Solutions manual to Tipler -Physics for Scientists and Engineers with Modern Physics(5th Edit)

Chapter 1 Systems of Measurement Conceptual Problems *1 • Determine the Concept The fundamental physical quantities in the SI system include mass, length, and time. Force, being the product of mass and acceleration, is not a fundamental quantity. (c) is correct. 2 • Picture the Problem We can express and simplify the ratio of m/s to m/s2 to determine the final units.

m 2 s = m ⋅ s = s and (d ) is correct. m m ⋅s 2 s

Express and simplify the ratio of m/s to m/s2:

3 • Determine the Concept Consulting Table 1-1 we note that the prefix giga means 109. (c ) is correct. 4 • Determine the Concept Consulting Table 1-1 we note that the prefix mega means 106. (d ) is correct. *5 • Determine the Concept Consulting Table 1-1 we note that the prefix pico means 10−12. (a ) is correct. 6 • Determine the Concept Counting from left to right and ignoring zeros to the left of the first nonzero digit, the last significant figure is the first digit that is in doubt. Applying this criterion, the three zeros after the decimal point are not significant figures, but the last zero is significant. Hence, there are four significant figures in this number.

(c) is correct.

1

2

Chapter 1

7 • Determine the Concept Counting from left to right, the last significant figure is the first digit that is in doubt. Applying this criterion, there are six significant figures in this number. (e) is correct. 8 • Determine the Concept The advantage is that the length measure is always with you. The disadvantage is that arm lengths are not uniform; if you wish to purchase a board of ″two arm lengths″ it may be longer or shorter than you wish, or else you may have to physically go to the lumberyard to use your own arm as a measure of length. 9 • (a) True. You cannot add ″apples to oranges″ or a length (distance traveled) to a volume (liters of milk). (b) False. The distance traveled is the product of speed (length/time) multiplied by the time of travel (time). (c) True. Multiplying by any conversion factor is equivalent to multiplying by 1. Doing so does not change the value of a quantity; it changes its units.

Estimation and Approximation *10 •• Picture the Problem Because θ is small, we can approximate it by θ ≈ D/rm provided that it is in radian measure. We can solve this relationship for the diameter of the moon. Express the moon’s diameter D in terms of the angle it subtends at the earth θ and the earth-moon distance rm:

D = θ rm

Find θ in radians:

θ = 0.524° ×

Substitute and evaluate D:

D = (0.00915 rad )(384 Mm )

2π rad = 0.00915 rad 360°

= 3.51 × 106 m

Systems of Measurement

3

*11 •• Picture the Problem We’ll assume that the sun is made up entirely of hydrogen. Then we can relate the mass of the sun to the number of hydrogen atoms and the mass of each. Express the mass of the sun MS as the product of the number of hydrogen atoms NH and the mass of each atom MH:

M S = NHM H

Solve for NH:

NH =

MS MH

Substitute numerical values and evaluate NH:

NH =

1.99 × 1030 kg = 1.19 × 1057 1.67 × 10 −27 kg

12 •• Picture the Problem Let P represent the population of the United States, r the rate of consumption and N the number of aluminum cans used annually. The population of the United States is roughly 3×108 people. Let’s assume that, on average, each person drinks one can of soft drink every day. The mass of a soft-drink can is approximately 1.8 ×10−2 kg. (a) Express the number of cans N used annually in terms of the daily rate of consumption of soft drinks r and the population P:

N = rP∆t

Substitute numerical values and approximate N:

⎛ 1can ⎞ ⎟⎟ 3 × 108 people N = ⎜⎜ ⋅ person d ⎝ ⎠ ⎛ d⎞ × (1 y )⎜⎜ 365.24 ⎟⎟ y⎠ ⎝

(

≈ 1011 cans (b) Express the total mass of aluminum used per year for soft drink cans M as a function of the number of cans consumed and the mass m per can:

M = Nm

)

4

Chapter 1

Substitute numerical values and evaluate M:

(c) Express the value of the aluminum as the product of M and the value at recycling centers:

(

)(

M = 1011 cans/y 1.8 × 10−2 kg/can

)

≈ 2 × 109 kg/y Value = ($1 / kg )M

(

= ($1 / kg ) 2 × 109 kg/y

)

= $2 × 10 / y 9

= 2 billion dollars/y 13 •• Picture the Problem We can estimate the number of words in Encyclopedia Britannica by counting the number of volumes, estimating the average number of pages per volume, estimating the number of words per page, and finding the product of these measurements and estimates. Doing so in Encyclopedia Britannica leads to an estimate of approximately 200 million for the number of words. If we assume an average word length of five letters, then our estimate of the number of letters in Encyclopedia Britannica becomes 109. (a) Relate the area available for one letter s2 and the number of letters N to be written on the pinhead to the area of the pinhead: Solve for s to obtain:

Substitute numerical values and evaluate s:

Ns 2 =

π 4

d 2 where d is the diameter of the

pinhead.

s=

πd 2 4N

⎡

s=

(b) Express the number of atoms per letter n in terms of s and the atomic spacing in a metal datomic:

n=

Substitute numerical values and evaluate n:

n=

⎣

2

cm ⎞⎤ ⎛ ⎟ in ⎠⎥⎦ ⎝ ≈ 10−8 m 9 4 10

π ⎢(161 in )⎜ 2.54

( )

s d atomic

10 −8 m ≈ 20 atoms 5 × 10 −10 atoms/m

*14 •• Picture the Problem The population of the United States is roughly 3 × 108 people. Assuming that the average family has four people, with an average of two cars per

Systems of Measurement

5

family, there are about 1.5 × 108 cars in the United States. If we double that number to include trucks, cabs, etc., we have 3 × 108 vehicles. Let’s assume that each vehicle uses, on average, about 12 gallons of gasoline per week. (a) Find the daily consumption of gasoline G: Assuming a price per gallon P = $1.50, find the daily cost C of gasoline: (b) Relate the number of barrels N of crude oil required annually to the yearly consumption of gasoline Y and the number of gallons of gasoline n that can be made from one barrel of crude oil: Substitute numerical values and estimate N:

(

)

G = 3×108 vehicles (2 gal/d ) = 6 ×108 gal/d

(

)

C = GP = 6 × 108 gal/d ($1.50 / gal) = $9 × 108 / d ≈ $1 billion dollars/d

N=

Y G∆t = n n

(6 ×10 N=

)

gal/d (365.24 d/y ) 19.4 gal/barrel 8

≈ 1010 barrels/y 15 •• Picture the Problem We’ll assume a population of 300 million (fairly accurate as of September, 2002) and a life expectancy of 76 y. We’ll also assume that a diaper has a volume of about half a liter. In (c) we’ll assume the disposal site is a rectangular hole in the ground and use the formula for the volume of such an opening to estimate the surface area required. (a) Express the total number N of disposable diapers used in the United States per year in terms of the number of children n in diapers and the number of diapers D used by each child in 2.5 y:

N = nD

Use the daily consumption, the number of days in a year, and the estimated length of time a child is in diapers to estimate the number of diapers D required per child:

D=

3 diapers 365.24 d × × 2.5 y d y

≈ 3 × 103 diapers/child

6

Chapter 1

Use the assumed life expectancy to estimate the number of children n in diapers:

⎛ 2 .5 y ⎞ ⎟⎟ 300 × 10 6 children n = ⎜⎜ ⎝ 76 y ⎠ ≈ 10 7 children

Substitute to obtain:

N = 107 children

(

(

(

)

× 3 × 10 diapers/child 3

)

)

≈ 3 × 1010 diapers (b) Express the required landfill volume V in terms of the volume of diapers to be buried: Substitute numerical values and evaluate V:

V = NVone diaper

(

)

V = 3 × 1010 diapers (0.5 L/diaper ) ≈ 1.5 × 107 m 3

(c) Express the required volume in terms of the volume of a rectangular parallelepiped:

V = Ah

Solve and evaluate h:

V 1.5 × 107 m 3 A= = = 1.5 × 106 m 2 10 m h

Use a conversion factor to express this area in square miles:

A = 1.5 × 106 m 2 ×

1 mi 2 2.590 km 2

≈ 0.6 mi 2 16 ••• Picture the Problem The number of bits that can be stored on the disk can be found from the product of the capacity of the disk and the number of bits per byte. In part (b) we’ll need to estimate (i) the number of bits required for the alphabet, (ii) the average number of letters per word, (iii) an average number of words per line, (iv) an average number of lines per page, and (v) a book length in pages. (a) Express the number of bits Nbits as a function of the number of bits per byte and the capacity of the hard disk Nbytes:

N bits = N bytes (8 bits/byte)

= (2 × 109 bytes)(8 bits/byte) = 1.60 × 1010 bits

Systems of Measurement (b) Assume an average of 8 letters/word and 8 bits/character to estimate the number of bytes required per word:

8

7

bits characters bits ×8 = 64 character word word bytes =8 word words bytes bytes ×8 = 4800 page word page

Assume 10 words/line and 60 lines/page:

600

Assume a book length of 300 pages and approximate the number bytes required:

300pages × 4800

Divide the number of bytes per disk by our estimated number of bytes required per book to obtain an estimate of the number of books the 2-gigabyte hard disk can hold:

N books =

bytes = 1.44 × 106 bytes page

2 × 109 bytes 1.44 × 106 bytes/book

≈ 1400 books

*17 •• Picture the Problem Assume that, on average, four cars go through each toll station per minute. Let R represent the yearly revenue from the tolls. We can estimate the yearly revenue from the number of lanes N, the number of cars per minute n, and the $6 toll per car C.

R = NnC = 14 lanes × 4

min h d $6 cars × 60 × 24 × 365.24 × = $177M min h d y car

Units 18 • Picture the Problem We can use the metric prefixes listed in Table 1-1 and the abbreviations on page EP-1 to express each of these quantities. (a)

(c)

1,000,000 watts = 10 watts 6

3 × 10 −6 meter = 3 µm

= 1 MW (d)

(b) −3

0.002gram = 2 × 10 g = 2 mg

30,000 seconds = 30 × 103 s = 30 ks

8

Chapter 1

19 • Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without prefixes. (c)

(a)

40 µW = 40 × 10 W = 0.000040 W

3 MW = 3 × 106 W = 3,000,000 W

(b)

(d)

−6

−9

4 ns = 4 × 10 s = 0.000000004 s

25 km = 25 × 103 m = 25,000 m

*20 • Picture the Problem We can use the definitions of the metric prefixes listed in Table 1-1 to express each of these quantities without abbreviations. (a) 10 −12 boo = 1 picoboo

(e) 106 phone = 1megaphone

(b) 10 9 low = 1 gigalow

(f) 10 −9 goat = 1 nanogoat

(c) 10 −6 phone = 1 microphone

(g) 1012 bull = 1 terabull

(d) 10 −18 boy = 1 attoboy

21 •• Picture the Problem We can determine the SI units of each term on the right-hand side of the equations from the units of the physical quantity on the left-hand side. (a) Because x is in meters, C1 and C2t must be in meters:

C1 is in m; C2 is in m/s

(b) Because x is in meters, ½C1t2 must be in meters:

C1 is in m/s 2

(c) Because v2 is in m2/s2, 2C1x must be in m2/s2:

C1 is in m/s 2

(d) The argument of trigonometric function must be dimensionless; i.e. without units. Therefore, because x

C1 is in m; C2 is in s −1

Systems of Measurement

9

is in meters: (e) The argument of an exponential function must be dimensionless; i.e. without units. Therefore, because v is in m/s:

C1 is in m/s; C2 is in s −1

22 •• Picture the Problem We can determine the US customary units of each term on the right-hand side of the equations from the units of the physical quantity on the left-hand side. (a) Because x is in feet, C1 and C2t must be in feet:

C1 is in ft; C2 is in ft/s

(b) Because x is in feet, ½C1t2 must be in feet:

C1 is in ft/s 2

(c) Because v2 is in ft2/s2, 2C1x must be in ft2/s2:

C1 is in ft/s 2

(d) The argument of trigonometric function must be dimensionless; i.e. without units. Therefore, because x is in feet:

C1 is in ft; C2 is in s −1

(e) The argument of an exponential function must be dimensionless; i.e. without units. Therefore, because v is in ft/s:

C1 is in ft/s; C2 is in s −1

Conversion of Units 23 • Picture the Problem We can use the formula for the circumference of a circle to find the radius of the earth and the conversion factor 1 mi = 1.61 km to convert distances in meters into distances in miles. (a) The Pole-Equator distance is one-fourth of the circumference:

c = 4 × 107 m

10

Chapter 1

(b) Use the formula for the circumference of a circle to obtain:

c 4 × 10−7 m R= = = 6.37 × 106 m 2π 2π

(c) Use the conversion factors 1 km = 1000 m and 1 mi = 1.61 km:

c = 4 × 107 m ×

1 km 1 mi × 3 10 m 1.61km

= 2.48 × 104 mi and

R = 6.37 × 106 m ×

1 km 1 mi × 3 10 m 1.61 km

= 3.96 × 103 mi 24 • Picture the Problem We can use the conversion factor 1 mi = 1.61 km to convert speeds in km/h into mi/h. Find the speed of the plane in km/s:

v = 2(340 m/s ) = 680 m/s m ⎞ ⎛ 1 km ⎞ ⎛ s⎞ ⎛ = ⎜ 680 ⎟ ⎜⎜ 3 ⎟⎟ ⎜ 3600 ⎟ s ⎠ ⎝ 10 m ⎠ ⎝ h⎠ ⎝ = 2450 km/h

Convert v into mi/h:

km ⎞ ⎛ 1 mi ⎞ ⎛ ⎟ v = ⎜ 2450 ⎟⎜ h ⎠ ⎜⎝ 1.61 km ⎟⎠ ⎝ = 1520 mi/h

*25 • Picture the Problem We’ll first express his height in inches and then use the conversion factor 1 in = 2.54 cm.

12 in + 10.5 in = 82.5 in ft

Express the player’s height into inches:

h = 6 ft ×

Convert h into cm:

h = 82.5 in ×

2.54 cm = 210 cm in

26 • Picture the Problem We can use the conversion factors 1 mi = 1.61 km, 1 in = 2.54 cm, and 1 m = 1.094 yd to complete these conversions.

Systems of Measurement (a)

100

11

km km 1 mi mi = 100 × = 62.1 h h 1.61km h 1in = 23.6 in 2.54 cm

(b)

60 cm = 60 cm ×

(c)

100 yd = 100 yd ×

1m = 91.4 m 1.094 yd

27 • Picture the Problem We can use the conversion factor 1.609 km = 5280 ft to convert the length of the main span of the Golden Gate Bridge into kilometers. Convert 4200 ft into km:

4200 ft = 4200 ft ×

1.609 km = 1.28 km 5280 ft

*28 • Picture the Problem Let v be the speed of an object in mi/h. We can use the conversion factor 1 mi = 1.61 km to convert this speed to km/h. Multiply v mi/h by 1.61 km/mi to convert v to km/h:

v

mi mi 1.61 km =v × = 1.61v km/h h h mi

29 • Picture the Problem Use the conversion factors 1 h = 3600 s, 1.609 km = 1 mi, and 1 mi = 5280 ft to make these conversions. (a) 1.296 × 105

km ⎛ km ⎞ ⎛ 1 h ⎞ km ⎟⎟ = 36.0 = ⎜1.296 × 105 2 ⎟ ⎜⎜ 2 h h ⎠ ⎝ 3600 s ⎠ h ⋅s ⎝

km ⎞ ⎛ 1 h ⎞ km ⎛ ⎟ (b) 1.296 × 10 2 = ⎜1.296 × 105 2 ⎟ ⎜⎜ h ⎠ ⎝ 3600 s ⎟⎠ h ⎝ 5

2

⎛ 103 m ⎞ m ⎜⎜ ⎟⎟ = 10.0 2 s ⎝ km ⎠

(c) 60

mi ⎛ mi ⎞ ⎛ 5280 ft ⎞ ⎛ 1 h ⎞ ft ⎟⎟ ⎜⎜ ⎟⎟ = 88.0 = ⎜ 60 ⎟ ⎜⎜ h ⎝ h ⎠ ⎝ 1 mi ⎠ ⎝ 3600 s ⎠ s

(d) 60

mi ⎛ mi ⎞ ⎛ 1.609 km ⎞ ⎛ 103 m ⎞ ⎛ 1 h ⎞ m ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = 26.8 = ⎜ 60 ⎟ ⎜⎜ h ⎝ h ⎠ ⎝ 1 mi ⎠ ⎝ km ⎠ ⎝ 3600 s ⎠ s

12

Chapter 1

30 • Picture the Problem We can use the conversion factor 1 L = 1.057 qt to convert gallons into liters and then use this gallons-to-liters conversion factor to convert barrels into cubic meters.

⎛ 4 qt ⎞ ⎛ 1 L ⎞ ⎟⎟ = 3.784 L ⎟⎟ ⎜⎜ ⎝ gal ⎠ ⎝ 1.057 qt ⎠

(a) 1gal = (1gal)⎜⎜

3 −3 ⎛ 42 gal ⎞ ⎛ 3.784 L ⎞ ⎛ 10 m ⎞ ⎟⎟ = 0.1589 m 3 ⎟⎟ ⎜⎜ ⎟ ⎜⎜ ⎝ barrel ⎠ ⎝ gal ⎠ ⎝ L ⎠

(b) 1 barrel = (1 barrel)⎜

31 • Picture the Problem We can use the conversion factor given in the problem statement and the fact that 1 mi = 1.609 km to express the number of square meters in one acre. Multiply by 1 twice, properly chosen, to convert one acre into square miles, and then into square meters:

⎛ 1mi 2 ⎞ ⎛ 1609 m ⎞ ⎟⎟ ⎜ 1acre = (1acre)⎜⎜ ⎟ ⎝ 640 acres ⎠ ⎝ mi ⎠

2

= 4050 m 2

32 •• Picture the Problem The volume of a right circular cylinder is the area of its base multiplied by its height. Let d represent the diameter and h the height of the right circular cylinder; use conversion factors to express the volume V in the given units. (a) Express the volume of the cylinder: Substitute numerical values and evaluate V:

V = 14 πd 2 h V = 14 π (6.8 in ) (2 ft ) 2

⎛ 1ft ⎞ ⎟⎟ = π (6.8 in ) (2 ft )⎜⎜ ⎝ 12 in ⎠

2

2

1 4

= 0.504 ft 3 3

(b) Use the fact that 1 m = 3.281 ft to convert the volume in cubic feet into cubic meters:

⎛ 1m ⎞ ⎟⎟ V = 0.504 ft ⎜⎜ ⎝ 3.281 ft ⎠

(c) Because 1 L = 10−3 m3:

⎛ 1L V = 0.0143m 3 ⎜⎜ − 3 3 ⎝ 10 m

(

3

)

= 0.0143 m 3

(

)

⎞ ⎟ = 14.3 L ⎟ ⎠

Systems of Measurement

13

*33 •• Picture the Problem We can treat the SI units as though they are algebraic quantities to simplify each of these combinations of physical quantities and constants. (a) Express and simplify the units of v2/x:

(m s )2 m

(b) Express and simplify the units of x a: (c) Noting that the constant factor 1 2 has no units, express and simplify the units of

1 2

=

m2 m = 2 2 m⋅s s

m = s2 = s m/s 2

( )

⎛m⎞ 2 ⎛m⎞ 2 ⎜ 2 ⎟(s ) = ⎜ 2 ⎟ s = m ⎝s ⎠ ⎝s ⎠

at 2 :

Dimensions of Physical Quantities 34 • Picture the Problem We can use the facts that each term in an equation must have the same dimensions and that the arguments of a trigonometric or exponential function must be dimensionless to determine the dimensions of the constants. (a) x = C1 + C2 t

L

L T T

L

(b)

x =

1 2

C1

t

2

L T2 T2

L

(d) x

=

C1 cos C2

L (e) v =

L T

L

C1

L T

t

1 T

T

exp( −C2 t)

1 T

T

(c)

v 2 = 2 C1

x

L T2

L

2

L T2

35 •• Picture the Problem Because the exponent of the exponential function must be dimensionl the dimension of λ must be T −1.

14

Chapter 1

*36 •• Picture the Problem We can solve Newton’s law of gravitation for G and substitute the dimensions of the variables. Treating them as algebraic quantities will allow us to express the dimensions in their simplest form. Finally, we can substitute the SI units for the dimensions to find the units of G.

Fr 2 m1m2

Solve Newton’s law of gravitation for G to obtain:

G=

Substitute the dimensions of the variables:

ML 2 ×L 2 L3 T G= = M2 MT 2

Use the SI units for L, M, and T:

Units of G are

m3 kg ⋅ s 2

37 •• Picture the Problem Let m represent the mass of the object, v its speed, and r the radius of the circle in which it moves. We can express the force as the product of m, v, and r (each raised to a power) and then use the dimensions of force F, mass m, speed v, and radius r to obtain three equations in the assumed powers. Solving these equations simultaneously will give us the dependence of F on m, v, and r. Express the force in terms of powers of the variables:

F = mavb r c

Substitute the dimensions of the physical quantities:

⎛L⎞ MLT −2 = M a ⎜ ⎟ Lc ⎝T ⎠

Simplify to obtain:

MLT −2 = M a Lb+cT − b

Equate the exponents to obtain:

a = 1, b + c = 1, and −b = −2

Solve this system of equations to obtain:

a = 1, b = 2, and c = −1

Substitute in equation (1):

F = mv 2 r −1 = m

b

v2 r

Systems of Measurement

15

38 •• Picture the Problem We note from Table 1-2 that the dimensions of power are ML2/T3. The dimensions of mass, acceleration, and speed are M, L/T2, and L/T respectively. Express the dimensions of mav:

[mav] = M ×

From Table 1-2:

[P ] = ML3

L L ML2 × = 3 T2 T T

2

T

Comparing these results, we see that the product of mass, acceleration, and speed has the dimensions of power. 39 •• Picture the Problem The dimensions of mass and velocity are M and L/T, respectively. We note from Table 1-2 that the dimensions of force are ML/T2. Express the dimensions of momentum:

[mv] = M × L = ML T

From Table 1-2:

T

[F ] = ML2 T

Express the dimensions of force multiplied by time:

[Ft ] = ML2 × T = ML T

T

Comparing these results, we see that momentum has the dimensions of force multiplied by time. 40 •• Picture the Problem Let X represent the physical quantity of interest. Then we can express the dimensional relationship between F, X, and P and solve this relationship for the dimensions of X. Express the relationship of X to force and power dimensionally: Solve for [ X ] :

[F ][X ] = [P]

[X ] = [P] [F ]

16

Chapter 1

Substitute the dimensions of force and power and simplify to obtain:

Because the dimensions of velocity are L/T, we can conclude that:

ML2 T3 = L [X ] = ML T 2 T

[P ] = [F ][v]

Remarks: While it is true that P = Fv, dimensional analysis does not reveal the presence of dimensionless constants. For example, if P = πFv , the analysis shown above would fail to establish the factor of π. *41 •• Picture the Problem We can find the dimensions of C by solving the drag force equation for C and substituting the dimensions of force, area, and velocity.

Fair Av 2

Solve the drag force equation for the constant C:

C=

Express this equation dimensionally:

[C ] = [Fair ]2 [A][v]

Substitute the dimensions of force, area, and velocity and simplify to obtain:

ML 2 [C ] = T 2 = M3 L ⎛L⎞ L2 ⎜ ⎟ ⎝T ⎠

42 •• Picture the Problem We can express the period of a planet as the product of these factors (each raised to a power) and then perform dimensional analysis to determine the values of the exponents.

T = Cr a G b M Sc

Express the period T of a planet as the product of r a , G b , and M Sc :

where C is a dimensionless constant.

Solve the law of gravitation for the constant G:

Fr 2 G= m1m2

Express this equation dimensionally:

2 F ][r ] [ [G ] = [m1 ][m2 ]

(1)

Systems of Measurement Substitute the dimensions of F, r, and m:

ML 2 × (L ) 2 L3 T = [G ] = M ×M MT 2

Noting that the dimension of time is represented by the same letter as is the period of a planet, substitute the dimensions in equation (1) to obtain:

⎛ L3 ⎞ ⎟ (M )c T = (L ) ⎜⎜ 2 ⎟ ⎝ MT ⎠

Introduce the product of M 0 and L0 in the left hand side of the equation and simplify to obtain:

M 0 L0T 1 = M c −b La +3bT −2b

Equate the exponents on the two sides of the equation to obtain:

0 = c – b, 0 = a + 3b, and 1 = –2b

Solve these equations simultaneously to obtain: Substitute in equation (1):

17

b

a

a = 32 , b = − 12 , and c = − 12

T = Cr 3 2G −1 2 M S−1 2 =

C r3 2 GM S

Scientific Notation and Significant Figures *43 • Picture the Problem We can use the rules governing scientific notation to express each of these numbers as a decimal number. (a) 3 × 10 4 = 30,000

(c) 4 × 10 −6 = 0.000004

(b) 6.2 × 10 −3 = 0.0062

(d) 2.17 × 105 = 217,000

44 • Picture the Problem We can use the rules governing scientific notation to express each of these measurements in scientific notation. (a) 3.1GW = 3.1 × 109 W

(c) 2.3 fs = 2.3 × 10 −15 s

18

Chapter 1

(b) 10 pm = 10 × 10 −12 m = 10 −11 m

(d) 4 µs = 4 × 10 −6 s

45 • Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions. (a) The number of significant figures in each factor is three; therefore the result has three significant figures: (b) Express both terms with the same power of 10. Because the first measurement has only two digits after the decimal point, the result can have only two digits after the decimal point: (c) We’ll assume that 12 is exact. Hence, the answer will have three significant figures: (d) Proceed as in (b):

(1.14)(9.99 × 104 ) =

1.14 × 105

(2.78 × 10 ) − (5.31 × 10 ) −8

−9

= (2.78 − 0.531) × 10−8

= 2.25 × 10−8

12π = 8.27 × 103 4.56 × 10 −3

(

)

27.6 + 5.99 × 10 2 = 27.6 + 599 = 627 = 6.27 × 10 2

46 • Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions. (a) Note that both factors have four significant figures. (b) Express the first factor in scientific notation and note that both factors have three significant figures.

(200.9)(569.3) =

1.144 × 105

(0.000000513)(62.3 × 107 )

(

)(

= 5.13 × 10 −7 62.3 × 107 = 3.20 × 10 2

)

Systems of Measurement

19

( ) = (2.841 × 10 ) + (5.78 × 10 )

(c) Express both terms in scientific notation and note that the second has only three significant figures. Hence the result will have only three significant figures.

28401 + 5.78 × 104

(d) Because the divisor has three significant figures, the result will have three significant figures.

63.25 = 1.52 × 104 −3 4.17 × 10

4

4

= (2.841 + 5.78) × 104 = 8.62 × 104

*47 • Picture the Problem Let N represent the required number of membranes and express N in terms of the thickness of each cell membrane. Express N in terms of the thickness of a single membrane:

N=

1in 7 nm

Convert the units into SI units and simplify to obtain:

N=

1in 2.54 cm 1m 1 nm × × × −9 7 nm in 100 cm 10 m

= 4 × 106 48 • Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions. (a) Both factors and the result have three significant figures: (b) Because the second factor has three significant figures, the result will have three significant figures: (c) Both factors and the result have three significant figures: (d) Write both terms using the same power of 10. Note that the result will have only three significant figures:

(2.00 × 10 )(6.10 × 10 ) = −2

4

(3.141592)(4.00 × 105 ) =

1.22 × 103

1.26 × 106

2.32 × 103 = 2.00 × 10−5 8 1.16 × 10

(5.14 × 10 ) + (2.78 × 10 ) = (5.14 × 10 ) + (0.278 × 10 ) 3

2

3

= (5.14 + 0.278) × 103 = 5.42 × 103

3

20

Chapter 1

(e) Follow the same procedure used in (d):

(1.99 × 10 ) + (9.99 × 10 ) = (1.99 × 10 ) + (0.000000999 × 10 ) −5

2

2

2

= 1.99 × 102 *49 • Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions. (a) The second factor and the result have three significant figures:

3.141592654 × (23.2 ) = 1.69 × 103 2

(b) We’ll assume that 2 is exact. Therefore, the result will have two significant figures:

2 × 3.141592654 × 0.76 = 4.8

(c) We’ll assume that 4/3 is exact. Therefore the result will have two significant figures:

4 π × (1.1)3 = 5.6 3

(d) Because 2.0 has two significant figures, the result has two significant figures:

(2.0)5 3.141592654

= 10

General Problems 50 • Picture the Problem We can use the conversion factor 1 mi = 1.61 km to convert 100 km/h into mi/h. Multiply 100 km/h by 1 mi/1.61 km to obtain:

100

km km 1 mi = 100 × h h 1.61km = 62.1 mi/h

*51 • Picture the Problem We can use a series of conversion factors to convert 1 billion seconds into years. Multiply 1 billion seconds by the appropriate conversion factors to convert into years:

Systems of Measurement 109 s = 109 s ×

21

1h 1day 1y × × = 31.7 y 3600 s 24 h 365.24 days

52 • Picture the Problem In both the examples cited we can equate expressions for the physical quantities, expressed in different units, and then divide both sides of the equation by one of the expressions to obtain the desired conversion factor.

3 × 108 m/s = 1.61 × 103 m/mi 5 1.86 × 10 mi/h

(a) Divide both sides of the equation expressing the speed of light in the two systems of measurement by 186,000 mi/s to obtain:

1=

(b) Find the volume of 1.00 kg of water:

Volume of 1.00 kg = 103 g is 103 cm3

Express 103 cm3 in ft3:

⎛ ⎞ ⎛ ⎞ (10 cm ) ⎜⎜ 1in ⎟⎟ ⎜⎜ 1ft ⎟⎟ ⎝ 2.54 cm ⎠ ⎝ 12 in ⎠ = 0.0353 ft 3

Relate the weight of 1 ft3 of water to the volume occupied by 1 kg of water:

1.00 kg lb = 62.4 3 3 0.0353 ft ft

Divide both sides of the equation by the left-hand side to obtain:

lb ft 3 = 2.20 lb/kg 1= 1.00 kg 0.0353 ft 3

m ⎞ ⎛ 1 km ⎞ ⎛ ⎟ = ⎜1.61 × 103 ⎟⎜ mi ⎠ ⎜⎝ 103 m ⎟⎠ ⎝ = 1.61 km/mi

3

3

3

62.4

53 •• Picture the Problem We can use the given information to equate the ratios of the number of uranium atoms in 8 g of pure uranium and of 1 atom to its mass. Express the proportion relating the number of uranium atoms NU in 8 g of pure uranium to the mass of 1 atom:

1atom NU = 8 g 4.0 × 10−26 kg

22

Chapter 1 ⎛ ⎞ 1atom ⎟⎟ N U = (8 g )⎜⎜ −26 ⎝ 4.0 × 10 kg ⎠

Solve for and evaluate NU:

= 2.0 × 10 23 54 •• Picture the Problem We can relate the weight of the water to its weight per unit volume and the volume it occupies. Express the weight w of water falling on the acre in terms of the weight of one cubic foot of water, the depth d of the water, and the area A over which the rain falls:

lb ⎞ ⎛ w = ⎜ 62.4 3 ⎟ Ad ft ⎠ ⎝

Find the area A in ft2:

⎛ 1 mi 2 ⎞ ⎛ 5280 ft ⎞ ⎟⎟ ⎜ A = (1acre)⎜⎜ ⎟ ⎝ 640 acre ⎠ ⎝ mi ⎠ = 4.356 × 10 4 ft 2

2

Substitute numerical values and evaluate w:

⎛ 1ft ⎞ lb ⎞ ⎛ ⎟⎟ = 3.17 × 105 lb w = ⎜ 62.4 3 ⎟ 4.356 × 10 4 ft 2 (1.4 in ) ⎜⎜ ft ⎠ ⎝ ⎝ 12 in ⎠

(

)

55 •• Picture the Problem We can use the definition of density and the formula for the volume of a sphere to find the density of iron. Once we know the density of iron, we can use these same relationships to find what the radius of the earth would be if it had the same mass per unit volume as iron.

m V

(a) Using its definition, express the density of iron:

ρ=

Assuming it to be spherical, express the volume of an iron nucleus as a function of its radius:

V = 43 π r 3

Substitute to obtain:

ρ=

3m 4π r 3

(1)

Systems of Measurement Substitute numerical values and evaluate ρ:

ρ=

( 4π (5.4 × 10

) m)

3 9.3 × 10 −26 kg −15

3

= 1.41 × 1017 kg/m 3 (b) Because equation (1) relates the density of any spherical object to its mass and radius, we can solve for r to obtain:

r=3

3m 4πρ

Substitute numerical values and evaluate r:

r=3

3 5.98 × 10 24 kg = 216 m 4π 1.41 × 1017 kg/m 3

( (

)

)

56 •• Picture the Problem Apply the general rules concerning the multiplication, division, addition, and subtraction of measurements to evaluate each of the given expressions. (a) Because all of the factors have two significant figures, the result will have two significant figures:

(5.6 × 10 ) (0.0000075) −5

2.4 × 10 −12

(5.6 × 10 ) (7.5 × 10 ) = −5

−6

2.4 × 10 −12

= 1.8 × 10 2 (b) Because the factor with the fewest significant figures in the first term has two significant figures, the result will have two significant figures. Because its last significant figure is in the tenth’s position, the difference between the first and second term will have its last significant figure in the tenth’s position: (c) Because all of the factors have two significant figures, the result will have two significant figures:

(14.2) (6.4 × 107 )(8.2 × 10−9 ) − 4.06 = 7.8 − 4.06 = 3.4

(6.1 × 10 ) (3.6 × 10 ) (3.6 × 10 ) −6 2

4 3

−11 1 2

= 2.9 × 108

23

24

Chapter 1

(d) Because the factor with the fewest significant figures has two significant figures, the result will have two significant figures.

(0.000064)1 3

(12.8 × 10 )(490 × 10 ) (6.4 × 10 ) = (12.8 × 10 ) (490 × 10 ) −1 1 2

−3

−5 1 3

−1 1 2

−3

= 0.45 *57 •• Picture the Problem We can use the relationship between an angle θ, measured in radians, subtended at the center of a circle, the radius R of the circle, and the length L of the arc to answer these questions concerning the astronomical units of measure.

S R

(a) Relate the angle θ subtended by an arc of length S to the distance R:

θ=

Solve for and evaluate S:

S = Rθ

(1)

⎛ 1 min ⎞ ⎟⎟ = (1 parsec)(1s )⎜⎜ 60 s ⎝ ⎠ ⎛ 1° ⎞ ⎛ 2π rad ⎞ ⎟⎟ ⎜ × ⎜⎜ ⎟ ⎝ 60 min ⎠ ⎝ 360° ⎠ = 4.85 × 10 −6 parsec (b) Solve equation (1) for and evaluate R:

R= =

S

θ 1.496 × 1011 m (1s ) ⎛⎜⎜ 1min ⎞⎟⎟ ⎛⎜⎜ 1° ⎞⎟⎟ ⎛⎜ 2π rad ⎞⎟ ⎝ 60 s ⎠ ⎝ 60 min ⎠ ⎝ 360° ⎠

= 3.09 × 1016 m (c) Relate the distance D light travels in a given interval of time ∆t to its speed c and evaluate D for ∆t = 1 y:

D = c∆t ⎛ m⎞ s⎞ ⎛ = ⎜ 3 × 108 ⎟ (1 y )⎜⎜ 3.156 × 107 ⎟⎟ s⎠ y⎠ ⎝ ⎝ = 9.47 × 1015 m

Systems of Measurement (d) Use the definition of 1 AU and the result from part (c) to obtain:

25

⎛ ⎞ 1 AU ⎟⎟ 1c ⋅ y = 9.47 × 1015 m ⎜⎜ 11 ⎝ 1.496 × 10 m ⎠

(

)

= 6.33 × 10 4 AU (e) Combine the results of parts (b) and (c) to obtain:

(

1 parsec = 3.08 × 1016 m

)

⎛ ⎞ 1c ⋅ y ⎟⎟ × ⎜⎜ 15 ⎝ 9.47 × 10 m ⎠ = 3.25 c ⋅ y

58 •• Picture the Problem Let Ne and Np represent the number of electrons and the number of protons, respectively and ρ the critical average density of the universe. We can relate these quantities to the masses of the electron and proton using the definition of density.

m N e me = V V

(a) Using its definition, relate the required density ρ to the electron density Ne/V:

ρ=

Solve for Ne/V:

Ne ρ = V me

Substitute numerical values and evaluate Ne/V:

6 × 10−27 kg/m 3 Ne = 9.11 × 10−31 kg/electron V

(1)

= 6.59 × 103 electrons/m 3 (b) Express and evaluate the ratio of the masses of an electron and a proton:

me 9.11 × 10−31 kg = = 5.46 × 10 −4 −27 mp 1.67 × 10 kg

Rewrite equation (1) in terms of protons:

Np

Divide equation (2) by equation (1) to obtain:

Np

V

=

ρ mp

(2)

V = me or N p = me ⎛ N e ⎞ ⎜ ⎟ Ne V mp ⎝ V ⎠ mp V

26

Chapter 1

Substitute numerical values and use the result from part (a) to evaluate Np/V:

Np V

(

= 5.46 × 10 −4

)

(

× 6.59 × 103 protons/m 3

)

= 3.59 protons/m 3 *59 •• Picture the Problem We can use the definition of density to relate the mass of the water in the cylinder to its volume and the formula for the volume of a cylinder to express the volume of water used in the detector’s cylinder. To convert our answer in kg to lb, we can use the fact that 1 kg weighs about 2.205 lb. Relate the mass of water contained in the cylinder to its density and volume:

m = ρV

Express the volume of a cylinder in terms of its diameter d and height h:

V = Abase h =

Substitute to obtain:

m=ρ

Substitute numerical values and evaluate m:

Convert 5.02 × 107 kg to tons:

π 4

π 4

d 2h

d 2h

⎛π ⎞ 2 m = 103 kg/m 3 ⎜ ⎟ (39.3 m ) (41.4 m ) ⎝4⎠ 7 = 5.02 × 10 kg

(

)

m = 5.02 × 107 kg ×

2.205 lb 1 ton × 2000 lb kg

= 55.4 × 103 ton The 50,000 − ton claim is conservative. The actual weight is closer to 55,000 tons. 60 ••• Picture the Problem We’ll solve this problem two ways. First, we’ll substitute two of the ordered pairs in the given equation to obtain two equations in C and n that we can solve simultaneously. Then we’ll use a spreadsheet program to create a graph of log T as a function of log m and use its curve-fitting capability to find n and C. Finally, we can identify the data points that deviate the most from a straight-line plot by examination of the graph.

Systems of Measurement

27

1st Solution for (a) (a) To estimate C and n, we can n apply the relation T = Cm to two arbitrarily selected data points. We’ll use the 1st and 6th ordered pairs. This will produce simultaneous equations that can be solved for C and n.

T1 = Cm1n and

T6 = Cm6n

Divide the second equation by the first to obtain:

T6 Cm6n ⎛ m6 ⎞ = =⎜ ⎟ T1 Cm12 ⎜⎝ m1 ⎟⎠

Substitute numerical values and solve for n to obtain:

1.75 s ⎛ 1 kg ⎞ ⎟ =⎜ 0.56 s ⎜⎝ 0.1kg ⎟⎠

n

n

or

3.125 = 10n ⇒ n = 0.4948 and so a ″judicial″ guess is that n = 0.5. Substituting this value into the second equation gives:

T5 = Cm50.5 so

1.75 s = C(1 kg )

0.5

Solving for C gives:

C = 1.75 s/kg 0.5

2nd Solution for (a) Take the logarithm (we’ll arbitrarily use base 10) of both sides of T = Cmn and simplify to obtain:

(

)

log(T ) = log Cm n = log C + log m n = n log m + log C which, we note, is of the form y = mx + b . Hence a graph of log T vs. log m should be linear with a slope of n and a log Tintercept log C.

The graph of log T vs. log m shown below was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.)

28

Chapter 1 0.4 0.3

log T = 0.4987log m + 0.2479

log T

0.2 0.1 0.0 -0.1 -0.2 -0.3 -1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

log m

n = 0.499

Comparing the equation on the graph generated by the Add Trendline function to log (T ) = n log m + log C , we

and

observe:

or

C = 10 0.2479 = 1.77 s/kg1 2

(

)

T = 1.77 s/kg1 2 m 0.499 (b) From the graph we see that the data points that deviate the most from a straight-line plot are:

m = 0.02 kg, T = 0.471 s, and m = 1.50 kg, T = 2.22 s

From the graph we see that the points generated using the data pairs (b) (0.02 kg, 0.471 s) and (0.4 kg, 1.05 s) deviate the most from the line representing the best fit to the points plotted on the graph. Remarks: Still another way to find n and C is to use your graphing calculator to perform regression analysis on the given set of data for log T versus log m. The slope yields n and the y-intercept yields log C. 61 ••• Picture the Problem We can plot log T versus log r and find the slope of the best-fit line to determine the exponent n. We can then use any of the ordered pairs to evaluate C. Once we know n and C, we can solve T = Crn for r as a function of T.

Systems of Measurement (a) Take the logarithm (we’ll arbitrarily use base 10) of both sides of T = Crn and simplify to obtain:

29

( )

log(T ) = log Cr n = log C + log r n = n log r + log C Note that this equation is of the form y = mx + b . Hence a graph of log T vs. log r should be linear with a slope of n and a log T -intercept log C.

The graph of log T versus log r shown below was created using a spreadsheet program. The equation shown on the graph was obtained using Excel’s ″Add Trendline″ function. (Excel’s ″Add Trendline″ function uses regression analysis to generate the trendline.)

1.0 0.8 y = 1.5036x + 1.2311

log T

0.6 0.4 0.2 0.0 -0.2 -0.4 -1.1

-1.0

-0.9

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

log r

From the regression analysis we observe that:

n = 1.50 and

C = 101.2311 = 17.0 y/(Gm )

32

(

or T = 17.0 y/(Gm )

32

(b) Solve equation (1) for the radius of the planet’s orbit:

⎛ ⎞ T ⎟ r = ⎜⎜ 32 ⎟ ( ) 17 . 0 y / Gm ⎝ ⎠

Substitute numerical values and evaluate r:

⎞ ⎛ 6.20 y ⎟ r = ⎜⎜ 32 ⎟ ⎝ 17.0 y/(Gm ) ⎠

)r

1.50

(1)

23

23

= 0.510 Gm

30

Chapter 1

*62 ••• Picture the Problem We can express the relationship between the period T of the pendulum, its length L, and the acceleration of gravity g as T = CLa g b and perform dimensional analysis to find the values of a and b and, hence, the function relating these variables. Once we’ve performed the experiment called for in part (b), we can determine an experimental value for C. (a) Express T as the product of L and g raised to powers a and b: Write this equation in dimensional form:

T = CLa g b

(1)

where C is a dimensionless constant.

[T ] = [L] a [g ] b b

Noting that the symbols for the dimension of the period and length of the pendulum are the same as those representing the physical quantities, substitute the dimensions to obtain:

⎛ L ⎞ T =L⎜ 2⎟ ⎝T ⎠

Because L does not appear on the left-hand side of the equation, we can write this equation as:

L0T 1 = La +bT −2b

Equate the exponents to obtain: Solve these equations simultaneously to find a and b: Substitute in equation (1) to obtain:

(b) If you use pendulums of lengths 1 m and 0.5 m; the periods should be about:

(c) Solve equation (2) for C:

a

a + b = 0 and − 2b = 1 a = 12 and b = − 12

T = CL1 2 g −1 2 = C

T (1 m ) = 2 s and T (0.5 m ) = 1.4 s

C =T

g L

L g

(2)

Systems of Measurement Evaluate C with L = 1 m and T = 2 s:

Substitute in equation (2) to obtain:

C = (2 s )

9.81 m/s 2 = 6.26 ≈ 2π 1m

T = 2π

L g

63 ••• Picture the Problem The weight of the earth’s atmosphere per unit area is known as the atmospheric pressure. We can use this definition to express the weight w of the earth’s atmosphere as the product of the atmospheric pressure and the surface area of the earth.

w A

Using its definition, relate atmospheric pressure to the weight of the earth’s atmosphere:

P=

Solve for w:

w = PA

Relate the surface area of the earth to its radius R:

A = 4π R 2

Substitute to obtain:

w = 4π R 2 P

Substitute numerical values and evaluate w: 2

⎛ 103 m ⎞ ⎛ 39.37 in ⎞ ⎛ lb ⎞ 19 ⎟⎟ ⎜ w = 4π (6370 km ) ⎜⎜ ⎟ ⎜14.7 2 ⎟ = 1.16 × 10 lb in ⎠ ⎝ km ⎠ ⎝ m ⎠ ⎝ 2

2

31

32

Chapter 1

Chapter 2 Motion in One Dimension Conceptual Problems 1 • Determine the Concept The "average velocity" is being requested as opposed to "average speed". The average velocity is defined as the change in position or displacement divided by the change in time. The change in position for any "round trip" is zero by definition. So the average velocity for any round trip must also be zero.

vav =

∆y ∆t

vav =

∆y 0 = = 0 ∆t ∆t

*2 • Determine the Concept The important concept here is that "average speed" is being requested as opposed to "average velocity". Under all circumstances, including constant acceleration, the definition of the average speed is the ratio of the total distance traveled (H + H) to the total time elapsed, in this case 2H/T. (d ) is correct. Remarks: Because this motion involves a round trip, if the question asked for "average velocity," the answer would be zero. 3 • Determine the Concept Flying with the wind, the speed of the plane relative to the ground (vPG) is the sum of the speed of the wind relative to the ground (vWG) and the speed of the plane relative to the air (vPG = vWG + vPA). Flying into or against the wind the speed relative to the ground is the difference between the wind speed and the true air speed of the plane (vg = vw – vt). Because the ground speed landing against the wind is smaller than the ground speed landing with the wind, it is safer to land against the wind. 4 • Determine the Concept The important concept here is that a = dv/dt, where a is the acceleration and v is the velocity. Thus, the acceleration is positive if dv is positive; the acceleration is negative if dv is negative. (a) Let’s take the direction a car is moving to be the positive direction:

Because the car is moving in the direction we’ve chosen to be positive, its velocity is positive (dx > 0). If the car is braking, then its velocity is decreasing (dv < 0) and its acceleration (dv/dt) is negative.

(b) Consider a car that is moving to

Because the car is moving in the direction

33

34

Chapter 2

the right but choose the positive direction to be to the left:

opposite to that we’ve chosen to be positive, its velocity is negative (dx < 0). If the car is braking, then its velocity is increasing (dv > 0) and its acceleration (dv/dt) is positive.

*5 • Determine the Concept The important concept is that when both the acceleration and the velocity are in the same direction, the speed increases. On the other hand, when the acceleration and the velocity are in opposite directions, the speed decreases. (a)

(b)

Because your velocity remains negative, your displacement must be negative. Define the direction of your trip as the negative direction. During the last five steps gradually slow the speed of walking, until the wall is reached.

(c) A graph of v as a function of t that is consistent with the conditions stated in the problem is shown below:

0

v (m/s)

-1

-2

-3

-4

-5 0

0.5

1

1.5

2

2.5

t (s)

6 • Determine the Concept True. We can use the definition of average velocity to express the displacement ∆x as ∆x = vav∆t. Note that, if the acceleration is constant, the average velocity is also given by vav = (vi + vf)/2. 7 • Determine the Concept Acceleration is the slope of the velocity versus time curve, a = dv/dt; while velocity is the slope of the position versus time curve, v = dx/dt. The speed of an object is the magnitude of its velocity.

Motion in One Dimension

35

(a) True. Zero acceleration implies that the velocity is constant. If the velocity is constant (including zero), the speed must also be constant. (b) True in one dimension. Remarks: The answer to (b) would be False in more than one dimension. In one dimension, if the speed remains constant, then the object cannot speed up, slow down, or reverse direction. Thus, if the speed remains constant, the velocity remains constant, which implies that the acceleration remains zero. (In more than one-dimensional motion, an object can change direction while maintaining constant speed. This constitutes a change in the direction of the velocity.) Consider a ball moving in a circle at a constant rotation rate. The speed (magnitude of the velocity) is constant while the velocity is tangent to the circle and always changing. The acceleration is always pointing inward and is certainly NOT zero. *8 •• Determine the Concept Velocity is the slope of the position versus time curve and acceleration is the slope of the velocity versus time curve. See the graphs below.

7 6

position (m)

5 4 3 2 1 0 0

5

10

15

20

25

15

20

25

time (s)

3

2

acceleration (m/s )

2 1 0 -1 -2 -3 -4 0

5

10 time (s)

36

Chapter 2 1.0 0.8

velocity (m/s)

0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 0

5

10

15

20

25

time (s)

9 • Determine the Concept False. The average velocity is defined (for any acceleration) as the change in position (the displacement) divided by the change in time vav = ∆x ∆t . It is always valid. If the acceleration remains constant the average velocity is also given by

vav =

vi + vf 2

Consider an engine piston moving up and down as an example of non-constant velocity. For one complete cycle, vf = vi and xi = xf so vav = ∆x/∆t is zero. The formula involving the mean of vf and vi cannot be applied because the acceleration is not constant, and yields an incorrect nonzero value of vi. 10 • Determine the Concept This can occur if the rocks have different initial speeds. Ignoring air resistance, the acceleration is constant. Choose a coordinate system in which the origin is at the point of release and upward is the positive direction. From the constant-acceleration equation

y = y0 + v0t + 12 at 2 we see that the only way two objects can have the same acceleration (–g in this case) and cover the same distance, ∆y = y – y0, in different times would be if the initial velocities of the two rocks were different. Actually, the answer would be the same whether or not the acceleration is constant. It is just easier to see for the special case of constant acceleration. *11 •• Determine the Concept Neglecting air resistance, the balls are in free fall, each with the same free-fall acceleration, which is a constant. At the time the second ball is released, the first ball is already moving. Thus, during any time interval their velocities will increase by exactly the same amount. What can be said about the speeds of the two balls? The first ball will always be moving faster than the second ball. This being the case, what happens to the separation of the two balls while they are both

Motion in One Dimension

37

falling? Their separation increases. (a ) is correct. 12 •• Determine the Concept The slope of an x(t) curve at any point in time represents the speed at that instant. The way the slope changes as time increases gives the sign of the acceleration. If the slope becomes less negative or more positive as time increases (as you move to the right on the time axis), then the acceleration is positive. If the slope becomes less positive or more negative, then the acceleration is negative. The slope of the slope of an x(t) curve at any point in time represents the acceleration at that instant. The slope of curve (a) is negative and becomes more negative as time increases.

Therefore, the velocity is negative and the acceleration is negative.

The slope of curve (b) is positive and constant and so the velocity is positive and constant.

Therefore, the acceleration is zero.

The slope of curve (c) is positive and decreasing.

Therefore, the velocity is positive and the acceleration is negative.

The slope of curve (d) is positive and increasing.

Therefore, the velocity and acceleration are positive. We need more information to conclude that a is constant.

The slope of curve (e) is zero.

Therefore, the velocity and acceleration are zero.

(d ) best shows motion with constant positive acceleration. *13 • Determine the Concept The slope of a v(t) curve at any point in time represents the acceleration at that instant. Only one curve has a constant and positive slope.

(b ) is correct.

14 • Determine the Concept No. The word average implies an interval of time rather than an instant in time; therefore, the statement makes no sense. *15 • Determine the Concept Note that the ″average velocity″ is being requested as opposed to the ″average speed.″

38

Chapter 2

Yes. In any roundtrip, A to B, and back to A, the average velocity is zero.

∆x ∆xAB + ∆xBA = ∆t ∆t ∆x + (− ∆xBA ) 0 = AB = ∆t ∆t

vav (A→B→A ) =

= 0 On the other hand, the average velocity between A and B is not generally zero.

vav (A→B ) =

∆xAB ≠ 0 ∆t

Remarks: Consider an object launched up in the air. Its average velocity on the way up is NOT zero. Neither is it zero on the way down. However, over the round trip, it is zero. 16 • Determine the Concept An object is farthest from the origin when it is farthest from the time axis. In one-dimensional motion starting from the origin, the point located farthest from the time axis in a distance-versus-time plot is the farthest from its starting point. Because the object’s initial position is at x = 0, point B represents the instant that the object is farthest from x = 0. (b) is correct. 17 • Determine the Concept No. If the velocity is constant, a graph of position as a function of time is linear with a constant slope equal to the velocity. 18 • Determine the Concept Yes. The average velocity in a time interval is defined as the displacement divided by the elapsed time vav = ∆x ∆t . The fact that vav = 0 for some time interval, ∆t, implies that the displacement ∆x over this interval is also zero. Because the instantaneous velocity is defined as v = lim ∆t →0 (∆x / ∆t ) , it follows that v must also be zero. As an example, in the following graph of x versus t, over the interval between t = 0 and t ≈ 21 s, ∆x = 0. Consequently, vav = 0 for this interval. Note that the instantaneous velocity is zero only at t ≈ 10 s.

Motion in One Dimension

39

600 500

x (m)

400 300 200 100 0 0

5

10

15

20

t (s)

19 •• Determine the Concept In the one-dimensional motion shown in the figure, the velocity is a minimum when the slope of a position-versus-time plot goes to zero (i.e., the curve becomes horizontal). At these points, the slope of the position-versus-time curve is zero; therefore, the speed is zero. (b) is correct. *20 •• Determine the Concept In one-dimensional motion, the velocity is the slope of a position-versus-time plot and can be either positive or negative. On the other hand, the speed is the magnitude of the velocity and can only be positive. We’ll use v to denote velocity and the word “speed” for how fast the object is moving. (a) curve a: v(t 2 ) < v(t1 )

(b) curve a: speed (t 2 ) < speed (t1 )

curve c: v(t 2 ) > v(t1 )

curve c: speed(t 2 ) < speed(t1 )

curve b: v(t 2 ) = v(t1 )

curve d: v(t 2 ) < v(t1 )

curve b: speed(t 2 ) = speed(t1 )

curve d: speed(t 2 ) > speed(t1 )

21 • Determine the Concept Acceleration is the slope of the velocity-versus-time curve, a = dv/dt, while velocity is the slope of the position-versus-time curve, v = dx/dt. (a) False. Zero acceleration implies that the velocity is not changing. The velocity could be any constant (including zero). But, if the velocity is constant and nonzero, the particle must be moving. (b) True. Again, zero acceleration implies that the velocity remains constant. This means that the x-versus-t curve has a constant slope (i.e., a straight line). Note: This does not necessarily mean a zero-slope line.

40

Chapter 2

22 • Determine the Concept Yes. If the velocity is changing the acceleration is not zero. The velocity is zero and the acceleration is nonzero any time an object is momentarily at rest. If the acceleration were also zero, the velocity would never change; therefore, the object would have to remain at rest. Remarks: It is important conceptually to note that when both the acceleration and the velocity have the same sign, the speed increases. On the other hand, when the acceleration and the velocity have opposite signs, the speed decreases. 23 • Determine the Concept In the absence of air resistance, the ball will experience a constant acceleration. Choose a coordinate system in which the origin is at the point of release and the upward direction is positive. The graph shows the velocity of a ball that has been thrown straight upward with an initial speed of 30 m/s as a function of time. Note that the slope of this graph, the acceleration, is the same at every point, including the point at which v = 0 (at the top of its flight). Thus, vtop of flight = 0 and atop of flight = − g . 30 20

v (m/s)

10 0 -10 -20 -30 0

1

2

3

4

5

6

t (s)

The acceleration is the slope (–g). 24 • Determine the Concept The "average speed" is being requested as opposed to "average velocity." We can use the definition of average speed as distance traveled divided by the elapsed time and expression for the average speed of an object when it is experiencing constant acceleration to express vav in terms of v0. The average speed is defined as the total distance traveled divided by the change in time:

total distance traveled total time H + H 2H = = T T

vav =

Motion in One Dimension

41

v0 + 0 H = 1 2 2T

Find the average speed for the upward flight of the object:

vav,up =

Solve for H to obtain:

H = 14 v0T

Find the average speed for the downward flight of the object:

vav,down =

Solve for H to obtain:

H = 14 v0T

Substitute in our expression for vav to obtain:

2( 14 v0T ) v = 0 T 2 Because v0 ≠ 0 , the average speed is not

0 + v0 H = 1 2 2T

vav =

zero. Remarks: 1) Because this motion involves a roundtrip, if the question asked for ″average velocity″, the answer would be zero. 2) Another easy way to obtain this result is take the absolute value of the velocity of the object to obtain a graph of its speed as a function of time. A simple geometric argument leads to the result we obtained above. 25 • Determine the Concept In the absence of air resistance, the bowling ball will experience constant acceleration. Choose a coordinate system with the origin at the point of release and upward as the positive direction. Whether the ball is moving upward and slowing down, is momentarily at the top of its trajectory, or is moving downward with ever increasing velocity, its acceleration is constant and equal to the acceleration due to gravity. (b) is correct. 26 • Determine the Concept Both objects experience the same constant acceleration. Choose a coordinate system in which downward is the positive direction and use a constantacceleration equation to express the position of each object as a function of time. Using constant-acceleration equations, express the positions of both objects as functions of time:

xA = x0, A + v0t + 12 gt 2 and

xB = x0, B + v0t + 12 gt 2 where v0 = 0.

Express the separation of the two objects by evaluating xB − xA:

xB − xA = x0,B − x0.A = 10 m and (d ) is correct.

*27 •• Determine the Concept Because the Porsche accelerates uniformly, we need to look for a graph that represents constant acceleration. We are told that the Porsche has a constant acceleration that is positive (the velocity is increasing); therefore we must look for a velocity-versus-time curve with a positive constant slope and a nonzero intercept.

42

Chapter 2

(c ) is correct. *28 •• Determine the Concept In the absence of air resistance, the object experiences constant acceleration. Choose a coordinate system in which the downward direction is positive. Express the distance D that an object, released from rest, falls in time t: Because the distance fallen varies with the square of the time, during the first two seconds it falls four times the distance it falls during the first second.

D = 12 gt 2

(a ) is correct.

29 •• Determine the Concept In the absence of air resistance, the acceleration of the ball is constant. Choose a coordinate system in which the point of release is the origin and upward is the positive y direction. The displacement of the ball halfway to its highest point is: Using a constant-acceleration equation, relate the ball’s initial and final velocities to its displacement and solve for the displacement:

∆y =

∆ymax 2

v 2 = v02 + 2a∆y = v02 − 2 g∆y

Substitute v0 = 0 to determine the maximum displacement of the ball:

∆ymax = −

Express the velocity of the ball at half its maximum height:

v 2 = v02 − 2 g∆y = v02 − 2 g

v02 v2 = 0 2(− g ) 2 g

= v02 − g∆ymax Solve for v:

∆ymax 2 2 v v2 = v02 − g 0 = 0 2g 2

2 v0 ≈ 0.707v0 2 and (c ) is correct. v=

30 • Determine the Concept As long as the acceleration remains constant the following constant-acceleration equations hold. If the acceleration is not constant, they do not, in general, give correct results except by coincidence.

x = x0 + v0t + 12 at 2

v = v0 + at

v 2 = v02 + 2a∆x

vav =

vi + vf 2

Motion in One Dimension

43

(a) False. From the first equation, we see that (a) is true if and only if the acceleration is constant. (b) False. Consider a rock thrown straight up into the air. At the "top" of its flight, the velocity is zero but it is changing (otherwise the velocity would remain zero and the rock would hover); therefore the acceleration is not zero. (c) True. The definition of average velocity, vav = ∆x ∆t , requires that this always be true. *31 • Determine the Concept Because the acceleration of the object is constant, the constantacceleration equations can be used to describe its motion. The special expression for average velocity for constant acceleration is vav =

vi + vf . (c ) is correct. 2

32 • Determine the Concept The constant slope of the x-versus-t graph tells us that the velocity is constant and the acceleration is zero. A linear position versus time curve implies a constant velocity. The negative slope indicates a constant negative velocity. The fact that the velocity is constant implies that the acceleration is also constant and zero.

(e ) is correct.

33 •• Determine the Concept The velocity is the slope of the tangent to the curve, and the acceleration is the rate of change of this slope. Velocity is the slope of the positionversus-time curve. A parabolic x(t) curve opening upward implies an increasing velocity. The acceleration is positive.

(a ) is correct.

34 •• Determine the Concept The acceleration is the slope of the tangent to the velocity as a function of time curve. For constant acceleration, a velocity-versus- time curve must be a straight line whose slope is the acceleration. Zero acceleration means that slope of v(t) must also be zero.

(c ) is correct.

35 •• Determine the Concept The acceleration is the slope of the tangent to the velocity as a function of time curve. For constant acceleration, a velocity-versus- time curve must be a straight line whose slope is the acceleration. The acceleration and therefore the slope can be positive, negative, or zero.

(d ) is correct.

36 •• Determine the Concept The velocity is positive if the curve is above the v = 0 line (the t axis), and the acceleration is negative if the tangent to the curve has a negative slope. Only graphs (a), (c), and (e) have positive v. Of these, only graph (e) has a negative slope.

(e ) is correct.

44

Chapter 2

37 •• Determine the Concept The velocity is positive if the curve is above the v = 0 line (the t axis), and the acceleration is negative if the tangent to the curve has a negative slope. Only graphs (b) and (d) have negative v. Of these, only graph (d) has a negative slope.

(d ) is correct.

38 •• Determine the Concept A linear velocity-versus-time curve implies constant acceleration. The displacement from time t = 0 can be determined by integrating vversus-t — that is, by finding the area under the curve. The initial velocity at t = 0 can be read directly from the graph of v-versus-t as the v-intercept; i.e., v(0). The acceleration of the object is the slope of v(t) . The average velocity of the object is given by drawing a horizontal line that has the same area under it as the area under the curve. Because all of these quantities can be determined

(e ) is correct.

*39 •• Determine the Concept The velocity is the slope of a position versus time curve and the acceleration is the rate at which the velocity, and thus the slope, changes.

Velocity

(a) Negative at t0 and t1. (b) Positive at t3, t4, t6, and t7. (c) Zero at t2 and t5.

Acceleration

(a) Negative at t4. (b) Positive at t2 and t6. (c) Zero at t0, t1, t3, t5, and t7.

The acceleration is positive at points where the slope increases as you move toward the right.

40 •• Determine the Concept Acceleration is the slope of a velocity-versus-time curve. (a) Acceleration is zero and constant while velocity is not zero. 3 2

v

1 0 -1 -2 -3 0

0.5

1

1.5 t

2

2.5

3

Motion in One Dimension (b) Acceleration is constant but not zero. 3 2

v

1 0 -1 -2 -3 0

0.5

1

1.5

2

2.5

3

2

2.5

3

2

2.5

3

t

(c) Velocity and acceleration are both positive. 3 2

v

1 0 -1 -2 -3 0

0.5

1

1.5 t

(d) Velocity and acceleration are both negative. 3 2

v

1 0 -1 -2 -3 0

0.5

1

1.5 t

45

46

Chapter 2

(e) Velocity is positive and acceleration is negative. 3 2

v

1 0 -1 -2 -3 0

0.5

1

1.5

2

2.5

3

2

2.5

3

t

(f) Velocity is negative and acceleration is positive. 3 2

v

1 0 -1 -2 -3 0

0.5

1

1.5 t

Motion in One Dimension

47

(g) Velocity is momentarily zero at the intercept with the t axis but the acceleration is not zero. 3 2

v

1 0 -1 -2 -3 0

0.5

1

1.5

2

2.5

3

t

41 •• Determine the Concept Velocity is the slope and acceleration is the slope of the slope of a position-versus-time curve. Acceleration is the slope of a velocity- versus-time curve. (a) For constant velocity, x-versus-t must be a straight line; v-versus-t must be a horizontal straight line; and a-versus-t must be a straight horizontal line at a = 0. (b) For velocity to reverse its direction x-versus-t must have a slope that changes sign and vversus-t must cross the time axis. The acceleration cannot remain zero at all times.

(a), (f), and (i) are the correct answers.

(c) and (d) are the correct answers.

(c) For constant acceleration, x-versus-t must be a straight line or a parabola, v-versus-t must be a straight line, and a-versus-t must be a horizontal straight line.

(a), (d), (e), (f), (h), and (i) are the correct answers.

(d) For non-constant acceleration, x-versus-t must not be a straight line or a parabola; v-versus-t must not be a straight line, or a-versus-t must not be a horizontal straight line.

(b), (c), and (g) are the correct answers.

48

Chapter 2

For two graphs to be mutually consistent, the curves must be consistent with the definitions of velocity and acceleration.

Graphs (a) and (i) are mutually consistent. Graphs (d) and (h) are mutually consistent. Graphs (f) and (i) are also mutually consistent.

Estimation and Approximation 42 • Picture the Problem Assume that your heart beats at a constant rate. It does not, but the average is pretty stable. (a) We will use an average pulse rate of 70 bpm for a seated (resting) adult. One’s pulse rate is defined as the number of heartbeats per unit time:

Pulse rate = and

# of heartbeats Time

# of heartbeats = Pulse rate × Time

The time required to drive 1 mi at 60 mph is (1/60) h or 1 min:

# of heartbeats = (70 beats/min )(1 min )

(b) Express the number of heartbeats during a lifetime in terms of the pulse rate and the life span of an individual:

# of heartbeats = Pulse rate × Time

= 70 beats

Assuming a 95-y life span, calculate the time in minutes:

Time = (95 y )(365.25 d/y )(24 h/d )(60 min/ h ) = 5.00 × 107 min Substitute numerical values and evaluate the number of heartbeats:

(

)

# of heartbeats = (70 beats / min ) 5.00 × 107 min = 3.50 ×109 beats *43 •• Picture the Problem In the absence of air resistance, Carlos’ acceleration is constant. Because all the motion is downward, let’s use a coordinate system in which downward is positive and the origin is at the point at which the fall began. (a) Using a constant-acceleration equation, relate Carlos’ final velocity to his initial velocity, acceleration, and distance fallen and solve for his final velocity: Substitute numerical values and evaluate v:

v 2 = v02 + 2a∆y and, because v0 = 0 and a = g,

v = 2 g∆y

v = 2(9.81 m/s 2 )(150 m ) = 54.2 m/s

Motion in One Dimension (b) While his acceleration by the snow is not constant, solve the same constant- acceleration equation to get an estimate of his average acceleration:

v 2 − v02 a= 2∆y

Substitute numerical values and evaluate a:

− 54 m/s 2 a= = −1.20 × 103 m/s 2 ( ) 2 1.22m

(

49

)

2

= − 123 g Remarks: The final velocity we obtained in part (a), approximately 121 mph, is about the same as the terminal velocity for an "average" man. This solution is probably only good to about 20% accuracy. 44 •• Picture the Problem Because we’re assuming that the accelerations of the skydiver and the mouse are constant to one-half their terminal velocities, we can use constantacceleration equations to find the times required for them to reach their ″upper-bound″ velocities and their distances of fall. Let’s use a coordinate system in which downward is the positive y direction. (a) Using a constant-acceleration equation, relate the upper-bound velocity to the free-fall acceleration and the time required to reach this velocity: Solve for ∆t:

Substitute numerical values and evaluate ∆t: Using a constant-acceleration equation, relate the skydiver’s distance of fall to the elapsed time ∆t: Substitute numerical values and evaluate ∆y: (b) Proceed as in (a) with vupper bound = 0.5 m/s to obtain:

vupper bound = v0 + g∆t or, because v0 = 0,

vupper bound = g∆t

∆t =

∆t =

vupper bound g

25 m/s = 2.55 s 9.81m/s 2

∆y = v0 ∆t + 12 a(∆t )

2

or, because v0 = 0 and a = g,

∆y = 12 g (∆t )

2

∆y =

∆t =

1 2

2

2

= 31.9 m

0.5 m/s = 0.0510 s 9.81m/s 2

and

∆y =

(9.81m/s ) (2.55 s)

1 2

(9.81m/s )(0.0510 s) 2

2

= 1.27 cm

50

Chapter 2

45 •• Picture the Problem This is a constant-acceleration problem. Choose a coordinate system in which the direction Greene is running is the positive x direction. During the first 3 s of the race his acceleration is positive and during the rest of the race it is zero. The pictorial representation summarizes what we know about Greene’s race.

Express the total distance covered by Greene in terms of the distances covered in the two phases of his race:

100 m = ∆x01 + ∆x12

Express the distance he runs getting to his maximum velocity:

∆x01 = v0 ∆t01 + 12 a01 (∆t01 ) = 12 a(3 s )

Express the distance covered during the rest of the race at the constant maximum velocity:

∆x12 = vmax ∆t12 + 12 a12 (∆t12 )

Substitute for these displacements and solve for a:

100 m = 12 a(3 s ) + a(3 s )(6.79 s )

2

2

2

= (a∆t01 )∆t12

= a(3 s )(6.79 s ) 2

and

a = 4.02 m/s 2 *46 •• Determine the Concept This is a constant-acceleration problem with a = −g if we take upward to be the positive direction. At the maximum height the ball will reach, its speed will be near zero and when the ball has just been tossed in the air its speed is near its maximum value. What conclusion can you draw from the image of the ball near its maximum height? To estimate the initial speed of the ball:

Because the ball is moving slowly its blur is relatively short (i.e., there is less blurring).

Motion in One Dimension

51

a) Estimate how far the ball being tossed moves in 1/30 s:

The ball moves about 3 ball diameters in 1/30 s.

b) Estimate the diameter of a tennis ball:

The diameter of a tennis ball is approximately 5 cm.

c) Now one can calculate the approximate distance the ball moved in 1/30 s:

Distance traveled = (3 diameters) × (5 cm/diameter ) = 15 cm

d) Calculate the average speed of the tennis ball over this distance:

Average speed =

e) Because the time interval is very short, the average speed of the ball is a good approximation to its initial speed: f) Finally, use the constantacceleration equation v 2 = v02 + 2a∆y to solve for and

15 cm = 450 cm/s 1 s 30 = 4.50 m/s

∴ v0 = 4.5 m/s

− v02 − (4.5 m/s ) = = 1.03 m 2a 2 − 9.81 m/s 2 2

∆y =

(

)

evaluate ∆y: Remarks: This maximum height is in good agreement with the height of the higher ball in the photograph. *47 •• Picture the Problem The average speed of a nerve impulse is approximately 120 m/s. Assume an average height of 1.7 m and use the definition of average speed to estimate the travel time for the nerve impulse. Using the definition of average speed, express the travel time for the nerve impulse:

∆t =

∆x vav

Substitute numerical values and evaluate ∆t:

∆t =

1.7 m = 14.2 ms 120 m/s

Speed, Displacement, and Velocity 48 • Picture the Problem Think of the electron as traveling in a straight line at constant speed and use the definition of average speed.

52

Chapter 2

(a) Using its definition, express the average speed of the electron:

Average speed =

Solve for and evaluate the time of flight:

∆t =

distance traveled time of flight ∆s = ∆t

∆s 0.16 m = Average speed 4 × 107 m s

= 4 × 10−9 s = 4.00 ns (b) Calculate the time of flight for an electron in a 16-cm long current carrying wire similarly.

∆t =

∆s 0.16 m = Average speed 4 × 10 −5 m s

= 4 × 103 s = 66.7 min *49 • Picture the Problem In this problem the runner is traveling in a straight line but not at constant speed - first she runs, then she walks. Let’s choose a coordinate system in which her initial direction of motion is taken as the positive x direction. (a) Using the definition of average velocity, calculate the average velocity for the first 9 min:

vav =

∆x 2.5 km = = 0.278 km / min ∆t 9 min

(b) Using the definition of average velocity, calculate her average speed for the 30 min spent walking:

vav =

∆x − 2.5 km = ∆t 30 min

= − 0.0833 km / min (c) Express her average velocity for the whole trip:

vav =

(d) Finally, express her average speed for the whole trip:

Average speed =

∆xround trip ∆t

=

0 = 0 ∆t

distance traveled elapsed time 2(2.5 km) = 30 min + 9 min = 0.128 km / min

50 • Picture the Problem The car is traveling in a straight line but not at constant speed. Let the direction of motion be the positive x direction. (a) Express the total displacement of the car for the entire trip:

∆x total = ∆x1 + ∆x2

Motion in One Dimension Find the displacement for each leg of the trip:

53

∆x1 = vav ,1∆t1 = (80 km/h )(2.5 h ) = 200 km and

∆x2 = vav , 2 ∆t2 = (40 km/h )(1.5 h ) = 60.0 km

Add the individual displacements to get the total displacement:

(b) As long as the car continues to move in the same direction, the average velocity for the total trip is given by:

∆xtotal = ∆x1 + ∆x2 = 200 km + 60.0 km = 260 km vav ≡

260 km ∆xtotal = ∆ttotal 2.5 h + 1.5 h

= 65.0 km h

51 • Picture the Problem However unlikely it may seem, imagine that both jets are flying in a straight line at constant speed. (a) The time of flight is the ratio of the distance traveled to the speed of the supersonic jet.

tsupersonic = =

sAtlantic speedsupersonic 5500 km 2(0.340 km/s )(3600 s/h )

= 2.25 h (b) The time of flight is the ratio of the distance traveled to the speed of the subsonic jet.

tsubsonic = =

sAtlantic speed subsonic 5500 km 0.9(0.340 km/s )(3600 s/h )

= 4.99 h (c) Adding 2 h on both the front and the back of the supersonic trip, we obtain the average speed of the supersonic flight. (d) Adding 2 h on both the front and the back of the subsonic trip, we obtain the average speed of the subsonic flight.

speed av, supersonic =

5500 km 2.25 h + 4.00 h

= 880 km h

speed av, subsonic =

5500 km 5.00 h + 4.00 h

= 611 km h

54

Chapter 2

*52 • Picture the Problem In free space, light travels in a straight line at constant speed, c. (a) Using the definition of average speed, solve for and evaluate the time required for light to travel from the sun to the earth:

average speed =

s t

and

t=

s 1.5 × 1011 m = average speed 3 × 108 m/s

= 500 s = 8.33 min (b) Proceed as in (a) this time using the moon-earth distance: (c) One light-year is the distance light travels in a vacuum in one year:

t=

3.84 ×108 m = 1.28 s 3×108 m/s

1 light - year = 9.48 ×1015 m = 9.48 ×1012 km

(

)

= 9.48 ×1012 km (1 mi/1.61 km ) = 5.89 ×10 mi 12

53 • Picture the Problem In free space, light travels in a straight line at constant speed, c. (a) Using the definition of average speed (equal here to the assumed constant speed of light), solve for the time required to travel the distance to Proxima Centauri: (b) Traveling at 10-4c, the delivery time (ttotal) will be the sum of the time for the order to reach Hoboken and the time for the pizza to be delivered to Proxima Centauri:

t=

distance traveled 4.1×1016 m = speed of light 3×108 m s

= 1.37 ×108 s = 4.33 y t total = torder to be sent to Hoboken + torder to be delivered 4.1×1013 km = 4.33 y + − 4 10 3 × 108 m s

(

)(

)

= 4.33 y + 4.33×106 y ≈ 4.33×106 y Since 4.33 × 10 6 y >> 1000 y, Gregor does not have to pay.

54 • Picture the Problem The time for the second 50 km is equal to the time for the entire journey less the time for the first 50 km. We can use this time to determine the average speed for the second 50 km interval from the definition of average speed. Using the definition of average speed, find the time required for the total journey:

t total =

total distance 100 km = = 2h average speed 50 km h

Motion in One Dimension Find the time required for the first 50 km:

t1st 50 km =

Find the time remaining to travel the last 50 km:

t2nd 50 km = t total − t1st 50 km = 2 h − 1.25 h

Finally, use the time remaining to travel the last 50 km to determine the average speed over this distance:

Average speed 2nd 50 km

55

50 km = 1.25 h 40 km h

= 0.75 h

=

distance traveled2nd 50 km time2nd 50 km

=

50 km = 66.7 km h 0.75 h

*55 •• Picture the Problem Note that both the arrow and the sound travel a distance d. We can use the relationship between distance traveled, the speed of sound, the speed of the arrow, and the elapsed time to find the distance separating the archer and the target. Express the elapsed time between the archer firing the arrow and hearing it strike the target: Express the transit times for the arrow and the sound in terms of the distance, d, and their speeds:

∆t = 1s = ∆tarrow + ∆tsound

∆tarrow =

varrow

=

d 40 m/s

=

d 340 m/s

and

∆tsound = Substitute these two relationships in the expression obtained in step 1 and solve for d:

d

d vsound

d d + = 1s 40 m/s 340 m/s and d = 35.8 m

56 •• Picture the Problem Assume both runners travel parallel paths in a straight line along the track. (a) Using the definition of average speed, find the time for Marcia:

distance run Marcia' s speed distance run = 1.15 (John' s speed ) 100 m = = 14.5 s 1.15 (6 m s )

tMarcia =

56

Chapter 2 xJohn = (6 m s )(14.5 s ) = 87.0 m

Find the distance covered by John in 14.5 s and the difference between that distance and 100 m:

and Marcia wins by

(b) Using the definition of average speed, find the time required by John to complete the 100-m run:

tJohn =

100 m − 87 m = 13.0 m

distance run 100 m = = 16.7 s John' s speed 6 m s

Marsha wins by 16.7 s – 14.5 s = 2.2 s Alternatively, the time required by John to travel the last 13.0 m is (13 m)/(6 m/s) = 2.17 s 57 • Picture the Problem The average velocity in a time interval is defined as the displacement divided by the time elapsed; that is vav = ∆x / ∆t . (a) ∆xa = 0

vav = 0

(b) ∆xb = 1 m and ∆tb = 3 s

vav = 0.333 m/s

(c) ∆xc = –6 m and ∆tc = 3 s

vav = − 2.00 m/s

(d) ∆xd = 3 m and ∆td = 3 s

vav = 1.00 m/s

58 •• Picture the Problem In free space, light travels in a straight line at constant speed c. We can use Hubble’s law to find the speed of the two planets.

(

)(

(a) Using Hubble’s law, calculate the speed of the first galaxy:

va = 5 × 10 22 m 1.58 × 10 −18 s −1

(b) Using Hubble’s law, calculate the speed of the second galaxy:

vb = 2 × 10 25 m 1.58 × 10 −18 s −1

(c) Using the relationship between distance, speed, and time for both galaxies, determine how long ago they were both located at the same place as the earth:

)

= 7.90 × 10 4 m/s

(

)(

= 3.16 × 107 m/s

r r 1 = = v rH H = 6.33 × 1017 s = 20.1× 109 y

t=

= 20.1 billion years

)

Motion in One Dimension

57

*59 •• Picture the Problem Ignoring the time intervals during which members of this relay time get up to their running speeds, their accelerations are zero and their average speed can be found from its definition. Using its definition, relate the average speed to the total distance traveled and the elapsed time:

vav =

Express the time required for each animal to travel a distance L:

tcheetah =

distance traveled elapsed time

tfalcon =

L vcheetah L vfalcon

,

,

and tsailfish = Express the total time, ∆t:

L vsailfish

⎛ 1 1 1 ⎞ ⎟⎟ ∆t = L⎜⎜ + + ⎝ vcheetah vfalcon vsailfish ⎠

Use the total distance traveled by the relay team and the elapsed time to calculate the average speed:

vav =

3L = 122 km/h ⎛ ⎞ 1 1 1 ⎟⎟ L⎜⎜ + + ⎝ 113 km/h 161 km/h 105 km/h ⎠

Calculate the average of the three speeds:

Averagethree speeds

=

113 km/h + 161 km/h + 105 km/h 3

= 126 km/h = 1.03vav

60 •• Picture the Problem Perhaps the easiest way to solve this problem is to think in terms of the relative velocity of one car relative to the other. Solve this problem from the reference frame of car A. In this frame, car A remains at rest.

Find the velocity of car B relative to car A:

vrel = vB – vA = (110 – 80) km/h = 30 km/h

Find the time before car B reaches car A:

∆t =

Find the distance traveled, relative to the road, by car A in 1.5 h:

d = (1.5 h )(80 km/h ) = 120 km

45 km ∆x = = 1.5 h vrel 30 km/h

58

Chapter 2

*61 •• Picture the Problem One way to solve this problem is by using a graphing calculator to plot the positions of each car as a function of time. Plotting these positions as functions of time allows us to visualize the motion of the two cars relative to the (fixed) ground. More importantly, it allows us to see the motion of the two cars relative to each other. We can, for example, tell how far apart the cars are at any given time by determining the length of a vertical line segment from one curve to the other.

(a) Letting the origin of our coordinate system be at the intersection, the position of the slower car, x1(t), is given by:

x1(t) = 20t where x1 is in meters if t is in seconds.

Because the faster car is also moving at a constant speed, we know that the position of this car is given by a function of the form:

x2(t) = 30t + b

We know that when t = 5 s, this second car is at the intersection (i.e., x2(5 s) = 0). Using this information, you can convince yourself that:

b = −150 m

Thus, the position of the faster car is given by:

x2 (t ) = 30t − 150

One can use a graphing calculator, graphing paper, or a spreadsheet to obtain the graphs of x1(t) (the solid line) and x2(t) (the dashed line) shown below: 350 300

x (m)

250 200 150 100 50 0 0

2

4

6

8

10

12

14

16

t (s)

(b) Use the time coordinate of the intersection of the two lines to determine the time at which the second car overtakes the first:

From the intersection of the two lines, one can see that the second car will "overtake" (catch up to) the first car at t = 15 s .

Motion in One Dimension

59

(c) Use the position coordinate of the intersection of the two lines to determine the distance from the intersection at which the second car catches up to the first car:

From the intersection of the two lines, one can see that the distance from the

(d) Draw a vertical line from t = 5 s to the red line and then read the position coordinate of the intersection of this line and the red line to determine the position of the first car when the second car went through the intersection:

From the graph, when the second car passes the intersection, the first car was

intersection is 300 m .

100 m ahead .

62 • Picture the Problem Sally’s velocity relative to the ground (vSG) is the sum of her velocity relative to the moving belt (vSB) and the velocity of the belt relative to the ground (vBG). Joe’s velocity relative to the ground is the same as the velocity of the belt relative to the ground. Let D be the length of the moving sidewalk.

Express D in terms of vBG (Joe’s speed relative to the ground): Solve for vBG:

Express D in terms of vBG + vSG (Sally’s speed relative to the ground):

Solve for vSG:

Express D in terms of vBG + 2vSB (Sally’s speed for a fast walk relative to the ground):

D = (2 min ) vBG

vBG =

D 2 min

D = (1 min )(vBG + vSG ) ⎛ D ⎞ = (1 min )⎜⎜ + vSG ⎟⎟ ⎝ 2 min ⎠

vSG =

D D D − = 1min 2 min 2 min

⎛ D 2D ⎞ ⎟⎟ + D = tf (vBG + 2vSB ) = tf ⎜⎜ ⎝ 2 min 2 min ⎠ ⎛ 3D ⎞ ⎟⎟ = tf ⎜⎜ ⎝ 2 min ⎠

Solve for tf as time for Sally's fast walk:

tf =

2 min = 40.0 s 3

60

Chapter 2

63 •• Picture the Problem The speed of Margaret’s boat relative to the riverbank ( vBR ) is the

sum or difference of the speed of her boat relative to the water ( vBW ) and the speed of the water relative to the riverbank ( vWR ), depending on whether she is heading with or against the current. Let D be the distance to the marina. Express the total time for the trip:

t tot = t1 + t 2

Express the times of travel with the motor running in terms of D, vWR

t1 =

and vBW :

and

t2 = Express the time required to drift distance D and solve for vWR :

t3 =

vBW

D = 4h − vWR

vBW

D + vWR

D = 8h vWR

and

From t1 = 4 h, find vBW :

Solve for t2:

Add t1 and t2 to find the total time:

vWR =

D 8h

vBW =

D D D 3D + vWR = + = 4h 4h 8h 8h

t2 =

vBW

D D = = 2h 3D D + vWR + 8h 8h

t tot = t1 + t 2 = 6 h

Acceleration 64 • Picture the Problem In part (a), we can apply the definition of average acceleration to find aav. In part (b), we can find the change in the car’s velocity in one second and add this change to its velocity at the beginning of the interval to find its speed one second later.

(a) Apply the definition of average acceleration:

∆v 80.5 km/h − 48.3 km/h = 3.7 s ∆t km = 8.70 h ⋅s

aav =

Motion in One Dimension Convert to m/s2:

m ⎞⎛ 1h ⎞ ⎛ ⎟ aav = ⎜ 8.70 × 103 ⎟⎜ h ⋅ s ⎠⎜⎝ 3600 s ⎟⎠ ⎝ = 2.42 m/s 2

(b) Express the speed of the car at the end of 4.7 s:

v(4.7 s ) = v(3.7 s ) + ∆v1s

Find the change in the speed of the car in 1 s:

km ⎞ ⎛ ∆v = aav ∆t = ⎜ 8.70 ⎟(1s ) h ⋅s ⎠ ⎝ = 8.70 km/h

Substitute and evaluate v(4.7 s):

v(4.7 s ) = 80.5 km/h + 8.7 km/h

= 80.5 km/h + ∆v1s

= 89.2 km/h 65 • Picture the Problem Average acceleration is defined as aav = ∆v/∆t.

The average acceleration is defined as the change in velocity divided by the change in time:

aav =

∆v (− 1m/s ) − (5m/s ) = (8s ) − (5s ) ∆t

= − 2.00 m/s 2 66 •• Picture the Problem The important concept here is the difference between average acceleration and instantaneous acceleration.

(a) The average acceleration is defined as the change in velocity divided by the change in time:

aav = ∆v/∆t

Determine v at t = 3 s, t = 4 s, and t = 5 s:

v(3 s) = 17 m/s v(4 s) = 25 m/s v(5 s) = 33 m/s

Find aav for the two 1-s intervals:

aav(3 s to 4 s) = (25 m/s – 17 m/s)/(1 s) = 8 m/s2 and aav(4 s to 5 s) = (33 m/s – 25 m/s)/(1 s) = 8 m/s2

The instantaneous acceleration is defined as the time derivative of the velocity or the slope of the velocityversus-time curve:

a=

dv = 8.00 m/s 2 dt

61

62

Chapter 2

(b) The given function was used to plot the following spreadsheet-graph of v-versus-t:

35 30 25

v (m/s)

20 15 10 5 0 -5 -10 0

1

2

3

4

5

t (s)

67 •• Picture the Problem We can closely approximate the instantaneous velocity by the average velocity in the limit as the time interval of the average becomes small. This is important because all we can ever obtain from any measurement is the average velocity, vav, which we use to approximate the instantaneous velocity v.

(a) Find x(4 s) and x(3 s):

x(4 s) = (4)2 – 5(4) + 1 = –3 m and x(3 s) = (3)2 – 5(3) + 1 = −5 m

Find ∆x:

∆x = x(4 s) – x(3 s) = (–3 m) – (–5 m) = 2m

Use the definition of average velocity:

vav = ∆x/∆t = (2 m)/(1 s) = 2 m/s

(b) Find x(t + ∆t):

x(t + ∆t) = (t + ∆t)2 − 5(t + ∆t) + 1 = (t2 + 2t∆t + (∆t)2) – 5(t + ∆t) + 1

Express x(t + ∆t) – x(t) = ∆x:

∆x =

(2t − 5)∆t + (∆t )2

where ∆x is in meters if t is in seconds. (c) From (b) find ∆x/∆t as ∆t → 0:

∆x (2t − 5)∆t + (∆t ) = ∆t ∆t = 2t − 5 + ∆t

2

and

Motion in One Dimension

63

v = lim ∆t →0 (∆x / ∆t ) = 2t − 5 where v is in m/s if t is in seconds. Alternatively, we can take the derivative of x(t) with respect to time to obtain the instantaneous velocity.

v(t ) = dx(t ) dt =

(

)

d at 2 + bt + 1 dt = 2at + b = 2t − 5

*68 •• Picture the Problem The instantaneous velocity is dx dt and the acceleration is dv dt .

Using the definitions of instantaneous velocity and acceleration, determine v and a:

v=

]

and

a=

Substitute numerical values for A and B and evaluate v and a:

[

dx d = At 2 − Bt + C = 2 At − B dt dt dv d = [2 At − B ] = 2 A dr dt

( ) (16 m/s ) t − 6m/s

v = 2 8m/s 2 t − 6 m/s = and

2

(

)

a = 2 8 m/s 2 = 16.0 m/s 2 69 •• Picture the Problem We can use the definition of average acceleration (aav = ∆v/∆t) to find aav for the three intervals of constant acceleration shown on the graph.

(a) Using the definition of average acceleration, find aav for the interval AB: Find aav for the interval BC:

Find aav for the interval CE:

aav, AB =

15 m/s − 5 m/s = 3.33 m/s 2 3s

aav, BC =

15 m/s − 15 m/s = 0 3s

aav, CE =

− 15 m/s − 15m/s = − 7.50m/s 2 4s

(b) Use the formulas for the areas of trapezoids and triangles to find the area under the graph of v as a function of t.

64

Chapter 2 ∆x = (∆x )A→B + (∆x )B→C + (∆x )C→D + (∆x )D→E =

1 2

(5 m/s + 15 m/s)(3 s ) + (15 m/s)(3 s) + 12 (15 m/s)(2 s) + 12 (−15 m/s)(2 s)

= 75.0 m (c) The graph of displacement, x, as a function of time, t, is shown in the following figure. In the region from B to C the velocity is constant so the x- versus-t curve is a straight line.

100

x (m)

80 60 40 20 0 0

2

4

6

8

10

t (s)

(d) Reading directly from the figure, we can find the time when the particle is moving the slowest.

At point D, t = 8 s, the graph crosses the time axis; therefore, v = 0.

Constant Acceleration and Free-Fall *70 • Picture the Problem Because the acceleration is constant (–g) we can use a constantacceleration equation to find the height of the projectile.

Using a constant-acceleration equation, express the height of the object as a function of its initial velocity, the acceleration due to gravity, and its displacement:

v 2 = v02 + 2a∆y

Solve for ∆ymax = h:

Because v(h) = 0,

h= From this expression for h we see that the maximum height attained is proportional to the square of the launch speed:

− v02 v2 = 0 2(− g ) 2 g

h ∝ v02

Motion in One Dimension Therefore, doubling the initial speed gives four times the height:

h2v 0

2 ( 2v0 ) =

and

(a ) is correct.

65

⎛ v2 ⎞ = 4⎜⎜ 0 ⎟⎟ = 4hv0 2g ⎝ 2g ⎠

71 • Picture the Problem Because the acceleration of the car is constant we can use constantacceleration equations to describe its motion.

(

)

(a) Uing a constant-acceleration equation, relate the velocity to the acceleration and the time:

v = v0 + at = 0 + 8 m s 2 (10 s )

(b) sing a constant-acceleration equation, relate the displacement to the acceleration and the time:

a ∆x = x − x0 = v0t + t 2 2

Substitute numerical values and evaluate ∆x:

∆x =

1 2 8 m s 2 (10 s ) = 400 m 2

vav =

∆x 400 m = = 40.0 m/s ∆t 10 s

(c) Use the definition of vav:

= 80.0 m s

(

)

Remarks: Because the area under a velocity-versus-time graph is the displacement of the object, we could solve this problem graphically. 72 • Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion.

Using a constant-acceleration equation, relate the velocity to the acceleration and the displacement: Solve for and evaluate the displacement:

v 2 = v02 + 2 a ∆x

∆x =

v 2 − v02 (152 − 52 )m 2 s 2 = 2a 2 (2 m s 2 )

= 50.0 m *73 • Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion.

Using a constant-acceleration equation, relate the velocity to the acceleration and the displacement:

v 2 = v02 + 2 a ∆x

66

Chapter 2

v 2 − v02 2 ∆x

Solve for the acceleration:

a=

Substitute numerical values and evaluate a:

(15 a=

2

)

− 102 m 2 s 2 = 15.6 m s 2 2(4 m )

74 • Picture the Problem Because the acceleration of the object is constant we can use constant-acceleration equations to describe its motion.

Using a constant-acceleration equation, relate the velocity to the acceleration and the displacement:

v 2 = v02 + 2 a ∆x

Solve for and evaluate v:

v=

(1 m s )2 + 2 (4 m

)

s 2 (1 m )

= 3.00 m/s Using the definition of average acceleration, solve for the time:

t=

∆v 3 m s − 1 m s = = 0.500 s aav 4 m s2

75 •• Picture the Problem In the absence of air resistance, the ball experiences constant acceleration. Choose a coordinate system with the origin at the point of release and the positive direction upward.

(a) Using a constant-acceleration equation, relate the displacement of the ball to the acceleration and the time: Setting ∆y = 0 (the displacement for a round trip), solve for and evaluate the time required for the ball to return to its starting position: (b) Using a constant-acceleration equation, relate the final speed of the ball to its initial speed, the acceleration, and its displacement: Solve for and evaluate H:

(c) Using the same constantacceleration equation with which we began part (a), express the displacement as a function of time:

∆y = v0t + 12 at 2

tround trip =

2v0 2(20 m/s ) = = 4.08 s g 9.81m/s 2

2 vtop = v02 + 2a∆y

or, because vtop = 0 and a = −g,

0 = v 02 + 2(− g )H

(20 m s ) = 20.4 m v02 = 2 g 2 9.81m s 2 2

H=

(

∆y = v0t + 12 at 2

)

Motion in One Dimension Substitute numerical values to obtain:

⎛ 9.81 m/s 2 ⎞ 2 ⎟⎟ t 15 m = (20 m/s )t − ⎜⎜ 2 ⎝ ⎠

Solve the quadratic equation for the times at which the displacement of the ball is 15 m:

The solutions are t = 0.991s

67

(this

corresponds to passing 15 m on the way up) and t = 3.09 s (this corresponds to passing 15 m on the way down).

76 •• Picture the Problem This is a multipart constant-acceleration problem using two different constant accelerations. We’ll choose a coordinate system in which downward is the positive direction and apply constant-acceleration equations to find the required times.

(a) Using a constant-acceleration equation, relate the time for the slide to the distance of fall and the acceleration: Solve for t1:

Substitute numerical values and evaluate t1:

∆y = y − y0 = h − 0 = v0t1 + 12 at12 or, because v0 = 0,

h = 12 at12 t1 =

2h g

t1 =

2(460 m ) = 9.68 s 9.81 m s 2

v1 = v0 + a1t1

(b) Using a constant-acceleration equation, relate the velocity at the bottom of the mountain to the acceleration and time:

or, because v0 = 0 and a1 = g,

Substitute numerical values and evaluate v1:

v1 = 9.81 m s 2 (9.68 s ) = 95.0 m s

(c) Using a constant-acceleration equation, relate the time required to stop the mass of rock and mud to its average speed and the distance it slides: Because the acceleration is constant:

Substitute to obtain:

v1 = gt1

(

)

∆t =

∆x vav

vav =

v1 + vf v1 + 0 v1 = = 2 2 2

∆t =

2∆x v1

68

Chapter 2

Substitute numerical values and evaluate ∆t:

∆t =

2(8000 m ) = 168 s 95.0 m s

*77 •• Picture the Problem In the absence of air resistance, the brick experiences constant acceleration and we can use constant-acceleration equations to describe its motion. Constant acceleration implies a parabolic position-versus-time curve.

y = y 0 + v 0 t + 12 (− g )t 2

(a) Using a constant-acceleration equation, relate the position of the brick to its initial position, initial velocity, acceleration, and time into its fall:

(

)

= 6 m + (5 m s ) t − 4.91 m s 2 t 2

(

)

The following graph of y = 6 m + (5 m s )t − 4.91 m s 2 t 2 was plotted using a spreadsheet program: 8 7 6

y (m)

5 4 3 2 1 0 0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

t (s)

(b) Relate the greatest height reached by the brick to its height when it falls off the load and the additional height it rises ∆ymax:

h = y0 + ∆ymax

Using a constant-acceleration equation, relate the height reached by the brick to its acceleration and initial velocity:

2 vtop = v02 + 2(− g )∆ymax

Solve for ∆ymax:

Substitute numerical values and evaluate ∆ymax:

or, because vtop = 0, 0 = v02 + 2(− g )∆ymax ∆ymax =

∆ymax =

v02 2g

(5 m s ) 2

(

2 9.81 m s 2

) = 1.27 m

Motion in One Dimension Substitute to obtain:

69

h = y0 + ∆ymax = 6 m + 1.27 m = 7.27 m

Note: The graph shown above confirms this result. (c) Using the quadratic formula, solve for t in the equation obtained in part (a):

⎛−g⎞ − v0 ± v02 − 4⎜ ⎟(− ∆y ) ⎝ 2 ⎠ t= ⎛−g⎞ 2⎜ ⎟ ⎝ 2 ⎠ ⎛ v ⎞⎛ 2 g (∆y ) ⎞⎟ = ⎜⎜ 0 ⎟⎟⎜⎜1 ± 1 − v02 ⎟⎠ ⎝ g ⎠⎝

With ybottom = 0 and yo = 6 m or ∆y = –6 m, we obtain:

t = 1.73 s and t = –0.708 s. Note: The

second solution is nonphysical.

(d) Using a constant-acceleration equation, relate the speed of the brick on impact to its acceleration and displacement, and solve for its speed:

v = 2 gh

Substitute numerical values and evaluate v:

v = 2 9.81 m s 2 (7.27 m ) = 11.9 m s

(

)

78 •• Picture the Problem In the absence of air resistance, the acceleration of the bolt is constant. Choose a coordinate system in which upward is positive and the origin is at the bottom of the shaft (y = 0).

(a) Using a constant-acceleration equation, relate the position of the bolt to its initial position, initial velocity, and fall time:

ybottom = 0

Solve for the position of the bolt when it came loose:

y0 = −v0t + 12 gt 2

Substitute numerical values and evaluate y0:

y0 = −(6 m s )(3 s ) + 12 (9.81 m s 2 )(3 s )

(b) Using a constant-acceleration equation, relate the speed of the bolt to its initial speed, acceleration, and fall time:

v = v0 + at

= y0 + v0t + 12 (− g )t 2

2

= 26.1 m

70

Chapter 2

Substitute numerical values and evaluate v :

(

)

v = 6 m s − 9.81 m s 2 (3s ) = −23.4 m s and

v = 23.4 m s *79 •• Picture the Problem In the absence of air resistance, the object’s acceleration is constant. Choose a coordinate system in which downward is positive and the origin is at the point of release. In this coordinate system, a = g and y = 120 m at the bottom of the fall.

Express the distance fallen in the last second in terms of the object’s position at impact and its position 1 s before impact:

∆ylast second = 120 m − y1s before impact (1)

Using a constant-acceleration equation, relate the object’s position upon impact to its initial position, initial velocity, and fall time:

y = y0 + v0t + 12 gt 2

Solve for the fall time:

Substitute numerical values and evaluate tfall:

or, because y0 = 0 and v0 = 0, 2 y = 12 gtfall

tfall =

2y g

tfall =

2(120 m ) = 4.95 s 9.81 m/s 2

(9.81 m/s )(3.95 s )

We know that, one second before impact, the object has fallen for 3.95 s. Using the same constantacceleration equation, calculate the object’s position 3.95 s into its fall:

y (3.95 s) =

Substitute in equation (1) to obtain:

∆ylast second = 120 m − 76.4 m = 43.6 m

1 2

2

2

= 76.4 m

80 •• Picture the Problem In the absence of air resistance, the acceleration of the object is constant. Choose a coordinate system with the origin at the point of release and downward as the positive direction.

Using a constant-acceleration equation, relate the height to the initial and final velocities and the acceleration; solve for the height:

vf2 = v02 + 2a∆y or, because v0 = 0,

h=

vf2 2g

(1)

Motion in One Dimension Using the definition of average velocity, find the average velocity of the object during its final second of fall:

vav =

vf -1s + vf 2

=

∆y 38 m = = 38 m s ∆t 1s

Express the sum of the final velocity and the velocity 1 s before impact:

vf -1s + vf = 2(38 m s ) = 76 m s

From the definition of acceleration, we know that the change in velocity of the object, during 1 s of fall, is 9.81 m/s:

∆v = vf − vf -1s = 9.81 m s

Add the equations that express the sum and difference of vf – 1 s and vf and solve for vf: Substitute in equation (1) and evaluate h:

76 m s + 9.81m s = 42.9 m s 2

vf =

h=

71

(42.9 m s )2

(

2 9.81 m s 2

)=

93.8 m

*81 • Picture the Problem In the absence of air resistance, the acceleration of the stone is constant. Choose a coordinate system with the origin at the bottom of the trajectory and the upward direction positive. Let vf -1 2 be the speed one-half second before impact

and vf the speed at impact. Using a constant-acceleration equation, express the final speed of the stone in terms of its initial speed, acceleration, and displacement:

vf2 = v02 + 2a∆y

Solve for the initial speed of the stone:

v0 = vf2 + 2 g∆y

Find the average speed in the last half second:

vav =

vf -1 2 + vf

2 = 90 m s

=

(1)

∆xlast half second 45 m = 0.5 s ∆t

and

vf -1 2 + vf = 2(90 m s ) = 180 m s

Using a constant-acceleration equation, express the change in speed of the stone in the last half second in terms of the acceleration and the elapsed time; solve for the change in its speed:

∆v = vf − vf -1 2 = g∆t

(

)

= 9.81 m s 2 (0.5 s ) = 4.91 m s

72

Chapter 2

Add the equations that express the sum and difference of vf – ½ and vf and solve for vf:

vf =

Substitute in equation (1) and evaluate v0:

v0 =

180 m s + 4.91m s = 92.5 m s 2

(92.5m s )2 + 2(9.81m

)

s 2 (− 200m )

= 68.1 m s

Remarks: The stone may be thrown either up or down from the cliff and the results after it passes the cliff on the way down are the same. 82 •• Picture the Problem In the absence of air resistance, the acceleration of the object is constant. Choose a coordinate system in which downward is the positive direction and the object starts from rest. Apply constant-acceleration equations to find the average velocity of the object during its descent.

Express the average velocity of the falling object in terms of its initial and final velocities: Using a constant-acceleration equation, express the displacement of the object during the 1st second in terms of its acceleration and the elapsed time:

vav =

v0 + vf 2

∆y1st second

gt 2 = = 4.91 m = 0.4 h 2

Solve for the displacement to obtain: Using a constant-acceleration equation, express the final velocity of the object in terms of its initial velocity, acceleration, and displacement:

h = 12.3 m

Substitute numerical values and evaluate the final velocity of the object:

vf = 2 9.81m s 2 (12.3 m ) = 15.5 m s

Substitute in the equation for the average velocity to obtain:

vf2 = v02 + 2 g∆y or, because v0 = 0,

vf = 2 g∆y

(

vav =

)

0 + 15.5 m s = 7.77 m s 2

83 •• Picture the Problem This is a three-part constant-acceleration problem. The bus starts from rest and accelerates for a given period of time, and then it travels at a constant velocity for another period of time, and, finally, decelerates uniformly to a stop. The pictorial representation will help us organize the information in the problem and develop our solution strategy.

Motion in One Dimension

73

(a) Express the total displacement of the bus during the three intervals of time.

∆xtotal = ∆x(0 → 12 s ) + ∆x(12 s → 37 s )

Using a constant-acceleration equation, express the displacement of the bus during its first 12 s of motion in terms of its initial velocity, acceleration, and the elapsed time; solve for its displacement:

∆x(0 → 12 s ) = v0t + 12 at 2

+ ∆x(37 s → end )

or, because v0 = 0,

∆x(0 → 12 s ) = 12 at 2 = 108 m

(

)

Using a constant-acceleration equation, express the velocity of the bus after 12 seconds in terms of its initial velocity, acceleration, and the elapsed time; solve for its velocity at the end of 12 s:

v12 s = v0 + a0→12 s ∆t = 1.5 m/s 2 (12 s )

During the next 25 s, the bus moves with a constant velocity. Using the definition of average velocity, express the displacement of the bus during this interval in terms of its average (constant) velocity and the elapsed time:

∆x(12 s → 37 s ) = v12 s ∆t = (18 m/s )(25 s )

Because the bus slows down at the same rate that its velocity increased during the first 12 s of motion, we can conclude that its displacement during this braking period is the same as during its acceleration period and the time to brake to a stop is equal to the time that was required for the bus to accelerate to its cruising speed of 18 m/s. Hence:

∆x(37 s → 49s ) = 108 m

Add the displacements to find the distance the bus traveled:

∆x total = 108 m + 450 m + 108 m

= 18 m/s

= 450 m

= 666 m

74

Chapter 2

(b) Use the definition of average velocity to calculate the average velocity of the bus during this trip:

vav =

∆xtotal 666 m = = 13.6 m s ∆t 49 s

Remarks: One can also solve this problem graphically. Recall that the area under a velocity as a function-of-time graph equals the displacement of the moving object. *84 •• Picture the Problem While we can solve this problem analytically, there are many physical situations in which it is not easy to do so and one has to rely on numerical methods; for example, see the spreadsheet solution shown below. Because we’re neglecting the height of the release point, the position of the ball as a function of time is given by y = v0t − 12 gt 2 . The formulas used to calculate the quantities in the columns are as follows:

Cell Content/Formula Algebraic Form B1 20 v0 B2 9.81 g B5 0 t B6 B5 + 0.1 t + ∆t C6 $B$1*B6 − 0.5*$B$2*B6^2 v0t − 1 gt 2 2

(a) A 1 2 3 4 5 6 7 44 45 46

B

t (s) 0.0 0.1 0.2

C m/s m/s^2 height (m) 0.00 1.95 3.80

3.9 4.0 4.1

3.39 1.52 −0.45

v0 = 20 g = 9.81

The graph shown below was generated from the data in the previous table. Note that the maximum height reached is a little more than 20 m and the time of flight is about 4 s.

Motion in One Dimension

75

25

height (m)

20

15

10

5

0 0

1

2

3

4

t (s )

(b) In the spreadsheet, change the value in cell B1 from 20 to 10. The graph should automatically update. With an initial velocity of 10 m/s, the maximum height achieved is approximately 5 m and the time-of-flight is approximately 2 s.

6

5

height (m)

4

3

2

1

0 0.0

0.5

1.0

1.5

2.0

t (s)

*85 •• Picture the Problem Because the accelerations of both Al and Bert are constant, constant-acceleration equations can be used to describe their motions. Choose the origin of the coordinate system to be where Al decides to begin his sprint.

(a) Using a constant-acceleration equation, relate Al's initial velocity, his acceleration, and the time to reach the end of the trail to his

∆x = v0t + 12 at 2

76

Chapter 2

displacement in reaching the end of the trail: Substitute numerical values to obtain:

35 m = (0.75 m/s)t + 12 (0.5 m/s 2 )t 2

Solve for the time required for Al to reach the end of the trail:

t = 10.4 s

(b) Using constant-acceleration equations, express the positions of Bert and Al as functions of time. At the instant Al turns around at the end of the trail, t = 0. Also, x = 0 at a point 35 m from the end of the trail:

x Bert = x Bert,0 + (0 .75 m/s ) t and

xAl = xAl,0 − (0.85 m/s ) t

= 35 m − (0.85 m/s ) t

Calculate Bert’s position at t = 0. At that time he has been running for 10.4 s:

xBert,0 = (0.75 m/s )(10.4 s ) = 7.80 m

Because Bert and Al will be at the same location when they meet, equate their position functions and solve for t:

7.80 m + (0.75 m/s )t = 35 m − (0.85 m/s )t and

t = 17.0 s

To determine the elapsed time from when Al began his accelerated run, we need to add 10.4 s to this time:

tstart = 17.0 s + 10.4 s = 27.4 s

(c) Express Bert’s distance from the end of the trail when he and Al meet:

d end of trail = 35 m − xBert,0

Substitute numerical values and evaluate dend of trail:

d end of trail = 35 m − 7.80 m

− d Bert runs until he meets Al − (17 s) (0.75 m/s )

= 14.5 m 86 •• Picture the Problem Generate two curves on one graph with the first curve representing Al's position as a function of time and the second curve representing Bert’s position as a function of time. Al’s position, as he runs toward the end of the trail, is given by xAl = v0t + 12 aAlt 2 and Bert’s position by xBert = x0, Bert + vBertt . Al’s position, once he’s

reached the end of the trail and is running back toward Bert, is given by xAl = xAl,0 + vAl (t − 10.5 s ) . The coordinates of the intersection of the two curves give the time and place where they meet. A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell

Content/Formula

Algebraic Form

Motion in One Dimension B1 B2 B3 B10 C10

0.75 0.50 −0.85 B9 + 0.25 $B$1*B10 + 0.5*$B$2*B10^2

C52

$C$51 + $B$3*(B52 − $B$51)

xAl,0 + vAl (t − 10.5 s )

F10

$F$9 + $B$1*B10

x0, Bert + vBertt

77

v0 aAl t t + ∆t

v0t + 12 aAlt 2

(b) and (c) 1 2 3 4 5 6 7 8 9 10 11

A B v0 = 0.75 a(Al) = 0.5 v(Al) = −0.85

C m/s m/s^2 m/s

D

E

F

t (s)

x (m)

x (m)

0.00 0.25 0.50

Al 0.00 0.20 0.44

Bert 0.00 0.19 0.38

49 50 51 52 53 54 55 56

10.00 10.25 10.50 10.75 11.00 11.25 11.50 11.75

32.50 33.95 35.44 35.23 35.01 34.80 34.59 34.38

7.50 7.69 7.88 8.06 8.25 8.44 8.63 8.81

119 120 121 122

27.50 27.75 28.00 28.25

20.99 20.78 20.56 20.35

20.63 20.81 21.00 21.19

127 128 129

29.50 29.75 30.00

19.29 19.08 18.86

22.13 22.31 22.50

*Al reaches end of trail and starts back toward Bert

The graph shown below was generated from the spreadsheet; the positions of both Al and Bert were calculated as functions of time. The dashed curve shows Al’s position as a function of time for the two parts of his motion. The solid line that is linear from the origin shows Bert’s position as a function of time.

78

Chapter 2 40

Position on trail (m)

35 30 25

Al

20

Bert

15 10 5 0 0

5

10

15

20

25

30

t (s)

Note that the spreadsheet and the graph (constructed from the spreadsheet data) confirm the results in Problem 85 by showing Al and Bert meeting at about 14.5 m from the end of the trail after an elapsed time of approximately 28 s. 87 •• Picture the Problem This is a two-part constant-acceleration problem. Choose a coordinate system in which the upward direction is positive. The pictorial representation will help us organize the information in the problem and develop our solution strategy.

(a) Express the highest point the rocket reaches, h, as the sum of its displacements during the first two stages of its flight:

h = ∆x1st stage + ∆x 2nd stage

Using a constant-acceleration equation, express the altitude reached in the first stage in terms of the rocket’s initial velocity, acceleration, and burn time; solve for the first stage altitude:

x1st stage = x0 + v0t + 12 a1st staget 2

Using a constant-acceleration equation, express the velocity of the rocket at the end of its first stage in terms of its initial velocity, acceleration, and displacement; calculate its end-of-first-stage velocity:

v1st stage = v0 + a1st staget

= 12 (20 m/s 2 )(25 s) 2 = 6250 m

= (20 m/s 2 )(25 s) = 500 m/s

Motion in One Dimension Using a constant-acceleration equation, express the final velocity of the rocket during the remainder of its climb in terms of its shut-off velocity, free-fall acceleration, and displacement; solve for its displacement:

79

2 2 vhighest point = vshutoff + 2a2 nd stage ∆y2nd stage

and, because vhighest point = 0,

2 − vshutoff (500 m s )2 = − 2g 2(9.81 m s 2 )

∆y 2 nd stage =

= 1.2742 ×10 4 m

Substitute in the expression for the total height to obtain:

h = 6250 m + 1.27 × 10 4 m = 19.0 km

(b) Express the total time the rocket is in the air in terms of the three segments of its flight:

∆ttotal = ∆tpowered climb + ∆t2nd segment + ∆tdescent

Express ∆t2nd segment in terms of the rocket’s displacement and average velocity: Substitute numerical values and evaluate ∆t2nd segment: Using a constant-acceleration equation, relate the fall distance to the descent time: Solve for ∆tdescent:

Substitute numerical values and evaluate ∆tdescent:

= 25 s + ∆t2nd segment + ∆tdescent

∆t2nd segment =

Displacement Average velocity

1.2742 × 10 4 m ∆t2nd segment = = 50.97 s ⎛ 0 + 500 m/s ⎞ ⎟ ⎜ 2 ⎝ ⎠ 2 1 ∆y = v0t + 2 g (∆tdescent ) or, because v0 = 0,

∆y = 12 g (∆tdescent )

2

∆tdescent =

∆t descent

2∆y g

(

)

2 1.90 × 10 4 m = = 62.2 s 9.81 m/s 2

Substitute and calculate the total time the rocket is in the air:

∆t = 25 s + 50.97 s + 62.2 s = 138 s

(c) Using a constant-acceleration equation, express the impact velocity of the rocket in terms of its initial downward velocity, acceleration under free-fall, and time of descent; solve for its impact velocity:

vimpact = v0 + g∆tdescent

= 2 min 18 s

and, because v0 = 0,

(

)

vimpact = g∆t = 9.81 m/s 2 (62.2 s ) = 610 m/s

80

Chapter 2

88

••

Picture the Problem In the absence of air resistance, the acceleration of the flowerpot is constant. Choose a coordinate system in which downward is positive and the origin is at the point from which the flowerpot fell. Let t = time when the pot is at the top of the window, and t + ∆t the time when the pot is at the bottom of the window. To find the distance from the ledge to the top of the window, first find the time ttop that it takes the pot to fall to the top of the window. Using a constant-acceleration equation, express the distance y below the ledge from which the pot fell as a function of time: Express the position of the pot as it reaches the top of the window: Express the position of the pot as it reaches the bottom of the window: Subtract ybottom from ytop to obtain an expression for the displacement ∆ywindow of the pot as it passes the window: Solve for ttop:

y = y0 + v0t + 12 at 2 Since a = g and v0 = y0 = 0,

y = 12 gt 2 2 ytop = 12 gt top

ybottom = 12 g (t top + ∆t window ) where ∆twindow = t top − tbottom

2

=

t top Substitute numerical values and evaluate ttop:

t top Substitute this value for ttop to obtain the distance from the ledge to the top of the window:

[ g [2t

2 ∆ywindow = 12 g (t top + ∆t window ) − t top 1 2

2

top

]

∆t window + (∆t window )

]

2∆ywindow 2 − (∆t window ) g = 2∆t window

2(4 m ) 2 − (0.2 s ) 2 9.81 m/s = = 1.839 s 2(0.2 s )

ytop = 12 (9.81 m/s 2 )(1.939 s) 2 = 18.4 m

*89 •• Picture the Problem The acceleration of the glider on the air track is constant. Its average acceleration is equal to the instantaneous (constant) acceleration. Choose a coordinate system in which the initial direction of the glider’s motion is the positive direction.

Using the definition of acceleration, express the average acceleration of the glider in terms of the glider’s velocity change and the elapsed time:

2

a = aav =

∆v ∆t

Motion in One Dimension Using a constant-acceleration equation, express the average velocity of the glider in terms of the displacement of the glider and the elapsed time: Solve for and evaluate the initial velocity:

vav =

∆x v0 + v = 2 ∆t

v0 =

2∆x 2(100 cm ) −v = − (−15 cm/s) ∆t 8s

= 40.0 cm/s Substitute this value of v0 and evaluate the average acceleration of the glider:

a=

− 15 cm/s − (40 cm/s) 8s

= − 6.88 cm/s 2

90 •• Picture the Problem In the absence of air resistance, the acceleration of the rock is constant and its motion can be described using the constant-acceleration equations. Choose a coordinate system in which the downward direction is positive and let the height of the cliff, which equals the displacement of the rock, be represented by h.

Using a constant-acceleration equation, express the height h of the cliff in terms of the initial velocity of the rock, acceleration, and time of fall:

81

∆y = v0t + 12 at 2 or, because v0 = 0, a = g, and ∆y = h,

h = 12 gt 2

Using this equation, express the displacement of the rock during the a) first two-thirds of its fall, and

2 3

b) its complete fall in terms of the time required for it to fall this distance.

h = 12 g (t + 1s )

Substitute equation (2) in equation (1) to obtain a quadratic equation in t:

t2 – (4 s)t – 2 s2 = 0

Solve for the positive root:

t = 4.45 s

Evaluate ∆t = t + 1 s:

∆t = 4.45 s + 1 s = 5.45 s

Substitute numerical values in equation (2) and evaluate h:

h=

h = 12 gt 2

(1) 2

1 2

(2)

(9.81m/s )(5.45 s ) 2

2

= 146 m

82

Chapter 2

91 ••• Picture the Problem Assume that the acceleration of the car is constant. The total distance the car travels while stopping is the sum of the distances it travels during the driver’s reaction time and the time it travels while braking. Choose a coordinate system in which the positive direction is the direction of motion of the automobile and apply a constant-acceleration equation to obtain a quadratic equation in the car’s initial speed v0.

(a) Using a constant-acceleration equation, relate the velocity of the car to its initial velocity, acceleration, and displacement during braking: Solve for the distance traveled during braking:

v 2 = v02 + 2a∆xbrk or, because the final velocity is zero,

0 = v02 + 2a∆x brk

∆xbrk = −

v02 2a

Express the total distance traveled by the car as the sum of the distance traveled during the reaction time and the distance traveled while slowing down:

∆xtot = ∆xreact + ∆xbrk

Rearrange this quadratic equation to obtain:

v02 − 2a∆treact v0 + 2a∆xtot = 0

Substitute numerical values and simplify to obtain:

v02 − 2 − 7 m/s 2 (0.5 s )v0

v02 = v0 ∆treact − 2a

(

) + 2(− 7 m/s )(4 m) = 0 2

or

v02 + (7 m/s )v0 − 56 m 2 / s 2 = 0

Solve the quadratic equation for the positive root to obtain:

v0 = 4.7613558 m/s

Convert this speed to mi/h::

⎛ 1 mi/h ⎞ ⎟⎟ v 0 = (4.7613558 m/s ) ⎜⎜ ⎝ 0.477 m/s ⎠ = 10.7 mi/h

(b) Find the reaction-time distance:

∆xreact = v0 ∆treact = (4.76 m/s)(0.5 s) = 2.38 m

Express and evaluate the ratio of the reaction distance to the total distance:

∆xreact 2.38 m = = 0.595 ∆xtot 4m

Motion in One Dimension

83

92 •• Picture the Problem Assume that the accelerations of the trains are constant. Choose a coordinate system in which the direction of the motion of the train on the left is the positive direction. Take xo = 0 as the position of the train on the left at t = 0.

( = (0.7 m s )t

)

Using a constant-acceleration equation, relate the distance the train on the left will travel before the trains pass to its acceleration and the time-to-passing: Using a constant-acceleration equation, relate the position of the train on the right to its initial velocity, position, and acceleration:

xL = 12 aLt 2 = 12 1.4m s 2 t 2

Equate xL and xR and solve for t:

0.7t 2 = 40 − 1.1t 2 and t = 4.71s

Find the position of the train initially on the left, xL, as they pass:

xL = 12 (1.4 m/s 2 )(4.71 s) 2 = 15.6 m

2

2

xR = 40 m − 12 aR t 2

(

)

= 40 m − 12 2.2m s 2 t 2

Remarks: One can also solve this problem by graphing the functions for xL and xR. The coordinates of the intersection of the two curves give one the time-to-passing and the distance traveled by the train on the left.

93 •• Picture the Problem In the absence of air resistance, the acceleration of the stones is constant. Choose a coordinate system in which the downward direction is positive and the origin is at the point of release of the stones.

Using constant-acceleration equations, relate the positions of the two stones to their initial positions, accelerations, and time-of-fall:

x1 = 12 gt 2 and

x2 = 12 g (t − 1.6 s) 2

Express the difference between x1 and x2:

x1 − x2 = 36 m

Substitute for x1 and x2 to obtain:

36 m = 12 gt 2 − 12 g (t − 1.6 s )

2

84

Chapter 2

Solve this equation for the time t at which the stones will be separated by 36 m:

t = 3.09 s

Substitute this result in the expression for x2 and solve for x2:

x2 =

1 2

(9.81 m s )(3.09 s − 1.6 s) 2

2

= 10.9 m

*94 •• Picture the Problem The acceleration of the police officer’s car is positive and constant and the acceleration of the speeder’s car is zero. Choose a coordinate system such that the direction of motion of the two vehicles is the positive direction and the origin is at the stop sign.

Express the velocity of the car in terms of the distance it will travel until the police officer catches up to it and the time that will elapse during this chase:

vcar =

d caught t car

Letting t1 be the time during which she accelerates and t2 the time of travel at v1 = 110 km/h, express the time of travel of the police officer:

tofficer = t1 + t2

Convert 110 km/h into m/s:

v1 = (110 km/h )(103 m/km) (1 h/3600 s) = 30.6 m/s

Express and evaluate t1:

t1 =

v1 amotorcycle

=

30.6 m/s = 4.94 s 6.2 m/s 2

Express and evaluate d1:

d1 = 12 v1t1 = 12 (30.6 m/s)(4.94 s) = 75.6 m

Determine d2:

d 2 = d caught − d1 = 1400 m − 75.6 m = 1324.4 m

Express and evaluate t2:

Express the time of travel of the car:

t2 =

d 2 1324.4 m = = 43.3 s v1 30.6 m/s

tcar = 2.0 s + 4.93 s + 43.3 s = 50.2 s

Motion in One Dimension Finally, find the speed of the car:

vcar =

d caught tcar

=

85

1400 m = 27.9 m/s 50.2 s

⎛ 1 mi/h ⎞ ⎟⎟ = (27.9 m/s )⎜⎜ ⎝ 0.447 m/s ⎠ = 62.4 mi/h 95 •• Picture the Problem In the absence of air resistance, the acceleration of the stone is constant. Choose a coordinate system in which downward is positive and the origin is at the point of release of the stone and apply constant-acceleration equations.

Using a constant-acceleration equation, express the height of the cliff in terms of the initial position of the stones, acceleration due to gravity, and time for the first stone to hit the water:

h = 12 gt12

Express the displacement of the second stone when it hits the water in terms of its initial velocity, acceleration, and time required for it to hit the water.

d 2 = v02t2 + 12 gt22

Because the stones will travel the same distances before hitting the water, equate h and d2 and solve for t.

where t2 = t1 – 1.6 s.

gt12 = v02t2 + 12 gt22

1 2

or 1 2

(9.81m/s )t = (32m/s)(t − 1.6s ) + (9.81m/s ) (t − 1.6s ) 2

2 1

1

1 2

2

2

1

Solve for t1 to obtain:

t1 = 2.37 s

Substitute for t1 and evaluate h:

h = 12 (9.81 m/s 2 )(2.37 s) 2 = 27.6 m

96 ••• Picture the Problem Assume that the acceleration of the passenger train is constant. Let xp = 0 be the location of the passenger train engine at the moment of sighting the freight train’s end; let t = 0 be the instant the passenger train begins to slow (0.4 s after the passenger train engineer sees the freight train ahead). Choose a coordinate system in which the direction of motion of the trains is the positive direction and use constantacceleration equations to express the positions of the trains in terms of their initial positions, speeds, accelerations, and elapsed time.

(a) Using constant-acceleration equations, write expressions for the positions of the front of the passenger train and the rear of the

xp = (29 m/s )(t + 0.4 s ) − 12 at 2

xf = (360 m ) + (6 m/s )(t + 0.4 s) where xp and xf are in meters if t is in

86

Chapter 2

freight train, xp and xf, respectively:

seconds.

at 2 − (23 m/s )t + 350.8 m = 0

Equate xf = xp to obtain an equation for t:

1 2

Find the discriminant D = B2 − 4AC of this equation:

⎛a⎞ 2 D = (23 m/s ) − 4⎜ ⎟(350.8 m ) ⎝2⎠

The equation must have real roots if it is to describe a collision. The necessary condition for real roots is that the discriminant be greater than or equal to zero:

If (23 m/s)2 – a (701.6 m) ≥ 0, then

(b) Express the relative speed of the trains:

vrel = vpf = vp − vf

Repeat the previous steps with a = 0.754 m/s2 and a 0.8 s reaction time. The quadratic equation that guarantees real roots with the longer reaction time is:

1 2

(

)

a ≤ 0.754 m/s 2

(1)

(0.754 m/s )t − (23 m/s)t 2

2

+ 341.6 m = 0

Solve for t to obtain the collision times:

t = 25.6 s and t = 35.4 s

Note that at t = 35.4 s, the trains have already collided; therefore this root is not a meaningful solution to our problem.

Note: In the graph shown below, you will see why we keep only the smaller of the two solutions.

Now we can substitute our value for t in the constant-acceleration equation for the passenger train and solve for the distance the train has moved prior to the collision:

xp = (29 m/s)(25.6 s + 0.8 s) – ( 0.377 m/s2)(25.6 s)2 = 518 m

Find the speeds of the two trains:

vp = vop + at = (29 m/s) + (–0.754 m/s2)(25.5 s) = 9.77 m/s and vf = vof = 6 m/s

Substitute in equation (1) and evaluate the relative speed of the trains:

vrel = 9.77 m/s - 6.00 m/s = 3.77 m/s

The graph shows the location of both trains as functions of time. The solid straight line is for the constant velocity freight train; the dashed curves are for the passenger train, with reaction times of 0.4 s for the lower curve and 0.8 s for the upper curve.

Motion in One Dimension

87

700 600

x (m)

500 400 0.4 s reaction time

300

Freight train 200

0.8 s reaction time

100 0 0

10

20

30

40

t (s)

Remarks: A collision occurs the first time the curve for the passenger train crosses the curve for the freight train. The smaller of two solutions will always give the time of the collision. 97 • Picture the Problem In the absence of air resistance, the acceleration of an object near the surface of the earth is constant. Choose a coordinate system in which the upward direction is positive and the origin is at the surface of the earth and apply constantacceleration equations.

Using a constant-acceleration equation, relate the velocity to the acceleration and displacement:

or, because v = 0 and a = −g,

Solve for the height to which the projectile will rise:

h = ∆y =

Substitute numerical values and evaluate h:

v 2 = v02 + 2a∆y 0 = v02 − 2 g∆y

h=

v02 2g

(300 m/s)2

(

2 9.81 m/s 2

)=

4.59 km

*98 • Picture the Problem This is a composite of two constant accelerations with the acceleration equal to one constant prior to the elevator hitting the roof, and equal to a different constant after crashing through it. Choose a coordinate system in which the upward direction is positive and apply constant-acceleration equations.

(a) Using a constant-acceleration equation, relate the velocity to the acceleration and displacement:

v 2 = v02 + 2 a ∆y or, because v = 0 and a = −g,

0 = v02 − 2 g ∆y

88

Chapter 2

Solve for v0:

v0 = 2 g∆y

Substitute numerical values and evaluate v0:

v0 = 2 9.81 m/s 2 10 4 m = 443 m/s

(b) Find the velocity of the elevator just before it crashed through the roof:

vf = 2 × 443 m/s = 886 m/s

Using the same constantacceleration equation, this time with v0 = 0, solve for the acceleration:

v 2 = 2a∆y

Substitute numerical values and evaluate a:

2 ( 886 m/s ) a= 2(150 m )

(

)(

)

= 2.62 × 103 m/s 2

= 267 g 99 •• Picture the Problem Choose a coordinate system in which the upward direction is positive. We can use a constant-acceleration equation to find the beetle’s velocity as its feet lose contact with the ground and then use this velocity to calculate the height of its jump.

Using a constant-acceleration equation, relate the beetle’s maximum height to its launch velocity, velocity at the top of its trajectory, and acceleration once it is airborne; solve for its maximum height: Because vhighest point = 0:

Now, in order to determine the beetle’s launch velocity, relate its time of contact with the ground to its acceleration and push-off distance: Substitute numerical values and 2 evaluate vlaunch : Substitute to find the height to which the beetle can jump:

2 2 vhighest point = vlaunch + 2a∆yfree fall 2 = vlaunch + 2(− g )h

h=

2 vlaunch 2g

2 vlaunch = v02 + 2a∆ylaunch

or, because v0 = 0, 2 vlaunch = 2a∆ylaunch

(

)(

2 = 2(400 ) 9.81 m/s 2 0.6 × 10 −2 m vlaunch

= 47.1 m 2 /s 2

h=

2 vlaunch 47.1m 2 /s 2 = = 2.40 m 2g 2 9.81m/s2

(

)

)

Motion in One Dimension Using a constant-acceleration equation, relate the velocity of the beetle at its maximum height to its launch velocity, free-fall acceleration while in the air, and time-to-maximum height: Solve for tmax height:

For zero displacement and constant acceleration, the time-of-flight is twice the time-to-maximum height:

89

v = v0 + at or vmax. height = vlaunch − gtmax. height and, because vmax height = 0,

0 = vlaunch − gtmax. height

tmax height =

vlaunch g

tflight = 2tmax. height = =

2vlaunch g

2(6.86 m/s ) = 1.40 s 9.81 m/s 2

100 • Picture the Problem Because its acceleration is constant we can use the constantacceleration equations to describe the motion of the automobile.

Using a constant-acceleration equation, relate the velocity to the acceleration and displacement:

v 2 = v02 + 2 a ∆x or, because v = 0,

0 = v02 + 2 a ∆x

Solve for the acceleration a:

− v02 a= 2∆x

Substitute numerical values and evaluate a:

− (98 km h ) (103 m km )(1 h 3600 s ) a= 2(50m )

[

]

= − 7.41 m s 2 Express the ratio of a to g and then solve for a:

a − 7.41m s 2 = = −0.755 g 9.81m s 2 and a = − 0.755 g

Using the definition of average acceleration, solve for the stopping time:

aav =

Substitute numerical values and evaluate ∆t:

∆t =

∆v ∆v ⇒ ∆t = aav ∆t

(− 98 km h )(103 m km )(1h

= 3.67 s

− 7.41m s

2

3600 s )

2

90

Chapter 2

*101 •• Picture the Problem In the absence of air resistance, the puck experiences constant acceleration and we can use constant-acceleration equations to describe its position as a function of time. Choose a coordinate system in which downward is positive, the particle starts from rest (vo = 0), and the starting height is zero (y0 = 0).

Using a constant-acceleration equation, relate the position of the falling puck to the acceleration and the time. Evaluate the y-position at successive equal time intervals ∆t, 2∆t, 3∆t, etc:

−g (∆t )2 = − g (∆t )2 2 2 −g (2∆t )2 = − g (4)(∆t )2 y2 = 2 2 −g (3∆t )2 = − g (9)(∆t )2 y3 = 2 2 −g (4∆t )2 = − g (16)(∆t )2 y4 = 2 2 etc.

Evaluate the changes in those positions in each time interval:

⎛−g⎞ 2 ∆y10 = y1 − 0 = ⎜ ⎟ (∆t ) ⎝ 2 ⎠ ⎛−g⎞ 2 ∆y21 = y2 − y1 = 3⎜ ⎟(∆t ) = 3∆y10 ⎝ 2 ⎠ ⎛−g⎞ 2 ∆y32 = y3 − y2 = 5⎜ ⎟(∆t ) = 5∆y10 ⎝ 2 ⎠ ⎛−g⎞ 2 ∆y43 = y4 − y3 = 7⎜ ⎟(∆t ) = 7 ∆y10 ⎝ 2 ⎠ etc.

y1 =

102 •• Picture the Problem Because the particle moves with a constant acceleration we can use the constant-acceleration equations to describe its motion. A pictorial representation will help us organize the information in the problem and develop our solution strategy.

Using a constant-acceleration equation, find the position x at t = 6 s. To find x at t = 6 s, we first need to find v0 and x0:

x = x0 + v0t + 12 at 2

Motion in One Dimension Using the information that when t = 4 s, x = 100 m, obtain an equation in x0 and v0:

x(4 s ) = 100 m

(

)

91

= x0 + v0 (4 s ) + 12 3 m/s 2 (4 s )

2

or

x0 + (4 s )v0 = 76 m

(

)

Using the information that when t = 6 s, v = 15 m/s, obtain a second equation in x0 and v0:

v(6 s ) = v0 + 3 m/s 2 (6 s )

Solve for v0 to obtain:

v0 = −3 m/s

Substitute this value for v0 in the previous equation and solve for x0:

x0 = 88 m

Substitute for x0 and v0 and evaluate x at t = 6 s:

x(6 s ) = 88 m + (− 3 m/s ) (6 s ) + 12 (3 m/s 2 ) (6 s ) = 124 m 2

*103 •• Picture the Problem We can use constant-acceleration equations with the final velocity v = 0 to find the acceleration and stopping time of the plane.

(a) Using a constant-acceleration equation, relate the known velocities to the acceleration and displacement: Solve for a:

Substitute numerical values and evaluate a: (b) Using a constant-acceleration equation, relate the final and initial speeds of the plane to its acceleration and stopping time: Solve for and evaluate the stopping time:

v 2 = v02 + 2 a ∆x

a=

v 2 − v02 − v02 = 2 ∆x 2 ∆x

a=

− (60 m s ) = − 25.7 m s 2 2(70 m ) 2

v = v0 + a∆t

∆t =

v − v0 0 − 60 m s = = 2.33 s a − 25.7 m/s 2

104 •• Picture the Problem This is a multipart constant-acceleration problem using three different constant accelerations (+2 m/s2 for 20 s, then zero for 20 s, and then –3 m/s2 until the automobile stops). The final velocity is zero. The pictorial representation will help us organize the information in the problem and develop our solution strategy.

92

Chapter 2

Add up all the displacements to get the total:

∆x 03 = ∆x01 + ∆x12 + ∆x 23

Using constant-acceleration formulas, find the first displacement:

∆x01 = v0t1 + 12 a01t12

The speed is constant for the second displacement. Find the displacement:

∆x12 = v1 (t2 − t1 )

= 0 + 12 (2 m/s 2 )(20 s) 2 = 400 m

where v1 = v0 + a01t1 = 0 + a01t1 and ∆x12 = a01t1 (t2 − t1 ) = (2 m/s 2 )(20 s)(20 s) = 800 m

Find the displacement during the braking interval:

v32 = v22 + 2a23∆x23 where v2 = v1 = a01t1 and v3 = 0 and

02 − (a01t1 ) − [(2 m/s) (20 s)] ∆x23 = = 2a23 2 − 3m s 2 2

(

2

)

= 267m ∆x03 = ∆x01 + ∆x12 + ∆x23 = 1467 m

Add the displacements to get the total:

= 1.47 km

Remarks: Because the area under the curve of a velocity-versus-time graph equals the displacement of the object experiencing the acceleration, we could solve this problem by plotting the velocity as a function of time and finding the area bounded by it and the time axis. *105 •• Picture the Problem Note: No material body can travel at speeds faster than light. When one is dealing with problems of this sort, the kinematic formulae for displacement, velocity and acceleration are no longer valid, and one must invoke the special theory of relativity to answer questions such as these. For now, ignore such subtleties. Although the formulas you are using (i.e., the constant- acceleration equations) are not quite correct, your answer to part (b) will be wrong by about 1%.

(a) This part of the problem is an exercise in the conversion of units. Make use of the fact that 1 c⋅y = 9.47×1015 m and 1 y = 3.16×107 s:

(

⎞ ⎛⎜ 3.16 × 10 7 s ⎛ 1c ⋅ y ⎟⎟ g = 9.81m/s 2 ⎜⎜ 15 9 . 47 10 m × (1 y )2 ⎠ ⎜⎝ ⎝

(

)

)

2

⎞ ⎟ = 1.03c ⋅ y / y 2 ⎟ ⎠

Motion in One Dimension (b) Let t1/2 represent the time it takes to reach the halfway point. Then the total trip time is:

t = 2 t1/2

Use a constant- acceleration equation to relate the half-distance to Mars ∆x to the initial speed, acceleration, and half-trip time t1/2 :

∆x = v0t + 12 at12 2

Because v0 = 0 and a = g:

The distance from Earth to Mars at closest approach is 7.8 × 1010 m. Substitute numerical values and evaluate t1/2 : Substitute for t1/2 in equation (1) to obtain:

93

(1)

t1 / 2 =

2∆x a

t1 / 2 =

2 3.9 × 1010 m = 8.92 × 104 s 2 9.81 m/s

(

)

t = 2(8.92 × 10 4 s ) = 1.78 × 105 s ≈ 2 d

Remarks: Our result in part (b) seems remarkably short, considering how far Mars is and how low the acceleration is. 106 • Picture the Problem Because the elevator accelerates uniformly for half the distance and uniformly decelerates for the second half, we can use constant-acceleration equations to describe its motion

Let t1/2 = 40 s be the time it takes to reach the halfway mark. Use the constant-acceleration equation that relates the acceleration to the known variables to obtain: Solve for a:

Substitute numerical values and evaluate a:

∆y = v0t + 12 at 2 or, because v0 = 0,

∆y = 12 at 2

a=

2 ∆y t12/ 2

a=

2( 12 )(1173 ft )(1 m/3.281ft ) = 0.223 m/s 2 2 (40 s )

= 0.0228 g 107 •• Picture the Problem Because the acceleration is constant, we can describe the motions of the train using constant-acceleration equations. Find expressions for the distances traveled, separately, by the train and the passenger. When are they equal? Note that the train is accelerating and the passenger runs at a constant minimum velocity (zero acceleration) such that she can just catch the train.

94

Chapter 2

1. Using the subscripts ″train″ and ″p″ to refer to the train and the passenger and the subscript ″c″ to identify ″critical″ conditions, express the position of the train and the passenger:

xtrain,c (tc ) =

Express the critical conditions that must be satisfied if the passenger is to catch the train:

vtrain,c = vp, c

2. Express the train’s average velocity. 3. Using the definition of average velocity, express vav in terms of xp,c and tc. 4. Combine steps 2 and 3 and solve for xp,c. 5. Combine steps 1 and 4 and solve for tc.

atrain 2 tc 2

and xp,c (tc ) = vp,c (tc − ∆t )

and

xtrain, c = xp, c vav (0 to tc ) = vav ≡ xp,c =

0 + vtrain, c vtrain, c = 2 2

∆x 0 + xp, c xp, c = = ∆t 0 + tc tc vtrain,ctc 2

vp,c (tc − ∆t ) =

vtrain,ctc 2

or tc − ∆t =

tc 2

and tc = 2 ∆t = 2 (6 s) = 12 s 6. Finally, combine steps 1 and 5 and solve for vtrain, c.

vp, c = vtrain,c = atrain tc = (0.4 m/s 2 )(12 s ) = 4.80 m/s

The graph shows the location of both the passenger and the train as a function of time. The parabolic solid curve is the graph of xtrain(t) for the accelerating train. The straight dashed line is passenger's position xp(t) if she arrives at ∆t = 6.0 s after the train departs. When the passenger catches the train, our graph shows that her speed and that of the train must be equal ( vtrain, c = vp, c ). Do you see why?

Motion in One Dimension

95

50 45 40

Train

35

Passenger

x (m)

30 25 20 15 10 5 0 0

4

8

12

16

t (s)

108 ••• Picture the Problem Both balls experience constant acceleration once they are in flight. Choose a coordinate system with the origin at the ground and the upward direction positive. When the balls collide they are at the same height above the ground.

Using constant-acceleration equations, express the positions of both balls as functions of time. At the ground y = 0. The conditions at collision are that the heights are equal and the velocities are related:

Express the velocities of both balls as functions of time:

Substituting the position and velocity functions into the conditions at collision gives:

yA = h − 12 gt 2 and yB = v0t − 12 gt 2 yA = yB and v A = − 2v B

vA = − gt and vB = v0 − gt h − 12 gtc2 = v0tc − 12 gtc2 and

− gtc = −2(v0 − gtc )

where tc is the time of collision. We now have two equations and two unknowns, tc and v0. Solving the equations for the unknowns gives:

tc =

Substitute the expression for tc into the equation for yA to obtain the height at collision:

⎛ 2h ⎞ 2h yA = h − 12 g ⎜⎜ ⎟⎟ = 3 ⎝ 3g ⎠

2h 3 gh and v0 = 3g 2

96

Chapter 2

Remarks: We can also solve this problem graphically by plotting velocity- versustime for both balls. Because ball A starts from rest its velocity is given by v A = − gt . Ball B initially moves with an unknown velocity vB0 and its velocity is given by v B = v B0 − gt . The graphs of these equations are shown below with T representing the time at which they collide.

The height of the building is the sum of the sum of the distances traveled by the balls. Each of these distances is equal to the magnitude of the area ″under″ the corresponding v-versus-t curve. Thus, the height of the building equals the area of the parallelogram, which is vB0T. The distance that A falls is the area of the lower triangle, which is (1/3) vB0T. Therefore, the ratio of the distance fallen by A to the height of the building is 1/3, so the collision takes place at 2/3 the height of the building. 109 ••• Picture the Problem Both balls are moving with constant acceleration. Take the origin of the coordinate system to be at the ground and the upward direction to be positive. When the balls collide they are at the same height above the ground. The velocities at collision are related by vA = 4vB.

Using constant-acceleration equations, express the positions of both balls as functions of time:

yA = h − 12 gt 2 and y B = v0t − 12 gt 2

The conditions at collision are that the heights are equal and the velocities are related:

Express the velocities of both balls as functions of time:

yA = yB and vA = 4vB

vA = − gt and v B = v0 − gt

Motion in One Dimension Substitute the position and velocity functions into the conditions at collision to obtain:

h − 12 gtc2 = v0t c − 12 gt c2 and

− gtc = 4(v0 − gtc )

where tc is the time of collision. We now have two equations and two unknowns, tc and v0. Solving the equations for the unknowns gives:

tc =

Substitute the expression for tc into the equation for yA to obtain the height at collision:

⎛ 4h ⎞ h yA = h − 12 g ⎜⎜ ⎟⎟ = 3 ⎝ 3g ⎠

4h 3gh and v 0 = 3g 4

*110 •• Determine the Concept The problem describes two intervals of constant acceleration; one when the train’s velocity is increasing, and a second when it is decreasing.

(a) Using a constant-acceleration equation, relate the half-distance ∆x between stations to the initial speed v0, the acceleration a of the train, and the time-to-midpoint ∆t: Solve for ∆t:

Substitute numerical values and evaluate the time-to-midpoint ∆t:

∆x = v0 ∆t + 12 a (∆t )

2

or, because v0 = 0,

∆x = 12 a (∆t )

2

∆t =

2∆x a

∆t =

2(450 m ) = 30.0 s 1 m/s 2

Because the train accelerates uniformly and from rest, the first part of its velocity graph will be linear, pass through the origin, and last for 30 s. Because it slows down uniformly and at the same rate for the second half of its journey, this part of its graph will also be linear but with a negative slope. The graph of v as a function of t is shown below.

97

98

Chapter 2 30 25

v (m/s)

20 15 10 5 0 0

10

20

30

40

50

60

t (s)

(b) The graph of x as a function of t is obtained from the graph of v as a function of t by finding the area under the velocity curve. Looking at the velocity graph, note that when the train has been in motion for 10 s, it will have traveled a distance of 1 2

(10 s )(10 m/s) = 50 m

and that this distance is plotted above 10 s on the following graph.

900 800 700

x (m)

600 500 400 300 200 100 0 0

10

20

30

40

50

60

t (s)

Selecting additional points from the velocity graph and calculating the areas under the curve will confirm the graph of x as a function of t that is shown. 111 •• Picture the Problem This is a two-stage constant-acceleration problem. Choose a coordinate system in which the direction of the motion of the cars is the positive direction. The pictorial representation summarizes what we know about the motion of the speeder’s car and the patrol car.

Motion in One Dimension

Convert the speeds of the vehicles and the acceleration of the police car into SI units:

km km 1h =8 × = 2.22 m/s 2 , h ⋅s h ⋅ s 3600 s km km 1h 125 = 125 × = 34.7 m/s , h h 3600 s 8

and

190

km km 1h = 190 × = 52.8 m/s h h 3600 s

(a) Express the condition that determines when the police car catches the speeder; i.e., that their displacements will be the same:

∆xP,02 = ∆xS,02

Using a constant-acceleration equation, relate the displacement of the patrol car to its displacement while accelerating and its displacement once it reaches its maximum velocity:

∆xP,02 = ∆xP,01 + ∆xP,12

Using a constant-acceleration equation, relate the displacement of the speeder to its constant velocity and the time it takes the patrol car to catch it:

∆xS,02 = vS,02 ∆t02

Calculate the time during which the police car is speeding up:

99

= ∆xP,01 + vP ,1 (t 2 − t1 )

= (34.7 m/s ) t 2

∆t P,01 = =

∆vP,01 vP ,1 − vP, 0 = aP,01 aP,01 52.8 m/s − 0 = 23.8 s 2.22 m/s 2

100 Chapter 2 Express the displacement of the patrol car:

∆xP,01 = vP,0 ∆tP,01 + 12 aP ,01∆tP2, 01

(

)

= 0 + 12 2.22 m/s 2 (23.8 s )

2

= 629 m ∆xP,02 = ∆xP,01 + ∆xP,12

Equate the displacements of the two vehicles:

= ∆xP,01 + vP ,1 (t 2 − t1 )

= 629 m + (52.8 m/s ) (t2 − 23.8 s) (34.7 m/s) t2 = 629 m + (52.8 m/s)(t2 – 23.8 s)

Solve for the time to catch up to obtain:

∴ t 2 = 34.7 s

∆xS,02 = vS,02 ∆t02 = (34.7 m/s )(34.7 s )

(b) The distance traveled is the displacement, ∆x02,S, of the speeder during the catch:

= 1.20 km

(c) The graphs of xS and xP are shown below. The straight line (solid) represents xS(t) and the parabola (dashed) represents xP(t). 1400 1200 Speeder

x (m)

1000

Officer

800 600 400 200 0 0

10

20

30

40

t (s)

112 •• Picture the Problem The accelerations of both cars are constant and we can use constant-acceleration equations to describe their motions. Choose a coordinate system in which the direction of motion of the cars is the positive direction, and the origin is at the initial position of the police car.

(a) The collision will not occur if, during braking, the displacements of the two cars differ by less than 100 m.

∆xP − ∆xS < 100 m.

Motion in One Dimension 101 Using a constant-acceleration equation, relate the speeder’s initial and final speeds to its displacement and acceleration and solve for the displacement:

or, because vs = 0,

∆xs =

− v02,s 2as

Substitute numerical values and evaluate ∆xs:

∆xS =

− (34.7 m/s ) = 100 m 2 − 6 m/s 2

vs2 = v02,s + 2as ∆xs

2

(

)

Using a constant-acceleration equation, relate the patrol car’s initial and final speeds to its displacement and acceleration and solve for the displacement:

or, assuming vp = 0,

Substitute numerical values and evaluate ∆xp:

− (52.8 m/s ) ∆xP = = 232 m 2 − 6 m/s 2

Finally, substitute these displacements into the inequality that determines whether a collision occurs:

232 m − 100 m = 132 m Because this difference is greater than

(b) Using constant-acceleration equations, relate the positions of both vehicles to their initial positions, initial velocities, accelerations, and time in motion:

xS = 100 m + (34.7 m/s )t − 3 m/s 2 t 2

Equate these expressions and solve for t:

vp2 = v02, p + 2ap ∆xp ∆xp =

− v02, p 2ap 2

(

)

100 m, the cars collide .

(

and

(

)

)

xP = (52.8 m/s )t − 3 m/s 2 t 2 100 m + (34.7 m/s) t – (3 m/s2) t2 = (52.8 m/s) t – (3m/s2) t2 and

t = 5.52 s

(c)

If you take the reaction time into account, the collision will occur sooner and be more severe.

113 •• Picture the Problem Lou’s acceleration is constant during both parts of his trip. Let t1 be the time when the brake is applied; L1 the distance traveled from t = 0 to t = t1. Let tfin be the time when Lou's car comes to rest at a distance L from the starting line. A pictorial representation will help organize the given information and plan the solution.

102 Chapter 2

(a) Express the total length, L, of the course in terms of the distance over which Lou will be accelerating, ∆x01, and the distance over which he will be braking, ∆x12:

L = ∆x01 + ∆x12

Express the final velocity over the first portion of the course in terms of the initial velocity, acceleration, and displacement; solve for the displacement:

v12 = v02 + 2a01∆x01

Express the final velocity over the second portion of the course in terms of the initial velocity, acceleration, and displacement; solve for the displacement: Substitute for ∆x01 and ∆x12 to obtain:

or, because v0 = 0, ∆x01 = L1, and a01 = a,

∆x01 = L1 =

v22 = v12 + 2a12 ∆x12 or, because v2 = 0 and a12 = −2a,

v12 L1 ∆x12 = = 4a 2 L = ∆x01 + ∆x12 = L1 + 12 L1 = 32 L1 and

L1 = (b) Using the fact that the acceleration was constant during both legs of the trip, express Lou’s average velocity over each leg: Express the time for Lou to reach his maximum velocity as a function of L1 and his maximum velocity:

2 3

L

vav, 01 = vav,12 =

∆t01 = and

t=

vmax 2

∆x01 2 L1 = vav,01 vmax

∆t01 ∝ L1 = Having just shown that the time required for the first segment of the trip is proportional to the length of the segment, use this result to express ∆t01 (= t1) in terms tfin:

2 v12 vmax = 2a 2a

2 3 fin

t

2 L 3

Motion in One Dimension 103 114 •• Picture the Problem There are three intervals of constant acceleration described in this problem. Choose a coordinate system in which the upward direction (shown to the left below) is positive. A pictorial representation will help organize the details of the problem and plan the solution.

(a)

The graphs of a(t) (dashed lines) and v(t) (solid lines) are shown below.

v (m/s) and a (m/s^2)

20

0

-20

-40 Velocity Acceleration

-60

-80 0

2

4

6

8

10

12

14

16

t (s)

(b) Using a constant-acceleration equation, express her speed in terms of her acceleration and the elapsed time; solve for her speed after 8 s of fall:

v1 = v0 + a01t1

(c) Using the same constantacceleration equation that you used in part (b), determine the duration of her constant upward acceleration:

v2 = v1 + a12 ∆t12 v −v − 5 m/s − (− 78.5 m/s ) ∆ t12 = 2 1 = a12 15 m/s 2

(

)

= 0 + 9.81 m/s 2 (8 s ) = 78.5 m/s

= 4.90 s (d) Find her average speed as she slows from 78.5 m/s to 5 m/s:

v1 + v2 78.5 m/s + 5 m/s = 2 2 = 41.8 m/s

vav =

104 Chapter 2 ∆y12 = vav ∆t12 = (41.8 m/s )(4.90 s)

Use this value to calculate how far she travels in 4.90 s:

= 204 m She travels 204 m while slowing down.

(e) Express the total time in terms of the times for each segment of her descent:

t total = ∆t 01 + ∆t12 + ∆t 23

We know the times for the intervals from 0 to 1 and 1 to 2 so we only need to determine the time for the interval from 2 to 3. We can calculate ∆t23 from her displacement and constant velocity during that segment of her descent.

∆y23 = ∆ytotal − ∆y01 − ∆y12

Add the times to get the total time:

t total = t01 + t12 + t23

⎛ 78.5 m/s ⎞ = 575 m − ⎜ ⎟(8 s ) − 204 m 2 ⎝ ⎠ = 57.0 m

= 8 s + 4.9 s +

57.0 m 5 m/s

= 24.3 s (f) Using its definition, calculate her average velocity:

vav =

∆x − 1500 m = = − 7.18 m/s ∆t 209 s

Integration of the Equations of Motion *115 • Picture the Problem The integral of a function is equal to the "area" between the curve for that function and the independent-variable axis.

(a) The graph is shown below: 35 30

v (m/s)

25 20 15 10 5 0 0

1

2

3

t (s)

4

5

Motion in One Dimension 105 The distance is found by determining the area under the curve. You can accomplish this easily because the shape of the area under the curve is a trapezoid. Alternatively, we could just count the blocks and fractions thereof. (b) To find the position function x(t), we integrate the velocity function v(t) over the time interval in question:

A = (36 blocks)(2.5 m/block) = 90 m or

⎛ 33 m/s + 3 m/s ⎞ A=⎜ ⎟(5 s − 0 s ) = 90 m 2 ⎝ ⎠ There are approximately 36 blocks each having an area of (5 m/s)(0.5 s) = 2.5 m. t

x(t ) = ∫ v(t ') dt ' 0 t

[(

)

]

= ∫ 6 m/s 2 t '+(3 m/s ) dt ' 0

and

(

)

x(t ) = 3 m/s 2 t 2 + (3 m/s )t

Now evaluate x(t) at 0 s and 5 s respectively and subtract to obtain ∆x:

∆x = x(5 s ) − x(0 s ) = 90 m − 0 m = 90.0 m

116 • Picture the Problem The integral of v(t) over a time interval is the displacement (change in position) during that time interval. The integral of a function is equivalent to the "area" between the curve for that function and the independent-variable axis. Count the grid boxes.

(a) Find the area of the shaded gridbox:

Area = (1 m/s )(1 s ) = 1 m per box

(b) Find the approximate area under curve for 1 s ≤ t ≤ 2 s:

∆x1 s to 2 s = 1.2 m

Find the approximate area under curve for 2 s ≤ t ≤ 3 s:

∆x2 s to 3 s = 3.2 m

(c) Sum the displacements to obtain the total in the interval 1 s ≤ t ≤ 3 s:

∆x1 s to 3 s = 1.2 m + 3.2 m = 4.4 m

∆x1 s to 3 s 4.4 m = = 2.20 m/s ∆t1 s to 3 s 2s

Using its definition, express and evaluate vav:

vav =

(d) Because the velocity of the particle is dx/dt, separate the

dx = 0.5m/s3 dt

(

so

)

106 Chapter 2 variables and integrate over the interval 1 s ≤ t ≤ 3 s to determine the displacement in this time interval:

(

x

∆x1s→3s = ∫ dx' = 0.5m/s

3s

3

x0

)∫ t '

2

dt '

1s

3s

⎡ t ′3 ⎤ = 0.5m/s ⎢ ⎥ = 4.33m ⎣ 3 ⎦1s

(

3

)

This result is a little smaller than the sum of the displacements found in part (b). Calculate the average velocity over the 2-s interval from 1 s to 3 s: Calculate the initial and final velocities of the particle over the same interval: Finally, calculate the average value of the velocities at t = 1 s and t = 3 s:

vav (1s− 3s) =

∆x1s− 3s ∆t1s− 3s

=

4.33m = 2.17m/s 2s

( ) v(3 s ) = (0.5 m/s )(3 s )

v(1s ) = 0.5 m/s 3 (1s ) = 0.5 m/s 2

2

3

= 4.5 m/s

v(1 s) + v(3 s) 0.5 m/s + 4.5 m/s = 2 2 = 2.50 m/s This average is not equal to the average velocity calculated above.

Remarks: The fact that the average velocity was not equal to the average of the velocities at the beginning and the end of the time interval in part (d) is a consequence of the acceleration not being constant. *117 •• Picture the Problem Because the velocity of the particle varies with the square of the time, the acceleration is not constant. The displacement of the particle is found by integration.

dx(t ) dt

Express the velocity of a particle as the derivative of its position function:

v(t ) =

Separate the variables to obtain:

dx(t ) = v(t )dt

Express the integral of x from xo = 0 to x and t from t0 = 0 to t: Substitute for v(t′) to obtain:

x(t ) =

x(t ) =

x (t )

t

t0 =0

t0 = 0

∫ dx' = ∫ v(t ') dt '

∫ [(7 m/s )t ' −(5 m/s)]dt ' t

3

t0 =0

=

(

7 3

)

2

m/s 3 t 3 − (5 m/s ) t

Motion in One Dimension 107 118 •• Picture the Problem The graph is one of constant negative acceleration. Because

vx = v(t) is a linear function of t, we can make use of the slope-intercept form of the equation of a straight line to find the relationship between these variables. We can then differentiate v(t) to obtain a(t) and integrate v(t) to obtain x(t). Find the acceleration (the slope of the graph) and the velocity at time 0 (the v-intercept) and use the slopeintercept form of the equation of a straight line to express vx(t): Find x(t) by integrating v(t):

a = −10 m/s 2 v x (t ) = 50 m/s + (−10 m/s 2 )t

[(

]

)

x(t ) = ∫ − 10 m/s 2 t + 50 m/s dt

(

)

= (50 m/s )t − 5 m/s 2 t 2 + C Using the fact that x = 0 when t = 0, evaluate C:

Substitute to obtain:

(

)

0 = (50 m/s )(0) − 5 m/s 2 (0) + C and C=0

(

2

)

x(t ) = (50 m/s ) t − 5 m/s 2 t 2 Note that this expression is quadratic in t and that the coefficient of t2 is negative and equal in magnitude to half the constant acceleration.

Remarks: We can check our result for x(t) by evaluating it over the 10-s interval shown and comparing this result with the area bounded by this curve and the time axis. 119 ••• Picture the Problem During any time interval, the integral of a(t) is the change in velocity and the integral of v(t) is the displacement. The integral of a function equals the "area" between the curve for that function and the independent-variable axis.

(a) Find the area of the shaded grid box in Figure 2-37:

Area = (0.5 m/s2)(0.5 s)

(b) We start from rest (vo = 0) at t = 0. For the velocities at the other times, count boxes and multiply by the 0.25 m/s per box that we found in part (a):

Examples: v(1 s) = (3.7 boxes)[(0.25 m/s)/box]

= 0.250 m/s per box

= 0.925 m/s v(2 s) = (12.9 boxes)[(0.25 m/s)/box] = 3.22 m/s and

108 Chapter 2 v(3 s) = (24.6 boxes)[(0.25 m/s)/box] = 6.15 m/s (c) The graph of v as a function of t is shown below: 7 6

v (m/s)

5 4 3 2 1 0 0

0.5

1

1.5

2

2.5

3

t (s)

Area = (1.0 m/s)(1.0 s) = 1.0 m per box Count the boxes under the v(t) curve to find the distance traveled:

x(3 s ) = ∆x(0 → 3 s )

= (7 boxes )[(1.0 m ) / box ] = 7.00 m

120 •• Picture the Problem The integral of v(t) over a time interval is the displacement (change in position) during that time interval. The integral of a function equals the "area" between the curve for that function and the independent-variable axis. Because acceleration is the slope of a velocity versus time curve, this is a non-constant-acceleration problem. The derivative of a function is equal to the "slope" of the function at that value of the independent variable.

(a) To obtain the data for x(t), we must estimate the accumulated area under the v(t) curve at each time interval: Find the area of a shaded grid box in Figure 2-38:

A = (1 m/s)(0.5 s) = 0.5 m per box.

We start from rest (vo = 0) at to= 0. For the position at the other times, count boxes and multiply by the 0.5 m per box that we found above. Remember to add the offset from the origin, xo = 5 m, and that boxes below the v = 0 line are counted as negative:

Examples:

⎛ 0.5 m ⎞ x(3 s ) = (25.8 boxes )⎜ ⎟ + 5m ⎝ box ⎠ = 17.9 m ⎛ 0.5 m ⎞ x(5 s ) = (48.0 boxes )⎜ ⎟ + 5m ⎝ box ⎠ = 29.0 m

Motion in One Dimension 109 ⎛ 0.5 m ⎞ x(10 s ) = (51.0 boxes )⎜ ⎟ ⎝ box ⎠ ⎛ 0.5 m ⎞ − (36.0 boxes )⎜ ⎟ + 5m ⎝ box ⎠ = 12.5 m A graph of x as a function of t follows: 35 30

x (m)

25 20 15 10 5 0 0

2

4

6

8

10

t (s)

(b) To obtain the data for a(t), we must estimate the slope (∆v/∆t) of the v(t) curve at each time. A good way to get reasonably reliable readings from the graph is to enlarge several fold:

Examples:

a (1s ) =

v(1.25 s ) − v(0.75 s ) 0 .5 s

4.9 m/s − 3.0 m/s = 3.8 m/s 2 0.5 s v(6.25 s ) − v(5.75 s ) a (6 s ) = 0 .5 s =

=

− 1.7 m/s − 0.4 m/s = −4.2 m/s 2 0.5 s

110 Chapter 2 A graph of a as a function of t follows: 6 4

a (m/s^2)

2 0 -2 -4 -6 0

2

4

6

8

10

t (s)

*121 •• Picture the Problem Because the position of the body is not described by a parabolic function, the acceleration is not constant.

Select a series of points on the graph of x(t) (e.g., at the extreme values and where the graph crosses the t axis), draw tangent lines at those points, and measure their slopes. In doing this, you are evaluating v = dx/dt at these points. Plot these slopes above the times at which you measured the slopes. Your graph should closely resemble the following graph.

8 6 4 2

v 0 -2 -4 -6 -8 0

0.2

0.4

0.6

0.8

1

1.2

1.4

t

Select a series of points on the graph of v(t) (e.g., at the extreme values and where the graph crosses the t axis), draw tangent lines at those points, and measure their slopes. In doing this, you are evaluating a = dv/dt at these points. Plot these slopes above the times at which you measured the slopes. Your graph should closely resemble the graph shown below.

Motion in One Dimension 111 15 10 5

a

0 -5 -10 -15 0

0.5

1

1.5

t

122 •• Picture the Problem Because the acceleration of the rocket varies with time, it is not constant and integration of this function is required to determine the rocket’s velocity and position as functions of time. The conditions on x and v at t = 0 are known as initial conditions.

(a) Integrate a(t) to find v(t):

v(t ) = ∫ a(t ) dt = b ∫ t dt = 12 bt 2 + C where C, the constant of integration, can be determined from the initial conditions.

Integrate v(t) to find x(t):

[

]

x(t ) = ∫ v(t ) dt = ∫ 12 bt 2 + C dt = 16 bt 3 + Ct + D where D is a second constant of integration.

Using the initial conditions, find the constants C and D:

v(0 ) = 0 ⇒ C = 0 and

x(0 ) = 0 ⇒ D = 0

∴ x(t ) = 16 bt 3 (b) Evaluate v(5 s) and x(5 s) with C = D = 0 and b = 3 m/s2:

v(5 s ) = and

x(5 s ) =

(

)

(

)

1 2 3m/s 2 (5s ) = 37.5 m/s 2 1 3 3m/s 2 (5s ) = 62.5 m 6

123 •• Picture the Problem The acceleration is a function of time; therefore it is not constant. The instantaneous velocity can be determined by integration of the acceleration and the average velocity from the displacement of the particle during the given time interval.

112 Chapter 2 v (t )

(a) Because the acceleration is the derivative of the velocity, integrate the acceleration to find the instantaneous velocity v(t).

dv a(t ) = ⇒ v(t ) = ∫ dv' = ∫ a(t ')dt ' dt v0 =0 t0 = 0

Calculate the instantaneous velocity using the acceleration given.

v(t ) = 0.2 m/s

(

t

) ∫ t ' dt ' t

3

t0 =0

and

v(t ) = (0.1 m/s3 )t 2

(b) To calculate the average velocity, we need the displacement: Because the velocity is the derivative of the displacement, integrate the velocity to find ∆x.

x (t )

t

dx v(t ) ≡ ⇒ x(t ) = ∫ dx' = ∫ v(t ')dt ' dt x0 = 0 t0 =0

(

x(t ) = 0.1m/s3

) ∫ t' t

2

(

dt ' = 0.1m/s3

t0 =0

)t3

3

and

∆x = x(7 s) − x(2 s) ⎡ (7 s )3 − (2 s)3 ⎤ = 0.1m/s3 ⎢ ⎥ 3 ⎣ ⎦ = 11.2 m

(

Using the definition of the average velocity, calculate vav.

)

vav =

∆x 11.2 m = = 2.23 m/s ∆t 5s

124 • Determine the Concept Because the acceleration is a function of time, it is not constant. Hence we’ll need to integrate the acceleration function to find the velocity as a function of time and integrate the velocity function to find the position as a function of time. The important concepts here are the definitions of velocity, acceleration, and average velocity.

(a) Starting from to = 0, integrate the instantaneous acceleration to obtain the instantaneous velocity as a function of time:

dv dt it follows that From a =

v

t

v0

0

∫ dv' = ∫ (a0 + bt ')dt '

and

v = v0 + a0t + 12 bt 2 (b) Now integrate the instantaneous velocity to obtain the position as a function of time:

From v =

dx dt

it follows that

Motion in One Dimension 113 x

t

x0

t0 = 0

∫ dx' = ∫ v(t ') dt ' b ⎞ ⎛ = ∫ ⎜ v9 + a0t '+ t '2 ⎟dt ' 2 ⎠ t0 ⎝ t

and

x = x0 + v0t + 12 a0t 2 + 16 bt 3 (c) The definition of the average velocity is the ratio of the displacement to the total time elapsed:

vav ≡

∆x x − x0 v0t + 12 a0t 2 + 16 bt 3 = = t − t0 t ∆t

and

vav = v0 + 12 a0t + 16 bt 2 Note that vav is not the same as that due to constant acceleration:

(v

)

constant acceleration av

v0 + v 2 v + v0 + a0t + 12 bt 2 = 0 2 1 = v0 + 2 a0t + 14 bt 2

=

(

)

≠ vav

General Problems 125 ••• Picture the Problem The acceleration of the marble is constant. Because the motion is downward, choose a coordinate system with downward as the positive direction. The equation gexp = (1 m)/(∆t)2 originates in the constant-acceleration equation

∆x = v0 ∆t + 12 a (∆t ) . Because the motion starts from rest, the displacement of the 2

marble is 1 m, the acceleration is the experimental value gexp, and the equation simplifies to gexp = (1 m)/(∆t)2. Express the percent difference between the accepted and experimental values for the acceleration due to gravity: Using a constant-acceleration equation, express the velocity of the marble in terms of its initial velocity, acceleration, and displacement:

% difference =

g accepted − g exp g accepted

vf2 = v02 + 2a∆y or, because v0 = 0 and a = g,

vf2 = 2 g∆y

Solve for vf:

vf = 2 g∆y

Let v1 be the velocity the ball has reached when it has fallen 0.5 cm,

v1 = 2 9.81 m/s 2 (0.005 m ) = 0.313 m/s

(

)

114 Chapter 2 and v2 be the velocity the ball has reached when it has fallen 0.5 m to obtain.

and

Using a constant-acceleration equation, express v2 in terms of v1, g and ∆t:

v 2 = v1 + g∆t

(

)

v2 = 2 9.81 m/s 2 (0.5 m ) = 3.13 m/s

∆t =

v2 − v1 g

Substitute numerical values and evaluate ∆t:

∆t =

3.13 m/s − 0.313 m/s = 0.2872 s 9.81m/s 2

Calculate the experimental value of the acceleration due to gravity from gexp = (1 m)/(∆t)2:

g exp =

Solve for ∆t:

Finally, calculate the percent difference between this experimental result and the value accepted for g at sea level.

1m = 12.13 m/s 2 2 (0.2872 s )

% difference =

9.81 m/s 2 − 12.13 m/s 2 9.81 m/s 2

= 23.6 %

*126 ••• Picture the Problem We can obtain an average velocity, vav = ∆x/∆t, over fixed time intervals. The instantaneous velocity, v = dx/dt can only be obtained by differentiation.

(a) The graph of x versus t is shown below:

8 6

v (m/s)

4 2 0 -2 -4 -6 0

5

10

15

20

25

30

35

t (s)

(b) Draw a tangent line at the origin and measure its rise and run. Use this ratio to obtain an approximate value for the slope at the origin:

The tangent line appears to, at least approximately, pass through the point (5, 4). Using the origin as the second point,

Motion in One Dimension 115 ∆x = 4 cm – 0 = 4 cm and Therefore, the slope of the tangent line and the velocity of the body as it passes through the origin is approximately:

∆t = 5 s – 0 = 5 s

v(0) =

rise ∆x 4 cm = = = 0.800 cm/s run ∆t 5s

(c) Calculate the average velocity for the series of time intervals given by completing the table shown below: t0 (s) 0 0 0 0 0 0

t (s) 6 3 2 1 0.5 0.25

∆t (s) 6 3 2 1 0.5 0.25

x0 (cm) 0 0 0 0 0 0

x (cm) 4.34 2.51 1.71 0.871 0.437 0.219

∆x (cm) 4.34 2.51 1.71 0.871 0.437 0.219

vav=∆x/∆t (m/s) 0.723 0.835 0.857 0.871 0.874 0.875

(d) Express the time derivative of the position:

dx = Aω cos ωt dt

Substitute numerical values and

dx = Aω cos 0 = Aω dt = (0.05 m ) 0.175 s −1

evaluate

dx at t = 0: dt

(

)

= 0.875 cm/s (e) Compare the average velocities from part (c) with the instantaneous velocity from part (d):

As ∆t, and thus ∆x, becomes small, the value for the average velocity approaches that for the instantaneous velocity obtained in part (d). For ∆t = 0.25 s, they agree to three significant figures.

127 ••• Determine the Concept Because the velocity varies nonlinearly with time, the acceleration of the object is not constant. We can find the acceleration of the object by differentiating its velocity with respect to time and its position function by integrating the velocity function. The important concepts here are the definitions of acceleration and velocity.

(a) The acceleration of the object is the derivative of its velocity with respect to time:

a=

dv d = [v max sin (ωt )] dt dt

= ω v max cos (ωt )

116 Chapter 2 Because a varies sinusoidally with time it is not constant. (b) Integrate the velocity with respect to time from 0 to t to obtain the change in position of the body:

x

t

x0

t0

∫ dx' = ∫ [v

max

sin (ωt ')]dt '

and

⎡− v ⎤ x − x0 = ⎢ max cos (ωt ')⎥ ⎦0 ⎣ ω t

=

− vmax

cos (ωt ) +

vmax

[1 − cos(ωt )]

ω

or

x = x0 +

ω

vmax

ω

Note that, as given in the problem statement, x(0 s) = x0. 128 ••• Picture the Problem Because the acceleration of the particle is a function of its position, it is not constant. Changing the variable of integration in the definition of acceleration will allow us to determine its velocity and position as functions of position.

(a) Because a = dv/dt, we must integrate to find v(t). Because a is given as a function of x, we’ll need to change variables in order to carry out the integration. Once we’ve changed variables, we’ll separate them with v on the left side of the equation and x on the right: Integrate from xo and vo to x and v:

a=

(

)

dv dv dx dv = = v = 2 s−2 x dt dx dt dx

or, upon separating variables,

(

)

vdv = 2 s −2 xdx

∫ v'dv' = ∫ (2 s )x'dx' v

x

v0 = 0

x0

and

−2

(

)(

v 2 − v02 = 2 s −2 x 2 − x02

)

Solve for v to obtain:

v = v02 + (2 s − 2 )(x 2 − x02 )

Now set vo = 0, xo = 1 m, x = 3 m, b =2 s–2 and evaluate the speed:

v = ± (2 s − 2 ) (3 m ) − (1m )

[

and

v = 4.00m/s

2

2

]

Motion in One Dimension 117 (b) Using the definition of v, separate the variables, and integrate to get an expression for t:

v( x ) =

dx dt

and t

x

0

x0

dx′

∫ dt ' = ∫ v(x′) To evaluate this integral we first must find v(x). Show that the acceleration is always positive and use this to find the sign of v(x).

a = (2 s–2 )x and x0 = 1 m. x0 is positive, so a0 is also positive. v0 is zero and a0 is positive, so the object moves in the direction of increasing x. As x increases the acceleration remains positive, so the velocity also remains positive. Thus,

(2 s )(x −2

v= Substitute

(2 s )(x −2

2

)

− x02 for v

and evaluate the integral. (It can be found in standard integral tables.)

t

∫ v(x′)

0

x0

dx′

x

=

∫ (2 s )(x′ −2

x0

= = Evaluate this expression with xo = 1 m and x = 3 m to obtain:

− x02 ) .

dx′

x

t = ∫ dt' =

2

x

1

(2 s

−2

)∫

x0

1

(2 s ) −2

2

− x02

)

dx′ x′2 − x02

⎛ x + x2 − x2 0 ln⎜ ⎜ x0 ⎝

⎞ ⎟ ⎟ ⎠

t = 1.25 s

129 ••• Picture the Problem The acceleration of this particle is not constant. Separating variables and integrating will allow us to express the particle’s position as a function of time and the differentiation of this expression will give us the acceleration of the particle as a function of time.

(a) Write the definition of velocity:

v=

dx dt

We are given that x = bv, where b = 1 s. Substitute for v and separate variables to obtain:

dx x dx = ⇒ dt = b dt b x

Integrate and solve for x(t):

t

⎛ x⎞ dx′ ⇒ (t − t0 ) = b ln⎜⎜ ⎟⎟ x′ ⎝ x0 ⎠ x0 x

∫ dt ' = b ∫

t0

and

118 Chapter 2 x(t ) = x0e (b) Differentiate twice to obtain v(t) and a(t):

v=

(t − t 0 ) / b

(t −t0 ) / b dx 1 = x0 e dt b

and

a= Substitute the result in part (a) to obtain the desired results:

(t −t0 ) / b dv 1 = 2 x0 e dt b

v(t ) =

1 x(t ) b

and 1 x(t ) b2

a(t ) = so

1 1 v(t ) = 2 x(t ) b b

a(t ) =

Because the numerical value of b, expressed in SI units, is one, the numerical values of a, v, and x are the same at each instant in time. 130 ••• Picture the Problem Because the acceleration of the rock is a function of time, it is not constant. Choose a coordinate system in which downward is positive and the origin at the point of release of the rock.

Separate variables in a(t) = dv/dt = ge−bt to obtain: Integrate from to = 0, vo = 0 to some later time t and velocity v:

dv = ge − bt dt v

t

0

0

v = ∫ dv' = ∫ ge −bt ' dt ' = =

(

)

(

g 1 − e −bt = vterm 1 − e −bt b

where

g b

vterm = Separate variables in

(

v = dy dt = vterm 1 − e

− bt

) to

(

)

dy = vterm 1 − e − bt dt

obtain: Integrate from to = 0, yo = 0 to some later time t and position y:

[ ]

g −bt ' e −b

−bt ' ∫ dy' = ∫ vterm (1 − e )dt ' y

t

0

0

)

t 0

Motion in One Dimension 119 t

⎡ 1 ⎤ y = vterm ⎢t '+ e −bt ' ⎥ ⎣ b ⎦0 = vtermt −

(

vterm 1 − e −bt b

)

This last result is very interesting. It says that throughout its free-fall, the object experiences drag; therefore it has not fallen as far at any given time as it would have if it were falling at the constant velocity, vterm. On the other hand, just as the velocity of the object asymptotically approaches vterm, the distance it has covered during its free-fall as a function of time asymptotically approaches the distance it would have fallen if it had fallen with vterm throughout its motion.

y (tlarge ) → vtermt −

v → vtermt b

This should not be surprising because in the expression above, the first term grows linearly with time while the second term approaches a constant and therefore becomes less important with time.

*131 ••• Picture the Problem Because the acceleration of the rock is a function of its velocity, it is not constant. Choose a coordinate system in which downward is positive and the origin is at the point of release of the rock.

Rewrite a = g – bv explicitly as a differential equation:

dv = g − bv dt

Separate the variables, v on the left, t on the right:

dv = dt g − bv

Integrate the left-hand side of this equation from 0 to v and the righthand side from 0 to t:

v

t

dv' ∫0 g − bv' = ∫0 dt ' and

1 ⎛ g − bv ⎞ ⎟=t − ln⎜⎜ b ⎝ g ⎟⎠ Solve this expression for v.

Finally, differentiate this expression with respect to time to obtain an expression for the acceleration and

(

)

v=

g 1 − e − bt b

a=

dv = ge −bt dt

120 Chapter 2 complete the proof. 132 ••• Picture the Problem The skydiver’s acceleration is a function of her velocity; therefore it is not constant. Expressing her acceleration as the derivative of her velocity, separating the variables, and then integrating will give her velocity as a function of time.

(a) Rewrite a = g – cv2 explicitly as a differential equation:

dv = g − cv 2 dt

Separate the variables, with v on the left, and t on the right:

dv = dt g − cv 2

Eliminate c by using c =

g : vT2

dv dv = g ⎡ ⎛ v g − 2 v2 ⎢1 − ⎜⎜ g vT ⎢⎣ ⎝ vT or dv ⎛ v 1 − ⎜⎜ ⎝ vT

⎞ ⎟⎟ ⎠

2

∫

The integral can be found in integral tables:

vT tanh −1 (v / vT ) = gt

0

t

dv' ⎛ v' 1− ⎜ ⎜v ⎝ T

2

⎤ ⎥ ⎥⎦

= gdt

Integrate the left-hand side of this equation from 0 to v and the righthand side from 0 to t:

v

⎞ ⎟⎟ ⎠

⎞ ⎟ ⎟ ⎠

2

= g ∫ dt ' = gt 0

or tanh -1 (v / vT ) = ( g / vT )t

Solve this equation for v to obtain:

⎛g v = vT tanh⎜⎜ ⎝ vT

Because c has units of m−1, and g has units of m/s2, (cg)−1/2 will have units of time. Let’s represent this expression with the time-scale factor T:

i.e., T = (cg)−1/2

⎞ ⎟⎟ t ⎠

= dt

Motion in One Dimension 121 The skydiver falls with her terminal velocity when a = 0. Using this definition, relate her terminal velocity to the acceleration due to gravity and the constant c in the acceleration equation:

0 = g − cvT2 and

vT =

Convince yourself that T is also equal to vT/g and use this relationship to eliminate g and vT in the solution to the differential equation:

g c

⎛t⎞ v(t ) = vT tanh⎜ ⎟ ⎝T ⎠

(b) The following table was generated using a spreadsheet and the equation we derived in part (a) for v(t). The cell formulas and their algebraic forms are: Cell Content/Formula Algebraic Form D1 56 vT D2 5.71 T B7 B6 + 0.25 t + 0.25 C7 $B$1*TANH(B7/$B$2) ⎛t⎞

vT tanh⎜ ⎟ ⎝T ⎠

A 1 2 3 4 5 6 7 8 9 10 54 55 56 57 58 59

B vT = 56 T= 5.71

C m/s s

time (s) 0.00 0.25 0.50 0.75 1.00

v (m/s) 0.00 2.45 4.89 7.32 9.71

12.00 12.25 12.50 12.75 13.00 13.25

54.35 54.49 54.61 54.73 54.83 54.93

D

E

122 Chapter 2 60 50

v (m/s)

40 30 20 10 0 0

2

4

6

8

10

12

14

t (s)

Note that the velocity increases linearly over time (i.e., with constant acceleration) for about time T, but then it approaches the terminal velocity as the acceleration decreases.

Chapter 3 Motion in Two and Three Dimensions Conceptual Problems *1 • Determine the Concept The distance traveled along a path can be represented as a sequence of displacements.

Suppose we take a trip along some path and consider the trip as a sequence of many very small displacements. The net displacement is the vector sum of the very small displacements, and the total distance traveled is the sum of the magnitudes of the very small displacements. That is,

r

r

r

r

total distance = ∆r0,1 + ∆r1, 2 + ∆r2,3 + ... + ∆rN −1, N where N is the number of very small displacements. (For this to be exactly true we have to take the limit as N goes to infinity and each displacement magnitude goes to zero.) Now, using ″the shortest distance between two points is a straight line,″ we have

r r r r r ∆r0, N ≤ ∆r0,1 + ∆r1, 2 + ∆r2,3 + ... + ∆rN −1, N , r

where ∆r0, N is the magnitude of the net displacement. Hence, we have shown that the magnitude of the displacement of a particle is less than or equal to the distance it travels along its path. 2 • Determine the Concept The displacement of an object is its final position vector minus r r r its initial position vector ( ∆r = rf − ri ). The displacement can be less but never more than the distance traveled. Suppose the path is one complete trip around the earth at the equator. Then, the displacement is 0 but the distance traveled is 2πRe.

123

124 Chapter 3 3 • Determine the Concept The important distinction here is that average velocity is being requested, as opposed to average speed. The average velocity is defined as the displacement divided by the elapsed time.

r r ∆r 0 vav = = =0 ∆t ∆t

The displacement for any trip around the track is zero. Thus we see that no matter how fast the race car travels, the average velocity is always zero at the end of each complete circuit. What is the correct answer if we were asked for average speed? The average speed is defined as the distance traveled divided by the elapsed time.

vav ≡

total distance ∆t

For one complete circuit of any track, the total distance traveled will be greater than zero and the average is not zero. 4 • False. Vectors are quantities with magnitude and direction that can be added and subtracted like displacements. Consider two vectors that are equal in magnitude and oppositely directed. Their sum is zero, showing by counterexample that the statement is false. 5 • Determine the Concept We can answer this question by expressing the relationship r between the magnitude of vector A and its component AS and then using properties of the cosine function.

Express AS in terms of A and θ :

AS = A cosθ

Take the absolute value of both sides of this expression:

⎜AS⎜ = ⎜A cosθ ⎜ = A⎜cosθ ⎜ and ⎜cosθ ⎜=

AS A

Motion in One and Two Dimensions 125 Using the fact that 0 < ⎟cosθ ⎜≤ 1, substitute for⎟cosθ ⎜to obtain:

0<

AS ≤ 1 or 0 < AS ≤ A A

No. The magnitude of a component of a vector must be less than or equal to the magnitude of the vector.

If the angle θ shown in the figure is equal to 0° or multiples of 180°, then the magnitude of the vector and its component are equal. *6 • Determine the Concept The diagram r shows a vector A and its components Ax and Ay. We can relate the magnitude of r A is related to the lengths of its components through the Pythagorean theorem.

r

Suppose that A is equal to zero. Then A2 = Ax2 + Ay2 = 0. But Ax2 + Ay2 = 0 ⇒ Ax = Ay = 0.

No. If a vector is equal to zero, each of its components must be zero too. 7 • r r Determine the Concept No. Consider the special case in which B = − A .

r

r

r

r

r

If B = − A ≠ 0, then C = 0 and the magnitudes of the components of A and B are

r

larger than the components of C . *8 • Determine the Concept The instantaneous acceleration is the limiting value, as ∆t r r approaches zero, of ∆v ∆t . Thus, the acceleration vector is in the same direction as ∆v .

False. Consider a ball that has been thrown upward near the surface of the earth and is slowing down. The direction of its motion is upward. The diagram shows the ball’s velocity vectors at two instants of r time and the determination of ∆v . r Note that because ∆v is downward so is the acceleration of the ball.

126 Chapter 3 9 • Determine the Concept The instantaneous acceleration is the limiting value, as ∆t r r approaches zero, of ∆v ∆t and is in the same direction as ∆v .

r

Other than through the definition of a, the instantaneous velocity and acceleration vectors are unrelated. Knowing the direction of the velocity at one instant tells one nothing about how the velocity is changing at that instant. (e) is correct. 10 • Determine the Concept The changing velocity of the golf ball during its flight can be understood by recognizing that it has both horizontal and vertical components. The nature of its acceleration near the highest point of its flight can be understood by analyzing the vertical components of its velocity on either side of this point. At the highest point of its flight, the ball is still traveling horizontally even though its vertical velocity is momentarily zero. The figure to the right shows the vertical components of the ball’s velocity just before and just after it has reached its highest point. The change in velocity during this short interval is a non-zero, downward-pointing vector. Because the acceleration is proportional to the change in velocity, it must also be nonzero.

(d ) is correct.

Remarks: Note that vx is nonzero and vy is zero, while ax is zero and ay is nonzero. 11 • Determine the Concept The change in the velocity is in the same direction as the acceleration. Choose an x-y coordinate system with east being the positive x direction and north the positive y direction.

r

r

Given our choice of coordinate system, the x component of a is negative and so v will r r decrease. The y component of a is positive and so v will increase toward the north.

(c) is correct. *12 • r Determine the Concept The average velocity of a particle, vav , is the ratio of the particle’s displacement to the time required for the displacement.

r

(a) We can calculate ∆r from the given information and ∆t is known. (a ) is correct.

r

(b) We do not have enough information to calculate ∆v and cannot compute the

Motion in One and Two Dimensions 127 particle’s average acceleration. (c) We would need to know how the particle’s velocity varies with time in order to compute its instantaneous velocity. (d) We would need to know how the particle’s velocity varies with time in order to compute its instantaneous acceleration. 13 •• Determine the Concept The velocity vector is always in the direction of motion and, thus, tangent to the path. (a)

The velocity vector, as a consequence of always being in the direction of motion, is tangent to the path.

(b) A sketch showing two velocity vectors for a particle moving along a path is shown to the right.

14 • Determine the Concept An object experiences acceleration whenever either its speed changes or it changes direction. The acceleration of a car moving in a straight path at constant speed is zero. In the other examples, either the magnitude or the direction of the velocity vector is changing and, hence, the car is accelerated. (b) is correct. *15 • r r Determine the Concept The velocity vector is defined by v = dr / dt , while the r r acceleration vector is defined by a = dv / dt. (a) A car moving along a straight road while braking. (b) A car moving along a straight road while speeding up. (c) A particle moving around a circular track at constant speed. 16 • Determine the Concept A particle experiences accelerated motion when either its speed or direction of motion changes. A particle moving at constant speed in a circular path is accelerating because the

128 Chapter 3 direction of its velocity vector is changing. If a particle is moving at constant velocity, it is not accelerating. 17 •• Determine the Concept The acceleration vector is in the same direction as the change in r velocity vector, ∆v . (a) The sketch for the dart thrown upward is shown to the right. The acceleration vector is in the direction of the change in the r velocity vector ∆v . (b) The sketch for the falling dart is shown to the right. Again, the acceleration vector is in the direction of the change in the r velocity vector ∆v . (c) The acceleration vector is in the direction of the change in the velocity vector … and hence is downward as shown the right: *18 •• Determine the Concept The acceleration vector is in the same direction as the change in r velocity vector, ∆v .

The drawing is shown to the right.

19 •• Determine the Concept The acceleration vector is in the same direction as the change in r velocity vector, ∆v . The sketch is shown to the right.

Motion in One and Two Dimensions 129 20 • Determine the Concept We can decide what the pilot should do by considering the speeds of the boat and of the current. Give up. The speed of the stream is equal to the maximum speed of the boat in still water. The best the boat can do is, while facing directly upstream, maintain its position relative to the bank. (d ) is correct. *21 • Determine the Concept True. In the absence of air resistance, both projectiles experience the same downward acceleration. Because both projectiles have initial vertical velocities of zero, their vertical motions must be identical. 22 • Determine the Concept In the absence of air resistance, the horizontal component of the projectile’s velocity is constant for the duration of its flight. At the highest point, the speed is the horizontal component of the initial velocity. The vertical component is zero at the highest point. (e) is correct. 23 • Determine the Concept In the absence of air resistance, the acceleration of the ball depends only on the change in its velocity and is independent of its velocity. As the ball moves along its trajectory between points A and C, the vertical component of its velocity decreases and the change in its velocity is a downward pointing vector. Between points C and E, the vertical component of its velocity increases and the change in its velocity is also a downward pointing vector. There is no change in the horizontal component of the velocity. (d ) is correct. 24 • Determine the Concept In the absence of air resistance, the horizontal component of the velocity remains constant throughout the flight. The vertical component has its maximum values at launch and impact. (a) The speed is greatest at A and E. (b) The speed is least at point C. (c) The speed is the same at A and E. The horizontal components are equal at these points but the vertical components are oppositely directed. 25 • Determine the Concept Speed is a scalar quantity, whereas acceleration, equal to the rate of change of velocity, is a vector quantity. (a) False. Consider a ball on the end of a string. The ball can move with constant speed

130 Chapter 3 (a scalar) even though its acceleration (a vector) is always changing direction. (b) True. From its definition, if the acceleration is zero, the velocity must be constant and so, therefore, must be the speed. 26 • r r Determine the Concept The average acceleration vector is defined by aav = ∆v / ∆t.

r

The direction of aav is that of

r r r ∆v = vf − vi , as shown to the right.

27 • r r r Determine the Concept The velocity of B relative to A is v BA = v B − v A .

r

r

r

The direction of v BA = v B − v A is shown to the right.

*28 ••

r

r

(a) The vectors A(t ) and A(t + ∆t ) are of equal length but point in slightly different

r

r

directions. ∆A is shown in the diagram below. Note that ∆A is nearly perpendicular

r

r

r

to A(t ) . For very small time intervals, ∆A and A(t ) are perpendicular to one another.

r

r

Therefore, dA / dt is perpendicular to A.

r

(b) If A represents the position of a particle, the particle must be undergoing circular motion (i.e., it is at a constant distance from some origin). The velocity vector is tangent to the particle’s trajectory; in the case of a circle, it is perpendicular to the circle’s radius. (c) Yes, it could in the case of uniform circular motion. The speed of the particle is constant, but its heading is changing constantly. The acceleration vector in this case is

Motion in One and Two Dimensions 131 always perpendicular to the velocity vector. 29 •• Determine the Concept The velocity vector is in the same direction as the change in the position vector while the acceleration vector is in the same direction as the change in the velocity vector. Choose a coordinate system in which the y direction is north and the x direction is east. (b)

(a) Path AB BC CD DE EF (c)

Direction of velocity vector north northeast east southeast south

Path AB BC CD DE EF

Direction of acceleration vector north southeast 0 southwest north

The magnitudes are comparable, but larger for DE since the radius of the path is smaller there.

*30 •• Determine the Concept We’ll assume that the cannons are identical and use a constantacceleration equation to express the displacement of each cannonball as a function of time. Having done so, we can then establish the condition under which they will have the same vertical position at a given time and, hence, collide. The modified diagram shown below shows the displacements of both cannonballs.

Express the displacement of the cannonball from cannon A at any time t after being fired and before any collision:

r r r ∆r = v0t + 12 gt 2

Express the displacement of the cannonball from cannon A at any time t′ after being fired and before any collision:

r r r ∆r ′ = v0′t ′ + 12 gt ′2

132 Chapter 3 If the guns are fired simultaneously, t = t ' and the balls are the same distance 1 2

gt 2 below the line of sight at all times. Therefore, they should fire the guns

simultaneously. Remarks: This is the ″monkey and hunter″ problem in disguise. If you imagine a monkey in the position shown below, and the two guns are fired simultaneously, and the monkey begins to fall when the guns are fired, then the monkey and the two cannonballs will all reach point P at the same time.

31 •• Determine the Concept The droplet leaving the bottle has the same horizontal velocity as the ship. During the time the droplet is in the air, it is also moving horizontally with the same velocity as the rest of the ship. Because of this, it falls into the vessel, which has the same horizontal velocity. Because you have the same horizontal velocity as the ship does, you see the same thing as if the ship were standing still. 32 • Determine the Concept

r r Because A and D are tangent to the path of the stone, either of them could (a) represent the velocity of the stone.

r r Let the vectors A(t ) and B (t + ∆t ) be of equal length but point in slightly different directions as the stone moves around the circle. These two r r vectors and ∆A are shown in the diagram above. Note that ∆A is nearly r r r (b) perpendicular to A(t ). For very small time intervals, ∆A and A(t ) are r r perpendicular to one another. Therefore, dA/dt is perpendicular to A and r only the vector E could represent the acceleration of the stone.

Motion in One and Two Dimensions 133 33 • Determine the Concept True. An object accelerates when its velocity changes; that is, when either its speed or its direction changes. When an object moves in a circle the direction of its motion is continually changing. 34 •• Picture the Problem In the diagram, (a) shows the pendulum just before it reverses direction and (b) shows the pendulum just after it has reversed its direction. The acceleration of the bob is in the direction of r r r the change in the velocity ∆v = v f − v i and is tangent to the pendulum trajectory at the point of reversal of direction. This makes sense because, at an extremum of motion, v = 0, so there is no centripetal acceleration. However, because the velocity is reversing direction, the tangential acceleration is nonzero. 35 • Determine the Concept The principle reason is aerodynamic drag. When moving through a fluid, such as the atmosphere, the ball's acceleration will depend strongly on its velocity.

Estimation and Approximation *36 •• Picture the Problem During the flight of the ball the acceleration is constant and equal to 9.81 m/s2 directed downward. We can find the flight time from the vertical part of the motion, and then use the horizontal part of the motion to find the horizontal distance. We’ll assume that the release point of the ball is 2 m above your feet. Make a sketch of the motion. Include coordinate axes, initial and final positions, and initial velocity components:

Obviously, how far you throw the ball will depend on how fast you can throw it. A major league baseball pitcher can throw a fastball at 90 mi/h or so. Assume that you can throw a ball at two-thirds that speed to obtain:

v0 = 60 mi/h ×

0.447 m/s = 26.8 m/s 1 mi/h

134 Chapter 3 There is no acceleration in the x direction, so the horizontal motion is one of constant velocity. Express the horizontal position of the ball as a function of time:

x = v0 xt

Assuming that the release point of the ball is a distance h above the ground, express the vertical position of the ball as a function of time:

y = h + v0 y t + 12 a y t 2

(a) For θ = 0 we have:

v0 x = v0 cosθ 0 = (26.8 m/s )cos 0°

(1)

(2)

= 26.8 m/s and

v0 y = v0 sin θ 0 = (26.8 m/s )sin 0° = 0

Substitute in equations (1) and (2) to obtain:

x = (26.8 m/s )t and

(

)

y = 2 m + 12 − 9.81 m/s 2 t 2 Eliminate t between these equations to obtain: At impact, y = 0 and x = R:

y = 2m −

4.91 m/s 2 2 x (26.8 m/s)2

0 = 2m −

4.91 m/s 2 2 R (26.8 m/s)2

Solve for R to obtain:

R = 17.1 m

(b) Using trigonometry, solve for v0x and v0y:

v0 x = v0 cos θ 0 = (26.8 m/s ) cos 45° = 19.0 m/s and

v0 y = v0 sin θ 0 = (26.8 m/s ) sin 45° = 19.0 m/s

Substitute in equations (1) and (2) to obtain:

x = (19.0 m/s )t and

(

)

y = 2 m + (19.0 m/s )t + 12 − 9.81 m/s 2 t 2

Eliminate t between these equations to obtain:

y = 2m + x −

4.905 m/s 2 2 x (19.0 m/s)2

Motion in One and Two Dimensions 135 At impact, y = 0 and x = R. Hence:

4.905 m/s 2 2 R 0 = 2m + R − (19.0 m/s)2 or

R 2 − (73.60 m )R − 147.2 m 2 = 0

Solve for R (you can use the ″solver″ or ″graph″ functions of your calculator) to obtain:

R = 75.6 m

(c) Solve for v0x and v0y:

v0 x = v0 = 26.8 m/s and

v0 y = 0 Substitute in equations (1) and (2) to obtain:

x = (26.8 m/s )t and

(

)

y = 14 m + 12 − 9.81 m/s 2 t 2 Eliminate t between these equations to obtain:

y = 14 m −

4.905 m/s 2 2 x (26.8 m/s)2

At impact, y = 0 and x = R:

4.905 m/s 2 2 R 0 = 14 m − (26.8 m/s)2

Solve for R to obtain:

R = 45.3 m

(d) Using trigonometry, solve for v0x and v0y:

v0x = v0 y = 19.0 m / s

Substitute in equations (1) and (2) to obtain:

x = (19.0 m/s )t

Eliminate t between these equations to obtain: At impact, y = 0 and x = R:

Solve for R (you can use the ″solver″ or ″graph″ function of your calculator) to obtain:

and

y = 14 m + (19.0 m/s ) t + 12 (− 9.81 m/s 2 )t 2 4.905 m/s 2 2 y = 14 m + x − x (19.0 m/s)2

0 = 14 m + R − R = 85.6 m

4.905 m/s 2 2 R (19.0 m/s)2

136 Chapter 3 37 •• Picture the Problem We’ll ignore the height of Geoff’s release point above the ground and assume that he launched the brick at an angle of 45°. Because the velocity of the brick at the highest point of its flight is equal to the horizontal component of its initial velocity, we can use constant-acceleration equations to relate this velocity to the brick’s x and y coordinates at impact. The diagram shows an appropriate coordinate system and the brick when it is at point P with coordinates (x, y).

Using a constant-acceleration equation, express the x coordinate of the brick as a function of time: Express the y coordinate of the brick as a function of time:

x = x0 + v0 x t + 12 ax t 2 or, because x0 = 0 and ax = 0,

x = v0 xt

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0 and ay = −g,

y = v0 y t − 12 gt 2 Eliminate the parameter t to obtain:

y = (tan θ 0 )x −

g 2 x 2v02x

Use the brick’s coordinates when it strikes the ground to obtain:

0 = (tan θ 0 )R −

g 2 R 2v02x

where R is the range of the brick. Solve for v0x to obtain:

Substitute numerical values and evaluate v0x:

v0 x =

v0 x =

gR 2 tan θ 0

(9.81m/s )(44.5 m ) = 2

2 tan 45°

14.8 m/s

Note that, at the brick’s highest point, vy = 0.

Vectors, Vector Addition, and Coordinate Systems 38 • Picture the Problem Let the positive y direction be straight up, the positive x direction r r be to the right, and A and B be the position vectors for the minute and hour hands. The

Motion in One and Two Dimensions 137 pictorial representation below shows the orientation of the hands of the clock for parts (a) through (d).

(a) The position vector for the minute hand at12:00 is:

r A12:00 =

(0.5 m ) ˆj

The position vector for the hour hand at 12:00 is:

r B12:00 =

(0.25 m ) ˆj

(b) At 3:30, the minute hand is positioned along the −y axis, while the hour hand is at an angle of (3.5 h)/12 h × 360° = 105°, measured clockwise from the top. The position vector for the minute hand is:

r A3:30 = − (0.5 m ) ˆj

Find the x-component of the vector representing the hour hand:

Bx = (0.25 m )sin 105° = 0.241m

Find the y-component of the vector representing the hour hand:

By = (0.25 m )cos105° = −0.0647 m

The position vector for the hour hand is:

r B3:30 =

(0.241m )iˆ − (0.0647 m ) ˆj

(c) At 6:30, the minute hand is positioned along the −y axis, while the hour hand is at an angle of (6.5 h)/12 h × 360° = 195°, measured clockwise from the top. The position vector for the minute hand is:

r A6:30 = − (0.5 m ) ˆj

Find the x-component of the vector representing the hour hand:

Bx = (0.25 m )sin 195° = −0.0647 m

Find the y-component of the vector representing the hour hand:

By = (0.25 m )cos195° = −0.241m

The position vector for the hour hand is:

r B6:30 = − (0.0647 m )iˆ − (0.241m ) ˆj

(d) At 7:15, the minute hand is positioned along the +x axis, while the hour hand is at an angle of (7.25 h)/12 h × 360° = 218°, measured clockwise from the top.

138 Chapter 3 The position vector for the minute hand is:

r A7:15 =

Find the x-component of the vector representing the hour hand:

Bx = (0.25 m )sin 218° = −0.154 m

Find the y-component of the vector representing the hour hand:

By = (0.25 m )cos 218° = −0.197 m

The position vector for the hour hand is:

r B7:15 = − (0.154 m ) iˆ − (0.197 m ) ˆj

r

r

(e) Find A − B at 12:00:

(0.5 m ) iˆ

r r A − B = (0.5 m ) ˆj − (0.25 m ) ˆj =

r

r

Find A − B at 3:30:

(0.25 m ) ˆj

r r A − B = −(0.5 m ) ˆj

[

− (0.241 m ) iˆ − (0.0647 m ) ˆj

]

= − (0.241m ) iˆ − (0.435 m ) ˆj

r

r

Find A − B at 6:30:

r r A − B = −(0.5 m ) ˆj

[

− (0.0647 m ) iˆ − (0.241 m ) ˆj

]

= − (0.0647 m ) iˆ − (0.259 m ) ˆj

r

r

Find A − B at 7:15:

r r A − B = (0.5 m ) ˆj

[

− − (0.152 m ) iˆ − (0.197 m ) ˆj =

(0.152 m ) iˆ + (0.697 m ) ˆj

*39 • Picture the Problem The resultant displacement is the vector sum of the individual displacements. The two displacements of the bear and its resultant displacement are shown to the right:

]

Motion in One and Two Dimensions 139 Using the law of cosines, solve for the resultant displacement:

R 2 = (12 m ) + (12 m ) 2

2

− 2(12 m )(12 m )cos135°

and

R = 22.2 m Using the law of sines, solve for α:

sin α sin 135° = 12 m 22.2 m ∴ α = 22.5° and the angle with the horizontal is 45° − 22.5° = 22.5°

40 • Picture the Problem The resultant displacement is the vector sum of the individual displacements. (a) Using the endpoint coordinates for her initial and final positions, draw the student’s initial and final position vectors and construct her displacement vector.

Find the magnitude of her displacement and the angle this displacement makes with the positive x-axis: (b)

Her displacement is 5 2 m @ 135°.

His initial and final positions are the same as in (a), so his displacement is also 5 2 @ 135°.

*41 • Picture the Problem Use the standard rules for vector addition. Remember that changing the sign of a vector reverses its direction. (a)

(b)

140 Chapter 3 (c)

(d)

(e)

42 • Picture the Problem The figure shows the paths walked by the Scout. The length of path A is 2.4 km; the length of path B is 2.4 km; and the length of path C is 1.5 km:

(a) Express the distance from the campsite to the end of path C:

2.4 km – 1.5 km = 0.9 km

(b) Determine the angle θ subtended by the arc at the origin (campsite):

θradians =

arc length 2.4 km = radius 2.4 km = 1 rad = 57.3° His direction from camp is 1 rad north of east.

(c) Express the total distance as the sum of the three parts of his walk:

dtot = deast + darc + dtoward camp

Substitute the given distances to find the total:

dtot = 2.4 km + 2.4 km + 1.5 km = 6.3 km

Motion in One and Two Dimensions 141 Express the ratio of the magnitude of his displacement to the total distance he walked and substitute to obtain a numerical value for this ratio:

Magnitude of his displacement 0.9 km = Total distance walked 6.3 km =

1 7

43 • Picture the Problem The direction of a vector is determined by its components. The vector is in the fourth quadrant and

⎛ − 3.5 m/s ⎞ ⎟⎟ = −32.5° θ = tan −1 ⎜⎜ ⎝ 5.5 m/s ⎠

(b) is correct.

44 • Picture the Problem The components of the resultant vector can be obtained from the components of the vectors being added. The magnitude of the resultant vector can then be found by using the Pythagorean Theorem. A table such as the one shown to the right is useful in organizing the r information in this problem. Let D

r r

Vector x-component y-component r 6 −3 A r 4 −3 B

r

be the sum of vectors A, B, and C .

r r r adding the components of A, B, r and C .

r C r D

2

Determine the components of D by

Dx = 5 and Dy = 6

Use the Pythagorean Theorem to r calculate the magnitude of D :

D = Dx2 + D y2 =

5

(5)2 + (6)2

= 7.81

and (d ) is correct. 45 • Picture the Problem The components of the given vector can be determined using righttriangle trigonometry. Use the trigonometric relationships between the magnitude of a vector and its components to calculate the x- and y-components of each vector. (a) (b) (c) (d)

A 10 m 5m 7 km 5 km

θ 30° 45° 60° 90°

Ax 8.66 m 3.54 m 3.50 km 0

Ay 5m 3.54 m 6.06 km 5 km

142 Chapter 3 (e) 15 km/s 150° −13.0 km/s 7.50 km/s (f) 10 m/s 240° −5.00 m/s −8.66 m/s (g) 8 m/s2 270° 0 −8.00 m/s2 *46 • Picture the Problem Vectors can be added and subtracted by adding and subtracting their components.

r

Write A in component form:

Ax = (8 m) cos 37° = 6.4 m Ay = (8 m) sin 37° = 4.8 m r ∴ A = (6.4 m ) iˆ + (4.8 m ) ˆj

(a), (b), (c) Add (or subtract) x- and y-components:

r D= r E= r F=

r

(d) Solve for G and add components to obtain:

(0.4m ) iˆ + (7.8m ) ˆj (− 3.4m ) iˆ − (9.8m ) ˆj (− 17.6m ) iˆ + (23.8m ) ˆj

(

)

r r 1 r r G = − A + B + 2C 2 = (1.3 m ) iˆ − (2.9 m ) ˆj

47 •• Picture the Problem The magnitude of each vector can be found from the Pythagorean theorem and their directions found using the inverse tangent function.

r

(a) A = 5 iˆ + 3 ˆj

r

(b) B = 10 iˆ − 7 ˆj

r

(c) C = −2 iˆ − 3 ˆj + 4kˆ

A=

Ax2 + Ay2 = 5.83 r and, because A is in the 1st quadrant, ⎛A ⎞ θ = tan −1 ⎜⎜ y ⎟⎟ = 31.0° ⎝ Ax ⎠ B = Bx2 + B y2 = 12.2 r and, because B is in the 4th quadrant, ⎛B ⎞ θ = tan −1 ⎜⎜ y ⎟⎟ = − 35.0° ⎝ Bx ⎠ C = C x2 + C y2 + C z2 = 5.39 ⎛ Cz ⎞ ⎟ = 42.1° ⎝C⎠

θ = cos −1 ⎜

where θ is the polar angle measured from the positive z-axis and

Motion in One and Two Dimensions 143 ⎛ Cx ⎞ −1 ⎛ − 2 ⎞ ⎟ = 112° ⎟ = cos ⎜ ⎝C ⎠ ⎝ 29 ⎠

φ = cos −1 ⎜

48 • Picture the Problem The magnitude and direction of a two-dimensional vector can be found by using the Pythagorean Theorem and the definition of the tangent function.

r

(a) A = −4iˆ − 7 ˆj

r B = 3iˆ − 2 ˆj

r r r C = A + B = − iˆ − 9 ˆj

(b) Follow the same steps as in (a).

A=

Ax2 + Ay2 = 8.06 r and, because A is in the 3rd quadrant, ⎛A ⎞ θ = tan −1 ⎜⎜ y ⎟⎟ = 240° ⎝ Ax ⎠ B = Bx2 + B y2 = 3.61 r and, because B is in the 4th quadrant, ⎛B ⎞ θ = tan −1 ⎜⎜ y ⎟⎟ = − 33.7° ⎝ Bx ⎠ C = C x2 + C y2 = 9.06 r and, because C is in the 3rd quadrant, ⎛C ⎞ θ = tan −1 ⎜⎜ y ⎟⎟ = 264° ⎝ Cx ⎠ A = 4.12 ; θ = − 76.0° B = 6.32 ; θ = 71.6° C = 3.61 ; θ = 33.7°

49 • Picture the Problem The components of these vectors are related to the magnitude of each vector through the Pythagorean Theorem and trigonometric functions. In parts (a) and (b), calculate the rectangular components of each vector and then express the vector in rectangular form.

144 Chapter 3 r

(a) Express v in rectangular form:

r v = v x iˆ + v y ˆj

Evaluate vx and vy:

vx = (10 m/s) cos 60° = 5 m/s and vy = (10 m/s) sin 60° = 8.66 m/s

Substitute to obtain:

r v = (5 m/s)iˆ + (8.66 m/s) ˆj

r

(b) Express v in rectangular form:

r A = Ax iˆ + Ay ˆj

Evaluate Ax and Ay:

Ax = (5 m) cos 225° = −3.54 m and Ay = (5 m) sin 225° = −3.54 m

Substitute to obtain:

r A = (− 3.54 m )iˆ + (− 3.54 m ) ˆj r r = (14m )iˆ − (6m ) ˆj

(c) There is nothing to calculate as we are given the rectangular components: 50

•

Picture the Problem While there are infinitely many vectors B that can be constructed such that A = B, the simplest are those which lie along the coordinate axes.

r

Determine the magnitude of A : Write three vectors of the same

r

magnitude as A : The vectors are shown to the right:

A=

Ax2 + Ay2 = 32 + 4 2 = 5

r r r B1 = 5iˆ, B2 = −5iˆ, and B3 = 5 ˆj

Motion in One and Two Dimensions 145 *51 •• Picture the Problem While there are several walking routes the fly could take to get from the origin to point C, its displacement will be the same for all of them. One possible route is shown in the figure.

Express the fly’s r displacement D during its trip from the origin to point C and find its magnitude:

r r r r D = A+ B +C = (3 m )iˆ + (3 m ) ˆj + (3 m )kˆ and

D=

(3 m )2 + (3 m )2 + (3 m )2

= 5.20 m *52 • Picture the Problem The diagram shows the locations of the transmitters relative to the ship and defines the distances separating the transmitters from each other and from the ship. We can find the distance between the ship and transmitter B using trigonometry.

Relate the distance between A and B to the distance from the ship to A and the angle θ: Solve for and evaluate the distance from the ship to transmitter B:

tan θ =

DSB =

DAB DSB

DAB 100 km = = 173 km tan θ tan 30°

146 Chapter 3

Velocity and Acceleration Vectors 53 • Picture the Problem For constant speed and direction, the instantaneous velocity is identical to the average velocity. Take the origin to be the location of the stationary radar and construct a pictorial representation.

Express the average velocity: Determine the position vectors:

r r ∆r vav = ∆t r r1 = (− 10 km ) ˆj and r r2 = (14.1 km )iˆ + (− 14.1 km ) ˆj

Find the displacement vector:

r r r ∆r = r2 − r1

= (14.1 km )iˆ + (− 4.1 km ) ˆj

r

Substitute for ∆r and ∆t to find the average velocity.

r (14.1km )iˆ + (− 4.1km ) ˆj vav = 1h = (14.1 km/h )iˆ + (− 4.1 km/h ) ˆj

54 • Picture the Problem The average velocity is the change in position divided by the elapsed time. (a) The average velocity is: Find the position vectors and the displacement vector:

vav =

∆r ∆t

r r0 = (2m )iˆ + (3m ) ˆj r r2 = (6m )iˆ + (7m ) ˆj and

r r r ∆r = r2 − r1 = (4 m )iˆ + (4 m ) ˆj

Find the magnitude of the displacement vector for the interval between t = 0 and t = 2 s:

∆r02 =

(4m )2 + (4m )2

= 5.66m

Motion in One and Two Dimensions 147 Substitute to determine vav:

vav =

5.66 m = 2.83 m/s 2s

and

⎛ 4m ⎞ ⎟⎟ = 45.0° measured 4 m ⎝ ⎠

θ = tan −1 ⎜⎜

from the positive x axis. (b) Repeat (a), this time using the displacement between t = 0 and t = 5 s to obtain:

r r5 = (13 m )iˆ + (14 m ) ˆj , r r r ∆r05 = r5 − r0 = (11 m )iˆ + (11 m ) ˆj ,

∆r05 = (11 m ) + (11 m ) = 15.6 m , 15.6 m vav = = 3.11 m/s , 5s 2

2

and

⎛ 11 m ⎞ ⎟⎟ = 45.0° measured ⎝ 11 m ⎠

θ = tan −1 ⎜⎜

from the positive x axis. *55 • Picture the Problem The magnitude of the velocity vector at the end of the 2 s of acceleration will give us its speed at that instant. This is a constant-acceleration problem. Find the final velocity vector of the particle:

r

Find the magnitude of v :

r v = vx iˆ + v y ˆj = vx 0 iˆ + a y tˆj

(

)

= (4.0 m/s ) iˆ + 3.0 m/s 2 (2.0 s ) ˆj = (4.0 m/s ) iˆ + (6.0 m/s ) ˆj v=

(4.0 m/s)2 + (6.0 m/s)2

= 7.21 m/s

and (b) is correct. 56 • Picture the Problem Choose a coordinate system in which north coincides with the positive y direction and east with the positive x direction. Expressing the west and north r r velocity vectors is the first step in determining ∆v and a av . (a) The magnitudes of r r v W and v N are 40 m/s and 30 m/s, respectively. The change in the magnitude of the particle’s velocity during this time is:

∆v = vN − vW = − 10 m/s

148 Chapter 3 (b) The change in the direction of the velocity is from west to north.

The change in direction is 90°

(c) The change in velocity is:

r r r ∆v = v N − v W = (30 m/s ) ˆj − (− 40 m/s ) iˆ = (40 m/s ) iˆ + (30 m/s ) ˆj

Calculate the magnitude and r direction of ∆v :

∆v =

(40 m/s )2 + (30 m/s)2

and

θ + x axis = tan −1 (d) Find the average acceleration during this interval:

30 m/s = 36.9° 40 m/s

r r (40 m/s ) iˆ + (30 m/s) ˆj aav ≡ ∆v ∆t = 5s 2 ˆ 2 ˆ = 8 m/s i + 6 m/s j

(

The magnitude of this vector is:

= 50 m/s

aav =

) (

)

(8 m/s ) + (6 m/s ) 2 2

2 2

= 10 m/s 2

and its direction is

⎛ 6 m/s 2 ⎞ ⎟ = 36.9° measured 2 ⎟ ⎝ 8 m/s ⎠

θ = tan −1 ⎜⎜

from the positive x axis. 57 • Picture the Problem The initial and final positions and velocities of the particle are given. We can find the average velocity and average acceleration using their definitions by first calculating the given displacement and velocities using unit vectors iˆ and ˆj. (a) The average velocity is:

r r vav ≡ ∆r ∆t

The displacement of the particle during this interval of time is:

r ∆r = (100 m )iˆ + (80 m ) ˆj

Substitute to find the average velocity:

r (100 m ) iˆ + (80 m ) ˆj vav = 3s

= (b) The average acceleration is:

(33.3 m/s ) iˆ + (26.7 m/s) ˆj

r r aav = ∆v ∆t

Motion in One and Two Dimensions 149 r r

v

Find v1 , v 2 , and ∆v :

r v1 = (28.3 m/s ) iˆ + (28.3 m/s ) ˆj and r v 2 = (19.3 m/s ) iˆ + (23.0 m/s ) ˆj r ∴ ∆v = (− 9.00 m/s ) iˆ + (− 5.30 m/s ) ˆj

Using ∆t = 3 s, find the average acceleration:

(− 3.00 m/s )iˆ + (− 1.77 m/s ) ˆj

r aav =

2

2

*58 •• Picture the Problem The acceleration is constant so we can use the constant-acceleration equations in vector form to find the velocity at t = 2 s and the position vector at t = 4 s. (a) The velocity of the particle, as a function of time, is given by: Substitute to find the velocity at t = 2 s:

r r r v = v0 + at r v = (2 m/s) iˆ + (−9 m/s) ˆj + (4 m/s 2 ) iˆ + (3 m/s 2 ) ˆj (2s )

[

]

= (10 m/s) iˆ + (−3 m/s) ˆj (b) Express the position vector as a function of time:

r r r r r = r0 + v0t + 12 at 2

Substitute and simplify:

r r = (4 m) iˆ + (3 m) ˆj + (2 m/s) iˆ + (-9 m/s) ˆj (4 s ) 2 + 1 (4 m/s 2 ) iˆ + (3 m/s 2 ) ˆj (4 s )

[ 2

]

[

]

= (44 m) iˆ + (−9 m) ˆj Find r the magnitude and direction of r at t = 4 s:

r (4 s) =

(44 m )2 + (− 9 m )2

= 44.9 m r and, because r is in the 4th quadrant, ⎛ −9m ⎞ ⎟⎟ = − 11.6° θ = tan −1 ⎜⎜ ⎝ 44 m ⎠

59 •• Picture the Problem The velocity vector is the time-derivative of the position vector and the acceleration vector is the time-derivative of the velocity vector.

r

Differentiate r with respect to time:

r r dr d (30t )iˆ + 40t − 5t 2 ˆj v= = dt dt = 30iˆ + (40 − 10t ) ˆj

[

(

)]

150 Chapter 3 r

where v has units of m/s if t is in seconds.

r

r r dv d a= = 30iˆ + (40 − 10t ) ˆj dt dt = − 10 m/s 2 ˆj

[

Differentiate v with respect to time:

(

]

)

60 •• Picture the Problem We can use the constant-acceleration equations in vector form to solve the first part of the problem. In the second part, we can eliminate the parameter t from the constant-acceleration equations and express y as a function of x.

r

r

r

r

[(

) (

) ] [(

) ]

)]

r v = 6m/s 2 iˆ + 4m/s 2 ˆj t

(a) Use v = v0 + at with v0 = 0 to

r find v :

r r r r r r Use r = r0 + v0t + 12 at 2 with r0 = (10 m )iˆ to find r :

[

(

r r = (10m ) + 3m/s 2 t 2 iˆ + 2m/s 2 t 2 ˆj (b) Obtain the x and y components of the path from the vector equation in (a): Eliminate the parameter t from these equations and solve for y to obtain:

x = 10 m + (3 m/s 2 )t 2 and

y = (2 m/s 2 )t 2 y=

20 2 x− m 3 3

Use this equation to plot the graph shown below. Note that the path in the xy plane is a straight line.

20 18 16 14

y (m)

12 10 8 6 4 2 0 0

10

20

x (m)

30

40

Motion in One and Two Dimensions 151 61 ••• Picture the Problem The displacements of the boat are shown in the figure. We need to determine each of the displacements in order to calculate the average velocity of the boat during the 30s trip.

(a) Express the average velocity of the boat:

r r ∆r vav = ∆t

Express its total displacement:

r ∆r =

r ∆rN

( )

= 12 a N ∆t N

r ∆rW

+ 2

v W = vN, f = aN ∆t N = 60 m/s

To calculate the displacement we first have to find the speed after the first 20 s:

so

Substitute to find the average velocity:

r r ∆r (600m ) − iˆ + ˆj vav = = 30s ∆t

r 2 ∆r = 12 aN (∆t N ) ˆj − (60 m/s )∆t W iˆ = (600m ) ˆj − (600m ) iˆ

(

=

(b) The average acceleration is given by:

(c) The displacement of the boat from the dock at the end of the 30-s trip was one of the intermediate results we obtained in part (a).

( )

ˆj + v ∆t − iˆ W W

)

(20m/s )(− iˆ + ˆj )

r r r r ∆r vf − vi aav = = ∆t ∆t (− 60 m/s) iˆ − 0 = = 30 s

(− 2 m/s )iˆ

r ∆r = (600m ) ˆj + (− 600m ) iˆ =

(600m )(− iˆ + ˆj )

2

152 Chapter 3 *62 ••• Picture the Problem Choose a coordinate system with the origin at Petoskey, the positive x direction to the east, and the positive y direction to the north. Let t = 0 at 9:00 a.m. and θ be the angle between Robert’s velocity vector and the easterly direction and let ″M″ and ″R″ denote Mary and Robert, respectively. You can express the positions of Mary and Robert as functions of time and then equate their north (y) and east (x) coordinates at the time they rendezvous. Express Mary’s position as a function of time:

r rM = vM t ˆj = (8t ) ˆj r where rM is in miles if t is in hours.

Note that Robert’s initial position coordinates (xi, yi) are:

(xi, yi) = (−13 mi, 22.5 mi)

Express Robert’s position as a function of time:

r

r rR = [ xi + (vR cosθ )(t − 1)]) iˆ + [ yi + (vR sinθ )(t − 1)] ˆj = [−13 + {6(t − 1) cos θ }] iˆ + [22.5 + {6(t − 1) sin θ }] ˆj

where rR is in miles if t is in hours. When Mary and Robert rendezvous, their coordinates will be the same. Equating their north and east coordinates yields: Solve equation (1) for cosθ :

Solve equation (2) for sinθ :

East:

–13 + 6t cosθ – 6 cosθ = 0

North: 22.5 + 6t sinθ – 6 sinθ = 8t (2)

cos θ =

13 6(t − 1)

(3)

sin θ =

8t − 22.5 6(t − 1)

(4)

Square and add equations (3) and (4) to obtain: 2

⎡ 8t − 22.5 ⎤ ⎡ 13 ⎤ sin θ + cos θ = 1 = ⎢ ⎥ +⎢ ⎥ ⎣ 6(t − 1) ⎦ ⎣ 6(t − 1) ⎦ 2

(1)

2

2

Simplify to obtain a quadratic equation in t:

28t 2 − 288t + 639 = 0

Solve (you could use your calculator’s ″solver″ function) this

t = 3.24 h = 3 h 15 min

Motion in One and Two Dimensions 153 equation for the smallest value of t (both roots are positive) to obtain: Now you can find the distance traveled due north by Mary:

rM = vM t = (8 mi/h )(3.24 h ) = 25.9 mi

Finally, solving equation (3) for θ and substituting 3.24 h for t yields:

⎡ ⎤ ⎡ 13 ⎤ 13 = cos −1 ⎢ ⎥ ⎥ = 14.7° ⎣ 6(t − 1) ⎦ ⎣ 6(3.24 − 1)⎦

θ = cos −1 ⎢

and so Robert should head 14.7° north of east. Remarks: Another solution that does not depend on the components of the vectors utilizes the law of cosines to find the time t at which Mary and Robert meet and then uses the law of sines to find the direction that Robert must head in order to rendezvous with Mary.

Relative Velocity 63 •• Picture the Problem Choose a coordinate system in which north is the positive y direction and east is the positive x direction. Let θ be the angle between north and the direction of the plane’s heading. The velocity of the plane relative r to the ground, v PG , is the sum of the velocity of the plane relative to the air, r v PA , and the velocity of the air relative to

r

the ground, vAG . i.e.,

r r r v PG = v PA + v AG The pilot must head in such a direction that r the east-west component of v PG is zero in order to make the plane fly due north. (a) From the diagram one can see that: Solve for and evaluate θ :

vAG cos 45° = vPA sinθ

⎛ 56.6 km/h ⎞ ⎟⎟ ⎝ 250 km/h ⎠

θ = sin −1 ⎜⎜

= 13.1° west of north (b) Because the plane is headed due north, add the north components of

r v PG = (250 km/h) cos 13.1° + (80 km/h) sin 45°

154 Chapter 3

r r v PA and v AG to determine the

= 300 km/h

plane’s ground speed: 64 •• r Picture the Problem Let vSB represent the velocity of the swimmer relative to the r bank; vSW the velocity of the swimmer

r

relative to the water; and v WB the velocity of the water relative to the shore; i.e.,

r r r vSB = vSW + v WB The current of the river causes the swimmer to drift downstream. (a) The triangles shown in the figure are similar right triangles. Set up a proportion between their sides and solve for the speed of the water relative to the bank: (b) Use the Pythagorean Theorem to solve for the swimmer’s speed relative to the shore:

vWB 40 m = vSW 80 m and

vWB =

1 2

(1.6 m/s ) =

0.800 m/s

2 2 vSB = vSW + v WS

=

(1.6 m/s)2 + (0.8 m/s)2

= 1.79 m/s (c) The swimmer should head in a direction such that the upstream component of her velocity is equal to the speed of the water relative to the shore:

Use a trigonometric function to evaluate θ:

⎛ 0.8 m/s ⎞ ⎟⎟ = 30.0° ⎝ 1.6 m/s ⎠

θ = sin −1 ⎜⎜

Motion in One and Two Dimensions 155 *65 •• Picture the Problem Let the velocity of the plane relative to the ground be r represented by v PG ; the velocity of the

r

plane relative to the air by v PA , and the velocity of the air relative to the ground by r v AG . Then

r r r v PG = v PA + v AG (1) Choose a coordinate system with the origin at point A, the positive x direction to the east, and the positive y direction to the north. θ is the angle between north and the direction of the plane’s heading. The pilot must head so that the east-west component r of vPG is zero in order to make the plane fly due north. Use the diagram to express the condition relating the eastward r component of v AG and the

(50 km/h) cos 45° = (240 km/h) sinθ

r

westward component of v PA . This must be satisfied if the plane is to stay on its northerly course. [Note: this is equivalent to equating the xcomponents of equation (1).]

⎡ (50 km/h )cos45° ⎤ ⎥ = 8.47° ⎦ ⎣ 240 km/h

Now solve for θ to obtain:

θ = sin −1 ⎢ r

Add the north components of v PA

r and v AG to find the velocity of the plane relative to the ground:

Finally, find the time of flight:

vPG + vAGsin45° = vPAcos8.47° and vPG = (240 km/h)cos 8.47° − (50 km/h)sin 45° = 202 km/h

t flight = =

distance travelled vPG 520 km = 2.57 h 202 km/h

156 Chapter 3 66

••

r

Picture the Problem Let v BS be the velocity of the boat relative to the shore; r v BW be the velocity of the boat relative to

r

the water; and v WS represent the velocity of the water relative to the shore. Independently of whether the boat is going upstream or downstream:

r r r v BS = v BW + v WS Going upstream, the speed of the boat relative to the shore is reduced by the speed of the water relative to the shore. Going downstream, the speed of the boat relative to the shore is increased by the same amount. For the upstream leg of the trip:

vBS = vBW − vWS

For the downstream leg of the trip:

vBS = vBW + vWS

Express the total time for the trip in terms of the times for its upstream and downstream legs:

ttotal = tupstream + tdownstream

Multiply both sides of the equation by (v BW − v WS )(v BW + v WS ) (the product of the denominators) and rearrange the terms to obtain: Solve the quadratic equation for vBW. (Only the positive root is physically meaningful.) 67 •• r Picture the Problem Let v pg be the velocity of the plane relative to the ground; r v ag be the velocity of the air relative to the

r

ground; and v pa the velocity of the plane

r

r

relative to the air. Then, v pg = v pa +

r v ag . The wind will affect the flight times

differently along these two paths.

=

L L + vBW − vWS vBW + vWS

2 vBW −

2L 2 vBW − vWS =0 t total

vBW = 5.18 km/h

Motion in One and Two Dimensions 157 The velocity of the plane, relative to the ground, on its eastbound leg is equal to its velocity on its westbound leg. Using the diagram, find the velocity of the plane relative to the ground for both directions:

vpg = vpa2 − vag2

Express the time for the east-west roundtrip in terms of the distances and velocities for the two legs:

troundtrip,EW = teastbound + t westbound

=

(15 m/s)2 − (5 m/s)2

=

radius of the circle vpg,eastbound +

=

= 14.1m/s

radius of the circle vpg,westbound

2 × 103 m = 141 s 14.1m/s

Use the distances and velocities for the two legs to express and evaluate the time for the north-south roundtrip:

troundtrip,NS = t northbound + tsouthbound =

radius of the circle radius of the circle + vpg,northbound vpg,southbound

103 m 103 m = + = 150 s (15 m/s) − (5 m/s) (15 m/s) + (5 m/s) Because troundtrip,EW < troundtrip,NS , you should fly your plane across the wind.

68 • Picture the Problem This is a relative velocity problem. The given quantities are the direction of the velocity of the plane relative to the ground and the velocity (magnitude and direction) of the air relative to the ground. Asked for is the direction of the velocity of the air relative to the r r r ground. Using v PG = v PA + v AG , draw a vector addition diagram and solve for the unknown quantity. Calculate the heading the pilot must take: Because this is also the angle of the plane's heading clockwise from north, it is also its azimuth or the required true heading:

θ = sin −1

30 kts = 11.5° 150 kts

Az = (011.5°)

158 Chapter 3 *69 •• Picture the Problem The position of B relative to A is the vector from A to B; i.e.,

r r r rAB = rB − rA The velocity of B relative to A is

r r v AB = drAB dt and the acceleration of B relative to A is

r r a AB = dv AB dt Choose a coordinate system with the origin at the intersection, the positive x direction to the east, and the positive y direction to the north.

r r

r

(a) Find rB , rA , and rAB :

[

]

r rB = 40m − 12 (2m/s 2 )t 2 ˆj r rA = [(20m/s )t ]iˆ and

r r r rAB = rB − rA

= [(− 20m/s ) t ] iˆ

[

(

) ]

+ 40m − 12 2m/s 2 t 2 ˆj r

Evaluate rAB at t = 6 s:

r

r

(b) Find v AB = drAB dt :

r rAB (6 s) = (120 m) iˆ + (4 m) ˆj r r r drAB d {(− 20 m/s)t}i v AB = = dt dt + {40 m − 12 (2 m/s 2 )t 2 }ˆj = (−20 m/s) iˆ + (−2 m/s 2 ) t ˆj

[

]

r

Evaluate v AB at t = 6 s:

r

r

(c) Find a AB = dv AB dt :

r v AB (6 s ) =

(− 20 m/s)iˆ − (12 m/s ) ˆj

[

r d (−20 m/s) iˆ + (−2 m/s 2 ) t ˆj aAB = dt = − 2 m/s 2 ˆj r Note that a AB is independent of time.

(

]

)

*70 ••• Picture the Problem Let h and h′ represent the heights from which the ball is dropped and to which it rebounds, respectively. Let v and v′ represent the speeds with which the ball strikes the racket and rebounds from it. We can use a constant-acceleration equation to relate the pre- and post-collision speeds of the ball to its drop and rebound heights.

Motion in One and Two Dimensions 159 (a) Using a constant-acceleration equation, relate the impact speed of the ball to the distance it has fallen: Relate the rebound speed of the ball to the height to which it rebounds:

v 2 = v02 + 2 gh or, because v0 = 0,

v = 2 gh

v 2 = v' 2 − 2 gh' or because v = 0,

v' = 2 gh' Divide the second of these equations by the first to obtain:

v' = v

2 gh'

Substitute for h′ and evaluate the ratio of the speeds:

v' = v

0.64h = 0.8 ⇒ v' = 0.8v h

2 gh

=

h' h

(b) Call the speed of the racket V. In a reference frame where the racket is unmoving, the ball initially has speed V, moving toward the racket. After it "bounces" from the racket, it will have speed 0.8 V, moving away from the racket. In the reference frame where the racket is moving and the ball initially unmoving, we need to add the speed of the racket to the speed of the ball in the racket's rest frame. Therefore, the ball's speed is:

(c)

v' = V + 0.8V = 1.8V = 45 m/s ≈ 100 mi/h This speed is close to that of a tennis pro’s serve. Note that this result tells us that the ball is moving significantly faster than the racket.

From the result in part (b), the ball can never move more than twice as fast as the racket.

Circular Motion and Centripetal Acceleration 71 • Picture the Problem We can use the definition of centripetal acceleration to express ac in terms of the speed of the tip of the minute hand. We can find the tangential speed of the tip of the minute hand by using the distance it travels each revolution and the time it takes to complete each revolution. Express the acceleration of the tip of the minute hand of the clock as a function of the length of the hand and the speed of its tip: Use the distance the minute hand travels every hour to express its speed:

ac =

v=

v2 R

2πR T

160 Chapter 3 Substitute to obtain:

Substitute numerical values and evaluate ac: Express the ratio of ac to g:

4π 2 R ac = T2 ac =

4π 2 (0.5 m ) = 1.52 × 10− 6 m/s 2 2 (3600 s )

ac 1.52 × 10−6 m/s 2 = = 1.55 × 10− 7 2 9.81m/s g

72 • Picture the Problem The diagram shows the centripetal and tangential accelerations experienced by the test tube. The tangential acceleration will be zero when the centrifuge reaches its maximum speed. The centripetal acceleration increases as the tangential speed of the centrifuge increases. We can use the definition of centripetal acceleration to express ac in terms of the speed of the test tube. We can find the tangential speed of the test tube by using the distance it travels each revolution and the time it takes to complete each revolution. The tangential acceleration can be found from the change in the tangential speed as the centrifuge is spinning up. (a) Express the acceleration of the centrifuge arm as a function of the length of its arm and the speed of the test tube: Use the distance the test tube travels every revolution to express its speed: Substitute to obtain:

Substitute numerical values and evaluate ac:

ac =

v=

v2 R

2πR T

ac = ac =

4π 2 R T2 4π 2 (0.15 m ) ⎛ 1 min 60 s ⎞ ⎜⎜ ⎟⎟ × ⎝ 15000 rev min ⎠

= 3.70 × 105 m/s 2

2

Motion in One and Two Dimensions 161 (b) Express the tangential acceleration in terms of the difference between the final and initial tangential speeds:

2πR −0 2πR vf − vi T = = at = ∆t ∆t T∆t

Substitute numerical values and evaluate aT:

at =

2π (0.15 m ) ⎛ 1 min 60 s ⎞ ⎜⎜ ⎟⎟(75 s ) × ⎝ 15000 rev min ⎠

= 3.14 m/s 2 73 • Picture the Problem The diagram includes a pictorial representation of the earth in its orbit about the sun and a force diagram showing the force on an object at the equator that is due to the earth’s rotation,

r FR , and the force on the object due to the

orbital motion of the earth about the sun, r Fo . Because these are centripetal forces, we can calculate the accelerations they require from the speeds and radii associated with the two circular motions. Express the radial acceleration due to the rotation of the earth: Express the speed of the object on the equator in terms of the radius of the earth R and the period of the earth’s rotation TR:

vR2 R 2πR vR = TR aR =

Substitute for vR in the expression for aR to obtain:

aR =

Substitute numerical values and evaluate aR:

aR =

4π 2 R TR2

(

4π 2 6370 × 103 m ⎡ ⎛ 3600 s ⎞⎤ ⎟⎟⎥ ⎢(24 h )⎜⎜ ⎝ 1 h ⎠⎦ ⎣

)

2

= 3.37 × 10−2 m/s 2 = 3.44 × 10−3 g Note that this effect gives rise to the wellknown latitude correction for g.

162 Chapter 3 Express the radial acceleration due to the orbital motion of the earth:

vo2 ao = r

Express the speed of the object on the equator in terms of the earth-sun distance r and the period of the earth’s motion about the sun To:

vo =

2π r To

Substitute for vo in the expression for ao to obtain:

ao =

4π 2 r To2

Substitute numerical values and evaluate ao:

ao =

(

4π 2 1.5 × 1011 m

)

⎡ ⎛ 24 h ⎞ ⎛ 3600 s ⎞⎤ ⎟⎟⎥ ⎟⎟ ⎜⎜ ⎢(365 d )⎜⎜ ⎝ 1d ⎠ ⎝ 1 h ⎠ ⎦ ⎣

2

= 5.95 × 10−3 m/s 2 = 6.07 × 10−4 g 74 •• Picture the Problem We can relate the acceleration of the moon toward the earth to its orbital speed and distance from the earth. Its orbital speed can be expressed in terms of its distance from the earth and its orbital period. From tables of astronomical data, we find that the sidereal period of the moon is 27.3 d and that its mean distance from the earth is 3.84×108 m. Express the centripetal acceleration of the moon:

ac =

Express the orbital speed of the moon:

v=

Substitute to obtain:

Substitute numerical values and evaluate ac:

v2 r

2πr T

4π 2 r ac = 2 T ac =

(

4π 2 3.84 × 108 m

)

24 h 3600 s ⎞ ⎛ × ⎜ 27.3 d × ⎟ d h ⎠ ⎝ = 2.72 × 10 −3 m/s 2 = 2.78 × 10 −4 g

2

Motion in One and Two Dimensions 163 Remarks: Note that

ac radius of earth = (ac is just the acceleration g distance from earth to moon

due to the earth’s gravity evaluated at the moon’s position). This is Newton’s famous ″falling apple″ observation. 75 • Picture the Problem We can find the number of revolutions the ball makes in a given period of time from its speed and the radius of the circle along which it moves. Because the ball’s centripetal acceleration is related to its speed, we can use this relationship to express its speed. Express the number of revolutions per minute made by the ball in terms of the circumference c of the circle and the distance x the ball travels in time t:

n=

x c

(1)

Relate the centripetal acceleration of the ball to its speed and the radius of its circular path:

v2 ac = g = R

Solve for the speed of the ball:

v = Rg

Express the distance x traveled in time t at speed v:

x = vt

Substitute to obtain:

x = Rg t

The distance traveled per revolution is the circumference c of the circle:

c = 2π R

Substitute in equation (1) to obtain:

Substitute numerical values and evaluate n:

n=

Rg t 1 = 2π R 2π

n=

1 2π

g t R

9.81m/s 2 (60 s ) = 33.4 min −1 0.8 m

Remarks: The ball will oscillate at the end of this string as a simple pendulum with a period equal to 1/n.

Projectile Motion and Projectile Range 76 • Picture the Problem Neglecting air resistance, the accelerations of the ball are constant and the horizontal and vertical motions of the ball are independent of each other. We can use the horizontal motion to determine the time-of-flight and then use this information to determine the distance the ball drops. Choose a coordinate system in which the origin is at the point of release of the ball, downward is the positive y direction, and the horizontal

164 Chapter 3 direction is the positive x direction. Express the vertical displacement of the ball:

∆y = v0 y ∆t + 12 a y (∆t )

2

or, because v0y = 0 and ay = g,

∆y = 12 g (∆t )

2

Find the time of flight from vx = ∆x/∆t:

∆t = =

Substitute to find the vertical displacement in 0.473 s:

∆y =

∆x vx

(18.4 m )(3600 s/h ) = 0.473 s (140 km/h )(1000 m/km) 1 2

(9.81m/s )(0.473 s ) 2

2

= 1.10 m

77 • Picture the Problem In the absence of air resistance, the maximum height achieved by a projectile depends on the vertical component of its initial velocity. The vertical component of the projectile’s initial velocity is:

v0y = v0 sinθ0

Use the constant-acceleration equation:

v 2y = v02 y + 2a y∆y

Set vy = 0, a = −g, and ∆y = h to obtain:

h=

(v0 sin θ 0 )2 2g

*78 •• Picture the Problem Choose the coordinate system shown to the right. Because, in the absence of air resistance, the horizontal and vertical speeds are independent of each other, we can use constant-acceleration equations to relate the impact speed of the projectile to its components.

The horizontal and vertical velocity components are:

v0x = vx= v0cosθ and v0y = v0sinθ

Using a constant-acceleration equation, relate the vertical

v y2 = v02y + 2a y ∆y or, because ay = −g and ∆y = −h,

Motion in One and Two Dimensions 165 component of the velocity to the vertical displacement of the projectile:

v y2 = (v0 sin θ ) + 2 gh

Express the relationship between the magnitude of a velocity vector and its components, substitute for the components, and simplify to obtain:

v 2 = vx2 + v y2 = (v0 cos θ ) + v y2

Substitute for v:

(1.2v0 )2 = v02 + 2 gh

Set v = 1.2 v0, h = 40 m and solve for v0:

v0 = 42.2 m/s

2

2

(

)

= v02 sin 2 θ + cos 2 θ + 2 gh = v + 2 gh 2 0

Remarks: Note that v is independent of θ. This will be more obvious once conservation of energy has been studied. 79 •• Picture the Problem Example 3-12 shows that the dart will hit the monkey unless the dart hits the ground before reaching the monkey’s line of fall. What initial speed does the dart need in order to just reach the monkey’s line of fall? First, we will calculate the fall time of the monkey, and then we will calculate the horizontal component of the dart’s velocity. Using a constant-acceleration equation, relate the monkey’s fall distance to the fall time:

h = 12 gt 2

Solve for the time for the monkey to fall to the ground:

t=

2h g

Substitute numerical values and evaluate t:

t=

2(11.2 m ) = 1.51s 9.81 m/s 2

Let θ be the angle the barrel of the dart gun makes with the horizontal. Then:

θ = tan −1 ⎜⎜

Use the fact that the horizontal velocity is constant to determine v0:

v0 =

⎛ 10 m ⎞ ⎟⎟ = 11.3° ⎝ 50 m ⎠

(50 m 1.51s ) = 33.8 m/s vx = cos θ cos11.3°

166 Chapter 3 80 •• Picture the Problem Choose the coordinate system shown in the figure to the right. In the absence of air resistance, the projectile experiences constant acceleration in both the x and y directions. We can use the constant-acceleration equations to express the x and y coordinates of the projectile along its trajectory as functions of time. The elimination of the parameter t will yield an expression for y as a function of x that we can evaluate at (R, 0) and (R/2, h). Solving these equations simultaneously will yield an expression for θ. Express the position coordinates of the projectile along its flight path in terms of the parameter t:

x = (v0 cos θ )t and

y = (v0 sin θ )t − 12 gt 2

Eliminate the parameter t to obtain:

y = (tan θ )x −

Evaluate equation (1) at (R, 0) to obtain:

R=

g x2 2 2v cos θ

(1)

2 0

2v02 sin θ cos θ g

Evaluate equation (1) at (R/2, h) to obtain:

2 ( v0 sin θ ) h=

Equate R and h and solve the resulting equation for θ :

θ = tan −1 (4) = 76.0°

2g

Remarks: Note that this result is independent of v0. 81 •• Picture the Problem In the absence of air resistance, the motion of the ball is uniformly accelerated and its horizontal and vertical motions are independent of each other. Choose the coordinate system shown in the figure to the right and use constant-acceleration equations to relate the x and y components of the ball’s initial velocity. Use the components of v0 to express θ in terms of v0x and v0y:

θ = tan −1

v0 y v0 x

(1)

Motion in One and Two Dimensions 167 Use the Pythagorean relationship between the velocity and its components to express v0:

v0 = v02x + v02y

Using a constant-acceleration equation, express the vertical speed of the projectile as a function of its initial upward speed and time into the flight:

vy= v0y+ ay t

Because vy = 0 halfway through the flight (at maximum elevation):

v0y = (9.81 m/s2)(1.22 s) = 12.0 m/s

Determine v0x:

Substitute in equation (2) and evaluate v0:

(2)

v0x =

∆x 40 m = = 16.4 m/s ∆t 2.44 s

v0 =

(16.4 m/s)2 + (12.0 m/s)2

= 20.3 m/s Substitute in equation (1) and evaluate θ :

⎛ 12.0 m/s ⎞ ⎟⎟ = 36.2° ⎝ 16.4 m/s ⎠

θ = tan −1 ⎜⎜

*82 •• Picture the Problem In the absence of friction, the acceleration of the ball is constant and we can use the constantacceleration equations to describe its motion. The figure shows the launch conditions and an appropriate coordinate system. The speeds v, vx, and vy are related through the Pythagorean Theorem.

The squares of the vertical and horizontal components of the object’s velocity are:

v y2 = v02 sin 2 θ − 2 gh

The relationship between these variables is:

v 2 = vx2 + v y2

Substitute and simplify to obtain:

v 2 = v02 − 2 gh

and vx2 = v02 cos 2 θ

Note that v is independent of θ ... as was to be shown.

168 Chapter 3 83 •• Picture the Problem In the absence of air resistance, the projectile experiences constant acceleration during its flight and we can use constant-acceleration equations to relate the speeds at half the maximum height and at the maximum height to the launch angle θ of the projectile.

The angle the initial velocity makes with the horizontal is related to the initial velocity components. Write the equation v y2 = v02y + 2a∆y, for ∆y = h and

tan θ =

v0 y v0 x

∆y = h ⇒ 0 = v02y − 2 gh

(1)

vy = 0: Write the equation v y2 = v02y + 2a∆y, for ∆y = h/2:

∆y =

We are given vy = (3/4)v0. Square both sides and express this using the components of the velocity. The x component of the velocity remains constant.

⎛3⎞ v + v = ⎜ ⎟ v02x + v02y ⎝4⎠ where we have used v x = v 0x .

h h ⇒ v y2 = v02y − 2 g 2 2 2

2 0x

2 y

(

(2)

)

(3)

(Equations 1, 2, and 3 constitute three equations and four unknowns v0x, v0y, vy, and h. To solve for any of these unknowns, we first need a fourth equation. However, to solve for the ratio (v0y/v0x) of two of the unknowns, the three equations are sufficient. That is 2 because dividing both sides of each equation by v 0x gives three equations and three unknowns vy/v0x, v0y/v0x, and h/ v 02x . Solve equation 2 for gh and substitute in equation 1: 2

Substitute for v y in equation 3:

2 0y

v

= 2(v − v )⇒ v = 2 0y

2 y

2 h

2

(

v02y 2

1 ⎛3⎞ v + v02y = ⎜ ⎟ v02x + v02y 2 ⎝ 4⎠ 2 0x

)

Motion in One and Two Dimensions 169 2

Divide both sides by v 0 x and solve for v0y/v0x to obtain:

2 2 1 v0 y 9 ⎛⎜ v0 y ⎞⎟ 1 = + 2 v02x 16 ⎜⎝ v02x ⎟⎠ and

1+

v0 y v0 x Using tan θ = v0y/v0x, solve for θ :

= 7

( )

⎛ v0 y ⎞ ⎟⎟ = tan −1 7 = 69.3° ⎝ v0 x ⎠

θ = tan −1 ⎜⎜

84 • Picture the Problem The horizontal speed of the crate, in the absence of air resistance, is constant and equal to the speed of the cargo plane. Choose a coordinate system in which the direction the plane is moving is the positive x direction and downward is the positive y direction and apply the constant-acceleration equations to describe the crate’s displacements at any time during its flight. (a) Using a constant-acceleration equation, relate the vertical displacement of the crate ∆y to the time of fall ∆t: Solve for ∆t:

∆y = v0 y ∆t + 12 g (∆t )

2

or, because v0y = 0,

∆y = 12 g (∆t )

2

∆t =

2∆y g

Substitute numerical values and evaluate ∆t:

∆t =

2(12 × 103 m ) = 49.5 s 9.81m/s 2

(b) The horizontal distance traveled in 49.5 s is:

R = ∆x = v0 x ∆t ⎛ 1h ⎞ ⎟⎟(49.5 s ) = (900 km/h )⎜⎜ ⎝ 3600 s ⎠ = 12.4 km

(c) Because the velocity of the plane is constant, it will be directly over the crate when it hits the ground; i.e., the distance to the aircraft will be the elevation of the aircraft.

∆y = 12.0 km

170 Chapter 3 *85 •• Picture the Problem In the absence of air resistance, the accelerations of both Wiley Coyote and the Roadrunner are constant and we can use constant-acceleration equations to express their coordinates at any time during their leaps across the gorge. By eliminating the parameter t between these equations, we can obtain an expression that relates their y coordinates to their x coordinates and that we can solve for their launch angles. (a) Using constant-acceleration equations, express the x coordinate of the Roadrunner while it is in flight across the gorge: Using constant-acceleration equations, express the y coordinate of the Roadrunner while it is in flight across the gorge:

Eliminate the parameter t to obtain:

Letting R represent the Roadrunner’s range and using the trigonometric identity sin2θ = 2sinθ cosθ, solve for and evaluate its launch speed:

x = x0 + v0 x t + 12 a xt 2 or, because x0 = 0, ax = 0 and v0x = v0 cosθ0,

x = (v0 cos θ 0 ) t

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, ay =−g and v0y = v0 sinθ0,

y = (v0 sin θ 0 ) t − 12 gt 2

y = (tan θ 0 )x −

g x2 2v cos 2 θ 0

Rg = v0 = sin 2θ 0

(1)

2 0

(16.5 m )(9.81m/s2 ) sin 30°

= 18.0 m/s

(b) Letting R represent Wiley’s range, solve equation (1) for his launch angle:

θ 0 = sin −1 ⎜⎜ 2 ⎟⎟ 2 ⎝ v0 ⎠

Substitute numerical values and evaluate θ0:

θ 0 = sin −1 ⎢

1

⎛ Rg ⎞

1 2

⎡ (14.5 m ) 9.81 m/s 2 ⎤ (18.0 m/s )2 ⎥⎦ ⎣

= 13.0°

(

)

Motion in One and Two Dimensions 171 86 • Picture the Problem Because, in the absence of air resistance, the vertical and horizontal accelerations of the cannonball are constant, we can use constantacceleration equations to express the ball’s position and velocity as functions of time and acceleration. The maximum height of the ball and its time-of-flight are related to the components of its launch velocity. (a) Using a constant-acceleration equation, relate h to the initial and final speeds of the cannonball:

or, because v = 0 and ay = −g,

Find the vertical component of the firing speed:

v0y = v0sinθ = (300 m/s)sin 45° = 212 m/s

Solve for and evaluate h:

(b) The total flight time is:

v 2 = v02y + 2a y ∆y 0 = v02y − 2 g∆y

h=

v02y 2g

=

(212 m/s)2

2(9.81m/s 2 )

= 2.29 km

∆t = tup + tdn = 2tup =2

v0 y g

=

2(212 m/s ) = 43.2 s 9.81 m/s 2

(c) Express the x coordinate of the ball as a function of time:

x = v0 x ∆t = (v0 cosθ )∆t

Evaluate x (= R) when ∆t = 43.2 s:

x = [(300 m/s )cos45°](43.2 s ) = 9.16 km

87 •• Picture the Problem Choose a coordinate system in which the origin is at the base of the tower and the x- and y-axes are as shown in the figure to the right. In the absence of air resistance, the horizontal speed of the stone will remain constant during its fall and a constant-acceleration equation can be used to determine the time of fall. The final velocity of the stone will be the vector sum of its x and y components.

172 Chapter 3 ∆y = v0 y ∆t + 12 a y (∆t )

(a) Using a constant-acceleration equation, express the vertical displacement of the stone (the height of the tower) as a function of the fall time:

or, because v0y = 0 and a = −g,

Solve for and evaluate the time of fall:

∆t = −

Use the definition of average velocity to find the velocity with which the stone was thrown from the tower:

2

∆y = − 12 g (∆t )

2

2∆y 2(− 24 m ) = − = 2.21s g 9.81m/s 2

vx = v0 x ≡

∆x 18 m = = 8.14 m / s ∆t 2.21s

(b) Find the y component of the stone’s velocity after 2.21 s:

v y = v0 y − gt

Express v in terms of its components:

v = vx2 + v y2

Substitute numerical values and evaluate v:

v=

= 0 − (9.81 m/s2)(2.21 s) = −21.7 m/s

(8.14 m/s)2 + (− 21.7 m/s)2

= 23.2 m/s 88 •• Picture the Problem In the absence of air resistance, the acceleration of the projectile is constant and its horizontal and vertical motions are independent of each other. We can use constant-acceleration equations to express the horizontal and vertical displacements of the projectile in terms of its time-of-flight. Using a constant-acceleration equation, express the horizontal displacement of the projectile as a function of time:

∆x = v0 x ∆t + 12 a x (∆t ) or, because v0x = v0cosθ and ax = 0, ∆x = (v0 cosθ )∆t

Using a constant-acceleration equation, express the vertical displacement of the projectile as a function of time:

∆y = v0 y ∆t + 12 a y (∆t ) or, because v0y = v0sinθ and ay = −g, 2 ∆y = (v0 cos θ )∆t − 12 g (∆t )

Substitute numerical values to obtain the quadratic equation:

− 200 m = (60m/s )(sin 60°)∆t

Solve for ∆t:

∆t = 13.6 s

2

2

− 12 (9.81 m/s 2 )(∆t )

2

Motion in One and Two Dimensions 173 Substitute for ∆t and evaluate the horizontal distance traveled by the projectile:

∆x = (60 m/s)(cos60°)(13.6 s) = 408 m

89 •• Picture the Problem In the absence of air resistance, the acceleration of the cannonball is constant and its horizontal and vertical motions are independent of each other. Choose the origin of the coordinate system to be at the base of the cliff and the axes directed as shown and use constant- acceleration equations to describe both the horizontal and vertical displacements of the cannonball. Express the direction of the velocity vector when the projectile strikes the ground: Express the vertical displacement using a constant-acceleration equation:

Set ∆x = −∆y (R = −h) to obtain: Solve for vx:

Find the y component of the projectile as it hits the ground: Substitute and evaluate θ :

90 • Picture the Problem In the absence of air resistance, the vertical and horizontal motions of the projectile experience constant accelerations and are independent of each other. Use a coordinate system in which up is the positive y direction and horizontal is the positive x direction and use constant-acceleration equations to describe the horizontal and vertical displacements of the projectile as functions of the time into the flight.

⎛ vy ⎞ ⎟⎟ ⎝ vx ⎠

θ = tan −1 ⎜⎜

∆y = v0 y ∆t + 12 a y (∆t )

2

or, because v0y = 0 and ay = −g,

∆y = − 12 g (∆t )

2

∆x = vx ∆t = 12 g (∆t )

2

vx =

∆x 1 = g∆t ∆t 2

v y = v0 y + a∆t = − g∆t = −2vx ⎛ vy ⎞ ⎟⎟ = tan −1 (− 2 ) = − 63.4° ⎝ vx ⎠

θ = tan −1 ⎜⎜

174 Chapter 3 (a) Use a constant-acceleration equation to express the horizontal displacement of the projectile as a function of time:

∆x = v0 x ∆t

Evaluate this expression when ∆t = 6 s:

∆x = (300 m/s )(cos60°)(6 s ) = 900 m

(b) Use a constant-acceleration equation to express the vertical displacement of the projectile as a function of time:

∆y = (v0 sin θ )∆t − 12 g (∆t )

= (v0 cosθ )∆t

2

Evaluate this expression when ∆t = 6 s:

(

)

∆y = (300 m/s )(sin60°)(6 s ) − 12 9.81 m/s 2 (6 s ) = 1.38 km 2

91 •• Picture the Problem In the absence of air resistance, the acceleration of the projectile is constant and the horizontal and vertical motions are independent of each other. Choose the coordinate system shown in the figure with the origin at the base of the cliff and the axes oriented as shown and use constant-acceleration equations to find the range of the cannonball. Using a constant-acceleration equation, express the horizontal displacement of the cannonball as a function of time:

∆x = v0 x ∆t + 12 a x (∆t ) or, because v0x = v0cosθ and ax = 0, ∆x = (v0 cosθ )∆t

Using a constant-acceleration equation, express the vertical displacement of the cannonball as a function of time:

∆y = v0 y ∆t + 12 a y (∆t )

2

2

or, because y = −40 m, a = −g, and v0y = v0sinθ,

− 40 m = (42.2 m/s )(sin 30°)∆t

(

)

− 12 9.81 m/s 2 (∆t )

2

Solve the quadratic equation for ∆t:

∆t = 5.73 s

Calculate the range:

R = ∆x = (42.2 m/s )(cos30°)(5.73 s ) = 209 m

Motion in One and Two Dimensions 175 *92 •• Picture the Problem Choose a coordinate system in which the origin is at ground level. Let the positive x direction be to the right and the positive y direction be upward. We can apply constantacceleration equations to obtain parametric equations in time that relate the range to the initial horizontal speed and the height h to the initial upward speed. Eliminating the parameter will leave us with a quadratic equation in R, the solution to which will give us the range of the arrow. In (b), we’ll find the launch speed and angle as viewed by an observer who is at rest on the ground and then use these results to find the arrow’s range when the horse is moving at 12 m/s.

(a) Use constant-acceleration equations to express the horizontal and vertical coordinates of the arrow’s motion: Solve the x-component equation for time: Eliminate time from the y-component equation:

Solve for the range to obtain:

R = ∆x = x − x0 = v0 xt and y = h + v0 y t + 12 (− g )t 2 where v0 x = v0 cosθ and v0 y = v0 sin θ t=

R R = v0 x v0 cosθ 2

R 1 ⎛ R ⎞ ⎟ − g⎜ y = h + v0 y v0 x 2 ⎜⎝ v0 x ⎟⎠ and, at (R, 0), g R2 0 = h + (tan θ )R − 2 2 2v0 cos θ R=

⎛ 2 gh v02 sin 2θ ⎜⎜1 + 1 + 2 2 2g v0 sin θ ⎝

Substitute numerical values and evaluate R: R=

2 ⎛ ⎞ ⎜1 + 1 + 2(9.81 m/s )(2.25 m ) ⎟ = 81.6 m ° sin 20 ⎜ 2(9.81 m/s 2 ) (45 m/s)2 (sin 2 10°) ⎟⎠ ⎝

(45 m/s)2

⎞ ⎟ ⎟ ⎠

176 Chapter 3 (b) Express the speed of the arrow in the horizontal direction:

v x = varrow + varcher

Express the vertical speed of the arrow:

v y = (45 m/s )sin10° = 7.81m/s

= (45 m/s )cos10° + 12 m/s = 56.3 m/s

Express the angle of elevation from the perspective of someone on the ground:

θ = tan −1 ⎜⎜

Express the arrow’s speed relative to the ground:

v0 = v x2 + v y2

⎛ vy ⎞ ⎛ 7.81 m/s ⎞ ⎟⎟ = tan −1 ⎜⎜ ⎟⎟ = 7.90° ⎝ 56.3 m/s ⎠ ⎝ vx ⎠

=

(56.3 m/s)2 + (7.81 m/s)2

= 56.8 m/s

Substitute numerical values and evaluate R: ⎛ 2(9.81 m/s 2 )(2.25 m ) ⎞⎟ ⎜ R= sin15.8° 1 + 1 + = 104 m ⎜ 2(9.81 m/s 2 ) (56.8 m/s)2 (sin 2 7.9°) ⎟⎠ ⎝

(56.8 m/s)2

Remarks: An alternative solution for part (b) is to solve for the range in the reference frame of the archer and then add to it the distance the frame travels, relative to the earth, during the time of flight. 93 • Picture the Problem In the absence of air resistance, the horizontal and vertical motions are independent of each other. Choose a coordinate system oriented as shown in the figure to the right and apply constant-acceleration equations to find the time-of-flight and the range of the spudplug.

∆y = v0 y ∆t + 12 a y (∆t )

(a) Using a constant-acceleration equation, express the vertical displacement of the plug:

or, because v0y = 0 and ay = −g,

Solve for and evaluate the flight time ∆t:

∆t = −

2

∆y = − 12 g (∆t )

2

2∆y 2(− 1.00 m ) = − g 9.81m/s 2

= 0.452 s

Motion in One and Two Dimensions 177 (b) Using a constant-acceleration equation, express the horizontal displacement of the plug: Substitute numerical values and evaluate R:

∆x = v0 x ∆t + 12 a x (∆t )

2

or, because ax = 0 and v0x = v0,

∆x = v0 ∆t

∆x = R = (50 m/s )(0.452 s ) = 22.6 m

94 •• Picture the Problem An extreme value (i.e., a maximum or a minimum) of a function is determined by setting the appropriate derivative equal to zero. Whether the extremum is a maximum or a minimum can be determined by evaluating the second derivative at the point determined by the first derivative. Evaluate dR/dθ0:

2 dR v02 d [sin (2θ 0 )] = 2v0 cos(2θ 0 ) = dθ 0 g dθ 0 g

Set dR/dθ0= 0 for extrema and solve for θ0:

2v02 cos(2θ 0 ) = 0 g and

θ 0 = 12 cos −1 0 = 45° Determine whether 45° is a maximum or a minimum:

d 2R 2 dθ 0

[

= − 4(v02 g )sin 2θ 0 θ0 =45°

<0

∴ R is a maximum at θ0 = 45° 95 • Picture the Problem We can use constantacceleration equations to express the x and y coordinates of a bullet in flight on the moon as a function of t. Eliminating this parameter will yield an expression for y as a function of x that we can use to find the range of the bullet. The necessity that the centripetal acceleration of an object in orbit at the surface of a body equal the acceleration due to gravity at the surface will allow us to determine the required muzzle velocity for orbital motion. (a) Using a constant-acceleration equation, express the x coordinate of a bullet in flight on the moon:

x = x0 + v0 xt + 12 a xt 2 or, because x0 = 0, ax = 0 and v0x = v0cosθ0,

x = (v0 cos θ 0 ) t

]

θ0 =45°

178 Chapter 3 Using a constant-acceleration equation, express the y coordinate of a bullet in flight on the moon:

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, ay = −gmoon and v0y = v0sinθ0,

y = (v0 sin θ 0 )t − 12 g moont 2

Eliminate the parameter t to obtain:

When y = 0 and x = R:

y = (tan θ 0 )x −

g moon x2 2 2v cos θ 0

0 = (tan θ 0 )R −

g moon R2 2 2v cos θ 0

2 0

2 0

and

R= Substitute numerical values and evaluate R:

v02 g moon

sin 2θ 0

2 ( 900 m/s ) R= sin90° = 4.85 × 105 m

1.67 m/s 2

= 485 km This result is probably not very accurate because it is about 28% of the moon’s radius (1740 km). This being the case, we can no longer assume that the ground is ″flat″ because of the curvature of the moon. (b) Express the condition that the centripetal acceleration must satisfy for an object in orbit at the surface of the moon:

ac = g moon

Solve for and evaluate v:

v = g moon r =

v2 = r

(1.67 m/s )(1.74 × 10 m ) 2

6

= 1.70 km/s 96 ••• Picture the Problem We can show that ∆R/R = –∆g/g by differentiating R with respect to g and then using a differential approximation. Differentiate the range equation with respect to g:

⎞ dR d ⎛ v02 v2 ⎜⎜ sin 2θ 0 ⎟⎟ = − 02 sin 2θ 0 = dg dg ⎝ g g ⎠ =−

R g

Motion in One and Two Dimensions 179 Approximate dR/dg by ∆R/∆g:

∆R R =− g ∆g

Separate the variables to obtain:

∆R ∆g =− R g i.e., for small changes in gravity ( g ≈ g ± ∆g ), the fractional change in R is linearly opposite to the fractional change in g.

Remarks: This tells us that as gravity increases, the range will decrease, and vice versa. This is as it must be because R is inversely proportional to g. 97 ••• Picture the Problem We can show that ∆R/R = 2∆v0/ v0 by differentiating R with respect to v0 and then using a differential approximation. Differentiate the range equation with respect to v0:

⎞ 2v dR d ⎛ v02 ⎜⎜ sin 2θ 0 ⎟⎟ = 0 sin 2θ 0 = dv0 dv0 ⎝ g g ⎠ =2

R v0

Approximate dR/dv0 by ∆R/∆v0:

∆R R =2 v0 ∆v0

Separate the variables to obtain:

∆R ∆v =2 0 R v0 i.e., for small changes in the launch velocity ( v0 ≈ v0 ± ∆v0 ), the fractional change in R is twice the fractional change in v0.

Remarks: This tells us that as launch velocity increases, the range will increase twice as fast, and vice versa. 98 ••• Picture the Problem Choose a coordinate system in which the origin is at the base of the surface from which the projectile is launched. Let the positive x direction be to the right and the positive y direction be upward. We can apply constant-acceleration equations to obtain parametric equations in time that relate the range to the initial horizontal speed and the height h to the initial upward speed. Eliminating the parameter will leave us with a quadratic equation in R, the solution to which is the result we are required to establish. Write the constant-acceleration equations for the horizontal and vertical parts of the projectile’s

x = v0 xt and

180 Chapter 3 motion:

y = h + v0 y t + 12 (− g ) t 2 where

v0 x = v0 cosθ and v0 y = v0 sin θ

Solve the x-component equation for time:

t=

x x = v0 x v0 cosθ

Using the x-component equation, eliminate time from the y-component equation to obtain:

y = h + (tan θ ) x −

g x2 2v cos 2 θ

When the projectile strikes the ground its coordinates are (R, 0) and our equation becomes:

0 = h + (tan θ )R −

g R2 2 2v cos θ

Using the plus sign in the quadratic formula to ensure a physically meaningful root (one that is positive), solve for the range to obtain:

⎛ 2 gh R = ⎜⎜1 + 1 + 2 2 v0 sin θ 0 ⎝

2 0

2 0

⎞ v02 ⎟ ⎟ 2 g sin 2θ 0 ⎠

*99 •• Picture the Problem We can use trigonometry to relate the maximum height of the projectile to its range and the sighting angle at maximum elevation and the range equation to express the range as a function of the launch speed and angle. We can use a constant-acceleration equation to express the maximum height reached by the projectile in terms of its launch angle and speed. Combining these relationships will allow us to conclude that tan φ = 12 tan θ . Referring to the figure, relate the maximum height of the projectile to its range and the sighting angle φ: Express the range of the rocket and use the trigonometric identity sin 2θ = 2 sin θ cos θ to rewrite the expression as: Using a constant-acceleration equation, relate the maximum height of a projectile to the vertical component of its launch speed: Solve for the maximum height h:

tan φ =

R=

h R2

v2 v2 sin( 2θ ) = 2 sin θ cosθ g g

v y2 = v02y − 2 gh or, because vy = 0 and v0y = v0sinθ,

v02 sin 2 θ = 2 gh h=

v2 sin 2 θ 2g

Motion in One and Two Dimensions 181 Substitute for R and h and simplify to obtain:

v2 2 sin 2 θ 2g tan φ = = 2 v 2 sin θ cos θ g

1 2

tan θ

100 • Picture the Problem In the absence of air resistance, the horizontal and vertical displacements of the projectile are independent of each other and describable by constant-acceleration equations. Choose the origin at the firing location and with the coordinate axes as shown in the figure and use constant-acceleration equations to relate the vertical displacement to vertical component of the initial velocity and the horizontal velocity to the horizontal displacement and the time of flight. (a) Using a constant-acceleration equation, express the vertical displacement of the projectile as a function of its time of flight: Solve for v0y:

Substitute numerical values and evaluate v0y:

∆y = v0 y ∆t + 12 a y (∆t )

2

or, because ay = −g,

∆y = v0 y ∆t − 12 g (∆t )

2

∆y + 12 g (∆t ) ∆t

2

v0 y =

450 m + 12 (9.81 m/s 2 )(20 s ) = 20 s

2

v0 y

= 121 m/s (b) The horizontal velocity remains constant, so: *101 •• Picture the Problem In the absence of air resistance, the acceleration of the stone is constant and the horizontal and vertical motions are independent of each other. Choose a coordinate system with the origin at the throwing location and the axes oriented as shown in the figure and use constant- acceleration equations to express the x and y coordinates of the stone while it is in flight.

v0 x = v x =

∆x 3000 m = = 150 m/s ∆t 20 s

182 Chapter 3 Using a constant-acceleration equation, express the x coordinate of the stone in flight: Using a constant-acceleration equation, express the y coordinate of the stone in flight:

x = x0 + v0 xt + 12 axt 2 or, because x0 = 0, v0x = v0 and ax = 0,

x = v0t

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, v0y = 0 and ay = g,

y = 12 gt 2

Referring to the diagram, express the relationship between θ, y and x at impact:

tan θ =

y x

Substitute for x and y and solve for the time to impact:

tan θ =

gt 2 g = t 2v0t 2v0

Solve for t to obtain:

Referring to the diagram, express the relationship between θ, L, y and x at impact: Substitute for y to obtain:

Substitute for t and solve for L to obtain:

t=

2v0 tan θ g

x = L cos θ =

y tan θ

gt 2 = L cosθ 2g

L=

2v02 tan θ g cosθ

102 ••• Picture the Problem The equation of a particle’s trajectory is derived in the text so we’ll use it as our starting point in this derivation. We can relate the coordinates of the point of impact (x, y) to the angle φ and use this relationship to eliminate y from the equation for the cannonball’s trajectory. We can then solve the resulting equation for x and relate the horizontal component of the point of impact to the cannonball’s range. The equation of the cannonball’s trajectory is given in the text:

⎛ ⎞ 2 g ⎟⎟ x y ( x) = (tan θ 0 ) x − ⎜⎜ 2 2 ⎝ 2v0 cos θ 0 ⎠

Relate the x and y components of a point on the ground to the angle φ:

y( x ) = (tan φ ) x

Express the condition that the cannonball hits the ground:

⎛

⎞ 2 g ⎟⎟ x 2 ⎝ 2v cos θ 0 ⎠

(tan φ )x = (tan θ 0 ) x − ⎜⎜

2 0

Motion in One and Two Dimensions 183 Solve for x to obtain:

2v02 cos 2 θ 0 (tan θ 0 − tan φ ) x= g

Relate the range of the cannonball’s flight R to the horizontal distance x:

x = R cos φ

Substitute to obtain:

Solve for R:

R cos φ =

R=

2v02 cos 2 θ 0 (tan θ 0 − tan φ ) g

2v02 cos 2 θ 0 (tan θ 0 − tan φ ) g cos φ

103 •• Picture the Problem In the absence of air resistance, the acceleration of the rock is constant and the horizontal and vertical motions are independent of each other. Choose the coordinate system shown in the figure with the origin at the base of the building and the axes oriented as shown and apply constant-acceleration equations to relate the horizontal and vertical displacements of the rock to its time of flight. Find the horizontal and vertical components of v0:

v0x = v0 cos53° = 0.602v0 v0y = v0 sin53° = 0.799v0

Using a constant-acceleration equation, express the horizontal displacement of the projectile:

∆x = 20 m = v0 x ∆t = (0.602v0 )∆t

Using a constant-acceleration equation, express the vertical displacement of the projectile:

∆y = −20 m = v0 y ∆t − 12 g (∆t )

2

= (0.799v0 )∆t − 12 g (∆t )

2

Solve the x-displacement equation for ∆t:

∆t =

Substitute ∆t into the expression for ∆y:

− 20 m = (0.799v 0 )∆t − 4.91 m/s 2 (∆t )

Solve for v0 to obtain:

v0 = 10.8 m/s

Find ∆t at impact:

20 m 0.602v0

(

∆t =

20 m = 3.08 s (10.8 m/s)cos53°

)

2

184 Chapter 3 Using constant-acceleration equations, find vy and vx at impact:

vx = v0 x = 6.50 m/s and

v y = v0 y − g∆t = −21 m/s Express the velocity at impact in vector form:

r v = (6.50 m/s) iˆ + (−21.6 m/s) ˆj

104 •• Picture the Problem The ball experiences constant acceleration, except during its collision with the wall, so we can use the constant-acceleration equations in the analysis of its motion. Choose a coordinate system with the origin at the point of release, the positive x axis to the right, and the positive y axis upward. Using a constant-acceleration equation, express the vertical displacement of the ball as a function of ∆t:

∆y = v0 y ∆t − 12 g (∆t )

When the ball hits the ground, ∆y = −2 m:

− 2 m = (10 m/s ) ∆t

Solve for the time of flight:

t flight = ∆t = 2.22 s

Find the horizontal distance traveled in this time:

∆x = (10 m/s) (2.22 s) = 22.2 m

The distance from the wall is:

∆x – 4 m = 18.2 m

2

(

Hitting Targets and Related Problems 105 • Picture the Problem In the absence of air resistance, the acceleration of the pebble is constant. Choose the coordinate system shown in the diagram and use constantacceleration equations to express the coordinates of the pebble in terms of the time into its flight. We can eliminate the parameter t between these equations and solve for the launch velocity of the pebble. We can determine the launch angle from the sighting information and, once the range is known, the time of flight can be found using the horizontal component of the initial velocity.

)

− 12 9.81 m/s 2 (∆t )

2

Motion in One and Two Dimensions 185 Referring to the diagram, express θ in terms of the given distances: Use a constant-acceleration equation to express the horizontal position of the pebble as a function of time:

Use a constant-acceleration equation to express the vertical position of the pebble as a function of time:

Eliminate the parameter t to obtain:

At impact, y = 0 and x = R:

Solve for v0 to obtain:

Substitute numerical values and evaluate v0: Substitute in equation (1) to relate R to tflight: Solve for and evaluate the time of flight:

⎛ 4.85 m ⎞ ⎟⎟ = 6.91° ⎝ 40 m ⎠

θ = tan −1 ⎜⎜

x = x0 + v0 xt + 12 axt 2 or, because x0 = 0, v0x = v0cosθ, and ax = 0, x = (v0 cosθ )t (1)

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, v0y = v0sinθ, and ay = −g,

y = (v0 sin θ )t − 12 gt 2

y = (tan θ )x −

g x2 2 2v cos θ

0 = (tan θ )R −

2 0

g R2 2v cos 2 θ 2 0

v0 =

Rg sin 2θ

v0 =

(40 m)(9.81 m/s 2 ) = 40.6 m/s sin 13.8°

R = (v0 cos θ ) t flight tflight =

40 m = 0.992 s (40.6 m/s)cos6.91°

*106 •• Picture the Problem The acceleration of the ball is constant (zero horizontally and – g vertically) and the vertical and horizontal components are independent of each other. Choose the coordinate system shown in the figure and assume that v and t are unchanged by throwing the ball slightly downward. Express the horizontal displacement of the ball as a function of time:

∆x = v0 x ∆t + 12 a x (∆t )

2

186 Chapter 3 or, because ax = 0,

∆x = v0 x ∆t

Solve for the time of flight if the ball were thrown horizontally:

∆t =

18.4 m ∆x = = 0.491 s v0 x 37.5 m/s

∆y = v0 y ∆t + 12 a y (∆t )

Using a constant-acceleration equation, express the distance the ball would drop (vertical displacement) if it were thrown horizontally:

or, because v0y = 0 and ay = −g,

Substitute numerical values and evaluate ∆y:

∆y = − 12 9.81 m/s 2 (0.491 s ) = −1.18 m

The ball must drop an additional 0.62 m before it gets to home plate.

y = (2.5 – 1.18) m = 1.32 m above ground

Calculate the initial downward speed the ball must have to drop 0.62 m in 0.491 s:

vy =

2

∆y = − 12 g (∆t )

2

(

Find the angle with horizontal:

)

2

− 0.62 m = −1.26 m 0.491s ⎛ vy ⎞ ⎛ − 1.26 m/s ⎞ ⎟⎟ = tan −1 ⎜⎜ ⎟⎟ ⎝ 37.5 m/s ⎠ ⎝ vx ⎠

θ = tan −1 ⎜⎜

= − 1.92° Remarks: One can readily show that

vx2 + v y2 = 37.5 m/s to within 1%; so the

assumption that v and t are unchanged by throwing the ball downward at an angle of 1.93° is justified. 107 •• Picture the Problem The acceleration of the puck is constant (zero horizontally and –g vertically) and the vertical and horizontal components are independent of each other. Choose a coordinate system with the origin at the point of contact with the puck and the coordinate axes as shown in the figure and use constant-acceleration equations to relate the variables v0y, the time t to reach the wall, v0x, v0, and θ0. Using a constant-acceleration equation for the motion in the y direction, express v0y as a function of the puck’s displacement ∆y:

v y2 = v02y + 2a y ∆y or, because vy= 0 and ay = −g,

0 = v02y − 2 g∆y

Motion in One and Two Dimensions 187 Solve for and evaluate v0y:

v0 y = 2 g∆y = 2(2.80 m )(9.81m/s 2 ) = 7.41 m/s

Find t from the initial velocity in the y direction:

t=

Use the definition of average velocity to find v0x:

v0 x = vx =

Substitute numerical values and evaluate v0:

v0 = v02x + v02y

v0 y

=

g

=

7.41 m/s = 0.756 s 9.81 m/s 2

∆x 12.0 m = = 15.9 m/s 0.756 s t

(15.9 m/s )2 + (7.41m/s )2

= 17.5 m/s Substitute numerical values and evaluate θ :

⎛ v0 y ⎞ ⎛ 7.41 m/s ⎞ ⎟⎟ = tan −1 ⎜⎜ ⎟⎟ ⎝ 15.9 m/s ⎠ ⎝ v0 x ⎠

θ = tan −1 ⎜⎜

= 25.0° 108 •• Picture the Problem In the absence of air resistance, the acceleration of Carlos and his bike is constant and we can use constant-acceleration equations to express his x and y coordinates as functions of time. Eliminating the parameter t between these equations will yield y as a function of x … an equation we can use to decide whether he can jump the creek bed as well as to find the minimum speed required to make the jump. (a) Use a constant-acceleration equation to express Carlos’ horizontal position as a function of time: Use a constant-acceleration equation to express Carlos’ vertical position as a function of time:

x = x0 + v0 xt + 12 axt 2 or, because x0 = 0, v0x = v0cosθ, and ax = 0,

x = (v0 cos θ ) t

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, v0y = v0sinθ, and ay = −g,

y = (v0 sin θ )t − 12 gt 2

188 Chapter 3 Eliminate the parameter t to obtain:

Substitute y = 0 and x = R to obtain:

Solve for and evaluate R:

y = (tan θ )x −

g x2 2v cos 2 θ

0 = (tan θ )R −

g R2 2 2v cos θ

2 0

2 0

(11.1m/s) sin 20° v02 sin (2θ 0 ) = 9.81 m/s 2 g = 4.30 m 2

R=

He should apply the brakes! (b) Solve the equation we used in the previous step for v0,min: Letting R = 7 m, evaluate v0,min:

v0, min =

v0,min =

Rg sin (2θ 0 )

(7m )(9.81m/s2 ) sin20°

= 14.2m/s = 51.0 km/h 109 ••• Picture the Problem In the absence of air resistance, the bullet experiences constant acceleration along its parabolic trajectory. Choose a coordinate system with the origin at the end of the barrel and the coordinate axes oriented as shown in the figure and use constant-acceleration equations to express the x and y coordinates of the bullet as functions of time along its flight path. Use a constant-acceleration equation to express the bullet’s horizontal position as a function of time:

x = x0 + v0 xt + 12 axt 2 or, because x0 = 0, v0x = v0cosθ, and ax = 0,

x = (v0 cosθ )t

Use a constant-acceleration equation to express the bullet’s vertical position as a function of time:

Eliminate the parameter t to obtain:

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, v0y = v0sinθ, and ay = −g,

y = (v0 sin θ )t − 12 gt 2

y = (tan θ )x −

g x2 2 2v cos θ 2 0

Motion in One and Two Dimensions 189 Let y = 0 when x = R to obtain:

Solve for the angle above the horizontal that the rifle must be fired to hit the target: Substitute numerical values and evaluate θ0:

0 = (tan θ )R −

g R2 2v cos 2 θ 2 0

⎛ Rg ⎞

θ 0 = 12 sin −1 ⎜⎜ 2 ⎟⎟ ⎝ v0 ⎠

(

)

⎡ (100 m ) 9.81 m/s 2 ⎤ (250 m/s)2 ⎥⎦ ⎣ = 0.450° Note: A second value for θ0, 89.6° is

θ 0 = 12 sin −1 ⎢

physically unreasonable. Referring to the diagram, relate h to θ0 and solve for and evaluate h:

tan θ 0 =

h 100 m

and

h = (100 m ) tan (0.450°) = 0.785 m

General Problems 110 • Picture the Problem The sum and difference of two vectors can be found from the components of the two vectors. The magnitude and direction of a vector can be found from its components.

(a) The table to the right r summarizes the components of A r and B .

(b) The table to the right shows the r components of S .

Vector

r A r B Vector

r A r Br S Determine the magnitude and r direction of S from its components:

x component (m) 0.707

y component (m) 0.707

0.866

−0.500

x component (m) 0.707

y component (m) 0.707

0.866

−0.500 0.207

1.57

S = S x2 + S y2 = 1.59 m r and, because S is in the 1st ⎛S ⎞ θ S = tan −1 ⎜⎜ y ⎟⎟ = 7.50° ⎝ Sx ⎠

190 Chapter 3 (c) The table to the right shows the

r

Vector

components of D :

r A r B r D Determine the magnitude and r direction of D from its components:

x component (m) 0.707

y component (m) 0.707

0.866

−0.500 1.21

−0.159

D = Dx2 + D y2 = 1.22 m r and, because D is in the 2nd quadrant, ⎛D ⎞ θ D = tan −1 ⎜⎜ y ⎟⎟ = 97.5° ⎝ Dx ⎠

*111 • Picture the Problem A vector quantity can be resolved into its components relative to any coordinate system. In this example, the axes are orthogonal and the components of the vector can be found using trigonometric functions.

r

The x and y components of g are related to g through the sine and cosine functions:

gx = gsin30° = 4.91 m/s 2 and gy = gcos30° = 8.50 m/s 2

112 • Picture the Problem The figure shows two arbitrary, co-planar vectors that (as drawn) do not satisfy the condition that A/B = Ax/Bx. Because Ax = A cos θ A and

Bx = B cosθ B ,

cosθ A = 1 for the cosθ B

condition to be satisfied.

r

r

∴ A/B = Ax/Bx if and only if A and B are parallel (θA = θB) or on opposite sides of the x-axis (θA = –θB). 113 • Picture the Problem We can plot the path of the particle by substituting values for t and r evaluating rx and ry coordinates of r . The velocity vector is the time derivative of the position vector. (a) We can assign values to t in the parametric equations x = (5 m/s)t and y = (10 m/s)t to obtain ordered pairs (x, y) that lie on the path of the particle. The path is shown in the following graph:

Motion in One and Two Dimensions 191 25

y (m)

20 15 10 5 0 0

2

4

6

8

10

12

x (m)

r

(b) Evaluate dr dt :

Use its components to find the r magnitude of v :

r r dr d (5 m/s)t iˆ + (10m/s)t ˆj v= = dt dt = (5 m/s ) iˆ + (10m/s ) ˆj

[

v = v x2 + v y2 = 11.2 m/s

114 •• Picture the Problem In the absence of air resistance, the hammer experiences constant acceleration as it falls. Choose a coordinate system with the origin and coordinate axes as shown in the figure and use constant-acceleration equations to describe the x and y coordinates of the hammer along its trajectory. We’ll use the equation describing the vertical motion to find the time of flight of the hammer and the equation describing the horizontal motion to determine its range. Using a constant-acceleration equation, express the x coordinate of the hammer as a function of time:

x = x0 + v0 xt + 12 axt 2 or, because x0 = 0, v0x = v0cosθ0, and ax = 0,

x = (v0 cosθ 0 )t

Using a constant-acceleration equation, express the y coordinate of the hammer as a function of time:

y = y0 + v0 y t + 12 a y t 2 or, because y0 = h, v0y = v0sinθ, and ay = −g,

y = h + (v0 sin θ ) t − 12 gt 2

]

192 Chapter 3 Substitute numerical values to obtain:

y = 10 m + (4 m/s )(sin 30°) t

Substitute the conditions that exist when the hammer hits the ground:

0 = 10 m − (4 m/s ) sin 30° t

Solve for the time of fall to obtain:

t = 1.24 s

Use the x-coordinate equation to find the horizontal distance traveled by the hammer in 1.24 s:

R = (4 m/s )(cos30°)(1.24 s )

(

)

− 12 9.81 m/s 2 t 2

− 12 (9.81 m/s 2 ) t 2

= 4.29 m

115 •• Picture the Problem We’ll model Zacchini’s flight as though there is no air resistance and, hence, the acceleration is constant. Then we can use constant- acceleration equations to express the x and y coordinates of Zacchini’s motion as functions of time. Eliminating the parameter t between these equations will leave us with an equation we can solve forθ. Because the maximum height along a parabolic trajectory occurs (assuming equal launch and landing elevations) occurs at half range, we can use this same expression for y as a function of x to find h.

Use a constant-acceleration equation to express Zacchini’s horizontal position as a function of time:

x = x0 + v0 xt + 12 axt 2 or, because x0 = 0, v0x = v0cosθ, and ax = 0,

x = (v0 cos θ ) t

Use a constant-acceleration equation to express Zacchini’s vertical position as a function of time:

Eliminate the parameter t to obtain:

Use Zacchini’s coordinates when he lands in a safety net to obtain:

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, v0y = v0sinθ, and ay = −g,

y = (v0 sin θ ) t − 12 gt 2 y = (tan θ )x −

g x2 2v cos 2 θ

0 = (tan θ )R −

g R2 2 2v cos θ

2 0

2 0

Motion in One and Two Dimensions 193 Solve for his launch angle θ :

⎛ Rg ⎞

θ = 12 sin −1 ⎜⎜ 2 ⎟⎟ ⎝ v0 ⎠

(

)

⎡ (53 m ) 9.81 m/s 2 ⎤ ⎥ = 31.3° 2 ⎣ (24.2 m/s ) ⎦

Substitute numerical values and evaluate θ :

θ = 12 sin −1 ⎢

Use the fact that his maximum height was attained when he was halfway through his flight to obtain:

R g ⎛R⎞ h = (tan θ ) − 2 ⎜ ⎟ 2 2 2v0 cos θ ⎝ 2 ⎠

2

Substitute numerical values and evaluate h: 2

53 m 9.81 m/s 2 ⎛ 53 m ⎞ h = (tan 31.3°) − ⎜ ⎟ = 8.06 m 2 2 2 2(24.2 m/s ) cos 31.3° ⎝ 2 ⎠ 116 •• Picture the Problem Because the acceleration is constant; we can use the constantacceleration equations in vector form and the definitions of average velocity and average (instantaneous) acceleration to solve this problem. (a) The average velocity is given by:

The average velocity can also be expressed as:

r r r r ∆r r2 − r1 = v av = ∆t ∆t ˆ = (3 m/s)i + (−2.5 m/s) ˆj r r r v1 + v 2 vav = 2 and

r r r v1 = 2vav − v 2

Substitute numerical values to obtain: (b) The acceleration of the particle is given by:

r v1 = (1 m/s) iˆ + (1 m/s) ˆj r r r r ∆v v 2 − v1 = a= ∆t ∆t = (2 m/s 2 ) iˆ + (−3.5 m/s 2 ) ˆj

(c) The velocity of the particle as a function of time is:

r r r v (t ) = v1 + at = [(1 m/s) + (2 m/s 2 )t ] iˆ + [(1 m/s) + (−3.5 m/s 2 )t ] ˆj (d) Express the position vector as a function of time:

r r r r r (t ) = r1 + v1t + 12 at 2

194 Chapter 3 r

Substitute numerical values and evaluate r (t ) :

(

)

r r (t ) = [(4 m) + (1 m/s)t + (1 m/s 2 )t 2 ] iˆ + [(3 m) + (1 m/s)t + − 1.75 m/s 2 t 2 ] ˆj *117 •• Picture the Problem In the absence of air resistance, the steel ball will experience constant acceleration. Choose a coordinate system with its origin at the initial position of the ball, the x direction to the right, and the y direction downward. In this coordinate system y0 = 0 and a = g. Letting (x, y) be a point on the path of the ball, we can use constant-acceleration equations to express both x and y as functions of time and, using the geometry of the staircase, find an expression for the time of flight of the ball. Knowing its time of flight, we can find its range and identify the step it strikes first. The angle of the steps, with respect to the horizontal, is: Using a constant-acceleration equation, express the x coordinate of the steel ball in its flight: Using a constant-acceleration equation, express the y coordinate of the steel ball in its flight:

The equation of the dashed line in the figure is: Solve for the flight time:

⎛ 0.18 m ⎞ ⎟⎟ = 31.0° ⎝ 0.3 m ⎠

θ = tan −1 ⎜⎜

x = x0 + v0t + 12 a y t 2 or, because x0 = 0 and ay = 0,

x = v0t

y = y0 + v0 y t + 12 a y t 2 or, because y0 = 0, v0y = 0, and ay = g,

y = 12 gt 2 y gt = tan θ = x 2v0 t=

2v0 tan θ g

Find the x coordinate of the landing position:

x=

2v 2 y = 0 tan θ tan θ g

Substitute the angle determined in the first step:

x=

2(3 m/s ) tan31° = 1.10 m 9.81 m/s 2 2

The first step with x > 1.10 m is the 4th step.

Motion in One and Two Dimensions 195 118 •• Picture the Problem Ignoring the influence of air resistance, the acceleration of the ball is constant once it has left your hand and we can use constant-acceleration equations to express the x and y coordinates of the ball. Elimination of the parameter t will yield an equation from which we can determine v0. We can then use the y equation to express the time of flight of the ball and the x equation to express its range in terms of x0, v0,θ and the time of flight. Use a constant-acceleration equation to express the ball’s horizontal position as a function of time:

Use a constant-acceleration equation to express the ball’s vertical position as a function of time:

Eliminate the parameter t to obtain:

For the throw while standing on level ground we have:

x = x0 + v0 xt + 12 axt 2 or, because x0 = 0, v0x = v0cosθ, and ax = 0, x = (v0 cosθ )t (1)

y = y0 + v0 y t + 12 a y t 2 or, because y0 = x0, v0y = v0sinθ, and ay = −g, y = x0 + (v0 sin θ )t − 12 gt 2 (2)

y = x0 + (tan θ )x −

0 = (tan θ )x0 −

g x2 2 2v cos θ 2 0

g x02 2v cos 2 θ 2 0

and

x0 =

v02 v2 v2 sin 2θ = 0 sin 2(45°) = 0 g g g

Solve for v0:

v0 = gx0

At impact equation (2) becomes:

0 = x0 +

(

tflight =

x0 sin θ + sin 2 θ + 2 g

Solve for the time of flight:

)

2 gx0 sin θ tflight − 12 gtflight

(

)

196 Chapter 3 Substitute in equation (1) to express the range of the ball when thrown from an elevation x0 at an angle θ with the horizontal:

( =(

R=

) x (sin θ + cosθ ) g

gx0 cosθ tflight gx0

0

(

sin 2 θ + 2

= x0 cosθ sin θ + sin 2 θ + 2

Substitute θ = 0°, 30°, and 45°:

)

)

x(0°) = 1.41x0 x(30°) = 1.73 x0 and

x(45°) = 1.62 x0 119 ••• Picture the Problem Choose a coordinate system with its origin at the point where the motorcycle becomes airborne and with the positive x direction to the right and the positive y direction upward. With this choice of coordinate system we can relate the x and y coordinates of the motorcycle (which we’re treating as a particle) using Equation 3-21. (a) The path of the motorcycle is given by: For the jump to be successful, h < y(x). Solving for v0, we find:

⎛ ⎞ g y ( x ) = (tanθ ) x − ⎜⎜ 2 2 ⎟⎟ x 2 ⎝ 2v 0 cos θ ⎠ vmin >

x cosθ

g 2( x tan θ − h)

(b) Use the values given to obtain:

vmin > 26.0 m/s or 58.0 mph

(c) In order for our expression for vmin to be real valued; i.e., to predict values for vmin that are physically meaningful, x tanθ − h > 0.

∴ hmax < x tanθ The interpretation is that the bike "falls away" from traveling on a straight-line path due to the free-fall acceleration downwards. No matter what the initial speed of the bike, it must fall a little bit before reaching the other side of the pit.

Motion in One and Two Dimensions 197 120 ••• Picture the Problem Let the origin be at the position of the boat when it was engulfed by the fog. Take the x and y directions to be east and north, r respectively. Let v BW be the velocity of

r

the boat relative to the water, v BS be the velocity of the boat relative to the shore, r and v WS be the velocity of the water with respect to the shore. Then

r r r v BS = v BW + v WS . r θ is the angle of v WS with respect to the x (east) direction. (a) Find the position vector for the boat at t = 3 h:

r rboat = {(32 km )(cos 135°)t}iˆ + {(32 km )(sin135°) t − 4 km} ˆj = {(− 22.6 km )t}iˆ + {(22.6 km ) t − 4 km} ˆj

Find the coordinates of the boat at t = 3 h:

rx = [(10 km/h ) cos135° + vWS cos θ ](3 h ) and

ry = [(10 km/h )sin 135° + vWS sin θ ](3 h )

Simplify the expressions involving rx and ry and equate these simplified expressions to the x and y components of the position vector of the boat: Divide the second of these equations by the first to obtain:

3vWS cosθ = −1.41 km/h and 3vWS sinθ = −2.586 km/h

tan θ =

− 2.586 km − 1.41 km

or

⎛ − 2.586 km ⎞ ⎟⎟ = 61.4° or 241.4° − 1.41 km ⎝ ⎠

θ = tan −1 ⎜⎜

Because the boat has drifted south, use θ = 241.4° to obtain:

vWS

1.41 km/h − vx 3 = = cos θ cos(241.4°) = 0.982 km/h at θ = 241.4°

198 Chapter 3 (b) Letting φ be the angle between east and the proper heading for the boat, express the components of the velocity of the boat with respect to the shore:

vBS,x = (10 km/h) cosφ + (0.982 km/h) cos(241.3°)

For the boat to travel northwest:

vBS,x = –vBS,y

Substitute the velocity components, square both sides of the equation, and simplify the expression to obtain the equations:

sinφ + cosφ = 0.133, sin2φ + cos2φ + 2 sinφ cosφ = 0.0177, and 1 + sin(2φ) = 0.0177

Solve for φ:

φ = 129.6° or 140.4°

Because the current pushes south, the boat must head more northerly than 135°:

Using 129.6°, the correct heading

(c) Find vBS:

vBS,x = –6.84 km/h and vBS = vBx /cos135° = 9.68 km/h

To find the time to travel 32 km, divide the distance by the boat’s actual speed:

t = (32 km)/(9.68 km/h)

vBS,y = (10 km/h) sinφ + (0.982 km/h) sin(241.3°)

is 39.6° west of north .

= 3.31 h = 3 h 18 min

*121 •• Picture the Problem In the absence of air resistance, the acceleration of the projectile is constant and the equation of a projectile for equal initial and final elevations, which was derived from the constant-acceleration equations, is applicable. We can use the equation giving the range of a projectile for equal initial and final elevations to evaluate the ranges of launches that exceed or fall short of 45° by the same amount. Express the range of the projectile as a function of its initial speed and angle of launch: Let θ0= 45° ± θ:

R=

v02 sin 2θ 0 g

R=

v02 sin (90° ± 2θ ) g

= Because cos(–θ) = cos(+θ) (the cosine function is an even function):

v02 cos(± 2θ ) g

R(45° + θ ) = R(45° − θ )

Motion in One and Two Dimensions 199 122 •• Picture the Problem In the absence of air resistance, the acceleration of both balls is that due to gravity and the horizontal and vertical motions are independent of each other. Choose a coordinate system with the origin at the base of the cliff and the coordinate axes oriented as shown and use constant-acceleration equations to relate the x and y components of the ball’s speed. Independently of whether a ball is thrown upward at the angle α or downward at β, the vertical motion is described by:

v y2 = v02y + 2a∆y

The horizontal component of the motion is given by:

vx = v0x

Find v at impact from its components:

v = v x2 + v y2 = v02x + v02y − 2 gh

= v02y − 2 gh

=

v02 − 2 gh

200 Chapter 3

Chapter 4 Newton’s Laws Conceptual Problems *1 •• Determine the Concept A reference frame in which the law of inertia holds is called an inertial reference frame. If an object with no net force acting on it is at rest or is moving with a constant speed in a straight line (i.e., with constant velocity) relative to the reference frame, then the reference frame is an inertial reference frame. Consider sitting at rest in an accelerating train or plane. The train or plane is not an inertial reference frame even though you are at rest relative to it. In an inertial frame, a dropped ball lands at your feet. You are in a noninertial frame when the driver of the car in which you are riding steps on the gas and you are pushed back into your seat. 2 •• Determine the Concept A reference frame in which the law of inertia holds is called an inertial reference frame. A reference frame with acceleration a relative to the initial frame, and with any velocity relative to the initial frame, is inertial. 3 • Determine the Concept No. If the net force acting on an object is zero, its acceleration is zero. The only conclusion one can draw is that the net force acting on the object is zero. *4 • Determine the Concept An object accelerates when a net force acts on it. The fact that an object is accelerating tells us nothing about its velocity other than that it is always changing. Yes, the object must have an acceleration relative to the inertial frame of reference. According to Newton’s 1st and 2nd laws, an object must accelerate, relative to any inertial reference frame, in the direction of the net force. If there is ″only a single nonzero force,″ then this force is the net force. Yes, the object’s velocity may be momentarily zero. During the period in which the force is acting, the object may be momentarily at rest, but its velocity cannot remain zero because it must continue to accelerate. Thus, its velocity is always changing. 5 • Determine the Concept No. Predicting the direction of the subsequent motion correctly requires knowledge of the initial velocity as well as the acceleration. While the acceleration can be obtained from the net force through Newton’s 2nd law, the velocity can only be obtained by integrating the acceleration. 6 • Determine the Concept An object in an inertial reference frame accelerates if there is a net force acting on it. Because the object is moving at constant velocity, the net force acting on it is zero. (c ) is correct.

201

202 Chapter 4 7 • Determine the Concept The mass of an object is an intrinsic property of the object whereas the weight of an object depends directly on the local gravitational field. Therefore, the mass of the object would not change and wgrav = mg local . Note that if the gravitational field is zero then the gravitational force is also zero. *8 • Determine the Concept If there is a force on her in addition to the gravitational force, she will experience an additional acceleration relative to her space vehicle that is proportional to the net force required producing that acceleration and inversely proportional to her mass. She could do an experiment in which she uses her legs to push off from the wall of her space vehicle and measures her acceleration and the force exerted by the wall. She could calculate her mass from the ratio of the force exerted by the wall to the acceleration it produced. *9 • Determine the Concept One’s apparent weight is the reading of a scale in one’s reference frame. Imagine yourself standing on a scale that, in turn, is on a platform accelerating upward with an acceleration a. The freebody diagram shows the force the r gravitational field exerts on you, mg, and

r

the force the scale exerts on you, wapp . The scale reading (the force the scale exerts on you) is your apparent weight.

Choose the coordinate system shown in the free-body diagram and apply

r

r

∑ F = ma

to the scale:

∑F

y

= wapp − mg = ma y

or

wapp = mg + ma y

So, your apparent weight would be greater than your true weight when observed from a reference frame that is accelerating upward. That is, when the surface on which you are standing has an acceleration a such that a y is positive: a y > 0 . 10 •• Determine the Concept Newton's 2nd law tells us that forces produce changes in the velocity of a body. If two observers pass each other, each traveling at a constant velocity, each will experience no net force acting on them, and so each will feel as if he or she is standing still. 11 • Determine the Concept Neither block is accelerating so the net force on each block is zero. Newton’s 3rd law states that objects exert equal and opposite forces on each other.

Newton’s Laws 203 (a) and (b) Draw the free-body diagram for the forces acting on the block of mass m1:

Apply

r

r

∑ F = ma to the block 1:

∑F

y

= Fn21 − m1 g = m1a1

or, because a1 = 0,

Fn21 − m1 g = 0

Therefore, the magnitude of the force that block 2 exerts on block 1 is given by:

Fn21 = m1 g

From Newton’s 3rd law of motion we know that the force that block 1 exerts on block 2 is equal to, but opposite in direction, the force that block 2 exerts on block 1.

r r Fn21 = − Fn12 ⇒ Fn12 = m1 g

(c) and (d) Draw the free-body diagram for the forces acting on block 2:

Apply

r

r

∑ F = ma to block 2:

∑F

2y

= FnT2 − Fn12 − m2 g = m2 a2

or, because a2 = 0,

FnT2 = FnT2 + m2 g = m1 g + m2 g = (m1 + m2 )g

and the normal force that the table exerts on body 2 is FnT2 =

From Newton’s 3rd law of motion we know that the force that block 2 exerts on the table is equal to, but opposite in direction, the force that the table exerts on block 2.

(m1 + m 2 )g

r r FnT2 = − Fn2T ⇒ Fn2T = (m1 + m2 ) g

204 Chapter 4 *12 • (a) True. By definition, action-reaction force pairs cannot act on the same object. (b) False. Action equals reaction independent of any motion of the two objects. 13 • Determine the Concept Newton’s 3rd law of motion describes the interaction between the man and his less massive son. According to the 3rd law description of the interaction of two objects, these are action-reaction forces and therefore must be equal in magnitude.

(b) is correct. 14 • Determine the Concept According to Newton’s 3rd law the reaction force to a force exerted by object A on object B is the force exerted by object B on object A. The bird’s weight is a gravitational field force exerted by the earth on the bird. Its reaction force is the gravitational force the bird exerts on the earth. (b) is correct. 15 • Determine the Concept We know from Newton’s 3rd law of motion that the reaction to the force that the bat exerts on the ball is the force the ball exerts on the bat and is equal in magnitude but oppositely directed. The action-reaction pair consists of the force with which the bat hits the ball and the force the ball exerts on the bat. These forces are equal in magnitude, act in opposite directions. (c) is correct. 16 • Determine the Concept The statement of Newton’s 3rd law given in the problem is not complete. It is important to remember that the action and reaction forces act on different bodies. The reaction force does not cancel out because it does not act on the same body as the external force. *17 • Determine the Concept The force diagrams will need to include the ceiling, string, object, and earth if we are to show all of the reaction forces as well as the forces acting on the object. (a) The forces acting on the 2.5-kg

r

object are its weight W, and the

r

tension T1, in the string. The reaction

r

forces are W ' acting on the earth and

r T1 ' acting on the string.

Newton’s Laws 205 (b) The forces acting on the string are its weight, the weight of the object,

r

and F, the force exerted by the ceiling. The reaction forces are

r r T1 acting on the string and F ' acting

on the ceiling.

18 • Determine the Concept Identify the objects in the block’s environment that are exerting forces on the block and then decide in what directions those forces must be acting if the block is sliding down the inclined plane. Because the incline is frictionless, the force the incline exerts on the block must be normal to the surface. The second object capable of exerting a force on the block is the earth and its force; the weight of the block acts directly downward. The magnitude of the normal force is less than that of the weight because it supports only a portion of the weight.

The forces shown in FBD (c) satisfy these conditions.

19 • Determine the Concept In considering these statements, one needs to decide whether they are consistent with Newton’s laws of motion. A good strategy is to try to think of a counterexample that would render the statement false. (a) True. If there are no forces acting on an object, the net force acting on it must be zero and, hence, the acceleration must be zero. (b) False. Consider an object moving with constant velocity on a frictionless horizontal surface. While the net force acting on it is zero (it is not accelerating), gravitational and normal forces are acting on it. (c) False. Consider an object that has been thrown vertically upward. While it is still rising, the direction of the gravitational force acting on it is downward. (d) False. The mass of an object is an intrinsic property that is independent of its location (the gravitational field in which it happens to be situated). 20 • Determine the Concept In considering these alternatives, one needs to decide which alternatives are consistent with Newton’s 3rd law of motion. According to Newton’s 3rd law, the magnitude of the gravitational force exerted by her body on the earth is equal and opposite to the force exerted by the earth on her. (a ) is correct.

206 Chapter 4 *21 • Determine the Concept In considering these statements, one needs to decide whether they are consistent with Newton’s laws of motion. In the absence of a net force, an object moves with constant velocity. (d ) is correct. 22 • Determine the Concept Draw the freebody diagram for the towel. Because the towel is hung at the center of the line, the

r

r

magnitudes of T1 and T2 are the same.

No. To support the towel, the tension in the line must have a vertical component equal to the towel’s weight. Thus θ > 0. 23 • Determine the Concept The free-body diagram shows the forces acting on a person in a descending elevator. The upward force exerted by the scale on the r person, wapp , is the person’s apparent weight.

Apply

∑F

y

= ma y to the person

and solve for wapp:

wapp – mg = may or wapp = mg + may = m(g + ay)

Because wapp is independent of v, the velocity of the elevator has no effect on the person' s apparent weight. Remarks: Note that a nonconstant velocity will alter the apparent weight.

Estimation and Approximation 24 •• Picture the Problem Assuming a stopping distance of 25 m and a mass of 80 kg, use Newton’s 2nd law to determine the force exerted by the seat belt. The force the seat belt exerts on the driver is given by:

Fnet = ma, where m is the mass of the driver.

Newton’s Laws 207 Using a constant-acceleration equation, relate the velocity of the car to its stopping distance and acceleration: Solve for a:

v 2 = v02 + 2a∆x or, because v = 0,

− v02 = 2a∆x a=

Substitute numerical values and evaluate a:

− v02 2∆x

⎛ km 1h 103 m ⎞ ⎜⎜ 90 ⎟ × × h 3600 s km ⎟⎠ ⎝ a=− 2(25 m )

2

= −12.5 m/s 2 Substitute for a and evaluate Fnet:

(

Fnet = (80 kg ) − 12.5 m/s 2

)

= − 1.00 kN Fnet is negative because it is opposite the direction of motion. *25 ••• Picture the Problem The free-body diagram shows the forces acting on you and your bicycle as you are either ascending or descending the grade. The magnitude of the normal force acting on you and your bicycle is equal to the component of your weight in the y direction and the magnitude of the tangential force is the x component of your weight. Assume a combined mass (you plus your bicycle) of 80 kg. (a) Apply

∑F

y

= ma y to you and

your bicycle and solve for Fn:

Fn – mg cosθ = 0, because there is no acceleration in the y direction. ∴ Fn = mg cosθ

Determine θ from the information concerning the grade:

tanθ = 0.08 and θ = tan−1(0.08) = 4.57°

Substitute to determine Fn:

Fn = (80 kg)(9.81 m/s2) cos4.57° = 782 N

Apply

∑F

x

= ma x to you and your

bicycle and solve for Ft, the tangential force exerted by the road on the wheels:

Ft – mg sinθ = 0, because there is no acceleration in the x direction.

208 Chapter 4 Evaluate Ft:

Ft = (80 kg)(9.81 m/s2) sin4.57° = 62.6 N

(b)

Because there is no acceleration, the forces are the same going up and going down the incline.

Newton’s First and Second Laws: Mass, Inertia, and Force 26 • Picture the Problem The acceleration of the particle can be found from the stopping distance by using a constant-acceleration equation. The mass of the particle and its acceleration are related to the net force through Newton’s second law of motion. Choose a coordinate system in which the direction the particle is moving is the v r positive x direction and apply Fnet = ma. Use Newton’s 2nd law to relate the mass of the particle to the net force acting on it and its acceleration: Because the force is constant, use a constant-acceleration equation with vx = 0 to determine a:

m=

Fnet ax

vx2 = v02x + 2ax ∆x and

− v02x ax = 2∆x

Substitute to obtain:

m=

2∆xFnet v02x

Substitute numerical values and evaluate m:

m=

2(62.5 m) (15.0 N) = 3.00 kg (25.0 m/s)2

and (b) is correct. 27 • Picture the Problem The acceleration of the object is related to its mass and the net force acting on it by Fnet = F0 = ma. (a) Use Newton’s 2nd law of motion to calculate the acceleration of the object:

a=

Fnet 2 F0 = m m

(

)

= 2 3 m/s 2 = 6.00 m/s 2

Newton’s Laws 209 (b) Let the subscripts 1 and 2 distinguish the two objects. The ratio of the two masses is found from Newton’s 2nd law:

1 m2 F0 a2 a1 3 m/s 2 = = = = 2 3 m1 F0 a1 a2 9 m/s

(c) The acceleration of the two-mass system is the net force divided by the total mass m = m1 + m2:

a= =

Fnet F0 = m m1 + m2 F0 m1 a1 = 1 + m2 m1 1 + 1 3

= 34 a1 = 2.25 m/s 2 28 • Picture the Problem The acceleration of an object is related to its mass and the net force acting on it by Fnet = ma . Let m be the mass of the ship, a1 be the acceleration of the ship when the net force acting on it is F1, and a2 be its acceleration when the net force is F1 + F2 . Using Newton’s 2nd law, express the net force acting on the ship when its acceleration is a1:

F1 = ma1

Express the net force acting on the ship when its acceleration is a2:

F1 + F2 = ma2

Divide the second of these equations by the first and solve for the ratio F2/F1:

F1 + F2 ma1 = F1 ma2 and

F2 a2 = −1 F1 a1 Substitute for the accelerations to determine the ratio of the accelerating forces and solve for F2:

F2 (16 km/h ) (10 s ) = −1 = 3 (4 km/h ) (10 s ) F1 or

F2 = 3F1 *29 •• Picture the Problem Because the deceleration of the bullet is constant, we can use a constant-acceleration equation to determine its acceleration and Newton’s 2nd law of motion to find the average resistive force that brings it to a stop. Apply

r

r

∑ F = ma

to express the

Fwood = ma

force exerted on the bullet by the wood: Using a constant-acceleration

v 2 = v02 + 2a∆x

210 Chapter 4 equation, express the final velocity of the bullet in terms of its acceleration and solve for the acceleration: Substitute to obtain:

Substitute numerical values and evaluate Fwood:

and

a=

v 2 − v02 − v02 = 2∆x 2∆x

Fwood = − Fwood = −

mv02 2∆x

(1.8 ×10

)

kg (500 m/s ) 2(0.06 m ) −3

2

= − 3.75 kN where the negative sign means that the direction of the force is opposite the velocity. *30 •• Picture the Problem The pictorial representation summarizes what we know about the motion. We can find the acceleration of the cart by using a constant-acceleration equation.

The free-body diagram shows the forces acting on the cart as it accelerates along the air track. We can determine the net force acting on the cart using Newton’s 2nd law and our knowledge of its acceleration.

(a) Apply

∑F

x

= ma x to the cart

F = ma

to obtain an expression for the net force F: Using a constant-acceleration equation, relate the displacement of the cart to its acceleration, initial speed, and travel time:

∆x = v0 ∆t + 12 a (∆t )

2

or, because v0 = 0,

Newton’s Laws 211 ∆x = 12 a (∆t )

2

2∆x (∆t )2

Solve for a:

a=

Substitute for a in the force equation to obtain:

F =m

Substitute numerical values and evaluate F:

F=

(b) Using a constant-acceleration equation, relate the displacement of the cart to its acceleration, initial speed, and travel time:

∆x = v0 ∆t + 12 a ' (∆t )

Solve for ∆t:

2(0.355 kg )(1.5 m ) = 0.0514 N (4.55 s )2 2

or, because v0 = 0,

∆x = 12 a ' (∆t )

2

∆t =

If we assume that air resistance is negligible, the net force on the cart is still 0.0514 N and its acceleration is: Substitute numerical values and evaluate ∆t:

2∆x 2m∆x = (∆t )2 (∆t )2

a' =

2∆x a'

0.0514 N = 0.0713 m/s 2 0.722 kg

∆t =

2(1.5 m ) = 6.49 s 0.0713 m/s 2

31 • Picture the Problem The acceleration of an object is related to its mass and the net force

r

r

acting on it according to Fnet = ma. Let m be the mass of the object and choose a coordinate system in which the direction of 2F0 in (b) is the positive x and the direction of the left-most F0 in (a) is the positive y direction. Because both force and acceleration are vector quantities, find the resultant force in each case and then find the resultant acceleration. (a) Calculate the acceleration of the object from Newton’s 2nd law of motion: Express the net force acting on the object:

Find the magnitude and direction of this net force:

r r Fnet a= m r Fnet = Fx iˆ + Fy ˆj = F0 iˆ + F0 ˆj and

Fnet = Fx2 + Fy2 = 2 F0 and

212 Chapter 4 ⎛ Fy ⎝ Fx

θ = tan −1 ⎜⎜ Use this result to calculate the magnitude and direction of the acceleration:

⎛F ⎞ ⎞ ⎟⎟ = tan −1 ⎜⎜ 0 ⎟⎟ = 45° ⎝ F0 ⎠ ⎠

Fnet 2 F0 = = 2 a0 m m = 2 (3 m/s 2 )

a =

4.24 m/s 2 @ 45.0° from each = force. (b) Calculate the acceleration of the object from Newton’s 2nd law of motion: Express the net force acting on the object:

r r a = Fnet /m r Fnet = Fx iˆ + Fy ˆj = (− F0 sin 45°)iˆ + (2 F0 + F0 cos 45°) ˆj

Find the magnitude and direction of this net force:

(− F0 sin 45°)2 + (2 F0 + F0 cos 45°)2

Fnet = Fx2 + Fy2 =

= 2.80 F0

and

⎛ Fy ⎝ Fx

⎞ ⎛ 2 F + F0 cos 45° ⎞ ⎟⎟ = tan −1 ⎜⎜ 0 ⎟⎟ = −75.4° − ° sin 45 F 0 ⎠ ⎝ ⎠ r = 14.6° from 2F0

θ = tan −1 ⎜⎜

Use this result to calculate the magnitude and direction of the acceleration:

Fnet F = 2.80 0 = 2.80a0 m m 2 = 2.80(3 m/s )

a=

r = 8.40 m/s 2 @ 14.6° from 2F0 32 • Picture the Problem The acceleration of an object is related to its mass and the net force

r

r

acting on it according to a = Fnet m .

r

r

Apply a = Fnet m to the object to obtain:

r r Fnet (6 N ) iˆ − (3 N ) ˆj a= = m 1.5 kg =

(4.00 m/s ) iˆ − (2.00 m/s ) ˆj 2

2

Newton’s Laws 213 r

Find the magnitude of a :

a = a x2 + a y2 =

(4.00 m/s ) + (2.00 m/s ) 2 2

2 2

= 4.47 m/s 2 33 • Picture the Problem The mass of the particle is related to its acceleration and the net force acting on it by Newton’s 2nd law of motion. Because the force is constant, we can use constant-acceleration formulas to calculate the acceleration. Choose a coordinate system in which the positive x direction is the direction of motion of the particle. The mass is related to the net force and the acceleration by Newton’s 2nd law:

r ∑ F Fx m= r = a ax ∆x = v0 x t + 12 a x (∆t ) , where v0 x = 0,

Because the force is constant, the acceleration is constant. Use a constant-acceleration equation to find the acceleration:

so

Substitute this result into the first equation and solve for and evaluate the mass m of the particle:

(12 N )(6 s ) F F (∆t ) = m= x = x 2∆x 2(18 m ) ax

2

ax =

2∆x (∆t )2 2

= 12.0 kg *34 • Picture the Problem The speed of either Al or Bert can be obtained from their accelerations; in turn, they can be obtained from Newtons 2nd law applied to each person. The free-body diagrams to the right show the forces acting on Al and Bert. The forces that Al and Bert exert on each other are action-and-reaction forces. (a) Apply ∑ Fx = ma x to Bert and solve for his acceleration:

− FAl on Bert = mBert aBert aBert =

− FAl on Bert mBert

=

− 20 N 100 kg

= −0.200 m/s 2 Using a constant-acceleration equation, relate Bert’s speed to his initial speed, speed after 1.5 s, and acceleration and solve for his speed at the end of 1.5 s:

v = v0 + a∆t = 0 + (−0.200 m/s2)(1.5 s) = − 0.300 m/s

2

214 Chapter 4 (b) From Newton's 3rd law, an equal but oppositely directed force acts on Al while he pushes Bert. Because the ice is frictionless, Al speeds off in the opposite direction. Apply Newton’s 2nd law to the forces acting on Al and solve for his acceleration: Using a constant-acceleration equation, relate Al’s speed to his initial speed, speed after 1.5 s, and acceleration; solve for his speed at the end of 1.5 s:

∑F

x , Al

= FBert on Al = mAl aAl

and

aAl =

FBert on Al 20 N = mAl 80 kg

= 0.250 m/s 2

v = v0 + a∆t = 0 + (0.250 m/s2)(1.5 s) = 0.375 m/s

35 • Picture the Problem The free-body diagrams show the forces acting on the two blocks. We can apply Newton’s second law to the forces acting on the blocks and eliminate F to obtain a relationship between the masses. Additional applications of Newton’s 2nd law to the sum and difference of the masses will lead us to values for the accelerations of these combinations of mass.

(a) Apply

∑F

x

= ma x to the two

blocks:

∑F

x ,1

and

∑F

x,2

Eliminate F between the two equations and solve for m2:

= F = m1 a1

m2 =

Express and evaluate the acceleration of an object whose mass is m2 – m1 when the net force acting on it is F:

a=

(b) Express and evaluate the acceleration of an object whose mass is m2 + m1 when the net force acting on it is F:

a=

= F = m2 a 2

a1 12 m/s 2 m1 = m1 = 4m1 a2 3 m/s 2

F F F = = m2 − m1 4m1 − m1 3m1

(

)

= 13 a1 = 13 12 m/s 2 = 4.00 m/s 2

=

F F = m2 + m1 4m1 + m1 F = 15 a1 = 15 (12 m/s 2 ) 5m1

= 2.40 m/s 2

Newton’s Laws 215 36 • Picture the Problem Because the velocity is constant, the net force acting on the log must be zero. Choose a coordinate system in which the positive x direction is the direction of motion of the log. The freebody diagram shows the forces acting on the log when it is accelerating in the positive x direction. (a) Apply

∑F

x

= ma x to the log

Fpull – Fres = max = 0

when it is moving at constant speed: Solve for and evaluate Fres:

(b) Apply

∑F

x

= ma x to the log

Fres = Fpull = 250 N Fpull – Fres = max

when it is accelerating to the right: Solve for and evaluate Fpull:

Fpull = Fres + max = 250 N + (75 kg) (2 m/s2) = 400 N

37 • Picture the Problem The acceleration can be found from Newton’s 2nd law. Because both forces are constant, the net force and the acceleration are constant; hence, we can use the constant-acceleration equations to answer questions concerning the motion of the object at various times. (a) Apply Newton’s 2nd law to the object to obtain:

r r r r Fnet F1 + F2 a= = m m ˆ (6 N ) i + (− 14 N ) ˆj = 4 kg

(

) (

)

= 1.50 m/s 2 iˆ + − 3.50 m/s 2 ˆj (b) Using a constant-acceleration equation, express the velocity of the object as a function of time and solve for its velocity when t = 3 s:

(c) Express the position of the object in terms of its average velocity and evaluate this expression at t = 3 s:

r r r v = v0 + at

[(

) (

)]

= 0 + 1.50 m/s 2 iˆ + − 3.50 m/s 2 ˆj (3 s ) =

(4.50 m/s) iˆ + (− 10.5 m/s) ˆj

r r r = vavt r = 12 v t =

(6.75 m ) iˆ + (− 15.8 m ) ˆj

216 Chapter 4

Mass and Weight *38 • Picture the Problem The mass of the astronaut is independent of gravitational fields and will be the same on the moon or, for that matter, out in deep space. Express the mass of the astronaut in terms of his weight on earth and the gravitational field at the surface of the earth:

m=

wearth 600 N = = 61.2 kg g earth 9.81 N/kg

and (c) is correct.

39 • Picture the Problem The weight of an object is related to its mass and the gravitational field through w = mg. (a) The weight of the girl is:

w = mg = (54 kg )(9.81 N/kg ) = 530 N

(b) Convert newtons to pounds:

w=

530 N = 119 lb 4.45 N/lb

40 • Picture the Problem The mass of an object is related to its weight and the gravitational field. Find the weight of the man in newtons: Calculate the mass of the man from his weight and the gravitational field:

165 lb = (165 lb )(4.45 N/lb ) = 734 N

m=

w 734 N = = 74.8 kg g 9.81 N/kg

Contact Forces *41 • Picture the Problem Draw a free-body diagram showing the forces acting on the

r

block. Fk is the force exerted by the spring,

r r r W = mg is the weight of the block, and Fn

is the normal force exerted by the horizontal surface. Because the block is resting on a surface, Fk + Fn = W.

(a) Calculate the force exerted by the spring on the block:

Fx = kx = (600 N/m )(0.1 m ) = 60.0 N

Newton’s Laws 217 (b) Choosing the upward direction to be positive, sum the forces acting on the block and solve for Fn:

r F ∑ = 0 ⇒ Fk + Fn − W = 0 and

Fn = W − Fk Substitute numerical values and evaluate Fn:

Fn = (12 kg)(9.81 N/kg) − 60 N = 57.7 N

42 • Picture the Problem Let the positive x direction be the direction in which the spring is stretched. We can use Newton’s 2nd law and the expression for the force exerted by a stretched (or compressed) spring to express the acceleration of the box in terms of its mass m, the stiffness constant of the spring k, and the distance the spring is stretched x. Apply Newton’s 2nd law to the box to obtain:

a=

∑F m

Express the force exerted on the box by the spring:

F = − kx

Substitute to obtain:

a=

Substitute numerical values and evaluate a:

a=−

− kx m

(800 N/m )(0.04 m ) 6 kg

= − 5.33 m/s 2 where the minus sign tells us that the box’s acceleration is toward its equilibrium position.

Free-Body Diagrams: Static Equilibrium 43 • Picture the Problem Because the traffic light is not accelerating, the net force acting on

r

r

r

it must be zero; i.e., T1 + T2 + mg = 0.

Construct a free-body diagram showing the forces acting on the knot and choose the coordinate system shown:

218 Chapter 4 Apply

∑F

x

= ma x to the knot:

Solve for T2 in terms of T1:

T1cos30° − T2cos60° = max = 0

cos 30° T1 = 1.73T1 cos 60° ∴ T2 is greater than T1 T2 =

44 • Picture the Problem Draw a free-body diagram showing the forces acting on the lamp and apply Fy = 0 .

∑

From the FBD, it is clear that T1 supports the full weight mg = 418 N.

Apply

∑F

y

= 0 to the lamp to

T1 − w = 0

obtain: Solve for T1: Substitute numerical values and evaluate T1:

T1 = w = mg

(

)

T1 = (42.6 kg ) 9.81 m/s 2 = 418 N and (b) is correct.

*45 •• Picture the Problem The free-body diagrams for parts (a), (b), and (c) are shown below. In both cases, the block is in equilibrium under the influence of the forces and we can use Newton’s 2nd law of motion and geometry and trigonometry to obtain relationships between θ and the tensions. (a) and (b)

(c)

(a) Referring to the FBD for part (a), use trigonometry to determine θ :

θ = cos −1

0.5 m = 36.9° 0.625 m

Newton’s Laws 219

2T sin θ − mg = 0 since a = 0

(b) Noting that T = T′, apply Fy = ma y to the 0.500-kg block

∑

and

and solve for the tension T:

T=

Substitute numerical values and evaluate T:

T=

mg 2 sin θ

(0.5 kg )(9.81m/s2 ) = 2sin36.9°

(c) The length of each segment is:

1.25 m = 0.417 m 3

Find the distance d:

d=

Express θ in terms of d and solve for its value:

4.08 N

1 m − 0.417m 2 = 0.2915 m

⎛

⎞ ⎛ 0.2915 m ⎞ d ⎟⎟ = cos −1 ⎜⎜ ⎟⎟ ⎝ 0.417 m ⎠ ⎝ 0.417 m ⎠

θ = cos −1 ⎜⎜ = 45.7°

Apply

∑F

y

= ma y to the 0.250-kg

block and solve for the tension T3:

T3 sin θ − mg = 0 since a = 0. and

T3 = Substitute numerical values and evaluate T3: Apply

∑F

x

= ma x to the 0.250-kg

T3 =

mg sin θ

(0.25 kg )(9.81m/s2 ) = sin45.7°

3.43 N

T3 cos θ − T2 = 0 since a = 0.

block and solve for the tension T2:

and

Substitute numerical values and evaluate T2:

T2 = (3.43 N ) cos 45.7° = 2.40 N

By symmetry:

T1 = T3 = 3.43 N

T2 = T3 cosθ

220 Chapter 4 46 • Picture the Problem The suspended body is in equilibrium under the influence of the r r r forces Th , T45 , and mg;

r

r

r

i.e., Th + T45 + mg = 0 Draw the free-body diagram of the forces acting on the knot just above the 100-N body. Choose a coordinate system with the positive x direction to the right and the positive y direction upward. Apply the conditions for translational equilibrium to determine the tension in the horizontal cord. If the system is to remain in static equilibrium, the vertical component of T45 must be exactly balanced by, and therefore equal to, the tension in the string suspending the 100-N body:

Tv = T45 sin45° = mg

Express the horizontal component of T45:

Th = T45 cos45°

Because T45 sin45° = T45 cos45°:

Th = mg = 100 N

47 • Picture the Problem The acceleration of any object is directly proportional to the net force acting on it. Choose a coordinate system in which the positive x direction is the

r

same as that of F1 and the positive y direction is to the right. Add the two forces to determine the net force and then use Newton’s 2nd law to find the acceleration of the

r

r

r

r

object. If F3 brings the system into equilibrium, it must be true that F3 + F1 + F2 = 0.

r

(a) Find the components of F1 and

r F2 :

r F1 = (20 N) iˆ r F2 = {(−30 N) sin 30°}iˆ + {(30 N) cos 30°} ˆj = (−15 N)iˆ + (26 N) ˆj

r

r

r

Add F1 and F2 to find Ftot :

r Ftot = (5 N) iˆ + (26 N) ˆj

Newton’s Laws 221 Apply

r r F ∑ = ma to find the

acceleration of the object:

r r Ftot a= m = (0.500 m/s 2 ) iˆ + (2.60 m/s 2 ) ˆj

(b) Because the object is in equilibrium under the influence of the three forces, it must be true that:

r r r F3 + F1 + F2 = 0 and

(

r r r F3 = − F1 + F2

)

(− 5.00 N ) iˆ + (− 26.0 N ) ˆj

=

*48 • r r Picture the Problem The acceleration of the object equals the net force, T − mg , divided by the mass. Choose a coordinate system in which upward is the positive y direction. Apply Newton’s 2nd law to the forces acting on this body to find the acceleration of the object as a function of T. (a) Apply

∑F

y

= ma y to the

T – w = T – mg = may

object: Solve this equation for a as a function of T:

ay =

T −g m

Substitute numerical values and evaluate ay:

ay =

5N − 9.81 m/s 2 = − 8.81 m/s 2 5 kg

(b) Proceed as in (a) with T = 10 N:

a = − 7.81m/s 2

(c) Proceed as in (a) with T = 100 N:

a = 10.2 m/s 2

49 •• Picture the Problem The picture is in equilibrium under the influence of the three forces r shown in the figure. Due to the symmetry of the support system, the vectors T and r T ' have the same magnitude T. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Apply the condition for translational equilibrium to obtain an expression for T as a function of θ and w. (a) Referring to Figure 4-37, apply the condition for translational equilibrium in the vertical direction and solve for T:

∑F

y

=2T sin θ − w = 0

and

T=

w 2 sin θ

222 Chapter 4 Tmin occurs when sinθ is a maximum: Tmax occurs when sinθ is a minimum. Because the function is undefined when sinθ = 0, we can conclude that: (b) Substitute numerical values in the result in (a) and evaluate T:

θ = sin −1 1 = 90° T → Tmax as θ → 0°

( 2 kg ) (9.81 m/s 2 ) T= = 2sin30°

19.6 N

Remarks: θ = 90° requires wires of infinite length; therefore it is not possible. As θ gets small, T gets large without limit. *50 ••• Picture the Problem In part (a) we can apply Newton’s 2nd law to obtain the given expression for F. In (b) we can use a symmetry argument to find an expression for tan θ0. In (c) we can use our results obtained in (a) and (b) to express xi and yi. (a) Apply

∑F

y

= 0 to the balloon:

F + Ti sin θ i − Ti −1 sin θ i −1 = 0

Solve for F to obtain:

F = Ti −1 sin θ i −1 − Ti sin θ i

(b) By symmetry, each support must balance half of the force acting on the entire arch. Therefore, the vertical component of the force on the support must be NF/2. The horizontal component of the tension must be TH. Express tanθ0 in terms of NF/2 and TH:

tan θ 0 =

NF 2 NF = TH 2TH

By symmetry, θN+1 = − θ0. Therefore, because the tangent function is odd:

tan θ 0 = − tan θ N +1 =

(c) Using TH = Ti cosθi = Ti−1cosθi−1, divide both sides of our result in (a) by TH and simplify to obtain:

F Ti −1 sin θ i −1 Ti sin θ i = − TH Ti −1 cos θ i −1 Ti cos θ i

Using this result, express tan θ1:

NF 2TH

= tan θ i −1 − tan θ i F tan θ1 = tan θ 0 − TH

Substitute for tan θ0 from (a):

tan θ1 =

NF F F − = (N − 2) 2TH TH 2TH

Newton’s Laws 223 Generalize this result to obtain:

tan θ i =

(N − 2i )

Express the length of rope between two balloons:

l between balloons =

Express the horizontal coordinate of the point on the rope where the ith balloon is attached, xi, in terms of xi−1 and the length of rope between two balloons:

xi = xi −1 +

Sum over all the coordinates to obtain:

Proceed similarly for the vertical coordinates to obtain:

F 2TH

L N +1

L cos θ i −1 N +1

xi =

L i −1 ∑ cosθ j N + 1 j =0

yi =

L i −1 ∑ sin θ j N + 1 j =0

(d) A spreadsheet program is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell C9

Content/Formula ($B$2-2*B9)/(2*$B$4)

D9

SIN(ATAN(C9))

E9

COS(ATAN(C9))

F10

F9+$B$1/($B$2+1)*E9

G10 G9+$B$1/($B$2+1)*D9

1 2 3 4 5 6 7 8 9 10 11 12 13

A L= N= F= TH=

B C 10 m 10 1 N 3.72 N

I 0 1 2 3 4

Algebraic Form

(N − 2i )

F 2TH

( ) cos(tan θ ) sin tan −1 θ i −1

i

L cos θ i −1 N +1 L yi −1 + cos θ i −1 N +1 xi −1 +

D

E

F

G

tan(thetai) sin(thetai) cos(thetai) xi yi 1.344 0.802 0.597 0.000 0.000 1.075 0.732 0.681 0.543 0.729 0.806 0.628 0.778 1.162 1.395 0.538 0.474 0.881 1.869 1.966 0.269 0.260 0.966 2.670 2.396

224 Chapter 4 14 15 16 17 18 19 20

5 6 7 8 9 10 11

0.000 −0.269 −0.538 −0.806 −1.075 −1.344

0.000 −0.260 −0.474 −0.628 −0.732 −0.802

1.000 0.966 0.881 0.778 0.681 0.597

3.548 4.457 5.335 6.136 6.843 7.462 8.005

2.632 2.632 2.396 1.966 1.395 0.729 0.000

(e) A horizontal component of tension 3.72 N gives a spacing of 8 m. At this spacing, the arch is 2.63 m high, tall enough for someone to walk through. 3.0

2.5

yi

2.0

1.5

1.0

0.5

0.0 0

1

2

3

4

5

6

7

8

xi

51 •• Picture the Problem We know, because the speed of the load is changing in parts (a) and (c), that it is accelerating. We also know that, if the load is accelerating in a particular direction, there must be a net force in that direction. A free-body diagram for part (a) is shown to the right. We can apply Newton’s 2nd law of motion to each part of the problem to relate the tension in the cable to the acceleration of the load. Choose the upward direction to be the positive y direction. (a) Apply

∑F

y

= ma y to the load

and solve for T:

Substitute numerical values and evaluate T:

(b) Because the crane is lifting the

T – mg = ma and

T = ma y + mg = m(a y + g )

(1)

T = (1000 kg )(2 m/s 2 + 9.81 m/s 2 ) = 11.8 kN T = mg = 9.81 kN

Newton’s Laws 225 load at constant speed, a = 0: (c) Because the acceleration of the load is downward, a is negative. Apply Fy = ma y to the load:

T – mg = may

Substitute numerical values in equation (1) and evaluate T:

T = (1000 kg)(9.81 m/s2 − 2 m/s2)

∑

= 7.81kN

52 •• Picture the Problem Draw a free-body diagram for each of the depicted situations and use the conditions for translational equilibrium to find the unknown tensions. (a)

ΣFx = T1cos60° − 30 N = 0 and T1 = (30 N)/cos60° = 60.0 N ΣFy = T1sin60° − T2 = 0 and T2 = T1sin60° = 52.0 N ∴ m = T2/g = 5.30 kg

(b)

ΣFx = (80 N)cos60° − T1sin60° = 0 and T1 = (80 N)cos60°/sin60° = 46.2 N ΣFy = (80 N)sin60° − T2 − T1cos60° = 0 T2 = (80 N)sin60° − (46.2 N)cos60° = 46.2 N m = T2/g = 4.71 kg

(c)

ΣFx = −T1cos60° + T3cos60° = 0 and T1 = T3 ΣFy = 2T1sin60° − mg = 0 and T1 = T3 = (58.9 N)/(2sin60°) = 34.0 N

226 Chapter 4 ∴ m = T1/g = 3.46 kg

53 •• Picture the Problem Construct the freebody diagram for that point in the rope at r which you exert the force F and choose the coordinate system shown on the FBD. We can apply Newton’s 2nd law to the rope to relate the tension to F.

(a) Noting that T1 = T2 = T, apply Fy = ma y to the car:

2Tsinθ − F = may = 0 because the car’s acceleration is zero.

Solve for and evaluate T:

T=

400 N F = = 3.82 kN 2 sin θ 2 sin 3°

(b) Proceed as in part (a):

T=

600 N = 4.30 kN 2 sin 4°

∑

Free-Body Diagrams: Inclined Planes and the Normal Force *54 • Picture the Problem The free-body diagram shows the forces acting on the box as the man pushes it across the frictionless floor. We can apply Newton’s 2nd law of motion to the box to find its acceleration.

Apply

∑F

x

= ma x to the box:

F cos θ = ma x

Solve for ax:

ax =

Substitute numerical values and evaluate ax:

ax =

F cos θ m

(250 N )cos35° = 20 kg

10.2 m/s 2

Newton’s Laws 227 55 • Picture the Problem The free-body diagram shows the forces acting on the box as the man pushes it up the frictionless incline. We can apply Newton’s 2nd law of motion to the box to determine the smallest force that will move it up the incline at constant speed.

Apply

∑F

x

= ma x to the box as it

Fmin cos(40° − θ ) − mg sin θ = 0

moves up the incline with constant speed: Solve for Fmin:

Fmin =

Substitute numerical values and evaluate Fmin:

Fmin =

mg sin θ cos (40° − θ )

(20 kg )(9.81m/s2 ) = cos25°

56.0 N

56 • Picture the Problem Forces always occur in equal and opposite pairs. If object A exerts a

r

r

r

force, FB , A on object B, an equal but opposite force, F A, B = − FB , A is exerted by object B on object A. The forces acting on the box are its weight,

r W, and the normal reaction force of the r inclined plane on the box, Fn . The reaction

forces are the forces the box exerts on the inclined plane and the gravitational force the box exerts on the earth. The reaction forces are indicated with primes.

57 • Picture the Problem Because the block whose mass is m is in equilibrium, the sum

r

r

r

of the forces Fn , T, and mg must be zero. Construct the free-body diagram for this object, use the coordinate system shown on the free-body diagram, and apply Newton’s 2nd law of motion. Apply

∑F

the incline:

x

= ma x to the block on

T – mgsin40° = max = 0 because the system is in equilibrium.

228 Chapter 4 Solve for m:

m=

The tension must equal the weight of the 3.5-kg block because that block is also in equilibrium:

T = (3.5 kg)g and

T g sin 40°

(3.5 kg) g 3.5 kg = g sin 40° sin 40°

m=

Because this expression is not included in the list of solution candidates,

(d ) is correct. Remarks: Because the object whose mass is m does not hang vertically, its mass must be greater than 3.5 kg. *58 • Picture the Problem The balance(s) indicate the tension in the string(s). Draw free-body diagrams for each of these systems and apply the condition(s) for equilibrium.

(a)

∑F

y

= T − mg = 0

and

(

)

T = mg = (10 kg ) 9.81 m/s 2 = 98.1 N

(b)

∑F

x

= T − T '= 0

or, because T ′= mg,

T = T ' = mg

(

)

= (10 kg ) 9.81m/s 2 = 98.1 N

(c)

∑F

y

= 2T − mg = 0

and

T = 12 mg =

(d)

1 2

∑F

x

(10 kg )(9.81m/s2 ) =

49.1 N

= T − mg sin 30° = 0

and

T = mg sin 30°

(

)

= (10 kg ) 9.81 m/s 2 sin 30° = 49.1 N

Newton’s Laws 229

Remarks: Note that (a) and (b) give the same answers … a rather surprising result until one has learned to draw FBDs and apply the conditions for translational equilibrium. 59 •• Picture the Problem Because the box is held in place (is in equilibrium) by the forces acting on it, we know that

r r r T + Fn + W = 0

Choose a coordinate system in which the positive x direction is in the direction of r T and the positive y direction is in the

r

direction of F n . Apply Newton’s 2nd law to the block to obtain expressions for

r r T and Fn .

(a) Apply

∑F

x

= ma x to the box:

T − mg sin θ = 0

Solve for T:

T = mg sin θ

Substitute numerical values and evaluate T:

T = (50 kg ) 9.81 m/s 2 sin 60° = 425 N

Apply

∑F

y

= ma y to the box:

Solve for Fn: Substitute numerical values and evaluate Fn:

(b) Using the result for the tension from part (a) to obtain:

(

)

Fn − mg cos θ = 0 Fn = mg cos θ

Fn = (50 kg )(9.81m/s 2 )cos 60° = 245 N T90° = mg sin 90° = mg and

T0° = mg sin 0° = 0

230 Chapter 4 60 •• Picture the Problem Draw a free-body diagram for the box. Choose a coordinate system in which the positive x-axis is parallel to the inclined plane and the positive y-axis is in the direction of the normal force the incline exerts on the block. Apply Newton’s 2nd law of motion to both the x and y directions.

(a) Apply

∑F

y

= ma y to the block:

Fn = mg cos 25° + (100 N )sin 25°

Solve for Fn: Substitute numerical values and evaluate Fn:

(b) Apply

Fn − mg cos 25° − (100 N )sin 25° = 0

∑F

x

= ma x to the block:

Solve for a:

Substitute numerical values and evaluate a:

Fn = (12 kg ) (9.81 m/s 2 )cos25° + (100 N )sin 25° = 149 N

(100 N )cos 25° − mg sin 25° = ma a=

(100 N )cos 25° − g sin 25°

a=

(100 N )cos 25° − (9.81m/s2 )sin 25°

m

12 kg

= 3.41 m/s 2 *61 •• Picture the Problem The scale reading (the boy’s apparent weight) is the force the scale exerts on the boy. Draw a free-body diagram for the boy, choosing a coordinate system in which the positive x-axis is parallel to and down the inclined plane and the positive y-axis is in the direction of the normal force the incline exerts on the boy. Apply Newton’s 2nd law of motion in the y direction.

Apply

∑F

y

= ma y to the boy to

find Fn. Remember that there is no acceleration in the y direction:

Fn − W cos 30° = 0

Newton’s Laws 231 Substitute for W to obtain:

Fn − mg cos 30° = 0

Solve for Fn:

Fn = mg cos 30°

Substitute numerical values and evaluate Fn:

(

)

Fn = (65 kg ) 9.81 m/s 2 cos 30° = 552 N

62 •• Picture the Problem The free-body diagram for the block sliding up the incline is shown to the right. Applying Newton’s 2nd law to the forces acting in the x direction will lead us to an expression for ax. Using this expression in a constantacceleration equation will allow us to express h as a function of v0 and g.

The height h is related to the distance ∆x traveled up the incline:

h = ∆xsin θ

Using a constant-acceleration equation, relate the final speed of the block to its initial speed, acceleration, and distance traveled:

v 2 = v02 + 2a x ∆x

Solve for ∆x to obtain:

Apply

∑F

x

= ma x to the block and

or, because v = 0,

0 = v02 + 2a x ∆x ∆x =

− v02 2a x

− mg sin θ = ma x

solve for its acceleration:

and ax = −g sinθ

Substitute these results in the equation for h and simplify:

⎛ v02 ⎞ ⎟⎟ sin θ h = ∆x sin θ = ⎜⎜ 2 g sin θ ⎠ ⎝ =

v02 2g

which is independent of the ramp’s angle θ.

232 Chapter 4

Free-Body Diagrams: Elevators 63 • Picture the Problem Because the elevator is descending at constant speed, the object

r

r

is in equilibrium and T + mg = 0. Draw a free-body diagram of the object and let the upward direction be the positive y direction. Apply Newton’s 2nd law with a = 0. Because the downward speed is constant, the acceleration is zero. Apply Fy = ma y and solve for

∑

T – mg = 0 ⇒ T = mg and

(a ) is correct.

T: 64 • Picture the Problem The sketch to the right shows a person standing on a scale in a descending elevator. To its right is a freebody diagram showing the forces acting on the person. The force exerted by the scale r on the person, wapp , is the person’s apparent weight. Because the elevator is slowing down while descending, the acceleration is directed upward.

Apply

∑F

y

= ma y to the person:

Solve for wapp:

wapp − mg = ma y wapp = mg + ma y > mg

The apparent weight will be higher. Because an upward acceleration is required to "slow" a downward velocity, the normal force exerted on you by the scale (your apparent weight ) must be greater than your weight. *65 • Picture the Problem The sketch to the right shows a person standing on a scale in the elevator immediately after the cable breaks. To its right is the free-body diagram showing the forces acting on the person. The force exerted by the scale on r the person, wapp , is the person’s apparent weight.

Newton’s Laws 233 r

r

r

r

From the free-body diagram we can see that wapp + mg = ma where g is the local

r

gravitational field and a is the acceleration of the reference frame (elevator). When the r r r r r r elevator goes into free fall ( a = g ), our equation becomes wapp + mg = ma = mg. This

r

tells us that wapp = 0. (e) is correct. 66 • Picture the Problem The free-body diagram shows the forces acting on the 10kg block as the elevator accelerates upward. Apply Newton’s 2nd law of motion to the block to find the minimum acceleration of the elevator required to break the cord. Apply

∑F

y

= ma y to the block:

T – mg = may

Solve for ay to determine the minimum breaking acceleration:

ay =

T − mg T = −g m m

Substitute numerical values and evaluate ay:

ay =

150 N − 9.81 m/s 2 = 5.19 m/s 2 10 kg

67 •• Picture the Problem The free-body diagram shows the forces acting on the 2-kg block as the elevator ascends at a constant velocity. Because the acceleration of the elevator is zero, the block is in equilibrium under the influence of

r r T and mg. Apply Newton’s 2nd law of

motion to the block to determine the scale reading. (a) Apply

∑F

y

= ma y to the block

T − mg = ma y

(1)

to obtain: For motion with constant velocity, ay = 0:

T − mg = 0 and T = mg

Substitute numerical values and evaluate T:

T = (2 kg ) 9.81m/s 2 = 19.6 N

(b) As in part (a), for constant velocity, a = 0:

T − mg = ma y

(

and

)

T = (2 kg ) (9.81 m/s 2 ) = 19.6 N

234 Chapter 4 (c) Solve equation (1) for T and simplify to obtain:

Because the elevator is ascending and its speed is increasing, we have ay = 3 m/s2. Substitute numerical values and evaluate T:

T = mg + ma y = m(g + a y )

T = (2 kg ) (9.81 m/s 2 + 3m/s 2 ) = 25.6 N

(d) For 0 < t < 5 s: ay = 0 and

T0→5 s = 19.6 N

Using its definition, calculate a for 5 s < t < 9 s:

a=

Substitute in equation (2) and evaluate T:

(2)

∆v 0 − 10 m/s = = −2.5 m/s 2 ∆t 4s

(

T5 s→9 s = (2 kg ) 9.81 m/s 2 − 2.5m/s 2

)

= 14.6 N

Free-Body Diagrams: Ropes, Tension, and Newton’s Third Law 68 • Picture the Problem Draw a free-body diagram for each object and apply Newton’s 2nd law of motion. Solve the resulting simultaneous equations for the ratio of T1 to T2. Draw the FBD for the box to the left and apply Fx = max :

∑

T1 = m1a1 Draw the FBD for the box to the Fx = ma x : right and apply

∑

T2 − T1 = m2a2 The two boxes have the same acceleration:

a1 = a2

Divide the second equation by the first:

T1 m = 1 T2 − T1 m2

Newton’s Laws 235 Solve for the ratio T1/T2 :

T1 m1 = and ( d ) is correct. T2 m1 + m2

69 •• Picture the Problem Call the common acceleration of the boxes a. Assume that box 1 moves upward, box 2 to the right, and box 3 downward and take this direction to be the positive x direction. Draw free-body diagrams for each of the boxes, apply Newton’s 2nd law of motion, and solve the resulting equations simultaneously. (a)

(c)

(b)

(a) Apply

∑F

x

= max to the box

T1 – w1 = m1a

whose mass is m1: Apply

∑F

x

= max to the box

T2 – T1 = m2a

whose mass is m2: Noting that T2 = T2' , apply

∑F

x

w3 – T2 = m3a

= ma x to the box whose mass

is m3: Add the three equations to obtain:

w3 − w1 = (m1 + m2 + m3)a

Solve for a:

a=

Substitute numerical values and evaluate a:

a=

(m3 − m1 )g m1 + m2 + m3

(2.5 kg − 1.5 kg )(9.81m/s2 ) 1.5 kg + 3.5 kg + 2.5 kg

= 1.31m/s 2 (b) Substitute for the acceleration in the equations obtained above to find the tensions:

T1 = 16.7 N and T2 = 21.3 N

236 Chapter 4 *70 •• Picture the Problem Choose a coordinate system in which the positive x direction is to the right and the positive y direction is

r

upward. Let F2,1 be the contact force

r

exerted by m2 on m1 and F1, 2 be the force exerted by m1 on m2. These forces are equal

r

r

and opposite so F2,1 = − F1, 2 . The freebody diagrams for the blocks are shown to the right. Apply Newton’s 2nd law to each block separately and use the fact that their accelerations are equal. (a) Apply

∑F

x

= ma x to the first

F − F2,1 = m1a1 = m1a

block: Apply

∑F

x

= ma x to the second

F1, 2 = m2 a2 = m2 a

(1)

block: Add these equations to eliminate F2,1 and F1,2 and solve for a = a1 = a2:

a=

Substitute your value for a into equation (1) and solve for F1,2:

F1,2 =

(b) Substitute numerical values in the equations derived in part (a) and evaluate a and F1,2:

a=

F m1 + m2 Fm2 m1 + m2

3.2 N = 0.400 m/s 2 2 kg + 6 kg

and

F1,2 =

(3.2 N )(6 kg ) = 2 kg + 6 kg

2.40 N

Remarks: Note that our results for the acceleration are the same as if the force F had acted on a single object whose mass is equal to the sum of the masses of the two blocks. In fact, because the two blocks have the same acceleration, we can consider them to be a single system with mass m1 + m2.

Newton’s Laws 237 71 • Picture the Problem Choose a coordinate system in which the positive x direction is to the right and the positive y direction is

r

upward. Let F2,1 be the contact force

r

exerted by m2 on m1 and F1, 2 be the force exerted by m1 on m2. These forces are equal

r

r

and opposite so F2,1 = − F1, 2 . The freebody diagrams for the blocks are shown. We can apply Newton’s 2nd law to each block separately and use the fact that their accelerations are equal. (a) Apply

∑F

x

= ma x to the first

F − F1, 2 = m2 a2 = m2 a

block: Apply

∑F

x

= ma x to the second

F2,1 = m1a1 = m1a

(1)

block: Add these equations to eliminate F2,1 and F1,2 and solve for a = a1 = a2: Substitute your value for a into equation (1) and solve for F2,1:

(b) Substitute numerical values in the equations derived in part (a) and evaluate a and F2,1:

F m1 + m2

a=

F2,1 =

a=

Fm1 m1 + m2

3.2 N = 0.400 m/s 2 2 kg + 6 kg

and

F2,1 =

(3.2 N )(2 kg ) = 2 kg + 6 kg

0.800 N

Remarks: Note that our results for the acceleration are the same as if the force F had acted on a single object whose mass is equal to the sum of the masses of the two blocks. In fact, because the two blocks have the same acceleration, we can consider them to be a single system with mass m1 + m2. 72 •• Picture the Problem The free-body diagrams for the boxes and the ropes are below. Because the vertical forces have no bearing on the problem they have not been included. Let the numeral 1 denote the 100-kg box to the left, the numeral 2 the rope connecting the boxes, the numeral 3 the box to the right and the numeral 4 the rope to which the force

r r r F is applied. F3, 4 is the tension force exerted by m3 on m4, F4,3 is the tension force r r exerted by m4 on m3, F2,3 is the tension force exerted by m2 on m3, F3, 2 is the tension

238 Chapter 4 r

r

force exerted by m3 on m2, F1, 2 is the tension force exerted by m1 on m2, and F2,1 is the

r

tension force exerted by m2 on m1. The equal and opposite pairs of forces are F2,1 =

r

r

r

r

r

− F1, 2 , F3, 2 = − F2,3 , and F4,3 = − F3, 4 . We can apply Newton’s 2nd law to each box and rope separately and use the fact that their accelerations are equal.

Apply

r

r

F2,1 = m1a1 = m1a

(1)

r

r

F3, 2 − F1, 2 = m2 a2 = m2 a

(2)

(3)

∑ F = ma to the box whose

mass is m1: Apply

∑ F = ma to the rope

whose mass is m2: Apply

r

r

F4,3 − F2,3 = m3a3 = m3a

r

r

F − F3, 4 = m4 a4 = m4 a

∑ F = ma to the box whose

mass is m3: Apply

∑ F = ma to the rope

whose mass is m4: Add these equations to eliminate F2,1, F1,2, F3,2, F2,3, F4,3, and F3,4 and solve for F:

F = (m1 + m2 + m3 + m4 )a

= (202 kg )(1.0 m/s 2 ) = 202 N

(

)

Use equation (1) to find the tension at point A:

F2,1 = (100 kg ) 1.0 m/s 2 = 100 N

Use equation (2) to find the tension at point B:

F3, 2 = F1, 2 + m2 a

(

= 100 N + (1 kg ) 1.0 m/s 2

)

= 101 N Use equation (3) to find the tension at point C:

F4,3 = F2,3 + m3a

(

= 101 N + (100 kg ) 1.0 m/s 2 = 201 N

)

Newton’s Laws 239 73 •• Picture the Problem Because the distribution of mass in the rope is uniform, we can express the mass m′ of a length x of the rope in terms of the total mass of the rope M and its length L. We can then express the total mass that the rope must support at a distance x above the block and use Newton’s 2nd law to find the tension as a function of x. Set up a proportion expressing the mass m′ of a length x of the rope as a function of M and L and solve for m′:

M m' M x = ⇒ m' = L x L

Express the total mass that the rope must support at a distance x above the block:

m + m' = m +

Apply

∑F

y

= ma y to the block

and a length x of the rope:

Solve for T to obtain:

M x L

M ⎞ ⎛ T − w = T −⎜m + x ⎟g L ⎠ ⎝ M ⎞ ⎛ = ⎜m + x⎟ a L ⎠ ⎝ T=

(a + g )⎛⎜ m + M ⎝

*74 •• Picture the Problem Choose a coordinate system with the positive y direction upward and denote the top link with the numeral 1, the second with the numeral 2, etc.. The free-body diagrams show the forces acting on links 1 and 2. We can apply Newton’s 2nd law to each link to obtain a system of simultaneous equations that we can solve for the force each link exerts on the link below it. Note that the net force on each link is the product of its mass and acceleration. (a) Apply

∑F

y

= ma y to the top

link and solve for F:

F − 5mg = 5ma and

F = 5m ( g + a )

⎞ x⎟ L ⎠

240 Chapter 4 Substitute numerical values and evaluate F:

(b) Apply

∑F

y

= ma y to a single

link:

(

F = 5(0.1kg ) 9.81 m/s 2 + 2.5 m/s 2 = 6.16 N

(

F1 link = m1 link a = (0.1 kg ) 2.5 m/s 2

)

= 0.250 N

(c) Apply

∑F

y

= ma y to the 1st

through 5th links to obtain:

Add equations (2) through (5) to obtain: Solve for F2 to obtain: Substitute numerical values and evaluate F2:

F − F2 − mg = ma , F2 − F3 − mg = ma ,

(1)

F3 − F4 − mg = ma , F4 − F5 − mg = ma , and F5 − mg = ma

(3)

(2) (4) (5)

F2 − 4mg = 4ma F2 = 4mg + 4ma = 4m(a + g )

(

F2 = 4(0.1kg ) 9.81 m/s 2 + 2.5 m/s 2 = 4.92 N

Substitute for F2 to find F3, and then substitute for F3 to find F4:

F3 = 3.69 N and F4 = 2.46 N

Solve equation (5) for F5:

F5 = m( g + a )

Substitute numerical values and evaluate F5:

(

F5 = (0.1 kg ) 9.81m/s 2 + 2.5 m/s 2 = 1.23 N

75 • Picture the Problem A net force is required to accelerate the object. In this problem the net force is the difference

r

r

r

between T and W (= mg ). The free-body diagram of the object is shown to the right. Choose a coordinate system in which the upward direction is positive. Apply

)

r

r

∑ F = ma to the object to

Fnet = T – W = T – mg

obtain: Solve for the tension in the lower portion of the rope:

T = Fnet + mg = ma + mg = m(a + g)

)

)

Newton’s Laws 241 Using its definition, find the acceleration of the object:

a ≡ ∆v/∆t = (3.5 m/s)/(0.7 s) = 5.00 m/s2

Substitute numerical values and evaluate T:

T = (40 kg)(5.00 m/s2 + 9.81 m/s2) = 592 N and (a ) is correct.

76 • Picture the Problem A net force in the downward direction is required to accelerate the truck downward. The net r r force is the difference between Wt and T . A free-body diagram showing these forces acting on the truck is shown to the right. Choose a coordinate system in which the downward direction is positive.

Apply

∑F

y

= ma y to the truck to

T − mt g = mt a y

obtain: Solve for the tension in the lower portion of the cable: Substitute to find the tension in the rope:

T = mt g + mt a y = mt (g + a y ) T = mt (g − 0.1g ) = 0.9mt g and (c) is correct.

77 •• Picture the Problem Because the string does not stretch or become slack, the two objects must have the same speed and therefore the magnitude of the acceleration is the same for each object. Choose a coordinate system in which up the incline is the positive x direction for the object of mass m1 and downward is the positive x direction for the object of mass m2. This idealized pulley acts like a piece of polished pipe; i.e., its only function is to change the direction the tension in the massless string acts. Draw a free-body diagram for each of the two objects, apply Newton’s 2nd law of motion to both objects, and solve the resulting equations simultaneously. (a) Draw the FBD for the object of mass m1:

Apply

∑F

x

= max to the object

whose mass is m1:

T – m1gsinθ = m1a

242 Chapter 4 Draw the FBD for the object of mass m2:

Apply

∑F

x

= max to the object

m2g − T = m2a

whose mass is m2: Add the two equations and solve for a:

a=

g (m2 − m1 sin θ ) m1 + m2

Substitute for a in either of the equations containing the tension and solve for T:

T=

gm1m2 (1 + sin θ ) m1 + m2

(b) Substitute the given values into the expression for a:

a = 2.45 m/s 2

Substitute the given data into the expression for T:

T = 36.8 N

78 • Picture the Problem The magnitude of the accelerations of Peter and the counterweight are the same. Choose a coordinate system in which up the incline is the positive x direction for the counterweight and downward is the positive x direction for Peter. The pulley changes the direction the tension in the rope acts. Let Peter’s mass be mP. Ignoring the mass of the rope, draw free-body diagrams for the counterweight and Peter, apply Newton’s 2nd law to each of them, and solve the resulting equations simultaneously. (a) Using a constant-acceleration equation, relate Peter’s displacement to her acceleration and descent time:

∆x = v0 ∆t + 12 a (∆t )

2

or, because v0 = 0,

∆x = 12 a (∆t )

2

Solve for the common acceleration of Peter and the counterweight:

a=

2∆x (∆t )2

Substitute numerical values and evaluate a:

a=

2(3.2 m ) = 1.32 m/s 2 (2.2 s )2

Newton’s Laws 243 Draw the FBD for the counterweight:

Apply

∑F

x

= max to the

T – mg sin50° = ma

counterweight: Draw the FBD for Peter:

Apply

∑F

x

= max to Peter:

mPg – T = mPa

Add the two equations and solve for m:

m=

Substitute numerical values and evaluate m:

m=

mP ( g − a ) a + g sin 50°

(50 kg )(9.81m/s2 − 1.32 m/s2 )

1.32 m/s 2 + (9.81m/s 2 ) sin50°

= 48.0 kg (b) Substitute for m in the force equation for the counterweight and solve for T:

T = m(a + g sin 50°)

(b) Substitute numerical values and evaluate T:

[

]

T = (48.0 kg ) 1.32 m/s 2 + (9.81m/s 2 ) sin50° = 424 N 79 •• Picture the Problem The magnitude of the accelerations of the two blocks are the same. Choose a coordinate system in which up the incline is the positive x direction for the 8-kg object and downward is the positive x direction for the 10-kg object. The peg changes the direction the tension in the rope acts. Draw free-body diagrams for each object, apply Newton’s 2nd law of motion to both of them, and solve the resulting equations simultaneously.

244 Chapter 4 (a) Draw the FBD for the 3-kg object:

Apply

∑F

x

= ma x to the 3-kg block:

T – m8g sin40° = m3a

Draw the FBD for the 10-kg object:

Apply

∑F

x

= ma x to the 10-kg block:

Add the two equations and solve for and evaluate a:

m10g sin50° − T = m10a

a=

g (m10 sin 50° − m8 sin 40°) m8 + m10

= 1.37 m/s 2 Substitute for a in the first of the two force equations and solve for T:

T = m8 g sin 40° + m8 a

Substitute numerical values and evaluate T:

T = (8 kg ) 9.81 m/s 2 sin 40°

[(

+ 1.37 m/s 2

]

)

= 61.4 N (b) Because the system is in equilibrium, set a = 0, express the force equations in terms of m1 and m2, add the two force equations, and solve for and evaluate the ratio m1/m2:

T – m1g sin40° = 0 m2g sin50°− T = 0 ∴ m2g sin50°– m1g sin40° = 0 and

m1 sin 50° = = 1.19 m2 sin 40°

Newton’s Laws 245 80 •• Picture the Problem The pictorial representations shown to the right summarize the information given in this problem. While the mass of the rope is distributed over its length, the rope and the 6-kg block have a common acceleration. Choose a coordinate system in which the direction of the 100-N force is the positive x direction. Because the surface is horizontal and frictionless, the only force that influences our solution is the 100-N force. (a) Apply

∑F

x

= ma x to the

100 N = (m1 + m2)a

objects shown for part (a): Solve for a to obtain:

a=

100 N m1 + m2

Substitute numerical values and evaluate a:

a=

100 N = 10.0 m/s 2 10 kg

(b) Let m represent the mass of a length x of the rope. Assuming that the mass of the rope is uniformly distributed along its length:

4 kg m m2 = = x Lrope 5 m and

⎛ 4 kg ⎞ ⎟⎟ x m = ⎜⎜ ⎝ 5m ⎠ Let T represent the tension in the rope at a distance x from the point at which it is attached to the 6-kg Fx = ma x to the block. Apply

∑

system shown for part (b) and solve for T:

T = (m1 + m)a

⎡ ⎛ 4 kg ⎞ ⎤ ⎟⎟ x ⎥ 10 m/s 2 = ⎢6 kg + ⎜⎜ ⎝ 5m ⎠ ⎦ ⎣

(

= 60 N + (8 N/m )x

*81 •• Picture the Problem Choose a coordinate system in which upward is the positive y direction and draw the free-body diagram for the frame-plus- painter. Noting that

r r F = −T, apply Newton’s 2nd law of

motion.

(a) Letting mtot = mframe + mpainter,

2T – mtotg = mtota and

)

246 Chapter 4 apply

∑F

y

= ma y to the frame-

T=

plus-painter and solve T: Substitute numerical values and evaluate T:

T=

mtot (a + g ) 2

(75 kg )(0.8 m/s 2 + 9.81m/s 2 ) 2

= 398 N Because F = T:

(b) Apply

F = 398 N

∑F

y

= ma y with a = 0

2T – mtotg = 0

to obtain: Solve for T:

T = 12 mtot g

Substitute numerical values and evaluate T:

T=

1 2

(75 kg )(9.81m/s 2 ) =

368 N

82 ••• Picture the Problem Choose a coordinate system in which up the incline is the positive x r r direction and draw free-body diagrams for each block. Noting that a20 = −a10 , apply Newton’s 2nd law of motion to each block and solve the resulting equations simultaneously. Draw a FBD for the 20-kg block:

Apply

∑F

x

= ma x to the block to

T – m20gsin20° = m20a20

obtain: Draw a FBD for the 10-kg block. Because all the surfaces, including the surfaces between the blocks, are frictionless, the force the 20-kg block exerts on the 10-kg block must be normal to their surfaces as shown to the right.

Apply obtain:

∑F

x

= ma x to the block to

T – m10gsin20° = m10a10

Newton’s Laws 247 Because the blocks are connected by a taut string:

a20 = −a10

Substitute for a20 and eliminate T between the two equations to obtain:

a10 = 1.12 m/s 2

and a20 = − 1.12 m/s 2 Substitute for either of the accelerations in the force equations and solve for T:

T = 44.8 N

83 ••• Picture the Problem Choose a coordinate system in which the positive x direction is to the right and draw free-body diagrams for each block. Because of the pulley, the string exerts a force of 2T. Apply Newton’s 2nd law of motion to both blocks and solve the resulting equations simultaneously. (a) Noting the effect of the pulley, express the distance the 20-kg block moves in a time ∆t:

∆x20 = 12 ∆x5 =

(b) Draw a FBD for the 20-kg block:

Apply

∑F

x

= ma x to the block to

2T = m20a20

obtain: Draw a FBD for the 5-kg block:

∑F

= ma x to the block to

m5g − T = m5a5

Using a constant-acceleration equation, relate the displacement of the 5-kg block to its acceleration

∆x5 = 12 a5 (∆t )

Apply

x

obtain: 2

1 2

(10 cm) =

5.00 cm

248 Chapter 4 and the time during which it is accelerated: Using a constant-acceleration equation, relate the displacement of the 20-kg block to its acceleration and the time during which it is accelerated: Divide the first of these equations by the second to obtain:

∆x20 = 12 a20 (∆t )

2

1 a (∆t ) ∆x5 a = 2 5 = 5 2 ∆x20 12 a20 (∆t ) a20 2

Use the result of part (a) to obtain:

a5 = 2a20

Let a20 = a. Then a5 = 2a and the force equations become:

2T = m20a and m5g – T = m5(2a)

Eliminate T between the two equations to obtain:

a = a20 =

Substitute numerical values and evaluate a20 and a5:

a20 = and

m5 g 2m5 + 12 m20

(5 kg )(9.81m/s2 ) = 2(5 kg ) + 12 (20 kg )

(

2.45 m/s 2

)

a5 = 2 2.45 m/s 2 = 4.91 m/s 2 Substitute for either of the accelerations in either of the force equations and solve for T:

T = 24.5 N

Free-Body Diagrams: The Atwood’s Machine *84 •• Picture the Problem Assume that m1 > m2. Choose a coordinate system in which the positive y direction is downward for the block whose mass is m1 and upward for the block whose mass is m2 and draw free-body diagrams for each block. Apply Newton’s 2nd law of motion to both blocks and solve the resulting equations simultaneously. Draw a FBD for the block whose mass is m2:

Newton’s Laws 249 Apply

∑F

y

= ma y to this block:

T – m2g = m2a2

Draw a FBD for the block whose mass is m1:

Apply

∑F

y

= ma y to this block:

Because the blocks are connected by a taut string, let a represent their common acceleration: Add the two force equations to eliminate T and solve for a:

Substitute for a in either of the force equations and solve for T:

m1g – T = m1a1 a = a1 = a2

m1 g − m2 g = m1 a + m2 a and

a=

m1 − m2 g m1 + m2

T=

2m1m2 g m1 + m2

85 •• Picture the Problem The acceleration can be found from the given displacement during the first second. The ratio of the two masses can then be found from the acceleration using the first of the two equations derived in Problem 89 relating the acceleration of the Atwood’s machine to its masses. Using a constant-acceleration equation, relate the displacement of the masses to their acceleration and solve for the acceleration:

∆y = v0t + 12 a (∆t )

2

or, because v0 = 0,

∆y = 12 a(∆t )

2

2∆y 2(0.3 m ) = = 0.600 m/s 2 2 2 (1s ) (∆t )

Solve for and evaluate a:

a=

Solve for m1 in terms of m2 using the first of the two equations given in Problem 84:

g + a 10.41 m/s 2 m1 = m2 = m2 = 1.13m2 g − a 9.21m/s 2

Find the second mass for m2 or m1 = 1.2 kg:

m2nd mass = 1.36 kg or 1.06 kg

250 Chapter 4 86 •• Picture the Problem Let Fnm be the force the block of mass m2 exerts on the pebble of mass m. Because m2 < m1, the block of mass m2 accelerates upward. Draw a freebody diagram for the pebble and apply Newton’s 2nd law and the acceleration equation given in Problem 84.

Apply

∑F

y

= ma y to the pebble:

Solve for Fnm:

Fnm – mg = ma

Fnm = m(a + g ) m1 − m 2 g m1 + m 2

From Problem 84:

a=

Substitute for a and simplify to obtain:

⎛ m − m2 ⎞ 2m1m Fnm = m⎜⎜ 1 g + g ⎟⎟ = g m1 + m2 ⎝ m1 + m2 ⎠

87 •• Picture the Problem Note from the free-body diagrams for Problem 89 that the net force exerted by the accelerating blocks is 2T. Use this information, together with the expression for T given in Problem 84, to derive an expression for F = 2T. From Problem 84 we have:

The net force, F, exerted by the Atwood’s machine on the hanger is:

If m1 = m2 = m, then:

If either m1 or m2 = 0, then:

T=

2m1m2 g m1 + m2

F = 2T =

F=

4m1m2 g m1 + m2

4m 2 g = 2mg … as expected. 2m

F = 0 … also as expected.

88 ••• Picture the Problem Use a constant-acceleration equation to relate the displacement of the descending (or rising) mass as a function of its acceleration and then use one of the results from Problem 84 to relate a to g. Differentiation of our expression for g will allow us to relate uncertainty in the time measurement to uncertainty in the measured value for g … and to the values of m2 that would yield an experimental value for g that is good to within 5%.

Newton’s Laws 251 m1 + m2 m1 − m2

(a) Use the result given in Problem 84 to express g in terms of a:

g=a

Using a constant-acceleration equation, express the displacement, L, as a function of t and solve for the acceleration:

∆y = L = v0 ∆t + 12 a (∆t )

Substitute this expression for a:

(b) Evaluate dg/dt to obtain:

2

or, because v0 = 0 and ∆t = t,

a=

2L t2

(2)

2 L ⎛ m1 + m2 ⎞ ⎜ ⎟ t 2 ⎜⎝ m1 − m2 ⎟⎠

g=

⎛ m + m2 ⎞ dg ⎟⎟ = −4 Lt −3 ⎜⎜ 1 dt ⎝ m1 − m2 ⎠ =

Divide both sides of this expression by g and multiply both sides by dt:

(1)

− 2 ⎡ 2 L ⎤⎛ m1 + m2 ⎞ − 2 g ⎜ ⎟= t ⎢⎣ t 2 ⎥⎦⎜⎝ m1 − m2 ⎟⎠ t

dg dt = −2 g t

(c) We have:

dg dt = ±0.05 and = ±0.025 g t

Solve the second of these equations for t to obtain:

t=

dt 1s = = 4s 0.025 0.025

Substitute in equation (2) to obtain:

a=

2(3 m ) = 0.375 m/s 2 2 (4 s )

Solve equation (1) for m2 to obtain:

m2 =

Evaluate m2 with m1 = 1 kg:

9.81m/s 2 − 0.375 m/s 2 (1kg ) m2 = 9.81 m/s 2 + 0.375 m/s 2 = 0.926 kg

Solve equation (1) for m1 to obtain:

m1 = m2

Substitute numerical values to obtain:

9.81 m/s2 + 0.375 m/s2 m1 = (0.926 kg ) 9.81 m/s2 − 0.375 m/s2 = 1.08 kg

g −a m1 g+a

g+a g −a

252 Chapter 4 Because the masses are interchangeable:

m2 = 0.926 kg or 1.08 kg

*89 •• Picture the Problem We can reason to this conclusion as follows: In the two extreme cases when the mass on one side or the other is zero, the tension is zero as well, because the mass is in free-fall. By symmetry, the maximum tension must occur when the masses on each side are equal. An alternative approach that is shown below is to treat the problem as an extreme-value problem. Express m2 in terms of M and m1:

m2 = M − m1

Substitute in the equation from Problem 84 and simplify to obtain:

T=

Differentiate this expression with respect to m1 and set the derivative equal to zero for extreme values:

dT ⎛ 2m ⎞ = 2 g ⎜1 − 1 ⎟ = 0 for extreme values dm1 M ⎠ ⎝

Solve for m1 to obtain:

m1 = 12 M

Show that m1 = M/2 is a maximum value by evaluating the second derivative of T with respect to m1 at m1 = M/2:

d 2T 4g =− < 0, independently of m1 2 dm1 M

⎛ m2 ⎞ 2 gm1 (M − m1 ) = 2 g ⎜⎜ m1 − 1 ⎟⎟ m1 + M − m1 M ⎠ ⎝

and we have shown that

T is a maximum when m1 = m2 = 12 M . Remarks: An alternative solution is to use a graphing calculator to show that T as a function of m1 is concave downward and has its maximum value when m1 = m2 = M/2. 90 ••• Picture the Problem The free-body diagrams show the forces acting on the objects whose masses are m1 and m2. The application of Newton’s 2nd law to these forces and the accelerations the net forces are responsible for will lead us to an expression for the tension in the string as a function of m1 and m2. Examination of this expression as for m2 >> m1 will yield the predicted result. (a) Apply

∑F

y

= ma y to the

objects whose masses are m1 and m2 to obtain:

T1 − m1 g = m1a1 and m2 g − T2 = m2 a2

Newton’s Laws 253 Assume that the role of the pulley is simply to change the direction the tension acts. Then T1 = T2 = T. Because the two objects have a common acceleration, let a = a1 = a2. Eliminate a between the two equations and solve for T to obtain:

T=

2m1m2 g m1 + m2

Divide the numerator and denominator of this fraction by m2:

T=

2m1 g m 1+ 1 m2

Take the limit of this fraction as m2 → ∞ to obtain:

T = 2m1 g

(b) Imagine the situation when m2 >> m1:

Under these conditions, the object whose mass is m2 is essentially in freefall, so the object whose mass is m1 is accelerating upward with an acceleration of magnitude g.

Under these conditions, the net force acting on the object whose mass is m1 is m1g and:

T – m1g = m1g ⇒ T = 2m1g. Note that this result agrees with that obtained using more analytical methods.

General Problems 91 • Picture the Problem Choose a coordinate system in which the force the tree exerts on the woodpecker’s head is in the negative-x direction and determine the acceleration of the woodpecker’s head from Newton’s 2nd law of motion. The depth of penetration, under the assumption of constant acceleration, can be determined using a constant- acceleration equation. Knowing the acceleration of the woodpecker’s head and the depth of penetration of the tree, we can calculate the time required to bring the head to rest. (a) Apply

∑F

x

= ma x to the

woodpecker’s head to obtain: (b) Using a constant-acceleration equation, relate the depth-ofpenetration into the bark to the acceleration of the woodpecker’s head: Solve for and evaluate ∆x:

ax =

−6N ∑ Fx = = − 100 m/s 2 m 0.060 kg

v 2 = v02 + 2a∆x or, because v = 0,

0 = v02 + 2a∆x − v02 − (3.5 m/s ) = = 6.13 cm 2a 2 − 100 m/s 2 2

∆x =

(

)

254 Chapter 4 (c) Use the definition of acceleration to express the time required for the woodpecker’s head to come to rest:

Substitute numerical values and evaluate ∆t:

∆t =

v − v0 a

or, because v = 0,

∆t =

v − v0 a

∆t =

− v0 − 3.5 m/s = = 35.0 ms a − 100 m/s 2

*92 •• Picture the Problem The free-body diagram shown to the right shows the forces acting on an object suspended from the ceiling of a car that is accelerating to the right. Choose the coordinate system shown and use Newton’s laws of motion and constant- acceleration equations in the determination of the influence of the forces on the behavior of the suspended object. The second free-body diagram shows the forces acting on an object suspended from the ceiling of a car that is braking while it moves to the right.

(a)

In accordance with Newton' s law of inertia, the object' s displacement will be in the direction opposite that of the acceleration.

(b) Resolve the tension, T, into its

r r components and apply ∑ F = ma to the object: Take the ratio of these two equations to eliminate T and m:

(c)

ΣFx = Tsinθ = ma and ΣFy = Tcosθ − mg = 0

T sin θ ma = T cos θ mg or a tanθ = ⇒ a = g tan θ g

Because the acceleration is opposite the direction the car is moving, the accelerometer will swing forward.

Newton’s Laws 255 Using a constant-acceleration equation, express the velocity of the car in terms of its acceleration and solve for the acceleration:

v 2 = v02 + 2a∆x or, because v = 0,

0 = v02 + 2a∆x a=

− v02 2∆x

Substitute numerical values and evaluate a:

a=

− (50 km/h ) = − 1.61 m/s 2 2(60 m )

Solve the equation derived in (b) for θ:

θ = tan −1 ⎜⎜ ⎟⎟

Substitute numerical values and evaluate θ :

⎛ 1.61m/s 2 ⎞ ⎟ = 9.32° θ = tan ⎜⎜ 2 ⎟ ⎝ 9.81 m/s ⎠

Solve for a:

2

⎛a⎞ ⎝g⎠

−1

93 •• Picture the Problem The free-body diagram shows the forces acting at the top of the mast. Choose the coordinate system shown and use Newton’s 2nd and 3rd laws of motion to analyze the forces acting on the deck of the sailboat.

Apply

∑F

x

= ma x to the top of the

TFsinθF − TBsinθB = 0

mast: Find the angles that the forestay and backstay make with the vertical:

⎛ 3.6 m ⎞ ⎟⎟ = 16.7° ⎝ 12 m ⎠

θ F = tan −1 ⎜⎜ and

⎛ 6.4 m ⎞ ⎟⎟ = 28.1° 12 m ⎝ ⎠

θ B = tan −1 ⎜⎜ Solve the x-direction equation for TB:

TB = TF

sin θ F sin 16.7° = (500 N ) sin θ B sin 28.1°

= 305 N Find the downward forces that TB and TF exert on the mast:

∑F

Solve for Fmast to obtain:

Fmast = TF cosθ F + TB cosθ B

y

= Fmast − TF cos θ F − TB cos θ B = 0

256 Chapter 4 Substitute numerical values and evaluate Fmast:

Fmast = (500 N ) cos16.7° + (305 N )cos 28.1° = 748 N The force that the mast exerts on the deck is the sum of its weight and the downward forces exerted on it by the forestay and backstay:

Fmast on the deck = 748 N + 800 N = 1.55 kN

94 •• Picture the Problem Let m be the mass of the block and M be the mass of the chain. The free-body diagrams shown below display the forces acting at the locations identified in the problem. We can apply Newton’s 2nd law with ay = 0 to each of the segments of the chain to determine the tensions. (a)

(a) Apply

(b)

∑F

y

= ma y to the block

(c)

Ta − mg = ma y

and solve for Ta:

or, because ay = 0,

Substitute numerical values and evaluate Ta:

Ta = (50 kg ) 9.81 m/s 2 = 491 N

(b) Apply

∑F

y

Ta = mg

= ma y to the block

and half the chain and solve for Tb:

(

)

M⎞ ⎛ Tb − ⎜ m + ⎟ g = ma y 2 ⎠ ⎝ or, because ay = 0,

M⎞ ⎛ Tb = ⎜ m + ⎟ g 2 ⎠ ⎝ Substitute numerical values and evaluate Tb: (c) Apply

∑F

y

= ma y to the block

and chain and solve for Tc:

(

)

Tb = (50 kg + 10 kg ) 9.81 m/s 2 = 589 N

Tc − (m + M )g = ma y or, because ay = 0,

Tc = (m + M )g

Newton’s Laws 257 Substitute numerical values and evaluate Tc:

Tc = (50 kg + 20 kg ) (9.81 m/s 2 ) = 687 N

*95 ••• Picture the Problem The free-body diagram shows the forces acting on the box as the man pushes it across a frictionless floor. Because the force is time-dependent, the acceleration will be, too. We can obtain the acceleration as a function of time from the application of Newton’s 2nd law and then find the velocity of the box as a function of time by integration. Finally, we can derive an expression for the displacement of the box as a function of time by integration of the velocity function. (a) The velocity is related to the acceleration according to: Apply

∑F

x

= ma x to the box and

solve for its acceleration:

dv = a(t ) dt F = ma and

a= Because the box’s acceleration is a function of time, separate variables in equation (1) and integrate to find v as a function of time:

Evaluate v at t = 3 s: (b) Integrate v = dx/dt between 0 and 3 s to find the displacement of the box during this time:

(1)

F (8 N/s )t = = (13 m/s3 )t m 24 kg t

v(t ) = ∫ a(t ')dt ' =

(

)

0

=

(

t

m/s3 ∫ t ' dt '

1 3

0

) t2 = (

)

2

1 3

m/s3

1 6

m/s3 t 2

v(3 s ) = (16 m/s3 )(3 s ) = 1.50 m/s 2

3s

∆x = ∫ v(t ')dt ' =

(

)∫ t ' 3s

1 6

m/s

3

2

dt '

0

0

3s

⎡ t '3 ⎤ = ⎢ 16 m/s3 ⎥ = 1.50 m 3 ⎦0 ⎣

(

(c) The average velocity is given by:

(d) Use Newton’s 2nd law to express the average force exerted on the box by the man:

vave =

)

∆x 1.5 m = = 0.500 m/s ∆t 3s

Fav = maav = m

∆v ∆t

258 Chapter 4 Substitute numerical values and evaluate Fav:

Fav = (24 kg )

1.5 m/s − 0 m/s = 12.0 N 3s

96 •• Picture the Problem The application of Newton’s 2nd law to the glider and the hanging weight will lead to simultaneous equations in their common acceleration a and the tension T in the cord that connects them. Once we know the acceleration of this system, we can use a constant-acceleration equation to predict how long it takes the cart to travel r r 1 m from rest. Note that the magnitudes of T and T ' are equal. (a) The free-body diagrams are shown to the right. m1 represents the mass of the cart and m2 the mass of the hanging weight.

(b) Apply

∑F

x

= ma x to the cart

and the suspended mass:

T − m1 g sin θ = m1a1 and m2 g − T = m2 a2

Letting a represent the common accelerations of the two objects, eliminate T between the two equations and solve a:

a=

m2 − m1 sin θ g m1 + m2

Substitute numerical values and evaluate a:

a=

0.075 kg − (0.270 kg )sin30° 0.075 kg + 0.270 kg

(

× 9.81 m/s 2

)

= − 1.71 m/s 2 i.e., the acceleration is down the incline. Substitute for a in either of the force equations to obtain:

T = 0.863 N

(c) Using a constant-acceleration equation, relate the displacement of the cart down the incline to its initial speed and acceleration:

∆x = v0 ∆t + 12 a (∆t )

Solve for ∆t:

Substitute numerical values and evaluate ∆t:

2

or, because v0 = 0,

∆x = 12 a (∆t )

2

∆t =

2∆x a

∆t =

2(1 m ) = 1.08 s 1.71 m/s 2

Newton’s Laws 259 97 •• Picture the Problem Note that, while the mass of the rope is distributed over its length, the rope and the block have a common acceleration. Because the surface is horizontal r and smooth, the only force that influences our solution is F . The figure misrepresents the situation in that each segment of the rope experiences a gravitational force; the combined effect of which is that the rope must sag.

r

r

(a) Apply a = Fnet / mtot to the ropeblock system to obtain: (b) Apply

r

r

∑ F = ma to the rope,

substitute the acceleration of the system obtained in (a), and simplify to obtain:

(c) Apply

r

r

∑ F = ma to the block,

substitute the acceleration of the system obtained in (a), and simplify to obtain:

a=

F m1 + m2

⎛ F ⎞ ⎟⎟ Fnet = m2 a = m2 ⎜⎜ ⎝ m1 + m2 ⎠ m2 F = m1 + m2 ⎛ F ⎞ ⎟⎟ T = m1a = m1 ⎜⎜ ⎝ m1 + m2 ⎠ m1 F = m1 + m2

r

(d) The rope sags and so F has both vertical and horizontal components; with its horizontal component being r less than F . Consequently, a will be somewhat smaller.

*98 •• Picture the Problem The free-body diagram shows the forces acting on the block. Choose the coordinate system shown on the diagram. Because the surface of the wedge is frictionless, the force it exerts on the block must be normal to its surface.

(a) Apply to obtain:

∑F

y

= ma y to the block

Fn sin 30° − w = ma y or, because ay = 0 and w = mg,

Fn sin 30° − mg = 0

or

260 Chapter 4

Apply

∑F

x

= ma x to the block:

Fn sin 30° = mg

(1)

Fn cos 30° = max

(2)

Divide equation (2) by equation (1) to obtain:

ax = cot 30° g

Solve for and evaluate ax:

a x = g cot 30° = 9.81 m/s 2 cot 30°

(

)

= 17.0 m/s 2 (b) An acceleration of the wedge greater than g cot30° would require that the normal force exerted on the body by the wedge be greater than that given in part (a); i.e., Fn > mg/sin30°.

Under this condition, there would be a net force in the y direction and the block would accelerate up the wedge.

99 •• Picture the Problem Because the system is initially in equilibrium, it follows that T0 = 5mg. When one washer is removed on the left side, the remaining washers will accelerate upward (and those on the right side downward) in response to the net force that results. The free-body diagrams show the forces under this unbalanced condition. Applying Newton’s 2nd law to each collection of washers will allow us to determine both the acceleration of the system and the mass of a single washer.

(a) Apply

∑F

y

= ma y to the rising

T − 4mg = (4m )a

(1)

5mg − T = (5m )a

(2)

masses: Apply

∑F

y

= ma y to the

descending masses: Eliminate T between these equations to obtain:

a = 19 g

Use this acceleration in equation (1) or equation (2) to obtain:

T=

Express the difference between T0 and T and solve for m:

T0 − T = 5mg − and

40 mg 9 40 mg = 0.3 N 9

Newton’s Laws 261 m = 0.0550 kg = 55.0 g (b) Proceed as in (a) to obtain:

T – 3mg = 3ma and 5mg – T = 5ma

Eliminate T and solve for a:

a = 14 g =

Eliminate a in either of the motion equations and solve for T to obtain:

T=

15 mg 4

Substitute numerical values and evaluate T:

T=

15 (0.0550 kg ) 9.81m/s2 4

1 4

(9.81m/s ) = 2

2.45 m/s 2

(

= 2.03 N 100 •• Picture the Problem The free-body diagram represents the Atwood’s machine with N washers moved from the left side to the right side. Application of Newton’s 2nd law to each collection of washers will result in two equations that can be solved simultaneously to relate N, a, and g. The acceleration can then be found from the given data. Apply

∑F

y

= ma y to the rising

T – (5 – N)mg = (5 – N)ma

= ma y to the

(5 +N)mg – T = (5 + N)ma

washers: Apply

∑F

y

descending washers: Add these equations to eliminate T:

(5 + N )mg − (5 − N )mg = (5 − N )ma + (5 + N )ma

Simplify to obtain:

2 Nmg = 10ma

Solve for N:

N = 5a/g

Using a constant-acceleration equation, relate the distance the washers fell to their time of fall:

∆y = v0 ∆t + 12 a(∆t )

2

or, because v0 = 0,

∆y = 12 a(∆t )

2

)

262 Chapter 4 Solve for the acceleration:

a=

2∆y (∆t )2

Substitute numerical values and evaluate a:

a=

2(0.471 m ) = 5.89 m/s 2 2 (0.40 s )

Substitute in the expression for N:

⎛ 5.89 m/s 2 ⎞ ⎟= 3 N = 5⎜⎜ 2 ⎟ ⎝ 9.81 m/s ⎠

101 •• Picture the Problem Draw the free-body diagram for the block of mass m and apply Newton’s 2nd law to obtain the acceleration of the system and then the tension in the rope connecting the two blocks. (a) Letting T be the tension in the connecting string, apply Fx = ma x to the block of

T – F1 = ma

∑

mass m: Apply

∑F

x

= ma x to both blocks

F2 – F1 = (m + 2m)a = (3m)a

to determine the acceleration of the system: Substitute and solve for a:

a = (F2 – F1)/3m

Substitute for a in the first equation and solve for T:

T=

(b) Substitute for F1 and F2 in the equation derived in part (a):

T = (2Ct +2Ct)/3 = 4Ct/3

Evaluate this expression for T = T0 and t = t0 and solve for t0:

t0 =

1 3

(F2 + 2F1 )

3T0 4C

Newton’s Laws 263 *102 ••• Picture the Problem Because a constantupward acceleration has the same effect as an increase in the acceleration due to gravity, we can use the result of Problem 89 (for the tension) with a replaced by a + g. The application of Newton’s 2nd law to the object whose mass is m2 will connect the acceleration of this body to tension from Problem 84. In Problem 84 it is given that, when the support pulley is not accelerating, the tension in the rope and the acceleration of the masses are related according to:

T=

2m1m2 g m1 + m2

Replace a with a + g:

T=

2m1m2 (a + g ) m1 + m2

Apply

∑F

y

= ma y to the object

whose mass is m2 and solve for a2:

Substitute for T and simplify to obtain:

The expression for a1 is the same as for a2 with all subscripts interchanged (note that a positive value for a1 represents acceleration upward):

T – m2g = m2a2 and

a2 =

a2 =

a1 =

T − m2 g m2

(m1 − m2 )g + 2m1a m1 + m2

(m2 − m1 )g + 2m2a m1 + m2

264 Chapter 4

Chapter 5 Applications of Newton’s Laws Conceptual Problems 1 • Determine the Concept Because the objects are speeding up (accelerating), there must be a net force acting on them. The forces acting on an object are the normal force exerted by the floor of the truck, the weight of the object, and the friction force; also exerted by the floor of the truck.

The force of friction between the object and the floor of the truck

Of these forces, the only one that acts in the direction of the acceleration (chosen to be to the right in the free-body diagram) is the friction force.

must be the force that causes the object to accelerate.

*2 • Determine the Concept The forces acting on an object are the normal force exerted by the floor of the truck, the weight of the object, and the friction force; also exerted by the floor of the truck. Of these forces, the only one that acts in the direction of the acceleration (chosen to be to the right in the free-body diagram) is the friction force. Apply Newton’s 2nd law to the object to determine how the critical acceleration depends on its weight. Taking the positive x direction to be to the right, apply ΣFx = max and solve for ax:

f = µsw = µsmg = max and ax = µsg

Because ax is independent of m and w, the critical accelerations are the same.

265

266 Chapter 5 3 • Determine the Concept The forces acting

r

on the block are the normal force Fn exerted by the incline, the weight of the r block mg exerted by the earth, and the

r

static friction force f s exerted by an external agent. We can use the definition of µs and the conditions for equilibrium to determine the relationship between µs and θ. Apply

∑F

= ma x to the block:

fs − mgsinθ = 0

(1)

Apply

∑F

= ma y in the y

Fn − mgcosθ = 0

(2)

x

y

direction: Divide equation (1) by equation (2) to obtain:

tan θ =

Substitute for fs (≤ µsFn):

tan θ ≤

fs Fn

µs Fn Fn

= µs

and (d ) is correct. *4 • Determine the Concept The block is in

r

r

equilibrium under the influence of Fn , mg,

r

and f s ; i.e.,

r r r Fn + mg + f s = 0 We can apply Newton’s 2nd law in the x direction to determine the relationship between fs and mg. Apply

∑F

x

Solve for fs:

= 0 to the block:

fs − mgsinθ = 0 fs = mgsinθ and (d ) is correct.

Applications of Newton’s Laws 267 5 •• Picture the Problem The forces acting on the car as it rounds a curve of radius R at maximum speed are shown on the free-body diagram to the right. The centripetal force is the static friction force exerted by the roadway on the tires. We can apply Newton’s 2nd law to the car to derive an expression for its maximum speed and then compare the speeds under the two friction conditions described. Apply

r

r

∑ F = ma to the car:

∑ Fx = fs, max = m

2 vmax R

and

∑F

y

= Fn − mg = 0

From the y equation we have:

Fn = mg

Express fs,max in terms of Fn in the x equation and solve for vmax:

vmax = µs gR or

vmax = constant µs Express v'max for µs' = 12 µs :

v'max = constant

µs 2

and (b) is correct. *6 •• Picture the Problem The normal reaction force Fn provides the centripetal force and the force of static friction, µsFn, keeps the cycle from sliding down the wall. We can apply Newton’s 2nd law and the definition of fs,max to derive an expression for vmin.

Apply

r

r

∑ F = ma to the motorcycle:

∑ Fx = Fn = m and

v2 R

= .707vmax ≈ 71%vmax

268 Chapter 5

∑F For the minimum speed: Substitute for fs, eliminate Fn between the force equations, and solve for vmin: Assume that R = 6 m and µs = 0.8 and solve for vmin:

y

= f s − mg = 0

fs = fs,max = µsFn

vmin =

vmin =

Rg

µs

(6 m )(9.81m/s2 ) 0.8

= 8.58 m/s = 30.9 km/h 7 •• Determine the Concept As the spring is extended, the force exerted by the spring on the block increases. Once that force is greater than the maximum value of the force of static friction on the block, the block will begin to move. However, as it accelerates, it will shorten the length of the spring, decreasing the force that the spring exerts on the block. As this happens, the force of kinetic friction can then slow the block to a stop, which starts the cycle over again. One interesting application of this to the real world is the bowing of a violin string: The string under tension acts like the spring, while the bow acts as the block, so as the bow is dragged across the string, the string periodically sticks and frees itself from the bow. 8 • True. The velocity of an object moving in a circle is continually changing independently of whether the object’s speed is changing. The change in the velocity vector and the acceleration vector and the net force acting on the object all point toward the center of circle. This center-pointing force is called a centripetal force. 9 • Determine the Concept A particle traveling in a vertical circle experiences a downward gravitational force plus an additional force that constrains it to move along a circular path. Because the net force acting on the particle will vary with location along its trajectory, neither (b), (c), nor (d) can be correct. Because the velocity of a particle moving along a circular path is continually changing, (a) cannot be correct. (e) is correct. *10 • Determine the Concept We can analyze these demonstrations by drawing force diagrams for each situation. In both diagrams, h denotes ″hand″, g denotes ″gravitational″, m denotes ″magnetic″, and n denotes ″normal″.

Applications of Newton’s Laws 269 Demonstration 2:

(a) Demonstration 1:

(b) Because the magnet doesn’t lift the iron in the first demonstration, the force exerted on the iron must be less than its (the iron’s) weight. This is still true when the two are falling, but the motion of the iron is not restrained by the table, and the motion of the magnet is not restrained by the hand. Looking at the second diagram, the net force pulling the magnet down is greater than its weight, implying that its acceleration is greater than g. The opposite is true for the iron: the magnetic force acts upwards, slowing it down, so its acceleration will be less than g. Because of this, the magnet will catch up to the iron piece as they fall. *11 ••• Picture the Problem The free-body diagrams show the forces acting on the two objects some time after block 2 is dropped.

r

r

Note that, while T1 ≠ T2 , T1 = T2.

The only force pulling block 2 to the left is the horizontal component of the tension. Because this force is smaller than the magnitude of the tension, the acceleration of block 1, which is identical to block 2, to the right (T1 = T2) will always be greater than the acceleration of block 2 to the left.

Because the initial distance from block 1 to the pulley is the same as the initial distance of block 2 to the wall, block 1 will hit the pulley before block 2 hits the wall. 12 • 1n True. The terminal speed of an object is given by vt = (mg b ) , where b depends on the shape and area of the falling object as well as upon the properties of the medium in which the object is falling. 13 • 1n Determine the Concept The terminal speed of a sky diver is given by vt = (mg b ) , where b depends on the shape and area of the falling object as well as upon the properties of the medium in which the object is falling. The sky diver’s orientation as she falls

270 Chapter 5 determines the surface area she presents to the air molecules that must be pushed aside.

(d ) is correct. 14 •• Determine the Concept In your frame of reference (the accelerating reference frame of the car), the direction of the force must point toward the center of the circular path along which you are traveling; that is, in the direction of the centripetal force that keeps you moving in a circle. The friction between you and the seat you are sitting on supplies this force. The reason you seem to be "pushed" to the outside of the curve is that your body’s inertia "wants" , in accordance with Newton’s law of inertia, to keep it moving in a straight line–that is, tangent to the curve. *15 • Determine the Concept The centripetal force that keeps the moon in its orbit around the earth is provided by the gravitational force the earth exerts on the moon. As described by Newton’s 3rd law, this force is equal in magnitude to the force the moon exerts on the earth.

(d ) is correct.

16 • Determine the Concept The only forces acting on the block are its weight and the force the surface exerts on it. Because the loop-the-loop surface is frictionless, the force it exerts on the block must be perpendicular to its surface. Point A: the weight is downward and the normal force is to the right.

Free-body diagram 3

Point B: the weight is downward, the normal force is upward, and the normal force is greater than the weight so that their difference is the centripetal force.

Free-body diagram 4

Point C: the weight is downward and the normal force is to the left.

Free-body diagram 5

Point D: both the weight and the normal forces are downward.

Free-body diagram 2

Applications of Newton’s Laws 271 17 •• Picture the Problem Assume that the drag force on an object is given by the Newtonian formula FD = 12 CAρv 2 , where A is the projected surface area, v is the object’s speed, ρ is the density of air, and C a dimensionless coefficient. Express the net force acting on the falling object:

Fnet = mg − FD = ma

Substitute for FD under terminal speed conditions and solve for the terminal speed:

mg − 12 CAρvT2 = 0 or

vT =

2mg CAρ

Thus, the terminal velocity depends on the ratio of the mass of the object to its surface area. For a rock, which has a relatively small surface area compared to its mass, the terminal speed will be relatively high; for a lightweight, spread-out object like a feather, the opposite is true. Another issue is that the higher the terminal velocity is, the longer it takes for a falling object to reach terminal velocity. From this, the feather will reach its terminal velocity quickly, and fall at an almost constant speed very soon after being dropped; a rock, if not dropped from a great height, will have almost the same acceleration as if it were in freefall for the duration of its fall, and thus be continually speeding up as it falls. An interesting point is that the average drag force acting on the rock will be larger than that acting on the feather precisely because the rock’s average speed is larger than the feather's, as the drag force increases as v2. This is another reminder that force is not the same thing as acceleration.

Estimation and Approximation *18 • Picture the Problem The free-body diagram shows the forces on the Tercel as it slows from 60 to 55 mph. We can use Newton’s 2nd law to calculate the average force from the rate at which the car’s speed decreases and the rolling force from its definition. The drag force can be inferred from the average and rolling friction forces and the drag coefficient from the defining equation for the drag force. (a) Apply

∑F

x

= max to the car to relate

the average force acting on it to its average velocity:

Fav = maav = m

∆v ∆t

272 Chapter 5 Substitute numerical values and evaluate Fav:

Fav = (1020 kg )

5

mi km 1h 1000 m × 1.609 × × h mi 3600 s km = 581 N 3.92 s

(b) Using its definition, express and evaluate the force of rolling friction:

f rolling = µ rolling Fn = µ rolling mg

(

= (0.02 )(1020 kg ) 9.81 m/s 2

)

= 200 N Assuming that only two forces are acting on the car in the direction of its motion, express their relationship and solve for and evaluate the drag force: (c) Convert 57.5 mi/h to m/s:

Fav = Fdrag + Frolling and

Fdrag = Fav − Frolling = 581 N − 200 N = 381 N 57.5

mi mi 1.609 km = 57.5 × h h mi 1h 103 m × × 3600 s km = 25.7 m/s

Using the definition of the drag force and its calculated value from (b) and the average speed of the car during this 5 mph interval, solve for C: Substitute numerical values and evaluate C:

Fdrag = 12 Cρ Av 2 ⇒ C =

C=

2 Fdrag

ρ Av 2

2(381 N ) 2 1.21 kg/m 1.91 m 2 (25.7 m/s )

(

3

)(

)

= 0.499 19 • Picture the Problem We can use the dimensions of force and velocity to determine the dimensions of the constant b and the dimensions of ρ, r, and v to show that, for n = 2, Newton’s expression is consistent dimensionally with our result from part (b). In parts (d) and (e), we can apply Newton’s 2nd law under terminal velocity conditions to find the terminal velocity of the sky diver near the surface of the earth and at a height of 8 km. (a) Solve the drag force equation for b with n = 1:

b=

Fd v

Applications of Newton’s Laws 273 Substitute the dimensions of Fd and v and simplify to obtain:

ML 2 [b] = TL = M T T and the units of b are kg/s

Fd v2

(b) Solve the drag force equation for b with n = 2:

b=

Substitute the dimensions of Fd and v and simplify to obtain:

ML 2 [b] = T 2 = M L ⎛L⎞ ⎜ ⎟ ⎝T ⎠ and the units of b are kg/m

(c) Express the dimensions of Newton’s expression:

[Fd ] = [12 ρπr 2v 2 ] = ⎛⎜ M3 ⎞⎟(L )2 ⎛⎜ L ⎞⎟ ⎝L ⎠

= From part (b) we have:

2

⎝T ⎠

ML T2

[Fd ] = [bv 2 ] = ⎛⎜ M ⎞⎟⎛⎜ L ⎞⎟

2

⎝ L ⎠⎝ T ⎠

= (d) Letting the downward direction be the positive y direction, apply Fy = ma y to the sky diver:

ML T2

mg − 12 ρπr 2 vt2 = 0

∑

Solve for and evaluate vt:

vt =

2mg = ρπ r 2

(

(

)

= 56.9 m/s (e) Evaluate vt at a height of 8 km:

vt =

(

)

2(56 kg ) 9.81m/s 2 2 π 0.514 kg/m 3 (0.3 m )

(

= 86.9 m/s

)

2(56 kg ) 9.81 m/s 2 2 π 1.2 kg/m 3 (0.3 m )

)

274 Chapter 5 20 •• Picture the Problem From Newton’s 2nd law, the equation describing the motion of falling raindrops and large hailstones is mg – Fd = ma where Fd = 12 ρπ r 2v 2 = bv 2 is the drag force. Under terminal speed conditions (a = 0), the drag force is equal to the weight of the falling object. Take the radius of a raindrop rr to be 0.5 mm and the radius of a golf-ball sized hailstone rh to be 2 cm. Using b = 12 πρ r 2 , evaluate br and bh:

(

)(

br = 12 π 1.2 kg/m 3 0.5 × 10 −3 m

)

2

= 4.71 × 10 −7 kg/m and

(

)(

bh = 12 π 1.2 kg/m 3 2 × 10 −2 m

)

2

= 7.54 × 10 −4 kg/m 4π r 3 ρ 3

Express the mass of a sphere in terms of its volume and density:

m = ρV =

Using ρr = 103 kg/m3 and ρh = 920 kg/m3, evaluate mr and mh:

4π (0.5 × 10 −3 m ) (103 kg/m 3 ) mr = 3 −7 = 5.24 × 10 kg 3

and

4π (2 × 10 −2 m ) (920 kg/m 3 ) 3 −2 = 3.08 × 10 kg 3

mh =

Express the relationship between vt and the weight of a falling object under terminal speed conditions and solve for vt: Use numerical values to evaluate vt,r and vt,h:

mg b

bvt2 = mg ⇒ vt =

vt,r =

(5.24 ×10

−7

)(

)

)(

)

kg 9.81 m/s 2 4.71× 10 −7 kg/m

= 3.30 m/s and

vt,h =

(3.08 ×10

−2

kg 9.81 m/s 2 7.54 × 10 −4 kg/m

= 20.0 m/s

Applications of Newton’s Laws 275

Friction *21 • Picture the Problem The block is in

r

equilibrium under the influence of Fn ,

r r mg, and f k ; i.e., r r r Fn + mg + f k = 0

We can apply Newton’s 2nd law to determine the relationship between fk, θ, and mg. Using its definition, express the coefficient of kinetic friction: Apply

∑F

x

µk =

= max to the block:

∑F

y

(1)

fk − mgsinθ = max = 0 because ax = 0 fk = mgsinθ

Solve for fk: Apply

fk Fn

= ma y to the block:

Fn − mgcosθ = may = 0 because ay = 0

Solve for Fn:

Fn = mgcosθ

Substitute in equation (1) to obtain:

µk =

mg sin θ = tan θ mg cos θ

and (b) is correct. 22 • Picture the Problem The block is in

r

equilibrium under the influence of Fn ,

r r r mg, Fapp , and f k ; i.e., r r r r Fn + mg + Fapp + f k = 0

We can apply Newton’s 2nd law to determine fk. Apply

∑F

Solve for fk:

x

= max to the block:

Fapp − fk = max = 0 because ax = 0 fk = Fapp = 20 N and (e) is correct.

276 Chapter 5 *23 • Picture the Problem Whether the friction force is that due to static friction or kinetic friction depends on whether the applied tension is greater than the maximum static friction force. We can apply the definition of the maximum static friction to decide whether fs,max or T is greater.

Calculate the maximum static friction force:

fs,max = µsFn = µsw = (0.8)(20 N) = 16 N

(a) Because fs,max > T:

f = fs = T = 15.0 N

(b) Because T > fs,max:

f = fk = µkw = (0.6)(20 N) = 12.0 N

24 • Picture the Problem The block is in equilibrium under the influence of the r r r forces T, f k , and mg; i.e.,

r r r T + f k + mg = 0

We can apply Newton’s 2nd law to determine the relationship between T and fk . Apply

∑F

x

= max to the block:

Solve for fk:

T cosθ − fk = max = 0 because ax = 0 fk = T cosθ and (b) is correct.

25 • Picture the Problem Whether the friction force is that due to static friction or kinetic friction depends on whether the applied tension is greater than the maximum static friction force.

Calculate the maximum static

fs,max = µsFn = µsw

Applications of Newton’s Laws 277 = (0.6)(100 kg)(9.81 m/s2) = 589 N

friction force:

Because fs,max > Fapp, the box does not move and :

Fapp = f s = 500 N

26 • Picture the Problem Because the box is moving with constant velocity, its acceleration is zero and it is in equilibrium

r

r

r

under the influence of Fapp , Fn , w , and

r f ; i.e.,

r r r r Fapp + Fn + w + f = 0

We can apply Newton’s 2nd law to determine the relationship between f and mg. The definition of µk is:

Apply

∑F

y

= ma y to the box:

Solve for Fn: Apply

∑F

x

µk =

fk Fn

Fn – w = may = 0 because ay = 0 Fn = w = 600 N

= max to the box:

ΣFx = Fapp – f = max = 0 because ax = 0

Solve for fk:

Fapp = fk = 250 N

Substitute to obtain µk:

µk = (250 N)/(600 N) = 0.417

27 • Picture the Problem Assume that the car is traveling to the right and let the positive x direction also be to the right. We can use Newton’s 2nd law of motion and the definition of µs to determine the maximum acceleration of the car. Once we know the car’s maximum acceleration, we can use a constant-acceleration equation to determine the least stopping distance.

278 Chapter 5 (a) Apply Apply

∑F

x

∑F

y

= max to the car:

= ma y to the car and

solve for Fn:

−fs,max = −µsFn = max

(1)

Fn − w = may = 0 or, because ay = 0, Fn = mg

(2)

Substitute (2) in (1) and solve for ax,max:

ax,max = µs g = (0.6)(9.81 m/s 2 )

(b) Using a constant-acceleration equation, relate the stopping distance of the car to its initial velocity and its acceleration and solve for its displacement:

v 2 = v02 + 2a∆x

Substitute numerical values and evaluate ∆x:

= − 5.89 m/s 2

or, because v = 0,

∆x =

− v02 2a

∆x =

− (30 m/s ) = 76.4 m 2 − 5.89 m/s 2 2

(

)

*28 • Picture the Problem The free-body diagram shows the forces acting on the drive wheels, the ones we’re assuming support half the weight of the car. We can use the definition of acceleration and apply Newton’s 2nd law to the horizontal and vertical components of the forces to determine the minimum coefficient of friction between the road and the tires. (a) Because µs > µ k , f will be greater if the wheels do not slip. (b) Apply Apply

∑F

x

∑F

y

= max to the car:

= ma y to the car and

fs = µsFn = max

(1)

Fn − 12 mg = ma y

solve for Fn:

Because ay = 0, Fn − 12 mg = 0 ⇒ Fn = 12 mg

Find the acceleration of the car:

ax =

∆v (90 km/h )(1000 m/km ) = 12 s ∆t

= 2.08 m/s 2

Applications of Newton’s Laws 279 Solve equation (1) for µs:

µs =

ma x 2a x = 1 g 2 mg

Substitute numerical values and evaluate ax:

µs =

2(2.08 m/s 2 ) = 0.424 9.81 m/s 2

29 • Picture the Problem The block is in equilibrium under the influence of the forces shown on the free-body diagram. We can use Newton’s 2nd law and the definition of µs to solve for fs and Fn.

(a) Apply

∑F

y

= ma y to the block

and solve for fs:

Solve for and evaluate fs:

f s − mg = ma y or, because ay = 0,

f s − mg = 0

f s = mg = (5 kg )(9.81 m/s 2 ) = 49.1 N

(b) Use the definition of µs to express Fn:

Fn =

Substitute numerical values and evaluate Fn:

Fn =

30 • Picture the Problem The free-body diagram shows the forces acting on the book. The normal force is the net force the student exerts in squeezing the book. Let the horizontal direction be the x direction and upward the y direction. Note that the normal force is the same on either side of the book because it is not accelerating in the horizontal direction. The book could be accelerating downward. We can apply Newton’s 2nd law to relate the minimum force required to hold the book in place to its mass and to the coefficients of static friction. In part (b), we can proceed similarly to relate the acceleration of the

f s,max

µs 49.1 N = 123 N 0.4

280 Chapter 5 book to the coefficients of kinetic friction. (a) Apply

r

r

∑ F = ma to the book:

∑F

= F2,min − F1,min = 0

x

and

∑F

= µ s ,1 F1',min + µ s,2 F2',min − mg = 0

Fmin =

mg µ,1 + µs,2

y

Noting that F1',min = F2' ,min , solve the y equation for Fmin: Substitute numerical values and evaluate Fmin: (b) Apply

∑F

y

= ma y with the

Fmin =

(10.2 kg )(9.81m/s2 ) =

∑F

= µ k ,1 F + µ k,2 F − mg = ma

y

0.32 + 0.16

208 N

book accelerating downward, to obtain: Solve for a to obtain:

Substitute numerical values and evaluate a:

a=

µk , + µk ,2 m

a =

F−g

0.2 + 0.09 (195 N ) − 9.81 m/s 2 10.2 kg

= − 4.27 m/s 2 31 • Picture the Problem A free-body diagram showing the forces acting on the car is shown to the right. The friction force that the ground exerts on the tires is the force fs shown acting up the incline. We can use the definition of the coefficient of static friction and Newton’s 2nd law to relate the angle of the incline to the forces acting on the car. Apply

r

r

∑ F = ma to the car:

∑F

x

= f s − mg sin θ = 0

(1)

= Fn − mg cos θ = 0

(2)

and

∑F

y

Solve equation (1) for fs and equation (2) for Fn:

f s = mg sin θ and

Applications of Newton’s Laws 281 Fn = mg cos θ

f s mg sin θ = = tan θ Fn mg cos θ

Use the definition of µs to relate fs and Fn:

µs =

Solve for and evaluate θ :

θ = tan −1 (µs ) = tan −1 (0.08) = 4.57°

*32 • Picture the Problem The free-body diagrams for the two methods are shown to the right. Method 1 results in the box being pushed into the floor, increasing the normal force and the static friction force. Method 2 partially lifts the box,, reducing the normal force and the static friction force. We can apply Newton’s 2nd law to obtain expressions that relate the maximum static r friction force to the applied force F . (a) Method 2 is preferable as it reduces Fn and, therefore, f s. (b) Apply

∑F

x

= max to the box:

∑F

= ma y to

F cosθ − fs = Fcosθ − µsFn = 0

the block and solve for Fn:

Fn – mg − Fsinθ = 0 ∴ Fn = mg + Fsinθ

Relate fs,max to Fn:

fs,max = µsFn = µs(mg + Fsinθ)

Method 1: Apply

y

∑F

= ma y to

(1)

the forces in the y direction and solve for Fn:

Fn – mg + Fsinθ = 0 and Fn = mg − Fsinθ

Relate fs,max to Fn:

fs,max = µsFn = µs(mg − Fsinθ)

(2)

Express the condition that must be satisfied to move the box by either method:

fs,max = Fcosθ

(3)

Method 1: Substitute (1) in (3) and solve for F:

F1 =

Method 2: Apply

y

µ s mg cosθ − µ s sin θ

(4)

282 Chapter 5

µ s mg cosθ + µ s sin θ

Method 2: Substitute (2) in (3) and solve for F:

F2 =

Evaluate equations (4) and (5) with θ = 30°:

F1 (30°) = 520 N

Evaluate (4) and (5) with θ = 0°:

F1 (0°) = F2 (0°) = µ s mg = 294 N

(5)

F2 (30°) = 252 N

33 • Picture the Problem Draw a free-body diagram for each object. In the absence of friction, the 3-kg box will move to the right, and the 2-kg box will move down.

r

The friction force is indicated by f without

r

r

subscript; it is f s for (a) and f k for (b). For values of µs less than the value found in part (a) required for equilibrium, the system will accelerate and the fall time for a given distance can be found using a constantacceleration equation. (a) Apply

∑F

x

= max to the 3-kg

T – fs = 0 because ax = 0

(1)

Fn,3 – m3g = 0 because ay = 0 and T – µs m3g = 0

(2)

m2g – T = 0 because ax = 0

(3)

box: Apply

∑F

y

= ma y to the 3-kg box,

solve for Fn,3, and substitute in (1):

Apply

∑F

x

= max to the 2-kg box:

Solve (2) and (3) simultaneously and solve for µs: (b) The time of fall is related to the acceleration, which is constant:

µs =

m2 = 0.667 m3

∆x = v0 ∆t + 12 a(∆t )

2

or, because v0 =0,

∆x = 12 a(∆t )

2

Solve for ∆t:

∆t =

2 ∆x a

(4)

Applications of Newton’s Laws 283 Apply

∑F

x

= max to each box:

Add equations (5) and (6) and solve for a: Substitute numerical values and evaluate a:

T – µk m3g = m3a and m2g – T = m2a

a=

(5) (6)

(m2 − µk m3 )g m2 + m3

[ 2 kg − 0.3(3 kg )](9.81m/s 2 ) a= 2 kg + 3 kg

= 2.16 m/s 2 Substitute numerical values in equation (4) and evaluate ∆t:

∆t =

2(2 m ) = 1.36 s 2.16 m/s 2

34 •• Picture the Problem The application of Newton’s 2nd law to the block will allow us to express the coefficient of kinetic friction in terms of the acceleration of the block. We can then use a constant-acceleration equation to determine the block’s acceleration. The pictorial representation summarizes what we know about the motion.

A free-body diagram showing the forces acting on the block is shown to the right.

Apply

∑F

= max to the block:

– fk = −µkFn = ma

(1)

Apply

∑F

= ma y to the block and

Fn – mg = 0 because ay = 0 and Fn = mg

(2)

x

y

solve for Fn:

284 Chapter 5 Substitute (2) in (1) and solve for µk:

µk = −a/g

Using a constant-acceleration equation, relate the initial and final velocities of the block to its displacement and acceleration:

v12 = v02 + 2a∆x

Solve for a to obtain:

Substitute for a in equation (3) to obtain:

(3)

or, because v1 = 0, v0 = v, and ∆x = d,

0 = v 2 + 2ad

a=

− v2 2d

µk =

v2 2 gd

*35 •• Picture the Problem We can find the speed of the system when it has moved a given distance by using a constantacceleration equation. Under the influence of the forces shown in the free-body diagrams, the blocks will have a common acceleration a. The application of Newton’s 2nd law to each block, followed by the elimination of the tension T and the use of the definition of fk, will allow us to determine the acceleration of the system. Using a constant-acceleration equation, relate the speed of the system to its acceleration and displacement; solve for its speed:

r

r

Apply Fnet = ma to the block whose mass is m1:

Using fk = µkFn, substitute (3) in (2) to obtain: Apply

∑F

x

= max to the block

v 2 = v02 + 2a∆x and, because v0 = 0,

v = 2a∆x

(1)

ΣFx = T – fk – m1gsin30° = m1a and ΣFy = Fn,1 – m1gcos30° = 0

(2) (3)

T – µk m1g cos30° – m1gsin30° = m1a

m2g – T = m2a

whose mass is m2: Add the last two equations to eliminate T and solve for a to

a=

(m2 − µk m1 cos 30° − m1 sin 30°)g m1 + m2

Applications of Newton’s Laws 285 obtain: Substitute numerical values and evaluate a:

a = 1.16 m/s 2

Substitute numerical values in equation (1) and evaluate v:

v = 2 1.16 m/s 2 (0.3 m ) = 0.835 m/s

(

)

and (a ) is correct.

36 •• Picture the Problem Under the influence of the forces shown in the free-body diagrams, the blocks are in static equilibrium. While fs can be either up or down the incline, the free-body diagram shows the situation in which motion is impending up the incline. The application of Newton’s 2nd law to each block, followed by the elimination of the tension T and the use of the definition of fs, will allow us to determine the range of values for m2. (a) Apply

r

r

∑ F = ma to the block

whose mass is m1: Using fs,max = µsFn, substitute (2) in (1) to obtain: Apply

∑F

x

= max to the block

ΣFx = T ± fs,max – m1gsin30° = 0 and ΣFy = Fn,1 – m1gcos30° = 0

T ± µ s m1 g cos 30° − m1 g sin 30° = m1a m2g – T = 0

(1) (2) (3)

(4)

whose mass is m2: Add equations (3) and (4) to eliminate T and solve for m2: Evaluate (5) denoting the value of m2 with the plus sign as m2,+ and the value of m2 with the minus sign as m2,- to determine the range of values of m2 for which the system is in static equilibrium:

m2 = m1 (± µs cos 30° + sin 30°)

= (4 kg )[± (0.4) cos 30° + sin 30°]

m2, + = 3.39 kg and m 2,- = 0.614 kg ∴ 0.614 kg ≤ m2 ≤ 3.39 kg

(5)

286 Chapter 5 (b) With m2 = 1 kg, the impending motion is down the incline and the static friction force is up the incline. Apply Fx = max to the block

T + fs – m1gsin30° = 0

(6)

∑

whose mass is m1: Apply

∑F

x

= max to the block

m2g – T = 0

(7)

whose mass is m2: Add equations (6) and (7) and solve for and evaluate fs:

fs = (m1sin30° – m2)g = [(4 kg)sin30° – 1 kg](9.81 m/s2) = 9.81 N

37 •• Picture the Problem Under the influence of the forces shown in the free-body diagrams, the blocks will have a common acceleration a. The application of Newton’s 2nd law to each block, followed by the elimination of the tension T and the use of the definition of fk, will allow us to determine the acceleration of the system. Finally, we can substitute for the tension in either of the motion equations to determine the acceleration of the masses. Apply

r

r

∑ F = ma to the block

whose mass is m1: Using fk = µkFn, substitute (2) in (1) to obtain: Apply

∑F

x

= max to the block

ΣFx = T – fk – m1gsin30° = m1a and ΣFy = Fn,1 – m1gcos30° = 0

T − µ k m1 g cos 30° − m1 g sin 30° = m1a m2g – T = m2a

(1) (2) (3)

(4)

whose mass is m2:

(m2 − µk m1 cos 30° − m1 sin 30°)g

Add equations (3) and (4) to eliminate T and solve for a to obtain:

a=

Substituting numerical values and evaluating a yields:

a = 2.36 m/s 2

m1 + m2

Applications of Newton’s Laws 287 T = 37.3 N

Substitute for a in equation (3) to obtain: *38 •• Picture the Problem The truck will stop in the shortest possible distance when its acceleration is a maximum. The maximum acceleration is, in turn, determined by the maximum value of the static friction force. The free-body diagram shows the forces acting on the box as the truck brakes to a stop. Assume that the truck is moving in the positive x direction and apply Newton’s 2nd law and the definition of fs,max to find the shortest stopping distance.

v 2 = v02 + 2a∆x

Using a constant-acceleration equation, relate the truck’s stopping distance to its acceleration and initial velocity; solve for the stopping distance:

r

or, since v = 0, ∆xmin =

r

Apply Fnet = ma to the block:

− v02 2amax

ΣFx = – fs,max = mamax and ΣFy = Fn – mg = 0

(2)

fs,max ≡ µsFn and amax = −µsg = − (0.3)(9.81 m/s2) = −2.943 m/s2

Using the definition of fs,max, solve equations (1) and (2) simultaneously for a:

Substitute numerical values and evaluate ∆xmin:

− (80 km/h ) (1000 km/m ) (1 h/3600 s ) = 9.16 m 2 − 2.943 m/s 2 2

∆xmin =

(1)

2

(

2

)

288 Chapter 5 39 •• Picture the Problem We can find the coefficient of friction by applying Newton’s 2nd law and determining the acceleration from the given values of displacement and initial velocity. We can find the displacement and speed of the block by using constant-acceleration equations. During its motion up the incline, the sum of the kinetic friction force and a component of the object’s weight will combine to bring the object to rest. When it is moving down the incline, the difference between the weight component and the friction force will be the net force. (a) Draw a free-body diagram for the block as it travels up the incline:

Apply

r

r

∑ F = ma to the block:

Substitute fk = µkFn and Fn from (2) in (1) and solve for µk:

ΣFx = – fk – mgsin37°= ma and ΣFy = Fn – mg cos37° = 0

µk =

− g sin 37° − a g cos 37°

a = − tan 37° − g cos 37° Using a constant-acceleration equation, relate the final velocity of the block to its initial velocity, acceleration, and displacement: Solving for a yields:

Substitute numerical values and evaluate a:

(1) (2)

(3)

v12 = v02 + 2a∆x

a=

v12 − v02 2∆x

2 2 ( 5.2 m/s ) − (14 m/s ) a= 2(8 m )

= −10.6 m/s 2

Applications of Newton’s Laws 289 Substitute for a in (3) to obtain:

− 10.6 m/s 2 µ k = − tan 37° − 9.81m/s 2 cos37°

(

)

= 0.599 (b) Use the same constantacceleration equation used above but with v1 = 0 to obtain:

0 = v02 + 2a∆x

∆x =

− v02 2a

Substitute numerical values and evaluate ∆x:

∆x =

− (14 m/s ) = 9.25 m 2 − 10.6 m/s 2

(c) When the block slides down the incline, fk is in the positive x direction:

ΣFx = fk – mgsin37°= ma and ΣFy = Fn – mgcos37° = 0

Solve for a as in part (a):

a = g (µ k cos 37° − sin 37°) = −1.21 m/s 2

Solve for ∆x to obtain:

2

(

)

Use the same constant-acceleration equation used in part (b) to obtain:

v 2 = v02 + 2a∆x

Set v0 = 0 and solve for v:

v = 2a∆x

Substitute numerical values and evaluate v:

v = 2 − 1.21 m/s 2 (− 9.25 m )

(

)

= 4.73 m/s

40 •• Picture the Problem We can find the stopping distances by applying Newton’s 2nd law to the automobile and then using a constant-acceleration equation. The friction force the road exerts on the tires and the component of the car’s weight along the incline combine to provide the net force that stops the car. The pictorial representation summarizes what we know about the motion of the car. We can use Newton’s 2nd law to determine the acceleration of the car and a constant-acceleration equation to obtain its stopping distance.

290 Chapter 5

(a) Using a constant-acceleration equation, relate the final speed of the car to its initial speed, acceleration, and displacement; solve for its displacement:

v12 = v02 + 2amax ∆xmin or, because v1 = 0, ∆xmin =

− v02 2amax

Draw the free-body diagram for the car going up the incline:

Apply

r

r

∑ F = ma to the car:

Substitute fs,max = µsFn and Fn from (2) in (1) and solve for a: Substitute numerical values in the expression for ∆xmin to obtain: (b) Draw the free-body diagram for the car going down the incline:

ΣFx = −fs,max – mgsin15° = ma and ΣFy = Fn – mgcos15° = 0

amax = − g (µ s cos 15° + sin 15°) = −9.17 m/s 2 − (30 m/s ) = 49.1 m 2 − 9.17 m/s 2 2

∆xmin =

(

)

(1) (2)

Applications of Newton’s Laws 291 Apply

r r F ∑ = ma to the car:

ΣFx = fs,max – mgsin15° = ma and ΣFy = Fn – mgcos15° = 0

Proceed as in (a) to obtain amax:

amax = g (µs cos15° − sin 15°) = 4.09 m/s 2

Again, proceed as in (a) to obtain the displacement of the car:

∆xmin =

(30 m/s ) = 110 m − v02 = 2amax 2 4.09 m/s 2 2

(

)

41 •• Picture the Problem The friction force the road exerts on the tires provides the net force that accelerates the car. The pictorial representation summarizes what we know about the motion of the car. We can use Newton’s 2nd law to determine the acceleration of the car and a constant-acceleration equation to calculate how long it takes it to reach 100 km/h.

(a) Because 40% of the car’s weight is on its two drive wheels and the accelerating friction forces act just on these wheels, the free-body diagram shows just the forces acting on the drive wheels. Apply

r r F ∑ = ma to the car:

Use the definition of fs,max in equation (1) and eliminate Fn between the two equations to obtain: (b) Using a constant-acceleration equation, relate the initial and final

ΣFx = fs,max = ma and ΣFy = Fn – 0.4mg = 0

(1) (2)

(

a = 0.4µs g = 0.4(0.7 ) 9.81 m/s 2 = 2.75 m/s 2

v1 = v0 + a∆t

)

292 Chapter 5 or, because v0 = 0 and ∆t = t1,

velocities of the car to its acceleration and the elapsed time; solve for the time:

t1 =

v1 a

Substitute numerical values and evaluate t1:

t1 =

(100 km/h )(1h/3600 s )(1000 m/km) = 2.75 m/s 2

10.1s

*42 •• Picture the Problem To hold the box in place, the acceleration of the cart and box must be great enough so that the static friction force acting on the box will equal the weight of the box. We can use Newton’s 2nd law to determine the minimum acceleration required. (a) Apply

r

r

∑ F = ma to the box:

ΣFx = Fn = mamin and ΣFy = fs,max – mg = 0

(1) (2)

Substitute µFn for fs,max in equation

µFn − mg = 0 , µ(ma min ) − mg = 0

(2), eliminate Fn between the two equations and solve for and evaluate amin:

and

(b) Solve equation (2) for fs,max, and substitute numerical values and evaluate fs,max:

fs,max = mg

(c) If a is twice that required to hold the box in place, fs will still have its maximum value given by:

fs,max = 19.6 N

amin =

g

µs

=

9.81 m/s 2 = 16.4 m/s 2 0.6

= (2 kg)(9.81 m/s2) = 19.6 N

(d) Because g µs is amin , the box will not fall if a ≥ g µs . 43

••

Picture the Problem Note that the blocks have a common acceleration and that the tension in the string acts on both blocks in accordance with Newton’s third law of motion. Let down the incline be the positive x direction. Draw the freebody diagrams for each block and apply Newton’s second law of motion and the definition of the kinetic friction force to each block to obtain simultaneous

Applications of Newton’s Laws 293 equations in ax and T. Draw the free-body diagram for the block whose mass is m1:

r

r

∑ F = ma to the upper block:

ΣFx = −fk,1 + T1 + m1gsinθ = m1ax and ΣFy = Fn,1 – m1gcosθ = 0

(2)

The relationship between fk,1 and Fn,1 is:

fk,1 = µk,1Fn,1

(3)

Eliminate fk,1 and Fn,1 between (1), (2), and (3) to obtain:

−µk,1m1gcosθ + T1 + m1gsinθ = m1ax

(4)

ΣFx = − fk,2 – T2 + m2gsinθ = m2ax and ΣFy = Fn,2 – m2gcosθ = 0

(5)

The relationship between fk,2 and Fn,2 is:

fk,2 = µk,2Fn,2

(7)

Eliminate fk,2 and Fn,2 between (5), (6), and (7) to obtain:

−µk,2m2gcosθ – T2 + m2gsinθ = m2ax (8)

Apply

(1)

Draw the free-body diagram for the block whose mass is m2:

Apply

r

r

∑ F = ma to the block:

(6)

294 Chapter 5 Noting that T2 = T1 = T, add equations (4) and (8) to eliminate T and solve for ax:

µ m + µ k,2 m2 ⎡ ⎤ a x = ⎢sin θ − k,1 1 cosθ ⎥ g m1 + m2 ⎣ ⎦

Substitute numerical values and evaluate ax to obtain:

a x = − 0.965 m/s 2 where the minus sign tells us that the acceleration is directed up the incline.

m1m2 (µ k,2 − µ k,1 )g cosθ m1 + m2

(b) Eliminate ax between equations (4) and (8) and solve for T = T1 = T2 to obtain:

T=

Substitute numerical values and evaluate T:

T = 0.184 N

*44 •• Picture the Problem The free-body diagram shows the forces acting on the two blocks as they slide down the incline. Down the incline has been chosen as the positive x direction. T is the force transmitted by the stick; it can be either tensile (T > 0) or compressive (T < 0). By applying Newton’s 2nd law to these blocks, we can obtain equations in T and a from which we can eliminate either by solving them simultaneously. Once we have expressed T, the role of the stick will become apparent. (a) Apply

r

r

∑ F = ma to block 1:

∑F

x

= T1 + m1 g sin θ − f k,1 = m1a

and

∑F

= Fn,1 − m1 g cos θ = 0

∑F

= m2 g sin θ − T2 − f k,2 = m2 a

y

Apply

r

r

∑ F = ma to block 2:

x

and

∑F

y

Letting T1 = T2 = T, use the definition of the kinetic friction force to eliminate fk,1 and Fn,1 between the equations for block 1 and fk,2 and Fn,1 between the equations for block 2 to obtain:

= Fn,2 − m2 g cos θ = 0

m1 a = m1 g sin θ + T − µ1 m1 g cos θ

(1)

and

m 2 a = m2 g sin θ − T − µ 2 m 2 g cos θ

(2)

Applications of Newton’s Laws 295 Add equations (1) and (2) to eliminate T and solve for a:

(b) Rewrite equations (1) and (2) by dividing both sides of (1) by m1 and both sides of (2) by m2 to obtain.

⎛ ⎞ µ m + µ 2 m2 a = g ⎜⎜ sin θ − 1 1 cosθ ⎟⎟ m1 + m2 ⎝ ⎠ a = g sin θ +

(3)

T − µ 2 g cos θ m2

(4)

and

a = g sin θ − Subtracting (4) from (3) and rearranging yields:

T − µ1 g cosθ m1

⎛ mm T = ⎜⎜ 1 2 ⎝ m1 − m2

⎞ ⎟⎟(µ1 − µ 2 )g cosθ ⎠

If µ1 = µ 2 , T = 0 and the blocks move down the incline with the same

acceleration of g (sinθ − µ cosθ ). Inserting a stick between them can' t change this; therefore, the stick must exert no force on either block. 45 •• Picture the Problem The pictorial representation shows the orientation of the two blocks on the inclined surface. Draw the free-body diagrams for each block and apply Newton’s 2nd law of motion and the definition of the static friction force to each block to obtain simultaneous equations in θc and T. (a) Draw the free-body diagram for the lower block:

Apply

r

r

∑ F = ma to the block:

The relationship between fs,1 and Fn,1 is:

ΣFx = m1gsinθc – fs,1 − T = 0 and ΣFy = Fn,1 – m1gcosθc = 0

(2)

fs,1 = µs,1Fn,1

(3)

(1)

296 Chapter 5

m1gsinθc − µs,1m1gcosθc − T = 0

(4)

ΣFx =T + m2gsinθc – fs,2 = 0 and ΣFy = Fn,2 – m2gcosθc = 0

(5)

The relationship between fs,2 and Fn,2 is:

fs,2 = µs,2Fn,2

(7)

Eliminate fs,2 and Fn,2 between (5), (6), and (7) to obtain:

T + m2gsinθc – µs,2m2gcosθc = 0

(8)

Eliminate fs,1 and Fn,1 between (1), (2), and (3) to obtain: Draw the free-body diagram for the upper block:

Apply

r r F ∑ = ma to the block:

⎡ µs,1m1 + µs,2 m2 ⎤ ⎥ ⎣ m1 + m2 ⎦

Add equations (4) and (8) to eliminate T and solve for θc:

(6)

θ c = tan −1 ⎢

⎡ (0.4)(0.2 kg ) + (0.6)(0.1 kg )⎤ = tan −1 ⎢ ⎥ 0.1 kg + 0.2 kg ⎣ ⎦ = 25.0° T = m1 g (sin θ C − µs,1 cos θ C )

(b) Because θc is greater than the angle of repose (tan−1(µs,1) = tan−1(0.4) = 21.8°) for the lower block, it would slide if T = 0. Solve equation (4) for T: Substitute numerical values and evaluate T:

(

)

T = (0.2 kg ) 9.81 m/s 2 [sin25° − (0.4 )cos25°] = 0.118 N

Applications of Newton’s Laws 297 46 •• Picture the Problem The pictorial representation shows the orientation of the two blocks with a common acceleration on the inclined surface. Draw the free-body diagrams for each block and apply Newton’s 2nd law and the definition of the kinetic friction force to each block to obtain simultaneous equations in a and T. (a) Draw the free-body diagram for the lower block:

r r F ∑ = ma to the lower

ΣFx = m1gsin20° − fk,1 – T = m1a and ΣFy = Fn,1 − m1gcos20° = 0

(2)

Express the relationship between fk,1 and Fn,1:

fk,1 = µk,1Fn,1

(3)

Eliminate fk,1 and Fn,1 between (1), (2), and (3) to obtain:

m1 g sin 20° − µ k,1m1 g cos 20°

Apply block:

− T = m1a

(1)

(4)

Draw the free-body diagram for the upper block:

Apply

r

r

∑ F = ma to the upper

block:

Express the relationship between fk,2 and Fn,2 :

ΣFx = T + m2gsin20° − fk,2 = m2a and ΣFy = Fn,2 – m2gcos20° = 0

(6)

fk,2 = µk,2Fn,2

(7)

(5)

298 Chapter 5 Eliminate fk,2 and Fn,2 between (5), (2), and (7) to obtain:

T + m2 g sin 20° − µ k,2 m2 g cos 20°

Add equations (4) and (8) to eliminate T and solve for a:

⎛ ⎞ µ m + µ 2 m2 a = g ⎜⎜ sin 20° − 1 1 cos 20° ⎟⎟ m1 + m2 ⎝ ⎠

Substitute the given values and evaluate a:

a = 0.944 m/s 2

(b) Substitute for a in either equation (4) or equation (8) to obtain:

T = − 0.426 N ; i.e., the rod is under

= m2 a

(8)

compression.

*47 •• Picture the Problem The vertical r component of F reduces the normal force; hence, the static friction force between the surface and the block. The horizontal component is responsible for any tendency to move and equals the static friction force until it exceeds its maximum value. We can apply Newton’s 2nd law to the box, under equilibrium conditions, to relate F to θ. (a) The static-frictional force opposes the motion of the object, and the maximum value of the static-frictional force is proportional to the normal force FN. The normal force is equal to the weight minus the vertical component FV of the force F. Keeping the magnitude F constant while increasing θ from zero results in a decrease in FV and thus a corresponding decrease in the maximum static-frictional force fmax. The object will begin to move if the horizontal component FH of the force F exceeds fmax. An increase in θ results in a decrease in FH. As θ increases from 0, the decrease in FN is larger than the decrease in FH, so the object is more and more likely to slip. However, as θ approaches 90°, FH approaches zero and no movement will be initiated. If F is large enough and if θ increases from 0, then at some value of θ the block will start to move. (b) Apply

r

r

∑ F = ma to the block:

Assuming that fs = fs,max, eliminate fs and Fn between equations (1) and (2) and solve for F:

ΣFx =Fcosθ – fs = 0 and ΣFy = Fn + Fsinθ – mg = 0

F=

µ s mg cosθ + µ s sin θ

(1) (2)

Applications of Newton’s Laws 299 Use this function with mg = 240 N to generate the table shown below:

θ F

(deg) (N)

0 240

10 220

20 210

30 206

40 208

50 218

60 235

The following graph of F(θ) was plotted using a spreadsheet program.

240 235

F (N)

230 225 220 215 210 205 0

10

20

30

40

50

60

theta (degrees)

From the graph, we can see that the minimum value for F occurs when θ ≈ 32°. Remarks: An alternative to manually plotting F as a function of θ or using a spreadsheet program is to use a graphing calculator to enter and graph the function. 48 ••• Picture the Problem The free-body diagram shows the forces acting on the block. We can apply Newton’s 2nd law, under equilibrium conditions, to relate F to θ and then set its derivative with respect to θ equal to zero to find the value of θ that minimizes F. (a) Apply

r r F ∑ = ma to the block:

ΣFx =Fcosθ – fs = 0 and ΣFy = Fn + Fsinθ – mg = 0

(1) (2)

300 Chapter 5 Assuming that fs = fs,max, eliminate fs and Fn between equations (1) and (2) and solve for F:

F=

µ s mg cosθ + µ s sin θ

(3)

To find θmin, differentiate F with respect to θ and set the derivative equal to zero for extrema of the function:

d d ( (cosθ + µs sin θ ) cos θ + µs sin θ ) (µs mg ) µs mg dF d d θ θ = − dθ (cosθ + µs sin θ )2 (cosθ + µs sin θ )2 µ mg (− sin θ + µs cosθ ) = s = 0 for extrema (cosθ + µs sin θ )2 Solve for θmin to obtain:

θ min = tan −1 µ s

(b) Use the reference triangle shown below to substitute for cosθ and sinθ in equation (3):

Fmin =

µ s mg 1 1+ µ

=

2 s

+ µs

µs 1 + µ s2

µ s mg 1 + µ s2 1 + µ s2

=

µs 1 + µ s2

mg

(c)

The coefficient of kinetic friction is less than the coefficient of static friction. An analysis identical to the one above shows that the minimum force one should apply to keep the block moving should be applied at an angle given by

θ min = tan −1 µ k . Therefore, once the block is moving the coefficient of friction will decrease, so the angle can be decreased. 49 •• r Picture the Problem The vertical component of F increases the normal force and the static friction force between the surface and the block. The horizontal component is responsible for any tendency to move and equals the static friction force until it exceeds its maximum value. We can apply Newton’s 2nd law to the box, under equilibrium conditions, to relate F to θ.

Applications of Newton’s Laws 301 (a) As θ increases from zero, F increases the normal force exerted by the surface and the static friction force. As the horizontal component of F decreases with increasing θ, one would expect F to continue to increase.

(b) Apply

r

r

∑ F = ma to the block:

ΣFx =Fcosθ – fs = 0 and ΣFy = Fn – Fsinθ – mg = 0

Assuming that fs = fs,max, eliminate fs and Fn between equations (1) and (2) and solve for F:

F=

(1) (2)

µ s mg cosθ − µ s sin θ

(3)

Use this function with mg = 240 N to generate the table shown below.

θ F

(deg) (N)

0 240

10 273

20 327

30 424

40 631

50 1310

60 very large

The graph of F as a function of θ, plotted using a spreadsheet program, confirms our prediction that F continues to increase with θ.

1400 1200

F (N)

1000 800 600 400 200 0 0

10

20

30

theta (degrees)

(a) From the graph we see that:

θ min = 0°

40

50

302 Chapter 5 (b) Evaluate equation (3) for θ = 0° to obtain:

F=

µs mg = µs mg cos 0° − µs sin 0°

You should keep the angle at 0°.

(c)

Remarks: An alternative to the use of a spreadsheet program is to use a graphing calculator to enter and graph the function. 50 •• Picture the Problem The forces acting on each of these masses are shown in the freebody diagrams below. m1 represents the mass of the 20-kg mass and m2 that of the 100-kg mass. As described by Newton’s 3rd law, the normal reaction force Fn,1 and the friction force fk,1 (= fk,2) act on both masses but in opposite directions. Newton’s 2nd law and the definition of kinetic friction forces can be used to determine the various forces and the acceleration called for in this problem. (a) Draw a free-body diagram showing the forces acting on the 20-kg mass:

Apply

r

r

∑ F = ma to this mass:

Solve equation (1) for fk,1:

ΣFx = fk,1 = m1a1 and ΣFy = Fn,1 – m1g = 0

(1) (2)

fk,1 = m1a1 = (20 kg)(4 m/s2) = 80.0 N

(b) Draw a free-body diagram showing the forces acting on the 100-kg mass:

Apply

∑F

x

= max to the 100-kg

object and evaluate Fnet:

Fnet = m2 a2

(

)

= (100 kg ) 6 m/s 2 = 600 N

Applications of Newton’s Laws 303 Express F in terms of Fnet and fk,2:

F = Fnet + fk,2 = 600 N + 80 N = 680 N

(c) When the 20-kg mass falls off, the 680-N force acts just on the 100-kg mass and its acceleration is given by Newton’s 2nd law:

a=

Fnet 680 N = = 6.80 m/s 2 m 100 kg

51 •• Picture the Problem The forces acting on each of these blocks are shown in the freebody diagrams to the right. m1 represents the mass of the 60-kg block and m2 that of the 100-kg block. As described by Newton’s 3rd law, the normal reaction force Fn,1 and the friction force fk,1 (= fk,2) act on both objects but in opposite directions. Newton’s 2nd law and the definition of kinetic friction forces can be used to determine the coefficient of kinetic friction and acceleration of the 100-kg block. (a) Apply

r

r

∑ F = ma to the 60-kg

block:

Apply

∑F

x

= max to the 100-kg

ΣFx = F − fk,1 = m1a1 and ΣFy = Fn,1 – m1g = 0

(2)

fk,2 = m2a2

(3)

(1)

block: Using equation (2), express the relationship between the kinetic r r friction forces f k ,1 and f k , 2 :

fk,1 = fk,2 = fk = µ kFn,1 = µ km1g (4)

Substitute equation (4) into equation (1) and solve for µ k:

µk =

F − m1a1 m1 g

Substitute numerical values and evaluate µ k:

µk =

320 N − (60 kg ) 3 m/s 2 = 0.238 (60 kg ) 9.81m/s2

(b) Substitute equation (4) into equation (3) and solve for a2:

a2 =

(

µ k m1 g m2

(

)

)

304 Chapter 5 Substitute numerical values and evaluate a2:

a2 =

(0.238)(60 kg )(9.81m/s 2 ) 100 kg

= 1.40 m/s 2 *52 •• Picture the Problem The accelerations of the truck can be found by applying Newton’s 2nd law of motion. The free-body diagram for the truck climbing the incline with maximum acceleration is shown to the right.

(a) Apply

r

r

∑ F = ma to the truck

when it is climbing the incline:

Solve equation (2) for Fn and use the definition of fs,max to obtain: Substitute equation (3) into equation (1) and solve for a: Substitute numerical values and evaluate a:

(b) When the truck is descending the incline with maximum acceleration, the static friction force points down the incline; i.e., its direction is reversed on the FBD. Apply Fx = max to the truck under

ΣFx = fs,max – mgsin12° = ma and ΣFy = Fn – mgcos12° = 0

(1)

fs,max = µsmgcos12°

(3)

(2)

a = g (µs cos12° − sin 12°) a = (9.81 m/s 2 )[(0.85) cos12° − sin 12°] = 6.12 m/s 2 – fs,max – mgsin12° = ma

(4)

∑

these conditions: Substitute equation (3) into equation (4) and solve for a: Substitute numerical values and evaluate a:

a = − g (µs cos12° + sin 12°)

(

a = − 9.81 m/s 2

)[(0.85)cos12° + sin 12°]

= − 10.2 m/s 2

Applications of Newton’s Laws 305 53 •• Picture the Problem The forces acting on each of the blocks are shown in the freebody diagrams to the right. m1 represents the mass of the 2-kg block and m2 that of the 4-kg block. As described by Newton’s 3rd law, the normal reaction force Fn,1 and the friction force fs,1 (= fs,2) act on both objects but in opposite directions. Newton’s 2nd law and the definition of the maximum static friction force can be used to determine the maximum force acting on the 4-kg block for which the 2-kg block does not slide. (a) Apply

r

r

∑ F = ma to the 2-kg

block:

Apply

r

r

∑ F = ma to the 4-kg block:

Using equation (2), express the relationship between the static

r

ΣFx = fs,1,max = m1amax and ΣFy = Fn,1 – m1g = 0

(1)

ΣFx = F – fs,2,max = m2amax and ΣFy = Fn,2 – Fn,1 - m2g = 0

(3) (4)

fs,1,max = fs,2,max = µs m1g

(5)

(2)

r

friction forces f s ,1,max and f s , 2,max : Substitute (5) in (1) and solve for amax:

amax = µsg = (0.3)g = 2.94 m/s2

Solve equation (3) for F = Fmax:

Fmax = m2 amax + µs m1 g

Substitute numerical values and evaluate Fmax:

(

)

Fmax = (4 kg ) 2.94 m/s 2 + (0.3)(2 kg )

(

× 9.81 m/s

2

)

= 17.7 N (b) Use Newton’s 2nd law to express the acceleration of the blocks moving as a unit:

a=

Substitute numerical values and evaluate a:

a=

F m1 + m2

1 2

(17.7 N )

2 kg + 4 kg

= 1.47 m/s 2

306 Chapter 5 Because the friction forces are an action-reaction pair, the friction force acting on each block is given by:

fs = m1a = (2 kg)(1.47 m/s2)

(c) If F = 2Fmax, then m1 slips on m2 and the friction force (now kinetic) is given by:

f = fk = µkm1g

Use

∑F

x

= max to relate the

acceleration of the 2-kg block to the net force acting on it and solve for a 1: Use

∑F

x

= max to relate the

= 2.94 N

fk = µkm1g = m1a1 and a1 = µkg = (0.2)g = 1.96 m/s 2

F − µkm1g = m2a2

acceleration of the 4-kg block to the net force acting on it: Solve for a2:

a2 =

F − µ k m1 g m2

Substitute numerical values and evaluate a2:

a2 =

2(17.7 N ) − (0.2 )(2 kg ) 9.81 m/s 2 4 kg

(

)

= 7.87 m/s 2 54 •• Picture the Problem Let the positive x direction be the direction of motion of these blocks. The forces acting on each of the blocks are shown, for the static friction case, on the free-body diagrams to the right. As described by Newton’s 3rd law, the normal reaction force Fn,1 and the friction force fs,1 (= fs,2) act on both objects but in opposite directions. Newton’s 2nd law and the definition of the maximum static friction force can be used to determine the maximum acceleration of the block whose mass is m1. (a) Apply block:

r

r

∑ F = ma to the 2-kg

ΣFx = fs,1,max = m1amax and

(1)

Applications of Newton’s Laws 307 ΣFy = Fn,1 – m1g = 0

(2)

ΣFx = T – fs,2,max = m2amax and ΣFy = Fn,2 – Fn,1 – m2g = 0

(3) (4)

Using equation (2), express the relationship between the static r r friction forces f s ,1,max and f s , 2,max :

fs,1,max = fs,2,max = µs m1g

(5)

Substitute (5) in (1) and solve for amax:

amax = µsg = (0.6)g = 5.89 m/s 2

Apply

r

r

∑ F = ma to the 4-kg

block:

(b) Use

∑F

x

= ma x to express the

T = (m1 + m2) amax

(6)

m3g – T = m3 amax

(7)

acceleration of the blocks moving as a unit: Apply

∑F

x

= max to the object

whose mass is m3:

µs (m1 + m2 ) (0.6)(10 kg + 5 kg ) = 1 − µs 1 − 0.6

Add equations (6) and (7) to eliminate T and then solve for and evaluate m3:

m3 =

(c) If m3 = 30 kg, then m1 will slide on m2 and the friction force (now kinetic) is given by:

f = fk = µkm1g

Use

∑F

x

= max to relate the

= 22.5 kg

m3g – T = m3a3

(8)

acceleration of the 30-kg block to the net force acting on it:

g (m3 − µ k m1 ) m2 + m3

Noting that a2 = a3 and that the friction force on the body whose mass is m2 is due to kinetic friction, add equations (3) and (8) and solve for and evaluate the common acceleration:

a2 = a3 =

With block 1 sliding on block 2, the

fk = µkm1g = m1a1

=

(9.81m/s )[30 kg − (0.4)(5 kg )] 2

10 kg + 30 kg

= 6.87 m/s 2 (1′)

308 Chapter 5 friction force acting on each is kinetic and equations (1) and (3) become: Solve equation (1′) for and evaluate a 1:

T – fk = T – µkm1g = m2a2

(

a1 = µ k g = (0.4 ) 9.81 m/s 2

(3′)

)

= 3.92 m/s 2 T = m2 a2 + µ k m1 g

Solve equation (3′) for T: Substitute numerical values and evaluate T:

T = (10 kg ) (6.87 m/s 2 ) + (0.4 )(5 kg ) (9.81 m/s 2 ) = 88.3 N 55 • Picture the Problem Let the direction of motion be the positive x direction. The free-body diagrams show the forces acting on both the block (M) and the

r

r

counterweight (m). While T1 ≠ T2 , T1 = T2. By applying Newton’s 2nd law to these blocks, we can obtain equations in T and a from which we can eliminate the tension. Once we know the acceleration of the block, we can use constant-acceleration equations to determine how far it moves in coming to a momentary stop. (a) Apply

r

r

∑ F = ma to the block

on the incline:

∑F

x

and

∑F

= Fn − Mg cosθ = 0

∑F

= mg − T2 = ma

y

Apply

r

= T1 − Mg sin θ − f k = Ma

r

∑ F = ma to the

x

(1)

counterweight: Letting T1 = T2 = T and using the definition of the kinetic friction force, eliminate fk and Fn between the equations for the block on the incline to obtain: Eliminate T from equations (1) and (2) by adding them and solve for a:

T − Mg sin θ − µ k Mg cos θ = Ma

a=

m − M (sin θ + µ k cos θ ) g m+M

(2)

Applications of Newton’s Laws 309 Substitute numerical values and evaluate a: a=

550 kg − (1600 kg ) (sin 10° + 0.15 cos10°) 9.81m/s 2 = 0.163 m/s2 550 kg + 1600 kg

(

(b) Using a constant-acceleration equation, relate the speed of the block at the instant the rope breaks to its acceleration and displacement as it slides to a stop. Solve for its displacement:

)

vf2 = vi2 + 2a∆x or, because vf = 0,

∆x =

− vi2 2a

(3)

The block had been accelerating up the incline for 3 s before the rope broke, so it has an initial speed of :

(0.163 m/s2)(3 s) = 0.489 m/s

From equation (2) we can see that, when the rope breaks (T = 0) and:

a = − g (sin θ + µ k cos θ )

(

)

= − 9.81 m/s 2 [sin 10° + (0.15)cos10°] = −3.15 m/s 2

where the minus sign indicates that the block is being accelerated down the incline, although it is still sliding up the incline. Substitute in equation (3) and evaluate ∆x:

− (0.489 m/s ) = 0.0380 m ∆x = 2 − 3.15 m/s 2

(c) When the block is sliding down the incline, the kinetic friction force will be up the incline. Express the block’s acceleration:

a = − g (sin θ − µ k cos θ )

2

(

(

)

)

= − 9.81 m/s 2 [sin 10° − (0.15) cos10°]

= − 0.254 m/s 2

56 ••• Picture the Problem If the 10-kg block is not to slide on the bracket, the maximum r value for F must be equal to the maximum value of fs and will produce the maximum acceleration of this block and the bracket. We can apply Newton’s 2nd law and the definition of fs,max to first calculate the maximum acceleration and then the maximum value of F. (a) and (b) Apply

r

r

∑ F = ma to the

10-kg block when it is experiencing its maximum acceleration:

ΣFx = fs,max – F = m2a2,max and ΣFy = Fn,2 – m2g = 0

(1) (2)

310 Chapter 5 Express the static friction force acting on the 10-kg block:

fs,max = µsFn,2

(3)

Eliminate fs,max and Fn,2 from equations (1), (2) and (3) to obtain:

µsm2g – F = m2a2,max

(4)

2F – µsm2g = m1a1,max

(5)

Apply

∑F

x

= max to the bracket

to obtain: Because a1,max = a2,max, denote this acceleration by amax. Eliminate F from equations (4) and (5) and solve for amax:

amax =

Substitute numerical values and evaluate amax:

amax =

µs m2 g m1 + 2m2

(0.4)(10 kg )(9.81m/s2 ) 5 kg + 2(10 kg )

= 1.57 m/s 2 Solve equation (4) for F = Fmax: Substitute numerical values and evaluate F:

F = µs m2 g − m2 amax = m2 (µs g − amax )

[

(

)

F = (10 kg ) (0.4) 9.81 m/s 2 − 1.57 m/s 2 = 23.5 N

*57 •• Picture the Problem The free-body diagram shows the forces acting on the block as it is moving up the incline. By applying Newton’s 2nd law, we can obtain expressions for the accelerations of the block up and down the incline. Adding and subtracting these equations, together with the data found in the notebook, will lead to values for gV and µk. Apply

r r F ∑i i = ma to the block when

it is moving up the incline:

∑F

x

and

∑F Using the definition of fk, eliminate Fn between the two equations to obtain:

= − f k − mg V sin θ = maup

y

= Fn − mg V cosθ = 0

aup = − µ k g V cos θ − g V sin θ

(1)

]

Applications of Newton’s Laws 311 When the block is moving down the incline, fk is in the positive x direction, and its acceleration is:

adown = µ k g V cos θ − g V sin θ

(2)

Add equations (1) and (2) to obtain:

aup + adown = −2 g V sin θ

(3)

Solve equation (3) for gV:

gV =

Determine θ from the figure:

aup + adown − 2 sin θ ⎡ 0.73 glapp ⎤ ⎥ = 10.8° ⎣ 3.82 glapp ⎦

θ = tan −1 ⎢

Substitute the data from the notebook in equation (4) to obtain:

1.73 glapp/plipp 2 + 1.42 glapp/plipp 2 gV = = − 8.41glapp/plipp 2 − 2 sin 10.8° Subtract equation (1) from equation (2) to obtain: Solve for µk:

adown − aup = 2µ k g V cos θ

µk =

adown − aup 2 g V cosθ

Substitute numerical values and evaluate µk:

µk =

− 1.42 glapp/plipp 2 − 1.73 glapp/plipp 2 = 0.191 2 − 8.41glapp/plipp 2 cos10.8°

(

)

*58 •• Picture the Problem The free-body diagram shows the block sliding down the incline under the influence of a friction force, its weight, and the normal force exerted on it by the inclined surface. We can find the range of values for m for the two situations described in the problem statement by applying Newton’s 2nd law of motion to, first, the conditions under which the block will not move or slide if pushed, and secondly, if pushed, the block will move up the incline. (a) Assume that the block is sliding down the incline with a constant velocity and with no hanging weight

r r (m = 0) and apply ∑ F = ma to

∑F

x

= − f k + Mg sin θ = 0

and

∑F

y

= Fn − Mg cosθ = 0

(4)

312 Chapter 5 the block: Using fk = µkFn, eliminate Fn between the two equations and solve for the net force acting on the block:

Fnet = − µ k Mg cosθ + Mg sin θ

If the block is moving, this net force must be nonnegative and:

(− µ k cosθ + sin θ )Mg ≥ 0

This condition requires that:

µ k ≤ tan θ = tan 18° = 0.325

Because µk = 0.2, this condition is satisfied and:

mmin = 0

To find the maximum value, note that the maximum possible value for the tension in the rope is mg. For the block to move down the incline, the component of the block’s weight parallel to the incline minus the frictional force must be greater than or equal to the tension in the rope:

Mgsinθ – µkMgcosθ ≥ mg

Solve for mmax:

mmax ≤ M (sin θ − µ k cos θ )

Substitute numerical values and evaluate mmax:

mmax ≤ (100 kg )[sin 18° − (0.2)cos18°]

The range of values for m is:

= 11.9 kg 0 ≤ m ≤ 11.9 kg

(b) If the block is being dragged up the incline, the frictional force will point down the incline, and:

Mg sinθ + µkMg cosθ < mg

Solve for and evaluate mmin:

mmin > M (sinθ + µk cosθ) = (100 kg)[sin18° + (0.2)cos18°] = 49.9 kg

If the block is not to move unless pushed:

Mg sinθ + µs Mg cosθ > mg

Solve for and evaluate mmax:

mmax < M (sinθ + µs cosθ) = (100 kg)[sin18° + (0.4)cos18°] = 68.9 kg

The range of values for m is:

49.9 kg ≤ m ≤ 68.9 kg

Applications of Newton’s Laws 313 59 ••• Picture the Problem The free-body diagram shows the forces acting on the 0.5 kg block when the acceleration is a minimum. Note the choice of coordinate system is consistent with the direction of r F . Apply Newton’s 2nd law to the block and solve the resulting equations for amin and amax.

r r F ∑ = ma to the 0.5-kg

ΣFx = Fnsinθ – fscosθ = ma and ΣFy = Fncosθ + fssinθ – mg = 0

(2)

Under minimum acceleration, fs = fs,max. Express the relationship between fs,max and Fn:

fs,max = µsFn

(3)

Substitute fs,max for fs in equation (2) and solve for Fn:

Fn =

Substitute for Fn in equation (1) and solve for a = amin:

amin = g

Substitute numerical values and evaluate amin:

amin = 9.81 m/s 2

(a) Apply block:

(1)

mg cosθ + µs sin θ

sin θ − µs cos θ cosθ + µs sin θ

(

° − (0.8)cos35° ) sin35 cos35° + (0.8)sin35°

= −0.627 m/s 2 Treat the block and incline as a single object to determine Fmin:

Fmin = mtotamin = (2.5 kg)( –0.627 m/s2)

To find the maximum acceleration, r reverse the direction of f s and apply

ΣFx = Fnsinθ + fscosθ = ma and ΣFy = Fncosθ – fssinθ – mg = 0

r r F ∑ = ma to the block:

= − 1.57 N (4) (5)

sin θ + µs cosθ cosθ − µs sin θ

Proceed as above to obtain:

amax = g

Substitute numerical values and evaluate amax:

amax = 9.81m/s 2

(

= 33.5 m/s 2

° + (0.8)cos35° ) sin35 cos35° − (0.8)sin35°

314 Chapter 5 Treat the block and incline as a single object to determine Fmax:

Fmax = mtotamax = (2.5 kg)(33.5 m/s2)

(b) Repeat (a) with µs = 0.4 to obtain:

Fmin = 5.75 N and Fmax = 37.5 N

= 83.8 N

60 • Picture the Problem The kinetic friction force fk is the product of the coefficient of sliding friction µk and the normal force Fn the surface exerts on the sliding object. By applying Newton’s 2nd law in the vertical direction, we can see that, on a horizontal surface, the normal force is the weight of the sliding object. Note that the acceleration of the block is opposite its direction of motion. (a) Relate the force of kinetic friction to µk and the normal force acting on the sliding wooden object: Substitute v = 10 m/s and evaluate fk :

(b) Substitute v = 20 m/s and evaluate fk:

f k = µ k Fn =

fk =

fk =

0.11

(1 + 2.3 ×10

−4

(

0.11(100 kg ) 9.81 m/s 2

(1 + 2.3 ×10

−4

(10 m/s)

(

(1 + 2.3 ×10

2

(20 m/s)

mg

)=

)

2 2

0.11(100 kg ) 9.81 m/s 2 −4

)

v2

103 N

)

)

2 2

= 90.5 N 61 •• Picture the Problem The pictorial representation shows the block sliding from left to right and coming to rest when it has traveled a distance ∆x. Note that the direction of the motion is opposite that of the block’s acceleration. The acceleration and stopping distance of the blocks can be found from constant-acceleration equations. Let the direction of motion of the sliding blocks be the positive x direction. Because the surface is horizontal, the normal force acting on the sliding block is the block’s weight.

Applications of Newton’s Laws 315 (a) Using a constant-acceleration equation, relate the block’s stopping distance to its initial speed and acceleration; solve for the stopping distance: Apply

∑F

x

= max to the sliding

block, introduce Konecny’s empirical expression, and solve for the block’s acceleration: Evaluate a with m = 10 kg:

v 2 = v02 + 2a∆x or, because v = 0,

− v02 ∆x = 2a

(1)

Fnet,x − f k 0.4 Fn0.91 = =− m m m 0.91 0.4(mg ) =− m

a=

0.91 ( 0.4)[(10 kg ) (9.81 m/s 2 )] a=−

10 kg

= − 2.60 m/s 2 Substitute in equation (1) and evaluate the stopping distance when v0 = 10 m/s:

∆x =

(b) Proceed as in (a) with m = 100 kg to obtain:

a=−

− (10 m/s ) = 19.2 m 2 − 2.60 m/s 2 2

(

)

(0.4)[(100 kg )(9.81m/s2 )] 0.91 100 kg

= − 2.11 m/s 2 Find the stopping distance as in (a):

− (10 m/s ) ∆x = = 23.7 m 2 − 2.11 m/s 2 2

(

)

*62 ••• Picture the Problem The kinetic friction force fk is the product of the coefficient of sliding friction µk and the normal force Fn the surface exerts on the sliding object. By applying Newton’s 2nd law in the vertical direction, we can see that, on a horizontal surface, the normal force is the weight of the sliding object. We can apply Newton’s 2nd law in the horizontal (x) direction to relate the block’s acceleration to the net force acting on it. In the spreadsheet program, we’ll find the acceleration of the block from this net force (which is velocity dependent), calculate the increase in the block’s speed from its acceleration and the elapsed time and add this increase to its speed at end of the previous time interval, determine how far it has moved in this time interval, and add this distance to its previous position to find its current position. We’ll also calculate the position of the block x2, under the assumption that µk = 0.11, using a constant-acceleration equation.

316 Chapter 5

The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content C9 C8+$B$6 D9 D8+F9*$B$6 E9 $B$5−($B$3)*($B$2)*$B$5/ (1+$B$4*D9^2)^2

Algebraic Form

F−

t + ∆t v + a∆t µ k mg

(1 + 2.34 ×10

F9

E10/$B$5

G9 K9

G9+D10*$B$6 0.5*5.922*I10^2

Fnet / m x + v∆t 2 1 2 at

L9

J10-K10

x − x2

A

1 2 3 4 5 6

B g= 9.81 Coeff1= 0.11 Coeff2= 2.30E04 Mass= 10 Applied 70 Force= Time 0.05 step=

7 8 9

C m/s^2

D

E

F

−4

v2

)

2

G

H

I

J

t

x

x2

x−x2

kg N s

Net force

10 11 12 13 14 15

t 0.00 0.05 0.10 0.15 0.20 0.25

v 0.00 0.30 0.59 0.89 1.18 1.48

a

59.22 59.22 59.22 59.22 59.23

205 206

9.75 9.80

61.06 61.40

66.84 6.68 292.37 66.88 6.69 295.44

5.92 5.92 5.92 5.92 5.92

x 0.00 0.01 0.04 0.09 0.15 0.22

0.00 0.05 0.10 0.15 0.20 0.25 9.75 9.80

mu=variable mu=constant 0.00 0.00 0.00 0.01 0.01 0.01 0.04 0.03 0.01 0.09 0.07 0.02 0.15 0.12 0.03 0.22 0.19 0.04 292.37 295.44

281.48 284.37

10.89 11.07

Applications of Newton’s Laws 317 207 208 209 210

9.85 9.90 9.95 10.00

61.73 62.07 62.40 62.74

66.91 66.94 66.97 67.00

6.69 6.69 6.70 6.70

298.53 301.63 304.75 307.89

9.85 9.90 9.95 10.00

298.53 301.63 304.75 307.89

287.28 290.21 293.15 296.10

11.25 11.42 11.61 11.79

The displacement of the block as a function of time, for a constant coefficient of friction (µk = 0.11) is shown as a solid line on the graph and for a variable coefficient of friction, is shown as a dotted line. Because the coefficient of friction decreases with increasing particle speed, the particle travels slightly farther when the coefficient of friction is variable.

300 mu = variable

250

mu = constant x (m)

200 150 100 50 0 0

2

4

6

8

10

t (s)

The velocity of the block, with variable coefficient of kinetic friction, is shown below.

70 60

v (m/s)

50 40 30 20 10 0 0

2

4

6 t (s)

8

10

318 Chapter 5 63 •• Picture the Problem The free-body diagram shows the forces acting on the block as it moves to the right. The kinetic friction force will slow the block and, eventually, bring it to rest. We can relate the coefficient of kinetic friction to the stopping time and distance by applying Newton’s 2nd law and then using constantacceleration equations. (a) Apply

r

r

∑ F = ma to the block

of wood:

∑F

x

= − f k = ma

and

∑F

y

= Fn − mg = 0

Using the definition of fk, eliminate Fn between the two equations to obtain:

a = −µ k g

Use a constant-acceleration equation to relate the acceleration of the block to its displacement and its stopping time:

∆x = v0 ∆t + 12 a(∆t )

(1)

2

v0 + v ∆t 2 = 12 v0 ∆t since v = 0.

(2)

Relate the initial speed of the block, v0, to its displacement and stopping distance:

∆x = vav ∆t =

Use this result to eliminate v0 in equation (2):

∆x = − 12 a(∆t )

Substitute equation (1) in equation (4) and solve for µk:

µk =

2∆x 2 g (∆t )

Substitute for ∆x = 1.37 m and ∆t = 0.97 s to obtain:

µk =

2(1.37 m ) = 0.297 2 9.81 m/s 2 (0.97 s )

(b) Use equation (3) to find v0:

2

v0 =

(

(3)

(4)

)

2∆x 2(1.37 m ) = = 2.82 m/s ∆t 0.97 s

Applications of Newton’s Laws 319 *64 •• Picture the Problem The free-body diagram shows the forces acting on the block as it slides down an incline. We can apply Newton’s 2nd law to these forces to obtain the acceleration of the block and then manipulate this expression algebraically to show that a graph of a/cosθ versus tanθ will be linear with a slope equal to the acceleration due to gravity and an intercept whose absolute value is the coefficient of kinetic friction. (a) Apply

r

r

∑ F = ma to the block

as it slides down the incline:

∑F

x

= mg sin θ − f k = ma

and

∑F

y

= Fn − mg cosθ = 0

Substitute µkFn for fk and eliminate Fn between the two equations to obtain:

a = g (sin θ − µ k cos θ )

Divide both sides of this equation by cosθ to obtain:

a = g tan θ − gµ k cos θ

Note that this equation is of the form y = mx + b:

Thus, if we graph a/cosθ versus tanθ, we should get a straight line with slope g and y-intercept −gµk.

(b) A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell C7 D7 E7

a TAN(C7*PI()/180)

F7

D7/COS(C7*PI()/180)

6 7 8 9

Formula/Content

Algebraic Form

θ

C theta 25 27 29

D a 1.691 2.104 2.406

π ⎞ ⎛ tan⎜θ × ⎟ ⎝ 180 ⎠ a π ⎞ ⎛ cos⎜θ × ⎟ ⎝ 180 ⎠ E tan(theta) 0.466 0.510 0.554

F a/cos(theta) 1.866 2.362 2.751

320 Chapter 5 10 11 12 13 14 15 16 17

31 33 35 37 39 41 43 45

2.888 3.175 3.489 3.781 4.149 4.326 4.718 5.106

0.601 0.649 0.700 0.754 0.810 0.869 0.933 1.000

3.370 3.786 4.259 4.735 5.338 5.732 6.451 7.220

A graph of a/cosθ versus tanθ is shown below. From the curve fit (Excel’s Trendline was used), g = 9.77 m/s2 and µ k =

2.62 m/s 2 = 0.268. 9.77 m/s 2

The percentage error in g from the commonly accepted value of 9.81 m/s2 is

⎛ 9.81 m/s 2 − 9.77 m/s 2 ⎞ ⎟⎟ = 0.408% 100⎜⎜ 9.81 m/s 2 ⎝ ⎠ 8 y = 9.7681x - 2.6154 R2 = 0.9981

7

a /cos(theta)

6 5 4 3 2 1 0 0.4

0.5

0.6

0.7 tan(theta)

0.8

0.9

1.0

Applications of Newton’s Laws 321

Motion Along a Curved Path 65 • Picture the Problem The free-body diagram showing the forces acting on the stone is superimposed on a sketch of the stone rotating in a horizontal circle. The only forces acting on the stone are the tension in the string and the gravitational force. The centripetal force required to maintain the circular motion is a component of the tension. We’ll solve the problem for the general case in which the angle with the horizontal is θ by applying Newton’s 2nd law of motion to the forces acting on the stone.

r

r

∑ F = ma to the stone:

ΣFx = Tcosθ = mac = mv2/r and ΣFy= Tsinθ – mg = 0

(2)

Use the right triangle in the diagram to relate r, L, and θ :

r = Lcosθ

(3)

Eliminate T and r between equations (1), (2) and (3) and solve for v2:

v 2 = gL cot θ cosθ

(4)

Express the velocity of the stone in terms of its period:

v=

Eliminate v between equations (4) and (5) and solve for θ :

θ = sin −1 ⎜⎜

Substitute numerical values and evaluate θ :

θ = sin −1 ⎢

Apply

2πr t1 rev

(1)

(5)

⎛ gt12rev ⎞ ⎟ 2 ⎟ 4 π L ⎠ ⎝

(

)

⎡ 9.81 m/s 2 (1.22 s )2 ⎤ ⎥ = 25.8° 2 ⎣ 4π (0.85 m ) ⎦

and (c) is correct.

322 Chapter 5 66 • Picture the Problem The free-body diagram showing the forces acting on the stone is superimposed on a sketch of the stone rotating in a horizontal circle. The only forces acting on the stone are the tension in the string and the gravitational force. The centripetal force required to maintain the circular motion is a component of the tension. We’ll solve the problem for the general case in which the angle with the horizontal is θ by applying Newton’s 2nd law of motion to the forces acting on the stone.

r r F ∑ = ma to the stone:

ΣFx = Tcosθ = mac = mv2/r and ΣFy= Tsinθ – mg = 0

(2)

Use the right triangle in the diagram to relate r, L, and θ:

r = Lcosθ

(3)

Eliminate T and r between equations (1), (2), and (3) and solve for v:

v = gL cot θ cosθ

Substitute numerical values and evaluate v:

v=

Apply

(1)

(9.81m/s )(0.8 m )cot 20° cos 20° 2

= 4.50 m/s

67 • Picture the Problem The free-body diagram showing the forces acting on the stone is superimposed on a sketch of the stone rotating in a horizontal circle. The only forces acting on the stone are the tension in the string and the gravitational force. The centripetal force required to maintain the circular motion is a component of the tension. We’ll solve the problem for the general case in which the angle with the vertical is θ by applying Newton’s 2nd law of motion to the forces acting on the stone. (a) Apply

r

r

∑ F = ma to the stone:

ΣFx = Tsinθ = mac = mv2/r and

(1)

Applications of Newton’s Laws 323 ΣFy= Tcosθ – mg = 0 Eliminate T between equations (1) and (2) and solve for v:

v = rg tan θ

Substitute numerical values and evaluate v:

v=

(b) Solve equation (2) for T:

T=

Substitute numerical values and evaluate T:

T=

(0.35 m )(9.81m/s 2 )tan30°

= 1.41 m/s mg cosθ

(0.75 kg )(9.81m/s2 ) = cos30°

*68 •• Picture the Problem The sketch shows the forces acting on the pilot when her plane is

r

at the lowest point of its dive. Fn is the force the airplane seat exerts on her. We’ll apply Newton’s 2nd law for circular motion to determine Fn and the radius of the circular path followed by the airplane. (a) Apply

∑F

y

(2)

= ma y to the pilot:

Fn − mg = mac

Solve for and evaluate Fn:

Fn = mg + mac = m(g + ac) = m(g + 8.5g) = 9.5mg = (9.5) (50 kg) (9.81 m/s2) = 4.66 kN

(b) Relate her acceleration to her velocity and the radius of the circular arc and solve for the radius:

ac =

v2 v2 ⇒ r= r ac

Substitute numerical values and evaluate r : 2 ( [ 345 km/h )(1h/3600 s )(1000 m/km )] r=

(

8.5 9.81 m/s 2

)

= 110 m

8.50 N

324 Chapter 5 69 •• Picture the Problem The diagram shows the forces acting on the pilot when her plane is at the lowest point of its dive.

r Fn is the force the airplane seat exerts on

her. We’ll use the definitions of centripetal acceleration and centripetal force and apply Newton’s 2nd law to calculate these quantities and the normal force acting on her. (a) Her acceleration is centripetal and given by: Substitute numerical values and evaluate ac:

ac =

ac

v2 , upward r

[(180 km/h )(1h/3600 s)(10 =

3

/km

300 m

= 8.33 m/s , upward 2

(b) The net force acting on her at the bottom of the circle is the force responsible for her centripetal acceleration: (c) Apply

∑F

y

= ma y to the pilot:

(

Fnet = mac = (65 kg ) 8.33 m/s 2

)

= 541 N, upward

Fn – mg = mac

Solve for Fn:

Fn = mg + mac = m(g + ac)

Substitute numerical values and evaluate Fn:

Fn = (65 kg)(9.81 m/s2 + 8.33 m/s2)

70 •• Picture the Problem The free-body diagrams for the two objects are shown to the right. The hole in the table changes the direction the tension in the string (which provides the centripetal force required to keep the object moving in a circular path) acts. The application of Newton’s 2nd law and the definition of centripetal force will lead us to an expression for r as a function of m1, m2, and the time T for one revolution.

= 1.18 kN, upward

)]

2

Applications of Newton’s Laws 325 Apply

∑F

x

= max to both objects

and use the definition of centripetal acceleration to obtain: Because F1 = F2 we can eliminate both of them between these equations to obtain: Express the speed v of the object in terms of the distance it travels each revolution and the time T for one revolution: Substitute to obtain:

m2g – F2 = 0 and F1 = m1ac = m1v2/r

m2 g − m1

v=

v2 =0 r

2πr T

m2 g − m1

4π 2 r 2 =0 rT 2

or

4π 2 r m2 g − m1 2 = 0 T Solve for r:

r=

m2 gT 2 4π 2 m1

*71 •• Picture the Problem The free-body diagrams show the forces acting on each block. We can use Newton’s 2nd law to relate these forces to each other and to the masses and accelerations of the blocks.

Apply

∑F

x

= max to the block

whose mass is m1: Apply

∑F

x

= max to the block

whose mass is m2: Relate the speeds of each block to their common period and their distance from the center of the

v12 T1 − T2 = m1 L1 T2 = m2

v1 =

v22 L1 + L2

2πL1 2π (L1 + L2 ) and v2 = T T

326 Chapter 5 circle: 2

Solve the first force equation for T2, substitute for v2, and simplify to obtain:

⎛ 2π ⎞ T2 = [m2 (L1 + L2 )] ⎜ ⎟ ⎝ T ⎠

Substitute for T2 and v1 in the first force equation to obtain:

⎛ 2π ⎞ T1 = [m2 (L1 + L2 ) + m1 L1 ] ⎜ ⎟ ⎝ T ⎠

2

*72 •• Picture the Problem The path of the particle and its position at 1-s intervals are shown. The displacement vectors are also shown. The velocity vectors for the average velocities in the first and second r r intervals are along r01 and r12 , respectively, and are shown in the lower diagram. r ∆ v points toward the center of the circle. Use the diagram to the right to find ∆r:

∆r = 2rsin22.5°= 2(4 cm) sin22.5° = 3.06 cm Find the average velocity of the particle along the chords:

vav = ∆r/∆t = (3.06 cm)/(1 s) = 3.06 cm/s

Using the lower diagram and the fact that the angle between r r v1 and v 2 is 45°, express ∆v in

∆v = 2v1sin22.5°

terms of v1 (= v2): Evaluate ∆v using vav as v1:

∆v = 2(3.06 cm/s)sin22.5° = 2.34 cm/s

Now we can determine a = ∆v/∆t:

a=

2.34 cm/s = 2.34 cm/s 2 1s

Applications of Newton’s Laws 327 2πr 2π(4 cm ) = = 3.14 cm/s T 8s

Find the speed v (= v1 = v2 …) of the particle along its circular path:

v=

Calculate the radial acceleration of the particle:

ac =

Compare ac and a by taking their ratio:

ac 2.46 cm/s 2 = = 1.05 a 2.34 cm/s 2

v 2 (3.14 cm/s) = = 2.46 cm/s 2 r 4 cm 2

or

ac = 1.05a 73 •• Picture the Problem The diagram to the right has the free-body diagram for the child superimposed on a pictorial representation of her motion. The force her r father exerts is F and the angle it makes with respect to the direction we’ve chosen as the positive y direction is θ. We can infer her speed from the given information concerning the radius of her path and the period of her motion. Applying Newton’s 2nd law as it describes circular motion will allow us to find both the direction and r magnitude of F . Apply

r

r

∑ F = ma to the child:

ΣFx = Fsinθ = mv2/r and ΣFy = Fcosθ − mg = 0

⎡ v2 ⎤ ⎥ ⎣ rg ⎦

Eliminate F between these equations and solve for θ :

θ = tan −1 ⎢

Express v in terms of the radius and period of the child’s motion:

v=

Substitute for v in the expression for θ to obtain:

θ = tan −1 ⎢

2πr T ⎡ 4π 2 r ⎤ 2 ⎥ ⎣ gT ⎦

328 Chapter 5 4π 2 (0.75 m ) ⎤ = 53.3° 2⎥ 2 ⎣ 9.81 m/s (1.5 s ) ⎦ ⎡

Substitute numerical values and evaluate θ :

θ = tan −1 ⎢

Solve the y equation for F:

F=

Substitute numerical values and evaluate F:

F=

(

)

mg cos θ

(25 kg )(9.81m/s2 ) = cos53.3°

410 N

74 •• Picture the Problem The diagram to the right has the free-body diagram for the bob of the conical pendulum superimposed on a pictorial representation of its motion. The r tension in the string is F and the angle it makes with respect to the direction we’ve chosen as the positive x direction isθ. We can findθ from the y equation and the information provided about the tension. Then, by using the definition of the speed of the bob in its orbit and applying Newton’s 2nd law as it describes circular motion, we can find the period T of the motion. Apply

r

r

∑ F = ma to the pendulum

bob:

ΣFx = Fcosθ = mv2/r and ΣFy = Fsinθ − mg = 0

⎛ mg ⎞ −1 ⎛ mg ⎞ ⎟⎟ = 9.59° ⎟ = sin ⎜⎜ ⎝ F ⎠ ⎝ 6mg ⎠

Using the given information that F = 6mg, solve the y equation for θ:

θ = sin −1 ⎜

With F = 6mg, solve the x equation for v:

v = 6rg cosθ

Relate the period T of the motion to the speed of the bob and the radius of the circle in which it moves:

T=

From the diagram, one can see that:

r =Lcosθ

2π r = v

2π r 6rg cosθ

Applications of Newton’s Laws 329 Substitute for r in the expression for the period to obtain:

T = 2π

L 6g

Substitute numerical values and evaluate T:

T = 2π

0.5 m = 0.579 s 6 9.81 m/s 2

(

)

75 •• Picture the Problem The static friction force fs is responsible for keeping the coin from sliding on the turntable. Using Newton’s 2nd law of motion, the definition of the period of the coin’s motion, and the definition of the maximum static friction force, we can find the magnitude of the friction force and the value of the coefficient of static friction for the two surfaces. (a) Apply

r r F ∑ = ma to the coin:

∑ Fx = f s = m

v2 r

and

∑F

y

= Fn − mg = 0

2πr T

If T is the period of the coin’s motion, its speed is given by:

v=

Substitute for v in the force equation and simplify to obtain:

fs =

Substitute numerical values and evaluate fs:

4π 2 ((0.1kg )(0.1 m ) fs = = 0.395 N (1s )2

(b) Determine Fn from the y equation: If the coin is about to slide at r = 16 cm, fs = fs,max. Solve for µs in terms of fs,max and Fn:

4π 2 mr T2

Fn = mg

µs =

f s ,max Fn

4π 2 mr 2 4π 2 r = T = mg gT 2

330 Chapter 5 Substitute numerical values and evaluate µs:

4π 2 (0.16 m ) µs = = 0.644 (9.81m/s2 )(1s)2

76 •• Picture the Problem The forces acting on the tetherball are shown superimposed on a pictorial representation of the motion. The r horizontal component of T is the centripetal force. Applying Newton’s 2nd law of motion and solving the resulting equations will yield both the tension in the cord and the speed of the ball. (a) Apply

r

r

∑ F = ma to the tetherball:

∑ Fx = T sin 20° = m

v2 r

and

∑F

y

Solve the y equation for T:

T=

Substitute numerical values and evaluate T:

T=

= T cos 20° − mg = 0

mg cos 20°

(0.25 kg )(9.81m/s2 ) = cos 20°

(b) Eliminate T between the force equations and solve for v:

v = rg tan 20°

Note from the diagram that:

r = Lsin20°

Substitute for r in the expression for v to obtain:

v = gL sin 20° tan 20°

Substitute numerical values and evaluate v:

v =

2.61 N

(9.81m/s )(1.2 m )sin 20° tan 20°

= 1.21 m/s

2

Applications of Newton’s Laws 331 *77 •• Picture the Problem The diagram includes a pictorial representation of the earth in its orbit about the sun and a force diagram showing the force on an object at the equator that is due to the earth’s

r

rotation, FR , and the force on the object due to the orbital motion of the earth about r the sun, Fo . Because these are centripetal forces, we can calculate the accelerations they require from the speeds and radii associated with the two circular motions.

aR =

vR2 R

vR =

2πR TR

Substitute for vR in the expression for aR to obtain:

aR =

4π 2 R TR2

Substitute numerical values and evaluate aR:

aR =

Express the radial acceleration due to the rotation of the earth: Express the speed of the object on the equator in terms of the radius of the earth R and the period of the earth’s rotation TR:

4π 2 (6370 km )(1000 m/km )

⎡ ⎛ 3600 s ⎞⎤ ⎟⎟⎥ ⎢(24 h )⎜⎜ 1 h ⎝ ⎠⎦ ⎣ −2 2 = 3.37 × 10 m/s = 3.44 × 10 −3 g

Express the radial acceleration due to the orbital motion of the earth: Express the speed of the object on the equator in terms of the earth-sun distance r and the period of the earth’s motion about the sun To:

ao =

vo2 r

vo =

2π r To

2

332 Chapter 5 Substitute for vo in the expression for ao to obtain:

4π 2 r ao = 2 To

Substitute numerical values and evaluate ac:

ao =

(

4π 2 1.5 × 1011 m

)

⎡ ⎛ 24 h ⎞ ⎛ 3600 s ⎞⎤ ⎟⎟ ⎜⎜ ⎟⎟⎥ ⎢(365 d )⎜⎜ 1 d 1 h ⎝ ⎠⎝ ⎠⎦ ⎣

2

= 5.95 × 10−3 m/s 2 = 6.07 × 10−4 g 78 • Picture the Problem The most significant force acting on the earth is the gravitational force exerted by the sun. More distant or less massive objects exert forces on the earth as well, but we can calculate the net force by considering the radial acceleration of the earth in its orbit. Similarly, we can calculate the net force acting on the moon by considering its radial acceleration in its orbit about the earth. (a) Apply

∑F

r

= mar to the earth:

Fon earth = m

Express the orbital speed of the earth in terms of the time it takes to make one trip around the sun (i.e., its period) and its average distance from the sun:

v=

Substitute for v to obtain:

v2 r

2π r T

Fon earth =

4π 2 mr T2

Substitute numerical values and evaluate Fon earth:

Fon earth =

(

)(

4π 2 5.98 × 10 24 kg 1.496 × 1011 m 24 h 3600 s ⎞ ⎛ × ⎜ 365.24 d × ⎟ d h ⎠ ⎝

2

)=

3.55 × 10 22 N

)=

2.00 × 10 20 N

(b) Proceed as in (a) to obtain:

Fon moon =

(

)(

4π 2 7.35 × 10 22 kg 3.844 × 108 m 24 h 3600 s ⎞ ⎛ × ⎜ 27.32 d × ⎟ d h ⎠ ⎝

2

Applications of Newton’s Laws 333 79 •• Picture the Problem The semicircular wire of radius 10 cm limits the motion of the bead in the same manner as would a 10-cm string attached to the bead and fixed at the center of the semicircle. The horizontal component of the normal force the wire exerts on the bead is the centripetal force. The application of Newton’s 2nd law, the definition of the speed of the bead in its orbit, and the relationship of the frequency of a circular motion to its period will yield the angle at which the bead will remain stationary relative to the rotating wire. Apply

r

r

∑ F = ma to the bead:

∑ Fx = Fn sin θ = m

v2 r

and

∑F

y

= Fn cosθ − mg = 0 v2 rg

Eliminate Fn from the force equations to obtain:

tan θ =

The frequency of the motion is the reciprocal of its period T. Express the speed of the bead as a function of the radius of its path and its period:

v=

Using the diagram, relate r to L and θ:

r = L sin θ

2πr T

⎡ gT 2 ⎤ 2 ⎥ ⎣ 4π L ⎦

Substitute for r and v in the expression for tanθ and solve for θ :

θ = cos −1 ⎢

Substitute numerical values and evaluate θ :

θ = cos −1 ⎢

(

)

⎡ 9.81 m/s 2 (0.5 s )2 ⎤ ⎥ = 51.6° 2 ( ) 4π 0.1 m ⎣ ⎦

334 Chapter 5 80 ••• Picture the Problem Note that the acceleration of the bead has two components, the radial component r perpendicular to v , and a tangential component due to friction that is opposite r to v . The application of Newton’s 2nd law will result in a differential equation with separable variables. Its integration will lead to an expression for the speed of the bead as a function of time. Apply

r

r

∑ F = ma to the bead in the

radial and tangential directions:

∑ Fr = Fn = m

v2 r

and

∑F = − f t

k

= mat = m

Express fk in terms of µk and Fn:

fk = µkFn

Substitute for Fn and fk in the tangential equation to obtain the differential equation:

µ dv = − k v2 dt r

Separate the variables to obtain:

µ dv = − k dt 2 v r

Express the integral of this equation with the limits of integration being from v0 to v on the left-hand side and from 0 to t on the right-hand side:

1 µ ∫v v' 2 dv' = − rk ∫0 dt' 0

Evaluate these integrals to obtain:

⎛1 1 ⎞ ⎛µ ⎞ − ⎜⎜ − ⎟⎟ = −⎜ k ⎟t ⎝ r ⎠ ⎝ v v0 ⎠

Solve this equation for v:

v

t

⎛ ⎞ ⎜ ⎟ 1 ⎜ ⎟ v = v0 ⎜ ⎛ µ k v0 ⎞ ⎟ ⎟t ⎟ ⎜1+ ⎜ ⎝ ⎝ r ⎠ ⎠

dv dt

Applications of Newton’s Laws 335 81 ••• Picture the Problem Note that the acceleration of the bead has two components−the radial component r perpendicular to v , and a tangential component due to friction that is opposite r to v . The application of Newton’s 2nd law will result in a differential equation with separable variables. Its integration will lead to an expression for the speed of the bead as a function of time. (a) In Problem 81 it was shown that:

⎛ ⎞ ⎜ ⎟ 1 ⎜ ⎟ v = v0 ⎜ ⎛ µ k v0 ⎞ ⎟ ⎟t ⎟ ⎜1+ ⎜ ⎝ ⎝ r ⎠ ⎠

Express the centripetal acceleration of the bead:

⎛ ⎞ ⎟ 2 2 ⎜ v v0 ⎜ 1 ⎟ ac = = r r ⎜ ⎛ µ k v0 ⎞ ⎟ ⎟t ⎟ ⎜1+ ⎜ ⎝ ⎝ r ⎠ ⎠

(b) Apply Newton’s 2nd law to the bead:

v2 ∑ Fr = Fn = m r

2

and

∑F = − f t

k

= mat = m

dv dt

v2 = − µ k ac r

Eliminate Fn and fk to rewrite the radial force equation and solve for at:

at = − µ k

(c) Express the resultant acceleration in terms of its radial and tangential components:

a = at2 + ac2 = = ac 1 + µ k2

(− µ k ac )2 + ac2

336 Chapter 5

Concepts of Centripetal Force *82 • Picture the Problem The diagram depicts a seat at its highest and lowest points. Let ″t″ denote the top of the loop and ″b″ the bottom of the loop. Applying Newton’s 2nd law to the seat at the top of the loop will establish the value of mv2/r; this can then be used at the bottom of the loop to determine Fn,b. Apply

∑F

r

= mar to the seat at the

mg +Fn,t = 2mg = mar = mv2/r

top of the loop: Apply

∑F

r

= mar to the seat at the

Fn,b – mg = mv2/r

bottom of the loop: Solve for Fn,b and substitute for mv2/r to obtain:

Fn,b = 3mg and (d ) is correct.

83 • Picture the Problem The speed of the roller coaster is imbedded in the expression for its radial acceleration. The radial acceleration is determined by the net radial force acting on the passenger. We can use Newton’s 2nd law to relate the net force on the passenger to the speed of the roller coaster. Apply

∑F

radial

= maradial to the

mg + 0.4mg = mv2/r

passenger: Solve for v:

v = 1.4 gr

Substitute numerical values and evaluate v:

v = 1.4 9.81 m/s 2 (12.0 m )

(

= 12.8 m/s

)

Applications of Newton’s Laws 337 84 • Picture the Problem The force F the passenger exerts on the armrest of the car door is the radial force required to maintain the passenger’s speed around the curve and is related to that speed through Newton’s 2nd law of motion. Apply

∑F

x

= max to the forces

acting on the passenger: Solve this equation for v:

Substitute numerical values and evaluate v:

F =m

v2 r

v=

rF m

v=

(80 m )(220 N ) = 15.9 m/s 70 kg

and (a ) is correct. *85 ••• Picture the Problem The forces acting on the bicycle are shown in the force diagram. The static friction force is the centripetal force exerted by the surface on the bicycle that allows it to move in a circular path.

r r Fn + f s makes an angle θ with the vertical

direction. The application of Newton’s 2nd law will allow us to relate this angle to the speed of the bicycle and the coefficient of static friction. (a) Apply

r

r

∑ F = ma to the bicycle:

∑ Fx = fs =

mv 2 r

and

∑F

y

Relate Fn and fs to θ :

= Fn − mg = 0

mv 2 f v2 tan θ = s = r = Fn mg rg

338 Chapter 5 Solve for v:

v = rg tan θ

Substitute numerical values and evaluate v:

v=

(b) Relate fs to µs and Fn:

fs =

Solve for µs and substitute for fs to obtain:

2 f s 2v 2 = µs = mg rg

Substitute numerical values and evaluate µs

2(7.25 m/s ) µs = = 0.536 (20 m ) 9.81m/s2

(20 m )(9.81m/s2 )tan15°

= 7.25 m/s 1 2

f s,max = 12 µs mg

2

(

)

86 •• Picture the Problem The diagram shows the forces acting on the plane as it flies in a horizontal circle of radius R. We can apply Newton’s 2nd law to the plane and eliminate the lift force in order to obtain an expression for R as a function of v and θ.

Apply

r

r

∑ F = ma to the plane:

∑ Fx = Flift sin θ = m

v2 R

and

∑F

y

Eliminate Flift between these equations to obtain: Solve for R:

Substitute numerical values and evaluate R:

= Flift cosθ − mg = 0

tan θ =

R=

v2 Rg

v2 g tan θ 2

⎛ km 1h ⎞ ⎜⎜ 480 ⎟ × h 3600 s ⎟⎠ ⎝ R= (9.81m/s2 )tan40° = 2.16 km

Applications of Newton’s Laws 339 87 • Picture the Problem Under the conditions described in the problem statement, the only forces acting on the car are the normal force exerted by the road and the gravitational force exerted by the earth. The horizontal component of the normal force is the centripetal force. The application of Newton’s 2nd law will allow us to express θ in terms of v, r, and g. Apply

r

r

∑ F = ma to the car:

∑ Fx = Fn sin θ = m

v2 r

and

∑F

y

Eliminate Fn from the force equations to obtain:

= Fn cosθ − mg = 0

tan θ =

Solve for θ :

v2 rg ⎡ v2 ⎤ ⎥ ⎣ rg ⎦

θ = tan −1 ⎢

Substitute numerical values and evaluate θ:

⎧ [(90 km/h )(1 h 3600 s )(1000 m/km )] 2 ⎫ ⎬ = 21.7° 2 ( ) 160 m 9.81 m/s ⎩ ⎭

θ = tan −1 ⎨

(

*88 •• Picture the Problem Both the normal force and the static friction force contribute to the centripetal force in the situation described in this problem. We can apply Newton’s 2nd law to relate fs and Fn and then solve these equations simultaneously to determine each of these quantities.

)

340 Chapter 5 (a) Apply

r

r

∑ F = ma to the car:

∑ Fx = Fn sin θ + f s cosθ = m

v2 r

and

∑F

y

Multiply the x equation by sinθ and the y equation by cosθ to obtain:

= Fn cosθ − f s sin θ − mg = 0

f s sin θ cosθ + Fn sin 2 θ = m

v2 sin θ r

and

Fn cos 2 θ − f s sin θ cosθ − mg cosθ = 0 Add these equations to eliminate fs:

Fn − mg cosθ = m

Solve for Fn:

Fn = mg cosθ + m

v2 sin θ r

v2 sin θ r ⎛ ⎞ v2 ⎜ = m⎜ g cosθ + sin θ ⎟⎟ r ⎝ ⎠

Substitute numerical values and evaluate Fn: 2 2 2 ⎡ ⎤ ( 85 km/h ) (1000 m/km ) (1h/3600 s ) 2 sin10°⎥ Fn = (800 kg ) ⎢ 9.81 m/s cos10° + 150 m ⎣ ⎦

(

)

= 8.25 kN (b) Solve the y equation for fs:

fs =

Fn cos θ − mg sin θ

Substitute numerical values and evaluate fs:

fs

( 8.25 kN ) cos10° − (800 kg ) (9.81m/s 2 ) = = sin10°

1.59 kN

(c) Express µs,min in terms of fs and Fn :

µs,min =

fs Fn

Substitute numerical values and evaluate µs,min:

µs,min =

1.59 kN = 0.193 8.25 kN

Applications of Newton’s Laws 341 89 •• Picture the Problem Both the normal force and the static friction force contribute to the centripetal force in the situation described in this problem. We can apply Newton’s 2nd law to relate fs and Fn and then solve these equations simultaneously to determine each of these quantities.

(a) Apply

r

r

∑ F = ma to the car:

v2 r ∑ Fy = Fn cosθ − f s sinθ − mg = 0

∑ Fx = Fn sin θ + f s cosθ = m

Multiply the x equation by sinθ and the y equation by cosθ :

v2 sin θ r Fn cos 2 θ − f s sin θ cosθ − mg cosθ = 0

Add these equations to eliminate fs:

Fn − mg cosθ = m

Solve for Fn:

Fn = mg cosθ + m

f s sin θ cosθ + Fn sin 2 θ = m

v2 sin θ r

v2 sin θ r ⎛ ⎞ v2 = m⎜⎜ g cosθ + sin θ ⎟⎟ r ⎝ ⎠

Substitute numerical values and evaluate Fn: 2 2 2 ⎡ ⎤ ( 38 km/h ) (1000 m/km ) (1 h/3600 s ) 2 sin10°⎥ Fn = (800 kg )⎢ 9.81 m/s cos10° + 150 m ⎣ ⎦

(

)

= 7.832 kN (b) Solve the y equation for fs:

Substitute numerical values and evaluate fs:

Fn cos θ − mg sin θ mg = Fn cot θ − sin θ

fs =

342 Chapter 5

( 800 kg ) (9.81 m/s 2 ) f s = (7.832 kN ) cot 10° − = sin10°

− 777 N

The negative sign tells us that fs points upward along the inclined plane rather than as shown in the force diagram. *90 ••• Picture the Problem The free-body diagram to the left is for the car at rest. The static friction force up the incline balances the downward component of the car’s weight and prevents it from sliding. In the free-body diagram to the right, the static friction force points in the opposite direction as the tendency of the moving car is to slide toward the outside of the curve.

r

r

∑ F = ma to the car that is

Apply at rest:

∑F

y

and

∑F

x

Substitute fs = fs,max = µsFn in equation (2) and solve for and evaluate the maximum allowable value of θ: Apply

r r F ∑ = ma to the car that is

moving with speed v:

Substitute numerical values into (5)

= Fn sin θ − f s cosθ = 0

(2)

θ = tan −1 (µs ) = tan −1 (0.08) = 4.57°

∑F

y

∑F

x

Substitute fs = µsFn in equations (3) and (4) and simplify to obtain:

= Fn cosθ + f s sin θ − mg = 0 (1)

= Fn cosθ − f s sin θ − mg = 0 (3)

= Fn sin θ + f s cosθ = m

Fn (cosθ − µs sin θ ) = mg Fn (µs cosθ + sin θ ) = m 0.9904Fn = mg

v2 r

(4) (5)

2

v r

(6)

Applications of Newton’s Laws 343 and (6) to obtain:

and

v2 0.1595Fn = m r Eliminate Fn and solve for r:

Substitute numerical values and evaluate r:

v2 r= 0.1610 g

r=

(60 km/h ×1h/3600 s ×1000 m/km)2

(

0.1610 9.81 m/s 2

)

= 176 m 91 ••• Picture the Problem The free-body diagram to the left is for the car rounding the curve at the minimum (not sliding down the incline) speed. The static friction force up the incline balances the downward component of the car’s weight and prevents it from sliding. In the free-body diagram to the right, the static friction force points in the opposite direction as the tendency of the car moving with the maximum safe speed is to slide toward the outside of the curve. Application of Newton’s 2nd law and the simultaneous solution of the force equations will yield vmin and vmax.

Apply

r

r

∑ F = ma to a car traveling

around the curve when the coefficient of static friction is zero:

∑ Fx = Fn sin θ = m and

∑F

y

Divide the first of these equations by the second to obtain:

2 vmin r

= Fn cosθ − mg =0

2 v2 −1 ⎛ v ⎞ tan θ = or θ = tan ⎜⎜ ⎟⎟ rg ⎝ rg ⎠

Substitute numerical values and evaluate the banking angle:

344 Chapter 5

(

)

⎡ (40 km/h )2 (1000 m/km )2 1h/3600 s 2 ⎤ ⎥ = 22.8° (30 m ) 9.81m/s2 ⎣ ⎦

θ = tan −1 ⎢ Apply

r r F ∑ = ma to the car

traveling around the curve at minimum speed:

(

2 vmin ∑ Fx = Fn sin θ − fs cosθ = m r

and

∑F

y

Substitute fs = fs,max = µsFn in the force equations and simplify to obtain:

Evaluate these equations for θ = 22.8° and µs = 0.3:

)

= Fn cos θ + f s sin θ − mg = 0

2 vmin Fn (µs cosθ − sin θ ) = m r

and

Fn (cosθ + µs sin θ ) = mg

0.1102Fn= m

2 vmin r

and 1.038Fn = mg Eliminate Fn between these two equations and solve for vmin:

vmin = 0.106rg

Substitute numerical values and evaluate vmin:

vmin = 0.106(30 m ) 9.81m/s 2

Apply

r

r

∑ F = ma to the car

traveling around the curve at maximum speed:

(

= 5.59 m/s = 20.1 km/h

∑ Fx = Fn sin θ + fs cosθ = m

Evaluate these equations for θ = 22.8° and µs = 0.3:

2 vmax r

and

∑F

y

Substitute fs = fs,max = µsFn in the force equations and simplify to obtain:

)

= Fn cosθ − f s sin θ − mg =0

Fn (µs cosθ + sin θ ) = m and

2 vmax r

Fn (cosθ − µ s sin θ ) = mg

0.6641Fn= m

2 vmax r

and 0.8056Fn = mg

Applications of Newton’s Laws 345 Eliminate Fn between these two equations and solve for vmax:

vmax = 0.8243rg

Substitute numerical values and evaluate vmax:

vmax =

(0.8243)(30 m )(9.81m/s2 )

= 15.6 m/s = 56.1 km/h

Drag Forces 92 • Picture the Problem We can apply Newton’s 2nd law to the particle to obtain its equation of motion. Applying terminal speed conditions will yield an expression for b that we can evaluate using the given numerical values. Apply

∑F

y

= ma y to the particle:

mg − bv = ma y

When the particle reaches its terminal speed v = vt and ay = 0:

mg − bvt = 0

Solve for b to obtain:

b=

Substitute numerical values and evaluate b:

b=

mg vt

(10

−13

)(

kg 9.81 m/s 2 3 × 10 −4 m/s

)

= 3.27 ×10 −9 kg/s 93 • Picture the Problem We can apply Newton’s 2nd law to the Ping-Pong ball to obtain its equation of motion. Applying terminal speed conditions will yield an expression for b that we can evaluate using the given numerical values. Apply

∑F

y

= ma y to the Ping-

mg − bv 2 = ma y

Pong ball: When the Ping-Pong ball reaches its terminal speed v = vt and ay = 0:

mg − bvt2 = 0

Solve for b to obtain:

b=

mg vt2

346 Chapter 5 Substitute numerical values and evaluate b:

b=

(2.3 ×10

−3

)(

kg 9.81 m/s 2 (9 m/s)2

)

= 2.79 × 10−4 kg/m *94 • Picture the Problem Let the upward direction be the positive y direction and apply Newton’s 2nd law to the sky diver. (a) Apply

∑F

y

= ma y to the sky

Fd − mg = ma y

diver:

or, because ay = 0,

Substitute numerical values and evaluate Fd:

Fd = (60 kg ) 9.81 m/s 2 = 589 N

(b) Substitute Fd = b vt2 in equation

bvt2 = mg

Fd = mg

(1)

(

)

(1) to obtain: Solve for b:

b=

mg Fd = 2 vt2 vt

Substitute numerical values and evaluate b:

b=

589 N = 0.942 kg/m (25 m/s)2

95 •• Picture the Problem The free-body diagram shows the forces acting on the car as it descends the grade with its terminal velocity. The application of Newton’s 2nd law with a = 0 and Fd equal to the given function will allow us to solve for the terminal velocity of the car. Apply

∑F

x

= max to the car:

mg sin θ − Fd = ma x or, because v = vt and ax = 0,

mg sin θ − Fd = 0

Substitute for Fd to obtain:

(

)

mg sin θ − 100 N − 1.2 N ⋅ s 2 / m 2 vt2 = 0

Applications of Newton’s Laws 347 Solve for vt:

vt =

Substitute numerical values and evaluate vt:

vt =

mg sin θ − 100 N 1.2 N ⋅ s 2 / m 2

(800 kg )(9.81m/s 2 )sin 6° − 100 N 1.2 N ⋅ s 2 / m 2

= 24.5 m/s = 88.2 km/h 96 ••• Picture the Problem Let the upward direction be the positive y direction and apply Newton’s 2nd law to the particle to obtain an equation from which we can find the particle’s terminal speed. (a) Apply

∑F

y

= ma y to a

mg − 6πηrv = ma y

pollution particle:

or, because ay = 0,

Solve for vt to obtain:

vt =

Express the mass of a sphere in terms of its volume:

⎛ 4π r 3 ⎞ ⎟⎟ m = ρV = ρ ⎜⎜ ⎝ 3 ⎠

Substitute for m to obtain:

Substitute numerical values and evaluate vt:

mg − 6πηrvt = 0

vt =

mg 6πηr

2r 2 ρg 9η

(

)( ( 2

)(

2 10 −5 m 2000 kg/m 3 9.81 m/s 2 vt = 9 1.8 ×10 −5 N ⋅ s/m 2

)

)

= 2.42 cm/s (b) Use distance equals average speed times the fall time to find the time to fall 100 m at 2.42 cm/s:

t=

10 4 cm = 4.13 × 103 s = 1.15 h 2.42 cm/s

*97 ••• Picture the Problem The motion of the centrifuge will cause the pollution particles to migrate to the end of the test tube. We can apply Newton’s 2nd law and Stokes’ law to derive an expression for the terminal speed of the sedimentation particles. We can then use this terminal speed to calculate the sedimentation time. We’ll use the 12 cm distance

348 Chapter 5 from the center of the centrifuge as the average radius of the pollution particles as they settle in the test tube. Let R represent the radius of a particle and r the radius of the particle’s circular path in the centrifuge. Express the sedimentation time in terms of the sedimentation speed vt: Apply

∑F

radial

= maradial to a

∆t sediment =

∆x vt

6πηRvt = mac

pollution particle: Express the mass of the particle in terms of its radius R and density ρ:

m = ρV = 43 π R 3 ρ

Express the acceleration of the pollution particles due to the motion of the centrifuge in terms of their orbital radius r and period T:

⎛ 2π r ⎞ ⎜ ⎟ 2 v 4π 2 r T ⎠ ⎝ = = 2 ac = r r T

Substitute for m and ac and simplify to obtain:

⎛ 4π 2 r ⎞ 16π 3 ρ rR 3 6πηRvt = π R ρ ⎜⎜ 2 ⎟⎟ = 3T 2 ⎝ T ⎠

Solve for vt:

Find the period T of the motion from the number of revolutions the centrifuge makes in 1 second:

2

4 3

3

vt =

8π 2 ρ rR 2 9ηT 2

T=

1 = 1.25 × 10 −3 min/rev 800 rev / min

= 1.25 × 10−3 min/rev × 60 s/min = 75.0 × 10-3 s/rev

Substitute numerical values and evaluate vt:

Find the time it takes the particles to move 8 cm as they settle in the test tube:

vt =

(

)

(

8π 2 2000 kg/m 3 (0.12 m ) 10−5 m

(

)(

9 1.8 × 10−5 N ⋅ s/m 2 75 × 10 −3 s = 2.08 m/s

∆tsediment =

8 cm ∆x = 208 cm/s v

= 38.5 ms

)

) 2

2

Applications of Newton’s Laws 349 8 cm ∆x = 2.42 cm/s v

In Problem 96 it was shown that the rate of fall of the particles in air is 2.42 cm/s. Find the time required to fall 8 cm in air under the influence of gravity:

∆tair =

Find the ratio of the two times:

∆tair/∆tsediment ≈ 100

= 3.31s

Euler’s Method 98 •• Picture the Problem The free-body diagram shows the forces acting on the baseball sometime after it has been thrown downward but before it has reached its terminal speed. In order to use Euler’s method, we’ll need to determine how the acceleration of the ball varies with its speed. We can do this by applying Newton’s 2nd law to the ball and using its terminal speed to express the constant in the acceleration equation in terms of the ball’s terminal speed. We can then use vn+1 = vn + an ∆t to find the speed of the ball at any given time. Apply Newton’s 2nd law to the ball to obtain: Solve for dv/dt to obtain:

When the ball reaches its terminal speed: Substitute to obtain:

Express the position of the ball to obtain: Letting an be the acceleration of the ball at time tn, express its speed when t = tn + 1:

mg − bv 2 = m

dv dt

dv b = g − v2 dt m 0=g−

b g b 2 = 2 vt ⇒ m vt m

⎛ v2 ⎞ dv = g ⎜⎜1 − 2 ⎟⎟ dt ⎝ vt ⎠

xn+1 = xn +

vn+1 + vn ∆t 2

vn+1 = vn + an ∆t where

⎛ v2 ⎞ an = g ⎜⎜1 − n2 ⎟⎟ ⎝ vt ⎠

350 Chapter 5 and ∆t is an arbitrarily small interval of time. A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A10 B10

Formula/Content B9+$B$1 B9+0.5*(C9+C10)*$B$1

C10 C9+D9*$B$1 D10 $B$4*(1−C10^2/$B$5^2)

A ∆t= x0= v0= a0= vt=

Algebraic Form t + ∆t

xn+1 = xn +

vn+1 + vn ∆t 2

vn+1 = vn+ an∆t

⎛ v2 ⎞ an = g ⎜⎜1 − n2 ⎟⎟ ⎝ vt ⎠

B

C

0.5 0 9.722 9.81 41.67

s m m/s m/s^2 m/s

D

1 2 3 4 5 6 7 8 9 10 11 12

t (s) 0.0 0.5 1.0 1.5

x (m) 0 6 14 25

v (m/s) 9.7 14.4 18.7 22.6

a (m/s^2) 9.28 8.64 7.84 6.92

28 29 30

9.5 10.0 10.5

317 337 358

41.3 41.4 41.5

0.17 0.13 0.10

38 39 40 41 42

14.5 15.0 15.5 16.0 16.5

524 545 566 587 608

41.6 41.7 41.7 41.7 41.7

0.01 0.01 0.01 0.01 0.00

From the table we can see that the speed of the ball after 10 s is approximately

41.4 m/s. We can estimate the uncertainty in this result by halving ∆t and recalculating the speed of the ball at t = 10 s. Doing so yields v(10 s) ≈ 41.3 m/s, a difference of about 0.02%. The graph shows the velocity of the ball thrown straight down as a function of time.

Applications of Newton’s Laws 351

v (m/s)

Ball Throw n Straight Dow n 45 40 35 30 25 20 15 10 5 0 0

5

10

15

20

t (s)

Reset ∆t to 0.5 s and set v0 = 0. Ninety-nine percent of 41.67 m/s is approximately 41.3 m/s. Note that the ball will reach this speed in about 10.5 s and that the distance it travels in this time is about 322 m. The following graph shows the distance traveled by the ball dropped from rest as a function of time. Ball Dropped From Rest 400 350 300 x (m)

250 200 150 100 50 0 0

2

4

6 t (s)

*99 •• Picture the Problem The free-body diagram shows the forces acting on the baseball after it has left your hand. In order to use Euler’s method, we’ll need to determine how the acceleration of the ball varies with its speed. We can do this by applying Newton’s 2nd law to the baseball. We can then use vn+1 = vn + an ∆t and

xn +1 = xn + vn ∆t to find the speed and

8

10

12

352 Chapter 5 position of the ball. Apply

∑F

y

= ma y to the baseball:

− bv v − mg = m

dv dt

where v = v for the upward part of the flight of the ball and v = −v for the downward part of the flight. Solve for dv/dt:

Under terminal speed conditions ( v = −vt ):

dv b = −g − v v dt m 0 = −g +

b 2 vt m

and

b g = 2 m vt Substitute to obtain:

⎛ vv dv g = − g − 2 v v = − g ⎜⎜1 + 2 dt vt vt ⎝

Letting an be the acceleration of the ball at time tn, express its position and speed when t = tn + 1:

yn+1 = yn + 12 (vn + vn−1 )∆t

⎞ ⎟ ⎟ ⎠

and

vn+1 = vn + an ∆t where

⎛ v v an = − g ⎜⎜1 + n 2 n vt ⎝

⎞ ⎟ ⎟ ⎠

and ∆t is an arbitrarily small interval of time. A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell D11 E10 E11 F10 F11

Formula/Content D10+$B$6 41.7 E10−$B$4* (1+E10*ABS(E10)/($B$5^2))*$B$6 0 F10+0.5*(E10+E11)*$B$6

G10 G11

0 $E$10*D11−0.5*$B$4*D11^2

Algebraic Form t + ∆t v0

vn+1 = vn + an ∆t y0

yn+1 = yn + 12 (vn + vn−1 )∆t y0

v0t − 12 gt 2

Applications of Newton’s Laws 353 D

E

F

G

t 0.0 0.1 0.2

v 41.70 39.74 37.87

y 0.00 4.07 7.95

y no drag 0.00 4.12 8.14

40 41 42 43 44 45 46

3.0 3.1 3.2 3.3 3.4 3.5 3.6

3.01 2.03 1.05 0.07 −0.91 −1.89 −2.87

60.13 60.39 60.54 60.60 60.55 60.41 60.17

81.00 82.18 83.26 84.25 85.14 85.93 86.62

78 79 80 81

6.8 6.9 7.0 7.1

−28.34 −28.86 −29.37 −29.87

6.26 3.41 0.49 −2.47

56.98 54.44 51.80 49.06

4 5 6 7 8 9 10 11 12

A B C g= 9.81 m/s^2 vt= 41.7 m/s ∆t= 0.1 s

From the table we can see that, after 3.5 s, the ball reaches a height of about 60.4 m. It reaches its peak a little earlier−at about 3.3 s, and its height at t = 3.3 s is 60.6 m. The ball hits the ground at about t = 7 s −so it spends a little longer coming down than going up. The solid curve on the following graph shows y(t) when there is no drag on the baseball and the dotted curve shows y(t) under the conditions modeled in this problem.

354 Chapter 5

90 80 70

y (m)

60 50 40

x with drag

30

x with no drag

20 10 0 0

1

2

3

4

5

6

7

t (s)

100 •• Picture the Problem The pictorial representation shows the block in its initial position against the compressed spring, later as the spring accelerates it to the right, and finally when it has reached its maximum speed at xf = 0. In order to use Euler’s method, we’ll need to determine how the acceleration of the block varies with its position. We can do this by applying Newton’s 2nd law to the box. We can then use vn+1 = vn + an ∆t and

xn +1 = xn + vn ∆t to find the speed and position of the block.

Apply

∑F

x

= max to the block:

Solve for an:

Express the position and speed of the block when t = tn + 1:

k (0.3 m − xn ) = man an =

k (0.3 m − xn ) m

xn +1 = xn + vn ∆t and

vn+1 = vn + an ∆t where

an =

k (0.3 m − xn ) m

and ∆t is an arbitrarily small interval of time. A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Applications of Newton’s Laws 355 Cell A10 B10

Formula/Content A9+$B$1 B9+C10*$B$1

Algebraic Form t + ∆t

C10

C9+D9*$B$1

vn + an ∆t k (0.3 − xn ) m

D10 ($B$4/$B$5)*(0.3−B10)

A ∆t= x0= v0= k= m=

B 0.005 0 0 50 0.8

xn + vn ∆t

C

D

1 2 3 4 5 6 7 8 9 10 11 12

t (s) 0.000 0.005 0.010 0.015

x (m) 0.00 0.00 0.00 0.00

v (m/s) 0.00 0.09 0.19 0.28

a (m/s^2) 18.75 18.72 18.69 18.63

45 46 47 48 49

0.180 0.185 0.190 0.195 0.200

0.25 0.27 0.28 0.29 0.30

2.41 2.42 2.43 2.44 2.44

2.85 2.10 1.34 0.58 −0.19

s m m/s N/m kg

From the table we can see that it took about 0.200 s for the spring to push the block 30 cm and that it was traveling about 2.44 m/s at that time. We can estimate the uncertainty in this result by halving ∆t and recalculating the speed of the ball at t = 10 s. Doing so yields v(0.200 s) ≈ 2.41 m/s, a difference of about 1.2%.

356 Chapter 5

General Problems 101 • Picture the Problem The forces that act on the block as it slides down the incline are shown on the free-body diagram to the right. The acceleration of the block can be determined from the distance-and-time information given in the problem. The application of Newton’s 2nd law to the block will lead to an expression for the coefficient of kinetic friction as a function of the block’s acceleration and the angle of the incline. Apply

r

r

∑ F = ma to the block:

ΣFx = mgsinθ − fk = ma and ΣFy = Fn − mg = 0

g sin θ − a g cosθ

Set fk = µkFn, Fn between the two equations, and solve for µk:

µk =

Using a constant-acceleration equation, relate the distance the block slides to its sliding time:

∆x = v0 ∆t + 12 a (∆t ) where v0 = 0

Solve for a:

a=

2∆x (∆t )2

Substitute numerical values and evaluate a:

a=

2(2.4 m ) = 0.1775 m/s 2 2 (5.2 s )

Find µk for a = 0.1775 m/s2 and θ = 28°:

2

µk

(9.81m/s ) sin28° − 0.1775 m/s = (9.81m/s ) cos28° 2

2

= 0.511

2

Applications of Newton’s Laws 357 102 • Picture the Problem The free-body diagram shows the forces acting on the model airplane. The speed of the plane can be calculated from the data concerning the radius of its path and the time it takes to make one revolution. The application of Newton’s 2nd law will give us the tension F in the string. (a) Express the speed of the airplane in terms of the circumference of the circle in which it is flying and its period:

v=

Substitute numerical values and evaluate v:

v=

(b) Apply

∑F

x

= max to the model

airplane: Substitute numerical values and evaluate F:

2πr T

2π(5.7 m ) = 10.7 m/s 4 s 1.2

F =m

v2 r

F = (0.4 kg )

(10.7 m/s)2 5.7 m

= 8.03 N

*103 •• Picture the Problem The free-body diagram shows the forces acting on the box. If the student is pushing with a force of 200 N and the box is on the verge of moving, the static friction force must be at its maximum value. In part (b), the motion is impending up the incline; therefore the direction of fs,max is down the incline. (a) Apply

r

r

∑ F = ma to the box:

∑F

x

= f s + F − mg sin θ = 0

and

∑F

y

= Fn − mg cosθ = 0

358 Chapter 5 F mg cosθ

Substitute fs = fs,max = µsFn, eliminate Fn between the two equations, and solve for µs:

µs = tan θ −

Substitute numerical values and evaluate µs:

µs = tan 30° −

200 N (800 N )cos30°

= 0.289 (b) Find fs,max from the x-direction force equation: Substitute numerical values and evaluate fs,max:

f s,max = mg sin θ − F f s,max = (800 N )sin30° − 200 N = 200 N

If the block is on the verge of sliding up the incline, fs,max must act down the incline. The x-direction force equation becomes:

− f s, max + F − mg sin θ = 0

Solve the x-direction force equation for F:

F = mg sin θ + f s,max

Substitute numerical values and evaluate F:

F = (800 N )sin30° + 200 N = 600 N

104 • Picture the Problem The path of the particle is a circle if r is a constant. Once we have shown that it is, we can calculate its value from its components. The direction of the particle’s motion can be determined by examining two positions of the particle at times that are close to each other. (a) and (b) Express the magnitude of r r in terms of its components:

r = rx2 + ry2

Evaluate r with rx = −10 m cos ωt and ry = 10 m sinωt:

r=

[(− 10 m )cos ωt ] 2 + [(10 m )sin ωt ] 2

(

)

= 100 cos 2ωt + sin 2 ωt m = 10.0 m

Applications of Newton’s Laws 359 (c) Evaluate rx and ry at t = 0 s:

Evaluate rx and ry at t = ∆t, where ∆t is small:

rx = −(10 m ) cos 0° = −10 m ry = (10 m )sin 0° = 0

rx = −(10 m )cos ω∆t ≈ −(10 m )cos 0° = −10 m ry = (10 m )sin ω∆t = ∆y where ∆y is positive the motion is clockwise

and

r

(d) Differentiate r with respect to r time to obtain v :

r

Use the components of v to find its speed:

r r v = dr / dt

= [(10ω sin ωt ) m] iˆ + [(10ω cos ωt )m] ˆj

v = vx2 + v y2

[(10ω sin ωt )m] 2 + [(10ω cos ωt )m] 2 = (10 m )ω = (10 m )(2 s −1 ) =

= 20.0 m/s (e) Relate the period of the particle’s motion to the radius of its path and its speed:

T=

2πr 2π (10 m ) = = πs v 20 m/s

105 •• Picture the Problem The free-body diagram shows the forces acting on the crate of books. The kinetic friction force opposes the motion of the crate up the incline. Because the crate is moving at constant speed in a straight line, its acceleration is zero. We can determine F by applying Newton’s 2nd law to the crate, substituting for fk, eliminating the normal force, and solving for the required force. Apply

r

r

∑ F = ma to the crate, with

both ax and ay equal to zero, to the crate:

∑F

x

= F cosθ − f k − mg sin θ = 0

and

∑F

y

= Fn − F sin θ − mg cosθ = 0

360 Chapter 5 Substitute µsFn for fk and eliminate Fn to obtain:

F=

mg (sin θ + µ k cos θ ) cos θ − µ k sin θ

Substitute numerical values and evaluate F:

F=

(100 kg )(9.81m/s2 )(sin30° + (0.5)cos30°) cos30° − (0.5)sin30°

= 1.49 kN

106 •• Picture the Problem The free-body diagram shows the forces acting on the object as it slides down the inclined plane. We can calculate its speed at the bottom of the incline from its acceleration and displacement and find its acceleration from Newton’s 2nd law. Using a constant-acceleration equation, relate the initial and final velocities of the object to its acceleration and displacement: solve for the final velocity: Apply

r

r

∑ F = ma to the sliding

v 2 = v02 + 2a∆x Because v0 = 0, v =

∑F

x

2a∆x

(1)

= − f k + mg sin θ = ma

and

object:

∑F

Solve the y equation for Fn and using fk = µkFn, eliminate both Fn and fk from the x equation and solve for a: Substitute equation (2) in equation (1) and solve for v:

y

= Fn − mg cosθ = 0

a = g (sin θ − µ k cosθ )

(2)

v = 2 g (sin θ − µ k cosθ )∆x

Substitute numerical values and evaluate v:

(

)

v = 2 9.81m/s 2 (sin 30° − (0.35)cos 30°)(72 m ) = 16.7 m/s and (d ) is correct.

Applications of Newton’s Laws 361 *107 •• Picture the Problem The free-body diagram shows the forces acting on the brick as it slides down the inclined plane. We’ll apply Newton’s 2nd law to the brick when it is sliding down the incline with constant speed to derive an expression for µk in terms of θ0. We’ll apply Newton’s 2nd law a second time for θ = θ1 and solve the equations simultaneously to obtain an expression for a as a function of θ0 and θ1. Apply

r

r

∑ F = ma to the brick

∑F

x

= − f k + mg sin θ 0 = 0

when it is sliding with constant speed:

and

Solve the y equation for Fn and using fk = µkFn, eliminate both Fn and fk from the x equation and solve for µk:

µ k = tan θ 0

Apply

r

r

∑ F = ma to the brick when

θ = θ1:

Solve the y equation for Fn, use fk = µkFn to eliminate both Fn and fk from the x equation, and use the expression for µk obtained above to obtain:

∑F

y

∑F

x

= Fn − mg cosθ 0 = 0

= − f k + mg sin θ 1 = ma

and

∑F

y

= Fn − mg cosθ 1 = 0

a = g (sin θ 1 − tan θ 0 cosθ 1 )

108 •• Picture the Problem The fact that the object is in static equilibrium under the influence r r r of the three forces means that F1 + F2 + F3 = 0. Drawing the corresponding force triangle will allow us to relate the forces to the angles between them through the law of sines and the law of cosines.

362 Chapter 5 (a) Using the fact that the object is in static equilibrium, redraw the force diagram connecting the forces head-to-tail:

Apply the law of sines to the triangle:

Use the trigonometric identity sin(π − α) = sinα to obtain: (b) Apply the law of cosines to the triangle: Use the trigonometric identity cos(π − α) = −cosα to obtain:

F3 F1 F2 = = sin (π − θ 23 ) sin (π − θ13 ) sin (π − θ12 ) F1 F2 F3 = = sin θ 23 sin θ13 sin θ12

F12 = F22 + F32 − 2 F2 F3 cos(π − θ 23 ) F12 = F22 + F32 + 2 F2 F3 cos θ 23

109 •• Picture the Problem We can calculate the acceleration of the passenger from his/her speed that, in turn, is a function of the period of the motion. To determine the longest period of the motion, we focus our attention on the situation at the very top of the ride when the seat belt is exerting no force on the rider. We can use Newton’s 2nd law to relate the period of the motion to the acceleration and speed of the rider. (a) Because the motion is at constant speed, the acceleration is entirely radial and is given by:

v2 ac = r

Express the speed of the motion of the ride as a function of the radius of the circle and the period of its motion:

v=

2π r T

Applications of Newton’s Laws 363 Substitute in the expression for ac to obtain:

4π 2 r ac = 2 T

Substitute numerical values and evaluate ac:

ac =

(b) Apply

r r F ∑ = ma to the

4π 2 (5 m ) = 49.3 m/s 2 2 (2 s )

∑F

r

= mg = ma c

passenger when he/she is at the top of the circular path and solve for ac:

and ac = g

Relate the acceleration of the motion to its radius and speed and solve for v:

v2 g = ⇒ v = gr r

Express the period of the motion as a function of the radius of the circle and the speed of the passenger and solve for Tm: Substitute numerical values and evaluate Tm:

Tm =

2π r r = 2π v g

Tm = 2π

5m = 4.49 s 9.81 m/s 2

Remarks: The rider is ″weightless″ under the conditions described in part (b). *110 •• Picture the Problem The pictorial representation to the right shows the cart and its load on the inclined plane. The load will not slip provided its maximum acceleration is not exceeded. We can find that maximum acceleration by applying Newton’s 2nd law to the load. We can then apply Newton’s 2nd law to the cart-plusload system to determine the tension in the rope when the system is experiencing its maximum acceleration.

364 Chapter 5 Draw the free-body diagram for the cart and its load:

Apply

∑F

x

= max to the cart plus

T − (m1 + m2 )g sin θ = (m1 + m2 )amax (1)

its load: Draw the free-body diagram for the load of mass m2 on top of the cart:

Apply

r r F ∑ = ma to the load on

top of the cart:

∑F

x

= f s,max − m2 g sin θ = m2 a max

and

∑F

y

= Fn , 2 − m2 g cosθ = 0

Using fs,max = µsFn,2, eliminate Fn,2 between the two equations and solve for the maximum acceleration of the load:

a max = g (µ s cosθ − sin θ )

Substitute equation (2) in equation (1) and solve for T :

T=

111 •• Picture the Problem The free-body diagram for the sled while it is held stationary by the static friction force is shown to the right. We can solve this problem by repeatedly applying Newton’s 2nd law under the conditions specified in each part of the problem.

(m1 + m2 )gµ s cosθ

(2)

Applications of Newton’s Laws 365 (a) Apply

∑F

y

= ma y to the sled:

Fn,1 − m1 g cosθ = 0

Solve for Fn,1:

Fn,1 = m1 g cosθ

Substitute numerical values and evaluate Fn,1:

Fn,1 = (200 N ) cos15° = 193 N

(b) Apply

∑F

x

= max to the sled:

f s − m1 g sin θ = 0

Solve for fs:

f s = m1 g sin θ

Substitute numerical values and evaluate fs:

f s = (200 N )sin 15° = 51.8 N

(c) Draw the free-body diagram for the sled when the child is pulling on the rope:

Apply

r

r

∑ F = ma to the sled to

∑F

x

= Fnet = F cos 30° − m1 g sin θ − f s,max

determine whether it moves: and

∑F

Solve the y-direction equation for Fn,1: Substitute numerical values and evaluate Fn,1:

y

= Fn,1 + F sin 30° − m1 g cosθ = 0

Fn,1 = − F sin 30° + m1 g cosθ Fn,1 = −(100 N )sin 30° + (200 N )cos15° = 143 N

Express fs,max:

fs,max = µsFn,1 = (0.5)(143 N) = 71.5 N

Use the x-direction force equation to evaluate Fnet:

Fnet = (100 N)cos30° − (200 N)sin15° − 71.5 N = −36.7 N

366 Chapter 5 Because the net force is negative, the sled does not move:

f k is undetermined

(d) Because the sled does not move:

µ k is undetermined

(e) Draw the FBD for the child:

Express the net force Fc exerted on the child by the incline:

Fc = Fn22 + f s,2max

Noting that the child is stationary, r r apply F = ma to the child:

∑F

∑

x

(1)

= f s,max − F cos 30° − m2 g sin 15° =0

and

∑F

y

Solve the x equation for fs,max and the y equation for Fn2:

= Fn2 − m2 g sin 15° − F sin 30° = 0

f s,max = F cos 30° + m2 g sin 15° and

Fn2 = m2 g sin 15° + F sin 30° Substitute numerical values and evaluate Fx and Fn2:

f s,max = (500 N ) cos 30° + (100 N ) sin 15° = 459 N and

Fn2 = (100 N )sin 15° + (500 N )sin 30° = 276 N

Substitute numerical values in equation (1) and evaluate F:

Fc =

(276 N )2 + (459 N )2

= 536 N

112 • Picture the Problem Let v represent the speed of rotation of the station, and r the distance from the center of the station. Because the O’Neill colony is, presumably, in deep space, the only acceleration one would experience in it would be that due to its rotation.

Applications of Newton’s Laws 367 (a) Express the acceleration of anyone who is standing inside the station:

a = v2/r

This acceleration is directed toward the axis of rotation. If someone inside the station drops an apple, the apple will not have any forces acting on it once released, but will move along a straight line at constant speed. However, from the point of view of our observer inside the station, if he views himself as unmoving, the apple is perceived to have an acceleration of mv2/r directed away from the axis of rotation (a "centrifugal" force). (b) Each deck must rotate the central axis with the same period T. Relate the speed of a person on a particular deck to his/her distance r from the center: Express the "acceleration of gravity" perceived by someone a distance r from the center:

(c) Relate the desired acceleration to the radius of Babylon 5 and its period: Solve for T:

v=

2π r T

v 2 4π 2 r = r T2 i.e., the " acceleration due to gravity" decreases as r decreases. a=

T=

Substitute numerical values and evaluate T:

T=

4π 2 r T2 4π 2 r a 1.609 km ⎞ ⎛ 4π 2 ⎜ 0.3 mi × ⎟ mi ⎠ ⎝ 9.8 m/s 2

= 44.1s = 0.735 min Take the reciprocal of this time to find the number of revolutions per minute Babylon 5 has to make in order to provide this ″earth-like″ acceleration:

T −1 = 1.36 rev / min

368 Chapter 5 113 •• Picture the Problem The free-body diagram shows the forces acting on the child as she slides down the incline. We’ll first use Newton’s 2nd law to derive an expression for µk in terms of her acceleration and then use Newton’s 2nd law to find her acceleration when riding the frictionless cart. Using a constantacceleration equation, we’ll relate these two accelerations to her descent times and solve for her acceleration when sliding. Finally, we can use this acceleration in the expression for µk. Apply

r

r

∑ F = ma to the child as

she slides down the incline:

Using fk = µkFn, eliminate fk and Fn between the two equations and solve for µk: Apply

∑F

x

= max to the child as

∑F

x

= mg sin 30° − f k = ma1

and

∑F

y

= Fn − mg cos 30° = 0

µ k = tan 30° −

a1 g cos 30°

mg sin 30° = ma2

she rides the frictionless cart down the incline and solve for her acceleration a2:

and

Letting s represent the distance she slides down the incline, use a constant-acceleration equation to relate her sliding times to her accelerations and distance traveled down the slide :

s = v0t1 + 12 a1t12 where v0 = 0

Equate these expressions, substitute t2 = 12 t1 and solve for a1:

a 2 = g sin 30° = 4.91 m/s 2

and

s = v0t2 + 12 a2t22 where v0 = 0

a1 = 14 a2 = 14 g sin 30° = 1.23 m/s2

(1)

Applications of Newton’s Laws 369 1.23 m/s 2 µ k = tan 30° − 9.81 m/s 2 cos 30°

Evaluate equation (1) with a1 = 1.23 m/s2:

(

)

= 0.433 *114 •• Picture the Problem The path of the particle is a circle if r is a constant. Once we have shown that it is, we can calculate its value from its components and determine the particle’s velocity and acceleration by differentiation. The direction of the net force acting on the particle can be determined from the direction of its acceleration.

r

(a) Express the magnitude of r in terms of its components:

r = rx2 + ry2

Evaluate r with rx = Rsinωt and ry = Rcosωt:

r=

[R sin ωt ] 2 + [R cos ωt ] 2

(

)

= R 2 sin 2ωt + cos 2 ωt = R = 4.0 m ∴the path of the particle is a circle centered at the origin.

r

(b) Differentiate r with respect to r time to obtain v :

r r v = dr / dt = [Rω cos ω t ] iˆ + [− Rω sin ω t ] ˆj =

Express the ratio

vx : vy

Express the ratio −

y : x

vx 8π cos ω t = = − cot ω t v y − 8π sin ω t −

∴ r

(c) Differentiate v with respect to r time to obtain a :

[(8π cos 2π t ) m/s] iˆ − [(8π sin 2π t ) m/s] ˆj

R cos ω t y =− = − cot ω t x R sin ω t vx y =− vy x

r r a = dv / dt

[(− 16π m/s )sin ω t ] iˆ + [(− 16π m/s )cos ω t ] ˆj 2

=

2

2

2

370 Chapter 5 r

Factor −4π2/s2 from a to obtain:

(

)[ )

(

circle in which the particle is traveling. Find the ratio

(d) Apply

v 2 (8π m/s ) = = 16π 2 m/s 2 = a r 4m

v2 : r r

2

r

∑ F = ma to the particle:

Fnet = ma = (0.8 kg )(16π 2 m/s 2 ) = 12.8π 2 N

r

Because the direction of Fnet is the

r

r Fnet is toward the center of the circle.

same as that of a : 115 •• Picture the Problem The free-body diagram showing the forces acting on a rider being held in place by the maximum static friction force is shown to the right. The application of Newton’s 2nd law and the definition of the maximum static friction force will be used to determine the period T of the motion. The reciprocal of the period will give us the minimum number of revolutions required per unit time to hold the riders in place. Apply

r

r

∑ F = ma to the riders

while they are held in place by friction:

Using f s,max = µ s Fn and v =

2π r , T

eliminate Fn between the force equations and solve for the period of the motion:

]

r a = − 4π 2 / s 2 (4 sin ω t ) iˆ + (4 cos ω t ) ˆj r = − 4π 2 / s 2 r r Because a is in the opposite direction from r r , it is directed toward the center of the

∑ Fx = Fn = m

v2 r

and

∑F

y

= f s,max − mg = 0

T = 2π

µs r g

Applications of Newton’s Laws 371 Substitute numerical values and evaluate T:

T = 2π

(0.4)(4 m ) 9.81 m/s 2

= 2.54 s = 0.00423 min The number of revolutions per minute is the reciprocal of the period in minutes:

23.6 rev/min

116 •• Picture the Problem The free-body diagrams to the right show the forces acting on the blocks whose masses are m1 and m2. The application of Newton’s 2nd law and the use of a constant-acceleration equation will allow us to find a relationship between the coefficient of kinetic friction and m1. The repetition of this procedure with the additional object on top of the object whose mass is m1 will lead us to a second equation that, when solved simultaneously with the former equation, leads to a quadratic equation in m1. Finally, its solution will allow us to substitute in an expression for µk and determine its value. Using a constant-acceleration equation, relate the displacement of the system in its first configuration as a function of its acceleration and fall time:

∆x = v0 ∆t + 12 a1 (∆t )

2

or, because v0 = 0,

∆x = 12 a1 (∆t )

2

Solve for a1:

a1 =

2∆x (∆t )2

Substitute numerical values and evaluate a1:

a1 =

2(1.5 m ) = 4.46 m/s 2 2 (0.82 s )

Apply

∑F

x

= max to the object

whose mass is m2 and solve for T1:

m2 g − T1 = m2 a1 and

T1 = m2 (g − a )

372 Chapter 5 Substitute numerical values and evaluate T1: Apply

r

r

∑ F = ma to the object

whose mass is m1:

(

T1 = (2.5 kg ) 9.81 m/s 2 − 4.46 m/s 2 = 13.375 N

∑F

= T1 − f k = m1 a1

x

and

∑F

y

= Fn,1 − m1 g = 0

Using fk = µkFn, eliminate Fn between the two equations to obtain:

T1 − µ k m1 g = m1 a1

Find the acceleration a2 for the second run:

a2 =

Evaluate T2:

T2 = m2 ( g − a )

Apply

∑F

x

)

(1)

2∆x 2(1.5 m ) = = 1.775 m/s 2 2 2 (∆t ) (1.3 s )

(

= (2.5 kg ) 9.81 m/s 2 − 1.775 m/s 2 = 20.1 N

= max to the 1.2-kg

T2 − µ k (m1 + 1.2 kg )g

(2)

= (m1 + 1.2 kg )a2

object in place:

)

T1 − m1a1 m1 g

Solve equation (1) for µk:

µk =

Substitute for µk in equation (2) and simplify to obtain the quadratic equation in m1:

2.685m12 + 9.947 m1 − 16.05 = 0

Solve the quadratic equation to obtain:

m1 = (− 1.85 ± 3.07 )kg ⇒ m1 = 1.22 kg

Substitute numerical values in equation (3) and evaluate µk:

µk =

(3)

(

13.375 N − (1.22 kg ) 4.66 m/s 2 (1.22 kg ) 9.81m/s2

= 0.643

(

)

)

Applications of Newton’s Laws 373 *117 ••• Picture the Problem The diagram shows a point on the surface of the earth at latitude θ. The distance R to the axis of rotation is given by R = rcosθ. We can use the definition of centripetal acceleration to express the centripetal acceleration of a point on the surface of the earth due to the rotation of the earth. (a) Referring to the figure, express ac for a point on the surface of the earth at latitude θ :

v2 ac = where R = rcosθ R

Express the speed of the point due to the rotation of the earth:

v=

Substitute for v in the expression for ac and simplify to obtain:

ac =

Substitute numerical values and evaluate ac:

4π 2 (6370 km )cosθ ac = [(24 h )(3600 s/h )] 2

2πR T where T is the time for one revolution.

=

4π 2 r cosθ T2

(3.37 cm/s )cosθ , toward the 2

earth' s axis.

(b)

A stone dropped from a hand at a location on earth. The effective weight of r r the stone is equal to mast, surf , where ast, surf is the acceleration of the falling stone (neglecting air resistance) relative to the local surface of the earth. The r r gravitational force on the stone is equal to mast,iner , where ast, iner is the acceleration of the local surface of the earth relative to the inertial frame (the acceleration of the surface due to the rotation of the earth). Multiplying r r r through this equation by m and rearranging gives mast, surf = mast, iner − masurf, iner , which relates the apparent weight to the acceleration due to gravity and the acceleration due to the earth' s rotation. A vector addition diagram can be used r r to show that the magnitude of mast, surf is slightly less than that of mast, iner .

374 Chapter 5 (c) At the equator, the gravitational acceleration and the radial acceleration are both directed toward the center of the earth. Therefore:

g = g eff + ac

(

)

= 978 cm/s2 + 3.37 cm/s 2 cos0° = 981.4 cm/s 2

At latitude θ the gravitational acceleration points toward the center of the earth whereas the centripetal acceleration points toward the axis of rotation. Use the law of cosines to relate geff, g, and a c:

2 g eff = g 2 + a c2 − 2 ga c cosθ

Substitute for θ, geff, and ac and simplify to obtain the quadratic equation:

g 2 − 4.75 cm/s 2 g − 962350 cm 2 /s 4 = 0

Solve for the physically meaningful (i.e., positive) root to obtain:

g = 983 cm/s 2

(

)

*118 ••• Picture the Problem The diagram shows the block in its initial position, an intermediate position, and as it is separating from the sphere. Because the sphere is frictionless, the only forces acting on the block are the normal and gravitational forces. We’ll apply Newton’s 2nd law and set Fn equal to zero to determine the angle θc at which the block leaves the surface. Taking the inward direction to be positive, apply Fr = mar to the

∑

v2 mg cosθ − Fn = m R

block: Apply the separation condition to obtain: Solve for cosθc:

Apply

∑F

t

= mat to the block:

mg cos θ c = m v2 cosθ c = gR

mg sin θ = mat or

v2 R (1)

Applications of Newton’s Laws 375 at =

dv = g sin θ dt

Note that a is not constant and, hence, we cannot use constant-acceleration equations. Multiply the left-hand side of the equation by one in the form of dθ/dθ and rearrange to obtain:

dv dθ = g sin θ dt dθ and

dθ dv = g sin θ dt dθ

Relate the arc distance s the block travels to the angle θ and the radius R of the sphere:

θ=

dθ 1 ds v s = = and R dt R dt R

where v is the block’s instantaneous speed.

Substitute to obtain:

v dv = g sin θ R dθ

Separate the variables and integrate from v′ = 0 to v and θ = 0 to θc:

v

θc

0

0

∫ v'dv' = gR ∫ sin θdθ or

v 2 = 2 gR(1 − cosθ c )

Substitute in equation (1) to obtain:

Solve for and evaluate θc:

2 gR(1 − cos θ c ) gR = 2(1 − cos θ c )

cos θ c =

⎛2⎞ ⎝3⎠

θ c = cos −1 ⎜ ⎟ = 48.2°

376 Chapter 5

Chapter 6 Work and Energy Conceptual Problems *1 • Determine the Concept A force does work on an object when its point of application moves through some distance and there is a component of the force along the line of motion. (a) False. The net force acting on an object is the vector sum of all the forces acting on the object and is responsible for displacing the object. Any or all of the forces contributing to the net force may do work. (b) True. The object could be at rest in one reference frame and moving in another. If we consider only the frame in which the object is at rest, then, because it must undergo a displacement in order for work to be done on it, we would conclude that the statement is true. (c) True. A force that is always perpendicular to the velocity of a particle changes neither it’s kinetic nor potential energy and, hence, does no work on the particle. 2 • Determine the Concept If we ignore the work that you do in initiating the horizontal motion of the box and the work that you do in bringing it to rest when you reach the second table, then neither the kinetic nor the potential energy of the system changed as you moved the box across the room. Neither did any forces acting on the box produce displacements. Hence, we must conclude that the minimum work you did on the box is zero. 3 • False. While it is true that the person’s kinetic energy is not changing due to the fact that she is moving at a constant speed, her gravitational potential energy is continuously changing and so we must conclude that the force exerted by the seat on which she is sitting is doing work on her. *4 • Determine the Concept The kinetic energy of any object is proportional to the square of its speed. Because K = 12 mv 2 , replacing v by 2v yields

(

)

K' = 12 m(2v ) = 4 12 mv 2 = 4 K . Thus doubling the speed of a car quadruples its kinetic 2

energy.

371

372 Chapter 6 5 • r Determine the Concept No. The work done on any object by any force F is defined as

r r r dW = F ⋅ dr . The direction of Fnet is toward the center of the circle in which the object r is traveling and dr is tangent to the circle. No work is done by the net force because r r Fnet and dr are perpendicular so the dot product is zero. 6 • Determine the Concept The kinetic energy of any object is proportional to the square of its speed and is always positive. Because K = 12 mv 2 , replacing v by 3v yields

(

)

K' = 12 m(3v ) = 9 12 mv 2 = 9 K . Hence tripling the speed of an object increases its 2

kinetic energy by a factor of 9 and (d ) is correct. *7 • Determine the Concept The work required to stretch or compress a spring a distance x is given by W = 12 kx 2 where k is the spring’s stiffness constant. Because W ∝ x2, doubling the distance the spring is stretched will require four times as much work. 8 • Determine the Concept No. We know that if a net force is acting on a particle, the particle must be accelerated. If the net force does no work on the particle, then we must conclude that the kinetic energy of the particle is constant and that the net force is acting perpendicular to the direction of the motion and will cause a departure from straight-line motion. 9 • Determine the Concept We can use the definition of power as the scalar product of force and velocity to express the dimension of power.

r r

Power is defined as:

P ≡ F ⋅v

Express the dimension of force:

[M][L/T 2]

Express the dimension of velocity:

[L/T]

Express the dimension of power in terms of those of force and velocity:

[M][L/T 2][L/T] = [M][L]2/[T]3 and (d ) is correct.

Work and Energy 373 10 • Determine the Concept The change in gravitational potential energy, over elevation changes that are small enough so that the gravitational field can be considered constant, is mg∆h, where ∆h is the elevation change. Because ∆h is the same for both Sal and Joe, their gains in gravitational potential energy are the same. (c) is correct. 11 • (a) False. The definition of work is not limited to displacements caused by conservative forces. (b) False. Consider the work done by the gravitational force on an object in freefall. (c) True. This is the definition of work done by a conservative force. *12 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x; i.e., Fx = − dU dx . (a) Examine the slopes of the curve at each of the lettered points, remembering that Fx is the negative of the slope of the potential energy graph, to complete the table:

(b) Find the point where the slope is steepest:

Point dU/dx Fx + A − 0 0 B + C − 0 0 D + E − 0 0 F

At point C Fx is greatest.

(c) If d2U/dx2 < 0, then the curve is concave downward and the equilibrium is unstable.

At point B the equilibrium is unstable.

If d2U/dx2 > 0, then the curve is concave upward and the equilibrium is stable.

At point D the equilibrium is stable.

Remarks: At point F, d2U/dx2 = 0 and the equilibrium is neither stable nor unstable; it is said to be neutral.

374 Chapter 6 13 • (a) False. Any force acting on an object may do work depending on whether the force produces a displacement … or is displaced as a consequence of the object’s motion. (b) False. Consider an element of area under a force-versus-time graph. Its units are N⋅s whereas the units of work are N⋅m. 14 • r r r Determine the Concept Work dW = F ⋅ ds is done when a force F produces a

(

)

r r r displacement ds . Because F ⋅ ds ≡ Fds cos θ = (F cos θ ) ds, W will be negative if the value of θ is such that Fcosθ is negative. (d ) is correct.

Estimation and Approximation *15 •• Picture the Problem The diagram depicts the situation when the tightrope walker is at the center of rope. M represents her mass and the vertical components of tensions

r r T1 and T2 , equal in magnitude, support her weight. We can apply a condition for static equilibrium in the vertical direction to relate the tension in the rope to the angle θ and use trigonometry to find s as a function of θ.

(a) Use trigonometry to relate the sag s in the rope to its length L and θ: Apply

∑F

y

= 0 to the tightrope

walker when she is at the center of the rope to obtain: Solve for θ to obtain:

Substitute numerical values and evaluate θ :

tan θ =

s L and s = tan θ 1 2 2 L

2T sin θ − Mg = 0 where T is the r r magnitude of T1 and T2 . ⎛ Mg ⎞ ⎟ ⎝ 2T ⎠

θ = sin −1 ⎜

(

)

⎡ (50 kg ) 9.81 m/s 2 ⎤ ⎥ = 2.81° 2(5000 N ) ⎣ ⎦

θ = sin −1 ⎢

Work and Energy 375 Substitute to obtain:

s=

(b) Express the change in the tightrope walker’s gravitational potential energy as the rope sags:

∆U = U at center − U end = Mg∆y

Substitute numerical values and evaluate ∆U:

10 m tan 2.81° = 0.245 m 2

(

)

∆U = (50 kg ) 9.81 m/s 2 (− 0.245 m ) = − 120 J

16 • Picture the Problem You can estimate your change in potential energy due to this change in elevation from the definition of ∆U. You’ll also need to estimate the height of one story of the Empire State building. We’ll assume your mass is 70 kg and the height of one story to be 3.5 m. This approximation gives us a height of 1170 ft (357 m), a height that agrees to within 7% with the actual height of 1250 ft from the ground floor to the observation deck. We’ll also assume that it takes 3 min to ride non-stop to the top floor in one of the high-speed elevators. (a) Express the change in your gravitational potential energy as you ride the elevator to the 102nd floor: Substitute numerical values and evaluate ∆U:

∆U = mg∆h

(

)

∆U = (70 kg ) 9.81m/s 2 (357 m ) = 245 kJ

(b) Ignoring the acceleration intervals at the beginning and the end of your ride, express the work done on you by the elevator in terms of the change in your gravitational potential energy:

W = Fh = ∆U

Solve for and evaluate F:

F=

∆U 245 kJ = = 686 N h 357 m

(c) Assuming a 3 minute ride to the top, express and evaluate the average power delivered to the elevator:

P=

∆U 245 kJ = (3 min )(60 s/min ) ∆t

= 1.36 kW

376 Chapter 6 17 • Picture the Problem We can find the kinetic energy K of the spacecraft from its definition and compare its energy to the annual consumption in the U.S. W by examining the ratio K/W.

(10000 kg )(3 ×107 m/s) 2

Using its definition, express and evaluate the kinetic energy of the spacecraft:

K = 12 mv 2 =

Express this amount of energy as a percentage of the annual consumption in the United States:

K 4.50 × 1018 J ≈ ≈ 1% E 5 × 10 20 J

1 2

= 4.50 × 1018 J

*18 •• Picture the Problem We can find the orbital speed of the Shuttle from the radius of its orbit and its period and its kinetic energy from K = 12 mv 2 . We’ll ignore the variation in the acceleration due to gravity to estimate the change in the potential energy of the orbiter between its value at the surface of the earth and its orbital value. (a) Express the kinetic energy of the orbiter:

K = 12 mv 2

Relate the orbital speed of the orbiter to its radius r and period T:

v=

Substitute and simplify to obtain:

2π 2 mr 2 ⎛ 2π r ⎞ K = m⎜ ⎟ = T2 ⎝ T ⎠

2π r T 2

1 2

Substitute numerical values and evaluate K:

K=

(

)

2π 2 8 × 10 4 kg [(200 mi + 3960 mi )(1.609 km/mi)] = 2.43 TJ [(90 min )(60 s / min )] 2

(b) Assuming the acceleration due to gravity to be constant over the 200 mi and equal to its value at the surface of the earth (actually, it is closer to 9 m/s2 at an elevation of 200 mi), express the change in gravitational potential energy of the orbiter, relative to the surface of the earth, as the Shuttle goes into orbit:

2

∆U = mgh

Work and Energy 377 Substitute numerical values and evaluate ∆U:

(

)(

)

∆U = 8 ×10 4 kg 9.81 m/s 2 × (200 mi )(1.609 km/mi) = 0.253 TJ

No, they shouldn' t be equal because there is more than just the force of gravity to consider here. When the shuttle is resting on the surface of (c)

the earth, it is supported against the force of gravity by the normal force the earth exerts upward on it. We would need to take into consideration the change in potential energy of the surface of earth in its deformation under the weight of the shuttle to find the actual change in potential energy.

19 • Picture the Problem Let’s assume that the width of the driveway is 18 ft. We’ll also assume that you lift each shovel full of snow to a height of 1 m, carry it to the edge of the driveway, and drop it. We’ll ignore the fact that you must slightly accelerate each shovel full as you pick it up and as you carry it to the edge of the driveway. While the density of snow depends on the extent to which it has been compacted, one liter of freshly fallen snow is approximately equivalent to 100 mL of water. Express the work you do in lifting the snow a distance h:

W = ∆U = mgh = ρ snowVsnow gh where ρ is the density of the snow.

Using its definition, express the densities of water and snow:

ρ snow =

Divide the first of these equations by the second to obtain:

ρ snow Vwater V or ρ snow = ρ water water = ρ water Vsnow Vsnow

Substitute and evaluate the ρsnow:

ρ snow = (103 kg/m 3 )

Calculate the volume of snow covering the driveway:

⎛ 10 ⎞ Vsnow = (50 ft )(18 ft )⎜ ft ⎟ ⎝ 12 ⎠ 28.32 L 10−3 m 3 = 750 ft 3 × × ft 3 L 3 = 21.2 m

Substitute numerical values in the expression for W to obtain an estimate (a lower bound) for the work you would do on the snow in removing it:

msnow m and ρ water = water Vsnow Vwater

(

)(

100 mL = 100 kg/m 3 L

)(

)

W = 100 kg/m 3 21.2 m 3 9.81 m/s 2 (1 m ) = 20.8 kJ

378 Chapter 6

Work and Kinetic Energy *20 • Picture the Problem We can use

1 2

mv 2 to find the kinetic energy of the bullet. K = 12 mv 2

(a) Use the definition of K:

=

1 2

(0.015 kg )(1.2 ×103 m/s) 2

= 10.8 kJ (b) Because K ∝ v2:

K' = 14 K = 2.70 kJ

(c) Because K ∝ v2:

K' = 4 K = 43.2 kJ

21 • Picture the Problem We can use

1 2

mv 2 to find the kinetic energy of the baseball and the

jogger. (a) Use the definition of K:

K = 12 mv 2 =

1 2

(0.145 kg )(45 m/s)2

= 147 J (b) Convert the jogger’s pace of 9 min/mi into a speed:

⎛ 1 mi ⎞ ⎛ 1 min ⎞ ⎛ 1609 m ⎞ ⎟⎟ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ v = ⎜⎜ ⎝ 9 min ⎠ ⎝ 60 s ⎠ ⎝ 1 mi ⎠ = 2.98 m/s

Use the definition of K:

K = 12 mv 2 =

1 2

(60 kg )(2.98 m/s)2

= 266 J 22 • Picture the Problem The work done in raising an object a given distance is the product of the force producing the displacement and the displacement of the object. Because the weight of an object is the gravitational force acting on it and this force acts downward, the work done by gravity is the negative of the weight of the object multiplied by its displacement. The change in kinetic energy of an object is equal to the work done by the net force acting on it. (a) Use the definition of W:

r r W = F ⋅ ∆y = F∆y

Work and Energy 379 = (80 N)(3 m) = 240 J (b) Use the definition of W:

r r r W = F ⋅ ∆y = −mg∆y, because F and r ∆y are in opposite directions. ∴ W = − (6 kg)(9.81 m/s2)(3 m) = − 177 J

(c) According to the work-kinetic energy theorem:

K = W + Wg = 240 J + (−177 J) = 63.0 J

23 • Picture the Problem The constant force of 80 N is the net force acting on the box and the work it does is equal to the change in the kinetic energy of the box.

(

Using the work-kinetic energy theorem, relate the work done by the constant force to the change in the kinetic energy of the box:

W = K f − K i = 12 m vf2 − vi2

Substitute numerical values and evaluate W:

W =

1 2

)

(5 kg )[(68 m/s )2 − (20 m/s )2 ]

= 10.6 kJ

*24 •• Picture the Problem We can use the definition of kinetic energy to find the mass of your friend. Using the definition of kinetic energy and letting ″1″ denote your mass and speed and ″2″ your girlfriend’s, express the equality of your kinetic energies and solve for your girlfriend’s mass as a function of both your masses and speeds: Express the condition on your speed that enables you to run at the same speed as your girlfriend:

1 2

m1v12 = 12 m2 v22

and

⎛v ⎞ m2 = m1 ⎜⎜ 1 ⎟⎟ ⎝ v2 ⎠

v2 = 1.25v1

2

(1)

(2)

380 Chapter 6 Substitute equation (2) in equation (1) to obtain:

2

⎛v ⎞ ⎛ 1 ⎞ m2 = m1 ⎜⎜ 1 ⎟⎟ = (85 kg )⎜ ⎟ ⎝ 1.25 ⎠ ⎝ v2 ⎠

2

= 54.4 kg

Work Done by a Variable Force 25 •• Picture the Problem The pictorial representation shows the particle as it moves along the positive x axis. The particle’s kinetic energy increases because work is done on it. We can calculate the work done on it from the graph of Fx vs. x and relate its kinetic energy when it is at x = 4 m to its kinetic energy when it was at the origin and the work done on it by using the work-kinetic energy theorem.

(a) Calculate the kinetic energy of the particle when it is at x = 0:

(b) Because the force and displacement are parallel, the work done is the area under the curve. Use the formula for the area of a triangle to calculate the area under the F as a function of x graph:

K 0 = 12 mv 2 =

1 2

(3 kg )(2 m/s) 2

= 6.00 J

(base)(altitude) = 12 (4 m )(6 N )

W0→4 =

1 2

= 12.0 J

(c) Express the kinetic energy of the particle at x = 4 m in terms of its speed and mass and solve for its speed:

v4 =

Using the work-kinetic energy theorem, relate the work done on the particle to its change in kinetic energy and solve for the particle’s kinetic energy at x = 4 m:

W0→4 = K4 – K0

2K 4 m

(1)

K 4 = K 0 + W0→4 = 6.00 J + 12.0 J = 18.0 J

Work and Energy 381 Substitute numerical values in equation (1) and evaluate v4:

2(18.0 J ) = 3.46 m/s 3 kg

v4 =

*26 •• Picture the Problem The work done by this force as it displaces the particle is the area under the curve of F as a function of x. Note that the constant C has units of N/m3. Because F varies with position nonlinearly, express the work it does as an integral and evaluate the integral between the limits x = 1.5 m and x = 3 m:

(

) ∫ x' 3m

W = C N/m

3

3

dx'

1.5 m

( ) [ x' ] (C N/m ) [(3 m) − (1.5 m) ] = = C N/m 3

3m

4

1 4

1.5 m

3

4

4

4

= 19C J 27 •• Picture the Problem The work done on the dog by the leash as it stretches is the area under the curve of F as a function of x. We can find this area (the work Lou does holding the leash) by integrating the force function. Because F varies with position nonlinearly, express the work it does as an integral and evaluate the integral between the limits x = 0 and x = x1:

x1

(

)

W = ∫ − kx'−ax'2 dx' 0

[

= − 12 kx'2 − 13 ax'3

]

x1 0

= − 12 kx12 − 13 ax13 28 •• Picture the Problem The work done on an object can be determined by finding the area bounded by its graph of Fx as a function of x and the x axis. We can find the kinetic energy and the speed of the particle at any point by using the work-kinetic energy theorem. (a) Express W, the area under the curve, in terms of the area of one square, Asquare, and the number of squares n:

W = n Asquare

Determine the work equivalent of one square:

W = (0.5 N)(0.25 m) = 0.125 J

382 Chapter 6 Estimate the number of squares under the curve between x = 0 and x = 2 m:

n ≈ 22

Substitute to determine W:

W = 22(0.125 J) = 2.75 J

(b) Relate the kinetic energy of the object at x = 2 m, K2, to its initial kinetic energy, K0, and the work that was done on it between x = 0 and x = 2 m:

K 2 = K 0 + W0→2 =

v=

(d) Estimate the number of squares under the curve between x = 0 and x = 4 m:

n ≈ 26

(e) Relate the kinetic energy of the object at x = 4 m, K4, to its initial kinetic energy, K0, and the work that was done on it between x = 0 and x = 4 m: Calculate the speed of the object at x = 4 m from its kinetic energy at the same location:

(3 kg )(2.40 m/s)2 + 2.75 J

= 11.4 J

(c) Calculate the speed of the object at x = 2 m from its kinetic energy at the same location:

Substitute to determine W:

1 2

2K 2 = m

2(11.4 J ) = 2.76 m/s 3 kg

W = 26(0.125 J ) = 3.25 J K 4 = K 0 + W0→4 =

1 2

(3 kg )(2.40 m/s )2 + 3.25 J

= 11.9 J

v=

2K 4 = m

2(11.9 J ) = 2.82 m/s 3 kg

*29 •• Picture the Problem We can express the mass of the water in Margaret’s bucket as the difference between its initial mass and the product of the rate at which it loses water and her position during her climb. Because Margaret must do work against gravity in lifting and carrying the bucket, the work she does is the integral of the product of the gravitational field and the mass of the bucket as a function of its position. (a) Express the mass of the bucket and the water in it as a function of

m( y ) = 40 kg − ry

Work and Energy 383 its initial mass, the rate at which it is losing water, and Margaret’s position, y, during her climb: Find the rate, r =

∆m , at which ∆y

r=

∆m 20 kg = = 1 kg/m ∆y 20 m

Margaret’s bucket loses water:

m( y ) = 40 kg − ry = 40 kg −

Substitute to obtain:

1 kg y m

(b) Integrate the force Margaret exerts on the bucket, m(y)g, between the limits of y = 0 and y = 20 m: 20 m

W=g

∫ 0

(

)[

1 kg ⎞ ⎛ y ' ⎟dy ' = 9.81 m/s 2 (40 kg ) y '− 12 (1 kg/m ) y '2 ⎜ 40 kg − m ⎠ ⎝

]

20 m 0

= 5.89 kJ

Remarks: We could also find the work Margaret did on the bucket, at least approximately, by plotting a graph of m(y)g and finding the area under this curve between y = 0 and y = 20 m.

Work, Energy, and Simple Machines 30 • Picture the Problem The free-body diagram shows the forces that act on the block as it slides down the frictionless incline. We can find the work done by these forces as the block slides 2 m by finding their components in the direction of, or opposite to, the motion. When we have determined the work done on the block, we can use the work-kinetic energy theorem or a constant-acceleration equation to calculate its kinetic energy and its speed at any given location.

384 Chapter 6 From the free - body diagram, we see that the forces acting on the block are (a) a gravitational force that acts downward and the normal force that the incline exerts perpendicularly to the incline. Identify the component of mg that acts down the incline and calculate the work done by it: Express the work done by this force: Substitute numerical values and evaluate W:

Fx = mg sin 60°

W = Fx ∆x = mg∆x sin 60°

(

)

W = (6 kg ) 9.81 m/s 2 (2 m )sin 60° = 102 J

Remarks: Fn and mgcos60°, being perpendicular to the motion, do no work on the block (b) The total work done on the block is the work done by the net force:

W = Fnet ∆x = mg∆x sin 60°

(

)

= (6 kg ) 9.81 m/s 2 (2 m )sin 60° = 102 J

(c) Express the change in the kinetic energy of the block in terms of the distance, ∆x, it has moved down the incline:

∆K = Kf – Ki = W = (mgsin60°)∆x or, because Ki = 0, Kf = W = (mgsin60°)∆x

Relate the speed of the block when it has moved a distance ∆x down the incline to its kinetic energy at that location:

v=

Determine this speed when ∆x = 1.5 m:

(d) As in part (c), express the change in the kinetic energy of the block in terms of the distance, ∆x, it has moved down the incline and

2K = m

2mg∆x sin 60° m

= 2 g∆x sin 60°

(

)

v = 2 9.81 m/s 2 (1.5 m )sin 60° = 5.05 m/s ∆K = Kf – Ki =W = (mg sin 60°)∆x and

Work and Energy 385 solve for Kf:

Kf = (mg sin 60°)∆x + Ki

Substitute for the kinetic energy terms and solve for vf to obtain:

vf = 2 g sin 60°∆x + vi2

Substitute numerical values and evaluate vf:

(

)

vf = 2 9.81 m/s 2 (1.5 m ) sin60° + (2 m/s ) = 5.43 m/s 2

31 • Picture the Problem The free-body diagram shows the forces acting on the object as in moves along its circular path on a frictionless horizontal surface. We can use Newton’s 2nd law to obtain an expression for the tension in the string and the definition of work to determine the amount of work done by each force during one revolution. (a) Apply

∑F

r

= mar to the 2-kg

object and solve for the tension:

(2.5 m/s) v2 T = m = (2 kg ) r 3m

2

= 4.17 N (b) From the FBD we can see that the forces acting on the object are:

r r r T , Fg , and Fn

Because all of these forces act perpendicularly to the direction of motion of the object, none of them do any work.

386 Chapter 6 *32 • Picture the Problem The free-body r diagram, with F representing the force required to move the block at constant speed, shows the forces acting on the block. We can apply Newton’s 2nd law to the block to relate F to its weight w and then use the definition of the mechanical advantage of an inclined plane. In the second part of the problem we’ll use the definition of work. (a) Express the mechanical advantage M of the inclined plane: Apply

∑F

x

= ma x to the block:

M =

w F

F − w sin θ = 0 because ax = 0.

Solve for F and substitute to obtain:

M =

Refer to the figure to obtain:

sin θ =

Substitute to obtain:

M =

w 1 = w sin θ sin θ H L

1 L = sin θ H

(b) Express the work done pushing the block up the ramp:

Wramp = FL = mgL sin θ

Express the work done lifting the block into the truck:

Wlifting = mgH = mgL sin θ and

Wramp = Wlifting 33 • Picture the Problem We can find the work done per revolution in lifting the weight and the work done in each revolution of the handle and then use the definition of mechanical advantage. Express the mechanical advantage of the jack:

M =

Express the work done by the jack in one complete revolution (the weight W is raised a distance p):

Wlifting = Wp

Express the work done by the force F in one complete revolution:

Wturning = 2π RF

W F

Work and Energy 387 Equate these expressions to obtain: Solve for the ratio of W to F:

Wp = 2π RF

M =

2π R W = F p

Remarks: One does the same amount of work turning as lifting; exerting a smaller force over a greater distance. 34 • r Picture the Problem The object whose weight is w is supported by two portions of the rope resulting in what is known as a mechanical advantage of 2. The work that is done in each instance is the product of the force doing the work and the displacement of the object on which it does the work. (a) If w moves through a distance h:

F moves a distance 2h

(b) Assuming that the kinetic energy of the weight does not change, relate the work done on the object to the change in its potential energy to obtain:

W = ∆U = wh cos θ = wh

(c) Because the force you exert on the rope and its displacement are in the same direction: Determine the tension in the ropes supporting the object:

W = F (2h )cosθ = F (2h )

∑F

vertical

= 2F − w = 0

and

F = 12 w Substitute for F:

W = F (2h ) = 12 w (2h ) = wh

(d) The mechanical advantage of the inclined plane is the ratio of the weight that is lifted to the force required to lift it, i.e.:

M =

w w = 1 = 2 F 2w

Remarks: Note that the mechanical advantage is also equal to the number of ropes supporting the load.

388 Chapter 6

Dot Products *35 • r r Picture the Problem Because A ⋅ B ≡ AB cos θ we can solve for cosθ and use the fact

r r

that A ⋅ B = − AB to find θ.

r r A⋅ B θ = cos AB

Solve for θ :

−1

r r

Substitute for A ⋅ B and evaluate θ :

θ = cos −1 (− 1) = 180°

36 • r r Picture the Problem We can use its definition to evaluate A ⋅ B .

r r

Express the definition of A ⋅ B : Substitute numerical values and r r evaluate A ⋅ B :

r r A ⋅ B = AB cos θ r r A ⋅ B = (6 m )(6 m )cos 60° = 18.0 m 2

37 • r r Picture the Problem The scalar product of two-dimensional vectors A and B is AxBx + A yB y.

r

(a) For A = 3 iˆ − 6 ˆj and

r B = −4 iˆ + 2 ˆj : r

(b) For A = 5 iˆ + 5 ˆj and

r B = 2 iˆ −4 ˆj :

r

r

(c) For A = 6 iˆ + 4 ˆj and B = 4 iˆ − 6 ˆj :

r r A ⋅ B = (3)( −4) + (−6)(2) = − 24

r r A ⋅ B = (5)(2) + (5)( −4) = − 10

r r A ⋅ B = (6)(4) + (4)( −6) = 0

Work and Energy 389 38 • r r Picture the Problem The scalar product of two-dimensional vectors A and B is AB cos θ r r = AxBx + AyBy. Hence the angle between vectors A and B is given by

θ = cos −1

Ax B x + Ay B y AB

. r r A ⋅ B = (3)( −4) + (−6)(2) = −24

r

(a) For A = 3 iˆ − 6 ˆj and

r B = −4 iˆ + 2 ˆj :

A= B=

(3)2 + (− 6)2 (− 4)2 + (2)2

and

θ = cos −1

= 20

− 24 = 143° 45 20

r r A ⋅ B = (5)(2) + (5)(−4) = -10

r

(b) For A = 5 iˆ + 5 ˆj and

r B = 2 iˆ − 4 ˆj :

A= B=

(5)2 + (5)2 = 50 (2)2 + (− 4)2 = 20

and

θ = cos −1 r

= 45

− 10 = 108° 50 20

r r A ⋅ B = (6)(4) + (4)( −6) = 0

r

(c) For A = 6 iˆ + 4 ˆj and B = 4 iˆ − 6 ˆj :

A= B=

(6)2 + (4)2 = 52 (4)2 + (− 6)2 = 52

and

θ = cos −1

0 = 90.0° 52 52

39 • r r Picture the Problem The work W done by a force F during a displacement ∆ s for

r

r

which it is responsible is given by F ⋅∆ s .

390 Chapter 6 r r W = F ⋅ ∆s = 2 N iˆ − 1 N ˆj + 1 N kˆ

(a) Using the definitions of work and the scalar product, calculate the work done by the given force during the specified displacement:

(

(

)

)

⋅ 3 m iˆ + 3 m ˆj − 2 m kˆ = [(2)(3) + (− 1)(3) + (1) (− 2)] N ⋅ m = 1.00 J W = F∆s cos θ = (F cos θ )∆s

(b) Using the definition of work that includes the angle between the force and displacement vectors, solve for r the component of F in the direction r of ∆ s :

and

F cosθ =

Substitute numerical values and evaluate Fcosθ :

F cos θ =

W ∆s

1J

(3 m )2 + (3 m )2 + (− 2 m )2

= 0.213 N 40 •• Picture the Problem The component of a vector that is along another vector is the scalar product of the former vector and a unit vector that is parallel to the latter vector.

r A Ax iˆ + Ay ˆj + Az kˆ = uˆ A = A Ax2 + Ay2 + Az2

(a) By definition, the unit vector r that is parallel to the vector A is:

r

r B uˆ B = = B

(b) Find the unit vector parallel to B :

r

r

3iˆ + 4 ˆj

(3)

(

2

+ (4 )

2

=

3ˆ 4 ˆ i+ j 5 5

)

r 4 ⎞ ⎛3 A ⋅ uˆ B = 2iˆ − ˆj − kˆ ⋅ ⎜ iˆ + ˆj ⎟ 5 ⎠ ⎝5 ⎛3⎞ ⎛4⎞ = (2)⎜ ⎟ + (− 1)⎜ ⎟ + (− 1)(0) ⎝5⎠ ⎝5⎠

The component of A along B is:

= 0.400 *41 •• Picture the Problem We can use the definitions of the magnitude of a vector and the dot

r

v

r

v

r

r

product to show that if A + B = A − B , then A ⊥ B .

Work and Energy 391

(

)

2

(

)

2

r

r2

r r2 r r A+ B = A+ B

Express A − B :

r

v

r r2 r r A− B = A− B

Equate these expressions to obtain:

(Ar + Bv ) = (Ar − Bv )

Express A + B :

Expand both sides of the equation to obtain:

2

2

r r r r A2 + 2 A ⋅ B + B 2 = A2 − 2 A ⋅ B + B 2

r r 4A ⋅ B = 0

Simplify to obtain:

or

r r A⋅ B = 0

From the definition of the dot product we have:

r

v

Because neither A nor B is the zero vector:

r r A ⋅ B = AB cos θ r v where θ is the angle between A and B. r r cos θ = 0 ⇒ θ = 90° and A ⊥ B.

42 •• Picture the Problem The diagram shows

ˆ and Bˆ arbitrarily the unit vectors A st located in the 1 quadrant. We can express these vectors in terms of the unit vectors iˆ and ˆj and their x and y components. We can then form the dot product of ˆ and Bˆ to show that A cos(θ1 − θ2) = cosθ1cosθ2 + sinθ1sinθ2.

ˆ in terms of the unit (a) Express A

ˆ = A iˆ + A ˆj A x y

vectors iˆ and ˆj :

where

Ax = cos θ1 and Ay = sin θ1 Proceed as above to obtain:

Bˆ = Bxiˆ + By ˆj where

Bx = cos θ 2

ˆ ⋅ Bˆ : (b) Evaluate A

(

and B y = sin θ 2

)

Aˆ ⋅ Bˆ = cos θ1iˆ + sin θ1 ˆj ⋅ cos θ 2 iˆ + sin θ 2 ˆj

(

)

= cos θ1 cos θ 2 + sin θ1 sin θ 2 From the diagram we note that:

Aˆ ⋅ Bˆ = cos(θ1 − θ 2 )

392 Chapter 6 Substitute to obtain:

cos(θ1 − θ 2 ) = cos θ1 cos θ 2 + sin θ1 sin θ 2

43 • r r Picture the Problem In (a) we’ll show that it does not follow that B = C by giving a counterexample.

r

r

Let A = iˆ , B = 3iˆ + 4 ˆj and

r r r r r C = 3iˆ − 4 ˆj. Form A ⋅ B and A ⋅ C :

( ) and r r A ⋅ C = iˆ ⋅ (3iˆ − 4 ˆj ) = 3

r r A ⋅ B = iˆ ⋅ 3iˆ + 4 ˆj = 3

No. We' ve shown by a counter r example that B is not necessarily r equal to C . 44 •• r r Picture the Problem We can form the dot product of A and r and require that r r r r lie on a straight line. A ⋅ r = 1 to show that the points at the head of all such vectors r We can use the equation of this line and the components of A to find the slope and intercept of the line.

r

(a) Let A = a x iˆ + a y ˆj . Then:

(

)(

r r A ⋅ r = a x iˆ + a y ˆj ⋅ x iˆ + y ˆj

)

= ax x + a y y = 1 Solve for y to obtain:

y= −

ax 1 x+ ay ay

which is of the form y = mx + b and hence is the equation of a line.

r

(b) Given that A = 2 iˆ − 3 ˆj :

m=−

ax 2 2 =− = ay 3 −3

and

b=

1 1 1 = = − ay − 3 3

Work and Energy 393 (c) The equation we obtained in (a) specifies all vectors whose component r parallel to A has constant magnitude; therefore, we can write such a vector as

r r r r A r = r 2 + B , where B is any vector A r perpendicular to A. This is shown

graphically to the right.

r

Because all possiblervectors B lie in a plane, the resultant r must lie in a plane as well, as is shown above. *45 •• Picture the Problem The rules for the differentiation of vectors are the same as those for the differentiation of scalars and scalar multiplication is commutative.

r r

(a) Differentiate r ⋅ r = r2 = constant:

r r

Because v ⋅ r = 0 :

r r

(b) Differentiate v ⋅ v = v2 = constant with respect to time:

r r

Because a ⋅ v = 0 :

r r d r r r dr dr r (r ⋅ r ) = r ⋅ + ⋅ r = 2vr ⋅ rr dt dt dt d = (constant ) = 0 dt r r v ⊥r

r r d r r r dv dv r (v ⋅ v ) = v ⋅ + ⋅ v = 2ar ⋅ vr dt dt dt d = (constant ) = 0 dt r r a ⊥v

The results of (a) and (b) tell us that r r a is perpendicular to r and and r parallel (or antiparallel) to r . r r

(c) Differentiate v ⋅ r = 0 with respect to time:

r r d r r r dr r dv (v ⋅ r ) = v ⋅ + r ⋅ dt dt dt r r d = v 2 + r ⋅ a = (0) = 0 dt

394 Chapter 6 r r

r r r ⋅ a = −v 2

Because v 2 + r ⋅ a = 0 : Express ar in terms of θ, where θ is r r the angle between r and a :

ar = a cos θ

Express r ⋅ a :

r r r ⋅ a = ra cos θ = rar

Substitute in equation (1) to obtain:

rar = −v 2

Solve for ar:

ar = −

r r

(1)

v2 r

Power 46 •• Picture the Problem The power delivered by a force is defined as the rate at which the force does work; i.e., P =

dW . dt

Calculate the rate at which force A does work:

PA =

5J = 0.5 W 10 s

Calculate the rate at which force B does work:

PB =

3J = 0.6 W and PB > PA 5s

47 • Picture the Problem The power delivered by a force is defined as the rate at which the force does work; i.e., P =

r r dW = F ⋅ v. dt

(a) If the box moves upward with a constant velocity, the net force acting it must be zero and the force that is doing work on the box is: The power input of the force is: Substitute numerical values and evaluate P:

F = mg

P = Fv = mgv

(

)

P = (5 kg ) 9.81 m/s 2 (2 m/s ) = 98.1 W

Work and Energy 395 (b) Express the work done by the force in terms of the rate at which energy is delivered:

W = Pt = (98.1 W) (4 s) = 392 J

48 • Picture the Problem The power delivered by a force is defined as the rate at which the force does work; i.e., P =

r r dW = F ⋅ v. dt

P 6W = = 2 m/s F 3N

(a) Using the definition of power, express Fluffy’s speed in terms of the rate at which he does work and the force he exerts in doing the work:

v=

(b) Express the work done by the force in terms of the rate at which energy is delivered:

W = Pt = (6 W) (4 s) = 24.0 J

49 • Picture the Problem We can use Newton’s 2nd law and the definition of acceleration to express the velocity of this object as a function of time. The power input of the force accelerating the object is defined to be the rate at which it does work; i.e.,

r r P = dW dt = F ⋅ v .

(a) Express the velocity of the object as a function of its acceleration and time: Apply

r

r

∑ F = ma to the object:

Substitute for a in the expression for v: (b) Express the power input as a function of F and v and evaluate P: (c) Substitute t = 3 s:

v = at

a = F/m

v=

F 5N t= t= m 8 kg

(

(

5 8

)

m/s 2 t

)

P = Fv = (5 N ) 85 m/s 2 t = 3.13t W/s P = (3.13 W/s )(3 s ) = 9.38 W

396 Chapter 6 50 • Picture the Problem The power delivered by a force is defined as the rate at which the force does work; i.e., P =

r r dW = F ⋅ v. dt

r

(a) For F = 4 N iˆ + 3 N kˆ and

r v = 6 m/s iˆ :

(

)(

r r P = F ⋅ v = 4 N iˆ + 3 N kˆ ⋅ 6 m/s iˆ

)

= 24.0 W

r

(b) For F = 6 N iˆ − 5 N ˆj and

r v = − 5 m/s iˆ + 4 m/s ˆj :

r r P = F ⋅v

(

)(

= 6 N iˆ − 5 N ˆj ⋅ − 5 m/s iˆ + 4 m/s ˆj

)

= − 50.0 W r

(c) For F = 3 N iˆ + 6 N ˆj

r and v = 2 m/s iˆ + 3 m/s ˆj :

r r P = F ⋅v = 3 N iˆ + 6 N ˆj ⋅ 2 m/s iˆ + 3 m/s ˆj

(

)(

)

= 24.0 W *51 • Picture the Problem Choose a coordinate system in which upward is the positive y direction. We can find Pin from the given information that Pout = 0.27 Pin . We can express

Pout as the product of the tension in the cable T and the constant speed v of the dumbwaiter. We can apply Newton’s 2nd law to the dumbwaiter to express T in terms of its mass m and the gravitational field g. Express the relationship between the motor’s input and output power:

Pout = 0.27 Pin or

Pin = 3.7 Pout Express the power required to move the dumbwaiter at a constant speed v: Apply

∑F

y

= ma y to the

Pout = Tv T − mg = ma y

dumbwaiter:

or, because ay = 0,

Substitute to obtain:

Pin = 3.7Tv = 3.7 mgv

Substitute numerical values and evaluate Pin:

T = mg

(

)

Pin = 3.7(35 kg ) 9.81 m/s 2 (0.35 m/s ) = 445 W

Work and Energy 397 52 •• Picture the Problem Choose a coordinate system in which upward is the positive y direction. We can express Pdrag as the product of the drag force Fdrag acting on the skydiver and her terminal velocity vt. We can apply Newton’s 2nd law to the skydiver to express Fdrag in terms of her mass m and the gravitational field g.

r r r Pdrag = Fdrag ⋅ v t r r or, because Fdrag and v t are antiparallel,

(a) Express the power due to drag force acting on the skydiver as she falls at her terminal velocity vt:

Pdrag = − Fdrag vt Apply

∑F

y

Fdrag − mg = ma y

= ma y to the skydiver:

or, because ay = 0,

Fdrag = mg

Pdrag = − mgvt

Substitute to obtain, for the magnitude of Pdrag:

(1)

Substitute numerical values and evaluate P:

Pdrag = − (55 kg) (9.81 m/s 2 ) (120

mi 1h 1.609 km × × ) = 2.89 × 10 4 W h 3600 s mi

(b) Evaluate equation (1) with v = 15 mi/h:

1h 1.609 km ⎛ mi ⎞ Pdrag = − (55 kg )(9.81 m/s 2 ) ⎜15 ⎟ × × ) = 3.62 kW mi ⎝ h ⎠ 3600 s *53 •• Picture the Problem Because, in the absence of air resistance, the acceleration of the cannonball is constant, we can use a constant-acceleration equation to relate its velocity to the time it has been in flight. We can apply Newton’s 2nd law to the cannonball to find

r

r

the net force acting on it and then form the dot product of F and v to express the rate at which the gravitational field does work on the cannonball. Integrating this expression over the time-of-flight T of the ball will yield the desired result. Express the velocity of the cannonball as a function of time while it is in the air: Apply

r

r

∑ F = ma to the

r v (t ) = 0iˆ + (v0 − gt ) ˆj

r F = − mg ˆj

cannonball to express the force acting on it while it is in the air:

r r

Evaluate F ⋅ v :

r r F ⋅ v = −mg ˆj ⋅ (v0 − gt ) ˆj

= −mgv0 + mg 2t

398 Chapter 6 r r

r r dW = F ⋅ v = − mgv0 + mg 2t dt

Relate F ⋅ v to the rate at which work is being done on the cannonball: Separate the variables and integrate over the time T that the cannonball is in the air:

(

T

)

W = ∫ − mgv0 + mg 2 t dt 0

(1)

= 12 mg 2T 2 − mgv0T v 2 = v02 + 2a∆y

Using a constant-acceleration equation, relate the speed v of the cannonball when it lands at the bottom of the cliff to its initial speed v0 and the height of the cliff H:

or, because a = g and ∆y = H,

v 2 = v02 + 2 gH

Solve for v to obtain:

v=

Using a constant-acceleration equation, relate the time-of-flight T to the initial and impact speeds of the cannonball:

v = v0 − gT

Solve for T to obtain:

T=

Substitute for T in equation (1) and simplify to evaluate W:

v0 + 2 gH 2

v0 − v g

W = 12 mg 2

v 02 − 2vv0 + v 2 g2

⎛v −v⎞ ⎟⎟ − mgv0 ⎜⎜ 0 ⎝ g ⎠ = 12 mv 2 − 12 mv02 = ∆K 54 •• Picture the Problem If the particle is acted on by a single force, that force is the net force acting on the particle and is responsible for its acceleration. The rate at which

r r

energy is delivered by the force is P = F ⋅ v . Express the rate at which this force

r

r

r r P = F ⋅v

does work in terms of F and v : The velocity of the particle, in terms of its acceleration and the time that the force has acted is:

r r v = at

Work and Energy 399

Using Newton’s 2nd law, substitute r for a :

r

Substitute for v in the expression for P and simplify to obtain:

r r F v= t m r r r r F F2 F ⋅F P=F⋅ t= t= t m m m

Potential Energy 55 • Picture the Problem The change in the gravitational potential energy of the earth-man system, near the surface of the earth, is given by ∆U = mg∆h, where ∆h is measured relative to an arbitrarily chosen reference position. Express the change in the man’s gravitational potential energy in terms of his change in elevation: Substitute for m, g and ∆h and evaluate ∆U:

∆U = mg∆h

∆U = (80 kg ) (9.81 m/s 2 ) (6 m ) = 4.71 kJ

56 • Picture the Problem The water going over the falls has gravitational potential energy relative to the base of the falls. As the water falls, the falling water acquires kinetic energy until, at the base of the falls; its energy is entirely kinetic. The rate at which energy is delivered to the base of the falls is given by P = dW / dt = − dU / dt. Express the rate at which energy is being delivered to the base of the falls; remembering that half the potential energy of the water is converted to electric energy: Substitute numerical values and evaluate P:

dW dU =− dt dt d dm = − 12 (mgh ) = − 12 gh dt dt

P=

( ) × (1.4 × 10 kg/s )

P = − 12 9.81 m/s 2 (− 128 m ) 6

= 879 MW

400 Chapter 6 57 • Picture the Problem In the absence of friction, the sum of the potential and kinetic energies of the box remains constant as it slides down the incline. We can use the conservation of the mechanical energy of the system to calculate where the box will be and how fast it will be moving at any given time. We can also use Newton’s 2nd law to show that the acceleration of the box is constant and constant-acceleration equations to calculate where the box will be and how fast it will be moving at any given time. (a) Express and evaluate the gravitational potential energy of the box, relative to the ground, at the top of the incline: (b) Using a constant-acceleration equation, relate the displacement of the box to its initial speed, acceleration and time-of-travel: Apply

∑F

x

= max to the box as it

Ui = mgh = (2 kg) (9.81 m/s2) (20 m) = 392 J

∆x = v0 ∆t + 12 a (∆t )

2

or, because v0 = 0,

∆x = 12 a (∆t )

2

mg sin θ = ma ⇒ a = g sin θ

slides down the incline and solve for its acceleration: Substitute for a and evaluate ∆x(t = 1 s):

∆x(1s ) =

(g sin θ )(∆t )2 2 = 12 (9.81 m/s 2 )(sin30°)(1s ) 1 2

= 2.45 m Using a constant-acceleration equation, relate the speed of the box at any time to its initial speed and acceleration and solve for its speed when t = 1 s:

v = v0 + at where v0 = 0 and

v(1s ) = a∆t = (g sin θ )∆t

(

)

= 9.81m/s 2 (sin 30°)(1s ) = 4.91 m/s

Work and Energy 401 (c) Calculate the kinetic energy of the box when it has traveled for 1 s:

Express the potential energy of the box after it has traveled for 1 s in terms of its initial potential energy and its kinetic energy: (d) Express the kinetic energy of the box at the bottom of the incline in terms of its initial potential energy and solve for its speed at the bottom of the incline: Substitute numerical values and evaluate v:

K = 12 mv 2 =

1 2

(2 kg )(4.91m/s )2

= 24.1 J

U = U i − K = 392 J − 24.1 J = 368 J

K = U i = 12 mv 2 = 392 J and

v=

2U i m

v=

2(392 J ) = 19.8 m/s 2 kg

58 • Picture the Problem The potential energy function U (x) is defined by the equation x

U ( x ) − U ( x0 ) = − ∫ Fdx. We can use the given force function to determine U(x) and then x0

the conditions on U to determine the potential functions that satisfy the given conditions. (a) Use the definition of the potential energy function to find the potential energy function associated with Fx:

x

U (x ) = U (x0 ) − ∫ Fx dx x0 x

= U (x0 ) − ∫ (6 N )dx' x0

= − (6 N )(x − x0 ) because U(x0) = 0. (b) Use the result obtained in (a) to find U (x) that satisfies the condition that U(4 m) = 0:

U (4 m ) = −(6 N )(4 m − x0 ) = 0 ⇒ x0 = 4 m and

U (x ) = −(6 N )(x − 4 m ) = 24 J − (6 N )x

402 Chapter 6 (c) Use the result obtained in (a) to find U that satisfies the condition that U(6 m) = 14 J:

U (6 m ) = −(6 N )(6 m − x0 ) = 14 J ⇒ x0 = 50 m and

25 ⎞ ⎛ U (x ) = −(6 N )⎜ x − m ⎟ 3 ⎠ ⎝ = 50 J − (6 N )x 59 • Picture the Problem The potential energy of a stretched or compressed ideal spring Us is related to its force (stiffness) constant k and stretch or compression ∆x by U s = 12 kx 2 . (a) Relate the potential energy stored in the spring to the distance it has been stretched:

U s = 12 kx 2

Solve for x:

x=

2U s k

Substitute numerical values and evaluate x:

x=

2(50 J ) = 0.100 m 10 4 N/m

(b) Proceed as in (a) with Us = 100 J:

x=

2(100 J ) = 0.141 m 10 4 N/m

*60 •• Picture the Problem In a simple Atwood’s machine, the only effect of the pulley is to connect the motions of the two objects on either side of it; i.e., it could be replaced by a piece of polished pipe. We can relate the kinetic energy of the rising and falling objects to the mass of the system and to their common speed and relate their accelerations to the sum and difference of their masses … leading to simultaneous equations in m1 and m2.

K=

Use the definition of the kinetic energy of the system to determine the total mass being accelerated:

and

In Chapter 4, the acceleration of the masses was shown to be:

a=

1 2

(m1 + m2 )v 2

m1 + m2 =

2K 2(80 J ) = = 10.0 kg (1) 2 v (4 m/s)2

m1 − m2 g m1 + m2

Work and Energy 403 Because v(t) = at, we can eliminate a in the previous equation to obtain: Solve for m1 − m2 :

v(t ) =

m1 − m2 gt m1 + m2

m1 − m2 =

(m1 + m2 )v(t ) gt

(10 kg )(4 m/s) = 1.36 kg (9.81m/s2 )(3 s )

Substitute numerical values and evaluate m1 − m2 :

m1 − m2 =

Solve equations (1) and (2) simultaneously to obtain:

m1 = 5.68 kg and m2 = 4.32 kg

(2)

61 •• Picture the Problem The gravitational potential energy of this system of two objects is the sum of their individual potential energies and is dependent on an arbitrary choice of where, or under what condition(s), the gravitational potential energy is zero. The best choice is one that simplifies the mathematical details of the expression of U. In this problem let’s choose U = 0 where θ = 0. (a) Express U for the 2-object system as the sum of their gravitational potential energies; noting that because the object whose mass is m2 is above the position we have chosen for U = 0, its potential energy is positive while that of the object whose mass is m1 is negative: (b) Differentiate U with respect toθ and set this derivative equal to zero to identify extreme values:

To be physically meaningful,

U (θ ) = U1 + U 2

= m2 gl 2 sin θ − m1 gl 1 sin θ =

(m2l 2 − m1l 1 )g sin θ

dU = (m2 l 2 − m1l 1 )g cosθ = 0 dθ from which we can conclude that cosθ = 0 and θ = cos−10.

∴θ = ± π 2

−π 2 ≤ θ ≤ π 2 :

Express the 2nd derivative of U with respect to θ and evaluate this derivative at θ = ± π 2 :

d 2U = −(m2l 2 − m1l 1 )g sin θ dθ 2

404 Chapter 6 If we assume, in the expression for U that we derived in (a), that m2l2 – m1l1 >0, then U(θ) is a sine function and, in the interval of interest, − π 2 ≤ θ ≤ π 2 , takes on its minimum value when θ = −π/2:

d 2U dθ 2

>0 −π 2

and U is a minimum at θ = − π 2

d 2U dθ 2

<0 π 2

and U is a maximum at θ = π 2 (c) If m1l1 = m2l2, then

(m2l2 − m1l1) = 0

and U = 0 independently of θ .

Remarks: An alternative approach to establishing the U is a maximum at θ = π/2 is to plot its graph and note that, in the interval of interest, U is concave downward with its maximum value at θ = π/2.

Force, Potential Energy, and Equilibrium 62 • Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, that is, Fx = − dU dx . Consequently, given U as a function of x, we can find Fx by differentiating U with respect to x. (a) Evaluate Fx = −

dU : dx

(b) Set Fx = 0 and solve for x:

Fx = −

( )

d Ax 4 = − 4 Ax 3 dx

Fx = 0 ⇒ x = 0

63 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, that is Fx = − dU dx . Consequently, given U as a function of x, we can find Fx by differentiating U with respect to x. (a) Evaluate Fx = −

(b) Because C > 0:

dU : dx

Fx = −

d ⎛C ⎞ C ⎜ ⎟= 2 dx ⎝ x ⎠ x

Fx is positive for x ≠ 0 and therefore r F is directed away from the origin.

Work and Energy 405 (c) Because U is inversely proportional to x and C > 0: (d) With C < 0:

Because U is inversely proportional to x and C < 0, U(x) becomes less negative as x increases:

U ( x ) decreases with increasing x.

Fx is negative for x ≠ 0 and therefore r F is directed toward from the origin. U ( x ) increases with increasing x.

*64 •• Picture the Problem Fy is defined to be the negative of the derivative of the potential function with respect to y, i.e. Fy = − dU dy . Consequently, we can obtain Fy by examining the slopes of the graph of U as a function of y. The table to the right summarizes the information we can obtain from Figure 6-40:

Slope Fy Interval (N) (N) 2 −2 A→B B→C transitional −2 → 1.4 1.4 −1.4 C→D

The graph of F as a function of y is shown to the right:

2.5 2.0 1.5

F (N)

1.0 0.5 0.0 0

1

2

3

4

5

6

-0.5 -1.0 -1.5

y (m)

65 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, i.e. Fx = − dU dx . Consequently, given F as a function of x, we can find U by integrating Fx with respect to x. Evaluate the integral of Fx with respect to x:

U ( x ) = − ∫ F ( x ) dx = − ∫ =

a dx x2

a +U0 x

where U0 is a constant determined by whatever conditions apply to U.

406 Chapter 6 66 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, that is, Fx = − dU dx . Consequently, given U as a function of x, we can find Fx by differentiating U with respect to x. To determine whether the object is in stable or unstable equilibrium at a given point, we’ll evaluate d 2U dx 2 at the point of interest. (a) Evaluate Fx = −

dU : dx

(b) We know that, at equilibrium, Fx = 0:

Fx = −

(

)

d 3x 2 − 2 x 3 = 6 x( x − 1) dx

When Fx =0, 6x(x – 1) = 0. Therefore, the object is in equilibrium at x = 0 and x = 1 m.

(c) To decide whether the equilibrium at a particular point is stable or unstable, evaluate the 2nd derivative of the potential energy function at the point of interest:

Evaluate

d 2U at x = 0: dx 2

(

)

dU d = 3x 2 − 2 x 3 = 6 x − 6 x 2 dx dx and

d 2U = 6 − 12 x dx 2 d 2U dx 2

=6>0 x =0

⇒ stable equilibrium at x = 0

Evaluate

d 2U at x = 1 m: dx 2

d 2U dx 2

= 6 − 12 < 0 x =1 m

⇒ unstable equilibrium at x = 1 m

67 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, i.e. Fx = − dU dx . Consequently, given U as a function of x, we can find Fx by differentiating U with respect to x. To determine whether the object is in stable or unstable equilibrium at a given point, we’ll evaluate d 2U dx 2 at the point of interest.

Work and Energy 407 (a) Evaluate the negative of the derivative of U with respect to x:

dU dx d = − (8 x 2 − x 4 ) = 4 x 3 − 16 x dx

Fx = −

= 4 x( x + 2)( x − 2) (b) The object is in equilibrium wherever Fnet = Fx = 0:

4 x( x + 2)( x − 2) = 0 ⇒ the equilibrium points are x = −2 m, 0, and 2 m.

(

)

(c) To decide whether the equilibrium at a particular point is stable or unstable, evaluate the 2nd derivative of the potential energy function at the point of interest:

d 2U d = 16 x − 4 x 3 = 16 − 12 x 2 2 dx dx

d 2U at x = −2 m: Evaluate dx 2

d 2U dx 2

= −32 < 0 x =−2 m

⇒

Evaluate

d 2U at x = 0: dx 2

d 2U dx 2

unstable equilibrium at x = −2 m

= 16 > 0 x =0

⇒ stable equilibrium at x = 0

Evaluate

d 2U at x = 2 m: dx 2

d 2U dx 2

= −32 < 0 x =2 m

⇒

unstable equilibrium at x = 2 m

Remarks: You could also decide whether the equilibrium positions are stable or unstable by plotting F(x) and examining the curve at the equilibrium positions. 68 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, i.e. Fx = − dU dx . Consequently, given F as a function of x, we can find U by integrating Fx with respect to x. Examination of d 2U dx 2 at extreme points will determine the nature of the stability at these locations.

408 Chapter 6 Determine the equilibrium locations by setting Fnet = F(x) = 0:

F(x) = x3 – 4x = x(x2 – 4) = 0 ∴ the positions of stable and unstable equilibrium are at

Evaluate the negative of the integral of F(x) with respect to x:

U (x ) = − ∫ F (x )

(

x = −2, 0 and 2 .

)

= − ∫ x 3 − 4 x dx =−

x4 + 2x2 + U 0 4

where U0 is a constant whose value is determined by conditions on U(x). Differentiate U(x) twice:

dU = − Fx = − x 3 + 4 x dx and

d 2U = −3 x 2 + 4 2 dx d 2U Evaluate at x = −2: dx 2

d 2U dx 2

= −8 < 0 x =−2

∴ the equilibrium is unstable at x = − 2

Evaluate

d 2U at x = 0: dx 2

d 2U dx 2

=4>0 x =0

∴ the equilibrium is stable at x = 0

Evaluate

d 2U at x = 2: dx 2

d 2U dx 2

= −8 < 0 x =2

∴ the equilibrium is unstable at x = 2 Thus U(x) has a local minimum at x = 0 and local maxima at x = ±2. 69 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, i.e. Fx = − dU dx . Consequently, given U as a function of x, we can find Fx by differentiating U with respect to x. To determine whether the object is in stable or unstable equilibrium at a given point, we can examine the graph of U.

Work and Energy 409 (a) Evaluate Fx = −

dU for x ≤ 3 m: dx

Set Fx = 0 to identify those values of x for which the 4-kg object is in equilibrium:

Evaluate Fx = −

dU for x > 3 m: dx

Fx = −

(

)

d 3x 2 − x 3 = 3x(2 − x ) dx

When Fx = 0, 3x(2 – x) = 0. Therefore, the object is in equilibrium at x = 0 and x = 2 m.

Fx = 0 because U = 0.

Therefore, the object is in neutral equilibrium for x > 3 m. (b) A graph of U(x) in the interval –1 m ≤ x ≤ 3 m is shown to the right:

4.0 3.5 3.0

U (J)

2.5 2.0 1.5 1.0 0.5 0.0 -1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

x (m)

(c) From the graph, U(x) is a minimum at x = 0:

∴ stable equilibrium at x = 0

From the graph, U(x) is a maximum at x = 2 m:

∴ unstable equilibrium at x = 2 m

(d) Relate the kinetic energy of the object to its total energy and its potential energy:

K = 12 mv 2 = E − U

Solve for v:

v=

2( E − U ) m

Evaluate U(x = 2 m):

U ( x = 2 m ) = 3(2) − (2) = 4 J

Substitute in the equation for v to obtain:

v=

2

3

2(12 J − 4 J ) = 2.00 m/s 4 kg

3.0

410 Chapter 6 70 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, that is Fx = − dU dx . Consequently, given F as a function of x, we can find U by integrating Fx with respect to x. (a) Evaluate the negative of the integral of F(x) with respect to x:

U ( x ) = − ∫ F ( x ) = − ∫ Ax −3 dx =

1 A + U0 2 x2

where U0 is a constant whose value is determined by conditions on U(x). For x > 0:

(b) As x → ∞,

U decreases as x increases

1 A → 0: 2 x2

∴ U0 = 0 and

U (x ) =

(c) The graph of U(x) is shown to the right:

1 A 1 8 N ⋅ m3 4 = = 2 N ⋅ m3 2 2 2x 2 x x

400 350 300 250 200 150 100 50 0 0.0

0.5

1.0

1.5

2.0

x (m)

*71 ••• Picture the Problem Let L be the total length of one cable and the zero of gravitational potential energy be at the top of the pulleys. We can find the value of y for which the potential energy of the system is an extremum by differentiating U(y) with respect to y and setting this derivative equal to zero. We can establish that this value corresponds to a minimum by evaluating the second derivative of U(y) at the point identified by the first derivative. We can apply Newton’s 2nd law to the clock to confirm the result we obtain by examining the derivatives of U(y). (a) Express the potential energy of the system as the sum of the potential energies of the clock and counterweights: Substitute to obtain:

U ( y ) = U clock ( y ) + U weights ( y )

(

U ( y ) = − mgy − 2 Mg L − y 2 + d 2

)

Work and Energy 411 (b) Differentiate U(y) with respect to y:

[

(

dU ( y ) d =− mgy + 2Mg L − y 2 + d 2 dy dy ⎡ = − ⎢mg − 2Mg ⎢⎣

⎤ ⎥ y 2 + d 2 ⎥⎦ y

or

y'

mg − 2Mg

Solve for y′ to obtain:

Find

d 2U ( y ) : dy 2

y' = d

m2 4M 2 − m 2

d 2U ( y ) d ⎡ = − ⎢mg − 2Mg dy 2 dy ⎢ ⎣ 2Mgd 2 = 32 y2 + d 2

(

d 2U ( y ) Evaluate at y = y′: dy 2

= 0 for extrema

y' 2 + d 2

)

2Mgd 2 d 2U ( y ) = 32 dy 2 y' y2 + d 2

(

=

⎤ ⎥ y 2 + d 2 ⎥⎦ y

)

y'

2Mgd

⎞ ⎛ m2 ⎜⎜ + 1⎟⎟ 2 2 ⎠ ⎝ 4M − m >0

32

and the potential energy is a minimum at

y= d

m2 4M 2 − m 2

(c) The FBD for the clock is shown to the right:

Apply

∑F

y

= 0 to the clock:

2Mg sin θ − mg = 0 and

sin θ =

m 2M

)]

412 Chapter 6 Express sinθ in terms of y and d:

Substitute to obtain:

sin θ =

m = 2M

y y2 + d 2 y y2 + d 2

which is equivalent to the first equation in part (b).

This is a point of stable equilibrium. Ifthe clock is displaced downward, θ increases, leading to a larger upward force on the clock. Similarly, if the clock is displaced upward, the net force from the cables decreases. Because of this, the clock will be pulled back toward the equilibrium point if it is displaced away from it. Remarks: Because we’ve shown that the potential energy of the system is a minimum at y = y′ (i.e., U(y) is concave upward at that point), we can conclude that this point is one of stable equilibrium.

General Problems *72 • Picture the Problem 25 percent of the electrical energy generated is to be diverted to do the work required to change the potential energy of the American people. We can calculate the height to which they can be lifted by equating the change in potential energy to the available energy. Express the change in potential energy of the population of the United States in this process:

∆U = Nmgh

Letting E represent the total energy generated in February 2002, relate the change in potential to the energy available to operate the elevator:

Nmgh = 0.25E

Solve for h:

h=

0.25 E Nmg

Work and Energy 413 Substitute numerical values and evaluate h:

h=

(0.25)(60.7 ×109 kW ⋅ h ) ⎛⎜⎜ 3600 s ⎞⎟⎟ ⎝ 1h ⎠ 287 × 10 (60 kg ) 9.81 m/s 2

(

6

)

(

)

= 323 km 73 • Picture the Problem We can use the definition of the work done in changing the potential energy of a system and the definition of power to solve this problem. (a) Find the work done by the crane in changing the potential energy of its load:

W = mgh = (6×106 kg) (9.81 m/s2) (12 m)

(b) Use the definition of power to find the power developed by the crane:

P≡

= 706 MJ

dW 706 MJ = = 11.8 MW dt 60 s

74 • Picture the Problem The power P of the engine needed to operate this ski lift is related to the rate at which it changes the potential energy U of the cargo of the gondolas according to P = ∆U/∆t. Because as many empty gondolas are descending as are ascending, we do not need to know their mass.

∆U ∆t

Express the rate at which work is done as the cars are lifted:

P=

Letting N represent the number of gondola cars and M the mass of each, express the change in U as they are lifted a vertical displacement ∆h:

∆U = NMg∆h

Substitute to obtain:

P≡

Relate ∆h to the angle of ascent θ and the length L of the ski lift:

∆h = Lsinθ

Substitute for ∆h in the expression for P:

P=

∆U NMg∆h = ∆t ∆t

NMgL sin θ ∆t

414 Chapter 6 Substitute numerical values and evaluate P:

P=

(

)

12(550 kg ) 9.81 m/s 2 (5.6 km )sin30° = 50.4 kW (60 min )(60 s/min )

75 • Picture the Problem The application of Newton’s 2nd law to the forces shown in the free-body diagram will allow us to relate R to T. The unknown mass and speed of the object can be eliminated by introducing its kinetic energy.

Apply

∑F

radial

= maradial the object

and solve for R:

mv 2 mv 2 and R = T= T R

Express the kinetic energy of the object:

K = 12 mv 2

Eliminate mv2 between the two equations to obtain:

R=

2K T

Substitute numerical values and evaluate R:

R=

2(90 J ) = 0.500 m 360 N

*76 • Picture the Problem We can solve this problem by equating the expression for the gravitational potential energy of the elevated car and its kinetic energy when it hits the ground. Express the gravitational potential energy of the car when it is at a distance h above the ground:

U = mgh

Express the kinetic energy of the car when it is about to hit the ground:

K = 12 mv 2

Equate these two expressions (because at impact, all the potential energy has been converted to kinetic energy) and solve for h:

h=

v2 2g

Work and Energy 415 Substitute numerical values and evaluate h:

2 ( [ 100 km/h )(1 h/3600 s )] h=

(

2 9.81 m/s 2

)

= 39.3 m

77 ••• Picture the Problem The free-body diagram shows the forces acting on one of the strings at the bridge. The force whose magnitude is F is one-fourth of the force (103 N) the bridge exerts on the strings. We can apply the condition for equilibrium in the y direction to find the tension in each string. Repeating this procedure at the site of the plucking will yield the restoring force acting on the string. We can find the work done on the string as it returns to equilibrium from the product of the average force acting on it and its displacement. (a) Noting that, due to symmetry, T′ = T, apply Fy = 0 to the string

∑

F − 2T sin 18° = 0

at the point of contact with the bridge: Solve for and evaluate T:

T=

1 (103 N ) = 41.7 N F = 4 2 sin 18° 2 sin 18°

(b) A free-body diagram showing the forces restoring the string to its equilibrium position just after it has been plucked is shown to the right: Express the net force acting on the string immediately after it is released: Use trigonometry to find θ:

Substitute and evaluate Fnet:

Fnet = 2T cos θ

⎛ 16.3 cm 10 mm ⎞ ⎟ = 88.6° × cm ⎟⎠ ⎝ 4 mm

θ = tan −1 ⎜⎜

Fnet = 2(34.4 N )cos88.6° = 1.68 N

416 Chapter 6 (c) Express the work done on the string in displacing it a distance dx′:

dW = Fdx'

If we pull the string out a distance x′, the magnitude of the force pulling it down is approximately:

F = (2T )

Substitute to obtain:

dW =

Integrate to obtain:

4T 2T 2 W= x'dx' = x ∫ L 0 L

x' 4T = x' L2 L

4T x' dx' L x

where x is the final displacement of the string. Substitute numerical values to obtain:

W=

2(41.7 N ) 4 × 10−3 m −2 32.6 × 10 m

(

)

2

= 4.09 mJ 78 •• Picture the Problem Fx is defined to be the negative of the derivative of the potential function with respect to x, that is Fx = − dU dx . Consequently, given F as a function of x, we can find U by integrating Fx with respect to x. Evaluate the integral of Fx with respect to x:

Apply the condition that U(0) = 0 to determine U0:

(

)

U (x ) = − ∫ F (x ) dx = − ∫ − ax 2 dx = 13 ax 3 + U 0 U(0) = 0 + U0 = 0 ⇒ U0 = 0

∴U (x ) =

1 3

ax 3

The graph of U(x) is shown to the right:

3 2

U (J)

1 0 -2.0

-1.5

-1.0

-0.5

0.0 -1 -2 -3

x (m)

0.5

1.0

1.5

2.0

Work and Energy 417 *79 •• Picture the Problem We can use the definition of work to obtain an expression for the position-dependent force acting on the cart. The work done on the cart can be calculated from its change in kinetic energy. (a) Express the force acting on the cart in terms of the work done on it:

F (x ) =

dW dx

Because U is constant:

F (x ) =

d 1 2 d ( 2 mv ) = dx dx

[ m(Cx ) ] 2

1 2

= mC 2 x (b) The work done by this force changes the kinetic energy of the cart:

W = ∆K = 12 mv12 − 12 mv02 = 12 mv12 − 0 = 12 m(Cx1 )

2

=

1 2

mC 2 x12

80 •• r Picture the Problem The work done by F depends on whether it causes a displacement in the direction it acts.

r

r

r

(a) Because F is along x-axis and the displacement is along y-axis:

W = ∫ F ⋅ ds = 0

(b) Calculate the work done by r F during the displacement from x = 2 m to 5 m:

r r W = ∫ F ⋅ ds =

(

= 2 N/m

2

)

∫ (2 N/m ) x dx

5m

2

2

2m

5m

⎡ x3 ⎤ ⎢ 3 ⎥ = 78.0 J ⎣ ⎦2 m

81 •• Picture the Problem The velocity and acceleration of the particle can be found by differentiation. The power delivered to the particle can be expressed as the product of its velocity and the net force acting on it, and the work done by the force and can be found from the change in kinetic energy this work causes. In the following, if t is in seconds and m is in kilograms, then v is in m/s, a is in m/s2, P is in W, and W is in J.

418 Chapter 6 (a) The velocity of the particle is given by:

v=

(

(6t

) dv d a= = (6t dt dt =

The acceleration of the particle is given by:

)

dx d = 2t 3 − 4t 2 dt dt 2

− 8t

2

− 8t

)

= (12t − 8)

(b) Express and evaluate the rate at which energy is delivered to this particle as it accelerates:

(c) Because the particle is moving in such a way that its potential energy is not changing, the work done by the force acting on the particle equals the change in its kinetic energy:

P = Fv = mav

(

) 8mt (9t − 18t + 8)

= m(12t − 8) 6t 2 − 8t =

2

W = ∆K = K1 − K 0

[

= 12 m (v(t1 )) − (v(0 )) 2

2

[(

= 12 m 6t12 − 8t1

)]

2

]

−0

= 2mt12 (3t1 − 4)

2

Remarks: We could also find W by integrating P(t) with respect to time. 82 •• Picture the Problem We can calculate the work done by the given force from its r r definition. The power can be determined from P = F ⋅ v and v from the change in kinetic energy of the particle produced by the work done on it. (a) Calculate the work done from its definition:

r r 3m W = ∫ F ⋅ ds = ∫ 6 + 4 x − 3x 2 dx

(

)

0

3m

⎡ 4 x 3x3 ⎤ = ⎢6 x + − = 9.00 J 2 3 ⎥⎦ 0 ⎣ 2

(b) Express the power delivered to the particle in terms of Fx=3 m and its velocity: Relate the work done on the particle to its kinetic energy and solve for its velocity:

r r P = F ⋅ v = Fx =3 m v

W = ∆K = K final = 12 mv 2 since v0 = 0

Work and Energy 419 Solve for and evaluate v:

v=

2(9 J ) = 2.45 m/s 3 kg

2K = m

Evaluate Fx=3 m:

Fx=3 m = 6 + 4(3) − 3(3) = −9 N

Substitute for Fx=3 m and v:

P = (− 9 N )(2.45 m/s ) = − 22.1 W

2

*83 •• Picture the Problem We’ll assume that the firing height is negligible and that the bullet lands at the same elevation from which it was fired. We can use the equation R = v02 g sin 2θ to find the range of the bullet and constant-acceleration equations to

(

)

find its maximum height. The bullet’s initial speed can be determined from its initial kinetic energy.

v02 sin 2θ g

Express the range of the bullet as a function of its firing speed and angle of firing:

R=

Rewrite the range equation using the trigonometric identity sin2θ = 2sinθ cosθ:

v02 sin 2θ 2v02 sin θ cos θ = R= g g

Express the position coordinates of the projectile along its flight path in terms of the parameter t:

Eliminate the parameter t and make use of the fact that the maximum height occurs when the projectile is at half the range to obtain:

x = (v0 cos θ )t and

y = (v0 sin θ )t − 12 gt 2

h=

(v0 sin θ )2 2g

Equate R and h and solve the resulting equation for θ:

tan θ = 4 ⇒ θ = tan −1 4 = 76.0°

Relate the bullet’s kinetic energy to its mass and speed and solve for the square of its speed:

K = 12 mv02 and v02 =

Substitute for v02 and θ and evaluate R:

R=

2K m

2(1200 J ) sin2(76°) (0.02 kg ) 9.81m/s2

= 5.74 km

(

)

420 Chapter 6 84 •• Picture the Problem The work done on the particle is the area under the force-versusdisplacement curve. Note that for negative displacements, F is positive, so W is negative for x < 0. (a) Use either the formulas for the areas of simple geometric figures or counting squares and multiplying by the work represented by one square to complete the table to the right:

(b) Choosing U(0) = 0, and using the definition of ∆U = −W, complete the third column of the table to the right:

x W (m) (J) −4 −11 −3 −10 −2 −7 −1 −3 0 0 1 1 2 0 3 −2 4 −3 ∆U (J) (J) −11 11 −10 10 −7 7 −3 3 0 0 1 −1 0 0 −2 2 −3 3

x

W

(m) −4 −3 −2 −1 0 1 2 3 4

12 10 8 6

U (J)

The graph of U as a function of x is shown to the right:

4 2 0 -4

-3

-2

-1

0 -2

x (m)

1

2

3

4

Work and Energy 421 85 •• Picture the Problem The work done on the particle is the area under the force-versusdisplacement curve. Note that for negative displacements, F is negative, so W is positive for x < 0. (a) Use either the formulas for the areas of simple geometric figures or counting squares and multiplying by the work represented by one square to complete the table to the right:

x (m) −4 −3 −2 −1 0 1 2 3 4

W (J) 6 4 2 0.5 0 0.5 1.5 2.5 3

(b) Choosing U(0) = 0, and using the definition of ∆U = −W, complete the third column of the table to the right:

x

W

(m) −4 −3 −2 −1 0 1 2 3 4

(J) 6 4 2 0.5 0 0.5 1.5 2.5 3

∆U (J) −6 −4 −2 −0.5 0 −0.5 −1.5 −2.5 −3

0 -4

-3

-2

-1

0 -1 -2

U (J)

The graph of U as a function of x is shown to the right:

-3 -4 -5 -6

x (m)

1

2

3

4

422 Chapter 6 86 •• Picture the Problem The pictorial representation shows the box at its initial position 0 at the bottom of the inclined plane and later at position 1. We’ll assume that the block is at position 0. Because the surface is frictionless, the work done by the tension will change both the potential and kinetic energy of the block. We’ll use Newton’s 2nd law to find the acceleration of the block up the incline and a constantacceleration equation to express v in terms of T, x, M, and θ. Finally, we can express the power produced by the tension in terms of the tension and the speed of the box. (a) Use the definition of work to express the work the tension T does moving the box a distance x up the incline: (b) Apply

∑F

x

= Max to the box:

W = Tx

T − Mg sin θ = Max T − Mg sin θ T = − g sin θ M M

Solve for ax:

ax =

Using a constant-acceleration equation, express the speed of the box in terms of its acceleration and the distance x it has moved up the incline:

v 2 = v02 + 2a x x

Substitute for ax to obtain:

(c) The power produced by the tension in the string is given by:

or, because v0 = 0,

v = 2a x x

v=

⎛T ⎞ 2⎜ − g sin θ ⎟ x ⎝M ⎠

⎛T ⎞ P = Tv = T 2⎜ − g sin θ ⎟ x ⎝M ⎠

Work and Energy 423 87 ••• Picture the Problem We can use the definition of the magnitude of vector to show that r the magnitude of F is F0 and the definition of the scalar product to show that its direction r is perpendicular to r . The work done as the particle moves in a circular path can be found from its definition.

r

r F = Fx2 + Fy2

(a) Express the magnitude of F :

2

⎛F ⎞ ⎛ F ⎞ = ⎜ 0 y⎟ + ⎜− 0 x⎟ ⎝ r ⎠ ⎝ r ⎠ F = 0 x2 + y2 r Because r =

r F F = 0 r

x2 + y2 :

r

r

Form the scalar product of F and r :

F0 r = F0 r

x2 + y2 =

(

2

)(

r r ⎛F ⎞ F ⋅ r = ⎜ 0 ⎟ y iˆ − x ˆj ⋅ x iˆ + y ˆj ⎝ r ⎠ ⎛F ⎞ = ⎜ 0 ⎟(xy − xy ) = 0 ⎝ r ⎠ r r

r

)

r

Because F ⋅ r = 0, F ⊥ r

r

r

r

(b) Because F ⊥ r , F is tangential to the circle and constant. At (5 m,

r 0), F points in the − ˆj direction. If r ds is in the − ˆj direction, dW > 0.

The work it does in one revolution is:

W = F0 (2π r ) = 2π (5 m )F0

= (10π m )F0 if the rotation is clockwise

and

W = (− 10π m )F0 if the rotation is counterclockwise.

W = (10π m) F0 if the rotation is clockwise, − (10π m) F0if the rotation is r counterclockwise. Because W ≠ 0 for a complete circuit, F is not conservative. *88 •••

r

Picture the Problem We can substitute for r and xiˆ + yˆj in F to show that the magnitude of the force varies as the inverse of the square of the distance to the origin, and that its direction is opposite to the radius vector. We can find the work done by this force by evaluating the integral of F with respect to x from an initial position x = 2 m, y = 0 m to a final position x = 5 m, y = 0 m. Finally, we can apply Newton’s 2nd law to the particle to relate its speed to its radius, mass, and the constant b.

424 Chapter 6 r ⎛ ⎞ 2 b ⎟ x + y 2 rˆ F = −⎜ 3 2 ⎜ x2 + y2 ⎟ ⎝ ⎠ where rˆ is a unit vector pointing from the

(a) Substitute for r and

r x iˆ + y ˆj in F to obtain:

(

)

origin toward the point of application of r F.

r ⎛ 1 ⎞ b ⎟rˆ = − 2 rˆ F = −b⎜⎜ 2 2 ⎟ r ⎝x +y ⎠

Simplify to obtain:

i.e., the magnitude of the force varies as the inverse of the square of the distance to the origin, and its direction is antiparallel

r (opposite) to the radius vector r = xiˆ + yˆj. (b) Find the work done by this force by evaluating the integral of F with respect to x from an initial position x = 2 m, y = 0 m to a final position x = 5 m, y = 0 m:

5m

5m

b ⎡1⎤ W = − ∫ 2 dx' = b ⎢ ⎥ x' ⎣ x' ⎦ 2 m 2m 1 ⎞ ⎛ 1 = 3 N ⋅ m2 ⎜ − ⎟ = − 0.900 J ⎝5m 2m⎠

(c) No work is done as the force is perpendicular to the velocity. (d) Because the particle is moving in a circle, the force on the particle must be supplying the centripetal acceleration keeping it moving in the circle. Apply Fr = mac to

b v2 m = r2 r

∑

the particle: Solve for v:

Substitute numerical values and evaluate v:

v=

b mr

v=

3 N ⋅ m2 = 0.463 m/s (2 kg )(7 m )

89 ••• Picture the Problem A spreadsheet program to calculate the potential is shown below. The constants used in the potential function and the formula used to calculate the ″6-12″ potential are as follows: Cell B2 B3

Content/Formula 1.09×10−7 6.84×10−5

Algebraic Form a b

Work and Energy 425 D8

$B$2/C8^12−$B$3/C8^6

C9

C8+0.1

a b − 6 12 r r r + ∆r

(a) A B C D 1 2 a = 1.09E-07 3 b = 6.84E-05 4 5 6 7 r U 8 3.00E-01 1.11E-01 9 3.10E-01 6.13E-02 10 3.20E-01 3.08E-02 11 3.30E-01 1.24E-02 12 3.40E-01 1.40E-03 13 3.50E-01 −4.95E-03 45 46 47 48

6.70E-01 6.80E-01 6.90E-01 7.00E-01

−7.43E-04 −6.81E-04 −6.24E-04 −5.74E-04

The graph shown below was generated from the data in the table shown above. Because the force between the atomic nuclei is given by F = −(dU dr ) , we can conclude that the shape of the potential energy function supports Feynman’s claim. "6-12" Potential 0.12 0.10

U (eV)

0.08 0.06 0.04 0.02 0.00 0.30 -0.02

0.35

0.40

0.45

0.50

0.55

0.60

0.65

0.70

r (nm)

(b) The minimum value is about −0.0107 eV, occurring at a separation of approximately 0.380 nm. Because the function is concave upward (a potential ″well″) at this separation,

426 Chapter 6 this separation is one of stable equilibrium, although very shallow.

dU d ⎡ a b⎤ = − ⎢ 12 − 6 ⎥ dr dr ⎣ r r ⎦ 12a 6b = 13 − 7 r r

(c) Relate the force of attraction between two argon atoms to the slope of the potential energy function:

F =−

Substitute numerical values and evaluate F(5 Å):

F=

(

) (

)

12 1.09 × 10−7 6 6.84 × 10 −5 1.6 × 10−19 J 1 nm −2 eV 4 . 18 10 − = − × × × −9 nm eV 10 m (0.5 nm )13 (0.5 nm )7

= − 6.69 × 10 −12 N where the minus sign means that the force is attractive. Substitute numerical values and evaluate F(3.5 Å):

F=

(

) (

)

12 1.09 × 10−7 6 6.84 × 10 −5 1.6 × 10−19 J 1 nm −1 eV 4 . 68 10 − = × × × −9 nm eV 10 m (0.35 nm )13 (0.35 nm )7

= 7.49 × 10−11 N where the plus sign means that the force is repulsive. *90 ••• Picture the Problem A spreadsheet program to plot the Yukawa potential is shown below. The constants used in the potential function and the formula used to calculate the Yukawa potential are as follows:

Cell Content/Formula Algebraic Form B1 4 U0 B2 2.5 a D8 −$B$1*($B$2/C9)*EXP(−C9/$B$2) ⎛ a ⎞ −r / a C10

− U 0 ⎜ ⎟e ⎝r⎠ r + ∆r

C9+0.1

(a) A 1 2 3 7 8 9 10

B U0= 4 a= 2.5

C

D

r 0.5 0.6

U −16.37 −13.11

pJ fm

Work and Energy 427 11 12 13 14

0.7 0.8 0.9 1

−10.80 −9.08 −7.75 −6.70

64 65 66 67 68 69 70

6 6.1 6.2 6.3 6.4 6.5 6.6

−0.15 −0.14 −0.14 −0.13 −0.12 −0.11 −0.11

U as a function of r is shown below. 0 -2 0

1

2

3

4

5

6

7

-4

U (pJ)

-6 -8 -10 -12 -14 -16 -18 r (fm)

(b) Relate the force between the nucleons to the slope of the potential energy function:

dU (r ) dr ⎤ d ⎡ ⎛a⎞ = − ⎢ − U 0 ⎜ ⎟e − r a ⎥ dr ⎣ ⎝r⎠ ⎦

F (r ) = −

⎛ a 1⎞ = − U 0 e −r / a ⎜ 2 + ⎟ r⎠ ⎝r (c) Evaluate F(2a):

⎛ a 1 ⎞ F (2a ) = −U 0 e −2 a / a ⎜⎜ + ⎟⎟ 2 ⎝ (2a ) 2a ⎠ ⎛ 3 ⎞ = −U 0 e −2 ⎜ ⎟ ⎝ 4a ⎠

428 Chapter 6 Evaluate F(a):

⎛ a 1⎞ F (a ) = −U 0 e −a / a ⎜⎜ 2 + ⎟⎟ a⎠ ⎝ (a ) ⎛1 1⎞ ⎛2⎞ = −U 0 e −1 ⎜ + ⎟ = −U 0 e −1 ⎜ ⎟ ⎝a a⎠ ⎝a⎠

Express the ratio F(2a)/F(a):

⎛ 3 ⎞ − U 0 e −2 ⎜ ⎟ F (2a ) ⎝ 4a ⎠ = 3 e −1 = F (a ) ⎛2⎞ 8 − U 0 e −1 ⎜ ⎟ ⎝a⎠ = 0.138

(d) Evaluate F(5a):

⎛ a 1 ⎞ F (5a ) = −U 0 e −5 a / a ⎜⎜ + ⎟⎟ 2 ⎝ (5a ) 5a ⎠ ⎛ 6 ⎞ = −U 0 e −5 ⎜ ⎟ ⎝ 25a ⎠

Express the ratio F(5a)/F(a):

⎛ 6 ⎞ − U 0 e −5 ⎜ ⎟ F (5a ) ⎝ 25a ⎠ = 3 e −4 = 25 F (a ) ⎛2⎞ − U 0 e −1 ⎜ ⎟ ⎝a⎠ = 2.20 × 10 −3

Chapter 7 Conservation of Energy Conceptual Problems *1 • Determine the Concept Because the peg is frictionless, mechanical energy is conserved as this system evolves from one state to another. The system moves and so we know that ∆K > 0. Because ∆K + ∆U = constant, ∆U < 0.

(a ) is correct.

2 • Determine the Concept Choose the zero of gravitational potential energy to be at ground level. The two stones have the same initial energy because they are thrown from the same height with the same initial speeds. Therefore, they will have the same total energy at all times during their fall. When they strike the ground, their gravitational potential energies will be zero and their kinetic energies will be equal. Thus, their speeds at impact will be equal. The stone that is thrown at an angle of 30° above the horizontal has a longer flight time due to its initial upward velocity and so they do not strike the ground at the same time. (c) is correct. 3 • (a) False. Forces that are external to a system can do work on the system to change its energy. (b) False. In order for some object to do work, it must exert a force over some distance. The chemical energy stored in the muscles of your legs allows your muscles to do the work that launches you into the air. 4 • Determine the Concept Your kinetic energy increases at the expense of chemical energy. *5 • Determine the Concept As she starts pedaling, chemical energy inside her body is converted into kinetic energy as the bike picks up speed. As she rides it up the hill, chemical energy is converted into gravitational potential and thermal energy. While freewheeling down the hill, potential energy is converted to kinetic energy, and while braking to a stop, kinetic energy is converted into thermal energy (a more random form of kinetic energy) by the frictional forces acting on the bike. *6 • Determine the Concept If we define the system to include the falling body and the earth, then no work is done by an external agent and ∆K + ∆Ug + ∆Etherm= 0. Solving for the change in the gravitational potential energy we find ∆Ug = −(∆K + friction energy).

437

438 Chapter 7 (b) is correct. 7 •• Picture the Problem Because the constant friction force is responsible for a constant acceleration, we can apply the constant-acceleration equations to the analysis of these statements. We can also apply the work-energy theorem with friction to obtain expressions for the kinetic energy of the car and the rate at which it is changing. Choose the system to include the earth and car and assume that the car is moving on a horizontal surface so that ∆U = 0. (a) A constant frictional force causes a constant acceleration. The stopping distance of the car is related to its speed before the brakes were applied through a constantacceleration equation. (b) Apply the work-energy theorem with friction to obtain: Express the rate at which K is dissipated:

v 2 = v02 + 2a∆s where v = 0. − v02 where a < 0. 2a Thus, ∆s ∝ v02 and statement (a) is false.

∴∆s =

∆K = −Wf = − µ k mg∆s

∆K ∆s = − µ k mg ∆t ∆t Thus,

∆K ∝ v and therefore not constant. ∆t

Statement (b) is false. (c) In part (b) we saw that:

K ∝ ∆s

Because ∆s ∝ ∆t:

K ∝ ∆t and statement (c) is false.

Because none of the above are correct:

(d ) is correct.

8 • Picture the Problem We’ll let the zero of potential energy be at the bottom of each ramp and the mass of the block be m. We can use conservation of energy to predict the speed of the block at the foot of each ramp. We’ll consider the distance the block travels on each ramp, as well as its speed at the foot of the ramp, in deciding its descent times. Use conservation of energy to find the speed of the blocks at the bottom of each ramp:

∆K + ∆U = 0 or

K bot − K top + U bot − U top = 0

Conservation of Energy 439 Because Ktop = Ubot = 0:

K bot − U top = 0

Substitute to obtain:

1 2

Solve for vbot:

vbot = 2 gH independently of the shape of

2 mvbot − mgH = 0

the ramp. Because the block sliding down the circular arc travels a greater distance (an arc length is greater than the length of the chord it defines) but arrives at the bottom of the ramp with the same speed that it had at the bottom of the inclined plane, it will require more time to arrive at the bottom of the arc. (b) is correct. 9 •• Determine the Concept No. From the work-kinetic energy theorem, no total work is being done on the rock, as its kinetic energy is constant. However, the rod must exert a tangential force on the rock to keep the speed constant. The effect of this force is to cancel the component of the force of gravity that is tangential to the trajectory of the rock.

Estimation and Approximation *10 •• Picture the Problem We’ll use the data for the "typical male" described above and assume that he spends 8 hours per day sleeping, 2 hours walking, 8 hours sitting, 1 hour in aerobic exercise, and 5 hours doing moderate physical activity. We can approximate his energy utilization using Eactivity = APactivity ∆tactivity , where A is the surface area of his body, Pactivity is the rate of energy consumption in a given activity, and ∆tactivity is the time spent in the given activity. His total energy consumption will be the sum of the five terms corresponding to his daily activities. (a) Express the energy consumption of the hypothetical male:

E = Esleeping + Ewalking + Esitting

Evaluate Esleeping:

Esleeping = APsleeping ∆tsleeping

+ Emod. act. + Eaerobic act.

(

)(

)

= 2 m 2 40 W/m 2 (8 h )(3600 s/h ) = 2.30 × 10 J 6

Evaluate Ewalking:

Ewalking = APwalking ∆t walking

(

)(

)

= 2 m 2 160 W/m 2 (2 h )(3600 s/h ) = 2.30 × 106 J Evaluate Esitting:

Esitting = APsitting ∆tsitting

(

)(

)

= 2 m 2 60 W/m 2 (8 h )(3600 s/h ) = 3.46 × 106 J

440 Chapter 7 Evaluate Emod. act.:

Emod. act. = APmod. act.∆tmod. act.

(

)(

)

= 2 m 2 175 W/m 2 (5 h )(3600 s/h ) = 6.30 × 106 J

Evaluate Eaerobic act.:

Eaerobic act. = APaerobic act.∆taerobic act.

(

)(

)

= 2 m 2 300 W/m 2 (1 h )(3600 s/h ) = 2.16 × 106 J

Substitute to obtain:

E = 2.30 × 106 J + 2.30 × 106 J + 3.46 × 106 J + 6.30 × 106 J + 2.16 × 106 J = 16.5 × 106 J

Express the average metabolic rate represented by this energy consumption:

Pav =

E 16.5 × 106 J = = 191 W ∆t (24 h )(3600 s/h )

or about twice that of a 100 W light bulb. (b) Express his average energy consumption in terms of kcal/day:

(c)

E=

16.5 × 106 J/day = 3940 kcal/day 4190 J/kcal

3940 kcal = 22.5 kcal/lb is higher than the estimate given in the statement of the 175 lb

problem. However, by adjusting the day's activities, the metabolic rate can vary by more than a factor of 2. 11 • Picture the Problem The rate at which you expend energy, i.e., do work, is defined as power and is the ratio of the work done to the time required to do the work. Relate the rate at which you can expend energy to the work done in running up the four flights of stairs and solve for your running time: Express the work done in climbing the stairs: Substitute for ∆W to obtain:

P=

∆W ∆W ⇒ ∆t = P ∆t

∆W = mgh

∆t =

mgh P

Conservation of Energy 441 Assuming that your weight is 600 N, evaluate ∆t:

∆t =

(600 N )(4 × 3.5 m ) = 250 W

33.6 s

12 • Picture the Problem The intrinsic rest energy in matter is related to the mass of matter through Einstein’s equation E0 = mc 2 . (a) Relate the rest mass consumed to the energy produced and solve for and evaluate m:

(b) Express the energy required as a function of the power of the light bulb and evaluate E:

Substitute in equation (1) to obtain:

E0 = mc 2 ⇒ m = m=

E0 c2

1J

(2.998 ×10

8

m/s

)

(1)

2

= 1.11× 10−17 kg

E = 3Pt

= 3(100 W )(10 y ) ⎛ 365.24 d ⎞ ⎛ 24 h ⎞ ⎛ 3600 s ⎞ ⎟⎟ ⎜ × ⎜⎜ ⎟⎜ ⎟ y ⎝ ⎠⎝ d ⎠⎝ h ⎠ = 9.47 × 1010 J

m=

9.47 × 1010 J

(2.998 ×10

8

m/s

)

2

= 1.05 µg

*13 • Picture the Problem There are about 3×108 people in the United States. On the assumption that the average family has 4 people in it and that they own two cars, we have a total of 1.5×108 automobiles on the road (excluding those used for industry). We’ll assume that each car uses about 15 gal of fuel per week. Calculate, based on the assumptions identified above, the total annual consumption of energy derived from gasoline:

(1.5 ×10

8

gal J ⎞ ⎛ ⎞ ⎛ weeks ⎞ ⎛ ⎟⎟ ⎜⎜ 2.6 × 108 ⎟⎟ = 3.04 × 1019 J/y auto ⎜15 ⎟ ⎜⎜ 52 y ⎠⎝ gal ⎠ ⎝ auto ⋅ week ⎠ ⎝

)

Express this rate of energy use as a fraction of the total annual energy use by the US:

3.04 × 1019 J/y ≈ 6% 5 × 1020 J/y

Remarks: This is an average power expenditure of roughly 9x1011 watt, and a total cost (assuming $1.15 per gallon) of about 140 billion dollars per year. 14 • Picture the Problem The energy consumption of the U.S. works out to an average power consumption of about 1.6×1013 watt. The solar constant is roughly 103 W/m2 (reaching

442 Chapter 7 the ground), or about 120 W/m2 of useful power with a 12% conversion efficiency. Letting P represent the daily rate of energy consumption, we can relate the power available at the surface of the earth to the required area of the solar panels using P = IA . Relate the required area to the electrical energy to be generated by the solar panels: Solve for and evaluate A:

P = IA where I is the solar intensity that reaches the surface of the Earth.

A=

(

P 2 1.6 × 1013 W = I 120 W/m 2

)

= 2.67 × 1011 m 2 where the factor of 2 comes from the fact that the sun is only up for roughly half the day. Find the side of a square with this area:

s = 2.67 × 1011 m 2 = 516 km

Remarks: A more realistic estimate that would include the variation of sunlight over the day and account for latitude and weather variations might very well increase the area required by an order of magnitude. 15 • Picture the Problem We can relate the energy available from the water in terms of its mass, the vertical distance it has fallen, and the efficiency of the process. Differentiation of this expression with respect to time will yield the rate at which water must pass through its turbines to generate Hoover Dam’s annual energy output. Assuming a total efficiencyη, use the expression for the gravitational potential energy near the earth’s surface to express the energy available from the water when it has fallen a distance h:

E = ηmgh

Differentiate this expression with respect to time to obtain:

P=

Solve for dV/dt:

dV P = dt ηρgh

Using its definition, relate the dam’s annual power output to the energy produced:

P=

Substitute numerical values to obtain:

4 × 109 kW ⋅ h P= = 4.57 × 108 W (365.24 d )(24 h/d )

d [ηmgh] = ηgh dm = ηρgh dV dt dt dt (1)

∆E ∆t

Conservation of Energy 443 Substitute in equation (1) and evaluate dV/dt:

4.57 × 108 W dV = dt 0.2(1kg/L ) 9.81 m/s 2 (211m )

(

)

= 1.10 × 106 L/s

The Conservation of Mechanical Energy 16 • Picture the Problem The work done in compressing the spring is stored in the spring as potential energy. When the block is released, the energy stored in the spring is transformed into the kinetic energy of the block. Equating these energies will give us a relationship between the compressions of the spring and the speeds of the blocks. Let the numeral 1 refer to the first case and the numeral 2 to the second case. Relate the compression of the spring in the second case to its potential energy, which equals its initial kinetic energy when released:

1 2

Relate the compression of the spring in the first case to its potential energy, which equals its initial kinetic energy when released:

1 2

kx22 = 12 m2 v22 =

1 2

(4m1 )(3v1 )2

= 18m1v12

kx12 = 12 m1v12

or

m1v12 = kx12 kx22 = 18kx12

Substitute to obtain:

1 2

Solve for x2:

x 2 = 6x1

17 • Picture the Problem Choose the zero of gravitational potential energy to be at the foot of the hill. Then the kinetic energy of the woman on her bicycle at the foot of the hill is equal to her gravitational potential energy when she has reached her highest point on the hill. Equate the kinetic energy of the rider at the foot of the incline and her gravitational potential energy when she has reached her highest point on the hill and solve for h: Relate her displacement along the

1 2

mv 2 = mgh ⇒ h =

d = h/sinθ

v2 2g

444 Chapter 7 incline d to h and the angle of the incline: Substitute for h to obtain:

d sin θ =

Solve for d:

d=

Substitute numerical values and evaluate d:

d=

v2 2g

v2 2 g sin θ

(10 m/s) 2

(

)

2 9.81 m/s 2 sin3°

and

= 97.4 m

(c) is correct.

*18 • Picture the Problem The diagram shows the pendulum bob in its initial position. Let the zero of gravitational potential energy be at the low point of the pendulum’s swing, the equilibrium position. We can find the speed of the bob at it passes through the equilibrium position by equating its initial potential energy to its kinetic energy as it passes through its lowest point.

Equate the initial gravitational potential energy and the kinetic energy of the bob as it passes through its lowest point and solve for v: Express ∆h in terms of the length L of the pendulum: Substitute and simplify:

mg∆h = 12 mv 2 and v = 2 g∆h

∆h =

v=

L 4 gL 2

19 • Picture the Problem Choose the zero of gravitational potential energy to be at the foot of the ramp. Let the system consist of the block, the earth, and the ramp. Then there are

Conservation of Energy 445 no external forces acting on the system to change its energy and the kinetic energy of the block at the foot of the ramp is equal to its gravitational potential energy when it has reached its highest point. Relate the gravitational potential energy of the block when it has reached h, its highest point on the ramp, to its kinetic energy at the foot of the ramp:

mgh = 12 mv 2

Solve for h:

v2 h= 2g

Relate the displacement d of the block along the ramp to h and the angle the ramp makes with the horizontal:

d = h/sinθ

Substitute for h:

v2 d sin θ = 2g

Solve for d:

d=

Substitute numerical values and evaluate d:

d=

v2 2 g sin θ

(

(7 m/s) 2

)

2 9.81 m/s 2 sin40°

= 3.89 m

20 • Picture the Problem Let the system consist of the earth, the block, and the spring. With this choice there are no external forces doing work to change the energy of the system. Let Ug = 0 at the elevation of the spring. Then the initial gravitational potential energy of the 3-kg object is transformed into kinetic energy as it slides down the ramp and then, as it compresses the spring, into potential energy stored in the spring. (a) Apply conservation of energy to relate the distance the spring is compressed to the initial potential energy of the block: Solve for x:

Wext = ∆K + ∆U = 0 and, because ∆K = 0,

− mgh + 12 kx 2 = 0

x=

2mgh k

446 Chapter 7 Substitute numerical values and evaluate x:

x=

(

)

2(3 kg ) 9.81 m/s 2 (5 m ) 400 N/m

= 0.858 m (b) The energy stored in the compressed spring will accelerate the block, launching it back up the incline:

The block will retrace its path, rising to a height of 5 m.

21 • Picture the Problem With Ug chosen to be zero at the uncompressed level of the spring, the ball’s initial gravitational potential energy is negative. The difference between the initial potential energy of the spring and the gravitational potential energy of the ball is first converted into the kinetic energy of the ball and then into gravitational potential energy as the ball rises and slows … eventually coming momentarily to rest. Apply the conservation of energy to the system as it evolves from its initial to its final state:

− mgx + 12 kx 2 = mgh

Solve for h:

h=

Substitute numerical values and evaluate h:

kx 2 −x 2mg

2 ( 600 N/m )(0.05 m ) h= − 0.05 m 2(0.015 kg ) (9.81 m/s 2 )

= 5.05 m 22 • Picture the Problem Let the system include the earth and the container. Then the work done by the crane is done by an external force and this work changes the energy of the system. Because the initial and final speeds of the container are zero, the initial and final kinetic energies are zero and the work done by the crane equals the change in the gravitational potential energy of the container. Choose Ug = 0 to be at the level of the deck of the freighter. Apply conservation of energy to the system: Because ∆K = 0:

Wext = ∆E sys = ∆K + ∆U

Wext = ∆U = mg∆h

Conservation of Energy 447 Evaluate the work done by the crane:

Wext = mg∆h

(

)

= (4000 kg ) 9.81 m/s 2 (− 8 m ) = − 314 kJ

23 • Picture the Problem Let the system consist of the earth and the child. Then Wext = 0. Choose Ug,i = 0 at the child’s lowest point as shown in the diagram to the right. Then the child’s initial energy is entirely kinetic and its energy when it is at its highest point is entirely gravitational potential. We can determine h from energy conservation and then use trigonometry to determine θ. Using the diagram, relate θ to h and L:

h⎞ ⎛ L−h⎞ −1 ⎛ ⎟ = cos ⎜1 − ⎟ ⎝ L⎠ ⎝ L ⎠

θ = cos −1 ⎜

mvi2 − mgh = 0

Apply conservation of energy to the system to obtain:

1 2

Solve for h:

h=

Substitute to obtain:

θ = cos −1 ⎜⎜1 −

vi2 2g ⎛ ⎝

Substitute numerical values and evaluate θ :

vi2 ⎞ ⎟ 2 gL ⎟⎠

2 ⎛ ⎞ ( 3.4 m/s ) ⎜ ⎟ θ = cos ⎜1 − 2 ⎟ ( ) 2 9.81 m/s 6 m ⎝ ⎠ −1

(

)

= 25.6° *24 •• Picture the Problem Let the system include the two objects and the earth. Then Wext = 0. Choose Ug = 0 at the elevation at which the two objects meet. With this choice, the initial potential energy of the 3-kg object is positive and that of the 2-kg object is negative. Their sum, however, is positive. Given our choice for Ug = 0, this initial potential energy is transformed entirely into kinetic energy. Apply conservation of energy:

Wext = ∆K + ∆U g = 0 or, because Wext = 0,

448 Chapter 7 ∆K = −∆Ug

mvf2 − 12 mvi2 = − ∆U g

Substitute for ∆K and solve for vf; noting that m represents the sum of the masses of the objects as they are both moving in the final state:

or, because vi = 0,

Express and evaluate ∆Ug:

∆U g = U g,f − U g,i

1 2

− 2∆U g

vf =

m

= 0 − (3 kg − 2 kg )(0.5 m )

(

× 9.81 m/s 2

)

= −4.91 J Substitute and evaluate vf:

− 2(− 4.91 J ) = 1.40 m/s 5 kg

vf =

25 •• Picture the Problem The free-body diagram shows the forces acting on the block when it is about to move. Fsp is the force exerted by the spring and, because the block is on the verge of sliding, fs = fs,max. We can use Newton’s 2nd law, under equilibrium conditions, to express the elongation of the spring as a function of m, k and θ and then substitute in the expression for the potential energy stored in a stretched or compressed spring. Express the potential energy of the spring when the block is about to move: Apply

r

r

∑ F = ma, under equilibrium

conditions, to the block:

U = 12 kx 2

∑F

x

and

∑F

y

Using fs,max = µsFn and Fsp = kx, eliminate fs,max and Fsp from the x equation and solve for x:

= Fsp − f s,max − mg sin θ = 0

x=

= Fn − mg cosθ = 0

mg (sin θ + µs cos θ ) k

Conservation of Energy 449

Substitute for x in the expression for U:

⎡ mg (sin θ + µs cosθ ) ⎤ U = k⎢ ⎥⎦ k ⎣

2

1 2

=

[mg (sin θ + µs cosθ )] 2 2k

26 •• Picture the Problem The mechanical energy of the system, consisting of the block, the spring, and the earth, is initially entirely gravitational potential energy. Let Ug = 0 where the spring is compressed 15 cm. Then the mechanical energy when the compression of the spring is 15 cm will be partially kinetic and partially stored in the spring. We can use conservation of energy to relate the initial potential energy of the system to the energy stored in the spring and the kinetic energy of block when it has compressed the spring 15 cm. Apply conservation of energy to the system:

∆U + ∆K = 0 or

U g,f − U g,i + U s,f − U s,i + K f − K i = 0 Because Ug,f = Us,I = Ki = 0:

− U g,i + U s,f + K f = 0

Substitute to obtain:

− mg (h + x ) + 12 kx 2 + 12 mv 2 = 0

Solve for v:

v = 2 g (h + x ) −

kx 2 m

Substitute numerical values and evaluate v:

(

)

v = 2 9.81 m/s 2 (5 m + 0.15 m ) −

(3955 N/m )(0.15 m )2 2.4 kg

= 8.00 m/s

450 Chapter 7 *27 •• Picture the Problem The diagram represents the ball traveling in a circular path with constant energy. Ug has been chosen to be zero at the lowest point on the circle and the superimposed free-body diagrams show the forces acting on the ball at the top and bottom of the circular path. We’ll apply Newton’s 2nd law to the ball at the top and bottom of its path to obtain a relationship between TT and TB and the conservation of mechanical energy to relate the speeds of the ball at these two locations. Apply

∑F

radial

= maradial to the ball

at the bottom of the circle and solve for TB:

vB2 TB − mg = m R and

vB2 TB = mg + m R Apply

∑F

radial

= maradial to the ball

at the top of the circle and solve for TT:

TT + mg = m

vT2 R

and

TT = −mg + m Subtract equation (2) from equation (1) to obtain:

(1)

vT2 R

TB − TT = mg + m

(2)

vB2 R

⎛ vT2 ⎞ ⎜ − ⎜ − mg + m ⎟⎟ R⎠ ⎝ 2 2 v v = m B − m T + 2mg R R Using conservation of energy, relate the mechanical energy of the ball at the bottom of its path to its mechanical energy at the top of the circle and solve for m

1 2

mvB2 = 12 mvT2 + mg (2 R )

m

vB2 v2 − m T = 4mg R R

vB2 v2 −m T : R R

Substitute in equation (3) to obtain:

TB − TT = 6mg

(3)

Conservation of Energy 451 28 •• Picture the Problem Let Ug = 0 at the lowest point in the girl’s swing. Then we can equate her initial potential energy to her kinetic energy as she passes through the low point on her swing to relate her speed v to R. The FBD show the forces acting on the girl at the low point of her swing. Applying Newton’s 2nd law to her will allow us to establish the relationship between the tension T and her speed.

Apply

∑F

radial

= maradial to the girl

T − mg = m

at her lowest point and solve for T:

v2 R

and T = mg + m

Equate the girl’s initial potential energy to her final kinetic energy and solve for

mg

v2 R

R 1 2 v2 = 2 mv ⇒ =g 2 R

v2 : R

Substitute for v2/R2 and simplify to obtain:

T = mg + mg = 2mg

29 •• Picture the Problem The free-body diagram shows the forces acting on the car when it is upside down at the top of the loop. Choose Ug = 0 at the bottom of the loop. We can express Fn in terms of v and R by apply Newton’s 2nd law to the car and then obtain a second expression in these same variables by applying the conservation of mechanical energy. The simultaneous solution of these equations will yield an expression for Fn in terms of known quantities. Apply

∑F

radial

= maradial to the car

at the top of the circle and solve for Fn :

Fn + mg = m

v2 R

and

Fn = m

v2 − mg R

(1)

452 Chapter 7 Using conservation of energy, relate the energy of the car at the beginning of its motion to its energy when it is at the top of the loop:

Solve for m

v2 : R

mgH = 12 mv 2 + mg (2 R )

m

Substitute equation (2) in equation (1) to obtain:

v2 ⎛H ⎞ = 2mg ⎜ − 2 ⎟ R ⎝R ⎠

⎛H ⎞ Fn = 2mg ⎜ − 2 ⎟ − mg ⎝R ⎠ ⎛ 2H ⎞ = mg ⎜ − 5⎟ ⎝ R ⎠

Substitute numerical values and evaluate Fn:

⎡ 2(23 m ) ⎤ − 5⎥ = 1.67 × 10 4 N ⇒ (c) is correct. Fn = (1500 kg ) 9.81 m/s 2 ⎢ ⎣ 7.5 m ⎦

(

)

30 • Picture the Problem Let the system include the roller coaster, the track, and the earth and denote the starting position with the numeral 0 and the top of the second hill with the numeral 1. We can use the workenergy theorem to relate the energies of the coaster at its initial and final positions.

(a) Use conservation of energy to relate the work done by external forces to the change in the energy of the system: Because the track is frictionless, Wext = 0:

Wext = ∆Esys = ∆K + ∆U

∆K + ∆U = 0 and

K1 − K 0 + U1 − U 0 = 0 mv12 − 12 mv02 + mgh1 − mgh0 = 0

Substitute to obtain:

1 2

Solve for v0:

v0 = v12 + 2 g (h1 − h0 )

If the coaster just makes it to the top of the second hill, v1 = 0 and:

v0 = 2 g (h1 − h0 )

(2)

Conservation of Energy 453 Substitute numerical values and evaluate v0:

(b)

(

)

v0 = 2 9.81 m/s 2 (9.5 m − 5 m ) = 9.40 m/s

No. Note that the required speed depends only on the difference in the heights of the two hills.

31 •• Picture the Problem Let the radius of the loop be R and the mass of one of the riders be m. At the top of the loop, the centripetal force on her is her weight (the force of gravity). The two forces acting on her at the bottom of the loop are the normal force exerted by the seat of the car, pushing up, and the force of gravity, pulling down. We can apply Newton’s 2nd law to her at both the top and bottom of the loop to relate the speeds at those locations to m and R and, at b, to F, and then use conservation of energy to relate vt and vb. Apply

∑F

radial

= ma radial to the

rider at the bottom of the circular arc: Solve for F to obtain:

Apply

∑F

radial

= ma radial to the

rider at the top of the circular arc:

F − mg = m

vb2 R

F = mg + m

vb2 R

mg = m

(1)

vt2 R

Solve for vt2 :

vt2 = gR

Use conservation of energy to relate the energies of the rider at the top and bottom of the arc:

Kb − Kt + U b − U t = 0 or, because Ub = 0,

Kb − Kt − U t = 0

mvb2 − 12 mvt2 − 2mgR = 0

Substitute to obtain:

1 2

Solve for vb2 :

vb2 = 5 gR

Substitute in equation (1) to obtain:

F = mg + m

5 gR = 6mg R

i.e., the rider will feel six times heavier than her normal weight.

454 Chapter 7 *32 •• Picture the Problem Let the system consist of the stone and the earth and ignore the influence of air resistance. Then Wext = 0. Choose Ug = 0 as shown in the figure. Apply the law of the conservation of mechanical energy to describe the energy transformations as the stone rises to the highest point of its trajectory.

Apply conservation of energy:

Wext = ∆K + ∆U = 0 and

K1 − K 0 + U1 − U 0 = 0 Because U0 = 0:

K1 − K 0 + U1 = 0

Substitute to obtain:

1 2

mvx2 − 12 mv 2 + mgH = 0

In the absence of air resistance, the r horizontal component of v is constant and equal to vx = vcosθ. Hence:

1 2

m(v cosθ ) − 12 mv 2 + mgH = 0

Solve for v:

Substitute numerical values and evaluate v:

2

v=

2 gH 1 − cos 2 θ

v=

2 9.81 m/s 2 (24 m ) = 27.2 m/s 1 − cos 2 53°

(

)

33 •• Picture the Problem Let the system consist of the ball and the earth. Then Wext = 0. The figure shows the ball being thrown from the roof of a building. Choose Ug = 0 at ground level. We can use the conservation of mechanical energy to determine the maximum height of the ball and its speed at impact with the ground. We can use the definition of the work done by gravity to calculate how much work was done by gravity as the ball rose to its maximum height. (a) Apply conservation of energy:

Wext = ∆K + ∆U = 0

Conservation of Energy 455 or

K 2 − K1 + U 2 − U 1 = 0 Substitute for the energies to obtain: Note that, at point 2, the ball is moving horizontally and: Substitute for v2 and h2:

1 2

mv 22 − 12 mv12 + mgh2 − mgh1 = 0

v 2 = v1 cosθ

1 2

m(v1 cosθ ) − 12 mv12 + mgH 2

− mgh1 = 0

(

)

v2 cos 2 θ − 1 2g

Solve for H:

H = h1 −

Substitute numerical values and evaluate H:

H = 12 m −

(30 m/s)2

(cos 2(9.81 m/s ) 2

2

)

40° − 1

= 31.0 m (b) Using its definition, express the work done by gravity:

Substitute numerical values and evaluate Wg:

(

Wg = − ∆U = − U H − U hi

)

= −(mgH − mghi ) = −mg (H − hi )

(

)

Wg = −(0.17 kg ) 9.81 m/s 2 (31 m − 12 m ) = − 31.7 J

mvi2 + mghi = 12 mvf2

(c) Relate the initial mechanical energy of the ball to its just-beforeimpact energy:

1 2

Solve for vf:

vf = vi2 + 2ghi

Substitute numerical values and evaluate vf

vf =

(30 m/s) 2 + 2(9.81m/s2 )(12 m )

= 33.7 m/s

456 Chapter 7 34 •• Picture the Problem The figure shows the pendulum bob in its release position and in the two positions in which it is in motion with the given speeds. Choose Ug = 0 at the low point of the swing. We can apply the conservation of mechanical energy to relate the two angles of interest to the speeds of the bob at the intermediate and low points of its trajectory. (a) Apply conservation of energy:

Wext = ∆K + ∆U = 0 or

Kf − Ki + U f − U i = 0

where U f and K i equal zero. ∴Kf − U i = 0 Express Ui:

U i = mgh = mgL(1 − cos θ 0 )

Substitute for Kf and Ui:

1 2

Solve for θ0:

⎛ v2 ⎞ ⎟⎟ θ 0 = cos ⎜⎜1 − ⎝ 2 gL ⎠

Substitute numerical values and evaluate θ0:

θ 0 = cos −1 ⎢1 −

mvf2 − mgL(1 − cos θ 0 ) = 0 −1

⎡ ⎣

⎤ (2.8 m/s)2 ⎥ 2 2(9.81m/s )(0.8 m ) ⎦

= 60.0° (b) Letting primed quantities describe the indicated location, use the law of the conservation of mechanical energy to relate the speed of the bob at this point to θ : Express U f' : Substitute for K f' , U f' and U i :

K f' − K i + U f' − U i = 0 where K i = 0. ∴K f ' + U f ' − U i = 0

U f' = mgh' = mgL(1 − cosθ ) 1 2

m(vf' ) + mgL(1 − cos θ ) 2

− mgL(1 − cos θ 0 ) = 0

Conservation of Energy 457 Solve for θ :

Substitute numerical values and evaluate θ :

⎡ (vf ')2

⎤ + cosθ 0 ⎥ ⎣ 2 gL ⎦

θ = cos −1 ⎢

⎤ (1.4 m/s) 2 + cos 60°⎥ 2 ⎣ 2(9.81m/s )(0.8 m ) ⎦ ⎡

θ = cos −1 ⎢

= 51.3°

*35 •• Picture the Problem Choose Ug = 0 at the bridge, and let the system be the earth, the jumper and the bungee cord. Then Wext = 0. Use the conservation of mechanical energy to relate to relate her initial and final gravitational potential energies to the energy stored in the stretched bungee, Us cord. In part (b), we’ll use a similar strategy but include a kinetic energy term because we are interested in finding her maximum speed. (a) Express her final height h above the water in terms of L, d and the distance x the bungee cord has stretched: Use the conservation of mechanical energy to relate her gravitational potential energy as she just touches the water to the energy stored in the stretched bungee cord:

h=L–d−x

(1)

Wext = ∆K + ∆U = 0 Because ∆K = 0 and ∆U = ∆Ug + ∆Us,

− mgL + 12 kx 2 = 0, where x is the maximum distance the bungee cord has stretched.

2mgL x2

Solve for k:

k=

Find the maximum distance the bungee cord stretches:

x = 310 m – 50 m = 260 m.

Evaluate k:

k=

2(60 kg ) (9.81m/s 2 )(310 m ) (260 m )2 = 5.40 N/m

458 Chapter 7 Fnet = kx − mg = 0

Express the relationship between the forces acting on her when she has finally come to rest and solve for x:

and

Evaluate x:

( 60 kg ) (9.81m/s 2 ) x= = 109 m

x=

mg k

5.40 N/m

Substitute in equation (1) and evaluate h: (b) Using conservation of energy, express her total energy E: Because v is a maximum when K is a maximum, solve for K::

h = 310 m − 50 m − 109 m = 151 m

E = K + U g + U s = Ei = 0 K = −U g − U s

(1)

= mg (d + x ) − 12 kx 2

Use the condition for an extreme value to obtain:

dK = mg − kx = 0 for extreme values dx

Solve for and evaluate x:

x=

From equation (1) we have:

1 2

Solve for v to obtain:

v = 2 g (d + x ) −

(

)

mg (60 kg ) 9.81m/s 2 = = 109 m k 5.40 N/m

mv 2 = mg (d + x ) − 12 kx 2 kx 2 m

Substitute numerical values and evaluate v for x = 109 m:

(

v = 2 9.81 m/s

Because

2

2 ( 5.4 N/m )(109 m ) )(50 m + 109 m) −

60 kg

= 45.3 m/s

d 2K = −k < 0, x = 109 m corresponds to Kmax and so v is a maximum. dx 2

Conservation of Energy 459 36 •• Picture the Problem Let the system be the earth and pendulum bob. Then Wext = 0. Choose Ug = 0 at the low point of the bob’s swing and apply the law of the conservation of mechanical energy to its motion. When the bob reaches the 30° position its energy will be partially kinetic and partially potential. When it reaches its maximum height, its energy will be entirely potential. Applying Newton’s 2nd law will allow us to express the tension in the string as a function of the bob’s speed and its angular position. (a) Apply conservation of energy to relate the energies of the bob at points 1 and 2:

Wext = ∆K + ∆U = 0 or K 2 − K1 + U 2 − U 1 = 0 1 2

Because U1 = 0:

mv22 − 12 mv12 + U 2 = 0

U 2 = mgL(1 − cos θ )

Express U2: Substitute for U2 to obtain:

1 2

mv22 − 12 mv12 + mgL(1 − cosθ ) = 0

v2 = v12 − 2 gL(1 − cos θ )

Solve for v2: Substitute numerical values and evaluate v2:

v2 =

(4.5 m/s )2 − 2(9.81m/s 2 )(3 m )(1 − cos30°) =

(b) From (a) we have: Substitute numerical values and evaluate U2: (c) Apply

∑F

obtain: Solve for T:

radial

= maradial to the bob to

3.52 m/s

U 2 = mgL(1 − cos θ )

(

)

U 2 = (2 kg ) 9.81 m/s 2 (3 m )(1 − cos30°) = 7.89 J v22 T − mg cos θ = m L ⎛ v22 ⎞ T = m⎜⎜ g cos θ + ⎟⎟ L⎠ ⎝

460 Chapter 7 Substitute numerical values and evaluate T: 2 ⎡ ( 3.52 m/s ) ⎤ 2 T = (2 kg )⎢ 9.81 m/s cos30° + ⎥ = 25.3 N 3m ⎣ ⎦

(

)

(d) When the bob reaches its greatest height:

Substitute for K1 and Umax Solve for θmax:

U = U max = mgL(1 − cos θ max ) and K1 + U max = 0 − 12 mv12 + mgL(1 − cos θ max ) = 0

θ max

Substitute numerical values and evaluate θmax:

⎛ v12 ⎞ ⎜ ⎟⎟ = cos ⎜1 − 2 gL ⎝ ⎠ −1

⎡

θ max = cos −1 ⎢1 − ⎣

= 49.0° 37 •• Picture the Problem Let the system consist of the earth and pendulum bob. Then Wext = 0. Choose Ug = 0 at the bottom of the circle and let points 1, 2 and 3 represent the bob’s initial point, lowest point and highest point, respectively. The bob will gain speed and kinetic energy until it reaches point 2 and slow down until it reaches point 3; so it has its maximum kinetic energy when it is at point 2. We can use Newton’s 2nd law at points 2 and 3 in conjunction with the law of the conservation of mechanical energy to find the maximum kinetic energy of the bob and the tension in the string when the bob has its maximum kinetic energy. (a) Apply

∑F

radial

= maradial to the

bob at the top of the circle and solve for v32 :

mg = m and

v32 = gL

v32 L

(4.5 m/s) 2 ⎤ ⎥ 2(9.81 m/s 2 )(3 m ) ⎦

Conservation of Energy 461 Use conservation of energy to express the relationship between K2, K3 and U3 and solve for K2:

K 3 − K 2 + U 3 − U 2 = 0 where U 2 = 0 Therefore,

K 2 = K max = K 3 + U 3

= 12 mv32 + mg (2 L )

Substitute for v32 and simplify to

K max = 12 m(gL ) + 2mgL =

5 2

mgL

obtain: (b) Apply

∑F

radial

= maradial to the

bob at the bottom of the circle and solve for T2:

Fnet = T2 − mg = m and

T2 = mg + m

v22 L

(1)

K 3 − K 2 + U 3 − U 2 = 0 where U 2 = 0

Use conservation of energy to relate the energies of the bob at points 2 and 3 and solve for K2:

K 2 = K3 + U 3

Substitute for v32 and K2 and solve

1 2

= 12 mv32 + mg (2 L )

mv22 = 12 m( gL ) + mg (2 L )

for v22 :

and

Substitute in equation (1) to obtain:

T2 = 6mg

38 •• Picture the Problem Let the system consist of the earth and child. Then Wext = 0. In the figure, the child’s initial position is designated with the numeral 1; the point at which the child releases the rope and begins to fall with a 2, and its point of impact with the water is identified with a 3. Choose Ug = 0 at the water level. While one could use the law of the conservation of energy between points 1 and 2 and then between points 2 and 3, it is more direct to consider the energy transformations between points 1 and 3. Given our choice of the zero of gravitational potential energy, the initial potential energy at point 1 is transformed into kinetic energy at point 3.

v22 L

v22 = 5 gL

462 Chapter 7

Apply conservation of energy to the energy transformations between points 1 and 3:

Wext = ∆K + ∆U = 0 K 3 − K1 + U 3 − U1 = 0

where U 3 and K1are zero. mv32 − mg [h + L(1 − cos θ )] = 0

Substitute for K3 and U1;

1 2

Solve for v3:

v3 = 2 g [h + L(1 − cos θ )]

Substitute numerical values and evaluate v3:

(

)

v3 = 2 9.81m/s 2 [3.2 m + (10.6 m )(1 − cos23°)] = 8.91m/s *39 •• Picture the Problem Let the system consist of you and the earth. Then there are no external forces to do work on the system and Wext = 0. In the figure, your initial position is designated with the numeral 1, the point at which you release the rope and begin to fall with a 2, and your point of impact with the water is identified with a 3. Choose Ug = 0 at the water level. We can apply Newton’s 2nd law to the forces acting on you at point 2 and apply conservation of energy between points 1 and 2 to determine the maximum angle at which you can begin your swing and then between points 1 and 3 to determine the speed with which you will hit the water. (a) Use conservation of energy to relate your speed at point 2 to your potential energy there and at point 1:

Wext = ∆K + ∆U = 0 or

K 2 − K1 + U 2 − U 1 = 0 1 2

Because K1 = 0:

Solve this equation for θ :

mv22 + mgh

− [mgL(1 − cos θ ) + mgh] = 0 ⎡

θ = cos −1 ⎢1 − ⎣

Apply

∑F

radial

= maradial to

T − mg = m

v22 ⎤ ⎥ 2 gL ⎦

v22 L

(1)

Conservation of Energy 463 yourself at point 2 and solve for T:

and

v22 T = mg + m L Because you’ve estimated that the rope might break if the tension in it exceeds your weight by 80 N, it must be that:

v22 m = 80 N L or (80 N )L v22 = m

Let’s assume your weight is 650 N. Then your mass is 66.3 kg and:

v22 =

Substitute numerical values in equation (1) to obtain:

θ = cos −1 ⎢1 −

(80 N )(4.6 m ) = 5.55 m 2 /s 2 66.3kg ⎡ ⎣

⎤ 5.55 m 2 /s 2 2 2 9.81 m/s (4.6 m )⎥⎦

(

)

= 20.2° (b) Apply conservation of energy to the energy transformations between points 1 and 3:

Wext = ∆K + ∆U = 0 K 3 − K1 + U 3 − U 1 = 0 where U 3 and K1are zero mv32 − mg [h + L(1 − cos θ )] = 0

Substitute for K3 and U1 to obtain:

1 2

Solve for v3:

v3 = 2 g [h + L(1 − cos θ )]

Substitute numerical values and evaluate v3:

(

)

v3 = 2 9.81 m/s 2 [1.8 m + (4.6 m )(1 − cos20.2°)] = 6.39 m/s

464 Chapter 7 40 •• Picture the Problem Choose Ug = 0 at point 2, the lowest point of the bob’s trajectory and let the system consist of the bob and the earth. Given this choice, there are no external forces doing work on the system. Because θ << 1, we can use the trigonometric series for the sine and cosine functions to approximate these functions. The bob’s initial energy is partially gravitational potential and partially potential energy stored in the stretched spring. As the bob swings down to point 2 this energy is transformed into kinetic energy. By equating these energies, we can derive an expression for the speed of the bob at point 2. Apply conservation of energy to the system as the pendulum bob swings from point 1 to point 2: Note, from the figure, that x ≈ Lsinθ when θ << 1: Also, when θ << 1: Substitute, simplify and solve for v2:

1 2

mv 22 = 12 kx 2 + mgL(1 − cosθ )

1 2

mv 22 = 12 k (L sin θ ) + mgL(1 − cosθ ) 2

sin θ ≈ θ and cosθ ≈ 1 − 12 θ 2 v 2 = Lθ

k g + m L

Conservation of Energy 465 41

•••

Picture the Problem Choose Ug = 0 at point 2, the lowest point of the bob’s trajectory and let the system consist of the earth, ceiling, spring, and pendulum bob. Given this choice, there are no external forces doing work to change the energy of the system. The bob’s initial energy is partially gravitational potential and partially potential energy stored in the stretched spring. As the bob swings down to point 2 this energy is transformed into kinetic energy. By equating these energies, we can derive an expression for the speed of the bob at point 2. 1 2

Apply conservation of energy to the system as the pendulum bob swings from point 1 to point 2:

mv 22 = 12 kx 2 + mgL(1 − cosθ ) (1)

Apply the Pythagorean theorem to the lower triangle in the diagram to obtain:

(x + 12 L )2 = L2 [sin 2 θ + ( 32 cosθ )2 ] = L2 [sin 2 θ + 94 − 3 cosθ + cos 2 θ ] = L2 (134 − 3 cosθ ) Take the square root of both sides of the equation to obtain:

x + 12 L = L

Solve for x:

x=L

[(

13 4

(134 − 3 cosθ )

− 3 cos θ ) − 12

]

Substitute for x in equation (1): 1 2

mv22 = 12 kL2

[(

13 4

− 3 cos θ ) − 12

]

2

+ mgL(1 − cosθ )

Solve for v22 to obtain:

v22 = 2 gL(1 − cos θ ) +

[ (

k 2 L m k ⎡ g = L2 ⎢2 (1 − cos θ ) + m ⎣ L

13 4

− 3 cosθ − 12

13 4

]

− 3 cosθ − 12

2

)

2

⎤ ⎥⎦

466 Chapter 7 Finally, solve for v2:

v2 = L 2

g (1 − cosθ ) + k L m

(

13 4

− 3 cos θ − 12

)

2

The Conservation of Energy 42 • Picture the Problem The energy of the eruption is initially in the form of the kinetic energy of the material it thrusts into the air. This energy is then transformed into gravitational potential energy as the material rises.

E = mg∆h

(a) Express the energy of the eruption in terms of the height ∆h to which the debris rises:

m V

Relate the density of the material to its mass and volume:

ρ=

Substitute for m to obtain:

E = ρVg∆h

Substitute numerical values and evaluate E:

(

)(

)(

)

E = 1600 kg/m 3 4 km 3 9.81 m/s 2 (500 m ) = 3.14 × 1016 J (b) Convert 3.13×1016 J to megatons of TNT:

3.14 × 1016 J = 3.14 × 1016 J ×

1Mton TNT = 7.48 Mton TNT 4.2 × 1015 J

43 •• Picture the Problem The work done by the student equals the change in his/her gravitational potential energy and is done as a result of the transformation of metabolic energy in the climber’s muscles. (a) The increase in gravitational potential energy is:

∆U = mg∆h

(

)

= (80 kg ) 9.81 m/s 2 (120 m ) = 94.2 kJ

Conservation of Energy 467 (b)

The energy required to do this work comes from chemical energy stored in the body.

(c) Relate the chemical energy expended by the student to the change in his/her potential energy and solve for E:

0.2 E = ∆U and

E = 5∆U = 5(94.2 kJ ) = 471 kJ

Kinetic Friction 44 • Picture the Problem Let the car and the earth be the system. As the car skids to a stop on a horizontal road, its kinetic energy is transformed into internal (i.e., thermal) energy. Knowing that energy is transformed into heat by friction, we can use the definition of the coefficient of kinetic friction to calculate its value. (a) The energy dissipated by friction is given by: Apply the work-energy theorem for problems with kinetic friction:

f∆s = ∆Etherm

Wext = ∆Emech + ∆Etherm = ∆Emech + f∆s or, because ∆Emech = ∆K = − K i and Wext = 0,

0 = − 12 mvi2 + f∆s Solve for f∆s to obtain:

f∆s = 12 mvi2

Substitute numerical values and evaluate f∆s:

f∆s =

(b) Relate the kinetic friction force to the coefficient of kinetic friction and the weight of the car and solve for the coefficient of kinetic friction:

f k = µ k mg ⇒ µ k =

Express the relationship between the energy dissipated by friction and the kinetic friction force and solve fk:

∆Etherm = f k ∆s ⇒ f k =

Substitute to obtain:

µk =

1 2

(2000 kg )(25 m/s)2 =

∆Etherm mg∆s

fk mg

∆Etherm ∆s

625 kJ

468 Chapter 7 Substitute numerical values and evaluate µk:

µk =

625 kJ (2000 kg ) 9.81 m/s 2 (60 m )

(

)

= 0.531 45 • Picture the Problem Let the system be the sled and the earth. Then the 40-N force is external to the system. The free-body diagram shows the forces acting on the sled as it is pulled along a horizontal road. The work done by the applied force can be found using the definition of work. To find the energy dissipated by friction, we’ll use Newton’s 2nd law to determine fk and then use it in the definition of work. The change in the kinetic energy of the sled is equal to the net work done on it. Finally, knowing the kinetic energy of the sled after it has traveled 3 m will allow us to solve for its speed at that location. (a) Use the definition of work to calculate the work done by the applied force: (b) Express the energy dissipated by friction as the sled is dragged along the surface: Apply

∑F

y

= ma y to the sled and

r r Wext ≡ F ⋅ s = Fs cosθ = (40 N )(3 m )cos 30° = 104 J

∆Etherm = f∆x = µ k Fn ∆x

Fn + F sin θ − mg = 0

solve for Fn:

and

Substitute to obtain:

∆Etherm = µ k ∆x(mg − F sin θ )

Substitute numerical values and evaluate ∆Etherm:

Fn = mg − F sin θ

[

(

∆Etherm = (0.4)(3 m ) (8 kg ) 9.81m/s 2 − (40 N )sin30°]

)

= 70.2 J (c) Apply the work-energy theorem

Wext = ∆Emech + ∆Etherm = ∆Emech + f∆s

Conservation of Energy 469 for problems with kinetic friction:

or, because ∆Emech = ∆K + ∆U and ∆U = 0,

Wext = ∆K + ∆Etherm

Solve for and evaluate ∆K to obtain:

∆K = Wext − ∆Etherm = 104 J − 70.2 J = 33.8 J

(d) Because Ki = 0:

K f = ∆K = 12 mvf2

Solve for vf:

vf =

2∆K m

Substitute numerical values and evaluate vf:

vf =

2(33.8 J ) = 2.91 m/s 8 kg

*46 • Picture the Problem Choose Ug = 0 at the foot of the ramp and let the system consist of the block, ramp, and the earth. Then the kinetic energy of the block at the foot of the ramp is equal to its initial kinetic energy less the energy dissipated by friction. The block’s kinetic energy at the foot of the incline is partially converted to gravitational potential energy and partially dissipated by friction as the block slides up the incline. The free-body diagram shows the forces acting on the block as it slides up the incline. Applying Newton’s 2nd law to the block will allow us to determine fk and express the energy dissipated by friction.

(a) Apply conservation of energy to the system while the block is moving horizontally:

Wext = ∆Emech + ∆Etherm = ∆K + ∆U + f∆s or, because ∆U = Wext = 0,

0 = ∆K + f∆s = K f − K i + f∆s

Solve for Kf:

K f = K i − f∆s

470 Chapter 7 mvf2 = 12 mvi2 − µ k mg∆x

Substitute for Kf, Ki, and f∆s to obtain:

1 2

Solving for vf yields:

vf = vi2 − 2µ k g∆x

Substitute numerical values and evaluate vf:

vf =

(b) Apply conservation of energy to the system while the block is on the incline:

Wext = ∆Emech + ∆Etherm

(7 m/s)2 − 2(0.3)(9.81m/s2 )(2 m )

= 6.10 m/s

= ∆K + ∆U + f∆s or, because Kf = Wext = 0,

0 = − K i + ∆U + f∆s

Apply

∑F

y

= ma y to the block

Fn − mg cos θ = 0 ⇒ Fn = mg cos θ

when it is on the incline: Express f∆s: The final potential energy of the block is: Substitute for Uf, Ui, and f∆s to obtain:

f∆s = f k L = µ k Fn L = µ k mgL cosθ U f = mgL sin θ 0 = − K i + mgL sin θ + µ k mgL cos θ

Solving for L yields:

L=

vi2 g (sin θ + µ k cos θ )

Substitute numerical values and evaluate L:

L=

(6.10 m/s)2 (9.81 m/s 2 )(sin40° + (0.3)cos40°)

1 2

1 2

= 2.17 m 47 • Picture the Problem Let the system include the block, the ramp and horizontal surface, and the earth. Given this choice, there are no external forces acting that will change the energy of the system. Because the curved ramp is frictionless, mechanical energy is conserved as the block slides down it. We can calculate its speed at the bottom of the ramp by using the law of the conservation of energy. The potential energy of the block at the top of the ramp or, equivalently, its kinetic energy at the bottom of the ramp is

Conservation of Energy 471 converted into thermal energy during its slide along the horizontal surface. (a) Choosing Ug = 0 at point 2 and letting the numeral 1 designate the initial position of the block and the numeral 2 its position at the foot of the ramp, use conservation of energy to relate the block’s potential energy at the top of the ramp to its kinetic energy at the bottom: Solve for v2 to obtain: Substitute numerical values and evaluate v2: (b) The energy dissipated by friction is responsible for changing the thermal energy of the system: Because ∆K = 0 for the slide: Substitute numerical values and evaluate Wf:

(c) The energy dissipated by friction is given by:

Wext = ∆Emech + ∆Etherm or, because Wext = Ki = Uf = ∆Etherm = 0,

0 = 12 mv22 − mg∆h = 0

v2 = 2 g∆h

(

)

v2 = 2 9.81 m/s 2 (3 m ) = 7.67 m/s Wf + ∆K + ∆U = ∆E therm + ∆K + ∆U = 0

Wf = −∆U = −(U 2 − U1 ) = U1

(

)

Wf = mg∆h = (2 kg ) 9.81 m/s 2 (3 m ) = 58.9 J ∆Etherm = f∆s = µ k mg∆x

Solve for µk:

µk =

∆Etherm mg∆x

Substitute numerical values and evaluate µk:

µk =

58.9 J = 0.333 (2 kg ) 9.81 m/s 2 (9 m )

(

)

48 •• Picture the Problem Let the system consist of the earth, the girl, and the slide. Given this choice, there are no external forces doing work to change the energy of the system. By the time she reaches the bottom of the slide, her potential energy at the top of the slide has been converted into kinetic and thermal energy. Choose Ug = 0 at the bottom of the slide and denote the top and bottom of the slide as shown in

472 Chapter 7 the figure. We’ll use the work-energy theorem with friction to relate these quantities and the forces acting on her during her slide to determine the friction force that transforms some of her initial potential energy into thermal energy.

Wext = ∆K + ∆U + ∆Etherm = 0

(a) Express the work-energy theorem:

0 = K 2 − U1 + ∆Etherm = 0

Because U2 = K1 = Wext = 0:

or ∆Etherm = U1 − K 2 = mg∆h − 12 mv22 Substitute numerical values and evaluate ∆Etherm:

(

)

∆Etherm = (20 kg ) 9.81 m/s 2 (3.2 m ) − 12 (20 kg )(1.3 m/s ) = 611 J (b) Relate the energy dissipated by friction to the kinetic friction force and the distance over which this force acts and solve for µk: Apply

∑F

y

= ma y to the girl and

2

∆Etherm = f∆s = µ k Fn ∆s and

µk =

∆Etherm Fn ∆s

Fn − mg cos θ = 0 ⇒ Fn = mg cosθ

solve for Fn: Referring to the figure, relate ∆h to ∆s and θ:

∆s =

Substitute for ∆s and Fn to obtain:

µk =

Substitute numerical values and evaluate µk:

∆h sin θ ∆Etherm ∆Etherm tan θ = ∆h mg∆h cos θ mg sin θ

Conservation of Energy 473

µk =

(611 J )tan20° = (20 kg )(9.81 m/s 2 )(3.2 m )

0.354

49 •• Picture the Problem Let the system consist of the two blocks, the shelf, and the earth. Given this choice, there are no external forces doing work to change the energy of the system. Due to the friction between the 4-kg block and the surface on which it slides, not all of the energy transformed during the fall of the 2-kg block is realized in the form of kinetic energy. We can find the energy dissipated by friction and then use the workenergy theorem with kinetic friction to find the speed of either block when they have moved the given distance.

∆Etherm = f∆s = µ k m1 gy

(a) The energy dissipated by friction when the 2-kg block falls a distance y is given by:

(

)

∆Etherm = (0.35)(4 kg ) 9.81 m/s 2 y

Substitute numerical values and evaluate ∆Etherm:

= (13.7 N ) y Wext = ∆Emech + ∆Etherm

(b) From the work-energy theorem with kinetic friction we have:

or, because Wext = 0,

∆Emech = −∆Etherm = − (13.7 N ) y

(m1 + m2 )v 2 − m2 gy = −∆Etherm

(c) Express the total mechanical energy of the system:

1 2

Solve for v to obtain:

v=

2(m2 gy − ∆E therm ) m1 + m2

(1)

Substitute numerical values and evaluate v: v=

[

(

)

]

2 (2 kg ) 9.81 m/s 2 (2 m ) − (13.73 N )(2 m ) = 1.98 m/s 4 kg + 2 kg

*50 •• Picture the Problem Let the system consist of the particle, the table, and the earth. Then Wext = 0 and the energy dissipated by friction during one revolution is the change in the thermal energy of the system. (a) Apply the work-energy theorem

Wext = ∆K + ∆U + ∆Etherm

474 Chapter 7 with kinetic friction to obtain:

or, because ∆U = Wext = 0,

0 = ∆K + ∆Etherm

Substitute for ∆Kf and simplify to obtain:

(

∆Etherm = − 12 mvf2 − 12 mvi2

[

= − 12 m( 12 v0 ) − 12 m(v0 ) =

(b) Relate the energy dissipated by friction to the distance traveled and the coefficient of kinetic friction: Substitute for ∆E and solve for µk to obtain:

(c)

)

3 8

2

2

]

mv02

∆Etherm = f∆s = µ k mg∆s = µ k mg (2πr )

µk =

3 mv 2 ∆Etherm 3v02 = 8 0 = 2πmgr 2πmgr 16πgr

Because it lost 34 K i in one revolution, it will only require another 1/3 revolution to lose the remaining 14 K i .

51 •• Picture the Problem The box will slow down and stop due to the dissipation of thermal energy. Let the system be the earth, the box, and the inclined plane and apply the work-energy theorem with friction. With this choice of the system, there are no external forces doing work to change the energy of the system. The freebody diagram shows the forces acting on the box when it is moving up the incline. Apply the work-energy theorem with friction to the system: Substitute for ∆K, ∆U, and ∆Etherm to obtain: Referring to the FBD, relate the normal force to the weight of the box and the angle of the incline: Relate ∆h to the distance L along the

Wext = ∆Emech + ∆Etherm

= ∆K + ∆U + ∆Etherm 0 = 12 mv12 − 12 mv02 + mg∆h + µ k Fn L Fn = mg cos θ ∆h = L sin θ

(1)

Conservation of Energy 475 incline: Substitute in equation (1) to obtain:

µ k mgL cos θ + 12 mv12 − 12 mv02 + mgL sin θ = 0

(2)

Solving equation (2) for L yields:

L=

v02 2 g (µ k cos θ + sin θ )

Substitute numerical values and evaluate L:

L=

(3.8 m/s)2 2(9.81 m/s 2 )[(0.3)cos37° + sin37°]

= 0.875 m Let vf represent the box’s speed as it passes its starting point on the way down the incline. For the block’s descent, equation (2) becomes: Set v1 = 0 (the block starts from rest at the top of the incline) and solve for vf :

µ k mgL cos θ + 12 mvf2 − 12 mv12 − mgL sin θ = 0

vf = 2 gL(sin θ − µk cosθ )

Substitute numerical values and evaluate vf:

(

]

)

vf = 2 9.81m/s2 (0.875 m)[sin37° − (0.3)cos37°] = 2.49 m/s 52 ••• Picture the Problem Let the system consist of the earth, the block, the incline, and the spring. With this choice of the system, there are no external forces doing work to change the energy of the system. The free-body diagram shows the forces acting on the block just before it begins to move. We can apply Newton’s 2nd law to the block to obtain an expression for the extension of the spring at this instant. We’ll apply the work-energy theorem with friction to the second part of the problem. (a) Apply

r

r

∑ F = ma to the block

∑F

x

= Fspring − f s,max − mg sin θ = 0

476 Chapter 7 when it is on the verge of sliding:

Eliminate Fn, fs,max, and Fspring between the two equations to obtain: Solve for and evaluate d:

(b) Begin with the work-energy theorem with friction and no work being done by an external force: Because the block is at rest in both its initial and final states, ∆K = 0 and: Let Ug = 0 at the initial position of the block. Then: Express the change in the energy stored in the spring as it relaxes to its unstretched length: The energy dissipated by friction is:

and

∑F

y

= Fn − mg cos θ = 0

kd − µ s mg cos θ − mg sin θ = 0

mg (sin θ + µ s cos θ ) k

d=

Wext = ∆Emech + ∆Etherm

= ∆K + ∆U g + ∆U s + ∆Etherm ∆U g + ∆U s + ∆Etherm = 0

(1)

∆U g = U g,final − U g,initial = mgh − 0 = mgd sin θ ∆U s = U s,final − U s,initial = 0 − 12 kd 2 = − 12 kd 2 ∆Etherm = f∆s = − f k d = − µ k Fn d = − µ k mgd cosθ

Substitute in equation (1) to obtain:

mgd sin θ − 12 kd 2 − µ k mgd cos θ = 0

Finally, solve for µk:

µk =

1 2

(tan θ − µs )

Mass and Energy 53 • Picture the Problem The intrinsic rest energy in matter is related to the mass of matter through Einstein’s equation E0 = mc 2 .

Conservation of Energy 477 (a) Relate the rest mass consumed to the energy produced and solve for and evaluate m:

(b) Express kW⋅h in joules:

E0 = mc 2

(

)(

= 1× 10 −3 kg 3 × 108 m/s

)

2

= 9.00 × 1013 J

(

)

1 kW ⋅ h = 1× 103 J/s (1 h )(3600 s/h ) = 3.60 × 10 J 6

Convert 9×1013 J to kW⋅h:

⎛ 1 kW ⋅ h ⎞ ⎟ 9 × 1013 J = 9 × 1013 J ⎜⎜ 6 ⎟ ⎝ 3.60 × 10 J ⎠ = 2.50 × 107 kW ⋅ h

Determine the price of the electrical energy:

⎛ $0.10 ⎞ Price = 2.50 × 107 kW ⋅ h ⎜ ⎟ ⎝ kW ⋅ h ⎠

(

)

(

)

= $2.5 × 106 (c) Relate the energy consumed to its rate of consumption and the time and solve for the latter:

E = Pt and

t=

E 9 × 1013 J = 100 W P

= 9 × 1011 s = 28,500 y 54 • Picture the Problem We can use the equation expressing the equivalence of energy and matter, E = mc2, to find the mass equivalent of the energy from the explosion. Solve E = mc2 for m:

Substitute numerical values and evaluate m:

m=

m=

E c2 5 × 1012 J

(2.998 ×10

8

m/s

)

2

= 5.56 × 10 −5 kg 55 • Picture the Problem The intrinsic rest energy in matter is related to the mass of matter through Einstein’s equation E0 = mc 2 . Relate the rest mass of a muon to its rest energy:

m0 =

E c2

478 Chapter 7

Express 1 MeV in joules:

1 MeV = 1.6×10−13 J

Substitute numerical values and evaluate m0:

m0 =

(105.7 MeV )(1.6 ×10−13 J/MeV )

(3 ×10

8

m/s

)

2

= 1.88 × 10 −28 kg *56 • Picture the Problem We can differentiate the mass-energy equation to obtain an expression for the rate at which the black hole gains energy. Using the mass-energy relationship, express the energy radiated by the black hole:

E = 0.01mc 2

Differentiate this expression to obtain an expression for the rate at which the black hole is radiating energy:

dm dE d = 0.01mc 2 = 0.01c 2 dt dt dt

Solve for dm/dt:

dm dE dt = dt 0.01c 2

Substitute numerical values and evaluate dm/dt:

dm 4 × 1031 watt = dt (0.01) 2.998 × 108 m/s

[

]

(

)

2

= 4.45 × 1016 kg/s 57 • Picture the Problem The number of reactions per second is given by the ratio of the power generated to the energy released per reaction. The number of reactions that must take place to produce a given amount of energy is the ratio of the energy per second (power) to the energy released per second. In Example 7-15 it is shown that the energy per reaction is 17.59 MeV. Convert this energy to joules:

The number of reactions per second is:

17.59 MeV = (17.59 MeV )

(

× 1.6 × 10 −19 J/eV = 28.1 × 10

−13

J

1000 J/s 28.1 × 10 −13 J/reaction = 3.56 × 1014 reactions/s

)

Conservation of Energy 479 58 • Picture the Problem The energy required for this reaction is the difference between the rest energy of 4He and the sum of the rest energies of 3He and a neutron.

He→3 He + n

Express the reaction:

4

The rest energy of a neutron (Table 7-1) is:

939.573 MeV

The rest energy of 4He (Example 7-15) is:

3727.409 MeV

The rest energy of 3He is:

2808.432 MeV

Substitute numerical values to find the difference in the rest energy of 4He and the sum of the rest energies of 3He and n:

E = [3727.409 − (2808.41 + 939.573)] MeV = 20.574 MeV 59 • Picture the Problem The energy required for this reaction is the difference between the rest energy of a neutron and the sum of the rest energies of a proton and an electron. The rest energy of a proton (Table 7-1) is:

938.280 MeV

The rest energy of an electron (Table 7-1) is:

0.511 MeV

The rest energy of a neutron (Table 7-1) is:

939.573 MeV

Substitute numerical values to find the difference in the rest energy of a neutron and the sum of the rest energies of a positron and an electron:

E = [939.573 − (938.280 + 0.511)] MeV

= 0.782 MeV

60 •• Picture the Problem The reaction is 2 H + 2 H→ 4 He + E . The energy released in this reaction is the difference between twice the rest energy of 2H and the rest energy of 4He.

480 Chapter 7 The number of reactions that must take place to produce a given amount of energy is the ratio of the energy per second (power) to the energy released per reaction. (a) The rest energy of 4He (Example 7-14) is:

3727.409 MeV

The rest energy of a deuteron, 2H, (Table 7-1) is:

1875.628 MeV

The energy released in the reaction is:

(b) The number of reactions per second is:

E = [2(1875.628) − 3727.409] MeV

= 23.847 MeV = 3.816 × 10 −12 J 1000 J/s 3.816 × 10 −12 J/reaction = 2.62 × 1014 reactions/s

61 •• Picture the Problem The annual consumption of matter by the fission plant is the ratio of its annual energy output to the square of the speed of light. The annual consumption of coal in a coal-burning power plant is the ratio of its annual energy output to energy per unit mass of the coal. (a) Express m in terms of E:

Assuming an efficiency of 33 percent, find the energy produced annually:

m=

E c2

(

)

E = 3P∆t = 3 3 × 109 J/s (1 y )

(

)

= 3 3 × 109 J/s (3600 s/h ) × (24 h/d )(365.24 d ) = 2.84 × 1017 J

Substitute to obtain:

(b) Assuming an efficiency of 38 percent, express the mass of coal required in terms of the annual energy production and the energy released per kilogram:

m=

mcoal

2.84 × 1017 J

(3 ×10

8

m/s )

2

= 3.16 kg

Eannual 9.47 × 1016 J = = 0.38(E / m ) 0.38 3.1× 107 J/kg

(

= 8.04 × 109 kg

)

Conservation of Energy 481

General Problems *62 •• Picture the Problem Let the system consist of the block, the earth, and the incline. Then the tension in the string is an external force that will do work to change the energy of the system. Because the incline is frictionless; the work done by the tension in the string as it displaces the block on the incline is equal to the sum of the changes in the kinetic and gravitational potential energies. Relate the work done by the tension force to the changes in the kinetic and gravitational potential energies of the block: Referring to the figure, express the change in the potential energy of the block as it moves from position 1 to position 2:

Wtension force = Wext = ∆U + ∆K

∆U = mg∆h = mgL sin θ

Because the block starts from rest:

∆K = K 2 = 12 mv 2

Substitute to obtain:

Wtension force = mgL sin θ + 12 mv 2 and (c) is correct.

63 •• Picture the Problem Let the system include the earth, the block, and the inclined plane. Then there are no external forces to do work on the system and Wext = 0. Apply the work-energy theorem with friction to find an expression for the energy dissipated by friction. Express the work-energy theorem with friction:

Wext = ∆K + ∆U + ∆Etherm = 0

482 Chapter 7 Because the velocity of the block is constant, ∆K = 0 and: In time ∆t the block slides a distance v∆t . From the figure: Substitute to obtain:

∆Etherm = − ∆U = −mg∆h ∆h = v∆t sin θ

∆Etherm = − mgv∆t sin θ and

(b) is correct.

64 • Picture the Problem Let the system include the earth and the box. Then the applied force is external to the system and does work on the system in compressing the spring. This work is stored in the spring as potential energy. Express the work-energy theorem:

Wext = ∆K + ∆U g + ∆U s + ∆Etherm

Because ∆K = ∆U g = ∆Etherm = 0 :

Wext = ∆U s

Substitute for Wext and ∆Us:

Fx = 12 kx 2

Solve for x:

x=

2F k

Substitute numerical values and evaluate x:

x=

2(70 N ) = 2.06 cm 6800 N/m

*65 • Picture the Problem The solar constant is the average energy per unit area and per unit time reaching the upper atmosphere. This physical quantity can be thought of as the power per unit area and is known as intensity.

P ∆E / ∆t = A A

Letting Isurface represent the intensity of the solar radiation at the surface of the earth, express Isurface as a function of power and the area on which this energy is incident:

I surface =

Solve for ∆E:

∆E = I surface A∆t

Conservation of Energy 483 Substitute numerical values and evaluate ∆E:

(

)(

)

∆E = 1 kW/m 2 2 m 2 (8 h )(3600 s/h ) = 57.6 MJ

66 •• Picture the Problem The luminosity of the sun (or of any other object) is the product of the power it radiates per unit area and its surface area. If we let L represent the sun’s luminosity, I the power it radiates per unit area (also known as the solar constant or the intensity of its radiation), and A its surface area, then L = IA. We can estimate the solar lifetime by dividing the number of hydrogen nuclei in the sun by the rate at which they are being transformed into energy. (a) Express the total energy the sun radiates every second in terms of the solar constant:

L = IA

Letting R represent its radius, express the surface area of the sun:

A = 4πR 2

Substitute to obtain:

L = 4πR 2 I

Substitute numerical values and evaluate L:

L = 4π (1.5 ×1011 m ) (1.35 kW/m 2 ) 2

= 3.82 ×10 26 watt Note that this result is in good agreement with the value given in the text of 3.9×1026 watt. (b) Express the solar lifetime in terms of the mass of the sun and the rate at which its mass is being converted to energy:

tsolar =

where M is the mass of the sun, m the mass of a hydrogen nucleus, and n is the number of nuclei used up.

Substitute numerical values to obtain:

tsolar

1.99 × 1030 kg 1.67 ×10 −27 kg/H nucleus = ∆n ∆t =

For each reaction, 4 hydrogen nuclei are "used up"; so:

N H nuclei M m = ∆n ∆t ∆n ∆t

1.19 ×1057 H nuclei ∆n ∆t

(

∆n 4 3.82 × 1026 J/s = ∆t 4.27 × 10−12 J = 3.57 × 1038 s −1

)

484 Chapter 7 ⎛ 1.19 × 1057 H nuclei ⎞ ⎟⎟ tsolar = 0.1⎜⎜ 38 −1 ⎝ 3.57 × 10 s ⎠

Because we’ve assumed that the sun will continue burning until roughly 10% of its hydrogen fuel is used up, the total solar lifetime should be:

= 3.33 × 1017 s = 1.06 × 1010 y

67 • Picture the Problem Let the system include the earth and the Spirit of America. Then there are no external forces to do work on the car and Wext = 0. We can use the workenergy theorem to relate the coefficient of kinetic friction to the given information. A constant-acceleration equation will yield the car’s velocity when 60 s have elapsed. (a) Apply the work-energy theorem with friction to relate the coefficient of kinetic friction µk to the initial and final kinetic energies of the car: Solve for µk:

Substitute numerical values and evaluate µk: (b) Express the kinetic energy of the car: Using a constant-acceleration equation, relate the speed of the car to its acceleration, initial speed, and the elapsed time: Express the braking force acting on the car:

1 2

mv 2 − 12 mv02 + µ k mg∆s = 0

or, because v = 0,

− 12 mv02 + µ k mg∆s = 0

µk =

v2 2 g∆s

2 [ ( 708 km/h )(1 h/3600 s )] µk = 2(9.81 m/s 2 )(9.5 km )

K = 12 mv 2

(1)

v = v0 + a∆t

Fnet = − f k = − µ k mg = ma

Solve for a:

a = −µk g

Substitute for a to obtain:

v = v0 − µ k g∆t

Substitute in equation (1) to obtain:

K = 12 m(v0 − µ k g∆t )

Substitute numerical values and evaluate K:

= 0.208

2

Conservation of Energy 485

K=

1 2

(1250 kg )[708 ×103 m/h − (0.208)(9.81m/s2 )(60 s )] 2 =

3.45 MJ

68 •• Picture the Problem The free-body diagram shows the forces acting on the skiers as they are towed up the slope at constant speed. Because the power r r required to move them is F ⋅ v , we need to find F as a function of mtot, θ, and µk. We can apply Newton’s 2nd law to obtain such a function. Express the power required as a function of force on the skiers and their speed: Apply

r r F ∑ = ma to the skiers:

P = Fv

∑F

x

(1)

= F − f k − mtot g sin θ = 0

and

∑F

y

Eliminate fk = µkFn and Fn between the two equations and solve for F: Substitute in equation (1) to obtain:

= Fn − mtot g cos θ = 0

F = mtot g sin θ + µ k mtot g cos θ P = (mtot g sin θ + µ k mtot g cos θ )v = mtot gv(sin θ + µ k cos θ )

Substitute numerical values and evaluate P:

(

)

P = 80(75 kg ) 9.81 m/s 2 (2.5 m/s )[sin 15° + (0.06 ) cos15°] = 46.6 kW

486 Chapter 7 69 •• Picture the Problem The free-body diagram for the box is superimposed on the pictorial representation shown to the right. The work done by friction slows and momentarily stops the box as it slides up the incline. The box’s speed when it returns to bottom of the incline will be less than its speed when it started up the incline due to the energy dissipated by friction while it was in motion. Let the system include the box, the earth, and the incline. Then Wext = 0. We can use the work-energy theorem with friction to solve the several parts of this problem.

From the FBD we can see that the forces acting on the box are the (a) normal force exerted by the inclined plane, a kinetic friction force, and the gravitational force (the weight of the box) exerted by the earth. (b) Apply the work-energy theorem with friction to relate the distance ∆x the box slides up the incline to its initial kinetic energy, its final potential energy, and the work done against friction:

− 12 mv12 + mg∆h + µ k mg∆x cos θ = 0

Referring to the figure, relate ∆h to ∆x to obtain:

∆h = ∆x sin θ

Substitute for ∆h to obtain:

− 12 mv12 + mg∆x sin θ + µ k mg∆x cos θ = 0

Solve for ∆x:

Substitute numerical values and evaluate ∆x:

∆x =

∆x =

v12 2 g (sin θ + µ k cos θ )

(

)

(3 m/s)2

2 9.81m/s 2 [sin60° + (0.3)cos60°]

= 0.451 m

Conservation of Energy 487 (c) Express and evaluate the energy dissipated by friction:

∆Etherm = f k ∆x = µ k mg∆x cosθ

(

)

= (0.3)(2 kg ) 9.81 m/s 2 (0.451 m )cos60° = 1.33 J (d) Use the work-energy theorem with friction to obtain:

Wext = ∆K + ∆U + ∆Etherm = 0 or

K1 − K 2 + U1 − U 2 + ∆Etherm = 0 Because K2 = U1 = 0 we have:

K1 − U 2 + ∆Etherm = 0 or 1 2

mv12 − mg∆x sin θ + µ k mg∆x cos θ = 0

Solve for v1:

v1 = 2 g∆x(sin θ − µ k cos θ )

Substitute numerical values and evaluate v1:

v1 = 2(9.81 m/s 2 )(0.451 m )[sin60° − (0.3)cos60°] = 2.52 m/s *70 • Picture the Problem The power provided by a motor that is delivering sufficient energy to exert a force F on a load which it is moving at a speed v is Fv. The power provided by the motor is given by: Because the elevator is ascending with constant speed, the tension in the support cable(s) is: Substitute for F to obtain: Substitute numerical values and evaluate P:

P = Fv

F = (melev + mload )g

P = (melev + mload )gv P = (2000 kg ) (9.81 m/s 2 )(2.3 m/s ) = 45.1 kW

488 Chapter 7 71 •• Picture the Problem The power a motor must provide to exert a force F on a load that it is moving at a speed v is Fv. The counterweight does negative work and the power of the motor is reduced from that required with no counterbalance. The power provided by the motor is given by: Because the elevator is counterbalanced and ascending with constant speed, the tension in the support cable(s) is: Substitute and evaluate P: Substitute numerical values and evaluate P:

Without a load:

P = Fv

F = (melev + mload − mcw )g

P = (melev + mload − mcw )gv

(

)

P = (500 kg ) 9.81 m/s 2 (2.3 m/s ) = 11.3 kW

F = (melev − mcw )g and

P = (melev − mcw )gv

(

)

= (− 300 kg ) 9.81 m/s 2 (2.3 m/s ) = − 6.77 kW

72 •• Picture the Problem We can use the work-energy theorem with friction to describe the energy transformation within the dart-spring-air-earth system. With this choice of the system, there are no external forces to do work on the system; i.e., Wext = 0. Choose Ug = 0 at the elevation of the dart on the compressed spring. The energy initially stored in the spring is transformed into gravitational potential energy and thermal energy. During the dart’s descent, its gravitational potential energy is transformed into kinetic energy and thermal energy. Apply conservation of energy during the dart’s ascent:

Wext = ∆K + ∆U + ∆Etherm = 0 or, because ∆K = 0,

U g,f − U g,i + U s,f − U s,i + ∆Etherm = 0

Because U g,i = U s,f = 0 :

U g,f − U s,i + ∆Etherm = 0

Conservation of Energy 489 Substitute for Ug,i and Us,f and solve for ∆Etherm:

∆Etherm = U s,i − U g,f = 12 kx 2 − mgh

Substitute numerical values and evaluate ∆Etherm:

∆Etherm =

(5000 N/m )(0.03 m )2 − (0.007 kg )(9.81 m/s 2 )(24 m )

1 2

= 0.602 J

Wext = ∆K + ∆U + ∆Etherm = 0

Apply conservation of energy during the dart’s descent:

or, because Ki = Ug,f = 0,

K f − U g,i + ∆Etherm = 0 1 2

Substitute for Kf and Ug,i to obtain: Solve for vf:

mvf2 − mgh + ∆Etherm = 0

vf =

2(mgh − ∆Etherm ) m

Substitute numerical values and evaluate vf:

vf =

[

(

]

)

2 (0.007 kg ) 9.81 m/s 2 (24 m ) − 0.602 J = 17.3 m/s 0.007 kg

*73 •• Picture the Problem Let the system consist of the earth, rock and air. Given this choice, there are no external forces to do work on the system and Wext = 0. Choose Ug = 0 to be where the rock begins its upward motion. The initial kinetic energy of the rock is partially transformed into potential energy and partially dissipated by air resistance as the rock ascends. During its descent, its potential energy is partially transformed into kinetic energy and partially dissipated by air resistance. (a) Using the definition of kinetic energy, calculate the initial kinetic energy of the rock: (b) Apply the work-energy theorem with friction to relate the energies of the system as the rock ascends: Because Kf = 0:

K i = 12 mvi2 =

1 2

(2 kg )(40 m/s )2

= 1.60 kJ

∆K + ∆U + ∆Etherm = 0

− K i + ∆U + ∆Etherm = 0 and

∆Etherm = K i − ∆U

490 Chapter 7 Substitute numerical values and evaluate ∆Etherm:

(c) Apply the work-energy theorem with friction to relate the energies of the system as the rock descends:

(

)

∆Etherm = 1600 J − (2 kg ) 9.81 m/s 2 (50 m ) = 619 J ∆K + ∆U + 0.7∆Etherm = 0

Because Ki = Uf = 0:

K f − U i + 0.7∆Etherm = 0

Substitute for the energies to obtain:

1 2

mvf2 − mgh + 0.7∆Etherm = 0 1.4∆Etherm m

Solve for vf:

vf = 2 gh −

Substitute numerical values and evaluate vf:

vf = 2 9.81 m/s 2 (50 m ) −

(

)

1.4(619 J ) 2 kg

= 23.4 m/s 74 •• Picture the Problem Let the distance the block slides before striking the spring be L. The pictorial representation shows the block at the top of the incline (1), just as it strikes the spring (2), and the block against the fully compressed spring (3). Let the block, spring, and the earth comprise the system. Then Wext = 0. Let Ug = 0 where the spring is at maximum compression. We can apply the work-energy theorem to relate the energies of the system as it evolves from state 1 to state 3.

Express the work-energy theorem:

∆K + ∆U g + ∆U s = 0 or

∆K + U g,3 − U g,1 + U s,3 − U s,1 = 0

Conservation of Energy 491 Because ∆K = Ug,3 = Us,1 = 0:

− U g,1 + U s,3 = 0

Substitute for each of these energy terms to obtain:

− mgh1 + 12 kx 2 = 0

Substitute for h3 and h1:

− mg (L + x )sin θ + 12 kx 2 = 0

Rewrite this equation explicitly as a quadratic equation:

x2 −

2mg sin θ 2mgL sin θ x− =0 k k

Solve this quadratic equation to obtain: 2

2mgL mg ⎛ mg ⎞ 2 sin θ + ⎜ sin θ x= ⎟ sin θ + k k ⎝ k ⎠ Note that the negative sign between the two terms leads to a non-physical solution. *75 • Picture the Problem We can find the work done by the girder on the slab by calculating the change in the potential energy of the slab. (a) Relate the work the girder does on the slab to the change in potential energy of the slab: Substitute numerical values and evaluate W:

W = ∆U = mg∆h

(

)(

)

W = 1.5 × 104 kg 9.81 m/s 2 (0.001 m ) = 147 J

The energy is transferred to the girder from its surroundings, which are (b)

warmer than the girder. As the temperature of the girder rises, the atoms in the girder vibrate with a greater average kinetic energy, leading to a larger average separation, which causes the girder' s expansion.

76 •• Picture the Problem The average power delivered by the car’s engine is the rate at which it changes the car’s energy. Because the car is slowing down as it climbs the hill, its potential energy increases and its kinetic energy decreases. Express the average power delivered by the car’s engine:

Pav =

∆E ∆t

492 Chapter 7 Express the increase in the car’s mechanical energy:

∆E = ∆K + ∆U = K top − K bot + U top − U bot 2 2 = 12 mvtop − 12 mvbot + mg∆h

(

2 2 = 12 m vtop − vbot + 2 g∆h

)

Substitute numerical values and evaluate ∆E:

∆E =

1 2

(1500 kg )[(10 m/s)2 − (24 m/s )2 + 2(9.81 m/s 2 )(120 m )]= 1.41 MJ v top + v bot

Assuming that the acceleration of the car is constant, find its average speed during this climb:

vav =

Using the vav, find the time it takes the car to climb the hill:

∆t =

∆s 2000 m = = 118 s vav 17 m/s

Substitute to determine Pav:

Pav =

1.41 MJ = 11.9 kW 118 s

2

= 17 m/s

*77 •• Picture the Problem Given the potential energy function as a function of y, we can find the net force acting on a given system from F = −dU / dy . The maximum extension of the spring; i.e., the lowest position of the mass on its end, can be found by applying the work-energy theorem. The equilibrium position of the system can be found by applying the work-energy theorem with friction … as can the amount of thermal energy produced as the system oscillates to its equilibrium position. (a) The graph of U as a function of y is shown to the right. Because k and m are not specified, k has been set equal to 2 and mg to 1. The spring is unstretched when y = y0 = 0. Note that the minimum value of U (a position of stable equilibrium) occurs near y = 5 m.

Conservation of Energy 493 1.0 0.8

U (J)

0.6 0.4 0.2 0.0 -0.2 -0.4 0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

y (m)

(b) Evaluate the negative of the derivative of U with respect to y:

F =−

dU d =− dy dy

(

1 2

ky 2 − mgy

)

= − ky + mg (c) Apply conservation of energy to the movement of the mass from y = 0 to y = ymax:

∆K + ∆U + ∆Etherm = 0

Because ∆K = 0 (the object starts from rest and is momentarily at rest at y = ymax) and ∆Etherm = 0 (no friction), it follows that:

∆U = U(ymax) – U(0) = 0

Because U(0) = 0:

U(ymax) = 0 ⇒

Solve for ymax:

y max =

(d) Express the condition of F at equilibrium and solve for yeq:

2 kymax − mgymax = 0

2mg k

Feq = 0 ⇒ − ky eq + mg = 0 and

yeq =

(e) Apply the conservation of energy to the movement of the mass from y = 0 to y = yeq and solve for ∆Etherm:

1 2

mg k

∆K + ∆U + ∆Etherm = 0 or, because ∆K = 0.

∆Etherm = −∆U = U i − U f

494 Chapter 7

(

Because U i = U (0 ) = 0 :

∆Etherm = −U f = − 12 kyeq2 − mgyeq

Substitute for yeq and simplify to obtain:

∆Etherm =

)

m2 g 2 2k

78 •• Picture the Problem The energy stored in the compressed spring is initially transformed into the kinetic energy of the signal flare and then into gravitational potential energy and thermal energy as the flare climbs to its maximum height. Let the system contain the earth, the air, and the flare so that Wext = 0. We can use the work-energy theorem with friction in the analysis of the energy transformations during the motion of the flare. (a) The work done on the spring in compressing it is equal to the kinetic energy of the flare at launch. Therefore: (b) Ignoring changes in gravitational potential energy (i.e., assume that the compression of the spring is small compared to the maximum elevation of the flare), apply the conservation of energy to the transformation that takes place as the spring decompresses and gives the flare its launch speed:

Ws = K i, flare =

or

K f − K i + U s,f − U s,i = 0

K f − U s,i = 0

Substitute for K f and U s,i :

1 2

(c) Apply the work-energy theorem with friction to the upward trajectory of the flare:

mv02

∆K + ∆U s = 0

Because Ki = ∆Ug = Us,f:

Solve for k to obtain:

1 2

mv02 − 12 kd 2 = 0

k=

mv02 d2

∆K + ∆U g + ∆Etherm = 0

Conservation of Energy 495 Solve for ∆Etherm:

∆Etherm = −∆K − ∆U g = Ki − Kf + U i − U f

Because Kf = Ui = 0:

∆Etherm =

1 2

mv02 − mgh

79 •• Picture the Problem Let UD = 0. Choose the system to include the earth, the track, and the car. Then there are no external forces to do work on the system and change its energy and we can use Newton’s 2nd law and the work-energy theorem to describe the system’s energy transformations to point G … and then the work-energy theorem with friction to determine the braking force that brings the car to a stop. The free-body diagram for point C is shown to the right. The free-body diagram for point D is shown to the right.

The free-body diagram for point F is shown to the right.

(a) Apply the work-energy theorem to the system’s energy transformations between A and B: If we assume that the car arrives at point B with vB = 0, then:

∆K + ∆U = 0 or

KB − KA +UB −UA = 0

− 12 mvA2 + mg∆h = 0 where ∆h is the difference in elevation between A and B.

496 Chapter 7 Solve for and evaluate ∆h:

(12 m/s) = 7.34 m v2 ∆h = A = 2 g 2 9.81 m/s 2

The height above the ground is:

h + ∆h = 10 m + 7.34 m = 17.3 m

(b) If the car just makes it to point B; i.e., if it gets there with vB = 0, then the force exerted by the track on the car will be the normal force: (c) Apply

∑F

x

= max to the car at

point C (see the FBD) and solve for a:

2

(

)

Ftrack on car = Fn = mg

(

= (500 kg ) 9.81 m/s 2 = 4.91 kN

mg sin θ = ma and

a = g sin θ = (9.81 m/s 2 )sin30° = 4.91 m/s 2

(d) Apply

∑F

y

= ma y to the car at

point D (see the FBD) and solve for Fn :

Fn − mg = m

vD2 R

and

Fn = mg + m

vD2 R

∆K + ∆U = 0

Apply the work-energy theorem to the system’s energy transformations between B and D:

or

Because KB = UD = 0:

KD −UB = 0

Substitute to obtain:

1 2

Solve for vD2 :

vD2 = 2 g (h + ∆h )

Substitute to find Fn:

KD − KB +UD −UB = 0

mv D2 − mg (h + ∆h ) = 0

vD2 R 2 g (h + ∆h ) = mg + m R ⎡ 2(h + ∆h ) ⎤ = mg ⎢1 + ⎥⎦ R ⎣

Fn = mg + m

)

Conservation of Energy 497 Substitute numerical values and evaluate Fn:

⎡ 2(17.3 m )⎤ Fn = (500 kg ) 9.81 m/s 2 ⎢1 + 20 m ⎥⎦ ⎣

(

)

= 13.4 kN, directed upward. (e) F has two components at point F; one horizontal (the inward force that the track exerts) and the other vertical (the normal force). Apply r r F = ma to the car at point F:

∑

Express the resultant of these two forces:

∑F

= Fn − mg = 0 ⇒ Fn = mg

y

and

∑ Fx = Fc = m F = Fc2 + Fn2 2

⎛ v2 ⎞ 2 = ⎜⎜ m F ⎟⎟ + (mg ) ⎝ R⎠ =m

Substitute numerical values and evaluate F:

v F2 R

vF4 + g2 R2

F = (500 kg )

(12 m/s)4 + (9.81m/s2 ) 2 (30 m )2

= 5.46 kN Find the angle the resultant makes with the x axis:

⎛ Fn ⎞ ⎛ gR ⎞ ⎟⎟ = tan −1 ⎜⎜ 2 ⎟⎟ ⎝ vF ⎠ ⎝ Fc ⎠

θ = tan −1 ⎜⎜

(

)

⎡ 9.81 m/s 2 (30 m ) ⎤ = 63.9° = tan ⎢ (12 m/s)2 ⎥⎦ ⎣ −1

(f) Apply the work-energy theorem with friction to the system’s energy transformations between F and the car’s stopping position: The work done by friction is also given by: Equate the two expressions for ∆Etherm and solve for Fbrake:

− K G + ∆Etherm = 0 and

∆Etherm = K G = 12 mvG2 ∆Etherm = f∆s = Fbrake d where d is the stopping distance.

Fbrake

mvF2 = 2d

498 Chapter 7 Substitute numerical values and evaluate Fbrake:

2 ( 500 kg )(12 m/s ) Fbrake = 2(25 m )

= 1.44 kN

*80 • Picture the Problem The rate of conversion of mechanical energy can be r r determined from P = F ⋅ v . The pictorial representation shows the elevator moving downward just as it goes into freefall as state 1. In state 2 the elevator is moving faster and is about to strike the relaxed spring. The momentarily at rest elevator on the compressed spring is shown as state 3. Let Ug = 0 where the spring has its maximum compression and the system consist of the earth, the elevator, and the spring. Then Wext = 0 and we can apply the conservation of mechanical energy to the analysis of the falling elevator and compressing spring. (a) Express the rate of conversion of mechanical energy to thermal energy as a function of the speed of the elevator and braking force acting on it: Because the elevator is moving with constant speed, the net force acting on it is zero and: Substitute for Fbraking and evaluate P:

P = Fbraking v0

Fbraking = Mg

P = Mgv0

(

)

= (2000 kg ) 9.81 m/s 2 (1.5 m/s ) = 29.4 kW

∆K + ∆U g + ∆U s = 0

(b) Apply the conservation of energy to the falling elevator and compressing spring:

or

Because K3 = Ug,3 = Us,1 = 0:

− 12 Mv02 − Mg (d + ∆y ) + 12 k (∆y ) = 0

K 3 − K 1 + U g,3 − U g,1 + U s,3 − U s,1 = 0 2

Conservation of Energy 499 Rewrite this equation as a quadratic equation in ∆y, the maximum compression of the spring: Solve for ∆y to obtain:

(∆y )2 − ⎛⎜ 2Mg ⎞⎟∆y − M (2 gd + v02 ) = 0 ⎝ k ⎠

∆y =

k

(

Mg M 2g2 M ± + 2 gd + v02 2 k k k

)

Substitute numerical values and evaluate ∆y:

∆y =

(2000 kg )(9.81m/s 2 ) 1.5 × 10 4 N/m +

(2000 kg )2 (9.81m/s 2 )2 +

(1.5 ×10

4

N/m

)

2

[(

)

2000 kg 2 2 9.81 m/s 2 (5 m ) + (1.5 m/s ) 4 1.5 × 10 N/m

]

= 5.19 m 81 • Picture the Problem We can use Newton’s 2nd law to determine the force of friction as a function of the angle of the hill for a given constant speed. The power output of the

r

r

engine is given by P = Ff ⋅ v . FBD for (a):

(a) Apply

FBD for (b):

∑F

x

= max to the car:

Evaluate Ff for the two speeds:

mg sin θ − Ff = 0 ⇒ Ff = mg sin θ

(

)

(

)

F20 = (1000 kg ) 9.81 m/s 2 sin2.87° = 491 N and

F30 = (1000 kg ) 9.81 m/s 2 sin5.74° = 981 N (b) Express the power an engine must deliver on a level road in order

P = Ff v P20 = (491 N )(20 m/s ) = 9.82 kW

500 Chapter 7 to overcome friction loss and evaluate this expression for v = 20 m/s and 30 m/s: (c) Apply

∑F

x

= max to the car:

and

P30 = (981 N )(30 m/s ) = 29.4 kW

∑F

x

= F − mg sin θ − F f = 0

P v

Relate F to the power output of the engine and the speed of the car:

Since P = Fv, F =

Substitute for F and solve for θ :

⎤ ⎡P − F20 ⎥ ⎢ θ = sin −1 ⎢ v ⎥ ⎢ mg ⎥ ⎦ ⎣

Substitute numerical values and evaluate θ :

⎡ 40 kW ⎢ 20 m/s − 491 N −1 θ = sin ⎢ 2 ⎢ (1000 kg ) 9.81 m/s ⎢⎣

(

⎤ ⎥ ⎥ ⎥ ⎥⎦

)

= 8.85° (d) Express the equivalence of the work done by the engine in driving the car at the two speeds: Let ∆V represent the volume of fuel consumed by the engine driving the car on a level road and divide both sides of the work equation by ∆V to obtain:

Solve for

(∆s )30

:

∆V

Substitute numerical values and evaluate

(∆s )30 ∆V

:

Wengine = F20 (∆s )20 = F30 (∆s )30

F20

(∆s )20 ∆V

(∆s )30 ∆V

(∆s )30 ∆V

= F30

(∆s )30 ∆V

=

F20 (∆s )20 F30 ∆V

=

491 N (12.7 km/L) 981 N

= 6.36 km/L

82 •• Picture the Problem Let the system include the earth, block, spring, and incline. Then Wext = 0. The pictorial representation to the left shows the block sliding down the incline

Conservation of Energy 501 and compressing the spring. Choose Ug = 0 at the elevation at which the spring is fully compressed. We can use the conservation of mechanical energy to determine the maximum compression of the spring. The pictorial representation to the right shows the block sliding up the rough incline after being accelerated by the fully compressed spring. We can use the work-energy theorem with friction to determine how far up the incline the block slides before stopping.

(a) Apply conservation of mechanical energy to the system as it evolves from state 1 to state 3:

∆K + ∆U g + ∆U s = 0 or

K 3 − K1 + U g,3 − U g,1 + U s,3 − U s,1 = 0

Because

K 3 = K 1 = U g,3 = U s,1 = 0 :

− U g,1 + U s,3 = 0 or

− mg∆h + 12 kx 2 = 0 Relate ∆h to L + x and θ and substitute to obtain:

∆h = (L + x )sin θ

∴ 12 kx 2 − mg (L + x )sin θ = 0 kx 2 − (mg sin θ )x − mgL sin θ = 0

Rewrite this equation in the form of an explicit quadratic equation:

1 2

Substitute for k, m, g, θ and L to obtain:

⎛ N⎞ 2 ⎜ 50 ⎟ x − (9.81 N )x − 39.24 J = 0 ⎝ m⎠

Solve for the physically meaningful (i.e., positive) root:

x = 0.989 m

(b) Proceed as in (a) but include energy dissipated by friction: The mechanical energy transformed to thermal energy is given by:

− U g,1 + U s,3 + ∆Etherm = 0 ∆Etherm = Ff (L + x ) = µ k Fn (L + x ) = µ k mg cos θ (L + x )

502 Chapter 7 Substitute for ∆h and ∆Etherm to obtain:

− mg (L + x ) sin θ + 12 kx 2

Substitute for k, m, g, θ, µk and L to obtain:

⎛ N⎞ 2 ⎜ 50 ⎟ x − (6.41 N )x − 25.65 J = 0 ⎝ m⎠

Solve for the positive root:

x = 0.783 m

(c) Apply the work-energy theorem with friction to the system as it evolves from state 3 to state 4: Because

K 4 = K 1 = U g,3 = U s,4 = 0 :

+ µ k mg cos θ (L + x ) = 0

K 4 − K 3 + U g,4 − U g,3 + U s,4 − U s,3 + ∆Etherm = 0

U g,4 − U s,3 + ∆Etherm = 0 or

− mg∆h'+ 12 kx 2 + ∆Etherm = 0 Substitute for ∆h′ and ∆Etherm to obtain:

− mg (L'+ x ) sin θ + 12 kx 2

Solve for L′ with x = 0.783 m:

L' = 1.54 m

+ µ k mg cos θ (L'+ x ) = 0

83 •• Picture the Problem The work done by the engines maintains the kinetic energy of the cars and overcomes the work done by frictional forces. Let the system include the earth, track, and the cars but not the engines. Then the engines will do external work on the system and we can use this work to find the power output of the train’s engines. (a) Use the definition of kinetic energy:

K = 12 mv 2 =

1 2

⎛ km 1 h ⎞ ⎟ ⋅ 2 × 10 kg ⎜⎜15 h 3600 s ⎟⎠ ⎝

(

6

)

2

= 17.4 MJ (b) The change in potential energy of the train is:

∆U = mg∆h

(

)(

= 1.39 × 1010 J (c) Express the energy dissipated by kinetic friction:

)

= 2 × 10 6 kg 9.81 m/s 2 (707 m )

∆Etherm = f∆s

Conservation of Energy 503 f = 0.008mg

Express the frictional force: Substitute for f and evaluate ∆Etherm:

(

)(

)

∆Etherm = 0.008mg∆s = 0.008 2 × 106 kg 9.81 m/s 2 (62 km ) = 9.73 × 109 J (d) Express the power output of the train’s engines in terms of the work done by them:

∆W ∆t

P=

Wext = ∆K + ∆U + ∆Etherm

Use the work-energy theorem with friction to find the work done by the train’s engines:

or, because ∆K = 0,

Find the time during which the engines do this work:

∆t =

Substitute in the expression for P to obtain:

P=

Wext = ∆U + ∆Etherm ∆s v

(∆U + ∆Etherm )v ∆s

Substitute numerical values and evaluate P:

⎛ km 1 h ⎞ ⎜⎜15 ⎟ ⋅ h 3600 s ⎟⎠ 10 9 ⎝ P = 1.39 × 10 J + 9.73 × 10 J = 1.59 MW 62 km

(

)

*84 •• Picture the Problem While on a horizontal surface, the work done by an automobile engine changes the kinetic energy of the car and does work against friction. These energy transformations are described by the work-energy theorem with friction. Let the system include the earth, the roadway, and the car but not the car’s engine. (a) The required energy equals the change in the kinetic energy of the car:

∆K = 12 mv 2 ⎛ km 1 h ⎞ ⎟ = (1200 kg )⎜⎜ 50 ⋅ h 3600 s ⎟⎠ ⎝ 1 2

= 116 kJ (b) The required energy equals the

∆Etherm = f∆s

2

504 Chapter 7 work done against friction: Substitute numerical values and evaluate ∆Etherm: (c) Apply the work-energy theorem with friction to express the required energy: Divide both sides of the equation by E to express the ratio of the two energies: Substitute numerical values and evaluate E′/E:

∆Etherm = (300 N )(300 m ) = 90.0 kJ

E ' = Wext = ∆K + ∆Etherm = ∆K + 0.75E

E ' ∆K = + 0.75 E E

E ' 116 kJ = + 0.75 = 2.04 90 kJ E

*85 ••• Picture the Problem Assume that the bob is moving with speed v as it passes the top vertical point when looping around the peg. There are two forces acting on the bob: the tension in the string (if any) and the force of gravity, Mg; both point downward when the ball is in the topmost position. The minimum possible speed for the bob to pass the vertical occurs when the tension is 0; from this, gravity must supply the centripetal force required to keep the ball moving in a circle. We can use conservation of energy to relate v to L and R. Express the condition that the bob swings around the peg in a full circle: Simplify to obtain: Use conservation of energy to relate the kinetic energy of the bob at the bottom of the loop to its potential energy at the top of its swing: Solve for v2: Substitute to obtain:

M

v2 > Mg R

v2 >g R 1 Mv 2 = Mg (L − 2 R ) 2

v 2 = 2 g (L − 2 R ) 2 g (L − 2 R ) >g R

Conservation of Energy 505 Solve for R:

R<

2 L 5

86 •• Picture the Problem If the wood exerts an average force F on the bullet, the work it does has magnitude FD. This must be equal to the change in the kinetic energy of the bullet, or because the final kinetic energy of the bullet is zero, to the negative of the initial kinetic energy. We’ll let m be the mass of the bullet and v its initial speed and apply the work-kinetic energy theorem to relate the penetration depth to v.

Wtotal = ∆K = K f − K i

Apply the work-kinetic energy theorem to relate the penetration depth to the change in the kinetic energy of the bullet:

or, because Kf = 0,

Substitute for Wtotal and Ki to obtain:

FD = − 12 mv 2

Solve for D to obtain:

Wtotal = − K i

D=−

mv 2 2F

For an identical bullet with twice the speed we have:

FD' = − 12 m(2v )

Solve for D′ to obtain:

⎛ mv 2 ⎞ ⎟⎟ = 4 D D' = 4⎜⎜ − 2 F ⎝ ⎠

2

and (c) is correct. 87 •• Picture the Problem For part (a), we’ll let the system include the glider, track, weight, and the earth. The speeds of the glider and the falling weight will be the same while they are in motion. Let their common speed when they have moved a distance Y be v and let the zero of potential energy be at the elevation of the weight when it has fallen the distance Y. We can use conservation of energy to relate the speed of the glider (and the weight) to the distance the weight has fallen. In part (b), we’ll let the direction of motion be the x direction, the tension in the connecting string be T, and apply Newton’s 2nd law to the glider and the weight to find their common acceleration. Because this acceleration is constant, we can use a constant-acceleration equation to find their common speed when they have moved a distance Y. (a) Use conservation of energy to relate the kinetic and potential energies of the system:

∆K + ∆U = 0 or

Kf − Ki + U f − U i = 0

Because the system starts from rest and Uf = 0:

Kf − U i = 0

Substitute to obtain:

1 2

mv 2 + 12 Mv 2 − mgY = 0

506 Chapter 7 Solve for v:

v=

2mgY M +m

(b) The free-body diagrams for the glider and the weight are shown to the right:

Apply Newton’s 3rd law to obtain:

r r T1 = T2 = T

Apply

∑F

= ma to the glider:

T = Ma

Apply

∑F

= ma to the weight:

mg − T = ma

x

x

Add these equations to eliminate T and obtain:

mg = Ma + ma

Solve for a to obtain:

a=g

Using a constant-acceleration equation, relate the speed of the glider to its initial speed and to the distance that the weight has fallen:

v 2 = v02 + 2aY or, because v0 = 0,

Substitute for a and solve for v to obtain:

v=

m m+M

v 2 = 2aY 2mgY , the same result we M +m

obtained in part (a). *88 •• Picture the Problem We’re given P = dW / dt and are asked to evaluate it under the assumed conditions. Express the rate of energy expenditure by the man:

Express the rate of energy expenditure P′ assuming that his

P = 3mv 2 = 3(10 kg )(3 m/s ) = 270 W

2

P = 15 P'

Conservation of Energy 507 muscles have an efficiency of 20%: Solve for and evaluate P′:

P' = 5 P = 5(270 W ) = 1.35 kW

89 •• Picture the Problem The pictorial representation shows the bob swinging through an angle θ before the thread is cut and it is launched horizontally. Let its speed at position 1 be v. We can use conservation of energy to relate v to the change in the potential energy of the bob as it swings through the angle θ . We can find its flight time ∆t from a constantacceleration equation and then express D as the product of v and ∆t. Relate the distance D traveled horizontally by the bob to its launch speed v and time of flight ∆t:

D = v∆t

Use conservation of energy to relate its launch speed v to the length of the pendulum L and the angle θ :

K1 − K 0 + U1 − U 0 = 0

(1)

or, because U1 = K0 = 0,

K1 − U 0 = 0

mv 2 − mgL(1 − cosθ ) = 0

Substitute to obtain:

1 2

Solving for v yields:

v = 2 gL(1 − cosθ )

In the absence of air resistance, the horizontal and vertical motions of the bob are independent of each other and we can use a constantacceleration equation to express the time of flight (the time to fall a distance H):

∆y = v0 y ∆t + 12 a y (∆t )

2

or, because ∆y = −H, ay = −g, and v0y = 0,

− H = − 12 g (∆t )

2

Solve for ∆t to obtain:

∆t = 2 H / g

Substitute in equation (1) and simplify to obtain:

D = 2 gL(1 − cosθ )

2H g

= 2 HL(1 − cosθ ) which shows that, while D depends on θ, it is independent of g.

508 Chapter 7 90 •• Picture the Problem The pictorial representation depicts the block in its initial position against the compressed spring (1), as it separates from the spring with its maximum kinetic energy (2), and when it has come to rest after moving a distance x + d. Let the system consist of the earth, the block, and the surface on which the block slides. With this choice, Wext = 0. We can use the work-energy theorem with friction to determine how far the block will slide before coming to rest.

(a) The work done by the spring on the block is given by:

Wspring = ∆U spring = 12 kx 2

Substitute numerical values and evaluate Wspring:

Wspring =

(b) The energy dissipated by friction is given by: Substitute numerical values and evaluate ∆Etherm: (c) Apply the conservation of energy between points 1 and 2:

1 2

(20 N/cm)(3 cm)2 =

∆Etherm = f∆s = µ k Fn ∆x = µ k mg∆x

(

= 0.294 J K 2 − K1 + U s,2 − U s,1 + ∆Etherm = 0

K 2 − U s,1 + ∆Etherm = 0

Substitute to obtain:

1 2

Substitute numerical values and evaluate v2:

)

∆Etherm = (0.2)(5 kg ) 9.81m/s 2 (0.03 m )

Because K1 = Us,2 = 0:

Solve for v2:

0.900 J

mv22 − 12 kx 2 + ∆Etherm = 0

v2 =

v2 =

kx 2 − 2∆Etherm m

(20 N/cm)(3 cm)2 − 2(0.294 J )

= 0.492 m/s

5 kg

Conservation of Energy 509 (d) Apply the conservation of energy between points 1 and 3: Because ∆K = Us,3 = 0:

∆K + U s,3 − U s,1 + ∆Etherm = 0

− U s,1 + ∆Etherm = 0 or

− 12 kx 2 + µ k mg ( x + d ) = 0

Solve for d:

Substitute numerical values and evaluate d:

kx 2 d= −x 2µ k mg 2 ( 20 N/cm )(3 cm ) d= − 0.03 m 2(0.2)(5 kg )(9.81m/s 2 )

= 6.17 cm 91 •• Picture the Problem The pictorial representation shows the block initially at rest at point 1, falling under the influence of gravity to point 2, partially compressing the spring as it continues to gain kinetic energy at point 3, and finally coming to rest at point 4 with the spring fully compressed. Let the system consist of the earth, the block, and the spring so that Wext = 0. Let Ug = 0 at point 3 for part (a) and at point 4 for part (b). We can use the work-energy theorem to express the kinetic energy of the system as a function of the block’s position and then use this function to maximize K as well as determine the maximum compression of the spring and the location of the block when the system has half its maximum kinetic energy.

∆K + ∆U g + ∆U s = 0

(a) Apply conservation of mechanical energy to describe the energy transformations between state 1 and state 3:

or

Because K1 = Ug,3 = Us,1 = 0:

K 3 − U g,1 + U s,3 = 0

K 3 − K 1 + U g,3 − U g,1 + U s,3 − U s,1 = 0

510 Chapter 7 and

K 3 = K = mg (h + x ) − 12 kx 2

Differentiate K with respect to x and set this derivative equal to zero to identify extreme values:

dK = mg − kx = 0 for extreme values. dx

Solve for x:

x=

Evaluate the second derivative of K with respect to x:

d 2K = −k < 0 dt 2 mg ⇒ x= maximizes K . k

Evaluate K for x = mg/k:

mg k

K max

⎛ mg ⎞ 1 ⎛ mg ⎞ = mgh + mg ⎜ ⎟ − 2 k⎜ ⎟ ⎝ k ⎠ ⎝ k ⎠

2

m2 g 2 = mgh + 2k (b) The spring will have its maximum compression at point 4 where K = 0:

Solve for x and keep the physically meaningful root:

2 mg (h + x max ) − 12 kx max =0

or 2 x max −

x max

2mg 2mgh x max − =0 k k

mg m 2 g 2 2mgh = + + k k k2

∆K + ∆U g + ∆U s = 0

(c) Apply conservation of mechanical energy to the system as it evolves from state 1 to the state in which K = 12 K max :

or

Because K1 = Ug,3 = Us,1 = 0:

K − U g,1 + U s,3 = 0

K − K 1 + U g,3 − U g,1 + U s,3 − U s,1 = 0

and

K = mg (h + x ) − 12 kx 2

Conservation of Energy 511 Substitute for K to obtain:

Express this equation in quadratic form:

Solve for the positive value of x:

1 2

⎛ m2 g 2 ⎞ ⎜⎜ mgh + ⎟⎟ = mg (h + x ) − 12 kx 2 2 k ⎝ ⎠

x2 −

⎛ m 2 g 2 mgh ⎞ 2mg ⎟=0 x + ⎜⎜ − 2 k k ⎟⎠ ⎝ 2k

x=

mg 2m 2 g 2 4mgh + + k k k2

92 ••• Picture the Problem The free-body diagram shows the forces acting on the pendulum bob. The application of Newton’s 2nd law leads directly to the required expression for the tangential acceleration. Recall that, provided θ is in radian measure, s = Lθ. Differentiation with respect to time produces the result called for in part (b). The remaining parts of the problem simply require following the directions for each part. (a) Apply

∑F

x

= max to the bob:

Solve for atan:

(b) Relate the arc distance s to the length of the pendulum L and the angle θ :

Ftan = − mg sin θ = matan a tan = dv / dt = − g sin θ s = Lθ

Differentiate with respect to time:

ds / dt = v = Ldθ / dt

dv dθ and by dt dθ dθ substitute for from part (b): dt

dv dv dθ dv dθ = = dt dt dθ dθ dt

(c) Multiply

=

dv ⎛ v ⎞ ⎜ ⎟ dθ ⎝ L ⎠

512 Chapter 7 (d) Equate the expressions for dv/dt from (a) and (c):

dv ⎛ v ⎞ ⎜ ⎟ = − g sin θ dθ ⎝ L ⎠ vdv = − gL sin θ dθ

Separate the variables to obtain: v

0

0

θ0

(e) Integrate the left side of the equation in part (d) from v = 0 to the final speed v and the right side from θ = θ0 to θ = 0:

∫ v' dv' = ∫ − gL sin θ ' dθ '

Evaluate the limits of integration to obtain:

1 2

Note, from the figure, that h = L(1 − cosθ0). Substitute and solve for v:

v=

v 2 = gL(1 − cosθ 0 )

2 gh

93 ••• Picture the Problem The potential energy of the climber is the sum of his gravitational potential energy and the potential energy stored in the spring-like bungee cord. Let θ be the angle which the position of the rock climber on the cliff face makes with a vertical axis and choose the zero of gravitational potential energy to be at the bottom of the cliff. We can use the definitions of Ug and Uspring to express the climber’s total potential energy. (a) Express the total potential energy of the climber:

U (s ) = U bungee cord + U g U ( s ) = 12 k (s − L ) + Mgy

Substitute to obtain:

2

= 12 k (s − L ) + MgH cos θ 2

⎛ s ⎞ 2 = 12 k (s − L ) + MgH cos ⎜ ⎟ ⎝H⎠ A spreadsheet solution is shown below. The constants used in the potential energy function and the formulas used to calculate the potential energy are as follows: Cell B3 B4 B5 B6 B7 D11 D12

Content/Formula 300 5 60 85 9.81 60 D11+1

Algebraic Form H k L M g s s+1

Conservation of Energy 513 E11

0.5*$B$4*(D11−$B$5)^2 +$B$6*$B$7*$B$3*(cos(D11/$B$3))

G11

E11−E61

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

B

H= k= L= m= g=

300 5 60 85 9.81

1 2

⎛ s ⎞ 2 k (s − L ) + MgH cos ⎜ ⎟ ⎝H⎠ U (60 m ) − U (110 m )

C

D

E

s 60 61 62 63 64

U(s) 2.45E+05 2.45E+05 2.45E+05 2.45E+05 2.45E+05

196 197 198 199 200

2.45E+05 2.45E+05 2.45E+05 2.45E+05 2.46E+05

m N/m m kg m/s^2

147 148 149 150 151

The following graph was plotted using the data from columns D (s) and E (U(s)).

246 245 244

U (kJ)

243 242 241 240 239 238 50

70

90

110

130 s (m)

150

170

190

210

514 Chapter 7 *94 ••• Picture the Problem The diagram shows the forces each of the springs exerts on the block. The change in the potential energy stored in the springs is due to the elongation of both springs when the block is displaced a distance x from its equilibrium position and

we can find ∆U using 12 k (∆L ) . We can find the magnitude of the force pulling the block back toward its equilibrium position by finding the sum of the magnitudes of the y components of the forces exerted by the springs. In Part (d) we can use conservation of energy to find the speed of the block as it passes through its equilibrium position. 2

(a) Express the change in the potential energy stored in the springs when the block is displaced a distance x:

[

]

∆U = 2 12 k (∆L ) = k (∆L ) 2

2

where ∆L is the change in length of a spring.

Referring to the force diagram, express ∆L:

∆L = L2 + x 2 − L

Substitute to obtain:

∆U = k

(b) Sum the forces acting on the block to express Frestoring:

Frestoring = 2 F cos θ = 2k∆L cos θ

( L + x − L) 2

2

x

= 2k∆L Substitute for ∆L to obtain:

Frestoring = 2k

2

L + x2 2

( L + x − L) 2

x

2

⎛ L = 2kx ⎜⎜1 − 2 L + x2 ⎝

L + x2 2

⎞ ⎟ ⎟ ⎠

(c) A spreadsheet program to calculate U(x) is shown below. The constants used in the potential energy function and the formulas used to calculate the potential energy are as follows: Cell B1 B2 B3 C8 D7

Content/Formula 1 1 1 C7+0.01 $B$2*((C7^2+$B$1^2)^0.5−$B$1)^2

Algebraic Form L k M x U(x)

Conservation of Energy 515

1 2 3 4 5 6 7 8 9 10 11 12

A B L = 0.1 k= 1 M= 1

23 24 25 26 27

C m N/m kg

D

x 0 0.01 0.02 0.03 0.04 0.05

U(x) 0 2.49E−07 3.92E−06 1.94E−05 5.93E−05 1.39E−04

0.16 0.17 0.18 0.19 0.20

7.86E−03 9.45E−03 1.12E−02 1.32E−02 1.53E−02

The following graph was plotted using the data from columns C (x) and D (U(x)).

16 14 12

U (mJ)

10 8 6 4 2 0 0.00

0.05

0.10

0.15

x (m)

(d) Use conservation of energy to relate the kinetic energy of the block as it passes through the equilibrium position to the change in its potential energy as it returns to its equilibrium position:

K equilibrium = ∆U or 1 2

Mv 2 = ∆U

0.20

516 Chapter 7 Solve for v to obtain:

v= =

2∆U = M

2k

( L + x − L) 2

2

( L + x − L) 2

2

M 2k M

Substitute numerical values and evaluate v:

v = ⎛⎜ ⎝

(0.1 m )2 + (0.1 m )2 − 0.1 m ⎞⎟ ⎠

2(1 N/m ) = 5.86 cm/s 1 kg

2

Chapter 8 Systems of Particles and Conservation of Momentum Conceptual Problems 1 • Determine the Concept A doughnut. The definition of the center of mass of an object does not require that there be any matter at its location. Any hollow sphere (such as a basketball) or an empty container with any geometry are additional examples of threedimensional objects that have no mass at their center of mass. *2 • Determine the Concept The center of mass is midway between the two balls and is in free-fall along with them (all forces can be thought to be concentrated at the center of mass.) The center of mass will initially rise, then fall. Because the initial velocity of the center of mass is half of the initial velocity of the ball thrown upwards, the mass thrown upwards will rise for twice the time that the center of mass rises. Also, the center of mass will rise until the velocities of the two balls are equal but opposite. (b) is correct. 3 • Determine the Concept The acceleration of the center of mass of a system of particles is

r

described by Fnet,ext =

r r F ∑ i,ext = Macm , where M is the total mass of the system. i

Express the acceleration of the center of mass of the two pucks:

acm =

Fnet,ext F1 = M m1 + m2

and (b) is correct. 4 • Determine the Concept The acceleration of the center of mass of a system of particles

r

is described by Fnet,ext =

r

∑F

i,ext

r = Macm , where M is the total mass of the system.

i

Express the acceleration of the center of mass of the two pucks:

acm =

Fnet,ext F1 = M m1 + m2

because the spring force is an internal force.

(b) is correct.

517

518 Chapter 8 *5 • Determine the Concept No. Consider a 1-kg block with a speed of 1 m/s and a 2- kg block with a speed of 0.707 m/s. The blocks have equal kinetic energies but momenta of magnitude 1 kg·m /s and 1.414 kg·m/s, respectively. 6 • (a) True. The momentum of an object is the product of its mass and velocity. Therefore, if we are considering just the magnitudes of the momenta, the momentum of a heavy object is greater than that of a light object moving at the same speed. (b) True. Consider the collision of two objects of equal mass traveling in opposite directions with the same speed. Assume that they collide inelastically. The mechanical energy of the system is not conserved (it is transformed into other forms of energy), but the momentum of the system is the same after the collision as before the collision, i.e., zero. Therefore, for any inelastic collision, the momentum of a system may be conserved even when mechanical energy is not. (c) True. This is a restatement of the expression for the total momentum of a system of particles. 7 • Determine the Concept To the extent that the system in which the rifle is being fired is an isolated system, i.e., the net external force is zero, momentum is conserved during its firing. Apply conservation of momentum to the firing of the rifle:

r r prifle + pbullet = 0 or

r r prifle = − pbullet

*8 • Determine the Concept When she jumps from a boat to a dock, she must, in order for momentum to be conserved, give the boat a recoil momentum, i.e., her forward momentum must be the same as the boat’s backward momentum. The energy she 2 imparts to the boat is Eboat = pboat 2mboat .

When she jumps from one dock to another, the mass of the dock plus the earth is so large that the energy she imparts to them is essentially zero. *9 •• Determine the Concept Conservation of momentum requires only that the net external force acting on the system be zero. It does not require the presence of a medium such as air.

Systems of Particles and Conservation of Momentum 519 10 • 2 . The kinetic Determine the Concept The kinetic energy of the sliding ball is 12 mvcm 2 energy of the rolling ball is 12 mvcm + K rel , where its kinetic energy relative to its center

of mass is K rel . Because the bowling balls are identical and have the same velocity, the rolling ball has more energy. 11 • Determine the Concept Think of someone pushing a box across a floor. Her push on the box is equal but opposite to the push of the box on her, but the action and reaction forces act on different objects. You can only add forces when they act on the same object. 12 • Determine the Concept It’s not possible for both to remain at rest after the collision, as that wouldn't satisfy the requirement that momentum is conserved. It is possible for one to remain at rest: This is what happens for a one-dimensional collision of two identical particles colliding elastically. 13 • Determine the Concept It violates the conservation of momentum! To move forward requires pushing something backwards, which Superman doesn’t appear to be doing when flying around. In a similar manner, if Superman picks up a train and throws it at Lex Luthor, he (Superman) ought to be tossed backwards at a pretty high speed to satisfy the conservation of momentum. *14 •• Determine the Concept There is only one force which can cause the car to move forward−the friction of the road! The car’s engine causes the tires to rotate, but if the road were frictionless (as is closely approximated by icy conditions) the wheels would simply spin without the car moving anywhere. Because of friction, the car’s tire pushes backwards against the road−from Newton’s third law, the frictional force acting on the tire must then push it forward. This may seem odd, as we tend to think of friction as being a retarding force only, but true. 15 •• Determine the Concept The friction of the tire against the road causes the car to slow down. This is rather subtle, as the tire is in contact with the ground without slipping at all times, and so as you push on the brakes harder, the force of static friction of the road against the tires must increase. Also, of course, the brakes heat up, and not the tires. 16 • Determine the Concept Because ∆p = F∆t is constant, a safety net reduces the force acting on the performer by increasing the time ∆t during which the slowing force acts. 17 • Determine the Concept Assume that the ball travels at 80 mi/h ≈ 36 m/s. The ball stops in a distance of about 1 cm. So the distance traveled is about 2 cm at an average speed of

520 Chapter 8 about 18 m/s. The collision time is

0.02 m ≈ 1 ms . 18 m/s

18 • Determine the Concept The average force on the glass is less when falling on a carpet because ∆t is longer. 19 • (a) False. In a perfectly inelastic collision, the colliding bodies stick together but may or may not continue moving, depending on the momentum each brings to the collision. (b) True. In a head-on elastic collision both kinetic energy and momentum are conserved and the relative speeds of approach and recession are equal. (c) True. This is the definition of an elastic collision. *20 •• Determine the Concept All the initial kinetic energy of the isolated system is lost in a perfectly inelastic collision in which the velocity of the center of mass is zero. 21 •• Determine the Concept We can find the loss of kinetic energy in these two collisions by finding the initial and final kinetic energies. We’ll use conservation of momentum to find the final velocities of the two masses in each perfectly elastic collision.

pbefore = pafter

(a) Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V:

or

Express the loss of kinetic energy for the case in which the two objects have oppositely directed velocities of magnitude v/2:

⎛ ⎛ v ⎞2 ⎞ ∆K = K f − K i = 0 − 2⎜ 12 m⎜ ⎟ ⎟ ⎜ ⎝2⎠ ⎟ ⎝ ⎠ 2 mv =− 4

Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V:

mv − mv = 2mV ⇒ V = 0

pbefore = pafter or

mv = 2mV ⇒ V = 12 v

Systems of Particles and Conservation of Momentum 521 Express the loss of kinetic energy for the case in which the one object is initially at rest and the other has an initial velocity v:

∆K = K f − K i 2

mv ⎛v⎞ = (2m )⎜ ⎟ − 12 mv 2 = − 4 ⎝2⎠

2

1 2

The loss of kinetic energy is the same in both cases. (b) Express the percentage loss for the case in which the two objects have oppositely directed velocities of magnitude v/2:

1 mv 2 ∆K = 14 2 = 100% K before 4 mv

Express the percentage loss for the case in which the one object is initially at rest and the other has an initial velocity v:

2 1 ∆K 4 mv = = 50% K before 12 mv 2

The percentage loss is greatest for the case in which the two objects have oppositely directed velocities of magnitude v/2. *22 •• Determine the Concept A will travel farther. Both peas are acted on by the same force, but pea A is acted on by that force for a longer time. By the impulse-momentum theorem, its momentum (and, hence, speed) will be higher than pea B’s speed on leaving the shooter. 23 •• Determine the Concept Refer to the particles as particle 1 and particle 2. Let the direction particle 1 is moving before the collision be the positive x direction. We’ll use both conservation of momentum and conservation of mechanical energy to obtain an expression for the velocity of particle 2 after the collision. Finally, we’ll examine the ratio of the final kinetic energy of particle 2 to that of particle 1 to determine the condition under which there is maximum energy transfer from particle 1 to particle 2. Use conservation of momentum to obtain one relation for the final velocities: Use conservation of mechanical energy to set the velocity of

m1v1,i = m1v1,f + m2 v 2,f

(1)

v 2,f − v1,f = −(v 2,i − v1,i ) = v1,i (2)

522 Chapter 8 recession equal to the negative of the velocity of approach: To eliminate v1,f, solve equation (2) for v1,f, and substitute the result in equation (1): Solve for v2,f:

Express the ratio R of K2,f to K1,i in terms of m1 and m2:

v1,f = v 2,f + v1,i

m1v1, i = m1 (v2, f − v1,i ) + m2v2, f

v 2,f =

2

R= =

Differentiate this ratio with respect to m2, set the derivative equal to zero, and obtain the quadratic equation: Solve this equation for m2 to determine its value for maximum energy transfer:

2m1 v1,i m1 + m2

−

K 2,f K1,i

⎛ 2m1 ⎞ 2 1 ⎜ ⎟⎟ v1,i 2 m2 ⎜ m m + 2 ⎠ ⎝ 1 = 2 1 m v 2 1 1,i

m2 4m12 m1 (m1 + m2 )2

m22 +1 = 0 m12

m2 = m1

(b) is correct because all of 1' s kinetic energy is transferred to 2 when m2 = m1. 24 • Determine the Concept In the center-of-mass reference frame the two objects approach with equal but opposite momenta and remain at rest after the collision. 25 • Determine the Concept The water is changing direction when it rounds the corner in the nozzle. Therefore, the nozzle must exert a force on the stream of water to change its direction, and, from Newton’s 3rd law, the water exerts an equal but opposite force on the nozzle. 26 • Determine the Concept The collision usually takes place in such a short period of time that the impulse delivered by gravity or friction is negligible.

Systems of Particles and Conservation of Momentum 523 27 • r r Determine the Concept No. Fext,net = dp dt defines the relationship between the net force acting on a system and the rate at which its momentum changes. The net external force acting on the pendulum bob is the sum of the force of gravity and the tension in the string and these forces do not add to zero. *28 •• Determine the Concept We can apply conservation of momentum and Newton’s laws of motion to the analysis of these questions. (a) Yes, the car should slow down. An easy way of seeing this is to imagine a "packet" of grain being dumped into the car all at once: This is a completely inelastic collision, with the packet having an initial horizontal velocity of 0. After the collision, it is moving with the same horizontal velocity that the car does, so the car must slow down. (b) When the packet of grain lands in the car, it initially has a horizontal velocity of 0, so it must be accelerated to come to the same speed as the car of the train. Therefore, the train must exert a force on it to accelerate it. By Newton’s 3rd law, the grain exerts an equal but opposite force on the car, slowing it down. In general, this is a frictional force which causes the grain to come to the same speed as the car. (c) No it doesn’t speed up. Imagine a packet of grain being "dumped" out of the railroad car. This can be treated as a collision, too. It has the same horizontal speed as the railroad car when it leaks out, so the train car doesn’t have to speed up or slow down to conserve momentum. *29 •• Determine the Concept Think of the stream of air molecules hitting the sail. Imagine that they bounce off the sail elastically−their net change in momentum is then roughly twice the change in momentum that they experienced going through the fan. Another way of looking at it: Initially, the air is at rest, but after passing through the fan and bouncing off the sail, it is moving backward−therefore, the boat must exert a net force on the air pushing it backward, and there must be a force on the boat pushing it forward.

Estimation and Approximation 30 •• Picture the Problem We can estimate the time of collision from the average speed of the car and the distance traveled by the center of the car during the collision. We’ll assume a car length of 6 m. We can calculate the average force exerted by the wall on the car from the car’s change in momentum and it’s stopping time. (a) Relate the stopping time to the assumption that the center of the car travels halfway to the wall with constant deceleration:

∆t =

d stopping vav

=

1 2

( 12 Lcar ) = vav

1 4

Lcar vav

524 Chapter 8 Express and evaluate vav:

vav =

vi + vf 2

km 1h 1000 m × × h 3600 s km = 2 = 12.5 m/s 0 + 90

Substitute for vav and evaluate ∆t:

∆t =

1 4

(6 m )

12.5 m/s

= 0.120 s

(b) Relate the average force exerted by the wall on the car to the car’s change in momentum:

⎛

Fav =

∆p = ∆t

(2000 kg ) ⎜⎜ 90 km × ⎝

h

1h 1000 m ⎞ ⎟ × 3600 s km ⎟⎠

0.120 s

= 417 kN

31 •• Picture the Problem Let the direction the railcar is moving be the positive x direction and the system include the earth, the pumpers, and the railcar. We’ll also denote the railcar with the letter c and the pumpers with the letter p. We’ll use conservation of momentum to relate the center of mass frame velocities of the car and the pumpers and then transform to the earth frame of reference to find the time of fall of the car. (a) Relate the time of fall of the railcar to the distance it falls and its velocity as it leaves the bank: Use conservation of momentum to find the speed of the car relative to the velocity of its center of mass:

Relate uc to up and solve for uc:

∆t =

∆y vc

r r pi = pf or mc uc + mp u p = 0 u c − u p = 4 m/s ∴ u p = u c − 4 m/s

Substitute for up to obtain:

mc uc + mp (uc − 4 m/s ) = 0

Systems of Particles and Conservation of Momentum 525 Solve for and evaluate uc:

Relate the speed of the car to its speed relative to the center of mass of the system:

Substitute and evaluate ∆t:

(b) Find the speed with which the pumpers hit the ground:

uc =

4 m/s 4 m/s = = 1.85 m/s m 350 kg 1+ c 1+ 4(75 kg ) mp

vc = uc + vcm m km 1h 1000 m + 32 × × s h 3600 s km = 10.74 m/s = 1.85

∆t =

25 m = 2.33 s 10.74 m/s

vp = vc − up = 10.74 m/s − 4 m/s = 6.74 m/s Hitting the ground at this speed, they may be injured.

*32 •• Picture the Problem The diagram depicts the bullet just before its collision with the melon and the motion of the melon-and-bullet-less-jet and the jet just after the collision. We’ll assume that the bullet stays in the watermelon after the collision and use conservation of momentum to relate the mass of the bullet and its initial velocity to the momenta of the melon jet and the melon less the plug after the collision.

Apply conservation of momentum to the collision to obtain: Solve for v2f:

Express the kinetic energy of the jet of melon in terms of the initial kinetic energy of the bullet:

m1v1i = (m2 − m3 + m1 ) v2f + 2m3 K 3

v2f =

m1v1i − 2m3 K 3 m2 − m3 + m1

K 3 = 101 K1 = 101

(

1 2

)

m1v1i2 =

1 20

m1v1i2

526 Chapter 8 Substitute and simplify to obtain:

v2f = =

m1v1i − 2m3 ( 201 m1v1i2 )

(

m2 − m3 + m1

v1i m1 − 0.1m1m3 m2 − m3 + m1

)

Substitute numerical values and evaluate v2f:

(

)

⎛ ft 1 m ⎞ 0.0104 kg − 0.1(0.0104 kg )(0.14 kg ) ⎟ = −0.386 m/s v2f = ⎜⎜1800 × s 3.281 ft ⎟⎠ 2.50 kg − 0.14 kg + 0.0104 kg ⎝ = − 1.27 ft/s Note that this result is in reasonably good agreement with experimental results.

Finding the Center of Mass 33 • Picture the Problem We can use its definition to find the center of mass of this system. Apply its definition to find xcm:

xcm =

m1 x1 + m2 x2 + m3 x3 (2 kg )(0) + (2 kg )(0.2 m ) + (2 kg )(0.5 m ) = = 0.233 m 2 kg + 2 kg + 2 kg m1 + m2 + m3

Because the point masses all lie along the x axis:

y cm = 0 and the center of mass of this system of particles is at

(0.233 m, 0)

.

*34 • Picture the Problem Let the left end of the handle be the origin of our coordinate system. We can disassemble the club-ax, find the center of mass of each piece, and then use these coordinates and the masses of the handle and stone to find the center of mass of the club-ax. Express the center of mass of the handle plus stone system: Assume that the stone is drilled and the stick passes through it. Use symmetry considerations to locate the center of mass of the stick:

xcm =

mstick xcm, stick + mstone xcm, stone mstick + mstone

xcm,stick = 45.0 cm

Systems of Particles and Conservation of Momentum 527 Use symmetry considerations to locate the center of mass of the stone: Substitute numerical values and evaluate xcm:

xcm,stone = 89.0 cm

xcm =

(2.5 kg )(45 cm) + (8 kg )(89 cm) 2.5 kg + 8 kg

= 78.5cm 35 • Picture the Problem We can treat each of balls as though they are point objects and apply the definition of the center of mass to find (xcm, ycm). Use the definition of xcm:

xcm = =

m A x A + m B x B + mC x C m A + m B + mC

(3 kg )(2 m ) + (1 kg )(1 m ) + (1 kg )(3 m ) 3 kg + 1 kg + 1 kg

= 2.00 m Use the definition of ycm:

ycm = =

m A y A + mB y B + mC yC m A + mB + mC

(3 kg )(2 m ) + (1 kg )(1 m ) + (1 kg )(0) 3 kg + 1 kg + 1 kg

= 1.40 m The center of mass of this system of particles is at:

(2.00 m,1.40 m )

36 • Picture the Problem The figure shows an equilateral triangle with its y-axis vertex above the x axis. The bisectors of the vertex angles are also shown. We can find x coordinate of the center-of-mass by inspection and the y coordinate using trigonometry. From symmetry considerations:

xcm = 0

528 Chapter 8 Express the trigonometric relationship between a/2, 30°, and ycm: Solve for ycm:

tan 30° =

ycm a2

ycm = 12 a tan 30° = 0.289a The center of mass of an equilateral triangle oriented as shown above is at

(0, 0.289a )

.

*37 •• Picture the Problem Let the subscript 1 refer to the 3-m by 3-m sheet of plywood before the 2-m by 1-m piece has been cut from it. Let the subscript 2 refer to 2-m by 1-m piece that has been removed and let σ be the area density of the sheet. We can find the centerof-mass of these two regions; treating the missing region as though it had negative mass, and then finding the center-of-mass of the U-shaped region by applying its definition. Express the coordinates of the center of mass of the sheet of plywood:

Use symmetry to find xcm,1, ycm,1, xcm,2, and ycm,2:

xcm = ycm =

m1 xcm,1 − m2 xcm, 2 m1 − m2 m1 ycm,1 − m2 ycm , 2 m1 − m2

xcm,1 = 1.5 m, ycm,1 = 1.5 m and xcm,2 = 1.5 m, ycm,2 = 2.0 m

Determine m1 and m2:

m1 = σA1 = 9σ kg and m2 = σA2 = 2σ kg

Substitute numerical values and evaluate xcm:

xcm =

(9σ kg )(1.5 m ) − (2σ kg )(1.5 kg ) 9σ kg − 2σ kg

= 1.50 m Substitute numerical values and evaluate ycm:

ycm =

(9σ kg )(1.5 m ) − (2σ kg )(2 m ) 9σ kg − 2σ kg

= 1.36 m The center of mass of the U-shaped sheet of plywood is at (1.50 m,1.36 m ) .

Systems of Particles and Conservation of Momentum 529 38 •• Picture the Problem We can use its definition to find the center of mass of the can plus water. By setting the derivative of this function equal to zero, we can find the value of x that corresponds to the minimum height of the center of mass of the water as it drains out and then use this extreme value to express the minimum height of the center of mass. (a) Using its definition, express the location of the center of mass of the can + water: Let the cross-sectional area of the cup be A and use the definition of density to relate the mass m of water remaining in the can at any given time to its depth x: Solve for m to obtain:

⎛H⎞ ⎛ x⎞ M ⎜ ⎟ + m⎜ ⎟ 2 ⎝2⎠ xcm = ⎝ ⎠ M +m M m ρ= = AH Ax

m=

Substitute to obtain:

xcm

x M H ⎛H⎞ ⎛ x ⎞⎛ x ⎞ M ⎜ ⎟ + ⎜ M ⎟⎜ ⎟ 2 H ⎠⎝ 2 ⎠ = ⎝ ⎠ ⎝ x M+ M H ⎛ ⎛ x ⎞2 ⎞ ⎜1+ ⎜ ⎟ ⎟ H ⎜ ⎝H⎠ ⎟ = 2 ⎜ 1+ x ⎟ ⎜ ⎟ ⎜ H ⎟⎠ ⎝

(b) Differentiate xcm with respect to x and set the derivative equal to zero for extrema:

2 ⎫ ⎧ ⎡ ⎛ x ⎞2 ⎤ d ⎛ x⎞ d ⎡ ⎛ x⎞ ⎤ x⎞ ⎛ ⎛ x ⎞2 ⎞ ⎛ ⎪ ⎢ ⎥ ⎢ ⎥ 1 + 1 + 1 + 1 + ⎜1+ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎟⎪ ⎜ ⎟ ⎜ ⎟ H dx H H dx H ⎪ ⎝ ⎠ ⎝ ⎠ ⎪⎪ ⎝ ⎠ ⎝ ⎠ ⎢ ⎥ ⎢ ⎥ dxcm H d ⎜ ⎝ H ⎠ ⎟ H ⎪ ⎦ ⎣ ⎦ −⎣ = = ⎨ ⎬ ⎜ ⎟ 2 2 x dx 2 dx ⎜ x⎞ x⎞ ⎪ ⎛ ⎛ ⎟ 2⎪ 1+ ⎜1 + ⎟ ⎜1 + ⎟ H ⎟ ⎜ ⎪ ⎪ ⎝ H⎠ ⎝ H⎠ ⎝ ⎠ ⎪⎭ ⎪⎩ ⎡ ⎛ x ⎞ 2 ⎤⎛ 1 ⎞ ⎫ ⎧⎛ x x 1 ⎪ ⎪ ⎜1 + ⎞⎟2⎛⎜ ⎞⎟⎛⎜ ⎞⎟ ⎢⎢1 + ⎜⎝ H ⎟⎠ ⎥⎥⎜⎝ H ⎟⎠ ⎪ H⎪ ⎪ H ⎠ ⎝ H ⎠⎝ H ⎠ ⎣ ⎦ = ⎨⎝ − ⎬=0 2 2 2 ⎪ x⎞ x⎞ ⎪ ⎛ ⎛ ⎜1 + ⎟ ⎜1 + ⎟ ⎪ ⎪ ⎝ H⎠ ⎝ H⎠ ⎩ ⎪⎭ Simplify this expression to obtain:

2

⎛ x⎞ ⎛ x⎞ ⎜ ⎟ + 2⎜ ⎟ − 1 = 0 ⎝H ⎠ ⎝H ⎠

530 Chapter 8 x=H

Solve for x/H to obtain:

(

)

2 − 1 ≈ 0.414 H

where we’ve kept the positive solution because a negative value for x/H would make no sense. Use your graphing calculator to convince yourself that the graph of xcm as a function of x is concave upward at x ≈ 0.414 H and that, therefore, the minimum value of xcm occurs at x ≈ 0.414 H . Evaluate xcm at x = H

(

)

2 − 1 to obtain:

= H

(

)

(

)

*39 •• Picture the Problem A semicircular disk and a surface element of area dA is shown in the diagram. Because the disk is a continuous object, we’ll use

r r Mrcm = ∫ r dm and symmetry to find its

center of mass.

xcm = 0 by symmetry. ycm =

)

2 −1

Finding the Center of Mass by Integration

Express the coordinates of the center of mass of the semicircular disk:

(

⎛ ⎛ H 2 − 1 ⎞2 ⎞ ⎜1+ ⎜ ⎟ ⎟ ⎟ ⎟ ⎜ ⎜ H H ⎠ ⎟ ⎝ xcm x= H ( 2 −1) = ⎜ 2⎜ H 2 −1 ⎟ ⎟⎟ ⎜⎜ 1 + H ⎠ ⎝

∫ yσ dA M

Express y as a function of r and θ :

y = r sin θ

Express dA in terms of r and θ :

dA = r dθ dr

Express M as a function of r and θ :

M = σAhalf disk = 12 σπR 2

Systems of Particles and Conservation of Momentum 531 Rπ

Substitute and evaluate ycm:

ycm = =

σ ∫ ∫ r 2 sin θ dθ dr 0 0

M

2σ 2 = r dr M ∫0 R

2σ 3 4 R = R 3M 3π

40 ••• Picture the Problem Because a solid hemisphere is a continuous object, we’ll use

r r Mrcm = ∫ r dm to find its center of mass. The volume element for a sphere is

dV = r2 sinθ dθ dφ dr, where θ is the polar angle and φ the azimuthal angle. Let the base of the hemisphere be the xy plane and ρ be the mass density. Then: Express the z coordinate of the center of mass:

Evaluate M =

∫ ρdV :

z = r cos θ

zcm =

∫ rρdV ∫ ρdV

M = ∫ ρdV = 12 ρVsphere

(

)

= 12 ρ 43 π R 3 = 23 πρR 3 Evaluate

∫ rρdV :

R π / 2 2π

∫ rρdV = ∫ 0

= Substitute and simplify to find zcm:

zcm =

1 4 2 3

∫ ∫r 0

πρR

3

sin θ cosθdθdφdr

0

4

2

πρR 4 = πρR 3

[ sin θ ] 2

1 2

3 8

π /2 0

=

πρR 4 4

R

41 ••• Picture the Problem Because a thin hemisphere shell is a continuous object, we’ll use r r Mrcm = r dm to find its center of mass. The element of area on the shell is dA = 2πR2

∫

sinθ dθ, where R is the radius of the hemisphere. Let σ be the surface mass density and express the z coordinate of the center of mass:

zcm =

∫ zσ dA ∫ σ dA

532 Chapter 8

∫

Evaluate M = σ dA :

M = ∫ σ dA = 12 σAspherical shell

(

)

= 12 σ 4π R 2 = 2πσR 2 Evaluate

∫ zσ dA :

π /2

∫ zσ dA = 2πR σ ∫ sin θ cos θ dθ 3

0

= πR 3σ

π /2

∫ sin 2θ dθ 0

= πR σ 3

Substitute and simplify to find zcm:

zcm =

πR 3σ = 2πσR 2

1 2

R

42 ••• Picture the Problem The parabolic sheet is shown to the right. Because the area of the sheet is distributed symmetrically with respect to the y axis, xcm = 0. We’ll integrate the element of area dA (= xdy) to obtain the total area of the sheet and yxdy to obtain the numerator of the definition of the center of mass. b

Express ycm:

ycm =

∫ xydy 0 b

∫ xdy 0

b

Evaluate

∫ xydy : 0

b

b

0

0

∫ xydy = ∫ =

b

Evaluate

∫ xdy : 0

2 5 a

b

b

0

0

∫ xdy = ∫ =

b

y1 2 1 ydy = y 3 2 dy ∫ a a0 b5 2 b

y1 2 1 dy = y1 2 dy ∫ a a0 2

3 a

b3 2

Systems of Particles and Conservation of Momentum 533 2

Substitute and simplify to determine ycm:

b5 2

= 3b ycm = 5 a 2 32 5 b 3 a Note that, by symmetry:

xcm = 0

(0, 35 b )

The center of mass of the parabolic sheet is at:

Motion of the Center of Mass 43 • Picture the Problem The velocity of the center of mass of a system of particles is related

r

to the total momentum of the system through P =

r

∑mv

i i

r = Mv cm .

i

r

Use the expression for the total momentum of a system to relate the velocity of the center of mass of the two-particle system to the momenta of the individual particles:

r v cm =

∑m v

i i

i

M

r r m1v1 + m2 v 2 = m1 + m2

r (3 kg )(vr1 + vr2 ) = 1 (vr + vr ) vcm = 1 2 2 6 kg = 1 (6 m/s ) iˆ − (3 m/s ) ˆj

Substitute numerical values and r evaluate v cm :

2

=

[

]

(3 m/s ) iˆ − (1.5 m/s ) ˆj

*44 • Picture the Problem Choose a coordinate system in which east is the positive x

r

direction and use the relationship P =

r

∑mv

i i

r = Mv cm to determine the velocity of the

i

center of mass of the system. Use the expression for the total momentum of a system to relate the velocity of the center of mass of the two-vehicle system to the momenta of the individual vehicles: Express the velocity of the truck:

r

r v cm =

∑m v

i i

i

M

r r mt v t + mc v c = mt + mc

r v t = (16 m/s ) iˆ

534 Chapter 8 r v c = (− 20 m/s ) iˆ

Express the velocity of the car:

r

Substitute numerical values and evaluate v cm :

r (3000 kg )(16 m/s) iˆ + (1500 kg )(− 20 m/s) iˆ = v cm = 3000 kg + 1500 kg

(4.00 m/s) iˆ

45 • Picture the Problem The acceleration of the center of mass of the ball is related to the

r

r

net external force through Newton’s 2nd law: Fnet,ext = Ma cm . Use Newton’s 2nd law to express the acceleration of the ball:

r Fnet,ext r acm = M

Substitute numerical values and r evaluate a cm :

r acm =

(12 N ) iˆ 3 kg + 1 kg + 1 kg

=

(2.4 m/s )iˆ 2

46 •• Picture the Problem Choose a coordinate system in which upward is the positive y r r direction. We can use Newton’s 2nd law Fnet,ext = Ma cm to find the acceleration of the

center of mass of this two-body system.

(a)

Yes; initially the scale reads ( M + m) g ; while m is in free fall, the reading is Mg.

(b) Using Newton’s 2nd law, express the acceleration of the center of mass of the system: Substitute to obtain:

(c) Use Newton’s 2nd law to express the net force acting on the scale while the object of mass m is falling: Substitute and simplify to obtain:

r Fnet,ext r acm = mtot r mg ˆ a cm = − j M +m

Fnet,ext = (M + m )g − ( M + m)acm

⎛ mg ⎞ Fnet,ext = (M + m )g − ( M + m) ⎜ ⎟ ⎝M +m⎠ = Mg

Systems of Particles and Conservation of Momentum 535 as expected, given our answer to part (a). *47 •• Picture the Problem The free-body diagram shows the forces acting on the platform when the spring is partially compressed. The scale reading is the force the scale exerts on the platform and is represented on the FBD by Fn. We can use Newton’s 2nd law to determine the scale reading in part (a) and the work-energy theorem in conjunction with Newton’s 2nd law in parts (b) and (c).

(a) Apply

∑F

y

= ma y to the

∑F

y

= Fn − mp g − Fball on spring = 0

spring when it is compressed a distance d: Solve for Fn:

Fn = mp g + Fball on spring ⎛m g⎞ = mp g + kd = mp g + k ⎜ b ⎟ ⎝ k ⎠

= mp g + mb g = (mp + mb )g (b) Use conservation of mechanical energy, with Ug = 0 at the position at which the spring is fully compressed, to relate the gravitational potential energy of the system to the energy stored in the fully compressed spring:

∆K + ∆U g + ∆U s = 0 Because ∆K = Ug,f = Us,i = 0,

U g,i − U s,f = 0

or

mb gd − 12 kd 2 = 0

2m b g k

Solve for d:

d=

Evaluate our force equation in (a)

Fn = mp g + Fball on spring

with d =

2m b g : k

⎛ 2m g ⎞ = mp g + kd = mp g + k ⎜ b ⎟ ⎝ k ⎠ = mp g + 2mb g = (mp + 2mb )g

536 Chapter 8 (c) When the ball is in its original position, the spring is relaxed and exerts no force on the ball. Therefore:

Fn = scale reading = mp g

*48 •• Picture the Problem Assume that the object whose mass is m1 is moving downward and take that direction to be the positive direction. We’ll use Newton’s 2nd law for a system of particles to relate the acceleration of the center of mass to the acceleration of the individual particles.

(a) Relate the acceleration of the center of mass to m1, m2, mc and their accelerations:

r r r r Ma cm = m1a1 + m2 a 2 + mc a c

m1 − m2 m1 + m2 + mc

Because m1 and m2 have a common acceleration a and ac = 0:

a cm = a

From Problem 4-81 we have:

a=g

Substitute to obtain:

⎛ m − m2 ⎞ ⎛ m1 − m2 ⎞ ⎟⎟ g ⎟⎟ ⎜⎜ acm = ⎜⎜ 1 ⎝ m1 + m2 ⎠ ⎝ m1 + m2 + mc ⎠ =

(b) Use Newton’s 2nd law for a system of particles to obtain:

Solve for F and substitute for acm from part (a):

m1 − m2 m1 + m2

(m1 − m2 )2 g (m1 + m2 )(m1 + m2 + mc )

F − Mg = − Macm where M = m1 + m2 + mc and F is positive upwards.

F = Mg − Macm 2 ( m1 − m2 ) g = Mg −

m1 + m2

⎡ 4m1 m2 ⎤ = ⎢ + mc ⎥ g ⎣ m1 + m2 ⎦ (c) From Problem 4-81:

T=

2m1 m2 g m1 + m2

Systems of Particles and Conservation of Momentum 537 Substitute in our result from part (b) to obtain:

⎡ 2m1m2 ⎤ F = ⎢2 + mc ⎥ g ⎣ m1 + m2 ⎦ ⎡ T ⎤ = ⎢2 + mc ⎥ g = 2T + mc g ⎣ g ⎦

49 •• Picture the Problem The free-body diagram shows the forces acting on the platform when the spring is partially compressed. The scale reading is the force the scale exerts on the platform and is represented on the FBD by Fn. We can use Newton’s 2nd law to determine the scale reading in part (a) and the result of Problem 7-96 part (b) to obtain the scale reading when the ball is dropped from a height h above the cup.

(a) Apply

∑F

y

= ma y to the spring

∑F

y

= Fn − mp g − Fball on spring = 0

when it is compressed a distance d: Solve for Fn:

Fn = mp g + Fball on spring ⎛m g⎞ = mp g + kd = mp g + k ⎜ b ⎟ ⎝ k ⎠

= mp g + mb g = (mp + mb )g (b) From Problem 7-96, part (b):

From part (a):

x max =

mb g ⎛ ⎜1 + 1 + 2kh k ⎜⎝ mb g

⎞ ⎟ ⎟ ⎠

Fn = mp g + Fball on spring = mp g + kxmax ⎛ 2kh = mp g + mb g ⎜⎜1 + 1 + mb g ⎝

⎞ ⎟ ⎟ ⎠

The Conservation of Momentum 50 • Picture the Problem Let the system include the woman, the canoe, and the earth. Then the net external force is zero and linear momentum is conserved as she jumps off the

538 Chapter 8 canoe. Let the direction she jumps be the positive x direction. Apply conservation of momentum to the system: Substitute to obtain:

r

Solve for v canoe :

r

∑m v

i i

r r = mgirl v girl + mcanoe v canoe = 0

(55 kg )(2.5 m/s ) iˆ + (75 kg ) vrcanoe = 0 r v canoe =

(− 1.83 m/s) iˆ

51 • Picture the Problem If we include the earth in our system, then the net external force is zero and linear momentum is conserved as the spring delivers its energy to the two objects.

Apply conservation of momentum to the system: Substitute numerical values to obtain:

r

Solve for v10 :

r

∑m v

i i

r r = m5 v 5 + m10 v10 = 0

(5 kg )(− 8 m/s ) iˆ + (10 kg ) vr10 = 0 r v10 =

(4 m/s) iˆ

*52 • Picture the Problem This is an explosion-like event in which linear momentum is conserved. Thus we can equate the initial and final momenta in the x direction and the initial and final momenta in the y direction. Choose a coordinate system in the positive x direction is to the right and the positive y direction is upward.

Equate the momenta in the y direction before and after the explosion:

∑p

y,i

= ∑ py,f = mv2 − 2mv1 = m(2v1 ) − 2mv1 = 0

We can conclude that the momentum was entirely in the x direction before the particle exploded.

∑p

= ∑ p x,f

Equate the momenta in the x direction before and after the explosion:

∴4mvi = mv3

Solve for v3:

vi = 14 v3 and (c) is correct.

x,i

Systems of Particles and Conservation of Momentum 539 53 • Picture the Problem Choose the direction the shell is moving just before the explosion to be the positive x direction and apply conservation of momentum.

Use conservation of momentum to relate the masses of the fragments to their velocities:

r

r r pi = pf or

r mviˆ = 12 mvˆj + 12 mv '

r v ' = 2viˆ − vˆj

Solve for v ' :

*54 •• Picture the Problem Let the system include the earth and the platform, gun and block. r Then Fnet,ext = 0 and momentum is conserved within the system.

(a) Apply conservation of momentum to the system just before and just after the bullet leaves the gun:

r

r

r r pbefore = pafter or r r 0 = pbullet + pplatform

Substitute for pbullet and pplatform and

r 0 = mb vb iˆ + mp v platform

solve for v platform :

and

r

r m v platform = − b vb iˆ mp (b) Apply conservation of momentum to the system just before the bullet leaves the gun and just after it comes to rest in the block: (c) Express the distance ∆s traveled by the platform: Express the velocity of the bullet relative to the platform:

r r p before = pafter or

r r 0 = p platform ⇒ v platform = 0 ∆s = vplatform ∆t

vrel = vb − vplatform = vb +

mb vb mp

⎛ m ⎞ m + mb vb = ⎜1 + b ⎟v b = p ⎜ m ⎟ mp p ⎠ ⎝ Relate the time of flight ∆t to L and vrel:

∆t =

L v rel

540 Chapter 8 Substitute to find the distance ∆s moved by the platform in time ∆t:

⎛m ⎞⎛ L ⎞ ⎟ ∆s = vplatform ∆t = ⎜ b vb ⎟⎜⎜ ⎜m ⎟ v ⎟ ⎝ p ⎠⎝ rel ⎠ ⎞ ⎛ ⎟ ⎜ ⎛ mb ⎞ ⎜ L ⎟ vb ⎟ ⎜ =⎜ ⎜m ⎟ m + mb ⎟ ⎝ p ⎠⎜ p vb ⎟ ⎟ ⎜ m p ⎠ ⎝ =

mb L mp + mb

55 •• Picture the Problem The pictorial representation shows the wedge and small object, initially at rest, to the left, and, to the right, both in motion as the small object leaves the wedge. Choose the direction the small object is moving when it leaves the wedge be the positive x direction and the zero of potential energy to be at the surface of the table. Let the speed of the small object be v and that of the wedge V. We can use conservation of momentum to express v in terms of V and conservation of energy to express v in terms of h.

Apply conservation of momentum to the small object and the wedge:

r r pi, x = pf, x or

r 0 = mviˆ + 2mV r

Solve for V :

r V = − 12 viˆ

(1)

and

V = 12 v Use conservation of energy to determine the speed of the small object when it exits the wedge: Because Uf = Ki = 0:

∆K + ∆U = 0 or Kf − Ki + U f − U i = 0 1 2

mv 2 + 12 (2m )V 2 − mgh = 0

Systems of Particles and Conservation of Momentum 541 Substitute for V to obtain:

1 2

mv 2 + 12 (2m )( 12 v ) − mgh = 0 2

Solve for v to obtain:

v=2

Substitute in equation (1) to r determine V :

r ⎛ gh ⎞ ˆ ⎟i = − V = − 12 ⎜⎜ 2 ⎟ 3 ⎝ ⎠

gh 3 gh ˆ i 3

i.e., the wedge moves in the direction opposite to that of the small object with a

gh . 3

speed of

*56 •• Picture the Problem Because no external forces act on either cart, the center of mass of the two-cart system can’t move. We can use the data concerning the masses and separation of the gliders initially to calculate its location and then apply the definition of the center of mass a second time to relate the positions X1 and X2 of the centers of the carts when they first touch. We can also use the separation of the centers of the gliders when they touch to obtain a second equation in X1 and X2 that we can solve simultaneously with the equation obtained from the location of the center of mass.

(a) Apply its definition to find the center of mass of the 2-glider system:

xcm = =

m1 x1 + m2 x2 m1 + m2

(0.1kg )(0.1m ) + (0.2 kg )(1.6 m ) 0.1 kg + 0.2 kg

= 1.10 m from the left end of the air track. Use the definition of the center of mass to relate the coordinates of the centers of the two gliders when they first touch to the location of the center of mass:

1.10 m = =

m1 X 1 + m2 X 2 m1 + m2

(0.1kg )X 1 + (0.2 kg )X 2

0.1 kg + 0.2 kg = X 1 + 23 X 2 1 3

(10 cm + 20 cm) = 0.15 m

Also, when they first touch, their centers are separated by half their combined lengths:

X 2 − X1 =

Thus we have:

0.333 X 1 + 0.667 X 2 = 1.10 m

1 2

and

X 2 − X 1 = 0.15 m

542 Chapter 8 Solve these equations simultaneously to obtain: (b)

X 1 = 1.00 m and X 2 = 1.15 m

No. The initial momentum of the system is zero, so it must be zero after the collision.

Kinetic Energy of a System of Particles *57 • Picture the Problem Choose a coordinate system in which the positive x direction is to the right. Use the expression for the total momentum of a system to find the velocity of the center of mass and the definition of relative velocity to express the sum of the kinetic energies relative to the center of mass.

(a) Find the sum of the kinetic energies:

K = K1 + K 2 = 12 m1v12 + 12 m2v22 =

1 2

(3 kg )(5 m/s )2 + 12 (3 kg )(2 m/s )2

= 43.5 J (b) Relate the velocity of the center of mass of the system to its total momentum:

r r r Mv cm = m1v1 + m2 v 2

Solve for vcm :

r r r m v + m2 v 2 v cm = 1 1 m1 + m2

Substitute numerical values and r evaluate vcm :

r (3 kg )(5 m/s ) iˆ − (3 kg )(2 m/s ) iˆ vcm = 3 kg + 3 kg

r

= (1.50 m/s ) iˆ (c) The velocity of an object relative to the center of mass is given by:

r r r v rel = v − v cm

Systems of Particles and Conservation of Momentum 543 Substitute numerical values to obtain:

r v1,rel = (5 m/s ) iˆ − (1.5 m/s ) iˆ

(3.50 m/s) iˆ r v 2,rel = (− 2 m/s ) iˆ − (1.5 m/s ) iˆ =

(− 3.50 m/s) iˆ

= (d) Express the sum of the kinetic energies relative to the center of mass:

K rel = K1,rel + K 2,rel = 12 m1v12,rel + 12 m2v22,rel

Substitute numerical values and evaluate Krel:

K rel =

1 2

(3 kg )(3.5 m/s )2 2 + 12 (3 kg )(− 3.5 m/s )

= 36.75 J (e) Find Kcm:

2 = K cm = 12 mtot vcm

1 2

(6 kg )(1.5 m/s )2

= 6.75 J = 43.5 J − 36.75 J = K − K rel 58 • Picture the Problem Choose a coordinate system in which the positive x direction is to the right. Use the expression for the total momentum of a system to find the velocity of the center of mass and the definition of relative velocity to express the sum of the kinetic energies relative to the center of mass.

(a) Express the sum of the kinetic energies:

K = K1 + K 2 = 12 m1v12 + 12 m2 v22

Substitute numerical values and evaluate K:

K=

(b) Relate the velocity of the center of mass of the system to its total momentum:

r

Solve for v cm :

1 2

(3 kg )(5 m/s )2 + 12 (5 kg )(3 m/s )2

= 6 0 .0 J r r r Mv cm = m1v1 + m2 v 2

r r r m1v1 + m2 v 2 v cm = m1 + m2

544 Chapter 8 Substitute numerical values and r evaluate v cm :

r (3 kg )(5 m/s) iˆ + (5 kg )(3 m/s) iˆ vcm = 3 kg + 5 kg

(3.75 m/s) iˆ

= (c) The velocity of an object relative to the center of mass is given by: Substitute numerical values and evaluate the relative velocities:

r r r v rel = v − v cm

r v1,rel = (5 m/s ) iˆ − (3.75 m/s ) iˆ = (1.25 m/s ) iˆ and

r v 2,rel = (3 m/s ) iˆ − (3.75 m/s ) iˆ

(− 0.750 m/s) iˆ

= (d) Express the sum of the kinetic energies relative to the center of mass: Substitute numerical values and evaluate Krel:

K rel = K1,rel + K 2,rel

= 12 m1v12,rel + 12 m2 v22,rel K rel =

(3 kg )(1.25 m/s)2 2 + 12 (5 kg )(− 0.75 m/s )

1 2

= 3.75 J (e) Find Kcm:

2 K cm = 12 mtot vcm =

1 2

(8 kg )(3.75 m/s )2

= 56.3 J = K − K rel

Impulse and Average Force 59 • Picture the Problem The impulse imparted to the ball by the kicker equals the change in the ball’s momentum. The impulse is also the product of the average force exerted on the ball by the kicker and the time during which the average force acts. I = ∆p = p f − pi (a) Relate the impulse delivered to the ball to its change in momentum: = mvf since vi = 0

Substitute numerical values and evaluate I:

I = (0.43 kg )(25 m/s ) = 10.8 N ⋅ s

Systems of Particles and Conservation of Momentum 545 (b) Express the impulse delivered to the ball as a function of the average force acting on it and solve for and evaluate Fav :

I = Fav ∆t and

Fav =

I 10.8 N ⋅ s = = 1.34 kN ∆t 0.008 s

60 • Picture the Problem The impulse exerted by the ground on the brick equals the change in momentum of the brick and is also the product of the average force exerted by the ground on the brick and the time during which the average force acts.

(a) Express the impulse exerted by the ground on the brick: Because pf,brick = 0:

I = ∆pbrick = pf,brick − pi,brick I = pi,brick = mbrick v

(1)

∆K + ∆U = 0

Use conservation of energy to determine the speed of the brick at impact:

or Kf − Ki + U f − U i = 0

Because Uf = Ki = 0:

Kf −Ui = 0

or 1 2

mbrick v 2 − mbrick gh = 0

Solve for v:

v = 2 gh

Substitute in equation (1) to obtain:

I = mbrick 2 gh

Substitute numerical values and evaluate I:

(c) Express the impulse delivered to the brick as a function of the average force acting on it and solve for and evaluate Fav :

(

)

I = (0.3 kg ) 2 9.81 m/s 2 (8 m ) = 3.76 N ⋅ s I = Fav ∆t and

Fav =

I 3.76 N ⋅ s = = 2.89 kN ∆t 0.0013 s

*61 • Picture the Problem The impulse exerted by the ground on the meteorite equals the change in momentum of the meteorite and is also the product of the average force exerted by the ground on the meteorite and the time during which the average force acts.

546 Chapter 8 I = ∆pmeteorite = pf − pi

Express the impulse exerted by the ground on the meteorite: Relate the kinetic energy of the meteorite to its initial momentum and solve for its initial momentum:

pi2 Ki = ⇒ pi = 2mK i 2m

Express the ratio of the initial and final kinetic energies of the meteorite:

pi2 p2 Ki = 2m2 = i2 = 2 Kf pf pf 2m

Solve for pf:

pf =

pi 2

pi ⎛ 1 ⎞ − pi = pi ⎜ − 1⎟ 2 ⎝ 2 ⎠ ⎛ 1 ⎞ = 2mK i ⎜ − 1⎟ ⎝ 2 ⎠

Substitute in our expression for I and simplify:

I=

Because our interest is in its magnitude, evaluate I :

I =

(

)(

)

⎛ 1 ⎞ 2 30.8 × 103 kg 617 ×106 J ⎜ − 1⎟ = 1.81 MN ⋅ s ⎝ 2 ⎠

Express the impulse delivered to the meteorite as a function of the average force acting on it and solve for and evaluate Fav :

I = Fav ∆t and

Fav =

I 1.81MN ⋅ s = = 0.602 MN ∆t 3s

62 •• Picture the Problem The impulse exerted by the bat on the ball equals the change in momentum of the ball and is also the product of the average force exerted by the bat on the ball and the time during which the bat and ball were in contact.

(a) Express the impulse exerted by the bat on the ball in terms of the change in momentum of the ball:

r r r r I = ∆pball = pf − pi = mv iˆ − − mv iˆ = 2mv iˆ f

(

where v = vf = vi

i

)

Systems of Particles and Conservation of Momentum 547 Substitute for m and v and evaluate I: (b) Express the impulse delivered to the ball as a function of the average force acting on it and solve for and evaluate Fav :

I = 2(0.15 kg )(20 m/s ) = 6.00 N ⋅ s

I = Fav ∆t and

Fav =

I 6.00 N ⋅ s = = 4.62 kN ∆t 1.3 ms

*63 •• Picture the Problem The figure shows the handball just before and immediately after its collision with the wall. Choose a coordinate system in which the positive x direction is to the right. The wall changes the momentum of the ball by exerting a force on it during the ball’s collision with it. The reaction to this force is the force the ball exerts on the wall. Because these action and reaction forces are equal in magnitude, we can find the average force exerted on the ball by finding the change in momentum of the ball.

Using Newton’s 3rd law, relate the average force exerted by the ball on the wall to the average force exerted by the wall on the ball: Relate the average force exerted by the wall on the ball to its change in momentum:

r

Express ∆v x for the ball:

r r Fav on wall = − Fav on ball and

Fav on wall = Fav on ball

(1)

r r r ∆p m∆v Fav on ball = = ∆t ∆t r ∆v x = vf , x iˆ − vi , x iˆ or, because vi,x = vcosθ and vf,x = −vcosθ,

r ∆v x = −v cos θ iˆ − v cos θ iˆ = −2v cos θ iˆ

Substitute in our expression for r Fav on ball :

r r m∆v 2mv cos θ ˆ Fav on ball = i =− ∆t ∆t

548 Chapter 8 r

2mv cosθ ∆t 2(0.06 kg )(5 m/s )cos40° = 2 ms = 230 N

Evaluate the magnitude of Fav on ball :

Fav on ball =

Substitute in equation (1) to obtain:

Fav on wall = 230 N

64 •• Picture the Problem The pictorial representation shows the ball during the interval of time you are exerting a force on it to accelerate it upward. The average force you exert can be determined from the change in momentum of the ball. The change in the velocity of the ball can be found by applying conservation of mechanical energy to its rise in the air once it has left your hand.

(a) Relate the average force exerted by your hand on the ball to the change in momentum of the ball: Letting Ug = 0 at the initial elevation of your hand, use conservation of mechanical energy to relate the initial kinetic energy of the ball to its potential energy when it is at its highest point: Substitute for Kf and Ui and solve for v2:

Fav =

∆p p2 − p1 mv2 = = ∆t ∆t ∆t

because v1 and, hence, p1 = 0.

∆K + ∆U = 0 or − Ki + U f = 0 since K f = U i = 0

− 12 mv 22 + mgh = 0 and v 2 = 2 gh

Relate ∆t to the average speed of the ball while you are throwing it upward:

∆t =

d d 2d = = vav v 2 v2 2

Systems of Particles and Conservation of Momentum 549 Substitute for ∆t and v2 in the expression for Fav to obtain:

Fav =

Substitute numerical values and evaluate Fav:

Fav =

mgh d

(0.15 kg )(9.81m/s 2 )(40 m ) 0.7 m

= 84.1 N (b) Express the ratio of the weight of the ball to the average force acting on it:

w mg (0.15 kg ) (9.81 m/s 2 ) = = < 2% Fav Fav 84.1 N

Because the weight of the ball is less than 2% of the average force exerted on the ball, it is reasonable to have neglected its weight. 65 •• Picture the Problem Choose a coordinate system in which the direction the ball is moving after its collision with the wall is the positive x direction. The impulse delivered to the wall or received by the player equals the change in the momentum of the ball. We can find the average forces from the rate of change in the momentum of the ball.

(a) Relate the impulse delivered to the wall to the change in momentum of the handball:

r r r r I = ∆p = mvf − mvi = (0.06 kg )(8 m/s ) iˆ

[

− − (0.06 kg )(10 m/s ) iˆ

]

= (1.08 N ⋅ s ) iˆ directed into wall. (b) Find Fav from the change in the ball’s momentum:

∆p 1.08 N ⋅ s = 0.003 s ∆t

Fav =

= 360 N, into wall. (c) Find the impulse received by the player from the change in momentum of the ball:

I = ∆pball = m∆v

= (0.06 kg )(8 m/s ) = 0.480 N ⋅ s, away from wall.

(d) Relate Fav to the change in the ball’s momentum:

Fav =

Express the stopping time in terms of the average speed vav of the ball

∆t =

∆p ball ∆t

d vav

550 Chapter 8 and its stopping distance d: Substitute to obtain:

Fav =

Substitute numerical values and evaluate Fav:

Fav =

vav ∆pball d

(4 m/s)(0.480 N ⋅ s ) 0.5 m

= 3.84 N, away from wall. 66 ••• Picture the Problem The average force exerted on the limestone by the droplets of water equals the rate at which momentum is being delivered to the floor. We’re given the number of droplets that arrive per minute and can use conservation of mechanical energy to determine their velocity as they reach the floor.

(a) Letting N represent the rate at which droplets fall, relate Fav to the change in the droplet’s momentum: Find the mass of the droplets:

Fav =

∆pdroplets ∆t

=N

m∆v ∆t

m = ρV = (1 kg/L )(0.03 mL ) = 3 × 10 −5 kg

Letting Ug = 0 at the point of impact of the droplets, use conservation of mechanical energy to relate their speed at impact to their fall distance:

∆K + ∆U = 0 or Kf − Ki + U f − U i = 0

mvf2 − mgh = 0

Because Ki = Uf = 0:

1 2

Solve for and evaluate v = vf:

v = 2 gh = 2(9.81 m/s 2 )(5 m ) = 9.90 m/s

Substitute numerical values and evaluate Fav:

⎛N⎞ Fav = ⎜ ⎟m∆v ⎝ ∆t ⎠ ⎛ droplets 1 min ⎞ ⎟ = ⎜⎜10 × min 60 s ⎟⎠ ⎝ × 3 × 10 −5 kg (9.90 m/s )

(

)

= 4.95 × 10 −5 N

Systems of Particles and Conservation of Momentum 551 (b) Calculate the ratio of the weight of a droplet to Fav:

w mg = Fav Fav =

(3 × 10

−5

)(

)

kg 9.81 m/s 2 ≈ 6 4.95 × 10 −5 N

Collisions in One Dimension *67 • Picture the Problem We can apply conservation of momentum to this perfectly inelastic collision to find the after-collision speed of the two cars. The ratio of the transformed kinetic energy to kinetic energy before the collision is the fraction of kinetic energy lost in the collision.

(a) Letting V be the velocity of the two cars after their collision, apply conservation of momentum to their perfectly inelastic collision: Solve for and evaluate V:

pinitial = pfinal or

mv1 + mv2 = (m + m )V

V=

v1 + v2 30 m/s + 10 m/s = 2 2

= 20.0 m/s (b) Express the ratio of the kinetic energy that is lost to the kinetic energy of the two cars before the collision and simplify:

∆K K − K initial = final K initial K initial = = =

Substitute numerical values to obtain:

K final −1 K initial 1 2 1 2

(2m )V 2

mv12 + 12 mv22

−1

2V 2 −1 v12 + v22

2(20 m/s ) ∆K = −1 K initial (30 m/s )2 + (10 m/s )2 = −0.200 2

20% of the initial kinetic energy is transformed into heat, sound, and the deformation of metal.

552 Chapter 8 68 • Picture the Problem We can apply conservation of momentum to this perfectly inelastic collision to find the after-collision speed of the two players.

Letting the subscript 1 refer to the running back and the subscript 2 refer to the linebacker, apply conservation of momentum to their perfectly inelastic collision:

pi = pf or

m1v1 = (m1 + m2 )V

Solve for V:

V=

m1 v1 m1 + m2

Substitute numerical values and evaluate V:

V=

85 kg (7 m/s) = 3.13 m/s 85 kg + 105 kg

69 • Picture the Problem We can apply conservation of momentum to this collision to find the after-collision speed of the 5-kg object. Let the direction the 5-kg object is moving before the collision be the positive direction. We can decide whether the collision was elastic by examining the initial and final kinetic energies of the system.

pi = pf

(a) Letting the subscript 5 refer to the 5-kg object and the subscript 2 refer to the 10-kg object, apply conservation of momentum to obtain:

or m5vi,5 − m10 vi,10 = m5vf,5

Solve for vf,5:

vf,5 =

Substitute numerical values and evaluate vf,5:

vf,5 =

m5vi,5 − m10 vi,10 m5

(5 kg )(4 m/s) − (10 kg )(3 m/s) 5 kg

= − 2.00 m/s where the minus sign means that the 5-kg object is moving to the left after the collision.

Systems of Particles and Conservation of Momentum 553 (b) Evaluate ∆K for the collision:

∆K = K f − K i =

1 2

(5 kg )(2 m/s )2 − [12 (5 kg )(4 m/s )2 + 12 (10 kg )(3 m/s )2 ] = −75.0 J

Because ∆K ≠ 0, the collision was inelastic. 70 • Picture the Problem The pictorial representation shows the ball and bat just before and just after their collision. Take the direction the bat is moving to be the positive direction. Because the collision is elastic, we can equate the speeds of recession and approach, with the approximation that vi,bat ≈ vf,bat to find vf,ball.

Express the speed of approach of the bat and ball: Because the mass of the bat is much greater than that of the ball:

vf, bat − vf, ball = −(vi, bat − vi, ball )

vi,bat ≈ vf,bat

Substitute to obtain:

vf, bat − vf, ball = −(vf, bat − vi, ball )

Solve for and evaluate vf,ball:

vf,ball = vf,bat + (vf,bat − vi,ball ) = −vi,ball + 2vf,bat = v + 2v = 3v

*71 •• Picture the Problem Let the direction the proton is moving before the collision be the positive x direction. We can use both conservation of momentum and conservation of mechanical energy to obtain an expression for velocity of the proton after the collision. r r r (a) Use the expression for the total P = mi vi = Mvcm i momentum of a system to find vcm:

∑

and r vcm = =

r mv p,i m + 12m

= 131 (300 m/s ) iˆ

(23.1m/s) iˆ

554 Chapter 8 (b) Use conservation of momentum to obtain one relation for the final velocities: Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach: To eliminate vnuc,f, solve equation (2) for vnuc,f, and substitute the result in equation (1): Solve for and evaluate vp,f:

mp v p,i = mp v p,f + mnuc v nuc,f

(1)

v nuc,f − v p,f = −(v nuc,i − v p,i ) = v p,i (2)

v nuc,f = v p,i + v p,f

mp v p,i = mp v p,f + mnuc (v p,i + v p,f )

vp,f = =

mp − mnuc mp + mnuc

vp,i

m − 12m (300 m/s) = − 254 m/s 13m

72 •• Picture the Problem We can use conservation of momentum and the definition of an elastic collision to obtain two equations in v2f and v3f that we can solve simultaneously.

Use conservation of momentum to obtain one relation for the final velocities: Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach: Solve equation (2) for v3f , substitute in equation (1) to eliminate v3f, and solve for and evaluate v2f:

Use equation (2) to find v3f:

m3 v3i = m3 v3f + m2 v 2f

(1)

v 2f − v3f = −(v 2i − v3i ) = v3i

v2f =

(2)

2m3v3i 2(3 kg )(4 m/s ) = m2 + m3 2 kg + 3 kg

= 4.80 m/s v3f = v2f − v3i = 4.80 m/s − 4.00 m/s = 0.800 m/s

Evaluate Ki and Kf:

K i = K 3i = 12 m3v3i2 = = 24.0 J

1 2

(3 kg )(4 m/s )2

Systems of Particles and Conservation of Momentum 555 and

K f = K 3f + K 2f = 12 m3v3f2 + 12 m2v2f2

(3 kg )(0.8 m/s)2 2 + 12 (2 kg )(4.8 m/s )

=

1 2

= 24.0 J Because K i = K f , we can conclude that the values obtained for v2f and v3f are consistent with the collision having been elastic. 73 •• Picture the Problem We can find the velocity of the center of mass from the definition of the total momentum of the system. We’ll use conservation of energy to find the maximum compression of the spring and express the initial (i.e., before collision) and final (i.e., at separation) velocities. Finally, we’ll transform the velocities from the center of mass frame of reference to the table frame of reference.

r r r P = ∑ mi v i = Mv cm

(a) Use the definition of the total momentum of a system to relate the initial momenta to the velocity of the center of mass:

or

Solve for vcm:

vcm =

Substitute numerical values and evaluate vcm:

vcm =

i

m1v1i = (m1 + m2 ) vcm

m1v1i + m2 v2i m1 + m2

(2 kg )(10 m/s) + (5 kg )(3 m/s) 2 kg + 5 kg

= 5.00 m/s (b) Find the kinetic energy of the system at maximum compression (u1 = u2 = 0):

2 K = K cm = 12 Mvcm

=

1 2

(7 kg )(5 m/s)2 = 87.5 J

∆K + ∆U s = 0

Use conservation of energy to relate the kinetic energy of the system to the potential energy stored in the spring at maximum compression:

or

Because Kf = Kcm and Usi = 0:

K cm − K i + 12 k (∆x ) = 0

K f − K i + U sf − U si = 0

2

556 Chapter 8 Solve for ∆x:

∆x =

2(K i − K cm ) k

[

=

2 12 m1v12i + 12 m2v22i − K cm k

=

m1v12i + m2v22i − 2 K cm k

]

Substitute numerical values and evaluate ∆x:

∆x =

(2 kg )(10 m/s)2 + (5 kg )(3 m/s)2 − 2(87.5 J ) ⎤ = 1120 N/m ⎥⎦

1120 N/m

(c) Find u1i, u2i, and u1f for this elastic collision:

0.250 m

u1i = v1i − vcm = 10 m/s − 5 m/s = 5 m/s, u 2i = v2i − vcm = 3 m/s − 5 m/s = −2 m/s, and u1f = v1f − vcm = 0 − 5 m/s = −5 m/s

Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach and solve for u2f:

Transform u1f and u2f to the table frame of reference:

u2f − u1f = −(u2i − u1i ) and

u2f = −u2i + u1i + u1f

= −(− 2 m/s ) + 5 m/s − 5 m/s = 2 m/s

v1f = u1f + vcm = −5 m/s + 5 m/s = 0 and

v 2f = u 2f + vcm = 2 m/s + 5 m/s = 7.00 m/s *74 •• Picture the Problem Let the system include the earth, the bullet, and the sheet of plywood. Then Wext = 0. Choose the zero of gravitational potential energy to be where the bullet enters the plywood. We can apply both conservation of energy and conservation of momentum to obtain the various physical quantities called for in this problem.

(a) Use conservation of mechanical energy after the bullet exits the sheet of plywood to relate its exit speed to the height to which it rises:

∆K + ∆U = 0 or, because Kf = Ui = 0,

− 12 mvm2 + mgh = 0

Systems of Particles and Conservation of Momentum 557 Solve for vm:

vm =

2 gh

Proceed similarly to relate the initial velocity of the plywood to the height to which it rises:

vM =

2 gH

r r pi = pf

(b) Apply conservation of momentum to the collision of the bullet and the sheet of plywood:

or

Substitute for vm and vM and solve for vmi:

vmi =

(c) Express the initial mechanical energy of the system (i.e., just before the collision):

Ei = 12 mvm2 i

Express the final mechanical energy of the system (i.e., when the bullet and block have reached their maximum heights): (d) Use the work-energy theorem with Wext = 0 to find the energy dissipated by friction in the inelastic collision:

mvmi = mvm + MvM 2 gh +

M m

2 gH

⎡ 2M = mg ⎢h + m ⎣⎢

2 ⎛M ⎞ ⎤ hH + ⎜ ⎟ H ⎥ ⎝ m ⎠ ⎥⎦

Ef = mgh + MgH = g (mh + MH )

Ef − Ei + Wfriction = 0 and

Wfriction = Ei − Ef ⎡ h M ⎤ = gMH ⎢2 + − 1⎥ ⎣ H m ⎦

75 •• Picture the Problem We can find the velocity of the center of mass from the definition of the total momentum of the system. We’ll use conservation of energy to find the speeds of the particles when their separation is least and when they are far apart.

(a) Noting that when the distance between the two particles is least, both move at the same speed, namely vcm, use the definition of the total momentum of a system to relate the initial momenta to the velocity of

r r r P = ∑ mi v i = Mv cm i

or

mp vpi = (mp + mα )vcm .

558 Chapter 8 the center of mass: Solve for and evaluate vcm:

vcm = v' =

mp vpi + mα vαi m1 + m2

=

mv0 + 0 m + 4m

= 0.200 v0 (b) Use conservation of momentum to obtain one relation for the final velocities: Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach: Solve equation (2) for vpf , substitute in equation (1) to eliminate vpf, and solve for vαf:

mp v0 = mpvpf + mα vαf

(1)

vpf − vαf = −(vpi − vαi ) = −vpi (2)

vαf =

2 m p v0 mp + mα

=

2mv0 = 0.400 v0 m + 4m

76 • Picture the Problem Let the numeral 1 denote the electron and the numeral 2 the hydrogen atom. We can find the final velocity of the electron and, hence, the fraction of its initial kinetic energy that is transferred to the atom, by transforming to the center-ofmass reference frame, calculating the post-collision velocity of the electron, and then transforming back to the laboratory frame of reference.

Express f, the fraction of the electron’s initial kinetic energy that is transferred to the atom:

f =

Ki − Kf K =1− f Ki Ki

⎛v m1v12f =1− = 1 − ⎜⎜ 1f 2 m1v1i ⎝ v1i 1 2 1 2

Find the velocity of the center of mass:

vcm =

⎞ ⎟⎟ ⎠

2

m1v1i m1 + m2

or, because m2 = 1840m1,

vcm = Find the initial velocity of the electron in the center-of-mass reference frame:

m1v1i 1 = v1i m1 + 1840m1 1841

u1i = v1i − vcm = v1i −

1 ⎞ ⎛ = ⎜1 − ⎟v1i ⎝ 1841 ⎠

1 v1i 1841

(1)

Systems of Particles and Conservation of Momentum 559 Find the post-collision velocity of the electron in the center-of-mass reference frame by reversing its velocity:

⎛ 1 ⎞ − 1⎟v1i u1f = −u1i = ⎜ ⎝ 1841 ⎠

To find the final velocity of the electron in the original frame, add vcm to its final velocity in the centerof-mass reference frame:

⎛ 2 ⎞ − 1⎟v1i v1f = u1f + vcm = ⎜ ⎝ 1841 ⎠

Substitute in equation (1) to obtain:

⎛⎛ 2 ⎞ ⎞ − 1⎟v1i ⎟ ⎜⎜ 2 1841 ⎠ ⎟ ⎛ 2 ⎞ = 1− ⎜ − 1⎟ f = 1− ⎜ ⎝ ⎟ ⎜ v1i ⎝ 1841 ⎠ ⎟ ⎜ ⎠ ⎝

2

= 2.17 × 10 −3 = 0.217%

77 •• Picture the Problem The pictorial representation shows the bullet about to imbed itself in the bob of the ballistic pendulum and then, later, when the bob plus bullet have risen to their maximum height. We can use conservation of momentum during the collision to relate the speed of the bullet to the initial speed of the bob plus bullet (V). The initial kinetic energy of the bob plus bullet is transformed into gravitational potential energy when they reach their maximum height. Hence we apply conservation of mechanical energy to relate V to the angle through which the bullet plus bob swings and then solve the momentum and energy equations simultaneously for the speed of the bullet.

Use conservation of momentum to relate the speed of the bullet just before impact to the initial speed of the bob plus bullet:

mvb = (m + M )V

Solve for the speed of the bullet:

⎛ M⎞ vb = ⎜ 1 + ⎟ V m⎠ ⎝

Use conservation of energy to relate

∆K + ∆U = 0

(1)

560 Chapter 8 or, because Kf = Ui = 0,

the initial kinetic energy of the bullet to the final potential energy of the system:

− Ki + U f = 0

− 12 (m + M )V 2

Substitute for Ki and Uf and solve for V:

+ (m + M )gL(1 − cos θ ) = 0

and

V = 2 gL(1 − cos θ )

M⎞ ⎛ vb = ⎜1 + ⎟ 2 gL(1 − cos θ ) m⎠ ⎝

Substitute for V in equation (1) to obtain: Substitute numerical values and evaluate vb:

⎛ 1.5 kg ⎞ ⎟⎟ 2 9.81 m/s 2 (2.3 m )(1 − cos60°) = 450 m/s vb = ⎜⎜1 + ⎝ 0.016 kg ⎠

(

)

*78 •• Picture the Problem We can apply conservation of momentum and the definition of an elastic collision to obtain equations relating the initial and final velocities of the colliding objects that we can solve for v1f and v2f.

Apply conservation of momentum to the elastic collision of the particles to obtain:

m1v1f + m2 v2f = m1v1i + m2 v2i

Relate the initial and final kinetic energies of the particles in an elastic collision:

1 2

Rearrange this equation and factor to obtain:

m2 v22f − v22i = m1 v12i − v12f

m1v12f + 12 m2 v22f = 12 m1v12i + 12 m2 v22i

(

)

(

)

or

m2 (v2 f − v2i )(v2f + v2i )

= m1 (v1i − v1f )(v1i + v1f )

Rearrange equation (1) to obtain:

m2 (v2f − v2i ) = m1 (v1i − v1f )

Divide equation (2) by equation (3) to obtain:

v2 f + v2i = v1i + v1f

Rearrange this equation to obtain equation (4):

v1f − v2f = v2i − v1i

Multiply equation (4) by m2 and add it to equation (1) to obtain:

(1)

(2)

(3)

(4)

(m1 + m2 )v1f = (m1 − m2 )v1i + 2m2v2i

Systems of Particles and Conservation of Momentum 561 Solve for v1f to obtain:

Multiply equation (4) by m1 and subtract it from equation (1) to obtain: Solve for v2f to obtain:

v1 f =

2m2 m1 − m2 v1i + v2 i m1 + m2 m1 + m2

(m1 + m2 )v2f = (m2 − m1 )v2i + 2m1v1i

v2 f =

2m1 m − m1 v1i + 2 v 2i m1 + m2 m1 + m2

Remarks: Note that the velocities satisfy the condition that v 2f − v1f = −(v 2i − v1i ) . This verifies that the speed of recession equals the speed of approach. 79 •• Picture the Problem As in this problem, Problem 78 involves an elastic, onedimensional collision between two objects. Both solutions involve using the conservation of momentum equation m1v1f + m2 v2 f = m1v1i + m2 v2i and the elastic collision

equation v1f − v2 f = v2i − v1i . In part (a) we can simply set the masses equal to each other and substitute in the equations in Problem 78 to show that the particles "swap" velocities. In part (b) we can divide the numerator and denominator of the equations in Problem 78 by m2 and use the condition that m2 >> m1 to show that v1f ≈ −v1i+2v2i and v2f ≈ v2i. (a) From Problem 78 we have:

m1 − m2 2m2 v1i + v2i m1 + m2 m1 + m2

(1)

v2f =

m − m1 2m1 v1i + 2 v 2i m1 + m2 m1 + m2

(2)

v1f =

2m v2 i = v2 i m+m

v1f = and

Set m1 = m2 = m to obtain:

and

v2 f =

2m v1i = v1i m+m

(b) Divide the numerator and denominator of both terms in equation (1) by m2 to obtain:

m1 −1 2 m2 v1f = v1i + v2i m1 m1 +1 +1 m2 m2

If m2 >> m1:

v1f ≈ −v1i +2v2i

562 Chapter 8 Divide the numerator and denominator of both terms in equation (2) by m2 to obtain:

If m2 >> m1:

2 v2f =

m1 m2

m1 +1 m2

1− v1i +

m1 m2

m1 +1 m2

v 2i

v2f ≈ v 2i

Remarks: Note that, in both parts of this problem, the velocities satisfy the condition that v 2f − v1f = −(v 2i − v1i ) . This verifies that the speed of recession equals the speed of approach.

Perfectly Inelastic Collisions and the Ballistic Pendulum 80 •• Picture the Problem Choose Ug = 0 at the bob’s equilibrium position. Momentum is conserved in the collision of the bullet with bob and the initial kinetic energy of the bob plus bullet is transformed into gravitational potential energy as it swings up to the top of the circle. If the bullet plus bob just makes it to the top of the circle with zero speed, it will swing through a complete circle.

Use conservation of momentum to relate the speed of the bullet just before impact to the initial speed of the bob plus bullet:

m1v = (m1 + m2 )V

Solve for the speed of the bullet:

⎛ m ⎞ v = ⎜⎜1 + 2 ⎟⎟V ⎝ m1 ⎠

Use conservation of energy to relate the initial kinetic energy of the bob plus bullet to their potential energy at the top of the circle:

∆K + ∆U = 0

(1)

or, because Kf = Ui = 0,

− Ki + U f = 0

Substitute for Ki and Uf:

− 12 (m1 + m2 )V 2 + (m1 + m2 )g (2 L ) = 0

Solve for V:

V = gL

Substitute for V in equation (1) and simplify to obtain:

⎛ m v = ⎜⎜1 + 2 m1 ⎝

⎞ ⎟⎟ gL ⎠

Systems of Particles and Conservation of Momentum 563 *81 •• Picture the Problem Choose Ug = 0 at the equilibrium position of the ballistic pendulum. Momentum is conserved in the collision of the bullet with the bob and kinetic energy is transformed into gravitational potential energy as the bob swings up to its maximum height.

Letting V represent the initial speed of the bob as it begins its upward swing, use conservation of momentum to relate this speed to the speeds of the bullet just before and after its collision with the bob:

m1v = m1 ( 12 v ) + m2V

m1 v 2m 2

Solve for the speed of the bob:

V =

Use conservation of energy to relate the initial kinetic energy of the bob to its potential energy at its maximum height:

∆K + ∆U = 0

(1)

or, because Kf = Ui = 0,

− Ki + U f = 0

Substitute for Ki and Uf:

− 12 m2V 2 + m2 gh = 0

Solve for h:

h=

Substitute V from equation (1) in equation (2) and simplify to obtain:

⎛ m1 ⎞ ⎜⎜ v⎟ 2 2m2 ⎟⎠ v 2 ⎛ m1 ⎞ ⎝ ⎜ ⎟ h= = 2g 8 g ⎜⎝ m2 ⎟⎠

V2 2g

(2)

2

82 • Picture the Problem Let the mass of the bullet be m, that of the wooden block M, the pre-collision velocity of the bullet v, and the post-collision velocity of the block+bullet be V. We can use conservation of momentum to find the velocity of the block with the bullet imbedded in it just after their perfectly inelastic collision. We can use Newton’s 2nd law to find the acceleration of the sliding block and a constant-acceleration equation to find the distance the block slides.

564 Chapter 8

0 = V 2 + 2a∆x

Using a constant-acceleration equation, relate the velocity of the block+bullet just after their collision to their acceleration and displacement before stopping:

because the final velocity of the block+bullet is zero.

Solve for the distance the block slides before coming to rest:

∆x = −

Use conservation of momentum to relate the pre-collision velocity of the bullet to the post-collision velocity of the block+bullet:

V2 2a

mv = (m + M )V

Solve for V:

V =

Substitute in equation (1) to obtain:

1 ⎛ m ⎞ ∆x = − ⎜ v⎟ 2a ⎝ m + M ⎠

Apply

r r F ∑ = ma to the

block+bullet (see the FBD in the diagram):

Use the definition of the coefficient of kinetic friction and equation (4) to obtain:

(1)

m v m+M

∑F

x

and

∑F

y

2

(2)

= − f k = (m + M ) a

(3)

=Fn − (m + M )g = 0

(4)

f k = µ k Fn = µ k (m + M )g

Substitute in equation (3):

− µ k (m + M )g = (m + M ) a

Solve for a to obtain:

a = −µk g

Substitute in equation (2) to obtain:

∆x =

1 ⎛ m ⎞ v⎟ ⎜ 2µ k g ⎝ m + M ⎠

2

Systems of Particles and Conservation of Momentum 565 Substitute numerical values and evaluate ∆x:

1 ∆x = 2(0.22 ) 9.81 m/s 2

(

2

)

⎛ ⎞ 0.0105 kg ⎜⎜ (750 m/s)⎟⎟ = 0.130 m ⎝ 0.0105 kg + 10.5 kg ⎠

83 •• Picture the Problem The collision of the ball with the box is perfectly inelastic and we can find the speed of the box-and-ball immediately after their collision by applying conservation of momentum. If we assume that the kinetic friction force is constant, we can use a constant-acceleration equation to find the acceleration of the box and ball combination and the definition of µk to find its value.

Using its definition, express the coefficient of kinetic friction of the table: Use conservation of momentum to relate the speed of the ball just before the collision to the speed of the ball+box immediately after the collision:

µk =

f k (M + m ) a a = = Fn (M + m )g g

MV = (m + M ) v

Solve for v:

v=

Use a constant-acceleration equation to relate the sliding distance of the ball+box to its initial and final velocities and its acceleration:

vf2 = vi2 + 2a∆x

Solve for a:

Substitute in equation (1) to obtain:

Use equation (2) to eliminate v:

(1)

MV m+M

(2)

or, because vf = 0 and vi = v,

0 = v 2 + 2a∆x a=−

v2 2∆x

µk =

v2 2 g∆x

µk =

1 ⎛ MV ⎞ ⎜ ⎟ 2 g∆x ⎝ m + M ⎠

⎛ ⎞ 1 ⎜ V ⎟ ⎜ ⎟ = 2 g∆x ⎜ m + 1 ⎟ ⎜ ⎟ ⎝M ⎠

2

2

566 Chapter 8 Substitute numerical values and evaluate µk: 2

⎛ ⎞ ⎜ ⎟ 1 1.3 m/s ⎟ ⎜ µk = = 0.0529 2 9.81 m/s 2 (0.52 m ) ⎜ 0.327 kg + 1 ⎟ ⎜ 0.425 kg ⎟ ⎝ ⎠

(

)

*84 •• Picture the Problem Jane’s collision with Tarzan is a perfectly inelastic collision. We can find her speed v1 just before she grabs Tarzan from conservation of energy and their speed V just after she grabs him from conservation of momentum. Their kinetic energy just after their collision will be transformed into gravitational potential energy when they have reached their greatest height h.

Use conservation of energy to relate the potential energy of Jane and Tarzan at their highest point (2) to their kinetic energy immediately after Jane grabbed Tarzan: Solve for h to obtain:

U 2 = K1 or

mJ+T gh = 12 mJ+TV 2

V2 h= 2g

Use conservation of momentum to relate Jane’s velocity just before she collides with Tarzan to their velocity just after their perfectly inelastic collision:

mJ v1 = mJ+TV

Solve for V:

V =

Apply conservation of energy to relate Jane’s kinetic energy at 1 to her potential energy at 0:

K1 = U 0

mJ v1 mJ+T

or 1 2

mJ v12 = mJ gL

(1)

(2)

Systems of Particles and Conservation of Momentum 567 Solve for v1:

v1 = 2 gL

Substitute in equation (2) to obtain:

V =

Substitute in equation (1) and simplify:

⎛ m ⎞ 1 ⎛ mJ ⎞ ⎜⎜ ⎟⎟ 2 gL = ⎜⎜ J ⎟⎟ L h= 2 g ⎝ mJ+T ⎠ ⎝ mJ+T ⎠

Substitute numerical values and evaluate h:

⎛ ⎞ 54 kg ⎟⎟ (25 m ) = 3.94 m h = ⎜⎜ ⎝ 54 kg + 82 kg ⎠

mJ 2 gL mJ+T 2

2

2

Exploding Objects and Radioactive Decay 85 •• Picture the Problem This nuclear reaction is 4Be → 2α + 1.5×10−14 J. In order to conserve momentum, the alpha particles will have move in opposite directions with the same velocities. We’ll use conservation of energy to find their speeds.

(

)

Letting E represent the energy released in the reaction, express conservation of energy for this process:

2 K α = 2 12 mα vα2 = E

Solve for vα:

vα =

E mα

Substitute numerical values and evaluate vα:

vα =

1.5 × 10 −14 J = 1.50 × 10 6 m/s − 27 6.68 × 10 kg

86 •• Picture the Problem This nuclear reaction is 5Li → α + p + 3.15 × 10−13 J. To conserve momentum, the alpha particle and proton must move in opposite directions. We’ll apply both conservation of energy and conservation of momentum to find the speeds of the proton and alpha particle.

Use conservation of momentum in this process to express the alpha particle’s velocity in terms of the proton’s:

pi = p f = 0 and

0 = mp vp − mα vα

568 Chapter 8 Solve for vα and substitute for mα to obtain: Letting E represent the energy released in the reaction, apply conservation of energy to the process:

vα =

mp mα

vp =

mp 4mp

vp = 14 vp

K p + Kα = E or 1 2

mp vp2 + 12 mα vα2 = E

Substitute for vα:

1 2

mp vp2 + 12 mα (14 vp ) = E

Solve for vp and substitute for mα to obtain:

vp =

32 E 32 E = 16mp + mα 16mp + 4mp

Substitute numerical values and evaluate vp:

vp =

32 3.15 × 10 −13 J 20 1.67 × 10 − 27 kg

2

( (

)

)

= 1.74 × 10 7 m/s Use the relationship between vp and vα to obtain vα:

vα = 14 vp =

1 4

(1.74 ×10

7

m/s

)

= 4.34 × 106 m/s

87 ••• Picture the Problem The pictorial representation shows the projectile at its maximum elevation and is moving horizontally. It also shows the two fragments resulting from the explosion. We chose the system to include the projectile and the earth so that no external forces act to change the momentum of the system during the explosion. With this choice of system we can also use conservation of energy to determine the elevation of the projectile when it explodes. We’ll also find it useful to use constant-acceleration equations in our description of the motion of the projectile and its fragments.

Systems of Particles and Conservation of Momentum 569 (a) Use conservation of momentum to relate the velocity of the projectile before its explosion to the velocities of its two parts after the explosion: The only way this equality can hold is if:

Express v3 in terms of v0 and substitute for the masses to obtain:

r r pi = pf r r r m3v 3 = m1v1 + m2 v 2 m3v3 iˆ = m1v x1iˆ + m1v y1 ˆj − m2 v y 2 ˆj

m3v3 = m1v x1 and m1v y1 = m2 v y 2 vx1 = 3v3 = 3v0 cosθ

= 3(120 m/s )cos30° = 312 m/s

and

v y1 = 2v y 2

Using a constant-acceleration equation with the downward direction positive, relate vy2 to the time it takes the 2-kg fragment to hit the ground: With Ug = 0 at the launch site, apply conservation of energy to the climb of the projectile to its maximum elevation: Solve for ∆y:

Substitute numerical values and evaluate ∆y: Substitute in equation (2) and evaluate vy2:

Substitute in equation (1) and evaluate vy1:

(1)

∆y = v y 2 ∆t + 12 g (∆t )

2

∆y − 12 g (∆t ) ∆t

2

vy2 =

(2)

∆K + ∆U = 0 Because Kf = Ui = 0, − K i + U f = 0 or

− 12 m3v y20 + m3 g∆y = 0

∆y =

∆y =

vy2

v y20 2g

=

(v0 sin 30°)2 2g

[(120 m/s)sin30°] 2

(

2 9.81 m/s 2

(

)

= 183.5 m

)

183.5 m − 12 9.81 m/s 2 (3.6 s ) = 3.6 s = 33.3 m/s

v y1 = 2(33.3 m/s ) = 66.6 m/s

2

570 Chapter 8 r

Express v1 in vector form:

r v1 = vx1iˆ + v y1 ˆj

= (b) Express the total distance d traveled by the 1-kg fragment: Relate ∆x to v0 and the time-toexplosion:

(312 m/s) iˆ + (66.6 m/s) ˆj

d = ∆x + ∆x'

(3)

∆x = (v0 cos θ )(∆t exp )

(4)

vy0

v0 sin θ g

Using a constant-acceleration equation, express ∆texp:

∆texp =

Substitute numerical values and evaluate ∆texp:

∆texp =

Substitute in equation (4) and evaluate ∆x:

∆x = (120 m/s )(cos30°)(6.12 s ) = 636.5 m

Relate the distance traveled by the 1-kg fragment after the explosion to the time it takes it to reach the ground: Using a constant-acceleration equation, relate the time ∆t′ for the 1-kg fragment to reach the ground to its initial speed in the y direction and the distance to the ground: Substitute to obtain the quadratic equation: Solve the quadratic equation to find ∆t′: Substitute in equation (3) and evaluate d:

=

g

(120 m/s)sin30° = 6.12 s 9.81 m/s 2

∆x' = vx1∆t'

∆y = v y1∆t' − 12 g (∆t' )

2

(∆t' )2 − (13.6 s )∆t' − 37.4 s 2 = 0 ∆t′ = 15.9 s

d = ∆x + ∆x' = ∆x + v x1∆t'

= 636.5 m + (312 m/s )(15.9 s ) = 5.61 km

Systems of Particles and Conservation of Momentum 571 (c) Express the energy released in the explosion: Find the kinetic energy of the fragments after the explosion:

Eexp = ∆K = K f − K i

(5)

K f = K1 + K 2 = 12 m1v12 + 12 m2v22

(1kg )([ 312 m/s)2 + (66.6 m/s)2 ] 2 + 12 (2 kg )(33.3 m/s )

=

1 2

= 52.0 kJ Find the kinetic energy of the projectile before the explosion:

K i = 12 m3v32 = 12 m3 (v0 cos θ )

2

=

1 2

(3 kg )[(120 m/s ) cos 30°] 2

= 16.2 kJ Substitute in equation (5) to determine the energy released in the explosion:

Eexp = K f − K i = 52.0 kJ − 16.2 kJ

= 35.8 kJ

*88 ••• Picture the Problem This nuclear reaction is 9B → 2α + p + 4.4×10−14 J. Assume that the proton moves in the –x direction as shown in the figure. The sum of the kinetic energies of the decay products equals the energy released in the decay. We’ll use conservation of momentum to find the angle between the velocities of the proton and the alpha particles. Note that vα = vα ' .

Express the energy released to the kinetic energies of the decay products:

Solve for vα:

K p + 2 Kα = Erel or 1 2

(

)

mp vp2 + 2 12 mα vα2 = Erel

vα =

Erel − 12 mp vp2 mα

572 Chapter 8 Substitute numerical values and evaluate vα: − 27 1 4.4 × 10 −14 J kg )(6 × 106 m/s ) 2 (1.67 × 10 = 1.44 × 106 m/s − 6.68 × 10 −27 kg 6.68 × 10 −27 kg 2

vα =

Given that the boron isotope was at rest prior to the decay, use conservation of momentum to relate the momenta of the decay products: Solve for θ :

r r pf = pi = 0 ⇒ p xf = 0 ∴ 2(mα vα cos θ ) − mp vp = 0 or

2(4mp vα cos θ ) − mp vp = 0 ⎡ vp ⎤ ⎥ ⎣ 8vα ⎦

θ = cos −1 ⎢

⎡ 6 × 106 m/s ⎤ = cos −1 ⎢ ⎥ = ±58.7° 6 ⎣ 8 1.44 ×10 m/s ⎦

(

Let θ′ equal the angle the velocities of the alpha particles make with that of the proton:

)

θ' = ±(180° − 58.7°) = ± 121°

Coefficient of Restitution 89 • Picture the Problem The coefficient of restitution is defined as the ratio of the velocity of recession to the velocity of approach. These velocities can be determined from the heights from which the ball was dropped and the height to which it rebounded by using conservation of mechanical energy.

vrec vapp

Use its definition to relate the coefficient of restitution to the velocities of approach and recession:

e=

Letting Ug = 0 at the surface of the steel plate, apply conservation of energy to express the velocity of approach:

∆K + ∆U = 0 Because Ki = Uf = 0,

Kf − Ui = 0

or 1 2

Solve for vapp:

2 mvapp − mghapp = 0

vapp = 2 ghapp

Systems of Particles and Conservation of Momentum 573 In like manner, show that:

vrec = 2 ghrec

Substitute in the equation for e to obtain:

e=

Substitute numerical values and evaluate e:

e=

2 ghrec 2 ghapp

=

hrec happ

2.5 m = 0.913 3m

*90 • Picture the Problem The coefficient of restitution is defined as the ratio of the velocity of recession to the velocity of approach. These velocities can be determined from the heights from which an object was dropped and the height to which it rebounded by using conservation of mechanical energy.

vrec vapp

Use its definition to relate the coefficient of restitution to the velocities of approach and recession:

e=

Letting Ug = 0 at the surface of the steel plate, apply conservation of energy to express the velocity of approach:

∆K + ∆U = 0 Because Ki = Uf = 0,

Kf − Ui = 0

or 1 2

2 mvapp − mghapp = 0

Solve for vapp:

vapp = 2 ghapp

In like manner, show that:

vrec = 2 ghrec

Substitute in the equation for e to obtain:

e=

Find emin:

emin =

173 cm = 0.825 254 cm

Find emax:

emax =

183 cm = 0.849 254 cm

2 ghrec 2 ghapp

=

hrec happ

574 Chapter 8 and 0.825 ≤ e ≤ 0.849 91 • Picture the Problem Because the rebound kinetic energy is proportional to the rebound height, the percentage of mechanical energy lost in one bounce can be inferred from knowledge of the rebound height. The coefficient of restitution is defined as the ratio of the velocity of recession to the velocity of approach. These velocities can be determined from the heights from which an object was dropped and the height to which it rebounded by using conservation of mechanical energy.

(a) We know, from conservation of energy, that the kinetic energy of an object dropped from a given height h is proportional to h:

K α h.

If, for each bounce of the ball, hrec = 0.8happ:

20% of its mechanical energy is lost.

vrec vapp

(b) Use its definition to relate the coefficient of restitution to the velocities of approach and recession:

e=

Letting Ug = 0 at the surface from which the ball is rebounding, apply conservation of energy to express the velocity of approach:

∆K + ∆U = 0 Because Ki = Uf = 0,

Kf − Ui = 0

or 1 2

2 mvapp − mghapp = 0

Solve for vapp:

vapp = 2 ghapp

In like manner, show that:

vrec = 2 ghrec

Substitute in the equation for e to obtain:

e=

Substitute for

hrec to obtain: happ

2 ghrec 2 ghapp

=

hrec happ

e = 0.8 = 0.894

Systems of Particles and Conservation of Momentum 575 92 •• Picture the Problem Let the numeral 2 refer to the 2-kg object and the numeral 4 to the 4-kg object. Choose a coordinate system in which the direction the 2-kg object is moving before the collision is the positive x direction and let the system consist of the earth, the surface on which the objects slide, and the objects. Then we can use conservation of momentum to find the velocity of the recoiling 4-kg object. We can find the energy transformed in the collision by calculating the difference between the kinetic energies before and after the collision and the coefficient of restitution from its definition.

r r pi = p f

(a) Use conservation of momentum in one dimension to relate the initial and final momenta of the participants in the collision:

or m2 v 2i = m4 v 4f − m2 v 2f

Solve for and evaluate the final velocity of the 4-kg object:

v4 f = =

m2v2i + m2v2 f m4

(2 kg )(6 m/s + 1 m/s) = 4 kg

Elost = K i − K f

(b) Express the energy lost in terms of the kinetic energies before and after the collision:

= 12 m2 v22i − =

1 2

[m (v 2

2 2i

(

1 2

3.50 m/s

m2 v22f + 12 m4 v42f

)

− v22f − m4 v42f

]

)

Substitute numerical values and evaluate Elost:

Elost =

1 2

[((2 kg ){(6 m/s) − (1m/s) })− (4 kg )(3.5 m/s) ] = 2

2

2

10.5 J

(c) Use the definition of the coefficient of restitution:

e=

vrec v4f − v2f 3.5 m/s − (− 1 m/s ) = = = 0.750 vapp v2 i 6 m/s

93 •• Picture the Problem Let the numeral 2 refer to the 2-kg block and the numeral 3 to the 3-kg block. Choose a coordinate system in which the direction the blocks are moving before the collision is the positive x direction and let the system consist of the earth, the surface on which the blocks move, and the blocks. Then we can use conservation of momentum find the velocity of the 2-kg block after the collision. We can find the coefficient of restitution from its definition.

576 Chapter 8

r r pi = p f

(a) Use conservation of momentum in one dimension to relate the initial and final momenta of the participants in the collision:

or m2 v 2i + m3 v3i = m2 v 2 f + m3 v3f

Solve for the final velocity of the 2-kg object:

v2 f =

m2v2i + m3v3i − m3v3f m2

Substitute numerical values and evaluate v2f:

v2 f =

(2 kg )(5 m/s) + (3 kg )(2 m/s − 4.2 m/s) = 2 kg

(b) Use the definition of the coefficient of restitution:

e=

1.70 m/s

vrec v3f − v2 f 4.2 m/s − 1.7 m/s = = 5 m/s − 2 m/s vapp v2i − v3i

= 0.833

Collisions in Three Dimensions *94 •• Picture the Problem We can use the definition of the magnitude of a vector and the definition of the dot product to establish the result called for in (a). In part (b) we can use the result of part (a), the conservation of momentum, and the definition of an elastic collision (kinetic energy is conserved) to show that the particles separate at right angles.

r

r

(a) Find the dot product of B + C with itself:

r

r

r

Because A = B + C : Substitute to obtain: (b) Apply conservation of momentum to the collision of the particles: Form the dot product of each side of this equation with itself to obtain:

(Br + Cr ) ⋅ (Br + Cr )

r r = B2 + C 2 + 2B ⋅ C

(

)(

r r2 r r r r A2 = B + C = B + C ⋅ B + C

r r A2 = B 2 + C 2 + 2 B ⋅ C r r r p1 + p2 = P r r

( pr1 + pr 2 ) ⋅ ( pr1 + pr 2 ) = P ⋅ P or

r r p12 + p22 + 2 p1 ⋅ p2 = P 2

Apply the definition of an elastic collision to obtain:

)

p12 p22 P 2 + = 2m 2m 2m

(1)

Systems of Particles and Conservation of Momentum 577 or

p12 + p22 = P 2 Subtract equation (1) from equation (2) to obtain:

r r 2 p1 ⋅ p2 = 0 or

(2)

r r p1 ⋅ p2 = 0

i.e., the particles move apart along paths that are at right angles to each other. 95 • Picture the Problem Let the initial direction of motion of the cue ball be the positive x direction. We can apply conservation of energy to determine the angle the cue ball makes with the positive x direction and the conservation of momentum to find the final velocities of the cue ball and the eight ball.

(a) Use conservation of energy to relate the velocities of the collision participants before and after the collision: This Pythagorean relationship tells r r r us that v ci , v cf , and v8 form a right triangle. Hence:

(b) Use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision: Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision: Solve these equations simultaneously to obtain:

1 2

mvci2 = 12 mvcf2 + 12 mv82

or

vci2 = vcf2 + v82

θ cf + θ 8 = 90° and

θ cf = 60° r r pxi = pxf or mvci = mvcf cos θ cf + mv8 cos θ 8

r r pyi = pyf

or 0 = mvcf sin θ cf + mv8 sin θ 8

vcf = 2.50 m/s and v8 = 4.33 m/s

578 Chapter 8 96 •• Picture the Problem We can find the final velocity of the object whose mass is M1 by using the conservation of momentum. Whether the collision was elastic can be decided by examining the difference between the initial and final kinetic energy of the interacting objects.

r r pi = pf

(a) Use conservation of momentum to relate the initial and final velocities of the two objects:

or

Simplify to obtain:

r v0 iˆ + v0 ˆj = 12 v0 iˆ + v1f

( )

( )

r mv0 iˆ + 2m 12 v0 ˆj = 2m 14 v0 iˆ + mv1f

r

r v1f =

Solve for v1f :

1 2

v0 iˆ + v0 ˆj

(b) Express the difference between the kinetic energy of the system before the collision and its kinetic energy after the collision:

[M v + M v − M v − 2mv ] = m[v + 2v − v − 2v ] − 2( v )] = mv

∆E = K i − K f = K1i + K 2i − (K1f + K 2f ) = =

1 2

=

1 2

[mv m[v

2 1i

+ 2mv − mv

2 0

+ 2 14 v02 − 54 v02

2 2i

( )

2 1f

2 2f

1 16

2 0

2 1 1i

1 2

2 1i

1 2

1 16

2 2 2i

2 2i

2 1f

2 1 1f

− M 2v22f

]

2 2f

2 0

Because ∆E ≠ 0, the collision is inelastic. *97 •• Picture the Problem Let the direction of motion of the puck that is moving before the collision be the positive x direction. Applying conservation of momentum to the collision in both the x and y directions will lead us to two equations in the unknowns v1 and v2 that we can solve simultaneously. We can decide whether the collision was elastic by either calculating the system’s kinetic energy before and after the collision or by determining whether the angle between the final velocities is 90°.

(a) Use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision:

pxi = pxf or mv = mv1 cos 30° + mv2 cos 60° or v = v1 cos 30° + v2 cos 60°

Systems of Particles and Conservation of Momentum 579 pyi = pyf

Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision:

or 0 = mv1 sin 30° − mv2 sin 60° or 0 = v1 sin 30° − v2 sin 60° v1 = 1.73 m/s and v2 = 1.00 m/s

Solve these equations simultaneously to obtain:

r

r

(b) Because the angle between v1 and v 2 is 90°, the collision was elastic. 98 •• Picture the Problem Let the direction of motion of the object that is moving before the collision be the positive x direction. Applying conservation of momentum to the motion in both the x and y directions will lead us to two equations in the unknowns v2 and θ2 that we can solve simultaneously. We can show that the collision was elastic by showing that the system’s kinetic energy before and after the collision is the same.

(a) Use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision:

Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision:

pxi = pxf

or 3mv0 = 5mv0 cos θ1 + 2mv2 cos θ 2 or 3v0 = 5v0 cos θ1 + 2v2 cos θ 2 pyi = pyf

or 0 = 5mv0 sin θ1 − 2mv2 sin θ 2 or 0 = 5v0 sin θ1 − 2v2 sin θ 2

Note that if tanθ1 = 2, then:

Substitute in the momentum equations to obtain:

cosθ 1 =

1

and sin θ 1 =

5

3v0 = 5v0

1 + 2v2 cos θ 2 5

or v0 = v2 cos θ 2 and

2 5

580 Chapter 8 2 − 2v2 sin θ 2 5

0 = 5v0

or 0 = v0 − v2 sin θ 2 Solve these equations simultaneously for θ2 :

θ 2 = tan −1 1 = 45.0°

Substitute to find v2:

v2 =

(b) To show that the collision was elastic, find the before-collision and after-collision kinetic energies:

v0 v0 = = cos θ 2 cos 45°

2v0

K i = 12 m(3v0 ) = 4.5mv02 2

and

(

)

(

K f = 12 m 5v0 + 12 (2m ) 2v0 2

)

2

= 4.5mv02 Because K i = K f , the collision is elastic. *99 •• Picture the Problem Let the direction of motion of the ball that is moving before the collision be the positive x direction. Let v represent the velocity of the ball that is moving before the collision, v1 its velocity after the collision and v2 the velocity of the initially-atrest ball after the collision. We know that because the collision is elastic and the balls have the same mass, v1 and v2 are 90° apart. Applying conservation of momentum to the collision in both the x and y directions will lead us to two equations in the unknowns v1 and v2 that we can solve simultaneously.

Noting that the angle of deflection for the recoiling ball is 60°, use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision: Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision:

pxi = pxf

or mv = mv1 cos 30° + mv2 cos 60° or v = v1 cos 30° + v2 cos 60° pyi = pyf

or 0 = mv1 sin 30° − mv2 sin 60° or 0 = v1 sin 30° − v2 sin 60°

Systems of Particles and Conservation of Momentum 581 Solve these equations simultaneously to obtain:

v1 = 8.66 m/s and v 2 = 5.00 m/s

100 •• Picture the Problem Choose the coordinate system shown in the diagram below with the x-axis the axis of initial approach of the first particle. Call V the speed of the target particle after the collision. In part (a) we can apply conservation of momentum in the x and y directions to obtain two equations that we can solve simultaneously for tanθ. In part (b) we can use conservation of momentum in vector form and the elastic-collision equation to show that v = v0cosφ.

(a) Apply conservation of momentum in the x direction to obtain:

v0 = v cos φ + V cos θ

(1)

Apply conservation of momentum in the y direction to obtain:

v sin φ = V sin θ

(2)

Solve equation (1) for Vcosθ :

V cos θ = v0 − v cos φ

(3)

Divide equation (2) by equation (3) to obtain:

V sin θ v sin φ = V cos θ v0 − v cos φ or

tan θ = (b) Apply conservation of momentum to obtain:

v sin φ v0 − v cos φ

r r r v0 = v + V

Draw the vector diagram representing this equation:

Use the definition of an elastic

v02 = v 2 + V 2

582 Chapter 8 collision to obtain: If this Pythagorean condition is to hold, the third angle of the triangle must be a right angle and, using the definition of the cosine function:

v = v0 cos φ

Center-of-Mass Frame 101 •• Picture the Problem The total kinetic energy of a system of particles is the sum of the kinetic energy of the center of mass and the kinetic energy relative to the center of mass. The kinetic energy of a particle of mass m is related to momentum according to K = p 2 2m .

Express the total kinetic energy of the system: Relate the kinetic energy relative to the center of mass to the momenta of the two particles:

K = K rel + K cm

K rel

(1)

p12 p12 p12 (m1 + m2 ) = + = 2m1 2m2 2m1m2

Express the kinetic energy of the center of mass of the two particles:

2 ( 2 p1 ) 2 p12 = K cm = 2(m1 + m2 ) m1 + m2

Substitute in equation (1) and simplify to obtain:

p12 (m1 + m2 ) 2 p12 K= + 2m1 m2 m1 + m2 =

In an elastic collision:

Simplify to obtain:

p12 ⎡ m12 + 6m1 m2 + m22 ⎤ ⎥ ⎢ 2 ⎣ m12 m2 + m1 m22 ⎦

Ki = Kf =

p12 ⎡ m12 + 6m1m2 + m22 ⎤ 2 ⎢⎣ m12 m2 + m1m22 ⎥⎦

=

p'12 ⎡ m12 + 6m1m2 + m22 ⎤ 2 ⎢⎣ m12 m2 + m1m22 ⎥⎦

(p ) = ( p ) ' 2 1

1

2

⇒ p1' = ± p1

and

If p1' = + p1 , the particles do not collide.

Systems of Particles and Conservation of Momentum 583 *102 •• Picture the Problem Let the numerals 3 and 1 denote the blocks whose masses are 3 kg r r mi v i = Mv cm to find the velocity of the center-ofand 1 kg respectively. We can use

∑ i

mass of the system and simply follow the directions in the problem step by step. (a) Express the total momentum of this two-particle system in terms of the velocity of its center of mass:

r r r r P = ∑ mi vi = m1v1 + m3v3 i

r r = Mvcm = (m1 + m3 )vcm

Solve for vcm :

r r r m v + m1v1 vcm = 3 3 m3 + m1

Substitute numerical values and r evaluate vcm :

r (3 kg )(− 5 m/s ) iˆ + (1 kg )(3 m/s) iˆ v cm = 3 kg + 1 kg

r

= (b) Find the velocity of the 3-kg block in the center of mass reference frame: Find the velocity of the 1-kg block in the center of mass reference frame: (c) Express the after-collision velocities of both blocks in the center of mass reference frame:

(d) Transform the after-collision velocity of the 3-kg block from the center of mass reference frame to the original reference frame: Transform the after-collision velocity of the 1-kg block from the center of mass reference frame to the original reference frame: (e) Express Ki in the original frame of

(− 3.00 m/s ) iˆ

r r r u3 = v3 − vcm = (− 5 m/s ) iˆ − (− 3 m/s ) iˆ =

(− 2.00 m/s) iˆ

r r r u1 = v1 − vcm = (3 m/s ) iˆ − (− 3 m/s ) iˆ =

(6.00 m/s) iˆ

r u3' =

(2.00 m/s) iˆ

and

r u1' =

(− 6.00 m/s) iˆ

r r r v3' = u3' + vcm = (2 m/s ) iˆ + (− 3 m/s ) iˆ =

(− 1.00 m/s ) iˆ

r r r v1' = u1' + v cm = (− 6 m/s ) iˆ + (− 3 m/s ) iˆ =

(− 9.00 m/s ) iˆ

K i = 12 m3v32 + 12 m1v12

584 Chapter 8 reference:

[(3 kg )(5 m/s) + (1kg )(3 m/s) ]

Substitute numerical values and evaluate Ki:

Ki =

Express Kf in the original frame of reference:

K f = 12 m3v'32 + 12 m1v'12

Substitute numerical values and evaluate Kf:

Kf =

1 2

2

2

= 42.0 J

1 2

[(3 kg )(1m/s) + (1kg )(9 m/s) ] 2

2

= 42.0 J

103 •• Picture the Problem Let the numerals 3 and 1 denote the blocks whose masses are 3 kg r r mi v i = Mv cm to find the velocity of the center-ofand 1 kg respectively. We can use

∑ i

mass of the system and simply follow the directions in the problem step by step. (a) Express the total momentum of this two-particle system in terms of the velocity of its center of mass:

r r r r P = ∑ mi vi = m3v3 + m5v5 i

r r = Mvcm = (m3 + m5 ) vcm

Solve for vcm :

r r r m3v3 + m5v5 vcm = m3 + m5

Substitute numerical values and r evaluate vcm :

r (3 kg )(− 5 m/s) iˆ + (5 kg )(3 m/s) iˆ v cm = 3 kg + 5 kg

r

= 0 (b) Find the velocity of the 3-kg block in the center of mass reference frame: Find the velocity of the 5-kg block in the center of mass reference frame:

(c) Express the after-collision velocities of both blocks in the center of mass reference frame:

r r r u3 = v3 − vcm = (− 5 m/s ) iˆ − 0 =

(− 5 m/s ) iˆ

r r r u5 = v5 − vcm = (3 m/s ) iˆ − 0 =

(3 m/s ) iˆ

r u3' =

(5 m/s) iˆ

and

Systems of Particles and Conservation of Momentum 585 u5' = 0.75 m/s

r r r v3' = u3' + vcm = (5 m/s ) iˆ + 0

(d) Transform the after-collision velocity of the 3-kg block from the center of mass reference frame to the original reference frame:

(5 m/s) iˆ

=

r r r v5' = u5' + vcm = (− 3 m/s ) iˆ + 0

Transform the after-collision velocity of the 5-kg block from the center of mass reference frame to the original reference frame:

(− 3 m/s) iˆ

=

(e) Express Ki in the original frame of reference:

K i = 12 m3v32 + 12 m5v52

Substitute numerical values and evaluate Ki:

Ki =

Express Kf in the original frame of reference:

K f = 12 m3v'32 + 12 m5v'52

1 2

[(3 kg )(5 m/s) + (5 kg )(3 m/s) ] 2

2

= 60.0 J

Substitute numerical values and evaluate Kf:

Kf =

1 2

[(3 kg )(5 m/s)

2

]

+ (5 kg )(3 m/s ) = 60.0 J 2

Systems With Continuously Varying Mass: Rocket Propulsion 104 •• Picture the Problem The thrust of a rocket Fth depends on the burn rate of its fuel dm/dt and the relative speed of its exhaust gases uex according to Fth = dm dt uex .

Using its definition, relate the rocket’s thrust to the relative speed of its exhaust gases: Substitute numerical values and evaluate Fth:

Fth =

dm uex dt

Fth = (200 kg/s )(6 km/s ) = 1.20 MN

586 Chapter 8 105 •• Picture the Problem The thrust of a rocket Fth depends on the burn rate of its fuel dm/dt and the relative speed of its exhaust gases uex according to Fth = dm dt uex . The final

velocity vf of a rocket depends on the relative speed of its exhaust gases uex, its payload to initial mass ratio mf/m0 and its burn time according to vf = −uex ln (mf m0 ) − gt b . (a) Using its definition, relate the rocket’s thrust to the relative speed of its exhaust gases:

Fth =

dm uex dt

Fth = (200 kg/s )(1.8 km/s ) = 360 kN

Substitute numerical values and evaluate Fth: (b) Relate the time to burnout to the mass of the fuel and its burn rate:

tb =

mfuel 0.8m0 = dm / dt dm / dt

Substitute numerical values and evaluate tb:

tb =

0.8(30,000 kg ) = 120 s 200 kg/s

(c) Relate the final velocity of a rocket to its initial mass, exhaust velocity, and burn time:

⎛m ⎞ vf = −uex ln⎜⎜ f ⎟⎟ − gtb ⎝ m0 ⎠

Substitute numerical values and evaluate vf:

(

)

⎛1⎞ vf = −(1.8 km/s ) ln⎜ ⎟ − 9.81 m/s 2 (120 s ) = 1.72 km/s ⎝5⎠ *106 •• Picture the Problem We can use the dimensions of thrust, burn rate, and acceleration to show that the dimension of specific impulse is time. Combining the definitions of rocket thrust and specific impulse will lead us to uex = gI sp .

(a) Express the dimension of specific impulse in terms of the dimensions of Fth, R, and g:

[I ] sp

M⋅L 2 [F ] = th = T = T [R][g ] M ⋅ L T T2

(b) From the definition of rocket thrust we have:

Fth = Ruex

Solve for uex:

uex =

Fth R

Systems of Particles and Conservation of Momentum 587 Substitute for Fth to obtain:

uex =

RgI sp

(c) Solve equation (1) for Isp and substitute for uex to obtain:

I sp =

Fth Rg

From Example 8-21 we have:

R = 1.384×104 kg/s and Fth = 3.4×106 N

Substitute numerical values and evaluate Isp:

3.4 × 106 N I sp = 1.384 × 104 kg/s 9.81m/s 2

R

= gI sp

(

(1)

)(

)

= 25.0 s *107 ••• Picture the Problem We can use the rocket equation and the definition of rocket thrust to show that τ 0 = 1+ a0 g . In part (b) we can express the burn time tb in terms of the initial and final masses of the rocket and the rate at which the fuel burns, and then use this equation to express the rocket’s final velocity in terms of Isp, τ0, and the mass ratio m0/mf. In part (d) we’ll need to use trial-and-error methods or a graphing calculator to solve the transcendental equation giving vf as a function of m0/mf.

(a) Express the rocket equation:

− mg + Ruex = ma

From the definition of rocket thrust we have:

Fth = Ruex

Substitute to obtain:

− mg + Fth = ma

Solve for Fth at takeoff:

Fth = m0 g + m0 a0

Divide both sides of this equation by m0g to obtain:

Fth a = 1+ 0 m0 g g

Because τ 0 = Fth /( m0 g ) :

(b) Use equation 8-42 to express the final speed of a rocket that starts from rest with mass m0: Express the burn time in terms of the burn rate R (assumed constant):

τ 0 = 1+

a0 g

vf = uex ln

m0 − gtb , mf

where tb is the burn time.

tb =

m0 − mf m0 ⎛ mf ⎞ ⎜1 − ⎟ = R R ⎜⎝ m0 ⎟⎠

(1)

588 Chapter 8 Multiply tb by one in the form gT/gT and simplify to obtain:

tb =

gFth m0 ⎛ mf ⎞ ⎜1 − ⎟ gFth R ⎜⎝ m0 ⎟⎠

gm0 Fth ⎛ mf ⎞ ⎜1 − ⎟ Fth gR ⎜⎝ m0 ⎟⎠ I ⎛ m ⎞ = sp ⎜⎜1 − f ⎟⎟ τ 0 ⎝ m0 ⎠ =

Substitute in equation (1):

m0 gI sp ⎛ mf ⎞ ⎜1 − ⎟ − τ 0 ⎜⎝ m0 ⎟⎠ mf

vf = uex ln

uex = gI sp ,

From Problem 32 we have:

where uex is the exhaust velocity of the propellant. Substitute and factor to obtain:

vf = gI sp ln

m0 gI sp ⎛ mf ⎞ ⎜1 − ⎟ − τ 0 ⎜⎝ m0 ⎟⎠ mf

⎡ ⎛ m ⎞ 1 ⎛ m ⎞⎤ = gI sp ⎢ln⎜⎜ 0 ⎟⎟ − ⎜⎜1 − f ⎟⎟⎥ ⎣ ⎝ mf ⎠ τ 0 ⎝ m0 ⎠⎦ (c) A spreadsheet program to calculate the final velocity of the rocket as a function of the mass ratio m0/mf is shown below. The constants used in the velocity function and the formulas used to calculate the final velocity are as follows:

Cell B1 B2 B3 D9 E8

1 2 3 4 5 6 7 8 9

Content/Formula 250 9.81 2 D8 + 0.25 $B$2*$B$1*(LOG(D8) − (1/$B$3)*(1/D8))

A B C Isp = 250 s g = 9.81 m/s^2 tau = 2

D

Algebraic Form Isp g

τ

m0/mf ⎡ ⎛ m ⎞ 1 ⎛ m ⎞⎤ gI sp ⎢ln⎜⎜ 0 ⎟⎟ − ⎜⎜1 − f ⎟⎟⎥ ⎣ ⎝ mf ⎠ τ 0 ⎝ m0 ⎠⎦

E

mass ratio vf 2.00 1.252E+02 2.25 3.187E+02

Systems of Particles and Conservation of Momentum 589 10 11 12

2.50 2.75 3.00

4.854E+02 6.316E+02 7.614E+02

36 9.00 2.204E+03 37 9.25 2.237E+03 38 9.50 2.269E+03 39 9.75 2.300E+03 40 10.00 2.330E+03 41 725.00 7.013E+03 A graph of final velocity as a function of mass ratio is shown below.

v f (km/s)

2

1

0 2

4

6

8

10

m 0/m f

(d) Substitute the data given in part (c) in the equation derived in part (b) to obtain:

⎛ m 1 ⎛ m ⎞⎞ 7 km/s = 9.81 m/s 2 (250 s )⎜⎜ ln 0 − ⎜⎜1 − f ⎟⎟ ⎟⎟ ⎝ mf 2 ⎝ m0 ⎠ ⎠

(

)

or

2.854 = ln x − 0.5 + Use trial-and-error methods or a graphing calculator to solve this transcendental equation for the root greater than 1:

0.5 where x = m0/mf. x x = 28.1 , a value considerably larger than the practical limit of 10 for single-stage rockets.

108 •• Picture the Problem We can use the velocity-at-burnout equation from Problem 106 to find vf and constant-acceleration equations to approximate the maximum height the rocket will reach and its total flight time.

(a) Assuming constant acceleration, relate the maximum height reached

2 h = 12 gt top

(1)

590 Chapter 8 by the model rocket to its time-totop-of-trajectory: From Problem 106 we have:

⎛ ⎛ m ⎞ 1 ⎛ m ⎞⎞ vf = gI sp ⎜⎜ ln⎜⎜ 0 ⎟⎟ − ⎜⎜1 − f ⎟⎟ ⎟⎟ ⎝ ⎝ mf ⎠ τ ⎝ m0 ⎠ ⎠

Evaluate the velocity at burnout vf for Isp = 100 s, m0/mf = 1.2, and τ = 5:

vf = 9.81 m/s 2 (100 s )

Assuming that the time for the fuel to burn up is short compared to the total flight time, find the time to the top of the trajectory:

t top =

vf 146 m/s = = 14.9 s g 9.81 m/s 2

Substitute in equation (1) and evaluate h:

h=

(9.81m/s )(14.9 s)

(b) Find the total flight time from the time it took the rocket to reach its maximum height:

tflight = 2t top = 2(14.9 s ) = 29.8 s

(c) Express and evaluate the fuel burn time tb:

(

)

⎡ 1⎛ 1 ⎞⎤ × ⎢ln (1.2) − ⎜1 − ⎟ 5 ⎝ 1.2 ⎠⎥⎦ ⎣ = 146 m/s

1 2

2

I sp ⎛ m f ⎜1 − τ ⎜⎝ m0 = 3.33 s

tb =

2

= 1.09 km

⎞ 100 s ⎛ 1 ⎞ ⎟⎟ = ⎜1 − ⎟ 5 ⎝ 1.2 ⎠ ⎠

Because this burn time is approximately 1/5 of the total flight time, we can' t expect the answer we obtained in Part (b) to be very accurate. It should, however, be good to about 30% accuracy, as the maximum distance the model rocket could possibly move in this time is 12 vtb = 243 m, assuming constant acceleration until burnout.

General Problems 109 • Picture the Problem Let the direction of motion of the 250-g car before the collision be the positive x direction. Let the numeral 1 refer to the 250-kg car, the numeral 2 refer to the 400-kg car, and V represent the velocity of the linked cars. Let the system include the earth and the cars. We can use conservation of momentum to find their speed after they have linked together and the definition of kinetic energy to find their initial and final kinetic energies.

Systems of Particles and Conservation of Momentum 591 Use conservation of momentum to relate the speeds of the cars immediately before and immediately after their collision:

pix = pfx or

m1v1 = (m1 + m2 )V

Solve for V:

V=

m1v1 m1 + m2

Substitute numerical values and evaluate V:

V =

(0.250 kg )(0.50 m/s ) =

Find the initial kinetic energy of the cars:

K i = 12 m1v12 =

Find the final kinetic energy of the coupled cars:

Kf =

0.250 kg + 0.400 kg 1 2

0.192 m/s

(0.250 kg )(0.50 m/s )2

= 31.3 mJ

(m1 + m2 )V 2 2 = 12 (0.250 kg + 0.400 kg )(0.192 m/s ) 1 2

= 12.0 mJ 110 • Picture the Problem Let the direction of motion of the 250-g car before the collision be the positive x direction. Let the numeral 1 refer to the 250-kg car and the numeral 2 refer to the 400-g car and the system include the earth and the cars. We can use conservation of momentum to find their speed after they have linked together and the definition of kinetic energy to find their initial and final kinetic energies.

(a) Express and evaluate the initial kinetic energy of the cars:

(b) Relate the velocity of the center of mass to the total momentum of the system:

K i = 12 m1v12 =

1 2

(0.250 kg )(0.50 m/s)2

= 31.3 mJ

r r r P = ∑ mi v i = mv cm i

Solve for vcm:

vcm =

Substitute numerical values and evaluate vcm:

vcm =

m1v1 + m2 v2 m1 + m2

(0.250 kg )(0.50 m/s) = 0.192 m/s 0.250 kg + 0.400 kg

592 Chapter 8 Find the initial velocity of the 250-g car relative to the velocity of the center of mass: Find the initial velocity of the 400-g car relative to the velocity of the center of mass:

u1 = v1 − vcm = 0.50 m/s − 0.192 m/s = 0.308 m/s u 2 = v2 − vcm = 0 m/s − 0.192 m/s = − 0.192 m/s

Express the initial kinetic energy of the system relative to the center of mass:

K i,rel = 12 m1u12 + 12 m2u 22

Substitute numerical values and evaluate Ki,rel:

K i,rel =

(0.250 kg )(0.308 m/s )2 2 + 12 (0.400 kg )(− 0.192 m/s )

1 2

= 19.2 mJ (c) Express the kinetic energy of the center of mass: Substitute numerical values and evaluate Kcm:

(d) Relate the initial kinetic energy of the system to its initial kinetic energy relative to the center of mass and the kinetic energy of the center of mass:

2 K cm = 12 Mvcm

K cm =

1 2

(0.650 kg )(0.192 m/s )2

= 12.0 mJ

K i = K i,rel + K cm = 19.2 mJ + 12.0 mJ = 31.2 mJ ∴ K i = K i,rel + K cm

*111 • Picture the Problem Let the direction the 4-kg fish is swimming be the positive x direction and the system include the fish, the water, and the earth. The velocity of the larger fish immediately after its lunch is the velocity of the center of mass in this perfectly inelastic collision.

Relate the velocity of the center of mass to the total momentum of the system:

r r r P = ∑ mi v i = mv cm i

Systems of Particles and Conservation of Momentum 593 Solve for vcm:

vcm =

Substitute numerical values and evaluate vcm:

vcm =

m4 v4 − m1.2 v1.2 m4 + m1.2

(4 kg )(1.5 m/s) − (1.2kg) (3 m/s) 4 kg + 1.2 kg

= 0.462 m/s 112 • Picture the Problem Let the direction the 3-kg block is moving be the positive x direction and include both blocks and the earth in the system. The total kinetic energy of the two-block system is the sum of the kinetic energies of the blocks. We can relate the momentum of the system to the velocity of its center of mass and use this relationship to find vcm. Finally, we can use the definition of kinetic energy to find the kinetic energy relative to the center of mass.

(a) Express the total kinetic energy of the system in terms of the kinetic energy of the blocks:

K tot = 12 m3v32 + 12 m6 v62

Substitute numerical values and evaluate Ktot:

K tot =

(b) Relate the velocity of the center of mass to the total momentum of the system:

1 2

(3 kg )(6 m/s )2 + 12 (6 kg )(3 m/s)2

= 81.0 J

r r r P = ∑ mi v i = mv cm i

Solve for vcm:

vcm =

Substitute numerical values and evaluate vcm:

vcm =

m3v3 + m6 v6 m1 + m2

(3 kg )(6 m/s) + (6 kg )(3 m/s) 3 kg + 6 kg

= 4.00 m/s (c) Find the center of mass kinetic energy from the velocity of the center of mass:

2 K cm = 12 Mvcm =

= 72.0 J

1 2

(9 kg )(4 m/s)2

594 Chapter 8 K rel = K tot − K cm

(d) Relate the initial kinetic energy of the system to its initial kinetic energy relative to the center of mass and the kinetic energy of the center of mass:

= 81.0 J − 72.0 J = 9.00 J

113 • Picture the Problem Let east be the positive x direction and north the positive y direction. Include both cars and the earth in the system and let the numeral 1 denote the 1500-kg car and the numeral 2 the 2000-kg car. Because the net external force acting on the system is zero, momentum is conserved in this perfectly inelastic collision.

r r r r r p = p1 + p2 = m1v1 + m2 v 2 = m v ˆj − m v iˆ

(a) Express the total momentum of the system:

1 1

2 2

r

Substitute numerical values and evaluate p :

r p = (1500 kg )(70 km/h ) ˆj − (2000 kg )(55 km/h ) iˆ

(

) (

)

= − 1.10 ×105 kg ⋅ km/h iˆ + 1.05 × 105 kg ⋅ km/h ˆj

r r r p v f = v cm = M

(b) Express the velocity of the wreckage in terms of the total momentum of the system:

r

Substitute numerical values and evaluate v f :

(

)

(

)

r − 1.10 × 105 kg ⋅ km/h iˆ 1.05 × 105 kg ⋅ km/h ˆj vf = + 1500 kg + 2000 kg 1500 kg + 2000 kg = −(31.4 km/h ) iˆ + (30.0 km/h ) ˆj

(31.4 km/h )2 + (30.0 km/h )2

Find the magnitude of the velocity of the wreckage:

vf =

Find the direction of the velocity of the wreckage:

θ = tan −1 ⎢

= 43.4 km/h ⎡ 30.0 km/h ⎤ ⎥ = −43.7° ⎣ − 31.4 km/h ⎦

The direction of the wreckage is 46.3° west of north.

Systems of Particles and Conservation of Momentum 595 *114 •• Picture the Problem Take the origin to be at the initial position of the right-hand end of raft and let the positive x direction be to the left. Let ″w″ denote the woman and ″r″ the raft, d be the distance of the end of the raft from the pier after the woman has walked to its front. The raft moves to the left as the woman moves to the right; with the center of mass of the woman-raft system remaining fixed (because Fext,net = 0). The diagram shows the initial (xw,i) and final (xw,f) positions of the woman as well as the initial (xr_cm,i) and final (xr_cm,f) positions of the center of mass of the raft both before and after the woman has walked to the front of the raft.

CM

x

×

xr_cm,i x w i =6 m , xC M

0

0.5 m

CM

x

×

xr_cm,f xr_cm,i

0

xw f ,

(a) Express the distance of the raft from the pier after the woman has walked to the front of the raft:

d

d = 0.5 m + xf,w

Express xcm before the woman has walked to the front of the raft:

xcm =

Express xcm after the woman has walked to the front of the raft:

xcm =

Because Fext,net = 0, the center of mass remains fixed and we can equate these two expressions for xcm to obtain:

P I E R

(1)

mw xw,i + mr xr_cm, i m w + mr mw xw,f + mr xr_cm,f m w + mr

mw xw ,i + mr xr_cm,i = mw xw,f + mr xr_cm,f

Solve for xw,f:

xw,f = xw,i −

From the figure it can be seen that xr_cm,f – xr_cm,i = xw,f. Substitute xw,f

xw,f =

mr (xr_cm,f − xr_cm,i ) mw

mw xw,i m w + mr

596 Chapter 8 for xr_cm,f – xr_cm,i and to obtain:

(60 kg )(6 m )

Substitute numerical values and evaluate xw,f:

xw,f =

Substitute in equation (1) to obtain:

d = 2.00 m + 0.5 m = 2.50 m

(b) Express the total kinetic energy of the system:

K tot = 12 mw vw2 + 12 mr vr2

Noting that the elapsed time is 2 s, find vw and vr:

vw =

60 kg + 120 kg

x w,f − x w,i ∆t

=

= 2.00 m

2m − 6m = −2 m/s 2s

relative to the dock, and

vr =

xr,f − xr,i ∆t

=

2.50 m − 0.5 m = 1 m/s , 2s

also relative to the dock. Substitute numerical values and evaluate Ktot:

K tot =

(60 kg )(− 2 m/s)2 2 + 12 (120 kg )(1 m/s )

1 2

= 180 J Evaluate K with the raft tied to the pier:

K tot = 12 mw vw2 =

1 2

(60 kg )(3 m/s )2

= 270 J

All the kinetic energy derives from the chemical energy of the woman and, (c) assuming she stops via static friction, the kinetic energy is transformed into her internal energy. After the shot leaves the woman' s hand, the raft - woman system constitutes an inertial reference frame. In that frame the shot has the same initial (d) velocity as did the shot that had a range of 6 m in the reference frame of the land. Thus, in the raft - woman frame, the shot also has a range of 6 m and lands at the front of the raft.

Systems of Particles and Conservation of Momentum 597 115 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation of the 1-kg block. We can use conservation of energy to find the speed of the bob just before its perfectly elastic collision with the block and conservation of momentum to find the speed of the block immediately after the collision. We’ll apply Newton’s 2nd law to find the acceleration of the sliding block and use a constant-acceleration equation to find how far it slides before coming to rest. ∆K + ∆U = 0 (a) Use conservation of energy to find the speed of the bob just before or its collision with the block: K − K +U −U = 0 f

Because Ki = Uf = 0:

1 2

i

f

i

2 mball vball + mball g∆h = 0

and vball = 2 g∆h Substitute numerical values and evaluate vball:

(

)

vball = 2 9.81 m/s 2 (2 m ) = 6.26 m/s

Because the collision is perfectly elastic and the ball and block have the same mass:

vblock = vball = 6.26 m/s

(b) Using a constant-acceleration equation, relate the displacement of the block to its acceleration and initial speed and solve for its displacement:

vf2 = vi2 + 2ablock ∆x

Apply

r

r

∑ F = ma to the sliding

block:

Since vf = 0, ∆x =

− vi2 − v2 = block 2ablock 2ablock

∑F

= − f k = mablock

x

and

∑F

y

= Fn − mblock g = 0

Using the definition of fk (µkFn) eliminate fk and Fn between the two equations and solve for ablock:

ablock = − µ k g

Substitute for ablock to obtain:

∆x =

2 − vblock v2 = block − 2µ k g 2µ k g

598 Chapter 8 Substitute numerical values and evaluate ∆x:

∆x =

(6.26 m/s)2 = 2(0.1) (9.81 m/s 2 )

20.0 m

*116 •• Picture the Problem We can use conservation of momentum in the horizontal direction to find the recoil velocity of the car along the track after the firing. Because the shell will neither rise as high nor be moving as fast at the top of its trajectory as it would be in the absence of air friction, we can apply the work-energy theorem to find the amount of thermal energy produced by the air friction.

(a)

No. The vertical reaction force of the rails is an external force and so the momentum of the system will not be conserved.

∆p x = 0

(b) Use conservation of momentum in the horizontal (x) direction to obtain:

or

Solve for and evaluate vrecoil:

vrecoil =

Substitute numerical values and evaluate vrecoil:

vrecoil =

mv cos 30° − Mvrecoil = 0 mv cos 30° M

(200 kg )(125 m/s)cos30° 5000 kg

= 4.33 m/s (c) Using the work-energy theorem, relate the thermal energy produced by air friction to the change in the energy of the system: Substitute for ∆U and ∆K to obtain:

Wext = Wf = ∆Esys = ∆U + ∆K

Wext = mgyf − mgyi + 12 mvf2 − 12 mvi2

(

= mg ( yf − yi ) + 12 m vf2 − vi2

)

Substitute numerical values and evaluate Wext:

(

)

[

]

Wext = (200 kg ) 9.81 m/s 2 (180 m ) + 12 (200 kg ) (80 m/s ) − (125 m/s ) = − 569 kJ 2

2

Systems of Particles and Conservation of Momentum 599 117 •• Picture the Problem Because this is a perfectly inelastic collision, the velocity of the block after the collision is the same as the velocity of the center of mass before the collision. The distance the block travels before hitting the floor is the product of its velocity and the time required to fall 0.8 m; which we can find using a constantacceleration equation.

Relate the distance D to the velocity of the center of mass and the time for the block to fall to the floor: Relate the velocity of the center of mass to the total momentum of the system and solve for vcm:

D = vcm ∆t

r r r P = ∑ mi v i = Mv cm i

and

vcm =

Substitute numerical values and evaluate vcm: Using a constant-acceleration equation, find the time for the block to fall to the floor:

vcm =

mbullet vbullet + mblock vblock mbullet + mblock

(0.015 kg )(500 m/s) = 9.20 m/s 0.015 kg + 0.8 kg

∆y = v0 ∆t + 12 a(∆t )

2

Because v0 = 0, ∆t =

2∆y g

2∆y g

Substitute to obtain:

D = vcm

Substitute numerical values and evaluate D:

D = (9.20 m/s )

2(0.8 m ) = 3.72 m 9.81 m/s 2

118 •• Picture the Problem Let the direction the particle whose mass is m is moving initially be the positive x direction and the direction the particle whose mass is 4m is moving r initially be the negative y direction. We can determine the impulse delivered by F and, hence, the change in the momentum of the system from the change in the momentum of r the particle whose mass is m. Knowing ∆p , we can express the final momentum of the

particle whose mass is 4m and solve for its final velocity. Express the impulse delivered by the r force F :

r r r r r I = FT = ∆p = pf − pi = m(4v ) iˆ − mv iˆ = 3mv iˆ

600 Chapter 8 r

Express p' 4 m :

r

Solve for v ' :

r r r r p' 4m = 4mv ' = p4 m (0 ) + ∆p = −4mv ˆj + 3mv iˆ

r v' =

3 4

v iˆ − v ˆj

119 •• Picture the Problem Let the numeral 1 refer to the basketball and the numeral 2 to the baseball. The left-hand side of the diagram shows the balls after the basketball’s elastic collision with the floor and just before they collide. The right-hand side of the diagram shows the balls just after their collision. We can apply conservation of momentum and the definition of an elastic collision to obtain equations relating the initial and final velocities of the masses of the colliding objects that we can solve for v1f and v2f.

(a) Because both balls are in freefall, and both are in the air for the same amount of time, they have the same velocity just before the basketball rebounds. After the basketball rebounds elastically, its velocity will have the same magnitude, but the opposite direction than just before it hit the ground.

The velocity of the basketball will be equal in magnitude but opposite in direction to the velocity of the baseball.

(b) Apply conservation of momentum to the collision of the balls to obtain:

m1v1f + m2 v2f = m1v1i + m2 v2i

Relate the initial and final kinetic energies of the balls in their elastic collision:

1 2

Rearrange this equation and factor to obtain:

m2 v22f − v22i = m1 v12i − v12f

m1v12f + 12 m2 v22f = 12 m1v12i + 12 m2 v22i

(

)

(

)

or

m2 (v2 f − v2i )(v2 f + v2i )

= m1 (v1i − v1f )(v1i + v1f )

Rearrange equation (1) to obtain: Divide equation (2) by equation (3) to obtain:

(1)

m2 (v2f − v2i ) = m1 (v1i − v1f ) v2 f + v2i = v1i + v1f

(2)

(3)

Systems of Particles and Conservation of Momentum 601 Rearrange this equation to obtain equation (4): Multiply equation (4) by m2 and add it to equation (1) to obtain: Solve for v1f to obtain:

v1f − v2f = v2i − v1i

(4)

(m1 + m2 )v1f = (m1 − m2 )v1i + 2m2v2i v1f =

m1 − m2 2m2 v1i + v2 i m1 + m2 m1 + m2

or, because v2i = −v1i,

v1f = = For m1 = 3m2 and v1i = v:

(c) Multiply equation (4) by m1 and subtract it from equation (1) to obtain: Solve for v2f to obtain:

v1f =

2m2 m1 − m2 v1i − v1i m1 + m2 m1 + m2 m1 − 3m2 v1i m1 + m2

3m2 − 3m2 v= 0 3m2 + m2

(m1 + m2 )v2f = (m2 − m1 )v2i + 2m1v1i

v2 f =

2m1 m − m1 v 2i v1i + 2 m1 + m2 m1 + m2

or, because v2i = −v1i,

v2 f = = For m1 = 3m2 and v1i = v:

v2 f =

2m1 m − m1 v1i − 2 v1i m1 + m2 m1 + m2

3m1 − m2 v1i m1 + m2 3(3m2 ) − m2 v = 2v 3m2 + m2

602 Chapter 8 120 ••• Picture the Problem In Problem 119 only two balls are dropped. They collide head on, each moving at speed v, and the collision is elastic. In this problem, as it did in Problem 119, the solution involves using the conservation of momentum equation

m1v1f + m2 v2f = m1v1i + m2 v2i and the elastic collision equation

v1f − v2 f = v2i − v1i , where the numeral 1 refers to the baseball, and the numeral 2 to the top ball. The diagram shows the balls just before and just after their collision. From Problem 119 we know that that v1i = 2v and v2i = −v.

m1 − m2 2m2 v1i + v2 i m1 + m2 m1 + m2

(a) Express the final speed v1f of the baseball as a function of its initial speed v1i and the initial speed of the top ball v2i (see Problem 78):

v1f =

Substitute for v1i and , v2i to obtain:

m1 − m2 (2v ) + 2m2 (− v ) m1 + m2 m1 + m2 m1 −1 m2 (2v ) + m 2 (− v ) v1f = m1 1 +1 +1 m2 m2

Divide the numerator and denominator of each term by m2 to introduce the mass ratio of the upper ball to the lower ball:

Set the final speed of the baseball v1f equal to zero, let x represent the mass ratio m1/m2, and solve for x:

v1f =

0=

x −1 (2v ) + 2 (− v ) x +1 x +1

and

x=

m1 1 = m2 2

(b) Apply the second of the two equations in Problem 78 to the collision between the top ball and the baseball:

v2 f =

2m1 m − m1 v1i + 2 v 2i m1 + m2 m1 + m2

Substitute v1i = 2v and are given that v2i = −v to obtain:

v2 f =

2m1 (2v ) + m2 − m1 (− v ) m1 + m2 m1 + m2

Systems of Particles and Conservation of Momentum 603 In part (a) we showed that m2 = 2m1. Substitute and simplify:

v3f =

2(2m1 ) (2v ) − 2m1 − m1 v m1 + 2m1 m1 + 2m1

=

4m1 (2v ) − m1 v = 83 v − 13 v 3m1 3m1

=

7 3

v

*121 •• Picture the Problem Let the direction the probe is moving after its elastic collision with Saturn be the positive direction. The probe gains kinetic energy at the expense of the kinetic energy of Saturn. We’ll relate the velocity of approach relative to the center of mass to urec and then to v.

(a) Relate the velocity of recession to the velocity of recession relative to the center of mass: Find the velocity of approach:

v = u rec + vcm

uapp = −9.6 km/s − 10.4 km/s = −20.0 km/s

Relate the relative velocity of approach to the relative velocity of recession for an elastic collision: Because Saturn is so much more massive than the space probe: Substitute and evaluate v:

u rec = −uapp = 20.0 km/s

vcm = vSaturn = 9.6 km/s

v = urec + vcm = 20 km/s + 9.6 km/s = 29.6 km/s

(b) Express the ratio of the final kinetic energy to the initial kinetic energy:

2 1 ⎛v ⎞ Kf 2 Mvrec = 1 = ⎜⎜ rec ⎟⎟ 2 Ki ⎝ vi ⎠ 2 Mvi

2

2

⎛ 29.6 km/s ⎞ ⎟⎟ = 8.10 = ⎜⎜ ⎝ 10.4 km/s ⎠

The energy comes from an immeasurably small slowing of Saturn.

604 Chapter 8 *122 •• Picture the Problem We can use the relationships P = c∆m and ∆E = ∆mc 2 to show that P = ∆E c . We can then equate this expression with the change in momentum of the flashlight to find the latter’s final velocity.

(a) Express the momentum of the mass lost (i.e., carried away by the light) by the flashlight:

P = c∆m

Relate the energy carried away by the light to the mass lost by the flashlight:

∆m =

∆E c2

P=c

∆E ∆E = 2 c c

Substitute to obtain:

(b) Relate the final momentum of the flashlight to ∆E:

∆E = ∆p = mv c because the flashlight is initially at rest.

∆E mc

Solve for v:

v=

Substitute numerical values and evaluate v:

1.5 × 103 J v= (1.5 kg ) 2.998 × 108 m/s

(

)

= 3.33 × 10 −6 m/s = 3.33 µm/s 123 • Picture the Problem We can equate the change in momentum of the block to the momentum of the beam of light and relate the momentum of the beam of light to the mass converted to produce the beam. Combining these expressions will allow us to find the speed attained by the block.

Relate the change in momentum of the block to the momentum of the beam: Express the momentum of the mass converted into a well-collimated beam of light: Substitute to obtain: Solve for v:

(M − m )v = Pbeam because the block is initially at rest.

Pbeam = mc

(M − m )v = mc v=

mc M −m

Systems of Particles and Conservation of Momentum 605 Substitute numerical values and evaluate v:

v=

(0.001kg )(2.998 × 108 m/s) 1kg − 0.001 kg

= 3.00 × 105 m/s 124 •• Picture the Problem Let the origin of the coordinate system be at the end of the boat at which your friend is sitting prior to changing places. If we let the system include you and your friend, the boat, the water and the earth, then Fext,net = 0 and the center of mass is at the same location after you change places as it was before you shifted.

Express the center of mass of the system prior to changing places:

xcm = =

Substitute numerical values and simplify to obtain an expression for xcm in terms of m:

xcm = =

mboat xboat + myou xyou + mxfriend mboat + myou + mfriend

xyou (mboat + myou ) + mxfriend mboat + myou + m

(2 m )(60 kg + 80 kg ) + (0) m 60 kg + 80 kg + m 280 kg ⋅ m 140 kg + m

Find the center of mass of the system after changing places:

x'cm =

mboat xboat + myou xyou + mxfriend mboat + myou + mfriend

=

(mboat + m )(2 m ± 0.2 m ) mboat + myou + m

+

myou (± 0.2 m ) mboat + myou + m

Substitute numerical values and simplify to obtain:

x'cm =

(60 kg + m )(2 m ± 0.2 m ) 60 kg + 80 kg + m +

+

(80 kg )(± 0.2 m ) 60 kg + 80 kg + m

(2 m )m ± 0.2m m ± 16 kg ⋅ m

140 kg + m Because Fext,net = 0, x'cm = xcm .

120 kg ⋅ m ± 12 kg ⋅ m 140 kg + m

m=

(160 ± 28) kg (2 ± 0.2)

m=

(160 + 28) kg = (2 − 0.2)

Equate the two expressions and solve for m to obtain: Calculate the largest possible mass for your friend:

=

104 kg

606 Chapter 8 Calculate the smallest possible mass for your friend:

m=

(160 − 28) kg = (2 + 0.2)

60.0 kg

125 •• Picture the Problem Let the system include the woman, both vehicles, and the earth. Then Fext,net = 0 and acm = 0. Include the mass of the man in the mass of the truck. We can use Newton’s 2nd and 3rd laws to find the acceleration of the truck and net force acting on both the car and the truck.

(a) Relate the action and reaction forces acting on the car and truck:

Fcar = Ftruck or

mcar acar = mtruck+ woman atruck Solve for the acceleration of the truck:

Substitute numerical values and evaluate atruck:

(b) Apply Newton’s 2nd law to either vehicle to obtain: Substitute numerical values and evaluate Fnet:

atruck =

atruck

mcar acar mtruck + woman

( 800 kg ) (1.2 m/s 2 ) = = 1600 kg

0.600 m/s 2

Fnet = mcar acar

(

)

Fnet = (800 kg ) 1.2 m/s 2 = 960 N

126 •• Picture the Problem Let the system include the block, the putty, and the earth. Then Fext,net = 0 and momentum is conserved in this perfectly inelastic collision. We’ll use conservation of momentum to relate the after-collision velocity of the block plus blob and conservation of energy to find their after-collision velocity.

Noting that, because this is a perfectly elastic collision, the final velocity of the block plus blob is the velocity of the center of mass, use conservation of momentum to relate the velocity of the center of mass to the velocity of the glob before the collision:

pi = p f or mgl vgl = Mvcm where M = mgl + mbl.

Systems of Particles and Conservation of Momentum 607 Solve for vgl to obtain:

Use conservation of energy to find the initial energy of the block plus glob: Use fk = µkMg to eliminate fk and solve for vcm: Substitute numerical values and evaluate vcm:

Substitute numerical values in equation (1) and evaluate vgl:

vgl =

M vcm mgl

(1)

∆K + ∆U + Wf = 0 Because ∆U = Kf = 0, 2 − 12 Mvcm + f k ∆x = 0

vcm = 2 µ k g∆x

(

)

vcm = 2(0.4) 9.81 m/s 2 (0.15 m ) = 1.08 m/s vgl =

13 kg + 0.4 kg (1.08 m/s) 0.4 kg

= 36.2 m/s *127 •• Picture the Problem Let the direction the moving car was traveling before the collision be the positive x direction. Let the numeral 1 denote this car and the numeral 2 the car that is stopped at the stop sign and the system include both cars and the earth. We can use conservation of momentum to relate the speed of the initially-moving car to the speed of the meshed cars immediately after their perfectly inelastic collision and conservation of energy to find the initial speed of the meshed cars.

Using conservation of momentum, relate the before-collision velocity to the after-collision velocity of the meshed cars: Solve for v1:

Using conservation of energy, relate the initial kinetic energy of the meshed cars to the work done by friction in bringing them to a stop: Substitute for Ki and, using fk = µkFn = µkMg, eliminate fk to

pi = pf or

m1v1 = (m1 + m2 )V

v1 =

⎛ m ⎞ m1 + m2 V = ⎜⎜1 + 2 ⎟⎟V m1 ⎝ m1 ⎠

∆K + ∆Ethermal = 0 or, because Kf = 0 and ∆Ethermal = f∆s,

− K i + f k ∆s = 0

− 12 MV 2 + µ k Mg∆x = 0

608 Chapter 8 obtain: Solve for V:

V = 2 µ k g∆x

Substitute to obtain:

⎛ m ⎞ v1 = ⎜⎜1 + 2 ⎟⎟ 2 µ k g∆x m1 ⎠ ⎝

Substitute numerical values and evaluate v1:

⎛ 900 kg ⎞ ⎟⎟ 2(0.92 ) (9.81 m/s 2 )(0.76 m ) = 6.48 m/s = 23.3 km/h v1 = ⎜⎜1 + 1200 kg ⎠ ⎝ The driver was not telling the truth. He was traveling at 23.3 km/h. 128 •• Picture the Problem Let the zero of gravitational potential energy be at the lowest point of the bob’s swing and note that the bob can swing either forward or backward after the collision. We’ll use both conservation of momentum and conservation of energy to relate the velocities of the bob and the block before and after their collision.

pm2 2m

Express the kinetic energy of the block in terms of its after-collision momentum:

Km =

Solve for m to obtain:

m=

Use conservation of energy to relate Km to the change in the potential energy of the bob:

∆K + ∆U = 0 or, because Ki = 0,

Solve for Km:

K m = −U f + U i

pm2 2K m

(1)

Km + Uf − Ui = 0

= mbob g [L(1 − cos θ i ) − L(1 − cos θ f )]

= mbob gL[cos θ f − cos θ i ] Substitute numerical values and evaluate Km:

(

)

K m = (0.4 kg ) 9.81 m/s 2 (1.6 m )[cos5.73° − cos53°] = 2.47 J

Systems of Particles and Conservation of Momentum 609 Use conservation of energy to find the velocity of the bob just before its collision with the block:

∆K + ∆U = 0 or, because Ki = Uf = 0,

Kf − Ui = 0

∴ 12 mbob v 2 − mbob gL(1 − cos θ i ) = 0 or v = 2 gL(1 − cos θ i ) Substitute numerical values and evaluate v:

(

)

v = 2 9.81 m/s 2 (1.6 m )(1 − cos53°) = 3.544 m/s ∆K + ∆U = 0

Use conservation of energy to find the velocity of the bob just after its collision with the block:

or, because Kf = Ui = 0,

Substitute for Ki and Uf to obtain:

− 12 mbob v'2 + mbob gL(1 − cos θ f ) = 0

Solve for v′:

v' = 2 gL(1 − cos θ f )

Substitute numerical values and evaluate v′:

− Ki + U f = 0

(

)

v' = 2 9.81 m/s 2 (1.6 m )(1 − cos5.73°) = 0.396 m/s pi = p f

Use conservation of momentum to relate pm after the collision to the momentum of the bob just before and just after the collision:

or mbob v = mbob v'± pm

Solve for and evaluate pm:

pm = mbob v ± mbob v'

= (0.4 kg )(3.544 m/s ± 0.396 m/s ) = 1.418 kg ⋅ m/s ± 0.158 kg ⋅ m/s

Find the larger value for pm:

pm = 1.418 kg ⋅ m/s + 0.158 kg ⋅ m/s = 1.576 kg ⋅ m/s

Find the smaller value for pm:

pm = 1.418 kg ⋅ m/s − 0.158 kg ⋅ m/s = 1.260 kg ⋅ m/s

Substitute in equation (1) to determine the two values for m:

m=

(1.576 kg ⋅ m/s) 2 2(2.47 J )

= 0.503 kg

610 Chapter 8 or

m=

(1.260 kg ⋅ m/s) 2 2(2.47 J )

= 0.321kg

129 •• Picture the Problem Choose the zero of gravitational potential energy at the location of the spring’s maximum compression. Let the system include the spring, the blocks, and the earth. Then the net external force is zero as is work done against friction. We can use conservation of energy to relate the energy transformations taking place during the evolution of this system.

Apply conservation of energy:

∆K + ∆U g + ∆U s = 0

Because ∆K = 0:

∆U g + ∆U s = 0 ∆U g = − mg∆h − Mgx sin θ

Express the change in the gravitational potential energy: Express the change in the potential energy of the spring:

∆U s = 12 kx 2

Substitute to obtain:

− mg∆h − Mgx sin θ + 12 kx 2 = 0

Solve for M:

M =

Relate ∆h to the initial and rebound positions of the block whose mass is m:

1 2

kx 2 − mg∆h kx 2m∆h = − gx sin 30° g x

∆h = (4 m − 2.56 m ) sin 30° = 0.720 m

Substitute numerical values and evaluate M:

M =

(11×10

)

N/m (0.04 m ) 2(1 kg )(0.72 m ) − = 8.85 kg 2 9.81 m/s 0.04 m 3

*130 •• Picture the Problem By symmetry, xcm = 0. Let σ be the mass per unit area of the disk. The mass of the modified disk is the difference between the mass of the whole disk and the mass that has been removed.

Systems of Particles and Conservation of Momentum 611 Start with the definition of ycm:

ycm = =

Express the mass of the complete disk: Express the mass of the material removed:

Substitute and simplify to obtain:

∑m y i

i

i

M − mhole mdisk ydisk − mhole y hole M − mhole

M = σA = σπ r 2 2

mhole

⎛r⎞ = σπ ⎜ ⎟ = 14 σπ r 2 = 14 M ⎝2⎠

ycm =

M (0) − ( 14 M )(− 12 r ) = M − 14 M

1 6

r

131 •• Picture the Problem Let the horizontal axis by the y axis and the vertical axis the z axis. By symmetry, xcm = ycm = 0. Let ρ be the mass per unit volume of the sphere. The mass of the modified sphere is the difference between the mass of the whole sphere and the mass that has been removed.

Start with the definition of ycm:

zcm = =

∑m y i

i

i

M − mhole msphere ysphere − mhole y hole M − mhole

Express the mass of the complete sphere:

M = ρV = 43 ρπ r 3

Express the mass of the material removed:

⎛r⎞ mhole = 43 ρπ ⎜ ⎟ = ⎝2⎠

Substitute and simplify to obtain:

zcm =

3

(

1 4 8 3

ρπ r 3 ) = 18 M

M (0) − ( 18 M )(− 12 r ) = M − 18 M

1 14

r

*132 •• Picture the Problem In this elastic head-on collision, the kinetic energy of recoiling nucleus is the difference between the initial and final kinetic energies of the neutron. We can derive the indicated results by using both conservation of energy and conservation of momentum and writing the kinetic energies in terms of the momenta of the particles before and after the collision.

612 Chapter 8 p2 pni2 p2 = nf + nucleus 2m 2m 2M

(1)

pni = pnf + pnucleus

(2)

Eliminate pnf in equation (1) using equation (2):

pnucleus pnucleus pni + − =0 2M 2m m

(3)

Use equation (3) to write pni2 2m in

(M + m ) p2 pni2 = K n = nucleus 2 2m 8M m

(4)

⎡ 4Mm ⎤ K nucleus = K n ⎢ 2⎥ ⎣ (M + m ) ⎦

(5)

(a) Use conservation of energy to relate the kinetic energies of the particles before and after the collision: Apply conservation of momentum to obtain a second relationship between the initial and final momenta:

terms of pnucleus: Use equation (4) to express 2 K nucleus = pnucleus 2M in terms of

2

Kn: (b) Relate the change in the kinetic energy of the neutron to the aftercollision kinetic energy of the nucleus: Using equation (5), express the fraction of the energy lost in the collision:

∆K n = − K nucleus

− ∆K n = Kn

m 4 4 Mm M = 2 (M + m ) ⎛ m ⎞ 2 ⎜1 + ⎟ ⎝ M⎠

133 •• Picture the Problem Problem 132 (b) provides an expression for the fractional loss of energy per collision.

(a) Using the result of Problem 132 (b), express the fractional loss of energy per collision: Evaluate this fraction to obtain:

Express the kinetic energy of one

K nf K ni − ∆K n (M − m ) = = K ni E0 (M + m )2 2

K nf (12m − m ) = = 0.716 E0 (12m + m )2 2

K nf = 0.716 N E0

Systems of Particles and Conservation of Momentum 613 neutron after N collisions: (b) Substitute for Knf and E0 to obtain:

0.716 N = 10 −8

Take the logarithm of both sides of the equation and solve for N:

N=

−8 ≈ 55 log 0.716

134 •• Picture the Problem We can relate the number of collisions needed to reduce the energy of a neutron from 2 MeV to 0.02 eV to the fractional energy loss per collision and solve the resulting exponential equation for N.

(a) Using the result of Problem 132 (b), express the fractional loss of energy per collision:

K nf K ni − ∆K n K ni − 0.63K ni = = K ni E0 K ni = 0.37

Express the kinetic energy of one neutron after N collisions:

K nf = 0.37 N E0

Substitute for Knf and E0 to obtain:

0.37 N = 10 −8

Take the logarithm of both sides of the equation and solve for N:

N=

(b) Proceed as in (a) to obtain:

−8 ≈ 19 log 0.37

K nf K ni − ∆K n K ni − 0.11K ni = = K ni E0 K ni = 0.89

Express the kinetic energy of one neutron after N collisions:

K nf = 0.89 N E0

Substitute for Knf and E0 to obtain:

0.89 N = 10 −8

Take the logarithm of both sides of the equation and solve for N:

N=

−8 ≈ 158 log 0.89

614 Chapter 8 135 •• Picture the Problem Let λ = M/L be the mass per unit length of the rope, the subscript 1 refer to the portion of the rope that is being supported by the force F at any given time, and the subscript 2 refer to the rope that is still on the table at any given time. We can find the height hcm of the center of mass as a function of time and then differentiate this expression twice to find the acceleration of the center of mass.

m1h1,cm + m2 h2,cm M

(a) Apply the definition of the center of mass to obtain:

hcm =

From the definition of λ we have:

M m1 M = ⇒ m1 = vt L vt L

h1,cm and h2,cm are given by :

h1,cm =

Substitute for m1, h1,cm, and h2,cm in equation (1) and simplify to obtain:

(b) Differentiate hcm twice to obtain acm:

hcm

(1)

1 vt and h2,cm = 0 2

⎛M ⎞ ⎜ vt ⎟h1,cm + m2 (0 ) v2 2 L ⎠ =⎝ = t M 2L

⎛ v2 ⎞ v2 dhcm = 2⎜⎜ ⎟⎟t = t dt L ⎝ 2L ⎠ and d 2 hcm v2 = = a cm dt 2 L

(c) Letting N represent the normal force that the table exerts on the rope, apply Fy = macm to the

F + N − Mg = Macm

∑

rope to obtain: Solve for F, substitute for acm and N to obtain:

Use the definition of λ again to obtain:

F = Mg + Macm − N v2 = Mg + M − m2 g L

m2 M ⎛ vt ⎞ ⇒ m2 = M ⎜1 − ⎟ = L − vt L ⎝ L⎠

Systems of Particles and Conservation of Momentum 615 Substitute for m2 and simplify:

F = Mg + M =

⎛ v2 v2 vt ⎛ vt ⎞ − M ⎜1 − ⎟ g = M ⎜⎜ g + − g + L L L ⎝ L⎠ ⎝

⎞ ⎛ v 2 vt ⎞ + ⎟⎟ g ⎟⎟ = Mg ⎜⎜ ⎠ ⎝ gL L ⎠

⎞ vt ⎛ v ⎜⎜ + 1⎟⎟ Mg L ⎝ gt ⎠

136 •• Picture the Problem The free-body diagram shows the forces acting on the platform when the spring is partially compressed. The scale reading is the force the scale exerts on the platform and is represented on the FBD by Fn. We can use Newton’s 2nd law to determine the scale reading in part (a). We’ll use both conservation of energy and momentum to obtain the scale reading when the ball collides inelastically with the cup.

(a) Apply

∑F

y

= ma y to the

Fn − mp g − Fball on spring = 0

spring when it is compressed a distance d: Solve for Fn:

Fn = mp g + Fball on spring = mp g + kd ⎛m g⎞ = mp g + k ⎜ b ⎟ ⎝ k ⎠

= mp g + mb g = (mp + mb )g

(b) Letting the zero of gravitational energy be at the initial elevation of the cup and vbi represent the velocity of the ball just before it hits the cup, use conservation of energy to find this velocity: Use conservation of momentum to

∆K + ∆U g = 0 where K i = U gf = 0 ∴ 12 mb vbi2 − mgh = 0 and vbi = 2 gh r r pi = pf

616 Chapter 8 find the velocity of the center of mass:

∴ vcm =

⎡ mb ⎤ mb vbi = 2 gh ⎢ ⎥ mb + mc ⎣ mb + mc ⎦

∆K cm + ∆U s = 0

Apply conservation of energy to the collision to obtain:

or, with Kf = Usi = 0,

Substitute for vcm and solve for kx2:

2 kx 2 = (mb + mc ) vcm

2 − 12 (mb + mc )vcm + 12 kx 2 = 0

⎡ mb ⎤ = 2 gh(mb + mc ) ⎢ ⎥ ⎣ mb + mc ⎦ 2 ghmb2 = mb + mc Solve for x:

From part (a):

x = mb

2

2 gh k (mb + mc )

Fn = mp g + kx = mp g + kmb

2 gh k (mb + mc )

⎞ ⎛ 2kh ⎟ = g ⎜⎜ mp + mb ⎟ ( ) g m m + b c ⎠ ⎝ (c) Because the collision is inelastic, the ball never returns to its original height. 137 •• Picture the Problem Let the direction that astronaut 1 first throws the ball be the positive direction and let vb be the initial speed of the ball in the laboratory frame. Note that each collision is perfectly inelastic. We can apply conservation of momentum and the definition of the speed of the ball relative to the thrower to each of the perfectly inelastic collisions to express the final speeds of each astronaut after one throw and one catch.

Use conservation of momentum to relate the speeds of astronaut 1 and the ball after the first throw: Relate the speed of the ball in the laboratory frame to its speed relative

m1v1 + mb v b = 0

(1)

v = v b − v1

(2)

Systems of Particles and Conservation of Momentum 617 to astronaut 1: Eliminate vb between equations (1) and (2) and solve for v1:

v1 = −

Substitute equation (3) in equation (2) and solve for vb:

vb =

Apply conservation of momentum to express the speed of astronaut 2 and the ball after the first catch:

mb v m1 + mb

m1 v m1 + mb

0 = mb v b = (m2 + mb )v 2

Solve for v2:

v2 =

mb vb m2 + mb

Express v2 in terms of v by substituting equation (4) in equation (6):

v2 =

mb m1 v m2 + mb m1 + mb

Use conservation of momentum to express the speed of astronaut 2 and the ball after she throws the ball: Relate the speed of the ball in the laboratory frame to its speed relative to astronaut 2:

⎡ ⎤ mb m1 =⎢ ⎥v ⎣ (m2 + mb )(m1 + mb ) ⎦

(3)

(4)

(5)

(6)

(7)

(m2 + mb )v2 = mb vbf + m2v2f

(8)

v = v 2f − v bf

(9)

Eliminate vbf between equations (8) and (9) and solve for v2f:

⎛ mb ⎞ ⎡ m1 ⎤ ⎟⎟ ⎢1 + v2 f = ⎜⎜ ⎥v ⎝ m2 + mb ⎠ ⎣ m1 + mb ⎦

(10)

Substitute equation (10) in equation (9) and solve for vbf:

⎡ mb ⎤ vbf = − ⎢1 − ⎥ ⎣ m2 + mb ⎦ ⎡ m1 ⎤ × ⎢1 + ⎥v ⎣ m1 + mb ⎦

(11)

(m1 + mb )v1f

(12)

Apply conservation of momentum to express the speed of astronaut 1 and the ball after she catches the ball:

= mb vbf + m1v1

618 Chapter 8 Using equations (3) and (11), eliminate vbf and v1 in equation (12) and solve for v1f:

m2 mb (2m1 + mb )

v1f = −

(m1 + mb )2 (m2 + mb )

v

*138 •• Picture the Problem We can use the definition of the center of mass of a system containing multiple objects to locate the center of mass of the earth−moon system. Any object external to the system will exert accelerating forces on the system.

(a) Express the center of mass of the earth−moon system relative to the center of the earth:

r r Mrcm = ∑ mi ri i

or

rcm = =

Substitute numerical values and evaluate rcm:

M e (0) + mm rem mm rem = M e + mm M e + mm rem Me +1 mm

3.84 × 105 km rcm = = 4670 km 81.3 + 1

Because this distance is less than the radius of the earth, the position of the center of mass of the earth − moon system is below the surface of the earth.

(b)

(c)

Any object not in the earth − moon system exerts forces on the system, e.g., the sun and other planets. Because the sun exerts the dominant external force on the earth − moon system, the acceleration of the system is toward the sun.

(d) Because the center of mass is at a fixed distance from the sun, the distance d moved by the earth in this time interval is:

d = 2rem = 2(4670 km ) = 9340 km

139 •• Picture the Problem Let the numeral 2 refer to you and the numeral 1 to the water leaving the hose. Apply conservation of momentum to the system consisting of yourself, the water, and the earth and then differentiate this expression to relate your recoil acceleration to your mass, the speed of the water, and the rate at which the water is

Systems of Particles and Conservation of Momentum 619 leaving the hose. Use conservation of momentum to relate your recoil velocity to the velocity of the water leaving the hose: Differentiate this expression with respect to t:

r r p1 + p 2 = 0

or m1v1 + m2 v 2 = 0 m1

dv1 dm dv dm2 + v1 1 + m2 2 + v 2 =0 dt dt dt dt

or

m1 a1 + v1 Because the acceleration of the water leaving the hose, a1, is zero … as is

dm2 , the rate at which you are dt

losing mass:

Substitute numerical values and evaluate a2:

dm1 dm2 + ma 2 + v 2 =0 dt dt

dm1 + m2 a 2 = 0 dt and v dm1 a2 = − 1 m 2 dt v1

a2 = −

30 m/s (2.4 kg/s) 75 kg

= − 0.960 m/s 2 *140 ••• Picture the Problem Take the zero of gravitational potential energy to be at the elevation of the pan and let the system include the balance, the beads, and the earth. We can use conservation of energy to find the vertical component of the velocity of the beads as they hit the pan and then calculate the net downward force on the pan from Newton’s 2nd law.

Use conservation of energy to relate the y component of the bead’s velocity as it hits the pan to its height of fall: Solve for vy: Substitute numerical values and evaluate vy: Express the change in momentum in the y direction per bead:

∆K + ∆U = 0 or, because Ki = Uf = 0, 1 2

mv y2 − mgh = 0

v y = 2 gh

(

)

v y = 2 9.81 m/s 2 (0.5 m ) = 3.13 m/s

∆p y = p yf − p yi = mv y − (− mv y ) = 2mv y

620 Chapter 8 Use Newton’s 2nd law to express the net force in the y direction exerted on the pan by the beads:

Fnet, y = N

∆p y ∆t

∆p y

Letting M represent the mass to be placed on the other pan, equate its weight to the net force exerted by the beads, substitute for ∆py, and solve for M:

and

Substitute numerical values and evaluate M:

M = (100 / s )

Mg = N

M =

∆t

N ⎛ 2mv y ⎞ ⎜ ⎟ ∆t ⎜⎝ g ⎟⎠

[2(0.0005 kg )(3.13 m/s)] 9.81m/s 2

= 31.9 g 141 ••• Picture the Problem Assume that the connecting rod goes halfway through both balls, i.e., the centers of mass of the balls are separated by L. Let the system include the dumbbell, the wall and floor, and the earth. Let the zero of gravitational potential be at the center of mass of the lower ball and use conservation of energy to relate the speeds of the balls to the potential energy of the system. By symmetry, the speeds will be equal when the angle with the vertical is 45°.

Use conservation of energy to express the relationship between the initial and final energies of the system: Express the initial energy of the system: Express the energy of the system when the angle with the vertical is 45°:

Ei = E f

Ei = mgL

Ef = mgL sin 45° + 12 (2m ) v 2

Substitute to obtain:

⎛ 1 ⎞ 2 gL = gL⎜ ⎟+v ⎝ 2⎠

Solve for v:

1 ⎞ ⎛ v = gL⎜1 − ⎟ 2⎠ ⎝

Systems of Particles and Conservation of Momentum 621 Substitute numerical values and evaluate v:

v=

(9.81m/s )L⎛⎜1 − 2

(

)

⎝

= 1.70 m 2 /s L 1

1 ⎞ ⎟ 2⎠

622 Chapter 8

Chapter 9 Rotation Conceptual Problems *1 • Determine the Concept Because r is greater for the point on the rim, it moves the greater distance. Both turn through the same angle. Because r is greater for the point on the rim, it has the greater speed. Both have the same angular velocity. Both have zero tangential acceleration. Both have zero angular acceleration. Because r is greater for the point on the rim, it has the greater centripetal acceleration. 2

•

⎡1⎤

(a) False. Angular velocity has the dimensions ⎢ ⎥ whereas linear velocity has ⎣T ⎦

⎡L⎤ ⎣ ⎦

dimensions ⎢ ⎥ . T (b) True. The angular velocity of all points on the wheel is dθ/dt. (c) True. The angular acceleration of all points on the wheel is dω/dt. 3 •• Picture the Problem The constant-acceleration equation that relates the given variables is ω 2 = ω02 + 2α∆θ . We can set up a proportion to determine the number of revolutions required to double ω and then subtract to find the number of additional revolutions to accelerate the disk to an angular speed of 2ω. Using a constant-acceleration equation, relate the initial and final angular velocities to the angular acceleration:

ω 2 = ω02 + 2α∆θ or, because ω02 = 0,

ω 2 = 2α∆θ

Let ∆θ10 represent the number of revolutions required to reach an angular velocity ω:

ω 2 = 2α∆θ10

(1)

Let ∆θ2ω represent the number of revolutions required to reach an angular velocity ω:

(2ω)2 = 2α∆θ2ω

(2)

Divide equation (2) by equation (1) and solve for ∆θ2ω:

∆θ2ω = 623

(2ω)2 ∆θ ω2

10

= 4∆θ10

624 Chapter 9 The number of additional revolutions is:

4∆θ10 − ∆θ10 = 3∆θ10 = 3(10 rev ) = 30 rev and (c) is correct.

*4

•

⎡ ML2 ⎤ . 2 ⎥ ⎣T ⎦

Determine the Concept Torque has the dimension ⎢

⎡ ML ⎤ . ⎣ T ⎥⎦ ⎡ ML2 ⎤ (b) Energy has the dimension ⎢ 2 ⎥ . ⎣T ⎦

(a) Impulse has the dimension ⎢

(b) is correct.

⎡ ML ⎤ . ⎣ T ⎥⎦

(c) Momentum has the dimension ⎢

5 • Determine the Concept The moment of inertia of an object is the product of a constant that is characteristic of the object’s distribution of matter, the mass of the object, and the square of the distance from the object’s center of mass to the axis about which the object is rotating. Because both (b) and (c) are correct (d ) is correct. *6 • Determine the Concept Yes. A net torque is required to change the rotational state of an object. In the absence of a net torque an object continues in whatever state of rotational motion it was at the instant the net torque became zero. 7 • Determine the Concept No. A net torque is required to change the rotational state of an object. A net torque may decrease the angular speed of an object. All we can say for sure is that a net torque will change the angular speed of an object. 8 • (a) False. The net torque acting on an object determines the angular acceleration of the object. At any given instant, the angular velocity may have any value including zero. (b) True. The moment of inertia of a body is always dependent on one’s choice of an axis of rotation. (c) False. The moment of inertia of an object is the product of a constant that is characteristic of the object’s distribution of matter, the mass of the object, and the square of the distance from the object’s center of mass to the axis about which the object is

Rotation 625 rotating. 9 • Determine the Concept The angular acceleration of a rotating object is proportional to the net torque acting on it. The net torque is the product of the tangential force and its lever arm. Express the angular acceleration of the disk as a function of the net torque acting on it: Because α ∝ d , doubling d will double the angular acceleration.

α=

τ net

=

I i.e., α ∝ d

Fd F = d I I

(b) is correct.

*10 • Determine the Concept From the parallel-axis theorem we know that I = I cm + Mh 2 , where Icm is the moment of inertia of the object with respect to an axis through its center of mass, M is the mass of the object, and h is the distance between the parallel axes. Therefore, I is always greater than Icm by Mh2.

(d ) is correct.

11 • Determine the Concept The power delivered by the constant torque is the product of the torque and the angular velocity of the merry-go-round. Because the constant torque causes the merry-go-round to accelerate, neither the power input nor the angular velocity of the merry-go-round is constant. (b) is correct. 12 • Determine the Concept Let’s make the simplifying assumption that the object and the surface do not deform when they come into contact, i.e., we’ll assume that the system is rigid. A force does no work if and only if it is perpendicular to the velocity of an object, and exerts no torque on an extended object if and only if it’s directed toward the center of the object. Because neither of these conditions is satisfied, the statement is false. 13 • Determine the Concept For a given applied force, this increases the torque about the hinges of the door, which increases the door’s angular acceleration, leading to the door being opened more quickly. It is clear that putting the knob far from the hinges means that the door can be opened with less effort (force). However, it also means that the hand on the knob must move through the greatest distance to open the door, so it may not be the quickest way to open the door. Also, if the knob were at the center of the door, you would have to walk around the door after opening it, assuming the door is opening toward you.

626 Chapter 9 *14 • Determine the Concept If the wheel is rolling without slipping, a point at the top of the wheel moves with a speed twice that of the center of mass of the wheel, but the bottom of the wheel is momentarily at rest. (c) is correct. 15 •• Picture the Problem The kinetic energies of both objects is the sum of their translational and rotational kinetic energies. Their speed dependence will differ due to the differences in their moments of inertia. We can express the total kinetic of both objects and equate them to decide which of their translational speeds is greater. Express the kinetic energy of the cylinder:

2 2 K cyl = 12 I cylω cyl + 12 mvcyl

=

(

1 1 2 2

mr

2

2 vcyl

)r

2

2 + 12 mvcyl

2 = 34 mvcyl

Express the kinetic energy of the sphere:

2 2 K sph = 12 I sphω sphl + 12 mvsph

=

(

1 2 2 5

mr

2

2 vsph

)r

2

2 + 12 mvsph

2 = 107 mvsph

Equate the kinetic energies and simplify to obtain:

vcyl =

v

14 15 sph

< vsph

and (b) is correct.

*16 • Determine the Concept You could spin the pipes about their center. The one which is easier to spin has its mass concentrated closer to the center of mass and, hence, has a smaller moment of inertia. 17 •• Picture the Problem Because the coin and the ring begin from the same elevation, they will have the same kinetic energy at the bottom of the incline. The kinetic energies of both objects is the sum of their translational and rotational kinetic energies. Their speed dependence will differ due to the differences in their moments of inertia. We can express the total kinetic of both objects and equate them to their common potential energy loss to decide which of their translational speeds is greater at the bottom of the incline.

Rotation 627 Express the kinetic energy of the coin at the bottom of the incline:

2 2 K coin = 12 I cylω coin + 12 mcoin vcoin

=

(

1 1 2 2

mcoin r 2

) vr

2 coin 2

2 + 12 mcoin vcoin

2 = 34 mcoin vcoin

Express the kinetic energy of the ring at the bottom of the incline:

2 2 K ring = 12 I ringω ring + 12 mring vring

=

1 2

(m

ring

r

2

2 vring

)r

2

2 + 12 mring vring

2 = mring vring

Equate the kinetic of the coin to its change in potential energy as it rolled down the incline and solve for vcoin: Equate the kinetic of the ring to its change in potential energy as it rolled down the incline and solve for vring:

3 4

2 mcoin v coin = mcoin gh

and 2 v coin = 43 gh

2 mring v ring = mring gh

and 2 v ring = gh

Therefore, vcoin > vring and (b) is correct. 18 •• Picture the Problem We can use the definitions of the translational and rotational kinetic energies of the hoop and the moment of inertia of a hoop (ring) to express and compare the kinetic energies. Express the translational kinetic energy of the hoop:

K trans = 12 mv 2

Express the rotational kinetic energy of the hoop:

K rot = 12 I hoopω 2 =

1 2

(mr ) vr 2

2 2

= 12 mv 2

Therefore, the translational and rotational kinetic energies are the same and

(c) is correct.

628 Chapter 9 19 •• Picture the Problem We can use the definitions of the translational and rotational kinetic energies of the disk and the moment of inertia of a disk (cylinder) to express and compare the kinetic energies. Express the translational kinetic energy of the disk:

K trans = 12 mv 2

Express the rotational kinetic energy of the disk:

K rot = 12 I hoopω 2 =

(

1 1 2 2

mr 2

) vr

2 2

= 14 mv 2

Therefore, the translational kinetic energy is greater and

(a ) is correct.

20 •• Picture the Problem Let us assume that f ≠ 0 and acts along the direction of motion. Now consider the acceleration of the center of mass and the angular acceleration about the point of contact with the plane. Because Fnet ≠ 0, acm ≠ 0. However, τ = 0 because l = 0, so α = 0. But α = 0 is not consistent with acm ≠ 0. Consequently, f = 0. 21 • Determine the Concept True. If the sphere is slipping, then there is kinetic friction which dissipates the mechanical energy of the sphere. 22 • Determine the Concept Because the ball is struck high enough to have topspin, the frictional force is forward; reducing ω until the nonslip condition is satisfied.

(a ) is correct.

Estimation and Approximation 23 •• Picture the Problem Assume the wheels are hoops, i.e., neglect the mass of the spokes, and express the total kinetic energy of the bicycle and rider. Let M represent the mass of the rider, m the mass of the bicycle, mw the mass of each bicycle wheel, and r the radius of the wheels. Express the ratio of the kinetic energy associated with the rotation of the wheels to that associated with the total kinetic energy of the bicycle and rider:

K rot K rot = K tot K trans + K rot

(1)

Rotation 629 Express the translational kinetic energy of the bicycle and rider:

K trans = K bicycle + K rider

Express the rotational kinetic energy of the bicycle wheels:

K rot = 2 K rot, 1 wheel = 2 12 I wω 2

= 12 mv 2 + 12 Mv 2

(

(

)

)

v2 = mw r 2 = mw v 2 r 2

Substitute in equation (1) to obtain:

K rot m v2 mw 2 = 1 2 1 w 2 = 1 = 2 1 m+M K tot 2 mv + 2 Mv + mw v 2 m + 2 M + mw 2+ mw K rot 2 Substitute numerical values and = 10.3 % = K tot 2 + 14 kg + 38 kg evaluate Krot/Ktot: 3 kg 24 •• Picture the Problem We can apply the definition of angular velocity to find the angular orientation of the slice of toast when it has fallen a distance of 0.5 m from the edge of the table. We can then interpret the orientation of the toast to decide whether it lands jellyside up or down. Relate the angular orientation θ of the toast to its initial angular orientation, its angular velocity ω, and time of fall ∆t:

θ = θ 0 + ω∆t

Use the equation given in the problem statement to find the angular velocity corresponding to this length of toast:

9.81 m/s 2 ω = 0.956 = 9.47 rad/s 0.1m

Using a constant-acceleration equation, relate the distance the toast falls ∆y to its time of fall ∆t:

∆y = v0 y ∆t + 12 a y (∆t )

Solve for ∆t:

Substitute numerical values and evaluate ∆t:

(1)

2

or, because v0y = 0 and ay = g,

∆y = 12 g (∆t )

2

∆t =

2∆y g

∆t =

2(0.5 m ) = 0.319 s 9.81 m/s 2

630 Chapter 9

(vf ')2 + cos θ

2 gL find θ :

0 Substitute in equation (1) to

θ=

π 6

+ (9.47 rad/s )(0.319 s )

= 3.54 rad ×

180° = 203° π rad

The orientation of the slice of toast will therefore be at an angle of 203° with respect to the ground, i.e. with the jelly - side down. *25 •• Picture the Problem Assume that the mass of an average adult male is about 80 kg, and that we can model his body when he is standing straight up with his arms at his sides as a cylinder. From experience in men’s clothing stores, a man’s average waist circumference seems to be about 34 inches, and the average chest circumference about 42 inches. We’ll also assume that about 20% of the body’s mass is in the two arms, and each has a length L = 1 m, so that each arm has a mass of about m = 8 kg. Letting Iout represent his moment of inertia with his arms straight out and Iin his moment of inertia with his arms at his side, the ratio of these two moments of inertia is:

I out I body + I arms = I in I in

Express the moment of inertia of the ″man as a cylinder″:

I in = 12 MR 2

Express the moment of inertia of his arms:

I arms = 2( 13 )mL2

Express the moment of inertia of his body-less-arms:

I body =

1 2

(M − m )R 2

Substitute in equation (1) to obtain:

I out = I in

1 2

(M − m )R 2 + 2( 13 )mL2

Assume the circumference of the cylinder to be the average of the average waist circumference and the average chest circumference: Find the radius of a circle whose circumference is 38 in:

cav =

1 2

MR 2

34 in + 42 in = 38 in 2

38 in ×

cav = 2π = 0.154 m

R=

Substitute numerical values and evaluate Iout/ Iin:

(1)

2.54 cm 1m × in 100 cm 2π

Rotation 631 I out = I in

1 2

(80 kg − 16 kg )(0.154 m )2 + 23 (8 kg )(1m )2 2 1 2 (80 kg )(0.154 m )

= 6.42

Angular Velocity and Angular Acceleration 26 • Picture the Problem The tangential and angular velocities of a particle moving in a circle are directly proportional. The number of revolutions made by the particle in a given time interval is proportional to both the time interval and its angular speed. (a) Relate the angular velocity of the particle to its speed along the circumference of the circle:

v = rω

Solve for and evaluate ω:

ω=

(b) Using a constant-acceleration equation, relate the number of revolutions made by the particle in a given time interval to its angular velocity:

⎛ 1 rev ⎞ rad ⎞ ⎛ ⎟⎟ ∆θ = ω ∆t = ⎜ 0.278 ⎟ (30 s )⎜⎜ s ⎠ ⎝ ⎝ 2π rad ⎠

v 25 m/s = = 0.278 rad/s r 90 m

= 1.33 rev

27 • Picture the Problem Because the angular acceleration is constant; we can find the various physical quantities called for in this problem by using constant-acceleration equations. (a) Using a constant-acceleration equation, relate the angular velocity of the wheel to its angular acceleration and the time it has been accelerating:

ω = ω 0 + α∆t or, when ω0 = 0,

ω = α∆t

Evaluate ω when ∆t = 6 s:

ω = ⎛⎜ 2.6 rad/s2 ⎞⎟ (6 s ) = 15.6 rad/s

(b) Using another constantacceleration equation, relate the angular displacement to the wheel’s angular acceleration and the time it

∆θ = ω 0 ∆t + 12 α (∆t )

⎝

⎠

or, when ω0 = 0,

∆θ = 12 α (∆t )

2

2

632 Chapter 9 has been accelerating: Evaluate ∆θ when ∆t = 6 s:

(c) Convert ∆θ (6 s ) from rad to revolutions:

∆θ (6 s ) =

1 2

(2.6 rad/s )(6 s )

2

2

∆θ (6 s ) = 46.8 rad ×

= 46.8 rad

1 rev = 7.45 rev 2π rad

(d) Relate the angular velocity of the particle to its tangential speed and evaluate the latter when ∆t = 6 s:

v = rω = (0.3 m )(15.6 rad/s ) = 4.68 m/s

Relate the resultant acceleration of the point to its tangential and centripetal accelerations when ∆t = 6 s:

a = at2 + ac2 =

Substitute numerical values and evaluate a:

a = (0.3 m ) (2.6 rad/s 2 ) + (15.6 rad/s )

(rα )2 + (rω 2 )2

= r α 2 + ω4

2

4

= 73.0 m/s 2

*28 • Picture the Problem Because we’re assuming constant angular acceleration; we can find the various physical quantities called for in this problem by using constant-acceleration equations. (a) Using its definition, express the angular acceleration of the turntable: Substitute numerical values and evaluate α:

α=

α=

∆ω ω − ω0 = ∆t ∆t

0 − 33 13

rev 2π rad 1 min × × min rev 60 s 26 s

= 0.134 rad/s 2

Rotation 633 (b) Because the angular acceleration is constant, the average angular velocity is the average of its initial and final values:

ωav =

=

ω0 + ω 2 33 13

rev 2π rad 1 min × × min rev 60 s 2

= 1.75 rad/s (c) Using the definition of ωav, find the number or revolutions the turntable makes before stopping:

∆θ = ωav ∆t = (1.75 rad/s )(26 s ) = 45.5 rad ×

1 rev = 7.24 rev 2π rad

29 • Picture the Problem Because the angular acceleration of the disk is constant, we can use a constant-acceleration equation to relate its angular velocity to its acceleration and the time it has been accelerating. We can find the tangential and centripetal accelerations from their relationships to the angular velocity and angular acceleration of the disk. (a) Using a constant-acceleration equation, relate the angular velocity of the disk to its angular acceleration and time during which it has been accelerating:

ω = ω 0 + α ∆t or, because ω0 = 0,

ω = α ∆t

Evaluate ω when t = 5 s:

ω (5 s ) = (8 rad/s 2 )(5 s ) = 40.0 rad/s

(b) Express at in terms of α:

a t = rα

Evaluate at when t = 5 s:

at (5 s ) = (0.12 m ) 8 rad/s 2

(

)

= 0.960 m/s 2 Express ac in terms of ω:

a c = rω 2

Evaluate ac when t = 5 s:

ac (5 s ) = (0.12 m )(40.0 rad/s )

2

= 192 m/s 2 30 • Picture the Problem We can find the angular velocity of the Ferris wheel from its definition and the linear speed and centripetal acceleration of the passenger from the relationships between those quantities and the angular velocity of the Ferris wheel.

634 Chapter 9 (a) Find ω from its definition:

(b) Find the linear speed of the passenger from his/her angular speed: Find the passenger’s centripetal acceleration from his/her angular velocity:

ω=

∆θ 2π rad = = 0.233 rad/s ∆t 27 s

v = rω = (12 m )(0.233 rad/s ) = 2.79 m/s ac = rω 2 = (12 m )(0.233 rad/s )

2

= 0.651 m/s 2

31 • Picture the Problem Because the angular acceleration of the wheels is constant, we can use constant-acceleration equations in rotational form to find their angular acceleration and their angular velocity at any given time. (a) Using a constant-acceleration equation, relate the angular displacement of the wheel to its angular acceleration and the time it has been accelerating:

∆θ = ω 0 ∆t + 12 α (∆t )

2

or, because ω0 = 0,

∆θ = 12 α (∆t )

2

2∆θ (∆t )2

Solve for α:

α=

Substitute numerical values and evaluate α:

⎛ 2π rad ⎞ 2(3 rev )⎜ ⎟ rev ⎠ ⎝ = 0.589 rad/s 2 α= 2 (8 s )

(b) Using a constant-acceleration equation, relate the angular velocity of the wheel to its angular acceleration and the time it has been accelerating:

ω = ω 0 + α∆t

Evaluate ω when ∆t = 8 s:

or, when ω0 = 0,

ω = α∆t

ω (8 s ) = (0.589 rad/s 2 )(8 s ) = 4.71 rad/s

Rotation 635 32 • Picture the Problem The earth rotates through 2π radians every 24 hours. Find ω using its definition:

ω≡

∆θ 2π rad = ∆t 24 h × 3600 s h

= 7.27 × 10 −5 rad/s 33 • Picture the Problem When the angular acceleration of a wheel is constant, its average angular velocity is the average of its initial and final angular velocities. We can combine this relationship with the always applicable definition of angular velocity to find the initial angular velocity of the wheel. Express the average angular velocity of the wheel in terms of its initial and final angular speeds:

ω0 + ω

ω av =

2

or, because ω = 0,

ω av = 12 ω 0

Express the definition of the average angular velocity of the wheel:

ω av ≡

Equate these two expressions and solve for ω0:

ω0 =

∆θ ∆t 2∆θ 2(5 rad ) = = 3.57 s and ∆t 2.8 s

(d ) is correct. 34 • Picture the Problem The tangential and angular accelerations of the wheel are directly proportional to each other with the radius of the wheel as the proportionality constant. Provided there is no slippage, the acceleration of a point on the rim of the wheel is the same as the acceleration of the bicycle. We can use its defining equation to determine the acceleration of the bicycle. Relate the tangential acceleration of a point on the wheel (equal to the acceleration of the bicycle) to the wheel’s angular acceleration and solve for its angular acceleration:

a = a t = rα and

α=

a r

636 Chapter 9 Use its definition to express the acceleration of the wheel:

a=

∆v v − v0 = ∆t ∆t

or, because v0 = 0,

a=

v ∆t

Substitute in the expression for α to obtain:

α=

v r∆t

Substitute numerical values and evaluate α:

⎛ km ⎞ ⎛ 1 h ⎞ ⎛ 1000 m ⎞ ⎟⎜ ⎜ 24 ⎟⎜ ⎟ h ⎠ ⎜⎝ 3600 s ⎟⎠ ⎝ km ⎠ ⎝ α= (0.6 m )(14.0 s ) = 0.794 rad/s 2

*35 •• Picture the Problem The two tapes will have the same tangential and angular velocities when the two reels are the same size, i.e., have the same area. We can calculate the tangential speed of the tape from its length and running time and relate the angular velocity to the constant tangential speed and the radius of the reels when they are turning with the same angular velocity. Relate the angular velocity of the tape to its tangential speed: Letting Rf represent the outer radius of the reel when the reels have the same area, express the condition that they have the same speed: Solve for Rf:

Substitute numerical values and evaluate Rf: Find the tangential speed of the tape from its length and running time:

ω=

v r

(1)

π Rf2 − π r 2 = 12 (π R 2 − π r 2 )

Rf =

Rf =

R2 + r 2 2

(45 mm)2 + (12 mm)2

L v= = ∆t

2

= 32.9 mm

100 cm m = 3.42 cm/s 3600 s 2h × h

246 m ×

Rotation 637 Substitute in equation (1) and evaluate ω:

ω=

3.42 cm/s 1 cm 32.9 mm × 10 mm

v = Rf

= 1.04 rad/s Convert 1.04 rad/s to rev/min:

1.04 rad/s = 1.04

rad 1 rev 60 s × × s 2π rad min

= 9.93 rev/min

Torque, Moment of Inertia, and Newton’s Second Law for Rotation 36 • Picture the Problem The force that the woman exerts through her axe, because it does not act at the axis of rotation, produces a net torque that changes (decreases) the angular velocity of the grindstone. (a) From the definition of angular acceleration we have:

Substitute numerical values and evaluate α:

∆ω ω − ω0 = ∆t ∆t or, because ω = 0, − ω0 α= ∆t

α=

730

α =−

rev 2π rad 1 min × × min rev 60 s 9s

= − 8.49 rad/s 2 where the minus sign means that the grindstone is slowing down. (b) Use Newton’s 2nd law in rotational form to relate the angular acceleration of the grindstone to the net torque slowing it:

τ net = Iα

Express the moment of inertia of disk with respect to its axis of rotation:

I = 12 MR 2

638 Chapter 9 Substitute to obtain: Substitute numerical values and evaluate τnet:

τ net = 12 MRα τ net = 12 (1.7 kg )(0.08 m )2 (8.49 rad/s 2 ) = 0.0462 N ⋅ m

*37 • Picture the Problem We can find the torque exerted by the 17-N force from the definition of torque. The angular acceleration resulting from this torque is related to the torque through Newton’s 2nd law in rotational form. Once we know the angular acceleration, we can find the angular velocity of the cylinder as a function of time. (a) Calculate the torque from its definition:

τ = Fl = (17 N )(0.11 m ) = 1.87 N ⋅ m

(b) Use Newton’s 2nd law in rotational form to relate the acceleration resulting from this torque to the torque:

α=

Express the moment of inertia of the cylinder with respect to its axis of rotation:

I = 12 MR 2

Substitute to obtain:

α=

2τ MR 2

Substitute numerical values and evaluate α:

α=

2(1.87 N ⋅ m ) = 124 rad/s 2 (2.5 kg )(0.11m )2

(c) Using a constant-acceleration equation, express the angular velocity of the cylinder as a function of time:

ω = ω0 + α t

Evaluate ω (5 s):

τ I

or, because ω0 = 0,

ω =αt

ω (5 s ) = (124 rad/s 2 )(5 s ) = 620 rad/s

38 •• Picture the Problem We can find the angular acceleration of the wheel from its definition and the moment of inertia of the wheel from Newton’s 2nd law.

Rotation 639 (a) Express the moment of inertia of the wheel in terms of the angular acceleration produced by the applied torque: Find the angular acceleration of the wheel:

I=

τ α

∆ω = ∆t

α=

600

rev 2π rad 1 min × × min rev 60 s 20 s

= 3.14 rad/s 2 50 N ⋅ m = 15.9 kg ⋅ m 2 3.14 rad/s 2

Substitute and evaluate I:

I=

(b) Because the wheel takes 120 s to slow to a stop (it took 20 s to acquire an angular velocity of 600 rev/min) and its angular acceleration is directly proportional to the accelerating torque:

τ fr = 16 τ =

1 6

(50 N ⋅ m ) =

39 •• Picture the Problem The pendulum and the forces acting on it are shown in the free-body diagram. Note that the tension in the string is radial, and so exerts no tangential force on the ball. We can use Newton’s 2nd law in both translational and rotational form to find the tangential component of the acceleration of the bob. (a) Referring to the FBD, express r the component of mg that is tangent

Ft = mg sin θ

to the circular path of the bob: Use Newton’s 2nd law to express the tangential acceleration of the bob:

at =

(b) Noting that, because the line-ofaction of the tension passes through the pendulum’s pivot point, its lever arm is zero and the net torque is due

∑τ

Ft = g sin θ m pivot point

= mgL sin θ

8.33 N ⋅ m

640 Chapter 9 to the weight of the bob, sum the torques about the pivot point to obtain: (c) Use Newton’s 2nd law in rotational form to relate the angular acceleration of the pendulum to the net torque acting on it:

τ net = mgL sin θ = Iα

Solve for α to obtain:

α=

Express the moment of inertia of the bob with respect to the pivot point:

I = mL2

Substitute to obtain:

α=

Relate α to at:

⎛ g sin θ ⎞ a t = rα = L ⎜ ⎟ = g sin θ ⎝ L ⎠

mgL sin θ I

mgL sin θ g sin θ = mL2 L

*40 ••• Picture the Problem We can express the velocity of the center of mass of the rod in terms of its distance from the pivot point and the angular velocity of the rod. We can find the angular velocity of the rod by using Newton’s 2nd law to find its angular acceleration and then a constant-acceleration equation that relates ω to α. We’ll use the impulsemomentum relationship to derive the expression for the force delivered to the rod by the pivot. Finally, the location of the center of percussion of the rod will be verified by setting the force exerted by the pivot to zero.

L ω 2

(a) Relate the velocity of the center of mass to its distance from the pivot point:

vcm =

Express the torque due to F0:

τ = F0 x = I pivotα

Solve for α:

α=

Express the moment of inertia of the rod with respect to an axis through

I pivot = 13 ML2

F0 x I pivot

(1)

Rotation 641 its pivot point:

3F0 x ML2

Substitute to obtain:

α=

Express the angular velocity of the rod in terms of its angular acceleration:

ω = α ∆t =

Substitute in equation (1) to obtain:

(b) Let IP be the impulse exerted by the pivot on the rod. Then the total impulse (equal to the change in momentum of the rod) exerted on the rod is:

vcm =

3F0 x∆t ML2

3F0 x∆t 2ML

I P + F0 ∆t = Mvcm and

I P = Mvcm − F0 ∆t

3F0 x∆t ⎛ 3x ⎞ − F0 ∆t = F0 ∆t ⎜ − 1⎟ 2L ⎝ 2L ⎠

Substitute our result from (a) to obtain:

IP =

Because I P = FP ∆t :

⎛ 3x ⎞ − 1⎟ FP = F0 ⎜ ⎝ 2L ⎠

In order for FP to be zero:

2L 3x −1 = 0 ⇒ x = 3 2L

41 ••• Picture the Problem We’ll first express the torque exerted by the force of friction on the elemental disk and then integrate this expression to find the torque on the entire disk. We’ll use Newton’s 2nd law to relate this torque to the angular acceleration of the disk and then to the stopping time for the disk. (a) Express the torque exerted on the elemental disk in terms of the friction force and the distance to the elemental disk:

dτ f = rdf k

(1)

Using the definition of the coefficient of friction, relate the

df k = µ k gdm

(2)

642 Chapter 9 force of friction to µk and the weight of the circular element: Letting σ represent the mass per unit area of the disk, express the mass of the circular element:

dm = 2π rσ dr

(3)

Substitute equations (2) and (3) in (1) to obtain:

dτ f = 2π µ kσ g r 2 dr

(4)

Because σ =

M : π R2

(b) Integrate dτ f to obtain the total torque on the elemental disk: (c) Relate the disk’s stopping time to its angular velocity and acceleration: Using Newton’s 2nd law, express α in terms of the net torque acting on the disk:

τf =

2 µk M g 2 ∫0 r dr = R2

∆t =

ω α

α=

R

2 3

MRµ k g

τf I

I = 12 MR 2

The moment of inertia of the disk, with respect to its axis of rotation, is: Substitute and simplify to obtain:

2 µk M g 2 r dr R2

dτ f =

∆t =

3 Rω 4µ k g

Calculating the Moment of Inertia 42 • Picture the Problem One can find the formula for the moment of inertia of a thin spherical shell in Table 9-1. The moment of inertia of a thin spherical shell about its diameter is:

I = 23 MR 2

Rotation 643 Substitute numerical values and evaluate I:

I=

(0.057 kg )(0.035 m )2

2 3

= 4.66 × 10 −5 kg ⋅ m 2

*43 • Picture the Problem The moment of inertia of a system of particles with respect to a given axis is the sum of the products of the mass of each particle and the square of its distance from the given axis. Use the definition of the moment of inertia of a system of particles to obtain: Substitute numerical values and evaluate I:

I = ∑ mi ri 2 i

= m1r12 + m2 r22 + m3 r32 + m4 r42

(

I = (3 kg )(2 m ) + (4 kg ) 2 2 m 2

)

2

+ (4 kg )(0) + (3 kg )(2 m ) 2

2

= 56.0 kg ⋅ m 2 44 • Picture the Problem Note, from symmetry considerations, that the center of mass of the system is at the intersection of the diagonals connecting the four masses. Thus the distance of each particle from the axis through the center of mass is 2 m . According to the parallel-axis theorem, I = I cm + Mh 2 , where Icm is the moment of inertia of the object with respect to an axis through its center of mass, M is the mass of the object, and h is the distance between the parallel axes. Express the parallel axis theorem:

I = I cm + Mh 2

Solve for Icm and substitute from Problem 44:

I cm = I − Mh 2

(

= 56.0 kg ⋅ m 2 − (14 kg ) 2 m

)

2

= 28.0 kg ⋅ m 2 Use the definition of the moment of inertia of a system of particles to express Icm: Substitute numerical values and evaluate Icm:

I cm = ∑ mi ri 2 i

= m1r12 + m2 r22 + m3 r32 + m4 r42

(

)

(

I cm = (3 kg ) 2 m + (4 kg ) 2 m

(

2

)

(

)

2

+ (4 kg ) 2 m + (3 kg ) 2 m = 28.0 kg ⋅ m 2

2

)

2

644 Chapter 9 45 • Picture the Problem The moment of inertia of a system of particles with respect to a given axis is the sum of the products of the mass of each particle and the square of its distance from the given axis. (a) Apply the definition of the moment of inertia of a system of particles to express Ix: Substitute numerical values and evaluate Ix:

I x = ∑ mi ri 2 i

= m1r12 + m2 r22 + m3 r32 + m4 r42 I x = (3 kg )(2 m ) + (4 kg )(2 m ) 2

2

+ (4 kg )(0 ) + (3 kg )(0)

= 28.0 kg ⋅ m 2 (b) Apply the definition of the moment of inertia of a system of particles to express Iy: Substitute numerical values and evaluate Iy:

I y = ∑ mi ri 2 i

= m1r12 + m2 r22 + m3 r32 + m4 r42 I y = (3 kg )(0) + (4 kg )(2 m )

2

+ (4 kg )(0) + (3 kg )(2 m )

2

= 28.0 kg ⋅ m 2 Remarks: We could also use a symmetry argument to conclude that Iy = Ix . 46 • Picture the Problem According to the parallel-axis theorem, I = I cm + Mh 2 , where Icm is the moment of inertia of the object with respect to an axis through its center of mass, M is the mass of the object, and h is the distance between the parallel axes. Use Table 9-1 to find the moment of inertia of a sphere with respect to an axis through its center of mass:

I cm = 52 MR 2

Express the parallel axis theorem:

I = I cm + Mh 2

Substitute for Icm and simplify to obtain:

I = 52 MR 2 + MR 2 =

7 5

MR 2

Rotation 645 47 •• Picture the Problem The moment of inertia of the wagon wheel is the sum of the moments of inertia of the rim and the six spokes. Express the moment of inertia of the wagon wheel as the sum of the moments of inertia of the rim and the spokes:

I wheel = I rim + I spokes

Using Table 9-1, find formulas for the moments of inertia of the rim and spokes:

I rim = M rim R 2 and

Substitute to obtain:

I wheel = M rim R 2 + 6 13 M spoke L2

I spoke = 13 M spoke L2

(

)

= M rim R 2 + 2M spoke L2 Substitute numerical values and evaluate Iwheel:

I wheel = (8 kg )(0.5 m ) + 2(1.2 kg )(0.5 m ) 2

2

= 2.60 kg ⋅ m 2

*48 •• Picture the Problem The moment of inertia of a system of particles depends on the axis with respect to which it is calculated. Once this choice is made, the moment of inertia is the sum of the products of the mass of each particle and the square of its distance from the chosen axis. (a) Apply the definition of the moment of inertia of a system of particles:

I = ∑ mi ri 2 = m1 x 2 + m2 (L − x )

(b) Set the derivative of I with respect to x equal to zero in order to identify values for x that correspond to either maxima or minima:

dI = 2m1 x + 2m2 (L − x )(− 1) dx = 2(m1 x + m2 x − m2 L )

If

dI = 0 , then: dx

2

i

= 0 for extrema m1 x + m2 x − m2 L = 0

646 Chapter 9 Solve for x:

x=

Convince yourself that you’ve found 2

a minimum by showing that

d I is dx 2

m2 L m1 + m2

x=

m2 L is, by definition, the m1 + m2

distance of the center of mass from m.

positive at this point.

49 •• Picture the Problem Let σ be the mass per unit area of the uniform rectangular plate. Then the elemental unit has mass dm = σ dxdy. Let the corner of the plate through which the axis runs be the origin. The distance of the element whose mass is dm from the corner r is related to the coordinates of dm through the Pythagorean relationship r2 = x2 + y2. (a) Express the moment of inertia of the element whose mass is dm with respect to an axis perpendicular to it and passing through one of the corners of the uniform rectangular plate: Integrate this expression to find I:

dI = σ (x 2 + y 2 )dxdy

a b

(

)

I = σ ∫ ∫ x 2 + y 2 dxdy 0 0

(

)

= 13 σ a 3b + ab 3 = (b) Letting d represent the distance from the origin to the center of mass of the plate, use the parallel axis theorem to relate the moment of inertia found in (a) to the moment of inertia with respect to an axis through the center of mass: Using the Pythagorean theorem, relate the distance d to the center of

1 3

(

m a 2 + b3

)

I = I cm + md 2 or

(

)

I cm = I − md 2 = 13 m a 2 + b 2 − md 2

d 2 = ( 12 a ) + ( 12 b ) = 2

2

1 4

(a

2

+ b2

)

Rotation 647 mass to the lengths of the sides of the plate: Substitute for d2 in the expression for Icm and simplify to obtain:

(

)

(

I cm = 13 m a 2 + b 2 − 14 m a 2 + b 2 =

1 12

(

m a2 + b2

)

)

2

*50 •• Picture the Problem Corey will use the point-particle relationship I app = mi ri 2 = m1r12 + m2 r22 for his calculation whereas Tracey’s calculation will take

∑ i

into account not only the rod but also the fact that the spheres are not point particles. (a) Using the point-mass approximation and the definition of the moment of inertia of a system of particles, express Iapp:

I app = ∑ mi ri 2 = m1r12 + m2 r22 i

Substitute numerical values and evaluate Iapp:

I app = (0.5 kg )(0.2 m ) + (0.5 kg )(0.2 m )

Express the moment of inertia of the two spheres and connecting rod system:

I = I spheres + I rod

Use Table 9-1 to find the moments of inertia of a sphere (with respect to its center of mass) and a rod (with respect to an axis through its center of mass):

I sphere = 52 M sphere R 2

Because the spheres are not on the axis of rotation, use the parallel axis theorem to express their moment of inertia with respect to the axis of rotation: Substitute to obtain: Substitute numerical values and evaluate I:

2

2

= 0.0400 kg ⋅ m 2

and

I rod = 121 M rod L2

I sphere = 52 M sphere R 2 + M sphere h 2 where h is the distance from the center of mass of a sphere to the axis of rotation.

{

}

I = 2 52 M sphere R 2 + M sphere h 2 + 121 M rod L2

648 Chapter 9

{ (0.5 kg )(0.05 m)

I =2

2 5

2

}

+ (0.5 kg )(0.2 m ) + 121 (0.06 kg )(0.3 m ) 2

2

= 0.0415 kg ⋅ m 2 Compare I and Iapp by taking their ratio:

(b)

I app

0.0400 kg ⋅ m 2 = = 0.964 I 0.0415 kg ⋅ m 2

The rotational inertia would increase because I cm of a hollow sphere is greater than I cm of a solid sphere.

51 •• Picture the Problem The axis of rotation passes through the center of the base of the tetrahedron. The carbon atom and the hydrogen atom at the apex of the tetrahedron do not contribute to I because the distance of their nuclei from the axis of rotation is zero. From the geometry, the distance of the three H nuclei from the rotation axis is a / 3 , where a is the length of a side of the tetrahedron. Apply the definition of the moment of inertia for a system of particles to obtain:

I = ∑ mi ri 2 = mH r12 + mH r22 + mH r32

Substitute numerical values and evaluate I:

I = 1.67 × 10−27 kg 0.18 × 10 −9 m

52 •• Picture the Problem Let the mass of the element of volume dV be dm = ρdV = 2πρhrdr where h is the height of the cylinder. We’ll begin by expressing the moment of inertia dI for the element of volume and then integrating it between R1 and R2.

i

2

⎛ a ⎞ = 3mH ⎜ ⎟ = mH a 2 ⎝ 3⎠

(

)(

= 5.41× 10−47 kg ⋅ m 2

)

2

Rotation 649 Express the moment of inertia of the element of mass dm: Integrate dI from R1 to R2 to obtain:

dI = r 2 dm = 2πρ hr 3 dr

R2

I = 2πρ h ∫ r 3dr = 12 πρ h(R24 − R14 ) R1

= 12 πρ h(R22 − R12 )(R22 + R12 ) The mass of the hollow cylinder is m = π ρ h R22 − R12 , so:

(

ρ=

)

m π h R22 − R12

(

)

Substitute for ρ and simplify to obtain:

⎛ m I = 12 π ⎜⎜ 2 2 ⎝ π h R2 − R1

(

⎞ ⎟⎟ h R22 − R12 R22 + R12 = ⎠

) (

)(

)

1 2

(

m R22 + R12

)

53 ••• Picture the Problem We can derive the given expression for the moment of inertia of a spherical shell by following the procedure outlined in the problem statement. Find the moment of inertia of a sphere, with respect to an axis through a diameter, in Table 9-1:

I = 52 mR 2

Express the mass of the sphere as a function of its density and radius:

m = 43 π ρ R 3

Substitute to obtain:

I = 158 π ρ R 5

Express the differential of this expression:

dI = 83 π ρ R 4 dR

(1)

Express the increase in mass dm as the radius of the sphere increases by dR:

dm = 4π ρ R 2 dR

(2)

Eliminate dR between equations (1) and (2) to obtain:

dI = 23 R 2 dm Therefore, the moment of inertia of the spherical shell of mass m is 23 mR 2 .

650 Chapter 9 *54 ••• Picture the Problem We can find C in terms of M and R by integrating a spherical shell of mass dm with the given density function to find the mass of the earth as a function of M and then solving for C. In part (b), we’ll start with the moment of inertia of the same spherical shell, substitute the earth’s density function, and integrate from 0 to R. (a) Express the mass of the earth using the given density function:

R

M = ∫ dm = ∫ 4π ρ r 2 dr 0

4π C 3 r dr = 4π C ∫ 1.22r dr − R ∫0 0 R

R

2

=

4π 1.22CR 3 − π CR 3 3 M R3

Solve for C as a function of M and R to obtain:

C = 0.508

(b) From Problem 9-40 we have:

dI = 83 π ρ r 4 dr

Integrate to obtain:

I = 83 π ∫ ρ r 4 dr

R

0

R R 8π (0.508)M ⎡ 1 5 ⎤ 4 − 1 . 22 r dr r dr ⎥ ⎢∫ ∫ 3R 3 R 0 0 ⎣ ⎦ 4.26M ⎡1.22 5 1 5 ⎤ = R − R ⎥ 6 ⎦ R 3 ⎢⎣ 5

=

= 0.329MR 2

Rotation 651 55 ••• Picture the Problem Let the origin be at the apex of the cone, with the z axis along the cone’s symmetry axis. Then the radius of the elemental ring, at a distance z from the apex, can be obtained from the

r R = . The mass dm of the z H elemental disk is ρdV = ρπr2dz. We’ll

proportion

integrate r2dm to find the moment of inertia of the disk in terms of R and H and then integrate dm to obtain a second equation in R and H that we can use to eliminate H in our expression for I. Express the moment of inertia of the cone in terms of the moment of inertia of the elemental disk:

I=

1 2

=

1 2

∫r

2

dm

R2 2 ⎛ R2 2 ⎞ ⎜ z ρπ ∫0 H 2 ⎜⎝ H 2 z ⎟⎟⎠dz

H

=

πρ R 4 2H 4

H

4 ∫ z dz =

πρ R 4 H 10

0

H

H

Express the total mass of the cone in terms of the mass of the elemental disk:

M = πρ ∫ r 2 dz = πρ ∫

Divide I by M, simplify, and solve for I to obtain:

I=

56 ••• Picture the Problem Let the axis of rotation be the x axis. The radius r of the elemental area is

R 2 − z 2 and its mass,

dm, is σ dA = 2σ R 2 − z 2 dz . We’ll integrate z2 dm to determine I in terms of σ and then divide this result by M in order to eliminate σ and express I in terms of M and R.

0

0

= 13 πρ R H 2

3 10

MR 2

R2 2 z dz H2

652 Chapter 9 Express the moment of inertia about the x axis:

I = ∫ z 2 dm = ∫ z 2σ dA R

=

∫z

2

(2σ

R 2 − z 2 dz

)

−R

= 14 σπR 4 The mass of the thin uniform disk is: Divide I by M, simplify, and solve for I to obtain:

M = σπR 2 I=

1 4

MR 2 , a result in agreement with

the expression given in Table 9-1 for a cylinder of length L = 0.

57 ••• Picture the Problem Let the origin be at the apex of the cone, with the z axis along the cone’s symmetry axis, and the axis of rotation be the x rotation. Then the radius of the elemental disk, at a distance z from the apex, can be obtained from the

r R = . The mass dm of the z H elemental disk is ρdV = ρπr2dz. Each

proportion

elemental disk rotates about an axis that is parallel to its diameter but removed from it by a distance z. We can use the result from Problem 9-57 for the moment of inertia of the elemental disk with respect to a diameter and then use the parallel axis theorem to express the moment of inertia of the cone with respect to the x axis. Using the parallel axis theorem, express the moment of inertia of the elemental disk with respect to the x axis: In Problem 9-57 it was established that the moment of inertia of a thin uniform disk of mass M and radius R rotating about a diameter is 14 MR 2 . Express this result in

dI x = dI disk + dm z 2

(1)

where

dm = ρ dV = ρπ r 2 dz dI disk =

1 4

(ρπ r dz )r 2

2 2

⎛ R2 ⎞ = ρπ ⎜⎜ 2 z 2 ⎟⎟ dz ⎝H ⎠ 1 4

Rotation 653 terms of our elemental disk: Substitute in equation (1) to obtain:

⎡ 1 ⎛ R 2 ⎞2 ⎤ dI x = πρ ⎢ ⎜⎜ 2 z 2 ⎟⎟ ⎥ dz ⎢⎣ 4 ⎝ H ⎠ ⎥⎦ ⎛ ⎛R + ⎜ πρ ⎜ ⎜ ⎝H ⎝

Integrate from 0 to H to obtain:

2 ⎞ ⎞ z ⎟ dz ⎟ z 2 ⎟ ⎠ ⎠

2 ⎡1 ⎛ R2 R2 4 ⎤ 2⎞ I x = πρ ∫ ⎢ ⎜⎜ 2 z ⎟⎟ + 2 z ⎥ dz 4 H H ⎥⎦ ⎠ 0 ⎢ ⎣ ⎝ ⎛ R4H R2H 3 ⎞ ⎟ = πρ ⎜⎜ + 5 ⎟⎠ ⎝ 20 H

H

H

R2 2 z dz H2

Express the total mass of the cone in terms of the mass of the elemental disk:

M = πρ ∫ r 2 dz = πρ ∫

Divide Ix by M, simplify, and solve for Ix to obtain:

⎛ H 2 R2 ⎞ ⎟ + I x = 3M ⎜⎜ 20 ⎟⎠ ⎝ 5

0

0

= πρ R H 1 3

2

Remarks: Because both H and R appear in the numerator, the larger the cones are, the greater their moment of inertia and the greater the energy consumption required to set them into motion.

Rotational Kinetic Energy 58 • Picture the Problem The kinetic energy of this rotating system of particles can be calculated either by finding the tangential velocities of the particles and using these values to find the kinetic energy or by finding the moment of inertia of the system and using the expression for the rotational kinetic energy of a system. (a) Use the relationship between v and ω to find the speed of each particle:

v3 = r3ω = (0.2 m )(2 rad/s ) = 0.4 m/s and

v1 = r1ω = (0.4 m )(2 rad/s ) = 0.8 m/s

654 Chapter 9 Find the kinetic energy of the system:

K = 2 K 3 + 2 K1 = m3v32 + m1v12 = (3 kg )(0.4 m/s ) + (1kg )(0.8 m/s ) 2

2

= 1.12 J (b) Use the definition of the moment of inertia of a system of particles to obtain: Substitute numerical values and evaluate I:

I = ∑ mi ri 2 i

= m1r12 + m2 r22 + m3 r32 + m4 r42 I = (1 kg )(0.4 m ) + (3 kg )(0.2 m ) 2

2

+ (1 kg )(0.4 m ) + (3 kg )(0.2 m ) 2

2

= 0.560 kg ⋅ m 2 Calculate the kinetic energy of the system of particles:

K = 12 Iω 2 =

1 2

(0.560 kg ⋅ m )(2 rad/s)

2

2

= 1.12 J

*59 • Picture the Problem We can find the kinetic energy of this rotating ball from its angular speed and its moment of inertia. We can use the same relationship to find the new angular speed of the ball when it is supplied with additional energy. (a) Express the kinetic energy of the ball:

K = 12 Iω 2

Express the moment of inertia of ball with respect to its diameter:

I = 52 MR 2

Substitute for I:

K = 15 MR 2ω 2

Substitute numerical values and evaluate K:

K=

1 5

(1.4 kg )(0.075 m )2

⎛ rev 2π rad 1 min ⎞ ⎟ × ⎜⎜ 70 × × rev 60 s ⎟⎠ ⎝ min = 84.6 mJ

(b) Express the new kinetic energy with K′ = 2.0846 J:

K ' = 12 Iω ' 2

Express the ratio of K to K′:

K' 12 Iω' 2 ⎛ ω' ⎞ = 1 =⎜ ⎟ 2 K ⎝ω ⎠ 2 Iω '

2

2

Rotation 655 Solve for ω′:

ω' = ω

Substitute numerical values and evaluate ω′:

K' K

ω' = (70 rev/min )

2.0846 J 0.0846 J

= 347 rev/min 60 • Picture the Problem The power delivered by an engine is the product of the torque it develops and the angular speed at which it delivers the torque. Express the power delivered by the engine as a function of the torque it develops and the angular speed at which it delivers this torque:

P =τω

Substitute numerical values and evaluate P:

⎛ rev 2π rad 1 min ⎞ ⎟ = 155 kW P = (400 N ⋅ m )⎜⎜ 3700 × × min rev 60 s ⎟⎠ ⎝ 61 •• Picture the Problem Let r1 and r2 be the distances of m1 and m2 from the center of mass. We can use the definition of rotational kinetic energy and the definition of the center of mass of the two point masses to show that K1/K2 = m2/m1.

Iω12 m1r12ω 2 m1r12 = = Iω 22 m2 r22ω 2 m2 r22

Use the definition of rotational kinetic energy to express the ratio of the rotational kinetic energies:

K1 = K2

Use the definition of the center of mass to relate m1, m2, r1, and r2:

r1 m1 = r2 m2

r Solve for 1 , substitute and r2

K1 m1 ⎛ m2 ⎞ m2 ⎜⎜ ⎟⎟ = = K 2 m2 ⎝ m1 ⎠ m1

simplify to obtain:

1 2 1 2

2

62 •• Picture the Problem The earth’s rotational kinetic energy is given by K rot = 12 Iω 2 where I is its moment of inertia with respect to its axis of rotation. The

656 Chapter 9 center of mass of the earth-sun system is so close to the center of the sun and the earthsun distance so large that we can use the earth-sun distance as the separation of their centers of mass and assume each to be point mass. Express the rotational kinetic energy of the earth:

K rot = 12 Iω 2

Find the angular speed of the earth’s rotation using the definition of ω:

ω=

From Table 9-1, for the moment of inertia of a homogeneous sphere, we find:

I = 25 MR 2

Substitute numerical values in equation (1) to obtain:

K rot =

(1)

2π rad ∆θ = ∆t 24 h × 3600 s h −5 = 7.27 × 10 rad/s

=

2 5

(6.0 ×10

24

)(

kg 6.4 × 106 m

)

2

= 9.83 × 1037 kg ⋅ m 2

(9.83 ×10 kg ⋅ m ) × (7.27 × 10 rad/s )

1 2

37

2

2

−5

= 2.60 × 10 29 J Express the earth’s orbital kinetic energy:

2 K orb = 12 Iω orb

Find the angular speed of the center of mass of the earth-sun system:

ω=

∆θ ∆t

(2)

2π rad

=

365.25 days × 24

h 3600 s × day h

= 1.99 × 10−7 rad/s Express and evaluate the orbital moment of inertia of the earth:

2 I = M E Rorb

(

)(

= 6.0 × 1024 kg 1.50 × 1011 m = 1.35 × 1047 kg ⋅ m 2

Substitute in equation (2) to obtain:

K orb =

(1.35 ×10 kg ⋅ m ) × (1.99 × 10 rad/s )

1 2

47

−7

= 2.67 × 1033 J

2

2

)

2

Rotation 657 Evaluate the ratio

K orb : K rot

K orb 2.67 × 1033 J = ≈ 10 4 29 K rot 2.60 × 10 J

*63 •• Picture the Problem Because the load is not being accelerated, the tension in the cable equals the weight of the load. The role of the massless pulley is to change the direction the force (tension) in the cable acts. (a) Because the block is lifted at constant speed:

(b) Apply the definition of torque at the winch drum: (c) Relate the angular speed of the winch drum to the rate at which the load is being lifted (the tangential speed of the cable on the drum): (d) Express the power developed by the motor in terms of the tension in the cable and the speed with which the load is being lifted:

(

T = mg = (2000 kg ) 9.81m/s 2

)

= 19.6 kN

τ = Tr = (19.6 kN )(0.30 m ) = 5.89 kN ⋅ m

ω=

v 0.08 m/s = = 0.267 rad/s r 0.30 m

P = Tv = (19.6 kN )(0.08 m/s )

= 1.57 kW

64 •• Picture the Problem Let the zero of gravitational potential energy be at the lowest point of the small particle. We can use conservation of energy to find the angular velocity of the disk when the particle is at its lowest point and Newton’s 2nd law to find the force the disk will have to exert on the particle to keep it from falling off. (a) Use conservation of energy to relate the initial potential energy of the system to its rotational kinetic energy when the small particle is at its lowest point: Solve for ωf:

∆K + ∆U = 0 or, because Uf = Ki = 0, 1 2

(I

disk

ωf =

+ I particle )ωf2 − mg∆h = 0

2mg∆h I disk + I particle

658 Chapter 9 Substitute for Idisk, Iparticle, and ∆h and simplify to obtain:

2mg (2 R ) = 2 2 1 MR + mR 2

ωf =

(b) The mass is in uniform circular motion at the bottom of the disk, so the sum of the force F exerted by the disk and the gravitational force must be the centripetal force:

F − mg = mRωf2

Solve for F and simplify to obtain:

F = mg + mRω f2

8mg R(2m + M )

⎛ 8mg ⎞ ⎟⎟ = mg + mR⎜⎜ ⎝ R(2mM ) ⎠ 8m ⎞ ⎛ = mg ⎜1 + ⎟ ⎝ 2m + M ⎠ 65 •• Picture the Problem Let the zero of gravitational potential energy be at the center of mass of the ring when it is directly below the point of support. We’ll use conservation of energy to relate the maximum angular velocity and the initial angular velocity required for a complete revolution to the changes in the potential energy of the ring. (a) Use conservation of energy to relate the initial potential energy of the ring to its rotational kinetic energy when its center of mass is directly below the point of support:

∆K + ∆U = 0 or, because Uf = Ki = 0, 1 2

2 I Pω max − mg∆h = 0

Use the parallel axis theorem and Table 9-1 to express the moment of inertia of the ring with respect to its pivot point P:

I P = I cm + mR 2

Substitute in equation (1) to obtain:

1 2

Solve for ωmax:

Substitute numerical values and evaluate ωmax:

(mR

2

(1)

)

2 + mR 2 ωmax − mgR = 0

ωmax =

g R

ωmax =

9.81 m/s 2 = 3.62 rad/s 0.75 m

Rotation 659 (b) Use conservation of energy to relate the final potential energy of the ring to its initial rotational kinetic energy: Noting that the center of mass must rise a distance R if the ring is to make a complete revolution, substitute for IP and ∆h to obtain: Solve for ωi:

Substitute numerical values and evaluate ωi:

∆K + ∆U = 0 or, because Ui = Kf = 0,

− 12 I Pω i2 + mg∆h = 0

− 12 (mR 2 + mR 2 )ωi2 + mgR = 0

g R

ωi =

9.81 m/s 2 ωi = = 3.62 rad/s 0.75 m

66 •• Picture the Problem We can find the energy that must be stored in the flywheel and relate this energy to the radius of the wheel and use the definition of rotational kinetic energy to find the wheel’s radius. Relate the kinetic energy of the flywheel to the energy it must deliver: Express the moment of inertia of the flywheel: Substitute for Icyl and solve for ω:

Substitute numerical values and evaluate R:

K rot = 12 I cylω 2 = (2 MJ/km )(300 km ) = 600 MJ I cyl = 12 MR 2

R=

2

ω

K rot M

2

R= 400

rev 2π rad × s rev

106 J 600 MJ × MJ 100 kg

= 1.95 m 67 •• Picture the Problem We’ll solve this problem for the general case of a ladder of length L, mass M, and person of mass m. Let the zero of gravitational potential energy be at floor level and include you, the ladder, and the earth in the system. We’ll use

660 Chapter 9 conservation of energy to relate your impact speed falling freely to your impact speed riding the ladder to the ground. Use conservation of energy to relate the speed with which a person will strike the ground to the fall distance L:

∆K + ∆U = 0 or, because Ki = Uf = 0, 1 2

mvf2 − mgL = 0

Solve for vf2 :

vf2 = 2 gL

Letting ωr represent the angular velocity of the ladder+person system as it strikes the ground, use conservation of energy to relate the initial and final momenta of the system:

∆K + ∆U = 0

Substitute for the moments of inertia to obtain: Substitute vr for Lωf and solve for vr2 :

Express the ratio

vr2 : vf2

Solve for vr to obtain:

or, because Ki = Uf = 0, 1 2

1 2

(I

person

L⎞ ⎛ + I ladder )ωr2 − ⎜ mgL + Mg ⎟ = 0 2⎠ ⎝

1 ⎞ 2 2 ⎛ L⎞ ⎛ ⎜ m + M ⎟ L ω f − ⎜ mgL + Mg ⎟ = 0 3 ⎠ 2⎠ ⎝ ⎝

M⎞ ⎛ 2 gL⎜ m + ⎟ 2 ⎠ ⎝ v r2 = M m+ 3 v r2 = vf2

M 2 M m+ 3 m+

vr = vf

6m + 3M 6m + 2 M

Unless M , the mass of the ladder, is zero, vr > vf . It is better to let go and fall to the ground.

Rotation 661

Pulleys, Yo-Yos, and Hanging Things *68 •• Picture the Problem We’ll solve this problem for the general case in which the mass of the block on the ledge is M, the mass of the hanging block is m, and the mass of the pulley is Mp, and R is the radius of the pulley. Let the zero of gravitational potential energy be 2.5 m below the initial position of the 2-kg block and R represent the radius of the pulley. Let the system include both blocks, the shelf and pulley, and the earth. The initial potential energy of the 2-kg block will be transformed into the translational kinetic energy of both blocks plus rotational kinetic energy of the pulley. (a) Use energy conservation to relate the speed of the 2 kg block when it has fallen a distance ∆h to its initial potential energy and the kinetic energy of the system: Substitute for Ipulley and ω to obtain:

Solve for v:

Substitute numerical values and evaluate v:

∆K + ∆U = 0 or, because Ki = Uf = 0, 1 2

(m + M )v 2 + 12 I pulleyω 2 − mgh = 0

1 2

(m + M )v 2 + 12 (12 MR 2 ) v 2 − mgh = 0 2

R

v=

2mgh M + m + 12 M p

v=

2(2 kg ) 9.81 m/s 2 (2.5 m ) 4kg + 2 kg + 12 (0.6 kg )

(

)

= 3.95 m/s (b) Find the angular velocity of the pulley from its tangential speed: 69 •• Picture the Problem The diagrams show the forces acting on each of the masses and the pulley. We can apply Newton’s 2nd law to the two blocks and the pulley to obtain three equations in the unknowns T1, T2, and a.

ω=

v 3.95 m/s = = 49.3 rad/s R 0.08 m

662 Chapter 9 Apply Newton’s 2nd law to the two blocks and the pulley:

∑F =T = m a , ∑τ = (T − T ) r = I α , x

p

1

(1)

4

2

1

(2)

p

and

∑F

x

= m2 g − T2 = m2 a

Eliminate α in equation (2) to obtain:

T2 − T1 = 12 M p a

Eliminate T1 and T2 between equations (1), (3) and (4) and solve for a:

a=

(3) (4)

m2 g m2 + m4 + 12 M p

Substitute numerical values and evaluate a:

( 2 kg ) (9.81 m/s 2 ) a= = 2 kg + 4 kg + 12 (0.6 kg )

Using equation (1), evaluate T1:

T1 = (4 kg ) 3.11 m/s 2 = 12.5 N

Solve equation (3) for T2:

T2 = m2 ( g − a )

Substitute numerical values and evaluate T2:

(

3.11 m/s 2

)

(

T2 = (2 kg ) 9.81 m/s 2 − 3.11 m/s 2

)

= 13.4 N

70 •• Picture the Problem We’ll solve this problem for the general case in which the mass of the block on the ledge is M, the mass of the hanging block is m, the mass of the pulley is Mp, and R is the radius of the pulley. Let the zero of gravitational potential energy be 2.5 m below the initial position of the 2-kg block. The initial potential energy of the 2-kg block will be transformed into the translational kinetic energy of both blocks plus rotational kinetic energy of the pulley plus work done against friction. (a) Use energy conservation to relate the speed of the 2 kg block when it has fallen a distance ∆h to its initial potential energy, the kinetic energy of the system and the work done against friction: Substitute for Ipulley and ω to obtain:

∆K + ∆U + Wf = 0 or, because Ki = Uf = 0, 1 2

(m + M )v 2 + 12 I pulleyω 2 − mgh + µ k Mgh = 0

(m + M )v 2 + 12 (12 M p ) v 2 2

1 2

R − mgh + µ k Mgh = 0

Rotation 663 Solve for v:

2 gh(m − µ k M ) M + m + 12 M p

v=

Substitute numerical values and evaluate v:

v=

(

)

2 9.81m/s 2 (2.5 m )[2 kg − (0.25)(4 kg )] = 2.79 m/s 4kg + 2 kg + 12 (0.6 kg )

(b) Find the angular velocity of the pulley from its tangential speed:

ω=

v 2.79 m/s = = 34.9 rad/s R 0.08 m

71 •• Picture the Problem Let the zero of gravitational potential energy be at the water’s surface and let the system include the winch, the car, and the earth. We’ll apply energy conservation to relate the car’s speed as it hits the water to its initial potential energy. Note that some of the car’s initial potential energy will be transformed into rotational kinetic energy of the winch and pulley. Use energy conservation to relate the car’s speed as it hits the water to its initial potential energy: Express ωw and ωp in terms of the speed v of the rope, which is the same throughout the system: Substitute to obtain:

Solve for v:

Substitute numerical values and evaluate v:

∆K + ∆U = 0 or, because Ki = Uf = 0, 1 2

mv 2 + 12 I w ω w2 + 12 I pω p2 − mg∆h = 0

v2 v2 and ω = p rw2 rp2

ωw =

1 2

v2 1 v2 mv + I w 2 + 2 I p 2 − mg∆h = 0 rw rp 2

v=

v=

1 2

2mg∆h I I m + w2 + p2 rw rp

(

)

2(1200 kg ) 9.81 m/s 2 (5 m ) 320 kg ⋅ m 2 4 kg ⋅ m 2 1200 kg + + (0.8 m )2 (0.3 m )2

= 8.21 m/s

664 Chapter 9 *72 •• Picture the Problem Let the system include the blocks, the pulley and the earth. Choose the zero of gravitational potential energy to be at the ledge and apply energy conservation to relate the impact speed of the 30-kg block to the initial potential energy of the system. We can use a constant-acceleration equations and Newton’s 2nd law to find the tensions in the strings and the descent time. (a) Use conservation of energy to relate the impact speed of the 30-kg block to the initial potential energy of the system: Substitute for ωp and Ip to obtain:

∆K + ∆U = 0 or, because Ki = Uf = 0, 1 2

m30 v 2 + 12 m20 v 2 + 12 I pω p2 + m20 g∆h − m30 g∆h = 0

1 2

m30 v + m20 v + 2

1 2

2

(

1 1 2 2

M pr

2

)

⎛ v2 ⎞ ⎜⎜ 2 ⎟⎟ ⎝r ⎠

+ m20 g∆h − m30 g∆h = 0 Solve for v:

Substitute numerical values and evaluate v:

2 g∆h(m30 − m20 ) m20 + m30 + 12 M p

v=

(

)

2 9.81 m/s 2 (2 m )(30 kg − 20 kg ) v= 20 kg + 30 kg + 12 (5 kg ) = 2.73 m/s v 2.73 m/s = = 27.3 rad/s r 0.1 m

(b) Find the angular speed at impact from the tangential speed at impact and the radius of the pulley:

ω=

(c) Apply Newton’s 2nd law to the blocks:

∑F ∑F

Using a constant-acceleration equation, relate the speed at impact to the fall distance and the

v 2 = v02 + 2a∆h

x

= T1 − m20 g = m20 a

(1)

x

= m30 g − T2 = m30 a

(2)

or, because v0 = 0,

Rotation 665 acceleration and solve for and evaluate a:

(2.73 m/s) = 1.87 m/s 2 v2 a= = 2∆h 2(2 m )

Substitute in equation (1) to find T1:

T1 = m20 ( g + a )

2

(

= (20 kg ) 9.81 m/s 2 + 1.87 m/s 2

)

= 234 N Substitute in equation (2) to find T2:

T2 = m30 (g − a )

(

= (30 kg ) 9.81m/s 2 − 1.87 m/s 2 = 238 N

(d) Noting that the initial speed of the 30-kg block is zero, express the time-of-fall in terms of the fall distance and the block’s average speed:

∆t =

∆h ∆h 2∆h = = vav 12 v v

Substitute numerical values and evaluate ∆t:

∆t =

2(2 m ) = 1.47 s 2.73 m/s

∑τ

= TR = I sphereα

(1)

= mg − T = ma

(2)

73 •• Picture the Problem The force diagram shows the forces acting on the sphere and the hanging object. The tension in the string is responsible for the angular acceleration of the sphere and the difference between the weight of the object and the tension is the net force acting on the hanging object. We can use Newton’s 2nd law to obtain two equations in a and T that we can solve simultaneously. (a)Apply Newton’s 2nd law to the sphere and the hanging object:

0

and

∑F

x

Substitute for Isphere and α in equation (1) to obtain:

TR =

(

2 5

MR 2

) Ra

(3)

)

666 Chapter 9 Eliminate T between equations (2) and (3) and solve for a to obtain:

a=

(b) Substitute for a in equation (2) and solve for T to obtain:

T=

g 2M 1+ 5m 2mMg 5m + 2 M

74 •• Picture the Problem The diagram shows the forces acting on both objects and the pulley. By applying Newton’s 2nd law of motion, we can obtain a system of three equations in the unknowns T1, T2, and a that we can solve simultaneously.

(a) Apply Newton’s 2nd law to the pulley and the two objects:

∑F =T −m g = m a, ∑ τ = (T − T )r = I α , x

0

1

2

1

1

(1)

1

0

(2)

and

∑F

x

= m2 g − T2 = m2 a

Substitute for I0 = Ipulley and α in equation (2) to obtain:

(T2 − T1 ) r = (12 mr 2 ) a

Eliminate T1 and T2 between equations (1), (3) and (4) and solve for a to obtain:

a=

Substitute numerical values and evaluate a:

a=

r

(3)

(4)

(m2 − m1 )g m1 + m2 + 12 m

(510 g − 500 g )(981cm/s2 ) 500 g + 510 g + 12 (50 g )

= 9.478 cm/s 2 (b) Substitute for a in equation (1) and solve for T1 to obtain:

T1 = m1 (g + a )

(

= (0.500 kg ) 9.81 m/s 2 + 0.09478 m/s 2 = 4.9524 N

)

Rotation 667 Substitute for a in equation (3) and solve for T2 to obtain:

T2 = m2 ( g − a )

(

= (0.510 kg ) 9.81m/s 2 − 0.09478 m/s 2 = 4.9548 N

∆T = T2 − T1 = 4.9548 N − 4.9524 N

Find ∆T:

= 0.0024 N (c) If we ignore the mass of the pulley, our acceleration equation is:

a=

(m2 − m1 )g

Substitute numerical values and evaluate a:

a=

(510 g − 500 g )(981cm/s2 )

m1 + m2

500 g + 510 g

= 9.713 cm/s 2 Substitute for a in equation (1) and solve for T1 to obtain:

T1 = m1 ( g + a )

Substitute numerical values and evaluate T1:

(

)

T1 = (0.500 kg ) 9.81 m/s 2 + 0.09713 m/s 2 = 4.9536 N From equation (4), if m = 0:

T1 = T2

*75 •• Picture the Problem The diagram shows the forces acting on both objects and the pulley. By applying Newton’s 2nd law of motion, we can obtain a system of three equations in the unknowns T1, T2, and α that we can solve simultaneously.

(a) Express the condition that the system does not accelerate:

τ net = m1 gR1 − m2 gR2 = 0

)

668 Chapter 9 R1 R2

Solve for m2:

m2 = m1

Substitute numerical values and evaluate m2:

m 2 = (24 kg )

(b) Apply Newton’s 2nd law to the objects and the pulley:

∑F ∑τ

1.2 m = 72.0 kg 0.4 m

x

= m1 g − T1 = m1 a ,

0

= T1 R1 − T2 R2 = I 0α , (2)

(1)

and

∑F

x

Eliminate a in favor of α in equations (1) and (3) and solve for T1 and T2:

Substitute for T1 and T2 in equation (2) and solve for α to obtain:

= T2 − m2 g = m2 a

T1 = m1 ( g − R1α )

(3) (4)

and

T2 = m2 ( g + R2α )

α=

(5)

(m1 R1 − m2 R2 )g m1 R12 + m2 R22 + I 0

Substitute numerical values and evaluate α:

α=

[(36 kg )(1.2 m ) − (72 kg )(0.4 m )](9.81 m/s 2 ) (36 kg )(1.2 m )2 + (72 kg )(0.4 m )2 + 40 kg ⋅ m 2

= 1.37 rad/s 2

Substitute in equation (4) to find T1:

[

(

)]

(

)]

T1 = (36 kg ) 9.81 m/s 2 − (1.2 m ) 1.37 rad/s 2 = 294 N Substitute in equation (5) to find T2:

[

T2 = (72 kg ) 9.81 m/s 2 + (0.4 m ) 1.37 rad/s 2 = 746 N

Rotation 669 76 •• Picture the Problem Choose the coordinate system shown in the diagram. By applying Newton’s 2nd law of motion, we can obtain a system of two equations in the unknowns T and a. In (b) we can use the torque equation from (a) and our value for T to findα. In (c) we use the condition that the acceleration of a point on the rim of the cylinder is the same as the acceleration of the hand, together with the angular acceleration of the cylinder, to find the acceleration of the hand. (a) Apply Newton’s 2nd law to the cylinder about an axis through its center of mass:

∑τ

0

= TR = I 0

a R

(1)

and

∑F

x

= Mg − T = 0

Solve for T to obtain:

T = Mg

(b) Rewrite equation (1) in terms of α:

TR = I 0α

Solve for α:

α=

TR I0

Substitute for T and I0 to obtain:

α=

MgR 2g = 2 1 R 2 MR

(c) Relate the acceleration a of the hand to the angular acceleration of the cylinder:

a = Rα

Substitute for α to obtain:

⎛ 2g ⎞ a = R⎜ ⎟ = 2g ⎝ R ⎠

(2)

670 Chapter 9 77 •• Picture the Problem Let the zero of gravitational potential energy be at the bottom of the incline. By applying Newton’s 2nd law to the cylinder and the block we can obtain simultaneous equations in a, T, and α from which we can express a and T. By applying the conservation of energy, we can derive an expression for the speed of the block when it reaches the bottom of the incline. (a) Apply Newton’s 2nd law to the cylinder and the block:

∑τ

0

x

(b) Substitute for a in equation (2) and solve for T:

(1)

and

∑F

Substitute for α in equation (1), solve for T, and substitute in equation (2) and solve for a to obtain:

= TR = I 0α

a=

T=

= m2 g sin θ − T = m2 a

g sin θ m 1+ 1 2m2 1 2

m1 g sin θ m 1+ 1 2m2

(c) Noting that the block is released from rest, express the total energy of the system when the block is at height h:

E = U + K = m2 gh

(d) Use the fact that this system is conservative to express the total energy at the bottom of the incline:

Ebottom = m2 gh

(e) Express the total energy of the system when the block is at the bottom of the incline in terms of its kinetic energies:

Ebottom = K tran + K rot = 12 m2 v 2 + 12 I 0ω 2

(2)

Rotation 671 Substitute for ω and I0 to obtain:

Solve for v to obtain:

(f) For θ = 0: For θ = 90°:

1 2

m2 v + 2

1 2

(

1 2

2

)

1 2

m1a ,

v2 m1r = m2 gh r2

2 gh m 1+ 1 2 m2

v=

a =T =0 g , m 1+ 1 2 m2

a=

T=

m1 g = m 1+ 1 2 m2 1 2

and

2 gh m 1+ 1 2 m2

v=

For m1 = 0:

a = g sin θ , T = 0 , and

v=

2 gh

*78 •• Picture the Problem Let r be the radius of the concentric drum (10 cm) and let I0 be the moment of inertia of the drum plus platform. We can use Newton’s 2nd law in both translational and rotational forms to express I0 in terms of a and a constantacceleration equation to express a and then find I0. We can use the same equation to find the total moment of inertia when the object is placed on the platform and then subtract to find its moment of inertia. (a) Apply Newton’s 2nd law to the platform and the weight:

∑τ ∑F

0

= Tr = I 0α

(1)

x

= Mg − T = Ma

(2)

672 Chapter 9 Substitute a/r for α in equation (1) and solve for T:

T=

I0 a r2

Substitute for T in equation (2) and solve for a to obtain:

I0 =

Mr 2 ( g − a ) a

Using a constant-acceleration equation, relate the distance of fall to the acceleration of the weight and the time of fall and solve for the acceleration:

(3)

∆x = v0 ∆t + 12 a (∆t )

2

or, because v0 = 0 and ∆x = D,

a=

2D (∆t )2

Substitute for a in equation (3) to obtain:

⎛ g (∆t )2 ⎞ ⎛g ⎞ − 1⎟⎟ I 0 = Mr 2 ⎜ − 1⎟ = Mr 2 ⎜⎜ ⎝a ⎠ ⎝ 2D ⎠

Substitute numerical values and evaluate I0:

I 0 = (2.5 kg )(0.1 m )

2

(

)

⎡ 9.81 m/s 2 (4.2 s )2 ⎤ ×⎢ − 1⎥ 2(1.8 m ) ⎣ ⎦ = 1.177 kg ⋅ m 2

(b) Relate the moments of inertia of the platform, drum, shaft, and pulley (I0) to the moment of inertia of the object and the total moment of inertia:

⎛g ⎞ I tot = I 0 + I = Mr 2 ⎜ − 1⎟ ⎝a ⎠

Substitute numerical values and evaluate Itot:

I tot = (2.5 kg )(0.1 m )

⎛ g (∆t )2 ⎞ = Mr ⎜⎜ − 1⎟⎟ ⎝ 2D ⎠ 2

2

(

)

⎡ 9.81m/s 2 (6.8 s )2 ⎤ ×⎢ − 1⎥ ( ) 2 1.8 m ⎣ ⎦ = 3.125 kg ⋅ m 2

Solve for and evaluate I:

I = I tot − I 0 = 3.125 kg ⋅ m 2 − 1.177 kg ⋅ m 2 = 1.948 kg ⋅ m 2

Rotation 673

Objects Rolling Without Slipping *79 •• Picture the Problem The forces acting on the yo-yo are shown in the figure. We can use a constant-acceleration equation to relate the velocity of descent at the end of the fall to the yo-yo’s acceleration and Newton’s 2nd law in both translational and rotational form to find the yo-yo’s acceleration. Using a constant-acceleration equation, relate the yo-yo’s final speed to its acceleration and fall distance:

v 2 = v02 + 2a∆h or, because v0 = 0,

v = 2a∆h

∑F

x

= mg − T = ma

(2)

0

= Tr = I 0α

(3)

Use Newton’s 2nd law to relate the forces that act on the yo-yo to its acceleration:

and

Use a = rα to eliminate α in equation (3)

Tr = I 0

Eliminate T between equations (2) and (4) to obtain:

mg −

Substitute

1 2

mR 2 for I0 in equation

(5):

∑τ

mg −

Solve for a:

a=

Substitute numerical values and evaluate a:

a=

Substitute in equation (1) and evaluate v:

(1)

a r

(4)

I0 a = ma r2 1 2

(5)

mR 2 a = ma r2

g R2 1+ 2 2r

9.81 m/s 2 = 0.0864 m/s 2 2 (1.5 m ) 1+ 2 2(0.1 m )

(

)

v = 2 0.0864 m/s 2 (57 m ) = 3.14 m/s

674 Chapter 9 80 •• Picture the Problem The diagram shows the forces acting on the cylinder. By applying Newton’s 2nd law of motion, we can obtain a system of two equations in the unknowns T, a, and α that we can solve simultaneously.

(a) Apply Newton’s 2nd law to the cylinder:

∑τ

0

= TR = I 0α

(1)

= Mg − T = Ma

(2)

and

∑F

x

(

)

⎛a⎞ MR 2 ⎜ ⎟ ⎝R⎠

Substitute for α and I0 in equation (1) to obtain:

TR =

Solve for T:

T = 12 Ma

Substitute for T in equation (2) and solve for a to obtain:

a=

(b) Substitute for a in equation (3) to obtain:

T = 12 M ( 23 g ) =

2 3

1 2

(3)

g

1 3

Mg

81 •• Picture the Problem The forces acting on the yo-yo are shown in the figure. Apply Newton’s 2nd law in both translational and rotational form to obtain simultaneous equations in T, a, and α from which we can eliminate α and solve for T and a.

Apply Newton’s 2nd law to the yo-yo:

∑F

x

= mg − T = ma

(1)

0

= Tr = I 0α

(2)

and

∑τ

Use a = rα to eliminate α in equation (2)

Tr = I 0

a r

(3)

Rotation 675 Eliminate T between equations (1) and (3) to obtain: Substitute

1 2

mR 2 for I0 in equation

(4):

mg −

(4)

mR 2 a = ma mg − r2 1 2

Solve for a:

a=

Substitute numerical values and evaluate a:

a=

Use equation (1) to solve for and evaluate T:

I0 a = ma r2

g R2 1+ 2 2r

9.81 m/s 2 = 0.192 m/s 2 2 (0.1 m ) 1+ 2 2(0.01 m )

T = m( g − a )

(

= (0.1 kg ) 9.81 m/s 2 − 0.192 m/s 2

)

= 0.962 N *82 • Picture the Problem We can determine the kinetic energy of the cylinder that is due to its rotation about its center of mass by examining the ratio K rot K .

(

) vr

2

Express the rotational kinetic energy of the homogeneous solid cylinder:

K rot = 12 I cylω 2 =

Express the total kinetic energy of the homogeneous solid cylinder:

K = K rot + K trans = 14 mv 2 + 12 mv 2 = 34 mv 2

Express the ratio

K rot : K

1 2

1 2

mr 2

2

= 14 mv 2

K rot 14 mv 2 1 = 3 2 = 3 and (b) is correct. K 4 mv

83 • Picture the Problem Any work done on the cylinder by a net force will change its kinetic energy. Therefore, the work needed to give the cylinder this motion is equal to its kinetic energy. Express the relationship between the work needed to stop the cylinder and its kinetic energy:

W = ∆K = 12 mv 2 + 12 Iω 2

676 Chapter 9 Because the cylinder is rolling without slipping, its translational and angular speeds are related according to:

v = rω

Substitute for I (see Table 9-1) and ω and simplify to obtain:

W = 12 mv 2 + 12 Iω 2 = 12 mv 2 + 12

(

1 2

mr 2

) vr

2 2

= 34 mv 2 Substitute for m and v to obtain:

W =

3 4

(60 kg )(5 m/s )2 =

1.13 kJ

84 • Picture the Problem The total kinetic energy of any object that is rolling without slipping is given by K = K trans + K rot . We can find the percentages associated with each motion by expressing the moment of inertia of the objects as kmr2 and deriving a general expression for the ratios of rotational kinetic energy to total kinetic energy and translational kinetic energy to total kinetic energy and substituting the appropriate values of k. Express the total kinetic energy associated with a rotating and translating object:

K = K trans + K rot = 12 mv 2 + 12 Iω 2

(

= 12 mv 2 + 12 kmr 2

) vr

2 2

= 12 mv 2 + 12 kmv 2 = 12 mv 2 (1 + k ) Express the ratio

K rot : K

Express the ratio

K trans : K

(a) Substitute k = 2/5 for a uniform sphere to obtain:

1 1 K rot kmv 2 k = 1 2 2 = = 1+ k 1+ 1 K 2 mv (1 + k ) k 2 1 1 K trans mv = 1 22 = K 1+ k 2 mv (1 + k )

1 K rot = = 0.286 = 28.6% 1 K 1+ 0.4 and

K trans 1 = = 0.714 = 71.4% K 1 + 0.4

Rotation 677 (b) Substitute k = 1/2 for a uniform cylinder to obtain:

1 K rot = = 33.3% 1 K 1+ 0.5 and

K trans 1 = = 66.7% K 1 + 0.5 (c) Substitute k = 1 for a hoop to obtain:

1 K rot = = 50.0% 1 K 1+ 1 and

K trans 1 = = 50.0% K 1+1 85 • Picture the Problem Let the zero of gravitational potential energy be at the bottom of the incline. As the hoop rolls up the incline its translational and rotational kinetic energies are transformed into gravitational potential energy. We can use energy conservation to relate the distance the hoop rolls up the incline to its total kinetic energy at the bottom of the incline. Using energy conservation, relate the distance the hoop will roll up the incline to its kinetic energy at the bottom of the incline: Express Ki as the sum of the translational and rotational kinetic energies of the hoop: When a rolling object moves with speed v, its outer surface turns with a speed v also. Hence ω = v/r. Substitute for I and ω to obtain:

∆K + ∆U = 0 or, because Kf = Ui = 0,

− Ki + Uf = 0

(1)

K i = K trans + K rot = 12 mv 2 + 12 Iω 2

K i = 12 mv 2 +

1 2

(mr ) vr 2

2 2

= mv 2

Letting ∆h be the change in elevation of the hoop as it rolls up the incline and ∆L the distance it rolls along the incline, express Uf:

U f = mg∆h = mg∆L sin θ

Substitute in equation (1) to obtain:

− mv 2 + mg∆L sin θ = 0

678 Chapter 9 v2 ∆L = g sin θ

Solve for ∆L:

Substitute numerical values and evaluate ∆L:

∆L =

(15 m/s)2

(9.81m/s ) sin30° = 2

45.9 m

*86 •• Picture the Problem From Newton’s 2nd law, the acceleration of the center of mass equals the net force divided by the mass. The forces acting on the sphere are its weight

r r mg downward, the normal force Fn that balances the normal component of the weight, r and the force of friction f acting up the incline. As the sphere accelerates down the incline, the angular velocity of rotation must increase to maintain the nonslip condition. We can apply Newton’s 2nd law for rotation about a horizontal axis through the center of mass of the sphere to find α, which is related to the acceleration by the nonslip condition.

r

r

r

The only torque about the center of mass is due to f because both mg and Fn act through the center of mass. Choose the positive direction to be down the incline.

r

Apply

∑ F = ma to the sphere:

r

mg sin θ − f = macm

Apply

∑τ = I

α to the sphere:

fr = I cmα

cm

Use the nonslip condition to eliminate α and solve for f:

fr = I cm

acm r

and

f =

I cm acm r2

Substitute this result for f in equation (1) to obtain:

mg sin θ −

From Table 9-1 we have, for a solid sphere:

I cm = 25 mr 2

I cm acm = macm r2

(1)

Rotation 679 mg sin θ − 25 acm = macm

Substitute in equation (1) and simplify to obtain: Solve for and evaluate θ :

⎛ 7 acm ⎞ ⎟⎟ ⎝ 5g ⎠

θ = sin −1 ⎜⎜

⎡ 7(0.2 g ) ⎤ = sin −1 ⎢ ⎥ = 16.3° ⎣ 5g ⎦ 87 •• Picture the Problem From Newton’s 2nd law, the acceleration of the center of mass equals the net force divided by the mass. The forces acting on the thin spherical shell are

r

r

its weight mg downward, the normal force Fn that balances the normal component of the

r

weight, and the force of friction f acting up the incline. As the spherical shell accelerates down the incline, the angular velocity of rotation must increase to maintain the nonslip condition. We can apply Newton’s 2nd law for rotation about a horizontal axis through the center of mass of the sphere to find α, which is related to the acceleration by the nonslip

r

r

condition. The only torque about the center of mass is due to f because both mg and

r Fn act through the center of mass. Choose the positive direction to be down the incline.

Apply

r

r

mg sin θ − f = macm

α to the thin

fr = I cmα

∑ F = ma to the thin

spherical shell: Apply

∑τ = I

cm

spherical shell: Use the nonslip condition to eliminate α and solve for f:

fr = I cm

Substitute this result for f in equation (1) to obtain:

mg sin θ −

From Table 9-1 we have, for a thin

I cm = 23 mr 2

I acm and f = cm acm r2 r I cm acm = macm r2

(1)

680 Chapter 9 spherical shell: Substitute in equation (1) and simplify to obtain: Solve for and evaluate θ :

mg sin θ − 23 acm = macm 5acm 3g 5(0.2 g ) = sin −1 = 19.5° 3g

θ = sin −1

Remarks: This larger angle makes sense, as the moment of inertia for a given mass is larger for a hollow sphere than for a solid one. 88 •• Picture the Problem The three forces acting on the basketball are the weight of the ball, the normal force, and the force of friction. Because the weight can be assumed to be acting at the center of mass, and the normal force acts through the center of mass, the only force which exerts a torque about the center of mass is the frictional force. We can use Newton’s 2nd law to find a system of simultaneous equations that we can solve for the quantities called for in the problem statement. (a) Apply Newton’s 2nd law in both translational and rotational form to the ball:

∑F ∑F

x

= mg sin θ − f s = ma ,

(1)

y

= Fn − mg cosθ = 0

(2)

and

∑τ

= f s r = I 0α

0

(3)

a r

Because the basketball is rolling without slipping we know that:

α=

Substitute in equation (3) to obtain:

fs r = I 0

From Table 9-1 we have:

I 0 = 23 mr 2

Substitute for I0 and α in equation (4) and solve for fs:

fs r =

(

2 3

a r

mr 2

(4)

) ar ⇒ f

s

= 23 ma

(5)

Rotation 681 g sin θ

Substitute for fs in equation (1) and solve for a:

a=

(b) Find fs using equation (5):

f s = 23 m( 53 g sin θ ) =

(c) Solve equation (2) for Fn:

Fn = mg cos θ

Use the definition of fs,max to obtain:

f s,max = µ s Fn = µ s mg cos θ max

Use the result of part (b) to obtain:

2 5

Solve for θmax:

3 5

2 5

mg sin θ

mg sin θ max = µs mg cos θ max

θ max = tan −1 ( 52 µs )

89 •• Picture the Problem The three forces acting on the cylinder are the weight of the cylinder, the normal force, and the force of friction. Because the weight can be assumed to be acting at the center of mass, and the normal force acts through the center of mass, the only force which exerts a torque about the center of mass is the frictional force. We can use Newton’s 2nd law to find a system of simultaneous equations that we can solve for the quantities called for in the problem statement. (a) Apply Newton’s 2nd law in both translational and rotational form to the cylinder:

∑F ∑F

x

= mg sin θ − f s = ma ,

(1)

y

= Fn − mg cosθ = 0

(2)

and

∑τ

0

= f s r = I 0α

(3)

a r

Because the cylinder is rolling without slipping we know that:

α=

Substitute in equation (3) to obtain:

fs r = I 0

From Table 9-1 we have:

I 0 = 12 mr 2

a r

(4)

682 Chapter 9 Substitute for I0 and α in equation (4) and solve for fs:

fs r =

(

1 2

2 3

g sin θ

mr 2

) ar ⇒ f

s

= 12 ma

Substitute for fs in equation (1) and solve for a:

a=

(b) Find fs using equation (5):

f s = 12 m( 23 g sin θ ) =

(c) Solve equation (2) for Fn:

Fn = mg cos θ

Use the definition of fs,max to obtain:

f s,max = µ s Fn = µ s mg cos θ max

Use the result of part (b) to obtain:

1 3

Solve for θmax:

θ max = tan −1 (3µs )

1 3

(5)

mg sin θ

mg sin θ max = µs mg cos θ max

*90 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation where the spheres leave the ramp. The distances the spheres will travel are directly proportional to their speeds when they leave the ramp. Express the ratio of the distances traveled by the two spheres in terms of their speeds when they leave the ramp:

L' v'∆t v' = = L v∆t v

Use conservation of mechanical energy to find the speed of the spheres when they leave the ramp:

∆K + ∆U = 0 or, because Ki = Uf = 0,

Express Kf for the spheres:

K f = K trans + K rot

(1)

Kf −Ui = 0

(2)

= 12 mv 2 + 12 I cmω 2

(

= 12 mv 2 + 12 kmR 2

) Rv

2 2

= 12 mv 2 + 12 kmv 2 = (1 + k ) 12 mv 2

where k is 2/3 for the spherical shell and 2/5 for the uniform sphere. Substitute in equation (2) to obtain:

(1 + k ) 12 mv 2 = mgH

Rotation 683 Solve for v:

2 gH 1+ k

v=

1 + 23 1+ k L' = = = 1.09 1 + k' 1 + 52 L

Substitute in equation (1) to obtain:

or

L' = 1.09 L 91 •• Picture the Problem Let the subscripts u and h refer to the uniform and thin-walled spheres, respectively. Because the cylinders climb to the same height, their kinetic energies at the bottom of the incline must be equal. Express the total kinetic energy of the thin-walled cylinder at the bottom of the inclined plane:

K h = K trans + K rot = 12 mh v 2 + 12 I hω 2

Express the total kinetic energy of the solid cylinder at the bottom of the inclined plane:

K u = K trans + K rot = 12 mu v'2 + 12 I uω 2

Because the cylinders climb to the same height:

3 4

= mh v + 2

1 2

1 2

= 12 mu v' 2 + 12 mu v' 2 = mu gh

and mh v 2 = mh gh mu v' 2 mu gh = mh v 2 mh gh

Divide the first of these equations by the second:

3 4

Simplify to obtain:

3v' 2 =1 4v 2

Solve for v′:

v' =

4 v 3

(

)

v2 mh r = mh v 2 2 r

(

2

1 2

mu r 2

) v'r

2

2

= 34 mu v' 2

684 Chapter 9 92 •• Picture the Problem Let the subscripts s and c refer to the solid sphere and thinwalled cylinder, respectively. Because the cylinder and sphere descend from the same height, their kinetic energies at the bottom of the incline must be equal. The force diagram shows the forces acting on the solid sphere. We’ll use Newton’s 2nd law to relate the accelerations to the angle of the incline and use a constant acceleration to relate the accelerations to the distances traveled down the incline. Apply Newton’s 2nd law to the sphere:

∑F ∑F

x

= mg sin θ − f s = ma s ,

(1)

y

= Fn − mg cosθ = 0 ,

(2)

= f s r = I 0α

(3)

and

∑τ

Substitute for I0 and α in equation (3) and solve for fs:

0

fsr =

(

2 5

) ar ⇒ f

mr 2

Substitute for fs in equation (1) and solve for a:

as = 75 g sin θ

Proceed as above for the thin-walled cylinder to obtain:

a c = 12 g sin θ

Using a constant-acceleration equation, relate the distance traveled down the incline to its acceleration and the elapsed time:

∆s = v0 ∆t + 12 a(∆t ) or, because v0 = 0,

Because ∆s is the same for both objects:

as t s2 = a c t c2

s

= 52 mas

2

∆s = 12 a(∆t )

2

(4)

where

t c2 = (t s + 2.4 ) = t s2 + 4.8t s + 5.76 2

provided tc and ts are in seconds. Substitute for as and ac to obtain the quadratic equation:

t s2 + 4.8t s + 5.76 = 107 t s2

Rotation 685 Solve for the positive root to obtain:

t s = 12.3 s

Substitute in equation (4), simplify, and solve for θ :

θ = sin −1 ⎢

Substitute numerical values and evaluate θ :

θ = sin −1 ⎢

⎡14∆s ⎤ 2 ⎥ ⎣ 5 gts ⎦ ⎤ 14(3 m ) 2⎥ 2 ⎣ 5 9.81 m/s (12.3 s ) ⎦ ⎡

(

)

= 0.324° 93 ••• Picture the Problem The kinetic energy of the wheel is the sum of its translational and rotational kinetic energies. Because the wheel is a composite object, we can model its moment of inertia by treating the rim as a cylindrical shell and the spokes as rods. K = K trans + K rot Express the kinetic energy of the wheel: = 1 M tot v 2 + 1 I cmω 2 2

2

= 12 M tot v 2 + 12 I cm

v2 R2

where Mtot = Mrim + 4Mspoke Express the moment of inertia of the wheel:

I cm = I rim + I spokes

(

= M rim R 2 + 4 13 M spoke R 2 = (M rim + 43 M spoke )R 2

Substitute for Icm in the equation for K:

[

K = 12 M tot v 2 + 12 (M rim + 43 M spoke )R 2 =

Substitute numerical values and evaluate K:

)

[ (M 1 2

tot

]

+ M rim ) + 23 M spoke v 2

] Rv

2

K = [ 12 (7.8 kg + 3 kg ) + 23 (1.2 kg )](6 m/s )

2

= 223 J

2

686 Chapter 9 94 ••• Picture the Problem Let M represent the combined mass of the two disks and their connecting rod and I their moment of inertia. The object’s initial potential energy is transformed into translational and rotational kinetic energy as it rolls down the incline. The force diagram shows the forces acting on this composite object as it rolls down the incline. Application of Newton’s 2nd law will allow us to derive an expression for the acceleration of the object. (a) Apply Newton’s 2nd law to the disks and rod:

∑F ∑F

x

= Mg sin θ − f s = Ma ,

(1)

y

= Fn − Mg cos θ = 0 ,

(2)

= f s r = Iα

(3)

and

∑τ

Eliminate fs and α between equations (1) and (3) and solve for a to obtain: Express the moment of inertia of the two disks plus connecting rod:

a=

0

Mg sin θ I M+ 2 r

(4)

I = 2 I disk + I rod

(

)

= 2 12 mdisk R 2 + 12 mrod r 2 = mdisk R 2 + 12 mrod r 2

Substitute numerical values and evaluate I:

I = (20 kg )(0.3 m ) + 12 (1 kg )(0.02 m )

Substitute in equation (4) and evaluate a:

a=

2

= 1.80 kg ⋅ m 2

(41kg )(9.81m/s 2 ) sin30° 41 kg +

1.80 kg ⋅ m 2 (0.02 m )2

= 0.0443 m/s 2 (b) Find α from a:

α=

a 0.0443 m/s 2 = = 2.21 rad/s 2 r 0.02 m

2

Rotation 687 (c) Express the kinetic energy of translation of the disks-plus-rod when it has rolled a distance ∆s down the incline:

K trans = 12 Mv 2

Using a constant-acceleration equation, relate the speed of the disks-plus-rod to their acceleration and the distance moved:

v 2 = v02 + 2a∆s or, because v0 = 0,

Substitute to obtain:

K trans = Ma∆s

v 2 = 2a∆s

(

)

= (41 kg ) 0.0443 m/s 2 (2 m ) = 3.63 J

(d) Express the rotational kinetic energy of the disks after rolling 2 m in terms of their initial potential energy and their translational kinetic energy: Substitute numerical values and evaluate Krot:

K rot = U i − K trans = Mgh − K trans

(

)

K rot = (41 kg ) 9.81m/s 2 (2 m )sin30° − 3.63 J = 399 J

95 ••• Picture the Problem We can express the coordinates of point P as the sum of the coordinates of the center of the wheel and the coordinates, relative to the center of the r wheel, of the tip of the vector r0 . Differentiation of these expressions with respect to time will give us the x and y components of the velocity of point P. (a) Express the coordinates of point P relative to the center of the wheel:

Because the coordinates of the center of the circle are X and R:

x = r0 cosθ and y = r0 sin θ

(x P , y P ) = ( X + r0 cosθ , R + r0 sin θ )

688 Chapter 9 (b) Differentiate xP to obtain:

Note that

d ( X + r0 cosθ ) dt dX dθ = − r0 sin θ ⋅ dt dt

v Px =

dX dθ V = V and = −ω = − so: dt dt R

v Px = V +

Differentiate yP to obtain:

v Py =

Because

dθ V = −ω = − : dt R r r

(c) Calculate v ⋅ r :

r0V sin θ R

d (R + r0 sin θ ) = r0 cos θ ⋅ dθ dt dt

vPy = −

r0V cosθ R

r r v ⋅ r = v Px rx + v Py ry

rV ⎛ ⎞ = ⎜V + 0 sin θ ⎟(r0 cos θ ) R ⎝ ⎠ ⎛rV ⎞ − ⎜ 0 cos θ ⎟(R + r0 sin θ ) ⎝ R ⎠ = 0 (d) Express v in terms of its components:

v = v x2 + v y2 2

rV ⎞ ⎛ ⎞ ⎛ rV = ⎜V + 0 sin θ ⎟ + ⎜ − 0 cosθ ⎟ R ⎠ ⎝ ⎠ ⎝ R = V 1+ 2 Express r in terms of its components:

r0 r2 sin θ + 02 R R

r = rx2 + ry2 =

(r0 cosθ )2 + (R + r0 sin θ )2

r0 r02 = R 1 + 2 sin θ + 2 R R Divide v by r to obtain:

ω=

v V = r R

2

Rotation 689 *96 ••• Picture the Problem Let the letter B identify the block and the letter C the cylinder. We can find the accelerations of the block and cylinder by applying Newton’s 2nd law and solving the resulting equations simultaneously. Apply

∑F

= ma x to the block:

F − f ' = maB

(1)

Apply

∑F

= ma x to the cylinder:

f = MaC ,

(2)

Apply

∑τ

fR = I CMα

(3)

Substitute for ICM in equation (3) and solve for f = f ′ to obtain:

f = 12 MRα

(4)

Relate the acceleration of the block to the acceleration of the cylinder:

aC = aB + aCB

x

x

CM

= I CMα to the cylinder:

or, because aCB = −Rα is the acceleration of the cylinder relative to the block,

aC = aB − Rα

and Equate equations (2) and (4) and substitute from (5) to obtain: Substitute equation (4) in equation (1) and substitute for aC to obtain: Solve for aB:

97 ••• Picture the Problem Let the letter B identify the block and the letter C the cylinder. In this problem, as in Problem 97, we can find the accelerations of the block and cylinder by applying Newton’s 2nd law and solving the resulting equations simultaneously.

Rα = aB − aC aB = 3aC

F − 13 Ma B = ma B

aB =

3F M + 3m

(5)

690 Chapter 9 Apply

∑F

= ma x to the block:

F − f = maB

(1)

Apply

∑F

= ma x to the cylinder:

f = MaC ,

(2)

Apply

∑τ

fR = I CMα

(3)

Substitute for ICM in equation (3) and solve for f:

f = 12 MRα

(4)

Relate the acceleration of the block to the acceleration of the cylinder:

aC = aB + aCB

x

x

CM

= I CMα to the cylinder:

or, because aCB = −Rα,

aC = aB − Rα

and

Rα = aB − aC

(a) Solve for α and substitute for aB to obtain:

(5)

aB − aC 3aC − aC 2aC = = R R R 2F = R(M + 3m )

α=

From the force diagram it is evident that the torque and, therefore, α is in the counterclockwise direction. (b) Equate equations (2) and (4) and substitute (5) to obtain:

aB = 3aC

From equations (1) and (4) we obtain:

F − 13 Ma B = ma B

Solve for aB:

aB =

Substitute to obtain the linear acceleration of the cylinder relative to the table:

3F M + 3m

aC = 13 a B =

F M + 3m

Rotation 691 (c) Express the acceleration of the cylinder relative to the block:

aCB = aC − aB = aC − 3aC = −2aC = −

2F M + 3m

98 ••• Picture the Problem Let the system include the earth, the cylinder, and the r block. Then F is an external force that changes the energy of the system by doing work on it. We can find the kinetic energy of the block from its speed when it has traveled a distance d. We can find the kinetic energy of the cylinder from the sum of its translational and rotational kinetic energies. In part (c) we can add the kinetic energies of the block and the cylinder to show that their sum is the work done by r F in displacing the system a distance d. (a) Express the kinetic energy of the block:

K B = Won block = 12 mvB2

Using a constant-acceleration equation, relate the velocity of the block to its acceleration and the distance traveled:

vB2 = v02 + 2aB d or, because the block starts from rest,

Substitute to obtain:

K B = 12 m(2aB d ) = maB d

vB2 = 2aB d (1)

Apply

∑F

= ma x to the block:

F − f = maB

(2)

Apply

∑F

= ma x to the cylinder:

f = MaC ,

(3)

Apply

∑τ

fR = I CMα

(4)

Substitute for ICM in equation (4) and solve for f:

f = 12 MRα

(5)

Relate the acceleration of the block to the acceleration of the cylinder:

aC = aB + aCB

x

x

CM

= I CMα to the

cylinder:

or, because aCB = −Rα,

692 Chapter 9 aC = aB − Rα and

Rα = aB − aC

Equate equations (3) and (5) and substitute in (6) to obtain:

aB = 3aC

Substitute equation (5) in equation (2) and use aB = 3aC to obtain:

F − MaC = maB

(6)

or

F − 13 MaB = maB Solve for aB:

Substitute in equation (1) to obtain:

(b) Express the total kinetic energy of the cylinder:

F m + 13 M

aB =

KB =

mFd m + 13 M

K cyl = K trans + K rot = 12 MvC2 + 12 I CMω 2 = 12 MvC2 + 12 I CM where vCB = vC − vB .

In part (a) it was established that:

aB = 3aC

Integrate both sides of the equation with respect to time to obtain:

vB = 3vC + constant

Substitute the initial conditions to obtain:

(7)

2 vCB R2

where the constant of integration is determined by the initial conditions that vC = 0 when vB = 0.

constant = 0 and

vB = 3vC Substitute in our expression for vCB to obtain: Substitute for ICM and vCB in equation (7) to obtain:

vCB = vC − vB = vC − 3vC = −2vC

K cyl = 12 MvC2 + 12 = 32 Mv

2 C

(

) (− 2Rv )

2

1 2

MR 2

C 2

(8)

Rotation 693 Because vC = 13 vB :

vC2 = 19 vB2

It part (a) it was established that:

vB2 = 2aB d and

Substitute to obtain:

aB =

F m + 13 M

vC2 =

1 9

=

⎛

F 1 ⎝m+ 3M

(2aB d ) = 92 ⎜⎜

⎞ ⎟⎟d ⎠

2 Fd 9(m + 13 M )

Substitute in equation (8) to obtain:

⎛ 2 Fd ⎞ ⎟⎟ K cyl = 32 M ⎜⎜ 1 ⎝ 9(m + 3 M ) ⎠ MFd = 3(m + 13 M )

(c) Express the total kinetic energy of the system and simplify to obtain:

K tot = K B + K cyl = =

mFd MFd + m + 13 M 3(m + 13 M )

(3m + M ) Fd = 3(m + 13 M )

Fd

99 •• Picture the Problem The forces responsible for the rotation of the gears are shown in the diagram to the right. The forces acting through the centers of mass of the two gears have been omitted because they produce no torque. We can apply Newton’s 2nd law in rotational form to obtain the equations of motion of the gears and the not slipping condition to relate their angular accelerations. (a) Apply

∑τ = Iα to the gears to

obtain their equations of motion:

Because the gears do not slip

2 N ⋅ m − FR1 = I1α1

(1)

and

FR2 = I 2α 2

(2) where F is the force keeping the gears from slipping with respect to each other.

R1α1 = R2α 2

694 Chapter 9 relative to each other, the tangential accelerations of the points where they are in contact must be the same:

or

Divide equation (1) by R1 to obtain:

2 N⋅m I − F = 1 α1 R1 R1

Divide equation (2) by R2 to obtain:

α2 =

F=

R1 α1 = 12 α1 R2

(3)

I2 α2 R2

Add these equations to obtain:

2 N ⋅ m I1 I = α1 + 2 α 2 R1 R1 R2

Use equation (3) to eliminate α2:

2 N ⋅ m I1 I = α1 + 2 α1 R1 R1 2 R2

Solve for α1 to obtain:

Substitute numerical values and evaluate α1:

α1 =

α1 =

2N⋅m R I1 + 1 I 2 2 R2 2N⋅m 0.5 m 1 kg ⋅ m 2 + 16 kg ⋅ m 2 2(1 m )

(

)

= 0.400 rad/s 2

(0.400 rad/s ) =

Use equation (3) to evaluate α2:

α2 =

(b) To counterbalance the 2-N·m torque, a counter torque of 2 N·m must be applied to the first gear. Use equation (2) with α1 = 0 to find F:

2 N ⋅ m − FR1 = 0 and

F=

1 2

2

0.200 rad/s 2

2N⋅m 2N⋅m = = 4.00 N R1 0.5 m

Rotation 695 *100 •• Picture the Problem Let r be the radius of the marble, m its mass, R the radius of the large sphere, and v the speed of the marble when it breaks contact with the sphere. The numeral 1 denotes the initial configuration of the sphere-marble system and the numeral 2 is configuration as the marble separates from the sphere. We can use conservation of energy to relate the initial potential energy of the marble to the sum of its translational and rotational kinetic energies as it leaves the sphere. Our choice of the zero of potential energy is shown on the diagram. (a) Apply conservation of energy:

∆U + ∆K = 0 or

U 2 − U 1 + K 2 − K1 = 0 Because U2 = K1 = 0:

− mg [R + r − (R + r ) cos θ ] + 12 mv 2 + 12 Iω 2 = 0 or

− mg [(R + r )(1 − cos θ )] + 12 mv 2 + 12 Iω 2 = 0

Use the rolling-without-slipping condition to eliminate ω:

− mg [(R + r )(1 − cosθ )] + 12 mv 2 + 12 I

From Table 9-1 we have: Substitute to obtain:

v2 =0 r2

I = 52 mr 2 − mg [(R + r )(1 − cosθ )] + 12 mv 2 + 12

(

2 5

mr 2

) vr

2 2

=0

or

− mg [(R + r )(1 − cos θ )] + 12 mv 2 + 15 mv 2 = 0

Solve for v2 to obtain:

Apply

∑F

r

= mar to the marble as

it separates from the sphere:

v2 =

10 g (R + r )(1 − cosθ ) 7

mg cos θ = m or

v2 R+r

696 Chapter 9 v2 cos θ = g (R + r ) Substitute for v2:

cos θ =

1 ⎡10 ⎤ g (R + r )(1 − cos θ )⎥ ⎢ g (R + r ) ⎣ 7 ⎦

⎡10 ⎤ = ⎢ (1 − cos θ )⎥ ⎣7 ⎦ Solve for and evaluate θ :

⎛ 10 ⎞ ⎟ = 54.0° ⎝ 17 ⎠

θ = cos −1 ⎜

The force of friction is always less than µs multiplied by the normal force on the marble. However, the normal force decreases to 0 at the (b) point where the ball leaves the sphere, meaning that the force of friction must be less than the force needed to keep the ball rolling without slipping before it leaves the sphere.

Rolling With Slipping 101 • Picture the Problem Part (a) of this problem is identical to Example 9-16. In part (b) we can use the definitions of translational and rotational kinetic energy to find the ratio of the final and initial kinetic energies. (a) From Example 9-16:

s1 =

12 v02 , 49 µ k g

t1 =

2 v0 , and 7 µk g

v1 = 52 µ k gt1 = (b) When the ball rolls without slipping, v1 = rω. Express the final kinetic energy of the ball:

5 v0 7

K f = K trans + K rot = 12 Mv12 + 12 Iω 2 = 12 Mv12 + 12

(

2 5

Mr 2

= 107 Mv12 = 145 Mv02

) vr

2 1 2

Rotation 697 Mv02 5 = 2 Mv0 7

Express the ratio of the final and initial kinetic energies:

Kf = Ki

(c) Substitute in the expressions in (a) to obtain:

(8 m/s) 12 s1 = = 26.6 m 49 (0.06) 9.81 m/s 2

5 14 1 2

2

(

)

t1 =

2 8 m/s = 3.88 s 7 (0.06) 9.81 m/s 2

v1 =

5 (8 m/s) = 5.71 m/s 7

(

)

*102 •• Picture the Problem The cue stick’s blow delivers a rotational impulse as well as a translational impulse to the cue ball. The rotational impulse changes the angular momentum of the ball and the translational impulse changes its linear momentum. Express the rotational impulse Prot as the product of the average torque and the time during which the rotational impulse acts: Express the average torque it produces about an axis through the center of the ball: Substitute in the expression for Prot to obtain: The translational impulse is also given by: Substitute to obtain: Solve for ω0:

Prot = τ av ∆t

τ av = P0 (h − r ) sin θ = P0 (h − r ) where θ (= 90°) is the angle between F and the lever arm h − r.

Prot = P0 (h − r )∆t = (P0 ∆t )(h − r ) = Ptrams (h − r ) = ∆L = Iω0

Ptrans = P0 ∆t = ∆p = mv0 mv0 (h − r ) = 52 mr 2ω0

ω0 =

5v0 (h − r ) 2r 2

698 Chapter 9 103 •• Picture the Problem The angular velocity of the rotating sphere will decrease until the condition for rolling without slipping is satisfied and then it will begin to roll. The force diagram shows the forces acting on the sphere. We can apply Newton’s 2nd law to the sphere and use the condition for rolling without slipping to find the speed of the center of mass when the sphere begins to roll without slipping. Relate the velocity of the sphere when it begins to roll to its acceleration and the elapsed time:

v = a∆t

(1)

Apply Newton’s 2nd law to the sphere:

∑F ∑F

x

= f k = ma ,

(2)

y

= Fn − mg = 0 ,

(3)

= f k r = I 0α

(4)

and

∑τ

0

Using the definition of fk and Fn from equation (3), substitute in equation (2) and solve for a:

a = µk g

Substitute in equation (1) to obtain:

v = a∆t = µ k g∆t

Solve for α in equation (4):

α=

Express the angular speed of the sphere when it has been moving for a time ∆t:

ω = ω0 − α ∆t = ω0 −

Express the condition that the sphere rolls without slipping:

v = rω

Substitute from equations (5) and (6) and solve for the elapsed time until the sphere begins to roll:

∆t =

(5)

fkr mar 5 µk g = 2 2 = I0 2 r 5 mr

2 rω 0 7 µk g

5µ k g ∆t 2r

(6)

Rotation 699 Use equation (4) to find v when the sphere begins to roll:

v = µ k g∆t =

2rω0 2 rω 0 µ k g = 7 µk g 7

104 •• Picture the Problem The sharp force delivers a rotational impulse as well as a translational impulse to the ball. The rotational impulse changes the angular momentum of the ball and the translational impulse changes its linear momentum. In parts (c) and (d) we can apply Newton’s 2nd law to the ball to obtain equations describing both the translational and rotational motion of the ball. We can then solve these equations to find the constant accelerations that allow us to apply constant-acceleration equations to find the velocity of the ball when it begins to roll and its sliding time. (a) Relate the translational impulse delivered to the ball to its change in its momentum:

Ptrans = Fav ∆t = ∆p = mv0

Solve for v0:

v0 =

Substitute numerical values and evaluate v0:

v0 =

Fav ∆t m

(20 kN )(2 ×10−4 s ) = 0.02 kg

200 m/s

(b) Express the rotational impulse Prot as the product of the average torque and the time during which the rotational impulse acts:

Prot = τ av ∆t

Letting h be the height at which the impulsive force is delivered, express the average torque it produces about an axis through the center of the ball:

τ av = Fl sin θ where θ is the angle between F and the lever arm l .

Substitute h − r for l and 90° for θ

τ av = F (h − r )

700 Chapter 9 to obtain: Substitute in the expression for Prot to obtain: Because Ptrans = F∆t:

Prot = F (h − r )∆t Prot = Ptrans (h − r ) = ∆L = Iω0 = 52 mr 2ω0

Express the translational impulse delivered to the cue ball:

Ptrans = P0 ∆t = ∆p = mv0

Substitute for Ptrans to obtain:

2 5

mr 2ω0 = mv0

Solve for ω0:

ω0 =

5v0 (h − r ) 2r 2

Substitute numerical values and evaluate ω0:

ω0 =

5(200 m/s )(0.09 m − 0.05 m ) 2 2(.05 m )

= 8000 rad/s (c) and (d) Relate the velocity of the ball when it begins to roll to its acceleration and the elapsed time:

v = a∆t

Apply Newton’s 2nd law to the ball:

∑F ∑F

(1)

x

= f k = ma ,

(2)

y

= Fn − mg = 0 ,

(3)

= f k r = I 0α

(4)

and

∑τ

0

Using the definition of fk and Fn from equation (3), substitute in equation (2) and solve for a:

a = µk g

Substitute in equation (1) to obtain:

v = a∆t = µ k g∆t

Solve for α in equation (4):

α=

fkr mar 5 µk g = 2 2 = I0 2 r 5 mr

(5)

Rotation 701 5µ k g ∆t 2r

Express the angular speed of the ball when it has been moving for a time ∆t:

ω = ω0 − α ∆t = ω0 −

Express the speed of the ball when it has been moving for a time ∆t:

v = v 0 + µ k g∆t

Express the condition that the ball rolls without slipping:

v = rω

Substitute from equations (6) and (7) and solve for the elapsed time until the ball begins to roll:

∆t =

2 rω 0 − v0 7 µk g

∆t =

2 ⎡ (0.05 m )(8000 rad/s ) − 200 m/s ⎤ ⎥ (0.5) 9.81m/s 2 7 ⎢⎣ ⎦

Substitute numerical values and evaluate ∆t:

(6)

(7)

(

)

= 11.6 s Use equation (4) to express v when the ball begins to roll: Substitute numerical values and evaluate v:

v = v0 + µ k g∆t

(

)

v = 200 m/s + (0.5) 9.81 m/s 2 (11.6 s ) = 257 m/s

105 •• Picture the Problem Because the impulse is applied through the center of mass, ω0 = 0. We can use the results of Example 9-16 to find the rolling time without slipping, the distance traveled to rolling without slipping, and the velocity of the ball once it begins to roll without slipping. (a) From Example 9-16 we have:

t1 =

2 v0 7 µk g

Substitute numerical values and evaluate t1:

t1 =

2 4 m/s = 0.194 s 7 (0.6) 9.81 m/s 2

s1 =

12 v02 49 µ k g

(b) From Example 9-16 we have:

(

)

702 Chapter 9 Substitute numerical values and evaluate s1:

(4 m/s) 12 s1 = = 0.666 m 49 (0.6) 9.81 m/s 2

(c) From Example 9-16 we have:

v1 =

5 v0 7

Substitute numerical values and evaluate v1:

v1 =

5 (4 m/s) = 2.86 m/s 7

2

(

)

106 •• Picture the Problem Because the impulsive force is applied below the center line, the spin is backward, i.e., the ball will slow down. We’ll use the impulsemomentum theorem and Newton’s 2nd law to find the linear and rotational velocities and accelerations of the ball and constantacceleration equations to relate these quantities to each other and to the elapsed time to rolling without slipping. (a) Express the rotational impulse delivered to the ball:

Prot = mv0 r = mv0 =

Solve for ω0:

(b) Apply Newton’s 2nd law to the ball to obtain:

ω0 =

∑τ ∑F

(

2 5

)

2R = I cmω0 3

mR 2 ω0

5 v0 3 R = f k R = I cmα ,

(1)

y

= Fn − mg = 0 ,

(2)

x

= − f k = ma

(3)

0

and

∑F

µ k mgR

µ k mgR

Using the definition of fk and Fn from equation (2), solve for α:

α=

Using a constant-acceleration equation, relate the angular speed of the ball to its acceleration:

ω = ω0 + α∆t = ω0 +

I cm

=

2 5

mR

2

=

5µ k g 2R

5µ k g ∆t 2R

Rotation 703 Using the definition of fk and Fn from equation (2), solve equation (3) for a:

a = −µ k g

Using a constant-acceleration equation, relate the speed of the ball to its acceleration:

v = v0 + a∆t = v0 − µ k g∆t

Impose the condition for rolling without slipping to obtain:

5µ g ⎞ ⎛ R⎜ ω 0 + k ∆t ⎟ = v0 − µ k g∆t 2R ⎝ ⎠

Solve for ∆t:

Substitute in equation (4) to obtain:

∆t =

(4)

16 v0 21 µ k g

⎛ 16 v0 ⎞ 5 v = v0 − µ k g ⎜⎜ g ⎟⎟ = v0 µ 21 k ⎠ 21 ⎝ = 0.238v0

(c) Express the initial kinetic energy of the ball:

K i = K trans + K rot = 12 mv02 + 12 Iω02 = mv + 1 2

2 0

(

2

)

⎛ 5v ⎞ 19 mR ⎜ 0 ⎟ = mv02 ⎝ 3R ⎠ 18 2

1 2 2 5

= 1.056mv02 (d) Express the work done by friction in terms of the initial and final kinetic energies of the ball:

Wfr = K i − K f

Express the final kinetic energy of the ball:

K f = 12 mv 2 + 12 I cmω 2

= 12 mv 2 +

1 2

(

2 5

mR 2

) Rv

2 2

= 107 mv 2

= 107 m(0.238v0 ) = 0.0397 mv02 2

Substitute to find Wfr:

Wfr = 1.056mv02 − 0.0397 mv 02

= 1.016mv02

704 Chapter 9 107 •• Picture the Problem The figure shows the forces acting on the bowling during the sliding phase of its motion. Because the ball has a forward spin, the friction force is in the direction of motion and will cause the ball’s translational speed to increase. We’ll apply Newton’s 2nd law to find the linear and rotational velocities and accelerations of the ball and constantacceleration equations to relate these quantities to each other and to the elapsed time to rolling without slipping. (a) and (b) Relate the velocity of the ball when it begins to roll to its acceleration and the elapsed time:

v = v0 + a∆t

(1)

Apply Newton’s 2nd law to the ball:

∑F ∑F

x

= f k = ma ,

(2)

y

= Fn − mg = 0 ,

(3)

= f k R = I 0α

(4)

and

∑τ

0

Using the definition of fk and Fn from equation (3), substitute in equation (2) and solve for a:

a = µk g

Substitute in equation (1) to obtain:

v = v 0 + a∆t = v 0 + µ k g∆t

Solve for α in equation (4):

α=

Relate the angular speed of the ball to its acceleration:

ω = ω0 −

Apply the condition for rolling without slipping:

5 µk g ⎞ ⎛ v = Rω = R⎜ ω0 − ∆t ⎟ 2 R ⎝ ⎠ 5 µk g ⎞ ⎛ 3v = R⎜ 0 − ∆t ⎟ ⎝ R 2 R ⎠

fk R maR 5 µk g = 2 = 2 I0 2 R 5 mR 5 µk g ∆t 2 R

(5)

Rotation 705 ∴ v = 3v0 −

5 µ k g∆t 2

Equate equations (5) and (6) and solve ∆t:

∆t =

Substitute for ∆t in equation (6) to obtain:

v=

(c) Relate ∆x to the average speed of the ball and the time it moves before beginning to roll without slipping:

∆x = vav ∆t =

(6)

4 v0 7 µk g

11 v0 = 1.57v0 7 1 2

(v0 + v )∆t

11 ⎞⎛ 4v0 ⎞ ⎛ ⎟ = 12 ⎜ v0 + v0 ⎟⎜⎜ 7 ⎠⎝ 7 µ k g ⎟⎠ ⎝ =

v2 36 v02 = 0.735 0 49 µ k g µk g

*108 •• Picture the Problem The figure shows the forces acting on the cylinder during the sliding phase of its motion. The friction force will cause the cylinder’s translational speed to decrease and eventually satisfy the condition for rolling without slipping. We’ll use Newton’s 2nd law to find the linear and rotational velocities and accelerations of the ball and constantacceleration equations to relate these quantities to each other and to the distance traveled and the elapsed time until the satisfaction of the condition for rolling without slipping. (a) Apply Newton’s 2nd law to the cylinder:

∑F ∑F

x

= − f k = Ma ,

(1)

y

= Fn − Mg = 0 ,

(2)

= f k R = I 0α

(3)

and

∑τ

Use fk = µkFn to eliminate Fn between equations (1) and (2) and solve for a:

0

a = −µ k g

706 Chapter 9 Using a constant-acceleration equation, relate the speed of the cylinder to its acceleration and the elapsed time:

v = v0 + a∆t = v0 − µ k g∆t

Similarly, eliminate fk between equations (2) and (3) and solve for α:

α=

Using a constant-acceleration equation, relate the angular speed of the cylinder to its acceleration and the elapsed time:

ω = ω0 + α∆t =

Apply the condition for rolling without slipping:

⎛ 2µ g ⎞ v = v0 − µ k g∆t = Rω = R⎜ k ∆t ⎟ ⎝ R ⎠ = 2 µ k g∆t

Solve for ∆t:

Substitute for ∆t in the expression for v: (b) Relate the distance the cylinder travels to its average speed and the elapsed time:

2µ k g R

∆t =

2µ k g ∆t R

v0

3µ k g

v = v0 − µ k g

∆x = vav ∆t =

v0

3µ k g 1 2

=

2 v0 3 ⎛ v0 ⎞ ⎟⎟ ⎝ 3µ k g ⎠

(v0 + 23 v0 )⎜⎜

5 v02 = 18 µ k g

(c) Express the ratio of the energy dissipated in friction to the cylinder’s initial mechanical energy:

Wfr K i − K f = Ki Ki

Express the kinetic energy of the cylinder as it begins to roll without slipping:

K f = 12 Mv 2 + 12 I cmω 2 = 12 Mv 2 + 12

(

1 2

MR 2

) Rv

2 2 2

=

3 3 ⎛2 ⎞ 1 Mv 2 = M ⎜ v0 ⎟ = Mv02 4 4 ⎝3 ⎠ 3

Rotation 707 Wfr 12 Mv02 − 13 Mv02 1 = = 2 1 Ki 3 2 Mv0

Substitute for Ki and Kf and simplify to obtain: 109 •• Picture the Problem The forces acting on the ball as it slides across the floor are its

r

r

weight mg, the normal force Fn exerted by

v

the floor, and the friction force f . Because the weight and normal force act through the center of mass of the ball and are equal in magnitude, the friction force is the net (decelerating) force. We can apply Newton’s 2nd law in both translational and rotational form to obtain a set of equations that we can solve for the acceleration of the ball. Once we have determined the ball’s acceleration, we can use constantacceleration equations to obtain its velocity when it begins to roll without slipping. (a) Apply

r

r

∑ F = ma to the ball:

∑F

x

= − f = ma

(1)

= Fn − mg = 0

(2)

and

∑F

y

From the definition of the coefficient of kinetic friction we have:

f = µ k Fn

Solve equation (2) for Fn:

Fn = mg

Substitute in equation (3) to obtain:

f = µ k mg

Substitute in equation (1) to obtain:

− µ k mg = ma

(3)

or

a = −µk g

Apply

∑τ = Iα to the ball:

Solve for α to obtain:

Assuming that the coefficient of kinetic friction is constant*, we can use constant-acceleration equations to describe how long it will take the ball to begin

fr = Iα

α=

fr µ k mgr = I I

vf − v = a∆t = − µ k g∆t and

ωf =

µ k gmr I

∆t

(4) (5)

708 Chapter 9 rolling without slipping: Once rolling without slipping has been established, we also have: Equate equations (5) and (6):

Solve for ∆t:

Substitute in equation (4) to obtain:

Solve for vf:

(b) Express the total kinetic energy of the ball:

ωf =

vf r

(6)

vf µ k gmr = ∆t r I ∆t =

vf I µ k gmr 2

⎛ vf I ⎞ ⎟ vf − v = − µ k g ⎜⎜ 2 ⎟ ⎝ µ k gmr ⎠ I = − 2 vf mr vf =

K=

1 I 1+ mr 2

v

1 2 1 2 mvf + Iωf 2 2

Because the ball is now rolling without slipping, v = rωf and: 2 2 2 2 1 ⎛ 1 1 1 2 ⎛⎜ 1 ⎞ 2 1 ⎛ ⎞ v ⎞ ⎞⎟ 2 ⎛ = mv 1 + I / mr ⎜ K = m⎜ ⎟ v + I⎜ ⎟ 2 ⎟ ⎜ 2 ⎝ 1 + I / mr 2 ⎠ 2 ⎝ 1 + I / mr 2 ⎠ r 2 2 ⎝ 1 + I / mr ⎠ ⎟⎠ ⎝

(

=

)

1 2⎛ 1 ⎞ mv ⎜ 2 ⎟ 2 ⎝ 1 + I / mr ⎠

* Remarks: This assumption is not necessary. One can use the impulse-momentum theorem and the related theorem for torque and change in angular momentum to prove that the result holds for an arbitrary frictional force acting on the ball, so long as the ball moves along a straight line and the force is directed opposite to the direction of motion of the ball.

General Problems *110 • Picture the Problem The angular velocity of an object is the ratio of the number of revolutions it makes in a given period of time to the elapsed time.

Rotation 709 The moon’s angular velocity is:

1rev 27.3 days 1rev 2π rad 1day 1h = × × × 27.3 days rev 24 h 3600 s

ω=

= 2.66 × 10−6 rad/s 111 • Picture the Problem The moment of inertia of the hoop, about an axis perpendicular to the plane of the hoop and through its edge, is related to its moment of inertia with respect to an axis through its center of mass by the parallel axis theorem. Apply the parallel axis theorem:

I = I cm + Mh 2 = MR 2 + MR 2 = 2mR 2

112 •• Picture the Problem The force you exert on the rope results in a net torque that accelerates the merry-go-round. The moment of inertia of the merry-go-round, its angular acceleration, and the torque you apply are related through Newton’s 2nd law. (a) Using a constant-acceleration equation, relate the angular displacement of the merry-go-round to its angular acceleration and acceleration time:

∆θ = ω 0 ∆t + 12 α (∆t )

2

or, because ω0 = 0,

∆θ = 12 α (∆t )

2

2∆θ 2(2π rad ) = = 0.0873 rad/s 2 2 2 (∆t ) (12 s )

Solve for and evaluate α:

α=

(b) Use the definition of torque to obtain:

τ = Fr = (260 N )(2.2 m ) = 572 N ⋅ m

(c) Use Newton’s 2nd law to find the moment of inertia of the merry-goround:

I=

τ net 572 N ⋅ m = α 0.0873 rad/s 2

= 6.55 × 103 kg ⋅ m 2

710 Chapter 9 113 • Picture the Problem Because there are no horizontal forces acting on the stick, the center of mass of the stick will not move in the horizontal direction. Choose a coordinate system in which the origin is at the horizontal position of the center of mass. The diagram shows the stick in its initial raised position and when it has fallen to the ice. Express the displacement of the right end of the stick ∆x as the difference between the position coordinates x2 and x2:

∆x = x2 − x1

Using trigonometry, find the initial coordinate of the right end of the stick:

x1 = l cos θ = (1 m ) cos30° = 0.866 m

Because the center of mass has not moved horizontally:

x2 = l = 1 m

Substitute to find the displacement of the right end of the stick:

∆x = 1 m − 0.866 m = 0.134 m

114 •• Picture the Problem The force applied to the string results in a torque about the center of mass of the disk that accelerates it. We can relate these quantities to the moment of inertia of the disk through Newton’s 2nd law and then use constant-acceleration equations to find the disk’s angular velocity the angle through which it has rotated in a given period of time. The disk’s rotational kinetic energy can be found from its definition. (a) Use the definition of torque to obtain:

τ ≡ FR = (20 N )(0.12 m ) = 2.40 N ⋅ m

(b) Use Newton’s 2nd law to express the angular acceleration of the disk in terms of the net torque acting on it and its moment of inertia:

α=

Substitute numerical values and evaluate α:

α=

(c) Using a constant-acceleration equation, relate the angular velocity of the disk to its angular

ω = ω 0 + α∆t

τ net I

=

τ net 1 2

MR 2

2(2.40 N ⋅ m ) = 66.7 rad/s 2 2 (5 kg )(0.12 m )

or, because ω0 = 0,

ω = α∆t

Rotation 711 acceleration and the elapsed time: Substitute numerical values and evaluate ω:

ω = (66.7 rad/s 2 )(5 s ) = 333 rad/s

(

(d) Use the definition of rotational kinetic energy to obtain:

K rot = 12 Iω 2 =

Substitute numerical values and evaluate Krot:

K rot =

(e) Using a constant-acceleration equation, relate the angle through which the disk turns to its angular acceleration and the elapsed time:

∆θ = ω 0 ∆t + 12 α (∆t )

1 4

1 1 2 2

)

MR 2 ω 2

(5 kg )(0.12 m )2 (333 rad/s)2

= 2.00 kJ 2

or, because ω0 = 0,

∆θ = 12 α (∆t )

2

(66.7 rad/s )(5 s)

Substitute numerical values and evaluate ∆θ :

∆θ =

(f) Express K in terms of τ and θ :

⎛τ ⎞ 2 2 K = 12 Iω 2 = 12 ⎜ ⎟(α∆t ) = 12 ατ (∆t ) ⎝α ⎠

1 2

2

= τ ∆θ 115 •• Picture the Problem The diagram shows the rod in its initial horizontal position and then, later, as it swings through its vertical position. The center of mass is denoted by the numerals 0 and 1. Let the length of the rod be represented by L and its mass by m. We can use Newton’s 2nd law in rotational form to find, first, the angular acceleration of the rod and then, from α, the acceleration of any point on the rod. We can use conservation of energy to find the angular velocity of the center of mass of the rod when it is vertical and then use this value to find its linear velocity. (a) Relate the acceleration of the center of the rod to the angular

a = lα =

L α 2

2

= 834 rad

712 Chapter 9 acceleration of the rod: Use Newton’s 2nd law to relate the torque about the suspension point of the rod (exerted by the weight of the rod) to the rod’s angular acceleration:

L 3g τ α = = 1 22 = I 3 ML 2L

Substitute numerical values and evaluate α:

α=

3 9.81 m/s 2 = 18.4 rad/s 2 2(0.8 m )

Substitute numerical values and evaluate a:

a=

1 2

(b) Relate the acceleration of the end of the rod to α:

Mg

(

)

(0.8 m )(18.4 rad/s 2 ) =

(

7.36 m/s 2

aend = Lα = (0.8 m ) 18.4 rad/s 2

)

= 14.7 m/s 2

(c) Relate the linear velocity of the center of mass of the rod to its angular velocity as it passes through the vertical:

v = ω∆h = 12 ωL

Use conservation of energy to relate the changes in the kinetic and potential energies of the rod as it swings from its initial horizontal orientation through its vertical orientation:

∆K + ∆U = K1 − K 0 + U 1 − U 0 = 0

Substitute to obtain: Substitute for ∆h and solve for ω:

Substitute to obtain:

Substitute numerical values and evaluate v:

or, because K0 = U1 = 0,

K1 − U 0 = 0

1 2

I Pω 2 = mg∆h 3g L

ω=

v = 12 L

v=

1 2

3g = L

(

1 2

3 gL

)

3 9.81 m/s 2 (0.8 m ) = 2.43 m/s

Rotation 713 116 •• Picture the Problem Let the zero of gravitational potential energy be at the bottom of the track. The initial potential energy of the marble is transformed into translational and rotational kinetic energy as it rolls down the track to its lowest point and then, because the portion of the track to the right is frictionless, into translational kinetic energy and, eventually, into gravitational potential energy.

∆K + ∆U = 0

Using conservation of energy, relate h2 to the kinetic energy of the marble at the bottom of the track:

or, because Kf = Ui = 0,

Substitute for Ki and Uf to obtain:

− 12 Mv 2 − Mgh2 = 0

Solve for h2:

Using conservation of energy, relate h1 to the kinetic energy of the marble at the bottom of the track:

− Ki + U f = 0

h2 =

v2 2g

(1)

∆K + ∆U = 0 or, because Ki = Uf = 0,

Kf −Ui = 0

Mv 2 + 12 Iω 2 − Mgh1 = 0

Substitute for Kf and Ui to obtain:

1 2

Substitute for I and solve for v2 to obtain:

v 2 = 107 gh1

Substitute in equation (1) to obtain:

h2 =

gh1 = 2g

10 7

5 7

h1

*117 •• Picture the Problem To stop the wheel, the tangential force will have to do an amount of work equal to the initial rotational kinetic energy of the wheel. We can find the stopping torque and the force from the average power delivered by the force during the slowing of the wheel. The number of revolutions made by the wheel as it stops can be found from a constant-acceleration equation. (a) Relate the work that must be done to stop the wheel to its kinetic energy:

W = 12 Iω 2 =

(

1 1 2 2

)

mr 2 ω 2 = 14 mr 2ω 2

714 Chapter 9 Substitute numerical values and evaluate W:

W=

1 4

(120 kg )(1.4 m )2

⎡ rev 2π rad 1min ⎤ × ⎢1100 × × min rev 60 s ⎥⎦ ⎣

2

= 780 kJ (b) Express the stopping torque is terms of the average power required:

Pav = τω av

Solve for τ :

τ=

Substitute numerical values and evaluate τ :

780 kJ (2.5 min )(60 s/min ) τ= (1100 rev/min )(2π rad/rev)(1 min/60 s ) 2

Pav

ωav

= 90.3 N ⋅ m

τ

90.3 N ⋅ m = 151 N 0.6 m

Relate the stopping torque to the magnitude of the required force and solve for F:

F=

(c) Using a constant-acceleration equation, relate the angular displacement of the wheel to its average angular velocity and the stopping time:

∆θ = ωav ∆t

Substitute numerical values and evaluate ∆θ:

⎛ 1100 rev/min ⎞ ∆θ = ⎜ ⎟ (2.5 min ) 2 ⎝ ⎠

R

=

= 1380 rev 118 •• Picture the Problem The work done by the four children on the merry-go-round will change its kinetic energy. We can use the work-energy theorem to relate the work done by the children to the distance they ran and Newton’s 2nd law to find the angular acceleration of the merry-go-round.

Rotation 715 (a) Use the work-kinetic energy theorem to relate the work done by the children to the kinetic energy of the merry-go-round: Substitute for I and solve for ∆s to obtain:

Substitute numerical values and evaluate ∆s:

Wnet force = ∆K = Kf or

4 F∆s = 12 Iω 2 ∆s =

Iω 2 12 mr 2ω 2 mr 2ω 2 = = 8F 8F 16 F

⎡ ⎤ (240 kg )(2 m ) ⎢1rev × 2π rad ⎥ rev ⎦ ⎣ 2.8 s ∆s = 16(26 N )

2

2

= 11.6 m

τ net

4 Fr 8F = 2 mr mr

(b) Apply Newton’s 2nd law to express the angular acceleration of the merry-go-round:

α=

Substitute numerical values and evaluate α:

α=

(c) Use the definition of work to relate the force exerted by each child to the distance over which that force is exerted:

W = F∆s = (26 N )(11.6 m ) = 302 J

(d) Relate the kinetic energy of the merry-go-round to the work that was done on it:

Wnet force = ∆K = K f − 0 = 4 F∆s

Substitute numerical values and evaluate Wnet force:

Wnet force = 4(26 N )(11.6 m ) = 1.21 kJ

I

=

1 2

8(26 N ) = 0.433 rad/s 2 (240 kg )(2 m )

119 •• Picture the Problem Because the center of mass of the hoop is at its center, we can use Newton’s second law to relate the acceleration of the hoop to the net force acting on it. The distance moved by the center of the hoop can be determined using a constantacceleration equation, as can the angular velocity of the hoop. (a) Using a constant-acceleration equation, relate the distance the

∆s = 12 a cm (∆t )

2

716 Chapter 9 center of the travels in 3 s to the acceleration of its center of mass: Relate the acceleration of the center of mass of the hoop to the net force acting on it: Substitute to obtain:

Substitute numerical values and evaluate ∆s:

a cm =

Fnet m

F (∆t ) 2m

2

∆s =

2 ( 5 N )(3 s ) ∆s = 2(1.5 kg )

= 15.0 m

(b) Relate the angular velocity of the hoop to its angular acceleration and the elapsed time:

ω = α ∆t

Use Newton’s 2nd law to relate the angular acceleration of the hoop to the net torque acting on it:

α=

Substitute to obtain:

ω=

F∆t mR

Substitute numerical values and evaluate ω:

ω=

(5 N )(3 s ) = (1.5 kg )(0.65 m )

τ net I

=

FR F = 2 mR mR

15.4 rad/s

120 •• Picture the Problem Let R represent the radius of the grinding wheel, M its mass, r the radius of the handle, and m the mass of the load attached to the handle. In the absence of information to the contrary, we’ll treat the 25-kg load as though it were concentrated at a point. Let the zero of gravitational potential energy be where the 25-kg load is at its lowest point. We’ll apply Newton’s 2nd law and the conservation of mechanical energy to determine the initial angular acceleration and the maximum angular velocity of the wheel. (a) Use Newton’s 2nd law to relate the acceleration of the wheel to the net torque acting on it:

α=

τ net I

=

1 2

mgr MR 2 + mr 2

Rotation 717 Substitute numerical values and evaluate α:

α=

(25 kg )(9.81m/s2 )(0.65 m ) 2 2 1 2 (60 kg )(0.45 m ) + (25 kg )(0.65 m )

= 9.58 rad/s 2 (b) Use the conservation of mechanical energy to relate the initial potential energy of the load to its kinetic energy and the rotational kinetic energy of the wheel when the load is directly below the center of mass of the wheel: Substitute and solve for ω:

∆K + ∆U = 0 or, because Ki = Uf = 0,

K f,trans + K f,rot − U i = 0 .

(

)

MR 2 ω 2 − mgr = 0 ,

1 2

mv 2 + 12

1 2

mr ω + MR 2ω 2 − mgr = 0 , 2

1 2

2

1 4

and

Substitute numerical values and evaluate ω:

ω=

4mgr 2mr 2 + MR 2

ω=

4(25 kg ) 9.81 m/s 2 (0.65 m ) 2 2 2(25 kg )(0.65 m ) + (60 kg )(0.45 m )

(

)

= 4.38 rad/s

*121 •• Picture the Problem Let the smaller block have the dimensions shown in the diagram. Then the length, height, and width of the larger block are Sl, Sh, and Sw, respectively. Let the numeral 1 denote the smaller block and the numeral 2 the larger block and express the ratios of the surface areas, masses, and moments of inertia of the two blocks. (a) Express the ratio of the surface areas of the two blocks:

A2 2(Sw)(Sl ) + 2(Sl )(Sh ) + 2(Sw)(Sh ) = A1 2wl + 2lh + 2wh =

S2 (2wl + 2lh + 2wh ) 2 wl + 2lh + 2 wh

= S2

718 Chapter 9 (b) Express the ratio of the masses of the two blocks:

M 2 ρV2 V2 (Sw)(Sl )(Sh ) = = = M 1 ρV1 V1 w lh =

(c) Express the ratio of the moments of inertia, about the axis shown in the diagram, of the two blocks:

I2 = I1 =

S3 (wlh ) = S3 wlh

1 12

[

M 2 (Sl ) + (Sh ) 2 2 1 12 M 1 l + h 2

[

2

[

]

]

]

M 2 S2 l 2 + h 2 ⎛ M 2 ⎞ 2 ⎟⎟ S = ⎜⎜ M1 l2 + h2 M ⎝ 1⎠

[

]

In part (b) we showed that:

M2 = S3 M1

Substitute to obtain:

I2 = (S3 )(S2 ) = S5 I1

( )

122 •• Picture the Problem We can derive the perpendicular-axis theorem for planar objects by following the step-by-step procedure outlined in the problem. (a) and (b)

(

)

I z = ∫ r 2 dm = ∫ x 2 + y 2 dm = ∫ x 2 dm + ∫ y 2 dm = Ix + Iy

(c) Let the z axis be the axis of rotation of the disk. By symmetry:

Ix = Iy

Express Iz in terms of Ix:

I z = 2I x

Letting M represent the mass of the disk, solve for Ix:

I x = 12 I z =

1 2

(

1 2

)

MR 2 =

1 4

MR 2

123 •• Picture the Problem Let the zero of gravitational potential energy be at the center of the disk when it is directly below the pivot. The initial gravitational potential energy of the disk is transformed into rotational kinetic energy when its center of mass is directly below the pivot. We can use Newton’s 2nd law to relate the force exerted by the pivot to the weight of the disk and the centripetal force acting on it at its lowest point.

Rotation 719 (a) Use the conservation of mechanical energy to relate the initial potential energy of the disk to its kinetic energy when its center of mass is directly below the pivot:

∆K + ∆U = 0 or, because Ki = Uf = 0,

K f,rot − U i = 0

Iω 2 − Mgr = 0

Substitute for K f,rot and U i :

1 2

Use the parallel-axis theorem to relate the moment of inertia of the disk about the pivot to its moment of inertia with respect to an axis through its center of mass:

I = I cm + Mh 2

Solve equation (1) for ω and substitute for I to obtain:

(1)

or

I = 12 Mr 2 + Mr 2 = 32 Mr 2

ω=

4g 3r

(b) Letting F represent the force exerted by the pivot, use Newton’s 2nd law to express the net force acting on the swinging disk as it passes through its lowest point:

Fnet = F − Mg = Mrω 2

Solve for F and simplify to obtain:

F = Mg + Mrω 2 = Mg + Mr = Mg + 43 Mg =

124 •• Picture the Problem The diagram shows a vertical cross-piece. Because we’ll need to take moments about the point of rotation (point P), we’ll need to use the parallelaxis theorem to find the moments of inertia of the two parts of this composite structure. Let the numeral 1 denote the vertical member and the numeral 2 the horizontal member. We can apply Newton’s 2nd law in rotational form to the structure to express its angular acceleration in terms of the net torque causing it to fall and its moment of inertia with respect to point P.

7 3

Mg

4g 3r

720 Chapter 9 (a) Taking clockwise rotation to be positive (this is the direction the structure is going to rotate), apply τ = I Pα :

⎛l ⎞ ⎛ w⎞ m2 g ⎜ 2 ⎟ − m1 g ⎜ ⎟ = I Pα ⎝ 2⎠ ⎝2⎠

∑

Solve for α to obtain:

α= or

α= Convert l 1 , l 2 , and w to SI units:

Using Table 9-1 and the parallelaxis theorem, express the moment of inertia of the vertical member about an axis through point P:

m2 gl 2 − m1 gw 2I P g (m2 l 2 − m1w) 2(I1P + I 2 P )

1m = 3.66 m , 3.281ft 1m l 2 = 6 ft × = 1.83 m , and 3.281ft 1m w = 2 ft × = 0.610 m 3.281ft

l 1 = 12 ft ×

⎛ w⎞ = m l + m1 ⎜ ⎟ ⎝2⎠ 2 = m1 13 l 1 + 14 w 2

I1P

2

2 1 1

1 3

(

)

[ (3.66 m) +

Substitute numerical values and evaluate I1P:

I1P = (350 kg )

Using the parallel-axis theorem, express the moment of inertia of the horizontal member about an axis through point P:

I 2 P = I 2,cm + m2 d 2

Solve for d:

(1)

2

1 3

1 4

(0.610 m )2 ]

= 1.60 × 103 kg ⋅ m 2 (2)

where

d 2 = (l 1 + 12 w) + ( 12 l 2 − w) 2

2

(l 1 + 12 w)2 + ( 12 l 2 − w)2

d=

Substitute numerical values and evaluate d:

d=

[3.66 m + 12 (0.610 m )] 2 + [12 (1.83 m ) − 0.610 m] 2

From Table 9-1 we have:

I 2,cm = 121 m2 l 22

Substitute in equation (2) to obtain:

I2P =

1 12

m2l 22 + m2 d 2

= m2

(

1 12

l 22 + d 2

)

= 3.86 m

Rotation 721

[

I 2 P = (175 kg ) 121 (1.83 m ) + (3.86 m )

Evaluate I2P:

2

2

]

= 2.66 × 103 kg ⋅ m 2 Substitute in equation (1) and evaluate α:

α=

(9.81m/s )[(175 kg )(1.83 m) − (350 kg )(0.61m)] = 2

2(1.60 + 2.66)× 10 kg ⋅ m 3

(b) Express the magnitude of the acceleration of the sparrow:

2

0.123 rad/s 2

a = αR where R is the distance of the sparrow from the point of rotation and

R 2 = (l 1 + w) + (l 2 − w) 2

Solve for R:

2

(l 1 + w)2 + (l 2 − w)2

R=

Substitute numerical values and evaluate R:

R=

(3.66 m + 0.610 m )2 + (1.83 m − 0.610 m )2

Substitute numerical values and evaluate a:

(

= 4.44 m

)

a = 0.123 rad/s 2 (4.44 m ) = 0.546 m/s 2

(c) Refer to the diagram to express ax in terms of a:

a x = a cos θ = a

Substitute numerical values and evaluate ax:

ax = 0.546 m/s 2

(

l1 + w R m + 0.61 m ) 3.664.44 m

= 0.525 m/s 2

722 Chapter 9 125 •• Picture the Problem Let the zero of gravitational potential energy be at the bottom of the incline. The initial potential energy of the spool is transformed into rotational and translational kinetic energy when the spool reaches the bottom of the incline. We can apply the conservation of mechanical energy to find an expression for its speed at that location. The force diagram shows the forces acting on the spool when there is enough friction to keep it from slipping. We’ll use Newton’s 2nd law in both translational and rotational form to derive an expression for the static friction force.

The spool will move down the plane

(a) In the absence of friction, the forces acting on the spool will be its weight, the normal force exerted by the incline, and the tension in the string. A component of its weight will cause the spool to accelerate down the incline and the tension in the string will exert a torque that will cause counterclockwise rotation of the spool. Use the conservation of mechanical energy to relate the speed of the center of mass of the spool at the bottom of the slope to its initial potential energy: Substitute for K f, trans , K f,rot and U i : Substitute for ω and solve for v to obtain:

at constant acceleration, spinning in a counterclockwise direction as string unwinds.

∆K + ∆U = 0 or, because Ki = Uf = 0,

K f,trans + K f,rot − U i = 0 .

1 2

1 2

Mv 2 + 12 Iω 2 − MgD sin θ = 0

v2 Mv + I 2 − MgD sin θ = 0 r 2

1 2

and

v=

2MgD sin θ I M+ 2 r

(1)

Rotation 723 (b) Apply Newton’s 2nd law to the spool:

Eliminate T between these equations to obtain:

∑F ∑τ fs =

x

= Mg sin θ − T − f s = 0

0

= Tr − f s R = 0 Mg sin θ , up the incline. R 1+ r

126 •• Picture the Problem While the angular acceleration of the rod is the same at each point along its length, the linear acceleration and, hence, the force exerted on each coin by the rod, varies along its length. We can relate this force the linear acceleration of the rod through Newton’s 2nd law and the angular acceleration of the rod.

Fnet = mg − F ( x ) = ma( x )

Letting x be the distance from the pivot, use Newton’s 2nd law to express the force F acting on a coin:

or

Use Newton’s 2nd law to relate the angular acceleration of the system to the net torque acting on it:

L τ 3g α = net = 1 22 = I 2L 3 ML

Relate a(x) and α:

a( x ) = xα = x

Substitute in equation (1) to obtain:

F ( x ) = m( g − gx ) = mg (1 − x )

Evaluate F(0.25 m):

F (0.25 m ) = mg (1 − 0.25 m ) = 0.75mg

Evaluate F(0.5 m):

F (0.5 m ) = mg (1 − 0.5 m ) = 0.5mg

Evaluate F(0.75 m):

F (0.75 m ) = mg (1 − 0.75 m ) = 0.25mg

Evaluate F(1 m):

F (1 m ) = F (1.25 m ) = F (1.5 m ) = 0

F ( x ) = m(g − a ( x ))

(1)

Mg

3g = gx 2(1.5 m )

*127 •• Picture the Problem The diagram shows the force the hand supporting the meterstick exerts at the pivot point and the force the earth exerts on the meterstick acting at the center of mass. We can relate the angular acceleration to the acceleration of the end of the meterstick using a = Lα and use Newton’s 2nd law in rotational form to relate α to the moment of inertia of the meterstick.

724 Chapter 9

(a) Relate the acceleration of the far end of the meterstick to the angular acceleration of the meterstick: Apply

∑τ

P

= I Pα to the

meterstick: Solve for α:

From Table 9-1, for a rod pivoted at one end, we have: Substitute to obtain:

a = Lα

(1)

⎛L⎞ Mg ⎜ ⎟ = I Pα ⎝2⎠

α=

MgL 2I P

1 I P = ML2 3

α=

3MgL 3g = 2ML2 2 L

a=

3g 2

Substitute numerical values and evaluate a:

a=

3(9.81 m/s 2 ) = 14.7 m/s 2 2

(b) Express the acceleration of a point on the meterstick a distance x from the pivot point:

a = αx =

Substitute in equation (1) to obtain:

Express the condition that the meterstick leaves the penny behind:

a>g

Substitute to obtain:

3g x>g 2L

Solve for and evaluate x:

x>

3g x 2L

2 L 2(1 m ) = = 66.7 cm 3 3

Rotation 725 128 •• Picture the Problem Let m represent the 0.2-kg mass, M the 0.8-kg mass of the cylinder, L the 1.8-m length, and x + ∆x the distance from the center of the objects whose mass is m. We can use Newton’s 2nd law to relate the radial forces on the masses to the spring’s stiffness constant and use the work-energy theorem to find the work done as the system accelerates to its final angular speed. (a) Express the net inward force acting on each of the 0.2-kg masses: Solve for k:

Substitute numerical values and evaluate k:

∑F

radial

= k∆x = m(x + ∆x )ω 2

m( x + ∆x )ω k= ∆x k=

2

(0.2 kg )(0.8 m )(24 rad/s)2 0.4 m

= 230 N/m (b) Using the work-energy theorem, relate the work done to the change in energy of the system:

W = K rot + ∆U spring

Express I as the sum of the moments of inertia of the cylinder and the masses:

I = I M + I 2m

From Table 9-1 we have, for a solid cylinder about a diameter through its center:

I = 14 mr 2 + 121 mL2

= 12 Iω 2 + 12 k (∆x )

2

(1)

= 12 Mr 2 + 121 ML2 + 2 I m

where L is the length of the cylinder.

For a disk (thin cylinder), L is small and:

I = 14 mr 2

Apply the parallel-axis theorem to obtain:

I m = 14 mr 2 + mx 2

Substitute to obtain:

I = 12 Mr 2 + 121 ML2 + 2(14 mr 2 + mx 2 )

Substitute numerical values and evaluate I:

= 12 Mr 2 + 121 ML2 + 2m(14 r 2 + x 2 )

726 Chapter 9 I=

1 2

(0.8 kg )(0.2 m )2 + 121 (0.8 kg )(1.8 m )2 + 2(0.2 kg )[14 (0.2 m )2 + (0.8 m )2 ]

= 0.492 N ⋅ m 2 Substitute in equation (1) to obtain:

W =

1 2

(0.492 N ⋅ m )(24 rad/s) 2

2

+ 12 (230 N/m )(0.4 m ) = 160 J 2

129 •• Picture the Problem Let m represent the 0.2-kg mass, M the 0.8-kg mass of the cylinder, L the 1.8-m length, and x + ∆x the distance from the center of the objects whose mass is m. We can use Newton’s 2nd law to relate the radial forces on the masses to the spring’s stiffness constant and use the work-energy theorem to find the work done as the system accelerates to its final angular speed. Using the work-energy theorem, relate the work done to the change in energy of the system:

W = K rot + ∆U spring

Express I as the sum of the moments of inertia of the cylinder and the masses:

I = I M + I 2m

From Table 9-1 we have, for a solid cylinder about a diameter through its center:

I = 14 mr 2 + 121 mL2

= 12 Iω 2 + 12 k (∆x )

(1)

2

= 12 Mr 2 + 121 ML2 + 2 I m

where L is the length of the cylinder.

For a disk (thin cylinder), L is small and:

I = 14 mr 2

Apply the parallel-axis theorem to obtain:

I m = 14 mr 2 + mx 2

Substitute to obtain:

I = 12 Mr 2 + 121 ML2 + 2 14 mr 2 + mx 2

(

(

= 12 Mr 2 + 121 ML2 + 2m 14 r 2 + x 2

)

Substitute numerical values and evaluate I:

I=

1 2

(0.8 kg )(0.2 m )2 + 121 (0.8 kg )(1.8 m )2 + 2(0.2 kg )[14 (0.2 m )2 + (0.8 m )2 ]

= 0.492 N ⋅ m 2

)

Rotation 727 Express the net inward force acting on each of the 0.2-kg masses: Solve for ω:

Substitute numerical values and evaluate ω: Substitute numerical values in equation (1) to obtain:

∑F

radial

= k∆x = m(x + ∆x )ω 2

ω=

k∆x m( x + ∆x )

ω=

(60 N/m )(0.4 m ) = 12.2 rad/s (0.2 kg )(0.8 m )

W=

1 2

(0.492 N ⋅ m )(12.2 rad/s)

2

2

+ 12 (60 N/m )(0.4 m )

2

= 41.4 J 130 •• Picture the Problem The force diagram shows the forces acting on the cylinder. Because F causes the cylinder to rotate clockwise, f, which opposes this motion, is to the right. We can use Newton’s 2nd law in both translational and rotational forms to relate the linear and angular accelerations to the forces acting on the cylinder.

τ net

FR 2F = 2 1 MR 2 MR

(a) Use Newton’s 2nd law to relate the angular acceleration of the center of mass of the cylinder to F:

α=

Use Newton’s 2nd law to relate the acceleration of the center of mass of the cylinder to F:

acm =

Express the rolling-without-slipping condition to the accelerations:

α' =

(b) Take the point of contact with the floor as the ″pivot″ point, express the net torque about that point, and solve for α:

τ net = 2 FR = Iα

I

=

Fnet F = M M

acm F = = 2α R MR

and

α=

2 FR I

728 Chapter 9 Express the moment of inertia of the cylinder with respect to the pivot point:

I = 12 MR 2 + MR 2 = 32 MR 2

Substitute to obtain:

α=

3 2

2 FR 4F = 2 MR 3MR 4F 3M

Express the linear acceleration of the cylinder:

acm = Rα =

Apply Newton’s 2nd law to the forces acting on the cylinder:

∑F

Solve for f:

f = Macm − F =

x

=

= F + f = Ma cm

1 3

4F −F 3

F in the positive x direction.

131 •• Picture the Problem As the load falls, mechanical energy is conserved. As in Example 9-7, choose the initial potential energy to be zero. Apply conservation of mechanical energy to obtain an expression for the speed of the bucket as a function of its position and use the given expression for t to determine the time required for the bucket to travel a distance y. Apply conservation of mechanical energy:

U f + Kf = U i + Ki = 0 + 0 = 0

Express the total potential energy when the bucket has fallen a distance y:

U f = U bf + U cf + U wf

(1)

⎛ y⎞ = −mgy − mc'g ⎜ ⎟ ⎝2⎠ where mc' is the mass of the hanging part of the cable.

Assume the cable is uniform and express mc' in terms of mc, y, and L:

mc' mc m = or mc' = c y y L L

Substitute to obtain:

mc gy 2 U f = −mgy − 2L

Rotation 729 K f = K bf + K cf + K wf

Noting that bucket, cable, and rim of the winch have the same speed v, express the total kinetic energy when the bucket is falling with speed v:

= 12 mv 2 + 12 mc v 2 + 12 Iωf2 = 12 mv 2 + 12 mc v 2 + 12 (12 MR 2 )

v2 R2

= 12 mv 2 + 12 mc v 2 + 14 Mv 2 Substitute in equation (1) to obtain:

− mgy −

Solve for v:

v=

mc gy 2 1 2 + 2 mv 2L + 12 mc v 2 + 14 Mv 2 = 0

2mc gy 2 L M + 2m + 2mc

4mgy +

A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content D9 0 D10 D9+$B$8 E9 0 E10 ((4*$B$3*$B$7*D10+2*$B$7*D10^2/(2*$B$5))/ ($B$1+2*$B$3+2*$B$4))^0.5

F10

F9+$B$8/((E10+E9)/2)

J9

0.5*$B$7*H9^2

1 2 3 4 5 6 7 8 9 10 11 12 13 15

A M= R= m= mc= L=

B 10 0.5 5 3.5 10

g= 9.81 dy= 0.1

C kg m kg kg m m/s^2 m

Algebraic Form y0 y + ∆y v0

2mc gy 2 L M + 2m + 2mc

4mgy +

⎛v +v ⎞ t n−1 + ⎜ n−1 n ⎟∆y 2 ⎝ ⎠ 1 2

D

E

F

y 0.0 0.1 0.2 0.3 0.4 0.5

v(y) 0.00 0.85 1.21 1.48 1.71 1.91

t(y) 0.00 0.23 0.33 0.41 0.47 0.52

G

gt 2

H

I

J

t(y) 0.00 0.23 0.33 0.41 0.47 0.52

y 0.0 0.1 0.2 0.3 0.4 0.5

1/2gt^2 0.00 0.27 0.54 0.81 1.08 1.35

730 Chapter 9 105 106 107 108 109

9.6 9.7 9.8 9.9 10.0

9.03 9.08 9.13 9.19 9.24

2.24 2.25 2.26 2.27 2.28

2.24 2.25 2.26 2.27 2.28

9.6 9.7 9.8 9.9 10.0

24.61 24.85 25.09 25.34 25.58

The solid line on the graph shown below shows the position y of the bucket when it is in free fall and the dashed line shows y under the conditions modeled in this problem. 20 18 16

y'

14

free fall

y (m)

12 10 8 6 4 2 0 0.0

0.4

0.8

1.2

1.6

t (s)

132 •• Picture the Problem The pictorial representation shows the forces acting on the cylinder when it is stationary. First, we note that if the tension is small, then there can be no slipping, and the system must roll. Now consider the point of contact of the cylinder with the surface as the “pivot” point. If τ about that point is zero, the system will not roll. This will occur if the line of action of the tension passes through the pivot point. From the diagram we see that:

⎛r⎞ ⎝R⎠

θ = cos −1 ⎜ ⎟

2.0

Rotation 731 *133 •• Picture the Problem Free-body diagrams for the pulley and the two blocks are shown to the right. Choose a coordinate system in which the direction of motion of the block whose mass is M (downward) is the positive y direction. We can use the given µ ∆θ relationship T 'max = Te s to relate the tensions in the rope on either side of the pulley and apply Newton’s 2nd law in both rotational form (to the pulley) and translational form (to the blocks) to obtain a system of equations that we can solve simultaneously for a, T1, T2, and M. (a) Use T 'max = Te s to evaluate the maximum tension required to prevent the rope from slipping on the pulley:

T 'max = (10 N ) e (0.3 )π = 25.7 N

(c) Given that the angle of wrap is π radians, express T2 in terms of T1:

T2 = T1e 0.3π = 2.57T1

µ ∆θ

Because the rope doesn’t slip, we can relate the angular acceleration, α, of the pulley to the acceleration, a, of the hanging masses by: Apply

∑F

y

= ma y to the two

blocks to obtain:

Apply

∑τ = Iα to the pulley to

obtain:

α=

(1)

a r

T1 − mg = ma

(2)

and

Mg − T2 = Ma

(3)

(T2 − T1 ) r = I a

(4)

r

Substitute for T2 from equation (1) in equation (4) to obtain:

(2.57T1 − T1 ) r = I a

Solve for T1 and substitute numerical values to obtain:

I 0.35 kg ⋅ m 2 T1 = a= a 2 (5) 1.57 r 2 1.57(0.15 m )

r

= (9.91 kg )a

Substitute in equation (2) to obtain:

(9.91 kg )a − mg = ma

732 Chapter 9 Solve for and evaluate a:

(b) Solve equation (3) for M:

mg g = 9.91 kg − m 9.91 kg − 1 m 2 9.81 m/s = = 1.10 m/s 2 9.91 kg −1 1 kg

a=

T2 g −a

M =

Substitute in equation (5) to find T1:

T1 = (9.91 kg ) (1.10 m/s 2 ) = 10.9 N

Substitute in equation (1) to find T2:

T2 = (2.57 )(10.9 N ) = 28.0 N

Evaluate M:

28.0 N = 3.21 kg 9.81 m/s 2 − 1.10 m/s 2

M =

134 ••• Picture the Problem When the tension is horizontal, the cylinder will roll forward and r the friction force will be in the direction of T . We can use Newton’s 2nd law to obtain equations that we can solve simultaneously for a and f. (a) Apply Newton’s 2nd law to the cylinder:

∑F

= T + f = ma

(1)

∑τ = Tr − fR = Iα

(2)

x

and

a 1 = 2 mRa (3) R

Substitute for I and α in equation (2) to obtain:

Tr − fR = 12 mR 2

Solve equation (3) for f:

f =

Tr 1 − ma R 2

(4)

Substitute equation (4) in equation (1) and solve for a:

a=

2T ⎛ r⎞ ⎜1 + ⎟ 3m ⎝ R ⎠

(5)

Substitute equation (5) in equation (4) to obtain:

f =

T ⎛ 2r ⎞ ⎜ − 1⎟ 3⎝ R ⎠

(b) Equation (4) gives the acceleration of the center of mass:

a=

2T ⎛ r⎞ ⎜1 + ⎟ 3m ⎝ R ⎠

Rotation 733 (c) Express the condition that a >

T : m

2T ⎛ r⎞ T 2⎛ r⎞ ⎜1 + ⎟ > ⇒ ⎜ 1 + ⎟ > 1 3m ⎝ R ⎠ m 3⎝ R⎠ or

r>

1 2

R

r f > 0, i.e., in the direction of T .

(d) If r > 12 R :

135 ••• Picture the Problem The system is shown in the drawing in two positions, with angles θ0 and θ with the vertical. The drawing also shows all the forces that act on the stick. These forces result in a rotation of the stick—and its center of mass—about the pivot, and a tangential acceleration of the center of mass. We’ll apply the conservation of mechanical energy and Newton’s 2nd law to relate the radial and tangential forces acting on the stick. Use the conservation of mechanical energy to relate the kinetic energy of the stick when it makes an angle θ with the vertical and its initial potential energy:

Kf − Ki + U f − U i = 0 or, because Kf = 0,

− 12 Iω 2 + Mg

L L cosθ − Mg cosθ 0 = 0 2 2

3g (cosθ − cosθ 0 ) L

Substitute for I and solve for ω2:

ω2 =

Express the centripetal force acting on the center of mass:

Fc = M

L 2 ω 2 L 3g (cosθ − cosθ 0 ) =M 2 L 3Mg (cosθ − cosθ 0 ) = 2

r

Express the radial component of Mg :

(Mg )radial = Mg cosθ

Express the total radial force at the hinge:

F|| = Fc + (Mg)radial

734 Chapter 9 = = Relate the tangential acceleration of the center of mass to its angular acceleration:

3Mg (cos θ − cos θ 0 ) + Mg cos θ 2 1 2

Mg (5 cos θ − 3 cos θ 0 )

a⊥= 12 Lα

L sin θ 3g sin θ 2 = 2 1 2L 3 ML

Use Newton’s 2nd law to relate the angular acceleration of the stick to the net torque acting on it:

α=

Express a⊥ in terms of α:

a⊥= 12 Lα =

Solve for F⊥ to obtain:

F⊥ = − 14 Mg sin θ where the minus sign

τ net I

=

Mg

3 4

gsinθ = gsinθ + F⊥/M

indicates that the force is directed oppositely to the tangential component of

r Mg.

Chapter 10 Conservation of Angular Momentum Conceptual Problems *1 • r r r r (a) True. The cross product of the vectors A and B is defined to be A × B = AB sin φ nˆ.

r

r

If A and B are parallel, sinφ = 0.

r

(b) True. By definition, ω is along the axis. (c) True. The direction of a torque exerted by a force is determined by the definition of the cross product. 2 • r r Determine the Concept The cross product of the vectors A and B is defined to be

r r A × B = AB sin φ nˆ. Hence, the cross product is a maximum when sinφ = 1. This r r condition is satisfied provided A and B are perpendicular. (c) is correct.

3 • r r r r r Determine the Concept L and p are related according to L = r × p. From this

r

r

definition of the cross product, L and p are perpendicular; i.e., the angle between them is 90°. 4 • r r r r r Determine the Concept L and p are related according to L = r × p. Because the motion is along a line that passes through point P, r = 0 and so is L. (b) is correct. *5 •• r r r r r Determine the Concept L and p are related according to L = r × p.

r

r r Doubling p doubles L.

r

r r Doubling r doubles L.

(a) Because L is directly proportional r to p : (b) Because L is directly proportional r to r :

735

736 Chapter 10 6 •• Determine the Concept The figure shows a particle moving with constant speed in a straight line (i.e., with constant velocity and constant linear momentum). The magnitude of L is given by rpsinφ = mv(rsinφ). Referring to the diagram, note that the distance rsinφ from P to the line along which the particle is moving is constant. Hence, mv(rsinφ) is constant and so

r L is constant.

7 • False. The net torque acting on a rotating system equals the change in the system’s angular momentum; i.e., τ net = dL dt , where L = Iω. Hence, if τ net is zero, all we can say for sure is that the angular momentum (the product of I and ω) is constant. If I changes, so mustω. *8 •• Determine the Concept Yes, you can. Imagine rotating the top half of your body with arms flat at sides through a (roughly) 90° angle. Because the net angular momentum of the system is 0, the bottom half of your body rotates in the opposite direction. Now extend your arms out and rotate the top half of your body back. Because the moment of inertia of the top half of your body is larger than it was previously, the angle which the bottom half of your body rotates through will be smaller, leading to a net rotation. You can repeat this process as necessary to rotate through any arbitrary angle. 9 • Determine the Concept If L is constant, we know that the net torque acting on the system is zero. There may be multiple constant or time-dependent torques acting on the system as long as the net torque is zero. (e) is correct. 10 •• Determine the Concept No. In order to do work, a force must act over some distance. In each ″inelastic collision″ the force of static friction does not act through any distance. 11 •• Determine the Concept It is easier to crawl radially outward. In fact, a radially inward force is required just to prevent you from sliding outward. *12 •• Determine the Concept The pull that the student exerts on the block is at right angles to r r r its motion and exerts no torque (recall that τ = r × F and τ = rF sin θ ). Therefore, we

Conservation of Angular Momentum 737 can conclude that the angular momentum of the block is conserved. The student does, however, do work in displacing the block in the direction of the radial force and so the block’s energy increases. (b) is correct. *13 •• Determine the Concept The hardboiled egg is solid inside, so everything rotates with a uniform velocity. By contrast, it is difficult to get the viscous fluid inside a raw egg to start rotating; however, once it is rotating, stopping the shell will not stop the motion of the interior fluid, and the egg may start rotating again after momentarily stopping for this reason. 14 • r r False. The relationship τ = dL dt describes the motion of a gyroscope independently of whether it is spinning. 15 • Picture the Problem We can divide the expression for the kinetic energy of the object by the expression for its angular momentum to obtain an expression for K as a function of I and L. Express the rotational kinetic energy of the object:

K = 12 Iω 2

Relate the angular momentum of the object to its moment of inertia and angular velocity:

L = Iω

Divide the first of these equations by the second and solve for K to obtain:

K=

L2 and so (b) is correct. 2I

16 • Determine the Concept The purpose of the second smaller rotor is to prevent the body of the helicopter from rotating. If the rear rotor fails, the body of the helicopter will tend to rotate on the main axis due to angular momentum being conserved. 17 •• Determine the Concept One can use a right-hand rule to determine the direction of the torque required to turn the angular momentum vector from east to south. Letting the fingers of your right hand point east, rotate your wrist until your fingers point south. Note that your thumb points downward. (b) is correct.

738 Chapter 10 18 •• Determine the Concept In turning east, the man redirects the angular momentum vector from north to east by exerting a clockwise torque (viewed from above) on the gyroscope. As a consequence of this torque, the front end of the suitcase will dip downward.

(d ) is correct. 19 •• (a) The lifting of the nose of the plane rotates the angular momentum vector upward. It veers to the right in response to the torque associated with the lifting of the nose. (b) The angular momentum vector is rotated to the right when the plane turns to the right. In turning to the right, the torque points down. The nose will move downward. 20 •• r Determine the Concept If L points up and the car travels over a hill or through a valley, the force on the wheels on one side (or the other) will increase and car will tend to r tip. If L points forward and car turns left or right, the front (or rear) of the car will tend to lift. These problems can be averted by having two identical flywheels that rotate on the same shaft in opposite directions. 21 •• Determine the Concept The rotational kinetic energy of the woman-plus-stool system is given by K rot = 12 Iω 2 = L2 2 I . Because L is constant (angular momentum is conserved) and her moment of inertia is greater with her arms extended, (b) is correct. *22 •• Determine the Concept Consider the overhead view of a tether pole and ball shown in the adjoining figure. The ball rotates counterclockwise. The torque about the center of the pole is clockwise and of magnitude RT, where R is the pole’s radius and T is the tension. So L must decrease and (e) is correct. 23 •• Determine the Concept The center of mass of the rod-and-putty system moves in a straight line, and the system rotates about its center of mass.

Conservation of Angular Momentum 739 24 • (a) True. The net external torque acting a system equals the rate of change of the angular

r r dL momentum of the system; i.e., ∑ τ i,ext = . dt i

(b) False. If the net torque on a body is zero, its angular momentum is constant but not necessarily zero.

Estimation and Approximation *25 •• Picture the Problem Because we have no information regarding the mass of the skater, we’ll assume that her body mass (not including her arms) is 50 kg and that each arm has a mass of 4 kg. Let’s also assume that her arms are 1 m long and that her body is cylindrical with a radius of 20 cm. Because the net external torque acting on her is zero, her angular momentum will remain constant during her pirouette. Express the conservation of her angular momentum during her pirouette:

Li = Lf or

I arms outω arms out = I arms inω arms in

Express her total moment of inertia with her arms out: Treating her body as though it is cylindrical, calculate its moment of inertia of her body, minus her arms:

(1)

I arms out = I body + I arms

I body = 12 mr 2 =

1 2

(50 kg )(0.2 m )2

= 1.00 kg ⋅ m 2

[

]

Modeling her arms as though they are rods, calculate their moment of inertia when she has them out:

I arms = 2 13 (4 kg )(1 m )

Substitute to determine her total moment of inertia with her arms out:

I arms out = 1.00 kg ⋅ m 2 + 2.67 kg ⋅ m 2

Express her total moment of inertia with her arms in:

I arms in = I body + I arms

= 2.67 kg ⋅ m

2

2

= 3.67 kg ⋅ m 2

[

= 1.00 kg ⋅ m 2 + 2 (4 kg )(0.2 m ) = 1.32 kg ⋅ m 2

2

]

740 Chapter 10 Solve equation (1) for ω arms in and substitute to obtain:

ω arms in = =

I arms out I arms in

ω arms out

3.67 kg ⋅ m 2 (1.5 rev/s) 1.32 kg ⋅ m 2

= 4.17 rev/s 26 •• Picture the Problem We can express the period of the earth’s rotation in terms of its angular velocity of rotation and relate its angular velocity to its angular momentum and moment of inertia with respect to an axis through its center. We can differentiate this expression with respect to I and then use differentials to approximate the changes in I and T.

2π

Express the period of the earth’s rotation in terms of its angular velocity of rotation:

T=

Relate the earth’s angular velocity of rotation to its angular momentum and moment of inertia:

ω=

L I

Substitute to obtain:

T=

2π I L

Find dT/dI:

dT 2π T = = dI L I

Solve for dT/T and approximate ∆T:

dT dI ∆I = or ∆T ≈ T T I I

Substitute for ∆I and I to obtain:

Substitute numerical values and evaluate ∆T:

ω

∆T ≈

mr 2 5m T= T 2 2 3M E 5 M E RE

∆T =

5 2.3 × 1019 kg (1d ) 3 6 × 10 24 kg

2 3

(

(

)

)

−6

= 6.39 × 10 d = 6.39 × 10 −6 d × = 0.552 s

24 h 3600 s × d h

Conservation of Angular Momentum 741 27 • Picture the Problem We can use L = mvr to find the angular momentum of the particle. In (b) we can solve the equation L = l(l + 1)h for l(l + 1) and the approximate value of

l. (a) Use the definition of angular momentum to obtain:

L = mvr

= (2 × 10−3 kg )(3 ×10 −3 m/s )(4 × 10−3 m )

= 2.40 ×10 −8 kg ⋅ m 2 /s L2 h2

(b) Solve the equation L = l(l + 1)h for l(l + 1) :

l(l + 1) =

Substitute numerical values and evaluate l(l + 1) :

⎛ 2.40 × 10 −8 kg ⋅ m 2 /s ⎞ ⎟⎟ l(l + 1) = ⎜⎜ −34 ⎝ 1.05 × 01 J ⋅ s ⎠

2

= 5.22 × 1052 Because l >>1, approximate its value with the square root of l(l + 1) :

l ≈ 2.29 × 10 26

The quantization of angular momentum is not noticed in macroscopic (c) physics because no experiment can differentiate between l = 2 × 10 26 and

l = 2 × 10 26 + 1. *28 •• Picture the Problem We can use conservation of angular momentum in part (a) to relate the before-and-after collapse rotation rates of the sun. In part (b), we can express the fractional change in the rotational kinetic energy of the sun as it collapses into a neutron star to decide whether its rotational kinetic energy is greater initially or after the collapse. (a) Use conservation of angular momentum to relate the angular momenta of the sun before and after its collapse:

I bωb = I aωa

Using the given formula, approximate the moment of inertia Ib of the sun before collapse:

2 I b = 0.059MRsun

(

(1)

)(

= 0.059 1.99 × 1030 kg 6.96 × 105 km = 5.69 × 10 46 kg ⋅ m 2

)

2

742 Chapter 10 Find the moment of inertia Ia of the sun when it has collapsed into a spherical neutron star of radius 10 km and uniform mass distribution:

I a = 52 MR 2 =

2 5

(1.99 ×10

30

)

kg (10 km )

2

= 7.96 ×1037 kg ⋅ m 2

Substitute in equation (1) and solve for ωa to obtain:

ωa =

Ib 5.69 × 10 46 kg ⋅ m 2 ωb = ωb Ia 7.96 × 1037 kg ⋅ m 2

= 7.15 × 108 ωb Given that ωb = 1 rev/25 d, evaluate ωa:

⎛ 1 rev ⎞ ⎟⎟ ⎝ 25 d ⎠

ωa = 7.15 × 108 ⎜⎜

= 2.86 × 10 7 rev/d The additional rotational kinetic energy comes at the expense of gravitational potential energy, which decreases as the sun gets smaller. Note that the rotational period decreases by the same factor of Ib/Ia and becomes:

Ta =

2π

ωa

=

2π = 3.02 × 10−3 s 2π rad 1d 1h 7 rev 2.86 × 10 × × × d rev 24 h 3600 s ∆K K a − K b K a = −1 = Kb Kb Kb

(b) Express the fractional change in the sun’s rotational kinetic energy as a consequence of its collapse and simplify to obtain:

I aωa2 = −1 I bωb2 1 2 1 2

=

I aωa2 −1 I bωb2

Substitute numerical values and evaluate ∆K/Kb: 2

7 1 ∆K ⎛ ⎞ ⎛ 2.86 × 10 rev/d ⎞ ⎟ − 1 = 7.15 × 108 (i.e., the rotational kinetic ⎜ =⎜ ⎟ K b ⎝ 7.15 × 108 ⎠ ⎜⎝ 1 rev/25 d ⎟⎠

energy increases by a factor of approximately 7×108.) 29 •• Picture the Problem We can solve I = CMR 2 for C and substitute numerical values in order to determine an experimental value of C for the earth. We can then compare this value to those for a spherical shell and a sphere in which the mass is uniformly distributed to decide whether the earth’s mass density is greatest near its core or near its crust.

Conservation of Angular Momentum 743 (a) Express the moment of inertia of the earth in terms of the constant C: Solve for C to obtain:

Substitute numerical values and evaluate C:

I = CMR 2 C=

I MR 2

C=

8.03 ×1037 kg ⋅ m 2 2 5.98 × 1024 kg (6370 km )

(

)

= 0.331 (b) If all of the mass were in the crust, the moment of inertia of the earth would be that of a thin spherical shell:

I spherical shell = 23 MR 2

If the mass of the earth were uniformly distributed throughout its volume, its moment of inertia would be:

I solid sphere = 25 MR 2

Because experimentally C < 2/5 = 0.4, the mass density must be greater near the center of the earth. *30 •• Picture the Problem Let’s estimate that the diver with arms extended over head is about 2.5 m long and has a mass M = 80 kg. We’ll also assume that it is reasonable to model the diver as a uniform stick rotating about its center of mass. From the photo, it appears that he sprang about 3 m in the air, and that the diving board was about 3 m high. We can use these assumptions and estimated quantities, together with their definitions, to estimate ω and L. Express the diver’s angular velocity ω and angular momentum L:

∆θ ∆t

(1)

L = Iω

(2)

ω= and

Using a constant-acceleration equation, express his time in the air:

∆t = ∆t rise 3 m + ∆tfall 6 m =

Substitute numerical values and evaluate ∆t: Estimate the angle through which he rotated in 1.89 s:

∆t =

2∆yup g

+

2∆ydown g

2(3 m ) 2(6 m ) + = 1.89 s 2 9.81 m/s 9.81 m/s 2

∆θ ≈ 0.5 rev = π rad

744 Chapter 10 Substitute in equation (1) and evaluate ω:

ω=

π rad 1.89 s

= 1.66 rad/s

Use the ″stick rotating about an axis through its center of mass″ model to approximate the moment of inertia of the diver:

I = 121 ML2

Substitute in equation (2) to obtain:

L = 121 ML2ω

Substitute numerical values and evaluate L:

L = 121 (80 kg )(2.5 m ) (1.66 rad/s ) 2

= 69.2 kg ⋅ m 2 /s ≈ 70 kg ⋅ m 2 /s

Remarks: We can check the reasonableness of this estimation in another way. Because he rose about 3 m in the air, the initial impulse acting on him must be about 600 kg⋅m/s (i.e., I = ∆p = Mvi). If we estimate that the lever arm of the force is roughly l = 1.5 m, and the angle between the force exerted by the board and a line running from his feet to the center of mass is about 5°, we obtain L = Ilsin5° ≈ 78 kg⋅m2/s, which is not too bad considering the approximations made here. 31 •• Picture the Problem First we assume a spherical diver whose mass M = 80 kg and whose diameter, when curled into a ball, is 1 m. We can estimate his angular velocity when he has curled himself into a ball from the ratio of his angular momentum to his moment of inertia. To estimate his angular momentum, we’ll guess that the lever arm l of the force that launches him from the diving board is about 1.5 m and that the angle between the force exerted by the board and a line running from his feet to the center of mass is about 5°. Express the diver’s angular velocity ω when he curls himself into a ball in mid-dive: Using a constant-acceleration equation, relate the speed with which he left the diving board v0 to his maximum height ∆y and our estimate of his angle with the vertical direction: Solve for v0:

Substitute numerical values and evaluate v0:

ω=

L I

(1)

0 = v02y + 2a y ∆y where

v0 y = v0 cos 5°

v0 =

2 g∆y cos 2 5°

v0 =

2 9.81m/s 2 (3 m ) = 7.70 m/s cos 5°

(

)

Conservation of Angular Momentum 745 Approximate the impulse acting on the diver to launch him with the speed v0:

I = ∆p = Mv0

Letting l represent the lever arm of the force acting on the diver as he leaves the diving board, express his angular momentum:

L = Il sin 5° = Mv0l sin 5°

Use the ″uniform sphere″ model to approximate the moment of inertia of the diver:

I = 52 MR 2

Substitute in equation (1) to obtain:

Substitute numerical values and evaluate ω:

ω=

Mv0l sin 5° 5v0l sin 5° = 2 MR 2 2R2 5

ω=

5(7.70 m/s )(1.5 m )sin 5° 2 2(0.5 m )

= 10.1 rad/s *32 •• Picture the Problem We’ll assume that he launches himself at an angle of 45° with the horizontal with his arms spread wide, and then pulls them in to increase his rotational speed during the jump. We’ll also assume that we can model him as a 2-m long cylinder with an average radius of 0.15 m and a mass of 60 kg. We can then find his take-off speed and ″air time″ using constant-acceleration equations, and use the latter, together with the definition of rotational velocity, to find his initial rotational velocity. Finally, we can apply conservation of angular momentum to find his initial angular momentum. Using a constant-acceleration equation, relate his takeoff speed v0 to his maximum elevation ∆y:

v 2 = v02y + 2a y ∆y or, because v0y = v0sin45°, v = 0, and ay = − g,

0 = v02 sin 2 45° − 2 g∆y Solve for v0 to obtain:

v0 =

2 g∆y 2 g∆y = 2 sin 45° sin 45°

Substitute numerical values and evaluate v0:

v0 =

2 9.81m/s 2 (0.6 m ) = 4.85 m/s sin45°

Use its definition to express Goebel’s angular velocity:

ω=

Use a constant-acceleration equation to express Goebel’s ″air time″ ∆t:

∆t = 2∆trise 0.6 m = 2

(

)

∆θ ∆t

2∆y g

746 Chapter 10 Substitute numerical values and evaluate ∆t:

∆t = 2

Substitute numerical values and evaluate ω:

ω=

Use conservation of angular momentum to relate his take-off angular velocity ω0 to his average angular velocity ω as he performs a quadruple Lutz: Assuming that he can change his angular momentum by a factor of 2 by pulling his arms in, solve for and evaluate ω0:

2(0.6 m ) = 0.699 s 9.81 m/s 2

4 rev 2π rad × = 36.0 rad/s 0.699 s rev

I 0ω0 = Iω

ω0 =

1 I ω = (36 rad/s ) = 18.0 rad/s I0 2

Express his take-off angular momentum:

L0 = I 0ω0

Assuming that we can model him as a solid cylinder of length l with an average radius r and mass m, express his moment of inertia with arms drawn in (his take-off configuration):

I 0 = 2 12 mr 2 = mr 2 where the factor of 2 represents our assumption that he can double his moment of inertia by extending his arms.

Substitute to obtain:

L0 = mr 2ω0

Substitute numerical values and evaluate L0:

L0 = (60 kg )(0.15 m ) (18 rad/s )

(

)

2

= 24.3 kg ⋅ m 2 /s

Vector Nature of Rotation 33 • r r Picture the Problem We can express F and r in terms of the unit vectors iˆ and ˆj and

r

then use the definition of the cross product to find τ .

r

r F = − Fiˆ

r

r r = Rˆj

Express F in terms of F and the unit vector iˆ :

Express r in terms of R and the unit

vector ˆj :

Conservation of Angular Momentum 747 r F:

(

r r r τ = r × F = FR ˆj × −iˆ

r

Calculate the cross product of r and

( )

)

= FR iˆ × ˆj = FRkˆ

34 • r r Picture the Problem We can find the torque is the cross product of r and F .

r

r

Compute the cross product of r and F :

(

)(

) ( )

r r r τ = r × F = xiˆ + yˆj − mgˆj = −mgx iˆ × ˆj − mgy ˆj × ˆj

( )

= − mgxkˆ 35 • r Picture the Problem The cross product of the vectors A = Ax iˆ + Ay ˆj

r

and B = B x iˆ + B y ˆj is given by

( )

( ) ( ) () ( ) ( )

( )

r r A × B = Ax B x iˆ × iˆ + Ax B y iˆ × ˆj + Ay B x ˆj × iˆ + Ay B y ˆj × ˆj = Ax B x (0) + Ax B y kˆ + Ay B x − kˆ + Ay B y (0) = A B kˆ + A B − kˆ x

r

y

()

r

y

x

r

(a) Find A × B for A = 4 iˆ and

r B = 6 iˆ + 6 ˆj :

(

) ( ) ( )

r r A × B = 4iˆ × 6iˆ + 6 ˆj = 24 iˆ × iˆ + 24 iˆ × ˆj

= 24(0) + 24kˆ = 24kˆ r

r

r

r

r r A × B = 2iˆ + 3 ˆj × 3iˆ + 2 ˆj = 6 iˆ × iˆ + 4 iˆ × ˆj + 9 ˆj × iˆ + 6 ˆj × ˆj

(c) Find A × B for A = 2 iˆ + 3 ˆj

r and B =3 iˆ + 2 ˆj :

) ( ) ( ) = 24(0) + 24(− ˆj ) = − 24 ˆj

r r A × B = 4iˆ × 6iˆ + 6kˆ = 24 iˆ × iˆ + 24 iˆ × kˆ

r B = 6 iˆ + 6 kˆ :

r

(

r

(b) Find A × B for A = 4 iˆ and

(

)( ) ( ) ( ) ( ) ( ) = 6(0 ) + 4(kˆ )+ 9(− kˆ ) + 6(0 ) = − 5kˆ

748 Chapter 10 *36 •

r

r

Picture the Problem The magnitude of A × B is given by AB sin θ .

r

r

AB sin θ = AB cos θ

Equate the magnitudes of A × B r r and A ⋅ B :

∴ sin θ = cos θ or

tan θ = ±1

Solve for θ to obtain:

θ = tan −1 ± 1 = ± 45° or ± 135°

37 •• r Picture the Problem Let r be in the xy r plane. Then ω points in the positive z direction. We can establish the results called for in this problem by forming the appropriate cross products and by r differentiating v .

r

r ω = ω kˆ

(a) Express ω using unit vectors:

r

r r = riˆ

Express r using unit vectors:

r

r

Form the cross product of ω and r :

( )

r r ω × r = ω kˆ × r iˆ = rω kˆ × iˆ = rω ˆj = vˆj r r r ∴ v = ω× r

r

(b) Differentiate v with respect to t to r express a :

r r dv d r r a= = (ω × r ) dt dt r r dω r r dr = × r + ω× dt dt r dω r r r = × r + ω× v dt r r r r = a t + ω × (ω × r ) r r = a t + ac r r r r where a c = ω × (ω × r ) r r and a t and a c are the tangential and

Conservation of Angular Momentum 749 centripetal accelerations, respectively. 38 •• r r Picture the Problem Because Bz = 0, we can express B as B = B x iˆ + B y ˆj and form its

r

cross product with A to determine Bx and By.

r

Express B in terms of its components:

r

r B = B x iˆ + B y ˆj

(1)

(

)

Express A × B :

r r A × B = 4iˆ × Bx iˆ + B y ˆj = 4 B y kˆ = 12kˆ

Solve for By:

By = 3

Relate B to Bx and By:

B 2 = B x2 + B y2

Solve for and evaluate Bx:

Bx = B 2 − B y2 = 52 − 32 = 4

Substitute in equation (1):

r B = 4iˆ + 3 ˆj

r

39 • r r Picture the Problem We can write B in the form B = B x iˆ + B y ˆj + B z kˆ and use the dot

r

r

product of A and B to find By and their cross product to find Bx and Bz.

r

r B = B x iˆ + B y ˆj + B z kˆ

r r

r r A ⋅ B = 3B y = 12

Express B in terms of its components: Evaluate A ⋅ B :

(1)

and By = 4

r

r

Evaluate A × B :

(

r r A × B = 3 ˆj × B x iˆ + 4 ˆj + B z kˆ = −3B kˆ + 3B iˆ x

r

r

Because A × B = 9 iˆ :

Bx = 0 and Bz = 3.

Substitute in equation (1) to obtain:

r B = 4 ˆj + 3kˆ

z

)

750 Chapter 10 40

••

r

r

r

Picture the Problem The dot product of A with the cross product of B and C is a scalar

ax

ay

az

quantity and can be expressed in determinant form as bx

by bz . We can expand this cx c y cz r r r r r r determinant by minors to show that it is equivalent to A ⋅ ( B × C ) , C ⋅ ( A × B ) , and r r r B ⋅ (C × A) .

r

The dot product of A with the cross

r r product of B and C is a scalar

quantity and can be expressed in determinant form as: Expand the determinant by minors to obtain:

ax r r r A ⋅ ( B × C ) = bx cx

ay

az

by cy

bz cz

ax

ay

az

bx cx

by cy

bz = a x b y c z − a x bz c y cz + a y bz c x − a y bx c z

(1)

+ a z bx c y − a z b y c x

r

Evaluate the cross product of B and

r C to obtain:

r r B × C = (by cz − bz c y ) iˆ

+ (bz cx − bx cz ) ˆj + (bx c y − by cx )kˆ

r

Form the dot product of A with

r r B × C to obtain:

(

)

r r r A ⋅ B × C = axby cz − axbz c y

+ a ybz cx − a ybx cz + az bx c y − az by cx

Because (1) and (2) are the same, we can conclude that:

Proceed as above to establish that:

ax r r r A ⋅ ( B × C ) = bx

ay

az

by

bz

cx

cy

cz

ax r r r C ⋅ ( A × B ) = bx

ay

az

by

bz

cx

cy

cz

and

(2)

Conservation of Angular Momentum 751 ax r r r B ⋅ (C × A) = bx cx

ay

az

by cy

bz cz

41 •• r Picture the Problem Let, without loss of generality, the vector C lie along the x axis and

r

the vector B lie in the xy plane as shown below to the left. The diagram to the right shows the parallelepiped spanned by the three vectors. We can apply the definitions of

r

(r ) r

the cross- and dot-products to show that A ⋅ B × C is the volume of the parallelepiped.

r

r

Express the cross-product of B and C :

( )

r r B × C = (BC sin θ ) − kˆ and

r r B × C = (B sin θ )C

= area of the parallelogram r r r cross-product of B and C to obtain:

Form the dot-product of A with the

(

)

r r r A ⋅ B × C = A(B sin θ )C cos φ

= (BC sin θ )( A cos φ )

= (area of base )(height ) = Vparallelepiped *42 •• Picture the Problem Draw the triangle using the three vectors as shown below.

r

r

r

Note that A + B = C . We can find the

r

magnitude of the cross product of A and

r r r B and of A and C and then use the cross r r r r r product of A and C , using A + B = C, to show that AC sin b = AB sin c or B/sin b = C/ sin c. Proceeding similarly, we can extend the law of sines to the third side of the triangle and the angle opposite it.

752 Chapter 10 Express the magnitude of the cross

r r A × B = AB sin c

Express the magnitude of the cross

r r A × C = AC sin b

r r product of A and B :

r r product of A and C :

r

Form the cross product of A with

r C to obtain:

r

r

r

r

(

)

r r r r r A× C = A× A + B r r r r = A× A + A× B r r = A× B r r because A × A = 0 . r r r r A× C = A× B

Because A × C = A × B :

and

AC sin b = AB sin c

Simplify and rewrite this expression to obtain:

B C = sin b sin c

Proceed similarly to extend this result to the law of sines:

A B C = = sin a sin b sin c

Angular Momentum 43 • r r r r r r Picture the Problem L and p are related according to L = r × p. If L = 0, then r r examination of the magnitude of r × p will allow us to conclude that sin φ = 0 and that the particle is moving either directly toward the point, directly away from the point, or through the point.

r r r r r r r × p = r × mv = mr × v = 0

r

Because L = 0:

or

r r r ×v = 0

r r

Express the magnitude of r × v :

r r r × v = rv sin φ = 0

Because neither r nor v is zero:

sin φ = 0 r r where φ is the angle between r and v .

Solve for φ:

φ = sin −1 0 = 0° or 180°

Conservation of Angular Momentum 753 44 • Picture the Problem We can use their definitions to calculate the angular momentum and moment of inertia of the particle and the relationship between L, I, and ω to determine its angular speed. (a) Express and evaluate the r magnitude of L :

L = mvr = (2 kg )(3.5 m/s )(4 m )

= 28.0 kg ⋅ m 2 /s

(b) Express the moment of inertia of the particle with respect to an axis through the center of the circle in which it is moving:

I = mr 2 = (2 kg )(4 m ) = 32 kg ⋅ m 2

(c) Relate the angular speed of the particle to its angular momentum and solve for and evaluate ω:

ω=

2

L 28.0 kg ⋅ m 2 /s = = 0.875 rad/s 2 32 kg ⋅ m 2 I

45 • Picture the Problem We can use the definition of angular momentum to calculate the angular momentum of this particle and the relationship between its angular momentum and angular speed to describe the variation in its angular speed with time. (a) Express the angular momentum of the particle as a function of its mass, speed, and distance of its path from the reference point:

L = rmv sin θ

(b) Because L = mr2ω:

ω∝

= (6 m )(2 kg )(4.5 m/s )sin90° = 54.0 kg ⋅ m 2 /s

1 and r2 ω increases as the particle approaches the point and decreases as it recedes.

*46 •• Picture the Problem We can use the formula for the area of a triangle to find the area swept out at t = t1, add this area to the area swept out in time dt, and then differentiate this expression with respect to time to obtain the given expression for dA/dt. Express the area swept out at t = t1:

A1 = 12 br1 cos θ1 = 12 bx1 r r where θl is the angle between r1 and v and

754 Chapter 10 r

x1 is the component of r1 in the direction of

r v.

Express the area swept out at t = t1 + dt: Differentiate with respect to t:

Because rsinθ = b:

A = A1 + dA = 12 b( x1 + dx ) = 12 b( x1 + vdt )

dA 1 dx 1 = 2b = 2 bv = constant dt dt 1 2

bv =

1 2

(r sin θ )v =

1 (rp sin θ ) 2m

L 2m

=

47 •• Picture the Problem We can find the total angular momentum of the coin from the sum of its spin and orbital angular momenta. Lspin = I cmω spin (a) Express the spin angular momentum of the coin: From Problem 9-44:

I = 14 MR 2

Substitute for I to obtain:

Lspin = 14 MR 2ωspin

Substitute numerical values and evaluate Lspin:

Lspin =

1 4

(0.015 kg )(0.0075 m )2

⎛ rev 2π rad ⎞ × ⎜10 × ⎟ s rev ⎠ ⎝ = 1.33 × 10−5 kg ⋅ m 2 /s

(b) Express and evaluate the total angular momentum of the coin:

L = Lorbit + Lspin = 0 + Lspin

(c) From Problem 10-14:

Lorbit = 0

= 1.33 × 10 −5 kg ⋅ m 2 /s

and

L = 1.33 × 10 −5 kg ⋅ m 2 /s (d) Express the total angular momentum of the coin:

L = Lorbit + Lspin

Conservation of Angular Momentum 755 Find the orbital momentum of the coin:

Lorbit = ± MvR

= ±(0.015 kg )(0.05 m/s )(0.1 m ) = ±7.50 × 10−5 kg ⋅ m 2 /s

where the ± is a consequence of the fact that the coin’s direction is not specified.

L = ±7.50 × 10 −5 kg ⋅ m 2 /s

Substitute to obtain:

+ 1.33 × 10 −5 kg ⋅ m 2 /s L = 8.83 × 10 −5 kg ⋅ m 2 /s

The possible values for L are:

or

L = − 6.17 × 10 −5 kg ⋅ m 2 /s 48 •• Picture the Problem Both the forces acting on the particles exert torques with respect to an axis perpendicular to the page and through point O and the net torque about this axis is their vector sum. Express the net torque about an axis perpendicular to the page and through point O:

r

r

r

Because r1 − r2 points along − F1 :

r r r r r r τ net = ∑ τ i = r1 × F1 + r2 × F2 i

r r r = (r1 − r2 )× F1 r r because F2 = − F1 r

(rr1 − rr2 ) × F1 = 0

Torque and Angular Momentum 49 • Picture the Problem The angular momentum of the particle changes because a net torque acts on it. Because we know how the angular momentum depends on time, we can find the net torque acting on the particle by differentiating its angular momentum. We can use a constant-acceleration equation and Newton’s 2nd law to relate the angular speed of the particle to its angular acceleration. (a) Relate the magnitude of the torque acting on the particle to the rate at which its angular momentum changes:

τ net =

dL d = [(4 N ⋅ m )t ] dt dt

= 4.00 N ⋅ m

756 Chapter 10 (b) Using a constant-acceleration equation, relate the angular speed of the particle to its acceleration and time-in-motion:

ω = ω0 + α t where ω0 = 0

Use Newton’s 2nd law to relate the angular acceleration of the particle to the net torque acting on it:

α=

τ net

Substitute to obtain:

ω=

τ net

Substitute numerical values and evaluate ω:

I

ω=

mr 2

=

τ net mr 2

t

(4 N ⋅ m )t (1.8 kg )(3.4 m )2

(

)

= 0.192 rad/s 2 t provided t is in seconds. 50 •• Picture the Problem The angular momentum of the cylinder changes because a net torque acts on it. We can find the angular momentum at t = 25 s from its definition and the net torque acting on the cylinder from the rate at which the angular momentum is changing. The magnitude of the frictional force acting on the rim can be found using the definition of torque. (a) Use its definition to express the angular momentum of the cylinder:

L = Iω = 12 mr 2ω

Substitute numerical values and evaluate L:

L=

1 2

(90 kg )(0.4 m )2

⎛ rev 2π rad 1 min ⎞ ⎟⎟ × ⎜⎜ 500 × × min rev 60 s ⎠ ⎝ = 377 kg ⋅ m 2 /s

(b) Express and evaluate

dL : dt

(

dL 377 kg ⋅ m 2 /s = dt 25 s

)

= 15.1 kg ⋅ m 2 /s 2 (c) Because the torque acting on the uniform cylinder is constant, the rate

τ=

dL = 15.1 kg ⋅ m 2 /s 2 dt

Conservation of Angular Momentum 757 of change of the angular momentum is constant and hence the instantaneous rate of change of the angular momentum at any instant is equal to the average rate of change over the time during which the torque acts: (d) Using the definition of torque that relates the applied force to its lever arm, express the magnitude of the frictional force f acting on the rim:

f =

τ l

=

15.1 kg ⋅ m 2 /s 2 = 37.7 N 0.4 m

*51 •• Picture the Problem Let the system include the pulley, string, and the blocks and assume that the mass of the string is negligible. The angular momentum of this system changes because a net torque acts on it. (a) Express the net torque about the center of mass of the pulley:

τ net = Rm2 g sin θ − Rm1 g = Rg (m2 sin θ − m1 ) where we have taken clockwise to be positive to be consistent with a positive upward velocity of the block whose mass is m1 as indicated in the figure.

(b) Express the total angular momentum of the system about an axis through the center of the pulley: (c) Express τ as the time derivative of the angular momentum:

L = Iω + m1vR + m2 vR ⎛ I ⎞ = vR⎜ 2 + m1 + m2 ⎟ ⎝R ⎠

τ=

dL d = dt dt

⎡ ⎛ I ⎞⎤ ⎢vR⎜ R 2 + m1 + m2 ⎟⎥ ⎠⎦ ⎣ ⎝

⎛ I ⎞ = aR⎜ 2 + m1 + m2 ⎟ ⎝R ⎠ Equate this result to that of part (a) and solve for a to obtain:

a=

g (m2 sin θ − m1 ) I + m1 + m2 R2

758 Chapter 10 52 •• Picture the Problem The forces resulting from the release of gas from the jets will exert a torque on the spaceship that will slow and eventually stop its rotation. We can relate this net torque to the angular momentum of the spaceship and to the time the jets must fire.

∆L

Relate the firing time of the jets to the desired change in angular momentum:

∆t =

Express the magnitude of the net torque exerted by the jets:

τ net = 2 FR

Letting ∆m/∆t′ represent the mass of gas per unit time exhausted from the jets, relate the force exerted by the gas on the spaceship to the rate at which the gas escapes:

F=

Substitute and solve for ∆t to obtain:

∆t =

τ net

=

I∆ω

τ net

∆m v ∆t '

I∆ω ∆m 2 vR ∆t '

Substitute numerical values and evaluate ∆t:

rev 2π rad 1 min ⎞ ⎟⎟ (4000 kg ⋅ m )⎛⎜⎜ 6 min × × rev 60 s ⎠= ⎝ ∆t = 2(10 kg/s )(800 m/s )(3 m ) 2

−2

52.4 s

53 •• Picture the Problem We can use constant-acceleration equations to express the projectile’s position and velocity coordinates as functions of time. We can use these r r coordinates to express the particle’s position and velocity vectors r and v . Using its

r

definition, we can express the projectile’s angular momentum L as a function of time and

r

then differentiate this expression to obtain dL dt . Finally, we can use the definition of the torque, relative to an origin located at the launch position, the gravitational force

r

r

r

exerts on the projectile to express τ and complete the demonstration that dL dt = τ . Using its definition, express the r angular momentum vector L of the projectile: Using constant-acceleration

r r r L = r × mv

x = v0 xt = (V cos θ )t

(1)

Conservation of Angular Momentum 759 equations, express the position coordinates of the projectile as a function of time:

and

Express the projectile’s position r vector r :

r r = [(V cos θ )t ]iˆ + (V sin θ )t − 12 gt 2 ˆj

Using constant-acceleration equations, express the velocity of the projectile as a function of time:

vx = v0 x = V cos θ

y = y0 + v0 y t + 12 a y t 2 = (V sin θ )t − 12 gt 2

[

]

and

v y = v0 y + a y t = V sin θ − gt

r v = [V cosθ ]iˆ + [V sin θ − gt ] ˆj

Express the projectile’s velocity r vector v :

{

]}

[

r L = [(V cos θ )t ]iˆ + (V sin θ )t − 12 gt 2 ˆj × m [V cos θ ]iˆ + [V sin θ − gt ] ˆj

Substitute in equation (1) to obtain:

(

{

}

)

= − 12 mgt 2V cos θ kˆ

r

Differentiate L with respect to t to obtain:

r dL d 1 = − 2 mgt 2V cosθ kˆ dt dt = (− mgtV cosθ ) kˆ

(

)

(2)

r τ = rr × (− mg ) ˆj = [(V cos θ )t ]iˆ + [(V sin θ )t − 12 gt 2 ] ˆj × (− mg ) ˆj or r τ = (− mgtV cosθ )kˆ (3)

Using its definition, express the torque acting on the projectile:

r dL r =τ dt

Comparing equations (2) and (3) we see that:

Conservation of Angular Momentum *54 • Picture the Problem Let m represent the mass of the planet and apply the definition of torque to find the torque produced by the gravitational force of attraction. We can use

r

r

r

Newton’s 2nd law of motion in the form τ = dL dt to show that L is constant and apply conservation of angular momentum to the motion of the planet at points A and B.

760 Chapter 10 (a) Express the torque produced by the gravitational force of attraction of the sun for the planet:

r

(b) Because τ = 0 :

Noting that at points A and B r r r × v = rv , express the relationship between the distances from the sun and the speeds of the planets:

r r r r τ = r × F = 0 because F acts along r the direction of r .

r r r r dL = 0 ⇒ L = r × mv = constant dt r1v1 = r2 v2 or

v1 r = 2 v2 r1

55 •• Picture the Problem Let the system consist of you, the extended weights, and the platform. Because the net external torque acting on this system is zero, its angular momentum remains constant during the pulling in of the weights. (a) Using conservation of angular momentum, relate the initial and final angular speeds of the system to its initial and final moments of inertia: Solve for ω f :

I iω i = I f ω f

ωf =

Ii ωi If

Substitute numerical values and evaluate ω f :

ωf =

6 kg ⋅ m 2 (1.5 rev/s) = 5.00 rev/s 1.8 kg ⋅ m 2

(b) Express the change in the kinetic energy of the system:

∆K = K f − K i = 12 I f ωf2 − 12 I iωi2

Substitute numerical values and evaluate ∆K:

∆K =

1 2

(

2

(

)

)

⎛ rev 2π rad ⎞ × 1.8 kg ⋅ m ⎜ 5 ⎟ rev ⎠ ⎝ s

2

⎛ rev 2π rad ⎞ − 12 6 kg ⋅ m 2 ⎜1.5 × ⎟ s rev ⎠ ⎝ = 622 J

2

Conservation of Angular Momentum 761 (c)

Because no external agent does work on the system, the energy comes from the internal energy of the man.

*56 •• Picture the Problem Let the system consist of the blob of putty and the turntable. Because the net external torque acting on this system is zero, its angular momentum remains constant when the blob of putty falls onto the turntable. (a) Using conservation of angular momentum, relate the initial and final angular speeds of the turntable to its initial and final moments of inertia and solve for ωf:

I 0ω i = I f ω f and

ωf =

I0 ωi If

Express the final rotational inertia of the turntable-plus-blob:

I f = I 0 + I blob = I 0 + mR 2

Substitute and simplify to obtain:

ωf =

I0 ωi = I 0 + mR 2

1 ωi mR 2 1+ I0

(b) If the blob flies off tangentially to the turntable, its angular momentum doesn’t change (with respect to an axis through the center of turntable). Because there is no external torque acting on the blob-turntable system, the total angular momentum of the system will remain constant and the angular momentum of the turntable will not change. Because the moment of inertia of the table hasn’t changed either, the turntable will continue to spin at ω' = ωf . 57 •• Picture the Problem Because the net external torque acting on the Lazy Susancockroach system is zero, the net angular momentum of the system is constant (equal to zero because the Lazy Susan is initially at rest) and we can use conservation of angular momentum to find the angular velocity ω of the Lazy Susan. The speed of the cockroach relative to the floor vf is the difference between its speed with respect to the Lazy Susan and the speed of the Lazy Susan at the location of the cockroach with respect to the floor. Relate the speed of the cockroach with respect to the floor vf to the speed of the Lazy Susan at the location of the cockroach:

vf = v − ωr

Use conservation of angular momentum to obtain:

LLS − LC = 0

(1)

762 Chapter 10 Express the angular momentum of the Lazy Susan:

LLS = I LSω = 12 MR 2ω

Express the angular momentum of the cockroach:

⎛v ⎞ LC = I CωC = mr 2 ⎜ − ω ⎟ ⎝r ⎠

Substitute to obtain:

1 2

Solve for ω to obtain:

⎛v ⎞ MR 2ω − mr 2 ⎜ − ω ⎟ = 0 ⎝r ⎠

ω=

Substitute in equation (1):

2mrv MR 2 + 2mr 2

vf = v −

2mr 2 v MR 2 + 2mr 2

Substitute numerical values and evaluate vf:

2(0.015 kg )(0.08 m ) (0.01 m/s ) = 9.67 mm/s (0.25 m )(0.15 m )2 + 2(0.015 kg )(0.08 m )2 2

vf = 0.01 m/s −

*58 •• Picture the Problem The net external torque acting on this system is zero and so we know that angular momentum is conserved as these disks are brought together. Let the numeral 1 refer to the disk to the left and the numeral 2 to the disk to the right. Let the angular momentum of the disk with the larger radius be positive. Using conservation of angular momentum, relate the initial angular speeds of the disks to their common final speed and to their moments of inertia: Solve for ωf:

Express I1 and I2:

I iω i = I f ω f or

I 1ω 0 − I 2ω 0 = (I 1 + I 2 )ω f

ωf =

I1 − I 2 ω0 I1 + I 2

I 1 = 12 m(2r ) = 2mr 2 2

and

I 2 = 12 mr 2 Substitute and simplify to obtain:

ωf =

2mr 2 − 12 mr 2 ω0 = 2mr 2 + 12 mr 2

3 5

ω0

Conservation of Angular Momentum 763 59 •• Picture the Problem We can express the angular momentum and kinetic energy of the block directly from their definitions. The tension in the string provides the centripetal force required for the uniform circular motion and can be expressed using Newton’s 2nd law. Finally, we can use the work-kinetic energy theorem to express the work required to reduce the radius of the circle by a factor of two. (a) Express the initial angular momentum of the block:

L0 = r0 mv0

(b) Express the initial kinetic energy of the block:

K0 =

(c) Using Newton’s 2nd law, relate the tension in the string to the centripetal force required for the circular motion:

v02 T = Fc = m r0

Use the work-kinetic energy theorem to relate the required work to the change in the kinetic energy of the block:

L2f L20 W = ∆K = K f − K 0 = − 2I f 2I0

Substitute the result from part (a) and simplify to obtain:

W = − 23 mv02

1 2

mv02

L20 L2 L2 ⎛ 1 ⎞ ⎟ − 0 = 0 ⎜⎜ 2If 2I0 2 ⎝ I f − I 0 ⎟⎠ ⎞ L2 ⎛ 1 2 L20 ⎟ = 0 ⎜⎜ = − 2 ⎝ m( 12 r0 )2 − mr02 ⎟⎠ 3 mr02 =

*60 •• Picture the Problem Because the force exerted by the rubber band is parallel to the position vector of the point mass, the net external torque acting on it is zero and we can use the conservation of angular momentum to determine the speeds of the ball at points B and C. We’ll use mechanical energy conservation to find b by relating the kinetic and elastic potential energies at A and B. (a) Use conservation of momentum to relate the angular momenta at points A, B and C: Solve for vB in terms of vA:

L A = LB = LC or

mv A rA = mv B rB = mvC rC vB = v A

rA rB

764 Chapter 10 Substitute numerical values and evaluate vB:

vB = (4 m/s )

Solve for vC in terms of vA:

vC = v A

Substitute numerical values and evaluate vC:

vC = (4 m/s )

(b) Use conservation of mechanical energy between points A and B to relate the kinetic energy of the point mass and the energy stored in the stretched rubber band:

E A = EB

Solve for b:

Substitute numerical values and evaluate b:

0.6 m = 2.40 m/s 1m

rA rC 0.6 m = 4.00 m/s 0.6 m

or 1 2

mv A2 + 12 brA2 = 12 mv B2 + 12 brB2

m(vB2 − v A2 ) b= 2 2 rA − rB 2 2 ( [ 0.2 kg )( 2.4 m/s ) − (4 m/s ) ] b= (0.6 m )2 − (1m )2

= 3.20 N/m

Quantization of Angular Momentum *61 • Picture the Problem The electron’s spin angular momentum vector is related to its z component as shown in the diagram.

Using trigonometry, relate the r magnitude of s to its z component:

θ = cos −1

1 2

h

0.75h

= 54.7°

62 •• Picture the Problem Equation 10-27a describes the quantization of rotational energy. We can show that the energy difference between a given state and the next higher state is proportional to l + 1 by using Equation 10-27a to express the energy difference.

Conservation of Angular Momentum 765 From Equation 10-27a we have: Using this equation, express the difference between one rotational state and the next higher state:

K l = l(l + 1)E 0 r ∆E = (l + 1)(l + 2)E0 r − l(l + 1)E0 r = 2(l + 1)E0 r

63 •• Picture the Problem The rotational energies of HBr molecule are related to l and E 0 r according to K l = l(l + 1)E 0 r where E0 r = h 2 2 I . (a) Express and evaluate the moment of inertia of the H atom:

I = mp r 2

(

)(

= 1.67 × 10 −27 kg 0.144 × 10 −9 m

)

2

= 3.46 × 10−47 kg ⋅ m 2 (b) Relate the rotational energies to l and E 0 r : Evaluate E 0 r :

K l = l(l + 1)E 0 r

( (

)

2

h2 1.05 × 10−34 J ⋅ s = 2 I 2 3.46 × 10−47 kg ⋅ m 2 1eV = 1.59 × 10−22 J × 1.60 × 10−19 J = 0.996 meV

E0 r =

)

Evaluate E for l = 1:

E1 = (1 + 1)(0.996 meV ) = 1.99 meV

Evaluate E for l = 2:

E2 = 2(2 + 1)(0.996 meV ) = 5.98 meV

Evaluate E for l = 3:

E3 = 3(3 + 1)(0.996 meV ) = 12.0 meV

64 •• Picture the Problem We can use the definition of the moment of inertia of point particles to calculate the rotational inertia of the nitrogen molecule. The rotational energies of nitrogen molecule are related to l and E 0 r according to K l = l(l + 1)E 0 r where E0 r = h 2 2 I .

766 Chapter 10 (a) Using a rigid dumbbell model, express and evaluate the moment of inertia of the nitrogen molecule about its center of mass: Substitute numerical values and evaluate I:

I = ∑ mi ri 2 = mN r 2 + mN r 2 i

= 2mN r 2

(

)(

I = 2(14) 1.66 × 10 −27 kg 5.5 × 10−11 m

)

2

= 1.41× 10−46 kg ⋅ m 2 (b) Relate the rotational energies to l and E 0 r : Evaluate E 0 r :

Substitute to obtain:

E l = l(l + 1)E 0 r

( (

)

2

h2 1.05 × 10−34 J ⋅ s = 2 I 2 1.41× 10−46 kg ⋅ m 2 1eV = 3.91× 10−23 J × 1.60 × 10−19 J = 0.244 meV

E0 r =

)

El = 0.244l(l + 1) meV

*65 •• Picture the Problem We can obtain an expression for the speed of the nitrogen molecule by equating its translational and rotational kinetic energies and solving for v. Because this expression includes the moment of inertia I of the nitrogen molecule, we can use the definition of the moment of inertia to express I for a dumbbell model of the nitrogen molecule. The rotational energies of a nitrogen molecule depend on the quantum number l according to El = L2 / 2 I = l(l + 1)h 2 / 2 I . Equate the rotational kinetic energy of the nitrogen molecule in its l = 1 quantum state and its translational kinetic energy:

E1 = 12 mN v 2

L2 l(l + 1) h 2 = 2I 2I

Express the rotational energy levels of the nitrogen molecule:

El =

For l = 1:

1(1 + 1) h 2 h 2 E1 = = 2I I

(1)

Conservation of Angular Momentum 767 Substitute in equation (1):

Solve for v to obtain:

Using a rigid dumbbell model, express the moment of inertia of the nitrogen molecule about its center of mass: Substitute in equation (2):

Substitute numerical values and evaluate v:

h2 1 = 2 mN v 2 I 2h 2 mN I

v=

(2)

I = ∑ mi ri 2 = mN r 2 + mN r 2 = 2mN r 2 i

and

mN I = 2mN2 r 2 v=

2h 2 h = 2 2 2mN r mN r

1.055 × 10 −34 J ⋅ s v= 14 1.66 ×10 −27 kg 5.5 ×10 −11 m

(

)(

)

= 82.5 m/s

Collision Problems 66 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation of the rod. Because the net external torque acting on this system is zero, we know that angular momentum is conserved in the collision. We’ll use the definition of angular momentum to express the angular momentum just after the collision and conservation of mechanical energy to determine the speed of the ball just before it makes its perfectly inelastic collision with the rod. Use conservation of angular momentum to relate the angular momentum before the collision to the angular momentum just after the perfectly inelastic collision:

Lf = Li

Use conservation of mechanical energy to relate the kinetic energy of the ball just before impact to its initial potential energy:

Kf − Ki + U f − U i = 0

Letting h represent the distance the

= mvr

or, because Ki = Uf = 0,

Kf −Ui = 0

v = 2 gh

768 Chapter 10 ball falls, substitute for K f and U i and solve for v to obtain: Substitute for v to obtain:

Lf = mr 2 gh

Substitute numerical values and evaluate Lf:

Lf = (3.2 kg )(0.9 m ) 2 9.81 m/s 2 (1.2 m )

(

)

= 14.0 J ⋅ s

*67 •• Picture the Problem Because there are no external forces or torques acting on the system defined in the problem statement, both linear and angular momentum are conserved in the collision and the velocity of the center of mass after the collision is the same as before the collision. Let the direction the blob of putty is moving initially be the positive x direction and toward the top of the page in the figure be the positive y direction. Using its definition, express the location of the center of mass relative to the center of the bar:

ycm =

Using its definition, express the velocity of the center of mass:

vcm =

Using the definition of L in terms of I and ω, express ω:

ω=

Express the angular momentum about the center of mass:

Using the parallel axis theorem, express the moment of inertia of the system relative to its center of mass: Substitute for ycm and simplify to obtain:

md below the center of the bar. M +m

mv M +m

Lcm I cm

(1)

Lcm = mv(d − ycm ) md ⎞ mMvd ⎛ = mv⎜ d − ⎟= M +m⎠ M +m ⎝ 2 I cm = 121 ML2 + Mycm + m(d − ycm )

2

Conservation of Angular Momentum 769 I cm = =

= =

2 2 md ⎞ ⎛ md ⎞ ⎛ 2 ML + M ⎜ ⎟ + m⎜ d − ⎟ M +m⎠ ⎝M +m⎠ ⎝ 2 2 2 ⎛ d (M + m ) − md ⎞ 1 ML2 + Mm d + m ⎜ ⎟ 12 (M + m )2 ⎝ M + m ⎠ 2 2 ( mM 2 d 2 M + m )mMd 2 2 1 ML2 + Mm d 1 + = 12 ML + 12 (M + m )2 (M + m )2 (M + m )2 2 1 ML2 + mMd 12 M +m 1 12

Substitute for Icm and Lcm in equation (1) and simplify to obtain:

ω=

mMvd 2 1 12 ML (M + m ) + Mmd 2

Remarks: You can verify the expression for Icm by letting m → 0 to obtain I cm = 121 ML2 and letting M → 0 to obtain Icm = 0. 68 •• Picture the Problem Because there are no external forces or torques acting on the system defined in the statement of Problem 67, both linear and angular momentum are conserved in the collision and the velocity of the center of mass after the collision is the same as before the collision. Kinetic energy is also conserved as the collision of the hard sphere with the bar is elastic. Let the direction the sphere is moving initially be the positive x direction and toward the top of the page in the figure be the positive y direction and v′ and V′ be the final velocities of the objects whose masses are m and M, respectively.

Apply conservation of linear momentum to obtain:

pi = pf or

mv = mv'+ MV ' Apply conservation of angular momentum to obtain:

(1)

Li = Lf or

mvd = mv' d + 121 ML2ω mv M

Set v′ = 0 in equation (1) and solve for V ′:

V' =

Use conservation of mechanical energy to relate the kinetic energies of translation and rotation before

Ki = Kf or 1 2

(2)

mv 2 = 12 MV '2 + 12

(3)

(

1 12

)

ML2 ω 2

(4)

770 Chapter 10 and after the elastic collision: Substitute (2) and (3) in (4) and simplify to obtain: Solve for d:

m 12m ⎛ d 2 ⎜ + 1= M M ⎜⎝ L2

d= L

⎞ ⎟⎟ ⎠

M −m 12m

69 •• Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot as shown in the diagram. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing.

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ: Apply conservation of momentum to the collision:

∆K + ∆U = 0 or, because Kf = Ui = 0,

− Ki + Uf = 0

and 1 2

d ⎛ ⎞ Iω 2 = ⎜ Mg + mgx ⎟(1 − cosθ ) (1) 2 ⎝ ⎠

Li = Lf or

Solve for ω to obtain:

Express the moment of inertia of the system about the pivot:

ω=

1 3

[ Md

]

2

+ (0.8d ) m ω

0.8dmv Md 2 + 0.64md 2

(2)

0.8dmv = Iω =

1 3

I = m(0.8d ) + 13 Md 2

2

2

= 0.64md 2 + 13 Md 2

(3)

Conservation of Angular Momentum 771

d ⎛ ⎞ ⎜ Mg + mgd ⎟(1 − cos θ ) 2 ⎝ ⎠

Substitute equations (2) and (3) in equation (1) and simplify to obtain:

0.32(dmv ) = 1 2 2 3 Md + 0.64md 2

Solve for v:

v=

(0.5 M + 0.8m ) (13 Md 2 + 0.64md 2 )g (1 − cosθ ) 0.32dm 2

Evaluate v for θ = 90° to obtain:

v=

(0.5 M + 0.8m)(13 Md 2 + 0.64md 2 )g 0.32dm 2

70 •• Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot as shown in the diagram. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing.

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ : Apply conservation of momentum to the collision:

Kf − Ki + U f − U i = 0 or, because Kf = Ui = 0,

− Ki + Uf = 0

and 1 2

d ⎛ ⎞ Iω 2 = ⎜ Mg + mgx ⎟(1 − cosθ ) (1) 2 ⎝ ⎠

Li = Lf or

0.8dmv = Iω

=

[ Md 1 3

2

]

+ (0.8d ) m ω 2

772 Chapter 10 Solve for ω to obtain:

ω=

Express the moment of inertia of the system about the pivot:

1 2

0.8dmv Md 2 + 0.64md 2

(2)

I = m(0.8d ) + 13 Md 2 2

= (0.64m + 13 M ) d 2

= [0.64(0.3 kg ) + 13 (0.8 kg )](1.2 m )

2

= 0.660 kg ⋅ m 2 Substitute equation (2) in equation (1) and simplify to obtain:

d ⎛ ⎞ ⎜ Mg + 0.8dmg ⎟(1 − cosθ ) 2 ⎝ ⎠

0.32(dmv ) I

2

= Solve for v:

g (0.5 M + 0.8m )(1 − cosθ )I 0.32dm 2

v=

Substitute numerical values and evaluate v for θ = 60° to obtain:

v=

(9.81m/s ) [0.5 (0.8 kg ) + 0.8 (0.3 kg )](0.5)(0.660 kg ⋅ m ) = 2

2

0.32 (1.2 m )(0.3 kg )

2

7.74 m/s

71 •• Picture the Problem Let the length of the uniform stick be l. We can use the impulsechange in momentum theorem to express the velocity of the center of mass of the stick. By expressing the velocity V of the end of the stick in terms of the velocity of the center of mass and applying the angular impulse-change in angular momentum theorem we can find the angular velocity of the stick and, hence, the velocity of the end of the stick.

(a) Apply the impulse-change in momentum theorem to obtain:

K = ∆p = p − p0 = p or, because p0 = 0 and p = Mvcm,

K = Mvcm

Solve for vcm to obtain:

(b) Relate the velocity V of the end of the stick to the velocity of the center of mass vcm: Relate the angular impulse to the change in the angular momentum of the stick:

vcm =

K M

V = vcm + vrel to c of m = vcm + ω ( 12 l ) K ( 12 l ) = ∆L = L − L0 = I cmω or, because L0 = 0,

K ( 12 l ) = I cmω

(1)

Conservation of Angular Momentum 773 Refer to Table 9-1 to find the moment of inertia of the stick with respect to its center of mass:

I cm = 121 Ml 2

Substitute to obtain:

K ( 12 l ) = 121 Ml 2ω

Solve for ω:

Substitute in equation (1) to obtain:

ω=

6K Ml

V=

K ⎛ 6K ⎞ l 4K +⎜ ⎟ = M ⎝ Ml ⎠ 2 M

(c) Relate the velocity V′ of the other end of the stick to the velocity of the center of mass vcm:

V = vcm − vrel to c of m = vcm − ω ( 12 l )

(d) Letting x be the distance from the center of mass toward the end not struck, express the condition that the point at x is at rest:

vcm − ωx = 0

Solve for x to obtain:

K 6K − x=0 M Ml

Solve for x to obtain:

K x= M = 6K Ml

=

K ⎛ 6K ⎞ l 2K −⎜ ⎟ = − M ⎝ Ml ⎠ 2 M

1 6

l

Note that for a meter stick struck at the 100-cm mark, the stationary point would be at the 33.3-cm mark. Remarks: You can easily check this result by placing a meterstick on the floor and giving it a sharp blow at the 100-cm mark. 72 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision.

(a) Use its definition to express the total angular momentum of the disk and projectile just before impact:

L0 = mp v0 b

774 Chapter 10 (b) Use conservation of angular momentum to relate the angular momenta just before and just after the collision:

L0 = L = Iω and ω =

Express the moment of inertia of the disk + projectile:

I = 12 MR 2 + mp b 2

2m p v 0 b

Substitute for I in the expression for ω to obtain:

ω=

(c) Express the kinetic energy of the system after impact in terms of its angular momentum:

Kf =

MR 2 + 2mp b 2

(m v b) L2 = 1 p 20 2 I 2 2 MR + mpb 2 2

(

(m v b)

=

(d) Express the difference between the initial and final kinetic energies, substitute, and simplify to obtain:

L0 I

2

p 0

MR 2 + 2mpb 2

∆E = K i − K f = mv − 2 p 0

1 2

=

)

1 2

(m v b)

2

p 0

MR 2 + 2mpb 2

⎛ ⎞ mpb 2 ⎟ mp v02 ⎜1 − 2 2 ⎜ MR + 2m b ⎟ p ⎝ ⎠

*73 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic of the aftercollision system is then transformed into gravitational potential energy as the rod-plusparticle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the masses M and m.

Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle:

Kf − Ki + U f − U i = 0 or, because Ki = 0,

Kf + U f − Ui = 0

Conservation of Angular Momentum 775 Substitute for Kf, Uf, and Ui to obtain: Solve for ω:

Letting ω′ represent the angular speed of the rod-and-particle system just after impact, use conservation of angular momentum to relate the angular momenta before and after the collision: Solve for ω′:

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax: Express the moment of inertia of the system with respect to the pivot: Substitute for θmax, I and ω′ in equation (1):

(

1 1 2 3

)

ML12 ω 2 + Mg

L1 − MgL1 = 0 2

3g L1

ω=

Li = Lf or

(

1 3

)

ML12 ω =

ω'=

1 3 1 3

(

1 3

)

ML12 + mL22 ω '

ML12

ML12 + mL22

ω

Kf − Ki + U f − U i = 0 or, because Kf = 0,

− 12 Iω '2 + Mg ( 12 L1 )(1 − cos θ max )

+ mgL2 (1 − cos θ max ) = 0

(1)

I = 13 ML12 + mL22

(

)

g 1 2 2 3 ML1 L1 = Mg ( 12 L1 ) + mgL2 2 2 1 ML + mL 1 2 3

3

m 2 m L1 L2 + 3L22 L1 + 6 L32 M M

Simplify to obtain:

L13 = 2

Simplify equation (2) by letting α = m/M and β = L2/L1 to obtain:

6α 2 β 3 + 3β 2 + 2αβ − 1 = 0

Substitute for α and simplify to obtain the cubic equation in β:

12 β 3 + 9β 2 + 4β − 3 = 0

Use the solver function* of your calculator to find the only real value

β = 0.349

(2)

776 Chapter 10 of β: *Remarks: Most graphing calculators have a ″solver″ feature. One can solve the cubic equation using either the ″graph″ and ″trace″ capabilities or the ″solver″ feature. The root given above was found using SOLVER on a TI-85. 74 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rodplus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the mass M to m.

(a) Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle: Substitute for Kf, Uf, and Ui to obtain: Solve for ω:

Letting ω′ represent the angular speed of the system after impact, use conservation of angular momentum to relate the angular momenta before and after the collision: Solve for ω′:

Kf − Ki + U f − U i = 0 or, because Ki = 0,

Kf + U f − Ui = 0

(

1 1 2 3

)

ML12 ω 2 + Mg

ω=

L1 − MgL1 = 0 2

3g L1

Li = Lf or

(

1 3

)

ML12 ω =

ω' =

(

1 3

)

ML12 + mL22 ω '

ML12 ω 2 2 1 3 ML1 + mL2 1 3

ML12 =1 2 2 3 ML1 + mL2 1 3

3g L1

(1)

Conservation of Angular Momentum 777 Substitute numerical values to obtain:

ω' =

(2 kg )(1.2 m )2 2 2 1 3 (2 kg )(1.2 m ) + m(0.8 m ) 1 3

×

(

3 9.81 m/s 2 1.2 m

)

4.75 kg ⋅ m 2 / s 0.960 kg ⋅ m 2 + 0.64 m 2 m 4.75 kg / s = 0.960 kg + 0.64m =

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax: Substitute for Ki, Uf, and Ui to obtain: Express the moment of inertia of the system with respect to the pivot: Substitute for θmax, I and ω′ in equation (1) and simplify to obtain:

(

)

Kf − Ki + U f − U i = 0 or, because Kf = 0,

− Ki + U f − U i = 0

− 12 Iω '2 + Mg ( 12 L1 )(1 − cos θ max )

+ mgL2 (1 − cos θ max ) = 0

I = 13 ML12 + mL22

1 2

(4.75 kg/s )2

0.960 kg + 0.64m

= 0.2 g (ML1 + mL2 )

Substitute for M, L1 and L2 and simplify to obtain:

m 2 + 3.00m − 8.901 = 0

Solve the quadratic equation for its positive root:

m = 1.84 kg

(b) The energy dissipated in the inelastic collision is:

∆E = U i − U f

Express Ui:

U i = Mg

Express Uf:

⎛ L ⎞ U f = (1 − cosθ max )g ⎜ M 1 + mL2 ⎟ ⎠ ⎝ 2

(2)

L1 2

778 Chapter 10 Substitute in equation (2) to obtain:

∆E = Mg

L1 2

⎛ L ⎞ − (1 − cosθ max )g ⎜ M 1 + mL2 ⎟ ⎝ 2 ⎠ Substitute numerical values and evaluate ∆E:

Uf =

(2 kg ) (9.81 m/s 2 )(1.2 m ) 2

⎞ ⎛ (2 kg )(1.2 m ) − (1 − cos37°) 9.81 m/s 2 ⎜ + (1.85 kg )(0.8 m )⎟ 2 ⎠ ⎝

(

)

= 6.51 J 75 •• Picture the Problem Let ωi and ωf be the angular velocities of the rod immediately before and immediately after the inelastic collision with the mass m. Let ω0 be the initial angular velocity of the rod. Choose the zero of gravitational potential energy be at a distance L1 below the pivot. We apply energy conservation to determine ωf and conservation of angular momentum to determine ωi. We’ll apply energy conservation to determine ω0. Finally, we’ll find the energies of the system immediately before and after the collision and the energy dissipated.

Express the energy dissipated in the inelastic collision:

∆E = U i − U f

Use energy conservation to relate the kinetic energy of the system immediately after the collision to its potential energy after a 180° rotation:

Kf − Ki + U f − U i = 0

Substitute for Kbottom, Utop, and Ubottom to obtain:

(1)

or, because Kf = Ktop = 0 and Ki = Kbottom,

− K bottom + U top − U bottom = 0

− 12 Iωf2 + 32 MgL1 + mg (L1 + L2 )

− 12 MgL1 − mg (L1 − L2 ) = 0

Simplify to obtain:

− 12 Iωf2 + MgL1 + 2mgL2 = 0

Express I:

I = 13 ML12 + mL22

Substitute for I in equation (2) and solve for ωf to obtain:

ωf =

2 g (ML1 + 2mL2 ) 2 2 1 3 ML1 + mL2

(2)

Conservation of Angular Momentum 779 Substitute numerical values and evaluate ωf:

ωf =

(

)

2 9.81 m/s 2 [(0.75 kg )(1.2 m ) + 2(0.4 kg )(0.8 m )] = 7.00 rad/s 2 2 1 3 (0.75 kg )(1.2 m ) + (0.4 kg )(0.8 m )

Use conservation of angular momentum to relate the angular momentum of the system just before the collision to its angular momentum just after the collision: Substitute for Ii and If and solve for ωi:

Li = Lf or

I iω i = I f ω f

(

1 3

)

ML12 ωi =

(

1 3

)

ML12 + mL22 ωf

and

⎡ 3m ⎛ L ⎞ 2 ⎤ ⎜⎜ 2 ⎟⎟ ⎥ ωf ωi = ⎢1 + ⎢⎣ M ⎝ L1 ⎠ ⎥⎦ Substitute numerical values and evaluate ωi:

⎡ 3(0.4 kg ) ⎛ 0.8 m ⎞ 2 ⎤ ⎟ ⎥ (7.00 rad/s ) ⎜ ωi = ⎢1 + 0.75 kg ⎜⎝ 1.2 m ⎟⎠ ⎥ ⎢⎣ ⎦ = 12.0 rad/s

Apply conservation of mechanical energy to relate the initial rotational kinetic energy of the rod to its rotational kinetic energy just before its collision with the particle:

Kf − Ki + U f − U i = 0

Substitute to obtain:

(

1 1 2 3

)

ML12 ωi2 − 12

(

1 3

)

ML12 ω02 + Mg

− MgL1 = 0 Solve for ω0:

Substitute numerical values and evaluate ω0:

ω 0 = ω i2 −

ω0 =

3g L1

(

(12 rad/s)2 − 3 9.81 m/s

= 10.9 rad/s

1.2 m

2

)

L1 2

780 Chapter 10 Substitute in equation (1) to express the energy dissipated in the collision:

∆E =

(

1 1 2 3

)

ML12 ωi2 − MgL1 + 2mgL2

Substitute numerical values and evaluate ∆E:

∆E =

1 6

(0.75 kg )(1.2 m )2 (12 rad/s )2 − (9.81m/s 2 ) [(0.75 kg )(1.2 m ) + 2(0.4 kg )(0.8 m )]

= 10.8 J

76 ••• Picture the Problem Let v be the speed of the particle immediately after the collision and ωi and ωf be the angular velocities of the rod immediately before and immediately after the elastic collision with the mass m. Choose the zero of gravitational potential energy be at a distance L1 below the pivot. Because the net external torque acting on the system is zero, angular momentum is conserved in this elastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the elastic collision with the particle and the kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the mass M to m.

Use energy conservation to relate the energies of the system immediately before and after the elastic collision: Substitute for Kf, Uf, and Ui to obtain:

Kf − Ki + U f − U i = 0 or, because Ki = 0,

Kf + U f − Ui = 0 1 2

mv 2 + Mg

L1 (1 − cosθ max ) − Mg L1 = 0 2 2

Solve for mv2:

mv 2 = MgL1 cos θ max

Apply conservation of energy to express the angular speed of the rod just before the collision:

Kf − Ki + U f − U i = 0

Substitute for Kf, Uf, and Ui to obtain:

Solve for ωi:

(1)

or, because Ki = 0,

Kf + U f − Ui = 0

(

1 1 2 3

)

ML12 ωi2 + Mg

ωi =

3g L1

L1 − MgL1 = 0 2

Conservation of Angular Momentum 781

(

Apply conservation of energy to the rod after the collision:

1 1 2 3

Solve for ωf:

)

ML12 ωf2 − Mg

L1 (1 − cosθ max ) = 0 2

0 .6 g L1

ωf = Li = Lf

Apply conservation of angular momentum to the collision:

or

(

Solve for mv:

1 3

)

ML12 ωi =

mv =

Substitute for ωf and ωI to obtain:

1 3

(

1 3

)

ML12 ωf + mvL2

ML12 (ωi − ωf ) L2

⎛ 3g 0.6 g ML12 ⎜⎜ − L1 ⎝ L1 mv = 3 L2

Divide equation (1) by equation (2) to eliminate m and solve for v:

v=

=

MgL1 cos θ max ⎛ 3g 0.6 g ML12 ⎜⎜ − L1 ⎝ L1 3L2

⎞ ⎟ ⎟ ⎠

(2)

⎞ ⎟ ⎟ ⎠

3gL2 cosθ max 3gL1 − 0.6 gL1

Substitute numerical values and evaluate v:

v=

(

(

)

3 9.81m/s 2 (0.8 m ) cos 37°

3 9.81m/s

2

) (1.2 m) −

(

)

0.6 9.81 m/s 2 (1.2 m )

= 5.72 m/s

Solve equation (1) for m:

m=

MgL1 cos θ max v2

Substitute for v in the expression for mv and solve for m:

m=

(2 kg )(9.81m/s2 )(1.2 m )cos 37° (5.72 m/s)2

= 0.575 kg Because the collision was elastic:

∆E = 0

782 Chapter 10 77 •• Picture the Problem We can determine the angular momentum of the wheel and the angular velocity of its precession from their definitions. The period of the precessional motion can be found from its angular velocity and the angular momentum associated with the motion of the center of mass from its definition.

w 2 Rω g

(a) Using the definition of angular momentum, express the angular momentum of the spinning wheel:

L = Iω = MR 2ω =

Substitute numerical values and evaluate L:

⎛ 30 N ⎞ ⎟(0.28 m )2 L = ⎜⎜ 2 ⎟ ⎝ 9.81 m/s ⎠ ⎛ rev 2π rad ⎞ × ⎜12 × ⎟ s rev ⎠ ⎝ = 18.1 J ⋅ s dφ MgD = dt L

(b) Using its definition, express the angular velocity of precession:

ωp =

Substitute numerical values and evaluate ωp:

ωp =

(c) Express the period of the precessional motion as a function of the angular velocity of precession:

T=

(d) Express the angular momentum of the center of mass due to the precession:

Lp = I cmωp = MD 2ωp

Substitute numerical values and evaluate Lp:

⎛ 30 N ⎞ ⎟(0.25 m )2 (0.414 rad/s ) Lp = ⎜⎜ 2 ⎟ ⎝ 9.81 m/s ⎠

(30 N )(0.25 m ) =

2π

ωp

18.1J ⋅ s

=

0.414 rad/s

2π = 15.2 s 0.414 rad/s

= 0.0791 J ⋅ s The direction of Lp is either up or down, depending on the direction of L. *78 •• Picture the Problem The angular velocity of precession can be found from its definition. Both the speed and acceleration of the center of mass during precession are related to the angular velocity of precession. We can use Newton’s 2nd law to find the vertical and

Conservation of Angular Momentum 783 horizontal components of the force exerted by the pivot. (a) Using its definition, express the angular velocity of precession:

2 gD dφ MgD MgD = = 1 = 2 2 dt I sωs R ωs 2 MR ωs

ωp =

Substitute numerical values and evaluate ωp:

ωp =

(

)

2 9.81 m/s 2 (0.05 m ) = 3.27 rad/s rev 2π rad 1 min ⎞ 2⎛ ⎟ (0.064 m ) ⎜⎜ 700 × × min rev 60 s ⎟⎠ ⎝ vcm = Dωp = (0.05 m )(3.27 rad/s )

(b) Express the speed of the center of mass in terms of its angular velocity of precession:

= 0.164 m/s

acm = Dωp2 = (0.05 m )(3.27 rad/s )

2

(c) Relate the acceleration of the center of mass to its angular velocity of precession:

= 0.535 m/s 2 Fv = Mg = (2.5 kg ) (9.81m/s 2 )

(d) Use Newton’s 2nd law to relate the vertical component of the force exerted by the pivot to the weight of the disk:

= 24.5 N

Fv = Macm = (2.5 kg ) (0.535 m/s 2 )

Relate the horizontal component of the force exerted by the pivot to the acceleration of the center of mass:

= 1.34 N

General Problems 79 • Picture the Problem While the 3-kg particle is moving in a straight line, it has angular

r

r

r

r

r

momentum given by L = r × p where r is its position vector and p is its linear

r

r

r

momentum. The torque due to the applied force is given by τ = r × F . (a) Express the angular momentum of the particle:

r

r

Express the vectors r and p :

r r r L=r×p r r = (12 m ) iˆ + (5.3 m ) ˆj

784 Chapter 10 and

r p = mviˆ = (3 kg )(3 m/s )iˆ = (9 kg ⋅ m/s ) iˆ

r

Substitute and simplify to find L :

[

]

r L = (12 m ) iˆ + (5.3 m ) ˆj × (9 kg ⋅ m/s ) iˆ = 47.7 kg ⋅ m 2 /s ˆj × iˆ

(

)(

(

)

)

= − 47.7 kg ⋅ m 2 /s kˆ (b) Using its definition, express the torque due to the force:

r

Substitute and simplify to find τ :

r r r τ = r ×F

[

] ( )

r τ = (12 m ) iˆ + (5.3 m ) ˆj × (− 3 N ) iˆ = −(15.9 N ⋅ m ) ˆj × iˆ

= (15.9 N ⋅ m ) kˆ 80 • Picture the Problem The angular momentum of the particle is given by

r r r r r L = r × p where r is its position vector and p is its linear momentum. The torque acting r r on the particle is given by τ = dL dt . Express the angular momentum of the particle:

Evaluate

r dr : dt

r r r r r r r L = r × p = r × mv = mr × v r r dr = mr × dt r dr = 6tˆj dt

r

Substitute and simplify to find L :

[

= Find the torque due to the force:

}]

{

r L = (3 kg ) (4 m ) iˆ + (3t 2 m/s 2 ) ˆj × (6t m/s ) ˆj

(72.0t J ⋅ s ) kˆ

r r dL d (72.0t J ⋅ s ) kˆ τ= = dt dt = (72.0 N ⋅ m ) kˆ

[

]

Conservation of Angular Momentum 785 81 •• Picture the Problem The ice skaters rotate about their center of mass; a point we can locate using its definition. Knowing the location of the center of mass we can determine their moment of inertia with respect to an axis through this point. The angular momentum of the system is then given by L = I cmω and its kinetic energy can be found

from K = L2 2 I cm . (a) Express the angular momentum of the system about the center of mass of the skaters:

L = I cmω

Using its definition, locate the center of mass, relative to the 85-kg skater, of the system:

xcm =

Calculate I cm :

I cm = (55 kg )(1.7 m − 0.668 m )

(55 kg )(1.7 m ) + (85 kg )(0)

55 kg + 85 kg = 0.668 m 2

+ (85 kg )(0.668 m )

2

= 96.5 kg ⋅ m 2 Substitute to determine L:

⎛ 1 rev 2π rad ⎞ ⎟⎟ L = 96.5 kg ⋅ m 2 ⎜⎜ × 2.5 s rev ⎝ ⎠

(

)

= 243 J ⋅ s (b) Relate the total kinetic energy of the system to its angular momentum and evaluate K: Substitute numerical values and evaluate K:

L2 K= 2 I cm

K=

(

(243 J ⋅ s )2

2 96.5 kg ⋅ m 2

)=

306 J

786 Chapter 10 *82 •• Picture the Problem Let the origin of the coordinate system be at the pivot (point P). The diagram shows the forces acting on the ball. We’ll apply Newton’s 2nd law to the ball to determine its speed. We’ll then use the derivative of its position vector to express its velocity and the definition of r angular momentum to show that L has both horizontal and vertical components. r We can use the derivative of L with respect to time to show that the rate at which the angular momentum of the ball changes is equal to the torque, relative to the pivot point, acting on it.

(a) Express the angular momentum of the ball about the point of support: Apply Newton’s 2nd law to the ball:

r r r r r L = r × p = mr × v

∑ Fx = T sin θ = m

(1)

v2 r sin θ

and

∑F

z

= T cos θ − mg = 0

Eliminate T between these equations and solve for v:

v = rg sin θ tan θ

Substitute numerical values and evaluate v:

v=

Express the position vector of the ball:

(1.5 m )(9.81m/s2 ) sin30°tan30°

= 2.06 m/s

(

r r = (1.5 m )sin 30° cos ωt iˆ + sin ωtˆj − (1.5 m ) cos 30°kˆ

)

where ω = ωkˆ. Find the velocity of the ball:

r r dr v= dt = (0.75ω m/s ) − sin ωt iˆ + cos ωt ˆj

(

Evaluate ω:

ω=

2.06 m/s = 2.75 rad/s (1.5 m )sin 30°

)

Conservation of Angular Momentum 787

(

r v = (2.06 m/s ) − sin ωt iˆ + cos ωtˆj

Substitute for ω to obtain:

)

r

Substitute in equation (1) and evaluate L :

[

(

)

r L = (2 kg )(1.5 m )sin 30° cos ωt iˆ + sin ωt ˆj − (1.5 m ) cos 30°kˆ × [(2.06 m/s ) − sin ωt iˆ + cos ωt ˆj

(

[ (

)

]

)]

]

= 5.36 cos ωt iˆ + sin ωt ˆj + 3.09 kˆ J ⋅ s

r

The horizontal component of L is:

r

The vertical component of L is:

(

)

5.36 cos ωt iˆ + sin ωt ˆj J ⋅ s 3.09 kˆ J ⋅ s

r dL : (b) Evaluate dt

r dL = 5.36ω − sin ωt iˆ + cos ωtˆj J dt

r dL Evaluate the magnitude of : dt

r dL = (5.36 N ⋅ m ⋅ s )(2.75 rad/s ) dt

[

(

)]

= 14.7 N ⋅ m Express the magnitude of the torque exerted by gravity about the point of support: Substitute numerical values and evaluate τ :

τ = mgr sin θ

τ = (2 kg ) (9.81m/s 2 )(1.5 m )sin 30° = 14.7 N ⋅ m

83 •• Picture the Problem In part (a) we need to decide whether a net torque acts on the object. In part (b) the issue is whether any external forces act on the object. In part (c) we can apply the definition of kinetic energy to find the speed of the object when the unwrapped length has shortened to r/2.

(a) Consider the overhead view of the cylindrical post and the object shown in the adjoining figure. The object rotates counterclockwise. The torque about the center of the cylinder is clockwise and of magnitude RT, where R is the radius of the cylinder and T is the tension. So

788 Chapter 10 L must decrease.

No, L decreases.

(b) Because, in this frictionless environment, no net external forces act on the object:

Its kinetic energy is constant.

(c) Express the kinetic energy of the object as it spirals inward:

K = 12 Iω 2 =

1 2

(mr ) vr 2

2 2

= 12 mv 2

v0 . (The kinetic energy remains constant.) 84 •• Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system.

Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

Li = Lf or

I iω i = I f ω f

Ii I ωi = i ω If If

Solve for ωi:

ωf =

Express Ii:

I i = 101 ML2 + 2 14 ml 2

Express If:

I f = 101 ML2 + 2 14 mL2

Substitute to express ω f in terms of ω :

ωf =

=

Express the initial kinetic energy of the system:

1 10 1 10

)

(

)

ML2 + 2(14 ml 2 ) ω ML2 + 2(14 mL2 ) l2 L2 ω M + 5m

M + 5m

K i = 12 I iω 2 = =

(

1 20

(ML

2

[

1 1 2 10

(

)]

ML2 + 2 14 ml 2 ω 2

)

+ 5ml 2 ω 2

Conservation of Angular Momentum 789

Express the final kinetic energy of the system and simplify to obtain:

K f = 12 I f ωf2 =

[

1 1 2 10

(

)]

ML2 + 2 14 mL2 ωf2 =

1 20

=

1 20

⎛ l2 ⎞ ⎜ M + 5m 2 ⎟ L ω⎟ = ML2 + 5mL2 ⎜ ⎜ M + 5m ⎟ ⎜ ⎟ ⎝ ⎠

(

1 20

)

(

2

2

2 f

2 ⎡⎛ l2 ⎞ ⎤ ⎢ ⎜⎜ ML + 5m ⎟⎟ ⎥ L⎠ ⎥ 2 1 ⎢⎝ 20 ⎢ ⎥ω M + 5m ⎢ ⎥ ⎢⎣ ⎥⎦

2

=

(ML + 5mL )ω

)

⎡ ML2 + 5ml 2 2 ⎤ 2 ω ⎢ 2 2 ⎥ ⎢⎣ ML + 5mL ⎥⎦

85 •• Determine the Concept Yes. The net external torque is zero and angular momentum is conserved as the system evolves from its initial to its final state. Because the disks come to the same final position, the initial and final configurations are the same as in Problem 84. Therefore, the answers are the same as for Problem 84. 86 •• Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system.

Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

Li = Lf or

I iω i = I f ω f

Ii I ωi = i ω If If

Solve for ωf:

ωf =

Relate the tension in the string to the angular speed of the system and solve for and evaluate ω:

l T = mrω 2 = m ω 2 2 and

ω=

2T = ml

2(108 N ) (0.4 kg )(0.6 m )

= 30.0 rad/s

(1)

790 Chapter 10 Express and evaluate Ii:

(

I i = 101 ML2 + 2 14 ml 2

)

= 101 (0.8 kg )(2 m ) + 12 (0.4 kg )(0.6 m ) 2

2

= 0.392 kg ⋅ m 2 Express and evaluate If:

(

I f = 101 ML2 + 2 14 mL2

)

= 101 (0.8 kg )(2 m ) + 12 (0.4 kg )(2 m ) 2

2

= 1.12 kg ⋅ m 2 Substitute in equation (1) and solve for ω f :

ωf =

0.392 kg ⋅ m 2 Ii (30.0 rad/s ) ω= 1.12 kg ⋅ m 2 If

= 10.5 rad/s Express and evaluate the initial kinetic energy of the system:

K i = 12 I iω 2 =

1 2

(0.392 kg ⋅ m )(30.0 rad/s)

2

2

= 176 J Express and evaluate the final kinetic energy of the system:

K f = 12 I f ωf2 =

(1.12 kg ⋅ m )(10.5 rad/s)

2

2

1 2

= 61.7 J 87 •• Picture the Problem Until the inelastic collision of the cylindrical objects at the ends of the cylinder, both angular momentum and energy are conserved. Let K’ represent the kinetic energy of the system just before the disks reach the end of the cylinder and use conservation of energy to relate the initial and final kinetic energies to the final radial velocity.

Using conservation of mechanical energy, relate the initial and final kinetic energies of the disks: Solve for vr:

Using conservation of angular momentum, relate the initial and final angular velocities to the initial

Ki = K ' or 1 2

I iω 2 = 12 I f ω f2 +

vr =

I iω 2 − I f ω f2 2m

Li = Lf or

1 2

(2mv ) 2 r

(1)

Conservation of Angular Momentum 791 and final moments of inertia: Solve for ω f :

I iω i = I f ω f

ωf =

Ii I ω= iω If If

Express Ii:

I i = 101 ML2 + 2 14 ml 2

)

Express If:

If =

(

Substitute to obtain ω f in terms of ω :

(

ωf = =

Substitute in equation (1) and simplify to obtain:

vr =

1 10

1 10 1 10

ML

+ 2

2

( (

1 4

mL

2

)

) )

ML2 + 2 14 ml 2 ω ML2 + 2 14 mL2

ML2 + 5ml 2 ω ML2 + 5mL2 lω 2L

(L

2

− l2 )

88 •• Picture the Problem Because the net torque acting on the system is zero, we can use conservation of angular momentum to relate the initial and final angular velocities and the initial and final kinetic energy of the system.

Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia: Solve for ω f :

Relate the tension in the string to the angular speed of the system: Solve for ω:

Substitute numerical values and evaluate ω:

Li = Lf or

I iω i = I f ω f

ωf =

Ii I ωi = i ω If If

(1)

l T = mrω 2 = m ω 2 2

ω=

2T ml

ω=

2(108 N ) = 30.0 rad/s (0.4 kg )(0.6 m )

792 Chapter 10 Express and evaluate Ii:

(

I i = 101 ML2 + 2 14 ml 2

)

= 101 (0.8 kg )(2 m ) + 12 (0.4 kg )(0.6 m ) 2

2

= 0.392 kg ⋅ m 2

(

Letting L′ represent the final separation of the disks, express and evaluate If:

I f = 101 ML2 + 2 14 mL'2

Substitute in equation (1) and solve for ω f :

ωf =

)

= 101 (0.8 kg )(2 m ) + 12 (0.4 kg )(1.6 m ) 2

2

= 0.832 kg ⋅ m 2

Ii 0.392 kg ⋅ m 2 (30.0 rad/s ) ω= 0.832 kg ⋅ m 2 If

= 14.1 rad/s Express and evaluate the initial kinetic energy of the system:

K i = 12 I iω 2 =

1 2

(0.392 kg ⋅ m )(30.0 rad/s) 2

2

= 176 J Express and evaluate the final kinetic energy of the system:

K f = 12 I f ωf2 =

1 2

(0.832 kg ⋅ m )(14.1rad/s) 2

= 82.7 J The energy dissipated in friction is:

∆E = K i − K f = 176 J − 82.7 J = 93.3 J

*89 •• Picture the Problem The drawing shows an elliptical orbit. The triangular element of the area is dA = 12 r (rdθ ) = 12 r 2 dθ .

Differentiate dA with respect to t to obtain:

dA 1 2 dθ 1 2 = r = r ω dt 2 dt 2

Because the gravitational force acts along the line joining the two objects, τ = 0 and:

L = mr 2ω = constant

2

Conservation of Angular Momentum 793 Eliminate r2ω between the two equations to obtain:

dA L = = constant dt 2m

90 •• Picture the Problem Let x be the radial distance each disk moves outward. Because the net torque acting on the system is zero, we can use conservation of angular momentum to relate the initial and final angular velocities to the initial and final moments of inertia. We’ll assume that the disks are thin enough so that we can ignore their lengths in expressing their moments of inertia.

Use conservation of angular momentum to relate the initial and final angular velocities of the disks: Solve for ωf:

Li = Lf or

I iω i = I f ω f

ωf =

Ii ωi If

(1)

Express the initial moment of inertia of the system:

I i = I cyl + 2I disk

Express the moment of inertia of the cylinder:

I cyl = 121 ML2 + 12 MR 2

(

= 121 M L2 + 6 R 2

[

)

= 121 (0.8 kg )(1.8 m ) + 6(0.2 m ) 2

2

]

= 0.232 kg ⋅ m 2 Letting l represent the distance of the clamped disks from the center of rotation and ignoring the thickness of each disk (we’re told they are thin), use the parallel-axis theorem to express the moment of inertia of each disk:

I disk = 14 mr 2 + ml 2

With the disks clamped:

I i = I cyl + 2 I disk

(

= 14 m r 2 + 4l 2 =

1 4

)

(0.2 kg )[(0.2 m )2 + 4(0.4 m )2 ]

= 0.0340 kg ⋅ m 2

(

= 0.232 kg ⋅ m 2 + 2 0.0340 kg ⋅ m 2 = 0.300 kg ⋅ m

2

)

794 Chapter 10

(

I disk = 14 m r 2 + 4l 2

With the disks unclamped, l = 0.6 m and:

=

1 4

)

(0.2 kg )[(0.2 m )2 + 4(0.6 m )2 ]

= 0.0740 kg ⋅ m 2

I f = I cyl + 2 I disk

Express and evaluate the final moment of inertia of the system:

(

= 0.232 kg ⋅ m 2 + 2 0.0740 kg ⋅ m 2

)

= 0.380 kg ⋅ m 2 0.300 kg ⋅ m 2 (8 rad/s ) ωf = 0.380 kg ⋅ m 2

Substitute in equation (1) to determine ωf:

= 6.32 rad/s Express the energy dissipated in friction:

∆E = Ei − E f

Apply Newton’s 2nd law to each disk when they are in their final positions:

∑F

Solve for k:

mrω 2 k= x

= 12 I iω i2 − radial

(

1 2

I f ω f2 + 12 kx 2

)

= kx = mrω 2

2 ( 0.2 kg )(0.6 m )(6.32 rad/s ) k=

Substitute numerical values and evaluate k:

0.2 m

= 24.0 N/m Express the energy dissipated in friction:

Wfr = Ei − Ef = 12 I iωi2 −

(

1 2

I f ωf2 + 12 kx 2

)

Substitute numerical values and evaluate Wfr:

Wfr =

1 2

(0.300 kg ⋅ m )(8 rad/s) 2

2

(

)

− 12 0.380 kg ⋅ m 2 (6.32 rad/s ) − 12 (24 N/m )(0.2 m ) 2

2

= 1.53 J 91 •• Picture the Problem Let the letters d, m, and r denote the disk and the letters t, M, and R the turntable. We can use conservation of angular momentum to relate the final angular speed of the turntable to the initial angular speed of the Euler disk and the moments of inertia of the turntable and the disk. In part (b) we’ll need to use the parallel-axis theorem

Conservation of Angular Momentum 795 to express the moment of inertia of the disk with respect to the rotational axis of the turntable. You can find the moments of inertia of the disk in its two orientations and that of the turntable in Table 9-1. (a) Use conservation of angular momentum to relate the initial and final angular momenta of the system:

I diωdi = I df ωdf + I tf ωtf

Because ωtf = ωdf:

I diωdi = I df ωtf + I tf ωtf

Solve for ωtf:

ωtf =

I di ωdi I df + I tf

Ignoring the negligible thickness of the disk, express its initial moment of inertia:

I di = 14 mr 2

Express the final moment of inertia of the disk:

I df = 12 mr 2

Express the final moment of inertia of the turntable:

I tf = 12 MR 2

Substitute in equation (1) to obtain:

ω tf = =

Express ωdi in rad/s:

Substitute numerical values in equation (2) and evaluate ωtf:

(1)

mr 2 ωdi 2 1 1 2 mr + 2 MR 1 4 2

(2)

1 ωdi MR 2 2+2 mr 2

ωdi = 30

rev 2π rad 1 min × × = π rad/s min rev 60 s

π rad/s

ω tf = 2+2

(0.735 kg )(0.25 m )2 (0.5 kg )(0.125 m )2

= 0.228 rad/s (b) Use the parallel-axis theorem to express the final moment of inertia of the disk when it is a distance L from the center of the turntable:

(

I df = 12 mr 2 + mL2 = m 12 r 2 + L2

)

796 Chapter 10 Substitute in equation (1) to obtain:

mr 2 ωtf = 1 2 2 1 ωdi m 2 r + L + 2 MR 2 1 4

(

=

)

1 2

L MR 2 2+4 2 +2 r mr 2

ωdi

Substitute numerical values and evaluate ωtf:

ωtf =

(0.1m )

π rad/s

(0.735 kg )(0.25 m )2 2+4 + 2 (0.125 m )2 (0.5 kg )(0.125 m )2 2

= 0.192 rad/s

92 •• Picture the Problem We can express the period of the earth’s rotation in terms of its angular velocity of rotation and relate its angular velocity to its angular momentum and moment of inertia with respect to an axis through its center. We can differentiate this expression with respect to T and then use differentials to approximate the changes in r and T.

(a) Express the period of the earth’s rotation in terms of its angular velocity of rotation:

T=

Relate the earth’s angular velocity of rotation to its angular momentum and moment of inertia:

ω=

Substitute and simplify to obtain:

T=

2π

ω L L = 2 2 I 5 mr 2π

(

)

mr 2 4π m 2 r = L 5L

2 5

(b) Find dT/dr:

2T dT ⎛T ⎞ ⎛ 4π m ⎞ = 2⎜ ⎟r = 2⎜ 2 ⎟r = dr r ⎝r ⎠ ⎝ 5L ⎠

Solve for dT/T:

dT dr ∆T ∆r ≈2 = 2 or T r T r

(c) Using the equation we just derived, substitute for the change in the period of the earth:

∆T 14 d ∆r 1y 1 = × = =2 y 365.24 d 1460 T r

Conservation of Angular Momentum 797 6.37 ×10 3 km r ∆r = = 2(1460) 2(1460)

Solve for and evaluate ∆r:

= 2.18 km *93 •• Picture the Problem Let ωP be the angular velocity of precession of the earth-asgyroscope, ωs its angular velocity about its spin axis, and I its moment of inertia with respect to an axis through its poles, and relate ωP to ωs and I using its definition.

τ

Use its definition to express the precession rate of the earth as a giant gyroscope:

ωP =

Substitute for I and solve for τ:

τ = LωP = IωωP

Express the angular velocity ωs of the earth about its spin axis:

ω=

L

2π where T is the period of rotation of T

the earth. Substitute to obtain:

τ=

2πIωP T

Substitute numerical values and evaluateτ:

τ=

(

)(

)

2π 8.03 × 1037 kg ⋅ m 2 7.66 × 10−12 s −1 = 4.47 × 1022 N ⋅ m 24 h 3600 s 1d × × d h

94 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied torque is removed, angular momentum is conserved.

Express the work done before the string breaks: Express the moment of inertia of the system (see Table 9-1):

W = ∆K = K f = 12 I f ωf2

(1)

I = I cyl + 2 I m = I ( x ) = 121 M cyl L2cyl + 2mx 2 = 121 (1.2 kg )(1.6 m ) + 2(0.4 kg )x 2 2

= 0.256 kg ⋅ m 2 + (0.8 kg )x 2

798 Chapter 10 Evaluate If = I(0.4 m):

I f = I (0.4 m ) = 0.256 kg ⋅ m 2 + (0.8 kg )(0.4 m )

2

= 0.384 kg ⋅ m 2 Using Newton’s 2nd law, relate the forces acting on a disk to its angular velocity: Solve for ωf:

Substitute numerical values and evaluate ωf:

∑F

= T = mrωf2

radial

where T is the tension in the string at which it breaks.

ωf =

T mr

ωf =

100 N = 25.0 rad/s (0.4 kg )(0.4 m )

Substitute in equation (1) to express the work done before the string breaks:

W = 12 I f ωf2

Substitute numerical values and evaluate W:

W=

With the applied torque removed, angular momentum is conserved and we can express the angular momentum as a function of x:

L = Ifωf

Solve for ω ( x ) :

Substitute numerical values to obtain:

1 2

(0.384 kg ⋅ m )(25 rad/s)

2

2

= 120 J

= I ( x )ω ( x )

ω (x ) = ω (x ) = =

I f ωf I (x )

(0.384 kg ⋅ m )(25 rad/s) 2

0.256 kg ⋅ m 2 + (0.8 kg )x 2 9.60 J ⋅ s 0.256 kg ⋅ m 2 + (0.8 kg )x 2

95 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied

Conservation of Angular Momentum 799 torque is removed, angular momentum is conserved.

W = ∆K = K f = 12 I f ωf2

Express the work done before the string breaks:

(1)

I = I cyl + 2 I m = I ( x ) = 121 M cyl L2cyl + 2mx 2

Express the moment of inertia of the system (see Table 9-1):

I = 121 (1.2 kg )(1.6 m ) + 2(0.4 kg )x 2 2

Substitute numerical values to obtain:

= 0.256 kg ⋅ m 2 + (0.8 kg )x 2

I f = I (0.4 m )

Evaluate If = I(0.4 m):

= 0.256 kg ⋅ m 2 + (0.8 kg )(0.4 m )

2

= 0.384 kg ⋅ m 2 Using Newton’s 2nd law, relate the forces acting on a disk to its angular velocity: Solve for ωf:

Substitute numerical values and evaluate ωf:

∑F

rad

where T is the tension in the string at which it breaks.

ωf =

T mr

ωf =

100 N = 25.0 rad/s (0.4 kg )(0.4 m )

L = Ifωf

With the applied torque removed, angular momentum is conserved and we can express the angular momentum as a function of x: Solve for ω ( x ) :

= T = mrω f2

= I ( x )ω ( x )

ω (x ) =

I f ωf I (x )

Substitute numerical values and simplify to obtain:

ω (x ) =

(0.384 kg ⋅ m )(25 rad/s) 2

0.256 kg ⋅ m + (0.8 kg )x

Evaluate ω (0.8 m ) :

2

2

=

9.60 J ⋅ s 0.256 kg ⋅ m 2 + (0.8 kg )x 2

800 Chapter 10

ω (0.8 m ) =

9.60 J ⋅ s = 12.5 rad/s 2 0.256 kg ⋅ m 2 + (0.8 kg )(0.8 m )

Remarks: Note that this is the angular velocity in both instances. Because the disks leave the cylinder with a tangential velocity of 12 Lω , the angular momentum of the system remains constant. 96 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied torque is removed, angular momentum is conserved.

Express the work done before the string breaks: Using the parallel axis theorem and treating the disks as thin disks, express the moment of inertia of the system (see Table 9-1): Substitute numerical values to obtain:

W = ∆K = K f = 12 I f ωf2

(1)

I (x ) = I cyl + 2 I m

(

= 121 ML2 + 12 MR 2 + 2 14 mR 2 + mx 2

(

)

(

= 121 M L2 + 6 R 2 + 2m 14 R 2 + x 2

[

I (x ) = 121 (1.2 kg ) (1.6 m ) + 6(0.4 m ) + 2(0.4 kg )

2

2

[ (0.4 m) + x ] 1 4

2

)

)

]

2

= 0.384 kg ⋅ m 2 + (0.8 kg )x 2 Evaluate If = I(0.4 m):

I f = I (0.4 m )

= 0.384 kg ⋅ m 2 + (0.8 kg )(0.4 m )

2

= 0.512 kg ⋅ m 2 Using Newton’s 2nd law, relate the forces acting on a disk to its angular velocity: Solve for ωf:

Substitute numerical values and evaluate ωf:

∑F

rad

= T = mrω f2

where T is the tension in the string at which it breaks.

ωf =

T mr

ωf =

100 N = 25.0 rad/s (0.4 kg )(0.4 m )

Conservation of Angular Momentum 801 Substitute in equation (1) to express the work done before the string breaks:

W = 12 I f ωf2

Substitute numerical values and evaluate W:

W=

With the applied torque removed, angular momentum is conserved and we can express the angular momentum as a function of x:

L = Ifωf

Solve for ω ( x ) :

Substitute numerical values to obtain:

1 2

(0.512 kg ⋅ m )(25 rad/s)

2

2

= 160 J

= I ( x )ω ( x )

ω (x ) =

I f ωf I (x )

(0.512 kg ⋅ m )(25 rad/s) ω (x ) = 0.384 kg ⋅ m + (0.8 kg )x 2

2

=

2

12.8 J ⋅ s 0.384 kg ⋅ m 2 + (0.8 kg )x 2

*97 ••• Picture the Problem Let the origin of the coordinate system be at the center of the pulley with the upward direction positive. Let λ be the linear density (mass per unit length) of the rope and L1 and L2 the lengths of the hanging parts of the rope. We can use conservation of mechanical energy to find the angular velocity of the pulley when the difference in height between the two ends of the rope is 7.2 m.

(a) Apply conservation of energy to relate the final kinetic energy of the system to the change in potential energy: Express the change in potential energy of the system:

∆K + ∆U = 0 or, because Ki = 0,

K + ∆U = 0

(1)

∆U = U f − U i

= − 12 L1f (L1f λ )g − 12 L2f (L2f λ )g

− [− 12 L1i (L1i λ )g − 12 L2i (L2i λ )g ]

(

)

(

) )]

= − 12 L1f2 + L22f λ g + 12 L1i2 + L22i λ g

[(

) (

= − 12 λ g L1f2 + L22f − L1i2 + L22i

802 Chapter 10 Because L1 + L2 = 7.4 m, L2i – L1i = 0.6 m, and L2f – L1f = 7.2 m, we obtain: Substitute numerical values and evaluate ∆U:

L1i = 3.4 m, L2i = 4.0 m, L1f = 0.1 m, and L2f = 7.3 m.

(

∆U = − 12 (0.6 kg/m ) 9.81 m/s 2

[

× (0.1 m ) + (7.3 m ) 2

2

− (3.4 m ) − (4 m ) = −75.75 J 2

Express the kinetic energy of the system when the difference in height between the two ends of the rope is 7.2 m: Substitute numerical values and simplify:

2

]

K = 12 I pω 2 + 12 Mv 2

( (

=

1 1 2 2

=

1 1 2 2

K=

1 1 2 2

)

M p R 2 ω 2 + 12 MR 2ω 2 M p + M )R 2ω 2

2

[ (2.2 kg ) + 4.8 kg ] ⎛⎜ 1.2 m ⎞⎟ ω 2

(

⎝ 2π ⎠

)

= 0.1076 kg ⋅ m ω Substitute in equation (1) and solve for ω:

)

2

(0.1076 kg ⋅ m )ω 2

2

2

− 75.75 J = 0

and

75.75 J = 26.5 rad/s 0.1076 kg ⋅ m 2

ω=

(b) Noting that the moment arm of each portion of the rope is the same, express the total angular momentum of the system:

L = Lp + Lr = I pω + M r R 2ω

Letting θ be the angle through which the pulley has turned, express U(θ):

U (θ ) = − 12 (L1i − Rθ ) + (L2i + Rθ ) λ g

Express ∆U and simplify to obtain:

( =( =

1 2 1 2

)

M p R2 + M r R2 ω M p + M r )R 2ω

[

(2)

2

2

∆U = U f − U i = U (θ ) − U (0)

[

]

]

= − 12 (L1i − Rθ ) + (L2i + Rθ ) λ g 2

(

)

+ 12 L + L λ g 2 1i

2 2i

2

= − R θ λ g + (L1i − L2i )Rθλ g 2

Assuming that, at t = 0, L1i ≈ L2i:

2

∆U ≈ − R 2θ 2λ g

Conservation of Angular Momentum 803

(0.1076 kg ⋅ m )ω

Substitute for K and ∆U in equation (1) to obtain:

2

Solve for ω:

− R 2θ 2λ g = 0

R 2θ 2λ g 0.1076 kg ⋅ m 2

ω=

Substitute numerical values to obtain:

2

2

(

)

⎛ 1.2 m ⎞ 2 ⎜ ⎟ (0.6 kg/m ) 9.81m/s 2π ⎠ ω= ⎝ θ 0.1076 kg ⋅ m 2

(

)

= 1.41s -1 θ Express ω as the rate of change of θ:

dθ dθ = 1.41s −1 θ ⇒ = 1.41s −1 dt dt θ

Integrate θ from 0 to θ to obtain:

ln θ = 1.41s −1 t

(

)

(

)

(

)

θ (t ) = e (1.41 s )t −1

Transform from logarithmic to exponential form to obtain: Differentiate to express ω as a function of time:

ω (t ) =

−1 dθ = 1.41s −1 e (1.41 s )t dt

(

)

(

)

−1 L = (12 M p + M r )R 2 1.41s −1 e (1.41 s )t

Substitute for ω in equation (2) to obtain: Substitute numerical values and evaluate L: 2

[(

)

] (

)

⎛ 1.2 m ⎞ −1 (1.41s−1 )t = 0.303 kg ⋅ m 2 / s e (1.41s−1 )t L = [ 12 (2.2 kg ) + (4.8 kg )]⎜ ⎟ 1.41s e ⎝ 2π ⎠

804 Chapter 10

Chapter 11 Gravity Conceptual Problems *1 • (a) False. Kepler’s law of equal areas is a consequence of the fact that the gravitational force acts along the line joining two bodies but is independent of the manner in which the force varies with distance. (b) True. The periods of the planets vary with the three-halves power of their distances from the sun. So the shorter the distance from the sun, the shorter the period of the planet’s motion. 2 • Determine the Concept We can apply Newton’s 2nd law and the law of gravity to the satellite to obtain an expression for its speed as a function of the radius of its orbit. Apply Newton’s 2nd law to the satellite to obtain:

∑ Fradial =

GMm v2 = m r r2

where M is the mass of the object the satellite is orbiting and m is the mass of the satellite. Solve for v to obtain:

v=

Thus the speed of the satellite is independent of its mass and:

GM r

(c) is correct.

3 •• Picture the Problem The acceleration due to gravity varies inversely with the square of the distance from the center of the moon. Express the dependence of the acceleration due to the gravity of the moon on the distance from its center:

a' ∝

1 r2

Express the dependence of the acceleration due to the gravity of the moon at its surface on its radius:

a∝

1 RM2

837

838 Chapter 11 Divide the first of these expressions by the second to obtain: Solve for a′:

a' RM2 = 2 a r a' =

RM2 RM2 a a = 161 a = 2 2 r (4 RM )

and (d ) is correct. 4 • Determine the Concept Measurement of G is difficult because masses accessible in the laboratory are very small compared to the mass of the earth. 5 • Determine the Concept The escape speed for a planet is given by ve =

2Gm R .

Between ve depends on the square root of M, doubling M increases the escape speed by a factor of

2 and (a ) is correct.

6 •• Determine the Concept We can take careful measurements of its position in order to determine whether its trajectory is an ellipse, a hyperbola, or a parabola. If the path is an ellipse, it will return; if its path is hyperbolic or parabolic, it will not return. 7 •• Determine the Concept The gravitational field is proportional to the mass within the sphere of radius r and inversely proportional to the square of r, i.e., proportional to r 3 r 2 = r. *8 • Determine the Concept Let m represent the mass of Mercury, MS the mass of the sun, v the orbital speed of Mercury, and R the mean orbital radius of Mercury. We can use Newton’s 2nd law of motion to relate the gravitational force acting on the Mercury to its orbital speed. Use Newton’s 2nd law to relate the gravitational force acting on Mercury to its orbital speed: Simplify to obtain:

Fnet

1 2

GM S m v2 = =m R2 R

GM Sm ⎛ GM S m ⎞ = − 12 ⎜ − ⎟ R R ⎠ ⎝ = − 12 U

mv 2 =

1 2

Gravity 839 or K = − 12 U 9 •• Picture the Problem We can use the definition of the gravitational field to express the ratio of the student’s weight at an elevation of two earth radii to her weight at the surface of the earth.

GM E m RE2

Express the weight of the student at the surface of the earth:

w = mg =

Express the weight of the student at an elevation of two earth radii:

w' = mg' =

Express the ratio of w′ to w:

GM E m 2 1 w' (3RE ) = = and (d ) is correct. w GM E m 9 RE2

GM E m (3RE )2

10 •• Determine the Concept One such machine would be a balance wheel with weights attached to the rim with half of them shielded using Cavourite. The weights on one side would be pulled down by the force of gravity, while the other side would not, leading to rotation, which can be converted into useful work, in violation of the law of the conservation of energy.

Estimation and Approximation 11 • Picture the Problem To approximate the mass of the galaxy we’ll assume the galactic center to be a point mass with the sun in orbit about it and apply Kepler’s 3rd law. Using Kepler’s 3rd law, relate the period of the sun T to its mean distance r from the center of the galaxy:

Solve for

r3 to obtain: T2

T2 =

3

r = T2

4π 2 r3 = GM galaxy

G

M galaxy Ms

4π Ms

2

G

4π 2 Ms r3 M galaxy Ms

M galaxy =

Ms 4π 2 GM s

840 Chapter 11 If we measure distances in AU and times in years: Substitute numerical values and evaluate Mgalaxy/Ms:

4π 2 r 3 M galaxy = 1 and 2 = T Ms GM s

M galaxy Ms

⎛ 6.3 × 104 AU ⎞ ⎜⎜ 3 × 104 LY × ⎟⎟ LY ⎠ =⎝ 2 6 250 × 10 y

(

3

)

= 1.08 × 10

11

or

M galaxy = 1.08 × 1011 M s *12 ••• Picture the Problem We can use Kepler’s 3rd law to find the size of the semi-major axis of the planet’s orbit and the conservation of momentum to find its mass. (a) Using Kepler’s 3rd law, relate the period of this planet T to the length r of its semi-major axis:

T2 =

4π 2 GM Iota Draconis 4π 2 Ms

= G

=

If we measure time in years, distances in AU, and masses in terms of the mass of the sun: Solve for r to obtain:

r3

M Iota Draconis Ms 4π 2 GM s

M Iota Draconis Ms

r3

r3

1 4π 2 = 1 and T 2 = r3 M MGs Iota Draconis Ms r=3

M Iota Draconis 2 T Ms

Substitute numerical values and evaluate r:

⎛ 1.05 M s ⎞ ⎟⎟(1.5 y )2 = 1.33 AU r = 3 ⎜⎜ ⎝ Ms ⎠

(b) Apply conservation of momentum to the planet (mass m and speed v) and the star (mass MIota Draconis and speed V) to obtain:

mv = M Iota DraconisV

Gravity 841 Solve for m to obtain:

Use its definition to find the speed of the orbiting planet:

m = M Iota Draconis

V v

∆d 2πr = T ∆t ⎛ 1.5 × 1011 m ⎞ ⎟⎟ ⎜ 2π ⎜1.33 AU × AU ⎠ ⎝ = 365.25 d 24 h 3600 s 1.50 y × × × y d h

v=

= 2.65 × 104 m/s Substitute numerical values and evaluate v:

⎛ 296 m/s ⎞ ⎟⎟ m = M Iota Draconis ⎜⎜ 4 ⎝ 2.65 × 10 m/s ⎠ = 0.0112 M Iota Draconis = 0.0112(1.05M sun )

(

= 0.0112(1.05) 1.99 × 1030 kg

)

= 2.34 × 10 28 kg Express m in terms of the mass MJ of Jupiter:

m 2.34 × 10 28 kg = = 12.3 M J 1.90 × 10 27 kg or

m = 12.3M J Remarks: A more sophisticated analysis, using the eccentricity of the orbit, leads to a lower bound of 8.7 Jovian masses. (Only a lower bound can be established, as the plane of the orbit is not known.) 13 ••• Picture the Problem We can apply Newton’s law of gravity to estimate the maximum angular velocity which the sun can have if it is to stay together and use the definition of angular momentum to find the orbital angular momenta of Jupiter and Saturn. In part (c) we can relate the final angular velocity of the sun to its initial angular velocity, its moment of inertia, and the orbital angular momenta of Jupiter and Saturn. (a) Gravity must supply the centripetal force which keeps an element of the sun’s mass m rotating around it. Letting the radius of the sun be R, apply Newton’s law of gravity to an element of mass m to obtain:

mω 2 R < or

ω 2R <

GMm R2

GM 2 R

where we’ve used the inequality because we’re estimating the maximum angular velocity which the sun can have if it is to stay together.

842 Chapter 11 Solve for ω:

ω<

GM R3

Substitute numerical values and evaluate ω:

ω<

(6.673 ×10

−11

)( m)

N ⋅ m 2 /kg 2 1.99 × 1030 kg

(6.96 ×10

8

Calculate the period of this motion from its angular velocity:

3

T=

2π

ω

=

)=

6.28 × 10 −4 rad/s

2π 6.28 × 10 −4 rad/s

= 1.00 × 10 4 s ×

1h = 2.78 h 3600 s

(b) Express the orbital angular momenta of Jupiter and Saturn:

LJ = mJ rJ vJ and LS = mS rSvS

Express the orbital speeds of Jupiter and Saturn in terms of their periods and distances from the sun:

vJ =

Substitute to obtain:

2π rJ 2π rS and vS = TJ TS

2π mJ rJ2 2π mS rS2 LJ = and LS = TJ TS

Substitute numerical values and evaluate LJ and LS:

(

)(

(

)(

2π (318M E )rJ2 2π (318) 5.98 × 1024 kg 778 × 109 m LJ = = 365.25 d 24 h 3600 s TJ 11.9 y × × × y d h

)

2

= 1.93 × 1043 kg ⋅ m 2 /s and

2π (95.1M E )rS2 2π (95.1) 5.98 × 10 24 kg 1430 × 109 m = LS = 365.25 d 24 h 3600 s TS 29.5 y × × × y d h

)

2

= 7.85 × 10 42 kg ⋅ m 2 /s Express the angular momentum of the sun as a fraction of the sum of the angular momenta of Jupiter and Saturn:

f = =

Lsun LJ + LS 1.91× 10 41 kg ⋅ m 2 /s (19.3 + 7.85)×1042 kg ⋅ m 2 /s

= 0.703 %

Gravity 843 (c) Relate the final angular momentum of the sun to its initial angular momentum and the angular momenta of Jupiter and Saturn: Solve for ωf to obtain:

Substitute for ωI and Isun:

Lf = Li + LJ + LS or

I sunωf = I sunωi + LJ + LS

ω f = ωi + ωf =

LJ + LS I sun

2π LJ + LS + 2 Tsun 0.059M sun Rsun

Substitute numerical values and evaluate ωf:

ωf =

(19.3 + 7.85)×1042 kg ⋅ m 2 /s 2π + 24 h 3600 s 0.059(1.99 × 1030 kg )(6.96 × 108 m ) 2 30 d × × d h

= 4.80 × 10 −4 rad/s Note that this result is about 76% of the maximum possible rotation allowed by gravity that we calculated in part (a).

Kepler’s Laws 14 • Picture the Problem We can use the relationship between the semi-major axis and the distances of closest approach and greatest separation, together with Kepler’s 3rd law, to find the greatest separation of Alex-Casey from the sun. Letting x represent the greatest distance from the sun, express the relationship between x, the distance of closest approach, and its semimajor axis R:

R=

x + 0.1AU 2

Solve for x to obtain:

x = 2 R − 0.1 AU

Apply Kepler’s 3rd law, with the period T measured in years and R in AU to obtain:

T 2 = R3

Solve for R:

R = 3 T2

Substitute numerical values and evaluate R:

R = 3 (127.4 y ) = 25.3 AU

Substitute in equation (1) and evaluate R:

x = 2(25.3 AU ) − 0.1 AU = 50.5 AU

(1)

2

844 Chapter 11 15 • Picture the Problem We can use Kepler’s 3rd law to relate the period of Uranus to its mean distance from the sun.

T 2 = Cr 3

Using Kepler’s 3rd law, relate the period of Uranus to its mean distance from the sun:

where C =

Solve for T:

4π 2 = 2.973 × 10 -19 s 2 /m 3 . GM s

T = Cr 3

Substitute numerical values and evaluate T:

T=

(2.973 ×10

−19

= 2.651× 109 s ×

s 2 /m3 ) (2.87 × 1012 m )

3

1h 1d 1y × × = 84.0 y 3600 s 24 h 365.25 d

16 • Picture the Problem We can use Kepler’s 3rd law to relate the period of Hektor to its mean distance from the sun.

T 2 = Cr 3

Using Kepler’s 3rd law, relate the period of Hektor to its mean distance from the sun:

where C =

Solve for T:

4π 2 = 2.973 × 10 -19 s 2 /m 3 . GM s

T = Cr 3

Substitute numerical values and evaluate T: 3

⎛ 1.50 × 1011 m ⎞ ⎟⎟ T = 2.973 × 10 s /m ⎜⎜ 5.16 AU × AU ⎝ ⎠ 1h 1d 1y = 3.713 × 108 s × × × = 11.8 y 3600 s 24 h 365.25 d

(

−19

2

3

)

17 •• Picture the Problem Kepler’s 3rd law relates the period of Icarus to the length of its semimajor axis. The aphelion distance ra is related to the perihelion distance rp and the semimajor axis by ra + rp = 2a.

Gravity 845 (a) Using Kepler’s 3rd law, relate the period of Icarus to the length of its semimajor axis: Solve for a:

Substitute numerical values and evaluate a:

T 2 = Ca 3

4π 2 = 2.973 × 10−19 s 2 /m3 . where C = GM s a=3

T2 C

⎛ 365.25 d 24 h 3600 s ⎞ ⎜⎜1.1y × ⎟ × × 3 y d h ⎟⎠ ⎝ a= 2.973 × 10 −19 s 2 /m 3

2

= 1.59 × 1011 m (b) Use the definition of the eccentricity of an ellipse to determine the perihelion distance of Icarus:

rp = a(1 − e )

(

)

= 1.59 × 1011 m (1 − 0.83) = 2.71× 1010 m

Express the relationship between rp and ra for an ellipse:

ra + rp = 2a

Solve for and evaluate ra:

ra = 2a − rp

(

)

= 2 1.59 × 1011 m − 2.71× 1010 m = 2.91× 1011 m 18 •• Picture the Problem The Hohmann transfer orbit is shown in the diagram. We can apply Kepler’s 3rd law to relate the time-in-orbit to the period of the spacecraft in its Hohmann Earth-to-Mars orbit. The period of this orbit is, in turn, a function of its semi-major axis which we can find from the average of the lengths of the semi-major axes of the Earth and Mars orbits.

846 Chapter 11 Using Kepler’s 3rd law, relate the period T of the spacecraft to the semi-major axis of its orbit:

T 2 = R3

Solve for T to obtain:

T = R3

Relate the transit time to the period of this orbit:

t trip = 12 T =

Express the semi-major axis of the Hohmann transfer orbit in terms of the mean sun-Mars and sun-Earth distances: Substitute numerical values and evaluate ttrip:

R=

1 2

R3

1.52 AU + 1.00 AU = 1.26 AU 2

t trip =

(1.26 AU)3

1 2

= 0.707 y ×

365.24 d = 258 d 1y

*19 •• Picture the Problem We can use a property of lines tangent to a circle and radii drawn to the point of contact to show that b = 90°. Once we’ve established that b is a right angle we can use the definition of the sine function to relate the distance from the sun to Venus to the distance from the sun to the earth. (a) The line from earth to Venus' orbit is tangent to the orbit of Venus at the point of maximum extension. Venus will appear closer to the sun in earth’s sky when it passes the line drawn from earth and tangent to its orbit. Hence:

b = 90°

(b) Using trigonometry, relate the distance from the sun to Venus dSV to the angle a:

sin a =

Solve for dSV:

dSV = dSE sin a

Substitute numerical values and evaluate dSV:

dSV = (1 AU )sin 47° = 0.731 AU

dSV dSE

Remarks: The correct distance from the sun to Venus is closer to 0.723 AU. 20 •• Picture the Problem Because the gravitational force the Earth exerts on the moon is along the line joining their centers, the net torque acting on the moon is zero and its angular momentum is conserved in its orbit about the Earth. Because energy is also conserved, we can combine these two expressions to solve for either vp or va initially and

Gravity 847 then substitute in the conservation of angular momentum equation to find the other. Letting m be the mass of the moon, apply conservation of angular momentum to the moon at apogee and perigee to obtain: Solve for va:

mv p rp = mva ra or

v p rp = v a ra va =

Apply conservation of energy to the moon-earth system to obtain:

rp ra

vp

(1)

1 2 GMm 1 2 GMm mv p − = mva − 2 rp 2 ra or

1 2 GM 1 2 GM vp − = va − 2 rp 2 ra Substitute for va to obtain:

2

1 2 GM 1 ⎛ rp ⎞ GM = ⎜⎜ v p ⎟⎟ − vp − 2 2 ⎝ ra ⎠ rp ra 2

1⎛r ⎞ GM = ⎜⎜ p ⎟⎟ v 2p − 2 ⎝ ra ⎠ ra Solve for vp to obtain:

2GM rp

vp =

⎛ ⎞ 1 ⎜ ⎟ ⎜1+ r r ⎟ p a ⎠ ⎝

Substitute numerical values and evaluate vp:

vp =

⎛ ⎞ ⎜ ⎟ 2 6.673 × 10 N ⋅ m /kg 5.98 × 10 kg ⎜ 1 ⎟ = 1.09 km/s 8 8 ⎜ 3.576 × 10 m ⎟ 3.576 × 10 m ⎜1+ ⎟ 8 ⎝ 4.064 × 10 m ⎠

(

−11

2

2

Substitute numerical values in equation (1) and evaluate va:

)(

)

24

va =

3.576 × 108 m (1.09 km/s) 4.064 × 108 m

= 0.959 km/s

Newton’s Law of Gravity *21 •• Picture the Problem We can use Kepler’s 3rd law to find the mass of Jupiter in part (a). In part (b) we can express the centripetal accelerations of Europa and Callisto and compare their ratio to the square of the ratio of their distances from the center of Jupiter

848 Chapter 11 to show that the given data is consistent with an inverse square law for gravity. (a) Assuming a circular orbit, apply Kepler’s 3rd law to the motion of Europa to obtain: Solve for the mass of Jupiter:

Substitute numerical values and evaluate MJ:

TE2 =

4π 2 3 RE GM J

MJ =

4π 2 3 RE GTE2

4π 2 MJ = 6.673 × 10 −11 N ⋅ m 2 /kg 2

(

×

(6.71× 10 m ) 8

)

3

24 h 3600 s ⎞ ⎛ × ⎜ 3.55 d × ⎟ d h ⎠ ⎝

2

= 1.90 × 10 27 kg , a result in excellent agreement with the accepted value of 1.902 × 10 27 kg. (b) Express the centripetal acceleration of both of the moons to obtain:

2

⎛ 2πR ⎞ ⎜ ⎟ 2 v 4π 2 R T ⎠ ⎝ = = R R T2 where R and T are the radii and periods of their motion.

Using this result, express the centripetal accelerations of Europa and Callisto:

aE =

4π 2 RE 4π 2 RC a and = C TE2 TC2

Substitute numerical values and evaluate aE:

aE =

4π 2 6.71× 108 m [(3.55 d )(24 h/d )(3600 s/h )]2

(

)

= 0.282 m/s 2 Substitute numerical values and evaluate aC:

aC =

(

= 0.0356 m/s 2 Evaluate the ratio of these accelerations:

)

4π 2 18.8 × 108 m [(16.7 d )(24 h/d )(3600 s/h )]2

aE 0.282 m/s 2 = = 7.91 aC 0.0356 m/s 2

Gravity 849 Evaluate the square of the ratio of the distance of Callisto divided by the distance of Europa to obtain:

2

2

⎛ RC ⎞ ⎛ 18.8 × 108 m ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ = 7.85 8 ⎝ RE ⎠ ⎝ 6.71× 10 m ⎠

The close agreement (within 1%) of our last two calculations strongly supports the conclusion that the gravitational force varies inversely with the square of the distance. *22 • Determine the Concept The weight of anything, including astronauts, is the reading of a scale from which the object is suspended or on which it rests. If the scale reads zero, then we say the object is ″weightless.″ The pull of the earth’s gravity, on the other hand, depends on the local value of the acceleration of gravity and we can use Newton’s law of gravity to find this acceleration at the elevation of the shuttle. (a) Apply Newton’s law of gravitation to an astronaut of mass m in a shuttle at a distance h above the surface of the earth: Solve for gshuttle:

mg shuttle =

gshuttle =

GmM E (h + RE )2

GM E (h + RE )2

Substitute numerical values and evaluate gshuttle:

g shuttle =

(6.673 × 10

−11

)(

)

N ⋅ m 2 /kg 2 5.98 × 10 24 kg = 8.71 m/s 2 2 (400 km + 6370 km )

Because they are in " free fall" everything on the shuttle is falling toward (b) the center of the earth with exactly the same acceleration, so the astronauts will seem to be " weightless." 23 • Picture the Problem We can use Kepler’s 3rd law to relate the periods of the moons of Saturn to their mean distances from its center. (a) Using Kepler’s 3rd law, relate the period of Mimas to its mean distance from the center of Saturn: Solve for TM:

TM2 =

TM =

4π 2 3 rM GM S 4π 2 3 rM GM S

850 Chapter 11 4π 2 3 T = rT GM S

(b) Using Kepler’s 3rd law, relate the period of Titan to its mean distance from the center of Saturn:

2 T

Substitute numerical values and evaluate TM:

TM =

( )(

)

3

4π 2 1.86 × 108 m = 8.18 × 104 s 5.69 × 1026 kg 6.6726 × 10 −11 N ⋅ m 2 /kg 2

(

Solve for rT:

)

rT = 3

TT2GM S 4π 2

Substitute numerical values and evaluate rT:

rT =

(1.38 × 10 s ) (6.6726 × 10 6

3

2

)(

−11

N ⋅ m 2 /kg 2 5.69 × 10 26 kg

4π

2

)=

1.22 × 109 m

24 • Picture the Problem We can use Kepler’s 3rd law to relate the period of the moon to the mass of the earth and the mean earth-moon distance. (a) Using Kepler’s 3rd law, relate the period of the moon to its mean orbital radius:

Tm2 =

Solve for ME:

4π 2 3 rm GM E

ME =

4π 2 3 rm GTm2

Substitute numerical values and evaluate ME:

ME =

(

4π 2 3.84 × 108 m

(6.6726 ×10

−11

)

)

3

24 h 3600 s ⎞ ⎛ N ⋅ m 2 /kg 2 ⎜ 27.3 d × × ⎟ d h ⎠ ⎝

2

= 6.02 × 1024 kg

Remarks: This analysis neglects the mass of the moon; consequently the mass calculated here is slightly too great. 25 • Picture the Problem We can use Kepler’s 3rd law to relate the period of the earth to the mass of the sun and the mean earth-sun distance.

Gravity 851 4π 2 3 T = rE GM S

(a) Using Kepler’s 3rd law, relate the period of the earth to its mean orbital radius:

2 E

Solve for MS:

MS =

4π 2 3 rE GTE2

Substitute numerical values and evaluate MS:

MS =

(

4π 2 1.496 × 1011 m

(6.6726 ×10

−11

)

3

⎛ 365.25 d 24 h 3600 s ⎞ ⎟ N ⋅ m /kg ⎜⎜1 y × × × y d h ⎟⎠ ⎝ 2

2

)

2

= 1.99 × 1030 kg *26 • Picture the Problem We can relate the acceleration of an object at any elevation to its acceleration at the surface of the earth through the law of gravity and Newton’s 2nd law of motion. Letting a represent the acceleration due to gravity at this altitude (RE) and m the mass of the object, apply Newton’s 2nd law and the law of gravity to obtain: Apply Newton’s 2nd law to the same object when it is at the surface of the earth:

∑F

radial

GmM E

(2 RE )2

= ma

and

a=

GM E

(1)

(2RE )2

∑F

radial

=

GmM E = mg RE2

and

g=

Divide equation (1) by equation (2) and solve for a:

=

GM E RE2

(2)

R2 a = E2 g 4RE and

a = 14 g =

1 4

(9.81 m/s ) = 2

2.45 m/s 2

27 • Picture the Problem Your weight is the local gravitational force exerted on you. We can use the definition of density to relate the mass of the planet to the mass of earth and the

852 Chapter 11 law of gravity to relate your weight on the planet to your weight on earth. Using the definition of density, relate the mass of the earth to its radius:

M E = ρ VE = 43 ρ πRE3

Relate the mass of the planet to its radius:

M P = ρ VP = 43 ρ πRP3

Divide the second of these equations by the first to express MP in terms of ME:

= 43 ρ π (10RE )

3

4 ρ π (10 RE ) MP =ρ3 4 3 ME 3 ρ πR E

3

and

M P = 103 M E Letting w′ represent your weight on the planet, use the law of gravity to relate w′ to your weight on earth:

(

GmM P Gm 103 M E w' = = RP2 (10 RE )2 = 10

)

GmM E = 10w RE2

where w is your weight on earth. 28 • Picture the Problem We can relate the acceleration due to gravity of a test object at the surface of the new planet to the acceleration due to gravity at the surface of the earth through use of the law of gravity and Newton’s 2nd law of motion. Letting a represent the acceleration due to gravity at the surface of this new planet and m the mass of a test object, apply Newton’s 2nd law and the law of gravity to obtain:

Simplify this expression to obtain:

∑F

radial

=

GmM E

( 12 RE )2

= ma

and

a=

GM E

( 12 RE )2

a=4

GM E = 4 g = 39.2 m/s 2 2 RE

29 • Picture the Problem We can use conservation of angular momentum to relate the planet’s speeds at aphelion and perihelion. Using conservation of angular

La = Lp

Gravity 853 momentum, relate the angular momenta of the planet at aphelion and perihelion:

or

mv p rp = mva ra

Solve for the planet’s speed at aphelion:

va =

Substitute numerical values and evaluate va:

va =

vp rp ra

(5 ×10

)(

m/s 1.0 × 1015 m 2.2 × 1015 m

4

)

= 2.27 × 10 4 m/s 30 • Picture the Problem We can use Newton’s law of gravity to express the gravitational force acting on an object at the surface of the neutron star in terms of the weight of the object. We can then simplify this expression be dividing out the mass of the object … leaving an expression for the acceleration due to gravity at the surface of the neutron star. Apply Newton’s law of gravity to an object of mass m at the surface of the neutron star to obtain:

GM Neutron Star m = mg 2 RNeutron Star where g represents the acceleration due to gravity at the surface of the neutron star.

Solve for g and substitute for the mass of the neutron star:

g=

GM Neutron Star G (1.60M sun ) = 2 2 RNeutron RNeutron Star Star

Substitute numerical values and evaluate g:

(

)(

)

1.60 6.673 × 10−11 N ⋅ m 2 /kg 2 1.99 × 1030 kg g= = 1.93 × 1012 m/s 2 2 (10.5 km ) *31 •• Picture the Problem We can use conservation of angular momentum to relate the asteroid’s aphelion and perihelion distances. Using conservation of angular momentum, relate the angular momenta of the asteroid at aphelion and perihelion: Solve for and evaluate the ratio of the asteroid’s aphelion and perihelion distances:

La = Lp or

mv p rp = mva ra ra vp 20 km/s = = = 1.43 rp va 14 km/s

854 Chapter 11 32 •• Picture the Problem We’ll use the law of gravity to find the gravitational force acting on the satellite. The application of Newton’s 2nd law will lead us to the speed of the satellite and its period can be found from its definition. (a) Letting m represent the mass of the satellite and h its elevation, use the law of gravity to express the gravitational force acting on it:

Fg =

Substitute numerical values and evaluate Fg:

Fg =

=

GmM E mRE2 g = (RE + h )2 (RE + h )2 mg ⎛ h ⎞ ⎜⎜1 + ⎟⎟ ⎝ RE ⎠

2

mg ⎛ h ⎞ ⎜⎜1 + ⎟⎟ ⎝ RE ⎠

2

=

(300 kg )(9.81 N/kg ) ⎛ 5 × 107 m ⎞ ⎜⎜1 + ⎟⎟ 6 ⎝ 6.37 × 10 m ⎠

2

= 37.6 N (b) Using Newton’s 2nd law, relate the gravitational force acting on the satellite to its centripetal acceleration: Solve for v:

Substitute numerical values and evaluate v:

Fg = m

v=

v=

v2 r

Fg r m

(37.6 N )(6.37 ×106 m + 5 ×107 m ) 300 kg

= 2.66 km/s

2π r v

(c) Express the period of the satellite:

T=

Substitute numerical values and evaluate T:

2π 6.37 × 10 6 m + 5 × 10 7 m T= 2.66 × 103 m/s 1h = 1.33 × 105 s × = 36.9 h 3600 s

(

)

Gravity 855 *33 •• Picture the Problem We can determine the maximum range at which an object with a given mass can be detected by substituting the equation for the gravitational field in the expression for the resolution of the meter and solving for the distance. Differentiating g(r) with respect to r, separating variables to obtain dg/g, and approximating ∆r with dr will allow us to determine the vertical change in the position of the gravity meter in the earth’s gravitational field is detectable.

GM E RE2

(a) Express the gravitational field of the earth:

gE =

Express the gravitational field due to the mass m (assumed to be a point mass) of your friend and relate it to the resolution of the meter:

g (r ) =

Solve for r:

Substitute numerical values and evaluate r:

r = RE

Gm GM E = 10 −11 g E = 10 −11 2 r RE2

1011 m ME

(

r = 6.37 × 10 6 m

)

1011 (80 kg ) 5.98 × 10 24 kg

= 7.37 m (b) Differentiate g(r) and simplify to obtain:

dg − 2Gm 2 ⎛ Gm ⎞ 2 = =− ⎜ 2 ⎟=− g 3 dr r r⎝ r ⎠ r

Separate variables to obtain:

dg dr = −2 = 10 −11 g r

Approximating dr with ∆r, evaluate ∆r with r = RE:

∆r = − 12 10−11 6.37 × 106 m

(

)(

)

= 3.19 × 10−5 m = 0.0319 mm

34 •• Picture the Problem We can use the law of gravity and Newton’s 2nd law to relate the force exerted on the planet by the star to its orbital speed and the definition of the period to relate it to the radius of the orbit.

856 Chapter 11 Using the law of gravity and Newton’s 2nd law, relate the force exerted on the planet by the star to its centripetal acceleration:

Fnet

KMm v2 = =m r r

Solve for v2 to obtain:

v 2 = KM

Express the period of the planet:

T=

2π r 2π r 2π r = = v KM KM

or

T ∝r *35 •• Picture the Problem We can use the definitions of the gravitational fields at the surfaces of the earth and the moon to express the accelerations due to gravity at these locations in terms of the average densities of the earth and the moon. Expressing the ratio of these accelerations will lead us to the ratio of the densities.

GM E Gρ EVE Gρ E 43 π RE3 = = RE2 RE2 RE2

Express the acceleration due to gravity at the surface of the earth in terms of the earth’s average density:

gE =

Express the acceleration due to gravity at the surface of the moon in terms of the moon’s average density:

g M = 43 Gρ Mπ RM

Divide the second of these equations by the first to obtain:

g M ρ M RM = gE ρ E RE

Solve for

ρM : ρE

Substitute numerical values and evaluate

ρM : ρE

= 43 Gρ Eπ RE

ρ M g M RE = ρ E g E RM ρ M (1.62 m/s 2 )(6.37 × 106 m ) = ρ E (9.81 m/s 2 )(1.738 × 106 m ) = 0.605

Gravity 857

Measurement of G 36 • Picture the Problem We can use the law of gravity to find the forces of attraction between the two masses and the definition of torque to determine the balancing torque required. (a) Use the law of gravity to express the force of attraction between the two masses:

F=

Gm1m2 r2

Substitute numerical values and evaluate F:

F=

(6.6726 ×10

−11

)

N ⋅ m 2 /kg 2 (10 kg )(0.01 kg ) = 1.85 × 10 −9 N 2 (0.06 m )

(b) Use its definition to find the torque exerted by the suspension to balance these forces:

τ = 2 Fr = 2(1.85 ×10 −9 N )(0.1 m ) = 3.70 × 10−10 N ⋅ m

Gravitational and Inertial Mass 37 • Picture the Problem Newton’s 2nd law of motion relates the masses and accelerations of these objects to their common accelerating force. (a) Apply Newton’s 2nd law to the standard object:

F = m1 a1

Apply Newton’s 2nd law to the object of unknown mass:

F = m2 a 2

Eliminate F between these two equations and solve for m2:

m2 =

a1 m1 a2

Substitute numerical values and evaluate m2:

m2 =

2.6587 m/s 2 (1 kg ) = 2.27 kg 1.1705 m/s 2

(b)

It is the inertial mass of m2 .

858 Chapter 11 38 • Picture the Problem Newton’s 2nd law of motion relates the weights of these two objects to their masses and the acceleration due to gravity. (a) Apply Newton’s 2nd law to the standard object:

Fnet = w1 = m1 g

Apply Newton’s 2nd law to the object of unknown mass:

Fnet = w2 = m2 g

Eliminate g between these two equations and solve for m2:

m2 =

w2 m1 w1

Substitute numerical values and evaluate m2:

m2 =

56.6 N (1 kg ) = 5.77 kg 9.81 N

(b)

Since this result is determined by the effect on m2 of the earth' s gravitational field, it is the gravitational mass of m2 .

*39 • Picture the Problem Noting that g1 ~ g2 ~ g, let the acceleration of gravity on the first object be g1, and on the second be g2. We can use a constant-acceleration equation to express the difference in the distances fallen by each object and then relate the average distance fallen by the two objects to obtain an expression from which we can approximate the distance they would have to fall before we might measure a difference in their fall distances greater than 1 mm. Express the difference ∆d in the distances fallen by the two objects in time t:

∆d = d1 − d 2

Express the distances fallen by each of the objects in time t:

d1 = 12 g1t 2 and

d 2 = 12 g 2t 2 Substitute to obtain:

∆d = 12 g1t 2 − 12 g 2t 2 =

Relate the average distance d fallen by the two objects to their time of fall:

d = 12 gt 2 or

t2 =

2d g

1 2

(g1 − g 2 ) t 2

Gravity 859 Substitute to obtain:

Solve for d to obtain:

Substitute numerical values and evaluate d:

∆d ≈ 12 ∆g

d = ∆d

2d ∆g =d g g

g ∆g

(

)( )

d = 10 −3 m 1012 = 109 m

Gravitational Potential Energy 40 • Picture the Problem Choosing the zero of gravitational potential energy to be at infinite separation yields, as the potential energy of a two-body system in which the objects are separated by a distance r, U (r ) = − GMm r , where M and m are the masses of the two bodies. In order for an object to just escape a gravitational field from a particular location, it must have enough kinetic energy so that its total energy is zero.

GM E m r

(a) Letting U(∞) = 0, express the gravitational potential energy of the earth-object system:

U (r ) = −

Substitute for GME and simplify to obtain:

U (RE ) = −

(1)

GM E m gR 2 m = − E = −mgRE RE RE

Substitute numerical values and evaluate U(RE):

U (RE ) = −(100 kg )(9.81 N / kg ) (6.37 × 106 m ) = − 6.25 × 109 J (b) Evaluate equation (1) with r = 2RE:

GM E m gRE2 m U (2 RE ) = − =− 2 RE 2 RE = − 12 mgRE

Substitute numerical values and evaluate U(2RE):

U (2 RE ) = − 12 (100 kg )(9.81 N / kg ) (6.37 × 106 m ) = − 3.12 × 109 J (c) Express the condition that an object must satisfy in order to escape from the earth’s gravitational

K esc (2 RE ) + U (2 RE ) = 0 or 1 2

2 mvesc + U (2 RE ) = 0

860 Chapter 11 field from a height RE above its surface: Solve for vesc:

Substitute numerical values and evaluate vesc:

vesc =

− 2U (2 RE ) m

vesc =

− 2 − 3.12 × 109 J = 7.90 km/s 100 kg

(

)

41 • Picture the Problem In order for an object to just escape a gravitational field from a particular location, an amount of work must be done on it that is equal to its potential energy in its initial position. Express the work needed to remove the point mass from the surface of the sphere to a point a very large distance away: Express the initial potential energy of the system: Substitute in equation (1) to obtain:

W = ∆U = U f − U i or, because Uf = 0,

W = ∆U = −U i

Ui = −

W=

(1)

GMm0 R

GMm0 R

42 • Picture the Problem Let the zero of gravitational potential energy be at infinity and let m represent the mass of the spacecraft. We’ll use conservation of energy to relate the initial kinetic and potential energies to the final potential energy of the earth-spacecraft system. Use conservation of energy to relate the initial kinetic and potential energies of the system to its final energy when the spacecraft is one earth radius above the surface of the planet:

Kf − Ki + U f − U i = 0 or, because Kf = 0,

− K (RE ) + U (2 RE ) − U (RE ) = 0

(1)

Gravity 861 GM E m r

Express the potential energy of the spacecraft-and-earth system when the spacecraft is at a distance r from the surface of the earth:

U (r ) = −

Substitute in equation (1) to obtain:

− 12 mv 2 −

Solve for v:

Substitute numerical values and evaluate v:

GM E m GM E m + =0 RE 2 RE gRE2 = gRE RE

v=

GM E = RE

v=

(9.81m/s )(6.37 ×10 m) 2

6

= 7.91 km/s *43 •• Picture the Problem Let the zero of gravitational potential energy be at infinity and let m represent the mass of the object. We’ll use conservation of energy to relate the initial potential energy of the object-earth system to the final potential and kinetic energies. Use conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth:

Express the potential energy of the object-earth system when the object is at a distance r from the surface of the earth: Substitute in equation (1) to obtain:

Solve for v:

Kf − Ki + U f − U i = 0 or, because Ki = 0,

K (RE ) + U (RE ) − U (RE + h ) = 0

where h is the initial height above the earth’s surface.

U (r ) = −

1 2

mv 2 −

GM E m r

GM E m GM E m + =0 RE RE + h

⎛ GM E GM E ⎞ ⎟ v = 2⎜⎜ − RE + h ⎟⎠ ⎝ RE ⎛ h ⎞ ⎟⎟ = 2 gRE ⎜⎜ ⎝ RE + h ⎠

(1)

862 Chapter 11 Substitute numerical values and evaluate v:

v=

(

)(

)(

)

2 9.81 m/s 2 6.37 × 106 m 4 × 106 m = 6.94 km/s 6.37 × 106 m + 4 × 106 m

44 •• Picture the Problem Let the zero of gravitational potential energy be at infinity, m represent the mass of the object, and h the maximum height reached by the object. We’ll use conservation of energy to relate the initial potential and kinetic energies of the objectearth system to the final potential energy. Use conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth:

Kf − Ki + U f − U i = 0 or, because Kf = 0,

K (RE ) + U (RE ) − U (RE + h ) = 0

(1)

where h is the initial height above the earth’s surface.

Express the potential energy of the object-earth system when the object is at a distance r from the surface of the earth: Substitute in equation (1) to obtain:

U (r ) = −

Solve for h:

h=

Substitute numerical values and evaluate h:

h=

1 2

mv 2 −

GM E m r

GM E m GM E m + =0 RE RE + h

RE 2 gRE −1 v2 6.37 × 106 m 2 9.81 m/s 2 6.37 × 106 m

(

)(

(4 ×10 m) 3

2

) −1

= 935 km 45 •• Picture the Problem When the point mass is inside the spherical shell, there is no mass between it and the center of the shell. On the other hand, when the point mass is outside the spherical shell we can use the law of gravity to express the force acting on it. In (b) we can derive U(r) from F(r).

Gravity 863 (a) The force exerted by the shell on a point mass m0 when m0 is inside the shell is: The force exerted by the shell on a point mass m0 when m0 is outside the shell is:

r Finside = 0

r r GMm0 Foutside = m0 g = − rˆ r2 where rˆ is radially outward from the center of the spherical shell.

(b) Use its definition to express U(r) for r > R:

r

r

∞

∞

U (r ) = − ∫ Fr dr = GMm0 ∫ r −2 dr = −

GMm0 r

When r = R:

U (R ) = −

GMm0 R

(c) For r < R, F = 0 and:

dU = 0 ⇒ U = constant dr

(d) Because U is continuous, then for r < R:

U (r ) = U (R ) = −

GMm0 R

(e) A sketch of U as a function of r/R (with GMm0 = 1) is shown below:

0.0

-0.2

-0.4

GMm 0 = 1

U -0.6 -0.8

-1.0

-1.2 0

1

2

r /R

3

4

864 Chapter 11 46 • Picture the Problem The escape speed from a planet is related to its mass according to ve = 2GM R , where M and R represent the mass and radius of the planet, respectively. Express the escape speed from Saturn:

Express the escape speed from Earth:

Divide equation (1) by equation (2) to obtain:

Substitute numerical values and

v evaluate e.S : ve.E Solve for and evaluate ve,S:

ve.S =

2GM S RS

(1)

ve.E =

2GM E RE

(2)

ve.S = ve.E

2GM S RS 2GM E RE

=

RE M S ⋅ RS M E

ve.S 1 95.2 = × = 3.17 ve.E 9.47 1 ve.S = 3.17ve.E = 3.17(11.2 km/s ) = 35.5 km/s

47 • Picture the Problem The escape speed from the moon or the earth is given by ve = 2GM R , where M and R represent the masses and radii of the moon or the earth. Express the escape speed from the moon:

Express the escape speed from earth:

Divide equation (1) by equation (2) to obtain:

ve.S =

2GM m = 2 g m Rm Rm

(1)

ve.E =

2GM E = 2 g E RE RE

(2)

ve.m = ve.E

g m Rm g E RE

=

g m Rm g E RE

Gravity 865 Solve for ve,m:

Substitute numerical values and evaluate ve,m:

ve.m =

g m Rm ve.E g E RE

ve.m =

(0.166)(0.273)(11.2 km/s )

= 2.38 km/s

*48 • Picture the Problem We’ll consider a rocket of mass m which is initially on the surface of the earth (mass M and radius R) and compare the kinetic energy needed to get the rocket to its escape velocity with its kinetic energy in a low circular orbit around the earth. We can use conservation of energy to find the escape kinetic energy and Newton’s law of gravity to derive an expression for the low earth-orbit kinetic energy. Apply conservation of energy to relate the initial energy of the rocket to its escape kinetic energy:

Kf − Ki + U f − U i = 0

Letting the zero of gravitational potential energy be at infinity we have Uf = Kf = 0 and:

− Ki − U i = 0 or

Apply Newton’s law of gravity to the rocket in orbit at the surface of the earth to obtain:

GMm v2 m = R2 R

Rewrite this equation to express the low-orbit kinetic energy Eo of the rocket: Express the ratio of Ko to Ke:

K e = −U i =

GMm R

K o = 12 mv 2 =

GMm 2R

GMm Ko 1 = 2 R = ⇒ K e = 2 K o , as K e GMm 2 R asserted by Heinlein.

49 •• Picture the Problem Let the zero of gravitational potential energy be at infinity, m represent the mass of the particle, and the subscript E refer to the earth. When the particle is very far from the earth, the gravitational potential energy of the earth-particle system will be zero. We’ll use conservation of energy to relate the initial potential and kinetic energies of the particle-earth system to the final kinetic energy of the particle. Use conservation of energy to relate the initial energy of the system to its energy when the particle is very

Kf − Ki + U f − U i = 0 or, because Uf = 0,

K (∞ ) − K (RE ) − U (RE ) = 0

(1)

866 Chapter 11 from the earth: Substitute in equation (1) to obtain:

1 2

mv∞2 − 12 m(2ve ) + 2

GM E m =0 RE

or, because GM E = gRE2 , 1 2

Solve for v∞:

mv∞2 − 12 mv 2 + mgRE = 0

(

v∞ = 2 2ve2 − gRE

)

Substitute numerical values and evaluate v∞:

[(

) (

)]

)(

2

v∞ = 2 2 11.2 × 103 m/s − 9.81 m/s 2 6.37 × 106 m = 19.4 km/s 50 •• Picture the Problem Let the zero of gravitational potential energy be at infinity, m represent the mass of the particle, and the subscript E refer to the earth. When the particle is very far from the earth, the gravitational potential energy of the earth-particle system will be zero. We’ll use conservation of energy to relate the initial potential and kinetic energies of the particle-earth system to the final kinetic energy of the particle.

Kf − Ki + U f − U i = 0

Use conservation of energy to relate the initial energy of the system to its energy when the particle is very far away:

or, because Uf = 0,

K (∞ ) − K (RE ) − U (RE ) = 0

Substitute in equation (1) to obtain:

1 2

mv∞2 − 12 mvi2 +

(1)

GM E m =0 RE

or, because GM E = gRE2 , 1 2

Solve for vi:

mv∞2 − 12 mvi2 + mgRE = 0

vi = v∞2 + 2gRE

Substitute numerical values and evaluate vi:

vi =

(11.2 ×10

3

)

2

(

)(

)

m/s + 2 9.81 m/s 2 6.37 × 106 m = 15.8 km/s

51 •• Picture the Problem We can use the definition of kinetic energy to find the energy necessary to launch a 1-kg object from the earth at escape speed.

Gravity 867 (a) Using the definition of kinetic energy, find the energy required to launch a 1-kg object from the surface of the earth at escape speed:

K = 12 mve2

(b) Using the conversion factor 1 kW⋅h = 3.6 MJ, convert 62.7 MJ to kW⋅h:

K = 62.7 MJ ×

(c) Express the cost of this project in terms of the mass of the astronaut:

Cost = rate ×

Substitute numerical values and find the cost:

Cost =

=

1 2

(1kg )(11.2 ×103 m/s)2

= 62.7 MJ

1 kW ⋅ h 3.6 MJ

= 17.4 kW ⋅ h required energy × mass kg

$0.10 17.4 kW ⋅ h (80 kg ) × kW ⋅ h kg

= $139 52 •• Picture the Problem Let m represent the mass of the body that is projected vertically from the surface of the earth. We’ll begin by using conservation of energy under the assumption that the gravitational field is constant to determine H ′. We’ll apply conservation of energy a second time, with the zero of gravitational potential energy at infinity, to express H. Finally, we’ll solve these two equations simultaneously to express H in terms of H ′. Kf − Ki + U f − U i = 0 Assuming the gravitational field to be constant and letting the zero of or, because Kf = Ui = 0, potential energy be at the surface of − Ki + U f = 0 the earth, apply conservation of energy to relate the initial kinetic energy and the final potential energy of the object-earth system: Substitute for Ki and Uf and solve for H ′:

− 12 mv 2 + mgH' = 0 and

H' =

Letting the zero of gravitational potential energy be at infinity, use conservation of energy to relate the initial kinetic energy and the final

v2 2g

Kf − Ki + U f − U i = 0 or, because Kf = 0,

− Ki + U f − U i = 0

(1)

868 Chapter 11 potential energy of the object-earth system: Substitute to obtain:

− 12 mv 2 −

GMm GMm + =0 RE + H RE

or

gRE2 gRE2 − v − + =0 RE + H RE 2

1 2

Solve for v2:

⎛ 1 1 ⎞ ⎟⎟ v 2 = 2 gRE2 ⎜⎜ − R R H + E ⎝ E ⎠ ⎛ H ⎞ ⎟⎟ = 2 gRE ⎜⎜ ⎝ RE + H ⎠

Substitute in equation (1) to obtain:

Solve for H:

⎛ H ⎞ ⎟⎟ H' = RE ⎜⎜ R + H ⎝ E ⎠ H=

H'RE RE − H'

Orbits 53 •• Picture the Problem We can use its definition to express the period of the spacecraft’s motion and apply Newton’s 2nd law to the spacecraft to determine its orbital velocity. We can then use this orbital velocity to calculate the kinetic energy of the spacecraft. We can relate the spacecraft’s angular momentum to its kinetic energy and moment of inertia. (a) Express the period of the spacecraft’s orbit about the earth:

T=

2π R 2π (3RE ) 6π RE = = v v v

where v is the orbital speed of the spacecraft. Use Newton’s 2nd law to relate the gravitational force acting on the spacecraft to its orbital speed: Solve for v to obtain:

Fradial =

v=

GM E m

(3RE )2

gRE 3

=m

v2 3RE

Gravity 869 Substitute for v in our expression for T to obtain:

T = 6 3π

RE g

Substitute numerical values and evaluate T:

T = 6 3π

6.37 × 10 6 m 9.81 m/s 2

= 2.631 × 10 4 s ×

1h = 7.31 h 3600 s

K = 12 mv 2 = 12 m( 13 gRE )

(b) Using its definition, express the spacecraft’s kinetic energy:

K=

Substitute numerical values and evaluate K:

1 6

(100 kg )(9.81m/s2 )(6.37 ×106 m )

= 1.04 GJ

(c) Express the kinetic energy of the spacecraft in terms of its angular momentum:

K=

L2 2I

Solve for L:

L = 2 IK

Express the moment of inertia of the spacecraft with respect to an axis through the center of the earth:

I = m(3RE )

Substitute and solve for L:

L = 18mRE2 K = 3RE 2mK

2

= 9mRE2

Substitute numerical values and evaluate L:

(

)

(

)

L = 3 6.37 × 106 m 2(100 kg ) 1.04 × 109 J = 8.72 × 1012 J ⋅ s *54 • Picture the Problem Let the origin of our coordinate system be at the center of the earth and let the positive x direction be toward the moon. We can apply the definition of center of mass to find the center of mass of the earth-moon system and find the ″orbital″ speed of the earth using xcm as the radius of its motion and the period of the moon as the period of this motion of the earth. (a) Using its definition, express the x coordinate of the center of mass of the earth-moon system:

xcm =

M E xE + mmoon xmoon M E + mmoon

870 Chapter 11 Substitute numerical values and evaluate xcm:

xcm =

(

)(

)

M E (0 ) + 7.36 × 10 22 kg 3.82 × 108 m = 4.64 × 106 m 5.98 × 10 24 kg + 7.36 × 10 22 kg

Note that, because the radius of the earth is 6.37×106 m, the center of mass is actually located about 1700 km below the surface of the earth. (b) Express the ″orbital″ speed of the earth in terms of the radius of its circular orbit and its period of rotation: Substitute numerical values and evaluate v:

v=

2πxcm T

v=

2π (4.64 × 10 6 m ) = 12.4 m/s 24 h 3600 s 27.3 d × × h d

55 •• Picture the Problem We can express the energy difference between these two orbits in terms of the total energy of a satellite at each elevation. The application of Newton’s 2nd law to the force acting on a satellite will allow us to express the total energy of each satellite as function of its mass, the radius of the earth, and its orbital radius. Express the energy difference:

∆E = E geo − E1000

Express the total energy of an orbiting satellite:

E tot = K + U = 12 mv 2 −

GM E m R

(1)

(2)

where R is the orbital radius. Apply Newton’s 2nd law to a satellite to relate the gravitational force to the orbital speed:

Fradial =

GM E m v2 m = R R2

or

gRE2 v 2 = R R2 Simplify and solve for v2:

Substitute in equation (2) to obtain:

v2 =

gRE2 R

E tot = 12 m

gRE2 gRE2 m mgRE2 − =− R R 2R

Gravity 871 Substitute in equation (1) and simplify to obtain:

∆E = −

mgRE2 mgRE2 + 2 Rgeo 2 R1000

mgRE2 = 2

⎛ 1 1 ⎞⎟ ⎜ − ⎜R ⎟ ⎝ 1000 Rgeo ⎠

Substitute numerical values and evaluate ∆E:

∆E =

1 2

⎛

⎞ 1 1 ⎟⎟ = 11.1GJ − 6 7 ⎝ 7.37 × 10 m 4.22 × 10 m ⎠

(500 kg )(9.81 N / kg ) (6.37 × 106 m )2 ⎜⎜

56 •• Picture the Problem We can use Kepler’s 3rd law to relate the periods of the moon and Earth, in their orbits about the earth and the sun, to their mean distances from the objects about which they are in orbit. We can solve these equations for the masses of the sun and the earth and then divide one by the other to establish a value for the ratio of the mass of the sun to the mass of the earth. Using Kepler’s 3rd law, relate the period of the moon to its mean distance from the earth:

Using Kepler’s 3rd law, relate the period of the earth to its mean distance from the sun:

Solve equation (1) for ME:

Solve equation (2) for Ms:

4π 2 3 T = rm GM E 2 m

(1)

where rm is the distance between the centers of the earth and the moon.

TE2 =

4π 2 3 rE GM s

(2)

where rE is the distance between the centers of the earth and the sun.

4π 2 3 ME = rm GTm2

(3)

4π 2 3 rE GTE2

(4)

Ms =

Divide equation (4) by equation (3) and simplify to obtain:

M s ⎛ rE ⎞ =⎜ ⎟ M E ⎜⎝ rm ⎟⎠

Substitute numerical values and evaluate Ms/ME:

M s ⎛ 1.5 × 1011 m ⎞ ⎟ =⎜ M E ⎜⎝ 3.82 × 108 m ⎟⎠

3

⎛ Tm ⎞ ⎜⎜ ⎟⎟ ⎝ TE ⎠

= 3.38 × 105

2

3

⎛ 27.3 d ⎞ ⎜⎜ ⎟⎟ ⎝ 365.24 d ⎠

2

872 Chapter 11 Express the difference between this value and the measured value of 3.33×105:

% diff =

3.38 × 105 − 3.33 × 105 3.33 × 105

= 1.50 %

The Gravitational Field 57 • r r Picture the Problem The gravitational field at any point is defined by g = F m . Using its definition, express the gravitational field at a point in space:

r r F (12 N ) iˆ = g= = m 3 kg

(4 N/kg ) iˆ

*58 • r r Picture the Problem The gravitational field at any point is defined by g = F m . Using its definition, express the gravitational field at a point in space:

r

Solve for F and substitute for m and r g to obtain:

r r F g= m r r F = mg

(

)

= (0.004 kg ) 2.5 × 10 −6 N/kg ˆj

(

)

= 10 −8 N ˆj 59 •• Picture the Problem We can use the definition of the gravitational field due to a point mass to find the x and y components of the field at the origin and then add these components to find the resultant field. We can find the magnitude of the field from its components using the Pythagorean theorem. (a) Express the gravitational field due to the point mass at x = L:

r Gm g x = 2 iˆ L

Express the gravitational field due to the point mass at y = L:

r Gm g y = 2 ˆj L

Add the two fields to obtain:

r r r Gm ˆ Gm ˆ g = gx + g y = i+ 2 j L2 L

Gravity 873

r

(b) Find the magnitude of g :

r Gm Gm + 2 g = g x2 + g y2 = L2 L

=

2

Gm L2

60 •• Picture the Problem We can find the net force acting on m by superposition of the forces due to each of the objects arrayed on the circular arc. Once we have expressed the net force, we can find the gravitational field at the center of curvature from its definition. (a) Express the net force acting on m:

r F = Fx iˆ + Fy ˆj

(1)

GMm GMm GMm cos 45° − + R2 R2 R2 GMm cos 45° − R2 =0

Express Fx:

Fx =

Express Fy:

Fy =

GMm GMm + sin 45° R2 R2 GMm + sin 45° R2 GMm (2 sin 45° + 1) = R2

Substitute numerical values and evaluate Fy:

Fy =

(6.673 ×10

−11

N ⋅ m 2 / kg 2 (0.1m )2

)

× (3 kg )(2 kg )(2 sin 45° + 1) = 9.67 × 10 −8 N Substitute in equation (1) to obtain:

r

(b) Using its definition, express g at the center of curvature of the arc:

r F = 0iˆ + 9.67 × 10 −8 N ˆj

(

)

r r F 0iˆ + 9.67 × 10 −8 N ˆj g= = 2 kg m

(

=

(4.83 × 10

−8

)

)

N/kg ˆj

874 Chapter 11 61 •• Picture the Problem The configuration of point masses is shown to the right. The gravitational field at any point can be found by superimposing the fields due to each of the point masses. (a) Express the gravitational field at x = 2 m as the sum of the fields due to the point masses m1 and m2:

r

r

r r r g = g1 + g 2

(1)

Express g1 and g2 :

r r Gm Gm g1 = − 2 1 iˆ and g 2 = 2 2 iˆ x1 x2

Substitute in equation (1) to obtain:

r Gm Gm g = − 2 1 iˆ + 2 2 iˆ x1 x2

Substitute numerical values and r evaluate g :

=−

Gm1 ˆ Gm2 ˆ i+ i x12 (2 x1 )2

=−

G (m1 − 14 m2 ) iˆ x12

r 6.6726 × 10 −11 N ⋅ m 2 /kg 2 g=− (2 m )2 × [2 kg − 1 (4 kg )] iˆ 4

= r

r

(− 1.67 ×10

−11

)

N/kg iˆ

(b) Express g1 and g 2 :

r r Gm Gm g1 = − 2 1 iˆ and g 2 = − 2 2 iˆ x1 x2

Substitute in equation (1) to obtain:

r Gm Gm g = − 2 1 iˆ − 2 2 iˆ x1 x2 =−

Gm1 ˆ Gm2 ˆ i− 2 i x2 (2 x2 )2

=−

G 1 ( 4 m1 + m2 ) iˆ x22

Gravity 875 r 6.6726 × 10 −11 N ⋅ m 2 /kg 2 g=− (6 m )2 × [ 1 (2 kg ) + 4 kg ] iˆ

Substitute numerical values and r evaluate g :

4

= r

(− 8.34 × 10

−12

)

N/kg iˆ

Gm1 Gm2 − =0 2 x (6 − x )2

(c) Express the condition that g = 0:

or

2 4 − =0 2 x (6 − x )2 Express this quadratic equation in standard form:

x 2 + 12 x − 36 = 0 , where x is in meters.

Solve the equation to obtain:

x = 2.48 m and x = −14.5 m

From the diagram it is clear that the physically meaningful root is the positive one at:

x = 2.48 m

62 •• Picture the Problem To show that the maximum value of g x for the field of Example 11-7 occurs at the points x = ± a

2 , we can differentiate gx with respect to x and set

the derivative equal to zero. From Example 11-7:

gx = −

(x

2GMx 2

+ a2

)

3/ 2

Differentiate gx with respect to x and set the derivative equal to zero to find extreme values:

[(

dg x = −2GM x 2 + a 2 dx Solve for x to obtain:

)

−3 / 2

(

− 3x 2 x 2 + a 2

x= ±

)

−5 / 2

a 2

] = 0 for extrema.

876 Chapter 11 Remarks: To establish that this value for x corresponds to a relative maximum, we need to either evaluate the second derivative of gx at x = ± a/ 2 or examine the graph of g x at x = ± a/ 2 for concavity downward. 63 •• Picture the Problem We can find the mass of the rod by integrating dm over its length. r The gravitational field at x0 > L can be found by integrating dg at x0 over the length of the rod. L

L

(a) Express the total mass of the stick:

M = ∫ λ dx = C ∫ xdx =

(b) Express the gravitational field due to an element of the stick of mass dm:

r Gdm ˆ Gλ dx ˆ dg = − i =− i 2 ( x0 − x ) (x0 − x )2

Integrate this expression over the length of the stick to obtain:

0

=−

1 2

CL2

0

GCxdx ˆ i (x0 − x )2

L r g = −GC ∫

xdx ˆ i 2 0 (x0 − x )

=

2GM L2

⎡ ⎛ x0 ⎞ ⎛ L ⎞⎤ ˆ ⎟⎟ − ⎜⎜ ⎟⎟⎥ i ⎢ln⎜⎜ ⎣ ⎝ x 0 − L ⎠ ⎝ x0 − L ⎠⎦

64 ••• Picture the Problem Choose a mass element dm of the rod of thickness dx at a distance x from the origin. All such elements produce a gravitational field at a point P located a distance x0 > 12 L from the origin. We can calculate the total field by integrating the magnitude of the field produced by dm from x = −L/2 to x = +L/2. (a) Express the gravitational field at P due to the element dm:

r Gdm dg x = − 2 iˆ r

Relate dm to dx:

dm =

Express the distance r between dm and point P in terms of x and x0:

r = x0 − x

Substitute these results to express r dg x in terms of x and x0:

⎫ ⎧ r GM dg x = ⎨− dx ⎬iˆ 2 ⎭ ⎩ L(x0 − x )

M dx L

Gravity 877 (b) Integrate to find the total field:

r GM gx = − L

L/2

∫ (x

−L / 2

dx 0

− x)

2

iˆ

⎧⎪ GM ⎡ 1 ⎤ L / 2 ⎫⎪ = ⎨− ⎬iˆ ⎢ ⎥ ⎪⎩ L ⎣ x0 − x ⎦ − L / 2 ⎪⎭ GM = − 2 1 2 iˆ x0 − 4 L

r

(c) Use the definition of g to

r express F :

r r GMm F = m0 g = − 2 1 0 2 iˆ x0 − 4 L

(d) Factor x02 from the denominator

r gx = −

r of our expression for g x to obtain:

For x0 >> L the second term in parentheses is very small and:

GM

⎛ L x 02 ⎜⎜1 − 2 ⎝ 4 x0 2

⎞ ⎟⎟ ⎠

iˆ

r GM g x ≈ − 2 iˆ x0 which is the gravitational field of a point mass M located at the origin.

r g due to Spherical Objects 65 • Picture the Problem The gravitational field inside a spherical shell is zero and the field at the surface of and outside the shell is given by g = GM r 2 . (a) Because 0.5 m < R:

g= 0

(b) Because 1.9 m < R:

g= 0

(c) Because 2.5 m > R:

g=

GM r2 6.6726 × 10−11 N ⋅ m 2 /kg 2 (300 kg ) = (2.5 m )2

(

= 3.20 × 10 −9 N/kg 66

•

)

878 Chapter 11 Determine the Concept The gravitational attraction is zero. The gravitational field inside the 2 m shell due to that shell is zero; therefore, it exerts no force on the 1 m shell, and, by Newton’s 3rd law, that shell exerts no force on the larger shell. *67 • Picture the Problem The gravitational field and acceleration of gravity at the surface of a sphere given by g = GM R 2 , where R is the radius of the sphere and M is its mass. Express the acceleration of gravity on the surface of S1:

g1 =

GM R2

Express the acceleration of gravity on the surface of S2:

g2 =

GM R2

Divide the second of these equations by the first to obtain:

GM 2 g2 = R = 1 or g1 = g 2 g1 GM R2

68 •• Picture the Problem The gravitational field and acceleration of gravity at the surface of a sphere given by g = GM R 2 , where R is the radius of the sphere and M is its mass. Express the acceleration of gravity on the surface of S1:

g1 =

GM R12

Express the acceleration of gravity on the surface of S2:

g2 =

GM R22

Divide the second of these equations by the first to obtain:

GM g2 R22 R12 = = g1 GM R22 R12

Solve for g2:

R12 g2 = g1 R22

Remarks: The accelerations depend only on the masses and radii because the points of interest are outside spherically symmetric distributions of mass. 69

••

Gravity 879 Picture the Problem The magnitude of the gravitational force is F = mg where g inside a spherical shell is zero and outside is given by g = GM r 2 . (a) At r = 3a, the masses of both spheres contribute to g:

F = mg = m

Gm(M 1 + M 2 ) 9a 2

= (b) At r = 1.9a, g due to M2 = 0:

(c) At r = 0.9a, g = 0:

G (M 1 + M 2 ) (3a )2

F = mg = m

GM 1 GmM 1 = 2 3.61a 2 (1.9a )

F= 0

70 •• Picture the Problem The configuration is shown on the right. The centers of the spheres are indicated by the center-lines. The x coordinates of the mass m for parts (a), (b), and (c) are indicated along the x axis. The magnitude of the gravitational force is F = mg where g inside a spherical shell is zero and outside is given by g =

GM . r2

(a) Express the force acting on the object whose mass is m:

F = m( g1x + g 2 x )

Find g1x at x = 3a:

g1x =

GM 1 GM 1 = (3a )2 9a 2

Find g2x at x = 3a:

g2x =

GM 2 GM 2 = 2 (3a − 0.8a ) 4.84a 2

Substitute to obtain:

GM 2 ⎞ ⎛ GM 1 F = m⎜ + ⎟ 2 4.84a 2 ⎠ ⎝ 9a =

Gm ⎛ M 1 M 2 ⎞ + ⎜ ⎟ 4.84 ⎠ a2 ⎝ 9

880 Chapter 11 GM 2 GM 2 = 2 (1.9a − 0.8a ) 1.21a 2

(b) Find g2x at x = 1.9a:

g2x =

Find g1x at x = 1.9a:

g1x = 0

Substitute to obtain:

(c) At x = 0.9a, g1x = g2x = 0:

F = mg =

GmM 2 1.21a 2

F= 0

r g Inside Solid Spheres *71 •• Picture the Problem The "weight" as measured by a spring scale will be the normal force which the spring scale presses up against you. There are two forces acting on you as you stand at a distance r from the center of the planet: the normal force (FN) and the force of gravity (mg). Because you are in equilibrium under the influence of these forces, your weight (the scale reading or normal force) will be equal to the gravitational force acting on you. We can use Newton’s law of gravity to express this force. (a) Express the force of gravity acting on you when you are a distance r from the center of the earth: Using the definition of density, express the density of the earth between you and the center of the earth and the density of the earth as a whole:

Because we’re assuming the earth to of uniform-density and perfectly spherical:

Fg =

ρ=

GM (r )m r2

(1)

M (r ) M (r ) = V (r ) 43 π r 3

and

ρ=

ME M = 4 E3 VE 3π R

M (r ) ME = 3 3 4 4 3π r 3π R or

⎛r⎞ M (r ) = M E ⎜ ⎟ ⎝R⎠ Substitute in equation (1) and simplify to obtain:

3

3

⎛r⎞ GM E ⎜ ⎟ m ⎝ R ⎠ = GM E m r Fg = r2 R2 R

Gravity 881 Apply Newton’s law of gravity to yourself at the surface of the earth to obtain:

mg =

GM E m R2

or

g=

GM E R2

where g is the magnitude of free-fall acceleration at the surface of the earth. Substitute to obtain:

Fg =

mg r R

i.e., the force of gravity on you is proportional to your distance from the center of the earth. (b) Apply Newton’s 2nd law to your body to obtain:

FN − mg

Solve for your ″effective weight″ (i.e., what a spring scale will measure) FN:

FN =

r = −mrω 2 R

mg ⎞ ⎛ mg r − mrω 2 = ⎜ − mω 2 ⎟r R ⎝ R ⎠

Note that this equation tells us that your effective weight increases linearly with distance from the center of the earth. The second term can be interpreted as a "centrifugal force" pushing out, which increases the farther you get from the center of the earth.

(c) We can decide whether the change in mass with distance from the center of the earth or the rotational effect is more important by examining the ratio of the two terms in the expression for your effective weight:

mg g r g gT 2 R = R = = 2 mrω 2 ω 2 4π 2 R ⎛ 2π ⎞ R⎜ ⎟ ⎝T ⎠

Substitute numerical values and evaluate this ratio:

3600 s ⎞ ⎛ 9.81 m/s ⎜ 24 h × ⎟ 2 gT h ⎠ ⎝ = 4π 2 R 4π 2 (6370 km )

(

2

)

= 291

The change in the mass between you and the center of the earth as you move away from the center is 291 times more important than the rotational effect.

2

882 Chapter 11 72 •• Picture the Problem We can find the loss in weight at this depth by taking the difference between the weight of the student at the surface of the earth and her weight at a depth d = 15 km. To find the gravitational field at depth d, we’ll use its definition and the mass of the earth that is between the bottom of the shaft and the center of the earth. We’ll assume (incorrectly) that the density of the earth is constant. Express the loss in weight:

∆w = w( RE ) − w(R )

(1)

Express the mass M inside R =RE – d:

M = ρ V = 43 ρπ (RE − d )

Express the mass of the earth:

M E = ρVE = 43 ρπRE3

Divide the first of these equations by the second to obtain:

4 ρπ (R − d ) (RE − d M = 3 4 E 3 = ME ρπ R RE3 E 3

Solve for M:

3

3

M = ME

)3

(RE − d )3 RE3

Express the gravitational field at R =RE – d:

GM GM E (RE − d ) g= 2 = R (RE − d )2 RE2

Express the gravitational field at R =RE:

gE =

Divide equation (2) by equation (3) to obtain:

GM E (RE − d ) (RE − d )2 RE2 = RE − d g = GM E gE RE 2 RE

Solve for g:

g=

Express the weight of the student at R =RE – d:

w (R ) = mg (R ) =

3

GM E RE2

(2)

(3)

3

RE − d gE RE

RE − d mg E RE

⎛ d ⎞ ⎟⎟mg E = ⎜⎜1 − ⎝ RE ⎠

Gravity 883 Substitute in equation (1) to obtain:

⎛ d ⎞ mg E d ⎟⎟mg E = ∆w = mg E − ⎜⎜1 − RE ⎝ RE ⎠

Substitute numerical values and evaluate ∆w:

∆w =

(800 N )(15 km ) = 6370 km

1.88 N

73 •• Picture the Problem We can use the hint to find the gravitational field along the x axis. Using the hint, express g ( x ) :

g ( x ) = g solid sphere + g hollow sphere

Substitute for gsolid sphere and ghollow sphere and simplify to obtain:

g (x ) =

GM solid sphere

+

GM hollow sphere

(x − 12 R )2 3 Gρ 0 ( 43 π R 3 ) Gρ 0 [− 43 π ( 12 R ) ] = + x2 (x − 12 R )2 x2

⎤ ⎛ 4πρ0 R 3 ⎞ ⎡ 1 1 ⎟⎟ ⎢ 2 − = G⎜⎜ 2⎥ 8(x − 12 R ) ⎦ ⎝ 3 ⎠⎣ x 74 ••• Picture the Problem The diagram shows the portion of the solid sphere in which the r hollow sphere is embedded. g1 is the field due to the solid sphere of radius R and r density ρ0 and g 2 is the field due to the sphere of radius ½R and negative density

ρ0 centered at ½R. We can find the resultant field by adding the x and y r r components of g1 and g 2 .

r

Use its definition to express g1 :

r

Find the x and y components of g1 :

r GM Gρ 0V 4πρ 0 r 3G g1 = 2 = = r r2 3r 2 4πρ 0 rG = 3

4πρ0Gx ⎛ x⎞ g1x = − g1 cos θ = − g1 ⎜ ⎟ = − 3 ⎝r⎠ and

884 Chapter 11 4πρ0Gy ⎛ y⎞ g1 y = − g1 sin θ = − g1 ⎜ ⎟ = − 3 ⎝r⎠ where the negative signs indicate that the field points inward.

r

r GM 2 Gρ 0V2 4πρ 0 r23G = g2 = 2 = 3r22 r r22

Use its definition to express g 2 :

=

4πρ 0 r2G 3

where r2 =

r

(x − 12 R )2 + y 2

Express the x and y components of g2 :

⎛ x − 12 R ⎞ 4πρ 0 G (x − 12 R ) ⎟⎟ = g 2 x = g 2 ⎜⎜ 3 ⎝ r2 ⎠ ⎛ y ⎞ 4πρ 0Gy g 2 y = g 2 ⎜⎜ ⎟⎟ = 3 ⎝ r2 ⎠

Add the x components to obtain the x component of the resultant field:

g x = g1x + g 2 x

4πρ 0Gx 4πρ 0G (x − 12 R ) + 3 3 2πρ 0GR =− 3 =−

where the negative sign indicates that the field points inward. Add the y components to obtain the y component of the resultant field:

r r evaluate g :

Express g in vector form and

g y = g1 y + g 2 y =−

4πρ 0Gy 4πρ 0Gy + =0 3 3

r ⎛ 2πρ0GR ⎞ ˆ g = g x iˆ + G y ˆj = ⎜ − ⎟i 3 ⎝ ⎠ and

r 2πρ0GR g = 3 75 ••• Picture the Problem The gravitational field will exert an inward radial force on the objects in the tunnel. We can relate this force to the angular velocity of the planet by using Newton’s 2nd law of motion.

Gravity 885 Letting r be the distance from the objects to the center of the planet, use Newton’s 2nd law to relate the gravitational force acting on the objects to their angular velocity: Solve for ω to obtain:

Use its definition to express g:

Fnet = Fg = mrω 2 or

mg = mrω 2

ω=

g r

(1)

GM Gρ 0V 4πρ 0 r 3G = = 3r 2 r2 r2 4πρ 0 rG = 3

g=

Substitute in equation (1) and simplify:

ω=

4πρ 0 rG 3 = r

4πρ0G 3

76 ••• Picture the Problem Because we’re given the mass of the sphere, we can find C by expressing the mass of the sphere in terms of C. We can use its definition to find the gravitational field of the sphere both inside and outside its surface. (a) Express the mass of a differential element of the sphere:

(

dm = ρ dV = ρ 4π r 2 dr

5m

(

) )

Integrate to express the mass of the sphere in terms of C:

M = 4π C ∫ rdr = 50 m 2 π C

Solve for C:

C=

Substitute numerical values and evaluate C:

C=

1011 kg = 6.436 kg/m 2 2 (50 m )π

(b) Use its definition to express the gravitational field of the sphere at a distance from its center greater than its radius:

g=

GM r2

0

M 50 m 2 π

(

)

886 Chapter 11 (1) For r > 5 m:

g=

= Use its definition to express the gravitational field of the sphere at a distance from its center less than its radius:

(6.6726 ×10

N ⋅ m 2 /kg 2 ) (1011 kg ) r2

6.75 × 10 −8 N/kg r2 r

g =G

2 ∫ 4π r ρ dr 0

r

2

r

=G

∫ 4π r

2

0

C dr r

r2

r

=G (2) For r < 5 m:

−11

4π C ∫ r dr 0 2

r

= 2π GC

( × (6.436 kg/m )

g = 2π 6.6726 × 10 −11 N ⋅ m 2 /kg 2

)

2

= 2.70 × 10 −9 N/kg Remarks: Note that g is continuous at r = 5 m. *77 ••• Picture the Problem We can use conservation of energy to relate the work done by the gravitational field to the speed of the small object as it strikes the bottom of the hole. Because we’re given the mass of the sphere, we can find C by expressing the mass of the sphere in terms of C. We can then use its definition to find the gravitational field of the sphere inside its surface. The work done by the field equals the negative of the change in the potential energy of the system as the small object falls in the hole. Use conservation of energy to relate the work done by the gravitational field to the speed of the small object as it strikes the bottom of the hole:

Solve for v:

Express the mass of a differential element of the sphere:

K f − K i + ∆U = 0 or, because Ki = 0 and W = −∆U,

W = 12 mv 2 where v is the speed with which the object strikes the bottom of the hole and W is the work done by the gravitational field.

v=

2W m

(1)

(

dm = ρ dV = ρ 4π r 2 dr

)

Gravity 887 5m

(

)

Integrate to express the mass of the sphere in terms of C:

M = 4π C ∫ rdr = 50 m 2 π C

Solve for and evaluate C:

C=

0

M 1011kg = 2 50 m π 50 m 2 π

(

)

= 6.436 kg/m Use its definition to express the gravitational field of the sphere at a distance from its center less than its radius:

Substitute in equation (1) and simplify to obtain:

)

r

g =G

2 ∫ 4π r ρ dr 0

r2

r

=G

∫ 4π r

2

0

C dr r

r2

r

=G Express the work done on the small object by the gravitational force acting on it:

(

2

4π C ∫ r dr 0

r2

= 2π GC

3m

W = − ∫ mgdr = (2 m )mg 5m

v=

2(2 m )m(2π GC ) = m

(8 m )π GC

Substitute numerical values and evaluate v:

v=

(8 m )π (6.6726 ×10−11 N ⋅ m 2 /kg 2 )(6.436 kg/m 2 ) =

0.104 mm/s

78 ••• Picture the Problem The spherical deposit of heavy metals will increase the gravitational field at the surface of the earth. We can express this increase in terms of the difference in densities of the deposit and the earth and then form the quotient ∆g/g.

G∆M r2

Express ∆g due to the spherical deposit:

∆g =

Express the mass of the spherical deposit:

M = ∆ρV = ∆ρ 43 π R 3 = 43 π ∆ρ R 3

Substitute in equation (1):

(1)

(

∆g =

4 3

Gπ ∆ρ R 3 r2

)

888 Chapter 11 Gπ ∆ρ R 3 4 Gπ ∆ρ R 3 ∆g r2 = = 3 g g gr 2

Express ∆g/g:

4 3

Substitute numerical values and evaluate ∆g/g:

(

)(

)

∆g 43 π 6.6726 ×10 −11 N ⋅ m 2 /kg 2 5000 kg/m 3 (1000 m ) = = 3.56 ×10 −5 2 g (9.81 N/kg )(2000 m ) 3

*79 ••• Picture the Problem The force of attraction of the small sphere of mass m to the lead r r sphere is the sum of the forces due to the solid sphere ( FS ) and the cavities ( FC ) of negative mass. (a) Express the force of attraction:

r r r F = FS + FC

Use the law of gravity to express the force due to the solid sphere:

r GMm FS = − 2 iˆ d

Express the magnitude of the force acting on the small sphere due to one cavity:

FC =

(1)

GM'm

⎛R⎞ d +⎜ ⎟ ⎝2⎠

2

2

where M′ is the negative mass of a cavity. Relate the negative mass of a cavity to the mass of the sphere before hollowing:

⎡ ⎛ R ⎞3 ⎤ M' = − ρ V = − ρ ⎢ 43 π ⎜ ⎟ ⎥ ⎣⎢ ⎝ 2 ⎠ ⎦⎥

Letting θ be the angle between the x axis and the line joining the center of the small sphere to the center of either cavity, use the law of gravity to express the force due to the two cavities:

r FC = 2

Express cosθ :

(

)

= − 18 43 πρ R 3 = − 18 M GMm ⎛ R2 8⎜⎜ d 2 + 4 ⎝

⎞ ⎟⎟ ⎠

cosθ iˆ

because, by symmetry, the y components add to zero.

cosθ =

d d2 +

R2 4

Gravity 889 Substitute to obtain:

r FC =

GMm

d

⎛ R ⎞ ⎟⎟ d 2 + R 4⎜⎜ d 2 + 4 ⎠ 4 ⎝ GMmd = iˆ 2 3/ 2 ⎛ R ⎞ ⎟ 4⎜⎜ d 2 + 4 ⎟⎠ ⎝

Substitute in equation (1) and simplify:

2

2

iˆ

r GMm GMmd F = − 2 iˆ + iˆ 3/ 2 d ⎛ 2 R2 ⎞ ⎟⎟ 4⎜⎜ d + 4 ⎝ ⎠

⎤ ⎡ d3 ⎥ ⎢ GMm ⎢ ⎥ˆ 4 = − 1− 2 3 / 2 ⎥i ⎢ 2 d ⎧ ⎫ R ⎥ ⎢ d2 + ⎬ ⎥ ⎢ ⎨⎩ 4 ⎭ ⎦ ⎣

r

(b) Evaluate F at d = R:

⎡ ⎤ R3 ⎢ ⎥ r GMm ⎢ ⎥ˆ 4 F (R ) = − 1− i 2 3/ 2 ⎥ ⎢ 2 R ⎧ ⎫ R 2 ⎢ ⎥ R + ⎬ ⎥ ⎢ ⎨⎩ 4 ⎭ ⎦ ⎣ = − 0.821

GMm ˆ i R2

80 •• Picture the Problem Let R be the size of the cluster, and N the total number of stars in it. We can apply Newton’s law of gravity and the 2nd law of motion to relate the net force (which depends on the number of stars N(r) in a sphere whose radius is equal to the distance between the star of interest and the center of the cluster) acting on a star at a distance r from the center of the cluster to its speed. We can use the definition of density, in conjunction with the assumption of uniform distribution of the starts within the cluster, to find N(r) and, ultimately, express the orbital speed v of a star in terms of the total mass of the cluster. Using Newton’s law of gravity and 2nd law, express the force acting on a star at a distance r from the center of the cluster:

GN (r )M 2 v2 F (r ) = =M r2 r where N(r) is the number of stars within a distance r of the center of the cluster and M is the mass of an individual star.

890 Chapter 11 Using the uniform distribution assumption and the definition of density, relate the number of stars N(r) within a distance r of the center of the cluster to the total number N of stars in the cluster:

ρ=

N (r )M NM = 4 3 3 4 3 πr 3 πR

or

N (r ) = N

Substitute to obtain:

r3 R3

GNM 2 r 3 v2 =M r 2 R3 r or

Solve for v to obtain:

GNM

r2 = v2 R3

v=r

GNM ⇒ v∝r R3

The mean velocity v of a star in a circular orbit around the center of the cluster increases linearly with distance r from the center.

General Problems *81 • Picture the Problem We can use Kepler’s 3rd law to relate Pluto’s period to its mean distance from the sun.

T 2 = Cr 3

Using Kepler’s 3rd law, relate the period of Pluto to its mean distance from the sun:

4π 2 = 2.973 × 10−19 s 2 /m 3 . where C = GM s

Solve for T:

T = Cr 3

Substitute numerical values and evaluate T:

⎛ 1.50 × 1011 m ⎞ ⎟⎟ T = (2.973 × 10 s /m )⎜⎜ 39.5 AU × AU ⎝ ⎠ 1h 1d 1y = 7.864 × 109 s × × × 3600 s 24 h 365.25 d −19

= 249 y

2

3

3

Gravity 891 82 • Picture the Problem Consider an object of mass m at the surface of the earth. We can relate the weight of this object to the gravitational field of the earth and to the mass of the earth. Using Newton’s 2nd law, relate the weight of an object at the surface of the earth to the gravitational force acting on it: Solve for ME:

Substitute numerical values and evaluate ME:

w = mg =

ME =

ME

GM E m RE2

gRE2 G

2 ( 9.81 N / kg ) (6.37 × 106 m ) =

6.6726 × 10 −11 N ⋅ m 2 /kg 2

= 5.97 × 10 24 kg 83 •• Picture the Problem The work you must do against gravity to move the particle from a distance r1 to r2 is the negative of the change in the particle’s gravitational potential energy. (a) Relate the work you must do to the change in the gravitational potential energy of the earth-particle system:

r2

r2

r1

r1

W = −∆U = − ∫ Fg dr = GM E m ∫

dr r2

⎛1 1⎞ = −GM E m⎜⎜ − ⎟⎟ ⎝ r2 r1 ⎠ ⎛1 1⎞ = GM E m⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠

(b) Substitute gRE2 for GME, RE for r1, and RE + h for r2 to obtain:

⎛ 1 1 ⎞ ⎟⎟ W = mgRE2 ⎜⎜ − ⎝ RE RE + h ⎠

(1)

892 Chapter 11 ⎛ R + h − RE ⎞ ⎟⎟ W = mgRE2 ⎜⎜ E ⎝ RE (RE + h ) ⎠

(c) Rewrite equation (1) with a common denominator and simplify to obtain:

⎛ ⎜ ⎛ RE ⎞ 1 ⎟⎟ = mgh⎜ = mgh⎜⎜ ⎜ h ⎝ RE + h ⎠ ⎜1+ R E ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

≈ mgh when h << RE. 84 •• Picture the Problem The gravitational field outside a uniform sphere is given by g = − GM r 2 and the field inside the sphere by g = − GM R 3 r.

(

)

GM r2

(a) Express g outside the sphere:

g=−

Find the mass of the sphere:

M = ρV = ρ

Substitute and simplify to obtain:

g=−

Gρ

(

4 3

(

4 3

π R3 )

π R3 )

r2

= − 43

Gρ R 3 r2

Substitute numerical values and evaluate g:

g=−

4 3

(6.673 ×10

−11

)(

)

N ⋅ m 2 / kg 2 2000 kg/m 3 (100 m ) 0.559 N ⋅ m 2 /kg = − r2 r2

(b) Express the gravitational field inside the uniform sphere:

3

(

)

Gρ 43 π R 3 GM g=− 3 r=− r R R3 = − 43 π ρ Gr

Substitute numerical values and evaluate g:

(

)(

)

(

)

g = − 43 π 2000 kg/m 3 6.6726 × 10 −11 N ⋅ m 2 /kg 2 r = − 5.59 × 10 −7 N/kg ⋅ m r 85 •• Picture the Problem We can use Kepler’s 3rd law to relate the period of the satellite to its mean distance from the center of Jupiter.

Gravity 893 4π 2 (RJ + h )3 T = GM J

Use Kepler’s 3rd law to relate the period of the satellite to its mean distance from the center of Jupiter:

2

Solve for h:

h=3

T 2 GM J − RJ 4π 2

(1)

Express the mass of Jupiter in terms of the mass of the earth:

M J = 320 M E

Express the volume of Jupiter in terms of the mass of the earth:

VJ = 1320VE

Express the volumes of Jupiter and Earth in terms of their radii and solve for RJ:

RJ = 3 1320RE

Substitute in equation (1) to obtain:

T 2 G{320M E } 3 h= − 1320 RE 4π 2

Express the period of the satellite in seconds:

T = 9 h + 50 min

3

60 s 3600 s + 50 min× min h 4 = 3.54 × 10 s = 9h ×

Substitute numerical values and evaluate h:

h=

(3.54 ×10 s ) (6.6726 ×10 4

3

(

2

− 3 1320 6.37 × 106 m

)

−11

){ (

N ⋅ m 2 /kg 2 320 5.98 × 1024 kg 4π 2

)}

= 8.96 × 107 m 86 •• Picture the Problem Let m represent the mass of the spacecraft. From Kepler’s 3rd law we know that its period will be a minimum when it is in orbit just above the surface of the moon. We’ll use Newton’s 2nd law to relate the angular velocity of the spacecraft to the gravitational force acting on it.

894 Chapter 11 2π

Relate the period of the spacecraft to its angular velocity:

T=

Using Newton’s 2nd law of motion, relate the gravitational force acting on the spacecraft when it is in orbit at the surface of the moon to the angular velocity of the spacecraft:

∑F

Solve for ω:

radial

=

GM M m = mRMω 2 RM2

G (43 π ρ RM3 ) GM M = RM3 RM3

ω= =

Substitute in equation (1) and simplify to obtain:

(1)

ω

4 3

Gπ ρ

Tmin =

4 3

2π = Gπ ρ

3π ρG

Substitute numerical values and evaluate Tmin:

Tmin =

(6.6726 ×10

−11

3π = 6503 s = 1 h 48 min N ⋅ m 2 /kg 2 )(3340 kg/m 3 )

87 •• Picture the Problem We can use conservation of energy to establish a relationship between the height h to which the projectile will rise and its initial speed. The application of Newton’s 2nd law will relate the orbital speed, which is equal to the initial speed of the projectile, to the mass and radius of the moon. Use conservation of energy to relate the initial energies of the projectile to its final energy:

Solve for h:

Use Newton’s 2nd law to relate velocity of the satellite to the

Kf − Ki + U f − U i = 0 or, because Kf = 0,

− 12 mv 2 −

GM M m GM M m + =0 RM + h RM

⎛ ⎞ ⎜ ⎟ 1 ⎜ − 1⎟⎟ h = R⎜ 2 v RM ⎜1− ⎟ ⎝ 2GM M ⎠

∑F

radial

GM M m v2 = =m RM RM2

(1)

Gravity 895 gravitational force acting on it: Solve for v2:

Substitute for v2 in equation (1) and simplify to obtain:

v2 =

GM M RM

⎛ ⎞ ⎜ 1 ⎟ h = R⎜ − 1⎟ = R = 1.70 Mm ⎜1− 1 ⎟ ⎜ ⎟ ⎝ 2 ⎠

*88 •• Picture the Problem If we assume the astronauts experience a constant acceleration in the barrel of the cannon, we can use a constant-acceleration equation to relate their exit speed (the escape speed from the earth) to the acceleration they would need to undergo in order to reach that speed. We can use conservation of energy to express their escape speed in terms of the mass and radius of the earth and then substitute in the constantacceleration equation to find their acceleration. To find the balance point between the earth and the moon we can equate the gravitational forces exerted by the earth and the moon at that point. (a) Assuming constant acceleration down the cannon barrel, relate the ship’s speed as it exits the barrel to the length of the barrel and the acceleration required to get the ship to escape speed: Solve for the acceleration:

ve2 = 2a∆l where l is the length of the cannon.

a=

ve2 2∆l

(1)

∆K + ∆U = 0

Use conservation of energy to relate the initial energy of astronaut’s ship to its energy when it has escaped the earth’s gravitational field:

or

When the ship has escaped the earth’s gravitational field:

Kf = U f = 0

Kf − Ki + U f − U i = 0

and

− Ki − U i = 0 or

⎛ GM E m ⎞ − 12 mve2 − ⎜ − ⎟=0 R ⎠ ⎝ where m is the mass of the spaceship. Solve for ve2 to obtain:

ve2 =

2GM E R

896 Chapter 11 Substitute in equation (1) to obtain:

Substitute numerical values and evaluate a:

a=

GM E ∆lR

(

a = 6.673 × 10 −11 N ⋅ m 2 /kg 2 ×

(5.98 × 10

24

)

)

kg (274 m )(6370 km )

= 2.29 × 105 m/s 2 ≈ 23,300 g Survival is extremely unlikely! (b) Let the distance from the center of the earth to the center of the moon be R, and the distance from the center of the spaceship to the earth be x. If M is the mass of the earth and m the mass of the moon, the forces will balance out when:

Solve for x to obtain:

GM Gm = 2 x ( R − x) 2 or

x R−x = M m

where we’ve ignored the negative solution, as it doesn't indicate a point between the two bodies.

R

x=

m M

1+ Substitute numerical values and evaluate x:

3.84 × 108 m

x= 1+

7.36 × 10 22 kg 5.98 × 10 24 kg

= 3.46 × 108 m

(c)

No it is not. During the entire trip, the astronauts would be in free - fall, and so would not seem to weigh anything.

89 •• Picture the Problem Let the origin of our coordinate system be at the center of mass of the binary star system and let the distances of the stars from their center of mass be r1 and r2. The period of rotation is related to the angular velocity of the star system and we can use Newton’s 2nd law of motion to relate this velocity to the separation of the stars. Relate the square of the period of the motion of the stars to their angular velocity:

T2 =

4π 2

ω2

(1)

Gravity 897 Using Newton’s 2nd law of motion, relate the gravitational force acting on the star whose mass is m2 to the angular velocity of the system: Solve for ω2:

Gm1 m2

∑F

=

ω2 =

Gm1

radial

(r1 + r2 )

r2 (r1 + r2 )

2

= m2 r2ω 2

(2)

2

From the definition of the center of mass we have:

m1 r1 = m2 r2 where r = r1 + r2

Eliminate r1 from equations (3) and (4) and solve for r2:

r2 =

rm1 m1 + m2

Eliminate r2 from equations (3) and (4) and solve for r1:

r1 =

rm2 m1 + m2

Substitute for r1 and r2 in equation (2) to obtain:

ω2 =

Finally, substitute in equation (1) and simplify:

T2 =

(3) (4)

G (m1 + m2 ) r3 4π 2 4π 2 r 3 = G (m1 + m2 ) G (m1 + m2 ) 3 r

90 •• Picture the Problem Because the two-particle system has zero initial energy and zero initial linear momentum; we can use energy and momentum conservation to obtain simultaneous equations in the variables r, v1 and v2. We’ll assume that initial separation distance of the particles and their final separation r is large compared to the size of the particles so that we can treat them as though they are point particles. Use conservation of energy to relate the speeds of the particles when their separation distance is r:

Use conservation of linear momentum to obtain a second relationship between the speeds of the particles and their masses:

Ei = E f or

0 = 12 m1v12 + 12 m2 v 22 −

Gm1 m2 r

(1)

pi = pf or

0 = m1v1 + m2 v 2

(2)

898 Chapter 11 Solve equation (2) for v1 and substitute in equation (1) to obtain: Solve equation (3) for v2:

Solve equation (2) for v1 and substitute for v2 to obtain:

⎛ m22 ⎞ 2Gm1 m2 ⎟= v ⎜⎜ m2 + m1 ⎟⎠ r ⎝ 2 2

v2 =

2Gm12 r (m1 + m2 )

v1 =

2Gm22 r (m1 + m2 )

(3)

*91 •• Picture the Problem We can find the orbital speeds of the planets from their distance from the center of mass of the system and the period of their motion. Application of Kepler’s 3rd law will allow us to express the period of their motion T in terms of the effective mass of the system … which we can find from its definition. Express the orbital speeds of the planets in terms of their period T:

v=

2πR T

where R is the distance to the center of mass of the four-planet system. Apply Kepler’s 3rd law to express the period of the planets:

T=

4π 2 3 R GM eff

where Meff is the effective mass of the four planets. Substitute to obtain:

The distance of each planet from the effective mass is: Find Meff from its definition:

v=

R=

2πR 4π R3 GM eff 2

=

GM eff R

a 2

1 1 1 1 1 = + + + M eff M M M M and

M eff = 14 M Substitute for R and Meff to obtain:

v=

2GM 4a

Gravity 899 92 •• Picture the Problem Let r represent the separation of the particle from the center of the earth and assume a uniform density for the earth. The work required to lift the particle from the center of the earth to its surface is the integral of the gravitational force function. This function can be found from the law of gravity and by relating the mass of the earth between the particle and the center of the earth to the earth’s mass. We can use the workkinetic energy theorem to find the speed with which the particle, when released from the surface of the earth, will strike the center of the earth. Finally, the energy required for the particle to escape the earth from the center of the earth is the sum of the energy required to get it to the surface of the earth and the kinetic energy it must have to escape from the surface of the earth. (a) Express the work required to lift the particle from the center of the earth to the earth’s surface:

Using the law of gravity, express the force acting on the particle as a function of its distance from the center of the earth: Express the ratio of M to ME:

Substitute for M in equation (2) to obtain:

Substitute for F in equation (1) and evaluate the integral:

W =

RE

∫ Fdr

(1)

0

where F is the gravitational force acting on the particle.

F=

GmM r2

(2)

where M is the mass of a sphere whose radius is r.

( (

ρ 4π r3 M = 43 M E ρ 3 π RE3

) )

⇒ M = ME

r3 RE3

F=

mg mgRE2 GmM E = r r= r 3 3 RE RE RE

W=

mg RE

RE

∫ rdr = 0

(b) Use the work-kinetic energy theorem to relate the kinetic energy of the particle as it reaches the center of the earth to the work done on it in moving it to the surface of the earth:

W = ∆K = 12 mv 2

Substitute for W and solve for v:

v=

gRE

gmRE 2

900 Chapter 11 (c) Express the total energy required for the particle to escape when projected from the center of the earth:

E esc = W + 12 mve2 2 = 12 mvesc

where ve is the escape speed from the surface of the earth.

Substitute for W and solve for vesc:

vesc = 3 gRE

Substitute numerical values and evaluate vesc:

vesc = 3(9.81 N/kg ) 6.37 × 106 m

(

)

= 13.7 km/s

93 •• Picture the Problem We need to find the gravitational field in three regions: r < R1, R1 < r < R2, and r > R2. For r < R1:

g= 0

For r > R2, g(r) is the field of a mass M centered at the origin:

g (r ) =

For R1 < r < R2, g(r) is determined by the mass within the shell of radius r:

Gm r2 where m = 43 πρ r 3 − R13

Express the density of the spherical shell:

ρ=

M M = 4 3 3 V 3 π R2 − R1

Substitute for ρ in equation (2) and simplify to obtain:

m=

M (r 3 − R13 ) R23 − R13

Substitute for m in equation (1) to obtain:

g (r ) =

GM r2

g (r ) =

(

(

(

) )

GM r 3 − R13 r 2 R23 − R13

(

)

(1)

)

(2)

Gravity 901

A graph of gr with R1 = 2, R2 = 3, and GM = 1 follows:

0.12 0.10 0.08 g r 0.06 0.04 0.02 0.00 0

2

4

6

8

r

94 •• Picture the Problem A ring of radius R is shown to the right. Choose a coordinate system in which the origin is at the center of the ring and x axis is as shown. An element of length dL and mass dm is responsible for the field dg at a distance x from the center of the ring. We can express the x component of dg and then integrate over the circumference of the ring to find the total field as a function of x. (a) Express the differential gravitational field at a distance x from the center of the ring in terms of the mass of elemental length dL: Relate the mass of the element to its length: Substitute to obtain: By symmetry, the y and z components of g vanish. Express the x component of dg:

dg =

Gdm R2 + x2

dm = λ dL where λ is the linear density of the ring. Gλ dL R2 + x2 dg x = dg cosθ

dg =

=

Gλ dL cosθ R2 + x2

902 Chapter 11 Referring to the figure, express cosθ :

cosθ =

Substitute to obtain:

dg x =

Because λ =

M : 2π R

R2 + x2

Gλ dL x Gλ xdL × = 3/ 2 2 2 2 2 R +x R2 + x2 R +x

(

)

GM xdL

dg x =

Integrate to find g(x):

x

2π R(R 2 + x 2 )

3/ 2

g (x ) =

GM x

(

2π R R 2 + x 2

=

(R

GM 2

+ x2

)

3/ 2

2πR

dL ) ∫ 3/ 2

0

x

A plot of gx is shown below. The curve is normalized for R = 1 and GM = 1.

0.40 0.35 0.30 0.25 g x 0.20 0.15 0.10 0.05 0.00 0

1

2

3

4

x

(b) Differentiate g(x) with respect to x and set the derivative equal to zero to identify extreme values:

(

)

⎡ x 2 + R 2 3 / 2 − x( 3 ) 2 dg 2 = GM ⎢ x + R2 2 2 3 dx R +x ⎢⎣ Simplify to obtain:

(

)

(

(x

2

⎤

) (2 x )⎥ = 0 for extrema 1/ 2

+ R2

⎥⎦

)

3/ 2

(

− 3x 2 x 2 + R 2

)

1/ 2

=0

Gravity 903 Solve for x to obtain:

x= ±

R 2

Because the curve is concave downward, we can conclude that this result corresponds to a maximum. Note that this result agrees with our graphical maximum. 95 ••• Picture the Problem The diagram shows a segment of the wire of length dx and mass dm = λdx at a distance x from the origin of our coordinate system. We can find the magnitude of the gravitational field at a distance r from the wire from the resultant gravitational force acting on a particle of mass m′ located at point P and then integrating over the length of the wire.

dF = m'dg

Express the gravitational force acting on a particle of mass m′ at a distance r from the wire due to the segment of the wire of length dx:

or

dg =

dF m'

Gm'λdx R2 or, because R 2 = x 2 + r 2 , Gm'λdx dF = 2 2 x +r

Using Newton’s law of gravity, express dF:

dF =

Substitute and simplify to express the gravitational field due to the segment of the wire of length dx:

dg =

Gλdx x2 + r 2

Gλdx cos θ x2 + r 2 Gλdx r = 2 2 2 x +r x + r2 Gλr = dx 32 2 x + r2

By symmetry, the segment on the opposite side of the origin at the same distance from the origin will cancel out all but the radial component of the field, so the gravitational field will be given by:

dg =

(

)

Integrate dg from x′ = −∞ to x′ = +∞ to obtain: ∞

g=

Gλr

∫ (x' + r )

−∞

2

2 3/ 2

∞

dx' = 2Gλ ∫ 0

∞

⎤ 2Gλ ⎡ 2Gλ x dx' = ⎢ ⎥ = 2 2 3/ 2 2 2 r ⎣ x' + r ⎦ 0 r x' + r

(

r

)

904 Chapter 11 96 ••• Picture the Problem We can use the relationship between the angular velocity of an orbiting object and its tangential velocity to express the speeds vin and vout of the innermost and outermost portions of the ring. In part (b) we can use Newton’s law of gravity, in conjunction with the 2nd law of motion, to relate the tangential speed of a chunk of the ring to the gravitational force acting on it. As in part (a), once we know vin and vout, we can express the difference between them to obtain the desired results. (a) Express the speed of a point in the ring at a distance R′ from the center of the planet under the assumption that the ring is solid and rotates with an angular velocity ω:

v(R') = ωR

Express the speeds vin and vout of the innermost and outermost portions of the ring:

vin = (R − 12 r )ω

Express the difference between vout and vin:

and

vout = (R + 12 r )ω vout − vin = (R + 12 r )ω − (R − 12 r )ω =ωr =

(b) Assume that a chunk of the ring is moving in a circular orbit around the center of the planet under the force of gravity. Then, we can find its velocity by equating the force of gravity to the centripetal force needed to keep it in orbit:

Substitute for R′ to express vout:

v r r= v R R

GMm mv 2 = R '2 R' or

GM R'

v=

where M is the mass of the planet and R' the distance from the center.

vout =

GM = R + 12 r

GM ⎛ 1 r⎞ R ⎜1 + ⎟ ⎝ 2 R⎠

GM ⎛ 1 r ⎞ = ⎜1 + ⎟ R ⎝ 2 R⎠ Expand binomially to obtain:

vout =

≈

−1 2

GM ⎛ 1 1 r ⎜1 − R ⎝ 22R + higher order terms ) GM ⎛ 1 r ⎞ ⎜1 − ⎟ R ⎝ 4 R⎠

Gravity 905 Proceed similarly to obtain, for vin:

vin ≈

GM ⎛ 1 r ⎞ ⎜1 + ⎟ R ⎝ 4 R⎠

Express the difference between vout and vin:

GM ⎛ 1 r ⎞ GM ⎛ 1 r ⎞ ⎜1 − ⎟− ⎜1 + ⎟= R ⎝ 4 R⎠ R ⎝ 4 R⎠ 1 GM , vout − vin ≈ − and, because v = 2 R

vout − vin ≈

GM ⎛ 1 r ⎞ ⎜− ⎟ R ⎝ 2 R⎠ r v R

97 ••• Picture the Problem Let U = 0 at x = ∞. The potential energy of an element of the stick dm and the point mass m0 is given by the definition of gravitational potential energy: dU = − Gm0 dm r where r is the separation of dm and m0.

Gm0 dm x0 − x

(a) Express the potential energy of the masses m0 and dm:

dU = −

The mass dm is proportional to the size of the element dx:

dm = λ dx

Substitute these results to express dU in terms of x:

where λ =

dU = −

M . L

Gm0 λ dx GMm0 dx = − x0 − x L ( x0 − x )

(b) Integrate to find the total potential energy for the system:

U =−

GMm0 L

= −

L/2

dx GMm0 ⎡ ⎛ L⎞ L ⎞⎤ ⎛ ln⎜ x0 − ⎟ − ln⎜ x0 + ⎟⎥ = ⎢ x −x L ⎣ ⎝ 2⎠ 2 ⎠⎦ ⎝ −L / 2 0

∫

GMm0 ⎛ x0 + L 2 ⎞ ⎟⎟ ln⎜⎜ L ⎝ x0 − L 2 ⎠

(c) Because x0 is a general point along the x axis:

⎡ ⎤ 1 ⎥ dU Gmm0 ⎢ 1 = − F ( x0 ) = − ⎢ ⎥ dx0 L ⎢x + L x − L ⎥ 0 0 2 2⎦ ⎣

906 Chapter 11 Simplify this expression to obtain:

F ( x0 ) = −

Gmm0 x 2 − L2 4

in agreement with the result of Example 11-8. *98 ••• Picture the Problem Choose a mass element dm of the rod of thickness dx at a distance x from the origin. All such elements of the rod experience a gravitational force dF due to presence of the sphere centered at the origin. We can find the total gravitational force of attraction experienced by the rod by integrating dF from x = a to x = a + L. Express the gravitational force dF acting on the element of the rod of mass dm:

dF =

GMdm x2

Express dm in terms of the mass m and length L of the rod:

dm =

m dx L

Substitute to obtain:

dF =

GMm dx L x2

Integrate dF from x = a to x = a + L to find the total gravitational force acting on the rod:

GMm F= L =

a+ L

a+ L

GMm ⎡ 1 ⎤ ∫a x dx = − L ⎢⎣ x ⎥⎦ a −2

GMm a(a + L )

99 ••• Picture the Problem The semicircular rod is shown in the figure. We’ll use an element of length Rdθ = mass dM is

M

π

L

π

dθ whose

dθ . By symmetry, Fy = 0 .

We’ll first find dFx and then integrate over θ from −π/2 to π/2. Express dFx:

dFx =

GmdM GMm = dθ cosθ 2 2 R ⎛L⎞ π⎜ ⎟ ⎝π ⎠

Gravity 907 Integrate dFx over θ from −π/2 to π/2:

Fx =

π /2

π GMm 2

L

∫ cos θ dθ = π

− /2

2π GMm L2

Substitute numerical values and evaluate Fx:

(

)

2π 6.6726 × 10−11 N ⋅ m 2 /kg 2 (20 kg )(0.1 kg ) Fx = = 33.5 pN (5 m )2 *100 ••• Picture the Problem We can begin by expressing the forces exerted by the sun and the moon on a body of water of mass m and taking the ratio of these forces. In (b) we’ll simply follow the given directions and in (c) we can approximate differential quantities with finite quantities to establish the given ratio. (a) Express the force exerted by the sun on a body of water of mass m:

FS =

GM S m rS2

Express the force exerted by the moon on a body of water of mass m:

Fm =

GM m m rm2

Divide the first of these equations by the second and simplify to obtain:

FS M S rm2 = Fm M m rS2

Substitute numerical values and evaluate this ratio:

FS 1.99 × 1030 kg 3.84 × 108 m = Fm 7.36 × 1022 kg 1.50 × 1011 m

( (

)( )(

= 177

(b) Find

dF : dr

Solve for the ratio

dF 2Gm1m2 F =− = −2 3 dr r r dF : F

(c) Express the change in force ∆F for a small change in distance ∆r:

dF dr = −2 F r

∆F = −2

F ∆r r

) )

2 2

908 Chapter 11 Express ∆FS :

GmM S rS2 ∆FS = −2 ∆rS rS GmM S ∆rS rS3 GmM m ∆Fm = −2 ∆rm rm3 = −2

Express ∆Fm :

Divide the first of these equations by the second and simplify:

MS ∆rS ∆FS rS3 M r 3 ∆r = S m3 S = ∆Fm M m ∆r M m rS ∆rm m 3 rm M S rm3 M m rS3

= because

Substitute numerical values and evaluate this ratio:

∆rS = 1. ∆rm

( (

)( )(

) )

∆FS 1.99 × 1030 kg 3.84 × 108 m = ∆Fm 7.36 × 1022 kg 1.50 × 1011 m

3 3

= 0.454 101 •• Picture the Problem Let MNS be the mass of the Neutron Star and m the mass of each robot. We can use Newton’s law of gravity to express the difference in the tidal-like forces acting on the coupled robots. Expanding the expression for the force on the robot further from the Neutron Star binomially will lead us to an expression for the distance at which the breaking tension in the connecting cord will be exceeded.

The gravitational force is greater on the lower robot, so if it were not for (a) the cable its acceleration would be greater than that of the upper robot, and they would separate. In opposing this separation the cable is stressed.

Gravity 909 (b) Letting the separation of the two robots be ∆r, and the distance from the center of the star to the lower robot be r, use Newton’s law of gravity to express the difference in the forces acting on the robots:

Ftide =

GM NS m GM NS m − r2 (r + ∆r )2

⎡ ⎤ ⎢ ⎥ 1 1 ⎢ ⎥ = GM NS m 2 − 2 ⎢r ⎛ ∆r ⎞ ⎥ r 2 ⎜1 + ⎟ ⎥ ⎢ r ⎠ ⎦ ⎝ ⎣ −2 GM NS m ⎡ ⎛ ∆r ⎞ ⎤ = ⎟ ⎥ ⎢1 − ⎜1 + r 2 ⎣⎢ ⎝ r ⎠ ⎥⎦

Expand the expression in the square brackets binomially to obtain:

⎛ ∆r ⎞ 1 − ⎜1 + ⎟ r ⎠ ⎝

Substitute to obtain:

Ftide ≈

Letting FB be the breaking tension of the cord, substitute for Ftide and solve for the value of r corresponding to the breaking strain being exceeded:

r=3

−2

∆r ⎞ ⎛ ≈ 1 − ⎜1 − 2 ⎟ r ⎠ ⎝ ∆r =2 r

2GM NS m ∆r r3

2GM NS m ∆r FB

Substitute numerical values and evaluate r:

r=3

(

)(

)

2 6.673 × 10 −11 N ⋅ m 2 /kg 2 1.99 × 1030 kg (1 kg ) (1m ) = 220 km 25 kN

910 Chapter 11

Chapter 12 Static Equilibrium and Elasticity Conceptual Problems 1 • (a) False. The conditions

r

∑F

i

i

= 0 and

r

∑τ

i i

= 0 must be satisfied.

(b) True. The necessary and sufficient conditions for static equilibrium are

r ∑iτ i = 0 . (c) True. The conditions

r

∑F i

i

= 0 and

r

∑τ

i i

r

∑F i

i

= 0 and

= 0 must be satisfied.

(d) False. An object is in equilibrium provided the conditions

r

∑F i

i

= 0 and

r

∑τ

i i

= 0 are

satisfied. 2 • False. The location of the center of gravity depends on the mass distribution. 3 • No. The definition of the center of gravity does not require that there be any material at its location. 4 • Determine the Concept When the acceleration of gravity is not constant over an object, the center of gravity is the pivot point for balance. 5 •• Determine the Concept This technique works because the center of mass must be directly under the balance point. Thus, a line drawn straight downward will pass through the center of mass, and another line drawn straight downward when the figure is hanging from another point will also pass through the center of mass. The center of mass is where the lines cross. *6 • Determine the Concept No. Because the floor can exert no horizontal force, neither can the wall. Consequently, the friction force between the wall and the ladder is zero regardless of the coefficient of friction between the wall and the ladder.

911

912 Chapter 12 7 • Determine the Concept We know that equal lengths of aluminum and steel wire of the same diameter will stretch different amounts when subjected to the same tension. Also, because we are neglecting the mass of the wires, the tension in them is independent of which one is closer to the roof and depends only on W. (b) is correct. 8 • Determine the Concept Yes; if it were otherwise, angular momentum conservation would depend on the choice of coordinates. *9 • Determine the Concept The condition that the bar is in rotational equilibrium is that the net torque acting on it be zero; i.e., R1M1 = R2M2. This condition is satisfied provided R1 = R2 and M1 = M2. (c) is correct. 10 •• Determine the Concept You cannot stand up because your body’s center of gravity must be above your feet. *11 •• Determine the Concept The tensile strengths of stone and concrete are at least an order of magnitude lower than their compressive strengths, so you want to build compressive structures to match their properties.

Estimation and Approximation 12 •• Picture the Problem The diagram to the right shows the forces acting on the crate as it is being lifted at its left end. Note that when the crowbar lifts the crate, only half the weight of the crate is supported by the bar. Choose the coordinate system shown and let the subscript ″pb″ refer to the pry bar. The diagram below shows the forces acting on the pry bar as it is being used to lift the end of the crate.

Static Equilibrium and Elasticity

913

Assume that the maximum force F ′ you can apply is 500 N (about 110 lb). Let l be the distance between the points of contact of the steel bar with the floor and the crate, and let L be the total length of the bar. Lacking information regarding the bend in pry bar at the fulcrum, we’ll assume that it is small enough to be negligible. We can apply the condition for rotational equilibrium to the pry bar and a condition for translational equilibrium to the crate when its left end is on the verge of lifting. Apply

∑F

Apply

∑ τ = 0 to the crate about

y

= 0 to the crate:

r

Fpb − W + Fn = 0

(1)

wFn − 12 wW = 0

an axis through point B and perpendicular to the plane of the page to obtain: Solve for Fn:

Fn = 12 W as noted in Picture the Problem.

Solve equation (1) for Fpb and substitute for Fn to obtain: Apply

r

∑ τ = 0 to the pry bar about

Fpb = W − 12 W = 12 W

F (L − l ) − lFpb = 0

an axis through point A and perpendicular to the plane of the page to obtain: Solve for L:

⎛ F ⎞ L = l ⎜⎜1 + pb ⎟⎟ F ⎠ ⎝

Substitute for Fpb to obtain:

⎛ W ⎞ L = l ⎜1 + ⎟ ⎝ 2F ⎠

914 Chapter 12 Substitute numerical values and evaluate L:

⎛ 4500 N ⎞ ⎟⎟ = 55.0 cm L = (0.1 m )⎜⎜1 + ⎝ 2(500 N ) ⎠

*13 •• Picture the Problem We can derive this expression by imagining that we pull on an area A of the given material, expressing the force each spring will experience, finding the fractional change in length of the springs, and substituting in the definition of Young’s modulus. (a) Express Young’s modulus:

Express the elongation ∆L of each spring: Express the force Fs each spring will experience as a result of a force F acting on the area A: Express the number of springs N in the area A: Substitute to obtain:

Substitute in equation (2) to obtain, for the extension of one spring:

F A ∆L L

Y=

(1)

Fs k F Fs = N

∆L =

N=

A a2

Fs =

Fa 2 A

∆L =

Fa 2 kA

(2)

Assuming that the springs extend/compress linearly, express the fractional extension of the springs:

∆Ltot ∆L 1 Fa 2 Fa = = = L a a kA kA

Substitute in equation (1) and simplify:

F k Y= A = Fa a kA

(b) From our result in part (a):

k = Ya

From Table 12-1:

Y = 200GN/m 2 = 2 × 1011 N/m 2

Assuming that a ~ 1 nm, evaluate k:

k = 2 × 1011 N/m 2 10 −9 m = 200 N/m

(

)(

)

Static Equilibrium and Elasticity

915

Conditions for Equilibrium 14 • Picture the Problem Let w1 represent the weight of the 28-kg child sitting at the left end of the board, w2 the weight of the 40kg child, and d the distance of the 40-kg child from the pivot point. We can apply the condition for rotational equilibrium to find d. Apply

r

∑ τ = 0 about an axis through

w1 (2 m ) − w2 d = 0

the pivot point P: Solve for and evaluate d:

d=

w1 (2 m ) (28 kg )g (2 m ) = = 1.4 m (40 kg )g w2

15 • Picture the Problem Let F1 represent the force exerted by the floor on Misako’s feet, F2 the force exerted on her hands, and m her mass. We can apply the condition for rotational equilibrium to find F2.

Apply

r

∑ τ = 0 about an axis

F2 (1.5 m ) − mg (0.9 m ) = 0

through point 0: Solve for F2:

F2 =

Substitute numerical values and evaluate F2:

F2 =

mg (0.9 m ) 1.5 m

(54 kg )(9.81m/s2 )(0.9 m ) 1.5 m

= 318 N *16 • Picture the Problem Let F represent the force exerted by Misako’s biceps. To find F we apply the condition for rotational equilibrium about a pivot chosen at the tip of her elbow. Apply

r

∑ τ = 0 about an axis

(5 cm )F − (28 cm)(18 N ) = 0

916 Chapter 12 through the pivot: Solve for F:

F=

(28 cm )(18 N ) = 5 cm

101 N

17 • Picture the Problem Choose a coordinate system in which upward is the positive y direction and to the right is the positive x direction and use the conditions for translational equilibrium. (a) Apply

r

∑ F = 0 to the forces

acting on the tip of the crutch:

∑F

x

= − f s + Fc sin θ = 0

(1)

= Fn − Fc cosθ = 0

(2)

and

∑F

y

Solve equation (2) for Fn and assuming that fs = fs,max, obtain:

f s = f s,max = µs Fn = µs Fc cosθ

Substitute in equation (1) and solve for µs:

µs = tan θ

(b)

Taking long strides requires a large coefficient of static friction because

θ is large for long strides.

(c) If µs is small, i.e., there is ice on the surface, θ must be small to avoid slipping.

The Center of Gravity 18 • Picture the Problem Let the weight of the automobile be w. Choose a coordinate system in which the origin is at the point of contact of the front wheels with the ground and the positive x axis includes the point of contact of the rear wheels with the ground. Apply the definition of the center of gravity to find its location. Use the definition of the center of gravity:

xcgW = ∑ wi xi i

= 0.58w (0) + 0.42w (2 m ) = (0.84 m ) w or, because W = w, xcg (w) = (0.84 m ) w

Static Equilibrium and Elasticity Solve for xcg:

xcg = 0.84 m

*19 • Picture the Problem The figures are shown on the right. The center of mass for each is indicated by a small +. At static equilibrium, the center of gravity is directly below the point of support.

20 •• Picture the Problem Using the coordinate system indicated in the figure, we can apply the definition of the center of gravity to determine xcg and ycg. Apply the definition of the center of gravity to find xcg:

xcgW = ∑ wi xi i

= (40 N )( 12 a ) + (60 N )( 12 a )

+ (30 N )( 32 a ) + (50 N )( 32 a )

= (170 N ) a

or, because W = 180 N,

xcg (180 N ) = (170 N ) a

Solve for xcg:

Apply the definition of the center of gravity to find ycg:

xcg =

170 N a = 0.944a 180 N

ycgW = ∑ wi yi i

= (40 N )( 12 a ) + (60 N )( 32 a )

+ (30 N )( 32 a ) + (50 N )( 12 a )

= (180 N ) a

or, because W = 180 N,

ycg (180 N ) = (180 N ) a

Solve for ycg: The coordinates of the center of gravity are:

ycg = a

(x

cg

, ycg ) =

(0.944a, a )

917

918 Chapter 12 21 •• Picture the Problem Let the origin of the coordinate system be at the lower left corner of the plate and the positive x direction be to the right. Let a and b be the length and width of the plate. Let σ be the mass per unit area of the plate. Then the weight of the plate is given by w = abσg and that of the matter missing from the hole is − πR 2σg . Noting that, by symmetry, ycg = b/2, we can apply the definition of the center of gravity to find xcg. Apply the definition of the center of gravity to find xcg:

xcgW = ∑ wi xi i

(

)

= (abσg )( 12 a ) − πR 2σg (a − R ) or, because

W = wplate − whole = abσg − πR 2σg ,

(

)

xcg abσg − πR 2σg = (abσg )( 12 a )

(

)

− πR 2σ (a − R ) Solve for xcg:

The coordinates of the center of gravity are:

xcg =

1 2

a 2b − πaR 2 + πR 3 ab − πR 2

⎛ 12 a 2b − πaR 2 + πR 3 1 ⎞ (xcg , ycg ) = ⎜⎜ , 2 b ⎟⎟ ab − πR 2 ⎝ ⎠

Some Examples of Static Equilibrium 22 • Picture the Problem We can use the given definition of the mechanical advantage of a lever and the condition for rotational equilibrium to show that M = x/X.

F f

(a) Express the definition of mechanical advantage for a lever:

M =

Apply the condition for rotational equilibrium to the lever:

xf − XF = 0

Solve for the ratio of F to f to obtain:

F x = f X

Substitute to obtain:

M =

x X

A shorter moment arm for the applied force is useful when one wishes to (b) move the load over a large distance using a short movement of the applied force.

Static Equilibrium and Elasticity 23 • Picture the Problem The force diagram r shows the tension in the forestay, TF , the

r

tension in the backstay, TB , the gravitational r force on the mast mg , and the force exerted

r

by the deck, FD . Let the origin of the coordinate system be at the foot of the mast with the positive x direction to the right and the positive y direction upward. Because the mast is in equilibrium, we can apply the conditions for both translational and rotational equilibrium to find the tension in the backstay and the force that the deck exerts on the mast. Apply

r

∑ τ = 0 to the mast about an axis

through its foot and solve for TB:

(4.88 m )(1000 N )sin θ F − (4.88 m )TB sin 45° = 0 and

(1000 N )sin θ F

TB =

sin 45°

Find θF, the angle of the forestay with the vertical:

θ F = tan −1 ⎜⎜

Substitute to obtain:

TB =

Apply the condition for translational equilibrium in the x direction to the mast:

⎛ 2.74 m ⎞ ⎟⎟ = 29.3° ⎝ 4.88 m ⎠

(1000 N )sin 29.3° = sin 45°

∑F

x

692 N

= FD cos θ + TB sin 45° − TF sin θ F = 0

or

FD cosθ = (1000 N )sin29.3°

− (692 N )sin45° ≈0

Apply the condition for translational equilibrium in the y direction to the mast:

∑F

y

= FD sin θ − TF cos θ F − TB cos 45° − mg = 0

or

919

920 Chapter 12 FD sin θ = (1000 N )cos29.3°

+ (692 N ) cos45°

(

+ (120 kg ) 9.81 m/s 2 = 2539 N Because FDcosθ = 0:

)

θ = 90° , FD = 2.54 kN and

no block is required to prevent the mast from moving. 24 •• Picture the Problem The diagram shows r r Mg, the weight of the beam, mg, the weight of the student, and the force the r ledge exerts F , acting on the beam. Because the beam is in equilibrium, we can apply the condition for rotational equilibrium to the beam to find the location of the pivot point P that will allow the student to walk to the end of the beam. Apply

r

∑ τ = 0 about an axis

Mg (5 m − x ) − mgx = 0

through the pivot point P: Solve for x:

x=

5M 5(300 kg ) = = 4.17 m M + m 300 kg + 60 kg

*25 •• Picture the Problem The diagram shows

r r w , the weight of the student, FP , the

force exerted by the board at the pivot, and r Fs , the force exerted by the scale, acting on the student. Because the student is in equilibrium, we can apply the condition for rotational equilibrium to the student to find the location of his center of gravity. Apply

r

∑ τ = 0 about an axis

Fs (2 m ) − wx = 0

Static Equilibrium and Elasticity through the pivot point P: Solve for x:

Substitute numerical values and evaluate x:

(2 m )Fs

x=

w

(2 m )(250 N ) = (70 kg )(9.81m/s2 )

x=

0.728 m

26 •• Picture the Problem The diagram shows

r r mg, the weight of the board, FH , the force r exerted by the hinge, Mg , the weight of r the block, and F , the force acting vertically at the right end of the board. Because the board is in equilibrium, we can apply the condition for rotational equilibrium to it to find the magnitude of

r F.

(a) Apply

r

∑ τ = 0 about an axis

F [(3 m ) cos 30°] − mg [(1.5 m ) cos 30°]

− Mg [(0.8 m ) cos 30°] = 0

through the hinge: Solve for F:

F=

Substitute numerical values and evaluate F:

F=

m(1.5 m ) + M (0.8 m ) g 3m

(5 kg )(1.5 m ) + (60 kg )(0.8 m )

(

× 9.81 m/s 2

3m

)

= 181 N (b) Apply

∑F

y

= 0 to the board:

Solve for and evaluate FH:

FH − Mg − mg + F = 0

FH = Mg + mg − F = (M + m )g − F

(

)

= (60 kg + 5 kg ) 9.81 m/s 2 − 181 N = 457 N

921

922 Chapter 12 (c) The force diagram showing the r force F acting at right angles to the board is shown to the right:

Apply

r

∑ τ = 0 about the hinge:

F (3 m ) − mg [(1.5 m ) cos 30°]

− Mg [(0.8 m ) cos 30°] = 0

Solve for F:

F=

Substitute numerical values and evaluate F:

F=

m(1.5 m ) + M (0.8 m ) g cos 30° 3m

(5 kg )(1.5 m ) + (60 kg )(0.8 m ) 3m

(

)

× 9.81m/s cos 30° 2

= 157 N Apply

∑F

y

= 0 to the board:

FH sin θ − Mg − mg + F cos 30° = 0 or

FH sin θ = (M + m )g − F cos 30° (1)

Apply

∑F

x

= 0 to the board:

FH cosθ − F sin 30° = 0 or

FH cos θ = F sin 30°

Divide the first of these equations by the second to obtain: Solve for θ:

Substitute numerical values and evaluate θ :

FH sin θ (M + m )g − F cos 30° = FH cosθ F sin 30°

⎡ (M + m )g − F cos 30° ⎤ ⎥⎦ F sin 30° ⎣

θ = tan −1 ⎢

(2)

Static Equilibrium and Elasticity

(

)

⎡ (65 kg ) 9.81 m/s 2 − (157 N )cos30° ⎤ ⎥ = 81.1° (157 N )sin30° ⎣ ⎦

θ = tan −1 ⎢

Substitute numerical values in equation (2) and evaluate FH:

FH =

(157 N )sin 30° = cos 81.1°

507 N

*27 • Picture the Problem The planes are frictionless; therefore, the force exerted by each plane must be perpendicular to that

r

plane. Let F1 be the force exerted by the 30°

r

plane, and let F2 be the force exerted by the 60° plane. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the cylinder is in equilibrium, we can use the conditions for translational

r

equilibrium to find the magnitudes of F1 and

r F2 .

Apply

∑F

= 0 to the cylinder:

F1 sin 30° − F2 sin 60° = 0

(1)

Apply

∑F

= 0 to the cylinder:

F1 cos 30° + F2 cos 60° − W = 0

(2)

x

y

Solve equation (1) for F1:

F1 = 3F2

3F2 cos 30° + F2 cos 60° − W = 0

Substitute in equation (2) to obtain: Solve for F2:

(3)

(

)

3 cos 30° + cos 60° F2 = W

or

F2 =

Substitute in equation (3):

W = 3 cos 30° + cos 60°

F1 = 3 ( 12 W ) =

3 2

W

1 2

W

923

924 Chapter 12 28 •• Picture the Problem The force diagram r r r shows the forces FH , T2 , and T1 acting on the strut. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the strut is in equilibrium, we can apply the conditions for translational and rotational equilibrium to it.

(a)

r r r The forces acting on the strut are the tensions T1 and T2 and FH , the force exerted on the strut by the hinge.

(b) Apply

r

∑ τ = 0 about an axis

T2l sin 30° − T1l = 0

through the hinge:

T2v = T2 sin 30° = T1

Solve for T1:

or, because T1 = 80 N,

T2v = 80 N (c) Apply

∑F

x

= 0 to the beam:

FH cosθ − T2 cos 30° = 0 or

FH cosθ = T2 cos 30°

Apply

∑F

y

= 0 to the beam:

(1)

FH sin θ + T2 sin 30° − T1 = 0 or

FH sin θ = T1 − T2 sin 30°

= 80 N − T2 sin 30° Divide equation (2) by equation (1) to obtain: Solve for θ :

Express T2 in terms of T2v:

tan θ =

80 N − T2 sin 30° T2 cos 30° ⎡ 80 N − T2 sin 30° ⎤ ⎥ ⎣ T2 cos 30° ⎦

θ = tan −1 ⎢

T2 =

T2v 80 N = = 160 N sin 30° sin 30°

(2)

Static Equilibrium and Elasticity

925

⎡ 80 N − (160 N )sin 30° ⎤ ⎥ = 0° ⎣ (160 N ) cos 30° ⎦

Evaluate θ:

θ = tan −1 ⎢

Substitute numerical values in equation (1) and evaluate FH:

FH =

(160 N )cos 30° = cos 0°

139 N to the

right. 29 •• Picture the Problem The force diagram r shows the weight of the pirate, Mg, the r weight of the victim, mg, and the force the deck exerts at the edge of the r ship, F acting at the fulcrum P. The diagram also shows, for part (b), the weight of the plank acting through the plank’s center of gravity. (a) Apply

r

∑ τ = 0 at the pivot point P:

Mg (8 m − x ) − mgx = 0 or

M (8 m − x ) − mx = 0

Solve for x:

(b) Apply

x= r

∑ τ = 0 about an axis

Mg (8 m - x ) + mp g (4 m − x ) − mgx = 0

through the pivot point P:

or

Solve for x:

x=

Substitute numerical values and evaluate x:

8M 8(105 kg ) = = 5.00 m M + m 105 kg + 63 kg

M (8 m − x ) + mp (4 m − x ) − mx = 0

x=

8M + 4mp M + m + mp

8(105 kg ) + 4(25 kg ) = 4.87 m 105 kg + 63 kg + 25 kg

926 Chapter 12 30 •• Picture the Problem The drawing shows the door and its two supports. The center of gravity of the door is 0.8 m above (and below) the hinge, and 0.4 m from the hinges horizontally. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Denote the horizontal and vertical components of the hinge force by FHh and FHv. Because the door is in equilibrium, we can use the conditions for translational and rotational equilibrium to determine the horizontal forces exerted by the hinges.

Apply

r

∑ τ = 0 about an axis through

FHh (1.6 m ) − mg (0.4 m ) = 0

the lower hinge: Solve for FHh:

FHh =

Substitute numerical values and evaluate FHh:

FHh =

mg (0.4 m ) 1.6 m

(18 kg )(9.81m/s2 )(0.4 m ) 1.6 m

= 44.1 N Apply

∑F

x

= 0 to the door and

solve for F 'Hh :

F ' Hh −FHh = 0 and

F 'Hh = 44.1 N Note that the upper hinge pulls on the door and the lower hinge pushes on it.

Static Equilibrium and Elasticity

927

31 •• Picture the Problem The figure shows the wheel on the verge of rolling over the edge of the step. Note that, under this condition, the normal force the floor exerts on the wheel is zero. Choose the coordinate system shown in the figure and apply the conditions for translational equilibrium and the result for F from Example 12-4 to the wheel.

Apply

r

∑ F = 0 to the wheel:

∑F

x

= F − F1x = 0

and

∑F

y

r

Write F1 in vector form:

= F1y − Mg = 0

r F1 = − F1x iˆ + F1y ˆj = − Fiˆ + Mgˆj

From Example 12-4 we have:

Substitute to obtain:

F=

Mg h(2 R − h ) R−h

r Mg h(2 R − h ) ˆ F1 = − i + Mgˆj R−h =

Mg h(2 R − h ) ˆ i + Mgˆj h−R

32 •• Picture the Problem The diagram shows

r

r

the forces F1 and F2 acting at the supports, r the weight of the board mg , acting at its center of gravity, and the weight of the r diver Mg acting at the end of the diving board. Because the board is in equilibrium, we can apply the condition for rotational equilibrium to find the forces at the supports. Apply

r

∑ τ = 0 about an axis through

the left support:

(1.2 m )F2 − (2.1m ) mg − (4.2 m )Mg = 0

928 Chapter 12 Solve for F2:

(2.1m ) m + (4.2 m )M (1.2 m )

F2 =

g

Substitute numerical values and evaluate F2:

(2.1m )(30 kg ) + (4.2 m )(70 kg) (9.81m/s2 ) = (1.2 m )

F2 =

Apply

r

∑ τ = 0 about an axis through

2.92 kN, compression

(1.2 m )F1 − (0.9 m ) mg − (3 m )Mg = 0

the right support: Solve for F1:

(0.9 m ) m + (3 m )M (1.2 m )

F1 =

g

Substitute numerical values and evaluate F1:

F1 =

(0.9 m )(30 kg ) + (3 m )(70 kg) (9.81m/s2 ) = (1.2 m )

1.94 kN, tension

33 •• Picture the Problem Let T be the tension in the line attached to the wall and L be the length of the strut. The figure includes w, the weight of the strut, for part (b). Because the strut is in equilibrium, we can use the conditions for both rotational and translational equilibrium to find the force exerted on the strut by the hinge.

(a) Express the force exerted on the strut at the hinge: Ignoring the weight of the strut, r apply τ = 0 at the hinge:

∑

Solve for the tension in the line:

Apply

r

∑ F = 0 to the strut:

r F = Fh iˆ + Fv ˆj

(1)

LT − (L cos 45°)W = 0

T = W cos 45° = (60 N ) cos 45° = 42.43 N

∑F

x

= Fh − T cos 45° = 0

Static Equilibrium and Elasticity

929

and

∑F

y

Solve for Fh:

= Fv + T cos 45° − Mg = 0

Th = T cos 45° = (42.43 N ) cos 45°

= 30.0 N Solve for Fv:

Fv = Mg − T cos 45°

= 60 N − (42.43 N )cos45°

= 30.0 N Substitute in equation (1) to obtain:

r F=

(b) Including the weight of the strut, r apply τ = 0 at the hinge:

⎛L ⎞ LT − (L cos 45°)W − ⎜ cos 45 ⎟ w = 0 ⎝2 ⎠

Solve for the tension in the line:

⎛1 ⎞ T = (cos 45°)W + ⎜ cos 45° ⎟ w ⎝2 ⎠ ⎛1 ⎞ = (cos 45°)(60 N ) + ⎜ cos 45° ⎟ (20 N ) ⎝2 ⎠ = 49.5 N

∑

Apply

r

∑ F = 0 to the strut:

(30.0 N ) iˆ + (30.0 N ) ˆj

∑F

x

= Fh − T cos 45° = 0

and

∑F

y

Solve for Fh:

= Fv + T cos 45° − W − w = 0

Th = T cos 45° = (49.5 N ) cos 45°

= 35.0 N Solve for Fv:

Fv = W + w − T cos 45°

= 60 N + 20 N − (49.5 N )cos45° = 45.0 N

Substitute in equation (1) to obtain:

r F=

(35.0 N ) iˆ + (45.0 N ) ˆj

930 Chapter 12 34 •• Picture the Problem Note that if the 60-kg mass is at the far left end of the plank, T1 and T2 are less than 1 kN. Let x be the distance of the 60-kg mass from T1. Because the plank is in equilibrium, we can apply the condition for rotational equilibrium to relate the distance x to the other distances and forces.

Apply

r

∑ τ = 0 about an axis

(5 m )T2 − (4 m ) mb g − (2.5 m ) mp g − mJ gx = 0

through the left end of the plank: Solve for x:

x=

(5 m )T2 − (4 m ) mb g − (2 m ) mp g mJ g

Substitute numerical values and simplify to obtain:

x=

(5 m )T2 − 3.63 kN ⋅ m

Set T2 = 1 kN and evaluate x:

x=

(5 m )(1kN ) − 3.63 kN ⋅ m = 2.33 m

0.5886 kN

0.5886 kN

and Julie is safe for 0 < x < 2.33 m.

Static Equilibrium and Elasticity 35 •• Picture the Problem The figure to the right shows the forces acting on the cylinder. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the cylinder is in equilibrium, we can apply the conditions for translational and rotational equilibrium to find Fn and the horizontal and vertical components of the force the corner of the step exerts on the cylinder.

(a) Apply

r

∑ τ = 0 to the cylinder

Mgl − Fn l − F (2 R − h ) = 0

about the step’s corner:

F (2 R − h ) l

Solve for Fn:

Fn = Mg −

Express l as a function of R and h:

l = R 2 − (R − h ) = 2 Rh − h 2 2

Fn = Mg −

F (2 R − h ) 2 Rh − h 2

= Mg − F

(b) Apply

∑F

x

= 0 to the cylinder:

∑F

Solve for Fc,v:

− Fc ,h + F = 0 Fc ,h = F

Solve for Fc,h:

(c) Apply

2R − h h

y

= 0 to the cylinder:

Fn − Mg + Fc ,v = 0 Fc,v = Mg − Fn

931

932 Chapter 12 Substitute the result from part (a):

⎧ 2R − h ⎫ Fc,v = Mg − ⎨Mg − F ⎬ h ⎭ ⎩ = F

2R − h h

36 •• Picture the Problem The figure to the right shows the forces acting on the cylinder. Because the cylinder is in equilibrium, we can use the condition for rotational equilibrium to express Fn in terms of F. Because, to roll over the step, the cylinder must lift off the floor, we can set Fn = 0 in our expression relating Fn and F and solve for F. Apply

r

∑ τ = 0 about the step’s corner:

Mgl − Fn l − F (2 R − h ) = 0

F (2 R − h ) l

Solve for Fn:

Fn = Mg −

Express l as a function of R and h:

l = R 2 − (R − h ) = 2 Rh − h 2

Substitute to obtain:

Fn = Mg −

2

F (2 R − h ) 2 Rh − h 2

= Mg − F

To roll over the step, the cylinder must lift off the floor, i.e., Fn = 0: Solve for F:

0 = Mg − F

F = Mg

2R − h h 2R − h h

h 2R − h

*37 •• Picture the Problem The diagram shows the forces F1 and F2 that the fencer’s hand exerts on the epee. We can use a condition for translational equilibrium to find the upward force the fencer must exert on the epee when it is in equilibrium and the definition of torque to determine the total torque exerted. In part (c) we can use the conditions for translational and rotational equilibrium to obtain two equations in F1 and

Static Equilibrium and Elasticity

933

F2 that we can solve simultaneously. In part (d) we can apply Newton’s 2nd law in rotational form and the condition for translational equilibrium to obtain two equations in F1 and F2 that, again, we can solve simultaneously.

(a) Letting the upward force exerted by the fencer’s hand be F, apply Fy = 0 to the epee to

F −W = 0

∑

obtain: Solve for and evaluate F:

F = mg = (0.7 kg ) (9.81 m/s 2 ) = 6.87 N

(b) Express the torque due to the weight about the left end of the epee:

τ = lw = (0.24 m )(6.87 N ) = 1.65 N ⋅ m

(c) Apply

∑F

y

= 0 to the epee to

− F1 + F2 − 6.87 N = 0

(1)

obtain: Apply

∑τ

0

= 0 to obtain:

Solve these equations simultaneously to obtain:

− (0.02 m )F1 + (0.12 m )F2 − 1.65 N ⋅ m = 0 F1 = 8.26 N and F2 = 15.1 N . Note that the force nearest the butt of the epee is directed downward and the force nearest the hand guard is directed upward.

38 •• Picture the Problem In the force diagram, the forces exerted by the hinges r r r are Fy , 2 , Fy ,1 , and Fx ,1 where the subscript 1 refers to the lower hinge. Because the gate is in equilibrium, we can apply the conditions for translational and rotational equilibrium to find the tension in the wire and the forces at the hinges.

934 Chapter 12

(a) Apply

r

∑ τ = 0 about an axis

l 1T sin θ + l 2T cos θ − l 1mg = 0

through the lower hinge and perpendicular to the plane of the page: Solve for T:

T=

l 1mg l 1 sin θ + l 2 cos θ

Substitute numerical values and evaluate T:

T=

(1.5 m )(200 N ) (1.5 m )sin 45° + (1.5 m )cos 45°

= 141 N (b) Apply

∑F

x

= 0 to the gate:

Solve for and evaluate Fx,1:

Fx ,1 − T cos 45° = 0 Fx ,1 = T cos 45° = (141 N )cos 45° = 99.7 N

(c) Apply

∑F

y

= 0 to the gate:

Because Fy,1 and Fy,2 cannot be determined independently, solve for and evaluate their sum:

Fy ,1 + Fy , 2 + T sin 45° − mg = 0 Fy ,1 + Fy , 2 = mg − T sin 45° = 200 N − 99.7 N = 100 N

39 ••• Picture the Problem Let T = the tension in the wire; Fn = the normal force of the surface; and fs,max = µsFn the maximum force of static friction. Letting the point at which

Static Equilibrium and Elasticity

935

the wire is attached to the log be the origin, the center of mass of the log is at (−1.838 m, −0.797 m) and the point of contact with the floor is at (−3.676 m, −1.594 m). Because the log is in equilibrium, we can apply the conditions for translational and rotational equilibrium.

Apply

∑F

x

= 0 to the log:

T sin θ − f s,max = 0 or

T sin θ = f s,max = µs Fn

Apply

∑F

y

= 0 to the log:

(1)

T cosθ + Fn − mg = 0 or

T cos θ = mg − Fn

Divide equation (1) by equation (2) to obtain:

µs Fn T sin θ = T cos θ mg − Fn or

θ = tan −1

Apply

r

∑ τ = 0 about an axis

µs mg −1 Fn

l 2 mg − l 1 Fn − l 3 µs Fn = 0

through the origin: Solve for Fn:

(2)

Fn =

l 2 mg l 1 + l 3 µs

(3)

936 Chapter 12

(

)

Substitute numerical values and evaluate Fn:

1.838(100 kg ) 9.81 m/s 2 Fn = = 389 N 3.676 + 1.594(0.6 )

Substitute in equation (3) and evaluate θ:

θ = tan −1

0.6 (100 kg ) 9.81m/s 2 − 1 389 N

(

)

= 21.5° Substitute numerical values in equation (1) and evaluate T:

T=

(0.6)(389 N ) = sin21.5°

636 N

40 ••• Picture the Problem Consider what happens just as θ increases beyond θ max. Because the top of the block is fixed by the cord, the block will in fact rotate with only the lower right edge of the block remaining in contact with the plane. It follows that just prior to this slipping, Fn and fs = µsFn act at the lower right edge of the block. Choose a coordinate system in which up the incline is the positive x r direction and the direction of Fn is the positive y direction. Because the block is in equilibrium, we can apply the conditions for translational and rotational equilibrium. Apply

∑F

= 0 to the block:

T + µs Fn − mg sin θ = 0

(1)

Apply

∑F

= 0 to the block:

Fn − mg cos θ = 0

(2)

Apply

∑ τ = 0 about an axis

x

y

r

1 2

a(mg cosθ ) + 12 b(mg sin θ ) − bT = 0 (3)

through the lower right edge of the block: Eliminate Fn between equations (1) and (2) and solve for T:

T = mg (sin θ − µs cos θ )

Static Equilibrium and Elasticity Substitute for T in equation (3):

1 2

a(mg cosθ ) + 12 b(mg sin θ )

Substitute 4a for b:

1 2

a(mg cos θ ) + 12 (4a )(mg sin θ )

Simplify to obtain:

(1 + 8µs )cosθ − 4 sin θ = 0

− b[mg (sin θ − µ s cosθ )] = 0

− (4a )[mg (sin θ − µs cos θ )] = 0

Solve for θ:

θ = tan −1

1 + 8µ s 4

Substitute numerical values and evaluate θ :

θ = tan −1

1 + 8(0.8) = 61.6° 4

*41 •• Picture the Problem The free-body diagram shown to the left below is for the weight and the diagram to the right is for the boat. Because both are in equilibrium under the influences of the forces acting on them, we can apply a condition for translational equilibrium to find the tension in the chain.

(a) Apply

∑F

x

= 0 to the boat:

Solve for T:

Apply

Fd − T cos θ = 0 T=

∑F

y

= 0 to the weight:

Substitute for T to obtain: Solve for θ :

Fd cos θ

2T sin θ − 100 N = 0 2 Fd tan θ − 100 N = 0

θ = tan −1

100 N 2 Fd

(1)

937

938 Chapter 12 Substitute for Fd and evaluate θ :

Solve equation (1) for T:

Substitute for θ and evaluate T:

θ = tan −1

100 N = 45° 2(50 N )

T=

100 N 2 sin θ

T=

100 N = 70.7 N 2 sin 45°

(b) Use the diagram to the right to relate the sag ∆y in the chain to the angle θ the chain makes with the horizontal:

∆y 1 2 L

sin θ =

where L is the length of the chain. Solve for ∆y:

∆y = 12 L sin θ

Because the horizontal and vertical forces in the chain are equal, θ = 45° and:

∆y =

(c) Relate the distance d of the boat from the dock to the angle θ the chain makes with the horizontal:

1 2

(5 m )sin 45° =

cos θ =

1 2 1 2

1.77 m

d d = L L

Solve for and evaluate d:

d = L cos θ = (5 m ) cos 45° = 3.54 m

(d) Relate the resultant tension in the chain to the vertical component of the tension Fv and the maximum drag force exerted on the boat by the water Fd,max:

Fv2 + Fd,2max = (500 N )

Solve for Fd,max:

Fd,max =

(500 N )2 − Fv2

Because the vertical component of the tension is 50 N:

Fd,max =

(500 N )2 − (50 N )2

2

= 497 N

42 •• Picture the Problem Choose a coordinate system in which the positive x axis is along the rod and the positive y direction is normal to the rod. The rod and the forces acting on

Static Equilibrium and Elasticity

939

it are shown in the free-body diagram. The forces acting at the supports are denoted by the numerals 1 and 2. The resultant forces at the supports are shown as dashed lines. We’ll assume that the rod is on the verge of sliding. Because the x components of the forces at the supports are friction forces, they are proportional to the normal, i.e., y, components of the forces at the supports. Because the rod is in equilibrium, we can apply the conditions for translational and rotational equilibrium.

Apply

r

∑ τ = 0 about an axis

l 2 F2, y − l 1mg cos θ = 0

through the support at x = 2 m: Solve for F2,y:

Substitute numerical values and evaluate F2,y:

F2, y =

F2, y

l 1mg cos θ l2

( 3 m )(20 kg ) (9.81m/s 2 )cos 30° = 4m

= 127.4 N Apply

r

∑ τ = 0 about an axis

(l 2 − l 1 ) mg cosθ − l 2 F1, y = 0

through the support at x = 6 m: Solve for F1,y:

Substitute numerical values and evaluate F1,y:

F1, y =

F1, y =

(l 2 − l 1 ) mg cosθ l2

(4 m − 3 m )(20 kg )(9.81m/s 2 ) 4m × cos 30°

= 42.48 N

940 Chapter 12 Apply

∑F

x

= 0 to the rail:

Assuming that the rod is on the verge of sliding and that the coefficient of static friction is the same for both supports:

F1, x + F2, x − mg sin 30° = 0

(1)

F1,x = µ s F1, y and

F2,x = µ s F2, y

Divide the first of these equations by the second and evaluate this ratio to obtain:

F1, x F1, y 42.48 N 1 = = = F2, x F2, y 127.4 N 3

Solve for F2,x:

F2, x = 3F1, x

Substitute in equation (1):

F1, x + 3F1, x − mg sin θ = 0

Solve for F1,x:

F1, x = 14 mg sin θ

Substitute numerical values and evaluate F1,x:

F1, x =

1 4

(20 kg )(9.81m/s 2 )sin 30°

= 24.53 N

F2, x = 3(24.53 N ) = 73.58 N

Evaluate F2,x: Find the angle θ1 the force at support 1 (x = 2 m) makes with the rod: Find the angle θ2 the force at support 2 makes with the rod:

r

Find the magnitude of F1 :

θ1 = tan −1

θ 2 = tan −1

F1, y F1, x

F2, y F2,x

= tan −1

42.48 N = 60.0° 24.53 N

= tan −1

127.4 N = 60.0° 73.58 N

F1 = F1,2x + F1,2y =

(24.53 N )2 + (42.48 N )2

= 49.1 N

Static Equilibrium and Elasticity

r

Find the magnitude of F2 :

941

F2 = F22,x + F22, y =

(73.58 N )2 + (127.4 N )2

= 147 N 43 • Picture the Problem The forces shown in the figure constitute a couple and will cause the plate to experience a counterclockwise angular acceleration. We can find this net torque by expressing the torque about either of the corners of the plate. Sum the torques about an axis through the upper left corner of the plate to obtain:

τ net = b[(80 N ) cos 30°] − a[(80 N )sin 30°] =

(69.3 N )b − (40.0 N ) a

44 • Picture the Problem We can use the condition for translational equilibrium and the definition of a couple to show that the force of static friction exerted by the surface and the applied force constitute a couple. We can use the definition of torque to find the torque exerted by the couple. We can use our result from (b) to find the effective point of application of the normal force when F = Mg/3 and the condition for rotational r equilibrium to find the greatest magnitude of F for which the cube will not tip. (a) Apply

r

∑F

x

= 0 to the stationary

r r F + fs = 0

cube:

r r ∴F = − f s and this pair of equal, parallel, and oppositely directed forces constitute a couple. The torque of the couple is:

τ couple = Fa

(b) Let x = the distance from the point of application of Fn to the center of the cube. Now, Fn = Mg, r τ = 0 to the cube so applying

Mgx − Fa = 0

∑

yields: Substitute for F = Mg/3 to obtain:

or

x=

Fa Mg

Mg a a x= 3 = Mg 3

(1)

942 Chapter 12 Mgx a

(c) Solve equation (1) for F:

F=

Noting that xmax = a/2, substitute to express the condition that the cube will tip:

Mgxmax F> = a

a 2 = Mg 2 a

Mg

45 •• Picture the Problem We can find the perpendicular distance between the lines of action of the two forces by following the outline given in the problem statement. Express the vertical components of the forces:

F cos 30° =

Express the horizontal components of the forces:

F sin 30° =

Express the net torque acting on the plate:

τ net =

Letting D be the moment arm of the couple, express the net torque acting on the plate:

τ net = FD

Equate these two expressions for τnet:

FD = 12 F

Solve for D:

D=

*46 •• Picture the Problem Choose the coordinate system shown in the diagram and let x be the coordinate of the thrust point. The diagram to the right shows the forces acting on the wall. The normal force must balance out the weight of the wall and the vertical component of the thrust from the arch and the frictional force must balance out the horizontal component of the thrust. We can apply the conditions for translational equilibrium to find f and Fn and the condition for rotational equilibrium to find the distance x from the origin of our coordinate system at which Fn acts.

3 F 2 F 2

3 1 1 Fb − Fa = F 2 2 2

1 2

(

(

3b − a

3b − a

)

)

(

3b − a

)

943

Static Equilibrium and Elasticity

∑F

(a) Apply the conditions for translational equilibrium to the wall to obtain:

x

= − f + Farch cos θ = 0

(1)

= Fn − mg − Farch sin θ = 0

(2)

and

∑F

y

Solve equation (1) for and evaluate f:

(

)

f = Farch cosθ = 2 × 10 4 N cos 30° = 17.3 kN

Solve equation (2) for Fn:

Fn = mg + Farch sin θ

Substitute numerical values and evaluate Fn:

Fn = 3 × 10 4 kg 9.81 m/s 2

(

(

)(

)

)

+ 2 × 10 4 N sin 30° = 304 kN

Apply

∑τ

z axis

= 0 to the to the wall:

Solve for x:

xFn − 12 wmg − hFarch cos θ = 0 x=

1 2

wmg + hFarch cosθ Fn

Substitute numerical values and evaluate x:

x=

1 2

(1.25 m ) (3 × 104 kg )(9.81 m/s 2 )+ (10 m ) (2 × 104 N )cos 30° = 304 kN

0.570 m

(b)

If there were no thrust on the side of the wall, the normal force would act through the center of mass, so making the weight larger compared to the thrust must move the point of action of the normal force closer to the center. 47 •• Picture the Problem Let h be the height of the structure, T be the thrust, θ the angle from the horizontal of the thrust, m′g the weight of the wall above height y, N(x) the normal force, f the friction force the lower part of the wall exerts on the upper part, and w the width of the structure. We can apply the conditions for translational and rotational equilibrium to the portion of the wall above the point at which the thrust is applied to obtain two equations that we can solve simultaneously for x.

944 Chapter 12 Apply

∑F

y

N ( x ) − T sin θ − m'g = 0

= 0 to that fraction of the

wall above height y:

m'g mg = h− y h

Assuming the wall is of uniform density, express m′g in terms of mg:

and

⎛ y⎞ m'g = mg ⎜1 − ⎟ ⎝ h⎠ Substitute to obtain:

⎛ y⎞ N ( x ) − T sin θ − mg ⎜1 − ⎟ = 0 ⎝ h⎠

Solve for N(x):

⎛ y⎞ N ( x ) = T sin θ + mg ⎜1 − ⎟ ⎝ h⎠

Apply

xN (x ) − (h − y )T cosθ

r

∑ τ = 0 about an axis

⎛ y⎞ − 12 mgw ⎜1 − ⎟ = 0 ⎝ h⎠

through (0,y) and perpendicular to the xy plane to obtain: Solve for x to obtain:

x=

Substitute for N(x) to obtain:

x=

1 2

mgw + hT cosθ ⎛ y ⎞ ⎜1 − ⎟ N (x ) ⎝ h⎠

( 12 mgw + hT cosθ )⎛⎜1 − y ⎞⎟ ⎝

⎛ y⎞ T sin θ + mg ⎜1 − ⎟ ⎝ h⎠

h⎠

Substitute numerical values and simplify to obtain:

[ (3 ×10 kg )(9.81m/s )(1.25 m ) + (10 m )(2 ×10 N )cos 30°]⎛⎜⎜1 − 10ym ⎞⎟⎟ 1 2

x=

=

4

2

4

⎝ ⎛ y ⎞ ⎟⎟ 2 × 10 4 N sin 30° + 3 × 10 4 kg 9.81 m/s 2 ⎜⎜1 − ⎝ 10 m ⎠

(

)

(

)(

)

35.71 m − 3.571y 30.43 − 2.943 m -1 y

Solve for y:

(

)

y=

35.71 m − 30.43x 3.571 − 2.943 m -1 x

(

)

⎠

Static Equilibrium and Elasticity The graph shown below was plotted using a spreadsheet program:

12 10

y (m)

8 6 4 2 0 0.0

0.2

0.4

0.6

0.8

1.0

1.2

x (m)

Ladder Problems *48 •• Picture the Problem The ladder and the forces acting on it at the critical moment of slipping are shown in the diagram. Use the coordinate system shown. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium. Using its definition, express µs:

Apply

r

∑ τ = 0 about the bottom of

the ladder:

µs =

f s,max Fn

[(9 m )cosθ ]Mg + [(5 m )cosθ ]mg − [(10 m )sin θ ] FW = 0 (9 m ) M + (5 m ) m g cosθ (10 m )sin θ

Solve for FW:

FW =

Find the angle θ:

θ = cos −1

2.8 m = 73.74° 10 m

(1)

945

946 Chapter 12 Evaluate FW:

FW =

(9 m )(70 kg ) + (5 m )(22 kg ) (10 m )sin 73.74°

(

)

× 9.81 m/s 2 cos 73.74° = 211.7 N Apply

∑F

x

= 0 to the ladder and

solve for fs,max:

Apply

∑F

y

= 0 to the ladder:

FW − f s, max = 0 and

f s,max = FW = 211.7 N

Fn − Mg − mg = 0

Solve for Fn:

Fn = (M + m )g

Substitute numerical values in equation (1) and evaluate µs:

µs =

= (70 kg + 22 kg ) (9.81m/s 2 ) = 902.5 N 211.7 N = 0.235 902.5 N

49 •• Picture the Problem The ladder and the forces acting on it are shown in the diagram. Because the wall is smooth, the force the wall exerts on the ladder must be horizontal. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium to it.

Apply

∑F

y

= 0 to the ladder and

Fn − Mg = 0 ⇒ Fn = Mg

= 0 to the ladder and

FW − f s, max = 0 ⇒ f s,max = FW

solve for Fn: Apply

∑F

x

solve for fs,max: Apply

r

∑ τ = 0 about the bottom of

the ladder:

Mgx cosθ − FW L sin θ = 0

Static Equilibrium and Elasticity

947

f L FW L sin θ = s,max tan θ Mg cosθ Mg µFL = s n tan θ = µs L tan θ Mg

Solve for x:

x=

Referring to the figure, relate x to h and solve for h:

sin θ =

h x

and

h = x sin θ = µs L tan θ sin θ 50 •• Picture the Problem The ladder and the forces acting on it are shown in the drawing. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the wall is smooth, the force the wall exerts on the ladder must be horizontal. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium. Apply

∑F

y

= 0 to the ladder and

and

solve for Fn:

Apply

∑F

x

Fn = 5mg = 0 to the ladder and

solve for fs,max:

Apply

Fn − mg − 4mg = 0

r

∑ τ = 0 about an axis

FW − f s, max = 0 and

f s,max = FW

mg

L cosθ + 4mgl cosθ − FW L sin θ = 0 2

l=

5µs mgL sin θ − 12 mgL cosθ 4mg cosθ

through the bottom of the ladder: Substitute for FW and then fs,max and solve for l:

948 Chapter 12 Simplify to obtain:

1⎞ ⎛ 5µ l = ⎜ s tan θ − ⎟ L 8⎠ ⎝ 4 1⎞ ⎛ 5(0.45) =⎜ tan 60° − ⎟ L 8⎠ ⎝ 4 = 0.849 L i.e., you can climb about 85% of the way to the top of the ladder.

51 •• Picture the Problem The ladder and the forces acting on it are shown in the figure. Because the ladder is separating from the wall, the force the wall exerts on the ladder is zero. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium.

To find the force required to pull the ladder away from the wall, apply r τ = 0 about an axis through the

∑

bottom of the ladder:

Solve for F:

L L cos θ − F sin θ = 0 2 2 L h or, because cos θ = , 2 tan θ mgh L − F sin θ = 0 tan θ 2 mg

F=

2mgh L tan θ sin θ

(1)

Apply

∑F

= 0 to the ladder:

F − f s,max = 0 ⇒ F = f s,max = µ s Fn (2)

Apply

∑F

= 0 to the ladder:

Fn − mg = 0 ⇒ Fn = mg

x

y

Equate equations (1) and (2) and substitute for Fn to obtain:

µs mg =

Solve for µs:

µs =

2mgh L tan θ sin θ

2h L tan θ sin θ

Static Equilibrium and Elasticity 52 •• Picture the Problem Assume that half the man’s weight acts on each side of the ladder. The force exerted by the frictionless floor must be vertical. D is the separation between the legs at the bottom and x is the distance of the cross brace from the apex. Because each leg of the ladder is in equilibrium, we can apply the condition for rotational equilibrium the right leg to relate the tension in the cross brace to its distance from the apex. (a) By symmetry, each leg carries half the total weight. So the force on each leg is: (b) Consider one of the ladder’s r legs and apply τ = 0 about the

∑

apex: Solve for T:

450 N

Fn

D − Tx = 0 2

T=

Fn D 2x D2 h

Using trigonometry, relate h and θ through the tangent function:

tan 12 θ =

Solve for D to obtain:

D = 2h tan 12 θ

Substitute and simplify to obtain:

T=

2 Fn h tan 12 θ Fn h tan 12 θ = x 2x

Substitute numerical values and evaluate T:

T=

Fn h tan 12 θ x

Apply

∑F

y

= 0 to the ladder and

Fn − 12 w = 0 and Fn = 12 w

solve for Fn: Substitute to obtain:

T=

wh tan 12 θ 2x

(1)

949

950 Chapter 12 Substitute numerical values and evaluate T:

T=

(900 N )(4 m ) tan 15° = 2(2 m )

241 N

T will decrease.

(c) From equation (1) we can see that, if x is increased, i.e., the brace moved lower: 53 •• Picture the Problem The figure shows the forces acting on the ladder. Because the wall is frictionless, the force the wall exerts on the ladder is perpendicular to the wall. Because the ladder is on the verge of slipping, the static friction force is fs,max. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium. Apply

∑F

= 0 to the ladder:

FW − f s,max = 0 ⇒ FW = f s,max = µs Fn

Apply

∑F

= 0 to the ladder:

Fn − mg = 0 ⇒ Fn = mg

Apply

∑ τ = 0 about an axis

x

y

r

mg

through the bottom of the ladder: Substitute for FW and Fn and simplify to obtain: Solve for and evaluate θ:

1 2

L cos θ − LFW sin θ = 0 2

cos θ − µs sin θ = 0

θ = tan −1

1 1 = tan −1 = 59.0° 2 µs 2(0.3)

Stress and Strain *54 • Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire ∆L is related to Young’s modulus by Y = (F A) (∆L L ). We can use Table 12-1 to find the numerical value of Young’s modulus for steel. Find the amount the wire is stretched from Young’s modulus:

Y=

F A ∆L L

951

Static Equilibrium and Elasticity

Solve for ∆L:

∆L =

FL YA

Substitute for F and A to obtain:

∆L =

mgL Yπ r 2

Substitute numerical values and evaluate ∆L:

∆L =

(50 kg )(9.81m/s 2 )(5 m )

(

2π × 1011 N/m 2 2 × 10 −3 m

)

2

= 0.976 mm 55 • Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire ∆L is related to Young’s modulus by

Y = stress strain = (F A) (∆L L ).

(a) Express the maximum load in terms of the wire’s breaking stress:

Fmax = breaking stress × A = breaking stress × π r 2

(

) (

Substitute numerical values and evaluate Fmax:

Fmax = 3 ×108 N/m 2 π 0.21× 10 −3 m

(b) Using the definition of Young’s modulus, express the fractional change in length of the copper wire:

∆L L =

)

2

= 41.6 N F A 1.5 × 108 N/m 2 = 1.1× 1011 N/m 2 Y

= 1.36 × 10−3 = 0.136%

56 • Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire ∆L is related to Young’s modulus by Y = (F A) (∆L L ). We can use Table 12-1 to find the numerical value of Young’s modulus for steel.

F A ∆L L

Find the amount the wire is stretched from Young’s modulus:

Y=

Solve for ∆L:

∆L =

FL YA

952 Chapter 12 Substitute for F and A to obtain:

∆L =

Substitute numerical values and evaluate ∆L:

∆L =

mgL Yπ r 2

(4 kg )(9.81m/s2 )(1.2 m )

(

2π × 1011 N/m 2 0.3 × 10 −3 m

)

2

= 0.833 mm *57 • Picture the Problem The shear stress, defined as the ratio of the shearing force to the area over which it is applied, is related to the shear strain through the definition of the shear modulus; M s =

shear stress Fs A = . shear strain tan θ Fs Ms A

Using the definition of shear modulus, relate the angle of shear, θ to the shear force and shear modulus:

tan θ =

Solve for θ :

θ = tan −1

Fs Ms A

Substitute numerical values and evaluate θ :

θ = tan −1

25 N 1.9 × 10 N/m 2 15 × 10−4 m 2

(

5

)(

)

= 5.01° 58 •• Picture the Problem The stretch in the wire ∆L is related to Young’s modulus by Y = (F A) (∆L L ) , where L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. For a composite wire, the length under stress is the unstressed length plus the sum of the elongations of the components of the wire. Express the length of the composite wire when it is supporting a mass of 5 kg: Express the change in length of the composite wire:

L = 3.00 m + ∆L

∆L = ∆Lsteel + ∆LAl =

F Lsteel F LAl + A Ysteel A YAl

=

F ⎛ Lsteel LAl ⎞ ⎟ ⎜ + A ⎜⎝ Ysteel YAl ⎟⎠

(1)

Static Equilibrium and Elasticity

Find the stress in each wire:

(

F (5 kg ) 9.81 m/s 2 = A π 0.5 × 10 −3 m 2

(

)

953

)

= 6.245 × 107 N/m 2 Substitute numerical values and evaluate ∆L:

⎛ ⎞ 1.5 m 1.5 m ⎟ = 1.81× 10 −3 m ∆L = (6.245 × 107 N/m 2 ) ⎜⎜ + 11 2 11 2 ⎟ ⎝ 2 × 10 N/m 0.7 × 10 N/m ⎠ Substitute in equation (1) and evaluate L:

L = 3.00 m + 1.81×10 −3 m = 3.0018 m

59 •• Picture the Problem We can use Hooke’s law and Young’s modulus to show that, if the wire is considered to be a spring, the force constant k is given by k = AY/L. By treating the wire as a spring we can show the energy stored in the wire is U = ½F∆L. Express the relationship between the stretching force, the stiffness constant, and the elongation of a spring:

F = k∆L or

k=

F ∆L

Using the definition of Young’s modulus, express the ratio of the stretching force to the elongation of the wire:

F AY = ∆L L

Equate these two expressions for F/∆L to obtain:

k=

Treating the wire as a spring, express its stored energy:

U = 12 k (∆L ) =

Solve equation (1) for F:

F=

AY∆L L

Substitute in our expression for U to obtain:

U=

1 2

(1)

AY L 2

1 2

AY∆L ∆L = L

AY (∆L ) 2 L

1 2

F∆L

954 Chapter 12 60 •• Picture the Problem Let L′ represent the stretched and L the unstretched length of the wire. The stretch in the wire ∆L is related to Young’s modulus by Y = (F A) (∆L L ) , where F is the force acting on it, and A is its cross-sectional area. In problem 58 we showed that the energy stored in the wire is U = ½F∆L, where Y is Young’s modulus and ∆L is the amount the wire has stretched. (a) Express the stretched length of the wire:

L' = L + ∆L

Using the definition of Young’s modulus, express ∆L:

∆L =

Substitute and simplify:

L' = L +

Solve for L:

L=

Substitute numerical values and evaluate L:

L=

LF AY LF F ⎞ ⎛ = L⎜1 + ⎟ AY ⎝ AY ⎠

L' F 1+ AY

1+

0.35 m 53 N

(

π 0.1× 10 −3 m

) (2 ×10 2

11

N/m 2

= 0.347 m (b) Using the expression from Problem 59, express the work done in stretching the wire:

*61 •• Picture the Problem The table to the right summarizes the ratios ∆L/F for the student’s data. Note that this ratio is constant, to three significant figures, for loads less than or equal to 200 g. We can use this ratio to calculate Young’s modulus for the rubber strip.

W = ∆U = 12 F∆L =

1 2

(53 N )(0.35 m − 0.347 m )

= 0.0795 J

Load (g) 100 200 300 400 500

F (N) 0.981 1.962 2.943 3.924 4.905

∆L (m) 0.006 0.012 0.019 0.028 0.05

∆L/F (m/N 6.12×10−3 6.12×10−3 6.46×10−3 7.14×10−3 10.2×10−3

)

Static Equilibrium and Elasticity (a) Referring to the table, we see that for loads ≤ 200 g:

∆L = 6.12 ×10 −3 m/N F

Use the definition of Young’s modulus to express Y:

Y=

FL L = A∆L A ∆L F

Substitute numerical values and evaluate Y:

Y=

5 ×10 −2 m = 1.82 ×106 N/m 2 −3 −3 −3 (3 ×10 m) (1.5 ×10 m ) (6.12 ×10 m/N )

(b) Interpolate to determine the stretch when the load is 150 g, and use the expression from Problem 58, to express the energy stored in the strip:

U = 12 F∆L =

1 2

(0.15 kg )(9.81m/s 2 )(9 ×10−3 m )

= 6.62 mJ

62 •• Picture the Problem The figure shows the forces acting on the wire where it passes over the nail. m represents the mass of the mirror and T is the tension in the supporting wires. The figure also shows the geometry of the right triangle defined by the support wires and the top of the mirror frame. The distance a is fixed by the geometry while h and L will change as the mirror is suspended from the nail. Express the distance between the nail and the top of the frame when the wire is under tension: Apply

∑F

y

= 0 to the wire where it

h' = h + ∆h = 0.4 m + ∆h

mg − 2T cosθ = 0

passes over the supporting nail: Solve for the tension in the wire:

T=

mg 2 cosθ

(1)

955

956 Chapter 12 Substitute numerical values and evaluate T:

Using its definition, find the stress in the wire:

( 2.4 kg ) (9.81 m/s 2 ) = 25.0 N T= ⎛ 0.4 m ⎞ ⎟⎟ 2⎜⎜ ⎝ 0.85 m ⎠

stress =

T 25.0 N = A π 0.1× 10 −3 m

(

)

2

= 7.96 × 108 N/m 2

∆L stress = L Y 7.96 × 108 N/m 2 = = 3.98 × 10 −3 11 2 2 × 10 N/m

Using the definition of Young’s modulus, find the strain in the hypotenuse of the right triangle shown in the figure:

strain =

Using the Pythagorean theorem, express the relationship between the sides of the right triangle in the figure:

a 2 + h 2 = L2

Express the differential of this equation:

2 a ∆a + 2 h∆h = 2 L ∆L or, because ∆a = 0,

h∆h = L∆L Solve for and evaluate ∆h:

L∆L L2 ∆L ∆h = = ⋅ h h L

Substitute numerical values and evaluate ∆h:

∆h =

Substitute in equation (1) to obtain:

h' = 0.4 m + 7.19 mm

(0.85 m ) 2 (3.98 ×10−3 ) = 7.19 mm 0.4 m

= 40.72 cm 63 •• Picture the Problem Let the numeral 1 denote the aluminum wire and the numeral 2 the steel wire. Because their initial lengths and amount they stretch are the same, we can use the definition of Young’s modulus to express the change in the lengths of each wire and then equate these expressions to obtain an equation solvable for the ratio M1/M2. Using the definition of Young’s modulus, express the change in

∆L1 =

M 1 gL1 A1YAl

957

Static Equilibrium and Elasticity length of the aluminum wire:

M 2 gL2 A2Ysteel

Using the definition of Young’s modulus, express the change in length of the steel wire:

∆L2 =

Because the two wires stretch by the same amount, equate ∆L1 and ∆L2 and simplify:

M1 M2 = A1YAl A2Ysteel

Solve for the ratio M1/M2:

M1 AY = 1 Al M 2 A2Ysteel

Substitute numerical values and evaluate M1/M2:

π

(

(0.7 mm) 0.7 ×1011 N/m 2 M1 4 = π M2 (0.5 mm)2 2 ×1011 N/m 2 4 (0.7 mm)2 0.7 ×1011 N/m 2 = (0.5 mm)2 2 ×1011 N/m 2 2

(

(

(

)

)

= 0.686 64 •• Picture the Problem The free-body diagram shows the forces acting on the ball as it rotates around the post in a horizontal plane. We can apply Newton’s 2nd law to find the tension in the wire and use the definition of Young’s modulus to find the amount by which the aluminum wire stretches. Express the length of the wire under tension to its unstretched length: Apply

∑F

y

= 0 to the ball:

Solve for the tension in the wire:

L = L0 + ∆L = 0.7 m + ∆L

T sin θ − mg = 0 T=

mg sin θ

(1)

)

)

958 Chapter 12 Substitute numerical values and evaluate T:

( 0.5 kg ) (9.81 m/s 2 ) T= = sin5°

Using the definition of Young’s modulus, express ∆L:

∆L =

Substitute numerical values and evaluate ∆L:

∆L =

56.3 N

FL AY

(56.3 N )(0.7 m )

π

(1.6 ×10 4

−3

m

) (0.7 ×10 2

11

N/m 2

)

= 0.280 mm Substitute in equation (1) to obtain:

L = 0.7 m + 0.280 mm = 70.03 cm

*65 •• Picture the Problem We can use the definition of stress to calculate the failing stress of the cable and the stress on the elevator cable. Note that the failing stress of the composite cable is the same as the failing stress of the test sample. Express the stress on the elevator cable:

Stress cable =

F 20 kN = A 1.2 × 10 − 6 m 2

= 1.67 × 1010 N/m 2 Express the failing stress of the sample:

Stress failing =

F 1 kN = A 0.2 ×10 − 6 m 2

= 0.500 × 1010 N/m 2

Because Stress failing < Stress cable , it will not support the elevator. *66 ••• Picture the Problem Let the length of the sides of the rectangle be x, y and z. Then the volume of the rectangle will be V = xyz and we can express the new volume V ′ resulting from the pulling in the x direction and the change in volume ∆V in terms of ∆x, ∆y, and ∆z. Discarding the higher order terms in ∆V and dividing our equation by V and using the given condition that ∆y/y = ∆z/z will lead us to the given expression for ∆y/y. Express the new volume of the rectangular box when its sides change in length by ∆x, ∆y, and ∆z:

V' = (x + ∆x )( y + ∆y )(z + ∆z ) = xyz + ∆x( yz ) + ∆y (xz ) + ∆z (xy ) + {z∆x∆y + y∆x∆z + x∆y∆z + ∆x∆y∆z} where the terms in brackets are very small (i.e., second order or higher).

Static Equilibrium and Elasticity

959

V' = V + ∆x( yz ) + ∆y ( xz ) + ∆z ( xy )

Discard the second order and higher terms to obtain:

or

Because ∆V = 0:

∆x( yz ) = −[∆y ( xz ) + ∆z ( xy )]

Divide both sides of this equation by V = xyz to obtain:

⎡ ∆y ∆z ⎤ ∆x = −⎢ + ⎥ x z ⎦ ⎣ y

Because ∆y/y = ∆z/z, our equation becomes:

∆y 1 ∆x ∆x ∆y or = − = −2 x y y 2 x

∆V = V' − V = ∆x( yz ) + ∆y ( xz ) + ∆z ( xy )

67 •• Picture the Problem We can evaluate the differential of the volume of the wire and, using the assumptions that the volume of the wire does not change under stretching and that the change in its length is small compared to its length, show that ∆r/r = −(1/2) ∆L/L. Express the volume of the wire:

V = π r2L

Evaluate the differential of V to obtain:

dV = π r 2 dL + 2π rLdr

Because dV = 0:

Because ∆L << L, we can approximate the differential changes dr and dL with small changes ∆r and ∆L to obtain:

0 = rdL + 2 Ldr ⇒

dr 1 dL =− r 2 L

∆r 1 ∆L = − r 2 L

*68 ••• Picture the Problem Because the volume of the thread remains constant during the stretching process, we can equate the initial and final volumes to express r0 in terms of r. We can also use Young’s modulus to express the tension needed to break the thread in terms of Y and r0. (a) Express the conservation of volume during the stretching of the spider’s silk: Solve for r:

Substitute for L to obtain:

π r 2 L = π r02 L0

r = r0

L0 L

r = r0

L0 = 0.316r0 10 L0

960 Chapter 12 (b) Express Young’s modulus in terms of the breaking tension T: Solve for T to obtain:

Because ∆L/L = 9:

T A T π r 2 10T π r02 Y= = = ∆L L ∆L L ∆L L T=

1 ∆L π r02Y 10 L

T=

9π r02Y 10

General Problems 69 • Picture the Problem Because the board is in equilibrium, we can apply the conditions for translational and rotational equilibrium to determine the forces exerted by the supports. Apply

r

∑τ i

i

= 0 about the right support:

Solve for and evaluate FL:

(2 m )(360 N ) + (5 m )(90 N ) − (10 m )FL = 0 FL =

(2 m )(360 N ) + (5 m )(90 N ) 10 m

= 117 N Apply

∑F

y

= 0 to the board:

Solve for and evaluate FR:

FL + FR − 90 N − 360 N = 0 FR = − FL + 90 N + 360 N

= −117 N + 90 N + 360 N = 333 N Remarks: We could just as easily found FR by applying

r

∑ τ = 0 about the left

support. 70 • Picture the Problem Because the man-and-board system is in equilibrium, we can apply the conditions for translational and rotational equilibrium to determine the forces exerted by the supports. Let d represent the distance from the man’s feet to his center of gravity. Apply

r

∑ τ = 0 about an axis

through the man’s feet and perpendicular to the page:

(845 N ) d − (1.88 m )(445 N ) = 0

Static Equilibrium and Elasticity Solve for and evaluate d:

d=

961

(1.88 m )(445 N ) = 0.990 m 845 N

= 99.0 cm No. Holding his head slightly above the board would not change the location of his center of mass and so the scale readings would not change. *71 • Picture the Problem We can apply the balance condition

r

∑ τ = 0 successively, starting

with the lowest part of the mobile, to find the value of each of the unknown weights. Apply

r

∑ τ = 0 about an axis

(3 cm)(2 N ) − (4 cm) w1 = 0

through the point of suspension of the lowest part of the mobile: Solve for and evaluate w1:

Apply

r

∑ τ = 0 about an axis

w1 =

(3 cm )(2 N ) = 4 cm

1.50 N

(2 cm) w2 − (4 cm)(2 N + 1.5 N ) = 0

through the point of suspension of the middle part of the mobile: Solve for and evaluate w2:

Apply

r

∑ τ = 0 about an axis

w2 =

(4 cm )(2 N + 1.5 N ) = 2 cm

7.00 N

(2 cm)(10.5 N ) − (6 cm) w3 = 0

through the point of suspension of the top part of the mobile: Solve for and evaluate w3:

w3 =

(2 cm )(10.5 N ) = 6 cm

3.50 N

72 • Picture the Problem We can determine the ratio of L to h by noting the number of ropes supporting the load whose mass is M. (a) Noting that three ropes support the pulley to which the object whose mass is M is fastened we can

L = 3 h

962 Chapter 12 conclude that: (b) Apply the work-energy principle to the block-tackle object to obtain:

Wext = ∆Esystem = ∆U block-tackle or

FL = Mgh 73 •• Picture the Problem The figure shows the equilateral triangle without the mass m, and then the same triangle with the mass m and rotated through an angle θ. Let the side length of the triangle to be 2a. Then the center of mass of the triangle is at a distance of

2a from each vertex. As the 3

triangle rotates, its center of mass shifts by

2a θ, for θ << 1. Also, the vertex to 3

which m is attached moves toward the plumb line by the distance d = 2aθ cos30° = 3aθ (see the drawing). Apply

r

∑ τ = 0 about an axis through

(

)

mg a − 3aθ − Mg

the point of suspension: Solve for m/M:

Substitute numerical values and evaluate m/M:

2θ m = M 3 1 − 3θ

(

2a θ =0 3

)

⎛ π rad ⎞ 2(6°)⎜ ⎟ m ⎝ 180° ⎠ = M ⎡ ⎛ π rad ⎞⎤ 3 ⎢1 − 3 (6°)⎜ ⎟⎥ ⎝ 180° ⎠⎦ ⎣ = 0.148

Static Equilibrium and Elasticity 74 •• Picture the Problem If the hexagon is to roll rather than slide, the incline’s angle must be such that the center of mass falls just beyond the support base. From the geometry of the hexagon, one can see that the critical angle is 30°. The free-body diagram shows the forces acting on the hexagonal pencil when it is on the verge of sliding. We can use Newton’s 2nd law to relate the coefficient of static friction to the angle of the incline for which rolling rather than sliding occurs. Apply

r

∑ F = 0 to the pencil:

∑F

x

= mg sin θ − f s,max = 0

(1)

= Fn − mg cos θ = 0

(2)

and

∑F

y

Substitute fs,max = µsFn in equation (1): Divide equation (3) by equation (2) to obtain: Thus, if the pencil is to roll rather than slide when the pad is inclined:

mg sin θ − µs Fn = 0

tan θ = µs

µs ≥ tan 30° = 0.577

75 •• Picture the Problem The box and the forces acting on it are shown in the figure. When the box is about to tip, Fn acts at its edge, as indicated in the drawing. We can use the definition of µs and apply the condition for rotational equilibrium in an accelerated frame to relate fs to the weight of the box and, hence, to the normal force. Using its definition, express µs:

µs ≥

fs Fn

(3)

963

964 Chapter 12 Apply

r

∑ τ = 0 about an axis through

wf s − 12 wFn = 0

the box’s center of mass: Solve for the ratio

fs : Fn

Substitute to obtain the condition for tipping: Therefore, if the box is to slide:

fs 1 = Fn 2

µs ≥ 0.500

µs < 0.500

76 •• Picture the Problem Because the balance is in equilibrium, we can use the condition for rotational equilibrium to relate the masses of the blocks to the lever arms of the balance in the two configurations described in the problem statement. Apply

r

∑ τ = 0 about an axis

(1.5 kg )L1 = (1.95 kg )L2

through the fulcrum: Solve for the ratio L1/L2:

Apply

r

∑ τ = 0 about an axis

L1 1.95 kg = = 1.30 L2 1.5 kg

(1.5 kg )L2 = ML1

through the fulcrum with 1.5 kg at L2 : Solve for and evaluate M:

M =

(1.5 kg )L2 L1

= 1.15 kg

=

1.5 kg 1.5 kg = L1 L2 1.30

Static Equilibrium and Elasticity *77 •• Picture the Problem The figure shows the location of the cube’s center of mass and the forces acting on the cube. The opposing couple is formed by the friction force fs,max and the force exerted by the wall. Because the cube is in equilibrium, we can use the condition for translational equilibrium to establish that f s,max = FW and Fn = Mg and the condition for rotational equilibrium to relate the opposing couples. Apply

r

∑ F = 0 to the cube:

∑F

y

= Fn − Mg = 0 ⇒ Fn = Mg

and

∑F

x

r

r

Noting that f s,max and FW form a

r r couple, as do Fn and Mg , apply r ∑ τ = 0 about an axis though the

= f s − FW = 0 ⇒ FW = f s

f s,max a sin θ − Mgd = 0

center of mass of the cube: Referring to the diagram to the right, note that d =

a sin (45° + θ ) . 2

Substitute for d and fs,max to obtain:

µs Mga sin θ − Mg

a sin (45° + θ ) = 0 2

or

µs sin θ − Solve for µs and simplify to obtain:

1 sin (45° + θ ) = 0 2

965

966 Chapter 12 1 sin (45° + θ ) = 2 sin θ

µs =

1 (sin 45° cosθ + cos 45° sin θ ) 2 sin θ

1 1 1 ⎛ 1 ⎞ (cot θ + 1) cosθ + sin θ ⎟ = ⎜ 2 2 sin θ ⎝ 2 2 ⎠

=

78 •• Picture the Problem Because the meter stick is in equilibrium, we can apply the condition for rotational equilibrium to find the maximum distance from the hinge at which the block can be suspended. Apply

r

∑ τ = 0 about an axis through the hinge to obtain:

(1 m )(75 N ) − (0.5 m )(5 kg ) (9.81 m/s 2 )cos 45° − d (10 kg ) (9.81 m/s 2 )cos 45° = 0

Solve for and evaluate d:

( 1m )(75 N ) − (0.5 m )(5 kg ) (9.81m/s 2 ) d= cos 45° = (10 kg )(9.81m/s2 )cos 45°

0.831 m

79 •• Picture the Problem Let m represent the mass of the ladder and M the mass of the person. The force diagram shows the forces acting on the ladder for part (b). From the condition for translational equilibrium, we can conclude that T = FW, a result we’ll need in part (b). Because the ladder is also in rotational equilibrium, summing the torques about the bottom of the ladder will eliminate both Fn and T. (a) Apply

r

∑τ i

i

= 0 about an axis

through the bottom of the ladder: Solve for and evaluate FW:

(5 m )FW − (0.75 m )(20 kg )(9.81m/s 2 ) − (0.75 m )(80 kg )(9.81 m/s 2 ) = 0 (0.75 m )(20 kg )(9.81m/s2 ) F = W

5m

+ = 147 N

(0.75 m )(80 kg )(9.81m/s2 ) 5m

Static Equilibrium and Elasticity (b) Solve for and evaluate f :

f =

(5 m )(200 N ) − (0.75 m )(20 kg )(9.81m/s2 ) = 0.724 (1.5 m )(80 kg )(9.81m/s2 )

Find the distance the 80-kg person can climb the ladder:

d = f (5 m ) = (0.724 )(5 m ) = 3.62 m

*80 •• Picture the Problem To ″roll″ the cube one must raise its center of mass from y = a/2 to y =

2a 2 , where a is the cube

length. During this process the work done is the change in the gravitational potential energy of the cube. No additional work is done on the cube as it ″flops″ down. We can also use the definition of work to express the work done in sliding the cube a distance a along a horizontal surface and then equate the two expressions to determine µk. Express the work done in moving the cube a distance a by raising its center of mass from y = a/2 to y = 2a 2 and then letting the

⎛ 2a a ⎞ mga − ⎟⎟ = W = mg ⎜⎜ 2 2⎠ 2 ⎝ = 0.207mga

cube flop down: Letting fk represent the kinetic friction force, express the work done in dragging the cube a distance a along the surface at constant speed:

W = f k a = µ k mga

Equate these two expressions to obtain:

µ k = 0.207

(

)

2 −1

967

968 Chapter 12 81 •• Picture the Problem The free-body diagram shows the forces acting on the block when it is on the verge of sliding. Because the block is in equilibrium, we can use the conditions for translational equilibrium to determine the minimum angle for which the block will slide. The diagram to the right of the FBD shows that the condition for tipping is that the plumb line from the center of mass pass outside of the base. We can determine the tipping angle from the geometry of the block under this condition. Apply

r

∑ F = 0 to the block:

∑F

x

= mg sin θ sliding − f s,max ≥ 0

if the block is to slide, and

∑F

y

Substitute for fs,max and eliminate Fn between these equations to obtain: Solve for the condition for sliding: Using the geometry of the block, express the condition on θ that must be satisfied if the block is to tip:

= Fn − mg cos θ sliding = 0

µs ≤ tan θ sliding θ sliding ≥ tan −1 (µs ) = tan −1 (0.4) = 21.8° ⎛ 0.5a ⎞ −1 ⎛ 1 ⎞ ⎟ = tan ⎜ ⎟ = 18.4° ⎝ 1.5a ⎠ ⎝ 3⎠

θ tipping ≥ tan −1 ⎜

Because θ tipping < θ sliding , the block will tip before it slides.

Static Equilibrium and Elasticity

969

82 •• Picture the Problem Let m represent the mass of the bar, M the mass of the suspended object, Fv the vertical component of the force the wall exerts on the bar, Fh the horizontal component of the force exerted the wall exerts on the bar, and T the tension in the cable. The free-body diagrams show these forces and their points of application on the bar for parts (a) and (b) of the problem. Because the bar is in equilibrium, we can apply the conditions for translational and rotational equilibrium to relate the various forces and distances.

r

∑τ

(a) Apply

i

i

= 0 about an axis

through the hinge: Solve for T:

(5 m )T − (7.5 m ) mg cos 30° − (15 m ) Mg cos 30° = 0 T=

Substitute numerical values and evaluate T:

T=

[(7.5 m ) m + (15 m ) M ] g cos 30° 5m

(7.5 m )(85 kg ) + (15 m )(360 kg )

(

)

5m

× 9.81 m/s cos 30° 2

= 10.3 kN Apply

r

∑F i

i

= 0 to the bar:

∑F

y

= Fv + T sin 60° − mg − Mg = 0

and

∑F

x

Solve the y equation for Fv:

= Fh − T cos 60° = 0

Fv = −T sin 60° + (m + M )g = −(10.3 kN )sin 60°

(

+ (85 kg + 360 kg ) 9.81 m/s 2 = −4.55 kN Solve the x equation for Fh:

Fh = T cos 60° = (10.3 kN )cos 60° = 5.15 kN

)

970 Chapter 12 Find the magnitude of the force exerted by the wall on the bar:

F = Fv2 + Fh2

(− 4.55 kN )2 + (5.15 kN )2

=

= 6.87 kN Find the direction of the force exerted by the wall on the bar:

⎛ Fv ⎞ ⎛ − 4.55 kN ⎞ ⎟⎟ = tan −1 ⎜⎜ ⎟⎟ ⎝ 5.15 kN ⎠ ⎝ Fh ⎠

θ = tan −1 ⎜⎜

= − 41.5° i.e., 41.5° below the horizontal. (b) Apply

r

∑ τ = 0 about the hinge:

[(10 m )sin 60°]T − (7.5 m ) mg cos 30° − (15 m ) Mg cos 30° = 0

Solve for T:

T=

(7.5 m )m + (15 m )M (10 m )sin 60°

Substitute numerical values and evaluate T:

T=

(7.5 m )(85 kg ) + (15 m )(360 kg ) (10 m )sin 60°

(

g cos 30°

)

× 9.81 m/s 2 cos 30° = 5.92 kN Apply

r

∑ F = 0 to the bar:

∑F

y

= Fv + T cos 60° − (85 kg )g − (360 kg )g = 0

and

∑F

x

= Fh − T sin 60° = 0

Solve the y equation for Fv:

Fv = −(5.92 kN ) cos 60°

Solve the x equation for Fh:

Fh = T sin 60° = (5.92 kN )sin 60°

(

+ (85 kg + 360 kg ) 9.81 m/s 2 = 1.41 kN

= 5.13 kN Find the magnitude of the force exerted by the wall on the bar:

F = Fv2 + Fh2 =

(1.41kN )2 + (5.13 kN )2

= 5.32 kN

)

Static Equilibrium and Elasticity Find the direction of the force exerted by the wall on the bar:

971

⎛ Fv ⎞ ⎛ 1.41 kN ⎞ ⎟⎟ = tan −1 ⎜⎜ ⎟⎟ ⎝ 5.13 kN ⎠ ⎝ Fh ⎠

θ = tan −1 ⎜⎜

= 15.4° i.e., 15.4° above the horizontal. 83 •• Picture the Problem The box and the forces acting on it are shown in the figure. When the box is about to tip, Fn acts at its edge, as indicated in the drawing. We can use the definition of µs and apply the condition for rotational equilibrium in an accelerated frame to relate fs to the weight of the box and, hence, to the normal force.

Using its definition, express µs: Apply

r

∑ τ = 0 about an axis

fs Fn wf s − 12 wFn = 0

µs ≥

through the box’s center of mass: Solve for the ratio

fs : Fn

Substitute to obtain the condition for tipping: Therefore, if the box is to slide:

fs 1 = Fn 2

µs ≥ 0.500

µs < 0.500 , as in Problem 75.

Remarks: The difference between problems 75 and 83 is that in 75 the maximum acceleration before slipping is 0.5g, whereas in 88 it is (0.5 cos9°− sin9°) = 0.337g. *84 •• Picture the Problem Let the mass of the rod be represented by M. Because the rod is in equilibrium, we can apply the condition for rotational equilibrium to relate the masses of the objects placed on it to its mass.

972 Chapter 12 Apply

r

∑ τ = 0 about an axis

through the pivot for the initial condition: Solve for and evaluate M:

(20 cm)(2m + 2 g ) − (40 cm) m − (10 cm ) M = 0

M =

(20 cm)(2m + 2 g ) − (40 cm ) m 10 cm

= 4.00 g Apply

r

∑ τ = 0 about an axis

(20 cm) m − (10 cm) M

=0

through the pivot for the second condition: Solve for and evaluate m:

m=

(10 cm) M 20 cm

= 12 M = 2.00 g

*85 •• Picture the Problem Let the distance from the center of the meterstick of either finger be x1 and x2 and W the weight of the stick. Because the meterstick is in equilibrium, we can apply the condition for rotational equilibrium to obtain expressions for the forces one’s fingers exert on the meterstick as functions of the distances x1 and x2 and the weight of the meterstick W. We can then explain the stop-and-start motion of one’s fingers as they are brought closer together by considering the magnitudes of these forces in relationship the coefficients of static and kinetic friction.

The stick remains balanced as long as the center of mass is between the two fingers. For a balanced stick the normal force exerted by the finger (a)

nearest the center of mass is greater than that exerted by the other finger. Consequently, a larger static - frictional force can be exerted by the finger closer to the center of mass, which means the slipping occurs at the other finger.

Static Equilibrium and Elasticity

(b) Apply

r

∑ τ = 0 about an axis through

973

F2 (x1 + x2 ) − Wx1 = 0

point 1 to obtain: Solve for F2 to obtain:

Apply

F2 = W

r

x1 x1 + x2

− F1 ( x1 + x2 ) + Wx2 = 0

∑ τ = 0 about an axis

through point 2 to obtain: Solve for F1 to obtain:

F1 = W

x2 x1 + x2

The finger farthest from the center of mass will slide inward until the normal force it exerts on the stick is sufficiently large to produce a kinetic - frictional force exceeding the maximum static - frictional force exerted by the other finger. At that point the finger that was not sliding begins to slide, the finger that was sliding stops sliding, and the process is reversed. When one finger is is slipping the other is not. 86 •• Picture the Problem The drawing shows a side view of the wall-and-picture system. Because the frame’s width is not specified, we assume it to be negligible. Note that 0.75, 0.4, and 0.85 form a Pythagorean triad. Thus, the nail will be at the same level as the top of the frame. We can apply the condition for rotational equilibrium to determine the force exerted by the wall.

(a) Because the center of gravity of the picture is in front of the wall, the torque due to mg about the nail must be balanced by an opposing torque due to the force of the wall on the Fx = 0 , the tension in the wire must have a picture, acting horizontally. So that

∑

horizontal component, and the picture must therefore tilt forward. (b) Apply

r

∑ τ = 0 about an axis

through the nail and parallel to the

(

− [(0.6 m )sin 5°](8 kg ) 9.81 m/s 2 + [(1.2 m )cos 5°] FW = 0

)

974 Chapter 12 wall to obtain: Solve for and evaluate FW:

FW =

[(0.6 m )sin 5°](8 kg )(9.81m/s2 ) (1.2 m )cos 5°

= 3.43 N 87 •• Picture the Problem The box car and rail are shown in the drawing. At the critical speed, the normal force is entirely on the outside rail. The center of gravity is 0.775 m from that rail and 2.15 m above it. Choose the coordinate system shown in the figure. To find the speed at which this situation prevails, we can apply the conditions for static equilibrium in an accelerated frame. Apply

r

∑ τ = 0 about an axis

(0.775 m )Fn − (2.15 m ) fs = 0

through the center of gravity of the box car: Apply

∑F

y

= 0 to the box car and

Fn − mg = 0 ⇒ Fn = mg

solve for Fn: Apply

∑F

x

= macm to the box:

Substitute in equation (1) to obtain:

fs = m

v2 R

(0.775 m ) mg − (2.15 m ) m v

2

R

Solve for v:

v = 0.360 Rg

(a) Evaluate v for R = 150 m:

v = 0.360(150 m ) 9.81m/s 2

(

= 23.0 m/s

=0

)

(1)

Static Equilibrium and Elasticity (b) Evaluate v for R = 240 m:

(

v = 0.360(240 m ) 9.81 m/s 2

)

= 29.1 m/s 88 •• Picture the Problem For neutral equilibrium, the center of mass of the system must be at the same height as the feet of the tightrope walker. The system is shown in the drawing. Let the origin of the coordinate system be at the rope. We’ll determine the distance d such that ycm = 0. We’ll then determine the angle θ subtended by one half the long rod.

(58 kg )(0.9 m ) − 2(20 kg ) d

Express the y coordinate of the center of mass of the system:

ycm =

Set ycm = 0 and solve for d:

d = 1.305 m

Relate the distances s and d and solve for s:

s = 0.65 m + d = 1.955 m

Relate s to R and θ:

s = R(1 − cos θ )

(1)

Relate R and θ to the half-length of the rod:

Rθ = 4 m

(2)

Substitute in equation (1) to obtain:

1.955 m = (4 m )

58 kg + 40 kg

1 − cos θ

θ

or

1 − cos θ

θ Use graphical or trial-and-error methods to solve for θ : Substitute in equation (2) to obtain:

= 0.489

θ = 1.08 rad

R=

4m = 3.70 m 1.08 rad

975

976 Chapter 12 *89 ••• Picture the Problem Let the mass of each brick be m and number them as shown in the diagrams for 3 bricks and 4 bricks below. Let l denote the maximum offset of the nth brick. Choose the coordinate system shown and apply the condition for rotational equilibrium about an axis parallel to the z axis and passing through the point P at the supporting edge of the nth brick.

(a) Apply

r

∑ τ = 0 about an axis

mg [L − ( 12 L + l )] − mgl = 0

through P and parallel to the z axis to bricks 1 and 2 for the 3-brick arrangement shown above on the left: Solve for l to obtain: (b) Apply

r

∑ τ = 0 about an axis

l=

1 4

L

mg [L − ( 12 L + l )] + mg [L − ( 34 L + l )] − mg ( 54 L + l − L ) = 0

through P and parallel to the z axis to bricks 1 and 2 for the 4-brick arrangement shown above on the right: Solve for l to obtain: Continuing in this manner we obtain, as the successive offsets, the sequence: (c) Express the offset of the (n +1)st brick in terms of the offset of the nth brick:

l = 16 L L L L L L , , , , ... 2 4 6 8 2n where n = 1, 2, 3, … N.

l n+1 = l n +

L 2n

Static Equilibrium and Elasticity

977

A spreadsheet program to calculate the sum of the offsets as a function of n is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form B5 B4+1 n+1 C5 C4+$B$1/(2*B5) L

ln +

A 1 2 3 4 5 6 7 8 9 10 11 12 13 98 99 100 101 102 103

B L= 1

2n

C

D

m n 1 2 3 4 5 6 7 8 9 10

offset 0.500 0.750 0.917 1.042 1.142 1.225 1.296 1.359 1.414 1.464

95 96 97 98 99 100

2.568 2.573 2.579 2.584 2.589 2.594

From the table we see that l5 = 1.142 m, l10 = 1.464 m, and l100 = 2.594 m. (d) Increasing N in the spreadsheet solution suggests that the sum of the individual offsets continues to grow as N increases without bound. The series is, in fact, divergent and the stack of bricks has no maximum offset or length.

978 Chapter 12 O ffset as a function of n 3.0 2.5

Offset

2.0 1.5 1.0 0.5 0.0 0

20

40

60

80

100

n

90 ••• Picture the Problem The four forces acting on the sphere: its weight, mg; the normal force of the plane, Fn; the frictional force, f, acting parallel to the plane; and the tension in the string, T, are shown in the figure. Choose the coordinate system shown. Because the sphere is in equilibrium, we can apply the conditions for translational and rotational equilibrium to find f, Fn, and T. (a) Apply

r

∑ τ = 0 about an axis

fR − TR = 0 ⇒ T = f

through the center of the sphere: Apply

∑F

x

= 0 to the sphere:

f + T cos θ − Mg sin θ = 0

Substitute for f and solve for T:

T=

Substitute numerical values and evaluate T:

T=

(b) Apply

∑F

y

= 0 to the sphere:

Mg sin θ 1 + cos θ

(3 kg ) (9.81 m/s2 )sin 30° = 1 + cos30°

Fn − T sin θ − Mg cos θ = 0

7.89 N

Static Equilibrium and Elasticity Solve for Fn: Substitute numerical values and evaluate Fn:

Fn = T sin θ + Mg cos θ

Fn = (7.89 N )sin30°

(

)

+ (3 kg ) 9.81 m/s 2 cos 30°

= 29.4 N (c) In part (a) we showed that f = T:

f = 7.89 N

91 ••• Picture the Problem Let L be the length of each leg of the tripod. Applying the Pythagorean theorem leads us to conclude that the distance a shown in the figure is

3 L and the distance b, the distance to the 2 2 3 L , and centroid of the triangle ABC is 3 2 L the distance c is . These results allow 3 L . Because us to conclude that cos θ = 3 the tripod is in equilibrium, we can apply the condition for translational equilibrium to find the compressional forces in each leg. Letting FC represent the compressional force in a leg of the tripod, apply r F = 0 to the apex of the tripod:

3FC cos θ − mg = 0

Solve for FC: Solve for FC:

mg 3 cos θ mg 3 FC = = mg 1 3 3× 3

Substitute numerical values and evaluate FC:

FC =

∑

FC =

(

)

3 (100 kg ) 9.81m/s2 = 566 N 3

979

980 Chapter 12 92 •• Picture the Problem The forces that act on the beam are its weight, mg; the force of the cylinder, Fc, acting along the radius of the cylinder; the normal force of the ground, Fn; and the friction force fs = µsFn. The forces acting on the cylinder are its weight, Mg; the force of the beam on the cylinder, Fcb = Fc in magnitude, acting radially inward; the normal force of the ground on the cylinder, Fnc; and the force of friction, fsc = µscFnc. Choose the coordinate system shown in the figure and apply the conditions for rotational and translational equilibrium.

fs Fn

Express µs,beam-floor in terms of fs and Fn:

µ s,beam−floor =

Express µs,cylinder-floor in terms of fsc and Fnc:

µ s,cylinder−floor =

Apply

r

∑ τ = 0 about an axis

(1)

f sc Fnc

(2)

[(10 cm)cosθ ]mg − (15 cm)Fc = 0

through the right end of the beam: Solve for and evaluate Fc:

Fc = =

[(10 cm )cosθ ]mg 15 cm

[10 cos 30°](5 kg )(9.81m/s2 ) 15

= 28.3N Apply

∑F

y

= 0 to the beam:

Fn = mg − Fc cos θ

Solve for Fn:

Apply

∑F

x

Fn + Fc cos(90° − θ ) − mg = 0

(

)

= (5 kg ) 9.81 m/s 2 − (28.3 N )cos30° = 24.5 N

= 0 to the beam:

− fs + Fc cos(90° − θ ) = 0

Static Equilibrium and Elasticity

981

f s = Fc cos(90° − θ ) = (28.3 N ) cos60°

Solve for and evaluate fs:

= 14.2 N r r Fcb is the reaction force to Fc : Apply

∑F

y

= 0 to the cylinder:

∑F

x

Fnc − Fcb cos θ − Mg = 0 Fnc = Fcb cos θ + Mg

Solve for and evaluate Fnc:

Apply

Fcb = Fc = 28.3 N radially inward.

(

= (28.3 N ) cos30° + (8 kg ) 9.81 m/s 2 = 103 N

= 0 to the cylinder:

fsc − Fcb cos(90° − θ ) = 0 f sc = Fcb cos(90° − θ ) = (28.3 N ) cos60°

Solve for and evaluate fsc:

= 14.2 N Substitute numerical values in equations (1) and (2) and evaluate µs,beam-floor and µs,cylinder-floor:

µ s,beam−floor =

14.2 N = 0.580 24.5 N

and

µ s,cylinder−floor =

14.2 N = 0.138 103 N

93 ••• Picture the Problem The geometry of the system is shown in the drawing. Let upward be the positive y direction and to the right be the positive x direction. Let the angle between the vertical center line and the line joining the two centers beθ. Then

sin θ =

R−r and tan θ = r

R−r . R(2r − R )

The force exerted by the bottom of the cylinder is just 2mg. Let F be the force that the top sphere exerts on the lower sphere. Because the spheres are in equilibrium, we can apply the condition for translational equilibrium. Apply

∑F

y

= 0 to the spheres:

Fn − mg − mg = 0

)

982 Chapter 12 Solve for Fn:

Fn = 2mg

Because the cylinder wall is smooth, Fcosθ = mg, and:

F=

Express the x component of F:

Fx = F sin θ = mg tan θ

Express the force that the wall of the cylinder exerts:

FW = mg

mg cosθ

R−r R(2r − R )

Remarks: Note that as r approaches R/2, Fw→∞. *94 ••• Picture the Problem Consider a small rotational displacement, δθ of the cube from equilibrium. This shifts the point of contact between cube and cylinder by Rδθ, where R = d/2. As a result of that motion, the cube itself is rotated through the same angle δθ, and so its center is shifted in the same direction by the amount (a/2) δθ, neglecting higher order terms in δθ .

If the displacement of the cube’s center of mass is less than that of the point of contact, the torque about the point of contact is a restoring torque, and the cube will return to its equilibrium position. If, on the other hand, (a/2) δθ > (d/2) δθ, then the torque about the point of contact due to mg is in the direction of δθ, and will cause the displacement from equilibrium to increase. We see that the minimum value of d/a for stable equilibrium is d/a = 1.

Chapter 13 Fluids Conceptual Problems 1 • Determine the Concept The absolute pressure is related to the gauge pressure according to P = Pgauge + Pat. While doubling the gauge pressure will increase the absolute pressure, we do not have enough information to say what the resulting absolute pressure will be.

(e ) is correct.

*2

•

Determine the Concept No. In an environment where g eff = Fg − m

v2 = 0 , there is no r

buoyant force; there is no ″up″ or ″down.″ 3 •• Determine the Concept As you lower the rock into the water, the upward force you exert on the rock plus the upward buoyant force on the rock balance its weight. When the thread breaks, there will be an additional downward force on the scale equal to the buoyant force on the rock (the water exerts the upward buoyant force on the rock and the reaction force is the force the rock exerts on the water … and hence on the scale). Let ρ represent the density of the water, V the volume of the rock, and wf the weight of the displaced water. Then the density of the rock is 3ρ. We can use Archimedes’ principle to find the additional force on the scale. Apply Archimedes’ principle to the rock:

B = wf = mf g = ρ f Vf g

Because Vf = Vrock:

B=ρ

M

ρ rock

g=ρ

M g = 13 Mg 3ρ

and (d ) is correct. 4 •• Determine the Concept The density of water increases with depth and the buoyant force on the rock equals the weight of the displaced water. Because the weight of the displaced water depends on the density of the water, it follows that the buoyant force on the rock increases as it sinks. (b) is correct. 5 •• Determine the Concept Nothing. The fish is in neutral buoyancy (that is, its density equals that water), so the upward acceleration of the fish is balanced by the downward

983

984

Chapter 13

acceleration of the displaced water. *6 •• Determine the Concept Yes. Because the volumes of the two objects are equal, the downward force on each side is reduced by the same amount when they are submerged, not in proportion to their masses. That is, if m1L1 = m2L2 and L1 ≠ L2, then (m1 − c)L1 ≠ (m2 − c)L2. 7 •• Determine the Concept The buoyant forces acting on these submerged objects are equal to the weight of the water each displaces. The weight of the displaced water, in turn, is directly proportional to the volume of the submerged object. Because ρPb > ρCu, the volume of the copper must be greater than that of the lead and, hence, the buoyant force on the copper is greater than that on the lead. (b) is correct. 8 •• Determine the Concept The buoyant forces acting on these submerged objects are equal to the weight of the water each displaces. The weight of the displaced water, in turn, is directly proportional to the volume of the submerged object. Because their volumes are the same, the buoyant forces on them must be the same. (c) is correct. 9 • Determine the Concept It blows over the ball, reducing the pressure above the ball to below atmospheric pressure. 10 • Determine the Concept From the equation of continuity (IV = Av = constant), we can conclude that, as the pipe narrows, the velocity of the fluid must increase. Using Bernoulli’s equation for constant elevation ( P + 12 ρv 2 = constant ), we can conclude that as the velocity of the fluid increases, the pressure must decrease. (b) is correct. *11 • Determine the Concept False. The buoyant force on a submerged object depends on the weight of the displaced fluid which, in turn, depends on the volume of the displaced fluid. 12 • Determine the Concept When the bottle is squeezed, the force is transmitted equally through the fluid, leading to a pressure increase on the air bubble in the diver. The air bubble shrinks, and the loss in buoyancy is enough to sink the diver.

Fluids

985

13 • Determine the Concept The buoyant force acting on the ice cubes equals the weight of the water they displace, i.e., B = wf = ρ f Vf g . When the ice melts the volume of water displaced by the ice cubes will occupy the space previously occupied by the submerged part of the ice cubes. Therefore, the water level remains constant. 14 • Determine the Concept The density of salt water is greater than that of fresh water and so the buoyant force exerted on one in salt water is greater than in fresh water. 15 •• Determine the Concept Because the pressure increases with depth, the object will be compressed and its density will increase. Its volume will decrease. Thus, it will sink to the bottom. 16 •• Determine the Concept The force acting on the fluid is the difference in pressure between the wide and narrow parts times the area of the narrow part. 17 •• Determine the Concept The drawing shows the beaker and a strip within the water. As is readily established by a simple demonstration, the surface of the water is not level while the beaker is accelerated, showing that there is a pressure gradient. That pressure gradient results in a net force on the small element shown in the figure. *18 •• Determine the Concept The water level in the pond will drop slightly. When the anchor is in the boat, the boat displaces enough water so that the buoyant force on it equals the sum of the weight of the boat, your weight, and the weight of the anchor. When you put the anchor overboard, it will displace its volume and the volume of water displaced by the boat will decrease. 19 •• Determine the Concept From Bernoulli's principle, the opening above which the air flows faster will be at a lower pressure than the other one, which will cause a circulation of air in the tunnel from opening 1 toward opening 2. It has been shown that enough air will circulate inside the tunnel even with the slightest breeze outside.

986

Chapter 13

*20 • Determine the Concept The diagram that follows shows the forces exerted by the pressure of the liquid on the two cups to the left.

Because the force is normal to the surface of the cup, there is a larger downward component to the net force on the cup on the left. Similarly, there will be less total force exerted by the fluid in the cup on the far right in the diagram in the problem statement.

Density 21 • Picture the Problem The mass of the cylinder is the product of its density and volume. The density of copper can be found in Figure 13-1. Using the definition of density, express the mass of the cylinder: Substitute numerical values and evaluate m:

m = ρV = ρ (π R 2 h )

( × (6 × 10

)(

m = π 8.93 × 103 kg/m 3 2 × 10 −2 m −2

m

)

)

2

= 0.673 kg 22 • Picture the Problem The mass of the sphere is the product of its density and volume. The density of lead can be found in Figure 13-1.

(

)

Using the definition of density, express the mass of the sphere:

m = ρV = ρ 43 π R 3

Substitute numerical values and evaluate m:

m = 43 π 11.3 ×10 3 kg/m 3 2 ×10 −2 m

(

= 0.379 kg

)(

)

3

Fluids

987

23 • Picture the Problem The mass of the air in the room is the product of its density and volume. The density of air can be found in Figure 13-1. Using the definition of density, express the mass of the air: Substitute numerical values and evaluate m:

m = ρV = ρLWH m = (1.293 kg/m 3 )(4 m )(5 m )(4 m ) = 103 kg

*24 • Picture the Problem Let ρ0 represent the density of mercury at 0°C and ρ′ its density at 80°C, and let m represent the mass of our sample at 0°C and m′ its mass at 80°C. We can use the definition of density to relate its value at the higher temperature to its value at the lower temperature and the amount spilled.

m' V

Using its definition, express the density of the mercury at 80°C:

ρ' =

Express the mass of the mercury at 80°C in terms of its mass at 0°C and the amount spilled at the higher temperature:

ρ' =

Substitute numerical values and evaluate ρ′:

ρ' = 1.3645 × 104 kg/m 3 −

m − ∆m m ∆m = − V V V ∆m = ρ0 − V 1.47 × 10 −3 kg 60 × 10 −6 m 3

= 1.3621× 104 kg/m 3

Pressure 25 • Picture the Problem The pressure due to a column of height h of a liquid of density ρ is given by P = ρgh. Letting h represent the height of the column of mercury, express the pressure at its base:

ρ Hg gh = 101kPa

Solve for h:

h=

101 kPa ρ Hg g

988

Chapter 13

Substitute numerical values and evaluate h:

1.01× 105 N/m 2 13.6 × 103 kg/m 3 9.81 m/s 2 1in = 0.757 m × 2.54 × 10−2 m

h=

(

)(

)

= 29.8 in of Hg 26 • Picture the Problem The pressure due to a column of height h of a liquid of density ρ is given by P = ρgh. (a) Express the pressure as a function of depth in the lake:

P = Pat + ρ water gh

Solve for and evaluate h:

h=

P − Pat 2 Pat − Pat Pat = = ρ water g ρ water g ρ water g

Substitute numerical values and evaluate h:

h=

1.01× 105 N/m 2 (103 kg/m3 )(9.81m/s2 )

= 10.3 m (b) Proceed as in (a) with ρwater replaced by ρHg to obtain:

h=

2 Pat − Pat P = at ρ Hg g ρ Hg g

Substitute numerical values and evaluate h:

h=

1.01× 105 N/m 2 13.6 × 103 kg/m 3 9.81 m/s 2

(

)(

)

= 75.7 cm *27 • Picture the Problem The pressure applied to an enclosed liquid is transmitted undiminished to every point in the fluid and to the walls of the container. Hence we can equate the pressure produced by the force applied to the piston to the pressure due to the weight of the automobile and solve for F. Express the pressure the weight of the automobile exerts on the shaft of the lift:

Pauto =

Express the pressure the force applied to the piston produces:

P=

wauto Ashaft

F Apiston

Fluids Because the pressures are the same, we can equate them to obtain: Solve for F:

Substitute numerical values and evaluate F:

989

wauto F = Ashaft Apiston

F = wauto

Apiston Ashaft

Apiston

= mauto g

Ashaft

⎛ 1cm ⎞ ⎟⎟ F = (1500 kg ) 9.81 m/s ⎜⎜ ⎝ 8 cm ⎠

(

2

)

2

= 230 N 28 •• Picture the Problem The pressure exerted by the woman’s heel on the floor is her weight divided by the area of her heel. Using its definition, express the pressure exerted on the floor by the woman’s heel:

P=

Substitute numerical values and evaluate P:

P=

F w mg = = A A A

(56 kg )(9.81m/s2 ) 10−4 m 2

= 5.49 × 106 N/m 2 ×

1atm 101.3 kPa

= 54.2 atm *29 • Picture the Problem The required pressure ∆P is related to the change in volume ∆V and the initial volume V through the definition of the bulk modulus B; B = −

∆P ∆V V

Using the definition of the bulk modulus, relate the change in volume to the initial volume and the required pressure:

B=−

Solve for ∆P:

∆P = − B

∆V V

∆P . ∆V V

990

Chapter 13

Substitute numerical values and evaluate ∆P:

⎛ − 0.01 L ⎞ ⎟⎟ ∆P = −2.0 × 109 Pa × ⎜⎜ ⎝ 1L ⎠ 1 atm = 2.00 × 10 7 Pa × 101.325 kPa = 198 atm

30 • Picture the Problem The area of contact of each tire with the road is related to the weight on each tire and the pressure in the tire through the definition of pressure. Using the definition of gauge pressure, relate the area of contact to the pressure and the weight of the car:

A=

Substitute numerical values and evaluate A:

A= =

1 4

w

Pgauge

1 4

(1500 kg ) (9.81 m/s 2 ) 200 kPa

1 4

(1500 kg ) (9.81 m/s 2 ) 200 × 103 N/m 2

= 1.84 × 10 −2 m 2 = 184 cm 2 31 •• Picture the Problem The force on the lid is related to pressure exerted by the water and the cross-sectional area of the column of water through the definition of density. We can find the mass of the water from the product of its density and volume. (a) Using the definition of pressure, express the force exerted on the lid:

F = PA

Express the pressure due to a column of water of height h:

P = ρ water gh

Substitute for P and A to obtain:

F = ρ water ghπ r 2

Substitute numerical values:

F = 103 kg/m 3 9.81 m/s 2

(

)(

× (12 m ) π (0.2 m )

2

= 14.8 kN

)

Fluids (b) Relate the mass of the water to its density and volume:

m = ρ waterV = ρ water hπ r 2

Substitute numerical values and evaluate m:

m = 103 kg/m 3 (12 m ) π 3 × 10 −3 m

(

)

(

991

)

2

= 0.339 kg

32 •• Picture the Problem The minimum elevation of the bag h that will produce a pressure of at least 12 mmHg is related to this pressure and the density of the blood plasma through P = ρ blood gh . Using the definition of the pressure due to a column of liquid, relate the pressure at its base to its height:

P = ρ blood gh

Solve for h:

h=

Substitute numerical values and evaluate h:

P

ρ blood g

133.32 Pa 1 mmHg h= 3 1.03 × 10 kg/m 3 9.81 m/s 2 12 mmHg ×

(

)(

)

= 0.158 m = 15.8 cm 33 •• Picture the Problem The depth h below the surface at which you would be able to breath is related to the pressure at that depth and the density of water ρw through P = ρ w gh . Express the pressure at a depth h and solve for h:

P = ρ w gh and

h=

Express the pressure at depth h in terms of the weight on your chest:

P=

Substitute to obtain:

h=

P

ρw g F A F Aρ w g

992

Chapter 13

Substitute numerical values and evaluate h:

h=

400 N 0.09 m 10 kg/m 3 9.81 m/s 2

(

2

)(

3

)(

)

= 0.453 m 34 •• Picture the Problem Let A1 and A2 represent the cross-sectional areas of the large piston and the small piston, and F1 and F2 the forces exerted by the large and on the small piston, respectively. The work done by the large piston is W1 = F1h1 and that done on the small piston is W2 = F2h2. We’ll use Pascal’s principle and the equality of the volume of the displaced liquid in both pistons to show that W1 and W2 are equal. Express the work done in lifting the car a distance h:

W1 = F1h1 where F is the weight of the car.

Using the definition of pressure, relate the forces F1 (= w) and F2 to the areas A1 and A2:

F1 F2 = A1 A2

Solve for F1:

F1 = F2

Equate the volumes of the displaced fluid in the two pistons:

h1 A1 = h2 A2

Solve for h1:

h1 = h2

A2 A1

Substitute in the expression for W1 and simplify to obtain:

W1 = F2

A1 A2 h2 = F2 h2 = W2 A2 A1

35 • Picture the Problem Because the pressure varies with depth, we cannot simply multiply the pressure times the half-area of a side of the cube to find the force exerted by the water. We therefore consider the force exerted on a strip of width a, height dh, and area dA = adh at a depth h and integrate from h = 0 to h = a/2. The water pressure at depth h is Pat + ρgh. We can

A1 A2

Fluids

993

omit the atmospheric pressure because it is exerted on both sides of the wall of the cube.

dF = PdA = ρghadh

Express the force dF on the element of length a and height dh in terms of the net pressure ρgh:

a 2

a 2

⎛ a2 ⎞ 1 F = ∫ dF = ρga ∫ hdh = ρga⎜⎜ ⎟⎟ 2 ⎝ 4 ⎠ 0 0

Integrate from h = 0 to h = a/2:

=

ρga 3 8

*36 ••• Picture the Problem The weight of the water in the vessel is the product of its mass and the gravitational field. Its mass, in turn, is related to its volume through the definition of density. The force the water exerts on the base of the container can be determined from the product of the pressure it creates and the area of the base. (a) Using the definition of density, relate the weight of the water to the volume it occupies:

w = mg = ρVg

Substitute for V to obtain:

w = 13 πρ r 2 hg

Substitute numerical values and evaluate w:

(

)(

)(

)(

2

)

w = 13 π 103 kg/m 3 15 ×10 −2 m 25 × 10 −2 m 9.81 m/s 2 = 57.8 N F = PA = ρghπ r 2

(b) Using the definition of pressure, relate the force exerted by the water on the base of the vessel to the pressure it exerts and the area of the base: Substitute numerical values and evaluate F:

(

)(

)(

) (

F = 103 kg/m 3 9.81m/s 2 25 × 10−2 m π 15 × 10−2 m

)

2

= 173 N

994

Chapter 13 This occurs in the same way that the force on Pascals barrel is much greater than the weight of the water in the tube. The downward force on the base is also the result of the downward component of the force exerted by the slanting walls of the cone on the water.

Buoyancy *37 • Picture the Problem The scale’s reading will be the difference between the weight of the piece of copper in air and the buoyant force acting on it. Express the apparent weight w′ of the piece of copper:

w' = w − B

Using the definition of density and Archimedes’ principle, substitute for w and B to obtain:

w' = ρ CuVg − ρ wVg

Express w in terms of ρCu and V and solve for Vg:

w = ρ CuVg ⇒ Vg =

Substitute to obtain:

w' = (ρ Cu − ρ w )

Substitute numerical values and evaluate w′:

= (ρ Cu − ρ w )Vg

w

ρ Cu

w

ρ Cu ⎛ ρ ⎞ = ⎜⎜1 − w ⎟⎟ w ⎝ ρ Cu ⎠

(

⎛ 1⎞ w' = ⎜1 − ⎟ (0.5 kg ) 9.81 m/s 2 ⎝ 9⎠

)

= 4.36 N 38 • Picture the Problem We can use the definition of density and Archimedes’ principle to find the density of the stone. The difference between the weight of the stone in air and in water is the buoyant force acting on the stone.

mstone Vstone

Using its definition, express the density of the stone:

ρ stone =

Apply Archimedes’ principle to obtain:

B = wf = mf g = ρ f Vf g

(1)

Fluids

995

B ρf g

Solve for Vf:

Vf =

Because Vf = Vstone and ρf = ρwater:

Vstone =

Substitute in equation (1) and simplify to obtain:

ρstone =

mstone g w ρ water = stone ρ water B B

Substitute numerical values and evaluate ρstone:

ρ stone =

60 N 103 kg/m 3 60 N − 20 N

B

ρ water g

(

)

= 3.00 ×103 kg/m 3 39 • Picture the Problem We can use the definition of density and Archimedes’ principle to find the density of the unknown object. The difference between the weight of the object in air and in water is the buoyant force acting on the object.

mobject

(a) Using its definition, express the density of the object:

ρ object =

Apply Archimedes’ principle to obtain:

B = wf = mf g = ρ f Vf g

Solve for Vf:

Vf =

Because Vf = Vobject and ρf = ρwater:

(1)

Vobject

B ρf g

Vobject =

B

ρ water g

Substitute in equation (1) and simplify to obtain:

ρ object =

mobject g w ρ water = object ρ water B B

Substitute numerical values and evaluate ρobject:

ρ object =

5N 10 3 kg/m 3 5 N − 4.55 N

(

)

= 11.1 × 10 3 kg/m 3

(b)

From Figure 13 - 1, we see that the unknown material has a density close to that of lead.

Chapter 13

996

40 • Picture the Problem We can use the definition of density and Archimedes’ principle to find the density of the unknown object. The difference between the weight of the object in air and in water is the buoyant force acting on it.

mmetal Vmetal

Using its definition, express the density of the metal:

ρ metal =

Apply Archimedes’ principle to obtain:

B = wf = mf g = ρ f Vf g

Solve for Vf:

Vf =

Because Vf = Vmetal and ρf = ρwater:

(1)

B ρf g

Vmetal =

B

ρ water g

Substitute in equation (1) and simplify to obtain:

ρ metal =

mmetal g w ρ water = metal ρ water B B

Substitute numerical values and evaluate ρmetal:

ρ metal =

90N 10 3 kg/m 3 90 N − 56.6 N

(

)

= 2.69 × 10 3 kg/m 3 41 •• Picture the Problem Let V be the volume of the object and V′ be the volume that is submerged when it floats. The weight of the object is ρVg and the buoyant force due to the water is ρwV′g. Because the floating object is translational equilibrium, we can use Fy = 0 to relate the buoyant forces acting on the object in the two liquids to its

∑

weight. Apply

∑F

y

= 0 to the object

ρ wV'g − mg = ρ wV'g − ρVg = 0 (1)

floating in water:

V' V

Solve for ρ:

ρ = ρw

Substitute numerical values and evaluate ρ:

ρ = (103 kg/m 3 )

0.8V = 800 kg/m 3 V

Fluids Apply

∑F

y

= 0 to the object

997

mg = 0.72Vρ L g

floating in the second liquid and solve for mg: Solve equation (1) for mg:

mg = 0.8ρ wVg

Equate these two expressions to obtain:

0.72 ρ L = 0.8ρ w

Substitute in the definition of specific gravity to obtain:

specific gravity =

ρ L 0.8 = = 1.11 ρ w 0.72

*42 •• Picture the Problem We can use Archimedes’ principle to find the density of the unknown object. The difference between the weight of the block in air and in the fluid is the buoyant force acting on the block. Apply Archimedes’ principle to obtain:

B = wf = mf g = ρ f Vf g

Solve for ρf:

ρf =

Because Vf = VFe block:

ρf =

B Vf g B VFe block g

=

B mFe block g

ρ Fe

Substitute numerical values and evaluate ρf:

ρf =

(5 kg )(9.81m/s2 ) − 6.16 N (7.96 ×103 kg/m3 ) = (5 kg )(9.81m/s2 )

6.96 × 103 kg/m 3

43 •• Picture the Problem The forces acting on the cork are B, the upward force due to the displacement of water, mg, the weight of the piece of cork, and Fs, the force exerted by the spring. The piece of cork is in equilibrium under the influence of these forces. Apply

∑F

y

= 0 to the piece of cork:

B − w − Fs = 0

(1)

or

B − ρ corkVg − Fs = 0

Express the buoyant force as a function of the density of water:

B = wf = ρ wVg

(2)

998

Chapter 13

Solve for Vg:

Vg =

Substitute for Vg in equation (2):

B

ρw

B − ρ cork

B

ρw

− Fs = 0

Solve equation (1) for B:

B = w + Fs

Substitute in equation (3) to obtain:

w + Fs − ρ cork or

w − ρ cork

w + Fs

ρw

w + Fs

ρw

(3)

− Fs = 0

=0

w w + Fs

Solve for ρcork:

ρ cork = ρ w

Substitute numerical values and evaluate ρcork:

ρ cork = (103 kg/m3 )

0.285 N 0.285 N + 0.855 N

= 250 kg/m 3 44 •• Picture the Problem Under minimum-volume conditions, the balloon will be in equilibrium. Let B represent the buoyant force acting on the balloon, wtot represent its total weight, and V its volume. The total weight is the sum of the weights of its basket, cargo, and helium in its balloon. Apply

∑F

y

= 0 to the balloon:

B − wtot = 0

Express the total weight of the balloon:

wtot = 2000 N + ρ HeVg

Express the buoyant force due to the displaced air:

B = wf = ρ airVg

Substitute to obtain:

ρ airVg − 2000 N − ρ HeVg = 0

Solve for V:

V=

2000 N (ρ air − ρ He )g

Fluids

999

Substitute numerical values and evaluate V:

V =

2000 N = 183 m 3 3 2 1.29 kg/m − 0.178 kg/m 9.81 m/s

(

)(

3

)

*45 •• Picture the Problem Let V = volume of diver, ρD the density of the diver, VPb the volume of added lead, and mPb the mass of lead. The diver is in equilibrium under the influence of his weight, the weight of the lead, and the buoyant force of the water. Apply

∑F

y

B − wD − wPb = 0

= 0 to the diver:

ρ wVD + Pb g − ρ DVD g − mPb g = 0

Substitute to obtain:

or

ρ wVD + ρ wVPb − ρ DVD − mPb = 0 mD

Rewrite this expression in terms of masses and densities:

ρw

Solve for the mass of the lead:

mPb =

ρD

+ ρw

mPb

ρ Pb

− ρD

mD

ρD

− mPb = 0

ρ Pb (ρ w − ρ D ) mD ρ D (ρ Pb − ρ w )

Substitute numerical values and evaluate mPb:

mPb =

(11.3 ×10 kg/m )(10 kg/m − 0.96 ×10 kg/m )(85 kg ) = (0.96 ×10 kg/m )(11.3 ×10 kg/m − 10 kg/m ) 3

3

3

3

3

3

3

3

3

3

3

3

3.89 kg

46 •• Picture the Problem The scale’s reading w′ is the difference between the weight of the aluminum block in air w and the buoyant force acting on it. The buoyant force is equal to the weight of the displaced fluid, which, in turn, is the product of its density and mass. We can apply a condition for equilibrium to relate the reading of the bottom scale to the weight of the beaker and its contents and the buoyant force acting on the block. Express the apparent weight w′ of the aluminum block:

w' = w − B

(1)

Letting F be the reading of the bottom scale and choosing upward to be the positive y direction, apply

F + w'− M tot g = 0

(2)

1000 Chapter 13

∑F

y

= 0 to the scale to obtain:

Using the definition of density and Archimedes’ principle, substitute for w and B in equation (1) to obtain:

w' = ρ AlVg − ρ wVg

Express w in terms of ρAl and V and solve for Vg:

w = ρ AlVg ⇒ Vg =

Substitute to obtain:

w' = (ρ Al − ρ w )

Substitute numerical values and evaluate w′:

= (ρ Al − ρ w )Vg

w

ρ Al

w

ρ Al ⎛ ρ ⎞ = ⎜⎜1 − w ⎟⎟ w ⎝ ρ Al ⎠

⎛ 103 kg/m 3 ⎞ ⎟ w' = ⎜⎜1 − 3 3 ⎟ ⎝ 2.7 × 10 kg/m ⎠ × (2 kg ) 9.81 m/s 2

(

)

= 12.4 N Solve equation (2) for F: Substitute numerical values and evaluate the reading of the bottom scale:

F = M tot g − w'

(

)

F = (5 kg ) 9.81m/s 2 − 12.4 N

= 36.7 N

47 ••• Picture the Problem Let V = displacement of ship in the two cases, m be the mass of ship without load, and ∆m be the load. The ship is in equilibrium under the influence of the buoyant force exerted by the water and its weight. We’ll apply the condition for floating in the two cases and solve the equations simultaneously to determine the loaded mass of the ship. Apply

∑F

= 0 to the ship in fresh water:

ρ wVg − mg = 0

Apply

∑F

= 0 to the ship in salt water:

ρ swVg − (m + ∆m )g = 0

y

y

Solve equation (1) for Vg:

Substitute in equation (2) to obtain:

Vg =

ρ sw

mg

ρw

mg

ρw

− (m + ∆m )g = 0

(1) (2)

Fluids 1001

ρ w ∆m ρ sw − ρ w

Solve for m:

m=

Add ∆m to both sides of the equation and simplify to obtain:

m + ∆m =

ρ w ∆m + ∆m ρ sw − ρ w

⎛ ρw ⎞ = ∆m⎜⎜ + 1⎟⎟ ⎝ ρ sw − ρ w ⎠ ∆mρ sw = ρ sw − ρ w Substitute numerical values and evaluate m + ∆m:

m + ∆m =

(6 ×10 kg )(1.025ρ ) 5

1.025ρ w − ρ w

w

(6 ×10 kg )(1.025) = 5

1.025 − 1

= 2.46 × 107 kg *48 ••• Picture the Problem For minimum liquid density, the bulb and its stem will be submerged. For maximum liquid density, only the bulb is submerged. In both cases the hydrometer will be in equilibrium under the influence of its weight and the buoyant force exerted by the liquids. (a) Apply

∑F

y

= 0 to the hydrometer:

B−w=0

ρ minVg − mtot g = 0

Using Archimedes’ principle to express B, substitute to obtain:

or

Solve for mPb:

mPb = ρ min (Vbulb + Vstem ) − mglass

ρ min (Vbulb + Vstem ) = mglass + mPb

Substitute numerical values and evaluate mPb:

⎡ 1 L ⎞⎤ π 2⎛ mPb = (0.9 kg/L ) ⎢0.020 L + (0.15 m )(0.005 m ) ⎜⎜ -3 3 ⎟⎟⎥ − 6 × 10−3 kg 4 ⎝ 10 m ⎠⎦ ⎣

(

= 14.7 g (b) Apply

∑F

y

= 0 to the hydrometer:

ρ maxVg − mtot g = 0 or

)

1002 Chapter 13

ρ maxVbulb = mglass + mPb Solve for ρmax:

ρ max =

Substitute numerical values and evaluate ρmax:

ρ max =

mglass + mPb Vbulb

6 g + 14.7 g = 1.04 kg/L 20 mL

49 • Picture the Problem We can relate the upward force exerted on the dam wall to the area over which it acts using F = Pg A and express Pg in terms of the depth of the water using Pg = ρgh . Using the definition of pressure, express the upward force exerted on the dam wall:

F = Pg A

Express the gauge pressure Pg of the water 5 m below the surface of the dam:

Pg = ρgh

Substitute to obtain:

F = ρghA

Substitute numerical values and evaluate F:

F = 103 kg/m 3 9.81m/s 2 (5 m ) 10 m 2

(

)(

)

= 491kN

50 •• Picture the Problem The forces acting on the balloon are the buoyant force B, its weight mg, and a drag force FD. We can find the initial upward acceleration of the balloon by applying Newton’s 2nd law at the instant it is released. We can find the terminal velocity of the balloon by recognizing that when ay = 0, the net force acting on the balloon will be zero. (a) Apply

∑F

y

= ma y to the

B − mballoon g = mballoon a y

balloon at the instant of its release to obtain: Solve for ay:

ay =

B − mballoon g B = −g mballoon mballoon

(

)

Fluids 1003 Using Archimedes principle, express the buoyant force B acting on the balloon:

B = wf = mf g = ρ f Vf g

Substitute to obtain:

4 3

Solve for ay:

⎞ ⎛ 4 πρ r 3 a y = ⎜⎜ 3 air − 1⎟⎟ g ⎠ ⎝ mballoon

Substitute numerical values and evaluate ay:

⎡ 4 π 1.29 kg/m 3 (2.5 m )3 ⎤ − 1⎥ ay = ⎢ 3 15 kg ⎣ ⎦ 2 × 9.81 m/s

= ρ airVballoon g = 43 πρair r 3 g

πρair r 3 g − mballoon g = mballoon a y

(

)

(

)

= 45.4 m/s 2 (b) Apply

∑F

y

B − mg − 12 π r 2 ρvt2 = 0

= ma y to the

balloon under terminal-speed conditions to obtain: Substitute for B:

4 3

Solve for vt:

πρair r 3 g − mg − 12 π r 2 ρvt2 = 0

(

)

2 43 πρ air r 3 − m g

vt =

πr ρ 2

Substitute numerical values and evaluate v:

vt =

[ (

)

](

)

2 43 π 1.29 kg/m3 (2.5 m ) − 15 kg 9.81 m/s 2 = 7.33 m/s π (2.5 m )2 1.29 kg/m3

(c) Relate the time required for the balloon to rise to 10 km to its terminal speed:

3

(

)

∆t =

h 10 km = = 1364 s vt 7.33 m/s

= 22.7 min

Continuity and Bernoulli's Equation *51 •• Picture the Problem Let J represent the flow rate of the water. Then we can use J = Av to relate the flow rate to the cross-sectional area of the circular tap and the velocity of the water. In (b) we can use the equation of continuity to express the diameter of the stream 7.5 cm below the tap and a constant-acceleration equation to find the velocity of the water at this distance. In (c) we can use a constant-acceleration equation to express the distance-to-turbulence in terms of the velocity of the water at turbulence vt and the definition of Reynolds number NR to relate vt to NR.

1004 Chapter 13 (a) Express the flow rate of the water in terms of the cross-sectional area A of the circular tap and the velocity v of the water: Solve for v:

Substitute numerical values and evaluate v: (b) Apply the equation of continuity to the stream of water:

J = Av = π r 2v = 14 π d 2v

v=

J πd 2

10.5 cm 3 / s v= = 9.28 cm/s 2 1 4 π (1.2 cm ) vf Af = vi Ai = vAi or

vf Solve for df:

1 4

π 4

d f2 = v

df =

π 4

d i2

v di vf

(2)

Use a constant-acceleration equation to relate vf and v to the distance ∆h fallen by the water:

vf2 = v 2 + 2 g∆h

Solve for vf to obtain:

vf = v 2 + 2 g∆h

Substitute numerical values and evaluate vf:

vf =

Substitute in equation (2) and evaluate df:

d f = (1.2 cm )

(c) Using a constant-acceleration equation, relate the fall-distance-toturbulence ∆d to its initial speed v and its speed vt when its flow becomes turbulent: Solve for ∆d to obtain: Express Reynolds number NR for turbulent flow:

(1)

(9.28 cm/s)2 + 2(981cm/s 2 )(7.5 cm )

= 122 cm/s

9.28 cm/s = 0.331 cm 122 cm/s

vt2 = v 2 + 2 g∆d

∆d =

NR =

vt2 − v 2 2g

2rρvt

η

(3)

Fluids 1005 From equation (1):

Substitute to obtain:

Solve for vt:

Substitute numerical values (see Figure 13-1 for the density of water and Table 13-1 for the coefficient of viscosity for water) and evaluate vt: Substitute in equation (3) and evaluate the fall-distance-to turbulence:

J π vt

r=

NR =

vt =

2 ρv t

J π vt

η

πN R2η 2 4ρ 2 J

π (2300)2 (1.8 × 10 −3 Pa ⋅ s )

2

vt =

(

)( 2

4 103 kg/m 3 10.5 cm 3 / s = 1.28 m/s

∆d =

)

(128 cm/s)2 − (9.28 cm/s)2

(

2 981 cm/s 2

)

= 8.31 cm in reasonable agreement with everyday experience. 52 • Picture the Problem Let A1 represent the cross-sectional area of the hose, A2 the crosssectional area of the nozzle, v1 the velocity of the water in the hose, and v2 the velocity of the water as it passes through the nozzle. We can use the continuity equation to find v2 and Bernoulli’s equation for constant elevation to find the pressure at the pump. (a) Using the continuity equation, relate the speeds of the water to the diameter of the hose and the diameter of the nozzle: Solve for v2:

A1v1 = A2 v2 or

πd12 4 v2 =

v1 =

πd 22 4

v2

d12 v1 d 22 2

Substitute numerical values and evaluate v2:

⎛ 3 cm ⎞ ⎟⎟ (0.65 m/s ) = 65.0 m/s v2 = ⎜⎜ ⎝ 0.3 cm ⎠

(b) Using Bernoulli’s equation for constant elevation, relate the pressure at the pump PP to the

PP + 12 ρv12 = Pat + 12 ρv22

1006 Chapter 13 atmospheric pressure and the velocities of the water in the hose and the nozzle:

(

PP = Pat + 12 ρ v22 − v12

Solve for the pressure at the pump:

)

Substitute numerical values and evaluate PP:

(

)[

PP = 101 kPa + 12 103 kg/m 3 (65 m/s ) − (0.65 m/s ) = 2.21× 106 Pa ×

2

2

]

1atm = 21.9 atm 101.325 kPa

53 • Picture the Problem Let A1 represent the cross-sectional area of the larger-diameter pipe, A2 the cross-sectional area of the smaller-diameter pipe, v1 the velocity of the water in the larger-diameter pipe, and v2 the velocity of the water in the smaller-diameter pipe. We can use the continuity equation to find v2 and Bernoulli’s equation for constant elevation to find the pressure in the smaller-diameter pipe. (a) Using the continuity equation, relate the velocities of the water to the diameters of the pipe:

A1v1 = A2 v2 or

πd12 4

Solve for and evaluate v2:

v2 =

v1 =

πd 22 4

v2

d12 v1 d 22 2

Substitute numerical values and evaluate v2:

⎛ d ⎞ v2 = ⎜⎜ 1 1 ⎟⎟ (3 m/s ) = 12.0 m/s ⎝ 2 d1 ⎠

(b) Using Bernoulli’s equation for constant elevation, relate the pressures in the two segments of the pipe to the velocities of the water in these segments:

P1 + 12 ρ w v12 = P2 + 12 ρ w v22

Solve for P2:

P2 = P1 + 12 ρ w v12 − 12 ρ w v22

(

= P1 + 12 ρ w v12 − v22

)

Fluids 1007 Substitute numerical values and evaluate P2:

P2 = 200 kPa + 12 (103 kg/m 3 )

[

× (3 m/s ) − (12 m/s ) 2

2

]

= 133 kPa (c) Using the continuity equation, evaluate IV1:

I V1 = A1v1 =

Using the continuity equation, express IV2:

I V2 = A2v2 =

Substitute numerical values and evaluate IV2:

Thus, as we expected would be the case:

π d12 4

π d 22 4

v1 =

π d12 4

(3 m/s)

v2

2

I V2

⎛d ⎞ π⎜ 1 ⎟ π d12 2 (3 m/s) = ⎝ ⎠ (12 m/s ) = 4 4

I V1 = I V2

54 • Picture the Problem Let A1 represent the cross-sectional area of the 2-cm diameter pipe, A2 the cross-sectional area of the constricted pipe, v1 the velocity of the water in the 2-cm diameter pipe, and v2 the velocity of the water in the constricted pipe. We can use the continuity equation to express d2 in terms of d1 and to find v1 and Bernoulli’s equation for constant elevation to find the velocity of the water in the constricted pipe. Using the continuity equation, relate the volume flow rate in the 2-cm diameter pipe to the volume flow rate in the constricted pipe: Solve for d2:

A1v1 = A2 v2 or

πd12 4

v1 =

d 2 = d1

πd 22 4

v2

v1 v2

2.80 L/s IV = = 8.91 m/s A1 π (0.02 m )2 4

Using the continuity equation, relate v1 to the volume flow rate IV:

v1 =

Using Bernoulli’s equation for constant elevation, relate the pressures in the two segments of the pipe to the velocities of the water in

P1 + 12 ρ w v12 = P2 + 12 ρ w v22

1008 Chapter 13 these segments: Solve for v2:

Substitute numerical values and evaluate v2:

2(P1 − P )

v2 =

ρw

+ v12

2(142 kPa − 101 kPa ) 2 + (8.91 m/s ) 3 3 10 kg/m

v2 =

= 12.7 m/s Substitute and evaluate d2:

d 2 = (2 cm )

8.91 m/s = 1.68 cm 12.7 m/s

*55 •• Picture the Problem We can use the definition of the volume flow rate to find the volume flow rate of blood in an aorta and to find the total cross-sectional area of the capillaries. (a) Use the definition of the volume flow rate to find the volume flow rate through an aorta: Substitute numerical values and evaluate IV:

I V = Av

(

)

I V = π 9 × 10 −3 m 3 (0.3 m/s ) = 7.63 × 10 −5

m 3 60 s 1L × × −3 3 s min 10 m

= 4.58 L/min (b) Use the definition of the volume flow rate to express the volume flow rate through the capillaries:

I V = Acap vcap

Solve for the total cross-sectional area of the capillaries:

Acap =

IV vcap

Substitute numerical values and evaluate Acap:

Acap =

7.63 × 10 −5 m 3 /s 0.001 m/s

= 7.63 × 10 −2 m 2 = 763 cm 2

Fluids 1009 56 •• Picture the Problem We can apply Bernoulli’s equation to points a and b to determine the rate at which the water exits the tank. Because the diameter of the small pipe is much smaller than the diameter of the tank, we can neglect the velocity of the water at the point a. The distance the water travels once it exits the pipe is the product of its speed and the time required to fall the distance H – h. Express the distance x as a function of the exit speed of the water and the time to fall the distance H – h:

x = vb ∆t

Apply Bernoulli’s equation to the water at points a and b:

Pa + ρ w gH + 12 ρ w va2 = Pb

(1)

+ ρ w g (H − h ) + 12 ρ w vb2

or, because va ≈ 0 and Pa = Pb = Pat,

gH = g (H − h ) + 12 vb2

Solve for vb:

vb = 2 gh

Using a constant-acceleration equation, relate the time of fall to the distance of fall:

∆y = v0y ∆t + 12 a(∆t )

Solve for ∆t:

Substitute in equation (1) to obtain:

2

or, because v0y = 0,

H − h = 12 g (∆t )

2

∆t =

2(H − h ) g

x = 2 gh

2(H − h ) = 2 h(H − h ) g

57 •• Picture the Problem Let the subscript 60 denote the 60-cm-radius pipe and the subscript 30 denote the 30-cm-radius pipe. We can use Bernoulli’s equation for constant elevation to express P30 in terms of v60 and v30, the definition of volume flow rate to find v60 and the continuity equation to find v30. Using Bernoulli’s equation for constant elevation, relate the pressures in the two pipes to the velocities of the oil:

2 2 P60 + 12 ρv60 = P30 + 12 ρv30

1010 Chapter 13

(

Solve for P30:

2 2 P30 = P60 + 12 ρ v60 − v30

Use the definition of volume flow rate to find v60:

v60 =

)

(1)

IV A60

m 3 1 day 1h × × day 24 h 3600 s = π (0.6 m )2 = 2.456 m/s 2.4 × 105

A60v60 = A30v30

Using the continuity equation, relate the velocity of the oil in the halfstandard pipe to its velocity in the standard pipe:

A60 π (0.6 m ) (2.456 m/s) v60 = A30 π (0.3 m ) 2 = 9.824 m/s 2

Solve for and evaluate v30:

v30 =

Substitute numerical values in equation (1) and evaluate P30:

(

)[

]

P30 = 180 kPa + 12 800 kg/m 3 (2.456 m/s ) − (9.824 m/s ) = 144 kPa 2

2

*58 •• Picture the Problem We’ll use its definition to relate the volume flow rate in the pipe to the velocity of the water and the result of Example 13-9 to find the velocity of the water.

I V = A1v1 = π r 2v1

Using its definition, express the volume flow rate: Using the result of Example 13-9, find the velocity of the water upstream from the Venturi meter:

2 ρ Hg gh

v1 =

⎛ R12 ⎞ − 1⎟⎟ 2 ⎝ R2 ⎠

ρ w ⎜⎜

Substitute numerical values and evaluate v1:

v1 =

(

)(

)

2 13.6 × 103 kg/m3 9.81m/s 2 (0.024 m ) = 1.847 m/s ⎡⎛ 0.095 m ⎞ 2 ⎤ ⎟⎟ − 1⎥ 103 kg/m3 ⎢⎜⎜ ⎢⎣⎝ 0.056 m ⎠ ⎥⎦

(

)

Fluids 1011 Substitute numerical values and evaluate IV:

IV =

π

(0.095 m )2 (1.847 m/s )

4 = 1.309 × 10 −2 m 3 /s = 13.1 L/s

59 •• Picture the Problem We can apply the definition of the volume flow rate to find the mass of water emerging from the hose in 1 s and the definition of momentum to find the momentum of the water. The force exerted on the water by the hose can be found from the rate at which the momentum of the water changes.

∆V ∆m = = Av ∆t ρ w ∆t

(a) Using its definition, express the volume flow rate of the water emerging from the hose:

IV =

Solve for ∆m:

∆m = Avρ w ∆t

Substitute numerical values and evaluate ∆m:

∆m = π (0.015 m ) (30 m/s ) 103 kg/m 3 (1s )

(b) Using its definition, express and evaluate the momentum of the water:

2

(

)

= 21.2 kg/s

p = ∆mv = (21.2 kg/s )(30 m/s ) = 636 kg ⋅ m/s

(c) The vector diagrams are to the right:

Express the change in momentum of the water: Substitute numerical values and evaluate ∆p:

r r r ∆p = pf − pi

(636 kg ⋅ m/s)2 + (636 kg ⋅ m/s)2 = (636 kg ⋅ m/s ) 2

∆p =

= 899 kg ⋅ m/s

1012 Chapter 13 Relate the force exerted on the water by the hose to the rate at which the water’s momentum changes and evaluate F:

F=

∆p 899 kg ⋅ m/s = = 899 N ∆t 1s

60 •• Picture the Problem Let the letter P denote the pump and the 2-cm diameter pipe and the letter N the 1-cm diameter nozzle. We’ll use Bernoulli’s equation to express the necessary pump pressure, the continuity equation to relate the velocity of the water coming out of the pump to its velocity at the nozzle, and a constant-acceleration equation to relate its velocity at the nozzle to the height to which the water rises. Using Bernoulli’s equation, relate the pressures, areas, and velocities in the pipe and nozzle:

PP + ρ w ghP + 12 ρ w vP2 = PN + ρ w ghN + 12 ρ w v N2 or, because PN = Pat and hP = 0,

PP + 12 ρ w vP2 = PN + ρ w ghN + 12 ρ w vN2

(

Solve for the pump pressure:

PP = Pat + ρ w ghN + 12 ρ w vN2 − vP2

Use the continuity equation to relate vP and vN to the cross-sectional areas of the pipe from the pump and the nozzle:

AP vP = AN vN

)

(1)

and 2

2 1 ⎛ 1 cm ⎞ AN 4 πd N ⎟⎟ vN vP = vN = 1 2 v N = ⎜⎜ AP π d 2 cm ⎝ ⎠ P 4 1 = 4 vN

v 2 = v 2N − 2 g∆h

Using a constant-acceleration equation, express the velocity of the water at the nozzle in terms of the desired height ∆h:

or, because v = 0,

v 2N = 2 g∆h

Substitute in equation (1) to obtain:

PP = Pat + ρ w ghN + 12 ρ w [2 g∆h − 161 (2 g∆h )] = Pat + ρ w ghN + 12 ρ w (158 g∆h ) 15 = Pat + ρ w g (hN + 16 ∆h )

Substitute numerical values and evaluate PP:

(

)(

)

15 (12 m )] = 241kPa PP = 101 kPa + 103 kg/m 3 9.81 m/s 2 [3 m + 16

Fluids 1013 61 ••• Picture the Problem We can apply Bernoulli’s equation to points a and b to determine the rate at which the water exits the tank. Because the diameter of the small pipe is much smaller than the diameter of the tank, we can neglect the velocity of the water at the point a. The distance the water travels once it exits the pipe is the product of its velocity and the time required to fall the distance H – h. That there are two values of h that are equidistant from the point h = 12 H can be shown by solving the quadratic equation that relates x to h and H. That x is a maximum for this value of h can be established by treating x = f(h) as an extreme-value problem. (a) Express the distance x as a function of the exit speed of the water and the time to fall the distance H – h: Apply Bernoulli’s equation to the water at points a and b:

x = vb ∆t

(1)

Pa + ρ w gH + 12 ρ w va2 = Pb + ρ w g (H − h ) + 12 ρ w vb2 or, because va ≈ 0 and Pa = Pb = Pat,

gH = g (H − h ) + 12 vb2

Solve for vb:

vb = 2 gh

Using a constant-acceleration equation, relate the time of fall to the distance of fall:

∆y = v0y ∆t + 12 a(∆t )

Solve for ∆t:

Substitute in equation (1) to obtain:

2

or, because v0y = 0,

H − h = 12 g (∆t )

2

2(H − h ) g

∆t =

x = 2 gh

2(H − h ) = 2 h(H − h ) g

(b) Square both sides of this equation and simplify to obtain:

x 2 = 4hH − 4h 2 or 4h 2 − 4 Hh + x 2 = 0

Solve this quadratic equation to obtain:

h=

1 2

H ± 12 H 2 − x 2

1014 Chapter 13 Find the average of these two values for h:

hav =

1 2

= (c) Differentiate

x = 2 h(H − h ) with respect to h:

Set the derivative equal to zero for extrema:

H + H 2 − x 2 + 12 H − H 2 − x 2 2 1 2

1 dx ⎛1⎞ − = 2⎜ ⎟[h(H − h )] 2 (H − 2h ) dh ⎝2⎠ H − 2h = h(H − h )

H − 2h

h(H − h )

Solve for h to obtain:

h=

Evaluate x = 2 h(H − h ) with

xmax = 2

h = 12 H :

H

1 2

=0

H 1 2

H (H − 12 H )

= H

Remarks: To show that this value for h corresponds to a maximum, one can either

d2x show that < 0 at h = 12 H or confirm that the graph of f(h) at h = 12 H is dh 2 concave downward. *62 •• Picture the Problem Let the numeral 1 denote the opening in the end of the inner pipe and the numeral 2 to one of the holes in the outer tube. We can apply Bernoulli’s principle at these locations and solve for the pressure difference between them. By equating this pressure difference to the pressure difference due to the height h of the liquid column we can express v as a function of ρ, ρg, g, and h. Apply Bernoulli’s principle at locations 1 and 2 to obtain:

P1 + 12 ρ g v12 = P2 + 12 ρ g v22 where we’ve ignored the difference in elevation between the two openings.

Solve for the pressure difference ∆P = P1 − P2:

∆P = P1 − P2 = 12 ρ g v22 − 12 ρ g v12

Express the velocity of the gas at 1:

v1 = 0 because the gas is brought to a halt (i.e., is stagnant) at the opening to the inner pipe.

Express the velocity of the gas at 2:

v2 = v because the gas flows freely past

Fluids 1015 the holes in the outer ring. Substitute to obtain: Letting A be the cross-sectional area of the tube, express the pressure at a depth h in the column of liquid whose density is ρ1:

∆P = 12 ρ g v 2

P1 = P2 +

wdisplaced liquid

−

B A

A where B = ρ g Ahg is the buoyant force acting on the column of liquid of height h.

Substitute to obtain:

P1 = P2 +

ρ ghA

−

ρ g ghA

A A = P2 + (ρ − ρ g )gh

or

∆P = P1 − P2 = (ρ − ρ g )gh

Equate these two expressions for ∆P: Solve for v2 to obtain:

1 2

ρ g v 2 = (ρ − ρ g )gh v2 =

2 gh(ρ − ρ g )

ρg

Note that the correction for buoyant force due to the displaced gas is very small and that, to a good approximation,

v=

2 ghρ

ρg

.

Remarks: Pitot tubes are used to measure the airspeed of airplanes. 63 •• Picture the Problem Let the letter ″a″ denote the entrance to the siphon tube and the letter ″b″ denote its exit. Assuming streamline flow between these points, we can apply Bernoulli’s equation to relate the entrance and exit speeds of the water flowing in the siphon to the pressures at either end, the density of the water, and the difference in elevation between the entrance and exit points. We can use the expression for the pressure as a function of depth in an incompressible fluid to find the pressure at the entrance to the tube in terms of its distance below the surface. We’ll also use the equation of continuity to argue that, provided the surface area of the beaker is large compared to the area of the opening of the tube, the entrance speed of the water is approximately zero. (a) Apply Bernoulli’s equation at the entrance to the siphon tube (point a) and at its exit (point b):

Pa + 12 ρva2 + ρg (H − h )

= Pb + 12 ρvb2 + ρg (H − h − d )

(1)

where H is the height of the containers.

1016 Chapter 13 Apply the continuity equation to a point at the surface of the liquid in the container to the left and to point a: Express the pressure at the inlet (point a) and the outlet (point b):

va Aa = vsurface Asurface or, because Aa << Asurface,

va = vsurface = 0

Pa = Patm + ρg (H − h ) and

Pb = Patm + ρg (H − h − d )

Letting vb = v, substitute in equation (1) to obtain:

Patm + ρg (H − h ) + ρgH

= Patm + ρg (H − h − d )

+ 12 ρv 2 + ρg (H − h − d )

or, upon simplification,

g (H − h ) + gH = g (H − h − d ) + 12 v 2 + g (H − h − d )

Solve for v:

v=

(b) Relate the pressure at the highest part of the tube Ptop to the pressure at point b:

Ptop + ρg (H − h ) + 12 ρvh2

2 gd

= Patm + ρg (H − h − d ) + 12 ρvb2 or, because vh = vb,

Ptop = Patm − ρgd Remarks: If we let Ptop = 0, we can use this result to find the maximum theoretical height a siphon can lift water.

Viscous Flow 64 • Picture the Problem The required pressure difference can be found by applying Poiseuille’s law to the viscous flow of water through the horizontal tube. Using Poiseuille’s law, relate the pressure difference between the two ends of the tube to its length, radius, and the volume flow rate of the water:

∆P =

Substitute numerical values and evaluate ∆P:

∆P =

8ηL IV π r4

8(1 mPa ⋅ s )(0.25 m )

π (0.6 ×10−3 m )

= 1.47 kPa

4

(0.3 mL/s)

Fluids 1017 65 • Picture the Problem Because the pressure difference is unchanged, we can equate the expressions of Poiseuille’s law for the two tubes and solve for the diameter of the tube that would double the flow rate. Using Poiseuille’s law, express the pressure difference required for the radius and volume flow rate of Problem 64:

∆P =

8ηL IV π r4

Express the pressure difference required for the radius r′ that would double the volume flow rate of Problem 57:

∆P =

8ηL (2 I ) π r' 4 V

Equate these equations and simplify to obtain:

8ηL (2 I V ) = 8ηL4 I V 4 π r' πr or

2 1 = 4 4 r' r Solve for r′:

r' = 4 2r

Express d ′:

d' = 2r' = 24 2r = 4 2d

Substitute numerical values and evaluate d ′:

d' = 4 2 (1.2 mm ) = 1.43 mm

*66 • Picture the Problem We can apply Poiseuille’s law to relate the pressure drop across the capillary tube to the radius and length of the tube, the rate at which blood is flowing through it, and the viscosity of blood. Using Poiseuille’s law, relate the pressure drop to the length and diameter of the capillary tube, the volume flow rate of the blood, and the viscosity of the blood:

∆P =

Solve for the viscosity of the blood:

η=

8ηL IV π r4

π r 4 ∆P 8 LI V

1018 Chapter 13 Using its definition, express the volume flow rate of the blood:

I V = Acap v = π r 2 v

Substitute and simplify:

η=

Substitute numerical values to obtain:

r 2 ∆P 8Lv

(3.5 ×10 m ) (2.60 kPa ) η= ⎛ 10 m ⎞ ⎟ 8(10 m )⎜⎜ 1s ⎟ 2

−6

−3

−3

⎝

⎠

= 3.98 mPa ⋅ s *67 • Picture the Problem We can use the definition of Reynolds number to find the velocity of a baseball at which the drag crisis occurs. Using its definition, relate Reynolds number to the velocity v of the baseball: Solve for v:

Substitute numerical values (see Figure 13-1 for the density of air and Table 13-1 for the coefficient of viscosity for air) and evaluate v:

NR =

v=

2rρv

η

ηN R 2 rρ

( 0.018 mPa ⋅ s ) (3 × 105 ) v= 2(0.05 m ) (1.293 kg/m 3 ) = 41.8 m/s ×

1 mi/h 0.447 m/s

= 93.4 mi/h Because most major league pitchers can throw a fastball in the low - to mid - 90s, this drag crisis may very well play a role in the game. Remarks: This is a topic which has been fiercely debated by people who study the physics of baseball.

Fluids

1019

68 ••• Picture the Problem Let the subscripts ″f″ refer to ″displaced fluid”, ″s″ to ″soda″, and ″g″ to the ″gas″ in the bubble. The free-body diagram shows the forces acting on the bubble prior to reaching its terminal velocity. We can apply Newton’s 2nd law, Stokes′ law, and Archimedes principle to express the terminal velocity of the bubble in terms of its radius, and the viscosity and density of water. Apply

∑F

y

= ma y to the bubble to

B − mg g − FD = ma y

obtain: Under terminal speed conditions:

B − mg g − FD = 0

Using Archimedes principle, express the buoyant force B acting on the bubble:

B = wf = mf g

Express the mass of the gas bubble:

mg = ρ gVg = ρ gVbubble

Substitute to obtain:

ρ wVbubble g − ρ gVbubble g − 6πηavt = 0

Solve for vt:

Substitute for Vbubble and simplify:

= ρf Vf g = ρsVbubble g

vt =

vt = ≈

Substitute numerical values and evaluate vt:

Vbubble g (ρ s − ρ g ) 6πηa 4 3

πa 3 g (ρ s − ρ g ) 2a 2 g (ρ s − ρ g ) = 6πηa 9η

2a 2 gρ s , since ρ s >> ρ g . 9η

(

)( 2

2 0.5 × 10−3 m 9.81 m/s 2 vt = 9 1.8 × 10−3 Pa ⋅ s

(

(

× 1.1× 103 kg/m 3 = 0.333 m/s Express the rise time ∆t in terms of the height of the soda glass h and the terminal speed of the bubble:

∆t =

h vt

)

)

)

1020 Chapter 13 Assuming that a "typical" soda glass has a height of about 15 cm, evaluate ∆t:

∆t =

0.15 m = 0.450 s 0.333 m/s

Remarks: About half a second seems reasonable for the rise time of the bubble.

General Problems *69 •• Picture the Problem We can solve the given equation for the coefficient of roundness C and substitute estimates/assumptions of typical masses and heights for adult males and females. Express the mass of a person as a function of C, ρ, and h: Solve for C:

M = Cρh 3 C=

M ρh 3

Assuming that a "typical" adult male stands 5' 10" (1.78 m) and weighs 170 lbs (77 kg), then:

C=

77 kg = 0.0137 3 10 kg/m 3 (1.78 m )

Assuming that a "typical" adult female stands 5' 4" (1.63 m) and weighs 110 lbs (50 kg), then:

C=

(

)

3

50 kg = 0.0115 3 10 kg/m 3 (1.63 m )

(

)

3

70 • Picture the Problem Let the letter ″s″ denote the shorter of the two men and the letter ″t″ the taller man. We can find the difference in weight of the two men using the relationship M = Cρh3 from Problem 69. Express the difference in weight of the two men:

∆w = wt − ws = M t g − M s g

Express the masses of the two men:

M s = Cρhs3

= (M t − M s )g

and

M t = Cρht3 Substitute to obtain:

( = (h

)

∆w = Cρht3 − Cρhs3 g 3 t

)

− hs3 Cρg

Assuming that a "typical" adult male stands 5' 10" (1.78 m) and weighs 170 lbs (77 kg), then:

C=

Express the heights of the two men in SI units:

ht = 72 in × 2.54 cm/in = 1.83 m and

77 kg = 0.0137 3 10 kg/m 3 (1.78 m )

(

3

)

Fluids

1021

hs = 69 in × 2.54 cm/in = 1.75 m Substitute numerical values (assume that ρ = 103 kg/m3) and evaluate ∆w:

[

∆w = (1.83 m ) − (1.75 m ) 3

3

(

]

)(

× (0.0137 ) 103 kg/m 3 9.81 m/s 2

= 103 N ×

)

1lb = 23.2 lb 4.4482 N

71 • Determine the Concept The net force is zero. Neglecting the thickness of the table, the atmospheric pressure is the same above and below the surface of the table. 72 • Picture the Problem The forces acting on the Ping-Pong ball, shown in the free-body diagram, are the buoyant force, the weight of the ball, and the tension in the string. Because the ball is in equilibrium under the influence of these forces, we can apply the condition for translational equilibrium to establish the relationship between them. We can also apply Archimedes’ principle to relate the buoyant force on the ball to its diameter. Apply

∑F

y

B − mg − T = 0

= 0 to the ball:

Using Archimedes’ principle, relate the buoyant force on the ball to its diameter:

B = wf = mf g = ρ wVball g = 16 πρ w d 3

Substitute to obtain:

1 6

Solve for d:

πρw d 3 − mg − T = 0

d =3

6(T + mg )

πρ w

Substitute numerical values and evaluate d:

d =3

[

(

)]

6 2.8 × 10 −2 N + (0.004 kg ) 9.81 m/s 2 = 5.05 cm π 103 kg/m 3

(

)

1022 Chapter 13 73 • Picture the Problem Let ρ0 represent the density of seawater at the surface. We can use the definition of density and the fact that mass is constant to relate the fractional change in the density of water to its fractional change in volume. We can also use the definition of bulk modulus to relate the fractional change in density to the increase in pressure with depth and solve the resulting equation for the density at the depth at which the pressure is 800 atm. Using the definition of density, relate the mass of a given volume of seawater to its volume:

m = ρV

Noting that the mass does not vary with depth, evaluate its differential:

ρdV + Vdρ = 0

Solve for dρ/ρ:

dρ

ρ

=−

∆V dV ∆ρ or ≈− V ρ V

∆P ∆P = ∆V V ∆ρ ρ 0

Using the definition of the bulk modulus, relate ∆P to ∆ρ/ρ0:

B=−

Solve ∆ρ:

∆ρ = ρ − ρ 0 =

Solve for ρ:

ρ = ρ0 +

ρ 0 ∆P B

ρ 0 ∆P B ⎛ ∆P ⎞ = ρ 0 ⎜1 + ⎟ B ⎠ ⎝

Substitute numerical values and evaluate ρ:

⎛ 1.01× 105 Pa ⎞ ⎟ ⎜ 800 atm × 1atm 3 ⎜ ⎟ = 1061 kg/m 3 ρ = 1025 kg/m ⎜1 + ⎟ 2.3 × 109 N/m 2 ⎟ ⎜ ⎠ ⎝

(

)

Fluids

1023

74 • Picture the Problem When it is submerged, the block is in equilibrium under the influence of the buoyant force due to the water, the force exerted by the spring balance, and its weight. We can use the condition for translational equilibrium to relate the buoyant force to the weight of the block and the definition of density to express the weight of the block in terms of its density. Apply

∑F

y

= 0 to the block:

B + 0.8mg − mg = 0 ⇒ B = 0.2mg

Substitute for B and m to obtain:

ρ wVblock g = 0.2 ρ blockVblock g

Solve for and evaluate ρblock:

ρ block =

ρw 0 .2

(

= 5 103 kg/m 3

)

= 5.00 × 10 3 kg/m 3 *75 • Picture the Problem When the copper block is floating on a pool of mercury, it is in equilibrium under the influence of its weight and the buoyant force acting on it. We can apply the condition for translational equilibrium to relate these forces. We can find the fraction of the block that is submerged by applying Archimedes’ principle and the definition of density to express the forces in terms of the volume of the block and the volume of the displaced mercury. Let V represent the volume of the copper block, V′ the volume of the displaced mercury. Then the fraction submerged when the material is floated on water is V′/ V. Choose the upward direction to be the positive y direction. Apply

∑F

y

= 0 to the block:

B − w = 0 , where B is the buoyant force and w is the weight of the block.

Apply Archimedes’ principle and the definition of density to obtain:

ρ HgV'g − ρ CuVg = 0

Solve for V′/ V:

V' ρ Cu = V ρ Hg

1024 Chapter 13 Substitute numerical values and evaluate V′/ V:

V' 8.93 × 103 kg/m3 = = 0.657 = 65.7% V 13.6 × 103 kg/m 3

76 • Picture the Problem When the block is floating on a pool of ethanol, it is in equilibrium under the influence of its weight and the buoyant force acting on it. We can apply the condition for translational equilibrium to relate these forces. We can find the fraction of the block that is submerged by applying Archimedes’ principle and the definition of density to express the forces in terms of the volume of the block and the volume of the displaced ethanol. Let V represent the volume of the copper block, V′ the volume of the displaced ethanol. Then the fraction of the volume of the block that will be submerged when the material is floated on water is V′/ V. Choose the upward direction to be the positive y direction. Apply

∑F

y

= 0 to the block

floating on ethanol:

Apply Archimedes’ principle to obtain: Apply

∑F

y

= 0 to the block

Beth − w = 0 , where Beth is the buoyant force due to the ethanol and w is the weight of the block.

w = ρ eth (0.9V )g

Bw − w = 0 , where Bw is the buoyant force

floating on water:

due to the water and w is the weight of the block.

Apply Archimedes’ principle to obtain:

w = ρ wV'g , where V′ is the volume of the displaced water.

Equate the two expressions for w and solve for V′/ V:

V' 0.9 ρ eth = V ρw

Substitute numerical values and evaluate V′/ V:

V' 0.9 0.806 × 103 kg/m 3 = V 103 kg/m3

(

)

= 0.725 = 72.5% 77 • Determine the Concept If you are floating, the density (or specific gravity) of the liquid in which you are floating is immaterial as you are in translational equilibrium under the influence of your weight and the buoyant force on your body. Thus, the buoyant force on your body is your weight in both (a) and (b).

Fluids

1025

78 • Picture the Problem Let m and V represent the mass and volume of your body. Because you are in translational equilibrium when you are floating, we can apply the condition for translational equilibrium and Archimedes’ principle to your body to express the dependence of the volume of water it displaces when it is fully submerged on your weight. Let the upward direction be the positive y direction. Apply

∑F

y

= 0 to your floating body:

Use Archimedes’ principle to relate the density of water to your volume: Substitute to obtain: Solve for V:

B − mg = 0 B = wf = mf g = ρ w (0.96V )g

ρ w (0.96V )g − mg = 0 V=

m 0.96 ρ w

79 •• Picture the Problem Let m and V represent the mass and volume of the block of wood. Because the block is in translational equilibrium when it is floating, we can apply the condition for translational equilibrium and Archimedes’ principle to express the dependence of the volume of water it displaces when it is fully submerged on its weight. We’ll repeat this process for the situation in which the lead block is resting on the wood block with the latter fully submerged. Let the upward direction be the positive y direction. Apply

∑F

y

= 0 to floating block:

Use Archimedes’ principle to relate the density of water to the volume of the block of wood: Using the definition of density, express the weight of the block in terms of its density: Substitute to obtain: Solve for and evaluate the density of the wood block:

B − mg = 0 B = wf = mf g = ρ w (0.68V )g

mg = ρ woodVg

ρ w (0.68V )g − ρ woodVg = 0 ρ wood = 0.68ρ w = 0.68(103 kg/m 3 ) = 680 kg/m 3

1026 Chapter 13 Use the definition of density to find the volume of the wood:

V =

mwood

ρ wood

=

1.5 kg 680 kg/m 3

= 2.206 × 10−3 m 3 Apply

∑F

y

= 0 to the floating

block when the lead block is placed on it:

Use Archimedes’ principle and the definition of density to obtain:

B' − m' g = 0 , where B′ is the new buoyant force on the block and m′ is the combined mass of the wood block and the lead block.

ρ wVg − (mPb + mblock )g = 0

Solve for the mass of the lead block:

mPb = ρ wV − mblock

Substitute numerical values and evaluate mPb:

mPb = 103 kg/m 3 2.206 × 10−3 m 3

(

)(

)

− 1.5 kg = 0.706 kg

*80 •• Picture the Problem The true mass of the Styrofoam cube is greater than that indicated by the balance due to the buoyant force acting on it. The balance is in rotational equilibrium under the influence of the buoyant and gravitational forces acting on the Styrofoam cube and the brass masses. Neglect the buoyancy of the brass masses. Let m and V represent the mass and volume of the cube and L the lever arm of the balance. Apply

r

∑ τ = 0 to the balance:

(mg − B )L − mbrass gL = 0

Use Archimedes’ principle to express the buoyant force on the Styrofoam cube as a function of volume and density of the air it displaces:

B = ρ airVg

Substitute and simplify to obtain:

m − ρ airV − mbrass = 0

Solve for m:

m = ρ airV + mbrass

Substitute numerical values and evaluate m:

m = 1.293 kg/m 3 (0.25 m ) + 20 × 10 −3 kg

(

)

3

= 4.02 × 10 −2 kg = 40.2 g

Fluids

1027

81 •• Picture the Problem Let din and dout represent the inner and outer diameters of the copper shell and V′ the volume of the sphere that is submerged. Because the spherical shell is floating, it is in translational equilibrium and we can apply a condition for translational equilibrium to relate the buoyant force B due to the displaced water and its weight w. Apply

∑F

y

= 0 to the spherical shell:

Using Archimedes’ principle and the definition of w, substitute to obtain:

B−w=0

ρ wV'g − mg = 0 or

ρ wV' − m = 0

(1)

1π 3 π 3 d out = d out 26 12

Express V′ as a function dout:

V' =

Express m in terms of din and dout:

m = ρ Cu (Vout − Vin )

⎛π 3 π 3 ⎞ = ρ Cu ⎜ d out − d in ⎟ 6 ⎠ ⎝6

π

⎛π 3 π 3 ⎞ 3 d out − ρ Cu ⎜ d out − d in ⎟ = 0 12 6 ⎠ ⎝6

Substitute in equation (1) to obtain:

ρw

Simplify:

1 3 3 ρ w d out − ρ Cu (d out − d in3 ) = 0 2

Solve for din:

Substitute numerical values and evaluate din:

d in = d out 3 1 −

ρw 2 ρ Cu

d in = (12 cm ) 3 1 −

1 = 11.8 cm 2(8.93)

82 •• Determine the Concept The additional weight on the beaker side equals the weight of the displaced water, i.e., 64 g. This is the mass that must be placed on the other cup to maintain balance. *83 •• Picture the Problem We can use the definition of Reynolds number and assume a value for NR of 1000 (well within the laminar flow range) to obtain a trial value for the radius

1028 Chapter 13 of the pipe. We’ll then use Poiseuille’s law to determine the pressure difference between the ends of the pipe that would be required to maintain a volume flow rate of 500 L/s.

2rρv

Use the definition of Reynolds number to relate NR to the radius of the pipe:

NR =

Use the definition of IV to relate the volume flow rate of the pipe to its radius:

I V = Av = π r 2v ⇒ v =

Substitute to obtain:

NR =

Solve for r:

r=

Substitute numerical values and evaluate r:

2 700 kg/m 3 0.500 m 3 /s r= = 27.9 cm π (0.8 Pa ⋅ s )(1000)

Using Poiseuille’s law, relate the pressure difference between the ends of the pipe to its radius:

∆P =

8ηL I π r4 V

Substitute numerical values and evaluate ∆P:

∆P =

8(0.8 Pa ⋅ s )(50 km ) 0.500m 3 /s 4 π (0.279 m )

η

IV π r2

2 ρI V ηπ r

2 ρI V ηπ N R

(

)(

)

(

)

= 8.41 × 10 6 Pa = 8.41 × 10 6 Pa ×

1 atm 1.01325 × 105 Pa

= 83.0 atm This pressure is too large to maintain in the pipe. Evaluate ∆P for a pipe of 50 cm radius:

∆P =

8(0.8 Pa ⋅ s )(50 km ) 0.500 m 3 /s 4 π (0.50 m )

(

)

= 8.15 × 105 Pa = 8.15 × 105 Pa × = 8.04 atm

1atm 1.01325 × 105 Pa

Fluids 1029 1 m is a reasonable diameter for the pipeline. 84 •• Picture the Problem We’ll measure the height of the liquid–air interfaces relative to the centerline of the pipe. We can use the definition of the volume flow rate in a pipe to find the speed of the water at point A and the relationship between the gauge pressures at points A and C to determine the level of the liquid-air interface at A. We can use the continuity equation to express the speed of the water at B in terms of its speed at A and Bernoulli’s equation for constant elevation to find the gauge pressure at B. Finally, we can use the relationship between the gauge pressures at points A and B to find the level of the liquid-air interface at B. Relate the gauge pressure in the pipe at A to the height of the liquid-air interface at A: Solve for hA:

Substitute numerical values and evaluate hA:

Pgauge,A = ρghA where hA is measured from the center of the pipe.

hA =

hA

Pgauge,A

ρg

( 1.22 atm ) (1.01 × 105 Pa/atm ) =

(10

3

)(

kg/m 3 9.81 m/s 2

)

= 12.6 m Determine the velocity of the water at A:

I V 0.8 × 10 −3 m 3 /s vA = = = 2.55 m/s π AA (0.02 m )2 4

Apply Bernoulli’s equation for constant elevation to relate PB and PA:

PA + 12 ρvA2 = PB + 12 ρvB2

Use the continuity equation to relate vB and vA:

AA vA = ABvB

Solve for vB:

(2 cm) v = 4v A d2 vB = A vA = A2 vA = A AB dB (1 cm )2 A

Substitute in equation (1) to obtain:

PA + 12 ρvA2 = PB + 8ρvA2

(1)

2

1030 Chapter 13 Solve for PB: Substitute numerical values and evaluate PB:

PB = PA − 152 ρvA2

(

PB = (2.22 atm ) 1.01× 105 Pa/atm

(

)

)

− 152 103 kg/m 3 (2.55 m/s )

2

= 1.75 × 105 Pa = 1.75 × 105 Pa ×

1atm 1.01× 105 Pa

= 1.733 atm Relate the gauge pressure in the pipe at B to the height of the liquid-air interface at B: Solve for hB:

Substitute numerical values and evaluate hB:

Pgauge,B = ρghB

hB =

Pgauge,B

ρg

[(1.733 − 1)atm] ⎛⎜1.01×105 hB =

(10

⎝

3

kg/m

3

Pa ⎞ ⎟ atm ⎠

)(9.81m/s ) 2

= 7.55 m 85 •• Picture the Problem We’ll measure the height of the liquid–air interfaces relative to the centerline of the pipe. We can use the definition of the volume flow rate in a pipe to find the speed of the water at point A and the relationship between the gauge pressures at points A and C to determine the level of the liquid-air interface at A. We can use the continuity equation to express the speed of the water at B in terms of its speed at A and Bernoulli’s equation for constant elevation to find the gauge pressure at B. Finally, we can use the relationship between the gauge pressures at points A and B to find the level of the liquid-air interface at B. Relate the gauge pressure in the pipe at A to the height of the liquid-air interface at A: Solve for hA:

Pgauge,A = ρghA where hA is measured from the center of the pipe.

hA =

Pgauge,A

ρg

Fluids 1031 Substitute numerical values and evaluate hA:

hA =

(1.22 atm )(1.01×105 Pa/atm )

(10

3

)(

kg/m 3 9.81 m/s 2

)

= 12.6 m Determine the velocity of the water at A:

vA =

I V 0.6 × 10 −3 m 3 /s = = 1.91 m/s π 2 AA (0.02 m ) 4

Use the continuity equation to relate vB and vA:

AA vA = ABvB

Solve for vB:

(2 cm ) v A d2 vB = A vA = A2 vA = AB dB (1 cm )2 A 2

= 4v A Apply Bernoulli’s equation for constant elevation to relate PB and PA:

PA + 12 ρvA2 = PB + 12 ρvB2

Substitute in equation (1) to obtain:

PA + 12 ρvA2 = PB + 8ρvA2

Solve for PB:

PB = PA − 152 ρvA2

Substitute numerical values and evaluate PB:

(

(1)

PB = (2.22 atm ) 1.01× 105 Pa/atm

(

)

)

− 152 103 kg/m 3 (1.91 m/s )

2

= 1.969 × 105 Pa = 1.969 × 105 Pa × = 1.95 atm Relate the gauge pressure in the pipe at B to the height of the liquid-air interface at B: Solve for hB:

Pgauge,B = ρghB

hB =

Pgauge,B

ρg

1atm 1.01× 105 Pa

1032 Chapter 13

[(1.95 − 1)atm] ⎛⎜1.01×105

Substitute numerical values and evaluate hB:

hB =

(10

⎝

3

kg/m

3

Pa ⎞ ⎟ atm ⎠

)(9.81m/s ) 2

= 9.78 m *86 •• Picture the Problem Because it is not given, we’ll neglect the difference in height between the centers of the pipes at A and B. We can use the definition of the volume flow rate to find the speed of the water at A and Bernoulli’s equation for constant elevation to find its speed at B. Once we know the speed of the water at B, we can use the equation of continuity to find the diameter of the constriction at B. Use the definition of the volume flow rate to find vA:

I V 0.5 × 10 −3 m 3 /s = = 1.59 m/s vA = π AA (0.02 m )2 4

Use Bernoulli’s equation for constant elevation to relate the pressures and velocities at A and B:

PB + 12 ρvB2 = PA + 12 ρvA2

Solve for vB2 :

vB2 =

2(PA − PB )

ρ

+ vA2

Substitute numerical values and evaluate vB2 :

vB2 =

[

)]

(

2 (1.187 − 0.1) atm 1.01× 105 Pa/atm 2 + (1.59 m/s ) = 222 m 2 /s 2 3 3 10 kg/m

Using the continuity equation, relate the volume flow rate to the radius at B: Solve for and evaluate rB:

I V = ABvB = π rB2vB

rB =

IV = π vB

0.5 × 10−3 m 3 /s = 3.27 mm π (14.9 m/s )

and

d B = 2rB = 6.54 mm

Fluids 1033 87 •• Picture the Problem Let V′ represent the volume of the buoy that is submerged and h′ the height of the submerged portion of the cylinder. We can find the fraction of the cylinder’s volume that is submerged by applying the condition for translational equilibrium to the buoy and using Archimedes’ principle. When the buoy is submerged it is in equilibrium under the influence of the tension T in the cable, the buoyant force due to the displaced water, and its weight. When the cable breaks, the net force acting on the buoy will accelerate it and we can use Newton’s 2nd law to find its acceleration. (a) Apply

∑F

y

= 0 to the cylinder:

Using Archimedes’ principle and the definition of weight, substitute for B and w:

B−w=0

ρswV'g − mg = 0 or

ρsw h'Ag − mg = 0

where A is the cross-sectional area of the buoy. Solve for and evaluate h′:

Substitute numerical values and evaluate h′:

h' =

h' =

m

ρsw A 600 kg

(1.025 ×10

3

kg/m 3

) π4 (0.9 m)

2

= 0.920 m Use h′ to find the height h of the buoy:

h − h' = 2.6 m − 0.920 m = 1.68 m

Express the fraction of the volume of the cylinder that is above water:

d 2 h' V − V' V' 4 = 1− = 1− π 2 V V d h 4 h' = 1− h

Substitute numerical values to obtain:

V − V' 0.920 m = 1− = 64.6% V 2.6 m

(b) Apply

∑F

y

submerged buoy:

= 0 to the

π

B −T − w = 0

1034 Chapter 13 Solve for T and substitute for B and w to obtain:

T = B − w = ρ swVg − mg

Substitute numerical values and evaluate T:

T = 1.025 × 103 kg/m 3

= (ρ swV − m )g

[(

) π4 (0.9 m ) (2.6 m) − 600 kg ](9.81 m/s ) 2

2

= 10.7 kN (c) Apply

∑F

y

= 0 to the buoy:

B − w = ma

Substitute for B − w and solve for a to obtain:

a=

B−w T = m m

Substitute numerical values and evaluate a:

a=

10.75 kN = 17.9 m/s 2 600 kg

88 •• Picture the Problem Because the floating object is in equilibrium under the influence of the buoyant force acting on it and its weight; we can apply the condition for translational equilibrium to relate B and w. Let ∆h represent the change in elevation of the liquid level and Vf the volume of the displaced fluid. Apply

∑F

y

= 0 to the floating

B−w=0

object: Using Archimedes’ principle and the definition of weight, substitute for B and w:

ρ 0 gVf − mg = 0

The volume of fluid displaced is the sum of the volume displaced in the two vessels:

Vf = ∆VA + ∆V3 A = A∆h + 3 A∆h

Substitute for Vf to obtain:

4 ρ 0 gA∆h − mg = 0

Solve for ∆h:

= 4 A∆h

∆h =

m 4 Aρ 0

Fluids 1035 89 •• Picture the Problem We can calculate the smallest pressure change ∆P that can be detected from the reading ∆h from ∆P = ρg∆h. Express and evaluate the pressure difference between the two columns of the manometer:

∆P = ρg∆h

(

)(

= 900 kg/m 3 9.81 m/s 2

(

× 0.05 × 10 −3 m

)

)

= 0.4415 Pa Express this pressure in mmHg and µmHg:

∆P = 0.4415 Pa × ×

1 atm 1.01325 × 105 Pa

760 mmHg 1 atm

= 3.31 × 10 −3 mmHg = 3.31 µmHg 90 •• Picture the Problem We can use the equality of the pressure at the bottom of the U-tube due to the water on one side and that due to the oil and water on the other to relate the various heights. Let h represent the height of the oil above the water. Then ho = h1w + h.

Using the constancy of the amount of water, express the relationship between h1w and h2w:

h1w + h2w = 56 cm

Find the height of the oil-water interface:

h1w = 56 cm − 34 cm = 22.0 cm

Express the equality of the pressure at the bottom of the two arms of the U tube:

ρ w g (34 cm ) = ρ w g (22 cm ) + 0.78ρ w ghoil

1036 Chapter 13 Solve for and evaluate hoil:

hoil = =

Find the height of the air-oil interface ho:

ρ w g (34 cm ) − ρ w g (22 cm ) 0.78ρ w g

(34 cm ) − (22 cm ) = 15.4 cm 0.78

ho = 22cm + 15.4 cm = 37.4 cm

91 •• Picture the Problem Let σL represent the specific gravity of the liquid. The specific gravity of the oil is σo = 0.8. We can use the equality of the pressure at the bottom of the U-tube due to the water on one side and that due to the oil and water on the other to relate the various heights. Express the equality of the pressure at the bottom of the two arms of the U tube: Solve for and evaluate σL:

σ L gh = σ L g (h − 7 cm ) + 0.8σ w g (12 cm )

σL =

0.8σ w (12 cm ) 0.8(12 cm ) = 7 cm 7 cm

= 1.37 92 •• Picture the Problem The block of wood is in translational equilibrium under the influence of the buoyant force due to the displaced water acting on it and on the lead block, its weight, and the weight of the lead block. We can use a condition for translational equilibrium and Archimedes’ principle to obtain a relationship between the mass of the lead block and the densities of water, wood, and lead and the mass of the wood block. Apply

∑F

y

= 0 to the block of wood:

Bwood + BPb − mwood g − mPb g = 0

Use Archimedes’ principle to express the buoyant force on the block of wood:

Bwood = ρ wVwood g

Use Archimedes’ principle to

BPb = ρ wVPb g

Fluids 1037 express the buoyant force on the lead block: Substitute and simplify to obtain:

ρ wVwood + ρ wVPb − mwood − mPb = 0

Express the volume of the wood block in terms of its density and mass:

Vwood =

Express the volume of the lead block in terms of its density and mass:

VPb =

Substitute for Vwood and VPb:

ρw

Solve for mPb:

⎛ ρw ⎞ ⎜⎜ − 1⎟⎟mwood ρ ⎠ mPb = ⎝ wood ρ 1− w

mwood

ρ wood

mPb

ρ Pb

mwood

ρ wood

+ ρw

mPb

ρ Pb

− mwood − mPb = 0

ρ Pb

Substitute numerical values and evaluate mPb:

⎛ 1 ⎞ − 1⎟(0.5 kg ) ⎜ 0.7 ⎠ mPb = ⎝ = 0.235 kg 1 1− 11.3

*93 •• Picture the Problem Because the balloon is in equilibrium under the influence of the buoyant force exerted by the air, the weight of its basket and load w, the weight of the skin of the balloon, and the weight of the helium. Choose upward to be the positive y direction and apply the condition for translational equilibrium to relate these forces. Archimedes’ principle relates the buoyant force on the balloon to the density of the air it displaces and the volume of the balloon. (a) Apply

∑F

y

= 0 to the balloon:

B − mskin g − mHe g − w = 0

Letting V represent the volume of the balloon, use Archimedes’ principle to express the buoyant force:

ρairVg − mskin g − mHe g − w = 0

Substitute for mHe:

ρairVg − mskin g − ρ HeVg − w = 0

1038 Chapter 13 Solve for V:

V=

mskin g + w (ρ air − ρ He )g

Substitute numerical values and evaluate V:

V=

(1.5 kg )(9.81m/s 2 ) + 750 N (1.293 − 0.1786)(kg/m 3 )(9.81m/s 2 )

= 70.0 m 3 (b) Apply

∑F

y

= ma to the balloon:

B − mtot g = mtot a B −g mtot

Solve for a:

a=

Assuming that the mass of the skin has not changed and letting V′ represent the doubled volume of the balloon, express mtot:

mtot = mload + mHe + mskin =

wload + ρ HeV '+ mskin g

Substitute numerical values and evaluate mtot:

mtot =

(

)(

)

900 N + 0.1786 kg/m 3 140 m 3 + 1.5 kg = 118 kg 2 9.81m/s

Express the buoyant force acting on the balloon:

B = wdisplaced fluid = ρ airV ' g

Substitute numerical values and evaluate B:

B = 1.293 kg/m 3 140 m 3 9.81 m/s 2 = 1.78 kN 1.78 kN a= − 9.81 m/s 2 = 5.27 m/s 2 118 kg

Substitute and evaluate a:

(

)(

)(

)

94 •• Picture the Problem When the hollow sphere is completely submerged but floating, it is in translational equilibrium under the influence of a buoyant force and its weight. The buoyant force is given by Archimedes’ principle and the weight of the sphere is the sum of the weights of the hollow sphere and the material filling its center. Apply

∑F

y

= 0 to the hollow sphere:

B−w=0

Fluids 1039

[

]

Express the buoyant force acting on the hollow sphere:

B = 2 ρ 0Vsphere g = 2 ρ 0 43 π (2 R ) g

Express the weight of the sphere when it’s hollow is filled with a material of density ρ′:

w = ρ 0Vhollow sphere g + ρ 'Vhollow g

=

64 3

3

ρ 0π R 3 g

[ {

}]

[

]

= ρ 0 43 π (2 R ) − R 3 g + ρ' 43 πR 3 g =

28 3

3

ρ 0π R g + 43 ρ'π R g 3

3

ρ0π R 3 g − 283 ρ 0π R 3 g − 43 ρ'π R 3 g = 0

Substitute to obtain:

64 3

Solve for ρ′:

ρ ' = 9 ρ0

*95 •• Picture the Problem We can differentiate the function P(h) to show that it satisfies the differential equation dP/P = −C dh and in part (b) we can use the approximation e−x ≈ 1 – x and ∆h << h0 to establish the given result. (a) Differentiate P(h) = P0 e−Ch:

Separate variables to obtain:

(b) Express P(h + ∆h):

dP = −CP0 e −Ch dh = − CP dP = −Cdh P

P(h + ∆h ) = P0e − C (h+∆h ) = P0e −Ch e −C∆h = P(h ) e −C∆h

For ∆h << h0:

∆h << 1 h0

Let h0 = 1/C. Then:

C∆h << 1 and

e −C∆h ≈ 1 − C∆h = 1 −

Substitute to obtain:

∆h h0

⎛ ∆h ⎞ ⎟ P(h + ∆h ) = P(h )⎜⎜1 − h0 ⎟⎠ ⎝

1040 Chapter 13 (c) Take the logarithm of both sides of the function P(h):

ln P = ln P0 e − Ch = ln P0 + ln e − Ch

Solve for C:

C=

1 ⎛ P0 ⎞ ln⎜ ⎟ h ⎝P⎠

C=

⎛ P ⎞ 1 1 ln⎜⎜ 1 0 ⎟⎟ = ln 2 5.5 km ⎝ 2 P0 ⎠ 5.5 km

Substitute numerical values and evaluate C:

= ln P0 − Ch

= 0.126 km −1 96 •• Picture the Problem Let V represent the volume of the submarine and V ′ the volume of seawater it displaces when it is on the surface. The submarine is in equilibrium in both parts of the problem. Hence we can apply the condition for translational equilibrium (neutral buoyancy) to the submarine to relate its weight to the buoyant force acting on it. We’ll also use Archimedes’ principle to connect the buoyant forces to the volume of seawater the submarine displaces. Let upward be the positive y direction. (a) Express f, the fraction of the submarine’s volume above the surface when the tanks are filled with air: Apply

∑F

y

= 0 to the submarine

f =

V − V' V' = 1− V V

B−w=0

when its tanks are full of air: Use Archimedes’ principle to express the buoyant force on the submarine in terms of the volume of the displaced water:

B = ρ swV'g

Substitute and solve for V′:

V' =

Substitute in equation (1) to obtain:

m

ρsw

f = 1−

m ρswV

(1)

Fluids 1041 Substitute numerical values and evaluate f:

f = 1−

2.4 × 106 kg 1.025 × 103 kg/m 3 2.4 × 103 m 3

(

)(

)

= 2.44 × 10−2 = 2.44 % (b) Express the volume of seawater in terms of its mass and density: Apply

∑F

y

= 0 , the condition for

Vsw =

msw

(2)

ρsw

B − wsub − wsw = 0

neutral buoyancy, to the submarine: Use Archimedes’ principle to express the buoyant force on the submarine in terms of the volume of the displaced water:

B = ρswVg

Substitute to obtain:

ρswVg − msub g − msw g = 0

Solve for msw:

msw = ρ swV − msub

Substitute for Vsw in equation (2) to obtain:

Vsw =

Substitute numerical values and evaluate Vsw:

Vsw = 2.4 × 103 m 3 −

ρ swV − msub m = V − sub ρ sw ρ sw 2.4 × 10 6 kg 1.025 × 103 kg/m 3

= 58.5 m 3 97 •• Picture the Problem While the loaded crate is under the surface, it is in equilibrium under the influence of the tension in the cable, the buoyant force acting on the gold, and the gravitational force acting on the gold. The empty crate has neutral buoyancy. When the crate is out of the water, the buoyant force of the air is negligible and the tension in the cable is the sum of the weights of the crate, the gold bullion, and the seawater. (a) Apply

∑F

y

= 0 to the crate

T + BAu − wAu = 0

while it is below the surface: Solve for the tension in the cable:

T = wAu − BAu

Using Archimedes’ principle, relate the buoyant force acting on the gold

BAu = ρswVAu g

1042 Chapter 13 to its density and volume: Substitute for BAu and simplify to obtain:

T = (ρ Au − ρsw )VAu g

Substitute numerical values and evaluate T:

[(

) (

)]

(

T = 19.3 × 103 kg/m 3 − 1.025 × 103 kg/m 3 (0.36)(1.4 m )(0.75 m )(0.5 m ) 9.81 m/s 2

)

= 33.9 kN (b) 1. Apply

∑F

y

= 0 to the crate

T − wAu − wcrate − wsw = 0

while it is being lifted to the deck of the ship with none of the seawater leaking out: Substitute for the weights of the gold, crate, and seawater and solve for the tension in the cable and express:

T = wAu + wcrate + wsw = ρ AuVAu g + mcrate g + ρswVsw g = (ρ AuVAu + mcrate + ρswVsw )g

Substitute numerical values and evaluate T:

[(

)

(

T = 19.3 × 103 kg/m 3 (0.36 )(1.4 m )(0.75 m )(0.5 m ) + 32 kg + 1.025 × 103 kg/m 3

(

× (0.64 )(1.4 m )(0.75 m )(0.5 m )] 9.81 m/s

2

)

)

= 39.8 kN 2. With the seawater term missing, the expression for the tension is:

T = wAu + wcrate = ρ AuVAu g + mcrate g

= (ρ AuVAu + mcrate )g Substitute numerical values and evaluate T:

[

T = (19.3 × 103 kg/m 3 )(0.36 )(1.4 m )(0.75 m )(0.5 m ) + 32 kg ](9.81 m/s 2 ) = 36.1 kN 98 ••• Picture the Problem In the three situations described in the problem the hydrometer will be in equilibrium under the influence of its weight and the buoyant force exerted by the liquids. We can use Archimedes’ principle to relate the buoyant force acting on the hydrometer to the density of the liquid in which it is floating and to its weight.

Fluids 1043 (a) Find the volume of the bulb:

Vbulb = 16 πd 3 = 16 π (2.4 cm ) = 7.238 cm 3

Find the volume of the tube:

Vtube = 14 πd 2 L = 14 π (0.75 cm ) (20 cm )

3

2

= 8.836 cm 2 Apply

∑F

y

= 0 to the hydrometer

B − whyd − mPb g = 0

just floating in the liquid: Substitute for B and wglass:

ρliqVhyd g − mglass g − mPb g = 0

Solve for mPb:

mPb = ρ liqVhyd − mhyd

Substitute numerical values and evaluate mPb:

mPb = 0.78 1 g/cm 3

(

(

)

× 7.238 cm 3 + 8.836 cm 3

)

− 7.28 g = 5.26 g (b) Letting V represent the volume of the hydrometer that is submerged, Fy = 0 to the hydrometer apply

ρ wVg − mg = 0

∑

just floating in the liquid: Solve for V:

Substitute numerical values and evaluate V: Relate the volume of the hydrometer that is submerged to the volume of the bulb and the volume of the tube that is submerged:

V=

V=

m

ρw

=

mhyd + mPb

ρw

7.28 g + 5.26 = 12.54 cm 3 3 1 g/cm

2 V = 14 π d tube h' + Vbulb

Solve for h′:

h' =

V − Vbulb 2 1 4 π d tube

Substitute numerical values and evaluate h′:

h' =

12.54 cm3 − 7.238 cm3 = 12.0 cm 2 1 ( ) 0 . 75 cm π 4

1044 Chapter 13 Find the length of the tube that shows above the surface of the water:

(c) Apply

∑F

y

= 0 to the

h = 20 cm − h' = 20 cm − 12.0 cm = 8.00 cm

ρ LVL g − mhyd g = 0

hydrometer floating in the liquid of unknown specific gravity: Solve for the density of the liquid:

Express the volume of the displaced liquid: Substitute numerical values and evaluate VL:

ρL =

mhyd VL

2 VL = Vbulb + 14 π d tube h'

VL = 7.238 cm 3 + 14 π (0.75 cm )

2

× (20 cm − 12.2 cm )

= 10.68 cm 3 12.54 g = 1.174 g/cm 3 3 10.68 cm

Substitute for VL and mhyd and evaluate ρL:

ρL =

Express and evaluate the specific gravity of the liquid:

specific gravityliquid =

ρw = 1.17 ρL

99 ••• Picture the Problem We can apply Bernoulli’s equation to the top of the keg and to the spigot opening to determine the rate at which the root beer exits the tank. Because the area of the spigot is much smaller than that of the keg, we can neglect the velocity of the root beer at the top of the keg. We’ll use the continuity equation to obtain an expression for the rate of change of the height of the root beer in the keg as a function of the its height and integrate this function to find h as a function of time. (a) Apply Bernoulli’s equation to the beer at the top of the keg and at the spigot:

P1 + ρ beer gh + 12 ρ beer v12 = P2 + ρ beer gh2 + 12 ρ beer v22 or, because v1 ≈ 0, h2 = 0, P1 = P2 = Pat, and h1 = h,

gh = 12 v22 Solve for v2:

v2 =

2 gh

Fluids 1045 (b) Use the continuity equation to relate v1 and v2:

A1v1 = A2v2

Substitute −dh/dt for v1:

− A1

Substitute for v2 and solve for dh/dt to obtain:

dh A = − 2 2 gh dt A1

(c) Separate the variables in the differential equation:

−

Express the integral from h′ = H to h and t′ = 0 to t:

A A − 1 2 2g

Evaluate the integral to obtain:

−

Solve for h:

⎛ ⎞ A h = ⎜⎜ H − 2 2 g t ⎟⎟ 2 A1 ⎝ ⎠

dh = A2v2 dt

A1 A2 dh = dt 2g h

A1 A2 2g

(d) Solve h(t) for the time-to-drain t′:

t' =

Substitute numerical values and evaluate t′

t' =

h

t

dh' ∫H h' = ∫0 dt'

(

A1 A2

)

H − h =t

2

2H g A1 −4

10 A1

2(2 m ) = 6.39 × 103 s 2 9.81 m/s

= 1 h 46 min

1046 Chapter 13

Chapter 14 Oscillations Conceptual Problems 1 • Determine the Concept The acceleration of an oscillator of amplitude A and frequency f is zero when it is passing through its equilibrium position and is a maximum when it is at its turning points. When v = vmax:

a= 0

When x = xmax:

a = ω 2 A = 4π 2 f 2 A

2 • Determine the Concept The condition for simple harmonic motion is that there be a linear restoring force; i.e., that F = −kx. Thus, the acceleration and displacement (when they are not zero) are always oppositely directed. v and a can be in the same direction, as can v and x. 3 • (a) False. In simple harmonic motion, the period is independent of the amplitude. (b) True. In simple harmonic motion, the frequency is the reciprocal of the period which, in turn, is independent of the amplitude. (c) True. The condition that the acceleration of a particle is proportional to the displacement and oppositely directed is equivalent to requiring that there be a linear restoring force; i.e., F = −kx ⇔ ma = −kx or a = − (k/m)x. *4 • Determine the Concept The energy of a simple harmonic oscillator varies as the square of the amplitude of its motion. Hence, tripling the amplitude increases the energy by a factor of 9. 5 •• Picture the Problem The total energy of an object undergoing simple harmonic motion is given by Etot = 12 kA2 , where k is the stiffness constant and A is the amplitude of the motion. The potential energy of the oscillator when it is a distance x from its equilibrium position is U ( x ) = 12 kx 2 .

1047

1048 Chapter 14 Express the ratio of the potential energy of the object when it is 2 cm from the equilibrium position to its total energy: Evaluate this ratio for x = 2 cm and A = 4 cm:

U (x ) 12 kx 2 x2 = 1 2 = 2 Etot A 2 kA

U (2 cm ) (2 cm ) 1 = = 2 Etot (4 cm ) 4 2

U (2 cm ) (2 cm ) 1 = = 2 Etot (4 cm ) 4 2

and (a ) is correct. 6 • (a) True. The factors determining the period of the object, i.e., its mass and the spring constant, are independent of the oscillator’s orientation. (b) True. The factors determining the maximum speed of the object, i.e., its amplitude and angular frequency, are independent of the oscillator’s orientation. 7 • False. In order for a simple pendulum to execute simple harmonic motion, the restoring force must be linear. This condition is satisfied, at least approximately, for small initial angular displacements. 8 • True. In order for a simple pendulum to execute periodic motion, the restoring force must be linear. This condition is satisfied for any initial angular displacement. *9 •• Determine the Concept Assume that the first cart is given an initial velocity v by the blow. After the initial blow, there are no external forces acting on the carts, so their center of mass moves at a constant velocity v/2. The two carts will oscillate about their center of mass in simple harmonic motion where the amplitude of their velocity is v/2. Therefore, when one cart has velocity v/2 with respect to the center of mass, the other will have velocity −v/2. The velocity with respect to the laboratory frame of reference will be +v and 0, respectively. Half a period later, the situation is reversed; one cart will move as the other stops, and vice-versa. *10 •• Determine the Concept The period of a simple pendulum depends on the reciprocal of the length of the pendulum. Increasing the length of the pendulum will decrease its period

Oscillations 1049 and the clock would run slow. 11 • True. The mechanical energy of a damped, undriven oscillator varies with time according to E = E0e − t τ where E0 is the oscillator’s energy at t = 0 and τ is the time constant. 12 • (a) True. The amplitude of the motion of a driven oscillator depends on the driving (ω) and natural (ω0) frequencies according to A = F0

(

)

2

m 2 ω02 − ω 2 + b 2ω 2 . When

ω = ω0, the amplitude of the motion is a maximum and is given by A = F0

b 2ω 2 .

(b) True. The width of the resonance curve (∆ω) depends on the Q value according to ∆ω ω0 = 1 Q . Thus when Q is large, ∆ω is small and the resonance is sharp. 13 • Determine the Concept Examples of driven oscillators include the pendulum of a clock, a bowed violin string, and the membrane of any loudspeaker. 14 • Determine the Concept The shattering of a crystal wineglass is a consequence of the glass being driven at or near its resonant frequency. (a ) is correct. *15 • Determine the Concept We can use the expression for the frequency of a spring-andmass oscillator to determine the effect of the mass of the spring. If m represents the mass of the object attached to the spring in a spring-and-mass oscillator, the frequency is given by: If the mass of the spring is taken into account, the effective mass is greater than the mass of the object alone. Divide the second of these equations by the first and simplify to obtain:

f =

1 2π

k m

f' =

1 2π

k meff

1 2π

k meff

f' = f

1 2π

k m

=

m meff

1050 Chapter 14 Solve for f ′:

f' = f

m meff

Because f' varies inversely with the square root of m, taking into account the effective mass of the spring predicts that the frequency will be reduced. 16 •• Determine the Concept The period of the lamp varies inversely with the square root of the effective value of the local gravitational field. 1. greater than T0 when

B. the train rounds a curve of radius R with speed v.

2. less than T0 when

D. the train goes over the crest of a hill of radius of curvature R with constant speed.

3. equal to T0 when

A. the train moves horizontally with constant velocity. C. the train climbs a hill of inclination θ at constant speed.

17

••

Picture the Problem We can use f =

1 2π

k to express the frequencies of the two M

mass-spring systems in terms of their masses. Dividing one of the equations by the other will allow us to express MA in terms of MB. Express the frequency of massspring system A as a function of its mass: Express the frequency of massspring system B as a function of its mass:

fA =

1 2π

k MA

fB =

1 2π

k MB

Divide the second of these equations by the first to obtain:

fB = fA

Solve for MA:

⎛ f ⎞ ⎛ f ⎞ M A = ⎜⎜ B ⎟⎟ M B = ⎜⎜ B ⎟⎟ M B = 14 M B ⎝ fA ⎠ ⎝ 2 fB ⎠

MA MB 2

2

Oscillations 1051 and (d ) is correct. 18 •• Picture the Problem We can relate the energies of the two mass-spring systems through either E = 12 kA2 or E = 12 Mω 2 A2 and investigate the relationship between their amplitudes by equating the expressions, substituting for MA, and expressing AA in terms of AB. Express the energy of mass-spring system A:

EA = 12 k A AA2 = 12 M AωA2 AA2

Express the energy of mass-spring system B:

EB = 12 k B AB2 = 12 M BωB2 AB2

Divide the first of these equations by the second to obtain:

1 EA M ω 2 A2 = 1 = 21 A A2 A2 EB 2 M BωB AB

Substitute for MA and simplify:

2M BωA2 AA2 2ωA2 AA2 = 2 2 1= M BωB2 AB2 ωB AB

Solve for AA:

AA =

ωB AB 2ωA

Without knowing how ωA and ωB, or kA and kB, are related, we cannot simplify this expression further. (d ) is correct. 19 •• Picture the Problem We can express the energy of each system using E = 12 kA2 and, because the energies are equal, equate them and solve for AA. Express the energy of mass-spring system A in terms of the amplitude of its motion:

EA = 12 k A AA2

Express the energy of mass-spring system B in terms of the amplitude of its motion:

EB = 12 kB AB2

1052 Chapter 14 Because the energies of the two systems are equal we can equate them to obtain: Solve for AA:

Substitute for kA and simplify to obtain:

1 2

kA AA2 = 12 kB AB2

AA =

kB AB kA

AA =

kB A AB = B 2k B 2

and (b) is correct. 20 •• Picture the Problem The period of a simple pendulum is independent of the mass of its bob and is given by T = 2π L g . Express the period of pendulum A:

Express the period of pendulum B:

TA = 2π

LA g

TB = 2π

LB g 2

Divide the first of these equations by the second and solve for LA/LB:

LA ⎛ TA ⎞ =⎜ ⎟ LB ⎜⎝ TB ⎟⎠

Substitute for TA and solve for LB to obtain:

⎛ 2T ⎞ LA = ⎜⎜ B ⎟⎟ LB = 4 LB ⎝ TB ⎠

2

and (c) is correct.

Estimation and Approximation 21 •• Picture the Problem The Q factor for this system is related to the decay constant τ through Q = ω0τ = 2πτ T and the amplitude of the child’s damped motion varies with time according to A = A0 e − t 2τ . We can set the ratio of two displacements separated by eight periods equal to 1/e to determine τ in terms of T. Express Q as a function of τ :

q = ω 0τ =

2πτ T

(1)

Oscillations 1053 The amplitude of the oscillations varies with time according to:

A = A0 e −t 2τ

The amplitude after eight periods is:

A8 = A0 e − (t +8T ) 2τ

Express and simplify the ratio A8/A:

A8 A0e − (t +8T ) 2τ = = e − 4T τ A A0 e −t 2τ

Set this ratio equal to 1/e and solve for τ :

e −4T τ = e −1 ⇒ τ = 4T

Substitute in equation (1) and evaluate Q:

Q=

2π (4T ) = 8π T

*22 •• Picture the Problem Assume that an average length for an arm is about 0.8 m, and that it can be treated as a uniform stick, pivoted at one end. We can use the expression for the period of a physical pendulum to derive an expression for the period of the swinging arm. When carrying a heavy briefcase, the mass is concentrated mostly at the end of the pivot (i.e., in the briefcase), so we can treat the arm-plus-briefcase as a simple pendulum. (a) Express the period of a uniform rod pivoted at one end:

T = 2π

I MgD

where I is the moment of inertia of the stick about an axis through one end, M is the mass of the stick, and D (= L/2) is the distance from the end of the stick to its center of mass. Express the moment of inertia of the stick with respect to an axis through its end:

I = 13 ML2

Substitute the values for I and D to find T:

T = 2π

ML2 2L = 2π 1 Mg ( 2 L ) 3g

Substitute numerical values and evaluate T:

T = 2π

2(0.8 m ) = 1.47 s 3 9.81 m/s 2

(b) Express the period of a simple pendulum:

T '= 2π

L' g

1 3

(

)

where L′ is slightly longer than the arm

1054 Chapter 14 length due to the size of the briefcase. Assuming L′ = 1 m, evaluate the period of the simple pendulum:

T' = 2π

1m = 2.01s 9.81 m/s 2

From observation of people as they walk, these estimates seem reasonable.

Simple Harmonic Motion 23 • Picture the Problem The position of the particle is given by x = A cos(ωt + δ ) where A is the amplitude of the motion, ω is the angular frequency, and δ is a phase constant. (a) Use the definition of ω to determine f:

ω 6π s −1 f = = = 3.00 Hz 2π 2π

(b) Evaluate the reciprocal of the frequency:

T=

(c) Compare x = (7 cm) cos 6π t to x = A cos(ω t + δ ) :

A = 7.00 cm

(d) x = 0 when cosω t = 0:

ωt = cos −1 0 =

Solve for t:

t=

Differentiate x to find v(t):

v=

1 1 = = 0.333 s f 3.00 Hz

π 2

π π = = 0.0833 s 2ω 2(6π )

d [(7 cm)cos 6π t ] dt = −(42π cm/s )sin 6π t

Evaluate v(0.0833 s):

v(0.0833 s ) = −(42π cm/s)sin 6π (0.0833 s ) < 0 Because v < 0, the particle is moving in the negative direction at t = 0.0833 s. 24 • Picture the Problem The initial position of the oscillating particle is related to the amplitude and phase constant of the motion by x0 = A cos δ where 0 ≤ δ < 2π.

Oscillations 1055 (a) For x0 = 0:

cos δ = 0 and

δ = cos −1 0 = (b) For x0 = −A:

π 3π 2

,

2

− A = A cos δ and

δ = cos −1 (− 1) = π

(c) For x0 = A:

A = A cos δ and

δ = cos −1 (1) = 0

(d) When x = A/2:

A = A cos δ 2 and

⎛1⎞ ⎝2⎠

δ = cos −1 ⎜ ⎟ =

π 3

*25 • Picture the Problem The position of the particle as a function of time is given by x = A cos(ωt + δ ) . Its velocity as a function of time is given by v = − Aω sin (ωt + δ )

and its acceleration by a = − Aω 2 cos(ωt + δ ) . The initial position and velocity give us two equations from which to determine the amplitude A and phase constantδ. (a) Express the position, velocity, and acceleration of the particle as a function of t:

x = A cos(ωt + δ ) v = − Aω sin (ωt + δ )

a = − Aω 2 cos(ωt + δ ) 2π 4π −1 = s = 4.19 s −1 3 T

Find the angular frequency of the particle’s motion:

ω=

Relate the initial position and velocity to the amplitude and phase constant:

x0 = A cos δ and

Divide these equations to eliminate A:

v0 − ωA sin δ = = −ω tan δ x0 A cos δ

v0 = −ωA sin δ

(1) (2) (3)

1056 Chapter 14 Solve for δ and substitute numerical values to obtain: Substitute in equation (1) to obtain:

⎛

⎛ 0 ⎞ v0 ⎞ ⎟⎟ = tan −1 ⎜⎜ − ⎟⎟ = 0 ⎝ x0ω ⎠ ⎝ x0ω ⎠

δ = tan −1 ⎜⎜ −

⎡⎛ 4π −1 ⎞ ⎤ x = (25 cm ) cos ⎢⎜ s ⎟t⎥ ⎠ ⎦ ⎣⎝ 3 =

(b) Substitute in equation (2) to obtain:

(25 cm)cos[(4.19 s −1 )t ]

⎛ 4π −1 ⎞ ⎡⎛ 4π −1 ⎞ ⎤ s ⎟ sin ⎢⎜ v = −(25 cm )⎜ s ⎟t⎥ ⎝ 3 ⎠ ⎣⎝ 3 ⎠ ⎦

[(

)]

= − (105 cm/s)sin 4.19 s −1 t (c) Substitute in equation (3) to obtain:

2

⎡⎛ 4π −1 ⎞ ⎤ ⎛ 4π −1 ⎞ a = −(25 cm )⎜ s ⎟ cos ⎢⎜ s ⎟t⎥ ⎝ 3 ⎠ ⎠ ⎦ ⎣⎝ 3

) [(

(

)]

= − 439 cm/s 2 cos 4.19 s −1 t

26 • Picture the Problem The maximum speed and maximum acceleration of the particle in are given by vmax = Aω and amax = Aω 2 . The particle’s position is given by

x = A cos(ωt + δ ) where A = 7 cm, ω = 6π s−1, and δ = 0, and its velocity is given by v = − Aω sin (ωt + δ ) .

(a) Express vmax in terms of A and ω:

(

vmax = Aω = (7 cm ) 6π s −1

)

= 42π cm/s = 1.32 m/s (b) Express amax in terms of A and ω:

(

amax = Aω 2 = (7 cm ) 6π s −1

)

2

= 252π 2 cm/s 2 = 24.9 m/s 2 cos ωt = 0

(c) When x = 0:

and

ωt = cos −1 0 = Evaluate v at ωt =

π 2

:

π 3π 2

,

2

⎛π ⎞ v = − Aω sin ⎜ ⎟ = − Aω ⎝2⎠ i.e., the particle is moving to the left.

Oscillations 1057 Evaluate v at ωt =

3π : 2

⎛ 3π v = − Aω sin ⎜ ⎝ 2

⎞ ⎟ = Aω ⎠

i.e., the particle is moving to the right. Solve for t:

t=

3π 3π = = 0.250 s 2ω 2 6π s −1

(

)

27 •• Picture the Problem The position of the particle as a function of time is given by x = A cos(ωt + δ ) . Its velocity as a function of time is given by v = − Aω sin (ωt + δ )

and its acceleration by a = − Aω 2 cos(ωt + δ ) . The initial position and velocity give us

two equations from which to determine the amplitude A and phase constant δ. (a) Express the position, velocity, and acceleration of the particle as functions of t:

x = A cos(ωt + δ ) v = − Aω sin (ωt + δ )

a = − Aω cos(ωt + δ ) 2

(2) (3)

2π 4π −1 = s = 4.19 s −1 T 3

Find the angular frequency of the particle’s motion:

ω=

Relate the initial position and velocity to the amplitude and phase constant:

x0 = A cos δ and

Divide these equations to eliminate A:

v0 − ωA sin δ = = −ω tan δ x0 A cos δ

Solve for δ:

(1)

v0 = −ωA sin δ

v0 ⎞ ⎟⎟ ω x ⎝ 0 ⎠ ⎛

δ = tan −1 ⎜⎜ − ⎛

⎞ 50 cm/s ⎟ −1 ⎟ ⎝ (25 cm )(4.19 2 s ) ⎠ = −0.445 rad

Substitute numerical values and evaluate δ:

δ = tan −1 ⎜⎜ −

Use either the x0 or v0 equation (x0 is used here) to find the amplitude:

A=

x0 25 cm = = 27.7 cm cos δ cos(− 0.445 rad )

Substitute in equation (1) to obtain:

x=

(27.7 cm )cos[(4.19 s −1 )t − 0.445]

1058 Chapter 14 ⎛ 4π −1 ⎞ v = −(27.7 cm )⎜ s ⎟ ⎝ 3 ⎠ ⎡⎛ 4π −1 ⎞ ⎤ × sin ⎢⎜ s ⎟ t − 0.445⎥ ⎠ ⎣⎝ 3 ⎦

(b) Substitute in equation (2) to obtain:

[(

]

)

= − (116 cm/s)sin 4.19 s −1 t − 0.445 (c) Substitute in equation (3) to obtain: 2

⎡⎛ 4π −1 ⎞ ⎤ ⎛ 4π −1 ⎞ a = −(27.7 cm )⎜ s ⎟ cos ⎢⎜ s ⎟ t − 0.445⎥ ⎝ 3 ⎠ ⎠ ⎣⎝ 3 ⎦

(

) [(

)

= − 486 cm/s 2 cos 4.19 s −1 t − 0.445

]

28 •• Picture the Problem The position of the particle as a function of time is given by x = A cos(ωt + δ ) . We’re given the amplitude A of the motion and can use the initial position of the particle to determine the phase constant δ. Once we’ve determined these quantities, we can express the distance traveled ∆x during any interval of time. Express the position of the particle as a function of t:

x = (12 cm ) cos(ωt + δ )

2π 2π π −1 = = s 8s 4 T

Find the angular frequency of the particle’s motion:

ω=

Relate the initial position of the particle to the amplitude and phase constant:

x0 = A cos δ

Solve for δ:

δ = cos−1

Substitute in equation (1) to obtain:

⎡⎛ π ⎞ π ⎤ x = (12 cm ) cos ⎢⎜ s −1 ⎟ t + ⎥ 2⎦ ⎣⎝ 4 ⎠

x0 0 π = cos−1 = A A 2

(1)

Oscillations 1059 Express the distance the particle travels in terms of tf and ti:

⎡⎛ π ⎞ π⎤ ∆x = (12 cm ) cos ⎢⎜ s −1 ⎟ tf + ⎥ 2⎦ ⎣⎝ 4 ⎠ ⎡⎛ π ⎞ π⎤ − (12 cm ) cos ⎢⎜ s −1 ⎟ ti + ⎥ 2⎦ ⎣⎝ 4 ⎠ ⎧ ⎡⎛ π ⎞ π⎤ = (12 cm )⎨cos ⎢⎜ s −1 ⎟ tf + ⎥ 2⎦ ⎩ ⎣⎝ 4 ⎠ ⎡⎛ π ⎞ π ⎤⎫ − cos ⎢⎜ s −1 ⎟ ti + ⎥ ⎬ 2 ⎦⎭ ⎣⎝ 4 ⎠

(a) Evaluate ∆x for tf = 2 s, ti = 1 s:

⎧ ⎡⎛ π ⎞ π⎤ ∆x = (12 cm )⎨cos ⎢⎜ s −1 ⎟ (2 s ) + ⎥ 2⎦ ⎩ ⎣⎝ 4 ⎠ ⎡⎛ π ⎞ π ⎤⎫ − cos ⎢⎜ s −1 ⎟ (0 ) + ⎥ ⎬ 2 ⎦⎭ ⎣⎝ 4 ⎠

= (12 cm ){− 1 − 0} = 12.0 cm (b) Evaluate ∆x for tf = 4 s, ti = 2 s:

⎧ ⎡⎛ π ⎞ π⎤ ∆x = (12 cm )⎨cos ⎢⎜ s −1 ⎟ (4 s ) + ⎥ 2⎦ ⎩ ⎣⎝ 4 ⎠ ⎡⎛ π ⎞ π ⎤⎫ − cos ⎢⎜ s −1 ⎟ (2 s ) + ⎥ ⎬ 2 ⎦⎭ ⎣⎝ 4 ⎠

= (12 cm ){0 − 1} = 12.0 cm (c) Evaluate ∆x for tf = 1 s, ti = 0:

⎧ ⎡⎛ π ⎞ π⎤ ∆x = (12 cm )⎨cos ⎢⎜ s −1 ⎟ (1s ) + ⎥ 2⎦ ⎩ ⎣⎝ 4 ⎠ ⎡⎛ π ⎞ π ⎤⎫ − cos ⎢⎜ s −1 ⎟ (0 ) + ⎥ ⎬ 2 ⎦⎭ ⎣⎝ 4 ⎠

= (12 cm ){− 0.7071 − 0} = 8.49 cm

1060 Chapter 14 (d) Evaluate ∆x for tf = 2 s, ti = 1 s:

⎧ ⎡⎛ π ⎞ π⎤ ∆x = (12 cm )⎨cos ⎢⎜ s −1 ⎟ (2 s ) + ⎥ 2⎦ ⎩ ⎣⎝ 4 ⎠ ⎡⎛ π ⎞ π ⎤⎫ − cos ⎢⎜ s −1 ⎟ (1s ) + ⎥ ⎬ 2 ⎦⎭ ⎣⎝ 4 ⎠

= (12 cm ){− 1 + 0.7071} = 3.51 cm

29 •• Picture the Problem The position of the particle as a function of time is given by x = (10 cm ) cos(ωt + δ ) . We can determine the angular frequency ω from the period of the motion and the phase constant δ from the initial position and velocity. Once we’ve determined these quantities, we can express the distance traveled ∆x during any interval of time. Express the position of the particle as a function of t:

x = (10 cm ) cos(ωt + δ )

2π 2π π −1 = = s T 8s 4

Find the angular frequency of the particle’s motion:

ω=

Find the phase constant of the motion:

δ = tan −1 ⎜⎜ −

Substitute in equation (1) to obtain:

(1)

⎛

⎛ 0 ⎞ v0 ⎞ ⎟⎟ = tan −1 ⎜⎜ − ⎟⎟ = 0 ⎝ x0ω ⎠ ⎝ x0ω ⎠

⎡⎛ π ⎞ ⎤ x = (10 cm ) cos ⎢⎜ s −1 ⎟ t ⎥ ⎣⎝ 4 ⎠ ⎦

Oscillations 1061 ⎡⎛ π −1 ⎞ ⎤ s ⎟ t ⎥ follows: ⎣⎝ 4 ⎠ ⎦

(a) A graph of x = (10 cm ) cos ⎢⎜ 10 8 6 4

x (cm)

2 0 -2 -4 -6 -8 -10 0

2

4

6

8

t (s)

(b) Express the distance the particle travels in terms of tf and ti:

⎡⎛ π ⎡⎛ π ⎞ ⎤ ⎞ ⎤ ∆x = (10 cm ) cos ⎢⎜ s −1 ⎟ t f ⎥ − (10 cm ) cos ⎢⎜ s −1 ⎟ ti ⎥ ⎣⎝ 4 ⎠ ⎦ ⎣⎝ 4 ⎠ ⎦ ⎧ ⎡⎛ π ⎡⎛ π ⎞ ⎤⎫ ⎞ ⎤ = (10 cm )⎨cos ⎢⎜ s −1 ⎟ t f ⎥ − cos ⎢⎜ s −1 ⎟ ti ⎥ ⎬ ⎣⎝ 4 ⎠ ⎦ ⎭ ⎩ ⎣⎝ 4 ⎠ ⎦ Substitute numerical values in equation (2) and evaluate ∆x in each of the given time intervals to obtain:

tf ti (s) (s) 1 0

∆x (cm)

2

1

7.07

3

2

7.07

4

3

2.93

(2)

2.93

*30 •• Picture the Problem We can use the expression for the maximum acceleration of an oscillator to relate the 10g military specification to the compliance frequency. Express the maximum acceleration of an oscillator:

amax = Aω 2

1062 Chapter 14 Express the relationship between the angular frequency and the frequency of the vibrations:

ω = 2πf

Substitute to obtain:

amax = 4π 2 Af 2

Solve for f:

Substitute numerical values and evaluate f:

f =

1 2π

f =

1 98.1 m/s 2 = 12.9 Hz 2π 1.5 × 10−2 m

amax A

31 •• Picture the Problem The maximum speed and acceleration of the particle are given by vmax = Aω and amax = Aω 2 . The velocity and acceleration of the particle are given by

v = − Aω sin ωt and a = − Aω 2 cos ωt. (a) Find vmax from A and ω:

vmax = Aω = (2.5 m )(π s −1 ) = 7.85 m/s

Find amax from A and ω:

(

amax = Aω 2 = (2.5 m ) π s −1

)

2

= 24.7 m/s 2 (b) Use the equation for the position of the particle to relate its position at x = 1.5 m to the time t′to reach this position:

1.5 m = (2.5 m ) cos πt'

Solve for π t′:

πt' = cos −1 0.6 = 0.9273 rad

Evaluate v when π t = π t′:

v = −(2.5 m )(π s −1 )sin (0.9273 rad ) = − 6.28 m/s where the minus sign indicates that the particle is moving in the negative direction.

Evaluate a when πt = πt′:

a = −(2.5 m )(π s −1 ) cos(0.9273 rad ) 2

= − 14.8 m/s 2

Oscillations 1063 where the minus sign indicates that the particle’s acceleration is in the negative direction. *32 •• Picture the Problem We can use the formula for the cosine of the sum of two angles to write x = A0 cos(ωt + δ) in the desired form. We can then evaluate x and dx/dt at t = 0 to relate Ac and As to the initial position and velocity of a particle undergoing simple harmonic motion. (a) Apply the trigonometric identity

cos(ωt + δ ) = cos ωt cos δ − sin ωt sin δ

to obtain:

x = A0 cos(ωt + δ ) = A0 [cos ωt cos δ − sin ωt sin δ ] = − A0 sin δ sin ωt

+ A0 cos δ cos ωt = As sin ωt + Ac cos ωt provided

As = − A0 sin δ and Ac = A0 cos δ

(b) At t = 0: Evaluate dx/dt:

Evaluate v(0) to obtain:

x(0) = A0 cos δ = Ac dx d = [ As sin ωt + Ac cos ωt ] dt dt = Asω cos ωt − Acω sin ωt

v=

v(0) = ωAs = − ωA0 sin δ

Simple Harmonic Motion and Circular Motion 33 • Picture the Problem We can find the period of the motion from the time required for the particle to travel completely around the circle. The frequency of the motion is the reciprocal of its period and the x-component of the particle’s position is given by x = A cos(ωt + δ ) . (b) Use the definition of speed to find the period of the motion:

T=

2πr 2π (0.4 m ) = = 3.14 s 0.8 m/s v

(a) Because the frequency and the period are reciprocals of each other:

f =

1 1 = = 0.318 Hz T 3.14 s

(c) Express the x component of the

x = A cos(ωt + δ )

1064 Chapter 14 position of the particle: Assuming that the particle is on the positive x axis at time t = 0: Substitute for A, ω, and δ to obtain:

A = A cos δ ⇒ δ = cos −1 1 = 0

x = A cos(2πft ) =

(40 cm)cos[(2 s −1 )t ]

*34 • Picture the Problem We can find the period of the motion from the time required for the particle to travel completely around the circle. The angular frequency of the motion is 2π times the reciprocal of its period and the x-component of the particle’s position is given by x = A cos(ωt + δ ) . (a) Use the definition of speed to express and evaluate the speed of the particle:

v=

2πr 2π (15 cm ) = = 31.4 cm/s T 3s

(b) Express the angular velocity of the particle:

ω=

2π 2π = rad/s T 3

(c) Express the x component of the position of the particle:

x = A cos(ωt + δ )

Assuming that the particle is on the positive x axis at time t = 0:

A = A cos δ ⇒ δ = cos −1 1 = 0

Substitute to obtain:

⎛ 2π −1 ⎞ x = (15 cm ) cos⎜ s ⎟t ⎝ 3 ⎠

Energy in Simple Harmonic Motion 35 • Picture the Problem The total energy of the object is given by Etot = 12 kA2 , where A is the amplitude of the object’s motion. Express the total energy of the system:

Etot = 12 kA2

Substitute numerical values and evaluate Etot:

Etot =

1 2

(4.5 kN/m )(0.1m )2 =

22.5 J

Oscillations 1065 36 • Picture the Problem The total energy of an oscillating object can be expressed in terms 2 of its kinetic energy as it passes through its equilibrium position: Etot = 12 mvmax . Its maximum speed, in turn, can be expressed in terms of its angular frequency and the amplitude of its motion. Express the total energy of the object in terms of its maximum kinetic energy:

2 E = 12 mvmax

Express vmax:

vmax = Aω = 2πAf

Substitute to obtain:

E = 12 m(2πAf ) = 2mA2π 2 f 2

Substitute numerical values and evaluate E:

E = 2(3 kg )(0.1 m ) π 2 2.4 s −1

2

2

(

)

2

= 3.41 J

37 • Picture the Problem The total mechanical energy of the oscillating object can be expressed in terms of its kinetic energy as it passes through its equilibrium position: 2 Etot = 12 mvmax . Its total energy is also given by Etot = 12 kA2 . We can equate these expressions to obtain an expression for A. (a) Express the total mechanical energy of the object in terms of its maximum kinetic energy:

2 E = 12 mvmax

Substitute numerical values and evaluate E:

E=

(b) Express the total energy of the object in terms of the amplitude of its motion:

Etot = 12 kA2

Solve for A:

Substitute numerical values and evaluate A:

1 2

(1.5 kg )(0.7 m/s )2 =

0.368 J

A=

2 Etot k

A=

2(0.368 J ) = 3.84 cm 500 N/m

1066 Chapter 14 38 • Picture the Problem The total energy of the oscillating object can be expressed in terms 2 of its kinetic energy as it passes through its equilibrium position: Etot = 12 mvmax . Its total energy is also given by Etot = 12 kA2 . We can solve the latter equation to find A and solve the former equation for vmax. (a) Express the total energy of the object as a function of the amplitude of its motion: Solve for A:

Substitute numerical values and evaluate A: (b) Express the total energy of the object in terms of its maximum speed: Solve for vmax:

Substitute numerical values and evaluate vmax:

Etot = 12 kA2

A=

2 Etot k

A=

2(0.9 J ) = 3.00 cm 2000 N/m

2 Etot = 12 mvmax

vmax =

2 Etot m

vmax =

2(0.9 J ) = 0.775 m/s 3 kg

39 • Picture the Problem The total energy of the object is given by Etot = 12 kA2 . We can solve this equation for the force constant k and substitute the numerical data to determine its value. Express the total energy of the oscillator as a function of the amplitude of its motion:

Etot = 12 kA2

Solve for k:

k=

2 Etot A2

Substitute numerical values and evaluate k:

k=

2(1.4 J ) = 1.38 kN/m (0.045 m )2

Oscillations 1067 *40 •• Picture the Problem The total energy of the object is given, in terms of its maximum 2 kinetic energy by Etot = 12 mvmax . We can express vmax in terms of A and ω and, in turn, express ω in terms of amax to obtain an expression for Etot in terms of amax. Express the total energy of the object in terms of its maximum kinetic energy:

2 Etot = 12 mvmax

Relate the maximum speed of the object to its angular frequency:

vmax = Aω

Substitute to obtain:

Etot = 12 m( Aω ) = 12 mA2ω 2

Relate the maximum acceleration of the object to its angular frequency:

amax = Aω 2

2

or

ω2 =

amax A

Substitute and simplify to obtain:

Etot = 12 mA2

Substitute numerical values and evaluate Etot:

Etot =

1 2

amax 1 = 2 mAamax A

(3 kg )(0.08 m )(3.50 m/s2 )

= 0.420 J

Springs 41

•

Picture the Problem The frequency of the object’s motion is given by f =

1 2π

k m.

Its period is the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are vmax = Aω and amax = Aω 2 , respectively. (a) The frequency of the motion is given by:

f =

1 2π

k m

Substitute numerical values and evaluate f:

f =

1 2π

4.5 kN/m = 6.89 Hz 2.4 kg

1068 Chapter 14 1 1 = = 0.145 s f 6.89 s −1

(b) The period of the motion to is the reciprocal of its frequency:

T=

(c) Because the object is released from rest after the spring to which it is attached is stretched 10 cm:

A = 0.100 m

(d) Express the object’s maximum speed:

vmax = Aω = 2πfA

Substitute numerical values and evaluate vmax:

vmax = 2π 6.89 s −1 (0.1 m ) = 4.33 m/s

(e) Express the object’s maximum acceleration:

amax = Aω 2 = ωvmax = 2πfvmax

Substitute numerical values and evaluate amax:

(

(

)

amax = 2π 6.89 s −1 (4.33 m/s ) = 187 m/s 2

(f) The object first reaches its equilibrium when:

t = 14 T =

Because the resultant force acting on the object as it passes through its equilibrium point is zero, the acceleration of the object is:

a= 0

42

)

1 4

(0.145 s ) =

36.3 ms

•

Picture the Problem The frequency of the object’s motion is given by f =

1 2π

k m.

Its period is the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are vmax = Aω and amax = Aω 2 , respectively. (a) The frequency of the motion is given by:

f =

1 2π

k m

Substitute numerical values and evaluate f:

f =

1 2π

700 N/m = 1.88 Hz 5 kg

Oscillations 1069 1 1 = = 0.531s f 1.88 s −1

(b) The period of the motion is the reciprocal of its frequency:

T=

(c) Because the object is released from rest after the spring to which it is attached is stretched 8 cm:

A = 0.0800 m

(d) Express the object’s maximum speed:

vmax = Aω = 2πfA

Substitute numerical values and evaluate vmax:

vmax = 2π 1.88 s −1 (0.08 m ) = 0.945 m/s

(e) Express the object’s maximum acceleration:

amax = Aω 2 = ω vmax = 2πfvmax

Substitute numerical values and evaluate amax:

(

)

(

)

amax = 2π 1.88 s −1 (0.945 m/s ) = 11.2 m/s 2

(f) The object first reaches its equilibrium when:

t = 14 T =

Because the resultant force acting on the object as it passes through its equilibrium point is zero, the acceleration of the object is:

a= 0

1 4

(0.531s ) =

0.133 s

43 • Picture the Problem The angular frequency, in terms of the force constant of the spring and the mass of the oscillating object, is given by ω 2 = k m . The period of the motion is the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are vmax = Aω and amax = Aω 2 , respectively. (a) Relate the angular frequency of the motion to the force constant of the spring:

ω2 =

k m

or

k = mω 2 = 4π 2 f 2 m Substitute numerical values to obtain:

(

k = 4π 2 2.4 s −1

) (3 kg ) = 2

682 N/m

1070 Chapter 14 1 1 = = 0.417 s f 2.4 s −1

(b) Relate the period of the motion to its frequency:

T=

(c) Express the maximum speed of the object:

vmax = Aω = 2πfA

Substitute numerical values and evaluate vmax:

vmax = 2π (2.4 s −1 )(0.1 m ) = 1.51 m/s

(d) Express the maximum acceleration of the object:

amax = Aω 2 = 4π 2 f 2 A

Substitute numerical values and evaluate amax:

amax = 4π 2 2.4 s −1

(

) (0.1m) = 2

22.7 m/s 2

*44 • Picture the Problem We can find the frequency of vibration of the car-and-passenger system using f =

1 2π

k , where M is the total mass of the system. The spring M

constant can be determined from the compressing force and the amount of compression. Express the frequency of the carand-passenger system:

f =

1 2π

Express the spring constant:

k=

F mg = ∆x ∆x

k M

where m is the person’s mass. Substitute to obtain:

Substitute numerical values and evaluate f:

f =

1 2π

mg M∆x

f =

1 2π

(85 kg )(9.81m/s 2 ) (2485 kg )(2.35 ×10−2 m )

= 0.601 Hz 45 • Picture the Problem We can relate the force constant k to the maximum acceleration by eliminating ω2 between ω 2 = k m and amax = Aω 2 . We can also express the frequency f

Oscillations 1071 of the motion by substituting mamax/A for k in f =

1 2π

k . m

k or k = ω 2 m m

(a) Relate the angular frequency of the motion to the force constant and the mass of the oscillator:

ω2 =

Relate the object’s maximum acceleration to its angular frequency and amplitude and solve for the square of the angular frequency:

amax = Aω 2 or

Substitute to obtain:

k=

Substitute numerical values and evaluate k:

k=

(b) Replace ω in equation (1) by 2πf and solve for f to obtain:

f =

1 2π

amax A

Substitute numerical values and evaluate f:

f =

1 2π

26 m/s 2 = 4.16 Hz 3.8 × 10−2 m

(c) The period of the motion is the reciprocal of its frequency:

T=

1 1 = = 0.240 s f 4.16 s −1

ω2 =

amax A

(1)

mamax A

(4.5 kg )(26 m/s2 ) = 3.8 × 10 −2 m

3.08 kN/m

46 • Picture the Problem We can find the frequency of the motion from its maximum speed and the relationship between frequency and angular frequency. The mass of the object can be found by eliminating ω between ω 2 = k m and vmax = Aω. (b) Express the object’s maximum speed as a function of the frequency of its motion:

vmax = Aω = 2πfA

Solve for f:

f =

vmax 2πA

Substitute numerical values and evaluate f:

f =

2.2 m/s = 6.04 Hz 2π 5.8 × 10−2 m

(

(1)

)

1072 Chapter 14 (a) Relate the square of the angular frequency of the motion to the force constant and the mass of the object: Eliminate ω between equations (1) and (2) to obtain: Substitute numerical values and evaluate m:

ω2 =

m=

k k ⇒ m= 2 m ω

(2)

kA2 2 vmax

(1.8 ×10 m=

3

)(

N/m 5.8 × 10 −2 m (2.2 m/s )2

)

2

= 1.25 kg (c) The period of the motion is the reciprocal of its frequency:

T=

1 1 = = 0.166 s f 6.04 s −1

47 •• Picture the Problem The maximum speed of the block is given by vmax = Aω and the angular frequency of the motion is ω = k m = 5.48 rad/s . We’ll assume that the position of the block is given by x = A cos ωt and solve for ωt for x = 4 cm and x = 0. We can use these values for ωt to find the time for the block to travel from x = 4 cm to its equilibrium position. (a) Express the maximum speed of the block as a function of the system’s angular frequency: Substitute numerical values and evaluate vmax: (b) Assuming that x = A cos ω t, evaluate ωt for x = 4 cm = A/2: Evaluate v for ωt = π 3 :

vmax = Aω

v max = (0.08 m )(5.48 rad/s ) = 0.438 m/s 1 π A = A cos ωt ⇒ ωt = cos −1 = 2 2 3 v = vmax sin ωt = (0.438 m/s )sin = (0.438 m/s )

Express a as a function of vmax and ω:

π 3

3 = 0.379 m/s 2

a = Aω 2 cos ωt = vmaxω cos ωt

Oscillations 1073 Substitute numerical values and evaluate a:

a = (0.438 m/s )(5.48 rad/s ) cos = 1.20 m/s

(c) Evaluate ωt for x = 0:

Let ∆t = time to go from ωt = π 3 to

ωt = π 2 . Then: Solve for and evaluate ∆t:

π 2

−

π 3

3

2

0 = A cos ωt ⇒ ωt = cos −1 0 =

ω∆t =

π

=

π 2

π 6

π π = = 95.5 ms 6ω 6(5.48 rad/s )

∆t =

*48 •• Picture the Problem Choose a coordinate system in which upward is the positive y direction. We can find the mass of the object using m = k ω 2 . We can apply a condition for translational equilibrium to the object when it is at its equilibrium position to determine the amount the spring has stretched from its natural length. Finally, we can use the initial conditions to determine A and δ and express x(t) and then differentiate this expression to obtain v(t) and a(t).

k k ⇒m= 2 m ω

(a) Express the angular frequency of the system in terms of the mass of the object fastened to the vertical spring and solve for the mass of the object:

ω2 =

Express ω2 in terms of f:

ω 2 = 4π 2 f 2

Substitute to obtain:

m=

k 4π 2 f 2 1800 N/m

Substitute numerical values and evaluate m:

m=

(b) Letting ∆x represent the amount the spring is stretched from its natural length when the object is in Fy = 0 to the equilibrium, apply

k∆x − mg = 0

∑

object when it is in equilibrium:

(

4π 2 5.5 s −1

)

2

= 1.51 kg

1074 Chapter 14 Solve for ∆x:

Substitute numerical values and evaluate ∆x: (c) Express the position of the object as a function of time: Use the initial conditions (x0 = −2.5 cm and v0 = 0) to find δ: Evaluate ω: Substitute to obtain:

mg k

∆x =

(1.51kg )(9.81m/s2 ) =

∆x =

1800 N/m

8.23 mm

x = A cos(ωt + δ )

⎛

v0 ⎞ ⎟⎟ = tan −1 0 = π ⎝ ω x0 ⎠

δ = tan −1 ⎜⎜ −

k 1800 N/m = = 34.5 rad/s m 1.51 kg x = (2.5 cm ) cos[(34.5 rad/s )t + π ]

ω=

= − (2.5 cm ) cos[(34.5 rad/s )t ] Differentiate x(t) to obtain v:

v = (86.4 cm/s )sin[(34.5 rad/s )t ]

Differentiate v(t) to obtain a:

a=

(29.8 m/s )cos[(34.5 rad/s)t ] 2

49 •• Picture the Problem Let the system include the object and the spring. Then, the net external force acting on the system is zero. Choose Ei = 0 and apply the conservation of mechanical energy to the system.

2π

Express the period of the motion in terms of its angular frequency:

T=

Apply conservation of energy to the system:

Ei = Ef or 0 = U g + U spring

Substitute for Ug and Uspring:

0 = − mg∆x + 12 k (∆x )

Solve for ω2 = k/m:

ω2 =

k 2g = m ∆x

Substitute numerical values and evaluate ω2:

ω2 =

2(9.81 m/s 2 ) = 574 rad/s 2 −2 3.42 × 10 m

ω

(1)

2

Oscillations 1075 Substitute in equation (1) to obtain:

T=

2π 574 rad/s 2

= 0.262 s

50 •• Picture the Problem Let the system include the object and the spring. Then the net external force acting on the system is zero. Because the net force acting on the object when it is at its equilibrium position is zero, we can apply a condition for translational equilibrium to determine the distance from the starting point to the equilibrium position. Letting Ei = 0, we can apply conservation of energy to the system to determine how far down the object moves before coming momentarily to rest. We can find the period of the motion and the maximum speed of the object from T = 2π m k and vmax = A k m . (a) Apply

∑F

y

= 0 to the object

ky0 − mg = 0

when it is at the equilibrium position: Solve for y0:

Substitute numerical values and evaluate y0: (b) Apply conservation of energy to the system:

y0 =

y0 =

mg k

(1 kg ) (9.81 m/s 2 ) = 250 N/m

3.92 cm

Ei = Ef or

0 = U g + U spring Substitute for Ug and Uspring:

0 = − mgyf + 12 kyf2

Solve for yf:

yf =

2mg k

yf =

2(1 kg )(9.81m/s 2 ) = 7.85 cm 250 N/m

Substitute numerical values and evaluate yf: (c) Express the period T of the motion in terms of the mass of the object and the spring constant:

T = 2π

m k

1076 Chapter 14 Substitute numerical values and evaluate T: (d) The object will be moving with its maximum speed when it reaches its equilibrium position: Substitute numerical values and evaluate vmax:

T = 2π

1 kg = 0.397 s 250 N/m

vmax = Aω = A

k m

vmax = (3.92 cm )

250 N/m 1 kg

= 62.0 cm/s (e) The time required for the object to reach equilibrium is one-fourth of its period:

t = 14 T =

1 4

(0.397 s ) =

99.3 ms

51 •• Picture the Problem The stunt woman’s kinetic energy, after 2 s of flight, is K 2 s = 12 mv22s . We can evaluate this quantity as soon as we know how fast she is moving after two seconds. Because her motion is oscillatory, her velocity as a function of time is v(t ) = − Aω sin (ωt + δ ). We can find the amplitude of her motion from her distance of fall and the angular frequency of her motion by applying conservation of energy to her fall to the ground. Express the kinetic energy of the stunt woman when she has fallen for 2 s: Express her velocity as a function of time:

K 2 s = 12 mv22s

(1)

v(t ) = − Aω sin (ωt + δ ) where δ = 0 (she starts from rest with positive displacement) and

A = 12 (192 m ) = 96 m ∴ v(t ) = −(96 m )ω sin (ωt ) 0 = U g + U elastic

Letting Ei = 0, use conservation of energy to find the force constant of the elastic band:

or

Solve for k:

k=

0 = − mgh + 12 kh 2 = 0 2mg h

(2)

Oscillations 1077 Substitute numerical values and evaluate k:

2(60 kg ) (9.81m/s 2 ) k= = 6.13 N/m 192 m

Express the angular frequency of her motion:

ω=

k m

Substitute numerical values and evaluate ω:

ω=

6.13 N/m = 0.320 rad/s 60 kg

Substitute in equation (2) to obtain:

v(t ) = −(96 m )(0.320 rad/s ) × sin[(0.320 rad/s )t ]

= (30.7 m/s )sin[(0.320 rad/s )t ]

v(2 s ) = (30.7 m/s )sin[(0.320 rad/s )(2 s )]

Evaluate v(2 s):

= 18.3 m/s Substitute in equation (1) and evaluate K(2 s):

K (2 s ) =

1 2

(60 kg )(18.3 m/s )2 =

*52 •• Picture the Problem The diagram shows the stretched bungie cords supporting the suitcase under equilibrium conditions. We can use f =

1 2π

k eff to express the M

frequency of the suitcase in terms of the effective ″spring″ constant keff and apply a condition for translational equilibrium to the suitcase to find keff. Express the frequency of the suitcase oscillator: Apply obtain:

∑F

y

= 0 to the suitcase to

f =

1 2π

k eff M

kx + kx − Mg = 0 or

2kx − Mg = 0

or

k eff x − Mg = 0 where keff = 2k

10.1 kJ

1078 Chapter 14 Solve for keff to obtain:

Substitute to obtain:

Substitute numerical values and evaluate f:

k eff =

Mg x

f =

1 2π

g x

f =

1 2π

9.81 m/s 2 = 2.23 Hz 0.05 m

53 •• Picture the Problem The frequency of the motion of the stone and block depends on the force constant of the spring and the mass of the stone plus block. The force constant can be determined from the equilibrium of the system when the spring is stretched additionally by the addition of the stone to the mass. When the block is at the point of maximum upward displacement, it is momentarily at rest and the net force acting on it is its weight. (a) Express the frequency of the motion in terms of k and m:

f =

1 2π

k mtot

where mtot is the total mass suspended from the spring. Apply

∑F

y

= 0 to the stone when

k∆y − mg = 0

it is at its equilibrium position: Solve for k:

k=

Substitute numerical values and evaluate k:

k=

Substitute and evaluate f:

(b) The time to travel from its lowest point to its highest point is one-half its period:

f =

mg ∆y

(0.03 kg )(9.81m/s2 ) = 5.89 N/m 0.05 m

1 2π

t = 12 T =

5.89 N/m = 0.997 Hz 0.15 kg

1 1 = = 0.502 s 2 f 2(0.997 s −1 )

Oscillations 1079 (c) When the stone is at a point of maximum upward displacement:

Fnet = mg = (0.03 kg ) (9.81 m/s 2 ) = 0.294 N

54 •• Picture the Problem We can use the maximum acceleration of the oscillator amax = Aω 2 to express amax in terms of A, k, and m. k can be determined from the equilibrium of the system when the spring is stretched additionally by the addition of the stone to the mass. If the stone is to remain in contact with the block, the block’s maximum downward acceleration must not exceed g. Express the maximum acceleration in terms of the angular frequency and amplitude of the motion:

amax = Aω 2

Relate ω2 to the force constant and the mass of the stone:

ω2 =

Substitute to obtain:

amax = A

Apply

∑F

y

= 0 to the stone when

k m k m

k∆y − mg = 0

it is at its equilibrium position: Solve for k:

k=

Substitute numerical values and evaluate k:

k=

Substitute numerical values to express amax in terms of A:

amax = A

Set amax = g and solve for Amax:

Amax =

g 39.3 s −2

Amax =

9.81 m/s 2 = 25.0 cm 39.3 s −2

Substitute for g and evaluate Amax:

mg ∆y

(0.03 kg )(9.81m/s2 ) = 5.89 N/m 0.05 m

(

)

5.89 N/m = 39.3 s −2 A 0.15 kg

1080 Chapter 14 55 •• Picture the Problem The maximum height above the floor to which the object rises is the sum of its initial distance from the floor and the amplitude of its motion. We can find the amplitude of its motion by relating it to the object’s maximum speed. Because the object initially travels downward, it will be three-fourths of the way through its cycle when it first reaches its maximum height. We can find the minimum initial speed the object would need to be given in order for the spring to become uncompressed by applying conservation of energy. (a) Relate h, the maximum height above the floor to which the object rises, to the amplitude of its motion:

h = A + 5.0 cm

Relate the maximum speed of the object to the angular frequency and amplitude of its motion and solve for the amplitude:

vmax = Aω

Using its definition, express and evaluate the force constant of the spring:

(1)

or

A = vmax

k=

m k

(2)

mg (2 kg )(9.81 m/s 2 ) = = 654 N/m ∆y 0.03 m

Substitute numerical values in equation (2) and evaluate A:

A = 0.3 m/s

Substitute in equation (1) to obtain:

h = 1.66 cm + 5.00 cm = 6.66 cm

(b) Express the time required for the object to reach its maximum height the first time:

t = 34 T

Express the period of the motion:

2 kg = 1.66 cm 654 N/m

T = 2π

m k

Substitute numerical values and evaluate T:

T = 2π

2 kg = 0.347 s 654 N/m

Substitute to obtain:

t=

3 4

(0.347 s ) =

0.261s

Oscillations 1081 (c) Because h < 8.0 cm: Using conservation of energy and letting Ug be zero 5 cm above the floor, relate the height to which the object rises, ∆y, to its initial kinetic energy: Because ∆y = L − yi :

the spring is never uncompressed.

∆K + ∆U g + ∆U s = 0 or, because Kf = Ui = 0, 1 2

mvi2 − mg∆y + 12 k (∆y )

2

− 12 k (L − yi ) = 0 2

1 2

mvi2 − mg∆y + 12 k (∆y ) − 12 k (∆y ) = 0 2

2

and 1 2

Solve for and evaluate vi for ∆y = 3 cm:

mvi2 − mg∆y = 0

(

)

vi = 2 g∆y = 2 9.81 m/s 2 (3 cm ) = 0.767 m/s i.e., the minimum initial velocity that must be given to the object for the spring to be uncompressed at some time is

0.767 m/s *56 •• Picture the Problem We can relate the elongation of the cable to the load on it using the definition of Young’s modulus and use the expression for the frequency of a spring and mass oscillator to find the oscillation frequency of the engine block at the end of the wire.

stress F A = strain ∆l l

(a) Using the definition of Young’s modulus, relate the elongation of the cable to the applied stress:

Y=

Solve for ∆l:

∆l =

Substitute numerical values and evaluate ∆l:

∆l =

Fl Mgl = AY AY

(950 kg )(9.81m/s2 )(2.5 m )

(1.5 cm )(150 GN/m ) 2

= 1.04 mm (b) Express the oscillation frequency of the wire-engine block system:

f =

1 2π

k eff M

2

1082 Chapter 14 Express the effective ″spring″ constant of the cable: Substitute to obtain:

Substitute numerical values and evaluate f:

k eff =

F Mg = ∆l ∆l

f =

1 2π

g ∆l

f =

1 2π

9.81 m/s 2 = 15.5 Hz 1.04 mm

Energy of an Object on a Vertical Spring 57 •• Picture the Problem Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is U s = 12 ky02 . A further extension of the spring by an amount y increases Us to

1 2

k ( y + y0 ) = 12 ky 2 + kyy0 + 12 ky02 = 12 ky 2 + mgy + 12 ky02 . Consequently, if we set 2

U = Ug + Us = 0, a further extension of the spring by y increases Us by ½ky2 + mgy while decreasing Ug by mgy. Therefore, if U = 0 at the equilibrium position, the change in U is 2 given by 12 k ( y ') , where y′ = y − y0. (a) Express the total energy of the system:

E = 12 kA2

Substitute numerical values and evaluate E:

E=

(b) Express and evaluate Ug when the object is at its maximum downward displacement:

U g = − mgA

(c) When the object is at its maximum downward displacement:

U s = 12 kA2 + mgA

1 2

(600 N/m )(0.03 m )2 =

(

0.270 J

)

= −(2.5 kg ) 9.81 m/s 2 (0.03 m ) = − 0.736 J

=

1 2

(600 N/m )(0.03 m )2 + (2.5 kg ) (9.81m/s 2 )(0.03 m )

= 1.01 J

Oscillations 1083 (d) The object has its maximum kinetic energy when it is passing through its equilibrium position:

K max = 12 kA2 =

1 2

(600 N/m )(0.03 m )2

= 0.270 J

58 •• Picture the Problem Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is U s = 12 ky02 . A further extension of the spring by an amount y increases Us to

1 2

k ( y + y0 ) = 12 ky 2 + kyy0 + 12 ky02 = 12 ky 2 + mgy + 12 ky02 . Consequently, if we set 2

U = Ug + Us = 0, a further extension of the spring by y increases Us by ½ky2 + mgy while decreasing Ug by mgy. Therefore, if U = 0 at the equilibrium position, the change in U is given by

1 2

k ( y ') , where y′ = y − y0. 2

(a) Express the total energy of the system:

E = 12 kA2

Letting ∆y represent the amount the spring is stretched from its natural length by the 1.5-kg object, apply Fy = ma y to the object when it is

k∆y − mg = 0

∑

in its equilibrium position: Solve for k:

Substitute for k to obtain:

Substitute numerical values and evaluate E:

k=

mg ∆y

E=

mgA2 2∆y

E=

(1.5 kg )(9.81m/s2 )(0.022 m )2 2(0.028 m )

= 0.127 J (b) Express Ug when the object is at its maximum downward displacement: Substitute numerical values and evaluate Ug:

U g = − mgA

(

)

U g = −(1.5 kg ) 9.81 m/s 2 (0.022 m ) = − 0.324 J

1084 Chapter 14 (c) When the object is at its maximum downward displacement:

U s = 12 kA2 + mgA

Substitute numerical values and evaluate Us:

Us =

1 2

(526 N/m )(0.022 m )2 + (1.5 kg ) (9.81 m/s 2 )(0.022 m )

= 0.451 J (d) The object has its maximum kinetic energy when it is passing through its equilibrium position:

K max = 12 kA2 =

(526 N/m )(0.022 m )2

1 2

= 0.127 J

*59 •• Picture the Problem We can find the amplitude of the motion by relating it to the maximum speed of the object. Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is U s = 12 ky02 . A further extension of the spring by an amount y increases Us to

1 2

k ( y + y0 ) = 12 ky 2 + kyy0 + 12 ky02 = 12 ky 2 + mgy + 12 ky02 . Consequently, if we set 2

U = Ug + Us = 0, a further extension of the spring by y increases Us by ½ky2 + mgy while decreasing Ug by mgy. Therefore, if U = 0 at the equilibrium position, the change in U is given by

1 2

k ( y ') , where y′ = y − y0. 2

(a) Relate the maximum speed of the object to the amplitude of its motion: Solve for A:

vmax = Aω

A=

vmax

ω

= vmax

Substitute numerical values and evaluate A:

A = (0.3 m/s )

(b) Express the energy of the object at maximum displacement:

E = 12 kA2

Substitute numerical values and evaluate E:

E=

1 2

m k

1.2 kg = 1.90 cm 300 N/m

(300 N/m )(0.019 m )2 =

0.0542 J

Oscillations 1085 (c) At maximum displacement from equilibrium: Substitute numerical values and evaluate Ug:

U g = −mgA

(

)

U g = −(1.2 kg ) 9.81 m/s 2 (0.019 m ) = − 0.224 J

(d) Express the potential energy in the spring when the object is at its maximum downward displacement:

U s = 12 kA2 + mgA

Substitute numerical values and evaluate Us:

Us =

(300 N/m )(0.019 m )2 + (1.2 kg ) (9.81 m/s 2 )(0.019 m )

1 2

= 0.278 J

Simple Pendulums 60 • Picture the Problem We can determine the required length of the pendulum from the expression for the period of a simple pendulum. Express the period of a simple pendulum:

Solve for L:

Substitute numerical values and evaluate L:

T = 2π

L=

L g

T 2g 4π 2

2 ( 5 s ) (9.81 m/s 2 ) L= =

4π 2

6.21 m

61 • Picture the Problem We can find the period of the pendulum from T = 2π L g moon where g moon = 16 g and L = 6.21 m. Express the period of a simple pendulum:

T = 2π

Substitute numerical values and evaluate T:

T = 2π

L g moon

1 6

6.21 m = 12.2 s 9.81 m/s 2

(

)

1086 Chapter 14 62 • Picture the Problem We can find the value of g at the location of the pendulum by solving the equation T = 2π L g for g and evaluating it for the given length and period. Express the period of a simple pendulum: Solve for g:

Substitute numerical values and evaluate g:

T = 2π

L g

4π 2 L g= T2 g=

4π 2 (0.7 m ) = 9.79 m/s 2 2 (1.68 s )

*63 • Picture the Problem We can use T = 2π L g to find the period of this pendulum. Express the period of a simple pendulum:

T = 2π

L g

Substitute numerical values and evaluate T:

T = 2π

34 m = 11.7 s 9.81 m/s 2

64 •• Picture the Problem The figure shows the simple pendulum at maximum angular displacement φ0. The total energy of the simple pendulum is equal to its initial gravitational potential energy. We can apply the definition of gravitational potential energy and use the small-angle approximation to show that E ≈ 12 mgLφ02 .

Express the total energy of the simple pendulum at maximum displacement:

E = U max displacement = mgh

For φ << 1:

cos φ ≈ 1 − 12 φ 2

= mgL[1 − cos φ0 ]

Oscillations 1087 Substitute and simplify to obtain:

[ (

)]

E = mgL 1 − 1 − 12 φ02 =

1 2

mgLφ02

65 •• Picture the Problem Because the cart is accelerating down the incline, the period of the simple pendulum will be given by T = 2π L g eff where geff is less than g by the acceleration of the cart. We can apply Newton’s 2nd law to the cart to find its acceleration down the incline and then subtract this acceleration from g to find geff. Express the period of a simple pendulum in terms of its length and the effective value of the acceleration of gravity: Relate geff to the acceleration of the cart: Apply

∑F

x

= ma x to the cart and

T = 2π

L g eff

g eff = g − a mg sin θ = ma

solve for its acceleration:

and

Substitute to obtain:

T = 2π

a = g sin θ

= 2π

L L = 2π g−a g − g sin θ L g (1 − sin θ )

1088 Chapter 14 66 •• Picture the Problem The figure shows the simple pendulum at maximum angular displacement φ0. We can express the angular position of the pendulum’s bob in terms of its initial angular position and time and differentiate this expression to find the maximum speed of the bob. We can use conservation of energy to find an exact value for vmax and the approximation cos φ ≈ 1 − 12 φ 2 to show that this value reduces to the former value for small φ.

dφ dt

(a) Relate the speed of the pendulum’s bob to its angular speed:

v=L

Express the angular position of the pendulum as a function of time:

φ = φ0 cos ωt

Differentiate this expression to express the angular speed of the pendulum:

dφ = −φ0ω sin ωt dt

Substitute in equation (1) to obtain:

v = − Lφ0ω sin ωt = −vmax sin ωt

Simplify vmax to obtain:

(b) Use conservation of energy to relate the potential energy of the pendulum at point 1 to its kinetic energy at point 2:

(1)

vmax = Lφ0

g = φ0 gL L

∆K + ∆U = 0 or, because K1 = U2 = 0,

K 2 −U1 = 0

mv22 − mgh = 0

Substitute for K2 and U1:

1 2

Express h in terms of L and φ0:

h = L(1 − cos φ0 )

Substitute for h and solve for v2 = vmax to obtain:

vmax =

2 gL(1 − cos φ0 )

(2)

Oscillations 1089 (c) For φ0 << 1:

1 − cos φ0 ≈ 12 φ02

Substitute in equation (2) to obtain:

vmax = 2 gL(12 φ02 ) = φ0 gL in agreement with our result in part (a).

(d) Express the difference in the results from (a) and (b):

∆v = vmax,a − vmax, b

Using φ0 = 0.20 rad and L = 1 m, evaluate the result in (b):

vmax,b = 2 9.81 m/s 2 (1 m )(1 − cos 0.2 )

Using φ0 = 0.20 rad and L = 1 m, evaluate the result in part (a):

vmax .a = (0.20 rad ) 9.81 m/s 2 (1 m )

Substitute in equation (3) to obtain:

∆v = 0.6264 m/s − 0.6254 m/s

(3)

(

)

= 0.6254 m/s

(

)

= 0.6264 m/s

= 0.001m/s = 1.00 mm/s

Physical Pendulums 67 • Picture the Problem The period of this physical pendulum is given by T = 2π I MgD where I is the moment of inertia of the thin disk with respect to an axis through its pivot point. We can use the parallel-axis theorem to express I in terms of the moment of inertia of the disk with respect to its center of mass and the distance from its center of mass to its pivot point. Express the period of physical pendulum: Using the parallel-axis theorem, find the moment of inertia of the thin disk about an axis through the pivot point: Substitute to obtain:

Substitute numerical values and evaluate T:

T = 2π

I MgD

I = I cm + MR 2 = 12 MR 2 + MR 2 = 32 MR 2

T = 2π

T = 2π

MR 2 3R = 2π MgR 2g

3 2

3(0.2 m ) = 1.10 s 2 9.81 m/s 2

(

)

1090 Chapter 14 68 • Picture the Problem The period of this physical pendulum is given by T = 2π I MgD where I is the moment of inertia of the circular hoop with respect to an axis through its pivot point. We can use the parallel-axis theorem to express I in terms of the moment of inertia of the hoop with respect to its center of mass and the distance from its center of mass to its pivot point. Express the period of the physical pendulum: Using the parallel-axis theorem, find the moment of inertia of the circular hoop about an axis through the pivot point: Substitute to obtain:

Substitute numerical values and evaluate T:

T = 2π

I MgD

I = I cm + MR 2 = MR 2 + MR 2 = 2MR 2

T = 2π

2 MR 2 2R = 2π MgR g

T = 2π

2(0.5 m ) = 2.01s 9.81 m/s 2

69 • Picture the Problem The period of a physical pendulum is given by T = 2π I MgD where I is its moment of inertia with respect to an axis through its pivot point. We can solve this equation for I and evaluate it using the given numerical data. Express the period of the physical pendulum: Solve for I:

Substitute numerical values and evaluate I:

T = 2π

I=

I MgD

MgDT 2 4π 2

2 ( 3 kg ) (9.81 m/s 2 )(0.1 m )(2.6 s ) I=

4π 2

= 0.504 kg ⋅ m 2

Oscillations 1091 *70 •• Picture the Problem We can use the expression for the period of a simple pendulum to find the period of the clock. (a) Express the period of a simple pendulum:

T = 2π

l g

Substitute numerical values and evaluate T:

T = 2π

4m = 4.01s 9.81 m/s 2

(b)

By effectively raising the center of mass of the pendulum, placing coins in the tray shortens the period.

71 •• Picture the Problem Let x be the distance of the pivot from the center of the rod, m the mass at each end of the rod, and L the length of the rod. We can express the period of the physical pendulum as a function of the distance x and then differentiate this expression with respect to x to show that, when x = L/2, the period is a minimum. (a) Express the period of a physical pendulum: Express the moment of inertia of the dumbbell with respect to an axis through its center of mass: Using the parallel-axis theorem, express the moment of inertia of the dumbbell with respect to an axis through the pivot point: Substitute in equation (1) to obtain:

I MgD

T = 2π

2

I cm

(1)

2

⎛L⎞ ⎛L⎞ = m⎜ ⎟ + m⎜ ⎟ = 12 mL2 ⎝2⎠ ⎝2⎠

I = I cm + 2mx 2 = 12 mL2 + 2mx 2

1 2

T = 2π =

2π g

mL2 + 2mx 2 2mgx 1 4

L2 + x 2 x

L2 + x 2 =C x 2π where C = g 1 4

(2)

1092 Chapter 14 Set dT/dx = 0 to find the condition for minimum T:

dT d =C⋅ dx dx

Evaluate the derivative to obtain:

2x2 −

( 1 4

x2 Because the denominator of this expression cannot be zero, it must be true that:

2 x2 −

Solve for x to obtain:

x=

(

1 2

1 4

1 4

L2 + x 2

L +x x 2

2

L2 + x 2 = 0 for extrema x

)=0 )

L2 + x 2 = 0

1 4

L

i.e., the period is a minimum when the pivot point is at one of the masses. (b) Substitute x = L/4 in equation (2) and simplify to obtain:

T = 2π

Substitute numerical values and evaluate T:

T =π

1 4

L2 + ( 14 L ) 5L =π 1 g 2 gL 2

5(2 m ) = 3.17 s 9.81 m/s 2

Remarks: In (a), we’ve shown that x = L/2 corresponds to an extreme value; i.e., to either a maximum or a minimum. To complete the demonstration that this value of x corresponds to a minimum, we can either (1) show that d2T/dx2 evaluated at x = L/2 is positive, or (2) graph T as a function of x and note that the graph is a minimum at x = L/2. 72 •• Picture the Problem Let x be the distance of the pivot from the center of the rod. We’ll express the period of the physical pendulum as a function of the distance x and then differentiate this expression with respect to x to find the location of the pivot point that minimizes the period of the physical pendulum. Express the period of a physical pendulum:

T = 2π

I MgD

(1)

Oscillations 1093 Express the moment of inertia of the dumbbell with respect to an axis through its center of mass:

Using the parallel-axis theorem, express the moment of inertia of the dumbbell with respect to an axis through the pivot point: Substitute in equation (1) to obtain:

2

I cm

2

⎛L⎞ ⎛L⎞ = m⎜ ⎟ + m⎜ ⎟ + 121 (2m )L2 ⎝2⎠ ⎝2⎠ = 23 mL2

I = I cm + 4mx 2 = 23 mL2 + 4mx 2

2 3

T = 2π =

π

mL2 + 4mx 2 4mgx 2 3

g

L2 + 4 x 2 x

or 2 3

T =C

L2 + 4 x 2 π where C = x g

Set dT/dx = 0 to find the condition for minimum T:

dT d = C× dx dx

Evaluate the derivative to obtain:

8 x 2 − ( 23 L2 + 4 x 2 ) 2 3

2x2

(

2 3

L2 + 4 x 2 = 0 for extrema x

L2 + 4 x 2 x

=0

)

Because the denominator of this expression cannot be zero, it follows that:

8x2 −

Solve for x to obtain:

x=

L 6

The distance to the pivot point from the nearer mass is:

d=

L L − = 0.0918L 2 6

Remarks: We’ve shown that x = L

2 3

L2 + 4 x 2 = 0

6 corresponds to an extreme value; i.e., to

either a maximum or a minimum. To complete the demonstration that this value of x corresponds to a minimum, we can either (1) show that d2T/dx2 evaluated at

1094 Chapter 14 x=L

6 is positive, or (2) graph T as a function of x and note that the graph is a

minimum at x = L

6.

*73 •• Picture the Problem Let x be the distance of the pivot from the center of the meter stick, m the mass of the meter stick, and L its length. We’ll express the period of the meter stick as a function of the distance x and then differentiate this expression with respect to x to determine where the hole should be drilled to minimize the period. Express the period of a physical pendulum: Express the moment of inertia of the meter stick with respect to its center of mass: Using the parallel-axis theorem, express the moment of inertia of the meter stick with respect to the pivot point: Substitute in equation (1) to obtain:

I MgD

T = 2π

(1)

I cm = 121 mL2 I = I cm + mx 2 = 121 mL2 + mx 2

1 12

T = 2π =

2π g

mL2 + mx 2 mgx

1 12

L2 + x 2 x

L2 + x 2 x 2π where C = g 1 12

=C

Set dT/dx = 0 to find the condition for minimum T:

dT d = C× dx dx

Evaluate the derivative to obtain:

2x2 − x2

Because the denominator of this expression cannot be zero, it follows

2 x2 −

(

1 12

1 12

(

L2 + x 2

L2 + x 2 x

1 12

1 12

L2 + x 2 = 0 for extrema x

)=0

)

L2 + x 2 = 0

Oscillations 1095 that:

L 100 cm = = 28.9 cm 12 12

Solve for and evaluate x to obtain:

x=

The hole should be drilled at a distance:

d = 50 cm − 28.9 cm = 21.1cm from the center of the meter stick.

74 •• Picture the Problem Let m represent the mass and r the radius of the uniform disk. We’ll use the expression for the period of a physical pendulum and the parallel-axis theorem to obtain a quadratic equation that we can solve for d. We will then treat our expression for the period of the pendulum as an extreme-value problem, setting its derivative equal to zero in order to determine the value for d that will minimize the period. (a) Express the period of a physical pendulum: Using the parallel-axis theorem, relate the moment of inertia with respect to an axis through the hole to the moment of inertia with respect to the disk’s center of mass: Substitute to obtain:

T = 2π

I = I cm + md 2 = 12 mR 2 + md 2

T = 2π = 2π

Square both sides of this equation, simplify, and substitute numerical values to obtain:

Solve the quadratic equation to obtain:

I mgd

d2 −

1 2

mR 2 + md 2 mgd

1 2

R2 + d 2 gd

(1)

gT 2 R2 d + =0 4π 2 2

or

d 2 − (1.553 m )d + 0.320 m 2 = 0

d = 0.245 m The second root, d = 1.31 m, is too large to be physically meaningful.

1096 Chapter 14 (b) Set the derivative of equation (1) equal to zero to find relative maxima and minima:

dT 2π d = ⋅ dd g dd

Evaluate the derivative to obtain:

2d 2 −

Solve this equation to obtain:

Evaluate equation (1) with

d=R

2 to obtain an expression

(

Remarks: We’ve shown that d = R

1 2

R2 + d 2

1 2

R2 + d 2 d

2d 2 −

(

d=

R 2

T = 2π

for the shortest possible period of this physical pendulum: Substitute numerical values and evaluate T:

R2 + d 2 d

= 0 for extrema

2d 2 Because the denominator of this fraction cannot be zero:

1 2

T = 2π

1 2

)=0

)

R2 + d 2 = 0

1 2

R 2 + 12 R 2 = 2π R g 2

2R g

2 (0.8 m ) = 2.13 s 9.81 m/s 2

2 corresponds to an extreme value; i.e., to

either a maximum or a minimum. To complete the demonstration that this value of d corresponds to a minimum, we can either (1) show that d2T/dd2 evaluated at d = R 2 is positive, or (2) graph T as a function of d and note that the graph is a minimum at d = R

2.

75 ••• Picture the Problem We can use the equation for the period of a physical pendulum and the parallel-axis theorem to show that h1 + h2 = gT 2/4π 2. Express the period of the physical pendulum: Using the parallel-axis theorem, relate the moment of inertia with respect to an axis through P1 to the

T = 2π

I mgd

I = I cm + mh12

Oscillations 1097 moment of inertia with respect to the disk’s center of mass: Substitute to obtain:

T = 2π

I cm + mh12 mgh1

Square both sides of this equation and rearrange to obtain:

mgT 2 I cm = + mh1 4π 2 h1

Because the period of oscillation is the same for point P2:

I I cm + mh1 = cm + mh2 h1 h2

Solve this equation for Icm:

I cm = mh1h2

Substitute in equation (1) to obtain:

mgT 2 mh1h2 = + mh1 4π 2 h1

(1)

or

h2 + h1 =

gT 2 4π 2

76 ••• Picture the Problem We can find the period of the physical pendulum in terms of the period of a simple pendulum by starting with T = 2π I mgL and applying the parallelaxis theorem. Performing a binomial expansion for r << L on the radicand of our expression for T will lead to T ≈ T0 (1 + r2/5L2). (a) Express the period of the physical pendulum: Using the parallel-axis theorem, relate the moment of inertia of the pendulum about an axis through its center of mass to its moment of inertia with respect to an axis through its point of support:

T = 2π

I mgL

I = I cm + mL2 = 52 mr 2 + mL2

1098 Chapter 14 Substitute and simplify to obtain:

T = 2π = 2π

2 5

mr 2 + mL2 = 2π mgL

12

⎛ 2r 2 ⎞ expand ⎜⎜1 + 2 ⎟⎟ ⎝ 5L ⎠

:

r 2 + L2 gL

L ⎛ 2r 2 ⎞ L 2r 2 ⎜⎜1 + 2 ⎟⎟ = 2π 1+ 2 g ⎝ 5L ⎠ g 5L

= T0 1 + (b) Using the binomial expansion,

2 5

12

⎛ 2r 2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ 5L ⎠

2r 2 5L2

1 ⎛ 2r 2 ⎞ 1 ⎛ 2r 2 ⎞ = 1 + ⎜⎜ 2 ⎟⎟ + ⎜⎜ 2 ⎟⎟ 2 ⎝ 5L ⎠ 8 ⎝ 5L ⎠

2

+ higher - order terms r2 ≈ 1+ 2 5L provided r << L

Substitute in our result from (a) to obtain:

⎛ r2 ⎞ T ≈ T0 ⎜⎜1 + 2 ⎟⎟ ⎝ 5L ⎠

(c) Express the fractional error when the approximation T = T0 is used for this pendulum:

∆T T − T0 T ≈ = −1 T T0 T0

Substitute numerical values and evaluate ∆T/T:

(2 cm ) = 0.008% ∆T ≈ 2 T 5(100 cm )

For an error of 1%:

r2 = 0.01 5L2

Solve for and evaluate r with L = 100 cm:

r = L 0.05 = (100 cm ) 0.05

r2 r2 = 1+ 2 −1 = 2 5L 5L 2

= 22.4 cm

77 ••• Picture the Problem The period of this physical pendulum is given by T = 2π I MgD . We can express its period as a function of the distance d by using the definition of the center of mass of the pendulum to find D in terms of d and the parallelaxis theorem to express I in terms of d. Solving the resulting quadratic equation yields d.

Oscillations 1099 In (b), because the clock is losing 5 minutes per day, one would reposition the disk so that the clock runs faster; i.e., so the pendulum has a shorter period. We can determine the appropriate correction to make in the position of the disk by relating the fractional time loss to the fractional change in its position. (a) Express the period of the physical pendulum: Solve for

T = 2π

I mtot gxcm

I T 2 gmtot = xcm 4π 2

I : xcm

(1)

Express the moment of inertia of the physical pendulum, relative to an axis through the pivot point, as a function of d:

I = I cm + Md 2 = 13 mL2 + 12 Mr 2 + Md 2

Substitute numerical values and evaluate I:

I=

(0.8 kg )(2 m )2 + 12 (1.2 kg )(0.15 m )2 + (1.2 kg )d 2 = 1.0802 kg ⋅ m 2 + (1.2 kg )d 2 1 3

(2 kg )xcm = (0.8 kg )(1 m ) + (1.2 kg )d

Locate the center of mass of the physical pendulum relative to the pivot point:

and

xcm = 0.4 m + 0.6d

Substitute in equation (1) to obtain:

(

)

1.0802 kg ⋅ m 2 + (1.2 kg ) d 2 T 2 9.81 m/s 2 (2 kg ) = = 0.49698 kg ⋅ m/s 2 T 2 2 0.4 m + 0.6d 4π Setting T = 2.5 s and solving for d yields:

(b) There are 1440 minutes per day. If the clock loses 5 minutes per day, then the period of the clock is related to the perfect period of the clock by:

(

)

(2)

d = 1.63572 m where we have kept more than three significant figures for use in part (b).

1435 T = 1440 Tperfect where Tperfect = 3.5 s.

1100 Chapter 14 1440 1440 (3.5 s ) Tperfect = 1435 1435 = 3.51220 s

Solve for and evaluate T:

T=

Substitute T = 3.51220 s in equation (2) and solve for d to obtain:

d = 3.40140 m

Substitute T = 3.50 s in equation (2) and solve for d ′ to obtain:

d' = 3.37825 m

Express the distance the disk needs to be moved upward to correct the period:

∆d = d − d' = 3.40140 m − 3.37825 m =

2.32 cm

*78 •• Picture the Problem The period of a simple pendulum depends on its amplitude φ0 according to T = 2π

2 ⎤ 1 1 ⎛3⎞ 1 L⎡ 2 1 1 + sin + sin 4 φ0 + ...⎥ . We can φ ⎢ 0 2 2 ⎜ ⎟ 2 2 ⎝4⎠ 2 g ⎢⎣ 2 ⎥⎦

approximate T to the second-order term and express ∆T/T = (Tslow – Taccurate)/T. Equating this expression to ∆T/T calculated from the fractional daily loss of time will allow us to solve for and evaluate the amplitude of the pendulum that corresponds to keeping perfect time. Express the fractional daily loss of time:

1h 48 ∆T 48 s 1day = × × = T day 24 h 3600 s 86400

Approximate the period of the clock to the second-order term:

T = 2π

Express the difference in the periods of the slow and accurate clocks:

L⎡ 1 ⎤ 1 1 + 2 sin 2 φ0 ⎥ ⎢ g⎣ 2 2 ⎦

∆T = Tslow − Taccurate 1 L ⎧⎡ ⎤ 2 1 ⎨⎢1 + 2 sin (8.4°)⎥ 2 g ⎩⎣ 2 ⎦

= 2π

1 1 ⎤⎞ ⎡ − ⎢1 + 2 sin 2 φ0 ⎥ ⎟⎟ 2 ⎦⎠ ⎣ 2 1 L⎡ 1 sin 2 (8.4°) 2 ⎢ 2 g ⎣2

= 2π −

1 1 ⎤ sin 2 φ0 ⎥ 2 2 2 ⎦

Oscillations 1101

Divide both sides of this equation by T to obtain: Substitute for

∆T and simplify to T

obtain:

∆T 1 2 1 1 = sin 4.2° − sin 2 φ0 T 4 4 2 1 2 1 1 48 sin 4.2° − sin 2 φ0 = 4 4 2 86400 and

1 sin φ0 = 0.05605 2 Solve for φ0:

φ0 = 6.43°

79 •• Picture the Problem The period of a simple pendulum depends on its amplitude φ0 2 ⎤ 1 1 ⎛3⎞ L⎡ 2 1 4 1 according to T = 2π ⎢1 + 2 sin φ0 + 2 ⎜ ⎟ sin φ0 + ...⎥ . We’ll approximate 2 2 ⎝4⎠ 2 g ⎣⎢ 2 ⎦⎥

T to the second-order term and express ∆T/T = (Tslow – Tcorrect)/T. Equating this expression to ∆T/T calculated from the fractional daily loss of time will allow us to solve for and evaluate the amplitude of the pendulum that corresponds to keeping correct time. Express the fractional daily loss of time:

1h 5 ∆T 5 min 1day = × × = T day 24 h 60 min 1440

Approximate the period of the clock to the second-order term:

T = 2π

L⎡ 1 1 ⎤ 1 + 2 sin 2 φ0 ⎥ ⎢ g⎣ 2 2 ⎦

Assuming that the amplitude of the slow-running clock’s pendulum is small enough to ignore, express the difference in the periods of the slow and corrected clocks:

∆T = Tslow − Tcorrect

Divide both sides of this expression by T to obtain:

∆T 1 1 = − sin 2 φ0 T 4 2

Substitute for obtain:

∆T and simplify to T

= 2π

L⎧ ⎡ 1 ⎤⎫ 2 1 ⎨1 − ⎢1 + 2 sin φ0 ⎥ ⎬ g⎩ ⎣ 2 2 ⎦⎭

= 2π

L⎡ 1 1 ⎤ − 2 sin 2 φ0 ⎥ ⎢ g⎣ 2 2 ⎦

−5 1 1 − sin 2 φ0 = 4 2 1440 and

1102 Chapter 14 1 sin φ0 = 0.1178 2 Solve for φ0:

φ0 = 13.5°

Damped Oscillations 80 • Picture the Problem We can use the definition of the damping constant and its dimensions to show that it has units of kg/s. Using its definition, relate the decay constant τ to the damping constant b: Substitute the units of m and τ to obtain:

81

τ=

m m ⇒ b= b τ

Dimensionally, b =

•

Picture the Problem For small damping, Q =

[M ] = [T ]

kg s

2π where ∆E/E is the fractional ( ∆E E )cycle

energy loss per cycle.

2π ( ∆E E )cycle

Relate the Q factor to the fractional energy loss per cycle:

Q=

Solve for and evaluate the fractional energy loss per cycle:

( ∆E E )

cycle

=

2π 2π = = 3.14% Q 200

82 • Picture the Problem We can find the period of the oscillator from T = 2π m k and its total initial energy from E0 = 12 kA2 . The Q factor can be found from its definition

Q = 2π ( ∆E E )cycle and the damping constant from Q = ω0 m b .

(a) The period of the oscillator is given by:

T = 2π

m k

Substitute numerical values and evaluate T:

T = 2π

2 kg = 0.444 s 400 N/m

(b) Relate the initial energy of the

E0 = 12 kA2

Oscillations 1103 oscillator to its amplitude: Substitute numerical values and evaluate E0:

E0 =

(c) Relate the fractional rate at which the energy decreases to the Q value and evaluate Q:

Q=

Express the Q value in terms of b:

Q=

Solve for the damping constant b: Substitute numerical values and evaluate b:

1 2

(400 N/m )(0.03 m )2 =

0.180 J

2π 2π = = 628 ( ∆E E )cycle 0.01

ω0 m b

2πm Q TQ 2π (2 kg ) b= = 0.0451kg/s (0.444 s )(628) b=

ω0 m

=

83 •• Picture the Problem The amplitude of the oscillation at time t is A(t ) = A0 e − t 2τ where

τ = m/b is the decay constant. We’ll express the amplitudes one period apart and then show that their ratio is constant. Relate the amplitude of a given oscillation peak to the time at which the peak occurs: Express the amplitude of the oscillation peak at t′ = t + T: Express the ratio of these consecutive peaks:

A(t ) = A0 e − t 2τ

A(t + T ) = A0 e − (t +T ) 2τ A(t ) A0e − t 2τ = = e −T 2τ A(t + T ) A0e −(t +T ) 2τ = constant

84 •• Picture the Problem We can relate the fractional change in the energy of the oscillator each cycle to the fractional change in its amplitude. Both the Q value and the decay constant τ can be found from their definitions. (a) Relate the energy of the oscillator to its amplitude:

E = 12 kA2

1104 Chapter 14 Take the differential of this relationship to obtain:

dE = kAdA

Divide both sides of this equation by E:

dE kAdA dA = 1 2 =2 E A 2 kA

Approximate dE and dA by ∆E and ∆A and evaluate ∆E/E:

∆E = 2(5%) = 10% E

(b) For small damping:

∆E T = τ E and

τ=

(c) Using its definition, express and evaluate Q:

T 3s = = 30 s ∆E E 0.01

A = ω0τ =

2π 2π (30 s ) = 62.8 τ= T 3s

85 •• Picture the Problem We can use the physical interpretation of Q for small damping

Q=

2π to find the fractional decrease in the energy of the oscillator each ( ∆E E )cycle

cycle. (a) Express the fractional decrease in energy each cycle as a function of the Q factor and evaluate ∆E E :

∆E 2π 2π = = = 0.314 20 E Q

(b) Using the definition of the Q factor, use Equation 14-35 to express ω′ as a function of Q:

⎡ 1 ⎛ b 2 ⎞⎤ ω' = ω0 ⎢1 − ⎜⎜ 2 2 ⎟⎟⎥ ⎣ 4 ⎝ m ω0 ⎠ ⎦ ⎡ 1 ⎤ = ω0 ⎢1 − 2⎥ ⎣ 4Q ⎦

Use the approximation (1 + x)½ ≈ 1 + ½x for small x to obtain:

⎡

ω' = ω0 ⎢1 − ⎣

1 ⎤ 8Q 2 ⎥⎦

12

12

Oscillations 1105 Express and evaluate ω′ − ω0:

⎡

ω' − ω0 = ω0 ⎢1 − ⎣

=−

1 ⎤ 1 − ω0 = − 2 2⎥ 8Q ⎦ 8Q

1 2 8(20 )

= − 3.13 × 10−2 percent 86 •• Picture the Problem The amplitude of the spring-and-mass oscillator varies with time according to A = A0 e − t 2τ and its energy according to E = E0 e − t τ . (a) Express the amplitude of the oscillations as a function of time: Evaluate the amplitude when t = 2 s:

A = (6 cm )e − t 4 s A(2 s ) = (6 cm )e −2 s 4 s = (6 cm )e −1 2 = 3.64 cm

Evaluate the amplitude when t = 4 s:

A(4 s ) = (6 cm )e −4 s 4 s = (6 cm )e −1 = 2.21cm

(b) Express the energy of the system at t = 0: Express the energy in the system at t =2 s: The energy dissipated in the first 2 s is:

E (0) = E0e −0 2 s = E0 = 60 J E (2 s ) = E0e −2 s 2 s = E0e −1

∆E0− 2 s = E (0) − E (2 s )

(

= E0 1 − e −1

(

)

= (60 J ) 1 − e −1

)

= 37.9 J The energy dissipated in the second 2-s interval is:

(

∆E2−4 s = E2 s 1 − e −2s 2 s

(

)

)

= (37.9 J ) 1 − e −1 = 24.0 J

*87 •• Picture the Problem We can find the fractional loss of energy per cycle from the physical interpretation of Q for small damping. We will also find a general expression for the earth’s vibrational energy as a function of the number of cycles it has completed. We can then solve this equation for the earth’s vibrational energy after any number of days.

1106 Chapter 14 (a) Express the fractional change in energy as a function of Q:

∆E 2π 2π = = = 1.57% E Q 400

(b) Express the energy of the damped oscillator after one cycle:

⎛ ∆E ⎞ E1 = E0 ⎜1 − ⎟ E ⎠ ⎝

Express the energy after two cycles:

⎛ ∆E ⎞ ⎛ ∆E ⎞ E2 = E1 ⎜1 − ⎟ = E0 ⎜ 1 − ⎟ E ⎠ E ⎠ ⎝ ⎝

Generalizing to n cycles:

⎛ ∆E ⎞ n En = E0 ⎜1 − ⎟ = E0 (1 − 0.0157 ) E ⎠ ⎝

2

n

= E0 (0.9843)

n

24 h 60 m × d h 1T = 2880 min× 54 min = 53.3T

(c) Express 2 d in terms of the number of cycles; i.e., the number of vibrations the earth will have experienced:

2d = 2d ×

Evaluate E(2 d):

E (2 d ) = E0 (0.9843)53.3 = 0.430 E0

88 •• Picture the Problem The diagram shows 1) the pendulum bob displaced through an angle θ0 and held in equilibrium by the force exerted on it by the air from the fan and 2) the bob accelerating, under the influence of gravity, tension force, and drag force, toward its equilibrium position. We can apply Newton’s 2nd law to the bob to obtain the differential equation of motion of the damped pendulum and then use its solution to find the decay time constant and the time required for the amplitude of oscillation to decay to 1°. (a) Apply to obtain:

∑τ = Iα to the pendulum

− mgl sin θ + lFd = I

d 2θ dt 2

Oscillations 1107 Express the moment of inertia of the pendulum with respect to an axis through its point of support: Substitute for I and Fd to obtain:

Because θ << 1 and v = lω = ldθ/dt:

I = ml 2

ml 2

d 2θ + lbv + mgl sin θ = 0 dt 2

d 2θ dθ ml + l 2b + mglθ = 0 2 dt dt 2

or

m The solution to this second-order homogeneous differential equation with constant coefficients is:

d 2θ dθ mg +b + θ =0 2 l dt dt

θ = θ 0 e −t 2τ cos(ω't + δ )

(1)

where θ0 is the maximum amplitude, τ = m/b is the time constant, and the frequency ω' = ω0 1 − (b 2mω0 ) . 2

Apply

r

r

∑ F = ma to the bob when

∑F

x

= Ffan − T sin θ 0 = 0

it is at its maximum angular displacement to obtain:

and

Divide the x equation by the y equation to obtain:

Ffan T sin θ 0 = = tan θ 0 mg T cos θ 0

∑F

y

= T cos θ 0 − mg = 0

or

Ffan = mg tan θ 0

When the bob is in equilibrium, the drag force on it equals Ffan:

bv = mg tan θ 0

Solve for m/b in the definition of τ to obtain:

τ=

m v = b g tan θ 0

Substitute numerical values and evaluate τ :

τ=

7 m/s = 8.16 s 9.81 m/s 2 tan 5°

(b) From equation (1) we have:

θ = θ 0e −t 2τ

When the amplitude has decreased to 1°:

5°e −t 2τ = 1° or e −t 2τ = 0.2

Take the natural logarithm of both sides of the equation to obtain:

−

(

)

t = ln (0.2) 2τ

1108 Chapter 14 Solve for t:

t = −2τ ln (0.2 )

Substitute for τ and evaluate t:

t = −2(8.16 s ) ln (0.2 ) = 26.3 s

Driven Oscillations and Resonance 89 • Picture the Problem The resonant frequency of a vibrating system depends on the mass of the system and on a “stiffness” constant according to f 0 =

1 2π

simple pendulum oscillating with small-amplitude vibrations, f 0 =

k or, in the case of a m 1 2π

g . L

(a) For this spring-and-mass oscillator we have:

f0 =

1 2π

400 N/m = 1.01 Hz 10 kg

(b) For this spring-and-mass oscillator we have:

f0 =

1 2π

800 N/m = 2.01 Hz 5 kg

f0 =

1 2π

9.81 m/s 2 = 0.352 Hz 2m

(c) For this simple pendulum we have:

90 • Picture the Problem We can use the physical interpretation of Q for small damping to find the Q factor for this damped oscillator. The width of the resonance curve depends on the Q factor according to ∆ω = ω0 Q . (a) Using the physical interpretation of Q for small damping, relate Q to the fractional loss of energy of the damped oscillator per cycle:

Q=

2π ( ∆E E )cycle

Evaluate this expression for ∆E E cycle = 2% :

Q=

2π = 314 0.02

(b) Relate the width of the resonance curve to the Q value of the oscillatory system:

∆ω =

(

)

ω0 Q

=

2πf 0 Q

Oscillations 1109

(

)

2π 300 s -1 ∆ω = = 6.00 rad/s 3.14

Substitute numerical values and evaluate ∆ω:

91 •• Picture the Problem The amplitude of the damped oscillations is related to the damping constant, mass of the system, the amplitude of the driving force, and the natural and driving frequencies through A =

(

F0

. Resonance occurs when

)

2

m 2 ω02 − ω 2 + b 2ω 2

ω = ω0 . At resonance, the amplitude of the oscillations is A = F0

b 2ω 2 and the

width of the resonance curve is related to the damping constant and the mass of the system according to ∆ω = b m . (a) Express the amplitude of the oscillations as a function of the driving frequency: Determine ω0:

Evaluate the radicand in the expression for A to obtain:

A=

ω0 =

F0

(

)

2

m 2 ω02 − ω 2 + b 2ω 2

k = m

400 N/m = 14.14 rad/s 2 kg

(2 kg )2 [(14.14 rad/s)2 − (10 rad/s)2 ] 2 2 + (2 kg/s ) (10 rad/s )

2

= 4.04 × 10 4 kg 2 / s 4

10 N

Substitute numerical values and evaluate A:

A=

(b) Resonance occurs when:

ω = ω0 = 14.1rad/s

(c) Express the amplitude of the motion at resonance:

A=

Substitute numerical values and evaluate A:

A=

(d) The width of the resonance curve is:

∆ω =

4.04 × 10 4 kg 2 / s 4

= 4.98 cm

F0 b 2ω02 10 N

(2 kg/s ) (14.14 rad/s)2 2

= 35.4 cm

b 2 kg/s = = 1.00 rad/s m 2 kg

1110 Chapter 14 92 •• Picture the Problem We’ll find a general expression for the damped oscillator’s energy as a function of the number of cycles it has completed. We can then solve this equation for the number of cycles corresponding to the loss of half the oscillator’s energy. The Q factor is related to the fractional energy loss per cycle through ∆E E = 2π Q and the width of the resonance curve is ∆ω = ω0 Q where ω0 is the oscillator’s natural angular frequency. (a) Express the energy of the damped oscillator after one cycle:

⎛ ∆E ⎞ E1 = E0 ⎜1 − ⎟ E ⎠ ⎝

Express the energy after two cycles:

⎛ ∆E ⎞ ⎛ ∆E ⎞ E2 = E1 ⎜1 − ⎟ = E0 ⎜ 1 − ⎟ E ⎠ E ⎠ ⎝ ⎝

Generalizing to n cycles:

⎛ ∆E ⎞ E n = E0 ⎜ 1 − ⎟ E ⎠ ⎝

Substitute numerical values:

0.5 E0 = E0 (1 − 0.035)

2

n

n

or

0.5 = (0.965)

n

Solve for n to obtain:

n=

ln 0.5 = 19.5 ln 0.965

≈ 20 complete cycles.

2π 2π = = 180 ∆E E 0.035

(b) Apply the physical interpretation of Q for small damping to obtain:

Q=

(c) The width of the resonance curve is given by:

∆ω =

ω0 Q

=

2πf 0 2π (100 Hz ) = Q 180

= 3.49 rad/s

Collisions 93 ••• Picture the Problem Let the system include the spring-and-mass oscillator and the second object of mass m. Because the net external force acting on this system is zero, momentum is conserved during the collision of the second object with the oscillator.

Oscillations 1111 Because the collision is elastic, we can also apply conservation of energy. Let the subscript 1 refer to the object attached to the spring and the subscript 2 identify the second object. mv1i + mv2i = mv2f (a) Using momentum conservation, relate the speeds of the objects or before and after their collision: v1i + v2i = v2f (1) 1 2

mv12i + 12 mv22i = 12 mv22f

Using conservation of energy, obtain a second relationship between the speeds of the objects before and after their collision:

or

Solve equation (2) for v22i :

v22i = v22f − v12i = (v2f + v1i )(v2f − v1i )

Substitute for v2f from equation (1):

v22i = (v1i + v2i + v1i )(v1i + v2i − v1i )

v12i + v22i = v22f

(2)

= (2v1i + v2i )(v2i ) = 2v1i v2i + v22i

or

2v1i v2i = 0 Because v1i ≠ 0 , it follows that:

v = v2i = 0 i.e., the second object must be initially at rest.

(b) Because v2i = 0, we have, from equation (1): Because the object connected to the spring was moving through its equilibrium position at the time of collision:

v2f = v1i

(

v1i = vmax = Aω = (0.1 m ) 40 s −1

)

= 4.00 m/s

94 ••• Picture the Problem Let the system include the spring-and-mass oscillator and the second object of mass m. Because the net external force acting on this system is zero, momentum is conserved during the collision of the second object with the oscillator. Because the collision is elastic, we can also apply conservation of energy. Let the subscript 1 refer to the object attached to the spring and the subscript 2 identify the second object. Using momentum conservation, relate the speeds of the objects

mv1i + mv2i = mv2f or

1112 Chapter 14 before and after their collision:

v1i + v2i = v2f

Using conservation of energy, obtain a second relationship between the speeds of the objects before and after their collision:

1 2

or

Solve equation (2) for v22i :

v22i = v22f − v12i = (v2f + v1i )(v2f − v1i )

Substitute for v2f from equation (1):

v22i = (v1i + v2i + v1i )(v1i + v2i − v1i )

(1)

mv12i + 12 mv22i = 12 mv22f

v12i + v22i = v22f

(2)

= (2v1i + v2i )(v2i ) = 2v1i v2i + v22i

or

2v1i v2i = 0 Because v1i ≠ 0 , it follows that:

v = v2i = 0 i.e., the second object must be initially at rest.

Because the object connected to the spring was moving through its equilibrium position at the time of collision:

(

v1i = vmax = Aω = (0.1 m ) 40 s −1

)

= 4 m/s

Express the total energy of the system just before the collision:

E = 12 mv12i

Solve for m:

m=

2E v12i

Substitute numerical values and evaluate m:

m=

2(8 J ) = 1.00 kg (4 m/s)2

Relate the spring constant to the angular frequency of the oscillator:

k = mω 2

Substitute numerical values and evaluate k:

k = (1 kg ) 40 s −1

(

)

2

= 1.60 kN/m

95 ••• Picture the Problem Let the system include the spring-and-mass oscillator and the 1-kg object. Because the net external force acting on this system is zero, momentum is

Oscillations 1113 conserved during the collision of the second object with the oscillator. Let the subscript 1 refer to the 1-kg object and the subscript 2 to the 2-kg object. We can relate the amplitude of the motion to the maximum speed of the oscillator (which we can find from conservation of momentum) and the angular frequency of the oscillator, which we can determine from its definition. Once we have found the amplitudes and angular frequencies for both collisions, we express the position of each as a function of time, using the initial conditions to find the phase constants. (a) Relate the amplitude of the motion to the angular frequency and maximum speed of the oscillator: Because the 2-kg object is initially at rest, the maximum speed of the oscillator will be its speed immediately after the collision. Use conservation of momentum to relate this maximum speed to the speed of the 1-kg object before the collision:

A=

vmax

(1)

ω

m1v1i = (m1 + m2 )vmax

Solve for vmax:

vmax =

m1 v1i m1 + m2

Substitute numerical values and evaluate vmax:

vmax =

1 kg (6 m/s) = 2 m/s 1 kg + 2 kg

Express the angular frequency of the oscillator:

ω=

k m1 + m2

Substitute numerical values and evaluate ω:

ω=

600 N/m = 14.14 rad/s 3 kg

Substitute in equation (1) and evaluate A:

A=

Express and evaluate the period of the oscillator’s period:

T=

(b) For an elastic collision:

vmax = v2f =

2 m/s = 14.1cm 14.14 s −1 2π

ω

=

2π = 0.444 s 14.14 s −1 2m1 v1i m1 + m2

1114 Chapter 14 Substitute numerical values and evaluate vmax:

vmax =

2(1 kg ) (6 m/s) = 4 m/s 3 kg

Using its definition, evaluate the angular frequency of the oscillator:

ω=

k = m2

Substitute in equation (1) and evaluate A:

A=

Express and evaluate the period of the oscillator’s period:

T=

(c) For the perfectly inelastic collision: Use the initial conditions to evaluate δ:

Substitute in equation (2) to obtain:

600 N/m = 17.32 rad/s 2 kg

4 m/s = 23.1cm 17.32 s −1 2π

ω

=

2π = 0.363 s 17.32 s −1

[(

)

x(t ) = (14.1cm )cos 14.1s −1 t + δ

]

(2)

⎛ v ⎞ v0 ⎞ π ⎟⎟ = tan −1 ⎜⎜ − 0 ⎟⎟ = − 2 ⎝ ω (0 ) ⎠ ⎝ ω x0 ⎠ ⎛

δ = tan −1 ⎜⎜ −

π⎤ ⎡ x(t ) = (14.1cm )cos ⎢ 14.1s −1 t − ⎥ 2⎦ ⎣

(

)

[(

)]

[(

)

= (14.1cm )sin 14.1s −1 t For the elastic collision: Use the initial conditions to evaluate δ:

Substitute in equation (3) to obtain:

x(t ) = (23.1cm ) cos 17.3 s −1 t + δ

]

(3)

⎛

⎛ v ⎞ v0 ⎞ π ⎟⎟ = tan −1 ⎜⎜ − 0 ⎟⎟ = − 2 ⎝ ω (0 ) ⎠ ⎝ ω x0 ⎠

δ = tan −1 ⎜⎜ −

π⎤ ⎡ x(t ) = (23.1cm ) cos ⎢ 17.3 s −1 t − ⎥ 2⎦ ⎣

(

=

)

(23.1cm )sin[(17.3 s −1 )t ]

General Problems 96 • Picture the Problem The particle’s displacement is of the form x = A cos(ωt + δ ) . Thus, we have A = 0.4 m, ω = 3 rad/s, and δ = π/4. We can find the frequency of the motion from its angular frequency and the period from the frequency. The particle’s position at t = 0 and t = 0.5 s can be found directly from its displacement function.

Oscillations 1115 (a) Express and evaluate the frequency of the particle’s motion:

f =

ω 3 rad/s = = 0.477 Hz 2π 2π

Use the relationship between the frequency and the period of the particle’s motion to find its period:

T=

1 1 = = 2.09 s f 0.477 s −1

(b) Using the expression for the particle’s displacement, find its position at t = 0:

π⎤ ⎡ x(0) = (0.4 m ) cos ⎢(3 rad/s )(0) + ⎥ 4⎦ ⎣ ⎡π ⎤ = (0.4 m ) cos ⎢ ⎥ = 0.283 m ⎣4⎦

(c) Using the expression for the particle’s displacement, find its position at t = 0.5 s:

π⎤ ⎡ x(0) = (0.4 m ) cos ⎢(3 rad/s )(0.5 s ) + ⎥ 4⎦ ⎣ = (0.4 m ) cos[2.29 rad] = − 0.264 m

97 • Picture the Problem We can express the velocity of the particle by differentiating its displacement with respect to time. (a) Differentiate the particle’s displacement to obtain:

dx dt d ⎧ π ⎤⎫ ⎡ = ⎨(0.4 m ) sin ⎢(3 rad/s)t + ⎥ ⎬ dt ⎩ 4 ⎦⎭ ⎣

v=

π⎤ ⎡ = − (1.2 m/s ) sin ⎢(3 rad/s )t + ⎥ 4⎦ ⎣ (b) Evaluate the result in part (a) at t = 0:

π⎤ ⎡ v(0) = −(1.2 m/s) sin ⎢(3 rad/s )(0) + ⎥ 4⎦ ⎣ ⎡π ⎤ = −(1.2 m/s) sin ⎢ ⎥ ⎣4⎦ = − 0.849 m/s

(c) By inspection of the result in part (a) (or from vmax = Aω ):

vmax = 1.20 m/s

(d) Substitute vmax for v to obtain:

π⎤ ⎡ 1.2 m/s = −(1.2 m/s ) sin ⎢(3 rad/s )t '+ ⎥ 4⎦ ⎣

1116 Chapter 14 or

(3 rad/s )t' + π

4

Solve for t′ to obtain:

= sin −1 (− 1) =

3π 2

t' = 1.31s

98 • Picture the Problem Let ∆y represent the amount by which the spring stretches. We’ll apply a condition for equilibrium to the object to relate the amount the spring has stretched to the angular frequency of its motion and then solve this equation for ∆y. Apply

∑F i

y

= 0 to the object

k∆y − mg = 0

when it is in its equilibrium position and solve for the elongation of the spring:

or

Relate the angular frequency of the object’s motion to its period:

ω=

Substitute to obtain:

⎛ T ⎞ ∆y = ⎜ ⎟ g ⎝ 2π ⎠

Substitute numerical values and evaluate ∆x:

⎛ 4.5 s ⎞ 2 ∆y = ⎜ ⎟ 9.81 m/s = 5.03 m ⎝ 2π ⎠

∆y =

m g g= 2 k ω

2π T 2

2

(

)

*99 •• Picture the Problem Compare the forces acting on the particle to the right in Figure 14-36 with the forces shown acting on the bob of the simple pendulum shown in the free-body diagram to the right. Because there is no friction, the only forces acting on the particle are mg and the normal force acting radially inward. In (b), we can think of the particles as the bobs of simple pendulums of equal length. (a) The normal force is identical to the tension in a string of length r that keeps the particle moving in a circular path and a component of mg provides, for small displacements θ0 or s2, the linear restoring force required for oscillatory motion.

Oscillations 1117 (b) The particles meet at the bottom. Because s1 and s2 are both much smaller than r, the particles behave like the bobs of simple pendulums of equal length; therefore they have the same periods.

100 •• Picture the Problem The diagram shows the ball when it is a horizontal distance x from the bottom of the bowl. Note that we’ve chosen the zero of gravitational potential energy to be at the bottom of the bowl. The total energy of the ball is the sum of its potential energy and kinetic energies due to translation and rotation. Once we’ve obtained an expression for the total energy of the rolling ball, we can require, because the surface is frictionless, that the total energy of the sliding object be the same as that of the rolling ball. Because the motion of the ball is simple harmonic motion, we can assume a solution to its differential equation of motion and express the total energy of the ball in terms of this assumed solution. Doing so will lead us to an expression that we can solve for the oscillation frequency of the ball.

(a) Express the total energy E of the ball:

E = U + K = U + K trans + K rot

Referring to the diagram shown above and assuming that R << r, express the potential energy of the ball when it is a horizontal distance x from the bottom of the bowl:

U ( x ) = mgr (1 − cos θ )

Express cosθ as a power series:

For θ << 1:

Substitute to obtain:

cos θ = 1 −

cos θ ≈ 1 −

θ2 2!

+

θ4 4!

(1)

+ ...

θ2 2!

⎡ ⎛ θ 2 ⎞⎤ U ( x ) ≈ mgr ⎢1 − ⎜⎜1 − ⎟⎟⎥ = 12 mgrθ 2 2! ⎠⎦ ⎣ ⎝

1118 Chapter 14 For R << r:

Substitute to obtain:

Substitute in equation (1):

Because the ball is rolling without slipping, v = Rω. Substitute for ω and I to obtain: Simplify to obtain:

θ≈

x r

mgx 2 U (x ) = 2r E=

mgx 2 1 2 1 2 + mv + Iω 2r 2 2

mgx 2 1 2 1 ⎛ 2 ⎞⎛ v ⎞ + mv + ⎜ mR 2 ⎟⎜ ⎟ E= 2r 2 2⎝5 ⎠⎝ R ⎠ E=

mgx 2 7 + mv 2 2r 10

(b) Because energy is conserved if the side of the bowl is frictionless:

mgx 2 7 + mv 2 = constant E= 2r 10

Because the motion is simple harmonic motion, assume a solution of the form:

x = x0. cos(ωt + δ )

Differentiate this assumed solution with respect to time to obtain:

v = −ωx0 sin (ωt + δ )

Substitute to obtain:

Express the condition the E = constant: Solve for ω to obtain:

mg (x0. cos(ωt + δ ))2 2r 7 2 + m(− ωx0 sin (ωt + δ )) 10 mgx02 = cos 2 (ωt + δ ) 2r 7 mω 2 x02 + sin 2 (ωt + δ ) 10

E=

mgx02 7 mω 2 x02 g 7ω 2 = or = r 2r 10 5

ω=

5g 7r

2

Oscillations 1119 101 •• Picture the Problem Assume that the plane is accelerating to the right with an acceleration a0. The free-body diagram shows the forces on the bob as seen in the accelerated frame of the airplane. Let g′ represent the effective value of the acceleration due to gravity. The period of the yo-yo is given by

T = 2π L g' where g′ is the effective value of the acceleration due to gravity. Express the period of your yo-yo pendulum as a function of the effective value for the acceleration due to gravity: Using the FBD, relate g′ and g:

Substitute to obtain:

Substitute numerical values and evaluate T:

T = 2π

L g'

mg = mg' cosθ ⇒ g' =

T = 2π

T = 2π

L cos θ g

(0.7 m )cos 22° 9.81 m/s 2

102 •• Picture the Problem The diagram shows the wire described in the problem statement with an object of moment of inertia I suspended from its end. We can apply Newton’s 2nd law to the suspended object to obtain its differential equation of motion. By comparing this equation to the equation of a simple harmonic oscillator, we can show that ω = κ I . Apply

∑τ = Iα to the object hung

from the wire to obtain:

g cos θ

− κθ = Iα = I

d 2θ dt 2

= 1.62 s

1120 Chapter 14 Divide both sides of this differential equation by I to obtain:

d 2θ κ + θ =0 dt 2 I

This equation can be written as:

d 2θ + ω 2θ = 0 where ω = 2 dt

κ I

103 •• Picture the Problem The diagram shows the torsion balance described in the problem statement. We can apply Newton’s 2nd law to the suspended object to obtain its differential equation of motion. By comparing this equation and its solution to that of a simple harmonic oscillator, we can obtain an equation that we can solve for the torsion constant κ. Apply

∑τ = Iα

to the torsion

pendulum:

− κθ = Iα = I

d 2θ dt 2

or

d 2θ κ + θ =0 dt 2 I The differential equation of simple harmonic motion is:

d 2x + ω2x = 0 2 dt where

x(t ) = x0 cos(ωt + δ ) and ω = The solution to equation (1) is:

θ (t ) = θ 0 cos(ωt + δ ) where

ω= Solve for κ to obtain:

(1)

κ I

κ = ω2I

2π T

Oscillations 1121 Express the moment of inertia of the torsion pendulum: Substitute to obtain:

Substitute numerical values and evaluate κ:

2

ml 2 ⎛l⎞ I = 2m⎜ ⎟ = 2 ⎝2⎠

κ=

ω 2 ml 2

κ=

2π 2 (0.050 kg )(0.05 m ) (80 s )2

2

4π 2 ml 2 2π 2 ml 2 = = 2T 2 T2 2

= 3.86 × 10 −7 N ⋅ m/rad *104 •• Picture the Problem Choose a coordinate system in which the direction the cube is initially displaced (downward) is the positive y direction. The figure shows the forces acting on the cube when it is in equilibrium floating in the water and when it has been pushed down a small distance y. We can find the period of its oscillatory motion from its angular frequency. By applying Newton’s 2nd law to the cube, we can obtain its equation of motion; from this equation we can determine the angular frequency of the cube’s small-amplitude oscillations.

Express the period of oscillation in terms of the angular frequency of the oscillations: Apply

∑F

y

= 0 to the cube when

T=

2π

ω

mg − FB = 0

it is floating in the water: Apply

∑T

y

= ma y to the cube

when it is pushed down a small distance y:

mg − FB' = ma y

(1)

1122 Chapter 14 Eliminate mg between these equations to obtain:

FB − FB' = ma y or

∆FB = FB − FB' = ma y For y << 1:

∆FB ≈ dFB = − ρVg = −a 2 ρgy = m

Rewrite the equation of motion as:

d2y m 2 = −a 2 ρgy dt

d2y dt 2

or

d2y a 2 ρg =− y = −ω 2 y 2 m dt a 2 ρg where ω 2 = m Solve for ω:

Substitute in equation (1) to obtain:

ω=a

ρg m 2π

T= a

ρg

=

2π a

m ρg

m

105 •• Picture the Problem Assume that the density of the earth ρ is constant and let m represent the mass of the clock. We can decide the question of where the clock is more accurate by applying the law of gravitation to the clock at a depth h below/above the surface of the earth and at the earth’s surface and expressing the ratios of the acceleration due to gravity below/above the surface of the earth to its value at the surface of the earth. Express the gravitational force acting on the clock when it is at a depth h in a mine:

Express the gravitational force acting on the clock at the surface of the earth:

mg' =

GM'm (RE − h )2

where M′ is the mass between the location of the clock and the center of the earth.

mg =

GM E m RE2

Oscillations 1123 Divide the first of these equations by the second to obtain:

GM' 2 g' (RE − h ) M' RE2 = = GM E g M E (RE − h )2 RE2

Express M ′:

M' = ρV' = 43 πρ (RE − h )

Express ME:

M E = ρV = 43 πρRE3

Substitute to obtain:

g' 43 πρ (RE − h ) RE2 = 3 4 g (RE − h )2 3 πρRE

Simplify and solve for g′:

⎛ R −h⎞ ⎟⎟ = g' = g ⎜⎜ E ⎝ RE ⎠

3

3

⎛ h ⎞ ⎟⎟ g ⎜⎜1 − ⎝ RE ⎠

or

⎛ h ⎞ ⎟⎟ g' = g ⎜⎜1 − ⎝ RE ⎠

(1)

Express the gravitational force acting on the clock when it is at an elevation h:

mg'' =

Express the gravitational force acting on the clock at the surface of the earth:

mg =

Divide the first of these equations by the second to obtain:

GM E 2 RE2 g'' (RE + h ) = = GM E g (RE − h )2 RE2 =

Solve for g′′:

GM E m (RE + h )2

GM E m RE2

1 ⎛ h ⎞ ⎜⎜1 + ⎟⎟ ⎝ RE ⎠

2

⎛ h ⎞ ⎟ g'' = g ⎜⎜1 + RE ⎟⎠ ⎝

−2

1124 Chapter 14 Comparing equations (1) and (2), we see that g' is closer to g than is g'' . Thus, the error is greater if the clock is elevated. 106 •• Picture the Problem The figure shows this system when it has an angular displacement θ. The period of the system is related to its angular frequency according to T = 2π/ω. We can find the equation of motion of the system by applying Newton’s 2nd law. By writing this equation in terms of θ and using a small-angle approximation, we’ll find an expression for ω that we can use to express T.

2π

(a) Express the period of the system in terms of its angular frequency:

T=

r r F ∑ = ma to the bob:

∑F

Apply

ω x

= − kx − T sin θ = Max

and

∑F

y

Eliminate T between the two equations to obtain: Noting that x = Lθ and

a x = Lα = L

d 2θ , dt 2

= T cosθ − Mg = 0

− kx − Mg tan θ = Max

ML

d 2θ = − kLθ − Mg tan θ dt 2

eliminate the variable x in favor of θ : For θ << 1, tanθ ≈ θ :

ML

d 2θ = − kLθ − Mgθ dt 2 = −(kL + Mg )θ

or

d 2θ ⎛ k g⎞ = −⎜ + ⎟θ = −ω 2θ 2 dt ⎝M L⎠ where

ω=

k g + M L

(1)

Oscillations 1125 Substitute in equation (1) to obtain:

2π k g + M L

T=

(b) When T = 2 s and M = 1 kg we have:

2=

When T = 1 s we have:

1=

2π g L 2π k+

Solve these equations simultaneously to obtain:

g L

k = 29.6 N/m

107 •• Picture the Problem Applying Newton’s 2nd law to the first object as it is about to slip will allow us to express µs in terms of the maximum acceleration of the system which, in turn, depends on the amplitude and angular frequency of the oscillatory motion. (a) Apply

∑F

x

= max to the

f s, max = m2 amax

second object as it is about to slip: Apply

∑F

y

= 0 to the second

Fn − m2 g = 0

object: Use f s, max = µs Fn to eliminate

f s, max and Fn between the two

µs m2 g = m2 amax and

amax g

equations:

µs =

Relate the maximum acceleration of the oscillator to its amplitude and angular frequency:

amax = Aω 2 = A

Substitute for amax to obtain:

µs =

k m1 + m2

Ak (m1 + m2 )g

1126 Chapter 14 (b)

A is unchanged. E is unchanged because E = 12 kA2 . ω is reduced by increasing the total mass of the system and T is increased.

108 •• Picture the Problem The diagram shows the box hanging from the stretched spring and the free-body diagram when the box is in equilibrium. We can apply Fy = 0 to

∑

the box to derive an expression for x. In (b) and (c), we can proceed similarly to obtain expressions for the effective spring constant, the new equilibrium position of the box, and frequency of oscillations when the box is released. (a) Apply

∑F

y

= 0 to the box to

k (x − x0 ) − mg = 0

obtain: Solve for x:

x=

Substitute numerical values and evaluate x:

x=

mg + x0 k

(100 kg )(9.81m/s2 ) + 0.5m 500 N/m

= 2.46 m (b) Draw the free-body diagram for the block with the two springs exerting equal upward forces on it:

Apply obtain:

∑F

y

= 0 to the box to

k (x − x0 ) + k (x − x0 ) − mg = 0 or

keff ( x − x0 ) − mg = 0

where

keff = 2k

(1)

Oscillations 1127 When the box is displaced from this equilibrium position and released, its motion is simple harmonic motion and its frequency is given by: Substitute numerical values and evaluate ω: (c) Solve equation (1) for x:

Substitute numerical values and evaluate x:

2k m

ω=

keff = m

ω=

2(500 N/m ) = 3.16 rad/s 100 kg

x=

mg + x0 2k

x=

(100 kg )(9.81m/s2 ) + 0.5 m 2(500 N/m )

= 1.48 m 109 •• Picture the Problem We’ll differentiate the expression for the period of simple

L with respect to g, separate the variables, and use a differential g ∆T 1 ∆g ≈− . approximation to establish that T 2 g pendulum T = 2π

(a) Express the period of a simple pendulum in terms of its length and the local value of the acceleration due to gravity:

T = 2π

L g

[

]

Differentiate this expression with respect to g to obtain:

dT d = 2π L g −1 2 = −π L g −3 2 dg dg T =− 2g

Separate the variables to obtain:

dT 1 dg =− T 2 g

Approximate dT and dg by ∆T and ∆g for ∆g << g:

∆T 1 ∆g ≈ − T 2 g

(b) Solve the result in part (a) for ∆g:

∆g = −2 g

∆T T

1128 Chapter 14 Express ∆T/T:

∆T s 1d 1h = −90 × × d 24 h 3600 s T = −1.04 × 10 −3

Substitute and evaluate ∆g:

(

)(

∆g = −2 9.81 m/s 2 − 1.04 × 10−3

)

= 0.0204 m/s 2 = 2.04 cm/s 2 110 •• Picture the Problem We can find the frequency of the vibrating system from its angular frequency; this depends on the spring constant and the total mass involved in the motion. The energy of the system can be found from the amplitude of its motion. (a) Relate the frequency of the vibrating system to its angular frequency: Substitute numerical values and evaluate f:

f =

ω 1 = 2π 2π

f =

1 2π

k 2m

240 N/m = 2.25 Hz 2(0.6 kg )

Express the total energy of the system:

E = 12 kA2

Substitute numerical values and evaluate E:

E=

1 2

Relate the frequency to the system’s new angular frequency:

f1 =

ω1 1 = 2π 2π

Substitute numerical values and evaluate f1:

f1 =

1 2π

240 N/m = 3.18 Hz 0.6 kg

A1 =

vmax

= vmax

(240 N/m )(0.6 m ) 2 =

43.2 J

(b) (1) The glue dissolves when the spring is at maximum compression:

Express the system’s new amplitude as a function of the oscillator’s maximum speed and its new angular frequency: Find the maximum speed of the oscillator:

ω1

k m

m k

(

)

vmax = Aω = 2πfA = 2π 2.25 s −1 (0.6 m ) = 8.48 m/s

Oscillations 1129 Substitute and evaluate A1:

A1 = (8.48 m/s )

0.6 kg 240 N/m

= 42.4 cm

(240 N/m )(0.424 m )2

Express and evaluate the energy of the system:

E1 = 12 kA12 =

(b) (2) The glue dissolves when the spring is at maximum extension and f2 is the same as f1:

f 2 = 3.18 Hz

Because the second object is at rest, the amplitude and energy of the system are unchanged:

A2 = A = 0.600 m

1 2

= 21.6 J

and

E2 = E = 43.2 J

111 •• Picture the Problem Choose a coordinate system in which the positive x direction is to the right and assume that the object is displaced to the right. In case (a), note that the two springs undergo the same displacement whereas in (b) they experience the same force. (a) Express the net force acting on the object:

(b) Express the force acting on each spring and solve for x2:

Fnet = − k1 x − k 2 x = −(k1 + k 2 )x = −keff x where keff = k1 + k 2

F = − k1 x1 = −k 2 x2 or

x2 =

k1 x1 k2

Express the total extension of the springs:

x1 + x2 = −

Solve for keff:

keff = − =

F keff

F − k1 x1 =− x1 + x2 x1 + x2

k1 x1 1 = k1 1 1 x1 + x1 + k2 k1 k 2

1130 Chapter 14 Take the reciprocal of both sides of the equation to obtain:

1 1 1 = + keff k1 k 2

*112 •• Picture the Problem If the displacement of the block is y = A sin ωt, its acceleration is a = −ω2Asinωt. (a) At maximum upward extension, the block is momentarily at rest. Its downward acceleration is g. The downward acceleration of the piston is ω 2A. Therefore, if ω 2A > g, the block will separate from the piston. (b) Express the acceleration of the small block:

a = − Aω 2 sin ωt

For ω 2 A = 3 g and A = 15 cm:

a = −3g sin ωt = − g

Solve for t:

Substitute numerical values and evaluate t:

t=

t=

A ⎛1⎞ ⎛1⎞ sin −1 ⎜ ⎟ = sin −1 ⎜ ⎟ ω 3g ⎝3⎠ ⎝3⎠ 1

0.15 m 1 sin −1 = 0.0243 s 2 3 9.81 m/s 3

(

)

113 •• Picture the Problem The plunger and ball are moving with their maximum speed as they pass through their equilibrium position (x = 0). Once it has passed its equilibrium position, the acceleration of the plunger becomes negative; therefore it begins to slow down and the ball, continuing with speed vs, separates from the plunger. We can find this separation speed by equating it to the maximum speed of the plunger. Application of conservation of energy to the motion of the plunger will allow us to express the distance at which the plunger comes momentarily to rest. (a) The ball will leave the plunger when the plunger is moving with its maximum speed; i.e., at its equilibrium position:

x= 0

(b) Express the speed of the ball upon separation in terms of the maximum speed of the plunger:

vs = vmax = Aω = x0ω

Oscillations 1131 The angular frequency is given by:

Substitute to obtain:

(c) Apply conservation of energy to the plunger:

ω=

k mb + mp

vs = x0

k mb + mp

K f − K i + U f,s − U i, s = 0 or, because Kf = Ui = 0,

− 12 mp vs2 + 12 kxf2 = 0 Solve for xf:

xf =

Substitute for vs and simplify to obtain:

mp k

xf = x0

vs

mp mb + mp

114 •• Picture the Problem Applying Newton’s 2nd law to the box as it is about to slip will allow us to express µs in terms of the maximum acceleration of the platform which, in turn, depends on the amplitude and angular frequency of the oscillatory motion. (a) Apply

∑F

x

= max to the box

f s,max = mamax

as it is about to slip: Apply

∑F

y

= 0 to the box:

Use f s, max = µs Fn to eliminate

f s, max and Fn between the two

Fn − mg = 0

µs mg = mamax and

amax g

equations:

µs =

Relate the maximum acceleration of the oscillator to its amplitude and angular frequency:

amax = Aω 2

Substitute for amax :

Aω 2 4π 2 A µs = = 2 g T g

1132 Chapter 14 4π 2 (0.4 m ) = 2.52 µs = (0.8 s )2 9.81m/s2

Substitute numerical values and evaluate µs:

(

(b) Solve the equation derived above for Amax:

Amax

Substitute numerical values and evaluate Amax:

Amax

)

µs g µs gT 2 = 2 = ω 4π 2 2 ( 0.4 ) (9.81 m/s 2 )(0.8 s ) =

4π 2

= 6.36 cm 115 ••• Picture the Problem In (b), we can use the condition Fnet = dU/dx = 0 for stable equilibrium to find the value of x = x0 at stable equilibrium. In (c) and (d), we can simply follow the outline provided in the problem statement. In (e), we can obtain the frequency from f =

1 2π

k using the value for k from the potential function. m

(a) A graph of U(x) follows: 10 9 8

U /U 0

7 6 5 4 3 2 1 0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

x /a

(b) Express the condition for equilibrium:

F=

Differentiate U with respect to x:

dU d ⎡ ⎛ x a ⎞⎤ U0⎜ + ⎟ = dx dx ⎢⎣ ⎝ a x ⎠⎥⎦

dU =0 dx

2 ⎡1 a ⎤ U ⎛ a ⎞ = U 0 ⎢ − 2 ⎥ = 0 ⎜⎜1 − 2 ⎟⎟ ⎣a x ⎦ a ⎝ x ⎠

Oscillations 1133 Set this derivative equal to zero and solve for x:

U0 ⎛ a2 ⎞ ⎜1 − 2 ⎟⎟ = 0 a ⎜⎝ x0 ⎠ and

x0 = a or α = 1 (c) Express U(x0 + ε):

⎡x +ε a ⎤ U ( x0 + ε ) = U 0 ⎢ 0 + ⎥ x0 + ε ⎦ ⎣ a ⎡ ⎤ ⎢ x0 ε 1 ⎥ = U0 ⎢ + + x0 ε ⎥ ⎢a a + ⎥ a a⎦ ⎣ or, because x0 = a,

⎤ ⎡ ⎢ ε 1 ⎥ U ( x0 + ε ) = U 0 ⎢1 + + ε⎥ ⎢ a 1+ ⎥ a⎦ ⎣

[

= U 0 1 + β + (1 + β ) where β = (d) Expand (1 + β ) to obtain: −1

]

ε a

(1 + β )−1 = 1 + (− 1)β + (− 1)(− 2) β 2 + ... 2 ×1

≈1− β + β Substitute in U(x0 + ε):

−1

2

[ = U [2 + β ]

U ( x0 + ε ) = U 0 1 + β + 1 − β + β 2 2

0

= 2U 0 + U 0

ε2 a2

= constant + U 0 (e) Express the potential energy of a simple harmonic oscillator:

U = constant + 12 kε 2

If the particle whose potential energy is given in part (d) is to undergo simple harmonic motion:

k=

2U 0 a2

ε2 a2

]

1134 Chapter 14 Express the frequency of the simple harmonic motion, substitute for k, and simplify to obtain:

f = =

1 2π

k 1 = m 2π

2U 0 a 2m

1 2U 0 2πa m

116 ••• Picture the Problem Let m represent the mass of the cylindrical drum, R its radius, and k the stiffness constant of the spring. We can find the angular frequency of the oscillations by equating the maximum kinetic energy of the drum and the maximum energy stored in the spring. We can then express the frequency of the system in terms of its angular frequency. The application of Newton’s 2nd law, under on-the-verge-of-sliding conditions, together with the introduction of the oscillator’s total energy, will lead us to an expression for the minimum value of the coefficient of static friction.

ω 2π

(a) Express the frequency of oscillation of the system for small displacements from equilibrium:

f =

Express the kinetic energy of the drum and simplify to obtain:

K = 12 Iω 2 + 12 mv 2

(1)

(

)

2

⎛v⎞ mR ⎜ ⎟ + 12 mv 2 = ⎝R⎠ = 34 mv 2 2

1 1 2 2

Apply conservation of energy to obtain:

2 K max = 34 mvmax = 12 kA2

Substitute Aω for vmax:

3 4

Solve for ω:

ω=

Substitute in equation (1) to obtain:

Substitute numerical values and evaluate f: (b) Apply

m( Aω ) = 12 kA2

∑F

x

= 0 to the drum to

2

2k 3m

f =

1 2π

2k 3m

f =

1 2π

2(4000 N/m ) = 3.36 Hz 3(6 kg )

kA − f s, max = 0 or

Oscillations 1135 establish the condition that governs slipping:

kA − µs Fn = 0

Using Fn = mg, solve for µs:

µs =

Express the oscillator’s total energy in terms of the amplitude of its motion:

E = 12 kA2 ⇒ kA = 2 Ek

Substitute in equation (2) to obtain:

Substitute numerical values and evaluate µs:

kA mg

(2)

µs =

2 Ek mg

µs =

2(5 J )(4000 N/m ) = 3.40 (6 kg ) 9.81m/s2

(

)

*117 ••• Picture the Problem The pictorial representation shows the two blocks connected by the spring and displaced from their equilibrium positions. We can apply Newton’s 2nd law to each of these coupled oscillators and solve the resulting equations simultaneously to obtain the differential equation of motion of the coupled oscillators. We can then compare this differential equation and its solution to the differential equation of motion of the simple harmonic oscillator and its solution to show that the oscillation frequency is ω = (k µ )1 2 where µ = m1m2/(m1 + m2) is the reduced mass of the system.

Apply

r

r

∑ F = ma

to the block

whose mass is m1 and solve for its acceleration:

k (x1 − x2 ) = m1a1 = m1 or

a1 = Apply

r

r

∑ F = ma

to the block

whose mass is m2 and solve for its

d 2 x1 dt 2

d 2 x1 k (x1 − x2 ) = 2 dt m1

− k (x1 − x2 ) = m2 a2 = m1

d 2 x2 dt 2

1136 Chapter 14 acceleration:

or

a2 = Subtract the first equation from the second to obtain:

d 2 x2 k (x2 − x1 ) = 2 dt m2

⎛ 1 d 2 ( x2 − x1 ) d 2 x 1 ⎞ ⎟⎟ x = 2 = −k ⎜⎜ + 2 dt dt ⎝ m1 m2 ⎠ where x = x2 − x1

The reduced mass of the system is:

1

µ

=

m1m2 1 1 or µ = + m1 + m2 m1 m2

Substitute to obtain:

d 2x k =− x 2 dt µ

Compare this differential equation with the differential equation of the simple harmonic oscillator:

d 2x k =− x 2 dt m

The solution to this equation is:

x = x0 cos(ωt + δ ) where ω =

Express the solution to equation (1):

(1)

k m

x = x0 cos(ωt + δ ) where ω =

k

µ

118 •• 12 Picture the Problem We can use ω = (k µ ) and µ = m1m2/(m1 + m2) from Problem 117 to find the spring constant for the HCl molecule. Use the result of Problem 118 to relate the oscillation frequency to the spring constant and reduced mass of the HCl molecule:

ω=

k

µ

Solve for k to obtain:

k = µω 2

Express the reduced mass of the HCl molecule:

µ=

Substitute to obtain:

k=

m1m2 m1 + m2 m1 m 2 ω 2 m1 + m 2

Oscillations 1137 Express the masses of the hydrogen and Cl atoms:

m1 = 1 amu = 1.67×10-27 kg and m2 = 35.45 amu = 5.92×10-26 kg

Substitute numerical values and evaluate k:

(1.67 ×10 k=

kg )(5.92 × 10 −26 kg )(8.969 × 1013 s -1 ) = 13.1 N/m 1.67 × 10 −27 kg + 5.92 × 10 −26 kg 2

−27

119 •• Picture the Problem In Problem 117, we derived an expression for the oscillation frequency of a spring-and-two-block system as a function of the stiffness constant of the spring and the reduced mass of the two blocks. We can solve this problem, assuming that the "spring constant" does not change, by using the result of Problem 117 and the reduced mass of a deuterium atom and a Cl atom in the equation for the oscillation frequency. Use the result of Problem 117 to relate the oscillation frequency to the spring constant and reduced mass of the HCl molecule: Express the reduced mass of the HCl molecule: Express the masses of the deuterium and Cl atoms: Evaluate the reduced mass of the molecule:

ω=

µ=

k

µ

m1m2 m1 + m2

m1 = 2 amu = 3.34×10-27 kg and m2 = 35.45 amu = 5.92×10-26 kg

µ=

(3.34 × 10

−27

)(

)

kg 5.92 × 10 −26 kg 3.34 × 10 −27 kg + 5.92 × 10 −26 kg

= 3.16 × 10 −27 kg Substitute numerical values and evaluate ω:

ω=

13.1 N/m 3.16 × 10 -27 kg

= 6.44 × 1013 rad/s 120 ••• Picture the Problem The pictorial representation shows the block moving from right to left with an instantaneous displacement x from its equilibrium position. The free-body diagram shows the forces acting on the block during the half-cycles that it moves from right to left. When the block is moving from left to right, the directions of the kinetic friction force and the restoring force exerted by the spring are reversed. We can apply Newton’s 2nd law to these motions to obtain the differential equations given in the problem statements and then use their solutions to plot the graph called for in (c).

1138 Chapter 14

(a) Apply

∑F

x

= max to the block

while it is moving to the left to obtain: Using fk = µkFn =µkmg, eliminate fk in the differential equation of motion:

d 2x f k − kx = m 2 dt

m

d 2x = −kx + µ k mg dt 2

or

d 2x µ mg ⎞ ⎛ m 2 = −k ⎜ x − k ⎟ dt k ⎠ ⎝ Let x0 =

µ k mg k

to obtain:

m

d 2x = − k ( x − x0 ) dt 2

or

d 2 x' k = − x' = −ω 2 x' 2 dt m provided x′ = x − x0 and

x0 = The solution to the differential equation is:

µ k mg k

=

µk g ω2

x' = x0' cos(ωt + δ ) and its derivative is

v' = −ωx0' sin (ωt + δ )

The initial conditions are:

x' (0) = x − x0 and v' (0 ) = 0

Apply these conditions to obtain:

x' (0) = x0' cos δ = x − x0 and

v' (0) = −ωx0' sin δ = 0

Solve these equations simultaneously to obtain:

δ = 0 and x0' = x − x0 and

x' = ( x − x0 ) cos ωt

Oscillations 1139 or

(b) Apply

r r F ∑ = ma to the block

while it is moving to the right to obtain: Using fk = µkFn =µkmg, eliminate fk in the differential equation of motion:

Let x0 =

µ k mg k

to obtain:

x = ( x − x0 ) cos ωt + x0

− f k − kx = m

m

(1)

d 2x dt 2

d 2x = − kx − µ k mg dt 2

or

m

d 2x µ mg ⎞ ⎛ = −k ⎜ x + k ⎟ 2 dt k ⎠ ⎝

m

d 2x = − k ( x + x0 ) dt 2

or

d 2 x" k = − x" = −ω 2 x" 2 dt m provided x″ = x + x0 and

x0 = The solution to the differential equation is:

µ k mg k

=

µk g . ω2

x" = x0" cos(ωt + δ ) and its derivative is

v" = −ωx0" sin (ωt + δ )

The initial conditions are:

x" (0) = x + x0 and v" (0) = 0

Apply these conditions to obtain:

x" (0) = x0" cos δ = x + x0 and

v" (0 ) = −ωx0" sin δ = 0

Solve these equations simultaneously to obtain:

δ = 0 and x0" = x + x0 and

x" = ( x + x0 ) cos ωt

or

x = (x + x0 )cos ω t − x0

(2)

(c) A spreadsheet program to calculate the position of the oscillator as a function of time (equations (1) and (2)) is shown below. The constants used in the position functions (x0 = 1 m and T = 2 s were used for simplicity) and the formulas used to calculate the positions are shown in the table. After each half-period, one must compute a new amplitude for the oscillation, using the final value of the position from the last half-period.

1140 Chapter 14 Cell B1 B2 C7 D7

Content/Formula 1 10 C6 + 0.1 ($B$2−$B$1)*COS(PI()*C7)+$B$1

( A − x0 ) cos πt + x0

D17

(ABS($D$6+$B$1))*COS(PI()*C17)−$B$1

x + x0 cos πt − x0

D27

(ABS($D$6−$B$1))*COS(PI()*C27)+$B$1

x − x0 cos πt + x0

D37 (ABS($D$36+$B$1))*COS(PI()*C37)−$B$1

x + x0 cos πt − x0

($D$46−$B$1)*COS(PI()*C47)+$B$1

D47

1 2 3 4 5 6 7 8 9 10

A B C x0= 1 m A= 10

Algebraic Form x0 A t + ∆t

(x − x0 ) cos πt + x0

D

t x (s) (m) 0.0 10.00 0.1 9.56 0.2 8.28 0.3 6.29 0.4 3.78

53 54 55 56

4.7 4.8 4.9 5.0

0.41 0.19 0.05 0.00

The graph shown below was plotted using the data from columns C (t) and

Note that the motion of the block ceases after five half - cycles.

10 8 6 4 2 x (m)

D (x).

0 -2 0

1

2

3

-4 -6 -8 -10 t (s)

4

5

Oscillations 1141 121 ••• Picture the Problem The diagram shows the half-cylinder displaced from its equilibrium position through an angle θ. The frequency of its motion will be found by expressing the mechanical energy E in terms of θ and dθ/dt. For small θ we will obtain an equation of 2

⎛ dθ ⎞ the form E = κθ + I ⎜ ⎟ . Differentiating both sides of this equation with respect ⎝ dt ⎠ ⎛ d 2θ ⎞ dθ to time will lead to 0 = ⎜⎜ κθ + I 2 ⎟⎟ , an equation that must be valid at all times. dt ⎠ dt ⎝ Because the situation of interest to us requires that dθ/dt is not always equal to zero, we d 2θ κ d 2θ have 0 = κθ + I 2 or + θ = 0 , the differential equation of simple harmonic dt 2 I dt motion with ω 2 = κ I . The distance from O to the center of mass D, where, from 1 2

2

1 2

Problem 8-39, D = (4/3π)R, and the distance from the contact point C to the center of mass is r. Finally, we’ll take the potential energy to be zero where θ is zero and assume that there is no slipping.

Apply conservation of energy to obtain:

E =U +K 1 ⎛ dθ ⎞ = Mg (h − D ) + I C ⎜ ⎟ 2 ⎝ dt ⎠

From Table 9-1, the moment of inertia of a solid cylinder about an axis perpendicular to its face and through its center is given by:

I 0, solid cylinder =

2

(1)

1 (2M )R 2 = MR 2 2

where M is the mass of the half-cylinder.

1142 Chapter 14 Express the moment of inertia of the half-cylinder about the same axis:

I 0, half cylinder = I 0 =

Use the parallel-axis theorem to relate Icm to I0:

I 0 = I cm + MD 2

Substitute for Icm and solve for Icm:

I cm = I 0 − D 2 M =

[

]

1 1 MR 2 = MR 2 2 2

1 MR 2 − D 2 M 2

1 MR 2 − D 2 M + Mr 2 2 ⎛1 ⎞ = M ⎜ R2 − D2 + r 2 ⎟ ⎝2 ⎠

Apply the parallel-axis theorem a second time to obtain an expression for IC:

IC =

Apply the law of cosines to obtain:

r 2 = R 2 + D 2 − 2 RD cosθ

Substitute for r2 in equation (2) to obtain:

D ⎛3 ⎞ ⎛1 ⎞ I C = M ⎜ R 2 − D 2 + R 2 + D 2 − 2 RD cosθ ⎟ = MR 2 ⎜ − 2 cos θ ⎟ R 2 2 ⎝ ⎠ ⎝ ⎠ Substitute for h and IC in equation (1):

1 D ⎞⎛ dθ ⎞ ⎛3 E = MgD(1 − cosθ ) + MR 2 ⎜ − 2 cos θ ⎟⎜ ⎟ 2 R ⎠⎝ dt ⎠ ⎝2

2

1 2

Use the small angle approximation cos θ ≈ 1 − θ 2 to obtain:

[

]

1 1 ⎛3 D ⎞⎛ dθ ⎞ E = MgDθ 2 + MR 2 ⎜ − 2 − θ 2 ⎟⎜ ⎟ 2 2 ⎝2 R ⎠⎝ dt ⎠

2

Because θ 2 << 2, we can neglect the θ 2 in the square brackets to obtain:

1 1 D ⎞⎛ dθ ⎞ ⎛3 E = MgDθ 2 + MR 2 ⎜ − 2 ⎟⎜ ⎟ 2 2 R ⎠⎝ dt ⎠ ⎝2

2

(2)

Oscillations 1143

Differentiating both sides with respect to time yields:

0 = MgDθ

dθ D ⎞⎛ dθ ⎞⎛ d 2θ ⎞ ⎛3 ⎟, + MR 2 ⎜ − 2 ⎟⎜ ⎟⎜ dt R ⎠⎝ dt ⎠⎜⎝ dt 2 ⎟⎠ ⎝2

D ⎞⎛ d 2θ ⎛3 R ⎜ − 2 ⎟⎜⎜ 2 R ⎠⎝ dt ⎝2 2

⎞ ⎟⎟ + gDθ = 0 , ⎠

and

d 2θ + dt 2

gD θ = 0, D⎞ 2⎛ 3 R ⎜ −2 ⎟ R⎠ ⎝2

the differential equation of simple harmonic motion with ω 2 =

Substitute for D to obtain:

ω2 =

Express the period of the motion in terms of ω and simplify to obtain:

T=

gD . D⎞ 2⎛ 3 R ⎜ −2 ⎟ R⎠ ⎝2

4 3π

g ⎛ 8 ⎞g =⎜ ⎟ ⎛ 3 8 ⎞ R ⎝ 9π − 16 ⎠ R ⎟ ⎜ − ⎝ 2 3π ⎠

2π

⎛ 9π − 16 ⎞ R = 2π ⎜ ⎟ ω ⎝ 8 ⎠g

= 7.78

R g

*122 ••• Picture the Problem The net force acting on the particle as it moves in the tunnel is the x-component of the gravitational force acting on it. We can find the period of the particle from the angular frequency of its motion. We can apply Newton’s 2nd law to the particle in order to express ω in terms of the radius of the earth and the acceleration due to gravity at the surface of the earth. (a) From the figure we see that:

Fx = Fr sin θ = − = −

GmM E x r RE3 r

GmM E x RE3

Because this force is a linear restoring force, the motion of the particle is simple

1144 Chapter 14 harmonic motion. (b) Express the period of the particle as a function of its angular frequency: Apply

∑F

x

= max to the particle:

Solve for a:

T=

−

2π

(1)

ω

GmM E x = ma RE3

a=−

GM E x = −ω 2 x RE3

where

Use GM E = gRE2 to simplify ω:

Substitute in equation (1) to obtain:

Substitute numerical values and evaluate T:

ω=

GM E RE2

ω=

gRE2 = RE3

T=

g RE

RE 2π = 2π g g RE

T = 2π

6.37 × 10 6 m = 5.06 × 103 s 2 9.81 m/s

= 84.4 min 123 ••• Picture the Problem The amplitude of a damped oscillator decays with time according 2

to A = A0e

− (b 2 m )t

⎛ b ⎞ ⎟⎟ and then substitute in . We can find b/2m from ω' = ω0 1 − ⎜⎜ 2 m ω 0 ⎠ ⎝

the amplitude equation to find the factor by which the amplitude is decreased during each oscillation. We’ll use our result from (a), together with the dependence of the energy of the oscillator on the square of its amplitude, to find the factor by which its energy is reduced during each oscillation. (a) Express the variation in amplitude with time:

A = A0e − (b 2 m )t

(1)

Oscillations 1145 Relate the damped and undamped frequencies of the oscillator:

⎛ b ⎞ ⎟⎟ ω' = ω0 1 − ⎜⎜ ⎝ 2mω0 ⎠

Solve for b 2m :

b ω '2 2 = ω0 1 − 2 = ω0 1 − (0.9) 2m ω0

2

(14-46)

= 0.436ω0 Find the period of the damped oscillations: Substitute in equation (1) with t = T to obtain:

T=

2π 2π = ω' 0.9ω0 ⎛ 2π ⎞

⎟⎟ −(0.436ω0 )⎜⎜ A ⎝ 0.9ω0 ⎠ =e = e −3.04 A0

= 0.0478 (b) Express the energy of the oscillator at time t = 0:

E0 = 12 kA02

Express the energy of the oscillator at time t = T:

E = 12 kA2

Divide the second of these equations by the first, simplify, and substitute to evaluate E/E0:

E A2 ⎛ A ⎞ 2 = 2 = ⎜⎜ ⎟⎟ = (0.0477 ) E0 A0 ⎝ A0 ⎠

2

= 0.00228

124 ••• Picture the Problem We can differentiate Equation 14-52 twice and substitute x and d2x/dt2 in Equation 14-51 to determine the condition that must be satisfied in order for Equation 14-52 to be a solution of Equation 14-51. The differential equation of motion is Equation 14-51: Its proposed solution is Equation 14-52: Obtain the first and second derivatives of x:

m

d 2x dx + b + mω02 x = F0 cos ωt 2 dt dt

x = A cos(ωt − δ )

dx = − Aω sin (ωt − δ ) dt and

1146 Chapter 14 d 2x = − Aω 2 cos(ωt − δ ) 2 dt Substitute in the differential equation to obtain:

− mAω 2 cos(ωt − δ ) − bAω sin (ωt − δ ) + mω02 A cos(ωt − δ ) = F0 cos ωt Using trigonometric identities, expand cos(ωt − δ ) and sin (ωt − δ ) to obtain:

− mAω 2 (cos ωt cos δ + sin ωt sin δ ) − bAω (sin ωt cos δ − cos ωt sin δ ) + mω02 A(cos ωt cos δ + sin ωt sin δ ) = F0 cos ωt Factor mA(cos ωt cos δ + sin ωt sin δ ) from the first and third terms to obtain:

(

)

mA ω02 − ω 2 (cos ωt cos δ + sin ωt sin δ ) − bAω (sin ωt cos δ − cos ωt sin δ ) = F0 cos ωt Factor cos ωt cos δ from the first term on the left-hand side of the equation and sin ωt cos δ from the 2nd term:

sin ωt sin δ ⎞ ⎛ ⎛ cos ωt sin δ ⎞ mA ω02 − ω 2 (cos ωt cos δ )⎜1 + ⎟ − bAω (sin ωt cos δ )⎜1 − ⎟ ⎝ cos ωt cos δ ⎠ ⎝ sin ωt cos δ ⎠ = F0 cos ωt

(

)

Simplify to obtain:

tan δ ⎞ ⎛ mA ω02 − ω 2 (cos ωt cos δ )(1 + tan ωt tan δ ) − bAω (sin ωt cos δ )⎜1 − ⎟ ⎝ tan ωt ⎠ = F0 cos ωt

(

)

(

)

Divide both sides of the equation by m ω 02 − ω 2 :

A(cos ωt cos δ )(1 + tan ωt tan δ ) − =

bAω (sin ωt cos δ )⎛⎜1 − tan δ ⎞⎟ 2 2 m ω0 − ω ⎝ tan ωt ⎠

(

)

F0 cos ωt m ω −ω2

(

2 0

)

The phase constant for a driven oscillator is given by Equation 14-54:

tan δ =

bω m ω02 − ω 2

(

)

Oscillations 1147 Substitute for tan δ :

⎛ ⎞ bω bAω ⎟− (sin ωt cos δ ) A(cos ωt cos δ )⎜⎜1 + tan ωt 2 2 ⎟ m ω0 − ω ⎠ m ω02 − ω 2 ⎝ bω ⎞ ⎛ ⎜ 2 2 ⎟ m ω0 − ω ⎟ F0 = cos ωt × ⎜1 − 2 ⎜ tan ωt ⎟ m ω0 − ω 2 ⎟ ⎜ ⎠ ⎝

(

(

)

)

(

(

)

)

Simplify to obtain:

(

)

A(cos ωt cos δ ) 1 + tan 2 δ =

Use the trigonometric identity 1 + tan 2 δ =

A(cos ωt cos δ )

Simplify to obtain:

Thus x = A cos(ωt − δ ) is a solution to Equation 14-51 provided:

F0 cos ωt m ω −ω2

(

)

2 0

1 : cos 2 δ

1 F0 = cos ωt 2 2 cos δ m ω0 − ω 2

(

)

A cos ωt =

A=

F0 cos δ cos ωt m ω02 − ω 2

(

F0 cos δ m ω02 − ω 2

(

)

)

*125 ••• Picture the Problem We can follow the step-by-step instructions provided in the problem statement to obtain the desired results. (a) Express the average power delivered by a driving force to a driven oscillator:

r r P = F ⋅ v = Fv cosθ or, because θ is 0°, P = Fv

Express F as a function of time:

F = F0 cos ωt

Express the position of the driven oscillator as a function of time:

x = A cos(ωt − δ )

1148 Chapter 14 Differentiate this expression with respect to time to express the velocity of the oscillator as a function of time: Substitute to express the average power delivered to the driven oscillator: (b) Expand sin (ωt − δ ) to obtain: Substitute in your result from (a) and simplify to obtain:

v = − Aω sin (ωt − δ )

P = (F0 cos ωt )[− Aω sin (ωt − δ )] = − AωF0 cos ωt sin (ωt − δ ) sin (ωt − δ ) = sin ωt cos δ − cos ωt sin δ P = − AωF0 cos ωt (sin ωt cos δ − cos ωt sin δ )

=

(c) Integrate sin θ cos θ over one period to determine sin θ cosθ :

AωF0 sin δ cos 2 ωt − AωF0 cos δ cos ωt sin ωt 2π ⎤ 1 ⎡ ⎢ ∫ sin θ cosθdθ ⎥ 2π ⎣ 0 ⎦ 2π ⎤ 1 ⎡1 2 = ⎢ sin θ ⎥ 2π ⎣⎢ 2 ⎥ 0 ⎦

sin θ cos θ =

=0 Integrate cos 2 θ over one period to determine cos 2 θ :

Substitute and simplify to express Pav:

1 2π

2π

=

1 2π

=

1 2π

=

1 2π

⎡ 1 2π ⎤ ⎢ ∫ (1 + cos 2θ )dθ ⎥ ⎣2 0 ⎦ 2π ⎡ 1 2π ⎤ 1 + d θ ⎢2 ∫ 2 ∫ cos 2θ dθ ⎥ 0 ⎣ 0 ⎦ (π + 0) = 1 2

cos 2 θ =

∫ cos θdθ 2

0

Pav = AωF0 sin δ cos 2 ωt − AωF0 cos δ cos ωt sin ωt

= 12 AωF0 sin δ − AωF0 cos δ (0 ) =

1 2

AωF0 sin δ

Oscillations 1149 (d) Construct a triangle that is consistent with

tan δ =

bω : m ω02 − ω 2

(

)

Using the triangle, express sinδ:

Using equation 14-53, reduce this expression to the simpler form:

(e) Solve sin δ =

bωA for ω: F0

Substitute in the expression for Pav to eliminate ω: Substitute for sin δ from (d) to obtain Equation 14-55:

sin δ =

sin δ =

ω=

bω

(

)

2

m 2 ω02 − ω 2 + b 2ω 2 bω A F0

F0 sin δ bA

Pav =

F02 sin 2 δ 2b

⎤ 1⎡ bω 2 F02 Pav = ⎥ ⎢ 2 2 2 ⎣⎢ m ω0 − ω 2 2 + b 2ω 2 ⎦⎥

(

)

126 ••• Picture the Problem We can follow the step-by-step instructions given in the problem statement to derive the given results. (a) Express the condition on the denominator of Equation 14-55 when the power input is half its maximum value: Factor the difference of two squares to obtain:

(

)

2

m 2 ω02 − ω 2 + b 2ω 2 = 2b 2ω02 and, for a sharp resonance,

(

m 2 ω02 − ω 2

)

2

≈ b 2ω02

m 2 [(ω0 − ω )(ω0 + ω )] ≈ b 2ω02 2

or

m 2 (ω0 − ω ) (ω0 + ω ) ≈ b 2ω02 2

(b) Use the approximation ω + ω0 ≈ 2ω0 to obtain:

2

m 2 (ω0 − ω ) (2ω0 ) ≈ b 2ω02 2

2

1150 Chapter 14 Solve for ω0 − ω:

(c) Using its definition, express Q:

Solve for b:

ω0 − ω = ±

Q=

ω0 m

b=

ω0 m

(1)

b

Q

(d) Substitute for b in equation (1) to obtain:

ω0 − ω = ±

Solve for ω:

ω = ω0 ±

Express the two values of ω:

b 2m

ω0 2Q

ω0 2Q

ω + = ω0 +

ω0 2Q

and

ω − = ω0 −

ω0 2Q

Remarks: Note that the width of the resonance at half-power is ∆ω = ω+ − ω− = ω0 Q , in agreement with Equation 14-49. 127 ••• Picture the Problem We can find the equilibrium separation for the Morse potential by setting dU/dr = 0 and solving for r. The second derivative of U will give the "spring constant" for small displacements from equilibrium. In (c), we can use ω = k µ , where k is our result from (b) and µ is the reduced mass of a homonuclear diatomic molecule, to find the oscillation frequency of the molecule. (a) A spreadsheet program to calculate the Morse potential as a function of r is shown below. The constants and cell formulas used to calculate the potential are shown in the table. Cell Content/Formula B1 5 B2 0.2 C9 C8 + 0.1 D8 $B$1*(1−EXP(−$B$2*(C8−$B$3)))^2

Algebraic Form D

β

[

r + ∆r

D 1 − e −β (r −r0 )

]

2

Oscillations 1151

1 2 3 4 5 6 7 8 9 10 11 12

A B C D= 5 eV Beta= 0.2 nm−1 r0= 0.75 nm

235 236 237 238 239

D

r (nm) 0.0 0.1 0.2 0.3 0.4

U(r) (eV) 0.13095 0.09637 0.06760 0.04434 0.02629

22.7 22.8 22.9 23.0 23.1

4.87676 4.87919 4.88156 4.88390 4.88618

The graph shown below was plotted using the data from columns C (r) and D (U(r)).

0.7 0.6

U (eV)

0.5 0.4 0.3 0.2 0.1 0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

r (nm)

{[

]} ]

(b) Differentiate the Morse potential with respect to r to obtain:

dU d = D 1 − e −β (r −r0 ) dr dr = −2 β D 1 − e −β ( r −r0 )

Set this derivative equal to zero for extrema:

− 2βD 1 − e − β ( r −r0 ) = 0

[

[

]

2

1152 Chapter 14 Solve for r to obtain:

r = r0

Evaluate the second derivative of U(r) to obtain:

d 2U d = − 2βD 1 − e −β ( r −r0 ) 2 dr dr = 2β 2 De −β ( r −r0 )

Evaluate this derivative at r = r0:

d 2U dr 2

{

= 2β 2 D

U = 12 kx 2

Differentiate this expression twice to obtain:

d 2U =k dx 2

By comparison with equation (1) we have:

k = 2β 2 D

ω=

]}

(1)

r =r0

Recall that the potential function for a simple harmonic oscillator is:

(c) Express the oscillation frequency of the diatomic molecule:

[

k

µ

where µ is the reduced mass of the molecule. Express the reduced mass of the homonuclear diatomic molecule:

µ=

Substitute and simplify to obtain:

ω=

m1m2 m2 m = = m1 + m2 2m 2 2β 2 D D = 2β m m 2

Remarks: An alternative approach in (b) is the expand the Morse potential in a Taylor series

U (r ) = U (r0 ) + (r − r0 )U' (r0 ) +

1 (r − r0 )2 U'' (r0 ) + higher order terms 2!

to obtain U(r) ≈ β 2 D (r − r0 ) . Comparing this expression to the energy of a spring2

and-mass oscillator we see that, as was obtained above, k = 2 β 2 D .

Chapter 15 Wave Motion Conceptual Problems *1 • Determine the Concept The speed of a transverse wave on a rope is given by v = F µ where F is the tension in the rope and µ is its linear density. The waves on the rope move faster as they move up because the tension increases due to the weight of the rope below. 2 • Determine the Concept The distance between successive crests is one wavelength and the time between successive crests is the period of the wave motion. Thus, T = 0.2 s and f = 1/T = 5 Hz. (b) is correct. 3 • Picture the Problem True. The energy per unit volume (the average energy density) is given by η av = 12 ρω 2 s02 where s0 is the displacement amplitude. 4 • Determine the Concept At every point along the rope the wavelength, speed, and frequency of the wave are related by λ = v f . The speed of the wave, in turn, is given by v =

F µ where F is the tension in the rope and µ is its linear density. Due to the

weight of the rope below, the tension is greater at the top and the speed of the wave is also greater at the top. Because λ ∝ v, the wavelength is greater at the top. *5 • Determine the Concept The speed of the wave v on the bullwhip varies with the tension F in the whip and its linear density µ according to v = F µ . As the whip tapers, the wave speed in the tapered end increases due to the decrease in the mass density, so the wave travels faster. 6 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10-12 W/m2 is defined to be the threshold of hearing. Express the intensity of the 60-dB sound:

60dB = (10 dB) log

1153

I 60 ⇒ I 60 = 106 I 0 I0

1154 Chapter 15 Express the intensity of the 30-dB sound: Because I60 = 103I30:

30 dB = (10 dB) log

I 30 ⇒ I 30 = 103 I 0 I0

The statement is false.

7 • Determine the Concept No. Because the source and receiver are at rest relative to each other, there is no relative motion of the source and receiver and there will be no Doppler shift in frequency. 8 • Determine the Concept Because there is no relative motion of the source and receiver, there will be no Doppler shift and the observer will hear sound of frequency f0.

(a ) is correct. *9 •• Determine the Concept The light from the companion star will be shifted about its mean frequency periodically due to the relative approach to and recession from the earth of the companion star as it revolves about the black hole. 10 • Determine the Concept In any medium, the wavelength, frequency, and speed of a wave are related through λ = v/f. Because the frequency of a wave is determined by its source and is independent of the nature of the medium, if v differs in the two media, the wavelengths will also differ. In this situation, the frequencies are the same but the speeds of propagation are different. 11 • (a) True. The particles that make up the string move at right angles to the direction the wave is propagating. (b) False. Sound waves in air are longitudinal waves of compression and rarefaction. (c) False. The speed of sound in air varies with the square root of the absolute temperature. The speed of sound at 20°C is 4% greater than that at 5°C. 12 • Determine the Concept In any medium, the wavelength, frequency, and speed of a sound wave are related through λ = v/f. Because the frequency of a wave is determined by its source and is independent of the nature of the medium, if v is greater in water than

Wave Motion 1155 in air, λ will be greater in water than in air. ( a ) is correct. *13 • Determine the Concept There was only one explosion. Sound travels faster in water than air. Abel heard the sound wave in the water first, then, surfacing, heard the sound wave traveling through the air, which took longer to reach him. 14 •• Determine the Concept The graph shown below shows the pulse at an earlier time (−∆t) and later time (∆t). One can see that at t = 0, the portion of the string between 1 cm and 2 cm is moving down, the portion between 2 cm and 3 cm is moving up, and the string at x = 2 cm is instantaneously at rest.

t=0 t>0

y

t<0

0

1

2

3

4

5

6

7

8

9

10

x , cm

vy

15 •• Determine the Concept The velocity of the string at t = 0 is shown. Note that the velocity is negative for 1 cm < x < 2 cm and is positive for 2 cm < x < 3 cm.

0

1

2

3

4

5

x , cm

6

7

8

9

10

1156 Chapter 15 16 •• Determine the Concept As the jar is evacuated, the speed of sound inside the jar decreases. Because of the mismatch between the speed of sound inside and outside of the jar, a larger fraction of the sound wave is reflected back into the jar, and a smaller fraction is transmitted through the glass of the bell jar. *17 •• Determine the Concept Path C. Because the wave speed is highest in the water, and more of path C is underwater than A or B, the sound wave will spend the least time on path C.

Estimation and Approximation 18 •• Picture the Problem The rate at which energy is delivered by sound waves is the product of its intensity and the area over which the energy is delivered. We can use the definition of the intensity level of the speech at 1 m to find the intensity of the sound and the formula for the area of a sphere to find the area over which the energy is distributed. Express the power of human speech as a function of its intensity:

P = IA

Express the area of a sphere of radius 1 m:

A = 4πr 2 = 4π (1 m ) = 4π m 2

Use β = (10 dB) log(I I 0 ) to solve for the intensity of the sound at the 65-dB level:

2

65 dB = (10 dB)log and

I I0

(

I = 106.5 I 0 = 106.5 10 −12 W/m 2

)

= 3.16 × 10 −6 W/m 2 Substitute and evaluate P:

(

)(

P = 3.16 × 10 −6 W/m 2 4π m 2

)

= 3.97 × 10 −5 W 19 •• Picture the Problem Let d represent the distance from the bridge to the water under the assumption that the time for the sound to reach the man is negligible; let t be the elapsed time between dropping the stone and hearing the splash. We’ll use a constantacceleration equation to find the distance to the water in all three parts of the problem, just improving our initial value with corrections taking into account the time required for the sound of the splash to reach the man on the bridge.

Wave Motion 1157 (a) Using a constant-acceleration equation, relate the distance the stone falls to its time-of-fall:

d = v0t + 12 gt 2 or, because v0 = 0,

d = 12 gt 2

(9.81m/s )(4 s)

Substitute numerical values and evaluate d:

d=

(b) Express the actual distance to the water d′ in terms of a time correction ∆t:

d' = 12 g (t − ∆t )

Express ∆t:

∆t =

Substitute to obtain:

⎛ d⎞ d ' = g ⎜⎜ t − ⎟⎟ ⎝ vs ⎠

Substitute numerical values and evaluate d′:

2

2

1 2

= 78.5 m

2

d vs 2

1 2

d' =

1 2

⎛ 78.5 m ⎞ ⎟ 9.81 m/s ⎜⎜ 4 s − 340 m/s ⎟⎠ ⎝

(

2

)

2

= 69.7 m (c) Express the total time for the rock to fall and the sound to return to the man:

∆t = ∆tfalling rock + ∆tsound =

2d d + g vs

Rewrite the equation in quadratic form:

⎛ 1 ∆t ⎞ 2 d 2 − 2vs2 ⎜⎜ + ⎟⎟d + vs2 (∆t ) = 0 ⎝ g vs ⎠

Substitute numerical values to obtain:

d 2 − 2.63 ×10 4 m d + 1.85 ×106 m 2 = 0

Solve for the positive value of d:

d = 70.5 m … about 1% larger than

(

)

our result in part (b) and 11% smaller than our first approximation in (a). *20 •• Picture the Problem You can use a protractor to measure the angle of the shock cone and then estimate the speed of the bullet using sin θ = v u . The speed of sound in helium at room temperature (293 K) is 977 m/s.

1158 Chapter 15 Relate the speed of the bullet u to the speed of sound v in helium and the angle of the shock cone θ : Solve for u:

sin θ =

u=

v u

v sin θ

Measure θ to obtain:

θ ≈ 70°

Substitute numerical values and evaluate u:

u=

977 m/s = 1.04 km/s sin 70°

21 •• Picture the Problem Let d be the distance to the townhouses. We can relate the speed of sound to the distance to the townhouses to the frequency of the clapping for which no echo is heard. Relate the speed of sound to the distance it travels to the townhouses and back to the elapsed time: Express d in terms of the number of strides and distance covered per stride:

v=

2d ∆t

d = (30 strides )(1.8 m/stride ) = 54 m

Relate the elapsed time ∆t to the frequency f of the clapping:

∆t =

Substitute numerical values and evaluate v:

v=

Express the percent difference between this result and 340 m/s:

340 m/s − 270 m/s = 20.6% 340 m/s

1 1 = = 0.4 s f 2.5 claps/s

2(54 m ) = 270 m/s 0.4 s

Speed of Waves 22 • Picture the Problem The speed of sound in a fluid is given by v =

B ρ where B is the

bulk modulus of the fluid and ρ is its density. (a) Express the speed of sound in water in terms of its bulk modulus:

v=

Substitute numerical values and evaluate v:

v=

B

ρ 2.0 × 10 9 N/m 2 = 1.41 km/s 10 3 kg/m 3

Wave Motion 1159 (b) Solve v =

B ρ for B:

Substitute numerical values and evaluate B:

B = ρv 2

(

)

B = 13.6 ×103 kg/m 3 (1410 m/s )

2

= 2.70 × 1010 N/m 2

*23 • Picture the Problem The speed of sound in a gas is given by v = γRT M where R is the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ is a constant that is characteristic of the particular molecular structure of the gas. Because hydrogen gas is diatomic, γ = 1.4. Express the dependence of the speed of sound in hydrogen gas on the absolute temperature: Substitute numerical values and evaluate v:

v=

v=

γRT M

1.4(8.314 J/mol ⋅ K )(300 K ) 2 × 10 −3 kg/mol

= 1.32 km/s 24 • Picture the Problem The speed of a transverse wave pulse on a wire is given by v = F µ where F is the tension in the wire, m is its mass, L is its length, and µ is its mass per unit length. Express the dependence of the speed of the pulse on the tension in the wire:

Substitute numerical values and evaluate v:

v=

F

µ

where µ is the mass per unit length of the wire.

v=

F = m L

900 N = 251 m/s 0.1 kg 7 m

25 • Picture the Problem The speed of transverse waves on a wire is given by v = F µ where F is the tension in the wire, m is its mass, L is its length, and µ is its mass per unit length.

1160 Chapter 15 Express the dependence of the speed of the pulse on the tension in the wire and the linear density of the wire:

v=

F m L

Solve for m:

m=

FL v2

Substitute numerical values and evaluate m:

m=

(550 N )(0.8 m ) = (150 m/s)2

19.6 g

*26 • Picture the Problem The speed of a wave pulse on a wire is given by v =

F µ where

F is the tension in the wire, m is its mass, L is its length, and µ is its mass per unit length. (a) Doubling the length while keeping the mass per unit length constant does not change the linear density: (b) Because v depends of

F,

v = 20 m/s

v = 2 (20 m/s ) = 28.3 m/s

doubling the tension increases v by a factor of 2 : (c) Because v depends on 1

µ,

doubling µ reduces v by a factor of 2:

v=

20 m/s 2

= 14.1 m/s

27 • Picture the Problem The speed of a transverse wave on the piano wire is given by v = F µ where F is the tension in the wire, m is its mass, L is its length, and µ is its mass per unit length. (a) The speed of transverse waves on the wire is given by:

v=

Substitute numerical values and evaluate v:

v=

F

µ

=

F m L

500 N = 265 m/s (0.005 kg ) (0.7 m )

Wave Motion 1161 (b) Letting m′ represent the mass of the wire when copper has been wrapped around the steel wire, express ∆m, the amount of copper wire required: Express the new wave speed v′:

∆m = m' − m

F m' L

v' =

Express the ratio of the speed of the waves in part (a) to the reduced wave speed:

v =2= v'

Solve for and evaluate m′:

m' = 4m

Substitute to obtain:

∆m = m' − m = 4m − m = 3(5 g )

F m L F m' L

=

m' m

= 15.0 g 28 •• Picture the Problem We can estimate the accuracy of this procedure by comparing the estimated distance to the actual distance. Whether a correction for the time it takes the light to reach you is important can be decided by comparing the times required for light and sound to travel a given distance.

m 1km × = 0.340 km/s s 1000 m

(a) Convert 340 m/s to km/s:

v = 340

(b) Express the fractional error in the procedure:

∆s s − s estimated 0.340t − 0.333t = = 0.340t s s 0.007 = = 2.06% 0.340

(c) Compare the time required for light to travel 1 km to the time required for sound to travel the same distance:

∆tlight ∆tsound

1 km v 340 m/s = c = = ≈ 10 −6 1 km c 3 × 108 m/s v

1162 Chapter 15 Because this fraction is so small, a correction for the time for light to reach you is not important. *29 •• Picture the Problem The speed of a transverse wave on a string is given by v = F µ where F is the tension in the wire and µ is its linear density. We can differentiate this expression with respect to F and then separate the variables to show that the differentials satisfy dv v = 12 dF F . We’ll approximate the differential quantities to determine by how much the tension must be changed to increase the speed of the wave to 312 m/s. (a) Evaluate dv/dF:

Separate the variables to obtain:

1 v dv d ⎡ F⎤ 1 1 = = ⋅ ⎢ ⎥= dF dF ⎣ µ ⎦ 2 Fµ 2 F dv 1 dF = v 2 F

(b) Solve for dF:

dF = 2 F

dv v

Approximate dF with ∆F and dv with ∆v to obtain:

∆F = 2 F

∆v v

Substitute numerical values and evaluate ∆F:

∆F = 2(500 N )

12 m/s = 40.0 N 300 m/s

30 •• Picture the Problem The speed of sound in a gas is given by v = γRT M where R is the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ is a constant that is characteristic of the particular molecular structure of the gas. We can differentiate this expression with respect to T and then separate the variables to show that the differentials satisfy dv v = 12 dT T . We’ll approximate the differential quantities to determine the percentage change in the velocity of sound when the temperature increases from 0 to 27°C. Lacking information regarding the nature of the gas, we can express the ratio of the speeds of sound at 300 K and 273 K to obtain an expression that involves just the temperatures.

Wave Motion 1163 (a) Evaluate dv/dT:

Separate the variables to obtain:

dv d ⎡ γRT = ⎢ dT dT ⎣ M 1v = 2T

⎤ 1 M ⎛ γR ⎞ ⎜ ⎟ ⎥= ⎦ 2 γRT ⎝ M ⎠

dv 1 dT = v 2 T

(b) Approximate dT with ∆T and dv with ∆v and substitute numerical values to obtain:

∆v 1 ∆T 1 ⎛ 27 K ⎞ = = ⎜ ⎟ = 4.95% v 2 T 2 ⎝ 273K ⎠

(c) Using a differential approximation, approximate the speed of sound at 300 K:

v300 K ≈ v273 K + v273 K

Substitute numerical values and evaluate v300 K:

v300 K = (331 m/s )(1 + 0.0495) = 347 m/s

Use v = γRT M to express the speed of sound at 300 K: Use v = γRT M to express the speed of sound at 273 K: Divide the first of these equations by the second and solve for and evaluate v300 K:

∆v v ⎛ ∆v ⎞ = v273 K ⎜1 + ⎟ v ⎠ ⎝

v300 K =

v273 K =

v300 K v273 K

γR(300 K ) M

γR(273 K ) M

γR(300 K ) =

300 M = 273 γR(273 K ) M

and

v300 K = (331 m/s )

300 = 347 m/s 273

Note that these two results agree to three significant figures. 31 ••• Picture the Problem The speed of sound in a gas is given by v = γRT M where R is the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ

1164 Chapter 15 is a constant that is characteristic of the particular molecular structure of the gas. Because T = t + 273 K, we can differentiate v with respect to t to show that dv dt = 12 (v T ). Evaluate dv/dt:

dv d ⎡ γR(t + 273 K ) ⎤ = ⎢ ⎥ dt dt ⎣ M ⎦ =

1 M ⎛ γR ⎞ ⎜ ⎟ 2 γR(t + 273 K ) ⎝ M ⎠

=

1v 2T

Substitute for t to obtain:

⎤ dv 1 ⎡ v = ⎢ dt 2 ⎣ t + 273 K ⎥⎦

Use the approximation

⎤ 1⎡ v v(t ) ≈ v(0°C ) + ⎢ t 2 ⎣ t + 273 K ⎥⎦ = v(0°C ) + ∆v

⎛ dv ⎞ v(T ) ≈ v(T0 ) + ⎜ ⎟ ∆T ⎝ dT ⎠T0 to write:

For t << 273 K:

Substitute in equation (1) to obtain:

(1)

where

∆v =

1 ⎡ 331 m/s ⎤ t 2 ⎢⎣ t + 273 K ⎥⎦

∆v ≈

1 ⎡ 331 m/s ⎤ t = (0.606 m/s ⋅ K )t 2 ⎢⎣ 273 K ⎥⎦

v(t ) = v(0°C ) + (0.606 m/s ⋅ K )t = 331 m/s + (0.606 m/s ⋅ K )t

32 •• Picture the Problem Let d be the distance to the munitions plant, v1 be the speed of sound in air, v2 be the speed of sound in rock, and ∆t be the difference in the arrival times of the sound at the man’s apartment. We can express ∆t in terms of t1 and t2 and then express these travel times in terms of the distance d and the speeds of the sound waves in air and in rock to obtain an equation we can solve for the distance from the man’s apartment to the munitions plant. Express the difference in travel times for the sound wave transmitted through air and the sound wave transmitted through the earth:

t1 − t 2 = ∆t

Wave Motion 1165 d d and t 2 = v1 v2

Express the transmission times in terms of the distance traveled and the speeds in the two media:

t1 =

Substitute to obtain:

d d − = ∆t v1 v2 or

⎛1 1⎞ ⎛v −v ⎞ d ⎜⎜ − ⎟⎟ = d ⎜⎜ 2 1 ⎟⎟ = ∆t ⎝ v1 v2 ⎠ ⎝ v1v2 ⎠ Solve for d:

d=

v1v2 ∆t v2 − v1

Substitute numerical values and evaluate d:

d=

(340 m/s)(3000 m/s) (3 s ) 3000 m/s − 340 m/s

= 1.15 km 33 ••• Picture the Problem The locations of the lightning strike (L), dorm room (R), and baseball park (P) are indicated on the diagram. We can neglect the time required for the electromagnetic pulse to reach the source of the radio transmission, which is the ballpark. The angle θ is related to the sides of the triangle through the law of cosines. We’re given the distance dPR and can find the other sides of the triangle using the speed of sound and the elapsed times. Using the law of cosines, relate the angle θ to the distances that make up the sides of the triangle: 2 2 2 2 2 d LR = d LP + d PR − 2d LP d PR cos(180° − θ ) = d LP + d PR + 2d LP d PR cos θ

Solve for θ:

2 2 2 ⎛ d LR − d LP − d PR 2d LP d PR ⎝

θ = cos −1 ⎜⎜

⎞ ⎟⎟ ⎠

1166 Chapter 15 Express the distance from the lightning strike to the ball park: Express the distance from the lightning strike to the dorm room:

d LP = vs ∆tLP = (340 m/s )(2 s ) = 680 m d LR = vs ∆tLR = (340 m/s )(6 s ) = 2040 m

Substitute and evaluate θ:

⎛ (2040 m )2 − (680 m )2 − (1600 m )2 ⎞ ⎟ = ±58.4° ⎟ ( )( ) 2 680 m 1600 m ⎝ ⎠

θ = cos −1 ⎜⎜

The lightning struck 680 m from the ballpark, 58.4° W (or E) of north. *34 ••• Picture the Problem Choose a coordinate system in which station Alpha is at the origin and the axes are oriented as shown in the pictorial representation. Because 0.75 mi = 1.21 km, Alpha’s coordinates are (0, 0), Beta’s are (1.21 km, 0), and those of the lightning strike are (x, y). We can relate the distances from the stations to the speed of sound in air and the times required to hear the thunder at the two stations. Relate the distance dα to the position coordinates of Alpha and the lightning strike:

x 2 + y 2 = dα2

Relate the distance dβ to the position coordinates of Beta and the lightning strike:

(x − 1.21 km )2 + y 2 = d β2

Relate the distance dα to the speed of sound in air v and the time that elapses between seeing the lightning at Alpha and hearing the thunder:

dα = v∆tα = (340 m/s )(3.4 s ) = 1156 m

Relate the distance dβ to the speed of sound in air v and the time that elapses between seeing the lightning at Beta and hearing the thunder:

d β = v∆t β = (340 m/s )(2.5 s ) = 850 m

Substitute in equations (1) and (2) to obtain:

x 2 + y 2 = (1156 m ) = 1.336 km 2

(1)

2

and

(2)

(3)

Wave Motion 1167

(x − 1.21 km )2 + y 2 = (850 m )2 = 0.7225 km 2 Subtract equation (4) from equation (3) to obtain:

(4)

x 2 − ( x − 1.21 km ) = 1.336 km 2 2

− 0.7225 km 2 or

(2.42 km )x − (1.21 km )2 = 0.6135 km 2

Solve for x to obtain:

x = 0.855 km

Substitute in equation (3) to obtain:

(0.855 km )2 + y 2 = 1.336 km 2

Solve for y, keeping the positive root because the lightning strike is to the north of the stations, to obtain:

y = 0.778 km

The coordinates of the lightning strike are:

(0.855 km, 0.778 km ) or

(0.531 mi, 0.484 mi)

35 ••• Picture the Problem The velocity of longitudinal waves on the Slinky is given by

v = B ρ where B is the bulk modulus of the material from which the slinky is constructed and ρ is its mass per unit volume. The velocity of transverse waves on the slinky is given by v = F µ where F is the tension in the slinky and µ is its mass per unit length. Substitution for B and ρ will lead to v = L k m in (a) and similar substitutions, together with the assumption that L0 << L will yield the same result for (b). (a) Express the velocity of longitudinal waves on the slinky:

v=

For the slinky:

ρ=

B

ρ

(1)

m V

and

B=−

Letting A be the cross-sectional area of the slinky:

ρ=

P ∆V V

∆L m and P = −k AL A

(2)

1168 Chapter 15 Substitute in equation (2) and simplify to obtain:

B=k

Substitute in equation (1):

v=

(b) Express the velocity of transverse waves on the slinky:

v=

For the slinky:

µ=

L A L A = L k m m AL

k

F

µ

(3)

m L

and

⎛ L ⎞ F = k∆L = k (L − L0 ) = kL⎜1 − 0 ⎟ L⎠ ⎝ ≈ kL if L0 << L Substitute in equation (3) to obtain:

v=

kL k = L m m L

The Wave Equation 36

•

Picture the Problem The general wave equation is

∂2 y 1 ∂2 y . To show that each of = ∂x 2 v 2 ∂t 2

the functions satisfies this equation, we’ll need to find their first and second derivatives with respect to x and t and then substitute these derivatives in the wave equation. (a) Find the first two spatial

derivatives of y ( x,t ) = k (x + vt ) : 3

∂y 2 = 3k (x + vt ) ∂x and

∂2 y = 6k (x + vt ) ∂x 2 Find the first two temporal

derivatives of y ( x,t ) = k (x + vt ) : 3

∂y 2 = 3kv(x + vt ) ∂t and

(1)

Wave Motion 1169

Express the ratio of equation (1) to equation (2):

∂2 y = 6kv 2 (x + vt ) (2) 2 ∂t ∂2 y ∂x 2 = 6k ( x + vt ) = 1 ∂ 2 y 6kv 2 ( x + vt ) v2 ∂t 2 3 confirming that y (x, t ) = k (x + vt ) satisfies the general wave equation.

(b) Find the first two spatial derivatives of y (x,t ) = Aeik ( x−vt ) :

∂y = ikAeik ( x−vt ) ∂x and

∂ 2 y 2 2 ik ( x−vt ) = i k Ae ∂x 2 or

∂2 y = −k 2 Aeik ( x−vt ) 2 ∂x Find the first two temporal derivatives of y (x,t ) = Aeik ( x−vt ) :

(3)

∂y = −ikvAeik ( x−vt ) ∂t and

∂ 2 y 2 2 2 ik ( x−vt ) = i k v Ae ∂t 2 or

∂2 y = −k 2v 2 Aeik ( x−vt ) ∂t 2 Express the ratio of equation (3) to equation (4):

(4)

∂2 y ik ( x − vt ) 2 1 ∂x 2 = − k Ae = 2 2 2 2 ik ( x − vt ) v ∂ y − k v Ae 2 ∂t confirming that y (x, t ) = Aeik ( x−vt ) satisfies the general wave equation.

(c) Find the first two spatial derivatives of y ( x, t ) = ln k ( x − vt ) :

k ∂y = ∂x x − vt and

∂2 y k2 =− ∂x 2 (x − vt )2

(5)

1170 Chapter 15 Find the first two temporal derivatives of y ( x, t ) = ln k ( x − vt ) :

vk ∂y =− ∂t x − vt and

∂2 y v2k 2 = − ∂t 2 (x − vt )2 Express the ratio of equation (5) to equation (6):

(6)

2 ∂2 y − k 2 ∂x 2 = (x − vt ) = 1 ∂2 y v2k 2 v2 − ∂t 2 (x − vt )2 confirming that y ( x, t ) = ln k ( x − vt )

satisfies the general wave equation. *37 • Picture the Problem The general wave equation is

∂2 y 1 ∂2 y = . To show that ∂x 2 v 2 ∂t 2

y = A sin kx cos ω t satisfies this equation, we’ll need to find the first and second

derivatives of y with respect to x and t and then substitute these derivatives in the wave equation. Find the first two spatial derivatives of y = A sin kx cos ω t :

∂y = Ak cos kx cos ωt ∂x and

∂2 y = − Ak 2 sin kx cos ωt 2 ∂x Find the first two temporal derivatives of y = A sin kx cos ω t :

(1)

∂y = −ωA sin kx sin ωt ∂t and

∂2 y = −ω 2 A sin kx cos ωt 2 ∂t Express the ratio of equation (1) to equation (2):

∂2 y 2 2 ∂x 2 = − Ak sin kx cos ωt = k ∂ 2 y − Aω 2 sin kx cos ωt ω 2 ∂t 2 1 = 2 v confirming that y = A sin kx cos ω t satisfies the general wave equation.

(2)

Wave Motion 1171

Harmonic Waves on a String 38 • Picture the Problem We can find the velocity of the waves from the definition of velocity and their wavelength from λ = v f . Express the wavelength of the waves:

λ=

v f

Using the definition of velocity, find the wave velocity:

v=

∆x 6 m = = 12 m/s ∆t 0.5 s

Substitute to obtain:

λ=

12 m/s = 20.0 cm 60 s −1

39 • Picture the Problem Equation 15-13, y ( x, t ) = A sin (kx − ωt ), describes a wave traveling in the positive x direction. For a wave traveling in the negative x direction we have y ( x, t ) = A sin (kx + ωt ). (a) Factor k from the argument of the sine function to obtain:

ω ⎞ ⎛ y ( x,t ) = A sin k ⎜ x − t ⎟ k ⎠ ⎝ = A sin k ( x − vt )

(b) Substitute k = 2π λ and ω = 2πf to obtain:

⎛ 2π ⎞ y (x,t ) = A sin ⎜ x − 2πft ⎟ ⎝ λ ⎠ ⎞ ⎛x = A sin 2π ⎜ − ft ⎟ ⎝λ ⎠

(c) Substitute k = 2π λ and

ω = 2π T to obtain:

⎛ 2π 2π ⎞ y (x,t ) = A sin ⎜ − t⎟ T ⎠ ⎝ λ ⎛x 1 ⎞ = A sin 2π ⎜ − t ⎟ ⎝λ T ⎠

1172 Chapter 15 (d) Substitute k = 2π λ to obtain:

⎛ 2π ⎞ y ( x,t ) = A sin ⎜ x − ωt ⎟ ⎝ λ ⎠ 2π ⎛ λω ⎞ = A sin t⎟ ⎜x− 2π ⎠ λ ⎝ 2π (x − vt ) = A sin

λ

(e) Substitute k = 2π λ and ω = 2πf to obtain:

y ( x,t ) = A sin (kx − 2πft ) ⎛ k ⎞ x − t ⎟⎟ = A sin 2πf ⎜⎜ ⎝ 2πf ⎠ ⎛x ⎞ = A sin 2πf ⎜ − t ⎟ ⎝v ⎠ For waves traveling in the negative x direction, we simply change the – signs to + signs.

*40 • Picture the Problem We can use f = c/λ to express the frequency of any periodic wave in terms of its wavelength and velocity. (a) Find the frequency of light of wavelength 4×10−7 m:

f =

Find the frequency of light of wavelength 7×10−7 m:

f =

Therefore the range of frequencies is: (b) Use the same relationship to calculate the frequency of these microwaves:

c

λ c

λ

=

2.998 × 108 m/s = 7.50 × 1014 Hz 4 × 10 −7 m

=

2.998 × 108 m/s = 4.28 × 1014 Hz 7 × 10 −7 m

4.28 × 1014 Hz ≤ f ≤ 7.50 × 1014 Hz

f =

c

λ

=

2.998 × 108 m/s 3 × 10 −2 m

= 1.00 × 1010 Hz

41 • Picture the Problem The average power propagated along the string by a harmonic wave is Pav = 12 µω 2 A2 v, where v is the speed of the wave, and µ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave,

Wave Motion 1173 respectively. Express and evaluate the power propagated along the string: The speed of the wave on the string is given by:

Pav = 12 µω 2 A2v

v=

Substitute for v to obtain:

F

µ F

Pav = 12 µω 2 A2

µ

Substitute numerical values and evaluate Pav:

Pav =

1 2

(4π )(0.05 kg/m ) (10 s ) (0.05 m ) 2

−1 2

2

80 N = 9.87 W 0.05 kg/m

42 • Picture the Problem The average power propagated along the rope by a harmonic wave is Pav = 12 µω 2 A2 v, where v is the velocity of the wave, and µ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. Rewrite the power equation in terms of the frequency of the wave: Solve for the frequency:

The wave velocity is given by:

Substitute for v and simplify to obtain:

Substitute numerical values and evaluate f:

Pav = 12 µω 2 A2 v = 2π 2 µf 2 A2 v

f =

v=

f =

Pav 2 Pav 1 = 2 2π µA v 2πA µv 2

F

µ 1 2πA

2 Pav 1 = 2πA F

µ

µ

2 Pav µF

1174 Chapter 15 f =

1 2π (0.01 m )

2(100 W ) ⎛ 0.1 kg ⎞ ⎜⎜ ⎟⎟(60 N ) ⎝ 2m ⎠

= 171 Hz

43 •• Picture the Problem Equation 15-13, y ( x, t ) = A sin (kx − ωt ), describes a wave traveling in the positive x direction. For a wave traveling in the negative x direction, we have y ( x, t ) = A sin (kx + ωt ) . We can determine A, k, and ω by examination of the wave function. The wavelength, frequency, and period of the wave can, in turn, be determined from k and ω.

(a)

Because the sign between the kx and ωt terms is positive, the wave is traveling in the negative x direction.

Find the speed of the wave:

(b) The coefficient of x is k and:

Solve for and evaluate λ:

The coefficient of t is ω and:

The period of the wave motion is the reciprocal of its frequency: (c) Express and evaluate the maximum speed of any string segment:

v=

k=

ω k

=

314 s −1 = 5.00 m/s 62.8 m −1

2π

λ

λ=

2π 2π = = 10.0 cm k 62.8 m −1

f =

ω 314 s −1 = = 50.0 Hz 2π 2π

T=

1 1 = = 0.0200 s f 50 s −1

vmax = Aω = (0.001m )(314 rad/s ) = 0.314 m/s

44 •• Picture the Problem Let the positive x direction be to the right. Then equation 15-13, y (x,t ) = A sin (kx − ωt ), describes a wave traveling in the positive x direction. We can find ω and k from the data included in the problem statement and substitute in the general equation. The maximum speed of a point on the string can be found from vmax = Aω and the maximum acceleration from amax = Aω 2 .

Wave Motion 1175 (a) Express the general form of the equation of a harmonic wave traveling to the right:

y ( x,t ) = A sin (kx − ω t )

ω = 2πf = 2π (80 s −1 ) = 503 s −1

Evaluate ω: Determine k:

k=

Substitute to obtain: y ( x,t ) =

ω v

=

503 s −1 = 41.9 m −1 12 m/s

(0.025 m )sin[(41.9 m −1 ) x − (503 s −1 )t ]

(b) Express and evaluate the maximum speed of a point on the string:

(

vmax = Aω = (0.025 m ) 503 s −1

)

= 12.6 m/s

(c) Express the maximum acceleration of a point on the string:

amax = Aω 2

Substitute numerical values and evaluate amax:

amax = (0.025 m ) 503 s −1

(

)

2

= 6.33 km/s 2

45 •• Picture the Problem The average total energy of waves on a string is given by ∆Eav = 12 µω 2 A2 ∆x, where µ is the linear density of the string, ω is its angular frequency, A the amplitude of the wave motion, and, in this problem, ∆x is the length of the string. The average power propagated along the string is Pav = 12 µω 2 A2 v. (a) ∆Eav is given by:

∆Eav = 12 µω 2 A2 ∆x = 2π 2 µf 2 A2 ∆x

Evaluate ∆Eav = 12 µω 2 A2 ∆x with ∆x = L = 20 m:

⎛ 0.06 kg ⎞ 2 ⎟⎟ (200 s −1 ) (0.012 m )2 (20 m ) = 6.82 J ∆Eav = 2π 2 ⎜⎜ ⎝ 20 m ⎠ (b) Express the power transmitted past a given point on the string:

Pav = 12 µω 2 A2v

1176 Chapter 15 The speed of the wave is given by:

v=

Substitute for v to obtain:

F

µ F

Pav = 12 µω 2 A2

µ

Substitute numerical values and evaluate Pav:

⎛ 0.06 kg ⎞ ⎟⎟ 200 s −1 Pav = 2π 2 ⎜⎜ 20 m ⎝ ⎠

(

) (0.012 m ) 2

50 N = 44.0 W 0.06 kg 20 m

2

*46 •• Picture the Problem The power propagated along the rope by a harmonic wave is P = 12 µω 2 A2v where v is the velocity of the wave, and µ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. We can use the wave function y = A0e − bx sin (kx − ωt ) to determine the

(

)

amplitude of the wave at x = 0 and at point x. (a) Express the power associated with the wave at the origin:

P = 12 µω 2 A2v

(

)

Evaluate the amplitude at x = 0:

A(0 ) = A0 e 0 = A0

Substitute to obtain:

P(0 ) =

(b) Express the amplitude of the wave at x: Substitute to obtain:

1 2

µω 2 A02v

(

A(x ) = A0e −bx

(

) )

P(x ) = 12 µω 2 A0e −bx v =

1 2

2

µω 2 A02ve −2bx

47 •• Picture the Problem The average power propagated along the rope by a harmonic wave is Pav = 12 µω 2 A2 v where v is the velocity of the wave, and µ, ω, and A are the linear density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively.

Wave Motion 1177 (a) Express the average power transmitted along the wire:

Pav = 12 µω 2 A2 v = 2π 2 µf 2 A2 v

Substitute numerical values and evaluate Pav:

Pav = 2π 2 (0.01 kg/m ) 400 s −1

(

× 0.5 × 10−3

( ) m ) (10 m/s ) 2

2

= 79.0 mW (b) Because Pav ∝ f 2 :

Increasing f by a factor of 10 would increase Pav by a factor of 100.

Because Pav ∝ A2 :

Increasing A by a factor of 10 would increase Pav by a factor of 100.

Because Pav ∝ v and v ∝

F:

Increasing F by a factor of 10 4 would increase v by a factor of 100 and Pav by a factor of 100.

(c)

Depending on the adjustability of the power source, increasing f or A would be the easiest.

*48 ••• Picture the Problem We can use the assumption that both the wave function and its first spatial derivative are continuous at x = 0 to establish equations relating A, B, C, k1, and k2. Then, we can solve these simultaneous equations to obtain expressions for B and C in terms of A, v1, and v2. (a) Let y1(x, t) represent the wave function in the region x < 0, and y2(x, t) represent the wave function in the region x > 0. Express the continuity of the two wave functions at x = 0:

Because the sine function is odd; i.e., sin (− θ ) = − sin θ :

y1 (0, t ) = y2 (0, t ) and

A sin [k1 (0) − ωt ] + B sin[k1 (0) + ωt ] = C sin[k 2 (0) − ωt ]

or

A sin (− ωt ) + B sin ωt = C sin (− ωt )

− A sin ωt + B sin ωt = −C sin ωt and

A− B =C

(1)

1178 Chapter 15 Differentiate the wave functions with respect to x to obtain:

∂y1 = Ak1 cos(k1 x − ωt ) ∂x + Bk1 cos(k1 x + ωt ) and

∂y2 = Ck 2 cos(k 2 x − ωt ) ∂x

Express the continuity of the slopes of the two wave functions at x = 0:

∂y 1 ∂y = 2 ∂x x = 0 ∂x x = 0 and

Ak1 cos[k1 (0 ) − ωt ] + Bk1 cos[k1 (0) + ωt ] = Ck 2 cos[k 2 (0) − ωt ]

or

Ak1 cos(− ωt ) + Bk1 cos ωt

= Ck 2 cos(− ωt )

Because the cosine function is even; i.e., cos(− θ ) = cos θ :

Ak1 cos ωt + Bk1 cos ωt = Ck 2 cos ωt and

k1 A + k1 B = k 2C Multiply equation (1) by k1 and add it to equation (2) to obtain: Solve for C:

2k1 A = (k1 + k 2 )C

C=

2k1 2 A= A k1 + k 2 1 + k 2 k1

Solve for C/A and substitute ω/v1 for k1 and ω/v2 for k2 to obtain:

C 2 2 = = A 1 + k 2 k1 1 + v1 v2

Substitute in equation (1) to obtain:

⎛ ⎞ 2 ⎟⎟ A A − B = ⎜⎜ + v v 1 1 2 ⎠ ⎝

Solve for B/A:

B 1 − v1 v2 = − A 1 + v1 v2

(2)

Wave Motion 1179 (b) We wish to show that

B2 +

B2 + (v1/v2)C2 = A2 Use the results of (a) to obtain the expressions B = −[(1 − α)/(1 + α)] A and C = 2A/(1 + α), where α = v1/v2. Substitute these expressions into

2

2

⎛1−α ⎞ 2 ⎛ 2 ⎞ 2 2 ⎟ A + α⎜ ⎜ ⎟ A =A + + α α 1 1 ⎠ ⎝ ⎝ ⎠ 2

2

⎛1−α ⎞ ⎛ 2 ⎞ ⎜ ⎟ + α⎜ ⎟ =1 ⎝1+ α ⎠ ⎝1+ α ⎠

(1 − α )2 + 4α (1 + α )2

B2 + (v1/v2)C2 = A2 and check to see if the resulting equation is an identity:

v1 2 C = A2 v2

=1

1 − 2α + α 2 + 4α =1 (1 + α )2 1 + 2α + α 2 =1 (1 + α )2

(1 + α )2 (1 + α )2

=1

1=1 The equation is an identity:

Therefore, B 2 +

v1 2 C = A2 v2

Remarks: Our result in (a) can be checked by considering the limit of B/A as v2/v1 → 0. This limit gives B/A = +1, telling us that the transmitted wave has zero amplitude and the incident and reflected waves superpose to give a standing wave with a node at x = 0.

Harmonic Sound Waves *49 • Picture the Problem The pressure variation is of the form p( x,t ) = p0 cos k ( x − vt ) where k =

π 2

and v = 340 m/s . We can find λ from k and f from ω and k.

(a) By inspection of the equation:

(b) Because k =

(c) Solve v =

ω k

2π

λ =

=

π 2

:

2πf for f to obtain: k

p0 = 0.750 Pa

λ = 4.00 m

π kv 2 = f = 2π

(340 m/s) 2π

= 85.0 Hz

1180 Chapter 15 v = 340 m/s

(d) By inspection of the equation:

50 • Picture the Problem The frequency, wavelength, and speed of the sound waves are related by v = fλ. (a) Express and evaluate the wavelength of middle C:

λ=

v 340 m/s = = 1.30 m f 262 s -1

(b) Double the frequency corresponding to middle C; solve for and evaluate λ:

λ=

v 340 m/s = = 0.649 m f 2 262 s −1

(

)

51 • Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave µ, the speed of the wave v, and the displacement amplitude s0 according to p0 = ρωvs0 . (a) Solve p0 = ρωvs0 for s0:

Substitute numerical values and evaluate s0:

s0 =

p0 ρωv

s0 =

(10 atm)(1.01325 × 10 Pa/atm) 2π (1.29 kg/m )(100 s )(340 m/s ) −4

5

3

−1

= 3.68 × 10 −5 m (b) Use p0 = ρωvs0 to find p0:

(

)(

)

(

)

p0 = 2π 1.29 kg/m 3 300 s −1 (340 m/s ) 10 −7 m = 8.27 ×10 −2 Pa 52 • Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave µ, the speed of the wave v, and the displacement amplitude s0 according to p0 = ρωvs0 . (a) Solve p0 = ρωvs0 for s0:

s0 =

p0 ρωv

Wave Motion 1181 Substitute numerical values and evaluate s0:

s0 =

29 Pa 2π 1.29 kg/m 1000 s −1 (340 m/s )

(

3

)(

)

= 2.10 × 10 −5 m (b) Proceed as in (a) with f = 1 kHz:

s0 =

29 Pa 2π 1.29 kg/m 1000 s −1 (340 m/s )

(

3

)(

)

= 1.05 × 10 −5 m 53 • Picture the Problem The pressure or density wave is 90° out of phase with the displacement wave. When the displacement is zero, the pressure and density changes are either a maximum or a minimum. When the displacement is a maximum or minimum, the pressure and density changes are zero. We can use p0 = ρωvs0 to find the maximum value of the displacement at any time and place. (a) If the pressure is a maximum at x1 when t = 0: (b) Solve p0 = ρωvs0 for s0:

Substitute numerical values and evaluate s0:

the displacement s is zero.

s0 =

p0 ρωv

s0 =

(10 atm)(1.01325 ×10 Pa/atm) 2π (1.29 kg/m )(1000 s )(340 m/s ) −4

5

3

−1

= 3.68 µm *54 • Picture the Problem A human can hear sounds between roughly 20 Hz and 20 kHz; a factor of 1000. An octave represents a change in frequency by a factor of 2. We can evaluate 2N = 1000 to find the number of octaves heard by a person who can hear this range of frequencies. Relate the number of octaves to the difference between 20 kHz and 20 Hz:

2 N = 1000

Take the logarithm of both sides of the equation to obtain:

log 2 N = log103 or

N log 2 = 3

Solve for and evaluate N:

N=

3 = 9.97 ≈ 10 log 2

1182 Chapter 15

Waves in Three Dimensions: Intensity 55 • Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave µ, the speed of the wave v, and the displacement amplitude s0 according to p0 = ρωvs0 . The intensity of the waves is given by

I = 12 ρω 2 s02 v =

1 2

p02 and the rate at which energy is delivered (power) is the product of ρv

the intensity and the surface area of the piston. (a) Using p0 = ρωvs0 , evaluate p0:

(

)(

p0 = 2π 1.29 kg/m 3 500 s −1

(

)

× (340 m/s ) 0.1× 10 m −3

)

= 138 Pa (b) Use I =

1 2

p02 to find the ρv

intensity of the waves:

(c) Using Pav = IA to find the power required to keep the piston oscillating:

(138 Pa ) 1 2 1.29 kg/m 3 (340 m/s ) 2

I=

(

)

= 21.7 W/m 2

(

)(

Pav = 21.6 W/m 2 10 −2 m 2

)

= 0.217 W

56 • Picture the Problem The intensity of the sound from the spherical source varies inversely with the square of the distance from the source. The power radiated by the source is the product of the intensity of the radiation and the surface area over which it is distributed. (a) Relate the intensity at 10 m to the distance from the source:

I=

Pav 4πr 2

or

Letting r′ represent the distance at which the intensity is 10−6 W/m2, express the intensity as in part (a):

10 − 4 W/m 2 =

Pav 2 4π (10 m )

10 −6 W/m 2 =

Pav 4πr' 2

Wave Motion 1183 Divide the first of these equations by the second to obtain:

Pav 2 10 W/m 4π (10 m ) = Pav 10 −6 W/m 2 4πr' 2

Solve for and evaluate r′:

r' =

(b) Solve I =

Pav for Pav: 4π r 2

Substitute numerical values and evaluate Pav:

−4

2

(10 )(10 m )

2

2

= 100 m

Pav = 4π r 2 I

(

)

Pav = 4π (10 m ) 10 −4 W/m 2 = 0.126 W 2

*57 • Picture the Problem Because the power radiated by the loudspeaker is the product of the intensity of the sound and the surface area over which it is distributed, we can use this relationship to find the average power, the intensity of the radiation, or the distance to the speaker for a given intensity or average power. (a) Use Pav = 4πr 2 I to find the total acoustic power output of the speaker:

(

Pav = 4π (20 m ) 10 − 2 W/m 2 2

)

= 50.3 W Pav 2 4π (20 m )

(b) Relate the intensity of the sound at 20 m to the distance from the speaker:

10 − 2 W/m 2 =

Relate the threshold-of-pain intensity to the distance from the speaker:

1 W/m 2 =

Divide the first of these equations by the second; solve for and evaluate r:

r = 10 −2 (20 m ) = 2.00 m

(c) Use I =

Pav to find the 4πr 2

intensity at 30 m:

Pav 4πr 2

2

I (30 m ) =

50.3 W 2 4π (30 m )

= 4.45 × 10 −3 W/m 2

1184 Chapter 15 58 •• Picture the Problem We can use conservation of energy to find the acoustical energy resulting from the dropping of the pin. The power developed can then be found from the given time during which the energy was transformed from mechanical to acoustical form. We can find the range at which the dropped pin can be heard from I = P/4π r2. (a) Assuming that I = P/4π r2, express the distance at which one can hear the dropped pin: Use conservation of energy to determine the sound energy generated when the pin falls:

r=

P 4π I

E = (0.0005)(mgh )

(

)(

= (0.0005) 0.1× 10 −3 kg 9.81 m/s 2 × (1 m )

)

= 4.91× 10 −7 J Express the power of the sound pulse:

P=

E 4.91 × 10 −7 J = ∆t 0.1s

= 4.91 × 10 −6 W Substitute numerical values and evaluate r: (b) Repeat the last step in (a) with I = 10-8 W/m2:

r=

4.91 × 10 −6 W = 198 m 4π 10 −11 W/m 2

r=

4.91 × 10 −6 W = 6.25 m 4π 10 −8 W/m 2

(

)

(

)

Intensity Level 59 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be the threshold of hearing. (a) Using its definition, calculate the intensity level of a sound wave whose intensity is 10-10 W/m2:

⎛ 10 −10 W/m 2 ⎞ ⎟ 2 ⎟ −12 ⎝ 10 W/m ⎠

β = (10 dB) log⎜⎜

= 10 log 10 2 = 20.0 dB

Wave Motion 1185 (b) Proceed as in (a) with I = 10−2 W/m2:

⎛ 10 −2 W/m 2 ⎞ ⎟ 2 ⎟ −12 ⎝ 10 W/m ⎠

β = (10 dB) log⎜⎜

= 10 log 1010 = 100 dB 60 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be the threshold of hearing. (a) Solve β = (10 dB) log(I I 0 ) for I

I = 10 β

(10 dB )

I0

to obtain: Evaluate I for β = 10 dB:

I = 10 (10 dB ) (10 dB ) I 0 = 10 I 0

(

)

= 10 10 −12 W/m 2 = 10 −11 W/m 2 (b) Proceed as in (a) with β = 3 dB:

I = 10 (3 dB ) (10 dB ) I 0 = 2 I 0

(

)

= 2 10 −12 W/m 2 = 2 × 10 −12 W/m 2 *61 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be the threshold of hearing.

⎛ I concert ⎞ ⎟⎟ I ⎝ 0 ⎠

Express the sound level of the rock concert:

β concert = (10 dB)log⎜⎜

Express the sound level of the dog’s bark:

⎛ I dog ⎞ ⎟⎟ 50 dB = (10 dB) log⎜⎜ ⎝ I0 ⎠

Solve for the intensity of the dog’s bark:

I dog = 105 I 0 = 105 10 −12 W/m 2

Express the intensity of the rock concert in terms of the intensity of the dog’s bark:

I concert = 10 4 I dog = 10 4 10 −7 W/m 2

(

(1)

)

= 10 −7 W/m 2

(

= 10 −3 W/m 2

)

1186 Chapter 15 Substitute in equation (1) and evaluate βconcert:

⎛ 10 −3 W/m 2 ⎞ ⎟ −12 2 ⎟ ⎝ 10 W/m ⎠

β concert = (10 dB)log⎜⎜

= (10 dB)log109 = 90.0 dB

62 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be the threshold of hearing.

⎛ IL ⎞ ⎟⎟ ⎝ I0 ⎠

Express the intensity level of the louder sound:

β L = (10 dB) log⎜⎜

Express the intensity level of the softer sound:

β S = (10 dB) log⎜⎜

Express the difference between the intensity levels of the two sounds:

β L − β S = 30 dB

⎛ IS ⎞ ⎟⎟ I ⎝ 0⎠

⎡ ⎛I ⎞ ⎛ I ⎞⎤ = (10 dB)⎢log⎜⎜ L ⎟⎟ − log⎜⎜ S ⎟⎟⎥ ⎝ I 0 ⎠⎦⎥ ⎣⎢ ⎝ I 0 ⎠ ⎛I I ⎞ = (10 dB)log⎜⎜ L 0 ⎟⎟ ⎝ IS I0 ⎠ ⎛I ⎞ = (10 dB)log⎜⎜ L ⎟⎟ ⎝ IS ⎠

Solve for and evaluate the ratio IL/IS:

IL = 103 and (a ) is correct. IS

63 • Picture the Problem We can use the definition of the intensity level to express the difference in the intensity levels of two sounds whose intensities differ by a factor of 2.

⎛ I ⎞ ⎟⎟ I ⎝ 0⎠

Express the intensity level before the intensity is doubled:

β1 = (10 dB) log⎜⎜

Express the intensity level with the intensity doubled:

β 2 = (10 dB) log⎜⎜

⎛ 2I ⎞ ⎟⎟ ⎝ I0 ⎠

Wave Motion 1187 Express and evaluate ∆β = β2 − β1:

∆β = β 2 − β1 ⎛ 2I ⎞ ⎛ I ⎞ = (10 dB) log⎜⎜ ⎟⎟ − (10 dB) log⎜⎜ ⎟⎟ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎛ 2I ⎞ = (10 dB) log⎜ ⎟ = (10 dB) log 2 ⎝ I ⎠ = 3.01 dB ≈ 3.0 dB

*64 • Picture the Problem We can express the intensity levels at both 90 dB and 70 dB in terms of the intensities of the sound at those levels. By subtracting the two expressions, we can solve for the ratio of the intensities at the two levels and then find the fractional change in the intensity that corresponds to a decrease in intensity level from 90 dB to 70 dB. Express the intensity level at 90 dB:

⎛I ⎞ 90 dB = (10 dB) log⎜⎜ 90 ⎟⎟ ⎝ I0 ⎠

Express the intensity level at 70 dB:

⎛I ⎞ 70 dB = (10 dB)log⎜⎜ 70 ⎟⎟ ⎝ I0 ⎠

Express ∆β = β90 − β70:

∆β = 20 dB ⎛I ⎞ ⎛I ⎞ = (10 dB) log⎜⎜ 90 ⎟⎟ − (10 dB) log⎜⎜ 70 ⎟⎟ ⎝ I0 ⎠ ⎝ I0 ⎠ ⎛I ⎞ = (10 dB) log⎜⎜ 90 ⎟⎟ ⎝ I 70 ⎠

Solve for I90:

I 90 = 100I 70

Express the fractional change in the intensity from 90 dB to 70 dB:

I 90 − I 70 100 I 70 − I 70 = = 99% 100 I 70 I 90

65 •• Picture the Problem The intensity at a distance r from a spherical source varies with distance from the source according to I = Pav 4πr 2 . We can use this relationship to relate the intensities corresponding to an 80-dB intensity level (I80) and the intensity corresponding to a 60-dB intensity level (I60) to their distances from the source. We can relate the intensities to the intensity levels through β = (10 dB) log(I I 0 ) .

1188 Chapter 15 (a) Express the intensity of the sound where the intensity level is 80 dB:

I10 =

Pav 4π r102

Express the intensity of the sound where the intensity level is 60 dB:

I 60 =

Pav 4π r 2

Divide the first of these equations by the second to obtain:

I 80 I 60

Pav 2 r2 4π (10 m ) = = Pav 100 m 2 2 4π r

Solve for r:

r = (10 m )

Find the intensity of the 80-dB sound level radiation:

⎛I ⎞ 80 dB = (10 dB) log⎜⎜ 80 ⎟⎟ ⎝ I0 ⎠

I 80 I 60

and

I 80 = 108 I 0 = 10 −4 W/m 2 Find the intensity of the 60-dB sound level radiation:

⎛I ⎞ 60 dB = (10 dB) log⎜⎜ 60 ⎟⎟ ⎝ I0 ⎠ and

I 60 = 10 6 I 0 = 10 −6 W/m 2 Substitute and evaluate r:

r = (10 m )

(b) Using the intensity corresponding to an intensity level of 80 dB, express and evaluate the power radiated by this source:

P = I 80 A

10 −4 W/m 2 = 100 m 10 −6 W/m 2

(

)[

= 10 − 4 W/m 2 4π(10 m )

2

]

= 0.126 W

66 •• Picture the Problem Let I1 and I2 be the intensities of the sound at distances r1 and r2. We can relate these intensities to the intensity levels through β = (10 dB) log(I I 0 ) and to the distances through I = Pav 4π r 2 .

Wave Motion 1189 Using β = (10 dB) log(I I 0 ) , express the ratio β1/β2:

β 2 (10 dB) log(I 2 I 0 ) = β1 (10 dB) log(I1 I 0 ) =

Express the ratio of the intensities at distances r1 and r2 from the source and solve for I2:

log I 2 − log I 0 log I1 − log I 0

I 2 r12 r2 = 2 and I 2 = 12 I1 I1 r2 r2

Substitute and simplify to obtain:

β2 = β1 =

r12 I1 − 10 log I 0 10 log I1 + 20 log(r1 r2 ) − 10 log I 0 r22 = log I1 − log I 0 10 log I1 − 10 log I 0

10 log

10 log(I1 I 0 ) + 20 log(r1 r2 ) β + 20 log(r1 r2 ) = 1 10 log(I1 I 0 ) β1

67 •• Picture the Problem We can use β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be the threshold of hearing, to find the intensity level at 20 m. Because the power radiated by the loudspeaker is the product of the intensity of the sound and the surface area over which it is distributed, we can use this relationship to find either the average power, the intensity of the radiation, or the distance to the speaker for a given intensity or average power. (a) Relate the intensity level to the intensity at 20 m:

⎛ I ⎞ ⎟⎟ ⎝ I0 ⎠

β = (10 dB) log⎜⎜

⎛ 10 −2 W/m 2 ⎞ ⎟ = (10 dB) log⎜⎜ −12 2 ⎟ ⎝ 10 W/m ⎠

( )

= (10 dB) log 1010 = 100 dB (b) Use Pav = 4πr 2 I to find the total acoustic power output of the speaker: (c) Relate the intensity of the sound at 20 m to the distance from the speaker:

Pav = 4π (20 m ) (10 − 2 W/m 2 ) 2

= 50.3 W 10 − 2 W/m 2 =

Pav 2 4π (20 m )

1190 Chapter 15 Pav 4π r 2

Relate the threshold-of-pain intensity to the distance from the speaker:

1 W/m 2 =

Divide the first of these equations by the second; solve for and evaluate r:

r = 10 −2 (20 m ) = 2.00 m

(d) Use I =

Pav to find the 4πr 2

2

I (30 m ) =

intensity at 30 m: Find the intensity level at 30 m:

50.3 W 2 4π (30 m )

= 4.45 × 10 −3 W/m 2

β (30 m ) = (10 dB) log

4.45 × 10 −3 W/m 2 10 −12 W/m 2

= (10 dB) log 4.45 × 109 = 96.5 dB 68 •• Picture the Problem Let I′ and I represent the sound in consecutive years. Then, we can use β = (10 dB) log(I I 0 ) to express the annual increase in intensity levels. (a) Express the annual change in intensity level:

Solve for I′/I:

∆β = β' − β = 1dB = (10 dB) log

I' I − (10 dB) log I0 I0

= (10 dB) log

I' I

I' = 10 0.1 = 1.26 I and the annual increase in intensity is 26%.

This is not a plausible annual

increase because, if it were true, the intensity level would increase by a factor of 10 in ten years. (b) Because doubling of the intensity corresponds to ∆β = 3 dB and the intensity is increasing 1 dB annually:

The intensity level will double in 3 years.

Wave Motion 1191 69 •• Picture the Problem We can find the intensities of the three sources from their intensity levels and, because their intensities are additive, find the intensity level when all three sources are acting. (a) Express the sound intensity level when the three sources act at the same time:

β3 sources = (10 dB) log

Find the intensities of each of the three sources:

70 dB = (10 dB) log

= (10 dB) log

I 3 sources I0 I 70 + I 73 + I80 I0

I 70 ⇒ I 70 = 107 I 0 I0 I 73 dB = (10 dB) log 73 ⇒ I 73 = 107.3 I 0 I0

and

80 dB = (10 dB) log

I 80 ⇒ I80 = 108 I 0 I0

Substitute and evaluate β3 sources:

β 3sources = (10 dB) log

10 7 I 0 + 10 7.3 I 0 + 108 I 0 = (10 dB) log 10 7 + 10 7.3 + 108 I0

(

)

= 81.1 dB (b) Find the intensity level with the two least intense sources eliminated:

β 80 = (10 dB) log

( )

108 I 0 = (10 dB) log 108 I0

= 80.0 dB Eliminating the two least intense sources does not reduce the intensity level significantly. *70 •• Picture the Problem Let P be the power radiated by the source of sound, and r be the initial distance from the source to the receiver. We can use the definition of intensity to find the ratio of the intensities before and after the distance is doubled and then use the definition of the decibel level to find the change in its level. Relate the change in decibel level to the change in the intensity level:

∆β = 10 log

I I'

1192 Chapter 15 Using its definition, express the intensity of the sound from the source as a function of P and r:

I=

P 4π r 2

Express the intensity when the distance is doubled:

I' =

P P = 2 4π (2r ) 16π r 2

Evaluate the ratio of I to I′:

P I 4π r 2 = =4 P I' 16π r 2

Substitute to obtain:

∆β = 10 log 4 = 6.02 dB and (c) is correct.

71 ••• Picture the Problem The sound level can be found from the intensity of the sound due to the talking people. When 38 people are talking, the intensities add. Express the sound level when all 38 people are talking:

β 38 = (10 dB) log

38I1 I0

= (10 dB) log 38 + (10 dB) log = (10 dB) log 38 + 72 dB = 87.8 dB An equivalent but longer solution:

38I1 I0

Express the sound level when all 38 people are talking:

β 38 = (10 dB) log

Express the sound level when only one person is talking:

β1 = 72 dB = (10 dB) log

Solve for and evaluate I1:

I1 = 10 7.2 I 0 = 10 7.2 10 −12 W/m 2

(

= 1.58 × 10 −5 W/m 2 Express the intensity when all 38 people are talking:

I 38 = 38 I1

I1 I0

)

I1 I0

Wave Motion 1193 The decibel level is:

(

38 1.58 × 10 −5 W/m 2 β 38 = (10 dB) log 10 −12 W/m 2

)

= 87.8 dB *72 ••• Picture the Problem Let η represent the efficiency with which mechanical energy is converted to sound energy. Because we’re given information regarding the rate at which mechanical energy is delivered to the string and the rate at which sound energy arrives at the location of the listener, we’ll take the efficiency to be the ratio of the sound power delivered to the listener divided by the power delivered to the string. We can calculate the power input directly from the given data. We’ll calculate the intensity of the sound at 35 m from its intensity level at that distance and use this result to find the power output. Express the efficiency of the conversion of mechanical energy to sound energy: Find the power delivered by the bow to the string: Using β = (10 dB) log(I I 0 ) , find the intensity of the sound at 35 m:

η=

Pout Pin

Pin = Fv = (0.6 N )(0.5 m/s ) = 0.3 W

60 dB = (10 dB) log

I 35 m I0

and

I 35 m = 10 6 I 0 = 10 −6 W/m 2

(

)

Find the power of the sound emitted:

Pout = IA = 4π 10 −6 W/m 2 (35 m )

Substitute numerical values and evaluate η:

η=

2

= 0.0154 W

0.0154 W = 5.13% 0.3 W

73 ••• Picture the Problem Because the sound intensities are additive, we’ll find the intensity due to one student by subtracting the background intensity from the intensity due to the students and dividing by 100. Then, we’ll use this result to calculate the intensity level due to 50 students. Express the intensity level due to 50 students:

β50 = (10 dB) log

50 I1 I0

1194 Chapter 15 Find the sound intensity when 100 students are writing the exam:

60 dB = (10 dB) log

I100 I0

and

I100 = 106 I 0 = 10−6 W/m 2 Find the sound intensity due to the background noise:

40 dB = (10 dB) log

I background I0

and

I background = 104 I 0 = 10−8 W/m 2 Express the sound intensity due to the 100 students:

I100 − I background = 10−6 W/m 2 − 10−8 W/m2

Find the sound intensity due to 1 student:

I100 − I background

Substitute numerical values and evaluate the intensity level due to 50 students:

β 50 = (10 dB) log

≈ 10− 6 W/m2

100

= 10−8 W/m 2

(

50 10−8 W/m 2 10−12 W/m 2

)

= 57.0 dB

The Doppler Effect 74

•

v±u ) to find the wavelength of fs v ± ur f s ) to find the the sound between the source and the listener and 15-35a ( f r = v ± us

Picture the Problem We can use equation 15-32 ( λ =

frequency heard by the listener. (a) Because the source is approaching the listener, use the minus sign in the numerator of Equation 15-32 to find the wavelength of the sound between the source and the listener:

λ=

340 m/s − 80 m/s = 1.30 m 200 s −1

Wave Motion 1195 (b) Because the listener is at rest and the source is approaching, ur = 0 and the denominator of Equation 15-35a is the difference between the two speeds:

fr = =

v fs v − us

(

340 m/s 200 s −1 340 m/s − 80 m/s

)

= 262 Hz

75 • Picture the Problem In the reference frame described, the speed of sound from the source to the listener is reduced by the speed of the wind. We can find the wavelength of the sound in the region between the source and the listener from v = fλ. Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than the frequency emitted by the source and can be calculated using f r =

v ± ur fs . v ± us

(a) The speed of sound in the reference frame of the source is:

v = 340 m/s − 80 m/s = 260 m/s

(b) Noting that the frequency is unchanged, express the wavelength of the sound:

λ=

(c) Apply Equation 15-35a to obtain:

⎛ v + ur ⎞ fr = ⎜ ⎟ fs ⎝ v ⎠

v 260 m/s = = 1.30 m 200 s −1 f

⎛ 260 m/s + 80 m/s ⎞ ⎟⎟ 200 s −1 = ⎜⎜ 260 m/s ⎝ ⎠

(

)

= 262 Hz 76 • Picture the Problem We can use λ = (v ± u ) f s to find the wavelength of the sound in the region between the source and the listener and f r =

v ± ur f s to find the frequency v ± us

heard by the listener. Because the sound waves in the region between the source and the listener will be spread out by the motion of the listener, the frequency of the sound heard by the listener will be lower than the frequency emitted by the source.

1196 Chapter 15 (a) Because the source is moving away from the listener, use the positive sign in the numerator of Equation 15-32 to find the wavelength of the sound between the source and the listener:

λ=

(b) Because the listener is at rest and the source is receding, ur = 0 and the denominator of Equation 15-35a is the sum of the two speeds:

fr =

340 m/s + 80 m/s 200 s −1

= 2.10 m

=

v fs v + us

(

340 m/s 200 s −1 340 m/s + 80 m/s

)

= 162 Hz 77 • Picture the Problem We can find the wavelength of the sound in the region between the source and the listener from v = fλ. Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than the frequency emitted by the source and can be calculated using f r =

v ± ur fs . v ± us

v 340 m/s = = 1.70 m 200 s −1 f

(a) Because the wavelength is unaffected by the motion of the observer:

λ=

(b) Apply Equation 15-35a to obtain:

⎛ v + ur ⎞ fr = ⎜ ⎟ f0 ⎝ v ⎠ ⎛ 340 m/s + 80 m/s ⎞ ⎟⎟ 200 s −1 = ⎜⎜ 340 m/s ⎝ ⎠

(

)

= 247 Hz 78 • Picture the Problem In this reference frame, the speed of sound will be increased by the speed of the listener. We can find the wavelength of the sound in the region between the source and the listener from v = fλ. Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than the frequency emitted by the source and can be calculated using v′ = frλ′.

Wave Motion 1197 (a) Moving at 80 m/s in still air:

The observer experiences a wind velocity of 80 m/s.

(b) Use the standard Galilean transformation to obtain:

v' = v + ur = 340 m/s + 80 m/s

(c) Because the distance between the wave crests is unchanged:

λ' =

v' 340 m/s = = 1.70 m 200 s −1 f

(d) Using the speed of sound in this reference frame, express and evaluate the frequency heard by the listener:

fr =

v' 420 m/s = = 247 Hz λ' 1.70 m

= 420 m/s

79 • Picture the Problem Because the listener is moving away from the source, we know that the frequency he/she will hear will be less than the frequency emitted by the source. We can use f r =

v ± ur f s , with us = 0 and the minus sign in the numerator, to determine its v ± us

value. Relate the frequency heard by the listener to that of the source:

⎛ v − ur ⎞ fr = ⎜ ⎟ fs ⎝ v ⎠ ⎛ 340 m/s − 80 m/s ⎞ ⎟⎟ 200 s −1 = ⎜⎜ 340 m/s ⎝ ⎠

(

= 153 Hz 80 • Picture the Problem The diagram shows the position of the supersonic plane at time t after it was directly over a person located at point P 5000 m below it. Let u represent the speed of the plane and v the speed of sound. We can use trigonometry to determine the angle of the shock wave as well as the location of the jet x when the person on the ground hears the shock wave.

)

1198 Chapter 15 (a) Referring to the diagram, express θ in terms of v, u, and t:

⎛ vt ⎞ ⎛ 1 ⎞ −1 ⎛ 1 ⎞ ⎟⎟ = sin −1 ⎜ ⎟ = sin ⎜⎜ ⎟ ⎝ ut ⎠ ⎝ 2 .5 ⎠ ⎝u v⎠

θ = sin −1 ⎜

= 23.6° (b) Using the diagram, relate the angle θ to the altitude h of the plane and the distance x and solve for x:

tan θ =

Substitute numerical values and evaluate x:

x=

h h ⇒ x= tan θ x

5000 m = 11.4 km tan23.6°

81 • Picture the Problem If both us and ur are much smaller than the speed of sound v, then the shift in frequency is given approximately by

∆f u = ± , where u = us ± ur is the fr v

relative speed of the source and receiver. For purposes of this problem, we’ll assume that you are an Olympics qualifier and can run at a top speed of approximately 10 m/s. Express the frequency, fr, you hear in terms of the frequency of the source fs and your running speed u:

⎛v+u⎞ fr = ⎜ ⎟ fs ⎝ v ⎠

Assuming that you can run 10 m/s , substitute numerical values and evaluate fr:

⎛ 340 m/s + 10 m/s ⎞ ⎟⎟(1000 Hz ) f r = ⎜⎜ 340 m/s ⎝ ⎠

Using the positive sign (you are approaching the source), express

10 m/s ∆f = = 2.94% f r 340 m/s

and evaluate the ratio

∆f : fr

= 1029 Hz

Because this fractional change in frequency is less than the 3% criterion for recognition of a change in frequency, it would be impossible to use your sense of pitch to estimate your running speed. 82 •• Picture the Problem Because the car is moving away from the radar device, the frequency fr it receives will be less than the frequency emitted by the device. The

Wave Motion 1199 microwaves reflected from the car, moving away from a stationary detector, will be of a still lower frequency fr ′. We can use the Doppler shift equations to derive an expression for the speed of the car in terms of difference of these frequencies. Express the frequency fr received by the moving car in terms of fs, u, and v:

⎛c−u ⎞ fr = ⎜ ⎟ fs ⎝ c ⎠

(1)

Express the frequency fr ′ received by the stationary source in terms of fr, u, and v:

⎛c−u ⎞ f r' = ⎜ ⎟ fr ⎝ c ⎠

(2)

Substitute equation (1) in equation (2) to eliminate fr:

⎛c−u ⎞ ⎛ 2u ⎞ f r' = ⎜ ⎟ f s ≈ ⎜1 − ⎟ f s c ⎠ ⎝ c ⎠ ⎝

2

provided u << c Express the frequency difference detected at the source:

⎛ 2u ⎞ ∆f = f s − f r ' = f s − ⎜1 − ⎟ f s c ⎠ ⎝ 2u = fs c

Solve for u:

u=

Substitute numerical values and evaluate u:

u=

c ∆f 2 fs

3 × 108 m/s (293 Hz ) 2(2 GHz ) m 1 km 3600 s = 22.0 × 3 × s 10 m h = 79.1 km/h

*83 •• Picture the Problem Because the radial component u of the velocity of the raindrops is small compared to the speed v = c of the radar pulse, we can approximate the fractional change in the frequency of the reflected radar pulse to find the speed of the winds carrying the raindrops in the storm system. Express the shift in frequency when the speed of the source (the storm system) u is much smaller than the wave speed v = c:

∆f u ≈ fs c

1200 Chapter 15 ∆f fs

Solve for u:

u=c

Substitute numerical values and evaluate u:

u = 2.998 × 108 m/s

(

= 156 m/s ×

325 Hz ) 625 MHz

0.6215 mi/h 0.2778 m/s

= 349 mi/h 84 •• Picture the Problem Let the depth of the submarine be represented by D and its vertical speed by u. The submarine acts as both a receiver and source. We can apply the definition of average speed to determine the depth of the submarine and use the Doppler shift equations to derive an expression for the vertical speed of the submarine in terms of the frequency difference. (a) Using the definition of average speed, relate the depth of the submarine to the time delay between the transmitted and reflected pulses:

2 D = v∆t ⇒ D = 12 v∆t

Substitute numerical values and evaluate D:

D=

(b) Express the frequency fr received by the submarine in terms of f0, u, and c:

⎛c±u ⎞ fr = ⎜ ⎟ fs ⎝ c ⎠

(1)

Express the frequency fr′ received by the destroyer in terms of fr, u, and c:

⎛c±u ⎞ f r' = ⎜ ⎟ fr ⎝ c ⎠

(2)

Substitute equation (1) in equation (2) to eliminate fr:

⎛c±u ⎞ ⎛ 2u ⎞ fr ' = ⎜ ⎟ f s ≈ ⎜1 ± ⎟ f s c ⎠ ⎝ c ⎠ ⎝

1 2

(1.54 km/s )(80 ms) =

61.6 m

2

provided u << c. Express the frequency difference detected by the destroyer:

⎛ 2u ⎞ ∆f = f s − f r' = f s − ⎜1 ± ⎟ f s c ⎠ ⎝ 2u = fs c

Wave Motion 1201 Solve for u:

u=

c ∆f 2 fs

Substitute numerical values and evaluate u:

u=

1.54 km/s (0.0420 MHz ) 2(40 MHz )

= 0.809 m/s where the positive speed indicates that the velocity of the submarine is downward. 85 •• Picture the Problem Because the car is moving away from the radar unit, the frequency fr it receives will be less than the frequency emitted by the unit. The radar waves reflected from the car, moving away from a stationary detector, will be of a still lower frequency fr′. Let u represent the relative speed of the police car and the receding car (140 km/h) and use the Doppler shift equations to derive an expression for the difference between fs, the transmitted signal, and fr′. Express the frequency fr received by the moving car in terms of fs, u, and c:

⎛c−u ⎞ fr = ⎜ ⎟ fs ⎝ c ⎠

(1)

Express the frequency fr′ received by the stationary source in terms of fr, u, and c:

⎛c−u ⎞ f' r = ⎜ ⎟ fr ⎝ c ⎠

(2)

Substitute equation (1) in equation (2) to eliminate fr:

⎛ u⎞ ⎛ 2u ⎞ f r' = ⎜1 − ⎟ f s ≈ ⎜1 − ⎟ f s c ⎠ ⎝ c⎠ ⎝

2

provided u << c Express the frequency difference detected at the source:

⎛ 2u ⎞ ∆f = f s − f' r = f s − ⎜1 − ⎟ f s c ⎠ ⎝ 2u = fs c

Substitute numerical values and evaluate ∆f:

⎡ ⎛ km 1h ⎞ ⎤ ⎟⎟ ⎥ × ⎢ 2⎜⎜140 h 3600 s ⎝ ⎠ ⎥ 3 × 1010 Hz ∆f = ⎢ 8 ⎢ ⎥ 3 × 10 m/s ⎢ ⎥ ⎣ ⎦

(

= 7.78 kHz

)

1202 Chapter 15 86 •• Picture the Problem Because the car is moving away from the radar unit, the frequency fr it receives will be less than the frequency emitted by the unit. The radar waves reflected from the car, moving away from a stationary detector, will be of a still lower frequency fr′. Let u represent the relative speed of the police car and the receding car (80 km/h) and use the Doppler shift equations to derive an expression for the difference between fs, the transmitted signal, and fr′. Express the frequency fr received by the moving car in terms of fs, u, and c:

⎛c−u ⎞ fr = ⎜ ⎟ fs ⎝ c ⎠

(1)

Express the frequency fr′ received by the stationary source in terms of fr, u, and c:

⎛c−u ⎞ f r' = ⎜ ⎟ fr ⎝ c ⎠

(2)

Substitute equation (1) in equation (2) to eliminate fr:

⎛ u⎞ ⎛ 2u ⎞ f r' = ⎜1 − ⎟ f s ≈ ⎜1 − ⎟ f s c ⎠ ⎝ c⎠ ⎝

2

provided u << c. Express the frequency difference detected at the source:

⎛ 2u ⎞ ∆f = f s − f r' = f s − ⎜1 − ⎟ fs 0 c ⎠ ⎝ 2u = fs c

Substitute numerical values and evaluate ∆f:

⎡ ⎛ km 1h ⎞ ⎤ ⎟⎥ × ⎢ 2⎜⎜ 80 h 3600 s ⎟⎠ ⎥ ⎝ ⎢ ∆f = 3 × 1010 Hz ⎢ ⎥ 3 × 108 m/s ⎢ ⎥ ⎣ ⎦

(

= 4.44 kHz

)

Wave Motion 1203 87 •• Picture the Problem The diagram shows the position of the supersonic plane flying due west at time t after it was directly over point P 12 km below it. Let x be measured from point P, u represent the speed of the plane, and v be the speed of sound. We can use trigonometry to determine the angle of the shock wave as well as the location of the jet x when the person on the ground hears the shock wave. Using the diagram to relate the distance x to the shock-wave angle θ and the elevation of the plane: Referring to the diagram, express θ in terms of v, u, and t and determine its value:

Substitute numerical values and evaluate x:

tan θ =

h h and x = x tan θ

⎛ vt ⎞ −1 ⎛ 1 ⎞ ⎟⎟ ⎟ = sin ⎜⎜ ⎝ ut ⎠ ⎝u v⎠

θ = sin −1 ⎜

⎛ 1 ⎞ = sin −1 ⎜ ⎟ = 38.7° ⎝ 1.6 ⎠ x=

12 km = 15.0 km west of P. tan38.7°

88 •• Picture the Problem The change in frequency of source as it oscillates on the air track is 3 Hz. We can use

∆f u ≈ ± to find the maximum speed of the vibrating mass-spring fs v

system in terms of this change in frequency and then use this speed to find the energy of the system. Knowing the energy of the system, we can find the amplitude of its motion. We can calculate the period of the motion from knowledge of the mass of the radio and the stiffness constant of the spring. (a) Express the energy of the vibrating mass-spring system in terms of its maximum speed: Relate the change in the frequency heard by the listener to the maximum speed of the oscillator:

2 E = 12 mumax

(1)

∆f u =± fs v where u = us ± ur is the relative speed of the source and receiver and v is the speed of

1204 Chapter 15 sound. Solve for and evaluate u = umax:

u max =

3 Hz ∆f (340 m/s) v= f0 800 Hz

= 1.275 m/s

(0.1kg )(1.275 m/s2 ) 2 =

Substitute numerical values in equation (1) and evaluate E:

E=

(b) Relate the energy of the oscillator to the amplitude of its motion:

E = 12 kA2

Solve for A to obtain:

Substitute numerical values and evaluate A:

1 2

A=

2E k

A=

2(81.3 mJ ) = 2.85 cm 200 N/m

81.3 mJ

89 •• Picture the Problem The received and transmitted frequencies are related through

fr =

v ± ur f s , where the variables have the meanings given in the problem statement. v ± us

Because the source and receiver are moving in the same direction, we use the minus signs in both the numerator and denominator. (a) Relate the received frequency f r to the frequency f0 of the source:

fr =

1 − ur v f0 1 − us v

= (1 − ur v )(1 − us v ) f 0 −1

(b) Apply the binomial expansion to (1 − us v )−1 : Substitute to obtain:

(1 − us v )−1 ≈ 1 + us

v

f r = (1 − ur v )(1 + us v ) f 0

= [1 + us v − ur v − (ur v )(us v )] f 0 ⎛ u −u ⎞ ≈ ⎜1 + s r ⎟ v ⎠ ⎝

because both us and ur are small compared to v.

Wave Motion 1205 Because urel = us − ur

⎛ u ⎞ f r' ≈ ⎜1 + rel ⎟ f 0 v ⎠ ⎝

90 •• Picture the Problem Because the students are walking away from each other, the frequency f ′ each receives will be less than the frequency fs = 440 Hz emitted by their tuning forks. Let u represent the speed of each student and use the Doppler shift equations to derive an expression for the difference between fs, the frequency of the tuning fork each carries, and the frequency heard from the other’s tuning fork. Because this equation will contain u, we’ll be able to solve for and evaluate each student’s walking speed.

1− u v fs 1+ u v

Using equation 15-35a, express the frequency fr received by either student, when they are walking away from each other, in terms of fs, u, and v:

fr =

Expand (1 + u v )

(1 + u v )−1 ≈ 1 − u

−1

binomially:

Substitute and simplify to obtain:

= (1 − u v )(1 + u v ) f s −1

f r = (1 − u v ) f s

v provided u << v.

2

(

)

= 1 − 2u v + u 2 v 2 f s ≈ (1 − 2u v ) f s for u << v. Express the frequency difference heard by each student:

2u ⎛ 2u ⎞ ∆f = f 0 − f r = f s − ⎜1 − ⎟ f s = fs v ⎠ v ⎝

Solve for u:

u=

∆f v 2 fs

Substitute numerical values and evaluate u:

u=

2 Hz (340 m/s) = 0.773 m/s 2(440 Hz )

91 •• Picture the Problem The student serves as a source moving toward the wall, and a moving receiver for the echo. The received and transmitted frequencies are related through f r =

v ± ur f s . We’ll express the frequency received by the wall and the v ± us

frequency it transmits back to the moving student in order to express the difference in the

1206 Chapter 15 frequency the student hears from the wall and frequency she hears directly from her tuning fork. Because this expression will contain the student’s walking speed, we’ll be able to solve for and evaluate this speed. Relate the frequency fr received by the wall to the frequency of the student’s tuning fork fs:

⎛ 2u ⎞ ∆f = f s − f r' = f s − ⎜1 − ⎟ f s c ⎠ ⎝ 2u fs = c

Because the source is moving toward a stationary receiver:

fr =

Apply the binomial expansion to (1 − us v )−1 :

(1 − us v )−1 ≈ 1 + us

Substitute to obtain: Relate the frequency fr′ reflected from the wall and received by the student to the frequency fr reflected from the wall: Substitute for fr to obtain:

1 −1 f s = (1 − us v ) f s 1 − us v v

because us << v.

f r = (1 + us v ) f 0 f r' =

1 ± ur v 1 fr = fr 1 ± us v 1 − us v

because the source (the wall) is at rest and the receiver is approaching.

f r' =

1 (1 + us v ) f s 1 − us v

= (1 + us v )(1 − us v ) f s −1

Expand (1 − us v ) binomially: −1

(1 − us v )−1 ≈ 1 + us

v

because us << v.

f r' = (1 + us v )(1 + us v ) f s ≈ (1 + 2us v ) f s

because us << v Express the difference between the frequency the student receives from the wall and the frequency of her tuning fork:

∆f = f r' − f s

= (1 + 2us v ) f s − f s =

2us fs v

Wave Motion 1207 Solve for us:

us =

∆f v 2 fs

Substitute numerical values and evaluate us:

us =

4 Hz (340 m/s) = 1.33 m/s 2(512 Hz )

*92 •• Picture the Problem The frequency heard by the stationary observer will vary with time as the speaker rotates on its support arm. We can use a Doppler equation to express the frequency heard by the observer as a function of the velocity of the source and find the velocity of the source from the expression for the tangential velocity of an object moving in a circular path.

1 −1 f s = (1 − us v ) f s 1 − us v

Express the frequency fr heard by a stationary observer:

fr =

Expand (1 − us v ) to obtain:

(1 − us v )−1 ≈ 1 + us

−1

v

because us/v << 1 Substitute in the expression for fr: Express the speed of the source as a function of time:

f r = (1 + us v ) f s

(1)

us = rω sin ω t

= (0.8 m )(4 rad/s ) sin[(4 rad/s )t ] = (3.2 m/s ) sin[(4 rad/s )t ]

Substitute in equation (1) to obtain:

⎛ 3.2 m/s ⎞ f r = ⎜1 + sin[(4 rad/s )t ]⎟ f s v ⎝ ⎠

Substitute for v and simplify:

⎛ ⎞ 3.2 m/s f r = ⎜⎜1 + sin[(4 rad/s )t ]⎟⎟ 1000 s −1 ⎝ 340 m/s ⎠

(

)

= 1000 Hz + (9.41 Hz )sin[(4 rad/s )t ]

93 •• Picture the Problem The simplest way to approach this problem is to transform to a reference frame in which the balloon is at rest. In that reference frame, the speed of sound is v = 340 m/s, and ur = 36 km/h = 10 m/s. Then, we can use the equations for a moving receiver and a moving source to find the frequencies heard at the window and on the balloon.

1208 Chapter 15 (a) Express the observed frequency in terms of the frequency of the source:

⎛ u ⎞ f r = ⎜1 + r ⎟ f s v⎠ ⎝

Substitute numerical values and evaluate fr:

⎛ 10 m/s ⎞ ⎟⎟(800 Hz ) = 824 Hz f r = ⎜⎜1 + ⎝ 340 m/s ⎠

(b) Treating the tall building as a moving source, express the frequency of the reflected sound heard by a person riding in the balloon:

⎛ ⎜ 1 f r' = ⎜ ⎜ 1 − us ⎜ v ⎝

Substitute numerical values and evaluate fr′:

⎛ ⎞ ⎜ ⎟ 1 ⎜ ⎟(824 Hz ) = 849 Hz f r' = 10 m/s ⎟ ⎜ ⎜ 1 − 340 m/s ⎟ ⎝ ⎠

⎞ ⎟ ⎟ fr ⎟ ⎟ ⎠

94 •• Picture the Problem We can relate the frequencies fr and fr′ heard by the stationary observer behind the car to the speed of the car u and the frequency of the car’s horn fs. Dividing these equations will eliminate the frequency of the car’s horn and allow us to solve for the speed of the car. We can then substitute to find the frequency of the car’s horn. We can find the frequency heard by the driver as a moving receiver by relating this frequency to the frequency reflected from the wall. (a) Relate the frequency heard by the observer directly from the car’s horn to the speed of the car:

fr =

1 fs 1+ u v

(1)

Relate the frequency reflected from the wall to the speed of the car:

f r' =

1 fs 1− u v

(2)

Divide equation (2) by equation (1) to obtain:

f r' 1 + u v = fr 1 − u v

Solve for u:

u=

f r' − f r v f r' + f r'

Wave Motion 1209 Substitute numerical values and evaluate u:

863 Hz − 745 Hz (340 m/s) 863 Hz + 745 Hz m 1 km 3600 s = 24.95 × 3 × s 10 m h

u=

= 89.8 km/h (b) Solve equation (1) for fs:

f s = (1 + u v ) f r

Substitute numerical values and evaluate fs:

⎛ 24.95 m/s ⎞ ⎟(745 Hz ) f s = ⎜⎜1 + 340 m/s ⎟⎠ ⎝ = 800 Hz

(c) The driver is a moving receiver and so we can relate the frequency heard by the driver to the frequency reflected by the wall (the frequency heard by the stationary observer):

⎛ u⎞ f driver = ⎜1 + ⎟ f r' ⎝ v⎠

Substitute numerical values and evaluate fdriver:

⎛ 24.95 m/s ⎞ ⎟(863 Hz ) f driver = ⎜⎜1 + 340 m/s ⎟⎠ ⎝ = 926 Hz

95 •• Picture the Problem Let t = 0 when the driver sounds her horn and let the distance to the cliff at that instant be d. The received and transmitted frequencies are related through f r =

v ± ur f s . Solving this equation for fs will allow us to determine the v ± us

frequency of the car horn. We can use the total distance the sound travels (car-to-cliff plus cliff back to car … now closer to the cliff) to determine the distance to the cliff when the horn was briefly sounded. Relate the frequency heard by the driver to her speed and to the frequency of her horn:

fr =

1 + ur v fs 1 − us v

Solve for fs:

fs =

1 − us v fr 1 + ur v

1210 Chapter 15 Substitute numerical values and evaluate fs:

km 1h × 27.78 m/s h 3600 s 1− 1− 340 m/s 340 m/s ( (840 Hz ) = 713 Hz f0 = 840 Hz ) = 27.78 m/s km 1h × 1+ 100 340 m/s h 3600 s 1+ 340 m/s 100

Relate the distance d to the cliff at t = 0 to the distance she travels in time ∆t = 1 s, her speed u, and the speed of sound v:

d + (d − u∆t ) = v∆t

Solve for d:

d=

1 2

(u + v )∆t

Substitute numerical values and evaluate d:

d=

1 2

(27.78 m/s + 340 m/s)(1s )

= 184 m

96 •• Picture the Problem You’ll year the sonic boom when the surface of its cone reaches your plane. In the diagram the Concorde is at C and your plane is at P. The distance h = 3 km. The distance between the planes when you hear the sonic boom is d. We can use trigonometry to determine the angle of the shock wave as well as the separation of the planes when you hear the sonic boom. Using the Pythagorean theorem, relate the separation of the planes d, to the distance h and the angle θ : Solve for d to obtain:

Express θ in terms of v, u, and t:

d 2 = h 2 + d 2 cos 2 θ

d =h

1 1 − cos 2 θ ⎛ vt ⎞ −1 ⎛ 1 ⎞ ⎟⎟ ⎟ = sin ⎜⎜ ⎝ ut ⎠ ⎝u v⎠

θ = sin −1 ⎜

Wave Motion 1211 ⎛ 1 ⎞ ⎟ = 38.7° ⎝ 1.6 ⎠

Substitute numerical values and evaluate θ :

θ = sin −1 ⎜

Substitute numerical values and evaluate h:

d = (3 km )

1 = 4.80 km 1 − cos 2 38.7°

*97 •• Picture the Problem The sun and Jupiter orbit about their effective mass located at their common center of mass. We can apply Newton’s 2nd law to the sun to obtain an expression for its orbital speed about the sun-Jupiter center of mass and then use this speed in the Doppler shift equation to estimate the maximum and minimum wavelengths resulting from the Jupiter-induced motion of the sun. Letting v be the orbital speed of the sun about the center of mass of the sun-Jupiter system, express the Doppler shift of the light due to this motion when the sun is approaching the earth: Solve for λ′:

f' =

c 1+ v / c c 1+ v / c = f = λ' 1− v / c λ 1− v / c

1− v / c 1+ v / c

λ' = λ

= λ (1 − v c )(1 + v c ) = λ (1 − v c )

12

Because v << c, we can expand

(1 − v c )

12

and (1 + v c )

−1 2

binomially to obtain:

(1 − v c )1 2 ≈ 1 −

−1

(1 + v c )−1 2

v 2c

and

(1 + v c )−1 2 ≈ 1 −

v 2c 2

Substitute to obtain:

1− v / c ⎛ v ⎞ v = ⎜1 − ⎟ ≈ 1 − 1 + v / c ⎝ 2c ⎠ c

When the sun is receding from the earth:

1+ v / c ⎛ v ⎞ v = ⎜1 + ⎟ ≈ 1 + 1 − v / c ⎝ 2c ⎠ c

Hence the motion of the sun will give an observed Doppler shift of:

2

⎛ ⎝

v⎞ c⎠

λ ' ≈ λ ⎜1 ± ⎟

(1)

1212 Chapter 15 Apply Newton’s 2nd law to the sun:

Solve for v to obtain:

GM S M eff v2 = MS 2 rcm rcm v=

GM eff rcm

(0)M S + rs-J M J

Measured from the center of the sun, the distance to the center of mass of the sun-Jupiter system is:

rcm =

The effective mass is related to the masses of the sun and Jupiter according to:

1 1 1 = + M eff M s M J

Ms + MJ

=

rs-J M J Ms + MJ

or

M eff = Substitute for Meff and rcm to obtain:

M sM J Ms + MJ MsM J GM s Ms + MJ = rs-J M J rs-J Ms + MJ

G v=

Using rs-J = 7.78 × 1011 m as the mean orbital radius of Jupiter, substitute numerical values and evaluate v:

v=

(6.673 ×10

−11

N ⋅ m 2 / kg 2 )(1.99 × 1030 kg ) = 1.306 × 104 m/s 11 7.78 × 10 m

Substitute in equation (1) to obtain:

⎛

λ ' ≈ (500 nm ) ⎜⎜1 ±

1.306 × 10 4 m/s ⎞ ⎟ 2.998 × 108 m/s ⎟⎠

⎝ = (500 nm ) 1 ± 4.36 × 10 −5

(

The maximum and minimum wavelengths are:

)

λmax = (500 nm ) (1 + 4.36 × 10 −5 ) and

λmin = (500 nm ) (1 − 4.36 × 10 −5 )

98 ••• Picture the Problem Choose a coordinate system in which downward is the positive y direction. Let d represent the distance the tuning fork has fallen when the student hears a

Wave Motion 1213 frequency of 400 Hz, t1 the time for the source to fall that distance, and t2 the time for the sound to travel back to the student. We can a constant-acceleration equation to express d in terms of the time that elapses between the dropping of the tuning fork and the return of the sound to the student. A Doppler-effect equation will allow us to solve for the speed of the tuning fork when the student hears a frequency of 400 Hz; we can use constantacceleration equations to find the fall time for the fork and the return time for the sound from the tuning fork. Using a constant-acceleration equation, relate the distance the source has fallen to the elapsed time:

d = v0t + 12 at 2 or, because v0 = 0 and a = g,

d = 12 gt 2

(1)

where t = t1 + t2.

1 fs 1 + us v

Relate the frequency fr heard by the student to the speed of the falling tuning fork:

fr =

Solve for us:

⎛f ⎞ us = ⎜⎜ s − 1⎟⎟v ⎝ fr ⎠

Substitute numerical values and evaluate us:

⎛ 440 Hz ⎞ us = ⎜⎜ − 1⎟⎟ (340 m/s ) = 34.0 m/s ⎝ 400 Hz ⎠

Letting y be the distance the fork has fallen when its speed is us, use a constant-acceleration equation to relate y and us:

us2 = v02 + 2 gy

Solve for y:

Substitute numerical values and evaluate y: Using a constant-acceleration equation, relate the speed of the falling tuning fork to its time of fall t1 : Solve for and evaluate t1:

or, because v0 = 0,

us2 = 2 gy

y=

y=

us2 2g

(34 m/s)2

2(9.81 m/s 2 )

= 58.92 m

us = v0 + gt1 or, because v0 = 0,

us = gt1 t1 =

us 34.0 m/s = = 3.466 s g 9.81 m/s 2

1214 Chapter 15 Using the relationship between distance traveled, time and average speed, find the time t2 for the sound to travel back to the student:

t2 =

Substitute in equation (1) and evaluate d:

d=

y 58.92 m = = 0.173 s v 340 m/s

1 2

(9.81 m/s )(3.466 s + 0.173 s )

2

2

= 65.0 m

99 •• Picture the Problem The angle θ of the Cerenkov shock wave is related to the speed of light in water v and the speed of light in a vacuum c according to sin θ = v c .

v c

Relate the speed of light in water v to the angle of the Cerenkov cone:

sin θ =

Solve for v:

v = c sin θ

Substitute numerical values and evaluate v:

v = 2.998 × 108 m/s sin 48.75°

(

)

= 2.25 × 108 m/s

General Problems 100 • Picture the Problem The equation of a wave traveling in the positive x direction is of the form y (x,t ) = f ( x − vt ) and that of a wave traveling in the negative x direction

is y (x,t ) = f ( x + vt ) .

Wave Motion 1215

(a) The pulse at t = 0 shown below was plotted using a spreadsheet program:

0.035 0.030

y (x ,0) (m)

0.025 0.020 0.015 0.010 0.005 0.000 -6

-4

-2

0

2

4

6

x (m)

(b) The wave function must be of the form: Replace x with x − (10 m/s ) t to obtain:

(c) The wave function must be of the form: Replace x with x + (10 m/s ) t to obtain:

y ( x,t ) = f ( x − vt ) = f [x − (10 m/s )t ] because v = 10 m/s

y ( x,t ) =

0.12 m 3 (2.00 m )2 + [x − (10 m/s)t ] 2

y ( x,t ) = f ( x + vt ) = f [x + (10 m/s )t ] because v = 10 m/s

0.12 m 3 y ( x, t ) = (2.00 m ) 2 + [x + (10 m/s)t ] 2

101 • Picture the Problem Let the subscript 1 refer to the initial situation−a tension of 800 N and a wavelength of 24 cm. Let the subscript 2 refer to the conditions that the tension is 600 N and the wavelength unknown. We can express the wavelengths of the waves on the wire in terms of the two tensions in the wire and then eliminate the constant frequency by expressing the ratio of the two wavelengths. Finally, we can solve this equation for λ2. Express the wavelength in terms of the frequency and speed of the

λ=

v f

1216 Chapter 15 wave: Express the speed of the wave as a function of the tension in the wire: Substitute to obtain:

T

v=

µ

λ=

1 f

T

Express the wavelength when the tension in the wire is 600 N:

λ2 =

1 f

T2

Express the wavelength when the tension in the wire is 800 N:

λ1 =

1 f

T1

Divide the first of these equations by the second and solve for λ2:

λ2 = λ1

T2 T1

Substitute numerical values and evaluate λ2:

λ2 = (24 cm )

µ

µ

µ

600 N = 20.8 cm 800 N

102 • Picture the Problem Let m represent the mass of the rubber tubing whose length is L. We can express the travel time for the pulse t in terms of the separation of the post and the pulley and its speed. The speed of the pulse, in turn, can be found from the tension in the rubber tubing and its linear density. Express the time required for the pulse to travel the length of the tubing in terms of its speed and the length of the rubber tubing:

t=

L v

Relate the speed of the pulses to the tension in the tubing:

v=

Substitute for v and simplify to obtain:

t=

Lm F

Substitute numerical values and evaluate t:

t=

(10 m )(0.7 kg ) =

F

µ

=

F = m L

110 N

FL m

0.252 s

Wave Motion 1217 103 • Picture the Problem The diagram shows the boat traveling on a still lake with a speed v. A bow wave generated a time t earlier is shown at an angle of θ with the direction of the boat’s motion. We can use trigonometry to relate the speed of the bow wave to the speed of the boat.

vt v = ut u

Using the diagram, relate u and v to the angle θ:

sin θ =

Solve for v:

v = u sin θ

Substitute numerical values and evaluate v:

v = (10 m/s )sin20° = 3.42 m/s

104 • Picture the Problem The frequencies and wavelengths of the sound waves are related to the speed of sound through f = v/λ. (a) Use f = v/λ to find f:

f =

340 m/s = 113 Hz 10(0.3 m )

(b) Proceed as in (a):

f =

340 m/s = 11.3 kHz 0.1(0.3 m )

(c) Proceed as in (a) and (b):

f =

340 m/s = 567 Hz 10(0.06 m )

f =

340 m/s = 56.7 kHz 0.1(0.06 m )

1218 Chapter 15 105 • Picture the Problem The diagram depicts the whistle traveling in a circular path of radius r = 1 m. The stationary listener will hear the maximum frequency when the whistle is at point 1 and the minimum frequency when it is at point 2. These maximum and minimum frequencies are determined by f0 and the tangential speed us = 2π r/T. We can relate the frequencies heard at point P to the speed of the approaching whistle at point 1 and the speed of the receding whistle at point 2.

1 fs 1 − us v

Relate the frequency heard at point P to the speed of the approaching whistle at point 1:

f max =

Use the relationship between translational velocity and angular velocity to find the speed us of the whistle:

⎛ rev 2π rad ⎞ us = rω = (1 m ) ⎜ 3 × ⎟ rev ⎠ ⎝ s = 18.85 m/s

Substitute numerical values and evaluate fmax:

f max =

1 (500 Hz ) 18.85 m/s 1− 340 m/s

= 529 Hz Relate the frequency heard at point P to the speed of the receding whistle at point 2:

f min =

Substitute numerical values and evaluate fmin:

f min =

1 fs 1 + us v 1 (500 Hz ) 18.85 m/s 1+ 340 m/s

= 474 Hz 106 • Picture the Problem The crest-to-crest separation of the waves is their wavelength. We can find the frequency of the waves from v = fλ. When you lift anchor and head out to sea

Wave Motion 1219 you’ll become a moving receiver and we can apply f r =

v ± ur f s to calculate the v ± us

frequency you’ll observe. (a) Express the frequency of the ocean waves in terms of their speed and wavelength:

f0 =

Substitute numerical values and evaluate fs:

f0 =

(b) Express the frequency of the waves in terms of their speed and the speed of a moving receiver: Substitute numerical values and evaluate fr:

v

λ

8.9 m/s = 0.593 Hz 15 m

f r = (1 + ur v ) f s

⎛ 15 m/s ⎞ ⎟⎟ (0.593 Hz ) = 1.59 Hz f r = ⎜⎜1 + ⎝ 8.9 m/s ⎠

107 •• Picture the Problem Let t be the time of travel of the lefthand pulse and the subscripts L and R refer to the pulse coming from the left and right, respectively. Because the pulse traveling from the right starts later than the pulse from the left, its travel time is t − ∆t, where ∆t = 25 ms. Both pulses travel at the same speed and the sum of the distances they travel is 12 m. Express the total distance the two pulses travel: Solve for vt to obtain: The speed of the pulse is given by:

Substitute for v to obtain:

d = dL + dR

= vt + v(t − ∆t ) vt = 12 (d + v∆t ) v=

F

µ

=

F m L

⎞ ⎛ F vt = 12 ⎜⎜ d + ∆t ⎟⎟ m L ⎠ ⎝

1220 Chapter 15

Substitute numerical values and evaluate vt:

⎡ vt = 12 ⎢12 m + ⎣

(180 N )(12 m ) (25 × 10 −3 s )⎤ = ⎥ ⎦

0.085 kg

7.99 m

*108 •• Picture the Problem Let the frequency of the car’s horn be fs, the frequency you hear as the car approaches fr, and the frequency you hear as the car recedes fr′. We can use

fr =

v ± ur f s to express the frequencies heard as the car approaches and recedes and v ± us

then use these frequencies to express the fractional change in frequency as the car passes you. Express the fractional change in frequency as the car passes you:

∆f = 0.1 fr

Relate the frequency heard as the car approaches to the speed of the car:

fr =

1 fs 1 − us v

Express the frequency heard as the car recedes in terms of the speed of the car:

f r' =

1 fs 1 + us v

Divide the second of these frequency equations by the first to obtain:

f r' 1 − us v = f r 1 + us v and

1 − us v f r f r' ∆f − = = 1− = 0.1 fr fr fr 1 + us v

Solve us: Substitute numerical values and evaluate us:

0.1 v 1.9 0.1 (340 m/s) us = 1.9 m 1 km 3600 s = 17.89 × 3 × s 10 h us =

= 64.4 km/h

Wave Motion 1221 109 •• Picture the Problem The pressure amplitude can be calculated directly from p0 = ρωvs0 , and the intensity from I = 12 ρω 2 s02 v. The power radiated is the intensity times the area of the driver. (a) Relate the pressure amplitude to the displacement amplitude, angular frequency, wave velocity, and air density: Substitute numerical values and evaluate p0:

p0 = ρωvs0

(

)[ ( )] × (340 m/s ) (0.025 × 10 m )

p0 = 1.29 kg/m 3 2π 800 s −1

−3

= 55.1 N/m 2 (b) Relate the intensity to these same quantities:

I = 12 ρω 2 s02 v

Substitute numerical values and evaluate I:

I=

(1.29 kg/m )[2π (800 s )] × (0.025 × 10 m ) (340 m/s ) −1

3

1 2

−3

2

2

= 3.46 W/m 2 (c) Express the power in terms of the intensity and the area of the driver:

P = IA = π r 2 I

Substitute numerical values and evaluate P:

P = π (0.1 m ) 3.46 W/m 2 = 0.109 W 2

(

)

110 •• Picture the Problem The frequency of the sound wave is related to the density of the air, displacement amplitude, and velocity by I = 12 ρω 2 s02 v. Relate the intensity of the sound wave to the density of the air, displacement amplitude, velocity, and angular frequency: Solve for the angular frequency:

I = 12 ρω 2 s02 v

ω=

1 s0

2I ρv

1222 Chapter 15 Solve for f:

f =

Substitute numerical values and evaluate f:

f =

1 2π s0

2I ρv

1 2π 10 −6 m

(

(

)

2 10 −2 W/m 2 1.29 kg/m 3 (340 m/s )

) (

)

= 1.07 kHz 111 •• Picture the Problem The force exerted on the plate is due to the change in momentum of the water. We can use Newton’s 2nd law in the form F = ∆p/∆t to relate F to the mass of water in a length of tube equal to vs∆t and to the speed of the water. This mass of water, in turn, is given by the product of its density and the volume of water in a length of the tube equal to vs∆t.

∆p ∆mv w = ∆t ∆t

Relate the force exerted on the plate to the change in momentum of the water:

F=

Express ∆m in terms of the mass of water in a length of tube equal to vs∆t:

∆m = ρ∆V = ρvs A∆t

Substitute to obtain:

F = ρvs Avw

Substitute numerical values and evaluate F:

(

)

[

]

F = 103 kg/m 3 (1.4 km/s ) π (0.05 m ) (7 m/s ) = 77.0 kN 2

112 •• Picture the Problem Let d be the horizontal distance from the soap bubble to the position of the microphone. The angle θ of the shock wave is related to the speed of sound in air u and the speed of the bullet c according to sin θ = u v . We can determine θ from the given information and then use this angle to find d.

0.35 m tan θ

Express d in terms of the angle of the shock wave and the distance from the soap bubble to the laboratory bench:

d=

Relate the speed of the bullet to the angle of the shock-wave cone:

sin θ =

u v

Wave Motion 1223 Solve for θ to obtain:

θ = sin −1

Substitute to obtain:

d=

Substitute numerical values and evaluate d:

d=

u v

0.35 m u⎤ ⎡ tan ⎢sin −1 ⎥ v⎦ ⎣ 0.35 m = 26.3 cm ⎡ −1 ⎛ u ⎞⎤ tan ⎢sin ⎜ ⎟⎥ ⎝ 1.25u ⎠⎦ ⎣

113 •• Picture the Problem The source of the problem is that it takes a finite time for the sound to travel from the front of the line of marchers to the back. We can use the given data to determine the time required for the beat to reach the marchers in the back of the column and then use this time and the speed of sound to find the length of the column. Express the length of the column in terms of the speed of sound and the time required for the beat to travel the length of the column:

L = v∆t

Calculate the time for the sound to travel the length of the column:

∆t =

Substitute and evaluate L:

L = (340 m/s )(0.6 s ) = 204 m

1 min = 0.6 s 100

114 •• Picture the Problem The interval between the arrival times of the echo pulses heard by the bat is the reciprocal of the frequency of the reflected pulses. We can use

fr =

v ± ur f s to relate the frequency of the reflected pulses to the speed of the bat and v ± us

the frequency it emits. Relate the interval between the arrival times of the echo pulses heard by the bat to frequency of the reflected pulses: Relate the frequency of the pulses received by the bat to its speed and the frequency it emits:

∆t =

fr =

1 fr

1 + ur v fs 1 − us v

1224 Chapter 15 1 − us v (1 + ur v ) fs

Substitute to obtain:

∆t =

Substitute numerical values and evaluate ∆t:

12 m/s 340 m/s = 11.6 ms ∆t = ⎛ 12 m/s ⎞ −1 ⎟⎟ 80 s ⎜⎜1 + ⎝ 340 m/s ⎠ 1−

(

)

*115 •• Picture the Problem Let d be the distance to the moon, h be the height of earth's atmosphere, and v be the speed of light in earth’s atmosphere. We can express d ′, the distance measured when the earth’s atmosphere is ignored, in terms of the time for a pulse of light to make a round-trip from the earth to the moon and solve this equation for the length of correction d ′ − d. Express the roundtrip time for a pulse of light to reach the moon and return: Express the "measured" distance d ′ when we do not account for the atmosphere:

t = t earth's atmosphere + tout of earth's atmosphere =2

h d −h +2 v c

d −h⎞ 1 1 ⎛ h ct = c⎜ 2 + 2 ⎟ c ⎠ 2 2 ⎝ v c = h+d −h v

d' =

Solve for the length of correction d ′ − d:

⎛c ⎞ d' − d = h⎜ − 1⎟ ⎝v ⎠

Substitute numerical values and evaluate d ′ − d:

c ⎛ ⎞ d' − d = (8 km ) ⎜ − 1⎟ ⎝ 0.99997c ⎠ = 24.0 cm

Remarks: This is larger than the accuracy of the measurements, which is about 3 to 4 cm. 116 •• Picture the Problem The frequency of the waves on the wire is the same as the frequency of the tuning fork and their period is the reciprocal of the frequency. We can find the speed of the waves from the tension in the wire and its linear density. The wavelength can be determined from the frequency and the speed of the waves and the wave number from its definition. The general form of the wave function for waves on a

Wave Motion 1225 wire is y (x,t ) = A sin (kx ± ωt ) , so, once we know k and ω, because A is given, we can write a suitable wave function for the waves on this wire. The maximum speed and acceleration of a point on the wire can be found from the angular frequency and amplitude of the waves. Finally, we can use Pav = 12 µω 2 A2 v to find the average rate at which energy must be supplied to the tuning fork to keep it oscillating with a steady amplitude. (a) The frequency of the waves on the wire is the same as the frequency of the tuning fork:

f = 400 Hz

The period of the waves on the wire is the reciprocal of their frequency:

T=

(b) Relate the speed of the waves to the tension in the wire and its linear density:

v=

(c) Use the relationship between the wavelength, speed and frequency of a wave to find λ:

λ=

Using its definition, express and evaluate the wave number:

k=

(d) Determine the angular frequency of the waves:

1 1 = = 2.50 ms f 400 s −1 F

µ

1 kN = 316 m/s 0.01 kg/m

=

v 316 m/s = = 79.0 cm f 400 s −1 2π

λ

=

2π = 7.95 m −1 −2 79 × 10 m

ω = 2πf = 2π (400 s −1 ) = 2.51× 10 3 s −1

Substitute for A, k, and ω in the general form of the wave function to obtain:

y (x,t ) =

(0.50 mm)sin[(7.95 m −1 ) x − (2.51×103 s −1 ) t ]

(e) Relate the maximum speed of a point on the wire to the amplitude of the waves and the angular frequency of the tuning fork:

vmax = Aω

(

)(

= 0.5 × 10 −3 m 2.51 × 103 s −1 = 1.26 m/s

)

1226 Chapter 15 Express the maximum acceleration of a point on the wire in terms of the amplitude of the waves and the angular frequency of the tuning fork:

amax = Aω 2

(f) Express the average power required to keep the tuning fork oscillating at a steady amplitude in terms of the linear density of the wire, the amplitude of its vibrations, and the speed of the waves on the wire:

Pav = 12 µω 2 A2 v

(

)(

= 0.5 × 10 −3 m 2.51 × 103 s −1 = 3.15 × 103 m/s 2

Substitute numerical values and evaluate Pav:

Pav =

1 2

(0.1 kg/m ) (2.51 × 103 s −1 )2 (0.5 × 10 −3 m )2 (316 m/s) =

117 ••• Picture the Problem Because the chain is rolling at high speed we can neglect the effect of gravity. The diagram shows a small portion of the chain. We’ll assume that the angle θ is small even though it shown as a large angle in the diagram. Let ∆m be the mass of the segment of the chain shown. We’ll apply Newton’s 2nd law to the segment in order to relate the tension in the chain to its linear density and speed. (a) Apply

∑ Fradial = m

v2 to a R

Fnet = ∆m

v02 R

segment of the chain whose mass is ∆m: Express ∆m in terms of µ, θ, and R: Express Fnet in terms of T and θ :

∆m = µdl = µRθ Fnet = 2F sin 12 θ

Substitute to obtain:

2 F sin 12 θ = µθ v02

24.9 W

)

2

Wave Motion 1227 Solve for F:

F=

µθv02

2 sin 12 θ

sin 12 θ ≈ 12 θ :

µθv02 F = 1 = µv02 2( 2 θ )

(b) The wave speed is the same as the speed at which the chain is moving:

v0 =

Apply the small angle approximation

(c)

F

µ

As seen by an observer at rest, the pulse remains at the same position because its speed along the chain is the same as the speed of the chain. With respect to a fixed point on the chain, the pulse travels through 360°.

118 ••• Picture the Problem Let ∆m represent the mass of the segment of length ∆x = 1 mm. We can find the wave speed from the given data for the tension in the rope and its linear density. The wavelength can be found from v = fλ. We’ll use the definition of linear momentum to find the maximum transverse linear momentum of the 1-mm segment and apply Newton’s 2nd law to the segment to find the maximum net force on it. (a) Find the wave speed from the tension and linear density:

v=

F

µ

=

10 N = 10.0 m/s 0.1 kg/m

v 10 m/s = = 2.00 m f 5 s −1

(b) Express the wavelength in terms of the speed and frequency of the wave:

λ=

(c) Relate the maximum transverse linear momentum of the 1-mm segment to the maximum transverse speed of the wave:

pmax = ∆mvmax = µ∆xAω = 2πfµ∆xA

Substitute numerical values and evaluate pmax:

( ) × (1 × 10 m )(0.04 m )

pmax = 2π 5 s −1 (0.1 kg/m ) −3

= 1.26 × 10 −4 kg ⋅ m/s

1228 Chapter 15 (d) Apply

∑F

radial

v2 =m to the r

v2 A2ω 2 = µ∆x = µ∆xAω 2 A A = ωpmax = 2πfpmax

Fmax = ∆m

1-mm segment and simplify to obtain: Substitute numerical values and evaluate Fmax:

(

)(

Fmax = 2π 5 s −1 1.26 × 10−4 kg ⋅ m/s

)

= 3.96 mN

*119 ••• Picture the Problem We can relate the speed of the pulse to the tension in the rope and its linear density. Because the rope hangs vertically, the tension in it varies linearly with the distance from its bottom. Once we’ve established the result in part (a), we can integrate the resulting velocity equation to find the time for the pulse to travel the length of the rope and then double this time to get the round-trip time. (a) Relate the speed of transverse waves to tension and linear density: Express the force acting on a segment of the rope of length y: Substitute to obtain:

v=

F

µ

F = mg = µyg

v=

µyg = µ

gy

(b) Because the speed of the pulse varies with the distance from the bottom of the rope, express v as dy/dt and solve for dt:

dy 1 dy = gy and dt = dt g y

Integrate the left side of the equation from 0 to t and the right side from 0 to 3 m:

t

1 ∫0 dt ' = g

3m

∫ 0

dy y

and

t=

(

1 2 y g

)

3m 0

=

2 3m 9.81m/s 2

= 1.106 s The time for the pulse to make the round-trip is:

tround- trip = 2t = 2(1.106 s ) = 2.21s

Wave Motion 1229 120 ••• Picture the Problem We can follow the step-by-step instructions outlined above to obtain the given expressions for ∆U. (a) Express the potential energy of a segment of the string: For ∆y/∆x << 1:

∆U = F (∆l − ∆x )

[

∆l = ∆x 1 + 12 (∆y ∆x )

2

and

]

[

]

∆l − ∆x = ∆x 1 + 12 (∆y ∆x ) − ∆x =

Substitute to obtain:

∆U = F =

1 2

(∆y

2

∆x ) ∆x 2

[ (∆y ∆x ) ]

1 2

1 2

2

F (∆y ∆x ) ∆x 2

(b) Differentiate

dy = kA cos(kx − ωt ) dx

Approximate ∆y/∆x by dy/dx and substitute in our result from part (a):

∆U = 12 F (kA cos(kx − ωt )) ∆x

y ( x,t ) = A sin (kx − ωt ) to obtain:

2

=

1 2

FA2 k 2 ∆x cos 2 (kx − ωt )

1230 Chapter 15

Chapter 16 Superposition and Standing Waves Conceptual Problems *1 •• Picture the Problem We can use the speeds of the pulses to determine their positions at the given times.

2 •• Picture the Problem We can use the speeds of the pulses to determine their positions at the given times.

3 • Determine the Concept Beats are a consequence of the alternating constructive and destructive interference of waves due to slightly different frequencies. The amplitudes of the waves play no role in producing the beats. (c) is correct. 4 • (a) True. The harmonics for a string fixed at both ends are integral multiples of the frequency of the fundamental mode (first harmonic). (b) True. The harmonics for a string fixed at both ends are integral multiples of the

1231

1232

Chapter 16

frequency of the fundamental mode (first harmonic). (c) True. If l is the length of the pipe and v the speed of sound, the excited harmonics are given by f n = n

v , where n = 1, 3, 5… 4l

5 •• Determine the Concept Standing waves are the consequence of the constructive interference of waves that have the same amplitude and frequency but are traveling in opposite directions. (b) is correct. *6 • Determine the Concept Our ears and brain find frequencies which are small-integer multiples of one another pleasing when played in combination. In particular, the ear hears frequencies related by a factor of 2 (one octave) as identical. Thus, a violin sounds much more "musical" than the sound of a drum. 7 • Picture the Problem The first harmonic displacement-wave pattern in an organ pipe open at both ends and vibrating in its fundamental mode is represented in part (a) of the diagram. Part (b) of the diagram shows the wave pattern corresponding to the fundamental frequency for a pipe of the same length L that is closed at one end. Letting unprimed quantities refer to the open pipe and primed quantities refer to the closed pipe, we can relate the wavelength and, hence, the frequency of the fundamental modes using v = fλ. Express the frequency of the first harmonic in the open pipe in terms of the speed and wavelength of the waves: Relate the length of the open pipe to the wavelength of the fundamental mode:

f1 =

v

λ1

λ1 = 2 L

Superposition and Standing Waves 1233 Substitute to obtain:

f1 =

v 2L

Express the frequency of the first harmonic in the closed pipe in terms of the speed and wavelength of the waves:

f 1' =

v

Relate the length of the closed pipe to the wavelength of the fundamental mode: Substitute to obtain:

Substitute numerical values and evaluate f 1' :

λ1'

λ1' = 4 L

f 1' =

v 1⎛ v ⎞ 1 = ⎜ ⎟ = f1 4L 2 ⎝ 2L ⎠ 2

1 (400 Hz ) = 200 Hz 2 and (a ) is correct. f 1' =

8 •• Picture the Problem The frequency of the fundamental mode of vibration is directly proportional to the speed of waves on the string and inversely proportional to the wavelength which, in turn, is directly proportional to the length of the string. By expressing the fundamental frequency in terms of the length L of the string and the tension F in it we can examine the various changes in lengths and tension to determine which would halve it. Express the dependence of the frequency of the fundamental mode of vibration of the string on its wavelength: Relate the length of the string to the wavelength of the fundamental mode:

f1 =

v

λ1

λ1 = 2 L

Substitute to obtain:

f1 =

v 2L

Express the dependence of the speed of waves on the string on the tension

v=

F

µ

1234 Chapter 16 in the string: Substitute to obtain:

f1 =

1 2L

F

µ

(a) Doubling the tension and the length would increase the frequency by a factor of 2 2 . (b) Halving the tension and keeping the length fixed would decrease the frequency by a factor of 1

2.

(c) Keeping the tension fixed and halving the length would double the frequency.

(c) is correct. 9 •• Determine the Concept We can relate the resonant frequencies of an organ pipe to the speed of sound in air and the speed of sound to the absolute temperature. Express the dependence of the resonant frequencies on the speed of sound:

f =

Relate the speed of sound to the temperature of the air:

v=

v

λ

γRT

M where γ and R are constants, M is the molar mass of the gas (air), and T is the absolute temperature.

Substitute to obtain:

f =

1 γRT λ M

Because v ∝ T , increasing the temperature increases the resonant frequencies. *10 • Determine the Concept Because the two waves move independently, neither impedes the progress of the other.

Superposition and Standing Waves 1235 11 • Determine the Concept No; the wavelength of a wave is related to its frequency and speed of propagation (λ = v/f). The frequency of the plucked string will be the same as the wave it produces in air, but the speeds of the waves depend on the media in which they are propagating. Because the velocities of propagation differ, the wavelengths will not be the same. 12 • Determine the Concept No; when averaged over a region in space including one or more wavelengths, the energy is unchanged. 13 • Determine the Concept When the edges of the glass vibrate, sound waves are produced in the air in the glass. The resonance frequency of the air columns depends on the length of the air column, which depends on how much water is in the glass. 14 •• Picture the Problem We can use v = fλ to relate the frequency of the sound waves in the organ pipes to the speed of sound in air, nitrogen, and helium. We can use v = γRT M to relate the speed of sound, and hence its frequency, to the properties of the three gases. Express the frequency of a given note as a function of its wavelength and the speed of sound:

f =

Relate the speed of sound to the absolute temperature and the molar mass of the gas used in the organ:

v=

v

λ

γRT

M where γ depends on the kind of gas, R is a constant, T is the absolute temperature, and M is the molar mass.

1 γRT λ M

Substitute to obtain:

f =

For air in the organ pipes we have:

f air =

1 γ air RT λ M air

(1)

When nitrogen is in the organ pipes:

f N2 =

1 γ N2 RT λ M N2

(2)

1236 Chapter 16 Express the ratio of equation (2) to equation (1) and solve for f N 2 :

f N2 f air

=

γ N M air γ air M N 2

2

and

f N2 = f air

γ N M air γ air M N 2

2

Because γ N2 = γ air and M air > M N2 :

f N2 > f air i.e.,

f will increase for each organ pipe. If helium were used, we’d have:

f He = f air

Because γ He > γ air and M air >> M He :

f He >> f air i.e.,

γ He M air γ air M He

the effect will be even more pronounced.

*15 •• Determine the Concept Increasing the tension on a piano wire increases the speed of the waves. The wavelength of these waves is determined by the length of the wire. Because the speed of the waves is the product of their wavelength and frequency, the wavelength remains the same and the frequency increases. (b) is correct. 16 •• Determine the Concept If connected properly, the speakers will oscillate in phase and interfere constructively. If connected incorrectly, they interfere destructively. It would be difficult to detect the interference if the wavelength is short, less than the distance between the ears of the observer. Thus, one should use bass notes of low frequency and long wavelength. 17 •• Determine the Concept The pitch is determined mostly by the resonant cavity of the mouth; the frequency of sounds he makes is directly proportional to their speed. Because vHe > vair (see Equation 15-5), the resonance frequency is higher if helium is the gas in the cavity.

Superposition and Standing Waves 1237 *18 •• Determine the Concept The light is being projected up from underneath the silk, so you will see light where there is a gap and darkness where two threads overlap. Because the two weaves have almost the same spatial period but not exactly identical (because the two are stretched unequally), there will be places where, for large sections of the cloth, the two weaves overlap in phase, leading to brightness, and large sections where the two overlap 90° out of phase (i.e., thread on gap and vice versa) leading to darkness. This is exactly the same idea as in the interference of two waves.

Estimation and Approximation 19 •• Determine the Concept Pianos are tuned by ringing the tuning fork and the piano note simultaneously and tuning the piano string until the beats are far apart; i.e., the time between beats is very long. If we assume that 2 s is the maximum detectable period for the beats, then one should be able to tune the piano string to at least 0.5 Hz. *20 • Picture the Problem We can use v = f1λ1 to express the resonance frequencies in the organ pipes in terms of their wavelengths and L = n

λn 2

, n = 1, 2, 3, ... to relate the length

of the pipes to the resonance wavelengths.

v

(a) Relate the fundamental frequency of the pipe to its wavelength and the speed of sound:

f1 =

Express the condition for constructive interference in a pipe that is open at both ends:

L=n

Solve for λ1:

λ1 = 2 L

Substitute and evaluate f1:

f1 =

(b) Relate the resonance frequencies of the pipe to their wavelengths and the speed of sound:

fn =

Solve equation (2) for λn:

λn =

λ1 λn 2

, n = 1, 2, 3, ...

(1)

340 m/s v = = 2.27 kHz 2 L 2(7.5 × 10 −2 m )

v

λn

2L n

1238 Chapter 16 Substitute to obtain:

fn = n

v 340 m/s =n 2L 2 7.5 × 10 −2 m

(

= n(2.27 kHz ) Set fn = 20 kHz and evaluate n:

n=

)

20 kHz = 8.81 2.27 kHz

The eighth harmonic is within the range defined as audible. The ninth harmonic might be heard by a person with very good hearing. 21 •• Picture the Problem Assume a pipe length of 5 m and apply the standing-wave resonance frequencies condition for a pipe that is open at both ends (the same conditions hold for a string that is fixed at both ends). Relate the resonance frequencies for a pipe open at both ends to the length of the pipe:

fn = n

Evaluate this expression for n = 1:

f1 =

Express the dependence of the speed of sound in a gas on the temperature:

v=

v , n = 1, 2, 3, ... 2L

340 m/s = 34.0 Hz 2(5 m )

γRT

M where γ and R are constants, M is the molar mass, and T is the absolute temperature.

Because v ∝ T , the frequency will be somewhat higher in the summer.

Superposition and Interference 22 • Picture the Problem We can use A = 2 y0 cos 12 δ to find the amplitude of the resultant wave. (a) Evaluate the amplitude of the resultant wave when δ = π/6:

1 ⎛π ⎞ A = 2 y0 cos 12 δ = 2(0.02 m ) cos ⎜ ⎟ 2⎝ 6⎠ = 3.86 cm

Superposition and Standing Waves 1239

1 ⎛π ⎞ A = 2 y0 cos 12 δ = 2(0.02 m ) cos ⎜ ⎟ 2⎝ 3⎠

(b) Proceed as in (a) with δ = π/3:

= 3.46 cm 23 • Picture the Problem We can use A = 2 y0 cos 12 δ to find the amplitude of the resultant wave.

1 ⎛π ⎞ A = 2 y0 cos 12 δ = 2(0.05 m ) cos ⎜ ⎟ 2⎝ 2⎠

Evaluate the amplitude of the resultant wave when δ = π/2:

= 7.07 cm *24 • Picture the Problem The phase shift in the waves generated by these two sources is due to their separation of λ/3. We can find the phase difference due to the path difference from δ = 2π

∆x

λ

and then the amplitude of the resultant wave from A = 2 y0 cos 12 δ .

∆x

λ3 2 = π λ 3

Evaluate the phase difference δ:

δ = 2π

Find the amplitude of the resultant wave:

1⎛2 ⎞ Ares = 2 y0 cos 12 δ = 2 A cos ⎜ π ⎟ 2⎝3 ⎠

λ

= 2π

= 2 A cos

π 3

= A

25 • Picture the Problem The phase shift in the waves generated by these two sources is due to a path difference ∆x = 5.85 m – 5.00 m = 0.85 m. We can find the phase difference due to this path difference from δ = 2π from A = 2 y 0 cos 12 δ .

∆x

λ

and then the amplitude of the resultant wave

(a) Find the phase difference due to the path difference:

δ = 2π

Calculate the wavelength of the sound waves:

λ=

∆x

λ

v 340 m/s = = 3.4 m 100 s −1 f

1240 Chapter 16 Substitute and evaluate δ:

δ = 2π

(b) Relate the amplitude of the resultant wave to the amplitudes of the interfering waves and the phase difference between them:

0.85 m π = rad = 90.0° 3.4 m 2

1 ⎛π ⎞ A = 2 y0 cos 12 δ = 2 A cos ⎜ ⎟ 2⎝ 2⎠ =

2A

*26 • Picture the Problem The diagram is shown below. Lines of constructive interference are shown for path differences of 0, λ, 2λ, and 3λ.

27 • Picture the Problem The intensity at the point of interest is dependent on whether the speakers are coherent and on the total phase difference in the waves arriving at the given point. We can use δ = 2π

∆x

λ

to determine the phase difference δ, A = 2 p0 cos 12 δ to

find the amplitude of the resultant wave, and the fact that the intensity I is proportional to the square of the amplitude to find the intensity at P for the given conditions.

λ =π λ

(a) Find the phase difference δ:

δ = 2π

Find the amplitude of the resultant wave:

A = 2 p0 cos 12 π = 0

Because the intensity is proportional to A2:

I= 0

(b) The sources are incoherent and

I = 2I0

1 2

Superposition and Standing Waves 1241 the intensities add:

δ tot = δ sources + δ path difference

(c) Express the total phase difference:

= π + 2π

∆x

⎛1⎞ = π + 2π ⎜ ⎟ λ ⎝2⎠

= 2π Find the amplitude of the resultant wave: Because the intensity is proportional to A2:

A = 2 p0 cos 12 (2π ) = 2 p0

(2 p0 ) I = 4 I A2 I = 0 0 2 0 p0 p02 2

I=

28 • Picture the Problem The intensity at the point of interest is dependent on whether the speakers are coherent and on the total phase difference in the waves arriving at the given point. We can use δ = 2π

∆x

λ

to determine the phase difference δ, A = 2 p0 cos 12 δ to

find the amplitude of the resultant wave, and the fact that the intensity I is proportional to the square of the amplitude to find the intensity at P for the given conditions. (a) Find the phase difference δ:

Find the amplitude of the resultant wave:

δ = 2π

λ = 2π λ

A = 2 p0 cos 12 (2π ) = 2 p0

(2 p0 ) I = 4 I A2 I = 0 0 2 0 p0 p02 2

Because the intensity is proportional to A2:

I=

(b) The sources are incoherent and the intensities add:

I = 2I 0

(c) Express the total phase difference:

δ tot = δ sources + δ path difference = π + 2π

⎛λ ⎞ = π + 2π ⎜ ⎟ λ ⎝λ ⎠

∆x

= 3π Find the amplitude of the resultant

A = 2 p0 cos 12 (3π ) = 0

1242 Chapter 16 wave: Because the intensity is proportional to A2:

I= 0

29 • Picture the Problem Let P be the point located a distance r1 from speaker 1 and a distance r2 from speaker 2. If the sound at point P is to be either a maximum or a minimum, the difference in the distances to the speakers will have to be such that this difference compensates for the 90° out-of-phase condition of the speakers.

δ sources

(a) Express the phase shift due to the speakers in terms of a path difference:

∆r =

Express the condition that r2 − r1 must

r2 − r1 =

1 4

λ

r1 − r2 =

1 4

λ

360°

λ=

90° λ = 14 λ 360°

satisfy in order to compensate for this path difference: (b) In this case, the smallest difference in path is again λ/4, but now: *30 •• Picture the Problem The drawing shows a generic point P located a distance r1 from source S1 and a distance r2 from source S2. The sources are separated by a distance d and we’re given that d < λ/2. Because the condition for destructive interference is that δ = nπ where n = 1, 2, 3,..., we’ll show that, with d < λ/2, this condition cannot be satisfied.

∆r

Relate the phase shift to the path difference and the wavelength of the sound:

δ = 2π

Relate ∆r to d and θ:

∆r < d sin θ ≤ d

Substitute to obtain:

δ < 2π

λ

d sin θ

λ

≤ 2π

d

λ

Superposition and Standing Waves 1243 Because d < λ 2 :

Express the condition for destructive interference:

δ < 2π

λ 2 =π λ

δ = nπ where n = 1, 2, 3,…

Because δ < π, there is no complete destructive interference in any direction. 31 •• Picture the Problem Let the positive x direction be the direction of propagation of the wave. We can express the phase difference in terms of the separation of the two points and the wavelength of the wave and solve for λ. In part (b) we can find the phase difference by relating the time between displacements to the period of the wave. I in part (c) we can use the relationship between the speed, frequency, and wavelength of a wave to find its velocity.

∆x

(a) Relate the phase difference to the wavelength of the wave:

δ = 2π

Solve for and evaluate λ:

λ = 2π

(b) Express and evaluate the period of the wave:

T=

Relate the time between the two displacements to the period of the wave:

1 5 ms = T 5

Express the phase difference corresponding to one-fifth of a period:

δ=

(c) Express the wave speed in terms of its frequency and wavelength:

λ ∆x

δ

= 2π

5 cm = 60.0 cm π 6

1 1 = = 25 ms f 40 s −1

2π 5

(

)

v = fλ = 40 s −1 (0.6 m ) = 24.0 m/s

32 •• Picture the Problem Assume a distance of about 20 cm between your ears. When you rotate your head through 90°, you introduce a path difference of 20 cm. We can apply the equation for the phase difference due to a path difference to determine the change in phase between the sounds received by your ears as you rotate your head through 90°.

1244 Chapter 16 Express the phase difference due to the rotation of your head through 90°:

δ = 2π

Find the wavelength of the sound waves:

λ=

Substitute to obtain:

δ = 2π

20 cm

λ

v 340 m/s = = 50 cm f 680 s −1 20 cm = 0.8π rad 50 cm

33 •• Picture the Problem Because the sound intensity diminishes as the observer moves, parallel to a line through the sources, away from her initial position, we can conclude that her initial position is one at which there is constructive interference of the sound coming from the two sources. We can apply the condition for constructive interference to relate the wavelength of the sound to the path difference at her initial position and the relationship between the velocity, frequency, and wavelength of the waves to express this path difference in terms of the frequency of the sources. Express the condition for constructive interference at (40 m, 0):

∆r = nλ , n = 1, 2, 3, ...

(1)

Express the path difference ∆r:

∆r = rB − rA

Using the Pythagorean theorem, find rB:

rB =

(40 m )2 + (2.4 m )2

Substitute for rB and evaluate ∆r:

∆r =

(40 m )2 + (2.4 m )2 − 40 m

= 0.07194 m 0.07194 m n

Substitute in equation (1) and solve for λ:

λ=

Using v = fλ, express f in terms of λ and n:

fn = n

Evaluate f for n = 1 and 2:

f1 = 4726 Hz and f 2 = 9452 Hz

v 340 m/s =n 0.07194 m 0.07194 m

= (4726 Hz )n

Superposition and Standing Waves 1245 34 •• Picture the Problem Because the sound intensity increases as the observer moves, parallel to a line through the sources, away from her initial position, we can conclude that her initial position is one at which there is destructive interference of the sound coming from the two sources. We can apply the condition for destructive interference to relate the wavelength of the sound to the path difference at her initial position and the relationship between the velocity, frequency, and wavelength of the waves to express this path difference in terms of the frequency of the sources.

λ

Express the condition for destructive interference at (40 m, 0):

∆r = n , n = 1, 3, 5, ... 2

Express the path difference ∆r:

∆r = rB − rA

Using the Pythagorean theorem, find rB:

rB =

(40 m )2 + (2.4 m )2

Substitute for rB and evaluate ∆r:

∆r =

(40 m )2 + (2.4 m )2 − 40 m

(1)

= 0.07194 m 2(0.07194 m ) 0.1439 m = n n

Substitute in equation (1) and solve for λ:

λ=

Using v = fλ, express f in terms of λ:

fn = n

Evaluate f for n = 1 and 3:

f1 = 2363 Hz

v 340 m/s =n 0.1439 m 0.1439 m = (2363 Hz )n

and

f 3 = 7089 Hz *35 •• Picture the Problem We can use the trigonometric identity

⎛ A+ B⎞ ⎛ A− B⎞ cos A + cos B = 2 cos⎜ ⎟ cos⎜ ⎟ to derive the expression given in (a) and the ⎝ 2 ⎠ ⎝ 2 ⎠ speed of the envelope can be found from the second factor in this expression; i.e., from cos[(∆k / 2)x − (∆ω / 2) t ] . (a) Express the amplitude of the resultant wave function y(x,t):

y ( x, t ) = A(cos(k1 x − ω1t )+ cos(k 2 x − ω2t ))

1246 Chapter 16 ⎛ A+ B⎞ ⎛ A− B⎞ ⎟ cos⎜ ⎟ to obtain: ⎝ 2 ⎠ ⎝ 2 ⎠

Use the trigonometric identity cos A + cos B = 2 cos⎜

k x − ω1t + k 2 x − ω 2 t k x − ω1t − k 2 x + ω 2 t ⎤ ⎡ cos 1 y ( x,t ) = 2 A⎢cos 1 ⎥ 2 2 ⎣ ⎦ ⎡ ⎛ k + k2 ω + ω 2 ⎞ ⎛ k1 − k 2 ω − ω1 ⎞⎤ x− 1 t ⎟cos⎜ x+ 2 t ⎟⎥ = 2 A⎢cos⎜ 1 2 2 ⎠ ⎝ 2 ⎠⎦ ⎣ ⎝ 2 Substitute ωave = (ω1 + ω2)/2, kave = (k1 + k2)/2, ∆ω = ω1 - ω2 and ∆k = k1 − k2 to obtain:

∆ω ⎞⎤ ⎛ ∆k y ( x,t ) = 2 A[cos(k ave x − ω ave t )cos⎜ x− t⎟ 2 ⎠⎥⎦ ⎝ 2 (b) A spreadsheet program to calculate y(x,t) between 0 m and 50 m at t = 0, 0.5 s, and 1 s follows. The constants and cell formulas used are shown in the table. Cell B11 C10

Content/Formula Algebraic Form B10+0.25 x + ∆x COS($B$3*B10−$B$5*$C$9) y ( x,0 ) + COS($B$4*B10−$B$6*$C$9) D10 COS($B$3*B10-$B$5*$D$9) y ( x,0.5 s ) + COS($B$4*B10−$B$6*$D$9) E10 COS($B$3*B10−$B$5*$E$9) y ( x,1s ) + COS($B$4*B10−$B$6*$E$9)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 206 207 208 209 210

A

B

k1= k2= w1= w2=

1 0.8 1 0.9 x (m)

C

D

E

m−1 m−1 rad/s rad/s y(x,0) y(x,0.5 s) y(x,1 s)

0.00 0.25 0.50 0.75 1.00

0.000 2.000 1.949 1.799 1.557 1.237

2.000 −0.643 −0.207 0.241 0.678 1.081

4.000 −1.550 −1.787 −1.935 −1.984 −1.932

49.00 49.25 49.50 49.75 50.00

0.370 0.397 0.397 0.364 0.298

−0.037 0.003 0.065 0.145 0.237

0.021 −0.024 −0.075 −0.124 −0.164

Superposition and Standing Waves 1247 The solid line is the graph of y(x,0), the dashed line that of y(x,0.5 s), and the dotted line is the graph of y(x,1 s). 2.0 1.5

f(x,0)

1.0

f(x,0.5 s)

y (x ,t )

0.5

f(x,1 s) 0.0 0

5

10

15

20

25

30

35

40

45

50

-0.5 -1.0 -1.5 -2.0

x (m)

(c) Express the speed of the envelope:

venvelope =

∆ω ω1 − ω 2 = ∆k k1 − k 2

Substitute numerical values and evaluate venvelope:

venvelope =

1 rad/s − 0.9 rad/s = 0.500 m/s 1 m −1 − 0.8 m −1

36 •• Picture the Problem The diagram shows the two sources separated by a distance d and the path difference ∆s. Because the lines from the sources to the distant point are approximately parallel, the triangle shown in the diagram is approximately a right triangle and we can use trigonometry to express ∆s in terms of d and θ. In the second part of the problem, we can apply a small-angle approximation to the larger triangle shown in Figure 16-29 to relate ym to D and θ and then use the condition for constructive interference to relate ym to D, λ, and d.

∆s and ∆s ≈ d sin θ d

(a) Using the diagram, relate ∆s to the separation of the sources and the angle θ:

sin θ ≈

(b) For θ << 1, we can approximate

∆s ≈ d tan θ

1248 Chapter 16 sinθ with tanθ to obtain:

ym D

Referring to Figure 16-29, express tanθ in terms of y and D:

tan θ ≈

Substitute to obtain:

∆s ≈

Express the condition on the phase difference for constructive interference:

δ = 2π

Substitute for ∆s:

2π

Simplify and solve for ym:

ym = m

dym D ∆s

λ

= 2πm, m = 1, 2, 3, ...

dy m = 2πm, m = 1, 2, 3, ... Dλ Dλ d

37 •• Picture the Problem Because a maximum is heard at 0° and the sources are in phase, we can conclude that the path difference is 0. Because the next maximum is heard at 23°, the path difference to that position must be one wavelength. We can use the result of part (a) of Problem 36 to relate the separation of the sources to the path difference and the angle θ. We’ll apply the condition for constructive interference to determine the angular locations of other points of maximum intensity in the interference pattern. Using the result of part (a) of Problem 36, express the separation of the sources in terms of ∆s and θ : Evaluate d with ∆s = λ and θ = 23°:

d=

∆s sin θ

λ

v sin 23° f sin 23° 340 m/s = = 1.81 m 480 s −1 sin23°

d=

=

(

Express the condition for additional intensity maxima:

)

d sin θ m = mλ where m = 1, 2, 3, …, or

⎡ mλ ⎤ ⎣ d ⎥⎦

θ m = sin −1 ⎢

Superposition and Standing Waves 1249 Evaluate this expression for m = 2:

2(340 m/s ) ⎤ ⎥ = 51.5° −1 ⎣ 480 s (1.81 m ) ⎦ ⎡

θ 2 = sin −1 ⎢

(

)

Remarks: It is easy to show that, for m > 2, the inverse sine function is undefined and that, therefore, there are no additional relative maxima at angles larger than 51.5°. *38 ••• Picture the Problem Because the speakers are driven in phase and the path difference is 0 at her initial position, the listener will hear a maximum at (D, 0). As she walks along a line parallel to the y axis she will hear a minimum wherever it is true that the path difference is an odd multiple of a half wavelength. She will hear an intensity maximum wherever the path difference is an integral multiple of a wavelength. We’ll apply the condition for destructive interference in part (a) to determine the angular location of the first minimum and, in part (b), the condition for constructive interference find the angle at which she’ll hear the first maximum after the one at 0°. In part (c), we can apply the condition for constructive interference to determine the number of maxima she can hear as keeps walking parallel to the y axis. (a) Express the condition for destructive interference:

d sin θ m = m

λ 2

where m = 1, 3, 5,…, or

⎛ mλ ⎞ ⎟ ⎝ 2d ⎠

θ m = sin −1 ⎜ Evaluate this expression for m = 1:

⎡ 340 m/s ⎤ ⎛ v ⎞ ⎟⎟ = sin −1 ⎢ ⎥ −1 ⎝ 2 fd ⎠ ⎣ 2 600 s (2 m ) ⎦

θ1 = sin −1 ⎜⎜

(

)

= 8.14° (b) Express the condition for additional intensity maxima:

d sin θ m = mλ where m = 0, 1, 2, 3,…, or

⎛ mλ ⎞ ⎟ ⎝ d ⎠

θ m = sin −1 ⎜ Evaluate this expression for m = 1:

⎡ 340 m/s ⎤ ⎛ v ⎞ ⎟⎟ = sin −1 ⎢ ⎥ −1 ⎝ fd ⎠ ⎣ 600 s (2 m ) ⎦

θ1 = sin −1 ⎜⎜

= 16.5°

(

)

1250 Chapter 16

λ

(c) Express the limiting condition on sinθ :

sin θ m = m

Solve for m to obtain:

600 s −1 (2 m ) fd = = 3.53 m≤ = 340 m/s λ v

Because m must be an integer:

m= 3

d

≤1

(

d

)

39 ••• Picture the Problem Let d be the separation of the two sound sources. We can express the wavelength of the sound in terms of the d and either of the angles at which intensity maxima are heard. We can find the frequency of the sources from its relationship to the speed of the waves and their wavelengths. Using the condition for constructive interference, we can find the angles at which intensity maxima are heard. Finally, in part (d), we’ll use the condition for destructive interference to find the smallest angle for which the sound waves cancel. (a) Express the condition for constructive interference: Solve for λ:

Evaluate λ for m = 1:

d sin θ m = mλ

(1)

where m = 0, 1, 2, 3,…

λ=

d sin θ m m

λ = (2 m )sin (0.140 rad ) = 0.279 m

(b) Express the frequency of the sound in terms of its wavelength and speed: (c) Solve equation (1) for θm:

f =

v

λ

=

340 m/s = 1.22 kHz 0.279 m ⎛ mλ ⎞ −1 ⎡ m(0.279 m ) ⎤ ⎟ = sin ⎢ ⎥ 2m ⎝ d ⎠ ⎦ ⎣

θ m = sin −1 ⎜

= sin −1 [(0.1395)m]

The table shows the values for θ as a function of m:

m

θm

3 4 5

(rad) 0.432 0.592 0.772

Superposition and Standing Waves 1251 6 7 8 (d) Express the condition for destructive interference:

d sin θ m = m

0.992 1.354 undefined

λ 2

where m = 1, 3, 5,… Solve for θm:

Evaluate this expression for m = 1:

⎛ ⎝

θ m = sin −1 ⎜ m

λ ⎞

⎟ 2d ⎠

⎡ 0.279 m ⎤ ⎥ = 0.0698 rad ⎣ 2(2 m ) ⎦

θ1 = sin −1 ⎢

40 ••• Picture the Problem The total phase shift in the waves arriving at the points of interest is the sum of the phase shift due to the difference in path lengths from the two sources to a given point and the phase shift due to the sources being out of phase by 90°. From Problem 39 we know that λ = 0.279 m. Using the conditions on the path difference ∆x for constructive and destructive interference, we can find the angles at which intensity maxima are heard. Letting the subscript ″pd ″ denote ″path difference″ and the subscript ″s″ the ″sources″, express the total phase shift δ:

δ = δ pd + δ s = 2π

∆x

λ

+

π 4

where ∆x is the path difference between the two sources and the points at which constructive or destructive interference is heard.

Express the condition for constructive interference:

δ = 2π

Solve for ∆x to obtain:

∆x =

∆x

λ

+

π 4

= 2π, 4π,6π , ...

(8m − 1) λ 7 15 23 λ, λ, λ,... = 8 8 8 8

where m = 1, 2, 3,… Relate ∆x to d to obtain:

∆x =

(8m − 1) λ = d sin θ 8

c

where the ″c″ denotes constructive interference.

1252 Chapter 16 Solve for θc:

⎡ (8m − 1)λ ⎤ ⎥⎦ , m = 1, 2, 3,… ⎣ 8d

θ c = sin −1 ⎢

The table shows the values for θc for m = 1 to 5:

∆x

θc

m 1

7.01°

2

15.2°

3

23.6°

4

35.1°

5

42.8°

π

Express the condition for destructive interference:

δ = 2π

Solve for ∆x to obtain:

(8m − 5) λ 3 11 19 ∆x = λ, λ, λ,... = 8 8 8 8

λ

+

4

= π,3π,5π, ...

where m = 1, 2, 3,… Letting ″d ″ denotes destructive interference, relate ∆x to d to obtain: Solve for θd:

∆x =

(8m − 5) λ = d sin θ 8

d

⎡ (8m − 5)λ ⎤ ⎥⎦ , m = 1, 2, 3,… ⎣ 8d

θ d = sin −1 ⎢

The table shows the values for θd for m = 1 to 5:

θd

m 1

3.00°

2

11.1°

3

19.3°

4

28.1°

5

37.6°

41 ••• Picture the Problem We can calculate the required phase shift from the path difference and the wavelength of the radio waves using δ = 2π

Express the phase delay as a function of the path difference and

∆s

δ = 2π

λ ∆s

λ

.

(1)

Superposition and Standing Waves 1253 the wavelength of the radio waves: Find the wavelength of the radio waves: Express the path difference for the signals coming from an angle θ with the vertical:

λ=

v 3 × 108 m/s = = 15 m f 20 × 106 s −1

∆s = d sin θ

Substitute numerical values and evaluate ∆s:

∆s = (200 m )sin10° = 34.73 m = 2.315λ = 2λ + 0.315λ

Substitute in equation (1) and evaluate δ:

δ = 2π

0.315λ

λ

= 1.98 rad = 113°

Beats 42 • Picture the Problem The beat frequency is the difference between the frequency of the tuning fork and the frequency of the violin string. Let f2 = 500 Hz. (a) Express the relationship between the beat frequency of the frequencies of the two tuning forks: Solve for f2:

(b)

f 2 = f1 ± ∆f = 500 Hz ± 4 Hz

f 2 = 504 Hz or 496 Hz

If the beat frequency is increased, then f 2 = 504 Hz; if it is diminished, f 2 = 496 Hz.

43 •• Picture the Problem The Doppler shift of the siren as heard by one of the drivers is given by the formula for source and receiver both moving and approaching each other f r = f s [(1 + u / v ) (1 − u / v )] , where u is the speed of the ambulance and v is the speed of sound. (a) Express the beat frequency:

f beat = f r − f s where fr is the frequency heard by either driver due to the other’s siren,

1254 Chapter 16 Express fr:

u v fr = fs u 1− v 1+

Substitute to obtain:

f beat

Substitute numerical values and evaluate fbeat:

(b)

u u ⎛ ⎞ ⎜1+ ⎟ v −f = f ⎜ v − 1⎟ = fs s s u u ⎜ ⎟ 1− ⎜1− ⎟ v v ⎝ ⎠ 2 = fs v −1 u 1+

f beat = (500 Hz )

2 = 70.5 Hz 340 m/s −1 22.4 m/s

The person on the street hears no beat frequency as the sirens of both ambulances are Doppler shifted up by the same amount (approx. 35 Hz).

Standing Waves *44 • Picture the Problem We can use v = fλ to relate the second-harmonic frequency to the wavelength of the standing wave for the second harmonic. Relate the speed of transverse waves on the string to their frequency and wavelength: Express λ2 in terms of the length L of the string: Substitute for λ2 and evaluate v:

v = f 2 λ2

λ2 = L v = f 2 L = (60 s −1 )(3 m ) = 180 m/s

45 • Picture the Problem We can find the wavelength of this standing wave from the standing-wave condition for a string fixed at both ends and its frequency from v = f3λ3. We can use the wave function for a standing wave on a string fixed at both ends ( y n ( x,t ) = An sin k n x cos ω n t )) to write the wave function for the wave described in this problem.

Superposition and Standing Waves 1255 (a) Using the standing-wave condition for a string fixed at both ends, relate the length of the string to the wavelength of the harmonic mode in which it is vibrating: Solve for λ3:

Express the frequency of the third harmonic in terms of the speed of transverse waves on the string and their wavelength: (b) Write the equation for a standing wave, fixed at both ends, in its third harmonic:

L=n

λn 2

, n = 1, 2, 3, ...

λ3 =

2 2 L = (3 m ) = 2.00 m 3 3

f3 =

v

λ3

=

50 m/s = 25.0 Hz 2m

y3 ( x, t ) = A3 sin k3 x cos ω3t

2π

2π = π m −1 2m

Evaluate k3:

k3 =

Evaluate ω3:

ω3 = 2πf 3 = 2π (25 s −1 ) = 50π s −1

Substitute to obtain:

λ3

=

y 3 ( x,t ) = (4 mm )sin kx cos ω t where k = π m−1 and

ω = 50π s−1. 46 • Picture the Problem The first harmonic displacement-wave pattern in an organ pipe open at both ends and vibrating in its fundamental mode is represented in part (a) of the diagram. Part (b) of the diagram shows the wave pattern corresponding to the fundamental frequency for a pipe of the same length L that is closed at one end. We can relate the wavelength to the frequency of the fundamental modes using v = fλ. (a) Express the dependence of the frequency of the fundamental mode of vibration in the open pipe on its

f1,open =

v

λ1,open

1256 Chapter 16 wavelength: Relate the length of the open pipe to the wavelength of the fundamental mode:

λ1,open = 2 L

Substitute and evaluate f1,open:

f1,open =

(b) Express the dependence of the frequency of the fundamental mode of vibration in the closed pipe on its wavelength:

f1,closed =

Relate the length of the closed pipe to the wavelength of the fundamental mode: Substitute to obtain:

v 340 m/s = = 17.0 Hz 2 L 2(10 m ) v

λ1,closed

λ1,closed = 4 L

f1,closed =

v 340 m/s = = 8.50 Hz 4 L 4(10 m )

47 • Picture the Problem We can find the speed of transverse waves on the wire using

v = F µ and the wavelengths of any harmonic from L = n

λn 2

, n = 1, 2, 3, ... . We can

use v = fλ to find the frequency of the fundamental. For a wire fixed at both ends, the higher harmonics are integer multiples of the first harmonic (fundamental). (a) Relate the speed of transverse waves on the wire to the tension in the wire and its linear density:

v=

F

Substitute numerical values and evaluate v:

v=

968 N = 521 m/s (0.005 kg ) (1.4 m )

(b) Using the standing-wave condition for a wire fixed at both ends, relate the length of the wire to the wavelength of the harmonic mode in which it is vibrating: Solve for λ1:

L=n

µ

λn 2

=

F m L

, n = 1, 2, 3, ...

λ1 = 2 L = 2(1.4 m ) = 2.80 m

Superposition and Standing Waves 1257

Express the frequency of the first harmonic in terms of the speed and wavelength of the waves: (c) Because, for a wire fixed at both ends, the higher harmonics are integer multiples of the first harmonic:

48

f1 =

v

=

λ1

521 m/s = 186 Hz 2.80 m

f 2 = 2 f1 = 2(186 Hz ) = 372 Hz and

f 3 = 3 f1 = 3(186 Hz ) = 558 Hz

•

Picture the Problem We can use Equation 16-13, f n = n

v = nf1,n = 1, 3, 5, ..., to find 4L

the resonance frequencies for a rope that is fixed at one end. (a) Using the resonance-frequency condition for a rope fixed at one end, relate the resonance frequencies to the speed of the waves and the length of the rope:

fn = n

Solve for f1:

f1 =

(b)

v = nf1 , n = 1, 3, 5, ... 4L

20 m/s = 1.25 Hz 4(4 m )

Because this rope is fixed at just one end, the system does not support a second harmonic.

(c) For the third harmonic, n = 3:

f 3 = 3 f1 = 3(1.25 Hz ) = 3.75 Hz

49 • Picture the Problem We can find the fundamental frequency of the piano wire using the general expression for the resonance frequencies of a wire fixed at both ends,

fn = n

v = nf1 , n = 1, 2, 3, ... , with n = 1. We can use v = F µ to express the 2L

frequencies of the fundamentals of the two wires in terms of their linear densities. Relate the fundamental frequency of the piano wire to the speed of transverse waves on it and its linear density:

f1 =

v 2L

1258 Chapter 16 Express the dependence of the speed of transverse waves on the tension and linear density:

v=

F

Substitute to obtain:

f1 =

1 2L

Doubling the linear density results in a new fundamental frequency f ′ given by:

f1' =

F 1 1 ⎛ 1 ⎜ = 2 L 2µ 2 ⎜⎝ 2 L

Substitute for f1 to obtain:

f1' =

µ

1 2

F

µ

(200 Hz ) =

F⎞ 1 ⎟= f1 µ ⎟⎠ 2

141 Hz

*50 • Picture the Problem Because the frequency and wavelength of sounds waves are inversely proportional, the greatest length of the organ pipe corresponds to the lowest frequency in the normal hearing range. We can relate wavelengths to the length of the pipes using the expressions for the resonance frequencies for pipes that are open at both ends and open at one end. Find the wavelength of a 20-Hz note:

(a) Relate the length L of a closedat-one-end organ pipe to the wavelengths of its standing waves: Solve for and evaluate λ1:

(b) Relate the length L of an open organ pipe to the wavelengths of its standing waves: Solve for and evaluate λ1:

v

λmax =

L=n

L=

λn 4

λmax 4

L=n

L=

f lowest

λn 2

λmax 2

=

340 m/s = 17 m 20 s −1

, n = 1, 3, 5, ...

=

17 m = 4.25 m 4

, n = 1, 2, 3, ...

=

17 m = 8.50 m 2

51 •• Picture the Problem We can find λ and f by comparing the given wave function to the general wave function for a string fixed at both ends. The speed of the waves can then be

Superposition and Standing Waves 1259 found from v = fλ. We can find the length of the string from its fourth harmonic wavelength.

2π

(a) Using the wave function, relate k and λ:

k=

Solve for λ:

λ=

Using the wave function, relate f and ω:

ω = 2πf = 300 s −1

Solve for f:

f =

(b) Express the speed of transverse waves in terms of their frequency and wavelength: (c) Relate the length of the string to the wavelengths of its standingwave patterns: Solve for L when n = 4:

λ

= 0.20 cm −1

2π = 10π cm = 31.4 cm 0.20 cm −1

300 s −1 = 47.7 Hz 2π

v = fλ = (47.7 Hz )(0.314 m )

= 15.0 m/s

L=n

λn 2

, n = 1, 2, 3, ...

L = 2λ4 = 2(31.4 cm ) = 62.8 cm

52 •• Picture the Problem We can find λ and f by comparing the given wave function to the general wave function for a string fixed at both ends. The speed of the waves can then be found from v = fλ. In a standing wave pattern, the nodes are separated by one-half wavelength. (a) Express the speed of the traveling waves in terms of their frequency and wavelength:

v = fλ

2π

Using the wave function, relate k and λ:

k=

Solve for λ:

λ=

Using the wave function, relate ω

ω = 2πf = 500 s −1

λ

= 2.5 m −1

2π = 0.8π m = 2.51m 2.5 m −1

1260 Chapter 16 and f: Solve for f:

Substitute to find v:

Express the amplitude of the standing wave in terms of the amplitude of the two traveling waves that result in the standing wave: Solve for and evaluate A:

(b) The distance between nodes is half the wavelength: (c) Because there is a standing wave on the string, the shortest possible length is:

f =

500 s −1 = 79.6 Hz 2π

(

)

v = 79.6 s −1 (2.51 m ) = 200 m/s

ASW = 2 A

A=

ASW 0.05 m = = 2.50 cm 2 2

λ

2.51m = 1.26 m 2

2

=

Lmin =

λ 2

= 1.26 m

Superposition and Standing Waves 1261 53 •• Picture the Problem We can evaluate the wave function of Problem 52 at the given times to obtain graphs of position as a function of x. We can find the period of the motion from its frequency f and find f from its angular frequency ω. (a) The function y(x,0) is shown below:

5 4 3 y (x ,0) (cm)

2 1 0 -1 -2 -3 -4 -5 0.0

0.5

1.0

1.5

2.0

2.5

x (m)

The functions y(x,T/4) and y(x,3T/4) are shown below. Because these functions are identical, only one graph is shown.

5 4

y (x ,T /4) (cm)

3 2 1 0 -1 -2 -3 -4 -5 0.0

0.5

1.0

1.5 x (m)

2.0

2.5

1262 Chapter 16 The function y(x,T/2) follows:

5 4

y (x ,T /2) (cm)

3 2 1 0 -1 -2 -3 -4 -5 0.0

0.5

1.0

1.5

2.0

2.5

x (m)

1 f

(b) Express the period in terms of the frequency:

T=

Using the wave function, relate ω and f:

ω = 2πf = 500 s −1

Solve for f:

500 s −1 = 79.6 Hz 2π 1 T= = 12.6 ms 79.6 s −1

Substitute for f and evaluate T:

(c)

f =

Because the string is moving either upward or downward when y ( x) = 0 for all x,the energy of the wave is entirely kinetic energy.

*54 •• Picture the Problem Whether these frequencies are for a string fixed at one end only rather than for a string fixed at both ends can be decided by determining whether they are integral multiples or odd-integral multiples of a fundamental frequency. The length of the string can be found from the wave speed and the wavelength of the fundamental frequency using the standing-wave condition for a string with one end free. (a) Letting the three frequencies be represented by f' , f'' , and f''' , find the ratio of the first two frequencies:

75 Hz f' 3 = = f'' 125 Hz 5

Superposition and Standing Waves 1263 Find the ratio of the second and third frequencies: (b)

There are no even harmonics, so the string must be fixed at one end only.

(c) Express the resonance frequencies in terms of the fundamental frequency: Noting that the frequencies are multiples of 25 Hz, we can conclude that:

(d)

f'' 125 Hz 5 = = f''' 175 Hz 7

f n = nf1 , n = 1, 3, 5, ...

f 3 75 Hz = = 25 Hz 3 3

f1 =

Because the frequencies are 3, 5, and 7 times the fundamental frequency, they are the third, fifth, and seventh harmonics.

(e) Express the length of the string in terms of the standing-wave condition for a string fixed at one end:

L=n

Using v = f1λ1, find λ1:

λ1 =

Evaluate L for λ1 = 16 m and n = 1:

L=

λn 4

, n = 1, 3, 5, ...

v 400 m/s = = 16 m 25 s −1 f1

λ1 4

=

16 m = 4.00 m 4

55 •• Picture the Problem The lowest resonant frequency in this closed-at-one-end tube is its fundamental frequency. This frequency is related to its wavelength through v = fminλmax. We can use the relationship between the nth harmonic and the fundamental frequency, f n = (2n + 1) f1 , n = 1, 2, 3, ... , to find the highest frequency less than or equal to 5000 Hz that will produce resonance. (a) Express the length of the space above the water in terms of the standing-wave condition for a closed pipe:

L=n

Solve for λn:

λn =

λn 4

, n = 1, 3, 5, ...

4L , n = 1, 3, 5, ... n

1264 Chapter 16 λmax corresponds to n = 1:

λmax = 4 L = 4(1.2 m ) = 4.8 m

Using v = fminλmax, find fmin:

f min =

(b) Express the nth harmonic in terms of the fundamental frequency (first harmonic): To find the highest harmonic below 5000 Hz, let fn = 5000 Hz: Solve for n (an integer) to obtain: Evaluate f34:

v

λmax

=

340 m/s = 70.8 Hz 4.8 m

f n = (2n + 1) f1 , n = 1, 2, 3, ...

5000 Hz = (2n + 1)(70.8 Hz )

n = 34

f 34 = 69 f1 = 69(70.8 Hz ) = 4.89 kHz

(c) There are 34 harmonics higher than the fundamental frequency so the total number is:

35

56 •• Picture the Problem Sound waves of frequency 460 Hz are excited in the tube, whose length L can be adjusted. Resonance occurs when the effective length of the tube Leff = L + ∆L equals 14 λ, 34 λ, 54 λ, and so on, where λ is the wavelength of the sound. Even though the pressure node is not exactly at the end of the tube, the wavelength can be found from the fact that the distance between water levels for successive resonances is half the wavelength. We can find the speed from v = fλ and the end correction from the fact that, for the fundamental, Leff = 14 λ = L1 + ∆L, where L1 is the distance from the top of the tube to the location of the first resonance. (a) Relate the speed of sound in air to its wavelength and the frequency of the tuning fork: Using the fact that nodes are separated by one-half wavelength, find the wavelength of the sound waves: Substitute and evaluate v:

v = fλ

λ = 2(55.8 cm − 18.3 cm ) = 75 cm

(

)

v = 460 s −1 (0.75 m ) = 345 m/s

Superposition and Standing Waves 1265 (b) Relate the end correction ∆L to the wavelength of the sound and effective length of the tube:

Leff = 14 λ

Solve for and evaluate ∆L:

∆L = 14 λ − L1 =

= L1 + ∆L

1 4

(75 cm ) − 18.3 cm

= 0.450 cm *57 •• Picture the Problem We can use v = fλ to express the fundamental frequency of the organ pipe in terms of the speed of sound and v =

γRT M

to relate the speed of sound and

the fundamental frequency to the absolute temperature. Express the fundamental frequency of the organ pipe in terms of the speed of sound:

f =

Relate the speed of sound to the temperature:

v=

v

λ

γRT

M where γ and R are constants, M is the molar mass, and T is the absolute temperature.

Substitute to obtain:

f =

1 γRT λ M

Using primed quantities to represent the higher temperature, express the new frequency as a function of T:

f' =

1 λ'

As we have seen, λ is proportional to the length of the pipe. For the first question, we assume the length of the pipe does not change, so λ = λ′. Then the ratio of f ′ to f is:

f' T' = f T

γRT' M

1266 Chapter 16 Solve for and evaluate f ′ with T ′ = 305 K and T = 289 K:

f' = f 305 K = f 289 K = (440.0 Hz )

305 K 289 K

305 K 289 K

= 452 Hz It would be better to have the pipe expand so that v/L, where L is the length of the pipe, is independent of temperature. 58 •• Picture the Problem We can express the wavelength of the fundamental in a pipe open at both ends in terms of the effective length of the pipe using λ = 2 Leff = 2(L + ∆L ) , where L is the physical length of the pipe and λ = v/f. Solving these equations simultaneously will lead us to an expression for L as a function of D. Express the wavelength of the fundamental in a pipe open at both ends in terms of the pipe’s effective length Leff: Solve for L to obtain:

λ = 2 Leff = 2(L + ∆L ) where L is its physical length.

L=

λ 2

− ∆L =

λ 2

− 0.3186 D

Express the wavelength of middle C in terms of its frequency f and the speed of sound v:

λ=

v f

Substitute to obtain:

L=

v − 0.3186 D 2f

Substitute numerical values to express L as a function of D:

L=

Evaluate L for D = 1 cm:

340 m/s − 0.3186 D 2(256 s -1 ) = 0.664 m − 0.3186 D

L = 0.664 m − 0.3186(0.01 m ) = 66.1cm

Evaluate L for D = 10 cm:

L = 0.664 m − 0.3186(0.1 m ) = 63.2 cm

Superposition and Standing Waves 1267 Evaluate L for D = 30 cm:

L = 0.664 m − 0.3186(0.3 m ) = 56.8 cm

59 •• Picture the Problem We know that, when a string is vibrating in its fundamental mode, its ends are one-half wavelength apart. We can use v = fλ to express the fundamental frequency of the organ pipe in terms of the speed of sound and v = F µ to relate the speed of sound and the fundamental frequency to the tension in the string. We can use this relationship between f and L, the length of the string, to find the length of string when it vibrates with a frequency of 650 Hz. (a) Express the wavelength of the standing wave, vibrating in its fundamental mode, to the length L of the string:

λ = 2 L = 2(40 cm ) = 80 cm

(b) Relate the speed of the waves combining to form the standing wave to its frequency and wavelength:

v = fλ

Express the speed of transverse waves as a function of the tension in the string:

v=

Substitute and solve for F to obtain:

F = f 2 λ2

F

µ m L

where m is the mass of the string and L is its length. Substitute numerical values and evaluate F:

(

F = 500 s −1 = 480 N

(c) Using v = fλ and assuming that the string is still vibrating in its fundamental mode, express its frequency in terms of its length:

f =

v

λ

=

v 2L

) (0.8 m ) 2

2

1.2 ×10 −3 kg 0.4 m

1268 Chapter 16 v 2f

Solve for L:

L=

Letting primed quantities refer to a second length and frequency, express L′ in terms of f ′:

L' =

Express the ratio of L′ to L and solve for L′:

L' f f = ⇒ L' = L L f' f'

Evaluate L650 Hz:

L650 Hz =

v 2 f'

=

500 Hz L500 Hz 650 Hz 500 Hz (40 cm ) = 30.77 cm 650 Hz

You should place your finger 9.23 cm from the scroll bridge. 60 •• Picture the Problem Let f′ represent the frequencies corresponding to the A, B, C, and D notes and x(f ′) represent the distances from the end of the string that a finger must be placed to play each of these notes. Then, the distances at which the finger must be placed are given by x( f ' ) = L( f G ) − L( f ' ) . Express the distances at which the finger must be placed in terms of the lengths of the G string and the frequencies f ′ of the A, B, C, and D notes:

x ( f ' ) = L ( f G ) − L ( f ')

v

Assuming that it vibrates in its fundamental mode, express the frequency of the G string in terms of its length:

fG =

Solve for LG:

LG =

v 2 fG

Letting primed quantities refer to the string lengths and frequencies of

L' =

v 2f'

λG

=

v 2LG

(1)

Superposition and Standing Waves 1269 the A, B, C, and D notes, express L′ in terms of f ′: Express the ratio of L′ to L and solve for L′:

L' f f = G ⇒ L' = G LG LG f' f'

Evaluate L′ = L(f′) for the notes A, B, C and D to complete the table:

Use equation (1) to evaluate x(f ′) and complete the table to the right:

Note Frequency (Hz) A 220 B 247 C 262 D 294

L(f ′) (cm) 26.73 23.81 22.44 20.00

Note Frequency L(f ′) (Hz) (cm) A 220 26.73

x(f ′) (cm)

3.27

B

247

23.81

6.19

C

262

22.44

7.56

D

294

20.00

10.0

61 •• Picture the Problem We can use the fact that the resonance frequencies are multiples of the fundamental frequency to find both the fundamental frequency and the harmonic numbers corresponding to 375 Hz and 450 Hz. We can find the length of the string by relating it to the wavelength of the waves on it and the wavelength to the speed and frequency of the waves. The speed of the waves is, in turn, a function of the tension in the string and its linear density, both of which we are given. (a) Express 375 Hz as an integer multiple of the fundamental frequency of the string: Express 450 Hz as an integer multiple of the fundamental frequency of the string: Solve equations (1) and (2) simultaneously for f1:

nf1 = 375 Hz

(1)

(n + 1) f1 = 450 Hz

(2)

f1 = 75.0 Hz

1270 Chapter 16 (b) Substitute in equation (1) to obtain:

n=5

The harmonics are the fifth and sixth.

λ

v 2 f1

(c) Express the length of the string as a function of the speed of transverse waves on it and its fundamental frequency:

L=

Express the speed of transverse waves on the string in terms of the tension in the string and its linear density:

v=

F

Substitute to obtain:

L=

1 2 f1

L=

1 2 75 s −1

Substitute numerical values and evaluate L:

2

=

µ

F

µ

(

360 N = 2.00 m 4 × 10 −3 kg/m

)

62 •• Picture the Problem We can use the fact that the resonance frequencies are multiples of the fundamental frequency and are expressible in terms of the speed of the waves and their wavelengths to find the harmonic numbers corresponding to wavelengths of 0.54 m and 0.48 m. We can find the length of the string by using the standing-wave condition for a string fixed at both ends.

v

(a) Express the frequency of the nth harmonic in terms of its wavelength:

nf1 =

Express the frequency of the (n + 1)th harmonic in terms of its wavelength:

(n + 1) f1 =

Solve these equations simultaneously for n:

n=8

λn

=

v 0.54 m

v

λn+1

=

v 0.48 m

The harmonics are the eighth and ninth.

Superposition and Standing Waves 1271

λn

(b) Using the standing-wave condition, both ends fixed, relate the length of the string to the wavelength of its nth harmonic:

L=n

Evaluate L for the eighth harmonic:

⎛ 0.54 m ⎞ L = 8⎜ ⎟ = 2.16 m ⎝ 2 ⎠

2

, n = 1, 2, 3, ...

63 •• Picture the Problem The linear densities of the strings are related to the transverse wave speed and tension through v =

F µ . We can use v = fλ = 2fL to relate the frequencies

of the violin strings to their lengths and linear densities. (a) Relate the speed of transverse waves on a string to the tension in the string and solve for the string’s linear density:

Express the dependence of the speed of the transverse waves on their frequency and wavelength:

F

v=

µ

and

µ=

F v2

v = fEλ = 2 fE L

Substitute to obtain:

µE =

Substitute numerical values and evaluate µE:

µE =

FE 4 f E2 L2 90 N

[ (

4 1.5 440 s −1

)] (0.3 m ) 2

2

= 5.74 × 10 −4 kg/m = 0.574 g/m (b) Evaluate µA:

µA =

(

90 N

4 440 s −1

) (0.3 m) 2

= 1.29 × 10 −3 kg/m = 1.29 g/m

2

1272 Chapter 16 Evaluate µD:

µD =

90 N

(

4 293 s −1

) (0.3 m) 2

2

= 2.91× 10 −3 kg/m = 2.91g/m Evaluate µG:

µG =

90 N

(

4 195 s −1

) (0.3 m) 2

2

= 6.57 × 10 −3 kg/m = 6.57 g/m 64 •• Picture the Problem The spatial period is one-half the wavelength of the standing wave produced by the sound and its reflection. Hence we can solve c = f ' λ ' for λ′ and use

f ' = f [1 (1 − v c )] to derive an expression for λ′/2 in terms of c, v, and f.

c f'

(a) Express the wavelength of the reflected sound as a function of its frequency and the speed of sound in air:

λ' =

Use the expression for the Dopplershift in frequency when to source is in motion to obtain:

f '= f

1 1−

v c

where c is the speed of sound. Substitute to obtain:

λ' 2

=

c = 2f'

c 2f

1

v c c ⎛ v⎞ c−v = ⎜1 − ⎟ = 2f ⎝ c⎠ 2f 1−

Substitute numerical values and evaluate the spatial period of the standing wave:

λ' 2

=

340 m/s − 22.4 m/s = 0.318 m 2 500 s -1

(

)

Superposition and Standing Waves 1273 As the ambulance moves closer to the wall, the sound waves from its siren (b)

will periodically move in and out of resonance (i.e., the reflected waves will sometimes interfere constructively and sometimes partially destructively) so the intensity will periodically get louder and softer.

65 •• Picture the Problem Beat frequencies are heard when the strings are vibrating with slightly different frequencies. To understand the beat frequency heard when the A and E strings are bowed simultaneously, we need to consider the harmonics of both strings. In part (c) we’ll relate the tension in the string to the frequency of its vibration and set up a proportion involving the frequencies corresponding to the two tensions that we can solve for the tension when the E string is perfectly tuned.

The two sounds produce a beat because the third harmonic of the A (a)

string equals the second harmonic of the E string, and the original frequency of the E string is slightly greater than 660 Hz. If f E = (660 + ∆f )Hz, a beat of 2∆f will be heard.

(b) Because fbeat increases with increasing tension, the frequency of the E string is greater than 660 Hz. Thus the frequency of the E string is:

f E = 660 Hz + 12 (3 Hz ) = 661.5 Hz

(c) Express the frequency of a string as a function of its tension:

f =

When the frequency of the E string is 660 Hz we have:

660 Hz =

When the frequency of the E string is 661.5 Hz we have:

661.5 Hz =

Divide the first of these equations by the second and solve for F660 Hz to obtain:

v

λ

=

1

F

λ

µ

1

F660 Hz

λ

µ 1 80 N

λ

µ 2

F660 Hz

⎛ 660 Hz ⎞ ⎟⎟ (80 N ) = 79.6 N = ⎜⎜ 661.5 Hz ⎠ ⎝

66 •• Picture the Problem We can use the condition for constructive interference of the waves reflected from the walls in front of and behind you to relate the path difference to the

1274 Chapter 16 wavelength of the sound. We can find the wavelength of the sound from its frequency and the speed of sound in air. Express the total path difference as you walk toward the far wall of the hall: Express the condition on the path difference for constructive interference:

∆x = ∆xnear wall + ∆xfar wall

nλ = ∆x where n = 1, 2, 3, …

The reduction in the distance to the nearer wall as you walk a distance d is:

∆xnear wall = 2d

The increase in the distance to the farther wall as you walk a distance d is:

∆xfar wall = 2d

Substitute in equation (1) to find the total path difference as you walk a distance d:

∆x = 2d + 2d = 4d

Relate λ to f and v:

λ=

Substitute in equation (2) to obtain:

n

Solve for and evaluate d for n = 1:

d=

(1)

(2)

v f

v = 4d f 340 m/s v = = 12.5 cm 4 f 4 680 s −1

(

)

*67 •• Picture the Problem Let the wave function for the wave traveling to the right be yR ( x, t ) = A sin (kx − ω t − δ ) and the wave function for the wave traveling to the left be yL ( x, t ) = A sin (kx + ω t + δ ) and use the identity

⎛α + β ⎞ ⎛α − β ⎞ sin α + sin β = 2 sin ⎜ ⎟ cos⎜ ⎟ to show that the sum of the wave functions ⎝ 2 ⎠ ⎝ 2 ⎠ can be written in the form y ( x, t ) = A' sin kx cos(ωt + δ ) .

Superposition and Standing Waves 1275 Express the sum of the traveling waves of equal amplitude moving in opposite directions:

y ( x, t ) = yR ( x, t ) + yL ( x, t ) = A sin (kx − ωt − δ ) + A sin (kx + ωt + δ ) Use the trigonometric identity to obtain:

⎛ kx − ωt − δ + kx + ωt + δ y (x, t ) = 2 A sin ⎜ 2 ⎝ = 2 A sin kx cos(− ωt − δ ) Because the cosine function is even; i.e., cos(−θ) = cosθ:

⎞ ⎛ kx − ωt − δ − kx − ωt − δ ⎞ ⎟ cos⎜ ⎟ 2 ⎠ ⎝ ⎠

y ( x, t ) = 2 A sin kx cos(ωt + δ ) = A' sin kx cos(ωt + δ ) where A′ = 2A.

Thus we have:

y ( x, t ) = A' sin kx cos(ωt + δ ) provided A′ = 2A.

68 •• Picture the Problem We can find ω3 and k3 from the given information and substitute to find the wave function for the 3rd harmonic. We can use the time-derivative of this expression (the transverse speed) to express the kinetic energy of a segment of mass dm and length dx of the string. Integrating this expression will give us the maximum kinetic energy of the string in terms of its mass. (a) Write the general form of the wave function for the 3rd harmonic: Evaluate ω3: Using the standing-wave condition for a string fixed at one end, relate the length of the string to its 3rd harmonic wavelength:

Evaluate k3:

y3 ( x, t ) = A3 sin k3 x cos ω3t

ω3 = 2πf 3 = 2π (100 s −1 ) = 200π s −1 L=3

λ3 4

and

λ3 = k3 =

4 4 8 L = (2 m ) = m 3 3 3 2π

λ3

=

2π 3π −1 = m (8 3) m 4

1276 Chapter 16

(

K max = 14 m 200π s −1

Substitute numerical values and evaluate Kmax:

) (0.03 m) 2

2

(88.8 J/kg )m

=

Substitute to obtain:

y3 ( x, t ) =

(0.03 m )sin ⎡⎢⎛⎜ 3π m −1 ⎞⎟ x⎤⎥ cos(200π s −1 )t

(b) Express the kinetic energy of a segment of string of mass dm: Express the mass of the segment in terms of its length dx and the linear density of the string:

⎣⎝ 4

⎠ ⎦

dK = 12 dmv y2

dm = µdx

Using our result in (a), evaluate vy:

vy =

⎤ ⎡⎛ 3π −1 ⎞ ⎤ ∂⎡ −1 ⎢(0.03 m )sin ⎢⎜ m ⎟ x ⎥ cos(200π s ) t ⎥ ∂t ⎣ ⎠ ⎦ ⎣⎝ 4 ⎦

⎡⎛ 3π ⎞ ⎤ = −(200π s −1 )(0.03 m )sin ⎢⎜ m −1 ⎟ x ⎥ sin (200π s −1 ) t ⎠ ⎦ ⎣⎝ 4 ⎡⎛ 3π ⎞ ⎤ = −(6π m/s )sin ⎢⎜ m −1 ⎟ x ⎥ sin (200π s −1 ) t ⎠ ⎦ ⎣⎝ 4 Substitute to obtain:

dK =

1 2

⎡ ⎡⎛ 3π −1 ⎞ ⎤ −1 ⎢(6π m/s )sin ⎢⎜ m ⎟ x ⎥ sin 200π s 4 ⎠ ⎦ ⎣⎝ ⎣

Express the condition on the time that dK is a maximum:

(

)

2

⎤ t ⎥ µdx ⎦

sin (200π s −1 )t = 1 or

(200π s )t = π2 , 32π ,... −1

Solve for and evaluate t:

t=

1 1 3π π , ,... −1 −1 200π s 2 200π s 2

= 2.50 ms,7.50 ms,...

Superposition and Standing Waves 1277 Because the string’s maximum kinetic energy occurs when y(x,t) = 0: (c) Integrate dK from (b) over the length of the string to obtain:

The string is a straight line. L

K max = ∫ 12 [ωA sin kx sin ωt ] µdx 2

0

L

= 12 µω 2 A2 ∫ sin 2 kxdx 0

= 12 µω 2 A2

1 1 [ 2 kx − 14 sin 2kx] 0L k

= 14 mω 2 A2 where m is the mass of the string. *69 •• Picture the Problem We can equate the expression for the velocity of a wave on a string and the expression for the velocity of a wave in terms of its frequency and wavelength to obtain an expression for the weight that must be suspended from the end of the string in order to produce a given standing wave pattern. By using the condition on the wavelength that must be satisfied at resonance, we can express the weight on the end of the string in terms of µ, f, L, and an integer n and then evaluate this expression for n = 1, 2, and 3 for the first three standing wave patterns. Express the velocity of a wave on the string in terms of the tension T in the string and its linear density µ:

T

v=

µ

=

mg

µ

where mg is the weight of the object suspended from the end of the string. Express the wave speed in terms of its wavelength λ and frequency f: Eliminate v to obtain:

v = fλ

fλ =

mg

µ

Solve for mg:

mg = µ f 2 λ2

Express the condition on λ that corresponds to resonance:

λ=

Substitute to obtain:

2L , n = 1, 2, 3, ... n 2

⎛ 2L ⎞ mg = µf 2 ⎜ ⎟ , n = 1, 2, 3, ... ⎝ n ⎠ or

mg =

4µf 2 L2 , n = 1, 2, 3, ... n2

1278 Chapter 16 Evaluate mg for n = 1:

(

4(0.415 g/m ) 80 s -1 mg = (1)2

) (0.2 m ) 2

2

= 0.425 N which corresponds, at sea level, to a mass of 43.3 g. Evaluate mg for n = 2:

(

4(0.415 g/m ) 80 s -1 mg = (2)2

) (0.2 m ) 2

2

= 0.106 N which corresponds, at sea level, to a mass of 10.8 g.

Wave Packets 70 • Picture the Problem We can find the maximum duration of each pulse under the conditions given in the problem from the reciprocal of frequency of the pulses and the range of frequencies from the wave packet condition on ∆ω and ∆t. (a) The maximum duration of each pulse is its period: (b) Express the wave packet condition on ∆ω and ∆t:

T=

1 1 = 7 −1 = 10−7 s = 0.100 µs f 10 s

∆ω∆t ≈ 1 or 2π∆f∆t ≈ 1

Solve for ∆f:

∆f ≈

1 T = 2π∆t 2π

Substitute numerical values and evaluate ∆f:

∆f ≈

107 s −1 = 1.59 MHz 2π

71 • Picture the Problem We can approximate the duration of the pulse from the product of the number of cycles in the interval and the period of each cycle and the wavelength from the number of complete wavelengths in ∆x. We can use its definition to find the wave number k from the wavelength λ. (a) Relate the duration of the pulse to the number of cycles in the interval and the period of each cycle:

∆t ≈ NT =

N f0

Superposition and Standing Waves 1279 (b) There are about N complete wavelengths in ∆x; hence:

λ≈

(c) Use its definition to express the wave number k:

k=

(d)

∆x N 2π

λ

=

2πN ∆x

N is uncertain because the waveform dies out gradually rather than stopping abruptly at some time; hence, where the pulse starts and stops is not well defined.

(e) Using our result in part (c), express the uncertainty in k:

∆k =

2π∆N 2π = ∆x ∆x

because ∆N = ±1.

General Problems 72 • Picture the Problem We can use v = fλ and v =

F µ to relate the tension in the piano

wire to its fundamental frequency. Relate the tension in the wire to the speed of transverse waves on it: Express the speed of the transverse in terms of their wavelength and frequency:

v=

F

µ

=

FL m

v = fλ

mf 2λ2 L

Equate these expressions and solve for F to obtain:

F=

Relate λ for the fundamental mode of vibration to the length of the piano wire:

λ = 2L

Substitute to obtain:

F = 4mf 2 L

Substitute numerical values and evaluate F:

F = 4 7 × 10 −3 kg 261.63 s −1

(

= 1.53 kN

)(

) (0.8 m) 2

1280 Chapter 16 73 • Picture the Problem We can use v = fnλn to express the resonance frequencies of the ear canal in terms of their wavelengths and L = n

λn 4

, n = 1, 3, 5, ... to relate the length of the

ear canal to its resonance wavelengths.

v

(a) Relate the resonance frequencies to the speed of sound and the wavelength of the compressional vibrations:

fn =

Express the condition for constructive interference in a pipe that is open at one end:

L=n

Solve for λn:

λn =

Substitute to obtain:

fn = n

Evaluate f1, f2, and f3:

f1 = 3.40 kHz ,

λn

λn 4

, n = 1, 3, 5, ...

4L n

v 340 m/s =n 4L 4 2.5 × 10−2 m = n(3.40 kHz )

(

)

f 3 = 3 × 3.40 kHz = 10.2 kHz , and

f 5 = 5 × 3.40 kHz = 17.0 kHz Frequencies near 3400 Hz will be

(b)

most readily perceived. 74

•

Picture the Problem We can use L = n

λn 4

, n = 1, 3, 5, ... to express the wavelengths of

the fundamental and next two harmonics in terms of the length of the rope and v = fnλn and v =

F

µ

to relate the resonance frequencies to their wavelengths.

(a) Express the condition for constructive interference on a rope

L=n

λn 4

, n = 1, 3, 5, ...

Superposition and Standing Waves 1281 that is fixed at one end:

4 L 4(4 m ) 16 m = = n n n

Solve for λn:

λn =

Evaluate λn for n = 1, 3, and 5:

λ1 = 16.0 m

λ3 =

16 m = 5.33 m 3

and

λ5 = (b) Relate the resonance frequencies to the speed and wavelength of the transverse waves:

fn =

Express the speed of the transverse waves as a function of the tension in the rope:

v=

Substitute to obtain:

v

λn

F

µ

FL m

=

where m and L are the mass and length of the rope.

fn = =

Evaluate fn for n = 1, 3, and 5:

16 m = 3.20 m 5

1

1 FL = λn m

λn

(400 N )(4 m ) 0.16 kg

100 m/s

λn

100 m/s = 6.25 Hz 16 m 100 m/s f3 = = 18.8 Hz 5.33 m f1 =

and

f5 =

100 m/s = 31.3 Hz 3.20 m

75 •• Picture the Problem The path difference at the point where the resultant wave an amplitude A is related to the phase shift between the interfering waves according to ∆x λ = δ 2π . We can use this relationship to find the phase shift and the expression for the amplitude resulting from the superposition of two waves of the same amplitude and frequency to find the phase shift.

1282 Chapter 16 Express the relation between the path difference and the phase shift at the point where the resultant wave has an amplitude A:

δ 2π

∆x = λ

A = 2 y0 cos 12 δ

Express the amplitude resulting from the superposition of two waves of the same amplitude and frequency: Solve for and evaluate δ:

δ = 2 cos −1

Substitute and simplify to obtain:

∆x = λ

A A 2π = 2 cos −1 = 2 y0 2A 3

2π 3 = 2π

1 3

λ

76 •• Picture the Problem We can use v = fnλn to express the resonance frequencies of the string in terms of their wavelengths and L = n

λn 2

, n = 1, 2, 3, ... to relate the length of the

string to the resonance wavelengths for a string fixed at both ends. Our strategy for part (b) will be the same … except that we’ll use the standing-wave condition

L=n

λn 4

, n = 1, 3, 5, ... for strings with one end free. v

(a) Relate the frequencies of the harmonics to their wavelengths and the speed of transverse waves on the string:

fn =

Express the standing-wave condition for a string with both ends fixed:

L=n

Solve for λn:

λn =

Substitute to obtain:

fn = n

Express the speed of the transverse waves as a function of the tension in the string:

v=

λn

λn 2

, n = 1, 2, 3, ...

2L n v 2L

F

µ

Superposition and Standing Waves 1283 Substitute to obtain:

fn = n =n

1 2L

F

µ

1 18 N 2(35 m ) 0.0085 kg/m

= n(0.657 Hz ) Calculate the 1st four harmonics:

f1 = 0.657 Hz f 2 = 2(0.657 Hz ) = 1.31 Hz f 3 = 3(0.657 Hz ) = 1.97 Hz and

f 4 = 4(0.657 Hz ) = 2.63 Hz

λn

(b) Express the standing-wave condition for a string fixed at one end:

L=n

Solve for λn:

λn =

The resonance frequencies equation becomes:

fn = n

4

, n = 1, 3, 5, ...

4L n

=n

1 4L

F

µ

1 18 N 4(35 m ) 0.0085 kg/m

= n(0.329 Hz ) Calculate the 1st four harmonics:

f1 = 0.329 Hz f 3 = 3(0.329 Hz ) = 0.987 Hz f 5 = 5(0.329 Hz ) = 1.65 Hz and

f 7 = 7(0.329 Hz ) = 2.30 Hz

77 •• Picture the Problem We’ll model the shaft as a pipe of length L with one end open. We can relate the frequencies of the harmonics to their wavelengths and the speed of sound using v = fnλn and the depth of the mine shaft to the resonance wavelengths using the

1284 Chapter 16 standing-wave condition for a pipe with one end open; L = n

λn 4

, n = 1, 3, 5, ... .

v

Relate the frequencies of the harmonics to their wavelengths and the speed of sound:

fn =

Express the standing-wave condition for a pipe with one end open:

L=n

Solve for λn:

λn =

Substitute to obtain:

fn = n

For fn = 63.58 Hz:

63.58 Hz = n

For fn+2 = 89.25 Hz:

89.25 Hz = (n + 2)

Divide either of these equations by the other and solve for n to obtain:

n = 4.95 ≈ 5

Substitute in the equation for fn = f5 = 63.58 Hz:

f5 =

5v 4L

Solve for and evaluate L:

L=

5v 5(340 m/s ) = = 6.68 m 4 f 5 4 63.58 s −1

λn λn 4

, n = 1, 3, 5, ...

4L n v 4L

(

v 4L v 4L

)

78 •• Picture the Problem We can use the standing-wave condition for a string with one end free to find the wavelength of the 5th harmonic and the definitions of the wave number and angular frequency to calculate these quantitities. We can then substitute in the wave function for a wave in the nth harmonic to find the wave function for this standing wave. (a) Express the standing-wave condition for a string with one end free:

L=n

λn 4

, n = 1, 3, 5, ...

Superposition and Standing Waves 1285 Solve for and evaluate λ5:

λ5 =

(b) Use its definition to calculate the wave number:

k5 =

(c) Using its definition, calculate the angular frequency: (d) Write the wave function for a standing wave in the nth harmonic:

4 L 4(5 m ) = = 4.00 m 5 5 2π

λ5

=

π −1 2π = m 4m 2

ω 5 = 2πf 5 = 2π (400 s −1 ) = 800π s −1 yn (x, t ) = A sin k n x cos ωnt

Substitute to obtain:

y5 (x, t ) = A sin (k5 x )cos(ω5t ) =

(0.03 m )sin ⎡⎢⎛⎜ π m −1 ⎞⎟ x ⎤⎥ cos(800π s −1 ) t ⎣⎝ 2

⎠ ⎦

79 •• Picture the Problem The coefficient of the factor containing the time dependence in the wave function is the maximum displacement of any point on the string. The time derivative of the wave function is the instantaneous speed of any point on the string and the coefficient of the factor containing the time dependence is the maximum speed of any point on the string. Differentiate the wave function with respect to t to find the speed of any point on the string:

(a) Referring to the wave function, express the maximum displacement of the standing wave: Evaluate equation (1) at x = 0.10 m:

∂ [0.02 sin 4π x cos 60π t ] ∂t = −(0.02)(60π )sin 4π x sin 60π t = −1.2π sin 4π x sin 60π t

vy =

[(

)]

ymax ( x ) = (0.02 m )sin 4π m −1 x

ymax (0.10 m ) = (0.02 m )

[(

(1)

]

)

× sin 4π m −1 (0.10 m )

= 1.90 cm Referring to the derivative of the wave function with respect to t, express the maximum speed of the

[(

)]

v y ,max (x ) = (1.2π m/s )sin 4π m −1 x (2)

1286 Chapter 16 standing wave: Evaluate equation (2) at x = 0.10 m:

v y ,max (0.10 m ) = (1.2π m/s )

[(

]

)

× sin 4π m −1 (0.10 m ) = 3.59 m/s (b) Evaluate equation (1) at x = 0.25 m:

ymax (0.25 m ) = (0.02 m )

[(

]

)

× sin 4π m −1 (0.25 m )

= 0 Evaluate equation (2) at x = 0.25 m:

v y ,max (0.25 m ) = (1.2π m/s )

[(

]

)

× sin 4π m −1 (0.25 m ) = 0 (c) Evaluate equation (1) at x = 0.30 m:

ymax (0.30 m ) = (0.02 m )

[(

]

)

× sin 4π m −1 (0.30 m )

= 1.18 cm Evaluate equation (2) at x = 0.30 m:

v y ,max (0.30 m ) = (1.2π m/s )

[(

]

)

× sin 4π m −1 (0.30 m ) = 2.22 m/s (d) Evaluate equation (1) at x = 0.50 m:

ymax (0.50 m ) = (0.02 m )

[(

]

)

× sin 4π m −1 (0.50 m )

= 0 Evaluate equation (2) at x = 0.50 m:

v y ,max (0.50 m ) = (1.2π m/s )

[(

)

]

× sin 4π m −1 (0.50 m ) = 0

80 •• Picture the Problem In part (a) we can use the standing-wave condition for a wire fixed at both ends and the fact that nodes are separated by one-half wavelength to find the harmonic number. In part (b) we can relate the resonance frequencies to their wavelengths and the speed of transverse waves and express the speed of the transverse

Superposition and Standing Waves 1287 waves in terms of the tension in the wire and its linear density. (a) Express the standing-wave condition for a wire fixed at both ends:

L=n

Solve for n:

n=

Solve for and evaluate λ1:

λn

, n = 1, 2, 3, ...

2

2L

λn

λ1 = 2 L = 2(2.5 m ) = 5 m 1 2

λn = 0.5 m

Relate the distance between nodes to the distance of the node closest to one end and solve for λn:

and

Substitute and evaluate n:

n=

(b) Express the resonance frequencies in terms of the their wavelengths and the speed of transverse waves on the wire:

fn =

Relate the speed of transverse waves on the wire to the tension in the wire:

v=

Substitute and simplify to obtain:

fn = n

λn = 1 m 2(2.5 m ) = 5 1m v

λn

=n

v

λ1

F

µ

1

λ1

FL 1 =n m 5m

(30 N )(2.5 m ) 0.1 kg

= n(5.48 Hz ) Evaluate fn for n = 1, 2, and 3:

f1 = 5.48 Hz f 2 = 2(5.48 Hz ) = 11.0 Hz and

f 3 = 3(5.48 Hz ) = 16.4 Hz

*81 •• Picture the Problem We can use v = fλ to relate the speed of sound in the gas to the distance between the piles of powder in the glass tube.

1288 Chapter 16 At resonance, standing waves are set up in the tube. At a displacement (a) antinode, the powder is moved about; at a node the powder is stationary, and so it collects at the nodes. (b) Relate the speed of sound to its frequency and wavelength: Letting D = distance between nodes, relate the distance between the nodes to the wavelength of the sound:

v = fλ

λ = 2D

Substitute to obtain:

v = 2 fD

(c) If we let the length L of the tube be 1.2 m and assume that vair = 344 m/s (the speed of sound in air at 20°C), then the 10th harmonic corresponds to D = 25.3 cm and a driving frequency of:

f air =

vair 344 m/s = = 680 Hz 2 D 2(0.253 m )

If f = 2 kHz and vHe = 1008 m/s (the speed of sound in helium at 20°C), then D for the 10 th harmonic in helium would 25.3 cm and D for the 10 th harmonic in air would be 8.60 cm. Hence, neglecting end effects at the (d) driven end, a tube whose length is the least common multiple of 8.60 cm and 25.3 cm (218 cm) would work well for the measurement of the speed of sound in either air or helium. 82 •• Picture the Problem We can use v =

F µ to express F as a function of v and

v = fλ to relate v to the frequency and wavelength of the string’s fundamental mode. Because, for a string fixed at both ends, fn = nf1, we can extend our result in part (a) to part (b). (a) Relate the speed of the transverse waves on the string to the tension in it:

v=

Solve for F:

F = µv 2

F

µ

(1)

Superposition and Standing Waves 1289 Relate the speed of the transverse waves on the string to their frequency and wavelength: Express the wavelength of the fundamental mode to the length of the string:

v = f1λ1

λ1 = 2 L

Substitute to obtain:

v = 2 fL

Substitute in equation (1) to obtain:

F = 4 f 2 L2 µ

Substitute numerical values and evaluate F:

F = 4 60 s −1

(b) For the nth harmonic, equation (2) becomes: Evaluate this expression for n = 2, 3, and 4:

(

(2)

) (2.5 m) (8 ×10 2

2

−3

kg/m

)

= 720 N

Fn = f n2 L2 µ = n 2 f12 L2 µ = n 2 (720 N ) F2 = 4(720 N ) = 2.88 kN F3 = 9(720 N ) = 6.48 kN and

F4 = 16(720 N ) = 11.5 kN

83 •• Picture the Problem We can use the conditions ∆f = f1 and f n = nf1 , where n is an integer, which must be satisfied if the pipe is open at both ends to decide whether the pipe is closed at one end or open at both ends. Once we have decided this question, we can use the condition relating ∆f and the fundamental frequency to determine the latter. In part (c) we can use the standing-wave condition for the appropriate pipe to relate its length to its resonance wavelengths.

∆f = f1

(a) Express the conditions on the frequencies for a pipe that is open at both ends:

and

Evaluate ∆f = f1:

∆f = 1834 Hz − 1310 Hz = 524 Hz

Using the 2nd condition, find n:

n=

f n = nf1

f n 1310 Hz = = 2.5 f1 524 Hz

1290 Chapter 16 The pipe is closed at one end. (b) Express the condition on the frequencies for a pipe that is open at both ends:

∆f = 2 f1

(524 Hz ) =

Solve for and evaluate f1:

f1 = 12 ∆f =

(c) Using the standing-wave condition for a pipe open at one end, relate the length of the pipe to its resonance wavelengths:

L=n

For n = 1 we have:

λ1 =

λ v v and L = 1 = f1 4 4 f1

Substitute numerical values and evaluate L:

L=

340 m/s = 32.4 cm 4 262 s −1

λn 4

1 2

262 Hz

, n = 1, 3, 5, ...

(

)

84 •• Picture the Problem We can relate the speed of sound in air to the frequency of the violin string and the wavelength of the sound in the open tube that is closed at one end by water. The wavelength of the sound, in turn, is a function of the length of the air column and so we can derive an expression for the speed of sound as a function of the frequency of the transverse waves on the violin string and the length of the air column above the water. Knowing that the violin string is vibrating in its fundamental mode, we can express this frequency in terms of the tension in the string and its linear density. Express the speed of sound in the tube in terms of its fundamental frequency and wavelength:

vs = f1λ1

λn

Using the standing-wave condition for a tube open at one end, relate the speed of sound to the length of the air column in the tube:

Lair column = n

Solve for λ1:

λ1 = 4Lair column

Substitute to obtain:

vs = 4 f1Lair column

4

, n = 1, 3, 5, ...

(1)

Superposition and Standing Waves 1291 Express the frequency of the transverse waves on the violin string in terms of their wavelength and the speed with which they propagate on the string:

f1 =

Relate the speed of the transverse waves on the string to the tension in it:

v=

Substitute to obtain:

f1 =

Substitute in equation (1) to obtain:

v

λ1

F

µ

=

v 2Lstring

=

FLstring

1 2 Lstring

m

FLstring

vs = 4 Lair column = 2 Lair column

Substitute numerical values and evaluate vs:

vs = 2(0.18 m )

m

=

F 4mLstring

F 4mLstring F mLstring

(10

(440 N ) kg )(0.5 m )

−3

= 338 m/s The method is not very accurate because it neglects end effects (see Problem 56). 85 •• Picture the Problem We know that the superimposed traveling waves have the same wave number and angular frequency as the standing-wave function, have equal amplitudes that are half that of the standing-wave function, and travel in opposite directions. From inspection of the standing-wave function we note that k = 12 π m −1 and ω = 40π s −1 . We can express the velocity of a segment of the rope by differentiating the standing-wave function with respect to time and the acceleration by differentiating the velocity function with respect to time. (a) Write the wave function for the wave traveling in the positive x direction:

y1 ( x, t ) =

(0.01m )sin ⎡⎢⎛⎜ π m −1 ⎞⎟ x − (40π s −1 )t ⎤⎥ ⎣⎝ 2

⎠

⎦

1292 Chapter 16 Write the wave function for the wave traveling in the negative x direction:

y 2 ( x, t ) =

(0.01m )sin ⎡⎢⎛⎜ π m −1 ⎞⎟ x + (40π s −1 )t ⎤⎥ ⎣⎝ 2

⎠

⎦

d = 12 λ

(b) Express the distance d between nodes in terms of the wavelength of the standing wave: Use the wave number to find the wavelength:

k = 12 π m −1 =

2π

λ

and

λ = 4m d=

Substitute and evaluate d:

1 2

(4 m ) =

2.00 m

(c) Differentiate the given wave function with respect to t to express the velocity of any segment of the rope:

v y ( x, t ) =

∂ ⎡ (0.02 m )sin⎛⎜ π m −1 ⎞⎟ x cos 40π s −1 ⎢ ∂t ⎣ ⎝2 ⎠

(

)t ⎤⎥ ⎦

⎛π ⎞ = −(0.8π m/s )sin ⎜ m −1 ⎟ x sin 40π s −1 t ⎝2 ⎠

(

)

Evaluate v y (1 m, t ) :

⎛π ⎞ v y (1 m, t ) = −(0.8π m/s ) sin ⎜ m −1 ⎟(1 m ) sin 40π s −1 t ⎝2 ⎠ −1 = −(0.8π m/s ) sin 40π s t

(

(

(

)

)

)

= − (2.51 m/s ) sin 40π s −1 t (d) Differentiate v y (x, t ) with respect to time to obtain a y ( x, t ) :

a y ( x, t ) =

∂ ⎡ ⎛π ⎞ − (0.8π m/s )sin ⎜ m −1 ⎟ x sin 40π s −1 ⎢ ∂t ⎣ ⎝2 ⎠

(

⎛π ⎞ = − 32π 2 m/s 2 sin ⎜ m −1 ⎟ x cos 40π s −1 t 2 ⎝ ⎠

(

)

(

)

)t ⎤⎥ ⎦

Superposition and Standing Waves 1293 Evaluate a y (1 m, t ) :

⎛π ⎞ a y (1 m, t ) = − 32π 2 m/s 2 sin ⎜ m −1 ⎟(1 m ) cos 40π s −1 t ⎝2 ⎠ −1 2 2 = − 32π m/s cos 40π s t

( (

=

) ) ( ) − (316 m/s )cos(40π s )t

(

)

−1

2

86 •• Picture the Problem We can use the definition of intensity to find the intensity of each speaker, the dependence of intensity on the square of the amplitude of the wave disturbance to express the amplitudes of the waves, and the dependence of the intensity on whether the speakers are coherent and their phase difference to find the intensity at the given point. (a) Express the intensity as a function of the distance of a point from the source:

I=

P 4πr 2

Evaluate I1:

I1 =

1 mW = 19.9 µW/m 2 2 4π (2 m )

Evaluate I2:

I2 =

1 mW = 8.84 µW/m 2 2 4π (3 m )

(b) Using v = fλ, find the wavelength of the sound:

λ=

v 340 m/s = = 0.5 m f 680 s −1

Express the path difference in terms of λ:

Express the intensity at point P due to the sound from source 1:

∆x = 2λ and so there is constructive interference at point P.

I1 = constant × A12 or

A1 = C I1 where C is a constant. Express the intensity at point P due the sound from source 2:

I 2 = constant × A22 or

A2 = C I 2

1294 Chapter 16 Express the square of the resultant amplitude at point P:

A2 = C 2

Solve for and evaluate I:

I= =

(

(

I1 + I 2

I1 + I 2

)

)

2

= C 2I

2

( 19.9 µW/m

2

+ 8.84 µW/m 2

)

2

2

− 8.84 µW/m 2

)

2

= 55.3 µW/m 2 (c) If they are driven coherently but are 180° out of phase we will have destructive interference at point P and the intensity is given by:

(d) Because the sources are incoherent, the intensities add arithmetically:

I= =

(

I1 − I 2

)

2

( 19.9 µW/m

= 2.21 µW/m 2 I = I1 + I 2 = 19.9 µW/m 2 + 8.84 µW/m 2 = 28.7 µW/m 2

87 •• Picture the Problem In Chapter 14, Section 14.1, it was shown that a harmonic function could be represented by a vector rotating at the angular frequency ω. The simplest way to do this problem is to use that representation. The vectors, of equal magnitude, are shown in the diagram. We can find the resultant wave function by finding the magnitude and direction of the resultant vector.

From the diagram it is evident that: Find the sum of the x components of the vectors: Relate the magnitude of the resultant vector to the sum of its x and y components:

∑v

y

=0

∑v

x

= A cos 60° + A cos 60° + A = 2 A

v= =

(∑ v ) + (∑ v ) 2

x

(2 A)2 + (0)2

2

y

= 2A

Superposition and Standing Waves 1295 Find the direction of the resultant vector: Express the resultant wave:

θ = tan −1 ⎜⎜ ∑

vy ⎞ ⎟ = tan −1 ⎛⎜ 0 ⎞⎟ = 0 ⎟ ⎝ 2A ⎠ ⎝ ∑ vx ⎠ yres (x, t ) = 2 A sin (kx − ω t ) ⎛

= 0.1sin (kx − ω t ) 88 •• Picture the Problem The diagram shows a two dimensional plane wave propagating at an angle θ with respect to the x axis. At a given point in time, the surface of constant phase for the wave is the line defined by kx x + k y y = φ , or y = − k x k y x + φ .

(

)

The wave itself moves in a direction perpendicular to the wavefront, i.e., in a direction specified by a line with slope ky/kx. Choose two points (x, y) and (x + ∆x, y + ∆y) that have a separation of 1 wavelength along such a line. Express the phase difference φ between the two points that have a separation of 1 wavelength along the line y = − k x k y x + φ in terms

(

)

φ ∆r

=

2π

λ

or φ =

where ∆r =

2π

λ

∆r

(∆x )2 + (∆y )2

of the spatial separation ∆r of the points: Substitute φ = 2π to obtain:

2π =

2π

λ

(∆x )2 + (∆y )2

or

λ= Express φ in terms of kx, ky, ∆x and ∆y:

(∆x )2 + (∆y )2

φ = k∆r = k x ∆x + k y ∆y or, because φ = 2π,

k x ∆x + k y ∆y = 2π

Because ∆y =

ky kx

∆x :

k x ∆x + or

∆x =

k y2 kx

∆x = 2π

2πk x k x2 + k y2

(1)

1296 Chapter 16 Similarly:

2πk y

∆y =

k x2 + k y2 2

Substitute in equation (1) to obtain:

⎛ 2πk ⎞ ⎛ 2πk ⎞ λ = ⎜⎜ 2 x 2 ⎟⎟ + ⎜⎜ 2 y 2 ⎟⎟ ⎝ kx + k y ⎠ ⎝ kx + k y ⎠ 2π = k x2 + k y2

Relate the wave velocity v to its angular frequency ω and wave number k:

v=

Substitute for λ to obtain:

Express the angle between the wave velocity and the x axis:

v=

ω

=ω

k

ω 2π

2

λ 2π

2π k +k 2 x

2 y

=

ω k + k y2 2 x

2πk y

θ = tan −1

k 2 + k y2 ∆y = tan −1 x 2πk x ∆x k x2 + k y2

⎛k ⎞ = tan −1 ⎜⎜ y ⎟⎟ ⎝ kx ⎠ *89 •• Picture the Problem We can express the fundamental frequency of the organ pipe as a function of the air temperature and differentiate this expression with respect to the temperature to express the rate at which the frequency changes with respect to temperature. For changes in temperature that are small compared to the temperature, we can approximate the differential changes in frequency and temperature with finite changes to complete the derivation of ∆f/f = ½∆T/T. In part (b) we’ll use this relationship and the data for the frequency at 20°C to find the frequency of the fundamental at 30°C. (a) Express the fundamental frequency of an organ pipe in terms of its wavelength and the speed of sound:

f =

Relate the speed of sound in air to the absolute temperature:

v=

v

λ

γRT M

=C T

Superposition and Standing Waves 1297 where

γR

C=

Defining a new constant C′, substitute to obtain:

f =

M

C

= constant

T = C' T

λ

because λ is constant for the fundamental frequency we ignore any change in the length of the pipe. Differentiate this expression with respect to T:

df 1 f = C 'T −1 2 = dT 2 2T

Separate the variables to obtain:

df 1 dT = f 2 T

For ∆T << T, we can approximate df by ∆f and dT by ∆T to obtain: (b) Express the fundamental frequency at 30°C in terms of its frequency at 20°C:

∆f 1 ∆T = f 2 T f 30 = f 20 + ∆f

∆T T

Solve our result in (a) for ∆f:

∆f =

Substitute numerical values and evaluate ∆f:

f 30 = 200 Hz + 12 (200 Hz )

1 2

f

10 K 293 K

= 203 Hz 90 •• Picture the Problem We’ll use a spreadsheet program to graph the wave functions individually and their sum as functions of x at t = 0 and at t = 1 s. In (c) and (d) we can add the wave functions algebraically to find the result wave function at t = 0 and at t = 1 s. (a) and (d) A spreadsheet program to calculate values for y1(x,t) and y2(x,t) between and plot their graphs is shown below. The constants and cell formulas used are shown in the table. Cell A5 A6

Content/Formula −5.0 A5+0.1

Algebraic Form x

x + ∆x

1298 Chapter 16 B5

0.05/(2+(A5−2*$B$1)^2)

C5

−0.05/(2+(A5+2*$B$1)^2)

D5

0.05/(2+(A5−2*$B$1)^2) −0.05/(2+(A5+2*$B$1)^2) 0.05/(2+(A5−2*$B$2)^2) −0.05/(2+(A5+2*$B$2)^2)

E5

A B t= 0 t= 1

C

y1 ( x,0 ) y 2 ( x ,0 ) y1 ( x,0 ) + y 2 ( x,0 ) y1 ( x,1) + y 2 (x,1)

D

E

1 2 3 4 5 6 7 8 9 10

x y1(x,0) y2(x,0) y1(x,0)+y2(x,0) y1(x,1)+y2(x,1) −5.0 0.001 −0.001 0.000 −0.001 −4.9 0.001 −0.001 0.000 −0.002 −4.8 0.001 −0.001 0.000 −0.002 −4.7 0.001 −0.001 0.000 −0.002 −4.6 0.001 −0.001 0.000 −0.002 −4.5 0.001 −0.001 0.000 −0.002

110 111 112 113

5.5 5.6 5.7 5.8

0.001 0.001 0.001 0.001

s

−0.001 −0.001 −0.001 −0.001

0.000 0.000 0.000 0.000

0.001 0.001 0.001 0.001

The four curves on the graph are identified in the legend. y1 is traveling from left to right and y2 from right to left. As time increases, y1 is farther to the right and y2 is farther to the left.

0.010 0.008 0.006 0.004 0.002

-5

-3

0.000 -1 -0.002

1

3

5

-0.004

y1(x,0)

-0.006

y2(x,0) y1(x,0)+y2(x,0)

-0.008 -0.010

y1(x,1)+y2(x,1)

Superposition and Standing Waves 1299 (b) Express the resultant wave function at t = 0:

y1 ( x,0 ) + y2 ( x,0) =

0.02 m 3 − 0.02 m 3 = 0 + 2 m2 + x2 2 m2 + x2

(c) Express the resultant wave function at t = 1 s:

y1 (x,1s ) + y2 (x,1s ) =

0.02 m 3 − 0.02 m 3 + 2 2 2 m 2 + (x − 2 s ) 2 m 2 + (x + 2 s )

91 •• Picture the Problem We can relate the frequency of the standing waves in the openended tube to its length and the speed of sound in air. (a) What you hear is the fundamental mode of the tube and its overtones. A more physical explanation is that the echo of the finger snap moves back and forth along the tube with a characteristic time of 2L/c, leading to a series of clicks from each echo. Because the clicks happen with a frequency of c/2L, the ear interprets this as a musical note of that frequency. v (b) Express the frequency of the f = 2L sound in terms of the length of the tube: Solve for L:

L=

v 2f

Substitute numerical values and evaluate L:

L=

340 m/s = 38.6 cm 2 440 s −1

(

)

92 •• Picture the Problem To find the total kinetic energy of the nth mode of vibration, we’ll need to differentiate yn (x, t ) = An sin k n x cos ωnt with respect to time, substitute in the expression for ∆K, and then integrate over the length of the string. (a) Write the wave function for a standing wave on a string fixed at both ends: Using the standing-wave condition for a string with both ends fixed, relate the length of the string to the

yn (x, t ) = An sin k n x cos ωnt 2π . where k n =

λn

L=n

λn 2

, n = 1, 2, 3, ...

1300 Chapter 16 wavelength of the nth harmonic: Solve for λn:

λn =

Substitute in the expression for kn to obtain:

kn = n

Differentiate this expression with respect to t:

Substitute in the given expression and simplify to obtain:

2L n

π L

∂y ∂ = [ An sin k n x cos ωnt ] ∂t ∂t = −ωn An sin k n x sin ωnt

∆K = 12 µ (− ωn An sin k n x sin ωnt ) ∆x 2

= 12 µωn2 An2 sin 2 k n x sin 2 ωnt∆x

Integrate this expression over the length of the string to find its total kinetic energy:

⎛ π ⎞ K = µω A sin ωnt ∫ sin 2 ⎜ n x ⎟dx ⎝ L ⎠ 0

(b) Express the condition that K = Kmax:

sin 2 ωnt = 1

Substitute to obtain:

K max =

(c) From equation (1), for K = Kmax:

sin 2 ωnt = 1 or ωnt =

Evaluate the wave function in (a) when ωnt =

π

2

:

L

2 n

1 2

=

1 4

2 n

2

mωn2 An2 sin 2 ωnt

1 4

(1)

mωn2 An2

π 2

⎛ π ⎞ π ⎟⎟ = An sin k n x cos = 0 yn ⎜⎜ x, 2 ⎝ 2ωn ⎠

(d) Using the result from part (b), express the maximum kinetic energy:

K max = 14 mωn2 An2

Relate ωn to ω1:

ωn = nω1

Substitute to obtain:

K max = n 2 14 mω12 An2

(

)

or, because m and ω1 are constants,

Superposition and Standing Waves 1301 K max ∝ n 2 An2 Remarks: Our result in part (b) is exactly the same result obtained in Problem 68 with ωn and An replacing ω and A. 93

••

Picture the Problem We can use f n = n

v , n = 1, 2, 3, ... to relate the resonant 2L

frequencies to the length of the string and the speed of transverse waves on the string and

v = F µ to express the speed of the transverse waves on the string in terms of the tension in the string. Differentiating of the resulting expression with respect to F will lead

df n 1 dF . For changes in f that are small compared to f, we can use a differential = fn 2 F ∆f 1 ∆F . approximation to obtain n = 2 F fn v (a) Using the standing-wave f n = n , n = 1, 2, 3, ... 2L condition for a string fixed at both to

ends, relate the resonant frequencies to the length of the string and the speed of transverse waves on the string: Express the speed of transverse waves on the string in terms of the tension in the string:

v=

F

Substitute to obtain:

fn =

n 2L

µ

F

µ

=C F

because n, L, and µ are constants. Differentiate fn with respect to F to obtain: Separate the variables to obtain:

df n C 1 1 fn = = dF 2 F 2 F df n 1 dF = fn 2 F

Because no conditions were placed on its derivation, this expression is valid for all harmonics.

1302 Chapter 16 (b) Because ∆f << f, one can approximate the differential quantitities in our result for part (a) to obtain:

∆f n 1 ∆F = fn 2 F

Solve for ∆F/F:

∆F ∆f =2 n F fn

Substitute numerical values and evaluate ∆F/F:

⎛ 2 Hz ⎞ ∆F ⎟⎟ = 1.54% = 2⎜⎜ F ⎝ 260 Hz ⎠

94 •• Picture the Problem Let the sources be denoted by the numerals 1 and 2. The phase difference between the two waves at point P is the sum of the phase difference due to the sources δ0 and the phase difference due to the path difference δ. (a) Write the wave function due to source 1: Write the wave function due to source 2:

f1 (x, t ) = A0 cos(kx1 − ω t ) f 2 (x, t ) = A0 cos(k (x1 + ∆x ) − ω t + δ s )

(b) Express the sum of the two wave functions:

f (x, t ) = f1 (x, t ) + f 2 (x, t ) = A0 cos(kx1 − ω t ) + A0 cos(k (x1 + ∆x ) − ω t + δ s ) = A0 [cos(kx1 − ω t )cos(k (x1 + ∆x ) − ω t + δ s )] ⎛α + β ⎞ ⎛α − β ⎞ ⎟ cos⎜ ⎟ to obtain: ⎝ 2 ⎠ ⎝ 2 ⎠

Use cos α + cos β = 2 cos⎜

⎡ ⎛ k∆x δ s ⎞ ⎛ ⎛ ∆x ⎞ δ ⎞⎤ f ( x, t ) = 2 A0 ⎢cos⎜ + ⎟cos⎜⎜ k ⎜ x + ⎟ − ω t + s ⎟⎟⎥ 2⎠ ⎝ ⎝ 2 ⎠ 2 ⎠⎦ ⎣ ⎝ 2 Express the phase difference δ in terms of the path difference ∆x and the wave number k:

δ ∆x

=

2π

λ

= k or k∆x = δ

Superposition and Standing Waves 1303 Substitute to obtain:

⎡ ⎛ δ + δs ⎞ ⎛ ⎛ ∆x ⎞ δ ⎞⎤ f ( x, t ) = 2 A0 ⎢cos⎜ ⎟cos⎜⎜ k ⎜ x + ⎟ − ω t + s ⎟⎟⎥ 2 ⎠ 2 ⎠⎦ ⎣ ⎝ 2 ⎠ ⎝ ⎝

A = 2 A0 cos 12 (δ + δ s )

The amplitude of the resultant wave function is the coefficient of the time-dependent factor:

I P = C ' A2

(c) Express the intensity at an arbitrary point P:

= C ' [2 A0 cos 12 (δ + δ s )]

2

[

= C ' 4 A02 cos 2

[

Evaluate I for δ = 0 and δs = Ct:

I = C ' 4 A02 cos 2

Because the average value of cos 2 θ over a complete period is ½:

I ave ∝ 2 A02 = 2 I 0 and

I ∝ 4 I 0 cos 2

1 2

1 2

1 2

(δ + δ s )]

(Ct )]

(Ct )

(d) Evaluate I for ∆x = 12 λ and

∆x = 12 λ ⇒ δ = π

δs = Ct:

∴ I = C ' 4 A02 cos 2

[

1 2

(π + Ct )]

and at t = 0, I = 0 . i.e., the waves interfere destructively. A spreadsheet program to calculate the intensity at point P as a function of time for a zero path difference and a path difference of λ is shown below. The constants and cell formulas used are shown in the table. Cell B1 B7 C6

Content/Formula 1 B6+0.1 COS($B$6*B6/2)^2

D6

COS($B$6*B6/2-PI()/2)^2 A

1 2 3 4 5 6

B C= 1

t (s) 0.00

Algebraic Form C

t + ∆t cos 2 12 (Ct )

cos 2

1 2

(π + Ct )

C

D

I (W/m^2) 1.000

I (W/m^2) 0.000

s−1

1304 Chapter 16 7 8 9

0.10 0.20 0.30

0.998 0.990 0.978

0.002 0.010 0.022

103 104 105 106

9.70 9.80 9.90 10.00

0.019 0.035 0.055 0.080

0.981 0.965 0.945 0.920

The solid curve is the graph of cos 2

cos

2 1 2

(π + Ct ) .

1 2

(Ct ) and the dashed curve is the graph of

1.0

I (W/m^2)

0.8 path diff = 0

0.6

path diff = lambda/2

0.4

0.2

0.0 0

2

4

6

8

10

t (s)

95 ••• Picture the Problem We can differentiate the sum of the two wave functions to find the velocity of a segment dx of the string. We can find the kinetic energy of this segment from dK = 12 v y2 dm = 12 µv y2 dx and integrate this expression from 0 to L to find the total kinetic energy of the resultant wave. (a) Express the resultant wave function:

yr (x, t ) = y1 (x, t ) + y2 (x, t ) = A1 cos ω1t sin k1 x + A2 cos ω2t sin k 2 x Differentiate this expression with respect to t to find vy:

v y ( x, t ) =

∂ [A1 cos ω1t sin k1 x + A2 cos ω2t sin k2 x] ∂t

= − ω1 A1 sin ω1t sin k1 x − ω2 A2 sin ω2t sin k 2 x

Superposition and Standing Waves 1305 (b) Express the kinetic energy of a segment of the string of length dx and mass dm:

dK = 12 v y2 dm = 12 µv y2 dx = 12 µ (ω1 A1 sin ω1t sin k1 x + ω2 A2 sin ω2t sin k 2 x ) dx 2

=

1 2

µ [ω12 A12 sin 2 ω1t sin 2 k1 x + 2ω1ω2 A1 A2 sin ω1t sin k1 x sin ω2t sin k2 x

]

+ ω22 A22 sin 2 ω2t sin 2 k 2 x dx

(c) Integrate dK from 0 to L to obtain: L

K = µ ∫ ω12 A12 sin 2 ω1t sin 2 k1 xdx 1 2

0

L

+ 12 µ ∫ 2ω1ω2 A1 A2 sin ω1t sin k1 x sin ω2t sin k 2 xdx 0

L

+ 12 µ ∫ ω22 A22 sin 2 ω2t sin 2 k 2 xdx 0

L

= 12 µω12 A12 sin 2 ω1t ∫ sin 2 n1 0

π L

xdx L

+ µω1ω2 A1 A2 sin ω1t sin ω2t ∫ sin n1 0

L

+ 12 µω22 A22 sin 2 ω2t ∫ sin 2 n2 =

(

0

1 2

L

L

x sin n2

π L

xdx

xdx

µω A sin ω1t )( L ) + (µω1ω2 A1 A2 sin ω1t sin ω2t )(0 )n ≠n 2 1

2 1

2

1 2

+ =

π

π

1 4

(

1 2

µω A sin ω2t )( L ) 2 2

2 2

2

1

2

1 2

mω12 A12 sin 2 ω1t + 14 mω22 A22 sin 2 ω2t

Note that, from Problem 92:

1 4

mω12 A12 sin 2 ω1t + 14 mω22 A22 sin 2 ω2t = K1 + K 2

96 ••• Picture the Problem We can use the relationship K max = 14 mω 2 A2 from Problem 92 to express the maximum kinetic energy of the wire and v = fλ and v =

F µ to find an

∆U 1 ⎛ ∂y ⎞ ≈ 2 F ⎜ ⎟ from Problem 15-120 to expression for ω. In part (d) we’ll use ∆x ⎝ ∂x ⎠ 2

determine where the potential energy per unit length has its maximum value. (a) From Problem 92 we have:

K max = 14 mω 2 A2

(1)

1306 Chapter 16 Express ω1 in terms of f1:

ω1 = 2πf1

Relate f1 to the speed of transverse waves on the wire and the wavelength of the fundamental mode:

f1 =

v

v 2L

=

λ1

where L is the length of the wire.

Express the speed of the transverse waves on the wire in terms of the tension in the wire:

v=

F

Substitute and simplify to obtain:

f1 =

1 2L

K max

⎡ F ⎤ 2 π 2F 2 A = m ⎢2π ⎥ A = 4 mL 4 L ⎣ ⎦

Substitute for ω1 and f1 in equation (1) to obtain:

Substitute numerical values and evaluate Kmax:

FL m

=

µ

FL = m

F 4mL 2

1 4

K max =

π 2 (40 N )

(2 ×10 4(2 m )

−2

m

)

2

= 19.7 mJ (b) Express the wave function for a standing wave in its first harmonic: At the instant the transverse displacement is given by (0.02 m) sin (πx/2):

y1 (x, t ) = A1 sin k1 x cos ω1t cos ω1t = 1 ⇒ ω1t = 0 and

K= 0

(2 m ) =

(c) dK is a maximum where the displacement of the wire is greatest; i.e., at its midpoint:

x = 12 L =

(d) From Problem 15-120:

∆U 1 ⎛ ∂y ⎞ ≈ 2 F⎜ ⎟ ∆x ⎝ ∂x ⎠

Express the condition on ∂y ∂x that

∂y ⎛ ∂y ⎞ =⎜ ⎟ ∂x ⎝ ∂x ⎠ max

maximizes ∆U/∆x:

1 2

2

1.00 m

(2)

Superposition and Standing Waves 1307 ∂y1 ∂ = ( A1 sin k1 x cos ω1t ) ∂x ∂x = k1 A1 cos k1 x cos ω1t

Differentiate

y1 (x, t ) = A1 sin k1 x cos ω1t with

respect to x and set the derivative equal to zero for extrema:

=0 or

cos k1 x = 0

Solve for k1x and then x:

k1 x = and

x= =

π 2

π 2k1 1 2

=

πλ = 14 λ = 2(2π )

(2 m ) =

1 4

(2 L )

1.00 m

i.e., the potential energy per unit length is a maximum at the midpoint of the wire. Remarks: In part (d) we’ve shown that ∆U/∆x has an extreme value at x = 1 m. To show that ∆U/∆x is a maximum at this location, you need to examine the sign of the 2nd derivative of y1(x,t) at this point. 97 ••• (a) A spreadsheet program to evaluate f(x) is shown below. Typical cell formulas used are shown in the table. Cell A6 B4 B5

Content/Formula A5+0.1 2*B3+1 (−1)^B$3*COS(B$4*$A5) /B$4*4/PI()

C5

B5+(−1)^C$3*COS(C$4*$A5) /C$4*4/PI()

1 2 3 4 5 6 7 8 9 10

Algebraic Form

x + ∆x 2n + 1 ∞ 4 (−1) n cos((2n + 1)x ) ∑ π n =0 2n + 1

4

(−1) n cos((2n + 1)x ) 2n + 1 n =0 ∞

∑ π

A

B

C

D

K

L

0.0 0.1 0.2 0.3 0.4 0.5

0 1 1.2732 1.2669 1.2479 1.2164 1.1727 1.1174

1 3 0.8488 0.8614 0.8976 0.9526 1.0189 1.0874

2 5 1.1035 1.0849 1.0352 0.9706 0.9130 0.8833

9 19 0.9682 1.0134 1.0209 0.9680 1.0057 1.0298

10 21 1.0289 0.9828 0.9912 1.0286 0.9742 1.0010

1308 Chapter 16 130 131 132 133 134 135

12.5 12.6 12.7 12.8 12.9 13.0

1.2704 1.2725 1.2619 1.2386 1.2030 1.1554

0.8544 0.8503 0.8711 0.9143 0.9740 1.0422

1.0952 1.1013 1.0710 1.0141 0.9493 0.8990

0.9924 0.9752 1.0287 1.0009 0.9691 1.0261

1.0031 1.0213 0.9714 1.0126 1.0146 0.9685

The solid curve is plotted from the data in columns A and B and is the graph of f(x) for 1 term. The dashed curve is plotted from the data in columns A and F and is the graph of f(x) for 5 terms. The dotted curve is plotted from the data in columns A and K and is the graph of f(x) for 10 terms. 1.5

n=1 n=5

1.0

n = 10

0.5

0.0 0

2

4

6

8

10

12

14

-0.5

-1.0 -1.5

(b) Evaluate f(2π) to obtain:

f (2π ) =

4 ⎛ cos 2π cos 3(2π ) − ⎜ 3 π⎝ 1 +

cos 5(2π ) ⎞ − ...⎟ 5 ⎠

4⎛ 1 1 1 ⎞ ⎜1 − + − + ... ⎟ π⎝ 3 5 7 ⎠ =1 =

which is equivalent to the Liebnitz formula. 98 ••• (a) A spreadsheet program to evaluate f(x) is shown below. Typical cell formulas used are shown in the table. Cell A6 B4 B5

Content/Formula A5+0.1 2*B3+1 (−1)^$B$3*sin($B$4*A5)/ ($B$4)^2*4/PI()

Algebraic Form

x + ∆x 2n + 1

(− 1)n sin (2n + 1)x ∑ (2n + 1)2 π n 4

Superposition and Standing Waves 1309 C5

(− 1)n sin (2n + 1)x ∑ (2n + 1)2 π n

B5+((−1)^$C$3*sin($C$4*A5)/ ($C$4)^2*4/PI()

4

A

B

C

D

K

L

1 2 3 4 5 6 7 8 9 10

0.0 0.1 0.2 0.3 0.4 0.5

0 1 0.0000 0.1271 0.2530 0.3763 0.4958 0.6104

1 3 0.0000 0.0853 0.1731 0.2654 0.3640 0.4693

2 5 0.0000 0.1097 0.2159 0.3163 0.4103 0.4998

9 19 0.0000 0.0986 0.2012 0.3004 0.3983 0.5011

10 21 0.0000 0.1011 0.1987 0.3005 0.4008 0.4985

72 73 74 75 76

6.7 6.8 6.9 7.0 7.1

0.5155 0.6291 0.7365 0.8365 0.9282

0.3812 0.4877 0.6005 0.7181 0.8380

0.4256 0.5146 0.6034 0.6963 0.7968

0.4153 0.5183 0.6171 0.7148 0.8183

0.4171 0.5154 0.6182 0.7166 0.8155

Graphs of f(x) for 1, 5, and 10 terms are shown below. Note that there is little difference between the graphs for 5 terms and 10 terms of this triangular wave function.

2.0 1.5

n=1

1.0

n=5 n = 10

0.5 0.0 -0.5 -1.0 -1.5 -2.0 0

1

2

3

4

5

6

7

99 ••• Picture the Problem From the diagram above, the nth echo will reflect n − 1 times going out, and the same number of times going back. If we "unfold" the ray into a straight line, we get the representation shown below. Using this figure we can express the distance dn traveled by the nth echo and then use this result to express the time delay between the nth and n + 1th echoes. The reciprocal of this time delay is the frequency corresponding to the nth echo.

1310 Chapter 16

(a) Apply the Pythagorean theorem to the right triangle whose base is L, whose height is 2(n − 1), and whose hypotenuse is dn to obtain:

d n = 2 4(n − 1) 2 r 2 + L2

Express the time delay between the nth and n + 1th echoes:

∆tn =

dn v

Substitute to obtain:

∆tn =

2⎛ ⎜ v⎝

(2n )2 r 2 + L2

−

[2(n − 1)] 2 r 2 + L2 ⎞⎟⎠

A spreadsheet program to calculate ∆tn as a function of n is shown below. The constants and cell formulas used are shown in the table. Cell B1 B2 B3 B8 C7

Content/Formula 90 1 340 B7+1 2/$B$3*((2*(B7−1) *$B$2)^2+$B$1^2)^0.5 A

1 2 3 4 5 6 7 8 9 10 11

Algebraic Form L r c n+1

∆tn

B L= 90 r= 1 c= 340

C

D

t(n) 0.5294 0.5295 0.5299 0.5306 0.5315

delta t(n) 0.0001 0.0004 0.0007 0.0009 0.0012

m m m/s

n 1 2 3 4 5

Superposition and Standing Waves 1311 202 203 204 205 206

196 197 198 199 200

2.3544 2.3659 2.3773 2.3888 2.4003

0.0115 0.0115 0.0115 0.0115 0.0115

The graph of ∆tn as a function of n shown below was plotted using the data from columns B and D.

0.012 0.010

Delta-t (n )

0.008 0.006 0.004 0.002 0.000 0

50

100

150

200

n

(c)

The frequency heard at any time is 1/∆tn , so because ∆tn increases over time, the frequency of the culvert whistler decreases.

The highest frequency corresponds to n = 1 and is given by: Substitute for ∆t1 to obtain:

Substitute numerical values and evaluate fhighest:

1 ∆t1 1 = = ∆t1 2⎛⎜ ⎝

f highest = f highest

f highest =

v

(2) r 2 + L2 − 2

L2

)

340 m/s 2 2⎛⎜ 4(1 m ) + (90 m ) − 90 m ) ⎝ 2

= 7.65 kHz The lowest frequency end can be found by examining the limit of ∆tn as n → ∞ :

⎡2 ⎛ L2 L2 2 ( ) − − + lim n→∞ ∆tn = lim n→∞ ⎢ ⎜ (2n ) r 2 + 2 1 n r (2n )2 (2(n − 1))2 ⎢⎣ v ⎜⎝ 2r 4r = (2n − 2n + 2) = v v

⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦

1312 Chapter 16 Express flowest in terms of ∆t∞:

Substitute numerical values and evaluate flowest:

f lowest =

1 v = ∆t∞ 4r

f lowest =

340 m/s = 85.0 Hz 4(1 m )

Chapter 17 Temperature and the Kinetic Theory of Gases Conceptual Problems *1 • (a) False. If two objects are in thermal equilibrium with a third, then they are in thermal equilibrium with each other. (b) False. The Fahrenheit and Celsius temperature scales differ in the number of intervals between the ice-point temperature and the steam-point temperature. (c) True. (d) False. The result one obtains for the temperature of a given system is thermometerdependent. 2 • Determine the Concept Put each in thermal equilibrium with a third body; e.g., a thermometer. If each body is in thermal equilibrium with the third, then they are in thermal equilibrium with each other. 3 • Picture the Problem We can decide which room was colder by converting 20°F to the equivalent Celsius temperature. Using the Fahrenheit-Celsius conversion, convert 20°F to the equivalent Celsius temperature:

tC =

5 9

(tF − 32°) = 59 (20° − 32°)

= −6.67°C so Mert' s room was colder.

4 •• Picture the Problem We can apply the ideal-gas law to the two vessels to decide which of these statements is correct. Apply the ideal-gas law to the particles in vessel 1: Apply the ideal-gas law to the particles in vessel 2: Divide the equation for vessel 1 by the equation for vessel 2:

P1V1 = N1kT1

P2V2 = N 2 kT2

P1V1 N kT = 1 1 P2V2 N 2 kT2

1313

Chapter 17

1314

Because the vessels are identical and are at the same temperature and pressure:

1=

N1 and N1 = N 2 N2

(a ) is correct.

5 •• Determine the Concept From the ideal-gas law, we have P = nR T V . In the process depicted, both the temperature and the volume increase, but the temperature increases faster than does the volume. Hence, the pressure increases. *6

••

Determine the Concept From the ideal-gas law, we have V = nR T P. In the process depicted, both the temperature and the pressure increase, but the pressure increases faster than does the temperature. Hence, the volume decreases. 7 • True. The kinetic energy of translation K for n moles of gas is directly proportional to the absolute temperature T of the gas (K = 32 nkT ) . 8 • Determine the Concept We can use vrms = 3RT M to relate the temperature of a gas to the rms speed of its molecules. Express the dependence of the rms speed of the molecules of a gas on their absolute temperature:

vrms =

3RT M

where R is the gas constant, M is the molar mass, and T is the absolute temperature.

Because vrms ∝ T , the temperature must be quadrupled in order to double the rms speed of the molecules. 9 • Picture the Problem The average kinetic energy of a molecule, as a function of the temperature, is given by K av = 32 kT and the pressure, volume, and temperature of an ideal gas are related according to PV = NkT . Express the average kinetic energy of a molecule in terms of its temperature:

K av = 32 kT

Temperature and the Kinetic Theory of Gases 1315 From the ideal-gas law we have:

PV = NkT

Eliminate kT between these equations to obtain:

K av =

3PV 2N

If P is doubled at constant V , K av increases by a factor of 2. If V is doubled at constant P, K av increases by a factor of 2. 10 •• Picture the Problem We can express the rms speeds of the helium atoms and the methane molecules using vrms = 3RT M .

3RT M He

Express the rms speed of the helium atoms:

vrms (He ) =

Express the rms speed of the methane molecules:

vrms (CH 4 ) =

3RT M CH4

Divide the first of these equations by the second to obtain:

vrms (He ) = vrms (CH 4 )

M CH4

Use Appendix C to find the molar masses of helium and methane:

vrms (He ) 16 g/mol = =2 vrms (CH 4 ) 4 g/mol

M He

and (b) is correct. 11 • False. Whether the pressure changes also depends on whether and how the volume changes. In an isothermal process, the pressure can increase while the volume decreases and the temperature is constant. 12 • Determine the Concept For the Celsius scale, the ice point (0°C) and the boiling point of water at 1 atm (100°C) are more convenient than 273 K and 373 K; temperatures in roughly this range are normally encountered. On the Fahrenheit scale, the temperature of warm-blooded animals is roughly 100°F; this may be a more convenient reference than approximately 300 K. Throughout most of the world, the Celsius scale is the standard for nonscientific purposes.

1316

Chapter 17

*13 • Determine the Concept Because 107 >> 273, it does not matter. 14 • Determine the Concept The average speed of the molecules in an ideal gas depends on the square root of the kelvin temperature. Because vav ∝ T , doubling the temperature while maintaining constant pressure increases the average speed by a factor of

2.

(d ) is correct. 15 • Determine the Concept From the ideal-gas law, we have PV = nRT . Halving both the temperature and volume of the gas leaves the pressure unchanged.

(b) is correct.

16 • Determine the Concept The average translational kinetic energy of the molecules of an ideal gas is given by K = 32 NkT = 32 nRT . The temperature of the ideal gas is related to the pressure of the gas.

(d ) is correct.

17 • Determine the Concept The only conclusion we can draw from the information that the vessel contains equal amounts, by weight, of helium and argon is that the temperatures of the helium and argon molecules are the same.

(d ) is correct.

18 •• Determine the Concept The two rooms are in thermal equilibrium and, because they are connected, the air in each is at the same pressure. Because P = NkT/V, and the volume of each room is identical, NATA = NBTB, so the cooler room (A) has more air in it. 19 • Determine the Concept The rms speed of an ideal gas is given by vrms = 3RT M and its average kinetic energy by K av = 32 kT . Because the gases are at the same temperature, their average kinetic energies are the same. Because vrms = 3RT M , the rms speeds are inversely proportional to the square root of the molecular masses. 20 •• Determine the Concept The pressure is a measure of the change in momentum per second of a gas molecule on collision with the wall of the container. When the gas is heated, the average velocity, the average momentum, and pressure of the molecules increase.

Temperature and the Kinetic Theory of Gases 1317 *21 •• Determine the Concept Because the temperature remains constant, the average speed of the molecules remains constant. When the volume decreases, the molecules travel less distance between collisions, so the pressure increases because the frequency of collisions increases. 22 •• Picture the Problem The average kinetic energies of the molecules are given by K av = 12 mv 2 av = 32 kT . Assuming that the room’s temperature distribution is uniform,

(

)

we can conclude that the oxygen and nitrogen molecules have equal average kinetic energies. Because the oxygen molecules are more massive, they must be moving slower than the nitrogen molecules.

(b) is correct.

23 •• Determine the Concept The average molecular speed of He gas at 300 K is about 1.4 km/s, so a significant fraction of He molecules have speeds in excess of earth’s escape velocity (11.2 km/s). Thus, they "leak" away into space. Over time, the He content of the atmosphere decreases to almost nothing.

Estimation and Approximation *24 •• Picture the Problem Assuming the steam to be an ideal gas at a temperature of 373 K, we can use the ideal-gas law to estimate the pressure inside the test tube when the water is completely boiled away.

NkT V

Using the ideal-gas law, relate the pressure inside the test tube to its volume and the temperature:

P=

Relate the number of particles N to the mass of water, its molar mass M, and Avogadro’s number NA: Solve for N:

m M = N NA N =m

NA M

(

)(

)

Relate the mass of 1 mL of water to its density:

m = ρV = 103 kg/m3 10−6 m3 = 1g

Substitute for m, NA, and M (18 g/mol) and evaluate N:

6.022 × 1023 particles/mol N = (1g ) 18 g/mol = 3.35 × 1022 particles

1318

Chapter 17

Substitute numerical values and evaluate P:

(3.35 × 10

)(

)

particles 1.381 × 10−23 J/K (373 K ) 10 × 10− 6 m3 1atm = 172 × 105 N/m 2 × 1.01 × 105 N/m 2

P=

22

= 171atm 25 ••• Picture the Problem We can find the escape temperatures for the earth and the moon by equating, in turn, 0.15ve and vrms of O2 and H2. We can compare these temperatures to explain the absence from the earth’s upper atmosphere and from the surface of the moon. (a) Express vrms for O2:

3RT M

vrms =

where R is the gas constant, T is the absolute temperature, and M is the molar mass of oxygen. Equate 0.15ve and vrms:

0.15 2 gRearth =

3RT M

0.045 gRearth M 3R

Solve for T to obtain:

T=

Evaluate T for O2:

0.045 9.81m/s 2 6.37 × 106 m T= 3(8.314 J/mol ⋅ K )

(

(1)

)(

(

× 32 × 10− 3 kg/mol

)

= 3.61 × 103 K (b) Substitute numerical values and evaluate T for H2:

T=

0.045(9.81m/s 2 )(6.37 × 106 m ) 3(8.314 J/mol ⋅ K ) × (2 × 10− 3 kg/mol)

= 225 K

(c)

)

If vrms > 15 ve or T ≥ 25Tatm , H 2 molecules escape. Therefore, the more energetic H 2 molecules escape from the upper atmosphere.

Temperature and the Kinetic Theory of Gases 1319 (d) Express equation (1) at the surface of the moon:

0.045 g moon Rmoon M 3R 0.045( 16 g earth )Rmoon M = 3R 0.0025 g earth Rmoon M = R

T=

Substitute numerical values and evaluate T for O2:

T=

0.0025(9.81m/s 2 )(1.738 × 106 m )(32 × 10−3 kg/mol) = 164 K 8.314 J/mol ⋅ K

Substitute numerical values and evaluate T for H2:

T=

0.0025(9.81m/s 2 )(1.738 × 106 m )(2 × 10−3 kg/mol) = 10.3 K 8.314 J/mol ⋅ K

If we assume that the temperature on the moon with an atmosphere would have been approximately 1000 K, then all O 2 and H 2 would have escaped during the time since the formation of the moon to the present. 26 •• Picture the Problem We can use vrms = 3RT M to calculate the rms speeds of H2, O2, and CO2 at 273 K and then compare these speeds to 20% of the escape velocity on Mars to decide the likelihood of finding these gases in the atmosphere of Mars.

3RT M

Express the rms speed of an atom as a function of the temperature:

vrms =

(a) Substitute numerical values and evaluate vrms for H2:

vrms, H 2 =

3(8.314 J/mol ⋅ K )(273 K ) 2 × 10− 3 kg/mol

= 1.85 km/s (b) Evaluate vrms for O2:

vrms,O 2 =

3(8.314 J/mol ⋅ K )(273 K ) 32 × 10 − 3 kg/mol

= 461m/s

1320

Chapter 17

(c) Evaluate vrms for CO2:

3(8.314 J/mol ⋅ K )(273 K ) 44 × 10 − 3 kg/mol

vrms,CO 2 =

= 393 m/s (d) Calculate 20% of vesc for Mars:

v = 15 vesc =

1 5

(5 km/s) = 1 km/s

Because v is greater than vrms for CO 2 and O 2 but less than vrms for H 2 O 2 and CO 2 , but not H 2 , should be present. *27 •• Picture the Problem We can use vrms = 3RT M to calculate the rms speeds of H2, O2, and CO2 at 123 K and then compare these speeds to 20% of the escape velocity on Jupiter to decide the likelihood of finding these gases in the atmosphere of Jupiter.

3RT M

Express the rms speed of an atom as a function of the temperature:

vrms =

(a) Substitute numerical values and evaluate vrms for H2:

vrms, H 2 =

3(8.314 J/mol ⋅ K )(123 K ) 2 × 10− 3 kg/mol

= 1.24 km/s (b) Evaluate vrms for O2:

vrms,O 2 =

3(8.314 J/mol ⋅ K )(123 K ) 32 × 10− 3 kg/mol

= 310 m/s (c) Evaluate vrms for CO2:

vrms,CO 2 =

3(8.314 J/mol ⋅ K )(123 K ) 44 × 10− 3 kg/mol

= 264 m/s (d) Calculate 20% of vesc for Jupiter:

v = 15 vesc =

1 5

(60 km/s) = 12 km/s

Because v is greater than vrms for O 2 , CO 2 , and H 2 , O 2 , CO 2 , and H 2 should be found on Jupiter.

Temperature and the Kinetic Theory of Gases 1321

Temperature Scales 28 • Picture the Problem We can convert both of these temperatures to the Fahrenheit scale and then express their difference to find the range of temperatures. Solve the Fahrenheit-Celsius conversion equation for the Fahrenheit temperature: Find the Fahrenheit equivalent of −12°C: Find the Fahrenheit equivalent of −7°C: The difference between these two temperatures is the range on the Fahrenheit scale:

tF = 95 tC + 32°

tF =

9 5

(− 12°) + 32° = 10.4°

tF =

9 5

(− 7°) + 32° = 19.4°F

Range = 19.4°F − 10.4°F

= 9.00F°

Remarks: We could take advantage of the fact that 9 F° = 5 C° to arrive at the aforementioned result in which the range of Celsius temperatures happens to be 5C°. If the temperature difference were other than 5C°, we could set up a proportion to quickly find the range on the Fahrenheit scale. 29 • Picture the Problem We can use the Fahrenheit-Celsius conversion equation to find this temperature on the Celsius scale. tC = 95 (t F − 32°) = 95 (1945.4° − 32°) Convert 1945.4°F to the equivalent Celsius temperature: = 1063°C *30 • Picture the Problem We can use the Fahrenheit-Celsius conversion equation to express the temperature of the human body on the Celsius scale. Convert 98.6°F to the equivalent Celsius temperature:

tC =

5 9

(tF − 32°) = 95 (98.6° − 32°)

= 37.0°C

1322

Chapter 17

31 • Picture the Problem While we could use Equation 17-1 to relate the Celsius temperature to the length of the column of mercury in the thermometer, an alternative solution is to use the diagram to the right to set up a proportion that will relate the Celsius temperature to the calibration temperatures and to the lengths of the mercury column. Using the diagram, set up a proportion relating the temperatures to the lengths of the column of mercury:

tc − 0°C L − L0 = t 100°C − 0°C L100 − L0

tc (L100 − L0 ) + L0 100° t (24.0 cm − 4.0 cm ) = c + 4.0 cm 100° t (20.0 cm ) = c + 4.0 cm 100°

Solve for and evaluate Lt:

Lt =

(a) Substitute 22.0°C for tc and evaluate Lt:

Lt =

(22.0°C)(20.0 cm ) + 4.0 cm 100°

= 8.40 cm (b) Substitute 25.4 cm for Lt and evaluate tc:

tc =

25.4 cm − 4.0 cm × 100° 20 cm

= 107°C 32 • Picture the Problem We can use the temperature conversion equations tF = 95 tC + 32° and tC = T − 273.15 K to convert 107 K to the Fahrenheit and Celsius temperatures. Express the kelvin temperature in terms of the Celsius temperature:

T = tC + 273.15 K

Temperature and the Kinetic Theory of Gases 1323 (a) Solve for and evaluate tC:

tC = T − 273.15 K = 107 K − 273.15 K ≈ 107 K

(b) Use the Celsius to Fahrenheit conversion equation to evaluate tF:

(

)

t F = 95 10 7 °C + 32° ≈ 1.80 × 10 7 °F

33 • Picture the Problem While we could convert 77.35 K to a Celsius temperature and then convert the Celsius temperature to a Fahrenheit temperature, an alternative solution is to use the diagram to the right to set up a proportion for the direct conversion of the kelvin temperature to its Fahrenheit equivalent. Use the diagram to set up the proportion:

32°F − t F 273.15 K − 77.35 K = 212°F − 32°F 373.15 K − 273.15 K or

32°F − tF 195 = 180 F° 100 Solve for and evaluate tF:

tF = 32°F −

195 × 180 F° = − 319 °F 100

34 • Picture the Problem We can use the fact that, for a constant-volume gas thermometer, the pressure and absolute temperature are directly proportional to calibrate the given thermometer; i.e., to find the constant of proportionality relating P and T. We can then use this equation to find the temperature corresponding to a given pressure or the pressure corresponding to a given temperature.

P = CT

Express the direct proportionality between the pressure and the temperature:

where C is a constant.

Use numerical values to evaluate C:

C=

P 0.400 atm = T 273.15 K

= 1.464 × 10 −3 atm/K

1324

Chapter 17

Substitute to obtain:

P = (1.464 × 10−3 atm/K )T

(1)

or

T = (682.9 K/atm )P

(a) Use equation (2) to find the temperature:

(b) Use equation (1) to find the boiling point of sulfur:

(2)

T = (682.9 K/atm )(0.1atm )

= 68.3 K

(

)

P = 1.464 × 10−3 atm/K × (444.6 + 273.15)K

= 1.05 atm *35 • Picture the Problem We can use the information that the thermometer reads 50 torr at the triple point of water to calibrate it. We can then use the direct proportionality between the absolute temperature and the pressure to either the pressure at a given temperature or the temperature for a given pressure. Using the ideal-gas temperature scale, relate the temperature to the pressure:

(a) Solve for and evaluate P when T = 300 K:

T=

273.16 K 273.16 K P= P P3 50 torr

= (5.463 K/torr )P

P = (0.1830 torr/K )T

= (0.1830 torr/K )(300 K ) = 54.9 torr

(b) Find T when the pressure is 678 torr:

T = (5.463 K/torr )(678 torr ) = 3704 K

36 • Picture the Problem We can use the equation for the ideal-gas temperature scale to express the temperature measured by this thermometer in terms of its pressure and the given data to calibrate the thermometer. Write the equation for the ideal-gas temperature scale:

T=

273.16 K P P3

17-4

Temperature and the Kinetic Theory of Gases 1325 (a) Solve for and evaluate the thermometer’s triple-point pressure:

P3 =

273.16 K 273.16 K (30 torr ) P= T 373 K

= 22.0 torr (b) Substitute for P3 in Equation 17-4:

T=

273.16 K 273.16 K (0.175 torr ) P= 22.0 torr 22.0 torr

= 2.17 K 37 • Picture the Problem We can find the temperature at which the Fahrenheit and Celsius scales give the same reading by setting tF = tC in the temperature-conversion equation. Set tF = tC in tC =

5 9

(tF − 32°) :

Solve for and evaluate tF:

tF =

5 9

(tF − 32°)

tC = tF = − 40.0°C = − 40.0°F

Remarks: If you’ve not already thought of doing so, you might use your graphing calculator to plot tC versus tF and tF = tC (a straight line at 45°) on the same graph. Their intersection is at (−40, −40). 38 • Picture the Problem We can use the Celsius-to-absolute conversion equation to find 371 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation to find the Fahrenheit temperature corresponding to 371 K. Express the absolute temperature as a function of the Celsius temperature: Solve for and evaluate tC:

T = tC + 273.15 K

tC = T − 273.15 K = 371K − 273.15 K = 97.9°C

Use the Celsius-to-Fahrenheit conversion equation to find tF:

tF = 95 tC + 32° =

9 5

(97.9°) + 32°

= 208°F

39 • Picture the Problem We can use the Celsius-to-absolute conversion equation to find 90.2 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation to find the

1326

Chapter 17

Fahrenheit temperature corresponding to 90.2 K. Express the absolute temperature as a function of the Celsius temperature: Solve for and evaluate tC:

T = tC + 273.15 K

tC = T − 273.15 K = 90.2 K − 273.15 K = − 183°C

Use the Celsius-to-Fahrenheit conversion equation to find tF:

tF = 95 tC + 32° =

9 5

(− 183°) + 32°

= − 297°F

40 •• Picture the Problem We can use the diagram to the right to set up proportions that will allow us to convert temperatures on the Réaumur scale to Celsius and Fahrenheit temperatures.

Referring to the diagram, set up a proportion to convert temperatures on the Réaumur scale to Celsius temperatures: Simplify to obtain:

Referring to the diagram, set up a proportion to convert temperatures on the Réaumur scale to Fahrenheit temperatures: Simplify to obtain:

tC − 0°C t − 0°R = R 100°C − 0°C 80°R − 0°R

tC t = R or tC = 1.25tR 100 80 t F − 32°F t − 0°R = R 212°F − 32°F 80°R − 0°R

t F − 32 t R or tF = = 180 80

t + 32

9 4 R

*41 ••• Picture the Problem We can use the temperature dependence of the resistance of the thermistor and the given data to determine R0 and B. Once we know these quantities, we

Temperature and the Kinetic Theory of Gases 1327 can use the temperature-dependence equation to find the resistance at any temperature in the calibration range. Differentiation of R with respect to T will allow us to express the rate of change of resistance with temperature at both the ice point and the steam point temperatures. (a) Express the resistance at the ice point as a function of temperature of the ice point:

7360 Ω = R0e B 273 K

(1)

Express the resistance at the steam point as a function of temperature of the steam point:

153 Ω = R0e B 373 K

(2)

Divide equation (1) by equation (2) to obtain:

7360 Ω = 48.10 = e B 273 K − B 373 K 153 Ω

Solve for B by taking the logarithm of both sides of the equation:

1 ⎞ −1 ⎛ 1 − ln 48.1 = B⎜ ⎟K ⎝ 273 373 ⎠ and

ln 48.1 = 3.94 × 103 K 1 ⎞ −1 ⎛ 1 − ⎟K ⎜ ⎝ 273 373 ⎠

B=

Solve equation (1) for R0 and substitute for B:

R0 =

7360 Ω = (7360 Ω )e − B 273 K B 273 K e

= (7360 Ω )e − 3.94×10

3

K 273 K

= 3.97 × 10− 3 Ω (b) From (a) we have:

R = (3.97 × 10−3 Ω )e3.94×10

Convert 98.6°F to kelvins to obtain:

T = 310 K

Substitute to obtain:

R = 3.97 × 10−3 Ω e3.94×10

(

)

3

K T

3

K 310 K

= 1.31kΩ (c) Differentiate R with respect to T to obtain:

(

)

dR d d ⎛B⎞ R0e B T = R0e B T = ⎜ ⎟ dT dT dT ⎝ T ⎠ −B RB = 2 R0e B T = − 2 T T

1328

Chapter 17

Evaluate dR/dT at the ice point:

(

(7360 Ω ) 3.94 × 103 K ⎛ dR ⎞ = − ⎟ ⎜ (273.16 K )2 ⎝ dT ⎠ice point

)

= − 389 Ω / K Evaluate dR/dT at the steam point:

(

(153 Ω ) 3.94 × 103 K ⎛ dR ⎞ = − ⎜ ⎟ (373.16 K )2 ⎝ dT ⎠steam point

)

= − 4.33 Ω / K

(d)

The thermistor is more sensitive; i.e., it has greater sensitivity at lower temperatures.

The Ideal-Gas Law 42 • Picture the Problem Let the subscript 1 refer to the gas at 50°C and the subscript 2 to the gas at 100°C. We can apply the ideal-gas law for a fixed amount of gas to find the ratio of the final and initial volumes. Apply the ideal-gas law for a fixed amount of gas:

P2V2 P1V1 = T2 T1 or, because P2 = P1,

V2 T2 = V1 T1 Substitute numerical values and evaluate V2/V1:

V2 (273.15 + 100)K = = 1.15 (273.15 + 50)K V1

43 • Picture the Problem We can use the ideal-gas law to find the number of moles of gas in the vessel and the definition of Avogadro’s number to find the number of molecules. Apply the ideal-gas law to the gas:

PV = nRT

Solve for the number of moles of gas in the vessel:

n=

PV RT

Temperature and the Kinetic Theory of Gases 1329 Substitute numerical values and evaluate n:

n=

(8.206 × 10

(4 atm )(10 L ) −2

)

L ⋅ atm/mol ⋅ K (273 K )

= 1.79 mol Relate the number of molecules N in the gas in terms of the number of moles n:

N = nN A

Substitute numerical values and evaluate N:

(

)

N = (1.79 mol) 6.022 × 10 23 molecules/mol = 1.08 × 10 24 molecules 44 •• Picture the Problem We can use the ideal-gas law to relate the number of molecules in the gas to its pressure, volume, and temperature. Solve the ideal-gas law for the number of molecules in a gas as a function of its pressure, volume, and temperature:

N=

Substitute numerical values and evaluate N:

N=

PV kT

(10

−8

)

(

torr (133.32 Pa/torr ) 10−6 m 3 1.381 × 10− 23 J/K (300 K )

(

)

= 3.22 × 108 45 •• Picture the Problem The pictorial representation to the right, in which T0 represents absolute zero, summarizes the information concerning the temperatures and pressures we are given. We know, from the ideal-gas law, that the pressure of a fixed volume of gas is proportional to its absolute temperature. We can use the diagram to set up a proportion relating the temperatures and pressures that we can solve for T0. Apply the ideal-gas law to obtain:

22 glips − T0 − 10 glips − T0 = 12.5 klads 8.7 klads

Solve for T0 to obtain:

T0 = − 83.2 glips

)

1330

Chapter 17

Remarks: Because the gas is ideal, its pressure is directly proportional to its temperature. Hence, a graph of P versus T will be linear and the linear equation relating P and T can be solved for the temperature corresponding to zero pressure. 46 •• Picture the Problem Let the subscript 1 refer to the tires when their pressure is 180 kPa and the subscript 2 to conditions when their pressure is 245 kPa. Assume that the air in the tires behaves as an ideal gas. Then, we can apply the ideal-gas law for a fixed amount of gas to relate the temperatures to the pressures and volumes of the tires. (a) Apply the ideal-gas law for a fixed amount of gas to the air in the tires:

P2V2 P1V1 = T2 T1

Solve for T2:

T2 = T1

Substitute numerical values to obtain:

T2 = (265 K )

(1)

P2 because V1 = V2. P1 245 kPa = 360.7 K 180 kPa

= 87.7°C (b) Use equation (1) with V2 = 1.07 V1. Solve for T2:

Substitute numerical values and evaluate T2:

T2 =

⎛P ⎞ P2V2 T1 = 1.07⎜⎜ 2 T1 ⎟⎟ P1V1 ⎝ P1 ⎠

T2 = 1.07(360.7 K ) = 385.9 K = 113°C

47 •• Picture the Problem We can apply the ideal-gas law to find the number of moles of air in the room as a function of the temperature. (a) Use the ideal-gas law to relate the number of moles of air in the room to the pressure, volume, and temperature of the air:

n=

PV RT

Substitute numerical values and evaluate n:

n=

(101.3 kPa )(90 m3 ) (8.314 J/mol ⋅ K )(300 K )

= 3.66 × 103 mol

(1)

Temperature and the Kinetic Theory of Gases 1331 (b) Letting n′ represent the number of moles in the room when the temperature rises by 5 K, express the number of moles of air that leave the room:

∆n = n − n'

Apply the ideal-gas law to obtain:

n' =

Divide equation (2) by equation (1) to obtain:

n' T T = and n' = n n T' T'

Substitute for n′ to obtain:

∆n = n − n

Substitute numerical values and evaluate ∆n:

⎛ 300 K ⎞ ⎟⎟ ∆n = 3.66 × 103 mol ⎜⎜1 − ⎝ 305 K ⎠

PV RT '

(2)

T ⎛ T⎞ = n⎜1 − ⎟ T' ⎝ T' ⎠

(

)

= 60.0 mol *48 •• Picture the Problem Let the subscript 1 refer to helium gas at 4.2 K and the subscript 2 to the gas at 293 K. We can apply the ideal-gas law to find the volume of the gas at 4.2 K and a fixed amount of gas to find its volume at 293 K.

nRT1 P1

(a) Apply the ideal-gas law to the helium gas and solve for its volume:

V1 =

Substitute numerical values to obtain:

V1 = n

Find the number of moles in 10 g of helium:

n=

Substitute for n to obtain:

V1 = (0.3447 L / mol)(2.5 mol)

(0.08206 L ⋅ atm/mol ⋅ K )(4.2 K )

= (0.3447 L / mol)n

10 g = 2.5 mol 4 g/mol

= 0.862 L (b) Apply the ideal-gas law for a fixed amount of gas and solve for

1atm

P2V2 P1V1 = T2 T1

1332

Chapter 17

the volume of the helium gas at 293 K:

and, because P1 = P2,

V2 =

T2 V1 T1

Substitute numerical values and evaluate V2:

V2 =

293 K (0.862 L ) = 60.1L 4.2 K

49 •• Picture the Problem Because the helium is initially in the liquid state, its temperature must be 4.2 K. Let the subscript 1 refer to helium gas at 4.2 K and the subscript 2 to the gas at 293 K. We can apply the ideal-gas law for a fixed volume of gas to relate the pressure at 293 K to the pressure at 4.2 K and use the ideal-gas law to find the pressure at 4.2 K. Apply the ideal-gas law for a fixed amount of gas:

P2V2 P1V1 = T2 T1

Solve for its pressure at 293 K:

P2 =

P1V1T2 T = P1 2 V2T1 T1

(1)

because V1 = V2

nRT1 V1

Apply the ideal-gas law to the helium gas at 4.2 K and solve for its pressure:

P1 =

Substitute numerical values to obtain:

P1 = n

Find the number of moles in 10 g of helium:

n=

Substitute for n to obtain:

P1 = (0.05744 atm / mol)(2.5 mol)

(0.08206 L ⋅ atm/mol ⋅ K )(4.2 K )

6L = (0.05744 atm / mol)n 10 g = 2.5 mol 4 g/mol

= 0.1436 atm Substitute in equation (1) and evaluate P2:

P2 = (0.1436 atm )

293 K = 10.0 atm 4.2 K

*50 •• Picture the Problem Let the subscript 1 refer to the tire when its temperature is 20°C and the subscript 2 to conditions when its temperature is 50°C. We can apply the ideal-

Temperature and the Kinetic Theory of Gases 1333 gas law for a fixed amount of gas to relate the temperatures to the pressures of the air in the tire. (a) Apply the ideal-gas law for a fixed amount of gas and solve for pressure at the higher temperature:

P2V2 P1V1 = T2 T1

(1)

and

P2 =

T2 P1 T1

because V1 = V2 Substitute numerical values to obtain:

323 K (200 kPa + 101 kPa ) 293 K = 332 kPa

P2 =

and

P2,gauge = 332 kPa − 101 kPa = 231 kPa

(b) Solve equation (1) for P2 with V2 = 1.1 V1 and evaluate P2:

P2 =

V1T2 P1 V2T1

323 K (200 kPa + 101 kPa ) 1.1(293 K ) = 302 kPa =

and

P2,gauge = 302 kPa − 101 kPa = 201 kPa 51 •• Picture the Problem Let ρ N2 and ρ 02 be the number densities (i.e., the number of particles per unit volume) of N2 and O2, respectively. We can express the density of air in terms of the densities of nitrogen and oxygen and their number densities as ρ air = mN2 ρ N2 + m02 ρ 02 . By applying the ideal-gas law, we can find the number density of air and, using the given composition of air, calculate the number densities of nitrogen and oxygen. Finally, we can find the masses of nitrogen and oxygen molecules from their atomic masses. Knowing ρ N2 , ρ O 2 , mN 2 , and mO 2 , we can calculate ρair. Express the density of air in terms of the densities of nitrogen and oxygen:

ρ air = mN ρ N + m0 ρ 0

Using the ideal-gas law, relate the number density of air N/V to its temperature and pressure:

PV = NkT and

2

2

2

2

N P = V kT

(1)

1334

Chapter 17

Substitute numerical values and evaluate the number density of air:

N 1.01 × 105 N/m 2 = V 1.381 × 10− 23 J/K (297 K )

(

)

= 2.46 × 10 m 25

Because air is approximately 74% N2 and 26% O2:

−3

2

and

Calculate the masses of N2 and O2 molecules:

)

(

)

N = 0.26 2.46 × 1025 m − 3 V = 6.40 × 1024 m − 3

ρ O = 0.26 2

(

N = 0.74 2.46 × 1025 m − 3 V = 1.82 × 1025 m − 3

ρ N = 0.74

(

mN 2 = (28 u ) 1.660 × 10−27 kg/u

)

= 4.65 × 10− 26 kg and

(

mO 2 = (32 u ) 1.660 × 10−27 kg/u

)

= 5.31 × 10− 26 kg Substitute in equation (1) and evaluate ρair:

ρ air = (4.65 × 10 −26 kg )(1.82 × 10 25 m −3 )

(

)(

+ 5.31× 10 −26 kg 6.40 × 10 24 m −3 = 1.19 kg/m 3

52 •• Picture the Problem Let the subscript 1 refer to the conditions at the bottom of the lake and the subscript 2 to the surface of the lake and apply the ideal-gas law for a fixed amount of gas. Apply the ideal-gas law for a fixed amount of gas:

P2V2 P1V1 = T2 T1

Solve for the volume of the bubble just before it breaks the surface:

V2 = V1

Find the pressure at the bottom of the lake:

T2 P1 T1P2

P1 = Patm + ρgh = 101.3 kPa

(

)(

)

+ 103 kg/m 3 9.81m/s 2 (40 m ) = 493.7 kPa

)

Temperature and the Kinetic Theory of Gases 1335 Substitute numerical values and evaluate V2:

(

V2 = 15 cm3

K )(493.7 kPa ) ) ((298 278 K )(101.3 kPa )

= 78.4 cm3 53 •• Picture the Problem Assume that the volume of the balloon is not changing. Then the air inside and outside the balloon must be at the same pressure of about 1 atm. The contents of the balloon are the air molecules inside it. We can use Archimedes principle to express the buoyant force on the balloon and we can find the weight of the air molecules inside the balloon Express the net force on the balloon and its contents:

Fnet = B − wair inside the balloon

Using Archimedes principle, express the buoyant force on the balloon:

B = wdisplaced fluid = mdisplaced fluid g

(1)

or

B = ρoVballoon g where ρo is the density of the air outside the balloon.

Express the weight of the air inside the balloon:

wair inside the balloon = ρiVballoon g where ρi is the density of the air inside the balloon.

Substitute in equation (1) for B and wair inside the balloon to obtain:

Fnet = ρ oVballoon g − ρiVballoon g

Express the densities of the air molecules in terms of their number densities, molecular mass, and Avogadro’s number:

ρ=

Using the ideal-gas law, relate the number density of air N/V to its temperature and pressure:

PV = NkT and

Substitute to obtain:

ρ=

Substitute in equation (2) and simplify to obtain:

= (ρ o − ρ i )Vballoon g M ⎛N⎞ ⎜ ⎟ NA ⎝ V ⎠

N P = V kT

M ⎛ P ⎞ ⎜ ⎟ N A ⎝ kT ⎠

Fnet =

MP ⎛ 1 1 ⎞ ⎜ − ⎟Vballoon g N A k ⎜⎝ To Ti ⎟⎠

(2)

1336

Chapter 17

Assuming that the average molecular weight of air is 29 g/mol, substitute numerical values and evaluate Fnet:

Fnet =

(29 g/mol)(1.01 × 105 N/m 2 )

(6.022 × 10 particles/mol)(1.381 × 10 × (1.5 m )(9.81m/s ) 23

3

− 23

⎛ 1 1 ⎞ ⎟ ⎜⎜ − J/K ⎝ 297 K 348 K ⎟⎠

)

2

= 2.56 N 54 ••• Picture the Problem We can find the number of moles of helium gas in the balloon by applying the ideal-gas law to relate n to the pressure, volume, and temperature of the helium and Archimedes principle to find the volume of the helium. In part (b), we can apply the result of Problem 13-95 to relate atmospheric pressure to altitude and use the ideal-gas law to determine the pressure of the gas when the balloon is fully inflated. In part (c), we’ll find the net force acting on the balloon at the altitude at which it is fully inflated in order to decide whether it can rise to that altitude. (a) Apply the ideal-gas law to the helium in the balloon and solve for n: Relate the net force on the balloon to its weight:

n=

PV RT

(1)

FB − wskin − wload − wHe = 30 N

Use Archimedes principle to express the buoyant force on the balloon in terms of the volume of the balloon: Substitute to obtain:

FB = wdisplaced air

Solve for the volume of the helium:

V=

Substitute numerical values and evaluate V:

V=

= ρ airVg

ρairVg − wskin − wload − ρ HeVg = 30 N 30 N + wskin + wload (ρair − ρ He )g 30 N + 50 N + 110 N 1.293 kg/m 3 − 0.179 kg/m 3 1 × 9.81m/s 2

(

(

= 17.39 m

)

3

)

Temperature and the Kinetic Theory of Gases 1337 1L ⎞ ⎟ 10− 3 m3 ⎟⎠ ⎝ n= 8.206 × 10− 2 L ⋅ atm/mol ⋅ K (273 K )

(1atm )(17.39 m3 )⎛⎜⎜

Substitute numerical values in equation (1) and evaluate n:

(

)

= 776 mol (b) Using the result of Problem 1395, express the variation in atmospheric pressure with altitude:

P(h ) = P0e − h h0 where h0 = 7.93 km

⎡ P ⎤ h = h0 ln ⎢ 0 ⎥ ⎣ P(h ) ⎦

Solve for h:

(2)

Neglecting changes in temperature with elevation, apply the ideal-gas law to find the pressure at which the balloon’s volume will be 32 m3:

P=

(

)

nRT (776 mol) 8.206 × 10 −2 L ⋅ atm/mol ⋅ K (273 K ) = = 0.543 atm 1L V 3 32 m × − 3 3 10 m

Substitute in equation (2) and evaluate h:

(c) Express the condition that must be satisfied if the balloon is to reach its fully inflated altitude: Express wtot:

⎛ 1atm ⎞ ⎟⎟ = 4.84 km h = (7.93 km )ln⎜⎜ ⎝ 0.543 atm ⎠

Fnet = FB − wtot ≥ 0

(3)

wtot = wload + wskin + wHe = 110 N + 50 N + wHe = 160 N + wHe

Express the weight of the helium:

wHe = ρ HeVg

Substitute for wHe and evaluate wtot:

wtot = 160 N + ρ HeVg

(

)(

= 160 N + 0.179 kg/m 3 17.38 m3

(

× 9.81m/s = 190.5 N

2

)

)

1338

Chapter 17

Determine the buoyant force on the balloon at h = 4.84 km: Express the dependence of the density of the air on atmospheric pressure:

FB = ρ air, hVg

(4)

P ρ air, h = P0 ρ air

(5)

or

ρair, h = Substitute and evaluate FB:

FB =

P ρair P0

P ρ airVg P0

( )( × (9.81m/s )

= 0.543 1.293 kg/m 3 32 m3

)

2

= 219.9 N Substitute in equation (3) and evaluate Fnet:

Fnet = 219.9 N − 190.5 N = 29.4 N ≥ 0

Because Fnet > 0, the balloon will rise higher than the altitude at which it is fully inflated. (d) The balloon will rise until the net force acting on it is zero. Because the buoyant force depends on the density of the air, the balloon will rise until the density of the air has decreased sufficiently for the buoyant force to just equal the total weight of the balloon. Substitute equation (5) in equation (2) to obtain:

h = h0 ln

Using equation (4), find the density of the air such that FB = 190.5 N:

ρ air, h =

ρair ρair, h

FB 190.5 N = Vg 32 m3 9.81m/s 2

(

)(

)

= 0.6068 kg/m3 Substitute numerical values and evaluate h:

⎛ 1.293 kg/m 3 ⎞ ⎟ h = (7.93 km )ln⎜⎜ 3 ⎟ ⎝ 0.6068 kg/m ⎠ = 6.00 km

Temperature and the Kinetic Theory of Gases 1339

Kinetic Theory of Gases *55 • Picture the Problem We can express the rms speeds of argon and helium atoms by combining PV = nRT and vrms = 3RT M to obtain an expression for vrms in terms of P, V, and M. Express the rms speed of an atom as a function of the temperature:

vrms =

From the ideal-gas law we have:

RT =

Substitute to obtain:

vrms =

3RT M

PV n 3PV nM

(a) Substitute numerical values and evaluate vrms for an argon atom:

vrms (Ar ) =

(

)

3(10 atm )(101.3 kPa/atm ) 10− 3 m3 = 276 m/s (1mol) 40 × 10−3 kg/mol

(

)

(b) Substitute numerical values and evaluate vrms for a helium atom:

vrms (He ) =

(

)

3(10 atm )(101.3 kPa/atm ) 10− 3 m3 = 872 m/s (1mol) 4 × 10−3 kg/mol

(

)

56 • Picture the Problem We can express the total translational kinetic energy of the oxygen gas by combining K = 32 nRT and the ideal-gas law to obtain an expression for K in terms of the pressure and volume of the gas. Relate the total translational kinetic energy of translation to the temperature of the gas: Using the ideal-gas law, substitute for nRT to obtain: Substitute numerical values and evaluate K:

K = 32 nRT

K = 32 PV K=

3 2

(101.3 kPa )(10−3 m3 ) =

152 J

1340

Chapter 17

57 • Picture the Problem Because we’re given the temperature of the hydrogen atom and know its molar mass, we can find its rms speed using vrms = 3RT M and its average kinetic energy from K av = 32 kT . Relate the rms speed of a hydrogen atom to its temperature and molar mass:

vrms =

3RT M

Substitute numerical values and evaluate vrms:

vrms =

3(8.314 J/mol ⋅ K ) 107 K 10− 3 kg/mol

(

)

)(

)

= 499 km/s Express the average kinetic energy of the hydrogen atom as a function of its temperature: Substitute numerical values and evaluate Kav:

K av = 32 kT

(

K av = 32 1.381 × 10 −23 J/K 107 K = 2.07 × 10 −16 J

*58 • Picture the Problem Because there are 6 squared terms in the expression for the total energy of an atom in this model, we can conclude that there are 6 degrees of freedom. Because the system is in equilibrium, we can conclude that there is energy of 12 kT per molecule or

1 2

RT per mole associated with each degree of freedom.

Express the average energy per atom in the solid in terms of its temperature and the number of degrees of freedom:

Eav = N ( 12 kT ) = 6( 12 kT ) = 3kT atom

Relate the total energy of one mole to its temperature and the number of degrees of freedom:

Etot = N ( 12 RT ) = 6( 12 RT ) = 3RT mole

59 • Picture the Problem We can combine λ =

1 and PV = nRT to express the 2nvπ d 2

mean free path for a molecule in an ideal gas in terms of the pressure and temperature.

Temperature and the Kinetic Theory of Gases 1341 Express the mean free path of a molecule in an ideal gas:

1 2nvπ d 2

λ= where

nv = N V = nN A V Solve the ideal-gas law for the volume of the gas:

V =

nRT P

Substitute in our expression for nv to obtain:

nv =

nN A P P= nRT kT

Substitute in the expression for the mean free path to obtain:

λ=

kT 2 Pπ d 2

60 •• Picture the Problem We can find the collision time from the mean free path and the average (rms) speed of the helium molecules. We can use the result of Problem 43 to find the mean free path of the molecules and vrms = 3RT M to find the average speed of the molecules.

λ

Express the collision time in terms of the mean free path for and the average speed of a helium molecule:

τ=

Use the result of Problem 43 to express the mean free path of the gas:

λ=

kT 2 Pπ d 2

λ=

(1.381× 10 J/K )(300 K ) 2 (7 × 10 Pa )π (10 m )

Substitute numerical values and evaluate λ:

(1)

vav

−23

−11

−10

2

= 1.332 × 109 m Express the average speed of the molecules:

vrms =

3RT M

Substitute numerical values and evaluate vrms:

vrms =

3(8.314 J/mol ⋅ K )(300 K ) 4 × 10− 3 kg/mol

= 1.368 × 103 m/s

1342

Chapter 17 1.332 × 109 m τ= = 9.74 × 105 s 3 1.368 × 10 m/s

Substitute in equation (1) and evaluate τ:

*61 •• Picture the Problem We can use K = 32 kT and ∆U = mgh = Mgh N A to express the ratio of the average kinetic energy of a molecule of the gas to the change in its gravitational potential energy if it falls from the top of the container to the bottom.

K = 32 kT

Express the average kinetic energy of a molecule of the gas as a function of its temperature: Letting h represent the height of the container, express the change in the potential energy of a molecule as it falls from the top of the container to the bottom:

∆U = mgh =

Mgh NA

3 K kT 3N A kT = 2 = ∆U Mgh 2Mgh NA

Express the ratio of K to ∆U and simplify to obtain:

Substitute numerical values and evaluate K/∆U:

(

)(

)

K 3 6.022 × 1023 1.381 × 10−23 J/K (300 K ) = = 7.95 × 104 2 −3 ∆U 2 32 × 10 kg 9.81m/s (0.15 m )

(

)(

)

The Distribution of Molecular Speeds 62 •• Picture the Problem Equation 17-37 gives the Maxwell-Boltzmann speed distribution. Setting its derivative with respect to v equal to zero will tell us where the function’s extreme values lie. 32 Differentiate Equation 17-37 with df d ⎡ 4 ⎛ m ⎞ 2 − mv 2 2 kT ⎤ = ⎜ ⎟ ve ⎢ ⎥ respect to v: dv dv 2kT

⎣⎢ π ⎝

=

4 ⎛ m ⎞ ⎜ ⎟ π ⎝ 2kT ⎠

⎠

32

⎦⎥

⎛ mv3 ⎞ − mv 2 ⎜⎜ 2v − ⎟e kT ⎟⎠ ⎝

2 kT

Temperature and the Kinetic Theory of Gases 1343 Set df/dv = 0 for extrema and solve for v:

2v −

2kT m

mv3 =0⇒ v= kT

Examination of the graph of f(v) makes it clear that this extreme value is, in fact, a maximum. See Figure 17-16 and note that it is concave downward at v =

2kT m .

Remarks: An alternative to the examination of f(v) in order to conclude that v = 2kT m maximizes the Maxwell-Boltzmann speed distribution function is to show that d2f/dv2 < 0 at v =

2kT m .

*63 •• Picture the Problem We can show that f(v) is normalized by using the given integral to integrate it over all possible speeds. Express the integral of Equation 17-37:

∞

∫ 0

Let a = m 2kT to obtain:

∞

∫

4 ⎛ m ⎞ f (v )dv = ⎜ ⎟ π ⎝ 2kT ⎠ f (v )dv =

0

Use the given integral to obtain:

∞

∫

f (v )dv =

0

4

π

3 2∞

∫v e

2 − mv 2 2 kT

dv

0

∞

a 3 2 ∫ v 2e − av dv 2

0

⎛ π −3 2 ⎞ a 3 2 ⎜⎜ a ⎟⎟ = 1 4 π ⎝ ⎠

4

i.e., f(v) is normalized. 64 •• Picture the Problem In Problem 63 we showed that f(v) is normalized. Hence we can ∞

evaluate vav using vf (v )dv .

∫ 0

Express the average speed of the molecules in the gas:

∞

vav = ∫ vf (v )dv 0

4 ⎛ m ⎞ = ⎜ ⎟ π ⎝ 2kT ⎠

3 2∞

∫v e 0

3 − mv 2 2 kT

dv

1344

Chapter 17

Substitute a = m 2kT :

Use the given integral to obtain:

vav =

vav =

∞

4

a

π

32

∫v e

3 − av 2

dv

0

⎛ a −2 ⎞ 2 1 ⎟⎟ = a 3 2 ⎜⎜ π π a ⎝ 2 ⎠

4

2

=

π

2kT m

*65 •• Picture the Problem Choose a coordinate system in which downward is the positive direction. We can use a constant-acceleration equation to relate the fall distance to the initial velocity of the molecule, the acceleration due to gravity, the fall time, and vrms = 3kT m to find the initial velocity of the molecules. (a) Using a constant-acceleration equation, relate the fall distance to the initial velocity of a molecule, the acceleration due to gravity, and the fall time:

y = v0t + 12 gt 2

Express the rms speed of the atom to its temperature and mass:

vrms =

Substitute numerical values and evaluate vrms:

vrms

(1)

3kT m

(

)

3 1.381 × 10− 23 J/K (120 nK ) = (85.47 u ) 1.660 × 10− 27 kg/u

(

)

= 5.92 × 10− 3 m/s

(

)

(

)

Letting vrms = v0, substitute in equation (1) to obtain:

0.1m = 5.92 × 10−3 m/s t + 12 9.81m/s 2 t 2

Solve this equation to obtain:

t = 0.142 s

(b) If the atom is initially moving upward:

vrms = v0 = −5.92 × 10−3 m/s

Substitute in equation (1) to obtain:

0.1 m = − 5.92 × 10 −3 m/s t

(

(

)

)

+ 12 9.81 m/s 2 t 2

Temperature and the Kinetic Theory of Gases 1345 Solve this equation to obtain:

t = 0.143 s

General Problems 66 • Picture the Problem We can use vrms = 3RT M to relate the temperature of the H2 molecule to its rms speed. Relate the rms speed of the molecule to its temperature:

vrms =

Solve for the temperature:

T=

Substitute numerical values and evaluate T:

3RT M

2 Mvrms 3R

(2 × 10 T=

)

kg/mol (331m/s ) 3(8.314 J/mol ⋅ K ) −3

2

= 8.79 K 67 •• Picture the Problem We can use the ideal-gas law to find the initial temperature of the gas and the ideal-gas law for a fixed amount of gas to relate the volumes, pressures, and temperatures resulting from the given processes. PV (a) Apply the ideal-gas law to T= nR express the temperature of the gas: Substitute numerical values and evaluate T: (b) Use the ideal-gas law for a fixed amount of gas to relate the temperatures and volumes:

Solve for and evaluate T2:

(c) Use the ideal-gas law for a fixed amount of gas to relate the

( 101.3 kPa )(10 × 10−3 m 3 ) T= = (1mol)(8.314 J/mol ⋅ K ) P1V1 P2V2 = T1 T2 or, because P1 = P2,

V1 V2 = T1 T2 V T2 = 2 T1 = 2(122 K ) = 244 K V1 P1V1 P2V2 = T1 T2

122 K

1346

Chapter 17

temperatures and pressures:

or, because V1 = V2,

P1 P2 = T1 T2 Solve for T2:

P2 =

T2 P1 T1

Substitute numerical values and evaluate P2:

P2 =

350 K (1atm ) = 1.43 atm 244 K

68 •• Picture the Problem We can use the definition of pressure to express the net force on each wall of the box in terms of its area and the pressure differential between the inside and the outside of the box. We can apply the ideal-gas law for a fixed amount of gas to find the pressure inside the box. Using the definition of pressure, express the net force on each wall of the box: Use the ideal-gas law for a fixed amount of gas to relate the initial and final pressures of the gas:

F = A∆P

= A(Pinside − Poutside )

P1V1 P2V2 = T1 T2 or, because V1 = V2,

P1 P2 = T1 T2 Solve for and evaluate Pinside:

P2 = Pinside =

T2 400 K (101.3 kPa ) P1 = T1 300 K

= 135.1 kPa Substitute and evaluate F:

F = (0.2 m ) (135.1 kPa − 101.3 kPa ) 2

= 1.35 kN *69 •• Picture the Problem We can use the molar mass of water to find the number of moles in 2 L of water. Because there are two hydrogen atoms in each molecule of water, there must be as many hydrogen molecules in the gas formed by electrolysis as there were molecules of water and, because there is one oxygen atom in each molecule of water, there must be half as many oxygen molecules in the gas formed by electrolysis as there were molecules of water.

Temperature and the Kinetic Theory of Gases 1347 Express the electrolysis of water into H2 and O2: Express the number of moles in 2 L of water: Because there is one hydrogen atom for each water molecule: Because there are two oxygen atoms for each water molecule:

70 •• Picture the Problem The diagram shows the cylinder before removal of the membrane. We’ll assume that the gases are at the same temperature. The approximate location of the center of mass (CM) is indicated. We can find the distance the cylinder moves by finding the location of the CM after the membrane is removed. Express the distance the cylinder will move in terms of the movement of the center of mass when the membrane is removed: Apply the ideal-gas law to both collections of molecules to obtain:

Divide the first of these equations by the second to obtain:

n(H 2 O ) → n(H 2 ) + 12 n(O 2 )

n(H 2O ) =

2000 g = 111 mol 18 g/mol

n(H 2 ) = 111 mol

n(O 2 ) = 12 n(H 2O ) =

1 2

(111mol)

= 55.5 mol

∆x = xcm,after − xcm,before

PN 2VN 2 = n(N 2 )kT and

PO2VO2 = n(O 2 )kT

PN 2 PO 2

=

n( N 2 ) n(O 2 )

or, because PN 2 = 2PO 2 ,

2 PO 2 PO 2 Express the mass of O2 in terms of its molar mass and the number of moles of oxygen:

=

n(N 2 ) ⇒ n(N2) = 2n(O2) n(O 2 )

m(O2) = n(O2)M(O2)

1348

Chapter 17

Express the mass of N2 in terms of its molar mass and the number of moles of nitrogen:

m(N2) = 2n(O2)M(N2).

Using its definition, express the center of mass before the membrane is removed:

∑xm = ∑m i

xcm, before

i

i

=

n(N 2 )M (N 2 )xcm, N 2 + n(O 2 )M (O 2 )xcm,O 2

i

n(N 2 )M (N 2 ) + n(O 2 )M (O 2 )

i

=

2n(O 2 )M (N 2 )xcm, N 2 + n(O 2 )M (O 2 )xcm,O 2

=

2 M (N 2 )xcm, N 2 + M (O 2 )xcm,O 2

2n(O 2 )M (N 2 ) + n(O 2 )M (O 2 ) 2 M (N 2 ) + M (O 2 )

Substitute numerical values and evaluate xcm,before:

xcm, before =

2(10 cm )(28 g ) + (30 cm )(32 g ) = 17.27 cm 2(28 g ) + 32 g

Locate the center of mass after the membrane is removed:

xcm,after =

2(20 cm )(28 g ) + (20 cm )(32 g ) 2(28 g ) + 32 g

= 20.0 cm Substitute to obtain:

∆x = 20.00 cm − 17.27 cm = 2.73 cm

Because momentum must be conserved during this process and the center of mass moved to the right, the cylinder moved 2.73 cm to the left. 71 •• Picture the Problem We can apply the ideal-gas law to the two processes to find the number of moles of hydrogen in terms of the number of moles of nitrogen in the gas. Using the definition of molar mass, we can relate the mass of each gas to the number of moles of each gas and their molar masses. Apply the ideal-gas law to the first case: Apply the ideal-gas law to the

P1V = [2n(N 2 ) + n(H 2 )]RT1 3P1V = [2n(N 2 ) + 2n(H 2 )]2 RT1

Temperature and the Kinetic Theory of Gases 1349 second case: Divide the second of these equations by the first and simplify to express n(H2) in terms of n(N2): Relate the mN to n(N2):

n(H 2 ) = 2n(N 2 )

(1)

mN = n(N 2 )M (N 2 )

= n(N 2 )(28 g/mol)

and

n(N 2 ) = Relate the mH to n(H2):

mN 28 g/mol

mH = n(H 2 ) M (H 2 )

= n(H 2 )(2 g/mol)

and

n(H 2 ) =

Substitute in equation (1) and solve for mN:

mH 2 g/mol

mH 2mN ⇒ mN = 7mH = 2 g/mol 28 g/mol

*72 •• Picture the Problem Initially, we have 3P0V = n0RT0. Later, the pressures in the three vessels, each of volume V, are still equal, but the number of moles is not. The total number of moles, however, is constant and equal to the number of moles in the three vessels initially. Applying the ideal-gas law to each of the vessels will allow us to relate the number of moles in each to the final pressure and temperature. Equating this sum n0 will leave us with an equation in P′ and P0 that we can solve for P′. Relate the number of moles of gas in the system in the three vessels initially to the number in each vessel when the pressure is P′:

n0 = n1 + n2 + n3

Relate the final pressure in the first vessel to its temperature and solve for n1:

P' =

n1R(2T0 ) P'V ⇒ n1 = V 2RT0

Relate the final pressure in the second vessel to its temperature and solve for n2:

P' =

n2 R(3T0 ) P'V ⇒ n2 = V 3RT0

1350

Chapter 17

Relate the final pressure in the third vessel to its temperature and solve for n3:

P' =

n3 RT0 P'V ⇒ n3 = V RT0

Substitute to obtain:

n0 =

P'V P'V P'V + + 2 RT0 3RT0 RT0

⎛ 1 1 ⎞ P'V 11 ⎛⎜ P'V = = ⎜ + + 1⎟ ⎝ 2 3 ⎠ RT0 6 ⎜⎝ RT0 Express the number of moles in the three vessels initially in terms of the initial pressure and total volume: Equate the two expressions for n0 and solve for P′ to obtain:

n0 =

P0 (3V ) RT0

P' =

18 P0 11

⎞ ⎟⎟ ⎠

73 •• Picture the Problem We can use the ideal-gas temperature scale to relate the temperature of the boiling substance to its pressure and the pressure at the triple point. If we assume a linear relationship between P/P3 and P3, we can calibrate this equation using the data from any two (or all) of the temperature measurements and then extrapolate this equation to zero gas pressure to find the ideal-gas temperature of the boiling substance. Using the ideal-gas temperature scale, relate the temperature of the boiling substance to its pressure and the pressure at the triple point:

⎛P⎞ T = 273.16 K⎜⎜ ⎟⎟ ⎝ P3 ⎠

Find the temperature of the first measurement:

⎛ 734 torr ⎞ ⎟⎟ T1 = 273.16 K⎜⎜ ⎝ 500 torr ⎠ = 273.16 K (1.4680) = 401.00 K

Find the temperature of the third measurement:

⎛ 146.65 torr ⎞ ⎟⎟ T3 = 273.16 K⎜⎜ ⎝ 100 torr ⎠ = 273.16 K (1.4655) = 400.59 K

(1)

Temperature and the Kinetic Theory of Gases 1351 Assume a linear relationship between P/P3 and P3:

P = a + bP3 P3 where a is the pressure ratio for P3 = 0.

Substitute using the data from the first measurement:

734 torr = a + b(500 torr ) 500 torr or

1.4680 = a + b(500 torr )

Substitute using the data from the third measurement:

146.65 torr = a + b(100 torr ) 100 torr or

1.4665 = a + b(100 torr ) Solve these equations simultaneously for a: Substitute in equation (1) to obtain:

a = 1.46613

T = 273.16 K (1.46613) = 400.49 K

*74 •• Picture the Problem Because the O2 molecule resembles 2 spheres stuck together, which in cross section look something like two circles, we can estimate the radius of the molecule from the formula for the area of a circle. We can express the area, and hence the radius, of the circle in terms of the mean free path and the number density of the molecules and use the ideal-gas law to express the number density. Express the area of two circles of diameter d that touch each other: Solve for d to obtain:

Relate the mean free path of the molecules to their number density and cross-sectional area: Solve for A to obtain:

Substitute in equation (1) to obtain:

⎛ πd 2 ⎞ πd 2 ⎟⎟ = A = 2⎜⎜ 2 ⎝ 4 ⎠ d=

2A

π

λ=

1 nv A

A=

1 nv λ

d=

2 πnv λ

(1)

1352

Chapter 17

Use the ideal-gas law to relate the number density of the O2 molecules to their temperature and pressure: Substitute to obtain:

Substitute numerical values and evaluate d:

PV = NkT or nv =

d=

N P = V kT

2kT πPλ

( (

)

2 1.381 × 10− 23 J/K (300 K ) d= π 1.01 × 105 Pa 7.1 × 10−8 m

)(

)

= 6.06 × 10−10 m = 0.606 nm 75 •• Picture the Problem We can use its definition to express the mean free path of the molecules and the ideal-gas law to obtain an expression for the number density of the hydrogen gas molecules. (a) Relate the mean free path of the molecules to their number density and cross-sectional area: Use the ideal-gas law to relate the number density of the H2 molecules to their temperature and pressure:

λ=

Express the effective cross-sectional area of a H2 molecule:

A = 14 πd 2

Substitute for nv and A to obtain:

Substitute numerical values and evaluate λ:

1 nv A

PV = NkT or nv =

λ= λ=

N P = V kT

4kT Pπd 2

(

)

4 1.381 × 10 −23 J/K (300 K )

π (1.01 × 105 N/m 2 )(1.6 × 10−10 m )

2

= 2.04 × 10 − 6 m (b) Relate the available volume per molecule to the number density nv:

V 1 kT = = N nv P

Substitute numerical values and evaluate V/N:

V 1.381 × 10−23 J/K (300 K ) = N 1.01 × 105 N/m 2

(

= 4.10 × 10− 26 m3 Express the volume of a spherical molecule:

V = 43 π r 3 = 16 π d 3

)

Temperature and the Kinetic Theory of Gases 1353 Solve for d:

Substitute numerical values and evaluate d:

d =3

d =3

6V

π

(

6 4.10 × 10 −26 m 3

π

)=

4.28 nm

The mean free path is larger by approximately a factor of 1000. 76 ••• Picture the Problem Let A be the cross-sectional area of the cylinder. We can use the ideal-gas law to find the height of the piston under equilibrium conditions. In (b), we can apply Newton’s 2nd law and the ideal-gas law for a fixed amount of gas to the show that, for small displacements from its equilibrium position, the piston executes simple harmonic motion. (a) Express the pressure inside the cylinder: Apply the ideal-gas law to obtain a second expression for the pressure of the gas in the cylinder:

Mg A nRT nRT Pin = = V hA

Equate these two expressions:

Patm +

Solve for h to obtain:

h=

Pin = Patm +

=

At STP, 0.1 mol of gas occupies 2.24 L. Therefore:

Substitute numerical values and evaluate h:

(1)

Mg nRT = A hA

(2.4 m )APatm nRT = APatm + Mg APatm + Mg 2.4 m Mg 1+ APatm

(2.4 m )A = 2.24 × 10−3 m3 and A = 9.33×10−4 m2

h=

2.4 m (1.4 kg ) 9.81m/s2 1+ 9.333 × 10− 4 m 2 (101.3 kPa )

(

= 2.096 m

(

)

)

1354

Chapter 17

(b) Relate the frequency of vibration of the piston to its mass and a ″stiffness″ constant:

Letting y be the displacement from Fy = ma y to equilibrium, apply

∑

f =

1 2π

k M

(2)

where M is the mass of the piston and k is a constant of proportionality.

Pin A − mg − Patm A = 0

the piston in its equilibrium position: For a small displacement y above equilibrium:

Pin'A − mg − Patm A = ma y or

Pin'A − Pin A = ma y Using the ideal-gas law for a fixed amount of gas and constant temperature, relate Pin' to Pin : Solve for Pin' :

(3)

Pin'V' = PinV or

Pin' (V + Ay ) = PinV V Pin' = Pin V + Ay

and

Pin'A = Pin A

Substitute in equation (3) to obtain:

1 Ah = Pin A y Ah + Ay 1+ h −1

y⎞ ⎛ Pin A⎜1 + ⎟ − Pin A = ma y ⎝ h⎠ or, for y << h,

y⎞ ⎛ Pin A⎜1 − ⎟ − Pin A ≈ ma y ⎝ h⎠ y ≈ ma y h

Simplify equation (4):

− Pin A

Substitute in equation (1) to obtain:

⎛ nRT ⎞ y −⎜ ⎟ A ≈ ma y ⎝ Ah ⎠ h or

⎛ nRT ⎞ − ⎜ 2 ⎟ y ≈ ma y ⎝ h ⎠

(4)

Temperature and the Kinetic Theory of Gases 1355 Solve for ay:

ay = −

nRT y mh 2

or

k y , the condition for SHM m k nRT where = m mh 2 ay = −

Substitute in equation (2) to obtain:

f =

nRT mh 2

1 2π

Substitute numerical values and evaluate f:

f =

*77

1 2π

(0.1mol)(8.314 J/mol ⋅ K )(300 K ) = (1.4 kg )(2.096 m )2

1.01Hz

••• V

Picture the Problem We can show that f (v )dv = I ( x ), where f(v) is the Maxwell-

∫ 0

Boltzmann distribution function, x = mV 2 2kT , and I(x) is the integral whose values are tabulated in the problem statement. Then, we can use this table to find the value of x corresponding to the fraction of the gas molecules with speeds less than v by evaluating I(x). (a) The Maxwell-Boltzmann speed distribution f(x) is given by:

32

4 ⎛ m ⎞ 2 −mv2 2 kT f (v ) = ⎜ ⎟ ve π ⎝ 2kT ⎠ which means that the fraction of particles with speeds between v and v + dv is f (v )dv.

Express the fraction F(V) of particles with speeds less than V = 400 m/s:

V

F (V ) = ∫ f (v )dv 0

4 ⎛ m ⎞ = ⎜ ⎟ π ⎝ 2kT ⎠ Change integration variables by letting z = v m 2kT so we can use the table of values to evaluate the integral. Then:

v=

32 V

2kT z ⇒ dv = m

∫v e

2 − mv 2 2 kT

0

2kT dz m

dv

1356

Chapter 17

Substitute in the integrand of F(V) to obtain:

12

2 −mv 2 2 kT

ve

2kT − z 2 ⎛ 2kT ⎞ dv = z e ⎜ ⎟ dz m ⎝ m ⎠ 2

32

⎛ 2kT ⎞ 2 − z 2 =⎜ ⎟ z e dz ⎝ m ⎠ Transform the integration limits to correspond to the new integration

When v = 0, z = 0, and

variable z = v m 2kT :

when v = V, z = V m 2kT

The new lower integration limit is 0. Evaluate z = V m 2kT to find the upper limit:

z = (400 m/s )

(32 u )(1.661 × 10− 27 kg ) = 1.06 2(1.381 × 10− 23 J / K )(273 K )

Evaluate F(400 m/s) to obtain: 400 m/s

4 ⎛ m ⎞ F (400 m/s ) = ∫ f (v )dv = ⎜ ⎟ π ⎝ 2kT ⎠ 0 = I (1.06 ) where I (x ) =

3 2 400 m/s 2 − mv 2 2 kT

∫v e

dv =

0

4

π

1.06

2 −z ∫ z e dz 2

0

x

4

2 −z ∫ z e dz 2

π

0

Letting r represent the fraction of the molecules with speeds less than 400 m/s, interpolate from the table to obtain:

(b) Express the fraction r of the molecules with speeds between V1 = 190 m/s and V2 = 565 m/s:

r − 0.438 0.788 − 0.438 = 1.06 − 1 1.5 − 1 and

r = 48.0% r = F (V2 ) − F (V1 ) = I (x 2 ) − I (x1 ) where

x1 = V1 m 2kT and x2 = V2 m 2kT

Evaluate x1 and x2 to obtain:

x1 = (190 m/s )

(32 u )(1.661 × 10− 27 kg ) = 0.504 2(1.381 × 10− 23 J / K )(273 K )

x2 = (565 m/s )

(32 u )(1.661×10−27 kg ) = 1.50 2(1.381× 10 −23 J / K )(273 K )

and

Substitute to obtain:

r = I (1.50 ) − I (0.504 )

(1)

Temperature and the Kinetic Theory of Gases 1357 Using the table, evaluate I(1.50):

I (1.50 ) = 0.788

Letting r represent the fraction of the molecules with speeds less than 190 m/s, interpolate from the table to obtain:

0.132 − 0.081 r − 0.081 = 0.504 − 0.5 0.6 − 0.5

Substitute in equation (1) to obtain:

and

r = 0.083 r = 0.788 − 0.083 = 70.5%

1358

Chapter 17

Chapter 18 Heat and the First Law of Thermodynamics Conceptual Problems 1 • Picture the Problem We can use the relationship Q = mc∆T to relate the temperature changes of bodies A and B to their masses, specific heats, and the amount of heat supplied to each. Express the change in temperature of body A in terms of its mass, specific heat, and the amount of heat supplied to it:

∆TA =

Q mA cA

Express the change in temperature of body B in terms of its mass, specific heat, and the amount of heat supplied to it:

∆TB =

Q mBcB

Divide the second of these equations by the first to obtain:

∆TB mA cA = ∆TA mBcB

Substitute and simplify to obtain:

∆TB (2mB )(2cB ) = =4 mBcB ∆TA or

∆TB = 4∆TA *2 • Picture the Problem We can use the relationship Q = mc∆T to relate the temperature changes of bodies A and B to their masses, specific heats, and the amount of heat supplied to each. Relate the temperature change of block A to its specific heat and mass:

∆TA =

Q M A cA

Relate the temperature change of block B to its specific heat and mass:

∆TB =

Q M BcB

1359

1360 Chapter 18 Equate the temperature changes to obtain:

1 1 = M BcB M A cA

Solve for cA:

cA =

MB cB MA

and (b) is correct. 3 • Picture the Problem We can use the relationship Q = mc∆T to relate the amount of energy absorbed by the aluminum and copper bodies to their masses, specific heats, and temperature changes. Express the energy absorbed by the aluminum object: Express the energy absorbed by the copper object: Divide the second of these equations by the first to obtain: Because the object’s masses are the same and they experience the same change in temperature:

QAl = mAlcAl ∆T

QCu = mCu cCu ∆T QCu mCu cCu ∆T = QAl mAlcAl ∆T QCu cCu = <1 QAl cAl or

QCu < QAl and (c) is correct. 4 • Determine the Concept Some examples of systems in which internal energy is converted into mechanical energy are: a steam turbine, an internal combustion engine, and a person performing mechanical work, e.g., climbing a hill. *5 • Determine the Concept Yes, if the heat absorbed by the system is equal to the work done by the system. 6 • Determine the Concept According to the first law of thermodynamics, the change in the internal energy of the system is equal to the heat that enters the system plus the work done on the system.

(b) is correct.

Heat and the First Law of Thermodynamics 1361 7 • Determine the Concept ∆Eint = Qin + Won . For an ideal gas, ∆Eint is a function of T only. Because Won = 0 and Qin = 0 in a free expansion, ∆Eint = 0 and T is constant. For a real gas, Eint depends on the density of the gas because the molecules exert weak attractive forces on each other. In a free expansion, these forces reduce the average kinetic energy of the molecules and, consequently, the temperature. 8 • Determine the Concept Because the container is insulated, no energy is exchanged with the surroundings during the expansion of the gas. Neither is any work done on or by the gas during this process. Hence, the internal energy of the gas does not change and we can conclude that the equilibrium temperature will be the same as the initial temperature. Applying the ideal-gas law for a fixed amount of gas we see that the pressure at equilibrium must be half an atmosphere.

(c) is correct.

9 • Determine the Concept The temperature of the gas increases. The average kinetic energy increases with increasing volume due to the repulsive interaction between the ions. *10 •• Determine the Concept The balloon that expands isothermally is larger when it reaches the surface. The balloon that expands adiabatically will be at a lower temperature than the one that expands isothermally. Because each balloon has the same number of gas molecules and are at the same pressure, the one with the higher temperature will be bigger. An analytical argument that leads to the same conclusion is shown below. 1γ

Letting the subscript ″a″ denote the adiabatic process and the subscript ″i″ denote the isothermal process, express the equation of state for the adiabatic balloon:

⎛P ⎞ P0V0 = Pf Vf,a ⇒ Vf,a = V0 ⎜⎜ 0 ⎟⎟ ⎝ Pf ⎠

For the isothermal balloon:

⎛P ⎞ P0V0 = Pf Vf,i ⇒ Vf,i = V0 ⎜⎜ 0 ⎟⎟ ⎝ Pf ⎠

Divide the second of these equations by the first and simplify to obtain:

⎛P ⎞ V0 ⎜⎜ 0 ⎟⎟ 1−1 λ Vf,i Pf ⎠ ⎛ P0 ⎞ ⎝ = = ⎜⎜ ⎟⎟ 1γ Vf,a ⎛P ⎞ ⎝ Pf ⎠ V0 ⎜⎜ 0 ⎟⎟ ⎝ Pf ⎠

γ

γ

1362 Chapter 18 Because P0/Pf > 1 and γ > 1:

Vf,i > Vf,a

11 • Determine the Concept The work done along each of these paths equals the area under its curve. The area is greatest for the path A→B→C and least for the path A→D→C.

(a ) is correct. 12 • Determine the Concept An adiabatic process is, by definition, one for which no heat enters or leaves the system.

(b) is correct.

13 • (a) False. The heat capacity of a body is the heat needed to raise the temperature of the body by one degree. (b) False. The amount of heat added to a system when it goes from one state to another is path dependent. (c) False. The work done on a system when it goes from one state to another is path dependent. (d) True. (e) True. (f) True. (g) True. *14 • Determine the Concept For a constant-volume process, no work is done on or by the gas. Applying the first law of thermodynamics, we obtain Qin = ∆Eint. Because the temperature must change during such a process, we can conclude that ∆Eint ≠ 0 and hence Qin ≠ 0. (b) and ( d ) are correct. 15 • Determine the Concept Because the temperature does not change during an isothermal process, the change in the internal energy of the gas is zero. Applying the first law of thermodynamics, we obtain Qin = −Won = Wby the system. Hence ( d ) is correct.

Heat and the First Law of Thermodynamics 1363 16 •• Determine the Concept The melting point of propane at 1 atm pressure is 83 K. Hence, at this low temperature and high pressure, C3H8 is a solid. 17 •• Picture the Problem We can use the given dependence of the pressure on the volume and the ideal-gas law to show that if the volume decreases, so does the temperature. We’re given that:

P V = constant

Because the gas is an ideal gas:

PV = P V

Solve for T:

T=

(

)

V = constant V = nRT

(constant )

V

nR

Because T varies with the square root of V , if the volume decreases, the temperature decreases. *18 •• Determine the Concept At room temperature, most solids have a roughly constant heat capacity per mole of 6 cal/mol-K (Dulong-Petit law). Because 1 mole of lead is more massive than 1 mole of copper, the heat capacity of lead should be lower than the heat capacity of copper. This is, in fact, the case. 19 •• Determine the Concept The heat capacity of a substance is proportional to the number of degrees of freedom per molecule associated with the molecule. Because there are 6 degrees of freedom per molecule in a solid and only 3 per molecule (translational) for a monatomic liquid, you would expect the solid to have the higher heat capacity.

Estimation and Approximation *20 •• Picture the Problem The heat capacity of lead is c = 128 J/kg⋅K. We’ll assume that all of the work done in lifting the bag through a vertical distance of 1 m goes into raising the temperature of the lead shot and use conservation of energy to relate the number of drops of the bag and the distance through which it is dropped to the heat capacity and change in temperature of the lead shot. (a) Use conservation of energy to relate the change in the potential energy of the lead shot to the change in its temperature:

Nmgh = mc∆T where N is the number of times the bag of shot is dropped.

1364 Chapter 18 Solve for ∆T to obtain:

∆T =

Substitute numerical values and evaluate ∆T:

Nmgh Ngh = mc c

50(9.81`m/s 2 )(1 m ) ∆T = = 3.83 K 128 J/kg ⋅ K

It is better to use a larger mass because the rate at which heat is lost by the lead shot is proportional to its surface area while the rate at which it gains (b) heat is proportional to its mass. The amount of heat lost varies as the

surface area of the shot divided by its mass (L2 /L3 = L−1 ); which decreases as the mass increases. 21 •• Picture the Problem Assume that the water is initially at 30°C and that the cup contains 200 g of water. We can use the definition of power to express the required time to bring the water to a boil in terms of its mass, heat capacity, change in temperature, and the rate at which energy is supplied to the water by the microwave oven. Use the definition of power to relate the energy needed to warm the water to the elapsed time: Solve for ∆t to obtain:

P=

∆W mc∆T = ∆t ∆t

∆t =

mc∆T P

Substitute numerical values and evaluate ∆t:

∆t =

(0.2 kg )(4.18 kJ/kg ⋅ K )(373 K − 303 K ) = 97.5 s = 600 W

1.63 min , an elapsed time

that seems to be consistent with experience. 22 • Picture the Problem The adiabatic compression from an initial volume V1 to a final volume V2 between the isotherms at temperatures T1 and T2 is shown to the right. We’ll assume a room temperature of 300 K and apply the equation for a quasi-static adiabatic process with γair = 1.4 to solve for the ratio of the initial to the final volume of the air. Express TV γ −1 = constant in terms of the initial and final values of T and V:

T1V1γ −1 = T2V2γ −1

Heat and the First Law of Thermodynamics Solve for V1/V2 to obtain:

Substitute numerical values and evaluate V1/V2:

V1 ⎛ T2 ⎞ =⎜ ⎟ V2 ⎜⎝ T1 ⎟⎠

1365

1

γ −1

1

V1 ⎛ 506 K ⎞ 1.4−1 ⎟ =⎜ = 3.69 V2 ⎜⎝ 300 K ⎟⎠

23 •• Picture the Problem We can use Q = mcp ∆T to express the specific heat of water during heating at constant pressure in terms of the required heat and the resulting change in temperature. Further, we can use the definition of the bulk modulus to express the work done by the water as it expands. Equating the work done by the water during its expansion and the heat gained during this process will allow us to solve for cp. Express the heat needed to raise the temperature of a mass m of a substance whose specific heat at constant pressure is cp by ∆T:

Q = mcp ∆T

Solve for cp to obtain:

cp =

Q m∆T

Use the definition of the bulk modulus to express the work done by the water as it expands:

B=

V∆P ∆P = ∆V ∆V V

or

W = V∆P = B∆V

B∆V m∆T

Assuming that the work done by the water in expanding equals the heat gained during the process, substitute to obtain:

cp =

Using the definition of the coefficient of volume expansion, express ∆V (see Chapter 20, Section 1):

∆V = βV∆T

Substitute to obtain:

cp =

BβV∆T Bβ V = m∆T m

1366 Chapter 18 Use the data given in the problem statement to find the average volume of 1 kg of water as it warms from 4°C to 100°C:

V =

m

ρ

1 kg 1.0000 g/cm + 0.9584 g/cm3 2 −3 3 = 1.02 × 10 m =

3

Substitute numerical values and evaluate cp:

cp =

(2 ×10

8

N/m 2 )(0.207 × 10 −3 K -1 )(1.02 × 10 −3 m 3 ) = 42.2 J/kg ⋅ K 1 kg

Express the ratio of cp to cwater:

cp cwater

=

42.2 J/kg ⋅ K = 1.01 × 10 −2 4184 J/kg ⋅ K

or

cp = (1.01% ) cwater

*24 •• Picture the Problem We can apply the condition for the validity of the equipartition theorem, i.e., that the spacing of the energy levels be large compared to kT, to find the critical temperature Tc: Express the failure condition for the equipartition theorem: Solve for Tc:

Substitute numerical values and evaluate Tc:

kTc ≈ 0.15 eV

Tc =

Tc =

0.15 eV k 1.602 × 10 −19 J eV = 1740 K −23 1.381× 10 J/K

0.15 eV ×

Heat Capacity; Specific Heat; Latent Heat *25 • Picture the Problem We can use the conversion factor 1 cal = 4.184 J to convert 2500 kcal into joules and the definition of power to find the average output if the consumed energy is dissipated over 24 h. (a) Convert 2500 kcal to joules:

2500 kcal = 2500 kcal × = 10.5 MJ

4.184 J cal

Heat and the First Law of Thermodynamics (b) Use the definition of average power to obtain:

1367

1.05 × 107 J ∆E Pav = = = 121 W ∆t 24 h × 3600 s h

Remarks: Note that this average power output is essentially that of a widely used light bulb. 26 • Picture the Problem We can use the relationship Q = mc∆T to calculate the amount of heat given off by the concrete as it cools from 25 to 20°C. Relate the heat given off by the concrete to its mass, specific heat, and change in temperature: Substitute numerical values and evaluate Q:

Q = mc∆T

(

)

Q = 105 kg (1 kJ/kg ⋅ K )(298 K − 293 K ) = 500 MJ

27 • Picture the Problem We can find the amount of heat that must be supplied by adding the heat required to warm the ice from −10°C to 0°C, the heat required to melt the ice, and the heat required to warm the water formed from the ice to 40°C. Express the total heat required:

Q = Qwarm ice + Qmelt ice + Qwarm water

Substitute for each term to obtain:

Q = mcice ∆Tice + mLf + mcwater ∆Twater = m(cice ∆Tice + Lf + cwater ∆Twater )

Substitute numerical values and evaluate Q:

Q = (0.06 kg )[(0.49 kcal/kg ⋅ K )(273 K − 263 K ) + 79.7 kcal/kg + (1 kcal/kg ⋅ K )(313 K − 273 K )] = 7.48 kcal 28 •• Picture the Problem We can find the amount of heat that must be removed by adding the heat that must be removed to cool the steam from 150°C to 100°C, the heat that must be removed to condense the steam to water, the heat that must be removed to cool the water from 100°C to 0°C, and the heat that must be removed to freeze the water.

1368 Chapter 18 Express the total heat that must be removed:

Q = Qcool steam + Qcondense steam + Qcool water + Qfreeze water

Q = mcsteam ∆Tsteam + mLv

Substitute for each term to obtain:

+ mcwater ∆Twater + mLf

= m(csteam ∆Tsteam + Lv + cwater ∆Twater + Lf )

Substitute numerical values and evaluate Q:

Q = (0.1 kg )[(2.01 kJ/kg ⋅ K )(423 K − 373 K ) + 2.26 MJ/kg

+ (4.18 kJ/kg ⋅ K )(373 K − 273 K ) + 333.5 kJ/kg ]

= 311.2 kJ ×

1 kcal 4.184 kJ

= 74.4 kcal 29 •• Picture the Problem We can find the amount of nitrogen vaporized by equating the heat gained by the liquid nitrogen and the heat lost by the piece of aluminum. Express the heat gained by the liquid nitrogen as it cools the piece of aluminum: Express the heat lost by the piece of aluminum as it cools: Equate these two expressions and solve for mN:

QN = mN LvN

QAl = mAlcAl ∆TAl

mN LvN = mAlcAl ∆TAl and

mN =

mAlcAl ∆TAl LvN

Substitute numerical values and evaluate mN:

mN =

(0.05 kg )(0.90 J/kg ⋅ K )(293 K − 77 K ) = 4.88 ×10−5 kg = 199 kJ/kg

48.8 mg

Heat and the First Law of Thermodynamics

1369

30 •• Picture the Problem Because the heat lost by the lead as it cools is gained by the block of ice (we’re assuming no heat is lost to the surroundings), we can apply the conservation of energy to determine how much ice melts. Apply the conservation of energy to this process:

∆Q = 0 or

− mPb (Lf,Pb + cPb ∆TPb ) + mw Lf,w = 0

Solve for mw: m

w

=

m

⎛L + c ∆T ⎞ Pb ⎜⎝ f, Pb Pb Pb ⎟⎠ L f, w

Substitute numerical values and evaluate mw:

mw =

(0.5 kg )(24.7 kJ/kg + (0.128 kJ/kg ⋅ K )(600 K − 273 K )) = 333.5 kJ/kg

99.8 g

*31 •• Picture the Problem The temperature of the bullet immediately after coming to rest in the block is the sum of its pre-collision temperature and the change in its temperature as a result of being brought to a stop in the block. We can equate the heat gained by the bullet and half its pre-collision kinetic energy to find the change in its temperature. Express the temperature of the bullet immediately after coming to rest in terms of its initial temperature and the change in its temperature as a result of being stopped in the block: Relate the heat absorbed by the bullet as it comes to rest to its kinetic energy before the collision: Substitute for Q and K to obtain: Solve for ∆T:

T = Ti + ∆T = 293 K + ∆T

Q = 12 K

mPb cPb ∆T = ∆T =

v2 4cPb

(

1 1 2 2

mPb v 2

)

1370 Chapter 18 Substitute to obtain:

Substitute numerical values and evaluate T:

v2 T = 293 K + 4cPb T = 293 K +

(420 m/s)2

4(0.128 kJ/kg ⋅ K )

= 638 K = 365°C 32 •• Picture the Problem We can find the heat available to warm the brake drums from the initial kinetic energy of the car and the mass of steel contained in the brake drums from Q = msteelcsteel∆T.

Q csteel ∆T

Express msteel in terms of Q:

msteel =

Find the heat available to warm the brake drums from the initial kinetic energy of the car:

Q = K = 12 mcar v 2

Substitute for Q to obtain:

msteel =

mcar v 2 csteel ∆T 1 2

2

Substitute numerical values and evaluate msteel:

msteel

(1400 kg )⎛⎜⎜ 80 km × 1h ⎞⎟⎟ h 3600 s ⎠ ⎝ = ⎛ kcal 4.186 kJ ⎞ ⎟ (120 K ) × 2⎜⎜ 0.11 kg ⋅ K kcal ⎟⎠ ⎝ = 6.26 kg

Calorimetry 33 • Picture the Problem Let the system consist of the piece of lead, calorimeter, and water. During this process the water will gain energy at the expense of the piece of lead. We can set the heat out of the lead equal to the heat into the water and solve for the final temperature of the lead and water. Apply conservation of energy to the system to obtain:

∆Q = 0 or Qin = Qout

Heat and the First Law of Thermodynamics Express the heat lost by the lead in terms of its specific heat and temperature change: Express the heat absorbed by the water in terms of its specific heat and temperature change: Substitute to obtain:

1371

Qout = mPb cPb ∆TPb

Qin = mw cw ∆Tw

mw cw ∆Tw = mPb cPb ∆TPb

Substitute numerical values:

(0.5 kg )(4.18 kJ/kg ⋅ K )(tf − 293 K ) = (0.2 kg )(0.128 kJ/kg ⋅ K )(363 K − tf ) Solve for tf to obtain:

tf = 293.8 K = 20.8°C

*34 • Picture the Problem During this process the water and the container will gain energy at the expense of the piece of metal. We can set the heat out of the metal equal to the heat into the water and the container and solve for the specific heat of the metal. Apply conservation of energy to the system to obtain: Express the heat lost by the metal in terms of its specific heat and temperature change: Express the heat gained by the water and the container in terms of their specific heats and temperature change:

∆Q = 0 or Qgained = Qlost Qlost = mmetal cmetal ∆Tmetal

Qgained = mw cw ∆Tw + mcontainer cmetal ∆Tw

Substitute to obtain:

mw cw ∆Tw + mcontainer cmetal∆Tw = mmetalcmetal∆Tmetal Substitute numerical values:

(0.5 kg )(4.18 kJ/kg ⋅ K )(294.4 K − 293 K ) + (0.2 kg )(294.4 K − 293 K ) cmetal = (0.1 kg )(373 K − 294.4 K ) cmetal

1372 Chapter 18 Solve for cmetal:

cmetal = 0.386 kJ/kg ⋅ K

35 •• Picture the Problem We can use Q = mc∆T to express the mass m of water that can be heated through a temperature interval ∆T by an amount of heat energy Q. We can then find the amount of heat energy expended by Armstrong from the definition of power. Express the amount of heat energy Q required to raise the temperature of a mass m of water by ∆T:

Q = mc∆T

Solve for m to obtain:

m=

Q c∆T

Use the definition of power to relate the heat energy expended by Armstrong to the rate at which he expended the energy:

P=

Q ⇒ Q = P∆t ∆t

Substitute to obtain:

m=

P∆t c∆T

Substitute numerical values and evaluate m:

m=

(400 J/s )(3600 s/h )(5 h/d )(20 d ) (4.184 kJ/kg ⋅ K )(373 K − 297 K )

= 453 kg 36 •• Picture the Problem During this process the ice and the water formed from the melted ice will gain energy at the expense of the glass tumbler and the water in it. We can set the heat out of the tumbler and the water that is initially at 24°C equal to the heat into the ice and ice water and solve for the final temperature of the drink. Apply conservation of energy to the system to obtain: Express the heat lost by the tumbler and the water in it in terms of their specific heats and common temperature change: Express the heat gained by the ice and the melted ice in terms of their specific heats and temperature

∆Q = 0 or Qgained = Qlost

Qlost = mglasscglass ∆T + mwater cwater ∆T

Qgained = mice cice ∆Tice + mice Lf + mice water cwater ∆Tice water

Heat and the First Law of Thermodynamics

1373

changes: Substitute to obtain:

mice cice ∆Tice + mice Lf + mice water c water ∆Tice water = mglass cglass ∆T + mwater c water ∆T Substitute numerical values:

(0.03 kg )(0.49 kcal/kg ⋅ K )(273 K − 270 K ) + (0.03 kg )(79.7 kcal/kg ) + (0.03 kg )(1 kcal/kg ⋅ K )tf = (0.025 kg )(0.2 kcal/kg ⋅ K )(297 K − tf ) + (0.2 kg )(1 kcal/kg ⋅ K )(297 K − tf ) Solve for tf:

tf = 283.6 K = 10.6°C

37 •• Picture the Problem Because we can not tell, without performing a couple of calculations, whether there is enough heat available in the 500 g of water to melt all of the ice, we’ll need to resolve this question first. (a) Determine the heat required to melt 200 g of ice:

Determine the heat available from 500 g of water:

Because Qwater < Qmelt ice:

Qmelt ice = mice Lf

= (0.2 kg )(79.7 kcal/kg ) = 15.94 kcal

Qwater = mwater cwater ∆Twater

= (0.5 kg )(1 kcal/kg ⋅ K ) × (293 K − 273 K ) = 10 kcal

The final temperature is 0°C.

(b) Equate the energy available from the water Qwater to miceLf and solve for mice:

mice =

Qwater Lf

Substitute numerical values and evaluate mice:

mice =

10 kcal = 125 g 79.7 kcal/kg

1374 Chapter 18 38 •• Picture the Problem Because the bucket contains a mixture of ice and water initially, we know that its temperature must be 0°C. We can equate the heat gained by the mixture of ice and water and the heat lost by the block of copper and solve for the amount of ice initially in the bucket. Apply conservation of energy to the system to obtain: Express the heat lost by the block of copper: Express the heat gained by the ice and the melted ice:

∆Q = 0 or Qgained = Qlost Qlost = mCu cCu ∆TCu

Qgained = mice Lf + mice water cwater ∆Tice water

mice Lf + mice water cwater ∆Tice water

Substitute to obtain:

− mCu cCu ∆TCu = 0 Solve for mice:

mice =

mCu cCu ∆TCu − mice water cwater ∆Tice water Lf

Substitute numerical values and evaluate mice:

mice =

(3.5 kg )(0.0923 kcal/kg ⋅ K )(353 K − 281 K ) 79.7 kcal/kg (1.2 kg )(1 kcal/kg ⋅ K )(281 K − 273 K ) − 79.7 kcal/kg

= 171 g 39 •• Picture the Problem During this process the ice and the water formed from the melted ice will gain energy at the expense of the condensing steam and the water from the condensed steam. We can equate these quantities and solve for the final temperature of the system. (a) Apply conservation of energy to the system to obtain: Express the heat required to melt the ice and raise the temperature of the

∆Q = 0 or Qgained = Qlost Qgained = mice Lf + mice water c water ∆Twater

Heat and the First Law of Thermodynamics

1375

ice water: Express the heat available from 20 g of steam and the cooling water formed from the condensed steam:

Qlost = msteam Lv + msteam cwater ∆Twater

Substitute to obtain:

mice Lf + mice water cwater ∆Twater = msteam Lv + msteam cwater ∆Twater Substitute numerical values:

(0.15 kg )(79.7 kcal/kg) + (0.15 kg )(1kcal/kg ⋅ K )(t f − 273 K ) = (0.02 kg )(540 kcal/kg ) + (0.02 kg )(1kcal/kg ⋅ K )(373 K − t f ) Solve for tf:

tf = 277.94 K = 4.94°C

(b) Because the final temperture is greater than 0°C, no ice is left. 40 •• Picture the Problem During this process the ice will gain heat and the water will lose heat. We can do a preliminary calculation to determine whether there is enough heat available to melt all of the ice and, if there is, equate the heat the heat lost by the water to the heat gained by the ice and resulting ice water as the system achieves thermal equilibrium. Apply conservation of energy to the system to obtain: Find the heat available to melt the ice:

Find the heat required to melt all of the ice:

Because Qavail > Qmelt ice, we know

∆Q = 0 or Qgained = Qlost

Qavail = mwater c water ∆Twater

= (1 kg )(1 kcal/kg ⋅ K ) × (303 K − 273 K ) = 30 kcal

Qmelt ice = mice Lf

= (0.05 kg )(79.7 kcal/kg ) = 3.985 kcal

Qlost = mwater cwater ∆Twater

1376 Chapter 18 that the final temperature will be greater than 273 K and we can express Qlost in terms of the change in temperature of the water: Express Qgained: Equate the heat gained and the heat lost to obtain:

Qgained = mice Lf + mice water cwater ∆Tice water mice Lf + mice water cwater ∆Tice water = mwater cwater ∆Twater

Substitute numerical values to obtain:

(0.05 kg )(79.7 kcal/kg) + (0.05 kg )(1kcal/kg ⋅ K )(Tf − 273 K ) = (1 kg )(1 kcal/kg ⋅ K )(303 K − Tf ) Solving for Tf yields:

Tf = 297.8 K = 24.8°C

Find the heat required to melt 500 g of ice:

Qmelt ice = mice Lf

= (0.5 kg )(79.7 kcal/kg ) = 39.85 kcal

Because the heat required to melt 500 g of ice is greater than the heat available, the final temperature will be 0°C. *41 •• Picture the Problem Assume that the calorimeter is in thermal equilibrium with the water it contains. During this process the ice will gain heat in warming to 0°C and melting, as will the water formed from the melted ice. The water in the calorimeter and the calorimeter will lose heat. We can do a preliminary calculation to determine whether there is enough heat available to melt all of the ice and, if there is, equate the heat the heat lost by the water to the heat gained by the ice and resulting ice water as the system achieves thermal equilibrium. Find the heat available to melt the ice:

Qavail = mwater cwater ∆Twater + mcalccal ∆Twater

= [(0.5 kg )(4.18 kJ/kg ⋅ K )+ (0.2 kg )(0.9 kJ/kg ⋅ K )](293 K − 273 K ) = 45.40 kJ

Heat and the First Law of Thermodynamics

1377

Find the heat required to melt all of the ice:

Qmelt ice = mice cice ∆Tice + mice Lf

= (0.1 kg )(2 kJ/kg ⋅ K )(273 K − 253 K ) + (0.1 kg )(333.5 kJ/kg ) = 37.35 kJ

(a) Because Qavail > Qmelt ice, we know that the final temperature will be greater than 0°C. Apply the conservation of energy to the system to obtain: Express Qlost in terms of the final temperature of the system: Express Qgained in terms of the final temperature of the system:

∆Q = 0 or Qgained = Qlost

Qlost = (mwater cwater + mcalccal )∆Twater+calorimeter

Qgained = mice cice ∆Tice + mice Lf

Substitute to obtain:

mice cice ∆Tice + mice Lf + mice water cwater ∆Tice water = (mwater cwater + mcalccal )∆Twater+calorimeter Substitute numerical values:

37.35 kJ + (0.1 kg )(4.18 kJ/kg ⋅ K )(tf − 273 K )

= [(0.5 kg )(4.18 kJ/kg ⋅ K ) + (0.2 kg )(0.9 kJ/kg ⋅ K )](293 K − tf )

Solving for tf yields:

tf = 276 K = 2.99°C

(b) Find the heat required to raise 200 g of ice to 0°C:

Qwarm ice = mice cice ∆Tice = (0.2 kg )(2 kJ/kg ⋅ K )(273 K − 253 K ) = 8.00 kJ Noting that there are now 600 g of water in the calorimeter, find the heat available from cooling the calorimeter and water from 3°C to 0°C:

Qavail = mwater cwater ∆Twater + mcalccal ∆Twater

= [(0.6 kg )(4.18 kJ/kg ⋅ K ) + (0.2 kg )(0.9 kJ/kg ⋅ K )](293 K − 273 K ) = 8.064 kJ

1378 Chapter 18 Qavail − Qwarm ice Lf

Express the amount of ice that will melt in terms of the difference between the heat available and the heat required to warm the ice:

mmelted ice =

Substitute numerical values and evaluate mmelted ice:

mmelted ice =

Find the ice remaining in the system:

mremaining ice = 200 g − 0.1919 g

8.064 kJ − 8 kJ 333.5 kJ/kg = 0.1919 g

= 199.8 g

(c)

Because the initial and final conditions are the same, the answer would be the same.

42 •• Picture the Problem Let the subscript B denote the block, w1 the water initially in the calorimeter, and w2 the 120 mL of water that is added to the calorimeter vessel. We can equate the heat gained by the calorimeter and its initial contents to the heat lost by the warm water and solve this equation for the specific heat of the block. Apply conservation of energy to the system to obtain: Express the heat gained by the block, the calorimeter, and the water initially in the calorimeter:

∆Q = 0 or Qgained = Qlost Qgained = mBcB ∆TB + mCu cCu ∆TCu

(

+ mw1 cw1 ∆Tw1

)

= mBcB + mCu cCu + mw1 cw1 ∆T because the temperature changes are the same for the block, calorimeter, and the water that is initially at 20°C.

Express the heat lost by the water that is added to the calorimeter:

Qlost = mw 2 cw 2 ∆Tw 2

Substitute to obtain:

(m c

B B

)

+ mCu cCu + mw1 cw1 ∆T = mw 2 cw 2 ∆Tw 2

Heat and the First Law of Thermodynamics

1379

Substitute numerical values to obtain:

[(0.1kg )cB + (0.025 kg )(0.386 kJ/kg ⋅ K )+ (0.06 kg )(4.18 kJ/kg ⋅ K )](327 K − 293 K ) = (120 × 10−3 kg )(4.18 kJ/kg ⋅ K )(353 K − 327 K ) Solve for cB to obtain:

cB = 1.23 kJ/kg ⋅ K = 0.294 cal/g ⋅ K

43 •• Picture the Problem We can find the temperature t by equating the heat gained by the warming water and calorimeter, and vaporization of some of the water. Apply conservation of energy to the system to obtain: Express the heat gained by the warming and vaporizing water:

Express the heat lost by the 100-g piece of copper as it cools:

∆Q = 0 or Qgained = Qlost

Qgained = mw, vaporized Lf, w + mw cw ∆Tw + mcalccal∆Tw

Qlost = mCu cCu ∆TCu

Substitute to obtain:

mw, vaporized Lf, w + mw cw ∆Tw + mcalccal∆Tw = mCu cCu ∆TCu Substitute numerical values:

(1.2 g )(540 cal/g ) + (200 g )(1cal/g ⋅ K )(311K − 289 K ) + (150 g )(0.0923 cal/g ⋅ K )(311 K − 289 K ) = (100 g )(0.0923 cal/g ⋅ K )(t − 311 K ) Solve for t to obtain:

t = 891 K = 618°C

44 •• Picture the Problem We can find the final temperature of the system by equating the heat gained by the calorimeter and the water in it to the heat lost by the cooling aluminum shot. In (b) we’ll proceed as in (a) but with the initial and final temperatures adjusted to minimize heat transfer between the system and its surroundings.

1380 Chapter 18 Apply conservation of energy to the system to obtain: (a) Express the heat gained by the warming water and the calorimeter: Express the heat lost by the aluminum shot as it cools: Substitute to obtain:

∆Q = 0 or Qgained = Qlost Qgained = mw cw ∆Tw + mcalcAl ∆Tw

Qlost = mshot cAl ∆TAl

(mw cw + mcalcAl )∆Tw = mshot cAl∆TAl

Substitute numerical values to obtain:

[(500 g )(1cal/g ⋅ K )+ (200 g )(0.0923 cal/g ⋅ K )](tf − 293 K ) = (300 g )(0.215 cal/g ⋅ K )(373 K − tf ) Solve for tf to obtain:

tf = 301.9 K = 28.9°C

(b) Let the initial and final temperatures of the calorimeter and its contents be:

(1) ti = 20°C – t0 and tf = 20°C + t0 where ti and tf are the temperatures above and below room temperature and t0 is the amount ti and tf must be below and above room temperature respectively.

Express and the heat gained by the water and calorimeter: Express the heat lost by the aluminum shot as it cools: Equate Qin and Qout to obtain:

Qin = mw cw ∆Tw + mcalcAl ∆Tw = (mw cw + mcalcAl )∆Tw

Qout = mshot cAl ∆TAl

(mw cw + mcalcAl )∆Tw

= mshot cAl ∆TAl

Substitute numerical values:

[(500 g )(1cal/g ⋅ K )+ (200 g )(0.215 cal/g ⋅ K )](293 K + t0 − 293 K + t0 ) = (300 g )(0.215 cal/g ⋅ K )(373 K − 293 K − t0 ) Solve for and evaluate t0:

t0 = 277.49 K = 4.49°C

Heat and the First Law of Thermodynamics Substitute in equation (1) to obtain:

1381

ti = 20°C − 4.49°C = 15.5°C

First Law of Thermodynamics 45 • Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas: The work done by the gas equals the negative of the work done on the gas. Substitute numerical values and evaluate ∆Eint:

∆Eint = Qin + Won

∆Eint = 600 cal ×

4.184 J − 300 J cal

= 2.21 kJ

*46 • Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process.

Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas: The work done by the gas is the negative of the work done on the gas. Substitute numerical values and evaluate ∆Eint:

∆Eint = Qin + Won

∆Eint = 400 kcal ×

4.184 J − 800 kJ cal

= 874 kJ

47 • Picture the Problem We can use the first law of thermodynamics to relate the change in the bullet’s internal energy to its pre-collision kinetic energy.

1382 Chapter 18 ∆Eint = Qin + Won

Using the first law of thermodynamics, relate the change in the internal energy of the bullet to the work done on it by the block of wood:

or, because Qin = 0,

Substitute for ∆Eint, Kf, and Ki to obtain:

mcPb (tf − ti ) = − 0 − 12 mv 2 = 12 mv 2

Solve for tf:

v2 t f = ti + 2cPb

Substitute numerical values and evaluate tf:

∆Eint = Won = ∆K = −(K f − K i )

(

t f = 293 K +

)

(200 m/s)2

2(0.128 kJ/kg ⋅ K )

= 449 K = 176°C 48 • Picture the Problem What is described above is clearly a limiting case because, as the water falls, it will, for example, collide with rocks and experience air drag; resulting in some of its initial potential energy being converted into internal energy. In this limiting case we can use the first law of thermodynamics to relate the change in the gravitational potential energy (take Ug = 0 at the bottom of the waterfalls) to the change in internal energy of the water and solve for the increase in temperature. (a) Using the first law of thermodynamics and noting that, because the gravitational force is conservative, Won = −∆U, relate the change in the internal energy of the water to the work done on it by gravity: Substitute for ∆Eint, Uf, and Ui to obtain: Solve for ∆T:

Substitute numerical values and evaluate ∆T:

∆Eint = Qin + Won or, because Qin = 0,

∆Eint = Won = −∆U = −(U f − U i )

mcw ∆T = −(0 − mg∆h ) = mg∆h

∆T =

g∆h cw

(9.81m/s )(50 m) = ∆T = 2

4.18 kJ/kg ⋅ K

0.117 K

Heat and the First Law of Thermodynamics (b) Proceed as in (a) with ∆h = 740 m:

(9.81m/s )(740 m) = ∆T =

1383

2

4.18 kJ/kg ⋅ K

1.74 K

49 • Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas: The work done by the gas is the negative of the work done on the gas. Substitute numerical values and evaluate ∆Eint:

∆Eint = Qin + Won

∆Eint = 20 cal ×

4.184 J − 30 J = 53.7 J cal

50 •• Picture the Problem We can use the definition of kinetic energy to express the speed of the bullet upon impact in terms of its kinetic energy. The heat absorbed by the bullet is the sum of the heat required to warm to bullet from 202 K to its melting temperature of 600 K and the heat required to melt it. We can use the first law of thermodynamics to relate the impact speed of the bullet to the change in its internal energy. Using the first law of thermodynamics, relate the change in the internal energy of the bullet to the work done on it by the target: Substitute for ∆Eint, Kf, and Ki to obtain:

∆Eint = Qin + Won or, because Qin = 0,

∆Eint = Won = ∆K = −(K f − K i )

(

or

mcPb (TMP − Ti ) + mLf,Pb = 12 mv 2

Solve for v to obtain:

)

mcPb ∆TPb + mLf,Pb = − 0 − 12 mv 2 = 12 mv 2

v = 2[cPb (TMP − Ti ) + Lf,Pb ]

Substitute numerical values and evaluate v:

v = 2{(0.128 kJ/kg ⋅ K )(600 K − 303 K ) + 24.7 kJ/kg} = 354 m/s

1384 Chapter 18 *51 •• Picture the Problem We can find the rate at which heat is generated when you rub your hands together using the definition of power and the rubbing time to produce a 5°C increase in temperature from ∆Q = (dQ dt )∆t and Q = mc∆T. (a) Express the rate at which heat is generated as a function of the friction force and the average speed of your hands:

dQ = P = f k v = µFn v dt

Substitute numerical values and evaluate dQ/dt:

dQ = 0.5(35 N )(0.35 m/s ) = 6.13 W dt

(b) Relate the heat required to raise the temperature of your hands 5 K to the rate at which it is being generated:

∆Q =

Solve for ∆t:

∆t =

Substitute numerical values and evaluate ∆t:

∆t =

dQ ∆t = mc∆T dt

mc∆T dQ dt

(0.35 kg )(4 kJ/kg ⋅ K )(5 K )

= 1143 s ×

6.13 W 1 min = 19.0 min 60 s

Work and the PV Diagram for a Gas 52 • Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat added to the system during this process.

Heat and the First Law of Thermodynamics 1385 (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram.

The work done by the gas equals the area under the shaded curve:

Wby gas

101.3 kPa ⎞ ⎛ 10 −3 m 3 ⎞ ⎛ ⎜ ⎟ = 608 J = P∆V = (3 atm )(2 L ) = ⎜ 3 atm × ⎟ 2L× atm ⎠ ⎜⎝ L ⎟⎠ ⎝

(b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain: Substitute numerical values and evaluate Qin:

Qin = ∆Eint − Won

= (Eint,2 − Eint,1 ) − (− Wby gas ) = (Eint,2 − Eint,1 ) + Wby gas

Qin = (912 J − 456 J ) + 608 J = 1.06 kJ

53 • Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat added to the system during this process (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram.

1386 Chapter 18 The work done by the gas equals the area under the curve:

Wby gas

101.3 kPa ⎞ ⎛ 10 −3 m 3 ⎞ ⎛ ⎟ = 405 J = P∆V = (2 atm )(2 L ) = ⎜ 2 atm × ⎟⎜2L× atm ⎠ ⎜⎝ L ⎟⎠ ⎝

(b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain: Substitute numerical values and evaluate Qin:

Qin = ∆Eint − Won

= (Eint,2 − Eint,1 ) − (− Wby gas ) = (Eint,2 − Eint,1 ) + Wby gas

Qin = (912 J − 456 J ) + 405 J = 861 J

*54 •• Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isothermal expansion. We can then use the first law of thermodynamics to find the heat added to the system during this process. (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram.

The work done by the gas equals the area under the curve:

V2

3L

dV V 1L

Wby gas = ∫ PdV = nRT1 ∫ V1

3L

dV 3L = P1V1 [ln V ]1 L V 1L

= P1V1 ∫

= P1V ln 3 Substitute numerical values and evaluate Wby gas:

Heat and the First Law of Thermodynamics 1387 101.3 kPa ⎞ ⎛ 10 −3 m 3 ⎞ ⎛ ⎟ ln 3 = 334 J Wby gas = ⎜ 3 atm × ⎟ ⎜⎜1 L × atm ⎠ ⎝ L ⎟⎠ ⎝ (b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain: Substitute numerical values and evaluate Qin:

Qin = ∆Eint − Won

= (Eint,2 − Eint,1 ) − (− Wby gas ) = (Eint,2 − Eint,1 ) + Wby gas

Qin = (912 J − 456 J ) + 334 J = 790 J

55 •• Picture the Problem We can find the work done by the gas during this process from the area under the curve. We can then use the first law of thermodynamics to find the heat added to the system during this process. (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram:

The work done by the gas equals the area under the curve:

(b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain: Substitute numerical values and evaluate Qin:

Wby gas = Atrapezoid = = 5 atm ⋅ L ×

1 2

(3 atm + 2 atm )(2 L)

101.3 J = 507 J atm ⋅ L

Qin = ∆Eint − Won

= (Eint,2 − Eint,1 ) − (− Wby gas ) = (Eint,2 − Eint,1 ) + Wby gas

Qin = (912 J − 456 J ) + 507 J = 963 J

1388 Chapter 18 Remarks: You could use the linearity of the path connecting the initial and final states and the coordinates of the endpoints to express P as a function of V. You could then integrate this function between 1 and 3 L to find the work done by the gas as it goes from its initial to its final state. 56 •• Picture the Problem We can find the work done by the gas during this process from the area under the curve. The path from the initial state i to the final state f is shown on the PV diagram:

The work done by the gas equals the area under the curve:

Wby gas = Atrapezoid =

1 2

(1atm + 3 atm )(50 L )

= 100 atm ⋅ L ×

101.3 J = 10.1 kJ atm ⋅ L

Remarks: You could use the linearity of the path connecting the initial and final states and the coordinates of the endpoints to express P as a function of V. You could then integrate this function between 1 and 3 L to find the work done by the gas as it goes from its initial to its final state. 57 •• Picture the Problem We can find the work done by the gas from the area under the PV curve provided we can find the pressure and volume coordinates of the initial and final states. We can find these coordinates by using the ideal gas law and the condition

T = AP 2 .

PV = RAP 2 ⇒ V = RAP

Apply the ideal-gas law with n = 1 mol and T = AP 2 to obtain:

(1) This result tells us that the volume varies linearly with the pressure.

Solve the condition on the temperature for the pressure of the gas:

P0 =

T0 A

Heat and the First Law of Thermodynamics 1389 Find the pressure when the temperature is 4T0:

P=

4T0 T = 2 0 = 2 P0 A A

(2V0 ,2P0 )

Using equation (1), express the coordinates of the final state: The PV diagram for the process is shown to the right:

Wby gas = Atrapezoid =

The work done by the gas equals the area under the curve:

=

3 2

1 2

(P0 + 3P0 )(2V0 − V0 )

P0V0

*58 • Picture the Problem From the ideal gas law, PV = NkT, or V = NkT/P. Hence, on a VT diagram, isobars will be straight lines with slope 1/P. A spreadsheet program was used to plot the following graph. The graph was plotted for 1 mol of gas. 300 P = 1 atm P = 0.5 atm

250

P = 0.1 atm

3

V (m )

200

150

100

50

0 0

50

100

150

200 T (K)

250

300

350

1390 Chapter 18 59 •• Picture the Problem The PV diagram shows the isothermal expansion of the ideal gas from its initial state 1 to its final state 2. We can use the ideal-gas law for a fixed amount of gas to find V2 and then evaluate

∫ PdV for an isothermal process to find the

work done by the gas. In part (b) of the problem we can apply the first law of thermodynamics to find the heat added to the gas during the expansion. V2

V2

(a) Express the work done by a gas during an isothermal process:

Wby gas = ∫ PdV = nRT ∫

Apply the ideal-gas law for a fixed amount of gas undergoing an isothermal process:

P1V1 = P2V2 or

Solve for and evaluate V2:

V2 =

Substitute numerical values and evaluate W:

Wby gas = (200 kPa )(4 L ) ∫

V

2 dV dV = P1V1 ∫ V V V1 V1

V1

V2 P1 = V1 P1

P1 200 kPa (4 L ) = 8 L V1 = P2 100 kPa 8L

dV V 4L

= (800 kPa ⋅ L ) [ln V ] 4 L 8L

⎛ 8L ⎞ ⎟⎟ = (800 kPa ⋅ L ) ln⎜⎜ ⎝ 4L⎠ 10 −3 m 3 = 800 kPa ⋅ L × 555 J L

(b) Apply the first law of thermodynamics to the system to obtain:

Qin = ∆Eint − Won or, because ∆Eint = 0 for an isothermal process,

Qin = −Won Because the work done by the gas is the negative of the work done on the gas:

Qin = −(− Wby gas ) = Wby gas

Heat and the First Law of Thermodynamics 1391 Substitute numerical values and evaluate Qin:

Qin = 555 J

Heat Capacities of Gases and the Equipartition Theorem 60 • Picture the Problem We can find the number of moles of the gas from its heat capacity at constant volume using CV = 32 nR . We can find the internal energy of the gas from

Eint = CVT and the heat capacity at constant pressure using CP = CV + nR . (a) Express CV in terms of the number of moles in the monatomic gas:

CV = 32 nR

Solve for n:

n=

2CV 3R

Substitute numerical values and evaluate n:

n=

2(49.8 J/K ) = 3.99 3(8.314 J/mol ⋅ K )

(b) Relate the internal energy of the gas to its temperature: Substitute numerical values and evaluate Eint: (c) Relate the heat capacity at constant pressure to the heat capacity at constant volume: Substitute numerical values and evaluate CP:

Eint = CVT Eint = (49.8 J/K )(300 K ) = 14.9 kJ

CP = CV + nR = 32 nR + nR = 52 nR

CP =

5 2

(3.99)(8.314 J/mol ⋅ K )

= 82.9 J/K

61 • Picture the Problem The Dulong-Petit law gives the molar specific heat of a solid, c′. The specific heat is defined as c = c′/M where M is the molar mass. Hence we can use this definition to find M and a periodic table to identify the element. (a) Apply the Dulong-Petit law:

c' = 3R or c =

3R M

1392 Chapter 18 Solve for M:

M =

3R c

Substitute numerical values and evaluate M:

M =

24.9 J/mol ⋅ K = 55.7 g/mol 0.447 kJ/kg ⋅ K

Consulting the periodic table of the elements we see that the element is most likely iron. *62 •• Picture the Problem The specific heats of air at constant volume and constant pressure are given by cV = CV/m and cP = CP/m and the heat capacities at constant volume and constant pressure are given by CV = 52 nR and CP = 72 nR , respectively. (a) Express the specific heats per unit mass of air at constant volume and constant pressure:

cV =

CV m

(1)

CP m

(2)

and

cP =

CV = 52 nR

Express the heat capacities of a diatomic gas in terms of the gas constant R, the number of moles n, and the number of degrees of freedom:

and

CP = 72 nR

Express the mass of 1 mol of air:

m = 0.74 M N 2 + 0.26M O2

Substitute in equation (1) to obtain:

cV =

5nR 2 0.74M N2 + 0.26M O2

(

)

Substitute numerical values and evaluate cV:

cV =

5(1 mol)(8.314 J/mol ⋅ K ) = 716 J/kg ⋅ K 2 0.74 28 × 10 −3 kg + 0.26 32 × 10 −3 kg

[ (

Substitute in equation (2) to obtain:

)

)]

(

cP =

7nR 2 0.74 M N 2 + 0.26M O2

(

)

Heat and the First Law of Thermodynamics 1393 Substitute numerical values and evaluate cP:

cP =

7(1mol)(8.314 J/mol ⋅ K ) = 1002 J/kg ⋅ K 2 0.74 28 × 10−3 kg + 0.26 32 × 10−3 kg

[ (

)

)]

(

(b) Express the percent difference between the value from the Handbook of Chemistry and Physics and the calculated value:

% difference =

1.032 J/g ⋅ K − 1.002J/g ⋅ K = 2.91% 1.032 J/g ⋅ K

63 •• Picture the Problem We know that, during a constant-volume process, no work is done and that we can calculate the heat added during this expansion from the heat capacity at constant volume and the change in the absolute temperature. We can then use the first law of thermodynamics to find the change in the internal energy of the gas. In part (b), we can proceed similarly; using the heat capacity at constant pressure rather than constant volume. (a) The increase in the internal energy of the ideal diatomic gas is given by: Substitute numerical values and evaluate ∆Eint:

∆Eint = 52 nR∆T

∆Eint =

5 2

(1mol)(8.315 J/mol ⋅ K )(300 K )

= 6.24 kJ

For a constant-volume process:

Won = 0

From the 1st law of thermodynamics we have:

Qin = ∆Eint − Won

Substitute numerical values and evaluate Qin:

Qin = 6.24 kJ − 0 = 6.24 kJ

(b) Because ∆Eint depends only on the temperature difference:

∆Eint = 6.24 kJ

Relate the heat added to the gas to its heat capacity at constant pressure and the change in its temperature:

Qin = CP ∆T = ( 52 nR + nR )∆T = 72 nR∆T

1394 Chapter 18 Substitute numerical values and evaluate Qin:

Apply the first law of thermodynamics to find W:

(c) Integrate dWon = P dV to obtain:

Qin =

7 2

(1mol)(8.314 J/mol ⋅ K )(300 K )

= 8.73 kJ Won = ∆Eint − Qin = 8.73 kJ − 6.24 kJ = 2.49 kJ Vf

Won = ∫ PdV = P(Vf − Vi ) = nR(Tf − Ti ) Vi

Substitute numerical values and evaluate Won:

Won = (1 mol)(8.314 J/mol ⋅ K )(300 K ) = 2.49 kJ

64 •• Picture the Problem Because this is a constant-volume process, we can use Q = CV ∆T to express Q in terms of the temperature change and the ideal-gas law for a fixed amount of gas to find ∆T. Express the amount of heat Q that must be transferred to the gas if its pressure is to triple: Using the ideal-gas law for a fixed amount of gas, relate the initial and final temperatures, pressures and volumes:

Q = CV ∆T

= 52 nR(Tf − T0 )

P0V 3P0V = T0 Tf

Solve for Tf:

Tf = 3T0

Substitute and simplify to obtain:

Q = 52 nR(2T0 ) = 5(nRT0 ) = 5 P0V

65 •• Picture the Problem Let the subscripts i and f refer to the initial and final states of the gas, respectively. We can use the ideal-gas law for a fixed amount of gas to express V′ in terms of V and the change in temperature of the gas when 13,200 J of heat are transferred to it. We can find this change in temperature using Q = C P ∆T . Using the ideal-gas law for a fixed amount of gas, relate the initial and

PiV Pf V' = Ti Tf

Heat and the First Law of Thermodynamics 1395 final temperatures, volumes, and pressures: Because the process is isobaric, we can solve for V′ to obtain:

Relate the heat transferred to the gas to the change in its temperature:

V' = V

⎛ ∆T ⎞ Tf T + ∆T ⎟⎟ =V i = V ⎜⎜1 + Ti Ti T i ⎠ ⎝

Q = CP ∆T = 72 nR∆T

Solve ∆T:

∆T =

Substitute to obtain:

⎛ 2Q ⎞ ⎟⎟ V' = V ⎜⎜1 + ⎝ 7 nRTi ⎠

2Q 7nR

One mol of gas at STP occupies 22.4 L. Substitute numerical values and evaluate V′:

⎛ ⎞ 2(13.2 kJ ) ⎟⎟ = 59.6 L V' = (22.4 × 10 −3 m 3 ) ⎜⎜1 + ⎝ 7(1 mol)(8.314 J/mol ⋅ K )(273 K ) ⎠ 66 •• Picture the Problem We can use the relationship between CP and CV ( C P = C V + nR ) to find the number of moles of this particular gas. In parts (b) and (c) we can use the number of degrees of freedom associated with monatomic and diatomic gases, respectively, to find CP and CV. (a) Express the heat capacity of the gas at constant pressure to its heat capacity at constant volume:

C P = C V + nR

Solve for n:

n=

C P − CV R

Substitute numerical values and evaluate n:

n=

29.1 J/K = 3.50 mol 8.314 J/mol ⋅ K

(b) CV for a monatomic gas is given by:

CV = 32 nR

1396 Chapter 18 Substitute numerical values and evaluate CV:

Express CP for a monatomic gas: Substitute numerical values and evaluate CP:

(c) If the diatomic molecules rotate but do not vibrate they have 5 degrees of freedom:

CV =

3 2

(3.5 mol)(8.314 J/mol ⋅ K )

= 43.6 J/K CP = 52 nR CP =

5 2

(3.5 mol)(8.314 J/mol ⋅ K )

= 72.7 J/K CV = 52 nR =

(3.5 mol)(8.314 J/mol ⋅ K )

5 2

= 72.7 J/K and

CP = 72 nR =

7 2

(3.5 mol)(8.314 J/mol ⋅ K )

= 102 J/K *67 •• Picture the Problem We can find the change in the heat capacity at constant pressure as CO2 undergoes sublimation from the energy per molecule of CO2 in the solid and gaseous states. Express the change in the heat capacity (at constant pressure) per mole as the CO2 undergoes sublimation: Express Cp,gas in terms of the number of degrees of freedom per molecule:

∆CP = CP,gas − CP,solid

CP,gas = f ( 12 Nk ) = 52 Nk because each molecule has three translational and two rotational degrees of freedom in the gaseous state.

We know, from the Dulong-Petit Law, that the molar specific heat of most solids is 3R = 3Nk. This result is essentially a per-atom result as it was obtained for a monatomic solid with six degrees of freedom. Use this result and the fact CO2 is triatomic to express CP,solid:

CP,solid =

Substitute to obtain:

∆CP = 52 Nk − 182 Nk = − 132 Nk

3Nk × 3 atoms = 9 Nk atom

Heat and the First Law of Thermodynamics 1397 68 •• Picture the Problem We can find the initial internal energy of the gas from U i = 32 nRT and the final internal energy from the change in internal energy resulting from the addition of 500 J of heat. The work done during a constant-volume process is zero and the work done during the constant-pressure process can be found from the first law of thermodynamics. (a) Express the initial internal energy of the gas in terms of its temperature: Substitute numerical values and evaluate Eint,i:

(b) Relate the final internal energy of the gas to its initial internal energy:

Eint,i = 32 nRT

Eint,i =

3 2

(1mol)(8.314 J/mol ⋅ K )(273 K )

= 3.40 kJ Eint,f = Eint,i + ∆Eint = Eint,i + CV ∆T

Qin CP

Express the change in temperature of the gas resulting from the addition of heat:

∆T =

Substitute to obtain:

Eint,f = Eint,i +

Substitute numerical values and evaluate Eint,f:

Eint,f = 3.40 kJ +

(c) Relate the final internal energy of the gas to its initial internal energy: Apply the first law of thermodynamics to the constantvolume process: Substitute numerical values and evaluate Eint,f:

CV Qin CP 3 2 5 2

nR (500 J ) = 3.70 kJ nR

Eint,f = Eint,i + ∆Eint

∆Eint = Qin + Won or, because Won = 0,

∆Eint = Qin = 500 J Eint,f = 3.40 kJ + 500 J = 3.90 kJ

1398 Chapter 18 69 •• Picture the Problem We can use CV,water = f ( 12 Nk ) to express CV,water and then count the number of degrees of freedom associated with a water molecule to determine f. Express CV,water in terms of the number of degrees of freedom per molecule:

CV,water = f ( 12 Nk ) where f is the number of degrees of freedom associated with a water molecule.

There are three translational degrees of freedom and three rotational degrees of freedom. In addition, each of the hydrogen atoms can vibrate against the oxygen atom, resulting in an additional 4 degrees of freedom (2 per atom). Substitute for f to obtain:

CV,water = 10( 12 Nk ) = 5 Nk

Quasi-Static Adiabatic Expansion of a Gas *70 •• Picture the Problem The adiabatic expansion is shown in the PV diagram. We can use the ideal-gas law to find the initial volume of the gas and the equation for a quasi-static adiabatic process to find the final volume of the gas. A second application of the ideal-gas law, this time at the final state, will yield the final temperature of the gas. In part (c) we can use the first law of thermodynamics to find the work done by the gas during this process. (a) Apply the ideal-gas law to express the initial volume of the gas:

Vi =

Substitute numerical values and evaluate Vi:

Vi =

nRTi Pi

(1mol)(8.314 J/mol ⋅ K )(273 K ) 10 atm ×

101.3 kPa atm

= 2.24 × 10 −3 m 3 = 2.24 L

Heat and the First Law of Thermodynamics Use the relationship between the pressures and volumes for a quasistatic adiabatic process to express Vf:

⎛P PiVi = Pf Vf ⇒ Vf = Vi ⎜⎜ i ⎝ Pf

Substitute numerical values and evaluate Vf:

⎛P Vf = Vi ⎜⎜ i ⎝ Pf

γ

γ

1γ

⎞ ⎟⎟ ⎠

1399

1γ

⎞ ⎟⎟ ⎠

⎛ 10 atm ⎞ ⎟⎟ = (2.24 L )⎜⎜ ⎝ 2 atm ⎠

35

= 5.88 L (b) Apply the ideal-gas law to express the final temperature of the gas:

Tf =

Substitute numerical values and evaluate Tf:

Tf =

Pf Vf nR

(2 atm)(5.88 L) 8.206 × 10 −2 L ⋅ atm/mol ⋅ K

= 143 K (c) Apply the first law of thermodynamics to express the work done on the gas: Substitute numerical values and evaluate Won: Because Wby the gas = −Won:

Won = ∆Eint − Qin or, because the process is adiabatic,

Won = ∆Eint = CV ∆T = 32 nR∆T Won =

3 2

(1mol)(8.314 J/mol ⋅ K )(− 130 K )

= −1.62 kJ Wby gas = 1.62 kJ

71 • Picture the Problem We can use the temperature-volume equation for a quasi-static adiabatic process to express the final temperature of the gas in terms of its initial temperature and the ratio of its heat capacitiesγ. Because C P = C V + nR , we can determine γ for each of the given heat capacities at constant volume. Express the temperature-volume relationship for a quasi-static adiabatic process:

TiVi γ -1 = Tf Vfγ -1

Solve for the final temperature:

⎛V Tf = Ti ⎜⎜ i ⎝ Vf

⎞ ⎟⎟ ⎠

γ −1

⎛ V ⎞ = Ti ⎜⎜ 1 i ⎟⎟ ⎝ 2 Vi ⎠

γ −1

= Ti (2 )

γ −1

1400 Chapter 18 (a) Evaluate γ for C V = 32 nR :

γ =

Evaluate Tf:

Tf = (293 K )(2 )3 = 465 K

(b) Evaluate γ for C V = 52 nR :

γ =

Evaluate Tf:

Tf = (293 K )(2 )5 = 387 K

CP 52 nR = = CV 32 nR 5

5 3

−1

CP 72 nR = = CV 52 nR 7

7 5

−1

72 • Picture the Problem We can use the temperature-volume and pressure-volume equations for a quasi-static adiabatic process to express the final temperature and pressure of the gas in terms of its initial temperature and pressure and the ratio of its heat capacities. Express the temperature-volume relationship for a quasi-static adiabatic process:

TiVi γ −1 = Tf Vfγ −1

Solve for the final temperature:

⎛V Tf = Ti ⎜⎜ i ⎝ Vf

Using γ = 5/3 for neon, evaluate Tf:

Tf = (293 K )(4 )3 = 738 K

Express the relationship between the pressures and volumes for a quasistatic adiabatic process:

PiVi γ = Pf Vfγ

Solve for Pf:

⎛ V ⎞ Pf = Pi ⎜⎜ 1 i ⎟⎟ ⎝ 4 Vi ⎠

Substitute numerical values and evaluate Pf:

⎞ ⎟⎟ ⎠

γ −1

⎛ V ⎞ = Ti ⎜⎜ 1 i ⎟⎟ ⎝ 4 Vi ⎠ 5

γ −1

= Ti (4 )

γ −1

−1

γ

Pf = (1atm )(4 )

53

= 10.1atm

*73 •• Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression (Equation 19-16) for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In part (b) we can use the relationship between the pressures

Heat and the First Law of Thermodynamics

1401

and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas. Use the ideal-gas law to express the initial volume of the gas:

Vi =

Substitute numerical values and evaluate Vi:

Vi =

nRTi Pi

(0.5 mol)(8.314 J/mol ⋅ K )(300 K ) 400 kPa −3

= 3.12 ×10 m 3 = 3.12 L (a) Because the process is isothermal:

Tf = Ti = 300 K

Use the ideal-gas law for a fixed amount of gas to express Vf:

PiVi Pf Vf = Ti Tf or, because T = constant,

Vf = Vi

Pi Pf

Substitute numerical values and evaluate Vf:

⎛ 400 kPa ⎞ ⎟⎟ = 7.80 L Vf = (3.12 L )⎜⎜ ⎝ 160 kPa ⎠

Express the work done by the gas during the isothermal expansion:

Wby gas = nRT ln

Substitute numerical values and evaluate Wby gas:

Vf Vi

Wby gas = (0.5 mol)(8.314 J/mol ⋅ K ) ⎛ 7.80 L ⎞ ⎟⎟ × (300 K ) ln⎜⎜ ⎝ 3.12 L ⎠ = 1.14 kJ

Noting that the work done by the gas during the process equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas:

Qin = ∆Eint − Won = 0 − (− 1.14 kJ ) = 1.14 kJ

1402 Chapter 18 (b) Using γ = 5/3 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf:

⎛P PiVi = Pf Vf ⇒ Vf = Vi ⎜⎜ i ⎝ Pf

Substitute numerical values and evaluate Vf:

⎛ 400 kPa ⎞ ⎟⎟ Vf = (3.12 L )⎜⎜ ⎝ 160 kPa ⎠

Apply the ideal-gas law to find the final temperature of the gas:

Tf =

Substitute numerical values and evaluate Tf:

γ

Tf

γ

1γ

⎞ ⎟⎟ ⎠

35

= 5.41L

Pf Vf nR

( 160 kPa ) (5.41× 10 −3 m 3 ) = (0.5 mol)(8.314 J/mol ⋅ K ) = 208 K

For an adiabatic process:

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic process: Substitute numerical values and evaluate Won:

Qin = 0

Won = ∆Eint − Qin = CV ∆T − 0 = 32 nR∆T

Won =

(0.5 mol)(8.314 J/mol ⋅ K ) × (208 K − 300 K )

3 2

= −574 J Because the work done by the gas equals the negative of the work done on the gas:

Wby gas = −(− 574 J ) = 574 J

74 •• Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression (Equation 19-16) for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In part (b) we can use the relationship between the pressures and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas.

Heat and the First Law of Thermodynamics

Use the ideal-gas law to express the initial volume of the gas:

Vi =

Substitute numerical values and evaluate Vi:

Vi =

nRTi Pi

(0.5 mol)(8.314 J/mol ⋅ K )(300 K ) 400 kPa −3

= 3.12 × 10 m 3 = 3.12 L (a) Because the process is isothermal:

Tf = Ti = 300 K

Use the ideal-gas law for a fixed amount of gas to express Vf:

PiVi Pf Vf = Ti Tf or, because T = constant,

Vf = Vi

Pi Pf

Substitute numerical values and evaluate Tf:

⎛ 400 kPa ⎞ ⎟⎟ = 7.80 L Vf = (3.12 L )⎜⎜ ⎝ 160 kPa ⎠

Express the work done by the gas during the isothermal expansion:

Wby gas = nRT ln

Substitute numerical values and evaluate Wby gas:

Vf Vi

Wby gas = (0.5 mol)(8.314 J/mol ⋅ K ) ⎛ 7.80 L ⎞ ⎟⎟ × (300 K ) ln⎜⎜ 3.12 L ⎝ ⎠ = 1.14 kJ

Noting that the work done by the gas during the isothermal expansion equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas: (b) Using γ = 1.4 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf:

Qin = ∆Eint − Won = 0 − (− 1.14 kJ ) = 1.14 kJ

⎛P PiVi = Pf Vf ⇒ Vf = Vi ⎜⎜ i ⎝ Pf γ

1403

γ

1γ

⎞ ⎟⎟ ⎠

1404 Chapter 18 1 1.4

Substitute numerical values and evaluate Vf:

⎛ 400 kPa ⎞ ⎟⎟ Vf = (3.12 L )⎜⎜ ⎝ 160 kPa ⎠

Apply the ideal-gas law to express the final temperature of the gas:

Tf =

Pf Vf nR

Substitute numerical values and evaluate Tf:

Tf =

(160 kPa )(6 ×10−3 m3 ) (0.5 mol)(8.314 J/mol ⋅ K )

= 6.00 L

= 231 K For an adiabatic process:

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion: Substitute numerical values and evaluate Won:

Qin = 0

Won = ∆Eint − Qin = CV ∆T − 0 = 52 nR∆T

Won =

(0.5 mol)(8.314 J/mol ⋅ K ) × (231 K − 300 K )

5 2

= −717 J Noting that the work done by the gas during the adiabatic expansion is the negative of the work done on the gas, we have:

Wby gas = −(− 717 J ) = 717 J

75 •• Picture the Problem We can eliminate the volumes from the equations relating the temperatures and volumes and the pressures and volumes for a quasi-static adiabatic process to obtain a relationship between the temperatures and pressures. We can find the initial volume of the gas using the ideal-gas law and the final volume using the pressurevolume relationship. In parts (d) and (c) we can find the change in the internal energy of the gas from the change in its temperature and use the first law of thermodynamics to find the work done by the gas during its expansion. (a) Express the relationship between temperatures and volumes for a quasi-static adiabatic process:

TiVi γ −1 = Tf Vfγ −1

Heat and the First Law of Thermodynamics Express the relationship between pressures and volumes for a quasistatic adiabatic process:

PiVi γ = Pf Vfγ

Eliminate the volume between these two equations to obtain:

⎛P Tf = Ti ⎜⎜ f ⎝ Pi

Substitute numerical values and evaluate Tf:

⎛ 1atm ⎞ ⎟⎟ Tf = (500 K )⎜⎜ ⎝ 5 atm ⎠

(b) Solve equation (1) for Vf:

⎛P Vf = Vi ⎜⎜ i ⎝ Pf

Apply the ideal-gas law to express Vi :

Vi =

Substitute numerical values and evaluate Vi:

Vi =

(1)

1− γ1

⎞ ⎟⎟ ⎠

1− 513

= 263 K

1

⎞γ ⎟⎟ ⎠

nRTi Pi

(0.5 mol)(8.314 J/mol ⋅ K )(500 K ) 5atm ×

101.3 kPa atm

= 4.10 L Substitute for Vi and evaluate Vf:

(d) Relate the change in the internal energy of the helium gas to the change in its temperature: Substitute numerical values and evaluate ∆Eint:

3

⎛ 5 atm ⎞ 5 ⎟⎟ = 10.8 L Vf = (4.10 L )⎜⎜ ⎝ 1atm ⎠

∆Eint = CV ∆T = 32 nR∆T

∆Eint =

(0.5 mol)(8.314 J/mol ⋅ K ) × (263 K − 500 K )

3 2

= − 1.48 kJ (c) Use the first law of thermodynamics to express the work done on the gas: Substitute numerical values and evaluate Won:

1405

Won = ∆Eint − Qin = ∆Eint − 0 = ∆Eint

Won = −1.48 kJ

1406 Chapter 18 Wby gas = −Won = −(− 1.48 kJ )

Because the work done by the gas equals the negative of Won:

= 1.48 kJ

*76 ••• Picture the Problem Consider the process to be accomplished in a single compression. The initial pressure is 1 atm = 101 kPa. The final pressure is (101 + 482) kPa = 583 kPa, and the final volume is 1 L. Because air is a mixture of diatomic gases, γair = 1.4. We can find the initial volume of the air using PiVi γ = Pf Vfγ and use Equation 19-39 to find the work done by the air.

PiVi − Pf Vf γ −1

Express the work done in an adiabatic process:

W=

Use the relationship between pressure and volume for a quasistatic adiabatic process to express the initial volume of the air:

⎛P PiVi = Pf Vf ⇒ Vi = Vf ⎜⎜ f ⎝ Pi

Substitute numerical values and evaluate Vi:

⎛ 583 kPa ⎞ 1.4 ⎟⎟ = 3.50 L Vi = (1 L )⎜⎜ ⎝ 101 kPa ⎠

γ

(1)

γ

1

⎞γ ⎟⎟ ⎠

1

Substitute numerical values in equation (1) and evaluate W:

W=

(101kPa )(3.5 ×10−3 m3 ) − (583 kPa )(10−3 m3 ) 1.4 − 1

= − 574 J

where the minus sign tells us that work is done on the gas. 77 ••• Picture the Problem We can integrate PdV using the equation of state for an adiabatic process to obtain Equation 18-39. V2

Express the work done by the gas during this adiabatic expansion:

Wby gas = ∫ PdV

For an adiabatic process:

PV γ = constant = C

V1

(1)

and

P = CV −γ Substitute and evaluate the integral to obtain:

V2

Wby gas = C ∫ V −γ dV = V1

(

C V21−γ − V11−γ 1− γ

)

Heat and the First Law of Thermodynamics From equation (1) we have: Substitute to obtain:

1407

CV21−γ = P2V2γ and CV11−γ = P1V1γ

Wby gas

P2V2γ − P1V1γ P1V1γ − P2V2γ = = , 1− γ γ −1

which is Equation 18-39.

Cyclic Processes 78 •• Picture the Problem To construct the PV diagram we’ll need to determine the volume occupied by the gas at the beginning and ending points for each process. Let these points be A, B, C, and D. We can apply the ideal-gas law to the starting point (A) to find VA. To find the volume at point B, we can use the relationship between pressure and volume for a quasi-static adiabatic process. We can use the ideal-gas law to find the volume at point C and, because they are equal, the volume at point D. We can apply the first law of thermodynamics to find the amount of heat added to or subtracted from the gas during the complete cycle. (a) Using the ideal-gas law, express the volume of the gas at the starting point A of the cycle:

VA =

Substitute numerical values and evaluate VA:

VA =

nRTA PA

(1mol)(8.314 J/mol ⋅ K )(293 K ) 5 atm ×

101.3 kPa atm

= 4.81L 1

Use the relationship between pressure and volume for a quasistatic adiabatic process to express the volume of the gas at point B; the end point of the adiabatic expansion:

⎛ PA ⎞ γ VB = VA ⎜⎜ ⎟⎟ ⎝ PB ⎠

Substitute numerical values and evaluate VB:

⎛ 5 atm ⎞ 1.4 ⎟⎟ = 15.2 L VB = (4.81 L )⎜⎜ ⎝ 1atm ⎠

Using the ideal-gas law for a fixed amount of gas, express the volume occupied by the gas at points C and D:

VC = VD =

1

nRTC PC

1408 Chapter 18 Substitute numerical values and evaluate VC:

VC =

(1mol)(8.314 J/mol ⋅ K )(293 K ) 1atm ×

101.3 kPa atm

= 24.0 L The complete cycle is shown in the diagram.

(b) Note that for the paths A→B and B→C, Wby gas, the work done by the gas, is positive. For the path D→A, Wby gas is negative, and greater in magnitude than WA→C. Therefore the total work done by the gas is negative. Find the area enclosed by the cycle by noting that each rectangle of dotted lines equals 5 atm⋅L and counting the rectangles:

⎛ 101.3 J ⎞ Wby gas ≈ −(13 rectangles)(5 atm ⋅ L/rectangle ) = (− 65 atm ⋅ L )⎜ ⎟ ⎝ atm ⋅ L ⎠ = − 6.58 kJ (c) The work done on the gas equals the negative of the work done by the gas. Apply the first law of thermodynamics to find the amount of heat added to or subtracted from the gas during the complete cycle: (d) Express the work done during the complete cycle: Because A→B is an adiabatic process:

Qin = ∆Eint − Won = 0 − (− 6.58 kJ ) = 6.58 kJ because ∆Eint = 0 for the complete cycle.

W = WA → B + WB→C + WC → D + WD → A

WA → B =

PAVAγ − PBVBγ γ −1

Heat and the First Law of Thermodynamics

1409

(5 atm )(4.82 L ) − (1atm )(15.2 L)

Substitute numerical values and evaluate WA→B:

WA → B =

B→C is an isobaric process:

WB→C = P∆V

1.4 − 1 ⎛ 101.3 J ⎞ = (22.3 atm ⋅ L )⎜ ⎟ ⎝ atm ⋅ L ⎠ = 2.25 kJ

= (1atm )(24.0 L − 15.2 L ) ⎛ 101.3 J ⎞ = (8.80 atm ⋅ L )⎜ ⎟ ⎝ atm ⋅ L ⎠ = 0.891kJ

C→D is a constant-volume process:

WC → D = 0

D→A is an isobaric process:

WD→A = P∆V = (5 atm )(5 L − 24 L ) ⎛ 101.3 J ⎞ = (− 95.0 atm ⋅ L )⎜ ⎟ ⎝ atm ⋅ L ⎠ = −9.62 kJ

Substitute to obtain:

W = 2.25kJ + 0.891 kJ + 0 − 9.62 kJ = − 6.48 kJ Note that our result in part (b) agrees with this more accurate value to within 2%.

*79 •• Picture the Problem The total work done as the gas is taken through this cycle is the area bounded by the two processes. Because the process from 1→2 is linear, we can use the formula for the area of a trapezoid to find the work done during this expansion. We can use Wisothermal process = nRT ln (Vf Vi ) to find the work done on the gas during the process 2→1. The total work is then the sum of these two terms. Express the net work done per cycle: Work is done by the gas during its expansion from 1 to 2 and hence is equal to the negative of the area of the trapezoid defined by this path and the vertical lines at V1 = 11.5 L and V2 = 23 L. Use the formula for the area of a trapezoid to express

Wtotal = W1→2 + W2→1

W1→2 = − Atrap

(1)

= − 12 (23 L − 11.5 L )(2 atm + 1atm )

= −17.3 L ⋅ atm

1410 Chapter 18 W1→2: Work is done on the gas during the isothermal compression from V2 to V1 and hence is equal to the area under the curve representing this process. Use the expression for the work done during an isothermal process to express W2→1:

⎛V W2→1 = nRT ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ ⎠

Apply the ideal-gas law at point 1 to find the temperature along the isotherm 2→1:

T=

(2 atm )(11.5 L) PV = = 280 K nR (1 mol) 8.206 × 10 −2 L ⋅ atm/mol ⋅ K

(

)

Substitute numerical values and evaluate W2→1:

⎛ 11.5 L ⎞ ⎟⎟ = 15.9 L ⋅ atm W2→1 = (1 mol) 8.206 × 10 −2 L ⋅ atm/mol ⋅ K (280 K ) ln⎜⎜ ⎝ 23 L ⎠

(

Substitute in equation (1) and evaluate Wnet:

)

Wnet = −17.3 L ⋅ atm + 15.9 L ⋅ atm = −1.40 L ⋅ atm ×

101.325 J L ⋅ atm

= − 142 J Remarks: The work done by the gas during each cycle is 142 J. 80 •• Picture the Problem We can apply the ideal-gas law to find the temperatures T1, T2, and T3. We can use the appropriate work and heat equations to calculate the heat added and the work done by the gas for the isothermal process (1→2), the constant-volume process (2→3), and the isobaric process (3→1).

Heat and the First Law of Thermodynamics

1411

(a) The cycle is shown in the diagram:

(c) Use the ideal-gas law to find T1:

T1 = =

P1V1 nR

(2 atm)(2 L )

(2 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 24.4 K Because the process 1→2 is isothermal:

T2 = 24.4 K

Use the ideal-gas law to find T3:

T3 = =

P3V3 nR

(2 atm)(4 L )

(2 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 48.7 K (b) Because the process 1→2 is isothermal, Qin,1→2 = Wby gas,1→2:

⎛V ⎞ Qin, 1→2 = Wby gas,1→2 = nRT ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

Substitute numerical values and evaluate Qin,1→2:

⎛ 4L ⎞ ⎟⎟ = 281 J Qin, 1→2 = (2 mol)(8.314 J/mol ⋅ K )(24.4 K )ln⎜⎜ 2 L ⎠ ⎝ Because process 2→3 takes place at constant volume: Because process 2→3 takes place at constant volume, Won,2→3 = 0, and:

W2→3 = 0

Qin,2→3 = ∆Eint,2→3 = C V ∆T = 32 nR(T3 − T2 )

Substitute numerical values and evaluate Qin,2→3:

1412 Chapter 18 Qin,2→3 =

3 2

(2 mol)(8.314 J/mol ⋅ K )(48.7 K − 24.4 K ) =

Because process 3→1 is isobaric:

606 J

Q3→1 = C P ∆T = 52 nR(T1 − T3 )

Substitute numerical values and evaluate Q3→1:

Q3→1 =

5 2

(2 mol)(8.314 J/mol ⋅ K )(24.4 K − 48.7 K ) =

The work done by the gas from 3 to 1 equals the negative of the work done on the gas: Substitute numerical values and evaluate Wby gas,3→2:

− 1.01 kJ

Wby gas,3→1 = − P1,3 ∆V = P1,3 (V1 − V3 )

Wby gas,3→1 = −(2 atm )(2 L − 4 L ) ⎛ 101.3 J ⎞ = −(− 4 atm ⋅ L )⎜ ⎟ ⎝ atm ⋅ L ⎠ = 405 J

81 ••• Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal monatomic gas (3 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle: 1. Use the ideal-gas law to find the volume of the gas at point D:

Wby gas,tot = WD→A + WA→B + WB→C + WC→D

VD = =

nRTD PD

(2 mol)(8.314 J/mol ⋅ K )(360 K ) (2 atm )(101.3 kPa/atm )

= 29.5 L 2. We’re given that the volume of the gas at point B is three times that at point D:

VB = VC = 3VD = 88.6 L

Use the ideal-gas law to find the pressure of the gas at point C:

Heat and the First Law of Thermodynamics

(

1413

)

nRTC (2 mol) 8.206 ×10 −2 L ⋅ atm/mol ⋅ K (360 K ) = = 0.667 atm PC = 88.6 L VC We’re given that the pressure at point B is twice that at point C: 3. Because path DC represents an isothermal process: Use the ideal-gas law to find the temperatures at points B and A:

PB = 2 PC = 2(0.667 atm ) = 1.33 atm

TD = TC = 360 K

PBVB nR (1.333 atm )(88.6 L ) = (2 mol) 8.206 ×10−2 L ⋅ atm/mol ⋅ K = 720 K

TA = TB =

(

Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that: The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right.

4. For the path D→A, WD→A = 0

PA = 2 PD = 4 atm

Point A B C D

P (atm) 4 1.33 0.667 2

V (L) 29.5 88.6 88.6 29.5

T (K) 720 720 360 360

QD→A = ∆Eint, D→A = 32 nR∆TD→A = 32 nR(TA − TD )

and: Substitute numerical values and evaluate QD→A:

QD → A = For the path A→B:

3 2

(2 mol)(8.314 J/mol ⋅ K )(720 K − 360 K ) = 8.98 kJ ⎛V WA→B = QA→B = nRTA,B ln⎜⎜ B ⎝ VA

Substitute numerical values and evaluate WA→B:

⎛ 88.6 L ⎞ ⎟⎟ = 13.2 kJ WA→B = (2 mol)(8.314 J/mol ⋅ K )(720 K )ln⎜⎜ 29.5 L ⎠ ⎝

⎞ ⎟⎟ ⎠

)

1414 Chapter 18 and, because process A→B is isothermal, ∆Eint, A→B = 0 For the path B→C, WB→C = 0 , and:

QB→C = ∆U B→C = C V ∆T = 32 nR(TC − TB )

Substitute numerical values and evaluate QB→C:

Q B→ C =

3 2

(2 mol)(8.314 J/mol ⋅ K )(360 K − 720 K ) = −8.98 kJ

For the path C→D:

⎛V ⎞ WC→D = nRTC,D ln⎜⎜ D ⎟⎟ ⎝ VC ⎠

Substitute numerical values and evaluate WC→D:

⎛ 2 9 .5 L ⎞ ⎟⎟ = −6.58 kJ WC→D = (2 mol)(8.314 J/mol ⋅ K )(360 K )ln⎜⎜ ⎝ 8 8 .6 L ⎠ Also, because process A→B is isothermal, ∆Eint,A→B = 0 , and

QC→D = WC→D = −6.58 kJ Qin, Won, and ∆Eint are summarized for each of the processes in the table to the right.

Referring to the table, find the total work done by the gas per cycle:

Won (kJ) 0

∆Eint (kJ) 8.98 0

− 8.98

−13. 2 0

− 6.58

6.58

Process

Qin (kJ)

D→A

8.98

A→B

13.2

B→C C→D

−8.9 8 0

Wtot = WD→A + WA →B + WB→C + WC→D = 0 + 13.2 kJ + 0 − 6.58 kJ = 6.62 kJ

Remarks: Note that, as it should be, ∆Eint is zero for the complete cycle. *82 ••• Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal diatomic gas (5 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for

Heat and the First Law of Thermodynamics

1415

each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle: 1. Use the ideal-gas law to find the volume of the gas at point D:

Wby gas,tot = WD→A + WA→B + WB→C + WC→D

VD = =

nRTD PD

(2 mol)(8.314 J/mol ⋅ K )(360 K ) (2 atm )(101.3 kPa/atm )

= 29.5 L 2. We’re given that the volume of the gas at point B is three times that at point D:

VB = VC = 3VD = 88.6 L

Use the ideal-gas law to find the pressure of the gas at point C:

PC =

(

)

nRTC (2 mol) 8.206 × 10 −2 L ⋅ atm/mol ⋅ K (360 K ) = = 0.667 atm 88.6 L VC

We’re given that the pressure at point B is twice that at point C: 3. Because path DC represents an isothermal process: Use the ideal-gas law to find the temperatures at points B and A:

PB = 2 PC = 2(0.667 atm ) = 1.33 atm

TD = TC = 360 K

PBVB nR (1.333 atm )(88.6 L ) = (2 mol) 8.206 ×10−2 L ⋅ atm/mol ⋅ K = 720 K

TA = TB =

(

Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that: The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right.

PA = 2 PD = 4 atm

Point A

P (atm) 4

V (L) 29.5

T (K) 720

)

1416 Chapter 18 B C D

1.33 0.667 2

88.6 88.6 29.5

720 360 360

4. For the path D→A, WD→A = 0 and:

QD→A = ∆U D→A = 52 nR∆TD→A = 52 nR(TA − TD ) =

5 2

(2 mol)(8.314 J/mol ⋅ K )(720 K − 360 K )

= 15.0 kJ For the path A→B:

⎛V WA→B = QA→B = nRTA,B ln⎜⎜ B ⎝ VA = 13.2 kJ

⎞ ⎛ 88.6 L ⎞ ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )(720 K )ln⎜⎜ ⎟⎟ ⎝ 29.5 L ⎠ ⎠

and, because process A→B is isothermal, ∆Eint,A→B = 0 For the path B→C, WB→C = 0 and:

QB→C = ∆U B→C = C V ∆T = 52 nR(TC − TB ) =

5 2

(2 mol)(8.314 J/mol ⋅ K )(360 K − 720 K )

= −15.0 kJ For the path C→D:

⎛V ⎞ ⎛ 2 9 .5 L ⎞ ⎟⎟ = −6.58 kJ WC→D = nRTC,D ln⎜⎜ D ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )(360 K )ln⎜⎜ ⎝ 8 8 .6 L ⎠ ⎝ VC ⎠ Also, because process A→B is isothermal, ∆Eint,A→B = 0 and

QC→D = WC→D = −6.58 kJ Qin, Won, and ∆Eint are summarized for each of the processes in the table to the right.

15.0

Won (kJ) 0

∆Eint (kJ) 15.0

A→B

13.2

−13.2

0

B→C

− 15.0

0

−15.0

C→D

− 6.58

6.58

0

Process

Qin (kJ)

D→A

Heat and the First Law of Thermodynamics

1417

Referring to the table and noting that the work done by the gas equals the negative of the work done on the gas, find the total work done by the gas per cycle:

Wby gas,tot = WD→A + WA→B + WB→C + WC→D = 0 + 13.2 kJ + 0 − 6.58 kJ = 6.62 kJ Remarks: Note that ∆Eint for the complete cycle is zero and that the total work done is the same for the diatomic gas of this problem and the monatomic gas of problem 81. 83 ••• Picture the Problem We can use the equations of state for adiabatic and isothermal processes to express the work done on or by the system, the heat entering or leaving the system, and the change in internal energy for each of the four processes making up the Carnot cycle. We can use the first law of thermodynamics and the definition of the efficiency of a Carnot cycle to show that the efficiency is 1 – Qc / Qh. (a) The cycle is shown on the PV diagram to the right:

(b) Because the process 1→2 is isothermal: Apply the first law of thermodynamics to obtain:

(c) Because the process 3→4 is isothermal:

∆Eint,1→2 = 0

⎛V ⎞ Qh = Q1→2 = W1→2 = nRTh ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ ∆U 3→4 = 0

1418 Chapter 18 Apply the first law of thermodynamics to obtain:

⎛V ⎞ Qc = Q3→4 = W3→4 = nRTc ln⎜⎜ 4 ⎟⎟ ⎝ V3 ⎠ ⎛V ⎞ = − nRTc ln⎜⎜ 3 ⎟⎟ ⎝ V4 ⎠ where the minus sign tells us that heat is given off by the gas during this process.

(d) Apply the equation for a quasistatic adiabatic process at points 4 and 1 to obtain: Solve for the ratio V1/V4:

Apply the equation for a quasi-static adiabatic process at points 2 and 3 to obtain: Solve for the ratio V2/V3:

Equate equations (1) and (2) and rearrange to obtain: (e) Express the efficiency of the Carnot cycle: Apply the first law of thermodynamics to obtain:

TcV4γ −1 = ThV1γ −1

1

γ −1 V1 ⎛ Tc ⎞ =⎜ ⎟ V4 ⎜⎝ Th ⎟⎠

(1)

ThV2γ −1 = TcV3γ −1

1

γ −1 V2 ⎛ Tc ⎞ ⎟ ⎜ = V3 ⎜⎝ Th ⎟⎠

(2)

V3 V2 = V4 V1

ε=

W Qh

Won = ∆Eint, cycle − Qin

= 0 − (Qh − Qc ) = −(Qh − Qc )

because Eint is a state function and ∆Eint, cycle = 0 . Substitute to obtain:

ε= =

Wby the gas Qh

=

− Won Qh

Qh − Qc Q = 1− c Qh Qh

Heat and the First Law of Thermodynamics (f) In part (b) we established that:

⎛V ⎞ Qh = nRTh ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

In part (c) we established that the heat leaving the system along the path 3→4 is given by:

⎛V ⎞ Qc = nRTc ln⎜⎜ 3 ⎟⎟ ⎝ V4 ⎠

Divide the second of these equations by the first to obtain:

⎛V ⎞ nRTc ln⎜⎜ 3 ⎟⎟ Qc ⎝ V4 ⎠ = Tc = Qh Th ⎛V ⎞ nRTh ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ because

1419

V3 V2 = . V4 V1

Remarks: This last result establishes that the efficiency of a Carnot cycle is also given by εC = 1 −

Tc . Th

General Problems 84 • Picture the Problem The isobaric process is shown on the PV diagram. We can express the heat that must be supplied to gas in terms of its heat capacity at constant pressure and the change in its temperature and then use the ideal-gas law for a fixed amount of gas to relate the final temperature to the initial temperature.

Relate Qin to CP and ∆T: Use the ideal-gas law for a fixed amount of gas to relate the initial and final volumes, pressures, and temperatures:

Solve for Tf:

Qin = CP ∆T = CP (Tf − Ti ) = 52 nR(Tf − Ti ) PiVi Pf Vf = Ti Tf or, because the process is isobaric,

Vi Vf = Ti Tf Tf =

Vf 200 L Ti = Ti = 4Ti Vi 50 L

1420 Chapter 18 Substitute to obtain: Substitute numerical values and evaluate Qin:

Qin = 52 nR(4Ti − 3Ti ) = 152 nRTi Qin = 152 (3 mol)(8.314 J/mol ⋅ K )(300 K ) = 56.1kJ

85 • Picture the Problem We can use the first law of thermodynamics to relate the heat removed from the gas to the work done on the gas.

Qin = ∆Eint − Won = −Won

Apply the first law of thermodynamics to this process:

because ∆Eint = 0 for an isothermal process.

Substitute numerical values to obtain:

Qin = −180 kJ

Because Qremoved = −Qin:

Qremoved = 180 kJ

*86 • Picture the Problem We can find the number of moles of the gas from the expression for the work done on or by a gas during an isothermal process. Express the work done on the gas during the isothermal process:

⎛V W = nRT ln⎜⎜ f ⎝ Vi

Solve for n:

n=

Substitute numerical values and evaluate n:

n=

W ⎛V RT ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠

− 180 kJ

(8.314 J/mol ⋅ K )(293 K )ln⎛⎜ 1 ⎞⎟ ⎝5⎠

= 45.9 mol 87 • Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process EDC is isobaric, we can find the area under this line geometrically and the first law of thermodynamics to find QAEC.

Heat and the First Law of Thermodynamics (a) Using the ideal-gas law, find the temperature at point A:

TA = =

PAVA nR

1421

(4 atm)(4.01L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 65.2 K Using the ideal-gas law, find the temperature at point C:

TC = =

PCVC nR

(1atm)(20 L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 81.2 K

(b) Express the work done by the gas along the path AEC:

WAEC = WAE + WEC = 0 + PEC ∆VEC = (1atm )(20 L − 4.01 L ) = 16.0 L ⋅ atm ×

101.3 J L ⋅ atm

= 1.62 kJ (c) Apply the first law of thermodynamics to express QAEC:

QAEC = WAEC + ∆Eint = WAEC + CV ∆T = WAEC + 32 nR∆T

= WAEC + 32 nR(TC − TA ) Substitute numerical values and evaluate QAEC:

QAEC = 1.62 kJ + 32 (3 mol)(8.314 J/mol ⋅ K )(81.2 K − 65.2 K ) = 2.22 kJ Remarks The difference between WAEC and QAEC is the change in the internal energy ∆Eint,AEC during this process. 88 •• Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process AB is isobaric, we can find the area under this line geometrically. We can use the expression for the work done during an isothermal expansion to find the work done between B and C and the first law of thermodynamics to find QABC.

1422 Chapter 18 (a) Using the ideal-gas law, find the temperature at point A:

TA = =

PAVA nR

(4 atm)(4.01L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 65.2 K Use the ideal-gas law to find the temperature at point C:

TC = =

PCVC nR

(1atm)(20 L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 81.2 K

(b) Express the work done by the gas along the path ABC:

WABC = WAB + WBC = PAB∆VAB + nRTB ln

VC VB

Use the ideal-gas law to find the volume of the gas at point B:

VB =

nRTB (3 mol) (8.206 × 10 −2 L ⋅ atm/mol ⋅ K )(81.2 K ) = = 5.00 L 4 atm PB

Substitute to obtain:

⎛ 20 L ⎞ ⎟⎟ WABC = (4 atm )(5 L − 4.01 L ) + (3 mol) (8.206 × 10 −2 L ⋅ atm/mol ⋅ K )(81.2 K )ln⎜⎜ ⎝ 5L ⎠ 101.3 J = 31.7 L ⋅ atm × = 3.21 kJ atm QABC = WABC + ∆Eint = WABC + CV ∆T (c) Apply the first law of thermodynamics to obtain:

= WAEC + 32 nR∆T

= WAEC + 32 nR(TC − TA ) Substitute numerical values and evaluate QABC:

QABC = 3.21 kJ + 32 (3 mol)(8.314 J/mol ⋅ K )(81.2 K − 65.2 K ) = 3.81 kJ Remarks: The difference between WABC and QABC is the change in the internal energy ∆Eint,ABC during this process.

Heat and the First Law of Thermodynamics 1423 *89 •• Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process DC is isobaric, we can find the area under this line geometrically. We can use the expression for the work done during an isothermal expansion to find the work done between A and D and the first law of thermodynamics to find QADC. (a) Using the ideal-gas law, find the temperature at point A:

TA = =

PAVA nR

(4 atm)(4.01L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 65.2 K Use the ideal-gas law to find the temperature at point C:

TC = =

PCVC nR

(1atm)(20 L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 81.2 K (b) Express the work done by the gas along the path ADC:

W ADC = W AD + W DC ⎛V ⎞ = nRTA ln⎜⎜ D ⎟⎟ + PDC ∆V DC ⎝ VA ⎠

Use the ideal-gas law to find the volume of the gas at point D:

VD =

(

)

nRTD (3 mol) 8.206 × 10−2 L ⋅ atm/mol ⋅ K (65.2 K ) = = 16.1 L 1atm PD

Substitute numerical values and evaluate WADC:

⎛ 16.1 L ⎞ ⎟⎟ WADC = (3 mol) 8.206 × 10 −2 L ⋅ atm/mol ⋅ K (65.2 K ) ln⎜⎜ ⎝ 4.01 L ⎠ + (1 atm )(20 L − 16.1 L ) 101.325 J = 26.2 L ⋅ atm × = 2.65 kJ L ⋅ atm

(

(c) Apply the first law of thermodynamics to obtain:

)

QADC = WADC + ∆Eint = WADC + CV ∆T = WADC + 32 nR∆T

= WADC + 32 nR(TC − TA )

1424 Chapter 18 Substitute numerical values and evaluate QADC:

QADC = 2.65 kJ + 32 (3 mol)(8.314 J/mol ⋅ K )(81.2 K − 65.2 K ) = 3.25 kJ 90 •• Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process AB is isobaric, we can find the area under this line geometrically. We can find the work done during the adiabatic expansion between B and C using WBC = −CV ∆TBC and the first law of thermodynamics to find QABC.

The work done by the gas along path ABC is given by:

WABC = WAB + WBC = PAB ∆VAB − C V ∆TBC = PAB ∆VAB − 32 nR∆TBC because, with Qin = 0, WBC = −∆Eint,BC.

Use the ideal-gas law to find TA:

TA = =

PAVA nR

(4 atm)(4.01L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 65.2 K Use the ideal-gas law to find TB:

TB = =

PBVB nR

(4 atm)(8.71L )

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 142 K Use the ideal-gas law to find TC:

TC = =

PCVC nR

(3 mol)(8.206 ×10−2 L ⋅ atm/mol ⋅ K )

= 81.2 K Apply the pressure-volume relationship for a quasi-static adiabatic process to the gas at points B and C to find the volume of the gas at point B:

PBVBγ = PCVCγ and

(1atm)(20 L )

Heat and the First Law of Thermodynamics 1425 1

3

⎛ PC ⎞ γ ⎛ 1atm ⎞ 5 ⎟ ⎜ ⎟⎟ (20 L ) VB = ⎜ ⎟ VC = ⎜⎜ 4 atm P ⎝ ⎠ ⎝ B⎠ = 8.71 L Substitute numerical values and evaluate WABC:

(

WABC = (4 atm )(8.71 L − 4.01 L ) − 32 (3 mol) 8.206 × 10 −2 L ⋅ atm/mol ⋅ K = 41.3 L ⋅ atm ×

× (81.2 K − 142 K )

)

101.325 J = 4.18 kJ atm QABC = WABC + ∆Eint = WABC + CV ∆T

Apply the 1st law of thermodynamics to obtain:

= WABC + 32 nR∆T

= WABC + 32 nR(TC − TA ) Substitute numerical values and evaluate QABC: QABC = 4.18 kJ + 32 (3 mol)(8.314 J/mol ⋅ K )(81.2 K − 65.2 K ) = 4.78 kJ 91 •• Picture the Problem We can find c at T = 4 K by direct substitution. Because c is a function of T, we’ll integrate dQ over the given temperature interval in order to find the heat required to heat copper from 1 to 3 K.

(

)

c = 0.0108 J/kg ⋅ K 2 T

(a) Substitute for a and b to obtain:

(

)

+ 7.62 × 10 −4 J/kg ⋅ K 4 T 3

(

)

c(4 K ) = 0.0108 J/kg ⋅ K 2 (4 K )

Evaluate c at T = 4 K:

(

)

+ 7.62 × 10 J/kg ⋅ K 4 (4 K ) −4

= 9.20 × 10 −2 J/kg ⋅ K (b) Express and evaluate the integral of Q: Tf

(

Q = ∫ c(T ) dT = 0.0108 J/kg ⋅ K 2 Ti

3K

) ∫ TdT + (7.62 ×10 3K

) ∫ T dT 3K

−4

J/kg ⋅ K 4

1K

3

1K

3K

⎡T 2 ⎤ ⎡T 4 ⎤ = 0.0108 J/kg ⋅ K 2 ⎢ ⎥ + 7.62 × 10 −4 J/kg ⋅ K 4 ⎢ ⎥ = 0.0584 J/kg ⎣ 2 ⎦1 K ⎣ 4 ⎦1 K

(

)

(

)

3

1426 Chapter 18 92 •• Picture the Problem We can use the first law of thermodynamics to relate the heat escaping from the system to the amount of work done by the gas and the change in its internal energy. We can use the expression for the work done during an isothermal process to find the temperature along the isotherm. Apply the first law of thermodynamics to this isothermal process:

Qin = ∆Eint − Won

For an isothermal process:

∆Eint = 0

Substitute to obtain:

4.184 J ⎞ ⎛ Won = −Qin = − ⎜ − 170 cal × ⎟ cal ⎠ ⎝ = 711 J

Because Wby gas = −Won:

Wby gas = − 711 J

Express the work done during an isothermal process:

⎛V ⎞ Wby gas = nRT ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

Solve for T = Ti = Tf:

T=

Substitute numerical values and evaluate T:

T=

Wby gas ⎛V ⎞ nR ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ − 711 J ⎛

⎞

(2 mol)(8.314 J/mol ⋅ K )ln⎜⎜ 8 L ⎟⎟ ⎝ 18 L ⎠

= 52.7 K 93 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final values of temperature, pressure, and volume. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path

P1V1 − P2V2 and find P1 by eliminating P2 using P1V1γ = P2V2γ , where, for a γ −1 diatomic gas, γ = 1.4. Once we’ve determined P1 we can use the ideal-gas law to find T1 using W =

and the first law of thermodynamics to find T2. Finally, we can apply the ideal-gas law a second time to determine P2.

Heat and the First Law of Thermodynamics P1V1 − P2V2 γ −1

Relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path:

Won =

Using the equation for a quasi-static adiabatic process, relate the initial and final pressures and volumes:

P ⎛V ⎞ P1V1 = P2V2 ⇒ 2 = ⎜⎜ 1 ⎟⎟ P1 ⎝ V2 ⎠

Substitute to obtain:

Solve for P1:

1427

⎛ P ⎞ P1 ⎜⎜V1 − 2 V2 ⎟⎟ P1 ⎠ = ⎝ γ −1 γ

γ

γ

γ ⎛ ⎛ V1 ⎞ ⎞⎟ ⎜ P1 V1 − ⎜⎜ ⎟⎟ V2 ⎜ ⎝ V2 ⎠ ⎟⎠ Won = ⎝ γ −1

P1 =

Substitute numerical values and evaluate P1:

P1 =

Use the ideal-gas law to find T1:

T1 =

W (γ − 1) γ

⎛V ⎞ V1 − ⎜⎜ 1 ⎟⎟ V2 ⎝ V2 ⎠

(− 820 J )(1.4 − 1) 1.4

⎛ 18 L ⎞ ⎟⎟ 18 L − ⎜⎜ ⎝ 8L ⎠

= 47.6 kPa

(8 L )

(47.6 kPa )(18 L ) P1V1 = nR (2 mol)(8.314 J/mol ⋅ K )

= 51.5 K Apply the first law of thermodynamics to obtain:

Solve for and evaluate T2:

∆Eint = Qin + Won or, because Qin = 0 for an adiabatic process,

∆Eint = Won = CV ∆T = 52 nR(T2 − T1 )

T2 = T1 −

Won 5 2 nR

= 51.5 K − = 71.2 K

5 2

− 820 J (2 mol)(8.314 J/mol ⋅ K )

1428 Chapter 18 Use the ideal-gas law to find P2:

P2 = =

nRT2 V2

(2 mol)(8.314 J/mol ⋅ K )(71.2 K ) 8L

= 148 kPa 94 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final state respectively. Because the gas is initially at STP, we know that V1 = 22.4 L, P1 = 1 atm, and T1 = 273 K. We can use W = −nRT ln (V2 V1 ) to find the work done on the gas during an isothermal compression. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path

P1V1 − P2V2 and find P1 by eliminating P2 using P1V1γ = P2V2γ , where, for a γ −1 diatomic gas, γ = 1.4. using W =

(a) Express the work done on the gas in compressing it isothermally:

⎛V ⎞ Won = −nRT ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

Find the number of moles in 30 g of CO (M = 28 g/mol):

n=

30 g = 1.07 mol 28 g/mol

Substitute numerical values and evaluate Won:

⎛1⎞ Won = −(1.07 mol)(8.314 J/mol ⋅ K )(273 K )ln⎜ ⎟ = 3.91kJ ⎝5⎠ (b) Express the work done on the gas in compressing it adiabatically:

Won = −

P1V1 − P2V2 γ −1

⎛ P ⎞ P1 ⎜⎜V1 − 2 V2 ⎟⎟ P1 ⎠ =− ⎝ γ −1 Using the equation for a quasi-static adiabatic process, relate the initial and final pressures and volumes:

P ⎛V ⎞ P1V1 = P2V2 ⇒ 2 = ⎜⎜ 1 ⎟⎟ P1 ⎝ V2 ⎠ γ

γ

γ

Heat and the First Law of Thermodynamics 1429

Substitute for P2/P1and simplify to obtain: γ γ γ ⎞ ⎛ ⎛ ⎛ ⎛ V1 ⎞ ⎛ V1 ⎞ V1 ⎞⎟ ⎛ V1 ⎞ ⎟ ⎜ ⎜ ⎜ P1 V1 − ⎜⎜ ⎟⎟ V2 P1 V1 − ⎜⎜ ⎟⎟ P1V1 1 − 0.2⎜⎜ ⎟⎟ ⎟ ⎟ ⎜ ⎜ ⎜ 5 V V ⎝ 2⎠ ⎝ 2⎠ ⎝ V2 ⎠ ⎠ ⎠ ⎝ ⎝ ⎝ =− =− Won = − γ −1 γ −1 γ −1

⎞ ⎟ ⎟ ⎠

Substitute numerical values and evaluate Won:

W =−

(101.3 kPa )(1.07 mol)(22.4 L/mol)(1 − 0.2(5)1.4 ) = 1.4 − 1

5.49 kJ

95 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final state respectively. Because the gas is initially at STP, we know that V1 = 22.4 L, P1 = 1 atm, and T1 = 273 K. We can use W = −nRT ln (V2 V1 ) to find the work done on the gas during an isothermal compression. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path using W =

P1V1 − P2V2 and find P1 by eliminating P2 using P1V1γ = P2V2γ . We can find γ γ −1

using the data in Table 19-3. (a) Express the work done on the gas in compressing it isothermally:

⎛V ⎞ Won = −nRT ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

Find the number of moles in 30 g of CO2 (M = 44 g/mol):

n=

30 g = 0.682 mol 44 g/mol

Substitute numerical values and evaluate Won:

⎛1⎞ Won = −(0.682 mol)(8.314 J/mol ⋅ K )(273 K )ln⎜ ⎟ = 2.49 kJ ⎝5⎠ (b) Express the work done on the gas in compressing it adiabatically:

Won = −

P1V1 − P2V2 γ −1

⎛ P ⎞ P1 ⎜⎜V1 − 2 V2 ⎟⎟ P1 ⎠ =− ⎝ γ −1

1430 Chapter 18 P ⎛V ⎞ P1V1 = P2V2 ⇒ 2 = ⎜⎜ 1 ⎟⎟ P1 ⎝ V2 ⎠

Using the equation for a quasi-static adiabatic process, relate the initial and final pressures and volumes:

γ

γ

γ

Substitute for P2/P1 and simplify to obtain: γ γ γ ⎛ ⎛ ⎞ ⎛ ⎛ V1 ⎞ ⎛ V1 ⎞ V1 ⎞⎟ ⎛ V1 ⎞ ⎜ ⎜ ⎟ ⎜ P1 V1 − ⎜⎜ ⎟⎟ V2 P1 V1 − ⎜⎜ ⎟⎟ P1V1 1 − 0.2⎜⎜ ⎟⎟ ⎜ ⎜ ⎟ ⎜ V2 ⎠ V2 ⎠ 5 ⎟ ⎝ V2 ⎠ ⎝ ⎝ ⎝ ⎠ =− ⎠ =− ⎝ Won = − ⎝ γ −1 γ −1 γ −1

⎞ ⎟ ⎟ ⎠

cV = 3.39 R

From Table 18-3 we have:

and

cP = (3.39 + 1.02 )R = 4.41R

Evaluate γ:

γ =

cP 4.41R = = 1.30 cV 3.39 R

Substitute numerical values and evaluate Won:

Won = −

(101.3 kPa )(0.682 mol)(22.4 L/mol)(1 − 0.2(5)1.3 ) = 1.3 − 1

3.20 kJ

96 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final states respectively. Because the gas is initially at STP, we know that V1 = 22.4 L, P1 = 1 atm, and T1 = 273 K. We can use W = −nRT ln (V2 V1 ) to find the work done on the gas during an isothermal compression. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path

P1V1 − P2V2 and find P1 by eliminating P2 using P1V1γ = P2V2γ , where, for a γ −1 monatomic gas, γ = 1.67. using W =

(a) Express the work done on the gas in compressing it isothermally:

⎛V ⎞ Won = −nRT ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠

Find the number of moles in 30 g of Ar (M = 40 g/mol):

n=

Substitute numerical values and evaluate Won:

30 g = 0.750 mol 40 g/mol

Heat and the First Law of Thermodynamics 1431 ⎛1⎞ Won = −(0.75 mol)(8.314 J/mol ⋅ K )(273 K )ln⎜ ⎟ = 2.74 kJ ⎝5⎠ (b) Express the work done on the gas in compressing it adiabatically:

Won = −

P1V1 − P2V2 γ −1

⎛ P ⎞ P1 ⎜⎜V1 − 2 V2 ⎟⎟ P1 ⎠ =− ⎝ γ −1

P ⎛V ⎞ P1V1 = P2V2 ⇒ 2 = ⎜⎜ 1 ⎟⎟ P1 ⎝ V2 ⎠

Using the equation for a quasi-static adiabatic process, relate the initial and final pressures and volumes:

γ

γ

γ

Substitute for P2/P1 and simplify to obtain: γ γ γ ⎛ ⎛ ⎞ ⎛ ⎛ V1 ⎞ ⎛ V1 ⎞ V1 ⎞⎟ ⎛ V1 ⎞ ⎜ ⎜ ⎟ ⎜ P1 V1 − ⎜⎜ ⎟⎟ V2 P1 V1 − ⎜⎜ ⎟⎟ P1V1 1 − 0.2⎜⎜ ⎟⎟ ⎜ ⎜ ⎟ ⎜ V2 ⎠ V2 ⎠ 5 ⎟ ⎝ V2 ⎠ ⎝ ⎝ ⎝ ⎠ ⎝ ⎠ ⎝ Won = − =− =− γ −1 γ −1 γ −1

⎞ ⎟ ⎟ ⎠

Substitute numerical values and evaluate Won:

Won = −

(101.3 kPa )(0.75 mol)(22.4 L/mol) (1 − 0.2(5)1.67 ) = 1.67 − 1

4.93 kJ

97 •• Picture the Problem We can use conservation of energy to relate the final temperature to the heat capacities of the gas and the solid. We can apply the Dulong-Petit law to find the heat capacity of the solid at constant volume and use the fact that the gas is diatomic to find its heat capacity at constant volume. Apply conservation of energy to this process:

Solve for Tf:

∆Q = 0 or

CV,gas (Tf − 100 K ) − CV,solid (200 K − Tf ) = 0

Tf =

(100 K )(CV,gas ) + (200 K )(CV,solid ) CV,gas + CV,solid

1432 Chapter 18 Using the Dulong-Petit law, determine the heat capacity of the solid at constant volume: Determine the heat capacity of the gas at constant volume:

CV,solid = 3nR

= 3(2 mol)(8.314 J/mol ⋅ K ) = 49.9 J/K

CV,gas = 52 nR =

5 2

(1mol)(8.314 J/mol ⋅ K )

= 20.8 J/K Substitute numerical values and evaluate Tf:

Tf =

(100 K )(20.8 J/K ) + (200 K )(49.9 J/K ) = 20.8 J/K + 49.9 J/K

171 K

*98 •• Picture the Problem We can express the work done during an isobaric process as the product of the temperature and the change in volume and relate Q to ∆T through the definition of CP. Finally, we can use the first law of thermodynamics to show that ∆Eint = Cv∆T.

For an ideal gas, the internal energy is the sum of the kinetic energies of the (a) gas molecules, which is proportional to kT . Consequently, U is a function of T only and ∆Eint = CV ∆T .

(b) Use the first law of thermodynamics to relate the work done on the gas, the heat entering the gas, and the change in the internal energy of the gas: At constant pressure:

∆Eint = Qin + Won

Wby gas = P(Vf − Vi ) = nR(Tf − Ti ) = nR∆T and

Won = −Wby gas = −nR∆T

Relate Qin to CP and ∆T:

Qin = CP ∆T

Substitute to obtain:

∆Eint = CP ∆T − nR∆T = (CP − nR )∆T = CV ∆T

Heat and the First Law of Thermodynamics 1433 99 •• Picture the Problem We can use Qin = CV ∆T = 32 nR∆T to find Qin for the constantvolume process and Qin = C P ∆T = 52 nR∆T to find Qin for the isobaric process. The Vf

∫

work done by the gas is given by W = PdV . Finally, we can apply the first law of Vi

thermodynamics to find the change in the internal energy of the gas from the work done on the gas and the heat that enters the gas. (a) The heat added to the gas during this process is given by: Substitute numerical values and evaluate Qin:

Qin = 32 nR∆T Qin =

3 2

(1mol)(8.314 J/mol ⋅ K )(300 K )

= 3.74 kJ

For a constant-volume process:

Wby the gas = 0

Apply the 1st law of thermodynamics to obtain:

∆Eint = Qin + Won

Substitute for Qin and Won in equation (1) and evaluate ∆Eint:

∆Eint = 3.74 kJ + 0 = 3.74 kJ

(b) Relate the heat absorbed by the gas to the change in its temperature:

For a constant-pressure process that begins at temperature Ti and ends at temperature Tf, the work done by the gas is given by: Substitute numerical values and evaluate Wby the gas:

Apply the 1st law of thermodynamics to express the change in the internal energy of the gas during this isobaric expansion:

(1)

Qin = CP ∆T = 52 nR∆T =

5 2

(1mol)(8.314 J/mol ⋅ K )(300 K )

= 6.24 kJ Vf

Wby the gas = ∫ PdV = P(Vf − Vi ) Vi

= nR(Tf − Ti ) Wby the gas = (1 mol)(8.314 J/mol ⋅ K )(300 K ) = 2.49 kJ ∆Eint = Won + Qin

1434 Chapter 18 Because the Wby gas = −Won:

∆Eint = −Wby the gas + Qin

Substitute numerical values and evaluate ∆Eint:

∆Eint = −2.49 kJ + 6.24 kJ = 3.75 kJ

Remarks: Because ∆Eint depends only on the initial and final temperatures of the gas, the values for ∆Eint for Part (a) and Part (b) should be they same. They differ slightly due to rounding. *100 •• Picture the Problem We can use Qin = CP∆T to find the change in temperature during this isobaric process and the first law of thermodynamics to relate W, Q, and ∆Eint. We can use ∆Eint = 52 nR∆T to find the change in the internal energy of the gas during the isobaric process and the ideal-gas law for a fixed amount of gas to express the ratio of the final and initial volumes. (a) Relate the change in temperature to Qin and CP and evaluate ∆T:

∆T = =

Qin Q = 7 in CP 2 nR 500 J (2 mol)(8.314 J/mol ⋅ K )

7 2

= 8.59 K (b) Apply the first law of thermodynamics to relate the work done on the gas to the heat supplied and the change in its internal energy: Substitute numerical values and evaluate Won:

Won = ∆Eint − Qin = CV ∆T − Qin = 52 nR∆T − Qin

Won =

5 2

(2 mol)(8.314 J/mol ⋅ K )(8.59 K )

− 500 J = −143 J

Because Wby gas = −Won:

Wby gas = 143 J

(c) Using the ideal-gas law for a fixed amount of gas, relate the initial and final pressures, volumes and temperatures:

PiVi Pf Vf = Ti Tf or, because the process is isobaric,

Vi Vf = Ti Tf

Heat and the First Law of Thermodynamics 1435 Solve for and evaluate Vf/Vi:

Vf Tf Ti + ∆T = = Vi Ti Ti =

293.15 K + 8.59 K = 1.03 293.15 K

101 •• Picture the Problem Knowing the rate at which energy is supplied, we can obtain the data we need to plot this graph by finding the time required to warm the ice to 0°C, melt the ice, warm the water formed from the ice to 100°C, vaporize the water, and warm the water to 110°C.

mcice ∆T P ( 0.1 kg )(2 kJ/kg ⋅ K )(10 K ) = 100 J/s = 20.0 s

Find the time required to warm the ice to 0°C:

∆t1 =

Find the time required to melt the ice:

∆t2 =

Find the time required to heat the water to 100°C:

∆t3 =

Find the time required to vaporize the water:

∆t4 =

Find the time required to heat the vapor to 110°C:

∆t5 =

mLf (0.1 kg )(333.5 kJ/kg ) = 100 J/s P = 333.5 s

mcw ∆T P (0.1kg )(4.18 kJ/kg ⋅ K )(100 K ) = 100 J/s = 418 s mLV (0.1 kg )(2257 kJ/kg ) = 100 J/s P = 2257 s

mcsteam ∆T P (0.1kg )(2 kJ/kg ⋅ K )(10 K ) = 100 J/s = 20 s

1436 Chapter 18 The temperature T as a function of time t is shown to the right:

*102 •• Picture the Problem We know that, for an adiabatic process, Qin = 0. Hence the work done by the expanding gas equals the change in its internal energy. Because we’re given the work done by the gas during the expansion, we can express the change in the temperature of the gas in terms of this work and CV. Express the final temperature of the gas as a result of its expansion: Apply the equation for adiabatic work and solve for ∆T:

Tf = Ti + ∆T

Wadiabatic = −CV ∆T and

∆T = −

Substitute and evaluate Tf:

Wadiabatic W = − 5adiabatic CV 2 nR

Tf = Ti −

Wadiabatic 5 2 nR

= 300 K − = 216 K 103 •• Picture the Problem Because PfVf = 4PiVi and Vf = Vi/2, the path for which the work done by the gas is a minimum while the pressure never falls below Pi is shown on the adjacent PV diagram. We can apply the first law of thermodynamics to relate the heat transferred to the gas to its change in internal energy and the work done on the gas.

5 2

3.5 kJ (2 mol)(8.314 J/mol ⋅ K )

Heat and the First Law of Thermodynamics 1437 Using the first law of thermodynamics, relate the heat transferred to the gas to its change in internal energy and the work done on the gas: Solve for Qin: Express the work done during this process:

∆Eint = Qin + Won

Qin = ∆Eint − Won Won = Wisobaric process + Wconstant volume = Pi ∆V + 0 = Pi ( 12 Vi − Vi ) = 12 PiVi = 12 nRT = 12 RT because n = 1 mol.

Express ∆Eint for the process:

∆Eint = CV ∆T = 32 nR∆T = 32 nR(3T ) = 92 RT because n = 1 mol.

Substitute to obtain:

Qin = 92 RT − 12 RT = 4 RT

104 •• Picture the Problem We can solve the ideal-gas law for the dilute solution for the increase in pressure and find the number of solute molecules dissolved in the water from their mass and molecular weight.

NkT V

Solve the ideal gas law for P to obtain:

P=

Express the number of solute molecules N in terms of the number of moles n and Avogadro’s number and then express the number of moles in terms of the mass of the salt and its molecular mass:

N = nN A =

Substitute to obtain:

P=

mN A M NaCl

mN A kT M NaClV

Substitute numerical values and evaluate P:

P=

(30 g )(6.022 ×1023 particles/mol)(1.381×10−23 J/K )(297 K ) = (58.4 g/mol)(10−3 m3 )

1.27 × 106 N/m 2

1438 Chapter 18 105 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final states in this adiabatic expansion. We can use an equation describing a quasi-static adiabatic process to express the final temperature as a function of the initial temperature and the initial and final volumes. Using the equation for a quasi-static adiabatic process, relate the initial and final volumes and temperatures:

T2V2γ −1 = T1V1γ −1

Solve for and evaluate T2:

⎛V ⎞ T2 = T1 ⎜⎜ 1 ⎟⎟ ⎝ V2 ⎠

γ −1

= (300 K )(2)

1.4−1

= 396 K 106 •• Picture the Problem We can simplify our calculations by relating Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas and solving for NAk. We can then calculate U300 K and U600 K and their difference. Express the increase in internal energy per mole resulting from the heating of diamond: Express the relationship between Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas: Substitute in the given equation to obtain: Determine U300 K:

∆U = U 600 K − U 300 K

nR = Nk ⇒ R =

U=

N k = NAk n

3RTE eTE T − 1

U 300 K =

3(8.314 J/mol ⋅ K )(1060 K ) e1060 K 300 K − 1

= 795 J Determine U600 K:

U 600 K =

3(8.314 J/mol ⋅ K )(1060 K ) e1060 K 600 K − 1

= 5.45 kJ

Heat and the First Law of Thermodynamics 1439 Substitute to obtain:

∆U = U 600 K − U 300 K = 5.45 kJ − 795 J = 4.66 kJ

*107 ••• Picture the Problem The isothermal expansion followed by an adiabatic compression is shown on the PV diagram. The path 1→2 is isothermal and the path 2→3 is adiabatic. We can apply the idealgas law for a fixed amount of gas and an isothermal process to find the pressure at point 2 and the pressure-volume relationship for a quasi-static adiabatic process to determineγ.

(a) Relate the initial and final pressures and volumes for the isothermal expansion and solve for and evaluate the final pressure: (b) Relate the initial and final pressures and volumes for the adiabatic compression:

P1V1 = P2V2 and

P2 = P1

V V1 = P0 1 = 2V1 V2

1 2

P0

P2V2γ = P3V3γ or 1 2

P0 (2V0 ) = 1.32 P0V0γ γ

which simplifies to

2γ = 2.64 Take the natural logarithm of both sides of this equation and solve for and evaluate γ :

(c)

γ ln 2 = ln 2.64 and

ln 2.64 = 1.40 ln 2 ∴ the gas is diatomic.

γ =

In the isothermal process, T is constant, and the translational kinetic energy is unchanged.

In the adiabatic process, T3 = 1.32T0 , and the translational kinetic energy increases by a factor of 1.32.

1440 Chapter 18 108 ••• Picture the Problem In this problem the specific heat of the combustion products depends on the temperature. Although CP increases gradually from (9/2)R per mol to (15/2)R per mol at high temperatures, we’ll assume that CP = 4.5R below T = 2000 K and CP = 7.5R above T = 2000 K. We’ll also use R = 2.0 cal/mol⋅K. We can find the final temperature following combustion from the heat made available during the combustion and the final pressure by applying the ideal-gas law to the initial and final states of the gases. (a) Relate the heat available in this combustion process to the change in temperature of the triatomic gases: Solve for Tf to obtain:

Express Q available to heat the gases above 2000 K:

Express the energy released in the combustion of 1 mol of benzene: Noting that there are 3 mol of H2O and 6 mol of CO2, find the heat required to form the products at 100°C:

Qavailable = nCP ∆T

= n(7.5R )(Tf − Ti )

Tf =

Qavailable + Ti 7.5nR

Qavailable = Qreleased − Q9 mol to 2000 K − Qheat CO 2 − Qsteam

Qreleased =

1 2

(1)

(2)

(1516 kcal) = 758 kcal

Qsteam = nM w c w ∆T + nM w LV = (3 mol)(18 g/mol)

× (1cal/g ⋅ K )(373K − 300 K )

+ (3 mol)(18 g/mol)(540 cal/g )

= 33.10 kcal and

Qheat CO2 = nCP ∆T = 4.5nR∆T

= 4.5(6 mol)(2 cal/mol ⋅ K ) × (373 K − 300K ) = 3.942 kcal

Find Q required to heat 9 mol of gas to 2000 K:

Q9 mol to 2000 K = nCP ∆T = 4.5nR∆T

= 4.5(9 mol)(2 cal/mol ⋅ K ) × (2000 K − 373K ) = 43.93 kcal

Heat and the First Law of Thermodynamics 1441 Substitute in equation (2) to obtain:

Qavailable = 758 kcal − 131.79 kcal − 3.94 kcal − 33.10 kcal = 589.2 kcal

Substitute in equation (1) and evaluate Tf:

Tf =

589.2 kcal + 2000 K 7.5(9 mol)(2 cal/mol ⋅ K )

= 6364 K Apply the ideal-gas law to express the final volume in terms of the final temperature and pressure:

Vf = =

nRTf Pf

(9 mol)(8.314 J/mol ⋅ K )(6364 K ) 101.3 kPa

= 4.70 m 3 (b) Apply the ideal-gas law to relate the final temperature, pressure, and volume to the number of moles in the final state: Apply the ideal-gas law to relate the initial temperature, pressure, and volume to the number of moles in the initial state: Divide the first of these equations by the second and solve for Pf:

Pf Vf = nf RTf

PiVi = ni RTi

Pf Vf nf RTf = PiVi ni RTi or, because Tf = Ti,

⎛ n ⎞⎛ V Pf = Pi ⎜⎜ f ⎟⎟⎜⎜ i ⎝ ni ⎠⎝ Vf Find the initial volume Vi occupied by 8.5 mol of gas at 300 K and 1 atm:

⎞ ⎟⎟ ⎠

(3)

⎛ 300 K ⎞ ⎟⎟ Vi = (22.4 L/mol)(8.5 mol)⎜⎜ ⎝ 273 K ⎠ = 209.2 L

1442 Chapter 18 Substitute numerical values in equation (3) and evaluate Vf:

⎛ 9 mol ⎞ ⎛ 209.2 L ⎞ ⎟⎟ ⎜⎜ ⎟⎟ Pf = (101.3 kPa )⎜⎜ ⎝ 8.5 mol ⎠ ⎝ 4700 L ⎠ 1atm = 4.774 kPa × 101.325 kPa = 0.0471atm

*109 ••• Picture the Problem In this problem the specific heat of the combustion products depends on the temperature. Although CP increases gradually from (9/2)R per mol to (15/2)R per mol at high temperatures, we’ll assume that CP = 4.5R below T = 2000 K and CP = 7.5R above T = 2000 K. We can find the final temperature following combustion from the heat made available during the combustion and the final pressure by applying the ideal-gas law to the initial and final states of the gases. (a) Apply the ideal-gas law to find the pressure due to 3 mol at 300 K in the container prior to the reaction:

Pi = =

nRTi Vi

(3 mol)(8.314 J/mol ⋅ K )(300 K ) 80 L

= 93.5 kPa (b) Relate the heat available in this adiabatic process to CV and the change in temperature of the gases:

∆Eint = Qavailable

= CV (Tf − Ti )

Because T > 2000 K:

CV = CP − nR = n(7.5R ) − nR = 6.5nR

Substitute to obtain:

Qavailable = 6.5nR(Tf − Ti )

Solve for Tf to obtain:

Tf =

Find Q required to raise 2 mol of CO2 to 2000 K:

Qheat CO 2 = CV ∆T

For T < 2000 K:

CV = CP − nR = n(4.5R ) − nR = 3.5nR

Qavailable + Ti 6.5nR

(1)

Heat and the First Law of Thermodynamics 1443 Substitute for CV and find the heat required to warm to CO2 to 2000 K:

Find Q available to heat 2 mol of CO2 above 2000 K: Substitute in equation (1) and evaluate Tf:

Qheat CO2 = 3.5nR∆T

= 3.5(2 mol)(8.314 J/mol ⋅ K ) × (2000 K − 300K ) = 98.94 kJ

Qavailable = 560 kJ − 98.94 kJ = 461.1 kJ Tf =

461.1kJ + 2000 K 6.5(2 mol)(8.314 J/mol ⋅ K )

= 6266 K Apply the ideal-gas law to relate the final temperature, pressure, and volume to the number of moles in the final state: Apply the ideal-gas law to relate the initial temperature, pressure, and volume to the number of moles in the initial state: Divide the first of these equations by the second and solve for Pf:

Pf Vf = nf RTf

PiVi = ni RTi

Pf Vf nf RTf = PiVi ni RTi or, because Vf = Vi,

⎛n Pf = Pi ⎜⎜ f ⎝ ni Substitute numerical values in equation (2) and evaluate Pf:

⎞⎛ Tf ⎟⎟⎜⎜ ⎠⎝ Ti

⎞ ⎟⎟ ⎠

(2)

⎛ 2 mol ⎞ ⎛ 6266 K ⎞ ⎟⎟ ⎟⎟ ⎜⎜ Pf = (93.53 kPa )⎜⎜ ⎝ 3 mol ⎠ ⎝ 300 K ⎠ = 1.30 MPa

(c) Substitute numerical values in equation (2) and evaluate Pf for Tf = 273 K:

⎛ 2 mol ⎞ ⎛ 273 K ⎞ ⎟⎟ ⎜⎜ ⎟⎟ Pf = (93.53 kPa )⎜⎜ ⎝ 3 mol ⎠ ⎝ 300 K ⎠ = 56.7 kPa

1444 Chapter 18 110 ••• Picture the Problem The molar heat capacity at constant volume is related to the internal energy per mole according to c'V =

1 dU . We can differentiate U with respect to n dT

temperature and use nR = Nk or R = NAk to establish the result given in the problem statement. From Problem 106 we have, for the internal energy per mol:

U=

3N A kTE eTE T − 1

Relate the molar heat capacity at constant volume to the internal energy per mol:

c'V =

1 dU n dT

Use c'V =

c'V =

1 dU to express c'V : n dT

⎡ 1 d ⎡ 3 N A kTE ⎤ d ⎡ 1 ⎤ − 1 ⎤ d TE T = RT e 3 RT = 3 −1 ⎢ E E 2⎥ T T T T T T dT ⎢⎣ e E − 1⎥⎦ n dT ⎢⎣ e E − 1 ⎥⎦ ⎢⎣ e E − 1 ⎥⎦ dT

(

(

)

)

2 ⎡ − 1 ⎤ ⎡ TE T ⎛ TE ⎞⎤ eTE T ⎛ TE ⎞ e 3 R = 3RTE ⎢ − = ⎜ ⎟ ⎜ ⎟ 2 ⎥⎢ 2 2 ⎥ T T ⎝ T ⎠⎦ ⎝ T ⎠ eTE T − 1 ⎢⎣ e E − 1 ⎥⎦ ⎣

(

)

(

)

111 ••• Picture the Problem We can rewrite our expression for c'V by dividing its numerator and T T

denominator by e E and then using the power series for ex to show that, for T > TE, c'V ≈ 3R . In part (b), we can use the result of Problem 103 to obtain values for c'V every 100 K between 300 K and 600 K and use this data to find ∆U numerically. (a) From Problem 110 we have:

Divide the numerator and T T denominator by e E to obtain:

2

eTE T ⎛ TE ⎞ c = 3R⎜ ⎟ ⎝ T ⎠ eTE T − 1 ' V

(

)

2

2

1 ⎛T ⎞ c'V = 3R⎜ E ⎟ 2TE T − 2eTE T + 1 ⎝T ⎠ e eTE T 2

1 ⎛T ⎞ = 3R⎜ E ⎟ TE T − 2 + e −TE T ⎝T ⎠ e

Heat and the First Law of Thermodynamics 1445 Apply the power series expansion to obtain:

e

TE T

−2+e

−TE T

T 1 ⎛T ⎞ = 1+ E + ⎜ E ⎟ T 2⎝ T ⎠

2

2

+ ... − 2 + 1 −

TE 1 ⎛ TE ⎞ + ⎜ ⎟ + ... T 2⎝ T ⎠

2

⎛T ⎞ ≈ ⎜ E ⎟ for T > TE ⎝T ⎠ 2

1 ⎛T ⎞ c ≈ 3 R⎜ E ⎟ = 3R 2 ⎝ T ⎠ ⎛ TE ⎞ ⎜ ⎟ ⎝T ⎠

Substitute to obtain:

' V

(b) Use the result of Problem 110 to verify the table to the right:

T (K) 300 400 500 600

cV (J/mol⋅K) 9.65 14.33 17.38 19.35

The following graph of specific heat as a function of temperature shown to the right was plotted using a spreadsheet program:

21 19

C V (J/mol-K)

17 15 13 11 9 7 5 300

350

400

450

T (K)

500

550

600

1446 Chapter 18

Integrate numerically, using the formula for the area of a trapezoid, to obtain:

∆U =

1 2

(100 K )(9.65 + 14.33) J/mol ⋅ K + 12 (100 K )(14.33 + 17.38) J/mol ⋅ K + 12 (100 K )(17.38 + 19.35) J/mol ⋅ K

= 4.62 kJ , a result in good agreement (< 1% difference) with the result of Problem 106. 112 ••• Picture the Problem In (a) we’ll assume that τ = f (A/V, T, k, m) with the factors dependent on constants a, b, c, and d that we’ll find using dimensional analysis. In (b) we’ll use our result from (a) and assume that the diameter of the puncture is about 2 mm, that the tire volume is 0.1 m3, and that the air temperature is 20°C. (a) Express τ = f (A/V, T, k, m):

⎛ A⎞ τ = ⎜ ⎟ (T ) b (k ) c (m ) d ⎝V ⎠

Rewrite this equation in terms of the dimensions of the physical quantities to obtain:

T = (L )

Simplify this dimensional equation to obtain:

a

−a

1

c

2 ⎛ ⎞ ⎟⎟ (M ) d (K ) ⎜⎜ ML 2 ⎝T K⎠

b

where K represents the dimension of temperature.

T1 = L− a K b M c L2c K − c T -2c M d or

T1 = L2 c−a K b−c M c+d T −2c Equate exponents to obtain:

T : − 2c = 1 , L : 2c − a = 0 , K : b − c = 0, and

M: c+d = 0

Solve these equations simultaneously to obtain:

(1)

c = − 12 , a = −1 , b = − 12 , and

d=

1 2

Heat and the First Law of Thermodynamics −1

1 1 1 ⎛ A⎞ τ = ⎜ ⎟ (T )− 2 (k )− 2 (m ) 2 ⎝V ⎠

Substitute in equation (1):

=

V m A kT

(b) Substitute numerical values and evaluate τ:

τ=

π 4

(1.293 kg/m )(0.1m )

0 .1 m 3

(2 ×10

−3

3

m

)

2

3

(8.314 J/mol ⋅ K )(293 K )

= 232 s = 3.87 min

1447

1448 Chapter 18

Chapter 19 The Second Law of Thermodynamics Conceptual Problems 1 • Determine the Concept Friction reduces the efficiency of the engine. *2 • Determine the Concept As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the room–the refrigerator actually heats the room it is in. The heating coils on an air conditioner are outside, so the waste heat is vented to the outside. 3 • Determine the Concept Increasing the temperature of the steam increases the Carnot efficiency, and generally increases the efficiency of any heat engine. 4 •• Determine the Concept To condense, water must lose heat. Because its entropy change is given by dS = dQrev/T and dQrev is negative, the entropy of the water decreases.

(c) is correct. *5 • Determine the Concept (a) Because the temperature changes during an adiabatic process, the internal energy of the system changes continuously during the process. (b) Both the pressure and volume change during an adiabatic process and hence work is done by the system. (c) ∆Q = 0 during an adiabatic process. Therefore ∆S = 0. (c ) is correct. (d) Because the pressure and volume change during an adiabatic process, so does the temperature. 6 •• (a) False. The complete conversion of mechanical energy into heat is not prohibited by either the 1st or 2nd laws of thermodynamics and is common place in energy transformations. (b) True. This is the heat-engine statement of the 2nd law of thermodynamics.

1449

1450

Chapter 19

(c) False. The efficiency of a heat engine is a function of the thermodynamic processes of its cycle. (d) False. With the input of sufficient energy, a heat pump can transfer a given quantity of heat from a cold reservoir to a hot reservoir. (e) False. The only restriction that the refrigerator statement of the 2nd law places on the COP is that it can not be infinite. (f) True. The Carnot engine, as a consequence of its thermodynamic processes, is reversible. (g) False. The entropy of one system can decrease at the expense of one or more other systems. (h) True. This is one statement of the 2nd law of thermodynamics. 7 •• Determine the Concept The two paths are shown on the PV diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem.

(a) Because Eint is a state function and the initial and final states are the same for the two paths and ∆Eint, A = ∆Eint, B . (b) and (c) S, like Eint, is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs and ∆S A = ∆S B . (d) (d ) is correct.

The Second Law of Thermodynamics 1451 *8 •• Determine the Concept The processes A→B and C→D are adiabatic; the processes B→C and D→A are isothermal. The cycle is therefore the Carnot cycle shown in the adjacent PV diagram.

9 •• Determine the Concept Note that A→B is an adiabatic expansion. B→C is a constant volume process in which the entropy decreases; therefore heat is released. C→D is an adiabatic compression. D→A is a constant volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). 10 •• Determine the Concept Refer to Figure 19-3. Here a→b is an adiabatic compression, so S is constant and T increases. Between b and c, heat is added to the system and both S and T increase. c→d is again isentropic, i.e., without change in entropy. d→a releases heat and both S and T decrease. The cycle on an ST diagram is sketched in the adjacent figure. 11 •• Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2→3 is adiabatic, so S is constant as V increases. Process 3→4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4→1 is adiabatic, i.e., isentropic, and S is constant while V decreases. The cycle is shown in the adjacent SV diagram.

1452

Chapter 19

12 •• Picture the Problem The SV diagram of the Otto cycle is shown in Figure 19-13. (see Problem 9) 13 •• Determine the Concept Process A→B is at constant entropy, i.e., an adiabatic process in which the pressure increases. Process B→C is one in which P is constant and S decreases; heat is exhausted from the system and the volume decreases. Process C→D is an adiabatic compression. Process D→A returns the system to its original state at constant pressure. The cycle is shown in the adjacent PV diagram. *14 • Picture the Problem Let ∆T be the change in temperature and ε = (Th − Tc)/Th be the initial efficiency. We can express the efficiencies of the Carnot engine resulting from the given changes in temperature and examine their ratio to decide which has the greater effect on increasing the efficiency. If Th is increased by ∆T , ε′, the new efficiency is:

ε' =

Th + ∆T − Tc Th + ∆T

If Tc is reduced by ∆T, the efficiency is:

ε'' =

Th − Tc + ∆T Th

Divide the second of these equations by the first to obtain:

Th − Tc + ∆T ε'' T + ∆T Th = = h >1 ε' Th + ∆T − Tc Th Th + ∆T

Therefore, a reduction in the temperature of the cold reservoir by ∆T increases the efficiency more than an equal increase in the temperature of the hot reservoir.

The Second Law of Thermodynamics 1453

Estimation and Approximation 15 •• Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasistatic adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio. Express the Carnot efficiency of an engine operating between the temperatures Tc and Th:

εC = 1 −

Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasistatic adiabatic compression from Vc to Vh:

TcVcγ −1 = ThVhγ −1

Solve for the ratio of Tc to Th:

Tc Vhγ −1 ⎛ Vh ⎞ = =⎜ ⎟ Th Vcγ −1 ⎜⎝ Vc ⎟⎠

Substitute to obtain:

⎛V ⎞ ε C = 1 − ⎜⎜ h ⎟⎟ ⎝ Vc ⎠

Express the compression ratio r:

r=

Substitute once more to obtain:

εC = 1 −

Substitute numerical values for r and γ (1.4 for diatomic gases) and evaluate εC:

Tc Th

γ −1

γ −1

Vc Vh

εC = 1 −

1 r

γ −1

1

(8)1.4 −1

= 0.565 = 56.5%

*16 •• Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0°C (273 K) and the room temperature to be about 30°C (303 K), then the refrigerator must be able to maintain a temperature difference of about 30 K. We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir and Qh and Qc to find an upper limit on the COP of a household refrigerator. In (b) we can solve the definition of COP for Qc and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator compartment. (a) Using its definition, express the COP of a household refrigerator:

COP =

Qc W

(1)

1454 Chapter 19 Apply the 1st law of thermodynamics to the refrigerator to obtain:

W + Qc = Qh

Substitute for W and simplify to obtain:

COP =

Qc Q h − Qc

=

1 Qh Qc

Assume, for the sake of finding the upper limit on the COP, that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to Qh and

−1

Qh Th = Qc Tc

Qc: Substitute to obtain:

COPmax =

1 Th −1 Tc

Substitute numerical values and evaluate COPmax:

COPmax =

(b) Solve equation (1) for Qc:

Qc = W (COP )

Differentiate equation (2) with respect to time to obtain:

dQc dW = (COP ) dt dt

Substitute numerical values and evaluate dQc/dt:

dQc = (9.10)(600 J/s ) = 5.46 kW dt

1 = 9.10 303 K −1 273 K (2)

17 •• Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting the earth and the definition of the change in entropy to find the changes in the entropy of the earth and the sun resulting from the radiation from the sun. (a) Using its definition, express the intensity of the sun’s radiation on the earth in terms of the power delivered to the earth P and the earth’s cross sectional area A:

I=

Solve for P and substitute for A to obtain:

P = IA = IπR 2

P A

where R is the radius of the earth.

The Second Law of Thermodynamics 1455 Substitute numerical values and evaluate P:

(

)(

P = π 1.3 kW/m 2 6.37 × 106 m

)

2

= 1.66 × 1017 W (b) Express dSearth/dt for the earth due to the flow of solar radiation:

dSearth P = dt Tearth

Substitute numerical values and evaluate dSearth/dt:

∆Searth =

1.66 × 1017 W 290 K

= 5.72 × 1014 J/K ⋅ s (c) Express dSsun/dt for the sun due to the outflow of solar radiation hitting the earth:

dSsun P = dt Tsun

Substitute numerical values and evaluate dSsun/dt:

dSsun 1.66 × 1017 W = 5400 K dt = 3.07 × 1013 J/K ⋅ s

18 •• Picture the Problem We can use the definition of intensity to find the total power radiated by the sun and the definition of the change in entropy to find the change in the entropy of the universe resulting from the radiation of 1011 stars in 1011 galaxies. (a) Using its definition, express the intensity of the sun’s radiation on the location of earth in terms of the total power it delivers to space P and the area of a sphere A whose radius is the distance from the sun to the earth:

I=

Solve for P and substitute for A to obtain:

P = IA = 4π IR 2

Substitute numerical values and evaluate P:

P A

where R is the distance from the sun to the earth.

(

)(

P = 4π 1.3 kW/m 2 1.5 × 1011 m = 3.68 × 1026 W

(b) Express ∆Suniverse:

∆S universe =

P Tuniverse

)

2

1456 Chapter 19 Substitute numerical values and evaluate ∆Suniverse:

∆S universe

(

1022 3.68 × 1026 W = 2.73 K

)

= 1.35 × 1048 J/K ⋅ s 19 •• Picture the Problem We can use the definition of entropy change to estimate the increase in entropy of the universe as a result of the heat produced by a typical human body. The entropy change is equivalent to the entropy change if the heat from the body were added to the universe reversibly.

∆Qday

∆Qnight

Express the increase in entropy of the universe as a result of the heat produced by a human body:

∆S u =

Using the definition of power, express the total heat produced by a human body:

∆Q = P∆t

Assume that half of the heat is produced during the day and half at night:

∆Qday = ∆Qnight = 12 P∆t

Substitute to obtain:

∆S u =

Tday

+

Tnight

P∆t 12 P∆t + Tday Tnight

1 2

⎛ 1 1 ⎞⎟ = 12 P∆t ⎜ + ⎜T ⎟ ⎝ day Tnight ⎠ Use T =

5 9

(tF − 32) + 273 to obtain:

Tday = 294 K and Tnight = 286 K

Substitute numerical values and evaluate ∆Su:

∆S u =

1 2

1 1 ⎞ ⎟⎟ = 29.8 kJ/K + 294 K 286 K ⎠ ⎝ ⎛

(100 J/s )(24 h/d )(3600 s/h )⎜⎜

*20 ••• Picture the Problem If you had one molecule in a box, it would have a 50% chance of being on one side or the other. We don’t care which side the molecules are on as long as they all are on one side, so with one molecule you have a 100% chance of it being on one side or the other. With two molecules, there are four possible combinations (both on one side, both on the other, one on one side and one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both being on a particular side, or a 50% chance of them both being on either side. Extending this logic, the probability of N molecules all being on one side of the box is P = 2/2N, which means that, if the molecules shuffle 100 times a second, the time it would take them to cover all the combinations and all get on one side

2N . In (e) we can apply the ideal gas law to find the number of or the other is t = 2(100 )

The Second Law of Thermodynamics 1457 molecules in 1 L of air at a pressure of 10−12 torr and an assumed temperature of 300 K.

(a) Evaluate t for N = 10 molecules:

(b) Evaluate t for N = 100 molecules:

t=

210 = 5.12 s 2(100)

t=

2100 = 6.34 × 1027 s 2(100)

= 2.01 × 1020 y (c) Evaluate t for N = 1000 molecules:

t=

21000 2(100 )

To evaluate 21000 let 10 x = 21000 and take the logarithm of both sides of the equation to obtain:

(1000)ln 2 = x ln10

Solve for x to obtain:

x = 301

Substitute to obtain:

t=

10301 = 0.5 × 10299 s 2(100)

= 1.58 × 10290 y (d) Evaluate t for N = 6.02×1023 molecules:

26.02×10 t= 2(100)

To evaluate 26.02×10

(6.02 × 10 )ln 2 = x ln10

23

let

23

23

10 x = 26.02×10 and take the 23

logarithm of both sides of the equation to obtain: Solve for x to obtain:

x ≈ 1023

Substitute to obtain:

23 1010 t≈ ≈ 1010 y 2(100)

(e) Solve the ideal gas law for the number of molecules N in the gas: Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N:

23

N=

PV kT

(10 N=

torr )(133.32 Pa/torr ) (10 −3 m 3 ) (1.381×10−23 J/K )(300 K )

−12

= 3.22 × 107 molecules

1458 Chapter 19 Evaluate T for N = 3.22×107 molecules:

23.22×10 t= 2(100)

To evaluate 23.22×10 let

(3.22 × 10 )ln 2 = x ln10

7

7

7

10 x = 23.22×10 and take the 7

logarithm of both sides of the equation to obtain: Solve for x to obtain:

x ≈ 107

Substitute to obtain:

7 1010 T= ≈ 1010 y 2(100)

Express the ratio of this waiting time to the lifetime of the universe Tuniverse:

7

7

T Tuniverse

7 1010 y = 10 ≈ 1010 10 y

or

T ≈ 1010 Tuniverse 7

Heat Engines and Refrigerators 21 • Picture the Problem We can use the definition of the efficiency of a heat engine to relate the work done W, the heat absorbed Qin, and the heat rejected each cycle Qout. (a) Express Qin in terms of W and ε :

(b) Solve the definition of efficiency for and evaluate Qout :

Qin =

W

ε

=

100 J = 500 J 0.2

Qout = Qin (1 − ε ) = (500 J )(1 − 0.2 ) = 400 J

22 • Picture the Problem We can use its definition to find the efficiency of a heat engine from the work done, the heat absorbed, and the heat rejected each cycle. (a) Use the definition of the efficiency of a heat engine: (b) Solve the definition of efficiency for and evaluate Qout :

ε≡

W 120 J = = 30% Qin 400 J

Qout = Qin (1 − ε ) = (400 J )(1 − 0.3) = 280 J

The Second Law of Thermodynamics 1459 23 • Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output. (a) Apply the definition of the efficiency of a heat engine:

ε =1−

(b) Use the definition of power to find the power output of this engine:

P=

Qout 60 J =1− = 40.0% Qin 100 J

∆W εQin 0.4(100 J ) = = = 80.0 W ∆t ∆t 0.5 s

*24 • Picture the Problem We can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine.

Qc W

(a) Using the definition of the COP, relate the heat absorbed from the cold reservoir to the work done each cycle:

COP =

Relate the work done per cycle to Qh and Qc:

W = Qh − Qc

Substitute to obtain:

COP =

Qc Qh − Qc

Substitute numerical values and evaluate COP:

COP =

5 kJ = 1.67 8 kJ − 5 kJ

(b) Use the definition of efficiency to relate the work done per cycle to the heat absorbed from the hightemperature reservoir:

ε=

W Qh

Substitute numerical values and evaluate ε :

ε=

3 kJ = 37.5% 8 kJ

1460 Chapter 19 25 •• Picture the Problem To find the heat added during each step we need to find the temperatures in states 1, 2, 3, and 4. We can then find the work done on or by the gas along each pass from the area under each straight-line segment and the heat that enters or leaves the system from Q = CV ∆T and Q = CP ∆T . We can find the efficiency of the cycle from the work done each cycle and the heat that enters the system each cycle. (a) The cycle is shown to the right:

Apply the ideal-gas law to state 1 to find T1:

T1 = =

P1V1 nR

(1atm )(24.6 L )

(1mol)(8.206 × 10− 2 L ⋅ atm/mol ⋅ K )

= 300 K T2 = 2T1 = 600 K

The pressure doubles while the volume remains constant between states 1 and 2. Hence:

T3 = 2T2 = 1200 K

The volume doubles while the pressure remains constant between states 2 and 3. Hence:

T4 = 12 T3 = 600 K

The pressure is halved while the volume remains constant between states 3 and 4. Hence: For path 1→2:

W1→ 2 = P∆V1→ 2 = 0 and

Q1→2 = ∆Eint,1→2 = CV ∆T1→2 = 32 R∆T1→2 = = 3.74 kJ

3 2

(8.314 J/mol ⋅ K )(600 K − 300 K )

The Second Law of Thermodynamics 1461 For path 2→3:

W2→3 = P∆V2→2 = (2 atm )(49.2 L − 24.6 L ) = 49.20 L ⋅ atm ×

101.325 J L ⋅ atm

= 4.99 kJ and

Q2→3 = CP ∆T2→3 = 52 R∆T2→3 =

5 2

(8.314 J/mol ⋅ K )(1200 K − 600 K )

= 12.5 kJ For path 3→4:

W3→4 = P∆V3→4 = 0 and

Q3→4 = ∆Eint,3→4 = CV ∆T3→4 = 32 R∆T3→4 =

3 2

(8.314 J/mol ⋅ K )(600 K − 1200 K )

= − 7.48 kJ For path 4→1:

W4→1 = P∆V4→1 = (1atm )(24.6 L − 49.2 L ) = −24.6 L ⋅ atm ×

101.3 J L ⋅ atm

= 2.49 kJ and

Q4→1 = CP ∆T4→1 = 52 R∆T4→1 =

5 2

(8.314 J/mol ⋅ K )(300 K − 600 K )

= − 6.24 kJ (b) Use its definition to find the efficiency of this cycle:

ε= =

W W2 → 3 + W4 →1 = Qin Q1→ 2 + Q2 →3 4.99 kJ − 2.49 kJ = 15.4% 3.74 kJ + 12.5 kJ

Remarks: Note that the work done per cycle is the area bounded by the rectangular path.

1462 Chapter 19 26 •• Picture the Problem The three steps in the process are shown on the PV diagram. We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.

Express the efficiency of the cycle:

Find the heat entering or leaving the system during the adiabatic expansion: Find the heat entering or leaving the system during the isobaric compression: Find the heat entering or leaving the system during the constant-volume process: Apply the 1st law of thermodynamics to the cycle ( ∆Eint, cycle = 0 ) to

ε=

Q1 = 0

Q2 = CV ∆T2 = 72 R∆T2 = 72 P∆V2 =

7 2

(1atm )(10 L − 20 L ) = −35 atm ⋅ L

Q3 = CV ∆T3 = 52 R∆T3 = 52 ∆PV3 =

5 2

(2.64 atm - 1atm )(10 L)

= 41atm ⋅ L Won = ∆Eint − Qin = −Qin = Q1 + Q2 + Q3 = 0 − 35atm ⋅ L + 41atm ⋅ L = 6 atm ⋅ L

obtain:

Substitute and evaluate ε :

W Qin

ε=

6 atm ⋅ L = 14.6% 41atm ⋅ L

27 •• Picture the Problem We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.

The Second Law of Thermodynamics 1463

Apply the ideal-gas law to states 1, 2, 3, and 4 to find the pressures at these points:

P1 =

nRT1 (1 mol) (8.206 × 10−2 L ⋅ atm/mol ⋅ K )(400 K ) = = 1.33 atm V1 24.6 K

Proceed as above to obtain the values shown in the table:

Point 1 2 3 4

P V (atm) (L) 1.330 24.6 0.667 49.2 0.500 49.2 1.000 24.6

T (K) 400 400 300 300

The PV diagram is shown to the right:

Express the efficiency of the cycle:

ε=

W Qin

(1)

Find the work done by the gas and the heat that enters the system during the isothermal expansion from 1 to 2:

⎛V ⎞ ⎛ 49.2 L ⎞ ⎟⎟ W1→2 = Q1→2 = nRT1 ln⎜⎜ 2 ⎟⎟ = (1 mol)(8.314 J/mol ⋅ K )(400 K ) ln⎜⎜ ⎝ 24.6 L ⎠ ⎝ V1 ⎠ = 2.305 kJ Find the work done by the gas and the heat that enters the system during the constant-volume compression from 2 to 3:

W2→3 = 0 and

Q2→3 = CV ∆T2→3 = (21 J/K )(300 K − 400 K ) = −2.10 kJ

1464 Chapter 19

Find the work done by the gas and the heat that enters the system during the isothermal expansion from 3 to 4:

⎛V ⎞ ⎛ 2 4 .6 L ⎞ ⎟⎟ W3→4 = Q3→4 = nRT3 ln⎜⎜ 4 ⎟⎟ = (1 mol)(8.314 J/mol ⋅ K )(300 K ) ln⎜⎜ 4 9 . 2 L V ⎠ ⎝ ⎝ 3⎠ = −1.729 kJ Find the work done by the gas and the heat that enters the system during the constant-volume process from 4 to 1:

W4→1 = 0 and

Q4→1 = CV ∆T4→1 = (21 J/K )(400 K − 300 K ) = 2.10 kJ

Evaluate the work done each cycle:

W = W1→2 + W2→3 + W3→4 + W4→1 = 2.305 kJ + 0 − 1.729 kJ + 0 = 0.5760 kJ

Find the heat that enters the system each cycle:

Qin = Q1→ 2 + Q4 →1 = 2.305 kJ + 2.100 kJ = 4.405 kJ

Substitute numerical values in equation (1) and evaluate ε :

ε=

0.5760 kJ = 13.1% 4.405 kJ

*28 •• Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas and the heat capacities at constant volume and constant pressure to find the heat flow for the constant-volume and isobaric processes. Because the change in internal energy is zero for the isothermal process, we can use the expression for the work done on or by a gas during an isothermal process to find the heat flow during such a process. Finally, we can find the efficiency of the cycle from its definition. (a) Use the ideal-gas law to find the temperature at point 1:

T1 =

(100 kPa )(25 L ) P1V1 = nR (1mol)(8.314 J/mol ⋅ K )

= 301K

The Second Law of Thermodynamics 1465 Use the ideal-gas law to find the temperatures at points 2 and 3:

T2 = T3 =

(200 kPa )(25 L ) P2V2 = nR (1mol)(8.314 J/mol ⋅ K )

= 601K (b) Find the heat entering or leaving the system for the constant-volume process from 1 to 2:

Q1→2 = CV ∆T1→2 = 32 R∆T1→2 =

3 2

(8.314 J/mol ⋅ K )(601K − 301K )

= 3.74 kJ Find the heat entering or leaving the system for the isothermal process from 2 to 3:

⎛V ⎞ ⎛ 50 L ⎞ ⎟⎟ = 3.46 kJ Q2→3 = nRT2 ln⎜⎜ 3 ⎟⎟ = (1 mol)(8.314 J/mol ⋅ K )(601 K )ln⎜⎜ ⎝ 25 L ⎠ ⎝ V2 ⎠ Find the heat entering or leaving the system during the isobaric compression from 3 to 1:

Q3→1 = CP ∆T3→1 = 52 R∆T3→1 =

5 2

(8.314 J/mol ⋅ K )(301K − 601K )

= − 6.24 kJ W W = Qin Q1→ 2 + Q2 →3

(c) Express the efficiency of the cycle:

ε=

Apply the 1st law of thermodynamics to the cycle:

W = ∑ Q = Q1→ 2 + Q2 →3 + Q3→1

(1)

= 3.74 kJ + 3.46 kJ − 6.24 kJ = 0.960 kJ because, for the cycle, ∆U = 0.

Substitute numerical values in equation (1) and evaluate ε :

ε=

0.960 kJ = 13.3% 3.74 kJ + 3.46 kJ

29 •• Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. We can find the efficiency of the cycle from its definition; using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin.

1466 Chapter 19 (a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1:

P1V1 P2V2 = T1 T2

Solve for and evaluate T2:

T2 = T1

(3 atm ) P P2V2 = T1 2 = (200 K ) (1atm ) P1V1 P1

= 600 K Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain:

T3 = T2

Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain:

T4 = T3

(b) Express the efficiency of the cycle:

ε=

Use the area of the rectangle to find the work done each cycle:

Apply the ideal-gas law to state 1 to find the product of n and R:

(300 L ) P3V3 V = T2 3 = (600 K ) (100 L) P2V2 V2

= 1800 K

(1atm ) P4V4 P = T3 4 = (1800 K ) (3 atm ) P3 P3V3

= 600 K W Qin

(1)

W = ∆P∆V

= (300 L − 100 L )(3 atm − 1atm ) = 400 atm ⋅ L

nR =

P1V1 (1atm )(100 L ) = T1 200 K

= 0.5 L ⋅ atm/K Noting that heat enters the system between states 1 and 2 and states 2 and 3, express Qin:

Qin = Q1→ 2 + Q2 →3 = CV ∆T1→ 2 + CP ∆T2 → 3 = 52 nR∆T1→ 2 + 72 nR∆T2 → 3 = ( 52 ∆T1→ 2 + 72 ∆T2 → 3 )nR

Substitute numerical values and evaluate Qin:

Qin = [ 52 (600 K − 200 K ) + 72 (1800 K − 600 K )](0.5 L ⋅ atm/K ) = 2600 atm ⋅ L

The Second Law of Thermodynamics 1467 Substitute numerical values in equation (1) and evaluate ε :

ε=

400 atm ⋅ L = 15.4% 2600 atm ⋅ L

30 ••• Picture the Problem To find the efficiency of the diesel cycle we can find the heat that enters the system and the heat that leaves the system and use the expression that gives the efficiency in terms of these quantities. Note that no heat enters or leaves the system during the adiabatic processes a→b and c→d. Express the efficiency of the cycle in terms of Qc and Qh: Express Q for the isobaric warming process b→c: Express Q for the constant-volume cooling process d→a: Substitute to obtain:

ε =1−

Qc Qh

Qb → c = Qh = CP (Tc − Tb ) Qd → a = Qc = CV (Td − Ta )

ε =1− =1−

CV (Td − Ta ) CP (Tc − Tb )

(Td − Ta ) γ (Tc − Tb )

Using the equation of state for an adiabatic process, relate the temperatures Ta and Tb:

TaVaγ −1 = TbVbγ −1

(1)

Proceeding similarly, relate the temperatures Tc and Td:

TcVcγ −1 = TdVdγ −1

(2)

Use equations (1) and (2) to eliminate Ta and Td:

⎛ Vcγ −1 V γ −1 ⎞ ⎜⎜ Tc γ −1 − Tb bγ −1 ⎟⎟ Va ⎠ Vd ε =1− ⎝ γ (Tc − Tb ) ⎛ ⎛ V ⎞γ −1 T ⎛ V ⎞γ −1 ⎞ ⎜⎜ c ⎟ − b ⎜ b ⎟ ⎟ ⎜ ⎜⎝ Va ⎟⎠ Tc ⎜⎝ Va ⎟⎠ ⎟ ⎝ ⎠ =1− ⎛ T ⎞ γ ⎜⎜1 − b ⎟⎟ ⎝ Tc ⎠ because Va = Vd.

1468 Chapter 19 Apply the ideal-gas law for a fixed amount of gas to relate Tb and Tc:

Tb Vb = Tc Vc because Pb = Pc.

Substitute and simplify to obtain:

⎛ ⎛ V ⎞γ −1 V ⎛ V ⎞γ −1 ⎞ ⎛ ⎛ V ⎞γ V ⎛ V ⎞γ −1 ⎞ ⎜ ⎜ c ⎟ − b ⎜ b ⎟ ⎟ Vc ⎜⎜ c ⎟ − b ⎜ b ⎟ ⎟ ⎜ ⎟ ⎜ ⎜⎝ Va ⎟⎠ ⎟ ⎜ ⎜⎝ Va ⎟⎠ Va ⎜⎝ Va ⎟⎠ ⎟ Vc ⎝ Va ⎠ Va ⎝ ⎠ ⎠ ⋅ =1− ⎝ ε =1− V ⎛ V ⎞ ⎛V V ⎞ c γ ⎜⎜1 − b ⎟⎟ γ ⎜⎜ c − b ⎟⎟ Va ⎝ Vc ⎠ ⎝ Va Va ⎠ ⎛ ⎛ V ⎞γ ⎛ V ⎞γ ⎞ ⎜⎜ c ⎟ − ⎜ b ⎟ ⎟ ⎜ ⎜⎝ Va ⎟⎠ ⎜⎝ Va ⎟⎠ ⎟ ⎠ = 1− ⎝ ⎛V V ⎞ γ ⎜⎜ c − b ⎟⎟ ⎝ Va Va ⎠ *31 •• Picture the Problem We can use the efficiency of a Carnot engine operating between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures. (a) Express the maximum efficiency of an engine operating between body temperature and 70°F: Use T =

5 9

(tF − 32) + 273 to obtain:

Substitute numerical values and evaluate ε C :

εC = 1 −

Tc Th

Tbody = 310 K and Troom = 294 K

εC = 1 −

294 K = 5.16% 310 K

The fact that this efficiency is considerably less than the actual efficiency of a human body does not contradict the second law of thermodynamics. The application of the second law to chemical reactions such as the ones that supply the body with energy have not been discussed in the text but we can note that don' t get our energy from heat swapping between our body and the environment. Rather, we eat food to get the energy that we need.

The Second Law of Thermodynamics 1469 Most warm − blooded animals survive under roughly the same conditions (b)

as humans. To make a heat engine work with appreciable efficiency, internal body temperatures would have to be maintained at an unreasonably high level.

32 ••• Picture the Problem The Carnot cycle’s four segments (shown to the right) are: (A) an isothermal expansion at T = Th from V1 to V2, (B) an adiabatic expansion from V2 to V3, (C) an isothermal compression from V3 to V4 at T = Tc, and (D) an adiabatic compression from V4 to V1. We can find the Carnot efficiency for a gas described by the Clausius equation by expressing the ratio of the work done per cycle to the heat entering the system per cycle.

Express the efficiency of the Carnot cycle in terms of the work done and the heat that enters the system per cycle: Apply the first law of thermodynamics to segment A:

ε=

W Qh

V2

QA = WA + ∆Eint, A = WA = ∫ PdV V1

⎛ V − bn ⎞ dV ⎟⎟ = nRTh ln⎜⎜ 2 V − bn ⎝ V1 − bn ⎠ V1

V2

= nRTh ∫ = Qh Follow the same procedure for segment C to obtain:

V4

QC = WC + ∆Eint, C = WC = ∫ PdV V3

⎛ V − bn ⎞ dV ⎟⎟ = nRTc ln⎜⎜ 4 V − bn ⎝ V3 − bn ⎠ V3

V4

= nRTc ∫ and

⎛ V − bn ⎞ ⎟⎟ Qc = nRTc ln⎜⎜ 3 ⎝ V4 − bn ⎠

1470 Chapter 19 Apply the first law of thermodynamics to the complete cycle ( ∆Eint, cycle = 0 ) to express W:

W = Qh − Qc

Substitute and simplify to obtain:

⎛ V − bn ⎞ ⎛ V − bn ⎞ ⎟⎟ − nRTc ln⎜⎜ 3 ⎟ nRTh ln⎜⎜ 2 V4 − bn ⎟⎠ V1 − bn ⎠ ⎝ ⎝ ε= ⎛ V − bn ⎞ ⎟⎟ nRTh ln⎜⎜ 2 ⎝ V1 − bn ⎠

⎛ V − bn ⎞ ⎛ V − bn ⎞ ⎟⎟ − nRTc ln⎜⎜ 3 = nRTh ln⎜⎜ 2 ⎝ V4 − bn ⎠ ⎝ V1 − bn ⎠

⎛ V − bn ⎞ ⎟⎟ Tc ln⎜⎜ 3 V bn − ⎝ 4 ⎠ = 1− ⎛ V2 − bn ⎞ ⎟⎟ Th ln⎜⎜ ⎝ V1 − bn ⎠ Apply the first law of thermodynamics to the adiabatic processes B and D: Separate variables and integrate to obtain:

dQB = 0 = dWB + dEint, B = PdV + CV dT =

dT

∫T

nRT dV + CV dT V − bn

=−

nT dV ∫ CV V − bn

= −(γ − 1)∫

dV V − bn

or

ln T = −(γ − 1)ln (V − bn ) + constant = ln (V − bn )

Simplify to obtain:

ln T + ln(V − bn )

γ −1

+ constant

γ −1

= constant

or

lnT (V − bn )

γ −1

= constant

and

T (V − bn )

Using this result, relate V2 and V3 to Th and Tc:

γ −1

Th (V2 − bn )

= constant

γ −1

= Tc (V3 − bn )

γ −1

= Tc (V4 − bn )

γ −1

Relate V1 and V4 to Th and Tc:

Th (V1 − bn )

Divide equation (1) by equation (2) and simplify to obtain:

Th (V2 − bn ) Tc (V3 − bn ) γ −1 = γ −1 Th (V1 − bn ) Th (V4 − bn ) γ −1

γ −1

γ −1

(1)

(2)

The Second Law of Thermodynamics 1471 or

V2 − bn V3 − bn = V1 − bn V4 − bn

Substitute in our expression for ε and simplify:

⎛ V − bn ⎞ ⎟ Tc ln⎜⎜ 2 V1 − bn ⎟⎠ T ⎝ = 1− c ε = 1− Th ⎛ V − bn ⎞ ⎟⎟ Th ln⎜⎜ 2 ⎝ V1 − bn ⎠ the same as for an ideal gas.

Second Law of Thermodynamics 33 •• Determine the Concept The relationship of the perfect engine and the refrigerator to each other and to the hot and cold reservoirs is shown below. To remove 500 J from the cold reservoir and reject 800 J to the hot reservoir, 300 J of work must be done on the system. Assuming that the heat-engine statement is false, one could use the 800 J rejected to the hot reservoir to do 300 J of work. Thus, running the refrigerator connected to the ″perfect″ heat engine would have the effect of transferring 500 J of heat from the cold to the hot reservoir without any work being done, in violation of the refrigerator statement of the second law.

*34 •• Determine the Concept The work done by the system is the area enclosed by the cycle, where we assume that we start with the isothermal expansion. It is only in this expansion that heat is extracted from a reservoir. There is no heat transfer in the adiabatic expansion or compression. Thus, we would completely convert heat to mechanical energy, without exhausting any heat to a cold reservoir, in violation of the second law.

1472 Chapter 19

Carnot Engines 35 • Picture the Problem We can find the efficiency of the Carnot engine using ε = 1 − Tc / Th and the work done per cycle from ε = W / Qh . We can apply conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle. We can find the COP of the engine working as a refrigerator from its definition. (a) Express the efficiency of the Carnot engine in terms of the temperatures of the hot and cold reservoirs: (b) Using the definition of efficiency, relate the work done each cycle to the heat absorbed from the hot reservoir:

εC = 1−

Tc 200 K = 1− = 33.3% 300 K Th

W = ε CQh = (0.333)(100 J ) = 33.3 J

(c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done:

Qc = Qh − W = 100 J − 33.3 J = 66.7 J

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

COP =

Qc 66.7 J = = 2.00 W 33.3 J

36 • Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from W = ε CQh , where εC is the Carnot efficiency. (a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir and the heat exhausted to the lowtemperature reservoir: (b) Express the additional work done if the engine is reversible:

ε=

W Qh − Qc Q = = 1− c Qh Qh Qh

=1−

200 J = 20.0% 250 J

∆W = WCarnot − Wpart a

The Second Law of Thermodynamics 1473 Relate the work done by a reversible engine to its Carnot efficiency:

⎛ T ⎞ W = ε CQh = ⎜⎜1 − c ⎟⎟Qh ⎝ Th ⎠ ⎛ 200 K ⎞ ⎟⎟ (250 J ) = 83.3 J = ⎜⎜1 − 300 K ⎝ ⎠

Substitute and evaluate ∆W:

∆W = 83.3 J − 50 J = 33.3 J

37 •• Determine the Concept Let the first engine be run as a refrigerator. Then it will remove 140 J from the cold reservoir, deliver 200 J to the hot reservoir, and require 60 J of energy to operate. Now take the second engine and run it between the same reservoirs, and let it eject 140 J into the cold reservoir, thus replacing the heat removed by the refrigerator. If ε2, the efficiency of this engine, is greater than 30%, then Qh2, the heat removed from the hot reservoir by this engine, is 140 J/(1 − ε2) > 200 J, and the work done by this engine is W = ε2Qh2 > 60 J. The end result of all this is that the second engine can run the refrigerator, replacing the heat taken from the cold reservoir, and do additional mechanical work. The two systems working together then convert heat into mechanical energy without rejecting any heat to a cold reservoir, in violation of the second law. 38 •• Determine the Concept If the reversible engine is run as a refrigerator, it will require 100 J of mechanical energy to take 400 J of heat from the cold reservoir and deliver 500 J to the hot reservoir. Now let the second engine, with ε2 > 0.2, operate between the same two heat reservoirs and use it to drive the refrigerator. Because ε2 > 0.2, this engine will remove less than 500 J from the hot reservoir in the process of doing 100 J of work. The net result is then that no net work is done by the two systems working together, but a finite amount of heat is transferred from the cold to the hot reservoir, in violation of the refrigerator statement of the second law. *39 •• Picture the Problem We can use the definition of efficiency to find the efficiency of the Carnot engine operating between the two reservoirs. (a) Use its definition to find the efficiency of the Carnot engine:

εC =

W 50 J = = 33.3% Qh 150 J

(b) If COP > 2, then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. So running engine (a) to operate the refrigerator with a COP > 2 will result in the transfer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law.

1474 Chapter 19 40 •• Picture the Problem We can use the definitions of the efficiency of a Carnot engine and the coefficient of performance of a refrigerator to find these quantities. The work done each cycle by the Carnot engine is given by W = ε CQh and we can use the conservation of energy to find the heat rejected to the low-temperature reservoir. (a) Use the definition of the efficiency of a Carnot engine to obtain: (b) Express the work done each cycle in terms of the efficiency of the engine and the heat absorbed from the high-temperature reservoir:

εC = 1 −

Tc 77 K =1− = 74.3% Th 300 K

W = ε CQh = (0.743)(100 J ) = 74.3 J

(c) Apply conservation of energy to obtain:

Qc = Qh − W = 100 J − 74.3 J = 25.7 J

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

COP =

Qc 25.7 J = = 0.346 W 74.3 J

41 •• Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures, volumes, and pressures at the end points of each process in the given cycle. We can use Q = C V ∆T and Q = C P ∆T to find the heat entering and leaving during the constantvolume and isobaric processes and the first law of thermodynamics to find the work done each cycle. Once we’ve calculated these quantities, we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures. (a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1:

Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1:

P1V1 P3V3 = T1 T3 or, because P1 = P3,

T3 = T1

V3 V1

P1V1 P2V2 = T1 T2 or, because V1 = V2,

(1)

The Second Law of Thermodynamics 1475 P2 = P1

Apply the state equation for an adiabatic process to relate the pressures and volumes at points 2 and 3: Noting that V1 = 22.4 L, solve for and evaluate V3:

Substitute in equation (1) and evaluate T3:

T2 423 K = (1atm ) = 1.55 atm T1 273 K

P1V1γ = P3V3γ

1

1

⎛ P1 ⎞ γ ⎛ 1.55 atm ⎞ 1.4 ⎜ ⎟ ⎟⎟ V3 = V1 ⎜ ⎟ = (22.4 L )⎜⎜ ⎝ 1atm ⎠ ⎝ P3 ⎠ = 30.6 L T3 = (273 K )

30.6 L = 373 K 22.4 L

and

t3 = T3 − 273 = 100°C (b) Process 1→2 takes place at constant volume (note that γ = 1.4 corresponds to a diatomic gas and that CP – CV = R): Process 2→3 takes place adiabatically: Process 3→1 is isobaric (note that CP = CV + R):

Q1→ 2 = C V ∆T1→ 2 = 52 R∆T1→ 2 =

5 2

(8.314 J/mol ⋅ K )(423 K − 273 K )

= 3.12 kJ Q2 → 3 = 0

Q3→1 = C P ∆T3→1 = 72 R∆T1→ 2 =

7 2

(8.314 J/mol ⋅ K )(273 K − 373 K )

= − 2.91kJ W Qin

(c) Use its definition to express the efficiency of this cycle:

ε=

Apply the first law of thermodynamics to the cycle:

∆K int = Qin + Won or, because ∆Eint, cycle = 0 (the system begins and ends in the same state) and Won = − Wby the gas = W, W = Qin .

1476 Chapter 19 Evaluate W:

W = ∑ Q = Q1→ 2 + Q2 →3 + Q3→1 = 3.12 kJ + 0 − 2.91kJ = 0.210 kJ

Substitute and evaluate ε :

ε=

(d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K:

εC = 1 −

0.210 kJ = 6.73% 3.12 kJ Tc 273 K =1− = 35.5% Th 423 K

42 •• Picture the Problem We can find the maximum efficiency of the steam engine by calculating the Carnot efficiency of an engine operating between the given temperatures. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1 h.

Tc 323 K =1− = 40.5% Th 543 K

(a) Find the efficiency of a Carnot engine operating between these temperatures:

ε max = 1 −

Find the efficiency of the steam engine as a percentage of the maximum possible efficiency:

ε steam engine =

(b) Relate the heat discharged to the engine’s surroundings to Qh and the efficiency of the engine:

0.30 ε max = 0.741ε max 0.405

Qc = (1 − ε )Qh

W

P∆t

Using its definition, relate the efficiency of the engine to the heat intake of the engine and the work it does each cycle:

Qh =

Substitute and evaluate Qc in 1 h:

Qc = (1 − ε )

ε

=

ε

P∆t

ε

= (1 − 0.3)

(200 kJ/s )(3600 s )

= 1.68 GJ

0.3

The Second Law of Thermodynamics 1477

Heat Pumps *43 • Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. We can apply equation 19-10 and the definition of power to find the minimum power needed to run the heat pump. (a) Express the COPHP in terms of Th and Tc:

COPHP =

Qh Qh = W Qh − Qc

=

1 1 = Q T 1− c 1− c Qh Th

=

Th Th − Tc 313 K = 6.26 313 K − 263 K

Substitute numerical values and evaluate the COPHP:

COPHP =

(b) Using its definition, express the power output of the engine:

P=

W ∆t

Use equation 19-10 to express the work done by the heat pump:

W=

Qh 1 + COPHP

Substitute and evaluate P:

P=

(c) Find the minimum power if the COP is 60% of the efficiency of an ideal pump:

Pmin =

Qh ∆t 20 kW = = 2.75 kW 1 + COPHP 1 + 6.26 Qc ∆t 20 kW = 1 + 0.6(COPHP, max ) 1 + 0.6(6.26 )

= 4.21kW

44 • Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and ∆t.

1478 Chapter 19 (a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP: Express the COP in terms of Th and Tc:

Qc = (COP )W

= (COP )P∆t

COP = =

=

Qc Q Q −W = c = h εQh W εQh 1− ε

ε

=

1

ε

−1 =

1 −1 Tc 1− Th

Tc Th − Tc

Substitute to obtain:

⎛ Tc ⎞ ⎟⎟ P∆t Qc = ⎜⎜ T T − ⎝ h c⎠

Substitute numerical values and evaluate Qc:

⎞ ⎛ 273 K ⎟⎟ (370 W )(60 s ) Qc = ⎜⎜ ⎝ 293 K − 273 K ⎠ = 303 kJ

(b) Find the heat removed if the COP is 70% of the efficiency of an ideal pump:

⎞ ⎛ 273 K ⎟⎟ (370 W )(60 s ) Qc' = (0.7 )⎜⎜ ⎝ 293 K − 273 K ⎠ = 212 kJ

45 • Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and ∆t. (a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP:

Qc = (COP )W

= (COP )P∆t

The Second Law of Thermodynamics 1479 Express the COP in terms of Th and Tc:

COP = = =

Qc Q Q −W = c = h W εQh εQh 1− ε

ε

=

1

ε

−1

1 Tc −1 = Tc Th − Tc 1− Th

Substitute to obtain:

⎛ Tc ⎞ ⎟⎟ P∆t Qc = ⎜⎜ ⎝ Th − Tc ⎠

Substitute numerical values and evaluate Qc:

⎞ ⎛ 273 K ⎟⎟ (370 W )(60 s ) Qc = ⎜⎜ ⎝ 308 K − 273 K ⎠ = 173 kJ

(b) Find the heat removed if the COP is 70% of the efficiency of an ideal pump:

⎞ ⎛ 273 K ⎟⎟ (370 W )(60 s ) Qc' = (0.7 ) ⎜⎜ ⎝ 308 K − 273 K ⎠ = 121kJ

Entropy Changes 46 • Picture the Problem We can use the definition of entropy change to find the change in entropy of the water as it freezes. Apply the definition of entropy change to obtain: Substitute numerical values and evaluate ∆S:

∆S =

∆Q − mLf = T T

∆S =

− (18 g )(333.5 J/g ) = − 22.0 J/K 273 K

*47 •• Picture the Problem The change in the entropy of the world resulting from the freezing of this water and the cooling of the ice formed is the sum of the entropy changes of the water-ice and the freezer. Note that, while the entropy of the water decreases, the entropy of the freezer increases. Express the change in entropy of the universe resulting from this freezing and cooling process:

∆S u = ∆S water + ∆S freezer

(1)

1480 Chapter 19 ∆S water = ∆S freezing + ∆S cooling

Express ∆Swater: Express ∆Sfreezing:

∆S freezing =

(2)

− Qfreezing

(3)

Tfreezing

where the minus sign is a consequence of the fact that heat is leaving the water as it freezes. Relate Qfreezing to the latent heat of fusion and the mass of the water:

Qfreezing = mLf − mLf Tfreezing

Substitute in equation (3) to obtain:

∆S freezing =

Express ∆Scooling:

⎛T ∆S cooling = mCp ln⎜⎜ f ⎝ Ti

Substitute in equation (2) to obtain:

∆S water =

Noting that the freezer gains heat (at 263 K) from the freezing water and cooling ice, express ∆Sfreezer:

∆S freezer =

⎞ ⎟⎟ ⎠

⎛T − mLf + mCp ln⎜⎜ f Tfreezing ⎝ Ti

=

∆Qice ∆Qcooling ice + Tfreezer Tfreezer mCp ∆T mLf + Tfreezer Tfreezer

Substitute for ∆Swater and ∆Sfreezer in equation (1):

∆S u =

⎛T − mLf + mCp ln⎜⎜ f Tfreezing ⎝ Ti

mCp ∆T ⎞ mLf ⎟⎟ + + Tfreezer ⎠ Tfreezer

⎡ − Lf ⎛T = m⎢ + Cp ln⎜⎜ f ⎢⎣ Tfreezing ⎝ Ti

C ∆T ⎤ ⎞ L ⎟⎟ + f + p ⎥ ⎠ Tfreezer Tfreezer ⎥⎦

⎞ ⎟⎟ ⎠

The Second Law of Thermodynamics 1481 Substitute numerical values and evaluate ∆Su:

⎡

3

⎣

273 K

333.5 × 10 ∆S u = (0.05 kg ) ⎢− +

J/kg

⎛

⎞

3

⎝ 273 K ⎠

263 K

263 K 333.5 × 10 ⎟⎟ + + (2100 J/kg ⋅ K )ln⎜⎜

J/kg

(2100 J/kg ⋅ K )(273 K − 263 K )⎤ 263 K

⎥ ⎦

= 2.40 J/K and, because ∆Su > 0, the entropy of the universe increases. 48 • Picture the Problem We can use the definition of entropy change and the first law of thermodynamics to express ∆S for the ideal gas as a function of its initial and final volumes.

∆Q T

(a) Use its definition to express the entropy change of the gas:

∆S =

Apply the first law of thermodynamics to the isothermal process:

⎡ ⎛V ∆Q = ∆Eint − Won = − ⎢− nRT ln⎜⎜ f ⎝ Vi ⎣

Substitute to obtain:

⎞⎤ ⎟⎟⎥ ⎠⎦

because ∆Eint = 0 for an isothermal process.

⎛V ∆S = nR ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ ⎠

⎛ 80 L ⎞ ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )ln⎜⎜ ⎝ 40 L ⎠ = 11.5 J/K (b) Because the process is reversible:

∆S u = 0

Remarks: The entropy change of the environment of the gas is −11.5 J/K. 49 • Picture the Problem We can use the definition of entropy change and the 1st law of thermodynamics to express ∆S for the ideal gas as a function of its initial and final volumes.

1482 Chapter 19 ∆Q T

(a) Use its definition to express the entropy change of the gas:

∆S =

Apply the first law of thermodynamics to the isothermal process:

⎡ ⎛V ∆Q = ∆Eint − Won = − ⎢− nRT ln⎜⎜ f ⎝ Vi ⎣

Substitute to obtain:

⎞⎤ ⎟⎟⎥ ⎠⎦

because ∆Eint = 0 for an isothermal process.

⎛V ∆S = nR ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ ⎠

⎛ 80 L ⎞ ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )ln⎜⎜ ⎝ 40 L ⎠ = 11.5 J/K (b) Because the process is not quasi-static, it is non-reversible:

∆S u > 0

50 • Picture the Problem We can use the definition of entropy change to find the change in entropy of the water as it changes to steam. Apply the definition of entropy change to obtain: Substitute numerical values and evaluate ∆S:

∆S =

∆Q mLv = T T

∆S =

(1kg )(2.26 MJ/kg ) = 373 K

6.06 kJ/K

51 • Picture the Problem We can use the definition of entropy change to find the change in entropy of the ice as it melts. Apply the definition of entropy change to obtain: Substitute numerical values and evaluate ∆S:

∆S =

∆Q mLf = T T

∆S =

(1kg )(333.5 kJ/kg ) = 273 K

1.22 kJ/K

52 •• Picture the Problem We can use the first law of thermodynamics to find the change in the internal energy of the system and the change in the entropy of the system from the

The Second Law of Thermodynamics 1483 change in entropy of the hot- and cold-reservoirs. (a) Apply the 1st law of thermodynamics to find the change in the internal energy of the system:

∆Eint = Qin + Won

(b) Express the change in entropy of the system as the sum of the entropy changes of the high- and lowtemperature reservoirs:

∆S = ∆S h − ∆Sc =

(c) Because the process is reversible:

∆S u = 0

(d) Because Ssystem is a state function:

∆Eint = 50 J , ∆S = 0.167 J/K ,

= (200 J − 100 J ) − 50 J = 50 J

=

Qh Qc − Th Tc

200 J 100 J − = 0.167 J/K 300 K 200 K

and

∆S u > 0 *53 •• Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. Because the entropy change for the complete cycle is the sum of the entropy changes for each process, we can find the temperature T from the entropy changes during the 1st two processes and the heat rejected during the third. (a) Because S is a state function of the system:

∆S complete cycle = 0

(b) Relate the entropy change for the complete cycle to the entropy change for each process:

Q1 Q2 Q3 + + =0 T1 T2 T

Substitute numerical values to obtain:

300 J 200 J − 400 J + + =0 300 K 400 K T

Solve for T:

T = 267 K

54 •• Picture the Problem We can use the definition of entropy change and the 1st law of

1484 Chapter 19 thermodynamics to express ∆S for the ideal gas as a function of its initial and final volumes.

∆Q T

(a) Use its definition to express the entropy change of the gas:

∆S =

Apply the first law of thermodynamics to the isothermal process:

⎡ ⎛V ∆Qin = ∆Eint − Won = − ⎢− nRT ln⎜⎜ f ⎝ Vi ⎣

Substitute to obtain:

⎞⎤ ⎟⎟⎥ ⎠⎦

because ∆Eint = 0 for free expansion.

⎛V ∆S = nR ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ ⎠

⎛ 80 L ⎞ ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )ln⎜⎜ 40 L ⎝ ⎠ = 11.5 J/K (b) Because the process is irreversible, Su > 0 and, because no heat is exchanged:

∆S u = 11.5 J/K

55 •• Picture the Problem Because the ice gains heat as it melts, its entropy change is positive and can be calculated from its definition. Because the temperature of the lake is just slightly greater than 0°C and the mass of water is so much greater than that of the block of ice, the absolute value of the entropy change of the lake will be approximately equal to the entropy change of the ice as it melts.

mLf (200 kg )(333.5 kJ/kg ) = T 273 K

(a) Use the definition of entropy change to find the entropy change of the ice:

∆Sice =

(b) Relate the entropy change of the lake to the entropy change of the ice:

∆S lake ≈ − ∆Sice = − 244 kJ/K

= 244 kJ/K

(c) Because the temperature of the lake is slightly greater than that of the ice, the magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy change of the universe is greater than zero. The melting of the ice is an irreversible process and ∆S u > 0 .

The Second Law of Thermodynamics 1485 56 •• Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a melting process and for constant-pressure processes to find the entropy change of the universe, i.e., the piece of ice and the water in the insulated container. (a) Apply conservation of energy to obtain:

Qlost = Qgained or

Qcooling water = Qmelting ice + Qwarming water

Substitute to relate the masses of the ice and water to their temperatures, specific heats, and the final temperature of the water:

(100 g )(1cal/g ⋅ C°)(100°C − t ) = (100 g )(79.7 cal/g) + (100 g )(1cal/g ⋅ C°)(t ) Solve for t to obtain:

(b) Express the entropy change of the universe: Using the expression for the entropy change for a constant-pressure process, express the entropy change of the melting ice and warming icewater:

t = 10.2°C

∆S u = ∆Sice + ∆S water

∆Sice = ∆S melting ice + ∆S warming water =

⎛T mLf + mcP ln⎜⎜ f Tf ⎝ Ti

⎞ ⎟⎟ ⎠

Substitute numerical values to obtain:

∆Sice =

(0.1kg )(333.5 kJ/kg ) + (0.1kg )(4.184 kJ/kg ⋅ K )ln⎛⎜ 283.2 K ⎞⎟ = 138 J/K ⎜ 273 K ⎟ ⎝ ⎠

273 K

Find the entropy change of the cooling water:

⎛ 283.2 K ⎞ ⎟⎟ = −115 J/K ∆S water = (0.1kg )(4.18 kJ/kg ⋅ K )ln⎜⎜ ⎝ 373 K ⎠ Substitute for ∆Sice and ∆Swater and evaluate the entropy change of the universe:

∆S u = 138 J/K − 115 J/K = 23.0 J/K

1486 Chapter 19 Remarks: The result that ∆Su > 0 tells us that this process is irreversible. *57 •• Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe. (a) Using the equation for the entropy change during a constantpressure process, express the entropy change of the copper block: Apply conservation of energy to obtain:

⎛T ∆S Cu = mCu cCu ln⎜⎜ f ⎝ Ti

⎞ ⎟⎟ ⎠

(1)

Qlost = Qgained or

Qcopper block = Qwarming water Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature Tf of the water:

(1kg )(0.386 kJ/kg ⋅ K )(373.15 K − Tf ) = (4 L )(1kg/L )(4.184 kJ/kg ⋅ K )(Tf − 273.15 K ) Tf = 275.40 K

Solve for Tf:

Substitute numerical values in equation (1) and evaluate ∆SCu:

⎛ 275.40 K ⎞ ⎟⎟ = − 117 J/K ∆S Cu = (1 kg )(0.386 kJ/kg ⋅ K )ln⎜⎜ ⎝ 373.15 K ⎠ (b) Express the entropy change of the water:

⎛T ∆S water = mwater c water ln⎜⎜ f ⎝ Ti

⎞ ⎟⎟ ⎠

Substitute numerical values and evaluate ∆Swater:

⎛ 275.40 K ⎞ ⎟⎟ = 137 J/K ∆S water = (4 kg )(4.184 kJ/kg ⋅ K ) ln⎜⎜ ⎝ 273.15 K ⎠

The Second Law of Thermodynamics 1487 (c) Substitute for ∆SCu and ∆Swater and evaluate the entropy change of the universe:

∆S u = ∆S Cu + ∆S water = −117 J/K + 137 J/K = 20.3 J/K

Remarks: The result that ∆Su > 0 tells us that this process is irreversible. 58 •• Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead, the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10°C. We can apply the equation for the entropy change during a constant pressure process to find the entropy changes of the piece of lead, the water in the lake, and the universe. Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake:

∆S u = ∆S Pb + ∆S w

Using the equation for the entropy change during a constant-pressure process, express and evaluate the entropy change of the lead:

⎛T ∆S Pb = mPb cPb ln⎜⎜ f ⎝ Ti

⎞ ⎛ 283.15 K ⎞ ⎟⎟ = (2 kg )(0.128 kJ/kg ⋅ K ) ln⎜⎜ ⎟⎟ = −70.66 J/K ⎝ 373.15 K ⎠ ⎠

Find the entropy change of the water in the lake:

∆S w = =

Qw QPb mPb cPb ∆TPb = = Tw Tw Tw

(2 kg )(0.128 kJ/kg ⋅ K )(90 K ) 283.15 K

= 81.37 J/K Substitute and evaluate ∆Su:

∆S u = −70.66 J/K + 81.37 J/K = 10.7 J/K

59 •• Picture the Problem Because the air temperature will not change appreciably as a result of this crash; we can assume that the kinetic energy of the car is transformed into heat at a temperature of 20°C. We can use the definition of entropy change to find the entropy change of the universe.

1488 Chapter 19 Express the entropy change of the universe as a consequence of the kinetic energy of the car being transformed into heat:

Q 12 mv 2 ∆S u = = T T

Substitute numerical values and evaluate ∆Su:

(1500 kg )⎛⎜⎜100 km × 1h ⎞⎟⎟ h 3600 s ⎠ ⎝ ∆S u = 293.15 K

2

1 2

= 1.97 kJ/K *60 •• Picture the Problem The total change in entropy resulting from the mixing of these gases is the sum of the changes in their entropies. (a) Express the total change in entropy resulting from the mixing of the gases:

∆S = ∆S A + ∆S B

Express the change in entropy of each of the gases:

⎛V ⎞ ∆S A = nR ln⎜⎜ fA ⎟⎟ ⎝ ViA ⎠ and

⎛V ⎞ ∆S B = nR ln⎜⎜ fB ⎟⎟ ⎝ ViB ⎠ Because the initial and final volumes of the gases are the same and both volumes double: Substitute numerical values and evaluate ∆S:

⎛V ∆S = 2nR ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ = 2nR ln (2 ) ⎠

∆S = 2(1mol)(8.314 J/mol ⋅ K )ln (2) = 11.5 J/K

Because the gas molecules are indistinguishable, the entropy doesn' t change. (b) A complete description of this phonomenon has been derived using quantum mechanics.

Entropy and Work Lost *61 •• Picture the Problem We can find the entropy change of the universe from the entropy changes of the high- and low-temperature reservoirs. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs.

The Second Law of Thermodynamics 1489 (a) Express the entropy change of the universe:

∆S u = ∆S h + ∆Sc = −

Q Q + Th Tc

⎛1 1⎞ = −Q⎜⎜ − ⎟⎟ ⎝ Th Tc ⎠ Substitute numerical values and evaluate ∆Su:

⎛ 1 1 ⎞ ⎟⎟ ∆S u = (− 500 J ) ⎜⎜ − ⎝ 400 K 300 K ⎠ = 0.417 J/K

(b) Express the heat that could have been converted into work in terms of the maximum efficiency of an engine operating between the two reservoirs:

W = ε max Qh

Tc Th

Express the maximum efficiency of an engine operating between the two reservoir temperatures:

ε max = ε C = 1 −

Substitute and evaluate W:

⎛ T ⎞ ⎛ 300 K ⎞ ⎟⎟ (500 J ) W = ⎜⎜1 − c ⎟⎟Qh = ⎜⎜1 − ⎝ 400 K ⎠ ⎝ Th ⎠ = 125 J

62 •• Picture the Problem Although in the adiabatic free expansion no heat is lost by the gas, the process is irreversible and the entropy of the gas increases. In the isothermal reversible process that returns the gas to its original state, the gas releases heat to the surroundings. However, because the process is reversible, the entropy change of the universe is zero. Consequently, the net entropy change is the negative of that of the gas in the isothermal compression. (a) Relate the entropy change of the universe to the entropy change of the gas during the isothermal compression:

⎛V ∆S u = − ∆S gas = −nR ln⎜⎜ f ⎝ Vi

⎞ ⎟⎟ ⎠

1490 Chapter 19 Substitute numerical values and evaluate ∆Su:

⎛ 12.3 L ⎞ ⎟⎟ = 5.76 J/K ∆S u = −(1 mol)(8.314 J/mol ⋅ K )ln⎜⎜ 24 . 6 L ⎝ ⎠

To the extent that the initial expansion was isothermal and (b) reversible, no work was done and none was wasted in the cycle. (c) Express the wasted work in terms of T and the entropy change of the universe:

Wlost = T∆S u = (300 K )(5.76 J/K ) = 1.73 kJ

General Problems 63 • Picture the Problem We can use the definition of power to find the work done each cycle and the definition of efficiency to find the heat that is absorbed each cycle. Application of the first law of thermodynamics will yield the heat given off each cycle. (a) Use the definition of power to relate the work done in each cycle to the period of each cycle:

Wcycle = P∆t = (200 W )(0.1s ) = 20.0J Wcycle

20 J = 66.7 J 0.3

(b) Express the heat absorbed in each cycle in terms of the work done and the efficiency of the engine:

Qh, cycle =

Apply the 1st law of thermodynamics to find the heat given off in each cycle:

Qc, cycle = Qh, cycle − W = 66.7 J − 20 J

ε

=

= 46.7 J

64 • Picture the Problem We can use their definitions to find the efficiency of the engine and that of a Carnot engine operating between the same reservoirs. (a) Apply the definition of efficiency:

ε=

Q W 125 J =1− c =1− = 16.7% Qh Qh 150 J

The Second Law of Thermodynamics 1491 (b) Find the efficiency of a Carnot engine operating between the same reservoirs: Express the ratio of the two efficiencies:

εC = 1 −

Tc 293.15 =1− = 21.4% Th 373.15

ε 16.7% = = 0.780 ε C 21.4%

65 • Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat exhausted in each cycle. (a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs:

(b) Use the definition of efficiency to find the work done in each cycle: (c) Apply the first law of thermodynamics to the cycle to obtain:

⎛

ε = 0.85ε C = 0.85⎜⎜1 − ⎝

Tc ⎞ ⎟ Th ⎟⎠

⎛ 200 K ⎞ ⎟⎟ = 51.0% = 0.85⎜⎜1 − ⎝ 500 K ⎠ W = εQh = 0.51(200 kJ ) = 102 kJ

Qc, cycle = Qh, cycle − W = 200 kJ − 102 kJ = 98.0 kJ

*66 •• Picture the Problem We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1 GW of power and, from this value, the power that is wasted. The rate at which heat is being delivered to the river is related to the requisite flow rate of the river by dQ dt = c∆Tρ dV dt .

Tc Th

(a) Express the Carnot efficiency of a plant operating between temperatures Tc and Th:

ε max = ε C = 1 −

Substitute numerical values and evaluate εC:

ε max = 1 −

298 K = 0.404 500 K

(c) Find the power that must be supplied, at 40.4% efficiency, to produce an output of 1 GW:

Psupplied =

Poutput

ε max

=

1 GW = 2.48 GW 0.404

1492 Chapter 19 (b) Relate the wasted power to the power generated and the power supplied:

Pwasted = Psupplied − Pgenerated

Substitute numerical values and evaluate Pwasted:

Pwasted = 2.48 GW − 1 GW = 1.48 GW

(d) Express the rate at which heat is being dumped into the river:

dQ dm d = c∆T = c∆T (ρV ) dt dt dt dV = c∆Tρ dt

Solve for the flow rate dV/dt of the river:

dV dQ dt = dt c∆Tρ

Substitute numerical values (see Table 19-1 for the specific heat of water) and evaluate dV/dt:

dV 1.48 × 109 J/s = dt (4180 J/kg )(0.5 K ) (103 kg/m 3 ) = 708 m 3 / s = 7.08 × 105 L/s

67 • Picture the Problem We can find the rate at which the house contributes to the increase in the entropy of the universe from the ratio of ∆S to ∆t. Using the definition of entropy change, express the rate of increase in the entropy of the universe: Substitute numerical values and evaluate ∆S/∆t:

∆S ∆Q T ∆Q ∆t = = T ∆t ∆t

∆S 30 kW = = 113 W/K ∆t 266 K

68 •• Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle, its efficiency is that of a Carnot engine operating between the temperatures of its isotherms. Express the Carnot efficiency of the cycle:

εC = 1−

Tc Th

Substitute numerical values and evaluate εC:

εC = 1−

300 K = 60.0% 750 K

The Second Law of Thermodynamics 1493 69 •• Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract amount of work from 1 kJ to find the energy that is lost. In part (b) we can use its definition to find the change in entropy for each process. (a) For process (2): Find the efficiency of a Carnot engine operating between 400 K and 300 K: Substitute to obtain:

W2,max = Wrecovered = ε CQin

εC = 1−

Tc 300 K = 1− = 0.25 Th 400 K

Wrecovered = 0.25(1 kJ ) = 250 J or 750 J are lost.

Process (1) is more wasteful of mechanical energy. Process (2) is more wasteful of total energy.

∆Q 500 J = = 1.67 J/K T 300 K

(b) Find the change in entropy of the universe for process (1):

∆S1 =

Express the change in entropy of the universe for process (2):

∆S 2 = ∆S h + ∆Sc = −

∆Q ∆Q + Th Tc

⎛1 1⎞ = ∆Q ⎜⎜ − ⎟⎟ ⎝ Tc Th ⎠ Substitute numerical values and evaluate ∆S2:

⎛ 1 1 ⎞ ⎟⎟ ∆S 2 = (1 kJ )⎜⎜ − ⎝ 300 K 400 K ⎠ = 0.833 J/K

70 •• Picture the Problem Denote the three states of the gas as 1, 2, and 3 with 1 being the initial state. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures, volumes, and pressures at points 1, 2, and 3. To find the work done during each cycle, we can use the equations for the work done during isothermal, isobaric, and adiabatic processes. Finally, we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion.

1494 Chapter 19 V1 1L = (16 atm ) = 4 atm V2 4L

(a) Apply the ideal-gas law to the isothermal expansion 1→2 to find P2:

P2 = P1

Apply the equation of state for an adiabatic process to relate the pressures and volumes at 1 and 3:

P1V1γ = P3V3γ and 1γ

⎛P⎞ V3 = V1 ⎜⎜ 1 ⎟⎟ ⎝ P3 ⎠ = 2.29 L

1 1.67

⎛ 16 atm ⎞ ⎟⎟ = (1L )⎜⎜ 4 atm ⎝ ⎠

The PV diagram is shown to the right:

(b) From (a) we have:

V3 = 2.29 L

Apply the equation of state for an adiabatic process (γ =1.67) to relate the temperatures and volumes at 1 and 3:

T3V3γ −1 = T1V1γ −1 and

⎛V ⎞ T3 = T1 ⎜⎜ 1 ⎟⎟ ⎝ V3 ⎠

γ −1

1.67−1

⎛ 1L ⎞ ⎟⎟ = (600 K )⎜⎜ ⎝ 2.29 L ⎠

= 344 K (c) Express the work done each cycle: For the process 1→2:

W = W1→2 + W2→3 + W3→1

⎛V ⎞ ⎛V ⎞ W1→2 = nRT1 ln⎜⎜ 2 ⎟⎟ = P1V1 ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ ⎝ V1 ⎠ ⎛ 4L ⎞ ⎟⎟ = (16 atm )(1 L )ln⎜⎜ ⎝ 1L ⎠ = 22.2 atm ⋅ L

For the process 2→3:

W2 →3 = P2 ∆V2 →3

= (4 atm )(2.29 L − 4 L ) = −6.84 atm ⋅ L

(1)

The Second Law of Thermodynamics 1495 For the process 3→1:

W3→1 = −CV ∆T3→1 = − 32 nR(T1 − T3 ) = − 32 (P1V1 − P3V3 )

= − 32 [(16 atm )(1L ) − (4 atm )(2.29 L )] = −10.3 atm ⋅ L Substitute numerical values in equation (1) and evaluate W:

W = 22.2 atm ⋅ L − 6.84 atm ⋅ L − 10.3 atm ⋅ L = 5.06 atm ⋅ L

(d) Using its definition, express and evaluate the efficiency of the cycle:

ε= =

W W W = = Qin Q1→2 W1→2 5.06 atm ⋅ L = 22.8% 22.2 atm ⋅ L

*71 •• Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and, given that the efficiency of the heat engine is half that of a Carnot engine, relate Tc to the work done by and the heat input to the real heat engine. Using its definition, relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir: Solve for Tc:

εC = 1−

Tc Th

Tc = Th (1 − ε C ) W 2W = 12 ε C or ε C = Qin Qin

Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures:

ε=

Substitute to obtain:

⎛ 2W Tc = Th ⎜⎜1 − ⎝ Qin

The work done by the gas in expanding the balloon is:

⎞ ⎟⎟ ⎠

W = P∆V = (1atm )(4 L ) = 4 atm ⋅ L

1496 Chapter 19

Substitute numerical values and evaluate Tc:

⎛ 101.325 J ⎞ ⎞ ⎛ ⎜ 2 ⎜ 4 atm ⋅ L × ⎟⎟ atm ⋅ L ⎠ ⎟ ⎝ ⎜ Tc = (393.15 K ) 1 − = 313 K ⎜ ⎟ 4 kJ ⎜ ⎟ ⎝ ⎠ 72 •• Picture the Problem We can use the definitions of the COP and εC to show that their relationship is COP = Tc / (εCTh). Using the definition of the COP, relate the heat removed from the cold reservoir to the work done each cycle: Apply energy conservation to relate Qc, Qh, and W:

COP =

Qc W

Qc = Qh − W Qh − W W

Substitute to obtain:

COP =

Divide numerator and denominator by Qh and simplify to obtain:

W Q −W Qh COP = h = W W Qh

Because ε C = W Qh :

⎛ T ⎞ Tc 1 − ⎜⎜1 − c ⎟⎟ Th ⎠ Th 1− εC COP = = = ⎝

1−

εC

=

εC

εC

Tc ε CTh

73 •• Picture the Problem We can use the definition of the COP to express the work the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer.

The Second Law of Thermodynamics 1497 Qc W

Using the definition of the COP, relate the heat that must be removed from the freezer to the work done by the motor:

COP =

Solve for W:

W =

Qc COP

Differentiate this expression with respect to time to express the power of the motor:

P=

dW dQc dt = dt COP

Express the maximum COP of the motor:

COPmax =

Substitute to obtain:

P=

Substitute numerical values and evaluate P:

⎛ 50 K ⎞ ⎟⎟ = 10.0 W P = (50 W )⎜⎜ 250 K ⎝ ⎠

Tc ∆T

dQc ∆T dt Tc

74 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric, adiabatic, and isothermal processes of the cycle. (a) Apply the ideal-gas law to find the volume of the gas at A:

VA = =

nRTA PA

(2 mol)(8.314 J/mol ⋅ K )(600 K ) 5 atm ×

101.325 kPa atm

= 19.7 L (b) We’re given that:

VB = 2VA = 2(19.7 L ) = 39.4 L

Apply the ideal-gas law to this isobaric process to obtain:

TB = TA

2V VB = (600 K ) A = 1200 K VA VA

1498 Chapter 19 (c) Because the process C→A is isothermal:

TC = TA = 600 K

(d) Apply the equation of state for an adiabatic process (γ = 1.4) to find the volume of the gas at C:

TBVBγ −1 = TCVCγ −1 and 1

1

⎛ T ⎞ γ −1 ⎛ 1200 K ⎞ 1.4−1 ⎟⎟ VC = VB ⎜⎜ B ⎟⎟ = (39.4 L )⎜⎜ ⎝ 600 K ⎠ ⎝ TC ⎠ = 223 L WA − B = PA (VB − VA ) = PA (2VA − VA )

(e) Express and evaluate the work done by the gas during the isobaric process AB:

= PAVA = (5 atm )(19.7 L ) = 98.50 atm ⋅ L ×

101.325 J atm ⋅ L

= 9.98 kJ Won, B−C = ∆Eint,B−C − Qin, B−C = ∆Eint, B−C − 0

Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC:

= ∆Eint, B−C = −ncV ∆TB−C = − 52 nR∆TB−C

Substitute numerical values and evaluate WB−C:

WB−C = − 52 (2 mol)(8.314 J/mol ⋅ K )(600 K − 1200 K ) = 24.9 kJ The work done by the gas during the isothermal compression CA is:

⎛V ⎞ ⎛ 19.7 L ⎞ ⎟⎟ = − 24.2 kJ WC−A = nRTC ln⎜⎜ A ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )(600 K )ln⎜⎜ V 223 L ⎝ ⎠ C ⎝ ⎠ (f) The heat absorbed during the isobaric expansion AB is:

QA−B = ncP ∆TA−B = 72 nR∆TA−B =

7 2

(2 mol)(8.314 J/mol ⋅ K )(1200 K − 600 K )

= 34.9 kJ The heat absorbed during the adiabatic expansion BC is:

QB−C = 0

The Second Law of Thermodynamics 1499 Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA:

QC−A = WC−A + ∆Eint, C−A = WC−A = − 24.2 kJ because ∆Eint,C−A = 0 for an isothermal process.

(g) The thermodynamic efficiency ε is:

ε=

W WA−B + WB−C + WC−A = Qin QA−B

Substitute numerical values and evaluate ε:

ε=

9.98 kJ + 24.9 kJ − 24.2 kJ 34.9 kJ

= 30.6% 75 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB:

PB = PA

VA V = (5 atm ) A VB 2VA

= 2.50 atm ×

101.325 kPa 1atm

= 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC:

TC = TB

Using the ideal-gas law, find the volume at B:

VB = =

PCVC PBVB

nRTB PB

(2 mol)(8.314 J/mol ⋅ K )(600 K ) 253 kPa

= 39.43 L Use the equation of state for an adiabatic process and γ = 1.4 to find the volume occupied by the gas at C:

1γ

⎛P ⎞ VC = VB ⎜⎜ B ⎟⎟ ⎝ PC ⎠ = 75.87 L

1 1.4

⎛ 2.5 atm ⎞ ⎟⎟ = (39.43 L )⎜⎜ ⎝ 1atm ⎠

1500 Chapter 19 Substitute and evaluate TC:

TC = (600 K )

(1atm )(75.87 L ) (2.5 atm )(39.43 L )

= 462 K (c) Express the work done by the gas in one cycle:

W = WA−B + WB−C + WC−D + WD−A

The work done during the isothermal expansion AB is:

⎛ 2V ⎛V ⎞ WA − B = nRTA ln⎜⎜ B ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )(600 K ) ln⎜⎜ A ⎝ VA ⎝ VA ⎠

⎞ ⎟ = 6.915 kJ ⎟ ⎠

The work done during the adiabatic expansion BC is:

WB−C = −CV ∆TB−C = − 52 nR∆TB−C = − 52 (2 mol)(8.314 J/mol ⋅ K )(462 K − 600 K ) = 5.737 kJ The work done during the isobaric compression CD is:

WC−D = PC (VD − VC ) = (1atm )(19.7 L − 75.87 L ) = −56.17 atm ⋅ L ×

101.325 J atm ⋅ L

= −5.690 kJ Express and evaluate the work done during the constant-volume process DA: Substitute numerical values and evaluate W:

Using its definition, express the thermodynamic efficiency of the cycle:

WD−A = 0

W = 6.915 kJ + 5.737 kJ − 5.690 kJ + 0 = 6.96 kJ

ε=

W W = Qin QA−B + QD−A

(1)

QA−B = WA−B + ∆Eint, A−B = WA−B = 6.915 kJ

Express and evaluate the heat entering the system during the isothermal process AB:

Because ∆Eint = 0 for an isothermal process.

Express the heat entering the system

QD−A = CV ∆TD−A = 52 nR∆TD−A

The Second Law of Thermodynamics 1501 during the constant-volume process DA: Apply the ideal-gas law to the constant-volume process DA to obtain:

TD = TA

PD 1atm = (600 K ) = 120 K PA 5 atm

The heat entering the system during the process DA is:

QD−A =

5 2

(2 mol)(8.314 J/mol ⋅ K )(600 K − 120 K ) = 20.0 kJ

Substitute numerical values in equation (1) and evaluate the efficiency of the cycle:

ε=

6.975 kJ = 25.9% 6.915 kJ + 20.0 kJ

76 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric, adiabatic, and isothermal processes of the cycle. (a) Apply the ideal-gas law to find the volume of the gas at A:

VA = =

nRTA PA

(2 mol)(8.314 J/mol ⋅ K )(600 K ) 5 atm ×

101.325 kPa atm

= 19.7 L (b) We’re given that:

VB = 2VA = 2(19.7 L ) = 39.4 L

Apply the ideal-gas law to this isobaric process to obtain:

TB = TA

(c) Because the process CA is isothermal:

TC = TA = 600 K

(d) Apply the equation of state for an adiabatic process (γ = 5/3) to find the volume of the gas at C:

TBVBγ −1 = TCVCγ −1 and

VB 2V = (600 K ) A = 1200 K VA VA

1502 Chapter 19 1

3

⎛ TB ⎞ γ −1 ⎛ 1200 K ⎞ 2 ⎟⎟ VC = VB ⎜⎜ ⎟⎟ = (39.4 L )⎜⎜ ⎝ 600 K ⎠ ⎝ TC ⎠ = 111L WA − B = PA (VB − VA ) = PA (2VA − VA )

(e) Express and evaluate the work done by the gas during the isobaric process AB:

= PAVA = (5 atm )(19.7 L ) = 98.50 atm ⋅ L ×

101.325 J atm ⋅ L

= 9.98 kJ

Won, B−C = ∆Eint, B−C − Qin, B−C

Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC:

= ∆Eint, B−C − 0

= ∆Eint, B−C = −(ncV ∆TB−C ) = − 32 nR∆TB−C

Substitute numerical values and evaluate WB−C:

Won, B−C = − 32 (2 mol)(8.314 J/mol ⋅ K )(600 K − 1200 K ) = 14.9 kJ The work done by the gas during the isothermal compression CA is:

⎛V ⎞ ⎛ 19.7 L ⎞ ⎟⎟ WC−A = nRTC ln⎜⎜ A ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )(600 K )ln⎜⎜ ⎝ 111 L ⎠ ⎝ VC ⎠ = − 17.2 kJ (f) The heat absorbed during the isobaric expansion AB is:

Qin, A−B = ncP ∆TA−B = 52 nR∆TA−B =

5 2

(2 mol)(8.314 J/mol ⋅ K )(1200 K − 600 K )

= 24.9 kJ Express and evaluate the heat absorbed during the adiabatic expansion BC:

QB−C = 0

The Second Law of Thermodynamics 1503

Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA:

QC− A = WC− A + ∆Eint, C−A = WC−A = − 17.2 kJ because ∆Eint = 0 for an isothermal process.

(g) The definition of thermodynamic efficiency ε is:

ε=

W WA−B + WB−C + WC−A = Qin QA−B

Substitute numerical values and evaluate ε:

ε=

9.98 kJ + 14.9 kJ − 17.2 kJ 24.9 kJ

= 30.8% 77 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB:

PB = PA

(b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC:

TC = TB

Using the ideal-gas law, find the volume at B:

VB =

VA V = (5 atm ) A VB 2VA

= 2.50 atm ×

=

101.3 kPa = 253 kPa 1atm

PCVC PBVB

nRTB PB

(2 mol)(8.314 J/mol ⋅ K )(600 K ) 253 kPa

= 39.43 L Use the equation of state for an adiabatic process and γ = 5/3 to find the volume occupied by the gas at C:

1γ

⎛P ⎞ VC = VB ⎜⎜ B ⎟⎟ ⎝ PC ⎠ = 68.33 L

⎛ 2.5 atm ⎞ ⎟⎟ = (39.43 L )⎜⎜ ⎝ 1atm ⎠

35

1504 Chapter 19 Substitute and evaluate TC:

TC = (600 K )

(1atm )(68.33 L ) (2.5 atm )(39.43 L )

= 416 K (c) Express the work done by the gas in one cycle:

W = WA−B + WB−C + WC−D + WD−A

(1)

The work done during the isothermal expansion AB is:

⎛ 2V ⎞ ⎛V ⎞ WA−B = nRTA ln⎜⎜ B ⎟⎟ = (2 mol)(8.314 J/mol ⋅ K )(600 K )ln⎜⎜ A ⎟⎟ = 6.915 kJ ⎝ VA ⎠ ⎝ VA ⎠ The work done during the adiabatic expansion BC is:

WB−C = −CV ∆TB−C = − 52 nR∆TB−C = − 32 (2 mol)(8.314 J/mol ⋅ K )(416 K − 600 K ) = 4.589 kJ The work done during the isobaric compression CD is:

WC−D = PC (VD − VC ) = (1atm )(19.7 L − 68.33 L ) = −48.63 atm ⋅ L ×

101.3 J atm ⋅ L

= −4.926 kJ The work done during the constantvolume process DA is: Substitute numerical values in equation (1) to obtain:

The thermodynamic efficiency of the cycle is given by: The heat entering the system during the isothermal process AB is:

WD−A = 0

W = 6.915 kJ + 4.589 kJ − 4.926 kJ + 0 = 6.58 kJ

ε=

W W = Qin QA−B + QD−A

(2)

QA−B = WA−B + ∆Eint, A−B = WA−B = 6.915 kJ because ∆Eint = 0 for an isothermal process.

The heat entering the system during the constant-volume process DA is:

QD−A = CV ∆TD−A = 32 nR∆TD−A

The Second Law of Thermodynamics 1505 Apply the ideal-gas law to the constant-volume process DA to obtain:

TD = TA

PD 1atm = (600 K ) = 120 K PA 5 atm

The heat entering the system during the process DA is:

QD−A =

3 2

(2 mol)(8.314 J/mol ⋅ K )(600 K − 120 K ) = 12.0 kJ

Substitute numerical values in equation (2) and evaluate the efficiency of the cycle:

ε=

6.58 kJ = 34.8% 6.915 kJ + 12.0 kJ

78 •• Picture the Problem We can express the efficiency of the Otto cycle using the result from Example 19-2. We can apply the relation TV γ −1 = constant to the adiabatic processes of the Otto cycle to relate the end-point temperatures to the volumes occupied by the gas at these points and eliminate the temperatures at c and d. We can use the idealgas law to find the highest temperature of the gas during its cycle and use this temperature to express the efficiency of a Carnot engine. Finally, we can compare the efficiencies by examining their ratio. The efficiency of the Otto engine is given in Example 19-2:

εO = 1 −

Td − Ta Tc − Tb

(1)

where the subscripts refer to the various points of the cycle as shown in Figure 193. Apply the relation

TV γ −1 = constant to the adiabatic process a→b to obtain: Apply the relation

TV γ −1 = constant to the adiabatic process c→d to obtain: Subtract the first of these equations from the second to obtain:

⎛V ⎞ Tb = Ta ⎜⎜ a ⎟⎟ ⎝ Vb ⎠ ⎛V Tc = Td ⎜⎜ d ⎝ Vc

γ −1

⎞ ⎟⎟ ⎠

γ −1

⎛V Tc − Tb = Td ⎜⎜ d ⎝ Vc

⎞ ⎟⎟ ⎠

γ −1

⎛V ⎞ − Ta ⎜⎜ a ⎟⎟ ⎝ Vb ⎠

γ −1

1506 Chapter 19 In the Otto cycle, Va = Vd and Vc = Vb. Substitute to obtain:

⎛V ⎞ Tc − Tb = Td ⎜⎜ a ⎟⎟ ⎝ Vb ⎠

γ −1

⎛V ⎞ − Ta ⎜⎜ a ⎟⎟ ⎝ Vb ⎠

⎛V ⎞ = (Td − Ta )⎜⎜ a ⎟⎟ ⎝ Vb ⎠ Substitute in equation (1) and simplify to obtain:

Td − Ta

εO = 1 −

⎛ ⎞ (Td − Ta )⎜⎜ Va ⎟⎟ ⎝ Vb ⎠

⎛V ⎞ = 1 − ⎜⎜ b ⎟⎟ ⎝ Va ⎠

γ −1

=1−

γ −1

γ −1

γ −1

Ta Tb

Note that, while Ta is the lowest temperature of the cycle, Tb is not the highest temperature. Apply the ideal-gas law to c and b to obtain an expression for the cycle’s highest temperature Tc:

P Pc Pb = ⇒ Tc = Tb c > Tb Pb Tc Tb

Express the efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle:

εC = 1−

Express the ratio of the efficiency of a Carnot engine to the efficiency of an Otto engine operating between the same temperatures:

Ta εC Tc = > 1 because Tc > Tb. ε O 1 − Ta Tb

Ta Tc

1−

*79 ••• Picture the Problem We can use nR = CP − CV , γ = CP CV , and TV γ −1 = a constant to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V1,T1) to state (V2,T2) is zero. Express the entropy change for a general process that proceeds from state 1 to state 2:

⎛V ⎞ ⎛T ⎞ ∆S = CV ln⎜⎜ 2 ⎟⎟ + nR ln⎜⎜ 2 ⎟⎟ ⎝ V1 ⎠ ⎝ T1 ⎠

The Second Law of Thermodynamics 1507 For an adiabatic process:

T2 ⎛ V1 ⎞ =⎜ ⎟ T1 ⎜⎝ V2 ⎟⎠

γ −1

Substitute and simplify to obtain: γ −1 ⎡ ⎛ V1 ⎞ ⎤ ⎢ CV ln⎜⎜ ⎟⎟ ⎥ γ −1 ⎛ V1 ⎞ ⎛ V2 ⎞ ⎛ V2 ⎞ ⎢ ⎝ V2 ⎠ ⎥ ∆S = CV ln⎜⎜ ⎟⎟ + nR ln⎜⎜ ⎟⎟ = ln⎜⎜ ⎟⎟ ⎢nR + ⎥ V ⎝ V2 ⎠ ⎝ V1 ⎠ ⎝ V1 ⎠ ⎢ ln 2 ⎥ V1 ⎢⎣ ⎥⎦ ⎡ ⎛ ⎞⎤ (γ − 1)CV ln⎜⎜ V1 ⎟⎟ ⎥ ⎢ ⎛V ⎞ ⎝ V2 ⎠ ⎥ = ln⎛⎜ V2 ⎞⎟ [nR − (γ − 1)C ] = ln⎜⎜ 2 ⎟⎟ ⎢nR + V ⎜V ⎟ V1 ⎥ ⎝ 1⎠ ⎝ V1 ⎠ ⎢ ln − ⎥ ⎢ V2 ⎥⎦ ⎢⎣

Use the relationship between CP and CV to obtain: Substitute for nR and γ and simplify:

nR = CP − CV

⎞ ⎤ ⎛C ⎛ V ⎞⎡ ∆S = ln⎜⎜ 2 ⎟⎟ ⎢CP − CV − ⎜⎜ p − 1⎟⎟CV ⎥ ⎝ V1 ⎠ ⎣ ⎠ ⎦ ⎝ CV = 0

80 ••• Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal mount of heat is transferred to the hot reservoir and W = 0. The entropy change of the universe is then ∆Su = Qc/Th − Qc/Tc. Because Th > Tc, Su < 0, i.e., the entropy of the universe would decrease. (b) In this case, is heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir, i.e., Qc = 0, then the entropy change of the universe is ∆Su = −Qh/Th + 0, which is negative. Again, the entropy of the universe would decrease. (c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir, but these statements do not specify the minimum amount of heat rejected or work that must be done. The statement ∆Su ≥ 0 is more restrictive. The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to ∆Su ≥ 0.

1508 Chapter 19 81 ••• Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Express the net efficiency of the two engines connected in series:

ε net =

Express the efficiencies of engines 1 and 2:

ε1 =

W1 + W2 Qh

W1 Qh

and

ε2 =

W2 Qm

ε1Qh + ε 2Qm

Solve for W1 and W2 and substitute to obtain:

ε net =

Express the efficiency of engine 1 in terms of Qm and Qh:

ε1 = 1 −

Solve for Qm/ Qh:

Qm = 1 − ε1 Qh

Substitute to obtain:

Qh

= ε1 +

Qm ε2 Qh

Qm Qh

ε net = ε1 + (1 − ε1 )ε 2

*82 ••• Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Finally, we can substitute the expressions for the efficiencies of the ideal reversible engines to obtain ε net = 1 − Tc Th . Express the efficiencies of ideal reversible engines 1 and 2:

Tm Th

(1)

Tc Tm

(2)

W1 + W2 Qh

(3)

ε1 = 1 − and

ε2 = 1− Express the net efficiency of the two engines connected in series:

ε net =

The Second Law of Thermodynamics 1509 W1 W and ε 2 = 2 Qh Qm

Express the efficiencies of engines 1 and 2:

ε1 =

Solve for W1 and W2 and substitute in equation (3) to obtain:

ε net =

Express the efficiency of engine 1 in terms of Qm and Qh:

ε1 = 1 −

Solve for Qm/ Qh:

Qm = 1 − ε1 Qh

ε 1Qh + ε 2Qm Qh

= ε1 +

Qm ε2 Qh

Qm Qh

Substitute to obtain:

ε net = ε1 + (1 − ε1 )ε 2

Substitute for ε1 and ε2 and simplify to obtain:

ε net = 1 − =1−

Tm ⎛ Tm ⎞ ⎛ Tc ⎞ + ⎜ ⎟ ⎜1 − ⎟ Th ⎜⎝ Th ⎟⎠ ⎜⎝ Tm ⎟⎠ Tm Tm Tc T + − = 1− c Th Th Th Th

83 ••• Picture the Problem There are 26 letters and four punctuation marks (space, comma, period, and exclamation point) used in the English language, disregarding capitalization, so we have a grand total of 30 characters to choose from. This fragment is 330 characters (including spaces) long; there are then 30330 different possible arrangements of the character set to form a fragment this long. We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey’s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare. Assuming the monkeys type at random, express the probability P that one monkey will write out this passage:

P=

Use the approximation

P=

30 ≈ 1000 = 10

1.5

to obtain:

Assuming the monkeys can type at a rate of 1 character per second, it would take about 330 s to write a passage of length equal to the quotation from Shakespeare. Find the time T required for a million monkeys to type this particular passage by accident:

T=

1 30330

1

10

(1.5 )(330 )

=

1 = 10 −495 495 10

(330 s )(10495 ) 106

⎛ ⎞ 1y ⎟ = 3.30 × 10 491 s ⎜⎜ 7 ⎟ ⎝ 3.16 × 10 s ⎠

(

≈ 10484 y

)

1510 Chapter 19 Express the ratio of T to Russell’s estimate:

T TRussell

10484 y = = 10478 6 10 y

or

T ≈ 10 478 TRussell

Chapter 20 Thermal Properties and Processes Conceptual Problems *1 • Determine the Concept The glass bulb warms and expands first, before the mercury warms and expands. 2 • Determine the Concept The heating of the sheet causes the average separation of its molecules to increase. The consequence of this increased separation is that the area of the hole always increases. (b) is correct. 3 • Determine the Concept Actually, it can be hard boiled, but it does take quite a bit longer than at sea level. (c) is the best response. 4 • Determine the Concept Gases that cannot be liquefied by applying pressure at 20°C are those for which Tc < 293 K. These are He, Ar, Ne, H2, O2, NO. *5 •• (a) With increasing altitude, P decreases; from curve OF, T of the liquid-gas interface diminishes, so the boiling temperature decreases. Likewise, from curve OH, the melting temperature increases with increasing altitude. (b) Boiling at a lower temperature means that the cooking time will have to be increased.

6 • Picture the Problem We can apply the Stefan-Boltzmann law to relate the rate at which an object radiates thermal energy to its environment. Using the Stefan-Boltzmann law, relate the power radiated by a body to its temperature:

Pr = eσAT 4 where A is the surface area of the body, σ is Stefan’s constant, and e is the emissivity of the object.

Because P varies with the fourth power of T, tripling the temperature increases the rate at which it radiates by a factor of 34 and (d ) is correct.

1511

1512

Chapter 20

*7 • Determine the Concept The thermal conductivity of metal and marble is much greater than that of wood; consequently, heat transfer from the hand is more rapid. 8 • (a) True (b) True (c) False. The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. (d) False. Water contracts on heating between 0°C and 4°C. (e) True 9 • Determine the Concept Because atoms are few and far between in space, the earth can not lose heat by conduction or convection. Thermal energy is radiated through space in the form of electromagnetic waves that move at the speed of light. (c ) is correct. 10 • Determine the Concept Because there is little, if any, molecule-to-molecule transportation of energy into a fireplace-heated room, the mechanisms are radiation and convection. 11 • Determine the Concept In the absence of matter to support conduction and convection, radiation is the only mechanism. 12 •• Determine the Concept Because the amount of heat lost by the house is proportional to the difference between the house temperature and that of the outside air, the rate at which the house loses heat (that must be replaced by the furnace) is greater at night when the temperature of the house is kept high than when it is allowed to cool down. 13 •• Picture the Problem The rate at which heat is conducted through a cylinder is given by I = dQ / dt = kA∆T / ∆x where A is the cross-sectional area of the cylinder. Express the rate at which heat is conducted through cylinder A:

I A = k Aπ d A2

∆T ∆x

Thermal Properties and Processes 1513 Express the rate at which heat is conducted through cylinder B:

I B = k Bπ d B2

Equate these expressions to obtain:

kAπ d A2

∆T ∆x

∆T ∆T = kBπ d B2 ∆x ∆x

or

kA d A2 = kB d B2 Because dA = 2dB:

k A (2d B ) = kB d B2 2

and

4k A = k B ⇒ (a ) is correct. 14 • Determine the Concept Most objects of everyday experience are at temperatures near the mean temperature of the earth, about 300 K. Their blackbody spectrum therefore has a peak near λmax = 2.898 mm K/ 300 K ≈ 0.01 mm = 10 µm = 10,000 nm. These wavelengths are in the infrared region of the spectrum, so the heat which most objects radiate away can be detected most easily in the infrared, which is the spectral region where most night-vision goggles and other types of optical "heat detectors" operate. However, if the temperature of the object increases, the wavelength decreases; so the peak radiation can be found in any spectral region, not just the infrared. *15 • Determine the Concept The temperature of an object is inversely proportional to the maximum wavelength at which the object radiates (Wein’s displacement law). Because blue light has a shorter wavelength than red light, an object for which the wavelength of the peak of thermal emission is blue is hotter than one that is red.

Estimation and Approximation 16 ••• Picture the Problem We can express the heat current through the insulation in terms of the rate of evaporation of the liquid helium and in terms of the temperature gradient across the superinsulation. Equating these equations will allow us to solve for the thermal conductivity k of the superinsulation. Express the heat current in terms of the rate of evaporation of the liquid helium:

I = Lv

dm dt

Express the heat current in terms of the temperature gradient across the superinsulation and the conductivity of the superinsulation:

I = kA

∆T ∆x

1514 Chapter 20 Equate these expressions and solve for k:

k=

dm dt A∆T

Lv ∆x

dm dV =ρ dt dt

Using the definition of density, express the rate of loss of liquid helium: Substitute to obtain:

k=

dV dt A∆T

Lv ∆xρ

Express the ratio of the area of the spherical container to its volume:

A 4πr 2 = V 43 πr 3

Solve for A:

A = 3 36πV 2

Substitute to obtain:

dV dt k=3 2 36πV ∆T Lv ∆xρ

Substitute numerical values and evaluate k:

⎛

k=

m3 ⎞ ⎟ 86400 s ⎟⎠

(21kJ/kg )(7 ×10−2 m )(125 kg/m3 )⎜⎜ 0.7 ×10 3

(

36π 200 × 10 −3 m

⎝

−3

) (288 K )

3 2

= 3.13 × 10 −6 W/m ⋅ K

17 •• Picture the Problem We can use the thermal current equation for the thermal conductivity of the skin. Use the thermal current equation to express the rate of conduction of thermal energy: Solve for k to obtain:

Substitute numerical values and evaluate k:

I = kA

k=

k=

∆T ∆x

I ∆T A ∆x 130 W = 18.1 mW/m ⋅ K 4K 2 1.8 m 10 −3 m

(

)

Thermal Properties and Processes 1515 *18 •• Picture the Problem The amount of heat radiated by the earth must equal the solar flux from the sun, or else the temperature on earth would continually increase. The emissivity of the earth is related to the rate at which it radiates energy into space by the StefanBoltzmann law Pr = eσAT 4 . Using the Stefan-Boltzmann law, express the rate at which the earth radiates energy as a function of its emissivity e and temperature T:

Pr = eσA'T 4 where A′ is the surface area of the earth.

Solve for the emissivity of the earth:

e=

Pr σA'T 4

Use its definition to express the intensity of the radiation received by the earth:

I=

Pabsorbed A

where A is the cross-sectional area of the earth.

For 70% absorption of the sun’s radiation incident on the earth:

I=

0.7 Pr A

Substitute for Pr and A and simplify to obtain:

e=

0.7 AI 0.7π R 2 I 0.7 I = = 4 2 4 σAT 4π R σT 4σT 4

Substitute numerical values and evaluate e:

e=

0.7 1370 W/m 2 4 4 5.670 × 10−8 W/m 2 ⋅ K 4 (288 K )

(

(

) )

= 0.615 19 •• Picture the Problem The wavelength at which maximum power is radiated by the gas falling into a black hole is related to its temperature by Wien’s displacement law. Express Wien’s displacement law:

Substitute for T and evaluate λmax:

λmax =

2.898 mm ⋅ K T

λmax =

2.898 mm ⋅ K = 2.90 nm 106 K

Thermal Expansion 20 • Picture the Problem We can find the length of the ruler at 100°C by adding its elongation due to the increase in temperature to its length at 20°C. We can find its elongation using the definition of the coefficient of linear expansion α = (∆L L ) ∆T .

1516 Chapter 20 Express the length of the ruler at 100°C in terms of its length at 20°C, its coefficient of linear expansion, and the change in its temperature: Substitute numerical values and evaluate L100°C:

L100°C = L20°C + ∆L

= L20°C + αL20°C ∆T = L20°C (1 + α∆T )

[ (

)

= 30.026 cm

21 •• Picture the Problem We can let the definition of the coefficient of linear expansion α = (∆L L ) ∆T , with ∆A replacing ∆L and A replacing L suggest a definition of the coefficient of area expansion. (a) Letting γ represent the coefficient of area expansion we have: (b) For a square:

γ ≡

∆A A ∆T

(1)

∆A = [L(1 + α∆T )] − L2 2

( = A(2α∆T + α

)

= L2 1 + 2α∆T + α 2 ∆T 2 − L2 2

∆T 2

)

Divide both sides of the equation by A to obtain:

∆A = 2α∆T + α 2 ∆T 2 A

Substitute in equation (1) to obtain:

γ =

Let ∆T→0 to obtain:

γ ≈ 2α∆T

For a circle:

∆A = π [R(1 + α∆T )] − π R 2

2α∆T + α 2 ∆T 2 = 2α + α 2 ∆T ∆T

2

(

)

= π R 2 1 + 2α∆T + α 2 ∆T 2 − π R 2

(

= A 2α∆T + α ∆T Divide both sides of the equation by A to obtain: Substitute in equation (1) to obtain:

]

L100°C = (30 cm ) 1 + 11× 10−6 /K (80 K )

2

∆A = 2α∆T + α 2 ∆T 2 A

γ =

2

)

2α∆T + α 2 ∆T 2 = 2α + α 2 ∆T ∆T

Thermal Properties and Processes 1517 Let ∆T→0 to obtain:

γ ≈ 2α∆T

22 •• Picture the Problem While the mass of a sample of aluminum will remain constant with increasing temperature, its volume will increase due to thermal expansion. Consequently, its density will decrease with increasing temperature. We can use the definition of density (mass/unit volume) to express the density when its volume has increased by ∆V and the definition of the coefficient of volume expansion to relate ∆V to the increase in temperature ∆T. The relationship β = 3α will allow us to relate the coefficient of volume expansion to the coefficient of linear expansion. Express the density of aluminum ρ′ when its volume has changed by ∆V:

ρ' =

m mV = V + ∆V 1 + ∆V V

Using the definition of the coefficient of volume expansion, substitute for ∆V/V to obtain:

ρ' =

ρ ρ = 1 + β ∆T 1 + 3α∆T

Substitute numerical values and evaluate ρ′:

because β = 3α.

ρ' =

2.70 × 103 kg/m 3 1 + 3 24 × 10−6 /K (200 K )

(

)

= 2.66 × 103 kg/m 3 23 •• Picture the Problem Because the temperature of the steel shaft does not change, we need consider just the expansion of the copper collar. We can express the required temperature in terms of the initial temperature and the change in temperature that will produce the necessary increase in the diameter D of the copper collar. This increase in the diameter is related to the diameter at 20°C and the increase in temperature through the definition of the coefficient of linear expansion. Express the temperature to which the copper collar must be raised in terms of its initial temperature and the increase in its temperature: Apply the definition of the coefficient of linear expansion to express the change in temperature required for the collar to fit on the

T = Ti + ∆T

⎛ ∆D ⎞ ⎜ ⎟ D ⎠ ⎝ ∆T =

α

1518 Chapter 20 shaft:

∆D αD

Substitute to obtain:

T = Ti +

Substitute numerical values and evaluate T:

T = 293 K +

0.02 cm 17 × 10 −6 /K (5.98 cm )

(

)

= 490 K = 217°C *24 •• Picture the Problem Because the temperatures of both the steel shaft and the copper collar change together, we can find the temperature change required for the collar to fit the shaft by equating their diameters for a temperature increase ∆T. These diameters are related to their diameters at 20°C and the increase in temperature through the definition of the coefficient of linear expansion. Express the temperature to which the collar and the shaft must be raised in terms of their initial temperature and the increase in their temperature: Express the diameter of the steel shaft when its temperature has been increased by ∆T: Express the diameter of the copper collar when its temperature has been increased by ∆T: If the collar is to fit over the shaft when the temperature of both has been increased by ∆T: Solve for ∆T to obtain: Substitute in equation (1) to obtain:

T = Ti + ∆T

(1)

Dsteel = Dsteel,20°C (1 + α steel ∆T )

DCu = DCu,20°C (1 + α Cu ∆T )

DCu,20°C (1 + α Cu ∆T )

= Dsteel,20°C (1 + α steel ∆T )

∆T =

Dsteel,20°C − DCu,20°C DCu,20°Cα Cu − Dsteel,20°Cα steel

T = Ti +

Dsteel,20°C − DCu,20°C DCu,20°Cα Cu − Dsteel,20°Cα steel

Thermal Properties and Processes 1519 Substitute numerical values and evaluate T:

T = 293 K +

6.0000 cm − 5.9800 cm = 854 K = 581°C (5.98 cm) 17 ×10−6 /K − (6.00 cm ) 11×10−6 /K

(

)

(

)

25 •• Picture the Problem The linear expansion coefficient of the container is one-third its coefficient of volume expansion. We can relate the changes in volume of the mercury and the container to their initial volumes, temperature change, and coefficients of volume expansion, and, because we know the amount of spillage, obtain an equation that we can solve for βc. Relate the linear expansion coefficient of the container to its coefficient of volume expansion: Express the difference in the change in the volume of the mercury and the container in terms of the spillage: Express ∆VHg using the definition

α c = 13 β c

(1)

∆VHg − ∆Vc = 7.5 mL

∆VHg = β HgVHg ∆T

of the coefficient of volume expansion: Express ∆Vc using the definition of

∆Vc = β cVc ∆T

the coefficient of volume expansion: Substitute to obtain: Solve for βc:

β HgVHg ∆T − β cVc ∆T = 7.5 mL βc =

β HgVHg ∆T − 7.5 mL Vc ∆T

or, because V = VHg = Vc,

βc =

β HgV∆T − 7.5 mL

V∆T 7.5 mL = β Hg − V∆T

1520 Chapter 20 7.5 mL 3V∆T 7.5 mL = α Hg − 3V∆T

Substitute in equation (1) to obtain:

α c = 13 β Hg −

α c = 13 (0.18 × 10−3 / K ) −

Substitute numerical values and evaluate αc:

7.5 mL 3(1.4 L )(40 K )

= 15.4 × 10−6 K −1 26 •• Picture the Problem We can use dFe,168°C = dFe,20°C(1+αFe∆T) to find the diameter of the hole in the aluminum sheet at 168°C and then dAl,20°C = dAl,168°C(1−αAl∆T) to find the diameter of the hole when the sheet has cooled to room temperature. dFe,168°C = dFe,20°C(1+αFe∆T)

Relate the diameter of the hole/steel drill bit at 168°C to its diameter at 20° C:

Substitute numerical values and evaluate dFe,168°C:

[

]

d Fe,168°C = (6.245 cm ) 1 + 11× 10 −6 K −1 (148 K ) = 6.255 cm Express the diameter of the hole in the plate at 20° C:

d Al,20° C = d Al,168° C (1 − α Al ∆T )

Substitute numerical values and evaluate dAl,20°C:

[ (

)

]

d Al,20°C = (6.255 cm ) 1 − 24 × 10 −6 K −1 (148 K ) = 6.233 cm Remarks: Note that the diameter of the hole in the plate at 20°C is less than the diameter of the drill bit at 20°C.

Thermal Properties and Processes 1521 *27 •• Picture the Problem Let L be the length of the rail at 20°C and L′ its length at 25°C. The diagram shows these distances and the height h of the buckle. We can use Pythagorean theorem to relate the height of the buckle to the distances L and L′ and the definition of the coefficient of linear expansion to relate L and L′. 2

2

Apply the Pythagorean theorem to obtain:

⎛ L' ⎞ ⎛ L ⎞ h = ⎜ ⎟ −⎜ ⎟ = ⎝ 2 ⎠ ⎝2⎠

Use the definition of the coefficient of linear expansion to relate L and L′:

L′2 = L2(1 + αsteel∆T )2 or, because (αsteel∆T )2 << 2αsteel∆T, L′2 ≈ L2(1 + 2αsteel∆T)

Substitute to obtain:

h=

Substitute numerical values and evaluate h:

h=

1 2

L' 2 − L2

1 2 L (1 + 2α steel ∆T ) − L2 2 L = 2α steel ∆T 2

(

)

1000 m 2 11× 10 −6 K −1 (5 K ) 2

= 5.24 m 28 •• Picture the Problem The amount of gas that spills is the difference between the change in the volume of the gasoline and the change in volume of the tank. We can find this difference by expressing the changes in volume of the gasoline and the tank in terms of their common volume at 10°C, their coefficients of volume expansion, and the change in the temperature. Express the spill in terms of the change in volume of the gasoline and the change in volume of the tank:

Vspill = ∆Vgas − ∆Vtank

Relate ∆Vgas to the coefficient of volume expansion for gasoline:

∆Vgas = βgasV∆T

1522 Chapter 20 Relate ∆Vtank to the coefficient of linear expansion for steel:

∆Vtank = βtankV∆T or, because βsteel = 3αsteel, ∆Vtank= 3αsteelV∆T

Substitute to obtain:

Vspill = βgasV∆T − 3αsteelV∆T = V∆T (βgas − 3αsteel)

Substitute numerical values and evaluate Vspill:

[

)]

(

Vspill = (60 L )(15 K ) 0.9 × 10 −3 K −1 − 3 11× 10 −6 K −1 = 0.780 L 29 •• Picture the Problem We can relate the diameter of the capillary tube to the height the mercury rises for a 1°C increase in temperature and to the difference in the volume changes of the mercury in the bulb and the glass bulb. These volume changes can, in turn, be expressed in terms of the coefficients of volume expansion of mercury and glass. Express the net change in volume of the mercury in the thermometer and the bulb and tube of the glass thermometer:

∆V = ∆VHg − ∆Vglass = A∆L where A = πd2/4 is the cross-sectional area of the capillary tube and d is its diameter.

Relate ∆VHg to the coefficient of linear expansion for mercury:

∆VHg = βHgV∆T

Relate ∆Vglass to the coefficient of linear expansion for glass:

∆Vglass = βglassV∆T or, because βglass = 3αglass, ∆Vglass = 3αglassV∆T

Substitute to obtain:

πd 2 4

Solve for d:

= β HgV∆T − 3α glassV∆T = V∆T (β Hg − 3α glass ) 4V∆T (β Hg − 3α glass ) π∆L

d=

Substitute numerical values and evaluate d:

d=

(

)

4 10 −6 m 3 (1K ) 0.18 × 10 −3 K −1 − 3 9 × 10 −6 K −1 = 0.255 mm π 3 × 10 −3 m

(

)(

(

))

Thermal Properties and Processes 1523 30 •• Picture the Problem We can relate the volume of the thermometer bulb to the height the mercury rises for the 8 C° increase in temperature and to the difference in the volume changes of the mercury in the bulb and the glass bulb. These volume changes can, in turn, be expressed in terms of the coefficients of volume expansion of mercury and glass. Express the net change in volume of the mercury in the thermometer and the bulb and tube of the glass thermometer:

∆V = ∆VHg − ∆Vglass = A∆L where A = πd2/4 is the cross-sectional area of the capillary tube and d is its diameter.

Relate ∆VHg to the coefficient of linear expansion for mercury:

∆VHg = βHgV∆T or, because βHg = 3αHg, ∆VHg= 3αHgV∆T

Relate ∆Vglass to the coefficient of linear expansion for glass:

∆Vglass = βglassV∆T or, because βglass = 3αglass, ∆Vglass = 3αglassV∆T

Substitute to obtain:

β HgV∆T − 3α glassV∆T = A∆L

Solve for V and substitute for A:

V= =

A∆L (β Hg − 3α glass )∆T

πd 2 ∆L 4(β Hg − 3α glass )∆T

Substitute numerical values and evaluate V:

π (0.4 × 10 −3 m ) (7.5 ×10 −2 m ) 2

V =

[

]

4 0.18 ×10 −3 K −1 − 3(9 × 10 −6 K −1 ) (8 K )

= 7.70 mL

31 ••• Picture the Problem We can determine whether the clock runs fast or slow from the expression for the period of a simple pendulum and the dependence of its length on the temperature. Letting TP represent the period of the pendulum and T the temperature, we can evaluate dTP/dT and use a differential approximation to find the time gained or lost in a 24-h period. (a) Express the period of the pendulum in terms of its length:

TP = 2π

L g

1524 Chapter 20 Because TP ∝ L and L is temperature dependent, the clock runs slow. ∆TP ∆t TP

(b) Because the clock runs slow at the higher temperature, we know that it will lose time. Express the loss in terms of the loss each period and the elapsed time ∆t:

Loss =

dTP as the product of dT dTP dL and : dL dT

dTP dTP dL = ⋅ dT dL dT

Write

Evaluate

dTP and simplify to obtain: dL

Express the dependence of the length of the pendulum on its calibration length L0 and the coefficient of linear expansion of brass α: Evaluate

dL : dT

Substitute to obtain:

(1)

dTP d ⎡ L ⎤ 1 ⎛ 2π ⎞⎛ L ⎞ ⎟⎜ ⎟ = ⎢2π ⎥= ⎜ dL dL ⎣ g ⎦ 2 ⎜⎝ g ⎟⎠⎜⎝ g ⎟⎠ 1 ⎛ 2π ⎞ g 1 ⎛ L⎞ ⎟ ⎜ 2π ⎟⎟ = = ⎜⎜ 2 ⎝ g ⎠ L 2 L ⎜⎝ g ⎟⎠ T = P 2L L = L0 (1 + α∆T )

dL d [L0 (1 + α∆T )] = αL0 = dT dT

α dTP ⎛ TP ⎞ ⎟⎟(αL0 ) = TP = ⎜⎜ dT ⎝ 2 L0 ⎠ 2

Use the differential approximation to obtain:

∆TP α ∆TP α = TP or = ∆T 2 2 ∆T TP

Substitute numerical values and evaluate ∆TP/TP:

∆TP = TP

1 2

(19 ×10

−6

= 9.50 × 10 −5

)

/K (10 K )

− 12

Thermal Properties and Processes 1525

(

)

3600 s ⎞ ⎛ Loss = 9.50 × 10−5 ⎜ 24 h × ⎟ h ⎠ ⎝

Substitute numerical values in equation (1) to obtain:

= 8.21s 32 ••• Picture the Problem The steel tube will fit inside the brass tube when its outside diameter equals the inside diameter of the brass tube. We can use the definition of the coefficient of linear expansion to express the diameters of the tubes when they fit in terms of the required temperature change and equate these expressions to find ∆T.

T = Ti + ∆T = 293 K + ∆T

Express the temperature at which the steel tube will fit inside the brass tube in terms of their initial temperature and the change in temperature:

(1)

d steel = d brass

Express the condition that the steel tube will fit inside the brass tube:

d steel = d 0,steel (1 + α steel ∆T )

Relate the diameter of the steel tube to its initial diameter, coefficient of linear expansion, and the change in temperature:

d brass = d 0,brass (1 + α brass ∆T )

Relate the diameter of the brass tube to its initial diameter, coefficient of linear expansion, and the change in temperature:

d 0,steel (1 + α steel ∆T ) = d 0,brass (1 + α brass ∆T )

Substitute to obtain: Solve for ∆T:

∆T =

d 0,steel − d 0,brass d 0,brassα brass − d 0,steelα steel

Substitute numerical values and evaluate ∆T:

∆T ==

3.000 cm − 2.997 cm = 125 K (3.000 cm ) 19 ×10−6 K −1 − (2.997 cm ) 11×10−6 K −1

(

Substitute in equation (1) to evaluate ∆T:

)

(

)

T = 293 K + 125 K = 418 K = 145°C

1526 Chapter 20 *33 ••• Picture the Problem We can use the definition of Young’s modulus to express the tensile stress in the copper in terms of the strain it undergoes as its temperature returns to 20°C. We can show that ∆L/L for the circumference of the collar is the same as ∆d/d for its diameter. Using Young’s modulus, relate the stress in the collar to its strain:

Stress = Y × Strain = Y

∆L L20°C

where L20°C is the circumference of the collar at 20°C. Express the circumference of the collar at the temperature at which it fits over the shaft: Express the circumference of the collar at 20°C: Substitute to obtain:

LT = πd T

L20°C = πd 20°C

Stress = Y =Y

Substitute numerical values and evaluate the stress:

(

πd T − πd 20°C πd 20°C d T − d 20°C d 20°C

Stress = 11× 10 −10 N/m 2

cm ) 0.02 5.98 cm

= 3.68 × 10 −12 N/m 2

The van der Waals Equation, Liquid-Vapor Isotherms, and Phase Diagrams 34 • Picture the Problem We can apply the ideal-gas law to find the volume of 1 mol of steam at 100°C and a pressure of 1 atm and then use the van der Waals equation to find the temperature at which the steam will this volume.

Thermal Properties and Processes 1527 nRT P (1mol)(8.314 J/mol ⋅ K )(373 K ) = 101.325 kPa 1atm × atom 1L = 3.06 × 10−2 m 3 × −3 3 10 m

(a) Use the ideal-gas law to find the volume:

V =

= 30.6 L ⎛ an 2 ⎞ ⎜⎜ P + 2 ⎟⎟(V − bn ) V ⎠ T=⎝ nR

(b) Solve van der Waals equation for T to obtain:

Substitute numerical values and evaluate T:

⎛ an 2 ⎞ ⎜⎜ P + 2 ⎟⎟(V − bn ) 2 ⎛ V ⎠ 0.55 Pa ⋅ m 6 /mol2 (1mol) ⎞⎟ ⎝ ⎜ = 101.3 kPa + T= 2 ⎜ ⎟ nR 3.06 × 10−2 m 3 ⎝ ⎠

(

×

(

)

(

)

)

3.06 × 10 m − 30 × 10 m /mol (1 mol) (1mol) (8.314 J/mol ⋅ K ) −2

3

−6

3

= 375 K 35 •• Picture the Problem We can find these temperatures and pressure by consulting Figure 20-3. (a) At 70 kPa, water boils at:

t ≈ 90°C

(b) At 0.5 atm (about 51 kPa):

t boil ≈ 82°C

(c) For tboil = 115°C:

P ≈ 170 kPa

*36 •• Picture the Problem Assume that a helium atom is spherical. Then we can find its radius from V = 43 π r 3 and its volume from the van der Waals equation.

1528 Chapter 20 Express the radius of a spherical atom in terms of its volume:

r=3

In the van der Waals equation, b is the volume of 1 mol of molecules. For He, 1 molecule = 1 atom. Use Avogadro’s number to express b in cm3/atom:

b=

Substitute numerical values and evaluate r:

r=3

3V 4π

(0.0237 L/mol)(103 cm3 /L ) 6.022 × 10 23 atoms/mol

= 3.94 × 10 −23 cm3 /atom

(

3b 3 3 3.94 × 10−23 cm3 = 4π 4π

)

= 2.11× 10−8 cm = 0.211 nm 37 ••• Picture the Problem Because, at the critical point, dP/dV = 0 and d2P/dV2 = 0, we can solve the van der Waals equation for P and set its first and second derivatives equal to zero to find Vc. We can then eliminate Vc between these equations to find Tc and then substitute in the van der Waals equation to express Pc. Finally, we can use their definitions to rewrite the van der Waals equation in terms of the reduced variables.

nRT an 2 − 2 V − bn V

(a) Solve the van der Waals equation for P:

P=

Evaluate dP/dV:

dP d ⎡ nRT an 2 ⎤ = − ⎥ ⎢ dV dV ⎣V − bn V 2 ⎦ 2an 2 nRT =− + (V − nb )2 V 3 = 0 for extrema

Evaluate

2an 2 ⎤ d 2P d ⎡ nRT = − + ⎥ ⎢ dV 2 dV ⎣ (V − nb )2 V3 ⎦ 2nRT 6an 2 = − (V − nb )3 V 4 = 0 for critical points

d 2P : dV 2

Solve equation (2) for

2an 2 : V3

2an 2 nRT = 3 V (V − nb )2

(1)

(2)

(3)

(4)

Thermal Properties and Processes 1529 6an 2 Solve equation (3) for : V4

6an 2 2nRT = 4 V (V − nb )3

Divide equation (4) by equation (5) and simplify to obtain:

1 3

Solve for V = Vc:

Vc = 3nb

Substitute in equation (4):

nRTc 2an 2 = 3 3 27n b (3nb − nb )2

Simplify and solve for Tc:

V =

Tc =

1 2

(V − nb )

8a 27 Rb

Substitute for Vc and Tc in equation (1) and simplify to obtain:

8a 2 a 27 Rb − an = Pc = 2 3bn − bn (3bn ) 27b 2

(b) Using the result for Vc from (a), express the reduced volume Vr:

Vr =

V V = and V = 3nbVr Vc 3nb

Using the result for Tc from (a), express the reduced temperature Tr:

Tr =

T 27 RbT = Tc 8a

nR

and

Using the result for Pc from (a), express the reduced pressure Pr:

T=

8a Tr 27 Rb

Pr =

P 27b 2 P = Pc a

and

P=

Substitute in the van der Waals equation to obtain:

a Pr 27b 2

⎛ a an 2 ⎞⎟ ⎜ P + ⎜ 27b 2 r (3nbV )2 ⎟(3nbVr − bn ) r ⎝ ⎠ 8a = nR Tr 27 Rb

(5)

1530 Chapter 20 Simplify to obtain:

⎛ 3 ⎞ ⎜⎜ Pr + 2 ⎟⎟(3Vr − 1) = 8Tr Vr ⎠ ⎝

Heat Conduction 38 • Picture the Problem We can use their definitions to find the thermal resistance of the bar, the thermal current in the bar, and the temperature gradient in the bar. Because the temperature varies linearly with distance along the bar, we can express the temperature in terms of the thermal gradient and evaluate this expression 25 cm from the hot end. (a) Using its definition, find the thermal resistance of the bar:

R= =

∆x ∆x = kA kπ r 2 2m (401 W/m ⋅ K ) π 10−4 m 2

[(

)]

= 15.9 K/W ∆T 100 K = = 6.29 W R 15.9 K/W

(b) Using its definition, find the thermal current in the bar:

I=

(c) Substitute numerical values and evaluate the temperature gradient:

∆T 100 K = = 50 K/m = 50 K/m ∆x 2m

(d) Express the linear dependence of the temperature in the bar on the distance from the cold end: Substitute numerical values and evaluate T(1.75 m):

T = T0 +

dT ∆x dx

T (1.75 m ) = 273 K + (50 K/m )(1.75 m ) = 360.5 K = 87.5°C

39 • Picture the Problem We can use its definition to express the thermal current in the slab in terms of the temperature differential across it and its thermal resistance and use the definition of the R factor to express I as a function of ∆T, the cross-sectional area of the slab, and Rf. Express the thermal current through the slab in terms of the temperature difference across it and its thermal

I=

∆T R

Thermal Properties and Processes 1531 resistance: Substitute to express R in terms of the insulation’s R factor:

I=

Substitute numerical values and evaluate I:

I=

A∆T ∆T = Rf / A Rf

(20 ft )(30 ft )(68°F − 30°F) 11 h ⋅ ft 2 ⋅ F°/Btu

= 2.07 kBtu/h 40 •• Picture the Problem We can use R = ∆x kA to find the thermal resistance of each cube and the fact that they are in series to find the thermal resistance of the two-cube system. We can use I = ∆T R to find the thermal current through the cubes and the temperature at their interface.

∆x kA

(a) Using its definition, express the thermal resistance of each cube:

R=

Substitute numerical values and evaluate the thermal resistance of the copper cube:

RCu =

Substitute numerical values and evaluate the thermal resistance of the aluminum cube:

RAl =

(b) Because the cubes are in series, their thermal resistances are additive:

(c) Using its definition, find the thermal current: (d) Express the temperature at the interface between the two cubes: Express the temperature differential across the copper cube:

3 cm (401 W/m ⋅ K )(3 cm )2

= 0.0831 K/W 3 cm (237 W/m ⋅ K )(3 cm)2

= 0.141 K/W R = RCu + RAl = 0.0831 K/W + 0.141 K/W = 0.224 K/W

I=

∆T 373 K − 293 K = = 357 W R 0.224 K/W

Tinterface = 373 K − ∆TCu

∆TCu = I Cu RCu = IRCu

1532 Chapter 20 Tinterface = 373 K − IRCu

Substitute numerical values and evaluate Tinterface:

= 373 K − (357 W )(0.0831 K/W ) = 343.3 K = 70.3°C

41 •• Picture the Problem We can use I = ∆T R and R = ∆x kA to find the thermal current in each cube. Because the currents are additive, we can find the equivalent resistance of the two-cube system from Req = ∆T I . (a) Using its definition, express the thermal current through each cube:

I=

∆T R

Using its definition, express the thermal resistance of each cube:

R=

∆x kA

Substitute to obtain:

I=

kA∆T ∆x

(1)

Substitute numerical values in equation (1) and evaluate the thermal current in the copper cube:

I Cu =

(401 W/m ⋅ K )(3 cm )2 (373 K − 293 K ) = 3 cm

962 W

Substitute numerical values in equation (1) and evaluate the thermal current in the aluminum cube:

I Al =

(237 W/m ⋅ K )(3 cm )2 (373 K − 293 K ) =

(b) Because the cubes are in parallel, their total thermal currents are additive: (c) Use the relationship between the thermal current, temperature differential and thermal resistance to find Req:

3 cm

569 W

I = I Cu + I Al = 962 W + 569 W = 1.53 kW

Req =

∆T 373 K − 293 K = I 1.53 kW

= 0.0523 K/W

Thermal Properties and Processes 1533 42 •• Picture the Problem The cost of operating the air conditioner is proportional to the energy used in its operation. We can use the definition of the COP to relate the rate at which the air conditioner removes heat from the house to rate at which it must do work to maintain a constant temperature differential between the interior and the exterior of the house. To obtain an expression for the minimum rate at which the air conditioner must do work, we’ll assume that it is operating with the maximum efficiency possible. Doing so will allow us to derive an expression for the rate at which energy is used by the air conditioner that we can integrate to obtain the energy (and hence the cost of operation) required. Relate the cost C of air conditioning the energy W required to operate the air conditioner: Express the rate dQ/dt at which heat flows into a house provided the house is maintained at a constant temperature:

Use the definition of the COP to relate the rate at which the air conditioner must remove heat dW/dt to maintain a constant temperature: Solve for dW/dt:

Express the maximum value of the COP:

C = uW (1) where u is the unit cost of the energy.

P=

dQ = k∆T dt

where ∆T is the temperature difference between the interior and exterior of the house.

COP =

dQ dt dW dt

dW dt =

COPmax =

dQ dt COP

Tc ∆T

where Tc is the temperature of the cold reservoir. Letting COP = COPmax, substitute to obtain an expression for the minimum rate at which the air conditioner must do work in order to maintain a constant temperature:

dW dQ dt = ∆T dt Tc

Substitute for dQ/dt to obtain:

dW k∆T k 2 = ∆T = (∆T ) dt Tc Tc

Separate variables and integrate this equation to obtain:

∆t

W =

k (∆T )2 ∫ dt' = k (∆T )2 ∆t Tc Tc 0

1534 Chapter 20 Substitute in equation (1) to obtain:

C =u

k (∆T )2 ∆t ∝ Tc

(∆T )2

43 ••• Picture the Problem We can follow the step-by-step instructions given in the problem statement to obtain the differential equation describing the variation of T with r. Integrating this equation will yield an equation that we can solve for the current I.

(a)

Conservation of energy requires that thermal current through each shell be the same.

(b) Express the thermal current I through such a shell element in terms of the area A = 4π r2, the thickness dr, and the temperature difference dT across the element:

I = −kA

dT dT = − 4π kr 2 dr dr

where the minus sign is a consequence of the heat current being opposite the temperature gradient.

(c) Separate the variables:

dT = −

Integrate from r = r1 to r = r2:

T2

I

dr 4π k r 2 I

r2

dr

∫ dT = − 4π k ∫ r

T1

2

r1

and r

I ⎡1 ⎤ 2 I ⎛1 1⎞ ⎜ − ⎟ = T2 − T1 = ⎢ ⎥ 4π k ⎣ r ⎦ r1 4π k ⎜⎝ r1 r2 ⎟⎠ Solve for I to obtain:

(d) When r2 − r1 << r1: Let r2 − r1 = ∆r and substitute to obtain:

I=

4π kr1r2 (T2 − T1 ) r2 − r1

r1 ≈ r2 = r I=

4π kr 2 (T2 − T1 ) = 4π kr 2 ∆T ∆r ∆r

which is Equation 20-7. *44 •• Picture the Problem We can use the expression for the thermal current to express the thickness of the walls in terms of the thermal conductivity of copper, the area of the walls, and the temperature difference between the inner and outer surfaces. Letting ∆A/∆x′

Thermal Properties and Processes 1535 represent the area per unit length of the pipe and L its length, we can eliminate the surface area and solve for and evaluate L.

∆T ∆x

Write the expression for the thermal current:

I = kA

Solve for A:

A=

I∆x k∆T

Express the total surface area of the pipe:

A=

∆A L ∆x'

Substitute for A and solve for L to obtain:

I∆x L = k∆T ∆A ∆x'

Substitute numerical values and evaluate L:

(

)

⎡ ⎤ (3 GW ) 4 ×10−3 m ⎢ (401 W/m ⋅ K )(873 K − 498 K ) ⎥ ⎦ L= ⎣ 0.12 m = 665 m

45 ••• Picture the Problem Consider an element with a cylindrical area of length L, radius r, and thickness dr. We can relate the heat current through this element to the conductivity of the walls of the pipe, its length and radius, and the temperature gradient across the wall. We can separate the variables in the resulting differential equation and integrate to find the rate of heat transfer. (a) Express the heat current through the cylindrical element:

I = −kA

dT dT = −2π kLr dr dr

where the minus sign is a consequence of the heat current being opposite the temperature gradient. Separate the variables:

dT = −

Integrate from r = r1 to r = r2 and T = T1 to T = T2:

T2

I

dr 2π kL r I

r2

dr r r1

∫ dT = − 2π kL ∫

T1

and

1536 Chapter 20 I

T2 − T1 = −

ln r ] r12 r

2π kL I r =− ln 2 2π kL r1 =

Solve for I to obtain:

I=

I 2π kL

ln

r1 r2

2π kL (T2 − T1 ) ln (r1 r2 )

Remarks: If we use the above result in Problem 44 (take 0.12 m2 to be the outside area per unit length of the pipe), then r1 = 1.91 cm and r2 = 1.51 cm. Solving for L one obtains 746 m.

Radiation 46 • Picture the Problem We can apply Wein’s displacement law to find the wavelength at which the power is a maximum. Use Wein’s displacement law to relate the wavelength at which the power is a maximum to the surface temperature of the skin:

λmax =

2.898 mm ⋅ K T

Substitute numerical values and evaluate λmax:

λmax =

2.898 mm ⋅ K = 9.47 µm 273 K + 33 K

47 • Picture the Problem We can apply the Stefan-Boltzmann law to find the net power radiated by the wires of its heater to the room.

(

Relate the net power radiated to the surface area of the heating wires, their temperature, and the room temperature:

Pnet = eσA T 4 − T04

Solve for A:

A=

Pnet eσ T 4 − T04

(

)

)

Thermal Properties and Processes 1537 Substitute numerical values and evaluate A:

A=

(1) (5.6703 ×10

1 kW −8

W/m ⋅ K 2

4

)[(1173 K ) − (293 K ) ] 4

4

= 9.35 ×10 − 3 m 2

48 •• Picture the Problem The rate at which the sphere absorbs radiant energy is given by dQ / dt = mcdT / dt and, from the Stephan-Boltzmann law, Pnet = eσA T 4 − T04 , where

(

)

A is the surface area of the sphere, T0 is its temperature, and T is the temperature of the walls. We can solve the first equation for dT/dt and substitute Pnet for dQ/dt in order to find the rate at which the temperature of the sphere changes.

dQ dT = mc dt dt

Relate the rate at which the sphere absorbs radiant energy to the rate at which its temperature changes:

Pnet =

Solve for dT/dt:

dT Pnet P P = = net = 4 net3 dt mc ρVc 3 π r ρ c

Apply the Stefan-Boltzmann law to relate the net power radiated to the sphere to the difference in temperature of the walls and the blackened copper sphere:

Pnet = eσA T 4 − T04

(

(

)

= 4π r 2eσ T 4 − T04

)

dT 4π r 2eσ (T 4 − T04 ) = 3 4 dt 3π r ρc

Substitute to obtain:

=

3eσ (T 4 − T04 ) rρ c

Substitute numerical values and evaluate dT/dt:

(

)[

]

dT 3(1) 5.6703 ×10 −8 W/m 2 ⋅ K 4 (293 K ) − (273 K ) =− = 2.24 × 10 −3 K/s 3 3 −2 dt 4 × 10 m 8.93 × 10 kg/m (0.386 kJ/kg ⋅ K )

(

)(

4

)

4

49 •• Picture the Problem We can apply the Stephan-Boltzmann law to express the net power radiated by the incandescent lamp to its surroundings.

1538 Chapter 20 Express the rate at which energy is radiated to the surroundings:

(

Pnet = eσA T 4 − T04

)

⎛ ⎛ T0 ⎞ 4 ⎞ = eσAT ⎜1 − ⎜ ⎟ ⎟ ⎜ ⎝T ⎠ ⎟ ⎝ ⎠ 4

Evaluate (T0 T ) : 4

4

⎛ T0 ⎞ ⎛ 273 K ⎞ ⎟⎟ ≈ 9 × 10−4 ⎜ ⎟ = ⎜⎜ ⎝ T ⎠ ⎝ 1573 K ⎠ 4

and, because this ratio is so small, we can neglect the temperature of the surroundings. Substitute to obtain:

Pnet ≈ eσAT 4

Solve for T:

⎛P ⎞ T = ⎜ net ⎟ ⎝ eσA ⎠

Express the temperature T ′when the electric power input is doubled:

⎛ 2P ⎞ T' = ⎜ net ⎟ ⎝ eσA ⎠

Divide the second of these equations by the first:

T' 14 = (2) T

Solve for T ′:

T' = (2) T

Substitute numerical values and evaluate T ′

14

14

14

T' = (2 ) (1573 K ) = 1871 K 14

= 1598°C

50 •• Picture the Problem We can differentiate Q = mL, where L is the latent heat of boiling for helium, with respect to time to obtain an expression for the rate at which the helium boils away. Express the rate at which the helium boils away in terms of the rate at which its container absorbs radiant energy:

(

)

dm Pnet eσA T 4 − T04 = = dt L L 2 4 eσπ d T − T04 = L 4 eσπ d 2 4 ⎛⎜ ⎛ T0 ⎞ ⎞⎟ T 1− ⎜ ⎟ = ⎜ ⎝T ⎠ ⎟ L ⎠ ⎝ 2 eσπ d 4 T ≈ L

(

)

Thermal Properties and Processes 1539 when T0 << T. Substitute numerical values and evaluate dm/dt:

(

)

dm (1) 5.6703 × 10−8 W/m 2 ⋅ K 4 π (0.3 m ) (77 K )4 = 2.68 ×10−5 kg × 3600 s ≈ dt s h 21 kJ/kg 2

= 9.66 × 10−2 kg/h = 96.6 g/h

General Problems *51 • Picture the Problem The distance by which the tape clears the ground equals the change in the radius of the circle formed by the tape placed around the earth at the equator. Express the change in the radius of the circle defined by the steel tape:

Substitute numerical values and evaluate ∆R.

∆R = Rα∆T where R is the radius of the earth, α is the coefficient of linear expansion of steel, and ∆T is the increase in temperature.

(

)(

)

∆R = 6.37 × 10 6 m 11 × 10 −6 K −1 (30 K ) = 2.10 × 103 m = 2.10 km

52 •• Picture the Problem We can differentiate the definition of the density of an isotropic material with respect to T and use the definition of the coefficient of volume expansion to express the rate at which the density of the material changes with respect to temperature. Once we have an expression for dρ in terms of dT, we can apply a differential approximation to obtain ∆ρ in terms of ∆T. Using its definition, relate the density of the material to its mass and volume: Using its definition, relate the volume of the material to its coefficient of volume expansion:

ρ=

m V

∆V = βV∆T

1540 Chapter 20 dρ dρ dV m = = − 2 βV dT dV dT V ρV = − 2 βV = − ρβ V

Differentiate ρ with respect to T and simplify to obtain:

or

dρ = − ρβdT ∆ρ = − ρβ ∆T

Use a differential approximation to obtain:

53 •• Picture the Problem We can apply the Stefan-Boltzmann law to express the effective temperature of the sun in terms of the total power it radiates. We can, in turn, use the intensity of the sun’s radiation in the upper atmosphere of the earth to approximate the total power it radiates.

Pr = eσAT 4

Apply the Stefan-Boltzmann law to relate the energy radiated by the sun to its temperature: Solve for T:

T =4

Pr eσA

Express the area of the sun:

A = 4πRS2

Relate the intensity of the sun’s radiation in the upper atmosphere to the total power radiated by the sun:

I=

where R is the earth-sun distance.

Solve for Pr:

Pr = 4π R 2 I

Substitute for Pr and A and simplify to obtain:

T=

Pr 4π R 2

4

4π R 2 I R2I 4 = eσ 4π RS2 eσRS2

Substitute numerical values and evaluate T:

T =4

(1.5 ×10 m) (1.35 kW/m ) (1)(5.67 ×10 W/m ⋅ K )(6.96 ×10 m ) 11

−8

2

2

2

8

2

= 5767 K

Thermal Properties and Processes 1541 54 •• Picture the Problem We can solve the thermal-current equation for the R factor of the material. Use the equation for the thermal current to express I in terms of the temperature gradient across the insulation:

I = kA

Rewrite this expression in terms of the R factor of the material:

I=

Solve for the R factor:

∆T ∆x

∆T ∆T A∆T = = ∆x Rf Rf A kA

Rf =

A∆T 6 Aone side ∆T = I I

Substitute numerical values and evaluate R: 2

⎛ 2.54 × 10−2 m ⎞ ⎟⎟ 6 ⎜⎜12 in × 2 in ⎝ ⎠ (363 K − 293 K ) = 0.390 K ⋅ m Rf = 100 W W K ⋅ m 2 9 F° 10.76 ft 2 1054 J 1h F° ⋅ ft 2 ⋅ h = 0.390 × × × × = 2.21 J 5K m2 Btu 3600 s Btu s 55 •• Picture the Problem Because the temperature of the copper-aluminum interface is (T1 + T2)/2, we can conclude that the temperature differences across the two sheets must be the same. We also know, because the sheets are in series, that the heat currents through them are equal. Express the thermal current through the aluminum sheet:

I Al = k Al AAl

Express the thermal current through the copper sheet:

I Cu = kCu ACu

Equate these currents and solve for ∆xAl:

k Al AAl and

∆TAl ∆xAl ∆TCu ∆xCu

∆T ∆TAl = k Cu ACu Cu ∆xCu ∆xAl

1542 Chapter 20 ∆xAl = ∆xCu

k Al k Cu

∆xAl = (2 cm )

Substitute numerical values and evaluate ∆xAl:

237 W/m ⋅ K = 1.18 cm 401 W/m ⋅ K

56 •• Picture the Problem We can relate the stress in the bar to the strain due to its elongation using the definition of Young’s modulus and express the strain in terms of the coefficient of linear expansion and the change in temperature of the bar.

F Y= A ∆L L

Using the definition of Young’s modulus, relate the force exerted by the bar on each wall to the strain in the bar due to the change in its length:

∆L = α∆T L

Using the definition of the coefficient of linear expansion, express the strain in the bar:

F αA∆T

Substitute to obtain:

Y=

Solve for F:

F = αAY∆T

Substitute numerical values and evaluate F:

(

)

(

)

F = 11× 10 −6 K −1 π (0.022 m ) 200 GN/m 2 (40 K ) = 1.34 × 105 N 2

57 •• Picture the Problem We can use the definition of the coefficient of volume expansion with the ideal-gas law to show that β = 1/T. (a) Use the definition of the coefficient of volume expansion to express β in terms of the rate of change of the volume with temperature:

β=

1 dV V dT

Thermal Properties and Processes 1543 For an ideal gas:

V=

nRT dV nR = and P dT P

Substitute to obtain:

β=

1 nR 1 = V P T

(b) Express the ratio of the experimental value to the theoretical value:

−1 −1 β exp − β th 0.003673 K − 273 K = 1 β th K −1

1

273

< 0.3% 58 •• Picture the Problem We can express L as the difference between LB and LA and express these lengths in terms of the coefficients of linear expansion brass and steel. Requiring that L be constant will lead us to the condition that LA/LB = αB/αA. (a) Express the condition that L does not change when the temperature of the materials changes: Using the definition of the coefficient of linear expansion, substitute for LB and LA:

L = LB − LA = constant

L = (LB + α B LB ∆T ) − (LA + α A LA ∆T ) = (LB − LA ) + (α B LB − α A LA )∆T = L + (α B LB − α A LA )∆T or

(α B LB − α A LA )∆T = 0

The condition that L remain constant when the temperature changes by ∆T is: Solve for the ratio of LA to LB:

α B LB − α A LA = 0

LA α B = LB α A

1544 Chapter 20 (b) From (a) we have:

LB = Lsteel = LA = (250 cm )

αA α = Lbrass brass αB α steel

19 × 10 −6 K −1 11× 10 −6 K −1

= 432 cm and

L = LB − LA = 432 cm − 250 cm = 182 cm 59 •• Picture the Problem We can apply the thermal-current equation to calculate the heat loss of the earth per second due to conduction from its core. We can also use the thermal-current equation to find the power per unit area radiated from the earth and compare this quantity to the solar constant. Express the heat loss of the earth per unit time as a function of the thermal conductivity of the earth and its temperature gradient:

I=

dQ ∆T = kA dt ∆x

(1)

or

dQ ∆T = 4πRE2 k dt ∆x

Substitute numerical values and evaluate dQ/dt:

(

dQ = 4π 6.37 × 106 m dt

1C° ⎞ ⎟= ) (0.74 J/m ⋅ s ⋅ K )⎛⎜⎜ 30 m⎟ 2

⎝

⎠

1.26 × 1010 kW

Rewrite equation (1) to express the thermal current per unit area:

I ∆T =k A ∆x

Substitute numerical values and evaluate I/A:

⎛ 1C° ⎞ I ⎟⎟ = (0.74 J/m ⋅ s ⋅ K )⎜⎜ A ⎝ 30 m ⎠ = 0.0247 W/m 2

Express the ratio of I/A to the solar constant:

I A 0.0247 W/m 2 = solar constant 1.35 kW/m 2 < 0.002%

Thermal Properties and Processes 1545 60 •• Picture the Problem We can find the temperature of the outside of the copper bottom by finding the temperature difference between the outside of the saucepan and the boiling water. This temperature difference is related to the rate at which the water is evaporating through the thermal-current equation.

Tout = Tin + ∆T

Express the temperature outside the pan in terms of the temperature inside the pan:

= 373 K + ∆T ∆Q ∆T = kA ∆t ∆x

Relate the thermal current through the bottom of the saucepan to its thermal conductivity, area, and the temperature gradient between its surfaces: Solve for ∆T:

∆T =

1 ∆Q ∆x kA ∆t

Because the water is boiling:

∆Q = mLv

Substitute to obtain:

∆T =

mLv ∆x kA∆t

Substitute numerical values and evaluate ∆T:

∆T =

(0.8 kg )(2.26 MJ/kg )(3 ×10−3 m ) = 1.28 K (401 W/m ⋅ K ) ⎡⎢ π (0.15 m )2 ⎤⎥ (600 s ) ⎣4

Substitute numerical values and evaluate Tout:

⎦

Tout = 373 K + 1.28K = 374.3 K = 101.3°C

*61 •• Picture the Problem We’ll do this problem twice. First, we’ll approximate the answer by disregarding the fact that the surrounding insulation is cylindrical. In the second solution, we’ll obtain the exact answer by taking into account the cylindrical insulation surrounding the side of the tank. In both cases, the power required to maintain the temperature of the water in the tank is equal to the rate at which thermal energy is conducted through the insulation.

1546 Chapter 20 1st solution:

∆T ∆x

Using the thermal current equation, relate the rate at which energy is transmitted through the insulation to the temperature gradient, thermal conductivity of the insulation, and the area of the insulation/tank:

I = kA

Letting d represent the inside diameter of the tank and L its inside height, express and evaluate its surface area:

A = Aside + Abases ⎛π d2 ⎞ ⎟⎟ = π dL + 2 ⎜⎜ ⎝ 4 ⎠ = π dL + 12 d 2

(

)

[

= π (0.55 m )(1.2 m ) + 12 (0.55 m )

2

]

= 2.55 m 2 Substitute numerical values and evaluate I:

⎛ 74 K ⎞ ⎟⎟ I = (0.035 W/m ⋅ K ) 2.55 m 2 ⎜⎜ ⎝ 0.05 m ⎠

(

)

= 132 W 2nd solution: Express the total heat loss as the sum of the losses through the top and bottom and the side of the hotwater tank: Express I through the top and bottom surfaces:

Substitute numerical values and evaluate Itop and bottom:

I = I top and bottom + I side

⎛ ∆T ⎞ I top and bottom = 2⎜ kA ⎟ ⎝ ∆x ⎠ ∆T = 12 π d 2 k ∆x 2 1 I top and bottom = 2 π (0.55 m ) ×

(0.035 W/m ⋅ K )(74 K ) 0.05 m

= 24.6 W Letting r represent the inside radius of the tank, express the heat current

I side = −kA

dT dT = −2π kLr dr dr

Thermal Properties and Processes 1547 through the cylindrical side:

where the minus sign is a consequence of the heat current being opposite the temperature gradient.

Separate the variables:

dT = −

Integrate from r = r1 to r = r2 and T = T1 to T = T2:

T2

I side dr 2π kL r r

I side 2 dr dT = − ∫ 2π kL ∫r1 r T1

and

T2 − T1 = − =−

Solve for Iside to obtain:

I side =

Substitute numerical values and evaluate Iside:

I side =

I side r ln r ] r12 2π kL I side r I r ln 2 = side ln 1 2π kL r1 2π kL r2

2π kL (T2 − T1 ) r ln 1 r2

2π (0.035 W/m ⋅ K )(1.2 m ) (74 K ) ⎛ 0.325 m ⎞ ⎟⎟ ln⎜⎜ ⎝ 0.275 m ⎠

= 117 W Substitute for Iside and evaluate I:

I = 24.6 W + 117 W = 142 W

62 ••• Picture the Problem We can use R = ∆T/I and I = −kAdT/dt to express dT in terms of the linearly increasing diameter of the rod. Integrating this expression will allow us to find ∆T and, hence, R.

∆T I

Express the thermal resistance of the rod in terms of the thermal current in it:

R=

Relate the thermal current in the rod to its thermal conductivity k, crosssectional area A, and temperature gradient:

I = −kA

(1)

dT dx

where the minus sign is a consequence of the heat current being opposite the temperature gradient.

1548 Chapter 20 Using the dependence of the diameter of the rod on x, express the area of the rod: Substitute to obtain:

A=

π d2 4

=

π 4

d 02 (1 + ax )

2

⎡π 2 ⎤ dT I = −k ⎢ d 02 (1 + ax ) ⎥ ⎣4 ⎦ dx

Separate variables to obtain:

dT = −

Integrate T from T1 to T2 and x from 0 to L:

T2

Idx

⎡π 2⎤ k ⎢ d 02 (1 + ax ) ⎥ ⎣4 ⎦ 4I dx =− 2 π kd0 (1 + ax )2

4I ∫T dT = − π kd02 1

L

dx

∫ (1 + ax )

2

0

and

T2 − T1 = ∆T = Substitute for ∆T and I in equation (1) and simplify to obtain:

4 IL π kd (1 + aL ) 2 0

4 IL π kd (1 + aL ) 4L R= = 2 I π kd 0 (1 + aL ) 2 0

63 ••• Picture the Problem Let ∆T = T2 – T1. We can apply Newton’s 2nd law to establish the relationship between L2 and L1 and angular momentum conservation to relate ω2 and ω1. We can express both E2 and E1 in terms of their angular momenta and rotational inertias and take their ratio to establish their relationship.

Apply

∆L

∑τ = ∆t

to the spinning

disk:

Apply conservation of angular momentum to relate the angular velocity of the disk at T2 to the angular velocity at T1:

Because

∑τ = 0 , ∆L = 0

and

L2 = L1 I 2ω 2 = I1ω1 and

ω2 =

I1 ω1 I2

Thermal Properties and Processes 1549 Express I2:

I 2 = mr22 = mr12 (1 + α∆T )

2

(

= I1 1 + 2α∆T + (α∆T )

2

≈ I1 (1 + 2α∆T )

)

because (α∆T)2 is small compared to α∆T. Substitute and apply the binomial expansion formula to obtain:

ω2 =

I1 ω1 I1 (1 + 2α∆T )

and, because 2α∆T << 1,

ω2 ≈ (1 − 2α∆T )ω1

Express E2 in terms of L2 and I2:

E2 =

L22 L2 = 1 2I 2 2I 2

because L2 = L1. Express E1 in terms of L1 and I1:

E1 =

L12 2I1

Express the ratio of E2 to E1:

L12 E 2 2 I 2 I1 = 2 = L1 I2 E1 2 I1

Solve for E2 and substitute for the ratio of I1 to I2:

E2 = E1

I1 = E1 (1 − 2α∆T ) I2

64 ••• Picture the Problem The amount of heat radiated by the earth must equal the solar flux from the sun, or else the temperature on earth would continually increase. The emissivity of the earth is related to the rate at which it radiates energy into space by the StefanBoltzmann law Pr = eσAT 4 . Using the Stefan-Boltzmann law, Pr = eσA'T 4 express the rate at which the earth where A′ is the surface area of the earth. radiates energy as a function of its emissivity e and temperature T: Use its definition to express the intensity of the radiation Pa absorbed by the earth:

For 70% absorption of the sun’s radiation incident on the earth:

I=

Pa or Pa = AI A

where A is the cross-sectional area of the earth.

Pa = 0.7 AI

1550 Chapter 20 0.7 AI = eσA'T 4

Equate Pr and Pa and simplify:

or

(

0.7π R 2 I = e 4π R 2σT 4 Solve for T to obtain:

Substitute numerical values for I and σ and simplify to obtain:

)

T =4

0 .7 I = Ce −1 4 4σe

T =4

0.7 1370 W/m 2 4 5.670 × 10 −8 W/m 2 ⋅ K 4 e

(1)

(

(

)

)

= (255 K ) e −1 4 A spreadsheet program to evaluate T as a function of e is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form B1 255 B4 0.4 e B5 B4+0.01 e + 0.1 C4 $B$1/(B4^0.25) (255 K )e −1 4 A 1 2 3 4 5 6 7 23 24 25 26

B T= 255

C K

e 0.40 0.41 0.42 0.43

T 321 319 317 315

0.59 0.60 0.61 0.62

291 290 289 287

D

Thermal Properties and Processes 1551 A graph of T as a function of e is shown below.

325 320 315 T (K)

310 305 300 295 290 285 0.40

0.45

0.50

0.55

0.60

e

Treating e as a variable, differentiate equation (1) to obtain: Divide equation (2) by equation (1) to obtain:

Use a differential approximation to obtain: Solve for ∆e:

Substitute numerical values (e ≈ 0.615 for Tearth = 288 K) and evaluate ∆e:

1 dT = − Ce −5 4 de 4 de dT = T

(2)

1 − Ce −5 4 de 1 de 4 =− −1 4 Ce 4 e

∆T 1 ∆e =− T 4 e ∆e = −4e

∆T T

∆e = −4(0.615)

1K = − 0.00854 288 K

or about a 1.39% change in e.

65 ••• Picture the Problem We can differentiate the expression for the heat that must be removed from water in order to form ice to relate dQ/dt to the rate of ice build-up dm/dt. We can apply the thermal-current equation to express the rate at which heat is removed from the water to the temperature gradient and solve this equation for dm/dt. In part (b) we can separate the variables in the differential equation relating dm/dt and ∆T and integrate to find out how long it takes for a 20-cm layer of ice to be built up. (a) Relate the heat that must be removed from the water to freeze it to its mass and heat of fusion:

Q = mLf

1552 Chapter 20 Differentiate this expression with respect to time: Using the definition of density, relate the mass of the ice added to the bottom of the layer to its density and volume:

dQ dm = Lf dt dt

m = ρV = ρAx

Differentiate with respect to time to express the rate of build-up of the ice:

dm dx = ρA dt dt

Substitute to obtain:

dQ dx = Lf ρA dt dt

Apply the thermal-current equation:

∆T dQ = kA dt x

Equate these expressions and solve for dx/dt:

Lf ρA

∆T dx = kA dt x

and

dx k ∆T = dt Lf ρ x

Substitute numerical values and evaluate dx/dt:

(0.592 W/m ⋅ K )(10 K ) dx = dt (333.5 kJ/kg ) 917 kg/m 3 (0.01 m ) = 1.94 µm/s

(

= 0.698 cm/h (b) Separate the variables in equation (1):

xdx =

Integrate x from xi to xf and t′ from 0 to t:

xf

k∆T dt Lf ρ

∫ xdx = xi

and 1 2

Solve for t to obtain:

(1)

(x

t=

2 f

t

k∆T dt' Lf ρ ∫0

)

− xi2 =

k∆T t ρLf

ρLf (xf2 − xi2 ) 2k∆T

)

Thermal Properties and Processes 1553 Substitute numerical values and evaluate t:

t=

(917 kg/m )(333.5 kJ/kg ) [(0.2 m) − (0.01m ) ] = 1.03 ×10 2(0.592 W/m ⋅ K )(10 K ) 3

2

2

6

s×

1h 1d × 3600 s 24 h

= 11.9 d *66 ••• Picture the Problem We can use the thermal current equation and the definition of heat capacity to obtain the differential equation describing the rate at which the temperature of the water in the 200-g container is changing. Integrating this equation will yield T = T0 e − t RC . Substituting for dT/dt in dQ/dt = −CdT/dt and integrating will lead

(

)

to Q = CT0 1 − e − t RC .

(a) Use the thermal current equation to express the rate at which heat is conducted from the water at 60°C by the rod: Using the definition of heat capacity, relate the thermal current to the rate at which the temperature of the water initially at 60°C is changing: Equate these two expressions to obtain:

I=

∆T T = R R

because the temperature of the second container is maintained at 0°C.

I=

C

dQ dT = −C dt dt

(1)

dT 1 = − T , the differential equation dt R

describing the rate at which the temperature of the water in the 200-g container is changing. Separate variables to obtain:

dT 1 =− dt T RC

Integrate dT from T0 to T and dt from 0 to t:

t ⎛T ⎞ 1 dT' 1 ∫T T ' = − RC ∫0 dt ' ⇒ ln⎜⎜⎝ T0 ⎟⎟⎠ = − RC t 0 T

1554 Chapter 20 T = T0 e − t RC

Transform from logarithmic to exponential form and solve for T to obtain:

(2)

(b) Use its definition to express the thermal resistance R:

R=

∆x kA

Substitute numerical values (see Table 20-8 for the thermal conductivity of copper) and evaluate R:

R=

0.1m (401 W/m ⋅ K ) 1.5 ×10−4 m 2

Use its definition to express the heat capacity of the water and the copper container:

C = mc cc + mw cw = mc cc + ρ wVw cw

Substitute numerical values (see Table 18-1 for the specific heats of water and copper) and evaluate C:

(

)

= 1.66 K/W

C = (0.2 kg )(386 kJ/kg ⋅ K )

(

)

+ 103 kg/m 3 (0.7 L )(4.18 kJ/kg ⋅ K )

= 3.00 kJ/K

Evaluate the product of R and C to find the ″time constant″ τ :

τ = RC = (1.66 K/W )(3.00 kJ/K )

(c) Solve equation (1) for dQ to obtain:

⎛ dT ⎞ dQ = −C ⎜ ⎟dt = −CdT ⎝ dt ⎠

Integrate dQ′ from Q = 0 to Q and dT from T0 to T:

Q

T

0

T0

Substitute (equation (2) for T(t) to obtain:

= 4985 s = 1.38 h

∫ dQ' = − ∫ CdT ⇒ Q = C (T0 − T (t ))

(

)

(

Q = C T0 − T0e −t RC = CT0 1 − e −t RC

A spreadsheet program to evaluate Q as a function of t is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell D1 D2 D3 A6 A7 B6

Formula/Content 1.35 60 3000 0 A6+0.1 $B$2*EXP(−A6/$B$1)

Algebraic Form

C7

$B$3*$B$2*(1−EXP(−A6/$B$1))

CT0 1 − e − t RC

τ

T0 C t t + ∆t

T0e − t RC

(

)

)

Thermal Properties and Processes 1555 A

B

C

tau= 1.35 T0= 60 C= 3000

D

1 2 3 4 5 6 7 8

t (hr) 0.0 0.1 0.2

T 60.00 55.72 51.74

Q 0.00E+00 1.29E+04 2.48E+04

Q/1000 0 13 24

13 14 15 16

0.7 0.8 0.9 1.0

35.72 33.17 30.81 28.61

7.28E+04 8.05E+04 8.76E+04 9.42E+04

71 79 86 92

33 34 35

2.7 2.8 2.9

8.12 7.54 7.00

1.56E+05 1.57E+05 1.59E+05

152 154 155

E

h deg-C J/K

From the table we can see that the temperature of the container drops to 30°C in a little more than 0.9 h. If we wanted to know this time to the nearest hundredth of an hour, we could change the step size in the spreadsheet program to 0.01 h. A graph of T as a function of t is shown in the following graph.

70 60

T (deg C)

50 40 30 20 10 0 0.0

0.5

1.0

1.5 t (h)

2.0

2.5

3.0

1556 Chapter 20 A graph of Q as a function of t follows.

160 140 120

Q (kJ)

100 80 60 40 20 0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

t (h)

67 ••• Picture the Problem We can use the Stefan-Boltzmann equation and the definition of heat capacity to obtain the differential equation expressing the rate at which the temperature of the copper block decreases. We can then approximate the differential equation with a difference equation for the purpose of solving for the temperature of the block as a function of time using Euler’s method.

(

(a) Express the rate at which heat is radiated away from the cube:

dQ = eσA T 4 − T04 dt

Using the definition of heat capacity, relate the thermal current to the rate at which the temperature of the cube is changing:

dQ dT = −C dt dt

Equate these expressions to obtain:

Approximate the differential equation by the difference equation: Solve for ∆T:

)

dT eσA 4 =− T − T04 dt C

(

∆T eσA 4 =− T − T04 ∆t C

(

∆T = −

)

)

eσA 4 T − T04 ∆t C

or

Tn+1 = Tn −

(

)

eσA 4 Tn − T04 ∆t C

(

)

(1)

Thermal Properties and Processes 1557 Use the definition of heat capacity to obtain:

C = mc = ρVc

Substitute numerical values (see Figure 13-1 for ρCu and Table 19-1 for cCu) and evaluate C:

C = 8.93 × 103 kg/m 3 10 −6 m 3 × (0.386 kJ/kg ⋅ K ) = 3.45 J/K

(

)(

)

(b) A spreadsheet program to calculate T as a function of t using equation (1) is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B4 B5 A9 B9

Formula/Content 5.67×10−8 6.00×10−4 3.45 273 10 A8+$B$5 B8-($B$1*$B$2/$B$3) *(B8^4−$B$4^4)*$B$5 A sigma= A= C= T0= dt=

Algebraic Form

σ

A C T0 ∆t t+∆t

Tn −

B 5.67E−08 6.00E−04 3.45 273 10

1 2 3 4 5 6 7 8 9 10 11

t (s) 0 10 20 30

T (K) 573.00 562.92 553.56 544.85

248 249 250 251

2400 2410 2420 2430

288.22 288.08 287.95 287.82

eσA 4 Tn − T04 ∆t C

(

)

C W/m^2⋅K^4 m^2 J/K K s

1558 Chapter 20 From the spreadsheet solution, the time to cool to 15°C (288 K) is about 2420 s or

40.5 min . A graph of T as a function of t follows.

570 520

T (K)

470 420 370 320 270 0

500

1000

1500 t (s)

2000

2500

Chapter 21 The Electric Field 1: Discrete Charge Distributions Conceptual Problems *1 •• Similarities:

Differences:

The force between charges and masses varies as 1/r2.

There are positive and negative charges but only positive masses.

The force is directly proportional to the product of the charges or masses.

Like charges repel; like masses attract.

The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k. 2 • Determine the Concept No. In order to charge a body by induction, it must have charges that are free to move about on the body. An insulator does not have such charges. 3 •• Determine the Concept During this sequence of events, negative charges are attracted from ground to the rectangular metal plate B. When S is opened, these charges are trapped on B and remain there when the charged body is removed. Hence B is negatively charged and (c) is correct. 4 •• (a) Connect the metal sphere to ground; bring the insulating rod near the metal sphere and disconnect the sphere from ground; then remove the insulating rod. The sphere will be negatively charged. (b) Bring the insulating rod in contact with the metal sphere; some of the positive charge on the rod will be transferred to the metal sphere. (c) Yes. First charge one metal sphere negatively by induction as in (a). Then use that negatively charged sphere to charge the second metal sphere positively by induction.

1

2

Chapter 21

*5 •• Determine the Concept Because the spheres are conductors, there are free electrons on them that will reposition themselves when the positively charged rod is brought nearby. (a) On the sphere near the positively charged rod, the induced charge is negative and near the rod. On the other sphere, the net charge is positive and on the side far from the rod. This is shown in the diagram. (b) When the spheres are separated and far apart and the rod has been removed, the induced charges are distributed uniformly over each sphere. The charge distributions are shown in the diagram. 6 • Determine the Concept The forces acting on +q are shown in the diagram. The force acting on +q due to −Q is along the line joining them and directed toward −Q. The force acting on +q due to +Q is along the line joining them and directed away from +Q. Because charges +Q and −Q are equal in magnitude, the forces due to these charges are equal and their sum (the net force on +q) will be to the right and so (e) is correct. Note that the vertical components of these forces add up to zero. *7 • Determine the Concept The acceleration of the positive charge is given by

r r F q0 r a= = E . Because q0 and m are both positive, the acceleration is in the same m m direction as the electric field. (d ) is correct.

*8 • r Determine the Concept E is zero wherever the net force acting on a test charge is zero. At the center of the square the two positive charges alone would produce a net electric field of zero, and the two negative charges alone would also produce a net electric field of zero. Thus, the net force acting on a test charge at the midpoint of the

The Electric Field 1: Discrete Charge Distributions

3

square will be zero. (b) is correct. 9 •• (a) The zero net force acting on Q could be the consequence of equal collinear charges being equidistant from and on opposite sides of Q. (b) The charges described in (a) could be either positive or negative and the net force on Q would still be zero. (c) Suppose Q is positive. Imagine a negative charge situated to its right and a larger positive charge on the same line and the right of the negative charge. Such an arrangement of charges, with the distances properly chosen, would result in a net force of zero acting on Q. (d) Because none of the above are correct, (d ) is correct. 10 • Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the sketch to the right we’ve assigned 2 field lines to each charge q.

*11 • Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right we’ve assigned 2 field lines to each charge q.

4

Chapter 21

*12 • Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right we’ve assigned 7 field lines to each charge q.

13 • Determine the Concept A positive charge will induce a charge of the opposite sign on the near surface of the nearby neutral conductor. The positive charge and the induced charge on the neutral conductor, being of opposite sign, will always attract one another.

(a) is correct. *14 • Determine the Concept Electric field lines around an electric dipole originate at the positive charge and terminate at the negative charge. Only the lines shown in (d) satisfy this requirement. (d ) is correct. *15 •• Determine the Concept Because θ ≠ 0, a dipole in a uniform electric field will experience a restoring torque whose magnitude is pE x sin θ . Hence it will oscillate about its equilibrium orientation, θ = 0. If θ << 1, sinθ ≈ θ, and the motion will be simple harmonic motion. Because the field is nonuniform and is larger in the x direction, the force acting on the positive charge of the dipole (in the direction of increasing x) will be greater than the force acting on the negative charge of the dipole (in the direction of decreasing x) and thus there will be a net electric force on the dipole in the direction of increasing x. Hence, the dipole will accelerate in the x direction as it oscillates about θ = 0. 16 •• (a) False. The direction of the field is toward a negative charge. (b) True. (c) False. Electric field lines diverge from any point in space occupied by a positive charge. (d) True

The Electric Field 1: Discrete Charge Distributions

5

(e) True 17 •• Determine the Concept The diagram shows the metal balls before they are placed in the water. In this situation, the net electric field at the location of the sphere on the left is due only to the charge –q on the sphere on the right. If the metal balls are placed in water, the water molecules around each ball tend to align themselves with the electric field. This is shown for the ball on the right with charge –q.

r

(a) The net electric field Enet that produces a force on the ball on the left is the

r

field E due to the charge –q on the ball on the right plus the field due to the layer of positive charge that surrounds the ball on the right. This layer of positive charge is due to the aligning of the water molecules in the electric field, and the amount of positive charge in the layer surrounding the ball on the left will be less than +q. Thus, Enet < E. Because Enet < E, the force on the ball on the left is less than it would be if the balls had not been placed in water. Hence, the force will

decrease when the balls are placed in the water. (b) When a third uncharged metal ball is placed between the first two, the net electric field at the location of the sphere on the right is the field due to the charge +q on the sphere on the left, plus the field due to the charge –Q and +Q on the sphere in the middle. This electric field is directed to the right. The field due to the charge –Q and +Q on the sphere in the middle at the location of the sphere on the right is to the right. It follows that the net electric field due to the charge +q on the sphere on the left, plus the field due to the charge –Q and +Q on the sphere in the middle is to the right and has a greater magnitude than the field due only to the charge +q on the sphere on the left. Hence, the force on either sphere will increase if a third uncharged metal ball is placed between them. Remarks: The reduction of an electric field by the alignment of dipole moments with the field is discussed in further detail in Chapter 24.

6

Chapter 21

*18 •• Determine the Concept Yes. A positively charged ball will induce a dipole on the metal ball, and if the two are in close proximity, the net force can be attractive. *19 •• Determine the Concept Assume that the wand has a negative charge. When the charged wand is brought near the tinfoil, the side nearer the wand becomes positively charged by induction, and so it swings toward the wand. When it touches the wand, some of the negative charge is transferred to the foil, which, as a result, acquires a net negative charge and is now repelled by the wand.

Estimation and Approximation 20 •• Picture the Problem Because it is both very small and repulsive, we can ignore the gravitational force between the spheres. It is also true that we are given no information about the masses of these spheres. We can find the largest possible value of Q by equating the electrostatic force between the charged spheres and the maximum force the cable can withstand. Using Coulomb’s law, express the electrostatic force between the two charged spheres:

F=

kQ 2 l2

Express the tensile strength Stensile of steel in terms of the maximum force Fmax in the cable and the crosssectional area of the cable and solve for F:

S tensile =

Equate these forces to obtain:

kQ 2 = AS tensile l2

Solve for Q:

Q=l

Fmax ⇒ Fmax = AS tensile A

AS tensile k

Substitute numerical values and evaluate Q:

Q = (1 m )

(1.5 × 10

−4

)(

)

m 2 5.2 × 108 N/m 2 = 2.95 mC 8.99 × 109 N ⋅ m 2 / C 2

21 •• Picture the Problem We can use Coulomb’s law to express the net force acting on the copper cube in terms of the unbalanced charge resulting from the assumed migration of half the charges to opposite sides of the cube. We can, in turn, find the unbalanced charge Qunbalanced from the number of copper atoms N and the number of electrons per atom.

The Electric Field 1: Discrete Charge Distributions (a) Using Coulomb’s law, express the net force acting on the copper rod due to the imbalance in the positive and negative charges:

2 kQunbalanced F= r2

Relate the number of copper atoms N to the mass m of the rod, the molar mass M of copper, and Avogadro’s number NA:

N m ρ CuVrod = = NA M M

Solve for N to obtain:

N=

ρ CuVrod

(1)

NA

M

Substitute numerical values and evaluate N:

(8.93 × 10 N=

3

)(

)(

)(

2

)

kg/m 3 0.5 × 10 −2 m 4 × 10 −2 m 6.02 × 10 23 atoms/mol 63.54 × 10 −3 kg/mol

= 8.461 × 10 22 atoms

Qunbalanced =

Because each atom has 29 electrons and protons, we can express Qunbalanced as:

1 2

(29) (10−7 )eN

Substitute numerical values and evaluate Qunbalanced:

Qunbalanced =

1 2

(29) (10−7 )(1.6 × 10−19 C)(8.461× 1022 ) = 1.963 × 10−2 C

Substitute for Qunbalanced in equation (1) to obtain:

(8.99 × 10 F=

9

)(

N ⋅ m 2 / C 2 1.963 × 10 −2 C (0.01 m )2

(b) Using Coulomb’s law, express the maximum force of repulsion Fmax in terms of the maximum possible charge Qmax: Solve for Qmax:

Express Fmax in terms of the tensile strength Stensile of copper:

Fmax =

Qmax =

)

2

= 3.46 × 1010 N

2 kQmax r2

r 2 Fmax k

Fmax = S tensile A where A is the cross sectional area of the cube.

7

8

Chapter 21

Substitute to obtain:

r 2 S tensile A k

Qmax =

Substitute numerical values and evaluate Qmax:

Qmax =

(0.01 m )2 (2.3 × 108 N/m 2 )(10−4 m 2 ) = 1.60 × 10−5 C 8.99 × 109 N ⋅ m 2 / C 2

Because Qunbalanced = 2Qmax :

(

Qunbalanced = 2 1.60 × 10 −5 C

)

= 32.0 µC Remarks: A net charge of −32 µC means an excess of 2.00×1014 electrons, so the net imbalance as a percentage of the total number of charges is 4.06×10−11 = 4×10−9 %. 22 ••• Picture the Problem We can use the definition of electric field to express E in terms of the work done on the ionizing electrons and the distance they travel λ between collisions. We can use the ideal-gas law to relate the number density of molecules in the gas ρ and the scattering cross-section σ to the mean free path and, hence, to the electric field. (a) Apply conservation of energy to relate the work done on the electrons by the electric field to the change in their kinetic energy:

W = ∆K = F ∆ s

From the definition of electric field we have:

F = qE

Substitute for F and ∆s to obtain:

W = qEλ , where the mean free path λ is the distance traveled by the electrons between ionizing collisions with nitrogen atoms.

Referring to pages 545-546 for a discussion on the mean-free path, use its definition to relate λ to the scattering cross-section σ and the number density for air molecules n: Substitute for λ and solve for E to obtain: Use the ideal-gas law to obtain:

λ=

E=

n=

1 σn

σ nW q N P = V kT

The Electric Field 1: Discrete Charge Distributions Substitute for n to obtain:

E=

σPW

9

(1)

qkT

Substitute numerical values and evaluate E:

E=

(10 m )(10 N/m )(1eV) (1.6 × 10 J/eV) = (1.6 × 10 C)(1.38 × 10 J/K )(300 K ) −19

2

5

−19

2

−19

(b) From equation (1) we see that:

−23

2.41 × 106 N/C

E ∝ P and E ∝ T −1 i.e., E increases linearly with pressure and varies inversely with temperature.

*23 •• Picture the Problem We can use Coulomb’s law to express the charge on the rod in terms of the force exerted on it by the soda can and its distance from the can. We can apply Newton’s 2nd law in rotational form to the can to relate its acceleration to the electric force exerted on it by the rod. Combining these equations will yield an expression for Q as a function of the mass of the can, its distance from the rod, and its acceleration. Use Coulomb’s law to relate the force on the rod to its charge Q and distance r from the soda can: Solve for Q to obtain:

Apply

∑τ

center of mass

= Iα to the

F=

kQ 2 r2

r 2F k

Q=

(1)

FR = Iα

can: Because the can rolls without slipping, we know that its linear acceleration a and angular acceleration α are related according to:

α=

a R

where R is the radius of the soda can.

I = MR 2

Because the empty can is a hollow cylinder:

where M is the mass of the can.

Substitute for I and α and solve for F to obtain:

F=

Substitute for F in equation (1):

Q=

MR 2 a = Ma R2

r 2 Ma k

10

Chapter 21

Substitute numerical values and evaluate Q:

Q=

(0.1 m )2 (0.018 kg ) (1 m/s 2 ) 8.99 × 109 N ⋅ m 2 / C 2

= 141 nC 24 •• Picture the Problem Because the nucleus is in equilibrium, the binding force must be equal to the electrostatic force of repulsion between the protons. Apply

r

∑ F = 0 to a proton:

Fbinding − Felectrostatic = 0

Solve for Fbinding:

Fbinding = Felectrostatic

Using Coulomb’s law, substitute for Felectrostatic:

Fbinding =

kq 2 r2

Substitute numerical values and evaluate Felectrostatic:

Fbinding

(8.99 × 10 =

9

)( m)

N ⋅ m 2 / C 2 1.6 × 10 −19 C

(10

−15

2

)

2

= 230 N

Electric Charge 25 • Picture the Problem The charge acquired by the plastic rod is an integral number of electronic charges, i.e., q = ne(−e).

q = ne (− e )

Relate the charge acquired by the plastic rod to the number of electrons transferred from the wool shirt: Solve for and evaluate ne:

ne =

q − 0.8 µC = = 5.00 × 1012 −19 − e − 1.6 × 10 C

26 • Picture the Problem One faraday = NAe. We can use this definition to find the number of coulombs in a faraday. Use the definition of a faraday to calculate the number of coulombs in a faraday:

(

)(

)

1 faraday = N A e = 6.02 × 10 23 electrons 1.6 × 10 −19 C/electron = 9.63 × 10 4 C

The Electric Field 1: Discrete Charge Distributions

11

*27 • Picture the Problem We can find the number of coulombs of positive charge there are in 1 kg of carbon from Q = 6nC e , where nC is the number of atoms in 1 kg of carbon and the factor of 6 is present to account for the presence of 6 protons in each atom. We can find the number of atoms in 1kg of carbon by setting up a proportion relating Avogadro’s number, the mass of carbon, and the molecular mass of carbon to nC.

Q = 6nC e

Express the positive charge in terms of the electronic charge, the number of protons per atom, and the number of atoms in 1 kg of carbon: Using a proportion, relate the number of atoms in 1 kg of carbon nC, to Avogadro’s number and the molecular mass M of carbon:

N m nC mC ⇒ nC = A C = M NA M

Substitute to obtain:

Q=

6 N A mC e M

Substitute numerical values and evaluate Q:

Q=

(

)

(

)

6 6.02 × 1023 atoms/mol (1 kg ) 1.6 × 10 −19 C = 4.82 × 107 C 0.012 kg/mol

Coulomb’s Law 28 • Picture the Problem We can find the forces the two charges exert on each by applying Coulomb’s law and Newton’s 3rd law. Note that rˆ1, 2 = iˆ because the vector pointing from q1 to q2 is in the positive x direction.

r kq q F1, 2 = 12 2 rˆ1, 2 r1, 2

(a) Use Coulomb’s law to express the force that q1 exerts on q2:

r

Substitute numerical values and evaluate F1, 2 :

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (4 µC)(6 µC) F1, 2 = i= (3 m)2

(24.0 mN) iˆ

12

Chapter 21 r r F2,1 = − F1, 2 = − (24.0 mN ) iˆ

(b) Because these are action-andreaction forces, we can apply Newton’s 3rd law to obtain: (c) If q2 is −6.0 µC:

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (4 µC)(− 6 µC) ˆ F1, 2 = i = − (24.0 mN ) iˆ 2 (3 m ) and

29

r r F2,1 = − F1, 2 = •

(24.0 mN )iˆ r

r

Picture the Problem q2 exerts an attractive force F2,1 on q1 and q3 a repulsive force F3,1 . We can find the net force on q1 by adding these forces.

Express the net force acting on q1:

r r r F1 = F2,1 + F3,1

Express the force that q2 exerts on q1:

r kq q F2,1 = 12 2 iˆ r2,1

Express the force that q3 exerts on q1:

r kq q F3,1 = 12 3 − iˆ r3,1

Substitute and simplify to obtain:

r kq q kq q F1 = 12 2 iˆ − 12 3 iˆ r2,1 r3,1

( )

⎛q q ⎞ = k q1 ⎜⎜ 22 − 23 ⎟⎟iˆ ⎝ r2,1 r3,1 ⎠

r

Substitute numerical values and evaluate F1 :

r ⎛ 4 µC 6 µC ⎞ ˆ ⎟ i = 1.50 ×10− 2 N iˆ F1 = 8.99 ×109 N ⋅ m 2 /C 2 (6 µC)⎜⎜ − 2 2 ⎟ ⎝ (3 m ) (6 m ) ⎠

(

)

(

)

The Electric Field 1: Discrete Charge Distributions

13

30 •• Picture the Problem The configuration of the charges and the forces on the fourth charge are shown in the figure … as is a coordinate system. From the figure it is evident that the net force on q4 is along the diagonal of the square and directed away from q3. We can apply Coulomb’s law to

r

r

r

express F1, 4 , F2, 4 and F3, 4 and then add them to find the net force on q4. Express the net force acting on q4:

r r r r F4 = F1, 4 + F2, 4 + F3, 4

Express the force that q1 exerts on q4:

r kq q F1, 4 = 12 4 ˆj r1, 4

r

Substitute numerical values and evaluate F1, 4 :

r ⎛ 3 nC ⎞ ˆ ⎟ j = 3.24 × 10 −5 N ˆj F1, 4 = 8.99 × 109 N ⋅ m 2 /C 2 (3 nC ) ⎜⎜ 2 ⎟ ⎝ (0.05 m ) ⎠

(

)

(

)

r kq q F2, 4 = 22 4 iˆ r2, 4

Express the force that q2 exerts on q4:

r

Substitute numerical values and evaluate F2, 4 :

r ⎛ 3 nC ⎞ ˆ ⎟ i = 3.24 × 10 −5 N iˆ F2, 4 = 8.99 × 109 N ⋅ m 2 /C 2 (3 nC) ⎜⎜ 2 ⎟ ⎝ (0.05 m ) ⎠

(

)

Express the force that q3 exerts on q4:

(

r kq q F3, 4 = 32 4 rˆ3, 4 , where rˆ3, 4 is a unit vector r3, 4 pointing from q3 to q4.

r

r

r

Express r3, 4 in terms of r3,1 and r1, 4 :

)

r r r r3, 4 = r3,1 + r1, 4 = (0.05 m ) iˆ + (0.05 m ) ˆj

14

Chapter 21

r

Convert r3, 4 to rˆ3, 4 :

r r3, 4 (0.05 m ) iˆ + (0.05 m ) ˆj = r = r3, 4 (0.05 m )2 + (0.05 m )2

rˆ3, 4

= 0.707 iˆ + 0.707 ˆj

r

Substitute numerical values and evaluate F3, 4 :

⎞ ⎛ r 3 nC ⎟ 0.707 iˆ + 0.707 ˆj F3, 4 = 8.99 × 109 N ⋅ m 2 /C 2 (− 3 nC )⎜ ⎜ 0.05 2 m 2 ⎟ ⎠ ⎝ −5 − 5 = − 1.14 × 10 N iˆ − 1.14 × 10 N ˆj

(

)

(

(

) (

)

)

(

)

r

Substitute and simplify to find F4 :

r F4 = 3.24 ×10−5 N ˆj + 3.24 ×10 −5 N iˆ − 1.14 ×10 −5 N iˆ − 1.14 ×10 −5 N ˆj

(

(

= 2.10 ×10 −5

) ( ) ( N )iˆ + (2.10 ×10 N ) ˆj

) (

)

−5

31 •• Picture the Problem The configuration of the charges and the forces on q3 are shown in the figure … as is a coordinate system. From the geometry of the charge distribution it is evident that the net force on the 2 µC charge is in the negative y direction. We can apply

r

r

Coulomb’s law to express F1,3 and F2,3 and then add them to find the net force on q3.

The net force acting on q3 is given by:

r r r F3 = F1,3 + F2,3

The Electric Field 1: Discrete Charge Distributions Express the force that q1 exerts on q3:

15

r F1,3 = F cos θ iˆ − F sin θ ˆj where

kq1q3 r2 8.99 × 109 N ⋅ m 2 / C 2 (5 µC )(2 µC ) = (0.03 m )2 + (0.08 m )2 = 12.3 N

F=

(

)

and

⎛ 3 cm ⎞ ⎟⎟ = 20.6° 8 cm ⎝ ⎠

θ = tan −1 ⎜⎜ Express the force that q2 exerts on q3:

r

r

Substitute for F1,3 and F2,3 and simplify to obtain:

r F2,3 = − F cos θ iˆ − F sin θ ˆj r F3 = F cos θ iˆ − F sin θ ˆj − F cos θ iˆ − F sin θ ˆj = −2 F sin θ ˆj

Substitute numerical values and r evaluate F3 :

r F3 = −2(12.3 N ) sin 20.6° ˆj = − (8.66 N ) ˆj

*32 •• Picture the Problem The positions of the charges are shown in the diagram. It is apparent that the electron must be located along the line joining the two charges. Moreover, because it is negatively charged, it must be closer to the −2.5 µC than to the 6.0 µC charge, as is indicated in the figure. We can find the x and y coordinates of the electron’s position by equating the two electrostatic forces acting on it and solving for its distance from the origin. We can use similar triangles to express this radial distance in terms of the x and y coordinates of the electron. Express the condition that must be

F1, e = F2, e

16

Chapter 21

satisfied if the electron is to be in equilibrium:

kq1e

Express the magnitude of the force that q1 exerts on the electron:

F1,e =

(r +

Express the magnitude of the force that q2 exerts on the electron:

F2,e =

k q2 e r2

Substitute and simplify to obtain:

Substitute for q1 and q2 and simplify: Solve for r to obtain:

(r +

1.25 m

q1

)

1.25 m

2

=

)

2

q2 r2

(− 1.4 m ) r + (2.2361m ) r −2

2

−1

+ 1.25 m = 0 r = 2.036 m and r = −0.4386 m Because r < 0 is unphysical, we’ll consider only the positive root.

Use the similar triangles in the diagram to establish the proportion involving the y coordinate of the electron: Solve for ye: Use the similar triangles in the diagram to establish the proportion involving the x coordinate of the electron: Solve for xe: The coordinates of the electron’s position are:

ye 2.036 m = 0.5 m 1.12 m

ye = 0.909 m xe 2.036 m = 1m 1.12 m

xe = 1.82 m

(xe , ye ) = (− 1.82 m, − 0.909 m )

The Electric Field 1: Discrete Charge Distributions

17

*33 •• Picture the Problem Let q1 represent the charge at the origin, q2 the charge at (0, 0.1 m), and q3 the charge at (0.2 m, 0). The diagram shows the forces acting on each of the charges. Note the action-and-reaction pairs. We can apply Coulomb’s law and the principle of superposition of forces to find the net force acting on each of the charges. Express the net force acting on q1:

r r r F1 = F2,1 + F3,1

Express the force that q2 exerts on q1:

r r kq2 q1 kq2 q1 r2,1 kq2 q1 r = 3 r2,1 F2,1 = 2 rˆ2,1 = 2 r2,1 r2,1 r2,1 r2,1 r

Substitute numerical values and evaluate F2,1 :

r (− 1 µC ) (− 0.1 m ) ˆj = (1.80 N ) ˆj F2,1 = 8.99 ×10 9 N ⋅ m 2 /C 2 (2 µC ) (0.1 m )3

(

)

r kq q r F3,1 = 33 1 r3,1 r3,1

Express the force that q3 exerts on q1:

r

Substitute numerical values and evaluate F3,1 :

r (− 1 µC ) (− 0.2 m ) iˆ = (0.899 N ) iˆ F3,1 = 8.99 × 109 N ⋅ m 2 /C 2 (4 µC ) (0.2 m )3

(

)

r

Substitute to find F1 : Express the net force acting on q2:

r F1 =

(0.899 N ) iˆ + (1.80 N ) ˆj

r r r F2 = F3, 2 + F1, 2 r r = F3, 2 − F2,1 r = F3, 2 − (1.80 N ) ˆj r r because F1, 2 and F2,1 are action-and-reaction forces.

18

Chapter 21 r kq q r F3, 2 = 33 2 r3, 2 r3, 2

Express the force that q3 exerts on q2:

=

[

kq3q2 (− 0.2 m ) iˆ + (0.1 m ) ˆj 3 r3, 2

]

r

Substitute numerical values and evaluate F3, 2 :

r F3, 2 = 8.99 × 109 N ⋅ m 2 /C 2 (4 µC )

(

)

(2 µC) [(− 0.2 m ) iˆ + (0.1 m ) ˆj ] (0.224 m )3

= (− 1.28 N ) iˆ + (0.640 N ) ˆj Find the net force acting on q2:

r r F2 = F3, 2 − (1.80 N ) ˆj = (− 1.28 N ) iˆ + (0.640 N ) ˆj − (1.80 N ) ˆj = r

(− 1.28 N ) iˆ − (1.16 N ) ˆj r

r

r

Noting that F1,3 and F3,1 are an action-and-reaction pair, as are F2,3 and F3, 2 , express the net force acting on q3:

[

r r r r r F3 = F1,3 + F2,3 = − F3,1 − F3, 2 = −(0.899 N ) iˆ − (− 1.28 N ) iˆ + (0.640 N ) ˆj =

(0.381 N ) iˆ − (0.640 N ) ˆj

34 •• Picture the Problem Let q1 represent the charge at the origin and q3 the charge initially at (8 cm, 0) and later at (17.75 cm, 0). The diagram shows the forces acting on q3 at (8 cm, 0). We can apply Coulomb’s law and the principle of superposition of forces to find the net force acting on each of the charges. Express the net force on q2 when it is at (8 cm, 0):

r r r F2 (8 cm,0 ) = F1,3 + F2,3 =

kq1q3 r kQ q r r1,3 + 32 3 r2,3 3 r1,3 r2,3

⎛q r Q r ⎞ = kq3 ⎜⎜ 31 r1,3 + 32 r2,3 ⎟⎟ r2,3 ⎝ r1,3 ⎠

]

The Electric Field 1: Discrete Charge Distributions

19

Substitute numerical values to obtain:

(− 19.7 N )iˆ =

(8.99 × 10

9

⎤ ⎡ 5 µC (0.08 m )iˆ + Q2 3 (0.04 m )iˆ⎥ N ⋅ m 2 /C 2 (2 µC ) ⎢ 3 (0.04 m ) ⎦ ⎣ (0.08 m )

)

Q2 = − 3.00 µC

Solve for and evaluate Q2:

Remarks: An alternative solution is to equate the electrostatic forces acting on q2 when it is at (17.75 cm, 0). 35 •• Picture the Problem By considering the symmetry of the array of charges we can see that the y component of the force on q is zero. We can apply Coulomb’s law and the principle of superposition of forces to find the net force acting on q. Express the net force acting on q:

r r r Fq = FQ on x axis, q + 2FQ at 45°, q

Express the force on q due to the charge Q on the x axis:

r kqQ FQ on x axis, q = 2 iˆ R

Express the net force on q due to the charges at 45°:

Substitute to obtain:

r kqQ 2FQ at 45°, q = 2 2 cos 45°iˆ R 2 kqQ ˆ = i 2 2 R r kqQ 2 kqQ ˆ Fq = 2 iˆ + i 2 R 2 R =

kqQ ⎛ 2 ⎞ˆ ⎜ ⎟i + 1 R 2 ⎜⎝ 2 ⎟⎠

36 ••• Picture the Problem Let the H+ ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0,

(

)

0), and H3 at a 2, a 3 2 ,0 . The N−3 ion, q4 in our notation, is then at

(a 2, a 2

)

3 , a 2 3 where a =1.64×10−10 m. To simplify our calculations we’ll set

ke a = C = 8.56 × 10 −9 N . We can apply Coulomb’s law and the principle of 2

2

superposition of forces to find the net force acting on each ion.

20

Chapter 21

Express the net force acting on q1:

r r r r F1 = F2,1 + F3,1 + F4,1

r

r kq q F2,1 = 12 2 rˆ2,1 = C − iˆ = −Ciˆ r2,1

r

r kq q F3,1 = 32 1 rˆ3,1 r3,1

( )

Find F2,1 :

Find F3,1 :

a⎞ ⎛ a 3⎞ˆ ⎛ ⎟j ⎜ 0 − ⎟ iˆ + ⎜⎜ 0 − 2 ⎟⎠ 2⎠ ⎝ ⎝ =C a ⎛1 3 ˆ⎞ j ⎟⎟ = −C ⎜⎜ iˆ + 2 2 ⎝ ⎠ Noting that the magnitude of q4 is three times that of the other charges and that it is

r

negative, express F4,1 :

a⎞ ⎛ a ⎞ˆ ⎛ a 2⎞ˆ ⎛ ⎟k ⎟ j + ⎜⎜ 0 − ⎜ 0 − ⎟iˆ + ⎜ 0 − ⎟ r 2 3 2 3 ⎠ ⎝ ⎝ ⎠ ⎝ ⎠ F4,1 = 3Crˆ4,1 = −3C 2 2 2 ⎛ a ⎞ ⎛ a ⎞ ⎛⎜ a 2 ⎞⎟ ⎟ + ⎜ ⎟ +⎜ ⎝ 2 ⎠ ⎝ 2 3 ⎠ ⎜⎝ 3 ⎟⎠ ⎛ a ⎞ ˆ ⎛ a ⎞ ˆ ⎛⎜ a 2 ⎞⎟ ˆ k ⎟j+ ⎜ ⎟i + ⎜ ⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞ ⎤ ⎝ 2 ⎠ ⎝ 2 3 ⎠ ⎜⎝ 3 ⎟⎠ ⎟kˆ ⎥ = 3C = 3C ⎢⎜ ⎟iˆ + ⎜ ⎟ ˆj + ⎜⎜ ⎟ 2 3 a ⎢⎣⎝ ⎠ ⎝ 2 3 ⎠ ⎝ ⎠ ⎥⎦

The Electric Field 1: Discrete Charge Distributions

r

21

r ⎛1 3 ˆ⎞ F1 = −Ciˆ − C ⎜⎜ iˆ + j⎟ 2 ⎟⎠ ⎝2

Substitute to find F1 :

⎡⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞ ⎤ ⎟kˆ ⎥ + 3C ⎢⎜ ⎟iˆ + ⎜ ⎟ ˆj + ⎜⎜ ⎟ ⎢⎣⎝ 2 ⎠ ⎝ 2 3 ⎠ ⎝ 3 ⎠ ⎥⎦ = C 6kˆ From symmetry considerations:

Express the condition that molecule is in equilibrium:

r

Solve for and evaluate F4 :

r r r F2 = F3 = F1 = C 6kˆ r r r r F1 + F2 + F3 + F4 = 0

(

)

r r r r r F4 = − F1 + F2 + F3 = −3F1 = − 3C 6kˆ

The Electric Field *37 • Picture the Problem Let q represent the charge at the origin and use Coulomb’s law for r E due to a point charge to find the electric field at x = 6 m and −10 m. (a) Express the electric field at a point P located a distance x from a charge q: Evaluate this expression for x = 6 m:

r kq E ( x ) = 2 rˆP,0 x

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (4 µC ) ˆ ( ) E 6m = i (6 m )2 =

(999 N/C)iˆ

r

(b) Evaluate E at x = −10 m:

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (4 µC ) ˆ E (− 10 m ) = −i = (10 m )2

( )

(− 360 N/C) iˆ

(c) The following graph was plotted using a spreadsheet program:

22

Chapter 21 500

Ex (N/C)

250

0

-250

-500 -2

-1

0

1

2

x (m)

*38 • Picture the Problem Let q represent the charges of +4 µC and use Coulomb’s law for r E due to a point charge and the principle of superposition for fields to find the electric field at the locations specified. Noting that q1 = q2, use Coulomb’s law and the principle of superposition to express the electric field due to the given charges at a point P a distance x from the origin:

r r r ⎛ 1 ⎞ 1 kq kq2 E ( x ) = E q1 ( x ) + E q2 ( x ) = 21 rˆq1 ,P + rˆ = kq1 ⎜⎜ 2 rˆq1 ,P + rˆ ⎟ 2 q2 ,P 2 q2 , P ⎟ x (8 m − x ) (8 m − x ) ⎝x ⎠ ⎛ 1 ⎞ 1 = 36 kN ⋅ m 2 /C ⎜⎜ 2 rˆq1 ,P + rˆ ⎟ 2 q2 , P ⎟ (8 m − x ) ⎝x ⎠

(

)

(a) Apply this equation to the point at x = −2 m:

( )

( )

()

( )

r ⎡ 1 ˆi + 1 ˆi ⎤ = − − E(− 2 m ) = 36 kN ⋅ m 2 /C ⎢ 2 (10 m )2 ⎥⎦ ⎣ (2 m )

(

)

(− 9.36 kN/C)ˆi

r

(b) Evaluate E at x = 2 m:

r ⎤ ⎡ 1 ˆ 1 E (2 m ) = 36 kN ⋅ m 2 /C ⎢ i + − iˆ ⎥ = (8.00 kN/C) iˆ 2 2 (6 m ) ⎦ ⎣ (2 m )

(

)

The Electric Field 1: Discrete Charge Distributions r

(c) Evaluate E at x = 6 m:

()

( )

r ⎤ ⎡ 1 ˆ 1 E (6 m ) = 36 kN ⋅ m 2 /C ⎢ i + − iˆ ⎥ = 2 2 (2 m ) ⎦ ⎣ (6 m )

(

)

(− 8.00 kN/C)iˆ

r

(d) Evaluate E at x = 10 m:

()

r ⎡ 1 E (10 m ) = 36 kN ⋅ m 2 /C ⎢ iˆ 2 ( ) 10 m ⎣

(

)

+

()

1 ˆ⎤ i = (9.35 kN/C) iˆ (2 m )2 ⎥⎦

E (4 m ) = 0

(e) From symmetry considerations:

(f) The following graph was plotted using a spreadsheet program:

100

Ex ( kN m2/C)

50

0

-50

-100 -4

0

4

8

12

x (m)

39 • Picture the Problem We can find the electric field at the origin from its definition and

r

r

the force on a charge placed there from F = qE . We can apply Coulomb’s law to find the value of the charge placed at y = 3 cm. (a) Apply the definition of electric field to obtain: (b) Express and evaluate the force on a charged body in an electric field:

r r F 8 × 10 −4 N ˆj E= = = q0 2 nC

(

)

(400 kN/C) ˆj

r r F = qE = (− 4 nC)(400 kN/C) ˆj =

(− 1.60 mN) ˆj

23

24

Chapter 21

(c) Apply Coulomb’s law to obtain:

Solve for and evaluate q:

kq (− 4 nC ) ˆ − j = (− 1.60 mN ) ˆj 2 (0.03 m )

( )

q=−

(1.60 mN )(0.03 m )2 (8.99 × 109 N ⋅ m 2 /C 2 )(4 nC)

= − 40.0 nC 40 • Picture the Problem We can compare the electric and gravitational forces acting on an electron by expressing their ratio. We can equate these forces to find the charge that would have to be placed on a penny in order to balance the earth’s gravitational force on it. (a) Express the magnitude of the electric force acting on the electron: Express the magnitude of the gravitational force acting on the electron:

Fe = eE

Fg = me g

Express the ratio of these forces to obtain:

Fe eE = Fg mg

Substitute numerical values and evaluate Fe/Fg:

Fe 1.6 × 10 −19 C (150 N/C ) = 9.11 × 10 −31 kg 9.81 m/s 2 Fg

(

(

) )(

)

= 2.69 × 1012 or

Fe =

(2.69 × 10 )F 12

g

, i.e., the electric

force is greater by a factor of 2.69×1012. (b) Equate the electric and gravitational forces acting on the penny and solve for q to obtain:

q=

Substitute numerical values and evaluate q:

q=

mg E

(3 × 10

−3

)(

kg 9.81m/s 2 150 N/C

= 1.96 × 10− 4 C

)

The Electric Field 1: Discrete Charge Distributions

25

41 •• Picture the Problem The diagram shows the locations of the charges q1 and q2 and the point on the x axis at which we are to

r

find E . From symmetry considerations we

r

can conclude that the y component of E at any point on the x axis is zero. We can use Coulomb’s law for the electric field due to point charges to find the field at any point

r

r

on the x axis and F = qE to find the force on a charge q0 placed on the x axis at x = 4 cm. (a) Letting q = q1 = q2, express the xcomponent of the electric field due to one charge as a function of the distance r from either charge to the point of interest:

r

Express E x for both charges:

r kq E x = 2 cosθ iˆ r

r kq E x = 2 2 cosθ iˆ r

Substitute for cosθ and r, substitute numerical values, and evaluate to obtain:

(

)

r kq 0.04 m ˆ 2kq(0.04 m ) ˆ 2 8.99 × 109 N ⋅ m 2 /C 2 (6 nC )(0.04 m ) ˆ Ex = 2 2 i= i= i 3 2 r r r3 (0.03 m )2 + (0.04 m )2

=

[

]

(34.5 kN/C)iˆ r

r

(b) Apply F = qE to find the force on a charge q0 placed on the x axis at x = 4 cm:

r F = (2 nC)(34.5 kN/C)iˆ =

(69.0 µN )iˆ

*42 •• Picture the Problem If the electric field at x = 0 is zero, both its x and y components must be zero. The only way this condition can be satisfied with the point charges of +5.0 µC and −8.0 µC are on the x axis is if the point charge of +6.0 µC is also on the x axis. Let the subscripts 5, −8, and 6 identify the point charges and their fields. We can use r Coulomb’s law for E due to a point charge and the principle of superposition for fields to determine where the +6.0 µC charge should be located so that the electric field at x = 0 is zero.

26

Chapter 21

Express the electric field at x = 0 in terms of the fields due to the charges of +5.0 µC, −8.0 µC, and +6.0 µC: Substitute for each of the fields to obtain:

r r r r E (0) = E5 µC + E−8 µC + E6 µC =0 kq5 kq kq rˆ5 + 26 rˆ6 + 2−8 rˆ−8 = 0 2 r5 r6 r−8 or

( )

( )

kq5 ˆ kq6 ˆ kq−8 ˆ i + 2 −i + 2 −i =0 r52 r6 r−8 Divide out the unit vector iˆ to obtain:

q5 q6 q−8 − − =0 r52 r62 r−28

Substitute numerical values to obtain:

−8 5 6 − 2− =0 2 (3 cm ) r6 (4 cm )2

Solve for r6:

r6 = 2.38 cm

43 •• Picture the Problem The diagram shows the electric field vectors at the point of interest r P due to the two charges. We can use Coulomb’s law for E due to point charges and the

r

r

r

superposition principle for electric fields to find E P . We can apply F = qE to find the force on an electron at (−1 m, 0).

(a) Express the electric field at (−1 m, 0) due to the charges q1 and q2:

r r r EP = E1 + E2

The Electric Field 1: Discrete Charge Distributions

r

Evaluate E1 :

(

)

r kq1 8.99 × 109 N ⋅ m 2 /C 2 (− 5 µC ) ⎛⎜ (− 5 m ) iˆ + (2 m ) ˆj ⎞⎟ ˆ E1 = 2 r1,P = ⎜ (5 m )2 + (2 m )2 ⎟ r1,P (5 m )2 + (2 m )2 ⎝ ⎠ 3 = − 1.55 × 10 N/C − 0.928 iˆ + 0.371 ˆj

)(

(

)

= (1.44 kN/C ) iˆ + (− 0.575 kN/C) ˆj

r

Evaluate E2 :

(

)

r kq 8.99 × 109 N ⋅ m 2 /C 2 (12 µC ) ⎛⎜ (− 2 m ) iˆ + (− 2 m ) ˆj ⎞⎟ E 2 = 2 2 rˆ2,P = ⎜ (2 m )2 + (2 m )2 ⎟ r2,P (2 m )2 + (2 m )2 ⎝ ⎠ 3 = 13.5 × 10 N/C − 0.707 iˆ − 0.707 ˆj

)(

(

)

= (− 9.54 kN/C ) iˆ + (− 9.54 kN/C) ˆj

r

r

r

Substitute for E1 and E 2 and simplify to find E P :

r EP = (1.44 kN/C) iˆ + (− 0.575 kN/C) ˆj + (− 9.54 kN/C) iˆ + (− 9.54 kN/C) ˆj = (− 8.10 kN/C) iˆ + (− 10.1kN/C) ˆj r

The magnitude of E P is:

EP =

(− 8.10 kN/C)2 + (− 10.1kN/C)2

= 12.9 kN/C

r

The direction of E P is:

⎛ − 10.1 kN/C ⎞ ⎟⎟ ⎝ − 8.10 kN/C ⎠

θ E = tan −1 ⎜⎜

= 231° Note that the angle returned by your

⎛ − 10.1 kN/C ⎞

−1 ⎟⎟ is the calculator for tan ⎜⎜ ⎝ − 8.10 kN/C ⎠

reference angle and must be increased by 180° to yield θE. (b) Express and evaluate the force on an electron at point P:

27

28

Chapter 21

)[

r r F = qEP = − 1.602 × 10−19 C (− 8.10 kN/C) iˆ + (− 10.1kN/C) ˆj = 1.30 × 10−15 N iˆ + 1.62 × 10−15 N ˆj

(

(

) (

r

Find the magnitude of F :

)

F=

(1.30 × 10

−15

]

) ( 2

N + 1.62 × 10−15 N

)

= 2.08 × 10−15 N r

Find the direction of F :

⎛ 1.62 × 10 −15 N ⎞ ⎟⎟ = 51.3° −15 ⎝ 1.3 × 10 N ⎠

θ F = tan −1 ⎜⎜

44 •• Picture the Problem The diagram shows the locations of the charges q1 and q2 and the r point on the x axis at which we are to find E . From symmetry considerations we can r conclude that the y component of E at any point on the x axis is zero. We can use Coulomb’s law for the electric field due to point charges to find the field at any point on the x axis. We can establish the results called for in parts (b) and (c) by factoring the radicand and using the approximation 1 + α ≈ 1 whenever α << 1.

(a) Express the x-component of the electric field due to the charges at y = a and y = −a as a function of the distance r from either charge to point P: Substitute for cosθ and r to obtain:

r kq E x = 2 2 cos θ iˆ r

r kq x 2kqx 2kqx E x = 2 2 iˆ = 3 iˆ = iˆ 32 2 2 r r r (x + a ) 2kqx iˆ = 2 (x + a 2 )3 2

2

The Electric Field 1: Discrete Charge Distributions

29

and

(b) For x << a, x2 + a2 ≈ a2, so:

For x >> a, x2 + a2 ≈ x2, so:

2kqx

Ex =

(x

Ex ≈

2kqx

Ex ≈

2

+ a2

)

32

(a )

=

2kqx a3

2kqx

=

2kq x2

2 32

(x )

2 32

For x >> a, the charges separated by a would appear to be a single charge (c) 2kq of magnitude 2q. Its field would be given by E x = 2 . x Factor the radicand to obtain:

For a << x:

⎡ ⎛ a 2 ⎞⎤ Ex = 2kqx ⎢ x 2 ⎜⎜1 + 2 ⎟⎟⎥ x ⎠⎦ ⎣ ⎝ a2 1+ 2 ≈1 x

−3 2

and

[ ]

E x = 2kqx x 2

−3 2

=

2kq x2

*45 •• Picture the Problem The diagram shows the electric field vectors at the point of interest r P due to the two charges. We can use Coulomb’s law for E due to point charges and the

r

r

r

superposition principle for electric fields to find E P . We can apply F = qE to find the force on a proton at (−3 m, 1 m).

30

Chapter 21

r r r E P = E1 + E 2

(a) Express the electric field at (−3 m, 1 m) due to the charges q1 and q2:

r

Evaluate E1 :

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (− 4 µC ) ⎛⎜ (− 5 m ) iˆ + (3 m ) ˆj ⎞⎟ kq1 ˆ E1 = 2 r1,P = ⎜ (5 m )2 + (3 m )2 ⎟ r1.P (5 m )2 + (3 m )2 ⎠ ⎝ = (− 1.06 kN/C ) − 0.857 iˆ + 0.514 ˆj = (0.908 kN/C ) iˆ + (− 0.544 kN/C ) ˆj

(

)

r

Evaluate E2 :

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (5 µC ) ⎛⎜ (− 4 m ) iˆ + (− 2 m ) ˆj ⎞⎟ kq E 2 = 2 2 rˆ2,P = ⎜ (4 m )2 + (2 m )2 ⎟ r2,P (4 m )2 + (2 m )2 ⎠ ⎝ = (2.25 kN/C ) − 0.894 iˆ − 0.447 ˆj = (− 2.01 kN/C ) iˆ + (− 1.01 kN/C ) ˆj

(

)

r

Substitute and simplify to find E P :

r EP = (0.908 kN/C) iˆ + (− 0.544 kN/C) ˆj + (− 2.01kN/C) iˆ + (− 1.01kN/C) ˆj = (− 1.10 kN/C) iˆ + (− 1.55 kN/C) ˆj r

The magnitude of E P is:

EP =

(1.10 kN/C)2 + (1.55 kN/C)2

= 1.90 kN/C

The Electric Field 1: Discrete Charge Distributions

r

⎛ − 1.55 kN/C ⎞ ⎟⎟ = 235° ⎝ − 1.10 kN/C ⎠

The direction of E P is:

θ E = tan −1 ⎜⎜

Note that the angle returned by your

⎛ − 1.55 kN/C ⎞

−1 ⎟⎟ is the calculator for tan ⎜⎜ ⎝ − 1.10 kN/C ⎠

reference angle and must be increased by 180° to yield θE. (b) Express and evaluate the force on a proton at point P:

)[ ) (

r r F = qE P = 1.6 ×10 −19 C (− 1.10 kN/C) iˆ + (− 1.55 kN/C) ˆj = − 1.76 ×10 −16 N iˆ + − 2.48 ×10 −16 N ˆj

(

(

)

]

r

The magnitude of F is:

F=

(− 1.76 × 10

−16

r

N

) + (− 2.48 × 10 2

−16

N

)

2

= 3.04 × 10 −16 N

⎛ − 2.48 × 10 −16 N ⎞ ⎟⎟ = 235° θ F = tan ⎜⎜ −16 ⎝ − 1.76 × 10 N ⎠

The direction of F is:

−1

where, as noted above, the angle returned by your calculator for

⎛ − 2.48 × 10 −16 N ⎞ ⎟⎟ is the reference tan −1 ⎜⎜ −16 ⎝ − 1.76 × 10 N ⎠ angle and must be increased by 180° to yield θE. 46 •• Picture the Problem In Problem 44 it is shown that the electric field on the x axis, due to equal positive charges located at (0, a) and (0,−a), is given by

E x = 2kqx (x 2 + a 2 )

−3 2

. We can identify the locations at which Ex has it greatest values

by setting dEx/dx equal to zero. (a) Evaluate

dEx : dx

31

32

Chapter 21

[

(

]

)

[(

−3 2 dEx d d 2kqx x 2 + a 2 = = 2kq x x2 + a2 dx dx dx −3 2 −3 2 ⎤ ⎡ d 2 = 2kq ⎢ x x + a2 + x2 + a2 ⎥⎦ ⎣ dx

(

)

⎡ ⎛ 3⎞ = 2kq ⎢ x⎜ − ⎟ x 2 + a 2 ⎣ ⎝ 2⎠

(

[

(

= 2kq − 3 x 2 x 2 + a 2

(

−5 2

−5 2

(

+ x2 + a2

2

+ a2

)

−3 2

− 3x 2 (x 2 + a 2 )

]

−5 2

Set this derivative equal to zero: Solve for x to obtain:

x= ±

−3 2

]

)

) (2 x ) + (x

)

)

)

−3 2

⎤ ⎥ ⎦

+ (x 2 + a 2 )

−3 2

=0

a 2

(b) The following graph was plotted using a spreadsheet program:

2kq = 1 and a = 1 0.4

0.2

Ex

0.0

-0.2

-0.4 -10

-5

0

5

10

x

47 ••• Picture the Problem We can determine the stability of the equilibrium in Part (a) and Part (b) by considering the forces the equal charges q at y = +a and y = −a exert on the test charge when it is given a small displacement along either the x or y axis. The application of Coulomb’s law in Part (c) will lead to the magnitude and sign of the charge that must be placed at the origin in order that a net force of zero is experienced by each of the three charges. (a) Because Ex is in the x direction, a positive test charge that is displaced from

The Electric Field 1: Discrete Charge Distributions

33

(0, 0) in either the +x direction or the −x direction will experience a force pointing away from the origin and accelerate in the direction of the force.

Consequently, the equilibrium at (0,0) is unstable for a small displacement along the x axis. If the positive test charge is displaced in the direction of increasing y (the positive y direction), the charge at y = +a will exert a greater force than the charge at y = −a, and the net force is then in the −y direction; i.e., it is a restoring force. Similarly, if the positive test charge is displaced in the direction of decreasing y (the negative y direction), the charge at y = −a will exert a greater force than the charge at y = −a, and the net force is then in the −y direction; i.e., it is a restoring force.

Consequently, the equilibrium at (0,0) is stable for a small displacement along the y axis. (b)

Following the same arguments as in Part (a ), one finds that, for a negative test charge, the equilibrium is stable at (0,0) for displacements along the x axis and unstable for displacements along the y axis.

kqq0 kq 2 = 2 + =0 a (2a )2

(c) Express the net force acting on the charge at y = +a:

∑F

Solve for q0 to obtain:

q0 = − 14 q0

q at y =+ a

Remarks: In Part (c), we could just as well have expressed the net force acting on the charge at y = −a. Due to the symmetric distribution of the charges at y = −a and y = +a, summing the forces acting on q0 at the origin does not lead to a relationship between q0 and q. *48 ••• Picture the Problem In Problem 44 it is shown that the electric field on the x axis, due to equal positive charges located at (0, a) and (0,−a), is given by

E x = 2kqx (x 2 + a 2 )

−3 2

. We can use T = 2π m k' to express the period of the motion in

terms of the restoring constant k′. (a) Express the force acting on the on the bead when its displacement from the origin is x:

Fx = −qEx = −

(x

2kq 2 x 2

+ a2

)

32

34

Chapter 21

Factor a2 from the denominator to obtain:

For x << a:

Fx = −

2kq 2 x ⎛ x2 ⎞ a ⎜⎜ 2 + 1⎟⎟ ⎝a ⎠

32

2

Fx = −

2kq 2 x a3

i.e., the bead experiences a linear restoring force. (b) Express the period of a simple harmonic oscillator:

T = 2π

Obtain k′ from our result in part (a):

k' =

Substitute to obtain:

m k'

2kq 2 a3

T = 2π

m ma3 2 = π 2kq 2 2kq 2 a3

Motion of Point Charges in Electric Fields 49 • Picture the Problem We can use Newton’s 2nd law of motion to find the acceleration of the electron in the uniform electric field and constant-acceleration equations to find the time required for it to reach a speed of 0.01c and the distance it travels while acquiring this speed. (a) Use data found at the back of your text to compute e/m for an electron:

1.6 × 10 −19 C e = me 9.11 × 10 −31 kg

(b) Apply Newton’s 2nd law to relate the acceleration of the electron to the electric field:

a=

Substitute numerical values and evaluate a:

a=

= 1.76 × 1011 C/kg Fnet eE = me me

(1.6 × 10

)

C (100 N/C ) 9.11 × 10 −31 kg −19

= 1.76 × 1013 m/s 2

The Electric Field 1: Discrete Charge Distributions The direction of the acceleration of an electron is opposite the electric field. (c) Using the definition of acceleration, relate the time required for an electron to reach 0.01c to its acceleration:

∆t =

v 0.01c = a a

Substitute numerical values and evaluate ∆t:

∆t =

0.01 3 × 108 m/s = 0.170 µs 1.76 × 1013 m/s2

(d) Find the distance the electron travels from its average speed and the elapsed time:

(

)

∆x = vav ∆t =

1 2

[0 + 0.01(3 × 10

8

)]

m/s (0.170 µs )

= 25.5 cm

*50 • Picture the Problem We can use Newton’s 2nd law of motion to find the acceleration of the proton in the uniform electric field and constant-acceleration equations to find the time required for it to reach a speed of 0.01c and the distance it travels while acquiring this speed. (a) Use data found at the back of your text to compute e/m for an electron:

1.6 × 10 −19 C e = m p 1.67 × 10 −27 kg

Apply Newton’s 2nd law to relate the acceleration of the electron to the electric field:

a=

Substitute numerical values and evaluate a:

a=

= 9.58 × 107 C/kg Fnet eE = mp mp

(1.6 × 10

)

C (100 N/C ) 1.67 × 10 −27 kg −19

= 9.58 × 109 m/s 2 The direction of the acceleration of a proton is in the direction of the electric field.

35

36

Chapter 21

(b) Using the definition of acceleration, relate the time required for an electron to reach 0.01c to its acceleration:

∆t =

v 0.01c = a a

Substitute numerical values and evaluate ∆t:

∆t =

0.01(3 × 108 m/s) = 313 µs 9.58 × 109 m/s2

51 • Picture the Problem The electric force acting on the electron is opposite the direction of the electric field. We can apply Newton’s 2nd law to find the electron’s acceleration and use constant acceleration equations to find how long it takes the electron to travel a given distance and its deflection during this interval of time. r r (a) Use Newton’s 2nd law to relate the r Fnet − eE = a= acceleration of the electron first to the me me net force acting on it and then the electric field in which it finds itself: Substitute numerical values and r evaluate a :

r 1.6 × 10 −19 C (400 N/C) ˆj a=− 9.11 × 10 −31 kg =

(b) Relate the time to travel a given distance in the x direction to the electron’s speed in the x direction: (c) Using a constant-acceleration equation, relate the displacement of the electron to its acceleration and the elapsed time:

∆t =

(− 7.03 × 10

)

m/s 2 ˆj

13

∆x 0.1 m = = 50.0 ns vx 2 × 106 m/s

r r 2 ∆y = 12 a y (∆t ) = =

1 2

(− 7.03 × 10

13

)

2 m/s 2 (50 ns ) ˆj

(− 8.79 cm ) ˆj

i.e., the electron is deflected 8.79 cm downward. 52 •• Picture the Problem Because the electric field is uniform, the acceleration of the electron will be constant and we can apply Newton’s 2nd law to find its acceleration and use a constant-acceleration equation to find its speed as it leaves the region in which there is a uniform electric field. Using a constant-acceleration

v 2 = v02 + 2a∆x

The Electric Field 1: Discrete Charge Distributions equation, relate the speed of the electron as it leaves the region of the electric field to its acceleration and distance of travel:

or, because v0 = 0,

Apply Newton’s 2nd law to express the acceleration of the electron in terms of the electric field:

a=

Fnet eE = me me

Substitute to obtain:

v=

2eE∆x me

Substitute numerical values and evaluate v:

v=

(

)(

37

v = 2 a ∆x

)

2 1.6 × 10 −19 C 8 × 10 4 N/C (0.05 m ) = 3.75 × 10 7 m/s −31 9.11 × 10 kg

Remarks: Because this result is approximately 13% of the speed of light, it is only an approximation. 53 •• Picture the Problem We can apply the work-kinetic energy theorem to relate the change in the object’s kinetic energy to the net force acting on it. We can express the net force acting on the charged body in terms of its charge and the electric field. Using the work-kinetic energy theorem, express the kinetic energy of the object in terms of the net force acting on it and its displacement: Relate the net force acting on the charged object to the electric field: Substitute to obtain:

W = ∆K = Fnet ∆x

Fnet = QE

∆K = K f − K i = QE∆x or, because Ki = 0,

K f = QE∆x

Solve for Q:

Q=

Kf E∆x

38

Chapter 21

Substitute numerical values and evaluate Q:

Q=

0.12 J = 800 µC (300 N/C )(0.50 m )

*54 •• Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the particle in terms of the parameter t and Newton’s 2nd law to express the constant acceleration in terms of the electric field. Eliminating the parameter will yield an equation for y as a function of x, q, and m that we can solve for Ey. Express the x and y coordinates of the particle as functions of time:

x = (v cos θ ) t and

y = (v sin θ )t − 12 a y t 2 Fnet, y

qE y

Apply Newton’s 2nd law to relate the acceleration of the particle to the net force acting on it:

ay =

Substitute in the y-coordinate equation to obtain:

y = (v sin θ )t −

Eliminate the parameter t between the two equations to obtain:

y = (tan θ )x −

Set y = 0 and solve for Ey:

Ey =

Substitute the non-particle specific data to obtain:

Ey =

m

=

m

qE y

2m

t2

qE y

2mv cos θ 2

2

x2

mv 2 sin 2θ qx

(

)

2

m 3 × 106 m/s sin 70° q(0.015 m ) m = 5.64 × 1014 m/s 2 q

(

)

(a) Substitute for the mass and charge of an electron and evaluate Ey:

9.11 × 10 −31 kg E y = 5.64 × 10 m/s 1.6 × 10 −19 C

(b) Substitute for the mass and charge of a proton and evaluate Ey:

1.67 × 10−27 kg E y = 5.64 × 10 m/s 1.6 × 10−19 C

(

14

2

)

= 3.21 kN/C

(

14

= 5.89 MN/C

2

)

The Electric Field 1: Discrete Charge Distributions 55 •• Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the electron in terms of the parameter t and Newton’s 2nd law to express the constant acceleration in terms of the electric field. Eliminating the parameter will yield an equation for y as a function of x, q, and m. We can decide whether the electron will strike the upper plate by finding the maximum value of its y coordinate. Should we find that it does not strike the upper plate, we can determine where it strikes the lower plate by setting y(x) = 0. Express the x and y coordinates of the electron as functions of time:

x = (v0 cos θ )t and

y = (v0 sin θ )t − 12 a y t 2

Fnet, y eE y = me me

Apply Newton’s 2nd law to relate the acceleration of the electron to the net force acting on it:

ay =

Substitute in the y-coordinate equation to obtain:

y = (v0 sin θ )t −

Eliminate the parameter t between the two equations to obtain:

y ( x ) = (tan θ )x −

To find ymax, set dy/dx = 0 for extrema:

eE y dy = tan θ − x' dx me v02 cos 2 θ

eE y 2me

t2

eE y x 2 (1) 2me v02 cos 2 θ

= 0 for extrema me v02 sin 2θ (See remark below.) 2eE y

Solve for x′ to obtain:

x' =

Substitute x′ in y(x) and simplify to obtain ymax:

ymax =

me v02 sin 2 θ 2eE y

Substitute numerical values and evaluate ymax:

ymax =

(9.11× 10 kg )(5 × 10 m/s) sin 45° = 1.02 cm 2(1.6 × 10 C )(3.5 × 10 N/C) −31

2

6

−19

2

3

and, because the plates are separated by 2 cm, the electron does not strike the upper plate.

39

40

Chapter 21 me v02 sin 2θ x= eE y

To determine where the electron will strike the lower plate, set y = 0 in equation (1) and solve for x to obtain: Substitute numerical values and evaluate x:

x=

(9.11× 10 kg )(5 × 10 m/s) sin 90° = (1.6 × 10 C)(3.5 × 10 N/C) −31

2

6

−19

3

4.07 cm

Remarks: x′ is an extremum, i.e., either a maximum or a minimum. To show that it is a maximum we need to show that d2y/dx2, evaluated at x′, is negative. A simple alternative is to use your graphing calculator to show that the graph of y(x) is a maximum at x′. Yet another alternative is to recognize that, because equation (1) is quadratic and the coefficient of x2 is negative, its graph is a parabola that opens downward. 56 •• Picture the Problem The trajectory of the electron while it is in the electric field is parabolic (its acceleration is downward and constant) and its trajectory, once it is out of the electric field is, if we ignore the small gravitational force acting on it, linear. We can use constant-acceleration equations and Newton’s 2nd law to express the electron’s x and y coordinates parametrically and then eliminate the parameter t to express y(x). We can find the angle with the horizontal at which the electron leaves the electric field from the x and y components of its velocity and its total vertical deflection by summing its deflections over the first 4 cm and the final 12 cm of its flight. (a) Using a constant-acceleration equation, express the x and y coordinates of the electron as functions of time: Because v0,y = 0:

x(t ) = v0t and

y (t ) = v0 , y t + 12 a y t 2

x(t ) = v0t and

y (t ) = 12 a y t 2

Using Newton’s 2nd law, relate the acceleration of the electron to the electric field:

ay =

Fnet − eE y = me me

(1)

The Electric Field 1: Discrete Charge Distributions

y (t ) = −

Substitute to obtain:

eE y 2me

y(x ) = −

Eliminate the parameter t between equations (1) and (2) to obtain:

t2

eE y 2 e 0

2m v

(2)

x2 = −

eE y 4K

x2

Substitute numerical values and evaluate y(4 cm):

y (0.04 m ) = −

(1.6 × 10

−19

C )(2 × 10 4 N/C )(0.04 m ) = − 6.40 mm 4(2 × 10 −16 J ) 2

v x = v0 cos θ

(b) Express the horizontal and vertical components of the electron’s speed as it leaves the electric field:

and

Divide the second of these equations by the first to obtain:

θ = tan −1

Using a constant-acceleration equation, express vy as a function of the electron’s acceleration and its time in the electric field:

v y = v0 sin θ vy v = tan −1 y vx v0

v y = v0, y + a y t or, because v0,y = 0

v y = a yt =

Fnet, y eE x t=− y me me v0

⎛ eE y x ⎞ ⎛ eE x ⎞ ⎟ = tan −1 ⎜⎜ − y ⎟⎟ 2 ⎟ ⎝ 2K ⎠ ⎝ mev0 ⎠

Substitute to obtain:

θ = tan −1 ⎜⎜ −

Substitute numerical values and evaluate θ :

⎡ (1.6 × 10 −19 C )(2 × 10 4 N/C)(0.04 m ) ⎤ ⎥ = − 17.7° 2(2 × 10 −16 J ) ⎣ ⎦

θ = tan −1 ⎢−

(c) Express the total vertical displacement of the electron: Relate the horizontal and vertical distances traveled to the screen to the horizontal and vertical components of its velocity:

ytotal = y4 cm + y12 cm x = v x ∆t and

y = v y ∆t

41

42

Chapter 21

Eliminate ∆t from these equations to obtain: Substitute numerical values and evaluate y: Substitute for y4 cm and y12 cm and evaluate ytotal:

y=

vy vx

x = (tan θ )x

y = [tan(− 17.7°)](0.12 m ) = −3.83 cm

ytotal = −0.640 cm − 3.83 cm = − 4.47 cm i.e., the electron will strike the fluorescent screen 4.47 cm below the horizontal axis.

57 • Picture the Problem We can use its definition to find the dipole moment of this pair of charges. (a) Apply the definition of electric dipole moment to obtain:

r r p = qL and

p = (2 pC )(4 µm ) = 8.00 × 10−18 C ⋅ m

(b) If we assume that the dipole is oriented as shown to the right, then r p is to the right; pointing from the negative charge toward the positive charge. *58 • Picture the Problem The torque on an electric dipole in an electric field is given by

r r r r r τ = p × E and the potential energy of the dipole by U = − p ⋅ E .

Using its definition, express the torque on a dipole moment in a uniform electric field:

r r r τ = p× E and

τ = pE sin θ

where θ is the angle between the electric dipole moment and the electric field. (a) Evaluate τ for θ = 0°:

τ = pE sin 0° = 0

The Electric Field 1: Discrete Charge Distributions

43

τ = (0.5 e ⋅ nm ) (4.0 × 10 4 N/C)sin 90°

(b) Evaluate τ for θ = 90°:

= 3.20 × 10 −24 N ⋅ m

τ = (0.5 e ⋅ nm ) (4.0 × 10 4 N/C)sin 30°

(c) Evaluate τ for θ = 30°:

= 1.60 × 10 −24 N ⋅ m r r U = − p ⋅ E = − pE cosθ

(d) Using its definition, express the potential energy of a dipole in an electric field:

U = −(0.5 e ⋅ nm )(4.0 × 104 N/C)cos 0°

Evaluate U for θ = 0°:

= − 3.20 × 10− 24 J

U = −(0.5 e ⋅ nm )(4.0 × 104 N/C)cos 90°

Evaluate U for θ = 90°:

= 0 U = −(0.5 e ⋅ nm )(4.0 × 104 N/C)cos 30°

Evaluate U for θ = 30°:

= − 2.77 × 10− 24 J *59 •• Picture the Problem We can combine the dimension of an electric field with the dimension of an electric dipole moment to prove that, in any direction, the dimension of

the far field is proportional to 1 [L] and, hence, the electric field far from the dipole falls 3

off as 1/r3. Express the dimension of an electric field: Express the dimension an electric dipole moment: Write the dimension of charge in terms of the dimension of an electric dipole moment: Substitute to obtain:

[E ] = [kQ2] [L]

[ p] = [Q][L] [Q] = [ p] [L]

[E ] = [k ][2 p] = [k ][ p3 ] [L] [L] [L] This shows that the field E due to a dipole

44

Chapter 21 p falls off as 1/r3.

60 •• Picture the Problem We can use its definition to find the molecule’s dipole moment. From the symmetry of the system, it is evident that the x component of the dipole moment is zero. Using its definition, express the molecule’s dipole moment: From symmetry considerations we have: The y component of the molecule’s dipole moment is:

r p = px iˆ + p y ˆj px = 0

p y = qL = 2eL

(

)

= 2 1.6 × 10 −19 C (0.058 nm ) = 1.86 × 10 −29 C ⋅ m

Substitute to obtain:

(

)

r p = 1.86 × 10 −29 C ⋅ m ˆj

61 •• Picture the Problem We can express the net force on the dipole as the sum of the forces acting on the two charges that constitute the dipole and simplify this expression r r to show that Fnet = Cpiˆ. We can show that, under the given conditions, Fnet is also given by (dE x dx ) piˆ by differentiating the dipole’s potential energy function with respect to x. (a) Express the net force acting on the dipole: Apply Coulomb’s law to express the forces on the two charges:

Substitute to obtain:

r r r Fnet = F−q + F+q r r F− q = −qE = −qC ( x1 − a ) iˆ and

r r F+ q = + qE = qC ( x1 + a ) iˆ r Fnet = −qC ( x1 − a ) iˆ + qC ( x1 + a ) iˆ

= 2aqC iˆ = Cp iˆ where p = 2aq.

The Electric Field 1: Discrete Charge Distributions (b) Express the net force acting on the dipole as the spatial derivative of U:

45

r dU ˆ d Fnet = − i = − [− px E x ]iˆ dx dx dE = px x iˆ dx

62 ••• Picture the Problem We can express the force exerted on the dipole by the electric field as −dU/dr and the potential energy of the dipole as −pE. Because the field is due to a point charge, we can use Coulomb’s law to express E. In the second part of the problem, we can use Newton’s 3rd law to show that the magnitude of the electric field of the dipole along the line of the dipole a distance r away is approximately 2kp/r3. (a) Express the force exerted by the electric field of the point charge on the dipole:

r dU F =− rˆ dr where rˆ is a unit radial vector pointing from Q toward the dipole.

kQ r2

Express the potential energy of the dipole in the electric field:

U = − pE = − p

Substitute to obtain:

r 2kQp d ⎡ kQ ⎤ F = − ⎢− p 2 ⎥ rˆ = − 3 rˆ dr ⎣ r ⎦ r

(b) Using Newton’s 3rd law, express the force that the dipole exerts on the charge Q at the origin:

Express Fon Q in terms of the field in

r r Fon Q = − F or Fon Q rˆ = − Frˆ and

Fon Q = F

Fon Q = QE

which Q finds itself: Substitute to obtain:

QE =

2kQp 2kp ⇒E = 3 r3 r

General Problems *63 • Picture the Problem We can equate the gravitational force and the electric force acting on a proton to find the mass of the proton under the given condition. (a) Express the condition that must be satisfied if the net force on the

Fg = Fe

46

Chapter 21

proton is zero: Use Newton’s law of gravity and Coulomb’s law to substitute for Fg and Fe:

Gm 2 ke 2 = 2 r2 r

Solve for m to obtain:

m=e

k G

Substitute numerical values and evaluate m:

(

m = 1.6 × 10 −19 C

)

8.99 × 10 9 N ⋅ m 2 / C 2 = 1.86 × 10 −9 kg 6.67 × 10 −11 N ⋅ m 2 / kg 2

(b) Express the ratio of Fe and Fg:

ke 2 2 r 2 = ke Gmp2 Gmp2 r2

Substitute numerical values to obtain:

(8.99 × 109 N ⋅ m 2 / C2 )(1.6 × 10−19 C) = 1.24 × 1036 ke 2 = Gmp2 (6.67 × 10 −11 N ⋅ m 2 / kg 2 )(1.67 × 10 −27 kg )2 2

64 •• Picture the Problem The locations of the charges q1, q2 and q2 and the points at which we are calculate the field are shown in the diagram. From the diagram it is evident that r r E along the axis has no y component. We can use Coulomb’s law for E due to a point r charge and the superposition principle to find E at points P1 and P2. Examining the distribution of the charges we can see that there are two points where E = 0. One is between q2 and q3 and the other is to the left of q1.

The Electric Field 1: Discrete Charge Distributions r r r r E P1 = E q1 + E q2 + E q3

Using Coulomb’s law, express the electric field at P1 due to the three charges:

=

kq1 ˆ kq2 ˆ kq3 ˆ i+ 2 i+ 2 i r12,P1 r2,P1 r3,P1

⎡q q q ⎤ = k ⎢ 21 + 22 + 23 ⎥ iˆ ⎣⎢ r1,P1 r2,P1 r3,P1 ⎦⎥

r

Substitute numerical values and evaluate E P1 :

r ⎡ − 5 µC 3 µC 5 µC ⎤ ˆ E P1 = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ i + + 2 2 2⎥ ⎣ (4 cm ) (3 cm ) (2 cm ) ⎦ = 1.14 × 108 N/C iˆ

(

)

(

)

r r r r E P2 = E q1 + E q2 + E q3

Express the electric field at P2:

⎡q q q ⎤ = k ⎢ 21 + 22 + 23 ⎥ iˆ ⎣⎢ r1,P2 r2,P2 r3,P2 ⎦⎥

r

Substitute numerical values and evaluate E P2 :

r ⎡ − 5 µC 3 µC 5 µC ⎤ ˆ E P2 = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ i + + 2 2 2⎥ ⎣ (16 cm ) (15 cm ) (14 cm ) ⎦ = 1.74 × 106 N/C iˆ

(

)

(

)

Letting x represent the x coordinate of a point where the magnitude of the electric field is zero, express the condition that E = 0 for the point between x = 0 and x = 1 cm:

⎡q q ⎤ q EP = k ⎢ 21 + 22 + 23 ⎥ = 0 ⎢⎣ r1,P r2,P r3,P ⎥⎦ or

3 µC 5 µC − 5 µC + 2 − =0 2 (x + 1cm ) x (1cm - x )2

Solve this equation to obtain:

x = 0.417 cm

For x < −1 cm, let y = −x to obtain:

5 µC 3 µC 5 µC − 2 − =0 2 ( y − 1cm ) y ( y + 1cm )2

Solve this equation to obtain:

x = 6.95 cm and y = − 6.95 cm

47

Chapter 21

48

65 •• Picture the Problem The locations of the charges q1, q2 and q2 and the point P2 at which we are calculate the field are shown in the diagram. The electric field on the x axis due to

r

r

r

the dipole is given by E dipole = 2kp x 3 where p = 2aq1iˆ . We can use Coulomb’s law

r

r

for E due to a point charge and the superposition principle to find E at point P2.

Express the electric field at P2 as the sum of the field due to the dipole and the point charge q2:

r r r E P2 = Edipole + E q 2

2kp ˆ kq2 ˆ i+ 2 i x3 x 2k (2q1a ) ˆ kq2 ˆ i+ 2 i = x3 x k ⎡ 4q a ⎤ = 2 ⎢ 1 + q2 ⎥ iˆ x ⎣ x ⎦ =

where a = 1 cm.

r Substitute numerical values and evaluate E P2 :

r ⎤ 8.99 × 109 N ⋅ m 2 / C 2 ⎡ 4(5 µC )(1 cm ) + 3µC⎥ iˆ = 1.73 × 10 6 N/C iˆ E P2 = ⎢ 2 −2 15 cm 15 × 10 m ⎣ ⎦

(

(

)

)

While the separation of the two charges of the dipole is more than 10% of the distance to the point of interest, i.e., x is not much greater than a, this result is in excellent agreement with that of Problem 64. *66 •• Picture the Problem We can find the percentage of the free charge that would have to be removed by finding the ratio of the number of free electrons ne to be removed to give the penny a charge of 15 µC to the number of free electrons in the penny. Because we’re assuming the pennies to be point charges, we can use Coulomb’s law to find the force of repulsion between them. (a) Express the fraction f of the free charge to be removed as the quotient of the number of electrons to be removed and the number of free

f =

ne N

The Electric Field 1: Discrete Charge Distributions

49

electrons: Relate N to Avogadro’s number, the mass of the copper penny, and the molecular mass of copper:

m N m ⇒ N = NA = M NA M

Relate ne to the free charge Q to be removed from the penny:

Q = ne [− e] ⇒ ne =

Q −e

Q QM f = −e =− m meN A NA M Substitute numerical values and evaluate f:

f =−

(− 15 µC )(63.5 g/mol)

(3 g ) (1.6 × 10

−19

C )(6.02 × 10 mol 23

(b) Use Coulomb’s law to express the force of repulsion between the two pennies:

−1

= 3.29 × 10 −9 = 3.29 × 10 −7 %

)

kq 2 k (ne e ) = r2 r2

2

F=

Substitute numerical values and evaluate F:

(8.99 × 10 F=

9

)(

N ⋅ m 2 /C 2 9.38 × 1013 (0.25 m )2

) (1.6 × 10 2

−19

C

)

2

= 32.4 N

67 •• Picture the Problem Knowing the total charge of the two charges, we can use Coulomb’s law to find the two combinations of charge that will satisfy the condition that both are positive and hence repel each other. If just one charge is positive, then there is just one distribution of charge that will satisfy the conditions that the force is attractive and the sum of the two charges is 6 µC. (a) Use Coulomb’s law to express the repulsive force each charge exerts on the other: Express q2 in terms of the total charge and q1:

F=

kq1q2 r12, 2

q2 = Q − q1

50

Chapter 21

Substitute to obtain:

F=

kq1 (Q − q1 ) r12, 2

Substitute numerical values to obtain:

8 mN =

(8.99 × 10

9

[

N ⋅ m 2 /C 2 ) (6 µC) q1 − q12 (3 m)2

]

Simplify to obtain:

q12 + (− 6 µC ) q1 + 8.01(µC ) = 0

Solve to obtain:

q1 = 3.99 µC and q2 = 2.01 µC

2

or

q1 = 2.01 µC and q2 = 3.99 µC (b) Use Coulomb’s law to express the attractive force each charge exerts on the other:

F =−

kq1q2 r12, 2

Proceed as in (a) to obtain:

q12 + (− 6 µC) q1 − 8.01(µC ) = 0

Solve to obtain:

q1 = 7.12 µC and q2 = − 1.12 µC

2

68 •• Picture the Problem The electrostatic forces between the charges are responsible for the tensions in the strings. We’ll assume that these are point charges and apply Coulomb’s law and the principle of the superposition of forces to find the tension in each string. Use Coulomb’s law to express the net force on the charge +q: Substitute and simplify to obtain:

Use Coulomb’s law to express the net force on the charge +4q: Substitute and simplify to obtain:

T1 = F2 q + F4 q kq(2q ) kq(4q ) 3kq 2 T1 = + = d2 d2 (2d )2 T2 = Fq + F2 q k (2q )(4q ) kq(4q ) 9kq 2 T2 = + = d2 d2 (2d )2

The Electric Field 1: Discrete Charge Distributions

51

*69 •• Picture the Problem We can use Coulomb’s law to express the force exerted on one charge by the other and then set the derivative of this expression equal to zero to find the distribution of the charge that maximizes this force. Using Coulomb’s law, express the force that either charge exerts on the other:

F=

kq1q2 D2

Express q2 in terms of Q and q1:

q2 = Q − q1

Substitute to obtain:

F=

Differentiate F with respect to q1 and set this derivative equal to zero for extreme values:

dF k d [q1 (Q − q1 )] = 2 dq1 D dq1

Solve for q1 to obtain:

kq1 (Q − q1 ) D2

k [q1 (− 1) + Q − q1 ] D2 = 0 for extrema =

q1 = 12 Q and

q2 = Q − q1 = 12 Q To determine whether a maximum or a minimum exists at q1 = 12 Q , differentiate F a second time and evaluate this derivative at q1 = 12 Q :

d 2F k d [Q − 2q1 ] = 2 2 dq1 D dq1 k (− 2) D2 < 0 independently of q1. =

∴ q1 = q2 = 12 Q maximizes F . *70 •• Picture the Problem We can apply Coulomb’s law and the superposition of forces to relate the net force acting on the charge q = −2 µC to x. Because Q divides out of our equation when F(x) = 0, we’ll substitute the data given for x = 8.0 cm. Using Coulomb’s law, express the net force on q as a function of x:

F (x ) = −

kqQ kq(4Q ) + 2 x (12 cm − x )2

52

Chapter 21

⎤ 4 F (x ) ⎡ 1 Q = ⎢− 2 + 2⎥ kq (12 cm − x ) ⎦ ⎣ x

Simplify to obtain:

Solve for Q:

Q=

F (x ) ⎡ 1 ⎤ 4 kq ⎢− 2 + 2⎥ (12 cm − x ) ⎦ ⎣ x

Evaluate Q for x = 8 cm:

Q=

126.4 N = 3.00 µC ⎡ ⎤ 1 4 8.99 × 109 N ⋅ m 2 /C 2 (2 µC ) ⎢− + 2 2⎥ ⎣ (8 cm ) (4 cm ) ⎦

(

)

71 •• Picture the Problem Knowing the total charge of the two charges, we can use Coulomb’s law to find the two combinations of charge that will satisfy the condition that both are positive and hence repel each other. If the spheres attract each other, then there is just one distribution of charge that will satisfy the conditions that the force is attractive and the sum of the two charges is 200 µC. (a) Use Coulomb’s law to express the repulsive force each charge exerts on the other:

F=

kq1q2 r12, 2

q2 = Q − q1

Express q2 in terms of the total charge and q1: Substitute to obtain:

F=

kq1 (Q − q1 ) r12, 2

Substitute numerical values to obtain:

80 N =

(8.99 × 10

9

)[

N ⋅ m 2 /C 2 (200 µC) q1 − q12 (0.6 m)2

]

Simplify to obtain the quadratic equation:

q12 + (− 0.2 mC) q1 + 3.20 × 10−3 (mC) = 0

Solve to obtain:

q1 = 17.5 µC and q2 = 183 µC

2

or

The Electric Field 1: Discrete Charge Distributions

53

q1 = 183 µC and q2 = 17.5 µC (b) Use Coulomb’s law to express the attractive force each charge exerts on the other:

F =−

kq1q2 r12, 2

Proceed as in (a) to obtain:

q12 + (− 0.2 mC) q1 − 3.20 × 10−3 (mC) = 0

Solve to obtain:

q1 = − 15.0 µC and q2 = 215 µC

2

72 •• Picture the Problem Choose the coordinate system shown in the diagram and let Ug = 0 where y = 0. We’ll let our system include the ball and the earth. Then the work done on the ball by the electric field will change the energy of the system. The diagram summarizes what we know about the motion of the ball. We can use the work-energy theorem to our system to relate the work done by the electric field to the change in its energy. Using the work-energy theorem, relate the work done by the electric field to the change in the energy of the system:

Welectricfield = ∆K + ∆U g = K 2 − K1 + U g,2 − U g,1 or, because K1 = Ug,2 = 0,

Welectricfield = K 2 − U g,1

Substitute for Welectric field, K2, and Ug,0 and simplify:

qEh = 12 mv12 − mgh

Solve for m:

m=

(

= 12 m 2 gh

)

2

− mgh = mgh

qE g

73 •• Picture the Problem We can use Coulomb’s law, the definition of torque, and the condition for rotational equilibrium to find the electrostatic force between the two charged bodies, the torque this force produces about an axis through the center of the

54

Chapter 21

meter stick, and the mass required to maintain equilibrium when it is located either 25 cm to the right or to the left of the mid-point of the rigid stick. (a) Using Coulomb’s law, express the electric force between the two charges:

F=

kq1q2 d2

Substitute numerical values and evaluate F:

(8.99 ×10 F=

9

N ⋅ m 2 / C 2 )(5 ×10 −7 C ) = 0.225 N (0.1m )2

(b) Apply the definition of torque to obtain: Substitute numerical values and evaluate τ:

(c) Apply

∑τ

center of the meter stick

=0

2

τ = Fl τ = (0.225 N )(0.5 m ) = 0.113 N ⋅ m, counterclockwise

τ − mgl' = 0

to the meterstick: Solve for m:

m=

Substitute numerical values and evaluate m: (d) Apply

∑τ

center of the meter stick

=0

m=

τ gl ' 0.113 N (9.81 m/s 2 )(0.25 m ) = 0.0461 kg

− τ + mgl' = 0

to the meterstick: Substitute for τ:

− Fl + mgl' = 0

Substitute for F:

−

kq1q2' + mgl' = 0 d2

where q′ is the required charge. Solve for q2′ to obtain:

q2 =

d 2 mgl' kq1l

Substitute numerical values and evaluate q2′:

q 2' =

(0.1 m )2 (0.0461kg ) (9.81 m/s 2 )(0.25 m ) = (8.99 × 109 N ⋅ m 2 / C 2 )(5 × 10 −7 C)(0.5 m )

5.03 × 10 −7 C

The Electric Field 1: Discrete Charge Distributions

55

74 •• Picture the Problem Let the numeral 1 refer to the charge in the 1st quadrant and the numeral 2 to the charge in the 4th quadrant. We can use Coulomb’s law for the electric field due to a point charge and the superposition of forces to express the field at the origin and use this equation to solve for Q. Express the electric field at the origin due to the point charges Q:

r r r kQ kQ E (0,0 ) = E1 + E 2 = 2 rˆ1, 0 + 2 rˆ2,0 r1,0 r2,0

kQ (− 4 m )iˆ + (− 2 m ) ˆj + kQ3 (− 4 m )iˆ + (2 m ) ˆj = − (8 m3)kQ iˆ 3 r r r ˆ = Ex i (8 m )kQ . where r is the distance from each charge to the origin and E x = − r3 =

[

]

[

]

Express r in terms of the coordinates (x, y) of the point charges:

r = x2 + y2

Substitute to obtain:

Ex = −

Solve for Q to obtain:

Substitute numerical values and evaluate Q:

(8 m )kQ

(x

2

(

+ y2

E x2 + y 2 Q= x k (8 m )

)

3 2

)

3 2

2 2 ( 4 kN/C)[(4 m ) + (2 m ) ] Q=− (8.99 × 109 N ⋅ m2 /C2 )(8 m)

3 2

= − 4.97 µC 75 •• Picture the Problem Let the numeral 1 denote one of the spheres and the numeral 2 the other. Knowing the total charge Q on the two spheres, we can use Coulomb’s law to find the charge on each of them. A second application of Coulomb’s law when the spheres carry the same charge and are 0.60 m apart will yield the force each exerts on the other. (a) Use Coulomb’s law to express the repulsive force each charge exerts on the other:

F=

kq1q2 r12, 2

56

Chapter 21

q2 = Q − q1

Express q2 in terms of the total charge and q1: Substitute to obtain:

F=

kq1 (Q − q1 ) r12, 2

Substitute numerical values to obtain:

120 N =

(8.99 × 10

9

[

N ⋅ m 2 /C 2 ) (200 µC) q1 − q12 (0.6 m )2

]

Simplify to obtain the quadratic equation:

q12 + (− 200 µC ) q1 + 4810 (µC ) = 0

Solve to obtain:

q1 = 28.0 µC and q2 = 172 µC

2

or

q1 = 172 µC and q2 = 28.0 µC (b) Use Coulomb’s law to express the repulsive force each charge exerts on the other when q1 = q2 = 100 µC:

F=

kq1q2 r12, 2

Substitute numerical values and evaluate F:

F = (8.99 × 109 N ⋅ m 2 /C 2 )

(100 µC)2 (0.6 m )2

= 250 N

76 •• Picture the Problem Let the numeral 1 denote one of the spheres and the numeral 2 the other. Knowing the total charge Q on the two spheres, we can use Coulomb’s law to find the charge on each of them. A second application of Coulomb’s law when the spheres carry the same charge and are 0.60 m apart will yield the force each exerts on the other. (a) Use Coulomb’s law to express the attractive force each charge exerts on the other: Express q2 in terms of the total charge and q1:

F =−

kq1q2 r12, 2

q2 = Q − q1

The Electric Field 1: Discrete Charge Distributions Substitute to obtain:

F =−

57

kq1 (Q − q1 ) r12, 2

Substitute numerical values to obtain:

120 N =

[

− (8.99 × 109 N ⋅ m 2 /C 2 ) (200 µC) q1 − q12 (0.6 m)2

]

Simplify to obtain the quadratic equation:

q12 + (− 200 µC) q1 − 4810 (µC ) = 0

Solve to obtain:

q1 = − 21.7 µC and q2 = 222 µC

2

or

q1 = 222 µC and q2 = − 21.7 µC (b) Use Coulomb’s law to express the repulsive force each charge exerts on the other when q1 = q2 = 100 µC:

F=

kq1q2 r12, 2

Substitute numerical values and evaluate F:

F = (8.99 × 10 N ⋅ m /C 9

2

2

2 ( 100 µC ) ) (0.6 m )2

= 250 N

77 •• Picture the Problem The charge configuration is shown in the diagram as are the approximate locations, labeled x1 and x2, where the electric field is zero. We can determine the charge Q by using Coulomb’s law and the superposition of forces to express the net force acting on q2. In part (b), by inspection, the points where E = 0 must be between the −3 µC and +4 µC charges. We can use Coulomb’s law for the field due to point charges and the superposition of electric fields to determine the coordinates x1 and x2.

58

Chapter 21

(a) Use Coulomb’s law to express the force on the 4.0-µC charge:

r r r F2 = F1, 2 + FQ , 2 =

( )

kq1q2 ˆ kQq2 ˆ i + 2 −i r12, 2 rQ , 2

⎡q Q ⎤ = kq2 ⎢ 21 − 2 ⎥ iˆ = F2 iˆ ⎣⎢ r1, 2 rQ , 2 ⎦⎥

⎡q F ⎤ Q = rQ2, 2 ⎢ 21 − 2 ⎥ ⎣⎢ r1, 2 kq2 ⎥⎦

Solve for Q:

Substitute numerical values and evaluate Q:

⎤ 240 N 2 ⎡ − 3 µC − Q = (0.12 m ) ⎢ ⎥ = − 97.2 µC 2 9 2 2 ( ) ( ) 8.99 10 N m /C 4 µ C × ⋅ ( ) 0.2 m ⎦ ⎣ (b) Use Coulomb’s law for electric fields and the superposition of fields to determine the coordinate x at which E = 0:

r E=−

kQ kq2 kq iˆ − iˆ + 21 iˆ = 0 2 2 x (0.32 m − x ) (0.2 m − x )

or

−

Q q2 q − + 12 = 0 2 2 (0.32 m − x ) (0.2 m − x ) x

Substitute numerical values to obtain:

−

− 3 µC − 97.2 µC 4 µC − + =0 2 2 x2 (0.32 m − x ) (0.2 m − x )

and

97.2 4 3 − − 2 =0 2 2 (0.32 m − x ) (0.2 m − x ) x Solve (preferably using a graphing calculator!) this equation to obtain:

x1 = 0.0508 m and x2 = 0.169 m

The Electric Field 1: Discrete Charge Distributions *78 •• Picture the Problem Each sphere is in static equilibrium under the influence of

r

r

the tension T , the gravitational force Fg ,

r

and the electric force FE . We can use Coulomb’s law to relate the electric force to the charge on each sphere and their separation and the conditions for static equilibrium to relate these forces to the charge on each sphere. (a) Apply the conditions for static equilibrium to the charged sphere:

∑F

x

= FE − T sin θ =

kq 2 − T sin θ = 0 r2

and

∑F

y

= T cos θ − mg = 0

Eliminate T between these equations to obtain:

tan θ =

kq 2 mgr 2

Solve for q:

q=r

mg tan θ k

Referring to the figure, relate the separation of the spheres r to the length of the pendulum L: Substitute to obtain:

r = 2L sin θ

q = 2 L sin θ

mg tan θ k

(b) Evaluate q for m = 10 g, L = 50 cm, and θ = 10°:

q = 2(0.5 m )sin 10°

(0.01kg )(9.81m/s 2 )tan 10° 8.99 × 109 N ⋅ m 2 /C 2

= 0.241 µC

59

60

Chapter 21

79 •• Picture the Problem Each sphere is in static equilibrium under the influence of

r

r

the tension T , the gravitational force Fg ,

r

and the electric force FE . We can use Coulomb’s law to relate the electric force to the charge on each sphere and their separation and the conditions for static equilibrium to relate these forces to the charge on each sphere. (a)Apply the conditions for static equilibrium to the charged sphere:

∑F

x

=FE − T sin θ =

kq 2 − T sin θ = 0 r2

and

∑F

y

Eliminate T between these equations to obtain:

=T cosθ − mg = 0

tan θ =

kq 2 mgr 2

r = 2L sin θ

Referring to the figure for Problem 80, relate the separation of the spheres r to the length of the pendulum L: Substitute to obtain:

tan θ =

kq 2 4mgL2 sin 2 θ

or

kq 2 sin θ tan θ = 4mgL2 2

Substitute numerical values and evaluate sin 2 θ tan θ :

sin 2 θ tan θ = Because sin 2 θ tan θ << 1 :

(8.99 × 10

)

N ⋅ m 2 /C2 (0.75 µC ) = 5.73 × 10− 3 2 2 4(0.01kg ) 9.81m/s (1.5 m ) 9

(

2

)

sin θ ≈ tan θ ≈ θ and

θ 3 ≈ 5.73 × 10 −3 Solve for θ to obtain:

θ = 0.179 rad = 10.3°

(1)

The Electric Field 1: Discrete Charge Distributions (b) Evaluate equation (1) with replacing q2 with q1q2:

sin 2 θ tan θ =

(8.99 × 10

N ⋅ m 2 /C2 )(0.5 µC)(1 µC) = 5.09 × 10− 3 ≈ θ 3 2 2 4(0.01kg )(9.81m/s )(1.5 m ) 9

Solve for θ to obtain:

θ = 0.172 rad = 9.86°

80 •• Picture the Problem Let the origin be at the lower left-hand corner and designate the charges as shown in the diagram. We can apply Coulomb’s law for point charges to find the forces exerted on q1 by q2, q3, and q4 and superimpose these forces to find the net force exerted on q1. In part (b), we’ll use Coulomb’s law for the electric field due to a point charge and the superposition of fields to find the electric field at point P(0, L/2). (a) Using the superposition of forces, express the net force exerted on q1:

r

Apply Coulomb’s law to express F2,1 :

r r r r F1 = F2,1 + F3,1 + F4,1 r kq q kq q r F2,1 = 22 1 rˆ2,1 = 23 1 r2,1 r2,1 r2,1 =

r

Apply Coulomb’s law to express F4,1 :

r

(

)

r kq q kq q r F4,1 = 42 1 rˆ4,1 = 43 1 r4,1 r4,1 r4,1 =

Apply Coulomb’s law to express F3,1 :

2 k (− q )q ˆj = kq ˆj − L L3 L2

2 k (− q )q ˆ = kq iˆ − L i L3 L2

(

)

r kq q kq q r F3,1 = 32 1 rˆ3,1 = 33 1 r3,1 r3,1 r3,1

(

kq 2 − L iˆ − L ˆj 2 3 2 L3 kq 2 = − 3 2 2 iˆ + ˆj 2 L =

(

)

)

61

62

Chapter 21

Substitute and simplify to obtain:

(

r kq 2 kq 2 ˆ ˆ F1 = 2 j − 3 2 2 i + L 2 L 2 kq kq 2 = 2 iˆ + ˆj − 3 2 2 L 2 L

(

)

)

( iˆ + ˆj )

(

kq 2 ⎛ 1 ⎞ˆ ˆ = 1− ⎟i+ j 2 ⎜ L ⎝ 2 2⎠ (b) Using superposition of fields, express the resultant field at point P:

r

Use Coulomb’s law to express E1 :

r

r kq kq ⎛ L E1 = 2 1 rˆ1, P = 3 ⎜ r1, P r1, P ⎝ 2

r

r

ˆj ⎞⎟ = 4kq ˆj 2 ⎠ L

− kq ⎛ L − 3 ⎜ ⎛L⎞ ⎝ 2 ⎜ ⎟ ⎝2⎠

ˆj ⎞⎟ ⎠

ˆj ⎞⎟ = 4kq ˆj 2 ⎠ L

r kq kq ⎛ L ⎞ E3 = 2 3 rˆ3, P = 3 ⎜ − Liˆ − ˆj ⎟ 2 ⎠ r3, P r3, P ⎝ =

Use Coulomb’s law to express E 4 :

kq ⎛ L 3 ⎜ ⎛L⎞ ⎝ 2 ⎜ ⎟ ⎝2⎠

(1)

ˆj ⎞⎟ ⎠

r kq k (− q ) ⎛ L E 2 = 2 2 rˆ2, P = 3 ⎜ r2, P r2, P ⎝ 2 =

Use Coulomb’s law to express E 3 :

)

r r r r r E P = E1 + E 2 + E 3 + E 4

=

Use Coulomb’s law to express E 2 :

2

ˆj + kq iˆ L2

8kq ⎛ ˆ 1 ⎜− i − 53 2 L2 ⎝ 2

ˆj ⎞⎟ ⎠

r kq k (− q ) ⎛ L ⎞ E 4 = 2 4 rˆ3, P = 3 ⎜ Liˆ − ˆj ⎟ 2 ⎠ r4, P r4, P ⎝ =

8kq ⎛ ˆ 1 ⎜i − 53 2 L2 ⎝ 2

ˆj ⎞⎟ ⎠

Substitute in equation (1) and simplify to obtain:

r 1 4kq 4kq 8kq ⎛ E P = 2 ˆj + 2 ˆj + 3 2 2 ⎜ − iˆ − L L 2 5 L ⎝ 81

••

ˆj ⎞⎟ + 8kq ⎛⎜ iˆ − 1 32 2 2 ⎠ 5 L ⎝

ˆj ⎞⎟ = 8kq ⎛⎜1 + 5 ⎞⎟ ˆj 25 ⎟⎠ L2 ⎜⎝ ⎠

The Electric Field 1: Discrete Charge Distributions

63

Picture the Problem We can apply Newton’s 2nd law in rotational form to obtain the differential equation of motion of the dipole and then use the small angle approximation sinθ ≈ θ to show that the dipole experiences a linear restoring torque and, hence, will experience simple harmonic motion. Apply

∑τ = Iα to the dipole:

− pE sin θ = I

d 2θ dt 2

where τ is negative because acts in such a direction as to decrease θ. For small values of θ, sinθ ≈ θ and: Express the moment of inertia of the dipole: Relate the dipole moment of the dipole to its charge and the charge separation: Substitute to obtain:

− pEθ = I

d 2θ dt 2

I = 12 ma 2

p = qa

1 2

ma 2

d 2θ = − qaEθ dt 2

or

d 2θ 2qE =− θ 2 dt ma the differential equation for a simple harmonic oscillator with angular frequency ω = 2qE ma . Express the period of a simple harmonic oscillator: Substitute to obtain:

T=

2π

ω

T = 2π

ma 2qE

82 •• Picture the Problem We can apply conservation of energy and the definition of the potential energy of a dipole in an electric field to relate q to the kinetic energy of the dumbbell when it is aligned with the field.

64

Chapter 21

Using conservation of energy, relate the initial potential energy of the dumbbell to its kinetic energy when it is momentarily aligned with the electric field: Express the change in the potential energy of the dumbbell as it aligns with the electric field in terms of its dipole moment, the electric field, and the angle through which it rotates:

∆K + ∆U = 0 or, because Ki = 0,

K + ∆U = 0 where K is the kinetic energy when it is aligned with the field.

∆U = U f − U i = − pE cos θ f + pE cos θ f = qaE (cos 60° − 1)

Substitute to obtain:

K + qaE (cos 60° − 1) = 0

Solve for q:

q=

Substitute numerical values and evaluate q:

5 × 10−3 J q= (0.3 m )(600 N/C)(1 − cos 60°)

K aE (1 − cos 60°)

= 55.6 µC *83 •• Picture the Problem The forces the electron and the proton exert on each other constitute an action-and-reaction pair. Because the magnitudes of their charges are equal and their masses are the same, we find the speed of each particle by finding the speed of either one. We’ll apply Coulomb’s force law for point charges and Newton’s 2nd law to relate v to e, m, k, and r. Apply Newton’s 2nd law to the positron:

Solve for v to obtain:

v2 ke 2 ke 2 = m = 2mv 2 ⇒ 1 r r2 r 2 v=

ke 2 2mr

84 •• Picture the Problem In Problem 81 it was established that the period of an electric dipole in an electric field is given by T = 2π ma 2qE . We can use this result to find the frequency of oscillation of a KBr molecule in a uniform electric field of 1000 N/C.

The Electric Field 1: Discrete Charge Distributions Express the frequency of the KBr oscillator:

f =

1 2π

2qE ma

Substitute numerical values and evaluate f:

f =

1 2π

2 1.6 × 10 −19 C (1000 N/C ) 1.4 × 10 −25 kg (0.282 nm )

(

(

) )

= 4.53 × 108 Hz 85 ••• Picture the Problem We can use Coulomb’s force law for point masses and the condition for translational equilibrium to express the equilibrium position as a function of k, q, Q, m, and g. In part (b) we’ll need to show that the displaced point charge experiences a linear restoring force and, hence, will exhibit simple harmonic motion. (a) Apply the condition for translational equilibrium to the point mass: Solve for y0 to obtain:

(b) Express the restoring force that acts on the point mass when it is displaced a distance ∆y from its equilibrium position:

kqQ − mg = 0 y02

kqQ mg

y0 =

F= ≈

kqQ kqQ − 2 2 ( y0 + ∆y ) y0 kqQ kqQ − 2 y + 2 y0 ∆y y0 2 0

because ∆y << y0. Simplify this expression further by writing it with a common denominator:

F =− =−

≈−

2 y0 ∆ykqQ y04 + 2 y03 ∆y 2 y0 ∆ykqQ ⎛ ∆y ⎞ y04 ⎜⎜1 + 2 ⎟⎟ y0 ⎠ ⎝ 2∆ykqQ y03

again, because ∆y << y0. From the 1st step of our solution:

kqQ = mg y 02

65

66

Chapter 21

Substitute to obtain:

F =−

Apply Newton’s 2nd law to the displaced point charge to obtain:

m

2mg ∆y y0

d 2 ∆y 2mg =− ∆y 2 dt y0

or

d 2 ∆y 2 g + ∆y = 0 dt 2 y0 the differential equation of simple harmonic motion with ω = 2 g y0 . 86 ••• Picture the Problem The free-body diagram shows the Coulomb force the positive charge Q exerts on the bead that is constrained to move along the x axis. The x component of this force is a restoring force, i.e., it is directed toward the bead’s equilibrium position. We can show that, for x << L, this restoring force is linear and, hence, that the bead will exhibit simple harmonic motion about its equilibrium position. Once we’ve obtained the differential equation of SHM we can relate the period of the motion to its angular frequency.

k (− q )Q kqQ =− 2 2 2 L +x L + x2

Using Coulomb’s law for point charges, express the force F that +Q exerts on −q:

F=

Express the component of this force along the x axis:

Fx = −

kqQ cos θ L + x2 kqQ x =− 2 2 L + x L2 + x 2 kqQ =− x 32 L2 + x 2 2

(

)

The Electric Field 1: Discrete Charge Distributions Factor L2 from the denominator of this equation to obtain:

kqQ

Fx = −

⎛ x2 ⎞ L3 ⎜⎜1 + 2 ⎟⎟ L ⎠ ⎝

32

x≈−

kqQ x L3

because x << L. Apply

∑F

x

= ma x to the bead to

m

obtain:

d 2x kqQ =− 3 x 2 dt L

or

d 2 x kqQ + x=0 dt 2 mL3 the differential equation of simple harmonic motion with ω =

kqQ mL3 .

2π

Express the period of the motion of the bead in terms of the angular frequency of the motion:

mL3 mL T= = 2π = 2πL kqQ kqQ ω

87 ••• Picture the Problem Each sphere is in static equilibrium under the influence of

r

r

the tension T , the gravitational force Fg ,

r r FCoulomb and the force FE exerted by the

electric field. We can use Coulomb’s law to relate the electric force to the charges on the spheres and their separation and the conditions for static equilibrium to relate these forces to the charge on each sphere. (a)Apply the conditions for static equilibrium to the charged sphere:

∑F

x

= FCoulomb − T sin θ =

kq 2 − T sin θ = 0 r2

and

∑F

y

Eliminate T between these equations to obtain: Referring to the figure for Problem 78, relate the separation of the

= T cos θ − mg − qE = 0

tan θ =

kq 2 (mg + qE )r 2

r = 2L sin θ

67

68

Chapter 21

spheres r to the length of the pendulum L: Substitute to obtain:

tan θ =

kq 2 4(mg + qE )L2 sin 2 θ

or

sin 2 θ tan θ =

kq 2 4(mg + qE )L2

Substitute numerical values and evaluate sin 2 θ tan θ to obtain:

sin 2 θ tan θ = 3.25 × 10 −3

Because sin 2 θ tan θ << 1 :

sin θ ≈ tan θ ≈ θ

(1)

and

θ 3 ≈ 3.25 × 10 −3 Solve for θ to obtain:

θ = 0.148 rad = 8.48°

(b) The downward electrical forces acting on the two spheres are no longer equal. Let the mass of the sphere carrying the charge of 0.5 µC be m1, and that of the sphere carrying the charge of 1.0 µC be m2. The free-body diagrams show the tension, gravitational, and electrical forces acting on each sphere. Because we already know from part (a) that the angles are small, we can use the small-angle approximation sinθ ≈ tanθ ≈θ. Apply the conditions for static equilibrium to the charged sphere whose mass is m1:

∑F

x ,1

=− =− ≈− =0

and

kq1q2 + T1 sin θ1 r2 kq1q2

(L sin θ1 + L sin θ 2 )2 kq1q2

L (θ1 + θ 2 ) 2

2

+ T1θ1

+ T1 sin θ1

The Electric Field 1: Discrete Charge Distributions

∑F

= T1, y − m1 g − q1 E = 0

∑F

=

y ,1

Apply the conditions for static equilibrium to the charged sphere whose mass is m2:

x,2

= ≈

kq1q2 − T2 sin θ 2 r2 kq1q2

(L sin θ1 + L sin θ 2 )2 kq1q2

L2 (θ1 + θ 2 )

2

69

+ T2 sin θ 2

+ T2θ 2

=0 and

∑F

Express θ1 and θ2 in terms of the

θ1 =

r r components of T1 and T2 :

y,2

= T2, y − m2 g − q2 E = 0

T1, x T1, y

(1)

and

θ2 =

Divide equation (1) by equation (2) to obtain:

T2, x T2, y

(2)

T1, x θ1 T1, y T2, y = = θ 2 T2, x T1, y T2, y

r

because the horizontal components of T1 and

r T2 are equal.

Substitute for T2,y and T1,y to obtain:

θ 1 m2 g + q 2 E = θ 2 m1 g + q1 E

Add equations (1) and (2) to obtain:

θ1 + θ 2 =

T1, x T2, x ⎡ ⎤ kq q 1 1 + = 2 1 2 2⎢ + T1, y T2, y L (θ1 + θ 2 ) ⎣ m1 g + q1E m2 g + q2 E ⎥⎦

Solve for θ1 + θ2:

θ1 + θ 2 = 3

⎤ kq1q2 ⎡ 1 1 + 2 ⎢ L ⎣ m1 g + q1E m2 g + q2 E ⎥⎦

70

Chapter 21

θ1 + θ 2 = 0.287 rad = 16.4°

Substitute numerical values and evaluate θ1 + θ2 and θ1/θ2:

and

Solve for θ1 and θ2 to obtain:

θ1 = 9.42° and θ1 = 6.98°

θ1 = 1.34 θ2

88 ••• Picture the Problem Each sphere is in static equilibrium under the influence of a tension, gravitational and Coulomb force. Let the mass of the sphere carrying the charge of 2.0 µC be m1 = 0.01 kg, and that of the sphere carrying the charge of 1.0 µC be m2 = 0.02 kg. We can use Coulomb’s law to relate the Coulomb force to the charge on each sphere and their separation and the conditions for static equilibrium to relate these forces to the charges on the spheres. Apply the conditions for static equilibrium to the charged sphere whose mass is m1:

∑F

x ,1

=− =− ≈−

kq1q2 + T1 sin θ1 r2 kq1q2

(L sin θ1 + L sin θ 2 )2 kq1q2

L (θ1 + θ 2 )

2

2

+ T1 sin θ1

+ T1θ1

=0 and

∑F Apply the conditions for static equilibrium to the charged sphere whose mass is m2:

y ,1

∑F

x,2

= T1, y − m1 g = 0

= = ≈

kq1q2 − T2 sin θ 2 r2 kq1q2

(L sin θ1 + L sin θ 2 )2 kq1q2

L (θ1 + θ 2 )

=0

2

2

+ T2θ 2

+ T2 sin θ 2

The Electric Field 1: Discrete Charge Distributions and

∑F Using the small-angle approximation sinθ ≈ tanθ ≈θ, express θ1 and θ2 in

θ1 =

terms of the components of T1 and

and

r

r T2 :

Divide equation (1) by equation (2) to obtain:

θ2 =

y ,2

= T2, y − m2 g = 0

T1, x

(1)

T1, y T2, x

(2)

T2, y T1, x θ1 T1, y T2, y = = θ 2 T2, x T1, y T2, y

because the horizontal components of

r r T1 and T2 are equal.

Substitute for T2,y and T1,y to obtain:

θ1 m2 = θ 2 m1

Add equations (1) and (2) to obtain:

θ1 + θ 2 = =

Solve for θ1 + θ2:

T1, x T2, x + T1, y T2, y kq1q2 2 L (θ1 + θ 2 )

θ1 + θ 2 = 3

2

⎡ 1 1 ⎤ ⎢m g + m g ⎥ 2 ⎦ ⎣ 1

1 ⎤ kq1q2 ⎡ 1 + ⎢ 2 L ⎣ m1 g m2 g ⎥⎦

θ1 + θ 2 = 0.496 rad = 28.4°

Substitute numerical values and evaluate θ1 + θ2 and θ1/θ2:

and

Solve for θ1 and θ2 to obtain:

θ1 = 9.47° and θ1 = 18.9°

θ1 1 = θ2 2

Remarks: While the small angle approximation is not as good here as it was in the preceding problems, the error introduced is less than 3%.

71

72

Chapter 21

89 ••• Picture the Problem We can find the effective value of the gravitational field by finding r r the force on the bob due to g and E and equating this sum to the product of the mass of

r

r

r

r

the bob and g ' . We can then solve this equation for E in terms of g , g ' , q, and M and r use the equation for the period of a simple pendulum to find the magnitude of g '

r r r r ⎛r q r⎞ F = Mg + q E = M ⎜ g + E ⎟ = Mg ' M ⎠ ⎝

Express the force on the bob due to r r g and E :

where

r r q r g' = g + E M

r

Solve for E to obtain:

r M r r ( g' − g ) E= q

Using the expression for the period of a simple pendulum, find the magnitude of g′:

T' = 2π and

g' =

L g'

4π 2 L 4π 2 (1 m ) = = 27.4 m/s 2 2 T2 (1.2 s )

r

Substitute numerical values and evaluate E :

[

]

r 5 × 10−3 kg ( E= 27.4 m/s2 ) ˆj − (9.81m/s2 ) ˆj = (− 1.10 × 104 N/C) ˆj − 8.0 µC *90 ••• Picture the Problem We can relate the force of attraction that each molecule exerts on the other to the potential energy function of either molecule using F = − dU dx. We can relate U to the electric field at either molecule due to the presence of the other through U = −pE. Finally, the electric field at either molecule is given by E = 2kp x 3 . Express the force of attraction between the dipoles in terms of the spatial derivative of the potential energy function of p1: Express the potential energy of the dipole p1:

F =−

dU 1 dx

(1)

U1 = − p1 E1 where E1 is the field at p1 due to p2.

The Electric Field 1: Discrete Charge Distributions Express the electric field at p1 due to p2:

E1 =

2kp2 x3

where x is the separation of the dipoles. Substitute to obtain:

U1 = −

Substitute in equation (1) and differentiate with respect to x:

F =−

Evaluate F for p1 = p2 = p and x = d to obtain:

F=

2kp1 p2 x3

d ⎡ 2kp1 p2 ⎤ 6kp1 p2 − = dx ⎢⎣ x 3 ⎥⎦ x4

6kp 2 d4

91 ••• Picture the Problem We can use Coulomb’s law for the electric field due to a point charge and superposition of fields to find the electric field at any point on the y axis. By applying Newton’s 2nd law, with the charge on the ring negative, we can show that the ring experiences a linear restoring force and, therefore, will execute simple harmonic motion. We can find ω from the differential equation of motion and use f = ω/2π to find the frequency of the motion. (a) Use Coulomb’s law for the electric field due to a point charge and superposition of fields, express the field at point P on the y axis:

r r r kq kq kQ r kQ r E P = E1 + E 2 = 2 1 rˆ1,P + 2 2 rˆ2,P = 3 r1,P + 3 r2,P r1,P r2,P r1,P r2,P

= =

(a

kQ 2

(a

+y

)

2 32

2kQy 2

+y

kQ ⎛ L ˆ ˆ⎞ ⎜ i + yj ⎟ + 2 ⎠ a + y2 ⎝2

)

2 32

(

)

32

⎛ L ˆ ˆ⎞ ⎜ − i + yj ⎟ ⎠ ⎝ 2

ˆj

where a = L/2.

(b) Relate the force on the charged ring to its charge and the electric field:

r r Fy = qE y =

(a

2kqQy 2

+y

)

2 32

ˆj

r

where q must be negative if Fy is to be a restoring force.

73

74

Chapter 21

(c) Apply Newton’s 2nd law to the ring to obtain:

d2y 2kqQ m 2 =− y 32 2 dt a + y2

(

)

or

d2y 2kqQ =− 2 dt m a2 + y2

(

Factor the radicand to obtain:

d2y =− dt 2

)

32

y

2kqQ

y 2 32 ⎛ ⎞ y ma 3 ⎜⎜1 + 2 ⎟⎟ ⎝ a ⎠ 2kqQ 16kqQ y=− y ≈− 3 ma mL3

provided y << a = L/2.

d2y 16kqQ =− y 2 dt mL3

Thus we have:

or

d 2 y 16kqQ + y=0 dt 2 mL3 the differential equation of simple harmonic motion.

ω 2π

Express the frequency of the simple harmonic motion in terms of its angular frequency:

f =

From the differential equation describing the motion we have:

ω2 =

16kqQ mL3

and

f =

1 2π

16kqQ mL3

Substitute numerical values and evaluate f:

f =

(

)

1 16 8.99 × 109 N ⋅ m 2 /C 2 (5 µC )(2 µC ) = 9.37 Hz 2π (0.03 kg )(0.24 m )3

The Electric Field 1: Discrete Charge Distributions

75

92 ••• Picture the Problem The free body diagram shows the forces acting on the microsphere of mass m and having an excess charge of q = Ne when the electric field is downward. Under terminal-speed conditions the sphere is in equilibrium r under the influence of the electric force Fe ,

r

r

its weight mg, and the drag force Fd . We can apply Newton’s 2nd law, under terminal-speed conditions, to relate the number of excess charges N on the sphere to its mass and, using Stokes’ law, find its terminal speed. (a) Apply

∑F

y

= ma y to the

Fe − mg − Fd = ma y

microsphere:

or, because ay = 0,

Substitute for Fe, m, and Fd,terminal to obtain:

qE − ρVg − 6πηrvt = 0

Fe − mg − Fd, terminal = 0

or, because q = Ne,

NeE − 43 πr 3 ρg − 6πηrvt = 0 Solve for N to obtain:

Substitute numerical values and evaluate 43 πr 3 ρg :

N= 4 3

4 3

πr 3 ρg + 6πηrvt eE

π r 3 ρg = 43 π (5.5 × 10 −7 m )

3

× (1.05 × 103 kg/m 3 )(9.81 m/s 2 )

= 7.18 × 10 −15 N Substitute numerical values and evaluate 6πηrvt :

( × (1.16 × 10

)( m/s )

6πηrvt = 6π 1.8 × 10−5 Pa ⋅ s 5.5 × 10−7 m −4

= 2.16 × 10−14 N

Substitute numerical values in equation (1) and evaluate N:

N=

7.18 × 10 −15 N + 2.16 × 10 −14 N 1.6 × 10 −19 C 6 × 10 4 V/m

(

)(

= 3 (b) With the field pointing upward, the electric force is downward and the application of Fy = ma y to

∑

Fd, terminal − Fe − mg = 0 or

6πηrvt − NeE − 43 πr 3 ρg − = 0

)

)

76

Chapter 21

the bead yields: Solve for vt to obtain:

vt =

NeE + 43 πr 3 ρg 6πηr

Substitute numerical values and evaluate vt:

(

)(

)

(

)(

)(

3 1.6 × 10 −19 C 6 × 10 4 V/m + 43 π 5.5 × 10 −7 m 1.05 × 103 kg/m 3 9.81 m/s 2 vt = 6π 1.8 × 10 −5 Pa ⋅ s 5.5 × 10 −7 m

(

3

)(

)

= 1.93 × 10 −4 m/s *93 ••• Picture the Problem The free body diagram shows the forces acting on the microsphere of mass m and having an excess charge of q = Ne when the electric field is downward. Under terminal-speed conditions the sphere is in equilibrium r under the influence of the electric force Fe ,

r

r

its weight mg, and the drag force Fd . We can apply Newton’s 2nd law, under terminal-speed conditions, to relate the number of excess charges N on the sphere to its mass and, using Stokes’ law, to its terminal speed. (a) Apply

∑F

y

= ma y to the

Fe − mg − Fd = ma y

microsphere when the electric field is downward:

or, because ay = 0,

Substitute for Fe and Fd,terminal to obtain:

qE − mg − 6πηrvu = 0

Fe − mg − Fd, terminal = 0

or, because q = Ne,

NeE − mg − 6πηrvu = 0

NeE − mg 6πηr

Solve for vu to obtain:

vu =

With the field pointing upward, the electric force is downward and the application of Fy = ma y to the

Fd, terminal − Fe − mg = 0

∑

microsphere yields: Solve for vd to obtain:

(1)

or

6πηrvd − NeE − mg = 0

vd =

NeE + mg 6πηr

(2)

)

The Electric Field 1: Discrete Charge Distributions Add equations (1) and (2) to obtain:

v = vu + vd =

=

77

NeE − mg 6πηr NeE + mg + 6πηr

NeE qE = 3πηr 3πηr

This has the advantage that you don' t need to know the mass of the microsphere. (b) Letting ∆v represent the change in the terminal speed of the microsphere due to a gain (or loss) of one electron we have: Noting that ∆v will be the same whether the microsphere is moving upward or downward, express its terminal speed when it is moving upward with N electronic charges on it:

∆v = vN +1 − vN

vN =

NeE − mg 6πηr

(N + 1)eE − mg

Express its terminal speed upward when it has N + 1 electronic charges:

vN +1 =

Substitute and simplify to obtain:

∆vN +1 =

6πηr

= Substitute numerical values and evaluate ∆v:

∆v =

(N + 1)eE − mg − NeE − mg 6πηr

6πηr

eE 6πηr

(1.6 × 10 6π (1.8 × 10

−19

)(

)

C 6 × 10 4 V/m −5 Pa ⋅ m 5.5 × 10 −7 m

= 5.15 × 10 −5 m/s

)(

)

78

Chapter 21

Chapter 22 The Electric Field 2: Continuous Charge Distributions Conceptual Problems *1 •• (a) False. Gauss’s law states that the net flux through any surface is given by φnet = En dA = 4πkQinside . While it is true that Gauss’s law is easiest to apply to

∫

S

symmetric charge distributions, it holds for any surface. (b) True 2 •• Determine the Concept Gauss’s law states that the net flux through any surface is given by φnet = En dA = 4πkQinside . To use Gauss’s law the system must display some

∫

S

symmetry. 3 ••• Determine the Concept The electric field is that due to all the charges, inside and outside the surface. Gauss’s law states that the net flux through any surface is given

∫

by φnet = En dA = 4πkQinside . The lines of flux through a Gaussian surface begin on S

charges on one side of the surface and terminate on charges on the other side of the surface. 4 •• Picture the Problem We can show that the charge inside a sphere of radius r is proportional to r3 and that the area of a sphere is proportional to r2. Using Gauss’s law, we can show that the field must be proportional to r3/r2 = r. Use Gauss’s law to express the electric field inside a spherical charge distribution of constant volume charge density:

4πkQinside A where A = 4πr 2 . E=

Express Qinside as a function of ρ and r:

Qinside = ρV = 43 πρr 3

Substitute to obtain:

E=

79

4πk 43 πρr 3 4kπρ = r 2 4πr 3

80

Chapter 22

*5 • (a) False. Consider a spherical shell, in which there is no charge, in the vicinity of an infinite sheet of charge. The electric field due to the infinite sheet would be non-zero everywhere on the spherical surface. (b) True (assuming there are no charges inside the shell). (c) True. (d) False. Consider a spherical conducting shell. Such a surface will have equal charges on its inner and outer surfaces but, because their areas differ, so will their charge densities. 6 • Determine the Concept Yes. The electric field on a closed surface is related to the net flux through it by Gauss’s law: φ = EdA = Qinside ∈0 . If the net flux through the closed

∫

S

surface is zero, the net charge inside the surface must be zero by Gauss’s law. 7 • Determine the Concept The negative point charge at the center of the conducting shell induces a charge +Q on the inner surface of the shell. (a ) is correct. 8 • Determine the Concept The negative point charge at the center of the conducting shell induces a charge +Q on the inner surface of the shell. Because a conductor does not have to be neutral, (d ) is correct. *9 •• Determine the Concept We can apply Gauss’s law to determine the electric field for r < R1 and r > R2. We also know that the direction of an electric field at any point is determined by the direction of the electric force acting on a positively charged object located at that point. From the application of Gauss’s law we know that the electric field in both of these regions is not zero and is given by:

En =

kQ r2

A positively charged object placed in either of these regions would experience an attractive force from the charge –Q located at the center of the shell. (b) is correct.

The Electric Field 2: Continuous Charge Distributions

81

*10 •• Determine the Concept We can decide what will happen when the conducting shell is grounded by thinking about the distribution of charge on the shell before it is grounded and the effect on this distribution of grounding the shell. The negative point charge at the center of the conducting shell induces a positive charge on the inner surface of the shell and a negative charge on the outer surface. Grounding the shell attracts positive charge from ground; resulting in the outer surface becoming electrically neutral. (b) is correct. 11 •• Determine the Concept We can apply Gauss’s law to determine the electric field for r < R1 and r > R2. We also know that the direction of an electric field at any point is determined by the direction of the electric force acting on a positively charged object located at that point. From the application of Gauss’s law we know that the electric field in the region r < R1 is given by E n =

kQ . A positively charged object placed in the region r < R1 will r2

experience an attractive force from the charge –Q located at the center of the shell. With the conducting shell grounded, the net charge enclosed by a spherical Gaussian surface of radius r > R2 is zero and hence the electric field in this region is zero.

(c) is correct. 12 •• Determine the Concept No. The electric field on a closed surface is related to the net

∫

flux through it by Gauss’s law: φ = EdA = Qinside ∈0 . φ can be zero without E being S

zero everywhere. If the net flux through the closed surface is zero, the net charge inside the surface must be zero by Gauss’s law. 13 •• False. A physical quantity is discontinuous if its value on one side of a boundary differs from that on the other. We can show that this statement is false by citing a counterexample. Consider the field of a uniformly charged sphere. ρ is discontinuous at the surface, E is not.

Estimation and Approximation *14 •• Picture the Problem We’ll assume that the total charge is spread out uniformly (charge density = σ) in a thin layer at the bottom and top of the cloud and that the area of each

82

Chapter 22

surface of the cloud is 1 km2. We can then use the definition of surface charge density and the expression for the electric field at the surface of a charged plane surface to estimate the total charge of the cloud.

Q = σA

Express the total charge Q of a thundercloud in terms of the surface area A of the cloud and the charge density σ :

σ

Express the electric field just outside the cloud:

E=

Solve for σ :

σ =∈0 E

Substitute for σ to obtain:

Q =∈0 EA

∈0

Substitute numerical values and evaluate Q:

(

)(

)(

)

Q = 8.85 × 10 −12 C 2 / N ⋅ m 2 3 × 10 6 V/m 1 km 2 = 26.6 C Remarks: This charge is in reasonably good agreement with the total charge transferred in a lightning strike of approximately 30 C. 15 •• Picture the Problem We’ll assume that the field is strong enough to produce a spark. Then we know that field must be equal to the dielectric strength of air. We can then use the relationship between the field and the charge density to estimate the latter. Suppose the field is large enough to produce a spark. Then:

E ≈ 3 × 10 6 V/m

Because rubbing the balloon leaves it with a surface charge density of +σ and the hair with a surface charge density of −σ, the electric field between the balloon and the hair is:

E=

Solve for σ :

σ = 2 ∈0 E

σ 2 ∈0

Substitute numerical values and evaluate σ :

σ = 2(8.85 × 10 −12 C 2 / N ⋅ m 2 )(3 × 106 V/m ) = 5.31 × 10 −5 C/m 2

The Electric Field 2: Continuous Charge Distributions

83

16 • Picture the Problem For x << r, we can model the disk as an infinite plane. For x >> r, we can approximate the ring charge by a point charge.

E x = 2πkσ

For x << r, express the electric field near an infinite plane of charge:

(

)(

(a) and (b) Because Ex is independent of x for x << r:

E x = 2π 8.99 × 109 N ⋅ m 2 /C 2 3.6 µC/m 2

For x >> r, use Coulomb’s law for the electric field due to a point charge to obtain:

E x (x ) =

= 2.03 × 10 5 N/C

kQ kπr 2σ = x2 x2

(c) Evaluate Ex at x = 5 m:

E x (5 m ) =

π (8.99 × 109 N ⋅ m 2 /C 2 )(2.5 cm )2 (3.6 µC/m 2 )

(5 m )2

= 2.54 N/C

(d) Evaluate Ex at x = 5 cm:

E x (5 cm ) =

π (8.99 × 109 N ⋅ m 2 /C 2 )(2.5 cm )2 (3.6 µC/m 2 )

(0.05 m )

2

= 2.54 × 10 4 N/C

Note that this is a very poor approximation because x = 2r is not much greater than r.

r Calculating E From Coulomb’s Law *17 • Picture the Problem We can use the definition of λ to find the total charge of the line of charge and the expression for the electric field on the axis of a finite line of charge to evaluate Ex at the given locations along the x axis. In part (d) we can apply Coulomb’s law for the electric field due to a point charge to approximate the electric field at x = 250 m. (a) Use the definition of linear charge density to express Q in terms of λ: Express the electric field on the axis of a finite line charge:

Q = λL = (3.5 nC/m )(5 m ) = 17.5 nC E x ( x0 ) =

kQ x0 ( x0 − L )

)

84

Chapter 22 E x (6 m ) =

(b) Substitute numerical values and evaluate Ex at x = 6 m:

(8.99 × 10 N ⋅ m /C )(17.5 nC) 9

2

2

(6 m )(6 m − 5 m )

= 26.2 N/C

E x (9 m ) =

(c) Substitute numerical values and evaluate Ex at x = 9 m:

(8.99 × 10 N ⋅ m /C )(17.5 nC) 9

2

(9 m )(9 m − 5 m )

= 4.37 N/C

(d) Substitute numerical values and evaluate Ex at x = 250 m:

E x (250 m ) =

(8.99 × 10 N ⋅ m /C )(17.5 nC) = 9

2

2

(250 m )(250 m − 5 m ) E x (x ) =

(e) Use Coulomb’s law for the electric field due to a point charge to obtain:

2.57 mN/C

kQ x2

Substitute numerical values and evaluate Ex(250 m):

(8.99 × 10 N ⋅ m /C )(17.5 nC) = E (250 m ) = 9

x

2

(250 m )2

2

2.52 mN/C

Note that this result agrees to within 2% with the exact value obtained in (d). 18 • Picture the Problem Let the charge densities on the two plates be σ1 and σ2 and denote the three regions of interest as 1, 2, and 3. Choose a coordinate system in which the positive x direction is to the right. We can apply the equation for r E near an infinite plane of charge and the superposition of fields to find the field in each of the three regions.

2

The Electric Field 2: Continuous Charge Distributions r r r r E1 = Eσ1 + Eσ 2 (a) Use the equation for E near an infinite plane of charge to express the field in region 1 when σ1 = σ2 = +3 µC/m2:

85

= −2πkσ 1iˆ − 2πkσ 2 iˆ = −4πkσ iˆ

r

Substitute numerical values and evaluate E1 :

r E1 = −4π 8.99 × 109 N ⋅ m 2 /C 2 3 µC/m 2 iˆ = − 3.39 × 105 N/C iˆ

(

)(

Proceed as above for region 2:

)

(

)

r r r E 2 = Eσ1 + Eσ 2 = 2πkσ 1iˆ − 2πkσ 2 iˆ = 2πkσiˆ − 2πkσiˆ = 0

Proceed as above for region 3:

r r r E3 = Eσ1 + Eσ 2 = 2πkσ 1iˆ + 2πkσ 2 iˆ = 4πkσiˆ

(

)(

)

= 4π 8.99 × 109 N ⋅ m 2 /C 2 3 µC/m 2 iˆ

(

)

= 3.39 × 105 N/C iˆ The electric field lines are shown to the right:

r

(b) Use the equation for E near an infinite plane of charge to express and evaluate the field in region 1 when σ1 = +3 µC/m2 and σ2 = −3 µC/m2: Proceed as above for region 2:

r r r E1 = Eσ1 + Eσ 2 = 2πkσ 1iˆ − 2πkσ 2 iˆ = 2πkσiˆ − 2πkσiˆ = 0

r r r E1 = Eσ1 + Eσ 2 = 2πkσ 1iˆ + 2πkσ 2 iˆ = 4πkσiˆ

= 4π (8.99 × 109 N ⋅ m 2 /C 2 )(3 µC ) iˆ = (3.39 × 105 N/C ) iˆ

86

Chapter 22 r r r E3 = Eσ1 + Eσ 2 = 2πkσ 1iˆ − 2πkσ 2 iˆ

Proceed as above for region 3:

= 2πkσiˆ − 2πkσiˆ = 0 The electric field lines are shown to the right:

19 • Picture the Problem The magnitude of the electric field on the axis of a ring of charge

(

is given by E x (x ) = kQx x 2 + a 2

)

32

where Q is the charge on the ring and a is the

radius of the ring. We can use this relationship to find the electric field on the x axis at the given distances from the ring.

r

Express E on the axis of a ring charge:

E x (x ) =

(x

kQx 2

)

32

+ a2

(a) Substitute numerical values and evaluate Ex for x = 1.2 cm:

E x (1.2 cm ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(2.75 µC )(1.2 cm )

[(1.2 cm)

]

+ (8.5 cm )

2 3 2

2

= 4.69 × 105 N/C

(b) Proceed as in (a) with x = 3.6 cm:

E x (3.6 cm ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(2.75 µC )(3.6 cm )

[(3.6 cm)

2

+ (8.5 cm )

]

2 3 2

= 1.13 × 10 6 N/C

(c) Proceed as in (a) with x = 4.0 m:

E x (4 m ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(2.75 µC )(4 m )

[(4 m)

(d) Using Coulomb’s law for the electric field due to a point charge, express Ex:

2

+ (8.5 cm )

]

2 3 2

Ex (x ) =

kQ x2

= 1.54 × 103 N/C

The Electric Field 2: Continuous Charge Distributions

87

Substitute numerical values and evaluate Ex at x = 4.0 m:

E x (4 m ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(2.75 µC ) = 1.55 × 103 N/C (4 m )2

This result agrees to within 1% with the result obtained in Part (c). It is slightly larger because the point charge is nearer x = 4 m than is the ring. 20

•

⎛

Picture the Problem We can use E x ( x ) = 2πkσ ⎜⎜1 −

⎝

⎞ ⎟ , the expression for the ⎟ x2 + R2 ⎠ x

electric field on the axis of a disk charge, to find Ex at x = 0.04 cm and 5 m.

⎛ x E x ( x ) = 2πkσ ⎜⎜1 − 2 x + R2 ⎝

Express the electric field on the axis of a disk charge:

⎞ ⎟ ⎟ ⎠

(a) Evaluate this expression for x = 0.04 cm:

⎛ E x = 2π 8.99 × 109 N ⋅ m 2 / C 2 3.6 µC/m 2 ⎜1 − ⎜ ⎝

(

)(

)

0.04 cm

(0.04 cm )2 + (2.5 cm )2

⎞ ⎟ ⎟ ⎠

= 2.00 × 105 N/C This value is about 1.5% smaller than the approximate value obtained in Problem 9. (b) Proceed as in (a) for x = 5 m:

⎛ E x = 2π 8.99 × 109 N ⋅ m 2 /C 2 3.6 µC/m 2 ⎜1 − ⎜ ⎝

(

)(

)

⎞ ⎟ = 2.54 N/C 2 2 ⎟ (5 m ) + (2.5 cm) ⎠ 5m

Note that the exact and approximate (from Problem 16) agree to within 1%. 21 • Picture the Problem We can use the definition of λ to find the total charge of the line of charge and the expression for the electric field on the perpendicular bisector of a finite line of charge to evaluate Ey at the given locations along the y axis. In part (e) we can apply Coulomb’s law for the electric field due to a point charge to approximate the electric field at y = 4.5 m. (a) Use the definition of linear

Q = λL = (6 nC/m )(5 cm ) = 0.300 nC

88

Chapter 22

charge density to express Q in terms of λ:

Ey (y) =

Express the electric field on the perpendicular bisector of a finite line charge:

2kλ y

1 2

L

( 12 L )2 + y 2

(b) Evaluate Ey at y = 4 cm:

E y (4 cm ) =

(

2 8.99 × 109 N ⋅ m 2 /C 2 0.04 m

)

(6 nC/m )(0.05 m ) (0.025 m )2 + (0.04 m )2 1 2

= 1.43 kN/C

(c) Evaluate Ey at y = 12 cm:

E y (12 cm ) =

(

)

(6 nC/m )(0.05 m ) (0.025 m )2 + (0.12 m )2

)

(6 nC/m )(0.05 m ) (0.025 m )2 + (4.5 m )2

2 8.99 × 109 N ⋅ m 2 /C 2 0.12 m

1 2

= 183 N/C

(d) Evaluate Ey at y = 4.5 m:

E y (4.5 m ) =

(

2 8.99 × 109 N ⋅ m 2 /C 2 4.5 m

(e) Using Coulomb’s law for the electric field due to a point charge, express Ey:

1 2

Ey (y) =

= 0.133 N/C

kQ y2

Substitute numerical values and evaluate Ey at y = 4.5 m:

E y (4.5 m ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(0.3 nC ) = 0.133 N/C (4.5 m )2

This result agrees to three decimal places with the value calculated in Part (d). 22 • Picture the Problem The electric field on the axis of a disk charge is given by

⎛ E x = 2πkq⎜⎜1 − ⎝

⎞ ⎟ . We can equate this expression and E x = 12 σ 2 ∈0 and ⎟ x2 + a2 ⎠ x

solve for x. Express the electric field on the axis of a disk charge:

⎛ E x = 2πkq⎜⎜1 − ⎝

⎞ ⎟ ⎟ x +a ⎠ x

2

2

The Electric Field 2: Continuous Charge Distributions

89

E x = 12 σ 2 ∈0

We’re given that: Equate these expressions:

⎛ σ x = 2πkσ ⎜⎜1 − 2 4ε 0 x + a2 ⎝

⎞ ⎟ ⎟ ⎠

Simplify to obtain:

⎛ σ x = 2πkσ ⎜⎜1 − 4ε 0 x2 + a2 ⎝

⎞ ⎟ ⎟ ⎠

or, because k = 1/4πε0,

1 = 1− 2 Solve for x to obtain:

23

x=

x x + a2 2

a 3

•

Picture the Problem We can use E x =

(x

kQx 2

+ a2

)

32

to find the electric field at the given

distances from the center of the charged ring. (a) Evaluate Ex at x = 0.2a:

E x (0.2a ) =

kQ(0.2a )

[(0.2a )

2

E x (0.5a ) =

kQ(0.5a )

[(0.5a )

E x (0.7a ) =

E x (a ) =

[a

]

kQ(0.7 a )

[(0.7a )

2

kQa 2

32

+a

]

3 2

kQ a2

= 0.385

(d) Evaluate Ex at x = a:

+ a2

2

= 0.358

(c) Evaluate Ex at x = 0.7a:

]

kQ a2

= 0.189

(b) Evaluate Ex at x = 0.5a:

+ a2

+ a2

]

3 2

kQ a2

2 3 2

= 0.354

kQ a2

90

Chapter 22 E x (2a ) =

(e) Evaluate Ex at x = 2a:

2kQa

[(2a )

2

+a

]

2 3 2

= 0.179

kQ a2

The field along the x axis is plotted below. The x coordinates are in units of x/a and E is in units of kQ/a2.

0.4

Ex

0.2

0.0

-0.2

-0.4 -3

-2

-1

0

1

2

3

x /a

24

•

⎛

⎞ ⎟ , where R is the radius of ⎟ x2 + R2 ⎠ x

Picture the Problem We can use E x = 2πkσ ⎜⎜1 −

⎝

the disk, to find the electric field on the axis of a disk charge. Express Ex in terms of ε0:

Ex = =

(a) Evaluate Ex at x = 0.2a:

2πσ 4π ∈0

σ ⎛

⎛ ⎜1 − ⎜ ⎝

⎜1 − 2 ∈0 ⎜⎝

E x (0.2a ) =

⎞ ⎟ ⎟ x +R ⎠ ⎞ x ⎟ 2 2 ⎟ x +R ⎠

σ ⎛⎜

2 ∈0 ⎜ ⎝

x

2

1−

= 0.804

2

⎞ ⎟ 2 2 ⎟ (0.2a ) + a ⎠

σ 2 ∈0

0.2a

The Electric Field 2: Continuous Charge Distributions (b) Evaluate Ex at x = 0.5a:

E x (0.5a ) =

σ ⎛⎜

2 ∈0 ⎜ ⎝

= 0.553

(c) Evaluate Ex at x = 0.7a:

E x (0.7a ) =

σ ⎛⎜

2 ∈0 ⎜ ⎝

E x (a ) =

2 ∈0 0.7a

(0.7a )2

⎞ ⎟ 2 ⎟ +a ⎠

σ 2 ∈0

a 1 − 2 ∈0 ⎜⎝ a2 + a2

E x (2a ) =

⎞ ⎟ 2 ⎟ +a ⎠

σ

σ ⎛⎜

= 0.293

(e) Evaluate Ex at x = 2a:

(0.5a )2

1−

= 0.427

(d) Evaluate Ex at x = a:

0.5a

1−

91

⎞ ⎟ ⎟ ⎠

σ 2 ∈0

σ ⎛⎜

2 ∈0 ⎜ ⎝

1−

= 0.106

2a

(2a )2

⎞ ⎟ 2 ⎟ +a ⎠

σ 2 ∈0

The field along the x axis is plotted below. The x coordinates are in units of x/a and E is in units of σ 2 ∈0 .

2.0

1.6

1.2 Ex 0.8

0.4

0.0 -3

-2

-1

0 x/R

1

2

3

92

Chapter 22

*25 •• Picture the Problem (a) The electric field on the x axis of a disk of radius r carrying a surface charge density σ is given by: (b) The electric field due to an infinite sheet of charge density σ is independent of the distance from the plane and is given by:

⎛ E x = 2πkσ ⎜⎜1 − ⎝

⎞ ⎟ ⎟ x +r ⎠ x

2

2

Eplate = 2πkσ

A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B3 B4 B5 A8 A9 B8

Content/Formula 9.00E+09 5.00E−10 0.3 0 0.01 2*PI()*$B$3*$B$4*(1−A8/ (A8^2+$B$5^2)^2)^0.5)

C8

2*PI()*$B$3*$B$4 A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 73 74 75 76 77

B

Algebraic Form k

σ

r x0 x0 + 0.01

⎛ x 2πkσ ⎜⎜1 − x2 + r 2 ⎝ 2πkσ C

k= 9.00E+09 Nm^2/C^2 sigma= 5.00E-10 C/m^2 r= 0.3 m x 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07

E(x) 28.27 27.33 26.39 25.46 24.54 23.63 22.73 21.85

E plate 28.3 28.3 28.3 28.3 28.3 28.3 28.3 28.3

0.65 0.66 0.67 0.68 0.69

2.60 2.53 2.47 2.41 2.34

28.3 28.3 28.3 28.3 28.3

⎞ ⎟ ⎟ ⎠

The Electric Field 2: Continuous Charge Distributions 78

0.70

2.29

93

28.3

The following graph shows E as a function of x. The electric field from an infinite sheet with the same charge density is shown for comparison – the magnitude of the electric fields differ by more than 10 percent for x = 0.03 m.

30

25

E

E (N/C

20

E plate 15

10

5

0 0 .0

0.1

0 .2

0 .3

0 .4

0.5

0 .6

0.7

x (m)

26 •• Picture the Problem Equation 22-10 expresses the electric field on the axis of a ring charge as a function of distance along the axis from the center of the ring. We can show that it has its maximum and minimum values at x = + a 2 and x = − a 2 by setting its first derivative equal to zero and solving the resulting equation for x. The graph of Ex will confirm that the maximum and minimum occur at these coordinates. Express the variation of Ex with x on the axis of a ring charge:

Ex =

(x

kQx

+ a2

2

)

32

Differentiate this expression with respect to x to obtain:

dE x d ⎡ x = kQ ⎢ 2 dx dx ⎢⎣ x + a 2

(x = kQ

(

2

+ a2

)

3 2

(

)

x2 + a2 ⎤ = kQ 3 2⎥ ⎥⎦

(

− x( 32 ) x 2 + a 2

(x

2

+ a2

)

Set this expression equal to zero for extrema and simplify:

3

)

( )

d 2 x + a2 dx 3 x2 + a2

3 2

−x

(

) (2 x ) = kQ (x 12

(x

2

+ a2

)

2

2

+ a2

)

32

( (x + a )

3 2

( (x + a ) ) − 3x (x 32

3 2

− 3x 2 x 2 + a 2 2 3

2

− 3x 2 x 2 + a 2

2

(x

+ a2

)

)

= 0,

)

= 0,

12

2 3

2

2

+ a2

12

)

12

94

Chapter 22 and

x 2 + a 2 − 3x 2 = 0 Solve for x to obtain:

x= ±

a 2

as our candidates for maxima or minima. A plot of E, in units of kQ/a2, versus x/a is shown to the right. This graph shows that E is a minimum at x = − a

2 and a maximum at x = a

2.

0.4

0.2

E x 0.0

-0.2

-0.4 -3

-2

-1

0

1

2

3

x/a

27 •• Picture the Problem The line charge and point (0, y) are shown in the diagram. Also shown is a line element of length dx and the r field dE its charge produces at (0, y). We

r

can find dEx from dE and then integrate from x = x1 to x = x2.

r

Express the x component of dE :

kλ sin θdx x + y2 kλ x =− 2 dx 2 2 x + y x + y2

dE x = −

=−

2

(x

kλx 2

+ y2

)

3 2

dx

The Electric Field 2: Continuous Charge Distributions Integrate from x = x1 to x2 to obtain:

x2

E x = − kλ ∫

x1

(x

x 2

+ y2

)

32

95

dx x2

⎡ = − kλ ⎢− ⎢⎣

⎤ ⎥ x 2 + y 2 ⎥⎦ x 1

1

⎡ ⎤ 1 1 ⎥ = − kλ ⎢− + ⎢⎣ x22 + y 2 x12 + y 2 ⎥⎦ ⎤ kλ ⎡ y y ⎢− ⎥ =− + 2 2 2 2 y ⎢ ⎥⎦ x y x y + + 2 1 ⎣ From the diagram we see that:

cos θ 2 =

y x22 + y 2

⎛ x2 ⎞ ⎟⎟ ⎝ y⎠

or θ 2 = tan −1 ⎜⎜

and

cos θ1 =

Substitute to obtain:

Ex = − =

y x +y 2 1

2

⎛ x1 ⎞ ⎟⎟ ⎝ y⎠

or θ1 = tan −1 ⎜⎜

kλ [− cos θ 2 + cos θ1 ] y kλ [cos θ 2 − cos θ1 ] y

28 •• Picture the Problem The diagram shows a segment of the ring of length ds that has a charge dq = λds. We can express the r electric field dE at the center of the ring due to the charge dq and then integrate this expression from θ = 0 to 2π to find the magnitude of the field in the center of the ring.

r

(a) and (b) The field dE at the center of the ring due to the charge dq is: The magnitude dE of the field at the center of the ring is:

r r r dE = dE x + dE y = −dE cos θ iˆ − dE sin θ ˆj

dE =

kdq r2

(1)

96

Chapter 22

Because dq = λds:

The linear charge density varies with θ according to λ(θ) = λ0 sin θ : Substitute rdθ for ds:

Substitute for dE in equation (1) to obtain:

r

Integrate dE from θ = 0 to 2π:

dE =

kλds r2

dE =

kλ0 sin θ ds r2

dE =

kλ0 sin θ rdθ kλ0 sin θ dθ = r2 r

r kλ sin θ cos θ dθ ˆ dE = − 0 i r kλ sin 2 θ dθ ˆ − 0 j r r kλ E=− 0 2r

2π

∫ sin 2θ dθ iˆ 0

kλ − 0 r =0−

2π

∫ sin

2

θ dθ ˆj

0

π kλ0 ˆ j

r π kλ0 ˆ = − j r

The field at the origin is in the negative y direction and its magnitude is

29 •• Picture the Problem The line charge and the point whose coordinates are (0, y) are shown in the diagram. Also shown is a segment of the line of length dx. r The field that it produces at (0, y) is dE .

r

We can find dEy from dE and then integrate from x = 0 to x = a to find the y component of the electric field at a point on the y axis. (a) Express the magnitude of the r field dE due to charge dq of the

dE =

kdq r2

π kλ0 r

.

The Electric Field 2: Continuous Charge Distributions element of length dx: Because dq = λdx :

Express the y component of dE:

where r 2 = x 2 + y 2

dE =

kλdx x2 + y 2

dE y =

kλ cos θ dx x + y2 2

Refer to the diagram to express cosθ in terms of x and y:

cos θ =

Substitute for cosθ in the expression for dEy to obtain:

dE y =

Integrate from x = 0 to x = a and simplify to obtain:

E y = kλy ∫

y x + y2 2

(x

kλy 2

+ y2

a

0

(x

)

32

dx

1 2

+ y2

)

32

dx a

⎡ ⎤ x = kλy ⎢ ⎥ ⎢⎣ y 2 x 2 + y 2 ⎥⎦ 0 ⎡ ⎤ a = kλ ⎢ ⎥ 2 2 ⎣⎢ y a + y ⎥⎦ kλ a = 2 y a + y2 *30 ••• Picture the Problem Consider the ring with its axis along the z direction shown in the diagram. Its radius is z = rcosθ and its width is rdθ. We can use the equation for the field on the axis of a ring charge and then integrate to express the field at the center of the hemispherical shell.

Express the field on the axis of the ring charge:

dE =

(r

kzdq

sin 2 θ + r 2 cos 2 θ kzdq = 3 r where z = rcosθ 2

)

32

97

98

Chapter 22

Express the charge dq on the ring:

dq = σdA = σ (2πr sin θ )rdθ = 2πσr 2 sin θdθ

Substitute to obtain:

Integrate dE from θ = 0 to π/2 to obtain:

k (r cosθ )2πσr 2 sin θdθ r3 = 2πkσ sin θ cosθdθ

dE =

π 2

E = 2πkσ ∫ sin θ cos θ dθ

[

0

= 2πkσ 12 sin 2 θ

]

π 2 0

= πkσ

Gauss’s Law 31 • r Picture the Problem The definition of electric flux is φ = E ⋅ nˆ dA . We can apply this

∫

S

definition to find the electric flux through the square in its two orientations. (a) Apply the definition of φ to find the flux of the field when the square is parallel to the yz plane:

φ = ∫ (2 kN/C)iˆ ⋅ iˆdA = (2 kN/C ) ∫ dA

(b) Proceed as in (a) with iˆ ⋅ nˆ = cos 30° :

φ = ∫ (2 kN/C ) cos 30°dA

S

S

= (2 kN/C)(0.1 m ) = 20.0 N ⋅ m 2 /C 2

S

= (2 kN/C ) cos 30°∫ dA S

= (2 kN/C )(0.1 m ) cos 30° 2

= 17.3 N ⋅ m 2 /C *32 • Determine the Concept While the number of field lines that we choose to draw radially outward from q is arbitrary, we must show them originating at q and, in the absence of other charges, radially symmetric. The number of lines that we draw is, by agreement, in proportion to the magnitude of q.

The Electric Field 2: Continuous Charge Distributions

99

(a) The sketch of the field lines and of the sphere is shown in the diagram to the right.

Given the number of field lines drawn from q, 3 lines enter the sphere. Had we chosen to draw 24 field lines, 6 would have entered the spherical surface. (b) The net number of lines crossing the surface is zero.

(c) The net flux is zero. 33 • Picture the Problem The field at both circular faces of the cylinder is parallel to the outward vector normal to the surface, so the flux is just EA. There is no flux through the r curved surface because the normal to that surface is perpendicular to E . The net flux through the closed surface is related to the net charge inside by Gauss’s law.

r

(a) Use Gauss’s law to calculate the flux through the right circular surface:

φright = E right ⋅ nˆ right A

Apply Gauss’s law to left circular surface:

φleft = E left ⋅ nˆ left A

2 = (300 N/C ) iˆ ⋅ iˆ(π )(0.04 m )

= 1.51 N ⋅ m 2 /C r

( )

2 = (− 300 N/C ) iˆ ⋅ − iˆ (π )(0.04 m )

= 1.51 N ⋅ m 2 /C

100 Chapter 22

φcurved = 0

(b) Because the field lines are parallel to the curved surface of the cylinder:

φnet = φright + φleft + φcurved

(c) Express and evaluate the net flux through the entire cylindrical surface:

= 1.51 N ⋅ m 2 /C + 1.51 N ⋅ m 2 /C + 0 = 3.02 N ⋅ m 2 /C

(d) Apply Gauss’s law to obtain:

φnet = 4πkQinside

Solve for Qinside:

Qinside =

Substitute numerical values and evaluate Qinside:

Qinside

φnet 4πk

3.02 N ⋅ m 2 /C = 4π 8.99 × 109 N ⋅ m 2 /C 2

(

)

= 2.67 × 10 −11 C 34 • Picture the Problem We can use Gauss’s law in terms of ε0 to find the net charge inside the box. (a) Apply Gauss’s law in terms of ε0 to find the net charge inside the box:

φnet =

1 Qinside ∈0

or

Qinside =∈0 φnet

Substitute numerical values and evaluate Qinside:

(b)

(

)(

Qinside = 8.85 × 10 −12 C 2 /N ⋅ m 2 6 kN ⋅ m 2 /C = 5.31 × 10 −8 C

You can only conclude that the net charge is zero. There may be an equal number of positive and negative charges present inside the box.

35 • Picture the Problem We can apply Gauss’s law to find the flux of the electric field through the surface of the sphere. (a) Use the formula for the surface area of a sphere to obtain:

A = 4πr 2 = 4π (0.5 m ) = 3.14 m 2 2

)

The Electric Field 2: Continuous Charge Distributions 101 (b) Apply Coulomb’s law to express and evaluate E:

E= =

1 q 4π ∈0 r 2 1 2 µC 2 2 −12 4π 8.85 × 10 C /N ⋅ m (0.5 m )2

(

)

= 7.19 × 10 4 N/C (c) Apply Gauss’s law to obtain:

r

φ = ∫ E ⋅ nˆ dA = ∫ EdA S

S

(

)(

= 7.19 × 10 N/C 3.14 m 2 4

)

= 2.26 × 10 5 N ⋅ m 2 /C

(d)

No. The flux through the surface is independent of where the charge is located inside the sphere.

(e) Because the cube encloses the sphere, the flux through the surface of the sphere will also be the flux through the cube:

φcube = 2.26 × 105 N ⋅ m 2 /C

*36 • Picture the Problem We’ll define the flux of the gravitational field in a manner that is analogous to the definition of the flux of the electric field and then substitute for the gravitational field and evaluate the integral over the closed spherical surface. Define the gravitational flux as:

r

φg = ∫ g ⋅ nˆ dA S

r Substitute for g and evaluate the integral to obtain:

Gm ⎛ Gm ˆ ⎞ ˆ r ⎟ ⋅ ndA = − 2 ∫ dA 2 r S ⎝ r ⎠ ⎛ Gm ⎞ = ⎜ − 2 ⎟ 4πr 2 = − 4πGm ⎝ r ⎠

φg = ∫ ⎜ − S

(

)

37 •• Picture the Problem We’ll let the square be one face of a cube whose side is 40 cm. Then the charge is at the center of the cube and we can apply Gauss’s law in terms of ε0 to find the flux through the square. Apply Gauss’s law to the cube to express the net flux:

φnet =

1 Qinside ∈0

102 Chapter 22 Express the flux through one face of the cube:

φsquare =

1 Qinside 6 ∈0

Substitute numerical values and evaluate φsquare:

φsquare =

2 µC 6 8.85 × 10 −12 C 2 /N ⋅ m 2

(

)

= 3.77 × 10 4 N ⋅ m 2 /C 38 •• Picture the Problem We can treat this portion of the earth’s atmosphere as though it is a cylinder with cross-sectional area A and height h. Because the electric flux increases with altitude, we can conclude that there is charge inside the cylindrical region and use Gauss’s law to find the charge and hence the charge density of the atmosphere in this region. The definition of volume charge density is: Express the charge inside a cylinder of base area A and height h for a charge density ρ: Taking upward to be the positive direction, apply Gauss’s law to the charge in the cylinder: Substitute to obtain:

ρ=

Q V

Q = ρAh

Q = −(Eh A − E0 A) ∈0 = (E0 A − Eh A) ∈0 where we’ve taken our zero at 250 m above the surface of a flat earth.

ρ=

(E0 A − Eh A) ∈0 = (E0 − Eh ) ∈0 Ah

h

Substitute numerical values and evaluate ρ:

ρ=

(150 N/C − 170 N/C) (8.85 × 10 −12 C 2 /N ⋅ m 2 ) = 250 m

− 7.08 × 10 −13 C/m 3

where we’ve been able to neglect the curvature of the earth because the maximum height of 400 m is approximately 0.006% of the radius of the earth.

Spherical Symmetry 39 • Picture the Problem To find En in these three regions we can choose Gaussian surfaces of appropriate radii and apply Gauss’s law. On each of these surfaces, Er is constant and

The Electric Field 2: Continuous Charge Distributions 103 Gauss’s law relates Er to the total charge inside the surface. (a) Use Gauss’s law to find the electric field in the region r < R1:

1

∫ E dA = ∈ n

S

Qinside

0

and

E r < R1 =

Qinside = 0 ∈0 A

because Qinside = 0.

q1 kq1 = 2 ∈0 (4πr ) r2

Apply Gauss’s law in the region R1 < r < R2 :

E R1 < r < R1 =

Using Gauss’s law, find the electric field in the region r > R2:

Er > R2 =

(b) Set Er >R2 = 0 to obtain:

q1 + q2 = 0

q1 + q2 k (q1 + q2 ) = 2 ∈0 (4πr ) r2

or

q1 = −1 q2 (c) The electric field lines for the situation in (b) with q1 positive is shown to the right.

40 • Picture the Problem We can use the definition of surface charge density and the formula for the area of a sphere to find the total charge on the shell. Because the charge is distributed uniformly over a spherical shell, we can choose a spherical Gaussian surface and apply Gauss’s law to find the electric field as a function of the distance from the center of the spherical shell. (a) Using the definition of surface charge density, relate the charge on the sphere to its area:

Q = σA = 4πσ r 2

= 4π (9 nC/m 2 )(0.06 m )

2

= 0.407 nC

104 Chapter 22 1

∫ E dA = ∈

Apply Gauss’s law to a spherical surface of radius r that is concentric the spherical shell to obtain:

S

n

Qinside

0

or

4πr 2 En =

Solve for En:

En =

Qinside ∈0

Qinside 1 kQinside = r2 4π ∈0 r 2

En (2 cm ) = 0

(b) Qinside a sphere whose radius is 2 cm is zero and hence:

En (5.9 cm ) = 0

(c) Qinside a sphere whose radius is 5.9 cm is zero and hence:

(d) Qinside a sphere whose radius is 6.1 cm is 0.407 nC and hence:

En (6.1 cm ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(0.407 nC ) = 983 N/C (0.061 m )2

(e) Qinside a sphere whose radius is 10 cm is 0.407 nC and hence:

En (10 cm ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(0.407 nC ) = 366 N/C (0.1 m )2

41 •• Picture the Problem We can use the definition of volume charge density and the formula for the volume of a sphere to find the total charge of the sphere. Because the charge is distributed uniformly throughout the sphere, we can choose a spherical Gaussian surface and apply Gauss’s law to find the electric field as a function of the distance from the center of the sphere. (a) Using the definition of volume charge density, relate the charge on the sphere to its volume:

Q = ρV = 43 πρr 3

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the spherical shell to obtain:

∫ E dA = ∈

= 43 π (450 nC/m 3 )(0.06 m )

3

= 0.407 nC 1

S

or

n

0

Qinside

The Electric Field 2: Continuous Charge Distributions 105 4πr 2 En =

Qinside ∈0

Qinside 1 kQinside = r2 4π ∈0 r 2

Solve for En:

En =

Because the charge distribution is uniform, we can find the charge inside the Gaussian surface by using the definition of volume charge density to establish the proportion:

Q Qinside = V V' where V′ is the volume of the Gaussian surface.

V' r3 =Q 3 V R

Solve for Qinside to obtain:

Qinside = Q

Substitute to obtain:

En (r < R ) =

Qinside 1 kQ r = 4π ∈0 r 2 R 3

(b) Evaluate En at r = 2 cm:

(8.99 × 10 E (2 cm ) =

9

n

N ⋅ m 2 /C 2 )(0.407 nC ) (0.02 m ) = 339 N/C (0.06 m )3

(c) Evaluate En at r = 5.9 cm:

En (5.9 cm ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(0.407 nC ) (0.059 m ) = 999 N/C (0.06 m )3

Apply Gauss’s law to the Gaussian surface with r > R:

4πr 2 En =

Solve for En to obtain:

En =

Qinside

ε0

kQinside kQ = 2 r2 r

(d) Evaluate En at r = 6.1 cm:

En (6.1 cm ) =

(8.99 × 10

(e) Evaluate En at r = 10 cm:

9

N ⋅ m 2 /C 2 )(0.407 nC ) = 983 N/C (0.061 m )2

106 Chapter 22 En (10 cm ) =

(8.99 × 10

9

N ⋅ m 2 /C 2 )(0.407 nC ) = 366 N/C (0.1 m )2

Note that, for r > R, these results are the same as those obtained for in Problem 40 for a uniform charge distribution on a spherical shell. This agreement is a consequence of the choices of σ and ρ so that the total charges on the two spheres is the same. *42 •• Determine the Concept The charges on a conducting sphere, in response to the repulsive Coulomb forces each experiences, will separate until electrostatic equilibrium conditions exit. The use of a wire to connect the two spheres or to ground the outer sphere will cause additional redistribution of charge. (a) Because the outer sphere is conducting, the field in the thin shell must vanish. Therefore, −2Q, uniformly distributed, resides on the inner surface, and −5Q, uniformly distributed, resides on the outer surface. (b) Now there is no charge on the inner surface and −5Q on the outer surface of the spherical shell. The electric field just outside the surface of the inner sphere changes from a finite value to zero. (c) In this case, the −5Q is drained off, leaving no charge on the outer surface and −2Q on the inner surface. The total charge on the outer sphere is then −2Q. 43 •• Picture the Problem By symmetry; the electric field must be radial. To find Er inside the sphere we choose a spherical Gaussian surface of radius r < R. On this surface, Er is constant. Gauss’s law then relates Er to the total charge inside the surface. Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the nonconducting sphere to obtain:

1

∫ E dA = ∈ S

r

Qinside

0

or

4πr 2 Er =

Qinside ∈0

Qinside 1 kQinside = r2 4π ∈0 r 2

Solve for Er:

Er =

Use the definition of charge density to relate Qinside to ρ and the volume defined by the Gaussian surface:

Qinside = ρVGaussian surface = 43 ρπr 3

The Electric Field 2: Continuous Charge Distributions 107 Substitute to obtain:

Er (r < R ) =

4 3

ρπkr 3 r2

= 43 ρπkr

Substitute numerical values and evaluate Er at r = 0.5R = 0.05 m:

Er (0.05 m ) = 43 π (2 nC/m 3 )(8.99 × 10 9 N ⋅ m 2 /C 2 )(0.05 m ) = 3.77 N/C 44 •• Picture the Problem We can find the total charge on the sphere by expressing the charge dq in a spherical shell and integrating this expression between r = 0 and r = R. By symmetry, the electric fields must be radial. To find Er inside the charged sphere we choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface. (a) Express the charge dq in a shell of thickness dr and volume 4πr2 dr:

dq = 4πr 2 ρdr = 4πr 2 ( Ar ) dr

= 4πAr 3 dr

[

R

Integrate this expression from r = 0 to R to find the total charge on the sphere:

Q = 4πA∫ r 3 dr = πAr 4

(b) Apply Gauss’s law to a spherical surface of radius r > R that is concentric with the nonconducting sphere to obtain:

∫ E dA = ∈

Solve for Er:

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the nonconducting sphere to obtain:

]

R 0

= πAR 4

0

1

S

r

Qinside

0

or

4πr 2 Er =

Qinside ∈0

Er (r > R ) =

Qinside 1 kQinside = 4π ∈0 r 2 r2

=

kAπR 4 AR 4 = r2 4 ∈0 r 2

1

∫ E dA = ∈ S

r

Qinside

0

or

4πr 2 Er =

Qinside ∈0

108 Chapter 22 Qinside Ar 2 πAr 4 Er (r < R ) = = = 4πr 2 ∈0 4πr 2 ∈0 4 ∈0

Solve for Er:

The graph of Er versus r/R, with Er in units of A/4∈0, was plotted using a spreadsheet program.

1.0

0.8

Er

0.6

0.4

0.2

0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

r/R

Remarks: Note that the results for (a) and (b) agree at r = R. 45 •• Picture the Problem We can find the total charge on the sphere by expressing the charge dq in a spherical shell and integrating this expression between r = 0 and r = R. By symmetry, the electric fields must be radial. To find Er inside the charged sphere we choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface. (a) Express the charge dq in a shell of thickness dr and volume 4πr2 dr:

dq = 4πr 2 ρdr = 4πr 2

Integrate this expression from r = 0 to R to find the total charge on the sphere:

R

= 4πBrdr

[

B dr r

Q == 4πB ∫ rdr = 2πBr 2 0

= 2πBR 2

]

R 0

The Electric Field 2: Continuous Charge Distributions 109 1

∫ E dA = ∈

(b) Apply Gauss’s law to a spherical surface of radius r > R that is concentric with the nonconducting sphere to obtain:

S

r

Qinside

0

or

4πr 2 Er =

Qinside ∈0

Er (r > R ) =

Solve for Er:

=

1

∫ E dA = ∈

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the nonconducting sphere to obtain:

S

r

Qinside 1 kQinside = 4π ∈0 r 2 r2 k 2πBR 2 BR 2 = r2 2 ∈0 r 2

Qinside

0

or

4πr 2 Er =

Qinside ∈0

Er (r < R ) =

Solve for Er:

=

Qinside 2πBr 2 = 4πr 2 ∈0 4πr 2 ∈0 B 2 ∈0

The graph of Er versus r/R, with Er in units of B/2∈0, was plotted using a spreadsheet program. 1.2 1.0

Er

0.8 0.6 0.4 0.2 0.0 0.0

0.5

1.0

1.5

2.0

2.5

r /R

Remarks: Note that our results for (a) and (b) agree at r = R.

3.0

110 Chapter 22 *46 •• Picture the Problem We can find the total charge on the sphere by expressing the charge dq in a spherical shell and integrating this expression between r = 0 and r = R. By symmetry, the electric fields must be radial. To find Er inside the charged sphere we choose a spherical Gaussian surface of radius r < R. To find Er outside the charged sphere we choose a spherical Gaussian surface of radius r > R. On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface. (a) Express the charge dq in a shell of thickness dr and volume 4πr2 dr:

dq = 4πr 2 ρdr = 4πr 2

= 4πCdr

C dr r2

R

Integrate this expression from r = 0 to R to find the total charge on the sphere:

Q = 4πC ∫ dr = [4πCr ] 0

(b) Apply Gauss’s law to a spherical surface of radius r > R that is concentric with the nonconducting sphere to obtain:

∫ E dA = ∈

Solve for Er:

R

0

= 4πCR

1

S

r

0

or

4πr 2 Er =

Qinside ∈0

Er (r > R ) =

Qinside 1 kQinside = 4π ∈0 r 2 r2 k 4πCR CR = 2 r ∈0 r 2

=

Apply Gauss’s law to a spherical surface of radius r < R that is concentric with the nonconducting sphere to obtain:

Solve for Er:

Qinside

1

∫ E dA = ∈ S

r

Qinside

0

or

4πr 2 Er =

Qinside ∈0

Er (r < R ) =

=

Qinside 4πCr = 2 4πr ∈0 4πr 2 ∈0 C ∈0 r

The graph of Er versus r/R, with Er in units of C / ∈0 R , was plotted using a spreadsheet

The Electric Field 2: Continuous Charge Distributions 111 program.

10

8

6 Er 4

2

0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

r/R

47 ••• Picture the Problem By symmetry, the electric fields resulting from this charge distribution must be radial. To find Er for r < a we choose a spherical Gaussian surface of radius r < a. To find Er for a < r < b we choose a spherical Gaussian surface of radius a < r < b. To find Er for r > b we choose a spherical Gaussian surface of radius r > b. On each of these surfaces, Er is constant. Gauss’s law then relates Er to the total charge inside the surface. (a), (b) Apply Gauss’s law to a spherical surface of radius r that is concentric with the nonconducting spherical shell to obtain:

1

∫ E dA = ∈ S

Qinside

r

0

or

4πr 2 Er =

Qinside ∈0

Solve for Er:

Er (r ) =

Evaluate Er(r < a):

Er (r < a ) =

Qinside 1 kQinside = r2 4π ∈0 r 2 Qinside 1 kQinside = = 0 4π ∈0 r 2 r2

because ρ(r < a) = 0 and, therefore, Qinside = 0.

112 Chapter 22 Integrate dq from r = a to r to find the total charge in the spherical shell in the interval a < r < b:

Qinside

2

=

Evaluate Er(a < r < b):

r

⎡ 4πCr '3 ⎤ = 4πρ ∫ r ' dr ' = ⎢ ⎥ ⎣ 3 ⎦a a r

4πρ 3 r − a3 3

(

Er (a < r < b ) =

kQinside r2 4πkρ 3 (r − a 3 ) = 3r 2

=

For r > b:

)

Qinside =

ρ 3 ∈0 r

4πρ 3 b − a3 3

(

and

Er (r > b ) =

(r

2

3

− a3 )

)

4πkρ 3 (b − a 3 ) 3r 2

ρ

=

3 ∈0 r

2

(b

3

− a3 )

Remarks: Note that E is continuous at r = b.

Cylindrical Symmetry 48 •• Picture the Problem From symmetry, the field in the tangential direction must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long, uniformly charged cylindrical shell. Apply Gauss’s law to the cylindrical surface of radius r and length L that is concentric with the infinitely long, uniformly charged cylindrical shell:

1

∫ E dA = ∈ S

n

Qinside

0

or

2πrLEn =

Qinside ∈0

where we’ve neglected the end areas because no flux crosses them. Solve for En:

En =

2kQinside Qinside = Lr 2πrL ∈0

The Electric Field 2: Continuous Charge Distributions 113 For r < R, Qinside = 0 and:

En (r < R ) = 0

For r > R, Qinside = λL and:

En (r > R ) =

2kλL 2kλ 2k (2πRσ ) = = Lr r r Rσ = ∈0 r

49 •• Picture the Problem We can use the definition of surface charge density to find the total charge on the shell. From symmetry, the electric field in the tangential direction must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the uniformly charged cylindrical shell.

Q = σA

(a) Using its definition, relate the surface charge density to the total charge on the shell:

= 2πRLσ

(

Q = 2π (0.06 m )(200 m ) 9 nC/m 2

Substitute numerical values and evaluate Q:

= 679 nC E (2 cm ) = 0

(b) From Problem 48 we have, for r = 2 cm:

E (5.9 cm ) = 0

(c) From Problem 48 we have, for r = 5.9 cm: (d) From Problem 48 we have, for r = 6.1 cm:

Er = and

E (6.1 cm ) =

σR ∈0 r

(9 nC/m )(0.06 m ) (8.85 × 10 C /N ⋅ m )(0.061 m ) = 2

−12

2

2

1.00 kN/C

(e) From Problem 48 we have, for r = 10 cm:

(9 nC/m )(0.06 m ) E (10 cm ) = (8.85 × 10 C /N ⋅ m )(0.1 m ) = 2

−12

2

2

610 N/C

)

114 Chapter 22 50 •• Picture the Problem From symmetry, the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylinder. Apply Gauss’s law to a cylindrical surface of radius r and length L that is concentric with the infinitely long nonconducting cylinder:

1

∫ E dA = ∈ S

n

Qinside

0

or

2πrLEn =

Qinside ∈0

where we’ve neglected the end areas because no flux crosses them.

2kQinside Qinside = Lr 2πrLε 0

Solve for En:

En =

Express Qinside for r < R:

Qinside = ρ (r )V = ρ0 πr 2 L

Substitute to obtain:

En (r < R ) =

(

(

)

)

ρ0 2k πρ 0 Lr 2 r = Lr 2 ∈0

or, because λ = ρπR 2

En (r < R ) =

λ 2π ∈0 R 2

r

Express Qinside for r > R:

Qinside = ρ (r )V = ρ 0 (πR 2 L )

Substitute to obtain:

2k (πρ0 LR En (r > R ) =

Lr

2

)=

ρ0 R 2 2 ∈0 r

or, because λ = ρπR 2

En (r > R ) =

λ 2π ∈0 r

51 •• Picture the Problem We can use the definition of volume charge density to find the total charge on the cylinder. From symmetry, the electric field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the uniformly charged cylinder.

The Electric Field 2: Continuous Charge Distributions 115

(

Qtot = ρV = ρ πR 2 L

(a) Use the definition of volume charge density to express the total charge of the cylinder:

(

)

)

Qtot = π 300 nC/m 3 (0.06 m ) (200 m )

Substitute numerical values to obtain:

2

= 679 nC

From Problem 50, for r < R, we have:

Er =

ρ 2 ∈0

r

(b) For r = 2 cm:

(300 nC/m )(0.02 m ) = E (2 cm ) = 2(8.85 × 10 C /N ⋅ m ) 3

−12

r

2

2

339 N/C

(c) For r = 5.9 cm:

Er (5.9 cm ) =

(300 nC/m )(0.059 m) = 2(8.85 × 10 C /N ⋅ m ) 3

−12

From Problem 50, for r > R, we have:

2

2

Er =

1.00 kN/C

ρR 2 2 ∈0 r

(d) For r = 6.1 cm:

Er (6.1 cm ) =

(300 nC/m )(0.06 m ) = 2(8.85 × 10 C /N ⋅ m )(0.061 m ) 2

3

−12

2

2

1.00 kN/C

(e) For r = 10 cm:

Er (10 cm ) =

(300 nC/m )(0.06 m ) = 2(8.85 × 10 C /N ⋅ m )(0.1 m ) 2

3

−12

2

2

610 N/C

Note that, given the choice of charge densities in Problems 49 and 51, the electric fields for r > R are the same. *52 •• Picture the Problem From symmetry; the field tangent to the surfaces of the shells must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from

116 Chapter 22 the centerline of the infinitely long, uniformly charged cylindrical shells. (a) Apply Gauss’s law to the cylindrical surface of radius r and length L that is concentric with the infinitely long, uniformly charged cylindrical shell:

1

∫ E dA = ∈ S

n

Qinside

0

or

2πrLEn =

Qinside ∈0

where we’ve neglected the end areas because no flux crosses them.

2kQinside Lr

Solve for En:

En =

For r < R1, Qinside = 0 and:

En (r < R1 ) = 0

Express Qinside for R1 < r < R2:

Qinside = σ 1 A1 = 2πσ 1R1 L

Substitute in equation (1) to obtain:

En (R1 < r < R2 ) =

Express Qinside for r > R2:

Qinside = σ 1 A1 + σ 2 A2

(1)

2k (2πσ 1 R1 L ) Lr σR = 1 1 ∈0 r

= 2πσ 1 R1 L + 2πσ 2 R2 L 2k (2πσ 1 R1 L + 2πσ 2 R2 L ) Lr σ R + σ 2 R2 = 1 1 ∈0 r

Substitute in equation (1) to obtain:

En (r > R2 ) =

(b) Set E = 0 for r > R2 to obtain:

σ 1 R1 + σ 2 R2 ∈0 r or

=0

σ 1 R1 + σ 2 R2 = 0

Solve for the ratio of σ1 to σ2:

R σ1 = − 2 σ2 R1

The Electric Field 2: Continuous Charge Distributions 117 Because the electric field is determined by the charge inside the Gaussian surface, the field under these conditions would be as given above:

σ 1 R1

En (R1 < r < R2 ) =

∈0 r

(c) Assuming that σ1 is positive, the field lines would be directed as shown to the right.

53 •• Picture the Problem The electric field is directed radially outward. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long, uniformly charged cylindrical shell. (a) Apply Gauss’s law to a cylindrical surface of radius r and length L that is concentric with the inner conductor:

1

∫ E dA = ∈ S

n

Qinside

0

or

2πrLEn =

Qinside ∈0

where we’ve neglected the end areas because no flux crosses them.

2kQinside Lr

Solve for En:

En =

For r < 1.5 cm, Qinside = 0 and:

En (r < 1.5 cm ) = 0

Letting R = 1.5 cm, express Qinside for 1.5 cm < r < 4.5 cm: Substitute in equation (1) to obtain:

(1)

Qinside = λL = 2πσRL

En (1.5 cm < r < 4.5 cm ) = =

Substitute numerical values and evaluate En(1.5 cm < r < 4.5 cm):

2kλ r

2k (λL ) Lr

118 Chapter 22

(

En (1.5 cm < r < 4.5 cm ) = 2 8.99 × 10 9 N ⋅ m 2 /C 2

) = (108 N ⋅ m/C) ) (6 nC/m r r

Qinside = 0

Express Qinside for 4.5 cm < r < 6.5 cm:

and

En (4.5 cm < r < 6.5 cm ) = 0

Qinside = σ 2 A2 = 2πσ 2 R2 L

Letting σ2 represent the charge density on the outer surface, express Qinside for r > 6.5 cm:

where R2 = 6.5 cm.

Substitute in equation (1) to obtain:

E n (r > R2 ) =

2k (2πσ 2 R2 L ) σ 2 R2 = Lr ∈0 r

In (b) we show that σ2 = 21.2 nC/m2. Substitute numerical values to obtain:

En (r > 6.5 cm ) =

(21.2 nC/m )(6.5 cm) = (8.85 × 10 C / N ⋅ m )r 2

−12

2

2

156 N ⋅ m/C r

(b) The surface charge densities on the inside and the outside surfaces of the outer conductor are given by:

σ1 =

−λ and σ 2 = −σ 1 2πR1

Substitute numerical values and evaluate σ1 and σ2:

σ1 =

− 6 nC/m = − 21.2 nC/m 2 2π (0.045 m )

and

σ 2 = 21.2 nC/m 2 54 •• Picture the Problem From symmetry considerations, we can conclude that the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylinder. 1 (a) Apply Gauss’s law to a En dA = Qinside S ∈0 cylindrical surface of radius r and length L that is concentric with the or infinitely long nonconducting Q 2πrLEn = inside cylinder: ∈

∫

0

where we’ve neglected the end areas

The Electric Field 2: Continuous Charge Distributions 119 because no flux crosses them.

Qinside 2πrL ∈0

Solve for En:

En =

Express dQinside for ρ(r) = ar:

dQinside = ρ (r )dV = ar (2πrL )dr

(1)

= 2πar 2 Ldr Integrate dQinside from r = 0 to R to obtain:

Divide both sides of this equation by L to obtain an expression for the charge per unit length λ of the cylinder:

R

⎡r3 ⎤ = 2πaL ∫ r dr = 2πaL ⎢ ⎥ ⎣ 3 ⎦0 0 2πaL 3 = R 3 R

Qinside

λ=

2

Qinside 2πaR 3 = L 3

(b) Substitute for Qinside in equation (1) to obtain:

2πaL 3 r a 2 3 En (r < R ) = = r 2π ∈0 Lr 3 ∈0

For r > R:

Qinside =

Substitute for Qinside in equation (1) to obtain:

2πaL 3 R aR 3 3 En (r > R ) = = 2πrL ∈0 3r ∈0

2πaL 3 R 3

55 •• Picture the Problem From symmetry; the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylinder. (a) Apply Gauss’s law to a cylindrical surface of radius r and length L that is concentric with the infinitely long nonconducting cylinder:

1

∫ E dA = ∈ S

or

n

0

Qinside

120 Chapter 22 2πrLEn =

Qinside ∈0

where we’ve neglected the end areas because no flux crosses them.

Qinside 2πrL ∈0

Solve for En:

En =

Express dQinside for ρ(r) = br2:

dQinside = ρ (r )dV = br 2 (2πrL )dr

(1)

= 2πbr 3 Ldr Integrate dQinside from r = 0 to R to obtain:

Divide both sides of this equation by L to obtain an expression for the charge per unit length λ of the cylinder: (b) Substitute for Qinside in equation (1) to obtain:

For r > R:

Substitute for Qinside in equation (1) to obtain:

R

⎡r4 ⎤ = 2πbL ∫ r dr = 2πbL ⎢ ⎥ ⎣ 4 ⎦0 0 πbL 4 = R 2 R

Qinside

λ=

3

Qinside πbR 4 = L 2

πbL En (r < R ) =

Qinside =

πbL 2

b 3 2 = r 2πLr ∈0 4 ∈0

R4

πbL En (r > R ) =

r4

R4

bR 4 2 = 2πrL ∈0 4r ∈0

56 ••• Picture the Problem From symmetry; the field tangent to the surface of the cylinder must vanish. We can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to find the electric field as a function of the distance from the centerline of the infinitely long nonconducting cylindrical shell. Apply Gauss’s law to a cylindrical surface of radius r and length L that

1

∫ E dA = ∈ S

n

0

Qinside

The Electric Field 2: Continuous Charge Distributions 121 is concentric with the infinitely long nonconducting cylindrical shell:

or

2πrLEn =

Qinside ∈0

where we’ve neglected the end areas because no flux crosses them.

Qinside 2πrL ∈0

Solve for En:

En =

For r < a, Qinside = 0:

E n (r < a ) = 0

Express Qinside for a < r < b:

Qinside = ρV = ρπr 2 L − ρπa 2 L

(

= ρπL r 2 − a 2 Substitute for Qinside to obtain:

En (a < r < b ) = =

Express Qinside for r > b:

ρπL(r 2 − a 2 ) 2π ∈0 Lr ρ (r 2 − a 2 ) 2 ∈0 r

Qinside = ρV = ρπb 2 L − ρπa 2 L

(

= ρπL b 2 − a 2 Substitute for Qinside to obtain:

)

En (r > b ) = =

)

ρπL(b 2 − a 2 ) 2π ∈0 rL ρ (b 2 − a 2 ) 2 ∈0 r

57 ••• Picture the Problem We can integrate the density function over the radius of the inner cylinder to find the charge on it and then calculate the linear charge density from its definition. To find the electric field for all values of r we can construct a Gaussian surface in the shape of a cylinder of radius r and length L and apply Gauss’s law to each region of the cable to find the electric field as a function of the distance from its centerline. R R (a) Find the charge Qinner on the C Qinner = ρ (r )dV = 2πrLdr inner cylinder: r

∫

∫

0

0

R

= 2πCL ∫ dr = 2πCLR 0

122 Chapter 22 Relate this charge to the linear charge density: Substitute numerical values and evaluate λinner: (b) Apply Gauss’s law to a cylindrical surface of radius r and length L that is concentric with the infinitely long nonconducting cylinder:

λinner =

Qinner 2πCLR = = 2πCR L L

λinner = 2π (200 nC/m )(0.015 m ) = 18.8 nC/m

∫

S

En dA =

1 Qinside ∈0

or

2πrLEn =

Qinside ∈0

where we’ve neglected the end areas because no flux crosses them.

Qinside 2πrL ∈0

Solve for En:

En =

Substitute to obtain, for r < 1.5 cm:

En (r < 1.5 cm ) =

Substitute numerical values and evaluate En(r < 1.5 cm):

200 nC/m 2 E n (r < 1.5 cm ) = 8.85 × 10 −12 C 2 /N ⋅ m 2

2πCLr C = 2π ∈0 Lr ∈0

= 22.6 kN/C Express Qinside for 1.5 cm < r < 4.5 cm: Substitute to obtain, for 1.5 cm < r < 4.5 cm:

Qinside = 2πCLR

En (1.5 cm < r < 4.5 cm ) = =

2CπRL 2π ∈0 rL CR ∈0 r

where R = 1.5 cm. Substitute numerical values and evaluate En(1.5 cm < r < 4.5 cm):

En (1.5 cm < r < 4.5 cm ) =

Because the outer cylindrical shell is a conductor:

(200 nC/m )(0.015 m ) = (8.85 × 10 C /N ⋅ m )r 2

−12

2

2

339 N ⋅ m/C r

E n (4.5 cm < r < 6.5 cm ) = 0

The Electric Field 2: Continuous Charge Distributions 123 For r > 6.5 cm, Qinside = 2πCLR and:

E n (r > 6.5 cm ) =

339 N ⋅ m/C r

Charge and Field at Conductor Surfaces *58 • Picture the Problem Because the penny is in an external electric field, it will have charges of opposite signs induced on its faces. The induced charge σ is related to the electric field by E = σ/ε0. Once we know σ, we can use the definition of surface charge density to find the total charge on one face of the penny.

σ

(a) Relate the electric field to the charge density on each face of the penny:

E=

Solve for and evaluate σ:

σ =∈0 E

∈0

(

)

= 8.85 × 10 −12 C 2 /N ⋅ m 2 (1.6 kN/C ) = 14.2 nC/m 2

Q Q = 2 A πr

(b) Use the definition of surface charge density to obtain:

σ=

Solve for and evaluate Q:

Q = σπr 2 = π (14.2 nC/m 2 )(0.01 m )

2

= 4.45 pC 59 • Picture the Problem Because the metal slab is in an external electric field, it will have charges of opposite signs induced on its faces. The induced charge σ is related to the electric field by E = σ / ∈0 .

σ

Relate the magnitude of the electric field to the charge density on the metal slab:

E=

Use its definition to express σ :

σ=

Q Q = A L2

Substitute to obtain:

E=

Q L ∈0

∈0

2

124 Chapter 22 Substitute numerical values and evaluate E:

E=

(0.12 m )

2

1.2 nC 8.85 ×10 −12 C 2 /N ⋅ m 2

(

)

= 9.42 kN/C 60 • Picture the Problem We can apply its definition to find the surface charge density of the nonconducting material and calculate the electric field at either of its surfaces from σ/2∈0. When the same charge is placed on a conducting sheet, the charge will distribute itself until half the charge is on each surface. (a) Use its definition to find σ :

σ=

(b) Relate the electric field on either side of the sheet to the density of charge on its surfaces:

E=

(c) Because the slab is a conductor the charge will distribute uniformly on its two surfaces so that:

σ=

(d) The electric field just outside the surface of a conductor is given by:

Q 6 nC = = 150 nC/m 2 2 A (0.2 m )

σ 2 ∈0

=

150 nC/m 2 2 8.85 × 10 −12 C 2 /N ⋅ m 2

(

)

= 8.47 kN/C

Q 6 nC = = 75.0 nC/m 2 2 2 A 2(0.2 m )

σ

75 nC/m 2 = E= ∈0 8.85 × 10 −12 C 2 /N ⋅ m 2 = 8.47 kN/C

61 • Picture the Problem We can construct a Gaussian surface in the shape of a sphere of radius r with the same center as the shell and apply Gauss’s law to find the electric field as a function of the distance from this point. The inner and outer surfaces of the shell will have charges induced on them by the charge q at the center of the shell. (a) Apply Gauss’s law to a spherical surface of radius r that is concentric with the point charge:

∫

S

En dA =

or

4πr 2 En =

Solve for En:

1 Qinside ∈0

En =

Qinside ∈0

Qinside 4πr 2 ∈0

(1)

The Electric Field 2: Continuous Charge Distributions 125 For r < a, Qinside = q. Substitute in equation (1) and simplify to obtain:

En (r < a ) =

q kq = 2 2 4πr ∈0 r

Qinside = 0

Because the spherical shell is a conductor, a charge –q will be induced on its inner surface. Hence, for a < r < b:

and

For r > b, Qinside = q. Substitute in equation (1) and simplify to obtain:

En (r > b ) =

E n (a < r < b ) = 0

q kq = 2 2 4πr ∈0 r

(b) The electric field lines are shown in the diagram to the right:

(c) A charge –q is induced on the inner surface. Use the definition of surface charge density to obtain:

σ inner =

A charge q is induced on the outer surface. Use the definition of surface charge density to obtain:

σ outer =

−q q = − 2 4πa 4πa 2

q 4πb 2

62 •• Picture the Problem We can construct a spherical Gaussian surface at the surface of the earth (we’ll assume the Earth is a sphere) and apply Gauss’s law to relate the electric field to its total charge. Apply Gauss’s law to a spherical surface of radius RE that is concentric with the earth:

∫

S

En dA =

or

4πRE2 En =

Solve for Qinside = Qearth to obtain:

1 Qinside ∈0

Qinside ∈0

Qearth = 4π ∈0 RE2 En =

RE2 En k

126 Chapter 22 Substitute numerical values and evaluate Qearth:

(6.37 × 10 m ) (150 N/C) = 6

Qearth

2

8.99 × 109 N ⋅ m 2 /C 2

= 6.77 × 105 C *63 •• Picture the Problem Let the inner and outer radii of the uncharged spherical conducting shell be a and b and q represent the positive point charge at the center of the shell. The positive point charge at the center will induce a negative charge on the inner surface of the shell and, because the shell is uncharged, an equal positive charge will be induced on its outer surface. To solve part (b), we can construct a Gaussian surface in the shape of a sphere of radius r with the same center as the shell and apply Gauss’s law to find the electric field as a function of the distance from this point. In part (c) we can use a similar strategy with the additional charge placed on the shell. (a) Express the charge density on the inner surface: Express the relationship between the positive point charge q and the charge induced on the inner surface qinner:

σ inner =

qinner A

q + qinner = 0

Substitute for qinner to obtain:

σ inner =

−q 4πa 2

Substitute numerical values and evaluate σinner:

σ inner =

− 2.5 µC = − 0.553 µC/m 2 2 4π (0.6 m )

Express the charge density on the outer surface:

σ outer =

qouter A

Because the spherical shell is uncharged: Substitute for qouter to obtain:

Substitute numerical values and evaluate σouter:

qouter + qinner = 0

σ outer =

− qinner 4πb 2

σ outer =

2.5 µC = 0.246 µC/m 2 2 4π (0.9 m )

The Electric Field 2: Continuous Charge Distributions 127

∫

(b) Apply Gauss’s law to a spherical surface of radius r that is concentric with the point charge:

S

En dA =

1 Qinside ∈0

or

4πr 2 En =

Solve for En:

Qinside ∈0

Qinside 4πr 2 ∈0

En =

(1)

For r < a = 0.6 m, Qinside = q. Substitute in equation (1) and evaluate En(r < 0.6 m) to obtain:

En (r < a ) = =

(

)

q kq 8.99 × 109 N ⋅ m 2 /C 2 (2.5 µC ) = = 4πr 2 ∈0 r 2 r2

(2.25 × 10

4

N ⋅ m 2 /C

) r1

2

Qinside = 0

Because the spherical shell is a conductor, a charge –q will be induced on its inner surface. Hence, for 0.6 m < r < 0.9 m:

and

E n (0.6 m < r < 0.9 m ) = 0

For r > 0.9 m, the net charge inside the Gaussian surface is q and:

En (r > 0.9 m ) =

(c) Because E = 0 in the conductor:

kq = r2

(2.25 × 10

4

N ⋅ m 2 /C

) r1

2

qinner = −2.5 µC and

σ inner = − 0.553 µC/m 2 as before. Express the relationship between the charges on the inner and outer surfaces of the spherical shell:

σouter is now given by:

qouter + qinner = 3.5 µC

and

qouter = 3.5 µC - qinner = 6.0 µC

σ outer =

6 µC = 0.589 µC/m 2 2 4π (0.9 m )

128 Chapter 22 For r < a = 0.6 m, Qinside = q and En(r < 0.6 m) is as it was in (a): Because the spherical shell is a conductor, a charge –q will be induced on its inner surface. Hence, for 0.6 m < r < 0.9 m:

En (r < a ) =

(2.25 × 10

4

N ⋅ m 2 /C

) r1

2

Qinside = 0 and

E n (0.6 m < r < 0.9 m ) = 0

For r > 0.9 m, the net charge inside the Gaussian surface is 6 µC and:

En (r > 0.9 m ) =

(

)

kq 1 = 8.99 ×10 9 N ⋅ m 2 /C 2 (6 µC ) 2 = 2 r r

(5.39 ×10

4

N ⋅ m 2 /C

)1

r2

64 •• Picture the Problem From Gauss’s law we know that the electric field at the surface of the charged sphere is given by E = kQ R 2 where Q is the charge on the sphere and R is its radius. The minimum radius for dielectric breakdown corresponds to the maximum electric field at the surface of the sphere.

kQ R2

Use Gauss’s law to express the electric field at the surface of the charged sphere:

E=

Express the relationship between E and R for dielectric breakdown:

E max =

Solve for Rmin:

Rmin =

Substitute numerical values and evaluate Rmin:

Rmin =

kQ 2 Rmin kQ E max

(8.99 × 10

= 23.2 cm

)

N ⋅ m 2 /C 2 (18 µC ) 3 × 10 6 N/C 9

The Electric Field 2: Continuous Charge Distributions 129 65 •• Picture the Problem We can use its definition to find the surface charge density just outside the face of the slab. The electric field near the surface of the slab is given by E = σ face ∈0 . We can find the electric field on each side of the slab by adding the fields due to the slab and the plane of charge. (a) Express the charge density per face in terms of the net charge on the slab:

σ face =

q 2 L2

Substitute numerical values to obtain:

σ face =

80 µC = 1.60 µC/m 2 2 2(5 m )

Express the electric field just outside one face of the slab in terms of its surface charge density:

Eslab =

Substitute numerical values and evaluate Eface:

Eslab =

σ face ∈0

1.60 µC/m 2 8.85 × 10 −12 C 2 /N ⋅ m 2

= 1.81 × 105 N/C (b) Express the total field on the side of the slab closest to the infinite charged plane:

r r r E near = E plane + E slab = Eplane rˆ − Eslab rˆ =

σ plane 2 ∈0

rˆ −

σ face ∈0

rˆ

where rˆ is a unit vector pointing away from the slab. Substitute numerical values and r evaluate E near :

r E near =

2 µC/m 2 rˆ 2 8.85 × 10 −12 C 2 /N ⋅ m 2

(

(

)

)

− 1.81 × 10 N/C rˆ =

5

(− 0.680 × 10

5

)

N/C rˆ

130 Chapter 22 Express the total field on the side of the slab away from the infinite charged plane: Substitute numerical values and r evaluate E far :

r σ σ E far = plane rˆ + face rˆ ∈0 2 ∈0 r E far =

2 µC/m 2 rˆ 2 8.85 × 10 −12 C 2 /N ⋅ m 2

(

(

)

)

+ 1.81 × 105 N/C rˆ =

(2.94 × 10

5

)

N/C rˆ

The charge density on the side of the slab near the plane is:

σ near =∈0 Enear = (8.85 × 10 −12 C 2 /N ⋅ m 2 )(0.680 × 105 N/C ) = 0.602 µC/m 2 The charge density on the far side of the slab is:

σ near =∈0 Enear = (8.85 × 10 −12 C 2 /N ⋅ m 2 )(2.94 × 105 N/C ) = 2.60 µC/m 2

General Problems 66 •• Determine the Concept We can determine the direction of the electric field between spheres I and II by imagining a test charge placed between the spheres and determining the direction of the force acting on it. We can determine the amount and sign of the charge on each sphere by realizing that the charge on a given surface induces a charge of the same magnitude but opposite sign on the next surface of larger radius. (a) The charge placed on sphere III has no bearing on the electric field between spheres I and II. The field in this region will be in the direction of the force exerted on a test charge placed between the spheres. Because the charge at the center is negative,

the field will point toward the center. (b) The charge on sphere I (−Q0) will induce a charge of the same magnitude but opposite sign on sphere II: + Q0 (c) The induction of charge +Q0 on the inner surface of sphere II will leave its outer surface with a charge of the same magnitude but opposite sign: − Q0 (d) The presence of charge −Q0 on the outer surface of sphere II will induce a charge of the same magnitude but opposite sign on the inner surface of sphere III: + Q0

The Electric Field 2: Continuous Charge Distributions 131 (e) The presence of charge +Q0 on the inner surface of sphere III will leave the outer surface of sphere III neutral: 0 (f) A graph of E as a function of r is shown to the right:

67 •• Picture the Problem Because the difference between the field just to the right of the origin E x ,right and the field just to the left of the origin E x ,left is the field due to the nonuniform surface charge, we can express E x ,left and the difference between E x ,right and

σ ∈0 . Express the electric field just to the left of the origin in terms of E x ,right

E x ,left = E x ,right −

σ ∈0

and σ ∈0 : Substitute numerical values and evaluate E x ,left :

E x ,left = 4.65 × 105 N/C −

3.10 µC/m 2 = 1.15 × 105 N/C 8.85 × 10 −12 C 2 /N ⋅ m 2

132 Chapter 22 68 •• Picture the Problem Let P denote the point of interest at (2 m, 1.5 m). The electric field at P is the sum of the electric fields due to the infinite line charge and the point charge.

r r r E = Eλ + Eq

Express the resultant electric field at P: Find the field at P due the infinite line charge:

(

)

r 2kλ 2 8.99 × 109 N ⋅ m 2 /C 2 (− 1.5 µC/m ) ˆ Eλ = rˆ = i = (− 6.74 kN/C ) iˆ 4m r

r kq E q = 2 rˆ r

Express the field at P due the point charge:

r = 1.12 m

Referring to the diagram above, determine r and rˆ :

and

rˆ = 0.893iˆ − 0.446 ˆj

r

Substitute and evaluate E q (2 m,1.5 m ) :

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (1.3 µC ) E q (2 m,1.5 m ) = 0.893iˆ − 0.446 ˆj (1.12 m )2 = (9.32 kN/C ) 0.893iˆ − 0.446 ˆj

(

= (8.32 kN/C ) iˆ − (4.16 kN/C ) ˆj Substitute to obtain:

(

)

)

The Electric Field 2: Continuous Charge Distributions 133

r E (2 m,1.5 m ) = (− 6.74 kN/C ) iˆ + (8.35 kN/C ) iˆ − (4.17 kN/C ) ˆj = (1.61 kN/C ) iˆ − (4.17 kN/C ) ˆj *69 •• Picture the Problem If the patch is small enough, the field at the center of the patch comes from two contributions. We can view the field in the hole as the sum of the field from a uniform spherical shell of charge Q plus the field due to a small patch with surface charge density equal but opposite to that of the patch cut out. (a) Express the magnitude of the electric field at the center of the hole:

E = Espherical shell + Ehole

Apply Gauss’s law to a spherical gaussian surface just outside the given sphere:

Espherical shell 4πr 2 =

Solve for Espherical shell to obtain:

Espherical shell =

The electric field due to the small hole (small enough so that we can treat it as a plane surface) is:

Ehole =

Substitute and simplify to obtain:

E=

(

= = (b) Express the force on the patch:

)

Qenclosed Q = ∈0 ∈0

Q 4π ∈0 r 2

−σ 2 ∈0

−σ Q + 2 4π ∈0 r 2 ∈0

Q Q − 2 4π ∈0 r 2 ∈0 (4π r 2 ) Q 8π ∈0 r 2

F = qE where q is the charge on the patch.

Assuming that the patch has radius a, express the proportion between its charge and that of the spherical shell:

q Q a2 = or q = Q π a 2 4π r 2 4r 2

Substitute for q and E in the expression for F to obtain:

⎞ ⎛ a2 ⎞ ⎛ Q Q 2a 2 ⎟ F = ⎜⎜ 2 Q ⎟⎟ ⎜⎜ = 2 ⎟ 32π ∈0 r 4 ⎝ 4r ⎠ ⎝ 8π ∈0 r ⎠

(c) The pressure is the force divided by the area of the patch:

P=

Q2a 2 32π ∈0 r 4

πa 2

=

Q2 32π 2 ∈0 r 4

134 Chapter 22 70 •• Picture the Problem The work done by the electrostatic force in expanding the soap

∫

bubble is W = PdV . From Problem 69:

P=

Q2 32π 2 ∈0 r 4

W = ∫ PdV = ∫ P 4πr 2 dr

Express W in terms of dr: Substitute for P and simplify:

W=

Q2 8π ∈0

0.2 m

dr r2 0.1 m

∫

Evaluating the integral yields:

Q2 W= 8π ∈0

2 ⎛ −1 ( 3 nC ) 1 ⎞ ⎡ 1⎤ ⎜ ⎟ = + 2 2 ⎜ −12 ⎢⎣− r ⎥⎦ 8π (8.85 × 10 C / N ⋅ m ) ⎝ 0.2 m 0.1 m ⎟⎠ 0.1 m 0.2 m

= 2.02 × 10 −7 J 71 •• Picture the Problem We can use E = kq/R2, where R is the radius of the droplet, to find the electric field at its surface. We can find R by equating the volume of the bubble at the moment it bursts to the volume of the resulting spherical droplet. Express the field at the surface of the spherical water droplet:

E=

kq R2

(1)

where R is the radius of the droplet and q is its charge. Express the volume of the bubble just before it pops:

V ≈ 4πr 2 t where t is the thickness of the soap bubble.

Express the volume of the sphere into which the droplet collapses:

4 V = πR 3 3

Because the volume of the droplet and the volume of the bubble are equal:

4 4π r 2t = πR 3 3

Solve for R:

R = 3 3r 2t

Assume a thickness t of 1 µm and evaluate R:

R = 3 3(0.2 m ) (1 µm ) = 4.93 × 10 −3 m 2

The Electric Field 2: Continuous Charge Distributions 135 Substitute numerical values in equation (1) and evaluate E:

E=

(8.99 × 10 N ⋅ m (4.93 × 10 9

2

)

/ C 2 (3 nC )

−3

m

)

2

= 1.11 × 10 6 N/C 72 •• Picture the Problem Let the numeral 1 refer to the infinite plane at x = −2 m and the numeral 2 to the plane at x = 2 m and let the letter A refer to the region to the left of plane 1, B to the region between the planes, and C to the region to the right of plane 2. We can use the expression for the electric field of in infinite plane of charge to express the electric field due to each plane of charge in each of the three regions. Their sum will be the resultant electric field in each region. Express the resultant electric field as the sum of the fields due to planes 1 and 2: (a) Express and evaluate the field due to plane 1 in region A:

r r r E = E1 + E 2

(1)

( )

r σ E1 ( A) = 1 − iˆ 2 ∈0

( )

− 3.5 µC/m 2 − iˆ 2(8.85 × 10 −12 C 2 /N ⋅ m 2 ) = (198 kN/C) iˆ =

Express and evaluate the field due to plane 2 in region A:

( )

r σ E 2 ( A) = 2 − iˆ 2 ∈0

)( )

6 µC/m 2 − iˆ 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (− 339 kN/C ) iˆ =

Substitute in equation (1) to obtain:

(

r E ( A) = (198 kN/C )iˆ + (− 339 kN/C )iˆ =

(− 141 kN/C) iˆ

136 Chapter 22 (b) Express and evaluate the field due to plane 1 in region B:

r σ E1 (B ) = 1 iˆ 2 ∈0 − 3.5 µC/m 2 iˆ 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (− 198 kN/C ) iˆ

=

Express and evaluate the field due to plane 2 in region B:

(

)

( )

r σ E 2 (B ) = 2 − iˆ 2 ∈0

)( )

6 µC/m 2 − iˆ 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (− 339 kN/C ) iˆ =

Substitute in equation (1) to obtain:

r E (B ) = (− 198 kN/C )iˆ + (− 339 kN/C )iˆ =

(c) Express and evaluate the field due to plane 1 in region C:

(

(− 537 kN/C)iˆ

r σ E1 (C ) = 1 iˆ 2 ∈0 − 3.5 µC/m 2 iˆ 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (− 198 kN/C ) iˆ =

Express and evaluate the field due to plane 2 in region C:

(

)

r σ E 2 (C ) = 2 iˆ 2 ∈0 6 µC/m 2 iˆ 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (339 kN/C ) iˆ =

Substitute in equation (1) to obtain:

(

)

r E (C ) = (− 198 kN/C ) iˆ + (339 kN/C ) iˆ = (141 kN/C ) iˆ

*73 •• Picture the Problem We can find the electric fields at the three points of interest by adding the electric fields due to the infinitely long cylindrical shell and the spherical shell. In Problem 42 it was established that, for an infinitely long cylindrical shell of radius R, Er (r < R ) = 0, and Er (r > R ) = σR ∈0 r . We know that, for a spherical shell of radius R, Er (r < R ) = 0, and E r (r > R ) = σR 2 ∈0 r 2 .

The Electric Field 2: Continuous Charge Distributions 137

Express the resultant electric field as the sum of the fields due to the cylinder and sphere: (a) Express and evaluate the electric field due to the cylindrical shell at the origin:

r r r E = E cyl + E sph

(1)

r E cyl (0,0) = 0 because the origin is inside the cylindrical shell.

Express and evaluate the electric field due to the spherical shell at the origin:

( ) ( (

)

2 r σR 2 − 12 µC/m 2 (0.25 m ) ˆ = Esph (0,0) = i − iˆ = (339 kN/C ) iˆ − 2 2 2 −12 ∈0 r 2 8.85 × 10 C /N ⋅ m (0.5 m )

Substitute in equation (1) to obtain:

)

( )

r E (0,0 ) = 0 + (339 kN/C ) iˆ =

(339 kN/C) iˆ

or

E (0,0 ) = 339 kN/C

and

θ = 0° (b) Express and evaluate the electric field due to the cylindrical shell at (0.2 m, 0.1 m):

138 Chapter 22

r (6 µC/m 2 )(0.15 m) σR ˆ ˆ ˆ E cyl (0.2 m,0.1 m ) = i= ∈0 r (8.85 × 10−12 C2 /N ⋅ m 2 )(0.2 m ) i = (508 kN/C) i r σR 2 Esph (r ) = rˆ ∈0 r 2 where rˆ is a unit vector pointing from (50

Express the electric field due to the charge on the spherical shell as a function of the distance from its center:

cm, 0) to (20 cm, 10 cm).

r = 0.316 m

Referring to the diagram shown above, find r and rˆ :

and

r r = −0.949iˆ + 0.316 ˆj

Substitute to obtain:

r Esph (0.2 m,0.1 m ) =

(− 12 µC/m )(0.25 m ) ( − 0.949iˆ + 0.316 ˆj ) (8.85 × 10 C /N ⋅ m )(0.316 m) 2

2

−12

2

2

2

(

= (− 849 kN/C ) − 0.949iˆ + 0.316 ˆj = (806 kN/C )iˆ + (− 268 kN/C ) ˆj

)

Substitute in equation (1) to obtain:

r E (0.2 m,0.1 m ) = (508 kN/C ) iˆ + (806 kN/C ) iˆ + (− 268 kN/C ) ˆj = (1310 kN/C ) iˆ + (− 268 kN/C ) ˆj or

E (0.2 m,0.1 m ) =

(1310 kN/C)2 + (− 268 kN/C)2

= 1340 kN/C

and

⎛ − 268 kN/C ⎞ ⎟⎟ = 348° ⎝ 1310 kN/C ⎠

θ = tan −1 ⎜⎜

(c) Express and evaluate the electric field due to the cylindrical shell at (0.5 m, 0.2 m):

r E cyl (0.5 m,0.2 m ) =

(6 µC/m )(0.15 m) ˆ ˆ (8.85 × 10 C /N ⋅ m )(0.5 m ) i = (203 kN/C)i

Express and evaluate the electric field due to the spherical shell at

2

−12

2

2

r E sph (0.5 m,0.2 m ) = 0

The Electric Field 2: Continuous Charge Distributions 139 (0.5 m, 0.5 m):

Substitute in equation (1) to obtain:

because (0.5 m, 0.2 m) is inside the spherical shell.

r E (0.5 m,0.2 m ) = (203 kN/C ) iˆ + 0 =

(203 kN/C) iˆ

or

E (0.5 m,0.2 m ) = 203 kN/C

and

θ = 0° 74 •• Picture the Problem Let the numeral 1 refer to the plane with charge density σ 1 and the numeral 2 to the plane with charge density σ 2. We can find the electric field at the two points of interest by adding the electric fields due to the charge distributions of the two infinite planes. Express the electric field at any point in space due to the charge distributions on the two planes:

r r r E = E1 + E 2

(1)

(a) Express the electric field at (6 m, 2 m) due to plane 1:

r 65 nC/m 2 σ1 ˆ ˆj = (3.67 kN/C ) ˆj E1 (6 m,2 m ) = j= 2 ∈0 2(8.85 × 10 −12 C 2 /N ⋅ m 2 ) Express the electric field at (6 m, 2 m) due to plane 2:

r σ 45 nC/m 2 E 2 (6 m,2 m ) = 2 rˆ = rˆ = (2.54 kN/C )rˆ 2 ∈0 2(8.85 × 10 −12 C 2 /N ⋅ m 2 ) where rˆ is a unit vector pointing from plane 2 toward the point whose coordinates are (6 m, 2 m). Refer to the diagram below to obtain:

rˆ = sin 30°iˆ − cos 30° ˆj

140 Chapter 22 Substitute to obtain:

(

)

r E 2 (6 m,2 m ) = (2.54 kN/C ) sin 30°iˆ − cos 30° ˆj = (1.27 kN/C ) iˆ + (− 2.20 kN/C ) ˆj Substitute in equation (1) to obtain:

r E (6 m,2 m ) = (3.67 kN/C ) ˆj + (1.27 kN/C ) iˆ + (− 2.20 kN/C ) ˆj = (1.27 kN/C ) iˆ + (1.47 kN/C ) ˆj

r

r

(b) Note that E1 (6 m,5 m ) = E1 (6 m,2 m ) so that:

r 65 nC/m 2 σ ˆ ˆj = (3.67 kN/C ) ˆj E1 (6 m, 5 m ) = j= 2 ∈0 2(8.85 × 10 −12 C 2 / N ⋅ m 2 ) r

r

Note also that E 2 (6 m,5 m ) = − E 2 (6 m,2 m ) so that:

r E 2 (6 m,5 m ) = (− 1.27 kN/C ) iˆ + (2.20 kN/C ) ˆj Substitute in equation (1) to obtain:

r E (6 m,5 m ) = (3.67 kN/C) ˆj + (− 1.27 kN/C ) iˆ + (2.20 kN/C) ˆj =

(− 1.27 kN/C)iˆ + (5.87 kN/C) ˆj

75 •• Picture the Problem Because the atom is uncharged, we know that the integral of the electron’s charge distribution over all of space must equal its charge e. Evaluation of this integral will lead to an expression for ρ0. In (b) we can express the resultant field at any point as the sum of the fields due to the proton and the electron cloud. (a) Because the atom is uncharged:

Substitute for ρ(r):

∞

∞

0

0

e = ∫ ρ (r )dV = ∫ ρ (r )4π r 2 dr ∞

e = ∫ ρ0e 0

Use integral tables or integration by parts to obtain:

∞

−2r a

4π r dr = 4πρ 0 ∫ r 2 e − 2 r a dr

2 −2r a ∫ r e dr = 0

∞

2

0

a3 4

The Electric Field 2: Continuous Charge Distributions 141 ⎛ a3 ⎞ e = 4πρ 0 ⎜⎜ ⎟⎟ = πa 3 ρ 0 ⎝ 4 ⎠

Substitute to obtain:

Solve for ρ0:

ρ0 =

e π a3

(b) The field will be the sum of the field due to the proton and that of the electron charge cloud:

E = Ep + Ecloud =

Express the field due to the electron cloud:

Ecloud (r ) =

kq + Ecloud r2

kQ(r ) r2

where Q(r) is the net negative charge enclosed a distance r from the proton. Substitute to obtain:

E (r ) =

As in (a), Q(r) is given by:

ke kQ(r ) + 2 r2 r r

Q(r ) = ∫ 4πr ' ρ (r ' )dr ' 0

Integrate to find Q(r) and substitute in the expression for E to obtain:

ke − 2 r / a ⎛ 2r 2r 2 ⎞ ⎜⎜1 + E (r ) = 2 e + 2 ⎟⎟ a r a ⎠ ⎝

*76 •• Picture the Problem We will assume that the radius at which they balance is large enough that only the third term in the expression matters. Apply a condition for equilibrium will yield an equation that we can solve for the distance r. Apply

∑ F = 0 to the proton:

2ke 2 − 2 r / a e − mg = 0 a2

To solve for r, isolate the exponential factor and take the natural logarithm of both sides of the equation:

r=

a ⎛ 2ke 2 ln⎜ 2 ⎜⎝ mga 2

⎞ ⎟⎟ ⎠

Substitute numerical values and evaluate r:

(

)(

)

2 0.0529 nm ⎡ 2 8.99 × 109 N ⋅ m 2 / C 2 1.60 × 10 −19 C ⎤ r= = 1.16 nm ln ⎢ 2⎥ 2 − 27 2 ⎣⎢ 1.67 × 10 kg 9.81 m/s (0.0529 nm ) ⎦⎥

(

)(

)

142 Chapter 22 Thus, even though the unscreened electrostatic force is 40 orders of magnitude larger than the gravitational force, screening reduces it to smaller than the gravitational force within a few nanometers. Remarks: Note that the argument of the logarithm contains the ratio between the gravitational potential energy of a mass held a distance a0 above the surface of the earth and the electrostatic potential energy for two unscreened charges a distance a0 apart. 77 •• Picture the Problem In parts (a) and (b) we can express the charges on each of the elements as the product of the linear charge density of the ring and the length of the segments. Because the lengths of the segments are the product of the angle subtended at P and their distances from P, we can express the charges in terms of their distances from P. By expressing the ratio of the fields due to the charges on s1 and s2 we can determine their dependence on r1 and r2 and, hence, the resultant field at P. We can proceed similarly in part (c) with E varying as 1/r rather than 1/r2. In part (d), with s1 and s2 representing areas, we’ll use the definition of the solid angle subtended by these areas to relate their charges to their distances from point P. (a) Express the charge q1 on the element of length s1:

Express the charge q2 on the element of length s2:

q1 = λs1 = λθr1 where θ is the angle subtended by the arcs of length s1 and s2.

q2 = λs2 = λθr2

Divide the first of these equations by the second to obtain:

q1 λθr1 r = = 1 q2 λθr2 r2

Express the electric field at P due to the charge associated with the element of length s1:

E1 =

kq1 kλs1 kλθr1 kλθ = 2 = 2 = r12 r1 r1 r1

Express the electric field at P due to the charge associated with the element of length s2:

E2 =

kλθ r2

Divide the first of these equations by the second to obtain:

kλθ E1 r r = 1 = 2 E2 kλθ r1 r2

The Electric Field 2: Continuous Charge Distributions 143 and, because r2 > r1,

E1 > E 2 (b) The two fields point away from their segments of arc.

(c) If E varies as 1/r:

Because E1 > E2 , the resultant field points toward s2 . E1 = and

E2 =

Therefore:

(d) Use the definition of the solid angle Ω subtended by the area s1 to obtain:

kq1 kλs1 kλθr1 = = = kλθ r1 r1 r1 kq2 kλs2 kλθr2 = = = kλθ r2 r2 r2

E1 = E 2

s Ω = 12 4π 4πr1 or

s1 = Ωr12

Express the charge q1 of the area s1:

q1 = σs1 = σΩr12

Similarly, for an element of area s2:

s 2 = Ωr22 and

q 2 = σΩr22 Express the ratio of q1 to q2 to obtain:

q1 σΩr12 r12 = = q 2 σΩr22 r22

Proceed as in (a) to obtain:

kq1 E1 r2 r 2q r 2σΩr12 = 1 = 22 1 = 22 =1 E2 kq2 r1 q2 r1 σΩr22 r22

144 Chapter 22

r E =0

Because the two fields are of equal magnitude and oppositely directed:

r If E ∝ 1/r , then s2 would produce the stronger field at P and E would point toward s1. 78 •• Picture the Problem We can apply the condition for translational equilibrium to the particle and use the expression for the electric field on the axis of a ring charge to obtain an expression for |q|/m. Doing so will lead us to the conclusion that |q|/m will be a minimum when Ez is a maximum and so we’ll use the result from Problem 26 that

z = −R

2 maximizes Ez.

(a) Apply

∑F

z

= 0 to the particle:

q E z − mg = 0

q

Solve for |q|/m:

m

g Ez

=

(1)

Note that this result tells us that the minimum value of |q|/m will be where the field due to the ring is greatest. Express the electric field along the z axis due to the ring of charge:

Ez =

(z

kQz 2

+ R2

)

32

Differentiate this expression with respect to z to obtain:

dE x d ⎡ x = kQ ⎢ dz dz ⎢⎣ z 2 + R 2

(z = kQ

(

2

+ R2

)

3 2

(

)

z 2 + R2 ⎤ = kQ 3 2⎥ ⎥⎦

(

− z ( 32 ) z 2 + R 2

(z

2

+ R2

)

3

)

(

−z

) (2 z ) = kQ (z 12

( )

d 2 z + R2 dx 3 z2 + R2

3 2

2

+ R2

)

)

3 2

( (z + R )

3 2

− 3z 2 z 2 + R 2

2

2 3

)

12

The Electric Field 2: Continuous Charge Distributions 145 Set this expression equal to zero for extrema and simplify:

(z

2

+ R2

)

( (z + R ) ) − 3z (z 3 2

− 3z 2 z 2 + R 2

2

+ R2

= 0,

)

= 0,

12

2 3

2

(z

)

32

2

2

+ R2

12

and

z 2 + R 2 − 3z 2 = 0 Solve for x to obtain:

z=±

R 2

as candidates for maxima or minima. You can either plot a graph of Ez or evaluate its second derivative at these points to show that it is a maximum at:

z=−

R 2

Substitute to obtain an expression Ez,max:

E z ,max =

Substitute in equation (1) to obtain:

q m

(b) If |q|/m is twice as great as in (a), then the electric field should be half its value in (a), i.e.:

=

⎛ ⎛ R ⎞2 ⎞ ⎜⎜ − ⎟ + R2 ⎟ ⎜⎝ ⎟ 2⎠ ⎝ ⎠

32

=

2kQ 27 R 2

27 gR 2 2kQ

kQ kQz = 2 27 R z2 + R2

(

)

32

or

1 = 27 R 4

Let a = z2/R2 and simplify to obtain:

⎛ R ⎞ kQ⎜ − ⎟ 2⎠ ⎝

z2 ⎛ z2 ⎞ ⎜ R ⎜1 + 2 ⎟⎟ ⎝ R ⎠

3

6

a 3 + 3a 2 − 24a + 1 = 0

The graph of f (a ) = a 3 + 3a 2 − 24a + 1 shown below was plotted using a spreadsheet program.

146 Chapter 22 20 15 10 5 f (a )

0 -5 -10 -15 -20 -25 -30 0.0

1.0

2.0

3.0

4.0

a

Use your calculator or trial-and-error methods to obtain:

a = 0.04188 and a = 3.596

The corresponding z values are:

z = −0.205 R and z = −1.90 R

The condition for a stable equilibrium position is that the particle, when displaced from its equilibrium position, experiences a restoring force, i.e. a force that acts toward the equilibrium position. When the particle in this problem is just above its equilibrium position the net force on it must be downward and when it is just below the equilibrium position the net force on it must be upward. Note that the electric force is zero at the origin, so the net force there is downward and remains downward to the first equilibrium position as the weight force exceeds the electric force in this interval. The net force is upward between the first and second equilibrium positions as the electric force exceeds the weight force. The net force is downward below the second equilibrium position as the weight force exceeds the electric force. Thus, the first (higher) equilibrium position is stable and the second (lower) equilibrium position is unstable. You might also find it instructive to use your graphing calculator to plot a graph of the electric force (the gravitational force is constant and only shifts the graph of the total force downward). Doing so will produce a graph similar to the one shown in the sketch to the right. Note that the slope of the graph is negative on both sides of −0.205R whereas it is positive on both sides of −1.90R; further evidence that −0.205R is a position of stable equilibrium and −1.90R a position of unstable equilibrium.

The Electric Field 2: Continuous Charge Distributions 147 79 •• Picture the Problem The loop with the small gap is equivalent to a closed loop and a charge of − Ql 2πR at the gap. The field at the center of a closed loop of uniform line charge is zero. Thus the field is entirely due to the charge − Ql 2πR . (a) Express the field at the center of the loop: Relate the field at the center of the loop to the charge in the gap: Use the definition of linear charge density to relate the charge in the gap to the length of the gap:

r r r E center = E loop + E gap

(1)

r kq E gap = − 2 rˆ R

λ=

q Q = l 2πR

or

q=

Ql 2πR

Substitute to obtain:

r kQl E gap = − rˆ 2πR 3

Substitute in equation (1) to obtain:

r kQl kQl E center = 0 − rˆ = − rˆ 3 2πR 2πR 3

If Q is positive, the field at the origin points radially outward. (b) From our result in (a) we see r that the magnitude of E center is:

Ecenter =

kQl 2πR 3

80 •• Picture the Problem We can find the electric fields at the three points of interest, labeled 1, 2, and 3 in the diagram, by adding the electric fields due to the charge distributions on the nonconducting sphere and the spherical shell.

148 Chapter 22

Express the electric field due to the nonconducting sphere and the spherical shell at any point in space: (a) Because (4.5 m, 0) is inside the spherical shell: Apply Gauss’s law to a spherical surface inside the nonconducting sphere to obtain:

r r r E = E sphere + E shell

(1)

r E shell (4.5 m,0 ) = 0

r 4π E sphere (r ) = kρriˆ 3

r

Evaluate E sphere (0.5 m ) :

r 4π E sphere (0.5 m ) = 8.988 × 10 9 N ⋅ m 2 /C 2 5 µC/m 2 (0.5 m ) iˆ = (94.1kN/C ) iˆ 3

(

Substitute in equation (1) to obtain:

)(

)

r E (4.5 m,0) = (94.1 kN/C ) iˆ + 0

= Find the magnitude and direction of

r E (4.5 m,0) :

(94.1 kN/C) iˆ

E (4.5 m,0 ) = 94.1 kN/C and

θ = 0° (b) Because (4 m, 1.1m) is inside the spherical shell:

r E shell (4 m,1.1m ) = 0

The Electric Field 2: Continuous Charge Distributions 149 r

Evaluate E sphere (1.1 m ) :

(

)(

)

3 r 4π 8.99 × 10 9 N ⋅ m 2 /C 2 5 µC/m 2 (0.6 m ) ˆ E sphere (1.1 m ) = j = (33.6 kN/C ) ˆj 2 3(1.1 m )

Substitute in equation (1) to obtain:

r E (4.5 m,0) = (33.6 kN/C ) ˆj + 0

= (33.6 kN/C ) ˆj Find the magnitude and direction of

r E (4.5 m,1.1m ) :

E (4.5 m,1.1 m ) = 33.6 kN/C and

θ = 90° r kQ E shell (r ) = shell rˆ r2 where rˆ is a unit vector pointing from

(c) Because (2 m, 3 m) outside the spherical shell:

(4 m, 0) to (2 m, 3 m).

(

)

Qshell = σAshell = 4π − 1.5 µC/m 2 (1.2 m )

Evaluate Qshell:

= −27.1 µC

Refer to the diagram below to find rˆ and r:

r = 3.61 m and

rˆ = −0.555iˆ + 0.832 ˆj

r

Substitute and evaluate E shell (2 m,3 m ) :

(

)

r 8.99 × 109 N ⋅ m 2 /C 2 (− 27.1 µC ) Eshell (3.61 m ) = rˆ (3.61 m )2 = (− 18.7 kN/C ) − 0.555iˆ + 0.832 ˆj

(

= (10.4 kN/C ) iˆ + (− 15.6 kN/C ) ˆj

)

2

150 Chapter 22 Express the electric field due to the charged nonconducting sphere at a distance r from its center that is greater than its radius:

r kQsphere E sphere (r ) = rˆ r2

Find the charge on the sphere:

Qsphere = ρVsphere =

4π 3 5 µC/m 2 (0.6 m ) 3

(

)

= 4.52 µC

r

Evaluate E sphere (3.61 m ) :

(

)

r 8.99 × 10 9 N ⋅ m 2 /C 2 (4.52 µC ) Esphere (2 m,3 m ) = rˆ = (3.12 kN/C )rˆ (3.61 m )2 = (3.12 kN/C ) − 0.555iˆ + 0.832 ˆj

(

)

= (− 1.73 kN/C ) iˆ + (2.59 kN/C ) ˆj Substitute in equation (1) to obtain:

r E (2 m,3 m ) = (10.4 kN/C ) iˆ + (− 15.6 kN/C ) ˆj + (− 1.73 kN/C ) iˆ + (2.59 kN/C ) ˆj

= (8.67 kN/C ) iˆ + (− 13.0 kN/C ) ˆj r

Find the magnitude and direction of E (2 m,3 m ) :

E (2 m,3 m ) =

(8.67 kN/C)2 + (− 13.0 kN/C)2

and

⎛ − 13.0 kN/C ⎞ ⎟⎟ = 304° ⎝ 8.67 kN/C ⎠

θ = tan −1 ⎜⎜

81 •• Picture the Problem Let the numeral 1 refer to the infinite plane whose charge density is σ 1 and the numeral 2 to the infinite plane whose charge density is σ 2. We can find the electric fields at the two points of interest by adding the electric fields due to the charge distributions on the infinite planes and the sphere.

= 15.6 kN/C

The Electric Field 2: Continuous Charge Distributions 151 r r r r Express the electric field due to the E = E sphere + E1 + E 2 (1) infinite planes and the sphere at any point in space: (a) Because (0.4 m, 0) is inside the sphere: Find the field at (0.4 m, 0) due to plane 1:

r E sphere (0.4 m,0) = 0 r σ E1 (0.4 m,0) = 1 ˆj 2 ∈0

3 µC/m 2 ˆj 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (169 kN/C ) ˆj =

(

)

Find the field at (0.4 m, 0) due to plane 2:

( ) (

)( )

r − 2 µC/m 2 σ2 ˆ E 2 (0.4 m,0) = −i = − iˆ = (113 kN/C ) iˆ 2 ∈0 2 8.85 × 10 −12 C 2 /N ⋅ m 2 Substitute in equation (1) to obtain:

r E (0.4 m,0 ) = 0 + (169 kN/C ) ˆj + (113 kN/C ) iˆ = (113 kN/C ) iˆ + (169 kN/C ) ˆj

Find the magnitude and direction of

r E (0.4 m,0) :

E (0.4 m,0) =

(113 kN/C)2 + (169 kN/C)2

= 203 kN/C and

⎛ 169 kN/C ⎞ ⎟⎟ = 56.2° ⎝ 113 kN/C ⎠

θ = tan −1 ⎜⎜ (b) Because (2.5 m, 0) is outside the sphere:

r kQ Esphere (0.4 m,0 ) = sphere rˆ r2 where rˆ is a unit vector pointing from (1 m, −0.6 m) to (2.5 m, 0).

152 Chapter 22 Qsphere = σAsphere = 4πσR 2

Evaluate Qsphere:

= 4π (− 3 µC/m 2 )(1 m )

2

= −37.7 µC

r = 1.62 m

Referring to the diagram above, determine r and rˆ :

and

rˆ = 0.928iˆ + 0.371 ˆj r

Substitute and evaluate Esphere (2.5 m,0 ) :

(

)

r 8.99 × 10 9 N ⋅ m 2 /C 2 (− 37.7 µC ) Esphere (2.5 m,0 ) = rˆ (1.62 m )2 = (− 129 kN/C ) 0.928iˆ + 0.371 ˆj

(

)

= (− 120 kN/C ) iˆ + (− 47.9 kN/C ) ˆj

r σ E1 (2.5 m,0) = 1 ˆj 2 ∈0

Find the field at (2.5 m, 0) due to plane 1:

3 µC/m 2 ˆj 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (169 kN/C ) ˆj =

(

)

r σ E 2 (2.5 m,0) = 2 iˆ 2 ∈0

Find the field at (2.5 m, 0) due to plane 2:

− 2 µC/m 2 iˆ = 2(8.85 × 10 −12 C 2 /N ⋅ m 2 ) = (− 113 kN/C )iˆ

Substitute in equation (1) to obtain:

r E (0.4 m,0) = (− 120 kN/C ) iˆ + (− 47.9 kN/C ) ˆj + (169 kN/C ) ˆj + (− 113 kN/C ) iˆ = (− 233 kN/C ) iˆ + (121 kN/C ) ˆj r

Find the magnitude and direction of E (2.5 m,0 ) :

The Electric Field 2: Continuous Charge Distributions 153 E (2.5 m,0 ) =

(− 233 kN/C )2 + (121 kN/C)2

= 263 kN/C

and

⎛ 121 kN/C ⎞ ⎟⎟ = 153° − 233 kN/C ⎝ ⎠

θ = tan −1 ⎜⎜

82 •• Picture the Problem Let P represent the point of interest at (1.5 m, 0.5 m). We can find the electric field at P by adding the electric fields due to the infinite plane, the infinite line, and the sphere. Once we’ve expressed the field at P in vector form, we can find its magnitude and direction.

r r r r E = E plane + E line + E sphere

Express the electric field at P:

r

r σ ˆ E plane = − i 2 ∈0

Find E plane at P:

2 µC/m 2 iˆ =− 2 8.85 × 10 −12 C 2 /N ⋅ m 2 = (− 113 kN/C ) iˆ

(

r

)

r 2kλ E line = rˆ r

Express E line at P:

Refer to the diagram to obtain:

r r = (0.5 m ) iˆ − (0.5 m ) ˆj

and

rˆ = (0.707 ) iˆ − (0.707 ) ˆj

Substitute to obtain:

(

)

r 2 8.99 × 10 9 N ⋅ m 2 /C 2 (4 µC/m ) (0.707 ) iˆ − (0.707 ) ˆj E line = 0.707 m = (102 kN/C ) (0.707 ) iˆ − (0.707 ) ˆj = (72.1 kN/C ) iˆ + (− 72.1 kN/C ) ˆj

[

Letting r′ represent the distance from the center of the sphere to P,

[

]

r 4π E sphere = kr'ρrˆ' 3

]

154 Chapter 22 where rˆ' is directed toward the center of the sphere.

apply Gauss’s law to a spherical surface of radius r′ centered at (1 m, 0) to obtain an expression r for E sphere at P:

r r ' = −(0.5 m ) iˆ − (0.5 m ) ˆj

Refer to the diagram used above to obtain:

and

rˆ' = −(0.707 ) iˆ − (0.707 ) ˆj

Substitute to obtain:

)[

r 4π Esphere = 8.99 × 109 N ⋅ m 2 /C 2 (0.707 m ) − 6 µC/m 3 (0.707 ) iˆ + (0.707 ) ˆj 3 = (− 113 kN/C) iˆ + ˆj = (− 113 kN/C ) iˆ + (− 113 kN/C) ˆj

(

)

(

(

)

r

Substitute and evaluate E :

r E = (− 113 kN/C ) iˆ + (72.1 kN/C ) iˆ + (− 72.1 kN/C ) ˆj + (− 113 kN/C )iˆ + (− 113 kN/C ) ˆj

= (− 154 kN/C ) iˆ + (− 185 kN/C ) ˆj Finally, find the magnitude and direction r of E :

E=

(− 154 kN/C)2 + (− 185 kN/C)2

= 241 kN/C and

⎛ − 154 kN/C ⎞ ⎟⎟ = 220° ⎝ − 185 kN/C ⎠

θ = tan −1 ⎜⎜ 83 •• Picture the Problem We can find the period of the motion from its angular frequency and apply Newton’s 2nd law to relate ω to m, q, R, and the electric field due to the infinite line charge. Because the electric field is given by E r = 2kλ r we can express ω and, hence, T as a function of m, q, R, and λ. Relate the period T of the particle to its angular frequency ω:

T=

2π

ω

(1)

]

The Electric Field 2: Continuous Charge Distributions 155 Apply Newton’s 2nd law to the particle to obtain: Solve for ω:

Express the electric field at a distance R from the infinite line charge: Substitute in the expression for ω:

Substitute in equation (1) to obtain:

∑F

= qEr = mRω 2

radial

ω=

qE r mR

E r = 2k

ω=

λ R

2kλq 1 = mR 2 R

T = 2πR

2kλq m

m 2kλq

*84 •• Picture the Problem Starting with the equation for the electric field on the axis of ring charge, we can factor the denominator of the expression to show that, for x << R, Ex is proportional to x. We can use Fx = qEx to express the force acting on the particle and apply Newton’s 2nd law to show that, for small displacements from equilibrium, the particle will execute simple harmonic motion. Finally, we can find the period of the motion from its angular frequency, which we can obtain from the differential equation of motion. (a) Express the electric field on the axis of the ring of charge:

Ex =

Factor R2 from the denominator of Ex to obtain:

Ex =

=

(x

kQx 2

+ R2

)

32

kQx

⎡ 2⎛ x 2 ⎞⎤ ⎜ ⎟ + R 1 ⎢ ⎜ 2 ⎟⎥ ⎣ ⎝ R ⎠⎦ kQx ⎛ x2 ⎞ R 3 ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

32

provided x << R. (b) Express the force acting on the particle as a function of its charge and the electric field:

Fx = qE x =

kqQ x R3

32

≈

kQ x R3

156 Chapter 22 (c) Because the negatively charged particle experiences a linear restoring force, we know that its motion will be simple harmonic. Apply Newton’s 2nd law to the negatively charged particle to obtain:

m

d 2x kqQ =− 3 x 2 dt R

or

d 2 x kqQ + x=0 dt 2 mR 3 the differential equation of simple harmonic motion.

2π

Relate the period T of the simple harmonic motion to its angular frequency ω:

T=

From the differential equation we have:

ω2 =

Substitute to obtain:

ω

kqQ mR 3

T = 2π

mR 3 kqQ

85 •• Picture the Problem Starting with the equation for the electric field on the axis of a ring charge, we can factor the denominator of the expression to show that, for x << R, Ex is proportional to x. We can use Fx = qEx to express the force acting on the particle and apply Newton’s 2nd law to show that, for small displacements from equilibrium, the particle will execute simple harmonic motion. Finally, we can find the angular frequency of the motion from the differential equation and use this expression to find its value when the radius of the ring is doubled and all other parameters remain unchanged. Express the electric field on the axis of the ring of charge:

Ex =

Factor R2 from the denominator of Ex to obtain:

Ex =

=

(x

kQx 2

+ R2

)

32

kQx ⎡ 2⎛ x 2 ⎞⎤ ⎟ ⎜ 1 R + ⎢ ⎜ 2 ⎟⎥ ⎣ ⎝ R ⎠⎦ kQx ⎛ x2 ⎞ R 3 ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

provided x << R.

32

32

≈

kQ x R3

The Electric Field 2: Continuous Charge Distributions 157 kqQ x R3

Express the force acting on the particle as a function of its charge and the electric field:

Fx = qE x =

Because the negatively charged particle experiences a linear restoring force, we know that its motion will be simple harmonic. Apply Newton’s 2nd law to the negatively charged particle to obtain:

d 2x kqQ m 2 =− 3 x dt R or

d 2 x kqQ + x=0 dt 2 mR 3 the differential equation of simple harmonic motion.

The angular frequency of the simple harmonic motion of the particle is given by:

ω=

kqQ mR 3

(1)

Express the angular frequency of the motion if the radius of the ring is doubled:

ω'=

kqQ 3 m(2 R )

(2)

Divide equation (2) by equation (1) to obtain:

Solve for and evaluate ω′:

kqQ 3 m(2 R )

ω' = ω

ω' =

=

kqQ mR 3

ω

=

8

1 8

21 rad/s = 7.42 rad/s 8

86 •• Picture the Problem Starting with the equation for the electric field on the axis of a ring charge, we can factor the denominator of the expression to show that, for x << R, Ex is proportional to x. We can use Fx = qEx to express the force acting on the particle and apply Newton’s 2nd law to show that, for small displacements from equilibrium, the particle will execute simple harmonic motion. Finally, we can find the angular frequency of the motion from the differential equation and use this expression to find its value when the radius of the ring is doubled while keeping the linear charge density on the ring constant. Express the electric field on the axis of the ring of charge:

Ex =

(x

kQx 2

+ R2

)

32

158 Chapter 22 Factor R2 from the denominator of Ex to obtain:

Ex =

=

kQx ⎡ 2⎛ x 2 ⎞⎤ ⎢ R ⎜⎜1 + 2 ⎟⎟⎥ ⎣ ⎝ R ⎠⎦ kQx ⎛ x ⎞ R 3 ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠ 2

32

32

≈

kQ x R3

provided x << R. Express the force acting on the particle as a function of its charge and the electric field: Because the negatively charged particle experiences a linear restoring force, we know that its motion will be simple harmonic. Apply Newton’s 2nd law to the negatively charged particle to obtain:

Fx = qE x =

m

kqQ x R3

d 2x kqQ =− 3 x 2 dt R

or

d 2 x kqQ + x =0, dt 2 mR 3 the differential equation of simple harmonic motion.

The angular frequency of the simple harmonic motion of the particle is given by:

ω=

kqQ mR 3

(1)

Express the angular frequency of the motion if the radius of the ring is doubled while keeping the linear charge density constant (i.e., doubling Q):

ω' =

kq(2Q ) 3 m(2 R )

(2)

Divide equation (2) by equation (1) to obtain:

Solve for and evaluate ω′:

kq(2Q ) 3 m(2 R )

ω' = ω

ω' =

kqQ mR 3

ω 2

=

=

1 2

21 rad/s = 10.5 rad/s 2

The Electric Field 2: Continuous Charge Distributions 159 87 •• r Picture the Problem We can apply Gauss’s law to express E as a function of r. We can use the hint to think of the fields at points 1 and 2 as the sum of the fields due to a sphere of radius a with a uniform charge distribution ρ and a sphere of radius b, centered at a/2 with uniform charge distribution −ρ. (a) The electric field at a distance r from the center of the sphere is given by:

r E = Erˆ (1) where rˆ is a unit vector pointing radially outward.

Apply Gauss’s law to a spherical surface of radius r centered at the origin to obtain:

Q ∫ E dA = E (4πr ) = ∈

Relate Qenclosed to the charge density ρ:

ρ=

Substitute for Qenclosed:

Solve for E to obtain:

Substitute for E in equation (1) to obtain:

(b) The electric field at point 1 is the sum of the electric fields due to the two charge distributions: Apply Gauss’s law to relate the magnitude of the field due to the positive charge distribution to the charge enclosed by the sphere: Solve for Eρ:

Proceed similarly for the spherical hole to obtain: Solve for E−ρ:

2

enclosed

n

S

0

Qenclosed ⇒ Qenclosed = 43 ρπr 3 3 4 π r 3

(

)

E 4πr 2 =

E=

ρπr 3

4 3

∈0

ρr 3 ∈0

r E=

ρ 3 ∈0

r rˆ

r r r E1 = E ρ + E − ρ = E ρ rˆ + E− ρ rˆ

(

)

E ρ 4πa 2 =

Eρ =

qencl 43 πa 3 ρ = ∈0 ∈0

aρ 2 ρb = 3 ∈0 3 ∈0

(

)

E− ρ 4πb 2 = E− ρ = −

ρb 3∈0

4 πb 3 ρ qencl =− 3 ∈0 ∈0

(2)

160 Chapter 22 Substitute in equation (2) to obtain:

The electric field at point 2 is the sum of the electric fields due to the two charge distributions:

r 2 ρb ρb ρb E1 = rˆ − rˆ = rˆ 3 ∈0 3 ∈0 3 ∈0

r r r E 2 = E ρ + E − ρ = E ρ rˆ + E− ρ rˆ

Because point 2 is at the center of the larger sphere:

Eρ = 0

The magnitude of the field at point 2 due to the negative charge distribution is:

E− ρ =

Substitute in equation (3) to obtain:

r ρb ρb E2 = 0 + rˆ = rˆ 3 ∈0 3 ∈0

(3)

ρb 3∈0

88 ••• Picture the Problem The electric field in the cavity is the sum of the electric field due to the uniform and positive charge distribution of the sphere whose radius is a and the electric field due to any charge in the spherical cavity whose radius is b. The electric field at any point inside the cavity is the sum of the electric fields due to the two charge distributions:

r r r E = E ρ + E charge inside = E ρ rˆ + Echarge inside rˆ where rˆ is a unit vector pointing radially outward.

Because there is no charge inside the cavity:

Echarge inside = 0

The magnitude of the field inside the cavity due to the positive charge distribution is:

Eρ =

r

Substitute in the expression for E to obtain:

ρb 3∈0

r ρb ρ E =0+ rˆ = brˆ 3 ∈0 3 ∈0

89 •• Picture the Problem We can use the hint given in Problem 87 to think of the fields at points 1 and 2 as the sum of the fields due to a sphere of radius a with a uniform charge distribution ρ and a sphere of radius b, centered at a/2 with charge Q spread uniformly throughout its volume. The electric field at point 1 is the sum of the electric fields due to the two charge distributions:

r r r E1 = E ρ + EQ = E ρ rˆ + EQ rˆ

(1)

where rˆ is a unit vector pointing radially outward.

The Electric Field 2: Continuous Charge Distributions 161

(

Apply Gauss’s law to relate the field due to the positive charge distribution to the charge of the sphere:

)

E ρ 4πa 2 =

qencl 43 πa 3 ρ = ∈0 ∈0

aρ 2 ρb = 3 ∈0 3 ∈0

Solve for Eρ:

Eρ =

Apply Gauss’s law to relate the field due to the negative charge distributed uniformly throughout the volume of the cavity :

EQ 4πb 2 =

(

)

qencl Q = ∈0 ∈0

where Q = ρ 'V = ρ ' 43 πb 3

(

Substitute for Q to obtain:

)

EQ 4πb 2 =

Solve for EQ:

EQ =

4 3

πρ ' b 3 ∈0

ρ'b 3 ∈0

r 2 ρb ρ'b E1 = rˆ + rˆ = 3 ∈0 3 ∈0

Substitute in equation (1) to obtain:

(2 ρ + ρ ')b rˆ 3 ∈0

r r r E 2 = E ρ + EQ = E ρ rˆ + EQ rˆ

The electric field at point 2 is the sum of the electric fields due to the two charge distributions: Because point 2 is at the center of the larger sphere:

Eρ = 0

The magnitude of the field at point 2 due to the uniformly distributed charge Q was shown above to be:

EQ =

Substitute in equation (2) to obtain:

r ρ'b ρ' E2 = 0 + rˆ = brˆ 3 ∈0 3 ∈0

(2)

ρ'b 3 ∈0

90 •• Picture the Problem Let the length of the cylinder be L, its radius R, and charge Q. Let P be a generic point of interest on the x axis. We can find the electric field at P by expressing the field due to an elemental disk of radius R, thickness dx, and charge dq and

(

then integrating E x = 2πkσ 1 − x

)

x2 + R2 .

162 Chapter 22

⎛ dE x = 2πkρ ⎜⎜1 − ⎝

Express the electric field on the x axis due to the charge carried by the disk of thickness dx:

⎞ ⎟dx ⎟ x2 + R2 ⎠ x

Integrate dEx for P beyond the end of the cylinder:

E x = 2πkρ

x+ L 2

⎛ ⎜1 − ⎜ x− L 2 ⎝

∫

⎞ ⎟dx ⎟ x2 + R2 ⎠ x

2 2 ⎤ ⎡ ⎞ ⎛L ⎞ ⎛L = 2πkρ ⎢ L − ⎜ + x ⎟ + R 2 + ⎜ − x ⎟ + R 2 ⎥ ⎥ ⎢ ⎠ ⎝2 ⎠ ⎝2 ⎦ ⎣

Integrate dEx for P inside the cylinder:

⎡ L 2+ x ⎛ E x = 2πkρ ⎢ ∫ ⎜⎜1 − ⎣⎢ 0 ⎝

L 2− x ⎛ ⎞ ⎟dx − ∫ ⎜1 − ⎜ 2 2 ⎟ x +R ⎠ 0 ⎝

x

⎞ ⎤ ⎟dx ⎥ 2 2 ⎟ x + R ⎠ ⎦⎥ x

2 2 ⎡ ⎤ L L ⎛ ⎞ ⎛ ⎞ 2 = 2πkρ ⎢2 x − ⎜ + x ⎟ + R + ⎜ − x ⎟ + R 2 ⎥ ⎢ ⎥ ⎝2 ⎠ ⎝2 ⎠ ⎣ ⎦

The effective charge density of the disk is given by:

ρ=

Q L πR 2

Substitute numerical values and evaluate ρ:

ρ=

50 µC = 5.53 µC/m 3 2 π (1.2 m ) (2 m )

Evaluate 2πkρ :

(

)(

)

2πkρ = 2π 8.99 × 109 N ⋅ m 2 /C 2 5.53 µC/m 3 = 3.12 × 105 N/C ⋅ m

The Electric Field 2: Continuous Charge Distributions 163 (a) Evaluate Ex(0.5 m):

(

E x (0.5 m ) = 3.12 × 105 N/C ⋅ m

)

2 2 ⎡ ⎤ 2m 2m ⎛ ⎞ ⎛ ⎞ 2 2 × ⎢2(0.5 m ) − ⎜ + 0.5 m ⎟ + (1.2 m ) + ⎜ − 0.5 m ⎟ + (1.2 m ) ⎥ ⎢ ⎥ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎣ ⎦

= 118 kN/C (b) Evaluate Ex(2 m):

(

E x (2 m ) = 3.12 × 105 N/C ⋅ m

)

2 2 ⎡ ⎤ ⎛ 2m ⎞ ⎛ 2m ⎞ 2 2 × ⎢2 m − ⎜ + 2 m ⎟ + (1.2 m ) + ⎜ − 2 m ⎟ + (1.2 m ) ⎥ ⎢ ⎥ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎣ ⎦

= 103 kN/C (c) Evaluate Ex(20 m):

(

E x (20 m ) = 3.12 × 105 N/C ⋅ m

)

2 2 ⎡ ⎤ ⎛ 2m ⎞ ⎛ 2m ⎞ 2 2 × ⎢2 m − ⎜ + 20 m ⎟ + (1.2 m ) + ⎜ − 20 m ⎟ + (1.2 m ) ⎥ ⎢ ⎥ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎣ ⎦

= 1.12 kN/C Remarks: Note that, in (c), the distance of 20 m is much greater than the length of the cylinder that we could have used Ex = kQ/x2. 91 •• Picture the Problem We can use E x = kQ [x0 (x0 − L )] to express the electric fields at x0 = 2L and x0 = 3L and take the ratio of these expressions to find the field at x0 = 3L. Express the electric field along the x axis due to a uniform line charge on the x axis:

E x ( x0 ) =

Evaluate Ex at x0 = 2L:

E x (2 L ) =

Evaluate Ex at x0 = 3L:

kQ x0 ( x0 − L )

kQ kQ = 2 2 L(2 L − L ) 2 L kQ kQ = 2 E x (3L ) = 3L(3L − L ) 6 L

(1) (2)

164 Chapter 22 Divide equation (2) by equation (1) to obtain:

Solve for and evaluate E x (3L ) :

kQ E x (3L ) 6 L2 1 = = E x (2 L ) kQ 3 2 L2 E x (3L ) =

1 1 E x (2 L ) = (600 N/C ) 3 3

= 200 N/C 92 ••• Picture the Problem Let the coordinates of one corner of the cube be (x,y,z), and assume that the sides of the cube are ∆x, ∆y, and ∆z and compute the flux through the faces of the cube that are parallel to the yz plane. The net flux of the electric field out of the gaussian surface is the difference between the flux out of the surface and the flux into the surface. The net flux out of the cube is given by:

φnet = φ ( x + ∆x ) − φ ( x )

Use a Taylor series expansion to express the net flux through faces of the cube that are parallel to the yz plane:

φnet = φ (x ) + (∆x )φ ' ( x ) + 12 (∆x )2 φ ' ' ( x ) + ... − φ ( x ) = (∆x )φ ' ( x ) + 12 (∆x )2 φ ' ' ( x ) + ... Neglecting terms higher than first order we have: Because the electric field is in the x direction, φ (x) is:

φnet = ∆xφ' ( x ) φ ( x ) = Ex ∆y∆z and

φ' (x ) = Substitute for φ ′(x) to obtain:

∂Ex ∆y∆z ∂x

φnet = ∆x

∂E x (∆y∆z ) ∂x

∂E x (∆x∆y∆z ) ∂x ∂E x ∆V = ∂x

=

The Electric Field 2: Continuous Charge Distributions 165 93 •• Picture the Problem We can use the definition of electric flux in conjunction with the

r

result derived in Problem 92 to show that ∇ ⋅ E = ρ / ∈0 .

q encl ρ V = ∈0 ∈0

From Gauss’s law, the net flux through any surface is:

φ net =

Generalizing our result from Problem 92 (see the remark following Problem 92):

φnet = ⎜⎜ x + y + z ⎟⎟V = ∇ ⋅ E V ∂y ∂z ⎠ ⎝ ∂x

∂E ⎞

(

Equate these two expressions to obtain:

(∇ ⋅ Er )V = ∈ρ V or ∇ ⋅ Er =

ρ

∂E

⎛ ∂E

0

r

)

∈0

*94 ••• Picture the Problem We can find the field due to the infinitely long line charge from E = 2kλ r and the force that acts on the dipole using F = p dE dr . Express the force acting on the dipole:

F=p

The electric field at the location of the dipole is given by:

E=

Substitute to obtain:

F=p

dE dr

2kλ r d ⎡ 2kλ ⎤ 2kλp = − 2 ⎥ ⎢ dr ⎣ r ⎦ r

where the minus sign indicates that the dipole is attracted to the line charge. 95 •• Picture the Problem We can find the distance from the center where the net force on either charge is zero by setting the sum of the forces acting on either point charge equal to zero. Each point charge experiences two forces; one a Coulomb force of repulsion due to the other point charge, and the second due to that fraction of the sphere’s charge that is between the point charge and the center of the sphere that creates an electric field at the location of the point charge. Apply

∑ F = 0 to either of the

FCoulomb − Ffield = 0

point charges: Express the Coulomb force on the proton:

FCoulomb =

The force exerted by the field E is:

Ffield = eE

ke 2 ke 2 = (2a )2 4a 2

(1)

166 Chapter 22

(

)

Apply Gauss’s law to a spherical surface of radius a centered at the origin:

E 4πa 2 =

Relate the charge density of the electron sphere to Qenclosed:

Qenclosed 2e 2ea 3 = ⇒ Q = enclosed 3 3 4 4 R3 3 πR 3 πa

(

Substitute for Qenclosed:

)

E 4πa 2 =

Solve for E to obtain:

E=

Substitute for FCoulomb and Ffield in equation (1):

Qenclosed ∈0

2ea 3 ∈0 R 3

ea e2a ⇒ F = field 2π ∈0 R 3 2π ∈0 R 3

ke 2 e2a − =0 4a 2 2π ∈0 R 3 or

ke 2 2ke 2 a − =0 4a 2 R3 Solve for a to obtain:

a=3

1 R = 0 .5 R 8

96 ••• Picture the Problem We can use the result of Problem 96 to express the force acting on both point charges when they are separated by 2a. We can then use this expression to write the force function when the point charges are each displaced a small distance x from their equilibrium positions and then expand this function binomially to show that each point charge experiences a linear restoring force. From Problem 95, the force function at the equilibrium position is:

F (a ) =

When the charges are displaced a distance x symmetrically from their equilibrium positions, the force function becomes:

F (a + x ) =

ke 2 2ke 2 a − =0 4a 2 R3 2 ke 2 (a + x )−2 − 2ke3 (a + x ) 4 R

Expand this function binomially to obtain:

(

)

2ke 2 2ke 2 ke 2 −2 a − 2a −3 x + ... − 3 a − 3 x 4 R R 2 2 2 2 2ke 2ke ke ke ≈ 2 − 3 x− 3 a− 3 x 4a 2a R R

F (a + x ) =

The Electric Field 2: Continuous Charge Distributions 167 Substitute for R using the result obtained in Problem 96 and simplify to obtain:

⎛ 3ke 2 ⎞ Frestoring = − ⎜⎜ 3 ⎟⎟ x ⎝ 4a ⎠ Hence, we’ve shown that, for a small displacement from equilibrium, the point charges experience a linear restoring force.

Remarks: An alternative approach that you might find instructive is to expand the force function using the Taylor series. 97 ••• Picture the Problem Because the restoring force found in Problem 96 is linear, we can express the differential equation of the proton’s motion and then identify ω2 from this equation. Apply

∑F

x

= ma to the displaced

−

proton to obtain:

3ke 2 d 2x x m = 4r 3 dt 2

or

3ke 2 d 2x = − x = −ω 2 x 4mr 3 dt 2 3ke 2 where ω 2 = 4mr 3 Solve for ω :

ω=

3ke 2 4mr 3

Substitute numerical values and evaluate ω:

(

)(

3 8.99 × 109 N ⋅ m 2 / C 2 1.6 × 10 −19 C ω= 3 4 1.67 × 10 −27 kg (0.08 nm )

(

)

)

2

= 4.49 × 1014 s −1

168 Chapter 22

Chapter 23 Electrical Potential Conceptual Problems *1 • Determine the Concept A positive charge will move in whatever direction reduces its potential energy. The positive charge will reduce its potential energy if it moves toward a region of lower electric potential. 2 •• Picture the Problem A charged particle placed in an electric field experiences an accelerating force that does work on the particle. From the work-kinetic energy theorem we know that the work done on the particle by the net force changes its kinetic energy and that the kinetic energy K acquired by such a particle whose charge is q that is accelerated through a potential difference V is given by K = qV. Let the numeral 1 refer to the alpha particle and the numeral 2 to the lithium nucleus and equate their kinetic energies after being accelerated through potential differences V1 and V2. Express the kinetic energy of the alpha particle when it has been accelerated through a potential difference V1: Express the kinetic energy of the lithium nucleus when it has been accelerated through a potential difference V2: Equate the kinetic energies to obtain:

K 1 = q1V1 = 2eV1

K 2 = q 2V2 = 3eV2

2eV1 = 3eV2 or

V2 = 23 V1 and (b ) is correct.

3 • r Determine the Concept If V is constant, its gradient is zero; consequently E = 0. 4

•

Determine the Concept No. E can be determined from either El = − known and differentiable or from El = −

dV provided V is dl

∆V provided V is known at two or more points. ∆l

169

170 Chapter 23 5 • Determine the Concept Because the field lines are always perpendicular to equipotential surfaces, you move always perpendicular to the field. 6 •• Determine the Concept V along the axis of the ring does not depend on the charge distribution. The electric field, however, does depend on the charge distribution, and the result given in Chapter 21 is valid only for a uniform distribution. *7 •• Picture the Problem The electric field lines, shown as solid lines, and the equipotential surfaces (intersecting the plane of the paper), shown as dashed lines, are sketched in the adjacent figure. The point charge +Q is the point at the right, and the metal sphere with charge −Q is at the left. Near the two charges the equipotential surfaces are spheres, and the field lines are normal to the metal sphere at the sphere’s surface. 8 •• Picture the Problem The electric field lines, shown as solid lines, and the equipotential surfaces (intersecting the plane of the paper), shown as dashed lines, are sketched in the adjacent figure. The point charge +Q is the point at the right, and the metal sphere with charge +Q is at the left. Near the two charges the equipotential surfaces are spheres, and the field lines are normal to the metal sphere at the sphere’s surface. Very far from both charges, the equipotential surfaces and field lines approach those of a point charge 2Q located at the midpoint.

Electric Potential 171 9 •• Picture the Problem The equipotential surfaces are shown with dashed lines, the field lines are shown in solid lines. It is assumed that the conductor carries a positive charge. Near the conductor the equipotential surfaces follow the conductor’s contours; far from the conductor, the equipotential surfaces are spheres centered on the conductor. The electric field lines are perpendicular to the equipotential surfaces. 10 •• Picture the Problem The equipotential surfaces are shown with dashed lines, the electric field lines are shown with solid lines. Near each charge, the equipotential surfaces are spheres centered on each charge; far from the charges, the equipotential is a sphere centered at the midpoint between the charges. The electric field lines are perpendicular to the equipotential surfaces. *11 • Picture the Problem We can use Coulomb’s law and the superposition of fields to find E at the origin and the definition of the electric potential due to a point charge to find V at the origin. Apply Coulomb’s law and the superposition of fields to find the electric field E at the origin:

Express the potential V at the origin:

r r r E = E +Q at −a + E +Q at a =

kQ ˆ kQ ˆ i − 2 i =0 a2 a

V = V+Q at −a + V+Q at a kQ kQ 2kQ + = a a a and (b) is correct. =

172 Chapter 23 12

•

r

Picture the Problem We can use E = −

∂V ˆ i to find the electric field corresponding the ∂x

given potential and then compare its form to those produced by the four alternatives listed. Find the electric field corresponding to this potential function:

r ∂V ˆ ∂ E =− i = − [4 x + V0 ]iˆ ∂x ∂x ⎡ 1if x ≥ 0 ⎤ ˆ ∂ = −4 [ x ]iˆ = −4 ⎢ ⎥i ∂x ⎣− 1 if x < 0⎦ ⎡− 4 if x ≥ 0⎤ ˆ =⎢ ⎥i ⎣ 4 if x < 0 ⎦

Of the alternatives provided above, only a uniformly charged sheet in the yz plane would produce a constant electric field whose direction changes at the origin. (c) is correct. 13 • Picture the Problem We can use Coulomb’s law and the superposition of fields to find E at the origin and the definition of the electric potential due to a point charge to find V at the origin. r r r Apply Coulomb’s law and the E = E +Q at −a + E −Q at a superposition of fields to find the 2kQ kQ kQ = 2 iˆ + 2 iˆ = 2 iˆ electric field E at the origin:

a

Express the potential V at the origin:

a

a

V = V+Q at −a + V−Q at a

kQ k (− Q ) + =0 a a and (c) is correct

=

14 •• (a) False. As a counterexample, consider two equal charges at equal distances from the origin on the x axis. The electric field due to such an array is zero at the origin but the electric potential is not zero. (b) True. (c) False. As a counterexample, consider two equal-in-magnitude but opposite-in-sign charges at equal distances from the origin on the x axis. The electric potential due to such an array is zero at the origin but the electric field is not zero.

Electric Potential 173 (d) True. (e) True. (f) True. (g) False. Dielectric breakdown occurs in air at an electric field strength of approximately 3×106 V/m. 15 •• (a) No. The potential at the surface of a conductor also depends on the local radius of the surface. Hence r and σ can vary in such a way that V is constant. (b) Yes; yes. *16 • Determine the Concept When the two spheres are connected, their charges will redistribute until the two-sphere system is in electrostatic equilibrium. Consequently, the entire system must be an equipotential. (c) is corrent.

Estimation and Approximation Problems 17 • Picture the Problem The field of a thundercloud must be of order 3×106 V/m just before a lightning strike. Express the potential difference between the cloud and the earth as a function of their separation d and electric field E between them:

V = Ed

Assuming that the thundercloud is at a distance of about 1 km above the surface of the earth, the potential difference is approximately:

V = 3 × 10 6 V/m 103 m

(

)(

)

= 3.00 × 109 V

Note that this is an upper bound, as there will be localized charge distributions on the thundercloud which raise the local electric field above the average value. *18 • Picture the Problem The potential difference between the electrodes of the spark plug is the product of the electric field in the gap and the separation of the electrodes. We’ll assume that the separation of the electrodes is 1 mm. Express the potential difference between the electrodes of the spark

V = Ed

174 Chapter 23 plug as a function of their separation d and electric field E between them:

(

)(

V = 2 × 10 7 V/m 10 −3 m

Substitute numerical values and evaluate V:

)

= 20.0 kV

19 •• Picture the Problem We can use conservation of energy to relate the initial kinetic energy of the protons to their electrostatic potential energy when they have approached each other to the given "radius".

Ki + U i = Kf + U f

(a) Apply conservation of energy to relate the initial kinetic energy of the protons to their electrostatic potential when they are separated by a distance r:

or, because Ui = Kf = 0,

Ki = U f

Because each proton has kinetic energy K:

2K =

e2 e2 ⇒ K= 4π ∈0 r 8π ∈0 r

Substitute numerical values and evaluate K:

(1.6 × 10

−19

)

2

C 1 eV = 1.15 × 10 −13 J × K= −12 2 2 −15 8π 8.85 × 10 C / N ⋅ m 10 m 1.6 × 10 −19 J

(

)(

)

= 0.719 MeV (b) Express and evaluate the ratio of the two energies:

f =

K 0.719 MeV = = 0.0767% Erest 938 MeV

20 •• Picture the Problem The magnitude of the electric field for which dielectric breakdown occurs in air is about 3 MV/m. We can estimate the potential difference between you and your friend from the product of the length of the spark and the dielectric constant of air. Express the product of the length of the spark and the dielectric constant of air:

V = (3 MV/m )(2 mm ) = 6000 V

Electric Potential 175

Potential Difference 21 • Picture the Problem We can use the definition of finite potential difference to find the potential difference V(4 m) − V(0) and conservation of energy to find the kinetic energy of the charge when it is at x = 4 m. We can also find V(x) if V(x) is assigned various values at various positions from the definition of finite potential difference. (a) Apply the definition of finite potential difference to obtain:

4m b r r V (4 m ) − V (0 ) = − ∫ E ⋅ d l = − ∫ Edl a

0

= −(2 kN/C )(4 m ) = − 8.00 kV (b) By definition, ∆U is given by:

∆U = q∆V = (3 µC )(− 8 kV ) = − 24.0 mJ ∆K + ∆U = 0

(c) Use conservation of energy to relate ∆U and ∆K:

or

Because K0 = 0:

K 4 m = −∆U = 24.0 mJ

Use the definition of finite potential difference to obtain: (d) For V(0) = 0:

K 4 m − K 0 + ∆U = 0

V ( x ) − V ( x0 ) = − E x ( x − x0 )

= −(2 kV/m )( x − x0 )

V ( x ) − 0 = −(2 kV/m )(x − 0 ) or

V (x ) = − (2 kV/m )x

(e) For V(0) = 4 kV:

V ( x ) − 4 kV = −(2 kV/m )( x − 0 ) or

V (x ) = 4 kV − (2 kV/m )x

(f) For V(1m) = 0:

V ( x ) − 0 = −(2 kV/m )( x − 1) or

V (x ) = 2 kV − (2 kV/m )x

176 Chapter 23 22 • Picture the Problem Because the electric field is uniform, we can find its magnitude from E = ∆V/∆x. We can find the work done by the electric field on the electron from the difference in potential between the plates and the charge of the electron and find the change in potential energy of the electron from the work done on it by the electric field. We can use conservation of energy to find the kinetic energy of the electron when it reaches the positive plate. (a) Express the magnitude of the electric field between the plates in terms of their separation and the potential difference between them:

E=

∆V 500 V = = 5.00 kV/m ∆x 0.1 m

Because the electric force on a test charge is away from the positive plate and toward the negative plate, the positive plate is at the higher potential. (b) Relate the work done by the electric field on the electron to the difference in potential between the plates and the charge of the electron: Convert 8.01×10−17 J to eV:

W = q∆V = (1.6 × 10 −19 C )(500V ) = 8.01 × 10 −17 J

⎞ ⎛ 1 eV ⎟ W = 8.01 × 10 −17 J ⎜⎜ −19 ⎟ ⎝ 1.6 × 10 J ⎠

(

)

= 500 eV (c) Relate the change in potential energy of the electron to the work done on it as it moves from the negative plate to the positive plate:

∆U = −W = − 500 eV

Apply conservation of energy to obtain:

∆K = −∆U = 500 eV

23 • Picture the Problem The Coulomb potential at a distance r from the origin relative to V = 0 at infinity is given by V = kq/r where q is the charge at the origin. The work that must be done by an outside agent to bring a charge from infinity to a position a distance r from the origin is the product of the magnitude of the charge and the potential difference due to the charge at the origin.

Electric Potential 177 kq r 8.99 × 109 N ⋅ m 2 /C 2 (2 µC ) = 4m

(a) Express and evaluate the Coulomb potential of the charge:

V=

(

)

= 4.50 kV

W = q∆V = (3 µC )(4.50 kV )

(b) Relate the work that must be done to the magnitude of the charge and the potential difference through which the charge is moved:

= 13.5 mJ

(c) Express the work that must be done by the outside agent in terms of the potential difference through which the 2-µC is to be moved:

W = q2 ∆V3 =

Substitute numerical values and evaluate W:

W=

kq2 q3 r

(8.99 × 10

9

N ⋅ m 2 /C 2 )(2 µC )(3 µC ) 4m

= 13.5 mJ 24 •• Picture the Problem In general, the work done by an external agent in separating the two ions changes both their kinetic and potential energies. Here we’re assuming that they are at rest initially and that they will be at rest when they are infinitely far apart. Because their potential energy is also zero when they are infinitely far apart, the energy Wext required to separate the ions to an infinite distance apart is the negative of their potential energy when they are a distance r apart.

Wext = ∆K + ∆U = 0 − U i

Express the energy required to separate the ions in terms of the work required by an external agent to bring about this separation:

=−

kq− q+ k (− e ) e ke 2 =− = r r r

Substitute numerical values and evaluate Wext:

Wext

(8.99 × 10 =

9

)(

N ⋅ m 2 /C 2 1.6 × 10 −19 C 2.80 × 10 −10 m

)

2

= 8.24 × 10 −19 J

178 Chapter 23 Convert Wext to eV:

⎛ ⎞ 1 eV ⎟ W = 8.24 × 10 −19 J ⎜⎜ −19 ⎟ ⎝ 1.6 × 10 J ⎠

(

)

= 5.14 eV 25 •• Picture the Problem We can find the final speeds of the protons from the potential difference through which they are accelerated and use E = ∆V/∆x to find the accelerating electric field. (a) Apply the work-kinetic energy theorem to the accelerated protons:

W = ∆K = K f or

e∆V = 12 mv 2 Solve for v to obtain:

Substitute numerical values and evaluate v:

v=

2e∆V m

v=

2 1.6 × 10 −19 C (5 MV ) 1.67 × 10 −27 kg

(

)

= 3.10 × 10 7 m/s (b) Assuming the same potential change occurred uniformly over the distance of 2.0 m, we can use the relationship between E, ∆V, and ∆x express and evaluate E:

E=

∆V 5 MV = = 2.50 MV/m ∆x 2m

*26 •• Picture the Problem The work done on the electrons by the electric field changes their kinetic energy. Hence we can use the work-kinetic energy theorem to find the kinetic energy and the speed of impact of the electrons. Use the work-kinetic energy theorem to relate the work done by the electric field to the change in the kinetic energy of the electrons: (a) Substitute numerical values and evaluate Kf:

W = ∆K = K f or

K f = e∆V

(1)

K f = (1 e )(30 kV ) = 3 × 10 4 eV

Electric Potential 179 (b) Convert this energy to eV:

⎛ 1.6 × 10 −19 J ⎞ ⎟⎟ K f = 3 × 10 4 eV ⎜⎜ eV ⎝ ⎠

(

)

= 4.80 × 10 −15 J (c) From equation (1) we have: Solve for vf to obtain:

Substitute numerical values and evaluate vf:

1 2

mvf2 = e∆V

vf =

2e∆V m

vf =

2 1.6 × 10 −19 C (30 kV ) 9.11 × 10 −31 kg

(

)

= 1.03 × 108 m/s Remarks: Note that this speed is about one-third that of light. 27 •• Picture the Problem We know that energy is conserved in the interaction between the α particle and the massive nucleus. Under the assumption that the recoil of the massive nucleus is negligible, we know that the initial kinetic energy of the α particle will be transformed into potential energy of the two-body system when the particles are at their distance of closest approach.

∆K + ∆U = 0

(a) Apply conservation of energy to the system consisting of the α particle and the massive nucleus:

or

Because Kf = Ui = 0 and Ki = E:

− E +Uf = 0

Kf − Ki + U f −Ui = 0

Letting r be the separation of the particles at closest approach, express Uf:

Uf =

kqnucleus qα k (Ze )(2e ) 2kZe 2 = = r r r

Substitute to obtain:

−E+

2kZe 2 =0 r

Solve for r to obtain:

2kZe 2 r= E

(b) For a 5.0-MeV α particle and a gold nucleus:

180 Chapter 23

(

)

(

)

= 4.55 × 10 −14 m = 45.4 fm

)

= 2.53 × 10 −14 m = 25.3 fm

2 8.99 × 109 N ⋅ m 2 /C 2 (79 ) 1.6 × 10 −19 C r5 = (5 MeV ) 1.6 × 10 −19 J/eV

(

)

2

For a 9.0-MeV α particle and a gold nucleus:

r9 =

(

)

(

2 8.99 × 109 N ⋅ m 2 /C 2 (79 ) 1.6 × 10 −19 C (9 MeV ) 1.6 ×10−19 J/eV

(

)

2

Potential Due to a System of Point Charges 28 • Picture the Problem Let the numerals 1, 2, 3, and 4 denote the charges at the four corners of square and r the distance from each charge to the center of the square. The potential at the center of square is the algebraic sum of the potentials due to the four charges.

kq1 kq 2 kq3 kq4 + + + r r r r k = (q1 + q 2 + q3 + q 4 ) r k 4 = ∑ qi r i =1

Express the potential at the center of the square:

V=

(a) If the charges are positive:

V=

8.99 × 109 N ⋅ m 2 /C 2 (4)(2 µC) 2 2m

= 25.4 kV (b) If three of the charges are positive and one is negative:

V=

8.99 × 109 N ⋅ m 2 /C 2 (2)(2 µC) 2 2m

= 12.7 kV (c) If two are positive and two are negative:

V= 0

29 • Picture the Problem The potential at the point whose coordinates are (0, 3 m) is the algebraic sum of the potentials due to the charges at the three locations given.

Electric Potential 181 3

Express the potential at the point whose coordinates are (0, 3 m):

V = k∑ i =1

⎛q q q ⎞ qi = k ⎜⎜ 1 + 2 + 3 ⎟⎟ ri ⎝ r1 r2 r3 ⎠

(a) For q1 = q2 = q3 = 2 µC:

⎛ 1 1 1 ⎞ ⎟ = 12.9 kV V = 8.99 × 109 N ⋅ m 2 /C 2 (2 µC )⎜⎜ + + ⎟ 3 m 3 2 m 3 5 m ⎝ ⎠

(

)

(b) For q1 = q2 = 2 µC and q3 = −2 µC:

⎛ 1 1 1 ⎞ ⎟ = 7.55 kV + − V = 8.99 × 10 9 N ⋅ m 2 /C 2 (2 µC ) ⎜⎜ ⎟ 3 m 3 2 m 3 5 m ⎝ ⎠

(

)

(c) For q1 = q3 = 2 µC and q2 = −2 µC:

⎛ 1 1 1 ⎞ ⎟ = 4.44 kV V = 8.99 × 109 N ⋅ m 2 /C 2 (2 µC )⎜⎜ − + ⎟ ⎝ 3m 3 2 m 3 5 m ⎠

(

)

30 • Picture the Problem The potential at point C is the algebraic sum of the potentials due to the charges at points A and B and the work required to bring a charge from infinity to point C equals the change in potential energy of the system during this process. (a) Express the potential at point C as the sum of the potentials due to the charges at points A and B:

⎛q q ⎞ VC = k ⎜⎜ A + B ⎟⎟ ⎝ rA rB ⎠

Substitute numerical values and evaluate VC:

⎛1 1⎞ ⎛ 1 1 ⎞ ⎟⎟ = 12.0 kV + VC = kq⎜⎜ + ⎟⎟ = 8.99 × 109 N ⋅ m 2 /C 2 (2 µC ) ⎜⎜ ⎝ 3m 3m ⎠ ⎝ rA rB ⎠

(

(b) Express the required work in terms of the change in the potential energy of the system: (c) Proceed as in (a) with qB = −2 µC:

)

W = ∆U = q5VC = (5 µC )(12.0 kV ) = 60.0 mJ

182 Chapter 23 ⎛q q ⎞ VC = k ⎜⎜ A + B ⎟⎟ = 8.99 × 109 N ⋅ m 2 /C 2 ⎝ rA rB ⎠

(

)⎛⎜⎜ 23µmC + − 32mµC ⎞⎟⎟ = ⎝

⎠

0

and W = ∆U = q5VC = (5 µC )(0 ) = 0 31 • Picture the Problem The electric potential at the origin and at the north pole is the algebraic sum of the potentials at those points due to the individual charges distributed along the equator. 6

qi q = 6k ri r

(a) Express the potential at the origin as the sum of the potentials due to the charges placed at 60° intervals along the equator of the sphere:

V = k∑

Substitute numerical values and evaluate V:

V = 6 8.99 × 109 N ⋅ m 2 /C 2

i =1

(

) 03.6µCm

= 270 kV (b) Using geometry, find the distance from each charge to the north pole:

r '= 0.6 2 m

Proceed as in (a) with r '= 0.6 2 m :

V = k∑

6

i =1

qi q = 6k ' ri r'

(

= 6 8.99 × 109 N ⋅ m 2 /C 2

)

3 µC 0.6 2 m

= 191kV *32 • Picture the Problem We can use the fact that the electric potential at the point of interest is the algebraic sum of the potentials at that point due to the charges q and q′ to find the ratio q/q'. Express the potential at the point of interest as the sum of the potentials due to the two charges:

kq kq' + =0 a 3 2a 3

Electric Potential 183 q' =0 2

Simplify to obtain:

q+

Solve for the ratio q/q':

q 1 = − q' 2

33 •• Picture the Problem For the two charges, r = x − a and x + a respectively and the electric potential at x is the algebraic sum of the potentials at that point due to the charges at x = +a and x = −a.

⎛ 1 1 V = kq⎜⎜ + ⎝ x−a x+a

(a) Express V(x) as the sum of the potentials due to the charges at x = +a and x = −a:

⎞ ⎟ ⎟ ⎠

(b) The following graph of V(x) versus x for kq = 1 and a = 1 was plotted using a spreadsheet program:

10

8

V (V)

6

4

2

0 -3

-2

-1

0

1

2

3

x (m)

(c) At x = 0:

dV dV = 0 and E x = − = 0 dx dx

*34 •• Picture the Problem For the two charges, r = x − a and x respectively and the electric potential at x is the algebraic sum of the potentials at that point due to the charges at x = a and x = 0. We can use the graph and the function found in part (a) to identify the points at which V(x) = 0. We can find the work needed to bring a third charge +e to the point

184 Chapter 23 x = 12 a on the x axis from the change in the potential energy of this third charge. V (x ) =

Express the potential at x:

k (3e ) k (− 2e ) + x x−a

The following graph of V(x) for ke = 1 and a =1 was plotted using a spreadsheet program.

25 20 15 V (V)

10 5 0 -5 -10 -15 -3

-2

-1

0

1

2

x (m)

(b) From the graph we can see that V(x) = 0 when:

x = ±∞

Examining the function, we see that V(x) is also zero provided:

3 2 − =0 x x−a

For x > 0, V(x) = 0 when:

x = 3a

For 0 < x < a, V(x) = 0 when:

x = 0 .6 a

(c) Express the work that must be done in terms of the change in potential energy of the charge: Evaluate the potential at x = 12 a :

W = ∆U = qV ( 12 a )

V ( 12 a ) = =

k (3e ) k (− 2e ) + 1 1 a 2 2a−a 6ke 4ke 2ke − = a a a

3

Electric Potential 185 Substitute to obtain:

2ke 2 ⎛ 2ke ⎞ W = e⎜ ⎟= a ⎝ a ⎠

Computing the Electric Field from the Potential 35 • Picture the Problem We can use the relationship Ex = − (dV/dx) to decide the sign of Vb − Va and E = ∆V/∆x to find E. (a) Because Ex = − (dV/dx), V is greater for larger values of x. So:

Vb − Va is positive.

(b) Express E in terms of Vb − Va and the separation of points a and b:

Ex =

∆V Vb − Va = ∆x ∆x

Substitute numerical values and evaluate Ex:

Ex =

105 V = 25.0 kV/m 4m

*36 • Picture the Problem Because Ex = −dV/dx, we can find the point(s) at which Ex = 0 by identifying the values for x for which dV/dx = 0. Examination of the graph indicates that dV/dx = 0 at x = 4.5 m. Thus Ex = 0 at:

x = 4 .5 m

37 • Picture the Problem We can use V(x) = kq/x to find the potential V on the x axis at x = 3.00 m and at x = 3.01 m and E(x) = kq/r2 to find the electric field at x = 3.00 m. In part (d) we can express the off-axis potential using V(x) = kq/r, where

r = x2 + y2 . (a) Express the potential on the x axis as a function of x and q: Evaluate V at x = 3 m:

V (x ) =

kq x

(8.99 × 10 V (3 m ) = = 8.99 kV

9

N ⋅ m 2 /C 2 )(3 µC ) 3m

186 Chapter 23

(8.99 × 10 V (3.01 m ) =

Evaluate V at x = 3.01 m:

9

N ⋅ m 2 /C 2 )(3 µC ) 3.01 m

= 8.96 kV (b) The potential decreases as x increases and:

−

∆V 8.96 kV − 8.99 kV =− ∆x 3.01 m − 3.00 m = 3.00 kV/m

(c) Express the Coulomb field as a function of x:

E (x ) =

Evaluate this expression at x = 3.00 m to obtain:

E (3 m ) =

kq x2

(8.99 × 10

9

)

N ⋅ m 2 /C 2 (3 µC ) (3 m )2

= 3.00 kV/m in agreement with our result in (b).

V ( x, y ) =

(d) Express the potential at (x, y) due to a point charge q at the origin:

kq x + y2 2

Evaluate this expression at (3.00 m, 0.01 m):

V (3.00 m,0.01 m ) =

(8.99 × 10

9

)

N ⋅ m 2 /C 2 (3 µC )

(3.00 m ) + (0.01 m )2 2

= 8.99 kV

For y << x, V is independent of y and the points (x, 0) and (x, y) are at the same potential, i.e., on an equipotential surface. 38 • Picture the Problem We can find the potential on the x axis at x = 3.00 m by expressing it as the sum of the potentials due to the charges at the origin and at x = 6 m. We can also express the Coulomb field on the x axis as the sum of the fields due to the charges q1 and q2 located at the origin and at x = 6 m. (a) Express the potential on the x axis as the sum of the potentials due to the charges q1 and q2 located at the origin and at x = 6 m:

⎛q q ⎞ V ( x ) = k ⎜⎜ 1 + 2 ⎟⎟ ⎝ r1 r2 ⎠

Electric Potential 187 Substitute numerical values and evaluate V(3 m):

V ( x ) = (8.99 × 109 N ⋅ m 2 /C 2 ) ⎛ 3 µC − 3 µC ⎞ ⎟ × ⎜⎜ + 3 m ⎟⎠ ⎝ 3m = 0 ⎛q q ⎞ kq1 kq2 + 2 = k ⎜⎜ 12 + 22 ⎟⎟ 2 r1 r2 ⎝ r1 r2 ⎠

(b) Express the Coulomb field on the x axis as the sum of the fields due to the charges q1 and q2 located at the origin and at x = 6 m:

Ex =

Substitute numerical values and evaluate E(3 m):

E x = 8.99 × 109 N ⋅ m 2 /C 2

(

)

⎛ 3 µC − 3 µC ⎞ ⎟ × ⎜⎜ − 2 2 ⎟ ⎝ (3 m ) (3 m ) ⎠ = 5.99 kV/m

(c) Express the potential on the x axis as the sum of the potentials due to the charges q1 and q2 located at the origin and at x = 6 m: Substitute numerical values and evaluate V(3.01 m):

⎛q q ⎞ V ( x ) = k ⎜⎜ 1 + 2 ⎟⎟ ⎝ r1 r2 ⎠

(

V (3.01 m ) = 8.99 × 109 N ⋅ m 2 /C 2

)

⎛ 3 µC − 3 µC ⎞ ⎟⎟ × ⎜⎜ + ⎝ 3.01 m 2.99 m ⎠ = − 59.9 V

Compute −∆V/∆x:

−

∆V − 59.9 V − 0 =− 3.01 m − 3.00 m ∆x = 5.99 kV/m = E x (3.00 m )

39 • Picture the Problem We can use the relationship Ey = − (dV/dy) to decide the sign of Vb − Va and E = ∆V/∆y to find E. (a) Because Ex = − (dV/dx), V is smaller for larger values of y. So:

Vb − Va is negative.

188 Chapter 23 (b) Express E in terms of Vb − Va and the separation of points a and b:

Ey =

∆V Vb − Va = ∆y ∆y

Substitute numerical values and evaluate Ey:

Ey =

2 × 10 4 V = 5.00 kV/m 4m

40 • Picture the Problem Given V(x), we can find Ex from −dV/dx. (a) Find Ex from −dV/dx:

Ex = −

d [2000 + 3000 x] dx

= − 3.00 kV/m (b) Find Ex from −dV/dx:

Ex = −

d [4000 + 3000 x] dx

= − 3.00 kV/m (c) Find Ex from −dV/dx:

Ex = −

d [2000 − 3000 x] dx

= 3.00 kV/m (d) Find Ex from −dV/dx:

Ex = −

d [− 2000] = 0 dx

41 •• Picture the Problem We can express the potential at a general point on the x axis as the sum of the potentials due to the charges at x = 0 and x = 1 m. Setting this expression equal to zero will identify the points at which V(x) = 0. We can find the electric field at any point on the x axis from Ex = −dV/dx.

kq k (− 3q ) + x x −1

(a) Express V(x) as the sum of the potentials due to the point charges at x = 0 and x = 1 m:

V (x ) =

(b) Set V(x) = 0:

⎛q 3q ⎞⎟ k ⎜⎜ − ⎟=0 ⎝ x x −1 ⎠

⎛q 3q ⎞⎟ = k ⎜⎜ − ⎟ ⎝ x x −1 ⎠

or

Electric Potential 189 1 3 − =0 x x −1 For x < 0:

1 3 − = 0 ⇒ x = − 0.500 m − x − ( x − 1)

For 0 < x < 1:

1 3 − = 0 ⇒ x = 0.250 m x − (x − 1)

Note also that:

V ( x ) → 0 as x → ±∞

(c) Evaluate V(x) for 0 < x < 1:

⎛ q 3q ⎞ V (0 < x < 1) = k ⎜ + ⎟ ⎝ x x −1⎠

Apply Ex = −dV/dx to find Ex in this region:

E x (0 < x < 1) = −

d ⎡ ⎛ q 3q ⎞⎤ k⎜ + ⎟ dx ⎢⎣ ⎝ x x − 1 ⎠⎥⎦

⎡1 3 ⎤ = kq ⎢ 2 + (x − 1)2 ⎥⎦ ⎣x Evaluate this expression at x = 0.25 m to obtain:

⎡ ⎤ 1 3 + E x (0.25 m ) = kq ⎢ ⎥ 2 (0.75 m )2 ⎦ ⎣ (0.25 m ) =

(21.3 m )kq −2

Evaluate V(x) for x < 0:

3 ⎞ ⎛1 V ( x < 0) = −kq⎜ + ⎟ ⎝ x 1− x ⎠

Apply Ex = −dV/dx to find Ex in this region:

E x (x < 0 ) = − kq

d ⎡1 3 ⎤ + ⎢ dx ⎣ x 1 − x ⎥⎦

⎡ 1 3 ⎤ = kq ⎢− 2 + (1 − x )2 ⎥⎦ ⎣ x Evaluate this expression at x = −0.5 m to obtain:

⎡ 1 3 ⎤ E x (− 0.5 m ) = kq ⎢− + = 2 2⎥ ⎣ (− 0.5 m ) (1.5 m ) ⎦

(− 2.67 m )kq −2

(d) The following graph of V(x) for kq = 1 and a = 1 was plotted using a spreadsheet

190 Chapter 23 program:

5 0

V (V)

-5 -10 -15 -20 -25 -2

-1

0

1

2

3

x (m)

*42 •• Picture the Problem Because V(x) and Ex are related through Ex = − dV/dx, we can find V from E by integration. Separate variables to obtain: Integrate V from V1 to V2 and x from 1 m to 2 m:

(

)

dV = − E x dx = − 2.0 x 3 kN/C dx V2

x2

∫ dV = −(2.0 kN/C)∫ x

V1

3

dx

x1

[ ]

= −(2.0 kN/C) 14 x 4 Simplify to obtain:

2m

1m

V2 −V1 = − 7.50 kV

43 •• Picture the Problem Let r1 be the distance from (0, a) to (x, 0), r2 the distance from (0, −a), and r3 the distance from (a, 0) to (x, 0). We can express V(x) as the sum of the potentials due to the charges at (0, a), (0, −a), and (a, 0) and then find Ex from −dV/dx. (a) Express V(x) as the sum of the potentials due to the charges at (0, a), (0, −a), and (a, 0):

V (x ) =

kq1 kq2 kq3 + + r1 r2 r3

where q1 = q2 = q3 = q

At x = 0, the fields due to q1 and q2 cancel, so Ex(0) = −kq/a2; this is also obtained from (b) if x = 0.

Electric Potential 191 As x→∞, i.e., for x >> a, the three charges appear as a point charge 3q, so Ex = 3kq/x2; this is also the result one obtains from (b) for x >> a. Substitute for the ri to obtain:

⎛ 1 V ( x ) = kq⎜⎜ + 2 2 ⎝ x +a

1 x2 + a2

(b) For x > a, x − a > 0 and:

+

1 x−a

⎞ ⎛ 2 1 ⎟ = kq⎜ ⎟ ⎜ x2 + a2 + x − a ⎠ ⎝

⎞ ⎟ ⎟ ⎠

x−a = x−a

Use Ex = −dV/dx to find Ex:

E x (x > a ) = −

d ⎡ ⎛ 2 1 ⎞⎤ ⎟⎥ = + ⎢kq⎜⎜ 2 dx ⎣⎢ ⎝ x + a 2 x − a ⎟⎠⎦⎥

For x < a, x − a < 0 and:

(x

2kqx +a

2

)

2 32

kq (x − a )2

+

x − a = −( x − a ) = a − x

Use Ex = −dV/dx to find Ex:

E x (x < a ) = −

d ⎡ ⎛ 2 1 ⎞⎤ ⎟⎥ = + ⎢kq⎜⎜ 2 dx ⎣⎢ ⎝ x + a 2 a − x ⎟⎠⎦⎥

(x

2kqx +a

2

)

2 32

−

kq (a − x )2

Calculations of V for Continuous Charge Distributions 44 • Picture the Problem We can construct Gaussian surfaces just inside and just outside the spherical shell and apply Gauss’s law to find the electric field at these locations. We can use the expressions for the electric potential inside and outside a spherical shell to find the potential at these locations. (a) Apply Gauss’s law to a spherical Gaussian surface of radius r < 12 cm:

r

r

∫ E ⋅ dA = S

Qenclosed

∈0

=0

because the charge resides on the outer surface of the spherical surface. Hence

r E (r < 12 cm ) = 0

Apply Gauss’s law to a spherical Gaussian surface of radius

(

)

E 4π r 2 =

q

∈0

192 Chapter 23 r > 12 cm:

and

E (r > 12 cm ) =

q kq = 2 2 4π r ∈0 r

Substitute numerical values and evaluate E (r > 12 cm ) :

E (r > 12 cm ) =

(8.99 ×10

)(

)

N ⋅ m 2 /C 2 10 −8 C = 6.24 kV/m (0.12 m )2

9

(b) Express and evaluate the potential just inside the spherical shell:

V (r ≤ R ) =

(

)(

)

kq 8.99 × 10 9 N ⋅ m 2 /C 2 10 −8 C = = 749 V R 0.12 m

Express and evaluate the potential just outside the spherical shell:

kq (8.99 × 109 N ⋅ m 2 /C 2 )(10 −8 C ) V (r ≥ R ) = = = 749 V r 0.12 m (c) The electric potential inside a uniformly charged spherical shell is constant and given by:

V (r ≤ R ) =

(

)(

)

kq 8.99 × 109 N ⋅ m 2 /C 2 10 −8 C = = 749 V R 0.12 m

In part (a) we showed that:

r E (r < 12 cm ) = 0

45 • Picture the Problem We can use the expression for the potential due to a line charge V = −2kλ ln

r , where V = 0 at some distance r = a, to find the potential at these a

distances from the line. Express the potential due to a line charge as a function of the distance from the line:

V = −2kλ ln

r a

Because V = 0 at r = 2.5 m:

0 = −2kλ ln

2.5 m , a

Electric Potential 193 0 = ln

2.5 m , a

and

2.5 m = ln −1 0 = 1 a Thus we have a = 2.5 m and:

⎛ r ⎞ ⎛ r ⎞ ⎟⎟ = − 2.70 × 10 4 N ⋅ m/C ln⎜⎜ ⎟⎟ V = −2 8.99 × 109 N ⋅ m 2 /C 2 (1.5 µC/m ) ln⎜⎜ ⎝ 2 .5 m ⎠ ⎝ 2 .5 m ⎠

(

)

(a) Evaluate V at r = 2.0 m:

(

)

⎛ 2m ⎞ ⎟⎟ V = − 2.70 × 10 4 N ⋅ m/C ln⎜⎜ ⎝ 2 .5 m ⎠

(

)

= 6.02 kV (b) Evaluate V at r = 4.0 m:

⎛ 4m ⎞ ⎟⎟ V = − 2.70 × 10 4 N ⋅ m/C ln⎜⎜ ⎝ 2 .5 m ⎠

(

)

= − 12.7 kV (c) Evaluate V at r = 12.0 m:

⎛ 12 m ⎞ ⎟⎟ V = − 2.70 × 10 4 N ⋅ m/C ln⎜⎜ 2 . 5 m ⎝ ⎠

(

)

= − 42.3 kV 46 •• Picture the Problem The electric field on the x axis of a disk charge of radius R is given

⎛ ⎜ ⎝

by E x = 2πkσ ⎜1 −

⎞ ⎟ . We’ll choose V(∞) = 0 and integrate from x′ = ∞ to x′ = ⎟ x2 + R2 ⎠ x

x to obtain Equation 23-21. Relate the electric potential on the axis of a disk charge to the electric field of the disk: Express the electric field on the x axis of a disk charge:

dV = − E x dx

⎛ E x = 2πkσ ⎜⎜1 − ⎝

⎞ ⎟ ⎟ x2 + R2 ⎠ x

194 Chapter 23 Substitute to obtain:

Let V(∞) = 0 and integrate from x′ = ∞ to x′ = x:

⎛ dV = −2πkσ ⎜⎜1 − ⎝

⎞ ⎟dx ⎟ x2 + R2 ⎠ x

x ⎛ x' V = −2πkσ ∫ ⎜⎜1 − x'2 + R 2 ∞⎝

= 2πkσ

(x

2

+ R2 − x

)

⎞ ⎟ dx' ⎟ ⎠

⎛ ⎞ R2 = 2πkσ x ⎜ 1 + 2 − 1⎟ ⎜ ⎟ x ⎝ ⎠ which is Equation 23-21. *47 •• Picture the Problem Let the charge per unit length be λ = Q/L and dy be a line element with charge λdy. We can express the potential dV at any point on the x axis due to λdy and integrate of find V(x, 0). (a) Express the element of potential dV due to the line element dy:

dV =

kλ dy r

where r = Integrate dV from y = −L/2 to y = L/2:

L2

kQ dy V ( x,0) = ∫ 2 L −L 2 x + y 2 =

(b) Factor x from the numerator and denominator within the parentheses to obtain:

Use ln

x2 + y2

2 2 kQ ⎛⎜ x + L 4 + L 2 ⎞⎟ ln L ⎜ x 2 + L2 4 − L 2 ⎟ ⎝ ⎠

2 ⎛ ⎞ ⎜ 1+ L + L ⎟ kQ ⎜ 4x 2 2x ⎟ V ( x,0 ) = ln⎜ ⎟ 2 L ⎜ 1+ L − L ⎟ ⎜ ⎟ 4x 2 2x ⎠ ⎝

a = ln a − ln b to obtain: b ⎛ kQ ⎧⎪ ⎛⎜ L2 L ⎞⎟ L2 L ⎞⎟⎫⎪ ⎜ − ln 1 + 2 − V (x,0 ) = ⎬ ⎨ln⎜ 1 + 2 + ⎜ L ⎪ ⎝ 4x 2 x ⎟⎠ 4x 2 x ⎟⎠⎪ ⎝ ⎭ ⎩

Electric Potential 195 L2 L2 12 1 1 2 Let ε = and use (1 + ε ) = 1 + 2 ε − 8 ε + ... to expand 1 + : 4x 2 4x 2 12

⎛ L2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ 4x ⎠

2

1 L2 1 ⎛ L2 ⎞ ⎟ + ... ≈ 1 for x >> L. = 1+ − ⎜ 2 4 x 2 8 ⎜⎝ 4 x 2 ⎟⎠

Substitute to obtain:

V ( x,0) =

Let δ =

kQ ⎧ ⎛ L⎞ L ⎞⎫ ⎛ ⎨ln⎜1 + ⎟ − ln⎜1 − ⎟⎬ L ⎩ ⎝ 2x ⎠ ⎝ 2 x ⎠⎭

L⎞ L ⎛ and use ln(1 + δ ) = δ − 12 δ 2 + ... to expand ln⎜1 ± ⎟: 2x ⎝ 2x ⎠ L⎞ L L2 L⎞ L L2 ⎛ ⎛ ln⎜1 + ⎟ ≈ − 2 and ln⎜1 − ⎟ ≈ − − 2 for x >> L. 2x 4x ⎝ 2x ⎠ 2x 4x ⎝ 2x ⎠

Substitute and simplify to obtain:

V ( x,0) =

kQ ⎧ L L2 ⎛ L L2 ⎞⎫ kQ − 2 ⎟⎟⎬ = ⎨ − 2 − ⎜⎜ − L ⎩ 2 x 4 x ⎝ 2 x 4 x ⎠⎭ x

48 •• Picture the Problem We can find Q by integrating the charge on a ring of radius r and thickness dr from r = 0 to r = R and the potential on the axis of the disk by integrating the expression for the potential on the axis of a ring of charge between the same limits. (a) Express the charge dq on a ring of radius r and thickness dr:

⎛ R⎞ dq = 2πrσdr = 2πr ⎜ σ 0 ⎟dr ⎝ r⎠ = 2πσ 0 Rdr

Integrate from r = 0 to r = R to obtain:

Q = 2πσ 0 R ∫ dr = 2πσ 0 R 2

R

0

196 Chapter 23 kdq 2πkσ 0 Rdr = r' x2 + r 2

(b) Express the potential on the axis of the disk due to a circular element of charge dq = 2πrσdr :

dV =

Integrate from r = 0 to r = R to obtain:

V = 2πkσ 0 R ∫

R

0

dr x2 + r 2

⎛ R + x2 + R2 = 2πkσ 0 R ln⎜ ⎜ x ⎝ 49 •• Picture the Problem We can find Q by integrating the charge on a ring of radius r and thickness dr from r = 0 to r = R and the potential on the axis of the disk by integrating the expression for the potential on the axis of a ring of charge between the same limits. (a) Express the charge dq on a ring of radius r and thickness dr:

⎛ r2 ⎞ dq = 2πrσdr = 2πr ⎜⎜ σ 0 2 ⎟⎟dr ⎝ R ⎠ 2πσ 0 3 = r dr R2

Integrate from r = 0 to r = R to obtain:

2πσ 0 R 3 Q= r dr = R 2 ∫0

(b)Express the potential on the axis of the disk due to a circular element of

dV =

charge dq =

2πσ 0 3 r dr : R2

kdq 2πkσ 0 = r' R2

1 2

πσ 0 R 2 r3 x2 + r 2

Integrate from r = 0 to r = R to obtain:

V=

2πkσ 0 R2

R

∫ 0

r 3 dr x2 + r 2

=

2πkσ 0 R2

⎛ R2 − 2x2 ⎜⎜ 3 ⎝

x2 + R2 +

2x3 ⎞ ⎟ 3 ⎟⎠

dr

⎞ ⎟ ⎟ ⎠

Electric Potential 197 50 •• Picture the Problem Let the charge per unit length be λ = Q/L and dy be a line element with charge λdy. We can express the potential dV at any point on the x axis due to λdy and integrate to find V(x, 0). Express the element of potential dV due to the line element dy:

dV =

kλ dy r

where r = Integrate dV from y = −L/2 to y = L/2:

V ( x,0) =

x2 + y2 L2

kQ dy ∫ L −L 2 x 2 + y 2

2 2 kQ ⎛⎜ x + L 4 + L 2 ⎞⎟ ln = L ⎜ x 2 + L2 4 − L 2 ⎟ ⎝ ⎠

*51 •• Picture the Problem The potential at any location on the axis of the disk is the sum of the potentials due to the positive and negative charge distributions on the disk. Knowing that the total charge on the disk is zero and the charge densities are equal in magnitude will allow us to find the radius of the region that is positively charged. We can then use the expression derived in the text to find the potential due to this charge closest to the axis and integrate dV from

r=R

2 to r = R to find the potential at x

due to the negative charge distribution. (a) Express the potential at a distance x along the axis of the disk as the sum of the potentials due to the positively and negatively charged regions of the disk: We know that the charge densities are equal in magnitude and that the

V ( x ) = V+ ( x ) + V- ( x )

Qr a or

198 Chapter 23 total charge carried by the disk is zero. Express this condition in terms of the charge in each of two regions of the disk:

σ 0πa 2 = σ 0πR 2 − σ 0πa 2

Solve for a to obtain:

a=

Use this result and the general expression for the potential on the axis of a charged disk to express V+(x):

⎛ ⎞ R2 V+ ( x ) = 2πkσ 0 ⎜ x 2 + − x⎟ ⎜ ⎟ 2 ⎝ ⎠

Express the potential on the axis of the disk due to a ring of charge a distance r > a from the axis of the ring:

dV− ( x ) = −2πkσ 0

Integrate this expression from r = R 2 to r = R to obtain:

R 2

where r ' =

r dr r'

x2 + r 2 .

R

V− ( x ) = −2πkσ 0

r

∫ R

x2 + r 2

2

dr

⎛ R 2 ⎞⎟ = −2πkσ 0 ⎜ x 2 + R 2 − x 2 + ⎜ 2 ⎟⎠ ⎝ Substitute and simplify to obtain:

⎛ ⎞ ⎛ R2 V ( x ) = 2πkσ 0 ⎜ x 2 + − x ⎟ − 2πkσ 0 ⎜ x 2 + R 2 − ⎜ ⎟ ⎜ 2 ⎝ ⎠ ⎝ ⎛ R2 R2 = 2πkσ 0 ⎜ x 2 + − x− x2 + R 2 + x 2 + ⎜ 2 2 ⎝

x2 +

R2 2

⎞ ⎟ ⎟ ⎠

⎞ ⎟ ⎟ ⎠

⎛ ⎞ R2 = 2πkσ 0 ⎜ 2 x 2 + − x2 + R2 − x ⎟ ⎜ ⎟ 2 ⎝ ⎠ (b) To determine V for x >> R, factor x from the square roots and expand using the binomial expansion:

12

⎛ R2 R2 ⎞ x + = x⎜⎜1 + 2 ⎟⎟ 2 ⎝ 2x ⎠ ⎛ R2 R4 ⎞ ⎜ ⎟ x 1 ≈ ⎜ + 2− 4 ⎟ 4 x 32 x ⎝ ⎠ 2

and

Electric Potential 199 12

⎛ R2 ⎞ x + R = x⎜⎜1 + 2 ⎟⎟ x ⎠ ⎝ ⎛ R2 R4 ⎞ ≈ x⎜⎜1 + 2 − 4 ⎟⎟ 8x ⎠ ⎝ 2x 2

2

Substitute to obtain:

⎞ ⎛ ⎛ πkσ 0 R 4 R2 R4 ⎞ ⎛ R2 R4 ⎞ ⎟ ⎜ ⎟ ⎟ V ( x ) ≈ 2πkσ 0 ⎜⎜ 2 x⎜⎜1 + 2 − x 1 x = − + − − ⎟ 32 x 4 ⎟⎠ ⎜⎝ 2 x 2 8 x 4 ⎟⎠ 8x3 ⎠ ⎝ ⎝ 4x 52 •• Picture the Problem Given the potential function

)

(

V ( x ) = 2πkσ 0 2 x 2 + R 2 2 − x 2 + R 2 − x found in Problem 51(a), we can find Ex from −dV/dx. In the second part of the problem, we can find the electric field on the axis of the disk by integrating Coulomb’s law for the oppositely charged regions of the disk and expressing the sum of the two fields. dV Relate Ex to dV/dx:

Ex = −

dx

From Problem 51(a) we have:

⎛ ⎞ R2 V ( x ) = 2πkσ 0 ⎜ 2 x 2 + − x2 + R2 − x ⎟ ⎜ ⎟ 2 ⎝ ⎠ Evaluate the negative of the derivative of V(x) to obtain:

⎞ d ⎛⎜ R2 2 − x2 + R2 − x ⎟ E x = −2πkσ 0 2 x + ⎟ 2 dx ⎜⎝ ⎠ ⎛ ⎜ 2x ⎜ = − 2πkσ 0 ⎜ − 2 R 2 ⎜⎜ x + 2 ⎝ Express the field on the axis of the disk as the sum of the field due to the positive charge on the disk and the field due to the negative charge

⎞ ⎟ x ⎟ − 1⎟ 2 2 x +R ⎟⎟ ⎠

E x = E x− + E x+

200 Chapter 23 on the disk: The field due to the positive charge (closest to the axis) is:

To determine Ex− we integrate the field due to a ring charge:

E x+

⎛ ⎜ ⎜ = 2πkσ 0 ⎜1 − ⎜⎜ ⎝

⎞ ⎟ x ⎟ ⎟ R2 ⎟ 2 x + ⎟ 2 ⎠

R

E x − = −2πkσ 0 R

∫ (x 2

rdr 2

+ r2

)

32

⎛ ⎜ x ⎜ = −2πkσ 0 ⎜ − 2 R 2 ⎜⎜ x + 2 ⎝

⎞ ⎟ x ⎟ 2 2 ⎟ x +R ⎟ ⎟ ⎠

Substitute and simplify to obtain:

⎛ ⎜ x ⎜ − E x = −2πkσ 0 ⎜ 2 R ⎜⎜ x 2 + 2 ⎝ ⎛ ⎜ 2x ⎜ = − 2πkσ 0 ⎜ − 2 R ⎜⎜ x 2 + 2 ⎝

⎞ ⎛ ⎟ ⎜ x ⎟ ⎜ + 2πkσ 0 ⎜1 − 2 2 ⎟ x +R ⎟ ⎜⎜ ⎟ ⎠ ⎝

⎞ ⎟ x ⎟ 2 ⎟ R ⎟ x2 + ⎟ 2 ⎠

⎞ ⎟ x ⎟ − 1⎟ 2 2 x +R ⎟⎟ ⎠

53 •• Picture the Problem We can express the electric potential dV at x due to an elemental charge dq on the rod and then integrate over the length of the rod to find V(x). In the second part of the problem we use a binomial expansion to show that, for x >> L/2, our result reduces to that due to a point charge Q.

Electric Potential 201 (a) Express the potential at x due to the element of charge dq located at u:

Integrate V from u = −L/2 to L/2 to obtain:

kdq kλdu = r x−u or, because λ = Q/L, kQ du dV = L x−u dV =

L2

kQ du V (x ) = ∫ L −L 2 x − u kQ L2 ln ( x − u ) − L 2 L ⎡ L⎞ L ⎞⎤ ⎛ ⎛ = ⎢− ln⎜ x − ⎟ + ln⎜ x + ⎟⎥ 2⎠ 2 ⎠⎦ ⎝ ⎝ ⎣ =

⎛ x+ kQ ⎜ ⎜ ln = L ⎜x− ⎜ ⎝ (b) Divide the numerator and denominator of the argument of the logarithm by x to obtain:

L⎞ ⎟ 2⎟ L⎟ ⎟ 2⎠

L ⎞ L⎞ ⎛ ⎛ ⎜1+ ⎟ ⎜ x+ ⎟ 2 ⎟ = ln⎜ 2 x ⎟ = ln⎛⎜ 1 + a ⎞⎟ ln⎜ ⎜1− L ⎟ ⎜ x− L ⎟ ⎝1− a ⎠ ⎜ ⎟ ⎜ ⎟ 2⎠ ⎝ ⎝ 2x ⎠ where a = L/2x.

Divide 1 + a by 1 − a to obtain:

⎛ 2a 2 ⎞ ⎛1+ a ⎞ ⎟ ln⎜ ⎟ = ln⎜⎜1 + 2a + 1 − a ⎟⎠ ⎝1− a ⎠ ⎝ ⎛ ⎜ = ln⎜1 + ⎜ ⎜ ⎝ ⎛ ≈ ln⎜1 + ⎝

L2 ⎞ ⎟ 2 L x ⎟ + x 2− L ⎟ x ⎟⎠ L⎞ ⎟ x⎠

provided x >> L/2. Expand ln(1 + L/x) binomially to obtain:

⎛ L⎞ L ln⎜1 + ⎟ ≈ x⎠ x ⎝ provided x >> L/2.

Substitute to express V(x) for x >> L/2:

V (x ) =

kQ L kQ = , the field due to a L x x

202 Chapter 23 point charge Q. 54 •• Picture the Problem The diagram is a cross-sectional view showing the charges on the sphere and the spherical conducting shell. A portion of the Gaussian surface over which we’ll integrate E in order to find V in the region r > b is also shown. For a < r < b, the sphere acts like point charge Q and the potential of the metal sphere is the sum of the potential due to a point charge at its center and the potential at its surface due to the charge on the inner surface of the spherical shell. (a) Express Vr > b: Apply Gauss’s law for r > b:

Vr >b = − ∫ Er >b dr

∫

S

r Q E r ⋅ nˆ dA = enclosed = 0

ε0

and Er>b = 0 because Qenclosed = 0 for r > b. Substitute to obtain: (b) Express the potential of the metal sphere:

Vr >b = − ∫ (0 )dr = 0

Va = VQ at its center + Vsurface

Express the potential at the surface of the metal sphere:

Vsurface =

Substitute and simplify to obtain:

Va =

k (− Q ) kQ =− b b

kQ kQ ⎛1 1⎞ − = kQ⎜ − ⎟ a b ⎝a b⎠

Electric Potential 203 55 •• Picture the Problem The diagram is a cross-sectional view showing the charges on the inner and outer conducting shells. A portion of the Gaussian surface over which we’ll integrate E in order to find V in the region a < r < b is also shown. Once we’ve determined how E varies with r, we

∫

can find Vb – Va from Vb − Va = − Er dr .

Express the potential difference Vb – Va:

Vb − Va = − ∫ Er dr

Apply Gauss’s law to cylindrical Gaussian surface of radius r and length L:

∫ E ⋅ nˆ dA = E (2πrL ) = ε

Solve for Er:

Er =

Substitute for Er and integrate from r = a to b:

r

q

r

S

0

q 2πε 0 rL

Vb − Va = −

b

q

dr r a

2πε L ∫ 0

= −

2kq ⎛ b ⎞ ln⎜ ⎟ L ⎝a⎠

56 •• Picture the Problem Let R be the radius of the sphere and Q its charge. We can express the potential at the two locations given and solve the resulting equations simultaneously for R and Q. Relate the potential of the sphere at its surface to its radius:

kQ = 450 V R

Express the potential at a distance of 20 cm from its surface:

kQ = 150 V R + 0.2 m

(1)

(2)

204 Chapter 23 Divide equation (1) by equation (2) to obtain:

kQ 450 V R = kQ 150 V R + 0.2 m or

R + 0.2 m =3 R

Solve for R to obtain:

R = 0.100 m

Solve equation (1) for Q:

Q = (450 V )

Substitute numerical values and evaluate Q:

Q = (450 V )

R k

(0.1m )

(8.99 ×10

9

N ⋅ m 2 /C 2

)

= 5.01 nC 57 •• Picture the Problem Let the charge density on the infinite plane at x = a be σ1 and that on the infinite plane at x = 0 be σ2. Call that region in space for which x < 0, region I, the region for which 0 < x < a region II, and the region for which a < x region III. We can integrate E due to the planes of charge to find the electric potential in each of these regions.

(a) Express the potential in region I in terms of the electric field in that region: Express the electric field in region I as the sum of the fields due to the charge densities σ1 and σ2:

x r r VI = − ∫ E I ⋅ dx 0

r σ σ σ ˆ σ ˆ E I = − 1 iˆ − 2 iˆ = − i− i 2 ∈0 2 ∈0 2 ∈0 2 ∈0

=−

σ ˆ i ∈0

Electric Potential 205 x ⎛ σ ⎞ σ VI = − ∫ ⎜⎜ − ⎟⎟dx = x + V (0) ∈0 ⎠ ∈0 0⎝

Substitute and evaluate VI:

=

σ σ x+0= x ∈0 ∈0

r r VII = − ∫ E II ⋅ dx + V (0)

Express the potential in region II in terms of the electric field in that region:

r σ σ σ ˆ σ ˆ E II = − 1 iˆ + 2 iˆ = − i+ i 2 ∈0 2 ∈0 2 ∈0 2 ∈0

Express the electric field in region II as the sum of the fields due to the charge densities σ1 and σ2:

=0

x

Substitute and evaluate VII:

VII = − ∫ (0 )dx = 0 + V (0 ) = 0 0

x r r VIII = − ∫ E III ⋅ dx

Express the potential in region III in terms of the electric field in that region:

a

r σ σ σ ˆ σ ˆ E III = 1 iˆ + 2 iˆ = i+ i 2 ∈0 2 ∈0 2 ∈0 2 ∈0

Express the electric field in region III as the sum of the fields due to the charge densities σ1 and σ2:

=

σ ˆ i ∈0

x ⎛σ ⎞ σ σ VIII = − ∫ ⎜⎜ ⎟⎟dx = − x + a ∈ ∈0 ∈0 a⎝ 0 ⎠

Substitute and evaluate VIII:

=

(b) Proceed as in (a) with σ1 = −σ and σ2 = σ to obtain:

σ (a − x ) ∈0

VI = 0 , VII = −

*58

σ σ x and VIII = − a ∈0 ∈0

••

Picture the Problem The potential on the axis of a disk charge of radius R and charge

[(

density σ is given by V = 2πkσ x 2 + R 2

)

12

]

−x .

206 Chapter 23 Express the potential on the axis of the disk charge:

[(

V = 2πkσ x 2 + R 2

)

12

−x

]

Factor x from the radical and use the binomial expansion to obtain:

(x

2

+R

)

2 12

12

⎡ ⎤ ⎛ R2 ⎞ R 2 ⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞ R 4 = x⎜⎜1 + 2 ⎟⎟ = x ⎢1 + 2 + ⎜ ⎟⎜ − ⎟⎜ ⎟ 4 + ...⎥ x ⎠ ⎣ 2 x ⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠ x ⎦ ⎝ 2 4 ⎡ R R ⎤ ≈ x ⎢1 + 2 − 4 ⎥ 8x ⎦ ⎣ 2x

Substitute for the radical term to obtain:

⎧ ⎡ ⎫ R2 R4 ⎤ V = 2πkσ ⎨ x ⎢1 + 2 − 4 ⎥ − x ⎬ 8x ⎦ ⎩ ⎣ 2x ⎭ ⎛ R2 R4 ⎞ = 2πkσ ⎜⎜ − 3 ⎟⎟ ⎝ 2 x 8x ⎠ ⎛ R2 ⎞ kQ ≈ 2πkσ ⎜⎜ ⎟⎟ = x ⎝ 2x ⎠ provided x >> R.

59 •• Picture the Problem The diagram shows a sphere of radius R containing a charge Q uniformly distributed. We can use the definition of density to find the charge q′ inside a sphere of radius r and the potential V1 at r due to this part of the charge. We can express the potential dV2 at r due to the charge in a shell of radius r′ and thickness dr′ at r′ > r using dV2 = kdq' r and then integrate this expression from r′ = r to r′ = R to find V2. (a) Express the potential V1 at r due to q′:

V1 =

kq' r

Use the definition of density and the fact that the charge density is uniform to relate q′ to Q:

ρ=

q' Q = 3 3 4 4 3π r 3π R

Electric Potential 207 Solve for q′:

q' =

r3 Q R3

Substitute to express V1:

V1 =

k ⎛ r3 ⎞ kQ 2 ⎜⎜ 3 Q ⎟⎟ = r r⎝R ⎠ R3

(b) Express the potential dV2 at r due to the charge in a shell of radius r′ and thickness dr′ at r′ > r:

dV2 =

Express the charge dq′ in a shell of radius r′ and thickness dr′ at r′ > r:

⎛ 3Q ⎞ dq' = 4πr' 2 ρdr ' = 4πr' 2 ⎜ dr' 3 ⎟ ⎝ 4πR ⎠ 3Q = 3 r' 2 dr' R

Substitute to obtain:

dV2 =

(c) Integrate dV2 from r′ = r to r′ = R to find V2:

V2 =

(d) Express the potential V at r as the sum of V1 and V2:

kdq' r

3kQ r'dr' R3

3kQ 3kQ 2 ( r'dr' = R − r2 ) 3 ∫ 3 R r 2R R

V = V1 + V2 =

(

kQ 2 3kQ 2 2 r + 3 R −r R3 2R

)

kQ ⎛ r2 ⎞ ⎜ 3 − 2 ⎟⎟ = 2 R ⎜⎝ R ⎠ 60 • Picture the Problem We can equate the expression for the electric field due to an infinite plane of charge and −∆V/∆x and solve the resulting equation for the separation of the equipotential surfaces.

σ 2 ∈0

Express the electric field due to the infinite plane of charge:

E=

Relate the electric field to the potential:

E=−

∆V ∆x

208 Chapter 23 Equate these expressions and solve for ∆x to obtain:

∆x =

Substitute numerical values and evaluate ∆x :

∆x =

2 ∈0 ∆V

σ

(

)

2 8.85 × 10 −12 C 2 /N ⋅ m 2 (100 V ) 3.5 µC/m 2

= 0.506 mm 61 • Picture the Problem The equipotentials are spheres centered at the origin with radii ri = kq/Vi. Evaluate r for V = 20 V:

r20 V =

(8.99 × 10

9

N ⋅ m 2 /C 2 20 V

9

N ⋅ m 2 /C 2 40 V

9

N ⋅ m 2 /C 2 60 V

9

N ⋅ m 2 /C 2 80 V

)(

1 9

× 10 −8 C

)

)(

1 9

× 10 −8 C

)

)(

1 9

× 10 −8 C

)

1 9

× 10 −8 C

)

= 0.499 m Evaluate r for V = 40 V:

r40 V

(8.99 × 10 = = 0.250 m

Evaluate r for V = 60 V:

r60 V

(8.99 × 10 = = 0.166 m

Evaluate r for V = 80 V:

r80 V =

(8.99 × 10

)(

= 0.125 m Evaluate r for V = 100 V:

r100 V =

(8.99 × 10

9

= 0.0999 m

N ⋅ m 2 /C 2 100 V

)(

1 9

× 10 −8 C

)

Electric Potential 209 The equipotential surfaces are shown in cross-section to the right:

The equipotential surfaces are not equally spaced. 62 • Picture the Problem We can relate the dielectric strength of air (about 3 MV/m) to the maximum net charge that can be placed on a spherical conductor using the expression for the electric field at its surface. We can find the potential of the sphere when it carries its maximum charge using V = kQmax R . (a) Express the dielectric strength of a spherical conductor in terms of the charge on the sphere:

Ebreakdown =

Solve for Qmax:

Qmax =

Substitute numerical values and evaluate Qmax:

Qmax =

kQmax R2

Ebreakdown R 2 k

(3 MV/m )(0.16 m )2 8.99 × 109 N ⋅ m 2 /C 2

= 8.54 µC (b) Because the charge carried by the sphere could be either positive or negative:

kQmax R 8.99 × 109 N ⋅ m 2 /C 2 (8.54 µC ) =± 0.16 m

Vmax = ±

(

)

= ± 480 kV *63 • Picture the Problem We can solve the equation giving the electric field at the surface of a conductor for the greatest surface charge density that can exist before dielectric breakdown of the air occurs. Relate the electric field at the surface of a conductor to the surface charge density:

E=

σ ∈0

210 Chapter 23 Solve for σ under dielectric breakdown of the air conditions: Substitute numerical values and evaluate σmax:

σ max =∈0 Ebreaddown σ max = (8.85 × 10 −12 C 2 /N ⋅ m 2 )(3 MV/m ) = 26.6 µC/m 2

64 •• Picture the Problem Let L and S refer to the larger and smaller spheres, respectively. We can use the fact that both spheres are at the same potential to find the electric fields near their surfaces. Knowing the electric fields, we can use σ =∈0 E to find the surface charge density of each sphere.

kQS kQ and EL = 2L 2 RS RL

Express the electric fields at the surfaces of the two spheres:

ES =

Divide the first of these equations by the second to obtain:

kQS ES R2 Q R2 = S = S L2 EL kQL QL RS RL2

Because the potentials are equal at the surfaces of the spheres:

Q R kQL kQS and S = S = RL RS QL RL

Substitute to obtain:

ES RS RL2 RL = = EL RL RS2 RS

Solve for ES:

ES =

RL 12 cm (200 kV/m ) EL = RS 5 cm

= 480 kV/m Use σ =∈ 0 E to find the surface charge density of each sphere:

σ 12 cm =∈0 E12 cm = (8.85 × 10−12 C 2 /N ⋅ m 2 )(200 kV/m ) = 1.77 µC/m 2 and

σ 5 cm =∈0 E5 cm = (8.85 × 10−12 C 2 /N ⋅ m 2 )(480 kV/m ) = 4.25 µC/m 2

Electric Potential 211 65 •• Picture the Problem The diagram is a cross-sectional view showing the charges on the concentric spherical shells. The Gaussian surface over which we’ll integrate E in order to find V in the region r ≥ b is also shown. We’ll also find E in the region for which a < r < b. We can then

∫

use the relationship V = − Edr to find Va and Vb and their difference.

Express Vb:

b

Vb = − ∫ E r ≥ a dr ∞

Apply Gauss’s law for r ≥ b:

∫

S

r Q E r ⋅ nˆ dA = enclosed = 0

∈0

and Er≥b = 0 because Qenclosed = 0 for r ≥ b. Substitute to obtain:

b

Vb = − ∫ (0)dr = 0 ∞

Express Va:

a

Va = − ∫ Er ≥a dr b

Apply Gauss’s law for r ≥ a:

(

)

Er ≥a 4π r 2 =

q

∈0

and

Er ≥a =

q 4π ∈0 r

2

=

kq r2

a

Substitute to obtain:

Va = − kq ∫

The potential difference between the shells is given by:

⎛1 1⎞ Va − Vb = Va = kq⎜ − ⎟ ⎝a b⎠

dr kq kq = − r2 a b b

*66 ••• Picture the Problem We can find the potential relative to infinity at the center of the sphere by integrating the electric field for 0 to ∞. We can apply Gauss’s law to find the

212 Chapter 23 electric field both inside and outside the spherical shell. R

∞

0

R

The potential relative to infinity the center of the spherical shell is:

V = ∫ Er < R dr + ∫ Er > R dr

Apply Gauss’s law to a spherical surface of radius r < R to obtain:

Q ∫ E dA = E (4πr ) = ∈ r
n

S

2

(1)

inside 0

Using the fact that the sphere is uniformly charged, express Qinside in terms of Q:

Qinside r3 Q = ⇒ Q = Q inside 3 3 4 4 R3 3 πr 3 πR

Substitute for Qinside to obtain:

Er < R 4πr 2 =

Solve for Er < R:

Er < R =

Apply Gauss’s law to a spherical surface of radius r > R to obtain:

Q ∫ E dA = E (4πr ) = ∈

Solve for Er>R to obtain:

Er > R =

Substitute for ErR in equation (1) and evaluate the resulting integral:

V=

(

)

r kQ Q= 3 r 3 R 4π ∈0 R

r >R

n

S

r3 Q ∈0 R 3

2

0

=

Q ∈0

Q kQ = 2 2 4π ∈0 r r ∞

R

kQ dr r dr + kQ ∫ 2 3 ∫ R 0 r R R

∞

kQ ⎡ r 2 ⎤ 3kQ ⎡ 1⎤ = 3 ⎢ ⎥ + kQ ⎢− ⎥ = R ⎣ 2 ⎦0 2R ⎣ r ⎦R

67 •• Picture the Problem (a) The field lines are shown on the figure. The charged spheres induce charges of opposite sign on the spheres near them so that sphere 1 is negatively charged, and sphere 2 is positively charged. The total charge of the system is zero.

(b)

inside

V1 = V2 because the spheres are connected. From the direction of the electric field lines it follows that V3 > V1.

Electric Potential 213 If 3 and 4 are connected, V3 = V4 and the conditions of part (b) can (c) only be satisfied if all potentials are zero. Consequently the charge

on each sphere is zero.

General Problems 68 • Picture the Problem Because the charges at either end of the electric dipole are point charges, we can use the expression for the Coulomb potential to find the field at any distance from the dipole charges.

kq+ kq− + d d k = (q+ + q− ) d

Using the expression for the potential due to a system of point charges, express the potential at the point 9.2×10−10 m from each of the two charges:

V=

Because q+ = −q−:

q+ + q− = 0 , V = 0 and (b) is correct.

69 • Picture the Problem The potential V at any point on the x axis is the sum of the Coulomb potentials due to the two point charges. Once we have found V, we can

r

use E = −grad V to find the electric field at any point on the x axis.

kqi ri

(a) Express the potential due to a system of point charges:

V =∑

Substitute to obtain:

V ( x ) = Vcharge at + a + Vcharge at -a

i

= =

kq x +a 2

2

+

2kq x2 + a2

kq x + a2 2

214 Chapter 23 (b) The electric field at any point on the x axis is given by:

r d ⎡ 2kq ⎤ ˆ E ( x ) = −grad V = − ⎢ ⎥i dx ⎣ x 2 + a 2 ⎦

=

(x

2kqx 2

+a

)

2 32

iˆ

70 • Picture the Problem The radius of the sphere is related to the electric field and the potential at its surface. The dielectric strength of air is about 3 MV/m.

V (r ) r

Relate the electric field at the surface of a conducting sphere to the potential at the surface of the sphere:

Er =

Solve for r:

r=

When E is a maximum, r is a minimum:

rmin =

Substitute numerical values and evaluate rmin:

rmin

V (r ) Er V (r ) Emax

10 4 V = = 3.33 mm 3 MV/m

*71 •• Picture the Problem The geometry of the wires is shown to the right. The potential at the point whose coordinates are (x, y) is the sum of the potentials due to the charge distributions on the wires. (a) Express the potential at the point whose coordinates are (x, y):

V ( x, y ) = Vwire at −a + Vwire at a ⎛r ⎛r ⎞ = 2kλ ln⎜⎜ ref ⎟⎟ + 2k (− λ ) ln⎜⎜ ref ⎝ r2 ⎝ r1 ⎠ ⎡ ⎛r ⎞ ⎛ r ⎞⎤ = 2kλ ⎢ln⎜⎜ ref ⎟⎟ − ln⎜⎜ ref ⎟⎟⎥ ⎝ r2 ⎠⎦ ⎣ ⎝ r1 ⎠ ⎛r ⎞ λ ln⎜⎜ 2 ⎟⎟ = 2π ∈0 ⎝ r1 ⎠ where V(0) = 0.

⎞ ⎟⎟ ⎠

Electric Potential 215 Because r1 =

r2 =

(x + a )2 + y 2 and

⎛ λ V ( x, y ) = ln⎜ 2π ∈0 ⎜

(x − a )2 + y 2 :

⎝

(x − a )2 + y 2 ⎞⎟ (x + a )2 + y 2 ⎟⎠

⎛ a2 + y2 λ ln⎜ V (0, y ) = 2π ∈0 ⎜ a 2 + y 2 ⎝

On the y-axis, x = 0 and:

= (b) Evaluate the potential at

( 14 a,0) = (1.25 cm, 0) :

V(

λ ln (1) = 0 2π ∈0 ⎛

1 4

λ ⎜ a ,0 ) = ln 2π ∈0 ⎜⎜ ⎝

= Equate V(x,y) and V ( 14 a,0 ) :

3 = 5

Solve for y to obtain:

⎞ ⎟ ⎟ ⎠

( 14 a − a )2 ⎞⎟ ( 14 a + a )2 ⎟⎟⎠

λ ⎛ 3⎞ ln⎜ ⎟ 2π ∈0 ⎝ 5 ⎠

(x − 5)2 + y 2 (x + 5)2 + y 2

y = ± 21.25 x − x 2 − 25

A spreadsheet program to plot y = ± 21.25 x − x 2 − 25 is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell A2

Content/Formula 1.25

Algebraic Form

A3 B2

A2 + 0.05 SQRT(21.25*A2 − A2^2 − 25)

x + ∆x

y = 21.25 x − x 2 − 25

B4

−SQRT(21.25*A2 − A2^2 − 25)

y = − 21.25 x − x 2 − 25

1 4

1 2 3 4 5 6 7

A x 1.25 1.30 1.35 1.40 1.45 1.50

B y_pos 0.00 0.97 1.37 1.67 1.93 2.15

C y_neg 0.00 −0.97 −1.37 −1.67 −1.93 −2.15

370 371 372

19.65 19.70 19.75

2.54 2.35 2.15

−2.54 −2.35 −2.15

a

216 Chapter 23 373 374 375 376

19.80 19.85 19.90 19.95

−1.93 −1.67 −1.37 −0.97

1.93 1.67 1.37 0.97

The following graph shows the equipotential curve in the xy plane for

V ( 14 a,0 ) =

λ ⎛3⎞ ln⎜ ⎟ . 2π ∈0 ⎝ 5 ⎠ 10 8 6

y (cm)

4 2 0 -2 -4 -6 -8 -10 0

5

10

15

20

x (cm)

72 •• Picture the Problem We can use the expression for the potential at any point in the xy plane to show that the equipotential curve is a circle. (a) Equipotential surfaces must satisfy the condition:

Solve for r2/r1:

V=

⎛r ⎞ λ ln⎜⎜ 2 ⎟⎟ 2π ∈0 ⎝ r1 ⎠

r2 =e r1

2π ∈0 V

λ

= C or r2 = Cr1

where C is a constant. Substitute for r1 and r2 to obtain:

(x − a )2 + y 2 = C 2 [(x + a )2 + y 2 ]

Expand this expression, combine like terms, and simplify to obtain:

C 2 +1 x + 2a 2 x + y 2 = −a 2 C −1 2

⎡ ⎛ C 2 + 1 ⎞2 ⎤ ⎟⎟ ⎥ to both sides of the equation: Complete the square by adding ⎢a 2 ⎜⎜ 2 ⎢⎣ ⎝ C − 1 ⎠ ⎥⎦

Electric Potential 217 ⎡ ⎛ C 2 +1 ⎞2 ⎤ ⎡ ⎛ C 2 +1⎞2 ⎤ C 2 +1 4a 2 C 2 2 2 x + 2a 2 x + ⎢a ⎜⎜ 2 ⎟⎟ ⎥ + y = ⎢a 2 ⎜⎜ 2 ⎟⎟ ⎥ − a 2 = 2 C −1 ⎢⎣ ⎝ C − 1 ⎠ ⎥⎦ ⎢⎣ ⎝ C − 1 ⎠ ⎥⎦ C 2 −1 2

Let α = 2a

(

C 2 +1 C and β = 2a 2 2 C −1 C −1

to obtain:

(x + α )2 + y 2 = β 2 ,

)

the equation of

circle in the xy plane with its center at (−α,0).

(b) The three - dimensional surfaces are cylinders parallel to the wires. 73 •• Picture the Problem Expressing the charge dq in a spherical shell of volume 4πr2dr within a distance r of the proton and setting the integral of this expression equal to e will allow us to solve for the value of ρ 0 needed for charge neutrality. In part (b), we can use the given charge density to express the potential function due to this charge and then integrate this function to find V as a function of r. Express the charge dq in a spherical shell of volume 4πr2dr within a distance r of the proton: Express the condition for charge neutrality:

(

)(

dq = ρdV = ρ 0 e −2 r a 4πr 2 dr

)

= 4πρ 0 r 2 e −2 r a dr ∞

e = 4πρ 0 ∫ r 2 e −2 r a dr 0

Integrate by parts twice to obtain:

Solve for ρ0:

e = 4πρ 0

ρ0 =

a3 = πρ 0 a 3 4

e πa 3

74 • Picture the Problem Let Q be the sphere’s charge, R its radius, and n the number of electrons that have been removed. Then Q = ne , where e is the electronic charge. We can use the expression for the Coulomb potential of the sphere to express Q and then Q = ne to find n. Letting n be the number of electrons that have been removed, express the sphere’s charge Q in terms of the electronic charge e: Solve for n:

Q = ne

n=

Q e

(1)

218 Chapter 23 Relate the potential of the sphere to its charge and radius:

V=

kQ R

Solve for the sphere’s charge:

Q=

VR k

Substitute in equation (1) to obtain:

n=

VR ke

Substitute numerical values and evaluate n:

n=

(8.99 × 10

(400 V )(0.05 m ) 9

)(

N ⋅ m 2 /C 2 1.6 × 10 −19 C

)=

1.39 × 1010

75 • Picture the Problem We can use conservation of energy to relate the change in the kinetic energy of the particle to the change in potential energy of the charge-and-particle system as the particle moves from x = 1.5 m to x = 1 m. The change in potential energy is, in turn, related to the change in electric potential.

Apply conservation of energy to the point charge Q and particle system:

∆K + ∆U = 0 or, because Ki = 0,

K f + ∆U if = 0

Solve for Kf: Relate the difference in potential between points i and f to the change in potential energy of the system as the body whose charge is q moves from i to f: Substitute to obtain:

K f = −∆U if ∆U if = − q∆Vif = − q(Vf − Vi ) ⎛ kQ kQ ⎞ ⎛1 1⎞ ⎟⎟ = − kqQ⎜⎜ − ⎟⎟ = − q⎜⎜ − xi ⎠ ⎝ xf ⎝ xf xi ⎠

⎛1 1⎞ K f = −kqQ⎜⎜ − ⎟⎟ ⎝ xf xi ⎠

Electric Potential 219 Solve for Q:

Q=−

Kf ⎛1 1⎞ kq⎜⎜ − ⎟⎟ ⎝ xf xi ⎠

Substitute numerical values and evaluate Q:

Q=−

0.24 J ⎛ 1 1 ⎞ ⎟⎟ 8.99 × 109 N ⋅ m 2 /C 2 (4 µC ) ⎜⎜ − ⎝ 1 m 1.5 m ⎠

(

)

= − 20.0 µC

*76 •• Picture the Problem We can use the definition of power and the expression for the work done in moving a charge through a potential difference to find the minimum power needed to drive the moving belt. Relate the power need to drive the moving belt to the rate at which the generator is doing work:

P=

dW dt

W = q∆V

Express the work done in moving a charge q through a potential difference ∆V: Substitute to obtain:

P=

d [q∆V ] = ∆V dq dt dt

P = (1.25 MV )(200 µC/s ) = 250 W

Substitute numerical values and evaluate P:

77 •• Picture the Problem We can use Wq→final position = q∆Vi→ f to find the work required to move these charges between the given points. (a) Express the required work in terms of the charge being moved and the potential due to the charge at x = +a:

W+Q→+ a = Q∆V∞→+ a

= Q[V (a ) − V (∞ )] kQ 2 ⎛ kQ ⎞ = QV (a ) = Q⎜ ⎟= 2a ⎝ 2a ⎠

220 Chapter 23 (b) Express the required work in terms of the charge being moved and the potentials due to the charges at x = +a and x = −a:

W−Q→0 = −Q∆V∞→0

= −Q[V (0) − V (∞ )] = −QV (0) = −Q Vcharge at -a + Vcharge at + a

[

]

− 2kQ 2 ⎛ kQ kQ ⎞ = −Q⎜ + ⎟= a ⎠ a ⎝ a (c) Express the required work in terms of the charge being moved and the potentials due to the charges at x = +a and x = −a:

W−Q→2 a = −Q∆V0→2 a

= −Q[V (2a ) − V (0)] = −Q Vcharge at -a + Vcharge at + a − V (0)

[

]

⎛ kQ kQ 2kQ ⎞ = −Q⎜ + − ⎟ a a ⎠ ⎝ 3a =

2kQ 2 3a

78 •• Picture the Problem Let q represent the charge being moved from x = 50 cm to the origin, Q the ring charge, and a the radius of the ring. We can use Wq→final position = q∆Vi→ f , where V is the expression for the axial field due to a ring charge, to find the work required to move q from x = 50 cm to the origin. Express the required work in terms of the charge being moved and the potential due to the ring charge at x = 50 cm and x = 0:

W = q∆V = q[V (0 ) − V (0.5 m )]

The potential on the axis of a uniformly charged ring is:

V (x ) =

Evaluate V(0):

V (0 ) =

kQ x + a2 2

kQ

a2 8.99 × 109 N ⋅ m 2 /C 2 (2 nC ) = 0.1 m

(

)

= 180 V Evaluate V(0.5 m):

V (0 ) =

(8.99 × 10

9

)

N ⋅ m 2 /C 2 (2 nC )

(0.5 m )2 + (0.1 m )2

= 35.3 V

Electric Potential 221 Substitute in the expression for W to obtain:

W = (1 nC )(180 V − 35.3 V ) = 1.45 × 10 −7 J = 1.45 × 10 −7 J ×

1 eV 1.6 × 10 −19 J

= 9.06 × 1011 eV 79 •• Picture the Problem We can find the speed of the proton as it strikes the negatively charged sphere from its kinetic energy and, in turn, its kinetic energy from the potential difference through which it is accelerated. Use the definition of kinetic energy to express the speed of the proton when it strikes the negatively charged sphere: Use the work-kinetic energy theorem to relate the kinetic energy of the proton to the potential difference through which it is accelerated: Express the work done on the proton in terms of its charge e and the potential difference ∆V between the spheres: Substitute to obtain: Substitute in equation (1) to obtain:

Substitute numerical values and evaluate v:

v=

2K p

(1)

mp

W = ∆K = K f − K i or, because Ki = 0 and Kf = Kp,

W = ∆K = K p

W = e∆V

K p = e∆V v=

2e∆V mp

v=

2 1.6 × 10 −19 C (100 V ) 1.67 × 10 −27 kg

(

= 1.38 × 105 m/s

)

222 Chapter 23 80

••

Picture the Problem Equation 23-20 is V = kQ

a2 + x2 .

(a) A spreadsheet solution is shown below for kQ = a = 1. The formulas used to calculate the quantities in the columns are as follows: Cell A4 B3

Content/Formula A3 + 0.1 1/(1+A3^2)^(1/2)

Algebraic Form x + ∆x

kQ a2 + x2

A

B

1 2 3 4 5 6 7 8 9

x −5.0 −4.8 −4.6 −4.4 −4.2 −4.0 −3.8

V(x) 0.196 0.204 0.212 0.222 0.232 0.243 0.254

49 50 51 52 53

4.2 4.4 4.6 4.8 5.0

0.232 0.222 0.212 0.204 0.196

The following graph shows V as a function of x: 1.0

V (V)

0.8

0.6

0.4

0.2 -5

-4

-3

-2

-1

0

1

x (arbit rary unit s)

2

3

4

5

Electric Potential 223 (b) Examining the graph we see that the maximum value of V occurs where:

x= 0

Because E = −dV/dx, examination of the graph tells us that:

E (0 ) = 0

81 •• Picture the Problem Let R2 be the radius of the second sphere and Q1 and Q2 the charges on the spheres when they have been connected by the wire. When the spheres are connected, the charge initially on the sphere of radius R1 will redistribute until the spheres are at the same potential. Express the common potential of the spheres when they are connected:

12 kV =

kQ1 R1

(1)

and

12 kV =

Express the potential of the first sphere before it is connected to the second sphere:

20 kV =

Solve equation (1) for Q1:

Q1 =

Solve equation (2) for Q2:

Substitute in equation (3) to obtain:

Q2 =

kQ2 R2

k (Q1 + Q2 ) R1

(3)

(12 kV )R1 k

(12 kV )R2 k

⎛ (12 kV )R1 (12 kV )R2 ⎞ + k⎜ ⎟ k k ⎝ ⎠ 20 kV = R1 ⎛R ⎞ = 12 kV + 12 kV⎜⎜ 2 ⎟⎟ ⎝ R1 ⎠ or

⎛R ⎞ 8 = 12⎜⎜ 2 ⎟⎟ ⎝ R1 ⎠ Solve for R2:

(2)

R2 =

2 R1 3

224 Chapter 23 *82 •• Picture the Problem We can use the definition of surface charge density to relate the radius R of the sphere to its charge Q and the potential function V (r ) = kQ r to relate Q to the potential at r = 2 m. Use its definition, relate the surface charge density σ to the charge Q on the sphere and the radius R of the sphere:

σ=

Q 4πR 2

Solve for R to obtain:

R=

Q 4πσ

Relate the potential at r = 2.0 m to the charge on the sphere:

V (r ) =

Solve for Q to obtain:

Q=

Substitute to obtain:

R= =

kQ r

rV (r ) k

rV (r ) = 4πkσ

4π ∈0 rV (r ) 4πσ

∈0 rV (r ) σ

Substitute numerical values and evaluate R:

R=

(8.85 × 10

−12

)

C 2 /N ⋅ m 2 (2 m )(500 V ) = 0.600 m 24.6 nC/m 2

83 •• Picture the Problem We can use the definition of surface charge density to relate the radius R of the sphere to its charge Q and the potential function V (r ) = kQ r to relate Q to the potential at r = 2 m.

Q πR 2

Use its definition, relate the surface charge density σ to the charge Q on the disk and the radius R of the disk:

σ=

Solve for Q to obtain:

Q = πσR 2

(1)

Electric Potential 225 Relate the potential at r to the charge on the disk: Substitute V(0.6 m) = 80 V: Substitute V(1.5 m) = 40 V:

V (r ) = 2πkσ

(x

2

+ R2 − x

2 80 V = 2πkσ ⎛⎜ (0.6 m ) + R 2 − 0.6 m ⎞⎟ ⎝ ⎠ 2 40 V = 2πkσ ⎛⎜ (1.5 m ) + R 2 − 1.5 m ⎞⎟ ⎝ ⎠

(0.6 m )2 + R 2 − 0.6 m (1.5 m )2 + R 2 − 1.5 m

Divide the first of these equations by the second to obtain:

2=

Solve for R to obtain:

R = 0.800 m

Express the electric field on the axis of a disk charge:

⎛ x E x = 2πkσ ⎜⎜1 − x2 + R2 ⎝

Solve for σ to obtain:

σ=

σ=

⎞ ⎟ ⎟ ⎠

Ex

⎛ x 2πk ⎜⎜1 − x2 + R2 ⎝ 2 ∈0 E x = x 1− x2 + R2

Evaluate σ using R = 0.8 m and E(1.5 m) = 23.5 V/m:

)

⎞ ⎟ ⎟ ⎠

(

)

2 8.85 × 10 −12 C 2 /N ⋅ m 2 (23.5 V/m ) 1.5 m 1− (1.5 m )2 + (0.8 m )2

= 3.54 nC/m 2 Substitute in equation (1) and evaluate Q:

Q = π (3.54 nC/m 2 )(0.8 m )

2

= 7.12 nC

84 •• Picture the Problem We can use U = kq1q2/R to relate the electrostatic potential energy of the particles to their separation. Express the electrostatic potential energy of the two particles in terms of their charge and separation:

U=

kq1q2 R

226 Chapter 23 Solve for R:

R=

Substitute numerical values and evaluate R:

(8.99 × 10 R=

)

kq1q2 U

(

N ⋅ m 2 /C 2 (2)(82) 1.6 × 10 −19 C 1.6 × 10 −19 C 5.30 MeV × eV

9

)

2

= 44.6 fm

85 •• Picture the Problem We can use ∆V = E∆l and the expression for the electric field due to a plane of charge to find the potential difference between the two planes. The conducting slab introduced between the planes in part (b) will have a negative charge induced on its surface closest to the plane with the positive charge density and a positive charge induced on its other surface. We can proceed as in part (a) to find the potential difference between the planes with the conducting slab in place. (a) Express the potential difference between the two planes:

∆V = E∆l = Ed

The electric field due to each plane is:

E =

Because the charge densities are of opposite sign, the fields are additive and the resultant electric field between the planes is: Substitute to obtain:

(b) The diagram shows the conducting slab between the two planes and the electric field lines in the region between the original two planes.

σ 2 ∈0

E = E plane 1 + E plane 2 =

σ σ σ + = 2∈ 0 2∈ 0 ∈ 0

∆V =

σd ∈0

Electric Potential 227 Express the new potential difference ∆V′ between the planes in terms of the potential differences ∆V1, ∆V2 and ∆V3: Express the electric fields in regions 1, 2 and 3:

∆V ' = ∆V1 + ∆V2 + ∆V3 = E1l 1 + E2 a + E3l 2

E1 = E3 =

σ and E2 = 0 ∈0

σ σ l1 + l 2 ∈0 ∈0 σ (l + l ) = ∈0 1 2

Substitute to obtain:

∆V ' =

Express l 1 + l 2 in terms of a and d:

l1 + l 2 = d − a

Substitute to obtain:

∆V ' =

σ (d − a ) ∈0

86 ••• Picture the Problem We need to consider three regions, as in Example 23-5. Region I, x > a; region II, 0 < x < a; and region III, x < 0. We can find V in each of these regions and then find E from E = − dV dl .

dVI dx

(a) Relate EI to VI:

EI = −

In region I we have:

VI =

Substitute and evaluate EI:

EI = −

Because x > 0:

x=x

For x > a:

x−a = x−a

Substitute to obtain:

EI = − =

kq1 kq2 + x x−a

d ⎡ kq1 kq2 ⎤ + ⎢ ⎥ dx ⎣⎢ x x − a ⎦⎥

d ⎡ kq1 kq2 ⎤ + dx ⎢⎣ x x − a ⎥⎦ kq1 kq2 + 2 x (x − a )2

228 Chapter 23

Proceed as above for regions II and III to obtain:

EII =

kq1 kq2 − 2 x (x − a )2

and

(b) The distance between q1 and a point on y axis is y and the distance between a point on the y axis and q2 is

EIII = −

kq1 kq2 − 2 x (x − a )2

V (y) =

kq1 + y

kq2 y2 + a2

y 2 + a 2 . Using these distances,

express the potential at a point on the y axis: (c) To obtain the y component of r E at a point on the y axis we take the derivative of V(y). For y > 0:

For y < 0:

d ⎡ kq1 kq2 ⎤ + ⎢ ⎥ 2 2 dy ⎢ y y a + ⎥⎦ ⎣ kq1 kq2 y = + 2 y (y 2 + a 2 )3 2

Ey = −

d ⎡ kq1 kq2 ⎤ + ⎢− ⎥ 2 2 dy ⎢ y y a + ⎥⎦ ⎣ kq kq2 y = − 21 + y (y 2 + a 2 )3 2

Ey = −

These are the components of the fields due to q1 and q2 that one obtains using Coulomb' s law. *87 ••• Picture the Problem We can consider the relationship between the potential and the electric field to show that this arrangement is equivalent to replacing the plane by a point charge of magnitude −q located a distance d beneath the plane. In (b) we can first find the field at the plane surface and then use σ =∈0 E to find the surface charge density. In (c) the work needed to move the charge to a point 2d away from the plane is the product of the potential difference between the points at distances 2d and 3d from −q multiplied by the separation ∆x of these points.

Electric Potential 229 The potential anywhere on the plane is 0 in either arrangement and the (a)

electric field is perpendicular to the plane in both arrangements, so they must give the same potential everywhere in the xy plane. Also, because the net charge is zero, the potential at infinity is zero.

(b) The surface charge density is given by: At any point on the plane, the electric field points in the negative x direction and has magnitude:

σ =∈0 E

(1)

kq cos θ d + r2 where θ is the angle between the horizontal E=

2

and a vector pointing from the positive charge to the point of interest on the xz plane and r is the distance along the plane from the origin (i.e., directly to the left of the charge). Because cos θ =

d d +r 2

2

:

Substitute for E in equation (1) to obtain:

E=

kq d + r2

=

(d

kqd

=

4π ∈0 (d 2 + r 2 )

σ=

d

2

2

d + r2 2

+ r2 ) qd

32

32

(

qd

4π d + r 2 2

)

3/ 2

88 ••• Picture the Problem We can express the potential due to the ring charges as the sum of the potentials due to each of the ring charges. To show that V(x) is a minimum at x = 0, we must show that the first derivative of V(x) = 0 at x = 0 and that the second derivative is positive. In part (c) we can use a Taylor expansion to show that, for x << L, the potential is of the form V(x) = V(0) + α x2. In part (d) we can obtain the potential energy function from the potential function and, noting that it is quadratic in x, find the ″spring″ constant and the angular frequency of oscillation of the particle provided its displacement from its equilibrium position is small. (a) Express the potential due to the ring charges as the sum of the

V ( x ) = Vring to the left + Vring to the right

230 Chapter 23 potentials due to each of their charges:

V (x ) =

The potential for a ring of charge is:

kQ x + R2 2

where R is the radius of the ring and Q is the charge of the ring. For the ring to the left we have:

Vring to the left =

For the ring to the right we have:

kQ

(x + L )2 + L2 kQ

Vring to the right =

(x − L )2 + L2

Substitute to obtain:

V (x ) =

kQ

(x + L )2 + L2

kQ

+

(x − L )2 + L2

(b) Evaluate dV/dx to obtain:

⎧⎪ dV L−x = kQ⎨ dx ⎪⎩ (L − x )2 + L2

[

− 3 2

⎫⎪ = 0 for extrema 3 2⎬ (L + x )2 + L2 ⎪⎭

] [

L+x

]

x=0

Solve for x to obtain: Evaluate d2V/dx2 to obtain: 2 ⎧⎪ 3(L − x ) d 2V kQ = ⎨ dx 2 ⎪⎩ (L − x )2 + L2

[

]

5 2

−

[(L − x )

2

+ L2 −

Evaluate this expression for x = 0 to obtain:

3(L + x )

2

1

]

3 2

+

[(L + x )

2

[(L − x )

1

2

]

2 3 2

+L

+ L2

]

5 2

⎫⎪ ⎬ ⎪⎭

d 2V (0 ) kQ = >0 2 dx 2 2 L3 Hence V ( x) is a maximum at x = 0.

Electric Potential 231 (c) The Taylor expansion of V(x) is:

V (x ) = V (0) + V ' (0)x + 12 V ' ' (0)x 2 + higher order terms

For x << L:

V ( x ) ≈ V (0 ) + V ' (0)x + 12 V ' ' (0 )x 2

Substitute our results from part (b) to obtain:

V (x ) =

2kQ ⎛ kQ ⎞ 2 + (0)x + 12 ⎜ ⎟x 3 L ⎝ 2 2L ⎠ 2kQ kQ + x2 3 L 4 2L

= or

V ( x ) = V (0 ) + αx 2

where

V (0 ) =

(d) Express the angular frequency of oscillation of a simple harmonic oscillator: From our result for part (c) and the definition of electric potential:

Substitute for k′ in the expression for ω:

89 ••• Picture the Problem The diagram shows part of the shells in a cross-sectional view under the conditions of part (a) of the problem. We can use Gauss’s law to find the electric field in the regions defined by the three surfaces and then find the electric potentials from the electric fields. In part (b) we can use the redistributed charges to find the charge on and potentials of the three surfaces.

ω=

2kQ kQ and α = L 4 2 L3

k' m

where k′ is the restoring constant.

1 ⎛ kqQ ⎞ 2 U ( x ) = qV (0 ) + ⎜ ⎟x 2 ⎝ 2 2 L3 ⎠ = qV (0 ) + 12 k ' x 2 kqQ where k ' = 2 2 L3

ω=

kqQ 2m 2 L3

232 Chapter 23

(a) Apply Gauss’s law to a spherical Gaussian surface of radius r ≥ c to obtain: Because Er(c) = 0: Apply Gauss’s law to a spherical Gaussian surface of radius b < r < c to obtain:

(

)

Er 4π r 2 =

Qenclosed

∈0

and Er = 0 because the net charge enclosed by the Gaussian surface is zero.

V (c ) = 0

(

)

Er 4π r 2 =

Q

∈0

and

Er (b < r < c ) = Use Er (b < r < c ) to find the potential difference between c and b:

=0

kQ r2 b

V (b ) − V (c ) = − kQ ∫ c

dr r2

⎛1 1⎞ = kQ⎜ − ⎟ ⎝b c⎠ Because V(c) = 0:

⎛1 1⎞ V (b ) = kQ⎜ − ⎟ ⎝b c⎠

The inner shell carries no charge, so the field between r = a and r = b is zero and:

⎛1 1⎞ V (a ) = V (b ) = kQ⎜ − ⎟ ⎝b c⎠

(b) When the inner and outer shells are connected their potentials become equal as a consequence of the redistribution of charge.

The charges on surfaces a and c are related according to:

Qa + Qc = −Q

(1)

Electric Potential 233 Qb does not change with the connection of the inner and outer shells: Express the potentials of shells a and c: In the region between the r = a and r = b, the field is kQa/r2 and the potential at r = b is then:

Qb = Q

V (a ) = V (c ) = 0

⎛1 1⎞ V (b ) = kQa ⎜ − ⎟ ⎝b a⎠

(2)

k (Qa + Q ) r2

The enclosed charge for b < r < c is Qa + Q, and by Gauss’s law the field in this region is:

Eb < r < c =

Express the potential difference between b and c:

⎛1 1⎞ V (c ) − V (b ) = k (Qa + Q ) ⎜ − ⎟ ⎝c b⎠ = −V (b ) because V(c) = 0.

Solve for V(b) to obtain:

⎛1 1⎞ V (b ) = k (Qa + Q ) ⎜ − ⎟ ⎝b c⎠

Equate equations (2) and (3) and solve for Qa to obtain:

Qa = − Q

a(c − b ) b(c − a )

(4)

Substitute equation (4) in equation (1) and solve for Qc to obtain:

Qc = − Q

c(b − a ) b(c − a )

(5)

Substitute (4) and (5) in (3) to obtain:

V (b ) = kQ

(c − b )(b − a ) b 2 (c − a )

(3)

234 Chapter 23 *90 ••• Picture the Problem The diagram shows a cross-sectional view of a portion of the concentric spherical shells. Let the charge on the inner shell be q. The dashed line represents a spherical Gaussian surface r over which we can integrate E ⋅ nˆ dA in order to find Er for r ≥ b. We can find V(b) from the integral of Er between r = ∞ and r = b. We can obtain a second expression for V(b) by considering the potential difference between a and b and solving the two equations simultaneously for the charge q on the inner shell.

(

)

Apply Gauss’s law to a spherical surface of radius r ≥ b:

Er 4πr 2 =

Solve for Er to obtain:

Er =

Use Er to find V(b):

Equate these expressions for V(b) to obtain: Solve for q to obtain:

ε0

k (Q + q ) r2 b

V (b ) = −k (Q + q )∫ =

We can also determine V(b) by considering the potential difference between a, i.e., 0 and b:

Q+q

k (Q + q ) b

dr r2 ∞

⎛1 1⎞ V (b ) = kq⎜ − ⎟ ⎝b a⎠ k (Q + q ) ⎛1 1⎞ = ka⎜ − ⎟ b ⎝b a⎠ a q= − Q b

91 ••• Picture the Problem We can use the hint to derive an expression for the electrostatic potential energy dU required to bring in a layer of charge of thickness dr and then integrate this expression from r = 0 to R to obtain an expression for the required work.

Electric Potential 235 If we build up the sphere in layers, then at a given radius r the net charge on the sphere will be given by:

⎛ r ⎞3 Q(r) = Q⎜ ⎟ ⎝ R⎠

When the radius of the sphere is r, the potential relative to infinity is:

V (r ) =

Q(r ) Q r2 = 4π ∈0 r 4π ∈0 R 3

dW = dU = V (r )dQ

Express the work dW required to bring in charge dQ from infinity to the surface of a uniformly charged sphere of radius r:

Integrate dW from 0 to R to obtain:

=

Q r2 ⎛ 3Q ⎞ dr ⎟ 4πr 2 3 ⎜ 3 4π ∈0 R ⎝ 4πR ⎠

=

3Q 2 r 4 dr 4π ∈0 R 6 R

3Q 2 W =U = r 4 dr 6 ∫ 4π ∈0 R 0 R

3Q 2 ⎡ r 5 ⎤ 3Q 2 = = ⎢ ⎥ 20π ∈0 R 4π ∈0 R 6 ⎣ 5 ⎦ 0 92 •• Picture the Problem We can equate the rest energy of an electron and the result of Problem 91 in order to obtain an expression that we can solve for the classical electron radius. From Problem 91 we have:

U=

3e 2 20π ∈0 R

The rest mass of the electron is given by:

E0 = m0 c 2

Equate these energies to obtain:

3e 2 = m0 c 2 20π ∈0 R

Solve for R:

R=

3e 2 20π ∈0 m0 c 2

Substitute numerical values and evaluate R:

(

)

2

3 1.6 × 10 −19 C R= 20π 8.85 × 10 −12 C 2 / N ⋅ m 2 5.11 × 10 5 eV 1.6 × 10 −19 J/eV

(

= 1.69 × 10 −15 m

)(

)(

)

236 Chapter 23 This model does not explain how the electron holds together against its own mutual repulsion. 93 •• Picture the Problem Because the post-fission volumes of the fission products are equal, we can express the post-fission radii in terms of the radius of the pre-fission sphere. (a) Relate the initial volume V of the uniformly charged sphere to the volumes V′ of the fission products: Substitute for V and V ′:

V = 2V '

4 3

πR 3 = 2( 43 πR'3 ) 1 R = 0.794 R 3 2

Solve for and evaluate R′:

R' =

(b) Express the difference ∆E in the total electrostatic energy as a result of fissioning:

∆E = E − E '

From Problem 91 we have:

After fissioning:

E=

3Q 2 20π ∈0 R

⎡ ⎤ 2 1 ⎢ ⎥ ⎛ 3Q' ⎞ 3( 2 Q ) ⎟⎟ = 2 ⎢ E ' = 2⎜⎜ ⎥ ⎢ 20π ∈0 1 R ⎥ ⎝ 20π ∈0 R' ⎠ 3 2 ⎥⎦ ⎣⎢ 2

= Substitute for E and E′ to obtain:

3

2 ⎛ 3Q 2 ⎞ ⎜ ⎟ = 0.630 E 2 ⎜⎝ 20π ∈0 R ⎟⎠

∆E = E − 0.630 E = 0.370 E

*94 ••• Picture the Problem We can use the definition of density to express the radius R of a nucleus as a function of its atomic mass N. We can then use the result derived in Problem 91 to express the electrostatic energies of the 235U nucleus and the nuclei of the fission fragments 140Xe and 94Sr. The energy released by this fission process is: Express the mass of a nucleus in terms of its density and volume:

∆E = U 235 U − (U 140 Xe + U 94 Sr ) Nm = 43 ρπR 3 where N is the nuclear number.

(1)

Electric Potential 237 Solve for R to obtain:

Substitute numerical values and evaluate R as a function of N:

R=3

3Nm 4πρ

R=3

3 1.660 × 10 −27 kg 1 3 N 4π 4 × 1017 kg/m 3

(

(

(

) )

)

= 9.97 × 10 −16 m N 1 3

(

)

RU = 9.97 × 10 −16 m (235)

The 'radius' of the 235U nucleus is therefore:

13

= 6.15 × 10 −15 m

From Problem 91 we have:

U=

3Q 2 20π ∈0 R

Substitute numerical values and evaluate the electrostatic energy of the 235U nucleus:

(

)

2

3 92 × 1.6 × 10 −19 C 20π 8.85 × 10 −12 C 2 / N ⋅ m 2 6.15 × 10 −15 m 1 eV = 1.91 × 10 −10 J × = 1189 MeV 1.6 × 10 −19 J/eV

U 235 U =

(

)(

)

Proceed as above to find the electrostatic energy of the fission fragments 140Xe and 94Sr:

U 140 Xe

(

and

U 94 Sr

(

)

2

3 54 × 1.6 × 10 −19 C = 20π 8.85 × 10 −12 C 2 / N ⋅ m 2 6.15 × 10 −15 m 1 eV = 6.57 × 10 −11 J × = 410 MeV 1.6 × 10 −19 J/eV

(

)(

)

)

2

3 38 × 1.6 × 10 −19 C = 20π 8.85 × 10 −12 C 2 / N ⋅ m 2 6.15 × 10 −15 m 1 eV = 3.25 × 10 −11 J × = 203 MeV 1.602 × 10 −19 J/eV

(

Substitute for U 235 U , U 140 Xe , and

U 94 Sr in equation (1) and evaluate

)(

)

∆E = 1189 MeV − (410 MeV + 203 MeV ) = 576 MeV

∆E: 95 ••• Picture the Problem The geometry of the point charge and the sphere is shown below. The charge is a distance R away from the center of a spherical shell of radius a.

238 Chapter 23

kdq kσdA =∫ sphere r sphere r

(a) The average potential over the surface of the sphere is given by:

Vav = ∫

Substitute for k, σ, and dA to obtain:

Vav =

1 4π ∈0

q (2πa sin θ )(adθ ) 4πa 2 r

π

∫ 0

Apply the law of cosines to the triangle to obtain:

r = R 2 + a 2 − 2aR cos θ

Substitute for r and simplify to obtain:

Vav =

q 8π ∈0

π

∫ (R

sin θdθ 2

+ a 2 − 2aR cos θ

2

+ a − 2aRu

0

and

−q Vav = 8π ∈0

−1

∫ (R 1

du 2

)

Then dv = − βdu and:

∫ (R 1

2

+ a 2 − 2aRu

12

(1)

α = R 2 + a 2 , β = 2aR , and v = α − βu

To simplify the integrand, let:

du

12

du = − sin θdθ

Change variables by letting u = cosθ. Then:

−1

)

)

12

=− =−

Substitute for α and β to obtain:

1

β

l2

∫

l1

1 aR

l2 −1 dv 2 1 v =− α − βu =− 1 l1 aR β v

[ α +β −

α −β

]

Electric Potential 239 −1

∫ (R 1

du 2

+ a − 2aRu 2

)

12

=−

1 aR

[ R + a + 2aR − 2

2

R 2 + a 2 − 2aR

]

1 ⎡ (R + a )2 − (R − a )2 ⎤⎥ ⎢ ⎣ ⎦ aR 1 =− [(R + a ) − (R − a )] = − 2 aR R

=−

Substitute in equation (1) to obtain:

Vav =

−q ⎛ 2 ⎞ q ⎜− ⎟ = 8π ∈0 ⎝ R ⎠ 4π ∈0 R

Note that this result is the potential at the center of the sphere due to the point charge. The superposition principle tells us that the potential at any point is the (b)

sum of the potentials due to any charge distributions in space. Because this result is independent of any properties of the sphere, this result must hold for any sphere and any configuration of charges outside of it.

240 Chapter 23

Chapter 24 Electrostatic Energy and Capacitance Conceptual Problems *1 • Determine the Concept The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, and the electrical properties of the matter between them. The capacitance is, therefore, independent of the voltage across the capacitor. (c ) is correct. 2 • Determine the Concept The capacitance of a parallel-plate capacitor is a function of the surface area of its plates, the separation of these plates, and the electrical properties of the matter between them. The capacitance is, therefore, independent of the charge of the capacitor. (c) is correct. 3 • Determine the Concept True. The energy density of an electrostatic field is given by ue = 12 ∈0 E 2 . 4 • Picture the Problem The energy stored in the electric field of a parallel-plate capacitor is related to the potential difference across the capacitor by U = 12 QV . Relate the potential energy stored in the electric field of the capacitor to the potential difference across the capacitor:

U = 12 QV

With Q constant, U is directly proportional to V . Hence, doubling V doubles U . *5 •• Picture the Problem The energy stored in a capacitor is given by U = 12 QV and the capacitance of a parallel-plate capacitor by C = ∈0 A d . We can combine these relationships, using the definition of capacitance and the condition that the potential difference across the capacitor is constant, to express U as a function of d. Express the energy stored in the capacitor:

U = 12 QV

241

242 Chapter 24 Use the definition of capacitance to express the charge of the capacitor:

Q = CV

Substitute to obtain:

U = 12 CV 2

Express the capacitance of a parallel-plate capacitor in terms of the separation d of its plates:

C=

Substitute to obtain:

Because U ∝

∈0 A d

where A is the area of one plate.

U=

∈0 AV 2 2d

1 , doubling the d

separation of the plates will reduce the energy stored in the capacitor to 1/2 its previous value:

(d ) is correct.

6 •• Picture the Problem Let V represent the initial potential difference between the plates, U the energy stored in the capacitor initially, d the initial separation of the plates, and V ′, U ′, and d ′ these physical quantities when the plate separation has been doubled. We can use U = 12 QV to relate the energy stored in the capacitor to the potential difference across it and V = Ed to relate the potential difference to the separation of the plates. Express the energy stored in the capacitor before the doubling of the separation of the plates: Express the energy stored in the capacitor after the doubling of the separation of the plates:

U = 12 QV

U' = 12 QV' because the charge on the plates does not change.

Express the ratio of U′ to U:

U' V' = U V

Express the potential differences across the capacitor plates before and after the plate separation in terms of the electric field E between the plates:

V = Ed and

V' = Ed' because E depends solely on the charge on the plates and, as observed above, the

Electrostatic Energy and Capacitance 243 charge does not change during the separation process. Substitute to obtain:

U' Ed' d' = = U Ed d

For d ′ = 2d:

U ' 2d = = 2 and (b) is correct U d

7 • Determine the Concept Both statements are true. The total charge stored by two capacitors in parallel is the sum of the charges on the capacitors and the equivalent capacitance is the sum of the individual capacitances. Two capacitors in series have the same charge and their equivalent capacitance is found by taking the reciprocal of the sum of the reciprocals of the individual capacitances. 8 •• (a) False. Capacitors connected in series carry the same charge. (b) False. The voltage across the capacitor whose capacitance is C0 is Q/C0 and that across the second capacitor is Q/2C0. (c) False. The energy stored by the capacitor whose capacitance is C0 is Q 2 2C0 and the energy stored by the second capacitor is Q 2 4C0 . (d) True 9 • Determine the Concept True. The capacitance of a parallel-plate capacitor filled with a dielectric of constant κ is given by C =

κ ∈0 A d

or C ∝ κ.

*10 •• Picture the Problem We can treat the configuration in (a) as two capacitors in parallel and the configuration in (b) as two capacitors in series. Finding the equivalent capacitance of each configuration and examining their ratio will allow us to decide whether (a) or (b) has the greater capacitance. In both cases, we’ll let C1 be the capacitance of the dielectric-filled capacitor and C2 be the capacitance of the air capacitor. In configuration (a) we have:

Ca = C1 + C2

244 Chapter 24 Express C1 and C2:

C1 =

κ ∈0 A1 d1

=

κ ∈0 12 A d

=

κ ∈0 A 2d

and

C2 =

∈0 A2 ∈0 12 A ∈0 A = = d2 d 2d

κ ∈0 A ∈0 A ∈0 A

(κ + 1)

Substitute for C1 and C2 and simplify to obtain:

Ca =

In configuration (b) we have:

1 1 1 C1C2 = + ⇒ Cb = Cb C1 C2 C1 + C2

Express C1 and C2:

C1 = and

C2 =

Substitute for C1 and C2 and simplify to obtain:

Divide Cb by Ca:

Because

4κ < 1 for κ > 1: (κ + 1)2

2d

+

2d

=

2d

∈0 A1 ∈0 A 2 ∈0 A = 1 = d1 d d 2

κ ∈0 A2 d2

=

κ ∈0 A 1 2

d

=

2κ ∈0 A d

⎛ 2 ∈0 A ⎞ ⎛ 2κ ∈0 A ⎞ ⎜ ⎟⎜ ⎟ d ⎠⎝ d ⎝ ⎠ Cb = 2 ∈0 A 2κ ∈0 A + d d ⎛ 2 ∈0 A ⎞ ⎛ 2κ ∈0 A ⎞ ⎜ ⎟⎜ ⎟ d ⎠⎝ d ⎝ ⎠ = 2 ∈0 A (κ + 1) d 2 ∈0 A ⎛ κ ⎞ = ⎜ ⎟ d ⎝ κ + 1⎠ 2 ∈0 A ⎛ κ ⎞ ⎜ ⎟ 4κ Cb d ⎝ κ + 1⎠ = = 2 ∈0 A Ca (κ + 1) (κ + 1) 2d Ca > Cb

Electrostatic Energy and Capacitance 245 11 • (a) False. The capacitance of a parallel-plate capacitor is defined to be the ratio of the charge on the capacitor to the potential difference across it. (b) False. The capacitance of a parallel-plate capacitor depends on the area of its plates A, their separation d, and the dielectric constant κ of the material between the plates according to C = κ ∈0 A d . (c) False. As in part (b), the capacitance of a parallel-plate capacitor depends on the area of its plates A, their separation d, and the dielectric constant κ of the material between the plates according to C = κ ∈0 A d . 12 •• Picture the Problem We can use the expression U = 12 CV 2 to express the ratio of the energy stored in the single capacitor and in the identical-capacitors-in-series combination. Express the energy stored in capacitors when they are connected to the 100-V battery:

U = 12 CeqV 2

Express the equivalent capacitance of the two identical capacitors connected in series:

Ceq =

C2 1 = 2C 2C

Substitute to obtain:

U=

( 12 C )V 2 = 14 CV 2

Express the energy stored in one capacitor when it is connected to the 100-V battery:

U 0 = 12 CV 2

Express the ratio of U to U0:

1 2

1 U CV 2 1 = 14 = U 0 2 CV 2 2

or

U = 12 U 0 and (d ) is correct

Estimation and Approximation 13 •• Picture the Problem The outer diameter of a "typical" coaxial cable is about 5 mm, while the inner diameter is about 1 mm. From Table 24-1 we see that a reasonable range of values for κ is 3-5. We can use the expression for the capacitance of a

246 Chapter 24 cylindrical capacitor to estimate the capacitance per unit length of a coaxial cable. The capacitance of a cylindrical dielectric-filled capacitor is given by:

C=

2πκ ∈0 L ⎛R ⎞ ln⎜⎜ 2 ⎟⎟ ⎝ R1 ⎠

where L is the length of the capacitor, R1 is the radius of the inner conductor, and R2 is the radius of the second (outer) conductor.

C 2πκ ∈0 κ = = L ⎛R ⎞ ⎛R ⎞ ln⎜⎜ 2 ⎟⎟ 2k ln⎜⎜ 2 ⎟⎟ ⎝ R1 ⎠ ⎝ R1 ⎠

Divide both sides by L to obtain an expression for the capacitance per unit length of the cable: If κ = 3:

C = L

3 ⎛ 2.5 mm ⎞ ⎟⎟ 2 8.99 × 10 N ⋅ m / C ln⎜⎜ ⎝ 0.5 mm ⎠

(

9

2

2

)

= 0.104 nF/m

If κ = 5:

C = L

5 ⎛ 2.5 mm ⎞ ⎟⎟ 2(8.99 × 10 N ⋅ m / C )ln⎜⎜ ⎝ 0.5 mm ⎠ 9

A reasonable range of values for C/L, corresponding to 3 ≤ κ ≤ 5, is:

2

= 0.173 nF/m

2

0.104 nF/m ≤

C ≤ 0.173 nF/m L

*14 •• Picture the Problem The energy stored in a capacitor is given by U = 12 CV 2 . Relate the energy stored in a capacitor to its capacitance and the potential difference across it:

U = 12 CV 2

Solve for C:

C=

The potential difference across the spark gap is related to the width of the gap d and the electric field E in the gap:

V = Ed

2U V2

Electrostatic Energy and Capacitance 247 Substitute for V in the expression for C to obtain:

C=

Substitute numerical values and evaluate C:

C=

2U E 2d 2

(3 × 10

2(100 J )

6

V/m ) (0.001m ) 2

2

= 22.2 µF 15 •• Picture the Problem Because ∆R << RE we can treat the atmosphere as a flat slab with an area equal to the surface area of the earth. Then the energy stored in the atmosphere can be estimated from U = uV, where u is the energy density of the atmosphere and V is its volume. Express the electric energy stored in the atmosphere in terms of its energy density and volume:

U = uV

Because ∆R << RE = 6370 km, we can consider the volume:

V = Asurface area of the earth ∆R

Express the energy density of the Earth’s atmosphere in terms of the average magnitude of its electric field:

u = 12 ∈0 E 2

Substitute for V and u to obtain:

U=

Substitute numerical values and evaluate U:

U=

= 4π RE2 ∆R

(

1 2

∈0 E 2 )(4π RE2 ∆R ) =

RE2 E 2 ∆R 2k

(6370 km )2 (200 V/m )2 (1km ) 2(8.99 × 109 N ⋅ m 2 /C 2 )

= 9.03 × 1010 J 16 •• Picture the Problem We’ll approximate the balloon by a sphere of radius R = 3 m and use the expression for the capacitance of an isolated spherical conductor. Relate the capacitance of an isolated spherical conductor to its radius:

C = 4π ∈0 R =

R k

248 Chapter 24 Substitute numerical values and evaluate C:

C=

3m = 0.334 nF 8.99 × 109 N ⋅ m 2 /C 2

U=

kq1q2 kq1q3 kq2 q3 + + r1, 2 r1,3 r2,3

Electrostatic Potential Energy 17 • The electrostatic potential energy of this system of three point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram. Express the work required to assemble this system of charges:

⎛qq qq q q ⎞ = k ⎜⎜ 1 2 + 1 3 + 2 3 ⎟⎟ r1,3 r2,3 ⎠ ⎝ r1, 2 Find the distances r1,2, r1,3, and r2,3:

r1, 2 = 3 m, r2,3 = 3 m, and r1,3 = 6 m

(a) Evaluate U for q1 = q2 = q3 = 2 µC:

⎡ (2 µC )(2 µC ) (2 µC )(2 µC ) (2 µC )(2 µC )⎤ + + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ ⎥ 3m 6m 3m ⎣ ⎦

(

)

= 30.0 mJ (b) Evaluate U for q1 = q2 = 2 µC and q3 = −2 µC:

⎡ (2 µC )(2 µC ) (2 µC )(− 2 µC ) (2 µC )(− 2 µC ) ⎤ + + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ ⎥ 3m 6m 3m ⎣ ⎦

(

)

= − 5.99 mJ (c) Evaluate U for q1 = q3 = 2 µC and q2 = −2 µC:

⎡ (2 µC )(− 2 µC ) (2 µC )(2 µC ) (− 2 µC )(2 µC ) ⎤ U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ + + ⎥ 3m 6m 3m ⎣ ⎦

(

= − 18.0 mJ

)

Electrostatic Energy and Capacitance 249 18 • Picture the Problem The electrostatic potential energy of this system of three point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram.

Express the work required to assemble this system of charges:

U=

kq1q2 kq1q3 kq2 q3 + + r1, 2 r1,3 r2,3

⎛qq qq q q ⎞ = k ⎜⎜ 1 2 + 1 3 + 2 3 ⎟⎟ r1,3 r2,3 ⎠ ⎝ r1, 2 r1, 2 = r2,3 = r1,3 = 2.5 m

Find the distances r1,2, r1,3, and r2,3: (a) Evaluate U for q1 = q2 = q3 = 4.2 µC:

⎡ (4.2 µC )(4.2 µC ) (4.2 µC )(4.2 µC ) + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ 2.5 m 2.5 m ⎣ (4.2 µC)(4.2 µC)⎤ + ⎥ 2.5 m ⎦

(

)

= 0.190 J (b) Evaluate U for q1 = q2 = 4.2 µC and q3 = −4.2 µC:

⎡ (4.2 µC )(4.2 µC ) (4.2 µC )(− 4.2 µC ) U = 8.988 × 109 N ⋅ m 2 /C 2 ⎢ + 2.5 m 2.5 m ⎣ (4.2 µC)(− 4.2 µC)⎤ + ⎥ 2.5 m ⎦

(

)

= − 63.4 mJ (c) Evaluate U for q1 = q2 = −4.2 µC and q3 = +4.2 µC:

250 Chapter 24

⎡ (− 4.2 µC )(− 4.2 µC ) (− 4.2 µC )(4.2 µC ) + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ 2.5 m 2.5 m ⎣ (− 4.2 µC)(4.2 µC)⎤ + ⎥ 2.5 m ⎦

(

)

= − 63.4 mJ *19 • Picture the Problem The potential of an isolated spherical conductor is given by V = kQ r ,where Q is its charge and r its radius, and its electrostatic potential energy by U = 12 QV . We can combine these relationships to find the sphere’s electrostatic potential energy. Express the electrostatic potential energy of the isolated spherical conductor as a function of its charge Q and potential V:

U = 12 QV

Express the potential of the spherical conductor:

V =

kQ r

Solve for Q to obtain:

Q=

rV k

Substitute to obtain:

rV 2 ⎛ rV ⎞ U = 12 ⎜ ⎟V = 2k ⎝ k ⎠

Substitute numerical values and evaluate U:

U=

(

(0.1m )(2 kV )2

2 8.99 × 109 N ⋅ m 2 /C 2

= 22.2 µJ 20 •• Picture the Problem The electrostatic potential energy of this system of four point charges is the work needed to bring the charges from an infinite separation to the final positions shown in the diagram. In part (c), depending on the configuration of the positive and negative charges, two energies are possible.

)

Electrostatic Energy and Capacitance 251 Express the work required to assemble this system of charges:

U=

kq1q2 kq1q3 kq1q4 kq2 q3 kq2 q4 kq3q4 + + + + + r1, 2 r1,3 r1, 4 r2,3 r2, 4 r3, 4

⎛qq qq qq q q q q q q ⎞ = k ⎜⎜ 1 2 + 1 3 + 1 4 + 2 3 + 2 4 + 3 4 ⎟⎟ r1,3 r1, 4 r2,3 r2, 4 r3, 4 ⎠ ⎝ r1, 2 Find the distances r1,2, r1,3, r1,4, r2,3, r2,4, and r3,4,:

r1, 2 = r2,3 = r3, 4 = r1, 4 = 4 m and

r1,3 = r2, 4 = 4 2 m (a) Evaluate U for q1 = q2 = q3 = q4 = −2 µC:

⎡ (− 2 µC )(− 2 µC ) (− 2 µC )(− 2 µC ) (− 2 µC )(− 2 µC ) + + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ 4m 4m 4 2m ⎣ (− 2 µC)(− 2 µC) + (− 2 µC)(− 2 µC) + (− 2 µC)(− 2 µC)⎤ + ⎥ 4m 4m 4 2m ⎦

(

)

= 48.7 mJ (b) Evaluate U for q1 = q2 = q3 = 2 µC and q4 = −2 µC:

⎡ (2 µC )(2 µC ) (2 µC )(2 µC ) (2 µC )(− 2 µC ) + + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ 4m 4m 4 2m ⎣ (2 µC)(2 µC) + (2 µC)(− 2 µC) + (2 µC)(− 2 µC)⎤ + ⎥ 4m 4m 4 2m ⎦

(

)

= 0 (c) Let q1 = q2 = 2 µC and q3 = q4 = −2 µC:

⎡ (2 µC )(2 µC ) (2 µC )(− 2 µC ) (2 µC )(− 2 µC ) + + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ 4m 4m 4 2m ⎣ (2 µC)(− 2 µC) + (2 µC)(− 2 µC) + (− 2 µC)(− 2 µC)⎤ + ⎥ 4m 4m 4 2m ⎦

(

)

= − 12.7 mJ Let q1 = q3 = 2 µC and q2 = q4 = −2 µC:

252 Chapter 24 ⎡ (2 µC )(− 2 µC ) (2 µC )(2 µC ) (2 µC )(− 2 µC ) + + U = 8.99 × 109 N ⋅ m 2 /C 2 ⎢ 4m 4m 4 2m ⎣ (− 2 µC)(2 µC) + (− 2 µC)(− 2 µC) + (2 µC)(− 2 µC)⎤ + ⎥ 4m 4m 4 2m ⎦

(

)

= − 23.2 mJ 21 •• Picture the Problem The diagram shows the four charges fixed at the corners of the square and the fifth charge that is released from rest at the origin. We can use conservation of energy to relate the initial potential energy of the fifth particle to its kinetic energy when it is at a great distance from the origin and the electrostatic potential at the origin to express Ui. Use conservation of energy to relate the initial potential energy of the particle to its kinetic energy when it is at a great distance from the origin: Express the initial potential energy of the particle to its charge and the electrostatic potential at the origin: Substitute for Kf and Ui to obtain:

∆K + ∆U = 0 or, because Ki = Uf = 0,

K f −U i = 0

U i = qV (0 )

1 2

mv 2 − qV (0) = 0 2qV (0) m

Solve for v:

v=

Express the electrostatic potential at the origin:

V (0 ) =

Substitute and simplify to obtain:

kq 2kq − 3kq 6kq + + + 2a 2a 2a 2a 6kq = 2a

v=

2q ⎛ 6kq ⎞ 6 2k ⎜ ⎟= q m ⎝ 2a ⎠ ma

Electrostatic Energy and Capacitance 253

Capacitance *22 • Picture the Problem The charge on the spherical conductor is related to its radius and potential according to V = kQ/r and we can use the definition of capacitance to find the capacitance of the sphere. (a) Relate the potential V of the spherical conductor to the charge on it and to its radius:

V =

kQ r

Solve for and evaluate Q:

Q=

rV k

=

(b) Use the definition of capacitance to relate the capacitance of the sphere to its charge and potential:

C=

(0.1m )(2 kV )

8.99 × 109 N ⋅ m 2 /C 2

= 22.2 nC

Q 22.2 nC = = 11.1pF V 2 kV

(c) It doesn' t. The capacitance of a sphere is a function of its radius. 23 • Picture the Problem We can use its definition to find the capacitance of this capacitor. Use the definition of capacitance to obtain:

C=

Q 30 µC = = 75.0 nF V 400 V

24 •• Picture the Problem Let the separation of the spheres be d and their radii be R. Outside the two spheres the electric field is approximately the field due to point charges of +Q and −Q, each located at the centers of spheres, separated by distance d. We can derive an expression for the potential at the surface of each sphere and then use the potential difference between the spheres and the definition of capacitance and to find the capacitance of the two-sphere system. The capacitance of the two-sphere system is given by:

C=

Q ∆V

where ∆V is the potential difference between the spheres.

254 Chapter 24 The potential at any point outside the two spheres is:

V=

k (+ Q ) k (− Q ) + r1 r2

where r1 and r2 are the distances from the given point to the centers of the spheres. For a point on the surface of the sphere with charge +Q: Substitute to obtain:

For δ << d:

r1 = R and r2 = d + δ where δ < R

V+ Q =

k (+ Q ) k (− Q ) + R d +δ

V+ Q =

kQ kQ − R d

and

V− Q = − The potential difference between the spheres is:

kQ kQ + R d

∆V = VQ − V− Q kQ kQ ⎛ − kQ kQ ⎞ − −⎜ + ⎟ R d ⎝ R d ⎠ ⎛1 1⎞ = 2kQ⎜ − ⎟ ⎝R d⎠ =

Q 2π ∈0 = ⎛1 1⎞ ⎛1 1⎞ 2kQ⎜ − ⎟ ⎜ − ⎟ ⎝R d⎠ ⎝R d⎠ 2π ∈0 R = R 1− d

Substitute for ∆V in the expression for C to obtain:

C=

For d very large:

C = 2π ∈0 R

The Storage of Electrical Energy 25 • Picture the Problem Of the three equivalent expressions for the energy stored in a charged capacitor, the one that relates U to C and V is U = 12 CV 2 . (a) Express the energy stored in the capacitor as a function of C and V:

U = 12 CV 2

Electrostatic Energy and Capacitance 255 Substitute numerical values and evaluate U: (b) Express the additional energy required as the difference between the energy stored in the capacitor at 200 V and the energy stored at 100 V:

U=

1 2

(3 µF)(100 V )2 =

15.0 mJ

∆U = U (200 V ) − U (100 V ) =

1 2

(3 µF)(200 V )2 − 15.0 mJ

= 45.0 mJ

26 • Picture the Problem Of the three equivalent expressions for the energy stored in a charged capacitor, the one that relates U to Q and C is U =

1 Q2 . 2 C

1 Q2 2 C

(a) Express the energy stored in the capacitor as a function of C and Q:

U=

Substitute numerical values and evaluate U:

1 (4 µC ) U= = 0.800 µJ 2 10 µF

(b) Express the energy remaining when half the charge is removed:

1 (2 µC ) U ( Q) = = 0.200 µJ 2 10 µF

2

2

1 2

27 • Picture the Problem Of the three equivalent expressions for the energy stored in a

1 Q2 . charged capacitor, the one that relates U to Q and C is U = 2 C 1 Q2 2 C

(a) Express the energy stored in the capacitor as a function of C and Q:

U=

Substitute numerical values and evaluate U:

1 (5 µC ) U (5 µC ) = = 0.625 J 2 20 pF

(b) Express the additional energy required as the difference between the energy stored in the capacitor when its charge is 5 µC and when its charge is 10 µC:

2

∆U = U (10 µC ) − U (5 µC ) 1 (10 µC ) = − 0.625 J 2 20 pF = 2.50 J − 0.625 J 2

= 1.88 J

256 Chapter 24 *28 • Picture the Problem The energy per unit volume in an electric field varies with the square of the electric field according to u = ∈0 E 2 2 . Express the energy per unit volume in an electric field:

u = 12 ∈0 E 2

Substitute numerical values and evaluate u:

u=

(8.85 × 10

1 2

−12

)

C 2 /N ⋅ m 2 (3 MV/m )

2

= 39.8 J/m 3

29 • Picture the Problem Knowing the potential difference between the plates, we can use E = V/d to find the electric field between them. The energy per unit volume is given by u = 12 ∈0 E 2 and we can find the capacitance of the parallel-plate capacitor using

C =∈0 A d . V d 100 V = = 100 kV/m 1 mm

(a) Express the electric field between the plates in terms of their separation and the potential difference between them:

E=

(b) Express the energy per unit volume in an electric field:

u = 12 ∈0 E 2

Substitute numerical values and evaluate u:

u=

(c) The total energy is given by:

1 2

(8.85 × 10

−12

)

C 2 /N ⋅ m 2 (100 kV/m )

2

= 44.3 mJ/m3 U = uV = uAd

(

)(

)

= 44.3 mJ/m3 2 m 2 (1 mm ) = 88.6 µJ

(d) The capacitance of a parallel-plate capacitor is given by:

∈0 A d 8.85 × 10 −12 C 2 /N ⋅ m 2 2 m 2 = 1mm

C=

(

= 17.7 nF

)(

)

Electrostatic Energy and Capacitance 257 (e) The total energy is given by:

U = 12 CV 2

Substitute numerical values and evaluate U:

U=

1 2

(17.7 nF)(100 V )2

= 88.5 µJ, in agreement with (c).

30 •• Picture the Problem The total energy stored in the electric field is the product of the energy density in the space between the spheres and the volume of this space. (a) The total energy U stored in the electric field is given by:

The energy density of the field is:

U = uV where u is the energy density and V is the volume between the spheres.

u=

1 ∈0 E 2 2

where E is the field between the spheres. The volume between the spheres is approximately: Substitute for u and V to obtain: The magnitude of the electric field between the concentric spheres is the sum of the electric fields due to each charge distribution: Because the two surfaces are so close together, the electric field between them is approximately the sum of the fields due to two plane charge distributions: Substitute for σQ to obtain:

Substitute for E in equation (1) and simplify:

V ≈ 4πr12 (r2 − r1 ) U = 2π ∈0 E 2 r12 (r2 − r1 )

(1)

E = EQ + E− Q

E=

E≈

σQ 2 ∈0

+

σ −Q 2 ∈0

=

σQ ∈0

Q 4π r12 ∈0 2

⎛ Q ⎞ 2 ⎟⎟ r1 (r2 − r1 ) U = 2π ∈0 ⎜⎜ 2 4 π r ∈ 1 0 ⎠ ⎝ 2 Q r2 − r1 = 8π ∈0 r12

258 Chapter 24 Substitute numerical values and evaluate U:

U=

(5 nC)2 (10.5 cm − 10.0 cm ) 2 8π (8.85 × 10−12 C 2 / N ⋅ m 2 )(10.0 cm )

(b) The capacitance of the twosphere system is given by:

C=

= 56.0 nJ

Q ∆V

where ∆V is the potential difference between the two spheres. The electric potentials at the surfaces of the spheres are:

V1 =

Q Q and V2 = 4π ∈0 r1 4π ∈0 r2

Substitute for ∆V and simplify to obtain:

C=

Q rr = 4π ∈0 1 2 Q Q r2 − r1 − 4π ∈0 r1 4π ∈0 r2

Substitute numerical values and evaluate C:

(

C = 4π 8.85 × 10−12 C 2 / N ⋅ m 2

Use ½ Q2/C to find the total energy stored in the electric field between the spheres:

10.0 cm )(10.5 cm ) ) (10 = .5 cm − 10.0 cm

0.234 nF

2 1 ⎡ (5 nC ) ⎤ U= ⎢ ⎥ = 53.4 nJ 2 ⎣ 0.234 nF ⎦

Note that our approximate result in (a ) is within 5% of our exact result obtained in (b). *31 •• Picture the Problem We can relate the charge Q on the positive plate of the capacitor to the charge density of the plate σ using its definition. The charge density, in turn, is related to the electric field between the plates according to σ =∈0 E and the electric field can be found from E = ∆V/∆d. We can use ∆U = 12 Q∆V in part (b) to find the increase in the energy stored due to the movement of the plates. (a) Express the charge Q on the positive plate of the capacitor in terms of the plate’s charge density σ and surface area A:

Q = σA

Electrostatic Energy and Capacitance 259 Relate σ to the electric field E between the plates of the capacitor:

σ =∈0 E

Express E in terms of the change in V as the plates are separated a distance ∆d:

E=

Substitute for σ and E to obtain:

Q =∈0 EA =∈0 A

∆V ∆d

∆V ∆d

Substitute numerical values and evaluate Q:

Q = (8.85 × 10−12 C2 /N ⋅ m 2 )(500 cm 2 )

(b) Express the change in the electrostatic energy in terms of the change in the potential difference: Substitute numerical values and evaluate ∆U:

100 V = 11.1nC 0.4 cm

∆U = 12 Q∆V

∆U =

1 2

(11.1 nC )(100 V ) =

0.553 µJ

32 ••• Picture the Problem By symmetry, the electric field must be radial. In part (a) we can find Er both inside and outside the ball by choosing a spherical Gaussian surface first inside and then outside the surface of the ball and applying Gauss’s law. (a) Relate the electrostatic energy density at a distance r from the center of the ball to the electric field due to the uniformly distributed charge Q: Relate the flux through the Gaussian surface to the electric field Er on the Gaussian surface at r < R: Using the fact that the charge is uniformly distributed, express the ratio of the charge enclosed by the Gaussian surface to the total charge of the sphere:

ue = 12 ∈0 E 2

Er (4π r 2 ) =

(1)

Qinside

∈0

Qinside ρVGaussian surface = Q ρVball =

4 3 4 3

π r3 r3 = π R3 R3

(2)

260 Chapter 24 r3 R3

Solve for Qinside to obtain:

Qinside = Q

Substitute in equation (2):

Er (4πr 2 ) =

Solve for Er < R:

Er < R =

Substitute in equation (1) to obtain:

Qr 3 ∈0 R 3

Qr 4π ∈0 R

kQ r R3

⎛ kQ ⎞ ue (r < R ) = ∈0 ⎜ 3 r ⎟ ⎝R ⎠

Er (4π r 2 ) =

∈ 0 k 2Q 2 2R

Qinside

6

=

∈0

r2

Q

∈0

Q = kQr − 2 4π r 2 ∈ 0

Solve for Er > R:

Er >R =

Substitute in equation (1) to obtain:

ue (r > R ) = 12 ∈0 kQr −2

(

= (b) Express the energy dU in a spherical shell of thickness dr and surface area 4π r2: For r < R:

2

1 2

=

Relate the flux through the Gaussian surface to the electric field Er on the Gaussian surface at r > R:

=

3

1 2

)

2

∈ 0 k 2Q 2 r − 4

dU shell = 4π r 2u (r ) dr

⎛ ∈ k 2Q 2 ⎞ dU shell (r < R ) = 4πr 2 ⎜⎜ 0 6 r 2 ⎟⎟dr ⎠ ⎝ 2R kQ 2 4 r dr = 2R6

For r > R:

(

= (c) Express the total electrostatic energy:

)

dU shell (r > R ) = 4πr 2 12 ∈0 k 2Q 2 r −4 dr 1 2

kQ 2 r − 2 dr

U = U (r < R ) + U (r > R )

(3)

Electrostatic Energy and Capacitance 261 Integrate Ushell(r < R) from 0 to R:

U shell (r < R ) = =

R

kQ 2 4 r dr 2 R 6 ∫0 kQ 2 10 R ∞

Integrate Ushell(r > R) from R to ∞:

kQ 2 U shell (r > R ) = kQ ∫ r dr = 2R R

Substitute in equation (3) to obtain:

U=

1 2

2

−2

kQ 2 kQ 2 3kQ 2 + = 10 R 2 R 5R

The field inside the shell is zero, so the first integral vanishes. The result is greater for the sphere because it includes the field energy within the sphere.

Combinations of Capacitors 33 • Picture the Problem We can apply the properties of capacitors connected in parallel to determine the number of 1.0-µF capacitors connected in parallel it would take to store a total charge of 1 mC with a potential difference of 10 V across each capacitor. Knowing that the capacitors are connected in parallel (parts (a) and (b)) we determine the potential difference across the combination. In part (c) we can use our knowledge of how potential differences add in a series circuit to find the potential difference across the combination and the definition of capacitance to find the charge on each capacitor. Q (a) Express the number of n= q capacitors n in terms of the charge q on each and the total charge Q: Relate the charge q on one capacitor to its capacitance C and the potential difference across it:

q = CV

Substitute to obtain:

n=

Q CV

Substitute numerical values and evaluate n:

n=

1mC = 100 (1µF)(10 V )

262 Chapter 24 (b) Because the capacitors are connected in parallel the potential difference across the combination is the same as the potential difference across each of them: (c) With the capacitors connected in series, the potential difference across the combination will be the sum of the potential differences across the 100 capacitors: Use the definition of capacitance to find the charge on each capacitor:

Vparallel combination = V = 10 V

Vseries combination = 100V

= 100(10 V ) = 1.00 kV

q = CV = (1 µF)(10 V ) = 10.0 µC

34 • Picture the Problem The capacitor array is shown in the diagram. We can find the equivalent capacitance of this combination by first finding the equivalent capacitance of the 3.0-µF and 6.0-µF capacitors in series and then the equivalent capacitance of this capacitor with the 8.0-µF capacitor in parallel. Express the equivalent capacitance for the 3.0-µF and 6.0-µF capacitors in series: Solve for C3+6: Find the equivalent capacitance of a 2-µF capacitor in parallel with an 8µF capacitor:

1 1 1 = + C3+6 3 µF 6 µF

C3+6 = 2 µF C2+8 = 2 µF + 8 µF = 10 µF

*35 • Picture the Problem Because we’re interested in the equivalent capacitance across terminals a and c, we need to recognize that capacitors C1 and C3 are in series with each other and in parallel with capacitor C2. Find the equivalent capacitance of C1 and C3 in series:

1 1 1 = + C1+3 C1 C3

Electrostatic Energy and Capacitance 263

C1C3 C1 + C3

Solve for C1+3:

C1+3 =

Find the equivalent capacitance of C1+3 and C2 in parallel:

Ceq = C2 + C1+3 = C2 +

C1C3 C1 + C3

36 • Picture the Problem Because the capacitors are connected in parallel we can add their capacitances to find the equivalent capacitance of the combination. Also, because they are in parallel, they have a common potential difference across them. We can use the definition of capacitance to find the charge on each capacitor. (a) Find the equivalent capacitance of the two capacitors in parallel:

Ceq = 10.0 µF + 20 µF = 30 µF

(b) Because capacitors in parallel have a common potential difference across them:

V = V10 + V20 = 6.00 V

(c) Use the definition of capacitance to find the charge on each capacitor:

Q10 = C10V = (10 µF)(6 V ) = 60.0 µC and

Q20 = C20V = (20 µF)(6 V ) = 120 µC

37 •• Picture the Problem We can use the properties of capacitors in series to find the equivalent capacitance and the charge on each capacitor. We can then apply the definition of capacitance to find the potential difference across each capacitor. Q10 = Q20 = CeqV (a) Because the capacitors are connected in series they have equal charges: Express the equivalent capacitance of the two capacitors in series:

1 1 1 = + Ceq 10 µF 20 µF

Solve for Ceq to obtain:

Ceq =

Substitute to obtain:

Q10 = Q20 = (6.67 µF)(6 V ) = 40.0 µC

(10 µF)(20 µF) = 6.67 µF 10 µF + 20 µF

264 Chapter 24 (b) Apply the definition of capacitance to find the potential difference across each capacitor:

Q10 40.0 µC = = 4.00 V 10 µF C10

V10 = and

Q20 40.0 µC = = 2.00 V 20 µF C20

V20 =

*38 •• Picture the Problem We can use the properties of capacitors connected in series and in parallel to find the equivalent capacitances for various connection combinations. (a) If their capacitance is to be a maximum, they must be connected in parallel. Find the capacitance of each capacitor:

Ceq = 3C = 15 µF and

C = 5 µF

(b) (1) Connect the three capacitors in series:

1 3 = and Ceq = 1.67 µF Ceq 5 µF

Ceq, two in parallel = 2(5 µF) = 10 µF

(2) Connect two in parallel, with the third in series with that combination:

and

Solve for Ceq:

Ceq =

(3) Connect two in series, with the third in parallel with that combination:

1 1 1 = + Ceq 10 µF 5 µF

(10 µF)(5 µF) = 10 µF + 5 µF

1 Ceq, two in series

=

3.33 µF

2 5 µF

or

Ceq, two in series = 2.5 µF

Find the capacitance equivalent to 2.5 µF and 5 µF in parallel:

Ceq = 2.5 µF + 5 µF = 7.50 µF

39 •• Picture the Problem We can use the properties of capacitors connected in series and in parallel to find the equivalent capacitance between the terminals and these properties and the definition of capacitance to find the charge on each capacitor.

Electrostatic Energy and Capacitance 265 (a) Relate the equivalent capacitance of the two capacitors in series to their individual capacitances:

1 C4+15

=

1

1 4 µF 15 µF +

(4 µF)(15 µF) = 3.16 µF

Solve for C4+15:

C4 +15 =

Find the equivalent capacitance of C4+15 in parallel with the 12-µF capacitor:

Ceq = 3.16 µF + 12 µF = 15.2 µF

(b) Using the definition of capacitance, express and evaluate the charge stored on the 12-µF capacitor: Because the capacitors in series have the same charge:

4 µF + 15 µF

Q12 = C12V12 = C12V

= (12 µF)(200 V ) = 2.40 mC

Q4 = Q15 = C4 +15V

= (3.16 µF)(200 V ) = 0.632 mC

(c) The total energy stored is given by:

1 U total = CeqV 2 2

Substitute numerical values and evaluate Utotal:

U total =

1 (15.2 µF)(200 V )2 = 0.304 J 2

40 •• Picture the Problem We can use the properties of capacitors in series to establish the results called for in this problem. (a) Express the equivalent capacitance of two capacitors in series: Solve for Ceq by taking the reciprocal of both sides of the equation to obtain:

1 1 1 C2 + C1 = + = Ceq C1 C2 C1C2

Ceq =

C1C2 C1 + C2

266 Chapter 24 (b) Divide numerator and denominator of this expression by C1 to obtain:

C2 < C2 C 1+ 2 C1 C because 1 + 2 > 1 . C1

Divide numerator and denominator of this expression by C2 to obtain:

C1 < C1 C1 1+ C2 C because 1 + 1 > 1 . C2

Using our result from part (a) for two of the capacitors, add a third capacitor C3 in series to obtain:

Take the reciprocal of both sides of the equation to obtain:

Ceq =

Ceq =

C + C2 1 1 = 1 + Ceq C1C2 C3 =

Ceq =

C1C3 + C2C3 + C1C2 C1C2C3 C1C2C3 C1C2 + C2C3 + C1C3

41 •• Picture the Problem Let Ceq1 represent the equivalent capacitance of the parallel combination and Ceq the total equivalent capacitance between the terminals. We can use the equations for capacitors in parallel and then in series to find Ceq. Because the charge on Ceq is the same as on the 0.3-µF capacitor and Ceq1, we’ll know the charge on the 0.3µF capacitor when we have found the total charge Qeq stored by the circuit. We can find the charges on the 1.0-µF and 0.25-µF capacitors by first finding the potential difference across them and then using the definition of capacitance.

(a) Find the equivalent capacitance for the parallel combination:

Ceq1 = 1 µF + 0.25 µF = 1.25 µF

Electrostatic Energy and Capacitance 267 The 0.30-µF capacitor is in series with Ceq1 … find their equivalent capacitance Ceq:

1 1 1 = + Ceq 0.3 µF 1.25 µF and Ceq = 0.242 µF

(b) Express the total charge stored by the circuit Qeq:

Qeq = Q0.3 = Q1.25 = CeqV = (0.242 µF )(10 V ) = 2.42 µC

The 1-µF and 0.25-µF capacitors, being in parallel, have a common potential difference. Express this potential difference in terms of the 10 V across the system and the potential difference across the 0.3µF capacitor: Using the definition of capacitance, find the charge on the 1-µF and 0.25µF capacitors:

V1.25 = 10 V − V0.3 = 10 V −

Q0.3 C0.3

= 10 V −

2.42 µC 0.3 µF

= 1.93 V Q1 = C1V1 = (1 µF)(1.93 V ) = 1.93 µC and

Q0.25 = C0.25V0.25 = (0.25 µF)(1.93 V ) = 0.483 µC

(c) The total stored energy is given by:

U = 12 CeqV 2

Substitute numerical values and evaluate U:

U=

1 2

(0.242 µF)(10 V )2 =

12.1 µJ

42 •• Picture the Problem Note that there are three parallel paths between a and b. We can find the equivalent capacitance of the capacitors connected in series in the upper and lower branches and then find the equivalent capacitance of three capacitors in parallel. (a) Find the equivalent capacitance of the series combination of capacitors in the upper and lower branch:

1 1 1 = + Ceq C0 C0 or Ceq =

C02 = 12 C0 2C0

268 Chapter 24 Now we have two capacitors with capacitance C0/2 in parallel with a capacitor whose capacitance is C0. Find their equivalent capacitance:

C'eq = 12 C0 + C0 + 12 C0 = 2C0

(b) If the central capacitance is 10C0, then:

C'eq = 12 C0 + 10C0 + 12 C0 = 11C0

43 •• Picture the Problem Place four of the capacitors in series. Then the potential across each is 100 V when the potential across the combination is 400 V. The equivalent capacitance of the series is 2/4 µF = 0.5 µF. If we place four such series combinations in parallel, as shown in the circuit diagram, the total capacitance between the terminals is 2 µF.

*44 •• Picture the Problem We can connect two capacitors in parallel, all three in parallel, two in series, three in series, two in parallel in series with the third, and two in series in parallel with the third. Connect 2 in parallel to obtain:

Ceq = 1 µF + 2 µF = 3 µF or

Ceq = 1 µF + 4 µF = 5 µF or

Ceq = 2 µF + 4 µF = 6 µF Connect all three in parallel to obtain:

Ceq = 1 µF + 2 µF + 4 µF = 7 µF

Connect two in series:

Ceq = or

Ceq = or

(1 µF)(2 µF) = 1 µF + 2 µF

(1 µF)(4 µF) = 1 µF + 4 µF

2 µF 3 4 µF 5

Electrostatic Energy and Capacitance 269 Ceq =

(2 µF)(4 µF) = 2 µF + 4 µF

4 µF 3

Connect all three in series:

Ceq =

(1µF)(2 µF)(4 µF) = (1 µF)(2 µF) + (2 µF)(4 µF) + (1 µF)(4 µF)

Connect two in parallel, in series with the third:

Ceq = or

Ceq = or

Connect two in series, in parallel with the third:

(4 µF)(1 µF + 2 µF) = 1 µF + 2 µF + 4 µF

(1 µF)(4 µF + 2 µF) = 1 µF + 2 µF + 4 µF

Ceq =

(2 µF)(4 µF + 1 µF) =

Ceq =

(1µF)(2 µF) + 4 µF =

or

Ceq = or

Ceq =

45 ••• Picture the Problem Let C be the capacitance of each capacitor in the ladder and let Ceq be the equivalent capacitance of the infinite ladder less the series capacitor in the first rung. Because the capacitance is finite and non-zero, adding one more stage to the ladder will not change the capacitance of the network. The capacitance of the two capacitor combination shown to the right is the equivalent of the infinite ladder, so it has capacitance Ceq also.

4 µF 7

1 µF + 2 µF + 4 µF

1 µF + 2 µF

(4 µF)(2 µF) + 1 µF = 4 µF + 2 µF

(1µF)(4 µF) + 2 µF = 1 µF + 4 µF

12 µF 7 6 µF 7 10 µF 7 14 µF 3 7 µF 3 14 µF 5

270 Chapter 24 (a) The equivalent capacitance of the parallel combination of C and Ceq is:

C + Ceq

The equivalent capacitance of the series combination of C and (C + Ceq) is Ceq, so:

1 1 1 = + Ceq C C + Ceq

Simply this expression to obtain a quadratic equation in Ceq:

Ceq2 + CCeq − C 2 = 0

Solve for the positive value of Ceq to obtain:

⎛ 5 −1⎞ ⎟ C = 0.618C Ceq = ⎜⎜ ⎟ 2 ⎝ ⎠

Because C = 1 µF:

Ceq = 0.618 µF

(b) The capacitance C′ required so that the combination has the same capacitance as the infinite ladder is:

C' = C + Ceq

Substitute for Ceq and evaluate C′:

C' = C + 0.618C = 1.618C

Because C = 1 µF:

C' = 1.618 µF

Parallel-Plate Capacitors 46 • Picture the Problem The potential difference V across a parallel-plate capacitor, the electric field E between its plates, and the separation d of the plates are related according to V = Ed. We can use this relationship to find Vmax corresponding to dielectric breakdown and the definition of capacitance to find the maximum charge on the capacitor. (a) Express the potential difference V across the plates of the capacitor in terms of the electric field between the plates E and their separation d: Vmax corresponds to Emax:

(b) Using the definition of capacitance, find the charge Q

V = Ed

Vmax = (3 MV/m )(1.6 mm ) = 4.80 kV

Q = CVmax = (2.0 µF)(4.80 kV ) = 9.60 mC

Electrostatic Energy and Capacitance 271 stored at this maximum potential difference: 47 • Picture the Problem The potential difference V across a parallel-plate capacitor, the electric field E between its plates, and the separation d of the plates are related according to V = Ed. In part (b) we can use the definition of capacitance and the expression for the capacitance of a parallel-plate capacitor to find the required plate radius. (a) Express the potential difference V across the plates of the capacitor in terms of the electric field between the plates E and their separation d: Substitute numerical values and evaluate V: (b) Use the definition of capacitance to relate the capacitance of the capacitor to its charge and the potential difference across it: Express the capacitance of a parallel-plate capacitor:

V = Ed

(

)

V = 2 × 10 4 V/m (2 mm ) = 40.0 V

C=

C=

Q V

∈0 A ∈0 π R 2 =

d

d

where R is the radius of the circular plates. Equate these two expressions for C:

∈0 π R 2 d

Solve for R to obtain:

R=

Substitute numerical values and evaluate R:

R=

=

Q V

Qd ∈0 πV

(10 µC )(2 mm )

π (8.85 × 10−12 C 2 /N ⋅ m 2 )(40 V )

= 4.24 m 48 •• Picture the Problem We can use the expression for the capacitance of a parallel-plate capacitor to find the area of each plate and the definition of capacitance to find the potential difference when the capacitor is charged to 3.2 µC. We can find the stored energy using U = 12 CV 2 and the definition of capacitance and the relationship between

272 Chapter 24 the potential difference across a parallel-plate capacitor and the electric field between its plates to find the charge at which dielectric breakdown occurs. Recall that Emax, air = 3 MV/m. (a) Relate the capacitance of a parallel-plate capacitor to the area A of its plates and their separation d:

C=

Solve for A:

A=

∈0 A d

Cd

∈0

(0.14 µF)(0.5 mm)

Substitute numerical values and evaluate A:

A=

(b) Using the definition of capacitance, express and evaluate the potential difference across the capacitor when it is charged to 3.2 µC:

V =

(c) Express the stored energy as a function of the capacitor’s capacitance and the potential difference across it:

U = 12 CV 2

Substitute numerical values and evaluate U:

U=

(d) Using the definition of capacitance, relate the charge on the capacitor to breakdown potential difference: Relate the maximum potential difference to the maximum electric field between the plates: Substitute to obtain: Substitute numerical values and evaluate Qmax:

8.85 × 10−12 C 2 /N ⋅ m 2

= 7.91m 2

Q 3.2 µC = = 22.9 V C 0.14 µF

1 2

(0.14 µF)(22.9 V )2 =

36.7 µJ

Qmax = CVmax

Vmax = Emax d

Qmax = CEmax d Qmax = (0.14 µF)(3 MV/m )(0.5 mm ) = 210 µC

Electrostatic Energy and Capacitance 273 *49 •• Picture the Problem The potential difference across the capacitor plates V is related to their separation d and the electric field between them according to V = Ed. We can use this equation with Emax = 3 MV/m to find dmin. In part (b) we can use the expression for the capacitance of a parallel-plate capacitor to find the required area of the plates. (a) Use the relationship between the potential difference across the plates and the electric field between them to find the minimum separation of the plates:

d min =

(b) Use the expression for the capacitance of a parallel-plate capacitor to relate the capacitance to the area of a plate:

C=

Solve for A:

A=

Substitute numerical values and evaluate A:

A=

V 1000 V = = 0.333 mm Emax 3 MV/m

∈0 A d

Cd ∈0

(0.1 µF)(0.333 mm) 8.85 × 10-12 C 2 /N ⋅ m 2

= 3.76 m 2

Cylindrical Capacitors 50 • Picture the Problem The capacitance of a cylindrical capacitor is given by C = 2πκ ∈0 L ln (r2 r1 ) where L is its length and r1 and r2 the radii of the inner and outer conductors. (a) Express the capacitance of the coaxial cylindrical shell:

C=

Substitute numerical values and evaluate C:

C=

2πκ ∈0 L ⎛r⎞ ln⎜ ⎟ ⎝R⎠

(

)

2π (1) 8.85 × 10−12 C 2 /N ⋅ m 2 (0.12 m ) ⎛ 1.5 cm ⎞ ⎟⎟ ln⎜⎜ ⎝ 0.2 mm ⎠

= 1.55 pF

274 Chapter 24 (b) Use the definition of capacitance to express the charge per unit length:

λ=

Q CV = L L

Substitute numerical values and evaluate λ:

λ=

(1.55 pF)(1.2 kV ) = 0.12 m

15.5 nC/m

51 •• Picture the Problem The diagram shows a partial cross-sectional view of the inner wire and the outer cylindrical shell. By symmetry, the electric field is radial in the space between the wire and the concentric cylindrical shell. We can apply Gauss’s law to cylindrical surfaces of radii r < R1, R1 < r < R2, and r > R2 to find the electric field and, hence, the energy density in these regions. (a) Apply Gauss’s law to a cylindrical surface of radius r < R1 and length L to obtain:

Er (2π rL ) =

Qinside =0 ∈0

and

Er < R1 = 0 Because E = 0 for r < R1:

ur < R1 = 0

Apply Gauss’s law to a cylindrical surface of radius R1 < r < R2 and length L to obtain:

Er (2π rL ) =

Qinside λL = ∈0 ∈0

where λ is the linear charge density.

λL 2kλ = 2π ∈0 r r

Solve for Er to obtain:

Er =

Express the energy density in the region R1 < r < R2:

⎛ 2kλ ⎞ u = ∈0 E = ∈0 ⎜ ⎟ ⎝ r ⎠ 1 2

2 r

2

1 2

2k 2 ∈0 Q 2 ⎛ 2kQ ⎞ = ∈0 ⎜ ⎟ = r 2 L2 ⎝ rL ⎠ 2

1 2

Electrostatic Energy and Capacitance 275 Apply Gauss’s law to a cylindrical surface of radius r > R2 and length L to obtain:

Er (2π rL ) =

Qinside =0 ∈0

and

Er > R2 = 0 Because E = 0 for r > R2:

(b) Express the energy residing in a cylindrical shell between the conductors of radius r, thickness dr, and volume 2π rL dr:

ur > R2 = 0

dU = 2π rLu (r )dr ⎛ 2k 2 ∈0 Q 2 ⎞ kQ 2 ⎟ = dr dr = 2π rL⎜⎜ 2 2 ⎟ r L rL ⎠ ⎝

dr kQ 2 ⎛ R2 ⎞ ln⎜ ⎟ = ∫r L ⎜⎝ R1 ⎟⎠ R1

R2

(c) Integrate dU from r = R1 to R2 to obtain:

kQ 2 U= L

Use U = 12 CV 2 and the expression

U = 12 CV 2

for the capacitance of a cylindrical capacitor to obtain:

Q2 Q2 = = C ⎞ ⎛ ⎟ ⎜ 2π ∈0 L ⎟ ⎜ 2 R2 ⎟ ⎜ ⎜ ln R ⎟ 1 ⎠ ⎝ 1 2

=

kQ 2 ⎛ R2 ⎞ ln⎜ ⎟ L ⎜⎝ R1 ⎟⎠

in agreement with the result from part (b). 52 ••• Picture the Problem Note that with the innermost and outermost cylinders connected together the system corresponds to two cylindrical capacitors connected in parallel. We can use C =

2π ∈0 κL to express the capacitance per unit length and then calculate and ln (Ro Ri )

add the capacitances per unit length of each of the cylindrical shell capacitors. Relate the capacitance of a cylindrical capacitor to its length L and inner and outer radii Ri and Ro: Divide both sides of the equation by L to express the capacitance per unit

C=

2π ∈0 κL ln (Ro Ri )

C 2π ∈0 κ = L ln (Ro Ri )

276 Chapter 24 length: Express the capacitance per unit length of the cylindrical system:

C ⎛C ⎞ ⎛C ⎞ +⎜ ⎟ =⎜ ⎟ L ⎝ L ⎠ outer ⎝ L ⎠inner

Find the capacitance per unit length of the outer cylindrical shell combination:

2π 8.85 × 10−12 C 2 /N ⋅ m 2 (1) ⎛C ⎞ = ⎜ ⎟ ln (0.8 cm 0.5 cm ) ⎝ L ⎠ outer = 118.3 pF/m

Find the capacitance per unit length of the inner cylindrical shell combination:

2π 8.85 × 10−12 C 2 /N ⋅ m 2 (1) ⎛C ⎞ = ⎜ ⎟ ln (0.5 cm 0.2 cm ) ⎝ L ⎠inner = 60.7 pF/m

Substitute in equation (1) to obtain:

(1)

(

)

(

)

C = 118.3 pF/m + 60.7 pF/m L = 179 pF/m

*53 •• Picture the Problem We can use the expression for the capacitance of a parallelplate capacitor of variable area and the geometry of the figure to express the capacitance of the goniometer.

∈0 ( A − ∆A) d

The capacitance of the parallel-plate capacitor is given by:

C=

The area of the plates is:

A = π R22 − R12

If the top plate rotates through an angle ∆θ, then the area is reduced by:

∆A = π R22 − R12

Substitute for A and ∆A in the expression for C to obtain:

(

(

C= =

) 2θπ = (R

− R12

2 2

) ∆2πθ = (R

2 2

)θ2

− R12

) ∆2θ

∈0 ⎡ 2 ∆θ ⎤ θ R2 − R12 − R22 − R12 ⎢ d ⎣ 2 2 ⎥⎦

(

) (

(

)

∈0 R22 − R12 (θ − ∆θ ) 2d

)

Electrostatic Energy and Capacitance 277 54 •• Picture the Problem Let C be the capacitance of the capacitor when the pressure is P and C′ be the capacitance when the pressure is P + ∆P. We’ll assume that (a) the change in the thickness of the plates is small, and (b) the total volume of material between the plates is conserved. We can use the expression for the capacitance of a dielectric-filled parallel-plate capacitor and the definition of Young’s modulus to express the change in the capacitance ∆C of the given capacitor when the pressure on its plates is increased by ∆P. Express the change in capacitance resulting from the decrease in separation of the capacitor plates by ∆d: Because the volume is constant:

∆C = C' − C =

κ ∈0 A' d − ∆d

−

κ ∈0 A d

A'd' = Ad or

⎛ d ⎞ ⎛d⎞ A' = ⎜ ⎟ A = ⎜ ⎟A ⎝ d − ∆d ⎠ ⎝ d' ⎠ Substitute for A′ in the expression for ∆C and simplify to obtain:

d ⎞ κ ∈0 A ⎟− ⎜ d − ∆d ⎝ d − ∆d ⎠ d κ ∈0 A κ ∈0 A = d2 − 2 d d (d − ∆d )

∆C =

κ ∈0 A ⎛

⎤ d2 − 1⎥ ⎢ 2 d ⎣ (d − ∆d ) ⎦ 2 ⎡ d ⎤ = C⎢ − 1⎥ 2 ⎣ (d − ∆d ) ⎦ =

κ ∈0 A ⎡

From the definition of Young’s modulus:

∆d P ⎛P⎞ = − ⇒ ∆d = −⎜ ⎟d d Y ⎝Y ⎠

Substitute for ∆d in the expression for ∆C to obtain:

⎤ ⎡ ⎥ ⎢ 2 κ ∈0 A ⎢ d ∆C = − 1⎥ 2 ⎥ ⎢ d ⎧ P ⎫ ⎥ ⎢ ⎨d + ⎛⎜ ⎞⎟d ⎬ ⎥⎦ ⎢⎣ ⎩ ⎝ Y ⎠ ⎭ ⎡ ⎧ ⎛ P ⎞ ⎫ −2 ⎤ = C ⎢⎨1 + ⎜ ⎟⎬ − 1⎥ ⎥⎦ ⎢⎣⎩ ⎝ Y ⎠⎭

⎛ ⎝

Expand ⎜1 − obtain:

−2

P⎞ ⎟ binomially to Y⎠

⎛ P⎞ ⎜1 − ⎟ ⎝ Y⎠

−2

2

= 1− 2

P ⎛P⎞ + 3⎜ ⎟ + ... Y ⎝Y ⎠

278 Chapter 24 −2

Provided P << Y:

⎛ P⎞ ⎜1 − ⎟ ⎝ Y⎠

Substitute in the expression for ∆C and simplify to obtain:

P ⎤ P ⎡ ∆C = C ⎢1 − 2 − 1⎥ = − 2 C Y ⎦ Y ⎣

≈ 1− 2

P Y

Spherical Capacitors *55 •• Picture the Problem We can use the definition of capacitance and the expression for the potential difference between charged concentric spherical shells to show that

C = 4π ∈0 R1 R2 (R2 − R1 ).

Q V

(a) Using its definition, relate the capacitance of the concentric spherical shells to their charge Q and the potential difference V between their surfaces:

C=

Express the potential difference between the conductors:

⎛1 1 ⎞ R − R1 V = kQ⎜⎜ − ⎟⎟ = kQ 2 R1R2 ⎝ R1 R2 ⎠

Substitute to obtain:

C=

=

(b) Because R2 = R1 + d:

Q R1R2 = R − R1 k (R2 − R1 ) kQ 2 R1R2 4π ∈0 R1R2 R2 − R1

R1R2 = R1 (R1 + d ) = R12 + R1d ≈ R12 = R 2 because d is small.

Substitute to obtain:

C≈

4π ∈0 R 2 ∈ A = 0 d d

Electrostatic Energy and Capacitance 279 56 •• Picture the Problem The diagram shows a partial cross-sectional view of the inner and outer spherical shells. By symmetry, the electric field is radial. We can apply Gauss’s law to spherical surfaces of radii r < R1, R1 < r < R2, and r > R2 to find the electric field and, hence, the energy density in these regions. (a) Apply Gauss’s law to a spherical surface of radius r < R1 to obtain:

(

)

Er 4π r 2 =

Qinside =0 ∈0

and

Er < R1 = 0 Because E = 0 for r < R1:

ur < R1 = 0

Apply Gauss’s law to a spherical surface of radius R1 < r < R2 to obtain:

Er 4π r 2 =

Solve for Er to obtain:

Er =

Express the energy density in the region R1 < r < R2:

⎛ kQ ⎞ u = ∈0 E = ∈0 ⎜ 2 ⎟ ⎝r ⎠

Apply Gauss’s law to a cylindrical surface of radius r > R2 to obtain:

(

)

Q kQ = 2 2 4π ∈0 r r 2 r

1 2

1 2

=

k 2 ∈0 Q 2 2r 4

(

)

Er 4π r 2 = and

Er > R2 = 0 Because E = 0 for r > R2:

Qinside Q = ∈0 ∈0

ur > R2 = 0

Qinside =0 ∈0

2

280 Chapter 24 (b) Express the energy in the electrostatic field in a spherical shell of radius r, thickness dr, and volume 4π r2dr between the conductors:

⎛ k 2 ∈0 Q 2 ⎞ ⎟⎟dr dU = 4π r 2u (r ) dr = 4πr 2 ⎜⎜ 4 ⎝ 2r ⎠

(c) Integrate dU from r = R1 to R2 to obtain:

kQ 2 U= 2

=

kQ 2 dr 2r 2

dr kQ 2 (R2 − R1 ) ∫ r 2 = 2 R1R2 R1

R2

1 2 ⎛ R2 − R1 ⎞ ⎟ Q ⎜ 2 ⎜⎝ 4π ∈0 R1R2 ⎟⎠

=

Note that the quantity in parentheses is 1/C , so we have U = 12 Q 2 C. 57 ••• Picture the Problem We know, from Gauss’s law, that the field inside the shell is zero. Applying Gauss’s law to a spherical surface of radius R > r will allow us to find the energy density in this region. We can then express the energy in the electrostatic field in a spherical shell of radius R, thickness dR, and volume 4π R2dR outside the spherical shell and find the total energy in the electric field by integrating from r to ∞. If we then integrate the same expression from r to R we can find the radius R of the sphere such that half the total electrostatic field energy of the system is contained within that sphere.

(

)

Apply Gauss’s law to a spherical shell of radius R > r to obtain:

Er 4π R 2 =

Solve for Er outside the spherical shell: Express the energy density in the region R > r:

Er =

Express the energy in the electrostatic field in a spherical shell of radius R, thickness dR, and volume 4πR2dR outside the spherical shell:

Integrate dU from r to ∞ to obtain:

Qinside Q = ∈0 ∈0

kQ R2

k 2 ∈0 Q 2 ⎛ kQ ⎞ u = 12 ∈0 ER2 = 12 ∈0 ⎜ 2 ⎟ = 2R 4 ⎝R ⎠ 2

dU = 4π R 2u (R )dR ⎛ k 2 ∈0 Q 2 ⎞ ⎟⎟dR = 4π R 2 ⎜⎜ 4 ⎠ ⎝ 2R 2 kQ = dR 2R 2 ∞

U tot

kQ 2 dR kQ 2 = = 2 ∫r R 2 2r

Electrostatic Energy and Capacitance 281 kQ 2 dR' kQ 2 ⎛ 1 1 ⎞ = ⎜ − ⎟ 2 ∫r R' 2 2 ⎝r R⎠ R

Integrate dU from r to R to obtain:

U=

Set U = 12 U tot to obtain:

kQ 2 ⎛ 1 1 ⎞ kQ 2 ⎜ − ⎟= 2 ⎝ r R ⎠ 4r

Solve for R:

R = 2r

Disconnected and Reconnected Capacitors 58 •• Picture the Problem Let C1 represent the capacitance of the 2.0-µF capacitor and C2 the capacitance of the 2nd capacitor. Note that when they are connected as described in the problem statement they are in parallel and, hence, share a common potential difference. We can use the equation for the equivalent capacitance of two capacitors in parallel and the definition of capacitance to relate C2 to C1 and to the charge stored in and the potential difference across the equivalent capacitor. Using the definition of capacitance, find the charge on capacitor C1:

Q1 = C1V = (2 µF)(12 V ) = 24 µC

Ceq = C1 + C2

Express the equivalent capacitance of the two-capacitor system and solve for C2:

and

Using the definition of capacitance, express Ceq in terms of Q2 and V2:

Ceq =

C2 = Ceq − C1 Q2 Q1 = V2 V2

where V2 is the common potential difference (they are in parallel) across the two capacitors and Q1 and Q2 are the (equal) charges on the two capacitors. Substitute to obtain:

C2 =

Q1 − C1 V2

Substitute numerical values and evaluate C2:

C2 =

24 µC − 2 µF = 4.00 µF 4V

282 Chapter 24 59 •• Picture the Problem Because, when the capacitors are connected as described in the problem statement, they are in parallel, they will have the same potential difference across them. In part (b) we can find the energy lost when the connections are made by comparing the energies stored in the capacitors before and after the connections. (a) Because the capacitors are in parallel:

(b) Express the energy lost when the connections are made in terms of the energy stored in the capacitors before and after their connection: Express and evaluate Ubefore:

V100 = V400 = 2.00 kV

∆U = U before − U after

U before = U100 + U 400 2 2 = 12 C100V100 + 12 C400V400

= 12 V 2 (C100 + C400 ) =

1 2

(2 kV )2 (500 pF)

= 1.00 mJ Express and evaluate Uafter:

U after = U100 + U 400 2 2 = 12 C100V100 + 12 C400V400

= 12 V 2 (C100 + C400 ) =

1 2

(2 kV )2 (500 pF)

= 1.00 mJ Substitute to obtain:

∆U = 1.00 mJ − 1.00 mJ = 0

*60 •• Picture the Problem When the capacitors are reconnected, each will have the charge it acquired while they were connected in series across the 12-V battery and we can use the definition of capacitance and their equivalent capacitance to find the common potential difference across them. In part (b) we can use U = 12 CV 2 to find the initial and final energy stored in the capacitors. (a) Using the definition of capacitance, express the potential difference across each capacitor when they are reconnected:

V =

2Q Ceq

(1)

where Q is the charge on each capacitor before they are disconnected.

Electrostatic Energy and Capacitance 283 Find the equivalent capacitance of the two capacitors after they are connected in parallel: Express the charge Q on each capacitor before they are disconnected: Express the equivalent capacitance of the two capacitors connected in series:

Ceq = C1 + C2 = 4 µF + 12 µF = 16 µF Q = C'eqV

C'eq =

(4 µF)(12 µF) = 3 µF C1C2 = C1 + C2 4 µF + 12 µF

Substitute to find Q:

Q = (3 µF)(12 V ) = 36 µC

Substitute in equation (1) and evaluate V:

V =

(b) Express and evaluate the energy stored in the capacitors initially:

U i = 12 C'eqVi 2 =

Express and evaluate the energy stored in the capacitors when they have been reconnected:

U f = 12 CeqVf2 =

2(36 µC ) = 4.50 V 16 µF 1 2

(3 µF)(12 V )2

1 2

(16 µF)(4.5 V )2

= 216 µJ

= 162 µJ

61 •• Picture the Problem Let C1 represent the capacitance of the 1.2-µF capacitor and C2 the capacitance of the 2nd capacitor. Note that when they are connected as described in the problem statement they are in parallel and, hence, share a common potential difference. We can use the equation for the equivalent capacitance of two capacitors in parallel and the definition of capacitance to relate C2 to C1 and to the charge stored in and the potential difference across the equivalent capacitor. In part (b) we can use U = 12 CV 2 to find the energy before and after the connection was made and, hence, the energy lost when the connection was made. (a) Using the definition of capacitance, find the charge on capacitor C1: Express the equivalent capacitance of the two-capacitor system and solve for C2:

Q1 = C1V = (1.2 µF)(30 V ) = 36 µC

Ceq = C1 + C2 and

284 Chapter 24 C2 = Ceq − C1 Using the definition of capacitance, express Ceq in terms of Q2 and V2:

Q2 Q1 = V2 V2

Ceq =

where V2 is the common potential difference (they are in parallel) across the two capacitors. Substitute to obtain:

C2 =

Q1 − C1 V2

Substitute numerical values and evaluate C2:

C2 =

36 µC − 1.2 µF = 2.40 µF 10 V

∆U = U before − U after

(b) Express the energy lost when the connections are made in terms of the energy stored in the capacitors before and after their connection:

= 12 C1V12 − 12 CeqVf2 =

1 2

(C V

2 1 1

− CeqVf2

)

Substitute numerical values and evaluate ∆U:

∆U =

1 2

[(1.2 µF)(30 V ) − (3.6 µF)(10 V ) ] = 2

2

360 µJ

62 •• Picture the Problem Because, when the capacitors are connected as described in the problem statement, they are in parallel, they will have the same potential difference across them. In part (b) we can find the energy lost when the connections are made by comparing the energies stored in the capacitors before and after the connections. (a) Using the definition of capacitance, express the charge Q on the capacitors when they have been reconnected:

Substitute numerical values to obtain: Using the definition of capacitance, relate the equivalent capacitance, charge, and final potential difference for the parallel connection:

Q = Q400 − Q100 = C400V400 − C100V100 = (C400 − C100 )V

where V is the common potential difference to which the capacitors have been charged.

Q = (400 pF − 100 pF)(2 kV ) = 600 nC Q = (C1 + C2 )Vf

Electrostatic Energy and Capacitance 285 Solve for and evaluate Vf:

600 nC Q = C1 + C2 100 pF + 400 pF

Vf =

= 1.20 kV across both capacitors. (b) Express the energy lost when the connections are made in terms of the energy stored in the capacitors before and after their connection:

∆U = U before − U after = 12 C1V12 + 12 C2V22 − 12 CeqVf 2 =

1 2

(C V

2 1 1

+ C2V22 − CeqVf2

)

Substitute numerical values and evaluate ∆U:

∆U =

1 2

[(100 pF)(2 kV ) + (400 pF)(2 kV ) − (500 pF)(1.2 kV ) ] = 2

2

2

0.640 mJ

63 •• Picture the Problem When the capacitors are reconnected, each will have a charge equal to the difference between the charges they acquired while they were connected in parallel across the 12-V battery. We can use the definition of capacitance and their equivalent capacitance to find the common potential difference across them. In part (b) we can use U = 12 CV 2 to find the initial and final energy stored in the capacitors. (a) Using the definition of capacitance, express the potential difference across the capacitors when they are reconnected: Express the final charge Qf on each capacitor: Use the definition of capacitance to substitute for Q2 and Q1:

Vf =

Qf Qf = C eq C1 + C 2

(1)

where Qf is the common charge on the capacitors after they are reconnected.

Qf = Q2 − Q1 Qf = C2V − C1V = (C2 − C1 )V

Substitute in equation (1) to obtain:

Vf =

C2 − C1 V C1 + C2

Substitute numerical values and evaluate Vf:

Vf =

12 µF − 4 µF (12 V ) = 6.00 V 12 µF + 4 µF

286 Chapter 24 (b) Express and evaluate the energy stored in the capacitors initially:

U i = 12 C1V 2 + 12 C2V 2 = 12 V 2 (C1 + C2 ) =

1 2

(12 V )2 (12 µF + 4 µF)

= 1.15 mJ Express and evaluate the energy stored in the capacitors when they have been reconnected:

U f = 12 C1Vf2 + 12 C2Vf2 = 12 Vf2 (C1 + C2 ) =

1 2

(6 V )2 (12 µF + 4 µF)

= 0.288 mJ *64 •• Picture the Problem Let the numeral 1 refer to the 20-pF capacitor and the numeral 2 to the 50-pF capacitor. We can use conservation of charge and the fact that the connected capacitors will have the same potential difference across them to find the charge on each capacitor. We can decide whether electrostatic potential energy is gained or lost when the two capacitors are connected by calculating the change ∆U in the electrostatic energy during this process. (a) Using the fact that no charge is lost in connecting the capacitors, relate the charge Q initially on the 20pF capacitor to the charges on the two capacitors when they have been connected:

Q = Q1 + Q2

Q1 Q2 = C1 C2

Because the capacitors are in parallel, the potential difference across them is the same:

V1 = V2 ⇒

Solve for Q1 to obtain:

Q1 =

C1 Q2 C2

Substitute in equation (1) and solve for Q2 to obtain:

Q2 =

Q 1 + C1 C2

Use the definition of capacitance to find the charge Q initially on the 20pF capacitor:

(1)

(2)

Q = C1V = (20 pF)(3 kV ) = 60 nC

Electrostatic Energy and Capacitance 287 60 nC = 42.9 nC 1 + 20 pF 50 pF

Substitute in equation (2) and evaluate Q2:

Q2 =

Substitute in equation (1) to obtain:

Q1 = Q − Q2 = 60nC − 42.9 nC = 17.1 nC

(b) Express the change in the electrostatic potential energy of the system when the two capacitors are connected:

∆U = U f − U i Q2 Q2 = − 2Ceq 2C1 =

Substitute numerical values and evaluate ∆U:

∆U =

Q 2 ⎛⎜ 1 1⎞ − ⎟ 2 ⎜⎝ Ceq C1 ⎟⎠

(60 nC)2 ⎛⎜ 2

1 1 ⎞ ⎜ 70 pF − 20 pF ⎟⎟ ⎝ ⎠

= −64.3 µJ Because ∆U < 0, electrostatic energy is lost when the two capacitors are connected. 65 ••• Picture the Problem Let upper case Qs refer to the charges before S3 is closed and lower case qs refer to the charges after this switch is closed. We can use conservation of charge to relate the charges on the capacitors before S3 is closed to their charges when this switch is closed. We also know that the sum of the potential differences around the circuit when S3 is closed must be zero and can use this to obtain a fourth equation relating the charges on the capacitors after the switch is closed to their capacitances. Solving these equations simultaneously will yield the charges q1, q2, and q3. Knowing these charges, we can use the definition of capacitance to find the potential difference across each of the capacitors. (a) With S1 and S2 closed, but S3 open, the charges on and the potential differences across the capacitors do not change and: (b) When S3 is closed, the charges can redistribute; express the conditions on the charges that must be satisfied as a result of this

V1 = V2 = V3 = 200 V

q2 − q1 = Q2 − Q1 , q3 − q2 = Q3 − Q2 , and

288 Chapter 24 redistribution:

q1 − q3 = Q1 − Q3 . V1 + V2 + V3 = 0

Express the condition on the potential differences that must be satisfied when S3 is closed:

where the subscripts refer to the three capacitors.

Use the definition of capacitance to eliminate the potential differences:

q1 q2 q3 + + =0 C1 C2 C3

Q1 = C1V = (2 µF)(200 V ) = 400 µC , Q2 = C2V = (4 µF)(200 V ) = 800 µC ,

Use the definition of capacitance to find the initial charge on each capacitor:

and

Let Q = Q1. Then:

Q2 = 2Q and Q3 = 3Q

Express q2 and q3 in terms of q1 and Q:

Q3 = C3V = (6 µF)(200 V ) = 1200 µC

q2 = Q + q1

(2)

and

q3 = q1 + 2Q Substitute in equation (1) to obtain:

(1)

(3)

q1 Q + q1 q1 + 2Q + + =0 C1 C2 C3 or

q1 Q + q1 q1 + 2Q + + =0 2 µF 4 µF 6 µF

Solve for and evaluate q1 to obtain:

q1 = − 117 Q = − 117 (400 µC ) = − 254 µC

Substitute in equation (2) to obtain:

q2 = 400 µC − 254 µC = 146 µC

Substitute in equation (3) to obtain:

q3 = −254 µC + 2(400 µC ) = 546 µC

(c) Use the definition of capacitance to find the potential difference across each capacitor with S3 closed:

q1 − 254 µC = = − 127 V , 2 µF C1 146 µC q = 36.5 V , V2 = 2 = 4 µF C2

V1 =

and

V3 =

q3 546 µC = = 91.0 V 6 µF C3

Electrostatic Energy and Capacitance 289 *66 •• Picture the Problem We can use the expression for the energy stored in a capacitor to express the ratio of the energy stored in the system after the discharge of the first capacitor to the energy stored in the system prior to the discharge. Express the energy U initially stored in the capacitor whose capacitance is C:

U=

Q2 2C 2

2

The energy U′ stored in the two capacitors after the first capacitor has discharged is:

⎛Q⎞ ⎛Q⎞ ⎜ ⎟ ⎜ ⎟ Q2 2⎠ 2 ⎝ +⎝ ⎠ = U' = 2C 2C 4C

Express the ratio of U′ to U:

Q2 U' 4C 1 = = ⇒ U '= U Q2 2 2C

1 2

U

Dielectrics 67 • Picture the Problem The capacitance of a parallel-plate capacitor filled with a dielectric of constant κ is given by C =

κ ∈0 A d

.

Relate the capacitance of the parallel-plate capacitor to the area of its plates, their separation, and the dielectric constant of the material between the plates:

C=

Substitute numerical values and evaluate C:

C=

κ ∈0 A d

(

)(

2.3 8.85 × 10−12 C 2 /N ⋅ m 2 400 cm 2 0.3 mm

)

= 2.71 nF 68 •• Picture the Problem The capacitance of a cylindrical capacitor is given by C = 2πκ ∈0 L ln (r2 r1 ) , where L is its length and r1 and r2 the radii of the inner and outer conductors. We can use this expression, in conjunction with the definition of capacitance, to express the potential difference between the wire and the cylindrical shell in the Geiger tube. Because the electric field E in the tube is related to the linear charge density λ on the wire according to E = 2kλ κr , we can use this expression to find 2kλ/κ for E = Emax. In part (b) we’ll use this relationship to find the charge per unit length λ on

290 Chapter 24 the wire. (a) Use the definition of capacitance and the expression for the capacitance of a cylindrical capacitor to express the potential difference between the wire and the cylindrical shell in the tube:

∆V =

Q Q = C 2πκ ∈0 L ln (R r )

⎛ R ⎞ 2kλ ⎛ R ⎞ ln⎜ ⎟ = ln⎜ ⎟ κ ⎝r⎠ 4π ∈0 κ ⎝ r ⎠ where λ is the linear charge density, κ is =

2λ

the dielectric constant of the gas in the Geiger tube, r is the radius of the wire, and R the radius of the coaxial cylindrical shell of length L. Express the electric field at a distance r greater than its radius from the center of the wire:

E=

Solve for 2kλ/κ :

2kλ

Noting that E is a maximum at r = 0.2 mm, evaluate 2kλ/κ :

κ 2kλ

κ

2kλ κr

= Er

(1)

= Emax r = (2 × 106 V/m )(0.2 mm ) = 400 V

Substitute and evaluate ∆Vmax:

⎛ 1.5 cm ⎞ ⎟⎟ = 1.73 kV ∆Vmax = (400 V ) ln⎜⎜ 0.2 mm ⎠ ⎝

(b) Solve equation (1) for λ:

λ=

Emaxκr 2k

Substitute numerical values and evaluate λ:

λ=

1.8(2 × 106 V/m )(0.2 mm) 2(8.99 × 109 N ⋅ m 2 /C 2 )

= 40.0 nC/m

Electrostatic Energy and Capacitance 291 69 •• Picture the Problem The diagram shows a partial cross-sectional view of the inner and outer spherical shells. By symmetry, the electric field is radial. We can apply Gauss’s law to spherical surfaces of radii r < R1, R1 < r < R2, and r > R2 to find the electric field and, hence, the energy density in these regions. (a) Apply Gauss’s law to a spherical surface of radius r < R1 to obtain:

(

)

Er 4π r 2 =

Qinside

=0

Qinside

=

κ ∈0

and

Er < R1 = 0 Because E = 0 for r < R1:

ur < R1 = 0

Apply Gauss’s law to a spherical surface of radius R1 < r < R2 to obtain:

Er 4π r 2 =

Solve for Er to obtain:

Er =

(

)

κ ∈0

Q 4πκ ∈0 r

2

=

Q

κ ∈0

kQ κr 2

Express the energy density in the region R1 < r < R2:

⎛ kQ ⎞ u = κ ∈0 E = κ ∈ 0 ⎜ 2 ⎟ ⎝ κr ⎠ k 2 ∈0 Q 2 = 2κr 4

Apply Gauss’s law to a cylindrical surface of radius r > R2 to obtain:

Er 4π r 2 =

2 r

1 2

(

)

and

Er > R2 = 0 Because E = 0 for r > R2:

ur > R2 = 0

1 2

Qinside

κ ∈0

=0

2

292 Chapter 24 (b) Express the energy in the electrostatic field in a spherical shell of radius r, thickness dr, and volume 4πr2dr between the conductors:

dU = 4π r 2u (r )dr ⎛ k 2κ ∈0 Q 2 ⎞ ⎟⎟dr = 4π r 2 ⎜⎜ 2 4 ⎝ 2κ r ⎠ kQ 2 dr 2κr 2

=

(c) Integrate dU from r = R1 to R2 to obtain:

kQ 2 U= 2κ

R2

dr

∫r

2

=

R1

kQ 2 (R2 − R1 ) = 2κR1 R2 =

1 2 ⎛ R2 − R1 ⎞ ⎟ Q ⎜ 2 ⎜⎝ 4πκε 0 R1 R2 ⎟⎠

Note that the quantity in parentheses is 1/C, so we have U = 12 Q 2 C. 70 •• Picture the Problem We can use the relationship between the electric field between the plates of a capacitor, their separation, and the potential difference between them to find the minimum plate separation. We can use the expression for the capacitance of a dielectric-filled parallel-plate capacitor to determine the necessary area of the plates. (a) Relate the electric field of the capacitor to the potential difference across its plates: Solve for d:

E=

V d

where d is the plate separation.

d=

V E

Noting that dmin corresponds to Emax, evaluate dmin:

d min =

(b) Relate the capacitance of a parallel-plate capacitor to the area of its plates:

C=

Solve for A:

A=

V 2000 V = = 50.0 µm Emax 4 × 107 V/m

κ ∈0 A d

Cd

κ ∈0

Electrostatic Energy and Capacitance 293 Substitute numerical values and evaluate A:

A=

(0.1 µF)(50 µm )

(

24 8.85 × 10-12 C 2 /N ⋅ m 2 −2

= 2.35 × 10 m

)

2

= 235 cm 2 71 •• Picture the Problem We can model this system as two capacitors in series, C1 of thickness d/4 and C2 of thickness 3d/4 and use the equation for the equivalent capacitance of two capacitors connected in series. Express the equivalent capacitance of the two capacitors connected in series:

1 1 1 = + Ceq C1 C2 or

Ceq =

Relate the capacitance of C1 to its dielectric constant and thickness:

C1 =

Relate the capacitance of C2 to its dielectric constant and thickness:

C2 =

C1C2 C1 + C2

κ1 ∈ 0 A 1 4

d

κ 2 ∈0 A 3 4

d

=

4κ1 ∈0 A d

=

4κ 2 ∈0 A 3d

Substitute and simplify to obtain:

⎛ 4κ1 ∈0 A ⎞⎛ 4κ 2 ∈0 A ⎞ ⎛ κ1 ⎞⎛ 4κ 2 ⎞ 4κ1κ 2 ⎟ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎜ d d d d 3 3 ⎠∈ A = ⎠ = ⎝ ⎠⎝ ⎠⎝ d Ceq = ⎝ ∈0 A 0 4κ1 ∈0 A 4κ 2 ∈0 A 3κ1 κ 2 + 3 κ κ 1 2 + + d 3d 3d 3d =

4κ1κ 2 ⎛ ∈0 A ⎞ ⎟= ⎜ 3κ1 + κ 2 ⎝ d ⎠

⎛ 4κ1κ 2 ⎞ ⎜⎜ ⎟⎟C0 ⎝ 3κ1 + κ 2 ⎠

*72 •• Picture the Problem Let the charge on the capacitor with the air gap be Q1 and the charge on the capacitor with the dielectric gap be Q2. If the capacitances of the capacitors were initially C, then the capacitance of the capacitor with the dielectric inserted is C' = κC. We can use the conservation of charge and the equivalence of the potential difference across the capacitors to obtain two equations that we can solve simultaneously for Q1 and Q2. Apply conservation of charge during

Q1 + Q2 = 2Q

(1)

294 Chapter 24 the insertion of the dielectric to obtain:

Q1 Q2 = C κC

Because the capacitors have the same potential difference across them: Solve equations (1) and (2) simultaneously to obtain:

2Q 1+ κ

Q1 =

(2)

and Q2 =

2Qκ 1+ κ

73 •• Picture the Problem We can model this system as two capacitors in series, C1 of thickness t and C2 of thickness d − t and use the equation for the equivalent capacitance of two capacitors connected in series.

1 1 1 = + Ceq C1 C2

Express the equivalent capacitance of the two capacitors connected in series:

or

Ceq =

Relate the capacitance of C1 to its dielectric constant and thickness:

C1 =

Relate the capacitance of C2 to its dielectric constant and thickness:

C2 =

C1C2 C1 + C2

κ ∈0 A t

∈0 A d −t

Substitute and simplify to obtain:

⎛ κ ∈0 A ⎞⎛ ∈0 A ⎞ ⎛ κ ⎞⎛ 1 ⎞ ⎛ κ ⎞⎛ 1 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ t ⎠⎝ d − t ⎠ ⎝ t ⎠⎝ d − t ⎠ t ⎠⎝ d − t ⎠ ⎝ ⎝ Ceq = = ∈0 A = ∈0 A 1 1 κ κ ∈0 A ∈0 A κ + + + t d −t t d −t t d −t =

κ

κ (d − t ) + t

⎛

κd

⎞ ⎟⎟C0 ⎝ κ (d − t ) + t ⎠

∈0 A = ⎜⎜

74 •• Picture the Problem Because d << r, we can model the membrane as a parallel-plate capacitor. We can use the definition of capacitance to find the charge on each side of the membrane in part (b) and the relationship between the potential difference across the

Electrostatic Energy and Capacitance 295 membrane, its thickness, and the electric field in it to find the electric field called for in part (c).

κ ∈0 A

(a) Express the capacitance of a parallel-plate capacitor:

C=

Substitute for the area of the plates:

C=

2πκ ∈0 rL κrL = d 2kd

Substitute numerical values and evaluate C:

C=

3 10 −5 m (0.1m ) 2 8.99 × 109 N ⋅ m 2 /C 2 10−8 m

d

(

(

)

)(

)

= 16.7 nF Q = CV = (16.7 nF)(70 mV ) = 1.17 nC

(b) Use the definition of capacitance to find the charge on each side of the membrane: (c) Express the electric field through the membrane as a function of its thickness d and the potential difference V across it:

E=

V d

Substitute numerical values and evaluate E:

E=

70 mV = 7.00 MV/m 10−8 m

*75 •• Picture the Problem The bound charge density is related to the dielectric constant and

⎛ ⎝

the free charge density according to σ b = ⎜1 −

1⎞ ⎟σ . κ⎠ f

Solve the equation relating σb, σf, and κ for κ to obtain:

κ=

(a) Evaluate this expression for σb/σf = 0.8:

κ=

1 = 5.00 1 − 0.8

(b) Evaluate this expression for σb/σf = 0.2:

κ=

1 = 1.25 1 − 0.2

(c) Evaluate this expression for σb/σf = 0.98:

κ=

1 = 50.0 1 − 0.98

1 1−σ b σ f

296 Chapter 24 76 •• Picture the Problem We can use the definition of the dielectric constant to find its value. In part (b) we can use the expression for the electric field in the space between the charged capacitor plates to find the area of the plates and in part (c) we can relate the surface charge densities to the induced charges on the plates. (a) Using the definition of the dielectric constant, relate the electric field without a dielectric E0 to the field with a dielectric E: Solve for and evaluate κ :

E=

E0

κ=

E0 2.5 × 105 V/m = = 2.08 E 1.2 × 105 V/m

κ

(b) Relate the electric field in the region between the plates to the surface charge density of the plates:

E0 =

Solve for A:

A=

Substitute numerical values and evaluate A:

A=

σ Q A = ∈0 ∈0

Q E0 ∈ 0 10 nC 2.5 ×10 V/m 8.85 × 10−12 C 2 /N ⋅ m 2

(

5

)(

= 4.52 × 10−3 m 2 = 45.2 cm 2 (c) Relate the surface charge densities to the induced charges on the plates:

⎛

1⎞

σ b = ⎜1 − ⎟σ f ⎝ κ⎠ or

σ b Qb 1 = = 1− σ f Qf κ Solve for Qb: Substitute numerical values and evaluate Qb:

⎛ 1⎞ Qb = ⎜1 − ⎟Qf ⎝ κ⎠ 1 ⎞ ⎛ Qb = ⎜1 − ⎟(10 nC ) = 5.19 nC ⎝ 2.08 ⎠

)

Electrostatic Energy and Capacitance 297 *77 •• Picture the Problem We can model this parallel-plate capacitor as a combination of two capacitors C1 and C2 in series with capacitor C3 in parallel. Express the capacitance of two series-connected capacitors in parallel with a third:

Express each of the capacitances C1, C2, and C3 in terms of the dielectric constants, plate areas, and plate separations:

C = C3 + Cs where

Cs =

1 2

C2 =

C3 =

Substitute in equation (1) to obtain:

C1C2 C1 + C2

κ1 ∈ 0

C1 =

and

Substitute in equation (2) to obtain:

(1)

(2)

( 12 A) = κ1 ∈0

d

κ 2 ∈0 1 2

d

d

,

( 12 A) = κ 2 ∈0

A

( 12 A) = κ 3 ∈0

A

d

κ 3 ∈0

A

d

,

2d

⎛ κ1 ∈0 A ⎞⎛ κ 2 ∈0 A ⎞ ⎜ ⎟⎜ ⎟ d ⎠⎝ d ⎝ ⎠ Cs = κ1 ∈ 0 A κ 2 ∈ 0 A + d d κ κ ⎛∈ A ⎞ = 1 2 ⎜ 0 ⎟ κ1 + κ 2 ⎝ d ⎠

C=

κ 3 ∈0 A 2d

+

κ1κ 2 ⎛∈0 A ⎞ ⎟ ⎜ κ1 + κ 2 ⎝ d ⎠

⎛ 2κ1κ 2 ⎞⎛∈0 A ⎞ ⎟⎜ = ⎜⎜ κ 3 + ⎟ κ1 + κ 2 ⎟⎠⎝ 2d ⎠ ⎝ 78 •• Picture the Problem The electric field E between the plates of a parallel-plate capacitor is related to the potential difference V between the plates and their separation d according to V = Ed and the electrostatic energy U depends on the electric field according to U = 12 ∈0 E02 Ad . We can use these relationships to find E, V, and U with and without the dielectric in place. (a) Relate the electric field E0 to the potential difference V between the plates and the plate separation d:

E0 =

V 100 V = = 25.0 kV/m d 4 mm

298 Chapter 24 Use the definition of energy density to relate the electrostatic energy U0 to the volume of the space between the plates:

U 0 = u0 Ad

Express the energy density in the electric field:

u0 = 12 ∈0 E02

Substitute to obtain:

U 0 = 12 ∈0 E02 Ad

Substitute numerical values and evaluate U0:

U0 =

(8.85 × 10 C /N ⋅ m )(25 kV/m ) × (600 cm )(4 mm ) −12

1 2

2

2

2

2

= 0.664 µJ (b) With the dielectric in place the electric field becomes: (c) Relate the potential difference V to the electric field E and the separation of the plates:

E=

E0

=

25 kV/m = 6.25 kV/m κ 4 V = Ed = (6.25 kV/m )(4 mm ) = 25.0 V

(d) Relate the new electrostatic energy U to the initial electrostatic energy U0 and the dielectric constant κ :

U=

U0

=

κ

0.664 µJ = 0.166 µJ 4

79 ••• Picture the Problem We can use the definition of capacitance and the relationship between the electric field in the capacitor and the potential difference across its plates to express C. In part (b) we can use σ b = (1 − 1 κ )σ f and κ = 1 + (3/y0)y to express the ratio σb/σf and evaluate it at y = 0 and y = y0. The application of Gauss’s law in part (c) will yield an expression for ρ (y) within the dielectric that we can integrate in part (d) to find the total induced bound charge. (a) Using its definition, express the capacitance of the parallel-plate capacitor: Express the potential difference V between the plates in terms of the electric field E between the plates:

C=

Q σA = V V

dV = Edy

(1)

Electrostatic Energy and Capacitance 299 E0 σ = κ ( y ) ∈0 κ ( y )

Express the electric field in the region between the plates:

E=

Substitute to obtain:

dV =

Integrate from y = 0 to y = y0:

Substitute in equation (1) and simplify to obtain:

(b) Relate σb to σf and κ :

σ dy = ∈0 κ ( y )

σ ⎧ ⎛ 3 ⎞ ⎫ ∈ 0 ⎨1 + ⎜⎜ ⎟⎟ y ⎬ ⎩ ⎝ y0 ⎠ ⎭

dy σ V= ∫ ∈0 0 1 + (3 y0 ) y σy y = 0 ln (1 + 3 y y0 ) 0 3 ∈0 σy = 0 ln (4) 3 ∈0 y0

C=

σA

σy0 ln (4) 3 ∈0 ⎛

=

0

3 ∈0 A y0 ln (4)

1⎞

σ b = ⎜1 − ⎟σ f ⎝ κ⎠ and

σb 1 = 1− σf κ

Substitute for κ to obtain:

σb 1 = 1− σf 1 + (3 y0 ) y

Evaluate σb/σf at y = 0:

σb σf

Evaluate σb/σf at y = y0:

(c) Consider a Gaussian surface of area A and width dy and recall that E into the surface is taken to be negative. Apply Gauss’s law to obtain:

σb σf

=1− y =0

1 = 0 1 + (3 y0 )(0)

= 1− y = y0

1 = 0.750 1 + (3 y0 ) y0

[E ( y ) − E ( y + dy )]A = Qinside ∈0

=

Adyρ ( y )

∈0

dy

300 Chapter 24 Divide both sides of the equation by dy:

[E ( y ) − E ( y + dy )] = ρ ( y ) or

− Solve for ρ (y) to obtain:

∈0

dy

dE ρ ( y ) = dy ∈0

ρ ( y ) = − ∈0 = − ∈0 = −σ

d ⎡ σ ⎤ ⎢ ⎥ dy ⎣∈0 κ ( y ) ⎦

⎤ d ⎡ 1 ⎢ ( ⎥ dy ⎣1 + 3 y0 ) y ⎦

3σ 2 y0 (1 + 3 y y0 )

[

=

(d) Integrate ρ(y) from y = 0 to y = y0 to obtain:

dE dy

y0

ρ = 3σ ∫ 0

]

dy 2 y0 (1 + 3 y y0 )

[

]

= − 34 σ , the charge per unit area in the dielectric, and just cancels out the induced surface charge density.

General Problems 80 •• Picture the Problem We can use the expression U 0 = 12 CeqV 2 to express the total energy stored in the combination of four capacitors in terms of their equivalent capacitance Ceq. The energy stored in one capacitor when it is connected to the 100-V battery is:

U 0 = 12 CV 2

When the four capacitors are connected to the battery in some combination, the total energy stored in them is:

U = 12 CeqV 2

Equate U and U0 and solve for Ceq to obtain:

Ceq = C

Electrostatic Energy and Capacitance 301 The equivalent capacitance C′ of two capacitors of capacitance C connected in series is their product divided by their sum: If we connect two of the capacitors in series in parallel with the other two capacitors connected in series, their equivalent capacitance will be:

C' =

C2 = 1C C +C 2

Ceq = C' + C' = 12 C + 12 C = C

Thus, a series combination of two of the capacitors in parallel with a series combination of the other two capacitors will result in total energy U 0 stored in all four capacitors. *81 • Picture the Problem We can use the equations for the equivalent capacitance of three capacitors connected in parallel and in series to find these equivalent capacitances. (a) Express the equivalent capacitance of three capacitors connected in parallel: Substitute numerical values and evaluate Ceq:

(b) Express the equivalent capacitance of the three capacitors connected in series:

Ceq = C1 + C2 + C3

Ceq = 2.0 µF + 4.0 µF + 8.0 µF = 14.0 µF Ceq =

C1C2C3 C1C2 + C2C3 + C1C3

Substitute numerical values and evaluate Ceq:

Ceq =

(2 µF)(4 µF)(8 µF) = (2 µF)(4 µF) + (4 µF)(8 µF) + (2 µF)(8 µF)

1.14 µF

82 • Picture the Problem We can first use the equation for the equivalent capacitance of two capacitors connected in parallel and then the equation for two capacitors connected in series to find the equivalent capacitance.

302 Chapter 24 Find the equivalent capacitance of a 1.0-µF capacitor connected in parallel with a 2.0-µF capacitor: Find the equivalent capacitance of a 3.0-µF capacitor connected in series with a 6.0-µF capacitor:

Ceq,1 = C1 + C2 = 1.0 µF + 2.0 µF = 3.0 µF C eq,2 =

C eq,1C 6 (3.0µF)(6.0µF) = C eq,1 + C 6 3.0µF + 6.0µF

= 2.00µF

83 • Picture the Problem The charge Q and the charge density σ are independent of the separation of the plates and do not change during the process described in the problem statement. Because the electric field E depends on σ, it too is constant. We can use U = 12 CV 2 and the relationship between V and E, together with the expression for the capacitance of a parallel-plate capacitor, to show that U ∝ d. Express the energy stored in the capacitor in terms of its capacitance C and the potential difference across its plates:

U = 12 CV 2

Express V in terms of E:

V = Ed where d is the separation of the plates.

Express the capacitance of a parallelplate capacitor:

C=

Substitute to obtain:

U=

Because U ∝ d, to double U one must double d. Hence:

κ ∈0 A d 1 2

κ ∈0 A d

(Ed )2 = (12 κ ∈0

)

AE 2 d

d f = 2d = 2(0.5 mm ) = 1.00 mm

84 •• Picture the Problem We can use the equations for the equivalent capacitance of capacitors connected in parallel and in series to find the single capacitor that will store the same amount of charge as each of the networks shown above. (a) Find the capacitance of the two capacitors in parallel:

Ceq,1 = C0 + C0 = 2C0

Electrostatic Energy and Capacitance 303 Find the capacitance equivalent to 2C0 in series with C0: (b) Find the capacitance of two capacitors of capacitance C0 in parallel: Find the capacitance equivalent to 2C0 in series with 2C0: (c) Find the equivalent capacitance of three equal capacitors connected in parallel:

Ceq,2 =

Ceq,1C0 Ceq,1 + C0

=

(2C0 )C0 2C0 + C0

=

2 3

C0

Ceq,1 = 2C0

Ceq,2 =

Ceq,1C0 Ceq,1 + C0

=

(2C0 )(2C0 ) = 2C0 + 2C0

C0

Ceq = C0 + C0 + C0 = 3C0

*85 •• Picture the Problem Note that with V applied between a and b, C1 and C3 are in series, and so are C2 and C4. Because in a series combination the potential differences across the two capacitors are inversely proportional to the capacitances, we can establish proportions involving the capacitances and potential differences for the left- and righthand side of the network and then use the condition that Vc = Vd to eliminate the potential differences and establish the relationship between the capacitances.

Q C1

Letting Q represent the charge on capacitors 1 and 2, relate the potential differences across the capacitors to their common charge and capacitances:

and

Divide the first of these equations by the second to obtain:

V1 C3 = V3 C1

(1)

Proceed similarly to obtain:

V2 C4 = V4 C2

(2)

Divide equation (1) by equation (2) to obtain:

V1V4 C3C2 = V3V2 C1C4

(3)

If Vc = Vd then we must have:

V1 = V2 and V3 = V4

Substitute in equation (3) and

V1 =

V3 =

Q C3

C2C3 = C1C4

304 Chapter 24 rearrange to obtain: 86 •• Picture the Problem Because the spheres are identical, each will have half the charge of the initially charged sphere when they are connected. We can find the fraction of the initial energy that is dissipated by finding the energy stored initially and the energy stored when the two spheres are connected. Express the fraction of the initial energy that is dissipated when the two spheres are connected: Express the initial energy of the sphere whose charge is Q: Relate the capacitance of a an isolated spherical conductor to its radius:

f =

Ui −Uf U = 1− f Ui Ui

Ui =

(1)

1 Q2 2 C

C = 4π ∈0 R

Substitute to obtain:

Ui =

1 Q2 1 kQ 2 = 2 4π ∈0 R 2 R

Express the energy of the connected spheres:

Uf =

1 k (Q 2) 1 k (Q 2) 1 kQ 2 + = 2 R 2 R 4 R

Substitute in equation (1) and simplify:

2

2

1 kQ 2 1 1 f = 1− 4 R2 = 1− = 1 kQ 2 2 2 R

87 •• Picture the Problem We can use the expression for the capacitance of a parallel-plate capacitor as a function of A and d to determine the effect on the capacitance of doubling the plate separation. We can use V = Ed to determine the effect on the potential difference across the capacitor of doubling the plate separation. Finally, we can use U = CV 2 2 to determine the effect of doubling the plate separation on the energy stored in the capacitor. (a) Express the capacitance of a capacitor whose plates are separated by a distance 2d:

Cnew =

∈0 A 2d

Electrostatic Energy and Capacitance 305 (b) Express the potential difference across a parallel-plate capacitor whose plates are separated by a distance d: Express the new potential difference across the plates resulting from the doubling of their separation:

V = Ed where the electric field E depends solely on the charge on the capacitor plates.

Vnew = E (2d ) = 2(Ed ) = 2V

1 1 ∈0 A 2 CV 2 = V 2 2 d

(c) Relate the energy stored in a parallel-plate capacitor to the separation of the plates:

U=

When the plate separation is doubled we have:

U new =

(d) Relate the work required to change the plate separation from d to 2d to the change in the electrostatic potential energy of the system:

2 1 ∈0 A (2V )2 = ∈0 AV 2 2d d

W = ∆U = U new − U i

∈0 AV 2 ∈0 AV 2

=

−

d

∈0 AV

=

2d

2

2d

88 •• Picture the Problem We can use the equation for the equivalent capacitance of two capacitors in series to relate C0 to C′ and the capacitance of the dielectric-filled parallelplate capacitor and then solve the resulting equation for C′. Express the equivalent capacitance of the system in terms of C′ and C, where C is the dielectric-filled capacitor:

C0 =

C'C C '+C

Solve for C′ to obtain:

C' =

C0C C − C0

Express the capacitance of the dielectric-filled capacitor:

C=

Substitute to obtain:

C' =

κ ∈0 A d

= κC0

C0 (κC0 ) κ = C κC0 − C0 κ −1 0

306 Chapter 24 89 •• Picture the Problem Modeling the Leyden jar as a parallel-plate capacitor, we can use the equation relating the capacitance of a parallel-plate capacitor to the area A and separation d of its plates to find the jar’s capacitance. To find the maximum charge the jar can carry without undergoing dielectric breakdown we can use the definition of capacitance to express Qmax in terms of Vmax … and then relate Vmax to Emax using Vmax = Emax d , where d is the thickness of the glass wall of the jar. (a) Treating it as a parallel-plate capacitor, express the capacitance of the Leyden jar

C=

κ ∈0 A d

=

κ ∈0 (2πRh )

d 4π ∈0 κRh κRh = = 2d 2kd

where h is the height of the jar and R is its inside radius. Substitute numerical values and evaluate C:

C=

5(0.04 m )(0.4 m ) 2 8.99 × 109 N ⋅ m 2 /C 2 2 × 10− 3 m

(

)(

)

(

)

= 2.22 nF (b) Using the definition of capacitance, relate the maximum charge of the capacitor to the breakdown voltage of the dielectric: Express the breakdown voltage in terms of the dielectric strength and thickness of the dielectric: Substitute to obtain: Substitute numerical values and evaluate Qmax:

Qmax = CVmax

Vmax = Emax d

Qmax = CEmax d Qmax = (2.22 nF)(15 MV/m ) 2 × 10−3 m = 66.6 µC

*90 •• Picture the Problem The maximum voltage is related to the dielectric strength of the medium according to Vmax = Emax d and we can use the expression for the capacitance of a parallel-plate capacitor to determine the required area of the plates. (a) Relate the maximum voltage that can be applied across this capacitor

Vmax = Emax d

Electrostatic Energy and Capacitance 307 to the dielectric strength of silicon dioxide:

(

)(

Substitute numerical values and evaluate Vmax:

Vmax = 8 × 106 V/m 5 × 10−6 m

(b) Relate the capacitance of a parallel-plate capacitor to area A of its plates:

C=

Solve for A to obtain:

A=

Substitute numerical values and evaluate A:

)

= 40.0 V

A=

κ ∈0 A d

Cd

κ ∈0

(10 pF)(5 × 10−6 m )

(

3.8 8.85 × 10−12 C 2 /N ⋅ m 2

)

= 1.49 × 10− 6 m 2 = 1.49 mm2 (c) Express the number of capacitors n in terms of the area of a square 1 cm by 1cm and the area required for each capacitor:

n=

(1cm)2 A

=

100 mm 2 ≈ 67 1.49 mm 2

91 •• Picture the Problem When the battery is removed, after having initially charged both capacitors, and the separation of one of the capacitors is doubled, the charge is redistributed subject to the condition that the total charge remains constant; i.e., Q = Q1 + Q2 where Q is the initial charge on both capacitors and Q2 is the charge on the capacitor whose plate separation has been doubled. We can use the conservation of charge during the plate separation process and the fact that, because the capacitors are in parallel, they share a common potential difference. Find the equivalent capacitance of the two 2-µF parallel-plate capacitors connected in parallel: Use the definition of capacitance to find the charge on the equivalent capacitor:

Ceq = 2 µF + 2 µF = 4 µF

Q = CeqV = (4 µF)(100 V ) = 400 µC

308 Chapter 24 Relate this total charge to charges distributed on capacitors 1 and 2 when the battery is removed and the separation of the plates of capacitor 2 is doubled: Because the capacitors are in parallel:

Q = Q1 + Q2

(1)

V1 = V2 and

Q1 Q2 Q 2Q2 = = 1 2 = C1 C2 ' 2 C2 C2 Solve for Q1 to obtain:

⎛C ⎞ Q1 = 2⎜⎜ 1 ⎟⎟Q2 ⎝ C2 ⎠

Substitute equation (2) in equation (1) and solve for Q2 to obtain:

Q2 =

Q 2(C1 C2 ) + 1

Substitute numerical values and evaluate Q2:

Q2 =

400 µC = 133 µC 2(2 µF 2 µF) + 1

Substitute in equation (1) or equation (2) and evaluate Q1:

Q1 = 267 µC

(2)

92 •• Picture the Problem We can relate the electric field in the dielectric to the electric field between the capacitor’s plates in the absence of a dielectric using E = E0/κ. In part (b) we can express the potential difference between the plates as the sum of the potential differences across the dielectrics and then express the potential differences in terms of the electric fields in the dielectrics. In part (c) we can use our result from (b) and the definition of capacitance to express the capacitance of the dielectric-filled capacitor. In part (d) we can confirm the result of part (c) by using the addition formula for capacitors in series. (a) Express the electric field E in a dielectric of constant κ in terms of the electric field E0 in the absence of the dielectric:

E=

E0

Express the electric field E0 in the absence of the dielectrics:

E0 =

σ Q = ∈0 ∈0 A

κ

Electrostatic Energy and Capacitance 309 Substitute to obtain:

Use this relationship to express the electric fields in dielectrics whose constants are κ1 and κ2: (b) Express the potential difference between the plates as the sum of the potential differences across the dielectrics: Relate the potential differences to the electric fields and the thicknesses of the dielectrics:

E=

E1 =

Q

κ ∈0 A Q

κ 1 ∈0 A

and

V =

d Qd = 2 2κ 2 ∈0 A

Qd 2κ1 ∈0 A

+

Qd 2κ 2 ∈0 A

Qd ⎛ 1 1 ⎞ ⎜ + ⎟ 2 ∈0 A ⎜⎝ κ1 κ 2 ⎟⎠

Q Qd ⎛ 1 1 ⎞ ⎜ + ⎟ 2 ∈0 A ⎜⎝ κ1 κ 2 ⎟⎠ 2 ∈0 A 2 ∈0 A ⎛ κ1κ 2 ⎞ ⎜ ⎟ = = d ⎜⎝ κ1 + κ 2 ⎟⎠ ⎛ κ1 + κ 2 ⎞ ⎟⎟ d ⎜⎜ ⎝ κ1κ 2 ⎠

C=

Q = V

⎛ κκ ⎞ = 2C0 ⎜⎜ 1 2 ⎟⎟ ⎝ κ1 + κ 2 ⎠ where C0 = ∈0 A d . (d) Express the equivalent capacitance C of capacitors C1 and C2 in series:

κ 2 ∈0 A

d Qd = 2 2κ1 ∈0 A

V1 = E1

=

(c) Use the definition of capacitance to obtain:

Q

V = V1 + V2

V2 = E2

Substitute and simplify to obtain:

and E2 =

C=

C1C2 C1 + C2

310 Chapter 24 Express C1:

Express C2:

Substitute to obtain:

κ1 ∈ 0 A

C1 =

d 2

C2 =

κ 2 ∈0 A d 2

=

2κ1 ∈0 A = 2κ1C0 d

=

2κ 2 ∈0 A = 2κ 2C0 d

(2κ1C0 )(2κ 2C0 ) =

C=

2κ1C0 + 2κ 2C0

⎛ κκ ⎞ 2C0 ⎜⎜ 1 2 ⎟⎟ , ⎝ κ1 + κ 2 ⎠

a result in agreement with part (c). 93 •• Picture the Problem Recall that within a conductor E = 0. We can use the definition of capacitance to express C in terms of the charge on the capacitor Q and the potential difference across the plates V. We can then express V in terms of E and the thickness of the air gap between the plates. Finally, we can express the electric field between the plates in terms of the charge on them and their area. Substitution in our expression for C will give us C in terms of d – t. In part (b) we can use the expression for the equivalent capacitance of two capacitors connected in series to derive the same expression for C. (a) Use its definition to express the capacitance of this parallel-plate capacitor: Relate the electric potential between the plates to the electric field between the plates:

Q V

C=

where Q is the charge on the capacitor.

V = E (d − t )

σ Q = ∈ 0 ∈0 A

Express the electric field E between the plates but outside the metal slab:

E=

Substitute and simplify to obtain:

C=

(b) Express the equivalent capacitance C of two capacitors C1 and C2 connected in series:

C=

Express the capacitances C1 and C2 of the plates separated by a and b, respectively:

C1 = and

∈ A Q Q = = 0 Q E (d − t ) (d − t ) d − t ∈0 A C1C2 C1 + C2

∈0 A a

Electrostatic Energy and Capacitance 311 C2 = Substitute and simplify to obtain:

∈0 A b

⎛∈0 A ⎞⎛∈0 A ⎞ ⎜ ⎟⎜ ⎟ a ⎠⎝ b ⎠ ∈0 A C=⎝ = ∈0 A ∈0 A a+b + a b

Solve the constraint that a + b + t = d for a + b to obtain:

a +b = d −t

Substitute for a + b to obtain:

C=

∈0 A d −t

*94 •• Picture the Problem We can express the ratio of Ceq to C0 to show that the capacitance with the dielectrics in place is (κ1 + κ2)/2 times greater than that of the capacitor in the absence of the dielectrics.

Because the capacitor plates are conductors, the potentials are (a) the same across the entire upper and lower plates. Hence, the system is equivalent to two capacitors, each of area A/2, in parallel. (b) Relate the capacitance C0, in the absence of the dielectrics, to the plate area and separation: Express the equivalent capacitance of capacitors C1 and C2, each with plate area A/2, connected in parallel:

Express the ratio of Ceq to C0 and simplify to obtain:

C0 =

∈0 A d

Ceq = C1 + C2

Ceq C0

( 12 A) + κ1 ∈0 ( 12 A)

=

κ1 ∈0

=

κ1 ∈0 A

d 2d

d

(κ1 + κ 2 )

κ1 ∈ 0 A =

2d

(κ1 + κ 2 )

∈0 A

=

1 2

(κ1 + κ 2 )

d 95 •• Picture the Problem We can use U = Q2/2C and the expression for the capacitance as a function of plate separation to express U as a function of x. Differentiation of this result

312 Chapter 24 with respect to x will yield dU. Because the work done in increasing the plate separation a distance dx equals the change in the electrostatic potential energy of the capacitor, we can evaluate F from dU/dx. Finally, we can express F in terms of Q and E by relating E to x using E = Vx and using the definition of capacitance and the expression for the capacitance of a parallel-plate capacitor. (a) Relate the electrostatic energy U stored in the capacitor to its capacitance C:

U=

Express the capacitance as a function of the plate separation:

C=

Substitute for C to obtain:

U=

(b) Use the result obtained in (a) to evaluate dU:

1 Q2 2 C

∈0 A x Q2 2 ∈0 A

dU =

x

⎤ dU d ⎡ Q2 x ⎥ dx dx = ⎢ dx dx ⎣ 2 ∈0 A ⎦

=

Q2 2 ∈0 A

dx

(c) Relate the work needed to move one plate a distance dx to the change in the electrostatic potential energy of the system:

W = dU = Fdx

Solve for and evaluate F:

F=

⎤ dU d ⎡ Q2 Q2 = = x ⎢ ⎥ dx dx ⎣ 2 ∈0 A ⎦ 2 ∈0 A

(d) Express the electric field between the plates in terms of their separation and their potential difference:

E=

V x

Use the definition of capacitance to eliminate V:

E=

Q Cx

Use the expression for the capacitance of a parallel-plate capacitor to eliminate C:

E=

Q Q = ∈0 A x ∈0 A x

Electrostatic Energy and Capacitance 313 Substitute in our result from part (c) to obtain:

F=

Q(∈0 AE ) = 2 ∈0 A

1 2

QE

The field E is due to the sum of the fields from charges +Q and –Q on the opposite plates of the capacitor. Each plate produces a field ½ E and the force is the product of charge Q and the field ½ E. 96 •• Picture the Problem We can model this capacitor as the equivalent of two capacitors connected in parallel. Let the numeral 1 denote the capacitor with the dielectric material whose constant is κ and the numeral 2 the air-filled capacitor. (a) Express the equivalent capacitance of the two capacitors in parallel:

C ( x ) = C1 + C2

Use the expression for the capacitance of a parallel-plate capacitor to express C1:

C1 =

Express the capacitance C0 of the capacitor with the dielectric removed, i.e., x = 0:

C0 =

Divide C1 by C0 to obtain:

κ ∈0 A1 d

κ ∈0 bx d

∈0 ab d

κ ∈0 bx κx C1 d = = ∈0 ab a C0 d or

C1 = Use the expression for the capacitance of a parallel-plate capacitor to express C2:

=

(1)

C2 =

κx a

C0

∈0 A2 d

=

∈0 b(a − x ) d

∈0 b(a − x )

Divide C2 by C0 to obtain:

C2 = C0 or

d ∈0 ab d

=

a−x a

314 Chapter 24 a−x C0 a

C2 = Substitute in equation (1) and simplify to obtain:

C (x ) =

κx

(b) Evaluate C for x = 0:

C (0) =

∈0 b

C0 +

a−x C0 a

a C0 = [a + (κ − 1)x] a ∈ b = 0 [a + (κ − 1)x ] d

d

[a] = ∈0 ab = d

C0

as expected. Evaluate C for x = a:

C (a ) = =

∈0 b d

[a + (κ − 1)a]

κ ∈0 ab d

as expected.

*97 ••• Picture the Problem We can model this capacitor as the equivalent of two capacitors connected in parallel, one with an air gap and other filled with a dielectric of constantκ. Let the numeral 1 denote the capacitor with the dielectric material whose constant is κ and the numeral 2 the air-filled capacitor. (a) Using the hint, express the energy stored in the capacitor as a function of the equivalent capacitance Ceq:

1 Q2 U= 2 Ceq

The capacitances of the two capacitors are:

C1 =

Because the capacitors are in parallel, Ceq is the sum of C1 and C2:

Ceq = C1 + C2 =

κ ∈0 ax d

and C2 =

κ ∈0 ax ∈0 a(a − x ) d

∈0 a (κx + a − x ) d ∈ a = 0 [(κ − 1)x + a ] d =

∈0 a(a − x ) d +

d

Electrostatic Energy and Capacitance 315 Substitute for Ceq in the expression for U and simplify to obtain:

Q 2d U= 2 ∈0 a[(κ − 1)x + a ]

(b) The force exerted by the electric field is given by:

F =−

(c) Rewrite our result in (b) to obtain:

dU dx ⎤ d ⎡1 Q 2d =− ⎢ ⎥ dx ⎣ 2 ∈0 a[(κ − 1)x + a ]⎦

{

=−

Q 2d d [(κ − 1)x + a ]−1 2 ∈0 a dx

=

(κ − 1)Q 2 d 2 2a ∈0 [(κ − 1)x + a ]

}

(κ − 1)Q 2 ⎛⎜ a ∈0 ⎞⎟ F=

⎝ d ⎠

⎛ a ∈0 ⎞ 2 2⎜ ⎟ [(κ − 1)x + a ] d ⎝ ⎠ (κ − 1)Q 2 ⎛⎜ a ∈0 ⎞⎟ ⎝ d ⎠ = 2 2Ceq =

2

(κ − 1)a ∈0 V 2 2d

Note that this expression is independent of x.

This force originates from the fringing fields around the edges of the (d) capacitor. The effect of the force is to pull the dielectric into the space between the capacitor plates. 98 •• Picture the Problem Because capacitors connected in series have a common charge, we can find the charge on each capacitor by finding the charge on the equivalent capacitor. We can also find the total energy stored in the capacitors, with and without the dielectric inserted in one of them, by using U = 12 CeqV 2 . In part (d) we can use our knowledge of the charge on each capacitor and the definition of capacitance to the potential differences across them. (a) Using the definition of capacitance, relate the charge on each capacitor to the equivalent

Q = Q1 = Q2 = CeqV

316 Chapter 24 capacitance: Express the equivalent capacitance of two capacitors in series:

Ceq =

Substitute numerical values and evaluate Ceq:

Ceq =

Substitute numerical values and evaluate Q: (b) Express the energy U stored in the capacitors as a function of Ceq and V: (c) Using the definition of capacitance, relate the charge on each capacitor to the new equivalent capacitance Ceq′: Express the new equivalent capacitance Ceq′ when the dielectric of constant κ has been inserted between the plates of one of the capacitors: Letting the capacitor with the dielectric between its plates be denoted by the numeral 1, express C1′ and C2′:

C1C2 C1 + C2

(4 µF)(4 µF) = 2 µF 4 µF + 4 µF

Q1 = Q2 = (2 µF)(24 V ) = 48.0 µC U = 12 CeqV 2 =

1 2

(2 µF)(24 V )2 =

Q' = Q1' = Q2' = Ceq'V

Ceq' =

576 µJ (1)

C1'C2' C1' + C2'

C1' = κC1 and

C2' = C2

Substitute to obtain:

Ceq' =

κC1C2 κC1 + C2

Substitute numerical values and evaluate Ceq′:

Ceq' =

4.2(4 µF)(4 µF) = 3.23 µF 4.2(4 µF) + 4 µF

Substitute in equation (1) to obtain:

Q1' = Q2' = (3.23 µF)(24 V ) = 77.5 µC

(d) Express the potential difference across each capacitor in terms of its

V1' =

77.5 µC Q1' Q1' = = = 4.61 V C1' κC1 4.2(4 µF)

Electrostatic Energy and Capacitance 317 charge and capacitance:

and

V2' =

Q2' Q2' 77.5 µC = = = 19.4 V 4 µF C2 ' C2

U = 12 Ceq'V 2

(e) Express the total stored energy in terms of the equivalent capacitance:

=

1 2

(3.23 µF)(24 V )2 =

930 µJ

99 •• Picture the Problem We can find the work required to pull the glass plate out of the capacitor by finding the change in the electrostatic energy of the system as a consequence of the removal of the dielectric plate.

W = ∆U = U f − U i

Express the change in the electrostatic energy of the system resulting from the removal of the glass plate:

=

1 Q2 1 Q2 − 2 C0 2 C

C = κC0

Express the capacitance C with the dielectric plate in place in terms of the dielectric constant κ and the aironly capacitance C0:

where C0 =

∈0 A d

.

1 Q2 1 Q2 Q2 ⎛ 1 ⎞ − = W= ⎜1 − ⎟ 2 C0 2 κC0 2C0 ⎝ κ ⎠

Substitute and factor to obtain:

Q = CV = κC0V

Use the definition of capacitance to relate the charge on the capacitor to the potential difference across its plates: Substitute to obtain:

W= =

κ 2C02V 2 ⎛

1 ⎞ κ 2C0V 2 ⎛ 1 ⎞ ⎜1 − ⎟ = ⎜1 − ⎟ 2 ⎝ κ⎠ ⎝ κ⎠

2C0

κ 2ε 0 AV 2 ⎛

1⎞ ⎜1 − ⎟ ⎝ κ⎠

2d

Substitute numerical values and evaluate W: 2 2 ( 5) (8.85 × 10−12 C 2 /N ⋅ m 2 )(1m 2 )(12 V ) ⎛ 1 ⎞ W= 1− =

(

2 0.5 × 10− 2 m

)

⎜ ⎝

⎟ 5⎠

2.55 µJ

318 Chapter 24 100 •• Picture the Problem The problem statement provides us with two conditions relating the potential between the plates of the capacitor and the charge on them. We can use the definition of capacitance to obtain simultaneous equations in Q and V and solve these equations to determine the capacitance of the capacitor and the initial and final voltages. Using the definition of capacitance, relate the initial potential between the plates of the capacitor to the charge carried by these plates: Again using the definition of capacitance, express the relationship between the charge on the capacitor and the increased voltage: Divide the second of these equations by the first to obtain:

15 µC = CVi

18 µC = CVf

= C (Vi + 6 V )

18 µC C (Vi + 6 V ) = 15 µC CVi or

6Vi = 5(Vi + 6 V )

Solve for V to obtain:

Vi = 30.0 V

and Vf = Vi + 6 V = 36.0 V Substitute in either of the first two equations to obtain:

C = 0.500 µF

101 •• Picture the Problem Let l be the variable separation of the plates. We can use the definition of the work done in charging the capacitor to relate the force on the upper plate to the energy stored in the capacitor. Solving this expression for the force and substituting for the energy stored in a parallel-plate capacitor will yield an expression that we can use to decide whether the balance is stable. We can use this same expression and a condition for equilibrium to find the voltage required to balance the object whose mass is M. (a) Express the work done in charging the capacitor (the energy stored in it) in terms of the force between the plates:

dW = dE = − Fdl or

F =−

dE dl

Electrostatic Energy and Capacitance 319 1 1 ⎛∈ A ⎞ CV 2 = ⎜ 0 ⎟V 2 2 2⎝ l ⎠

The energy stored in the capacitor is given by:

E=

Differentiate E with respect to l to obtain:

F =−

d ⎡ 1 ⎛ ∈0 A ⎞ 2 ⎤ ⎛ ∈0 A ⎞ 2 ⎜ ⎟V = ⎜ ⎟V dl ⎢⎣ 2 ⎝ l ⎠ ⎥⎦ ⎝ 2l 2 ⎠

Because F increases as l decreases, a decrease in plate separation will unbalance the system and the balance is unstable. (b) Apply

∑ F = 0 to the object whose

mass is M to obtain: Solve for V:

⎛∈ A ⎞ Mg − ⎜ 0 2 ⎟V 2 = 0 ⎝ 2l ⎠ V = l

2 Mg ∈0 A

*102 ••• Picture the Problem Recall that the dielectric strength of air is 3 MV/m. We can express the maximum energy to be stored in terms of the capacitance of the air-gap capacitor and the maximum potential difference between its plates. This maximum potential can, in turn, be expressed in terms of the maximum electric field (dielectric strength) possible in the air gap. We can solve the resulting equation for the volume of the space between the plates. In part (b) we can modify the equation we derive in part (a) to accommodate a dielectric with a constant other than 1. (a) Express the energy stored in the capacitor in terms of its capacitance and the potential difference across it: Express the capacitance of the airgap parallel-plate capacitor: Relate the maximum potential difference across the plates to the maximum electric field between them: Substitute to obtain:

2 U max = 12 CVmax

C=

∈0 A d

Vmax = Emax d

⎛∈ A ⎞ 2 2 U max = 12 ⎜ 0 ⎟(Emax d ) = 12 ∈0 ( Ad )Emax ⎝ d ⎠ 1 = 2 ∈0 υE 2 where υ = Ad is the volume between the

320 Chapter 24 plates. Solve for υ:

υ=

2U max 2 ∈0 Emax

Substitute numerical values and evaluate υ:

υ=

(8.85 × 10

(1)

−12

2(100 kJ ) 2 C 2 /N ⋅ m 2 (3 MV/m )

)

= 2.51 × 103 m3 (b) With the dielectric in place equation (1) becomes:

υ=

Evaluate equation (2) with κ = 5 and Emax = 3×108 V/m:

υ=

2U max 2 κ ∈0 Emax

(2)

2(100 kJ )

(

5 8.85 × 10

−12

)(

C 2 /N ⋅ m 2 3 × 108 V/m

)

2

= 5.02 × 10− 2 m 3 103 ••• Picture the Problem We can use the definition of capacitance to find the charge on each capacitor in part (a). In part (b) we can express the total energy stored as the sum of the energy stored on the two capacitors … using our result from (a) for the charge on each capacitor. When the dielectric is removed in part (c) each capacitor will carry half the charge carried by the capacitor system previously and we can proceed as in (b). Knowing the total charge stored by the capacitors, we can use the definition of capacitance to find the final voltage across the two capacitors in part (d). (a) Use the definition of capacitance to express the charge on each capacitor as a function of its capacitance: (b) Express the total stored energy of the capacitors as the sum of stored energy in each capacitor:

Q1 = C!V =

(200 V )C1

and

Q2 = C2V = κC1V =

(200 V )κC1

U = U1 + U 2 = 12 C1V 2 + 12 C2V 2 = 12 C1V 2 + 12 κC1V 2 = 12 C1V 2 (1 + κ ) = =

1 2

(200 V )2 C1 (1 + κ ) (2 × 104 V 2 )(1 + κ )C1

Electrostatic Energy and Capacitance 321 (c) With the dielectric removed, each capacitor carries charge Q/2. Express the final energy stored by the capacitors under this condition: Using the definition of capacitance, express the total charge carried by the capacitors with the dielectric in place in C2: Substitute to obtain:

Uf = =

1 Q12 1 Q22 1 Q 2 1 Q 2 + = + 2 C1 2 C2 2 4C1 2 4C1 Q2 4C1

Q = Q1 + Q2 = C1V + C2V

= C1V + κC1V = C1V (1 + κ ) = (200 V )C1 (1 + κ )

Uf =

[(200 V )C1 (1 + κ )]2

(

4C1

)

= 104 V 2 C1 (1 + κ ) (d) Use the definition of capacitance to express the final voltage across the capacitors:

Vf =

2

(200 V )C1 (1 + κ ) Q = Ceq 2C1

= 100(1 + κ ) V

104 ••• Picture the Problem We can use the definition of capacitance and the expression for the capacitance of a cylindrical capacitor to find the potential difference between the cylinders. In part (b) we can apply the definition of surface charge density to find the density of the free charge σf on the inner and outer cylindrical surfaces. We can use the fact that that Q and Qb are proportional to E and Eb to express Qb at a and b and then apply the definition of surface charge density to express σb(a) and σb(b). In part (d) we can use U = 12 QV to find the total stored electrostatic energy and in (e) find the mechanical work required from the change in electrostatic energy of the system resulting from the removal of the dielectric cylindrical shell. (a) Using the definition of capacitance, relate the potential difference between the cylinders to their charge and capacitance:

V =

Q C

Express the capacitance of a cylindrical capacitor as a function of its radii a and b and length L:

C=

2π ∈0 κL ln (b a )

Substitute to obtain:

V =

Q ln (b a ) 2kQ ln (b a ) = κL 2π ∈0 κL

322 Chapter 24 (b) Apply the definition of surface charge density to obtain:

σ f (a ) =

Q 2πaL

and

σ f (b ) = (c) Noting that Q and Qb are proportional to E and Eb, express Qb at a and b:

Qb (a ) = and

Qb (b ) = Apply the definition of surface charge density to express σb(a) and σb(b):

−Q 2πbL

− Q(κ − 1)

κ Q(κ − 1)

κ − Q(κ − 1)

Qb (a ) κ = A 2πaL − Q(κ − 1) = 2πaLκ

σ b (a ) =

and

Q(κ − 1)

Qb (b ) κ = 2πbL A Q(κ − 1) = 2πbLκ

σ b (b ) =

(d) Express the total stored electrostatic energy in terms of the charge stored and the potential difference between the cylinders: (e) Express the work required to remove the dielectric cylindrical shell in terms of the change in the electrostatic potential energy of the system: Substitute for U and U′ to obtain:

⎡ 2kQ ln(b a ) ⎤ U = 12 QV = 12 Q ⎢ ⎥⎦ κL ⎣ kQ 2 ln (b a ) = κL W = ∆U = U' − U where U ′ = κU is the electrostatic potential energy of the system with the dielectric shell in place.

W = κU − U = U (κ − 1) =

kQ 2 (κ − 1)ln (b a ) κL

Electrostatic Energy and Capacitance 323 105 ••• Picture the Problem Let the numeral 1 denote the 35-µF capacitor and the numeral 2 the 10-µF capacitor. We can use U = 12 CeqV 2 to find the energy initially stored in the system and the definition of capacitance to find the charges on the two capacitors. When the dielectric is removed from the capacitor the two capacitors will share the total charge stored equally. Finally, we can find the final stored energy from the total charge stored and the equivalent capacitance of the two equal capacitors in parallel. (a) Express the stored energy of the system in terms of the equivalent capacitance and the charging potential:

U = 12 CeqV 2

Express the equivalent capacitance:

Ceq = C1 + C2

Substitute to obtain:

U=

1 2

(C1 + C2 )V 2

Substitute numerical values and evaluate U:

U=

1 2

(35 µF + 10 µF)(100 V )2

(b) Use the definition of capacitance to find the charges on the two capacitors:

(c) Because the capacitors are connected in parallel, when the dielectric is removed their charges will be equal; as will be their capacitances and: (d) Express the final stored energy in terms of the total charge stored and the equivalent capacitance: Substitute numerical values and evaluate Uf:

= 0.225 J Q1 = C1V = (35 µ F )(100 V ) = 3.50 mC and

Q2 = C2V = (10 µ F )(100 V ) = 1.00 mC

Q1 = Q2 = 12 Q =

1 2

(3.5 mC + 1mC)

= 2.25 mC

2 1 Qtot Uf = 2 Qeq

1 (4.5 mC) = 0.506 J 2 2(10 µF) 2

Uf =

324 Chapter 24 *106 ••• Picture the Problem We can express the two conditions on the voltage in terms of the charges Q1 and Q2 and the capacitances C1 and C2 and solve the equations simultaneously to find Q1 and Q2. We can then use the definition of capacitance to find the initial voltages V1 and V2. Express the condition for the series connection:

V1 +V2 = 80 V or

Q1 Q2 + = 80 V C1 C2 Substitute numerical values to obtain:

Q1 Q2 + = 80 V 0.4 µF 1.2 µF or

3Q1 + Q2 = 96 µC

Use the definition of capacitance to express the condition for the parallel connection: Because the capacitors are now connected in parallel: Substitute to obtain:

(1)

Q1 + Q2 = 20 V Ceq

Ceq = C1 + C2 = 0.4 µF + 1.2 µF = 1.6 µF Q1 + Q2 = 20 V 1.6 µF or

Q1 + Q2 = 32 µC

Solve equations (1) and (2) simultaneously to obtain: Use the definition of capacitance to obtain:

(2)

Q1 = 32 µC and Q2 = 0

V1 =

Q1 32 µC = = 80.0 V C1 0.4 µF

and

V2 =

0 Q2 = = 0 C2 0.4 µF

107 ••• Picture the Problem Note that, with switch S closed, C1 and C2 are in parallel and we can use U closed = 12 CeqV 2 and Ceq = C1 + C2 to obtain an equation we can solve for C2.

Electrostatic Energy and Capacitance 325 We can use the definition of capacitance to express Q2 in terms of V2 and C2 and U open = 12 C1V12 + 12 C2V22 to obtain an equation from which we can determine V2. Express the energy stored in the capacitors after the switch is closed: Express the equivalent capacitance of C1 and C2 in parallel:

U closed = 12 CeqV 2

Ceq = C1 + C2

(C1 + C2 )V 2

Substitute to obtain:

U closed =

Solve for C2:

C2 =

2U closed − C1 V2

Substitute numerical values and evaluate C2:

C2 =

2(960 µJ ) − 0.2 µF = 0.100 µF (80 V )2

Express the charge on C2 when the switch is open: Express the energy stored in the capacitors with the switch open: Solve for V2 to obtain:

Substitute in equation (1) to obtain:

1 2

Q2 = C2V2

(1)

U open = 12 C1V12 + 12 C2V22

V2 =

2U open − C1V12

Q2 = C2

C2 2U open − C1V12

(

C2

= C2 2U open − C1V12

)

Substitute numerical values and evaluate Q2:

Q2 =

(0.1 µF)[2(1440 µJ ) − (0.2 µF)(40 V )2 ] =

16.0 µC

108 ••• Picture the Problem We can express the electric fields in the dielectric and in the free space in terms of the charge densities and then use the fact that the electric field has the same value inside the dielectric as in the free space between the plates to establish that σ1 = 2σ2. In part (c) we can model the system as two capacitors in parallel to show that

326 Chapter 24 the equivalent capacitance is 3ε0A/2d and then use the definition of capacitance to show that the new potential difference is 23 V .

The potential difference between the plates is the same for both halves (the (a) plates are equipotential surfaces). Therefore, E = V/d must be the same everywhere between the plates.

σ κ ∈0

(b) Relate the electric field in each region to σ and κ :

E=

Solve for σ :

σ = κ ∈0 E

Express σ1 and σ2:

σ 1 = κ1 ∈0 E1 = 2 ∈0 E1 and

σ 2 = κ 2 ∈0 E2 =∈0 E1

Divide the 1st of these equations by the 2nd and simplify to obtain: (c) Model the partially dielectricfilled capacitor as two capacitors in parallel to obtain:

σ 1 = 2σ 2 Ceq = C1 + C2 where

C1 = and

C2 = Substitute and simplify to obtain:

κ ∈0

( 12 A) = κ ∈0

2d

d

∈0

( 12 A) = ∈0

A 2d

d

Ceq =

A

κ ∈0 A ∈0 A +

2d 3 ∈0 A = 2d

2d

=

2 ∈0 A ∈0 A + 2d 2d

Qf Cf

Use the definition of capacitance to relate Vf, Qf, and Cf:

Vf =

Because the capacitors are in parallel:

Qf = Qi = VCi =

V ∈0 A d

Electrostatic Energy and Capacitance 327 Substitute to obtain:

Vf =

V ∈0 A V ∈0 A = = Cf d ⎛ 3 ∈0 A ⎞ ⎜ ⎟d ⎝ 2d ⎠

2 3

V

109 ••• Picture the Problem Note that when the capacitors are connected in the manner described they are in parallel with each other. Let the numeral one refer to the capacitor with the air gap and the numeral 2 to the capacitor that receives the dielectric and let primes denote physical quantitities after the insertion of the dielectric. We can find the energy stored in the system from our knowledge of the charge on and capacitance of each capacitor. In part (b) we can find the final charges on the two capacitors by first finding the equivalent capacitance and the potential difference across the modified system of capacitors. We can use the final potential difference across the system and our knowledge of the stored charge to find the final stored energy of the system. (a) Express the stored energy in the system as the sum of the energy stored in the two capacitors: Substitute numerical values and evaluate U:

U=

1 Q12 1 Q12 Q12 + = 2 C1 2 C1 C1

2 ( 100 µC ) U=

10 µF

= 1.00 mJ

(b) Relate the final charges Q1′ and Q2′ to the total charge stored by the capacitors:

Q1' + Q2' = 200 µC

Express the common potential difference across the capacitors:

V=

Express the equivalent capacitance when the dielectric is inserted between the plates of capacitor 2:

Q1' + Q2' Ceq

Ceq = C1 + C2 = C1 + κC1 = C1 (1 + κ )

Substitute to obtain:

V=

Q1' + Q2' C1 (1 + κ )

Substitute numerical values and evaluate V:

V=

200 µC = 4.76 V (10 µF)(1 + 3.2)

Use the definition of capacitance to find Q1′ and Q2′:

Q1' = VC1' = (4.76 V )(10 µF) = 47.6 µC and

328 Chapter 24 Q2' = VC2' = VκC2 = (4.76 V )(3.2)(10 µF) = 152 µC (c) Express the final stored energy of the system in terms of the total charge stored and the final potential difference across the capacitors connected in parallel:

U f = 12 QV =

1 2

(200 µC)(4.76 V )

= 0.476 mJ

*110 ••• Picture the Problem Choose a coordinate system in which the positive x direction is the right and the origin is at the left edge of the capacitor. We can express an element of capacitance dC and then integrate this expression to find C for this capacitor. Express an element of capacitance dC of length b, width dx and separation d = y0 + (y0/a)x: These elements are all in parallel, so the total capacitance is obtained by integration:

dC =

C=

∈0 b d

∈0 b y0

y0

dx =

1

∈0 b

y0 (1 + x a )

∫ 1 + x a dx = 0

dx

∈0 ab y0

ln 2

111 ••• Picture the Problem The diagram to the left shows the dielectric-filled parallel-plate capacitor before compression and the diagram to the right shows the capacitor when the plate separation has been reduced to x. We can use the definition of capacitance and the expression for the capacitance of a parallel-plate capacitor to derive an expression for the capacitance as a function of voltage across the capacitor. We can find the maximum voltage that can be applied from the dielectric strength of the dielectric and the separation of the plates. In part (c) we can find the fraction of the total energy that is electrostatic field energy and the fraction that is mechanical stress energy by expressing either of these as a fraction of their sum.

Electrostatic Energy and Capacitance 329 (a) Use its definition to express the capacitance as a function of the voltage across the capacitor:

C (V ) =

The limiting value of the capacitance is:

C0 =

Substitute numerical values and evaluate C0:

C0 =

Q V

(1)

κ ∈0 A d

(

)

3 8.85 × 10−12 C 2 /N ⋅ m 2 A 0.2 mm

= 0.133 A C 2 /N ⋅ m 3 Let x be the variable separation. Because κ is independent of x:

C (x ) =

κ ∈0 A x

and

κ ∈0 A

Q( x ) = C ( x )V =

κ ∈0 A

Substitute in equation (1) to obtain:

C (V ) = =

x V

x

V

=

V

κ ∈0 A x

(2)

κ ∈0 A d − ∆x

Q 2 (x ) 2κ ∈0 A

The force of attraction between the plates is given in Problem 95c:

F=

Substitute to obtain:

⎛ κ ∈0 A ⎞ V⎟ ⎜ 2 x ⎝ ⎠ = − κ ∈0 AV F= 2κ ∈0 A 2x2 2

where the minus sign is used to indicate that the force acts to decrease the plate separation x. Apply Hooke’s law to relate the stress to the strain:

Y=

F A ∆x x

or

∆x F = x YA

Substitute for F to obtain:

κ ∈0 V 2 ∆x =− x 2Yx2

330 Chapter 24 and

∆x = −

κ ∈0 V 2 2Yx

=−

κ ∈0 V 2 2Yd

(3)

provided ∆x << d

V = Emax d = (40 kV/mm )(0.2 mm )

The voltage across the capacitor is:

= 8.00 kV 3(8.85 × 10 −12 C 2 / N ⋅ m 2 )(8 kV ) 2(5 × 106 N/m 2 )(0.2 mm )

2

Substitute numerical values in equation (3) and evaluate ∆x:

∆x = −

= 8.50 × 10− 7 m = 8.50 × 10 − 4 mm Substitute in equation (2) to obtain:

C (V ) =

−1

⎛ κ ∈0 V 2 ⎞ ⎛ κ ∈0 V 2 ⎞ κ ∈0 A κ ∈0 A ⎜ ⎟ = C ≈ = − 1 C 0⎜ 0⎜ ⎜1 + 2Yd 2 ⎟⎟ κ ∈0 V 2 2Yd 2 ⎟⎠ ⎛ κ ∈0 V 2 ⎞ ⎝ ⎝ ⎠

d−

2Yd

⎟ d ⎜⎜1 − 2Yd 2 ⎟⎠ ⎝

provided ∆x << d. (b) Express the maximum voltage that can be applied in terms of the maximum electric field:

(

)

Vmax = Emax (d − ∆x ) = (40 kV/mm ) 0.2 mm − 8.5 × 10-4 mm = 7.97 kV (c) The fraction of the total energy of the capacitor that is mechanical stress energy is: Express the maximum electric field energy:

f =

UM UM +UE

(4)

2 U E, max = 12 C (Vmax )Vmax

Evaluate C(Vmax):

[ (

)

]

C (7.97 kV ) = (0.133 A ) 1 + 6.64 × 10 −11 (7.97 kV ) µF / m 2 = 0.134 A µF / m 2 Substitute for C(Vmax) and evaluate UE,max:

2

U E,max =

1 2

(0.134 A µF/m )(7.96 kV )

= 4.25 J/m 2

2

2

Electrostatic Energy and Capacitance 331 1 (∆x ) Y 2 d 2

The mechanical stress energy is given by:

UM =

Substitute numerical values and evaluate UM:

1 8.5 × 10 −4 mm 5 × 106 N/m 2 UM = 2 0.2 mm

(

)( 2

)

= 8.92 mJ Substitute numerical values in equation (4) and evaluate f:

f =

8.92 mJ = 0.209% 8.92 mJ + 4.25 J/m 2

and the fraction of the total energy that is electrostatic field energy is

1 − f = 1 − 0.209% = 99.8 % 112 ••• Picture the Problem Note that, due to symmetry, the electric field, wherever it exists, will be radial. We can integrate the electric flux over spherical Gaussian surfaces with radii r < R1, R1 < r < R2, and r > R2 to find the electric field everywhere in space. Once we know the electric field everywhere we can find the potential of the conducting sphere by using dV = −Edr and integrating E in the regions R1 < r < R2 and r > R2. Finally, knowing the electrostatic potential at the surface of the conducting sphere we can use U tot = 12 QV (R1 ) to find the total electrostatic potential energy of the system. (a) Integrate the electric flux over a spherical Gaussian surface with radius r < R1 to obtain:

Solve for Er(r < R1) to obtain:

(

)

Er 4π r 2 =

Qinside

∈0

because Qinside the conducting surface is zero.

Er (r < R1 ) = 0

(

)

Qinside

Integrate the electric flux over a spherical Gaussian surface with radius R1 < r < R2 to obtain:

Er 4π r 2 =

Solve for Er(R1 < r < R2) to obtain:

Er (R1 < r < R2 ) =

Integrate the electric flux over a spherical Gaussian surface with radius r > R2 to obtain:

=0

(

)

Er 4π r 2 =

κ ∈0

Qinside

∈0

=

Q

κ ∈0

Q 4π ∈0 κr =

Q

∈0

2

=

kQ κr 2

332 Chapter 24 Solve for Er(r > R2) to obtain:

(b) Express the potential at the surface of the conducting sphere in terms of the electric fields Er(R1 < r < R2) and Er(r > R2):

Er (r > R2 ) =

4π ∈0 r

2

=

kQ r2

R1

V (R1 ) = − ∫ Edr ∞

R2

R

kQ 1 1 1 dr = −kQ ∫ 2 dr − r κ R∫2 r 2 ∞ =

(c) Express the total electrostatic potential energy of the system in terms of V(R1) and Q:

Q

kQ ⎛ R1 (κ − 1) + R2 ⎞ ⎜ ⎟⎟ R1 R2 κ ⎜⎝ ⎠

U tot = 12 QV (R1 ) ⎛ kQ ⎛ R1 (κ − 1) + R2 ⎞ ⎞ ⎜⎜ ⎟⎟ ⎟ = 12 Q⎜⎜ ⎟ R1R2 ⎠⎠ ⎝ κ ⎝ =

kQ 2 ⎛ R1 (κ − 1) + R2 ⎞ ⎜ ⎟⎟ 2κ ⎜⎝ R1R2 ⎠

Chapter 25 Electric Current and Direct-Current Circuits Conceptual Problems *1 • Determine the Concept When current flows, the charges are not in equilibrium. In that case, the electric field provides the force needed for the charge flow. 2 • Determine the Concept Water, regarded as a viscous liquid flowing from a water tower through a pipe to ground is another mechanical analog of a simple circuit. 3 • Picture the Problem The resistances of the wires are given by R = ρL A , where L is the length of the wire and A is its cross-sectional area. We can express the ratio of the resistances and use our knowledge of their lengths and diameters to find the resistance of wire A.

Express the resistance of wire A:

RA =

ρLA AA

where ρ is the resistivity of the wire. Express the resistance of wire B:

Divide the first of these equations by the second to obtain:

R=

ρLB AB

ρLA RA L A A = A = A⋅ B ρLB LB AA R AB or, because LA = LB,

RA = Express the area of wire A in terms of its diameter:

AB R AA

AA = 14 πd A2

333

(1)

334 Chapter 25 Express the area of wire B in terms of its diameter: Substitute in equation (1) to obtain:

AB = 14 πd B2

RA =

d B2 R d A2

or, because dA = 2dB,

RA =

d B2 R = 14 R 2 (2d B )

and (e) is correct. 4 •• Determine the Concept An emf is a source of energy that gives rise to a potential difference between two points and may result in current flow if there is a conducting path whereas a potential difference is the consequence of two points in space being at different potentials. *5 •• Picture the Problem The resistance of the metal bar varies directly with its length and inversely with its cross-sectional area. Hence, to minimize the resistance of the bar, we should connect to the surface for which the ratio of the length to the contact area is least. Denoting the surfaces as a, b, and c, complete the table to the right:

Surface a b c

Because connecting to surface c minimizes R:

(c) is correct.

L 10 4 2

A 8 20 40

L/A 0.8 0.2 0.05

6 •• Picture the Problem The resistances of the wires are given by R = ρL A , where L is the length of the wire and A is its cross-sectional area. We can express the ratio of the resistances and use the definition of density to eliminate the cross-sectional areas of the wires in favor of the ratio of their lengths.

`Electric Current and Direct-Current Circuits 335

Express the resistance of wire A:

RA =

ρLA AA

where ρ is the resistivity of copper. Express the resistance of wire B:

Divide the first of these equations by the second to obtain:

RB =

ρLB AB

ρLA A L A RA = A = A B RB ρLB LB AA AB or, because LA = 2LB,

RA = 2

AB RB AA

(1)

mA = ρ'VA = ρ'LA AA

Using the definition of density, express the mass of wire A:

where ρ′ is the density of copper.

Express the mass of wire B

mB = ρ'VB = ρ'LB AB

Because the masses of the wires are equal:

ρ'LA AA = ρ'LB AB or

AB LA = AA LB Substitute in equation (1) to obtain:

RA = 2

LA RB = 2(2)RB = 4 RB LB

and (b) is correct. 7 • Picture the Problem The power dissipated in the resistor is given by P = I2R. We can express the power dissipated when the current is 3I and, assuming that the resistance does not change, express the ratio of the two rates of energy dissipation to find the power dissipated when the current is 3I. Express the power dissipated in the

P = I 2R

336 Chapter 25 resistor when the current in it is I: Express the power dissipated in the resistor when the current in it is 3I:

P' = (3I ) R = 9 I 2 R

Divide the second of these equations by the first to obtain:

P' 9 I 2 R = 2 =9 P I R

2

or

P' = 9 P and (d ) is correct. 8 • Picture the Problem Assuming the current (which depends on the resistance) to be constant, the power dissipated in a resistor is directly proportional to the voltage drop across it. Express the power dissipated in the resistor when the voltage drop across it is V: Express the power dissipated in the resistor when the voltage drop across it is increased to 2V: Divide the second of these equations by the first to obtain:

P=

V2 R

P' =

(2V )2 R

=

4V 2 R

4V 2 P' = R2 = 4 ⇒ P' = 4 P V P R (c) is correct.

9 • Determine the Concept You should decrease the resistance. Because the voltage across the resistor is constant, the heat out is given by P = V 2 R . Hence, decreasing the resistance will increase P. *10 • Picture the Problem We can find the equivalent resistance of this two-resistor combination and then apply the condition that R1 >> R2. Express the equivalent resistance of R1 and R2 in parallel:

1 1 1 = + Req R1 R2

`Electric Current and Direct-Current Circuits 337

R1 R2 R1 + R2

Solve for Req to obtain:

Req =

Factor R1 from the denominator and simplify to obtain:

Req =

If R1 >> R2, then:

Req = Reff ≈ R2 and (b) is correct

R1R2 R2 = R2 ⎛ R ⎞ R1 ⎜⎜1 + 2 ⎟⎟ 1 + R1 ⎝ R1 ⎠

11 • Picture the Problem We can find the equivalent resistance of this two-resistor combination and then apply the condition that R1 >> R2. Express the equivalent resistance of R1 and R2 in series:

Req = R1 + R2

Factor R1 to obtain:

⎛ R ⎞ Req = R1 ⎜⎜1 + 2 ⎟⎟ ⎝ R1 ⎠

If R1 >> R2, then:

Req = Reff ≈ R1 and (a ) is correct

12 • Picture the Problem Because the potential difference across resistors connected in parallel is the same for each resistor; we can use Ohm’s law to relate the currents through the resistors to their resistances. Using Ohm’s law, express the current carried by resistor A:

IA =

V V = RA 2RB

Using Ohm’s law, express the current carried by resistor B:

IB =

V RB

Divide the second of these equations by the first to obtain:

V IB R = B =2 V IA 2 RB and

I B = 2 I A and (b) is correct.

338 Chapter 25 *13 • Determine the Concept In a series circuit, because there are no alternative pathways, all resistors carry the same current. The potential difference across each resistor, keeping with Ohm’s law, is given by the product of the current and the resistance and, hence, is not the same across each resistor unless the resistors are identical. (a ) is correct. 14 •• Picture the Problem Because the potential difference across the two combinations of resistors is constant, we can use P = V 2 R to relate the power delivered by the battery to the equivalent resistance of each combination of resistors. Express the power delivered by the battery when the resistors are connected in series: Letting R represent the resistance of the identical resistors, express Req: Substitute to obtain:

Express the power delivered by the battery when the resistors are connected in parallel: Express the equivalent resistance of the identical resistors connected in parallel: Substitute to obtain:

Divide equation (2) by equation (1) to obtain:

Solve for and evaluate Pp:

Ps =

V2 Req

Req = R + R = 2 R

Ps =

V2 2R

Pp =

V2 Req

Req =

Pp =

(1)

(R )(R ) = 1 R R+R

2

V 2 2V 2 = 1 R 2 R

(2)

2V 2 Pp = R2 = 4 V Ps 2R

Pp = 4 Ps = 4(20 W ) = 80 W and (e) is correct.

`Electric Current and Direct-Current Circuits

339

15 • Determine the Concept While Kirchhoff’s loop rule is a statement about potential differences around a closed loop in a circuit, recall that electric potential at a point in space is the work required to bring a charged object from infinity to the given point. Hence, the loop rule is actually a statement that energy is conserved around any closed path in an electric circuit. (b) is correct. 16 • Determine the Concept An ideal voltmeter would have infinite resistance. A voltmeter consists of a galvanometer movement connected in series with a large resistance. The large resistor accomplishes two purposes; 1) it protects the galvanometer movement by limiting the current drawn by it, and 2) minimizes the loading of the circuit by the voltmeter by placing a large resistance in parallel with the circuit element across which the potential difference is being measured. (a ) is correct. *17 • Determine the Concept An ideal ammeter would have zero resistance. An ammeter consists of a very small resistance in parallel with a galvanometer movement. The small resistance accomplishes two purposes: 1) It protects the galvanometer movement by shunting most of the current in the circuit around the galvanometer movement, and 2) It minimizes the loading of the circuit by the ammeter by minimizing the resistance of the ammeter. (b) is correct. 18 • Determine the Concept An ideal voltage source would have zero internal resistance. The terminal potential difference of a voltage source is given by V = ε − Ir, where ε is the emf of the source, I is the current drawn from the source, and r is the internal resistance of the source. (b) is correct. 19 • Determine the Concept If we apply Kirchhoff’s loop rule with the switch closed, we obtain ε − IR – VC = 0. Immediately after the switch is closed, I = 0 and we have ε = VC.

(b) is correct. 20 •• Determine the Concept The energy stored in the fully charged capacitor is U = 12 Cε 2 . During the charging process, a total charge Qf = εC flows through the battery. The battery therefore does work W = Qfε = Cε 2. The energy dissipated in the resistor is the difference between W and U. (b) is correct.

340 Chapter 25 *21 •• Determine the Concept Applying Kirchhoff’s loop rule to the circuit, we obtain ε − VR − VC = 0 , where VR is the voltage drop across the resistor. Applying Ohm’s law to the resistor, we obtain VR = IR. Because I decreases as the capacitor is charged, VR decreases with time.

(e) is correct.

22 •• Picture the Problem We can express the variation of charge on the discharging capacitor as a function of time to find the time T it takes for the charge on the capacitor to drop to half its initial value. We can also express the energy remaining in the electric field of the discharging capacitor as a function of time and find the time t′ for the energy to drop to half its initial value in terms of T. Express the dependence of the charge stored on a capacitor on time: For Q(t) = ½ Q0:

Q(t ) = Q0e −t τ where τ = RC. 1 2

Q0 = Q0 e −T τ

or

= e −T τ

1 2

Take the natural logarithm of both sides of the equation and solve for T to obtain: Express the dependence of the energy stored in a capacitor on the potential difference VC across its terminals:

T = τ ln 2

U (t ) = 12 CVC2

Express the potential difference across a discharging capacitor as a function of time:

VC = V0e −t RC

Substitute to obtain:

U (t ) = 12 C V0e −t RC

(

= U 0e −2t RC For U(t) = ½ U0:

1 2

U 0 = U 0 e −2t' RC

or 1 2

= e −2t ' RC

)

2

= 12 CV02e −2t RC

`Electric Current and Direct-Current Circuits

Take the natural logarithm of both sides of the equation and solve for t′ to obtain:

t' = 12 τ ln 2 =

1 2

341

T

23 • Determine the Concept A small resistance because P = ε2/R. *24 • Determine the Concept The potential difference across an external resistor of resistance R is given by

R V, where r is the internal resistance and V the voltage supplied by the r+R

source. The higher R is, the higher the voltage drop across R. Put differently, the higher the resistance a voltage source sees, the less its own resistance will change the circuit. 25 • Determine the Concept Yes. Kirchhoff’s rules are statements of the conservation of energy and charge and hence apply to all circuits. 26 •• Determine the Concept All of the current provided by the battery passes through R1, whereas only half this current passes through R2 and R3. Because P = I 2R, the power dissipated in R1 will be four times that dissipated in R2 and R3. (c) is correct.

Estimation and Approximation 27 •• Picture the Problem We can use Ohm’s law and the definition of resistivity to find the maximum voltage that can be applied across 40 m of the 16-gauge copper wire. In part (b) we can find the electric field in the wire using E = V/L. In part (c) we can use P = I 2 R to find the power dissipated in the wire when it carries 6 A. (a) Use Ohm’s law to relate the potential difference across the wire to its maximum current and its resistance:

Vmax = I max R

Use the definition of resistivity to relate the resistance of the wire to its length and cross-sectional area:

R=ρ

Substitute to obtain:

Vmax = I max ρ

L A

L A

342 Chapter 25 Substitute numerical values (see Tables 25-1 and 25-2) for the resistivity of copper and the crosssectional area of 16-gauge wire:

(

Vmax = (6 A ) 1.7 × 10 −8 Ω ⋅ m

)

⎛ 40 m ⎞ ⎟ × ⎜⎜ 2 ⎟ ⎝ 1.309 mm ⎠ = 3.12 V V 3.12 V = = 78.0 mV/m L 40 m

(b) Relate the electric in the wire to the potential difference between its ends and the length of the wire:

E=

(c) Relate the power dissipated in the wire to the current in and the resistance of the wire:

P = I 2R

Substitute for R to obtain:

P = I 2ρ

Substitute numerical values and evaluate P:

⎛ 40 m ⎞ 2 ⎟ P = (6 A ) 1.7 × 10 −8 Ω ⋅ m ⎜⎜ 2 ⎟ ⎝ 1.309 mm ⎠

L A

(

)

= 18.7 W 28 •• Picture the Problem We can use the definition of resistivity to find the resistance of the jumper cable. In part (b), the application of Ohm’s law will yield the potential difference across the jumper cable when it is starting a car, and, in part (c), we can use the expression for the power dissipated in a conductor to find the power dissipation in the jumper cable.

L A

(a) Noting that a jumper cable has two leads, express the resistance of the cable in terms of the wire’s resistivity and the cable’s length, and cross-sectional area:

R=ρ

Substitute numerical values (see Table 25-1 for the resistivity of copper) and evaluate R:

R = 1.7 × 10−8 Ω ⋅ m

(b) Apply Ohm’s law to the cable to obtain:

V = IR = (90 A )(0.0102 Ω ) = 0.918 V

(

m )106mm

2

= 0.0102 Ω

`Electric Current and Direct-Current Circuits 343

(c) Use the expression for the power dissipated in a conductor to obtain:

P = IV = (90 A )(0.918 V ) = 82.6 W

29 •• Picture the Problem We can combine the expression for the rate at which energy is delivered to the water to vaporize it (P = ε 2/R) and the expression for the resistance of a conductor ( R = ρL A ) to obtain an expression for the required length L of wire. Use an expression for the power dissipated in a resistor to relate the required resistance to rate at which energy is delivered to generate the steam: Relate the resistance of the wire to its length, cross-sectional area, and resistivity: Equate these two expressions and solve for L to obtain: Express the power required to generate the steam in terms of the rate of energy delivery: Substitute to obtain:

Substitute numerical values (see Table 25-1 for the resistivity of Nichrome and Table 18-2 for the latent heat of vaporization of water) and evaluate L:

R=

ε2 P

R=ρ

L=

P=

L=

L A

ε 2A ρP ∆E ∆(mLv ) ∆m = = Lv ∆t ∆t ∆t

ε 2A ρLv

∆m ∆t

(120 V )2 π (1.80 mm)2 4 L= (10−6 Ω ⋅ m)(2257 kJ/kg )(8 g/s) = 2.03 m

*30 •• Picture the Problem We can find the annual savings by taking into account the costs of the two types of bulbs, the rate at which they consume energy and the cost of that energy, and their expected lifetimes. Express the yearly savings:

∆$ = Cost incandescent − Cost fluorescent

(1)

344 Chapter 25 Express the annual cost with the incandescent bulbs:

Cost incandescent = Cost bulbs + Cost energy

Express and evaluate the annual cost of the incandescent bulbs:

Cost bulbs = number of bulbs in use × annual consumption of bulbs × cost per bulb 24 h ⎞ ⎛ ⎟ ⎜ 365.24 d × d ⎟ ($1.50 ) = $65.74 = (6 ) ⎜ 1200 h ⎟ ⎜ ⎟ ⎜ ⎠ ⎝ Find the cost of operating the incandescent bulbs for one year:

Cost energy = energy consumed × cost per unit of energy

= 6(75 W )(365.25 d )(24 h/d )($0.115 / kW ⋅ h ) = $453.64

Express the annual cost with the fluorescent bulbs:

Cost fluorescent = Cost bulbs + Cost energy

Express and evaluate the annual cost of the fluorescent bulbs:

Cost bulbs = number of bulbs in use × annual consumption of bulbs × cost per bulb 24 h ⎞ ⎛ ⎜ 365.24 d × ⎟ d ⎟ ($6 ) = $39.45 = (6 ) ⎜ 8000 h ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ Find the cost of operating the fluorescent bulbs for one year:

Cost energy = energy consumed × cost per unit of energy 24 h ⎞ ⎛ = 6(20 W ) ⎜ 365.24 d × ⎟ ($0.115 / kW ⋅ h ) d ⎠ ⎝ = $120.97 Substitute in equation (1) and evaluate the cost savings ∆$:

∆$ = Cost incandescent − Cost fluorescent = ($65.74 + $453.64) − ($39.45 + $120.97 ) = $358.96

`Electric Current and Direct-Current Circuits 345

31 •• Picture the Problem We can use an expression for the power dissipated in a resistor to relate the Joule heating in the wire to its resistance and the definition of resistivity to relate the resistance to the length and cross-sectional area of the wire. Express the power the wires must dissipate in terms of the current they carry and their resistance:

P = I 2R

Divide both sides of the equation by L to express the power dissipation per unit length:

P I 2R = L L

Using the definition of resistivity, relate the resistance of the wire to its resistivity, length and crosssectional area:

R=ρ

Substitute to obtain:

P 4 ρI 2 = L πd 2

Solve for d to obtain:

Substitute numerical values (see Table 25-1 for the resistivity of copper wire) and evaluate d:

d = 2I

4 ρL L L =ρπ 2 = A πd 2 4 d

ρ

π (P / L )

d = 2(20 A )

1.7 × 10 −8 Ω ⋅ m π (2 W/m )

= 2.08 mm

*32 •• Picture the Problem Let r be the internal resistance of each battery and use Ohm’s law to express the current in laser diode as a function of the potential difference across r. We can find the power of the laser diode from the product of the potential difference across the internal resistance of the batteries and the current delivered by them I and the time-todischarge from the combined capacities of the two batteries and I. (a) Use Ohm’s law to express the current in the laser diode: The potential difference across the internal resistance is:

I=

Vinternal resistance 2r

Vinternal resistance = ε − 2.3 V

346 Chapter 25 Substitute to obtain:

Assuming that r = 125 Ω:

(b) The power delivered by the batteries is given by:

I=

I=

ε − 2.3 V 2r 2(1.55) − 2.3 V = 3.20 mA 2(125 Ω )

P = IV = (3.2 mA )(2.3 V ) = 7.36 mW

(7.36 mW ) =

The power of the laser is half this value:

Plaser = 12 P =

Express the ratio of Plaser to Pquoted:

Plaser 3.68 mW = = 1.23 Pquoted 3 mW

1 2

3.68 mW

or

Plaser = 123% Pquoted (c) Express the time-to-discharge:

∆t =

Capacity I

Because each battery has a capacity of 20 mA⋅h, the series combination has a capacity of 40 mA⋅h and:

∆t =

40 mA ⋅ h = 12.5 h 3.20 mA

Current and the Motion of Charges 33 • Picture the Problem We can relate the drift velocity of the electrons to the current density using I = nevd A . We can find the number density of charge carriers n using

n = ρN A M , where ρ is the mass density, NA Avogadro’s number, and M the molar mass. We can find the cross-sectional area of 10-gauge wire in Table 25-2. Use the relation between current and drift velocity to relate I and n:

I = nevd A

Solve for vd:

vd =

I neA

`Electric Current and Direct-Current Circuits 347

The number density of charge carriers n is related to the mass density ρ, Avogadro’s number NA, and the molar mass M: For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Substitute and evaluate n:

n=

ρN A

n=

(8.93 g/cm )(6.02 × 10

M

3

23

atoms/mol

)

63.5 g/mol

= 8.47 × 10 28 atoms/m 3

Using Table 25-2, find the crosssectional area A of 10-gauge wire:

A = 5.261 mm 2

Substitute and evaluate vd:

vd =

(8.47 × 10

28

m

−3

20 A = 0.281 mm/s 1.60 × 10 −19 C 5.261 mm 2

)(

)(

)

34 • Picture the Problem Note that, while the positive and negative charges flow in opposite directions, the total current is their sum. Express the total current I in the tube as the sum of the electron current and the ion current:

I = I electron + I ion

The electron current is the product of the number of electrons through the cross-sectional area each second and the charge of each electron:

I electron = ne

Proceed in the same manner to find the ion current:

I ion = nion qion

(

= 2 × 1018 electrons/s

(

)

× 1.60 × 10 −19 C/electron

)

= 0.320 A

(

= 0.5 × 1018 electrons/s

(

)

× 1.60 × 10 −19 C/electron

)

= 0.0800 A Substitute to obtain:

I = 0.320 A + 0.0800 A = 0.400 A

35 • Picture the Problem We can solve K = 12 me v 2 for the velocity of an electron in the beam and use the relationship between current and drift velocity to find the beam current.

348 Chapter 25 K = 12 me v 2

(a) Express the kinetic energy of the beam: Solve for v:

Substitute numerical values and evaluate v:

v=

2K me

v=

2(10 keV ) 1.60 × 10 −19 J/eV 9.11 × 10 −31 kg

(

)

= 5.93 × 10 7 m/s (b) Use the relationship between current and drift velocity (here the velocity of an electron in the beam) to obtain:

I = nevd A

Express the cross-sectional area of the beam in terms of its diameter D:

A = 14 πD 2

Substitute to obtain:

I = 14 πnevd D 2

Substitute numerical values and evaluate I: Substitute numerical values and evaluate I:

(

)(

)(

)(

)

2

I = 14 π 5 × 10 6 cm −3 1.60 × 10 −19 C 5.93 × 10 7 m/s 10 −3 m = 37.3 µA 36 •• Picture the Problem We can use the definition of current, the definition of charge density, and the relationship between period and frequency to derive an expression for the current as a function of a, λ, and ω. Use the definition of current to relate the charge ∆Q associated with a segment of the ring to the time ∆t it takes the segment to pass a given point:

I=

∆Q ∆t

Because each segment carries a charge ∆Q and the time for one

I=

∆Q = ∆Qf T

(1)

`Electric Current and Direct-Current Circuits 349

revolution is T:

∆Q 2πa

Use the definition of the charge density λ to relate the charge ∆Q to the radius a of the ring:

λ=

Solve for ∆Q to obtain:

∆Q = 2πaλ

Substitute in equation (1) to obtain:

I = 2πaλf

Because ω = 2πf we have:

I = aλω

*37 •• Picture the Problem The current will be the same in the two wires and we can relate the drift velocity of the electrons in each wire to their current densities and the crosssectional areas of the wires. We can find the number density of charge carriers n using n = ρN A M , where ρ is the mass density, NA Avogadro’s number, and M the molar mass. We can find the cross-sectional area of 10- and 14-gauge wires in Table 25-2. Relate the current density to the drift velocity of the electrons in the 10gauge wire: Solve for vd:

The number density of charge carriers n is related to the mass density ρ, Avogadro’s number NA, and the molar mass M: For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Substitute and evaluate n: Use Table 25-2 to find the crosssectional area of 10-gauge wire: Substitute numerical values and evaluate vd.10:

I10 gauge

= nevd

A10 gauge

vd, 10 =

n=

I10 gauge neA10 gauge

ρN A M

(8.93 g/cm )(6.02 × 10 n= 3

23

63.5 g/mol

= 8.47 × 10 28 atoms/m 3 A10 = 5.261 mm 2

atoms/mol)

350 Chapter 25 vd, 10 =

(8.47 × 10

28

m

−3

15 A = 0.210 mm/s 1.60 × 10 −19 C 5.261 mm 2

)(

Express the continuity of the current in the two wires:

)(

)

I10 gauge = I14 gauge or

nevd,10 A10 gauge = nevd,14 A14 gauge Solve for vd,14 to obtain:

vd,14 = vd,10

A10 gauge A14 gauge

Use Table 25-2 to find the crosssectional area of 14-gauge wire:

A14 = 2.081 mm 2

Substitute numerical values and evaluate vd,14:

vd,14 = (0.210 mm/s)

5.261 mm 2 2.081 mm 2

= 0.531 mm/s 38 •• Picture the Problem We can use I = neAv to relate the number n of protons per unit volume in the beam to current I. We can find the speed of the particles in the beam from their kinetic energy. In part (b) we can express the number of protons N striking the target per unit time as the product of the number of protons per unit volume n in the beam and the volume of the cylinder containing those protons that will strike the target in an elapsed time ∆t and solve for N. Finally, we can use the definition of current to express the charge arriving at the target as a function of time. (a) Use the relation between current and drift velocity to relate I and n:

I = neAv

Solve for n to obtain:

n=

Express the kinetic energy of the protons and solve for v:

K = 12 mp v 2 ⇒ v =

Relate the cross-sectional area A of the beam to its diameter D:

I eAv

A = 14 πD 2

2K mp

`Electric Current and Direct-Current Circuits 351

Substitute for v and A to obtain:

I

n= 1 4

πeD 2

2K mp

=

4I πeD 2

mp 2K

Substitute numerical values and evaluate n:

n=

4(1 mA ) π (1.60 × 10 −19 C )(2 mm )2

1.67 × 10 −27 kg = 3.21 × 1013 mm −3 −19 2(20 MeV ) (1.60 × 10 J/eV )

(b) Express the number of protons N striking the target per unit time as the product of the number n of protons per unit volume in the beam and the volume of the cylinder containing those protons that will strike the target in an elapsed time ∆t and solve for N: Substitute for v and A to obtain:

N = n(vA) ⇒ N = nvA∆t ∆t

N = 14 πD 2 n∆t

2K mp

Substitute numerical values and evaluate N:

N = 14 π (2 mm ) (3.21 × 1013 m −3 )(1 min ) 2

(c) Using the definition of current, express the charge arriving at the target as a function of time:

2(20 MeV ) (1.60 × 10 −19 J/eV ) = 3.75 × 1017 − 27 1.67 × 10 kg

Q = It = (1 mA ) t = (1 mC/s ) t

*39 •• Picture the Problem We can relate the number of protons per meter N to the number n of free charge-carrying particles per unit volume in a beam of cross-sectional area A and then use the relation between current and drift velocity to relate n to I. (a) Express the number of protons per meter N in terms of the number n of free charge-carrying particles per unit volume in a beam of cross-

N = nA

(1)

352 Chapter 25 sectional area A: Use the relation between current and drift velocity to relate I and n:

I = enAv

Solve for n to obtain:

n=

Substitute to obtain:

N=

IA I = eAv ev

Substitute numerical values and evaluate N:

N=

5 mA 1.60 × 10 C 3 × 108 m/s

I eAv

(

−19

)(

)

= 1.04 × 108 m −1 (b) From equation (1) we have:

n=

N 1.04 × 108 m −1 = A 10−6 m 2

= 1.04 × 1014 m −3

Resistance and Ohm’s Law 40 • Picture the Problem We can use Ohm’s law to find the potential difference between the ends of the wire and V = EL to find the magnitude of the electric field in the wire. (a) Apply Ohm’s law to obtain:

V = RI = (0.2 Ω )(5 A ) = 1.00 V

(b) Relate the electric field to the potential difference across the wire and the length of the wire:

E=

V 1V = = 0.100 V/m L 10 m

41 • Picture the Problem We can apply Ohm’s law to both parts of this problem, solving first for R and then for I. (a) Apply Ohm’s law to obtain:

R=

V 100 V = = 33.3 Ω I 3A

`Electric Current and Direct-Current Circuits 353

(b) Apply Ohm’s law a second time to obtain:

I=

V 25 V = = 0.751 A R 33.3 Ω

42 • Picture the Problem We can use R = ρL A to find the resistance of the block and Ohm’s law to find the current in it for the given potential difference across its length.

L A

(a) Relate the resistance of the block to its resistivity ρ, cross-sectional area A, and length L:

R=ρ

Substitute numerical values (see Table 2625-1 for the resistivity of carbon) and evaluate R:

R = 3500 × 10−8 Ω ⋅ m

(b) Apply Ohm’s law to obtain:

I=

(

)

3 cm (0.5 cm )2

= 42.0 mΩ V 8.4 V = = 200 A R 42.0 mΩ

43 • Picture the Problem We can solve the relation R = ρL A for L to find the length of the carbon rod that will have a resistance of 10 Ω. Relate the resistance of the rod to its resistivity ρ, cross-sectional area A, and length L: Solve for L to obtain:

Substitute numerical values (see Table 2625-1 for the resistivity of carbon) and evaluate L:

R=ρ

L=

L=

L A

AR

ρ

=

πr 2 R ρ

π (0.1 mm )2 (10 Ω ) 3500 × 10 −8 Ω ⋅ m

= 8.98 mm

*44 • Picture the Problem We can use R = ρL A to find the resistance of the track. (a) Relate the resistance of the track to its resistivity ρ, cross-sectional area A, and length L:

R=ρ

L A

354 Chapter 25 Substitute numerical values and evaluate R:

(

R = 10 − 7 Ω ⋅ m

10 km ) 55 cm

2

= 0.182 Ω

45 • Picture the Problem We can use Ohm’s law in conjunction with R = ρL A to find the potential difference across one wire of the extension cord. Using Ohm’s law, express the potential difference across one wire of the extension cord:

V = IR

Relate the resistance of the wire to its resistivity ρ, cross-sectional area A, and length L:

R=ρ

L A

Substitute to obtain:

V =ρ

LI A

Substitute numerical values (see Table 25-1 for the resistivity of copper and Table 25-2 for the crosssectional area of 16-gauge wire) and evaluate V:

V = (1.7 × 10 −8 Ω ⋅ m )

(30 m )(5 A ) 1.309 mm 2

= 1.95 V

46 • Picture the Problem We can use R = ρL A to find the length of a 14-gauge copper wire that has a resistance of 2 Ω. (a) Relate the resistance of the wire to its resistivity ρ, cross-sectional area A, and length L:

R=ρ

Solve for L to obtain:

L=

Substitute numerical values (see Table 2625-1 for the resistivity of copper and Table 2625-2 for the cross-sectional area of 14-gauge wire) and evaluate L:

L A

RA

ρ

( 2 Ω ) (2.081 mm 2 ) L= = 1.7 × 10 −8 Ω ⋅ m

245 m

`Electric Current and Direct-Current Circuits 355

47 •• Picture the Problem We can use R = ρL A to express the resistances of the glass cylinder and the copper wire. Expressing their ratio will eliminate the common crosssectional areas and leave us with an expression we can solve for the length of the copper wire. Relate the resistance of the glass cylinder to its resistivity, crosssectional area, and length:

Rglass = ρ glass

Relate the resistance of the copper wire to its resistivity, crosssectional area, and length:

RCu = ρ Cu

Divide the second of these equations by the first to obtain:

RCu = Rglass

Lglass Aglass

LCu ACu

ρ Cu ρ glass

LCu L ρ ACu = Cu ⋅ Cu Lglass ρ glass Lglass Aglass

or, because Aglass = ACu and RCu = Rglass,

1=

Solve for LCu to obtain:

Substitute numerical values (see Table 251 for the resistivities of glass and copper) and evaluate LCu:

ρ Cu LCu ⋅ ρ glass Lglass

LCu =

ρ glass L ρ Cu glass

1012 Ω ⋅ m (1cm ) 1.7 ×10 −8 Ω ⋅ m 1c ⋅ y = 5.88 ×1017 m × 9.461× 1015 m

LCu =

= 62.2 c ⋅ y

356 Chapter 25 48 •• Picture the Problem We can use Ohm’s law to relate the potential differences across the two wires to their resistances and R = ρL A to relate their resistances to their lengths, resistivities, and cross-sectional areas. Once we’ve found the potential differences across each wire, we can use E = V L to find the electric field in each wire. (b) Apply Ohm’s law to express the potential drop across each wire:

VCu = IRCu and

VFe = IRFe Relate the resistances of the wires to their resistivity, cross-sectional area, and length:

and

RFe = ρ Fe Substitute to obtain:

VCu = and

VFe = Substitute numerical values (see Table 2625-1 for the resistivities of copper and iron) and evaluate the potential differences:

LCu ACu

RCu = ρ Cu

VCu =

LFe AFe

ρ Cu LCu

I

ACu

ρ Fe LFe

I

AFe

(1.7 × 10 1 4

)

Ω ⋅ m (80 m ) (2 A ) π (1 mm )2 −8

= 3.46 V and

VFe =

(10 × 10 1 4

)

Ω ⋅ m (49 m ) (2 A ) π (1 mm )2 −8

= 12.5 V (a) Express the electric field in each conductor in terms of its length and the potential difference across it:

Substitute numerical values and evaluate the electric fields:

ECu =

VCu LCu

and

EFe =

VFe LFe

ECu =

3.46 V = 43.3 mV/m 80 m

`Electric Current and Direct-Current Circuits 357

and

EFe =

12.5 V = 255 mV/m 49 m

*49 •• Picture the Problem We can use Ohm’s law to express the ratio of the potential differences across the two wires and R = ρL A to relate the resistances of the wires to their lengths, resistivities, and cross-sectional areas. Once we’ve found the ratio of the potential differences across the wires, we can use E = V L to decide which wire has the greater electric field. (a) Apply Ohm’s law to express the potential drop across each wire:

VCu = IRCu and

VFe = IRFe Divide the first of these equations by the second to express the ratio of the potential drops across the wires:

VCu IRCu RCu = = VFe IRFe RFe

Relate the resistances of the wires to their resistivity, cross-sectional area, and length:

RCu = ρ Cu

LCu ACu

and

RFe = ρ Fe Divide the first of these equations by the second to express the ratio of the resistances of the wires:

(1)

RCu RFe

LFe AFe

LCu ACu ρ Cu = = L ρ Fe ρ Fe Fe AFe

ρ Cu

because LCu = LFe and ACu = AFe. Substitute in equation (1) to obtain:

VCu ρ Cu = VFe ρ Fe

Substitute numerical values (see Table 2625-1 for the resistivities of copper and iron) and evaluate the ratio of the potential differences:

VCu 1.7 × 10−8 Ω ⋅ m = = 0.170 VFe 10 × 10−8 Ω ⋅ m

358 Chapter 25 (b) Express the electric field in each conductor in terms of its length and the potential difference across it: Divide the first of these equations by the second to obtain:

ECu =

VCu V and EFe = Fe LCu LFe

VCu V L = Cu = Cu = 0.170 VCu VFe LCu

ECu EFe or

EFe = Because EFe = 5.88ECu:

ECu = 5.88 ECu 0.17

E is greater in the iron wire.

50 •• Picture the Problem We can use R = ρL A to relate the resistance of the salt solution to its length, resistivity, and cross-sectional area. To find the resistance of the filled tube when it is uniformly stretched to a length of 2 m, we can use the fact that the volume of the solution is unchanged to relate the new cross-sectional area of the solution to its original cross-sectional area.

L A

(a) Relate the resistance of the filled tube to the resistivity, crosssectional area, and length of the salt solution:

R=ρ

Substitute numerical values and evaluate R:

R = 10−3 Ω ⋅ m

(b) Relate the resistance of the stretched tube to the resistivity, cross-sectional area, and length of the salt solution:

R' = ρ

Letting V represent volume, express the relationship between the beforestretching volume V and the afterstretching volume V′:

V = V'

Solve for A′ to obtain:

(

)

1m = 79.6 Ω π (2 mm )2

L' 2L =ρ A' A'

or

LA = L'A'

A' =

L A = 12 A L'

(1)

`Electric Current and Direct-Current Circuits 359

Substitute in equation (1) to obtain:

R' = ρ

2L ⎛ L⎞ = 4⎜ ρ ⎟ A ⎝ A⎠

1 2

= 4(79.6 Ω ) = 318 Ω 51 •• Picture the Problem We can use R = ρL A to relate the resistance of the wires to their lengths, resistivities, and cross-sectional areas. To find the resistance of the stretched wire, we can use the fact that the volume of the wire does not change during the stretching process to relate the new cross-sectional area of the wire to its original cross-sectional area. Relate the resistance of the unstretched wire to its resistivity, cross-sectional area, and length:

R=ρ

L A

Relate the resistance of the stretched wire to its resistivity, cross-sectional area, and length:

R' = ρ

L' A'

Divide the second of these equations by the first to obtain:

L' ρ L' A R' = A' = L L A' R ρ A or

R' = 2 Letting V represent volume, express the relationship between the beforestretching volume V and the afterstretching volume V′:

A R A'

(1)

V = V' or

LA = L'A'

Solve for A/A′ to obtain:

A L' = =2 A' L

Substitute in equation (1) to obtain:

R' = 2(2 )R = 4 R = 4(0.3 Ω ) = 1.20 Ω

360 Chapter 25 52 •• Picture the Problem We can use R = ρL A to find the resistance of the wire from its length, resistivity, and cross-sectional area. The electric field can be found using E = V/L and Ohm’s law to eliminate V. The time for an electron to travel the length of the wire can be found from L = vd∆t, with vd expressed in term of I using I = neAvd .

L A

(a) Relate the resistance of the unstretched wire to its resistivity, cross-sectional area, and length:

R=ρ

Substitute numerical values (see Table 25-1 for the resistivity of copper and Table 25-2 for the crosssectional area of 10-gauge wire) and evaluate R:

R = 1.7 × 10−8 Ω ⋅ m

(b) Relate the electric field in the wire to the potential difference between its ends:

E=

V L

Use Ohm’s law to obtain:

E=

IR L

Substitute numerical values and evaluate E:

E=

(30 A )(0.323 Ω ) =

(c) Express the time ∆t for an electron to travel a distance L in the wire in terms of its drift speed vd:

∆t =

L vd

Relate the current in the wire to the drift speed of the charge carriers:

I = neAvd

Solve for vd to obtain:

vd =

I neA

Substitute to obtain:

∆t =

neAL I

(

100 m ) 5.261 mm

2

= 0.323 Ω

100 m

96.9 mV/m

Substitute numerical values (in Example 25-1 it is shown that n = 8.47×1028 m−3) and evaluate ∆t:

`Electric Current and Direct-Current Circuits

(8.47 × 10 ∆t =

28

361

m −3 )(1.6 × 10 −19 C )(5.261 mm 2 )(100 m ) = 2.38 × 105 s 30 A

53 •• Picture the Problem We can use R = ρL A to find express the resistance of the wire from in terms of its length, resistivity, and cross-sectional area. The fact that the volume of the copper does not change as the cube is drawn out to form the wire will allow us to eliminate either the length or the cross-sectional area of the wire and solve for its resistance.

L A

Express the resistance of the wire in terms of its resistivity, crosssectional area, and length:

R=ρ

Relate the volume V of the wire to its length and cross-sectional area:

V = AL

Solve for L to obtain:

L=

Substitute to obtain:

R=ρ

V A

(

Substitute numerical values (see Table 25-1 for the resistivity of copper and Table 25-2 for the crosssectional area of 14-gauge wire) and evaluate R:

V A2 −8

R = 1.7 × 10 Ω ⋅ m

)

(2 cm )3

(2.081mm )

2 2

= 0.0314 Ω

*54 •• A spreadsheet program to plot I as a function of V is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 A5 A6 B5

1 2 3

Content/Formula 2 −200 A5 + 25 $B$1*(EXP(A5/25) − 1) A I_0= 2

B

V

I

Algebraic Form I0 V (mV) V + ∆V

(

)

I 0 eV / 25 mV − 1 C nA

362 Chapter 25 4 5 6 7

(mV) −200.0 −175.0 −150.0

(nA) −2.00 −2.00 −2.00

15 16 17

50.0 75.0 100.0

12.78 38.17 107.20

The following graph was plotted using the data in spreadsheet table shown above. 110 90

I (nA)

70 50 30 10 -10 -200

-150

-100

-50

0

50

100

V (mV)

A spreadsheet program to plot ln(I) as a function of V for V > 0.3 V follows. The formulas used to calculate the quantities in the columns are as follows: Cell B1 A5 A6 B5

Content/Formula 2 300 A5 + 10 LN($B$1*(EXP(A5/25) − 1))

A I_0= 2

B

C nA

V (mV) 300 310 320 330 340 350

ln(I) 12.69 13.09 13.49 13.89 14.29 14.69

970

39.49

Algebraic Form 2 nA V V + ∆V

[ (

)]

ln I 0 eV / 25 mV − 1

`Electric Current and Direct-Current Circuits

980 990 1000

363

39.89 40.29 40.69

A graph of ln(I) as a function of V follows. Microsoft Excel’s Trendline feature was used to obtain the equation of the line. 45 40 35

ln(I ) = 0.04V + 0.6931

ln(I )

30 25 20 15 10 5 0 300

400

500

600

700

800

900

1000

V (mV)

For V >> 25 mV:

eV / 25 mV − 1 ≈ eV / 25 mV and

I ≈ I 0eV / 25 mV Take the natural logarithm of both sides of the equation to obtain:

(

)

= ln (I 0 ) +

1 V 25 mV

ln (I ) = ln I 0eV / 25 mV

which is of the form y = mx + b, where

m=

1 −1 = 0.04 (mV ) 25 mV

in agreement with our graphical result. 55 •• Picture the Problem We can use the first graph plotted in Problem 55 to conclude that, if V < 0.5 V, then the diode’s resistance is effectively infinite. We can use Ohm’s law to estimate the current through the diode. (a)

From the first graph plotted in Problem 55 we see that, if V < 0.5 V, then the current is negligible and the diode has essentially infinite resistance.

(b) Use Ohm’s law to express the current flowing through the diode:

I=

Vresistor R

364 Chapter 25 For a potential difference across the diode of approximately 0.5 V:

I=

5 V - 0.5 V = 90.0 mA 50 Ω

56 ••• Picture the Problem We can use, as our element of resistance, a semicircular strip of width t, radius r, and thickness dr. Then dR = (πrρ t )dr . Because the strips are in parallel, integrating over them will give us the reciprocal of the resistance of half ring. b

Integrate dR from r = a to r = b to obtain:

1 t dr t ⎛b⎞ ln⎜ ⎟ = = ∫ R πρ a r πρ ⎝ a ⎠

Take the reciprocal of both sides of the equation to obtain:

R=

ρπ ⎛b⎞ t ln⎜ ⎟ ⎝a⎠

57 ••• Picture the Problem The element of resistance we use is a segment of length dx and cross-sectional area π[a + (b − a)x/L]2. Because these resistance elements are in series, integrating over them will yield the resistance of the wire.

Express the resistance of the chosen element of resistance:

dR = ρ

Integrate dR from x = 0 to x = L and simplify to obtain:

ρ R= π

dx ρ dx = A π [a + (b − a )( x L )] 2

L

dx

∫ [a + (b − a )(x L )]

2

0

ρL

⎛1 ⎞ 1 ⎜⎜ − ⎟ π (b − a ) ⎝ a a + (b − a ) ⎟⎠ ρL ⎛ 1 1 ⎞ = ⎜ − ⎟ π (b − a ) ⎝ a b ⎠

=

= =

ρL

⎛b−a⎞ ⎜ ⎟ π (b − a ) ⎝ ab ⎠

ρL πab

`Electric Current and Direct-Current Circuits

*58 ••• Picture the Problem The diagram shows a cross-sectional view of the concentric spheres of radii a and b as well as a spherical-shell element of radius r. Using the Hint we can express the resistance dR of the spherical-shell element and then integrate over the volume filled with the material whose resistivity ρ is given to find the resistance between the conductors. Note that the elements of resistance are in series.

dr dr =ρ A 4πr 2

Express the element of resistance dR:

dR = ρ

Integrate dR from r = a to r = b to obtain:

R=

ρ b dr ρ ⎛ 1 1 ⎞ = ⎜ − ⎟ 4π ∫a r 2 4π ⎝ a b ⎠

Substitute numerical values and evaluate R:

R=

109 Ω ⋅ m ⎛ 1 1 ⎞ ⎜⎜ ⎟ − 4π ⎝ 1.5 cm 5 cm ⎟⎠

= 3.71× 109 Ω 59 ••• Picture the Problem The diagram shows a cross-sectional view of the coaxial cylinders of radii a and b as well as a cylindrical-shell element of radius r. We can express the resistance dR of the cylindrical-shell element and then integrate over the volume filled with the material whose resistivity ρ is given to find the resistance between the two cylinders. Note that the elements of resistance are in series. Express the element of resistance dR:

dR = ρ

Integrate dR from r = a to r = b to obtain:

R=

dr dr =ρ A 2πrL

ρ b dr ρ ⎛b⎞ ln⎜ ⎟ = ∫ 2πL a r 2πL ⎝ a ⎠

365

366 Chapter 25 (b) Apply Ohm’s law to obtain:

I=

Substitute numerical values and evaluate I:

I=

V V 2πLV = = ρ ⎛b⎞ R ⎛b⎞ ln⎜ ⎟ ρ ln⎜ ⎟ 2πL ⎝ a ⎠ ⎝a⎠ 2π (50 cm )(10 V ) = 2.05 A ⎛ 2.5 cm ⎞ ⎟⎟ (30 Ω ⋅ m ) ln⎜⎜ ⎝ 1.5 cm ⎠

Temperature Dependence of Resistance *60 • Picture the Problem We can use R = ρL/A to find the resistance of the rod at 20°C. Ignoring the effects of thermal expansion, we can we apply the equation defining the temperature coefficient of resistivity, α, to relate the resistance at 40°C to the resistance at 20°C.

L A

(a) Express the resistance of the rod at 20°C as a function of its resistivity, length, and crosssectional area:

R20 = ρ 20

Substitute numerical values and evaluate R20:

R20 = 5.5 × 10−8 Ω ⋅ m

(

)

0.5 m (1mm)2

= 27.5 mΩ (b) Express the resistance of the rod at 40°C as a function of its resistance at 20°C and the temperature coefficient of resistivity α:

R40 = ρ 40

L A

= ρ 20 [1 + α (tC − 20C°)]

L A

L L + ρ 20 α (tC − 20C°) A A = R20 [1 + α (tC − 20C°)] = ρ 20

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of tungsten) and evaluate R40:

[

]

R40 = (27.5 mΩ ) 1 + (4.5 × 10 −3 K −1 )(20 C°) = 30.0 mΩ

`Electric Current and Direct-Current Circuits 367

61 • Picture the Problem The resistance of the copper wire increases with temperature according to RtC = R20 [1 + α (tC − 20C°)] . We can replace RtC by 1.1R20 and solve for tC to find the temperature at which the resistance of the wire will be 110% of its value at 20°C. Express the resistance of the wire at 1.1R20: Simplify this expression to obtain:

1.1R20 = R20 [1 + α (tC − 20C°)] 1.1R20 = R20 + R20α (tC − 20C°) or

0.1 = α (tC − 20C°)

Solve to tC to obtain:

tC =

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of copper) and evaluate tC:

tC =

0.1

α

+ 20C°

0.1 + 20C° 3.9 × 10−3 K −1

= 45.6°C

62 •• Picture the Problem Let the primed quantities denote the current and resistance at the final temperature of the heating element. We can express R′ in terms of R20 and the final temperature of the wire tC using R' = R20 [1 + α (tC − 20C°)] and relate I′, R′, I20, and R20 using Ohm’s law. Express the resistance of the heating element at its final temperature as a function of its resistance at 20°C and the temperature coefficient of resistivity for Nichrome:

R' = R20 [1 + α (t C − 20C°)]

Apply Ohm’s law to the heating element when it is first turned on:

V = I 20 R20

Apply Ohm’s law to the heating element when it has reached its final temperature:

V = I'R'

Because the voltage is constant, we have:

I'R' = I 20 R20 or

(1)

368 Chapter 25 R' = Substitute in equation (1) to obtain:

I 20 R20 I'

I 20 R20 = R20 [1 + α (t C − 20C°)] I' or

I 20 = 1 + α (tC − 20C°) I' Solve for tC to obtain:

I 20 −1 I' tC = + 20C°

α

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of Nichrome) and evaluate tC:

1.5 A −1 1.3 A tC = + 20C° = 405°C 0.4 × 10−3 K −1

63 •• Picture the Problem We can apply Ohm’s law to find the initial current drawn by the cold heating element. The resistance of the wire at 1000°C can be found using R1000 = R20 [1 + α (tC − 20C°)] and the power consumption of the heater at this temperature from P = V2/R1000. (a) Apply Ohm’s law to find the initial current I20 drawn by the heating element: (b) Express the resistance of the heating element at 1000°C as a function of its resistance at 20°C and the temperature coefficient of resistivity for Nichrome: Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of Nichrome) and evaluate R1000: (c) Express and evaluate the operating wattage of this heater at 1000°C:

I=

V 120 V = = 15.0 A R20 8Ω

R1000 = R20 [1 + α (tC − 20C°)]

[ (

R1000 = (8 Ω ) 1 + 0.4 × 10 −3 K −1 × (1000°C − 20C°)]

)

= 11.1 Ω

(120 V ) = 1.30 kW V2 = R1000 11.1Ω 2

P=

`Electric Current and Direct-Current Circuits 369

64 •• Picture the Problem We can find the resistance of the copper leads using RCu = ρ Cu L A and express the percentage error in neglecting the resistance of the leads as the ratio of RCu to RNichrome. In part (c) we can express the change in resistance in the Nichrome wire corresponding to a change ∆tC in its temperature and then find ∆tC by substitution of the resistance of the copper wires in this equation.

L A

(a) Relate the resistance of the copper leads to their resistivity, length, and cross-sectional area:

RCu = ρ Cu

Substitute numerical values (see Table 25-1 for the resistivity of copper) and evaluate RCu:

RCu = 1.7 × 10−8 Ω ⋅ m

(b) Express the percentage error as the ratio of RCu to RNichrome:

% error =

(

)

1 4

50 cm 2 π (0.6 mm )

= 30.1 mΩ

=

RCu RNichrome 30.1 mΩ = 0.301% 10 Ω

(c) Express the change in the resistance of the Nichrome wire as its temperature changes from tC to tC′:

∆R = R' − R = R20 [1 + α (t C' − 20C°)]

Solve for ∆tC to obtain:

∆tC =

∆R R20α

Set ∆R equal to the resistance of the copper wires (see Table 25-1 for the temperature coefficient of resistivity of Nichrome wire) and evaluate ∆tC:

∆t C =

30.1 mΩ = 7.53 C° (10 Ω ) 0.4 × 10 −3 K −1

− R20 [1 + α (t C − 20C°)]

= R20α∆t C

(

)

*65 ••• Picture the Problem Expressing the total resistance of the two current-carrying (and hence warming) wires connected in series in terms of their resistivities, temperature coefficients of resistivity, lengths and temperature change will lead us to an expression in which, if ρ1L1α1 + ρ2L2α2 = 0, the total resistance is temperature independent. In part (b) we can apply the condition that ρ1L1α1 + ρ2L2α2 = 0 to find the ratio of the lengths of the carbon and copper wires.

370 Chapter 25 (a) Express the total resistance of these two wires connected in series:

R = R1 + R2 = ρ1

L1 (1 + α1∆T ) + ρ 2 L2 (1 + α 2 ∆T ) + 1 [ρ1L1 (1 + α1∆T ) + ρ 2 L2 (1 + α 2 ∆T )] A A A

Expand and simplify this expression to obtain:

R=

1 [ρ1L1 + ρ 2 L2 + (ρ1L1α1 + ρ1L1α 2 )∆T ] A

If ρ1L1α1 + ρ2L2α2 = 0, then:

1 [ρ1 L1 + ρ 2 L2 ] independently of A the temperature.

R=

(b) Apply the condition for temperature independence obtained in (a) to the carbon and copper wires:

ρ C LCα C + ρ Cu LCuα Cu = 0

Solve for the ratio of LCu to LC:

ρα LCu =− C C LC ρ Cuα Cu

Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of carbon and copper) and evaluate the ratio of LCu to LC:

(

)( )(

)

LCu 3500 × 10 −8 Ω ⋅ m − 0.5 × 10 −3 K −1 =− = 264 LC 1.7 × 10 −8 Ω ⋅ m 3.9 × 10 −3 K −1

(

)

66 ••• Picture the Problem We can use the relationship between the rate at which energy is transformed into heat and light in the filament and the resistance of and potential difference across the filament to estimate the resistance of the filament. The linear dependence of the resistivity on temperature will allow us to find the resistivity of the filament at 2500 K. We can then use the relationship between the resistance of the filament, its resistivity, and cross-sectional area to find its diameter. (a) Express the wattage of the lightbulb as a function of its resistance R and the voltage V supplied by the source:

P=

V2 R

`Electric Current and Direct-Current Circuits 371

Solve for R to obtain:

Substitute numerical values and evaluate R: (b) Relate the resistance R of the filament to its resistivity ρ, radius r, and length l: Solve for r to obtain:

V2 R= P R= R=

(100 V )2 40 W

= 250 Ω

ρl π r2

r=

ρl πR

and the diameter d of the filament is

d =2 Because the resistivity varies linearly with temperature, we can use a proportion to find its value at 2500 K:

Solve for ρ2500 K to obtain:

ρl πR

(1)

ρ 2500 K − ρ 293 K 2500 K − 293 K = ρ3500 K − ρ 293 K 3500 K − 293 K =

ρ 2500 K =

2207 3207

2207 (ρ3500 K − ρ293 K ) + ρ293 K 3207

Substitute numerical values and evaluate ρ2500 K:

ρ 2500 K =

2207 (1.1 µΩ ⋅ m − 56 nΩ ⋅ m ) + 56 nΩ ⋅ m = 774.5 nΩ ⋅ m 3207

Substitute numerical values in equation (1) and evaluate d:

d =2

(774.5 nΩ ⋅ m )(0.5 cm ) π (250 Ω )

= 4.44 µm 67 ••• Picture the Problem We can use the relationship between the rate at which an object radiates and its temperature to find the temperature of the bulb. (a) At a temperature T, the power emitted by a perfect blackbody is:

Solve for T:

P = σAT 4

where σ = 5.67×10−8 W/m2⋅K4 is the Stefan-Boltzmann constant.

T =4

P 4 P = σA σπdL

372 Chapter 25 or, because P = V2/R,

T =4 Relate the resistance R of the filament to its resistivity ρ:

R=

Substitute for R in the expression for T to obtain:

T=

V2 σπdLR

ρL A

4

=

4 ρL πd 2

V2 V 2d =4 4 ρL 4σL2 ρ σπdL 2 πd

Substitute numerical values and evaluate T:

T=

4

(5 V )2 (40 ×10−6 m ) 2 4(5.67 ×10 −8 W/m 2 ⋅ K 4 )(0.03 m ) (3 ×10 −5 Ω ⋅ m )

= 636 K

As the filament heats up, its resistance increases, leading to more power (b) being dissipated, leading to further heat, leading to a higher temperature, etc. This thermal runaway can burn out the filament if not controlled.

Energy in Electric Circuits *68 • Picture the Problem We can use P = V 2 R to find the power dissipated by the two resistors. Express the power dissipated in a resistor as a function of its resistance and the potential difference across it:

V2 P= R

(a) Evaluate P for V = 120 V and R = 5 Ω:

P=

(b) Evaluate P for V = 120 V and R = 10 Ω:

P=

(120 V )2 5Ω

(120 V )2 10 Ω

= 2.88 kW

= 1.44 kW

69 • 2 Picture the Problem We can solve Pmax = I max R for the maximum current the resistor can carry and apply Ohm’s law to find the maximum voltage that can be placed across the resistor.

`Electric Current and Direct-Current Circuits

(a) Express the maximum power the resistor can dissipate in terms of the current flowing through it: Solve for Imax to obtain:

Substitute numerical values and evaluate Imax:

373

2 Pmax = I max R

I max =

Pmax R

I max =

0.25 W = 5.00 mA 10 kΩ

(b) Apply Ohm’s law to relate the maximum voltage across this resistor to the maximum current through it:

Vmax = I max R

Substitute numerical values and evaluate Vmax:

Vmax = (5 mA )(10 kΩ ) = 50.0 V

70 • Picture the Problem We can use P = V 2 R to find the resistance of the heater and Ohm’s law to find the current it draws. (a) Express the power output of the heater in terms of its resistance and its operating voltage:

P=

V2 R

Solve for and evaluate R:

V 2 (240 V ) = = 57.6 Ω R= 1 kW P

Apply Ohm’s law to find the current drawn by the heater:

I=

(b) Evaluate the power output of the heater operating at 120 V:

P=

2

V 240 V = = 4.17 A R 57.6 Ω

(120 V )2 57.6 Ω

= 250 W

71 • Picture the Problem We can use the definition of power and the relationship between the battery’s power output and its emf to find the work done by it under the given conditions. Use the definition of power to relate the work done by the battery to the

P=

∆W ∆t

374 Chapter 25 time current is drawn from it: Solve for the work done in time ∆t: Express the power output of the battery as a function of the battery’s emf: Substitute to obtain: Substitute numerical values and evaluate ∆W:

∆W = P∆t

P = εI

∆W = εI∆t

∆W = (12 V )(3 A )(5 s ) = 180 J

72 • Picture the Problem We can relate the terminal voltage of the battery to it emf, internal resistance, and the current delivered by it and then solve this relationship for the internal resistance. Va − Vb = ε − Ir Express the terminal potential difference of the battery in terms of its emf and internal resistance: Solve for r:

Substitute numerical values and evaluate r:

r=

r=

ε − (Va − V )b I

12 V − 11.4 V = 0.0300 Ω 20 A

*73 • Picture the Problem We can find the power delivered by the battery from the product of its emf and the current it delivers. The power delivered to the battery can be found from the product of the potential difference across the terminals of the starter (or across the battery when current is being drawn from it) and the current being delivered to it. In part (c) we can use the definition of power to relate the decrease in the chemical energy of the battery to the power it is delivering and the time during which current is drawn from it. In part (d) we can use conservation of energy to relate the energy delivered by the battery to the heat develope the battery and the energy delivered to the starter (a) Express the power delivered by the battery as a function of its emf

P = εI = (12 V )(20 A ) = 240 W

`Electric Current and Direct-Current Circuits

375

and the current it delivers: (b) Relate the power delivered to the starter to the potential difference across its terminals:

Pstarter = Vstarter I

(c) Use the definition of power to express the decrease in the chemical energy of the battery as it delivers current to the starter:

∆E = P∆t

(d) Use conservation of energy to relate the energy delivered by the battery to the heat developed in the battery and the energy delivered to the starter:

Edelivered by battery = Etransformed into heat

Express the energy delivered by the battery and the energy delivered to the starter in terms of the rate at which this energy is delivered:

P∆t = Q + Ps ∆t

Solve for Q to obtain: Substitute numerical values and evaluate Q:

= (11.4 V )(20 A ) = 228 W

= (240 W )(3 min ) = 43.2 kJ

+ Edelivered to starter = Q + Edelivered to starter

Q = (P − Ps )∆t Q = (240 W − 228 W )(3 min ) = 2.16 kJ

74 • Picture the Problem We can use conservation of energy to relate the emf of the battery to the potential differences across the variable resistor and the energy converted to heat within the battery. Solving this equation for I will allow us to find the current for the four values of R and we can use P = I 2R to find the power delivered the battery for the four values of R. Apply conservation of energy (Kirchhoff’s loop rule) to the circuit to obtain: Solve for I to obtain:

ε = IR + Ir

I=

ε R+r

376 Chapter 25 Express the power delivered by the battery as a function of the current drawn from it:

P = I 2R

(a) For R = 0:

I=

ε

6V

=

R+r

0.3 Ω

= 20 A

and

P = (20 A ) (0 ) = 0 2

(b) For R = 5 Ω:

I=

ε R+r

=

6V = 1.13 A 5 Ω + 0.3 Ω

and

P = (1.13 A ) (5 Ω ) = 6.38 W 2

(c) For R = 10 Ω:

I=

ε R+r

=

6V = 0.583 A 10 Ω + 0.3 Ω

and

P = (0.583A ) (10 Ω ) = 3.40 W 2

(d) For R = ∞:

I=

ε R+r

= lim R→∞

6V = 0 R + 0.3 Ω

and

P= 0 75 •• Picture the Problem We can express the total stored energy ∆U in the battery in terms of its emf and the product I∆t of the current it can deliver for a period of time ∆t. We can apply the definition of power to relate the lifetime of the battery to the rate at which it is providing energy to the pair of headlights (a) Express ∆U in terms of ε and the product I∆t: Substitute numerical values and evaluate ∆U:

∆U = εI∆t

∆U = (12 V )(160 A ⋅ h ) = 1.92 kW ⋅ h = 1.92 kW ⋅ h × = 6.91 MJ

3.6 MJ kW ⋅ h

`Electric Current and Direct-Current Circuits 377

(b) Use the definition of power to relate the lifetime of the battery to the rate at which it is providing energy to the pair of headlights:

∆t =

∆U P

Substitute numerical values and evaluate ∆t:

∆t =

1.92 kW ⋅ h = 12.8 h 150 W

76 •• Picture the Problem We can use conservation of energy (aka Kirchhoff’s loop Rule) to relate the emf at the fuse box and the voltage drop in the wires to the voltage at the outlet box (delivered to the space heater). We can find the number of 60-W li bulbs that could be supplied by the this line without blowing the fuse by dividing the current available by the current drawn by each 60-W bulb.

ε − Vwires − Voutlet = 0

(a) Apply Kirchhoff’s loop rule to the circuit to obtain:

or

Solve for Voutlet to obtain:

Voutlet = ε − IRwires

ε − IRwires − Voutlet = 0

Relate the resistance of the copper wires to the resistivity of copper, the length of the wires, and the crosssectional area of 12-gauge wire:

Rwires = ρ Cu

L A

Substitute to obtain:

Voutlet = ε −

Iρ Cu L A

Substitute numerical values (see Table 25 -1 for the resistivity of copper and Table 25-2 for the crosssectional area of 12-gauge wire) and evaluate Voutlet:

Voutlet = 120 V

(b) Relate the number of bulbs N to the maximum current available and the current drawn by each 60-W bulb:

N=

−

(12.5 A ) (1.7 × 10 −8 Ω ⋅ m )(60 m ) 3.309 mm 2

= 116 V I max − 12.5 A I bulb

(1)

378 Chapter 25 Use Ohm’s law to relate the current drawn by each bulb to the potential difference across it and it’s resistance: Express the resistance of each 60-W bulb:

I bulb =

Rbulb =

Substitute to obtain:

I bulb =

Substitute in equation (1) to obtain:

N=

V Rbulb

ε2 P PV

ε2

I max − 12.5 A PV

ε2

= Substitute numerical values and evaluate N:

N=

ε 2 (I − 12.5 A ) max

PV

(120 V )2 (20 A − 12.5 A ) (60 W )(116 V )

= 15 bulbs *77 •• Picture the Problem We can use P = fv to find the power the electric motor must develop to move the car at 80 km/h against a frictional force of 1200 N. We can find the total charge that can be delivered by the 10 batteries using ∆Q = NI∆t . The total electrical energy delivered by the 10 batteries before recharging can be found using the definition of emf. We can find the distance the car can travel from the definition of work and the cost per kilometer of driving the car this distance by dividing the cost of the required energy by the distance the car has traveled. (a) Express the power the electric motor must develop in terms of the speed of the car and the friction force: (b) Use the definition of current to express the total charge that can be delivered before charging:

P = fv = (1200 N )(80 km/h ) = 26.7 kW

⎛ 3600 s ⎞ ∆Q = NI∆t = 10(160 A ⋅ h ) ⎜ ⎟ ⎝ h ⎠ = 5.76 MC where N is the number of batteries.

`Electric Current and Direct-Current Circuits 379

(c) Use the definition of emf to express the total electrical energy available in the batteries:

W = Qε = (5.76 MC )(12 V ) = 69.1 MJ

(d) Relate the amount of work the batteries can do to the work required to overcome friction:

W = fd

Solve for and evaluate d:

d=

W 69.1 MJ = = 57.6 km f 1200 N

(e) Express the cost per kilometer as the ratio of the ratio of the cost of the energy to the distance traveled before recharging:

Cost/km =

($0.09/kW ⋅ h )εIt = ($0.09/kW ⋅ h )(120 V )(160 A ⋅ h ) = d

57.6 km

$0.03 / km

78 ••• Picture the Problem We can use the definition of power to find the current drawn by the heater and Ohm’s law to find its resistance. In part (b) we’ll use the hint to show that ∆P/P ≈ 2∆V/V and in part (c) use this result to find the approximate power dissipated in the heater if the potential difference is decreased to 115 V. (a) Use the definition of power to relate the current I drawn by the heater to its power rating P and the potential difference across it V:

I=

P V

Substitute numerical values and evaluate I:

I=

100 W = 0.833 A 120 V

Apply Ohm’s law to relate the resistance of the heater to the voltage across it and the current it draws:

R=

V 120 V = = 144 Ω I 0.833 A

(b) Approximate dP/dV by differentials:

dP ∆P dP ≈ or ∆P ≈ ∆V dV ∆V dV

Express the dependence of P on V:

P=

V2 R

380 Chapter 25 Assuming R to be constant, evaluate dP/dV: Substitute to obtain:

Divide both sides of the equation by P to obtain:

dP d ⎡V 2 ⎤ 2V = ⎢ ⎥= dV dV ⎣ R ⎦ R

∆P ≈

2V V 2 ∆V ∆V ∆V = 2 = 2P R R V V

∆P ∆V =2 P V

(c) Express the approximate power dissipated in the heater as the sum of its power consumption and the change in its power dissipation when the voltage is decreased by ∆V:

P ≈ P0 + ∆P

Substitute numerical values and evaluate P:

⎛ ⎛ − 5 V ⎞⎞ ⎟⎟ ⎟⎟ = 91.7 W P ≈ (100 W ) ⎜⎜1 + 2⎜⎜ ⎝ 120 V ⎠ ⎠ ⎝

∆V V ∆V ⎞ ⎛ = P0 ⎜1 + 2 ⎟ V ⎠ ⎝ = P0 + 2 P0

Combinations of Resistors *79 • Picture the Problem We can either solve this problem by using the expression for the equivalent resistance of three resistors connected in parallel and then using Ohm’s law to find the current in each resistor, or we can apply Ohm’s law first to find the current through each resistor and then use Ohm’s law a second time to find the equivalent resistance of the parallel combination. We’ll follow the first procedure. (a) Express the equivalent resistance of the three resistors in parallel and solve for Req:

1 1 1 1 = + + Req 4 Ω 3 Ω 6 Ω and

Req = 1.33 Ω (b) Apply Ohm’s law to each of the resistors to find the current flowing through each:

V 12 V = = 3.00 A 4Ω 4Ω 12 V V I3 = = = 4.00 A 3Ω 3Ω

I4 =

and

`Electric Current and Direct-Current Circuits 381

I6 =

V 12 V = = 2.00 A 6Ω 6Ω

Remarks: You would find it instructive to use Kirchhoff’s junction rule (conservation of charge) to confirm our values for the currents through the three resistors. 80 • Picture the Problem We can simplify the network by first replacing the resistors that are in parallel by their equivalent resistance. We can then add that resistance and the 3-Ω resistance to find the equivalent resistance between points a and b. In part (b) we’ll denote the currents through each resistor with subscripts corresponding to the resistance through which the current flows and apply Ohm’s law to find those currents. (a) Express the equivalent resistance of the two resistors in parallel and solve for Req,1:

1 1 1 1 1 = + = + Req,1 R6 R2 6 Ω 2 Ω and

Req,1 = 1.50 Ω Because the 3-Ω resistor is in series with Req,1:

Req = R3 + Req,1

(b) Apply Ohm’s law to the network to find I3:

I3 =

Find the potential difference across the parallel resistors:

V6 & 2 = Vab − V3

Use the common potential difference across the resistors in parallel to find the current through each of them:

I6 =

= 3 Ω + 1.5 Ω = 4.50 Ω Vab 12 V = = 2.67 A , Req 4.5 Ω

= 12 V − (2.67 A )(3 Ω ) = 4 V

V6 4 V = = 0.667 A R6 6Ω

and

I2 =

V6 & 2 R2

=

4V = 2.00 A 2Ω

81 • Picture the Problem Note that the resistors between a and c and between c and b are in series as are the resistors between a and d and between d and b. Hence, we have two branches in parallel, each branch consisting of two resistors R in series. In part (b) it will be important to note that the potential difference between point c and point d is zero.

382 Chapter 25 (a) Express the equivalent resistance between points a and b in terms of the equivalent resistances between acb and adb:

1 1 1 1 1 = + = + Req Racb Radb 2 R 2 R

Solve for Req to obtain:

Req = R

(b)

Because the potential difference between points c and d is zero, no current would flow through the resistor connected between these two points, and the addition of that resistor would not change the network.

82 •• Picture the Problem Note that the 2-Ω resistors are in parallel with each other and with the 4-Ω resistor. We can Apply Kirchhoff’s loop rule to relate the current I3 drawn from the battery to the emf of the battery and equivalent resistance Req of the resistor network. We can find the current through the resistors connected in parallel by applying Kirchhoff’s loop rule a second time. In part (b) we can find the power delivered by the battery from the product of its emf and the current it delivers to the circuit. (a) Apply Kirchhoff’s loop rule to obtain: Solve for I3:

Find the equivalent resistance of the three resistors in parallel:

ε − I 3 Req = 0 I3 =

ε Req

(1)

1 1 1 1 1 1 1 = + + = + + Req,1 R2 R2 R4 2 Ω 2 Ω 4 Ω and

Req,1 = 0.8 Ω Find the equivalent resistance of Req,1 and R3 in series:

Req = R3 + Req,1 = 3 Ω + 0.8 Ω = 3.8 Ω

Substitute in equation (1) and evaluate I3:

I3 =

6V = 1.58 A 3.8 Ω

Express the current through each of the parallel resistors in terms of their common potential difference V:

I2 =

V V and I 4 = R2 R4

`Electric Current and Direct-Current Circuits 383

Apply Kirchhoff’s loop rule to obtain:

ε − I 3 R3 − V = 0

Solve for V:

V = ε − I 3 R3

Substitute numerical values and evaluate I2 and I4:

I2 =

= 6 V − (1.58 A )(3 Ω ) = 1.26 V 1.26 V = 0.630 A 2Ω

and I4 = (b) Express P in terms of ε and I3:

1.26V = 0.315 A 4Ω

P = εI 3 = (6 V )(1.58 A ) = 9.48 W

Remarks: Note that the currents I3, I2, and I4 satisfy Kirchhoff’s junction rule. *83 •• Picture the Problem Let r represent the resistance of the internal resistance of the power supply, ε the emf of the power supply, R the resistance of the external resistor to be placed in series with the power supply, and I the current drawn from the power supply. We can use Ohm’s law to express the potential difference across R and apply Kirchhoff’s loop rule to express the current through R in terms of ε, r, and R. Express the potential difference across the resistor whose resistance is R: Apply Kirchhoff’s loop rule to the circuit to obtain: Solve for I to obtain:

Substitute in equation (1) to obtain:

VR = IR

ε − Ir − IR = 0 I=

ε r+R

⎛ ε ⎞ VR = ⎜ ⎟R ⎝r+R⎠

(1)

384 Chapter 25 Solve for R to obtain:

R=

VR r ε − VR

Substitute numerical values and evaluate R:

R=

(4.5 V )(50 Ω ) = 5 V − 4.5 V

450 Ω

84 •• Picture the Problem We can apply Kirchhoff’s loop rule to the two circuits described in the problem statement and solve the resulting equations simultaneously for r and ε. (a) and (b) Apply Kirchhoff’s loop rule to the two circuits to obtain:

Substitute numerical values to obtain:

ε − I1r − I1R5 = 0 and

ε − I 2 r − I 2 R11 = 0 ε − (0.5 A ) r − (0.5 A )(5 Ω ) = 0

or

ε − (0.5 A ) r = 2.5 V

(1)

and

ε − (0.25 A ) r − (0.25 A )(11 Ω ) = 0

or

ε − (0.25 A ) r = 2.75 V

Solve equations (1) and (2) simultaneously to obtain:

ε=

(2)

3.00 V and r = 1.00 Ω

85 •• Picture the Problem We can use the formula for the equivalent resistance for two resistors in parallel to show that Req = R1x/(1 + x).

R1R2 R1 + R2

(a) Express the equivalent resistance of R1 and R2 in parallel:

Req =

Let x = R2/R1 to obtain:

xR12 x Req = = R1 R1 + xR1 1+ x

(b) The following graph of Req /R1 versus x was plotted using a spreadsheet program.

`Electric Current and Direct-Current Circuits 385

1.0 0.8 0.6 R eq/R 1 0.4 0.2 0.0 0

2

4

6

8

10

x

86 •• Picture the Problem We can use Kirchhoff’s loop rule to relate the required resistance to the emf of the source and the desired current. We can apply Kirchhoff’s rule a second time to the circuit that includes the load resistance r to establish the largest value it can have if it is to change the current drawn from the source by at most 10 percent. (a) Apply Kirchhoff’s loop rule to the circuit that includes the source and the resistance R to obtain: Solve for R to obtain:

Substitute numerical values and evaluate R: (b) Letting I′ represent the current in the loaded circuit, express the condition that the current drops by less than 10%: Letting r represent the load resistance, apply Kirchhoff’s loop rule to the circuit to obtain: Solve for I′ to obtain:

Substitute for I and I′ in equation (1):

ε − IR = 0

R= R=

ε I 5V = 500 Ω 10 mA

I − I' I' = 1 − < 0.1 I I

(1)

ε − I'r − I'R = 0

I' =

ε r+R

ε

R 1 − r + R < 0.1 ⇒ 1 − < 0.1 ε r+R R

386 Chapter 25 Solve for r to obtain: Substitute the numerical value for R to obtain:

0.1R 0.9 0.1(500 Ω ) r< = 55.6 Ω 0.9

r<

87 •• Picture the Problem We can simplify the network by first replacing the resistors that are in parallel by their equivalent resistance. We’ll then have a parallel network with two resistors in series in each branch and can use the expressions for resistors in series to simplify the network to two resistors in parallel. The equivalent resistance between points a and b will be the single resistor equivalent to these two resistors. In part (b) we’ll use the fact that the potential difference across the upper branch is the same as the potential difference across the lower branch, in conjunction with Ohm’s law, to find the currents through each resistor. (a) Express and evaluate the equivalent resistance of the two 6-Ω resistors in parallel and solve for Req,1:

Req,1 =

R6 R6 (6 Ω )2 = = 3Ω R6 + R6 6 Ω + 6 Ω

Find the equivalent resistance of the 6-Ω resistor is in series with Req,1:

Req,2 = R6 + Req,1

Find the equivalent resistance of the 12-Ω resistor in series with the 6-Ω resistor:

Req,3 = R6 + R12

Finally, find the equivalent resistance of Req,3 in parallel with Req,2:

Req =

(b) Apply Ohm’s law to the upper branch to find the current Iu = I12 = I6 :

= 6Ω + 3Ω = 9 Ω

= 6 Ω + 12 Ω = 18 Ω

=

Req,2 Req,3 Req,2 + Req,3

(9 Ω )(18 Ω ) = 9 Ω + 18 Ω

I u = I12 = I 6 =

6.00 Ω

Vab Req,3

12 V = = 0.667 A 18 Ω

,

`Electric Current and Direct-Current Circuits 387

Apply Ohm’s law to the lower branch to find the current Il = I6-Ω resistor in series:

I l = I 6-Ω resistor in series =

Express the current through the 6-Ω resistors in parallel:

I 6-Ω resistors in parallel = 12 I l

=

Vab Req,2

12 V = 1.33 A 9Ω

=

1 2

(1.33 A )

= 0.667 A 88 •• Picture the Problem Assign currents in each of the resistors as shown in the diagram. We can simplify the network by first replacing the resistors that are in parallel by their equivalent resistance. We’ll then have a parallel network with two resistors in series in each branch and can use the expressions for resistors in series to simplify the network to two resistors in parallel. The equivalent resistance between points a and b will be the single resistor equivalent to these two resistors. In part (b) we’ll use the fact that the potential difference across the upper branch is the same as the potential difference across the lower branch, in conjunction with Ohm’s law, to find the currents through each resistor.

(R2 + R4 )R4 (R2 + R4 ) + R4 (6 Ω )(4 Ω ) =

(a) Express and evaluate the equivalent resistance of the resistors in parallel in the upper branch and solve for Req,1:

Req,1 =

Find the equivalent resistance of the 6-Ω resistor is in series with Req,1:

Req,2 = R6 + Req,1

Express and evaluate the equivalent resistance of the resistors in parallel in the lower branch and solve for

Req,2 =

2Ω + 4Ω + 4Ω

= 2.4 Ω

= 6 Ω + 2.4 Ω = 8.4 Ω

R8 R8 = 12 R8 = R8 + R8

1 2

(8 Ω ) = 4 Ω

388 Chapter 25 Req,2: Find the equivalent resistance of the 4-Ω resistor is in series with Req,2:

Req,3 = R4 + Req,2

Finally, find the equivalent resistance of Req,2 in parallel with Req,3:

Req =

= 4Ω + 4Ω = 8Ω

=

Req,2 Req,3 Req,2 + Req,3

(8.4 Ω )(8 Ω ) = 8.4 Ω + 8 Ω

4.10 Ω

Vab 12 V = = 1.43 A Req,2 8.4 Ω

(b) Apply Ohm’s law to the upper branch to find the current I1:

I1 =

Find the potential difference across the 4-Ω and 6-Ω parallel combination in the upper branch:

V4&6 = 12 V − V6 = 12 V − I u R6

Apply Ohm’s law to find the current I4 :

I4 =

V6 3.43 V = = 0.572 A R6 6Ω

Apply Ohm’s law to find the current I3 :

I3 =

V4 3.43 V = = 0.858 A R4 4Ω

Apply Ohm’s law to the lower branch to find the current I2:

I2 =

Vab 12 V = = 1.50 A Req,2 8 Ω

Find the potential difference across the 8-Ω and 8-Ω parallel combination in the lower branch:

V8&8 = 12 V − I 3 R4

Apply Ohm’s law to find I5 = I6:

I5 = I6 =

= 12 V − (1.43 A )(6 Ω ) = 3.43 V

= 12 V − (1.5 A )(4 Ω ) = 6V

V8&8 6 V = = 0.750 A 8Ω 8Ω

*89 •• Picture the Problem We can use the equation for N identical resistors connected in parallel to relate N to the resistance R of each piece of wire and the equivalent resistance Express the resistance of the N pieces connected in parallel:

1 N = Req R where R is the resistance of one of the N

`Electric Current and Direct-Current Circuits 389

pieces.

Rwire N

Relate the resistance of one of the N pieces to the resistance of the wire:

R=

Substitute to obtain:

1 N2 = Req Rwire

Solve for N:

Substitute numerical values and evaluate N:

N=

Rwire Req

N=

120 Ω = 8 1.875 Ω

90 •• Picture the Problem We can assign currents as shown in the diagram of the first arrangement of resistors and apply Kirchhoff’s loop rule to obtain an expression for I8,1.

Assign currents as shown in the diagram below for the second arrangement of the resistors and apply Kirchhoff’s loop rule to obtain an expression for I8,2 that we can equate to I8,1 and solve for R.

Apply Kirchhoff’s loop rule to the first arrangement of the resistors:

ε − I1 Req,1 = 0 where I1 is the current drawn from the battery.

390 Chapter 25 Solve for I1 to obtain:

Find the equivalent resistance of the first arrangement of the resistors:

ε

I1 =

Req,1

Req,1 = =

Substitute to obtain:

I1 =

(8 Ω )R 8Ω + R

+ 16 Ω

(24 Ω )R + 128 Ω 2 R + 8Ω

ε

(24 Ω )R + 128 Ω 2 R + 8Ω

=

Apply Kirchhoff’s loop rule to the loop containing R and the 8-Ω resistor: Solve for I8,1 to obtain:

ε (R + 8 Ω )

(24 Ω )R + 128 Ω 2

− I 8 (8 Ω ) + R(I1 − I 8 ) = 0

I 8,1 =

R I1 R + 8Ω

⎛ R ⎞⎛ ε (R + 8 Ω ) ⎟⎟⎜⎜ = ⎜⎜ 2 R 8 + Ω ⎝ ⎠⎝ (24 Ω )R + 128 Ω εR = (24 Ω )R + 128 Ω 2 Express I8,2 in terms of I1 and I8,1:

I 8, 2 = I1 − I 8,1

Substitute for I1 and I8,1 and simplify to obtain:

I 8, 2 =

ε (R + 8 Ω )

(24 Ω )R + 128 Ω 2 −

εR

(24 Ω )R + 128 Ω 2 (8 Ω )ε = (24 Ω )R + 128 Ω 2 Apply Kirchhoff’s loop rule to the second arrangement of the resistors:

ε − I 2 Req,2 = 0 where I2 is the current drawn from the battery.

⎞ ⎟⎟ ⎠

`Electric Current and Direct-Current Circuits 391

Solve for I2 (= I8,2) to obtain:

I 2 = I 8, 2 =

ε Req,2

Find the equivalent resistance of the second arrangement of the resistors:

Req,2 = R + 24 Ω

Substitute to obtain:

I 8, 2 =

Equate I8,1 and I8,2:

Solve for R to obtain:

ε R + 24 Ω

εR

(24 Ω )R + 128 Ω

2

=

(8 Ω )ε (24 Ω )R + 128 Ω 2

R = 8.00 Ω

91 •• Picture the Problem We can find the equivalent resistance Rab between points a and b and then set this resistance equal, in turn, to R1, R3, and R1 and solve for R3, R2, and R1, respectively. (a) Express the equivalent resistance between points a and b:

Rab =

R1 R2 + R3 R1 + R2

Equate this expression to Rl:

R1 =

R1 R2 + R3 R1 + R2

Solve for R3 to obtain:

R12 R3 = R1 + R2

(b) Set R3 equal to Rab:

R3 =

Solve for R2 to obtain:

R2 = 0

(c) Set R1 equal to Rab:

R1 =

R1 R2 + R3 R1 + R2

R1 R2 + R3 R1 + R2

or

R12 − R3 R1 − R2 R3 = 0

392 Chapter 25 Solve the quadratic equation for R1 to obtain:

R3 + R32 + 4 R2 R3 R1 = 2 where we’ve used the positive sign because resistance is a non-negative quantity.

92 •• Picture the Problem We can substitute the given resistances in the equations derived in Problem 91 to check our results from Problem 78. (a) For R1 = 4 Ω and R2 = 6 Ω:

(4 Ω ) = 1.60 Ω R12 = R1 + R2 4 Ω + 6 Ω 2

R3 = and

Rab = = (b) For R1 = 4 Ω and R3 = 3 Ω:

R1 R2 + R3 R1 + R2

(4 Ω )(6 Ω ) + 1.6 Ω =

4.00 Ω

4Ω + 6Ω

R2 = 0 and

Rab =

R1 (0) + R3 = 0 + 3 Ω R1 + 0

= 3.00 Ω (c) For R2 = 6 Ω and R3 = 3 Ω:

R1 = =

3Ω +

(3 Ω )2 + 4(6 Ω )(3 Ω ) 2

3Ω + 9 Ω = 6.00 Ω 2

and

Rab = =

R1 R2 + R3 R1 + R2

(6 Ω )(6 Ω ) + 3 Ω = 6Ω + 6Ω

6.00 Ω

Kirchhoff’s Rules *93 • Picture the Problem We can relate the current provided by the source to the rate of Joule heating using P = I 2 R and use Ohm’s law and Kirchhoff’s rules to find the potential difference across R and the value of r.

`Electric Current and Direct-Current Circuits 393

(a) Relate the current I in the circuit to rate at which energy is being dissipated in the form of Joule heat:

P = I 2 R or I =

P R

Substitute numerical values and evaluate I:

I=

(b) Apply Ohm’s law to find VR:

VR = IR = (4 A )(0.5 Ω ) = 2.00 V

(c) Apply Kirchhoff’s loop rule to obtain: Solve for r:

Substitute numerical values and evaluate r: 94

8W = 4.00 A 0 .5 Ω

ε − Ir − IR = 0 r=

r=

ε − IR = ε I

I

−R

6V − 0.5 Ω = 1.00 Ω 4A

•

Picture the Problem Assume that the current flows clockwise in the circuit and let ε1

represent the 12-V source and ε2 the 6-V source. We can apply Kirchhoff’s loop rule

(conservation of energy) to this series circuit to relate the current to the emfs of the sources and the resistance of the circuit. In part (b) we can find the power delivered or

absorbed by each source using P = εI and in part (c) the rate of Joule heating using P = I 2R. (a) Apply Kirchhoff’s loop rule to the circuit to obtain: Solve for I:

ε 1 − IR2 − ε 2 − IR4 = 0 I=

ε1 − ε 2 R2 + R4 12 V − 6 V = 1.00 A 2Ω + 4Ω

Substitute numerical values and evaluate I:

I=

(b) Express the power delivered/absorbed by each source in terms of its emf and the current drawn from or forced through it:

P12 = ε 12 I = (12 V )(1 A ) = 12 W and

P6 = ε 6 I = (− 6 V )(1 A ) = − 6 W

394 Chapter 25 where the minus sign means that this source is absorbing power. (c) Express the rate of Joule heating in terms of the current through and the resistance of each resistor:

95

P2 = I 2 R2 = (1 A ) (2 Ω ) = 2.00 W 2

and

P4 = I 2 R4 = (1 A ) (4 Ω ) = 4.00 W 2

••

Picture the Problem The circuit is shown in the diagram for part (a). ε1 and r1 denote

the emf of the ″sick″ battery and its internal resistance, ε2 and r2 the emf of the second

battery and its internal resistance, and R is the load resistance. Let I1, I2, and IR be the currents. We can apply Kirchhoff’s rules to determine the unknown currents. In part (c)

we can use P = εI to find the power delivered or absorbed by each battery and P = I2R to

find the power dissipated in the internal and load resistors. (a) The circuit diagram is shown to the right:

(b) Apply Kirchhoff’s junction rule to junction 1 to obtain: Apply Kirchhoff’s loop rule to loop 1 to obtain:

Apply Kirchhoff’s loop rule to loop 2 to obtain:

Solve equations (1), (2) and (3) simultaneously to obtain:

I1 + I 2 = I R

(1)

ε 1 − r1 I1 − RI R = 0 or

11.4 V − (0.01 Ω )I1 − (2 Ω )I R = 0 (2)

ε 2 − r2 I 2 − RI R = 0 or

12.6 V − (0.01 Ω )I 2 − (2 Ω )I R = 0 (3)

I1 = − 57.0 A , I 2 = 63.0 A , and

`Electric Current and Direct-Current Circuits 395

I R = 6.00 A where the minus sign for I1 means that the current flows in the direction opposite to the direction we arbitrarily chose, i.e., the battery is being charged. (c) Express the power delivered by the second battery in terms of its emf and the current drawn from it:

P2 = ε 2 I 2 = (12.6 V )(63.0 A ) = 794 W

Express the power absorbed by the second battery in terms of its emf and the current forced through it:

P1 = ε 1 I1 = (11.4 V )(57.0 A ) = 650 W

Find the power dissipated in the internal resistance r1:

Pr1 = I12 r1 = (57 A ) (0.01 Ω ) = 32.5 W

Find the power dissipated in the internal resistance r2:

Pr2 = I 22 r2 = (63 A ) (0.01 Ω ) = 39.7 W

Find the power dissipated in the load resistance R:

PR = I R2 R = (6 A ) (2 Ω ) = 72.0 W

2

2

2

Remarks: Note that the sum of the power dissipated in the internal and load resistances and that absorbed by the second battery is the same as that delivered by the first battery … just as we would expect from conservation of energy. 96 •• Picture the Problem Note that when both switches are closed the 50-Ω resistor is shorted. With both switches open, we can apply Kirchhoff’s loop rule to find the current I in the 100-Ω resistor. With the switches closed, the 100-Ω resistor and R are in parallel. Hence, the potential difference across them is the same and we can express the current I100 in terms of the current Itot flowing into the parallel branch whose resistance is R, and the resistance of the 100-Ω resistor. Itot, in turn, depends on the equivalent resistance of the closed-switch circuit, so we can express I100 = I in terms of R and solve for R. Apply Kirchhoff’s loop rule to a loop around the outside of the circuit with both switches open:

ε − (300 Ω )I − (100 Ω )I − (50 Ω )I = 0

396 Chapter 25 Solve for I to obtain: Relate the potential difference across the 100-Ω resistor to the potential difference across R when both switches are closed: Apply Kirchhoff’s junction rule at the junction to the left of the 100-Ω resistor and R:

ε

1.5 V = 3.33 mA 450 Ω 450 Ω (100 Ω )I100 = RI R

I=

=

I tot = I100 + I R or

I R = I tot − I100 where Itot is the current drawn from the source when both switches are closed.

Substitute to obtain:

(100 Ω )I100 = R(I tot − I100 ) or

I100 = Express the current Itot drawn from the source with both switches closed:

I tot =

Express the equivalent resistance when both switches are closed:

Req =

Substitute to obtain:

I tot =

Substitute in equation (1) to obtain:

I100

Solve for and evaluate R:

RI tot R + 100 Ω

(1)

ε Req

(100 Ω )R R + 100 Ω

+ 300 Ω

1.5 V (100 Ω )R + 300 Ω R + 100 Ω

⎛ ⎞ ⎜ ⎟ R 1.5 V ⎜ ⎟ = ⎜ ⎟ ( ) R Ω 100 R + 100 Ω ⎜ R + 100 Ω + 300 Ω ⎟ ⎝ ⎠ (1.5 V )R = (400 Ω )R + 30,000 Ω 2 = 3.33 mA

R = 600 Ω

`Electric Current and Direct-Current Circuits 397

Remarks: Note that we can also obtain the result in the third step by applying Kirchhoff’s loop rule to the parallel branch of the circuit. *97 •• Picture the Problem Let I1 be the current delivered by the left battery, I2 the current delivered by the right battery, and I3 the current through the 6-Ω resistor, directed down. We can apply Kirchhoff’s rules to obtain three equations that we can solve simultaneously for I1, I2, and I3. Knowing the currents in each branch, we can use Ohm’s law to find the potential difference between points a and b and the power delivered by both the sources. (a) Apply Kirchhoff’s junction rule at junction a: Apply Kirchhoff’s loop rule to a loop around the outside of the circuit to obtain: Apply Kirchhoff’s loop rule to a loop around the left-hand branch of the circuit to obtain: Solve these equations simultaneously to obtain:

I1 + I 2 = I 3 12 V − (4 Ω )I1 + (3 Ω )I 2 − 12 V = 0 or

− (4 Ω )I1 + (3 Ω )I 2 = 0

12 V − (4 Ω )I1 − (6 Ω )I 3 = 0

I1 = 0.667 A , I 2 = 0.889 A , and

I 3 = 1.56 A (b) Apply Ohm’s law to find the potential difference between points a and b: (c) Express the power delivered by the 12-V battery in the left-hand branch of the circuit: Express the power delivered by the 12-V battery in the right-hand branch of the circuit:

Vab = (6 Ω )I 3 = (6 Ω )(1.56 A ) = 9.36 V Pleft = εI1 = (12 V )(0.667 A ) = 8.00 W Pright = εI 2 = (12 V )(0.889 A ) = 10.7 W

398 Chapter 25 98 •• Picture the Problem Let I1 be the current delivered by the 7-V battery, I2 the current delivered by the 5-V battery, and I3, directed up, the current through the 1-Ω resistor. We can apply Kirchhoff’s rules to obtain three equations that we can solve simultaneously for I1, I2, and I3. Knowing the currents in each branch, we can use Ohm’s law to find the potential difference between points a and b and the power delivered by both the sources. (a) Apply Kirchhoff’s junction rule at junction a: Apply Kirchhoff’s loop rule to a loop around the outside of the circuit to obtain: Apply Kirchhoff’s loop rule to a loop around the left-hand branch of the circuit to obtain: Solve these equations simultaneously to obtain:

I1 = I 2 + I 3 7 V − (2 Ω )I1 − (1 Ω )I 3 = 0

7 V − (2 Ω )I1 − (3 Ω )I 2 + 5 V = 0 or

(2 Ω )I1 + (3 Ω )I 2 = 12 V I1 = 3.00 A , I 2 = 2.00 A ,

and

I 3 = 1.00 A (b) Apply Ohm’s law to find the potential difference between points a and b:

Vab = −5 V + (3 Ω )I 2

= −5 V + (3 Ω )(2 A ) = 1.00 V

(c) Express the power delivered by the 7-V battery:

P7 = εI1 = (7 V )(3 A ) = 21.0 W

(c) Express the power delivered by the 5-V battery:

P5 = εI 2 = (5 V )(2 A ) = 10.0 W

99 •• Picture the Problem We can apply Kirchhoff’s rules to the two circuits to determine the current, and hence the power, supplied to the load resistor R for the two connections of the batteries. Differentiation, with respect to the load resistor, of the expressions for the power delivered to the load resistor will allow us to

`Electric Current and Direct-Current Circuits 399

identify the conditions under which the power delivered is a maximum and to decide whether the power supplied to R greater when R < r or when R > r. The series connection of the batteries is shown to the right:

Express the power supplied to R: Apply Kirchhoff’s loop rule to obtain:

Ps = I s2 R − rI s + ε − rI s + ε − RI s = 0 2ε 2r + R

Solve for Is to obtain:

Is =

Substitute to obtain:

4ε 2 R ⎛ 2ε ⎞ Ps = ⎜ ⎟ R= (2r + R )2 ⎝ 2r + R ⎠

Set the derivative, with respect to R, of equation (1) equal to zero for extrema:

dPs d ⎡ 4ε 2 R ⎤ = ⎢ ⎥ dR dR ⎣ (2r + R )2 ⎦

2

(1)

2 ( 2r + R ) 4ε 2 − 4ε 2 R(2)(2r + R ) = (2r + R )4

= 0 for extrema. Solve for R to obtain:

R = 2r

Examination of the second derivative of Ps at R = 2r shows that R = 2r corresponds to a maximum value of Pp and hence, for the series combination,

the power delivered to the load is greater if R > r and is greatest when R = 2r.

400 Chapter 25 The parallel connection of the batteries is shown to the right:

Express the power supplied to R:

Pp = I p2 R

Apply Kirchhoff’s junction rule to point a to obtain:

I p = I1 + I 2

Apply Kirchhoff’s loop rule to loop 1 to obtain:

− rI1 + ε − ε

+ rI 2 = 0

or

I1 = I 2 = 12 I p Apply Kirchhoff’s loop rule to the outer loop to obtain:

Solve for Ip to obtain:

Substitute to obtain:

ε − RI p − rI1 = 0 or

ε − RI p − 12 rI p = 0 Ip =

ε 1 2

r+R

⎛ ε ⎞ ⎟⎟ R = Pp = ⎜⎜ 1 ⎝ 2r+R⎠ 2

Evaluate equation (1) when r = R:

Ps (r = R ) =

Evaluate equation (2) when r = R:

Pp (r = R ) =

ε 2R

( 12 r + R )2

(2)

4ε 2 R 4ε2 = (2 R + R )2 9 R

ε 2R

( 12 R + R )2

=

4ε2 9 R

Thus, we see that if r = R, both arrangements provide the same power to the load.

`Electric Current and Direct-Current Circuits 401

Set the derivative, with respect to R, of equation (2) equal to zero for extrema:

dPp dR

= =

d ⎡ ε 2R ⎤ ⎢ ⎥ dR ⎣ ( 12 r + R )2 ⎦

( 12 r + R )2 ε 2 − ε 2 R(2) ( 12 r + R ) ( 12 r + R )4

= 0 for extrema. Solve for R to obtain:

R = 12 r

Examination of the second derivative of Pp at R = 12 r shows that R = 12 r corresponds to a maximum value of Pp and hence, for the parallel combination, the power delivered to the load is greater if R < r and a maximum when R = 12 r. *100 •• Picture the Problem Let the current drawn from the source be I. We can use Ohm’s law in conjunction with Kirchhoff’s loop rule to express the output voltage as a function of V, R1, and R2. In (b) we can use the result of (a) to express the condition on the output voltages in terms of the effective resistance of the loaded output and the resistances R1 and R2. (a) Use Ohm’s law to express Vout in terms of R2 and I:

Vout = IR2

Apply Kirchhoff’s loop rule to the circuit to obtain:

V − IR1 − IR2 = 0

Solve for I:

I=

V R1 + R2

Substitute for I in the expression for Vout to obtain:

⎛ R2 ⎞ ⎛ V ⎞ ⎟⎟ ⎟⎟ R2 = V ⎜⎜ Vout = ⎜⎜ R R R R + + 2 ⎠ 2 ⎠ ⎝ 1 ⎝ 1

(b) Relate the effective resistance of the loaded circuit Reff to R2 and Rload:

1 1 1 = + Reff R2 Rload

Solve for Rload:

Letting V 'out represent the output voltage under load, express the condition that Vout drops by less than 10 percent of its unloaded

Rload =

R2 Reff R2 − Reff

Vout − V'out V' = 1 − out < 0.1 Vout Vout

(1)

(2)

402 Chapter 25 value: Using the result from (a), express V'out in terms of the effective output load Reff:

⎛ Reff V' out = V ⎜⎜ ⎝ R1 + Reff

Substitute for Vout and V' out in equation (2) and simplify to obtain:

Reff R + Reff 1− 1 < 0.1 R2 R1 + R2 or

1− Solve for Reff:

⎞ ⎟⎟ ⎠

Reff (R1 + R2 ) < 0.1 R2 (R1 + Reff )

Reff >

0.9 R1R2 R1 + 0.1R2

Substitute numerical values and evaluate Reff:

Reff >

0.9(10 kΩ )(10 kΩ ) = 8.18 kΩ 10 kΩ + 0.1(10 kΩ )

Finally, substitute numerical values in equation (1) and evaluate Rload:

Rload <

(10 kΩ )(8.18 kΩ ) = 10 kΩ − 8.18 kΩ

44.9 kΩ

101 •• Picture the Problem In the equivalent Thevenin circuit shown to the right, R2 is in parallel with Rload. We can apply Ohm’s law to express Vout in terms of Reff and I and then use Kirchhoff’s loop rule to express I in terms of V, R1, and Reff. Simplification of the resulting equation will yield both of the indicated results.

(a) and (b) Use Ohm’s law to express Vout in terms of Reff and I:

Vout = IReff

Apply Kirchhoff’s loop rule to the circuit to obtain:

V − IR1 − IReff = 0

Solve for I:

Substitute for I in the expression for Vout to obtain:

I=

V R1 + Reff

⎛ V Vout = ⎜⎜ ⎝ R1 + Reff

⎞ ⎛ Reff ⎟⎟ Reff = V ⎜⎜ ⎠ ⎝ R1 + Reff

⎞ ⎟⎟ ⎠

`Electric Current and Direct-Current Circuits 403

Express the effective resistance Reff in terms of Rload and R2: Substitute for Reff in the expression for Vout to obtain:

Simplify to obtain:

Reff =

Vout

R2 Rload R2 + Rload

R2 Rload ⎛ ⎜ R2 + Rload =V⎜ ⎜ R2 Rload ⎜ R1 + R + R 2 load ⎝

Vout = V

R1 Rload

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

R2 Rload + R2 Rload + R1 R2

⎛ ⎜ ⎛ Rload R2 ⎞ ⎜ ⎟⎟ = ⎜⎜V ⎝ R1 + R2 ⎠ ⎜ Rload + R1 R2 ⎜ R1 + R2 ⎝ ⎛ R2 ⎞ ⎛ Rload ⎞ ⎟⎟ ⎟⎟ ⎜⎜ = ⎜⎜V ⎝ R1 + R2 ⎠ ⎝ Rload + R' ⎠ Rload = V' Rload + R'

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

where

R' =

R1 R2 R2 and V' = V R1 + R2 R1 + R2

102 •• Picture the Problem Let I1 be the current in the 1-Ω resistor, directed to the right; let I2 be the current, directed up, in the middle branch; and let I3 be the current in the 6-Ω resistor, directed down. We can apply Kirchhoff’s rules to find these currents, the power supplied by each source, and the power dissipated in each resistor. (a) Apply Kirchhoff’s junction rule at the top junction to obtain: Apply Kirchhoff’s loop rule to the outside loop of the circuit to obtain:

Apply the loop rule to the inside loop at the left-hand side of the circuit to obtain:

I1 + I 2 = I 3

(1)

8 V − (1 Ω )I1 + 4 V − (2 Ω )I1 − (6 Ω )I 3 = 0 or

(3 Ω )I1 + (6 Ω )I 3 = 12 V

(2)

8 V − (1 Ω )I1 + 4 V − (2 Ω )I1

+ (2 Ω )I 2 − 4 V = 0

or

8 V − (3 Ω )I1 + (2 Ω )I 2 = 0

(3)

404 Chapter 25 Solve equations (1), (2), and (3) simultaneously to obtain:

I1 = 2.00 A , I 2 = − 1.00 A , and

I 3 = 1.00 A where the minus sign indicates that the current flows downward rather than upward as we had assumed. (b) Express the power delivered by the 8-V source:

P8 = ε 8 I1 = (8 V )(2 A ) = 16.0 W

Express the power delivered by the 4-V source:

P4 = ε 4 I 2 = (4 V )(− 1 A ) = − 4.00 W where the minus sign indicates that this source is having current forced through it.

(c) Express the power dissipated in the 1-Ω resistor:

P1Ω = I12 R1Ω

Express the power dissipated in the 2-Ω resistor in the left branch:

P2 Ω ,left = I12 R2 Ω

Express the power dissipated in the 2-Ω resistor in the middle branch:

P2 Ω ,middle = I 22 R2 Ω

Express the power dissipated in the 6-Ω resistor:

P6 Ω = I 32 R6 Ω

= (2 A ) (1 Ω ) = 4.00 W 2

= (2 A ) (2 Ω ) = 8.00 W 2

= (1 A ) (2 Ω ) = 2.00 W 2

= (1 A ) (6 Ω ) = 6.00 W 2

103 •• Picture the Problem Let I1 be the current in the left branch resistor, directed up; let I3 be the current, directed down, in the middle branch; and let I2 be the current in the right branch, directed up. We can apply Kirchhoff’s rules to find I3 and then the potential difference between points a and b. Relate the potential at a to the potential at b:

Va − R4 I 3 − 4 V = Vb or

`Electric Current and Direct-Current Circuits 405

Va − Vb = R4 I 3 + 4 V Apply Kirchhoff’s junction rule at a to obtain: Apply the loop rule to a loop around the outside of the circuit to obtain:

I1 + I 2 = I 3

(1)

2 V − (1 Ω )I1 + (1 Ω )I 2 − 2 V

+ (1 Ω )I 2 − (1 Ω )I1 = 0

or

I1 − I 2 = 0 Apply the loop rule to the left side of the circuit to obtain:

(2)

2 V − (1 Ω )I1 − (4 Ω )I 3 − 4 V

− (1 Ω )I1 = 0

or

− (1 Ω )I1 − (2 Ω )I 3 = 1 V

Solve equations (1), (2), and (3) simultaneously to obtain:

(3)

I1 = −0.200 A , I 2 = −0.200 A , and

I 3 = −0.400 A where the minus signs indicate that all the directions we chose for the currents were wrong. Substitute to obtain:

Va − Vb = (4 Ω )(− 0.4 A ) + 4 V = 2.40 V

Remarks: Note that point a is at the higher potential. 104 •• Picture the Problem Let ε1 be the emf of the 9-V battery and r1 its internal resistance of 0.8 Ω, and ε2 be the emf of the 3-V battery and r2 its internal resistance of 0.4 Ω. The series connection is shown to the right. We can apply Kirchhoff’s rules to both connections to find the currents Is and Ip delivered to the load resistor in the series and parallel connections.

406 Chapter 25 (a) Apply Kirchhoff’s loop rule to the batteries connected in series: Solve for Is to obtain:

ε 1 − r2 I s + ε 2 − RI s − r1 I s = 0 Is =

ε1 + ε 2 r1 + r2 + R

=

12 V 1.2 Ω + R

Suppose the two batteries are connected in parallel and their terminals are then connected to R. Let I1 be the current delivered by ε1, I2 be the current delivered by ε2, and Ip the current through the load resistor R in the parallel connection. Apply the junction rule at a to obtain: Apply the loop rule to a loop around the outside of the circuit:

Apply the loop rule to loop 2 to obtain:

I1 + I 2 = I p

(1)

ε 1 − RI p − r1I1 = 0 or

9 V − RI p − (0.8 Ω )I1 = 0

(2)

ε 1 − ε 2 + r2 I 2 − r1 I1 = 0 or

9 V − 3 V + (0.4 Ω )I 2 − (0.8 Ω )I 1 = 0

or

6 V + (0.4 Ω )I 2 − (0.8 Ω )I1 = 0

Eliminate I2 between equations (1) and (3) to obtain:

I1 = 5 A + 13 I p

(4)

7.5 V 1.5R + 0.4 Ω

Substitute equation (4) in equation (2) and solve for Ip to obtain:

Ip =

(b) Evaluate Is and Ip for R = 0.2 Ω:

I s (R = 0.2 Ω ) =

12 V 1.2 Ω + 0.2 Ω

= 8.57 A and

(3)

`Electric Current and Direct-Current Circuits 407

I p (R = 0.2 Ω ) =

7.5 V 1.5(0.2 Ω ) + 0.4 Ω

= 10.7 A (c), (d), and (e) Proceed as in (b) to complete the table to the right:

R (Ω) 0.6 1.0 1.5

(c) (d) (e)

Is (A) 6.67 5.45 4.44

Ip (A) 5.77 3.95 2.83

Note that for R = 0.4 Ω, I s = I p = 7.5 A. When R > 0.4 Ω, the series connection gives the larger current through R.

Ammeters and Voltmeters *105 •• Picture the Problem Let I be the current drawn from source and Req the resistance equivalent to R and 10 MΩ connected in parallel and apply Kirchhoff’s loop rule to express the measured voltage V across R as a function of R. The voltage measured by the voltmeter is given by:

V = IReq

Apply Kirchhoff’s loop rule to the circuit to obtain:

10 V − IReq − I (2 R ) = 0

Solve for I:

Express Req in terms of R and 10-MΩ resistance in parallel with it: Solve for Req:

Substitute for I in equation (1) and simplify to obtain:

Substitute for Req and simplify to obtain:

I=

(1)

10 V Req + 2 R

1 1 1 = + Req 10 MΩ R

Req =

(10 MΩ )R R + 10 MΩ

⎛ 10 V ⎞ ⎟ R = 10 V V =⎜ ⎜ R + 2 R ⎟ eq 2R ⎝ eq ⎠ 1+ Req V =

(10 V )(5 MΩ ) R + 15 MΩ

(2)

408 Chapter 25

(10 V )(5 MΩ ) =

(a) Evaluate equation (2) for R = 1 kΩ:

V =

(b) Evaluate equation (2) for R =10 kΩ:

V =

(c) Evaluate equation (2) for R =1 MΩ:

V =

(d) Evaluate equation (2) for R =10 MΩ:

V =

(e) Evaluate equation (2) for R =100 MΩ:

V =

(f) Express the condition that the measured voltage to be within 10 percent of the true voltage Vtrue:

Vtrue − V V = 1− < 0.1 Vtrue Vtrue

3.33 V

1 kΩ + 15 MΩ

(10 V )(5 MΩ )

= 3.33 V

10 kΩ + 15 MΩ

(10 V )(5 MΩ ) 1 MΩ + 15 MΩ

= 3.13 V

(10 V )(5 MΩ ) 10 MΩ + 15 MΩ

= 2.00 V

(10 V )(5 MΩ ) 100 MΩ + 15 MΩ

= 0.435 V

(10 V )(5 MΩ )

Substitute for V and Vtrue to obtain:

R + 15 MΩ < 0.1 IR

1−

or, because I = 10 V/3R,

(10 V )(5 MΩ )

1−

Solve for R to obtain:

R<

R + 15 MΩ < 0.1 10 V 3 1.5 MΩ = 1.67 MΩ 0.9

106 •• Picture the Problem The diagram shows a voltmeter connected in parallel with a galvanometer movement whose internal resistance is R. We can apply Kirchhoff’s loop rule to express R in terms of I and V.

Apply Kirchhoff’s loop rule to the loop that includes the galvanometer movement and the voltmeter: Solve for R:

V − IR = 0

R=

V I

`Electric Current and Direct-Current Circuits 409

Substitute numerical values and evaluate R:

R=

0.25 V = 5.00 kΩ 50 µA

107 •• Picture the Problem When there is a voltage drop of 0.25 V across this galvanometer, the meter reads full scale. The diagram shows the galvanometer movement with a resistor of resistance r in parallel. The purpose of this resistor is to limit the current through the movement to Ig = 50 µA. We can apply Kirchhoff’s loop rule to the circuit fragment containing the galvanometer movement and the shunt resistor to derive an expression for r. Apply Kirchhoff’s loop rule to the circuit fragment to obtain:

− RI g + rI r = 0

Apply Kirchhoff’s junction rule at point a to obtain:

Ir = I − Ig

Substitute for Ir in the loop equation:

− RI g + r (I − I g ) = 0

Solve for r:

r=

RI g I − Ig

where RIg = 0.25 V Substitute numerical values and evaluate r:

r=

0.25 V = 2.50 Ω 100 mA − 50 µA

108 •• Picture the Problem The circuit diagram shows the ammeter connected in series with a 100-Ω resistor and a 10-V power supply. We can apply Kirchhoff’s rules to obtain an expression for I as a function of r, Ig, the potential difference provided by the source, and the resistance of the series resistor.

410 Chapter 25 (a) Apply Kirchhoff’s loop rule to the inner loop of the circuit to obtain:

10 V − (100 Ω )I − rI r = 0

Apply Kirchhoff’s junction rule at point a to obtain:

Ir = I − Ig

Substitute for Ir in the loop equation:

10 V − (100 Ω )I − r (I − I g ) = 0

Solve for I:

I=

In Problem 107 it was established that r = 2.50 Ω. Substitute numerical values and evaluate I:

I=

(b) Under these conditions, equation (1) becomes:

I=

Substitute numerical values and evaluate I:

I=

10 V + rI g

(1)

100 Ω + r

10 V + (2.50 Ω )(50 µA ) 100 Ω + 2.50 Ω

= 97.6 mA 1 V + rI g 10 Ω + r

1 V + (2.50 Ω )(50 µA ) 10 Ω + 2.50 Ω

= 80.0 mA Remarks: Our result in (b) differs from that obtained in (a) by about 18 percent. *109 •• Picture the Problem The circuit diagram shows a fragment of a circuit in which a resistor of resistance r is connected in series with the meter movement of Problem 106. The purpose of this resistor is to limit the current through the galvanometer movement to 50 µA and to produce a deflection of the galvanometer movement that is a measure of the potential difference V. We can apply Kirchhoff’s loop rule to express r in terms of Vg, Ig, and R. Apply Kirchhoff’s loop rule to the circuit fragment to obtain: Solve for r:

V − rI g − RI g = 0 r=

V − RI g Ig

=

V −R Ig

(1)

`Electric Current and Direct-Current Circuits 411

Use Ohm’s law to relate the current Ig through the galvanometer movement to the potential difference Vg across it:

Ig =

Vg

⇒ R=

R

Vg Ig

Use the values for Vg and Ig given in Problem 106 to evaluate R:

R=

0.25 V = 5000 Ω 50 µA

Substitute numerical values in equation (1) and evaluate r:

r=

10 V − 5000 Ω = 195 kΩ 50 µA

Remarks: The total series resistance is the sum of r and R or 200 kΩ. 110 •• Picture the Problem The voltmeter shown in Figure 25-64 is equivalent to a resistor of resistance Rg = 200 kΩ as shown in the circuit diagram to the right. The voltage reading across R1 is given by V1 = IR1 . We can use Kirchhoff’s loop rule and the expression for the equivalent resistance of two resistors in parallel to find I.

The voltage reading across R1 is given by: Apply Kirchhoff’s loop rule to the loop including the source, R1, and R2 : Solve for I to obtain:

V1 = IR1

ε − IReq − IR2 = 0

I=

ε Req + R2

εR1

Substitute for I in the expression for V1 :

V1 =

Express Req in terms of R1 and Rg:

Req =

Substitute numerical values and evaluate Req:

Req + R2

Req =

(1)

R1Rg R1 + Rg

(200 kΩ )(200 kΩ ) = 100 kΩ 200 kΩ + 200 kΩ

412 Chapter 25 Substitute numerical values in equation (1) and evaluate V1:

(10 V )(200 kΩ )

V1 =

200 kΩ + 200 kΩ

= 5.00 V

RC Circuits 111 • Picture the Problem We can use the definition of capacitance to find the initial charge on the capacitor and Ohm’s law to find the initial current in the circuit. We can find the time constant of the circuit using its definition and the charge on the capacitor after 6 ms using Q(t ) = Q0e −t τ . (a) Use the definition of capacitance to find the initial charge on the capacitor:

Q0 = CV0 = (6 µF)(100 V ) = 600 µC

(b) Apply Ohm’s law to the resistor to obtain:

I0 =

(c) Use its definition to find the time constant of the circuit:

τ = RC = (500 Ω )(6 µF) = 3.00 ms

V0 100 V = = 0.200 A R 500 Ω

Q(t ) = Q0e −t τ

(d) Express the charge on the capacitor as a function of time:

Q(6 ms ) = (600 µC ) e −6 ms 3 ms = 81.2 µC

Substitute numerical values and evaluate Q(6 ms):

112 • Picture the Problem We can use U 0 = 12 CV02 to find the initial energy stored in the

capacitor and U (t ) = 12 C (VC (t )) with VC (t ) = V0 e −t τ to show that U = U0e −2t /τ. 2

(a) The initial energy stored in the capacitor is given by:

U 0 = 12 CV02

(b) Express the energy stored in the discharging capacitor as a function of time:

U (t ) = 12 C (VC (t ))

=

1 2

(6 µF)(100 V )2 = 2

where

VC (t ) = V0e −t τ

30.0 mJ

`Electric Current and Direct-Current Circuits 413

(

U (t ) = 12 C V0 e − t τ

Substitute to obtain:

)

2

= 12 CV02 e − 2 t τ = U 0 e − 2 t τ (c) A graph of U versus t is shown below. U is in units of U0 and t is in units of τ.

1.0

0.8

0.6

U 0.4

0.2

0.0 0

0.5

1

1.5

2

t

*113 •• Picture the Problem We can find the resistance of the circuit from its time constant and use Ohm’s law and the expression for the current in a charging RC circuit to express τ as a function of time, V0, and V(t). Express the resistance of the resistor in terms of the time constant of the circuit: Using Ohm’s law, express the voltage drop across the resistor as a function of time: Express the current in the circuit as a function of the elapsed time after the switch is closed: Substitute to obtain:

R=

τ C

(1)

V (t ) = I (t )R

I (t ) = I 0e −t τ

V (t ) = I 0 e − t τ R = (I 0 R ) e − t τ = V0 e − t τ

414 Chapter 25 Take the natural logarithm of both sides of the equation and solve for τ to obtain:

τ =−

Substitute in equation (1) to obtain:

R=−

Substitute numerical values and evaluate R using the data given for t = 4 s:

R=−

t ⎡V (t )⎤ ln ⎢ ⎥ ⎣ V0 ⎦ t ⎡V (t ) ⎤ C ln ⎢ ⎥ ⎣ V0 ⎦

4s = 2.18 MΩ ⎛ 20 V ⎞ ⎟⎟ (2 µF) ln⎜⎜ ⎝ 50 V ⎠

*114 •• Picture the Problem We can find the resistance of the circuit from its time constant and use the expression for the charge on a discharging capacitor as a function of time to express τ as a function of time, Q0, and Q(t). Express the effective resistance across the capacitor in terms of the time constant of the circuit:

R=

τ

(1)

C

Q(t ) = Q0e −t τ

Express the charge on the capacitor as a function of the elapsed time after the switch is closed: Take the natural logarithm of both sides of the equation and solve for τ to obtain:

τ =−

Substitute in equation (1) to obtain:

R=−

Substitute numerical values and evaluate R:

R=−

t Q(t ) ln Q0

t Q(t ) C ln Q0 4s

(0.12 µF)ln Q0 Q0 1 2

= 48.1 MΩ

115 •• Picture the Problem We can use the definition of capacitance to find the final charge on the capacitor and Q(t ) = Qf 1 − e −t τ to express the charge on the capacitor as a function

(

)

of time. In part (b) we can let Q(t) = 0.99Qf and solve for t to find the time required for the

`Electric Current and Direct-Current Circuits 415

capacitor to reach 99% of its final charge. (a) After a very long time has elapsed, the capacitor will be fully charged. Use the definition of capacitance to find its charge:

Qf = CV = (1.6 µF)(5 V ) = 8.00 µC

(

Q(t ) = Qf 1 − e −t τ

)

(b) Express the charge on the capacitor as a function of time:

where τ = RC.

When Q = 0.99Qf:

0.99Qf = Qf 1 − e −t τ

(

)

or

0.01 = e −t τ Take the natural logarithm of both sides of the equation and solve for t to obtain: Substitute numerical values and evaluate t:

t = − RC ln (0.01)

t = −(10 kΩ )(1.6 µF) ln (0.01) = 73.7 ms

116 •• Picture the Problem We can use Kirchhoff’s loop rule (conservation of energy) to find both the initial and steady-state currents drawn from the battery and Ohm’s law to find the maximum voltage across the capacitor. (a) Apply Kirchhoff’s loop rule to a loop around the outside of the circuit to obtain: Because the capacitor initially is uncharged:

ε − (1.2 MΩ)I 0 − VC0 = 0

VC0 = 0 and

I0 = (b) When a long time has passed: Apply Kirchhoff’s loop rule to a loop that includes the source and both resistors to obtain:

ε 1.2 MΩ

=

120 V = 0.100 mA 1.2 MΩ

I C∞ = 0

ε − (1.2 MΩ )I ∞ − (600 kΩ )I ∞ = 0

416 Chapter 25 Solve for and evaluate I∞:

I∞ = =

ε 1.2 MΩ + 600 kΩ 120 V = 66.7 µA 1.2 MΩ + 600 kΩ

VC∞ = I ∞ R600 kΩ

(c) The maximum voltage across the capacitor equals the potential difference across the 600-kΩ under steady-state conditions. Apply Ohm’s law to obtain:

= (66.7 µA )(600 kΩ ) = 40.0 V

117 •• Picture the Problem We can use Q(t ) = Qf 1 − e −t τ = Cε 1 − e −t τ to find the charge

(

)

(

)

on the capacitor at t = τ and differentiate this expression with respect to time to find the rate at which the charge is increasing (the current). The power supplied by the battery is given by Pτ = Iτ ε and the power dissipated in the resistor by PR ,τ = Iτ2 R . In part (f) we

can differentiate U (t ) = Q 2 (t ) 2C with respect to time and evaluate the derivative at t = τ

to find the rate at which the energy stored in the capacitor is increasing.

(

)

(

Q(t ) = Qf 1 − e −t τ = Cε 1 − e −t τ

)

(1)

(a) Express the charge Q on the capacitor as a function of time:

where τ = RC.

Evaluate Q(τ) to obtain:

Q(τ ) = (1.5 µF)(6 V ) (1 − e −1 ) = 5.69 µC

(b) and (c) Differentiate equation (1) with respect to t to obtain: Apply Kirchhoff’s loop rule to the circuit just after the circuit is completed to obtain: Because VC0 = 0 we have:

dQ(t ) = I = I 0 e −t τ dt ε − RI 0 − VC0 = 0

I0 =

ε R

Substitute to obtain:

dQ(t ) ε = I (t ) = e −t τ dt R

Substitute numerical values and evaluate I(τ):

I (τ ) =

6 V −1 e = 1.10 µA 2 MΩ

`Electric Current and Direct-Current Circuits 417

(d) Express the power supplied by the battery as the product of its emf and the current drawn from it at t = τ: (e) The power dissipated in the resistor is given by:

(f) Express the energy stored in the capacitor as a function of time: Differentiate this expression with respect to time to obtain:

Evaluate this expression when t = τ to obtain:

P(τ ) = I (τ )ε = (1.10 µC/s )(6 V ) = 6.60 µW

PR (τ ) = I 2 (τ )R = (1.10 µA ) (2 MΩ ) = 2.42 µW 2

U (t ) =

Q 2 (t ) 2C

dU (t ) 1 d 2 = Q (t ) dt 2C dt 1 (2Q(t )) dQ(t ) = dt 2C Q(t ) I (t ) = C

[

]

dU (τ ) Q(τ ) = I (τ ) dt C 5.69 µC (1.10 µA ) = 1.5 µF = 4.17 µW

Remarks: Note that our answer for part (f) is the difference between the power delivered by the battery at t = τ and the rate at which energy is dissipated in the resistor at the same time. 118 •• Picture the Problem We can apply Kirchhoff’s junction rule to find the current in each branch of this circuit and then use the loop rule to obtain equations solvable for R1, R2, and R3 . (a) Apply Kirchhoff’s junction rule at the junction of the 5-µF capacitor and the 10-Ω and 50-Ω resistors under steady-state conditions:

I bat = I10 Ω + 5 A

Because the potential differences across the 5-µF capacitor and the 10-Ω resistor are the same:

I10 Ω =

V10 Ω V = C 10 Ω 10 Ω

(1)

418 Chapter 25 Qf C

Express the potential difference across the capacitor to its steadystate charge:

VC =

Substitute to obtain:

I10 Ω =

Substitute in equation (1) to obtain:

I bat =

Qf + 5A (10 Ω )C

Substitute numerical values and evaluate Ibat:

I bat =

1000 µC + 5 A = 25.0 A (10 Ω )(5 µF)

(b) Use Kirchhoff’s junction rule to find the currents I5 Ω, IR3, and IR1:

I 5 Ω = 10 A ,

Qf (10 Ω )C

I R 3 = 15 A , and

I R1 = I bat = 25 A Apply the loop rule to the loop that includes the battery, R1, and the 50Ω and 5-Ω resistors: Solve for R1 to obtain: Apply the loop rule to the loop that includes the battery, R1, the 10-Ω resistor and R3: Solve for R3 to obtain: Apply the loop rule to the loop that includes the 10-Ω and 50-Ω resistors and R2: Solve for R2 to obtain:

310 V − (25 A )R1 − (5 A )(50 Ω ) − (10 A )(5 Ω ) = 0

R1 = 0.400 Ω

310 V − (25 A )(0.4 Ω ) − (20 A )(10 Ω ) − (15 A )R3 = 0 R3 = 6.67 Ω

− (20 A )(10 Ω ) − (5 A )R2 + (5 A )(50 Ω ) = 0

R2 = 10.0 Ω

119 •• Picture the Problem We can solve Equation 25-35 for dQ/dt and separate the variables in order to obtain the equation given above. Integrating this differential equation will yield Equation 25-36.

`Electric Current and Direct-Current Circuits 419

Solve Equation 25-35 for dQ/dt to obtain:

dQ εC − Q = dt RC dQ dt = εC − Q RC

Separate the variables to obtain:

Integrate dQ′ from 0 to Q and dt′ from 0 to t:

Q

t

1 dQ' ∫0 εC − Q' = RC ∫0 dt'

and

⎛ εC ⎞ t ⎟⎟ = ln⎜⎜ ⎝ εC − Q ⎠ RC

Transform from logarithmic to exponential form to obtain: Solve for Q to obtain Equation 2536:

εC

εC − Q

t

= e RC

(

)

(

Q = εC 1 − e −t RC = Qf 1 − e −t RC

)

*120 ••• Picture the Problem We can find the time-to-discharge by expressing the voltage across the capacitor as a function of time and solving for t. We can use U (t ) = 12 CV 2 (t ) to find the energy released/stored in the capacitor when the lamp flashes. In part (c) we can integrate dUbat = εdI(t) to find the energy supplied by the battery during the charging cycle.

Q(t ) Qf = 1 − e −t RC C C −t RC = Vf 1 − e

( )

)

(a) Express the voltage across the capacitor as a function of time:

V (t ) =

Solve for t to obtain:

⎛ V (t ) ⎞ ⎟ t = − RC ln⎜⎜1 − Vf ⎟⎠ ⎝

Substitute numerical values and evaluate t:

⎛ 7V⎞ ⎟⎟ t = −(18 kΩ )(0.15 µF) ln⎜⎜1 − ⎝ 9V ⎠

(

= 4.06 ms (b) Express the energy stored in the capacitor as a function of time:

U (t ) = 12 CV 2 (t )

420 Chapter 25

(

U (t ) = 12 CVf2 1 − e −t RC

Substitute for V(t) to obtain:

)

2

Substitute numerical values and evaluate U(4.06 ms):

U (4.06 ms ) =

1 2

(0.15 µF)(9 V )2 (1 − e −4.06 ms (18 kΩ )(0.15 µF ) )2 =

3.67 µJ

2 t ε U bat (t ) = ε ∫ I (t' )dt' = e −t' RC dt' R ∫ t

(c) Relate the energy provided by the battery to its emf and the current it delivers:

0

=

0

ε [RC (1 − e −t RC )] 2

R = Cε 2 1 − e −t RC

(

)

Substitute numerical values and evaluate Ubat(4.06 ms):

(

)

U bat (4.06 ms ) = (0.15 µF)(9 V ) 1 − e −4.06 ms (18 kΩ )(0.15 µF ) = 9.45 µJ 2

Express the fraction f of the energy supplied by the battery during the charging cycle that is dissipated in the resistor: Use conservation of energy to relate the energy supplied by the battery to the energy dissipated in the resistor and the energy released when the lamp flashes:

f =

UR U bat

U bat = U R + U flash or

U R = U bat − U flash

U bat − U flash U = 1 − flash U bat U bat

Substitute to obtain:

f =

Substitute numerical values and evaluate f:

f = 1−

3.67 µJ = 61.2% 9.45 µJ

*121 ••• Picture the Problem Let R1 = 200 Ω, R2 = 600 Ω, I1 and I2 their currents, and I3 the current into the capacitor. We can apply Kirchhoff’s loop rule to find the initial battery current I0 and the battery current I∞ a long time after the switch is closed. In part (c) we can apply both the loop and junction rules to obtain equations that we can use to obtain a linear differential equation with constant coefficients describing the current in the 600-Ω

`Electric Current and Direct-Current Circuits 421

resistor as a function of time. We can solve this differential equation by assuming a solution of a given form, differentiating this assumed solution and substituting it and its derivative in the differential equation. Equating coefficients, requiring the solution to hold for all values of the assumed constants, and invoking an initial condition will allow us to find the constants in the assumed solution. (a) Apply Kirchhoff’s loop rule to the circuit at the instant the switch is closed:

ε − (200 Ω)I 0 − VC0 = 0

Because the capacitor is initially uncharged:

VC0 = 0

Solve for and evaluate I0:

I0 =

(b) Apply Kirchhoff’s loop rule to the circuit after a long time has passed:

ε 200 Ω

=

50 V = 0.250 A 200 Ω

50 V − (200 Ω )I ∞ − (600 Ω )I ∞ = 0

50 V = 62.5 mA 800 Ω

Solve for I∞ to obtain:

I∞ =

(c) Apply the junction rule at the junction between the 200-Ω resistor and the capacitor to obtain: Apply the loop rule to the loop containing the source, the 200-Ω resistor and the capacitor to obtain:

I1 = I 2 + I 3

(1)

ε − R1I1 − Q = 0

(2)

Apply the loop rule to the loop containing the 600-Ω resistor and the capacitor to obtain:

Q − R2 I 2 = 0 C

(3)

Differentiate equation (2) with respect to time to obtain:

d ⎡ ε − R1I1 − Q ⎤⎥ = 0 − R1 dI1 − 1 dQ ⎢ dt ⎣ C⎦ dt C dt dI 1 = − R1 1 − I 3 = 0 dt C

C

or

R1

dI1 1 = − I3 dt C

(4)

422 Chapter 25

Differentiate equation (3) with respect to time to obtain:

d ⎡Q dI 2 ⎤ 1 dQ R I R − = − =0 2 2 2 ⎥⎦ C dt dt ⎢⎣ C dt or

R2

dI 2 1 = I3 dt C

(5)

Using equation (1), substitute for I3 in equation (5) to obtain:

dI 2 1 (I1 − I 2 ) = dt R2C

Solve equation (2) for I1:

I1 =

Substitute for I1 in equation (6) and simplify to obtain the differential equation for I2:

ε −Q R1

Differentiate I2(t) with respect to time to obtain: Substitute for I2 and dI2/dt to obtain:

Equate coefficients of e −t τ to obtain:

=

ε − R2 I 2 R1

⎞ dI 2 1 ⎛ ε − R2 I 2 ⎜⎜ = − I 2 ⎟⎟ dt R2C ⎝ R1 ⎠ =

To solve this linear differential equation with constant coefficients we can assume a solution of the form:

C

(6)

ε

⎛ R + R2 ⎞ ⎟I2 − ⎜⎜ 1 R1R2C ⎝ R1R2C ⎟⎠

I 2 (t ) = a + be −t τ

[

(7)

]

dI 2 d b = a + be −t τ = − e −t τ τ dt dt ε − ⎛⎜ R1 + R2 ⎞⎟ a + be−t τ b − e −t τ = R1R2C ⎜⎝ R1R2C ⎟⎠ τ

(

τ=

R1R2C R1 + R2

ε

Requiring the equation to hold for all values of a yields:

a=

If I2 is to be zero when t = 0:

0 = a+b

R1 + R2

or

b = −a = −

ε R1 + R2

)

`Electric Current and Direct-Current Circuits 423

Substitute in equation (7) to obtain:

I 2 (t ) = =

ε R1 + R2

ε

R1 + R2

−

ε R1 + R2

e −t τ

(1 − e ) −t τ

where

τ=

R1 R2C (200 Ω )(600 Ω )(5 µF) = R1 + R2 200 Ω + 600 Ω

= 0.750 ms Substitute numerical values and evaluate I2(t):

I 2 (t ) = =

(

50 V 1 − e −t 0.750 ms 200 Ω + 600 Ω

)

(62.5 mA ) (1 − e −t 0.750 ms )

122 ••• Picture the Problem Let R1 represent the 1.2-MΩ resistor and R2 the 600-kΩ resistor. Immediately after switch S is closed, the capacitor has zero charge and so the potential difference across it (and the 600 kΩ-resistor) is zero. A long time after the switch is closed, the capacitor will be fully charged and the potential difference across it will be given by both Q/C and I∞R2. When the switch is opened after having been closed for a long time, both the source and the 1.2-MΩ resistor will be out of the circuit and the fully charged capacitor will discharge through R1. We can use Kirchhoff’s loop to find the currents drawn from the source immediately after the switch is closed and a long time after the switch is closed, as well as the current in the RC circuit when the switch is again opened and the capacitor discharges through R2. (a) Apply Kirchhoff’s loop rule to the circuit immediately after the switch is closed to obtain: Solve for and evaluate I0:

(b) Apply Kirchhoff’s loop rule to the circuit a long time after the switch is closed to obtain:

ε − I 0 R1 − VC0 = 0 or, because VC0 = 0,

ε − I 0 R1 = 0 I0 =

ε R1

=

50 V = 41.7 µA 1.2 MΩ

ε − I ∞ R1 − I ∞ R2 = 0

424 Chapter 25 Solve for and evaluate I∞:

I∞ = =

ε R1 + R2 50 V = 27.8 µA 1.2 MΩ + 600 kΩ

VC (t ) − R2 I (t ) = 0

(c) Apply Kirchhoff’s loop rule to the RC circuit sometime after the switch is opened and solve for I(t) to obtain:

or

I (t ) =

VC (t ) R2

Substitute for VC(t):

I (t ) =

VC∞ −t τ e = I ∞ e −t τ R2

where τ = R2C. Substitute numerical values to obtain:

I (t ) = (27.8 µA ) e −t (600 kΩ )(2.5 µF ) =

(27.8 µA ) e −t 1.5s

123 ••• Picture the Problem In part (a) we can apply Kirchhoff’s loop rule to the circuit immediately after the switch is closed in order to find the initial current I0. We can find the time at which the voltage across the capacitor is 24 V by again applying Kirchhoff’s loop rule to find the voltage across the resistor when this condition is satisfied and then using the expression I (t ) = I 0e −t τ for the current through the resistor as a function of time and solving for t. (a) Apply Kirchhoff’s loop rule to the circuit immediately after the switch is closed: Solve for and evaluate I0:

ε − 12 V − I 0 R = 0 I0 =

ε − 12 V

R 36 V − 12 V = = 48.0 µA 0.5 MΩ

(b) Apply Kirchhoff’s loop rule to the circuit when VC = 24 V and solve for VR: Express the current through the resistor as a function of I0 and τ :

36 V − 24 V − I (t )R = 0 and

I (t )R = 12 V

I (t ) = I 0e −t τ where τ = RC.

`Electric Current and Direct-Current Circuits 425

Substitute to obtain:

RI 0e −t τ = 12 V or

e −t τ = t

12 V RI 0 12 V RI 0

Take the natural logarithm of both sides of the equation to obtain:

−

Solve for t:

⎛ 12 V ⎞ ⎛ 12 V ⎞ ⎟⎟ = − RC ln⎜⎜ ⎟⎟ t = −τ ln⎜⎜ RI RI ⎝ 0 ⎠ ⎝ 0 ⎠

τ

= ln

Substitute numerical values and evaluate t:

⎡ ⎤ 12 V t = −(0.5 MΩ )(2.5 µF) ln ⎢ ⎥ = 0.866 s ⎣ (0.5 MΩ )(48 µA ) ⎦ 124 ••• Picture the Problem In part (a) we can apply Kirchhoff’s loop rule to the circuit immediately after the switch is closed in order to find the initial current I0. We can find the time at which the voltage across the capacitor is 24 V by again applying Kirchhoff’s loop rule to find the voltage across the resistor when this condition is satisfied and then using the expression I (t ) = I 0 e −t τ for the current through the resistor as a function of time and solving for t. (a) Apply Kirchhoff’s loop rule to the circuit immediately after the switch is closed: Solve for and evaluate I0:

ε + 12 V − I 0 R = 0

I0 =

ε + 12 V

R 36 V + 12 V = = 96.0 µA 0.5 MΩ

(b) Apply Kirchhoff’s loop rule to the circuit when VC = 24 V and solve for VR: Express the current through the resistor as a function of I0 and τ :

36 V − 24 V − I (t )R = 0 and

I (t )R = 12 V

I (t ) = I 0e −t τ where τ = RC.

426 Chapter 25 Substitute to obtain:

RI 0e −t τ = 12 V or

e −t τ = t

12 V RI 0 12 V RI 0

Take the natural logarithm of both sides of the equation to obtain:

−

Solve for t:

⎛ 12 V ⎞ ⎛ 12 V ⎞ ⎟⎟ = − RC ln⎜⎜ ⎟⎟ t = −τ ln⎜⎜ RI RI ⎝ 0 ⎠ ⎝ 0 ⎠

τ

= ln

Substitute numerical values and evaluate t:

⎡ ⎤ 12 V t = −(0.5 MΩ )(2.5 µF) ln ⎢ ⎥ = 1.73 s ⎣ (0.5 MΩ )(96 µA ) ⎦

General Problems *125 •• Determine the Concept Because all of the current drawn from the battery passes through R1, we know that I1 is greater than I2 and I3. Because R2 ≠ R3, I2 ≠ I3 and so (b) is false. Because R3 > R2, I3 < I2 and so (c) is false. (a ) is correct. 126 •• A 25-W lightbulb is connected in series with a 100-W lightbulb and a voltage V is placed across the combination. Which lightbulb is brighter? Explain. Determine the Concept The 25-W bulb will be brighter. The brightness of a bulb is proportional to the power it dissipates. The resistance of the 25-W bulb is greater than that of the 100-W bulb, and in the series combination, the same current I flows through the bulbs. Hence, I 2R25 > I 2R100. 127 • Picture the Problem We can apply Ohm’s law to find the current drawn from the battery and use Kirchhoff’s loop rule to find the current in the 6-Ω resistor.

ε

Using Ohm’s law, express the current I1 drawn from the battery:

I1 =

Find Req:

Req = 4 Ω +

Req

(6 Ω )(12 Ω ) = 8 Ω 6 Ω + 12 Ω

`Electric Current and Direct-Current Circuits 427

Substitute and evaluate I1:

Apply Kirchhoff’s loop rule to a loop that includes the battery and the 4-Ω and 6-Ω resistors: Solve for I2 to obtain:

I1 =

24 V = 3A 8Ω

24 V − (4 Ω )(3 A ) − (6 Ω )I 2 = 0

I 2 = 2 A and (b) is correct.

128 • 2 R to find the maximum current the resistor Picture the Problem We can use P = I max can tolerate and Ohm’s law to find the voltage across the resistor that will produce this current. (a) Relate the maximum current the resistor can tolerate to its power and resistance: Solve for and evaluate Imax:

(b) Use Ohm’s law to relate the voltage across the resistor to this maximum current:

2 P = I max R

I max =

P 5W = = 0.707 A R 10 Ω

V = I max R = (0.707 A )(10 Ω ) = 7.07 V

129 • Picture the Problem We can use Ohm’s law to find the short-circuit current drawn from the battery and the relationship between the terminal potential difference, the emf of the battery, and the current being drawn from it to find the terminal voltage when the battery is delivering a current of 20 A. (a) Apply Ohm’s law to the shorted battery to find the short-circuit current: (b) Express the terminal voltage as the difference between the emf of the battery and the current being drawn from it:

I sc =

ε r

=

12 V = 30.0 A 0.4 Ω

Vterm = ε − Ir

428 Chapter 25 Substitute numerical values and evaluate Vterm:

Vterm = 12 V − (20 A )(0.4 Ω ) = 4.00 V

130 •• Picture the Problem We can use Kirchhoff’s loop rule to obtain two equations relating ε and r that we can solve simultaneously to find these quantities. Use Kirchhoff’s loop rule to relate the emf of the battery to the current drawn from it and the internal and external resistance: When I = 1.80 A and a 7.0-Ω resistor is connected across the battery terminals equation (1) becomes: When I = 2.20 A and a 12-Ω resistor is connected in parallel with the 7.0Ω resistor: Find the equivalent resistance:

Substitute to obtain:

ε − IR − Ir = 0

(1)

ε − (1.8 A )(7 Ω ) − (1.8 A ) r = 0 or

ε − 12.6 V − (1.8 A ) r = 0

(2)

ε − (2.2 A )Req − (2.2 A ) r = 0

Req =

(7 Ω )(12 Ω ) = 4.42 Ω 7 Ω + 12 Ω

ε − (2.2 A )(4.42 Ω ) − (2.2 A )r = 0 or

ε − 9.72 V − (2.2 A ) r = 0

Solve equations (2) and (3) simultaneously to obtain:

ε=

(3)

25.5 V and r = 7.19 Ω

*131 •• Picture the Problem We can apply Kirchhoff’s loop rule to the circuit that includes the box and the 21-V source to obtain two equations in the unknowns ε and R that we can solve simultaneously.

`Electric Current and Direct-Current Circuits 429

Apply Kirchhoff’s loop rule to the circuit when the polarity of the 21V source and the direction of the current are as shown in the diagram: Apply Kirchhoff’s loop rule to the circuit when the polarity of the source is reversed and the current is 2 A in the opposite direction: Solve these equations simultaneously to obtain:

21 V + ε − (1 A )R = 0

− 21 V + ε

+ (2 A )R = 0

R = 14.0 Ω and

ε=

− 7.00 V

132 •• Picture the Problem When the switch is closed, the initial potential differences across the capacitors are zero (they have no charge) and the resistors in the bridge portion of the circuit are in parallel. When a long time has passed, the current through the capacitors will be zero and the resistors will be in series. In both cases, the application of Kirchhoff’s loop rule to the entire circuit will yield the current in the circuit. To find the final charges on the capacitors we can use the definition of capacitance and apply Kirchhoff’s loop rule to the loops containing two resistors and a capacitor to find the potential differences across the capacitors. (a) Apply Kirchhoff’s loop rule to the circuit immediately after the switch is closed:

50 V − I 0 (10 Ω ) − I 0 Req = 0

50 V 10 Ω + Req

Solve for I0:

I0 =

Find the equivalent resistance of 15 Ω, 12 Ω, and 15 Ω in parallel:

1 1 1 1 = + + Req 15 Ω 12 Ω 15 Ω and

Req = 4.62 Ω Substitute for Req and evaluate I0:

(b) Apply Kirchhoff’s loop rule to the circuit a long time after the switch is closed:

I0 =

50 V = 3.42 A 10 Ω + 4.62 Ω

50 V − I ∞ (10 Ω ) − I ∞ Req = 0

430 Chapter 25 50 V 10 Ω + Req

Solve for I∞:

I∞ =

Find the equivalent resistance of 15 Ω, 12 Ω, and 15 Ω in series:

Req = 15 Ω + 12 Ω + 15 Ω = 42 Ω

Substitute for Req and evaluate I∞:

50 V = 0.962 A 10 Ω + 42 Ω Q10 µF = C10 µFV10 µF

(c) Using the definition of capacitance, express the charge on the capacitors in terms of their final potential differences: Apply Kirchhoff’s loop rule to the loop containing the 15-Ω and 12-Ω resistors and the 10 µF capacitor to obtain: Solve for V10 µF: Substitute in equation (1) and evaluate Q10 µF:

I∞ =

(1)

and

Q5 µF = C5 µFV5 µF

(2)

V10 µF − (15 Ω )I ∞ − (12 Ω )I ∞ = 0

V10 µF = (27 Ω )I ∞ Q10 µF = C10 µF (27 Ω )I ∞

= (10 µF)(27 Ω )(0.962 A ) = 260 µC

Apply Kirchhoff’s loop rule to the loop containing the 15-Ω and 12-Ω resistors and the 5 µF capacitor to obtain: Solve for V5 µF: Substitute in equation (2) and evaluate Q5 µF:

V5 µF − (15 Ω )I ∞ − (12 Ω )I ∞ = 0

V5 µF = (27 Ω )I ∞ Q5 µF = C5 µF (27 Ω )I ∞

= (5 µF)(27 Ω )(0.962 A ) = 130 µC

*133 •• Picture the Problem Let the current flowing through the galvanometer by IG. By applying Kirchhoff’s rules to the loops including 1) R1, the galvanometer, and Rx, and 2) R2, the galvanometer, and R0, we can obtain two equations relating the unknown

`Electric Current and Direct-Current Circuits 431

resistance to R1, R2 and R0. Using R = ρL/A will allow us to express Rx in terms of the length of wire L1 that corresponds to R1 and the length of wire L2 that corresponds to R2. Apply Kirchhoff’s loop rule to the loop that includes R1, the galvanometer, and Rx to obtain: Apply Kirchhoff’s loop rule to the loop that includes R2, the galvanometer, and R0 to obtain:

− R1 I1 + Rx I 2 = 0

− R2 (I1 − I G ) + R0 (I 2 + I G ) = 0

R1I1 = Rx I 2

(1)

(2)

(3)

When the bridge is balanced, IG = 0 and equations (1) and (2) become:

and

Divide equation (3) by equation (4) and solve for x to obtain:

Rx = R0

Express R1 and R2 in terms of their lengths, cross-sectional areas, and the resistivity of their wire:

R1 = ρ

Substitute in equation (5) to obtain:

Rx = R0

(a) When the bridge balances at the 18-cm mark, L1 = 18 cm, L2 = 82 cm and:

Rx = (200 Ω )

18 cm = 43.9 Ω 82 cm

(b) When the bridge balances at the 60-cm mark, L1 = 60 cm, L2 = 40 cm and:

Rx = (200 Ω )

60 cm = 300 Ω 40 cm

(c) When the bridge balances at the 95-cm mark, L1 = 95 cm, L2 = 5 cm and:

Rx = (200 Ω )

95 cm = 3.80 kΩ 5 cm

R2 I1 = R0 I 2

(4)

R1 R2

(5)

L1 L and R2 = ρ 2 A A

L1 L2

134 •• Picture the Problem Let the current flowing through the galvanometer by IG. By applying Kirchhoff’s rules to the loops including 1) R1, the galvanometer, and Rx, and 2) R2, the galvanometer, and R0, we can obtain two equations relating the unknown

432 Chapter 25 resistance to R1, R2 and R0. Using R = ρL/A will allow us to express Rx in terms of the length of wire L1 that corresponds to R1 and the length of wire L2 that corresponds to R2. To find the effect of an error of 2 mm in the location of the balance point we can use the relationship ∆Rx = (dRx dL )∆L to determine ∆Rx and then divide by Rx = R0 L (1 − L ) to find the fractional change (error) in Rx resulting from a given error in the determination of the balance point. Apply Kirchhoff’s loop rule to the loop that includes R1, the galvanometer, and Rx to obtain: Apply Kirchhoff’s loop rule to the loop that includes R2, the galvanometer, and R0 to obtain:

− R1 I1 + Rx I 2 = 0

(1)

− R2 (I1 − I G ) + R0 (I 2 + I G ) = 0

R1I1 = Rx I 2

When the bridge is balanced, IG = 0 and equations (1) and (2) become:

and

Divide equation (3) by equation (4) and solve for x to obtain:

Rx = R0

Express R1 and R2 in terms of their lengths, cross-sectional areas, and the resistivity of their wire:

R1 = ρ

Substitute in equation (5) to obtain:

Rx = R0

(a) When the bridge balances at the 98-cm mark, L1 = 98 cm, L2 = 2 cm and:

Rx = (200 Ω )

(b) Express Rx in terms of the distance to the balance point:

Rx = R0

Express the error ∆Rx in Rx resulting from an error ∆L in L:

∆Rx =

(2)

(3)

R2 I1 = R0 I 2

(4)

R1 R2

(5)

L1 L and R2 = ρ 2 A A

L1 L2

(6)

98 cm = 9.80 kΩ 2 cm

L 1− L

dRx d ⎡ L ⎤ ∆L ∆L = R0 dL dL ⎢⎣1 − L ⎥⎦ 1 = R0 ∆L (1 − L )2

`Electric Current and Direct-Current Circuits 433

Divide ∆Rx by Rx to obtain:

∆Rx = Rx

R0

1 ∆L (1 − L )2 = 1 ∆L L 1− L L R0 1− L

Evaluate ∆Rx/Rx for L = 98 cm and ∆L = 2 mm:

∆Rx 1m 2 mm = = 10.0% Rx 1m − 0.98 m 1 m

(c) Solve equation (6) for the ratio of L1 to L2:

L1 Rx = L2 R0

For L1 = 50 cm, L2 = 50 cm, and R0 = Rx = 9.80 kΩ. Hence, a resistor of approximately 10 kΩ will cause the bridge to balance near the 50 - cm mark. 135 •• Picture the Problem Knowing the beam current and charge per proton, we can use I = ne to determine the number of protons striking the target per second. The energy deposited per second is the power delivered to the target and is given by P = IV. We can find the elapsed time before the target temperature rises 300C° using ∆Q = P∆t = mcCu∆T. (a) Relate the current to the number of protons per second n arriving at the target:

I = ne

Solve for and evaluate n:

n=

(b) Express the power of the beam in terms of the beam current and energy:

P = IV = (3.5 µA )(60 MeV ) = 210 J/s

(c) Relate the energy delivered to the target to its heat capacity and temperature change:

∆Q = P∆t = CCu ∆T = mcCu ∆T

Solve for ∆t:

∆t =

I 3.50 µA = = 2.19 × 1013 / s e 1.60 × 10 −19 C

mcCu ∆T P

434 Chapter 25 Substitute numerical values (see Table 19-1 for the specific heat of copper)and evaluate ∆t:

(50 g )(0.386 kJ/kg ⋅ K )(300C°)

∆t =

210 J/s

= 27.6 s

136 •• Picture the Problem We can use the definition of current to express the current delivered by the belt in terms of the surface charge density, width, and speed of the belt. The minimum power needed to drive the belt can be found from P = IV. (a) Use its definition to express the current carried by the belt: Substitute numerical values and evaluate I:

I=

dQ dx = σw = σwv dt dt

(

)

I = 5 mC/m 2 (0.5 m )(20 m/s ) = 50.0 mA

(b) Express the minimum power of the motor in terms of the current delivered and the potential of the charge:

P = IV

Substitute numerical values and evaluate P:

P = (50 mA )(100 kV ) = 5.00 kW

137 •• Picture the Problem We can differentiate the expression relating the amount of heat required to produce a given temperature change with respect to time to express the mass flow-rate required to maintain the temperature of the coils at 50°C. We can then use the definition of density to find the necessary volume flow rate. Express the heat that must be dissipated in terms of the specific heat and mass of the water and the desired temperature change of the water:

Q = mcwater ∆T

Differentiate this expression with respect to time to obtain an expression for the power dissipation:

P=

dQ dm = cwater ∆T dt dt

`Electric Current and Direct-Current Circuits 435

Solve for dm/dt:

dm P = dt cwater ∆T

Substitute for the power dissipated to obtain:

dm IV = dt cwater ∆T (100 A )(240 V ) dm = dt (4.18 kJ/kg ⋅ K )(50°C − 15°C )

Substitute numerical values and evaluate dm/dt:

= 0.164 kg/s Using the definition of density, express the volume flow rate in terms of the mass flow rate to obtain:

dV 1 dm 0.164 kg/s = = dt ρ dt 103 kg/m 3 ⎛ 1L ⎞ = 0.164 × 10 −3 m 3 /s ⎜⎜ −3 3 ⎟⎟ ⎝ 10 m ⎠

(

)

= 0.164 L/s 138 •• Picture the Problem We can use the expressions for the capacitance of a dielectricfilled parallel-plate capacitor and the resistance of a conductor to show that RC = ∈0ρκ. Express the capacitance of the dielectric-filled parallel-plate capacitor:

C=

Express the resistance of a conductor with the same dimensions:

R=

The product of C and R is:

κ ∈0 A d

ρd A

RC =

ρd κ ∈0 A A

d

= ∈0 ρκ

139 •• Picture the Problem We can use the expressions for the capacitance of a dielectric-filled cylindrical capacitor and the resistance of a cylindrical conductor to show that RC = ∈0ρκ. Express the capacitance of the dielectric-filled cylindrical capacitor whose inner and outer radii are r1 and r2, respectively:

C=

2πlκ ∈0 ⎛r ⎞ ln⎜⎜ 2 ⎟⎟ ⎝ r1 ⎠

where l is the length of the capacitor. Express the resistance of a cylindrical resistor with the same dimensions:

R=

⎛ r2 ⎞ ⎟ ⎟ r ⎝ 1⎠ 2πl

ρ ln⎜⎜

436 Chapter 25 The product of C and R is:

⎛ r2 ⎞ ⎟ r1 ⎟⎠ 2πlκ ∈0 ⎝ RC = = ∈0 ρκ 2πl ⎛ r2 ⎞ ln⎜⎜ ⎟⎟ ⎝ r1 ⎠ This result holds independently of the geometries of the capacitor and

ρ ln⎜⎜

the resistor. *140 •• Picture the Problem We’ll assume that the capacitor is fully charged initially and apply Kirchhoff’s loop rule to the circuit fragment to obtain the differential equation describing the discharge of the leaky capacitor. We’ll show that the solution to this equation is the familiar expression for an exponential decay with time constant τ = ∈0ρκ.

If we think of the leaky capacitor as a resistor/capacitor combination, (a) the voltage drop across the resistor must be the same as voltage drop across the capacitor. Hence, they must be in parallel. (b) Assuming that the capacitor is initially fully charged, apply Kirchhoff’s loop rule to the circuit fragment to obtain:

Q − RI = 0 C or, because I = −

dQ , dt

Q dQ +R =0 C dt Separate variables in this differential equation to obtain:

dQ 1 =− dt Q RC

From Problems 138 and 139 we have:

RC =∈0 ρκ

Substitute for RC in the differential equation to obtain:

1 dQ =− dt ∈0 ρκ Q

Integrate this equation from Q′ = Q0 to Q to obtain:

Q = Q0 e −t τ where

τ = ∈0 ρκ

`Electric Current and Direct-Current Circuits 437

(c) Because Q/Q0 = 0.1:

e −t τ = 0.1

Solve for t by taking the natural logarithm of both sides of the equation:

−

t

τ

= ln 0.1 ⇒ t = − ∈0 ρκ ln 0.1

Substitute numerical values and evaluate t:

t = −(8.85 × 10 −12 C 2 / N ⋅ m 2 )(9 × 1013 Ω ⋅ m )(5) ln 0.1 = 9.17 × 103 s = 2.55 h 141 ••• Picture the Problem We can use its definition to find the time constant of the charging circuit in part (a). In part (b) we can use the expression for the potential difference as a function of time across a charging capacitor and utilize the hint given in the problem statement to show that the voltage across the capacitor increases almost linearly over the time required to bring the potential across the switch to its critical value. In part (c) we can use the result derived in part (b) to find the value of R1 such that C charges from 0.2 V to 4.2 V in 0.1 s. In part (d) we can use the expression for the potential difference as a function of time across a discharging capacitor to find the discharge time. Finally, in part (e) we can integrate I 2R1 over the discharge time to find the rate at which energy is dissipated in R1 during the discharge of the capacitor and use the difference in the energy stored in the capacitor initially and when the switch opens to find the rate of energy dissipation in resistance of the capacitor. (a) When the capacitor is charging, the switch is open and the resistance in the charging circuit is R1. Hence: (b) Express the voltage across the charging capacitor as a function of time:

τ = R1C = (0.5 MΩ )(0.02 µF) = 10.0 ms

(

V (t ) = ε 1 − e −t τ

)

V (t )

Solve for the exponential term to obtain:

e −t τ = 1 −

Noting that V(t) << ε, let η = V(t)/ε

e −t τ = 1 − η

ε

or

et τ = (1 − η ) ≈ 1 + η −1

because η << 1.

(1)

438 Chapter 25 Use the power series for ex to expand et τ :

e

tτ

2

1⎛t⎞ = 1 + + ⎜ ⎟ + ... τ 2! ⎝ τ ⎠ 1 ≈ 1+ t t

τ

provided t τ << 1 . Substitute in equation (1) to obtain:

1 V (t ) 1+ t ≈ 1 +η = 1 +

ε

τ

Solve for t to obtain the linear relationship: (c) Using the result derived in (b), relate the time ∆t required to change the voltage across the capacitor by an amount ∆V to ∆V:

εt

V (t ) =

ε ∆t

∆V (t ) =

τ

or

τ = R1C =

Solve for R1:

R1 =

Substitute numerical values and evaluate R1:

R1 =

(1)

τ

ε

∆V (t )

ε

C∆V (t )

∆t

∆t

(800 V )(0.1s ) (0.02 µF)(4.2 V − 0.2 V )

= 1.00 GΩ (d) Express the potential difference across the capacitor as a function of time:

VC (t ) = VC0 e −t τ' where τ′ = R2C.

Solve for t to obtain:

⎛ V (t ) ⎞ ⎛ V (t ) ⎞ t = −τ' ln⎜⎜ C ⎟⎟ = − R2C ln⎜⎜ C ⎟⎟ ⎝ VC0 ⎠ ⎝ VC0 ⎠

Substitute numerical values and evaluate t:

⎛ 0 .2 V ⎞ ⎟⎟ t = −(0.001 Ω )(0.02 µF) ln⎜⎜ ⎝ 4 .2 V ⎠ = 60.9 ps

(e) Express the rate at which energy is dissipated in R1 as a function of its

P1 =

∆E1 = I 2 R1 ∆t

`Electric Current and Direct-Current Circuits 439

resistance and the current through it: Because the current varies with time, we need to integrate over time to find ∆E1:

⎛ V (t ) ⎞ ⎟⎟ R1dt ∆E1 = ∫ I R1dt = ∫ ⎜⎜ ⎝ R1 ⎠ 2

2

⎛εt ⎞ ⎟⎟ R1dt = ∫ ⎜⎜ R τ ⎝ 1⎠ 2

⎛ε = ⎜⎜ ⎝τ

2

0.105 s

⎞ 1 ⎟⎟ t 2 dt ∫ ⎠ R1 0.005 s 0.105 s

2

⎛ 800 V ⎞ 1 ⎡ t 3 ⎤ ⎟⎟ = ⎜⎜ ⎢ ⎥ ⎝ 20 s ⎠ 1 GΩ ⎣ 3 ⎦ 0.005s = 6.17 × 10 −10 J Substitute and evaluate P1:

P1 =

6.17 × 10 −10 J = 6.17 nW 0.1s

∆U C U Ci − U Cf = ∆t ∆t 2 1 1 CV − CV 2 1 C Vi 2 − Vf2 = 2 i 2 f = 2 ∆t ∆t

Express the rate at which energy is dissipated in the switch resistance:

P2 =

Substitute numerical values and evaluate P2:

P2 =

(

1 2

)

(0.02 µF)([ 4.2 V )2 − (0.2 V )2 ] 60.9 ps

= 2.89 kW 142 ••• Picture the Problem We can apply both the loop and junction rules to obtain equations that we can use to obtain a linear differential equation with constant coefficients describing the current in R2 as a function of time. We can solve this differential equation by assuming a solution of an appropriate form, differentiating this assumed solution and substituting it and its derivative in the differential equation. Equating coefficients, requiring the solution to hold for all values of the assumed constants, and invoking an initial condition will allow us to find the constants in the assumed solution. Once we know how the current varies with time in R2, we can express the potential difference across it (as well as across C because they are in parallel). To find the voltage across the capacitor at t = 8 s, we can express the dependence of the voltage on time for a discharging capacitor (C is discharging after t = 2 s) and evaluate this function, with a time constant differing from that found in (a), at t = 6 s.

440 Chapter 25 (a) Apply the junction rule at the junction between the two resistors to obtain:

I1 = I 2 + I 3

(1)

Apply the loop rule to the loop containing the source, R1, and the capacitor to obtain:

ε − R1I1 − Q = 0

(2)

Apply the loop rule to the loop containing R2 and the capacitor to obtain:

Q − R2 I 2 = 0 C

(3)

Differentiate equation (2) with respect to time to obtain:

d ⎡ ε − R1I1 − Q ⎤⎥ = 0 − R1 dI1 − 1 dQ ⎢ dt ⎣ C⎦ dt C dt dI 1 = − R1 1 − I 3 = 0 dt C

C

or

R1 Differentiate equation (3) with respect to time to obtain:

dI1 1 = − I3 dt C

(4)

d ⎡Q dI ⎤ 1 dQ − R2 2 = 0 − R2 I 2 ⎥ = ⎢ dt ⎣ C dt ⎦ C dt or

R2

dI 2 1 = I3 dt C

(5)

Using equation (1), substitute for I3 in equation (5) to obtain:

dI 2 1 (I1 − I 2 ) = dt R2C

Solve equation (2) for I1:

I1 =

Substitute for I1 in equation (6) and simplify to obtain the differential equation for I2:

ε −Q R1

=

ε − R2 I 2 R1

⎞ dI 2 1 ⎛ ε − R2 I 2 ⎜⎜ = − I 2 ⎟⎟ dt R2C ⎝ R1 ⎠ =

To solve this linear differential equation with constant coefficients we can assume a solution of the form:

C

(6)

ε

⎛ R + R2 ⎞ ⎟I2 − ⎜⎜ 1 R1R2C ⎝ R1R2C ⎟⎠

I 2 (t ) = a + be −t τ

(7)

`Electric Current and Direct-Current Circuits 441

[

]

Differentiate I2(t) with respect to time to obtain:

dI 2 d b = a + be −t τ = − e −t τ τ dt dt

Substitute for I2 and dI2/dt to obtain:

b − e −t τ =

τ

Equate coefficients of e −t τ to obtain:

τ=

Requiring the equation to hold for all values of a yields:

a=

If I2 is to be zero when t = 0:

0 = a+b

⎛ R + R2 ⎞ ⎟⎟ a + be −t τ − ⎜⎜ 1 R1R2C ⎝ R1R2C ⎠

(

R1R2C R1 + R2

ε R1 + R2

or

b = −a = −

Substitute in equation (7) to obtain:

ε

I 2 (t ) = =

ε R1 + R2

ε R1 + R2

ε

R1 + R2

ε

−

R1 + R2

e −t τ

(1 − e ) −t τ

where

τ=

R1 R2C R1 + R2

Substitute numerical values and evaluate τ :

τ=

(2 MΩ )(5 MΩ )(1 µF) = 1.43 s

Substitute numerical values and evaluate I2(t):-

I 2 (t ) =

2 MΩ + 5 MΩ

(

10 V 1 − e −t 1.43s 2 MΩ + 5 MΩ

(

= (1.43 µA ) 1 − e −t 1.43s Because C and R2 are in parallel, they have a common potential difference given by:

Evaluate VC at t = 2 s:

VC (t ) = V2 (t ) = I 2 (t )R2

(

)

)

= (1.43 µA )(5 MΩ ) 1 − e −t 1.43s

(

= (7.15 V ) 1 − e − t 1.43s

(

)

)

)

VC (2 s ) = (7.15 V ) 1 − e −2 s 1.43s = 5.38 V

)

442 Chapter 25 The voltage across the capacitor as a function of time is shown in the figure. The current through the 5-MΩ resistor R2 follows the same time course, its value being VC/(5×106) A.

6 5 4 V C (V) 3 2 1 0 0

2

4

6

8

10

t (s)

(b) The value of VC at t = 2 s has already been determined to be: When S is opened at t = 2 s, C discharges through R2 with a time constant given by: Express the potential difference across C as a function of time: Evaluate VC at t = 8 s to obtain:

VC (2 s ) = 5.38 V

τ ' = R2C = (5 MΩ )(1 µF) = 5 s

VC (t ) = VC0 e −t τ ' = (5.38 V ) e −t 5 s VC (8 s ) = (5.38 V ) e −6 s 5 s = 1.62 V in good agreement with the graph.

`Electric Current and Direct-Current Circuits 443

143 ••• Picture the Problem Let I1 be the current delivered by ε1, I2 the current delivered by ε2, and I3 the current through the resistor R. We can apply Kirchhoff’s rules to obtain three equations in the unknowns I1, I2¸and I3 that we can solve simultaneously to find I3. We can then express the power delivered by the sources to R. Setting the derivative of this expression equal to zero will allow us to solve for the value of R that maximizes the power delivered by the sources. Apply Kirchhoff’s junction rule at a to obtain: Apply the loop rule around the outside of the circuit to obtain: Apply the loop rule around the inside of the circuit to obtain: Eliminate I1 from equations (1) and (2) to obtain: Solve equation (3) for I2 to obtain:

Substitute for I2 in equation (4) to obtain: Solve for I3 to obtain:

I1 + I 2 = I 3

(1)

ε 1 − I 3 R − r1I1 = 0

(2)

ε 2 − I 3 R − r2 I 2 = 0

(3)

ε 1 − I 3 R − r1 (I 3 − I 2 ) = 0

(4)

I2 =

ε 2 − I3R r2

ε 1 − I 3 R − r1 ⎛⎜⎜ I 3 − ε 2 − I 3 R ⎞⎟⎟ = 0 ⎝

I3 =

ε 1r2 + ε 2 r1

r1r2 + R(r1 + r2 )

r2

⎠

444 Chapter 25 Express the power delivered to R:

⎛ ε 1r2 + ε 2 r1 ⎞ ⎟⎟ R P = I R = ⎜⎜ ( ) + + r r R r r 1 2 ⎠ ⎝ 12 2

2 3

⎛ ε r + ε 2 r1 ⎞ ⎡ R ⎤ ⎟⎟ ⎢ = ⎜⎜ 1 2 2⎥ ⎝ r1 + r2 ⎠ ⎣ (R + A) ⎦ 2

where

A=

Noting that the quantity in parentheses is independent of R and that therefore we can ignore it, differentiate P with respect to R and set the derivative equal to zero:

r1r2 r1 + r2

dP d ⎡ R ⎤ = ⎢ ⎥ dR dR ⎣ (R + A)2 ⎦ (R + A)2 − R d (R + A)2 dR = (R + A)4 2 ( R + A) − 2 R (R + A) = (R + A)4

= 0 for extrema Solve for R to obtain:

To establish that this value for R corresponds to a maximum, we need to evaluate the second derivative of P with respect to R at R = A and show that this quantity is negative, i.e., concave downward:

We can conclude that:

R= A=

r1r2 r1 + r2

2 d 2 P d ⎡ (R + A) − 2 R (R + A) ⎤ = ⎢ ⎥ dR 2 dR ⎣ (R + A)4 ⎦ 2R − 4 A = (R + A)4

and

− 2A d 2P = <0 2 dR R = A (R + A)4 R=

r1r2 maximizes the power r1 + r2

delivered by the sources. *144 ••• Picture the Problem Let Q1 and Q2 represent the final charges on the capacitors C1 and C2. Knowing that charge is conserved as it is redistributed to the two capacitors and that the final-state potential differences across the two capacitors will be the same, we can obtain two equations in the unknowns Q1 and Q2 that we can solve simultaneously. We

`Electric Current and Direct-Current Circuits 445

can compare the initial and final energies stored in this system by expressing and simplifying their ratio. We can account for any difference between these energies by considering the role of the resistor in the circuit. (a) Relate the total charge stored initially to the final charges Q1 and Q2 on C1 and C2:

Q = C1V0 = Q1 + Q2

(1)

Because, in their final state, the potential differences across the two capacitors will be the same:

Q1 Q2 = C1 C2

(2)

Solve equation (2) for Q2 and substitute in equation (1) to obtain:

C1 Q2 + Q2 = C1V0 C2 Q2 =

C1C2 V0 C1 + C2

Substitute in either (1) or (2) and solve for Q1 to obtain:

Q1 =

C12 V0 C1 + C2

(b) Express the ratio of the initial and final energies of the system:

Ui = Uf

Solve for Q2 to obtain:

=

Simplify this expression further to obtain:

2 1 2 1 0 2 1 1 2 1

CV

1 2

Q Q22 + C C2 C1V02 2

⎛ C1C2 ⎞ ⎛ C12 ⎞ ⎜⎜ ⎜⎜ V0 ⎟⎟ V0 ⎟⎟ ⎝ C1 + C2 ⎠ + ⎝ C1 + C2 ⎠ C1 C2

Ui C = 1+ 2 Uf C1 or Ui is greater than Uf by a factor of 1 + C2/C1.

(c)

The decrease in energy equals the energy dissipated as Joule heat in the resistor connecting the two capacitors.

2

446 Chapter 25 145 ••• Picture the Problem Let q1 and q2 be the time-dependent charges on the two capacitors after the switches are closed. We can use Kirchhoff’s loop rule and the conservat charge to a obtain a first-order linear differential equation describing the current I2 through R after the switches are closed. We can solve this differential equation assuming a solution of the form q2 (t ) = a + be −t τ and requiring that the solution satisfy the boundary condition that q2(0) = 0 and the differential equation be satisfied for all values of t. Once we know I2, we can find the energy dissipated in the resistor as a function of time and the total energy dissipated in the resistor. (a) Apply Kirchhoff’s loop rule to the circuit to obtain:

q1 q − IR − 2 = 0 C1 C2 or, because I = dq2/dt,

q1 dq q −R 2 − 2 =0 C1 dt C2 Apply conservation of charge during the redistribution of charge to obtain:

q1 = Q − q2 = C1V0 − q2

Substitute for q1 to obtain:

V0 −

Rearrange to obtain the first- order differential equation:

R

Assume a solution of the form:

q2 (t ) = a + be −t τ

Differentiate the assumed solution with respect to time to obtain:

dq2 (t ) d b = a + be −t τ = − e −t τ dt dt τ

Substitute for dq2/dt and q2 in the differential equation to obtain:

q2 dq q −R 2 − 2 =0 C1 dt C2

dq2 ⎛ C1 + C2 ⎞ ⎟q2 = V0 +⎜ dt ⎜⎝ C1C2 ⎟⎠

[

(1)

]

⎛ b ⎞ R ⎜ − e −t τ ⎟ ⎝ τ ⎠ ⎛ C + C2 ⎞ ⎟⎟(a + be −t τ ) = V0 + ⎜⎜ 1 ⎝ C1C2 ⎠

`Electric Current and Direct-Current Circuits 447

Rearrange to obtain:

⎡ R −t τ ⎤ ⎛⎜ C1 + C2 ⎞⎟ ⎢⎣− τ e ⎥⎦b + ⎜ C C ⎟a ⎝ 1 2 ⎠ ⎡⎛ C + C2 ⎞ −t τ ⎤ ⎟⎟e ⎥b = V0 + ⎢⎜⎜ 1 ⎣⎝ C1C2 ⎠ ⎦

If this equation is to be satisfied for all values of t:

a=

C1C2 V0 = CeqV0 C1 + C2

and

⎡⎛ C1 + C2 ⎞ −t τ ⎤ ⎡ R −t τ ⎤ ⎢⎣− τ e ⎥⎦b + ⎢⎜⎜ C C ⎟⎟e ⎥b = 0 ⎣⎝ 1 2 ⎠ ⎦ or

−

R

τ

+

C1 + C2 =0 C1C2 C1C2 = RCeq C1 + C2

Solve for τ to obtain:

τ =R

Substitute the boundary condition q2(0) = 0 in equation (1):

0= a+b

Substitute for a and b in equation (1) to obtain: Differentiate q2(t) with respect to time to find the current:

or

b = −a = −CeqV0 q2 (t ) = CeqV0 − CeqV0 e −t τ

(

= CeqV0 1 − e −t τ

dq2 (t ) d 1 − e −t τ = CeqV0 dt dt ⎛ 1⎞ = CeqV0 − e −t τ ⎜ − ⎟ ⎝ τ⎠ C V V = eq 0 e −t τ = 0 e −t τ R τ

(

I (t ) =

(

(b) Express the energy dissipated in the resistor as a function of time:

)

2

⎛V ⎞ P(t ) = I R = ⎜ 0 e −t τ ⎟ R ⎝R ⎠ 2

= (c) The energy dissipated in the resistor is the integral of P(t)

)

V02 −2t τ e R ∞

E=

V02 −2t' RCeq e dt' = R ∫0

1 2

V02Ceq

)

448 Chapter 25 between t = 0 and t =∞:

This is exactly the difference between the initial and final stored energies found in the preceding problem, which confirms the statement at the end of that problem that the difference in the stored energies equals the energy dissipated in the resistor. 146 ••• Picture the Problem We can apply Kirchhoff’s loop rule to find the initial current drawn from the battery and the current drawn from the battery a long time after Sl is closed. We can also use the loop rule to find the final voltages across the capacitors and the current in the 150-Ω resistor when S2 is opened after having been closed for a long time. (a) Apply Kirchhoff’s loop rule to the loop that includes the source, the 100-Ω resistor, and the capacitor immediately after S1 is closed to obtain: Because the capacitor is initially uncharged:

Solve for and evaluate Ibat(0):

(b) Apply Kirchhoff’s loop rule to the loop that includes the source, the 100-Ω, 50-Ω, and 150-Ω resistor a long time after S1 is closed to obtain: Solve for and evaluate I∞:

(c) Apply Kirchhoff’s loop rule to the loop that includes the source, the 100-Ω resistor, and C1 a long time after both switches are closed to obtain:

12 V − I bat (0 )(100 Ω ) − VC0 = 0

VC0 = 0 and

12 V − I bat (0)(100 Ω ) = 0

I bat (0) =

12 V = 0.120 A 100 Ω

12 V − I ∞ (100 Ω ) − I ∞ (50 Ω )

− I ∞ (150 Ω ) = 0

I∞ =

12 V = 40.0 mA 100 Ω + 50 Ω + 150 Ω

12 V − I bat (100 Ω ) − VC1 = 0

`Electric Current and Direct-Current Circuits 449

Solve for and evaluate VC1: (d) Apply Kirchhoff’s loop rule to the loop that includes the 150-Ω resistor and C2 a long time after both switches are closed to obtain: Solve for and evaluate VC2:

VC1 = 12 V − (40 mA )(100 Ω ) = 8.00 V

− VC2 + I bat (150 Ω ) = 0

VC2 = I bat (150 Ω ) = (40 mA )(150 Ω ) = 6.00 V VC2 (t ) − I (t )(150 Ω ) = 0

(e) Apply Kirchhoff’s loop rule to the loop that includes the 150-Ω resistor and C2 after S2 is opened to obtain:

or

Solve for I(t) to obtain:

I (t ) =

VC2 (0 ) e −t τ − I (t )(150 Ω ) = 0

VC2 (0) 150 Ω

e −t τ =

6 V −t (150 Ω )(50 µF ) e 150 Ω

(40 mA ) e −t 7.50 ms

=

*147 ••• Picture the Problem We can use the definition of differential resistance and the expression for the diode current given in problem 54 to express Rd and establish the required results. The differential resistance Rd is given by:

dV ⎛ dI ⎞ Rd = =⎜ ⎟ dI ⎝ dV ⎠

From Problem 54, the current in the diode is given by:

I = I 0 eV

Substitute for I to obtain:

⎫ ⎧ d Rd = ⎨ I 0 eV 25 mV − 1 ⎬ ⎭ ⎩ dV 25 mV −V 25 mV = e I0

(

25 mV

−1

)

−1

(1)

[ (

For V > 0.6 V, equation (1) becomes:

I ≈ I 0 eV

Solve for the exponential factor to obtain:

eV

25 mV

)]

−1

(2)

25 mV

≈

I ⇒ e −V I0

25 mV

≈

I0 I

450 Chapter 25 Substitute in equation (2) to obtain:

Rd ≈

25 mV I 0 25 mV = I0 I I

Examination of equation (2) shows that, for V < 0, Rd increases exponentially. This result, together with that for V > 0.6 V, justifies the assumptions made in Problem 55. 148 •• Picture the Problem We can approximate the slope of the graph in Figure 25-77 and take its reciprocal to obtain values for Rd that we can plot as a function of V. Use the graph in Figure 25-77 to complete the table to the right.

V (V) Rd (Ω) 0 6.67 0.1 17.9 0.3 −75.2 0.4 42.9 0.5 8

The following graph was plotted using a spreadsheet program.

40

R d (ohms)

20 0 -20 -40 -60 -80 0.0

0.1

0.2

0.3

0.4

0.5

V (V)

The differential resistance becomes negative at approximately 0.14 V. 149 ••• Picture the Problem We can use the definition of current to find the number of electrons accelerated in each pulse and the average current in the beam. The average and peak power of the accelerator can be found using Pav = I avV and

`Electric Current and Direct-Current Circuits 451

Ppeak = I peakV and the duty factor from its definition. (a) Use the definition of current to relate the number of electrons accelerated in each pulse to the duration of the pulse: Solve for and evaluate n:

I pulse =

∆Q ne = ∆t ∆t

where n is the number of electrons in each pulse.

n=

I pulse ∆t

e (1.6 A )(0.1 µs ) = 9.99 × 1011 ≈ 1012 = 1.602 × 10 −19 C

(b) Using the definition of current we have:

I av = =

Qpulse ∆t between pulses

(

=

ne 10 −3 s

1012 1.60 × 10 −19 C 10 -3 s

)

= 0.160 mA (c) Express the average power output in terms of the average current: (d) Express the peak power output in terms of the pulse current:

(e) The duty factor is defined to be:

Pav = I avV = (0.160 mA )(400 MV ) = 64.0 kW Ppeak = I pulseV = (1.6 A )(400 MV ) = 640 MW ∆t time between pulses 0.1 µs = −3 = 10 −4 10 s

duty factor =

452 Chapter 25 150 ••• Picture the Problem Let R be the resistance of each resistor in the ladder and let Req be the equivalent resistance of the infinite ladder. If the resistance is finite and non-zero, then adding one or more stages to the ladder will not change the resistance of the network. We can apply the rules for resistance combination to the diagram shown to the right to obtain a quadratic equation in Req that we can solve for the equivalent resistance between points a and b. The equivalent resistance of the series combination of R and (R || Req) is Req, so:

Req = R + R Req = R +

Simplify to obtain:

Req2 − RReq − R 2 = 0

Solve for Req to obtain:

⎛1+ 5 ⎞ ⎟R Req = ⎜⎜ ⎟ 2 ⎝ ⎠

For R = 1Ω:

⎛1+ 5 ⎞ ⎟ (1 Ω ) = 1.62 Ω Req = ⎜⎜ ⎟ 2 ⎝ ⎠

RReq R + Req

*151 ••• Picture the Problem Let Req be the equivalent resistance of the infinite ladder. If the resistance is finite and non-zero, then adding one or more stages to the ladder will not change the resistance of the network. We can apply the rules for resistance combination to the diagram shown to the right to obtain a quadratic equation in Req that we can solve for the equivalent resistance between points a and b. The equivalent resistance of the series combination of R1 and (R2 || Req) is Req, so:

Req = R1 + R2 Req = R1 +

Simplify to obtain:

Req2 − R1 Req − R1 R2 = 0

Solve for the positive value of Req to obtain:

R1 + R12 + 4 R1 R2 Req = 2

R2 Req R2 + Req

Chapter 26 The Magnetic Field Conceptual Problems *1 • Determine the Concept Because the electrons are initially moving at 90° to the magnetic field, they will be deflected in the direction of the magnetic force acting on them. Use the right-hand rule based on the expression for the magnetic force acting on a moving charge

r r r F = qv × B , remembering that, for a negative charge, the force is in the direction

opposite that indicated by the right-hand rule, to convince yourself that the particle will follow the path whose terminal point on the screen is 2. (b) is correct. 2 • Determine the Concept One cannot define the direction of the force arbitrarily. By r r experiment, F is perpendicular to B . 3 • Determine the Concept False. An object experiences acceleration if either its speed changes or the direction it is moving changes. The magnetic force, acting perpendicular to the direction a charged particle is moving, changes the particle’s velocity by changing the direction it is moving and hence accelerates the particle. 4 • Determine the Concept Yes; it will be deflected upward. Because the beam passes through undeflected when traveling from left to right, we know that the upward electric force must be balanced by a downward magnetic force. Application of the right-hand rule tells us that the magnetic field must be out of the page. When the beam is reversed, the magnetic force (as well as the electric force) acting on it is upward. *5 • Determine the Concept The alternating current running through the filament is changing direction every 1/60 s, so in a magnetic field the filament experiences a force which alternates in direction at that frequency. 6 • Determine the Concept The magnitude of the torque on a current loop is given by τ = µB sin θ , where θ is the angle between the magnetic field and a normal to the surface of the loop. To maximizeτ , sinθ = 1 and θ = 90°. Hence the normal to the plane of the r loop should be perpendicular to B .

453

454 Chapter 26 7 • (a) True. This is an experimental fact and is the basis for the definition of the magnetic force on a moving charged particle being expressed in terms of the cross product of

v v r r r v and B ; i.e. F = qv × B .

(b) True. This is another experimental fact. The torque on a magnet is a restoring torque, i.e., one that acts in such a direction as to align the magnet with magnetic field. (c) True. We can use a right-hand rule to relate the direction of the magnetic field around the loop to the direction of the current. Doing so indicates that one side of the loop acts like a north pole and the other like a south pole. (d) False. The period of a particle moving in a circular path in a magnetic field is given by T = 2π mr qvB and, hence, is proportional to the square root of the radius of the circle. (e) True. The drift velocity is related to the Hall voltage according to vd = VH/Bw where w is the width of the Hall-effect material. *8 • . Determine the Concept The direction in which a particle is deflected by a magnetic field will be unchanged by any change in the definition of the direction of the magnetic field. Since we have reversed the direction of the field, we must define the direction in which particles are deflected by a "left-hand" rule instead of a "right-hand" rule. 9 • Determine the Concept Choose a right-handed coordinate system in which east is the positive x direction and north is the positive y direction. Then the magnetic force acting

r

( )

on the particle is given by F = qviˆ × Bˆj = qvB iˆ × ˆj = qvBkˆ . Hence, the magnetic force is upward. 10 • Determine the Concept Application of the right-hand rule tells us that this positively charged particle would have to be moving in the northwest direction with the magnetic field upward in order for the magnetic force to be toward the northeast. The situation described cannot exist. (e) is correct. 11 • Picture the Problem We can use Newton’s 2nd law for circular motion to express the radius of curvature R of each particle in terms of its charge, momentum, and the magnetic field. We can then divide the proton’s radius of curvature by that of the 7Li nucleus to decide which of these alternatives is correct.

The Magnetic Field 455 Apply

∑F

radial

= mac to the lithium

nucleus to obtain:

v2 qvB = m R mv qB

Solve for r:

R=

For the 7Li nucleus this becomes:

RLi =

For the proton we have:

Rp =

Divide equation (2) by equation (1) and simplify to obtain:

Because the momenta are equal:

Rp

pLi 3eB pp eB pp

p = eB = 3 p pLi RLi pLi 3eB

Rp RLi

= 3 and (a ) is correct.

*12 • Determine the Concept Application of the right-hand rule indicates that a positively charged body would experience a downward force and, in the absence of other forces, be deflected downward. Because the direction of the magnetic force on an electron is opposite that of the force on a positively charged object, an electron will be deflected upward.

(c) is correct.

13 •• Determine the Concept From relativity; this is equivalent to the electron moving from right to left at velocity v with the magnet stationary. When the electron is directly over the magnet, the field points directly up, so there is a force directed out of the page on the electron. 14

•

Similarities Magnetic field lines are similar to electric field lines in that their density is a measure of the strength of the field; the lines point in the direction of the field; also, magnetic field lines do not cross.

Differences They differ from electric field lines in that magnetic field lines must close on themselves (there are no isolated magnetic poles), and the force on a charge depends on the velocity of the charge and is perpendicular to the magnetic field lines.

456 Chapter 26 15 • r Determine the Concept If only F and I are known, one can only conclude that the r r r magnetic field B is in the plane perpendicular to F . The specific direction of B is undetermined.

Estimation and Approximation *16 •• Picture the Problem If the electron enters the magnetic field in the coil with speed v, it will travel in a circular path under the influence of the magnetic force acting on. We can apply Newton’s 2nd law to the electron in this field to obtain an expression for the magnetic field. We’ll assume that the deflection of the electron is small over the distance it travels in the magnetic field, but that, once it is through the region of the magnetic field, it travels at an angle θ with respect to the direction it was originally traveling. Apply

∑ F = ma

c

to the electron

in the magnetic field to obtain: Solve for B:

The kinetic energy of the electron is: Solve for v to obtain:

Substitute for v in the expression for r: Because θ << 1:

evB = m

B=

v2 r

mv er

K = eV = 12 mv 2 v=

B=

2eV m m er

2eV 1 2mV = m r e

d ≈ r sin θ ⇒ r ≈

d sin θ

Substitute for r in the expression for B to obtain:

B=

sin θ d

For maximum deflection, θ ≈ 45°. Substitute numerical values and evaluate B:

B=

sin 45° 2 9.11× 10 −31 kg (15 kV ) 0.05 m 1.60 × 10 −19 C

2mV e

(

)

= 5.84 mT 17 •• Picture the Problem Let h be the height of the orbit above the surface of the earth, m the mass of the micrometeorite, and v its speed. We can apply Newton’s 2nd law to the orbiting micrometeorite with Fmag = qvB to derive an expression for the charge-to-mass ratio of the micrometeorite.

The Magnetic Field 457 (a) Apply

∑ F = ma to the

v2 qvB = m h + Rearth

c

micrometeorite orbiting under the influence of the magnetic force:

q v = m B(h + Rearth )

Solve for q/m to obtain:

Substitute numerical values and evaluate q/m:

30 km/s q = = 88.6 C/kg −5 m 5 × 10 T (400 km + 6370 km )

(

)

q = (88.6 C/kg )m

(b) Solve the result for q/m obtained in (a) for q to obtain:

(

q = (88.6 C/kg ) 3 × 10 −10 kg

Substitute numerical values and evaluate q:

)

= 26.6 nC

Force Exerted by a Magnetic Field 18 • r r r Picture the Problem The magnetic force acting on a charge is given by F = qv × B .

r r scalar q to find F .

r

We can express v and B , form their vector (a.k.a. ″cross″) product, and multiply by the

r r r F = qv × B

Express the force acting on the proton:

r

r v = (4.46 Mm/s) iˆ

r

r B = (1.75 T )kˆ

Express v : Express B :

r

Substitute numerical values and evaluate F :

)[

]

r F = 1.60 × 10−19 C (4.46 Mm/s) iˆ × (1.75 T ) kˆ = − (1.25 pN ) ˆj

(

19 • r r r Picture the Problem The magnetic force acting on the charge is given by F = qv × B .

r

r

We can express v and B , form their vector (a.k.a. ″cross″) product, and multiply by the

458 Chapter 26

r

scalar q to find F .

r r r F = qv × B

Express the force acting on the charge:

[(

r r F = (− 3.64 nC) 2.75 × 106 m/s iˆ × B

Substitute numerical values to obtain:

r

)

]

r

(a) Evaluate F for B = 0.38 T ˆj :

[(

]

r F = (− 3.64 nC) 2.75 × 106 m/s iˆ × (0.38 T ) ˆj = − (3.80 mN )kˆ

r

)

r

(b) Evaluate F for B = 0.75 T iˆ + 0.75 T ˆj :

}]

{

[

r F = (− 3.64 nC) (2.75 × 106 m/s )iˆ × (0.75 T ) iˆ + (0.75 T ) ˆj = − (7.51mN )kˆ

r

r

(c) Evaluate F for B = 0.65 T iˆ :

[(

]

r F = (− 3.64 nC ) 2.75 × 106 m/s iˆ × (0.65 T ) iˆ = 0

)

r r (d) Evaluate F for B = 0.75 T iˆ + 0.75 T kˆ :

[(

]

r F = (− 3.64 nC ) 2.75 × 106 m/s iˆ × (0.75 T ) iˆ + (0.75 T ) kˆ =

)

(7.51mN) ˆj

20 • r r r Picture the Problem The magnetic force acting on the proton is given by F = qv × B .

r r scalar q to find F .

r

We can express v and B , form their vector (a.k.a. ″cross″) product, and multiply by the

r r r F = qv × B

Express the force acting on the proton:

r

r (a) Evaluate F for v = 2.7 Mm/s iˆ :

)[(

]

r F = 1.60 × 10−19 C 2.7 × 106 m/s iˆ × (1.48 T ) kˆ = − (0.640 pN ) ˆj

(

)

r

r (b) Evaluate F for v = 3.7 Mm/s ˆj :

)[(

]

r F = 1.60 × 10 −19 C 3.7 × 106 m/s ˆj × (1.48 T ) kˆ =

(

)

(0.876 pN ) iˆ

The Magnetic Field 459 r

r (c) Evaluate F for v = 6.8 Mm/s kˆ :

)[(

]

r F = 1.60 × 10 −19 C 6.8 × 106 m/s kˆ × (1.48 T ) kˆ = 0

(

r

)

r

(d) Evaluate F for v = 4.0 Mm/s iˆ + 3.0 Mm/s ˆj :

r F = 1.60 × 10−19 C

(

=

)[{(4.0 Mm/s)iˆ + (3.0 Mm/s) ˆj}× (1.48 T )kˆ ]

(0.710 pN ) iˆ − (0.947 pN ) ˆj

21 • Picture the Problem The magnitude of the magnetic force acting on a segment of wire is given by F = IlB sin θ where l is the length of the segment of wire, B is the magnetic field, and θ is the angle between the segment of wire and the direction of the magnetic field.

F = IlB sin θ

Express the magnitude of the magnetic force acting on the segment of wire:

F = (2.6 A )(2 m )(0.37 T )sin 30°

Substitute numerical values and evaluate F:

= 0.962 N

*22 • r r r Picture the Problem We can use F = IL × B to find the force acting on the wire segment.

r r r F = IL × B

Express the force acting on the wire segment:

[

]

r F = (2.7 A ) (3 cm )iˆ + (4 cm ) ˆj × (1.3 T ) iˆ

Substitute numerical values and r evaluate F :

= − (0.140 N )kˆ

23 • r r r Picture the Problem The magnetic force acting on the electron is given by F = qv × B .

r

r

We can form the vector product of v and B and multiply by the charge of the electron to

r

find F and obtain its magnitude using F =

(

Fx2 + Fy2 + Fz2 . The direction angles are

)

given by θ x = tan −1 (Fx F ) , θ y = tan −1 Fy F , and θ z = tan −1 (Fz F ) .

460 Chapter 26 r r r F = qv × B

Express the force acting on the proton:

r

Express the magnitude of F in terms of its components:

F = Fx2 + Fy2 + Fz2

(1)

r

Substitute numerical values and evaluate F :

)[{

)]

}(

r F = − 1.60 × 10 −19 C (2 Mm/s ) iˆ − (3 Mm/s ) ˆj × 0.8 iˆ + 0.6 ˆj − 0.4 kˆ T = (− 0.192 pN ) kˆ + (− 0.128 pN ) ˆj + (− 0.384 pN ) kˆ + (− 0.192 pN ) iˆ

(

= − (0.192 pN ) iˆ − (0.128 pN ) ˆj − (0.576 pN ) kˆ Substitute in equation (1) to obtain:

F=

(− 0.192 pN )2 + (− 0.128 pN )2 + (− 0.576 pN )2

Express and evaluate the angle r F makes with the x axis:

= 0.621pN

⎛ Fx ⎞ −1 ⎛ − 0.192 pN ⎞ ⎟⎟ ⎟ = cos ⎜⎜ ⎝F⎠ ⎝ 0.621 pN ⎠

θ x = cos −1 ⎜ = 108°

Express and evaluate the angle r F makes with the y axis:

⎛ − 0.128 pN ⎞ ⎛ Fy ⎞ ⎟⎟ ⎟⎟ = cos −1 ⎜⎜ ⎝F ⎠ ⎝ 0.621 pN ⎠

θ y = cos −1 ⎜⎜ = 102°

Express and evaluate the angle r F makes with the z axis:

⎛ Fz ⎞ −1 ⎛ − 0.576 pN ⎞ ⎟⎟ ⎟ = cos ⎜⎜ ⎝F⎠ ⎝ 0.621 pN ⎠

θ z = cos −1 ⎜ = 158°

24 •• r r r Picture the Problem We can use F = I l × B to find the force acting on the segments of the wire as well as the magnetic force acting on the wire if it were a straight segment from a to b. Express the magnetic force acting on the wire:

r r r F = F3 cm + F4 cm

The Magnetic Field 461

[

r

r F3 cm = (1.8 A ) (3 cm ) iˆ × (1.2 T ) kˆ = (0.0648 N ) − ˆj

Evaluate F3 cm :

( )

]

= −(0.0648 N ) ˆj

[

r

]

Evaluate F4 cm :

r F4 cm = (1.8 A ) (4 cm ) ˆj × (1.2 T ) kˆ = (0.0864 N ) iˆ

Substitute to obtain:

r F = −(0.0648 N ) ˆj + (0.0864 N ) iˆ

=

(0.0864 N )iˆ − (0.0648 N ) ˆj

r l = (3 cm ) iˆ + (4 cm ) ˆj

If the wire were straight from a to b: The magnetic force acting on the wire is:

[

]

r F = (1.8 A ) (3 cm )iˆ + (4 cm ) ˆj × (1.2 T )kˆ = −(0.0648 N ) ˆj + (0.0864 N ) iˆ

(0.0864 N )iˆ − (0.0648 N ) ˆj

=

in agreement with the result obtained above when we treated the two straight segments of the wire separately. 25 •• Picture the Problem Because the magnetic field is horizontal and perpendicular to the wire, the force it exerts on the current-carrying wire will be vertical. Under equilibrium conditions, this upward magnetic force will be equal to the downward gravitational force acting on the wire. Apply

∑F

vertical

= 0 to the wire:

Express Fmag:

Fmag − w = 0 Fmag = IlB because θ = 90°.

Substitute to obtain:

IlB − mg = 0

Solve for I:

I=

mg lB

I=

(50 g )(9.81m/s2 ) = (25 cm )(1.33 T )

Substitute numerical values and evaluate I:

1.48 A

462 Chapter 26 *26 •• Picture the Problem The diagram shows the gaussmeter displaced from equilibrium under the influence of the gravitational and magnetic forces acting on it. We can apply the condition for translational equilibrium in the x direction to find the equilibrium angular displacement of the wire from the vertical. In part (b) we can solve the equation derived in part (a) for B and evaluate this expression for the given data to find the horizontal magnetic field sensitivity of this gaussmeter. (a) Apply

∑F

x

= 0 to the wire to

mg sin θ − F cosθ = 0

obtain: Substitute for F and solve for θ to obtain:

mg sin θ − IlB cosθ = 0

(1)

and

⎛ IlB ⎞ ⎟⎟ ⎝ mg ⎠

θ = tan −1 ⎜⎜ Substitute numerical values and evaluate θ :

⎡ (0.2 A )(0.5 m )(0.04 T )⎤ 2 ⎥ ⎣ (0.005 kg ) 9.81 m/s ⎦

θ = tan −1 ⎢

(

)

= 4.66°

mg tan θ Il

(b) Solve equation (1) for B to obtain:

B=

For a displacement from vertical of 0.5 mm:

tan θ ≈ sin θ =

0.5 mm = 0.001 0.5 m

and

θ = 0.001rad

Substitute numerical values and evaluate B:

B=

(0.005 kg )(9.81m/s 2 )(0.001rad ) (20 A )(0.5 m )

= 4.91 µT 27

••

The Magnetic Field 463 Picture the Problem With the current in the direction indicated and the magnetic field in the z direction, pointing out of the plane of the page, the force is in the radial direction and we can integrate the element of force dF acting on an element of length dℓ between θ = 0 and π to find the force acting on the semicircular portion of the loop and use the expression for the force on a current-carrying wire in a uniform magnetic field to find the force on the straight segment of the loop. Express the net force acting on the semicircular loop of wire: Express the force acting on the straight segment of the loop:

F = Fsemicircular loop + Fstraight segment

r r r Fstraight segment = I l × B = −2 RIB

Express the force dF acting on the element of the wire of length dℓ:

dF = IdlB = IRBdθ

Express the x and y components of dF:

dFx = dF cosθ and

dFy = dF sin θ

Because, by symmetry, the x component of the force is zero, we can integrate the y component to find the force on the wire:

dFy = IRB sin θ dθ and π

Fy = RIB ∫ sin θ dθ = 2 RIB 0

Substitute in equation (1) to obtain:

F = 2 RIB − 2 RIB = 0

(1)

464 Chapter 26 28 •• Picture the Problem We can use the information given in the 1st and 2nd sentences to r obtain an expression containing the components of the magnetic field B . We can then use the information in the 1st and 3rd sentences to obtain a second equation in these r components that we can solve simultaneously for the components of B .

r

Express the magnetic field B in terms of its components:

r

r

Express F in terms of B :

r B = Bx iˆ + By ˆj + Bz kˆ

(1)

r r r F = Il × B

[

] ( ) = (0.4 A ⋅ m )kˆ × (B iˆ + B ˆj + B kˆ )

= (4 A )(0.1 m )kˆ × Bx iˆ + By ˆj + Bz kˆ x

y

z

= (0.4 A ⋅ m )By ˆj − (0.4 A ⋅ m )By iˆ Equate the components of this r expression for F with those given in the second sentence of the statement of the problem to obtain: Noting that Bz is undetermined, solve for Bx and By: When the wire is rotated so that the current flows in the positive x direction:

(0.4 A ⋅ m )By = 0.2 N and

(0.4 A ⋅ m )Bx = 0.2 N Bx = 0.5 T and By = 0.5 T r r r F = Il × B

[

]( ) = (0.4 A ⋅ m ) iˆ × (B iˆ + B ˆj + B kˆ )

= (4 A )(0.1 m ) iˆ × Bx iˆ + By ˆj + Bz kˆ x

y

z

= −(0.4 A ⋅ m )Bz ˆj + (0.4 A ⋅ m )By kˆ Equate the components of this r expression for F with those given in the third sentence of the problem statement to obtain: Solve for Bz and By to obtain:

− (0.4 A ⋅ m )Bz = 0 and

(0.4 A ⋅ m )By = 0.2 N Bz = 0

and, in agreement with our results above,

By = 0.5 T

Substitute in equation (1) to obtain:

r B=

(0.5 T )iˆ + (0.5 T ) ˆj

The Magnetic Field 465 29 •• Picture the Problem We can use the information given in the 1st and 2nd sentences to r obtain an expression containing the components of the magnetic field B . We can then use the information in the 1st and 3rd sentences to obtain a second equation in these r components that we can solve simultaneously for the components of B .

r

Express the magnetic field B in terms of its components:

r

r

Express F in terms of B :

r B = Bx iˆ + By ˆj + Bz kˆ

(1)

r r r F = Il × B

[

] ( ) = (0.2 A ⋅ m ) iˆ × (B iˆ + B ˆj + B kˆ )

= (2 A )(0.1 m ) iˆ × Bx iˆ + B y ˆj + Bz kˆ x

y

z

= −(0.2 A ⋅ m )Bz ˆj + (0.2 A ⋅ m )B y kˆ Equate the components of this r expression for F with those given in the second sentence of the statement of the problem to obtain: Noting that Bx is undetermined, solve for Bz and By:

− (0.2 A ⋅ m )Bz = 3 N and

(0.2 A ⋅ m )By = 2 N Bz = −15 T

and

By = 10 T

When the wire is rotated so that the current flows in the positive y direction:

r r r F = Il × B

[

] ( ) = (0.2 A ⋅ m ) ˆj × (B iˆ + B ˆj + B kˆ ) = (2 A )(0.1 m ) ˆj × Bx iˆ + By ˆj + Bz kˆ x

y

z

= (0.2 A ⋅ m )Bz iˆ − (0.2 A ⋅ m )Bx kˆ Equate the components of this r expression for F with those given in the third sentence of the problem statement to obtain: Solve for Bx and Bz to obtain:

(0.2 A ⋅ m )Bx = −3 N and

− (0.2 A ⋅ m )Bz = −2 N

Bx = −15 T and, in agreement with our results above,

Bz = 10 T

466 Chapter 26 Substitute in equation (1) to obtain:

r B = (10 T ) iˆ + (10 T ) ˆj − (15 T )kˆ

30 ••• r Picture the Problem We can integrate the expression for the force dF acting on an

r

r

r

r

element of the wire of length dL from a to b to show that F = IL × B.

r

Express the force dF acting on the

r

r r r dF = IdL × B

element of the wire of length dL : Integrate this expression to obtain:

r Because B and I are constant:

r b r r F = ∫ IdL × B a

r r r ⎛ b r⎞ r F = I ⎜⎜ ∫ dL ⎟⎟ × B = IL × B ⎝a ⎠ r where L is the vector from a to b.

Motion of a Point Charge in a Magnetic Field *31 • Picture the Problem We can apply Newton’s 2nd law to the orbiting proton to relate its speed to its radius. We can then use T = 2πr/v to find its period. In Part (b) we can use the relationship between T and v to determine v. In Part (c) we can use its definition to find the kinetic energy of the proton. (a) Relate the period T of the motion of the proton to its orbital speed v: Apply

∑F

radial

= mac to the proton

T=

2πr v

qvB = m

to obtain:

(1)

v2 r

Solve for v/r to obtain:

v qB = r m

Substitute to obtain:

T=

2πm qB

Substitute numerical values and evaluate T:

T=

2π 1.67 × 10−27 kg = 87.4 ns 1.60 × 10−19 C (0.75 T )

(

(

)

)

The Magnetic Field 467 (b) From equation (1) we have:

v=

2πr 2π (0.65 m ) = T 87.4 ns

= 4.67 × 107 m/s (c) Using its definition, express and evaluate the kinetic energy of the proton:

(

)(

)

2

K = 12 mv 2 = 12 1.67 × 10− 27 kg 4.67 × 107 m/s = 1.82 × 10−12 J ×

1eV 1.60 × 10−19 J

= 11.4 MeV 32 • Picture the Problem We can apply Newton’s 2nd law to the orbiting electron to obtain an expression for the radius of its orbit as a function of its mass m, charge q, speed v, and the magnetic field B. Using the definition of its kinetic energy will allow us to express r in terms of m, q, B, and its kinetic energy K. We can use T = 2πr/v to find the period of the motion and calculate the frequency from the reciprocal of the period of the motion. (a) Apply

∑F

radial

= mac to the

qvB = m

proton to obtain:

v2 r

mv qB

Solve for r:

r=

Express the kinetic energy of the electron:

K = 12 mv 2

Solve for v to obtain:

v=

2K m

(1)

(2)

Substitute in equation (1) to obtain:

r=

m 2K 1 = 2 Km qB m qB

Find the frequency from the reciprocal of the period:

f =

1 1 = = 9.10 GHz T 0.110 ns

Substitute numerical values and evaluate r:

468 Chapter 26 ⎛ 1.6 × 10−19 J ⎞ −31 ⎜ ⎟⎟ 2(45 keV )⎜ 9.11× 10 kg × eV ⎝ ⎠ = 2.20 mm 1.6 × 10−19 C (0.325 T )

r=

(

)

(b) Relate the period of the electron’s motion to the radius of its orbit and its orbital speed:

T=

Substitute equation (2) to obtain:

T=

2πr v

2πr 2m = πr K 2K m

Substitute numerical values and evaluate T:

T = π (2.20 mm )

(

)

2 9.11× 10−31 kg = 0.110 ns 1.6 × 10−19 J 45 keV × eV

33 • Picture the Problem We can apply Newton’s 2nd law to the orbiting electron to obtain an expression for the radius of its orbit as a function of its mass m, charge q, speed v, and the magnetic field B. (a) Apply

∑F

radial

= mac to the

proton to obtain:

v2 qvB = m r mv qB

Solve for r:

r=

Substitute numerical values and evaluate r:

(9.11×10 kg )(10 m/s) r= (1.60 ×10 C)(4 ×10 T ) −31

7

−19

−7

= 142 m (b) For B = 2×10−5 T:

(9.11×10 kg )(10 m/s) r= (1.60 ×10 C)(2 ×10 T ) −31

−19

= 2.84 m

7

−5

The Magnetic Field 469 34 •• Picture the Problem We can apply Newton’s 2nd law to an orbiting particle to obtain an expression for the radius of its orbit R as a function of its mass m, charge q, speed v, and the magnetic field B. Apply

∑F

radial

= mac to an orbiting

qvB = m

particle to obtain:

v2 r

mv qB

Solve for r:

r=

Express the kinetic energy of the particle:

K = 12 mv 2

Solve for v to obtain:

2K m

v=

m 2K 1 = 2 Km qB m qB

Substitute in equation (1) to obtain:

r=

Using equation (1), express the ratio Rd/Rp:

1 2 Kmd q Rd qd B = = p 1 Rp 2 Kmp qd qp B =

Using equation (1), express the ratio Rα /Rp:

e 2mp = e mp

md mp

2

1 2 Kmα q Rα qα B = = p 1 Rp 2 Kmp qα qp B =

(1)

mα mp

e 4mp = 1 2e mp

35 •• Picture the Problem We can apply Newton’s 2nd law to the orbiting particles to derive an expression for their velocities as a function of their charge, their mass, the magnetic field in which they are moving, and the radii of their orbits. We can then compare their velocities by expressing their ratio. In parts (b) and (c) we can proceed similarly starting

470 Chapter 26 with the definitions of kinetic energy and angular momentum. (a) Apply

∑F

radial

= mac to an

qvB = m

orbiting particle to obtain:

v2 r

qBr m

Solve for v:

v=

Express the velocities of the particles:

vp =

Divide the second of these equations by the first to obtain:

qα Br q m 2emp vα 1 m = α = α p = = qp Br qp mα e(4mp ) vp 2 mp

(b) Express the kinetic energy of an orbiting particle:

q 2 B 2r 2 ⎛ qBr ⎞ K = mv = m⎜ ⎟ = 12 m ⎝ m ⎠

Using this relationship, express the ratio of Kα to Kp:

qα2 B 2 r 2 qα2 mp Kα mα = = K p 1 qp2 B 2 r 2 qp2 mα 2 mp

qp Br mp

Express the ratio of Lα to Lp:

qα Br mα

2

2

1 2

1 2

1 2

=

(c) Express the angular momenta of the particles:

and vα =

(2e )2 mp

e 2 (4mp )

= 1

Lα = mα vα r and Lp = mp vp r Lα mα vα r (4mp )(12 vp ) = = = 2 Lp mp vp r mp vp

36 •• Picture the Problem We can use the definition of momentum to express p in terms of v and apply Newton’s 2nd law to the orbiting particle to express v in terms of q, B, R, and m. In part (b) we can express the particle’s kinetic energy in terms of its momentum and use our result from part (a) to show that K = 12 B 2 q 2 R 2 m.

The Magnetic Field 471 (a) Express the momentum of the particle: Apply

∑F

radial

= mac to the

p = mv

qvB = m

orbiting particle to obtain:

(1)

v2 R

qBR m

Solve for v:

v=

Substitute in equation (1) to obtain:

⎛ qBR ⎞ p = m⎜ ⎟ = qBR ⎝ m ⎠

(b) Express the kinetic energy of the orbiting particle as a function of its momentum:

K=

Substitute our result from part (a) to obtain:

K=

p2 2m

(qBR )2 2m

=

q2B2 R2 2m

*37 •• r Picture the Problem The particle’s velocity has a component v1 parallel to B and a r component v2 normal to B . v1 = v cosθ and is constant, whereas v2 = v sinθ , being r normal to B , will result in a magnetic force acting on the beam of particles and circular r motion perpendicular to B . We can use the relationship between distance, rate, and time and Newton’s 2nd law to express the distance the particle moves in the direction of the field during one period of the motion. Express the distance moved in the r direction of B by the particle during one period: Express the period of the circular motion of the particles in the beam: Apply

∑F

radial

= mac to a particle

x = v1T

T=

2πr v2

qv2 B = m

in the beam to obtain: Solve for v2:

v2 =

(1)

qBr m

v22 r

472 Chapter 26 2πr 2πm = qBr qB m

Substitute to obtain:

T=

Because v1 = v cosθ, equation (1) becomes:

⎛ m⎞ ⎛ 2πm ⎞ ⎟⎟ = 2π ⎜⎜ ⎟⎟v cosθ x = (v cosθ )⎜⎜ ⎝ qB ⎠ ⎝ qB ⎠

38 •• Picture the Problem The trajectory of the proton is shown to the right. We know that, because the proton enters the field perpendicularly to the field, its trajectory while in the field will be circular. We can use symmetry considerations to determine φ. The application of Newton’s 2nd law to the proton while it is in the magnetic field and of trigonometry will allow us to conclude that r = d and to determine their value. From symmetry, it is evident that the angle θ in Figure 26-35 equals the angle φ:

φ = 60.0°

Use trigonometry to obtain:

sin (90° − θ ) = sin 30° =

1 d 2 = r 2

or r = d. Apply

∑F

radial

= mac to the proton

qvB = m

while it is in the magnetic field to obtain:

v2 r

mv qB

Solve for r:

r=

Substitute numerical values and evaluate r = d:

d =r=

(1.67 ×10 kg )(10 m/s) (1.60 ×10 C)(0.8 T )

= 0.130 m

−27

−19

7

The Magnetic Field 473 39 •• Picture the Problem The trajectory of the proton is shown to the right. We know that, because the proton enters the field perpendicularly to the field, its trajectory while in the field will be circular. We can use symmetry considerations to determine φ. The application of Newton’s 2nd law to the proton while it is in the magnetic field and of trigonometry will allow us to conclude that r = d and to determine their value. (a) From symmetry, it is evident that the angle θ in Figure 26-33 equals the angle φ:

φ = 24.0°

Use trigonometry to obtain:

sin (90° − θ ) = sin 24° =

d 2 rp

or

rp = Apply

∑F

radial

= mac to the proton

while it is in the magnetic field to obtain: Solve for and evaluate vp:

Substitute numerical values and evaluate vp:

d 0.4 m = = 0.492 m 2 sin 24° 2 sin 24°

qp vp B = mp

vp =

vp =

vp2 rp

qp rp B mp

(1.60 ×10

C )(0.492 m )(0.6 T ) 1.67 × 10−27 kg −19

= 2.83 × 107 m/s (b) Express vd:

vd =

qd rd B qp rp B = md 2mp

474 Chapter 26 Substitute numerical values and evaluate vd:

(1.60 ×10 C)(0.492 m)(0.6 T ) = 2(1.67 × 10 kg ) −19

vd

− 27

= 1.41× 107 m/s 40 •• Picture the Problem We can apply Newton’s 2nd law of motion to express the orbital speed of the particle and then find the period of the dust particle from this orbital speed. The period of the dust particle’s motion is given by: Apply

∑ F = ma to the particle: c

T=

2πr v

qvB = m

v2 r

Solve for v to obtain:

v=

qBr m

Substitute for v in the expression for T and simplify:

T=

2πrm 2πm = qBr qB

Substitute numerical values and evaluate T:

T=

(

2π 10 × 10 −6 g × 10 −3 kg/g (0.3 nC) 10−9 T 1y = 2.094 × 1011 s × 31.56 Ms

(

)

)

= 6.64 × 103 y

The Velocity Selector *41 • Picture the Problem Suppose that, for positively charged particles, their motion is from left to right through the velocity selector and the electric field is upward. Then the magnetic force must be downward and the magnetic field out of the page. We can apply the condition for translational equilibrium to relate v to E and B. In (b) and (c) we can use the definition of kinetic energy to find the energies of protons and electrons that pass through the velocity selector undeflected. (a) Apply to obtain:

∑F

y

= 0 to the particle

Felec − Fmag = 0 or

qE − qvB = 0

The Magnetic Field 475 Solve for v to obtain:

v=

E B

Substitute numerical values and evaluate v:

v=

0.46 MV/m = 1.64 × 106 m/s 0.28 T

(b) Express and evaluate the kinetic energy of protons passing through the velocity selector undeflected:

K p = 12 mp v 2 =

1 2

(1.67 ×10

− 27

kg

= 2.26 × 10−15 J ×

)(1.64 ×10

6

m/s

)

2

1eV 1.60 × 10 −19 J

= 14.0 keV K e = 12 me v 2

(c) The kinetic energy of electrons passing through the velocity selector undeflected is given by:

=

1 2

(9.11×10

−31

kg

= 1.23 × 10−18 J ×

)(1.64 ×10

6

m/s

)

2

1eV 1.60 × 10−19 J

= 7.66 eV 42 • Picture the Problem Because the beam of protons is not deflected; we can conclude that the electric force acting on them is balanced by the magnetic force. Hence, we can find the magnetic force from the given data and use its definition to express the electric field.

r r Felec Eelec = q

(a) Use the definition of electric field to relate it to the electric force acting on the beam of protons:

r Fmag = qviˆ × Bˆj = qvBkˆ

Express the magnetic force acting on the beam of protons:

Because the electric force must be equal in magnitude but opposite in direction:

r Felec = − qvBkˆ = − 1.60 × 10−19 C (12.4 km/s )(0.85 T ) kˆ = − 1.69 × 10 −15 N kˆ

(

Substitute in the equation for the electric field to obtain:

)

(

(

)

)

r − 1.69 × 10 −15 N kˆ Eelec = 1.6 × 10 −19 C = − (10.5 kV/m ) kˆ

476 Chapter 26 r

r

(b) Because both Fmag and Felec are reversed, electrons are not deflected.

Thomson’s Measurement of q/m for Electrons and the Mass Spectrometer *43 •• Picture the Problem Figure 26-18 is reproduced below. We can express the total deflection of the electron beam as the sum of the deflections while the beam is in the field between the plates and its deflection while it is in the field-free space. We can, in turn, use constant-acceleration equations to express each of these deflections. The resulting equation is in terms of v0 and E. We can find v0 from the kinetic energy of the beam and E from the potential difference across the plates and their separation. In part (b) we can equate the electric and magnetic forces acting on an electron to express B in terms of E and v0.

(a) Express the total deflection ∆y of the electrons:

Use a constant-acceleration equation to express ∆y1:

∆y = ∆y1 + ∆y2

(1)

where ∆y1 is the deflection of the beam while it is in the electric field and ∆y2 is the deflection of the beam while it travels along a straight-line path outside the electric field.

∆y1 = 12 a y (∆t )

2

(2)

where ∆t = x1/v0 is the time an electron is in the electric field between the plates.

Apply Newton’s 2nd law to an electron between the plates to obtain:

qE = ma y

Solve for ay and substitute into equation (2) to obtain:

ay = and

qE m

The Magnetic Field 477 ⎛ qE ⎞ ⎛⎜ x1 ⎞⎟ ∆y1 = ⎜ ⎟ ⎝ m ⎠ ⎜⎝ v0 ⎟⎠

2

1 2

Express the vertical deflection ∆y2 of the electrons once they are out of the electric field: Use a constant-acceleration equation to find the vertical speed of an electron as it leaves the electric field:

(3)

∆y2 = v y ∆t2

(4)

v y = v0 y + a y ∆t1 =0+

qE ⎛ x1 ⎞ ⎜ ⎟ m ⎜⎝ v0 ⎟⎠

qE ⎛ x1 ⎞ ⎛ x2 ⎞ qEx1 x2 ⎜ ⎟⎜ ⎟ = m ⎜⎝ v0 ⎟⎠ ⎜⎝ v0 ⎟⎠ mv02

Substitute in equation (4) to obtain:

∆y 2 =

Substitute equations (3) and (5) in equation (1) to obtain:

⎛ qE ⎞ ⎛⎜ x1 ⎞⎟ qEx1 x2 + ∆y = ⎜ ⎟ mv02 ⎝ m ⎠ ⎜⎝ v0 ⎟⎠

(5)

2

1 2

or

∆y =

Use the definition of kinetic energy to find the speed of the electrons:

qEx1 ⎛ x1 ⎞ + x2 ⎟ 2 ⎜ mv0 ⎝ 2 ⎠

(6)

K = 12 mv02 and

2K 2(2.8 keV ) = m 9.11 × 10 − 31 kg

v0 =

= 3.14 × 107 m/s Express the electric field between the plates in terms of their potential difference:

E=

V d

Substitute numerical values and evaluate E:

E=

25 V V = = 2.08 kV/m d 1.2 cm

Substitute numerical values in equation (6) and evaluate ∆y:

(1.60 ×10 C)(2.08 kV/m )(6 cm) ⎛⎜ 6 cm + 30 cm ⎞⎟ = ∆y = (9.11×10 kg )(31.4 Mm/s) ⎝ 2 ⎠ −19

−31

2

7.34 mm

478 Chapter 26 (b) Because the electrons are deflected upward, the electric field must be downward and the magnetic Fy = 0 to an field upward. Apply

∑

Fmag − Felec = 0 or

qvB = qE

electron to obtain: Solve for B:

B=

E v

Substitute numerical values and evaluate B:

B=

2.08 kV/m = 66.2 µT 3.14 × 107 m/s

44 •• Picture the Problem The diagram below represents the paths of the two ions entering the magnetic field at the left. The magnetic force acting on each causes them to travel in circular paths of differing radii due to their different masses. We can apply Newton’s 2nd law to an ion in the magnetic field to obtain an expression for its radius and then express their final separation in terms of these radii that, in turn, depend on the energy with which the ions enter the field. We can connect their energy to the potential through which they are accelerated using the work-kinetic energy theorem and relate their separation ∆s to the accelerating potential difference ∆V.

Express the separation ∆s of the chlorine ions: Apply

∑F

radial

= mac to an ion in

the magnetic field of the mass spectrometer: Solve for r to obtain:

∆s = 2(r37 − r35 )

(1)

v2 qvB = m r

r=

mv qB

(2)

The Magnetic Field 479 Relate the speed of an ion as it enters the magnetic field to the potential difference through which it has been accelerated: Solve for v to obtain:

q∆V = 12 mv 2

2q∆V m

v=

m 2q∆V = qB m

2m∆V qB 2

Substitute in equation (2) to obtain:

r=

Use this equation to express the radii of the paths of the two chlorine isotopes to obtain:

r35 =

Substitute in equation (1) to obtain:

⎛ 2m35 ∆V 2m35∆V ∆s = 2⎜⎜ − 2 qB qB 2 ⎝

2m35∆V 2m37 ∆V and r37 = 2 qB qB 2

⎛ 1 2∆V = 2⎜⎜ q ⎝B Solve for ∆V:

Substitute numerical values and evaluate ∆V:

(

⎞ ⎟ ⎟ ⎠

)

⎞ m37 − m35 ⎟⎟ ⎠

qB 2 (2∆s )

2

∆V =

(

2 m37 − m35

(

)

)

2

⎛ 1.4 cm ⎞ 1.60 × 10 C (1.2 T ) ⎜ ⎟ 2 ⎠ ⎝ ∆V = 2 2 37 u − 35 u −19

(

=

(

2

2

)

5.65 × 10− 24 C ⋅ T 2 ⋅ m 2

)( 2

37 − 35 1.66 × 10− 27 kg

)

= 122 kV 45 •• Picture the Problem We can apply Newton’s 2nd law to an ion in the magnetic field to obtain an expression for r as a function of m, v, q, and B and use the work-kinetic energy theorem to express the kinetic energy in terms of the potential difference through which the ion has been accelerated. Eliminating v between these equations will allow us to express r in terms of m, q, B, and ∆V.

480 Chapter 26 Apply

∑F

radial

= mac to an ion in

the magnetic field of the mass spectrometer:

v2 qvB = m r

mv qB

Solve for r to obtain:

r=

Apply the work-kinetic energy theorem to relate the speed of an ion as it enters the magnetic field to the potential difference through which it has been accelerated:

q∆V = 12 mv 2

Solve for v to obtain:

(1)

2 q ∆V m

v=

m 2q∆V = qB m

Substitute in equation (1) to obtain:

r=

(a) Substitute numerical values and evaluate equation (2) for 24Mg :

r24 =

(

2m∆V qB 2

(2)

)

2 3.983 × 10− 26 kg (2.5 kV )

(1.60 × 10

−19

)(

C 557 × 10− 4 T

)

2

= 63.3 cm (b) Express the difference in the radii for 24Mg and 26Mg: Substitute numerical values and evaluate equation (2) for 26Mg :

∆r = r26 − r24

r26 =

(

)

⎛ 26 ⎞ 2⎜ ⎟ 3.983 × 10 − 26 kg (2.5 kV ) ⎝ 24 ⎠ 2 1.60 × 10 −19 C 557 × 10 − 4 T

(

)(

)

= 65.9 cm Substitute to obtain:

∆r = 65.9 cm − 63.3 cm = 2.60 cm

*46 •• Picture the Problem We can apply Newton’s 2nd law to an ion in the magnetic field of the spectrometer to relate the diameter of its orbit to its charge, mass, velocity, and the magnetic field. If we assume that the velocity is the same for the two ions, we can then express the ratio of the two diameters as the ratio of the masses of the ions and solve for

The Magnetic Field 481 the diameter of the orbit of 7Li. Apply

∑F

radial

= mac to an ion in

qvB = m

the field of the spectrometer:

v2 r

Solve for r to obtain:

r=

mv qB

Express the diameter of the orbit:

d=

2mv qB

Express the diameters of the orbits for 6Li and 7Li:

d6 =

Assume that the velocities of the two ions are the same and divide the 2nd of these diameters by the first to obtain:

2m7 v d7 m qB = = 7 d 6 2m6v m6 qB

Solve for and evaluate d7:

d7 =

2m6v 2m7 v and d 7 = qB qB

m7 7u (15 cm) = 17.5 cm d6 = m6 6u

The Cyclotron 47 •• Picture the Problem The time required for each of the ions to complete its semicircular paths is half its period. We can apply Newton’s 2nd law to an ion in the magnetic field of the spectrometer to obtain an expression for r as a function of the charge and mass of the ion, its velocity, and the magnetic field. Express the time for each ion to complete its semicircular path: Apply

∑F

radial

= mac to an ion in the

∆t = 12 T =

qvB = m

field of the spectrometer: Solve for r to obtain:

r=

mv qB

πr

v2 r

v

482 Chapter 26 Substitute to obtain:

Substitute numerical values and evaluate ∆t58 and ∆t60:

∆t =

πm qB

∆t58 =

(

)

58π 1.66 × 10−27 kg 1.60 × 10−19 C (0.12 T )

(

)

= 15.8 µs and

(

)

60π 1.66 × 10−27 kg ∆t60 = 1.60 × 10−19 C (0.12 T )

(

)

= 16.3 µs 48 •• Picture the Problem We can apply a condition for equilibrium to ions passing through the velocity selector to obtain an expression relating E, B, and v that we can solve for v. We can, in turn, express E in terms of the potential difference V between the plates of the selector and their separation d. In (b) we can apply Newton’s 2nd law to an ion in the bending field of the spectrometer to relate its diameter to its mass, charge, velocity, and the magnetic field. (a) Apply

∑F

y

= 0 to the ions in

Felec − Fmag = 0

the crossed fields of the velocity selector to obtain:

or

Solve for v to obtain:

v=

E B

Express the electric field between the plates of the velocity selector in terms of their separation and the potential difference across them:

E=

V d

Substitute to obtain:

v=

V dB

Substitute numerical values and evaluate v:

v=

160 V = 1.90 × 105 m/s (2 mm)(0.42 T )

(b) Express the difference in the diameters of the orbits of singly

qE − qvB = 0

∆d = d 238 − d 235

(1)

The Magnetic Field 483 ionized 238U and 235U: Apply

∑F

radial

= mac to an ion in

qvB = m

the spectrometer’s magnetic field: Solve for the radius of the ion’s orbit:

r=

mv qB

Express the diameter of the orbit:

d=

2mv qB

Express the diameters of the orbits for 238U and 235U:

d 238 =

Substitute in equation (1) to obtain:

∆d =

v2 r

2m238v 2m235v and d 235 = qB qB

2m238v 2m235v − qB qB 2v (m238 − m235 ) = qB

Substitute numerical values and evaluate ∆d:

⎛ 1.66 × 10−27 kg ⎞ ⎟⎟ 2 1.90 × 105 m/s (238 u − 235 u )⎜⎜ u ⎝ ⎠ = 9.86 mm ∆d = −19 1.60 × 10 C (1.2 T )

(

)

(

)

*49 •• Picture the Problem We can express the cyclotron frequency in terms of the maximum orbital radius and speed of the protons/deuterons. By applying Newton’s 2nd law, we can relate the radius of the particle’s orbit to its speed and, hence, express the cyclotron frequency as a function of the particle’s mass and charge and the cyclotron’s magnetic field. In part (b) we can use the definition of kinetic energy and their maximum speed to find the maximum energy of the emerging protons. (a) Express the cyclotron frequency in terms of the proton’s orbital speed and radius: Apply

∑F

radial

= mac to a proton in

the magnetic field of the cyclotron:

f =

1 1 v = = T 2πr v 2πr

qvB = m

v2 r

(1)

484 Chapter 26 Solve for r to obtain:

r=

mv qB

Substitute to obtain:

f =

qBv qB = 2πmv 2πm

Substitute numerical values and evaluate f:

f =

(1.60 ×10 C)(1.4 T ) = 2π (1.67 × 10 kg )

(b) Express the maximum kinetic energy of a proton:

(2)

−19

−27

21.3 MHz

2 K max = 12 mvmax

qBrmax m

Solve equation (1) for vmax to obtain:

vmax =

Substitute to obtain:

⎛ q2B2 ⎞ 2 ⎛ qBrmax ⎞ ⎟⎟rmax (3) K = m⎜ ⎟ = 12 ⎜⎜ ⎝ m ⎠ ⎝ m ⎠

Substitute numerical values and evaluate K:

⎛ 1.60 ×10 −19 C 2 (1.4 T )2 ⎞ ⎟(0.7 m )2 K= ⎜ 27 − ⎜ ⎟ 1.67 ×10 kg ⎝ ⎠ 1eV = 7.36 × 10 −12 J × 1.60 ×10-19 J

2

1 2

1 2

(

)

= 46.0 MeV (c) From equation (2) we see that doubling m halves f:

f deuterons =

From equation (3) we see that doubling m halves K:

K deuterons = 12 K protons = 23.0 MeV

1 2

f protons = 10.7 MHz

50 •• Picture the Problem We can express the cyclotron frequency in terms of the maximum orbital radius and speed of the protons be accelerated in the cyclotron. By applying Newton’s 2nd law, we can relate the radius of the proton’s orbit to its speed and, hence, express the cyclotron frequency as a function of the its mass and charge and the cyclotron’s magnetic field. In part (b) we can use the definition of kinetic energy express the minimum radius required to achieve the desired emergence energy. In part (c) we can find the number of revolutions required to achieve this emergence energy from the

The Magnetic Field 485 energy acquired during each revolution. (a) Express the cyclotron frequency in terms of the proton’s orbital speed and radius: Apply

∑F

radial

= mac to a proton in

1 1 v = = T 2πr v 2πr

f =

qvB = m

the magnetic field of the cyclotron:

v2 r

Solve for r to obtain:

r=

mv qB

Substitute to obtain:

f =

qBv qB = 2πmv 2πm

Substitute numerical values and evaluate f:

(1.60 ×10 C)(1.8 T ) f = 2π (1.67 × 10 kg )

(1)

−19

− 27

= 27.4 MHz (b) Using the definition of kinetic energy, relate emergence energy of the protons to their velocity: Solve for v to obtain:

K = 12 mv 2

2K m

v=

Substitute in equation (1) and simplify to obtain:

r=

Substitute numerical values and evaluate rmin:

r=

m 2K = qB m

2 Km qB

(

2(25 MeV ) 1.67 × 10 −27 kg 1.60 × 10 −19 C (1.8 T )

(

)

= 0.401 m (c) Express the required number of revolutions N in terms of the energy gained per revolution: Because the beam is accelerated through a potential difference of 50

N=

25 MeV Erev

Erev = 2q∆V = 100 keV

)

486 Chapter 26 kV twice during each revolution: Substitute and evaluate N:

25 MeV = 250 rev 100 keV/rev

N=

51 •• Picture the Problem We can express the cyclotron frequency in terms of the maximum orbital radius and speed of a particle being accelerated in the cyclotron. By applying Newton’s 2nd law, we can relate the radius of the particle’s orbit to its speed and, hence, express the cyclotron frequency as a function of its charge-to-mass ratio and the cyclotron’s magnetic field. We can then use data for the relative charges and masses of deuterons, alpha particles, and protons to establish the ratios of their cyclotron frequencies. Express the cyclotron frequency in terms of a particle’s orbital speed and radius: Apply

∑F

radial

= mac to a particle

1 1 v = = T 2πr v 2πr

f =

qvB = m

in the magnetic field of the cyclotron:

v2 r

Solve for r to obtain:

r=

mv qB

Substitute to obtain:

f =

qBv B q = 2πmv 2π m

Evaluate equation (1) for deuterons:

fd =

B qd B e = 2π md 2π md

Evaluate equation (1) for alpha particles:

fα =

B qα B 2e B e = = 2π mα 2π 2md 2π md

(1)

and

f d = fα Evaluate equation (1) for protons:

fp =

B qp B = 2π mp 2π

= 2 fd and

⎛ B e ⎞ e ⎜⎜ ⎟⎟ = 2 1 ⎝ 2π md ⎠ 2 md

The Magnetic Field 487 1 2

f p = f d = fα

52 •• Picture the Problem We can apply Newton’s 2nd law to the orbiting charged particle to obtain an expression for its radius as a function of its particle’s kinetic energy. Because the energy gain per revolution is constant, we can express this kinetic energy as the product of the number of orbits completed and the energy gained per revolution and, hence, show that the radius is proportional to the square root of the number of orbits completed. Apply

∑F

radial

= mac to a particle

qvB = m

in the magnetic field of the cyclotron:

v2 r

mv qB

Solve for r to obtain:

r=

Express the kinetic energy of the particle in terms of its speed and solve for v:

K = 12 mv 2 ⇒ v =

Noting that the energy gain per revolution is constant, express the kinetic energy in terms of the number of orbits N completed by the particle and energy Erev gained by the particle each revolution: Substitute equations (2) and (3) in equation (1) to obtain:

(1)

2K m

(2)

K = NErev

r= =

(3)

1 m 2K 2mK = qB m qB 2mErev 1 2 N qB

1 2mNErev = qb

or r ∝ N 1 2

Torques on Current Loops and Magnets 53 • Picture the Problem We can use the definition of the magnetic moment of a coil to

r

r

r

evaluate µ and the expression for the torque exerted on the coil τ = µ × B to find the magnitude of τ.

488 Chapter 26 (a) Using its definition, express the magnetic moment of the coil: Substitute numerical values and evaluate µ:

(b) Express the magnitude of the torque exerted on the coil: Substitute numerical values and evaluate τ :

µ = NIA = NIπr 2 µ = (20)(3 A )π (0.04 m )2 = 0.302 A ⋅ m 2

τ = µB sin θ τ = (0.302 A ⋅ m 2 )(0.5 T )sin 60° = 0.131 N ⋅ m

54 • Picture the Problem The coil will experience the maximum torque when the plane of the r coil makes an angle of 90° with the direction of B . The magnitude of the maximum torque is then given by τ max = µB . Express the maximum torque acting on the coil:

τ max = µB

Use its definition to express the magnetic moment of the coil:

µ = NIA = NIπr 2

Substitute to obtain:

τ max = NIπr 2 B

Substitute numerical values and evaluate τ :

τ max = (400 )(1.6 mA )π (0.75 cm )2 (0.25 T ) = 2.83 × 10− 5 N ⋅ m

*55 • r r r Picture the Problem We can use τ = µ × B to find the torque on the coil in the two orientations of the magnetic field. Express the torque acting on the coil: Express the magnetic moment of the coil:

r r r τ = µ× B r µ = ± IAkˆ = ± IL2 kˆ

The Magnetic Field 489 r

r

(a) Evaluate τ for B in the z direction:

r

r

(b) Evaluate τ for B in the x direction:

r τ = ± IL2 kˆ × Bkˆ = ± IL2 B kˆ × kˆ = 0

(

)

( )

r τ = ± IL2 kˆ × Biˆ = ± IL2 B kˆ × iˆ

= ±(2.5 A )(0.06 m ) (0.3 T ) ˆj 2

(

)

= ± 2.70 × 10 −3 N ⋅ m ˆj 56 • r r r Picture the Problem We can use τ = µ × B to find the torque on the equilateral triangle in the two orientations of the magnetic field. Express the torque acting on the coil: Express the magnetic moment of the coil: Relate the area of the equilateral triangle to the length of its side:

Substitute to obtain:

r

r

(a) Evaluate τ for B in the z direction:

r r r τ = µ× B r µ = ± IAkˆ A = 12 base × altitude =

1 ⎛ 3L ⎞ 3 2 ⎟= (L )⎜⎜ L ⎟ 2 ⎝ 2 ⎠ 4

r 3L2 I ˆ µ=± k 4 r 3L2 I ˆ τ =± k × Bkˆ 4 3L2 IB ˆ ˆ =± k×k = 0 4

(

r

r

(b) Evaluate τ for B in the x direction:

)

( )

r 3L2 I ˆ 3L2 IB ˆ ˆ τ =± k × Biˆ = ± k×i 4 4 2 3 (0.08 m ) (2.5 A )(0.3 T ) ˆ =± j 4 = ± (2.08 × 10 −3 N ⋅ m ) ˆj

490 Chapter 26 57 •• Picture the Problem The loop will start to lift off the table when the magnetic torque equals the gravitational torque. Express the magnetic torque acting on the loop: Express the gravitational torque acting on the loop: Because the loop is in equilibrium under the influence of the two torques: Solve for B to obtain:

τ mag = µB = IπR 2 B

τ grav = mgR IπR 2 B = mgR

B=

mg IπR

58 •• Picture the Problem The diagram to the right shows the coil as it would appear from along the positive z axis. The righthand rule for determining the direction of nˆ has been used to establish nˆ as shown. We can use the geometry of this figure to determine θ and to express the unit normal vector nˆ . The magnetic moment of the r coil is given by µ = NIAnˆ and the torque

r

r

r

exerted on the coil by τ = µ × B . Finally, we can find the potential energy of the coil

r r

in this field from U = − µ ⋅ B . (a) Noting that θ and the angle whose measure is 37° have their right and left sides mutually perpendicular, we can conclude that:

θ = 37°

(b) Use the components of nˆ to

nˆ = nx iˆ + n y ˆj = cos 37°iˆ − sin 37° ˆj

express nˆ in terms of iˆ and ˆj :

(c) Express the magnetic moment of

= 0.799iˆ − 0.602 ˆj r µ = NIAnˆ

The Magnetic Field 491 the coil:

r

Substitute numerical values and evaluate µ :

)(

(

) (

) (

)

r µ = (50)(1.75 A ) 48 cm 2 0.799iˆ − 0.602 ˆj = 0.336 A ⋅ m 2 iˆ − 0.253 A ⋅ m 2 ˆj (d) Express the torque exerted on the coil:

r r r τ = µ× B

r

r

Substitute for µ and B to obtain:

{(

) (

)}

r τ = 0.336 A ⋅ m 2 iˆ − 0.253 A ⋅ m 2 ˆj × (1.5 T ) ˆj

( )

( )

= (0.504 N ⋅ m ) iˆ × ˆj − (0.380 N ⋅ m ) ˆj × ˆj =

(0.504 N ⋅ m ) kˆ

r r U = −µ⋅ B

(e) Express the potential energy of the coil in terms of its magnetic moment and the magnetic field:

r

r

Substitute for µ and B and evaluate U:

{(

)}

) (

U = − 0.336 A ⋅ m 2 iˆ − 0.253 A ⋅ m 2 ˆj ⋅ (1.5 T ) ˆj = −(0.504 N ⋅ m ) iˆ ⋅ ˆj + (0.380 N ⋅ m ) ˆj ⋅ ˆj = 0.380 J

( )

( )

59 •• Picture the Problem We can use the right-hand rule for determining the direction of nˆ

r

r

r

to establish the orientation of the coil for value of nˆ and τ = µ × B to find the torque exerted on the coil in each orientation. (a) The orientation of the coil is shown to the right:

r

r Evaluate τ for B = 2.0 T ˆj and

nˆ = iˆ :

r r r r τ = µ × B = NIAnˆ × B

(

)

= (50 )(1.75 A ) 48 cm 2 iˆ × (2 T ) ˆj = (0.840 N ⋅ m ) iˆ × ˆj

( )

=

(0.840 N ⋅ m ) kˆ

492 Chapter 26 (b) The orientation of the coil is shown to the right:

r

r Evaluate τ for B = 2.0 T ˆj and

r r r r τ = µ × B = NIAnˆ × B

(

)

= (50)(1.75 A ) 48 cm 2 ˆj × (2 T ) ˆj = (0.840 N ⋅ m ) ˆj × ˆj

nˆ = ˆj :

( )

= 0 (c) The orientation of the coil is shown to the right:

r

r Evaluate τ for B = 2.0 T ˆj and

r r r r τ = µ × B = NIAnˆ × B

(

)

= −(50 )(1.75 A ) 48 cm 2 ˆj × (2 T ) ˆj = (− 0.840 N ⋅ m ) ˆj × ˆj

nˆ = − ˆj :

( )

= 0 (d) The orientation of the coil is shown to the right:

r

r Evaluate τ for B = 2.0 T ˆj and

nˆ = ( iˆ + ˆj )/ 2 :

r r r r τ = µ × B = NIAnˆ × B =

(50)(1.75 A )(48 cm2 ) (iˆ + ˆj )× (2 T ) ˆj 2

( ) ( )

= (0.594 N ⋅ m ) iˆ × ˆj + (0.594 N ⋅ m ) ˆj × ˆj =

(0.594 N ⋅ m ) kˆ

Magnetic Moments *60 •• Picture the Problem Because the small magnet can be modeled as a magnetic dipole; we can use the equation for the torque on a current loop to find its magnetic moment. Express the magnitude of the torque

τ = µB sin θ

The Magnetic Field 493 acting on the magnet: Solve for µ to obtain:

µ=

Substitute numerical values and evaluate µ:

µ=

τ B sin θ 0.10 N ⋅ m = 2.89 A ⋅ m 2 (0.04 T )sin 60°

61 •• Picture the Problem We can use the definition of the magnetic moment to find the magnetic moment of the given current loop and a right-hand rule to find its direction.

µ = IA

Using its definition, express the magnetic moment of the current loop: Express the area bounded by the loop:

A=

Substitute to obtain:

µ=

Substitute numerical values and evaluate µ:

µ=

1 2

(πR

2 outer

πI

(R 2

2 outer

π (1.5 A ) 2

)

2 − πRinner =

2 − Rinner

π 2

(R

2 outer

2 − Rinner

)

[(0.5 m)

2

− (0.3 m )

2

]

= 0.377 A ⋅ m 2 Apply the right-hand rule for determining the direction of the unit normal vector (the

r

direction of µ) to conclude that µ points into the page. 62 •• Picture the Problem We can use the definition of the magnetic moment of a coil to find the magnetic moment of a wire of length L that is wound into a circular coil of N loops. We can find the area of the coil from its radius R and we can find R by dividing the length of the wire by the number of turns. Use its definition to express the magnetic moment of the coil: Express the circumference of each loop:

µ = NIA

L = 2πR N

(1)

)

494 Chapter 26 where R is the radius of a loop.

L 2πN

Solve for R to obtain:

R=

Express the area of the coil:

L2 ⎛ L ⎞ A = πR = π ⎜ ⎟ = 4πN 2 ⎝ 2πN ⎠

Substitute in equation (1) and simplify to obtain:

⎛ L2 ⎞ IL2 ⎟= µ = NI ⎜⎜ 2 ⎟ 4πN ⎝ 4πN ⎠

2

2

63 •• Picture the Problem We can use the definition of current and the relationship between the frequency of the motion and its period to show that I = qω/2π . We can use the definition of angular momentum and the moment of inertia of a point particle to show that the magnetic moment has the magnitude µ = 12 qωr 2 . Finally, we can express the

r

r

r

ratio of µ to L and the fact that µ and L are both parallel to ω to conclude that

r r µ = (q/2m) L .

(a) Using its definition, relate the average current to the charge passing a point on the circumference of the circle in a given period of time:

I=

∆q q = = qf ∆t T

Relate the frequency of the motion to the angular frequency of the particle:

f =

ω 2π

Substitute to obtain:

I=

qω 2π

From the definition of the magnetic moment we have:

µ = IA = ⎜

(b) Express the angular momentum of the particle: The angular momentum of the particle is:

⎛ qω ⎞ 2 ⎟ πr = ⎝ 2π ⎠

L = Iω

I = mr 2

( )

1 2

qωr 2

The Magnetic Field 495

( )

Substitute to obtain:

L = mr 2 ω = mr 2ω

Express the ratio of µ to L and simplify to obtain:

µ

r

r

Because µ and L are both parallel

r to ω :

L

=

r µ=

q qωr 2 q = ⇒ µ= L 2 mr ω 2m 2m

1 2

q r L 2m

*64 ••• Picture the Problem We can express the magnetic moment of an element of charge dq in a cylinder of length L, radius r, and thickness dr, relate this charge to the length, radius, and thickness of the cylinder, express the current due to this rotating charge, substitute for A and dI in our expression for µ and then integrate to complete our derivation for the magnetic moment of the rotating cylinder as a function of its angular velocity. Express the magnetic moment of an element of charge dq in a cylinder of length L, radius r, and thickness dr:

dµ = AdI where A = πr2.

Relate the charge dq in the cylinder to the length of the cylinder, its radius, and thickness:

dq = 2πLρrdr

Express the current due to this rotating charge:

dI =

Substitute to obtain:

dµ = πr 2 (Lωρrdr ) = Lωρπr 3dr

Integrate r from Ri to R0 to obtain:

ω ω (2πLρrdr ) = Lωρrdr dq = 2π 2π

R0

µ = Lωρπ ∫ r 3dr =

1 4

(

Lωρπ R04 − Ri4

Ri

65 ••• Picture the Problem We can follow the step-by-step outline provided in the problem statement to establish the given results. (a) Express the magnetic moment of the rotating element of charge: The area enclosed by the rotating

dµ = AdI

A = πx 2

(1)

)

496 Chapter 26 element of charge:

dI =

dq λdx = ∆t ∆t

where ∆t is the time required for one revolution. Express the time ∆t required for one revolution:

∆t =

1 2π = ω f

Substitute to obtain:

dI =

λω dx 2π

Substitute in equation (1) and simplify to obtain:

⎛ λω ⎞ dµ = πx 2 ⎜ dx ⎟ = ⎝ 2π ⎠

(b) Integrate dµ from x = 0 to x= l:

µ = 12 λω ∫ x 2 dx =

(c) Express the angular momentum of the rod:

( ) l

1 6

1 2

λωx 2 dx

λωl 3

0

L = Iω where L is the angular momentum of the rod and I is the moment of inertia of the rod with respect to the point about which it is rotating.

I = 13 mL2

Express the moment of inertia of the rod with respect to an axis through its end:

where L is now the length of the rod.

Substitute to obtain:

L = 13 mL2ω

Divide our expression for µ by L to obtain:

µ L

=

1 6 1 3

λωL3 λL = mL2ω 2m

or, because Q = λL,

µ= r

r

r

Because ω and L = Iω point in the same direction:

r µ=

Q L 2m Q r L 2M

The Magnetic Field 497 66 ••• Picture the Problem We can express the magnetic moment of an element of current dI due to a ring of radius r, and thickness dr with charge dq. Integrating this expression from r = 0 to r = R will give us the magnetic moment of the disk. We can integrate the charge on the ring between these same limits to find the total charge on the disk and divide µ by Q to establish the relationship between them. In part (b) we can find the angular momentum of the disk by first finding the moment of inertia of the disk by integrating r2dm between the same limits used above. (a) Express the magnetic moment of an element of the disk:

dµ = AdI

The area enclosed by the rotating element of charge is:

A = πx 2

Express the element of current dI:

dI =

Substitute and simplify to obtain:

dµ = πr 2

Integrate dµ from r = 0 to r = R to obtain: Express the charge dq within a distance r of the center of the disk:

Integrate dq from r = 0 to r = R to obtain:

dq σdA = = fσdA ∆t ∆t ω ⎛ r⎞ σω 2 = ⎜ σ 0 ⎟(2πrdr ) = 0 r dr 2π ⎝ R ⎠ R

µ=

σ 0ω R

σ 0πω R R

r 2 dr =

∫ r dr = 4

σ 0πω

1 5

R

r 4 dr

σ 0πωR 4

(1)

0

⎛ r⎞ dq = 2πrσdr = 2πr ⎜ σ 0 ⎟dr ⎝ R⎠ 2πσ 0 2 = r dr R 2πσ 0 2 r dr = 23 πσ 0 R 2 ∫ R 0 R

Q=

(2)

498 Chapter 26 Divide equation (1) by Q to obtain:

σ 0πωR 4 3ωR 2 = 2 = 2 10 Q 3 πσ 0 R µ

1 5

and

µ=

3 10

QωR 2

(b) Express the moment of inertia of an element of mass dm of the disk:

dI = r 2 dm = r 2σ m dA

Integrate dI from r = 0 to r = R to obtain:

2πmσ 0 4 2πmσ 0 4 I= r dr = R ∫ 5Q QR 0

Divide I by equation (2) and simplify to obtain:

2πmσ 0 4 R I 3m 2 5Q = 2 = R 2 πσ 0 R 5Q Q 3

(3)

⎛m ⎞ = r 2 ⎜⎜ σ ⎟⎟(2πrdr ) ⎝Q ⎠ ⎛r ⎞ 2πm⎜ σ 0 ⎟ ⎝ R ⎠ r 3dr = Q 2πmσ 0 4 = r dr QR R

and

I=

3m 2 R 5

Express the angular momentum of the disk:

L = Iω = 53 mR 2ω

Divide equation (3) by L and simplify to obtain:

µ L

=

3 10 3 5

QωR 2 Q = 2 mR ω 2m

and

µ= r

Because µ is in the same direction

r as ω :

r µ=

Q L 2m Q r L 2m

67 ••• Picture the Problem We can use the general result from Example 26-11 and Problem 63 to express µ as a function of Q, M, and L. We can then use the definitions of surface

The Magnetic Field 499 charge density and angular momentum to substitute for Q and L to obtain the magnetic moment of the rotating sphere.

Q L 2M

Express the magnetic moment of the spherical shell in terms of its mass, charge, and angular momentum:

µ=

Use the definition of surface charge density to express the charge on the spherical shell is:

Q = σA = 4πσR 2

Express the angular momentum of the spherical shell:

L = Iω = 23 MR 2ω

Substitute to obtain:

⎛ 4πσR 2 ⎞⎛ 2 ⎞ ⎟⎟⎜ MR 2ω ⎟ = µ = ⎜⎜ ⎠ ⎝ 2 M ⎠⎝ 3

4 3

πσR 4ω

68 ••• Picture the Problem We can use the general result from Example 26-11 and Problem 63 to express µ as a function of Q, M, and L. We can then use the definitions of volume charge density and angular momentum to substitute for Q and L to obtain the magnetic moment of the rotating sphere.

Q L 2M

Express the magnetic moment of the solid sphere in terms of its mass, charge, and angular momentum:

µ=

Use the definition of volume charge density to express the charge of the sphere:

Q = ρV = 43 πρR 3

Express the angular momentum of the solid sphere:

L = Iω = 52 MR 2ω

Substitute to obtain:

⎛ 4 πρR 3 ⎞⎛ 2 ⎞ ⎟⎟⎜ MR 2ω ⎟ = µ = ⎜⎜ 3 ⎠ ⎝ 2 M ⎠⎝ 5

4 15

πρR 5ω

*69 ••• Picture the Problem We can use its definition to express the torque acting on the disk and the definition of the precession frequency to find the precession frequency of the disk.

500 Chapter 26 (a) The magnitude of the net torque acting on the disk is:

τ = µB sin θ where µ is the magnetic moment of the disk.

1 4

From example 26-11:

µ = πσr 4ω

Substitute for µ in the expression for τ to obtain:

τ=

1 πσr 4ωB sin θ 4

Ω=

τ Iω

I=

1 2 mr 2

(b) The precession frequency Ω is equal to the ratio of the torque divided by the spin angular momentum: For a solid disk, the moment of inertia is given by: Substitute for τ and I to obtain:

Ω=

1 4

πσr 4ωB sin θ πσr 2 B = sin θ 2 1 2m 2 mr ω

Remarks: It’s interesting that the precession frequency is independent of ω.

The Hall Effect 70 • Picture the Problem We can use the Hall effect equation to find the drift velocity of the electrons and the relationship between the current and the number density of charge carriers to find n. In (c) we can use a right-hand rule to decide whether a or b is at the higher potential. (a) Express the Hall voltage as a function of the drift velocity of the electrons in the strip:

VH = vd Bw

Solve for vd:

vd =

VH Bw

Substitute numerical values and evaluate vd:

vd =

4.27 µV = 0.107 mm/s (2 T )(2 cm)

(b) Express the current as a function of the number density of charge carriers:

I = nAqvd

The Magnetic Field 501 Solve for n:

n=

I Aqvd

Substitute numerical values and evaluate n:

n=

20 A = 5.84 × 10 28 m −3 (2 cm)(0.1cm) 1.60 ×10−19 C (0.107 mm/s)

(

)

r

r

(c) Apply a right-hand rule to I l and B to conclude that positive charge will accumulate

at a and negative charge at b and therefore Va > Vb .

71 •• Picture the Problem We can use I = nqvd A to find the drift velocity and VH = vd Bw to find the Hall voltage.

I = nqvd A

(a) Express the current in the metal strip in terms of the drift velocity of the electrons: Solve for vd:

vd =

I nqA

Substitute numerical values and evaluate vd:

vd =

(8.47 ×10

22

cm

−3

10 A = 3.69 × 10 −5 m/s −19 1.60 × 10 C (2 cm )(0.1cm )

)(

(b) Relate the Hall voltage to the drift velocity and the magnetic field: Substitute numerical values and evaluate VH:

)

VH = vd Bw

(

)

VH = 3.69 ×10 −5 m/s (2 T )(2 cm ) = 1.48 µV

*72 •• Picture the Problem We can use VH = vd Bw to express B in terms of VH and

I = nqvd A to eliminate the drift velocity vd and derive an expression for B in terms of VH, n, and t. Relate the Hall voltage to the drift velocity and the magnetic field:

VH = vd Bw

502 Chapter 26 Solve for B to obtain:

B=

Express the current in the metal strip in terms of the drift velocity of the electrons:

VH vd w

I = nqvd A

Solve for vd to obtain:

vd =

Substitute and simplify to obtain:

B=

I nqA

VH nqAVH nqwtVH = = I Iw Iw w nqA nqt = VH I

Substitute numerical values and simplify to obtain:

B=

(8.47 × 10

22

)(

)

cm −3 1.60 × 10 −19 C (0.1 cm )VH = 6.78 × 105 s/m 2 VH 20 A

(a) Evaluate B for VH = 2.00 µV:

(

)

B = (6.78 ×10 5 s/m 2 )(2.00 µV ) = 1.36 T

(b) Evaluate B for VH = 5.25 µV:

B = (6.78 × 10 5 s/m 2 )(5.25 µV ) = 3.56 T

(c) Evaluate B for VH = 8.00 µV:

(

)

B = 6.78 × 10 5 s/m 2 (8.00 µV ) = 5.42 T

73 •• Picture the Problem We can use VH = vd Bw to find the Hall voltage developed across the diameter of the artery. Relate the Hall voltage to the flow speed of the blood vd, the diameter of the artery w, and the magnetic field B:

VH = vd Bw

The Magnetic Field 503 Substitute numerical values and evaluate VH:

VH = (0.6 m/s )(0.2 T )(0.85 cm ) = 1.02 mV

74 •• Picture the Problem Let the width of the slab be w and its thickness t. We can use the definition of the Hall electric field in the slab, the expression for the Hall voltage across it, and the definition of current density to show that the Hall coefficient is also given by 1/(nq). Express the Hall coefficient: Using its definition, express the Hall electric field in the slab:

R=

Ey

J x Bz V Ey = H w

I = nqvd wt

Express the current density in the slab:

Jx =

Substitute to obtain:

VH VH R= w = nqvd Bz nqvd wBz

Express the Hall voltage in terms of vd, B, and w: Substitute and simplify to obtain:

VH = vd Bz w

R=

1 vd Bz w = nqvd wBz nq

*75 •• Picture the Problem We can determine the number of conduction electrons per atom from the quotient of the number density of charge carriers and the number of charge carriers per unit volume. Let the width of a slab of aluminum be w and its thickness t. We can use the definition of the Hall electric field in the slab, the expression for the Hall voltage across it, and the definition of current density to find n in terms of R and q and na = ρN A M , to express na. Express the number of electrons per atom N:

N=

n na

(1)

where n is the number density of charge carriers and na is the number of atoms per unit volume.

504 Chapter 26 Ey

From the definition of the Hall coefficient we have:

R=

Express the Hall electric field in the slab:

Ey =

VH w

Express the current density in the slab:

Jx =

I = nqvd wt

Substitute to obtain:

VH VH R= w = nqvd Bz nqvd wBz

J x Bz

VH = vd Bz w

Express the Hall voltage in terms of vd, B, and w: Substitute and simplify to obtain:

R=

1 vd Bz w = nqvd wBz nq

Solve for and evaluate n:

n=

1 Rq

Express the number of atoms na per unit volume:

na = ρ

Substitute equations (2) and (3) in equation (1) to obtain:

N=

(2)

NA M

(3)

M qRρN A

Substitute numerical values and evaluate N:

N=

(− 1.60 ×10

−19

)(

C − 0.3 × 10

−10

27 g/mol m /C 2.7 × 103 kg/m 3 6.02 × 1023 atoms/mol 3

)(

)(

)

= 3.46

General Problems 76 • Picture the Problem We can use the expression for the magnetic force acting on a wire r r r ( F = I l × B ) to find the force per unit length on the wire. Express the magnetic force on the

r r r F = Il × B

The Magnetic Field 505 wire:

r

r

Substitute for I l and B to obtain:

r F = (6.5 A )liˆ × (1.35 T ) ˆj and

r F = (6.5 A ) iˆ × (1.35 T ) ˆj l

Simplify to obtain:

r F = (8.78 N/m ) iˆ × ˆj = (8.78 N/m )kˆ l

( )

77 • Picture the Problem We can express the period of the alpha particle’s motion in terms of its orbital speed and use Newton’s 2nd law to express its orbital speed in terms of known quantities. Knowing the particle’s period and the radius of its motion we can find its speed and kinetic energy. (a) Relate the period of the alpha particle’s motion to its orbital speed: Apply

∑F

radial

= mac to the alpha

T=

2πr v

qvB = m

particle to obtain: Solve for v to obtain:

v=

Substitute and simplify to obtain:

T=

Substitute numerical values and evaluate T:

(1)

v2 r

qBr m 2πr 2πm = qBr qB m

T=

(

)

2π 6.65 × 10−27 kg = 0.131 µs 2 1.60 × 10−19 C (1T )

(

)

(b) Solve equation (1) for v:

v=

2πr T

Substitute numerical values and evaluate v:

v=

2π (0.5 m ) = 2.40 × 107 m/s 0.131 µs

(c) Express the kinetic energy of the alpha particle:

K = 12 mv 2

506 Chapter 26 K=

Substitute numerical values and evaluate K:

1 2

(6.65 ×10

−27

)(

kg 2.41× 107 m/s

= 1.93 × 10 −12 J ×

)

2

1eV 1.60 × 10 −19 J

= 12.0 MeV 78 •• Picture the Problem The configuration of the magnet and field are shown in the

r

figure. We’ll assume that a force + qm B is exerted on the north pole and a force r − qm B is exerted on the south pole and show that this assumption leads to the familiar expression for the torque acting on a magnetic dipole.

r

Assuming that a force + qm B is

− Bq m L Bqm L sin θ − sin θ 2 2 = Bq m L sin θ

τ=

exerted on the north pole and a force

r − qm B is exerted on the south pole,

express the net torque acting on the bar magnet: Substitute for qm to obtain:

r µ τ = B L sin θ = µB sin θ L or

r r r τ = µ× B

*79 •• r r r Picture the Problem We can use F = qv × B to show that motion of the particle in the x direction is not affected by the magnetic field. The application of Newton’s 2nd law to motion of the particle in yz plane will lead us to the result that r = mv0y /qB. By expressing the period of the motion in terms of v0y we can show that the time for one complete orbit around the helix is t = 2πm/qB. (a) Express the magnetic force acting on the particle:

r r r F = qv × B

The Magnetic Field 507 r

r

Substitute for v and B and simplify to obtain:

(

)

r F = q v0 x iˆ + v0 y ˆj × Biˆ

( )

( )

= qv0 x B iˆ × iˆ + qv0 y B ˆj × iˆ = 0 − qv0 y Bkˆ = −qv0 y Bkˆ

i.e., the motion in the direction of the magnetic field (the x direction) is not affected by the field. Apply

∑F

radial

= mac to the motion

of the particle in the plane

qv0 y B = m

v02y r

(1)

perpendicular to iˆ (i.e., the yz plane): Solve for r:

mv0 y

r=

qB 2πr v0 y

(b) Relate the time for one orbit around the helix to the particle’s orbital speed:

t=

Solve equation (1) for v0y:

v0 y =

Substitute and simplify to obtain:

t=

qBr m

2πr 2πm = qBr qB m

*80 •• Picture the Problem We can use a constant-acceleration equation to relate the velocity of the crossbar to its acceleration and Newton’s 2nd law to express the acceleration of the crossbar in terms of the magnetic force acting on it. We can determine the direction of r r r motion of the crossbar using a right-hand rule or, equivalently, by applying F = I l × B . We can find the minimum field B necessary to start the bar moving by applying a condition for static equilibrium to it. (a) Using a constant-acceleration equation, express the velocity of the bar as a function of its acceleration and the time it has been in motion:

v = v0 + at or, because v0 = 0,

v = at

508 Chapter 26 Use Newton’s 2nd law to express the acceleration of the rail:

a=

F m

where F is the magnitude of the magnetic force acting in the direction of the crossbar’s motion.

F t m

Substitute to obtain:

v=

Express the magnetic force acting on the current-carrying crossbar:

F = ILB

Substitute to obtain:

v=

(b)

ILB t m

Apply to conclude that the magnetic force is to the right and so the motion of the crossbar will also be to the right.

(c) Apply

∑F

x

= 0 to the crossbar:

ILBmin − f s,max = 0 or

ILBmin − µs mg = 0

Solve for Bmin to obtain:

Bmin =

µs mg IL

81 •• Picture the Problem Note that with the r rails tilted, F still points horizontally to

r

the right (I, and hence l , is out of the page). Choose a coordinate system in which down the incline is the positive x direction. Then we can apply a condition for translational equilibrium to find the r vertical magnetic field B is needed to keep the bar from sliding down the rails. In part (b) we can apply Newton’s 2nd law to find the acceleration of the crossbar when B is twice its value found in (a). (a) Apply

∑F

x

= 0 to the crossbar

mg sin θ − IlB cos θ = 0

The Magnetic Field 509 to obtain:

r mg mg tan θ uˆ v tan θ and B = − Il Il where uˆ v is a unit vector in the vertical

Solve for B:

B=

direction. (b) Apply

∑F

x

= ma to the

IlB' cos θ − mg sin θ = ma

crossbar to obtain:

IlB' cos θ − g sin θ m

Solve for a:

a=

Substitute B′ = 2B and simplify to obtain:

mg tan θ l I a= cos θ − g sin θ m = 2 g sin θ − g sin θ 2 Il

= g sin θ Note that the direction of the acceleration is up the slope. 82 •• Picture the Problem We’re being asked to show that, for small displacements from equilibrium, the bar magnet executes simple harmonic motion. To show its motion is SHM we need to show that the bar magnet experiences a linear restoring torque when displaced from equilibrium. We can accomplish this by applying Newton’s 2nd law in rotational form and using a small angle approximation to obtain the differential equation for simple harmonic motion. Once we have the DE we can identify ω and express f. Apply

∑τ = Iα to the bar magnet:

d 2θ − µB sin θ = I 2 dt where the minus sign indicates that the torque acts in such a manner as to align the magnet with the magnetic field and I is the moment of inertia of the magnet.

For small displacements from equilibrium, θ << 1 and:

sin θ ≈ θ

510 Chapter 26 Hence our differential equation of motion becomes:

d 2θ I 2 = − µBθ dt

Thus for small displacements from equilibrium we see that the differential equation describing the motion of the bar magnet is the differential equation of simple harmonic motion. Solve this equation for d2θ/dt2 to obtain:

d 2θ µB =− θ = −ω 2θ 2 dt I µB where ω = I

Relate f to ω to obtain:

f =

1 ω = 2π 2π

µB I

83 •• r r r Picture the Problem We can use F = qv × B to find the magnitude and direction of the magnetic force experienced by an electron in the conducting wire. In (b) we can use a r r condition for translational equilibrium to relate E to F . In (c) we can apply the definition of electric field in terms of potential difference to evaluate the difference in potential between the ends of the moving wire.

r r r F = qv × B = qviˆ × Bkˆ = qvB iˆ × kˆ = −qvBˆj

(a) Express the magnetic force on an electron in the conductor:

( )

r

Substitute numerical values and evaluate F :

r F = − − 1.60 × 10 −19 C (20 m/s )(0.5 T ) ˆj = 1.60 × 10 −18 N ˆj

(

(b) Sum the forces acting on an electron under steady-state conditions to obtain:

)

(

)

r r qE + F = 0

Solve for E :

r r F E=− q

Substitute our result in part (a) to obtain:

r 1.60 × 10−18 N ˆj E=− = (10.0 V/m ) ˆj − 1.60 × 10−19 C

r

(

)

The Magnetic Field 511 (c) The potential difference between the ends of the wire is:

∆V = E∆x = (10.0 V/m )(2 m ) = 20.0 V

84 ••• Picture the Problem We can use T = 2π I MgD to find the period of smalldisplacement oscillations with no current flowing in the frame. With a current flowing, the frame will experience an additional restoring torque that will reduce its period. In part (c) we can apply the condition for rotational equilibrium to find the magnitude of the current that will put the frame in equilibrium. (a) Express the period of a physical pendulum:

I MgD

T = 2π

(1)

where D is the distance from the pivot to the center of mass of the pendulum. Express the moment of inertia of the frame:

I = I hor. segment + 2 I vert. segment

(

= mhor. segment h 2 + 2 13 mver. segment h 2

)

where h = 10 cm. Using the linear density of the frame, calculate mhor. segment and mver. segment:

mhor. segment = λw

= (20 g/cm )(6 cm ) = 0.12 kg

and

mver. segment = λh

= (20 g/cm )(10 cm ) = 0.2 kg

Substitute and evaluate I:

I = (0.12 kg )(0.1m )

2

[

+ 2 13 (0.2 kg )(0.1m )

2

−3

= 2.53 × 10 kg ⋅ m

]

2

2(0.05 m )(0.2 kg ) + (0.1 m )(0.12 kg ) 0.12 kg + 0.2 kg + 0.2 kg = 6.15 cm

Evaluate the distance D to the center of mass from the A-A axis:

D=

Substitute in equation (1) and evaluate T:

2.53 × 10 −3 kg ⋅ m 2 T = 2π (0.52 kg ) 9.81m/s 2 (6.15 cm )

(

= 0.564 s

)

512 Chapter 26 (b) Express the restoring torque with r B and I as shown:

τ = (MgD + BIA)θ where A is the area of the loop and provided θ << 1 rad.

I MgD + BIA

Rewrite equation (1) with this restoring torque:

T' = 2π

Evaluate BIA:

BIA = (0.2 T )(8 A )(10 cm )(6 cm ) = 9.60 × 10− 3 N ⋅ m

Substitute numerical values and evaluate T′:

T = 2π

2.53 × 10 −3 kg ⋅ m 2 0.314 N ⋅ m + 9.60 × 10 −3 N ⋅ m

= 0.556 s (c) Apply

∑τ = 0 to the frame

MgD sin θ − BIA sin θ = 0

when it is in equilibrium to obtain:

MgD BA

Solve for I:

I=

Substitute numerical values and evaluate I:

( 0.52 kg ) (9.81 m/s 2 )(6.15 cm ) I= (0.2 T )(10 cm )(6 cm ) = 262 A

*85 ••• Picture the Problem We can use a constant-acceleration equation to express the height to which the wire rises in terms of its initial speed and the acceleration due to gravity. We can then use the impulse-change in momentum equation to express the initial speed of the wire in terms of the impulsive magnetic force acting on it. Finally, we can use the definition of current to relate the charge delivered by the battery to the time during which the impulsive force acts. Using a constant-acceleration equation, relate the height h to the initial and final speeds and the acceleration of the wire: Solve for h:

v 2 = v02 + 2a y h or, because v = 0 and ay = g,

0 = v02 − 2 gh

h=

v02 2g

(1)

The Magnetic Field 513 Use the impulse-momentum equation to relate the change in momentum of the wire to the impulsive force accelerating it: Express the impulsive (magnetic) force acting on the wire: Substitute to obtain: Solve for v0 and substitute in equation (1):

Use the definition of current to relate the charge delivered by the battery to the time during which it delivers the current: Substitute to obtain:

∆p = F∆t or pf − pi = F∆t and, because pi = 0, mv0 = F∆t

F = IlB

mv0 = IlB∆t ⎛ IlB∆t ⎞ ⎜ ⎟ (IlB∆t )2 m ⎠ h=⎝ = 2g 2m 2 g 2

∆Q = I∆t

2 ( lB∆Q ) h=

2m 2 g

Substitute numerical values and evaluate h:

2 ( [ 0.25 m )(0.4 T )(2 C )] h= = 2 2(0.02 kg ) (9.81 m/s 2 )

5.10 m

86 ••• Picture the Problem We’re being asked to show that, for small displacements from equilibrium, the circular loop executes simple harmonic motion. To show its motion is SHM we must show that the loop experiences a linear restoring torque when displaced from equilibrium. We can accomplish this by applying Newton’s 2nd law in rotational form and using a small angle approximation to obtain the differential equation for simple harmonic motion. Once we have the DE we can identify ω and express the period of the motion T. Apply

∑τ = Iα to the loop:

− IAB sin θ = I inertia

d 2θ dt 2

where the minus sign indicates that the torque acts in such a manner as to align the loop with the magnetic field and Iinertia is the moment of inertia of the loop.

514 Chapter 26 For small displacements from equilibrium, θ << 1 and:

sin θ ≈ θ

d 2θ = − IABθ dt 2

Hence, our differential equation of motion becomes:

I inertia

Thus for small displacements from equilibrium we see that the differential equation describing the motion of the current loop is the differential equation of simple harmonic motion. Solve this equation for d2θ/dt2 to obtain:

d 2θ IAB θ =− 2 dt I inertia .

Noting that the moment of inertia of a hoop about its diameter is 12 mR 2 ,

d 2θ IπR 2 B 2 IπB θ =− θ = −ω 2θ =− 1 2 2 dt m 2 mR

substitute for Iinertia and simplify to obtain:

where ω =

Relate the period T of the motion to ω and substitute to obtain:

T=

2π

ω

2πIB m

= 2π

m 2πIB

87 ••• r Picture the Problem We can express µ in terms of its components and calculate U from

r r r r r µ and B using U = − µ ⋅ B . Knowing U we can calculate the components of F using Fx

= −dU/dx and Fy = −dU/dy. Express the net force acting on the magnet in terms of its components:

r

Express µ in terms of its

r F = Fx iˆ + Fy ˆj

(1)

r µ = µ x iˆ + µ y ˆj + µ z kˆ

components: Express the potential energy of the bar magnetic in the nonuniform magnetic field:

r

Because µ is constant but

r B depends on x and y:

r r U = −µ⋅ B = − µ x iˆ + µ y ˆj + µ z kˆ ⋅ Bx (x ) iˆ + By ( y ) ˆj

(

)(

= − µ x Bx (x ) − µ y By ( y )

Fx = −

dU ⎛ ∂B ⎞ = µx ⎜ x ⎟ dx ⎝ ∂x ⎠

)

The Magnetic Field 515 and

Fy = −

Substitute in equation (1) to obtain:

⎛ ∂B ⎞ dU = µ y ⎜⎜ y ⎟⎟ dy ⎝ ∂y ⎠

r ∂B ∂B F = µ x x iˆ + µ y y ˆj ∂x ∂y

*88 ••• Picture the Problem We can apply Newton’s 2nd law to the particle to derive an expression for the radius of its orbit and then express its period in terms of its orbital speed and radius.

r

(a) Because B is perpendicular to r v , the magnitude of force on the particle is given by: Apply

∑ F = ma to the orbiting

particle to obtain: Solve for r:

F = qvB

v qvB = m(v )

2

r

r=

The period T of the particle’s motion is related to the radius r of its orbit and its orbital speed v:

T=

Substitute for r and simplify to obtain:

T=

v = γ (v )m

2

r

γ (v )mv qB 2πr v 2πγ (v )m qB

(b) A spreadsheet program to calculate r and T as functions of ln(γ) follows. The formulas used to calculate the quantities in the columns are given in the table. Cell B1 B2 B3 B4 A7 A8 B7 C7 D7

Content/Formula 9.11E−31 1.60E−19 10 3.00E+08 0.100 0.101 1/SQRT(1 − (A7)^2) LN(B7) B7*$B$1*A7*$B$4/($B$2*$B$3)

Algebraic Form m e B c v/c v/c + 0.001

E7

D7*10^8

106r

γ ln(γ) γ mv qB

516 Chapter 26 F7

A

2πγm × 1012 qB

(2*PI()*A7*$B$1/($B$2*$B$3))*10^12

B m= 9.11E−31 e= 1.60E−19 B= 10 c= 3.00E+08

1 2 3 4 5 6 7 8 9 10 11

v/c 0.100 0.101 0.102 0.103 0.104

903 904 905 906

0.996 0.997 0.998 0.999

C

D

E

F

gamma 1.0050 1.0051 1.0052 1.0053 1.0055

ln(gamma) 0.005 0.005 0.005 0.005 0.005

r 1.72E-05 1.73E-05 1.75E-05 1.77E-05 1.79E-05

r (microns) 17.2 17.3 17.5 17.7 17.9

T (ps) 0.358 0.361 0.365 0.368 0.372

11.1915 12.9196 15.8193 22.3663

2.415 2.559 2.761 3.108

1.90E-03 2.20E-03 2.70E-03 3.82E-03

1904.0 2200.2 2696.7 3816.6

3.563 3.567 3.570 3.574

kg C T m/s

The following graph of r as a function of ln(γ) was plotted using the data in columns C and E.

4000

r (microns)

3000

2000

1000

0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

ln(gamma)

The following graph of T as a function of ln(γ) was plotted using the data in columns C and F.

The Magnetic Field 517 4

T (ps)

3

2

1

0 0.0

0.5

1.0

1.5

2.0

ln(gamma)

2.5

3.0

3.5

518 Chapter 26

Chapter 27 Sources of the Magnetic Field Conceptual Problems *1 • Picture the Problem The electric forces are described by Coulomb’s law and the laws of attraction and repulsion of charges and are independent of the fact the charges are moving. The magnetic interaction is, on the other hand, dependent on the motion of the charges. Each moving charge constitutes a current that creates a magnet field at the location of the other charge. (a) The electric forces are repulsive; the magnetic forces are attractive (the two charges moving in the same direction act like two currents in the same direction). (b) The electric forces are again repulsive; the magnetic forces are also repulsive.

2 • No. The magnitude of the field depends on the location within the loop. 3 • Picture the Problem The field lines for the electric dipole are shown in the sketch to the left and the field lines for the magnetic dipole are shown in the sketch to the right. Note that, while the far fields (the fields far from the dipoles) are the same, the near fields (the fields between the two charges and inside the current loop/magnetic dipole) are not, and that, in the region between the two charges, the electric field is in the opposite direction to that of the magnetic field at the center of the magnetic dipole. It is especially important to note that while the electric field lines begin and terminate on electric charges, the magnetic field lines are continuous, i.e., they form closed loops.

4 • Determine the Concept Applying the right-hand rule to the wire to the left we see that the magnetic field due to its current is out of the page at the midpoint. Applying the righthand rule to the wire to the right we see that the magnetic field due to its current is out of

519

520 Chapter 27 the page at the midpoint. Hence, the sum of the magnetic fields is out of the page as well.

(c) is correct. 5 • Determine the Concept While we could express the force wire 1 exerts on wire 2 and compare it to the force wire 2 exerts on wire 1 to show that they are the same, it is simpler to recognize that these are action and reaction forces. (a ) is correct. *6 • Determine the Concept Applying the right-hand rule to the wire to the left we see that the magnetic field due to the current points to west at all points north of the wire.

(c) is correct. 7 • Determine the Concept At points to the west of the vertical wire, the magnetic field due to its current exerts a downward force on the horizontal wire and at points to the east it exerts an upward force on the horizontal wire. Hence, the net magnetic force is zero and

(e) is correct. 8 • Picture the Problem The field-line sketch follows. An assumed direction for the current in the coils is shown in the diagram. Note that the field is stronger in the region between the coaxial coils and that the field lines have neither beginning nor ending points as do electric-field lines. Because there are an uncountable infinity of lines, only a representative few have been shown.

*9 • Picture the Problem The field-line sketch is shown below. An assumed direction for the current in the coils is shown in the diagram. Note that the field lines never begin or end and that they do not touch or cross each other. Because there are an uncountable infinity of lines, only a representative few have been shown.

Sources of the Magnetic Field 521

10 • Determine the Concept Because all of these statements regarding Ampère’s law are true,

(e) is correct. 11 • (a) True (b) True *12 • Determine the Concept The magnetic susceptibility χm is defined by the

r r r r Bapp equation M = χ m , where M is the magnetization vector and Bapp is the applied

µ0

magnetic field. For paramagnetic materials, χm is a small positive number that depends on temperature, whereas for diamagnetic materials, it is a small negative constant independent of temperature. (a ) is correct. 13 • (a) False. The magnetic field due to a current element is perpendicular to the current element. (b) True (c) False. The magnetic field due to a long wire varies inversely with the distance from the wire. (d) False. Ampère’s law is easier to apply if there is a high degree of symmetry, but is valid in all situations. (e) True

522 Chapter 27 14 • Determine the Concept Yes. The classical relation between magnetic moment and

r

angular momentum is µ =

q r L . Thus, if its charge density is zero, a particle with 2m

angular momentum will not have a magnetic moment. 15 • Determine the Concept No. The classical relation between magnetic moment and

r

angular momentum is µ =

q r L . Thus, if the angular momentum of the particle is zero, 2m

its magnetic moment will also be zero. 16 • Determine the Concept Yes, there is angular momentum associated with the magnetic r moment. The magnitude of L is extremely small, but very sensitive experiments have demonstrated its presence (Einstein-de Haas effect). 17 • Determine the Concept From Ampère’s law, the current enclosed by a closed path within the tube is zero, and from the cylindrical symmetry it follows that B = 0 everywhere within the tube. *18 • Determine the Concept The force per unit length experienced by each segment of the wire, due to the currents in the other segments of the wire, will be equal. These equal forces will result in the wire tending to form a circle. 19 • Determine the Concept H2, CO2, and N2 are diamagnetic (χm < 0); O2 is paramagnetic (χm > 0).

Estimation and Approximation 20 •• Picture the Problem We can use the definition of the magnetization of the earth’s core to find its volume and radius. (a) Express the magnetization of the earth’s core in terms of the magnetic moment of the earth and the volume of the core:

M =

µ V

Sources of the Magnetic Field 523 Solve for and evaluate V:

µ

V =

M

=

9 × 1022 A ⋅ m 2 1.5 × 109 A/m

= 6.00 × 1013 m3 (b)Assuming a spherical core centered with the earth:

V = 43 πr 3

Solve for r:

r=3

3V 4π

Substitute numerical values and evaluate r:

r=3

3(6 × 1013 m 3 ) = 2.43 × 104 m 4π

*21 •• Picture the Problem We can model the lightning bolt as a current in a long wire and use the expression for the magnetic field due to such a current to estimate the transient magnetic field 100 m from the lightning bolt. The magnetic field due to the current in a long, straight wire is:

B=

µ0 2I 4π r

where r is the distance from the wire. Assuming that the height of the cloud is 1 km, the charge transfer will take place in roughly 10−3 s and the current associated with this discharge is: Substitute numerical values and evaluate B:

I=

∆Q 30 C = −3 = 3 × 10 4 A ∆t 10 s

B=

4π × 10 −7 N/A 2 2(3 × 10 4 A ) 4π 100 m

= 60.0 µT *22 •• Picture the Problem A rotating disk with total charge Q and surface charge density σ is shown in the diagram. We can find Q by deriving an expression for the magnetic field B at the center of the disk due to its rotation. We’ll use Ampere’s law to express the field dB at the center of the disk due to the element of current dI and then integrate over r to find B.

524 Chapter 27

µ0 I

Applying Ampere’s law to a circular current loop of radius r we obtain:

B=

The B field at the center of an annular ring on a rotating disk of radius r and thickness dr is:

dB =

If σ represents the surface charge density, then the current in the annular ring is given by: Because T =

2π

ω

2r

µ0

dI

2r

(1)

σ (2π r )

dI =

T

dr, where σ =

Q πR 2

dI = σω rdr

:

Substitute for dI in equation (1) to obtain:

dB =

Integrate from r = 0 to R to obtain:

B=

Substitution for σ yields:

B=

Solve for Q to obtain:

Q=

µ0 2r

σω rdr =

µ 0σω 2

R

∫ dr = 0

µ 0σω 2

dr

µ 0σωR 2

⎛ Q ⎞ ωR 2 ⎟ µ Qω ⎝ πR ⎠ = 0 2 2π R

µ0 ⎜

2π RB

µ 0ω

Substitute numerical values and evaluate Q:

Q=

(

)

2π 10 7 m (0.1 T ) = 5.00 × 1014 C −7 −2 2 4π × 10 N/A 10 rad/s

(

)(

)

The electric field above the sunspot is given by:

E=

Substitute numerical values and evaluate E:

E=

σ 2 ∈0

(

=

Q 2π ∈0 R 2 5.00 × 1014 C

)(

2π 8.85 × 10 −12 C 2 / N ⋅ m 2 10 7 m

= 90.0 GN/C

)

2

Sources of the Magnetic Field 525

The Magnetic Field of Moving Point Charges 23 • r Picture the Problem We can substitute for v and q in the equation describing the

r µ0 qvr × rˆ ), evaluate r and rˆ for magnetic field of the moving charged particle ( B = 4π r 2 r each of the given points of interest, and substitute to find B . Express the magnetic field of the moving charged particle:

r µ 0 qvr × rˆ B= 4π r 2

(

)

= 10 −7 N/A 2 (12 µC )

(

= 36.0 pT ⋅ m 2 (a) Find r and rˆ for the particle at (0, 2 m) and the point of interest at the origin:

r

Substitute and evaluate B(0,0 ) :

(30 m/s) iˆ × rˆ r2

) i r× rˆ ˆ

2

r r = −(2 m ) ˆj , r = 2 m , and rˆ = − ˆj

) ( )

r iˆ × − ˆj B (0,0 ) = 36.0 pT ⋅ m 2 (2 m )2

(

= − (9.00 pT )kˆ (b) Find r and rˆ for the particle at (0, 2 m) and the point of interest at (0, 1 m):

r

Substitute and evaluate B (0,1 m ) :

r r = −(1 m ) ˆj , r = 1 m , and rˆ = − ˆj

) ( )

r iˆ × − ˆj B (0,1 m ) = 36.0 pT ⋅ m 2 (1m )2

(

= − (36.0 pT )kˆ (c) Find r and rˆ for the particle at (0, 2 m) and the point of interest at (0, 3 m):

r

Substitute and evaluate B (0,3 m ) :

r r = (1m ) ˆj , r = 1m , and rˆ = ˆj

r iˆ × ˆj B (0,3 m ) = 36.0 pT ⋅ m 2 (1m )2

(

=

)

(36.0 pT )kˆ

526 Chapter 27 (d) Find r and rˆ for the particle at (0, 2 m) and the point of interest at (0, 4 m):

r

Substitute and evaluate B (0,4 m ) :

r r = (2 m ) ˆj , r = 2 m , and rˆ = ˆj

r iˆ × ˆj B (0,4 m ) = 36.0 pT ⋅ m 2 (2 m )2

(

)

(9.00 pT )kˆ

=

24 • r Picture the Problem We can substitute for v and q in the equation describing the

r µ0 qvr × rˆ ), evaluate r and rˆ for magnetic field of the moving charged particle ( B = 4π r 2 r each of the given points of interest, and substitute to find B . The magnetic field of the moving charged particle is given by:

r µ 0 qvr × rˆ B= 4π r 2

= (10 −7 N/A 2 )(12 µC ) = (36.0 pT ⋅ m 2 )

(30 m/s) iˆ × rˆ r2

iˆ × rˆ r2

(a) Find r and rˆ for the particle at (0, 2 m) and the point of interest at (1 m, 3 m):

r r = (1 m ) iˆ + (1 m ) ˆj , r = 2 m , and 1 ˆ 1 ˆ rˆ = i+ j 2 2

Substitute for rˆ and evaluate

r B (1 m,3 m ) = 36.0 pT ⋅ m 2

r B (1m,3 m ) :

(

)

⎛ 1 ˆ 1 iˆ × ⎜ i+ 2 2 ⎝ × 2 2m

(

=

(36.0 pT ⋅ m )

)

2

2

ˆj ⎞⎟ ⎠

(

kˆ 2m

= (12.7 pT )kˆ (b) Find r and rˆ for the particle at (0, 2 m) and the point of interest at (2 m, 2 m):

r r = (2 m ) iˆ , r = 2 m , and rˆ = iˆ

)

2

Sources of the Magnetic Field 527 r iˆ × iˆ B (2 m,2 m ) = 36.0 pT ⋅ m 2 (2 m )2

(

Substitute for rˆ and evaluate r B (2 m,2 m ) :

)

= 0 r r = (2 m ) iˆ + (1 m ) ˆj , r = 5 m , and 2 ˆ 1 ˆ rˆ = i+ j 5 5

(c) Find r and rˆ for the particle at (0, 2 m) and the point of interest at (2 m, 3 m):

r

Substitute for rˆ and evaluate B (2 m,3 m ) :

⎛ 2 ˆ 1 iˆ × ⎜ i+ r 5 5 ⎝ 2 B (2 m,3 m ) = 36.0 pT ⋅ m 2 5m

(

)

(

)

ˆj ⎞⎟ ⎠=

(3.22 pT )kˆ

25 • r Picture the Problem We can substitute for v and q in the equation describing the

r µ0 qvr × rˆ ), evaluate r and rˆ for each of the magnetic field of the moving proton ( B = 4π r 2 r given points of interest, and substitute to find B. The magnetic field of the moving proton is given by:

) [(

) (

)]

r µ 0 qvr × rˆ 10 4 m/s iˆ + 2 × 10 4 m/s ˆj × rˆ 2 −7 −19 B= = 10 N/A 1.60 × 10 C r2 4π r 2 iˆ + 2 ˆj × rˆ = 1.60 × 10 −22 T ⋅ m 2 r2

(

)(

)(

(

)

r r = −(1 m ) iˆ − (2 m ) ˆj , r = 5 m , and 1 ˆ 2 ˆ rˆ = − i− j 5 5

(a) Find r and rˆ for the proton at (3 m, 4 m) and the point of interest at (2 m, 2 m):

r

Substitute for rˆ and evaluate B (1 m,3 m ) :

r B (1 m,3 m ) = 1.60 × 10 −22 T ⋅ m 2

(

=

(1.60 × 10

− 22

5

T ⋅ m2

)

(iˆ + 2 ˆj )× ⎛⎜ −

)

⎝

1 ˆ 2 ˆ⎞ i− j⎟ 5 5 ⎠

r2 ⎡ − 2kˆ + 2kˆ ⎤ ⎢ ⎥= 0 2 ⎢⎣ 5 m ⎥⎦

(

)

528 Chapter 27 r r = (3 m ) iˆ , r = 3 m , and rˆ = iˆ

(b) Find r and rˆ for the proton at (3 m, 2 m) and the point of interest at (6 m, 4 m):

r

Substitute for rˆ and evaluate B (6 m,4 m ) :

r B (6 m,4 m ) = 1.60 × 10 −22 T ⋅ m 2

(

(

) (i + 2 j )× i = (1.60 × 10 ˆ

ˆ

ˆ

(3 m ) 2

)

− 22

⎛ − 2kˆ ⎞ ⎟ T ⋅ m 2 ⎜⎜ 2 ⎟ ⎝9m ⎠

)

= − 3.56 × 10 −23 T kˆ r r = (2 m ) ˆj , r = 2 m , and rˆ = ˆj

(c) Find r and rˆ for the proton at (3 m, 4 m) and the point of interest at the (3 m, 6 m):

r

Substitute for rˆ and evaluate B (3 m,6 m ) :

)(

)

r iˆ + 2 ˆj × ˆj B (3 m,6 m ) = 1.60 × 10 −22 T ⋅ m 2 = 1.60 × 10 −22 T ⋅ m 2 2 (2 m )

(

=

(4.00 × 10

− 23

(

)

ˆ ⎞ ⎟ 2 ⎟ ⎠

)⎛⎜⎜ 4 mk ⎝

T kˆ

26 • Picture the Problem The centripetal force acting on the orbiting electron is the Coulomb force between the electron and the proton. We can apply Newton’s 2nd law to the electron to find its orbital speed and then use the expression for the magnetic field of a moving charge to find B. Express the magnetic field due to the motion of the electron: Apply

∑F

radial

= mac to the

electron:

B=

µ0 ev 4π r 2

v2 ke 2 = m r r2

Solve for v to obtain:

v=

ke 2 mr

Substitute and simplify to obtain:

B=

µ0 e 4π r 2

ke 2 µ 0 e 2 = mr 4πr 2

k mr

Sources of the Magnetic Field 529 Substitute numerical values and evaluate B:

(10 B=

−7

)(

N/A 2 1.6 × 10 −19 C

(5.29 × 10

−11

m

)

2

)

2

8.99 × 109 N ⋅ m 2 / C 2 = 12.5 T 9.11 × 10 −31 kg 5.29 × 10 −11 m

(

)(

)

*27 •• Picture the Problem We can find the ratio of the magnitudes of the magnetic and electrostatic forces by using the expression for the magnetic field of a moving charge and r r Coulomb’s law. Note that v and r , where r is the vector from one charge to the other, r are at right angles. The field B due to the charge at the origin at the location (0, b, 0) is r perpendicular to v and r . Express the magnitude of the magnetic force on the moving charge at (0, b, 0):

Express the magnitude of the repulsive electrostatic interaction between the two charges: Express the ratio of FB to FE and simplify to obtain:

FB = qvB =

µ0 q 2v 2 4π b 2

and, applying the right hand rule, we find that the direction of the force is toward the charge at the origin; i.e., the magnetic force between the two moving charges is attractive.

1 q2 FE = 4πε 0 b 2

µ0 q 2v 2 v2 FB 4π b 2 2 v = = ε µ = 0 0 2 2

FE

1 q 4πε 0 b 2

c

where c is the speed of light in a vacuum.

The Magnetic Field of Currents: The Biot-Savart Law 28 • r Picture the Problem We can substitute for v and q in the Biot-Savart relationship

r r µ0 Id l × rˆ ), evaluate r and rˆ for each of the points of interest, and substitute to ( dB = 4π r 2 r find dB .

530 Chapter 27 Express the Biot-Savart law for the given current element:

r r µ 0 Id l × rˆ dB = 4π r 2

(

= 10 −7 N/A 2

) (2 A )(2rmm)k × rˆ

(

= 0.400 nT ⋅ m 2 (a) Find r and rˆ for the point whose coordinates are (3 m, 0, 0):

r

Evaluate dB at (3 m, 0, 0):

r

Evaluate dB at (−6 m, 0, 0):

) kr× rˆ ˆ

2

r r = (3 m ) iˆ , r = 3 m , and rˆ = iˆ

r kˆ × iˆ dB (3 m,0,0) = 0.400 nT ⋅ m 2 (3 m )2

(

)

(44.4 pT ) ˆj

= (b) Find r and rˆ for the point whose coordinates are (−6 m, 0, 0):

ˆ

2

r r = −(6 m ) iˆ , r = 6 m , and rˆ = −iˆ

) ( )

r kˆ × − iˆ dB (− 6 m,0,0) = 0.400 nT ⋅ m 2 (6 m )2

(

= − (11.1pT ) ˆj (c) Find r and rˆ for the point whose coordinates are (0, 0, 3 m):

r

Evaluate dB at (0, 0, 3 m):

r r = (3 m )kˆ , r = 3 m , and rˆ = kˆ

r kˆ × kˆ dB (0,0, 3 m ) = 0.400 nT ⋅ m 2 (3 m )2

(

)

= 0 (d) Find r and rˆ for the point whose coordinates are (0, 3 m, 0):

r r = (3 m ) ˆj , r = 3 m , and rˆ = ˆj

Sources of the Magnetic Field 531 r

Evaluate dB at (0, 3 m, 0):

r kˆ × ˆj dB (0,3 m,0) = 0.400 nT ⋅ m 2 (3 m )2

(

)

= − (44.4 pT ) iˆ 29 • r Picture the Problem We can substitute for v and q in the Biot-Savart relationship

r r µ0 Id l × rˆ r ), evaluate r and rˆ for (0, 3 m, 4 m), and substitute to find dB. ( dB = 2 4π r Express the Biot-Savart law for the given current element:

r r µ 0 Id l × rˆ dB = 4π r 2

(

= 10 −7 N/A 2

) (2 A )(2rmm)k × rˆ ˆ

2

(

= 0.400 nT ⋅ m 2 Find r and rˆ for the point whose coordinates are (0, 3 m, 4 m):

) kr× rˆ ˆ

2

r r = (3 m ) ˆj + (4 m )kˆ , r = 5m , and

rˆ =

3ˆ 4ˆ j+ k 5 5

r

Evaluate dB at (3 m, 0, 0):

4 ⎞ ⎛3 kˆ × ⎜ ˆj + kˆ ⎟ r 5 5 ⎠ dB (3 m,0,0) = (0.400 nT ⋅ m 2 ) ⎝ = − (9.60 pT ) iˆ 2 (5 m ) *30 • r Picture the Problem We can substitute for v and q in the Biot-Savart relationship

r r r µ0 Id l × rˆ ˆ r ), evaluate r and for the given points, and substitute to find d B . ( dB = 4π r 2 Express the Biot-Savart law for the given current element:

r r µ 0 Id l × rˆ dB = 4π r 2

(

= 10 −7 N/A 2

(

) (2 A )(2rmm)k × rˆ

= 0.400 nT ⋅ m 2

ˆ

2

) kr× rˆ ˆ

2

532 Chapter 27

r r = (2 m ) iˆ + (4 m ) ˆj ,

(a) Find r and rˆ for the point whose coordinates are (2 m, 4 m, 0):

r = 2 5m, and

rˆ =

2 ˆ 4 ˆ 1 ˆ 2 ˆ i+ j= i+ j 2 5 2 5 5 5

r

Evaluate dB at (2 m, 4 m, 0):

r dB (2 m,4 m,0) = 0.400 nT ⋅ m 2

(

)

⎛ 1 ˆ 2 kˆ × ⎜ i+ 5 ⎝ 5

(2

5m

)

2

ˆj ⎞⎟ ⎠ = − (17.9 pT ) iˆ + (8.94 pT ) ˆj

The diagram is shown to the right:

r r = (2 m ) iˆ + (4 m )kˆ ,

(b) Find r and rˆ for the point whose coordinates are (2 m, 0, 4 m):

r = 2 5m, and

rˆ =

2 ˆ 4 ˆ 1 ˆ 2 ˆ i+ k= i+ k 2 5 2 5 5 5

r

Evaluate dB at (2 m, 0, 4 m):

r dB (2 m,0,4 m ) = (0.400 nT ⋅ m 2 )

⎛ 1 ˆ 2 ˆ⎞ kˆ × ⎜ i+ k⎟ 5 ⎠ ⎝ 5 = (8.94 pT ) ˆj 2 2 5m

(

)

The diagram is shown to the right:

r B Due to a Current Loop 31

•

Picture the Problem We can use Bx =

µ0 2πR 2 I to find B on the axis of the 4π (x 2 + R 2 )3 2

Sources of the Magnetic Field 533 current loop. Express B on the axis of a current loop: Substitute numerical values to obtain:

Bx =

µ0 2πR 2 I 4π (x 2 + R 2 )3 2

(

−7

Bx = 10 N/A =

2

)

2π (0.03 m ) (2.6 A ) 2

(x

+ (0.03 m )

−9

1.47 × 10 T ⋅ m

(x

2

)

2 32

2

3

)

+ (0.03 m )

2 32

1.47 × 10−9 T ⋅ m 3

(a) Evaluate B at the center of the loop:

B(0) =

(b) Evaluate B at x = 1 cm:

B(0.01 m ) =

(0 + (0.03 m) )

2 32

= 54.5 µT

1.47 × 10 −9 T ⋅ m 3

((0.01m) + (0.03 m) ) 2

2 32

= 46.5 µT (c) Evaluate B at x = 2 cm:

B(0.02 m ) =

1.47 × 10 −9 T ⋅ m 3

((0.02 m) + (0.03 m) ) 2

2 32

= 31.4 µT (d) Evaluate B at x = 35 cm:

B(0.35 m ) =

1.47 × 10 −9 T ⋅ m 3

((0.35 m) + (0.03 m) ) 2

2 32

= 33.9 nT *32 •

µ0 2πR 2 I for I with x = 0 and substitute Picture the Problem We can solve Bx = 4π (x 2 + R 2 )3 2 the earth’s magnetic field at the equator to find the current in the loop that would produce a magnetic field equal to that of the earth. Express B on the axis of the current loop:

µ0 2πR 2 I Bx = 4π (x 2 + R 2 )3 2

Solve for I with x = 0:

I=

4π R B µ 0 2π x

534 Chapter 27 Substitute numerical values and evaluate I: 3 ⎞ ⎛ ( 1 0 .1 m ) (0.7 G )⎜⎜ 14T ⎟⎟ = 11.1 A I= 2 2 −7 (10 N/A ) 2π (0.1 m) ⎝ 10 G ⎠

The orientation of the loop and current is shown in the sketch:

33

••

Picture the Problem We can solve Bx =

µ0 2πR 2 I for B0, express 4π (x 2 + R 2 )3 2

the ratio of Bx to B0, and solve the resulting equation for x. Express B on the axis of the current loop:

Bx =

µ 0 2πR 2 I 4π (x 2 + R 2 )3 2

Evaluate Bx for x = 0:

B0 =

µ 0 2πI 4π R

Express the ratio of Bx to B0:

Solve for x to obtain:

µ 0 2πR 2 I 32 Bx 4π (x 2 + R 2 ) R3 = = µ 0 2πI B0 (x 2 + R 2 )3 2 4π R ⎛B ⎞ x = R ⎜⎜ 0 ⎟⎟ ⎝ Bx ⎠

23

−1

(1)

(a) Evaluate equation (1) for Bx = 0. 1B0:

⎛ B ⎞ x = 10 cm ⎜⎜ 0 ⎟⎟ ⎝ 0.1B0 ⎠

23

(b) Evaluate equation (1) for Bx = 0. 01B0:

⎛ B0 x = 10 cm ⎜⎜ ⎝ 0.01B0

⎞ ⎟⎟ ⎠

− 1 = 19.1 cm

23

− 1 = 45.3 cm

Sources of the Magnetic Field 535 (a) Evaluate equation (1) for Bx = 0. 001B0:

34

⎛ B0 ⎞ ⎟⎟ x = 10 cm ⎜⎜ 0 . 001 B 0 ⎠ ⎝

23

− 1 = 99.5 cm

••

Picture the Problem We can solve Bx =

µ0 2πR 2 I for I with x = 0 and substitute 4π (x 2 + R 2 )3 2

the earth’s magnetic field at the equator to find the current in the loop that would produce a magnetic field equal to that of the earth. Express B on the axis of the current loop:

Bx =

Solve for I with x = 0 and Bx = BE:

I=

µ0 2πR 2 I 4π (x 2 + R 2 )3 2

4π R BE µ 0 2π

Substitute numerical values and evaluate I:

I=

⎛ ⎞ 1 0.085 m (0.7 G )⎜⎜ 14T ⎟⎟ = 9.47 A 2 10 N/A 2π ⎝ 10 G ⎠

(

−7

)

The normal to the plane of the loop must be in the direction of the earth’s field, and the current must be counterclockwise as seen from

r

above. Here BE denotes the earth’s

r

field and BI the field due to the current in the coil. 35 •• Picture the Problem We can use the expression for the magnetic field on the axis of a current loop and the expression for the electric field on the axis of ring of charge Q to plot graphs of Bx/B0 and E(x)/(kQ/R2) as functions of x/R. (a) Express Bx on the axis of a current loop:

Bx =

µ 0 2πR 2 I 4π (x 2 + R 2 )3 2

Express B0 at the center of the loop:

B0 =

µ 0 2πR 2 I µ 0 I = 4π (R 2 )3 2 2 R

536 Chapter 27 Bx 1 = 32 B0 ⎛ x2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

Express the ratio of Bx to B0 and simplify to obtain:

The graph of Bx/B0 as a function of x/R shown below was plotted using a spreadsheet program:

1.0

0.8

Bx /B 0

0.6

0.4

0.2

0.0 -5

-4

-3

-2

-1

0

1

2

3

4

5

x /R

Express Ex on the axis due to a ring of radius R carrying a total charge Q:

Divide both sides of this equation by kQ/R2 to obtain:

E (x ) =

(R

kQx 2

+ x2

)

32

x kQ R = 2 32 R ⎛ x2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

x E (x ) R = 32 kQ ⎛ x2 ⎞ ⎜⎜1 + 2 ⎟⎟ R2 ⎝ R ⎠

The graph of Ex as a function of x/R shown below was plotted using a spreadsheet program. Here E(x) is normalized, i.e., we’ve set kQ/R2 = 100.

Sources of the Magnetic Field 537 1.0 0.8 0.6 0.4 0.2

E x 0.0 -0.2 -0.4 -0.6 -0.8 -1.0 -5

-4

-3

-2

-1

0

1

2

3

4

5

x /R

(b) Express the magnetic field on the x axis due to the loop centered at x = 0:

B1 ( x ) = =

(

µ0 R 2 I

2 x2 + R2 µ0 I

)

3 2

⎛ x2 ⎞ 2 R⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

3 2

where N is the number of turns. Because B0 =

µ0 I 2R

:

B1 (x ) =

B0 ⎛ x2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

3 2

or 2 B1 ( x ) ⎡ ⎛ x ⎞ ⎤ = ⎢1 + ⎜ ⎟ ⎥ B0 ⎣⎢ ⎝ R ⎠ ⎦⎥

Express the magnetic field on the x axis due to the loop centered at x = R:

B2 (x ) =

[

−3 2

µ0 R 2 I

2 (R − x ) + R 2 2

]

3 2

538 Chapter 27 B2 ( x ) =

Simplify this expression to obtain:

=

=

[

µ0 R 2 I

2 (R − x ) + R 2 µ0 I 2

]

3 2

2 ⎡⎛ ⎤ x⎞ 2 R ⎢⎜1 − ⎟ + 1⎥ ⎣⎢⎝ R ⎠ ⎦⎥

32

B0 2 ⎡⎛ ⎤ x⎞ ⎢⎜1 − ⎟ + 1⎥ ⎥⎦ ⎣⎢⎝ R ⎠

32

or 2 ⎤ B2 ( x ) ⎡⎛ x⎞ = ⎢⎜1 − ⎟ + 1⎥ B0 ⎢⎣⎝ R ⎠ ⎥⎦

−3 2

The graphs of B1/B0, B2/B0, and B1/B0 + B2/B0 as functions of x/R with the second loop displaced by d = R from the center of the first loop along the x axis shown below were plotted using a spreadsheet program.

1.6 1.4

B1/B0

1.2

B2/B0 B1/B0 + B2/B0

1.0 0.8 0.6 0.4 0.2 0.0 -3

-2

-1

0

1

2

3

x /R

Note that, midway between the two loops, dB(x)/dx = 0. Also, when d = R, B(x) is nearly flat at the midpoint which shows that in the region midway between the two coils B(x) is nearly constant. 36 •• Picture the Problem Let the origin be midway between the coils so that one of them is centered at x = −r/2 and the other is centered at x = r/2. Let the numeral 1 denote the coil centered at x = −r/2 and the numeral 2 the coil centered at x = r/2. We can express the magnetic field in the region between the coils as the sum of the magnetic fields B1 and B2 due to the two coils.

Sources of the Magnetic Field 539 B1 ( x ) =

Express the magnetic field on the x axis due to the coil centered at x = −r/2:

µ 0 Nr 2 I 2 ⎡⎛ r ⎤ ⎞ 2 ⎢⎜ + x ⎟ + r 2 ⎥ ⎠ ⎣⎢⎝ 2 ⎦⎥

32

where N is the number of turns.

B2 (x ) =

Express the magnetic field on the x axis due to the coil centered at x = r/2:

µ 0 Nr 2 I 2 ⎡⎛ r ⎤ ⎞ 2 ⎢⎜ − x ⎟ + r 2 ⎥ ⎠ ⎣⎢⎝ 2 ⎦⎥

32

Add these equations to express the total magnetic field along the x axis:

Bx ( x ) = B1 ( x ) + B2 ( x ) =

=

µ 0 Nr 2 I 2 ⎡⎛ r ⎤ ⎞ 2 ⎢⎜ + x ⎟ + r 2 ⎥ ⎠ ⎣⎢⎝ 2 ⎦⎥

µ 0 Nr 2 I ⎛⎜ ⎡⎛ r 2

2 ⎤ ⎞ 2 x r + + ⎜ ⎟ ⎢ ⎥ ⎜⎜ ⎝ 2 ⎠ ⎢ ⎥⎦ ⎣ ⎝

−3 2

32

+

µ 0 Nr 2 I 2 ⎡⎛ r ⎤ ⎞ 2 ⎢⎜ − x ⎟ + r 2 ⎥ ⎠ ⎣⎢⎝ 2 ⎦⎥

2 ⎡⎛ r ⎤ ⎞ + ⎢⎜ − x ⎟ + r 2 ⎥ ⎠ ⎢⎣⎝ 2 ⎥⎦

−3 2

32

⎞ ⎟ ⎟⎟ ⎠

The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B3 B5

Formula/Content 1.13×10−7 0.30 250 15 0.5*$B$1*$B$3*($B$2^2)*$B$4

A8 B8

−0.30 $B$5*(($B$2/2+A8)^2+$B$2^2)^(−3/2)

Algebraic Form

µ0 r N I

Coeff =

−r

$B$5* (($B$2/2−A8)^2+$B$2^2)^(−3/2)

1 2 3

C N/A^2 m turns

2 ⎤ ⎞ 2 ⎢⎜ − x ⎟ + r ⎥ ⎠ ⎣⎢⎝ 2 ⎦⎥

Bx = 10 4 (B1 + B2 )

10^4(B8+C8) A B mu_0= 1.26E-06 r= 0.3 N= 250

2 ⎤ ⎞ 2 ⎢⎜ + x ⎟ + r ⎥ ⎠ ⎣⎢⎝ 2 ⎦⎥

µ 0 Nr 2 I ⎡⎛ r 2

D8

2

µ 0 Nr 2 I ⎡⎛ r 2

C8

µ 0 Nr 2 I

D

−3 2

−3 2

540 Chapter 27 4 I= 15 A 5 Coeff= 2.13E−04 6 7 x B_1 B_2 B(x) 8 −0.30 5.63E−03 1.34E−03 70 9 −0.29 5.86E−03 1.41E−03 73 10 −0.28 6.08E−03 1.48E−03 76 11 −0.27 6.30E−03 1.55E−03 78 12 −0.26 6.52E−03 1.62E−03 81 13 −0.25 6.72E−03 1.70E−03 84 14 −0.24 6.92E−03 1.78E−03 87 15 −0.23 7.10E−03 1.87E−03 90 61 62 63 64 65 66 67 68

0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30

1.87E−03 1.78E−03 1.70E−03 1.62E−03 1.55E−03 1.48E−03 1.41E−03 1.34E−03

7.10E−03 6.92E−03 6.72E−03 6.52E−03 6.30E−03 6.08E−03 5.86E−03 5.63E−03

90 87 84 81 78 76 73 70

The following graph of Bx as a function of x was plotted using the data in the above table. 120 100

B x (G)

80 60 40 20 0 -0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

x (m)

The maximum value of Bx is 113 G. Twenty percent of this maximum value is 23 G. Referring to the table of values we see that the field is within 20 percent of 113 G in the interval − 0.23 m < x < 0.23 m. 37 ••• Picture the Problem Let the numeral 1 denote the coil centered at the origin and the numeral 2 the coil centered at x = R. We can express the magnetic field in the region between the coils as the sum of the magnetic fields due to the two coils and then evaluate

Sources of the Magnetic Field 541 the derivatives at x = R/2.

B1 ( x ) =

Express the magnetic field on the x axis due to the coil centered at x = 0:

(

µ 0 NR 2 I

2 x2 + R2

)

32

where N is the number of turns.

B2 ( x ) =

Express the magnetic field on the x axis due to the coil centered at x = R:

µ0 NR 2 I

[

2 (x − R ) + R 2 2

]

32

Add these equations to express the total magnetic field along the x axis:

Bx ( x ) = B1 ( x ) + B2 ( x ) = =

µ0 NR 2 I ⎛⎜

(

µ0 NR 2 I

2 x2 + R2 1

(

⎜ x2 + R2 ⎝

2

+

)

32

)

32

+

[

2 (x − R ) + R 2 2

]

32

⎞ ⎟ 2 2 32 ⎟ (x − R ) + R ⎠ 1

[

x1 ( 12 R ) =

Evaluate x1 and x2 at x = R/2:

µ0 NR 2 I

]

1 4

R2 + R2 =

(

5 4

R2

and

x2 ( 12 R ) =

( 12 R − R )2 + R 2

)

12

= (54 R 2 )

dBx µ 0 NR 2 I d ⎛ 1 1 ⎞ ⎜ + ⎟ = dx dx ⎜⎝ x13 x23 ⎟⎠ 2

Differentiate Bx with respect to x to obtain:

=

µ 0 NR 2 I ⎛ x 2

x−R⎞ ⎜⎜ 5 + 5 ⎟⎟ x2 ⎠ ⎝ x1

Evaluate dBx/dx at x = R/2 to obtain:

dBx dx

= x = 12 R

µ 0 NR 2 I ⎛⎜ 2

⎜ ⎝

(

1 2 5 4

R

R

)

2 52

+

(

− 12 R ⎞⎟ = 0 2 52 ⎟ 5 R 4 ⎠

)

Differentiate dBx/dx with respect to x to obtain: 2 d 2 Bx µ0 NR 2 I d ⎛ x x − R ⎞ µ0 NR 2 I ⎛ 1 1 5 x 2 5( x − R ) ⎞ ⎟ ⎟ ⎜ ⎜ = + − − = + ⎜ x5 x5 x7 ⎟ dx 2 2 dx ⎜⎝ x15 x25 ⎟⎠ 2 x27 2 1 ⎝ 1 ⎠

Evaluate d2Bx/dx2 at x = R/2 to obtain:

12

542 Chapter 27 d 2 Bx dx 2

=

µ0 NR 2 I ⎛⎜ ⎜ ⎝

2

x = 12 R

(

1 5 4

R

)

2 52

+

(

1 5 4

R

)

2 52

−

(

5 4 5 4

R2

R

)

2 72

−

(

5 4 5 4

R2

R

)

2 72

⎞ ⎟= 0 ⎟ ⎠

Differentiate d2Bx/dx2 with respect to x to obtain: 2 d 3 Bx µ0 NR 2 I d ⎛ 1 1 5 x 2 5( x − R ) ⎞ ⎟ ⎜ = + − − ⎟ 2 dx 3 dx ⎜⎝ x15 x25 x17 x27 ⎠

=

µ0 NR 2 I ⎛ 35 x 3 15 x 15( x − R ) 35( x − R )3 ⎞⎟ ⎜⎜ 9 − 7 − x1 ⎝ x1

2

−

x27

⎟ ⎠

x29

Evaluate d3Bx/dx3 at x = R/2 to obtain:

d 3 Bx dx 3

= x = 12 R

µ0 NR 2 I ⎛⎜ 2

⎜ ⎝

(

35 8 5 4

R3

R2

)

92

−

(

15 2 5 4

R

R2

)

72

−

(

− 152 R 5 4

R2

)

72

−

− 358 R 3 ⎞⎟ = 0 2 92 ⎟ 5 R 4 ⎠

(

)

*38 ••• Picture the Problem Let the origin be midway between the coils so that one of them is centered at x = −r 3 / 2 and the other is centered at x = r 3 / 2 . Let the numeral 1 denote the coil centered at x = −r 3 / 2 and the numeral 2 the coil centered at

x = r 3 / 2 . We can express the magnetic field in the region between the coils as the difference of the magnetic fields B1 and B2 due to the two coils. Express the magnetic field on the x axis due to the coil centered at x = −r 3 / 2 :

B1 ( x ) =

µ 0 Nr 2 I 2 ⎡⎛ r 3 ⎤ ⎞ 2 ⎢⎜⎜ + x ⎟⎟ + r 2 ⎥ ⎢⎝ 2 ⎥ ⎠ ⎣ ⎦

32

where N is the number of turns. Express the magnetic field on the x axis due to the coil centered at x = r 3 /2:

B2 (x ) =

µ0 Nr 2 I 2 ⎡⎛ r 3 ⎤ ⎞ 2 ⎜ ⎟ ⎢ 2 ⎜ − x⎟ + r ⎥ ⎢⎝ 2 ⎥ ⎠ ⎣ ⎦

Subtract these equations to express the total magnetic field along the x axis:

32

Sources of the Magnetic Field 543 Bx (x ) = B1 (x ) − B2 (x ) =

µ0 Nr 2 I 2 ⎡⎛ r 3 ⎤ ⎞ + x ⎟⎟ + r 2 ⎥ 2⎢⎜⎜ ⎢⎝ 2 ⎥ ⎠ ⎣ ⎦

32

−

µ0 Nr 2 I 2 ⎡⎛ r 3 ⎤ ⎞ − x ⎟⎟ + r 2 ⎥ 2⎢⎜⎜ ⎢⎝ 2 ⎥ ⎠ ⎣ ⎦

32

−3 2 −3 2 2 2 ⎛⎡ ⎞ ⎤ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎟ r 3 µ0 Nr I ⎜ ⎜ r 3 2 2 ⎟ ⎜ ⎟ ⎢ ⎥ ⎢ ⎥ = + x⎟ + r − ⎜ − x⎟ + r ⎜ ⎟ ⎜ 2 ⎜ ⎢⎝ 2 ⎥ ⎢⎝ 2 ⎥ ⎟ ⎠ ⎠ ⎦ ⎣ ⎦ ⎠ ⎝⎣ 2

The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B3 B5

Formula/Content 1.26×10−6 0.30 250 15 0.5*$B$1*$B$3*($B$2^2)*$B$4

A8 B8

−0.30 $B$5*(($B$2*SQRT(3)/2+A8)^2 +$B$2^2)^(−3/2)

Algebraic Form

µ0 r N I

Coeff =

−r

$B$5* (($B$2*SQRT(3)/2−A8)^2 +$B$2^2)^(−3/2)

−3 2

µ0 Nr 2 I ⎡⎛⎜ r 3

−3 2

2

D8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

10^4*(B8−C8) A mu_0= r= N= I= Coeff=

B 1.26E−06 0.3 250 15 2.13E−04

C N/A^2 m turns A

x −0.30 −0.29 −0.28 −0.27 −0.26 −0.25 −0.24 −0.23

B_1 5.63E−03 5.86E−03 6.08E−03 6.30E−03 6.52E−03 6.72E−03 6.92E−03 7.10E−03

B_2 1.34E−03 1.41E−03 1.48E−03 1.55E−03 1.62E−03 1.70E−03 1.78E−03 1.87E−03

D

B(x) 68.4 68.9 69.2 69.2 68.9 68.4 67.5 66.4

2

µ0 Nr 2 I ⎡⎛⎜ r 3 2

C8

µ 0 Nr 2 I

2 ⎤ ⎞ ⎟ + r2⎥ ⎢ + x ⎟ ⎢⎜⎝ 2 ⎥ ⎠ ⎣ ⎦ 2 ⎤ ⎞ ⎟ + r2⎥ ⎢ − x ⎟ ⎢⎜⎝ 2 ⎥ ⎠ ⎣ ⎦ Bx = B1 − B2

544 Chapter 27 61 62 63 64 65 66 67 68

0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30

1.87E−03 1.78E−03 1.70E−03 1.62E−03 1.55E−03 1.48E−03 1.41E−03 1.34E−03

7.10E−03 6.92E−03 6.72E−03 6.52E−03 6.30E−03 6.08E−03 5.86E−03 5.63E−03

−66.4 −67.5 −68.4 −68.9 −69.2 −69.2 −68.9 −68.4

The following graph of Bx as a function of x was plotted using the data in the above table. 60 40

B x (G)

20 0 -20 -40 -60 -0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

x (m)

39 •• Picture the Problem The diagram shows the two coils of radii r1 and r2 with the currents flowing in the directions given. We can use the expression for B on the axis of a current loop to express the difference of the fields due to the two loops at a distance x from their common center. We’ll denote each field by the subscript identifying the radius of the current loop. The magnitude of the field on the x axis due to the current in the inner loop is:

µ 0 2π r12 I µ 0 r12 I B1 = = 4π (x 2 + r12 ) 3 2 2(x 2 + r12 ) 3 2

The magnitude of the field on the x axis due to the current in the outer loop is:

B2 =

µ 0 2π r22 I µ 0 r22 I = 4π (x 2 + r22 ) 3 2 2(x 2 + r22 ) 3 2

The resultant field at x is the difference between B1 and B2:

Sources of the Magnetic Field 545 Bx ( x ) = B1 ( x ) − B2 ( x ) =

(

µ 0 r12 I

2 x 2 + r12

)

32

−

(

µ 0 r22 I

2 x 2 + r22

)

32

(a) The spreadsheet program to calculate Bx as a function of x, for r2 = 10.1 cm, is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content B1 1.26×10−6 B2 0.1 B3 1 B4 0.101 A7 0 B7 0.5*$B$1*$B$2^2*$B$3/(A7^2+$B$2^2)^(3/2)

Algebraic Form

µ0 r1 I r2 x

(

µ 0 r12 I

2 x 2 + r12 C7

)

32

)

32

0.5*$B$1*$B$4^2*$B$3/(A7^2+$B$4^2)^(3/2)

µ 0 r22 I

10^4*(B7−C7)

2 x 2 + r22 Bx ( x ) = B1 ( x ) − B2 ( x )

(

D7

The spreadsheet for Bx when r = 10.1 cm follows. The other three tables are similar.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 56 57 58 59 60

A mu_0= r_1= I= r_2= r_2= r_2= r_2=

B 1.26E−06 0.1 1 0.101 0.11 0.15 0.2

C N/A^2 m A m m m m

D

x 0.00 0.01 0.02 0.03 0.04 0.05

B_1 6.30E−06 6.21E−06 5.94E−06 5.54E−06 5.04E−06 4.51E−06

B_2 6.24E−06 6.15E−06 5.89E−06 5.49E−06 5.01E−06 4.49E−06

B_x 6.24E−04 5.97E−04 5.21E−04 4.14E−04 2.95E−04 1.81E−04

0.46 0.47 0.48 0.49 0.50

6.04E−08 5.68E−08 5.34E−08 5.04E−08 4.75E−08

6.15E−08 5.78E−08 5.45E−08 5.13E−08 4.84E−08

−1.13E−05 −1.07E−05 −1.01E−05 −9.51E−06 −8.99E−06

The following graph shows B(x) for r = 10.1 cm, 11 cm, 15 cm, and 20 cm.

546 Chapter 27

B (G)

0.035

r = 10.1 cm

0.030

r = 11 cm

0.025

r = 15 cm

0.020

r = 20 cm

0.015 0.010 0.005 0.000 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 -0.005 -0.010 x (m)

40 ••• Picture the Problem We can approximate B(x) by using the result from Problem 39 for the field due to a single coil of radius r and evaluating B( x ) ≈

∂B ∆r at ∂r

r = r1. The magnetic field at a distance x on the axis of a coil of radius r is given by:

B(x ) =

µ0 I 2πr 2 4π (x 2 + r 2 ) 3 2

Express B(x) in terms of the rate of change of B with respect to r:

B( x ) ≈

∂B ∆r ∂r

Evaluate the partial derivative of B with respect to r:

(1)

Sources of the Magnetic Field 547 ∂ ⎡ µ0 I ⎢ ∂r ⎢⎣ 4π

⎛ 2πr 2 ⎜ ⎜ x2 + r 2 ⎝

(

⎞⎤ µ 0 I ∂ ⎡ 2πr 2 ⎤ ⎟ = ⎢ 2 ⎥ 3 / 2 ⎟⎥ 2 3/ 2 4 π r ∂ x r + ⎥ ⎢ ⎥⎦ ⎠⎦ ⎣ 3/ 2 ∂ ∂ 2 ⎡ 2 x + r2 2πr 2 − 2πr 2 x + r2 ⎢ µ0 I r r ∂ ∂ = ⎢ 3 4π ⎢ x 2 + r12 ⎣

)

(

)

(

)

(

[(

)

(

⎡ 2 x + r2 µ0 I ⎢ ⎢ = 4π ⎢ ⎢⎣

(

(

⎡ 2 2 ⎢ x +r = µ 0 Ir ⎢ ⎢ ⎣

)

) (4πr ) − 2πr ⎡⎢ 32 (x ⎣ (x + r ) 3/ 2

2

−

(

(

( )

)

( )

)

3 2 2 r x + r2 2 3 x2 + r 2

3/ 2

)

⎡ 2 x 2 + r 2 3 / 2 − 3r 2 x 2 + r 2 = µ 0 Ir ⎢ 3 2 x2 + r 2 ⎣⎢

(

= µ 0 Ir x 2 + r 2 =

)

1/ 2

(

(

+ r2

2

)

1/ 2

1/ 2

3/ 2

]⎤⎥ ⎥ ⎥ ⎦

) (2r )⎤⎥ ⎤⎥ 1/ 2

⎦⎥ ⎥ ⎥⎦

2 3 1

2

)

)

⎤ ⎥ ⎥ ⎥ ⎦ ⎤ ⎥ ⎦⎥

⎡ 2 x 2 + r 2 − 3r 2 ⎤ ⎢ ⎥ 2 2 3 ⎣⎢ 2 x + r ⎦⎥

(

)

µ0 I ⎡ 2 x 2 r − r 3 ⎤ ⎢ 2 ⎢⎣ x 2 + r 2

(

Evaluate ∂B ∂x at r = r1 to obtain:

)

52

⎥ ⎥⎦

µ I ⎡ 2 x 2 r1 − r13 ⎤ ∂B = 0 ⎢ ⎥ ∂r r =r1 2 ⎢⎣ x 2 + r12 5 2 ⎦⎥

(

Substitute in equation (1) to obtain:

)

2 3 ⎛ µ I∆r ⎞ ⎡ 2 x r1 − r1 ⎤ B(x ) ≈ ⎜ 0 ⎟⎢ 2 52⎥ ⎝ 2 ⎠ ⎢⎣ x + r12 ⎥⎦

(

)

Remarks: This solution shows that the field due to two coils separated by ∆r can be approximated by the given expression. 41

•••

Picture the Problem We can factor x from the denominator of the equation from

⎛ µ 0 I∆r ⎞ ⎛ 2r1 ⎞ ⎟⎜ 3 ⎟ . ⎝ 2 ⎠⎝ x ⎠

Problem 40 to show that B ( x) ≈ ⎜ From Problem 40:

2 3 ⎛ µ I∆r ⎞ ⎡ 2 x r1 − r1 ⎤ B(x ) ≈ ⎜ 0 ⎟⎢ 2 52⎥ ⎝ 2 ⎠ ⎣⎢ x + r12 ⎦⎥

(

)

548 Chapter 27 Factor x2 from the denominator of the expression to obtain:

⎤ ⎡ ⎥ ⎢ 2 3 ⎛ µ 0 I∆r ⎞ ⎢ 2r1 x − r1 ⎥ B( x) ≈ ⎜ ⎟ 52 ⎝ 2 ⎠ ⎢⎢ 5 ⎛ r12 ⎞ ⎥⎥ x ⎜1 + 2 ⎟⎟ ⎢⎣ ⎜⎝ x ⎠ ⎥⎦

For x >> r1:

⎛ r12 ⎞ x ⎜1 + 2 ⎟ ⎜ x ⎟⎠ ⎝ 5

5/ 2

≈ x5

2 3 ⎛ µ I∆r ⎞ ⎛ 2r1 x − r1 ⎞ ⎟⎟ B ( x) ≈ ⎜ 0 ⎟ ⎜⎜ x5 ⎝ 2 ⎠⎝ ⎠

Substitute and simplify to obtain:

2 r13 ⎞ ⎛ µ I∆r ⎞ ⎛ 2r1 x ⎟ ⎜ =⎜ 0 − ⎟⎜ 5 x 5 ⎟⎠ ⎝ 2 ⎠⎝ x 3 ⎛ µ I∆r ⎞ ⎛ 2r1 r1 ⎞ =⎜ 0 ⎟ ⎜⎜ 3 − 5 ⎟⎟ x ⎠ ⎝ 2 ⎠⎝ x

⎛ µ I∆r ⎞ ⎛ 2r1 ⎞ B ( x) ≈ ⎜ 0 ⎟⎜ 3 ⎟ ⎝ 2 ⎠⎝ x ⎠

For x >> r:

The spreadsheet-generated graph that follows provides a comparison of the exact and approximate fields. Note that the two solutions agree for large values of x.

0.001

B (G)

0.000 -0.001 Exact solution -0.002 A pproximate solution

-0.003 -0.004 0.00

0.10

0.20

0.30 x (m)

0.40

0.50

Sources of the Magnetic Field 549

Straight-Line Current Segments 42 • Picture the Problem The magnetic field due to the current in a long straight wire is given by B =

µ0 2 I where I is the current in the wire and R is the distance from the wire. 4π R µ0 2 I 4π R

Express the magnetic field due to a long straight wire:

B=

Substitute numerical values to obtain:

B = 10−7 T ⋅ m/A

(

=

) 2(10R A )

2.00 × 10−6 T ⋅ m R

(a) Evaluate B at R = 10 cm:

B(10 cm ) =

2.00 ×10 −6 T ⋅ m = 20.0 µT 0.1 m

(b) Evaluate B at R = 50 cm:

B(50 cm ) =

2.00 × 10−6 T ⋅ m = 4.00 µT 0.5 m

(c) Evaluate B at R = 2 m:

B(2 m ) =

2.00 × 10−6 T ⋅ m = 1.00 µT 2m

Problems 43 to 48 refer to Figure 27-45, which shows two long straight wires in the xy plane and parallel to the x axis. One wire is at y = −6 cm and the other is at y = +6 cm. The current in each wire is 20 A.

Figure 27-45 Problems 43-48 *43 • Picture the Problem Let + denote the wire (and current) at y = +6 cm and − the wire (and current) at y = −6 cm. We can use B =

µ0 2 I to find the magnetic field due to each of 4π R

the current carrying wires and superimpose the magnetic fields due to the currents in the

550 Chapter 27 wires to find B at the given points on the y axis. We can apply the right-hand rule to find r the direction of each of the fields and, hence, of B . (a) Express the resultant magnetic field at y = −3 cm: Find the magnitudes of the magnetic fields at y = −3 cm due to each wire:

r r r B (− 3 cm ) = B+ (− 3 cm ) + B− (− 3 cm )

(

B+ (− 3 cm ) = 10 −7 T ⋅ m/A

A) ) 20(.20 09 m

= 44.4 µT and

(

B− (− 3 cm ) = 10−7 T ⋅ m/A

A) ) 20(.20 03 m

= 133 µT Apply the right-hand rule to find the

r B+ (− 3 cm ) = (44.4 µT )kˆ

r r directions of B+ and B− :

and

Substitute to obtain:

r B (− 3 cm ) = (44.4 µT )kˆ − (133 µT )kˆ

r B− (− 3 cm ) = −(133 µT )kˆ

= − (88.6 µT )kˆ (b) Express the resultant magnetic field at y = 0:

r

r

r r r B (0) = B+ (0) + B− (0)

Because B+ (0 ) = − B− (0 ) :

r B (0 ) = 0

(c) Proceed as in (a) to obtain:

r B+ (3 cm ) = (133 µT )kˆ , r B− (3 cm ) = −(44.4 µT )kˆ , and

r B (3 cm ) = (133 µT )kˆ − (44.4 µT )kˆ = − (88.6 µT )kˆ

(d) Proceed as in (a) with y = 9 cm to obtain:

r B+ (9 cm ) = −(133 µT )kˆ , r B− (9 cm ) = −(26.7 µT )kˆ , and

Sources of the Magnetic Field 551 r B (9 cm ) = −(133 µT )kˆ − (26.7 µT )kˆ = − (160 µT )kˆ 44 •• Picture the Problem The diagram shows the two wires with the currents flowing in the negative x direction. We can use the expression for B due to a long, straight wire to express the difference of the fields due to the two currents. We’ll denote each field by the subscript identifying the position of each wire.

µ0 2I 4π 0.06 m − y

The field due to the current in the wire located at y = 6 cm is:

B6 =

The field due to the current in the wire located at y = −6 cm is:

B−6 =

µ0 2I 4π 0.06 m + y

The resultant field Bz is the difference between B6 and B−6:

Bz = B6 − B−6 =

⎞ µ0 µ µ I⎛ I I 1 1 ⎟ − 0 = 0 ⎜⎜ − 4π 0.06 m − y 4π 0.06 m − y 4π ⎝ 0.06 m − y 0.06 m + y ⎟⎠

The following graph of Bz as a function of y was plotted using a spreadsheet program:

552 Chapter 27 4

Bz (G)

2

0

-2

-4 -0.10

-0.08

-0.06

-0.04

-0.02

0.00

0.02

0.04

0.06

0.08

0.10

y (m)

45 • Picture the Problem Let + denote the wire (and current) at y = +6 cm and − the wire (and current) at y = −6 cm. We can use B =

µ0 2 I to find the magnetic field due to each of 4π R

the current carrying wires and superimpose the magnetic fields due to the currents in the wires to find B at the given points on the y axis. We can apply the right-hand rule to find r the direction of each of the fields and, hence, of B. (a) Express the resultant magnetic field at y = −3 cm: Find the magnitudes of the magnetic fields at y = −3 cm due to each wire:

r r r B (− 3 cm ) = B+ (− 3 cm ) + B− (− 3 cm )

(

B+ (− 3 cm ) = 10 −7 T ⋅ m/A

A) ) 20(.20 09 m

= 44.4 µT and

(

B− (− 3 cm ) = 10−7 T ⋅ m/A

A) ) 20(.20 03 m

= 133 µT Apply the right-hand rule to find the

r B + (− 3 cm ) = −(44.4 µT )kˆ

directions of B+ and B− :

and

Substitute to obtain:

r B (− 3 cm ) = −(44.4 µT )kˆ − (133 µT )kˆ

r

r

r B − (− 3 cm ) = −(133 µT )kˆ

= − (177 µT )kˆ

(b) Express the resultant magnetic field at y = 0: Find the magnitudes of the magnetic fields at y = 0 cm due to each wire:

Sources of the Magnetic Field 553 r r r B (0) = B+ (0) + B− (0)

(

B+ (0) = 10 −7 T ⋅ m/A

A) ) 20(.20 06 m

= 66.7 µT and

(

B− (0) = 10 −7 T ⋅ m/A

A) ) 20(.20 06 m

= 66.7 µT Apply the right-hand rule to find the

r B + (0) = −(66.7 µT )kˆ

directions of B+ and B− :

and

Substitute to obtain:

r B (0) = −(66.7 µT )kˆ − (66.7 µT )kˆ

r

r

r B − (0) = −(66.7 µT )kˆ

= − (133 µT )kˆ (c) Proceed as in (a) with y = +3 cm to obtain:

r B + (3 cm ) = −(133 µT )kˆ , r B − (3 cm ) = −(44.4 µT )kˆ , and

r B (3 cm ) = −(133 µT ) kˆ − (44.4 µT ) kˆ = − (177 µT ) kˆ

(d) Proceed as in (a) with y = +9 cm to obtain:

r B+ (9 cm ) = (133 µT ) kˆ , r B− (9 cm ) = −(26.7 µT ) kˆ , and

r B (9 cm ) = (133 µT ) kˆ − (26.7 µT ) kˆ = (106 µT ) kˆ

46 •• Picture the Problem The diagram shows the two wires with the currents flowing in the negative x direction. We can use the expression for B due to a long, straight wire to express the difference of the fields due to the two currents. We’ll denote each field by the subscript identifying the position of each wire.

554 Chapter 27

µ0 2I 4π 0.06 m − y

The field due to the current in the wire located at y = 6 cm is:

B6 =

The field due to the current in the wire located at y = −6 cm is:

B−6 =

µ0 2I 4π 0.06 m + y

The resultant field Bz is the sum of B6 and B−6:

Bz = B6 − B−6 =

⎞ µ0 µ µ I⎛ I I 1 1 ⎟ + 0 = 0 ⎜⎜ + 4π 0.06 m − y 4π 0.06 m − y 4π ⎝ 0.06 m − y 0.06 m + y ⎟⎠

The following graph of Bz as a function of y was plotted using a spreadsheet program:

4

Bz (G)

2

0

-2

-4 -0.10

-0.05

0.00

0.05

0.10

y (m)

47 • Picture the Problem Let + denote the wire (and current) at y = +6 cm and − the wire (and current) at y = −6 cm. We can use B =

µ0 2 I to find the magnetic field due to each 4π R

of the current carrying wires and superimpose the magnetic fields due to the currents in the wires to find B at the given points on the z axis.

Sources of the Magnetic Field 555 (a) Apply the right-hand rule to show that, for the currents parallel and in the negative x direction, the directions of the fields are as shown to the right:

Express the magnitudes of the magnetic fields at z = +8 cm due to the current-carrying wires at y = −6 cm and y = +6 cm:

Noting that the z components add to zero, express the resultant magnetic field at z = +8 cm:

(

B z − = B z + = 10 −7 T ⋅ m/A

×

2(20 A )

)

(0.06 m )2 + (0.08 m )2

= 40.0 µT r B ( z = 8 cm ) = 2(40.0 µT ) sin θ ˆj = 2(40.0 µT )(0.8) ˆj =

(64.0 µT ) ˆj

(b) Apply the right-hand rule to show that, for the currents antiparallel with the current in the wire at y = −6 cm in the negative x direction, the directions of the fields are as shown to the right: Noting that the y components add to zero, express the resultant magnetic field at z = +8 cm:

r B (z = 8 cm ) = −2(40.0 µT ) cos θ kˆ = −2(40.0 µT )(0.6) kˆ = − (48.0 µT ) kˆ

48 • Picture the Problem Let + denote the wire (and current) at y = +6 cm and − the wire (and current) at y = −6 cm. The forces per unit length the wires exert on each other are action and reaction forces and hence are equal in magnitude. We can use F = IlB to express the force on either wire and B =

µ 0 2I to express the magnetic field at the location of either 4π R

wire due to the current in the other. Express the force exerted on either wire:

F = IlB

556 Chapter 27

µ 0 2I 4π R

Express the magnetic field at either location due to the current in the wire at the other location:

B=

Substitute to obtain:

⎛ µ 2 I ⎞ 2lµ 0 I 2 lµ 0 I 2 = F = Il⎜ 0 ⎟= 4π R 2π R ⎝ 4π R ⎠

Divide both sides of the equation by l to obtain:

F 2µ 0 I 2 = l 4π R

Substitute numerical values and evaluate F/ l :

F 2 10 −7 T ⋅ m/A (20 A ) = 0.12 m l

(

)

2

= 667 µN/m 49

•

Picture the Problem We can use

F 2µ 0 I 2 = to relate the force per unit length each l 4π R

current-carrying wire exerts on the other to their common current. (a)

Because the currents repel, they are antiparallel.

(b)Express the force per unit length experienced by each wire:

F 2µ 0 I 2 = l 4π R

Solve for I:

I=

Substitute numerical values and evaluate I:

I=

4πR F 2µ 0 l

(

(8.6 cm )

2 10 −7 T ⋅ m/A

) (3.6 nN/m )

= 39.3 mA 50 •• Picture the Problem Note that the current segments a-b and e-f do not contribute to the magnetic field at point P. The current in the segments b-c, c-d, and d-e result in a magnetic field at P that points into the plane of the paper. Note that the angles bPc and ePd are 45° and use the expression for B due to a straight wire segment to find the contributions to the field at P of segments bc, cd, and de.

Sources of the Magnetic Field 557

Express the resultant magnetic field at P:

B = Bbc + Bcd + Bde

µ0 I (sin θ1 + sin θ 2 ) 4π R

Express the magnetic field due to a straight line segment:

B=

Use equation (1) to express Bbc and Bde:

Bbc =

Use equation (1) to express Bcd:

Bcd =

Substitute to obtain:

B=

µ0 4π µ = 0 4π

(1)

I (sin 45° + sin 0°) R I sin 45° R

µ0 I (sin 45° + sin 45°) 4π R µ I = 2 0 sin 45° 4π R

+

µ0 I µ0 I sin 45° + 2 sin 45° 4π R 4π R µ0 I

sin 45° 4π R µ0 I =4 sin 45° 4π R Substitute numerical values and evaluate B:

(

B = 4 10 −7 T ⋅ m/A

) 0.801Am sin 45°

= 226 µT 51 •• Picture the Problem The forces acting on the wire are the upward magnetic force FB and the downward gravitational force mg, where m is the mass of the wire. We can use a condition for translational equilibrium and the expression for the force per unit length between parallel current-carrying wires to relate the required current to the mass of the wire, its length, and the separation of the two wires. Apply

∑F

y

= 0 to the floating

FB − mg = 0

558 Chapter 27 wire to obtain: Express the repulsive force acting on the upper wire: Substitute to obtain:

FB = 2

2

Solve for I:

µ0 I 2l 4π R

µ0 I 2l − mg = 0 4π R 4πmgR 2µ 0 l

I=

Substitute numerical values and evaluate I:

I=

(14 × 10

kg )(9.81 m/s 2 )(1.5 × 10 −3 m ) = 80.2 A 2(10 −7 T ⋅ m/A )(0.16 m ) −3

*52 •• Picture the Problem Note that the forces on the upper wire are away from and directed along the lines to the lower wire and that their horizontal components cancel. We can

µ0 I 2 F use =2 to find the resultant force in the upward direction (the y direction) l 4π R acting on the top wire. In part (b) we can use the right-hand rule to determine the directions of the magnetic fields at the upper wire due to the currents in the two lower wires and use B =

µ 0 2I to find the magnitude of the resultant field due to these 4π R

currents. (a) Express the force per unit length each of the lower wires exerts on the upper wire: Noting that the horizontal components add up to zero, express the net upward force per unit length on the upper wire:

µ I2 F =2 0 l 4π R

∑

µ0 I 2 cos 30° 4π R l µ0 I 2 +2 cos 30° 4π R µ I2 =4 0 cos 30° 4π R

Fy

=2

Sources of the Magnetic Field 559 Substitute numerical values and evaluate

∑

Fy

l

∑

:

= 4(10 −7 T ⋅ m/A )

Fy

l

(15 A )2 cos 30° 0.1 m

= 7.79 × 10 −4 N/m

(b) Noting, from the geometry of the wires, the magnetic field vectors both are at an angle of 30° with the horizontal and that their y components cancel, express the resultant magnetic field:

r µ 2I cos 30°iˆ B=2 0 4π R

(

Substitute numerical values and evaluate B:

B = 2 10 −7 T ⋅ m/A

A) ) 20(15 cos 30° .1 m

= 52.0 µT 53 •• Picture the Problem Note that the forces on the upper wire are away from the lower left hand wire and toward the lower right hand wire and that, due to symmetry, their vertical

µ0 I 2 F =2 to find the resultant force in the x components cancel. We can use l 4π R direction (to the right) acting on the top wire. In part (b) we can use the right-hand rule to determine the directions of the magnetic fields at the upper wire due to the currents in the two lower wires and use B =

µ 0 2I to find the magnitude of the resultant field due 4π R

to these currents. (a) Express the force per unit length each of the lower wires exerts on the upper wire:

µ I2 F =2 0 l 4π R

Noting that the vertical components add up to zero, express the net force per unit length acting to the right on the upper wire:

Fx µ0 I 2 = 2 ∑ l 4π R cos 60° µ I2 +2 0 cos 60° 4π R µ I2 =4 0 cos 60° 4π R

560 Chapter 27 F (15 A ) ∑ lx = 4 10−7 T ⋅ m/A 0.1 m cos 60°

(

Substitute numerical values and evaluate

F ∑ lx :

)

2

= 4.50 × 10 −4 N/m

r µ 2I sin 30° ˆj B = −2 0 4π R

(b) Noting, from the geometry of the wires, that the magnetic field vectors both are at an angle of 30° with the horizontal and that their x components cancel, express the resultant magnetic field:

(

Substitute numerical values and evaluate B:

B = 2 10 −7 T ⋅ m/A

A) ) 20(15 sin 30° .1 m

= 30.0 µT 54 •• Picture the Problem Let the numeral 1 denote the current flowing in the positive x direction and the magnetic field resulting from it and the numeral 2 denote the current flowing in the positive y direction and the magnetic field resulting from it. We can express the magnetic field anywhere in the xy plane using B =

r

µ 0 2I and the right-hand 4π R

rule and then impose the condition that B = 0 to determine the set of points that satisfy this condition. Express the resultant magnetic field due to the two current-carrying wires: Express the magnetic field due to the current flowing in the positive x direction: Express the magnetic field due to the current flowing in the positive y direction: Substitute to obtain:

r r r B = B1 + B2

r µ 2I B1 = 0 1 kˆ 4π y r µ 2I B2 = − 0 2 kˆ 4π x

r µ 2I µ 2I ˆ B= 0 kˆ − 0 k 4π y 4π x ⎛ µ 2I µ 0 2I ⎞ ˆ ⎟⎟k = ⎜⎜ 0 − y x 4 4 π π ⎝ ⎠

Sources of the Magnetic Field 561 because I = I1 = I2.

r

µ0 2I µ 0 2I − = 0 ⇒ x = y. 4π y 4π x

For B = 0 :

r Hence, B = 0 along a line that makes an angle of 45° with the x axis.

55 •• Picture the Problem Let the numeral 1 denote the current flowing along the positive z axis and the magnetic field resulting from it and the numeral 2 denote the current flowing in the wire located at x = 10 cm and the magnetic field resulting from it. We can express the magnetic field anywhere in the xy plane using B =

r

µ 0 2I and the right-hand rule 4π R

and then impose the condition that B = 0 to determine the current that satisfies this condition. (a) Express the resultant magnetic field due to the two current-carrying wires: Express the magnetic field at x = 2 cm due to the current flowing in the positive z direction: Express the magnetic field at x = 2 cm due to the current flowing in the wire at x = 10 cm: Substitute to obtain:

r r r B = B1 + B2

r µ 2 I1 ˆ B1 ( x = 2 cm ) = 0 j 4π 2 cm

r µ 2I 2 ˆ B2 ( x = 2 cm ) = − 0 j 4π 8 cm r µ 2 I1 ˆj − µ 0 2 I 2 ˆj B= 0 4π 2 cm 4π 8 cm ⎛ µ 2I1 µ 0 2 I 2 ⎞ ˆ ⎟⎟ j = ⎜⎜ 0 − ⎝ 4π 2 cm 4π 8 cm ⎠

r

For B = 0 :

µ 0 2 I1 µ 0 2 I 2 − =0 4π 2 cm 4π 8 cm or

I1 I − 2 =0 2 cm 8 cm

562 Chapter 27 Solve for and evaluate I2:

(b) Express the magnetic field at x = 5 cm:

Substitute numerical values and

r evaluate B (x = 5 cm ) :

I 2 = 4 I 1 = 4(20 A ) = 80.0 A

r µ 2 I1 ˆj − µ 0 2 I 2 ˆj B= 0 4π 5 cm 4π 5 cm 2µ 0 (I1 − I 2 ) ˆj = 4π (5 cm ) r 2(10 −7 T ⋅ m/A ) (20 A − 80 A ) ˆj B= 5 cm = − (0.240 mT ) ˆj

56 •• Picture the Problem Choose a coordinate system with its origin at the lower left-hand corner of the square, the positive x axis to the right and the positive y axis upward. We can use B =

µ 0 2I and the right-hand rule to find the magnitude and direction of the 4π R

magnetic field at the unoccupied corner due to each of the currents, and superimpose these fields to find the resultant field. (a) Express the resultant magnetic field at the unoccupied corner:

r r r r B = B1 + B2 + B3

(1)

When all the currents are into the paper their magnetic fields at the unoccupied corner are as shown to the right: Express the magnetic field at the unoccupied corner due to the current I1 : Express the magnetic field at the unoccupied corner due to the current I2 :

Express the magnetic field at the unoccupied corner due to the current I3 :

r µ 2I ˆ B1 = − 0 j 4π L

(

r µ 2I B2 = 0 cos 45° iˆ − ˆj 4π L 2 µ 2I ˆ ˆ = 0 i−j 4π 2 L

(

r µ 2I ˆ B3 = 0 i 4π L

)

)

Sources of the Magnetic Field 563 Substitute in equation (1) and simplify to obtain:

( )

( )

r µ 2I ˆ µ0 2I ˆ ˆ µ0 2I ˆ µ0 2I ⎛ ˆ 1 ˆ ˆ ˆ ⎞ B=− 0 j+ i−j + i= ⎜− j + i − j + i ⎟ 4π L 4π 2 L 4π L 4π L ⎝ 2 ⎠ 3µ 0 I ˆ ˆ µ 2 I ⎡⎛ 1 ⎞ ˆ ⎛ 1⎞ ⎤ = 0 i−j ⎜1 + ⎟i + ⎜ − 1 − ⎟ ˆj ⎥ = ⎢ 4π L ⎣⎝ 2 ⎠ ⎝ 2⎠ ⎦ 4πL

[ ]

(b) When I2 is out of the paper the magnetic fields at the unoccupied corner are as shown to the right:

Express the magnetic field at the unoccupied corner due to the current I2 :

(

r µ 2I B2 = 0 cos 45° − iˆ + ˆj 4π L 2 µ 2I ˆ ˆ = 0 −i + j 4π 2 L

(

)

)

Substitute in equation (1) and simplify to obtain:

(

)

(

)

r µ 2I ˆ µ0 2I ˆ ˆ µ0 2I ˆ µ0 2I ⎛ ˆ 1 ˆ ˆ ˆ ⎞ B=− 0 j+ −i + j + i= ⎜− j + − i + j + i ⎟ 4π L 4π 2 L 4π L 4π L ⎝ 2 ⎠ µ 2 I ⎡⎛ 1 ⎞ ˆ ⎛ µ0 I ˆ ˆ 1 ⎞ ⎤ µ 2 I ⎡ 1 ˆ 1 ˆ⎤ = 0 i − j⎥ = i−j ⎜1 − ⎟i + ⎜ − 1 + ⎟ ˆj ⎥ = 0 ⎢ ⎢ 4π L ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ 4π L ⎣ 2 2 ⎦ 4πL

[ ]

(c) When I1 and I2 are in and I3 is out of the paper the magnetic fields at the unoccupied corner are as shown to the right: From (a) or (b) we have:

From (a) we have:

r µ 2I ˆ B1 = − 0 j 4π L

(

r µ 2I B2 = 0 cos 45° iˆ − ˆj 4π L 2 µ 2I ˆ ˆ = 0 i−j 4π 2 L

(

Express the magnetic field at the unoccupied corner due to the current I3 :

r µ 2I ˆ B3 = − 0 i 4π L

)

)

564 Chapter 27 Substitute in equation (1) and simplify to obtain:

( )

( )

r µ 2I ˆ µ0 2I ˆ ˆ µ0 2I ˆ µ0 2I ⎛ ˆ 1 ˆ ˆ ˆ ⎞ B=− 0 j+ i−j − i= ⎜− j + i − j − i ⎟ 4π L 4π 2 L 4π L 4π L ⎝ 2 ⎠ µ 2 I ⎡⎛ µ0 I ˆ ˆ 1⎞ ⎛ 1⎞ ⎤ = 0 −i −3j ⎜ − 1 + ⎟iˆ + ⎜ − 1 − ⎟ ˆj ⎥ = ⎢ 4π L ⎣⎝ 2⎠ ⎝ 2⎠ ⎦ 4πL

[

]

*57 •• Picture the Problem Choose a coordinate system with its origin at the lower left-hand corner of the square, the positive x axis to the right and the positive y axis upward. Let the numeral 1 denote the wire and current in the upper left-hand corner of the square, the numeral 2 the wire and current in the lower left-hand corner (at the origin) of the square, and the numeral 3 the wire and current in the lower right-hand corner of the square. We can use B =

µ 0 2I and the right-hand rule to find the magnitude and direction of the 4π R

magnetic field at, say, the upper right-hand corner due to each of the currents, superimpose these fields to find the resultant field, and then use F = IlB to find the force per unit length on the wire. (a) Express the resultant magnetic field at the upper right-hand corner:

r r r r B = B1 + B2 + B3

(1)

When all the currents are into the paper their magnetic fields at the upper right-hand corner are as shown to the right: Express the magnetic field due to the current I1: Express the magnetic field due to the current I2:

r µ 2I ˆ B1 = − 0 j 4π a

(

r µ 2I B2 = 0 cos 45° iˆ − ˆj 4π a 2 µ 2I ˆ ˆ = 0 i−j 4π 2a

(

Express the magnetic field due to the current I3:

r µ 2I ˆ B3 = 0 i 4π a

Substitute in equation (1) and simplify to obtain:

)

)

Sources of the Magnetic Field 565

( )

r µ 2I ˆ µ0 2I ˆ ˆ B=− 0 j+ i−j 4π a 4π 2a

( )

µ0 2 I ˆ µ0 2 I ⎛ ˆ 1 ˆ ˆ ˆ ⎞ i= ⎜− j + i − j + i ⎟ 4π a 4π a ⎝ 2 ⎠ µ 2 I ⎡⎛ 1 ⎞ ˆ ⎛ 1 ⎞ ⎤ 3µ I = 0 ⎜1 + ⎟i + ⎜ − 1 − ⎟ ˆj ⎥ = 0 iˆ − ˆj ⎢ 4π a ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ 4πa +

[ ]

F = BI l

Using the expression for the magnetic force on a currentcarrying wire, express the force per unit length on the wire at the upper right-hand corner:

(2)

r F 3µ 0 I 2 ˆ ˆ = i−j 4πa l

[ ]

Substitute to obtain:

and 2

⎛ 3µ I 2 ⎞ ⎛ 3µ I 2 ⎞ F = ⎜⎜ 0 ⎟⎟ + ⎜⎜ 0 ⎟⎟ l ⎝ 4πa ⎠ ⎝ 4πa ⎠ =

2

3 2µ0 I 2 4πa

(b) When the current in the upper right-hand corner of the square is out of the page, and the currents in the wires at adjacent corners are oppositely directed, the magnetic fields at the upper right-hand are as shown to the right:

(

r µ 2I cos 45° − iˆ + ˆj B2 = 0 4π a 2 µ 2I ˆ ˆ = 0 −i + j 4π 2a

Express the magnetic field at the upper right-hand corner due to the current I2:

r

(

)

)

r

Using B1 and B3 from (a), substitute in equation (1) and simplify to obtain:

(

)

(

r µ 2I ˆ µ0 2I ˆ ˆ µ0 2I ˆ µ0 2I ⎛ ˆ 1 ˆ −i + j + B=− 0 j+ i= ⎜− j + − i + 4π a 4π 2a 4π a 4π a ⎝ 2 µ 2 I ⎡⎛ 1 ⎞ ˆ ⎛ 1 ⎞ ⎤ µ 2 I ⎡ 1 ˆ 1 ˆ⎤ µ0 I ˆ = 0 i − j⎥ = i− ⎜1 − ⎟i + ⎜ − 1 + ⎟ ˆj ⎥ = 0 ⎢ 2 ⎠ ⎦ 4π a ⎢⎣ 2 2 ⎦ 4πa 4π a ⎣⎝ 2 ⎠ ⎝

)

ˆj + iˆ ⎞⎟ ⎠

[ ˆj ]

566 Chapter 27 r F µ0 I 2 ˆ ˆ = i−j 4πa l

[ ]

Substitute in equation (2) to obtain:

and 2

⎛ µ I2 ⎞ ⎛ µ I2 ⎞ F = ⎜⎜ 0 ⎟⎟ + ⎜⎜ 0 ⎟⎟ l ⎝ 4πa ⎠ ⎝ 4πa ⎠ =

2

2µ0 I 2 4πa

58 •• Picture the Problem The configuration is shown in the adjacent figure. Here the z axis points out of the plane of the paper, the x axis points to the right, the y axis points up. We can use B =

µ 0 2I and the 4π R

right-hand rule to find the magnetic field due to the current in each wire and add these magnetic fields vectorially to find the resultant field.

r r r r r r B = B1 + B2 + B3 + B4 + B5

Express the resultant magnetic field on the z axis:

r B1 is given by:

r B1 = Bˆj

r B2 is given by:

r B2 = (B cos 45°)iˆ + (B sin 45°) ˆj

r B3 is given by:

r B3 = Biˆ

r B4 is given by:

r B4 = (B cos 45°)iˆ − (B sin 45°) ˆj

r B5 is given by:

r B5 = − Bˆj

r

r

r

r

r

Substitute for B1 , B2 , B3 , B4 , and B5 and simplify to obtain:

r B = Bˆj + (B cos 45°)iˆ + (B sin 45°) ˆj + Biˆ + (B cos 45°)iˆ − (B sin 45°) ˆj − Bˆj = (B cos 45°)iˆ + Biˆ + (B cos 45°)iˆ = (B + 2 B cos 45°)iˆ = 1 + 2 Biˆ

(

)

Sources of the Magnetic Field 567

µ 0 2I 4π R

Express B due to each current at z = 0:

B=

Substitute to obtain:

r µI B = 1 + 2 0 iˆ 2πR

(

)

r B Due to a Current in a Solenoid 59

•

⎛ b a + ⎜ 2 2 2 a + R2 ⎝ b +R

Picture the Problem We can use B x = 12 µ 0 nI ⎜

⎞ ⎟ to find B at ⎟ ⎠

any point on the axis of the solenoid. Note that the number of turns per unit length for this solenoid is 300 turns/0.3 m = 1000 turns/m.

⎛ b a + B x = 12 µ 0 nI ⎜⎜ 2 2 a2 + R2 ⎝ b +R

⎞ ⎟ ⎟ ⎠

⎛ b a + T ⋅ m/A (1000 )(2.6 A ) ⎜ 2 ⎜ b 2 + (0.012 m )2 a 2 + (0.012 m ) ⎝

⎞ ⎟ ⎟ ⎠

Express the magnetic field at any point on the axis of the solenoid: Substitute numerical values to obtain:

Bx =

1 2

(4π × 10

−7

)

⎛ b a = (1.63 mT ) ⎜ + 2 2 ⎜ b 2 + (0.012 m ) a 2 + (0.012 m ) ⎝

⎞ ⎟ ⎟ ⎠

(a) Evaluate Bx for a = b = 0.15 m:

⎛ Bx = (1.63 mT ) ⎜ ⎜ ⎝

0.15 m

(0.15 m )2 + (0.012 m )2

+

0.15 m

(0.15 m )2 + (0.012 m )2

⎞ ⎟ = 3.25 mT ⎟ ⎠

(b) Evaluate Bx for a = 0.1 m and b = 0.2 m:

⎛ Bx (0.2 m ) = (1.63 mT ) ⎜ ⎜ ⎝

0.2 m

(0.2 m )2 + (0.012 m )2

= 3.25 mT (c) Evaluate Bx (= Bend) for a = 0 and b = 0.3 m:

+

0.1 m

(0.1 m )2 + (0.012 m )2

⎞ ⎟ ⎟ ⎠

568 Chapter 27

Note that Bend

⎛ Bx = (1.63 mT ) ⎜ ⎜ ⎝ 1 = 2 Bcenter .

⎞ ⎟ = 1.63 mT ⎟ ⎠

0.3 m

(0.3 m )2 + (0.012 m )2

*60 • Picture the Problem We can use Bx = µ 0 nI to find the approximate magnetic field on the axis and inside the solenoid. Express Bx as a function of n and I:

Bx = µ 0 nI

Substitute numerical values and evaluate Bx:

⎛ 600 ⎞ ⎟⎟(2.5 A ) Bx = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 2.7 m ⎠

(

)

= 0.698 mT 61 ••• Picture the Problem The solenoid, extending from x = − l 2 to x = l 2 , with the origin at its center, is shown in the diagram. To find the field at the point whose coordinate is x outside the solenoid we can determine the field at x due to an infinitesimal segment of the solenoid of width dx′ at x′, and then integrate from x = − l 2 to x = l 2 . The segment may be considered as a coil ndx′ carrying a current I. Express the field dB at the axial point whose coordinate is x:

dBx =

µ0 2πR 2 I 4π ( x − x')2 + R 2

[

]

3 2

dx'

Integrate dBx from x = − l 2 to x = l 2 to obtain:

Bx =

µ 0 nIR 2 2

l2

dx'

∫ [(x − x' )

−l 2

2

+ R2

]

3 2

Refer to the diagram to express cosθ1 and cosθ2:

=

µ 0 nI ⎛⎜ 2 ⎜ ⎝

x+l 2

(x + l 2)2 + R 2

cos θ1 = and

[R

x + 12 l

2

⎞ ⎟ 2 2 ⎟ (x − l 2) + R ⎠ x−l 2

−

+ (x + 12 l )

]

2 12

Sources of the Magnetic Field 569 cos θ 2 =

Substitute to obtain:

B=

1 2

[R

x − 12 l

2

]

+ (x − 12 l )

2 12

µ 0 nI (cos θ1 − cos θ 2 )

62 ••• Picture the Problem We can use Equation 27-35, together with the small angle approximation for the cosine and tangent functions, to show that θ1 and θ2 are as given and that B is given by Equation 27-37. (a) The angles θ1 and θ2 are shown in the diagram. Note that tan θ1 = R ( x + l 2 ) and

tan θ 2 = R ( x − l 2 ) .

Apply the small angle approximation tanθ ≈ θ to obtain:

θ1 ≈

R x + 12 l

and

θ2 ≈

(b) Express the magnetic field outside the solenoid: Apply the small angle approximation for the cosine function to obtain:

R x − 12 l

B = 12 µ 0 nI (cosθ1 − cosθ 2 )

⎛ R ⎞ ⎟ cos θ1 = 1 − ⎜⎜ 1 ⎟ + x l 2 ⎠ ⎝

2

1 2

and

⎛ R ⎞ ⎟ cos θ 2 = 1 − 12 ⎜⎜ 1 ⎟ − x l 2 ⎠ ⎝

2

Substitute and simplify to obtain: 2 2 ⎡ ⎡ ⎤ ⎛ R ⎞ ⎤ 1 ⎛ R ⎞ 1 1 1 1 ⎟⎟ ⎥ = 4 µ 0 nIR 2 ⎢ ⎟⎟ − 1 + 2 ⎜⎜ B = µ 0 nI ⎢1 − 2 ⎜⎜ − ⎥ 2 2 1 1 1 1 ⎢⎣ ⎢⎣ (x − 2 l ) (x + 2 l ) ⎦ ⎝ x − 2 l ⎠ ⎥⎦ ⎝ x+ 2l⎠ 1 2

Let r1 = x − 12 l be the distance to the near end of the solenoid, r1 = x + 12 l the distance to the far

B=

µ0 4π

⎛ qm q m ⎞ ⎜⎜ 2 − 2 ⎟⎟ r2 ⎠ ⎝ r1

570 Chapter 27 end, and q m = nIπR 2 = µ l , where

µ = nIπR2 is the magnetic moment of the solenoid to obtain:

Ampère’s Law *63 • Picture the Problem We can apply Ampère’s law to a circle centered on the axis of the cylinder and evaluate this expression for r < R and r > R to find B inside and outside the cylinder. Apply Ampère’s law to a circle centered on the axis of the cylinder:

r

r

∫ B ⋅ dl = µ I

0 C

C

Note that, by symmetry, the field is the same everywhere on this circle.

r r Binside ⋅ d l = µ 0 (0 ) = 0

Evaluate this expression for r < R:

∫

Solve for Binside to obtain:

Binside = 0

C

r r Boutside ⋅ d l = B(2πR ) = µ 0 I

Evaluate this expression for r > R:

∫

Solve for Boutside to obtain:

Boutside =

C

µ0 I 2πR

64 • r r Picture the Problem We can use Ampère’s law, B ⋅ d l = µ 0 I C , to find the line integral

∫

C

r r B ∫ ⋅ d l for each of the three paths. C

(a) Evaluate

Evaluate

Evaluate

r

r

∫ B ⋅ d l for C : 1

C

r r B ∫ ⋅ dl for C2: C

r

r

∫ B ⋅ dl for C : C

3

r r B ∫ ⋅ dl = µ0 (8 A ) C1

r

r

∫ B ⋅ dl = µ (8 A − 8 A ) = C2

0

r r B ∫ ⋅ dl = − µ0 (8 A ) C1

0

because the field

is opposite the direction of integration.

Sources of the Magnetic Field 571 (b)

None of the paths can be used to find B at a general point because there the current configuration does not have cylindrical symmetry.

65 • Picture the Problem Let the current in the wire and outer shell be I. We can apply Ampère’s law to a circle, concentric with the inner wire, of radius r to find B at points between the wire and the shell far from the ends (r < R), and outside the cable (r > R).

r r Br < R ⋅ d l = Br < R (2πr ) = µ 0 I

(a) Apply Ampère’s law for r < R:

∫

Solve for Br
Br < R =

C

µ0 I 2πr

r r Br > R ⋅ dl = µ 0 (0 )

(b) Apply Ampère’s law for r > R:

∫

Solve for Br>R to obtain:

Br > R = 0

C

66 •• Picture the Problem. Let the radius of the wire be a. We can apply Ampère’s law to a circle, concentric with the center of the wire, of radius r to find B at various distances from the center of the wire. Express Ampère’s law:

r r B ∫ ⋅ d l = µ0 I C C

Using the fact that the current is uniformly distributed over the crosssectional area of the wire, relate the current enclosed by a circle of radius r to the total current I carried by the wire:

IC I = 2 2 πr πa or

IC = I

r2 a2

Substitute and evaluate the integral to obtain:

B r (2πr ) =

Solve for Br
Br
For r ≥ a:

∫

C

µ0r 2 a2

µ0 r I 2πa 2

I

(1)

r r Br ≥a ⋅ d l = Br ≥a (2πr ) = µ 0 I

572 Chapter 27 Solve for Br ≥ a:

Br ≥a =

µ0 I 2πr

(2)

(a) Use equation (1) to evaluate B(0.1 cm):

B(0.1 cm ) =

(4π × 10

)

N/A 2 (0.001 m ) (100 A ) = 8.00 × 10 −4 T 2 2π (0.005 m ) −7

(b) Use either equation to evaluate B at the surface of the wire:

B(0.005 cm ) =

(4π × 10

)

N/A 2 (0.005 m ) (100 A ) = 4.00 × 10 −3 T 2 2π (0.005 m ) −7

(4π × 10 B(0.007 m ) =

)

N/A 2 (100 A ) 2π (0.007 m )

(c) Use equation (2) to evaluate B(0.7 cm):

−7

= 2.86 × 10 −3 T (d) A graph of B as a function of r follows:

1.0

B (arbitrary units)

0.8 0.6 0.4 0.2 0.0 0

1

2

3

4

5

r (arbitrary units)

*67 •• Determine the Concept The contour integral consists of four portions, two horizontal r r portions for which B ⋅ d l = 0 , and two vertical portions. The portion within the

∫

C

magnetic field gives a nonvanishing contribution, whereas the portion outside the field gives no contribution to the contour integral. Hence, the contour integral has a finite value. However, it encloses no current; thus, it appears that Ampère’s law is violated. What this demonstrates is that there must be a fringing field so that the contour integral does vanish.

Sources of the Magnetic Field 573 68 •• Picture the Problem Let r1 = 1 mm, r2 = 2 mm, and r3 = 3 mm and apply Ampère’s law in each of the three regions to obtain expressions for B in each part of the coaxial cable and outside the coaxial cable. Apply Ampère’s law to a circular path of radius r < r1 to obtain:

Br
Because the current is uniformly distributed over the cross section of the inner wire:

r2 IC I = ⇒ IC = 2 I r1 π r 2 π r12

Substitute for IC to obtain:

Br
Solve for Br
B r < r1 =

Apply Ampère’s law to a circular path of radius r1 < r < r2 to obtain: Solve for Br1
Apply Ampère’s law to a circular path of radius r2 < r < r3 to obtain:

r2 I r12

2µ 0 I r 4π r12

Br1
Br1
2µ0 I 1 4π r

(2)

Br2
Because the current is uniformly distributed over the cross section of the outer conductor:

I' I = 2 2 π r − π r2 π r3 − π r22

Solve for I′:

I '=

Substitute for I′ to obtain:

⎛ r2 − r2 Br2
Solve for Br1
(1)

2

r 2 − r22 I r32 − r22

Br2
2 µ0 I ⎛ r 2 − r22 ⎞ ⎜1 − ⎟ 4π ⎜⎝ r32 − r22 ⎟⎠

⎞ I ⎟⎟ ⎠

(3)

574 Chapter 27 A spreadsheet program was used to plot the following graph of equations (1), (2), and (3).

3.0 2.5

B (G)

2.0 1.5 1.0 0.5 0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

r (mm)

Apply Ampère’s law to a circular path of radius r > r3 to obtain:

Br >r3 (2π r ) = µ 0 I C

= µ 0 (I − I ) = 0

and Br >r3 = 0 69 •• Picture the Problem We can use Ampère’s law to calculate B because of the high degree of symmetry. The current through C depends on whether r is less than or the inner radius a, greater than the inner radius a but less than the outer radius b, or greater than the outer radius b. (a) Apply Ampère’s law to a circular path of radius r < a to obtain:

∫

C

r r B r < a ⋅ d l = µ 0 I C = µ 0 (0 ) = 0

and

Br < a = 0

(b) Use the uniformity of the current over the cross-section of the conductor to express the current I′ enclosed by a circular path whose radius satisfies the condition a < r < b:

I' I = 2 2 π r −a π b − a2

Solve for IC = I′:

I C = I' = I

(

2

)

(

r 2 − a2 b2 − a2

)

Sources of the Magnetic Field 575

∫

Substitute in Ampère’s law to obtain:

C

r r Ba
Solve for Ba
Ba
r 2 − a2 b2 − a2

µ0 I r 2 − a 2 2πr b 2 − a 2

IC = I

(c) Express IC for r > b:

r r Br >b ⋅ d l = Br >b (2πr ) = µ 0 I

Substitute in Ampère’s law to obtain:

∫

Solve for Br>b:

Br >b =

C

µ0 I 2πr

70 •• Picture the Problem The number of turns enclosed within the rectangular area is na. Denote the corners of the rectangle, starting in the lower left-hand corner and proceeding counterclockwise, as 1, 2, 3, and 4. We can apply Ampère’s law to each side of this r r rectangle in order to evaluate B ⋅ d l .

∫

C

r r r r r r r r ⋅ d = ⋅ d + ⋅ d + l l l B B B B ∫C ∫1→2 ∫2→3 ∫3→4⋅ d l r r + ∫ B ⋅ dl

Express the integral around the closed path C as the sum of the integrals along the sides of the rectangle: Evaluate

r

4→1

r

r

r

∫ B ⋅ d l = aB

∫ B ⋅ dl : 1→2

1→2

r

For the paths 2 → 3 and 4 → 1, B is either zero (outside the solenoid) or r is perpendicular to d l and so:

r

For the path 3 → 4, B =0 and:

Substitute in Ampère’s law to obtain:

r

r

r

r

∫ B ⋅ dl = ∫ B ⋅ dl = 0 2→3

r

4→1

r

∫ B ⋅ dl = 0 3→4

r r B ∫ ⋅ d l = aB + 0 + 0 + 0 = aB C

= µ 0 I C = µ 0 naI

576 Chapter 27 B = µ 0 nI

Solve for B to obtain:

71 •• Picture the Problem The magnetic field inside a tightly wound toroid is given by B = µ 0 NI (2πr ) , where a < r < b and a and b are the inner and outer radii of the toroid. Express the magnetic field of a toroid:

B=

µ 0 NI 2πr

(a) Substitute numerical values and evaluate B(1.1 cm):

B(1.1 cm ) =

(4π × 10

−7

)

N/A 2 (1000)(1.5 A ) = 27.3 mT 2π (1.1 cm )

(b) Substitute numerical values and evaluate B(1.5 cm):

(4π × 10 B(1.5 cm ) =

−7

)

N/A 2 (1000)(1.5 A ) = 20.0 mT 2π (1.5 cm )

*72 •• Picture the Problem In parts (a), (b), and (c) we can use a right-hand rule to determine the direction of the magnetic field at points above and below the infinite sheet of current. r r In part (d) we can evaluate B ⋅ d l around the specified path and equate it to µ0IC and

∫

C

solve for B.

(

)

(a)

At P the magnetic field points to the right i.e., in the − iˆ direction since its vertical components cancel.

(b)

Because the sheet is infinite, the same argument used in (a) applies; B is in the − iˆ direction.

(c)

Below the sheet the magnetic field points to the left, i.e., in the iˆ direction. The vertical components cancel.

(d) Express

r r B ∫ ⋅ d l , in the C

counterclockwise direction, for the given path:

r r B ∫ ⋅ dl = 2 C

r r r r B ⋅ d l + 2 B ∫ ∫ ⋅ dl

parallel

⊥

Sources of the Magnetic Field 577 r r ∫ B ⋅ dl = 0

For the paths perpendicular to the r r sheet, B and d l are perpendicular to each other and:

⊥

r

r

∫ B ⋅ d l = Bw

For the paths parallel to the sheet,

r r B and d l are in the same direction

parallel

and:

r

r

r

r

∫ B ⋅ d l = 2 ∫ B ⋅ d l = 2 Bw

Substitute to obtain:

C

parallel

= µ 0 I C = µ 0 (λw)

r B = 12 µ 0 λ and Babove = − 12 µ 0 λiˆ

Solve for B:

Magnetization and Magnetic Susceptibility 73 • Picture the Problem We can use B = Bapp = µ0 nI to find B and Bapp at the center when there is no core in the solenoid and B = Bapp + µ 0 M when there is an iron core with a magnetization M = 1.2×106 A/m.

B = Bapp = µ0 nI

(a) Express the magnetic field, in the absence of a core, in the solenoid :

⎛ 400 ⎞ ⎟⎟(4 A ) B = Bapp = 4π × 10 −7 N/A 2 ⎜⎜ 0 . 2 m ⎝ ⎠

(

Substitute numerical values and evaluate B and Bapp:

)

= 10.1 mT Bapp = 10.1 mT

(b) With an iron core with a magnetization M = 1.2×106 A/m present:

and

(

)(

)

B = Bapp + µ 0 M = 10.1 mT + 4π × 10 −7 N/A 2 1.2 × 10 6 A/m = 1.52 T 74 • Picture the Problem We can use B = Bapp = µ0 nI to find B and Bapp at the center when there is no core in the solenoid and B = Bapp + µ 0 M when there is an aluminum core. We

578 Chapter 27 can use M = χ m

Bapp

µ0

to find the magnetization of the core with the aluminum present.

Express the magnetic field, in the absence of a core, in the solenoid : Substitute numerical values and evaluate B and Bapp:

B = Bapp = µ0 nI

⎛ 400 ⎞ ⎟⎟ (4 A ) B = Bapp = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0 .2 m ⎠

(

)

= 10.1 mT

Bapp

Express the magnetization in the core with the aluminum present:

M = χm

Use Table 27-1 to find the value of χm for aluminum:

χ m, Al = 2.3 × 10−5

Substitute numerical values and evaluate M:

M = 2.3 × 10 − 5

µ0

10.1 mT 4π × 10 − 7 N/A 2

= 0.185 A/m 75 • Picture the Problem We can use Bapp = µ 0 nI to find Bapp at the center of the tungsten core in the solenoid. The magnetization is related to Bapp and χm according to M = χ m Bapp µ0 = χ m nI and we can use B = Bapp (1 + χ m ) to find B. Express the magnetic field, for a tungsten core, in the solenoid : Substitute numerical values and evaluate Bapp:

Bapp = µ0 nI

⎛ 400 ⎞ ⎟⎟ (4 A ) Bapp = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0 .2 m ⎠

(

)

= 10.053 mT

Bapp

Express the magnetization in the core with the aluminum present:

M = χm

Use Table 27-1 to find the value of χm for tungsten:

χ m, tungstenl = 6.8 × 10 −5

µ0

= χ m nI

Sources of the Magnetic Field 579

Substitute numerical values and evaluate M:

⎛ 400 ⎞ ⎟⎟(4 A ) M = 6.8 × 10 −5 ⎜⎜ ⎝ 0 .2 m ⎠

(

)

= 0.544 A/m Express B in terms of Bapp and χm: Substitute numerical values and evaluate B:

B = Bapp (1 + χ m ) B = (10.053 mT ) (1 + 6.8 × 10 −5 ) = 10.054 mT

76 • Picture the Problem We can use B = Bapp (1 + χ m ) to relate B and Bapp to the magnetic susceptibility of tungsten. Dividing both sides of this equation by Bapp and examining the value of χm, tungsten will allow us to decide whether the field inside the solenoid decreases or increases when the core is removed. Express the magnetic field inside the solenoid with the tungsten core present B in terms of Bapp and χm:

B = Bapp (1 + χ m ) where Bapp is the magnetic field in the absence of the tungsten core.

Express the ratio of B to Bapp:

B = 1 + χm Bapp

(a) Because χm, tungsten > 0:

B > Bapp

(1)

and

B will decrease when the tungsten core is removed. (b) From equation (1) the fractional change is:

χ m = 6.8 × 10 −5 = 6.8 × 10−3%

77 • Picture the Problem We can use B = Bapp (1 + χ m ) to relate B and Bapp to the magnetic susceptibility of liquid sample. Express the magnetic field inside the solenoid with the liquid sample present B in terms of Bapp and χm,

B = Bapp (1 + χ m, sample ) where Bapp is the magnetic field in the absence of the liquid sample.

580 Chapter 27 sample:

The fractional change in the magnetic field in the core is:

∆B = χ m, sample Bapp

Substitute numerical values and evaluate χm, sample:

χ m, sample =

∆B = −0.004% Bapp

= − 4.00 × 10 −5 78 • Picture the Problem We can use B = Bapp = µ0 nI to find B and Bapp at the center when

there is no core in the solenoid and B = Bapp (1 + χ m ) when there is an aluminum or silver

core.

B = Bapp = µ0 nI

(a) Express the magnetic field, in the absence of a core, in the solenoid:

Substitute numerical values and evaluate B and Bapp:

⎛ 50 ⎞ ⎟⎟ (10 A ) = 62.8 mT B = Bapp = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0.01 m ⎠

(

(b) With an aluminum core: Use Table 27-1 to find the value of χm for aluminum:

)

B = Bapp (1 + χ m )

χ m, Al = 2.3 × 10−5 and

1 + χ m, Al = 1 + 2.3 × 10 −5 ≈ 1 Substitute numerical values and evaluate B and Bapp:

⎛ 50 ⎞ ⎟⎟ (10 A ) = 62.8 mT B = Bapp = (4π × 10 −7 N/A 2 )⎜⎜ ⎝ 0.01 m ⎠ (c) With a silver core: Use Table 27-1 to find the value of χm for silver:

B = Bapp (1 + χ m )

χ m, Ag = −2.6 × 10−5 and

1 + χ m, Ag = 1 − 2.6 × 10 −5 ≈ 1

Sources of the Magnetic Field 581 Substitute numerical values and evaluate B and Bapp:

⎛ 50 ⎞ ⎟⎟ (10 A ) = 62.8 mT B = Bapp = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0.01 m ⎠

(

)

*79 •• Picture the Problem We can use the data in the table and Bapp = µ 0 nI to plot B versus Bapp. We can find Km using B = K m Bapp .

Bapp = µ0 nI

We can find the applied field Bapp for a long solenoid using: Km can be found from Bapp and B using:

Km =

B Bapp

The following graph was plotted using a spreadsheet program. The abscissa values for the graph were obtained by multiplying nI by µ0. B initially rises rapidly, and then becomes nearly flat. This is characteristic of a ferromagnetic material.

2.0 1.6

B (T)

1.2 0.8 0.4 0.0 0.000

0.002

0.004

0.006

0.008

0.010

0.012

0.014

B app (T)

The graph of Km versus nI shown below was also plotted using a spreadsheet program. Note that Km becomes quite large for small values of nI but then diminishes. A more revealing graph would be to plot B/(nI), which would be quite large for small values of nI and then drop to nearly zero at nI = 10,000 A/m, corresponding to saturation of the magnetization.

582 Chapter 27 6000 5000

Km

4000 3000 2000 1000 0 0

2000

4000

6000

8000

10000

nI (A/m)

80 •• Picture the Problem We can use the definition of the magnetization of a sample to find M and the relationship between the Bohr magneton and the magnetic moment of the sample to find the number of electrons aligned in the sample. In part (c) we can express the magnetic moment of the disk in terms of the amperian surface current and solve for the latter. (a) Express the magnetization of the sample in terms of its magnetic moment and volume:

M =

Substitute numerical values and evaluate M:

M=

µ V

=

µ πr 2 d

1.5 × 10 −2 A ⋅ m 2

π (1.4 cm )2 (0.3 cm )

= 8.12 × 10 3 A/m (b) Relate the magnetic moment of the sample to the Bohr magneton: Solve for and evaluate N:

µ = Nµ B

N=

1.5 × 10 −2 A ⋅ m 2 µ = µ B 9.27 × 10 −24 A ⋅ m 2

= 1.62 × 10 21 (c) Express the magnetic moment of the disk in terms of the amperian

µ = AI

Sources of the Magnetic Field 583 surface current: Solve for I and substitute for µ to obtain:

I=

µ A

=

MV MAt = = Mt A A

where t is the thickness of the disk.

I = (8.12 × 103 A/m )(0.3 cm )

Substitute numerical values and evaluate I:

= 24.4 A

81 •• Picture the Problem We can imagine the cylinder with the hole cut out as the superposition of two uniform cylinders with radii r and R, respectively, and magnetization −M and M, respectively. We can use the expression for B on the axis of a current loop to express the difference of the fields due to the two cylinders at a distance x from their common center. We’ll denote each field by the subscript identifying the radius of the current loop. From Problem 39 we have:

µ0 2π r 2 I µ0 r 2 I = 4π (x 2 + r 2 ) 3 2 2(x 2 + r 2 ) 3 2

Br = and

BR =

µ0 2π R 2 I µ0 R 2 I = 4π (x 2 + R 2 ) 3 2 2(x 2 + R 2 ) 3 2

The resultant field at x is the difference between BR and Br:

Bx ( x ) = BR (x ) − Br (x ) = =

µ0 I ⎡

R2

⎢ 2 ⎢⎣ x 2 + R 2

(

)

(

µ0 R 2 I

2 x2 + R2

32

−

(x

)

32

r2 2

+ r2

)

−

32

(

µ0 r 2 I

2 x2 + r 2

)

32

⎤ ⎥ ⎥⎦

The resultant magnetization of the disks is M = B/µ0:

M (x ) =

⎤ I⎡ R2 r2 − ⎢ 2 ⎥ 3 2 3 2 2 ⎢⎣ (x + R 2 ) (x 2 + r 2 ) ⎥⎦

The magnetization current is the product of M and the thickness of the disks:

584 Chapter 27 The magnetization is related to the amperian current:

M =

dI amperian dl

t

∫

⇒ I amperian = Mdl 0

Substitute for M to obtain:

I⎡ R2 =∫ ⎢ 2 ⎣⎢ x 2 + R 2 0 t

I amperian

(

)

32

−

(x

r2 2

+ r2

)

⎤ It ⎡ R2 dl = ⎢ 32⎥ 2 ⎢⎣ x 2 + R 2 ⎦⎥

(

)

32

−

(x

r2 2

+ r2

)

⎤ 32⎥ ⎦⎥

Atomic Magnetic Moments *82 •• Picture the Problem We can find the magnetic moment of a nickel atom µ from its relationship the saturation magnetization MS using M S = nµ where n is the number of molecules. n, in turn, can be found from Avogadro’s number, the density of nickel, and its molar mass using n =

NA ρ . Μ

M S = nµ

Express the saturation magnetic field in terms of the number of molecules per unit volume and the magnetic moment of each molecule:

or

µ=

MS n

Express the number of molecules per unit volume in terms of Avogadro’s number NA, the molecular mass M, and the density ρ:

n=

NA ρ Μ

Substitute and simplify to obtain:

µ=

MS µM µM M = 0 S = 0 S N A ρ µ0 N A ρ µ0 N A ρ Μ Μ

Substitute numerical values and evaluate µ:

µ=

(4π × 10

(0.61 T ) (58.7 × 10 −3 kg/mol) −7

N/A

2

)(6.02 × 10

Express the value of 1 Bohr magneton:

23

)(

atoms/mol 8.7 g/cm

3

)

= 5.44 × 10 −24 A ⋅ m 2

µ B = 9.27 ×10−24 A ⋅ m 2

Sources of the Magnetic Field 585

µ 5.44 ×10 −24 A ⋅ m 2 = = 0.587 µ B 9.27 ×10 −24 A ⋅ m 2

Divide µ by µB to obtain:

or

µ = 0.587 µ B 83 •• Picture the Problem We can find the magnetic moment of a cobalt atom µ from its relationship to the saturation magnetization MS using M S = nµ, where n is the number of molecules. n, in turn, can be found from Avogadro’s number, the density of cobalt, and its molar mass using n =

NA ρ . Μ

M S = nµ

Express the saturation magnetic field in terms of the number of molecules per unit volume and the magnetic moment of each molecule:

or

µ=

MS n

Express the number of molecules per unit volume in terms of Avogadro’s number NA, the molecular mass M, and the density ρ:

n=

NA ρ Μ

Substitute and simplify to obtain:

µ=

MS µM µM M = 0 S = 0 S N A ρ µ0 N A ρ µ0 N A ρ Μ Μ

Substitute numerical values and evaluate µ:

µ=

(4π × 10

(1.79 T ) (58.9 × 10 −3 kg/mol) −7

N/A

2

)(6.02 × 10

23

)(

atoms/mol 8.9 g/cm

3

)

= 1.57 × 10 −23 A ⋅ m 2

Express the value of 1 Bohr magneton:

µ B = 9.27 ×10−24 A ⋅ m 2

Divide µ by µB to obtain:

µ 1.57 ×10 −23 A ⋅ m 2 = = 1.69 µ B 9.27 ×10−24 A ⋅ m 2 or

µ = 1.69µ B

586 Chapter 27

Paramagnetism 84 • Picture the Problem We can show that χm = µµ0Ms/3kT by equating Curie’s law and the equation that defines χm ( M = χ m

Bapp

µ0

Express Curie’s law:

) and solving for χm.

M =

1 µBapp MS 3 kT

where MS is the saturation value. Express the magnetization of the substance in terms of its magnetic susceptibility χm:

M = χm

Equate these expressions to obtain:

χm

Bapp

µ0

Bapp

µ0

=

1 µBapp MS 3 kT

or

χm 1 µ = M µ0 3 kT S

Solve for χm to obtain:

χm =

µ0 µM S 3kT

85 •• Picture the Problem We can use the assumption that M = fM S and Curie’s law to solve these equations simultaneously for the fraction f of the molecules have their magnetic moments aligned with the external magnetic field. (a) Assume that some fraction f of the molecules have their magnetic moments aligned with the external magnetic field and that the rest of the molecules are randomly oriented and so do not contribute to the magnetic field: From Curie’s law we have:

M = fM S

M =

1 µBapp MS 3 kT

Sources of the Magnetic Field 587 Equate these expressions and solve for f to obtain:

fM S =

1 µBapp MS 3 kT

and

µB

f =

3kT

because B given in the problem statement is the external magnetic field Bapp.

(9.27 × 10 f = 3(1.381 × 10

)

A ⋅ m 2 (1 T ) − 23 J/K (300 K )

−24

(b) Substitute numerical values and evaluate f:

)

= 7.46 × 10 −4 *86 •• Picture the Problem In (a) we can express the saturation magnetic field in terms of the number of molecules per unit volume and the magnetic moment of each molecule and use n = N A ρ Μ to express the number of molecules per unit volume in terms of Avogadro’s number NA, the molecular mass M, and the density ρ. We can use χ m = µ0 µM S 3kT from Problem 84 to calculate χm. (a) Express the saturation magnetic field in terms of the number of molecules per unit volume and the magnetic moment of each molecule:

M S = nµ B

NA ρ Μ

Express the number of molecules per unit volume in terms of Avogadro’s number NA, the molecular mass M, and the density ρ:

n=

Substitute to obtain:

MS =

NA ρ µB Μ

Substitute numerical values and evaluate MS:

MS =

(6.02 ×10

23

)(

)(

atoms/mol 2.7 × 103 kg/m 3 9.27 × 10−24 A ⋅ m 2 27 g/mol

= 5.58 × 105 A/m and

BS = µ 0 M S = (4π × 10 −7 N/A 2 )(5.58 × 105 A/m ) = 0.701 T

)

588 Chapter 27

(b) From Problem 84 we have:

χm =

µ0 µM S 3kT

Substitute numerical values and evaluate χm:

χm =

(4π ×10

−7

)(

)(

)

N/A 2 9.27 × 10−24 A ⋅ m 2 5.58 × 105 A/m = 5.23 × 10−4 3 1.381× 10−23 J/K (300 K )

(

)

(c) In calculating χ m in (b ) we neglected any diamagnetic effects. 87 •• Picture the Problem We can use Equation 27-17 to express Bapp and Equation 27-21 to express B in terms of Bapp and M.

µ0 NI for R − r < a < R + r 2πa

Express Bapp inside a tightly wound toroid:

Bapp =

The resultant field B in the ring is the sum of Bapp and µ0M:

B = Bapp + µ 0 M =

µ 0 NI + µ0 M 2πa

88 •• Picture the Problem We can find the magnetization using M = χ m Bapp µ0 and the magnetic field using B = Bapp (1 + χ m ) .

(a) Using Equation 27-22, express the magnetization M in terms of χm and Bapp:

M = χm

Express Bapp inside a tightly wound toroid:

Bapp =

Substitute to obtain:

Substitute numerical values and evaluate M:

Bapp

µ0

µ0 NI 2πrmean

µ0 NI NI 2πrmean M = χm = χm µ0 2πrmean

(4 × 10 )(2000)(15 A ) M= −3

2π (0.2 m )

= 95.5 A/m

Sources of the Magnetic Field 589

B = Bapp (1 + χ m )

(b) Express B in terms of Bapp and χm: Substitute for Bapp to obtain:

B=

µ0 NI (1 + χ m ) 2πrmean

Substitute numerical values and evaluate B:

B=

(4π × 10

−7

)

N/A 2 (2000)(15 A ) 1 + 4 × 10 −3 = 30.1 mT 2π (0.2 m )

(

)

∆B B − Bapp = B B Bapp (1 + χ m ) − Bapp χ m Bapp = = B B 1 χm = = 1 1 + χm +1

(c) Express the fractional increase in B produced by the liquid oxygen:

χm

∆B = B

Substitute numerical values and evaluate ∆B/B:

1 1 +1 4 × 10 −3

= 3.98 × 10 −3

= 0.398% 89

••

Picture the Problem We can use B = Bapp (1 + χ m ) and Bapp = within the substance and M = χ m

Bapp

µ0

µ0 NI = µ0 nI to find B 2πrmean

to find the magnitude of the magnetization.

(a) Express the magnetic field B within the substance in terms of Bapp and χm:

B = Bapp (1 + χ m )

µ0 NI = µ0 nI 2πrmean

Express Bapp inside the toroid:

Bapp =

Substitute to obtain:

B = µ0 nI (1 + χ m )

Substitute numerical values and evaluate B:

590 Chapter 27

(

)(

)

(

)

B = 4π × 10 −7 N/A 2 60 × 10 2 m −1 (4 A ) 1 + 2.9 × 10 −4 = 30.2 mT (b) Express the magnetization M in terms of χm and Bapp:

M = χm

Substitute for Bapp to obtain:

M = χm

Substitute numerical values and evaluate M:

(c) If there were no paramagnetic core present:

Bapp

µ0 µ0 nI = χ m nI µ0

(

)(

)

M = 2.9 × 10 −4 6000 m −1 (4 A ) = 6.96 A/m B = Bapp = 30.2 mT

Ferromagnetism *90 • Picture the Problem We can use B = K m Bapp to find B and M = (K m − 1) Bapp µ 0 to find M. Express B in terms of M and Km: Substitute numerical values and evaluate B:

Relate M to Km and Bapp:

Substitute numerical values and evaluate M:

B = K m Bapp

(

B = (5500 ) 1.57 × 10 −4 T

)

= 0.864 T M = (K m − 1)

M =

Bapp

µ0

≈

K m Bapp

µ0

(5500) (1.57 × 10 −4 T ) 4π × 10 −7 N/A 2

= 6.87 × 105 A/m 91 •• Picture the Problem We can relate the permeability µ of annealed iron to χm using

µ = (1 + χ m )µ0 , find χm using Equation 27-22 ( M = χ m ( K m = 1 + χ m ) to evaluate Km.

Bapp

µ0

), and use its definition

Sources of the Magnetic Field 591 Express the permeability µ of annealed iron in terms of its magnetic susceptibility χm:

µ = (1 + χ m )µ0

(1)

Bapp

Using Equation 27-22, express the magnetization M in terms of χm and Bapp:

M = χm

Solve for and evaluate χm (see Table 27-2 for the product of µ0 and M):

χm =

Use its definition to express and evaluate the relative permeability Km:

K m = 1 + χ m = 1 + 10.75 = 11.75

Substitute numerical values in equation (1) and evaluate µ:

µ0

µ0 M Bapp

=

2.16 T = 10.75 0.201 T

µ = (1 + 10.75) (4π × 10 −7 N/A 2 ) = 1.48 × 10 −5 N/A 2

92 •• Picture the Problem We can use the relationship between the magnetic field on the axis of a solenoid and the current in the solenoid to find the minimum current is needed in the solenoid to demagnetize the magnet. Relate the magnetic field on the axis of a solenoid to the current in the solenoid:

Bx = µ0 nI

Solve for I to obtain:

I=

Bx µ0 n

Let Bapp = Bx to obtain:

I=

Bapp

Substitute numerical values and evaluate I:

I=

µ0 n 5.53 × 10 −2 T ⎛ 600 ⎞ ⎟⎟ 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0.15 m ⎠

(

= 11.0 A

)

592 Chapter 27 93 •• Picture the Problem We can use the equation describing the magnetic field on the axis of a solenoid, as a function of the current in the solenoid, to find Bapp. We can then use B = Bapp + µ0 M to find M and B = K m Bapp to evaluate Km. (a) Relate the magnetic field on the axis of a solenoid to the current in the solenoid: Substitute numerical values to obtain:

Bx = µ0 nI

(

)(

)

Bapp = 4π × 10 −7 N/A 2 50 cm −1 (2 A ) = 12.6 mT

(b) Relate M to B and Bapp:

B = Bapp + µ0 M

Solve for and evaluate M:

M =

B − Bapp

µ0

=

1.72 T − 12.6 mT 4π × 10 −7 N/A 2

= 1.36 × 10 6 A/m (c) Express B in terms of Km and Bapp: Solve for and evaluate Km:

B = K m Bapp

Km =

B 1.72 T = = 137 Bapp 12.6 mT

94 •• Picture the Problem We can use the equation describing the magnetic field on the axis of a solenoid, as a function of the current in the solenoid, to find Bapp. We can then use B = Bapp + µ0 M to find M and B = K m Bapp to evaluate Km. (a) Relate the magnetic field on the axis of the solenoid to the current in the solenoid: Substitute numerical values and evaluate Bapp:

(b) Relate M to B and Bapp:

Bx = µ0 nI

(

)(

)

Bapp = 4π × 10 −7 N/A 2 50 cm −1 (0.2 A ) = 1.26 mT

B = Bapp + µ0 M

Sources of the Magnetic Field 593 Solve for M:

B − Bapp

M =

Substitute numerical values and evaluate M:

M =

µ0 1.58 T − 1.26 mT 4π × 10−7 N/A 2

= 1.26 × 106 A/m B = K m Bapp

(c) Express B in terms of Km and Bapp: Solve for and evaluate Km:

Km =

B 1.58 T = = 1.25 × 103 Bapp 1.26 mT

95 •• Picture the Problem The magnetic field in the core of a hollow solenoid is related to the current in its coils according to Bx = Bapp = µ0 nI . The presence of the iron increases the magnetic field by a factor of Km. In part (b), requiring that the magnetic field be unchanged when the iron core is removed will allow us to find the current that will produce the same field within the solenoid.

Bx = Bapp = µ0 nI

(a) Relate the magnetic field on the axis of the solenoid to the current in the solenoid: Express B in terms of Bapp:

B = K m Bapp

Substitute to obtain:

B = K m µ0 nI

Substitute numerical values and evaluate B:

(

)(

)

B = 1200 4π × 10 −7 N/A 2 2000 m −1 (20 mA ) = 60.3 mT (b) We require, that with the iron core removed, the magnetic field is unchanged: Solve for and evaluate I0:

B = K m µ0 nI = µ0 nI 0

I 0 = K m I = 1200(20 mA ) = 24.0 A

594 Chapter 27 *96 •• Picture the Problem Because the wires carry equal currents in opposite directions, the magnetic field midway between them will be twice that due to either current alone and will be greater, by a factor of Km, than it would be in the absence of the insulator. We can use Ampère’s law to find the field, due to either current, at the midpoint of the plane of r r r the wires and dF = Id l × B to find the force per unit length on either wire. (a) Relate the magnetic field in the insulator to the magnetic field in its absence: Apply Ampère’s law to a closed circular path a distance r from a current-carrying wire to obtain:

B = K m Bapp

r r B ∫ ⋅ d l = Bapp (2πr ) = µ0 I C = µ0 I C

µ0 I 2πr

Solve for Bapp to obtain:

Bapp =

Because there are two current carrying wires, with their currents in opposite directions, the fields are additive and:

B = 2K m

Substitute numerical values and evaluate B:

B=

µ0 I K m µ0 I = 2πr πr

(

)

120 4π × 10 −7 N/A 2 (40 A ) π (0.02 m )

= 96.0 mT (b) Express the force per unit length experienced by either wire due to the current in the other: Apply Ampère’s law to obtain:

F = BI l r

r

∫ B ⋅ d l = B(2πr ) = µ I

0 C

C

= µ0 I

where r is the separation of the wires.

µ0 I K µI and Bapp = m 0 2πr 2πr

Solve for B:

B=

Substitute to obtain:

F K m µ0 I 2 = l 2πr

Sources of the Magnetic Field 595 Substitute numerical values and evaluate

F : l

F 120(4π × 10 −7 N/A 2 )(40 A ) = l 2π (0.04 m )

2

= 0.960 N/m

97 •• Picture the Problem We can use B = Bapp + µ 0 M and the expression for the magnetic field inside a tightly wound toroid to find the magnetization M. We can find Km from its definition, µ = K m µ 0 to find µ, and K m = 1 + χ m to find χm for the iron sample. (a) Relate the magnetization to B and Bapp: Solve for M:

Express the magnetic field inside a tightly wound toroid:

B = Bapp + µ0 M

M =

µ0

Bapp =

Substitute and simplify to obtain:

M =

Substitute numerical values and evaluate M:

B − Bapp

M =

µ0 NI 2πr

B−

µ0 NI 2πr = B − NI µ0 µ0 2πr

1.8 T 2000(10 A ) − −7 2 4π ×10 N/A 2π (0.2 m )

= 1.42 ×106 A/m (b) Use its definition to express Km:

Km =

Substitute numerical values and evaluate Km:

Km =

2πrB B B = = Bapp µ0 NI µ0 NI 2πr 2π (0.2 m )(1.8 T ) 4π × 10 −7 N/A 2 (2000)(10 A )

(

)

= 90.0 Now that we know Km we can find µ using:

Relate χm to Km:

µ = K m µ0 = 90(4π × 10−7 N/A 2 ) = 1.13 × 10 −4 T ⋅ m/A Km = 1+ χm

596 Chapter 27 Solve for and evaluate χm:

χ m = K m − 1 = 89.0

98 •• Picture the Problem We can substitute the expression for applied magnetic field ( Bapp =

µ0 NI ) in the defining equation for Km ( B = K m Bapp ) to obtain an expression 2πr

for the magnetic field B in the toroid. Relate the magnetic field in the toroid to the relative permeability of its core:

B = K m Bapp

Express the applied magnetic field in the toroid in terms of the current in its winding:

Bapp =

Substitute to obtain:

B=

Express the number of turns N of wire in terms of the number of turns per unit length n: Substitute to obtain: Substitute numerical values and evaluate B:

µ0 NI 2πr

K m µ0 NI 2πr

N = 2πrn

B = K m µ0 nI

(

)(

)

B = 500 4π × 10 −7 N/A 2 60 cm −1 (0.2 A ) = 0.754 T

99 •• Picture the Problem We can use Ampère’s law to obtain expressions for the magnetic field inside the wire, inside the ferromagnetic material, and in the region outside the insulating ferromagnetic material. (a) Apply Ampère’s law to a circle of radius r < 1 mm and concentric with the center of the wire: Assuming that the current is distributed uniformly over the crosssectional area of the wire (uniform current density), express IC in terms

r r B ∫ ⋅ d l = B(2πr ) = µ0 I C C

IC I = 2 2 πr πR or

Sources of the Magnetic Field 597 r2 I R2

of the total current I:

IC =

Substitute to obtain:

B(2πr ) =

Solve for B:

B=

Substitute numerical values and evaluate B:

B=

µ0 Ir 2 R2

µ0 I r 2πR 2

(4π × 10

)

N/A 2 (40 A ) r 2 2π (1 mm ) −7

= (8.00 T/m ) r (b) Relate the magnetic field inside the ferromagnetic material to the magnetic field due to the current in the wire: Apply Ampère's law to a circle of radius 1 mm < r < 4 mm and concentric with the center of the wire:

B = K m Bapp

r r B ∫ ⋅ d l = Bapp (2πr ) = µ0 I C = µ0 I C

Solve for Bapp:

Bapp =

Substitute to obtain:

B=

Substitute numerical values and evaluate B:

B=

(c) Apply Ampère’s law to a circle of radius r > 4 mm and concentric with the center of the wire: Solve for B:

µ0 I 2πr

K m µ0 I 2πr

400(4π × 10 −7 N/A 2 )(40 A ) 2πr 1 = (3.20 × 10 −3 T ⋅ m ) r

r

r

∫ B ⋅ d l = B(2πr ) = µ I

0 C

C

B=

µ0 I 2πr

= µ0 I

598 Chapter 27

Substitute numerical values and evaluate B:

N/A 2 )(40 A ) 2πr 1 = (8.00 × 10 −6 T ⋅ m ) r

B=

(4π × 10

−7

(d) Note that the field in the ferromagnetic region is that which would be produced in a nonmagnetic region by a current of 400I = 1600 A. The ampèrian current on the inside of the surface of the ferromagnetic material must therefore be (1600 − 40) A = 1560 A in the direction of I. On the outside surface there must then be an ampèrian current of 1560 A in the opposite direction.

General Problems 100 • Picture the Problem Because point P is on the line connecting the straight segments of the conductor, these segments do not contribute to the magnetic field at P. Hence, we can use the expression for the magnetic field at the center of a current loop to find BP. Express the magnetic field at the center of a current loop:

B=

µ0 I 2R

where R is the radius of the loop. Express the magnetic field at the center of half a current loop:

B=

Substitute numerical values and evaluate B:

B=

1 µ0 I µ0 I = 2 2R 4R

(4π × 10

)

N/A 2 (15 A ) 4(0.2 m ) −7

= 2.36 × 10 −5 T *101 • Picture the Problem Let out of the page be the positive x direction. Because point P is on the line connecting the straight segments of the conductor, these segments do not contribute to the magnetic field at P. Hence, the resultant magnetic field at P will be the sum of the magnetic fields due to the current in the two semicircles, and we can use the

r

expression for the magnetic field at the center of a current loop to find BP . Express the resultant magnetic field at P:

r r r BP = B1 + B2

Sources of the Magnetic Field 599 Express the magnetic field at the center of a current loop:

B=

µ0 I 2R

where R is the radius of the loop. Express the magnetic field at the center of half a current loop:

r

r

Express B1 and B2 :

B=

1 µ0 I µ0 I = 2 2R 4R

r µI B1 = 0 iˆ 4 R1 and

r µI B2 = − 0 iˆ 4 R2

Substitute to obtain:

102

r µ I µ I µ I⎛ 1 1 ⎞ˆ ⎟i B P = 0 iˆ − 0 iˆ = 0 ⎜⎜ − 4 R1 4 R2 4 ⎝ R1 R2 ⎟⎠

••

Picture the Problem We can express B as a function of N, I, and R using B = eliminate R by relating l to R through l = 2πRN .

µ0 NI 2R

and

µ0 NI

Express the magnetic field at the center of a coil of N turns and radius R:

B=

Relate l to the number of turns N:

l = 2πRN

Solve for R to obtain:

R=

Substitute to obtain:

B=

2R

l 2πN

µ 0 NI 2

l 2πN

=

µ 0πN 2 I l

103 •• Picture the Problem The magnetic field at P (which is out of the page) is the sum of the magnetic fields due to the three parts of the wire. Let the numerals 1, 2, and 3 denote the left-hand, center (short), and right-hand wires. We can then use the expression for B due to a straight wire segment to find each of these fields and their sum.

600 Chapter 27 Express the resultant magnetic field at point P:

BP = B1 + B2 + B3

Because B1 = B3:

BP = 2 B1 + B2

Express the magnetic field due to a straight wire segment:

B=

For wires 1 and 3 (the long wires), θ1 = 90° and θ2 = 45°:

B1 =

For wire 2, θ1 = θ2 = 45°:

Substitute and simplify to obtain:

µ0 I (sin θ1 + sin θ 2 ) 4π R

µ0 4π µ = 0 4π

I (sin 90° + sin 45°) a 1 ⎞ I⎛ ⎜1 + ⎟ a⎝ 2⎠

µ0 4π µ = 0 4π

I (sin 45° + sin 45°) a I⎛ 2 ⎞ ⎜ ⎟ a⎝ 2⎠

B2 =

⎡µ I ⎛ 1 ⎞⎤ µ 0 I ⎛ 2 ⎞ BP = 2 ⎢ 0 ⎜ 1 + ⎟⎥ + ⎜ ⎟ 2 ⎠⎦ 4π a ⎝ 2 ⎠ ⎣ 4π a ⎝ µ I⎛ 1 1 ⎞ = 0 ⎜1 + + ⎟ 2π a ⎝ 2 2⎠ =

(

µ0 I ⎛ µ0 I 2 ⎞ 1+ 2 ⎜1 + ⎟= 2π a ⎝ 2π a 2⎠

)

*104 •• Picture the Problem Depending on the direction of the wire, the magnetic field due to its current (provided this field is a large enough fraction of the earth’s magnetic field) will either add to or subtract from the earth’s field and moving the compass over the ground in the vicinity of the wire will indicate the direction of the current. Apply Ampère’s law to a circle of radius r and concentric with the center of the wire: Solve for B to obtain:

r r B ∫ ⋅ d l = Bwire (2πr ) = µ0 I C = µ0 I C

Bwire =

µ0 I 2πr

Sources of the Magnetic Field 601

(4π × 10

)

N/A 2 (50 A ) 2π (2 m ) = 0.0500 G −7

Substitute numerical values and evaluate Bwire:

Bwire =

Express the ratio of Bwire to Bearth:

Bwire 0.05 G = ≈ 7% Bearth 0.7 G Thus, the field of the current-carrying wire should be detectable with a good compass.

If the cable runs east-west, its magnetic field is in the north-south direction and thus either adds to or subtracts from the earth’s field, depending on the current direction and location of the compass. Moving the compass over the region one should be able to detect the change. If the cable runs north-south, its magnetic field is perpendicular to that of the earth, and moving the compass about one should observe a change in the direction of the compass needle. 105 •• r r r Picture the Problem Let I1 and I2 represent the currents of 20 A and 5 A, F1 , F2 , F3 ,

r

and F4 the forces that act on the horizontal wire at the top of the loop, and the other

r

r

r

r

wires following the current in a counterclockwise direction, and B1 , B2 , B3 , and B4 the magnetic fields at these wires due to I1. Let the positive x direction be to the right and the positive y direction be upward. Note that only the components into or out of the paper r r r r r r r r of B1 , B2 , B3 , and B4 contribute to the forces F1 , F2 , F3 , and F4 , respectively. (a) Express the forces F2 and F4 in

r

r r r F2 = I 2 l 2 × B2

terms of I2 and B2 and B4 :

and

r

r

r

Express B2 and B4 :

r

r

r r r F4 = I 2 l 4 × B4 r µ 2I B2 = − 0 1 kˆ 4π R1

and

r µ 2I B4 = − 0 1 kˆ 4π R4

602 Chapter 27

r ⎛ µ 2I ⎞ F2 = − I 2l 2 ˆj × ⎜⎜ − 0 1 kˆ ⎟⎟ ⎝ 4π R1 ⎠ µl II = 0 2 1 2 iˆ 2πR2

Substitute to obtain:

and

r ⎛ µ 2I ⎞ F4 = I 2 l 4 ˆj × ⎜⎜ − 0 1 kˆ ⎟⎟ ⎝ 4π R4 ⎠ µl II = − 0 4 1 2 iˆ 2πR4 r

r

Substitute numerical values and evaluate F2 and F4 :

r (4π × 10 −7 N/A 2 )(0.1 m )(20 A )(5 A ) F2 = iˆ = (1.00 × 10 −4 N ) iˆ 2π (0.02 m ) and

r (4π × 10−7 N/A 2 )(0.1 m )(20 A )(5 A ) iˆ = F4 = − 2π (0.07 m )

−4

)

N iˆ

r r r r r Fnet = F1 + F2 + F3 + F4

(b) Express the net force acting on the coil:

(1)

r r F1 + F3 = 0

Because the lengths of segments 1 and 3 are the same and the currents in these segments are in opposite directions:

r

(− 0.286 × 10

and

r r r Fnet = F2 + F4

r

Substitute for F2 and F4 in equation (1) and simplify to obtain:

r Fnet = (− 0.250 × 10 −4 N ) ˆj + (1.00 × 10 −4 N ) iˆ + (0.250 × 10 −4 N ) ˆj + (− 0.286 × 10 −4 N ) iˆ

(

)

= 0.714 × 10 −4 N iˆ 106 •• Picture the Problem Let out of the page be the positive x direction and the numerals 40 and 60 refer to the circular arcs whose radii are 40 cm and 60 cm. Because point P is on the line connecting the straight segments of the conductor, these segments do not contribute to the magnetic field at P. Hence the resultant magnetic field at P will be the sum of the magnetic fields due to the current in the two circular arcs and we can use the

Sources of the Magnetic Field 603 r expression for the magnetic field at the center of a current loop to find BP . r r r BP = B40 + B60

Express the resultant magnetic field at P: Express the magnetic field at the center of a current loop:

B=

µ0 I 2R

where R is the radius of the loop. Express the magnetic field at the center of one-sixth of a current loop:

r

B=

r

1 µ0 I µ0 I = 6 2 R 12 R

r µI B40 = − 0 iˆ 12 R40

Express B40 and B60 :

and

r µI B60 = 0 iˆ 12 R60

r µI µI BP = − 0 iˆ + 0 iˆ 12 R40 12 R60

Substitute to obtain:

=

µ0 I ⎛ 1

1 ⎞ˆ ⎜⎜ ⎟i − 12 ⎝ R60 R40 ⎟⎠

r

Substitute numerical values and evaluate BP :

(

)

r 4π × 10 −7 N/A 2 (8 A ) ⎛ 1 1 ⎞ˆ ⎜⎜ ⎟⎟ i = BP = − 12 ⎝ 0.6 m 0.4 m ⎠

(− 6.98 × 10 T )iˆ −7

107 •• Picture the Problem Let the positive x direction be into the page and the numerals 20 and 40 refer to the circular arcs whose radii are 20 cm and 40 cm. Because point P is on the line connecting the straight segments of the conductor, these segments do not contribute to the magnetic field at P and the resultant field at P is the sum of the fields due to the two semicircular current loops. Express the resultant magnetic field at P: Express the magnetic field at the center of a circular current loop:

r r r BP = B20 + B40

B=

µ0 I 2R

604 Chapter 27 where R is the radius of the loop. Express the magnetic field at the center of half a circular current loop:

r

r

B=

1 µ0 I µ0 I = 2 2R 4R

Express B20 and B40 :

r r µI µI B20 = 0 iˆ and B40 = 0 iˆ 4 R20 4 R40

Substitute to obtain:

r µI µI BP = 0 iˆ + 0 iˆ 4 R20 4 R40 =

µ0 I ⎛ 1

1 ⎞ˆ ⎜⎜ ⎟i + 4 ⎝ R20 R40 ⎟⎠

Substitute numerical values and evaluate BP:

(

)

r 4π × 10 −7 N/A 2 (3 A ) ⎛ 1 1 ⎞ˆ ⎜⎜ ⎟⎟ i = BP = + 4 ⎝ 0.2 m 0.4 m ⎠

(7.07 µT ) iˆ

*108 •• Picture the Problem Chose the coordinate system shown to the right. Then the current is in the positive z direction. Assume that the electron is at (1 cm, 0, 0). We can use

r r r µ 2I r r r ˆj to F = qv × B to relate the magnetic force on the electron to v and B and B = 0 4π r r express the magnetic field at the location of the electron. We’ll need to express v for r r r each of the three situations described in the problem in order to evaluate F = qv × B .

Express the magnetic force acting on the electron:

r r r F = qv × B

Sources of the Magnetic Field 605

r µ 2I ˆj B= 0 4π r

Express the magnetic field due to the current in the wire as a function of distance from the wire:

( )

r r µ 2 I ˆ 2qµ0 I r ˆ F = qv × 0 j= v× j 4π r 4πr

Substitute to obtain:

(1)

r v = viˆ

(a) Express the velocity of the electron when it moves directly away from the wire:

(

)

r 2qµ0 I 2qµ0 Iv ˆ viˆ × ˆj = F= k 4πr 4πr

Substitute to obtain:

r

Substitute numerical values and evaluate F :

(

)(

)(

)

r 2 4π × 10 −7 N/A 2 − 1.6 × 10 −19 C 5 × 10 6 m/s (20 A ) kˆ F= 4π (0.01 m )

=

(− 3.20 × 10

−16

N )kˆ

r

r v = vkˆ

(b) Express v when the electron is traveling parallel to the wire in the direction of the current:

(

)

r 2qµ0 I 2qµ0 Iv ˆ vkˆ × ˆj = − F= i 4πr 4πr

Substitute in equation (1) to obtain:

r

Substitute numerical values and evaluate F :

(

)(

)(

)

r 2 4π × 10 −7 N/A 2 − 1.6 × 10 −19 C 5 × 10 6 m/s (20 A ) iˆ F =− = 3.20 × 10 −16 N iˆ 4π (0.01 m ) r

(c) Express v when the electron is traveling perpendicular to the wire and tangent to a circle around the wire: Substitute in equation (1) to obtain:

(

r v = vˆj

(

)

r 2 qµ 0 I F= vˆj × ˆj = 0 4πr

)

606 Chapter 27 109 •• Picture the Problem We can apply Ampère's law to derive expressions for the magnetic field as a function of the distance from the center of the wire. Apply Ampère's law to a closed circular path of radius r < r0 to obtain:

Br
Because the current is uniformly distributed over the cross section of the wire:

IC I r2 = ⇒ I = I C π r 2 π r02 r02

Substitute to obtain:

Br
Solve for Br
Br
Apply Ampère's law to a closed circular path of radius r > r0 to obtain: Solve for Br >r0 :

µ0 r 2 I r02

µ 0 rI µ 0 2 I = r 2πr02 4π r02

(1)

Br >r0 (2πr ) = µ 0 I C = µ 0 I

Br >r0 =

µ0 2I 4π r

(2)

The spreadsheet program to calculate B as a function of r in the interval 0 ≤ r ≤ 10r0 is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1

Formula/Content 1.00E−07

Algebraic Form

B2 B3 A6 B6 C6

5 1 2.55E−03 0.00E+00 10^4*$B$1*2*$B$2*A6/$B$3^2

I I r (m) r (mm)

C17

10^4*$B$1*2*$B$2*A6/A17

1

A B mu/4pi= 1.00E−07

µ0 4π

µ0 2 I r 4π r02 µ0 2 I 4π r C N/A^2

Sources of the Magnetic Field 607 2 3 4 5 6 7 8 9 10

I= 5 r_0= 2.55E−03

A m

r (m) 0.00E+00 2.55E−04 5.10E−04 7.65E−04 1.02E−03

r (mm) 0.00E+00 2.55E−01 5.10E−01 7.65E−01 1.02E+00

B (T) 0.00E+00 3.92E−01 7.84E−01 1.18E+00 1.57E+00

102 103 104 105 106

2.45E−02 2.47E−02 2.50E−02 2.52E−02 2.55E−02

2.45E+01 2.47E+01 2.50E+01 2.52E+01 2.55E+01

4.08E−01 4.04E−01 4.00E−01 3.96E−01 3.92E−01

A graph of B as a function of r follows. 4

B (G)

3

2

1

0 0

4

8

12

16

20

24

r (mm)

110 •• Picture the Problem We can use

r

r r r τ = µ × B to find the torque exerted on the small coil r

(magnetic moment = µ ) by the magnetic field B due to the current in the large coil. Relate the torque exerted by the large coil on the small coil to the r magnetic moment µ of the small

r

coil and the magnetic field B due to the current in the large coil: Express the magnetic moment of the small coil:

r r r τ = µ× B or, because the planes of the two coils are perpendicular,

τ = µB

µ = NIA where I is the current in the coil, N is the number of turns in the coil, and A is the

608 Chapter 27 cross-sectional area of the coil. Express the magnetic field at the center of the large coil:

B=

N'µ 0 I' 2R

where I′is the current in the large coil, N′ is the number of turns in the coil, and R is its radius. Substitute to obtain:

τ=

NN'II'Aµ 0 2R

Substitute numerical values and evaluate τ :

τ=

(50)(20)(4 A )(1 A )π (0.5 cm )2 (4π × 10 −7 N/A 2 ) = 2(10 cm )

1.97 µN ⋅ m

*111 •• Picture the Problem We can apply Newton’s 2nd law for rotational motion to obtain the differential equation of motion of the bar magnet. While this equation is not linear, we can use a small-angle approximation to render it linear and obtain an expression for the square of the angular frequency that we can solve for κ when there is an external field and for the period T in the absence of an external field. Apply

∑τ = Iα to the bar magnet

when B ≠ 0 to obtain the differential equation of motion for the magnet:

For small displacements from equilibrium (θ << 1): Rewrite the differential equation as:

− κθ − µB sin θ = I

d 2θ dt 2

where I is the moment of inertia of the magnet about an axis through its point of suspension.

− κθ − µBθ ≈ I

I

d 2θ dt 2

d 2θ + (κ + µB )θ = 0 dt 2

or

d 2θ ⎛ κ + µB ⎞ +⎜ ⎟θ = 0 dt 2 ⎝ I ⎠ Because the coefficient of the linear term is the square of the angular frequency, we have:

ω2 =

κ + µB I

(1)

Sources of the Magnetic Field 609

Express the moment of inertia (see Table 9-1) of the bar magnet about an axis through its center:

I = 121 mL2

Substitute to obtain:

ω2 =

κ + µB 1 12

mL2 ⎛ 4π 2 ⎞ ⎟ − µB 2 ⎟ ⎝T ⎠

Solve for κ to obtain:

κ = 121 mL2ω 2 − µB = 121 mL2 ⎜⎜ =

π 2 mL2 3T 2

− µB

Substitute numerical values and evaluate κ :

κ=

π 2 (0.8 kg )(0.16 m )2 3(0.5 s )

2

(

)

− 0.12 A ⋅ m 2 (0.2 T ) = 0.246 N ⋅ m/rad

Substitute B = 0 and ω = 2π/T in equation (1) to obtain:

4π 2 κ = T2 I

Solve for T:

T = 2π

Substitute numerical values and evaluate T:

I

κ

= 2π

T = π (0.16 m )

mL2 m = πL 12κ 3κ

0.8 kg 3(0.246 N ⋅ m/rad )

= 0.523 s 112 •• Picture the Problem We can apply Newton’s 2nd law for rotational motion to obtain the differential equation of motion of the bar magnet. While this equation is not linear, we can use a small-angle approximation to render it linear and obtain an expression for the square of the angular frequency that we can solve for the frequency f of the motion. Apply

∑τ = Iα to the bar magnet

to obtain the differential equation of motion for the magnet:

− µB sin θ = I

d 2θ dt 2

where I is the moment of inertia of the magnet about an axis through its point of suspension.

610 Chapter 27

For small displacements from equilibrium (θ << 1):

d 2θ − µBθ ≈ I 2 dt

Rewrite the differential equation as:

d 2θ I 2 + µBθ = 0 dt or

d 2θ µB + θ =0 dt 2 I Because the coefficient of the linear term is the square of the angular frequency, we have: Solve for ω to obtain:

ω2 =

ω=

µB I

µB I

113 •• Picture the Problem We can use the potential energy of the displaced bar magnet to find the force acting on it to return it to its equilibrium position. While this restoring force is not, in general, linear, we can use a binomial expansion to show that for displacements that are small compared to the radius of the coil, the restoring force is linear and, hence, the motion of the bar magnet is simple harmonic motion. We can then apply Newton’s 2nd law to obtain the differential equation of motion of the bar magnet and use the coefficient of the linear term to express the period of the motion. Express the potential energy of the displaced bar magnet: Express the magnetic field on the axis of the current loop:

U = − µB

B=

µ0 2πNR 2 I 4π (x 2 + R 2 )3 2

where I is the current in the loop and R is its radius. Substitute to obtain:

µ0 2πµNR 2 I U =− 4π (x 2 + R 2 )3 2

Sources of the Magnetic Field 611 Differentiate U with respect to x to find the restoring force acting on the bar magnet:

2

Factor R from the radical to obtain:

Fx = −

dU dx

[(

)

(

)

⎤ ⎡ ⎥ ⎢ 3µ0 µNR 2 I ⎢ 1 ⎥x Fx = − 5 2 5 ⎥ ⎢ 2 2R ⎢ ⎛⎜1 + x ⎞⎟ ⎥ ⎢⎣ ⎜⎝ R 2 ⎟⎠ ⎥⎦ 3µ µNI ⎛ x2 ⎞ = − 0 3 ⎜⎜1 + 2 ⎟⎟ 2R ⎝ R ⎠

Expand the radical factor to obtain:

]

−3 2 d 2 x + R2 dx 2 ⎡ ⎤ 3µ µNR I 1 =− 0 ⎢ 2 ⎥x 2 52 2 ⎣⎢ x + R ⎦⎥

= 12 µ 0 µNR 2 I

⎛ x2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

−5 2

= 1−

−5 2

x

5 x2 + higher order 2 R2

terms For x << R:

⎛ x2 ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ R ⎠

Substitute in Fx to obtain:

Fx = −

−5 2

≈1

3µ0 µNI x 2R3

Thus, we' ve shown that the bar magnet experiences a linear restoring force and, hence, its motion will be simple harmonic motion. Apply

r r F ∑ = ma to the bar

−

magnet to obtain:

3µ0 µNI d 2x x = m 2R3 dt 2

or

d 2 x 3µ 0 µNI + x=0 dt 2 2mR 3 Because the coefficient of the linear term is the square of the angular frequency we have:

ω2 =

4π 2 3µ0 µNI = T2 2mR 3

612 Chapter 27 Solve for T to obtain:

2mR 3 3µ0 µNI

T = 2π

Substitute numerical values and evaluate T:

2(0.1 kg )(0.1 m ) T = 2π = 10.2 s −7 3 4π × 10 N/A 2 0.04 A ⋅ m 2 (100)(5 A ) 3

(

)(

)

114 •• Picture the Problem We can apply Newton’s 2nd law for rotational motion to obtain the differential equation of motion of the bar magnet. While this equation is not linear, we can use a small-angle approximation to render it linear and obtain an expression for the square of the angular frequency that we can solve for the frequency f of the motion. Apply

∑τ = Iα to the bar magnet

to obtain the differential equation of motion for the magnet:

For small displacements from equilibrium (θ << 1): Rewrite the differential equation as:

− µB sin θ = I

d 2θ dt 2

where I is the moment of inertia of the magnet about an axis through its point of suspension.

− µBθ ≈ I

I

d 2θ dt 2

d 2θ + µBθ = 0 dt 2

or

d 2θ µB + θ =0 dt 2 I Because the coefficient of the linear term is the square of the angular frequency, we have: Solve for f to obtain:

ω 2 = 4π 2 f 2 =

µB I

where f is the frequency of oscillation.

f =

1 2π

µB

I or, because µ = 2.2NµB where N is the number of iron atoms in the bar magnet,

f =

1 2π

2.2 Nµ B B I

Sources of the Magnetic Field 613 I = 121 mL2 = 121 ρVL2

From Table 9-1 we have: Express the number of iron atoms in terms of Avogadro’s number and the atomic weight of iron M:

Substitute for I and N and simplify to obtain:

N m ρV = = NA M M and

N=

N A ρV M

f =

1 2π

=

2.2 N A ρVµ B B 2 1 12 ρVL M

1 6.6 N A µ B B πL M

Substitute numerical values and evaluate f:

f =

(

)(

)(

1 6.6 6.02 × 10 23 / mol 9.27 × 10 −24 A ⋅ m 2 0.5 × 10 −4 T π (0.08 m ) 55.85 g/mol

)

= 0.723 Hz 115 •• Picture the Problem We can solve the equation for the frequency f of the compass needle given in Problem 112 for magnetic dipole moment of the needle. In Parts (b) and (c) we can use their definitions to find the magnetization M and the amperian current Iamperian. (a) In Problem 112 it is established that the frequency of the compass needle is:

f =

1 2π

µB I

where I is the moment of inertia of the needle. Solve for µ to obtain:

4π 2 f 2 I µ= B

Express the moment of inertia of the needle:

I = 121 mL2 = 121 ρVL2 = 121 ρπr 2 L3

Substitute to obtain:

µ=

π 3 f 2 ρr 2 L3 3B

614 Chapter 27 Substitute numerical values and evaluate µ:

π 3 (1.4 s -1 ) (7.96 × 103 kg/m 3 )(0.85 × 10 −3 m ) (0.03 m )3 2

µ=

2

3(0.6 × 10 −4 T )

(b) Use its definition to express the magnetization M:

M =

Substitute to obtain:

M =

= 5.24 × 10 −2 A ⋅ m 2

µ V

µ V

=

π 3 f 2 ρr 2 L3 3BV

=

π 2 f 2 ρL2 3B

Substitute numerical values and evaluate M:

π 2 (1.4 s -1 ) (7.96 × 103 kg/m 3 )(0.03 m )2 2

M =

(

3 0.6 × 10 −4 T

)

= 7.70 × 105 A/m

(c) Express and evaluate the amperian current on the surface of the needle:

(

)

I amperian = ML = 7.70 × 105 A/m (0.03 m ) = 2.31 × 10 4 A *116 •• Picture the Problem We can use the definition of angular momentum and Equation 2727, together with the definition of the magnetization M of the iron bar, to derive an expression for the rotational angular velocity of the bar just after it has been demagnetized. Assuming its angular momentum to be conserved, use the definition of L to express the angular momentum of the iron bar just after it has been demagnetized:

L = Iω

Solve for the angular velocity ω:

ω=

Assuming that Equation 27-27 holds yields:

2m 2m 2m µ = e MV = e Mπr 2l q e e where r is the radius of the bar and l its

L I

L=

length.

Sources of the Magnetic Field 615 Modeling the bar as a cylinder, express its moment of inertia with respect to its axis:

I = 12 mr 2 = 12 ρVr 2 = 12 ρπr 4l

Substitute to obtain:

2me Mπr 2l 4me M e ω= 1 = 4 eρ r 2 2 ρπr l

Substitute numerical values (see Table 13-1 for the density of iron) and evaluate ω:

ω=

(

)(

)

4 9.11 × 10 −31 kg 1.72 × 10 6 A/m = 4.92 × 10 −5 rad/s 2 3 3 −19 1.6 × 10 C 7.96 × 10 kg/m (0.01 m )

(

)(

)

117 •• Picture the Problem The dipole moment of the bar is given by µ = 2.219 Nµ B , where N is the number of atoms in the bar. We can express N in terms of Avogadro’s number, the density of iron, the volume of the bar, and the atomic weight of iron. We can use the definition of torque to find the torque that must be supplied to hold the iron bar perpendicular to the given magnetic field.

µ = 2.219 Nµ B

(a) Express the magnetic dipole moment of the magnetized iron bar:

where N is the number of iron atoms in the bar.

N m ρV = = NA M M

Express the number of iron atoms in terms of Avogadro’s number and the atomic weight of iron M:

and

Substitute to obtain:

N=

N A ρV M

µ=

2.219 N A ρVµ B 2.219 N A ρlAµ B = M M

Substitute numerical values and evaluate µ:

µ=

(

)(

)

2.219 6.02 × 10 23 mol-1 7.96 × 103 kg/m 3 (0.2 m ) 55.85 × 10 -3 kg/mol

(

)(

× 2 × 10 −4 m 2 9.27 × 10 −24 A ⋅ m 2 = 70.6 A ⋅ m 2

)

616 Chapter 27 (b) Express the torque required to hold the iron bar perpendicular to the magnetic field: Substitute numerical values and evaluate τ:

τ = µB sin θ = µB sin 90° = µB

τ = (70.6 A ⋅ m 2 )(0.25 T ) = 17.7 N ⋅ m

*118 •• Picture the Problem Note that Be and Bc are perpendicular to each other and that the resultant magnetic field is at an angle θ with north. We can use trigonometry to relate Bc and Be and express Bc in terms of the geometry of the coil and the current flowing in it. Express Bc in terms of Be:

Bc = Be tan θ where θ is the angle of the resultant field from north.

Express the field Bc due to the current in the coil:

Bc =

Nµ 0 I 2R

where N is the number of turns. Substitute to obtain:

Solve for I:

Nµ 0 I = Be tan θ 2R I=

2 RBe tan θ µ0 N

119 •• Picture the Problem Let the positive x direction be out of the page. We can use the expressions for the magnetic fields due to an infinite straight line and a circular loop to express the net magnetic field at the center of the circular loop. We can set this net field to zero and solve for r. Express the net magnetic field at the center of circular loop:

r r r B = Bloop + Bline

Letting R represent the radius of the r loop, express Bloop :

r µI Bloop = − 0 iˆ 2R

Express the magnetic field due to the current in the infinite straight line:

r µI Bline = 0 iˆ 2πr

Sources of the Magnetic Field 617 Substitute to obtain:

r

If B = 0 , then:

r µI µI ⎛ µ I µ I⎞ B = − 0 iˆ + 0 iˆ = ⎜ − 0 + 0 ⎟ iˆ 2R 2πr ⎝ 2 R 2πr ⎠ −

µ0 I

+

2R

µ0 I =0 2πr

or

−

Solve for r:

Substitute numerical values and evaluate r:

1 1 + =0 R πr

r=

r=

R

π 10 cm

π

= 3.18 cm

120 •• Picture the Problem Note that only the current in the section of wire of length 2a contributes to the field at P. Hence, we can use the expression for B due to a straight wire segment to find the magnetic field at P. In Part (b) we can use our result from (a), together with the value for θ when the polygon has N sides, to obtain an expression for B at the center of a polygon of N sides. Express the magnetic field at P due to a straight wire segment: Because θ1 = θ2 = θ :

Refer to the figure to obtain:

Substitute to obtain:

BP =

µ0 I (sin θ1 + sin θ 2 ) 4π R

BP =

µ0 I (2 sin θ ) = ⎛⎜ µ0 I ⎞⎟ sin θ 4π R ⎝ 2π R ⎠

sin θ =

BP =

a a + R2 2

µ 0 aI 2πR a 2 + R 2

π

(b) Express θ for an N-sided polygon:

θ=

Because each side of the polygon contributes to B an amount equal to that obtained in (a):

⎛ Nµ 0 I ⎞ ⎛ π ⎞ B= ⎜ ⎟ sin ⎜ ⎟ ⎝ 2πR ⎠ ⎝ N ⎠

N

618 Chapter 27 As N → ∞:

π ⎛π ⎞ sin ⎜ ⎟ → N ⎝N⎠ and

µI ⎛ Nµ0 I ⎞⎛ π ⎞ B→⎜ ⎟⎜ ⎟ = 0 , the 2R ⎝ 2πR ⎠⎝ N ⎠ expression for the magnetic field at the center of a current-carrying circular loop. 121 •• Picture the Problem We can use Ampère’s law to derive expressions for B(r) for r < R, r = R, and r > R that we can evaluate for the given distances from the center of the cylindrical conductor. Apply Ampère’s law to a closed circular path a distance r < R from the center of the cylindrical conductor to obtain:

r

r

∫ B ⋅ d l = B(r )(2πr ) = µ I

= µ0 I (r )

0 C

C

Solve for B(r) to obtain:

B(r ) =

µ0 I (r ) 2πr

Substitute for I(r):

B(r ) =

µ 0 (50 A/m) r µ 0 (50 A/m) = 2π r 2π

(a) and (b) Noting that B is independent of r, substitute numerical values and evaluate B(5 cm) and B(10 cm):

(c) Apply Ampère’s law to a closed circular path a distance r > R from the center of the cylindrical conductor to obtain: Solve for B(r):

B(5 cm ) = B(10 cm ) =

(4π × 10

−7

)

N/A 2 (50 A/m ) 2π

= 10.0 µT r

r

∫ B ⋅ d l = B(r )(2πr ) = µ I

0 C

C

B(r ) =

Substitute numerical values and evaluate B(20 cm):

µ 0 I (R ) 2πr

= µ 0 I (R )

Sources of the Magnetic Field 619 B(20 cm ) =

(4π × 10

−7

)

N/A 2 (50 A/m )(0.1 m ) = 5.00 µT 2π (0.2 m )

122 •• r Picture the Problem The field B due to the 10-A current is in the yz plane. The net force on the wires of the square along the y direction cancel and do not contribute to a net r r r r r r torque or force. We can use τ = l × F , F = I l × B , and the expression for the magnetic field due to an infinite straight wire to express the torque acting on each of the wires and hence, the net torque acting on the loop.

r r r τ = l ×r F

(a) Express the torque on the loop:

where l is the lever arm.

r r r F = Il × B

Express the magnetic force on a current element:

(

r µ 2I 1 B y =10 = 0 − ˆj − kˆ 4π R 2

Express the magnetic field at the wire at y = 10 cm:

)

where

(0.1 m )2 + (0.1 m )2

R=

= 0.141 m .

r

Substitute numerical values and evaluate B y =10 :

r 4π × 10 −7 N/A 2 2(10 A ) ˆ ˆ B y =10 = − j − k = (1.00 × 10 −5 T ) − ˆj − kˆ 0.141 m 4π 2

(

)

(

)

(

)

r B y =−10 = (1.42 × 10 −5 T ) − ˆj + kˆ

Proceed similarly to obtain:

r

Evaluate Fy =10 :

(

r r r Fy =10 = I l × B y =10 = (5 A )(0.2 m ) iˆ × (1.00 × 10 −5 T ) − ˆj − kˆ

(

)[ (

)] (

)(

)

= 1.00 × 10 −5 N iˆ × − ˆj − kˆ = 1.00 × 10 −5 N − kˆ + ˆj

)

r

Evaluate Fy =−10 :

(

r Fy =−10 = (5 A )(− 0.2 m ) iˆ × (1.00 × 10 −5 T ) − ˆj + kˆ

(

)[ (

)] (

)

)(

= − 1.00 × 10 −5 N iˆ × − ˆj + kˆ = 1.00 × 10 −5 N kˆ + ˆj

)

620 Chapter 27 Express and evaluate the net force acting on the loop:

(

)

(

r r r F = Fy =10 + Fy =−10 = (1.00 × 10 −5 N ) − kˆ + ˆj + (1.00 × 10 −5 N ) kˆ + ˆj

(

)

)

= 2.00 × 10 −5 N ˆj Express and evaluate the torque about the x axis acting on the loop:

(b) The net force acting on the loop is the sum of the forces acting on the four sides (see the next to last step in (a)):

τ = (0.1 m ) (2.00 × 10 −5 N ) = 2.00 × 10 −6 N ⋅ m

r r r F = Fy =10 + Fy =−10 =

(2.00 × 10

−5

)

N ˆj

123 •• Picture the Problem The force acting on the lower wire is given by Flower wire = IlB, where I is the current in the lower wire, l is the length of the wire on the balance, and B is the magnetic field at the location of the lower wire due to the current in the upper wire. We can apply Ampère’s law to find B at the location of the wire on the pan of the balance. The force experienced by the lower wire is given by: Apply Ampère’s law to a closed circular path of radius r centered on the upper wire to obtain: Solve for B to obtain:

Substitute for B in the expression for the force on the lower wire:

Solve for I to obtain:

Noting that the force on the lower wire is the increase in the reading of the balance, substitute numerical values and evaluate I:

Flower wire = IlB B(2πr ) = µ 0 I C = µ 0 I

B=

µ0 I 2π r

Flower wire

⎛ µ 0 I ⎞ µ 0 lI 2 ⎟⎟ = = Il⎜⎜ 2 π r ⎝ ⎠ 2π r

I=

2π rFlower wire µ0l

I=

2π (2 cm ) 5 × 10 −6 kg 4π × 10 −7 N/A 2 (10 cm )

(

(

)

)

= 2.24 A

124 •• Picture the Problem We can use a proportion to relate minimum current detectible using

Sources of the Magnetic Field 621 this balance to its sensitivity and to the current and change in balance reading from Problem 123. The minimum current Imin detectible is to the sensitivity of the balance as the current in Problem 123 is to the change in the balance reading in Problem 123:

2.24 A I min = 0.1 mg 5.0 mg

Solve for and evaluate Imin:

⎛ 2.24 A ⎞ ⎟⎟ = 44.8 mA I min = (0.1 mg ) ⎜⎜ ⎝ 5.0 mg ⎠

The " standard" current balance can be made very sensitive by increasing the length (i.e., moment arm) of the wire balance, which one cannot do with this kind; however, this is compensated somewhat by the high sensitivity of the electronic balance. *125 ••• Picture the Problem The diagram shows the rotating disk and the circular strip of radius r and width dr with charge dq. We can use the definition of surface charge density to express dq in terms of r and dr and the definition of current to show that dI = ωσr dr. We can then use this current and expression for the magnetic field on the axis of a current loop to obtain the results called for in (b) and (c).

(a) Express the total charge dq that passes a given point on the circular strip once each period: Using its definition, express the current in the element of width dr:

dq = σdA = 2πσrdr

dI =

dq 2πσrdr = = ωσrdr 2π dt

ω

622 Chapter 27 (c) Express the magnetic field dBx at a distance x along the axis of the disk due to the current loop of radius r and width dr:

Integrate from r = 0 to r = R to obtain:

µ 0 2πr 2 dI dBx = 4π (x 2 + r 2 )3 2 =

Bx = =

(b) Evaluate Bx for x = 0:

(

µ 0ωσr 3

2 x2 + r 2

µ 0ωσ

32

R

∫ (x

2

0

dr

)

r3 2

+ r2

)

32

dr

⎞ ⎟ 2 x − ⎟ ⎜ 2 2 R x + ⎠ ⎝

µ 0ωσ ⎛ R 2 + 2 x 2 ⎜

Bx (0 ) =

2

µ 0ωσ ⎛ R 2 ⎞ 2

⎜ ⎟= ⎜ 2 ⎟ ⎝ R ⎠

1 2

µ 0σωR

126 ••• Picture the Problem From the symmetry of the system it is evident that the fields due to each segment of length l are the same in magnitude. We can express the magnetic field at (x,0,0) due to one side (segment) of the square, find its component in the x direction, and then multiply by four to find the resultant field.

Express B due to a straight wire segment:

B=

µ0 I (sin θ1 + sin θ 2 ) 4π R

where R is the perpendicular distance from the wire segment to the field point. Use θ1 = θ2 and R =

x 2 + l 2 4 to

B1 ( x,0,0 ) =

express B due to one side at (x,0,0):

=

µ0 4π µ0 2π

I l2 x + 4 I

(2 sin θ1 )

2

l2 x + 4 2

(sin θ1 )

Sources of the Magnetic Field 623 Referring to the diagram, express sinθ1:

Substitute to obtain:

l sin θ1 = 2 = d

l 2

x2 +

µ B1 ( x,0,0) = 0 2π

l2 2

l 2

I x2 +

l2 4

x2 +

µ0 I

=

4π x 2 +

l2 2

l l2 4

x2 +

l2 2

By symmetry, the sum of the y and z components of the fields due to the four segments must vanish, whereas the x components will add. The diagram to the right is a view of the xy plane showing the relationship

r

between B1 and the angle β it makes with the x axis. Express B1x:

B1x = B1 cos β

Substitute and simplify to obtain:

µ0 I

B1x =

l 2

l l2 l2 x2 + 4 2 2 µ 0 Il

4π x 2 +

=

The resultant magnetic field is the sum of the fields due to the 4 wire segments (sides of the square):

x2 +

⎛ l2 ⎞ l2 8π ⎜⎜ x 2 + ⎟⎟ x 2 + 4⎠ 2 ⎝

r B = 4 B1x iˆ

=

µ 0 Il 2 ⎛ l ⎞ l 2π ⎜⎜ x 2 + ⎟⎟ x 2 + 4⎠ 2 ⎝ 2

2

iˆ

l2 4

624 Chapter 27 Factor x2 from the two factors in the denominator to obtain:

r B=

=

For x >> l:

µ 0 Il 2 ⎛ ⎛ l ⎞ l ⎞ 2πx 2 ⎜⎜1 + 2 ⎟⎟ x 2 ⎜⎜1 + 2 ⎟⎟ ⎝ 4x ⎠ ⎝ 2x ⎠ 2

2

µ 0 Il 2 ⎛ l ⎞ 2πx 3 ⎜⎜1 + 2 ⎟⎟ ⎝ 4x ⎠ 2

⎛ l ⎞ ⎜⎜1 + 2 ⎟⎟ ⎝ 2x ⎠

r µ 0 Il 2 µ0 µ ˆ B≈ iˆ = i 3 2πx 2πx 3 where µ = Il2.

2

iˆ

iˆ

Chapter 28 Magnetic Induction Conceptual Problems *1 • Determine the Concept We know that the magnetic flux (in this case the magnetic field because the area of the conducting loop is constant and its orientation is fixed) must be changing so the only issues are whether the field is increasing or decreasing and in which direction. Because the direction of the magnetic field associated with the clockwise current is into the page, the changing field that is responsible for it must be either increasing out of the page (not included in the list of possible answers) or a decreasing field directed into the page. (d ) is correct. 2 • Determine the Concept Note that when R is constant, B in the loop to the right points out of the paper. (a) If R increases, I decreases and so does B. By Lenz’s law, the induced current is counterclockwise. (b) If R decreases, the induced current is clockwise. 3 •• Determine the Concept If the counterclockwise current in loop A increases, so does the magnetic flux through B. To oppose this increase in flux, the induced current in loop B will by clockwise. If the counterclockwise current in loop A decreases, so does the magnetic flux through B. To oppose this decrease in flux, the induced current in loop B r r r will be counterclockwise. We can use F = I l × B to determine the direction of the forces on each loop and, hence, whether they will attract or repel each other. (a) If the current in B is clockwise the loops repel one another. (b) If the current in B is counterclockwise the loops attract one another. 4 •• Determine the Concept We know that, as the magnet moves to the right, the flux through the loop first increases until the magnet is half way through the loop and then decreases. Because the flux first increases and then decreases, the current will change directions, having its maximum values when the flux is changing most rapidly. (a) and (b) The following graph shows the flux and the induced current as a function of

625

626 Chapter 28 time as the bar magnet passes through the coil. When the center of the magnet passes through the plane of the coil dφm /dt = 0 and the current is zero.

1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6

Flux Current

-0.8 -1.0 -3

-2

-1

0

1

2

3

time (arbitrary units)

5 •• Determine the Concept Because the magnet moves with simple harmonic motion, the flux and the induced current will vary sinusoidally. The current will be a maximum wherever the flux is changing most rapidly and will be zero wherever the flux is momentarily constant. (a), (b) The following graph shows the flux, φm , and the induced current (proportional to −dφm/dt) in the loop as a function of time.

1.0 Flux

0.8

Current

0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0 -5

0

5 time (arbitrary units)

10

15

Magnetic Induction 627 *6 • Determine the Concept The magnetic energy stored in an inductor is given by

U m = 12 LI 2 . Doubling I quadruples Um. (c) is correct. 7 • Determine the Concept The protection is needed because if the current is suddenly interrupted, the resulting emf generated across the inductor due to the large flux change can blow out the inductor. The diode allows the current to flow (in a loop) even when the switch is opened. 8 • Determine the Concept The inductance of a coil depends on the product n2l, where n is the number of turns per unit length and l is the length of the coil. If n increases by a factor of 3, l will decrease by the same factor, because the inductors are made from the same length of wire. Hence, the inductance increases by a factor of (3) (1 3) = 3 . 2

9 • (a) False. The induced emf in a circuit is proportional to the rate of change of the magnetic flux through the circuit. (b) True. (c) True. (d) False. The inductance of a solenoid is determined by its length, cross-sectional area, number of turns per unit length, and the permeability of the matter in its core. (e) True. *10 • Determine the Concept In the configuration shown in (a), energy is dissipated by eddy currents from the emf induced by the pendulum movement. In the configuration shown in (b), the slits inhibit the eddy currents and the braking effect is greatly reduced. 11 • Determine the Concept The time varying magnetic field of the magnet sets up eddy currents in the metal tube. The eddy currents establish a magnetic field with a magnetic moment opposite to that of the moving magnet; thus the magnet is slowed down. If the tube is made of a nonconducting material, there are no eddy currents. 12 •• Determine the Concept When the current is turned on, the increasing magnetic field in the coil induces a large emf in the ring. As described by Lenz's law, the direction of the

628 Chapter 28 current resulting from this induced emf is in such a direction that its magnetic field opposes the changing flux in the coil, i.e., the current induced in the ring will be in such a direction that the magnetic field in the coil will repel it. The demonstration will not work if a slot is cut in the ring, because the emf will not be able to induce a current in the ring.

Estimation and Approximation *13 •• Picture the Problem We can use Faraday’s law to relate the induced emf to the angular velocity with which the students turn the jump rope. (a) It seems unlikely that the students could turn the ″jump rope″ wire faster than 5 revolutions per second. This corresponds to a maximum angular velocity of: (b) The magnetic flux φm through the rotating circular loop of wire varies sinusoidally with time according to:

ω =5

rev 2π rad × = 31.4 rad/s. s rev

φm = BA sin ωt and

dφm = BAω cos ωt dt

Because the average value of the cosine function, over one revolution, is ½, the average rate at which the flux changes through the circular loop is:

dφ m dt

From Faraday’s law, the magnitude of the induced emf in the loop is:

ε = dφm

= 12 BAω = 12 π r 2 Bω av

dt

= 12 π r 2 Bω

Substitute numerical values and evaluate ε:

ε (c)

(d)

⎛ 1.5 m ⎞ = π⎜ ⎟ ⎝ 2 ⎠ 1 2

2

⎛ 1T ⎞ ⎜⎜ 0.7 G × 4 ⎟⎟ (31.4 rad/s ) = 1.94 mV 10 G ⎠ ⎝

No. To generate an emf of 1 V, the students would have to rotate the jump rope about 500 times faster.

The use of multiple strands of lighter wire (so that the composite wire could be rotated at the same angular speed) looped several times around would increase the induced emf.

14 • Picture the Problem We can compare the energy density stored in the earth’s electric field to that of the earth's magnetic field by finding their ratio.

Magnetic Induction 629 The energy density in an electric field E is given by:

ue =

1 ∈0 E 2 2

The energy density in a magnetic field B is given by:

um =

B2 2µ 0 B2 2µ0

Express the ratio of um to ue to obtain:

B2 um = = ue 1 ∈ E 2 µ 0 ∈0 E 2 0 2

Substitute numerical values and evaluate um/ ue:

(

)

2

5 × 10 −5 T um == = 2.25 × 10 4 2 2 2 2 −7 −12 ue 4π × 10 N/A 8.85 × 10 C / N ⋅ m (100 V/m )

(

or

um =

)(

(2.25 × 10 )u

)

4

e

15 •• Picture the Problem We can apply Faraday’s law to estimate the maximum emf induced by the lightning strike in the antenna. Use Faraday’s law to express the magnitude of the induced emf in antenna:

ε = dφm dt

=

d [BA] dt

where A is the area of the antenna. Because the lightning strike has such a short duration:

ε ≈ BA

The magnetic field induced in the loop is given by:

B=

∆t

µ0 2I µ0 I = 4π r 2π r

where r is the distance from the antenna to the lightning strike. Substitute for B to obtain:

ε=

Substitute numerical values and evaluate ε:

ε=

µ 0 IA 2π r∆t

(4π × 10

⎛ 30 C ⎞ ⎟⎟ 0.1 m 2 N/A 2 ⎜⎜ 1 s µ ⎝ ⎠ 2π (300 m )(1 µs )

−7

= 2.00 kV

)

(

)

630 Chapter 28

Magnetic Flux 16 • r Picture the Problem Because the surface is a plane with area A and B is constant in magnitude and direction over the surface and makes an angle θ with the unit normal vector, we can use φm = BA cos θ to find the magnetic flux through the coil.

1T ⎞

⎛

φm = ⎜⎜ 2000 G ⋅ 4 ⎟⎟(5 × 10 −2 m ) cosθ 10 G

Substitute for B and A to obtain:

2

⎠ ⎝ −4 = 5.00 × 10 Wb cos θ

(

)

φm = (5.00 × 10−4 Wb )cos 0°

(a) For θ = 0°:

= 5.00 × 10 −4 Wb = 0.500 mWb

φm = (5.00 × 10 −4 Wb )cos30°

(b) For θ = 30°:

= 4.33 × 10 −4 Wb = 0.433 mWb

φm = (5.00 × 10 −4 Wb )cos60°

(c) For θ = 60°:

= 2.50 × 10 −4 Wb = 0.250 mWb

φm = (5.00 × 10 −4 Wb )cos90°

(d) For θ = 90°:

= 0 *17 • r Picture the Problem Because the coil defines a plane with area A and B is constant in magnitude and direction over the surface and makes an angle θ with the unit normal vector, we can use φm = NBA cos θ to find the magnetic flux through the coil. Substitute for N, B, and A to obtain:

⎛

φm = NBπ r 2 cos θ = 25⎜⎜ 0.7 G ⋅ ⎝ −5 = 1.37 × 10 Wb cos θ

(

)

2 1T ⎞ ⎟⎟π 5 × 10 −2 m cosθ 4 10 G ⎠

(

)

Magnetic Induction 631 (a) When the plane of the coil is horizontal, θ = 90°:

(b) When the plane of the coil is vertical with its axis pointing north, θ = 0°: (c) When the plane of the coil is vertical with its axis pointing east, θ = 90°: (d) When the plane of the coil is vertical with its axis making an angle of 30° with north, θ = 30°:

φm = (1.37 ×10−5 Wb )cos 90° = 0

φm = (1.37 × 10 −5 Wb )cos 0° = 1.37 × 10 −5 Wb

φm = (1.37 ×10−5 Wb )cos 90° = 0

φm = (1.37 × 10−5 Wb )cos 30° = 1.19 × 10 −5 Wb

18 • r Picture the Problem Because the square coil defines a plane with area A and B is constant in magnitude and direction over the surface and makes an angle θ with the unit normal vector, we can use φm = NBA cos θ to find the magnetic flux through the coil. Substitute for N, B, and A to obtain:

φm = NBA cos θ

(

)

= 14(1.2 T ) 5 × 10 −2 m cosθ = (42.0 mWb) cos θ (a) For θ = 0°:

2

φm = (42.0 mWb)cos 0° = 42.0 mWb

(b) For θ = 60°:

φm = (42.0 mWb) cos 60° = 21.0 mWb

19 • Picture the Problem Noting that the flux through the base must also penetrate the spherical surface, we can apply its definition to express φm. Apply the definition of magnetic flux to obtain:

φm = AB = πR 2 B

632 Chapter 28 20 •• Picture the Problem We can use φm = NBA cos θ to express the magnetic flux through

the solenoid and B = µ 0 nI to relate the magnetic field in the solenoid to the current in its coils. Express the magnetic flux through a coil with N turns:

φm = NBA cosθ

Express the magnetic field inside a long solenoid:

B = µ0 nI where n is the number of turns per unit length.

φm = Nµ0 nIA cosθ

Substitute to obtain:

or, because n = N/L and θ = 0°,

φm =

N 2 µ0 IA N 2 µ0 Iπ r 2 = L L

Substitute numerical values and evaluate φm:

φm =

(400)2 (4π × 10 −7 N/A 2 )(3 A )π(0.01 m )2 0.25 m

= 7.58 × 10 −4 Wb

21 •• Picture the Problem We can use φm = NBA cos θ to express the magnetic flux through

the solenoid and B = µ 0 nI to relate the magnetic field in the solenoid to the current in its coils. Express the magnetic flux through a coil with N turns:

φm = NBA cosθ

Express the magnetic field inside a long solenoid:

B = µ0 nI

Substitute to obtain:

where n is the number of turns per unit length.

φm = Nµ0 nIA cosθ or, because n = N/L and θ = 0°,

φm =

N 2 µ0 IA N 2 µ0 Iπ r 2 = L L

Magnetic Induction 633 Substitute numerical values and evaluate φm: 2 2 ( 800) (4π × 10 −7 N/A 2 )(2 A )π (0.02 m ) φm =

0.3 m

= 6.74 × 10 −3 Wb

22 •• Picture the Problem We can apply the definitions of magnet flux and of the dot product to find the flux for the given unit vectors.

r

φm = N ∫ B ⋅ nˆ dA

Apply the definition of magnetic flux to the coil to obtain:

S

r

(

r

r

)

φm = NB ⋅ nˆ ∫ dA = N B ⋅ nˆ A

Because B is constant:

(

S

)

r = N B ⋅ nˆ π r 2

r

r B = (0.4 T ) iˆ

Evaluate B :

φm = (15)[(0.4 T )]π (0.04 m )2

Substitute numerical values and simplify to obtain:

( ) = (0.0302 T ⋅ m )iˆ ⋅ iˆ = = 0.0302 T ⋅ m 2 iˆ ⋅ nˆ

(a) Evaluate φm for nˆ = iˆ :

φm

(b) Evaluate φm for nˆ = ˆj :

φ m = (0.0302 T ⋅ m 2 ) iˆ ⋅ ˆj = 0

( )

(c) Evaluate φm for nˆ = iˆ + ˆj

2

φm = (0.0302 T ⋅ m 2 )iˆ ⋅

2:

=

0.0302 Wb

(iˆ + ˆj ) 2

0.0302 T ⋅ m = 0.0213 Wb 2 2

φ m = (0.0302 T ⋅ m 2 ) iˆ ⋅ kˆ = 0

(d) Evaluate φm for nˆ = kˆ : (e) Evaluate φm for nˆ = 0.6iˆ + 0.8 ˆj :

(

)

φm = (0.0302 T ⋅ m 2 ) iˆ ⋅ 0.6iˆ + 0.8 ˆj = 0.6(0.0302 T ⋅ m 2 )iˆ ⋅ iˆ

(

)

+ 0.8 0.0302 T ⋅ m 2 iˆ ⋅ ˆj

(

)

= 0.6 0.0302 T ⋅ m 2 = 0.0181 Wb

634 Chapter 28 23 •• Picture the Problem The magnetic field outside the solenoid is, to a good approximation, zero. Hence, the flux through the loop is the flux in the core of the solenoid. The magnetic field inside the solenoid is uniform. Hence, the flux through this small loop is given by the same expression with R3 replacing R1: (a) Express the flux through the large circular loop outside the solenoid:

φm = NBA = µ 0 nINπR12

(b) Express the flux through the small loop inside the solenoid:

φm = NBA = µ 0 nINπR32

*24 •• Picture the Problem We can use the hint to set up the element of area dA and express the flux dφm through it and then carry out the details of the integration to express φm.

dφm = BdA

(a) Express the flux through the strip of area dA:

where dA = bdx.

µ0 2 I µ0 I = 4π x 2π x

Express B at a distance x from a long, straight wire:

B=

Substitute to obtain:

dφm =

Integrate from x = d to x = d + a:

φm =

µ0 I µ Ib dx bdx = 0 2π x 2π x

µ0 Ib d + a dx µ Ib d + a = 0 ln ∫ d 2π d x 2π

(b) Substitute numerical values and evaluate φm:

φm =

(4π × 10

−7

)

N/A 2 (20 A )(0.1 m ) ⎛ 7 cm ⎞ ⎟⎟ = 5.01 × 10 −7 Wb ln⎜⎜ 2π 2 cm ⎠ ⎝

25 ••• Picture the Problem Consider an element of area dA = Ldr where r ≤ R. We can use its definition to express dφm through this area in terms of B and Ampere’s law to express B as a function of I. The fact that the current is uniformly distributed over the crosssectional area of the conductor allows us to set up a proportion from which we can obtain I as a function of r. With these substitutions in place we can integrate dφm to obtain φm/L.

Magnetic Induction 635 Express the flux dφm through an area Ldr: Apply Ampere’s law to the current contained inside a cylindrical region of radius r < R:

Using the fact that the current I is uniformly distributed over the crosssectional area of the conductor, express its variation with distance r from the center of the conductor: Substitute and simplify to obtain:

Substitute in equation (1):

Integrate dφm from r = 0 to r = R to obtain: Divide both sides of this equation by L to express the magnetic flux per unit length:

dφm = BdA = BLdr

(1)

r r B ∫ ⋅ d l = 2πrB = µ0 IC C

and

µ0 I C 2πr

B=

I (r ) πr 2 = 2 I πR or

r2 I (r ) = I C = I 2 R B=

µ0 I r 2 µI = 0 2r 2 2πr R 2πR

dφ m =

φm =

φm L

=

µ 0 LI rdr 2πR 2

µ0 LI R µ LI rdr = 0 2 ∫ 2πR 0 4π

µ0 I 4π

26 ••• Picture the Problem We can use its definition to express the flux through the rectangular region and Ampere’s law to relate the magnetic field to the current in the wire and the position of the long straight wire. (a) Note that for 0 ≤ x ≤ b, B is symmetric about the wire, into the paper for the region below the wire and out of the paper for the region above the wire. Thus, for the area 2(b − x)a:

φm,net = 0

To find the flux through the

dφm = BdA

636 Chapter 28 remaining area of the rectangle, express the flux through a strip of area dA:

where dA = adx.

Using Ampere’s law, express B at a distance x from a long, straight wire:

B=

Substitute to obtain:

dφm =

For 0 ≤ x ≤ b, integrate from x′ = b − x to x′ = x :

µ Ia φm (0 ≤ x ≤ b ) = 0 ∫ 2π

µ0 2 I µ0 I = 4π x 2π x µ0 I µ Ia dx adx = 0 2π x 2π x x

dx' x' b− x

=

For x ≥ b, integrate from x′ = x to x′ = x + b:

φm ( x ≥ b ) = =

µ0 Ia ⎛ x ⎞ ln⎜ ⎟ 2π ⎝ b − x ⎠

µ 0 Ia x+b dx' 2π ∫x x' µ 0 Ia ⎛ x + b ⎞ ln⎜ ⎟ 2π ⎝ x ⎠

(b) From the expressions derived in (a) we see that φm → ∞ as:

x→0

The flux is a minimum (φm = 0) for:

x = 12 b as expected from symmetry.

Induced EMF and Faraday’s Law *27 • Picture the Problem We can find the induced emf by applying Faraday’s law to the loop. The application of Ohm’s law will yield the induced current in the loop and we can find the rate of joule heating using P = I 2 R .

dφm d dB dB = ( AB ) = A = πR 2 dt dt dt dt

(a) Apply Faraday’s law to express the induced emf in the loop in terms of the rate of change of the magnetic field:

ε

=

Substitute numerical values and

ε

= π (0.05 m ) (40 mT/s ) = 0.314 mV 2

Magnetic Induction 637 evaluate

ε

:

ε

0.314 mV = 0.785 mA 0.4 Ω

(b) Using Ohm’s law, relate the induced current to the induced voltage and the resistance of the loop and evaluate I:

I=

(c) Express the rate at which power is dissipated in a conductor in terms of the induced current and the resistance of the loop and evaluate P:

P = I 2 R = (0.785 mA ) (0.4 Ω )

R

=

2

= 0.247 µW

28 • Picture the Problem Given φm as a function of time, we can use Faraday’s law to express ε as a function of time. (a) Apply Faraday’s law to express the induced emf in the loop in terms of the rate of change of the magnetic field:

ε = − dφm = − d [(t 2 − 4t )× 10−1 Wb]

(b) Evaluate φm at t = 0:

φm (0 s ) = (0)2 − 4(0) × 10−1 Wb = 0

Evaluate

ε

at t = 0:

dt dt = −(2t − 4) × 10−1 Wb/s = − (0.2t − 0.4) V

[

]

ε (0 s ) = −[0.2(0) − 0.4]V = 0.400 V

Proceed as above to complete the table to the right:

ε φm t (s) (Wb) (V) 0 0 0 2 −0.400 0 4 0 −0.400 6 1.20 −0.800

29 • Picture the Problem We can find the time at which the flux is a minimum by looking for the lowest point on the graph of ε versus t and the emf at this time by determining the value of V at this time from the graph. We can interpret the graphs to find the times at which the flux is zero and the corresponding values of the emf.

638 Chapter 28 (a) The flux, φm, and the induced emf, ε , are shown as functions of t in the following graph. The solid curve represents φm, the dashed curve represents ε .

1.5

flux

1.0

emf

0.5

0.0

-0.5

-1.0 0

1

2

3

4

5

6

t (s)

(b)

(c)

Referring to the graph, we see that the flux is a minimum at t = 2 s and that V = 0 at this instant. The flux is zero at t = 0 and t = 4 s. At these times, ε = 0.4 V and − 0.4 V, respectively.

30 • Picture the Problem We can use its definition to find the magnetic flux through the solenoid and Faraday’s law to find the emf induced in the solenoid when the external field is reduced to zero in 1.4 s. (a) Express the magnetic flux through the solenoid in terms of N, B, A, and θ :

φm = NBA cosθ

Substitute numerical values and evaluate φm:

φm = (400 )(0.06 T )π (0.008 m )2 cos 50°

= NBπR 2 cosθ

= 3.10 mWb

Magnetic Induction 639 (b) Apply Faraday’s law to obtain:

ε = − dφm = − 0 − 3.10 mWb dt

1.4 s

= 2.22 mV *31 •• Picture the Problem We can use the definition of average current to express the total charge passing through the coil as a function of Iav. Because the induced current is proportional to the induced emf and the induced emf, in turn, is given by Faraday’s law, we can express ∆Q as a function of the number of turns of the coil, the magnetic field, the resistance of the coil, and the area of the coil. Knowing the reversal time, we can find the average current from its definition and the average emf in the coil from Ohm’s law. (a) Express the total charge that passes through the coil in terms of the induced current:

∆Q = I av ∆t

Relate the induced current to the induced emf:

I = I av =

Using Faraday’s law, express the induced emf in terms of φm:

ε = − ∆φm

Substitute and simplify to obtain:

− ε ∆Q = ∆t =

ε R

∆t

∆φm ∆t ∆t = − 2φm R R R ⎛π ⎞ 2 NB⎜ d 2 ⎟ 2 NBA ⎝4 ⎠ =− =− R R 2 NBπd =− 2R

where d is the diameter of the coil. Substitute numerical values and evaluate ∆Q:

2 ( 100 )(1 T )π (0.02 m ) ∆Q = − 2(50 Ω )

= − 1.26 mC (b) Apply the definition of average current to obtain:

I av =

∆Q 1.26 mC = = 12.6 mA ∆t 0.1s

640 Chapter 28

ε av = I av R = (12.6 mA )(50 Ω )

(c) Using Ohm’s law, relate the average emf in the coil to the average current:

= 630 mV

32 •• Picture the Problem We can use the definition of average current to express the total charge passing through the coil as a function of Iav. Because the induced current is proportional to the induced emf and the induced emf, in turn, is given by Faraday’s law, we can express ∆Q as a function of the number of turns of the coil, the magnetic field, the resistance of the coil, and the area of the coil. Express the total charge that passes through the coil in terms of the induced current:

∆Q = I av ∆t

Relate the induced current to the induced emf:

I = I av =

Using Faraday’s law, express the induced emf in terms of φm:

ε = − ∆φm

Substitute to obtain:

− ε ∆Q = ∆t =

ε R

∆t

∆φm ∆t ∆t R R 2φ 2 NBA =− m =− R R

Substitute numerical values and evaluate ∆Q:

∆Q = −

(

)(

)

2(1000 ) 0.7 × 10 −4 T 300 × 10 −4 m 2 = 0.280 mC 15 Ω

33 •• Picture the Problem We can use Faraday’s law to express the earth’s magnetic field at this location in terms of the induced emf and Ohm’s law to relate the induced emf to the charge that passes through the current integrator. Using Faraday’s law, express the induced emf in terms of the change in the magnetic flux as the coil is rotated through 90°:

ε = − ∆φm ∆t

=

NBA NBπ r 2 = ∆t ∆t

Magnetic Induction 641 Solve for B:

Using Ohm’s law, relate the induced emf to the induced current:

B=

ε∆t Nπ r 2

ε = IR = ∆Q R ∆t

where ∆Q is the charge that passes through the current integrator. Substitute to obtain:

∆Q R∆t ∆QR B = ∆t 2 = Nπ r Nπ r 2

Substitute numerical values and evaluate B:

B=

(9.4 µC)(20 Ω ) (300)π (0.05 m )2

= 79.8 µT

34 •• Picture the Problem We can use Faraday’s law to express the induced emf in the coil in terms of the rate of change of the magnetic flux. We can use its definition to express the magnetic flux through the rectangular region and Ampere’s law to relate the magnetic field to the current in the wire and the position of the long straight wire. (a) Apply Faraday’s law to relate the induced emf to the changing magnetic flux:

ε = − dφm

Note that for 0 ≤ x ≤ b, B is symmetric about the wire, into the paper for the region below the wire and out of the paper for the region above the wire. Thus, for the area 2(b − x)a:

φm,net = 0

To find the flux through the remaining area of the rectangle, express the flux through a strip of area dA:

dφm = BdA

Using Ampere’s law, express B at a distance x from a long, straight wire:

dt

where dA = adx.

B=

µ0 2 I µ0 t = π x 4π x

(1)

642 Chapter 28

µ0 t µ ta dx adx = 0 π x π x

Substitute to obtain:

dφ m =

For 0 ≤ x ≤ b, integrate from x′ = b − x to x′ = x :

φm (0 ≤ x ≤ b ) = =

µ 0ta x dx' π b∫− x x' µ 0ta ⎛ x ⎞ ln⎜ ⎟ π ⎝b− x⎠

Differentiate this expression with respect to time to obtain:

dφm d ⎡ µ 0ta ⎛ x ⎞⎤ = ⎢ ln⎜ ⎟⎥ dt dt ⎣ π ⎝ b − x ⎠⎦ µa ⎛ x ⎞ = 0 ln⎜ ⎟ π ⎝b− x⎠

Substitute in equation (1) and evaluate ε for x = b/4:

ε = − µ0 a ln⎛⎜⎜

µ a ⎛1⎞ b4 ⎞ ⎟⎟ = − 0 ln⎜ ⎟ π ⎝3⎠ ⎝b−b 4⎠

π

= 1.10

(b) Using Ohm’s law, express and evaluate R:

ε

µ0 a π

1.10 µ 0 a I πI 1.10 4π × 10 −7 N/A 2 (1.5 m ) = π (0.1 A )

R=

=

(

)

= 6.60 µΩ Because the magnetic flux due to I is increasing into the page, the induced current will be in such a direction that its magnetic field will oppose this increase; i.e, it will be out of the page. Thus the induced current is counterclockwise. 35 •• Picture the Problem We can use Faraday’s law to express the induced emf in the coil in terms of the rate of change of the magnetic flux. We can use its definition to express the magnetic flux through the rectangular region and Ampere’s law to relate the magnetic field to the current in the wire and the position of the long straight wire. (a) Apply Faraday’s law to relate the induced emf to the changing magnetic flux:

ε = − dφm dt

(1)

Magnetic Induction 643 Note that for 0 ≤ x ≤ b, B is symmetric about the wire, into the paper for the region below the wire and out of the paper for the region above the wire. Thus, for the area 2(b − x)a:

φm,net = 0

To find the flux through the remaining area of the rectangle, express the flux through a strip of area dA:

dφm = BdA where dA = adx.

µ0 2 I µ0 t = π x 4π x

Using Ampere’s law, express B at a distance x from a long, straight wire:

B=

Substitute to obtain:

dφ m =

For 0 ≤ x ≤ b, integrate from x′ = b − x to x′ = x :

µ 0 ta x dx' φm (0 ≤ x ≤ b ) = π b∫− x x'

µ0 t µ ta dx adx = 0 π x π x

=

µ 0 ta ⎛ x ⎞ ln⎜ ⎟ π ⎝b− x⎠

Differentiate this expression with respect to time to obtain:

dφm d ⎡ µ 0ta ⎛ x ⎞⎤ = ⎢ ln⎜ ⎟⎥ dt dt ⎣ π ⎝ b − x ⎠⎦ µa ⎛ x ⎞ = 0 ln⎜ ⎟ π ⎝b− x⎠

Substitute in equation (1) and evaluate ε for x = b/3:

ε = − µ0 a ln⎛⎜⎜

µ a ⎛1⎞ b3 ⎞ ⎟⎟ = − 0 ln⎜ ⎟ π ⎝2⎠ ⎝b −b 3⎠

π

= 0.693

(b) Using Ohm’s law, express and evaluate R:

ε

µ0 a π

0.693µ 0 a πI I 0.693 4π × 10 −7 N/A 2 (1.5 m ) = π (0.1 A )

R=

=

(

= 4.16 µΩ

)

644 Chapter 28 Because the magnetic flux due to I is increasing into the page, the induced current will be in such a direction that its magnetic field will oppose this increase, i.e, it will be out of the page. Thus, the induced current is counterclockwise.

Motional EMF *36 • Picture the Problem We can apply the equation for the force on a charged particle moving in a magnetic field to find the magnetic force acting on an electron in the rod. We r r r can use E = v × B to find E and V = El , where l is the length of the rod, to find the potential difference between its ends. (a) Relate the magnetic force on an electron in the rod to the speed of the rod, the electronic charge, and the magnetic field in which the rod is moving: Substitute numerical values and evaluate F:

(b) Express the electrostatic field r E in the rod in terms of the r magnetic field B : Substitute numerical values and evaluate B:

r r r F = qv × B and

F = qvB sin θ

F = (1.6 × 10 −19 C )(8 m/s )(0.05 T ) sin90° = 6.40 × 10 −20 N r r r E = v×B and

E = vB sin θ

E = (8 m/s )(0.05 T )sin90° = 0.400 V/m

(c) Relate the potential difference between the ends of the rod to its length l and the electric field E:

V = El

Substitute numerical values and evaluate V:

V = (0.4 V/m )(0.3 m ) = 0.120 V

37 • r r r Picture the Problem We can use E = v × B to relate the speed of the rod to the electric field in the rod and magnetic field in which it is moving and V = El to relate the electric field in the rod to the potential difference between its ends.

Magnetic Induction 645

r

Express the electrostatic field E in the rod in terms of the magnetic r field B and solve for v:

r r r E = v×B and

v=

E B sin θ V l

Relate the potential difference between the ends of the rod to its length l and the electric field E and solve for E:

V = El ⇒ E =

Substitute for E to obtain:

v=

V Bl sin θ

Substitute numerical values and evaluate v:

v=

6V = 400 m/s (0.05 T )(0.3 m )

38 • Picture the Problem Because the speed of the rod is constant, an external force must act on the rod to counter the magnetic force acting on the induced current. We can use the motional-emf equation ε = vBl to evaluate the induced emf, Ohm’s law to find the current in the circuit, Newton’s 2nd law to find the force needed to move the rod with constant velocity, and P = Fv to find the power input by the force. (a) Relate the induced emf in the circuit to the speed of the rod, the magnetic field, and the length of the rod: (b) Using Ohm’s law, relate the current in the circuit to the induced emf and the resistance of the circuit:

(c) Because the rod is moving with constant velocity, the net force acting on it must be zero. Apply

ε = vBl = (10 m/s)(0.8 T )(0.2 m ) = 1.60 V

I=

ε R

=

1.6 V = 0.800 A 2Ω

Note that, because the rod is moving to the right, the flux in the region defined by the rod, the rails, and the resistor is increasing. Hence, in accord with Lenz’s law, the current must be counterclockwise if its magnetic field is to counter this increase in flux.

∑F

x

and

= F − Fm = 0

646 Chapter 28 Newton’s 2nd law to relate F to the magnetic force Fm:

F = Fm = BIl

(d) Express the power input by the force in terms of the force and the velocity of the rod:

P = Fv = (0.128 N )(10 m/s ) = 1.28 W

(e) The rate of Joule heat production is given by:

P = I 2 R = (0.8 A ) (2 Ω ) = 1.28 W

= (0.8 T )(0.8 A )(0.2 m ) = 0.128 N

2

39 •• Picture the Problem We’ll need to determine how long it takes for the loop to completely enter the region in which there is a magnetic field, how long it is in the region, and how long it takes to leave the region. Once we know these times, we can use its definition to express the magnetic flux as a function of time. We can use Faraday’s law to find the induced emf as a function of time. (a) Find the time required for the loop to enter the region where there is a uniform magnetic field: Letting w represent the width of the loop, express and evaluate φm for 0 < t < 4.17 s : Find the time during which the loop is fully in the region where there is a uniform magnetic field:

Express φm for 4.17 s < t < 8.33 s :

The left-end of the loop will exit the field when t = 12.5 s. Express φm for 8.33 s < t < 12.5 s : For t = 8.33 s and φm = 8.50 mWb:

t=

l side of loop v

=

10 cm = 4.17 s 2.4 cm/s

φm = NBA = NBwvt

= (1.7 T )(0.05 m )(0.024 m/s ) t = (2.04 mWb/s ) t

t=

l side of loop v

=

10 cm = 4.17 s 2.4 cm/s

i.e., the loop will begin to exit the region when t = 8.33 s.

φ m = NBA = NBlw

= (1.7 T )(0.1 m )(0.05 m )

= 8.50 mWb φm = mt + b where m is the slope of the line and b is the φm-intercept.

8.50 mWb = m(8.33 s ) + b

(1)

Magnetic Induction 647 0 = m(12.5 s ) + b

For t = 12.5 s and φm = 0:

(2)

φm = −(2.04 mWb/s) t + 25.5 mWb

Solve equations (1) and (2) simultaneously to obtain:

φm = 0

The loop will be completely out of the magnetic field when t > 12.5 s and:

The following graph of φm (t ) was plotted using a spreadsheet program.

9 8

Magnetic flux (mWb)

7 6 5 4 3 2 1 0 -1 0

2

4

6

8

10

12

14

t (s)

(b) Using Faraday’s law, relate the induced emf to the magnetic flux:

ε = − dφm

During the interval 0 < t < 4.17 s :

ε = − d [(2.04 mWb/s)t ] = −2.04 mV

dt

dt

During the interval

4.17 s < t < 8.33 s :

During the interval

8.33 s < t < 12.5 s :

For t > 12.5 s:

ε = − d [8.50 mWb] = 0 dt

ε = − d [(- 2.04 mWb/s)t + 25.5 mWb] dt = 2.04 mV

ε =0

648 Chapter 28 The following graph of ε(t) was plotted using a spreadsheet program.

2.5 2.0 1.5

emf (V)

1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 -2.5 0

2

4

6

8

10

12

14

t (s)

40 •• Picture the Problem The rod is executing simple harmonic motion in the xy plane, i.e., in a plane perpendicular to the magnetic field. The emf induced in the rod is a consequence of its motion in this magnetic field and is given by ε = vBl . Because we’re given the position of the oscillator as a function of time, we can differentiate this expression to obtain v.

dx dt

Express the motional emf in terms of v, B, and l :

ε

Evaluate dx/dt:

dx d = [(2 cm ) cos 120π t ] dt dt = −(2 cm ) (120 s −1 )π sin 120π t = −(7.54 m/s ) sin 120π t

Substitute numerical values and evaluate

ε

ε

= vBl = Bl

:

= −(1.2 T )(0.15 m )(7.54 m/s ) sin 120π t = − (1.36 V ) sin 120π t

41 •• Picture the Problem Let m be the mass of the rod and F be the net force acting on it due to the current in it. We can obtain the equation of motion of the rod by applying

Magnetic Induction 649 Newton’s 2nd law to relate its acceleration to B, I, and l . The net emf that drives I in this circuit is the emf of the battery minus the emf induced in the rod as a result of its motion.

dv dt

(a) Letting the direction of motion of the rod be the positive x Fx = ma x to direction, apply

where

the rod:

I=

Substitute to obtain:

dv Bl (ε − Blv ) = dt mR

(b) Express the condition on dv/dt when the rod has achieved its terminal speed:

Bl (ε − Blvt ) = 0 mR

∑

Solve for vt to obtain:

(c) Substitute vt for v in equation (2) to obtain:

BIl = m

ε − Blv

(2)

R

ε

vt =

I=

(1)

Bl

ε − Bl R

ε Bl = 0

*42 •• Picture the Problem In Example 28-9 it is shown that the speed of the rod is given by v = v0 e − (B l

2 2

) . We can use the definition of power and the expression for a motional

mR t

emf to express the power dissipated in the resistance in terms of B, l , v, and R. We can then separate the variables and integrate over all time to show that the total energy dissipated is equal to the initial kinetic energy of the rod. Express the power dissipated in terms of ε and R: Express ε as a function of B, l , and v:

2 ε P=

R

ε = Blv where

v = v0e − (B l

2 2

Substitute to obtain:

P=

(Blv )2 R

)

mR t

650 Chapter 28 The total energy dissipated as the rod comes to rest is obtained by integrating dE = P dt:

∞

E=∫ 0

∞

=∫

(Blv )2 dt R

(Blv e ( 0

)

− B 2l 2 mR t

R

0

) dt 2

∞

B 2l 2v02 −2 (B2l2 mR )t = e dt R ∫0 Evaluate the integral (by changing variables to u = −

2 2

2B l ) to mR

E=

B 2 l 2 v02 ⎛ mR ⎞ ⎜ ⎟= R ⎝ 2 B 2l 2 ⎠

1 2

mv02

obtain: 43 •• Picture the Problem In Example 28-9 it is shown that the speed of the rod is given by v = v0 e − (B l

2 2

) . We can write v as dx/dt, separate the variables and integrate to find

mR t

the total distance traveled by the rod. Apply the result from Example 28-9 to obtain:

dx = v0e −Ct dt where

C=

B 2l 2 mR ∞

x

Separate variables and integrate x′ from 0 to x and t′ from 0 to ∞:

∫ dx' = v ∫ e

Evaluate the integrals to obtain:

x=

Substitute for C and simplify:

0

0

x=

0

v0 C mv0 R B 2l 2

− Ct

dt

Magnetic Induction 651 44 •• Picture the Problem Let m be the mass of the rod. The net force acting on the rod is due to the current in it. We can obtain the equation of motion for the rod by applying Newton’s 2nd law to relate its acceleration to B, I, and l . The net emf that drives I in this circuit is the emf of the capacitor minus the emf induced in the rod as a result of its motion. (a) Letting the direction of motion of the rod be the positive x direction, apply Fx = ma x to the

∑

rod:

Solve equation (1) for I:

BIl = m

dv dt

(1)

where

Q − Blv I= C R I=

(2)

m dv Bl dt

or, because the capacitor is discharging,

− Simplify to obtain:

Integrate Q′ from Q0 to Q and v′ from 0 to v:

dQ m dv = dt Bl dt

dQ = −

m dv Bl

Q

v

m ∫Q dQ' = − Bl ∫0 dv' 0 and

Q = Q0 − Substitute in equation (2) to obtain:

m v Bl

m v Bl − Blv C I= R m Q0 − v Bl − Blv = R CR Q0 −

652 Chapter 28 Substitute in equation (1) to obtain the equation of motion of the rod:

m ⎞ ⎛ Q − v ⎟ dv Bl ⎜ 0 Bl ⎜ = − Blv ⎟ dt mR ⎜ C ⎟ ⎟ ⎜ ⎠ ⎝ =

(b) When the rod has achieved its terminal speed:

BlQ0 ⎛ 1 B 2l 2 ⎞ ⎟v − ⎜⎜ + mRC ⎝ RC mR ⎟⎠

BIl = m

dv =0 dt

and

Qf − Blvt I= C =0 R Solve for vt to obtain:

vt =

Qf CBl

*45 •• Picture the Problem The free-body diagram shows the forces acting on the rod as it slides down the inclined plane. The retarding force is the component of Fm acting up the incline, i.e., in the −x direction. We can express Fm using the expression for the force acting on a conductor moving in a magnetic field. Recognizing that only the horizontal r component of the rod’s velocity v produces an induced emf, we can apply the expression for a motional emf in conjunction with Ohm’s law to find the induced current in the rod. In part (b) we can apply Newton’s 2nd law to obtain an expression for dv/dt and set this expression equal to zero to obtain vt. (a) Express the retarding force acting on the rod:

F = Fm cosθ

(1)

where

Fm = IlB and I is the current induced in the rod as a consequence of its motion in the magnetic

Magnetic Induction 653 field. Express the induced emf due to the motion of the rod in the magnetic field:

ε = Blv cosθ

ε

Blv cos θ R

Using Ohm’s law, relate the current I in the circuit to the induced emf:

I=

Substitute in equation (1) to obtain:

⎛ Blv cosθ ⎞ F =⎜ ⎟lB cosθ R ⎝ ⎠ =

(b) Apply

∑F

x

= max to the rod:

R

=

B 2l 2v cos 2 θ R

B 2l 2v dv mg sin θ − cos 2 θ = m R dt and

dv B 2l 2v = g sin θ − cos 2 θ dt mR When the rod reaches its terminal velocity vt, dv/dt = 0 and: Solve for vt to obtain:

0 = g sin θ −

vt =

B 2 l 2 vt cos 2 θ mR

mgR sin θ B 2l 2 cos 2 θ

46 •• Picture the Problem The diagram shows the square loop being pulled from the r magnetic field B by the constant force

r F . The time required to pull the loop out of

the magnetic field depends on the terminal speed of the loop. We can apply Newton’s 2nd law and use the expressions for the magnetic force on a moving wire in a magnetic field to obtain the equation of motion for the loop and, from this equation, an expression for the terminal speed of the loop. Apply

r

r

∑ F = ma to the square

loop to obtain:

F − Fm = m

dv dt

(1)

654 Chapter 28 The magnetic force is given by:

Fm = IlB =

ε lB R

where R is the resistance of the loop. Substitute for Fm in equation (1) to obtain: The induced emf ε is related to the speed of the loop: Substitute for in equation (2) to obtain the equation of motion of the loop: When the loop reaches its terminal speed, dv/dt = 0 and:

F−

ε l B = m dv R

dt

(2)

ε = vBl F−

dv l2 B2 v=m R dt

R l2 B2 F− vt = 0 ⇒ v t = 2 2 F R l B

This result tells us that doubling F doubles the terminal speed vt. Hence, doubling F will halve the time required to pull the loop from the magnetic field and (c) is correct. 47 •• Picture the Problem The diagram shows the square loop being pulled from the r magnetic field B by the constant force

r F . The time required to pull the loop out of

the magnetic field depends on the terminal speed of the loop. We can apply Newton’s 2nd law and use the expressions for the magnetic force on a moving wire in a magnetic field to obtain the equation of motion for the loop and, from this equation, an expression for the terminal speed of the loop. Apply

r

r

∑ F = ma to the square

loop to obtain: The magnetic force is given by:

F − Fm = m Fm = IlB =

dv dt

(1)

ε lB R

where R is the resistance of the loop. Substitute for Fm in equation (1) to obtain: The induced emf ε is related to the speed of the loop:

F−

ε l B = m dv R

ε = vBl

dt

(2)

Magnetic Induction 655 Substitute for in equation (2) to obtain the equation of motion of the loop:

dv l2 B2 F− v=m R dt

When the loop reaches its terminal speed, dv/dt = 0 and:

F−

R l2 B2 vt = 0 ⇒ v t = 2 2 F R l B

This result tells us that halving R halves the terminal speed vt. Hence, halving R will double the time required to pull the loop from the magnetic field and (b) is correct. 48 •• Picture the Problem The diagram shows the initial position of the sphere and its position at t = 3 s. We can find the velocity of the sphere and the magnetic field when t r r r r = 3 s and use E = v × B to find E . We can find the voltage across the sphere at this time from the electric field at its center and its diameter.

(a) Relate the electric field at the center of the sphere to the magnetic field at that location:

r r r E = v×B

( )

Express the magnetic field as a function of the distance y from the current-carrying wire:

r µ 2I µ I B= 0 − iˆ = − 0 iˆ 4π y 2π y

Using a constant-acceleration equation, find the position of the sphere at t = 3 s:

y = y0 + v0, y ∆t + 12 a(∆t )

2

or, because y0 = h, v0,y = 0, and a = −g,

y = h − 12 g (∆t )

2

(

)

= 45 m − 12 9.81 m/s 2 (3 s )

2

= 0.855 m

r

Substitute and evaluate B :

r µ I B = − 0 iˆ 2π y

4π × 10 −7 N/A 2 20 A ˆ i 2π 0.855 m = (− 4.68 × 10 −6 T )iˆ =−

656 Chapter 28 Using a constant-acceleration equation, find the velocity of the sphere at t = 3 s:

r r r v = v0 + a∆t r r or, because v 0 = 0 and a = − g ˆj r v = − g∆t ˆj = − 9.81 m/s 2 (3 s ) ˆj

(

)

= (− 29.4 m/s ) ˆj

r

Substitute and evaluate E :

r E = (− 29.4 m/s ) ˆj × − 4.68 × 10 −6 T iˆ

(

=

)

(− 0.138 mV/m) kˆ

(b) The potential difference across the sphere depends on the electric field at the center of the sphere and the diameter of the sphere:

V = 2 RE

Substitute numerical values and evaluate V:

V = 2(0.02 m )(0.138 mV/m ) = 5.52 µV

49 •• Picture the Problem The free-body diagram shows the forces acting on the rod as it slides down the inclined plane. The retarding force is the component of Fm acting up the incline; i.e., in the −x direction. We can express Fm using the expression for the force acting on a conductor moving in a magnetic field. We can apply the expression for a motional emf in conjunction with Ohm’s law to find the induced current in the rod. In part (b) we can apply Newton’s 2nd law to obtain an expression for dv/dt and set this expression equal to zero to obtain vt. (a) Noting that only the horizontal component of the rod’s velocity r v produces an induced emf, express ε due to the motion of the rod in the magnetic field:

ε = Blv cosθ

Magnetic Induction 657 Substitute numerical values and evaluate ε : (b) Apply Newton’s 2nd law to the rod:

ε = (1.2 T )(15 m ) v (cos30°) = (15.6 T ⋅ m ) v

mg sin θ − Fm cos θ = m

dv dt

and

dv F = g sin θ − m cosθ dt m

(1)

where

Fm = IlB and I is the current induced in the rod as a consequence of its motion in the magnetic field. Using Ohm’s law, relate the current I in the circuit to the induced emf:

I=

ε R

=

Blv cos θ R

and

Fm =

B 2l 2v cos θ R

Substitute in equation (1) to obtain the equation of motion of the rod:

dv B 2l 2v = g sin θ − cos 2 θ dt mR

When the rod reaches its terminal velocity vt, dv/dt = 0:

B 2 l 2 vt 0 = g sin θ − cos 2 θ mR

Solve for vt:

vt =

mgR sin θ B 2l 2 cos 2 θ

Substitute numerical values and evaluate vt:

vt =

(0.4 kg ) (9.81 m/s 2 )(2 Ω )sin30° (1.2 T )2 (15 m )2 cos 2 30°

= 1.61 cm/s 50 ••• Picture the Problem Let Ff be the friction force between the rails and cylinder, Fm the magnetic force on the cylinder, and Im the cylinder’s moment of inertia. Because the current through the rod is uniformly distributed, we can treat the current as though it were r concentrated at the center of the rod. We can find the magnitude of B by applying Newton’s 2nd law to the cylinder. The application of Ohm’s law to the circuit will allow us to express the net force acting on the cylinder in terms of its speed. Setting this net

658 Chapter 28 force equal to zero will lead us to a value for the terminal velocity of the cylinder. We can use the definition of kinetic energy (both translational and rotational) to find the kinetic energy of the cylinder when it has reached its terminal velocity. (a) Apply

∑F

x

= max to the

Fm − Ff = ma x or

cylinder:

BIa − Ff = m Apply

∑τ = Iα to the cylinder:

Solve for Ff and substitute to obtain:

Solve for r

dω : dt

dω dt

Ff r = I m BIa −

dv dω = mr dt dt

I m dω dω = mr r dt dt

dω BIa BIa 2 BIa = = = 2 1 mr dt m + I m 3m m+ 2 2 2 r r

r or

dv 2 BIa = dt 3m

(1)

Solve for B:

dv dt B= 2 Ia

Apply Ohm’s law to the circuit to find I:

I=

Substitute numerical values and evaluate B:

3(4 kg ) 0.1 m/s 2 B= = 0.750 T 2(2A )(0.4 m )

r

r r r direction of B :

Apply F = I l × B to determine the

3m

ε R

=

12V = 2A 6Ω

(

r B is downward.

(b) Multiply both sides of equation (1) by m to express the net force acting on the cylinder:

Fnet = m

Use Ohm’s law to express the current as a function of the emf of

I=

dv 2 BIa = 3 dt

ε − Bav R

)

Magnetic Induction 659 the battery and the induced emf in the cylinder: Substitute to express the net force acting on the cylinder as a function of the velocity of the cylinder:

Fnet

⎛ ε − Bav ⎞ 2 B⎜ ⎟a R ⎠ ⎝ = 3

2 Baε 2 B 2 a 2 − v 3R 3R

=

(c) Set Fnet = 0 and solve for the terminal velocity of the cylinder:

2 Baε 2 B 2 a 2 − vt = 0 3R 3R and

vt = (d) Express the total kinetic energy of the cylinder when it has reached its terminal velocity:

ε Ba

=

12 V = 40.0 m/s (0.75 T )(0.4 m )

K = 12 mvt2 + 12 I mω t2

= 12 mvt2 + 12

(

1 2

mr 2

) vr

2 t 2

= 34 mvt2 Substitute numerical values and evaluate K:

K=

3 4

(4 kg )(40 m/s )2 =

4.80 kJ

*51 ••• Picture the Problem We can use the expression for a motional emf and Ampere’s law to express the net emf induced in the moving loop. We can also use express the magnetic flux through the loop and apply Faraday’s law to obtain the same result. (a) Express the motional emf induced in the segments parallel to the current-carrying wire: Using Ampere’s law, express B(d + vt) and B(d + a + vt):

ε = B(x ) vb

B(d + vt ) =

µ0 I

2π (d + vt )

and

B(d + a + vt ) = Substitute to express ε1 for the near wire and ε2 for the far wire:

ε1 =

µ 0 Ivb 2π (d + vt )

µ0 I

2π (d + a + vt )

660 Chapter 28 and

µ 0 Ivb

ε1 = Noting that the emfs both point upward and hence oppose one another, express the net emf induced in the loop:

2π (d + a + vt )

ε = ε1 − ε 2 =

µ 0 Ivb µ 0 Ivb − 2π (d + vt ) 2π (d + a + vt )

=

µ 0 Ivb ⎛ 1 1 ⎞ − ⎜ ⎟ 2π ⎝ d + vt d + a + vt ⎠

The motion of the segments perpendicular to the long wire does not change the flux through the rectangular loop. Consequently, these segments do not contribute to the the induced emf. (b) From Faraday’s law we have:

ε = − dφm dt

Express the magnetic flux in an area of length b and width vdt:

dφm = B( x ) dA = B( x ) b dx where, from Ampere’s law,

B(x ) = Substitute and integrate from x = d + vt to d + a + vt :

µ0 I 2πx

d + a + vt

µ 0 Ib d +a +vt dx φm = ∫ B( x ) dx = 2π d +∫vt x d + vt =

µ 0 Ib ⎡ d + a + vt ⎤ ln ⎢ 2π ⎣ d + vt ⎥⎦

Differentiate with respect to time and simplify to obtain:

µ Ib d ⎡ d + a + vt ⎤ ⎡ µ 0 Ib d + a + vt ⎤ =− 0 ln ln ⎢ ⎥ dt ⎣ 2π d + vt ⎦ d + vt ⎥⎦ 2π dt ⎢⎣

ε =− d =−

µ 0 Ib ⎡⎛ d + vt ⎞⎛ (d + vt )v − (d + a + vt ) v ⎞⎤ ⎟⎟⎥ ⎜ ⎟⎜ 2π ⎢⎣⎝ d + a + vt ⎠⎜⎝ (d + vt )2 ⎠⎦⎥

=−

µ 0 Ibv ⎡ (d + vt ) − (d + a + vt ) ⎤ µ Ibv ⎡ 1 1 ⎤ =− 0 − ⎢ ⎥ ⎢ 2π ⎣ (d + vt )(d + a + vt ) ⎦ 2π ⎣ d + a + vt d + vt ⎥⎦

=

µ 0 Ibv ⎡ 1 1 ⎤ − ⎢ 2π ⎣ d + vt d + a + vt ⎥⎦

Magnetic Induction 661 52 ••• r r r Picture the Problem We can use F = qv × B to express the magnetic force acting on the moving charged body. Expressing the emf induced in a segment of the rod of length dr and integrating this expression over the length of the rod will lead us to an expression for the induced emf. (a) Using the equation for the magnetic force on a moving charged body, express the force acting on the charged body a distance r from the pivot:

r

r

Because v ⊥ B and v = rω: (b) Use the motional emf equation to express the emf induced in a segment of the rod of length dr and at a distance r from the pivot: Integrate this expression from r = 0 to r = l to obtain:

r r r F = qv × B and

F = qvB sin θ

F = qBrω dε = Brdv = Brωdr

ε

l

0

0

∫ dε' = Bω ∫ rdr and

ε=

1 2

Bωl 2

dφm dt

(c) Using Faraday’s law, relate the induced emf to the rate at which the flux changes:

ε

Express the area dA, for any value of θ, between r and r + dr:

dA = rθdr

=

l

Integrate from r = 0 to r = l to obtain:

A = θ ∫ rdr = 12 θl 2

Using its definition, express the magnetic flux through this area:

φm = BA =

Differentiate φm with respect to time to obtain:

ε

0

=

d dt

[

1 2

1 2

Bl 2θ

]

Bl 2θ = 12 Bl 2

dθ = dt

1 2

Bl 2ω

662 Chapter 28

Inductance 53 • Picture the Problem We can use φm = LI and the dependence of I on t to find the magnetic flux through the coil. We can apply Faraday’s law to find the induced emf in the coil. (a) Use the definition of selfinductance to express φm:

φm = LI

Express I as a function of time:

I = 3 A + (200 A/s ) t

Substitute to obtain:

φm = L[3 A + (200 A/s) t ]

Substitute numerical values and express φm:

φm = (8 H )[3 A + (200 A/s) t ] = 24 Wb + (1600 H ⋅ A/s ) t

(b) Use Faraday’s law to relate ε, L, and dI/dt:

ε = − L dI

Substitute numerical values and evaluate ε :

ε = −(8 H )(200 A/s) =

dt − 1.60 kV

*54 • Picture the Problem We can apply φm = LI to find φm and Faraday’s law to find the self-induced emf as functions of time. Use the definition of self-inductance to express φm:

φm = LI = LI 0 sin 2πft

The graph of the flux φm as a function of time shown below was plotted using a spreadsheet program. The maximum value of the flux is LI0 and we have chosen 2πf = 1 rad/s.

Magnetic Induction 663 1.0

flux

0.5

0.0

-0.5

-1.0 0

1

2

3

4

5

6

t (s)

Apply Faraday’s law to relate ε , L, and dI/dt:

ε = − L dI dt

= −L

d [I 0 sin 2πft ] dt

= − 2πfLI 0 cos 2πft The graph of the emf ε as a function of time shown below was plotted using a spreadsheet program. The maximum value of the induced emf is 2πfLI0 and we have chosen 2πf = 1 rad/s.

1.0

emf

0.5

0.0

-0.5

-1.0 0

1

2

3

t (s)

4

5

6

664 Chapter 28 55 •• Picture the Problem We can use B = µ 0 nI to find the magnetic field on the axis at the center of the solenoid and the definition of magnetic flux to evaluate φm. We can use the definition of magnetic flux in terms of L and I to find the self-inductance of the solenoid. Finally, we can use Faraday’s law to find the induced emf in the solenoid when the current changes at 150 A/s.

B = µ 0 nI

(a) Apply the expression for B inside a long solenoid to express and evaluate B:

⎛ 400 ⎞ ⎟⎟(3 A ) = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0.25 m ⎠

(

)

= 6.03 mT (b) Apply the definition of magnetic flux to obtain:

φm = NBA = (400 )(6.03 mT )π (0.01 m )

2

= 7.58 × 10 −4 Wb (c) Relate the self-inductance of the solenoid to the magnetic flux through it and its current: (d) Apply Faraday’s law to obtain:

L=

φm I

=

7.58 × 10 −4 Wb = 0.253 mH 3A

ε = − L dI dt

= −(0.253 mH )(150 A/s )

= − 38.0 mV 56 •• Picture the Problem We can find the mutual inductance of the two coaxial solenoids using M 2,1 =

φm2 I1

= µ 0 n2 n1lπr12 .

Substitute numerical values and evaluate M2,1:

⎛ 300 ⎞ ⎛ 1000 ⎞ ⎟⎟(0.25 m )π (0.02 m )2 = 1.89 mH ⎟⎟ ⎜⎜ M 2,1 = 4π × 10 −7 N/A 2 ⎜⎜ ⎝ 0.25 m ⎠ ⎝ 0.25 m ⎠

(

)

*57 •• Picture the Problem Note that the current in the two parts of the wire is in opposite directions. Consequently, the total flux in the coil is zero. We can find the resistance of the wire-wound resistor from the length of wire used and the resistance per unit length.

Magnetic Induction 665 Because the total flux in the coil is zero:

L= 0

Express the total resistance of the wire:

⎛ Ω⎞ ⎛ Ω⎞ R = ⎜18 ⎟ L = ⎜18 ⎟(9 m ) = 162 Ω ⎝ m⎠ ⎝ m⎠

58 ••• Picture the Problem We can apply Kirchhoff’s loop rule to the galvanometer circuit to relate the potential difference across L2 to the potential difference across R2. Integration of this equation over time will yield an equation that relates the mutual inductance between the two coils to the steady-state current in circuit 1 and the charge that flows through the galvanometer. Apply Kirchhoff’s loop rule to the galvanometer circuit:

dI1 dI + L2 2 − R2 I 2 = 0 dt dt

M or

MdI1 + L2 dI 2 − R2 I 2 dt = 0 Integrate each term from t = 0 to t = ∞:

∞

∞

∞

0

0

0

M ∫ dI1 + L2 ∫ dI 2 − R2 ∫ I 2 dt = 0 and

MI1∞ + L2 I 2∞ − R2Q = 0 Because I2∞ = 0:

MI1∞ − R2Q = 0

Solve for M:

M=

Substitute numerical values and evaluate M:

M

R2Q I1∞

( 300 Ω ) (2 × 10 −4 C ) = = 5A

12.0 mH

59 ••• Picture the Problem We can use Ampere’s law to express the magnetic field inside the rectangular toroid and the definition of magnetic flux to express φm through the toroid. We can then use the definition of self-inductance of a solenoid to express L. Using the definition of the selfinductance of a solenoid, express L in terms of φm, N, and I:

L=

Nφ m I

(1)

666 Chapter 28 Apply Ampere’s law to a closed path or radius a < r < b:

r r B ∫ ⋅ d l = B 2πr = µ0 I C C

or, because IC = NI,

B 2πr = µ 0 NI

µ 0 NI 2πr

Solve for B to obtain:

B=

Express the flux in a strip of height H and width dr:

dφm = BHdr

Substitute for B and integrate dφm from r = a to r = b to obtain:

µ0 NIH b dr µ0 NIH ⎛ b ⎞ = φm = ln⎜ ⎟ 2π ∫a r 2π ⎝a⎠

Substitute in equation (1) and simplify to obtain:

L=

µ0 N 2 H ⎛ b ⎞ ln⎜ ⎟ 2π ⎝a⎠

Magnetic Energy 60 • Picture the Problem The current in an LR circuit, as a function of time, is given by I = I f 1 − e −t τ , where If = ε0/R and τ = L/R. The energy stored in the inductor under

(

)

steady-state conditions is stored in its magnetic field and is given by U m = 12 LI f2 . (a) Express and evaluate If:

(b) Express and evaluate the energy stored in an inductor:

If =

ε0 R

=

24 V = 2.00 A 12 Ω

U m = 12 LI f2 =

1 2

(2 H )(2 A )2 =

*61 •• Picture the Problem We can examine the ratio of um to uE with E = cB and c = 1 ε 0 µ0 to show that the electric and magnetic energy densities are equal. Express the ratio of the energy density in the magnetic field to the energy density in the electric field:

B2 2µ um B2 = 1 02 = uE 2 ε 0 E µ 0ε 0 E 2

Substitute E = cB:

1 um B2 = = 2 2 u E µ 0ε 0 c B µ 0ε 0 c 2

4.00 J

Magnetic Induction 667 Substitute for c:

u m µ 0ε 0 = = 1 ⇒ um = uE u E µ 0ε 0

62 •• Picture the Problem We can use L = µ 0 n 2 Al to find the inductance of the solenoid and

B = µ 0 nI to find the magnetic field inside it.

(a) Express the magnetic energy stored in the solenoid:

U m = 12 LI 2

Relate the inductance of the solenoid to its dimensions and properties:

L = µ 0 n 2 Al

Substitute to obtain:

U m = 12 µ 0 n 2 AlI 2

Substitute numerical values and evaluate Um:

Um =

1 2

⎛ 2000 ⎞ ⎟⎟ 4π × 10 N/A ⎜⎜ ⎝ 0.3 m ⎠

(

−7

(

2

)

2

)

× 4 × 10− 4 m 2 (0.3 m )(4 A )

2

= 53.6 mJ (b) The magnetic energy per unit volume in the solenoid is:

53.6 mJ Um Um = = 4 × 10 −4 m 2 (0.3 m ) V Al

(

)

= 447 J/m 3 (c) Express the magnetic field in the solenoid in terms of n and I:

N I l 4π × 10 −7 N/A 2 (2000 )(4 A ) = 0.3 m

B = µ 0 nI = µ 0

(

)

= 33.5 mT (d) The magnetic energy density is:

(33.5 mT ) B2 = 2µ 0 2(4π × 10 −7 N/A 2 ) 2

um =

= 447 J/m 3 in agreement with our result in Part (b).

668 Chapter 28 63 •• Picture the Problem Consider a cylindrical annulus of thickness dr at a radius r < a. We can use its definition to express the total magnetic energy dU m inside the cylindrical annulus and divide both sides of this expression by the length of the wire to express the magnetic energy per unit length dU' m . Integration of this expression will give us the magnetic energy per unit length within the wire. Express the magnetic energy within the cylindrical annulus:

B2 B2 dU m = Vannulus = 2πrldr 2µ0 2µ0 =

Divide both sides of the equation by l to express the magnetic energy per unit length dU' m : Use Ampere’s law to express the magnetic field inside the wire at a distance r < a from its center:

dU' m =

B2

µ0 B2

µ0

πrldr

πrdr

(1)

2πrB = µ 0 I C and

B=

µ0 I C 2πr

where IC is the current inside the cylinder of radius r. Because the current is uniformly distributed over the cross-sectional area of the wire:

I C πr 2 r2 = ⇒ IC = 2 I a I πa 2

Substitute to obtain:

B=

Substitute for B in equation (1) to obtain:

⎛ µ 0 rI ⎞ ⎜ ⎟ µ I2 2πa 2 ⎠ ⎝ dU 'm = πrdr = 0 4 r 3 dr µ0 4πa

µ 0 rI 2πa 2 2

Magnetic Induction 669 Integrate dU' m from r = 0 to r = a:

U' m =

µ0 I 2 a 3 µ0 I 2 a 4 µ0 I 2 r dr = ⋅ = 4πa 4 ∫0 4πa 4 4 16π

Remarks: Note that the magnetic energy per unit length is independent of the radius of the cylinder and depends only on the total current. *64 •• Picture the Problem The wire of length d and radius a is shown in the diagram, as is the inductor constructed with this wire and whose inductance L is to be found. We can use the equation for the self-inductance of a cylindrical inductor to derive an expression for L.

The self-inductance of an inductor with length l, cross-sectional area A, and number of turns per unit length n is:

L = µ 0 n 2 Al

The number of turns N is given by:

l 2a 1 N n= = l 2a

The number of turns per unit length n is:

(1)

N=

Assuming that a << r, the length of the wire d is related to n and r:

πr ⎛ l ⎞ d = N (2π r ) = ⎜ ⎟2π r = l a ⎝ 2a ⎠

Solve for l to obtain:

l=

Substitute for l, A, and n in equation (1) to obtain:

⎛ 1 ⎞ L = µ0 ⎜ ⎟ π r 2 ⎝ 2a ⎠

ad πr 2

(

)⎛⎜⎜ πadr ⎞⎟⎟ = ⎝

⎠

⎛ rd ⎞ ⎟ ⎝ 4a ⎠

µ0 ⎜

65 • Picture the Problem We can substitute numerical values in the expression derived in Problem 64 to find the self-inductance of the inductor.

670 Chapter 28 From Problem 64 we have:

L=

µ 0 rR 4a

Substitute numerical values and evaluate L:

L=

(4π × 10

−7

N/A 2 )(0.25 cm )(10 cm ) = 0.157 µH 4(0.5 mm )

66 •• Picture the Problem We can find the number of turns on the coil from the length of the superconducting wire and the cross-sectional radius of the coil. We can use B = (µ 0 NI ) (2πrmean ) to find the magnetic field at the mean radius. We can find the

energy density in the magnetic field from u m = B 2 (2 µ 0 ) and the total energy stored in

the toroid by multiplying u m by the volume of the toroid. (a) Express the number of turns in terms of the length of the wire L and length required per turn 2πr:

N=

1000 m L = = 7958 2πr 2π (0.02 m )

(b) Use the equation for B inside a tightly wound toroid to find the magnetic field at the mean radius:

B=

µ 0 NI 2πrmean

=

(4π × 10

−7

)

N/A 2 (7958)(400 A ) 2π (0.25 m )

= 2.55 T (c) Express and evaluate the energy density in the magnetic field:

(2.55 T ) B2 = 2 µ 0 2(4π × 10 −7 N/A 2 ) 2

um =

= 2.59 × 10 6 J/m 3 Relate the total energy stored in the toroid to the energy density in its magnetic field and the volume of the toroid: Think of the toroid as a cylinder of radius r and height 2πrmean to obtain: Substitute for Vtoroid to obtain:

U m = u mVtoroid

Vtoroid = πr 2 (2πrmean ) = 2π 2 r 2 rmean

U m = 2π 2 r 2 rmeanum

Magnetic Induction 671 Substitute numerical values and evaluate Um:

U m = 2π 2 (0.02 m ) (0.25 m ) (2.59 × 10 6 J/m 3 ) = 5.11 kJ 2

RL Circuits 67 • Picture the Problem We can find the current using I = I f 1 − e −t τ where If = ε0/R and

(

)

τ = L/R and its rate of change by differentiating this expression with respect to time. Express the dependence of the current on If and τ: Evaluate If and τ:

I = I f (1 − e −t τ )

If =

ε0 R

=

100 V = 12.5 A 8Ω

and

τ= Substitute to obtain:

L 4H = = 0.5 s R 8Ω

(

I = (12.5 A ) 1 − e −t 0.5 s

(

= (12.5 A ) 1 − e −2t s Express dI/dt:

(

)

)

)(

−1 dI = (12.5 A ) − e −2t s − 2 s −1 dt

= (25 A/s) e −2t s

(a) When t = 0:

−1

)

−1

I = (12.5 A ) (1 − e 0 ) = 0 and

dI = (25 A/s) e 0 = 25.0 A/s dt (b) When t = 0.1 s:

(

)

I = (12.5 A ) 1 − e −0.2 = 2.27 A and

dI = (25 A/s) e −0.2 = 20.5 A/s dt (c) When t = 0.5 s:

(

)

I = (12.5 A ) 1 − e −1 = 7.90 A and

672 Chapter 28 dI = (25 A/s) e −1 = 9.20 A/s dt (d) When t = 1.0 s:

(

)

I = (12.5 A ) 1 − e −2 = 10.8 A and

dI = (25 A/s) e −2 = 3.38 A/s dt

68 • Picture the Problem We can find the current using I = I 0 e −t τ , where I0 is the current at time t = 0 and τ = L/R. Express the current as a function of time:

I = I 0 e −t τ = (2 A ) e −t τ L 1 mH = = 10 −4 s R 10 Ω

Evaluate τ:

τ=

Substitute to obtain:

I = (2 A ) e −10

(a) When t = 0.5 ms:

4 −3 −1 I = (2 A ) e −10 (0.5×10 s )s = (2 A ) e −5

4

t s −1

= 13.5 mA (b) When t = 10 ms:

4 −3 −1 I = (2 A ) e −10 (10×10 s )s = (2 A ) e −100

= 7.44 × 10 −44 A ≈ 0 *69 •• Picture the Problem We can find the current using I = I f 1 − e −t τ , where

(

)

If = ε0/R,and τ = L/R, and its rate of change by differentiating this expression with respect to time. Express the dependence of the current on If and τ: Evaluate If and τ:

I = I f (1 − e −t τ )

If = and

ε0 R

=

12 V = 4A 3Ω

Magnetic Induction 673

τ=

L 0.6 H = = 0.2 s R 3Ω

(

(

)

I = (4 A ) 1 − e −t 0.2 s = (4 A ) 1 − e −5t s

Substitute to obtain:

(

)(

−1 dI = (4 A ) − e −5t s − 5 s −1 dt

Express dI/dt:

= (20 A/s ) e −5t s

)

−1

(

I (0.5 s ) = (4 A ) 1 − e −5(0.5 s )s

(a) Find the current at t = 0.5 s:

−1

−1

)

= 3.67 A

P(0.5 s ) = I (0.5 s )ε

The rate at which the battery supplies power at t = 0.5 s is:

= (3.67 A )(12 V ) = 44.0 W

PJ (0.5 s ) = [I (0.5 s )] R 2

(b) The rate of joule heating is:

= (3.67 A ) (3 Ω ) 2

= 40.4 W (c) Using the expression for the magnetic energy stored in an inductor, express the rate at which energy is being stored:

dU L d = dt dt

[

1 2

LI 2 = LI

]

dI dt

Substitute for L, I, and dI/dt to obtain:

dU L d = dt dt

[

1 2

LI 2 = LI

]

dI dt

Substitute numerical values and evaluate

(

dU L : dt

)

(

)

−1 −1 −1 −1 dU L = (0.6 H )(4 A ) 1 − e −5t s (20 A/s ) e −5t s = (48 W ) 1 − e −5t s e −5t s dt

Evaluate this expression for t = 0.5 s:

(

)

−1 −1 dU L = (48 W ) 1 − e −5(0.5 s )s e −5(0.5 s )s dt = (48 W ) 1 − e −2.5 e −2.5

(

= 3.62 W

)

)

674 Chapter 28 Remarks: Note that, to a good approximation, dUL/dt = P − PJ. 70 •• Picture the Problem We can find the current using I = I f 1 − e −t τ , where If = ε0/R and

(

)

τ = L/R, and its rate of change by differentiating this expression with respect to time. Express the dependence of the current on If and τ: Evaluate If and τ:

(

I = I f 1 − e −t τ

If =

)

ε 0 = 12 V = 4 A 3Ω

R

and

τ= Substitute to obtain:

L 0.6 H = = 0.2 s 3Ω R

(

I = (4 A ) 1 − e −t 0.2 s

(

= (4 A ) 1 − e −5t s Express dI/dt:

−1

)

(

)(

−1 dI = (4 A ) − e −5t s − 5 s −1 dt

= (20 A/s ) e −5t s

(a) Find the current at t = 1 s:

) )

−1

(

I (1 s ) = (4 A ) 1 − e −5(1 s ) s

−1

)

= 3.97 A The rate at which the battery supplies power at t = 1 s:

Find the current at t = 100 s:

P(1 s ) = I (1 s )ε = (3.97 A )(12 V )

= 47.7 W

(

I (100 s ) = (4 A ) 1 − e −5(100 s ) s

−1

)

= 4.00 A The rate at which the battery supplies power at t = 100 s:

(b) The rate of joule heating at t = 1 s is:

P(100 s ) = I (100 s )ε = (4 A )(12 V )

= 48.0 W PJ (1s ) = [I (1s )] R 2

= (3.97 A ) (3 Ω ) 2

= 47.3 W

Magnetic Induction 675 The rate of joule heating at t = 100 s is:

PJ (100 s ) = (4 A ) (3Ω ) = 48.0 W

Using the expression for the magnetic energy stored in an inductor, express the rate at which energy is being stored:

dU L d = dt dt

[

Substitute for L, I and dI/dt to obtain:

dU L d = dt dt

[

2

1 2

1 2

LI 2 = LI

]

dI dt

]

dI dt

LI 2 = LI

(

= (0.6 H )(4 A ) 1 − e −5t s × (20 A/s ) e −5t s

(

−1

)

−1

)

= (48 W ) 1 − e −5t s e −5t s −1

(

−1

)

−1 −1 dU L = (48 W ) 1 − e −5(1s )s e −5(1s )s dt = (48 W ) 1 − e −5 e −5

Evaluate dUL/dt for t = 1 s:

(

)

= 0.321 W

(

)

−1 −1 dU L = (48 W ) 1 − e −5(100 s )s e −5(100 s )s dt = (48 W ) 1 − e −500 e −500

Evaluate dUL/dt for t = 100 s:

(

)

= 0 Remarks: Note that, to a good approximation, dUL/dt = P − PJ. 71 •• Picture the Problem If the current is initially zero in an LR circuit, its value at some later time t is given by I = I f 1 − e −t τ , where If = ε0/R and τ = L/R is the time constant

(

)

for the circuit. We can find the time constant of the circuit from the given information and then use the definition of the time constant to find the self-inductance. (a) Express the current in the circuit as a function of time: Express the current when t = 4 s:

(

)

I = I f 1 − e −t τ where τ =

(

0.5I f = I f 1 − e −4 s τ

)

L R

or

0.5 = 1 − e −4 s τ ⇒ 0.5 = e −4 s τ

(1)

676 Chapter 28 4s

Take logarithms of both sides of this equation to obtain:

ln 12 = −

Solve for and evaluate τ:

τ=

(b) Solve equation (1) for and evaluate L:

L = Rτ = (5 Ω )(5.77 s ) = 28.9 H

τ

4s = 5.77 s ln2

72 •• Picture the Problem If the current is initially zero in an LR circuit, its value at some later time t is given by I = I f 1 − e −t τ , where If = ε0/R and τ = L/R is the time constant

(

)

for the circuit. We can find the number of time constants that must elapse before the current reaches any given fraction of its final value by solving this equation for t/τ . Express the fraction of its final value to which the current has risen as a function of time:

I = 1 − e −t τ If

Solve for t/τ :

⎛ I = − ln⎜⎜1 − τ ⎝ If

(a) Evaluate t/τ for I/If = 0.9:

t

t

τ (b) Evaluate t/τ for I/If = 0.99:

= − ln (1 − 0.9 ) = 2.30 90%

t

τ (c) Evaluate t/τ for I/If = 0.999:

= − ln (1 − 0.99 ) = 4.61 99%

t

τ

⎞ ⎟⎟ ⎠

= − ln (1 − 0.999 ) = 6.91 99.9%

73 •• Picture the Problem If the current is initially zero in an LR circuit, its value at some later time t is given by I = I f 1 − e −t τ , where If = ε0/R and τ = L/R is the time constant

(

)

for the circuit. We can find the rate of increase of the current by differentiating I with respect to time and the time for the current to reach any given fraction of its initial value by solving for t. (a) Express the current in the circuit as a function of time:

I=

ε 0 (1 − e−t τ ) R

Magnetic Induction 677 dI ε 0 d 1 − e −t τ = dt R dt

(

Express the initial rate of increase of the current by differentiating this expression with respect to time:

)

ε ⎛ 1 ⎞ ε0 = = 0 (− e −t τ ) − ⎟ ⎜ ⎝ τ⎠

R

ε 0 e − RL t

= dI dt

Evaluate dI/dt at t = 0 to obtain:

τR

e

R − t L

L

=

ε 0 e0 =

t =0

L

12 V = 3.00 kA/s 4 mH

(b) When I = 0.5If:

0.5 = 1 − e −t τ ⇒ e −t τ = 0.5

Evaluate dI/dt with e −t τ = 0.5 to obtain:

dI dt

= 0 .5 e−t τ =0.5

ε0

⎛ 12 V ⎞ ⎟⎟ = 0.5⎜⎜ L ⎝ 4 mH ⎠

= 1.50 kA/s (c) Calculate If from ε and R:

If =

ε0 = R

12 V = 80.0mA 150 Ω

(d) When I = 0.99If:

0.99 = 1 − e −t τ ⇒ e −t τ = 0.01

Solve for and evaluate t:

t = −τ ln (0.01) = − =−

L ln (0.01) R

4 mH ln (0.01) = 0.123 ms 150 Ω

74 •• Picture the Problem If the current is initially zero in an LR circuit, its value at some later time t is given by I = I f 1 − e −t τ , where If = ε0/R and τ = L/R is the time constant

(

)

for the circuit. We can find the time for the current to reach any given value by solving this equation for t. Evaluate If and τ :

If =

ε 0 = 250 V = 31.25 A R

8Ω

and

τ=

L 50 H = = 6.25 s R 8Ω

678 Chapter 28

(

)

Solve I = I f 1 − e −t τ for t:

⎛ I ⎞ t = −τ ln⎜⎜1 − ⎟⎟ ⎝ If ⎠ ⎛ ⎞ I ⎟⎟ = −(6.25 s )ln⎜⎜1 − 31 . 25 A ⎝ ⎠

(a) Evaluate t for I = 10 A:

⎛ 10 A ⎞ ⎟⎟ t 10 A = −(6.25 s ) ln⎜⎜1 − ⎝ 31.25 A ⎠ = 2.41s

(b) Evaluate t for I = 30 A:

⎛ 30 A ⎞ ⎟⎟ t 30 A = −(6.25 s ) ln⎜⎜1 − ⎝ 31.25 A ⎠ = 20.1s

*75 ••• Picture the Problem The self-induced emf in the inductor is proportional to the rate at which the current through it is changing. Under steady-state conditions, dI/dt = 0 and so the self-induced emf in the inductor is zero. We can use Kirchhoff’s loop rule to obtain the current through and the voltage across the inductor as a function of time. (a) Because, under steady-state conditions, the self-induced emf in the inductor is zero and because the inductor has negligible resistance, we can apply Kirchhoff’s loop rule to the loop that includes the source, the 10-Ω resistor, and the inductor to find the current drawn from the battery and flowing through the inductor and the 10-Ω resistor: By applying Kirchhoff’s junction rule at the junction between the resistors, we can conclude that:

10 V − (10 Ω )I = 0 and

I=

10 V = 1.00 A 10 Ω

I100-Ω resistor = I battery − I inductor = 0

(b) When the switch is closed, the current cannot immediately go to zero in the circuit because of the inductor. For a time, a current will circulate in the circuit loop between the inductor and the 100-Ω resistor. Because the current flowing through this circuit is initially 1 A, the voltage drop across the 100-Ω resistor is initially

100 V. Conservation of energy (Kirchhoff’s loop rule) requires that the voltage drop

Magnetic Induction 679 across the inductor is also 100 V.

dI + IR = 0 dt

(c) Apply Kirchhoff’s loop rule to the RL circuit to obtain:

L

The solution to this differential equation is:

I (t ) = I 0 e

R − t L

−

t

= I 0e τ 2H L = 0.02 s where τ = = R 100 Ω

A spreadsheet program to generate the data for graphs of the current and the voltage across the inductor as functions of time is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form B1 2 L B2 100 R B3 1 I0 A6 0 t0 R B6 $B$3*EXP((−$B$2/$B$1)*A6) − t

I 0e

A B L= 2 R= 100 I_0= 1

L

C H ohms A

1 2 3 4 5 6 7 8 9 10 11 12

t 0.000 0.005 0.010 0.015 0.020 0.025 0.030

I(t) V(t) 1.00E+00 100.00 7.79E−01 77.88 6.07E−01 60.65 4.72E−01 47.24 3.68E−01 36.79 2.87E−01 28.65 2.23E−01 22.31

32 33 34 35 36

0.130 0.135 0.140 0.145 0.150

1.50E−03 1.17E−03 9.12E−04 7.10E−04 5.53E−04

0.15 0.12 0.09 0.07 0.06

The following graph of the current in the inductor as a function of time was plotted using the data in columns A and B of the spreadsheet program.

680 Chapter 28 1.0 0.8

I (A)

0.6 0.4 0.2 0.0 0.00

0.03

0.06

0.09

0.12

0.15

t (s)

The following graph of the voltage across the inductor as a function of time was plotted using the data in columns A and C of the spreadsheet program. 100

V (V)

80 60 40 20 0 0.00

0.03

0.06

0.09

0.12

0.15

t (s)

76 •• Picture the Problem We can evaluate the derivative of Equation 28-26 with respect to time at t = 0 to find the slope of the linear function of current as a function of time. Because the I-intercept of this equation is I0, we can evaluate I(t) at t = τ to show that the current is zero after one time constant. Equation 28-26 describes the current in an LR circuit from which the source has been removed:

I = I 0e

−

t

τ

Magnetic Induction 681 Differentiate this expression with respect to t to obtain:

t

dI dt

t

I0

e

=−

I0

=− Evaluate dI/dt at t = 0:

t

− ⎛ 1⎞ dI d − = I 0 e τ = I 0e τ ⎜ − ⎟ dt dt ⎝ τ⎠

τ

t =0

Assuming that the current decreases steadily at this rate, express I as a linear function of t to obtain:

I (t ) = −

I0

Evaluate this function when t=τ:

I (τ ) = −

I0

τ

τ

−

τ

τ

t + I0

τ + I0 = 0

as was to have been shown. 77 •• Picture the Problem The current in an initially energized but source-free RL circuit is given by I = I 0 e −t τ . We can find τ from this equation and then use its definition to evaluate L. (a) Express the current in the RL circuit as a function of time:

I = I 0 e −t τ

Solve for and evaluate τ :

τ =−

t

⎛ I ⎞ ln⎜⎜ ⎟⎟ ⎝ I0 ⎠

=−

45 ms = 88.1 ms ⎛ 1.5 A ⎞ ⎟⎟ ln⎜⎜ ⎝ 2.5 A ⎠

(b) Using the definition of the inductive time constant, relate L to R:

L = τR

Substitute numerical values and evaluate L:

L = (0.0881s )(0.4 Ω ) = 35.2 mH

78 • Picture the Problem We can model this coil as a resistance-free inductor in series with an inductance-free resistor and express the potential difference across the coil as the sum of the potential differences across the inductor and the resistor. We can then use the given data to obtain two equations in the unknowns R and L and solve these equations

682 Chapter 28 simultaneously for the resistance and self-inductance of the coil. Express the potential difference across the coil as the sum of the potential difference across a resistor and the potential difference across an inductor: When I = 5 A and dI/dt = 10 A/s: When I = 5 A and dI/dt = −10 A/s: Add these equations to obtain:

V = VR + VL

= IR + L

dI dt

140 V = (5 A )R + (10 A/s )L 60 V = (5 A )R − (10 A/s)L 200 V = (10 A )R and

R=

Substitute in either of the equations to obtain:

200V = 20.0 Ω 10 A

L = 4.00 H

79 •• Picture the Problem We can use the definition of inductance to express the rate at which the current changes through the inductors and the resistor and the result of Problem 88 to find the effective inductance in the circuit. We can find the final/steadystate current by applying Ohm’s law. (a) Express the rate of change of the current through the resistor:

ε dI R = dt Leff

Using the result given in Problem 88, find Leff:

1 1 1 = + Leff 8 mH 4 mH and

Leff = 2.67 mH

Substitute numerical values and evaluate dIR/dt at t = 0:

dI R dt

= t =0

24 V = 9.00 kA/s 2.67 mH

Magnetic Induction 683

ε

Express the rate of change of the current through the 8-mH inductor:

dI 8 mH

Express the rate of change of the current through the 4-mH inductor:

dI 4 mH

Because IR = 0 when t = 0:

VLeff = V8 mH = V4 mH = 24 V

Substitute numerical values in equation (1) and evaluate dI 8 mH dt :

dI 8 mH

Substitute numerical values in equation (2) and evaluate dI 4 mH dt :

dI 4 mH

(b) After a long time has passed, the inductors will act as a short and the final current will be determined solely by the resistance in the circuit:

If =

=

dt

ε

=

dt

dt

dt

ε R

(1)

L8 mH

(2)

L4 mH

=

24 V = 3.00 kA/s 8 mH

=

24 V = 6.00 kA/s 4 mH

=

24 V = 1.60 A 15 Ω

*80 •• Picture the Problem If the current is initially zero in an LR circuit, its value at some later time t is given by I = I f 1 − e −t τ , where If = ε0/R and τ = L/R is the time constant

(

)

for the circuit. We can find the time at which the power dissipation in the resistor equals the rate at which magnetic energy is stored in the inductor by equating expressions for these rates and using the expression for I and its rate of change. Express the rate at which magnetic energy is stored in the inductor:

dU L d = dt dt

Express the rate at which power is dissipated in the resistor:

P = I 2R

Equate these expressions to obtain:

I 2 R = LI

Simplify to obtain:

I =τ

dI dt

[

1 2

]

LI 2 = LI

dI dt

dI dt (1)

684 Chapter 28

(

I = I f 1 − e −t τ

Express the current and its rate of change:

)

and

dI d ⎛ 1⎞ = I f (1 − e − t τ ) = − I f e − t τ ⎜ − ⎟ dt dt ⎝ τ⎠ I = f e−t τ

τ

(

)

⎛I ⎞ I f 1 − e −t τ = τ ⎜ f e −t τ ⎟ τ ⎝ ⎠

Substitute in equation (1) to obtain:

or

1 − e −t τ = e −t τ ⇒ 1 = 2e −t τ Solve for t:

t = −τ ln 12

Using τ = 333 µs from Example 2811, evaluate t to obtain:

t = −(333 µs )ln 12 = 231 µs

81 ••• Picture the Problem We can integrate dE dt = ε 0 I, where I = I f 1 − e −t τ , to find the

(

)

energy supplied by the battery, dEJ dt = I R to find the energy dissipated in the 2

resistor, and U L (τ ) = 12 L(I (τ )) to express the energy that has been stored in the 2

inductor when t = τ. (a) Express the rate at which energy is supplied by the battery:

dE = ε0I dt

Express the current in the circuit as a function of time:

I=

Substitute to obtain:

dE ε 02 = 1 − e −t τ dt R

Separate variables and integrate from t = 0 to t = τ to obtain:

ε 0 (1 − e−t τ ) R

(

)

2 τ ε 0 (1 − e −t τ )dt E=

R

∫ 0

=

ε [τ − (− τ e −1 + τ )]

=

ε 02 τ = ε 02 L

2 0

R

R e

R 2e

Magnetic Induction 685 Substitute numerical values and evaluate E:

2 ( 12 V ) (0.6 H ) = E= (3 Ω )2 e

(b) Express the rate at which energy is being dissipated in the resistor:

dEJ ⎡ε ⎤ = I 2 R = ⎢ 0 (1 − e −t τ )⎥ R dt ⎣R ⎦ 2

ε 02 (1 − 2e −t τ + e −2t τ )

= Separate variables and integrate from t = 0 to t = τ to obtain:

3.53 J

EJ =

R

ε 02 τ (1 − 2e −t τ + e −2t τ )dt R

ε

∫ 0

τ ⎞ ⎛ 2τ τ ⎜ − − 2⎟ R ⎝ e 2 2e ⎠ 2 ε 1 1 ⎞ 0L⎛2 = 2 ⎜ − − 2⎟ R ⎝ e 2 2e ⎠ =

Substitute numerical values and evaluate EJ:

EJ =

2 0

(12 V )2 (0.6 H ) ⎛ 2 − 1 − 1 ⎞ ⎜ ⎟ (3 Ω )2 ⎝ e 2 2e 2 ⎠

= 1.61J (c) Express the energy stored in the inductor when t = τ :

U L (τ ) = 12 L(I (τ ))

2

⎛ε ⎞ = L⎜ 0 1 − e −1 ⎟ ⎝R ⎠ 2 2 Lε 0 = 1 − e −1 2 2R

(

1 2

(

Substitute numerical values and evaluate EL:

U L (τ ) =

)

2

)

(0.6 H )(12 V )2 (1 − e −1 )2 2 2(3 Ω )

= 1.92 J Remarks: Note that, as we would expect from energy conservation, E = EJ + EL.

General Problems 82 • Picture the Problem We can apply the definition of magnetic flux to find the flux through the coil in its two orientations with respect to the magnetic field.

686 Chapter 28 (a) Using its definition, express the magnetic flux through the coil:

φm = NBA cos θ = NBπr 2 cos θ = (6)(0.5 T )π (0.03 m ) cos 0° 2

= 8.48 mWb (b) Proceed as in (a) with θ = 20°:

φm = NBA cos θ = NBπr 2 cos θ = (6)(0.5 T )π (0.03 m ) cos 20° 2

= 7.97 mWb 83 • Picture the Problem We can apply the definition of magnetic flux to find the flux through the coil in its two orientations with respect to the magnetic field and then use Faraday’s law to find the emfs induced in the coil. Using Faraday’s law, express the emf induced in the coil:

ε = − ∆φm = −

φm,f − φm,i ∆t

∆t because φm,f = 0

=

φm,i ∆t

(a) Using its definition, express the magnetic flux through the coil:

φm = NBA cosθ = NBπr 2 cosθ

Substitute to obtain:

ε = NBπr

2

cos θ

∆t

Substitute numerical values and evaluate ε:

ε = (6)(0.5 T )π (0.03 m )

2

cos 0°

1.2 s

= 7.07 mV (b) Proceed as in (a) with θ = 20°:

ε = (6)(0.5 T )π (0.03 m )

2

cos 20°

1.2 s

= 6.64 mV 84 • Picture the Problem We can apply Faraday’s and Ohm’s laws to obtain expressions for the induced emf that we can equate and solve for the rate at which the perpendicular magnetic field must change to induce a current of 4.0 A in the coil. Using Faraday’s law, relate the induced emf in the coil to the

ε

=

dφm dB = NA dt dt

Magnetic Induction 687 changing magnetic flux: Using Ohm’s law, relate the induced emf to the resistance of the coil and the current in it: Equate these expressions and solve for dB/dt:

ε

= IR

NA

dB = IR dt

and

dB IR IR = = dt NA Nπr 2 Substitute numerical values and evaluate dB/dt:

(4 A )(25 Ω ) = 199 T/s dB = dt (100)π (0.04 m )2

*85 •• Picture the Problem We can apply Faraday’s law and the definition of magnetic flux to derive an expression for the induced emf in the coil (potential difference between the slip rings). In part (b) we can solve this equation for ω under the given conditions. (a) Use Faraday’s law to express the induced emf: Using the definition of magnetic flux, relate the magnetic flux through the loop to its angular velocity: Substitute to obtain:

ε = − dφm dt

φm (t ) = NBA cos ωt

ε = − d [NBA cos ωt ]

dt = − NBabω (− sin ωt )

= NBabω sin ωt (b) Express the condition under which ε = εmax:

sin ωt = 1

688 Chapter 28 Solve for and evaluate ω under this condition:

ω= =

ε max NBab 110 V (1000)(2 T )(0.01 m )(0.02 m )

= 275 rad/s 86 •• Picture the Problem We can apply Faraday’s law and the definition of magnetic flux to derive an expression for the induced emf in the rotating coil gaussmeter.

ε = − dφm

Use Faraday’s law to express the induced emf:

dt

φm (t ) = NBA cos ωt

Using the definition of magnetic flux, relate the magnetic flux through the loop to its angular velocity:

ε = − d [NBA cos ωt ]

Substitute to obtain:

dt = − NBAω (− sin ωt )

= NBAω sin ωt = ε max sin ωt where

ε max = NBAω

Substitute numerical values and evaluate εmax:

ε max = (400)(0.45 T ) (1.4 × 10 −4 m 2 )⎛⎜⎜180 rev × 2π rad × 1 min ⎞⎟⎟ = ⎝

min

rev

60 s ⎠

0.475 V

The maximum induced emf occurs at the moment the plane of the coil is r parallel to the magnetic field B. At this instant, φ m is zero, but ε is a maximum. 87 •• Picture the Problem We can use the equality of the currents in the inductors connected in series and the additive nature of the total induced emf across the inductors to show that the inductances are additive.

Magnetic Induction 689 Relate the total induced emf to the effective inductance Leff and the rate at which the current is changing in the inductors:

ε = Leff dI

Because the inductors L1 and L2 are in series:

I1 = I 2 = I

dt

and

dI1 dI 2 dI = = dt dt dt Express the total induced emf:

ε = ε1 + ε 2 = L1 dI + L2 dI dt

= (L1 + L2 ) Substitute in equation (1) and simplify to obtain:

dt

dI dt

Leff = L1 + L2

*88 •• Picture the Problem We can use the common potential difference across the parallel combination of inductors and the fact that the current into the parallel combination is the sum of the currents through each inductor to find an expression of the equivalent inductance. Define Leff by:

Leff =

ε dI dt

or

dI 1 =ε dt Leff Relate the common potential difference across the inductors to their inductances and the rate at which the current is changing in each: Because the current divides at the parallel junction:

ε1 = L1 dI1 dt

(1)

(2)

and

ε 2 = L2 dI 2 dt

I = I1 + I 2 and

dI dI1 dI 2 = + dt dt dt

(3)

690 Chapter 28 Solve equations (2) and (3) for dI1/dt and dI2/dt and substitute to obtain: Express the relationship between an emf ε applied across the parallel combination of inductors and the emfs ε1 and ε2 across the individual inductors: Substitute to obtain:

Substitute in equation (1) and solve for 1/ Leff:

dI ε 1 ε 2 = + dt L1 L2

ε = ε1 = ε 2

⎛1 1 ⎞ dI ε ε = + = ε ⎜⎜ + ⎟⎟ dt L1 L2 ⎝ L1 L2 ⎠

1 1 1 = + Leff L1 L2

*89 •• Picture the Problem

As the magnet passes through the coil, it induces an emf because of the (a) changing flux through the coil. This allows the coil to " sense" when the magnet is passing through it.

(b)

One cannot use a cylinder made of conductive material because eddy currents induced in it by a falling magnet would slow the magnet.

As the magnet approaches the loop, the flux increases, resulting in the increasing voltage signal. When the magnet is passing the coil, the flux (c) goes from increasing to decreasing, so the induced emf becomes zero and then negative. The time at which the induced emf is zero is the time at which the magnet is at the center of the coil. (d) Each time represents a point when the distance has increased by 10 cm. The following graph of distance versus time was plotted using a spreadsheet program. The regression curve, obtained using Excel’s ″Add Trendline″ feature, is shown as a dashed line.

Magnetic Induction 691 1.4 1.2

y = 4.9257t 2 + 1.3931t + 0.0883

y (m)

1.0 0.8 0.6 0.4 0.2 0.0 0

0.1

0.2

0.3

0.4

t (s)

The coefficient of the second-degree term is

(

)

1 2

g. Consequently,

g = 2 4.9257 m/s 2 = 9.85 m/s 2 90 •• Picture the Problem The current equals the induced emf divided by the resistance. We can calculate the emf induced in the circuit as the coil moves by calculating the rate of change of the flux through the coil. The flux is proportional to the area of the coil in the magnetic field. We can find the direction of the current from Lenz’s law.

ε

(a) and (c) Express the magnitude of the induced current:

I=

Using Faraday’s law, express the magnitude of the induced emf:

ε

=

dφm dt

When the coil is moving to the right (or to the left), the flux does not change (until the coil leaves the region of magnetic field). Thus:

ε

=

dφm =0 dt

(b) and (d) Letting x represent the length of the side of the rectangular coil that is in the magnetic field, express the magnetic flux through the coil:

and

I=

(1)

R

ε R

= 0

φm = NBwx

692 Chapter 28 Compute the rate of change of the flux when the coil is moving up or down:

dφm dx = NBw dt dt = (80)(1.4 T )(0.25 m )(2 m/s ) = 56.0 V

Substitute in equation (1) to obtain:

I=

56 V = 2.33 A 24 Ω

When the coil is moving upward, the outward flux increases and the (b) induced current will be in the sense as to produce inward flux. I is clockwise. When the coil is moving downward, the outward flux decreases and the (d) induced current will be in the sense as to produce outward flux. I is counterclockwise. *91 •• Picture the Problem We can apply Faraday’s law and the definition of magnetic flux to derive an expression for the induced emf in the coil. We can then apply Ohm’s law to find the induced current as a function of time. Note that only half of the loop is in the magnetic field. Apply Ohm’s law to relate the induced current to the induced emf: Use Faraday’s law to express the induced emf: Using the definition of magnetic flux, relate the magnetic flux through the loop to its angular velocity: Substitute to obtain:

I (t ) =

ε (t )

(1)

R

ε (t ) = − dφm (t ) dt

φm (t ) = NBA cos ωt

ε (t ) = − d [NBA cos ωt ]

dt = − NBAω (− sin ωt ) = NBAω sin ωt

Substitute in equation (1) to obtain:

I (t ) =

NBAω sin ωt R

Magnetic Induction 693 Substitute numerical values and evaluate I(t):

I (t ) = =

(80)(1.4 T )(0.25 m )(0.15 m )(2 rad/s) sin (2 rad/s)t 24Ω

(0.350 A )sin (2 rad/s)t

92 •• Picture the Problem We can use the laws of Ohm and Faraday to express the charge dQ passing through the coil in time dt and integrate this expression to show that Q = N(φm1 − φm2)/R. Use Ohm’s law to express the induced current in terms of the induced emf: Apply Faraday’s law to obtain:

Integrate dQ from 0 to Q and dφm = φm1 to φm2 to obtain:

ε dQ ε = ⇒ dQ = dt dt R R dQ = −

N N dφm dt = − dφm R dt R

Q

N ∫0 dQ' = − R

φm2

∫ dφ

m

φm1

and

Q=

N (φm1 − φm2 ) R

93 •• Picture the Problem We can apply Faraday’s law to relate the induced electric field E to the rates at which the magnetic flux is changing at distances r < R and r > R from the axis of the solenoid. (a) Apply Faraday’s law to relate the induced electric field to the magnetic flux in the solenoid within a cylindrical region of radius r < R:

r r dφ E ∫C ⋅ d l = − dtm or

E (2πr ) = −

dφm dt

Express the field within the solenoid:

B = µ0 nI

Express the magnetic flux through an area for which r < R:

φm = BA = π r 2 µ0 nI

(1)

694 Chapter 28 Substitute in equation (1) to obtain:

Because I = I 0 sin ωt :

[

d π r 2 µ0 nI dt dI = −π r 2 µ0 n dt

E (2πr ) = −

E = − 12 rµ0 n

]

d [I 0 sin ωt ] dt

= − 12 rµ0 nI 0ω cos ωt (b) Proceed as in (a) with r > R to obtain:

Solve for E to obtain:

[

]

d π R 2 µ0 nI dt dI = −π R 2 µ0 n dt 2 = −π R µ0 nI 0ω cos ωt

E (2πr ) = −

E= −

µ0 nR 2 I 0ω 2r

cos ωt

94 ••• Picture the Problem The system exhibits cylindrical symmetry, so one can use Ampère’s law to determine B inside the inner cylinder, between the cylinders, and outside the outer cylinder. We can use um = B 2 2µ 0 and the expression for B from part (a) to express the magnetic energy density in the region between the cylinders. We can integrate this expression for um over the volume between the cylinders to find the total magnetic energy in a volume of length l . Finally, we can use our result in part (c) and U m = 12 LI 2 to find the self-inductance of the cylinders per unit length. (a) For r < r1 and for r > r2 the net enclosed current is zero; consequently, in these regions:

B= 0

For r1 < r < r2:

2πrB = µ0 I C ⇒ B =

(b) Express the magnetic energy density in the region between the cylinders:

um =

B2 2µ0

µ0 I 2πr

Magnetic Induction 695 2

Substitute for B and simplify to obtain:

⎛ µ0 I ⎞ ⎜ ⎟ µ0 I 2 2πr ⎠ ⎝ = um = 2µ0 8π 2 r 2

(c) Express the magnetic energy dUm in the cylindrical element of volume dV:

µ0 I 2 dU m = um dV = 2 2 (l 2πrdr ) 8π r 2 µ I l dr = 0 ⋅ 4π r

Integrate this expression from r = r1 to r = r2 to obtain:

µ0 I 2l Um = 4π

(d) Express the energy in the magnetic field in terms of L and I:

U m = 12 LI 2

Solve for L:

L=

From our result in (c):

U m µ0 r = l ln 2 2 I 4π r1

Substitute to obtain:

⎛µ r ⎞ µ r L = 2⎜⎜ 0 l ln 2 ⎟⎟ = 0 l ln 2 r1 ⎠ 2π r1 ⎝ 4π

Express the ratio L/ l :

µ0 r2 L = ln l 2π r1

r2

∫

r1

µ0 2 r dr = I l ln 2 r1 4π r

2U m I2

95 ••• Picture the Problem We can use its definition to express the magnetic flux through a rectangular element of area dA and then integrate from r = r1 to r = r2 to express the total flux through the region. Substituting in L = φm/I will yield the same result found in Part (d) of Problem 94.

φm

Use the definition of self-inductance to relate the magnetic flux through the region of interest to the current I:

L=

Consider a strip of unit length l and width dr at a distance r from the

dφm = BdA = Bldr = Bdr because l = 1.

I

(1)

696 Chapter 28 axis. The flux through this area is given by: Apply Ampere’s law to express the magnetic field at a distance r from the axis:

2πrB = µ0 I ⇒ B =

Substitute to obtain:

dφm =

Integrate from r = r1 to r = r2 to obtain:

µ I dr φm = 0 ∫ 2π r r

µ0 I 2πr

µ0 I dr 2π r r2

1

and

Substitute in equation (1) to obtain:

φm =

µ0 I r2 ln 2π r1

L=

µ0 r2 ln 2π r1

*96 ••• Picture the Problem We can use I = ε/R and ε = Bv l to find the current induced in the loop and Lenz’s law to determine its direction. We can apply the equation for the force on a current-carrying wire to find the net magnetic force acting on the loop and then sum the forces to find the net force on the loop. Separating the variables in the differential equation and integrating will lead us to an expression for v(t) and a second integration to an expression for y(t). We can solve the latter equation for y = 1.40 m to find the time it takes the loop to exit the magnetic field and our expression for v(t) to find its exit speed. Finally, we can use a constant-acceleration equation to find its exit speed in the absence of the magnetic field. (a) Relate the magnitude of the induced current to the induced emf and the resistance of the loop: Relate the induced emf to the motion of the loop: Substitute for ε to obtain:

I=

ε R

ε = Bvl I=

Bl v R

Magnetic Induction 697 As the loop falls, the flux into the page decreases. The direction of the induced current is such that its magnetic field opposes this decrease, i.e., clockwise. (b) Express the velocity-dependent force that acts on the loop in terms of the current in the loop:

Fv = BIl

Substitute for I to obtain:

B 2l 2 ⎛ Bl ⎞ Fv = B⎜ ⎟vl = v R ⎝ R⎠

r r r Apply dF = Id l × B to the horizontal portion of the loop that is in the magnetic field to conclude that the net magnetic force is upward. Note that the magnetic force on the left side of the loop is to the left and the magnetic force on the right side of the loop is to the right. (c) The net force acting on the loop is the difference between the downward gravitational force and the upward magnetic force: (d) Apply Newton’s 2nd law of motion to the loop to obtain its equation of motion:

Fnet = mg − Fv = mg −

mg −

B 2l 2 v R

B 2l 2 dv v=m R dt

or

dv B 2l 2 =g− v dt mR Factor g to obtain an alternate form of the equation of motion:

(e) Separate the variables to obtain:

⎛ ⎛ B 2l 2 ⎞ dv v = g ⎜⎜1 − v ⎟⎟ = g ⎜⎜1 − dt ⎝ mgR ⎠ ⎝ vt mgR where vt = 2 2 B l dv = dt B 2l 2 g− v mR or

dv = dt a − bv

⎞ ⎟⎟ ⎠

698 Chapter 28 B 2l 2 where a = g and b = mR v

t

Integrate v ′ from 0 to v and t ′ from 0 to t:

dv' 1 ⎛ a − bv ⎞ ∫0 a - bv' = ∫0 dt' ⇒ − b ln⎜⎝ a ⎟⎠ = t

Transform from logarithmic to exponential form and solve for v to obtain:

v(t ) =

Noting that vt =

(

a 1 − e − bt b

)

v(t ) = vt (1 − e −t τ )

a , we have: b

vt vt = . a g

where τ =

(

)

dy = vt 1 − e −t τ dt

(f) Write v as dy/dt and separate variables to obtain: Integrate y′ from 0 to y and t′ from 0 to t: y

t

0

0

(

)

−t' τ ∫ dy' = vt ∫ 1 − e dt'

and

]

y (t ) = vt [t − τ (1 − e −t τ )

(g) A spreadsheet program to generate the data for graphs of position y as a function of time t is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B4 B5

Formula/Content 0.05 0.2 0.4 0.3 $B$1*$B$7*$B$2/($B$3^2*$B$4^2)

B6 $B$5/$B$7 B7 9.81 A10 0.00 B10 $B$5*(A10−$B$6*(1−EXP(−A10/$B$6))) C10 0.5*$B$7*A10^2

Algebraic Form m R B L

vt τ g t y 1 2

gt 2

Magnetic Induction 699 A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

m= R= B= L= vt= tau= g= t 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

B 0.05 0.2 0.4 0.3 6.813 0.694 9.81 y 0.000 0.012 0.047 0.103 0.179 0.273 0.384 0.511 0.654 0.809 0.978 1.159 1.351 1.553 1.764 1.985 2.214 2.451 2.695 2.946 3.202

C kg ohms T m m/s s m/s^2 y (no B) 0.000 0.012 0.049 0.110 0.196 0.307 0.441 0.601 0.785 0.993 1.226 1.484 1.766 2.072 2.403 2.759 3.139 3.544 3.973 4.427 4.905

Examining the table, we see that y = 1.4 m when t ≈ 0.60 s. The following graph shows y as a function of t for B ≠ 0 (solid curve) and B = 0 (dashed curve).

700 Chapter 28 5 4

B>0

y (m)

B=0 3 2 1 0 0.0

0.2

0.4

0.6

0.8

1.0

t (s)

97 ••• Picture the Problem We can use the expression for the period of this spring-and-mass oscillator to find the spring constant κ. We can express the induced current in the loop by relating it to the induced emf and relating the induced emf to the velocity of the loop. Knowing that the loop is executing SHM, we can express its velocity as a sinusoidal function of time. We can use the expression for the magnetic force on a current-carrying wire in a magnetic field to express the damping force acting on the loop. (a) Express the period of the massspring system:

T = 2π

m

κ

κ=

4π 2 m T2

Substitute numerical values and evaluate κ :

κ=

4π 2 (0.5 kg ) = 30.8 N/m (0.8 s )2

(b) Express the current in the loop in terms of its resistance and the induced emf:

I=

Solve for κ to obtain:

Relate the induced emf in the wire to the motion of the wire:

ε R

ε = Bvl or, because l = w (where w is the width of the loop),

ε = Bvw

Express the position of the massspring system as a function of time:

y = y0 sin ωt

Magnetic Induction 701 Differentiate this expression with respect to time to express the velocity of the system: Substitute in our expression for I to obtain:

v=

dy = y0ω cos ωt dt

I=

By0ω w cos ωt R

(c) Express the damping force Fd acting on the loop:

Fd = BIw

Substitute for I and simplify to obtain:

Fd = − Bw

Because v = y0ωcosωt:

By0ωw cos ωt R B 2 w2 =− y0ω cos ωt R

Fd = −

B 2 w2 v = − βv R

where β =

(d) Choosing the static equilibrium position of the coil as the origin, apply

r r F ∑ = ma to the coil when

it is displaced slightly from this equilibrium position to obtain: Substituting for Fr and Fd yields the differential equation describing the motion of the coil:

B 2 w2 . R

− Fd − Fr = m

d2y dt 2

where Fr is the restoring force exerted by the plastic spring.

−β

dy d2y − κy = m 2 dt dt

or

d 2 y β dy κ + + y=0 dt 2 m dt m Note: compare this equation to Equation 14-35 on page 446 of Volume 1 of your textbook. For weak damping, the solution to this differential equation is:

y (t ) = ( y0 cos ωt ) e − ( β

2 m )t

Note: see Equation 14-36 on page 447 of your textbook.

702 Chapter 28 Differentiate y(t) with respect to time to obtain the velocity of the coil:

⎛ β ⎞

βy ⎛ ⎞ −⎜ ⎟ t v(t ) = −⎜ ωy0 sin ωt + 0 cos ωt ⎟ e ⎝ 2 m ⎠ 2m ⎝ ⎠

A spreadsheet program to generate the data for graphs of position y and velocity v as functions of time t is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B4 B5 B6 B7 B8 A11 B11 C11

Formula/Content Algebraic Form 0.05 y0 0.8 T 0.4 B 0.2 R 0.3 w 0.05 m 2*PI()/$B$2 ω $B$3^2*$B$5^2/$B$4 β 0.00 t $B$1*COS($B$7*A11)* y(t) EXP((-$B$8/(2*$B$6))*A11) v(t) −($B$1*$B$7*SIN($B$7*A11) +($B$8*$B$1/(2*$B$6))*COS($B$7*A11)) *EXP((−$B$8)/(2*$B$6))*A11) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

A y_0= T= B= R= w= m= omega= beta=

B 0.05 0.8 0.4 0.2 0.3 0.05 7.85 0.072

C m s T ohms m kg s^-1 kg/s

t 0.00 0.01 0.02 0.03 0.04 0.05

y 0.050 0.049 0.049 0.048 0.046 0.045

v −0.036 −0.066 −0.096 −0.124 −0.151 −0.177

235 236 237 238 239

2.24 2.25 2.26 2.27 2.28

0.003 0.004 0.004 0.005 0.006

0.072 0.069 0.066 0.062 0.057

Magnetic Induction 703

The graph of y(t) follows: 0.06 0.04

y (m)

0.02 0.00 -0.02 -0.04 -0.06 0.0

0.5

1.0

1.5

2.0

2.5

1.5

2.0

2.5

t (s)

The graph of v(t) follows: 0.3 0.2

v (m/s)

0.1 0.0 -0.1 -0.2 -0.3 -0.4 0.0

0.5

1.0 t (s)

98 ••• Picture the Problem If the coil is twisted through an angle θ, a restoring torque equal to κθ acts on it return it its equilibrium position. However, if it rotates back with angular speed ω = dθ dt, there will be an emf induced in the coil. The direction of the current resulting from this induced emf will be such that its magnetic field will oppose the change in flux resulting from the rotation of the coil. The net effect is that the motion of the coil is damped. We can apply Newton’s 2nd law to relate the net restoring torque to the moment of inertia of the coil and its angular acceleration and use the laws of Faraday and Ohm to find the emf and current induced in the coil. Apply

∑τ = Iα to the rotating

coil to obtain:

τ restoring − τ retarding = I

d 2θ dt 2

704 Chapter 28 The magnitude of the retarding (damping) torque is given by:

Substitute for τrestoring and τretarding to obtain:

τ retarding = NiBA cosθ where i is the current induced in the coil whose cross-sectional area is A.

− κθ − NiBA cosθ = I

d 2θ dt 2

(1)

Apply Faraday’s law to express the emf induced in the coil:

ε = − d (NBA sin θ ) = −(NBA cosθ ) dθ

From Ohm’s law, the magnitude of the induced current i in the coil is:

i=

Substitute for the induced current i in equation (1) to obtain:

N 2 B 2 A2 cos 2 θ dθ d 2θ − κθ − =I 2 R dt dt

For small displacements from equilibrium, cosθ ≈ 1 and:

− κθ −

Rearrange terms to obtain the differential equation of motion of the coil:

d 2θ N 2 B 2 A2 dθ κ + + θ ≈0 dt 2 RI dt I

Let β =

N 2 B 2 A2 κ and ω = to RI I

dt

ε R

=

dt

NBA cos θ dθ R dt

N 2 B 2 A2 dθ d 2θ ≈I 2 R dt dt

d 2θ dθ +β + ω 2θ ≈ 0 2 dt dt

obtain: The solution to this second-order, homogeneous, linear differential equation with constant coefficients is:

θ (t ) = θ 0 e −( β

2 )t

cos ωt

Chapter 29 Alternating-Current Circuits Conceptual Problems *1 • Determine the Concept Because the rms current through the resistor is given by

I rms = ε rms R and both εrms and R are independent of frequency, (b) is correct.

2 • Picture the Problem We can use the relationship between V and Vpeak to decide the effect of doubling the rms voltage on the peak voltage.

Vmax 2

Express the initial rms voltage in terms of the peak voltage:

Vrms =

Express the doubled rms voltage in terms of the new peak voltage V' max :

2Vrms =

Divide the second of these equations by the first and simplify to obtain:

V'max V' 2Vrms = 2 or 2 = max Vmax Vmax Vrms 2

Solve for V' max :

V'max = 2Vmax and (a ) is correct.

V'max 2

3 • Determine the Concept The inductance of an inductor is determined by the details of its construction and is independent of the frequency of the circuit. The inductive reactance, on the other hand, is frequency dependent. (b) is correct. 4 • Determine the Concept The inductive reactance of an inductor varies with the frequency according to X L = ωL. Hence, doubling ω will double XL. (a ) is correct.

705

706 Chapter 29 *5 • Determine the Concept The capacitive reactance of an capacitor varies with the frequency according to X C = 1 ωC . Hence, doubling ω will halve XC. (c) is correct. 6 • Determine the Concept Yes to both questions. While the current in the inductor is increasing, the inductor absorbs power from the generator. When the current in the inductor reverses direction, the inductor supplies power to the generator. 7 • Determine the Concept Yes to both questions. While charge is accumulating on the capacitor, the capacitor absorbs power from the generator. When the capacitor is discharging, it supplies power to the generator. 8 • Picture the Problem We can use the definitions of the capacitive reactance and inductive reactance to find the SI units of LC. Use its definition to express the inductive reactance:

X L = 2πfL

XL 2πf

Solve for L:

L=

Use its definition to express the capacitive reactance:

XC =

Solve for C:

C=

Express the product of L and C:

LC =

1 2πfC

1 2πfX C XL 1 XL = 2 2 2πf 2πfX C 4π f X C

Because the units of XL and XC cancel, the units of LC are those of 1/f 2 or s2.

(a ) is correct. *9 •• Determine the Concept To make an LC circuit with a small resonance frequency requires a large inductance and large capacitance. Neither is easy to construct.

Alternating-Current Circuits

707

10 • (a) True. The Q factor and the width of the resonance curve at half power are related according to Q = ω0 ∆ω ; i.e., they are inversely proportional to each other. (b) True. The impedance of an RLC circuit is given by Z =

R 2 + ( X L − X C ) . At 2

resonance XL = XC and so Z = R. (c) True. The phase angle δ is related to XL and XC according to δ = tan −1

X L − XC . At R

resonance XL = XC and so δ = 0. 11 • Determine the Concept Yes. The power factor is defined to be cos δ = R Z and, because Z is frequency dependent, so is cosδ. *12 • Determine the Concept Yes; the bandwidth must be wide enough to accommodate the modulation frequency. 13 • Determine the Concept Because the power factor is defined to be cos δ = R Z , if R = 0, then the power factor is zero. 14 • Determine the Concept A transformer is a device used to raise or lower the voltage in a circuit without an appreciable loss of power. (c) is correct. 15 • True. If energy is to be conserved, the product of the current and voltage must be constant. 16 •• Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. Assuming no loss of power in the transformer, we can equate the power in the primary circuit to the power in the secondary circuit and solve for the current in the primary windings. Assuming no loss of power in the transformer:

P1 = P2

708 Chapter 29 Substitute for P1 and P2 to obtain:

I1V1 = I 2V2

Solve for I1:

I1 = I 2 and

V2 I 2V2 P2 = = V1 V1 V1

(b ) is correct

17 • (a) False. The effective (rms) value of the current is not zero. (b) True. The reactance of a capacitor goes to zero as f approaches very high frequencies.

Estimation and Approximation *18 •• Picture the Problem We can find the resistance and inductive reactance of the plant’s total load from the impedance of the load and the phase constant. The current in the power lines can be found from the total impedance of the load the potential difference across it and the rms voltage at the substation by applying Kirchhoff’s loop rule to the substation-transmission wires-load circuit. The power lost in transmission can be found 2 from Ptrans = I rms Rtrans . We can find the cost savings by finding the difference in the power lost in transmission when the phase angle is reduced to 18°. Finally, we can find the capacitance that is required to reduce the phase angle to 18° by first finding the capacitive reactance using the definition of tanδ and then applying the definition of capacitive reactance to find C.

(a) Relate the resistance and inductive reactance of the plant’s total load to Z and δ: Express Z in terms of the current I in the power lines and voltage εrms at the plant: Express the power delivered to the plant in terms of εrms, Irms, and δ and

R = Z cos δ and

X L = Z sin δ

Z=

ε rms I

Pav = ε rms I rms cos δ and

Alternating-Current Circuits solve for Irms:

I rms =

Substitute to obtain:

Z=

Substitute numerical values and evaluate Z:

Substitute numerical values and evaluate R and XL:

Z=

Pav ε rms cos δ

(1)

2 ε rms cos δ

Pav

(40 kV )2 cos 25° = 630 Ω 2.3 MW

R = (630 Ω ) cos 25° = 571 Ω and

X L = (630 Ω ) sin 25° = 266 Ω

(b) Use equation (1) to find the current in the power lines: Apply Kirchhoff’s loop rule to the circuit:

I rms =

2.3 MW = 63.4 A (40 kV ) cos 25°

ε sub − I rms Rtrans − IZ tot = 0

Solve for εsub:

ε sub = I rms (Rtrans + Z tot )

Evaluate Ztot:

Z tot = R 2 + X L2 =

Substitute numerical values and evaluate εsub:

(c) The power lost in transmission is:

(d) Express the cost savings ∆C in terms of the difference in energy consumption (P25° − P18°)∆t and the per-unit cost u of the energy: Express the power list in

(571Ω )2 + (266 Ω )2

= 630 Ω

ε sub = (63.4 A )(5.2 Ω + 630 Ω ) = 40.3 kV 2 Ptrans = I rms Rtrans = (63.4 A ) (5.2 Ω ) 2

= 20.9 kW

∆C = (P25° − P18° )∆tu

P18° = I182 ° Rtrans

709

710 Chapter 29 transmission when δ = 18°:

2.3 MW = 60.5 A (40 kV ) cos18°

Find the current in the transmission lines when δ = 18°:

I18° =

Evaluate P18° :

P18° = (60.5 A ) (5.2 Ω ) = 19.0 kW 2

Substitute numerical values and evaluate ∆C:

∆C = (20.9 kW − 19.0 kW )(16 h/d )(30 d/month )($0.07 / kW ⋅ h ) = $63.84 Relate the new phase angle δ to the inductive reactance XL, the reactance due to the added capacitance XC, and the resistance of the load R:

tan δ =

XL − XC R

Solve for and evaluate XC:

X C = X L − R tan δ

Substitute numerical values and evaluate C:

C=

= 266 Ω − (571 Ω ) tan 18° = 80.5 Ω

(

2π 60 s

1 −1

)(80.5 Ω) =

33.0 µF

Alternating Current Generators 19 • Picture the Problem We can use the relationship

ε max = 2πNBAf between the

maximum emf induced in the coil and its frequency to find f when εmax is given and εmax when f is given . (a) Relate the induced emf to the angular frequency of the coil:

Solve for f:

Substitute numerical values and evaluate f:

ε = ε max cos ωt where

ε max = NBAω = 2πNBAf f =

f =

ε max

2πNBA 10 V 2π (200)(0.5 T ) 4 × 10 −4 m 2

= 39.8 Hz

(

)

Alternating-Current Circuits (b) From (a) we have: Substitute numerical values and evaluate εmax:

711

ε max = NBAω = 2πNBAf ε max = 2π (200)(0.5 T )(4 ×10−4 m 2 )(60 s −1 ) = 15.1 V

20 • Picture the Problem We can use the relationship

ε max = 2πNBAf between the

maximum emf induced in the coil and the magnetic field in which it is rotating to find B required to generate a given emf at a given frequency. Relate the induced emf to the magnetic field in which the coil is rotating:

ε max = NBAω = 2πNBAf

Solve for B:

Substitute numerical values and evaluate B:

B=

B=

ε max

2πNfA 10 V 2π (200 ) 60 s −1 4 × 10 − 4 m 2

(

)(

)

= 0.332 T *21 • Picture the Problem We can use the relationship

ε max = 2πNBAf to relate the

maximum emf generated to the area of the coil, the number of turns of the coil, the magnetic field in which the coil is rotating, and the frequency at which it rotates. (a) Relate the induced emf to the magnetic field in which the coil is rotating:

ε max = NBAω = 2πNBAf

(1)

Substitute numerical values and evaluate εmax:

ε max = 2π (300)(0.4 T ) (2 ×10−2 m )(1.5 ×10−2 m )(60 s −1 ) = (b) Solve equation (1) for f:

Substitute numerical values and evaluate f:

f =

ε max

2πNBA

13.6 V

712 Chapter 29 f =

110 V = 486 Hz 2π (300)(0.4 T ) 2 × 10 − 2 m 1.5 × 10 − 2 m

(

)(

22 • Picture the Problem We can use the relationship

)

ε max = 2πNBAf to relate the

maximum emf generated to the area of the coil, the number of turns of the coil, the magnetic field in which the coil is rotating, and the frequency at which it rotates. Relate the induced emf to the magnetic field in which the coil is rotating: Solve for B:

ε max = NBAω = 2πNBAf B=

ε max

2πNfA

Substitute numerical values and evaluate B:

B=

(

2π (300 ) 60 s

−1

24 V = 0.707 T 2 × 10 − 2 m 1.5 ×10 − 2 m

)(

)(

)

Alternating Current in a Resistor *23 • Picture the Problem We can use Pav = ε rms I rms to find Irms, I max =

ε

and Pmax = I max

max to

2I rms to find Imax,

find Pmax.

(a) Relate the average power delivered by the source to the rms voltage across the bulb and the rms current through it:

Pav = ε rms I rms

Pav

100 W = 0.833 A 120 V

Solve for and evaluate Irms:

I rms =

(b) Express Imax in terms of Irms:

I max = 2I rms

Substitute for Irms and evaluate Imax:

I max = 2 (0.833 A ) = 1.18 A

(c) Express the maximum power in terms of the maximum voltage and maximum current:

ε rms

=

Pmax = I maxε max

Alternating-Current Circuits Substitute numerical values and evaluate Pmax: 24 • Picture the Problem We can I max =

Pmax = (1.18 A ) 2 (120 V ) = 200 W

2I rms to find the largest current the breaker can

carry and Pav = I rmsVrms to find the average power supplied by this circuit. (a) Express Imax in terms of Irms:

(b) Relate the average power to the rms current and voltage:

I max = 2 I rms = 2 (15 A ) = 21.2 A

Pav = I rmsVrms = (15 A )(120 V ) = 1.80 kW

Alternating Current in Inductors and Capacitors 25 • Picture the Problem We can use X L = ωL to find the reactance of the inductor at any frequency. Express the inductive reactance as a function of f:

X L = ωL = 2πfL

(

)

(a) At f = 60 Hz:

X L = 2π 60 s −1 (1 mH ) = 0.377 Ω

(b) At f = 600 Hz:

X L = 2π 600 s −1 (1 mH ) = 3.77 Ω

(c) At f = 6 kHz:

X L = 2π 6000 s −1 (1 mH ) = 37.7 Ω

26 • (a) Relate the reactance of the inductor to its inductance:

X L = ωL = 2πfL

Solve for and evaluate L:

L=

(b) At 160 Hz:

X L = 2π 160 s −1 (0.199 H ) = 200 Ω

(

)

(

)

100 Ω XL = = 0.199 H 2πf 2π 80 s −1

(

(

)

)

713

714 Chapter 29 27 • Picture the Problem We can equate the reactances of the capacitor and the inductor and then solve for the frequency. Express the reactance of the inductor:

X L = ωL = 2πfL

1 1 = ωC 2πfC

Express the reactance of the capacitor:

XC =

Equate these reactances to obtain:

2πfL =

Solve for f to obtain:

f =

1 2π

1 LC

Substitute numerical values and evaluate f:

f =

1 2π

1 = 1.59 kHz (10 µF)(1mH )

1 2πfC

28 • Picture the Problem We can use X C = 1 ωC to find the reactance of the capacitor at any frequency. Express the capacitive reactance as a function of f:

XC =

1 1 = ωC 2πfC

(a) At f = 60 Hz:

XC =

1 = 2.65 MΩ 2π 60 s −1 (1 nF)

(b) At f = 6 kHz:

XC =

1 = 26.5 kΩ 2π 6000 s −1 (1 nF)

(c) At f = 6 MHz:

XC =

1 = 26.5 Ω 2π 6 × 106 s −1 (1 nF)

(

(

(

)

)

)

*29 • Picture the Problem We can use Imax = εmax/XC and XC = 1/ωC to express Imax as a function of εmax, f, and C. Once we’ve evaluate Imax, we can use Irms = Imax/ 2 to find Irms.

Alternating-Current Circuits Express Imax in terms of εmax and XC:

I max =

ε max XC 1 1 = ωC 2πfC

Express the capacitive reactance:

XC =

Substitute to obtain:

I max = 2πfCε max

(a) Substitute numerical values and evaluate Imax: (b) Express Irms in terms of Imax:

715

(

)

I max = 2π 20 s −1 (20 µF)(10 V ) = 25.1 mA I rms =

I max 25.1 mA = = 17.8 mA 2 2

30 • Picture the Problem We can use X C = 1 ωC = 1 2πfC to relate the reactance of the capacitor to the frequency.

1 1 = ωC 2πfC

Using its definition, express the reactance of a capacitor:

XC =

Solve for f to obtain:

f =

1 2πCX C

(a) Find f when XC = 1 Ω:

f =

1 = 15.9 kHz 2π (10 µF)(1Ω )

(b) Find f when XC = 100 Ω:

f =

1 = 159 Hz 2π (10 µF)(100 Ω )

(c) Find f when XC = 0.01 Ω:

f =

1 = 1.59 MHz 2π (10 µF)(0.01Ω )

31 •• Picture the Problem We can use the trigonometric identity cos θ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ ) to find the sum of the phasors V1 and V2 and then use this sum to express I as a function of time. In (b) we’ll use a phasor diagram to obtain the same result and in (c) we’ll use the phasor diagram appropriate to the given voltages to express the current as a function of time.

716 Chapter 29 (a) Express the current in the resistor:

I=

V V1 + V2 = R R

Use the trigonometric identity cos θ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ ) to find V1 + V2:

V1 + V2 = (5.0 V )[cos(ωt − α ) + cos(ωt + α )] = (5 V )[2 cos 12 (2ωt )cos 12 (− 2α )] = (10 V ) cos

π 6

cos ωt = (8.66 V ) cos ωt

Substitute to obtain:

(b) Express the magnitude of the current in R:

I=

(8.66 V )cos ωt = (0.346 A )cos ωt

I =

V R

25 Ω

The phasor diagram for the voltages is shown to the right.

Use vector addition to find V :

V = 2 V1 cos 30° = 2(5 V ) cos 30° = 8.66 V

Substitute to obtain:

I = and

I=

8.66 V 25 Ω

= 0.346 A

(0.346 A )cos ωt

(c) The phasor diagram is shown to the right. Note that the phase angle between V1 and V2 is now 90°.

Use the Pythagorean theorem to find V :

V = V1 + V2 2

= 8.60 V

2

=

(5 V )2 + (7 V )2

Alternating-Current Circuits Express I as a function of t:

I=

717

V cos(ωt + δ ) R

where

δ = 45° − (90° − α ) = α − 45° ⎛7V ⎞ ⎟⎟ − 45° = 9.46° = 0.165 rad = tan −1 ⎜⎜ 5 V ⎝ ⎠

Substitute numerical values and evaluate I:

I= =

8.60 V cos(ωt + 0.165 rad ) 25 Ω

(0.344 A )cos(ωt + 0.165 rad )

LC and RLC Circuits without a Generator *32 • Picture the Problem We can use XL = ωL and XC = 1/ωC to show the 1

LC has the

–1

unit s . Alternatively, we can use the dimensions of C and L to establish this result. Substitute the units for L and C in the expression 1

LC to obtain:

1 = H⋅F

1

(Ω ⋅ s )⎛⎜ s ⎞⎟

=

1 s

2

= s −1

⎝Ω⎠

Alternatively, use the defining equation (C = Q/V) for capacitance to obtain the dimension of C: Solve the defining equation ( V = L dI dt ) for inductance to obtain the dimension of L: Express the dimension of 1

[C ] = [Q] [V ] [V ]

2 V ] [V ][T ] [ [L] = = [Q] = [Q] ⎡ dI ⎤ ⎢⎣ dt ⎥⎦ [T ] 2

LC : ⎡ 1 ⎤ ⎡ ⎢ LC ⎥ = ⎢ ⎣ ⎦ ⎣⎢ ⎡ =⎢ ⎢⎣

⎤ ⎡ ⎥ ⎢ 1 ⎤ ⎢ 1 ⎥ ⎥=⎢ 2 [L][C ] ⎥⎦ ⎢ [V ][T ] [Q] ⎥⎥ ⎢⎣ [Q] [V ] ⎥⎦ 1 ⎤ 1 ⎥= 2 [T ] ⎥⎦ [T ]

Because the SI unit of time is the second, we’ve shown that 1

LC has units of s −1.

718 Chapter 29 33 • Picture the Problem We can use T = 2π/ω and ω = 1

LC to relate T (and hence f) to

L and C.

2π

(a) Express the period of oscillation of the LC circuit:

T=

For an LC circuit:

ω=

Substitute to obtain:

T = 2π LC

Substitute numerical values and evaluate T:

T = 2π

(b) Solve equation (1) for L to obtain:

L=

Substitute numerical values and evaluate L:

L=

ω 1 LC (1)

(2 mH )(20 µF) =

1.26 ms

T2 1 = 2 2 2 4π C 4π f C

(

4π 60 s 2

1

) (80 µF)

−1 2

= 88.0 mH

34 •• Picture the Problem We can use the expression f 0 = 1 2π LC for the resonance frequency of an LC circuit to show that each circuit oscillates with the same frequency. In (b) we can use I max = ωQ0 , where Q0 is the charge of the capacitor at time zero, and the definition of capacitance Q0 = CV to express Imax in terms of ω, C and V.

1

Express the resonance frequency for an LC circuit:

f0 =

(a) Express the product of L and C for each circuit:

Circuit 1 : L1C1 , Circuit 2 : L2C2 = (2 L1 )( 12 C1 ) = L1C1 ,

2π LC

and

Circuit 3 : L3C3 = ( 12 L1 )(2C1 ) = L1C1

Because L1C1 = L2C2 = L3C3 , the resonance frequencies of the three circuits are the same. (b) Express Imax in terms of the

I max = ωQ0

Alternating-Current Circuits charge stored in the capacitor: Express Q0 in terms of the capacitance of the capacitor and the potential difference across the capacitor: Substitute to obtain:

Q0 = CV

I max = ωCV or, for ω and V constant,

I max ∝ C

The circuit with C = C3 has the greatest I max . 35 •• Picture the Problem We can use U = 12 CV 2 to find the energy stored in the electric field of the capacitor, ω0 = 2πf 0 = 1

LC to find f0, and I max = ωQ0 and Q0 = CV to

find Imax. (a) Express the energy stored in the system as a function of C and V:

U = 12 CV 2

Substitute numerical values and evaluate U:

U=

(b) Express the resonance frequency of the circuit in terms of L and C:

ω0 = 2πf 0 =

Solve for f0:

f0 =

Substitute numerical values and evaluate f0: (c) Express Imax in terms of the charge stored in the capacitor: Express Q0 in terms of the capacitance of the capacitor and the potential difference across the capacitor:

f0 =

1 2

(5 µF)(30 V )2 =

2.25 mJ

1 LC

1 2π LC 1 = 712 Hz 2π (10 mH )(5 µF)

I max = ωQ0 Q0 = CV

719

720 Chapter 29 Substitute to obtain: Substitute numerical values and evaluate Imax:

I max = ωCV

(

)

I max = 2π 712 s −1 (5 µF)(30 V )

= 0.671 A

36 • Picture the Problem We can use its definition to find the power factor of the circuit and 2 Irms = ε/Z to find the rms current in the circuit. In (c) we can use Pav = I rms R to find the average power supplied to the circuit.

R Z

(a) Express the power factor of the circuit:

cos δ =

Express Z for the circuit:

Z = R 2 + X L2

Substitute to obtain:

cos δ =

Substitute numerical values and evaluate cosδ:

cos δ =

R R 2 + X L2

=

R R 2 + (2πfL )

2

100 Ω

(100 Ω )2 + [2π (60 s −1 )(0.4 H )] 2

= 0.553 (b) Express the rms current in terms of the rms voltage and the impedance of the circuit:

I rms =

Substitute numerical values and evaluate Irms:

I rms =

ε Z

=

ε R + (2πfL ) 2

2

120 V

(100 Ω )2 + [2π (60 s −1 )(0.4 H )] 2

= 0.663 A (c) Express the average power supplied to the circuit in terms of the rms current and the resistance of the inductor: Substitute numerical values and evaluate Pav:

2 Pav = I rms R

Pav = (0.663 A ) (100 Ω ) = 44.0 W 2

Alternating-Current Circuits

721

*37 •• Picture the Problem Let Q represent the instantaneous charge on the capacitor and apply Kirchhoff’s loop rule to obtain the differential equation for the circuit. We can then solve this equation to obtain an expression for the charge on the capacitor as a function of time and, by differentiating this expression with respect to time, an expression for the current as a function of time. We’ll use a spreadsheet program to plot the graphs. Apply Kirchhoff’s loop rule to a clockwise loop just after the switch is closed:

Q dI +L =0 C dt

Because I = dQ dt :

L

d 2Q Q d 2Q 1 + = 0 or + Q=0 dt 2 LC dt 2 C

Q(t ) = Q0 cos(ωt − δ )

The solution to this equation is:

where ω =

1 LC

Because Q(0) = Q0, δ = 0 and:

Q(t ) = Q0 cos ωt

The current in the circuit is the derivative of Q with respect to t:

I=

dQ d = [Q0 cos ωt ] = −ωQ0 sin ωt dt dt

(a) A spreadsheet program was used to plot the following graph showing both the charge on the capacitor and the current in the circuit as functions of time. L, C, and Q0 were all arbitrarily set equal to one to obtain these graphs. Note that the current leads the charge by one-fourth of a cycle or 90°. 1 Charge Current

0

-1 0

2

4

6

8

10

t (s)

(b) The equation for the current is:

I = −ωQ0 sin ωt

The sine and cosine functions are related through the identity:

π⎞ ⎛ − sin θ = cos⎜θ + ⎟ 2⎠ ⎝

(1)

722 Chapter 29 Use this identity to rewrite equation (1):

π⎞ ⎛ I = −ωQ0 sin ωt = ωQ0 cos⎜ ωt + ⎟ 2⎠ ⎝ showing that the current leads the charge by 90°.

RL Circuits with a Generator 38 •• Picture the Problem We can express the ratio of VR to VL and solve this expression for the resistance R of the circuit. In (b) we can use the fact that, in an LR circuit, VL leads VR by 90° to find the ac input voltage. (a) Express the potential differences across R and L in terms of the common current through these components:

VL = IX L = IωL and

VR = IR

Divide the second of these equations by the first to obtain:

VR IR R = = VL IωL ωL

Solve for R:

⎛V ⎞ R = ⎜⎜ R ⎟⎟ωL ⎝ VL ⎠

Substitute numerical values and evaluate R:

⎛ 30 V ⎞ ⎟⎟2π 60 s −1 (1.4 H ) = 396 Ω R = ⎜⎜ ⎝ 40 V ⎠

(b) Because VR leads VL by 90° in an LR circuit:

V = VR2 + VL2

Substitute numerical values and evaluate V:

V=

(

)

(30 V )2 + (40 V )2

= 50.0 V

39 •• Picture the Problem We can solve the expression for the impedance in an LR circuit for the inductive reactance and then use the definition of XL to find L. Express the impedance of the coil in terms of its resistance and inductive reactance:

Z = R 2 + X L2

Solve for XL to obtain:

X L = Z 2 − R2

Alternating-Current Circuits Express XL in terms of L:

X L = 2πfL

Equate these two expressions to obtain:

2πfL = Z 2 − R 2

Solve for L:

L=

Z 2 − R2 2πf

L=

(200 Ω )2 − (80 Ω )2 2π (1 kHz )

Substitute numerical values and evaluate L:

723

= 29.2 mH

40 •• Picture the Problem We can express the two output voltage signals as the product of the current from each source and R = 1 kΩ. We can find the currents due to each source using the given voltage signals and the definition of the impedance for each of them. (a) Express the voltage signals observed at the output side of the transmission line in terms of the potential difference across the resistor: Express I1 and I2:

V1, out = I1 R and

V2, out = I 2 R

(10 V )cos100t (103 Ω)2 + [(100 s-1 )(1H )]2 = (9.95 mA )cos100t

I1 =

V1 = Z1

and

I2 =

V2 = Z2

(10 V )cos104 t

(10 Ω) + [(10 3

2

4

)

= (0.995 mA ) cos104 t Substitute for I1 and I2 to obtain:

(

)

V1, out = 103 Ω (9.95 mA ) cos100t

(9.95 V )cos100t

= and

(

)

V2, out = 103 Ω (0.995 mA ) cos10 4 t =

(0.995 V )cos104 t

]

s −1 (1H )

2

724 Chapter 29 (b) Express the ratio of V1,out to V2,out:

V1, out 9.95 V = = 10.0 V2, out 0.995 V

41 •• Picture the Problem The average power supplied to coil is related to the power factor by 2 Pav = ε rms I rms cos δ . In (b) we can use Pav = I rms R to find R. Because the inductance L is related to the resistance R and the phase angle δ according to X L = ωL = R tan δ , we can use this relationship to find the resistance of the coil. Finally, we can decide whether the current leads or lags the voltage by noting whether XL is less than or greater than R. (a) Express the average power supplied to the coil in terms of the power factor of the circuit:

Pav = ε rms I rms cos δ

Solve for the power factor:

cos δ =

Pav ε rms I rms

Substitute numerical values and evaluate cosδ:

cos δ =

60 W = 0.333 (120 V )(1.5 A )

(b) Express the power supplied by the source in terms of the resistance of the coil: Solve for and evaluate R:

(c) Relate the inductive reactance to the resistance and phase angle: Solve for L:

2 Pav = I rms R

R=

Pav 60 W = = 26.7 Ω 2 I rms (1.5 A )2

X L = ωL = R tan δ

L=

R tan δ

=

ω

(

R tan cos −1 0.333 2πf

(26.7 Ω ) tan 70.5° =

)

Substitute numerical values and evaluate L:

L=

(d) Evaluate XL:

X L = (26.7 Ω ) tan 70.5° = 75.4 Ω

Because XL > R, the circuit is inductive and:

(

2π 60 s −1

)

I lags ε by 70.5° .

0.200 H

Alternating-Current Circuits 42

725

••

Picture the Problem We can use I max = ε max

R 2 + (ωL ) and 2

VL ,max = I max X L = ωLI max to find the maximum current in the circuit and the maximum voltage across the inductor. Once we’ve found VL,max we can find VL,rms using 2 VL ,rms = VL ,max 2 . We can use Pav = 12 I max R to find the average power dissipation, and 2 to find the maximum energy stored in the magnetic field of the inductor. U L ,max = 12 LI max

The average energy stored in the magnetic field of the inductor can be found from U L ,av = Pav dt .

∫

Express the maximum current in the circuit:

I max =

Substitute numerical values and evaluate Imax:

I max =

ε max Z

ε max

=

R 2 + (ωL )

2

345 V

(40 Ω )2 + [(150π s −1 )(36 mH )]2

= 7.94 A Relate the maximum voltage across the inductor to the current flowing through it: Substitute numerical values and evaluate VL,max:

VL,rms is related to VL,max according to: Relate the average power dissipation to Imax and R: Substitute numerical values and evaluate Pav: The maximum energy stored in the magnetic field of the inductor is:

The definition of UL,av is:

VL ,max = I max X L = ωLI max

(

)

VL ,max = 150π s −1 (36 mH )(7.94 A ) = 135 V

VL ,rms =

VL ,max 2

=

135 V = 95.5 V 2

2 Pav = 12 I max R

Pav =

(7.94 A )2 (40 Ω ) =

1 2

2 U L ,max = 12 LI max =

= 1.13 J T

U L ,av =

1 U (t )dt T ∫0

1 2

1.26 kW

(36 mH )(7.94 A )2

726 Chapter 29 U(t) is given by:

Substitute for U(t) to obtain:

U (t ) =

U L ,av

1 2 L[I (t )] 2

L = 2T

T

∫ [I (t )] dt 2

0

Evaluating the integral yields:

U L ,av =

L ⎡1 2 ⎤ 1 2 I max ⎥T = LI max ⎢ 2T ⎣ 2 4 ⎦

Substitute numerical values and evaluate UL,av:

U L ,av =

1 (36 mH )(7.94 A )2 = 0.567 J 4

43 •• Picture the Problem We can use the definition of the power factor to find the relationship between XL and R when f = 60 Hz and then use the definition of XL to relate the inductive reactance at 240 Hz to the inductive reactance at 60 Hz. We can then use the definition of the power factor to determine its value at 240 Hz.

R = Z

R

Using the definition of the power factor, relate R and XL:

cos δ =

Square both sides of the equation to obtain:

cos 2 δ =

Solve for X L2 (60 Hz ) :

⎞ ⎛ 1 X L2 (60 Hz ) = R 2 ⎜ − 1⎟ 2 ⎠ ⎝ cos δ ⎛ ⎞ 1 − 1⎟⎟ = 13 R 2 X L2 (60 Hz ) = R 2 ⎜⎜ 2 ⎝ (0.866 ) ⎠

Substitute for cosδ and simplify to obtain: Use the definition of XL to obtain:

R + X L2 2

R2 R 2 + X L2

X L2 ( f ) = 4πf 2 L2 and

X L2 ( f' ) = 4πf' 2 L2

Divide the second of these equations by the first to obtain:

X L2 ( f' ) 4πf' 2 L2 f' 2 = = 2 X L2 ( f ) 4πf 2 L2 f X L2 ( f' ) 4πf' 2 L2 f' 2 = = 2 X L2 ( f ) 4πf 2 L2 f or

(1)

Alternating-Current Circuits

727

2

⎛ f' ⎞ X ( f' ) = ⎜⎜ ⎟⎟ X L2 ( f ) ⎝ f ⎠ 2 L

2

Substitute numerical values to obtain:

⎛ 240 s −1 ⎞ 2 ⎟ X L (60 Hz ) X (240 Hz ) = ⎜⎜ −1 ⎟ ⎝ 60 s ⎠ ⎛ 1 ⎞ 16 = 16⎜ R 2 ⎟ = R 2 ⎝3 ⎠ 3

Substitute in equation (1) to obtain:

(cos δ )240 Hz =

2 L

=

R 16 R2 + R2 3 3 = 0.397 19

*44 •• Picture the Problem We can apply Kirchhoff’s loop rule to obtain expressions for IR and IL and then use trigonometric identities to show that I = IR + IL = Imax cos (ωt − δ), where tan δ = R/XL and Imax = εmax/Z with Z −2 = R −2 + X L−2 . (a) Apply Kirchhoff’s loop rule to a clockwise loop that includes the source and the resistor: Solve for IR:

(b) Apply Kirchhoff’s loop rule to a clockwise loop that includes the source and the inductor: Solve for IL:

(c) Express the current drawn from the source in terms of Imax and the phase constant δ: Use a trigonometric identity to expand cos(ωt − δ):

ε max cos ωt − I R R = 0

IR =

ε max cos ωt R

ε max cos(ωt − 90°) − I L X L = 0 because the current lags the potential difference across the inductor by 90°.

IL =

ε max cos(ωt − 90°) XL

I = I R + I L = I max cos(ωt − δ )

I = I max (cos ωt cos δ + sin ωt sin δ ) = I max cos ωt cos δ + I max sin ωt sin δ

728 Chapter 29 From our results in (a):

I = IR + IL = + =

A useful trigonometric identity is:

ε max cos ωt R

ε max cos(ωt − 90°) XL

ε max cos ωt + ε max sin ωt R

XL

A cos ωt + B sin ωt = A2 + B 2 cos(ωt − δ ) where

δ = tan −1 Apply this identity to obtain:

B A

⎛ε ⎞ ⎛ε ⎞ I = ⎜ max ⎟ + ⎜⎜ max ⎟⎟ cos(ωt − δ ) (1) ⎝ R ⎠ ⎝ XL ⎠ 2

2

and

⎛ ε max ⎜ −1 ⎜ X L δ = tan ⎜ ε max ⎜ R ⎝

Simplify equation (1) and rewrite equation (2) to obtain:

⎞ ⎟ ⎟ = tan −1 ⎛⎜ R ⎞⎟ ⎜X ⎟ ⎟ ⎝ L⎠ ⎟ ⎠

(2)

⎛ε ⎞ ⎛ε I = ⎜ max ⎟ + ⎜⎜ max ⎝ R ⎠ ⎝ XL

⎞ ⎟⎟ cos(ωt − δ ) ⎠

= ε max

⎛1⎞ ⎛ 1 ⎜ ⎟ + ⎜⎜ ⎝ R⎠ ⎝ XL

⎞ ⎟⎟ cos(ωt − δ ) ⎠

= ε max

⎛1⎞ ⎜ ⎟ cos(ωt − δ ) ⎝Z ⎠

2

=

2

2

2

2

ε max cos(ωt − δ ) Z

where

tan δ =

R 1 1 1 and 2 = 2 + 2 XL Z R XL

45 •• Picture the Problem We can use the complex numbers method to find the impedances of the parallel portion of the circuit and the total impedance of the circuit. We can then use

Alternating-Current Circuits

729

Kirchhoff’s loop rule to obtain an expression for the current drawn form the source. Knowing the current drawn from the source, we can find the potential difference across the parallel portion of the circuit and then use this information to find the currents drawn by the load and the inductor. (a) Express the rms currents in R, C, and RL:

I R , rms = I L , rms =

Express the total impedance of the circuit:

Use complex numbers to relate Zp to RL and XC:

ε rms , I

Z Vp,rms

RL , rms

=

Vp,rms RL

,and

XL

Z = R + Zp where Zp is the impedance of the parallel branch of the circuit.

R + iX L 1 1 1 = + = L Z p RL iX L iRL X L or

Multiple the numerator and denominator of this fraction by the complex conjugate of RL + iXL and simplify to obtain: Substitute numerical values and evaluate XL: Substitute numerical values and evaluate Zp:

Zp =

iRL X L RL + iX L

Zp =

iRL X L RL − iX L RL + iX L RL − iX L RL X L2 RL2 X L i + RL2 + X L2 RL2 + X L2

=

X L = ωL = 2πfL

(

)

= 2π 500 s −1 (3.2 mH ) = 10.1Ω

2 ( 20 Ω )(10.1Ω ) Zp = (20 Ω )2 + (10.1Ω )2 (20 Ω )2 (10.1Ω ) +i (20 Ω )2 + (10.1Ω )2 = 4.06 Ω + i (8.05 Ω )

and

Zp = Substitute to evaluate Z:

(4.06 Ω )2 + (8.05 Ω )2

Z = 4 Ω + 4.03 Ω + i (8.05 Ω ) = 8.03 Ω + i (8.05 Ω )

= 9.02 Ω

730 Chapter 29 and

(8.03 Ω )2 + (8.05 Ω )2

Z = Express and evaluate the power factor: Apply Kirchhoff’s loop rule to obtain: Solve for and evaluate IR, rms:

Express and evaluate Vp, rms:

= 11.4 Ω

R 8.05 Ω = = 0.706 Z 11.4 Ω

cos δ =

ε rms − I R ,rms Z

ε rms

I R , rms =

Z

=0

=

100 V 2 = 6.20 A 11.4 Ω

Vp, rms = I RL ,rms Z p

= (6.20 A )(9 Ω ) = 55.8 V

Substitute numerical values and evaluate IR, rms:

I RL , rms =

Substitute numerical values and evaluate IL, rms:

I L , rms =

(b) Proceed as in (a) with f = 2000 Hz. Substitute numerical values and evaluate XL: Substitute numerical values and evaluate Zp:

55.8 V = 2.79 A 20 Ω 55.8 V = 5.52 A 10.1Ω

X L = ωL = 2πfL

(

)

= 2π 2000 s −1 (3.2 mH ) = 40.2 Ω

2 ( 20 Ω )(40.2 Ω ) Zp = (20 Ω )2 + (40.2 Ω )2 (20 Ω )2 (40.2 Ω ) +i (20 Ω )2 + (40.2 Ω )2 = 16.0 Ω + i (7.98 Ω )

and

Zp = Substitute to evaluate Z:

(16.0 Ω )2 + (7.98 Ω )2

= 17.9 Ω

Z = 4 Ω + 16.0 Ω + i (7.97 Ω ) = 20.0 Ω + i (7.98 Ω ) and

Z =

(20.0 Ω )2 + (7.98 Ω )2

= 21.5 Ω

Alternating-Current Circuits Find the power factor:

Apply Kirchhoff’s loop rule to obtain: Solve for and evaluate IR, rms:

Express and evaluate Vp, rms:

cos δ =

R 20.0 Ω = = 0.930 Z 21.5 Ω

ε rms − I R ,rms Z I R , rms =

731

ε rms Z

=0

=

100 V 2 = 3.29 A 21.5 Ω

Vp, rms = I RL ,rms Z p

= (3.29 A )(17.9 Ω ) = 58.9 V

Substitute numerical values and evaluate IR, rms:

I RL , rms =

58.9 V = 2.95 A 20 Ω

Substitute numerical values and evaluate IL, rms:

I L , rms =

58.9 V = 1.47 A 40.2 Ω

(c) Express the fraction of the power dissipated in the resistor:

I R2L , rm RL PL ,rms = Ptot ε rms I R , rms cos δ

Evaluate this fraction for f = 500 Hz:

PL ,rms Ptot

2 ( 2.79A ) (20 Ω ) = ⎛ 100 V ⎞ f =500 Hz ⎜ ⎟ (6.20 A )(0.706)

⎝

2 ⎠

= 0.503 = 50.3% When f = 2000 Hz:

PL ,rms Ptot

= f =2000 Hz

(2.95 A )2 (20 Ω ) ⎛ 100 V ⎞ ⎜ ⎟ (3.29 A )(0.930) ⎝

2 ⎠

= 0.804 = 80.4% 46 •• Picture the Problem We can treat the ac and dc components separately. For the dc component, L acts like a short circuit. For convenience we let ε1 denote the maximum value of the ac emf. We can use P = ε 12 R1, 2 to find the power dissipated in the resistors due to the dc source. We’ll apply Kirchhoff’s loop rule the loop including L, R1, and R2 to derive an expression for the power dissipated in the resistors due to the ac source. Note that only the power dissipated in the resistor R2 due to the ac source is frequency

732 Chapter 29 dependent. (a) Express the total power dissipated in R1 and R2: Express and evaluate the dc power dissipated in R1 and R2:

P = Pdc + Pac

P1,dc = and

P2,dc =

Express and evaluate the average ac power dissipated in R1: Apply Kirchhoff’s loop rule to a clockwise loop that includes R1, L, and R2:

(1)

ε 22 = (16 V )2

= 25.6 W

ε 22 = (16 V )2

= 32.0 W

10 Ω

R1

8Ω

R2

1 ε12 1 (20 V ) = = = 20.0 W 2 R1 2 10 Ω 2

P1,ac

R1 I1 − Z 2 I 2 = 0

I2 =

ε R1 R ε I1 = 1 1 = 1 Z2 Z 2 R1 Z 2

Express the average ac power dissipated in R2:

P2, ac

⎛ε ⎞ 1 ε 12 R2 = I R2 = ⎜⎜ 1 ⎟⎟ R2 = 2 Z 22 ⎝ Z2 ⎠

Substitute numerical values and evaluate P2, ac:

P2, ac =

Solve for I2:

2

1 2

2 2

1 2

(20 V ) (8 Ω ) 1 2 2 (8 Ω ) + 2π 100 s −1 {6 mH} 2

[

2

( {

}

)]

= 20.5 W Substitute in equation (1) to obtain:

P1 = 25.6 W + 20.0 W = 45.6 W P2 = 32.0 W + 20.5 W = 52.5 W and

P = P1 + P2 = 98.1 W (b) Proceed as in (a) to evaluate P2, ac with f = 200 Hz:

P2, ac =

(20 V ) (8 Ω ) 1 2 (8 Ω )2 + 2π 200 s −1 {6 mH} 2

[

= 13.2 W

2

( {

}

)]

Alternating-Current Circuits Substitute in equation (1) to obtain:

733

P1 = 25.6 W + 20.0 W = 45.6 W P2 = 32.0 W + 13.2 W = 45.2 W and

P = P1 + P2 = 90.8 W (c) Proceed as in (a) to evaluate P2, ac with f = 800 Hz:

P2, ac

(20 V ) (8 Ω ) 1 = 2 2 (8 Ω ) + (2π {800 s −1 }{6 mH})2

[

2

]

= 1.64 W Substitute in equation (1) to obtain:

P1 = 25.6 W + 20.0 W = 45.6 W P2 = 32.0 W + 1.64 W = 33.6 W and

P = P1 + P2 = 79.2 W 47 •• Picture the Problem We can use the phasor diagram for an RC circuit to find the voltage across the resistor. Sketch the phasor diagram for the voltages in the circuit:

Use the Pythagorean theorem to express VR:

VR =

2 ε rms − VC2

Substitute numerical values and evaluate VR:

VR =

(100 V ) 2 − (80 V ) 2

= 60.0 V

Filters and Rectifiers *48 •• Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. We’ll then assume a solution to this equation that is a linear combination of sine and cosine terms with coefficients that we can find by substitution in the differential equation. Repeating this process for the output side of the filter will yield the desired equation.

734 Chapter 29 Apply Kirchhoff’s loop rule to the input side of the filter to obtain:

Vin − V − IR = 0 where V is the potential difference across the capacitor.

Substitute for Vin and I to obtain:

Vpeak cos ωt − V − R

Because Q = CV:

dQ d dV = [CV ] = C dt dt dt

Substitute for dQ/dt to obtain:

Vpeak cos ωt − V − RC

dQ =0 dt

dV =0 dt

the differential equation describing the potential difference across the capacitor. Assume a solution of the form:

V = Vc cos ωt + Vs sin ωt

Substitution of this assumed solution and its first derivative in the differential equations, followed by equating the coefficients of the sine and cosine terms, yields two coupled linear equations:

Vc + ωRCVs = Vpeak

Solve these equations simultaneously to obtain:

and

Vs − ωRCVc = 0

Vc = and

Vs =

1 V 2 peak 1 + (ωRC )

ωRC V 2 peak 1 + (ωRC )

Vout = VH cos(ωt − δ )

Note that the output voltage is the voltage across the resistor and that it is phase shifted relative to the input voltage:

where VH is the amplitude of the signal.

Assume that VH is of the form:

VH (t ) = vc cos ωt + vs sin ωt

The input, output, and capacitor voltages are related according to: Substitute for VH (t ) , Vpeak (t ) , and

V (t ) and use the previously

established values for Vc and Vs to obtain: Substitute for Vc and Vs to obtain:

VH (t ) = Vin (t ) − V (t )

vc = Vpeak − Vc and

vs = −Vs

( ωRC )2 vc = V 2 peak 1 + (ωRC )

Alternating-Current Circuits and

vs = −

735

ωRC V 2 peak 1 + (ωRC )

VH, vc, and vs are related according to the Pythagorean relationship:

VH = vc2 + vs2

Substitute for vc and vs to obtain:

VH =

ωRC 1 + (ωRC )

2

Vpeak

Vpeak

=

⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

2

49 •• Picture the Problem We can use some of the intermediate results from Problem 48 to express the tangent of the phase constant. (a) Because, as was shown in Problem 48, VH =

v +v : 2 c

Also from Problem 48:

2 s

tan δ =

( ωRC )2 vc = V 2 peak 1 + (ωRC ) and

vs = − Substitute to obtain:

(b) Solve for δ:

vs vc

ωRC V 2 peak 1 + (ωRC )

ωRC V 2 peak 1 1 + (ωRC ) tan δ = = − 2 ωRC (ωRC ) V 2 peak 1 + (ωRC ) −

1 ⎤ ⎡ ⎣ ωRC ⎥⎦

δ = tan −1 ⎢−

As ω → 0:

δ → − 90°

(c) As ω → ∞:

δ→ 0

50 •• Picture the Problem We can use the results obtained in Problems 48 and 49 to find f3 dB and to plot graphs of log(Vout) versus log(f) and δ versus log(f).

736 Chapter 29 (a) Express the ratio Vout/Vin:

Vpeak 2

Vout Vin

⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠ = = Vpeak

When Vout = Vin / 2 :

1 ⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

2

=

Square both sides of the equation and solve for ωRC to obtain:

ωRC = 1 ⇒ ω =

Substitute numerical values and evaluate f3 dB:

f 3 dB =

(b) From Problem 48 we have:

Vout =

1 ⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

2

1 2

1 1 ⇒ f 3 dB = RC 2πRC

1 = 531Hz 2π (20 kΩ )(15 nF) Vpeak ⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

2

1 ⎤ ⎡ ⎣ ωRC ⎥⎦

From Problem 49 we have:

δ = tan −1 ⎢−

Rewrite these expressions in terms of f3 dB to obtain:

Vout =

Vpeak ⎛ 1 ⎞ ⎟⎟ 1 + ⎜⎜ ⎝ 2πfRC ⎠

2

=

Vpeak ⎛f ⎞ 1 + ⎜⎜ 3 dB ⎟⎟ ⎝ f ⎠

2

and

⎡

⎡ f ⎤ 1 ⎤ = tan −1 ⎢− 3 dB ⎥ ⎥ f ⎦ ⎣ 2πfRC ⎦ ⎣

δ = tan −1 ⎢−

A spreadsheet program to generate the data for a graph of Vout versus f and δ versus f is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B4 A8

Formula/Content 2.00E+03 1.50E−08 1 531 53

Algebraic Form R C Vpeak f3 dB 0.1f3 dB

Alternating-Current Circuits C8

$B$3/SQRT(1+(1($B$4/A8))^2)

Vpeak ⎛f ⎞ 1 + ⎜⎜ 3 dB ⎟⎟ ⎝ f ⎠

D8 E8

LOG(C8) ATAN(−$B$4/A8)

F8

E8*180/PI() A

2

log(Vout)

⎡ f ⎤ tan −1 ⎢− 3 dB ⎥ ⎣ f ⎦ δ in degrees

1 2 3 4 5 6 7 8 9 10 11

R= C= V_peak= f_3 dB=

B 2.00E+04 1.50E−08 1 531

C ohms F V Hz

D

E

f 53 63 73 83

log(f) 1.72 1.80 1.86 1.92

V_out 0.099 0.118 0.136 0.155

55 56 57

523 533 543

2.72 2.73 2.73

0.702 0.709 0.715

−0.154 −0.150 −0.146

−0.793 −0.783 −0.774

−45.4 −44.9 −44.3

531 532 533 534

5283 5293 5303 5313

3.72 3.72 3.72 3.73

0.995 0.995 0.995 0.995

−0.002 −0.002 −0.002 −0.002

−0.100 −0.100 −0.100 −0.100

−5.7 −5.7 −5.7 −5.7

log(V_out) delta(rad) −1.003 −1.471 −0.928 −1.453 −0.865 −1.434 −0.811 −1.416

F

delta(deg) −84.3 −83.2 −82.2 −81.1

The following graph of log(Vout) versus log(f) was plotted for Vpeak = 1 V. 0.0

log(V out)

-0.2 -0.4 -0.6 -0.8 -1.0 1.5

2.0

2.5

3.0 log(f )

A graph of δ in degrees as a function of log(f) follows.

3.5

4.0

737

738 Chapter 29 0 -10 delta (degrees)

-20 -30 -40 -50 -60 -70 -80 -90 1.5

2.0

2.5

3.0

3.5

4.0

log(f )

Referring to the spreadsheet program, we see that when f = f3 dB, δ ≈ − 44.9°. This result is in good agreement with its calculated value of −45.0°. 51 ••• Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. Because the voltage drop across the resistor is small compared to the voltage drop across the capacitor, we can express the voltage drop across the capacitor in terms of the input voltage. Apply Kirchhoff’s loop rule to the input side of the filter to obtain:

Vin − VC − IR = 0 where VC is the potential difference across the capacitor.

Substitute for Vin and I to obtain:

Vpeak cos ωt − Vc − R

Because Q = CVC:

dQ d dV = [CVC ] = C C dt dt dt

Substitute for dQ/dt to obtain:

Vpeak cos ωt − VC − RC

dQ =0 dt

dVC =0 dt

the differential equation describing the potential difference across the capacitor.

Vpeak cos ωt − VC ≈ 0

Because the voltage drop across the resistor is very small compared to the voltage drop across the capacitor:

and

Consequently, the potential difference across the resistor is given by:

VR = RC

VC ≈ Vpeak cos ωt

[

dVC d ≈ RC Vpeak cos ωt dt dt

]

Alternating-Current Circuits

739

52 •• Picture the Problem We can use the expression for VH from Problem 48 and the definition of β given in the problem to show that every time the frequency is halved, the output drops by 6 dB. From Problem 48:

VH =

Vpeak ⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

2

or

VH = Vpeak

Express this ratio in terms of f and f3 dB:

For f << f3dB:

From the definition of β we have:

Substitute for VH/Vpeak to obtain:

VH = Vpeak

VH ≈ Vpeak

1 ⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠ 1 ⎛f ⎞ 1 + ⎜⎜ 3 dB ⎟⎟ ⎝ f ⎠

2

2

f ⎛ f2 ⎞ f 32dB ⎜⎜1 + 2 ⎟⎟ f 3 dB ⎠ ⎝

β = 20 log10

VH Vpeak

β = 20 log10

f

Doubling the frequency yields:

β' = 20 log10

The change in decibel level is:

∆β = β' − β

f

= f

=

2 3 dB

⎛ f2 ⎞ ⎜⎜1 + 2 ⎟⎟ f 3 dB ⎠ ⎝

f f 3 dB

f 3 dB

= 20 log10

2f f 3 dB

2f f − 20 log10 f 3 dB f 3 dB

= 20 log10 2 = 6.02 dB *53 •• Picture the Problem We can express the instantaneous power dissipated in the resistor and then use the fact that the average value of the square of the cosine function over one cycle is ½ to establish the given result.

740 Chapter 29 The instantaneous power P(t) dissipated in the resistor is: The output voltage Vout is: From Problem 48:

Substitute in the expression for P(t) to obtain:

2 Vout P(t ) = R

Vout = VH cos(ωt − δ ) Vpeak

VH =

⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

P(t ) = =

Because the average value of the square of the cosine function over one cycle is ½:

Simplify this expression to obtain:

Pave =

2

VH2 cos 2 (ωt − δ ) R 2 Vpeak ⎡ ⎛ 1 ⎞ ⎤ R ⎢1 + ⎜ ⎟ ⎥ ⎣⎢ ⎝ ωRC ⎠ ⎥⎦ 2

cos 2 (ωt − δ )

2 Vpeak

⎡ ⎛ 1 ⎞2 ⎤ 2 R ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ ωRC ⎠ ⎥⎦

2 ⎛ (ωRC )2 ⎞ Vpeak ⎜ ⎟ Pave = 2 R ⎜⎝ 1 + (ωRC )2 ⎟⎠

54 •• Picture the Problem We can solve the expression for VH from Problem 48 for the required capacitance of the capacitor. From Problem 48:

We require that:

Vpeak

VH =

⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

VH = Vpeak

2

1 ⎛ 1 ⎞ 1+ ⎜ ⎟ ⎝ ωRC ⎠

2

=

1 10

or 2

⎛ 1 ⎞ 1+ ⎜ ⎟ = 10 ⎝ ωRC ⎠ Solve for C to obtain:

C=

1 1 = 99ωR 2π 99 Rf

Alternating-Current Circuits Substitute numerical values and evaluate C:

C=

1 = 13.3 nF 2π 99 (20 kΩ )(60 Hz )

55 •• Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. We’ll then assume a solution to this equation that is a linear combination of sine and cosine terms with coefficients that we can find by substitution in the differential equation. The solution to these simultaneous equations will yield the amplitude of the output voltage. Apply Kirchhoff’s loop rule to the input side of the filter to obtain:

Vin − IR − V = 0 where V is the potential difference across the capacitor.

Substitute for Vin and I to obtain:

Vpeak cos ωt − R

Because Q = CV:

dQ d dV = [CV ] = C dt dt dt

Substitute for dQ/dt to obtain:

Vpeak cos ωt − RC

dQ −V = 0 dt

dV −V = 0 dt

the differential equation describing the potential difference across the capacitor. Assume a solution of the form:

V = Vc cos ωt + Vs sin ωt

Substitution of this assumed solution and its first derivative in the differential equation, followed by equating the coefficients of the sine and cosine terms, yields two coupled linear equations:

Vc + ωRCVs = Vpeak

Solve these equations simultaneously to obtain:

and

Vs − ωRCVc = 0

Vc = and

Vs =

1 V 2 peak 1 + (ωRC )

ωRC V 2 peak 1 + (ωRC )

Vout = VL cos(ωt − δ )

Note that the output voltage is the voltage across the capacitor and that it is phase shifted relative to the input voltage:

where VL is the amplitude of the signal.

VL, Vc, and Vs are related according to the Pythagorean relationship:

VL = Vc2 + Vs2

741

742 Chapter 29 Substitute for Vc and Vs to obtain: 2

⎡ ⎤ ⎡ ωRC ⎤ 1 VL = ⎢ V +⎢ V 2 peak ⎥ 2 peak ⎥ ⎣1 + (ωRC ) ⎦ ⎣1 + (ωRC ) ⎦ Simplify algebraically to obtain:

2

Vpeak

VL =

1 + (ωRC )

2

As f → 0, VL → Vpeak . As f → ∞, VL → 0. 56 •• Picture the Problem We can use some of the intermediate results from Problem 55 to express the tangent of the phase constant. From Problem 55:

VL = Vc2 + Vs2 where tan δ =

Also from Problem 55:

Vc = and

Vs =

Vs Vc

1 V 2 peak 1 + (ωRC )

ωRC V 2 peak 1 + (ωRC )

Substitute to obtain:

tan δ =

ωRC V 2 peak 1 + (ωRC ) 1 V 2 peak 1 + (ωRC )

Solve for δ:

δ = tan −1 (ωRC )

As ω → 0:

δ→ 0

(c) As ω → ∞:

δ → 90°

= ωRC

Remarks: See the spreadsheet solution in the following problem for additional evidence that our answer for Part (c) is correct. *57 •• Picture the Problem We can use the expressions for VL and δ derived in Problem 56 to plot the graphs of VL versus f and δ versus f for the low-pass filter of Problem 55. We’ll simplify the spreadsheet program by expressing both VL and δ as functions of f3 dB.

Alternating-Current Circuits From Problem 56 we have:

VL =

Vpeak 1 + (ωRC )

2

and

δ = tan −1 (ωRC )

Rewrite each of these expressions in terms of f3 dB to obtain:

VL =

Vpeak 1 + (2πfRC )

2

=

Vpeak ⎛ f ⎞ ⎟⎟ 1 + ⎜⎜ ⎝ f 3 dB ⎠

2

and

⎛ f ⎞

⎟ δ = tan −1 (2πfRC ) = tan −1 ⎜⎜ ⎟ ⎝ f 3 dB ⎠ A spreadsheet program to generate the data for graphs of VL versus f and δ versus f for the low-pass filter is shown below. Note that Vpeak has been arbitrarily set equal to 1 V. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form B1 2.00E+03 R B2 C 5.00E−09 B3 1 Vpeak B4 f3 dB (2*PI()*$B$1*$B$2)^−1 B8 $B$3/SQRT(1+((A8/$B$4)^2)) Vpeak

⎛ f ⎞ ⎟⎟ 1 + ⎜⎜ ⎝ f 3 dB ⎠ C8

ATAN(A8/$B$4)

D8

C8*180/PI() A

2

⎛ f ⎞ ⎟⎟ tan −1 ⎜⎜ ⎝ f 3 dB ⎠ δ in degrees

1 2 3 4 5 6 7 8 9 10

R= C= V_peak= f_3 dB=

B 1.00E+04 5.00E−09 1 3.183

C ohms F V kHz

D

f(kHz) 0 1 2 3

V_out 1.000 0.954 0.847 0.728

delta(rad) 0.000 0.304 0.561 0.756

delta(deg) 0.0 17.4 32.1 43.3

54 55 56

47 48 49

0.068 0.066 0.065

1.503 1.505 1.506

86.1 86.2 86.3

743

744 Chapter 29 57

50

0.064

1.507

86.4

A graph of Vout as a function of f follows: 1.0

V out (V)

0.8 0.6 0.4 0.2 0.0 0

10

20

30

40

50

30

40

50

f (kHz)

A graph of δ as a function of f follows: 100

delta (degrees)

80 60 40 20 0 0

10

20 f (kHz)

58 ••• Picture the Problem We can use Kirchhoff’s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. We’ll then assume a solution to this equation that is a linear combination of sine and cosine terms with coefficients that we can find by substitution in the differential equation. The solution to these simultaneous equations will yield the amplitude of the output voltage. Apply Kirchhoff’s loop rule to the input side of the filter to obtain: Substitute for Vin and I to obtain:

Vin − IR − VC = 0 where VC is the potential difference across the capacitor.

Vpeak cos ωt − R

dQ − VC = 0 dt

Alternating-Current Circuits Because Q = CVC:

dQ d dV = [CVC ] = C C dt dt dt

Substitute for dQ/dt to obtain:

Vpeak cos ωt − RC

dVC − VC = 0 dt

the differential equation describing the potential difference across the capacitor. The output voltage is the voltage across the capacitor. Because this voltage is small :

Vpeak cos ωt − RC

Separate the variables in this differential equation and solve for VC:

VC =

dVC ≈0 dt

1 Vpeak cos ωtdt RC ∫

*59 ••• Picture the Problem We can apply Kirchhoff’s loop rule to both the input side and output side of the trap filter to obtain an expression for the impedance of the trap. Requiring that the impedance of the trap be zero will yield the frequency at which the circuit rejects signals. Defining the bandwidth as ∆ω = ω − ωtrap and requiring that

Ztrap = R will yield an expression for the bandwidth and reveal its dependence on R. Apply Kirchhoff’s loop rule to the output of the trap circuit to obtain: Solve for Vout:

Vout − IX L − IX C = 0 Vout = I ( X L + X C ) = IZ trap where Z trap = X L + X C

Apply Kirchhoff’s loop rule to the input of the trap circuit to obtain:

Vin − IR − IX L − IX C = 0

Solve for I:

I=

Substitute for I in equation (1) to obtain: Because X L = iωL and

XC =

−i : ωC

Note that Ztrap = 0 and Vout = 0 provided:

Vin Vin = R + X L + X C R + Z trap

Vout = Vin

Z trap R + Z trap

1 ⎞ ⎛ Z trap = i⎜ ωL − ⎟ ωC ⎠ ⎝

ω=

1 LC

(1)

745

746 Chapter 29 Let the bandwidth ∆ω be:

∆ω = ω − ωtrap

Let the frequency bandwidth to be defined by the frequency at

ωL −

which Z trap = R . Then:

or

1 =R ωC

ω 2 LC − 1 = ωRC 2

1 : LC

Because ωtrap =

(2)

⎛ ω ⎞ ⎜ ⎟ − 1 = ωRC ⎜ω ⎟ ⎝ trap ⎠

For ω ≈ ωtrap:

2 ⎛ ω 2 − ωtrap ⎞ ⎜ ⎟ ≈ ωtrap RC ⎜ ω ⎟ trap ⎝ ⎠

2 Solve for ω 2 − ωtrap :

2 ω 2 − ωtrap = (ω − ωtrap )(ω + ωtrap )

Because ω ≈ ωtrap,

2 ω 2 − ωtrap ≈ 2ωtrap (ω − ωtrap )

ω − ωtrap ≈ 2ωtrap :

Substitute in equation (2) to obtain:

∆ω = ω − ωtrap =

2 RCωtrap

2

=

R 2L

60 •• Picture the Problem For voltages greater than 0.6 V, the output voltage will mirror the input voltage minus a 0.6 V drop. But when the voltage swings below 0.6 V, the output voltage will be 0. A spreadsheet program was used to plot the following graph. The angular frequency and the peak voltage were both arbitrarily set equal to one.

1.0 0.8

V_in

0.6

V_out

V in, V out (V)

0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0 0

2

4

6 t (s)

8

10

12

Alternating-Current Circuits

747

61 •• Picture the Problem We can use the decay of the potential difference across the capacitor to relate the time constant for the RC circuit to the frequency of the input signal. Expanding the exponential factor in the expression for VC will allow us to find the approximate value for C that will limit the variation in the output voltage by less than 50 percent (or any other percentage). The voltage across the capacitor is given by:

VC = Vin e −t RC

Expand the exponential factor to obtain:

e −t RC ≈ 1 −

For a decay of less than 50 percent:

1−

Solve for C to obtain:

Because the voltage goes positive every cycle, t = 1/60 s and:

1 t RC

1 t ≤ 0.5 RC 2 C≤ t R C≤

2 ⎛ 1 ⎞ ⎜ s ⎟ = 33.3 µF 1 kΩ ⎝ 60 ⎠

LC Circuits with a Generator 62 •• Picture the Problem We know that the current leads the voltage across and capacitor and lags the voltage across an inductor. We can use I L ,max = ε max X L and

I C ,max = ε max X C to find the amplitudes of these currents. The current in the generator

will vanish under resonance conditions, i.e., when I L = I C . To find the currents in the inductor and capacitor at resonance, we can use the common potential difference across them and their reactances … together with our knowledge of the phase relationships mentioned above. (a) Express the amplitudes of the currents through the inductor and the capacitor:

I L ,max = and

I C ,max =

Express XL and XC:

ε max XL

ε max XC

X L = ωL and X C =

1 ωC

748 Chapter 29 Substitute to obtain:

I L ,max =

100 V (4 H )ω , lagging ε by 90°

=

25 V/H

I C ,max =

100 V 1 (25 µF)ω

ω

and

=

(b) Express the condition that I = 0:

(2.5 ×10

−3

leading ε

)

V ⋅ F ω, by 90°

I L = IC or

ε

ωL

ε

=

1 ωC

= ωCε

Solve for ω to obtain:

ω=

1 LC

Substitute numerical values and evaluate ω:

ω=

1 = 100 rad/s (4 H )(25 µF)

(c) Express the current in the inductor at ω = ω0:

⎛ 25 V/H ⎞ ⎟ cos[(100 rad/s )t − 90°] I L = ⎜⎜ −1 ⎟ ⎝ 100 s ⎠

(0.250 A )sin[(100 s −1 )t ]

= Express the current in the capacitor at ω = ω0:

(

)(

)

[(

)]

I C = 2.5 × 10 −3 V ⋅ F 100 s −1

× cos[(100 rad/s )t + 90°]

= − (0.25 A )sin 100 s −1 t (d) The phasor diagram is shown to the right.

Alternating-Current Circuits

749

63 •• Picture the Problem We can differentiate Q with respect to time to find I as a function of time. In (b) we can find C by using ω = 1

LC . The energy stored in the magnetic

field of the inductor is given by U m = LI and the energy stored in the electric field of 1 2

the capacitor by U e =

1 2

2

Q2 . C

(a) Use the definition of current to obtain:

π ⎞⎤ dQ d ⎡ ⎛ = ⎢(15 µC ) cos⎜1250 t + ⎟⎥ 4 ⎠⎦ dt dt ⎣ ⎝

I=

π⎞ ⎛ = −(15 µC ) 1250 s −1 sin ⎜1250 t + ⎟ 4⎠ ⎝

(

)

π⎞ ⎛ = − (18.75 mA ) sin ⎜1250 t + ⎟ 4⎠ ⎝ (b) Relate C to L and ω:

ω=

1 LC

Solve for C to obtain:

C=

1

Substitute numerical values and evaluate C:

C=

ω2L 1

(1250 s ) (28 mH) −1 2

= 22.86 µF

(c) Express and evaluate the magnetic energy Um:

U m = 12 LI 2 = =

1 2

(28 mH )(18.75 mA )2 sin 2 ⎛⎜1250t + π ⎞⎟ ⎝

4⎠

(4.92 µJ )sin 2 ⎛⎜1250t + π ⎞⎟

Express and evaluate the electrical energy Ue:

⎝

4⎠

Q2 C 2 ( 15 µF) π⎞ ⎛ 1 =2 cos 2 ⎜1250 t + ⎟ 22.86 µF 4⎠ ⎝

Ue =

=

1 2

(4.92 µJ ) cos 2 ⎛⎜1250 t + π ⎞⎟

The total energy stored in the electric and magnetic fields is:

⎝

4⎠

750 Chapter 29

π⎞ π⎞ ⎛ ⎛ U = U m + U e = (4.92 µJ )sin 2 ⎜1250 t + ⎟ + (4.92 µJ )cos 2 ⎜1250 t + ⎟ = 4.92 µJ 4⎠ 4⎠ ⎝ ⎝ *64 ••• Picture the Problem We can use the definition of the capacitance of a dielectric-filled capacitor and the expression for the resonance frequency of an LC circuit to derive an expression for the fractional change in the thickness of the dielectric in terms of the resonance frequency and the frequency of the circuit when the dielectric is under compression. We can then use this expression for ∆t/t to calculate the value of Young’s modulus for the dielectric material.

stress ∆P = strain ∆t t

Use its definition to express Young’s modulus of the dielectric material:

Y=

Letting t be the initial thickness of the dielectric, express the initial capacitance of the capacitor:

C0 =

Express the capacitance of the capacitor when it is under compression:

Cc =

Express the resonance frequency of the capacitor before the dielectric is compressed:

ω0 =

1 = C0 L

1 κ ∈0 AL t

When the dielectric is compressed:

ωc =

1 = Cc L

1 κ ∈0 AL t − ∆t

Express the ratio of ωc to ω0 and simplify to obtain:

κ ∈0 AL ωc ∆t t = = 1− t ω0 κ ∈0 AL

(1)

κ ∈0 A t

κ ∈0 A t − ∆t

t − ∆t

Expand the radical binomially to obtain:

ωc ⎛ ∆t ⎞ = ⎜1 − ⎟ ω0 ⎝ t ⎠

12

provided ∆t << t.

≈ 1−

∆t 2t

Alternating-Current Circuits Solve for ∆t/t:

⎛ ω ⎞ ∆t = 2⎜⎜1 − c ⎟⎟ t ⎝ ω0 ⎠

Substitute in equation (1) to obtain:

Y=

Substitute numerical values and evaluate Y:

Y=

751

∆P ⎛ ω ⎞ 2⎜⎜1 − c ⎟⎟ ⎝ ω0 ⎠

(800 atm)(101.325 kPa/atm ) ⎛ 116 MHz ⎞ ⎟⎟ 2⎜⎜1 − ⎝ 120 MHz ⎠

= 1.22 × 109 N/m 2 65 ••• Picture the Problem We can model this capacitor as the equivalent of two capacitors connected in parallel. Let C1 be the capacitance of the dielectric-filled capacitor and C2 be the capacitance of the air-filled capacitor. We’ll derive expressions for the capacitances of the parallel capacitors and add these expressions to obtain C(x). We can then use the given resonance frequency when x = w/2 and the given value for L to evaluate C0. In Part (b) we can use our result for C(x) and the relationship between f, L, and C(x) at resonance to express f(x).

κ ∈0 A1 ∈0 A2

(a) Express the equivalent capacitance of the two capacitors in parallel:

C (x ) = C1 + C2 =

Express A2 in terms of the total area of a capacitor plate A, w, and the distance x:

A2 x x = ⇒ A2 = A A w w

Express A1 in terms of A and A2:

x⎞ ⎛ A1 = A − A2 = A⎜1 − ⎟ ⎝ w⎠

Substitute in equation (1) and simplify to obtain:

C (x ) =

d

κ ∈0 A ⎛

+

d

x⎞ ∈ A x ⎜1 − ⎟ + 0 d ⎝ w⎠ d w x⎞ x⎤ ∈ A⎡ ⎛ = 0 ⎢κ ⎜1 − ⎟ + ⎥ d ⎣ ⎝ w ⎠ w⎦ ⎡ κ −1 ⎤ x = κC0 ⎢1 − κw ⎥⎦ ⎣

(1)

752 Chapter 29 where C0 =

∈0 A d

⎛ w⎞ ⎡ κ −1 w ⎤ C ⎜ ⎟ = κC0 ⎢1 − κw 2 ⎥⎦ ⎝2⎠ ⎣ ⎡ κ − 1⎤ = κC0 ⎢1 − 2κ ⎥⎦ ⎣ κ +1 = C0 2

Find C(w/2):

Express the resonance frequency of the circuit in terms of L and C(x):

f (x ) =

Evaluate f(w/2):

⎛ w⎞ f⎜ ⎟= ⎝2⎠

1 2π LC (x )

=

Solve for C0 to obtain:

C0 =

Substitute numerical values and evaluate C0:

C0 =

(2)

1 2π LC0 1 2π

κ +1 2

2 (κ + 1)LC0

1 ⎛ w⎞ 2π 2 f 2 ⎜ ⎟ L(κ + 1) ⎝2⎠ 1

2π (90 MHz ) (2 mH )(4.8 + 1) 2

2

= 5.39 ×10 −16 F f (x ) =

(b) Substitute for C(x) in equation (2) to obtain:

1 ⎡ κ −1 ⎤ x 2π LκC0 ⎢1 − κw ⎥⎦ ⎣

Substitute numerical values and evaluate f(x):

f (x ) =

1 2π

(2 mH )(4.8)(5.39 ×10−16 F)⎡⎢1 − ⎣

4.8 − 1 ⎤ x 4.8(0.2 m ) ⎥⎦

=

70.0 MHz

(

)

1 − 3.96 m −1 x

Alternating-Current Circuits

RLC Circuits with a Generator 66 • Picture the Problem We can use the expression for the resonance frequency of a series RLC circuit to obtain an expression for C as a function of f. Express the resonance frequency as a function of L and C:

ω = 2πf =

Solve for C to obtain:

C=

1 4π f 2 L 2

1 2 4π (1600 kHz ) (1 µH ) = 9.89 nF

Substitute numerical values and evaluate the smallest value for C:

C smallest =

Substitute numerical values and evaluate the largest value for C:

Clargest =

Therefore:

1 LC

2

1 2 4π (500 kHz ) (1 µH ) = 101 nF 2

9.89 nF ≤ C ≤ 101 nF

67 • Picture the Problem The diagram shows the relationship between δ, XL, XC, and R. We can use this reference triangle to express the power factor for the circuit in Example 29-5. In (b) we can use the reference triangle to relate ω to tanδ.

R = Z

R

(a) Express the power factor for the circuit:

cos δ =

Evaluate XL and XC:

X L = ωL = 400 s −1 (2 H ) = 800 Ω

R2 + (X L − X C )

2

(

)

and

XC =

1 1 = = 1250 Ω −1 ωC 400 s (2 µF)

(

)

753

754 Chapter 29 Substitute numerical values and evaluate cosδ:

20 Ω

cos δ =

(20 Ω )2 + (800 Ω − 1250 Ω )2

= 0.0444 (b) Express tanδ:

X − XC tan δ = L = R

ωL −

1 ωC

R

LCω 2 − CR tan δω − 1 = 0

Rewrite this equation explicitly as a quadratic equation in ω: Substitute numerical values to obtain:

[(2 H )(2 µF)]ω 2 − [(2 µF)(20 Ω ) tan (cos −1 0.5)]ω − 1 = 0 or

(4 × 10

−6

)

(

)

s 2 ω 2 ± 69.3 × 10 −6 s ω − 1 = 0

Solve for ω to obtain:

ω = 491 rad/s or ω = 509 rad/s

68 • Picture the Problem The diagram shows the relationship between δ, XL, XC, and R. We can use this reference triangle to express the power factor for the given circuit. In (b) we can find the rms current from the rms potential difference and the impedance of the circuit. We can find the average power delivered by the source from the rms current and the resistance of the resistor. (a) The power factor is defined to be:

cos δ =

With no inductance in the circuit:

XL = 0

R = Z

R R2 + (X L − X C )

2

and

cos δ =

R R 2 + X C2

=

R R2 +

1 ω C2 2

Alternating-Current Circuits Substitute numerical values and evaluate cosδ:

755

80 Ω

cos δ =

(80 Ω )2 +

1

(400 s ) (20 µF) −1 2

2

= 0.539 (b) Express the rms current in the circuit:

I rms =

ε rms Z

ε max =

2 R + X C2

ε max

=

2 R2 +

Insert numerical values and evaluate Irms:

2

1 ω C2 2

20 V

I rms = 2

(80 Ω )2 +

1

(400 s ) (20 µF) −1 2

2

= 95.3 mA (c) Express and evaluate the average power delivered by the generator:

2 Pav = I rms R = (95.3 mA ) (80 Ω ) 2

= 0.727 W

*69 •• Picture the Problem The impedance of an ac circuit is given by

Z = R 2 + ( X L − X C ) . We can evaluate the given expression for Pav first for 2

XL = XC = 0 and then for R = 0. (a) For X = 0, Z = R and:

(b), (c) If R = 0, then:

2 2 Rε rms Rε rms = = Pav = Z2 R2

Pav =

2 ε rms

R

2 2 (0)ε rms Rε rms = = 0 Z2 ( X L − X C )2

Remarks: Recall that there is no energy dissipation in an ideal inductor or capacitor. 70 •• Picture the Problem We can use ω0 = 1

LC to find the resonant frequency of the

circuit, I rms = ε rms R to find the rms current at resonance, the definitions of XC and XL to

756 Chapter 29 find these reactances at ω = 8000 rad/s, the definitions of Z and Irms to find the impedance and rms current at ω = 8000 rad/s, and the definition of the phase angle to find δ. (a) Express the resonant frequency ω0 in terms of L and C:

ω0 =

1 LC

Substitute numerical values and evaluate ω0:

ω0 =

1 (10 mH )(2 µF)

= 7.07 × 103 rad/s (b) Relate the rms current at resonance to εrms and the impedance of the circuit at resonance:

I rms =

(c) Express and evaluate XC and XL at ω = 8000 rad/s:

XC =

ε rms = ε max R

2R

100 V 2 (5 Ω )

=

= 14.1 A 1 1 = = 62.5 Ω −1 ωC 8000 s (2 µF)

(

and

)

(

)

X L = ωL = 8000 s −1 (10 mH ) = 80.0 Ω

(d) Express the impedance in terms of the reactances, substitute the results from (c), and evaluate Z:

Z = R2 + (X L − X C )

2

(5 Ω )2 + (80 Ω − 62.5 Ω )2

=

= 18.2 Ω

ε rms = ε max

Relate the rms current at ω = 8000 rad/s to εrms and the impedance of the circuit at this frequency:

I rms =

(e) Using its definition, express and evaluate δ:

δ = tan −1 ⎜

Z

2Z

=

100 V 2 (18.2 Ω )

= 3.89 A ⎛ XL − XC ⎞ ⎟ R ⎝ ⎠ ⎛ 80 Ω − 62.5 Ω ⎞ ⎟⎟ = 74.1° = tan −1 ⎜⎜ 5Ω ⎝ ⎠

71 •• Picture the Problem We can use f 0 = 1 2π LC to find the resonant frequency of the circuit, the definitions of XC and XL to find these reactances at f = 1000 Hz, the definitions of Z and Irms to find the impedance and rms current at f= 1000 Hz, and the definition of

Alternating-Current Circuits

757

the phase angle to find δ.

1

(a) Express the resonant frequency f0 in terms of L and C:

f0 =

Substitute numerical values and evaluate f0:

f0 =

(b) Express and evaluate XC and XL at f = 1000 Hz:

XC =

2π LC 1 = 1.13 kHz 2π (10 mH )(2 µF)

1 1 = 2πfC 2π 1000 s −1 (2 µF)

(

)

= 79.6 Ω and

X L = 2πfL = 2π (1000 s −1 )(10 mH ) = 62.8 Ω

(c) Express the impedance in terms of the reactances, substitute the results from (b), and evaluate Z:

Z = R2 + (X L − X C )

2

(5 Ω )2 + (62.8 Ω − 79.6 Ω )2

=

= 17.5 Ω

ε rms = ε max

Relate the rms current at f = 1000 Hz to εrms and the impedance of the circuit at this frequency:

I rms =

(d) Using its definition, express and evaluate δ:

δ = tan −1 ⎜

Z

2Z

=

100 V 2 (17.5 Ω )

= 4.04 A ⎛ XL − XC ⎞ ⎟ R ⎝ ⎠ ⎛ 62.8 Ω − 79.6 Ω ⎞ ⎟⎟ = − 73.4° = tan −1 ⎜⎜ 5Ω ⎝ ⎠

72 •• Picture the Problem Note that the reactances and, hence, the impedance of an ac circuit are frequency dependent. We can use the definitions of XL, XC, and Z, δ, and cosδ to find the phase angle and the power factor of the circuit at the given frequencies. Express the phase angle δ and the power factor cosδ for the circuit:

⎛ X L − XC ⎞ ⎟ R ⎝ ⎠

δ = tan −1 ⎜ and

(1)

758 Chapter 29 cos δ =

(a) Evaluate XL, XC, and Z at f = 900 Hz:

R Z

(2)

X L = 2πfL

= 2π (900 s −1 )(10 mH ) = 56.5 Ω, 1 1 = XC = 2πfC 2π 900 s −1 (2 µF) = 88.4 Ω,

(

)

and

Z = R2 + (X L − X C )

2

=

(5 Ω )2 + (56.5 Ω − 88.4 Ω )2

= 32.3 Ω Substitute in equations (1) and (2) to obtain:

⎛ 56.5 Ω − 88.4 Ω ⎞ ⎟⎟ = − 81.1° Ω 5 ⎝ ⎠

δ = tan −1 ⎜⎜ and

cos δ =

(b) Evaluate XL, XC, and Z at f = 1.1 kHz:

5Ω = 0.155 32.3 Ω

X L = 2πfL

(

)

= 2π 1100 s −1 (10 mH ) = 69.1Ω, 1 1 = XC = 2πfC 2π 1100 s −1 (2 µF) = 72.3 Ω,

(

)

and

Z = R2 + (X L − X C )

2

=

(5 Ω )2 + (69.1Ω − 72.3 Ω )2

= 5.94 Ω Substitute in equations (1) and (2) to obtain:

⎛ 69.1 Ω − 72.3 Ω ⎞ ⎟⎟ = − 32.6° 5Ω ⎝ ⎠

δ = tan −1 ⎜⎜ and

cos δ =

5Ω = 0.842 5.94 Ω

Alternating-Current Circuits (c) Evaluate XL, XC, and Z at f = 1.3 kHz:

X L = 2πfL

(

759

)

= 2π 1300 s −1 (10 mH ) = 81.7 Ω, 1 1 = XC = 2πfC 2π 1300 s −1 (2 µF) = 61.2 Ω,

(

)

and

Z = R2 + (X L − X C )

2

(5 Ω )2 + (81.7 Ω − 61.2 Ω )2

=

= 21.1 Ω Substitute in equations (1) and (2) to obtain:

⎛ 81.7 Ω − 61.2 Ω ⎞ ⎟⎟ = 76.3° 5Ω ⎝ ⎠

δ = tan −1 ⎜⎜ and

5Ω = 0.237 21.1 Ω

cos δ =

73 •• Picture the Problem The Q factor of the circuit is given by Q = ω0 L R , the resonance width by ∆f = f 0 Q = ω0 2πQ , and the power factor by cos δ = R Z . Because Z is frequency dependent, we’ll need to find XC and XL at ω = 8000 rad/s in order to evaluate cosδ. Using their definitions, express the Q factor and the resonance width of the circuit:

(a) Express and evaluate the resonance frequency for the circuit:

Q=

ω0 L

and

(1)

R

∆f =

f0 ω = 0 Q 2πQ

ω0 =

1 = LC

(2)

1 (10 mH )(2 µF)

= 7.07 × 103 rad/s Substitute numerical values in equation (1) and evaluate Q: (b) Substitute numerical values in equation (2) and evaluate ∆f:

(7.07 × 10 Q=

3

)

rad/s (10 mH ) = 14.1 5Ω

7.07 × 103 rad/s ∆f = = 79.8 Hz 2π (14.1)

760 Chapter 29 (c) Express the power factor of the circuit: Evaluate XL, XC, and Z at ω = 8000 rad/s:

R Z X L = ωL

cos δ =

(

)

= 8000 s −1 (10 mH ) = 80.0 Ω, 1 1 = XC = −1 ωC 8000 s (2 µF) = 62.5 Ω,

(

)

and

Z = R2 + (X L − X C )

2

(5 Ω )2 + (80 Ω − 62.5 Ω )2

=

= 18.2 Ω Substitute numerical values and evaluate cosδ:

cos δ =

5Ω = 0.275 18.2 Ω

*74 •• Picture the Problem We can use its definition, Q = f 0 ∆f to find the Q factor for the circuit. Express the Q factor for the circuit:

Q=

f0 ∆f

Substitute numerical values and evaluate Q:

Q=

100.1 MHz 0.05 MHz

= 2002

75 •• Picture the Problem We can use I = ε max Z to find the current in the coil and the definition of the phase angle to evaluate δ. We can equate XL and XC to find the capacitance required so that the current and the voltage are in phase. Finally, we can find the voltage measured across the capacitor by using VC = IX C . (a) Express the current in the coil in terms of the potential difference across it and its impedance:

I=

Substitute numerical values and evaluate I:

I=

ε max Z

100 V = 10.0 A 10 Ω

Alternating-Current Circuits R X = sin −1 Z Z ⎛ 8Ω ⎞ ⎟⎟ = 53.1° = sin −1 ⎜⎜ ⎝ 10 Ω ⎠

(b) Express and evaluate the phase angle δ:

δ = cos −1

(c) Express the condition on the reactances that must be satisfied if the current and voltage are to be in phase:

X L = X C or X L =

Solve for C to obtain:

C=

1 1 = ωX L 2πfX L

Substitute numerical values and evaluate C:

C=

1 = 332 µF 2π 60 s −1 (8 Ω )

(d) Express the potential difference across the capacitor:

761

(

1 ωC

)

VC = IX C

V R

Relate the current I in the circuit to the impedance of the circuit when XL = XC:

I=

Substitute to obtain:

VC =

Relate the impedance of the circuit to the resistance of the coil:

Z = R2 + X 2

Solve for and evaluate the resistance of the coil:

R = Z2 − X 2 =

Substitute numerical values and evaluate VC:

VC =

VX C V = R 2πfCR

(10 Ω )2 − (8 Ω )2

= 6Ω 100 V = 133 V 2π 60 s (332 µF)(6 Ω )

(

−1

)

76 •• Picture the Problem We can find C using VC = I rms X C and Irms from the potential difference across the inductor. In the absence of resistance in the circuit, the measured rms voltage across both the capacitor and inductor is V = VL − VC .

762 Chapter 29 I rms 2πfC

(a) Relate the capacitance C to the potential difference across the capacitor:

VC = I rms X C =

Solve for C to obtain:

C=

Use the potential difference across the inductor to express the rms current in the circuit:

I rms =

Substitute to obtain:

C=

VL (2πf )2 LVC

Substitute numerical values and evaluate C:

C=

[2π (60 s )] (0.25 H )(75 V )

I rms 2πfVC VL V = L X L 2πfL

50 V

−1

2

= 18.8 µF (b) Express the measured rms voltage V across both the capacitor and the inductor when R = 0:

V = VL − VC

Substitute numerical values and evaluate V:

V = 50 V − 75 V = 25.0 V

77 •• Picture the Problem We can rewrite Equation 29-51 in terms of ω, L, and C and factor L from the resulting expression to obtain the given equation. In (b) and (c) we can use the expansions for cot−1x and tan−1x to approximate δ at very low and very high frequencies. (a) From Equation 29-51:

ω 2L −1 C tan δ = = R ωR 2 L(ω 2 − ω02 ) L(ω − 1 LC ) = = ωR ωR

(b) Rewrite tanδ as:

tan δ =

For ω << 1:

ωL − 1 ωC

ωL R

tan δ ≈ −

−

1 ωRC

1 ωRC

(1)

Alternating-Current Circuits

763

and

cot δ = −ωRC or δ = cot −1 (− ωRC )

Use the expansion for cot−1x to obtain: Recall that, for negative values of the argument, the angle approaches −π/2*, to obtain:

cot −1 x = ±

−

π 2

2

−x

− δ = −ωRC

or

δ= −

π 2

+ ωRC

ωL

(c) For ω >> 1, equation (1) becomes:

tan δ ≈

Use the expansion for tan−1x to obtain:

tan −1 x =

*

π

R

or δ ≈ tan −1

π

−

2

ωL R

π R 1 or δ = − x 2 ωL

You can easily confirm this using your graphing calculator.

78 •• Picture the Problem We can use the definition of the power factor to express cosδ in the absence of an inductor and simplify the resulting equation to obtain the equation given above. (a) Express the power factor for a series RLC circuit:

cos δ =

R = Z

R R2 + (X L − X C )

With no inductance in the circuit, XL = 0 and:

cos δ =

R = Z

R + (− X C )

Substitute for XC and simplify to obtain:

cos δ =

2

R

R R2 +

=

2

2

1 (ωC )2

R

= R 1+

1 (ωRC )2

ωRC 1 + (ωRC )

2

(b)A spreadsheet program to generate the data for a graph of cosδ versus ωRC is shown below. The formulas used to calculate the quantities in the columns are as follows:

764 Chapter 29 F o r m u l a / C o n t e n t

A l g e b r a i c

A 2

0 . 0

ω

A 3

A 2

ω

C e l l

F o r m

R C R C

+ + 0 . 5 B 2

0 . 5

ωRC

A 2 / ( 1

1 + (ωRC )

2

+ A 2 ^ 2 ) ^ ( 0 . 5 ) A 1 2

B R= 1 C= 1

C ohm F

Alternating-Current Circuits 3 4 5 6 7

omega 0.0 0.5 1.0

cos(delta) 0.000 0.447 0.707

13 14 15

4.0 4.5 5.0

0.970 0.976 0.981

765

The following graph of cos δ as a function of ω was plotted using the data in the above table. Note that both R and C were set equal to 1. 1.0

cosine(delta)

0.8

0.6

0.4

0.2

0.0 0

1

2

3

4

5

omega

*79 •• Picture the Problem We can find the rms current in the circuit and then use it to find the potential differences across each of the circuit elements. We can use phasor diagrams and our knowledge the phase shifts between the voltages across the three circuit elements to find the voltage differences across their combinations. (a) Express the potential difference between points A and B in terms of Irms and XL: Express Irms in terms of ε and Z:

V AB = I rms X L

I rms =

ε Z

=

(1)

ε R2 + (X L − X C )

2

766 Chapter 29 Evaluate XL and XC to obtain:

(

)

X L = 2πfL = 2π 60 s −1 (137 mH ) = 51.6 Ω and

1 1 = −1 2πfC 2π 60 s (25 µF) = 106.1Ω

XC =

Substitute numerical values and evaluate Irms:

I rms =

(

)

115 V

(50 Ω )2 + (51.6 Ω − 106.1Ω )2

= 1.55 A Substitute numerical values in equation (1) and evaluate VAB: (b) Express the potential difference between points B and C in terms of Irms and R: (c) Express the potential difference between points C and D in terms of Irms and XC:

VAB = (1.55 A )(51.6 Ω ) = 80.0 V

VBC = I rms R = (1.55 A )(50 Ω ) = 77.5 V VCD = I rms X C = (1.55 A )(106.1Ω ) = 164 V

(d) The voltage across the inductor lags the voltage across the resistor as shown in the phasor diagram to the right:

Use the Pythagorean theorem to find VAC:

(e) The voltage across the inductor lags the voltage across the resistor as shown in the phasor diagram to the right:

2 2 VAC = VAB + VBC

=

(80.0 V )2 + (77.5 V )2

= 111 V

Alternating-Current Circuits Use the Pythagorean theorem to find VBD:

767

2 2 VBD = VCD + VBC

=

(164 V )2 + (77.5 V )2

= 181V

80 •• Picture the Problem We can use Pav = ε rms I rms cos δ to find the power supplied to the 2 circuit and Pav = I rms R to find the resistance. In (c) we can relate the capacitive reactance

to the impedance, inductive reactance, and resistance of the circuit and solve for the capacitance C. We can use the condition on XL and XC at resonance to find the capacitance or inductance you would need to add to the circuit to make the power factor equal to 1. (a) Express the power supplied to the circuit in terms of εrms, Irms, and the power factor cosδ: Substitute numerical values and evaluate Pav:

Pav = ε rms I rms cos δ

Pav = (120 V )(11 A ) cos 45° = 933 W

Pav 2 I rms

(b) Relate the power dissipated in the circuit to the resistance of the resistor:

2 Pav = I rms R or R =

Substitute numerical values and evaluate R:

R=

933 W = 7.71Ω (11A )2

(c) Express the capacitance of the capacitor in terms of its reactance:

C=

1 1 = ωX C 2πfX C

Relate the capacitive reactance to the impedance, inductive reactance, and resistance of the circuit: Express the impedance of the circuit in terms of the rms emf ε and the rms current Irms: Substitute to obtain:

(1)

Z 2 = R2 + (X L − X C )

2

Z2 =

ε2 I

2 rms

ε2 2 I rms

= R 2 + (X L − X C )

2

768 Chapter 29 Solve for X L − X C :

Note that because I leads ε, the circuit is capacitive and XC > XL. Hence:

ε2

X L − XC =

X L − X C = −( X L − X C ) and

ε2

XC = X L +

I

2 rms

− R2

ε2

= 2πfL +

Substitute numerical values and evaluate XC:

− R2

2 I rms

I

(

2 rms

− R2

)

X C = 2π 60 s −1 (0.05 H ) +

(120 V )2 − (7.71Ω )2 (11A )2

= 26.6 Ω Substitute in equation (1) and evaluate C: (d) Express the relationship between XL and XC when cosδ = 1: Because XL = 18.1 Ω, we could make X L = X C by adding 7.75 Ω of

C=

2π (60 s

1 −1

)(26.6 Ω) =

99.7 µF

X L = XC

L=

7.75 Ω XL = = 20.6 mH 2πf 2π (60 s −1 )

inductive reactance to the circuit. Find the series inductance equivalent to 7.75 Ω of inductive reactance: Alternatively, we could make X L = X C by reducing the capacitive

1 1 1 = + 18.1Ω 26.6 Ω X C

reactance by 7.75 Ω. Find the capacitive reactance that you have to added in parallel to the existing capacitive reactance to reduce the equivalent capacitive reactance by 7.75 Ω:

and

X C = 56.6 Ω

Alternating-Current Circuits Find the capacitance corresponding to a capacitive reactance of 56.6 Ω:

C=

769

1 1 = −1 2πfX C 2π 60 s (56.6 Ω )

(

)

= 46.9 µF 81 •• Picture the Problem We can find XC using the equation relating XC, XL, R, and tanδ and then solve the defining equation for XC for C.

1 1 = ωX C 2πfX C

Express the capacitance of the circuit in terms of its capacitive reactance:

C=

Express the phase angle δ in terms of XL, XC, and R:

tan δ =

Solve for XC to obtain:

X C = 2πfL − R tan δ

Substitute numerical values and evaluate XC: Substitute numerical values and evaluate C:

X L − XC R

X C = 2π (500 s −1 )(0.15 H ) − (35 Ω ) tan 75° = 341 Ω C=

1 = 0.933 µF 2π 500 s −1 (341Ω )

(

)

82 •• Picture the Problem We can use the condition on XL and XC at resonance to find f0. By expressing the phase angle δ in terms of XL, XC, and R we can obtain a quadratic equation in ω that we can solve for the frequencies corresponding to the given phase angles. We can then use these frequencies to express the ratios of f to f0 for the given phase angles. (a) Relate XC and XL at resonance:

X L = XC or

2πf 0 L =

1 2πf 0C

Solve for f0:

f0 =

1 2π

1 LC

Substitute numerical values and evaluate f0:

f0 =

1 2π

1 = 120 Hz (0.35 H )(5 µF)

770 Chapter 29 (b) Express the phase angle δ in terms of XL, XC, and R:

tan δ =

X L − XC R

or

R tan δ = ωL −

Rewrite this equation explicitly as a quadratic equation to obtain: Substitute numerical values and simplify to obtain:

1 ωC

LCω 2 − (RC tan δ )ω − 1 = 0

(1.75 ×10

−6

)

F⋅H ω2

[(

)

]

− 2 × 10 −3 Ω ⋅ H tan δ ω − 1 = 0 or

(1F ⋅ H )ω 2 − [(1.14 ×103 Ω ⋅ H )tan δ ]ω − 5.71× 105 = 0

For δ = 60°:

(1F ⋅ H )ω 2 − (1.97 ×103 Ω ⋅ H )ω − 5.71× 105 = 0

Solve for the positive value of ω:

ω = 2.23 ×103 s −1 and

f =

Calculate the ratio f/f0:

For δ = −60°:

ω 2.23 ×103 s −1 = = 355 Hz 2π 2π

f 355 Hz = = 2.96 f 0 120 Hz

(1F ⋅ H )ω 2 + (1.97 ×103 Ω ⋅ H )ω − 5.71× 105 = 0

Solve for the positive value of ω and then for f:

ω = 256 s −1 and

f =

Calculate the ratio f/f0:

ω 256 s −1 = = 40.7 Hz 2π 2π

f 40.7 Hz = = 0.339 f 0 120 Hz

Remarks: Note that these ratios are reciprocals of each other. 83

••

Alternating-Current Circuits

771

Picture the Problem The impedance for the three circuits as functions of the angular frequency is shown in the three figures below. Also shown in each figure (dashed line) is the asymptotic approach for large angular frequencies. (a)

(c)

(b)

*84 •• Picture the Problem We can substitute for XL and XC in Equation 29-48 and simplify the resulting equation to obtain the given equation for Imax. Equation 29-48 is:

I max =

Substitute for XL and XC to obtain:

I max =

ε max

R2 + (X L − X C )

2

ε max 1 ⎞ ⎛ R + ⎜ ωL − ⎟ ωC ⎠ ⎝

2

2

Simplify algebraically to obtain:

ε max

I max =

=

1 ⎞ ⎛ R 2 + ω 2 L2 ⎜1 − 2 ⎟ ⎝ ω LC ⎠

ε max

1

ω

ω 2 R 2 + L2 (ω 2 − ω

2

)

2 2 0

=

ε max ⎛ ω2 ⎞ R + ω L ⎜⎜1 − 02 ⎟⎟ ⎝ ω ⎠ 2

=

2

ε max

=

2

R2 +

2

L2

ω

2

(ω

2

− ω02 )

2

ωε max

ω 2 R 2 + L2 (ω 2 − ω02 )

2

85 •• Picture the Problem We can use the constraints on L and C at resonance and the given values for XL and XC to obtain simultaneous equations that we can solve for L and C. In (b) we can find Q from its definition and in (c) we can calculate Imax from εmax and Z. (a) Relate XL and XC at resonance:

X L = X C or ω0 L =

1 ω0C

772 Chapter 29 Solve for the product of L and C:

Express XC and XL:

1

LC =

ω

=

2 0

(10

1

4

rad/s )

2

= 10 −8 s 2

(1)

1 = 16 Ω ωC

XC = and

X L = ωL = 4 Ω

Eliminate ω between these equations to obtain:

L = 64 Ω 2 C

Solve equations (1) and (2) simultaneously to obtain:

L = 0.800 mH and C = 12.5 µF

(b) Express Q in terms of R, L, and ω0:

Q=

ω0 L

Substitute numerical values and evaluate Q:

Q=

(10

(c) Relate the maximum current in the circuit to εmax and Z:

I max =

Substitute numerical values and evaluate Imax:

I max =

(2)

R 4

rad/s )(0.800 mH ) = 1.60 5Ω

ε max Z

ε max

=

R + (X L − X C ) 2

2

26 V

(5 Ω ) + (4 Ω − 16 Ω )2 2

= 2.00 A 86 •• Picture the Problem We can find the maximum current in the circuit from the maximum voltage across the capacitor and the reactance of the capacitor. To find the range of 2 inductance that is safe to use we can express Z2 for the circuit in terms of ε max and 2 I max and solve the resulting quadratic equation for L.

(a) Express the maximum current in terms of the maximum potential difference across the capacitor and its reactance:

I max =

VC ,max = ωCVC ,max XC

Alternating-Current Circuits Substitute numerical values and evaluate Imax:

I max = (2500 rad/s )(8 µF)(150 V ) = 3.00 A 2 ε max or Z 2 = ε max

(b) Relate the maximum current in the circuit to the emf of the source and the impedance of the circuit:

I max =

Express Z2 in terms of R, XL, and XC:

Z 2 = R2 + (X L − X C )

Substitute to obtain:

2 ε max

Substitute numerical values to obtain:

2 I max

Z

2

I Evaluate XC:

2 max

XC =

= R2 + (X L − X C )

2

1 1 = = 50 Ω ωC (2500 rad/s )(8 µF)

(200 V )2 = (60 Ω )2 (3 A )2 2 + ((2500 rad/s )L − 50 Ω ) or

[(

)

844 Ω 2 = 2500 s −1 L − 50 Ω Solve for L to obtain:

Denoting the solutions as L+ and L−, find the values for the inductance: Express the ranges for L:

773

L=

]

2

50 Ω ± 844 Ω 2 2500 s −1

L+ = 31.6 mH and L− = 8.38 mH

8.00 mH < L < 8.38 mH and

31.6 mH < L < 40.0 mH 87 •• Picture the Problem We can find the impedance of the circuit from the applied emf and 2 the current drawn by the device. In (b) we can use Pav = I rms R to find R and the definition of the impedance of a series RLC circuit to find X = XL − XC. ε (a) Express the impedance of the Z= I device in terms of the current it

774 Chapter 29 draws and the emf provided by the power line: Substitute numerical values to obtain: (b) Use the relationship between the average power supplied to the device and the rms current it draws to find R:

Express the impedance of a series RLC circuit:

Z=

120 V = 12.0 Ω 10 A

2 Pav = I rms R

and

R=

Pav 720 W = = 7.20 Ω 2 I rms (10 A )2

Z = R2 + (X L − X C )

2

or

Z 2 = R2 + (X L − X C )

2

Solve for X L − X C :

X = X L − X C = Z 2 − R2

Substitute numerical values and evaluate X:

X =

(12 Ω )2 − (7.2 Ω )2

= 9.60 Ω

(c) If the current leads the emf, the reactance is capacitive. *88 •• Picture the Problem We can use the fact that when the current is a maximum, XL = XC, to find the inductance of the circuit. In (b), we can find Imax from εmax and the impedance of the circuit at resonance.

1 ω0C

(a) Relate XL and XC at resonance:

X L = X C or ω0 L =

Solve for L to obtain:

L=

1 ω02C

Substitute numerical values and evaluate L:

L=

(5000 s ) (10 µF)

(b) Noting that, at resonance, X = 0, express Imax in terms of the applied emf and the impedance of the circuit at resonance:

I max =

1

−1 2

ε max Z

=

= 4.00 mH

10 V = 0.100 A 100 Ω

Alternating-Current Circuits

775

89 •• Picture the Problem We can use Ohm’s law to express the current through the resistor as a function of time. Because the resistor and capacitor are in parallel they have the same potential difference across them … the emf of the source. We can relate the charge on the capacitor as a function of time to its capacitance and the potential difference across it and differentiate this expression with respect to time to express IC(t). We can then apply Kirchhoff’s junction rule to express the total current drawn from the source. Using the results of (a) and (b) we can show that I = IR + IC = Imax cos (ωt + δ), where tan δ = R/XC and Imax = εmax/Z with Z−2 = R−2 + XC−2. (a) Apply Ohm’s law to obtain:

I R (t ) = =

(b) Express the potential difference across the capacitor in terms of the instantaneous charge on the capacitor:

VC (t ) =

Differentiate q(t) to express the current to the capacitor:

I C (t ) =

Use the definition of XC and the trigonometric identity cos(α + 90°) = − sin α to obtain:

I C (t ) =

(c) Apply Kirchhoff’s junction rule to obtain:

We know that the current is also expressible in the form:

V (t ) ε max cos ωt = R R

ε max cos ωt R

q(t ) or q (t ) = CVC (t ) C

dq(t ) d = C (VC (t )) dt dt d = C (ε max cos ωt ) dt = −ωCε max sin ωt

ε max cos(ωt + 90°) XC

I = I R + IC =

ε max cos ωt + ε max cos(ωt + 90°)

=

ε max cos ωt − ε max sin ωt

R

XC

R

XC

I = I max cos(ωt + δ )

776 Chapter 29 Expand this expression, using the formula for the cosine of the sum of two angles, to obtain: Equate these expressions and rewrite the resulting equation to obtain:

Express the conditions that must be satisfied if this equation is to be true for all values of t:

I = I max cos ωt cos δ − I max sin ωt sin δ

⎛ ε max ⎞ − I max cos δ ⎟ cos ωt ⎜ ⎝ R ⎠

⎛ε ⎞ − ⎜⎜ max − I max sin δ ⎟⎟ = 0 ⎝ XC ⎠

ε max − I R

cos δ = 0

max

sin δ = 0

and

ε max − I XC

Rewrite these equations as:

max

I max sin δ =

ε max

and

I max cos δ = Divide equation (1) by equation (2) and simplify to obtain: Square equations (1) and (2) and add to obtain:

tan δ =

(1)

XC

ε max

(2)

R

R XC

2 2 I max sin 2 δ + I max cos 2 δ

(

2 sin 2 δ + cos 2 δ = I max

)

⎛ε ⎞ ⎛ε ⎞ = ⎜⎜ max ⎟⎟ + ⎜ max ⎟ ⎝ XC ⎠ ⎝ R ⎠ 2

=I

2 max

=ε or

I max =

2 max

2

2 ⎛ 1 1 ⎞ ε max ⎜⎜ 2 + 2 ⎟⎟ = 2 ⎝ XC R ⎠ Z

ε max Z

where Z −2 = X C−2 + R −2 *90 •• Picture the Problem Because we’ll need to use it repeatedly in solving this problem, we’ll begin by using complex numbers to derive an expression for the impedance Zp of

Alternating-Current Circuits

777

the parallel combination of C with L and RL in series. The total impedance of the circuit is then Z = R + Zp. We can apply Kirchhoff’s loop rule to obtain expressions for the voltages across the load resistor with S either open or closed. Use complex numbers to relate Zp to RL, XL, and XC:

1 1 1 = + Z p − iX C RL + iX L =

RL + i ( X L − X C ) X C X L − iRL X C

Zp =

X C X L − iRL X C RL + i ( X L − X C )

Multiple the numerator and denominator of this fraction by the complex conjugate of RL + i( XL − XC):

Zp =

X C X L − iRL X C RL − i ( X L − X C ) RL + i ( X L − X C ) RL − i ( X L − X C )

Simplify to obtain:

Zp =

RL X C2 2 RL2 + ( X L − X C )

or

[

]

X C RL2 + X L ( X L − X C ) −i 2 RL2 + ( X L − X C ) (a) S is closed. Because L is shorted: Evaluate XC:

Substitute numerical values in equation (1) and evaluate Zp, Z, Z , and δ:

XL = 0

1 1 = 2πfC 2π 10 s −1 (8 µF) = 1.99 kΩ

XC =

(

)

Z p = 30 Ω − i (0.452 Ω ) , Z = 40 Ω − i (0.452 Ω ) ,

and

Z = In Problem 29-77 we showed that for a parallel combination of a resistor and capacitor, the phase angle δ is given by:

(1)

(40 Ω )2 + (0.452 Ω )2 ⎛ R ⎝ XC

δ = tan −1 ⎜⎜

⎞ ⎟⎟ ⎠

= 40.0 Ω

778 Chapter 29 Substitute numerical values and evaluate δ:

⎛

40 Ω ⎞ ⎟⎟ = − 89.4° ⎝ − 0.452 Ω ⎠

δ = tan −1 ⎜⎜

No phasor diagram is shown because it is impossible to represent it to scale. (b) S is open; i.e., the inductor is in the circuit. Find XL:

(

)

X L = ωL = 2πfL = 2π 10 s −1 (0.15 H ) = 9.42 Ω Z p = 30.3 Ω + i (9.01 Ω ) ,

Substitute numerical values in equation (1) and evaluate Zp, Z, Z ,

Z = 40.3 Ω + i (9.01 Ω ) ,

and δ:

Z =

(40.3 Ω )2 + (9.01 Ω )2

= 41.3 Ω

and

⎛X⎞ −1 ⎛ 9.01 Ω ⎞ ⎟⎟ ⎟ = tan ⎜⎜ ⎝R⎠ ⎝ 40.3 Ω ⎠

δ = tan −1 ⎜

= 12.6° The phasor diagram for this case is shown to the right.

(c) S is closed. Apply Kirchhoff’s loop rule to a loop including the source, R, and RL:

ε − IR − VR

L

Solve for VRL :

VRL = ε − IR

Express the current I in the circuit:

I=

Substitute and simplify to obtain:

From (a) we have:

=0

ε Z

VRL = ε −

εR = ⎛⎜1 − Z

⎜ ⎝

R Z

⎞ ⎟ε max cos(ωt − δ ) ⎟ ⎠

Z p = 30 Ω − i (0.452 Ω ) , Z = 40 Ω − i (0.452 Ω ) , Z = 40.0 Ω , and

Alternating-Current Circuits ⎛ − 0.452 ⎞ ⎟ = −0.647° ≈ 0° ⎝ 40 ⎠

δ = tan −1 ⎜ Substitute numerical values to obtain:

⎛ 10 Ω ⎞ ⎟⎟ (100 V ) cos 20 s −1 πt VRL = ⎜⎜1 − 40 Ω ⎠ ⎝

[(

= S is open. Apply Kirchhoff’s loop rule to a loop including the source, R, L, and RL when S is open:

) ]

(75 V ) cos[(20 s −1 )πt ]

ε − IR − IX L − VR

L

=0

Solve for VRL :

VRL = ε − IR − IX L = ε − I (R + X L )

Express the current I in the circuit:

I=

Substitute to obtain:

Substitute numerical values and evaluate Zp and Z:

ε Z

⎛ R + XL ⎞ ⎟ε max cos(ωt − δ ) VRL = ⎜⎜1 − ⎟ Z ⎝ ⎠ Z p = 30.3 Ω + i (9.01 Ω ) ,

Z = 40.3 Ω + i (9.01 Ω ) , Z = 41.3 Ω , and

⎛ XL ⎞ ⎛ 9.42 Ω ⎞ ⎟⎟ = tan −1 ⎜⎜ ⎟⎟ ⎝ 40.3 Ω ⎠ ⎝ R + RL ⎠

δ = tan −1 ⎜⎜ = 13.2° Substitute numerical values and evaluate VRL :

⎛ 10 Ω + 9.42 Ω ⎞ ⎟⎟ VRL = ⎜⎜1 − 41.3 Ω ⎝ ⎠ × (100 V )cos 20 s −1 πt − 13.2° =

(d) Find XL and XC when f = 1000 Hz:

[( ) ] (53.0 V )cos[(20 s )πt − 13.2°] −1

(

)

X L = 2π 1000 s −1 (0.15 H ) = 942 Ω and

XC =

1 = 19.9 Ω 2π 1000 s −1 (8 µF)

(

)

779

780 Chapter 29 S is closed. XL = 0, and Zp simplifies to:

RL X C2 RL2 X C Zp = 2 −i 2 RL + X C2 RL + X C2 Z p = 9.17 Ω − i (13.8 Ω ) ,

Substitute numerical values in equation (1) and evaluate Zp, Z, Z ,

Z = 19.17 Ω − i (13.8 Ω ) ,

and δ:

Z =

(19.17 Ω )2 + (13.8 Ω )2

= 23.6 Ω

and

⎛ − 13.8 Ω ⎞ ⎟⎟ = − 35.7° ⎝ 19.17 Ω ⎠

δ = tan −1 ⎜⎜ A phasor diagram for this circuit is shown to the right,

S is open. Substitute numerical values in equation (1) and evaluate Zp, Z, Z , and δ:

Z p = 0.0140 Ω − i (20.3 Ω ) ,

Z = 10.0 Ω − i (20.3 Ω ) , Z =

(10.0 Ω )2 + (20.3 Ω )2

= 22.6 Ω

and Find the total impedance, its magnitude, and phase angle for the circuit:

Z = 10.0 Ω − i (20.4 Ω ) ,

Z=

(10.0 Ω )2 + (20.4 Ω )2

= 22.7 Ω

and

⎛ − 20.4 Ω ⎞ ⎟⎟ = − 63.9° ⎝ 10 Ω ⎠

δ = tan −1 ⎜⎜ The phasor diagram is shown to the right.

(e)

The load voltage at the higher frequency is much more attenuated with S open, while opening S does not reduce the low frequency load voltage significantly. Therefore, S open is the better arrangement for a low - pass filter. 91

••

Alternating-Current Circuits

781

Picture the Problem We can find the resonant frequency of any parallel ac circuit by setting the imaginary part of the reciprocal of the impedance equal to zero. In (b) we can use complex numbers to find the impedance of each branch of the circuit and then relate the common potential difference across each branch to its impedance and the current in the resistors. (a) Express the reciprocal of the impedance of the circuit: Rewrite this expression with a common denominator and simplify to obtain:

1 1 1 = + Z R1 − iX C R2 + iX L

(R1 + R2 ) + i( X L − X C ) 1 = Z (R1 R2 + X C X L ) + i (R1 X L − R2 X C )

Multiply this expression by 1 in the form of the complex conjugate of the denominator divided by itself and simplify (separate the real part of the expression from the imaginary part) to obtain:

1 (R1 + R2 )(R1R2 + X C X L ) + ( X L − X C )(R1 X L − R2 X C ) = Z (R1R2 + X C X L )2 + (R1 X L − R2 X C )2 +i

( X L − X C )(R1R2 + X C X L ) − (R1 + R2 )(R1 X L − R2 X C ) (R1R2 + X C X L )2 + (R1 X L − R2 X C )2

Set the imaginary part of 1/Z equal to zero to obtain:

( X L − X C )(R1R2 + X C X L ) − (R1 + R2 )(R1 X L − R2 X C ) = 0 Substitute numerical values for R1 and R2 (suppress the units to save space and make the resulting equation more readable) to obtain:

Simplify this equation by clearing the fractions and combining like terms to obtain: Solve for ω0:

Substitute numerical values for L and C and evaluate ω0:

⎛ 1 ⎞⎛ L ⎜⎜ ω0 L − ⎟⎟⎜ 8 + ⎞⎟ ω0C ⎠⎝ C ⎠ ⎝ ⎛ 4 ⎞ ⎟=0 − 6⎜⎜ 2ω0 L − ω0C ⎟⎠ ⎝

(8LC ω0 =

2

)

+ L2C − 12 LC 2 ω02 = L − 16C

L − 16C 8 LC + L2C − 12 LC 2 2

1.15 × 10 −2 ω0 = = 1.64 × 103 rad/s −9 4.28 × 10

782 Chapter 29

ε

(b) Express the currents in each branch at resonance:

IC =

Evaluate ZC,res and Z C ,res :

Z C ,res = 2 Ω − i

ZC

and I L =

ε ZL

1

(1.64 ×10 s )(30 ×10 F) 3

−1

−6

= 2 Ω − i (20.3 Ω ),

Substitute to obtain:

Z C ,res =

(2 Ω )2 + (20.3 Ω )2

I C ,rms =

40 V = 1.39 A 2 (20.4 Ω )

= 20.4 Ω ,

and

⎛ − 20.3 Ω ⎞ ⎟⎟ = − 84.4° ⎝ 2Ω ⎠

δ C = tan −1 ⎜⎜ Evaluate ZL,res and Z L ,res :

Substitute to obtain:

(

)(

Z L ,res = 4 Ω + i 1.64 × 103 s −1 12 × 10 −3 H = 4 Ω + i (19.7 Ω ),

Z L ,res =

(4 Ω )2 + (19.7 Ω )2

I L ,rms =

40 V = 1.41 A 2 (20.1 Ω )

)

= 20.1 Ω ,

and

⎛ 19.7 Ω ⎞ ⎟⎟ = 78.5° ⎝ 4Ω ⎠

δ L = tan −1 ⎜⎜ Express and evaluate the rms current supplied by the source:

I rms = I L ,rms cos δ L + I C ,rms cos δ C = (1.41 A ) cos 78.5° + (1.39 A ) cos(− 84.4°) = 0.417 A

92 •• Picture the Problem We can use its definition to express Q in terms of ω0 and ∆ω. By expressing the current drawn from the source we can obtain an expression for the energy stored in the system each cycle and then use this result to establish the relationship between ω, R, L, and C when the energy stored per cycle is at half-maximum. Finally, we

Alternating-Current Circuits can solve the resulting equation for the values of ω that will allow us to determine ∆ω. The definition of Q is:

ω0 ∆ω

Q=

where ∆ω is the width of the resonance at half maximum. Express the resonance frequency of the circuit:

ω0 =

Substitute to obtain:

Q=

Express the current to the capacitor:

IC =

1 LC 1 LC ∆ω

(1)

V = ωCV XC

with IC leading V by 90°. Express the current in the inductor:

IL =

V V = X L ωL

with IL lagging V by 90°. Express the current in the resistor:

IR =

V R

with IR in phase with V. Express the total current drawn from the source:

2

V ⎛1⎞ ⎛ 1 ⎞ I = =V ⎜ ⎟ +⎜ − ωC ⎟ Z ⎝ R ⎠ ⎝ ωL ⎠ V ⎛ 1 ⎞ = 1 + R2 ⎜ − ωC ⎟ R ⎝ ωL ⎠

At resonance, the reactive term is zero and the total current is the current in the resistor: Substitute to obtain:

Express the total energy stored in the circuit per cycle:

I0 =

V R

I = I0

U tot

2

⎛ 1 ⎞ 1+ R ⎜ − ωC ⎟ ⎝ ωL ⎠

Q02 = 2C

2

2

2

783

784 Chapter 29 where Q0 is the maximum charge on the capacitor. Relate the maximum value of the current to the maximum value of the charge:

I max = ωQ0

2 I max 1 V2 = 2ω 2C 2ω 2C R 2 I 02 1 = 2 2ω 2C ⎞ 2⎛ 1 1+ R ⎜ − ωC ⎟ ⎝ ωL ⎠

Substitute to obtain:

U tot =

At resonance we have:

U tot,res =

I 02

2ω 2C

I2 1 U tot,res = 02 2 4ω C 1 = 2ω 2C

At half Utot,res:

I 02 ⎛ 1 ⎞ 1+ R ⎜ − ωC ⎟ ⎝ ωL ⎠ 2

or

1 = 2

⎛ ⎝

Solve for R⎜ ωC −

1 ⎞ ⎟ to obtain: ωL ⎠

1 ⎛ 1 ⎞ 1+ R ⎜ − ωC ⎟ ⎝ ωL ⎠

2

2

1 ⎞ ⎛ R ⎜ ωC − ⎟ = ±1 ωL ⎠ ⎝

(2)

Rewrite equation (2) explicitly as a quadratic equation:

RLCω 2 ± Lω − R = 0

Letting + denote the roots with a positive coefficient of ω and − the roots with a negative coefficient, solve this equation for ω+ and ω−:

1 1 ⎛ 1 ⎞ ω+ = ± ⎜ ⎟ + 2 RC LC ⎝ 2 RC ⎠

2

and 2

1 1 ⎛ 1 ⎞ ω− = − ± ⎜ ⎟ + 2 RC LC ⎝ 2 RC ⎠

2

Alternating-Current Circuits Express ∆ω:

Substitute in equation (1) to obtain:

∆ω = ω+ − ω− =

785

1 RC

RC C = R L LC

Q=

93 •• Picture the Problem We can use the expression for the resonance frequency derived by equating the capacitive and inductive reactances at resonance to express ω0 in terms of L and C. In (b) we can use the result derived in Problem 92 to find R from Q, L, and C. (a) Express the resonance frequency ω0 in terms of L and C:

ω0 =

Solve for C to obtain:

C=

Substitute numerical values and evaluate C:

C=

1 LC 1

ω L 2 0

=

1 4π f 2 L

(

2

1

4π 4 × 10 s −1 2

3

) (4 mH) 2

= 0.396 µF (b) From Problem 92 we have:

Q=R

C L

Solve for R to obtain:

R=Q

L C

Substitute numerical values and evaluate R:

R =8

4 mH = 804 Ω 0.396 µF

786 Chapter 29 94 •• Picture the Problem We can use the expression for the resonance frequency derived by equating the capacitive and inductive reactances at resonance to express ω0 in terms of L and C. We can use the result derived in Problem 92 to find the Q-value resulting from halving the capacitance and to find the resistance necessary to give Q = 8. Express the resonance frequency ω0 in terms of L and C:

ω0 =

Substitute numerical values and evaluate f0:

f0 =

1 1 or f 0 = LC 2π LC

2π

1 2

1 (4 mH )(0.396 µF)

= 5.66 kHz From Problem 92 we have:

Q=R

C L

Letting C′represent the halved capacitance, express Q′:

Q' = R

C' L

Divide Q′ by Q and simplify to obtain:

C' Q' L = C' = C Q C R L

Because C ' = 12 C :

(1)

R

Q' =

Q 8 = = 5.66 2 2

Solve equation (1) for R to obtain:

R=Q

Substitute numerical values and evaluate R:

R =8

L C

1 2

4 mH = 1.14 kΩ (0.396 µF)

95 •• Picture the Problem We can use its definition to find the resonance frequency of this series RLC circuit and the fact that, at resonance, Z = R, to find the resonance current. Because, at resonance VL = VC, we can find the voltage across either element from the product of the current and its reactance. In (c) we can use the definition of the Q factor to find the angular frequency corresponding to f = f 0 + 12 ∆f and then use this result to

Alternating-Current Circuits

787

find XL, XC, and Z at this frequency. Finally, we can use these values for XL, XC, and Z to find the rms current and the rms voltages across the inductor and capacitor. (a) Express the resonance frequency f0 in terms of L and C:

f0 =

Substitute numerical values and evaluate f0:

f0 =

(b) At resonance, Z = R and:

I =

Express and evaluate the equal (at resonance) rms voltages across the capacitor and the inductor:

1 2π LC 1 = 13.26 kHz 2π (36 mH )(4 nF)

ε R

=

VC = VL = IX L = ω0 IL = 2πf 0 IL

= 2π (13.26 kHz )(0.2 A )(36 mH ) = 600 V

(c) Express the rms current in the circuit and the rms voltages across the inductor and capacitor:

I=

ε ,V

Express the Q factor for an RLC circuit:

Q=

ω0 L

Solve for ∆f:

∆f =

Express f:

f = f0 +

Substitute numerical values and evaluate f and ω:

20 V = 200 mA 100 Ω

Z

R

L

≈

= IX L , and VC = IX C

ω0 f = 0 ∆ω ∆f

R f ω0 L 0 ⎛ R ⎞ ⎟⎟ f 0 = f 0 ⎜⎜1 + 2ω0 L ⎝ 2ω0 L ⎠ R

f = (13.26 kHz ) ⎛ ⎞ 100 Ω ⎟⎟ × ⎜⎜1 + ( )( ) π 4 13 . 26 kHz 36 mH ⎝ ⎠ = 13.48 kHz and

ω = 2πf = 2π (13.48 kHz ) = 84.7 krad/s

Calculate XL and XC at 84.7 krad/s:

X L = ωL = (84.7 krad/s)(36 mH ) = 3.05 kΩ

788 Chapter 29 and

1 1 = ωC (84.7 krad/s)(4 nF) = 2.95 kΩ

XC =

Now we can find Z:

Z = R2 + (X L − X C )

2

=

(100 Ω )2 + (3.05 kΩ - 2.95 kΩ )2

= 141 Ω Substitute numerical values and evaluate I, VL, and VC:

I=

20 V = 142 mA , 141 Ω

VL = (142 mA )(3.05 kΩ ) = 433 V , and

VC = (142 mA )(2.95 kΩ ) = 419 V

96 ••• Picture the Problem We can use complex numbers to find the impedance in the branches of the given circuit. We can then use Kirchhoff’s loop rule to find the currents in the branches and a current phasor diagram to find the total current and its phase relative to the applied voltage. (a) Use complex numbers to find ZL, Z L , and δL:

Z L = R2 + iX L = 40 Ω + i (30 Ω ) ,

Z L = R22 + X L2 =

(40 Ω )2 + (30 Ω )2

= 50.0 Ω

and

⎛ XL ⎞ −1 ⎛ 30 Ω ⎞ ⎟⎟ = 36.9° ⎟ = tan ⎜⎜ ⎝ R ⎠ ⎝ 40 Ω ⎠

δ L = tan −1 ⎜

Use complex numbers to find ZC, Z C , and δC:

Z C = R1 + iX C = 10 Ω − i (10 Ω ) , Z C = R12 + X C2 =

(10 Ω )2 + (10 Ω )2

= 14.1 Ω

and

⎛ XC ⎞ −1 ⎛ − 10 Ω ⎞ ⎟⎟ ⎟ = tan ⎜⎜ ⎝ R ⎠ ⎝ 10 Ω ⎠

δ C = tan −1 ⎜

= − 45.0°

Alternating-Current Circuits (b) Apply Kirchhoff’s loop rule to the source and the inductive branch to obtain:

789

V − IL X L = 0 or

V 110 V = Z L 50 Ω

IL =

= 2.20 A lagging the voltage by 36.9° Apply Kirchhoff’s loop rule to the source and the capacitive branch to obtain:

V − IC X C = 0 or

IC =

110 V V = Z C 1 4 .1 Ω

= 7.80 A leading the voltage by 45.0° (c) The current phasor diagram is shown to the right.

Express the total current in terms of its horizontal and vertical components:

2 2 I = I hor + I vert

and

⎛ I vert ⎝ I hor

δ = tan −1 ⎜⎜ Find the horizontal component Ihor of the total current:

⎞ ⎟⎟ ⎠

I hor = I C cos δ C + I L cos δ L

= (7.80 A ) cos 45° + (2.20 A ) cos 36.9° = 7.27 A

Find the vertical component Ivert of the total current:

I vert = I C sin δ C − I L sin δ L

= (7.80 A )sin 45° − (2.20 A )sin 36.9° = 4.19 A

Substitute numerical values and evaluate I and δ:

I=

(4.19 A )2 + (7.27 A )2

= 8.39 A

and

⎛ 4.19 A ⎞ ⎟⎟ = 30.0° ⎝ 7.27 A ⎠

δ = tan −1 ⎜⎜

790 Chapter 29 Remarks: The total current leads the applied voltage by 30.0°. *97 ••• Picture the Problem We can manipulate Equation 29-47 into a form that has the ratio of L to R in it and then use the definition of Q to eliminate L and R. In (b) we can approximate ω 2 − ω02 , near resonance, as 2ω0 ∆ω and substitute in the result from (a) to obtain the desired result. (a) From Equation 29-47:

Express Q in terms of ω0, L and R:

Solve for L/R to obtain:

Substitute to obtain:

(b) Near resonance:

Substitute in equation (1) to obtain:

ω2L −1 C R ωR 2 L(ω − 1 LC ) L(ω 2 − ω02 ) = = ωR ωR

tan δ =

Q=

ωL − 1 ωC

=

ω0 L R

L Q = R ω0 tan δ =

(

Q ω 2 − ω02

ωω0

)

(1)

ω 2 − ω02 = (ω + ω0 )(ω − ω0 ) ≈ 2ω0 ∆ω tan δ =

Q(2ω0 ∆ω )

ωω0

=

2Q(ω − ω0 )

ω

(c) A following graph of δ as a function of x = ω/ω0 was plotted using a spreadsheet program. The solid curve is for a high-Q circuit and the dashed curve is for a low-Q circuit.

Alternating-Current Circuits

791

2.0 1.5 Q=10 Q=2

1.0

delta

0.5 0.0 -0.5 -1.0 -1.5 -2.0 0.0

0.5

1.0

1.5

2.0

x

98 ••• Picture the Problem We can rewrite Equation 29-45 in terms of the current and then differentiate Equation 29-46. Substituting for I, dI/dt, XL, and XC will allow us to use the trigonometric identities for the sine and cosine of the sum of two angles to rewrite the equation in such a form that we can equate the coefficients of sinωt and cosωt to obtain Equation 29-47 and an equation that is satisfied provided Z is given by Equation 29-49.

1 dI + RI + ∫ Idt = ε max cos ωt dt C

Rewrite Equation 29-45 in terms of the current:

L

Equation 29-46 is:

I = I max cos(ωt − δ )

Differentiate Equation 29-50 with respect to time to obtain:

dI = −ωI max sin (ωt − δ ) dt

Evaluate

∫ Idt :

∫ Idt = I ∫ cos(ωt − δ ) dt max

=

Substitute to obtain:

I max

ω

sin (ωt − δ )

L(− ωI max sin (ωt − δ ))

+ RI max cos(ωt − δ ) +

1 ⎛ I max ⎞ sin (ωt − δ )⎟ = ε max cos ωt ⎜ C⎝ ω ⎠

792 Chapter 29 Divide through by Imax to obtain:

L(− ω sin (ωt − δ ))

+ R cos(ωt − δ ) +

Use the definitions of XL, XC, and Z to obtain:

1⎛1 ⎞ ε ⎜ sin (ωt − δ )⎟ = max cos ωt C ⎝ω ⎠ I max

− X L sin (ωt − δ ) + R cos(ωt − δ )

+ X C sin (ωt − δ ) = Z cos ωt

Use the trigonometric identities for sin (α + β ) and cos(α + β ) to obtain:

− X L (sin ωt cos δ − cos ωt sin δ ) + R(cos ωt cos δ + sin ωt sin δ ) + X C (sin ωt cos δ − cos ωt sin δ ) = Z cos ωt

Collect the terms in sinωt and cosωt:

(− X L cos δ + R sin δ + X C cos δ ) sin ωt + (R cos δ − X C sin δ + X L sin δ ) cos ωt = Z cos ωt Equate the coefficients of sinωt and cosωt to obtain:

− X L cos δ + R sin δ + X C cos δ = 0 and

R cos δ − X C sin δ + X L sin δ = Z

Solve the first of these equations for tanδ: Rewrite the second equation as:

tan δ =

X L − XC R

Equation 29-47

R − X C tan δ + X L tan δ =

Z cos δ

or

( X L − X C ) tan δ + R = Simplify this equation to obtain Equation 29-49:

Z=

Z cos δ

R2 + (X L − X C )

2

99 ••• Picture the Problem In (a) we can apply Kirchhoff’s loop rule to obtain the 2nd order differential equation relating the charge on the capacitor to the time. In (b) we’ll assume a solution of the form Q = Qmax cos ωt , differentiate it twice, and substitute for d2Q/dt2 and Q to show that the assumed solution satisfies the DE provided

Alternating-Current Circuits Qmax = −

ε max . In (c) we’ll use our results from (a) and (b) to establish the for L(ω 2 − ω 02 )

Imax given in the problem statement. (a) Apply Kirchhoff’s loop rule to obtain: Substitute for ε and rearrange the differential equation to obtain:

ε − Q − L dI C

L

dt

=0

dI Q + = ε max cos ωt dt C

Because I = dQ dt :

L

d 2Q Q + = ε max cos ωt dt 2 C

(b) Assume that the solution is:

Q = Qmax cos ωt

Differentiate the assumed solution twice to obtain:

dQ = −ωQmax sin ωt dt and

d 2Q = −ω 2Qmax cos ωt 2 dt Qmax cos ωt C = ε max cos ωt

Substitute in the differential equation to obtain:

− ω 2 LQmax cos ωt +

Factor cosωt from the left-hand side of the equation:

Q ⎞ ⎛ 2 ⎜ − ω LQmax + max ⎟ cos ωt C ⎠ ⎝ = ε max cos ωt Q − ω 2 LQmax + max = ε max C

If this equation is to hold for all values of t it must be true that:

or

Qmax =

Factor L from the denominator and substitute for 1/LC to obtain:

Qmax =

=

ε max − ω 2L +

1 C

ε max 1 ⎞ ⎛ L⎜ − ω 2 + ⎟ LC ⎠ ⎝

− ε max

(

L ω 2 − ω02

)

793

794 Chapter 29 (c) From (a) and (b) we have:

I=

dQ = −ωQmax sin ωt dt

ωε max sin ωt L(ω 2 − ω02 ) = I max sin ωt =

where

I max =

ωε max ε max = 2 2 L 2 L ω − ω0 ω − ω02 ω

ε max

=

ωL −

=

1 ωC

ε max

X L − XC

If ω > ω0, XL > XC and the current lags the voltage by 90°. Therefore:

I = I max sin ωt = I max cos(ωt − 90°)

If ω < ω0, XL < XC and the current leads the voltage by 90°. Therefore:

I = − I max sin ωt = I max cos(ωt + 90°)

100 ••• Picture the Problem We can use the condition determining the half-power points to obtain a quadratic equation that we can solve for the frequencies corresponding to the half-power points. Expanding these solutions binomially will lead us to the result that ∆ω = ω2 −ω1 ≈ R/L. We can then use the definition of Q to complete the proof that Q ≈ ω0 /∆ω. Equation 29-58 is:

Pav =

(

2 ε rms Rω 2

L2 ω 2 − ω02

)

2

+ ω 2R2

The half-power points will occur when the denominator is twice the value near resonance, that is, when:

L2(ω2 −ω0 2)2 = ω2R2 ≈ ω02R2 or

Let ω1 and ω2 be the solutions of this equation. Then:

⎛L⎞ 2 2 ⎜ ⎟ ω1 − ω0 R ⎝ ⎠

2

(

)

= ω02

(

)

= ω02

(

)

= ω02

⎛L⎞ 2 2 ⎜ ⎟ ω − ω0 ⎝R⎠ 2

2

2

and 2

⎛L⎞ 2 2 ⎜ ⎟ ω 2 − ω0 R ⎝ ⎠

2

Alternating-Current Circuits Solve these equations for ω1 and ω2 to obtain:

795

12

⎛ R ⎞ ⎟⎟ ω1 = ω0 ⎜⎜1 − ω L 0 ⎝ ⎠ and

12

⎛ R ⎞ ⎟⎟ ω2 = ω0 ⎜⎜1 + ⎝ ω0 L ⎠ Expand these solutions binomially to obtain:

⎛

ω1 = ω0 ⎜⎜1 − ⎝

⎞ + higher order terms ⎟⎟ 2ω0 L ⎠ R

and

⎛

ω2 = ω0 ⎜⎜1 + ⎝

For R << XL (a condition that holds for a sharply peaked resonance):

⎛

ω1 ≈ ω0 ⎜⎜1 −

⎞ + higher order terms ⎟⎟ 2ω0 L ⎠ R

R ⎞ ⎟, 2ω0 L ⎟⎠

⎝ ⎛ R ⎞ ⎟⎟ , ω2 ≈ ω0 ⎜⎜1 + ⎝ 2ω0 L ⎠

and

∆ω = ω2 − ω1 ≈ From the definition of Q:

R ω0 = L Q

Substitute to obtain:

∆ω ≈

Solve for Q:

Q≈

R . L

ω0 Q

ω0 ∆ω

101 ••• Picture the Problem We’ll differentiate Q = Q0 e − Rt 2 L cos ω ' t twice and substitute this function and both its derivatives in the differential equation of the circuit. Rewriting the resulting equation in the form Acosω′t + Bsinω′t = 0 will reveal that B vanishes. Requiring that Acosω′t = 0 hold for all values of t will lead to ω ' = Equation 29-47b is:

(1 LC ) − (R

d 2Q Q dQ L 2 + +R =0 dt C dt

2L) . 2

796 Chapter 29 Q = Q0 e − Rt 2 L cos ω ' t

Assume a solution of the form: Differentiate Q(t) twice to obtain:

dQ R ⎤ cos ω't ⎥ = −Q0e − Rt 2 L [ω' sin ω't + 2L dt ⎦ and

⎡ 2 d 2Q Rω' ⎤ − Rt 2 L ⎛ R 2⎞ = Q e sin ω't ⎥ ⎢⎜⎜ 2 − ω' ⎟⎟ cos ω't + 0 2 dt L ⎦ ⎠ ⎣⎝ 4 L Substitute these derivatives in the differential equation and simplify to obtain:

LQ0e

− Rt 2 L

⎡⎛ R 2 Rω' ⎤ Q 2⎞ sin ω't ⎥ + 0 e − Rt 2 L cos ω't ⎢⎜⎜ 2 − ω' ⎟⎟ cos ω't + L ⎦ C ⎠ ⎣⎝ 4 L − RQ0e − Rt 2 L [ω' sin ω't +

R ⎤ cos ω't ⎥ = 0 2L ⎦

or

⎡⎛ R 2 ⎞ Rω' R ⎤ 1 ⎤ L ⎢⎜⎜ 2 − ω' 2 ⎟⎟ cos ω't + sin ω't ⎥ + cos ω't − R[ω' sin ω't + cos ω't ⎥ = 0 L 2L ⎦ C ⎦ ⎠ ⎣⎝ 4 L Rewrite this equation in the form Acosω′t + Bsinω′t = 0:

⎡ ⎛

2

⎞

⎤ ⎥ cos ω't = 0 2L ⎦

(Rω' − Rω' )sin ω't + ⎢ L⎜⎜ R 2 − ω' 2 ⎟⎟ + 1 − R ⎣ ⎝ 4L

⎠ C

2

or

⎡⎛ R 2 1 R 2 ⎞⎤ ⎟⎥ cos ω't = 0 − Lω' 2 + − ⎢⎜⎜ C 2 L ⎟⎠⎦ ⎣⎝ 4 L If this equation is to hold for all values of t, its coefficient must vanish: Solve for ω′:

R2 1 R2 − Lω' 2 + − =0 4L C 2L

ω' =

1 ⎛ R ⎞ −⎜ ⎟ LC ⎝ 2 L ⎠

2

,

the condition that must be satisfied if Q = Q0e − Rt 2 L cos ω't is the solution to Equation 29-47b.

Alternating-Current Circuits

797

*102 ••• Picture the Problem We can use L = µ 0 n 2 Al to determine the inductance of the empty solenoid and the resonance condition to find the capacitance of the sample-free circuit when the resonance frequency of the circuit is 6.0000 MHz. By expressing L as a function of f0 and then evaluating df0/dL and approximating the derivative with ∆f0/∆L , we can evaluate χ from its definition.

L = µ 0 n 2 Al

(a) Express the inductance of an air-core solenoid: Substitute numerical values and evaluate L: 2

⎛ 400 ⎞ π ⎟⎟ (0.003 m )2 (0.04 m ) = 35.5 µH L = 4π × 10 N/A ⎜⎜ ⎝ 0.04 m ⎠ 4

(

−7

2

)

(b) Express the condition for resonance in the LC circuit:

X L = XC or

2πf 0 L =

1 2πf 0C

(1)

C=

1 4π f 02 L

Substitute numerical values and evaluate C:

C=

1 = 119 µF 4π (6 MHz )(35.5 µH )

(c) Express the sample’s susceptibility in terms of L and ∆L:

χ=

∆L L

Solve equation (1) for f0:

f0 =

Solve for C to obtain:

2

2

(2)

1 2π LC

Differentiate f0 with respect to L:

df 0 1 d −1 2 1 = L =− L−3 2 dL 2π C dL 4π C f 1 =− =− 0 2L 4πL LC

Approximate df0/dL by ∆f0/∆L:

f ∆f 0 ∆f ∆L = − 0 or 0 = − 2L f0 2L ∆L

798 Chapter 29 ∆f 0 f0

Substitute in equation (2) to obtain:

χ = −2

Substitute numerical values and evaluate χ:

χ = −2⎜⎜

⎛ 5.9989 MHz - 6.0000 MHz ⎞ ⎟⎟ 6.0000 MHz 0 ⎝ ⎠

= 3.67 × 10 −4 103 ••• Picture the Problem We can find the angular frequency ω for the circuit in Problem 91 such that the magnitudes of the reactances of the two parallel branches are equal by 2 equating the reactances in the two branches. We can use P = 12 I 2 R = (ε Z ) (R 2 ) , where Z is, in turn, ZL and ZC, to find the power dissipated in each resistor.

1 LC

(a) When the reactances of the parallel branches are equal:

X L = X C and ω =

Substitute numerical values to obtain:

ω=

(b) Express the power dissipation in a resistor in an ac circuit:

⎛ε ⎞ P = 12 I 2 R = 12 ⎜ ⎟ R ⎝Z⎠

Find ZL and Z L at 1.67 krad/s:

Z L = R2 + iωL

1 = 1.67 krad/s (12 mH)(30 µF) 2

= 4 Ω + i (1.67 krad/s )(12 mH ) = 4 Ω + i (20.0 Ω )

and

ZL = Find ZC and Z C at 1.67 krad/s:

(4 Ω )2 + (20 Ω )2

Z C = R1 − i

= 20.4 Ω

1 ωC

1 (1.67 krad/s)(30 µF) = 2 Ω − i (20.0 Ω ) = 2Ω − i

and

ZC =

(2 Ω )2 + (20.0 Ω )2

= 20.1 Ω

Alternating-Current Circuits Evaluate the power dissipated in R1 and R2:

1⎛ ε P1 = ⎜⎜ 2 ⎝ ZC

2

799

2

⎞ 1 ⎛ 40 V ⎞ ⎟ R1 = ⎜⎜ ⎟⎟ (2 Ω ) ⎟ Ω 2 20 . 1 ⎝ ⎠ ⎠

= 3.96 W and

1⎛ ε ⎞ 1 ⎛ 40 V ⎞ ⎟ (4 Ω ) P2 = ⎜⎜ ⎟⎟ R2 = ⎜⎜ 2 ⎝ ZL ⎠ 2 ⎝ 20.4 Ω ⎟⎠ 2

2

= 7.69 W 104 ••• Picture the Problem We can equate the power dissipated in the two resistors to obtain a relationship between the currents in and the resistances of the two branches. Expressing the currents in terms of the impedances of the two branches and the common potential difference across them will lead us to an equation that is quadratic in ω2 that we can solve for ω. In (b) we can use complex numbers to find the reactances of each of the two parallel branches and then use these results to draw the phasor diagram of (c). We can use the results of (b) to find the impedance of the circuit in (d). (a) Express the condition under which the power dissipation in the two resistors is the same: Express the ratio of the squares of the currents in the two resistors:

Equate these expressions to obtain:

I12 R1 = I 22 R2 or

I12 R2 = I 22 R1

⎛ ε ⎞ ⎜ ⎟ 2 Z L2 I1 ⎜ Z C ⎟ = = 2 ZC I12 ⎜ ε ⎟ ⎜Z ⎟ ⎝ L ⎠ 2

R2 Z L2 R22 + X L2 = = R1 Z C2 R12 + X C2 or

(

)

R2 2 R1 + X C2 = R22 + X L2 R1 Substitute for XL, XC, and the ratio of R2 to R1 to obtain:

1 ⎞ ⎛ 2⎜ 4 Ω 2 + 2 2 ⎟ = 16 Ω 2 + ω 2 L2 ω C ⎠ ⎝ or

8 Ω2 +

2 = 16 Ω 2 + ω 2 L2 2 ω C 2

800 Chapter 29 Combine like terms and clear fractions to obtain: Substitute numerical values to obtain: Use the ″solver″ capability of your calculator to solve for ω2:

(

)

L2C 2ω 4 + 8 Ω 2 C 2ω 2 − 2 = 0

(1.30 ×10

−13

(

)

s4 ω 4

)

+ 7.20 × 10 −9 s 2 ω 2 − 2 = 0

ω 2 = 3.89 × 10 6 (rad/s )2 and

ω = 1.97 × 103 rad/s (b) Express and evaluate ZC, Z C , and δC:

Z C = R1 − i

1 ωC

1 1.97 × 10 rad/s (30 µF) = 2 Ω − i (16.9 Ω ) = 2Ω − i

ZC =

(

3

(2 Ω )2 + (16.9 Ω )2

)

= 17.0 Ω

and

⎛ XC ⎝ R

δ C = tan −1 ⎜

⎞ −1 ⎛ − 16.9 Ω ⎞ ⎟⎟ ⎟ = tan ⎜⎜ ⎠ ⎝ 2Ω ⎠

= − 83.3° Express and evaluate ZL, Z L , and

Z L = R2 + iωL

(

)

= 4 Ω + i 1.97 × 103 rad/s (12 mH ) = 4 Ω + i (23.6 Ω )

δL:

ZL =

(4 Ω )2 + (23.6 Ω )2

= 23.9 Ω

and

⎛ XL ⎞ −1 ⎛ 23.9 Ω ⎞ ⎟⎟ ⎟ = tan ⎜⎜ ⎝ R ⎠ ⎝ 4Ω ⎠

δ L = tan −1 ⎜

= 80.5° (c) The applied voltage and the currents in the two branches are shown on the phasor diagram to the right.

Alternating-Current Circuits (d) Express the impedance of the circuit and simplify to obtain:

Multiply Z by 1 in the form of the complex conjugate of 6 Ω + i(6.70Ω) divided by itself and simplify to obtain:

Z=

Z L ZC Z L + ZC

=

[4 Ω + i(23.6 Ω )][2 Ω − i(16.9 Ω )] 4 Ω + i (23.6 Ω ) + 2 Ω − i (16.9 Ω )

=

407 Ω 2 − i 20.4 Ω 2 6 Ω + i (6.70 Ω )

(

(

)

⎛ 407 Ω 2 − i 20.4 Ω 2 Z = ⎜⎜ ⎝ 6 Ω + i (6.70 Ω ) ⎛ 6 Ω − i (6.70 Ω ) ⎞ ⎟⎟ × ⎜⎜ ⎝ 6 Ω − i (6.70 Ω ) ⎠

) ⎞⎟ ⎟ ⎠

(

2.58 × 103 Ω3 − i 2.85 × 103 Ω3 80.9 Ω 2 = 31.9 Ω − i (35.2 Ω ) =

Find the magnitude of the circuit’s impedance and the phase angle for the circuit:

Z =

(31.9 Ω )2 + (35.2 Ω )2

)

= 47.5 Ω

and

XC R ⎛ − 35.2 Ω ⎞ ⎟⎟ = − 47.8° = tan −1 ⎜⎜ ⎝ 31.9 Ω ⎠

δ = tan −1

The Transformer *105 • Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can use V2 N1 = V1 N 2 and N1 I1 = N 2 I 2 to find the turn ratio and the primary current when the transformer connections are reversed. (a) Relate the number of primary and secondary turns to the primary and secondary voltages:

V2 N1 = V1 N 2

Solve for and evaluate the ratio N2/N1:

24 V 1 N 2 V2 = = = 5 N1 V1 120 V

(b) Relate the current in the primary to the current in the secondary and

I1 =

N2 I2 N1

(1)

801

802 Chapter 29 to the turns ratio: Express the current in the primary winding in terms of the voltage across it and its impedance:

I2 =

V2 Z2

Substitute to obtain:

I1 =

N 2 V2 N1 Z 2

Substitute numerical values and evaluate I1:

⎛ 5 ⎞⎛ 120 V ⎞ ⎟⎟ = 50.0 A I1 = ⎜ ⎟⎜⎜ ⎝ 1 ⎠⎝ 12 Ω ⎠

106 • Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can decide whether the transformer is a step-up or step-down transformer by examining the ratio of the number of turns in the secondary to the number of terms in the primary. We can relate the open-circuit voltage in the secondary to the primary voltage and the turns ratio.

(a)

Because there are fewer turns in the secondary than in the primary it is a step - down transformer.

(b) Relate the open-circuit voltage V2 in the secondary to the voltage V1 in the primary: Substitute numerical values and evaluate V2:

V2 =

N2 V1 N1

V2 =

8 (120 V ) = 2.40 V 400

(c) Because there are no losses:

V1I1 = V2 I 2

Solve for and evaluate I2:

I2 =

V1 120 V (0.1A ) = 5.00 A I1 = V2 2.40 V

107 • Picture the Problem Let the subscript 1 denote the primary and the subscript 2 the secondary. We can use I1V1 = I 2V2 to find the current in the primary and V2 N1 = V1 N 2 to find the number of turns in the secondary. (a) Because we have 100 percent efficiency:

I1V1 = I 2V2

Alternating-Current Circuits Solve for and evaluate I1:

(b) Relate the number of primary and secondary turns to the primary and secondary voltages:

I1 = I 2

803

V2 9V = (20 A ) = 1.50 A V1 120 V

V2 N1 = V1 N 2

Solve for the ratio N2/N1:

N2 =

V2 N1 V1

Substitute numerical values and evaluate N2/N1:

N2 =

9V (250) = 18.8 ≈ 19 120 V

108 • Picture the Problem We can relate the input and output voltages to the number of turns in the primary and secondary using V2 N1 = V1 N 2 .

N2 V1 N1

Relate the output voltages V2 to the input voltage V1 and the number of turns in the primary N1 and secondary N2:

V2 =

Solve for N2:

N 2 = N1

Evaluate N2 for V2 = 2.5 V:

⎛ 2.5 V ⎞ ⎟⎟ = 10.4 N 2 = (500 )⎜⎜ ⎝ 120 V ⎠

Evaluate N2 for V2 = 7.5 V:

⎛ 7.5 V ⎞ ⎟⎟ = 31.3 N 2 = (500 )⎜⎜ 120 V ⎝ ⎠

Evaluate N2 for V2 = 9 V:

⎛ 9V ⎞ ⎟⎟ = 37.5 N 2 = (500 )⎜⎜ ⎝ 120 V ⎠

V2 V1

109 • Picture the Problem We can relate the input and output voltages to the number of turns in the primary and secondary using V2 N1 = V1 N 2 . Relate the output voltages V2 to the input voltage V1 and the number of

V2 =

N2 V1 N1

804 Chapter 29 turns in the primary N1 and secondary N2:

V1 V2

Solve for N1:

N1 = N 2

Substitute numerical values and evaluate N1:

⎛ 2000 V ⎞ ⎟⎟ = 3333 N1 = (400 )⎜⎜ ⎝ 240 V ⎠

*110 •• Picture the Problem Note: In a simple circuit maximum power transfer from source to load requires that the load resistance equals the internal resistance of the source. We can use Ohm’s law and the relationship between the primary and secondary currents and the primary and secondary voltages and the turns ratio of the transformer to derive an expression for the turns ratio as a function of the effective resistance of the circuit and the resistance of the speaker(s).

V1 I1

Express the effective loudspeaker resistance at the primary of the transformer:

Reff =

Relate V1 to V2, N1, and N2:

V1 = V2

N1 N2

Express I1 in terms of I2, N1, and N2:

I1 = I 2

N2 N1

Substitute to obtain:

N1 2 N 2 ⎛ V2 ⎞⎛ N1 ⎞ ⎟ = = ⎜ ⎟⎜ N 2 ⎜⎝ I 2 ⎟⎠⎜⎝ N 2 ⎟⎠ I2 N1 V2

Reff

Solve for N1/N2:

N1 = N2

I 2 Reff = V2

Evaluate N1/N2 for Reff = Rint:

N1 = N2

2000 Ω = 15.8 8Ω

Express the power delivered to the two speakers connected in parallel:

Psp = I12 Reff

Reff R2

(1)

(2)

Alternating-Current Circuits Find the equivalent resistance Rsp of the two 8-Ω speakers in parallel:

805

1 1 1 2 1 = + = = Rsp 8 Ω 8 Ω 8 Ω 4 Ω and

Rsp = 4 Ω Solve equation (1) for Reff to obtain:

Substitute numerical values and evaluate Reff: Find the current drawn from the source: Substitute numerical values in equation (2) and evaluate the power delivered to the parallel speakers:

Reff

⎛N ⎞ = R2 ⎜⎜ 1 ⎟⎟ ⎝ N2 ⎠

2

Reff = (4 Ω )(15.8) = 999 Ω 2

I1 =

V 12 V = = 4.00 mA Rtot 2000 Ω + 999 Ω

Psp = (4 mA ) (999 Ω ) = 16.0 mW 2

111 •• Picture the Problem We can substitute I2 = V2/Z in Equation 29-62 to show that I1 = ε/[(N1/N2)2Z] and then use this result in Zeff = ε/I1 to show that Zeff = (N1/N2)2Z. From Equation 29-62 we have:

I1 =

N2 I2 N1

or, because I2 = V2/Z,

I1 =

N 2 V2 N1 Z

From Equation 29-61 we have:

V2 =

N2 N V1 = 2 ε N1 N1

Substitute to obtain:

N2 ε 2 ⎛ N2 ⎞ ε N 2 N1 ⎟⎟ = ⎜⎜ I1 = N1 Z ⎝ N1 ⎠ Z =

Express the effective impedance Zeff of the speaker in terms of ε and I1:

Z eff =

( N1

ε I1

ε

N2 ) Z 2

806 Chapter 29 Substitute for I1 to obtain:

Z eff =

( N1

ε ε

=

( N1

N2 ) Z 2

N2 ) Z 2

General Problems 112 • Picture the Problem We can use Pav = ε rms I rms to find the rms current and

I max = 2I rms to find the maximum current drawn by the dryer. (a) Express the average power delivered by the source in terms of εrms and Irms: Solve for and evaluate Irms:

(b) Relate the maximum current Imax to the rms current Irms: (c) Proceed as in (a) and (b) to obtain:

Pav = ε rms I rms

I rms =

Pav

ε rms

=

5 kW = 20.8 A 240 V

I max = 2 I rms = 2 (20.8 A ) = 29.5 A I rms = 41.6 A and I max = 59.0 A

113 • Picture the Problem We can use its definition to find the reactance of the capacitor at the given frequencies. Express the reactance of a capacitor:

XC =

1 1 = ωC 2πfC

(a) Evaluate XC at f = 60 Hz:

XC =

1 = 265 Ω 2π (60 Hz )(10 µF)

(b) Evaluate XC at f = 6 kHz:

XC =

1 = 2.65 Ω 2π (6 kHz )(10 µF)

(a) Evaluate XC at f = 6 MHz:

XC =

1 = 2.65 mΩ 2π (6 MHz )(10 µF)

Alternating-Current Circuits 114 •• Picture the Problem We can use its definition, I rms =

( )

to the current carried by the resistor and find I 2

av by

I rms =

(a) Express the rms current in terms of the I 2 av :

( )

(I ) 2

av

807

to relate the rms current

integrating I2.

(I ) 2

av

Evaluate I2:

I 2 = [(5 A ) sin 120πt + (7 A ) sin 240πt ]

(

)

(

2

)

(

)

= 25 A 2 sin 2 120πt + 70 A 2 sin 120πt sin 240πt + 49 A 2 sin 2 240πt

( )

Find I 2

av by

integrating I2 from t = 0 to t = T = 2π/ω and dividing by T:

(I ) 2

av

=

ω 2π

∫ {(25 A )sin 2π ω

0

2

2

120πt + (70 A 2 )sin 120πt sin 240πt

+ (49 A 2 )sin 2 240πt }dt

Use the trigonometric identity sin 2 x = 12 (1 − cos 2 x ) to simplify

(I ) 2

av

= 12.5 A 2 + 0 + 24.5 A 2 = 37.0 A 2

and evaluate the 1st and 3rd integrals and recognize that the middle term is of the form sinxsin2x to obtain: Substitute to obtain:

(b) Relate the power dissipated in the resistor to its resistance and the rms current in it: Substitute numerical values and evaluate P: (c) Express the rms voltage across the resistor in terms of R and Irms: Substitute numerical values and evaluate Vrms:

I rms = 37.0 A 2 = 6.08 A 2 P = I rms R

P = (6.08 A ) (12 Ω ) = 444 W 2

Vrms = I rms R Vrms = (6.08 A )(12 Ω ) = 73.0 V

808 Chapter 29 *115 •• Picture the Problem The average of any quantity over a time interval ∆T is the integral of the quantity over the interval divided by ∆T. We can use this definition to find both the average of the voltage squared, V 2 av and then use the definition of the rms voltage.

( )

(V )

(a) From the definition of Vrms we have:

Vrms =

Noting that − V02 = V02 , evaluate

Vrms = V02 = V0 = 12.0 V

2 0 av

Vrms: (b) Noting that the voltage during the second half of each cycle is now zero, express the voltage during the first half cycle of the time interval 1 2 ∆T : Express the square of the voltage during this half cycle: Calculate V 2 av by integrating V2

( )

from t = 0 to t =

1 2

∆T and dividing

V = V0

V 2 = V02

(V ) 2

av

=

V02 ∆T

∫

1 ∆T 2

0

dt =

V02 12 ∆T 1 2 [t ]0 = 2 V0 ∆T

by ∆T: Substitute to obtain:

Vrms =

1 2

V02 =

V0 12 V = = 8.49 V 2 2

116 •• Picture the Problem We can use the definitions of Irms and Vrms to find the rms value of the waveform and the average power delivered by the pulse generator. (a) From the definition of Irms we have:

( )

Evaluate I 2

av over

1 s:

(I )

I rms =

(I ) 2

av

2

av

= =

(0.1s )I 2 + (0.9 s )(0) 1s

(0.1s )I 2 = (0.1)(15 A )2 1s

= 22.5 A 2

Alternating-Current Circuits Substitute to obtain: (b) Express the average power delivered by the pulse generator in terms of Irms and Vrms: From the definition of Vrms we have:

( )

Evaluate V 2

av

over 1 s:

809

I rms = 22.5 A 2 = 4.74 A Pav = I rmsVrms

(V )

Vrms =

(V ) 2

av

2

av

( 0.1s )V 2 + (0.9 s )(0) = 1s

=

(0.1s )V

2

1s

= (0.1)(100 V )

2

= 1000 V 2 Evaluate Vrms: Substitute numerical values and evaluate Pav:

Vrms = 1000 V 2 = 31.6 V Pav = (4.74 A )(31.6 V ) = 150 W

117 •• Picture the Problem We can use the definition of capacitance to find the charge on each capacitor and the definition of current to find the steady-state current in the circuit. We can find the maximum and minimum energy stored in the capacitors using U = 12 CeqV 2 , where V is either the maximum or the minimum potential difference across the capacitors. (a) Use the definition of capacitance to express the charge on each capacitor:

Substitute to obtain:

Q1 = C1V1 and Q2 = C2V2 or, because the capacitors are in parallel, Q1 = C1V and Q2 = C2V where V = ε + 24 V

Q1 = C1 (ε + 24 V )

= (3 µF)[(20 V )cos(120πt ) + 24 V ] =

(60 µF)cos(120πt ) + 72 µF

and

Q2 = C2 (ε + 24 V )

= (1.5 µF)[(20 V )cos(120πt ) + 24 V ] =

(30 µF)cos(120πt ) + 36 µF

810 Chapter 29 (b) Express the steady-state current as the rate at which charge is being delivered to the capacitors:

I=

Substitute for Q1 and Q2 and evaluate I:

I=

dQ d = (Q1 + Q2 ) dt dt

d [(60 µF) cos(120πt ) + 72 µF dt + (30 µF) cos(120πt ) + 36 µF] = −120π (60 µF) sin (120πt ) − 120π (30 µF) sin (120πt )

= − (33.9 mA ) sin (120πt ) (c) Express Umax in terms of the maximum potential difference across the capacitors: Because Vmax = 44 V and Ceq = C1 + C2 = 3 µF + 1.5 µF = 4.5 µF: (d) Express Umin in terms of the minimum potential difference across the capacitors: The minimum energy stored in the capacitors occurs when Vmin = 24 V − εmax = 4 V:

2 U max = 12 CeqVmax

U max =

(4.5 µF)(44 V )2 =

1 2

4.36 mJ

2 U min = 12 CeqVmin

U min =

1 2

(4.5 µF)(4 V )2 =

36.0 µJ

118 •• Picture the Problem The average of any quantity over a time interval ∆T is the integral of the quantity over the interval divided by ∆T. We can use this definition to find both the average current Iav, and the average of the current squared, I 2 av

( )

From the definition of Iav and Irms we have:

I av =

(a) Express the current during the first half cycle of time interval ∆T:

I=

1 ∆T

∫

∆T

0

Idt and I rms =

(I ) 2

4 t ∆T

where I is in A when t and T are in seconds.

av

Alternating-Current Circuits Evaluate Iav:

Express the square of the current during this half cycle: Noting that the average value of the squared current is the same for each time interval ∆T, calculate I 2 av by

I av =

1 ∆T

=

4

I2 =

(I )

=

2

av

1 ∆T

from t = 0 to t =

1 2

∆T and dividing

by ∆T:

Substitute in the expression for Irms to obtain:

∆T

⎡t 2 ⎤ ⎢ ⎥ = 2.00 A ⎣ 2 ⎦0

∆T

16

∫ (∆T ) 0

2

t 2 dt

I = 4A

4 A 12 ∆T 4 A 12 ∆T [t ]0 = 2.00 A dt = ∫ ∆T 0 ∆T

I 2 = 16 A 2

integrating I2

tdt

16 2 A = 2.31 A 3

I rms =

Express the square of the current during this half cycle: av by

0

∆T

I av =

( )

∆T

16 ⎡ t 3 ⎤ 16 = = ⎥ 3 ⎢ (∆T ) ⎣ 3 ⎦ 0 3

Evaluate Iav:

Calculate I 2

0

∫

16 2 t (∆T )2

integrating I2 from t = 0 to t = ∆T and dividing by ∆T:

(b) Noting that the current during the second half of each cycle is zero, express the current during the first half cycle of the time interval 12 ∆T :

4 4 tdt = ∆T (∆T )2

∆T

(∆T )2

( )

Substitute in the expression for Irms to obtain:

∫

811

(I ) 2

av

16 A 2 12 ∆T dt = ∆T ∫0 16 A 2 12 ∆T = [t ]0 = 8 A 2 ∆T

I rms = 8 A 2 = 2.83 A

119 •• Picture the Problem We can apply Kirchhoff’s loop rule to express the current in the circuit in terms of the emfs of the sources and the resistance of the resistor. We can then

812 Chapter 29 find Imax and Imin by considering the conditions under which the time-dependent factor in I will be a maximum or a minimum. The average of any quantity over a time interval ∆T is the integral of the quantity over the interval divided by ∆T. We can use this definition to find average of the current squared, I 2 av and then Irms.

( )

ε 1 + ε 2 − IR = 0

Apply Kirchhoff’s loop rule to obtain: Solve for I:

Substitute numerical values to obtain:

I=

ε1 + ε 2

I=

(20 V )cos(2π (180 s −1 )t ) + 18 V

R

36 Ω

(

)

= 0.5 A + (0.556 A ) cos 1131s −1 t

(

)

Express the condition that must be satisfied if the current is to be a maximum:

cos 1131s −1 t = 1

Evaluate Imax:

I max = 0.5 A + 0.556 A = 1.06 A

cos(1131s −1 )t = −1

Express the condition that must be satisfied if the current is to be a minimum: Evaluate Imin:

I min = 0.5 A − 0.556 A = − 0.0560 A

Because the average value of cosωt = 0:

I av = 0.500 A

Express and evaluate the average current delivered by the source whose emf is ε2:

I2 =

(

ε2 R

=

18 V = 0.5 A 36 Ω

)

Because I1 = (0.556 A )cos 1131s −1 t :

(I )

2 1 av

=

(

)

5.56 ms 1 (0.556 A )2 cos 2 1131s −1 tdt ∫ 5.56 ms 0

Use the trigonometric identity cos 2 x =

1 2

(1 + cos 2 x ) to obtain:

Alternating-Current Circuits

(I )

2 1 av

(

))

0.309 A 2 5.56 ms (1 + cos 2 1131s −1 t dt ∫ 0 2(5.56 ms )

=

⎡ 1 = 27.8 A / s ⎢t + sin 2262 s −1 -1 ⎣ 2262 s

(

( )

Evaluate I12

(I )

2 1 av

av

813

2

)

(

5.56 ms

⎤ t⎥ ⎦0

)

:

⎡ ⎤ 1 = 27.8 A 2 / s ⎢5.56 ms + sin 2262 s -1 (5.56 ms )⎥ = 0.1543 A 2 −1 2262 s ⎣ ⎦

(

)

( )

Express I 2

av :

(

(I ) 2

av

)

( ) + (I )

= I12

2 2 av

av

= 0.1543 A 2 + (0.5 A )

2

= 0.4043 A 2 Evaluate Irms:

I rms =

(I ) 2

av

= 0.4043 A 2

= 0.636 A *120 •• Picture the Problem We can apply Kirchhoff’s loop rule to obtain an expression for charge on the capacitor as a function of time. Differentiating this expression with respect to time will give us the current in the circuit. We can then find Imax and Imin by considering the conditions under which the time-dependent factor in I will be a maximum or a minimum. We can use the maximum value of the current to find Irms. Apply Kirchhoff’s loop rule to obtain: Substitute numerical values and solve for q(t):

ε 1 + ε 2 − q(t ) = 0 C

(

)

q (t ) = (2 µF)(20 V ) cos 1131s −1 t + (2 µF)(18 V )

(

)

= (40 µC ) cos 1131s −1 t + 36 µC Differentiate this expression with respect to t to obtain the current as a function of time:

I=

[

(

)

dq d (40 µC)cos 1131s −1 t = dt dt + 36 µC]

(

)

= − (45.2 mA )sin 1131s −1 t Express the condition that must be

sin (1131s −1 )t = 1

814 Chapter 29 satisfied if the current is to be a minimum:

and

Express the condition that must be satisfied if the current is to be a maximum:

sin 1131s −1 t = −1

I min = − 45.2 mA

(

)

and

I max = 45.2 mA

Because the dc source sees the capacitor as an open circuit and the average value of the sine function over a period is zero:

I av = 0

Because the peak current is 45.2 mA:

I rms =

I max 45.2 mA = = 32.0 mA 2 2

21 •• Picture the Problem The inductance acts as a short circuit to the constant voltage source. The current is infinite at all times. Consequently, Imax = Irms = ∞; there is no minimum current.

Alternating-Current Circuits

815

Chapter 30 Maxwell’s Equations and Electromagnetic Waves Conceptual Problems *1 • (a) False. Maxwell’s equations apply to both time-independent and time-dependent fields. (b) True (c) True (d) True (e) False. The magnitudes of the electric and magnetic field vectors are related according to E = cB. (f) True 2 •• Determine the Concept Two changes would be required. Gauss’s law for magnetism would become Bn dA = µ 0 qm and Faraday’s law would

∫

S

r r d I become ∫ E ⋅ d l = − ∫ Bn dA − m , where Im is the current associated with the motion C S dt ∈0 of the magnetic poles. 3 • Determine the Concept X rays have greater frequencies whereas light waves have longer wavelengths (see Table 30-1). *4 • Determine the Concept The frequencies of ultraviolet radiation are greater than those of infrared radiation (see Table 30-1). 5 • Determine the Concept Consulting Table 30-1 we see that FM radio and televisions waves have wavelengths of the order of a few meters. 6 • Determine the Concept The dipole antenna detects the electric field, the loop antenna detects the magnetic field of the wave.

815

816 Chapter 30 7 • Determine the Concept The dipole antenna should be in the horizontal plane and normal to the line from the transmitter to the receiver. *8 • Determine the Concept A red plastic filter absorbs all the light incident on it except for the red light and a green plastic filter absorbs all the light incident on it except for the green light. If the red beam is incident on a red filter it will pass through, whereas, if it is incident on the green filter it will be absorbed. Because the green filter absorbs more energy than does the red filter, the laser beam will exert a greater force on the green filter.

Estimation and Approximation 9 •• Picture the Problem We’ll assume that the plastic bead has the same density as water. Applying a condition for translational equilibrium to the bead will allow us to relate the gravitational force acting on it to the force exerted by the laser beam. Because the force exerted by the laser beam is related to the radiation pressure and the radiation pressure to the intensity of the beam, we’ll be able to find the beam’s intensity. Knowing the beam’s intensity, we find the total power needed to lift the bead. Apply

∑F

y

= 0 to the bead:

Fby laser beam − mg = 0

Relate the force exerted by the laser beam to the radiation pressure exerted by the beam:

1 Fby laser beam = Pr A = πd 2 Pr 4

Substitute to obtain:

1 2 πd Pr − mg = 0 4

The radiation pressure Pr is the quotient of the intensity I and the speed of light c:

Pr =

Substitute for Pr to obtain:

1 πd 2 I − mg = 0 4 c

Express the mass of the bead:

1 m = ρV = πρ d 3 6

Substitute for m in equation (1) to obtain:

1 πd 2 I 1 − πρ d 3 g = 0 4 c 6

I c

(1)

Maxwell’s Equations and Electromagnetic Waves Solve for I:

I=

817

2 cρ dg 3

Substitute numerical values and evaluate I:

I=

2 ( 3 × 108 m/s )(103 kg/m 3 )(15 µm ) (9.81 m/s 2 ) = 2.94 × 10 7 W/m 2 3

The power needed is the product of the beam intensity and the crosssectional area of the bead:

1 P = IAbead = π d 2 I 4

Substitute numerical values and evaluate P:

1 2 P = π (15 µm ) 2.94 × 10 7 W/m 2 4

(

)

= 5.20 mW 10 ••• Picture the Problem The net force acting on the spacecraft is the difference between the repulsive force due to radiation pressure and the attractive gravitational force. We can apply Newton’s 2nd law to the spacecraft and solve the resulting equation for the acceleration of the spacecraft. Because the acceleration turns out to be a function of r, we’ll need to integrate a to find v2. We’ll assume that the sail absorbs all of the radiation incident on it. Apply Newton’s 2nd law to the spacecraft (including sail) to obtain:

Fr − Fg = ma

Solve for a:

a=

Assuming that the sail absorbs all of the incident solar radiation:

Fr − Fg m

Fr = Pr A =

IA c

where A is the area of the sail. Because I =

Ps : 4πr 2

Fr =

Ps A 4π r 2 c

818 Chapter 30 Substitute for Fr and Fg to obtain:

Ps A GM s m − 2 Ps A GM 4π r c r2 a= = − 2s 2 4π r mc m r Ps A − GM s m 4π c = mr 2

Neglecting the gravitational term:

a=

(b) Because a is a function of r, the velocity must be found by integration. Note that:

a=

Ps A 4π r 2 mc

dv dv dr dv = = v ⇒ vdv = adr dt dr dt dr

Substitute for a and integrate v′ from v0 to v and r′ from r0 to r:

⎞ ⎛ Ps A ⎞ ⎛ Ps A − GM s m ⎟ − GM s m ⎟ r ⎜ ⎜ 4π c ⎟⎛⎜ 1 − 1 ⎞⎟ ⎟ dr' = ⎜ 4π c v 2 − v02 = ⎜ 2 ∫ ⎟⎜⎝ r0 r ⎟⎠ ⎜ ⎟r r' ⎜ m m ⎟ ⎜ ⎟0 ⎜ ⎠ ⎝ ⎠ ⎝

∫ v'dv' = ( v

1 2

v0

)

Solve for v2 to obtain:

⎞ ⎛ Ps A − GM s m ⎟ ⎜ 4π c ⎟ ⎛⎜ 1 − 1 ⎞⎟ v 2 = v02 + 2⎜ ⎟ ⎜⎝ r0 r ⎟⎠ ⎜ m ⎟ ⎜ ⎠ ⎝ Ignore the gravitational term to obtain:

⎛ P A ⎞⎛ 1 1 ⎞ v 2 = v02 + ⎜ s ⎟ ⎜⎜ − ⎟⎟ ⎝ 2πmc ⎠ ⎝ r0 r ⎠

(c)

This scheme is not likely to work effectively. For any reasonable mass, the surface mass density of the sail would have to be extremely small and the sail would have to be huge. Additionally, unless struts are built into the sail, it would collapse during the acceleration of the spacecraft.

Maxwell’s Equations and Electromagnetic Waves

819

11 •• Picture the Problem We can use I = ErmsBrms/µ0 and Brms = Erms/c to express Erms in terms of I. We can then use Brms = Erms/c to find Brms. The average power output of the sun is given by Pav = 4πR 2 I where R is the earth-sun distance. The intensity and the radiation pressure at the surface of the sun can be found from the definitions of these physical quantities.

Erms Brms

(a) Express the intensity I of the radiation as a function of its average power and the distance r from the station:

I=

Solve for Erms:

Erms = cµ0 I

µ0

=

2 Erms cµ0

Substitute numerical values and evaluate Erms:

Erms =

(3 × 10

8

)(

)(

)

m/s 4π × 10 −7 N/A 2 1.37 kW/m 2 = 719 V/m

Use Brms = Erms/c to evaluate Brms:

Brms =

(b) Express the average power output of the sun in terms of the solar constant:

Pav = 4π R 2 I

719 V/m = 2.40 µT 3 × 10 8 m/s

where R is the earth-sun distance.

(

)(

Substitute numerical values and evaluate Pav:

Pav = 4π 1.5 × 1011 m 1.37 kW/m 2

(c) Express the intensity at the surface of the sun in terms of the sun’s average power output and radius r:

I=

Substitute numerical values and evaluate I at the surface of the sun:

2

= 3.87 × 10 26 W

I=

Pav 4π r 2

3.87 × 10 26 W

(

4π 6.96 × 108 m

)

2

= 6.36 × 10 7 W/m 2 Express the radiation pressure in terms of the intensity:

Pr =

I c

Substitute numerical values and evaluate Pr:

Pr =

6.36 × 10 7 W/m 2 = 0.212 Pa 3 × 108 m/s

)

820 Chapter 30 *12 •• Picture the Problem We can find the radiation pressure force from the definition of pressure and the relationship between the radiation pressure and the intensity of the radiation from the sun. We can use Newton’s law of gravitation to find the gravitational force the sun exerts on the earth. The radiation pressure exerted on the earth is given by:

Pr =

Fr ⇒ Fr = Pr A A

where A is the cross-sectional area of the earth. Express the radiation pressure in terms of the intensity of the radiation I from the sun:

Pr =

Substituting for Pr and A yields:

Iπ R 2 Fr = c

Substitute numerical values and evaluate Fr:

Fr =

I c

π (1370 W/m 2 )(6370 km )2 3 × 108 m/s

= 5.82 × 108 N The gravitational force exerted on the earth by the sun is given by:

F=

Gmsun mearth r2

where r is the radius of the earth’s orbit. Substitute numerical values and evaluate F:

F=

(6.67 × 10

−11

)(

)(

N ⋅ m 2 / kg 2 1.99 × 10 30 kg 5.98 × 10 24 kg

(1.5 × 10

11

Express the ratio of the force due radiation pressure Fr to the gravitational force F:

m

)

2

) = 3.53 × 10

22

N

Fr 5.82 × 108 N = = 1.65 × 10 −14 F 3.53 × 10 22 N

The gravitational force is greater by a factor of approximately 1014. *13 •• Picture the Problem We can find the radiation pressure force from the definition of pressure and the relationship between the radiation pressure and the intensity of the radiation from the sun. We can use Newton’s law of gravitation to find the gravitational force the sun exerts on Mars. The radiation pressure exerted on Mars is given by:

Pr =

Fr ⇒ Fr = Pr A A

where A is the cross-sectional area of Mars.

Maxwell’s Equations and Electromagnetic Waves Express the radiation pressure on Mars in terms of the intensity of the radiation IMars from the sun:

Pr =

Substituting for Pr and A yields:

Fr =

821

I Mars c

2 I Marsπ RMars c

Express the ratio of the solar constant at the earth Iearth to the solar constant IMars at Mars:

⎛r ⎞ I Mars ⎛ rearth ⎞ ⎟⎟ ⇒ I Mars = I earth ⎜⎜ earth ⎟⎟ = ⎜⎜ I earth ⎝ rMars ⎠ ⎝ rMars ⎠

Substitute for IMars to obtain:

I π R2 ⎛ r ⎞ Fr = earth Mars ⎜⎜ earth ⎟⎟ c ⎝ rMars ⎠

2

2

2

Substitute numerical values and evaluate Fr:

Fr =

π (1370 W/m 2 )(3395 km )2 ⎛ 1.50 × 1011 m ⎞

2

⎜⎜ ⎟⎟ = 7.09 × 10 7 N 11 ⎝ 2.29 × 10 m ⎠

3 × 108 m/s

The gravitational force exerted on Mars by the sun is given by:

F=

Gmsun mMars Gmsun (0.11mearth ) = r2 r2

where r is the radius of Mars’ orbit. Substitute numerical values and evaluate F:

F=

(6.67 × 10

−11

N ⋅ m 2 / kg 2 )(1.99 × 1030 kg )(0.11) (5.98 × 10 24 kg )

(2.29 × 10

11

Express the ratio of the force due radiation pressure Fr to the gravitational force F:

m)

2

= 1.66 × 10 21 N

Fr 7.09 × 10 7 N = 4.27 × 10 −14 = F 1.66 × 10 21 N

Because the ratio of these forces is 1.65 × 10 −14 for the earth and 4.27 × 10 −14 for Mars, Mars has the larger ratio. The reason that the ratio is higher for Mars is that the dependence of the radiation pressure on the distance from the Sun is the same for both forces (r −2 ), whereas the dependence on the radii of the planets is different. Radiation pressure varies as R 2 , whereas the gravitational force varies as R 3 (assuming that the two planets have the same density, an assumption that is nearly true). Consequently, the ratio of the forces goes as R 2 / R 3 = R −1. Because Mars is smaller than earth, the ratio is larger.

822 Chapter 30 *14 •• Picture the Problem We can use Newton’s 2nd law to express the acceleration of an atom in terms of the net force acting on the atom and the relationship between radiation pressure and the intensity of the beam to find the net force. Once we know the acceleration of an atom, we can use the definition of acceleration to find the stopping time for a rubidium atom at room temperature. (a) Apply

∑ F = ma to the atom to

F = ma

obtain:

where F is the force exerted by the laser beam.

The radiation pressure Pr and intensity of the beam I are related according to:

Pr =

F I = A c

F=

IA Iλ2 = c c

Solve for F to obtain:

Substitute for F in the expression of Newton’s 2nd law to obtain:

Iλ2 = ma c

Solve for a:

Iλ2 a= mc

Substitute numerical values and evaluate a:

a=

(10 W/m )(780 nm) 2

2

⎛ ⎞ g 1 mol ⎜⎜ 85 ⎟⎟ 3 × 108 m/s × 23 ⎝ mol 6.02 × 10 particles ⎠

(b) Using the definition of acceleration, express the stopping time ∆t of the atom: Because vfinal ≈ 0:

(

)

= 1.44 × 105 m/s 2

∆t =

vfinal − vinitial a

∆t ≈

− v initial a

Using the rms speed as the initial speed of an atom, relate vinitial to the temperature of the gas:

vinitial = vrms =

Substitute in the expression for the stopping time to obtain:

∆t = −

1 3kT a m

3kT m

Maxwell’s Equations and Electromagnetic Waves

823

Substitute numerical values and evaluate ∆t:

∆t = −

1 − 1.44 × 105 m/s 2

3(1.38 × 10 −23 J/K )(300 K ) = 2.06 ms ⎛ ⎞ g 1 mol ⎜⎜ 85 ⎟⎟ × 23 ⎝ mol 6.02 × 10 particles ⎠

Maxwell’s Displacement Current 15 • Picture the Problem We can differentiate the expression for the electric field between the plates of a parallel-plate capacitor to find the rate of change of the electric field and the definitions of the conduction current and electric flux to compute Id.

Q

(a) Express the electric field between the plates of the parallelplate capacitor:

E=

Differentiate this expression with respect to time to obtain an expression for the rate of change of the electric field:

1 dQ dE d ⎡ Q ⎤ I = = ⎢ ⎥= dt dt ⎣∈0 A ⎦ ∈0 A dt ∈0 A

∈0 A

Substitute numerical values and evaluate dE/dt:

dE 5A = = 3.40 × 1014 V/m ⋅ s 2 −12 2 2 dt 8.85 × 10 C / N ⋅ m π (0.023 m )

(

)

(b) Express the displacement current Id :

I d =∈0

dφe dt

Substitute for the electric flux to obtain:

I d =∈0

d [EA] =∈0 A dE dt dt

Substitute numerical values and evaluate Id:

(

)

(

)

I d = 8.85 × 10 −12 C 2 / N ⋅ m 2 π (0.023 m ) 3.40 × 1014 V/m ⋅ s = 5.00 A 2

824 Chapter 30 16 • Picture the Problem We can express the displacement current in terms of the electric flux and differentiate the resulting expression to obtain Id in terms dE/dt. Express the displacement current Id:

I d =∈0

dφe dt

Substitute for the electric flux to obtain:

I d =∈0

d [EA] =∈0 A dE dt dt

Because E = (0.05 N/C )sin 2000t :

d [(0.05 N/C)sin 2000t ] dt = 2000 s -1 ∈ 0 A(0.05 N/C ) cos 2000t

I d =∈ 0 A

(

Id will have its maximum value when cos 2000t = 1. Hence:

)

(

)

I d,max = 2000 s -1 ∈0 A(0.05 N/C )

Substitute numerical values and evaluate Id,max:

(

)(

)(

)

I d = 2000 s −1 8.85 × 10 −12 C 2 / N ⋅ m 2 1 m 2 (0.05 N/C ) = 8.85 × 10 −10 A 17 •• Picture the Problem We can use Ampere’s law to a circular path of radius r between the plates and parallel to their surfaces to obtain an expression relating B to the current enclosed by the amperian loop. Assuming that the displacement current is uniformly distributed between the plates, we can relate the displacement current enclosed by the circular loop to the conduction current I.

r

r

Apply Ampere’s law to a circular path of radius r between the plates and parallel to their surfaces to obtain:

∫ B ⋅ d l = 2πrB = µ I

Assuming that the displacement current is uniformly distributed:

Id I r2 = ⇒ I = 2 Id π r 2 π R2 R

0 enclosed

C

= µ0 I

where R is the radius of the circular plates. Substitute to obtain:

2π rB =

µ0 r 2 R2

Id

Maxwell’s Equations and Electromagnetic Waves

825

µ0 r Id 2π R 2

Solve for B:

B=

Substitute numerical values and evaluate B:

B(r ) =

(4π × 10

N / A 2 )(5 A ) r 2 2π (0.023 m ) −7

= (1.89 × 10 −3 T/m ) r

18 •• Picture the Problem We can use the definitions of the displacement current and electric flux, together with the expression for the capacitance of an air-core-parallel-plate capacitor to show that Id = C dV/dt. (a) Use its definition to express the displacement current Id:

I d =∈0

dφe dt

Substitute for the electric flux to obtain:

I d =∈0

d [EA] =∈0 A dE dt dt

Because E = V/d:

I d =∈0 A

The capacitance of an air-coreparallel-plate capacitor whose plates have area A and that are separated by a distance d is given by:

C=

Substitute to obtain:

Id = C

(b) Substitute in the expression derived in (a) to obtain:

I d = (5 nF)

d ⎡V ⎤ ∈0 A dV = dt ⎢⎣ d ⎥⎦ d dt

∈0 A d

dV dt

d [(3 V ) cos 500π t ] dt = −(5 nF)(3 V ) 500π s −1 sin 500π t

(

)

= − (23.6 µA ) sin 500π t *19 •• Picture the Problem We can use the conservation of charge to find Id, the definitions of the displacement current and electric flux to find dE/dt, and Ampere’s law to evaluate r r B ⋅ d l around the given path.

826 Chapter 30 (a) From conservation of charge we know that:

I d = I = 10.0 A

(b) Express the displacement current Id :

I d =∈0

Substitute for dE/dt:

I dE = d dt ∈0 A

Substitute numerical values and evaluate dE/dt:

10 A dE = −12 8.85 × 10 C 2 / N ⋅ m 2 0.5 m 2 dt

dφe d dE =∈0 [EA] =∈0 A dt dt dt

(

= 2.26 × 1012

)(

)

V m ⋅s

r r B ∫ ⋅ d l = µ0 I enclosed

(c) Apply Ampere’s law to a circular path of radius r between the plates and parallel to their surfaces to obtain:

C

I enclosed I d π r2 = ⇒ I = Id enclosed A π r2 A

Assuming that the displacement current is uniformly distributed:

where R is the radius of the circular plates.

r r µ 0π r 2 B ∫C ⋅ d l = A I d

Substitute for Ienclosed to obtain:

Substitute numerical values and evaluate

(

r

r

∫ B ⋅ dl : C

)

r r 4π × 10 −7 N / A 2 π (0.1 m )2 (10 A ) B = 7.90 × 10 −7 T ⋅ m 2 ∫C ⋅ d l = 0.5 m 20 ••• Picture the Problem If Q = Q0 e −t τ is the charge on the capacitor plates, then the conduction current I = dQ/dt. We can use I d =∈0 and I b =

dφe to find the displacement current dt

dQb to find the current due to the rate of change of the bound charges. The dt

total current is the sum of I, Id, and Ib.

Maxwell’s Equations and Electromagnetic Waves

827

dQ dt

(a) The conduction current is given by:

I=

The charge on the capacitor varies with time according to:

Q = Q0 e −t τ , where τ = RC

Substitute for Q to obtain:

I=

[

]

Q0 −t τ d Q0 e −t τ = e τ dt

This current is in the direction of the electric field, which is from the positive plate to the negative plate. By choosing the positive sign for this current we define this to be the positive direction.

dφe d dE =∈0 [EA] =∈0 A dt dt dt

(b) The displacement current is given by:

I d =∈0

Relate the electric field E to the potential difference V between the plates and the separation of the plates d:

E=

Substitute to obtain:

⎡V ⎤ ∈0 A dV ⎢⎣ d ⎥⎦ = d dt κ ∈0 A , or, because C = d C dV Id = κ dt

V varies with time according to:

V = V0 e −t τ =

Substituting in the expression for Id yields:

Id =

V d

I d =∈0 A

d dt

Q0 −t τ e C

Q C d ⎡ Q0 −t τ ⎤ e ⎥ = − 0 e −t τ ⎢ κ dt ⎣ C κτ ⎦ 1 = − I

κ

(c) As the voltage across the dielectric decreases the magnitude of the bound charges also decreases. The current Ib due to the flow of these bound charges though a

Ib =

dQb where Qb is the bound charge dt

on the surface of the dielectric next to the plate with charge Q.

828 Chapter 30 stationary surface is given by: It follows that Q and Qb are opposite in sign and are related by Equation 24-27:

⎛ 1⎞ Qb = −⎜1 − ⎟Q ⎝ κ⎠

Substitute in the expression for Ib and carry out the differentiation to obtain:

Ib =

(d) Add the currents found in (a), (b), and (c) to obtain:

d ⎡ ⎛ 1⎞ ⎤ ⎛ 1 ⎞ dQ − ⎜1 − ⎟Q ⎥ = −⎜1 − ⎟ ⎢ dt ⎣ ⎝ κ ⎠ ⎦ ⎝ κ ⎠ dt

⎛ 1⎞ = − ⎜1 − ⎟ I ⎝ κ⎠ I total = I + I d + I b =I−

⎛ 1⎞ I − ⎜1 − ⎟ I κ ⎝ κ⎠

1

= 0 Remarks: In more sophisticated treatments of electrodynamics it is conventional to refer to the sum Id + Ib as the displacement current. 21 ••• Picture the Problem We can find the conduction current as a function of time using I = V(t)/R and substituting for V(t). We can use I d =∈0 φe to obtain an expression for the displacement current Id as a function of time. Finally, equating the conduction and displacement currents will yield an expression for the time at which they are equal. (a) Express the conduction current in terms of the potential difference between the plates of the capacitor:

I=

V (t ) AV (t ) = R ρd

Substitute for V(t) to obtain:

I=

(0.01 V/s )A t

(b) The displacement current is given by:

ρd

d (EA) =∈0 d ⎛⎜ V A ⎞⎟ dt dt ⎝ d ⎠ ∈ A dV = 0 d dt

I d =∈0

Maxwell’s Equations and Electromagnetic Waves ∈0 A d [(0.01 V/s )t ] d dt (0.01 V/s )∈0 A = d

Substitute for V and simplify to obtain:

Id =

(c) Set Id = I to obtain:

(0.01 V/s )∈0 d

••

Picture the Problem We can use I d =∈0

A

=

A(0.01 V/s ) t ρd

t = ∈0 ρ

Solve for t:

22

829

dφe and the relationship between the voltage dt

across the plates and the electric field between them to find the displacement current. The conduction current between the plates is given by I =

V AV = where A is the area of R ρd

the plates and d is their separation.

dφe d dE =∈0 [EA] =∈0 A dt dt dt

(a) The displacement current is given by:

I d =∈0

Relate the electric field E to the potential difference V between the plates and the separation of the plates d:

E=

Substitute to obtain:

I d =∈0 A

V d

d ⎡V ⎤ ∈0 A dV = dt ⎢⎣ d ⎥⎦ d dt

V varies with time according to:

V = V0 cos ωt

Substituting in the expression for Id yields:

Id =

Substitute numerical values and evaluate Id:

∈0 A d [V0 cos ωt ] d dt ∈ π r 2V0 sin ωt =− 0 ωd

830 Chapter 30 Id

(8.85 × 10 =− (

C 2 / N ⋅ m 2 )π (20 cm ) (40 V ) sin (120π rad/s ) t (120π rad/s)(1 mm)

−12

2

)

= − 1.18 × 10 −10 A sin (120π rad/s ) t (b) The conduction current between the plates is given by:

I=

Substitute numerical values and simplify to obtain:

I=

V AV AV0 = = cos ωt ρd R ρd

π (0.2 m )2 (40 V )

(10

4

)(

Ω ⋅ m 10 −3 m

) cos(120π rad/s)t

(0.503 A ) cos(120π rad/s)t

=

*23 ••• Picture the Problem We can follow the step-by-step instructions in the problem statement to show that Equation 30-4 gives the same result for B as that given in Part (a). (a) Express the magnetic field at P using the expression for B due to a straight wire segment:

BP =

µ0 I (sin θ1 + sin θ 2 ) 4π R

where

sin θ1 = sin θ 2 = Substitute for sinθ1 and sinθ2 to obtain:

(b) Express the electric flux through the circular strip of radius r and width dr in the yz plane: The electric field due to the dipole is:

a R + a2 2

µ0 I 2a 4π R R 2 + a 2 1 µ Ia = 0 2 2π R R + a 2

BP =

dφe = E x dA = E x (2π rdr )

Ex =

2kQ 2kQa cos θ1 = 32 2 r +a r 2 + a2 2

(

)

Maxwell’s Equations and Electromagnetic Waves Substitute for Ex to obtain:

dφe = E x dA =

(c) Multiply both sides of the expression for φe by ∈0:

(r

2kQa + a2

2

)

32

2Qa

=

4π ∈0 (r 2 + a 2 )

=

Qa

32

∈0 (r 2 + a 2 )

∈0 dφe =

32

(r

Qa 2

+ a2

)

32

831

(2π rdr )

(2π rdr )

rdr

rdr

Integrate r from 0 to R to obtain: R

∈0 φe = Qa ∫ 0

(r

rdr 2

+a

)

2 32

⎛ ⎛ 1⎞ a −1 = Qa⎜⎜ + ⎟⎟ = Q⎜⎜1 − 2 2 a⎠ R2 + a2 ⎝ R +a ⎝

⎞ ⎟ ⎟ ⎠

dφe d ⎡ ⎛ a = ⎢Q⎜⎜1 − 2 dt dt ⎢⎣ ⎝ R + a2 ⎛ ⎞ dQ a ⎟ = ⎜⎜1 − 2 2 ⎟ dt R +a ⎠ ⎝ ⎛ ⎞ a ⎟ = − I ⎜⎜1 − 2 2 ⎟ R +a ⎠ ⎝

(d) The displacement current is defined to be:

I d =∈0

The total current is the sum of I and Id :

⎛ a I + I d = I − I ⎜⎜1 − R2 + a2 ⎝ a = I 2 R + a2

(e) Apply Equation 30-4 (the generalized form of Ampere’s law) to obtain:

∫

Solve for B:

B=

C

⎞ ⎟ ⎟ ⎠

r r B ⋅ d l = 2πRB = µ 0 (I + I d )

µ0 (I + I d ) 2π R

⎞⎤ ⎟⎥ ⎟ ⎠⎥⎦

832 Chapter 30 Substitute for I + Id from (d) to obtain:

B=

a µ0 ⎛ ⎜I ⎜ 2π R ⎝ R2 + a2

⎞ ⎟ ⎟ ⎠

µ 0 Ia 1 2 2π R R + a 2

=

Maxwell’s Equations and the Electromagnetic Spectrum 24 •• Picture the Problem The figure shows the end view of a pillbox surrounding a small area dA of the surface. The normal components of the magnetic field, v v Bn, top and B n, bottom , are shown with different magnitudes. When performing the surface integral the normal to the surface is outward, as shown in the figure. Apply Gauss’s law for magnetism to the pillbox to obtain:

r B ∫ ⋅ nˆ dA = S

r B ∫ ⋅ nˆ dA +

bottom surface

r B ∫ ⋅ nˆ dA +

lateral surface

r

r B ∫ ⋅ nˆ dA = 0

top surface

r B ∫ ⋅ nˆ dA = 0 , and:

Because the horizontal component of B is zero,

lateral surface

r B ∫ ⋅ nˆ dA = S

r B ∫ ⋅ nˆ dA +

bottom surface

r

Because B and nˆ are oppositely directed at the bottom surface:

r

Because B and nˆ are parallel at the top surface: Substitute in equation (1) to obtain: Solve for Bn,top:

r B ∫ ⋅ nˆ dA = 0

(1)

top surface

r B ∫ below ⋅ nˆ dA = − Bn,below A

bottom surface

r B ∫ below ⋅ nˆ dA = Bn,above A

top surface

− Bn, below A + Bn, above A = 0 Bn, above = Bn, below ; i.e., the normal r component of B is continuous across the surface.

Maxwell’s Equations and Electromagnetic Waves

833

*25 • Picture the Problem We can use c = fλ to find the wavelengths corresponding to the given frequencies. Solve c = fλ for λ:

λ=

c f

(a) For f = 1000 kHz:

λ=

3 × 108 m/s = 300 m 1000 × 103 s −1

(b) For f = 100 MHz:

λ=

3 × 108 m/s = 3.00 m 100 × 10 6 s −1

*26 • Picture the Problem We can use c = fλ to find the frequency corresponding to the given wavelength. Solve c = fλ for f:

f =

Substitute numerical values and evaluate f:

f =

c

λ 3 × 108 m/s = 1010 Hz = 10.0 GHz −2 3 × 10 m

27 • Picture the Problem We can use c = fλ to find the frequency corresponding to the given wavelength. Solve c = fλ for f:

Substitute numerical values and evaluate f:

f =

f =

c

λ 3 × 108 m/s = 3.00 × 1018 Hz −9 0.1 × 10 m

Electric Dipole Radiation 28 •• Picture the Problem We can use the intensity I1 at a distance r = 10 m and at an angle θ = 90° to find the proportionality constant in the expression for the intensity of radiation from an electric dipole and then use the resulting equation to find the intensity at the given distances and angles.

834 Chapter 30

Express the intensity of radiation as a function of r and θ :

I (θ , r ) =

C sin 2 θ r2

(1)

where C is a constant of proportionality. Express I(90°,10 m):

I (90°,10 m ) = I1 = =

(

C 100 m 2

)

Solve for C:

C = 100 m 2 I1

Substitute in equation (1) to obtain:

I (θ , r ) =

(a) Evaluate equation (2) for r = 30 m and θ = 90°:

(100 m )I 2

r

I (90°,30 m ) = =

(b) Evaluate equation (2) for r = 10 m and θ = 45°:

I (45°,10 m ) = =

(c) Evaluate equation (2) for r = 20 m and θ = 30°:

C sin 2 90° 2 (10 m )

1

2

sin 2 θ

(100 m ) I 2

(30 m )2

1 9

1

sin 2 90°

1

sin 2 45°

I1

(100 m )I 2

(10 m )

1 2

2

I1

(100 m ) I I (30°,20 m ) = 2

(20 m )

=

(2)

1 16

2

1

sin 2 30°

I1

29 •• Picture the Problem We can use the intensity I1 at a distance r = 10 m and at an angle θ = 90° to find the proportionality constant in the expression for the intensity of radiation from an electric dipole and then use the resulting equation to find the angle for a given intensity and distance and the distance corresponding to a given intensity and angle. Express the intensity of radiation as a function of r and θ :

I (θ , r ) =

C sin 2 θ 2 r

(1)

where C is a constant of proportionality.

Maxwell’s Equations and Electromagnetic Waves Express I(90°,10 m):

I (90°,10 m ) = I1 =

)

Solve for C:

C = 100 m 2 I1

Substitute in equation (1) to obtain:

I (θ , r ) =

(a) For r = 5 m and I(θ,r) = I1:

(100 m )I 2

r

(100 m )I =

2

2

I1

C sin 2 90° 2 (10 m )

C 100 m 2

=

(

835

(5 m )

1

2

1

sin 2 θ

(2)

sin 2 θ

or

sin 2 θ = Solve for θ to obtain: (b) For θ = 45° and I(θ,r) = I1:

θ = sin −1 I1 = r2 = r=

1 2

= 30.0°

(100 m ) I 2

1

r2

or

Solve for r to obtain:

1 4

1 2

sin 2 45°

(100 m )

1 2

2

(100 m ) = 2

7.07 m

30 •• Picture the Problem We can use the intensity I at a distance r = 4000 m and at an angle θ = 90° to find the proportionality constant in the expression for the intensity of radiation from an electric dipole and then use the resulting equation to find the intensity at sea level and 1.5 km from the transmitter. Express the intensity of radiation as a function of r and θ :

I (θ , r ) =

C sin 2 θ 2 r

(1)

where C is a constant of proportionality. Use the given data to obtain:

C sin 2 90° (4 km )2 C = (4 km )2

4 × 10 −12 W/m 2 =

836 Chapter 30

(

C = (4 km ) 4 × 10 −12 W/m 2 2

Solve for C:

)

−5

= 6.40 × 10 W Substitute in equation (1) to obtain:

I (θ , r ) =

6.40 × 10 −5 W 2 sin θ r2

For a point at sea level and 1.5 km from the transmitter:

θ = tan −1

2 km = 53.1° 1.5 km

(2)

Evaluate I(53.1°,1.5 km):

6.40 × 10 −5 W 2 I (53.1°,1.5 km ) = sin 53.1° = 18.2 pW/m 2 2 (1.5 km ) 31 ••• Picture the Problem The intensity of radiation from an electric dipole is equal to I0(sin2θ)/r2, where θ is the angle between the electric dipole moment and the position r vector r . We can integrate the intensity to express the total power radiated by the antenna and use this result to evaluate I0. Knowing I0 we can find the intensity at a horizontal distance of 120 km directly in front of the station.

sin 2 θ r2

Express the intensity of the signal as a function of r and θ :

I (r ,θ ) = I 0

At a horizontal distance of 120 km from the station and directly in front of it:

I (120 km,90°) = I 0

From the definition of intensity we have:

dP = IdA

sin 2 90° (120 km )2 I0 = (120 km )2

and

Ptot = ∫∫ I (r , θ ) dA

where, in polar coordinates,

dA = r 2 sin θ dθ dφ Substitute for dA to obtain:

2π π

Ptot =

∫ ∫ I (r ,θ ) r 0 0

2

sin θ dθ dφ

(1)

Maxwell’s Equations and Electromagnetic Waves Substitute for I(r,θ):

837

2π π

Ptot = I 0 ∫ ∫ sin 3 θ dθ dφ 0 0

From integral tables we find that:

π

∫ sin

3

θ dθ = − 13 cos θ (sin 2 θ + 2 )]0 = π

0

4 3

2π

Substitute and integrate with respect to φ to obtain:

4 4 8π 2π Ptot = I 0 ∫ dφ = I 0 [φ ] 0 = I0 3 0 3 3

Solve for I0:

I0 =

3 Ptot 8π

Substitute for Ptot and evaluate I0:

I0 =

3 (500 kW ) = 59.7 kW 8π

Substitute for I0 in equation (1) and evaluate I(120 km,90°):

I (120 km,90°) =

59.7 kW (120 km )2

= 4.15 µW/m 2 Express the number of photons incident on an area A in time ∆t:

N N NI = = A∆t (P I )∆t P∆t =

Substitute numerical values and evaluate I/hf:

NI I I = = E E N hf

I 4.15 µW/m 2 = hf 6.63 × 10 -34 J ⋅ s (1.20 MHz ) photons = 5.21 × 10 21 m2 ⋅ s photons = 5.22 × 1017 cm 2 ⋅ s

(

)

*32 ••• Picture the Problem The intensity of radiation from an electric dipole is given by I0(sin2θ)/r2, where θ is the angle between the electric dipole moment and the position r vector r . We can integrate the intensity to express the total power radiated by the antenna and use this result to evaluate I0. Knowing I0 we can find the total power radiated by the station. From the definition of intensity we

dP = IdA

838 Chapter 30 have:

and

Ptot = ∫∫ I (r , θ ) dA

where, in polar coordinates,

dA = r 2 sin θ dθ dφ Substitute for dA to obtain:

2π π

Ptot =

∫ ∫ I (r ,θ ) r

2

sin θ dθ dφ

0 0

Express the intensity of the signal as a function of r and θ : Substitute for I(r,θ):

sin 2 θ I (r ,θ ) = I 0 2 r

(1)

2π π

Ptot = I 0 ∫ ∫ sin 3 θ dθ dφ 0 0

From integral tables we find that:

π

∫ sin

3

θ dθ = − 13 cos θ (sin 2 θ + 2 )]0 = π

0

2π

Substitute and integrate with respect to φ to obtain:

4 4 8π 2π Ptot = I 0 ∫ dφ = I 0 [φ ] 0 = I0 3 0 3 3

From equation (1) we have:

I0 =

Substitute to obtain:

Substitute numerical values and evaluate Ptot:

4 3

I (r ,θ )r 2 sin 2 θ

8π I (r ,θ )r 2 3 sin 2 θ or, because θ = 90°, 8π Ptot = I (r )r 2 3 Ptot =

Ptot =

8π 2 2 ×10 −13 W/m 2 (30 km ) 3

(

)

= 1.51 mW 33 ••• Picture the Problem The intensity of radiation from the airport’s vertical dipole antenna is given by I0(sin2θ)/r2, where θ is the angle between the electric dipole moment and the r position vector r . We can integrate the intensity to express the total power radiated by the antenna and use this result to evaluate I0. Knowing I0 we can find the intensity of the

Maxwell’s Equations and Electromagnetic Waves

839

signal at the plane’s elevation and distance from the airport. Express the intensity of the signal as a function of r and θ :

I (r ,θ ) = I 0

From the definition of intensity we have:

dP = IdA

sin 2 θ r2

(1)

and

Ptot = ∫∫ I (r , θ ) dA

where, in polar coordinates,

dA = r 2 sin θ dθ dφ Substitute for dA to obtain:

2π π

Ptot =

∫ ∫ I (r ,θ ) r

2

sin θ dθ dφ

0 0

Substitute for I(r,θ):

2π π

Ptot = I 0 ∫ ∫ sin 3 θ dθ dφ 0 0

From integral tables we find that:

π

∫ sin

3

θ dθ = − 13 cos θ (sin 2 θ + 2)]0 = π

0

4 3

2π

Substitute and integrate with respect to φ to obtain:

4 4 8π 2π Ptot = I 0 ∫ dφ = I 0 [φ ] 0 = I0 3 0 3 3

Solve for I0:

I0 =

Substitute for I0 in equation (1):

I (r ,θ ) =

At the elevation of the plane:

θ = tan −1 ⎜⎜

3 Ptot 8π 3Ptot sin 2 θ 8π r 2 ⎛ 2500 m ⎞ ⎟⎟ = 32.0° ⎝ 4000 m ⎠

and

r= Substitute numerical values and evaluate I(4717 m,32°):

(2500 m )2 + (4000 m )2

I (4717 m,32°) =

= 4717 m

3(100 W ) sin 2 32° 8π (4717 m )2

= 0.151 µW/m 2

840 Chapter 30

Energy and Momentum in an Electromagnetic Wave 34 • Picture the Problem We can use Pr = I/c to find the radiation pressure. The intensity of the electromagnetic wave is related to the rms values of its electric and magnetic fields according to I = ErmsBrms/µ0, where Brms = Erms/c.

I c

(a) Express the radiation pressure in terms of the intensity of the wave:

Pr =

Substitute numerical values and evaluate Pr:

100 W/m 2 Pr = = 0.333 µPa 3 × 108 m/s

(b) Relate the intensity of the electromagnetic wave to Erms and Brms:

I=

Erms Brms

µ0

or, because Brms = Erms/c,

I=

Erms Erms c

µ0

=

2 Erms µ0c

Erms = µ0cI

Solve for Erms: Substitute numerical values and evaluate Erms:

Erms =

(4π × 10

−7

)(

)(

)

N/A 2 3 × 108 m/s 100 W/m 2 = 194 V/m

(c) Express Brms in terms of Erms:

Brms =

Erms c

Substitute numerical values and evaluate Brms:

Brms =

194 V/m = 0.647 µT 3 × 108 m/s

35 • Picture the Problem The rms values of the electric and magnetic fields are found from their amplitudes by dividing by the square root of two. The rms values of the electric and magnetic fields are related according to Brms = Erms/c. We can find the intensity of the radiation using I = ErmsBrms/µ0 and the radiation pressure using Pr = I/c. (a) Relate Erms to E0:

E rms =

E0 2

=

400 V/m 2

= 283 V/m

Maxwell’s Equations and Electromagnetic Waves (b) Find Brms from Erms:

841

Erms 283 V/m = c 3 × 108 m/s

Brms =

= 0.943 µT (c) The intensity of an electromagnetic wave is given by:

I=

Erms Brms

Substitute numerical values and evaluate I:

I=

(283 V/m )(0.943 µT ) =

(d) Express the radiation pressure in terms of the intensity of the wave:

Pr =

I c

Substitute numerical values and evaluate Pr:

Pr =

212 W/m 2 = 0.707 µPa 3 × 108 m/s

µ0

4π × 10 N/A −7

2

212 W/m 2

36 • Picture the Problem Given Erms, we can find Brms using Brms = Erms/c. The average energy density of the wave is given by uav = ErmsBrms/µ0c and the intensity of the wave by I = uavc . (a) Express Brms in terms of Erms:

Brms =

Erms c

Substitute numerical values and evaluate Brms:

Brms =

400 V/m = 1.33 µT 3 × 108 m/s

(b) The average energy density uav is given by:

uav =

Substitute numerical values and evaluate uav:

uav =

Erms Brms µ0c

(400 V/m )(1.33 µT )

(4π × 10

−7

= 1.41 µJ/m 3 (c) Express the intensity as the product of the average energy density and the speed of light in a vacuum:

I = uav c

)(

N/A 2 3 × 108 m/s

)

842 Chapter 30 Substitute numerical values and evaluate I:

(

)(

I = 1.41 µJ/m 3 3 × 108 m/s

)

= 423 W/m 2

37 • Picture the Problem We can simplify the units of cB to show that this product has the same units as E. Express the units of cB and simplify:

m m N m N N N m J V ×T = × = × = = × = = C s s A⋅m s m ⋅m C C m C⋅m s *38 • Picture the Problem Given Brms, we can find Erms using Erms = cBrms. The average energy density of the wave is given by uav = ErmsBrms/µ0c and the intensity of the wave by I = uavc. (a) Express Erms in terms of Brms: Substitute numerical values and evaluate Erms:

Erms = cBrms

(

)

Erms = 3 × 108 m/s (0.245 µT ) = 73.5 V/m

(b) The average energy density uav is given by:

uav =

Substitute numerical values and evaluate uav:

uav =

Erms Brms µ0c

(73.5 V/m )(0.245 µT )

(4π × 10

−7

)(

N/A 2 3 × 108 m/s

= 47.8 nJ/m 3 (c) Express the intensity as the product of the average energy density and the speed of light in a vacuum: Substitute numerical values and evaluate I:

I = uav c

(

)(

I = 47.8 nJ/m 3 3 × 108 m/s = 14.3 W/m 2

)

)

Maxwell’s Equations and Electromagnetic Waves

843

39 •• Picture the Problem We can find the force exerted on the card using the definition of pressure and the relationship between radiation pressure and the intensity of the electromagnetic wave. Note that, when the card reflects all the radiation incident on it, conservation of momentum requires that the force is doubled. (a) Using the definition of pressure, express the force exerted on the card by the radiation:

F = Pr A

Relate the radiation pressure to the intensity of the wave:

Pr =

I c

Substitute to obtain:

F=

IA c

Substitute numerical values and evaluate F:

F=

(200 W/m )(0.2 m )(0.3 m) 2

3 × 108 m/s

= 40.0 nN (b) If the card reflects all of the radiation incident on it, the force exerted on the card is doubled:

F = 80.0 nN

40 •• Picture the Problem Only the normal component of the radiation pressure exerts a force on the card. (a) Using the definition of pressure, express the force exerted on the card by the radiation:

F = 2 Pr A cosθ where the factor of 2 is a consequence of the fact that the card reflects the radiation incident on it.

Relate the radiation pressure to the intensity of the wave:

Pr =

I c

Substitute to obtain:

F=

2 IA cosθ c

844 Chapter 30

(

)

2 200 W/m 2 (0.2 m )(0.3 m ) cos 30° F= 3 × 108 m/s

Substitute numerical values and evaluate F:

= 69.3 nN *41 •• Picture the Problem We can use I = Pav/4πr2 and I = ErmsBrms/µ0 to express Erms in terms of Pav and the distance r from the station. Express the intensity I of the radiation as a function of its average power and the distance r from the station: The intensity is also given by:

Equate these expressions to obtain:

Solve for Emax:

I=

Pav 4π r 2

I=

Erms Brms

µ0

2 2 Erms Emax = = cµ0 2cµ0

2 Pav Emax = 4π r 2 2cµ0

cµ0 Pav ⎛ 1 ⎞ ⎜ ⎟ 2π ⎝ r ⎠

Emax =

(a) Substitute numerical values and evaluate Emax for r = 500 m:

Emax (500 m ) =

(3 × 10

8

)(

)

m/s 4π × 10 −7 N/A 2 (50 kW ) ⎛ 1 ⎞ ⎟⎟ = 3.46 V/m ⎜⎜ 2π ⎝ 500 m ⎠

Use Bmax = Emax/c to evaluate Bmax:

Bmax =

3.46 V/m = 11.5 nT 3 × 108 m/s

(b) Substitute numerical values and evaluate Emax for r = 5 km:

Emax (5 km ) =

(3 × 10

8

)(

)

m/s 4π × 10 −7 N/A 2 (50 kW ) ⎛ 1 ⎞ ⎟⎟ = 0.346 V/m ⎜⎜ 2π ⎝ 5 km ⎠

Use Bmax = Emax/c to evaluate Bmax:

Bmax =

0.346 V/m = 1.15 nT 3 × 108 m/s

(c) Substitute numerical values and evaluate Emax for r = 50 km:

Maxwell’s Equations and Electromagnetic Waves Emax (500 m ) =

(3 × 10

8

)(

845

)

m/s 4π × 10 −7 N/A 2 (50 kW ) ⎛ 1 ⎞ ⎟⎟ = 0.0346 V/m ⎜⎜ 2π ⎝ 50 km ⎠

Use Bmax = Emax/c to evaluate Bmax:

Bmax =

0.0346 V/m = 0.115 nT 3 × 108 m/s

42 •• Picture the Problem We can use I = Pav/A to express Erms in terms of I. We can then use Brms = Erms/c to find Brms. The average power output of the sun is given by Pav = 4πR 2 I, where R is the earth-sun distance. The intensity and the radiation pressure at the surface of the sun can be found from the definitions of these physical quantities.

Pav 4 Pav = A π d2

(a) From the definition of intensity we have:

I=

Substitute numerical values and evaluate I:

I=

π (10 m )

(b) Express the intensity I of the radiation as a function of its average power and the distance r from the station:

I=

Erms Brms

Solve for Erms:

Erms = cµ0 I

4(1.5 mW ) −3

µ0

2

=

= 1.91kW/m 2

2 Erms cµ0

Substitute numerical values and evaluate Erms:

Erms =

(3 × 10

8

)(

)(

)

m/s 4π × 10 −7 N/A 2 1.91 kW/m 2 = 849 V/m

Use Brms = Erms/c to evaluate Brms:

Brms =

(d) Express the radiation pressure in terms of the intensity:

Pr =

I c

Substitute numerical values and evaluate Pr:

Pr =

1.91 × 103 W/m 2 = 6.37 µPa 3 × 108 m/s

849 V/m = 2.83 µT 3 × 108 m/s

846 Chapter 30 *43 •• Picture the Problem We can use I = ErmsBrms/µ0 and Brms = Erms/c to express Erms in terms of I. We can then use Brms = Erms/c to find Brms.

Erms Brms

Express the intensity I of the radiation as a function of its average power and the distance r from the station:

I=

Solve for Erms:

Erms = cµ0 I

Use the definition of intensity to relate the intensity of the electromagnetic wave to the power in the beam:

I=

Substitute for I to obtain:

Erms =

µ0

=

2 Erms cµ0

P I trans. lineV = A A

cµ0 I trans. lineV A

Substitute numerical values and evaluate Erms:

Erms =

(3 × 10

8

)(

)(

)

Brms =

75.2 kV/m = 0.251 mT 3 × 108 m/s

m/s 4π × 10 −7 N/A 2 103 A (750 kV ) = 75.2 kV/m 50 m 2

Use Brms = Erms/c to evaluate Brms:

44 •• Picture the Problem The spatial length L of the pulse is the product of its speed c and duration ∆t. We can find the energy density within the pulse using its definition (u = U/V). The electric amplitude of the pulse is related to the energy density in the beam according to u =∈0 E 2 and we can find B from E using B = E/c. (a) The spatial length L of the pulse is the product of its speed c and duration ∆t: Substitute numerical values and evaluate L:

L = c∆t

(

)

L = 3 × 108 m/s (10 ns ) = 3.00 m

Maxwell’s Equations and Electromagnetic Waves (b) The energy density within the pulse is the energy of the beam per unit volume:

u=

U U = V π r 2L

Substitute numerical values and evaluate u:

u=

20 J = 531kJ/m 3 2 π (2 mm ) (3.00 m )

(c) E is related to u according to: Solve for E0 to obtain:

Substitute numerical values and evaluate E0:

847

2 u =∈0 Erms = 12 ∈0 E02

2u

E0 =

∈0

(

)

2 531 kJ/m 3 8.85 × 10 −12 C 2 / N ⋅ m 2

E0 =

= 346 MV/m Use B0 = E0/c to find B0:

B0 =

346 MV/m = 1.15 T 3 × 108 m/s

*45 •• Picture the Problem We can determine the direction of propagation of the wave, its wavelength, and its frequency by examining the argument of the cosine function. We can

r

find E from S = E 2 µ 0 c and B from B = E/c. Finally, we can use the definition of the

r

r

r

Poynting vector and the given expression for S to find E and B .

(a)

Because the argument of the cosine function is of the form kx − ωt , the wave propagates in the positive x direction.

2π

(b) Examining the argument of the cosine function, we note that the wave number k of the wave is:

k=

Solve for and evaluate λ:

λ=

Examining the argument of the cosine function, we note that the angular frequency ω of the wave is:

ω = 2πf = 3 ×109 s −1

λ

= 10 m −1

2π = 0.628 m 10 m −1

848 Chapter 30 3 × 109 s −1 f = = 477 MHz 2π

Solve for and evaluate f to obtain:

r

(c) Express the magnitude of S in terms of E:

r E2 S = µ0c

Solve for E:

r E = µ0c S

Substitute numerical values and evaluate E:

(3 × 10

E=

8

)(

)(

)

m/s 4π × 10 −7 N/A 2 100 W/m 2 = 194 V/m

[

)]

r r 1 r r E×B: Because S ( x, t ) = 100 W/m 2 cos 2 10 x − 3 × 109 t iˆ and S =

(

)

(

[

µ0

)]

r E ( x, t ) = (194 V/m ) cos 10 x − 3 × 109 t ˆj Use B = E/c to evaluate B:

r

Because S =

B=

(

194 V/m = 0.647 µT 3 × 108 m/s

r r 1 r r E × B , the direction of B must be such that the cross product of E

µ0

r with B is in the positive x direction:

[

)]

r B (x, t ) = (0.647 µT ) cos 10 x − 3 × 109 t kˆ

(

46 •• Picture the Problem We can use the definition of the electric field between the plates of the parallel-plate capacitor and the definition of the displacement current to show that the displacement current in the capacitor is equal to the conduction current in the capacitor leads. In (b) we can use the definition of the Poynting vector and the directions of the electric and magnetic fields to determine the direction of the Poynting vector between the

r

capacitor plates. In (c), we’ll demonstrate that the flux of S into the region between the plates is equal to the rate of change of the energy stored in the capacitor by evaluating these quantities separately and showing that they are equal.

(

V (t ) V = 1 − e −t / RC d d

(a) The electric field between the plates of the capacitor is given by:

E=

The displacement current is proportional to the rate at which the

I D (t ) =∈0

)

dφe d dE =∈0 ( AE ) =∈0 A dt dt dt

Maxwell’s Equations and Electromagnetic Waves flux is changing between the plates: Substitute for E and carry out the details of the differentiation to obtain:

(

)

[(

)]

d ⎡V ⎤ 1 − e −t / RC ⎥ ⎢ dt ⎣ d ⎦ ∈ AV d = 0 1 − e −t / RC d dt ∈ AV d = 0 − e −t / RC d dt ∈ AV −t / RC = 0 e dRC

I D (t ) =∈0 A

[

Because the capacitance of an airfilled-parallel-plate capacitor is given by C =

I D (t ) =

∈0 A : d

(b) Apply Ampere’s law to a closed circular path of radius r (the radius of the capacitor plates) to obtain:

]

CV −t / RC e = I (t ) RC

B(2π r ) = µ 0 I C = µ 0 I D

Substitute for ID from (a):

B(2π r ) = µ 0 ∈0

Solve for B to obtain:

B = µ 0 ∈0

πr 2V

d (RC )

e −t / RC

rV e −t / RC 2d (RC )

r r Because E is perpendicular to the plates of the capacitor and B is tangent to circles that are concentric and whose center is through the middle of the r capacitor plates, S points radially inward toward the center of the capacitor. (c) The magnitude of the Poynting vector is:

r BE S =I=

µ0

Substitute for B and E and simplify to obtain:

I=

∈0 V 2 r −t / RC e 1 − e −t / RC 2 2 d RC

The total power is:

P=

dE = 2π rdI dt

(

)

849

850 Chapter 30 Substitute for I to obtain:

dE V 2πr 2 −t / RC ( =∈0 e 1 − e −t / RC ) dt dRC

Because the capacitance of an airfilled-parallel-plate capacitor is

dE V 2 −t / RC = e ( 1 − e −t / RC ) dt R

∈0 π r 2 : given by C = d

(1)

1 2 C [V (t )] 2

The energy in the capacitor at any time is:

E=

Differentiate E with respect to time to obtain:

dE d ⎡ 1 dV (t ) 2⎤ = ⎢ C (V (t )) ⎥ = CV (t ) dt dt ⎣ 2 dt ⎦

Substitute for V(t) and complete the differentiation to obtain:

dE V 2 −t / RC = e ( 1 − e −t / RC ) dt R

(2)

r The equivalence of equations (1) and (2) proves that the flux of S into this region is equal to the rate of change of the energy stored in the capacitor. 47 •• Picture the Problem The diagram shows the displacement of the pendulum bob, through an angle θ, as a consequence of the complete absorption of the radiation incident on it. We can use conservation of energy (mechanical energy is conserved after the collision) to relate the maximum angle of deflection of the pendulum to the initial momentum of the pendulum bob. Because the displacement of the bob during the absorption of the pulse is negligible, we can use conservation of momentum (conserved during the collision) to equate the momentum of the electromagnetic pulse to the initial momentum of the bob. Apply conservation of energy to obtain:

Kf − Ki + U f − U i = 0 or, since Ui = Kf = 0 and K i = pi2 2m ,

Maxwell’s Equations and Electromagnetic Waves

851

pi2 − + Uf = 0 2m Uf is given by:

U f = mgh = mgL(1 − cos θ )

Substitute for Uf:

pi2 − + mgL(1 − cos θ ) = 0 2m

Solve for θ to obtain:

⎛ p2 ⎞ θ = cos −1 ⎜⎜1 − 2i ⎟⎟ ⎝ 2m gL ⎠

Use conservation of momentum to relate the momentum of the electromagnetic pulse to the initial momentum pi of the pendulum bob:

pem wave =

Substitute for pi:

U P∆t = = pi c c

where ∆t is the duration of the pulse.

⎛

θ = cos −1 ⎜⎜1 − ⎝

2 P 2 (∆t ) ⎞ ⎟ 2m 2 c 2 gL ⎟⎠

Substitute numerical values and evaluate θ : 2 2 ⎛ ⎞ ( 1000 MW ) (200 ns ) ⎜ ⎟ = 6.10 × 10 −3 degrees θ = cos 1 − ⎜ 2(10 mg )2 (3 × 108 m/s)2 (9.81 m/s 2 )(0.04 m ) ⎟ ⎝ ⎠

−1

Remarks: The solution presented here is valid only if the displacement of the bob during the absorption of the pulse is negligible. (Otherwise, the horizontal component of the momentum of the pulse-bob system is not conserved during the collision.) We can show that the displacement during the pulse-bob collision is small by solving for the speed of the bob after absorbing the pulse. Applying conservation of momentum (mv = P(∆t)/c) and solving for v gives v = 6.67×10−7 m/s. This speed is so slow compared to c, we can conclude that the duration of the collision is extremely close to 200 ns (the time for the pulse to travel its own length). Traveling at 6.67×10−7 m/s for 200 ns, the bob would travel 1.33×10−13 m—a distance 1000 times smaller that the diameter of a hydrogen atom. (Since 6.67×10−7 m/s is the maximum speed of the bob during the collision, the bob would actually travel less than 1.33×10−13 m during the collision.) 48 •• Picture the Problem We can use the definitions of pressure and the relationship between radiation pressure and the intensity of the radiation to find the force due to radiation pressure on one of the mirrors.

852 Chapter 30 (a) Because only about 0.01 percent of the energy inside the laser "leaks out", the average power of the radiation incident on one of the mirrors is:

P=

(b) Use the definition of radiation pressure to obtain:

Pr =

15 W = 1.50 × 105 W 10 −4

F A

where F is the force due to radiation pressure and A is the area of the mirror on which the radiation is incident. The radiation pressure is also related to the intensity of the radiation:

Pr =

2I 2P = c Ac

where P is the power of the laser and the factor of 2 is due to the fact that the mirror is essentially totally reflecting. Equate the two expression for the radiation pressure and solve for F:

F 2P 2P ⇒ F= = A Ac c

Substitute numerical values and evaluate F:

F=

(

)

2 1.50 × 105 W = 1.00 mN 3 × 108 m/s

49 •• Picture the Problem The card, pivoted at point P, is shown in the diagram. Note that the force exerted by the radiation acts along the dashed line. Let the length of the card be l, the width of the card be w, and the force acting on an area dA = w dx be dFradiation. We can find the total torque exerted on the card due to radiation pressure by integrating dτradiation over the length l of the card and then relate the intensity of the light to the angle θ by applying the condition for rotational equilibrium to the card. Express the torque, due to F, acting at a distance x from P:

dτ radiation = xdFradiation

Maxwell’s Equations and Electromagnetic Waves Relate dFradiation to the intensity of the light:

dFradiation =

853

2I cos θ dA c

where the factor of 2 arises from the total reflection of the radiation incident on the mirror.

2I cos θ xdA c 2I = cos θ xwdx c

Substitute to obtain:

dτ radiation =

Integrate x from 0 to l:

τ radiation =

l

2 Iw cos θ ∫ xdx c 0

⎛ l 2 ⎞ IAl 2 Iw = cos θ ⎜⎜ ⎟⎟ = cos θ c c ⎝2⎠ Apply

∑τ

P

= 0 to the card:

Solve for I to obtain:

IAl cos θ − ( 12 l sin θ )mg = 0 c mgc tan θ 2A

I=

Substitute numerical values and evaluate I:

I=

(2 g ) (9.81 m/s 2 )(3 × 108 m/s) tan 1° = 2(0.1 m )(0.15 m )

3.42 MW/m 2

The Wave Equation for Electromagnetic Waves 50 • Picture the Problem We can show that Equation 30-17a is satisfied by the wave function Ey by showing that the ratio of ∂2Ey/∂x2 to ∂2Ey/∂t2 is 1/c2 where c = ω/k. Differentiate

E y = E0 sin (kx − ωt ) with respect

to x: Evaluate the second partial derivative of Ey with respect to x:

∂E y

∂ [E0 sin(kx − ωt )] ∂x ∂x = kE0 cos(kx − ωt ) =

∂2Ey ∂x

2

∂ [kE0 cos(kx − ωt )] ∂x = − k 2 E0 sin(kx − ωt )

=

(1)

854 Chapter 30 Differentiate E y = E0 sin (kx − ωt )

∂E y

∂ [E0 sin(kx − ωt )] ∂t ∂t = −ωE0 cos(kx − ωt )

with respect to t:

=

Evaluate the second partial derivative of Ey with respect to t:

∂2Ey

Divide equation (1) by equation (2) to obtain:

∂2 Ey 2 2 ∂x 2 = − k E0 sin (kx − ωt ) = k ∂ 2 E y − ω 2 E0 sin (kx − ωt ) ω 2 ∂t 2

∂t

2

∂ [− ωE0 cos(kx − ωt )] (2) ∂t = −ω 2 E 0 sin(kx − ωt ) =

or 2 2 1 ∂ Ey k 2 ∂ Ey = 2 = 2 ω ∂t 2 c ∂t 2 ∂x 2 provided c = ω/k.

∂2Ey

51 • Picture the Problem Substitute numerical values and evaluate c:

c=

(4π ×10

1

−7

N/A

2

)(8.85 ×10

−12

C / N⋅m 2

2

)

= 3.00 ×108 m/s

*52 ••• Picture the Problem We can use Figures 30-10 and 30-11and a derivation similar to that in the text to obtain the given results.

∂E z ∆x ∂x

In Figure 30-11, replace Bz by Ez. For ∆x small:

E z ( x2 ) = E z ( x1 ) +

Evaluate the line integral of r E around the rectangular area ∆x∆z:

r r ∂E E ∫ ⋅ d l ≈ − ∂xz ∆x∆z

Express the magnetic flux through the same area: Apply Faraday’s law to obtain:

(1)

∫ B dA = B ∆x∆z S

n

r

y

r

∂

∂

∫ E ⋅ d l ≈ − ∂ t ∫ B dA = − ∂ t (B ∆x∆z ) S

y

n

=−

∂B y ∂t

∆x∆z

Maxwell’s Equations and Electromagnetic Waves Substitute in equation (1) to obtain:

855

∂B ∂E z ∆x∆z = − y ∆x∆z ∂x ∂t

− or

∂E z ∂B y = ∂t ∂x In Figure 30-10, replace Ey by By and evaluate the line integral of r B around the rectangular area ∆x∆z: Evaluate these integrals to obtain:

r

r

∫ B ⋅ dl = µ

∈0

∂B y

= µ0 ∈0

S

En dA

∂E z ∂t

∂ ⎛ ∂E z ⎞ ∂ ⎛ ∂B y ⎜ ⎟= ⎜ ∂x ⎝ ∂x ⎠ ∂x ⎜⎝ ∂ t

⎞ ⎟⎟ ⎠

or

∂ 2 Ez ∂ ⎛ ∂B y = ⎜⎜ 2 ∂x ∂ t ⎝ ∂x Use the second result obtained in (a) to obtain:

∫

provided there are no conduction currents.

∂x (b) Using the first result obtained in (a), find the second partial derivative of Ez with respect to x:

0

⎞ ⎟⎟ ⎠

∂ 2 Ez ∂ ⎛ ∂E z ⎞ ∂ 2 Ez ⎜ ⎟ = µ ε µ ∈ = 0 0 0 0 ∂x 2 ∂ t ⎜⎝ ∂ t ⎟⎠ ∂t2 or, because µ0∈0 = 1/c2,

∂ 2 Ez 1 ∂ 2 Ez = . ∂x 2 c2 ∂ t 2 Using the second result obtained in (a), find the second partial derivative of By with respect to x:

∂ ⎛ ∂B y ⎜ ∂x ⎜⎝ ∂x or

∂ 2 By ∂x Use the second result obtained in (a) to obtain:

⎞ ∂ ⎛ ∂E z ⎞ ⎟⎟ = µ 0 ∈0 ⎜ ⎟ ∂x ⎜⎝ ∂ t ⎟⎠ ⎠

2

∂ 2 By ∂x 2

= µ 0 ∈0

∂ ⎛ ∂E z ⎞ ⎜ ⎟ ∂ t ⎝ ∂x ⎠

= µ 0 ∈0

∂ ⎛ ∂B y ⎜ ∂ t ⎜⎝ ∂ t

or, because µ0∈0 = 1/c2,

∂ 2 By ∂x 2

=

2 1 ∂ By . c2 ∂ t 2

∂ 2 By ⎞ ⎟⎟ = µ 0 ∈0 ∂t2 ⎠

856 Chapter 30 53 ••• Picture the Problem We can show that these functions satisfy the wave equations by differentiating them twice (using the chain rule) with respect to x and t and equating the expressions for the second partial of f with respect to u. Let u = x − vt. Then:

∂ f ∂u ∂ f ∂ f = = ∂ x ∂ x ∂u ∂u and

∂ f ∂u ∂ f ∂f = = −v ∂t ∂u ∂u ∂t

Express the second derivatives of f with respect to x and t to obtain:

∂2 f ∂2 f = ∂ x2 ∂ u 2 and 2 ∂2 f 2 ∂ f = v ∂t2 ∂ u2

Thus, for any f(u):

Let u = x + vt. Then:

∂2 f 1 ∂2 f = ∂ x2 v2 ∂ t 2 ∂ f ∂u ∂ f ∂ f = = ∂ x ∂ x ∂u ∂u and

∂ f ∂u ∂ f ∂f = =v ∂t ∂u ∂u ∂t

Express the second derivatives of f with respect to x and t to obtain:

∂2 f ∂2 f = ∂ x2 ∂ u 2 and 2 ∂2 f 2 ∂ f =v ∂t2 ∂ u2

Thus, for any f(u):

∂2 f 1 ∂2 f = ∂ x2 v2 ∂ t 2

Maxwell’s Equations and Electromagnetic Waves

857

General Problems 54 • Picture the Problem We can substitute the appropriate units and simplify to show that the units of the Poynting vector are watts per square meter and that those of radiation pressure are newtons per square meter.

r

(a) Express the units of S and simplify:

(b) Express the units of Pr and simply:

J N × V C⋅m C⋅ m ×T s m = N N A2 A2 J J =C = = W s s C J W N⋅m 2 2 m = s ⋅ m = m2 = N m m m m2 s s

55 •• Determine the Concept The current induced in a loop antenna is proportional to the time-varying magnetic field. For maximum signal, the antenna’s plane should make an angle θ = 0° with the line from the antenna to the transmitter. For any other angle, the induced current is proportional to cos θ. The intensity of the signal is therefore proportional to cos θ. 56 •• Picture the Problem We can use c = fλ to find the wavelength. Examination of the argument of the cosine function will reveal the direction of propagation of the wave. We can find the magnitude, wave number, and angular frequency of the electric vector from r the given information and the result of (a) and use these results to obtain E (z, t). Finally, we can use its definition to find the Poynting vector. (a) Relate the wavelength of the wave to its frequency and the speed of light:

λ=

c f

Substitute numerical values and evaluate λ:

λ=

3 × 108 m/s = 3.00 m 100MHz

858 Chapter 30 From the sign of the argument of the cosine function and the spatial dependence on z , we can conclude that the wave propagates in the z direction.

(

r

)(

E = cB = 3 × 108 m/s 10 −8 T = 3.00 V/m

(b) Express the amplitude of E :

)

ω = 2π f = 2π (100 MHz ) = 6.28 × 108 s −1

Find the angular frequency and wave number of the wave:

and

k= r

2π

λ

=

2π = 2.09 m −1 3.00 m

r

Because S is in the positive z direction, E must be in the negative y direction in order to satisfy the Poynting vector expression:

[

]

r E ( z , t ) = −(3.00 V/m ) cos (2.09 m −1 ) z − (6.28 × 108 s −1 )t ˆj (c) Use its definition to express the Poynting vector:

r 1 r r − (3.00 V/m ) (10 −8 T ) 2 S= E×B = cos (2.09 m −1 ) z − (6.28 × 108 s −1 )t ˆj × iˆ 2 −7 µ0 4π × 10 N/A

](

[

or

[

)

]

r S = (23.9 mW/m 2 )cos 2 (2.09 m −1 ) z − (6.28 × 108 s −1 )t kˆ

The intensity of the wave is the average magnitude of the Poynting vector. The average value of the square of the cosine function is 1/2:

r I= S =

1 2

(23.9 mW/m ) 2

= 12.0 mW/m 2

*57 •• Picture the Problem The maximum rms voltage induced in the loop is given by ε rms = AωB0 2 , where A is the area of the loop, B0 is the amplitude of the magnetic field, and ω is the angular frequency of the wave. We can use the definition of density and the expression for the intensity of an electromagnetic wave to derive an expression for B0. The maximum induced rms emf occurs when the plane of the loop is r perpendicular to B :

ε rms =

AωB0 2

=

πR 2ωB0 2

(1)

where R is the radius of loop of wire.

Maxwell’s Equations and Electromagnetic Waves From the definition of intensity we have:

I=

859

P 4π r 2

where r is the distance from the transmitter.

E0 B0 B02 c = 2µ0 2µ0

The intensity is also given by:

I=

Substitute to obtain:

B02 c P = 2µ 0 4π r 2

Solve for B0:

B0 =

1 µ0 P r 2π c

2 ε rms = π R (2π f )

Substitute in equation (1) to obtain:

2r

=

R2 f 2r

µ0 P 2π c

2π 3 µ 0 P c

Substitute numerical values and evaluate εrms:

ε rms

2 ( 0.3 m ) (100 MHz ) =

(

5

2 10 m

)

(

)

2π 3 4π × 10 −7 N/A 2 (50 kW ) = 7.25 mV 3 × 108 m/s

58 •• Picture the Problem The voltage induced in the piece of wire is the product of the electric field and the length of the wire. The maximum rms voltage induced in the loop is given by ε = AωB0 , where A is the area of the loop, B0 is the amplitude of the magnetic field, and ω is the angular frequency of the wave. (a) Because E is independent of x:

V = El where l is the length of the wire.

Substitute numerical values and evaluate V:

(b) The voltage induced in a loop is given by:

[(

)

]

V = 10−4 N/C cos106 t (0.5 m ) =

(50.0 µV )cos106 t

ε = ωB0 A where A is the area of the loop and B0 is the amplitude of the magnetic field.

860 Chapter 30

ωE0π R 2

Eliminate B0 in favor of E0 and substitute for A to obtain:

ε=

Substitute numerical values and

ε = (10

evaluate ε:

c 6

)(

)

s −1 10 −4 N/C π (0.2 m ) 3 × 108 m/s

2

= 41.9 nV 59 •• Picture the Problem Some of the charge entering the capacitor passes through the resistive wire while the rest of it accumulates on the upper plate. The total current is the rate at which the charge passes through the resistive wire plus the rate at which it accumulates on the upper plate. The magnetic field between the capacitor plates is due to both the current in the resistive wire and the displacement current though a surface bounded by a circle a distance r from the resistive wire. The phase difference between the supplied current and the applied voltage may be calculated using a phasor diagram.

(a) The current drawn by the capacitor is the sum of the conduction current through the resistance wire and dQ/dt, where Q is the charge on the upper plate of the capacitor:

I = Ic +

Express the conduction current Ic in terms of the potential difference between the plates and the resistance of the wire:

Ic =

Express the displacement current between the capacitor plates. Let C be the capacitance of the capacitor:

dQ dt

V V0 = sin ωt R R

Q = CV so

dQ dV =C = ωCV0 cos ωt dt dt

(1)

Maxwell’s Equations and Electromagnetic Waves Substitute in equation (1):

I=

V0 sin ωt + ωCV 0 cos ωt R

861

(2)

∈0 A ∈0 π a 2

Using Equation 24-10 for the capacitance of a parallel-plate capacitor with plate area A and plate separation d we have:

C=

Substituting for C equation 2 gives:

⎛1 ⎞ ω ∈0 π a 2 I = V0 ⎜⎜ sin ωt + cos ωt ⎟⎟ d ⎝R ⎠

(b) Apply the generalized form of Ampere’s law to a circular path of radius r centered within the plates of the capacitor, where I'd is the

d

=

d

r r B ∫ ⋅ d l = µ0 (I c + I'd ) C

displacement current through the flat surface S bounded by the path and Ic is the conduction current through the same surface: By symmetry the line integral is B times the circumference of the circle of radius r: In the region between the capacitor plates there is a uniform electric field due to the surface charges +Q and –Q. The associated displacement current through S is: To evaluate the displacement current we first must evaluate E everywhere on S. Near the surface of a conductor E = σ ∈0 (Equation 22-25), where σ is the surface charge density:

B(2π r ) = µ 0 (I c + I'd )

(3)

dφe d =∈0 ( A'E ) dt dt dE dE =∈0 A' =∈0 π r 2 dt dt provided (r ≤ a ) I'd =∈0

(

E = σ ∈0 , where σ = Q A = Q π a 2 so

E=

Q

∈0 π a 2

)

862 Chapter 30

dE d ⎛ Q ⎞ ⎟ =∈0 π r 2 ⎜⎜ dt dt ⎝∈0 π a 2 ⎟⎠ r 2 dQ r 2 d (V0 sin ωt ) = 2 = d dt d 2 dt r2 = ω 2 V0 cos ωt d

Substituting for E in the equation for I'd gives:

I'd =∈0 π r 2

Substituting for Ic and I'd in

B(r ) =

equation (3) and solving for B gives:

(c) Both the charge Q and the conduction current Ic are in phase with V. However, dQ/dt, which is equal to the displacement current Id through S for r ≥ a, lags V by 90°. (Mathematically, cos ωt lags behind sin ωt by 90°.) The voltage V leads the current I = Ic + Id by phase angle δ. The current relation is expressed in terms of the current amplitudes: The values of the conduction and displacement current amplitudes are obtained by comparison with the answer to part (a):

A phasor diagram for adding the currents Ic and Id is shown to the right. The conduction current Ic is in phase with the voltage V across the resistor and Id lags behind it by 90°:

µ 0 (I c + I'd ) 2π r

=

⎞ µ 0 ⎛ V0 r2 ⎜⎜ sin ωt + ω 2 V0 cos ωt ⎟⎟ 2π r ⎝ R a ⎠

=

⎞ µ 0V0 ⎛ 1 r2 ⎜⎜ sin ωt + ω 2 cos ωt ⎟⎟ 2π r ⎝ R a ⎠

I = Ic + Id or

I max sin (ωt + δ ) = I c,max sin ωt + I d, max cos ωt

I c,max =

V0 R

and

I d,max =

ω ∈0 π a 2V0 d r Ic

Ic,max δ

I d,max r Id

I max r I

r V

Maxwell’s Equations and Electromagnetic Waves From the phasor diagram we have:

tan δ =

=

I d,max I c,max

=

V0

863

ω ∈0 π a 2 d V0 R

Rω ∈0 π a 2 d

so

⎛ Rω ∈ 0 π a 2 ⎞ ⎟⎟ d ⎝ ⎠

δ = tan −1 ⎜⎜

Remarks: The capacitor and the resistive wire are connected in parallel. The potential difference across each of them is the applied voltage V0 sin ωt. 60 •• Picture the Problem The total force on the surface is the sum of the force due to the reflected radiation and the force due to the absorbed radiation. From the conservation of momentum, the force due to the 10 kW that are reflected is twice the force due to the 10 kW that are absorbed. Express the total force on the surface:

Ftot = Fr + Fa

Substitute for Fr and Fa to obtain:

Ftot =

2( 12 P ) 12 P 3P + = c c 2c

Substitute numerical values and evaluate Ftot:

Ftot =

3(20 kW ) = 0.100 mN 2 3 × 108 m/s

(

)

*61 •• Picture the Problem We can use the definition of the Poynting vector and the r r relationship between B and E to find the instantaneous Poynting vectors for each of the resultant wave motions and the fact that the time average of the cross product term is zero for ω1 ≠ ω2, and ½ for the square of cosine function to find the time-averaged Poynting vectors.

r

r

(a) Because E1 and E 2 propagate in

r r r E × B = µ 0 Siˆ ⇒ B = Bkˆ

the x direction: Express B in terms of E1 and E2: Substitute for E1 and E2 to obtain:

B=

1 (E1 + E2 ) c

864 Chapter 30

r 1 B = [E1, 0 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω2t + δ )]kˆ c Express the instantaneous Poynting vector for the resultant wave motion:

r 1 (E1,0 cos(k1 x − ω1t ) + E2,0 cos(k2 x − ω2t + δ )) ˆj S=

µ0

× =

1

µ0c

1 (E1,0 cos(k1 x − ω1t ) + E2,0 cos(k2 x − ω2t + δ ))kˆ c

(

(E

2 cos(k1 x − ω1t ) + E2,0 cos(k 2 x − ω2t + δ )) ˆj × kˆ

1, 0

[E µc 1

=

2 1, 0

)

cos 2 (k1 x − ω1t ) + 2 E1, 0 E2, 0 cos(k1 x − ω1t )

0

]

× cos(k 2 x − ω2t + δ ) + E22, 0 cos 2 (k 2 x − ω2t + δ ) iˆ

(b) The time average of the cross product term is zero for ω1 ≠ ω2, and the time average of the square of the cosine terms is ½:

r S av =

[E 2µ c 1

2 1, 0

]

+ E22, 0 iˆ

0

r

(c) In this case B2 = − Bkˆ because the wave with k = k2 propagates in the − iˆ direction. The magnetic field is then:

r 1 B = [E1, 0 cos(k1 x − ω1t ) − E2,0 cos(k 2 x + ω2t + δ )]kˆ c Express the instantaneous Poynting vector for the resultant wave motion:

r 1 (E1,0 cos(k1 x − ω1t ) + E2,0 cos(k2 x − ω2t + δ )) ˆj S=

µ0

× =

1 (E1,0 cos(k1 x − ω1t ) − E2,0 cos(k2 x + ω2t + δ ))kˆ c

[E µc 1

2 1, 0

]

cos 2 (k1 x − ω1t ) − E22, 0 cos 2 (k 2 x + ω2t + δ ) iˆ

0

(d) The time average of the square of the cosine terms is ½:

r S av =

[E 2µ c 1

2 1, 0

]

− E22,0 iˆ

0

*62 •• Picture the Problem We can use the definitions of power and intensity to express the area of the surface as a function of P, I, and the efficiency ε.

Maxwell’s Equations and Electromagnetic Waves Use the definition of power to relate the required surface area to the intensity of the solar radiation:

E ε = IAε t where ε is the efficiency of the system.

Solve for A to obtain:

A=

P Iε

Substitute numerical values and evaluate A:

A=

25 kW = 111 m 2 2 0.3 0.75 kW/m

865

P=

(

)

63 •• Picture the Problem We can use the relationship between the average value of the Poynting vector (the intensity), E0, and B0 to find B0. The application of Faraday’s law will allow us to find the emf induced in the antenna. The emf induced in a 2-m wire oriented in the direction of the electric field can be found using ε = El and the relationship between E and B. (a) The intensity of the signal is related the amplitude of the magnetic field in the wave:

S av = I =

Solve for B0:

B0 =

E0 B0 cB02 = 2µ0 2µ0

2µ0 I c

Substitute numerical values and evaluate B0:

B0 =

(

)(

)

2 4π × 10 −7 N/A 2 10 −14 W/m 2 = 9.15 × 10 −15 T 3 × 108 m/s

ε

(b) Apply Faraday’s law to the antenna coil to obtain:

Substitute numerical values and evaluate

ε

(

ε

d (BA) = A d (NK m B0 sin ωt ) dt dt = NK m AB0ω cos ωt =

:

)

= 2000(200)π (0.01 m ) 9.15 × 10 −15 T [2π (140 kHz )]cos[2π (140 kHz )]t 2

(

)

= (1.01 µV ) cos 8.80 × 105 s −1 t (c) The voltage induced in the wire

ε = El

866 Chapter 30 is the product of its length l and the amplitude of electric field E0: Relate E to B:

E = cB = cB0 sin ωt

Substitute for E to obtain:

ε = clB0 sin ωt

Substitute numerical values and evaluate

ε

:

ε = (3 × 108 m/s)(2 m ) (9.15 × 10 −15 T )sin[2π (140 kHz )]t =

(5.49 µV )sin (8.80 × 105 s −1 )t

64 •• Picture the Problem We’ll choose the curve with sides ∆x and ∆z in the xy plane shown in the diagram and apply Equation 30-6d to show that

∂ Ey ∂ Bz = − µ 0 ∈0 . ∂x ∂t

Because ∆x is very small, we can approximate the difference in Bz at the points x1 and x2 by: Then:

The flux of the electric field through this curve is approximately: Apply Faraday’s law to obtain:

Bz ( x2 ) − Bz ( x1 ) = ∆B ≈

∂Bz ∆x ∂x

r r ∂E y B ∫C ⋅ d l ≈ µ0 ∈0 ∂t ∆x∆z

∫

S

En dA = E y ∆x∆y

∂E y ∂Bz ∆x∆z = − µ 0 ∈0 ∆x∆z ∂x ∂t or

∂E y ∂Bz = − µ 0 ∈0 ∂x ∂t *65 ••• Picture the Problem We can use Ohm’s law to relate the electric field E in the conductor to I, ρ, and a and Ampere’s law to find the magnetic field B just outside the conductor.

Maxwell’s Equations and Electromagnetic Waves

r

r

867

r

Knowing E and B we can find S and, using its normal component, show that the rate of energy flow into the conductor equals I2R, where R is the resistance. (a) Apply Ohm’s law to the cylindrical conductor to obtain:

V = IR =

Solve for E:

E=

(b) Apply Ampere’s law to a circular path of radius a at the surface of the cylindrical conductor: Solve for B to obtain:

(c) The electric field at the surface of the conductor is in the direction of the current and the magnetic field at the surface is tangent to the surface. Use the results of (a) and (b) and the right-hand rule to r evaluate S :

IρL IρL = = EL A π a2

Iρ π a2

r r B ∫ ⋅ d l = B(2π a ) = µ0 I enclosed = µ0 I C

B=

µ0 I 2π a

r 1 r r S= E×B

µ0

=

⎛µ I ⎞ 1 ⎛ Iρ ⎞ ⎜⎜ ⎟uˆ × ⎜ 0 ⎟⎟uˆ tangent 2 ⎟ parallel ⎜ µ0 ⎝ π a ⎠ ⎝ 2π a ⎠

= −

I 2ρ rˆ 2π 2 a 3

where rˆ is a unit vector directed radially outward from the cylindrical conductor. (d) The flux through the surface of the conductor into the conductor is: Substitute for Sn, the inward

r component of S , and simplify to

∫ S dA =S (2π aL ) n

∫ Sn dA =

I 2ρ I 2 ρL ( ) 2 = π aL π a2 2π 2 a 3

obtain: Since R =

ρL A

=

ρL : πa 2

∫ S dA = n

I 2R

66 ••• Picture the Problem We can use Faraday’s law to express the induced electric field at a distance r < R from the solenoid axis in terms of the rate of change of magnetic flux and B = nµ 0 at to express B in terms of the current in the windings of the solenoid. We can

r

use the results of (a) to find the magnitude and direction of the Poynting vector S at the

868 Chapter 30 cylindrical surface r = R just inside the solenoid windings. In part (c) we’ll use the definition of flux and the expression for the magnetic energy in a given region to show

r

that the flux of S into the solenoid equals the rate of increase of the magnetic energy inside the solenoid. (a) Apply Faraday’s law to a circular path of radius r < R:

∫ E ⋅ dl = E (2π r ) = −

Solve for E to obtain:

E=−

C

Express the magnetic field inside a long solenoid:

1 dφm 2π r dt

φm = BA = nµ 0 atπ r 2

Substitute in equation (1) to obtain:

E=−

r

At the cylindrical surface just inside the windings: Substitute to obtain:

r

(1)

B = nµ 0 I = nµ 0 at

The magnetic flux through a circle of radius r is:

(b) Express the magnitude of S at r = R:

dφm dt

S=

nµ a r d nµ 0 atπ r 2 = − 0 2π r dt 2 1

[

]

EB

µ0

B = nµ 0 at ⎛ nµ 0 a R ⎞ ⎜ ⎟(nµ 0 at ) n 2 µ 0 a 2 Rt 2 ⎠ ⎝ S= = µ0 2

Because the field E is tangential and directed so as to give an induced current that opposes the increase in r r r B , E × B is a vector that points toward the axis of the solenoid. Hence:

r n 2 µ 0 a 2 Rt S= − rˆ 2 where rˆ is a unit vector that points radially

(c) Consider a cylindrical surface of length L and radius R. Because r S points inward, the energy flowing

⎛ n 2 µ 0 a 2 Rt ⎞ ⎜⎜ ⎟⎟ = = S dA π RLS π RL 2 2 ∫ n 2 ⎝ ⎠ = n 2π µ 0 R 2 La 2t

outward.

Maxwell’s Equations and Electromagnetic Waves

869

into the solenoid per unit time is: Express the magnetic energy in the solenoid:

U B = u mV =

(

B2 π R2L 2µ0

)

( µ 0 nat )2 (π R 2 L ) = 2µ0

=

Evaluate dUB/dt:

n 2π µ0 R 2 La 2t 2 2

dU B d ⎡ n 2π µ 0 R 2 La 2t 2 ⎤ = ⎢ ⎥ dt dt ⎣ 2 ⎦ = n 2π µ 0 R 2 La 2t = ∫ S n dA

*67 ••• Picture the Problem We can use a condition for translational equilibrium to obtain an expression relating the forces due to gravity and radiation pressure that act on the particles. We can express the force due to radiation pressure in terms of the radiation pressure and the effective cross sectional area of the particles and the radiation pressure in terms of the intensity of the solar radiation. We can solve the resulting equation for r. Apply the condition for translational equilibrium to the particle:

Fr − Fg = 0 or, since Fr = PrA and Fg = mg,

Pr A − The radiation pressure Pr depends on the intensity of the radiation I:

Pr =

The intensity of the solar radiation at a distance R is:

I=

Substitute to obtain:

Pr =

Substitute for Pr, A, and m in equation (1):

GM s m =0 R2

I c

P 4π R 2 P 4π R 2 c

3 4 P 2 3 π r ρ GM s (π r ) − R 2 = 0 4π R 2 c

(1)

870 Chapter 30 Solve for R to obtain:

r=

3P 16π ρ c GM s

Substitute numerical values and evaluate r:

r=

(

16π 1 g/cm

3

( )(

)

3 3.83 × 10 26 W 3 × 108 m/s 6.67 × 10 −11 N ⋅ m 2 / kg 2 1.99 × 10 30 kg

)(

)(

)

= 0.574 µm 68 ••• Picture the Problem

r

(a) At a perfectly conducting surface E = 0 . Therefore, the sum of the electric fields of

r

r

the incident and reflected wave must add to zero, and so E i = − E r . (b) Let the incident and reflected waves be described by:

Ei = E 0 y cos(ωt − kx ) and

E r = − E 0 y cos(ωt + kx )

Use the trigonometric identity cos(α + β) = cosαcosβ − sinαsinβ to obtain:

Ei + E r = E 0 y cos(ωt − kx ) − E 0 y cos(ωt + kx ) = E 0 y [cos(ωt − kx ) − cos(ωt + kx )] = E 0 y [cos ωt cos(− kx ) − sin ωt sin (− kx ) − cos ωt cos kx + sin ωt sin (kx )]

= E 0 y [cos ωt cos kx + sin ωt sin kx − cos ωt cos kx + sin ωt sin kx] = 2 E 0 y sin ωt sin kx , the equation of a standing wave. r

r

r

r

(c) Because E × B = µ 0 S and S is in the direction of propagation of the wave, we see

r

that for the incident wave Bi = Bz cos(ωt − kx ) . Since both S and Ey are reversed for the

reflected wave, Br = B z cos(ωt + kx ) . So the magnetic field vectors are in the direction

r

r

at the reflecting surface and add at that surface. Hence B = 2 Br . *69 ••• Picture the Problem Let the point source be a distance a above the plane. Consider a ring of radius r and thickness dr in the plane and centered at the point directly below the light source. Express the force of force on this ring and integrate the resulting expression to obtain F.

Maxwell’s Equations and Electromagnetic Waves The intensity anywhere along this infinitesimal ring is P/4π (r2 + a2) and the element of force dF on this ring of area 2π rdr is given by:

dF =

P rdr c r 2 + a2

=

Pardr

( (

c r 2 + a2

)

871

a r 2 + a2

)

32

where we have taken into account that only the normal component of the incident radiation contributes to the force on the plane, and that the plane is a perfectly reflecting plane. Integrate dF from r = 0 to r = ∞:

From integral tables:

∞

F=

∞

∫ (r

Pa rdr ∫ c 0 (r 2 + a 2 )3 2 rdr

2

0

+ a2

)

32

∞

⎤ 1 = ⎥ = 2 2 r + a ⎦0 a −1

Substitute to obtain:

F=

Pa ⎛ 1 ⎞ P ⎜ ⎟= c ⎝a⎠ c

Substitute numerical values and evaluate F:

F=

1 MW = 3.33 mN 3 × 108 m/s

872 Chapter 30

Chapter 31 Properties of Light Conceptual Problems 1 • Determine the Concept The population inversion between the state E2,Ne and the state 1.96 eV below it (see Figure 31-9) is achieved by inelastic collisions between neon atoms and helium atoms excited to the state E2,He. 2 •• Determine the Concept Although the excited atoms emit the light of the same frequency on returning to the ground state, the light is emitted in a random direction, not exclusively in the direction of the incident beam. Consequently, the beam intensity is greatly diminished. 3 • Determine the Concept The layer of water greatly reduces the light reflected back from the car’s headlights, but increases the light reflected by the road of light from the headlights of oncoming cars. 4 • Determine the Concept When light passes from air into water its wavelength changes ( λ water = λair n water ), its speed changes ( v water = c n water ), and the direction of its propagation changes in accordance with Snell’s law. (c ) is correct. *5 •• Determine the Concept The change in atmospheric density results in refraction of the light from the sun, bending it toward the earth. Consequently, the sun can be seen even after it is just below the horizon. Also, the light from the lower portion of the sun is refracted more than that from the upper portion, so the lower part appears to be slightly higher in the sky. The effect is an apparent flattening of the disk into an ellipse. 6 • Determine the Concept (a) Yes. (b) Her procedure is based on Fermat’s principle in that, since the ball presumably travels at constant speed, the path that requires the least time of travel corresponds to the shortest distance of travel. 7 • Determine the Concept Because she can run faster than she can swim, she should choose the path that will maximize her running distance. Path LES is the path that satisfies this criterion.

873

874

Chapter 31

8 • Picture the Problem The intensity of the light transmitted by the second polarizer is given by I trans = I 0 cos 2 θ, where I 0 = 12 I . Therefore, I trans = 12 I cos 2 θ and

(b) is correct. 9 • Picture the Problem Polarized light can be produced from unpolarized light by absorption, reflection, birefringence, and scattering. Therefore, (d ) is correct. *10 •• Determine the Concept The diagram shows that the radiated intensity for a dipole is zero in the direction of the dipole moment. Because the dipole axis is in the same direction as the polarization, for light polarized parallel to plane of incidence, the dipole axis will point in the same direction as the reflected wave, i.e., in the direction described by Brewster’s law. As the diagram indicates, there is zero field in the direction of the refracted ray. On the other hand, if the incoming wave is polarized perpendicular to the plane of incidence, the dipole axis will never point along the direction of propagation for the reflected or refracted wave.

11 •• Determine the Concept The diagram shows unpolarized light from the sun incident on the smooth surface at the polarizing angle for that particular surface. The reflected light is polarized perpendicular to the plane of incidence, i.e., in the horizontal direction. The sunglasses are shown in the correct orientation to pass vertically polarized light and block the reflected sunlight.

Properties of Light

875

12 • (a) True. (b) False. Most of the light incident normally on an air–glass interface is transmitted. (c) False. The relationship between the angles of incidence and refraction depends on the indices of refraction on both sides of the interface. (d) False. The index of refraction of water is a function of the wavelength of light. (e) True. 13 •• Picture the Problem Because the speed of light in a given medium is inversely proportional to the index of refraction of the medium, we can decide which of the statements are true by referring to Figure 31-26. (a) The graphs of n vs. λ are not horizontal lines and so the speed of light is a function of its wavelength. (b) Because the index of refraction decreases with wavelength, violet light has the lowest speed and red light the highest speed. (c) Because the index of refraction decreases with wavelength, violet light has the lowest speed and red light the highest speed. (c) is correct. (d) Examination of Figure 31-26 tells us that this statement is false.

876

Chapter 31

(e) Examination of Figure 31-26 tells us that this statement is false. *14 •• Picture the Problem The sound is reflected specularly from the surface of the water (we assume it is calm). It is then refracted back toward the water in the region above the water because the speed of sound depends on the temperature of the air and is greater at the higher temperature. The pattern of the sound wave is shown schematically below.

15 • Determine the Concept In resonance absorption, the molecules respond to the frequency of the light through the Einstein photon relation E = hf. Thus, the color appears to be the same in spite of the fact that the wavelength has changed.

Estimation and Approximation 16 • Picture the Problem We can use the distance, rate, and time relationship to estimate the time required to travel 6 km (see Problem 31-14). Express the distance D to light traveled in terms of its speed c and the elapsed time ∆t:

D = c∆t

Solve for ∆t:

∆t =

D c

Substitute numerical values and evaluate ∆t:

∆t =

6 km = 20.0 µs 2.998 × 10 8 m/s

17 • Picture the Problem We can use the period of Io’s motion and the position of the earth at B to find the number of eclipses of Io during the earth’s movement and then use this information to find the number of days before a night-time eclipse. During the 42.5 h between eclipses of Jupiter’s moon, the earth moves from A to B, increasing the distance from Jupiter by approximately the distance from the earth to the Sun, making the path for the light longer and introducing a delay in the onset of the eclipse.

Properties of Light

877

Tearth 4 365.24 d 24 h = × 4 d = 2191 h

(a) Find the time it takes the earth to travel from point A to point B:

t A→B =

Because there are 42.5 h between eclipses of Io, the number of eclipses N occurring in the time it takes for the earth to move from A to B is:

N=

t A→B 2191 h = = 51.55 42.5 h TIo

Hence, in one-fourth of a year, there will be 51.55 eclipses. Because we want to find the next occurrence that happens in the evening hours, we’ll use 52 as the number of eclipses. We’ll also assume that Jupiter is visible so that the eclipse of Io can be observed at the time we determine. Relate the time t(N) at which the Nth eclipse occurs to N and the period TIo of Io: Evaluate t(52) to obtain:

Subtract the number of whole days to find the clock time t:

t ( N ) = NTIo

⎛ 1d ⎞ ⎟ t (52 ) = (52 )⎜⎜ 42.5 h × 24 h ⎟⎠ ⎝ = 92.083 d

t = t (52 ) − 92 d = 92.083 d − 92 d 24 h = 0.083 d × = 1.992 h d ≈ 2 am

Because June, July, and August have 30, 31, and 31 d, respectively, the date is: (b) Express the time delay ∆t in the arrival of light from Io due to the earth’s location at B:

September 1

∆t =

rearth-sun c

878

Chapter 31 1.5 × 1011 m ∆t = = 500 s 2.998 × 108 m/s = 8.33 min

Substitute numerical values and evaluate ∆t:

Hence, the eclipse will actually occur at 2 : 08 pm. 18 •• Picture the Problem We can express the relative error in using the small angle approximation and then either 1) use trial-and-error methods, 2) use a spreadsheet program, or 3) use the Solver capability of a scientific calculator to solve the transcendental equation the results from setting the error function equal to 0.01. Express the relative error δ in using the small angle approximation:

δ (θ ) =

θ − sin θ θ = −1 sin θ sin θ

A spreadsheet program was used to plot the following graph of δ (θ ). 0.016 0.014

delta(theta)

0.012 0.010 0.008 0.006 0.004 0.002 0.000 0.00

0.05

0.10

0.15

0.20

0.25

0.30

theta (radians)

From the graph, we can see that δ (θ ) < 1% for θ ≤ 0.24 radians. In degree measure,

θ ≤ 14° Remarks: Using the Solver program on a TI-85 gave θ = 0.244 radians.

Properties of Light

879

Sources of Light 19 • Picture the Problem We can use the definition of power to find the total energy of the pulse. The ratio of the total energy to the energy per photon will yield the number of photons emitted in the pulse. (a) Use the definition of power to obtain: Substitute numerical values and evaluate E:

E = P∆t

E = (10 MW )(1.5 ns ) = 15.0 mJ

E

(b) Relate the number of photons N to the total energy in the pulse and the energy of a single photon Ephoton:

N=

The energy of a photon is given by:

E photon =

Substitute for Ephoton to obtain:

N=

E photon

hc

λ

λE hc

Substitute numerical values (the wavelength of light emitted by a ruby laser is 694.3 nm) and evaluate N:

N=

(694.3 nm )(15.0 mJ ) 1240 eV ⋅ nm

1 eV = 5.25 × 1016 −19 1.60 × 10 J

20 • Picture the Problem We can express the number of photons emitted per second as the ratio of the power output of the laser and energy of a single photon.

P

Relate the number of photons per second n to the power output of the pulse and the energy of a single photon Ephoton:

n=

The energy of a photon is given by:

E photon =

Ephoton

hc

λ

880

Chapter 31

Substitute for Ephoton to obtain:

n=

λP hc

Substitute numerical values and evaluate n:

n=

(632.8 nm )(4 mW ) 1240 eV ⋅ nm

1 eV = 1.28 × 1016 photons/s −19 1.60 × 10 J

21 • Picture the Problem We can use the Einstein equation for photon energy to find the wavelength of the radiation for resonance absorption. We can use the same relationship, with ERaman = Einc − ∆E where ∆E is the energy for resonance absorption, to find the wavelength of the Raman scattered light. (a) Use the Einstein equation for photon energy to relate the wavelength of the radiation to energy of the first excited state:

λ=

hc E

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 435 nm 2.85 eV

(b) The wavelength of the Raman scattered light is given by:

λRaman =

Relate the energy of the Raman scattered light ERaman to the energy of the incident light Einc:

Substitute numerical values and evaluate λRaman:

1240 eV ⋅ nm E Raman

E Raman = Einc − ∆E 1240 eV ⋅ nm − 2.85 eV 320 nm = 1.025 eV =

λRaman =

1240 eV ⋅ nm = 1210 nm 1.025 eV

22 •• Picture the Problem The incident radiation will excite atoms of the gas to higher energy states. The scattered light that is observed is a consequence of these atoms returning to their ground state. The energy difference between the ground state and the atomic state excited by the irradiation is given by ∆E = hf =

hc

λ

.

Properties of Light The energy difference between the ground state and the atomic state excited by the irradiation is given by:

∆E = hf =

Substitute 368 nm for λ and evaluate ∆E:

∆E =

hc

λ

=

881

1240 eV ⋅ fm

λ

1240 eV ⋅ fm = 3.37 eV 368 nm

23 •• Picture the Problem The ground state and the three excited energy levels are shown in the diagram to the right. Because the wavelength is related to the energy of a photon by λ = hc/∆E, longer wavelengths correspond to smaller energy differences.

(a) The maximum wavelength of radiation that will result in resonance fluorescence corresponds to an excitation to the 3.2 eV level followed by decays to the 2.11 eV level and the ground state:

λmax =

The fluorescence wavelengths are:

λ21 = and

1240 eV ⋅ fm = 387.5 nm 3.2 eV

1240 eV ⋅ fm = 1138 nm 3.2 eV − 2.11 eV

λ10 =

1240 eV ⋅ fm = 587.7 nm 2.11 eV − 0

(b) For excitation:

λ03 =

1240 eV ⋅ fm = 285.1 nm 4.35 eV

The fluorescence wavelengths corresponding to the possible transitions are:

λ32 =

1240 eV ⋅ fm = 1078 nm 4.35 eV − 3.2 eV 1240 eV ⋅ fm λ21 = = 1138 nm 3.2 eV − 2.11 eV 1240 eV ⋅ fm λ10 = = 587.7 nm 2.11 eV − 0 1240 eV ⋅ fm λ31 = = 553.6 nm 4.35 eV − 2.11 eV

882

Chapter 31 and

λ20 =

1240 eV ⋅ fm = 387.5 nm 3.2 eV − 0

*24 •• Determine the Concept The energy difference between the ground state and the first excited state is 3E0 = 40.8 eV, corresponding to a wavelength of 30.4 nm. This is in the far ultraviolet, well outside the visible range of wavelengths. There will be no dark lines in the transmitted radiation.

The Speed of Light 25 • Picture the Problem We can use the distance, rate, and time relationship to find the distance to the spaceship. Relate the distance D to the spaceship to the speed of electromagnetic radiation in a vacuum and to the time for the message to reach the astronauts: Noting that the time for the message to reach the astronauts is half the time for Mission Control to hear their response, substitute numerical values and evaluate D:

D = c∆t

(

)

D = 2.998 × 10 8 m/s (2.5 s ) = 7.50 × 10 m 8

and

(a ) is correct.

26 • Picture the Problem We can use the conversion factor, found in EP-3, to convert a distance in km into c⋅y: Convert D = 2×1019 km into lightyears:

D = 2 × 1019 km ×

1c ⋅ y 9.46 × 1015 m

= 2.11 × 10 6 c ⋅ y 27 • Picture the Problem We can use the distance, rate, and time relationship to find the time delay between sending the signal from the earth and receiving it on Mars. Relate the distance D to Mars to the

D = c∆t

Properties of Light

883

speed of electromagnetic radiation in a vacuum and to the travel time for the signal: Solve for ∆t:

∆t =

D c

Substitute numerical values and evaluate ∆t:

∆t =

9.7 × 1010 m = 324 s 2.998 × 10 8 m/s

= 5 min 23 s 28 • Picture the Problem We can use the given information that the uncertainty in the measured distance ∆x is related to the uncertainty in the time ∆t by ∆x = c∆t to evaluate ∆x. The uncertainty in the distance is: Substitute numerical values and evaluate ∆t:

∆x = ± c∆t

(

)

∆x = ± 2.998 × 108 m/s (1.0 ns ) = ± 30.0 cm

*29 •• Picture the Problem We can use the distance, rate, and time relationship to find the time difference Galileo would need to be able to measure the speed of light successfully. (a) Relate the distance separating Galileo and his assistant to the speed of light and the time required for it travel to the assistant and back to Galileo:

D = c∆t

Solve for ∆t:

∆t =

D c

Substitute numerical values and evaluate ∆t:

∆t =

2(3 km ) = 20.0 µs 3 × 108 m/s

(b) Express the ratio of the human reaction time to the transit time for the light:

∆t reaction 0.2 s = = 10 4 ∆t 20 µs or

884

Chapter 31 ∆t reaction = 10 4 ∆t

Reflection and Refraction 30 • Picture the Problem Let the subscript 1 refer to air and the subscript 2 to water and use the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface. 2

Express the intensity I of the light reflected from an air-water interface at normal incidence in terms of the indices of refraction and the intensity I0 of the incident light:

⎛ n − n2 I = ⎜⎜ 1 ⎝ n1 + n2

⎞ ⎟⎟ I 0 ⎠

Solve for the ratio I/I0:

I ⎛ n1 − n2 =⎜ I 0 ⎜⎝ n1 + n2

Substitute numerical values and evaluate I/I0:

I ⎛ 1 − 1.33 ⎞ =⎜ ⎟ = 0.0201 = 2.01% I 0 ⎝ 1 + 1.33 ⎠

⎞ ⎟⎟ ⎠

2

2

*31 •• Picture the Problem The diagram shows ray 1 incident on the vertical surface at an angle θ1, reflected as ray 2, and incident on the horizontal surface at an angle of incidence θ3. We’ll prove that rays 1 and 3 are parallel by showing that θ1 = θ4, i.e., by showing that they make equal angles with the horizontal. Note that the law of reflection has been used in identifying equal angles of incidence and reflection.

θ 2 + 90° + 90° − θ1 = 180°

We know that the angles of the right triangle formed by ray 2 and the two mirror surfaces add up to 180°:

or

The sum of θ2 and θ3 is 90°:

θ 3 = 90° − θ 2

Because θ1 = θ 2 :

θ 3 = 90° − θ1

θ1 = θ 2

Properties of Light The sum of θ4 and θ3 is 90°:

θ 3 + θ 4 = 90°

Substitute for θ3 to obtain:

90° − θ1 + θ 4 = 90° ⇒ θ1 = θ 4

885

32 •• Picture the Problem Diagrams showing the light rays for the two cases are shown below. In (a) the light travels from air into water and in (b) it travels from water into air. (a) (b)

(a) Apply Snell’s law to the airwater interface to obtain:

Solve for θ2:

n1 sin θ1 = n2 sin θ 2 where the angles of incidence and refraction are θ1 and θ2, respectively.

⎞ ⎛ n1 sin θ1 ⎟⎟ ⎠ ⎝ n2

θ 2 = sin −1 ⎜⎜

A spreadsheet program to graph θ2 as a function of θ1 is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 A6 A7 B6

Content/Formula 1 1.33333 0 A6 + 5 A6*PI()/180

C6

ASIN(($B$1/$B$2)*SIN(B6))

D6

C6*180/PI()

Algebraic Form n1 n2 θ1 (deg) θ1 + ∆θ

θ1 ×

π

180

⎞ ⎛n sin −1 ⎜⎜ 1 sin θ1 ⎟⎟ ⎠ ⎝ n2 180 θ2 ×

π

886

Chapter 31 A B n1= 1 n2= 1.33333

1 2 3 4 5 6 7 8 9

theta1 (deg) 0 1 2 3

21 22 23 24

87 88 89 90

C

D

theta1 (rad) 0.00 0.02 0.03 0.05

theta2 (rad) 0.000 0.013 0.026 0.039

theta2 (deg) 0.00 0.75 1.50 2.25

1.52 1.54 1.55 1.57

0.847 0.847 0.848 0.848

48.50 48.55 48.58 48.59

A graph of θ2 as a function of θ1 follows: 50 45

Angle of refraction (deg)

40 35 30 25 20 15 10 5 0 0

10

20

30

40

50

60

70

80

90

Angle of incidence (deg)

(b) Change the contents of cell B1 to 1.33333 and the contents of cell B2 to 1 to obtain the following graph:

Properties of Light

887

90 Angle of refraction (deg)

80 70 60 50 40 30 20 10 0 0

10

20

30

40

50

Angle of incidence (deg)

Note that as the angle of incidence approaches the critical angle for a water-air interface (48.6°), the angle of refraction approaches 90°. No light will be refracted into the air if the angle of incidence is greater than 48.6°. 33 • Picture the Problem We can use the definition of the index of refraction to find the speed of light in water and in glass. The definition of the index of refraction is:

n=

c v

Solve for v to obtain:

v=

c n

Substitute numerical values and evaluate vwater:

vwater =

3 × 108 m/s = 2.25 × 108 m/s 1.33

Substitute numerical values and evaluate vglass:

vglass =

3 × 108 m/s = 2.00 × 108 m/s 1.5

34 • Picture the Problem Let the subscript 1 refer to the air and the subscript 2 to the silicate glass and apply Snell’s law to the air-glass interface. Apply Snell’s law to the air-glass interface to obtain: Solve for θ2:

n1 sin θ1 = n 2 sin θ 2

⎞ ⎛ n1 sin θ 1 ⎟⎟ ⎠ ⎝ n2

θ 2 = sin −1 ⎜⎜

888

Chapter 31

⎛ 1 ⎞ sin 45° ⎟ = 25.2° ⎝ 1.66 ⎠

Substitute numerical values for the light of wavelength 400 nm and evaluate θ2, 400 nm:

θ 2, 400 nm = sin −1 ⎜

Substitute numerical values for the light of wavelength 700 nm and evaluate θ2, 700 nm:

θ 2, 700 nm = sin −1 ⎜

⎞ ⎛ 1 sin 45° ⎟ = 26.1° ⎝ 1.61 ⎠

35 •• Picture the Problem Let the subscript 1 refer to the water and the subscript 2 to the glass and apply Snell’s law to the water-glass interface. Apply Snell’s law to the water-glass interface to obtain: Solve for θ2:

(a) Evaluate θ2 for θ1 = 60°:

(b) Evaluate θ2 for θ1 = 45°:

(c) Evaluate θ2 for θ1 = 30°:

n1 sin θ1 = n 2 sin θ 2

⎛ n1 ⎞ sin θ 1 ⎟⎟ ⎝ n2 ⎠

θ 2 = sin −1 ⎜⎜

⎛ 1.33 ⎞ sin 60° ⎟ = 50.2° ⎠ ⎝ 1.5

θ 2 = sin −1 ⎜

⎛ 1.33 ⎞ sin 45° ⎟ = 38.8° ⎠ ⎝ 1.5

θ 2 = sin −1 ⎜

⎛ 1.33 ⎞ sin 30° ⎟ = 26.3° ⎠ ⎝ 1.5

θ 2 = sin −1 ⎜

36 •• Picture the Problem Let the subscript 1 refer to the glass and the subscript 2 to the water and apply Snell’s law to the glass-water interface. Apply Snell’s law to the water-glass interface to obtain: Solve for θ2:

n1 sin θ1 = n 2 sin θ 2

⎛ n1 ⎞ sin θ 1 ⎟⎟ ⎝ n2 ⎠

θ 2 = sin −1 ⎜⎜

Properties of Light (a) Evaluate θ2 for θ1 = 60°:

(b) Evaluate θ2 for θ1 = 45°:

(c) Evaluate θ2 for θ1 = 30°:

889

⎛ 1.5 ⎞ sin 60° ⎟ = 77.6° ⎝ 1.33 ⎠

θ 2 = sin −1 ⎜

⎛ 1.5 ⎞ sin 45° ⎟ = 52.9° ⎝ 1.33 ⎠

θ 2 = sin −1 ⎜

⎛ 1.5 ⎞ sin 30° ⎟ = 34.3° 1 . 33 ⎝ ⎠

θ 2 = sin −1 ⎜

*37 •• Picture the Problem Let the subscript 1 refer to the medium to the left (air) of the first interface, the subscript 2 to glass, and the subscript 3 to the medium (air) to the right of the second interface. Apply the equation relating the intensity of reflected light at normal incidence to the intensity of the incident light and the indices of refraction of the media on either side of the interface to both interfaces. We’ll neglect multiple reflections at glass-air interfaces. Express the intensity of the transmitted light in the second medium:

Express the intensity of the transmitted light in the third medium:

I 2 = I 1 − I r,1

⎛ n − n2 = I 1 − ⎜⎜ 1 ⎝ n1 + n 2

⎡ ⎛n −n 2 = I 1 ⎢1 − ⎜⎜ 1 n n + ⎢⎣ ⎝ 1 2

⎞ ⎟⎟ ⎠

2

2

⎞ ⎟⎟ I 1 ⎠

⎤ ⎥ ⎥⎦ 2

I 3 = I 2 − I r,2

⎛ n − n3 ⎞ ⎟⎟ I 2 = I 2 − ⎜⎜ 2 ⎝ n 2 + n3 ⎠

⎡ ⎛ n − n ⎞2 ⎤ 3 ⎟⎟ ⎥ = I 2 ⎢1 − ⎜⎜ 2 ⎢⎣ ⎝ n 2 + n3 ⎠ ⎥⎦

Substitute for I2 to obtain:

⎡ ⎛n −n 2 I 3 = I 1 ⎢1 − ⎜⎜ 1 ⎢⎣ ⎝ n1 + n2

Solve for the ratio I3/I1:

I 3 ⎡ ⎛ n1 − n 2 = ⎢1 − ⎜ I 1 ⎢ ⎜⎝ n1 + n2 ⎣

⎞ ⎟⎟ ⎠

⎞ ⎟⎟ ⎠ 2

2

⎤⎡ ⎛ n − n ⎞2 ⎤ 3 ⎟⎟ ⎥ ⎥ ⎢1 − ⎜⎜ 2 ⎥⎦ ⎢⎣ ⎝ n 2 + n3 ⎠ ⎥⎦

⎤⎡ ⎛ n − n ⎞2 ⎤ 3 ⎟⎟ ⎥ ⎥ ⎢1 − ⎜⎜ 2 ⎥⎦ ⎢⎣ ⎝ n 2 + n3 ⎠ ⎥⎦

890

Chapter 31

Substitute numerical values and evaluate I3/I1:

2 2 I 3 ⎡ ⎛ 1 − 1.5 ⎞ ⎤ ⎡ ⎛ 1.5 − 1 ⎞ ⎤ = ⎢1 − ⎜ − 1 ⎟ ⎥⎢ ⎜ ⎟ ⎥ I 1 ⎢⎣ ⎝ 1 + 1.5 ⎠ ⎥⎦ ⎢⎣ ⎝ 1.5 + 1 ⎠ ⎥⎦

= 0.922 = 92.2% 38 •• Picture the Problem As the line enters the muddy field, its speed is reduced by half and the direction of the forward motion of the line is changed. In this case, the forward motion in the muddy field makes an angle of 14.5o with respect to the normal of the boundary line. Note that the separation between successive lines in the muddy field is half that in the dry field. Picture the Problem As the line enters the muddy field, its speed is reduced by half and the direction of the forward motion of the line is changed. In this case, the forward motion in the muddy field makes an angle of 14.5o with respect to the normal of the boundary line. Note that the separation between successive lines in the muddy field is half that in the dry field. 39 •• Picture the Problem We can apply Snell’s law consecutively, first to the n1-n2 interface and then to the n2-n3 interface. Apply Snell’s law to the n1-n2 interface: Apply Snell’s law to the n2-n3 interface: Equate the two expressions for n 2 sin θ 2 to obtain:

n1 sin θ1 = n 2 sin θ 2

n2 sin θ 2 = n3 sin θ 3

n1 sin θ1 = n3 sin θ 3

Properties of Light

891

*40 ••• Picture the Problem Let x be the perpendicular separation between the two rays and let l be the separation between the points of emergence of the two rays on the glass surface. We can use the geometry of the refracted and reflected rays to express x as a function of l, d, θr, and θi. Setting the derivative of the resulting equation equal to zero will yield the value of θi that maximizes x. (a) Express l in terms of d and the angle of refraction θr:

l = 2d tan θ r

Express x as a function of l, d, θr, and θi:

x = 2d tan θ r cos θ i

Differentiate x with respect toθi:

⎛ ⎞ dx d (tan θ r cos θ i ) = 2d ⎜⎜ − tan θ r sin θ i + sec 2 θ r cos θ i dθ r ⎟⎟ = 2d dθ i dθ i dθ i ⎠ ⎝ Apply Snell’s law to the air-glass interface:

n1 sin θ i = n 2 sin θ r

(1)

(2)

or, since n1 = 1 and n2 = n,

sin θ i = n sin θ r

Differentiate implicitly with respect toθI to obtain:

cos θ i dθ i = n cos θ r dθ r or

dθ r 1 cos θ i = dθ i n cos θ r

Substitute in equation (1) to obtain:

⎛ 1 cos 2 θ i sin θ r sin θ i ⎛ sin θ r dx 1 cos θ i cos θ i ⎞ ⎜⎜ ⎟ d 2 = − sin θ i + = 2d ⎜⎜ − 3 dθ i n cos 2 θ r cos θ r ⎟⎠ cos θ r ⎝ n cos θ r ⎝ cos θ r

⎞ ⎟⎟ ⎠

892

Chapter 31

Substitute 1 − sin 2 θ i for

cos 2 θ i and

⎛ 1 − sin 2 θ i sin 2 θ i dx = 2d ⎜⎜ − 3 dθ i ⎝ n cos θ r n cos θ r

1 sin θ i for sin θ r to n

⎞ ⎟⎟ ⎠

obtain: Multiply the second term in parentheses by cos 2 θ r cos 2 θ r and simplify to obtain:

⎛ 1 − sin 2 θ i sin 2 θ i cos 2 θ r dx = 2d ⎜⎜ − 3 dθ i n cos 3 θ r ⎝ n cos θ r

⎞ 2d ⎟⎟ = 1 − sin 2 θ i − sin 2 θ i cos 2 θ r 3 ⎠ n cos θ r

(

Substitute 1 − sin 2 θ r for cos 2 θ r :

[

(

2d dx = 1 − sin 2 θ i − sin 2 θ i 1 − sin 2 θ r 3 dθ i n cos θ r Substitute

)]

1 sin θ i for sin θ r to obtain: n 2d dx = dθ i n cos 3 θ r

⎡ 1 ⎛ ⎞⎤ 2 2 2 ⎢1 − sin θ i − sin θ i ⎜1 − n 2 sin θ i ⎟⎥ ⎝ ⎠⎦ ⎣

Factor out 1/n2, simplify, and set equal to zero to obtain:

[

]

2d dx = 3 sin 4 θ i − 2n 2 sin 2 θ i + n 2 = 0 for extrema dθ i n cos 3 θ r If dx/dθ1 = 0, then it must be true that: Solve this quartic equation for θi to obtain:

(b) Evaluate θI for n = 1.60:

sin 4 θ i − 2n 2 sin 2 θ i + n 2 = 0

⎛

θ i = sin − 1 ⎜ n 1 − 1 − ⎜ ⎝

⎛

θ i = sin −1 ⎜1.6 1 − 1 − ⎜ ⎝

= 48.5° In (a) we showed that:

x = 2d tan θ r cos θ i

1 ⎞⎟ n 2 ⎟⎠ 1

(1.6)2

⎞ ⎟ ⎟ ⎠

)

Properties of Light Solve equation (2) for θr:

Substitute numerical values and evaluate θr: Substitute numerical values and evaluate x:

893

⎛ n1 ⎞ sin θ i ⎟⎟ ⎝ n2 ⎠

θ r = sin −1 ⎜⎜

⎞ ⎛ 1 sin 48.5° ⎟ = 27.9° ⎝ 1.6 ⎠

θ r = sin −1 ⎜

x = 2(4 cm ) tan 27.9° cos 48.5° = 2.81cm

Total Internal Reflection 41 • Picture the Problem Let the subscript 1 refer to the glass and the subscript 2 to the water and use Snell’s law under total internal reflection conditions. Use Snell’s law to obtain:

n1 sin θ 1 = n 2 sin θ 2

When there is total internal reflection:

θ1 = θ c and θ 2 = 90°

Substitute to obtain:

n1 sin θ c = n2 sin 90° = n2

Solve for θc:

Substitute numerical values and evaluate θc:

θ c = sin −1

n2 n1

θ c = sin −1

1.33 = 62.5° 1.5

42 •• Picture the Problem Let the index of refraction of glass be represented by n1 and the index of refraction of water by n2 and apply Snell’s law to the glass-water interface under total internal reflection conditions.

Apply Snell’s law to the glass-water interface: At the critical angle, θ1 = θc and

n1 sin θ1 = n2 sin θ 2

n1 sin θ c = n2 sin 90°

894

Chapter 31

θ2 = 90°: Solve for θc:

Substitute numerical values and evaluate θc:

⎡ n2 ⎤ sin 90°⎥ ⎣ n1 ⎦

θ c = sin −1 ⎢

⎡1.33 ⎤ sin 90°⎥ = 62.5° ⎣ 1.5 ⎦

θ c = sin −1 ⎢

43 •• Picture the Problem We can apply Snell’s law to the water-air interface to express the critical angle θc in terms of the indices of refraction of water (n1) and air (n2) and then relate the radius of the circle to the depth d of the point source and θc.

Express the area of the circle whose radius is r: Relate the radius of the circle to the depth d of the point source and the critical angle θc: Apply Snell’s law to the water-air interface to obtain: Solve for θc:

Substitute for r and θc to obtain:

A = π r2

r = d tan θ c

n1 sin θ c = n2 sin 90° = n2

θ c = sin −1

n2 n1

A = π [d tan θ c ]

2

⎡ ⎛ n ⎞⎤ = π ⎢d tan⎜⎜ sin −1 2 ⎟⎟⎥ n1 ⎠⎦ ⎝ ⎣

2

Properties of Light Substitute numerical values and evaluate A:

⎡ 1 ⎞⎤ ⎛ A = π ⎢(5 m ) tan⎜ sin −1 ⎟ 1.33 ⎠⎥⎦ ⎝ ⎣

895

2

= 102 m 2 44 •• Picture the Problem We can use the definition of the index of refraction to express the speed of light in the prism in terms of the index of refraction n1 of the prism. The application of Snell’s law at the glass-air interface will allow us to relate the index of refraction of the prism to the critical angle for total internal reflection. Finally, we can use the geometry of the isosceles-right-triangle prism to conclude that θc = 45°. Express the speed of light v in the prism in terms of its index of refraction n1: Apply Snell’s law to the glass-air interface to obtain:

c n1

v=

n1 sin θ c = n2 sin 90° = 1

1 sin θ c

Solve for n1:

n1 =

Substitute to obtain:

v = c sin θ c

Substitute numerical values and evaluate v:

(

)

v = 2.998 × 108 m/s sin 45° = 2.12 × 108 m/s

45 •• Picture the Problem The observer above the surface of the fluid will not see any light until the angle of incidence of the light at the fluid-air interface is less than or equal to the critical angle for the two media. We can use Snell’s law to express the index of refraction of the fluid in terms of the critical angle and use the geometry of card and light source to express the critical angle.

896

Chapter 31

Apply Snell’s law to the fluid-air interface to obtain: Light is seen by the observer when θ1 = θc and θ2 = 90°:

n1 sin θ1 = n2 sin θ 2

n1 sin θ c = n2 sin 90° = n2

1 sin θ c

Because the medium above the interface is air, n2 = 1. Solve for n1 to obtain:

n1 =

From the geometry of the diagram:

tan θ c =

Substitute to obtain:

n1 =

Substitute numerical values and evaluate n1:

n1 =

r r ⇒ θ c = tan −1 d d

1 r⎤ ⎡ sin ⎢ tan −1 ⎥ d⎦ ⎣ 1 ⎡ 6 cm ⎤ sin ⎢ tan −1 5 cm ⎥⎦ ⎣

= 1.30

*46 •• Picture the Problem We can use the geometry of the figure, the law of refraction at the air-n1 interface, and the condition for total internal reflection at the n1-n2 interface to show that the numerical aperture is given by

n22 − n32 .

Referring to the figure, note that:

sin θ c = and

n3 a = n2 c

Properties of Light sin θ 2 = Apply the Pythagorean theorem to the right triangle to obtain:

897

b c

a 2 + b2 = c2 or

a 2 b2 + =1 c2 c2 Solve for

b : c

Substitute for

b a2 = 1− 2 c c b a and to obtain: c c

Use the law of refraction to relate θ1 and θ2: Substitute for sinθ2 and let n1 = 1 (air) to obtain:

n32 sin θ 2 = 1 − 2 n2 n1 sin θ1 = n2 sin θ 2

sin θ1 = n2 1 −

n32 = n22

n22 − n32

47 • Picture the Problem We can use the result of Problem 46 to find the maximum angle of incidence under the given conditions. From Problem 46:

sin θ 0 = n12 − n22

Solve for θ0:

θ 0 = sin −1 n12 − n22

Substitute numerical values and evaluate θ0:

θ 0 = sin −1 ⎛⎜ (1.492)2 − (1.489)2 ⎞⎟

(

⎝

) ⎠

= 5.43° 48 •• Picture the Problem Examination of the figure reveals that, if the length of the tube is L, the distance traveled by the pulse that enters at an angle θ0 is the ratio of a to b multiplied by L. Let the subscripts 1 and 2 denote the pulses entering the tube normally and at an angle θ0, respectively.

898

Chapter 31

Express the difference in time ∆t needed for the two pulses to travel a distance L:

b L ∆t = t2 − t1 = a − c c n1 n1

Substitute for t2 and t1 and simplify to obtain:

b L n1L ⎛ b ⎞ ∆t = a − = ⎜ − 1⎟ c c c ⎝a ⎠ n1 n1

Referring to the figure, note that:

sin θ c =

b a

From Snell’s law, the sine of the critical angle is also given by:

sin θ c =

n2 b n ⇒ = 2 n1 a n1

Substitute for b/a in equation (1) and simplify to obtain:

∆t =

n1L ⎛ n2 ⎞ L ⎜ − 1⎟ = (n2 − n1 ) c ⎜⎝ n1 ⎟⎠ c

Substitute numerical values and evaluate ∆t:

∆t =

15 km (1.492 − 1.489) 3 ×108 m/s

L

L

= 150 ns 49 ••• Picture the Problem Let the index of refraction of glass be represented by n1, the index of refraction of water by n2, and the index of refraction of air by n3. We can apply Snell’s law to the glass-water interface under total internal reflection conditions to find the critical angle for total internal reflection. The application of Snell’s law to glass-air and glass-water interfaces will allow us to decide whether there are angles of incidence greater than θc for glass-to-air refraction for which light rays will leave the glass and the water and pass into the air. (a) Apply Snell’s law to the glasswater interface:

n1 sin θ1 = n2 sin θ 2

(1)

Properties of Light At the critical angle, θ1 = θc and θ2 = 90°: Solve for θc:

Substitute numerical values and evaluate θc: (b) Apply Snell’s law to a glass-air interface:

899

n1 sin θ c = n2 sin 90°

⎡ n2 ⎤ sin 90°⎥ ⎣ n1 ⎦

θ c = sin −1 ⎢

⎡1.33 ⎤ sin 90°⎥ = 62.5° ⎣ 1.5 ⎦

θ c = sin −1 ⎢

n1 sin θ c = n3 sin 90° or

1.5 sin θ c = sin 90° = 1

Solve for θc:

Apply Snell’s law to a ray incident at the critical angle for a glass-water interface: Solve for θ2:

⎛ 1 ⎞ ⎟ = 41.8° ⎝ 1.5 ⎠

θ c = sin −1 ⎜

n1 sin θ1 = n2 sin θ 2 or

1.5 sin 41.8° = 1.33 sin θ 2

⎛ 1.5 sin 41.8° ⎞ ⎟ = 48.7° 1.33 ⎝ ⎠

θ 2 = sin −1 ⎜

Note that θ 2 equals the critical angle for a water - air interface. Therefore, the ray will not leave the water for θ1 ≥ 41.8°. 50 ••• Picture the Problem The situation is shown in the adjacent figure. We can use the geometry of the diagram and trigonometric relationships to derive an expression for d in terms of the angles of incidence and refraction. Applying Snell’s law will yield θr.

Express the distance x in terms of t and θr:

x = 2t tan θ r

900

Chapter 31

The separation of the reflected rays is: Substitute to obtain: Apply Snell’s law at the air-glass interface to obtain: Solve for θr:

Substitute in equation (1) to obtain: Substitute numerical values and evaluate d:

d = x cosθ i d = 2t tan θ r cos θ i

(1)

sin θ i = n sin θ r

⎛ sin θ i ⎞ ⎟ ⎝ n ⎠

θ r = sin −1 ⎜

⎡ ⎛ sin θ i ⎞⎤ d = 2t tan ⎢sin −1 ⎜ ⎟⎥ cosθ i ⎝ n ⎠⎦ ⎣ ⎡ ⎛ sin 40° ⎞⎤ d = 2(3 cm ) tan ⎢sin −1 ⎜ ⎟⎥ cos 40° ⎝ 1.5 ⎠⎦ ⎣ = 2.18 cm

Dispersion *51 •• Picture the Problem We can apply Snell’s law of refraction to express the angles of refraction for red and violet light in silicate flint glass. Express the difference between the angle of refraction for violet light and for red light: Apply Snell’s law of refraction to the interface to obtain: Solve for θr:

∆θ = θ r,red − θ r,violet

(1)

sin 45° = n sin θ r

⎛ 1 ⎞ ⎟ ⎝ 2n ⎠

θ r = sin −1 ⎜

Substitute in equation (1):

⎛ 1 ⎞ ⎛ 1 ⎞ ⎟ − sin −1 ⎜ ⎟ ∆θ = sin −1 ⎜⎜ ⎟ ⎜ ⎟ ⎝ 2nred ⎠ ⎝ 2nviolet ⎠

Substitute numerical values and evaluate ∆θ :

⎞ ⎞ ⎛ ⎛ 1 1 ⎟⎟ ⎟⎟ − sin −1 ⎜⎜ ∆θ = sin −1 ⎜⎜ ( ) ( ) 2 1 . 60 2 1 . 66 ⎠ ⎠ ⎝ ⎝ = 26.23° − 25.21° = 1.02°

Properties of Light

901

52 •• Picture the Problem The transit times will be different because the speed with which light of various wavelengths propagates in silicate crown glass is dependent on the index of refraction. We can use Table 31-26 to estimate the indices of refraction for pulses of wavelengths 500 and 700 nm. Express the difference in time needed for two short pulses of light to travel a distance L in the fiber:

∆t =

L L − v500 v700

Substitute for L, v500, and v700 and simplify to obtain:

∆t =

n500 L n700 L L − = (n500 − n700 ) c c c

n500 ≈ 1.55

Use Table 31-26 to find the indices of refraction of silicate crown glass for the two wavelengths:

and

Substitute numerical values and evaluate ∆t:

∆t =

n700 ≈ 1.50

15 km (1.55 − 1.50) 2.998 × 108 m/s

= 2.50 µs

Polarization 53 • Picture the Problem The polarizing angle is given by Brewster’s law: tan θ p = n2 n1 where n1 and n2 are the indices of refraction on the near and far sides of the interface, respectively.

⎛ n2 ⎞ ⎟⎟ ⎝ n1 ⎠

Use Brewster’s law to obtain:

θ p = tan −1 ⎜⎜

(a) For n1 = 1 and n2 = 1.33:

θ p = tan −1 ⎜

(b) For n1 = 1 and n2 = 1.50:

θ p = tan −1 ⎜

⎛ 1.33 ⎞ ⎟ = 53.1° ⎝ 1 ⎠ ⎛ 1.50 ⎞ ⎟ = 56.3° ⎝ 1 ⎠

54 • Picture the Problem The intensity of the transmitted light I is related to the intensity of the incident light I0 and the angle the transmission axis makes with the horizontal θ

902

Chapter 31

according to I = I 0 cos 2 θ . Express the intensity of the transmitted light in terms of the intensity of the incident light and the angle the transmission axis makes with the horizontal: Solve for θ :

Substitute numerical values and evaluate θ :

I = I 0 cos 2 θ

θ = cos −1

I I0

θ = cos −1 0.15 = 67.2° and ( d ) is correct.

55 • Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the intensity of the light transmitted through all three sheets for

θ = 45° and θ = 30°. (a) Express the intensity of the light between the first and second sheets:

I1 = 12 I 0

Express the intensity of the light between the second and third sheets:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 45° = 14 I 0

Express the intensity of the light that has passed through the third sheet:

I 3 = I 2 cos 2 θ 2,3 = 14 I 0 cos 2 45° =

(b) Express the intensity of the light between the first and second sheets:

1 8

I0

I1 = 12 I 0

Express the intensity of the light between the second and third sheets:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 30° = 83 I 0

Express the intensity of the light that has passed through the third sheet:

I 3 = I 2 cos 2 θ 2,3 = 83 I 0 cos 2 60° =

3 32

I0

56 •• Picture the Problem Because the light is polarized in the vertical direction and the first polarizer is also vertically polarized, no loss of intensity results from the first

Properties of Light

903

transmission. We can use Malus’s law to find the intensity of the light after it has passed through the second polarizer. The intensity of the beam is the ratio of its power to cross-sectional area: Express the intensity of the light between the first and second polarizers:

I=

P A

I1 = I 0 and P1 = P0

Express Malus’s law in terms of the power of the beam:

P P0 = cos 2 θ ⇒ P = P0 cos 2 θ A A

Express the power of the beam after the second transmission:

P2 = P1 cos 2 θ1, 2 = P0 cos 2 θ12

Substitute numerical values and evaluate I2:

P2 = (5 mW )cos 2 27° = 3.97 mW

57 •• Picture the Problem Assume that light is incident in air (n1 = 1). We can use the relationship between the polarizing angle and the angle of refraction to determine the latter and Brewster’s law to find the index of refraction of the substance. (a) At the polarizing angle, the sum of the angles of polarization and refraction is 90°:

θ p + θ r = 90°

Solve for θr:

θ r = 90° − θ p

Substitute for θp to obtain:

θ r = 90° − 60° = 30.0°

(b) From Brewster’s law we have:

tan θ p =

n2 n1

or, because n1 = 1,

n2 = tan θ p

Substitute for θp and evaluate n2:

n2 = tan 60° = 1.73

904

Chapter 31

58 •• Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the intensity of the light transmitted through the three sheets. Express the intensity of the light between the first and second sheets:

I1 = 12 I 0

Express the intensity of the light between the second and third sheets:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 θ

Express the intensity of the light that has passed through the third sheet and simplify to obtain:

I 3 = I 2 cos 2 θ 2,3 = 12 I 0 cos 2 θ cos 2 (90° − θ ) = 12 I 0 cos 2 θ sin 2 θ

= 18 I 0 (2 cosθ sin θ )

2

= Because the sine function is a maximum when its argument is 90°, the maximum value of I3 occurs when:

1 8

I 0 sin 2 2θ

θ = 45.0°

59 •• Picture the Problem Let In be the intensity after the nth polarizing sheet, use I = I 0 cos 2 θ to find the intensity of the light transmitted through each sheet, and replace

θ with ωt. Express the intensity of the light between the first and second sheets:

I1 = 12 I 0

Express the intensity of the light between the second and third sheets:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 ωt

Express the intensity of the light that has passed through the third sheet and simplify to obtain:

I 3 = I 2 cos 2 θ 2,3 = 12 I 0 cos 2 ωt cos 2 (90° − ωt ) = 12 I 0 cos 2 ωt sin 2 ωt

= 18 I 0 (2 cos ωt sin ωt )

2

=

1 8

I 0 sin 2 2ωt

Properties of Light *60 •• Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the ratio of In+1 to In. (a) Find the ratio of In+1 to In:

I n+1 π = cos 2 In 2N

Because there are N such reductions of intensity:

I N +1 I N +1 ⎛ π ⎞ = = cos 2 N ⎜ ⎟ I1 I0 ⎝ 2N ⎠ and

⎛ π ⎞ I N +1 = I 0 cos 2 N ⎜ ⎟ ⎝ 2N ⎠ (b) A spreadsheet program to graph IN+1/I0 as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A2 A3 B2

Content/Formula 2 A2 + 1 (cos(PI()/(2*A2))^(2*A2)

Algebraic Form N N+1

⎛ π ⎞ cos 2 N ⎜ ⎟ ⎝ 2N ⎠

1 2 3 4 5

A N 2 3 4 5

B I/I0 0.250 0.422 0.531 0.605

95 96 97 98 99 100

95 96 97 98 99 100

0.974 0.975 0.975 0.975 0.975 0.976

A graph of I/I0 as a function of N follows.

905

906

Chapter 31 1.0 0.9 0.8 0.7

I /I 0 0.6 0.5 0.4 0.3 0.2 0

10

20

30

40

50

60

70

80

90

100

N

(c)

In each case, the polarization of the transmitted beam is perpendicular to that of the incident beam.

61 •• Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the ratio of In+1 to In. Because each sheet introduces a 2% loss of intensity, the net transmission after N sheets (0.98)N. Find the ratio of In+1 to In:

I n+1 π = (0.98) cos 2 In 2N

Because there are N such reductions of intensity:

I N +1 ⎛ π ⎞ N = (0.98) cos 2 N ⎜ ⎟ I0 ⎝ 2N ⎠

(b) A spreadsheet program to graph IN+1/I0 for an ideal polarizer as a function of N, the percent transmission, and IN+1/I0 for a real polarizer as a function of N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A3 B2

Content/Formula 1 (cos(PI()/(2*A2))^(2*A2)

C3

(0.98)^A3

D4

B3*C3

Algebraic Form N

⎛ π ⎞ cos 2 N ⎜ ⎟ ⎝ 2N ⎠ (0.98)N

(0.98)N cos2 N ⎛⎜

π ⎞

⎟ ⎝ 2N ⎠

Properties of Light A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

B Ideal Polarizer 0.000 0.250 0.422 0.531 0.605 0.660 0.701 0.733 0.759 0.781 0.798 0.814 0.827 0.838 0.848 0.857 0.865 0.872 0.878 0.884

N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

C Percent Transmission 0.980 0.960 0.941 0.922 0.904 0.886 0.868 0.851 0.834 0.817 0.801 0.785 0.769 0.754 0.739 0.724 0.709 0.695 0.681 0.668

907

D Real Polarizer 0.000 0.240 0.397 0.490 0.547 0.584 0.608 0.624 0.633 0.638 0.639 0.638 0.636 0.632 0.626 0.620 0.613 0.606 0.598 0.590

A graph of I/I0 as a function of N for the quantities described above follows: 1.0

0.8

0.6 I /I 0

Ideal Polarizer

0.4

Percent Transmission 0.2

Real Polarizer

0.0 0

5

10

15

20

Number of sheets (N )

Inspection of the table, as well as of the graph, tells us that the optimum number of sheets is 13. *62 •• Picture the Problem A circularly polarized wave is said to be right circularly polarized if the electric and magnetic fields rotate clockwise when viewed along the direction of propagation and left circularly polarized if the fields rotate counterclockwise.

908

Chapter 31

For a circularly polarized wave, the x and y components of the electric field are given by:

E x = E0 cos ωt and

E y = E0 sin ωt or E y = − E0 sin ωt

for left and right circular polarization, respectively. For a wave polarized along the x axis:

r r E right + Eleft = E0 cos ωt iˆ + E0 cos ωt iˆ = 2 E0 cos ωt iˆ

63 •• Picture the Problem Let In be the intensity after the nth polarizing sheet and use I = I 0 cos 2 θ to find the intensity of the light transmitted by the four sheets. (a) Express the intensity of the light between the first and second sheets:

I1 = 12 I 0

Express the intensity of the light between the second and third sheets:

I 2 = I1 cos 2 θ1, 2 = 12 I 0 cos 2 30° = 83 I 0

Express the intensity of the light between the third and fourth sheets:

I 3 = I 2 cos 2 θ 2,3 = 83 I 0 cos 2 30° =

Express the intensity of the light to the right of the fourth sheet:

I 4 = I 3 cos 2 θ 3, 4 =

9 32

9 32

I0

27 I 0 cos 2 30° = 128 I0

= 0.211I 0

Note that, for the single sheet between the two end sheets at θ = 45o, I = 0.125I0. Using two sheets at relative angles of 30o increases the transmitted intensity. Remarks: We could also apply the result obtained in Problem 60(a) to solve this problem. *64 •• r r Picture the Problem We can use the components of E to show that E is constant in time and rotates with angular frequency ω.

r

Express the magnitude of E in terms of its components:

E = E x2 + E y2

Properties of Light

909

Substitute for Ex and Ey to obtain:

E=

[E0 sin (kx − ωt )] 2 + [E0 cos(kx − ωt )] 2

= E0

[

]

= E02 sin 2 (kx − ωt ) + cos 2 (kx − ωt )

r

and the E vector rotates in the yz plane with angular frequency ω. 65 •• Picture the Problem We can apply the given definitions of right and left circular polarization to the electric field and magnetic fields of the wave. The electric field of the wave in Problem 64 is:

r E = E0 sin (kx − ωt ) ˆj + E0 cos(kx − ωt )kˆ The corresponding magnetic field is:

r B = B0 sin (kx − ωt )kˆ − B0 cos(kx − ωt ) ˆj Because these fields rotate clockwise when viewed along the direction of propagation, the wave is right circularly polarized. For a left circularly polarized wave traveling in the opposite direction:

r E = E0 sin (kx + ωt ) ˆj − E0 cos(kx + ωt )kˆ

General Problems 66 • Picture the Problem We can use v = fλ and the definition of the index of refraction to relate the wavelength of light in a medium whose index of refraction is n to the wavelength of light in air.

v c λ = = 0 f nf n

(a) The wavelength λn of light in a medium whose index of refraction is n is given by:

λn =

Substitute numerical values and evaluate λwater:

λwater =

700 nm 700 nm = = 526 nm nwater 1.33

910 (b)

Chapter 31 Because the color observed depends on the frequency of the light, a swimmer observes the same color in air and in water.

67 •• Picture the Problem We can use Snell’s law, under critical angle and polarization conditions, to relate the polarizing angle of the substance to the critical angle for internal reflection. Apply Snell’s law, under critical angle conditions, to the interface: Apply Snell’s law, under polarization conditions, to the interface:

Solve for θp:

Solve equation (1) for the ratio of n2 to n1: Substitute for n2/n1 in equation (2) to obtain: Substitute numerical values and evaluate θp: *68 •• Picture the Problem Angle ADE is the angle between the direction of the incoming ray and that reflected by the two mirror surfaces. Note that triangle ABC is isosceles and that angles CAB and ABC are equal and their sum equals θ. Also from the law of reflection, angles CAD and CBD equal angle ABC. Because angle BAD is twice BAC and angle DBA is twice CBA, angle ADE is twice the angle θ.

n1 sin θ c = n2

(1)

n1 sin θ p = n2 sin (90° − θ p ) = n2 cosθ p or

tan θ p =

n2 n1 ⎛ n2 ⎞ ⎟⎟ ⎝ n1 ⎠

θ p = tan −1 ⎜⎜

(2)

n2 = sin θ c n1

θ p = tan −1 (sin θ c ) θ p = tan −1 (sin 45°) = 35.3°

Properties of Light

911

69 •• Picture the Problem The sketch shows the ray from the coin passing through the water to the eye of the observer. We can use trigonometry to express the apparent depth d in terms of the depth h of the water, the 20° angle, and the angle of incidence θi. The application of Snell’s law at the interface will yield an expression for θi. Express the apparent depth d in terms of the distance x: Relate the distance x to the depth of the water and the angle θi: Substitute for x in equation (1) to obtain: Apply Snell’s law to the water-air interface: Solve for θi:

d = x tan 20°

(1)

x = h tan θ i d = h tan θ i tan 20°

(2)

n1 sin θ i = n2 sin θ r

⎛ n2 ⎞ sin θ r ⎟⎟ ⎝ n1 ⎠

θ i = sin −1 ⎜⎜

Substitute for θI in equation (2) to obtain:

⎡ ⎛n ⎞⎤ d = h tan ⎢sin −1 ⎜⎜ 2 sin θ r ⎟⎟⎥ tan 20° ⎝ n1 ⎠⎦ ⎣

Substitute numerical values and evaluate d:

⎡ ⎛ 1 ⎞⎤ d = (4 m ) tan ⎢sin −1 ⎜ sin 70° ⎟⎥ tan 20° ⎝ 1.33 ⎠⎦ ⎣ = 1.45 m

70 •• Picture the Problem Assume that the sound source is the voice of the fisherman and that the fisherman’s mouth is 2 m from the surface of the water as shown below. We can apply Snell’s law at the air-water interface to find θc and use trigonometry to find θ1. If we can show that θ1 > θc, then we can conclude that the noise on shore cannot possibly be sensed by fish 20 m from shore.

912

Chapter 31

Apply Snell’s law at the air-water interface 20 m from the shore: For θ1 = θc:

n1 sin θ1 = n2 sin θ 2

⎛ n2 ⎞ ⎛v ⎞ ⎟⎟ = sin −1 ⎜⎜ 1 ⎟⎟ ⎝ n1 ⎠ ⎝ v2 ⎠

θ c = sin −1 ⎜⎜

⎛ 330 m/s ⎞ ⎟⎟ = 13.2° ⎝ 1450 m/s ⎠

Substitute numerical values and evaluate θc:

θ c = sin −1 ⎜⎜

Relate θ1 to the distance from the shore and the distance from the surface of the water to the fisherman’s mouth:

tan (90° − θ1 ) =

Solve for and evaluate θ1:

θ1 = 90° − tan −1 (0.1) = 84.3°

2m 20 m

Because θ1 > θ c , all the sound is reflected at air - water interface. *71 •• Picture the Problem We can apply Snell’s law to the water-air interface to express the critical angle θc in terms of the indices of refraction of water (n1) and air (n2) and then relate the radius of the circle to the depth d of the swimmer and θc.

Properties of Light

Relate the radius of the circle to the depth d of the point source and the critical angle θc: Apply Snell’s law to the water-air interface to obtain: Solve for θc:

913

r = d tan θ c

n1 sin θ c = n2 sin 90° = n2

⎛ n2 ⎞ ⎟⎟ ⎝ n1 ⎠

θ c = sin −1 ⎜⎜

Substitute for θc to obtain:

⎡ ⎛ n ⎞⎤ r = d tan ⎢sin −1 ⎜⎜ 2 ⎟⎟⎥ ⎝ n1 ⎠⎦ ⎣

Substitute numerical values and evaluate r:

⎡ ⎛ 1 ⎞⎤ r = (3 m ) tan ⎢sin −1 ⎜ ⎟⎥ = 3.42 m ⎝ 1.33 ⎠⎦ ⎣

72 •• Picture the Problem Let φ be the initial angle of incidence. Since the angle of reflection with the normal to the mirror is alsoφ, the angle between incident and reflected rays is 2φ. If the mirror is now rotated by a further angle θ, the angle of incidence is increased by θ to φ +θ, and so is the angle of reflection. Consequently, the reflected beam is rotated by 2θ relative to the incident beam.

73 •• Picture the Problem We can apply Snell’s law at the glass-air interface to express θc in terms of the index of refraction of the glass and use Figure 31-25 to find the index of refraction of the glass for the given wavelengths of light. Apply Snell’s law at the glass-air interface:

n1 sin θ1 = n2 sin θ 2

914

Chapter 31

If θ1 = θc and n2 = 1:

n1 sin θ c = sin 90° = 1 and

⎛1⎞ ⎟⎟ ⎝ n1 ⎠

θ c = sin −1 ⎜⎜

⎛ 1 ⎞ ⎟ = 36.8° ⎝ 1.67 ⎠

(a) For violet light of wavelength 400 nm, n2 = 1.67:

θ c = sin −1 ⎜

(b) For red light of wavelength 700 nm, n2 = 1.60:

θ c = sin −1 ⎜

⎛ 1 ⎞ ⎟ = 38.7° ⎝ 1.60 ⎠

74 •• Picture the Problem We’ll neglect multiple reflections at the glass-air interfaces. We can use the expression (Equation 31-11) for the reflected intensity at an interface to express the intensity of the light in the glass slab as the difference between the intensity of the incident beam and the reflected beam. Repeating this analysis at the glass-air interface will lead to the desired result. Express the intensity of the light transmitted into the glass:

The intensity of the light reflected at the air-glass interface is: Substitute and simplify to obtain:

Express the intensity of the light transmitted at the glass-air interface:

I glass = I 0 − I R,1 where IR,1 is the intensity of the light reflected at the air-glass interface.

⎛1− n ⎞ =⎜ ⎟ I0 ⎝1+ n ⎠ 2

I R,1

⎛1− n ⎞ = I0 − ⎜ ⎟ I0 ⎝1+ n ⎠ ⎡ ⎛ 1 − n ⎞2 ⎤ = I 0 ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 1 + n ⎠ ⎥⎦ ⎡ 4n ⎤ = I0 ⎢ 2⎥ ⎣ (1 + n ) ⎦ 2

I glass

I T = I glass − I R,2 where IR,2 is the intensity of the light reflected at the glass-air interface.

Properties of Light

915

⎛1− n ⎞ =⎜ ⎟ I glass ⎝1+ n ⎠ 2

The intensity of the light reflected at the glass-air interface is:

I R,2

⎛ 1 − n ⎞ ⎡ 4n ⎤ =⎜ I ⎟ ⎢ 2⎥ 0 ⎝ 1 + n ⎠ ⎣ (1 + n ) ⎦ 2

⎡ 4n ⎤ ⎛ 1 − n ⎞ ⎡ 4n ⎤ −⎜ IT = I0 ⎢ I ⎟ ⎢ 2⎥ 2⎥ 0 ⎣ (1 + n ) ⎦ ⎝ 1 + n ⎠ ⎣ (1 + n ) ⎦ ⎡ ⎛ 1 − n ⎞ 2 ⎤ ⎡ 4n ⎤ = I 0 ⎢1 − ⎜ ⎟ ⎥⎢ 2⎥ ⎢⎣ ⎝ 1 + n ⎠ ⎥⎦ ⎣ (1 + n ) ⎦ ⎡ 4n ⎤ ⎡ 4 n ⎤ = I0 ⎢ 2 ⎥⎢ 2⎥ ⎣ (1 + n ) ⎦ ⎣ (1 + n ) ⎦ 2

Substitute and simplify to obtain:

⎡ 4n ⎤ = I0 ⎢ 2⎥ ⎣ (1 + n ) ⎦

2

75 •• Picture the Problem We can write an expression for the total distance traveled by the light as a function of x and set the derivative of this expression equal to zero to find the value of x that minimizes the distance traveled by the light. The adjacent figure shows the two points and the reflecting surface. The x and y coordinates are in meters.

D = d1 + d 2

(a) Express the total distance D traveled by the light:

=

(x + 2)2 + 4 + (2 − x )2 + 36

Differentiate D with respect to x:

dD d ⎡ = dx dx ⎢⎣ =

1 2

(x + 2)2 + 4 + (2 − x )2 + 36 ⎤⎥

[(x + 2)

2

+4

]

− 12

[

⎦

2(x + 2 ) + 12 (2 − x ) + 36

Simplify this expression to obtain:

2

]

− 12

2(2 − x )(− 1) = 0 for extrema

916

Chapter 31 x+2

(x + 2)2 + 4

−

2− x

=0

(2 − x )2 + 36

Solve for x to obtain:

x = − 1.00 m

(b) With x = −1 m:

θ i = tan −1 ⎢

⎡ − 2 − (− 1) ⎤ ⎣ 0 − 2 ⎥⎦ ⎛1⎞ = tan −1 ⎜ ⎟ = 26.6° ⎝2⎠

and

⎡ − 1 − (2) ⎤ ⎣ 0 − 6 ⎥⎦ ⎛3⎞ = tan −1 ⎜ ⎟ = 26.6° ⎝6⎠

θ r = tan −1 ⎢

*76 •• Picture the Problem Let the angle of refraction at the first interface by θ1 and the angle of refraction at the second interface be θ2. We can apply Snell’s law at each interface and eliminate θ1 and n2 to show that θ2 = θP2. Apply Snell’s Brewster’s law at the n1-n2 interface:

tan θ P1 =

n2 n1

Draw a reference triangle consistent with Brewster’s law:

Apply Snell’s law at the n1-n2 interface: Solve for θ1 to obtain:

Referring to the reference triangle we note that:

n1 sin θ P1 = n2 sin θ1

⎛ n1 ⎞ sin θ P1 ⎟⎟ ⎝ n2 ⎠

θ1 = sin −1 ⎜⎜

⎛n

⎞ ⎟ ⎜ n2 n 2 + n 2 ⎟ 1 2 ⎠ ⎝ ⎛ ⎞ n1 ⎟ = sin −1 ⎜ ⎜ n2 + n2 ⎟ 2 ⎠ ⎝ 1

θ1 = sin −1 ⎜

1

n2

Properties of Light

917

i.e., θ1 is the complement of θp1. Apply Snell’s law at the n2-n1 interface: Solve for θ2 to obtain:

n2 sin θ1 = n1 sin θ 2

⎛ n2 ⎞ sin θ1 ⎟⎟ ⎝ n1 ⎠

θ 2 = sin −1 ⎜⎜

⎛n

⎞ ⎟ ⎜ n1 n 2 + n 2 ⎟ 1 2 ⎠ ⎝ ⎛ ⎞ n2 ⎟= θ = sin −1 ⎜ P2 ⎜ n2 + n2 ⎟ 2 ⎠ ⎝ 1

Refer to the reference triangle again to obtain:

θ 2 = sin −1 ⎜

Equate these expressions for n2 sin θ1 to obtain:

n1 sin θ P = n1 sin θ 2 ⇒ θ 2 = θ P

2

n1

77 •• Picture the Problem We can use Brewster’s law in conjunction with index of refraction data from Figure 31-29 to calculate the polarization angles for the air-glass interface. From Brewster’s law we have:

⎛ n2 ⎞ ⎟⎟ ⎝ n1 ⎠

θ p = tan −1 ⎜⎜

or, for n1 = 1,

θ p = tan −1 n2

For silicate flint glass, n2 ≈ 1.62 and:

θ p = tan −1 (1.62) = 58.3°

For borate flint glass, n2 ≈ 1.57 and:

θ p = tan −1 (1.57 ) = 57.5°

For quartz glass, n2 ≈ 1.54 and:

θ p = tan −1 (1.54) = 57.0°

For silicate crown glass, n2 ≈ 1.51 and:

θ p = tan −1 (1.51) = 56.5°

918

Chapter 31

78 ••• Picture the Problem The diagram to the right shows the angles of incidence, refraction, and deviation at the first interface. We can use the geometry of this symmetric passage of the light to express θr in terms of α and δ1 in terms of θr and α. We can then use a symmetry argument to express the deviation at the second interface and the total deviation δ. Finally, we can apply Snell’s law at the first interface to complete the derivation of the given expression. (a) With respect to the normal to the left face of the prism, let the angle of incidence be θi and the angle of refraction be θr. From the geometry of the figure, it is evident that: Express the angle of deviation at the refracting surface: By symmetry, the angle of deviation at the second refracting surface is also of this magnitude. Thus: Solve for θi: Apply Snell’s law, with n1 = 1 and n2 = n, to the first interface: Substitute for θI to obtain:

(b) The angular separation is: Solve equation (1) for δ:

θ r = 12 α

δ1 = θ i − θ r = θ i − 12 α δ = 2δ1 = 2θ i − α

θi =

1 2

(α + δ )

sin θ i = n sin 12 α

sin

α +δ 2

= n sin

α 2

∆δ = δ violet − δ red

α⎤ ⎡ δ = 2 sin −1 ⎢n sin ⎥ − α 2⎦ ⎣

(1)

Properties of Light

919

Substitute to obtain:

⎧ ⎫ α⎤ α⎤ ⎡ ⎡ ∆δ = 2 sin −1 ⎢nviolet sin ⎥ − α − ⎨2 sin −1 ⎢nred sin ⎥ − α ⎬ 2⎦ 2⎦ ⎣ ⎣ ⎩ ⎭ α⎤ α⎤ ⎡ ⎡ = 2 sin −1 ⎢nviolet sin ⎥ − 2 sin −1 ⎢nred sin ⎥ 2⎦ 2⎦ ⎣ ⎣ Substitute numerical values and evaluate ∆δ:

60° ⎤ 60° ⎤ ⎡ ⎡ ∆δ = 2 sin −1 ⎢1.52 sin − 2 sin −1 ⎢1.48 sin = 3.47° ⎥ 2 ⎦ 2 ⎥⎦ ⎣ ⎣ *79 •• Picture the Problem We can apply Snell’s law at the critical angle and the polarizing angle to show that tan θp = sin θc. (a) Apply Snell’s law at the medium-vacuum interface:

n1 sin θ1 = n2 sin θ r

For θ1 = θc, n1 = n, and n2 = 1:

n sin θ c = sin 90° = 1

For θ1 = θp, n1 = n, and n2 = 1:

tan θ p =

Because both expressions equal one: (b) For any value of θ :

n2 1 = ⇒ n tan θ p = 1 n1 n

tan θ p = sin θ c tan θ > sin θ ⇒ θ p > θ c

80 •• Picture the Problem Let the numeral 1 refer to the side of the interface from which the light is incident and the numeral 2 to the refraction side of the interface. We can apply Snell’s law, under the conditions described in the problem statement, at the interface to derive an expression for n as a function of the angle of incidence (also the polarizing angle). (a) Apply Snell’s law at the airmedium interface: Because the reflected and refracted rays are mutually perpendicular:

sin θ1 = n sin θ 2

θ1 + θ 2 = 90° ⇒ θ 2 = 90° − θ1

920

Chapter 31

Substitute for θ2 to obtain:

sin θ1 = n sin (90° − θ1 ) = n cos θ1 or

n = tan θ1 = tan θ p

Substitute for θp and evaluate n: (b) Apply Snell’s law at the interface under conditions of total internal reflection: Because n1 = 1:

Substitute for n and evaluate θc:

n = tan 58° = 1.60

n2 sin θ c = n1 sin 90° = n1

⎛1⎞ ⎛1⎞ ⎟⎟ = sin −1 ⎜ ⎟ ⎝n⎠ ⎝ n2 ⎠

θ c = sin −1 ⎜⎜

⎛ 1 ⎞ ⎟ = 38.7° ⎝ 1.6 ⎠

θ c = sin −1 ⎜

81 •• Picture the Problem We can apply Snell’s law at the glass–liquid and liquid–air interfaces to find the refractive index of the unknown liquid, the angle of incidence (glass-air interface) for total internal reflection, and the angle of refraction of a ray into the liquid film. (a) Apply Snell’s law, under critical-angle conditions, at the glass–liquid interface: Solve for nliquid: Substitute numerical values and evaluate nliquid:

sin θ c =

nliquid nglass

nliquid = nglass sin θ c nliquid = (1.655) sin 53.7° = 1.33

⎛ 1 ⎞ ⎟ ⎟ n glass ⎝ ⎠

(b) With the liquid removed:

θ c = sin −1 ⎜⎜

Substitute numerical values and evaluate θc:

θ c = sin −1 ⎜

(c) Apply Snell’s law at the glass−liquid interface:

⎛ 1 ⎞ ⎟ = 37.2° ⎝ 1.655 ⎠

nglass sin θ1 = nliquid sin θ 2

Properties of Light ⎡ nglass

Solve for θ2:

θ 2 = sin −1 ⎢

⎢⎣ nliquid

921

⎤ sin θ1 ⎥ ⎥⎦

⎤ ⎡1.655 sin 37.2°⎥ = 48.8° ⎦ ⎣ 1.33

Substitute numerical values and evaluate θ2:

θ 2 = sin −1 ⎢

Because θ 2 is also the angle of incidence at the liquid − air interface and because it is larger than the critical angle for total internal reflection at this interface, no light will emerge. 82 •• Picture the Problem We can use Equation 31-18 and the result of Problem 86 to find the angular separation of these colors in the primary rainbow. Express the angular separation ∆φ of the colors:

∆φ = φd,blue − φd,red

From Equation 31-18, with nair = 1 and nwater = n:

φd = π + 2θ1 − 4 sin −1 ⎜

From Problem 86:

cosθ1m =

(1)

⎛ sin θ1 ⎞ ⎟ ⎝ n ⎠

n2 −1 3

or

⎡ n2 − 1 ⎤ ⎥ 3 ⎥⎦ ⎢⎣

θ1m = cos −1 ⎢

Substitute to obtain:

⎡

φd = π + 2 cos −1 ⎢ ⎢⎣

2 ⎛ ⎧⎪ ⎫⎞ ⎜ sin cos −1 ⎡⎢ n − 1 ⎤⎥ ⎪ ⎟ ⎬⎟ ⎨ ⎜ ⎪ 3 ⎢ ⎥ n2 − 1 ⎤ ⎣ ⎦ ⎪⎭ ⎟ −1 ⎩ ⎥ − 4 sin ⎜ ⎟ ⎜ 3 ⎥⎦ n ⎟ ⎜ ⎟ ⎜ ⎠ ⎝

Evaluate φd for blue light in water:

922

Chapter 31

⎡

φd,blue = π + 2 cos −1 ⎢ ⎢⎣

2 ⎛ ⎧ ⎫⎞ ⎜ sin ⎪cos −1 ⎡⎢ (1.3435) − 1 ⎤⎥ ⎪ ⎟ ⎬ ⎜ ⎨ 3 ⎥⎦ ⎪⎭ ⎟⎟ ⎢⎣ (1.3435)2 − 1 ⎤⎥ − 4 sin −1 ⎜ ⎪⎩ ⎜ ⎟ 3 1.3435 ⎥⎦ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠

= 139.42° Evaluate φd for red light in water:

⎡

φd,red = π + 2 cos −1 ⎢ ⎢⎣

2 ⎛ ⎧ ⎫⎞ ⎜ sin ⎪cos −1 ⎡⎢ (1.3318) − 1 ⎤⎥ ⎪ ⎟ ⎬ ⎨ ⎜ 3 ⎢⎣ ⎥⎦ ⎪⎭ ⎟⎟ (1.3318)2 − 1 ⎤⎥ − 4 sin −1 ⎜ ⎪⎩ ⎟ ⎜ 3 1.3318 ⎥⎦ ⎟ ⎜ ⎟ ⎜ ⎠ ⎝

= 137.75° Substitute in equation (1) and evaluate ∆φ:

∆φ = 139.42 − 137.75° = 1.67°

83 •• Picture the Problem We can use the result, obtained in Problem 74, that each slab 2

⎡ 4n ⎤ reduces the intensity of the transmitted light by ⎢ , to find the ratio of the 2⎥ ( ) n + 1 ⎣ ⎦ transmitted intensity to the incident intensity through N parallel slabs of glass for light of normal incidence. 2

(a) From Problem 74, each slab reduces the intensity by the factor:

⎡ 4n ⎤ ⎢ 2⎥ ⎣ (n + 1) ⎦

For N slabs:

⎡ 4n ⎤ I t = I0 ⎢ 2⎥ ⎣ (n + 1) ⎦

2N

and

(b) Evaluate equation (1) with N = 3 and n = 1.5:

⎡ 4n ⎤ It = ⎢ 2⎥ I0 ⎣ (n + 1) ⎦

2N

I t ⎡ 4(1.5) ⎤ =⎢ ⎥ I 0 ⎣ (1.5 + 1)2 ⎦

2 (3 )

(1)

= 0.783

Properties of Light (c) Begin the solution of equation (1) for N by taking the logarithm (arbitrarily to base 10) of both sides of the equation:

⎡ 4n ⎤ ⎛I ⎞ log⎜⎜ t ⎟⎟ = log ⎢ 2⎥ ⎝ I0 ⎠ ⎣ (n + 1) ⎦

Solve for N:

⎛I ⎞ log⎜⎜ t ⎟⎟ ⎝ I0 ⎠ N= ⎡ 4n ⎤ 2 log ⎢ 2⎥ ⎣ (n + 1) ⎦

Substitute numerical values and evaluate N:

N=

923

2N

⎡ 4n ⎤ = 2 N log ⎢ 2⎥ ⎣ (n + 1) ⎦

log(0.1) = 28.2 ≈ 28 ⎡ 4(1.5) ⎤ 2 log ⎢ 2⎥ ⎣ (1.5 + 1) ⎦

84 •• Picture the Problem We can apply Snell’s law at the air-slab interface to express the index of refraction n in terms of θ1 and θ2 and then use the geometry of the figure to relate θ2 to t and d.

Apply Snell’s law to the first interface:

sin θ1 = n sin θ 2

sin θ1 sin θ 2

Solve for n:

n=

From the diagram:

⎛d ⎞ d = t tan θ 2 ⇒ θ 2 = tan −1 ⎜ ⎟ ⎝t⎠

Substitute to obtain:

n=

sin θ1 ⎡ ⎛ d ⎞⎤ sin ⎢ tan −1 ⎜ ⎟⎥ ⎝ t ⎠⎦ ⎣

*85 •• Picture the Problem The angle that the rain appears to make with the vertical, according to the marathoner, is the angle whose tangent is the ratio of vrunner to vrain. The circular

924

Chapter 31

motion of the star is analogous to the circular motion of the cloud with vrunner = vearth and vrain = c.

⎛ vrunner ⎞ ⎟⎟ v ⎝ rain ⎠

(a) The angle that the rain appears to make with the vertical to the marathoner is given by:

θ = tan −1 ⎜⎜

Substitute numerical values and evaluate θ :

θ = tan −1 ⎜⎜

⎛ 4 m/s ⎞ ⎟⎟ = 24.0° ⎝ 9 m/s ⎠

R = H tan θ

(b) The cloud moves in a circle whose radius is given by:

R = (10 km ) tan 24° = 4.45 km

Substitute numerical values and evaluate R: (c) Here vrunner = vearth and vrain = c:

⎛ vearth ⎞ (1) ⎟ ⎝ c ⎠ where θ = 12 (angular diameter )

(d) From equation (1):

c=

Convert 20.6″ to degrees:

20.6" = 20.6" ×

Substitute numerical values and evaluate c:

c=

θ = tan −1 ⎜

vearth 2πRearth-sun = tan θ Tearth tan θ 1' 1° × = 5.722 × 10 −3° 60" 60'

2π (1.5 × 1011 m ) (1 y ) (3.156 × 107 s/y ) tan(20.6")

= 2.99 × 108 m/s Substitute numerical values and evaluate c:

c=

(

)

2π 1.5 × 1011 m = 2.99 × 108 m/s (1 y ) 3.156 × 107 s/y tan 5.722 × 10 −3°

(

) (

)

86 ••• Picture the Problem We can follow the directions given in the problem statement and use the hint to establish the given result.

Properties of Light ⎛ nair sin θ1 ⎞ ⎟⎟ ⎝ nwater ⎠

(a) Equation 31-18 is:

φd = π + 2θ1 − 4 sin −1 ⎜⎜

For nair = 1and nwater = n:

φd = π + 2θ1 − 4 sin −1 ⎜

Use the hint to differentiate φd with respect to θ1:

dφd d = dθ1 dθ1

(b) Set dφd/dθ1 = 0:

2−

⎛ sin θ1 ⎞ ⎟ ⎝ n ⎠

⎡ −1 ⎛ sin θ1 ⎞ ⎤ ⎢π + 2θ1 − 4 sin ⎜ n ⎟⎥ ⎝ ⎠⎦ ⎣ 4 cos θ1 = 2− n 2 − sin 2 θ1 4 cos θ1 n 2 − sin 2 θ1

= 0 for extrema

(

Simplify to obtain:

16 cos 2 θ1 = 4 n 2 − sin 2 θ1

Replace sin2θ1 with 1 − cos2θ1 and simplify:

12 cos 2 θ1 = 4n 2 − 4

Solve for cosθ1 = cosθ1m:

)

n2 − 1 3

cos θ1m = and

θ1m Evaluate θ1m for n = 1.33:

87 ••• Picture the Problem Let the thickness of the slab be t and the separation of the incident and emerging rays be ds. We can apply Snell’s law at both interfaces and use the geometry of the diagram and trigonometric relationships to show that the emerging ray and incident ray are parallel and to derive an expression for d.

925

⎡ n2 − 1 ⎤ = cos ⎢ ⎥ 3 ⎦⎥ ⎣⎢ −1

⎡ (1.33)2 − 1 ⎤ ⎥ = 59.6° 3 ⎢⎣ ⎥⎦

θ1m = cos −1 ⎢

926

Chapter 31

Apply Snell’s law at the two interfaces to obtain:

sin θ1 = n sin θ 2

(1)

and

n sin θ 3 = sin θ 4

sin θ1 = n sin θ 3

Because θ2 and θ3 are equal (they are alternate interior angles formed by parallel lines and a transversal):

and

Substitute for n sin θ 3 in the first of

sin θ1 = sin θ 4 ⇒ θ1 = θ 4 and

these equations to obtain:

n sin θ 3 = sin θ 4

the emerging ray and incident ray are parallel.

Express the distance dBD in terms of t and θ1: The distance dBC is: Use the distances dBD and dBC to express the distance dCD:

d BD = t tan θ1

d BC = t tan θ 2 d CD = d BD − d BC = t (tan θ1 − tan θ 2 )

Because α and θ1 have their right and left sides mutually perpendicular, they are equal and:

s = t (tan θ1 − tan θ 2 ) cos α

Substitute for tanθ1 and tanθ2 and simplify to obtain:Solve equation (1) for θ2:

⎛ sin θ1 sin θ 2 ⎞ ⎟⎟ cos θ1 s = t ⎜⎜ − cos θ cos θ 2 ⎠ 1 ⎝ ⎛ sin θ 2 cos θ1 ⎞ ⎟ = t ⎜⎜ sin θ1 − cos θ 2 ⎟⎠ ⎝ t (sin θ1 cos θ 2 − sin θ 2 cos θ1 ) = cos θ 2

= t (tan θ1 − tan θ 2 ) cos θ1

=

(2)

t sin (θ1 − θ 2 ) cos θ 2

Remarks: One can also derive this expression using the law of sines. 88 •• Picture the Problem We can use Snell’s law to determine θ2 and then apply the result of Problem 87 to find s.

Properties of Light From Problem 87 we have:

Apply Snell’s law to the first interface to obtain: Solve for θ2:

s=

927

t sin (θ1 − θ 2 ) cos θ 2

sin θ1 = n sin θ 2

⎛ sin θ1 ⎞ ⎟ ⎝ n ⎠

θ 2 = sin −1 ⎜

⎛ sin 30° ⎞ ⎟ = 19.47° ⎝ 1.5 ⎠

Substitute numerical values and evaluate θ2:

θ 2 = sin −1 ⎜

Substitute numerical values and evaluate s:

s=

(15 mm)sin (30° − 19.47°) cos(19.47°)

= 2.91 mm 89 ••• Picture the Problem The figure below shows the prism and the path of the ray through it. The dashed lines are the normals to the prism faces. The triangle formed by the interior ray and the prism faces has interior angles of α, 90° − θ2, and 90° − θ3. Consequently, θ 2 + θ 3 = α . We can apply Snell’s law at both interfaces to express the angle of deviation δ as a function of θ3 and then set the derivative of this function equal to zero to find the conditions on θ3 and θ2 that result in δ being a minimum.

Express the angle of deviation: Apply Snell’s law to relate θ1 to θ2 and θ3 to θ4:

δ = θ1 + θ 2 − α

(1)

sin θ1 = n sin θ 2

(2)

and

n sin θ 3 = sin θ 4

Solve equation (2) for θ1 and

θ1 = sin −1 (n sin θ 2 )

(3)

928

Chapter 31

equation (3) for θ4:

and

θ 4 = sin −1 (n sin θ 3 )

Substitute in equation (1) to obtain:

δ = sin −1 (n sin θ 2 ) + sin −1 (n sin θ 3 ) − α = sin −1 [n sin (θ 3 − α )] + sin −1 (n sin θ 3 ) − α Note that the only variable in this expression is θ3. To determine the condition that minimizesδ, take the derivative of δ with respect to θ3 and set it equal to zero.

dδ d = sin −1 [n sin (θ 3 − α )] + sin −1 (n sin θ 3 ) − α dθ 3 dθ 3

{

=−

n cos(α − θ 3 )

1 − [n sin (α − θ 3 )]

2

+

n cos θ 3

1 − (n sin θ 3 )

2

}

= 0 for extrema

This equation is satisfied provided:

α − θ 3 = θ 3 ⇒ θ 3 = 12 α

Because θ 2 = α − θ 3 :

θ 2 = α − 12 α = 12 α

Because θ 2 = θ 3 , we can conclude that the deviation angle is a minimum if the ray passes through the prism symmetrically. Remarks: Setting dδ/dθ3 = 0 establishes the condition on θ3 that δ is either a maximum or a minimum. To establish that δ is indeed a minimum when θ3 = θ2 = 12 α, we can either show that d 2 δ dθ 32 , evaluated at θ3 = 12 α , is positive or, alternatively, plot a graph of δ (θ3) to show that it is concave upward at θ3 = 12 α .

Chapter 32 Optical Images Conceptual Problems 1 • Determine the Concept Yes. Note that a virtual image is ″seen″ because the eye focuses the diverging rays to form a real image on the retina. Similarly, the camera lens can focus the diverging rays onto the film. 2 • Determine the Concept Yes; the mirror image is a left-handed coordinate system. 3 •• (a) False. The virtual image formed by a concave mirror when the object is between the focal point and the vertex of the mirror depends on the distance of the object from the vertex. (b) False. When the object is outside the focal point, the image is real. (c) True. (d) False. When the object is between the center of curvature and the focal point, the image is enlarged and real. *4 •• Determine the Concept Let s be the object distance and f the focal length of the mirror. (a) If s < f, the image is virtual, upright, and larger than the object. (b) If s < f, the image is virtual, upright, and larger than the object. (c) If s > 2f, the image is real, inverted, and smaller than the object. (d) If f < s < 2f, the image is real, inverted, and larger than the object. 5 •• Determine the Concept A convex mirror always produces a virtual, upright image that is s than the object. It never produces an enlarged image.

929

930

Chapter 32

6 •• Determine the Concept They appear more distant because the images are smaller than they would be in a flat mirror. 7 •• Picture the Problem The ray diagram shows three object positions 1, 2, and 3 as the object is moved from a great distance toward the focal point F of a concave mirror. The real images corresponding to each of these object positions are labeled with the same numeral. (b) is correct.

8 • Picture the Problem The diagram shows two rays (from the bundle of rays) of light refracted at the air-water interface. Because the index of refraction of water is greater than that of air, the rays are bent toward the normal. The diver will, therefore, think that the rays are diverging from a point above the bird and so the bird appears to be farther from the surface than it actually is. *9 • Determine the Concept (a) The lens will be positive if its index of refraction is greater than that of the surrounding medium and the lens is thicker in the middle than at the edges. Conversely, if the index of refraction of the lens is less than that of the surrounding medium, the lens will be positive if it is thinner at its center than at the edges. (b) The lens will be negative if its index of refraction is greater than that of the surrounding medium and the lens is thinner at the center than at the edges. Conversely, if the index of refraction of the lens is less than that of the surrounding medium, the lens will be negative if it is thicker at the center than at the edges.

Optical Images

931

10 • Determine the Concept The focal length depends on the index of refraction, and n is a function of wavelength. 11 •• Picture the Problem We can use a ray diagram to determine the general features of the image. In the diagram shown, the parallel ray and central ray have been used to locate the image.

From the diagram, we see that the image is virtual (only one of the rays from the head of the object actually pass through the head of the image), upright, and diminished.

(d ) is correct. 12 •• Picture the Problem We can use a ray diagram to determine the general features of the image. In the diagram shown, the ray parallel to the principle axis and the central ray have been used to locate the image.

From the diagram, we see that the image is virtual (neither ray from the head of the object passes through the head of the image), upright, and enlarged. (c) is correct. 13 • Determine the Concept The muscles in the eye change the thickness of the lens and thereby change the focal length of the lens to accommodate objects at different distances. A camera, on the other hand, has a fixed focal length so that focusing is accomplished by varying the distance between the lens and the film.

932

Chapter 32

*14 • Determine the Concept The eye muscles of a farsighted person lack the ability to shorten the focal length of the lens in the eye sufficiently to form an image on the retina of the eye. A convex lens (a lens that is thicker in the middle than at the circumference) will bring the image forward onto the retina. (a ) is correct. 15 •• Determine the Concept Refraction of light at the water-cornea interface is less than at the air-cornea interface and so an image that would normally (that is, without a corrective lens) be in front of the retina, is formed on the retina.

(b) is correct.

16 •• Determine the Concept A nearsighted person’s lenses form sharp images (unless the person is also astigmatic) of nearby object’s on the retinas of her eyes. The corrective lenses (convex) give a reduced image of the object and, therefore, should be removed.

(b) is correct. *17 • Determine the Concept Referring to the ray diagram show below we note that the image is always virtual and diminished. (d ) is correct.

18 • Determine the Concept Converging lenses can form real or virtual images that can be enlarged or reduced. (c) is correct. 19 • Picture the Problem We can apply the lens maker’s equation to the air-glass lens and to the water-glass lens to find the ratio of their focal lengths. Apply the lens maker’s equation to the air-glass interface:

⎛1 1⎞ 1 = (1.6 − 1)⎜⎜ − ⎟⎟ f air ⎝ r1 r2 ⎠

Optical Images Apply the lens maker’s equation to the water-glass interface: Divide the first of these equations by the second to obtain:

1 f water

f water f air

933

⎛1 1⎞ = (1.6 − 1.33)⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠

⎛1 1⎞ (1.6 − 1)⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠ = 2.22 = ⎛1 1⎞ (1.6 − 1.33)⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠

or f water = 2.22 f air and (a ) is correct. 20 •• (a) True. (b) True. (c) False. Where the rays intersect the axis of a spherical mirror depends on how far from the axis they are reflected from the mirror. (d) True. (e) False. The image distance for a virtual image is negative. *21 • Determine the Concept Microscopes ordinarily produce images (either the intermediate one produced by the objective or the one viewed through the eyepiece) that are larger than the object being viewed. A telescope, on the other hand, ordinarily produces images that are much reduced compared to the object. The object is normally viewed from a great distance and the telescope magnifies the angle subtended by the object.

Estimation and Approximation 22 •• Picture the Problem We can use the lens-maker’s equation to obtain a relationship between the two radii of curvature of the lenses we are to design. For a thin lens of focal length 27 cm and index of refraction of 1.6:

⎛1 1⎞ 1 = (1.6 − 1)⎜⎜ − ⎟⎟ 27 cm ⎝ r1 r2 ⎠ or

1 1 1 − = r1 r2 16.2 cm

934

Chapter 32

One solution is a plano-convex lens (one with a flat surface and a convex surface). Let r2 = ∞. Then

r1 = 16.2 cm and r2 = ∞ .

Another design is a double convex lens (one with both surfaces convex and radii of curvature that are equal in magnitude) obtained by letting r2 = −r1. Then r1 = 32.4 cm

and

r2 = − 32.4 cm . A third possibility is a double convex lens with unequal curvature, e.g., let r2 = 12 cm. Then

r1 = 6.89 cm and r2 = 12.0 cm .

23 •• Picture the Problem We can use the lens-maker’s equation to obtain a relationship between the two radii of curvature of the lenses we are to design. For a thin lens of focal length −27 cm and index of refraction of 1.6:

⎛1 1⎞ 1 = (1.6 − 1) ⎜⎜ − ⎟⎟ − 27 cm ⎝ r1 r2 ⎠ or

1 1 1 − = r1 r2 − 16.2 cm One solution is a plano-concave lens (one with a flat surface and a concave surface), Let r2 = ∞. Then

r1 = − 16.2 cm and r2 = ∞ .

Another design is a biconcave lens

Optical Images

935

(one with both surfaces concave) by letting r2 = −r1. Then

r1 = − 32.4 cm and r2 = 32.4 cm . A third possibility is a lens with r2 = 8.1 cm. Then r1 = 5.40 cm and r2 = 16.2 cm .

*24 •• Picture the Problem Because the focal length of a spherical lens depends on its radii of curvature and the magnification depends on the focal length, there is a practical upper limit to the magnification.

xnp

Use equation 32-20 to relate the magnification M of a simple magnifier to its focal length f:

M=

Use the lens-maker’s equation to relate the focal length of a lens to its radii of curvature and the index of refraction of the material from which it is constructed:

⎛1 1⎞ 1 = (n − 1) ⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠

For a plano-convex lens, r2 = ∞. Hence:

r 1 n −1 = ⇒ f = 1 f r1 n −1

Substitute in the expression for M and simplify to obtain:

M=

f

(n − 1)xnp r1

Note that the smallest reasonable value for r1 will maximize M. A reasonable smallest value for the radius of a magnifier is 1 cm. Use this value and n = 1.5 to estimate Mmax:

M max =

(1.5 − 1)(25 cm ) = 1 cm

12.5

936

Chapter 32

Plane Mirrors 25 • Determine the Concept Rays from the source and reflected by the mirror are shown. The reflected rays appear to diverge from the image. The eye can see the image if it is in the region between rays 1 and 2.

26 • Determine the Concept The mirror must be half the height of the person, i.e., 81 cm. The top of the mirror must be 7.5 cm below the top of the head, or 154.5 cm above the floor. The bottom of the mirror must be 73.5 cm above the floor. A ray diagram showing rays from the person’s feet and the top of her head reaching her eyes is shown to the right. *27 • Determine the Concept Draw rays of light from the object that satisfy the law of reflection at the two mirror surfaces. Three virtual images are formed, as shown in the adjacent figure. The eye should be to the right and above the mirrors in order to see these images.

28 • Determine the Concept Draw rays of light from the object that satisfy the law of reflection at the two mirror surfaces. The images are located at the intersection of the

Optical Images

937

dashed lines (extensions of the reflected rays). (a) The diagram to the right shows selected rays emanating from a point object (P) that form the two virtual images directly below the horizontal mirror:

The diagram to the right shows selected rays emanating from the point object (P) that form the image that lies on the bisector of the angle. There are two additional virtual images to the left of the mirror that is at 60° with the horizontal. Hence, the total number of images formed when a point object is on the bisector of the 60° angle is five. (b) The diagram to the right shows selected rays emanating from a point object (P) that form the two virtual images at the intersection of the dashed lines (extensions of the reflected rays):

29 •• Determine the Concept (a) The first image in the mirror on the left is 10 cm behind the mirror. The mirror on the right forms an image 20 cm behind that mirror or 50 cm from the left mirror. This image will result in a second image 50 cm behind the left mirror. The first image in the left mirror is 40 cm from the right mirror and forms an image 40 cm behind the right mirror or 70 cm from the left mirror. That image gives an image 70 cm behind the left mirror. The fourth image behind the left mirror is 110 cm behind that mirror.

938

Chapter 32

(b) Proceeding as in Part (a) for the mirror on the right, one finds the location of the images to be 20 cm, 40 cm, 80 cm, and 100 cm behind the right-hand mirror.

Spherical Mirrors *30 •• Picture the Problem The easiest rays to use in locating the image are 1) the ray parallel to the principal axis and passes through the focal point of the mirror, the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and 2) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object. (a) The ray diagram is shown to the right. The image is real, inverted, and reduced.

The image is real, inverted, and reduced. (b) The ray diagram is shown to the right.

The image is real, inverted, and the same size as the object.

Optical Images

939

(c) The ray diagram is shown to the right. The object is at the focal point of the mirror.

The emerging rays are parallel and do not form an image. (d) The ray diagram is shown to the right.

The image is virtual, erect, and enlarged. 31 • Picture the Problem In describing the images, we must indicate where they are located, how large they are in relationship to the object, whether they are real or virtual, and whether they are upright or inverted. The object distance s, the image distance s′, and the focal length of a mirror are related according to

1 1 1 + = , where f = 12 r and r is the s s' f

radius of curvature of the mirror. In this problem, f = 20 cm because r is positive for a concave mirror. Solve the mirror equation for s′:

s' =

fs s− f

(a) When s = 50 cm:

s' =

(12 cm )(50 cm) =

The lateral magnification of the image is:

50 cm − 12 cm

m=−

15.8 cm

s' 15.8 cm =− = −0.316 s 50 cm

940

Chapter 32 Because the image distance is positive and the lateral magnification is less than one and negative, we can conclude that the image real, inverted, and reduced.

(b) When s = 24 cm:

The lateral magnification of the image is:

s' =

(12 cm )(24 cm) =

24.0 cm

24 cm − 12 cm

m=−

s' 24 cm =− = −1 s 24 cm

Because the image distance is positive and the lateral magnification is one and negative, we can conclude that the image real, inverted, and the same size as the object. (c) When s = 12 cm:

s' =

(12 cm )(12 cm) = 12 cm − 12 cm

∞

and there is no image. (d) When s = 8 cm:

The lateral magnification of the image is:

s' =

(12 cm)(8 cm ) = 8 cm − 12 cm

m=−

− 24.0 cm

− 24 cm s' =− =3 s 8 cm

Because the image distance is negative and the lateral magnification is three and positive, we can conclude that the image virtual, erect, and three times the size of the object. 32 •• Picture the Problem The easiest rays to use in locating the image are 1) the ray parallel to the principal axis and passes through the focal point of the mirror, the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and 2) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object.

Optical Images

941

(a) The ray diagram is shown to the right.

The image is virtual, upright, and reduced. (b) The ray diagram is shown to the right.

The image is virtual, upright, and reduced. (c) The ray diagram is shown to the right.

The image is virtual, upright, and reduced. (d) The ray diagram is shown to the right.

The image is virtual, upright, and reduced. 33 • Picture the Problem In describing the images, we must indicate where they are located, how large they are in relationship to the object, whether they are real or virtual, and whether they are upright or inverted. The object distance s, the image distance s′, and the focal length of a mirror are related according to

1 1 1 + = , where f = 12 r and r is the s s' f

radius of curvature of the mirror. In this problem, f = −20 cm because r is negative for a

942

Chapter 32

convex mirror. Solve the mirror equation for s′:

s' =

fs s− f

(a) When s = 55 cm:

s' =

(− 12 cm )(55 cm ) = 55 cm − (− 12 cm )

The lateral magnification of the image is:

m=−

− 9.85 cm

− 9.85 cm s' =− = 0.179 s 55 cm

Because the image distance is negative and the lateral magnification is less than one in magnitude and positive, we can conclude that the image is virtual, upright, and reduced.

(− 12 cm)(24 cm) = 24 cm − (− 12 cm )

(b) When s = 24 cm:

s' =

The lateral magnification of the image is:

m=−

− 8.00 cm

− 8 cm s' =− = 0.333 s 24 cm

Because the image distance is negative and the lateral magnification is less than one in magnitude and positive, we can conclude that the image is virtual, upright, and reduced.

(− 12 cm )(12 cm ) = 12 cm − (− 12 cm )

(c) When s = 12 cm:

s' =

The lateral magnification of the image is:

m=−

− 6.00 cm

− 6 cm 1 s' = =− 12 cm 2 s

Because the image distance is negative and the lateral magnification is one − half in magnitude and positive, we can conclude that the image virtual, upright, and half the size of the object. (d) When s = 8 cm:

s' =

(− 12 cm )(8 cm ) = 8 cm − (− 12 cm )

− 4.80 cm

Optical Images The lateral magnification of the image is:

m=−

− 4.80 cm s' =− = 0.600 s 8 cm

Because the image distance is negative and the lateral magnification is less than one and positive, we can conclude that the image is virtual, upright, and reduced. 34 • Picture the Problem We can solve the mirror equation for 1/s′ and then examine the implications of f < 0 and s > 0. Solve the mirror equation for 1/s′:

1 1 1 s− f = − = s' f s sf

For a convex mirror:

f <0

With s > 0, the numerator is positive and the denominator negative. Consequently:

1 < 0 ⇒ s' < 0 s'

*35 • Picture the Problem We can use the mirror equation and the definition of the lateral magnification to find the radius of curvature of the mirror. (a) Express the mirror equation:

1 1 1 2 + = = s s' f r

Solve for r:

r=

The lateral magnification of the mirror is given by:

m=−

Solve for s′:

s' = −ms

Substitute for s′ in equation (1) to obtain:

r=

− 2ms 1− m

Substitute numerical values and evaluate r:

r=

− 2(5.5)(2.1 cm ) = 5.13 cm 1 − 5.5

2ss' s'+ s

(1)

s' s

943

944 (b)

Chapter 32 The mirror must be concave. A convex mirror always produces a diminished virtual image.

36 •• Picture the Problem We can use the mirror equation and the relationship between the focal length of a mirror and its radius of curvature to find the location of the image. We can then use the definition of the lateral magnification of the mirror to find the height of the image formed in the mirror. (a) and (b) Solve the mirror equation for for s′: Relate the focal length of the mirror to its radius of curvature:

s' =

fs s− f

f = 12 r

1 2

Substitute to obtain:

s' =

rs rs = 1 s − 2 r 2s − r

Substitute numerical values and evaluate s′:

s' =

(− 1.2 m )(10 m ) = 2(10 m ) − (− 1.2 m )

− 0.566 m

and

the image is 56.6 cm behind the mirror.

y' s' =− y s

(c) Express the lateral magnification of the mirror:

m=

Solve for y′:

y' = −

s' y s

Substitute numerical values and evaluate y′:

y' = −

− 0.566 m (2 m ) = 11.3 cm 10 m

37 •• Picture the Problem We can use the mirror equation to locate the image formed in this mirror and the expression for the lateral magnification of the mirror to find the diameter of the image. Solve the mirror equation for the location of the image of the moon:

s' =

fs f −s

Optical Images Because f = 12 r :

s' =

1 2

rs rs = 1 r − 2s 2 r −s

(8 m ) (3.8 × 108 m )

Substitute numerical values and evaluate s′:

s' =

8 m − 2(3.8 × 108 m )

Express the lateral magnification of the mirror:

m=

y' s' =− y s

Solve for y′:

y' = −

s' y s

Substitute numerical values and evaluate y′:

y' = −

− 4m 3.5 × 10 6 m 8 3.8 × 10 m

= 3.68 cm 38 •• Picture the Problem The rays from the point object are shown in the diagram to the right. Note that the rays that reflect from the mirror far from the axis do not converge at the same point as those that reflect from the mirror close to the mirror axis. For the small-angle rays, the point of convergence is 4.5 cm from the mirror. The 60° ray crosses the axis at 3 cm from the mirror. Consequently, the image extends from 4.5 cm to 3.0 cm, or about 1.5 cm along the axis.

945

(

= − 4.00 m

)

946

Chapter 32

*39 •• Picture the Problem (a) The figure to the right shows the mirror and the four rays drawn to scale. Using a calibrated ruler, the spread of the crossing points is δx ≈ 1.0 cm. Note that the triangles formed by the center of curvature, the point of reflection on the mirror, and the point of intersection of the reflected ray and the mirror axis are isosceles triangles. Express the equal angles of the isosceles triangles:

⎛ y⎞ ⎝R⎠

θ r = sin −1 ⎜ ⎟ where y is the distance of the incoming ray from the mirror axis and R is the radius of curvature of the mirror.

Using the law of cosines, the distance between the point of intersection and the mirror is given by: Evaluate d for y/R = 2/3:

Evaluate d for y/R = 1/12:

Express the spread δx:

−1 ⎧⎪ ⎡ ⎛ −1 ⎛ y ⎞ ⎞⎤ ⎫⎪ d = R ⎨1 − ⎢2 cos⎜⎜ sin ⎜ ⎟ ⎟⎟⎥ ⎬ ⎝ R ⎠ ⎠⎦ ⎪⎭ ⎝ ⎪⎩ ⎣

−1 ⎧⎪ ⎡ ⎛ −1 ⎛ 2 ⎞ ⎞⎤ ⎫⎪ d = (6 cm ) ⎨1 − ⎢2 cos⎜⎜ sin ⎜ ⎟ ⎟⎟⎥ ⎬ ⎝ 3 ⎠ ⎠⎦ ⎪⎭ ⎝ ⎪⎩ ⎣ = 1.975 cm

−1 ⎧⎪ ⎡ ⎛ −1 ⎛ 1 ⎞ ⎞⎤ ⎫⎪ d = (6 cm ) ⎨1 − ⎢2 cos⎜⎜ sin ⎜ ⎟ ⎟⎟⎥ ⎬ ⎝ 12 ⎠ ⎠⎦ ⎪⎭ ⎝ ⎪⎩ ⎣ = 2.990 cm

δx = 2.990 cm − 1.975 cm = 1.01 cm in good agreement with the result obtained above.

Optical Images (b) Evaluate d for y/R = 1/3:

947

−1 ⎧⎪ ⎡ ⎛ −1 ⎛ 1 ⎞ ⎞⎤ ⎫⎪ d = (6 cm ) ⎨1 − ⎢2 cos⎜⎜ sin ⎜ ⎟ ⎟⎟⎥ ⎬ ⎝ 3 ⎠ ⎠⎦ ⎪⎭ ⎝ ⎪⎩ ⎣ = 2.818 cm

Express the new spread δx′:

δx' = 2.990 cm − 2.818 cm = 0.172 cm

Express the ratio of δx′ to δx:

δx' 0.172 cm = = 17.0% δx 1.01 cm

By blocking off the edges of the mirror so that only paraxial rays within 2 cm of the mirror axis are reflected, the spread is reduced by 83.0%. 40 •• Picture the Problem We can use the mirror equation to find the focal length of the mirror and then apply it a second time to find the object position after the mirror has been moved. Solve the mirror equation for f:

f =

ss' s' + s

Substitute numerical values and evaluate f:

f =

(100 cm )(75 cm) = 42.86 cm

Solve the mirror equation for s:

s=

fs' s' − f

Find s for f = −42.86 cm and s′ = − 35 cm:

s=

(− 42.86 cm )(− 35 cm ) = 190.9 cm − 35 cm − (− 42.86 cm )

The distance d the mirror moved is:

d = 190.9 cm − 100 cm = 90.9 cm

75 cm + 100 cm

41 •• Picture the Problem We can use the mirror equation, with s = ∞, to find the image distance in the large mirror. Because this image serves as a virtual object for the small mirror, we can use the mirror equation a second time to find the focal length and, hence, the radius of curvature of the small mirror. (a) Express the mirror equation:

1 1 2 + = s s' r

948

Chapter 32

Because s = ∞:

Substitute numerical values and evaluate s′:

1 2 = and s' = 12 r s' r s' =

1 2

(5 m ) = 2.5 m

This image serves as a virtual object for the small mirror at s = −0.5 m. Solve the mirror equation for the focal length of the small mirror:

f small =

ss' s '+ s

Substitute numerical values and evaluate fsmall:

f small =

(− 0.5 m )(2 m ) = −0.667 m 2 m + (− 0.5 m )

The radius of curvature is twice the focal length:

rsmall = 2 f small = 2(− 0.667 m ) = − 1.33 m

(b) Because f small < 0, the small mirror is convex.

Images Formed by Refraction 42 • Picture the Problem The diagram shows two rays (from the bundle of rays) of light refracted at the glass-air interface. Because the index of refraction of air is less than that of water, the rays are bent away from the normal. The writing on the paper will, therefore, appear to be closer than it actually is. We can use the equation for refraction at a single surface to find the distance s′. Use the equation for refraction at a single surface to relate the image and object distances: Here we have n1 = n, n2 = 1, and r = ∞. Therefore:

n1 n2 n2 − n1 + = s s' r

n 1 + =0 s s'

Optical Images

Solve for s′:

s' = −

s n

Substitute numerical values and evaluate s′:

s' = −

2 cm = − 1.33 cm 1.5

949

where the minus sign tells us that the image is 1.33 cm below the glass surface. 43 • Picture the Problem The diagram shows two rays (from the bundle of rays) of light refracted at the water-air interface. Because the index of refraction of air is less than that of water, the rays are bent away from the normal. The fish will, therefore, appear to be closer than it actually is. We can use the equation for refraction at a single surface to find the distance s′. We’ll assume that the glass bowl is thin enough that we can ignore the refraction of the light passing through it. (a) Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Here we have n1 = n and n2 = 1. Therefore:

n 1 1− n + = s s' r

Solve for s′:

s' =

rs s (1 − n ) − nr

Substitute numerical values and evaluate s′:

s' =

(− 20 cm )(10 cm ) (10 cm )(1 − 1.33) − (1.33)(− 20 cm )

= − 8.54 cm where the minus sign tells us that the image is 8.54 cm from the front surface of the bowl.

950

Chapter 32

(b) Repeat (a) with s = 30 cm:

s' =

(− 20 cm )(30 cm) (30 cm )(1 − 1.33) − (1.33)(− 20 cm )

= − 35.9 cm where the minus sign tells us that the image is 35.9 cm from the front surface of the bowl. *44 •• Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

(1)

Here we have n1 = 1 and n2 = n = 1.5. Therefore:

1 n n −1 + = s s' r

Solve for s′:

s' =

nrs s (n − 1) − r

(a) Substitute numerical values (s = 35 cm and r = 7.2 cm) and evaluate s′:

s' =

(1.5)(7.2 cm )(35 cm ) (35 cm )(1.5 − 1) − (7.2 cm )

= 36.7 cm where the positive distance tells us that the image is 36.7 cm in back of the surface and is real.

(b) Substitute numerical values (s = 6.5 cm and r = 7.2 cm) and evaluate s′:

s' =

(1.5)(7.2 cm )(6.5 cm ) (6.5 cm )(1.5 − 1) − (7.2 cm )

= − 17.8 cm where the minus sign tells us that the image

Optical Images

951

is 17.8 cm in front of the surface and is

virtual.

(c) When s = ∞, equation (1) becomes:

n n −1 = s' r

Solve for s′:

s' =

nr n −1

Substitute numerical values and evaluate s′:

s' =

(1.5) (7.2 cm) = 1.5 − 1

21.6 cm

i.e., the image is at the focal point, is

real, and of zero size.

45 •• Picture the Problem We can use the equation for refraction at a single surface to find the image distance that corresponds to parallel light rays in the rod. Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Parallel rays imply that s′ = ∞. Therefore:

1 n −1 = s r

Solve for s:

s=

r n −1

Substitute numerical values and

s=

7.2 cm = 14.4 cm 1.5 − 1

(1)

952

Chapter 32

evaluate s: The ray diagram is shown below:

46 •• Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

(1)

Here we have n1 = 1 and n2 = n = 1.5. Therefore:

1 n n −1 + = s s' r

Solve for s′:

s' =

nrs s (n − 1) − r

(a) Substitute numerical values (s = 35 cm and r = −7.2 cm) and evaluate s′:

s' =

(1.5)(− 7.2 cm )(35 cm ) (35 cm )(1.5 − 1) − (− 7.2 cm )

= − 15.3 cm where the minus sign tells us that the image is 15.3 cm in front of the surface of the rod and is virtual.

Optical Images

(b) Substitute numerical values (s = 6.5 cm and r = −7.2 cm) and evaluate s′:

s' =

953

(1.5)(− 7.2 cm )(6.5 cm ) (6.5 cm )(1.5 − 1) − (− 7.2 cm )

= − 6.72 cm where the minus sign tells us that the image is 6.72 cm in front of the surface of the rod (located at the object) and is virtual.

(c) When s = ∞, equation (1) becomes:

n n −1 = s' r

Solve for s′:

s' =

Substitute numerical values and evaluate s′:

s' =

nr n −1

(1.5) (− 7.2 cm ) 1 .5 − 1

= − 21.6 cm where the minus sign tells us that the image is 21.6 cm in front of the surface of the rod and is virtual.

954

Chapter 32

47 •• Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

(1)

Solve for s′:

s' =

n2 rs s (n2 − n1 ) − n1r

(a) Substitute numerical values (s = 35 cm, n1 = 1.33, n2 = 1.5, and r = 7.2 cm) and evaluate s′:

s' =

(1.5)(7.2 cm)(35 cm ) (35 cm)(1.5 − 1.33) − (1.33)(7.2 cm )

= − 104 cm where the negative distance tells us that the image is 104 cm in front of the surface and is virtual.

(b) Substitute numerical values (s = 6.5 cm) and evaluate s′:

s' =

(1.5)(7.2 cm)(6.5 cm ) (6.5 cm )(1.5 − 1.33) − (1.33)(7.2 cm )

= − 8.29 cm where the minus sign tells us that the image is 8.29 cm in front of the surface and is

virtual.

Optical Images

(c) When s = ∞, equation (1) becomes:

n2 n2 − n1 = s' r

Solve for s′:

s' = −

Substitute numerical values and evaluate s′:

s' =

955

n2 r n2 − n1

1.5 (7.2 cm ) = 63.5 cm 1.5 − 1.33

i.e., the image is 63.5 cm to the right of the surface (at the focal point) and is real.

48 •• Picture the Problem We can use the equation for refraction at a single surface to find the images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

(1)

Solve for s′:

s' =

n2 rs s(n2 − n1 ) − n1r

(a) Substitute numerical values (s = 35 cm) and evaluate s′:

s' =

(1.5)(− 7.5 cm )(35 cm) (35 cm )(1.5 − 1.33) − (1.33)(− 7.5 cm)

= − 24.7 cm where the minus sign tells us that the image is 24.7 cm in front of the surface and is

virtual.

956

Chapter 32

(b) Substitute numerical values (s = 6.5 cm) and evaluate s′:

s' =

(1.5)(− 7.5 cm )(6.5 cm ) (6.5 cm )(1.5 − 1.33) − (1.33)(− 7.5 cm )

= − 6.60 cm where the minus sign tells us that the image is 6.60 cm in front of the surface and is

virtual.

(c) When s = ∞, equation (1) becomes:

n2 n2 − n1 = s' r

Solve for s′:

s' =

n2 r n2 − n1

Substitute numerical values and evaluate s′:

s' =

(1.5) (− 7.5 cm ) = 1.5 − 1.33

− 66.2 cm

i.e., the image is at the focal point, is

virtual, and of zero size.

Optical Images

957

*49 •• Picture the Problem We can use the equation for refraction at a single surface to find the images due to refraction at the ends of the glass rod. The image formed by the refraction at the first surface will serve as the object for the second surface. The sign of the final image distance will tell us whether the image is real or virtual. (a) Use the equation for refraction at a single surface to relate the image and object distances at the first surface:

n1 n2 n2 − n1 + = s s' r

Solve for s′:

s' =

n2 rs s (n2 − n1 ) − n1r

Substitute numerical values and evaluate s′:

s' =

(1.6)(8 cm)(20 cm) (20 cm )(1.6 − 1) − (8 cm)

(1)

= 64.0 cm (b) The object for the second surface is 96 cm − 64 cm = 32 cm from the surface whose radius is 16 cm. Substitute numerical values and evaluate s′:

(c)

s' =

(1)(− 16 cm )(32 cm ) (32 cm )(1 − 1.6) − (1.6)(− 16 cm )

= − 80.0 cm

The final image is 96 cm − 80 cm = 16 cm from the surface whose radius is 8 cm and is virtual.

50 •• Picture the Problem We can use the equation for refraction at a single surface to find the images due to refraction at the ends of the glass rod. The image formed by the refraction at the first surface will serve as the object for the second surface. The sign of the final image distance will tell us whether the image is real or virtual.

958

Chapter 32

(a) Use the equation for refraction at a single surface to relate the image and object distances at the first surface:

n1 n2 n2 − n1 + = s s' r

(1)

Solve for s′:

s' =

n2 rs s (n2 − n1 ) − n1r

Substitute numerical values and evaluate s′:

s' =

(1.6)(16 cm )(20 cm ) (20 cm )(1.6 − 1) − (16 cm )

= − 128 cm (b) The object for the second surface is 96 cm + 128 cm = 224 cm from the surface whose radius is 8 cm. Substitute numerical values and evaluate s′:

s' =

(1)(− 8 cm )(224 cm) (224 cm )(1 − 1.6) − (1.6)(− 8 cm )

= 14.7 cm

(c) The final image is 14.7 cm from the far end of the rod and is real.

Thin Lenses 51 • Picture the Problem We can use the lens-maker’s equation to find the focal length of each of the lenses. The lens-maker’s equation is:

⎛1 1⎞ 1 = (n − 1) ⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively.

(a) For r1 = 15 cm and r2 = −26 cm:

⎛ 1 1 1 ⎞ ⎟⎟ = (1.5 − 1) ⎜⎜ − f ⎝ 15 cm − 26 cm ⎠ and

f = 19.0 cm

Optical Images A double convex lens is shown to the right:

(b) For r1 = ∞ and r2 = −15 cm:

⎛1 1 1 ⎞ ⎟⎟ = (1.5 − 1) ⎜⎜ − f ⎝ ∞ − 15 cm ⎠ and

f = 30.0 cm A plano-convex lens is shown to the right:

(c) For r1 = −15 cm and r2 = +15 cm:

⎛ 1 1 1 ⎞ ⎟⎟ = (1.5 − 1) ⎜⎜ − f ⎝ − 15 cm 15 cm ⎠ and

f = − 15.0 cm A double concave lens is shown to the right:

(d) For r1 = ∞ and r2 = +26 cm:

⎛1 1 1 ⎞ ⎟⎟ = (1.5 − 1) ⎜⎜ − f ⎝ ∞ 26 cm ⎠ and

f = − 52.0 cm A plano-concave lens is shown to the right:

959

960

Chapter 32

52 • Picture the Problem We can use the lens-maker’s equation to find the focal length of the lens. The lens-maker’s equation is:

⎛1 1⎞ 1 = (n − 1) ⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively.

Substitute numerical values to obtain:

⎛ 1 1 1 ⎞ ⎟⎟ = (1.62 − 1) ⎜⎜ − f ⎝ − 100 cm − 40 cm ⎠

Solve for f:

f = 108 cm

*53 • Picture the Problem We can use the lens-maker’s equation to find the focal length of the lens and the thin-lens equation to locate the image. We can use m = −

s' to find the s

lateral magnification of the image. (a) The lens-maker’s equation is:

⎛1 1⎞ 1 = (n − 1) ⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively.

Substitute numerical values to obtain:

⎛ 1 1 1 ⎞ ⎟⎟ = (1.45 − 1) ⎜⎜ − f ⎝ − 30 cm 25 cm ⎠

Solve for f:

f = − 30.3 cm

(b) Use the thin-lens equation to relate the image and object distances:

1 1 1 + = s s' f

Solve for s′:

s' =

fs s− f

Substitute numerical values and evaluate s′:

s' =

(− 30.3 cm )(80 cm ) = 80 cm − (− 30.3 cm )

− 22.0 cm

Optical Images (c) The lateral magnification of the image is given by:

m=−

s' s

Substitute numerical values and evaluate m:

m=−

− 22 cm = 0.275 80 cm

961

(d) Because s' < 0 and m > 0, the image is virtual and upright. 54 • Picture the Problem We can use the lens-maker’s equation to find the focal length of each of the lenses described in the problem statement. The lens-maker’s equation is:

⎛1 1⎞ 1 = (n − 1) ⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠ where the numerals 1 and 2 denote the first and second surfaces, respectively.

(a) For r1 = 20 cm, r2 = 10 cm:

⎛ 1 1 1 ⎞ ⎟⎟ = (1.6 − 1) ⎜⎜ − f 20 cm 10 cm ⎝ ⎠ and

f = − 33.3 cm A sketch of the lens is shown to the right:

(b) For r1 = 10 cm, r2 = 20 cm:

⎛ 1 1 1 ⎞ ⎟⎟ = (1.6 − 1) ⎜⎜ − f ⎝ 10 cm 20 cm ⎠ and

f = 33.3 cm A sketch of the lens is shown to the right:

962

Chapter 32 ⎛ 1 1 1 ⎞ ⎟⎟ = (1.6 − 1)⎜⎜ − f ⎝ − 10 cm − 20 cm ⎠

(c) For r1 = −10 cm, r2 = −20 cm:

and

f = − 33.3 cm A sketch of the lens is shown to the right:

Remarks: Note that the lenses that are thicker on their axis than on their circumferences are positive (converging) lenses and those that are thinner on their axis are negative (diverging) lenses. *55 • Picture the Problem The parallel and central rays were used to locate the image in the diagram shown below. The power P of the lens, in diopters, can be found from P = 1/f and the size of the image from m =

y' s' =− . y s

The image is real, inverted, and diminished. The thin-lens equation is:

1 1 1 + = s s' f

Solve for s′:

s' =

fs s− f

Use the definition of the power of the lens to find its focal length:

f =

1 1 = = 0.1 m = 10 cm P 10 m −1

Substitute numerical values and evaluate s′:

s' =

(10 cm )(25 cm ) = 25 cm − 10 cm

16.7 cm

Optical Images

963

y' s' =− y s

Use the lateral magnification equation to relate the height of the image y′ to the height y of the object and the image and object distances:

m=

Solve for y′:

y' = −

s' y s

Substitute numerical values and evaluate y′:

y' = −

16.7 cm (3 cm ) = − 2.00 cm 25 cm

Because s' > 0 and y' = −2.00 cm, the image is real, inverted, and diminished in agreement with the ray diagram. 56 • Picture the Problem The parallel and central rays were used to locate the image in the diagram shown below. The power P of the lens, in diopters, can be found from P = 1/f and the size of the image from m =

y' s' =− . y s

The image is real and inverted and appears to be the same size as the object. The thin-lens equation is:

1 1 1 + = s s' f

Solve for s′:

s' =

fs s− f

Use the definition of the power of the lens to find its focal length:

f =

1 1 = = 0.1 m = 10 cm P 10 m −1

Substitute numerical values and evaluate s′:

s' =

(10 cm )(20 cm) = 20 cm − 10 cm

20.0 cm

964

Chapter 32 y' s' =− y s

Use the lateral magnification equation to relate the height of the image y′ to the height y of the object and the image and object distances:

m=

Solve for y′:

y' = −

s' y s

Substitute numerical values and evaluate y′:

y' = −

20 cm (1cm ) = − 1.00 cm 20 cm

Because s' > 0 and y' = −1 cm, the image is real, inverted, and the same size as the object in agreement with the ray diagram. 57 • Picture the Problem The parallel and central rays were used to locate the image in the diagram shown below. The power P of the lens, in diopters, can be found from P = 1/f and the size of the image from m =

y' s' =− . y s

The image is virtual, upright, and diminished. The thin-lens equation is:

1 1 1 + = s s' f

Solve for s′:

s' =

fs s− f

Use the definition of the power of the lens to find its focal length:

f =

1 1 = = −0.1 m = − 10 cm P − 10 m −1

Substitute numerical values and evaluate s′:

s' =

(− 10 cm )(20 cm ) = 20 cm − (− 10 cm )

− 6.67 cm

Optical Images

965

y' s' =− y s

Use the lateral magnification equation to relate the height of the image y′ to the height y of the object and the image and object distances:

m=

Solve for y′:

y' = −

s' y s

Substitute numerical values and evaluate y′:

y' = −

− 6.67 cm (1.5 cm ) = 0.500 cm 20 cm

Because s' < 0 and y' = 0.500 cm, the image is virtual, erect, and about one - third the size of the object in agreement with the ray diagram. 58 •• Picture the Problem The parallel and central rays were used to locate the image in the diagram shown below. The power P of the lens, in diopters, can be found from P = 1/f and the size of the image from m = −

s' . s

(a) A negative object distance implies that the object is a virtual object, i.e., that light rays converge on the object rather than diverge from the object. A virtual object can occur in a two-lens system when the first lens forms an image that is at a distance − s from the second lens. (b) The thin-lens equation is:

1 1 1 + = s s' f

Solve for s′:

s' =

fs s− f

Substitute numerical values and evaluate s′:

s' =

(20 cm)(− 20 cm) = − 20 cm − (20 cm )

The lateral magnification is:

m=−

10.0 cm

s' 10 cm =− = 0.500 s − 20 cm

The parallel and central rays were used to locate the image in the ray diagram shown below:

966

Chapter 32

Because s' > 0 and m > 0, the image is real, erect, and one - half the size of the virtual object. (c) Proceed as in (b) with s = −10 cm and f = −30 cm:

s' =

(− 30 cm)(− 10 cm) = − 10 cm − (− 30 cm )

15.0 cm

and

m=−

s' 15 cm =− = 1.500 s − 10 cm

The parallel and central rays were used to locate the image in the ray diagram shown below:

Because s' > 0 and m = 1.5, the image is real, erect, and one and one - half times the size of the virtual object. *59 •• Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens. (a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.

Optical Images f1s1 s1 − f1

Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

Find the lateral magnification of the first image:

m1 = −

Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm − 20 cm = 15 cm. Equation (1) applied to the second lens is:

s2' =

Substitute numerical values and evaluate s2' :

s2' =

967

(1)

(10 cm)(20 cm ) = 20 cm 20 cm − 10 cm

s1' 20 cm =− = −1 s 20 cm

f 2 s2 s2 − f 2

(10 cm )(15 cm ) = 30 cm 15 cm − 10 cm

and the final image is 85.0 cm from the object. Find the lateral magnification of the second image:

m2 = −

s2' 30 cm =− = −2 s 15 cm

Because s' 2 > 0 and m = m1m2 = 2, the image is real, erect, and twice the size of the object. The overall lateral magnification of the image is the product of the magnifications of each image:

m = m1m2 = (− 1)(− 2 ) = 2.00

60 •• Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens. (a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.

968

Chapter 32

f1s1 s1 − f1

Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

Find the lateral magnification of the first image:

m1 = −

Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm − 20 cm = 15 cm. Equation (1) applied to the second lens is:

s2' =

f 2 s2 s2 − f 2

Substitute numerical values and evaluate s2' :

s2' =

(− 15 cm )(15 cm ) = −7.5 cm 15 cm − (− 15 cm )

(1)

(10 cm)(20 cm ) = 20 cm 20 cm − 10 cm

s'1 20 cm =− = −1 s 20 cm

and the final image is

47.5 cm from the

object. Find the lateral magnification of the second image:

m2 = −

s2' − 7.5 cm =− = 0.5 s 15 cm

Because s' 2 < 0 and m = m1m2 = −0.5, the image is virtual, inverted, and half as large as the object. The overall lateral magnification of the image is the product of the magnifications of each image:

m = m1m2 = (− 1)(0.5) = − 0.500

61 •• Picture the Problem We can use the thin-lens equation and the definition of the lateral magnification to show that s = (m − 1)f/m.

Optical Images

(a) Express the thin-lens equation:

1 1 1 + = s s' f

Express the lateral magnification of the image and solve for s′:

m=−

Substitute to obtain:

1 1 1 + = s − ms f

Solve for s:

s=

969

s' ⇒ s' = − ms s

(m − 1) f m

(b) The magnification m is:

m=−

Substitute numerical values and evaluate s:

s=

y' 24 mm =− = −0.0137 y 1.75 m

(− 0.0137 − 1)(50 mm) = − 0.0137

3.70 m

62 •• Picture the Problem We can plot the first graph by solving the thin-lens equation for the image distance s′ and the second graph by using the definition of the magnification of the image.

fs s− f

(a) and (b) Solve the thin-lens equation for s′ to obtain:

s' =

The magnification of the image is given by:

m=−

s' s

A spreadsheet program to calculate s′as a function of s is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 A4 A5 B4

Content/Formula 12 13.2 A4 + 1 $B$1*A4/(A4 − $B$1)

C5

−B4/A4

Algebraic Form f s s + ∆s

fs s− f s' − s

Chapter 32

970

A

B

C

1 2 3 4 5 6 7 8 9

s 13.2 14.2 15.2 16.2 17.2 18.2

s' 132.00 77.45 57.00 46.29 39.69 35.23

m −10.00 −5.45 −3.75 −2.86 −2.31 −1.94

108 109 110 111

117.2 118.2 119.2 120.2

13.37 13.36 13.34 13.33

−0.11 −0.11 −0.11 −0.11

f= 12

cm

A graph of s′ as a function of s follows. 140 120

s' (cm)

100

m

80 60 40 20 0 -20 0

20

40

60

80

100

120

s (cm)

(c) The images are real and inverted for this range of object distances.

The asymptotes of the graph of s' versus s correspond to the focal length of the lens. The horizontal asymptote of the graph of m versus (d) s indicates the fact that, as the object moves away from the lens, the image formed by the lens approaches the far focal point and its size approaches zero.

Optical Images

971

63 •• Picture the Problem We can plot the first graph by solving the thin-lens equation for the image distance s′ and the second graph by using the definition of the magnification of the image.

fs s− f

(a) and (b) Solve the thin-lens equation for s′ to obtain:

s' =

The magnification of the image is given by:

m=−

s' s

A spreadsheet program to calculate s′as a function of s is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 A4 A5 B4

Content/Formula 12 0.12 A4 + 0.1 $B$1*A4/(A4 − $B$1)

C5

−B4/A4

A

Algebraic Form f s s + ∆s

fs s− f s' − s

B

C

1 2 3 4 5 6 7 8 9

s 0.12 0.22 0.32 0.42 0.52 0.62

s' −0.12 −0.22 −0.33 −0.44 −0.54 −0.65

m 1.01 1.02 1.03 1.04 1.05 1.05

108 109 110 111

10.52 10.62 10.72 10.82

−85.30 −92.35 −100.50 −110.03

8.11 8.70 9.37 10.17

f= 12

cm

972

Chapter 32

A graph of s′ as a function of s follows. 10 0 -10 -20 -30 -40

s' (cm)

-50

m

-60 -70 -80 -90 -100 -110 0

2

4

6

8

10

12

s (cm)

(c) The images are virtual and erect for this range of object distances.

The asymptote of the graph of s ' versus s corresponds to the image approaching infinity as the object distance approaches the focal (d) length of the lens. The horizontal asymptote of the graph of m versus s indicates that, as the object moves toward the lens, the height of the image formed by the lens approaches the height of the object. *64 •• Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.

f1s1 s1 − f1

Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

With s1' = ∞, the thin-lens equation

1 1 ⇒ s 2' = f 2 = 15.0 cm = s2' f 2

applied to the second lens becomes:

(1)

(15 cm )(15 cm) = ∞ 15 cm − 15 cm

Optical Images

973

A ray diagram is shown below:

The final image is 50 cm from the object, real, inverted, and the same size as the object. 65 •• Picture the Problem We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.

f1s1 s1 − f1

Apply the thin-lens equation to express the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

With s'1 = ∞, the thin-lens equation

1 1 ⇒ s2' = f 2 = 15.0 cm = s2' f 2

applied to the second lens becomes:

(1)

(15 cm )(15 cm) = ∞ 15 cm − 15 cm

A ray diagram is shown below:

The final image is 50 cm from the object, real, inverted, and the same size as the object. 66 ••• Picture the Problem We can substitute x = s − f and x′ = s′ − f in the thin-lens equation and the equation for the lateral magnification of an image to obtain Newton’s equations.

974

Chapter 32

Express the thin-lens equation:

1 1 1 + = s s' f

If x = s − f and x′ = s′ − f:

1 1 1 + = x + f x' + f f

Expand this expression to obtain:

f ( x' + x + 2 f ) = ( x + f )( x' + f ) = xx' + xf + x'f + f 2 or, simplifying,

The lateral magnification is:

m=−

xx' = f 2

(1)

s' s

or, because x = s − f and x′ = s′ − f,

m=−

x' + f x+ f

f2 x'

Solve equation (1) for x:

x=

Substitute for x and simplify to obtain:

m=−

The lateral magnification is also given by:

m=−

From equation (1) we have:

f2 x' = x

Substitute to obtain:

f2 + f m=− x =− x+ f

x' + f x' + f =− 2 f ( f + x' ) f +f x' x' x' = − f

x'+ f x+ f

⎛f ⎞ f ⎜ + 1⎟ ⎝x ⎠= − f f⎞ x ⎛ x⎜1 + ⎟ x⎠ ⎝

Optical Images

975

The variables x, f, s,and s′ are shown in the sketch below:

67 ••• Picture the Problem The ray diagram shows the two lens positions and the corresponding image and object distances (denoted by the numerals 1 and 2). We can use the thin-lens equation relate the two sets of image and object distances to the focal length of the lens and then use the hint to express the relationships between these distances and the distances D and L to eliminate s1, s1′, s2, and s2′ and obtain an expression relating f, D, and L. Relate the image and object distances for the two lens positions to the focal length of the lens:

1 1 1 1 1 1 + = and + = s1 s1' f s2 s2' f

Solve for f to obtain:

f =

The distances D and L can be expressed in terms of the image and object distances: Substitute for the sums of the image and object distances in equation (1) to obtain:

s1s1' s s' = 2 2 s1 + s1' s2 + s2'

D = s1 + s1' = s2 + s2' and

L = s2 − s1 = s1' − s2'

f =

s1s1' s2 s2' = D D

From the hint:

s1 = s2' and s1' = s2

Hence D = s1 + s2 and:

D − L = 2 s1 and D + L = 2 s2

Take the product of D − L and D + L to obtain:

(1)

(D − L )(D + L ) = D 2 − L2 = 4s1s2 = 4s1s1'

From the thin-lens equation:

4 s1s2 = rs1s1' = 4 fD

Substitute to obtain:

4 fD = D 2 − L2

976

Chapter 32

Solve for f:

f =

D 2 − L2 4D

68 •• Picture the Problem We can use results obtained in Problem 67 to find the focal length of the lens and the two locations of the lens with respect to the object. (a) From Problem 77 we have:

f =

2 2 ( 1.7 m ) − (0.72 m ) = 4(1.7 m )

Substitute numerical values and evaluate f:

f

(b) Solve the thin-lens equation for the image distance to obtain:

s' =

In Problem 77 it was established that:

D 2 − L2 4D

fs f −s

= 34.9 cm

(1)

D − L = 2 s1 and D + L = 2 s2

Solve for s1 and s2:

s1 =

D−L D+L and s 2 = 2 2

Substitute numerical values and evaluate s1 and s2:

s1 =

170 cm − 72 cm = 49.0 cm 2

and

s2 =

170 cm + 72 cm = 121cm 2

69 ••• Picture the Problem The ray diagram shows four rays from the head of the object that locate images I1 and I2. We can use the thin-lens equation to find the location of the image formed in the positive lens and then, knowing the separation of the two lenses, determine the object distance for the second lens and apply the thin lens a second time to find the location of the final image.

Optical Images (a) Express the object-to-image distance d:

d = s1 + 5 cm + s2'

Apply the thin-lens equation to the positive lens:

1 1 1 + = s1 s1' f1

Solve for s1' :

s1' =

Substitute numerical values and evaluate s1' :

s1' =

Find the object distance for the negative lens:

977

(1)

f1s1 s1 − f1

(8.5 cm)(17.5 cm) = 16.53 cm 17.5 cm − 8.5 cm

s2 = 5 cm − s1' = 5 cm − 16.53 cm = −11.53 cm

The image distance s2′ is given by:

s2' =

f 2 s2 s2 − f 2

Substitute numerical values and evaluate s2′:

s2' =

(− 30 cm )(− 11.53 cm ) = 18.7 cm − 11.53 cm − (− 30 cm )

Substitute numerical values in equation (1) and evaluate d:

(b) The overall lateral magnification is given by:

d = 17.5 cm + 5 cm + 18.7 cm = 41.2 cm m = m1m2

s1' s' and m2 = − 2 s1 s2

Express m1 and m2:

m1 = −

Substitute to obtain:

⎛ s ' ⎞⎛ s ' ⎞ s 's ' m = ⎜⎜ − 1 ⎟⎟⎜⎜ − 2 ⎟⎟ = 1 2 ⎝ s1 ⎠⎝ s2 ⎠ s1s2

Substitute numerical values and evaluate m:

m=

(16.53 cm)(18.7 cm ) = (17.5 cm )(− 11.53 cm )

− 1.53

Because m < 0, the image is inverted. Because s2' > 0, the image is real.

978

Chapter 32

Aberrations *70 • Determine the Concept Chromatic aberrations are a consequence of the differential refraction of light of differing wavelengths by lenses. (a ) is correct. 71 • (a) False. Aberrations are a consequence of imperfections in lenses. (b) True. 72 • Picture the Problem We can use the lens-maker’s equation to find the focal length the this lens for the two colors of light. The lens-maker’s equation relates the radii of curvature and the index of refraction to the focal length of the lens:

⎛1 1⎞ 1 = (n − 1) ⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠

(a) For red light:

⎛ 1 1 1 ⎞ ⎟⎟ = (1.47 − 1) ⎜⎜ − f red ⎝ 10 cm − 10 cm ⎠ and

f red = 10.6 cm (b) For blue light:

1

f blue

⎛ 1 1 ⎞ ⎟⎟ = (1.53 − 1) ⎜⎜ − ⎝ 10 cm − 10 cm ⎠

and

f blue = 9.43 cm

The Eye *73 •• Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens. (a) Use the thin-lens equation to relate the image and object distances to the power of the lens:

1 1 1 + = =P s s' f

Optical Images Because s′ = d and, for a distance object, s = ∞: (b) If

xnp is the closest distance an

Pmin =

Pmax =

object could be and still remain in clear focus on the screen, equation (1) becomes: (c) Use our result in (a) to obtain:

979

1 1 = s' d

1 1 + xnp d

Use the results of (a) and (b) to express the accommodation of the model eye:

1 = 40.0 D 2.5 cm 1 1 1 1 + − = A = Pmax − Pmin = xnp d d xnp

Substitute numerical values and evaluate A:

A=

Pmin =

1 = 4.00 D 25 cm

74 •• Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens. (a) Use the thin-lens equation to relate the image and object distances to the power of the lens:

1 1 1 + = =P s s' f

Because s′ = d and s = xfp:

Pmin =

(b) To correct for the nearsightedness of this eye, we need a lens that will form an image 25 cm in front of the eye of an object at the eye’s far point:

Pmin =

1 1 + xfp d 1 1 + = − 2.00 D 50 cm − 25 cm

75 •• Picture the Problem The thin-lens equation relates the image and object distances to the power of a lens. (a) Use the thin-lens equation to relate the image and object distances to the power of the lens:

1 1 1 + = =P s s' f

980

Chapter 32 1 1 + x'np d

Because s′ = d and s = x′np:

P'max =

(b) For a normal eye:

Pmax =

The amount by which the power of the lens is too small is the difference between equations (2) and (1):

Pmax − P' max =

1 1 + xnp d

(2)

1 1 ⎛ 1 1⎞ + −⎜ + ⎟ xnp d ⎜⎝ x' np d ⎟⎠ 1 1 − xnp x' np

=

(c) For xnp = 15 cm and x′np = 150 cm:

Pmax − P' max =

(1)

1 1 − 15 cm 150 cm

= 6.00 D 76 • Picture the Problem We can use the thin-lens equation to find the distance from the lens to the image and then take their difference to find the distance the lens would have to be moved. Express the distance d that the lens would have to move:

d = s' − f

Solve the thin-lens equation for s′:

s' =

fs s− f

Substitute to obtain:

d=

fs −f s− f

Substitute numerical values and evaluate d:

d=

(2.5 cm )(25 cm ) − 2.5 cm 25 cm − 2.5 cm

= 0.278 cm That is, the lens would have to move 0.278 cm toward the object. 77 • Picture the Problem We can apply the thin-lens equation for the two values of s to find ∆f.

Optical Images Express the change ∆f in the focal length:

∆f = f s=3 m − f s=0.3 m

Solve the thin-lens equation for s:

ss' s' + s s3 m s'3 m s0.3 m s' 0.3 m ∆f = − s'3 m + s3 m s' 0.3 m + s0.3 m

Substitute to obtain:

981

f =

or, because s′3 m = s′0.3 m,

⎡ s3 m ⎤ s0.3 m ∆f = s'3 m ⎢ − ⎥ ⎣⎢ s'3 m + s3 m s' 0.3 m + s0.3 m ⎦⎥ Substitute numerical values and evaluate ∆f:

⎡ ⎤ 300 cm 30 cm ∆f = (2.5 cm ) ⎢ − ⎥ = 0.172 cm = 1.72 mm ⎣ 2.5 cm + 300 cm 2.5 cm + 30 cm ⎦ 78 • Picture the Problem We can use the thin-lens equation and the definition of the power of a lens to express the near point distance as a function of P and s. From the thin-lens equation we have:

1 1 1 + = =P s s' f

Solve for s′:

s' =

s Ps − 1

Substitute numerical values and evaluate s′:

s' =

25 cm = −44.4 cm 1.75 m (0.25 m ) − 1

(

−1

)

The person' s near point with lenses is 44.4 cm. *79 • Picture the Problem We can use the relationship between a distance measured along the arc of a circle and the angle subtended at its center to approximate the smallest angle the two points can subtend and the separation of the two points 20 m from the eye. (a) Relate θmin to the diameter of the eye and the distance between the activated cones:

d eyeθ min ≈ 2 µm

982

Chapter 32

Solve for θmin:

θ min =

2 µm d eye

Substitute numerical values and evaluate θmin:

θ min =

2 µm = 80.0 µrad 2.5 cm

(b) Let D represent the separation of the points R = 20 m from the eye to obtain:

D = Rθ min = (20 m )(80 µrad ) = 1.60 mm

80 •• Picture the Problem We can use the thin-lens equation to find f and the definition of the power of a lens to find P. (a) Solve the thin-lens equation for f:

f =

ss' s' + s

Noting that s′ < 0, substitute numerical values and evaluate f:

f =

(45 cm )(− 80 cm ) =

(b) Use the definition of the power of a lens to obtain:

P=

1 1 = = 0.971diopters f 1.03 m

− 80 cm + 45 cm

103 cm

81 •• Picture the Problem We can use the thin-lens equation to find f and the definition of the power of a lens to find P.

1 f

Express the required power of the lens:

P=

The thin-lens equation is:

1 1 1 + = s s' f

For s = ∞:

1 1 = ⇒ f = s' s' f

Substitute for f to obtain:

P=

1 s'

Substitute for s′ and evaluate P:

P=

1 = 0.444 diopters 2.25 m

Optical Images

983

82 •• Picture the Problem We can use the lens-maker’s equation with s = ∞ to find the radius of the cornea modeled as a homogeneous sphere with an index of refraction of 1.4. Use the lens-maker’s equation to relate the radius of the cornea to its index of refraction and that of air:

n1 n2 n2 − n1 + = s s' r

Because n2 = n, n1 = 1, and s = ∞:

n n −1 = s' r

Solve for r:

r=

Substitute numerical values and evaluate r:

1 ⎞ ⎛ r = ⎜1 − ⎟ (2.5 cm ) = 0.714 cm ⎝ 1.4 ⎠

s' (n − 1) ⎛ 1 ⎞ = ⎜1 − ⎟ s' n ⎝ n⎠

The eye is not a homogeneous sphere. It is filled with a transparent liquid (vitreous humor) which has an index of refraction that is not known. If that index of refraction differs from 1.4, there is refraction at the inner surface of the cornea which will result in the formation of the image nearer the cornea' s surface if n > 1.4 and farther if n < 1.4, where n is the index of refraction of the vitreous humor. If n < 1.4, then r as calculated above is too small. 83 •• Picture the Problem We can use the definition of the power of a lens and the thin-lens equation to find the power of the lens that should be used in the glasses. Express the power of the lens that should be used in the glasses:

P = Peye + Plens =

1 1 + f eye f glasses

s' = −80 cm + 2 cm = −78 cm

Because the glasses are 2 cm from the eye:

and

Apply the thin-lens equation to the eye with s′ = ∞:

1 1 ⇒ f eye = s = s f eye

s = 25 cm − 2 cm = 23 cm

(1)

984

Chapter 32

Apply the thin-lens equation to the glasses with s = ∞:

1 1 ⇒ f glasses = s' = s' f glasses

Substitute for feye and fglasses in equation (1) to obtain:

P=

1 1 + s s'

Substitute numerical values and evaluate P:

P=

1 1 + = 3.07 D 0.23 m − 0.78 m

84 ••• Picture the Problem We can use the thin-lens equation and the distance from her eyes to her glasses to derive an expression for the location of her near point. (a) Express her near point, xnp, at age 45 in terms of the location of her glasses: Because the glasses are 2.2 cm from her eye:

xnp = s' + 2.2 cm

(1)

s = 25 cm − 2.2 cm = 22.8 cm

Apply the thin-lens equation to the glasses:

1 1 1 + = =P s s' f glasses

Solve for s′:

s' =

1 s = Ps − 1 P − 1 s

Substitute in equation (1) to obtain:

xnp =

Substitute numerical values and evaluate xnp:

xnp =

1 1 P− s

+ 2.2 cm

1 2.1 m −1 −

1 0.228 m

(2)

+ 2.2 cm

= 45.9 cm (b) At age 55:

s = 40 cm − 2.2 cm = 37.8 cm

Optical Images Substitute numerical values in equation (2) and evaluate s′:

1

xnp =

2.1 m −1 −

1 0.378 m

985

+ 2.2 cm

= 185 cm (c) Solve the thin-lens equation for f: From the definition of P:

For s = 22.8 cm and s′ = 183.3 cm:

ss' s' + s 1 s' + s P= = f s's f =

P=

183.3 cm + 22.8 cm = 4.93 D (183.3 cm )(22.8 cm )

The Simple Magnifier *85 • Picture the Problem We can use the definitions of the magnifying power of a lens ( M = xnp f ) and of the power of a lens ( P = 1 f ) to find the magnifying power of the given lens. The magnifying power of the lens is given by:

M =

xnp f

= Pxnp

where P is the power of the lens. Substitute numerical values and evaluate M:

(

)

M = 20 m −1 (0.3 m ) = 6.00

86 • Picture the Problem We can use the definition of the magnifying power of a lens to find the required focal length so that this person’s lens will have magnification power of 5.

xnp

The magnifying power of the lens is given by:

M =

Solve for f:

f =

xnp

f =

25 cm = 5.00 cm 5

Substitute numerical values and evaluate f:

f

M

986

Chapter 32

87 • Picture the Problem We can use the definition of the magnifying power of a lens to find the magnifying power of this lens. The magnifying power of the lens is given by:

M =

xnp

Substitute numerical values and evaluate M:

M =

35 cm = 5.00 7 cm

f

88 •• Picture the Problem Let the numerals 1 and 2 denote the 1st and 2nd persons, respectively. We can use the definition of magnifying power to find the effective magnifying power of the lens for each person. The relative height of the images on the retinas of the two persons is given by the ratio of the effective magnifying powers. The magnifying power of the lens is given by:

M =

xnp

Substitute numerical values and evaluate M1 and M2:

M1 =

25 cm = 4.17 6 cm

f

and

M2 =

40 cm = 6.67 6 cm

From the definition of magnifying power we have:

y1 M1 y f = = 1 M 2 y2 y2 f

Substitute for M1 and M2 and evaluate the ratio of y1 to y2:

y1 4.17 = = 0.625 y2 6.67

89 •• Picture the Problem We can use the definition of angular magnification to find the expected angular magnification if the final image is at infinity and the thin-lens equation and the expression for the magnification of a thin lens to find the angular magnification when the final image is at 25 cm.

Optical Images (a) Express the angular magnification when the final image is at infinity: Substitute numerical values and evaluate M:

M =

xnp f

where P is the power of the lens.

(

)

M = (25 cm ) 12 m −1 = 3.00

(b) Express the magnification of the lens when the final image is at 25 cm:

m=−

Solve the thin-lens equation for s:

s=

Substitute to obtain:

m=−

Substitute numerical values and evaluate m:

= xnp P

s' s

fs' s' − f

s' s' − f s' = − +1 =− fs' f f s' − f = 1 − s'P

(

)

m = 1 − (− 0.25 m ) 12 m −1 = 4

*90 •• Picture the Problem We can use the definition of the angular magnification of a lens and the thin-lens equation to show that M =

xnp f

+1.

θ θ0

(a) Express the angular magnification of the simple magnifier in terms of the angles subtended by the object and the image:

M =

Solve the thin-lens equation for s:

fs' s' − f s' = − xnp

Because the image is virtual: Substitute to obtain:

(1)

s=

s=

f (− x np ) − x np − f

=

fx np x np + f

987

988

Chapter 32

Express the angle subtended by the object:

y xnp

θ0 =

where y is the height of the object. Express the angle subtended by the image:

θ=

Substitute for s to obtain:

θ=

y s y fxnp

=

y (xnp + f ) fxnp

xnp + f Substitute in equation (1) and simplify:

(b) In terms of the power of the magnifying lens: The magnification of a 20-D lens for a person with a near point of 30 cm and the final image at the near point is:

y (xnp + f ) fxnp y xnp

M =

=

xnp + f f

=

xnp f

+1

M = xnp P + 1

(

)

M = (0.3 m ) 20 m −1 + 1 = 7.00

A ray diagram for this situation is shown to the right:

91 •• Picture the Problem We can use the definitions of lateral and angular magnification and the result given in Problem 82 to show that, when the image of a simple magnifier is viewed at the near point, the lateral and angular magnifications are equal. Express the lateral magnification of the lens:

M =

xnp f

Optical Images Because the image is viewed at the near point, f = s and:

M =

xnp

From Problem 32-82:

M =

xnp

989

s

f

+1

and

xnp s

=

xnp f

+ 1 or M lateral = M angular

The Microscope 92 •• Picture the Problem We can use the thin-lens equation to find the location of the object and the expression for the magnifying power of a microscope to find the magnifying power of the given microscope for a person whose near point is at 25 cm. (a) Using the thin-lens equation, relate the object distance s to the focal length of the objective lens f0:

1 1 1 + = s s' f 0

Solve for s to obtain:

s=

From Figure 32-48, the image distance for the image formed by the objective lens is:

f 0 s' s' − f 0

s' = f 0 + L = 1.7 cm + 16 cm = 17.7 cm

(1.7 cm)(17.7 cm) =

Substitute numerical values and evaluate s:

s=

(b) Express the magnifying power of a microscope:

M =−

L xnp f0 fe

Substitute numerical values and evaluate M:

M =−

16 cm 25 cm = − 46.1 1.7 cm 5.1 cm

17.7 cm − 1.7 cm

1.88 cm

*93 •• Picture the Problem The lateral magnification of the objective is mo = − L f o and the magnifying power of the microscope is M = mo M e .

990

Chapter 32

(a) The lateral magnification of the objective is given by:

mo = −

L fo

Substitute numerical values and evaluate mo:

mo = −

16 cm = − 1.88 8.5 mm

(b) The magnifying power of the microscope is given by:

Substitute numerical values and evaluate M:

M = mo M e where Me is the angular magnification of the lens.

M = (− 1.88)(10 ) = − 18.8

94 •• Picture the Problem We can find the tube length from the length of the tube to which the lenses are fastened and the focal lengths of the objective and eyepiece. We can use their definitions to find the lateral magnification of the objective and the magnifying power of the microscope. The distance of the object from the objective can be found using the thin-lens equation. (a) The tube length L is given by:

L = D − fo − fe = 0.30 m −

2 = 20.0 cm 20 m −1

L 20 cm =− = − 4.00 fo 5 cm

(b) The lateral magnification of the objective mo is given by:

mo = −

(c) The magnifying power of the microscope is given by:

M = mo M e = mo

Substitute numerical values and evaluate M:

M = (− 4)

(d) From the thin-lens equation we have:

1 1 1 + = so so' f o where so' = f o + L

Substitute to obtain:

1 1 1 + = so f o + L f o

xnp fe

25 cm = − 20.0 5 cm

Optical Images Solve for so:

so =

Substitute numerical values and evaluate so:

so =

991

f o ( f o + L) L

(5 cm )(5 cm + 20 cm ) = 20 cm

6.25 cm

*95 •• Picture the Problem The magnifying power of a compound microscope is the product of the magnifying powers of the objective and the eyepiece. Express the magnifying power of the microscope in terms of the magnifying powers of the objective and eyepiece:

M = mo me

xnp

(1)

From Problem 82, the magnification of the eyepiece is given by:

me =

The magnification of the objective is given by:

L fo where L = D − f o − f e

Substitute to obtain:

mo = −

Substitute for me and mo in equation (1) to obtain:

⎛ D − fo − fe ⎞ ⎟⎟ M = (Pe xnp + 1)⎜⎜ − f o ⎝ ⎠

fe

+ 1 = Pe xnp + 1

mo = −

D − fo − fe fo

Substitute numerical values and evaluate M:

⎛ 28 cm − 2.22 cm − 1.25 cm ⎞ ⎟⎟ = − 232 M = [(80 D )(0.25 m ) + 1] ⎜⎜ − 2.22 cm ⎝ ⎠ 96 ••• Picture the Problem We can find the focal length of the eyepiece from its angular magnification and the near point of a normal eye. The location of the object such that it is in focus for a normal relaxed eye can be found from the lateral magnification of the eyepiece and the magnifying power of the microscope. Finally, we can use the thin-lens equation to find the focal length of the objective lens.

992

Chapter 32 xnp

(a) Relate the focal length of the eyepiece to its angular magnifying power:

Me =

Substitute numerical values and evaluate fe:

fe =

(b) Relate s to s′ through the lateral magnification of the objective:

mo = −

Relate the magnifying power of the microscope M to the lateral magnification of its objective m0 and the angular magnification of its eyepiece Me:

⇒ fe =

fe

xnp Me

25 cm = 1.67 cm 15 s' s' ⇒ s=− s mo

M = mo M e

Solve for mo:

mo =

M Me

Substitute to obtain:

s=−

s'M e M

Evaluate s′:

s' = 22 cm − f e = 22 cm − 1.67 cm = 20.33 cm

Substitute numerical values and evaluate s:

s=−

(20.33 cm )(15) =

(c) Solve the thin-lens equation for fo :

fo =

ss' s' + s

Substitute numerical values and evaluate fo:

fo =

− 600

0.508 cm

(0.508 cm )(20.33 cm ) 20.33 cm + 0.508 cm

= 0.496 cm

The Telescope 97 • Picture the Problem Because of the great distance to the moon, its image formed by the objective lens is at the focal point of the objective lens and we can use D = f oθ to find

Optical Images

993

the diameter D of the image of the moon. Because angle subtended by the final image at infinity is given by θ e = Mθ o = Mθ, we can solve (b) and (c) together by first using M = −fo/fe to find the magnifying power of the telescope. (a) Relate the diameter D of the image of the moon to the image distance and the angle subtended by the moon: Because the image of the moon is at the focal point of the objective lens:

D = so'θ

so' = f o and

D = f oθ

Substitute numerical values and evaluate D: (b) and (c) Relate the angle subtended by the final image at infinity to the magnification of the telescope and the angle subtended at the objective:

D = (100 cm )(0.009 rad ) = 9.00 mm

θ e = Mθ o = Mθ

Express the magnifying power of the telescope:

M =−

fo fe

Substitute numerical values and evaluate M and θe:

M =−

100 cm = − 20.0 5 cm

and

θ e = (− 20)(0.009 rad ) = − 0.180 rad

98 • Picture the Problem Because of the great distance to the moon, its image formed by the objective lens is at the focal point of the objective lens and we can use D = f oθ to find the diameter D of the image of the moon. Relate the diameter D of the image of the moon to the image distance and the angle subtended by the moon: Because the image of the moon is at

D = so'θ

so' = f o

994

Chapter 32

the focal point of the objective lens: Substitute numerical values and evaluate D:

and

D = f oθ D = (19.5 m )(0.009 rad ) = 17.6 cm

*99 •• Picture the Problem Because the light-gathering power of a mirror is proportional to its area, we can compare the light-gathering powers of these mirrors by finding the ratio of their areas. We can use the ratio of the focal lengths of the objective and eyepiece lenses to find the magnifying power of the Palomar telescope. (a) Express the ratio of the lightgathering powers of the Palomar and Yerkes mirrors:

Substitute numerical values and evaluate PPalomar/PYerkes:

π

d2 PPalomar APalomar mirror 4 Palomar mirror = = π 2 PYerkes AYerkes mirror d Yerkes mirror 4 d2 mirror = Palomar 2 d Yerkes mirror PPalomar (200 in ) = = 25.0 PYerkes (40 in )2 2

or

PPalomar =

(25.0)PYerkes

(b) Express the magnifying power of the Palomar telescope:

M =−

fo fe

Substitute numerical values and evaluate M:

M =−

1.68 m = − 134 1.25 cm

100 •• Picture the Problem We can use the expression for the magnifying power of a telescope and the fact that the length of a telescope is the sum of focal lengths of its objective and eyepiece lenses to obtain simultaneous equations in fo and fe. The magnifying power of the telescope is given by: The length of the telescope is the sum of the focal lengths of the objective and eyepiece lenses:

M =−

fo =7 fe

L = f o + f e = 32 cm

Optical Images Solve these equations simultaneously to obtain:

995

f o = 28.0 cm and f e = 4.00 cm

101 •• Picture the Problem The magnification of a telescope is the ratio of the angle subtended at the eyepiece lens to the angle subtended at the objective lens. We can use the geometry of the ray diagram to express both θe and θo. (b) The ray diagram is shown below:

(a) Express the magnifying power M of the telescope:

M =

θe θo

Because the image formed by the objective lens is at the focal point, F ′ 1:

θo =

h fo

where we have assumed that θo << 1 so that tan θo ≈ θo.

h where fe is negative. fe

Express the angle subtended by the eyepiece:

θe =

Substitute to obtain:

h f f M = e = o h fe fo and M = −

fo is positive. fe

Remarks: Because the object for the eyepiece is at its focal point, the image is at infinity. As is also evident from the ray diagram, the image is virtual and upright. 102 •• Picture the Problem We can use the thin-lens equation to find the image distance for the objective lens and the object distance for the eyepiece lens. The separation of the lenses is

996

Chapter 32

the sum of these distances. We can use the definition of the angular magnification and the angles subtended at the objective and eyepiece lenses to find the height of the final image. (a) Solve the thin-lens equation for so' :

so' =

Substitute numerical values and evaluate so' :

so' =

f o so so − f o

(1m )(30 m ) = 30 m − 1m

103.45 cm

where we have kept more than three significant figures in the answer for use in (c) and (d). (b) Solve the thin-lens equation for se:

se =

f e se' se' − f e

Noting that se′ = −25 cm, substitute numerical values and evaluate se:

se =

(− 5 cm )(− 25 cm ) − 25 cm − (− 5 cm )e

= − 6.25 cm

where the minus sign tells us that the object of the eyepiece is virtual. (c) Express the separation D of the lenses:

D = so' + se

Substitute numerical values and evaluate D:

D = 103.45 cm − 6.25 cm = 97.2 cm

(d) Express the height h′ of the final image in terms of the magnification M of the telescope:

h' = Mh

The magnification of the telescope is the product of the magnifications of the objective and eyepiece lenses:

M = mo me =

Substitute to obtain:

h' =

so' se' h so se

so' se' so se

Optical Images Substitute numerical values and evaluate h′:

997

⎛ 103.45 cm ⎞ ⎛ − 25 cm ⎞ ⎟⎟ ⎜⎜ ⎟⎟(1.5 m ) h' = ⎜⎜ ⎝ 3000 cm ⎠ ⎝ − 6.25 cm ⎠ = 20.7 cm

Express the angular magnification of the telescope:

M =

θe θo

The angle subtended by the object is:

θo =

h so

The angle subtended by the image is:

θ e = tan −1 ⎜⎜

Substitute to obtain:

⎛ h' ⎞ tan −1 ⎜⎜ ⎟⎟ ⎝ se ⎠ = so tan −1 ⎛⎜ h' ⎞⎟ M = ⎜s ⎟ h h ⎝ e⎠ so

Substitute numerical values and evaluate M:

M =

⎛ h' ⎞ ⎟⎟ ⎝ se ⎠

⎛ 20.7 cm ⎞ 30 m ⎟⎟ = 25.6 tan −1 ⎜⎜ 1.5 m ⎝ 6.25 cm ⎠

103 ••• Picture the Problem The roles of the objective and eyepiece lenses are reversed. Express the magnifying power of the ″wrong end″ telescope:

M =−

fe fo

Substitute numerical values and evaluate M:

M =−

1.5 cm = −6.67 × 10 −3 2.25 m

= − 1 / 150

General Problems 104 • Picture the Problem We can solve the thin-lens equation for s′ and then argue that the signs of the numerator and denominator are such that their quotient is always negative.

998

Chapter 32 fs s− f

Solve the thin-lens equation for s′:

s' =

For a diverging lens:

f < 0 and s > 0 for a real object.

Consequently, the denominator is positive and the numerator is negative, so s' must always be negative. *105 • Picture the Problem We can express the distance ∆s that the lens must move as the difference between the image distances when the object is at 30 m and when it is at infinity and then express these image distances using the thin-lens equation. Express the distance ∆s that the lens must move to change from focusing on an object at infinity to one at a distance of 30 m:

∆s = s'30 − s' ∞

Solve the thin-lens equation for s′:

fs s− f fs30 fs∞ ∆s = − s30 − f s∞ − f

Substitute and simplify to obtain:

s' =

=

fs30 f − s30 − f 1 − f s∞

⎡ s ⎤ = f ⎢ 30 − 1⎥ ⎣ s30 − f ⎦ Substitute numerical values and evaluate ∆s:

⎤ ⎡ 30 m ∆s = (200 mm ) ⎢ − 1⎥ ⎣ 30 m − 0.2 m ⎦ = 1.34 mm

106 • Picture the Problem We can express the distance ∆s that the lens must move as the difference between the image distances when the object is at 30 m and when it is at infinity and then express these image distances using the thin-lens equation. Express the distance ∆s that the lens must move to change from focusing on an object at infinity to one at a

∆s = s'5 − s' ∞

Optical Images

999

distance of 5 m:

fs s− f

Solve the thin-lens equation for s′:

s' =

Substitute and simplify to obtain:

∆s = =

fs5 fs∞ − s5 − f s∞ − f fs5 f − s5 − f 1 − f s∞

⎡ s ⎤ = f ⎢ 5 − 1⎥ ⎣ s5 − f ⎦ Substitute numerical values and evaluate ∆s:

⎤ ⎡ 5m ∆s = (28 mm ) ⎢ − 1⎥ ⎣ 5 m − 0.028 m ⎦ = 0.158 mm

107 • Picture the Problem We can use the thin-lens and magnification equations to obtain simultaneous equations that we can solve to find the image and object distances for the two situations described in the problem statement. (a) Use the thin-lens equation to relate the image and object distances to the focal length of the lens:

1 1 1 + = s s' f

Because the image is twice as large as the object:

m=−

Substitute to obtain:

1 1 1 + = s − 2s f

Solve for s: Substitute numerical values and evaluate s and s′:

s=

1 2

s=

1 2

and

s' ⇒ s' = −2 s s

f

(10 cm ) =

5.00 cm

s' = −2(5 cm ) = − 10.0 cm

1000 Chapter 32 (b) If the image is inverted, then:

s' = 2 s and

Solve for s:

s=

Substitute numerical values and evaluate s and s′:

3 2

1 1 1 + = s 2s f

f

s = 12 3(10 cm ) = 15.0 cm and

s' = 2(15 cm ) = 30.0 cm

The ray diagrams for (a) (left) and (b) (right) are shown below:

108 •• (a) In an astronomical telescope the eyepiece (short focal length) and objective (long focal length) lenses are separated by the sum of their focal lengths. Given these two lenses, we’ll use the 25 mm lens as the eyepiece lens and the 75 mm lens as the objective lens and mount them 100 mm apart. The angular magnification is then M =

f 0 75 mm = = 3 . f e 25 mm

(b) A ray diagram showing how rays from a distant object are magnified by an astronomical telescope follows. A real and inverted image of the distant object is formed by the objective lens near its second focal point. The eyepiece lens forms an enlarged and inverted image of the image formed by the objective lens.

Optical Images 1001 109 •• Determine the Concept (a) Because the focal lengths appear in the magnification formula as a product, it would appear that it does not matter in which order we use them. The usual arrangement would be to use the shorter focal length lens as the objective but we get the same magnification in the reverse order. What difference does it make then? None in this problem. However, it is generally true that the smaller the focal length of a lens, the smaller its diameter. This condition makes it harder to use the shorter focal length lens, with its smaller diameter, as the eyepiece lens. If we separate the objective and eyepiece lenses by L + fe + fo = 16 cm + 7.5 cm + 2.5 cm = 26.0 cm, the overall magnification will be

M = m0 M e = −

L xnp 16 cm 25 cm =− = − 21.3 . f0 fe 7.5 cm 2.5 cm

In a compound microscope, the lenses are separated by:

δ = L + fe + f0

Substitute numerical values and evaluate δ:

δ = 16 cm + 7.5 cm + 2.5 cm = 26.0 cm

The overall magnification of a compound microscope is given by:

M = m0 M e = −

Substitute numerical values and evaluate M:

M =−

L xnp f0 fe

16 cm 25 cm = − 21.3 7.5 cm 2.5 cm

(b) A ray diagram showing how rays from a near-by object are magnified by a compound microscope follows. A real and inverted image of the near-by object is formed by the objective lens at the first focal point of the eyepiece lens. The eyepiece lens forms an inverted and virtual image of this image at infinity.

1002 Chapter 32 *110 •• Picture the Problem We can use the equation for refraction at a single surface to locate the image of the fish and the expression for the magnification due to refraction at a spherical surface to find the magnification of the image. (a) Use the equation describing refraction at a single surface to relate the image and object distances:

n1 n2 n2 − n1 + = s s' r

Solve for s′:

s' =

n2 rs (n2 − n1 )s − n1r

Substitute numerical values and evaluate s′:

s' =

(1)(0.5 m )(2.5 m ) (1 − 1.33)(2.5 m ) − (1.33)(0.5 m )

= − 0.839 m Note that the fish appears to be much closer to the diver than it actually is. (b) Express the magnification due to refraction at a spherical surface:

m=−

n1s' n2 s

Substitute numerical values and evaluate m:

m=−

(1.33)(− 0.839 m ) = (1)(2.5 m )

0.446

Note that the fish appears to be smaller than it actually is. 111 •• Picture the Problem We can use the thin-lens equation and the definition of the magnification of an image to determine where the person should stand. Use the thin-lens equation to relate s and s′:

1 1 1 + = s s' f

The magnification of the image is given by:

m=−

s' 2.4 cm =− = −1.37 × 10−2 s 175 cm

and

s' = − ms

Optical Images 1003 Substitute to obtain:

1 1 1 − = s ms f

Solve for s:

⎛ 1⎞ s = ⎜1 − ⎟ f ⎝ m⎠

Substitute numerical values and evaluate s:

1 ⎛ ⎞ (50 mm) s = ⎜1 − −2 ⎟ ⎝ − 1.37 × 10 ⎠ = 3.70 m

112 •• Picture the Problem We can use the thin-lens equation and the definition of the magnification of an image to determine the ideal focal length of the lens. Use the thin-lens equation to relate s and s′:

1 1 1 + = s s' f

The magnification of the image is given by:

m=−

s' 3.6 cm =− = −1.80 × 10−2 s 200 cm

and

s' = − ms Substitute to obtain:

1 1 1 − = s ms f

Solve for f:

f =

s 1−

Substitute numerical values and evaluate f:

f =

1 m

30 m = 0.530 m 1 1− − 1.80 × 10−2

113 •• Picture the Problem Let the numeral 1 refer to the first lens and the numeral 2 to the second lens. We apply the thin-lens equation twice; once to locate the image formed by the first lens and a second time to find the image formed by the second lens. The magnification of the image is the product of the magnifications produced by the two lenses.

1004 Chapter 32 (a) Solve the thin-lens equation for the location of the image formed by the first lens:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

Because the second lens is 20 cm to the right of the first lens:

f1s1 s1 − f1

(10 cm)(12 cm) = 60.0 cm 12 cm − 10 cm

s2 = 20 cm − 60 cm = −40 cm

Solve the thin-lens equation for the location of the image formed by the second lens:

s2' =

Substitute numerical values and evaluate s2' :

s2' =

f 2 s2 s2 − f 2

(12.5 cm )(− 40 cm) = − 40 cm − 12.5 cm

9.52 cm

i.e., the final image is 9.52 cm to the right of the second lens. (b) Express the magnification of the final image:

⎛ s ' ⎞⎛ s ' ⎞ s 's ' m = m1m2 = ⎜⎜ − 1 ⎟⎟⎜⎜ − 2 ⎟⎟ = 1 2 ⎝ s1 ⎠⎝ s2 ⎠ s1s2

Substitute numerical values and evaluate m:

m=

(60 cm )(9.52 cm ) = (12 cm )(− 40 cm )

− 1.19

i.e., the final image is about 20% larger than the object and is inverted. (c) The ray diagram is shown in the figure. The enlarged, inverted image formed by the first lens serves as a virtual object for the second lens. The image formed from this virtual object is the real, inverted image shown in the ray diagram.

Optical Images 1005 114 •• Picture the Problem We can apply the equation for refraction at a surface to both surfaces of the lens and add the resulting equations to obtain an equation relating the image and object distances to the indices of refraction. We can then use the lens maker’s equation to complete the derivation of the given relationship between f ′ and f. (a) Relate s and s′ at the water-lens interface:

nw n n − nw + = s s1' r1

Relate s and s′ at the lens-water interface:

n n n −n + = w r2 − s1' s'

Add these equations to obtain:

Let

1 1 1 = + to obtain: f' s s'

The lens-maker’s equation is:

⎛1 1⎞ ⎛1 1 ⎞ nw ⎜ + ⎟ = (n − nw )⎜⎜ − ⎟⎟ ⎝ s s' ⎠ ⎝ r1 r2 ⎠ ⎛1 1⎞ nw = (n − nw ) ⎜⎜ − ⎟⎟ f' ⎝ r1 r2 ⎠ ⎛1 1⎞ 1 = (n − 1)⎜⎜ − ⎟⎟ f ⎝ r1 r2 ⎠ and

1 1 1 − = r1 r2 (n − 1) f Substitute to obtain:

Solve for f ′:

(b) Use the lens-maker’s equation to find the focal length of the lens in air:

⎛ 1 nw = (n − nw ) ⎜⎜ f' ⎝ (n − 1) f f'=

⎞ ⎟⎟ ⎠

nw (n − 1) f n − nw

⎛ 1 1 1 ⎞ ⎟⎟ = (1.5 − 1)⎜⎜ − f ⎝ − 30 cm 35 cm ⎠ and

f = − 32.3 cm Use the result derived in (a) to find f ′:

f '=

(1.33)(1.5 − 1) (− 32.3 cm ) 1.5 − 1.33

= − 126 cm

1006 Chapter 32 *115 •• Picture the Problem Here we must consider refraction at each surface separately. To find the focal length we imagine the object at s = ∞, and find the image from the first refracting surface at s′1. That image serves as the object for the second refracting surface. We’ll find that this is a virtual image for the second refracting surface, i.e., s2 is negative. Using the equation for refraction at a single surface a second time, we can locate the image formed by the second refracting surface by the virtual object at s2. The location of that image is then the focal point of the thick lens. We’ll let the numeral 1 denote the first surface and the numeral 2 the second surface. In part (b) we can proceed as in part (a) (except that now n1 = 1.33 for the first refraction and n2 = 1.33 for the second refraction) to determine the focal length in water, which we denote by fw. (a) Use the equation for refraction at a single surface to relate s1 and s1′:

n1 n2 n2 − n1 + = s1 s1' r1

For s1 = ∞:

n2 n2 − n1 = s1' r1

Solve for s1′:

s1' =

Substitute numerical values and evaluate s1' :

s1' =

The object distance s2 for the second lens is:

n2 r1 n2 − n1

(1)

(1.5)(20 cm ) = 60.0 cm 1.5 − 1

s2 = −(s1' − 4 cm ) = −(60 cm − 4 cm ) = −56 cm

Solve the equation for refraction at a single surface for s2' :

s2' =

n2 r2 s2 (n2 − n1 )s2 − n1r2

Substitute numerical values and evaluate s2' :

s2' =

(1)(− 20 cm )(− 56 cm ) (1 − 1.5)(− 56 cm ) − (1.5)(− 20 cm )

= 19.3 cm Because f is measured from the center of the lens:

(b) Substitute numerical values in equation (1) and evaluate s1' :

f = s2' + 2 cm = 19.3 cm + 2 cm = 21.3 cm s1' =

(1.5)(20 cm) = 176 cm 1.5 − 1.33

(2)

Optical Images 1007

The object distance s2 for the second lens is:

s2 = −(s1' − 4 cm ) = −(176 cm − 4 cm ) = −172 cm

Substitute numerical values in equation (2) and evaluate s′2:

s2' =

(1.33)(− 20 cm )(− 172 cm) = 77.2 cm (1.33 − 1.5)(− 172 cm ) − (1.5)(− 20 cm)

Because fw is measured from the center of the lens:

f w = s2' + 2 cm = 77.2 cm + 2 cm = 79.2 cm

Remarks: Note that if we use the expression given in Problem 114 we obtain fw = 83.3 cm, in only moderate agreement with the exact result given above. 116 •• Picture the Problem Let the numeral 1 denote the CCl4-H2O interface and the numeral 2 the H2O-air interface. We can locate the final image by applying the equation for refraction at a single surface to both interfaces. The ray diagram shown below shows a spot at the bottom of the tank and the rays of light emanating from it that form the intermediate and final images. Use the equation for refraction at a single surface to relate s and s′ at the CCl4-H2O interface:

nCCl4 s1

s1

Substitute numerical values and evaluate s1′:

nH2O s1'

=

nH2O − nCCl4 r

or, because r = ∞,

nCCl4

Solve for s1′:

+

+

s1' = −

s1' = −

nH2O s1'

=0

nH2O s1 nCCl4

(1)

(1.33)(4 cm ) = −3.64 cm 1.46

1008 Chapter 32 The depth of this image, as viewed from the H2O-air interface is:

s2 = 2 cm − s1' = 2 cm − (− 3.64 cm ) = 5.64 cm

At the H2O-air interface equation (1) becomes:

s2' = −

Substitute numerical values and evaluate s2′:

s2' = −

nair s2 n H 2O

(1)(5.64 cm ) = −4.24 cm

1.33 The apparent depth is 4.24 cm.

117 •• Picture the Problem The speed of the jogger as seen in the mirror is v' = ds' dt. We can use the mirror equation to derive an expression for v′ in terms of f and ds/dt. Solve the mirror equation for s′:

Differentiate s′ with respect to time to obtain:

⎛ 1 1⎞ s' = ⎜⎜ − ⎟⎟ ⎝ f s⎠

−1

(1)

ds' d ⎛ 1 1 ⎞ = ⎜ − ⎟ v' = dt dt ⎜⎝ f s ⎟⎠

−1

−2

⎛ 1 1 ⎞ ⎛ 1 ⎞ ds = −⎜⎜ − ⎟⎟ ⎜ 2 ⎟ ⎝ f s ⎠ ⎝ s ⎠ dt Simplify this result to obtain:

2

⎛ s' ⎞ v' = −⎜ ⎟ v ⎝s⎠

(2)

−1

Rewrite equation (1) in terms of r:

⎛ 2 1⎞ s' = ⎜ − ⎟ ⎝r s⎠

Find s′ when s = 5 m:

⎛ 2 1 ⎞ ⎟⎟ = −0.833 m − s' = ⎜⎜ ⎝ − 2m 5m ⎠

Use equation (2) to find v' when

⎛ − 0.833 m ⎞ ⎟⎟ (3.5 m/s ) v' = − ⎜⎜ ⎝ 5m ⎠

v = 3.5 m/s:

−1

2

= 0.0971 m/s 118 •• Picture the Problem Let the numerals 1 and 2 denote to the first and second refracting surfaces of the spherical lens, respectively, and follow the steps given in the hint.

Optical Images 1009 Use the equation for refraction at a single surface to relate s1 and s1′:

n1 n2 n2 − n1 + = s1 s1' r1

When s1 = ∞:

n2 n2 − n1 = s1' r1

Solve for s1′:

s1' =

Substitute numerical values and evaluate s1′:

s1' =

Because the thickness of the glass sphere is 4 mm: Use the equation for refraction at a single surface to relate s2 and s2′:

n2 r1 n2 − n1

(1.5)(2 mm) = 6.00 mm 1.5 − 1

s2 = 4 mm − s1' = 4 mm − 6 mm = −2 mm

n2 n1 n1 − n2 + = s2 s2' r2

Solve for s2′:

s2' =

n1r2 s2 (n1 − n2 ) s2 − n2 r2

Substitute numerical values and evaluate s2′:

s2' =

(1)(− 2 mm)(− 2 mm) (1 − 1.5)(− 2 mm) − [1.5](− 2 mm)

= 1.00 mm Because s2' = 1.00 mm = r/2, f = 1.00 mm. 119 ••• Picture the Problem We can use the thin-lens equation to locate the first image formed by the lens, the mirror equation to locate the image formed in the mirror, and the thin-lens equation a second time to locate the final image formed by the lens as the rays pass back through it. (b) and (c) The ray diagram is shown below. The numeral 1 represents the object. The parallel and central rays from 1 are shown; one passes through the center of the lens, the other is paraxial and then passes through the focal point F′. The two rays intersect behind the mirror, and the image formed there, identified by the numeral 2, serves as a virtual object for the mirror. Two rays are shown emanating from this virtual image, one through the center of the mirror, the other passing through its focal point (halfway between C and the mirror surface) and then continuing as a paraxial ray. These two rays intersect in front

1010 Chapter 32 of the mirror, forming a real image, identified by the numeral 3. Finally, the image 3 serves as a real object for the lens; again we show two rays, a paraxial ray that then passes through the focal point F and a ray through the center of the lens. These two rays intersect to form the final real, upright, and diminished image, identified as 4. To see this image the eye must be to the left of the image 4.

(a) Solve the thin-lens equation for s1′:

s1' =

Substitute numerical values and evaluate s1′:

s1' =

Because the image formed by the lens is behind the mirror:

fs1 s1 − f

(10 cm)(15 cm ) = 30 cm 15 cm − 10 cm

s2 = 25 cm − 30 cm = −5 cm

Solve the mirror equation for s2′:

s2' =

Substitute numerical values and evaluate s2′:

s2' =

fs2 s2 − f

(5 cm )(− 5 cm ) = 2.50 cm and the − 5 cm − 5 cm

image is 22.5 cm from the lens; i.e., s3 = 22.5 cm. Solve the thin-lens equation for s3′:

s3' =

Substitute numerical values and evaluate s3′:

s3' =

fs3 s3 − f

(10 cm )(22.5 cm ) = 22.5 cm − 10 cm

18.0 cm

*120 ••• Picture the Problem The mirror surfaces must be concave to create inverted images on reflection. Therefore, the lens is a diverging lens. Let the numeral 1 denote the lens in its initial orientation and the numeral 2 the lens in its second orientation. We can use the mirror equation to find the magnitudes of the radii of the lens’ surfaces, the thin-lens equation to find its focal length, and the lens maker’s equation to find its index of refraction.

Optical Images 1011 Solve the mirror equation for r1 :

r1 =

2s1s1' s1' + s1

Substitute numerical values and evaluate r1 :

r1 =

2(30 cm )(6 cm ) = 10.0 cm 6 cm + 30 cm

Solve the mirror equation for r2 :

r2 =

2s2 s2' s2' + s2

Substitute numerical values and evaluate r2 :

r2 =

2(30 cm )(10 cm ) = 15.0 cm 10 cm + 30 cm

Solve the thin-lens equation for f:

f =

ss' s' + s

Substitute numerical values and evaluate f:

f =

(30 cm )(− 7.5 cm) = −10.0 cm

Solve the lens-maker’s equation for n to obtain:

n=

Because the lens is a diverging lens, r1 = −10 cm and r2 = 15 cm. Substitute numerical values and evaluate n:

n=

− 7.5 cm + 30 cm 1

⎛1 1⎞ f ⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠

+1

1 +1 ⎛ 1 1 ⎞ ⎟⎟ (− 10 cm)⎜⎜ − ⎝ − 10 cm 15 cm ⎠

= 1.60

121 ••• Picture the Problem Assume that the object is very small compared to r so that all incident and reflected rays traverse 1 cm of water. The problem involves two refractions at the air−water interface and one reflection at the mirror. Let the numeral 1 refer to the first refraction at the air−water interface, the numeral 2 to the reflection in the mirror surface, and the numeral 3 to the second refraction at the water−air interface. Use the equation for refraction at a single surface to relate s1 and s1′:

n1 n2 n2 − n1 + = s1 s1' r or, because r = ∞,

n1 n2 + =0 s1 s1'

1012 Chapter 32 Solve for s1′:

Let n2 = n. Because n1 = 1: Find the object distance for the mirror:

s1' = −

n2 s1 n1

s1' = −ns1 s2 = 1 − s1' = 1 + ns1 where 1 has units of cm. −1

Solve the mirror equation for s2′:

⎛2 1 ⎞ s2' = ⎜⎜ − ⎟⎟ ⎝ r s2 ⎠

Substitute for s2:

⎛2 1 ⎞ ⎟⎟ s2' = ⎜⎜ − r + ns 1 1⎠ ⎝

Find the object distance s3 for the water−air interface:

⎛2 1 ⎞ ⎟⎟ s3 = 1 − s2' = 1 − ⎜⎜ − ⎝ r 1 + ns1 ⎠

Use the equation for refraction at a single surface to relate s3 and s3′:

n1 n2 n2 − n1 + = s3 s3' r

−1

−1

or, because r = ∞,

n1 n2 + =0 s3 s3' Solve for s3′:

s3' = −

Because n2 = 1 and n1 = n:

⎛2 1 ⎞ ⎟⎟ 1 − ⎜⎜ − + 1 r ns s3 1⎠ s3' = − = − ⎝ n n

Equate s3′ and s1:

⎛2 1 ⎞ ⎟ 1 − ⎜⎜ − r 1 + ns1 ⎟⎠ ⎝ s1 = − n

Simplify to obtain:

s12 +

n2 s3 n1

−1

2−r 1− r s1 + 2 = 0 n n

−1

Optical Images 1013 Substitute numerical values and simplify:

2 cm − 50 cm 1cm − 50 cm s1 + =0 1.33 (1.33)2

s12 + or

s12 − 36.09s1 − 27.70 = 0 where s1 is in cm. Solve for the positive value of s1:

s1 = 36.8 cm

122 ••• Picture the Problem We can use the lens maker’s equation, in conjunction with the result given in Problem 114, to find the index of refraction of the liquid. Solve the lens-maker’s equation for n:

n=

Substitute numerical values and evaluate n:

n=

1 ⎛1 1⎞ f ⎜⎜ − ⎟⎟ ⎝ r1 r2 ⎠

+1

1 +1 ⎛ 1 1 ⎞ ⎟⎟ (27.5 cm )⎜⎜ − ⎝ − 17 cm − 8 cm ⎠

= 1.55 From Problem 114, the focal length of the lens in the liquid, fL, is related to the focal length of the lens in air, f, according to:

fL =

nL (n − 1) f n − nL

Solve for nL:

nL =

nf L (n − 1) f + f L

Substitute numerical values and evaluate nL:

nL =

(1.55)(109 cm ) (1.55 − 1)(27.5 cm ) + 109 cm

= 1.36 123 ••• Picture the Problem The problem involves two refractions and one reflection. We can use the refraction at spherical surface equation and the mirror equation to find the images formed in the two refractions and one reflection. Let the numeral 1 refer to the first refraction at the air-glass interface, the numeral 2 to the reflection from the silvered surface, and the numeral 3 refer to the refraction at the glass-air interface.

1014 Chapter 32 (a) The image and object distances for the first refraction are related according to:

n1 n2 n2 − n1 + = s1 s1' r

Solve for s1′ to obtain:

s1' =

n2 rs1 (n2 − n1 )s1 − n1r

Substitute numerical values and evaluate s1′:

s1' =

(1.5)(10 cm)(30 cm ) (1.5 − 1)(30 cm ) − (1)(10 cm )

= 90.0 cm The object for the mirror surface is behind the mirror and its distance from the surface of the mirror is: Use the mirror equation to relate s2 and s2′: Solve for s2′:

Substitute numerical values and evaluate s2′: The object for the second refraction at the glass-air interface is in front of the mirrored surface and its distance from the glass-air interface is: The image and object distances for the second refraction are related according to:

s2 = 20 cm − 90 cm = −70 cm

1 1 2 + = s2 s2' r s 2' =

rs 2 2s 2 − r

s2' =

(10 cm )(− 70 cm ) = 4.67 cm 2(− 70 cm ) − 10 cm

s3 = 20 cm − 4.67 cm = 15.3 cm

n1 n2 n2 − n1 + = s3 s3' r

Solve for s3′ to obtain:

s3' =

n2 rs3 (n2 − n1 )s3 − n1r

Noting that r = −10 cm, substitute numerical values and evaluate s3′:

s3' =

(1)(− 10 cm)(15.3 cm ) (1 − 1.5)(15.3 cm) − (1.5)(− 10 cm)

= −20.8 cm

Optical Images 1015 The final image is − 20.8 cm + 20 cm = 0.8 cm behind the mirror surface.

(b)

Proceed as in (a ) with s1 = 20 cm to obtain s3 = − 20 cm and the final image to be at the mirror surface.

124 ••• Picture the Problem We can solve the lens maker’s equation for f and then differentiate with respect to n and simplify to obtain df/f = −dn/(n − 1).

⎛1 1⎞ 1 1 = (n − 1) ⎜⎜ − ⎟⎟ = (n − 1) f ⎝ r1 r2 ⎠ C 1 1 1 = − where C r1 r2

(a) The lens maker’s equation is:

f = C (n − 1)

−1

Solve for f: Differentiate f with respect to n and simplify:

df d f (n − 1) f −1 −2 C (n − 1) = −C (n − 1) = − =− = −1 dn dn n −1 (n − 1)

[

]

Solve for df/f :

(b) Express the focal length for blue light in terms of the focal length for red light:

−2

df dn = − f n −1

f blue = f red + ∆f

Approximate df/f by ∆f/f and dn by ∆n to obtain:

∆f ∆n ≈− f n −1

Solve for ∆f :

∆f = −

Substitute for ∆f in equation (1) to obtain:

(1)

f ∆n n −1

f blue = f red −

⎛ f red ∆n ∆n ⎞ ⎟⎟ = f red ⎜⎜1 − n −1 ⎝ nred − 1 ⎠

1016 Chapter 32 Substitute numerical values and evaluate fblue:

⎛ 1.53 − 1.47 ⎞ f blue = (20 cm )⎜1 − ⎟ 1.47 − 1 ⎠ ⎝ = 17.4 cm

*125 ••• Picture the Problem We examine the amount by which the image distance s′ changes due to a change in s. Solve the thin-lens equation for s′:

⎛ 1 1⎞ s' = ⎜⎜ − ⎟⎟ ⎝ f s⎠

−1

Differentiate s′ with respect to s: −1 1 1 ds' d ⎡⎛ 1 1 ⎞ ⎤ s' 2 = ⎢⎜⎜ − ⎟⎟ ⎥ = − = − = −m 2 2 2 2 ds ds ⎢⎝ f s ⎠ ⎥ s s ⎛ 1 1⎞ ⎣ ⎦ ⎜⎜ − ⎟⎟ ⎝ f s⎠

The image of an object of length ∆s will have a length − m 2 ∆s.

Chapter 33 Interference and Diffraction Conceptual Problems *1 • Determine the Concept The energy is distributed nonuniformly in space; in some regions the energy is below average (destructive interference), in others it is higher than average (constructive interference). 2 • Determine the Concept Coherent sources have a constant phase difference. The pairs of light sources that satisfy this criterion are (b), (c), and (e). 3 • Determine the Concept The thickness of the air space between the flat glass and the lens is approximately proportional to the square of d, the diameter of the ring. Consequently, the separation between adjacent rings is proportional to 1/d. 4 •• Determine the Concept The distance between adjacent fringes is so small that the fringes are not resolved by the eye. 5 •• Determine the Concept If the film is thick, the various colors (i.e., different wavelengths) will give constructive and destructive interference at that thickness. Consequently, what one observes is the reflected intensity of white light. *6 • (a) The phase change on reflection from the front surface of the film is 180°; the phase change on reflection from the back surface of the film is 0°. As the film thins toward the top, the phase change associated with the film’s thickness becomes negligible and the two reflected waves interfere destructively. (b) The first constructive interference will arise when t = λ/4. Therefore, the first band will be violet (shortest visible wavelength). (c) When viewed in transmitted light, the top of the film is white, since no light is reflected. The colors of the bands are those complementary to the colors seen in reflected light; i.e., the top band will be red.

1017

1018

Chapter 33

7 • Determine the Concept The first zeroes in the intensity occur at angles given by sin θ = λ a. Hence, decreasing a increases θ and the diffraction pattern becomes wider. 8 • Determine the Concept Equation 33-2 expresses the condition for an intensity maximum i slit interference. Here d is the slit separation, λ the wavelength of the light, m an integer, a the angle at which the interference maximum appears. Equation 33-11 expresses the condition for the first minimum in single-slit diffraction. Here a is the width of the slit, λ the wavelength of the light, and θ the angle at which the first minimum appears, assuming m = 1. 9 • Picture the Problem We can solve d sin θ = mλ for θ with m = 1 to express the location of the first-order maximum as a function of the wavelength of the light. The interference maxima in a diffraction pattern are at angles θ given by:

d sin θ = mλ where d is the separation of the slits and m = 0, 1, 2, …

Solve for the angular location θ1 of the first-order maximum :

⎛λ⎞ θ1 = sin −1 ⎜ ⎟

Because λgreen light < λred light:

θ green light < θ red light and (a ) is correct.

⎝d ⎠

*10 • Determine the Concept The distance on the screen to mth bright fringe is given by

ym = m

λL d

, where L is the distance from the slits to the screen and d is the separation of

the slits. Because the index of refraction of air is slightly larger than the index of refraction of a vacuum, the introduction of air reduces λ to λ/n and decreases ym. Because the separation of the fringes is ym − ym−1, the separation of the fringes decreases and (b) is correct. 11 • (a) False. When destructive interference of light waves occurs, the energy is no longer distributed evenly. For example, light from a two-slit device forms a pattern with very bright and very dark parts. There is practically no energy at the dark fringes and a great deal of energy at the bright fringe. The total energy over the entire pattern equals the

Interference and Diffraction 1019 energy from one slit plus the energy from the second slit. Interference re-distributes the energy. (b) True (c) True (d) True (e) True

Estimation and Approximation *12 • Picture the Problem We’ll assume that the diameter of the pupil of the eye is 5 mm and that the wavelength of light is 600 nm. Then we can use the expression for the minimum angular separation of two objects than can be resolved by the eye and the relationship between this angle and the width of an object and the distance from which it is viewed to support the claim. Relate the width w of an object that can be seen at a height h to the critical angular separation αc:

tan α c =

Solve for w:

w = h tan α c

The minimum angular separation αc of two point objects that can just be resolved by an eye depends on the diameter D of the eye and the wavelength λ of light: Substitute for αc in the expression for w to obtain:

w h

α c = 1.22

λ D

λ⎞ ⎛ w = h tan⎜1.22 ⎟ D⎠ ⎝

In low-earth orbit:

⎛ 600 nm ⎞ ⎟ = 58.6 m w = (400 km ) tan⎜⎜1.22 5 mm ⎟⎠ ⎝ Because the width of the Great Wall is about 5 m, a naked eye would not be able to see it from the moon. At a distance equal to that of the distance of the moon from earth:

Chapter 33

1020

⎛ 600 nm ⎞ ⎟ = 56.2 km w = 3.84 × 108 m tan⎜⎜1.22 5 mm ⎟⎠ ⎝

(

)

Because the width of the Great Wall is about 5 m, a naked eye would not be able to see it from the moon. 13

•

Picture the Problem We can use sin θ = 1.22

λ D

to relate the diameter D of the opaque-

disk water droplets to the angular diameter θ of a coronal ring and to the wavelength of light. We’ll assume a wavelength of 500 nm. The angle θ subtended by the first diffraction minimum is related to the wavelength λ of light and the diameter D of the opaque-disk water droplet:

sin θ = 1.22

Because of the great distance to the cloud of water droplets, θ << 1 and:

θ ≈ 1.22

Solve for D to obtain:

D=

Substitute numerical values and evaluate D:

14

D=

•

Picture the Problem We can use sin θ = 1.22

λn D

λ D

λ D

1.22λ

θ 1.22(500 nm ) = 3.50 µm π rad 10° × 180°

to relate the diameter D of a

microsphere to the angular diameter θ of a coronal ring and to the wavelength of light in water. The angle θ subtended by the first diffraction minimum is related to the wavelength λn of light in water and the diameter D of the microspheres: Because θ << 1:

sin θ = 1.22

θ ≈ 1.22

λn

λ nD

D

= 1.22

λ nD

Interference and Diffraction 1021 Substitute numerical values and evaluate θ :

θ≈

1.22(632.8 nm ) = 0.116 rad 1.33(5 µm )

= 6.65° 15

•

Picture the Problem We can use sin θ = 1.22

λ D

to relate the diameter D of a pollen

grain to the angular diameter θ of a coronal ring and to the wavelength of light. We’ll assume a wavelength of 450 nm for blue light and 650 nm for red light. The angle θ subtended by the first diffraction minimum is related to the wavelength λ of light and to the diameter D of the microspheres: Because θ << 1:

Substitute numerical values and evaluate θ for red light:

sin θ = 1.22

θ ≈ 1.22 θ red ≈

λ D

λ nD

1.22(650 nm ) = 3.17 × 10 −2 rad 25 µm

= 1.82° Substitute numerical values and evaluate θ for blue light:

θ blue ≈

1.22(450 nm ) = 2.20 × 10 −2 rad 25 µm

= 1.26° *16 •• Picture the Problem The diagram shows the hair whose diameter d = a, the screen a distance L from the hair, and the separation ∆y of the first diffraction peak from the center. We can use the geometry of the experiment to relate ∆y to L and a and the condition for diffraction maxima to express θ in terms of the diameter of the hair and the wavelength of the light illuminating the hair.

Relate θ to ∆y:

tan θ =

∆y L

1022

Chapter 33

Solve for ∆y:

∆y = L tan θ

Diffraction maxima occur where:

a sin θ = (m + 12 )λ where m = 1, 2, 3, …

Solve for θ to obtain:

Substitute for θ in the expression for ∆y to obtain:

⎡ (m + 12 )λ ⎤ ⎥⎦ ⎣ a

θ = sin −1 ⎢

⎧ ⎡ (m + 12 )λ ⎤ ⎫ ∆y = L tan ⎨sin −1 ⎢ ⎥⎦ ⎬ ⎣ a ⎩ ⎭

For the first peak, m = 1. Substitute numerical values and evaluate ∆y:

⎧ ⎡ (1 + 12 )(632.8 nm ) ⎤ ⎫ ∆y = (10 m ) tan ⎨sin −1 ⎢ ⎥ ⎬ = 13.6 cm 70 µm ⎣ ⎦⎭ ⎩

Phase Difference and Coherence 17 • Picture the Problem A path difference ∆r contributes a difference δ given by δ =

∆r

λ

360° . ∆r

(a) Relate a path difference ∆r to a phase shift δ :

δ=

Solve for∆r:

∆r =

λ

Substitute numerical values and evaluate ∆r:

∆r =

(b) Substitute numerical values in equation (1) and evaluate δ:

δ=

360°

(1)

δλ 360°

(180°)(600 nm ) = 360°

300 nm

300 nm 360° = 135° 800 nm

18 • Picture the Problem The wavelength of light in a medium whose index of refraction is n is the ratio of the wavelength of the light in air divided by n. The number of wavelengths of light contained in a given distance is the ratio of the distance to the wavelength of light in the given medium. The difference in phase between the two waves is the sum of a π phase shift in the reflected wave and a phase shift due to the additional distance traveled by the wave reflected from the bottom of the water−air interface.

Interference and Diffraction 1023

λ

500 nm = 376 nm 1.33

(a) Express the wavelength of light in water in terms of the wavelength of light in air:

λn =

(b) Relate the number of wavelengths N to the thickness t of the film and the wavelength of light in water:

2 × 10 −4 cm = = 5.32 N= 376 nm λn

(c) Express the phase difference as the sum of the phase shift due to reflection and the phase shift due to the additional distance traveled by the wave reflected from the bottom of the water−air interface: Substitute for N and evaluate δ:

=

n

2t

δ = δ reflection + δ additional distance traveled =π +

2t

λn

2π = π + 2π N

δ = π rad + 2π (5.32 rad ) = 11.6π rad or, subtracting 11.6π rad from 12π rad,

δ = 0.4π rad *19 •• Picture the Problem The difference in phase depends on the path difference according to δ =

∆r

λ

360°. The path difference is the difference in the distances of (0, 15 cm) and

(3 cm, 14 cm) from the origin.

∆r

Relate a path difference ∆r to a phase shift δ:

δ=

The path difference ∆r is:

∆r = 15 cm −

λ

360°

(3 cm )2 + (14 cm )2

= 0.682 cm Substitute numerical values and evaluate δ:

δ=

0.682 cm 360° = 164° 1.5 cm

1024

Chapter 33

Interference in Thin Films 20 • Picture the Problem Because the mth fringe occurs when the path difference 2t equals m wavelengths, we can express the additional distance traveled by the light in air as an mλ. The thickness of the wedge, in turn, is related to the angle of the wedge and the distance from its vertex to the mth fringe.

(a)

The first band is dark because the phase difference due to reflection by the back surface of the top plate and the top surface of the bottom plate is 180°

(b) The mth fringe occurs when the path difference 2t equals m wavelengths: Relate the thickness of the air wedge to the angle of the wedge:

2t = mλ

θ=

t ⇒ t = xθ x

where we’ve used a small-angle approximation to replace an arc length by the length of a chord. Substitute to obtain:

2 xθ = mλ

mλ 1 m λ = 2x 2 x

Solve for θ :

θ=

Substitute numerical values and evaluate θ :

θ= ⎜

1⎛ 5 ⎞ −4 ⎟ (700 nm ) = 1.75 × 10 rad 2 ⎝ cm ⎠

*21 •• Picture the Problem The condition that one sees m fringes requires that the path difference between light reflected from the bottom surface of the top slide and the top surface of the bottom slide is an integer multiple of a wavelength of the light. The mth fringe occurs when the path difference 2d equals m wavelengths:

Because the nineteenth (but not the twentieth) bright fringe can be seen,

2 d = mλ ⇒ d =

mλ 2

(m − 12 ) λ < d < (m + 12 ) λ 2

2

Interference and Diffraction 1025 the limits on d must be:

where m = 19

Substitute numerical values to obtain:

(19 − 12 ) 590 nm < d < (19 + 12 ) 590 nm 2

2

or

5.46 µm < d < 5.75 µm 22 •• Picture the Problem The light reflected from the top surface of the bottom plate (wave 2 in the diagram) is phase shifted relative to the light reflected from the bottom surface of the top plate (wave 1 in the diagram). This phase difference is the sum of a phase shift of π (equivalent to a λ/2 path difference) resulting from reflection plus a phase shift due to the additional distance traveled.

Relate the extra distance traveled by wave 2 to the distance equivalent to the phase change due to reflection and to the condition for constructive interference:

2t + 12 λ = λ , 2λ , 3λ , ... or

2t = 12 λ , 32 λ , 52 λ , ...

and

2t = (m + 12 )λ where m = 0, 1, 2, … and λ

is the wavelength of light in air. Solve for m:

m=

2t

λ

−

1 2(2r ) 1 4r 1 = − = − 2 λ 2 λ 2

where r is the radius of the wire. Substitute numerical values and evaluate m:

m=

4(0.025 mm ) 1 − = 166 600 nm 2

23 •• Picture the Problem We can use the condition for destructive interference in a thin film to find its thickness. Once we’ve found the thickness of the film, we can use the condition for constructive interference to find the wavelengths in the visible portion of the spectrum that will be brightest in the reflected interference pattern and the condition for destructive interference to find the wavelengths of light missing from the reflected light when the film is placed on glass with an index of refraction greater than that of the

1026

Chapter 33

film. (a) Express the condition for destructive interference in the thin film:

2t + 12 λ' = 12 λ' , 32 λ' , 52 λ' , ... or

2t = λ' ,2λ' ,3λ' , ...

or

2t = mλ' = m

λ n

(1)

where m = 1, 2, 3, … and λ′ is the wavelength of the light in the film. Solve for λ:

λ=

Substitute for the missing wavelengths to obtain:

450 nm =

Divide the first of these equations by the second and simplify to obtain:

2nt 450 nm m +1 = m = 2 nt 360 nm m m +1

2nt m 2nt 2nt and 360 nm = m m +1

Solve for m:

m = 4 for λ = 450 nm

Solve equation (1) for t:

t=

mλ 2n

Substitute numerical values and evaluate t:

t=

4(450 nm ) = 600 nm 2(1.5)

(b) Express the condition for constructive interference in the thin film:

2t + 12 λ' = λ' , 2λ' , 3λ' ,... or

2t = 12 λ' , 32 λ' , 52 λ' ,... = (m + 12 )λ'

(1)

where λ′ is the wavelength of light in the oil and m = 0, 1, 2, … Substitute for λ′ to obtain:

2t = (m + 12 )

λ n

where n is the index of refraction of the film.

Interference and Diffraction 1027 Solve for λ:

λ=

2nt m + 12

Substitute numerical values and simplify to obtain:

λ=

2(1.5)(600 nm ) 1800 nm = m + 12 m + 12

Substitute for m and evaluate λ to obtain the following table: m λ (nm)

0 3600

1 1200

2 720

3 514

4 400

5 327

From the table, we see that the only wavelengths in the visible spectrum are 720 nm, 514 nm, and 400 nm.

2t = 12 λ' , 32 λ' , 52 λ' , ...

(c) Because the index of refraction of the glass is greater than that of the film, the light reflected from the film-glass interface will be shifted by 12 λ (as is the wave reflected

or

2t = (m + 12 )

λ n

where n is the index of refraction of the film and m = 0, 1, 2, …

from the top surface) and the condition for destructive interference becomes: Solve for λ:

λ=

2nt m + 12

Substitute numerical values and simplify to obtain:

λ=

2(1.5)(600 nm ) 1800 nm = m + 12 m + 12

Substitute for m and evaluate λ to obtain the following table: m λ (nm)

0 3600

1 1200

2 720

3 514

4 400

5 327

From the table we see that the missing wavelengths in the visible spectrum are 720 nm, 514 nm, and 400 nm. 24 •• Picture the Problem Because there is a

1 2

λ phase change due to reflection at both the

air-oil and oil-water interfaces, the condition for constructive interference is that twice

1028

Chapter 33

the thickness of the oil film equal an integer multiple of the wavelength of light in the film.

2t = λ' , 2λ' , 3λ' ,...

Express the condition for constructive interference:

or

2t = mλ' (1) where λ′ is the wavelength of light in the oil a = 1, 2, 3, … Substitute for λ′ to obtain:

2t = m

λ n

Solve for t:

t=

mλ 2n

Substitute numerical values and evaluate t:

t=

(2)(650 nm ) = 2(1.22)

25 •• Picture the Problem Because there is a

1 2

533 nm

λ phase change due to reflection at both the

air-oil and oil-glass interfaces, the condition for constructive interference is that twice the thickness of the oil film equal an integer multiple of the wavelength of light in the film. Express the condition for constructive interference:

Substitute for λ′ to obtain:

2t = λ' , 2λ' , 3λ' ,... = mλ'

(1)

where λ′ is the wavelength of light in the oil and m = 0, 1, 2, …

2t = m

λ n

where n is the index of refraction of the oil. Solve for λ:

λ=

Substitute for the predominant wavelengths to obtain:

690 nm =

Divide the first of these equations by the second and simplify to obtain:

2nt 690 nm m +1 = m = 2nt 460 nm m m +1

2nt m 2nt 2nt and 460 nm = m m +1

Interference and Diffraction 1029 Solve for m:

m = 2 for λ = 690 nm

Solve equation (1) for t:

t=

mλ 2n

Substitute numerical values and evaluate t:

t=

(2)(690 nm ) = 2(1.45)

476 nm

*26 •• Picture the Problem Because the index of refraction of air is less than that of the oil, there is a phase shift of π rad ( 12 λ ) in the light reflected at the air-oil interface. Because the index of refraction of the oil is greater than that of the glass, there is no phase shift in the light reflected from the oil-glass interface. We can use the condition for constructive interference to determine m for λ = 700 nm and then use this value in our equation describing constructive interference to find the thickness t of the oil film. Express the condition for constructive interference between the waves reflected from the air-oil interface and the oil-glass interface:

2t + 12 λ' = λ' , 2λ' , 3λ' ,... or

2t = 12 λ' , 32 λ' , 52 λ' ,... = (m + 12 )λ'

where λ′ is the wavelength of light in the oil and m = 0, 1, 2, …

Substitute for λ' and solve for λ to obtain:

λ=

Substitute the predominant wavelengths to obtain:

700 nm =

Divide the first of these equations by the second to obtain:

2nt 700 nm m + 12 m + 32 = = 2nt 500 nm m + 12 m + 32

2nt m + 12 2nt 2nt and 500 nm = 1 m+ 2 m + 32

Solve for m:

m = 2 for λ = 700 nm

Solve equation (1) for t:

t = (m + 12 )

Substitute numerical values and evaluate t:

(1)

t = (2 + 12 )

λ 2n

700 nm = 603 nm 2(1.45)

1030

Chapter 33

Newton’s Rings *27 •• Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is air. A phase change of 180° ( 12 λ ) occurs at the top of the flat glass plate. We can use the condition for constructive interference to derive the result given in (a) and use the geometry of the lens on the plate to obtain the result given in (b). We can then use these results in the remaining parts of the problem. (a) The condition for constructive interference is:

2t + 12 λ = λ , 2λ , 3λ ,... or

2t = 12 λ , 32 λ , 52 λ ,... = (m + 12 )λ

where λ is the wavelength of light in air and m = 0, 1, 2, … Solve for t:

(b) From Figure 33-39 we have:

t=

(m + 12 ) λ , m = 0,1, 2, ... 2

(1)

r 2 + (R − t ) = R 2 2

or

R 2 = r 2 + R 2 − 2 Rt + t 2 For t << R we can neglect the last term to obtain:

R 2 ≈ r 2 + R 2 − 2 Rt

Solve for r:

r=

2 Rt

(2)

(c) The transmitted pattern is complementary to the reflected pattern. (d) Square equation (2) and substitute for t from equation (1) to obtain:

r 2 = (m + 12 )Rλ

Solve for m:

m=

r2 1 − Rλ 2

Substitute numerical values and evaluate m:

m=

(2 cm )2 − 1 = 67 (10 m )(590 nm ) 2

and so there will be 68 bright fringes.

Interference and Diffraction 1031 (e) The diameter of the mth fringe is: Noting that m = 5 for the sixth fringe, substitute numerical values and evaluate D:

D = 2r = 2 (m + 12 )Rλ D = 2 (5 + 12 )(10 m )(590 nm ) = 1.14 cm

The wavelength of the light in the film becomes λair / n = 444 nm. The (f) separation between fringes is reduced and the number of fringes that will

be seen is increased by the factor n = 1.33. 28 •• Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is air. A phase change of 180° ( 12 λ ) occurs at the top of the flat glass plate. We can use the condition for constructive interference and the results of Problem 27(b) to determine the radii of the first and second bright fringes in the reflected light. The condition for constructive interference is:

2t + 12 λ = λ , 2λ , 3λ ,... or

2t = 12 λ , 32 λ , 52 λ ,... = (m + 12 )λ

where λ is the wavelength of light in air and m = 0, 1, 2, …

λ

Solve for t:

t = (m + 12 ) , m = 0,1, 2, ... 2

From Problem 27(b):

r = 2tR

Substitute for t to obtain:

r=

(m + 12 )λR

The first fringe corresponds to m = 0:

r=

1 2

(520 nm )(2 m ) =

0.721 mm

The second fringe corresponds to m = 1:

r=

3 2

(520 nm )(2 m ) =

1.25 mm

1032

Chapter 33

29 •• Picture the Problem This arrangement is essentially identical to a ″thin film″ configuration, except that the ″film″ is oil. A phase change of 180° ( 12 λ ) occurs at lensoil interface. We can use the condition for constructive interference and the results from Problem 27(b) to determine the radii of the first and second bright fringes in the reflected light. The condition for constructive interference is:

2t + 12 λ' = λ' , 2λ' , 3λ' ,... or

2t = 12 λ' , 32 λ' , 52 λ' ,... = (m + 12 )λ'

where λ′ is the wavelength of light in the oil and m = 0, 1, 2, … Substitute for λ′ and solve for t:

t = (m + 12 )

λ 2n

, m = 0,1, 2, ...

where λ is the wavelength of light in air. From Equation 33-29:

r = 2tR

Substitute for t to obtain:

r=

(m + 12 ) λR

The first fringe corresponds to m = 0:

r=

1 (520 nm )(2 m ) = 0.535 mm 2 1.82

The second fringe corresponds to m = 1:

r=

3 (520 nm )(2 m ) = 0.926 mm 2 1.82

n

Two-Slit Interference Pattern *30 • Picture the Problem The number of bright fringes per unit distance is the reciprocal of the separation of the fringes. We can use the expression for the distance on the screen to the mth fringe to find the separation of the fringes.

1 ∆y

Express the number N of bright fringes per centimeter in terms of the separation of the fringes:

N=

Express the distance on the screen to the mth and (m + 1)st bright fringe:

ym = m

(1)

λL d

and ym+1 = (m + 1)

λL d

Interference and Diffraction 1033

λL

Subtract the second of these equations from the first to obtain:

∆y =

Substitute in equation (1) to obtain:

N=

d λL

Substitute numerical values and evaluate N:

N=

1 mm = 8.33 cm −1 (600 nm )(2 m )

d

31 • Picture the Problem We can use the expression for the distance on the screen to the mth and (m + 1)st bright fringes to obtain an expression for the separation ∆y of the fringes as a function of the separation of the slits d. Because the number of bright fringes per unit length N is the reciprocal of ∆y, we can find d from N, λ, and L. Express the distance on the screen to the mth and (m + 1)st bright fringe:

ym = m

Subtract the second of these equations from the first to obtain:

∆y =

Solve for d:

d=

Because the number of fringes per unit length N is the reciprocal of ∆y: Substitute numerical values and evaluate d:

λL d

λL d

λL d

λL ∆y

d = Nλ L

(

)

d = 28 cm −1 (589 nm )(3 m ) = 4.95 mm

32 • Picture the Problem We can use the geometry of the setup, represented to the right, to find the separation of the slits. To find the number of interference maxima that can be observed we can apply the equation describing two-slit interference maxima and require that sinθ ≤ 1. Because d << L, we can approximate sinθ1 as:

and ym+1 = (m + 1)

sin θ1 ≈

λ d

1034

Chapter 33

λ sin θ1

Solve for d to obtain:

d≈

From the right triangle whose sides are L and y1 we have:

sin θ1 =

Substitute numerical values in equation (1) and evaluate d:

d≈

(b) The equation describing two-slit interference maxima is:

(1)

0.82 m

(12 m )2 + (0.82 m )2

= 0.06817

633 nm = 9.29 µm 0.06817

d sin θ = mλ, m = 0,1, 2, ...

Because sinθ ≤ 1 determines the maximum number of interference fringes that can be seen:

d = mmax λ

Solve for mmax:

mmax =

Substitute numerical values and evaluate mmax:

mmax =

d

λ 9.29 µm = 14 because m must be 633 nm

an integer. Because there are 14 fringes on either side of the central maximum:

N = 2mmax + 1 = 2(14 ) + 1 = 29

33 •• Picture the Problem We can use the equation for the distance on a screen to the mth bright fringe to derive an expression for the spacing of the maxima on the screen. In (c) we can use this same relationship to express the slit separation d. (a) Express the distance on the screen to the mth and (m + 1)st bright fringe:

ym = m

Subtract the second of these equations from the first to obtain:

∆y =

Substitute numerical values and evaluate ∆y:

∆y =

λL d

and ym+1 = (m + 1)

λL

λL d

(1)

d

(500 nm )(1m ) = 1cm

50.0 µm

Interference and Diffraction 1035 (b)

Not with the unaided eye. The separation is too small to be observed with the naked eye.

(c) Solve equation (1) for d:

Substitute numerical values and evaluate d:

d=

d=

λL ∆y

(500 nm )(1m ) = 1 mm

0.500 mm

34 •• Picture the Problem Let the separation of the slits be d. We can find the total path difference when the light is incident at an angle φ and set this result equal to an integer multiple of the wavelength of the light to obtain the given equation. Express the total path difference:

∆l = d sin φ + d sin θ m ∆l = mλ

The condition for constructive interference is:

where m is an integer.

Substitute to obtain:

d sin φ + d sin θ m = mλ

Divide both sides of the equation by d to obtain:

sin φ + sin θ m =

mλ d

*35 •• Picture the Problem Let the separation of the slits be d. We can find the total path difference when the light is incident at an angle φ and set this result equal to an integer multiple of the wavelength of the light to relate the angle of incidence on the slits to the direction of the transmitted light and its wavelength. Express the total path difference:

∆l = d sin φ + d sin θ ∆l = mλ

The condition for constructive interference is:

where m is an integer.

Substitute to obtain:

d sin φ + d sin θ = mλ

Divide both sides of the equation by d to obtain:

sin φ + sin θ =

mλ d

1036

Chapter 33

Set θ = 0 and solve for λ:

λ=

d sin φ m

Substitute numerical values and simplify to obtain:

λ=

(2.5 µm )sin 30° = 1.25 µm m

m

Evaluate λ for positive integral values of m: m 1 2 3 4

λ (nm) 1250 625 417 313

From the table we can see that 625 nm and 417 nm are in the visible portion of the electromagnetic spectrum. 36 •• Picture the Problem The diagram shows the two speakers, S1 and S2, the centralbright image and the first-order image to the left of the central-bright image. The distance y is measured from the center of the central-bright image. We can apply the conditions for constructive and destructive interference from two sources and use the geometry of the speakers and microphone to find the distance to the first interference minimum and the distance to the first interference maximum. Relate the distance ∆y to the first minimum from the center of the central maximum to θ and the distance L from the speakers to the plane of the microphone:

tan θ =

y L

Solve for y to obtain:

y = L tan θ

Interference minima occur where:

d sin θ = (m + 12 )λ where m = 0, 1, 2, 3, …

Solve for θ to obtain:

⎡ (m + 12 )λ ⎤ ⎥⎦ ⎣ d

θ = sin −1 ⎢

(1)

Interference and Diffraction 1037 Relate the wavelength λ of the sound waves to the speed of sound v and the frequency f of the sound:

λ=

Substitute for λ in the expression for θ to obtain:

θ = sin −1 ⎢

v f ⎡ (m + 12 )v ⎤ ⎥ ⎣ df ⎦

⎧ ⎡ (m + 12 )v ⎤ ⎫ y = L tan ⎨sin −1 ⎢ ⎥⎬ ⎣ df ⎦ ⎭ ⎩

Substitute for θ in equation (1):

(2)

Noting that the first minimum corresponds to m = 0, substitute numerical values and evaluate ∆y:

⎧ ⎡ ( 1 ) (343 m/s ) ⎤ ⎫ y1st min = (1 m ) tan ⎨sin −1 ⎢ 2 ⎥ ⎬ = 0.365 m ⎣ (5 cm )(10 kHz ) ⎦ ⎭ ⎩ d sin θ = mλ

The maxima occur where:

where m = 1, 2, 3, …

⎧ ⎡ mv ⎤ ⎫ ∆y = L tan ⎨sin −1 ⎢ ⎥ ⎬ ⎣ af ⎦ ⎭ ⎩

For diffraction maxima, equation (2) becomes:

Noting that the first maximum corresponds to m = 1, substitute numerical values and evaluate ∆y:

⎧ ⎡ (1)(343 m/s ) ⎤ ⎫ y1st max = (1m ) tan ⎨sin −1 ⎢ ⎥ ⎬ = 0.943 m ( )( ) 5 cm 10 kHz ⎦⎭ ⎣ ⎩

Diffraction Pattern of a Single Slit 37 • Picture the Problem We can use the expression locating the first zeroes in the intensity to find the angles at which these zeroes occur as a function of the slit width a.

λ

The first zeroes in the intensity occur at angles given by:

sin θ =

Solve for θ :

⎛λ⎞ θ = sin −1 ⎜ ⎟

a

⎝a⎠

(a) For a = 1 mm:

⎛ 600 nm ⎞ ⎟⎟ = 0.600 mrad ⎝ 1 mm ⎠

θ = sin −1 ⎜⎜

1038

Chapter 33 ⎛ 600 nm ⎞ ⎟⎟ = 6.00 mrad ⎝ 0.1 mm ⎠

(b) For a = 0.1 mm:

θ = sin −1 ⎜⎜

(c) For a = 0.01 mm:

θ = sin −1 ⎜⎜

⎛ 600 nm ⎞ ⎟⎟ = 60.0 mrad ⎝ 0.01 mm ⎠

38 • Picture the Problem We can use the expression locating the first zeroes in the intensity to find the wavelength of the radiation as a function of the angle at which the first diffraction minimum is observed and the width of the plate.

λ

The first zeroes in the intensity occur at angles given by:

sin θ =

Solve for λ:

λ = a sin θ

Substitute numerical values and evaluate λ:

a

λ = (5 cm )sin 37° = 3.01cm

*39 •• Picture the Problem The diagram shows the beam expanding as it travels to the moon and that portion of it that is reflected from the mirror on the moon expanding as it returns to earth. We can express the diameter of the beam at the moon as the product of the beam divergence angle and the distance to the moon and use the equation describing diffraction at a circular aperture to find the beam divergence angle. We can follow this same procedure to find the diameter of the beam when it gets back to the earth. In Parts (c) and (d) we can use the dependence of the power in a beam on its cross-sectional area to find the fraction of the power of the beam that is reflected back to earth and the fraction of the original beam energy that is recaptured upon return to earth.

(a) Relate the diameter D of the beam at the moon to the distance to the moon L and the beam divergence angle θ :

D ≈θ L

Interference and Diffraction 1039 The angle θ subtended by the first diffraction minimum is related to the wavelength λ of the light and the diameter of the telescope opening dtelescope by: Because θ << 1, sinθ ≈ θ and:

sin θ = 1.22

θ ≈ 1.22

Substitute for θ in equation (1) to obtain:

D=

λ d telescope

λ d telescope

1.22 Lλ d telescope

Substitute numerical values and evaluate D:

⎤ ⎡ ⎥ ⎢ 1.22(500 nm ) D = 3.82 × 108 m ⎢ ⎥ = 1.53 km ⎢ 6 in × 2.54 cm × 1 m ⎥ ⎢⎣ in 10 2 cm ⎥⎦

(

)

λ

(b) The portion of the beam reflected back to the earth will be that portion incident on the mirror, so the diffraction angle is:

θ ≈ 1.22

The beam will expand back to:

⎡ λ ⎤ D' = L ⎢1.22 ⎥ d mirror ⎦ ⎣

d mirror

Substitute numerical values and evaluate D′:

⎡ ⎤ ⎢ ⎥ 1.22(500 nm ) D' = 3.82 × 108 m ⎢ ⎥ = 459 m ⎢ 20 in × 2.54 cm × 1 m ⎥ ⎢⎣ in 10 2 cm ⎥⎦

(

)

(c) Because the power of the beam is proportional to its cross-sectional area, the fraction of the power that is reflected back to the earth is the ratio of the area of the mirror to the area of the expanded beam at the moon:

π

2 d2 P' Amirror 4 mirror ⎛ d mirror ⎞ = = =⎜ ⎟ π 2 P Abeam D ⎝ ⎠ D 4

1040

Chapter 33

Substitute for D to obtain:

⎛ ⎜ P' ⎜ d mirror = P ⎜ 1.22 Lλ ⎜ ⎝ d telescope

Substitute numerical values and evaluate P′/P:

2 ⎡ ⎛ 2.54 cm ⎞ ⎤ ( )( ) 20 in 6 in ⎜ ⎟ ⎥ ⎢ P' ⎢ ⎝ in ⎠ ⎥ = P ⎢1.22 3.82 × 108 m (500 nm ) ⎥ ⎥ ⎢ ⎦ ⎣

2

⎞ ⎟ 2 ⎟ = ⎛⎜ d mirror d telescope ⎞⎟ (1) ⎟ ⎜ 1.22 Lλ ⎟ ⎠ ⎟ ⎝ ⎠

(

2

)

= 1.10 × 10 −7

λ

(d) The angular spread of the beam from reflection from the 20-in mirror is given by:

θ ≈ 1.22

The diameter D′ of the beam on return to earth will be:

D' ≈ 1.22 L

d mirror

λ d mirror

π

Letting P′′ represent the power intercepted by the telescope, we have:

2 d telescope P'' 4 = = π 2 P' Abeam D' 4

Atelescope

⎛d ⎞ = ⎜⎜ telescope ⎟⎟ ⎝ D' ⎠ Substitute for D′ and simplify:

2

P'' ⎛ d telescoped mirror ⎞ ⎟ =⎜ P' ⎜⎝ 1.22 Lλ ⎟⎠

2

(2)

Multiply equation (2) by equation (1) and simplify to obtain:

P'' P' P'' ⎛ d telescoped mirror ⎞ ⎟ = =⎜ P' P P ⎜⎝ 1.22 Lλ ⎟⎠ Substitute numerical values and evaluate P′′/P:

2

2

⎛ d mirror d telescope ⎞ ⎛ d mirror d telescope ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ 1.22 Lλ ⎠ ⎝ 1.22 Lλ ⎠

4

2 ⎡ ⎛ 2.54 cm ⎞ ⎤ ( )( ) 20 in 6 in ⎜ ⎟ ⎥ ⎢ P'' ⎢ ⎝ in ⎠ ⎥ = P ⎢1.22 3.82 × 108 m (500 nm ) ⎥ ⎥ ⎢ ⎦ ⎣

(

= 1.21 × 10 −14

)

4

Interference and Diffraction 1041

Interference-Diffraction Pattern of Two Slits 40 • Picture the Problem We need to find the value of m for which the mth interference maximum coincides with the first diffraction minimum. Then there will be N = 2m − 1 fringes in the central maximum. The number of fringes N in the central maximum is:

N = 2m − 1

Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating:

sin θ1 =

Express the angle θm corresponding to the mth interference maxima in terms of the separation d of the slits:

sin θ m =

Because we require that θ1 = θm, we can equate these expressions to obtain:

mλ λ = d a

Solve for and evaluate m:

m=

Substitute in equation (1) to obtain:

N = 2(5) − 1 = 9

If d = na:

m=

(1)

λ a

mλ d

d 5a = =5 a a

d na = =n a a

and

N = 2n − 1 41 •• Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the fifth interference maximum occurs to find a. We can then find the number of bright interference fringes seen in the central diffraction maximum using N = 2m − 1. (a) Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating:

sin θ1 =

λ a

1042

Chapter 33 5λ d

Express the angle θ5 corresponding to the mth fifth interference maxima maximum in terms of the separation d of the slits:

sin θ 5 =

Because we require that θ1 = θm5, we can equate these expressions to obtain:

5λ λ = d a

Solve for and evaluate ma:

a=

(b) Because m = 5:

N = 2m − 1 = 2(5) − 1 = 9

d 0.1 mm = = 20.0 µm 5 5

42 •• Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the mth interference maximum occurs to find m. We can then find the number of bright interference fringes seen in the central diffraction maximum using N = 2m − 1. The number of fringes N in the central maximum is:

N = 2m − 1

Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating:

sin θ1 =

Express the angle θm corresponding to the mth interference maxima in terms of the separation d of the slits:

sin θ m =

Because we require that θ1 = θm, we can equate these expressions to obtain:

mλ λ = d a

Solve for m:

m=

d a

Substitute in equation (1) to obtain:

N=

2d −1 a

λ a

mλ d

(1)

Interference and Diffraction 1043 Substitute numerical values and evaluate N:

N=

2(0.2 mm ) − 1 = 39 0.01 mm

*43 •• Determine the ConceptPicture the Problem There are 8 interference fringes on each side of the central maximum. The secondary diffraction maximum is half as wide as the central one. It follows that it will contain 8 interference maxima. 44 •• Picture the Problem We can equate the sine of the angle at which the first diffraction minimum occurs to the sine of the angle at which the mth interference maximum occurs to find m. We can then find the number of bright interference fringes seen in the central diffraction maximum using N = 2m − 1. In (b) we can use the expression relating the intensity in a single-slit diffraction pattern to phase constant φ =

2π

λ

a sin θ to find the

ratio of the intensity of the third interference maximum to the side of the centerline to the intensity of the center interference maximum. (a) The number of fringes N in the central maximum is:

N = 2m − 1

Relate the angle θ1 of the first diffraction minimum to the width a of the slits of the diffraction grating:

sin θ1 =

Express the angle θm corresponding to the mth interference maxima in terms of the separation d of the slits:

sin θ m =

Because we require that θ1 = θm, we can equate these expressions to obtain:

mλ λ = d a

Solve for m:

m=

d a

Substitute in equation (1) to obtain:

N=

2d −1 a

Substitute numerical values and evaluate N:

N=

2(0.15 mm ) −1 = 9 0.03 mm

λ a

mλ d

(1)

1044

Chapter 33

(b) Express the intensity for a single-slit diffraction pattern as a function of the phase difference φ:

2

⎛ sin 1 φ ⎞ I = I 0 ⎜⎜ 1 2 ⎟⎟ ⎝ 2φ ⎠ 2π where φ = a sin θ

(2)

λ

For m = 3:

sin θ 3 = and

φ=

2π

λ

3λ d

a sin θ 3 =

2π ⎛ 3λ ⎞ ⎛a⎞ a⎜ ⎟ = 6π ⎜ ⎟ λ ⎝ d ⎠ ⎝d ⎠

⎛ 0.03 mm ⎞ 6π ⎟⎟ = 0 . 15 mm ⎝ ⎠ 5

Substitute numerical values and evaluate φ:

φ = 6π ⎜⎜

Solve equation (2) for the ratio of I3 to I0:

I ⎛ sin 12 φ ⎞ ⎟ =⎜ I 0 ⎜⎝ 12 φ ⎟⎠

Substitute numerical values and evaluate I3/I0:

⎡ 1 ⎛ 6π ⎞ ⎤ sin ⎜ ⎟ I3 ⎢ 2 ⎝ 5 ⎠ ⎥ ⎥ = 0.255 =⎢ I 0 ⎢ 1 ⎛ 6π ⎞ ⎥ ⎢⎣ 2 ⎜⎝ 5 ⎟⎠ ⎥⎦

2

2

Using Phasors to Add Harmonic Waves 45 • Picture the Problem Chose the coordinate system shown in the phasor diagram. We can use the standard methods of vector addition to find the resultant of the two waves.

The resultant of the two waves is of the form:

r

Express R in vector form:

r

Find the magnitude of R :

E = R sin (ωt + δ )

r R = 2iˆ − 3 ˆj R=

(2)2 + (− 3)2

= 3.61

Interference and Diffraction 1045 ⎛ −3⎞ ⎟ = −56.3° ⎝ 2 ⎠

Find the phase angle δ between r r R and E1 :

δ = tan −1 ⎜

Substitute to obtain:

E = 3.61sin (ωt − 56.3°)

*46 • Picture the Problem Chose the coordinate system shown in the phasor diagram. We can use the standard methods of vector addition to find the resultant of the two waves.

The resultant of the two waves is of the form:

E = R sin (ωt + δ )

r

Rx = 4 + 3 cos 60° = 5.50

r

R y = 0 + 3 sin 60° = 2.60

Express the x component of R : Express the y component of R :

r

(5.50)2 + (2.60)2

Find the magnitude of R :

R=

= 6.08

Find the phase angle δ between r r R and E1 :

δ = tan −1 ⎜⎜

Substitute to obtain:

E = 6.08 sin (ωt + 25.3°)

⎛ Ry ⎞ 2.60 ⎞ ⎟⎟ = tan −1 ⎛⎜ ⎟ = 25.3° 5 . 50 R ⎝ ⎠ ⎝ x⎠

Remarks: We could have used the law of cosines to find R and the law of sines to find δ. 47 •• Picture the Problem We can evaluate the expression for the intensity for a single-slit diffraction pattern at the second secondary maximum to express I2 in terms of I0. The intensity at the second secondary maximum is given by:

⎡ sin 1 φ ⎤ I2 = I0 ⎢ 1 2 ⎥ ⎣ 2φ ⎦

2

1046

Chapter 33 where

φ= At this second secondary maximum:

λ

a sin θ

a sin θ = and

φ= Substitute for φ and evaluate I2:

2π

5 λ 2

2π ⎛ 5λ ⎞ ⎜ ⎟ = 5π λ ⎝ 2 ⎠

⎡ ⎛ 5π ⎢ sin ⎜ 2 I2 = I0 ⎢ ⎝ ⎢ 5π ⎢⎣ 2

2

⎞⎤ ⎟⎥ ⎠ ⎥ = 0.0162 I 0 ⎥ ⎥⎦

48 •• Picture the Problem We can use phasor concepts to find the phase angle δ in terms of the number of phasors N (three in this problem) forming a closed polygon of N sides at the minima and then use this information to express the path difference ∆r for each of these locations. Applying a small angle approximation, we can obtain an expression for y that we can evaluate for enough of the path differences to establish the pattern given in the problem statement.

2π N

Express the phase angle δ in terms of the number of phasors N forming a closed polygon of N sides at the first minimum:

δ=

Express the path difference ∆r in terms of sinθ and the separation d of the slits:

∆r = d sin θ or, provided the small angle approximation is valid,

∆r =

yd L

where L is the distance to the screen. Solve for y:

y=

L L ∆r y = δ d 2d

For three equally spaced sources, the phase angle corresponding to the first minimum is:

δ=

λ 2π and ∆r = δ = 13 λ 3 2π

Interference and Diffraction 1047

λL L⎛λ⎞ ⎜ ⎟ = (1) d⎝3⎠ 3d

Substitute to obtain:

y1 =

The phase angle corresponding to the second minimum is:

δ= ⎜

Substitute to obtain:

y2 =

1 ⎛ 2π ⎞ λ δ = 23 λ ⎟ and ∆r = 2⎝ 3 ⎠ 2π

λL L ⎛ 2λ ⎞ ⎜ ⎟ = (2) d⎝ 3 ⎠ 3d

When the path difference is λ, we have an interference maximum. The path difference corresponding to the fourth minimum is: Substitute to obtain:

Continue in this manner to obtain:

(b) For L = 1 m, λ = 5×10–7 m, and d = 0.1 mm:

∆r = 43 λ

y2 =

λL L ⎛ 4λ ⎞ ⎜ ⎟ = (4) d⎝ 3 ⎠ 3d

ymin =

2 ymin =

nλ L , n = 1, 2, 4, 5 ,7, 8, ... 3d

2(500 nm )(1 m ) = 3.33 mm 3(0.1 mm )

49 •• Picture the Problem We can use phasor concepts to find the phase angle δ in terms of the number of phasors N (four in this problem) forming a closed polygon of N sides at the minima and then use this information to express the path difference ∆r for each of these locations. Applying a small angle approximation, we can obtain an expression for y that we can evaluate for enough of the path differences to establish the pattern given in the problem statement.

2π N

Express the phase angle δ in terms of the number of phasors N forming a closed polygon of N sides at the first minimum:

δ=

Express the path difference ∆r in terms of sinθ and the separation d of the slits:

∆r = d sin θ or, provided the small angle approximation is valid,

1048

Chapter 33 yd L

∆r =

where L is the distance to the screen.

L ∆r d

Solve for y:

y=

For four equally spaced sources, the phase angle corresponding to the first minimum is:

δ=

Substitute to obtain:

y1 =

The phase angle corresponding to the second minimum is:

δ = π and ∆r =

Substitute to obtain:

y2 =

The phase angle angle corresponding to the third minimum is:

δ=

Substitute to obtain:

y3 =

Continue in this manner to obtain:

π 2

and ∆r =

λ δ = 14 λ 2π

L⎛λ⎞ λL ⎜ ⎟ = (1) d ⎝4⎠ 4d

λ δ = 12 λ 2π

λL L⎛λ⎞ ⎜ ⎟ = (2 ) d⎝2⎠ 4d

λ ⎛ 3π ⎞ 3λ 3π and ∆r = ⎜ ⎟= 2 2π ⎝ 2 ⎠ 4 λL L ⎛ 3λ ⎞ ⎜ ⎟ = (3) d⎝ 4 ⎠ 4d

ymin =

nλ L , n = 1, 2, 3, 5 ,6, 7, 9,... 4d

(b) For L = 2 m, λ = 6×10–7 m, d = 0.1 mm, and n = 1:

2 ymin =

2(600 nm )(2 m ) = 6.00 mm 4(0.1 mm )

For two slits:

2 ymin =

2(m + 12 )λL d

For L = 2 m, λ = 6×10–7 m, d = 0.1 mm, and m = 0:

2 y min =

(600 nm )(2 m ) = 12.0 mm 0.1mm

The width for four sources is half the width for two sources.

Interference and Diffraction 1049 50 •• Picture the Problem We can use sin θ = λ a to find the first zeros in the intensity pattern. The four-slit interference maxima occur at angles given by d sin θ = mλ, m = 0,1, 2, ... . In (c) we can use the result of Problem 49 to find the angular spread between the central interference maximum and the first interference minimum on either side of it. In (d) we’ll proceed as in Example 33-6, using a phasor diagram for a four-slit grating, to find the resultant amplitude at a given point in the intensity pattern as a function of the phase constant δ, that, in turn, is a function of the angle θ that determines the location of a point in the interference pattern.

λ

(a) The first zeros in the intensity occur at angles given by:

sin θ =

Solve for θ :

⎛λ⎞ θ = sin −1 ⎜ ⎟

a

⎝a⎠

Substitute numerical values and evaluate θ :

(b) The four-slit interference maxima occur at angles given by: Solve for θm:

⎛ 480 nm ⎞ ⎟⎟ = 0.242 rad ⎝ 2 µm ⎠

θ = sin −1 ⎜⎜

d sin θ = mλ, m = 0,1, 2, ...

⎡ mλ ⎤ ⎣ d ⎥⎦

θ m = sin −1 ⎢

⎡ m(480 nm ) ⎤ −1 ⎥ = sin (0.08m ) 6 m µ ⎣ ⎦

Substitute numerical values to obtain:

θ m = sin −1 ⎢

Evaluate θm for m = 0, 1, 2, and 3:

θ 0 = sin −1 [0(0.08)] = 0 θ1 = sin −1 [1(0.08)] = 80.1 mrad θ 2 = sin −1 [2(0.08)] = 0.161 rad

θ 3 = sin −1 [3(0.08)] = 0.242 rad

where θ3 will not be seen as it coincides with the first minimum in the diffraction pattern. (c) From Problem 49 we have:

θ min =

nλ 4d

1050

Chapter 33

For n = 1:

θ min =

480 nm = 0.0200 rad 4(6 µm )

(d) Use the phasor method to show the superposition of four waves of the same amplitude A0 and constant phase difference δ =

Express A in terms of δ ′ and δ ′′: Because the sum of the external angles of a polygon equals 2π: Examining the phasor diagram we see that: Eliminate α and solve for δ '' to obtain: Because the sum of the internal angles of a polygon of n sides is (n − 2)π : From the definition of a straight angle we have: Eliminate φ between these equations to obtain:

2π

λ

d sin θ .

A = 2( A0 cos δ '' + A0 cos δ ' ) (1) 2α + 3δ = 2π

α + δ '' = π δ '' = 32 δ 3φ + 2δ '' = 3π

φ −δ ' +δ = π

δ ' = 12 δ

Interference and Diffraction 1051 A = 2 A0 (cos 32 δ + cos 12 δ )

Substitute for δ '' and δ ' in equation (1) to obtain:

I = 4 I 0 (cos 32 δ + cos 12 δ )

2

Because the intensity is proportional to the square of the amplitude of the resultant wave:

The following graph of I/I0 as a function of sinθ was plotted using a spreadsheet 2

⎛ sin 1 φ ⎞ I program. The diffraction envelope was plotted using = 42 ⎜⎜ 1 2 ⎟⎟ , where I0 ⎝ 2φ ⎠

φ=

2π

λ

a sin θ . Note the excellent agreement with the results calculated in (a), (b), and

(c).

18

int ensit y

16 diffract ion envelope

14 12 10 I /I 0 8 6 4 2 0 -2 -0.3

-0.2

-0.1

0 sin(t het a)

51 ••• Picture the Problem We can find the phase constant δ from the geometry of the diagram to the right. Using the value of δ found in this fashion we can express the intensity at the point 1.72 cm from the centerline in terms of the intensity on the centerline. On the centerline, the amplitude of the resultant wave is 3 times that of each individual wave and the intensity is 9 times that of each source acting separately.

0.1

0.2

0.3

1052

Chapter 33

(a) Express δ for the adjacent slits: For small angles, sinθ ≈ tanθ :

δ=

2π

λ

d sin θ

sin θ ≈ tan θ =

Substitute to obtain:

δ=

Substitute numerical values and evaluate δ :

δ=

y L

2πdy λL

2π (0.06 mm )(1.72 cm ) (550 nm )(2.5 m ) 3π = rad = 270° 2

The three phasors, 270° apart, are shown in the diagram to the right. Note that they form three sides of a square. Consequently, their sum, shown as the resultant R, equals the magnitude of one of the phasors. (b) Express the intensity at the point 1.72 cm from the centerline:

I ∝ R2

Because I0 ∝ 9R2:

I I R2 ⇒ I= 0 = 2 9 I 0 9R

Substitute for I0 and evaluate I:

0.05 W/m 2 I= = 5.56 mW/m2 9

*52 ••• Picture the Problem We can use the phasor diagram shown in Figure 33-26 to determine the first three values of φ that produce subsidiary maxima. Setting the derivative of Equation 33-19 equal to zero will yield a transcendental equation whose roots are the values of φ corresponding to the maxima in the diffraction pattern. (a) Referring to Figure 33-26 we see that the first subsidiary maximum occurs when: A minimum occurs when:

φ = 3π

φ = 4π

Interference and Diffraction 1053 Another maximum occurs when: Thus, subsidiary maxima occur when:

(b) The intensity in the single-slit diffraction pattern is given by:

φ = 5π φ = (2n + 1)π, n = 1, 2, 3, ... and the first three subsidiary maxima are at φ = 3π, 5π, and 7π.

⎛ sin 1 φ ⎞ I = I 0 ⎜⎜ 1 2 ⎟⎟ ⎝ 2φ ⎠

2

Set the derivative of this expression equal to zero for extrema:

⎛ sin 1 φ ⎞ ⎡ 1 φ cos 12 φ − 12 sin 12 φ ⎤ dI = 2 I 0 ⎜⎜ 1 2 ⎟⎟ ⎢ 4 ⎥ = 0 for relative maxima and minima dφ ( 12 φ )2 ⎥⎦ ⎝ 2 φ ⎠ ⎢⎣ Simplify to obtain the transcendental equation: Solve this equation numerically (use the ″Solver″ function of your calculator) to obtain:

tan 12 φ = 12 φ

φ = 2.86π, 4.92π, and 6.94π

Remarks: Note that our results in (b) are smaller than the approximate values found in (a) by 4.80%, 1.63%, and 0.865% and that the agreement improves as n increases.

Diffraction and Resolution 53 • Picture the Problem We can use

θ = 1.22

λ

D

to find the angle between the

central maximum and the first diffraction minimum for a Fraunhofer diffraction pattern and the diagram to the right to find the distance between the central maximum and the first diffraction minimum on a screen 8 m away from the pinhole. (a) The angle between the central maximum and the first diffraction

θ = 1.22

λ D

1054

Chapter 33

minimum for a Fraunhofer diffraction pattern is given by: Substitute numerical values and evaluate θ : (b) Referring to the diagram, we see that: Substitute numerical values and evaluate y:

θ = 1.22

700 nm = 8.54 mrad 0.1 mm

y = L tan θ

y = (8 m ) tan (8.54 mrad ) = 6.83 cm

54 • Picture the Problem We can apply Rayleigh’s criterion to the overlapping diffraction patterns and to the diameter D of the pinhole to obtain an expression that we can solve for ∆y.

Rayleigh’s criterion is satisfied provided:

α c = 1.22

Relate αc to the separation ∆y of the light sources:

αc ≈

Equate these expressions to obtain: Solve for ∆y:

Substitute numerical values and evaluate ∆y:

λ D

∆y provided αc << 1. L

∆y λ = 1.22 L D λL ∆y = 1.22 D ∆y = 1.22

(700 nm )(10 m ) = 0.1 mm

8.54 cm

Interference and Diffraction 1055 *55 • Picture the Problem We can use Rayleigh’s criterion for slits and the geometry of the diagram to the right showing the overlapping diffraction patterns to express x in terms of λ, L, and the width a of the slit.

Referring to the diagram, relate αc, L, and x:

αc ≈

For slits, Rayleigh’s criterion is:

αc =

x L

λ a

Equate these two expressions to obtain:

x λ = L a

Solve for x:

x=

Substitute numerical values and evaluate x:

λL

x =

a

(700 nm )(5 m ) 0.5 mm

= 7.00 mm

56 • Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to express L in terms of λ, x, and the diameter D of your pupil.

Referring to the diagram, relate αc, L, and x:

αc ≈

x L

1056

Chapter 33

λ

For circular apertures, Rayleigh’s criterion is:

α c = 1.22

Equate these two expressions to obtain:

x λ = 1.22 L D

Solve for L:

L=

xD 1.22λ

Substitute numerical values and evaluate L:

L=

(112 cm )(5 mm ) = 1.22(550 nm )

D

8.35 km

57 • Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to express L in terms of λ, x, and the diameter D of your pupil.

Referring to the diagram, relate αc, L, and x:

αc ≈

For circular apertures, Rayleigh’s criterion is:

α c = 1.22

Equate these two expressions to obtain:

x L

λ D

x λ = 1.22 L D

Solve for L:

L=

xD 1.22λ

Substitute numerical values and evaluate L:

L=

(6.5 cm )(5 mm) = 1.22(550 nm )

484 m

Interference and Diffraction 1057 58 •• Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to the right showing the overlapping diffraction patterns to express L in terms of λ, x, and the diameter D of your pupil.

x provided α << 1 L

(a) Referring to the diagram, relate αc, L, and x:

αc ≈

For circular apertures, Rayleigh’s criterion is:

α c = 1.22

Equate these two expressions to obtain:

λ D

x λ = 1.22 L D

Solve for L:

L=

xD 1.22λ

Substitute numerical values and evaluate L:

L=

(6 mm )(5 mm) = 1.22(500 nm )

(b)

49.2 m

Because L is inversely proportional to λ, the holes can be resolved better with violet light which has a shorter wavelength.

59 •• Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to obtain an expression we can solve for the minimum separation ∆x of the stars.

1058

Chapter 33

(a) Rayleigh’s criterion is satisfied provided:

α c = 1.22

Relate αc to the separation ∆x of the light sources:

αc ≈

λ D

∆x because αc << 1 L

∆x λ = 1.22 L D λL ∆x = 1.22 D

Equate these expressions to obtain: Solve for ∆x:

Substitute numerical values and evaluate ∆x:

⎛

∆x = 1.22

m⎞ ⎟⎟ 1c ⋅ y ⎝ ⎠ = 5.00 × 109 m 2.54 cm 200 in × 1in

(550 nm )⎜⎜ 4 c ⋅ y × 9.461 × 10

15

*60 •• Picture the Problem We can use Rayleigh’s criterion for circular apertures and the geometry of the diagram to obtain an expression we can solve for the minimum diameter D of the pupil that allows resolution of the binary stars.

(a) Rayleigh’s criterion is satisfied provided:

α c = 1.22

Solve for D:

D = 1.22

λ D

λ αc

Interference and Diffraction 1059 Substitute numerical values and evaluate D:

D = 1.22

550 nm π rad 1° × 14'' × 3600'' 180°

= 9.89 mm ≈ 1 cm

Diffraction Gratings 61 • Picture the Problem We can solve d sin θ = mλ for θ with m = 1 to express the location of the first-order maximum as a function of the wavelength of the light. The interference maxima in a diffraction pattern are at angles θ given by: Solve for the angular location θm of the maxima : Relate the number of slits N per centimeter to the separation d of the slits:

d sin θ = mλ where d is the separation of the slits and m = 0, 1, 2, …

⎛ mλ ⎞ ⎟ ⎝ d ⎠

θ m = sin −1 ⎜

N=

1 d

Substitute to obtain:

θ m = sin −1 (mNλ )

Evaluate θ1 for λ = 434 nm:

θ1 = sin −1 [(2000 cm −1 )(434 nm )] = 86.9 mrad

Evaluate θ1 for λ = 410 nm:

θ1 = sin −1 [(2000 cm −1 )(410 nm )] = 82.1 mrad

*62 • Picture the Problem We can solve d sin θ = mλ for λ with m = 1 to express the location of the first-order maximum as a function of the angles at which the first-order images are found. The interference maxima in a diffraction pattern are at angles θ given by:

d sin θ = mλ where d is the separation of the slits and m = 0, 1, 2, …

1060

Chapter 33

Solve for λ:

λ=

d sin θ m

Relate the number of slits N per centimeter to the separation d of the slits:

N=

1 d

Let m =1 and substitute for d to obtain:

λ=

d sin θ N

Substitute numerical values and evaluate λ1 for θ1 = 9.72 ×10–2 rad:

λ1 =

sin (9.72 ×10 −2 rad ) = 485 nm 2000 cm −1

Substitute numerical values and evaluate λ1 for θ 2 = 1.32 ×10–1 rad:

λ1 =

sin 1.32 × 10−1 rad = 658 nm 2000 cm −1

(

)

63 • Picture the Problem We can solve d sin θ = mλ for θ with m = 1 to express the location of the first-order maximum as a function of the wavelength of the light. The interference maxima in a diffraction pattern are at angles θ given by: Solve for the angular location θm of the maxima : Relate the number of slits N per centimeter to the separation d of the slits:

d sin θ = mλ where d is the separation of the slits and m = 0, 1, 2, …

⎛ mλ ⎞ ⎟ ⎝ d ⎠

θ m = sin −1 ⎜

N=

1 d

Substitute to obtain:

θ m = sin −1 (mNλ )

Evaluate θ1 for λ = 434 nm:

θ 1 = sin −1 [(15000 cm −1 )(434 nm )] = 0.7089 rad = 40.6°

Evaluate θ1 for λ = 410 nm:

θ 1 = sin −1 [(15000 cm −1 )(410 nm )] = 0.6624 rad = 38.0°

Interference and Diffraction 1061 64 • Picture the Problem We can use the grating equation with sinθ = 1 and m = 5 to find the longest wavelength that can be observed in the fifth-order spectrum with the given grating spacing. The interference maxima are at angles θ given by:

d sin θ = mλ, m = 1, 2, 3, ...

d sin θ m

Solve for λ:

λ=

Evaluate λ for sinθ = 1 and m = 5:

1 d 4000 cm −1 λ= = = 500 nm 5 5

65 • Picture the Problem We can use the grating equation to find the angle at which normally incident blue light will be diffracted by the Morpho’s wings. The grating equation is:

d sin θ = mλ where m = 1, 2, 3, …

Solve for θ to obtain:

Substitute numerical values and evaluate θ1:

⎡ mλ ⎤ ⎣ d ⎥⎦

θ = sin −1 ⎢

⎡ (1)(440 nm )⎤ ⎥ = 30.0° ⎣ 880 nm ⎦

θ = sin −1 ⎢

66 •• Picture the Problem We can use the grating equation to find the angular separation of the first-order spectrum of the two lines. In (b) we can apply the definition of the resolving power of the grating to find the width of the grating that must be illuminated for the lines to be resolved. (a) Express the angular separation in the first-order spectrum of the two lines: Solve the grating equation for θ :

Substitute to obtain:

∆θ = θ 579 − θ 577

⎛ mλ ⎞ ⎟ ⎝ d ⎠

θ = sin −1 ⎜

1062

Chapter 33 ⎡ ⎤ ⎡ ⎤ ⎢ m(579 nm ) ⎥ ⎢ m(577 nm ) ⎥ ⎥ − sin −1 ⎢ ⎥ ∆θ = sin −1 ⎢ 1 1 ⎢ ⎥ ⎢ ⎥ ⎢ 2000 cm −1 ⎥ ⎢ 2000 cm −1 ⎥ ⎣ ⎦ ⎣ ⎦

For m = 1:

⎡ ⎤ ⎡ ⎤ ⎢ (1)(579 nm ) ⎥ ⎢ (1)(577 nm ) ⎥ ⎥ − sin −1 ⎢ ⎥ = 0.0231° ∆θ = sin −1 ⎢ 1 1 ⎢ ⎥ ⎢ ⎥ ⎢ 2000 cm −1 ⎥ ⎢ 2000 cm −1 ⎥ ⎣ ⎦ ⎣ ⎦ (b) Express the width of the beam necessary for these lines to be resolved: Relate the resolving power of the diffraction grating to the number of slits N that must be illuminated in order to resolve these wavelengths in the mth order:

w = Nd

λ = mN ∆λ

For m = 1:

N=

λ ∆λ

Substitute in equation (1) to obtain:

w=

λd ∆λ

Letting λ be the average of the two wavelengths, substitute numerical values and evaluate w:

(1)

⎛

w=

⎞ 1 ⎟ −1 ⎟ 2000 cm ⎝ ⎠

(578 nm ) ⎜⎜

2 nm

= 1.45 mm

*67 •• Picture the Problem We can use the grating equation d sin θ = mλ, m = 1, 2, 3, ... to express the order number in terms of the slit separation d, the wavelength of the light λ, and the angle θ. The interference maxima in the diffraction pattern are at angles θ

d sin θ = mλ, m = 1, 2, 3, ...

Interference and Diffraction 1063 given by: Solve for m: If one is to see the complete spectrum:

d sin θ

m=

λ

sin θ ≤ 1 and m ≤

Evaluate mmax:

m max

1 4800 cm −1

λ max

d

λ 1 4800 cm −1 = = 2.98 700 nm

Because m max = 2.98, one can see the complete spectrum only for m = 1 and 2. Express the condition for overlap:

m1 λ1 ≥ m 2 λ 2

Because 700 nm < 2 × 400 nm, there is no overlap of the second - order spectrum into the first - order spectrum; however, there is overlap of long wavelengths in the second order with short wavelengths in the third - order spectrum. 68 •• Picture the Problem We can use the grating equation and the resolving power of the grating to derive an expression for the angle at which you should look to see a wavelength of 510 nm in the fourth order. The interference maxima in the diffraction pattern are at angles θ given by: The resolving power R is given by:

d sin θ = mλ, m = 1, 2, 3, ...

(1)

R = mN where N is the number of slits and m is the order number.

Relate d to the width w of the grating:

d=

w N

Substitute for N to obtain:

d=

mw R

1064

Chapter 33

Substitute for d in equation (1) to obtain: Solve for θ :

Substitute numerical values and evaluate θ :

mw sin θ = mλ R ⎛ Rλ ⎞ ⎟ ⎝ w ⎠

θ = sin −1 ⎜

⎡ (22,000 )(510 nm ) ⎤ ⎥ = 13.0° 5 cm ⎣ ⎦

θ = sin −1 ⎢

69 •• Picture the Problem The distance on the screen to the mth bright fringe can be found using ym = mλL d , where d is the slit separation. We can use θ min = λ Nd = ∆y 2 L to find the width of the central maximum and the R = mN, where N is the number of slits in the grating, to find the resolution in the first order. (a) The distance on the screen to the mth bright fringe is given by:

ym = m

λL d

or, because d = n−1,

y m = mnλL

Substitute numerical values to obtain: Evaluate y1 and y2:

(

)

y m = m 4000 cm −1 (589 nm )(1.5 m ) = (0.353 m )m

y1 = (0.353 m )(1) = 0.353 m and

y 2 = (0.353 m )(2 ) = 0.706 m

(b) The angle θmin that locates the first minima in the diffraction pattern is given by:

θ min =

λ Nd

=

∆y 2L

where ∆y is the width of the central maximum.

Solve for ∆y:

∆y =

Substitute numerical values and evaluate ∆y:

∆y =

2 Lλ Nd 2(1.5 m )(589 nm ) ⎞ 1 ⎟ −1 ⎟ 4000 cm ⎠ ⎝ ⎛

(8000 lines) ⎜⎜ = 88.4 µm

Interference and Diffraction 1065

(c) The resolution R in the mth order is given by: Substitute numerical values and evaluate R:

R = mN

R = (1)(8000 ) = 8000

70 •• Picture the Problem The width of the grating w is the product of its number of lines N and the separation of its slits d. Because the resolution of the grating is a function of the average wavelength, the difference in the wavelengths, and the order number, we can express w in terms of these quantities. Express the width w of the grating as a function of the number of lines N and the slit separation d:

w = Nd

The resolving power R of the grating is given by:

R=

Solve for N to obtain:

λ = mN ∆λ

N=

Substitute for N in the expression for w to obtain:

w=

λ m∆λ

λd m∆λ

Letting λ be the average of the given wavelengths, substitute numerical values and evaluate w:

1 2

w=

⎞ 1 ⎟ −1 ⎟ ⎝ 8400 cm ⎠ = 3.43 cm 2(519.322 nm − 519.313 nm ) ⎛

(519.313 nm + 519.322 nm )⎜⎜

*71 •• Picture the Problem We can use the expression for the resolving power of a grating to find the resolving power of the grating capable of resolving these two isotopic lines in the third-order spectrum. Because the total number of the slits of the grating N is related to width w of the illuminated region and the number of lines per centimeter of the grating and the resolving power R of the grating, we can use this relationship to find the number of lines per centimeter of the grating

1066

Chapter 33

The resolving power of a diffraction grating is given by:

R=

λ = mN ∆λ

Substitute numerical values and evaluate R:

R=

546.07532 546.07532 − 546.07355

(1)

= 3.09 × 105 Express n, be the number of lines per centimeter of the grating, in terms of the total number of slits N of the grating and the width w of the grating:

n=

N w

From equation (1) we have:

N=

R m

Substitute to obtain:

n=

R mw

Substitute numerical values and evaluate n:

n=

3.09 × 105 = 5.15 × 10 4 cm −1 (3)(2 cm )

72 •• Picture the Problem We can differentiate the grating equation implicitly to obtain an expression for the number of lines per centimeter n as a function of cosθ and dθ /dλ. We can use the Pythagorean identity sin2θ + cos2θ = 1 and the grating equation to write cosθ in terms of n, m, and λ. Making this substitution and approximating dθ /dλ by ∆θ /∆λ will yield an expression for n in terms of m, λ, ∆λ, and ∆θ. (a) The grating equation is:

d sin θ = mλ, m = 0, 1, 2, ...

Differentiate both sides of this equation with respect to λ:

d (d sin θ ) = d (mλ ) dλ dλ or

d cos θ

Because n = 1/d:

cos θ

dθ =m dλ

dθ = nm dλ

(1)

Interference and Diffraction 1067 Solve for n to obtain:

Approximate dθ /dλ by ∆θ /∆λ:

Substitute for cosθ :

n=

1 dθ cos θ m dλ

n=

1 ∆θ cos θ m ∆λ

n=

1 ∆θ 1 − sin 2 θ m ∆λ

From equation (1):

sin θ =

Substitute to obtain:

n=

Solve for n:

n=

mλ = nmλ d

1 ∆θ 1 − n 2 m 2 λ2 m ∆λ 1 ⎛ ∆λ ⎞ m λ2 + ⎜ ⎟ ⎝ ∆θ ⎠

2

Substitute numerical values and evaluate n:

1

n=

⎛ ⎞ ⎜ ⎟ ⎛ 480 nm + 500 nm ⎞ ⎜ 500 nm − 480 nm ⎟ 3 ⎜ ⎟ + 2 ⎠ ⎜ 12° × π rad ⎟ ⎝ ⎜ ⎟ 180° ⎝ ⎠

2

= 6.677 × 10 5 m −1

2

= 6677 cm −1 (b) Express mmax in terms of d and λmax:

m max =

Substitute numerical values and evaluate mmax:

m max =

d

λ max

=

1 nλ max 1

(6677 cm )(500 nm) −1

= 3

1068

Chapter 33

73 •• Picture the Problem We can use the grating equation and the geometry of the diagram to derive an expression for the separation ∆y = y2 − y1 of the spectral lines in terms of the distance L to the screen, the wavelengths of the resolved lines, and the number of grating slits per centimeter n. We will assume that the angle θ 2 is small and then verify that this is a justified assumption. (a) The grating equation is:

d sin θ = mλ, m = 0, 1, 2, ...

Assuming that θ2 << 1 and m = 2:

sin θ 2 ≈ tan θ 2 =

Substitute to obtain:

d

Solve for y:

y=

Letting the numerals 1 and 2 refer to the spectral lines, express y2 – y1:

∆y = y2 − y1 =

Solve for d to obtain:

d=

mL (λ2 − λ1 ) y2 − y1

The number of lines per centimeter n is the reciprocal of d:

n=

y2 − y1 mL(λ2 − λ1 )

Substitute numerical values and evaluate n:

n=

8.4 cm (2)(8 m )(590 nm − 520 nm )

y L

y = mλ L mLλ d mL (λ2 − λ1 ) d

= 750 cm −1 To confirm our assumption that θ2 << 1, solve the grating equation for θ2:

⎛ 2λ ⎞ −1 ⎟ = sin (2λn ) ⎝ d ⎠

θ 2 = sin −1 ⎜

Interference and Diffraction 1069

θ 2 = sin −1 [2(590 nm ) (750 cm −1 )]

Substitute numerical values and evaluate θ2:

= 8.86 × 10 −2 <<1 sin θ 2 ≈ tan θ 2 ≈ θ 2 , as was assumed

Because θ2 << 1:

above. (b) The separation of the wavelengths is given by:

∆y =

mL (λ2 − λ1 ) = mLn(λ2 − λ1 ) d

For m = 1:

(

)

(

)

∆y = (1)(8 m ) 750 cm −1 (590 nm − 520 nm ) = 4.20 cm For m = 3:

∆y = (3)(8 m ) 750 cm −1 (590 nm − 520 nm ) = 12.6 cm 74 ••• Picture the Problem We can differentiate the grating equation implicitly and approximate dθ /dλ by ∆θ /∆λ to obtain an expression ∆θ as a function of m, n, ∆λ, and cosθ. We can use the Pythagorean identity sin2θ + cos2θ = 1 and the grating equation to write cosθ in terms of n, m, and λ. Making these substitutions will yield the given equation. The grating equation is:

d sin θ = mλ, m = 0, 1, 2, ...

Differentiate both sides of this equation with respect to λ:

d (d sin θ ) = d (mλ ) dλ dλ or

d cos θ

dθ =m dλ

dθ = nm dλ

Because n = 1/d:

cos θ

Solve for n to obtain:

n=

1 dθ cos θ m dλ

Approximate dθ /dλ by ∆θ /∆λ:

n=

1 ∆θ cos θ m ∆λ

(1)

1070

Chapter 33

Solve for ∆θ :

∆θ =

Substitute for cosθ :

∆θ =

nm∆λ cos θ

From equation (1):

sin θ =

Substitute to obtain:

∆θ =

nm∆λ 1 − sin 2 θ

mλ = nmλ d nm∆λ 1 − n 2 m 2λ2

Simplify by dividing the numerator and denominator by nm:

∆θ =

∆λ 1 1 − n 2 m 2λ2 nm

=

∆λ 1− n m λ n2m2 2

2 2

∆λ 1 − λ2 2 2 nm

=

75 ••• Picture the Problem We can use the grating equation and the geometry of the grating to derive an expression for φm in terms of the order number m, the wavelength of the light λ, and the groove separation a. (a) The grating equation is: Because φ and θI have their left and right sides mutually perpendicular: Substitute to obtain: Solve for φm:

(b) For m = 2:

d sin θ = mλ, m = 0, 1, 2, ...

(1)

θ i = φm d sin φm = mλ ⎛ mλ ⎞ ⎟ ⎝ d ⎠

φm = sin −1 ⎜

⎛ ⎞ ⎜ ⎟ −1 ⎜ (2 )(450 nm ) ⎟ φ2 = sin = 64.2° ⎜ ⎟ 1 ⎜ 10,000 cm −1 ⎟ ⎝ ⎠

Interference and Diffraction 1071 76 ••• Picture the Problem We can follow the procedure outlined in the problem statement to obtain R = λ/∆λ = mN. (a) Express the relationship between the phase difference φ and the path difference ∆r: Because ∆r = dsinθ :

(b) Differentiate this expression with respect to θ to obtain: Solve for dφ:

(c) From (b):

Substitute 2π/N for dφ to obtain:

(d) Equation 33-27 is: Differentiate this expression implicitly with respect to λ to obtain:

Solve for dθ to obtain:

(e) Equate the two expressions for dθ obtained in (c) and (d): Solve for R = λ/∆λ:

2π∆r φ ∆r = ⇒φ= λ λ 2π

2πd

φ=

λ

sin θ

dφ d ⎡ 2πd ⎤ 2πd = sin θ ⎥ = cos θ ⎢ dθ dθ ⎣ λ λ ⎦ 2πd

dφ =

dθ =

λ

cos θ dθ

λdφ 2πd cos θ

λ

dθ =

Nd cos θ

33-30

d sin θ = mλ, m = 0, 1, 2, ... d [d sin θ ] = d [mλ ] dλ dλ or

d cos θ

dθ =m dλ

dθ =

mdλ d cos θ

λ Nd cos θ R=

=

mdλ d cos θ

λ = mN dλ

33-31

1072

Chapter 33

General Problems *77 • Picture the Problem We can apply the condition for constructive interference to find the angular position of the first maximum on the screen. Note that, due to reflection, the wave from the image is 180o out of phase with that from the source. (a) Because y0 << L, the distance from the mirror to the first maximum is given by: Express the condition for constructive interference: Solve for θ :

For the first maximum, m = 0 and:

Substitute in equation (1) to obtain: Because the image of the slit is as far behind the mirror’s surface as the slit is in front of it, d = 2 mm. Substitute numerical values and evaluate y0:

y 0 = Lθ 0

(1)

d sin θ = (m + 12 )λ, m = 0,1, 2, ...

λ⎤ ⎡ θ = sin −1 ⎢(m + 12 ) ⎥ d⎦ ⎣ ⎡ λ⎤ θ 0 = sin −1 ⎢( 12 ) ⎥ ⎣ d⎦ ⎡ λ⎤ y0 = L sin −1 ⎢( 12 ) ⎥ ⎣ d⎦ ⎡ 600 nm ⎤ y0 = (1 m ) sin −1 ⎢( 12 ) 2 mm ⎥⎦ ⎣ = 0.150 mm

λL

(b) The separation of the fringes on the screen is given by:

∆y =

The number of dark bands per centimeter is the reciprocal of the fringe separation:

n=

1 d = ∆y λL

Substitute numerical values and evaluate n:

n=

2 mm = 3.33 × 103 m −1 (600 nm )(1m )

d

78 •• Picture the Problem The light from the radio galaxy reaches the radio telescope by two paths; one coming directly from the galaxy and the other reflected from the surface of the lake. The latter is phase shifted 180°, relative to the former, by reflection from the surface

Interference and Diffraction 1073 of the lake. We can use the condition for constructive interference of two waves to find the angle above the horizon at which the light from the galaxy will interfere constructively.

Because the reflected light is phase shifted by 180°, the condition for constructive interference at point P is: Referring to the figure, note that:

Substitute for ∆r to obtain:

Noting that m = 0 for the first interference maximum, substitute numerical values and evaluate θ0:

∆r = (m +

1 2

)λ

where m = 0, 1, 2, …

sin θ ≈

∆r ⎡ ∆r ⎤ ⇒ θ = sin −1 ⎢ ⎥ d ⎣d ⎦ ⎡ (m + 12 )λ ⎤ ⎥ ⎣ d ⎦

θ = sin −1 ⎢

⎡ 1 (20 cm )⎤ = 5.00 × 10 −3 rad ⎥ ⎣ 20 m ⎦

θ 0 = sin −1 ⎢ 2

= 0.286° 79 • Picture the Problem We can use the condition determining the location of points of zero intensity in a diffraction pattern to express the location of the first zero in terms of y and L. The width of the central maximum can then be found from ∆y = 2y. Express the horizontal length of the principal diffraction maximum on the screen: Referring to the diagram, relate the angle θ to the distances y and L:

∆y = 2 y

tan θ =

(1)

y L

or, because θ << 1, tanθ ≈ sinθ and

1074

Chapter 33 sin θ =

The points of zero intensity for a single-slit diffraction pattern are determined by the condition:

y L

a sin θ = mλ, m = 1, 2, ...

Substitute for sinθ to obtain:

ay = mλ L

Solve for y:

y=m

Substitute for y in equation (1):

At the first diffraction minimum, m = 1. Substitute numerical values and evaluate ∆y:

λL a

∆y = 2m

λL

∆y = 2(1)

a

(700 nm )(6 m ) = 0.5 mm

1.68 cm

80 • Picture the Problem We can use the Rayleigh criterion to express αc in terms of λ and the diameter of the opera glasses lens D and the geometry of the problem to relate αc to separation y of the singer’s eyelashes and the observation distance L.

λ

The critical angular separation, according to Rayleigh’s criterion, is:

α c = 1.22

Given that αc << 1, it is also given by:

αc ≈

Equating these two expressions yields:

y λ = 1.22 L D

D

y L

Interference and Diffraction 1075 Solve for D to obtain:

Substitute numerical values and evaluate D:

D = 1.22

D = 1.22

λL y

(550 nm )(25 m ) = 0.5 mm

33.6 mm

81 • Picture the Problem The resolving power of a telescope is the ability of the instrument to resolve two objects that are close together. Hence we can use Rayleigh’s criterion as the resolving power of the Arecibo telescope. Rayleigh’s criterion for resolution is:

α c = 1.22

Substitute numerical values and evaluate αc:

α c = 1.22

λ D 3.2 cm = 0.130 mrad 300 m

*82 •• Picture the Problem Note that reflection at both surfaces involves a phase shift of π rad. We can apply the condition for destructive interference to find the thickness t of the nonreflective coating.

The condition for destructive interference is:

2t = (m + 12 )λcoating = (m + 12 )

Solve for t:

t = (m + 12 )

Evaluate t for m = 0:

t = ( 12 )

λair 2ncoating

600 nm = 115 nm 2(1.30)

λair ncoating

1076

Chapter 33

83 •• Picture the Problem The Fabry-Perot interferometer is shown in the figure. For constructive interference in the transmitted light the path difference must be an integral multiple of the wavelength of the light. This path difference can be found using the geometry of the interferometer. Express the path difference between the two rays that emerge from the interferometer: For constructive interference we Require that: Equate these expressions to obtain: Solve for a to obtain:

∆r =

2a cos θ

∆r = mλ, m = 0,1, 2, ...

2a cos θ mλ a= cos θ 2

mλ =

84 •• Picture the Problem The gaps in the spectrum of the visible light are the result of destructive interference between the incident light and the reflected light. Noting that there is a π rad phase shift at the first air-mica interface, we can use the condition for destructive interference to find the index of refraction n of the mica sheet. Because there is a π rad phase shift at the first air-mica interface, the condition for destructive interference is:

2t = mλmica = m

Solve for n:

n=m

For λ = 474 nm:

λair 2t

2t = (474 nm ) m

λair n

, m = 1, 2, 3, ...

(1)

Interference and Diffraction 1077 For λ = 421 nm:

2t = (421 nm )(m + 1)

Equate these two expressions for 2t and solve for m to obtain:

m = 8 for λ = 474 nm

Substitute numerical values in equation (1) and evaluate n:

n=8

474 nm = 1.58 2(1.2 µm )

85 •• Picture the Problem Note that the light reflected at both the air-film and film-lens interfaces undergoes a π rad phase shift. We can use the condition for destructive interference between the light reflected from the air-film interface and the film-lens interface to find the thickness of the film. In (c) we can find the factor by which light of the given wavelengths is reduced by this film from I ∝ cos 2 12 δ . (a) Express the condition for destructive interference between the light reflected from the air-film interface and the film-lens interface: Solve for t:

2t = (m + 12 )λfilm = (m + 12 )

λair

(1)

n

where m = 0, 1, 2, …

t = (m + 12 )

λair 2n

Evaluate t for m = 0:

⎛ 1 ⎞ 540 nm t =⎜ ⎟ = 97.8 nm ⎝ 2 ⎠ 2(1.38)

(b) Solve equation (1) for λair:

λair =

2tn m + 12

Evaluate λair for m = 1:

λair =

2(97.8 nm )(1.38) = 180 nm 1 + 12

No; because 180 nm is not in the visible portion of the spectrum. (c) Express the reduction factor f as

f = cos 2 12 δ

(2)

1078

Chapter 33

a function of the phase difference δ between the two reflected waves: Relate the phase difference to the path difference ∆r: Because ∆r = 2t:

⎛ ∆r ⎞ δ ∆r ⎟⎟ = ⇒ δ = 2π ⎜⎜ 2π λfilm λ ⎝ film ⎠ ⎛ 2t ⎞ ⎟⎟ ⎝ λfilm ⎠

δ = 2π ⎜⎜

Substitute in equation (2) to obtain:

⎡ ⎛ 2t ⎞⎤ ⎡ 2π t ⎤ ⎟⎟⎥ = cos 2 ⎢ f = cos 2 ⎢ 12 2π ⎜⎜ ⎥ ⎝ λfilm ⎠⎦ ⎣ λfilm ⎦ ⎣ ⎡ 2π nt ⎤ = cos 2 ⎢ ⎥ ⎣ λair ⎦

Evaluate f for λ = 400 nm:

⎡ 2π (1.38)(97.8 nm ) ⎤ f 400 = cos 2 ⎢ ⎥ 400 nm ⎣ ⎦ = 0.273

Evaluate f for λ = 700 nm:

⎡ 2π (1.38)(97.8 nm ) ⎤ f 400 = cos 2 ⎢ ⎥ 700 nm ⎦ ⎣ = 0.124

86 •• Picture the Problem As indicated in the problem statement, we can find the optimal size of the pinhole by equating the angular width of the object at the film and the angular width of the diffraction pattern.

D D ⇒θ = L 2L

Express the angular width of the a distant object at the film in terms of the diameter D of the pinhole and the distance L from the pinhole to the object:

2θ =

Using Rayleigh’s criterion, express the angular width of the diffraction

θ diffraction = 1.22

λ D

Interference and Diffraction 1079 pattern: Equate these two expressions to obtain:

D λ = 1.22 D 2L

Solving for D yields:

D = 2.44λL

Substitute numerical values and evaluate D:

D = 2.44(550 nm )(10 cm ) = 0.366 mm

*87 •• Picture the Problem We can use the geometry of the dots and the pupil of the eye and Rayleigh’s criterion to find the greatest viewing distance that ensures that the effect will work for all visible wavelengths.

d L

Referring to the diagram, express the angle subtended by the adjacent dots:

θ≈

Letting the diameter of the pupil of the eye be D, apply Rayleigh’s criterion to obtain:

α c = 1.22

Set θ = αc to obtain:

d λ = 1.22 L D

Solve for L:

L=

Dd 1.22λ

Evaluate L for the shortest wavelength light in the visible portion of the spectrum:

L=

(3 mm)(2 mm) = 1.22(400 nm )

λ D

12.3 m

1080

Chapter 33

*88 ••• Picture the Problem It is given that with one tube evacuated and one full of air at 1-atm pressure, there are 198 more wavelengths of light in the tube full of air than in the evacuated tube of the same length. We can use this condition to obtain an equation that expresses this difference in terms of L, λn, and λ0. We can obtain a second equation relating λn, n, and λ0 ( λn =

λ0 n

) and solve the two equations simultaneously to find n.

(a) The wavelengths are related by:

λn = L

The number of wavelengths in length L is the length L divided by the wavelength. Thus:

λn

Substitute for λn:

nL

λ0

Solve for λn to obtain:

−

−

λ0 n L

λ0

L

λ0

= 198

= 198

198λ0 L ⎛ 589 nm ⎞ ⎟⎟ = 1.0002916 n = 1 + 198⎜⎜ ⎝ 0 .4 m ⎠

n =1+

Substitute numerical values and evaluate n:

(b) Replace 198 with 198 ± 0.25 and assume that the uncertainties in L and λ0 are negligible:

n =1+

λ0 L

(198 ± 0.25) =

1.0002916 ± 0.0000004

89 ••• Picture the Problem We can use the condition that determines points of zero intensity for a single slit diffraction pattern and the geometry of the slit and screen shown in the diagram to derive the given width of the central maximum on the screen. (a) The points of zero intensity for a single-slit diffraction pattern are given by:

a sin θ = mλ, m = 1, 2, 3, ...

(1)

Interference and Diffraction 1081

Relate the half-width y of the diffraction pattern to θ and L:

tan θ =

y L

Because θ is very small, tanθ ≈ sinθ and:

sin θ ≈

y L

Substitute for sinθ in equation (1) to obtain:

a

Solve for y:

y≈m

The width of the central maximum (m = 1) is: (b) Set a = obtain:

2 Lλ and simplify to a

y ≈ mλ L

2y ≈

2y ≈

λL a 2λ L a

2λL = a 2 Lλ a

1082

Chapter 33

Chapter 34 Wave-Particle Duality and Quantum Physics Conceptual Problems *1 • Determine the Concept The Young double-slit experiment, the diffraction of light by a small aperture, and the J.J. Thomson cathode-ray experiment all demonstrated the wave nature of electromagnetic radiation. Only the photoelectric effect requires an explanation based on the quantization of electromagnetic radiation.

(c ) is correct.

2 •• Determine the Concept Since the power radiated by a source is the energy radiated per unit area and per unit time, it is directly proportional to the energy. The energy radiated varies inversely with the wavelength ( E = hc λ ); i.e., the longer the wavelength, the less energy is associated with the electromagnetic radiation.

(b ) is correct.

3 • (a) True (b) False. The work function of a metal is a property of the metal and is independent of the frequency of the incident light. (c) True (d) True 4 • Determine the Concept In the photoelectric effect, the number of electrons emitted per second is a function of the light intensity, proportional to the light intensity, independent of the work function of the emitting surface and independent of the frequency of the light.

(b ) is correct.

*5 • Determine the Concept The threshold wavelength for emission of photoelectrons is related to the work function of a metal through φ = hc λt . Hence λt = hc φ and

(a ) is correct. 1083

1084

Chapter 34

6 •• Determine the Concept In order for electrons to be emitted hc/λ must be greater thanφ. Evidently, hc/λ1 < φ, but hc/λ2 > φ. 7 • (a) True (b) True (c) True (d) False. Electrons are too small to be resolved by an electron microscope. 8 • Determine the Concept If the de Broglie wavelengths of an electron and a proton are equal, their momenta must be equal. Since mp > me, vp < ve. Response (c) is correct. 9 • Picture the Problem The kinetic energy of a particle can be expressed, in terms of its momentum, as K =

p2 . We can use the equality of the kinetic energies and the fact that 2m

me < mp to determine the relative sizes of their de Broglie wavelengths.

pp2

Express the equality of the kinetic energies of the proton and electron in terms of their momenta and masses:

pe2 = 2mp 2me

Use the de Broglie relation for the wavelength of matter waves to obtain:

h2 h2 = 2mp λ2p 2me λ2e or

mp λ2p = me λ2e Since me < mp:

λ2p < λ2e and λe > λp and

(c ) is correct.

10 • Determine the Concept Yes. <x> can equal a value for which P(x) is zero. An example is the asymmetric well for all even numbered states.

Wave-Particle Duality and Quantum Physics 1085 *11 • Determine the Concept In the photoelectric effect, an electron absorbs the energy of a single photon. Therefore, Kmax = hf − φ, independently of the number of photons incident on the surface. However, the number of photons incident on the surface determines the number of electrons that are emitted. 12 •• Picture the Problem The probability of a particular event occurring is the number of ways that event can occur divided by the number of possible outcomes. The expectation value, on the other hand, is the average value of the experiment. (a) Find the probability of a 1 coming up when the die is thrown:

P(1) =

(b) Find the average value of a large number of throws of the die:

n =

3 1 = 6 2

3 ×1 + 3 × 2 = 1.5 6

13 •• Determine the Concept According to quantum theory, the average value of many measurements of the same quantity will yield the expectation value of that quantity. However, any single measurement may differ from the expectation value.

Estimation and Approximation 14 •• Picture the Problem From Einstein’s photoelectric equation we have K max = hf − φ , which is of the form y = mx + b , where the slope is h and the Kmax-intercept is the work function. Hence we should plot a graph of Kmax versus f in order to obtain a straight line whose slope will be an experimental value for Planck’s constant. (a) The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form A3 544 λ (nm) B3 0.36 Kmax (eV) C3 A3*10^−19 λ (m) D3 3*10^8/C3 c/λ E3 B3*1.6*10^−19 Kmax (J) lambda Kmax (nm) (eV)

lambda (m)

f=c/lambda (Hz)

Kmax (J)

1086 Chapter 34 544 594 604 612 633

0.36 0.199 0.156 0.117 0.062

5.44E−07 5.94E−07 6.04E−07 6.12E−07 6.33E−07

5.51E+14 5.05E+14 4.97E+14 4.90E+14 4.74E+14

5.76E−20 3.18E−20 2.50E−20 1.87E−20 9.92E−21

The following graph was plotted from the data shown in the above table. Excel’s ″Add Trendline″ was used to fit a linear function to the data. 7.E-20 6.E-20 K max = 6.19E-34f - 2.83E-19 K max (J)

5.E-20 4.E-20 3.E-20 2.E-20 1.E-20 0.E+00 4.6E+14

4.8E+14

5.0E+14

5.2E+14

5.4E+14

5.6E+14

f (Hz)

(b) From the regression line we note that the experimental value for Planck’s constant is:

hexp = 6.19 × 10 −34 J ⋅ s

(c) Express the percent difference between hexp and h:

% diff =

h − hexp

= 1−

hexp

h h −34 6.19 × 10 J ⋅ s = 1− = 6.64% 6.63 × 10 −34 J ⋅ s

15 •• Picture the Problem From Einstein’s photoelectric equation we have K max = hf − φ , which is of the form y = mx + b , where the slope is h and the Kmax-intercept is the work function. Hence we should plot a graph of Kmax versus f in order to obtain a straight line whose intercept will be an experimental value for the work function. (a) The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form A3 544 λ (nm)

Wave-Particle Duality and Quantum Physics 1087 B3 C3 D3 E3

0.36 A3*10^−19 3*10^8/C3 B3*1.6*10^−19

Kmax (eV) λ (m)

c/λ

Kmax (J)

lambda Kmax lambda f=c/lambda Kmax (nm) (eV) (m) (Hz) (J) 544 0.36 5.44E−07 5.51E+14 5.76E−20 594 0.199 5.94E−07 5.05E+14 3.18E−20 604 0.156 6.04E−07 4.97E+14 2.50E−20 612 0.117 6.12E−07 4.90E+14 1.87E−20 633 0.062 6.33E−07 4.74E+14 9.92E−21 The following graph was plotted from the data shown in the above table. Excel’s ″Add Trendline″ was used to fit a linear function to the data. 7.E-20 6.E-20 K max = 6.19E-34f - 2.83E-19 K max (J)

5.E-20 4.E-20 3.E-20 2.E-20 1.E-20 0.E+00 4.6E+14

4.8E+14

5.0E+14

5.2E+14

5.4E+14

5.6E+14

f (Hz)

(b) From the regression line we note that the experimental value for the work function φ is:

φexp = 2.83 × 10 −19 J ×

1 eV 1.6 × 10 −19 J

= 1.77 eV

(c) The value of φexp = 1.77 eV is closest to the work function for cesium. *16 •• Picture the Problem From the Compton-scattering equation we have λ2 − λ1 = λC (1 − cosθ ) , where λC = h mec is the Compton wavelength. Note that this equation is of the form y = mx + b provided we let y = λ2 − λ1 and x = 1 − cosθ. Thus, we can linearize the Compton equation by plotting ∆λ = λ2 − λ1 as a function of 1− cos θ . The slope of the resulting graph will yield an experimental value for the Compton wavelength.

1088 Chapter 34 (a) The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form A3 45 θ (deg) B3 1 − cos(A3*PI()/180) 1 − cosθ C3 6.47E^−13 ∆λ = λ2 − λ1

θ

1− cosθ

λ2−λ1

(deg) 45 75 90 135 180

0.293 0.741 1.000 1.707 2.000

6.47E−13 1.67E−12 2.45E−12 3.98E−12 4.95E−12

The following graph was plotted from the data shown in the above table. Excel’s ″Add Trendline″ was used to fit a linear function to the data. The regression line is

∆λ = 2.48 × 10 −12 (1 − cos θ ) − 1.03 × 10 −13 6.0E-12

delta lambda = 2.48E-12[1-cos(theta)]- 1.03E-13

delta lambda

5.0E-12 4.0E-12 3.0E-12 2.0E-12 1.0E-12 0.0E+00 0.0

0.5

1.0

1.5

2.0

1-cos(theta)

From the regression line we note that the experimental value for the Compton wavelength λC,exp is:

λC,exp = 2.48 × 10−12 m

The Compton wavelength is given by:

λC =

h hc = me c me c 2

Substitute numerical values and evaluate λC:

λC =

1240 eV ⋅ nm = 2.43 × 10 −12 m 5.11× 105 eV

Wave-Particle Duality and Quantum Physics 1089 Express the percent difference between λC and λC,exp:

% diff = =

λC,exp − λexp λC,exp = −1 λexp λexp 2.48 × 10−12 m − 1 = 2.06 % 2.43 × 10−12 m

*17 •• Picture the Problem The de Broglie wavelength of an object is given by λ = h/p, where p is the momentum of the object. The de Broglie wavelength of an object, in terms of its mass m and speed v, is:

λ=

h mv

The values in the following table were obtained using the internet:

Type of ball Baseball Tennis Golf Soccer

The de Broglie wavelength of a baseball, moving with its maximum speed, is: Proceed as above to obtain the values shown in the table:

λ=

vmax m (g) (m/s) 142 44 57 54 57 42 250 31

6.63 × 10−34 J ⋅ s = 1.06 × 10−34 m (0.142 kg )(44 m/s)

Type of ball Baseball

m

vmax

(g) (m/s) 142 44

Tennis

57

54

Golf

57

42

Soccer

250

31

λ (m)

1.06 ×10 −34 2.15 × 10−34 2.77 × 10−34 0.855 × 10−34

Examination of the table indicates that the soccer ball has the shortest de Broglie wavelength.

The Particle Nature of Light: Photons 18 • Picture the Problem We can find the photon energy for an electromagnetic wave of a given frequency f from E = hf where h is Planck’s constant.

1090 Chapter 34 (a) For f = 100 MHz:

E = hf

(

)

= 6.63 × 10 −34 J ⋅ s (100 MHz ) = 6.63 × 10 −26 J = 6.63 × 10 −26 J ×

1 eV 1.60 × 10 −19 J

= 4.14 × 10 −7 eV (b) For f = 900 kHz:

E = hf

(

)

= 6.63 × 10 −34 J ⋅ s (900 kHz ) = 5.96 × 10 −28 J = 5.96 × 10 −28 J ×

1 eV 1.60 × 10 −19 J

= 3.73 × 10 −9 eV 19 • Picture the Problem The energy of a photon, in terms of its frequency, is given by E=hf. (a) Express the frequency of a photon in terms of its energy and evaluate f for E = 1 eV:

f =

(b) For E = 1 keV:

f =

1eV E = h 4.14 × 10 −15 eV ⋅ s

= 2.42 × 1014 Hz 1 keV 4.14 × 10 −15 eV ⋅ s

= 2.42 × 1017 Hz (c) For E = 1 MeV:

f =

1 MeV 4.14 × 10 −15 eV ⋅ s

= 2.42 × 10 20 Hz *20 • Picture the Problem We can use E = hc/λ to find the photon energy when we are given the wavelength of the radiation. (a) Express the photon energy as a function of wavelength and evaluate E for λ = 450 nm:

E=

hc

λ

=

1240 eV ⋅ nm = 2.76 eV 450 nm

Wave-Particle Duality and Quantum Physics 1091 (b) For λ = 550 nm:

(c) For λ = 650 nm:

E=

1240 eV ⋅ nm = 2.25 eV 550 nm

E=

1240 eV ⋅ nm = 1.91 eV 650 nm

21 • Picture the Problem We can use E = hc/λ to find the photon energy when we are given the wavelength of the radiation. (a) Express the photon energy as a function of wavelength and evaluate E for λ = 0.1 nm: (b) For λ =1 fm:

E=

E=

hc

λ

=

1240 eV ⋅ nm = 12.4 keV 0.1 nm

1240 eV ⋅ nm = 1.24 GeV 10 −6 nm

22 •• Picture the Problem We can express the density of photons in the beam as the number of photons per unit volume. The number of photons per unit volume is, in turn, the ratio of the power of the laser to the energy of the photons and the volume occupied by the photons emitted in one second is the product of the cross-sectional area of the beam and the speed at which the photons travel, i.e., the speed of light. Express the density of photons in the beam as a function of the number of photons emitted per second and the volume occupied by those photons:

ρ=

N V

Relate the number of photons emitted per second to the power of the laser and the energy of the photons:

N=

P Pλ = E hc

Express the volume containing the photons emitted in one second as a function of the cross sectional area of the beam:

V = Ac

1092 Chapter 34 Substitute to obtain:

Pλ hc 2 A

ρ=

Substitute numerical values and evaluate ρ:

ρ=

(3 mW )(632 nm )

(6.63 × 10

−34

2⎛ π 2⎞ J ⋅ s 3 × 108 m/s ⎜ (1 mm ) ⎟ ⎝4 ⎠

)(

)

= 4.05 × 1013 m −3

*23 • Picture the Problem The number of photons per unit volume is, in turn, the ratio of the power of the laser to the energy of the photons and the volume occupied by the photons emitted in one second is the product of the cross-sectional area of the beam and the speed at which the photons travel; i.e., the speed of light. Relate the number of photons emitted per second to the power of the laser and the energy of the photons:

N=

Substitute numerical values and evaluate N:

N=

P Pλ = E hc

(2.5 mW )(1.55 µm )

(6.63 × 10

−34

J ⋅ s )(3 × 108 m/s )

= 1.95 × 1016 s −1

The Photoelectric Effect 24 • Picture the Problem The threshold wavelength and frequency for emission of photoelectrons is related to the work function of a metal through φ = hf t = hc λt . We can use Einstein’s photoelectric equation K max =

hc

λ

− φ to find the maximum kinetic

energy of the electrons for the given wavelengths of the incident light. (a) Express the threshold frequency in terms of the work function for tungsten and evaluate ft:

Using v = fλ, express the threshold wavelength in terms of the threshold

ft =

φ h

=

4.58 eV 4.14 × 10−15 eV ⋅ s

= 1.11× 1015 Hz

λt =

3 × 108 m/s v = = 270 nm f t 1.11 × 1015 Hz

Wave-Particle Duality and Quantum Physics 1093 frequency and evaluate λt: (b) Using Einstein’s photoelectric equation, relate the maximum kinetic energy of the electrons to their wavelengths and evaluate Kmax:

K max = E − φ = hf − φ =

=

hc

λ

−φ

1240 eV ⋅ nm − 4.58 eV 200 nm

= 1.62 eV (c) Evaluate Kmax for λ = 250 nm:

K max =

1240 eV ⋅ nm − 4.58 eV 250 nm

= 0.380 eV 25 • Picture the Problem We can use the Einstein equation for photon energy to find the energy of an incident photon and his photoelectric equation to relate the work function for potassium to the maximum energy of the photoelectrons. The threshold wavelength can be found from λt = hc φ .

hc

1240 eV ⋅ nm = 4.13 eV 300 nm

(a) Use the Einstein equation for photon energy to relate the energy of the incident photon to its wavelength:

E=

(b) Using Einstein’s photoelectric equation, relate the work function for potassium to the maximum kinetic energy of the photoelectrons:

K max = E − φ

Solve for and evaluate φ:

φ = E − K max = 4.13 eV − 2.03 eV

λ

=

= 2.10 eV (c) Proceed as in (b) with E = hc λ :

K max = =

hc

λ

−φ

1240 eV ⋅ nm − 2.10 eV 430 nm

= 0.784 eV

1094 Chapter 34 (d) Express the threshold wavelength as a function of potassium’s work function and evaluate λt:

λt =

hc

=

φ

1240 eV ⋅ nm = 590 nm 2.10 eV

26 • Picture the Problem We can find the work function for silver using φ = hc λt and the maximum kinetic energy of the electrons using Einstein’s photoelectric equation.

hc

1240 eV ⋅ nm = 4.73 eV 262 nm

(a) Express the work function for silver as a function of the threshold wavelength:

φ=

(b) Using Einstein’s photoelectric equation, relate the work function for silver to the maximum kinetic energy of the photoelectrons:

K max = E − φ =

=

λt

=

hc

λ

−φ

1240 eV ⋅ nm − 4.73 eV 175 nm

= 2.36 eV 27 • Picture the Problem We can find the threshold frequency and wavelength for cesium using φ = hf t = hc λt and the maximum kinetic energy of the electrons using Einstein’s photoelectric equation. (a) Use the Einstein equation for photon energy to express and evaluate the threshold wavelength for cesium: Use v = fλ to find the threshold frequency:

λt =

ft =

hc

φ

v

λt

=

1240 eV ⋅ nm = 653 nm 1.9 eV

=

3 × 108 m/s 653 nm

= 4.59 × 1014 Hz (b) Using Einstein’s photoelectric equation, relate the maximum kinetic energy of the photoelectrons to the wavelength of the incident light and evaluate Kmax for λ = 250 nm:

K max = =

hc

λ

−φ

1240 eV ⋅ nm − 1.90 eV 250 nm

= 3.06 eV

Wave-Particle Duality and Quantum Physics 1095 (c) Proceed as above with λ = 350 nm:

K max =

1240 eV ⋅ nm − 1.90 eV 350 nm

= 1.64 eV *28 •• Picture the Problem We can use Einstein’s photoelectric equation to find the work function of this surface and then apply it a second time to find the maximum kinetic energy of the photoelectrons when the surface is illuminated with light of wavelength 365 nm.

hc

Use Einstein’s photoelectric equation to relate the maximum kinetic energy of the emitted electrons to their total energy and the work function of the surface:

K max =

Using Einstein’s photoelectric equation, find the work function of the surface:

φ = E − K max =

Substitute for φ and λ and evaluate Kmax:

λ

−φ

hc

λ

− K max

1240 eV ⋅ nm − 0.37 eV 780 nm = 1.22 eV

=

K max =

1240 eV ⋅ nm − 1.22 eV 410 nm

= 1.80 eV

Compton Scattering 29 • Picture the Problem We can calculate the shift in wavelength using the Compton relationship ∆λ =

h (1 − cosθ ) . me c

The shift in wavelength is given by:

Substitute numerical values and evaluate ∆λ:

∆λ =

h (1 − cos θ ) me c

1096 Chapter 34 6.63 × 10 −34 J ⋅ s ∆λ = (9.11× 10−31 kg )(3 × 108 m/s) (1 − cos60°) = 1.21 pm 30 • Picture the Problem We can calculate the scattering angle using the Compton relationship ∆λ =

h (1 − cosθ ) . me c

Using the Compton scattering equation, relate the shift in wavelength to the scattering angle:

∆λ =

h (1 − cosθ ) me c

⎛ ⎝

Solve for θ :

θ = cos −1 ⎜1 −

me c ⎞ ∆λ ⎟ h ⎠

Substitute numerical values and evaluate θ :

(

)(

)

⎛ ⎞ 9.11 × 10 −31 kg 3 × 108 m/s ⎜ (0.33 pm )⎟⎟ = 30.2° θ = cos ⎜1 − −34 6.63 × 10 J ⋅ s ⎝ ⎠ −1

31 • Picture the Problem We can calculate the shift in wavelength using the Compton relationship ∆λ =

h (1 − cosθ ) . me c

Express the wavelength of the incident photons in terms of the fractional change in wavelength:

∆λ

Using the Compton scattering equation, relate the shift in wavelength to the scattering angle:

∆λ =

λ

= 2.3% ⇒ λ =

∆λ 0.023

h (1 − cosθ ) me c

Substitute numerical values and evaluate λ:

λ=

6.63 × 10 −34 J ⋅ s (1 − cos135°) = 180 pm 0.023(9.11 × 10 −31 kg )(3 × 108 m/s )

Wave-Particle Duality and Quantum Physics 1097 *32 • Picture the Problem We can use the Einstein equation for photon energy to find the energy of both the incident and scattered photon and the Compton scattering equation to find the wavelength of the scattered photon.

hc

1240 eV ⋅ nm = 17.4 keV 0.0711 nm

(a) Use the Einstein equation for photon energy to obtain:

E=

(b) Express the wavelength of the scattered photon in terms of its prescattering wavelength and the shift in its wavelength during scattering:

λ2 = λ1 + ∆λ = λ1 +

λ1

=

h (1 − cos θ ) me c

Substitute numerical values and evaluate λ2:

6.63 × 10 −34 J ⋅ s λ2 = 0.0711 nm + (9.11× 10−31 kg )(3 × 108 m/s) (1 − cos180°) = 0.0760 nm (c) Use the Einstein equation for photon energy to obtain:

E=

hc

λ2

=

1240 eV ⋅ nm = 16.3 keV 0.0760 nm

33 • Picture the Problem Compton used X rays of wavelength 71.1 pm. Let the direction the incident photon (and the recoiling electron) is moving be the positive direction. We can use p = h/λ to find the momentum of the incident photon and the conservation of momentum to find its momentum after colliding with the electron. Use the expression for the momentum of a photon to find the momentum of Compton’s photons:

Using the Compton scattering equation, relate the shift in wavelength to the scattering angle:

6.63 × 10 −34 J ⋅ s p1 = = 71.1 pm λ1 h

= 9.32 × 10 −24 kg ⋅ m/s

λ2 = λ1 + λC (1 − cos θ )

Substitute numerical values and evaluate λ2:

λ2 = 71.1 pm + (2.43 × 10 −12 m )(1 − cos180°) = 76.0 pm

1098 Chapter 34 Apply conservation of momentum to obtain:

p1 = pe − p2 ⇒ pe = p1 − p2

Substitute for p1 and p2 and evaluate pe:

⎛ 6.63 × 10 −34 J ⋅ s ⎞ ⎟⎟ = 1.80 × 10 −23 kg ⋅ m/s pe = 9.32 × 10 −24 kg ⋅ m/s − ⎜⎜ − 76.0 pm ⎝ ⎠ 34 •• Picture the Problem We can calculate the shift in wavelength using the Compton relationship ∆λ =

h (1 − cosθ ) = λC (1 − cosθ ) and use conservation of energy to find me c

the kinetic energy of the scattered electron. (a) Use the Compton scattering equation to find the change in wavelength of the photon:

∆λ = λC (1 − cos θ )

= (2.43 × 10 −12 m )(1 − cos90°) = 2.43 pm hc

hc

(b) Use conservation of energy to relate the change in the kinetic energy of the electron to the energies of the incident and scattered photon:

∆Ee =

Find the wavelength of the scattered photon:

λ2 = λ1 + ∆λ = 6 pm + 2.43 pm

Substitute and evaluate the kinetic energy of the electron (equal to the change in its energy since it was stationary prior to the collision with the photon):

∆Ee =

λ1

−

λ2

= 8.43 pm hc

λ1

−

hc

λ2

⎛ 1 1 ⎞ ⎟⎟ = 1240 eV ⋅ nm⎜⎜ − ⎝ 6 pm 8.43 pm ⎠ = 59.6 keV

35 •• Picture the Problem We can find the number of head-on collisions required to double the wavelength of the incident photon by dividing the required change in wavelength by the change in wavelength per collision. The change in wavelength per collision can be found using the Compton scattering equation.

Wave-Particle Duality and Quantum Physics 1099 Express the number of collisions required in terms of the change in wavelength per collision: Using the Compton scattering equation, express the wavelength shift per collision:

N=

∆λ ∆λ collision

∆λ = λC (1 − cos θ )

Substitute numerical values and evaluate ∆λ:

∆λ = (2.43 × 10 −12 m )(1 − cos180°) = 4.86 pm

Substitute and evaluate N:

N=

200 pm = 42 4.86 pm

Electrons and Matter Waves 36

•

Picture the Problem From Equation 34-16 we have λ =

1.226 nm provided K is in K

electron volts. (a) For K = 2.5 eV:

λ=

1.226 nm = 0.775 nm 2.5

(b) For K = 250 eV:

λ=

1.226 nm = 0.0775 nm 250

(c) For K = 2.5 keV:

λ=

1.226 nm = 0.0245 nm 2500

(d) For K = 25 keV:

λ=

1.226 nm = 7.75 pm 25000

37 • Picture the Problem We can use its definition to find the de Broglie wavelength of this electron.

1100 Chapter 34 h p h = me v

Use its definition to express the de Broglie wavelength of the electron in terms of its momentum:

λ=

Substitute numerical values and evaluate λ:

λ=

6.63 × 10 −34 J ⋅ s (9.11× 10−31 kg )(2.5 × 105 m/s)

= 2.91 nm 38 • Picture the Problem We can find the momentum of the electron from the de Broglie equation and its kinetic energy from λ =

1.226 nm , where K is in eV. K h

6.63 × 10−34 J ⋅ s 200 nm

(a) Use the de Broglie relation to express the momentum of the electron:

p=

(b) Use the electron wavelength equation to relate the electron’s wavelength to its kinetic energy:

λ=

Solve for and evaluate K:

⎛ 1.226 eV1 2 nm ⎞ ⎟⎟ = 3.76 × 10 −5 eV K = ⎜⎜ ⎝ 200 nm ⎠

*39

λ

=

= 3.31× 10 −27 kg ⋅ m/s 1.226 nm K

2

••

Picture the Problem The momenta of these particles can be found from their kinetic energies and speeds. Their de Broglie wavelengths are given by λ = h/p. (a) The momentum of a particle p, in terms of its kinetic energy K, is given by: Substitute numerical values and evaluate pe:

p = 2mK

Wave-Particle Duality and Quantum Physics 1101 pe = 2me K = 2(9.11 × 10

−31

⎛ 1.6 × 10 −19 C ⎞ ⎜ ⎟⎟ kg )⎜150 keV × eV ⎝ ⎠

= 2.09 × 10 −22 N ⋅ s Substitute numerical values and evaluate pp:

pp = 2mp K = 2(1.67 × 10

− 27

⎛ 1.6 × 10 −19 C ⎞ ⎟⎟ kg )⎜⎜150 keV × eV ⎝ ⎠

= 8.95 × 10 −21 N ⋅ s Substitute numerical values and evaluate pα:

⎛ 1.66 × 10 −27 kg ⎞ ⎛ 1.6 × 10 −19 C ⎞ ⎟⎟ ⎜⎜150 keV × ⎟⎟ pα = 2mα K = 2⎜⎜ 4 u × u eV ⎝ ⎠⎝ ⎠ = 1.79 × 10 −20 N ⋅ s (b) The de Broglie wavelengths of the particles are given by:

λ=

Substitute numerical values and evaluate λp:

λp =

h p

=

Substitute numerical values and evaluate λe:

λe =

h pp 6.63 × 10 −34 J ⋅ s = 7.41 × 10 −14 m 8.95 × 10 −21 N ⋅ s h pe

6.63 × 10 −34 J ⋅ s = = 3.17 × 10 −12 m − 22 2.09 × 10 N ⋅ s Substitute numerical values and evaluate λα:

λα = =

h pα 6.63 × 10 −34 J ⋅ s = 3.70 × 10 −14 m − 20 1.79 × 10 N ⋅ s

1102 Chapter 34 40 • Picture the Problem The wavelength associated with a particle of mass m and kinetic energy K is given by Equation 34-15 as λ =

1240 eV ⋅ nm 2mc 2 K

Substitute numerical data in Equation 34-15 to obtain:

λ=

.

1240 eV ⋅ nm

2(940 MeV )(0.02 eV )

= 0.202 nm 41 • Picture the Problem The wavelength associated with a particle of mass m and kinetic energy K is given by Equation 34-15 as λ =

1240 eV ⋅ nm

Substitute numerical data in Equation 34-15 to obtain:

2mc 2 K

λ=

.

1240 eV ⋅ nm 2(938 MeV )(2 MeV )

= 2.02 × 10 −5 nm = 20.2 fm *42 • Picture the Problem We can use its definition to calculate the de Broglie wavelength of this proton. Use its definition to express the de Broglie wavelength of the proton:

λp =

h h = pp mp vp

Substitute numerical values and evaluate λp:

λ=

43

6.63 × 10 −34 J ⋅ s (1.67 × 10−27 kg ) 0.003(3 × 108 m/s) = 0.441 pm

[

]

•

Picture the Problem We can solve Equation 34-15 ( λ =

1240 eV ⋅ nm 2mc 2 K

) for the kinetic

energy of the proton and use the rest energy of a proton mc2 = 938 MeV to simplify our computation. Solve Equation 34-15 for the kinetic energy of the proton:

2 ( 1240 eV ⋅ nm ) K=

2mc 2λ2

Wave-Particle Duality and Quantum Physics 1103 (a) Substitute numerical values and evaluate K for λ = 1 nm:

2 ( 1240 eV ⋅ nm ) K= 2 2(938 MeV )(1 nm )

= 0.820 meV (b) Evaluate K for λ = 1 nm:

2 ( 1240 eV ⋅ nm ) K= 2 2(938 MeV ) (10 −6 nm )

= 820 MeV 44 • Picture the Problem We’ll need to convert oz and mph into SI units. Then we can use its definition to calculate the de Broglie wavelength of the baseball. Use its definition to express the de Broglie wavelength of the baseball:

λbaseball =

h pbaseball

=

h mbaseballvbaseball

Substitute numerical values and evaluate λbaseball:

λbaseball =

6.63 × 10 −34 J ⋅ s = 1.10 × 10 −34 m m⎞ ⎛ 0.447 ⎟ ⎛ 1 lb 1 kg ⎞ ⎜ mi s ⎟ ⎟⎟ ⎜ 95 × ⎜⎜ 5 oz × × mi 16 oz 2.20 lb ⎠ ⎜ h ⎟ ⎝ ⎜ ⎟ h ⎝ ⎠

For the tennis ball:

λtennis ball = =

h ptennis ball h mtennis ball vtennis balll

Substitute numerical values and evaluate λtennis ball:

λ tennis balll

6.63 × 10 −34 J ⋅ s = 2.01 × 10 −34 m = m⎞ ⎛ 0.447 ⎟ ⎜ ⎛ ⎞ 1lb 1 kg ⎜ mi s ⎟ ⎜⎜ 2 oz × ⎟⎟ 130 × × mi 16 oz 2.20 lb ⎠ ⎜ h ⎟ ⎝ ⎟ ⎜ h ⎠ ⎝ The tennis ball has the longer de Broglie wavelength.

1104 Chapter 34 Remarks: Because λ = h/p, we could have solved the problem by determining which ball has the smaller momentum. 45 • Picture the Problem If K is in electron volts, the wavelength of a particle is given by λ =

1.226 nm provided K is in eV. K

Evaluate λ for K = 54 eV:

46

λ=

1.226 1.226 nm = nm = 0.167 nm 54 K

•

Picture the Problem We can use λ =

1.226 nm , where K is in eV, to find the energy of K

electrons whose wavelength isλ.

1.226 nm K

Relate the wavelength of the electrons to their kinetic energy:

λ=

Solve for K:

⎛ 1.226 nm ⋅ eV1 2 ⎞ ⎟⎟ K = ⎜⎜ λ ⎝ ⎠

Substitute numerical values and evaluate K:

⎛ 1.226 nm ⋅ eV1 2 ⎞ ⎟⎟ = 22.8 eV K = ⎜⎜ 0 . 257 nm ⎝ ⎠

2

2

*47 • Picture the Problem We can use λ =

1.226 nm , where K is in eV, to find the K

wavelength of 70-keV electrons. Relate the wavelength of the electrons to their kinetic energy:

λ=

Substitute numerical values and evaluate λ:

λ=

1.23 nm K

1.226 70 × 103 eV

nm = 4.63 pm

48 • Picture the Problem We can use its definition to calculate the de Broglie wavelength of a neutron with speed 106 m/s.

Wave-Particle Duality and Quantum Physics 1105 Use its definition to express the de Broglie wavelength of the neutron:

λn =

h h = pn mn vn

Substitute numerical values and evaluate λn:

λn =

6.63 × 10 −34 J ⋅ s 1.67 × 10 −27 kg 106 m/s

(

)(

)

= 0.397 pm

Wave-Particle Duality 49 • Picture the Problem In order for diffraction to occur, the diameter of the aperture d must be approximately equal to the de Broglie wavelength of the spherical object. We can use the de Broglie relationship to find the size of the aperture necessary for this object to show diffraction. Express the de Broglie wavelength of the spherical object:

λobject =

Substitute numerical values and evaluate λobject:

λobject =

h pobject

=

h mobject vobject

6.63 × 10 −34 J ⋅ s 4 × 10 −3 kg (100 m/s )

(

)

= 1.66 × 10 −33 m This is many orders of magnitude smaller than even the diameter of a proton and so no common objects would be able to squeeze through such an aperture. 50 • Picture the Problem In order for diffraction to occur, the size of the object must be approximatelyλ. The wavelength of the neutron is given by λ =

1240 eV ⋅ nm 2mc 2 K

. The rest

energy of the neutron is mc2 = 940 MeV. Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm 2(940 MeV )(10 MeV )

= 9.04 × 10 −6 nm ≈ 10 fm This wavelength is of the same order - of - magnitude as a nuclear diameter and so nuclei would be suitable targets to demonstrate the wave nature of neutrons with this energy.

1106 Chapter 34 51

•

Picture the Problem We can use λ =

1.226 nm , where K is in eV, to find the K

wavelength of 200-eV electrons. In order for diffraction to occur, the size of the target must be approximatelyλ. Relate the wavelength of the electrons to their kinetic energy:

λ=

Substitute numerical values and evaluate λ:

λ=

1.226 nm K 1.226 nm = 0.0867 nm 200 eV

This distance is of the order of the size of an atom and so atoms would be suitable targets to demonstrate the wave nature of electrons with this energy.

A Particle in a Box *52 ••

nπx 2 sin . The L L following graphs were plotted using a spreadsheet program. The graph of ψ(x) for Picture the Problem The wave function for state n isψ n (x ) =

n = 1 is shown below:

Wave function

1.0

0.5

0.0 0.0

0.2

0.4

0.6

x /L

0.8

1.0

Wave-Particle Duality and Quantum Physics 1107 The graph of ψ 2(x) for n = 1 is shown below:

1.0

Probability density

0.8

0.6

0.4

0.2

0.0 0.0

0.2

0.4

0.6

0.8

1.0

0.6

0.8

1.0

x /L

The graph of ψ(x) for n = 2 is shown below:

1.0

Wave function

0.5

0.0

-0.5

-1.0 0.0

0.2

0.4

x /L

1108 Chapter 34 The graph of ψ 2(x) for n = 2 is shown below:

1.0

Probability density

0.8

0.6

0.4

0.2

0.0 0.0

0.2

0.4

0.6

0.8

1.0

0.6

0.8

1.0

x /L

The graph of ψ(x) for n = 3 is shown below:

1.0

Wave function

0.5

0.0

-0.5

-1.0 0.0

0.2

0.4

x /L

Wave-Particle Duality and Quantum Physics 1109 The graph of ψ 2(x) for n = 3 is shown below:

1.0

Probability density

0.8

0.6

0.4

0.2

0.0 0.0

0.2

0.4

0.6

0.8

1.0

x /L

53

••

Picture the Problem We can find the ground-state energy using E1 =

h2 and the 8mL2

energies of the excited states using En = n 2 E1 . The wavelength of the electromagnetic radiation emitted when the proton transitions from one state to another is given by the Einstein equation for photon energy ( E =

hc

λ

).

h2 E1 = 8mp L2

(a) Express the ground-state energy:

Substitute numerical values and evaluate E1:

E1 =

(6.63 × 10

−34

J ⋅s

)

2

1 eV = 206 MeV −19 8 1.67 × 10 −27 kg 10 −15 m 1.6 × 10 J

(

Find the energies of the first two excited states:

)(

)

2

E2 = 2 2 E1 = 4(206 MeV ) = 824 MeV and

E3 = 32 E1 = 9(206 MeV ) = 1.85 GeV

1110 Chapter 34 The energy-level diagram for this system is shown to the right:

hc ∆E 1240 eV ⋅ nm = ∆E

(b) Relate the wavelength of the electromagnetic radiation emitted during a proton transition to the energy released in the transition:

λ=

For the n = 2 to n = 1 transition:

∆E = E2 − E1 = 4 E1 − E1 = 3E1

Substitute numerical values and evaluate λ2→1:

λ2→1 =

1240 eV ⋅ nm 1240 eV ⋅ nm = 3E1 3(206 MeV )

= 2.01 fm (c) For the n = 3 to n = 2 transition:

∆E = E3 − E2 = 9 E1 − 4 E1 = 5E1

Substitute numerical values and evaluate λ3→2:

λ3→2 =

1240 eV ⋅ nm 1240 eV ⋅ nm = 5E1 5(206 MeV )

= 1.20 fm (d) For the n = 3 to n = 1 transition:

∆E = E3 − E1 = 9 E1 − E1 = 8E1

Substitute numerical values and evaluate λ3→1:

λ3→1 =

1240 eV ⋅ nm 1240 eV ⋅ nm = 8E1 8(206 MeV )

= 0.752 fm 54

••

Picture the Problem We can find the ground-state energy using E1 =

h2 and the 8mL2

energies of the excited states using En = n 2 E1 . The wavelength of the electromagnetic radiation emitted when the proton transitions from one state to another is given by the

Wave-Particle Duality and Quantum Physics 1111 Einstein equation for photon energy ( E =

hc

λ

(a) Express the ground-state energy:

).

E1 =

h2 8mp L2

Substitute numerical values and evaluate E1:

(6.63 × 10

−34

)

2

J ⋅s 1 eV E1 = = 5.14 meV 2 −19 − 27 8 1.67 × 10 kg (0.2 nm ) 1.60 × 10 J

(

Find the energies of the first two excited states:

)

E2 = 2 2 E1 = 4(5.14 meV ) = 20.6 meV and

E3 = 32 E1 = 9(5.14 meV ) = 46.3 meV hc ∆E 1240 eV ⋅ nm = ∆E

(b) Relate the wavelength of the electromagnetic radiation emitted during a proton transition to the energy released in the transition:

λ=

For the n = 2 to n = 1 transition:

∆E = E2 − E1 = 4 E1 − E1 = 3E1

Substitute numerical values and evaluate λ2→1:

λ2→1 =

1240 eV ⋅ nm 1240 eV ⋅ nm = 3E1 3(5.14 meV )

= 80.4 µm (c) For the n = 3 to n = 2 transition:

∆E = E3 − E2 = 9 E1 − 4 E1 = 5E1

Substitute numerical values and evaluate λ3→2:

λ3→2 =

1240 eV ⋅ nm 1240 eV ⋅ nm = 5E1 5(5.14 meV )

= 48.2 µm (d) For the n = 3 to n = 1 transition:

∆E = E3 − E1 = 9 E1 − E1 = 8E1

Substitute numerical values and evaluate λ3→1:

λ3→1 =

1240 eV ⋅ nm 1240 eV ⋅ nm = 8E1 8(5.14 meV )

= 30.2 µm

1112 Chapter 34

Calculating Probabilities and Expectation Values 55 •• Picture the Problem The probability of finding the particle in some range ∆x is ψ2dx. The interval ∆x = 0.002L is so small that we can neglect the variation in ψ(x) and just compute ψ2∆x. Express the probability of finding the particle in the interval ∆x:

P = P( x )∆x = ψ 2 ( x )∆x

πx 2 sin L L

Express the wave function for a particle in the ground state:

ψ 1 (x ) =

Substitute to obtain:

P=

(a) Evaluate P at x = L/2:

P = 0.004 sin 2

2 2 πx sin ∆x L L πx ⎞ 2⎛ = ⎜ sin 2 ⎟(0.002 L ) L⎝ L⎠ πx = 0.004 sin 2 L

πL 2L

= 0.004 sin 2

π 2

= 0.004 (b) Evaluate P at x = 2L/3:

P = 0.004 sin 2

2πL 2π = 0.004 sin 2 3L 3

= 0.003 (c) Evaluate P at x = L:

P = 0.004 sin 2

πL L

= 0.004 sin 2 π

= 0 *56 •• Picture the Problem The probability of finding the particle in some range ∆x is ψ 2dx. The interval ∆x = 0.002L is so small that we can neglect the variation in ψ(x) and just compute ψ 2∆x. Express the probability of finding the particle in the interval ∆x:

P = P( x )∆x = ψ 2 ( x )∆x

Wave-Particle Duality and Quantum Physics 1113 2 2πx sin L L

Express the wave function for a particle in its first excited state:

ψ 2 (x ) =

Substitute to obtain:

P=

(a) Evaluate P at x = L/2:

P = 0.004 sin 2

2 2 2πx sin ∆x L L 2⎛ 2πx ⎞ = ⎜ sin 2 ⎟(0.002 L ) L⎝ L ⎠ 2πx = 0.004 sin 2 L

2πL = 0.004 sin 2 π 2L

= 0 (b) Evaluate P at x = 2L/3:

P = 0.004 sin 2

4πL 4π = 0.004 sin 2 3L 3

= 0.003 (c) Evaluate P at x = L:

P = 0.004 sin 2

2πL = 0.004 sin 2 2π L

= 0 57

••

Picture the Problem We’ll use f ( x ) = (a) Express ψ(x) for the n = 2 state:

Express x using the n = 2 wave function: Change variables by letting θ =

2πx . Then: L

∫ f (x )ψ (x ) dx withψ (x ) = 2

n

ψ 2 (x ) =

nπx 2 sin . L L

2 2πx sin L L

2 x 2 2πx sin dx L L 0

L

x =∫

L θ, 2π 2π dθ = dx, and L L dx = dθ 2π and the limits on θ are 0 and 2π. x=

1114 Chapter 34 L

2 ⎛ L ⎞ ⎛ L ⎞ x = ∫ ⎜ θ ⎟ sin 2 θ ⎜ dθ ⎟ L 0 ⎝ 2π ⎠ ⎝ 2π ⎠

Substitute to obtain:

=

2π

L 2π

∫ θ sin

2

2

θ dθ

0

2π

Using a table of integrals, evaluate the integral:

L ⎡θ 2 θ sin 2θ cos 2θ ⎤ x = − − 2π 2 ⎢⎣ 4 4 8 ⎥⎦ 0 =

L ⎡ 2 1 1⎤ L π − + ⎥= 2 ⎢ 2π ⎣ 8 8⎦ 2

2x2 2πx sin 2 =∫ dx L L 0 L

(b) Express x 2 using the n = 2

x

2

x

2

wave function:

L

Change variables as in (a) and substitute to obtain:

2 = ∫ L0

2

⎞ ⎛ L ⎞ 2 ⎛ L dθ ⎟ ⎜ θ ⎟ sin θ ⎜ ⎝ 2π ⎠ ⎝ 2π ⎠ 2π

=

L2 θ 2 sin 2 θ dθ 3 ∫ 4π 0

Using a table of integrals, evaluate the integral: 2π

x

58

2

L2 ⎡θ 3 ⎛ θ 2 1 ⎞ L2 ⎡ 4π 3 π ⎤ θ cos 2θ ⎤ ⎜ ⎟ = − − θ = − ⎥ sin 2 − ⎥ ⎢ 4π 3 ⎣ 6 ⎜⎝ 4 8 ⎟⎠ 4 ⎦0 4π 3 ⎢⎣ 3 2⎦ 1 ⎞ ⎛1 = L2 ⎜ − 2 ⎟ = 0.321L2 ⎝ 3 8π ⎠

••

Picture the Problem We’ll use f ( x ) =

∫ f (x )ψ (x ) dx with ψ (x ) = 2

n

nπx 2 . sin L L

In Part (c) we’ll use P ( x ) = ψ 22 ( x ) to determine the probability of finding the particle in some small region dx centered at x = ½L. (a) Express the wave function for a particle in its first excited state:

ψ 2 (x ) =

2 2πx sin L L

Square both sides of the equation to obtain:

ψ 22 ( x ) =

2 2 2πx sin L L

Wave-Particle Duality and Quantum Physics 1115 The graph of ψ 2(x) as a function of x is shown to the right:

(b) Express x using the n = 2 wave function: Change variables by

2πx letting θ = . Then: L

Substitute to obtain:

2 x 2 2πx sin dx L L 0

L

x =∫

L θ, 3π 3π dθ = dx, and L L dx = dθ 3π and the limits on θ are 0 and 2π. x=

L

2 ⎛ L ⎞ ⎛ L ⎞ x = ∫ ⎜ θ ⎟ sin 2 θ ⎜ dθ ⎟ L 0 ⎝ 2π ⎠ ⎝ 2π ⎠ =

Using a table of integrals, evaluate the integral:

2π

2

∫ θ sin

2

θ dθ

0

2π

L ⎡θ 2 θ sin 2θ cos 2θ ⎤ − − x = 2π 2 ⎢⎣ 4 4 8 ⎥⎦ 0 =

(c) Express P(x):

2π

L

L ⎡ 2 1 1⎤ L π − + ⎥= 2 ⎢ 2π ⎣ 8 8⎦ 2

P(x ) = ψ 22 (x ) =

2 2 2πx dx sin L L

Evaluate P(L/2):

2π L ⎛L⎞ 2 P⎜ ⎟ = sin 2 ⋅ L 2 ⎝2⎠ L 2 = sin 2 π = 0 L

Because P(L/2) = 0:

⎛L⎞ P⎜ ⎟dx = 0 ⎝2⎠

1116 Chapter 34

(d)

The answers to Parts (b) and (c) are not contradictory. (b) states that the average value of measurements of the position of the particle will yield L/2, even though the probability that any one measurement of position will yield L/2 is zero.

59 •• Picture the Problem We can find the constant A by applying the normalization ∞

condition ψ 2 ( x ) dx = 1 and finding the value for A that satisfies this condition. As soon

∫

−∞

as we have found the normalization constant, we can calculate the probability of the a

finding the particle in the region −a ≤ x ≤ a using P = ψ 2 ( x ) dx .

∫

−a

(a) Express the normalization condition:

∞

∫ψ (x ) dx = 1 2

−∞

Substitute ψ (x ) = Ae

−x a

:

∫ (Ae

∞

) dx = 2 A ∫ e

−∞

From integral tables:

∞

∫e

−αx

Solve for A:

(b) Express the normalized wave function: The probability of finding the particle in the region −a ≤ x ≤ a is:

2

−2 x a

dx

0

dx =

0

Therefore:

∞

−x a 2

1

α

∞

⎛a⎞ 2 A2 ∫ e −2 x a dx = 2 A2 ⎜ ⎟ = aA2 = 1 ⎝2⎠ 0

1

A=

a

ψ (x ) =

1 −x e a

a

a

a

1 P = ∫ψ ( x ) dx = 2∫ e −2 x a dx a −a 0 2

a

=

2 −2 x a e dx = 1 − e −2 = 0.865 a ∫0

Wave-Particle Duality and Quantum Physics 1117 60 •• Picture the Problem The probability density for the particle in its ground state is given by P(x ) =

2 2π sin x . We’ll evaluate the integral of P(x) between the limits specified in L L

(a), (b), and (c). Express P(x) for 0 < x < d:

P (x ) =

Change variables by

x=

letting θ =

π

L

x . Then:

L

π

dθ = dx =

Substitute to obtain:

2 π sin 2 xdx ∫ L0 L d

θ,

π L L

π

P(x ) = =

dx, and dθ θ'

2 ⎛L ⎞ sin 2 θ ⎜ dθ ⎟ ∫ L0 ⎝π ⎠ 2

θ'

sin π∫

2

θdθ

0

Using a table of integrals, evaluate θ'

∫ sin

2

θ dθ :

P(x ) =

θ'

2 ⎡θ sin 2θ ⎤ − 4 ⎥⎦ 0 π ⎢⎣ 2

(1)

0

π 2

(a) Noting that the limits on θ are 0 and π/2, evaluate equation (1) over the interval 0 < x < ½L:

2 ⎡θ sin 2θ ⎤ 2 ⎡π ⎤ = ⎢ ⎥ P(x ) = ⎢ − ⎥ π ⎣2 π ⎣4⎦ 4 ⎦0

(b) Noting that the limits on θ are 0 and π/3, Evaluate equation (1) over the interval 0 < x < L/3:

2 ⎡θ sin 2θ ⎤ P(x ) = ⎢ − π ⎣2 4 ⎥⎦ 0 2 ⎡ π sin 2π 3 ⎤ = ⎢ − π ⎣6 4 ⎥⎦

= 0.500 π 3

=

1 3 − = 0.196 3 4π

1118 Chapter 34 (c) Noting that the limits on θ are 0 and 3π/4, Evaluate equation (1) over the interval 0 < x < 3L/4:

3π 4

2 ⎡θ sin 2θ ⎤ P(x ) = ⎢ − π ⎣2 4 ⎥⎦ 0 2 ⎡ 3π sin 6π 4 ⎤ = ⎢ − ⎥⎦ π⎣8 4 3 1 = + = 0.909 4 2π

61 •• Picture the Problem The probability density for the particle in its first excited state is given by P ( x ) =

2 2 2π sin x . We’ll evaluate the integral of P(x) between the limits L L

specified in (a), (b), and (c).

2 2π xdx sin 2 ∫ L0 L d

Express P(x) for 0 < x < d:

P(x ) =

Change variables by

L θ, 2π 2π dθ = dx, and L L dx = dθ 2π

letting θ =

2π x . Then: L

Substitute to obtain:

x=

P(x ) = =

θ'

2 ⎛ L ⎞ sin 2 θ ⎜ dθ ⎟ ∫ L0 ⎝ 2π ⎠ 1

θ'

sin π∫

2

θdθ

0

Using a table of integrals, evaluate θ'

∫ sin

2

θ dθ :

P(x ) =

θ'

1 ⎡θ sin 2θ ⎤ − 4 ⎥⎦ 0 π ⎢⎣ 2

(1)

0

(a) Noting that the limits on θ are 0 and π, evaluate equation (1) over the interval 0 < x < ½L:

π

1 ⎡θ sin 2θ ⎤ 1 ⎡π ⎤ = ⎢ ⎥ P(x ) = ⎢ − ⎥ π ⎣2 4 ⎦0 π ⎣ 2 ⎦

= 0.500

Wave-Particle Duality and Quantum Physics 1119 2π 3

(b) Noting that the limits on θ are 0 and 2π/3, evaluate equation (1) over the interval 0 < x < L/3:

1 ⎡θ sin 2θ ⎤ P(x ) = ⎢ − π ⎣2 4 ⎥⎦ 0 =

1 ⎡ 2π sin 4π 3 ⎤ − π ⎢⎣ 6 4 ⎥⎦

=

1 3 2 + = 0.402 3 4π 3π 2

(c) Noting that the limits on θ are 0 and 3π/2, evaluate equation (1) over the interval 0 < x < 3L/4:

1 ⎡θ sin 2θ ⎤ P(x ) = ⎢ − π ⎣2 4 ⎥⎦ 0

=

1 ⎡ 3π ⎤ π ⎢⎣ 4 ⎥⎦

= 0.750

62 •• Picture the Problem Classically, x = xP( x ) dx and x 2 = x 2 P( x ) dx .

∫

∫

Evaluate x with P(x) = 1/L:

Evaluate x 2 with P(x) = 1/L:

63

L

⎡ x2 ⎤ x L x = ∫ dx = ⎢ ⎥ = L 2 ⎣ 2L ⎦ 0 0 L

L

⎡ x3 ⎤ x2 L2 = ∫ dx = ⎢ ⎥ = L 3 ⎣ 3L ⎦ 0 0 L

x

2

••

Picture the Problem We’ll use f ( x ) =

∫ f (x )ψ (x ) dx with ψ (x ) = 2

n

2 nπx sin to L L

L L2 L2 2 show that x = = and x = . 2 3 2n 2π 2 (a) Express x for a particle in the nth state:

0

Change variables by letting θ =

L

x =∫

nπx . Then: L

2 x 2 nπx sin dx L L

L θ, nπ nπ dθ = dx, and L L dx = dθ nπ and the limits on θ are 0 and nπ. x=

1120 Chapter 34 Substitute to obtain:

2 x = L =

nπ

⎛ L

⎞

∫ ⎜⎝ nπ θ ⎟⎠ sin

2

⎛ L ⎞ dθ ⎟ ⎝ nπ ⎠

θ⎜

0

nπ

2L θ sin 2 θ dθ 2 2 ∫ nπ 0

Using a table of integrals, evaluate the integral: nπ

2 L ⎡θ 2 θ sin 2θ cos 2θ ⎤ − x = 2 2⎢ − nπ ⎣4 4 8 ⎥⎦ 0

2 L ⎡ n 2π 2 nπ sin 2nπ cos 2nπ 1 ⎤ 2 L ⎡ n 2π 2 1 1 ⎤ = 2 2⎢ − − + ⎥= 2 2 ⎢ − + ⎥ nπ ⎣ 4 4 8 8⎦ n π ⎣ 4 8 8⎦ L = 2 x

2

nth state: Change variables by letting θ =

nπx . Then: L

2x2 nπx sin 2 dx L L 0

L

Express x 2 for a particle in the

=∫

L θ, nπ nπ dθ = dx, and L L dx = dθ nπ and the limits on θ are 0 and nπ. x=

Substitute to obtain:

x

2

2 = L

nπ

∫ 0

2

⎛ L ⎞ ⎛ L ⎞ θ ⎟ sin 2 θ ⎜ dθ ⎟ ⎜ ⎝ nπ ⎠ ⎝ nπ ⎠ nπ

=

2 L2 θ 2 sin 2 θ dθ 3 3 ∫ nπ 0

Using a table of integrals, evaluate the integral: nπ

x

2

2 L2 ⎡θ 3 ⎛ θ 2 1 ⎞ 2 L2 ⎡ n 3π 3 nπ ⎤ θ cos 2θ ⎤ = 3 3 ⎢ − ⎜⎜ − ⎟⎟ sin 2θ − = − ⎥ 4 ⎦0 4 ⎥⎦ n π ⎣ 6 ⎝ 4 8⎠ n 3π 3 ⎢⎣ 6

=

L2 L2 − 2 2 3 2n π

Wave-Particle Duality and Quantum Physics 1121 (b)

For large values of n, the result agrees with the classical value of L2 / 3 given in Problem 62.

*64 •• Picture the Problem From Problem 63 we have x =

L L2 L2 and x 2 = − 2 2 .A 2 3 2n π

spreadsheet program was used to plot the following graphs of <x> and <x2> as a function of n.

0.55

0.50

0.45

<x> <x^2>

0.40

0.35

0.30

0.25 1

10

19

28

37

46

55

64

73

82

91

100

n

As n → ∞, x 2 →

L2 3

65 •• Picture the Problem For the ground state, n = 1 and so we’ll evaluate

f ( x ) = ∫ f ( x )ψ 2 ( x ) dx using ψ 1 (x ) = Because ψ 12 ( x ) is an even function of x, xψ

2 1

(x ) is an odd function of

x. It follows that the integral of xψ 12 ( x ) between −L/2 and L/2 is zero. Thus:

πx 2 cos . L L x = 0 for all values of n.

1122 Chapter 34 L2

Express x 2 :

x

Change variables by letting θ = Then:

πx L

.

π 2 = x 2 cos 2 xdx ∫ L −L 2 L

2

x=

L

π

dθ = dx =

θ,

π L L

π

dx, and dθ

and the limits on θ are −π/2 and π/2. π 2

Substitute to obtain:

x

2

=

Use a trigonometric identity to rewrite the integrand:

x

2

2

2 ⎛L ⎞ ⎞ 2 ⎛ L = ⎜ θ ⎟ cos θ ⎜ dθ ⎟ ∫ L −π 2 ⎝ π ⎠ ⎝π ⎠

= =

2 L2

π3 2 L2

π

π 2

∫θ π

−

2

cos 2 θ dθ

2

π 2

∫ θ (1 − sin θ )dθ π 2

3

−

2

2

π 2 π 2 ⎤ 2 L2 ⎡ 2 θ d θ θ 2 sin 2 θ dθ ⎥ − 3 ⎢ ∫ ∫ π ⎢⎣−π 2 ⎥⎦ −π 2

Evaluate the second integral by looking it up in the tables: π 2

x2

2 L2 ⎡θ 3 ⎧θ 3 ⎛ θ 2 1 ⎞ θ cos 2θ ⎫⎤ = 3 ⎢ − ⎨ − ⎜⎜ − ⎟⎟ sin 2θ − ⎬⎥ 4 ⎭⎦⎥ π ⎣⎢ 3 ⎩ 6 ⎝ 4 8 ⎠ −π π 2

= =

=

2 L2 ⎡θ 3 ⎛ θ 2 1 ⎞ θ cos 2θ ⎤ + ⎜⎜ − ⎟⎟ sin 2θ + ⎥ 3 ⎢ 4 ⎦ −π π ⎣ 6 ⎝ 4 8⎠ 2 L2 ⎡ π 3 ⎛ π 2 1 ⎞ π cos π + ⎜⎜ − ⎟⎟ sin π + 3 ⎢ 8 π ⎣ 48 ⎝ 16 8 ⎠ π3 ⎛π 2 1⎞ π cos π ⎤ + + ⎜⎜ − ⎟⎟ sin π + ⎥ 48 ⎝ 16 8 ⎠ 8 ⎦ 2 L2 ⎡ π 3 π ⎤ 1 ⎤ ⎡1 − ⎥ = L2 ⎢ − 2 ⎥ 3 ⎢ π ⎣ 24 4 ⎦ ⎣12 2π ⎦

2

2

Wave-Particle Duality and Quantum Physics 1123 Remarks: The result differs from that of Example 34-8. Since we have shifted the origin by ∆x = L/2, we could have arrived at the above result, without performing the integration, by subtracting (∆x)2 = L2/4 from x 2 as given in Example 34-8. 66 •• Picture the Problem For the first excited state, n = 2, and so we’ll evaluate

f ( x ) = ∫ f ( x )ψ 2 ( x ) dx using ψ 2 (x ) = Since ψ 22 ( x ) is an even function of

(x ) is an odd function of x. It follows that the integral of xψ 22 ( x ) x, xψ

2 2πx sin . L L x = 0

2 2

between −L/2 and L/2 is zero. Thus: Express x 2 :

Change variables by letting

θ=

2πx . Then: L

Substitute to obtain:

L2

x

2

2 2π = x 2 sin 2 xdx ∫ L −L 2 L

L θ, 2π 2π dθ = dx, and L L dx = dθ 2π and the limits on θ are −π and π. x=

π

x

2

2

2 ⎛ L ⎞ ⎛ L ⎞ = ∫ ⎜ θ ⎟ sin 2 θ ⎜ dθ ⎟ L −π ⎝ 2π ⎠ ⎝ 2π ⎠ π

L2 = θ 2 sin 2 θ dθ 3 ∫ 4π −π

1124 Chapter 34

Evaluate the integral by looking it up in the tables: π

x

2

θ cos 2θ ⎤ L2 ⎡θ 3 ⎛ θ 2 1 ⎞ = − ⎜⎜ − ⎟⎟ sin 2θ − ⎥ 3 ⎢ 4π ⎣ 6 ⎝ 4 8 ⎠ 4 ⎦ −π =

π cos 2π L2 ⎡ π 3 ⎛ π 2 1 ⎞ − ⎜⎜ − ⎟⎟ sin 2π − 3 ⎢ 4π ⎣ 6 ⎝ 4 8 ⎠ 4 +

π cos 2π ⎤ 1⎞ − ⎜⎜ − ⎟⎟ sin 2π − ⎥ 6 ⎝ 4 8⎠ 4 ⎦

π3 ⎛π 2

=

L2 ⎡ π 3 π π 3 π ⎤ − + − 4π 3 ⎢⎣ 6 4 6 4 ⎥⎦

=

L2 ⎡ π 3 π ⎤ 1 ⎤ ⎡1 − ⎥ = L2 ⎢ − 2 ⎥ 3 ⎢ 4π ⎣ 3 2 ⎦ ⎣12 8π ⎦

Remarks: The result differs from that of Example 34-8. Since we have shifted the origin by ∆x = L/2, we could have arrived at the above result, without performing the integration, by subtracting (∆x)2 = L2/4 from x 2 as given in Example 34-8.

General Problems *67 • Picture the Problem We can use the Einstein equation for photon energy to find the energy of each photon in the beam. The intensity of the energy incident on the surface is the ratio of the power delivered by the beam to its delivery time. Hence, we can express the energy incident on the surface in terms of the intensity of the beam. (a) Use the Einstein equation for photon energy to express the energy of each photon in the beam: Substitute numerical values and evaluate Ephoton:

Ephoton = hf =

(b) Relate the energy incident on a surface of area A to the intensity of the beam:

E = IA∆t

Ephoton =

hc

λ

1240 eV ⋅ nm = 3.10 eV 400 nm

Wave-Particle Duality and Quantum Physics 1125 Substitute numerical values and evaluate E:

(

)(

)

E = 100 W/m 2 10 −4 m 2 (1s ) 1 eV = 0.01 J × 1.60 × 10 −19 J = 6.25 × 1016 eV

(c) Express the number of photons striking this area in 1 s as the ratio of the total energy incident on the surface to the energy delivered by each photon: 68

N=

E Ephoton

=

6.25 × 1016 ev 3.10 eV

= 2.02 × 1016

•

Picture the Problem The particle’s nth-state energy is En = n 2

h2 . We can find n by 8mL2

solving this equation for n and substituting the particle’s kinetic energy for En.

h2 8mL2

Express the energy of the particle when it is in its nth state:

En = n 2

Solve for n:

n=

Express the energy (kinetic) of the particle:

En = 12 mv 2

Substitute to obtain:

n=

2mvL h

Substitute numerical values and evaluate n:

n=

2 10 −9 kg 10−3 m/s 10−2 m 6.63 × 10 −34 J ⋅ s

L 8mEn h

(

)(

)(

)

= 3.02 × 1019 ≈ 3 × 1019 69 • Picture the Problem We can use the fact that the uncertainties are given by ∆x/L = 0.01 percent and ∆p/p = 0.01 percent to find ∆x and ∆p. (a) Assuming that ∆x/L = 0.01 percent, find ∆x:

∆x = 10 −4 (L ) = 10 −4 (10 −2 m ) = 1.00 µm

1126 Chapter 34 Assuming that ∆p/p = 0.01 percent, find ∆p:

(b) Evaluate (∆x∆p)/ h :

(

)(

∆p = 10 −4 mv = 10 −4 10 −9 kg 10 −3 m/s

)

= 10 −16 kg ⋅ m/s

(

∆x∆p (1 µm ) 10 −16 kg ⋅ m/s = 1.054 × 10 −34 h

)

= 0.949 × 1012 70 • Picture the Problem We can estimate the number of emitted photons from the ratio of the total energy in the flash to the energy of a single photon.

E

Letting N be the number of emitted photons, express the ratio of the total energy in the flash to the energy of a single photon:

N=

Relate the energy in the flash to the power produced:

E = P∆t

Express the energy of a single photon as a function of its wavelength:

Ephoton =

Substitute to obtain:

N=

Substitute numerical values and evaluate N:

(5 ×10 W )(10 s )(400 ×10 m) N= (6.63 ×10 J ⋅ s)(3 ×10 m/s)

Ephoton

hc

λ

P∆tλ hc −12

15

−34

−9

8

= 1.01× 1022 71

•

Picture the Problem We can use the electron wavelength equation λ = K is in eV to find the minimum energy required to see an atom. Relate the energy of the electron to the size of an atom (the wavelength of the electron):

λ=

1.23 nm K

provided K is in eV.

1.23 nm, where K

Wave-Particle Duality and Quantum Physics 1127 Solve for K:

(1.23 eV K=

12

⋅ nm

λ2

Substitute numerical values and evaluate K:

(1.23 eV K=

⋅ nm (0.1 nm )2 12

)

2

)

2

= 151eV

72 • Picture the Problem The flea’s de Broglie wavelength is λ = h p , where p is the flea’s momentum immediately after takeoff. We can use a constant acceleration equation to find the flea’s speed and, hence, momentum immediately after takeoff.

h h = p mv0

Express the de Broglie wavelength of the flea immediately after takeoff:

λ=

Using a constant acceleration equation, express the height the flea can jump as a function of its takeoff speed:

v 2 = v02 + 2a∆y or, since v = 0 and a = −g,

Substitute to obtain:

λ=

Substitute numerical values and evaluate λ:

λ=

v0 = 2 g∆y h m 2 g∆y

(8 ×10

6.63 × 10 −34 J ⋅ s −6

) (

)

kg 2 9.81 m/s 2 (0.2 m )

= 4.18 × 10 −29 m *73 •• Picture the Problem We can relate the fraction of the photons entering the eye to ratio of the area of the pupil to the area of a sphere of radius R. We can find the number of photons emitted by the source from the rate at which it emits and the energy of each photon which we can find using the Einstein equation. Letting r be the radius of the pupil, Nentering eye the number of photons per second entering the eye, and Nemitted the number of photons emitted by the source per second, express the fraction of the light energy entering the eye at a distance R from the

N entering eye N emitted

=

Aeye 4πR 2

=

πr 2 4πR 2

=

r2 4R 2

1128 Chapter 34 source: Solve for R to obtain:

R=

r 2

N emitted N entering eye

Find the number of photons emitted by the source per second:

N emitted =

Using the Einstein equation, express the energy of the photons:

Ephoton =

Substitute numerical values and evaluate Ephoton:

Ephoton =

Substitute and evaluate Nemitted:

N emitted =

(1)

P Ephoton

hc

λ 1240 eV ⋅ nm = 2.07 eV 600 nm 100 W (2.07 eV ) 1.60 ×10−19 J/eV

(

)

= 3.02 × 1020 s −1 Substitute for Nemitted in equation (1) and evaluate R:

R=

3.5 mm 3.02 × 10 20 s −1 2 20 s −1

= 6.80 × 103 km 74 •• Picture the Problem The intensity of the light such that one photon per second passes through the pupil is the ratio of the energy of one photon to the product of the area of the pupil and time interval during which the photon passes through the pupil. We’ll use the Einstein equation to express the energy of the photon. Use its definition to relate the intensity of the light to the energy of a 600-nm photon:

I1 photon =

Using the Einstein equation, express the energy of a 600-nm photon:

E1 photon =

Substitute for E1 photon to obtain:

I1 photon =

P E1 photon = A A∆t hc

λ hc λA∆t

Wave-Particle Duality and Quantum Physics 1129 Substitute numerical values and evaluate I1 photon:

I1 photon =

(1240 eV ⋅ nm )(1.602 ×10−19 J/eV ) = (600 nm ) ⎡⎢ π (5 ×10−3 m )2 ⎤⎥(1s ) ⎣4

1.69 × 10−14 W/m 2

⎦

75 •• Picture the Problem We can find the intensity at a distance of 1.5 m directly from its definition. The number of photons striking the surface each second can be found from the ratio of the energy incident on the surface to the energy of a 650-nm photon. (a) Use its definition to express the intensity of the light as a function of distance from the light bulb:

I=

P P = A 4πR 2

Substitute numerical data to obtain:

I=

90 W = 3.18 W/m 2 2 4π (1.5 m )

(b) Express the number of photons per second that strike the surface as the ratio of the energy incident on the surface to the energy of a 650nm photon:

N=

where A is the area of the surface.

Use the Einstein equation to express the energy of the 650-nm photons:

E=

Substitute to obtain:

N=

IAλ hc

Substitute numerical values and evaluate N:

N=

(3.18 W/m )(10 m )(650 nm ) (1240 eV ⋅ nm ) (1.60 × 10 J/eV )

IA Ephoton

hc

λ

2

−4

2

−19

= 1.04 × 1015 76 •• Picture the Problem The maximum kinetic energy of the photoelectrons is related to the frequency of the incident photons and the work function of the cathode material through the Einstein equation. We can apply this equation to the two sets of data and solve the resulting equations simultaneously for the work function.

1130 Chapter 34 Using the Einstein equation, relate the maximum kinetic energy of the emitted electrons to the frequency of the incident photons and the work function of the cathode material:

K max = hf − φ

Substitute numerical data for the light of wavelength λ1:

1.8 eV =

Substitute numerical data for the light of wavelength λ1/2:

5.5 eV =

Solve these equations simultaneously for φ to obtain:

φ = 1.90 eV

=

hc

λ

−φ

hc

λ1

−φ

hc 2hc −φ = −φ λ1 2 λ1

77 •• Picture the Problem We can use the Einstein equation to express the energy of the scattered photon in terms of its wavelength and the Compton scattering equation to relate this wavelength to the scattering angle and the pre-scattering wavelength. Express the energy of the scattered photon E′ as a function of their wavelength λ′:

E' =

hc λ'

Express the wavelength of the scattered photon as a function of the scattering angle θ :

λ' =

h (1 − cos θ ) + λ me c

Substitute and simplify to obtain:

where λ is the wavelength of the incident photon.

E' =

hc h (1 − cos θ ) + λ me c hc

=

=

λ

hc (1 − cos θ ) + 1 me c 2 λ E E (1 − cos θ ) + 1 me c 2

Wave-Particle Duality and Quantum Physics 1131 78 •• Picture the Problem While we can work with either of the transitions described in the problem statement, we’ll use the first transition in which radiation of wavelength 114.8 nm is emitted. We can express the energy released in the transition in terms of the difference between the energies in the two states and solve the resulting equation for n. Express the energy of the emitted radiation as the particle goes from the nth to n – 1 state:

∆E = En − En −1

Express the energy of the particle in nth state:

En = n 2 E1

Express the energy of the particle in the n – 1 state:

En −1 = (n − 1) E1

Substitute and simplify to obtain:

∆E = n 2 E1 − (n − 1) E1

2

2

= (2n − 1)E1 =

hc

λ

Solve for n:

n=

hc 1 + 2λE1 2

Substitute numerical values and evaluate n:

n=

1240 eV ⋅ nm 1 + = 5 2(114.8 nm )(1.2 eV ) 2

*79 •• Picture the Problem We can use the expression for the energy of a particle in a well to find the energy of the most energetic electron in the uranium atom. Relate the energy of an electron in ⎛ h2 ⎞ ⎟ En = n 2 ⎜⎜ 2 ⎟ the uranium atom to its quantum ⎝ 8mL ⎠ number n: Substitute numerical values and evaluate E92: 2 ⎡ 6.63 ×10 −34 J ⋅ s ) 1eV ⎤ ( E92 = (92) ⎢ = 1.28 MeV × 2 −19 ⎥ −31 ⎢⎣ 8(9.11× 10 kg )(0.05 nm ) 1.6 × 10 J ⎥⎦ 2

The rest energy of an electron is:

1132 Chapter 34 ⎞ 2⎛ 1 eV ⎟ = 0.512 MeV me c 2 = (9.11 × 10 −31 kg )(3 × 108 m/s ) ⎜⎜ −19 ⎟ ⎝ 1.6 × 10 J ⎠ Express the ratio of E92 to mec2:

E92 1.28 MeV = = 2.50 2 me c 0.512 MeV

The energy of the most energetic electron is approximately 2.5 times the rest - mass energy of an electron. 80 •• Picture the Problem We can express the kinetic energy of an electron in the beam in terms of its momentum. We can use the de Broglie relationship to relate the electron’s momentum to its wavelength and use the condition for constructive interference to find λ.

Express the kinetic energy of an electron in terms of its momentum:

K=

p2 2m

(1)

h

Using the de Broglie relationship, relate the momentum of an electron to its momentum:

p=

Substitute for p in equation (1) to obtain:

h2 K= 2mλ2

The condition for constructive interference is:

d sin θ = nλ

λ

(2)

where d is the slit separation and n = 0, 1, 2, …

d sin θ n

Solve for λ:

λ=

For θ << 1, sinθ is also given by:

sin θ ≈

∆y L

Wave-Particle Duality and Quantum Physics 1133 Substitute for sinθ to obtain:

Substitute for λ in equation (2) to obtain:

λ=

d∆y nL

K=

n 2 L2 h 2 2 2md 2 (∆y )

Substitute numerical values (n = 1) and evaluate K: 2 2 2 ( 1) (1.5 m ) (6.63 × 10 −34 J ⋅ s ) 1 eV K= = 2 2 −19 −31 2(9.11 × 10 kg )(54 nm ) (0.68 mm ) 1.6 × 10 J

2.52 keV

81 •• Picture the Problem The maximum kinetic energy of the photoelectrons is related to the frequency of the incident photons and the work function of the illuminated surface through the Einstein equation. We can apply this equation to either set of data and solve the resulting equations simultaneously for the work function of the surface and the wavelength of the incident photons. Using the Einstein equation, relate the maximum kinetic energy of the emitted electrons to the frequency of the incident photons and the work function of the cathode material:

K max = hf − φ

Substitute numerical data for the light of wavelength λ:

1.2 eV =

Substitute numerical values for the light of wavelength λ′:

1.76 eV =

Solve these equations simultaneously for φ to obtain:

φ = 1.04 eV

Substitute in either of the equations and solve for λ:

λ = 554 nm

=

hc

λ

−φ

hc

λ

−φ

hc hc −φ = −φ λ' 0.8λ

1134 Chapter 34 82 •• Picture the Problem The diagram shows the pendulum with an angular displacement θ. The energy of the oscillator is equal to its initial potential energy mgh = mgL(1 − cosθ). We can find n by equating this initial energy to En = (n + 12 )hf 0 and solving for n. In part (b) we’ll express the ratio of ∆En to En and solve for ∆n. (a) Express the nth-state energy as a function of the frequency of the pendulum:

h 2π

g L

h 2π

g L

En = (n + 12 )hf 0 = (n + 12 )

En = mgL(1− cos θ )

Express the energy of the pendulum: Substitute to obtain:

mgL(1 − cos θ ) = (n + 12 )

Solve for n:

2πm g L3 2 (1 − cosθ ) 1 n= − h 2

Substitute numerical values and evaluate n:

2π (0.3 kg ) 9.81 m/s 2 (1m ) (1 − cos10°) 1 − = 1.35 × 1032 −34 6.63 × 10 J ⋅ s 2 32

n=

(b) Express the ratio of ∆En to En:

∆En (n + ∆n + 12 )hf 0 − (n + 12 )hf 0 = (n + 12 )hf 0 En =

Solve for and evaluate ∆n:

∆n = 10−4 1 n+ 2

∆n = 10 −4 (n + 12 ) ≈ 10 −4 n = 1.35 × 10 28

*83 •• Picture the Problem We can use the fact that the energy of the nth state is related to the energy of the ground state according to En = n 2 E1 to express the fractional change in energy in terms of n and then examine this ratio as n grows without bound.

Wave-Particle Duality and Quantum Physics 1135 En+1 − En (n + 1) − n 2 2n + 1 = = En n2 n2 2

(a) Express the ratio (En + 1 − En)/En:

=

2 1 2 + 2 ≈ n n n

for n >> 1. (b) Evaluate

(c)

E1001 − E1000 : E1000

2 E1001 − E1000 ≈ = 0.2% 1000 E1000

Classically, the energy is continuous. For very large values of n, the energy difference between adjacent levels is infinitesimal.

84 •• Picture the Problem We can apply the definition of power in conjunction with the de Broglie equation for the energy of a photon to derive an expression for the average power produced by the laser.

∆E ∆t

The average power produced by the laser is:

P=

Use the de Broglie equation to express the energy of the emitted photons:

∆E = Nhf =

Nhc

λ

where N is number of photons in each pulse.

Substitute for ∆E to obtain:

P=

Nhc λ∆t

Substitute numerical values and evaluate P:

P=

(5 ×10 )(6.63 ×10 9

J ⋅ s )(3 × 108 m/s ) = 117 mW (850 nm )(10−8 s ) −34

Remarks: Note that the pulse length has no bearing on the solution. 85 •• Picture the Problem We can find the rate at which energy is delivered to the atom using the definitions of power and intensity. We can also use the definition of power to determine how much time is required for an amount of energy equal to the work function to fall on one atom.

1136 Chapter 34 (a) Relate the energy per second (power) falling on an atom to the intensity of the incident radiation: Substitute numerical values and evaluate P:

P=

∆E = IA ∆t

P = (0.01 W/m 2 )(0.01 × 10 −18 m 2 ) J 1 eV = 10 −22 × s 1.60 × 10 −19 J = 6.25 × 10 −4 eV/s

(b) Classically:

∆t =

∆E φ = P P

Substitute numerical values and evaluate ∆t:

∆t =

2 eV = 3200 s 6.25 × 10 −4 eV/s

= 53.3 min

Chapter 35 Applications of the Schrödinger Equation Conceptual Problems 1 • True 2 • Determine the Concept Looking at the graphs in the text for the n = 1, 2, and 3 states, we note that the n = 4 state graph of the wave function must have four extrema in the region 0 < x < L and decay in toward zero in the regions x < 0 and x > L. (a)

(b)

3 • Determine the Concept Looking at the graphs in the text for the n = 1, 2, and 3 states, we note that the n = 5 state graph of the wave function must have five extrema in the region 0 < x < L and decay in toward zero in the regions x < 0 and x > L. (a)

(b)

1137

1138

Chapter 35

Estimation and Approximation *4 • Picture the Problem Assume a mass of 150 g for the baseball, 30 cm for the width of the locker, and 1 cm/s for the speed of the ball, and equate the kinetic energy of the ball and the quantum-mechanical energy and solve for the quantum number n. The allowed energy states of a particle of mass m in a 1-dimensional infinite potential well of width L are given by:

⎛ h2 ⎞ ⎟ En = n ⎜⎜ 2 ⎟ ⎝ 8mL ⎠

The kinetic energy of the ball is:

K=

For En = K:

⎛ h2 ⎞ 1 2 ⎟ = mv n 2 ⎜⎜ 2 ⎟ ⎝ 8mL ⎠ 2

Solve for the quantum number n:

n=

2mvL h

Substitute numerical values and evaluate n:

n=

2(0.15 kg )(0.01 m/s )(0.3 m ) 6.63 × 10 −34 J ⋅ s

2

1 2 mv 2

= 1.36 × 1030 ≈ 1030

The Schrödinger Equation 5 •• Picture the Problem We can show that ψ3(x) is a solution to the time-independent Schrödinger equation by differentiating it twice and substituting in Equation 35-4.

h 2 d 2ψ ( x ) + U (x )ψ (x ) = Eψ ( x ) 2m dx 2

Equation 35-4 is:

−

Because ψ1 (x) and ψ 2 (x) are

h 2 d 2ψ 1 (x ) − + U ( x )ψ 1 ( x ) = Eψ 1 (x ) 2m dx 2

solutions of Equation 35-4:

and

h 2 d 2ψ 2 ( x ) − + U ( x )ψ 2 ( x ) = Eψ 2 ( x ) 2m dx 2

Applications of the Schrödinger Equation 1139 Add these equations to obtain:

−

h 2 ⎡ d 2ψ 1 ( x) d 2ψ 2 ( x) ⎤ + + U (x )[ψ 1 ( x) + ψ 2 ( x)] = E [ψ 1 ( x) + ψ 2 ( x)] (1) 2m ⎢⎣ dx 2 dx 2 ⎥⎦

Differentiate ψ 3 ( x) = ψ 1 ( x) + ψ 2 ( x) twice with respect to x to obtain:

dψ 3 ( x) dψ 1 ( x) dψ 2 ( x) = + dx dx dx and

d 2ψ 3 ( x) d 2ψ 1 ( x) d 2ψ 2 ( x) = + dx 2 dx 2 dx 2 Substitute in equation (1) to obtain:

h 2 d 2ψ 3 ( x) + U ( x )ψ 3 ( x) = Eψ 3 ( x) 2m dx 2 which shows that ψ 3 ( x) = ψ 1 ( x) + ψ 2 ( x) satisfies Equation 35-4.

−

The Harmonic Oscillator 6 •• Picture the Problem We can relate the spring constant to the mass of the hydrogen atom and its angular frequency and then use the relationship between the allowed energy levels and the angular frequency ω to derive an expression for the spring constant k. The spring constant k is related to the mass m of the hydrogen molecule and its angular frequency ω:

k = mω 2

Relate the energy spacing ∆E to the angular frequency ω:

∆E = hf =

Solve for ω:

ω=

Substitute for ω in equation (1) to obtain:

⎛ ∆E ⎞ k = m⎜ ⎟ ⎝ h ⎠

Substitute numerical values and evaluate k:

(1)

hω = hω 2π

∆E h 2

1140 Chapter 35 2

⎛ 1.66 × 10 −27 kg ⎞ ⎛ 8.7 × 10 −20 J ⎞ ⎟⎟ ⎜⎜ ⎟⎟ = 1.14 kN/m k = ⎜⎜1 u × −34 u ⎝ ⎠ ⎝ 1.05 × 10 J ⋅ s ⎠ Remarks: Our result is very similar to the stiffness constant of typical macroscopic springs. Note that strictly speaking one should use the reduced mass of a hydrogen molecule rather than the simpler model of a single atom attached to a fixed point. 7

•• 2

∫

Determine the Concept The integral x = xψ dx = 0 because the integrand is an odd function of x for the ground state as well as any excited state of the harmonic oscillator. *8 •• Picture the Problem We can differentiate ψ(x) twice and substitute in the Schrödinger equation for the harmonic oscillator. Substitution of the given value for a will lead us to an expression for E1.

ψ 1 ( x ) = A1 xe − ax

The wave function for the first excited state of the harmonic oscillator is:

2

[

]

2 2 dψ 1 ( x ) d = A1 xe −ax = A1e −ax dx dx

Compute dψ1(x)/dx:

Compute d2ψ1(x)/dx2:

[

]

d 2ψ 1 ( x ) d −ax 2 − ax 2 −ax 2 −ax 2 2 3 = A e = − 2 axA e − 4 axA e + 4 a x A e 1 1 1 1 dx 2 dx = (4a 2 x 3 − 6ax )A1e −ax

2

Substitute in the Schrödinger equation:

[

]

2 2 2 h2 − ( 4a 2 x 3 − 6ax )A1e −ax + 12 mω02 x 2 A1 xe −ax = E1 A1 xe −ax 2m

Divide out A1e − ax to obtain: 2

− or

[(

)]

h2 4a 2 x 3 − 6ax + 12 mω02 x 3 = E1 x 2m

Applications of the Schrödinger Equation 1141 −

(

)

h2 h2 (6ax ) + 12 mω02 x 3 = E1 x 4a 2 x 3 + 2m 2m

Substitute for a to obtain:

h 2 ⎛ mω0 ⎞ 3 h 2 ⎛ mω0 ⎞ 2 3 4⎜ 6⎜ ⎟ x + 12 mω0 x = E1 x ⎟ x + 2m ⎝ 2h ⎠ 2m ⎝ 2h ⎠ 2

−

Solve for E1 to obtain:

E1 =

3 2

hω0 = 3E0

9 ••• Picture the Problem We must show that, with A0 = (2mω0/h)1/4, the normalization 2

∞

condition

∞

∫ ψ (x ) dx = ∫ A e 0

2

− ax 2

0

−∞

dx = 1 is satisfied.

−∞

∞

We need to show that:

∫ ψ (x )

2

dx =

0

−∞

∞

With

⎛ 2mω0 ⎞ A0 = ⎜ ⎟ ⎝ h ⎠

14

14

⎛ mω0 ⎞ ⎟⎟ , the = ⎜⎜ ⎝ πh ⎠

∫ ψ (x )

2

∞

∫ Ae

− ax 2

0

dx = 1

−∞

2

0

−∞

12 ∞

⎛ mω0 ⎞ ⎟⎟ dx = ⎜⎜ ⎝ πh ⎠

∫e

− 2 ax 2

dx

−∞

normalization condition becomes: In Example 35-1 it is shown that:

a=

Substitute to obtain:

∞

mω0 2h

∫ ψ 0 (x )

2

−∞

⎛ mω0 ⎞ Let s = ⎜ ⎟ x. Then: ⎝ h ⎠ 12

12 ∞

⎛ mω0 ⎞ ⎟⎟ dx = ⎜⎜ ⎝ πh ⎠

∫e

−∞

⎛ mω0 ⎞ ds = ⎜ ⎟ dx ⎝ h ⎠ 12

and

⎛ mω0 ⎞ dx = ⎜ ⎟ ⎝ h ⎠ Substitute for dx and simplify to obtain:

∞

2

−1 2

∫ ψ 0 (x ) dx =

−∞

ds

1

π

∞

∫e

−∞

−s2

ds

−

mω0 2 x h

dx

1142 Chapter 35 From integral tables (see Table D-5):

∞

∫e

−s2

ds = π

−∞

∞

Therefore:

∫ ψ (x )

2

0

−∞

10

⎛ 2mω0 ⎞ dx = 1 provided A0 = ⎜ ⎟ ⎝ h ⎠

14

•••

Picture the Problem We are required to evaluate x 2 = x 2 ψ 0 ( x ) with 2

∫

ψ 0 ( x ) = A0 e − ax , where a = 2

mω0 . We can then use U av = 12 mω02 x 2 to find the 2h

average potential energy of the harmonic oscillator. We need to evaluate:

+∞

x

∫x

=

2

2

ψ dx 2

−∞

For the ground state of the harmonic oscillator: Substitute for ψ

2

to obtain:

ψ

2

= A02 e −2 ax

2

+∞

x 2 = A02 ∫ x 2 e −2 ax dx 2

−∞

+∞

= 2A

2 0

∫x e

2 − 2 ax 2

dx

0

Use the appropriate integral from the inside of the back cover of the text to obtain: The normalization condition is:

x 2 = 2 A02

1 4

π

(2a )

2

∫ψ

2

dx = A

2 0

∫e

− 2 ax

−∞

−∞

+∞

= 2A

∫e

2 0

− 2 ax 2

dx

0

Again, use the appropriate integral from the inside of the back cover of the text to obtain: Solve for A02 to obtain:

1 = 2 A02

A02 =

π

A02 4a

2a

+∞

+∞

1=

=

π 1 π = A02 2 2a 2a

2a

π

2

dx

(1)

Applications of the Schrödinger Equation 1143 Substitute in equation (1) to obtain: From Example 36-1:

a=

Substitute for a in equation (2) to obtain:

1 4a

x2 =

2a

π

π

2a

=

(2)

mω0 2h h 2h = 4mω0 2mω0

x2 =

The average potential energy of the oscillator is:

U av = 12 mω02 x 2

Substitute for x 2 and simplify:

⎛ h ⎞ ⎟⎟ = U av = 12 mω02 ⎜⎜ ⎝ 2mω0 ⎠ = 12 E0

11

1 4a

1 4

hω 0

••

x 2 from Problem 10 and the

Picture the Problem We can combine the result for

x2 − x

result for x from Problem 7 to obtain an expression for

energy of the electron in an infinite potential well is given by E1 =

(a) From Problem 10 we have:

The ground-state energy is given by: Solve for ω0 to obtain:

h 2mω0

x2 =

x =0

obtain:

2

to

(1)

h2 4mE0

From Problem 7 we have:

x2 − x

h2 . 8mL2

1 hω0 2 2E ω0 = 0 h x2 =

the expression

. The lowest

E0 =

Substitute in equation (1) and simplify to obtain:

Substitute equations (1) and (2) in

2

x

2

− x

(2) 2

=

h2 h 1 −0 = 4mE0 2 mE0

1144 Chapter 35 x2 − x

Substitute numerical values and evaluate

x2 − x

2

=

1.05 × 10 −34 J ⋅ s 2

2

:

1

(9.11× 10

−31

⎛ 1.6 × 10 −19 J ⎞ ⎟⎟ kg ⎜⎜ 2.1 × 10 −4 eV × eV ⎠ ⎝

)

= 9.49 nm

h2 E1 = 8mL2

(b) The lowest energy of an electron trapped in an infinite potential well is: Letting L =

x2 − x

2

E1 =

yields:

h2 8m

x2 − x

2

Substitute numerical values and evaluate E1:

(6.63 × 10 J ⋅ s ) = 8(9.11 × 10 kg )(9.49 × 10 −34

E1

−31

2

1 eV = 4.19 meV −19 −9 m 1.6 × 10 J

)

2

12 ••• Picture the Problem We can begin by equating the average kinetic energy of the harmonic oscillator and its average potential energy and solving for p 2 and then evaluating and substituting for

x2 .

According to the problem statement:

p2

2m Solve for p 2 :

=

1 k x2 2

p 2 = mk x 2 or, because ω02 =

k , m

p 2 = m 2ω02 x 2 We need to evaluate:

2

(1) ∞

x 2 = ∫ x 2 ψ = A02 ∫ x 2 e −2 ax dx mω0 where a = 2h

2

−∞

(2)

Applications of the Schrödinger Equation 1145 Let y2 = 2ax2. Then x 2 =

y2 and: 2a

dx =

ydy 2ax

With appropriate substitutions, the integral becomes: ∞

∞

∞

1 y 2 − y 2 ydy y 3 − y 2 2a dy e = ∫−∞ 2a 2ax −∫∞ 2a e 2ay = (2a )3 2 From integral tables (see Table D5): In Problem 35-9 it was given that:

Substitute in equation (2) to obtain:

Substitute for a and simplify:

Substitute for x 2 in equation (1):

∞

∫y e

2 − y2

dy =

−∞

∫y e

2 − y2

dy

−∞

1 π 2

⎛ 2mω0 ⎞ A0 = ⎜ ⎟ ⎝ h ⎠

14

1 ⎛ 2mω0 ⎞ ⎛ 1 ⎞ π⎟ = ⎟ ⎜ 32 ⎜ (2a ) ⎝ h ⎠ ⎝ 2 ⎠ 12

x

2

x

2

=

⎛ mω0 ⎞ ⎜ ⎟ ⎝ h ⎠ h = 2mω0

⎛ 2mω0 ⎞ ⎛ 1 ⎞ π⎟ ⎜ ⎟ ⎜ ⎠ ⎝ h ⎠ ⎝2 12

1 32

⎛ h ⎞ ⎟⎟ = p 2 = m 2ω02 ⎜⎜ 2 ω m 0 ⎠ ⎝

1 2

hmω0

13 ••• Picture the Problem We can use the definition of the standard deviation of ∆x and ∆p and the results of Problems 7, 10, and 12 to determine the uncertainty product ∆x ∆p for the ground state of the harmonic oscillator. Express the standard deviation of ∆p (see Equation 17-35a):

Because pav = 0:

(∆p )2 = [( p − pav )2 ]av

[

= p 2 − 2 ppav − p av2

(∆p )2 = ( p 2 )av

]

av

1146 Chapter 35 Express the standard deviation of ∆x (see Equation 17-35a):

(∆x )2 = [(x − xav )2 ]av

[

= x 2 − 2 xxav − x av2

]

av

(∆x )2 = (x 2 )av

Because xav = 0: We have, from Problems 10 and 12, for the ground state of the harmonic oscillator:

x 2 = (∆x ) = 2

and

p 2 = (∆p ) = 2

Express the product of (∆x ) and 2

(∆p )2 :

Take the square root of both sides of the equation to obtain:

h 2mω0 hmω0 2

⎛

h ⎞ ⎛ hmω0 ⎞ h 2 ⎟⎟ ⎜ ⎟= ⎝ 2mω0 ⎠ ⎝ 2 ⎠ 4

(∆x )2 (∆p )2 = ⎜⎜ ∆x∆p =

h 2

Reflection and Transmission of Electron Waves: Barrier Penetration *14 •• Picture the Problem We can use the total energy of the particle in the region x > 0 to express k2 in terms of α and k1. Knowing k2 in terms of k1, we can use 2 ( k1 − k 2 ) R= (k1 + k 2 )2

to find R and T = 1 − R to determine the transmission coefficient T.

(a) Using conservation of energy, express the energy of the particle in the region x > 0: Solve for k2:

From the equation for the total energy of the particle:

h 2 k 22 + U 0 = αU 0 2m

k2 =

2mU 0 (α − 1) h

k1 =

2mαU 0 h

Applications of the Schrödinger Equation 1147 Express the ratio of k2 to k1:

k2 = k1

2mU 0 (α − 1) α −1 h = α 2mαU 0 h

and k 2 =

(b) The reflection coefficient R is given by:

R=

α −1 k α 1

(k1 − k 2 )2 (k1 + k 2 )2 2

Factor k1 from the numerator and denominator to obtain:

⎛ k2 ⎞ ⎜⎜1 − ⎟⎟ k1 ⎠ R=⎝ 2 ⎛ k2 ⎞ ⎜⎜1 + ⎟⎟ k1 ⎠ ⎝

Substitute our result from (a) for k2/k1:

2 ⎛ α − 1 ⎞⎟ ⎛ ⎞ − α 1 ⎜1 − ⎜ ⎟ ⎜ α ⎟⎠ ⎜ 1 − α ⎟ ⎝ = R= 2 ⎜ ⎟ ⎛ ⎞ α −1 ⎟ ⎜1+ α −1 ⎟ ⎜1 + ⎜ α ⎠ α ⎟⎠ ⎝ ⎝

The transmission coefficient is given by:

⎛ ⎞ ⎜1− α −1 ⎟ α ⎟ T = 1− R = 1− ⎜ ⎜ α −1 ⎟ ⎜1+ ⎟ α ⎠ ⎝

2

2

A spreadsheet program to plot R and T as functions of α is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A2 B2

Content/Formula 1.0 (1−SQRT((A2−1)/A2))/ (1+SQRT((A2−1)/A2))^2

Algebraic Form

α

⎛ α − 1 ⎞⎟ ⎜1− α ⎟ ⎜ ⎜ α −1 ⎟ ⎜1+ ⎟ α ⎠ ⎝

2

1148 Chapter 35 C2

1−B2

⎛ ⎞ ⎜1− α −1 ⎟ α ⎟ 1− ⎜ ⎜ α −1 ⎟ ⎜1+ ⎟ α ⎠ ⎝

1 2 3 4 5

A alpha 1.0 1.2 1.4 1.6

B R 1.000 0.298 0.198 0.149

C T 0.000 0.702 0.802 0.851

18 19 20 21 22

4.2 4.4 4.6 4.8 5.0

0.036 0.034 0.032 0.031 0.029

0.964 0.966 0.968 0.969 0.971

2

The following graph was plotted using the data in the above table:

1.0

0.8

0.6

R T

0.4

0.2

0.0 1

2

2

3

3

4

4

5

5

alpha

15 •• Picture the Problem We can use the total energy of the particle in the region x > 0 to find k2. Knowing k2, we can use R = determine the transmission coefficient T.

(k1 − k 2 )2 (k1 + k 2 )2

to find R and T = 1 − R to

Applications of the Schrödinger Equation 1149 (a) Using conservation of energy, express the particle in the region x > 0: Solve for k2:

From the equation for the total energy of the particle:

h 2 k 22 − U 0 = 2U 0 2m

k2 =

6mU 0 h

k1 =

4mU 0 h

Express the ratio of k2 to k1:

k2 = k1

(b) The reflection coefficient R is given by:

Substitute for k2/k1 and evaluate R:

6mU 0 h = 4mU 0 h

3 ⇒ k2 = 2

3 k1 2

2

R=

⎛ k2 ⎞ ⎜⎜1 − ⎟⎟ k1 ⎠ =⎝ 2 ⎛ k2 ⎞ ⎜⎜1 + ⎟⎟ k1 ⎠ ⎝

(k1 − k 2 )2 (k1 + k 2 )2

⎛ ⎜1 − ⎜ R=⎝ ⎛ ⎜1 + ⎜ ⎝

3⎞ ⎟ 2 ⎟⎠

2

3⎞ ⎟ 2 ⎟⎠

2

= 0.0102

(c) Because R + T = 1:

T = 1 − R = 1 − 0.0102 = 0.990

(d) If we let N0 represent the number of particles incident upon the potential step, then the number that continue beyond is:

N 0T = 10 6 × 0.990 = 9.90 × 105

Classically, all 10 6 would continue to move past the potential step.

16 •• Picture the Problem We can use the energies in the regions U = 0 and U = U0 to express the ratio of the potential energy to the total energy in terms of the ratio of the wave numbers. We can also express this ratio in terms of the reflection coefficient R to obtain an expression for the ratio of E to U in terms of R.

1150 Chapter 35 In the region U = 0:

In the region U = U0:

E=

h 2 k12 ⇒ k1 = 2m

E −U0 =

k r= 2 = k1

2 m( E − U 0 ) U h2 = 1− 0 E 2mE 2 h

U =1− r2 E

k2 : k1

Solve for r to obtain:

Substitute for r in equation (1):

(1)

2

Write the reflection coefficient R as a function of r =

2 m(E − U 0 ) h2

h 2 k 22 ⇒ k2 = 2m

Let r equal the ratio of k2 to k1:

Letting U0 = U, solve for U/E:

2mE h2

2 ( k1 − k 2 ) R= (k1 + k 2 )2

r=

⎛ k2 ⎞ ⎜⎜1 − ⎟⎟ 2 k1 ⎠ ( 1− r) ⎝ = = 2 (1 + r )2 ⎛ k2 ⎞ ⎜⎜1 + ⎟⎟ k1 ⎠ ⎝

1− R 1+ R

⎛1− R ⎞ U ⎟ = 1 − ⎜⎜ ⎟ E ⎝1+ R ⎠

2

and 2 E ⎡ ⎛1− R ⎞ ⎤ ⎟ ⎥ = ⎢1 − ⎜ U ⎢ ⎜⎝ 1 + R ⎟⎠ ⎥ ⎣ ⎦

Substitute a numerical value for R and evaluate E/U:

−1

−1

2 E ⎡ ⎛ 1 − 0.5 ⎞ ⎤ ⎟ ⎥ = 1.03 = ⎢1 − ⎜ U ⎢ ⎜⎝ 1 + 0.5 ⎟⎠ ⎥ ⎣ ⎦

Applications of the Schrödinger Equation 1151 17 •• Picture the Problem The probability that a proton will tunnel out of a nucleus in one collision with a nuclear barrier if it has a given energy is given by Equation 35-29.

T = e −2αa

Equation 35-29 is:

where

Multiply the numerator and denominator of α by c to obtain:

α=

2m(U 0 − E ) = h2

α=

2mc 2 (U 0 − E ) hc

2m(U 0 − E ) h

where

hc = 1.974 × 10 −13 MeV ⋅ m Using mp c 2 = 938 MeV, evaluate T:

⎧⎪ 2(938 MeV )(6 MeV ) ⎫⎪ T = exp⎨− 2 10 −15 m ⎬ = 0.341 1.974 × 10 −13 MeV ⋅ m ⎪⎭ ⎪⎩

(

)

*18 •• Picture the Problem The probability that the electron with a given energy will tunnel through the given barrier is given by Equation 35-29.

T = e −2αa

(a) Equation 35-29 is:

where

Multiply the numerator and denominator of α by c to obtain:

α=

2m(U 0 − E ) = h2

α=

2mc 2 (U 0 − E ) hc

2m(U 0 − E ) h

where Using me c 2 = 511 keV, evaluate T:

⎧⎪ T = exp⎨− 2 10 −9 m ⎪⎩

(

(b) Repeat with a = 10−10 m:

hc = 1.974 × 10 −13 MeV ⋅ m

keV )(25 eV − 10 eV ) ⎫⎪ ) 2(1511 ⎬= .974 × 10 MeV ⋅ m −13

⎪⎭

5.91 × 10 −18

1152 Chapter 35

⎧⎪ 2(511 keV )(25 eV − 10 eV ) ⎫⎪ −2 T = exp⎨− 2(10 −10 m ) ⎬ = 1.89 × 10 −13 1.974 × 10 MeV ⋅ m ⎪⎭ ⎪⎩ 19 ••• Picture the Problems We can find the distance of closest approach by equating the kinetic energy of the alpha particle and the Coulomb potential energy. The probability that the electron with a given energy will tunnel through the given barrier is given by T = e −2αa , where α is the transmission coefficient and depends on ∆E.

k 2eZe r1

(a) The distance of closest approach is related to the kinetic energy E of the alpha particles:

E=

Solve for r1:

2kZe 2 r1 = E

For E = 4 MeV:

r1, 4 MeV

(

)

(

)

(

)

(

)

2 8.99 × 109 N ⋅ m 2 / C 2 (92) 1.6 × 10 −19 C = 1.6 × 10 −19 J 4 MeV × eV

2

= 6.62 × 10 −14 m

For K = 7 MeV:

r1, 7 MeV

2 8.99 × 109 N ⋅ m 2 / C 2 (92) 1.6 × 10−19 C = = 3.78 × 10−14 m −19 1.6 × 10 J 7 MeV × eV

(b) The transmission coefficient T is given by:

2

T = e −2αa

(1)

where

α=

2m∆E = h2

2m∆E h

Evaluate α4 MeV for ∆E = 4 MeV:

α 4 MeV =

⎛ 1.66 × 10− 27 kg ⎞ ⎛ 1.6 × 10−19 J ⎞ ⎟⎟ ⎟⎟ ⎜⎜ 4 MeV × 2⎜⎜ 4 u × u eV ⎠ ⎝ ⎠⎝ = 8.78 × 1014 m −1 − 34 1.05 × 10 J ⋅ s

Applications of the Schrödinger Equation 1153 Evaluate α7 MeV for ∆E = 7 MeV:

α 7 MeV =

⎛ 1.66 × 10− 27 kg ⎞ ⎛ 1.6 × 10−19 J ⎞ ⎟⎟ ⎟⎟ ⎜⎜ 7 MeV × 2⎜⎜ 4 u × u eV ⎠ ⎝ ⎠⎝ = 1.16 × 1015 m −1 − 34 1.05 × 10 J ⋅ s T4 MeV = e −2 (8.78×10

Substitute numerical values in equation (1) and evaluate T4 MeV:

14

)(

m −1 6.62×10 −14 m

)

= 3.27 × 10− 51 T7 MeV = e −2 (1.16×10

15

Substitute numerical values in equation (1) and evaluate T7 MeV:

)(

m −1 3.78×10 −14 m

)

= 8.21 × 10 −39

The Schrödinger Equation in Three Dimensions 20

•

Picture the Problem We can use En1 ,n2 ,n3

h 2π 2 = 2m

⎛ n12 n22 n32 ⎞ ⎜⎜ 2 + 2 + 2 ⎟⎟ with the given sides ⎝ L1 L2 L3 ⎠

of the box to find the quantum numbers n1, n2, n3 that correspond to the lowest ten quantum states of this box. The energies of the quantum states are given by Equation 35-34: For a box with sides L1, L2 = 2L1, and L3 = 3L1:

En1 ,n2 ,n3 =

h 2π 2 2m

⎛ n12 n22 n32 ⎞ ⎜⎜ 2 + 2 + 2 ⎟⎟ ⎝ L1 L2 L3 ⎠

⎛ n12 n22 n2 ⎞ ⎜⎜ 2 + 2 + 32 ⎟⎟ ⎝ L1 4 L1 9 L1 ⎠ h 2 ⎛ 2 n22 n32 ⎞ ⎜ n1 + = + ⎟⎟ 8mL12 ⎜⎝ 4 9 ⎠ h2 = 36n12 + 9n22 + n32 2 288mL1

En1 ,n2 ,n3 =

h 2π 2 2m

(

The energies in units of

h2 are listed in the following table: 288mL12 n1 n2 n3 1 1 1 1 1 2

E 49 61

)

1154 Chapter 35 1 1 1 1 1 1 1 1

2 1 2 2 1 3 3 2

1 3 2 3 4 1 2 4

76 81 88 108 109 121 133 136

21 • Picture the Problem The wave functions are of the form

⎛ n1π ⎝ L1

ψ = A sin ⎜⎜ 22

⎞ ⎛n π ⎞ ⎛nπ x ⎟⎟ sin ⎜⎜ 2 y ⎟⎟ sin ⎜⎜ 3 ⎠ ⎝ 2 L1 ⎠ ⎝ 3L1

⎞ z ⎟⎟ ⎠

•

Picture the Problem We can use En1 ,n2 ,n3 =

h 2π 2 2m

⎛ n12 n22 n32 ⎞ ⎜⎜ 2 + 2 + 2 ⎟⎟ with the given sides ⎝ L1 L2 L3 ⎠

of the box to find the quantum numbers n1, n2, n3 that correspond to the lowest ten quantum states of this box. (a) The energies of the quantum states are given by Equation 35-34: For a box with sides L1, L2 = 2L1, and L3 = 4L1:

En1 ,n2 ,n3

h 2π 2 = 2m

En1 ,n2 ,n3 =

h 2π 2 2m

h2 = 8mL12 =

The energies in units of

⎛ n12 n22 n32 ⎞ ⎜⎜ 2 + 2 + 2 ⎟⎟ ⎝ L1 L2 L3 ⎠

⎛ n12 n22 n2 ⎞ ⎜⎜ 2 + 2 + 3 2 ⎟⎟ ⎝ L1 4 L1 16 L1 ⎠ ⎛ 2 n22 n32 ⎞ ⎜⎜ n1 + + ⎟ 4 16 ⎟⎠ ⎝

(

h2 16n12 + 4n22 + n32 128mL12

h2 are listed in the following table: 128mL12 n1 n2 n3 1 1 1 1 1 2 1 1 3 1 2 1

E 21 24 29 33

)

Applications of the Schrödinger Equation 1155 1 1 1 1 1 1 1 1

1 2 2 1 2 3 1 3

4 2 3 5 4 1 6 2

36 36 41 45 48 53 56 56

(1,1,4) and(1,2,2 )

Referring to the table, we see that there are two degenerate levels:

and

(1,1,6) and(1,3,2 )

*23 • Picture the Problem The wave functions are of the form

⎛ n1π ⎝ L1

ψ = A sin ⎜⎜

⎞ ⎛n π x ⎟⎟ sin ⎜⎜ 2 ⎠ ⎝ 2 L1

⎞ ⎛nπ y ⎟⎟ sin ⎜⎜ 3 ⎠ ⎝ 4 L1

⎞ z ⎟⎟ ⎠

24 • Picture the Problem The boundary conditions in the y and z directions are the same those in Figure 35-1. In the x direction, we’ll require the ψ = 0 at −L/2 and L/2. (a) The boundary conditions in the x direction are: The general solution of the timeindependent Schrödinger equation is: Apply the boundary conditions to obtain:

ψ (− 12 L ) = ψ ( 12 L ) = 0 ψ ( x ) = A sin kx + B cos kx

ψ (− 12 L ) = − A sin

kL kL + B cos =0 2 2

and

ψ ( 12 L ) = A sin

kL kL + B cos =0 2 2

kL =0 2

Eliminate the terms in B by subtracting the equations:

A sin

For A ≠ 0:

kL = sin −1 0 = 0, π , 2π , ... 2

1156 Chapter 35 or

k=

n1π , n1 = 0, 2, 4, ... L kL =0 2

Eliminate the terms in A by adding the equations:

B sin

For B ≠ 0:

kL π 3π 5π = cos −1 0 = , , , ... 2 2 2 2 or

k=

n2π , n2 = 1, 3, 5, ... L

Thus:

⎛ n1π ⎞ ⎛ n2π ⎞ ⎛ n3π ⎞ x ⎟ sin ⎜ y ⎟ sin ⎜ z ⎟, n1 = 2n + 1 ⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠

ψ ( x,y,z ) = B cos⎜ and

⎛ n1π ⎝ L

ψ ( x,y,z ) = A sin ⎜

⎞ ⎛n π ⎞ ⎛nπ x ⎟ sin ⎜ 2 y ⎟ sin ⎜ 3 ⎠ ⎝ L ⎠ ⎝ L

⎞ z ⎟, n1 = 2n ⎠

The ground-state wave function is:

⎛π ⎞ ⎛π ⎞ ⎛π ⎞ x ⎟ sin ⎜ y ⎟ cos⎜ z ⎟ ⎝L ⎠ ⎝L ⎠ ⎝L ⎠

ψ (1,1,1) = A cos⎜

(b) The allowed energies are the same as those for a well with U = 0 for 0 < x < L. 25 •• Picture the Problem We can apply the solution to the time-independent Schrödinger equation in three dimensions to obtain the wave function and the allowed energies for the given two-dimensional region. In (c), we must find three different sets of quantum numbers (m,n) for which the sum of the squares are the same. (a) The solution to the timeindependent Schrödinger equation in two-dimensions is:

ψ ( x, y ) = A sin k1 x sin k 2 y = A sin

nπ mπ x sin y L L

where n and m are integers.

Applications of the Schrödinger Equation 1157 (b) The energy is quantized to the values: (c) The lowest two states that are degenerate are:

E n ,m

(

h2 = n2 + m2 2 8mL

E1, 2 = E2,1 =

5h 2 8mL2 n m En,m 1 7 50 7 1 50 5 5 50

(d) The energies of three lowest states that have the same energies (in units of

)

h2 ) are listed in the 8mL2

table to the right: The quantum numbers for the three states are:

(1,7 ), (7,1),

and

(5,5)

and their energies are

E = (50 )

h2 25h 2 = 8mL2 4mL2

The Schrödinger Equation for Two Identical Particles 26 • Picture the Problem We must differentiate Equation 35-37 twice and substitute these derivatives in this equation to show that it is a solution. With U = 0, Equation 35-35 becomes:

−

h 2 ∂ 2ψ ( x1 , x2 ) h 2 ∂ 2ψ ( x1 , x2 ) − (1) 2m ∂x12 2m ∂x22

= Eψ ( x1 , x2 )

Differentiate Equation 35-7 with respect to x1:

∂ψ ∂ = ∂x1 ∂x1

πx1 2πx2 ⎤ ⎡ ⎢⎣ A sin L sin L ⎥⎦ Aπ πx 2πx2 cos 1 sin = L L L

Compute the second derivative with respect to x1:

πx 2πx2 ⎤ ∂ 2ψ ∂ ⎡ Aπ cos 1 sin = 2 ⎢ L L ⎥⎦ ∂x1 ∂x1 ⎣ L =−

πx 2πx2 Aπ 2 sin 1 sin 2 L L L

1158 Chapter 35 Differentiate Equation 35-7 with respect to x2:

∂ψ ∂ = ∂x2 ∂x2

πx1 2πx2 ⎤ ⎡ ⎢⎣ A sin L sin L ⎥⎦ 2 Aπ πx 2πx2 sin 1 cos = L L L

Compute the second derivative with respect to x2:

∂ 2ψ ∂ = 2 ∂x2 ∂x2 =−

2πx2 ⎤ πx1 ⎡ 2 Aπ ⎢⎣ L sin L cos L ⎥⎦

4 Aπ 2 πx 2πx2 sin 1 sin 2 L L L

Substitute in equation (1) to obtain:

−

πx1 2πx2 ⎤ h 2 ⎡ 4 Aπ 2 πx 2πx2 ⎤ h 2 ⎡ Aπ 2 sin sin sin 1 sin − − − ⎢ ⎥ ⎢ 2 2 2m ⎣ L L L ⎦ 2m ⎣ L L L ⎥⎦ = EA sin

πx1 L

sin

2πx2 L

Solve for E to obtain:

5h 2π 2 E= 2mL2

Thus we' ve shown that Equation 35 - 37 satisfies Equation 35 - 35 provided E=

5h 2π 2 . 2mL2

27 • Picture the Problem Because bosons have symmetric wave functions and do not obey the Pauli exclusion principle, they can occupy the same ground state. The ground-state energy of a single particle in a one-dimensional box of length L is: For 10 bosons:

E0,1 particle

h2 = 8mL2

E0,10 bosons

10h 2 5h 2 = = 8mL2 4mL2

*28 • Picture the Problem For fermions, such as neutrons for which the spin quantum number is ½, two particles can occupy the same spatial state.

Applications of the Schrödinger Equation 1159

(

E = 2 E1 12 + 2 2 + 32 + 4 2 + 52

The lowest total energy for the 10 fermions is:

)

⎛ h2 ⎞ ⎟(55) = 2⎜⎜ 2 ⎟ ⎝ 8mL ⎠ 55h 2 4mL2

=

Orthogonality of Wave Functions 29

•• ∞

Picture the Problem We need to show that

∫ψ (x )ψ (x ) dx = 0 , where ψ (x) and ψ (x) 0

0

1

1

−∞

are given by Equations 35-23 and 35-25, respectively. Equations 35-23 and 35-25 are:

ψ 0 ( x ) = A0 e − ax

2

35-23

and

ψ 1 (x ) = A1 xe − ax Note that ψ1(x) is antisymmetric, whereas ψ0(x) is symmetric. Because the product of an antisymmetric function and a symmetric function is antisymmetric:

35-25

ψ 0 ( x )ψ 1 ( x ) is antisymmetric

∞

Because the integral of an antisymmetric function over symmetric limits is zero: 30

2

∫ψ (x )ψ (x ) dx = 0 0

1

−∞

•• ∞

Picture the Problem We need to show that

∫ψ (x )ψ (x ) dx = 0 , where ψ (x) is given 1

2

2

−∞

in the problem statement and ψ 1 (x ) = A1 xe − ax . 2

Note that ψ1(x) is antisymmetric, whereas ψ2(x) is symmetric. Because the product of a symmetric function and an antisymmetric function is antisymmetric:

ψ 1 (x )ψ 2 ( x ) is antisymmetric.

1160 Chapter 35 ∞

Because the integral of an antisymmetric function over symmetric limits is zero: 31

∫ψ (x )ψ (x ) dx = 0 1

2

−∞

•• L

Picture the Problem We need to show that

⎛ nπx ⎞ ⎛ mπx ⎞ ⎟ sin ⎜ ⎟dx = 0. L ⎠ ⎝ L ⎠

∫ sin⎜⎝ 0

Use the trigonometric identity (sin aα )(sin bα ) =

1 2

{cos[(a − b )α ] − cos[(a + b )α ] }

to rewrite the product of the two sine functions as the difference of two cosine functions:

πx ⎤ πx ⎤ ⎫ ⎛ nπx ⎞ ⎛ mπx ⎞ 1 ⎧ ⎡ ⎡ sin ⎜ ⎟ sin ⎜ ⎟ = ⎨cos ⎢(n − m ) ⎥ − cos ⎢(n + m ) ⎥ ⎬ L ⎦⎭ L⎦ ⎝ L ⎠ ⎝ L ⎠ 2⎩ ⎣ ⎣ ⎛ nπx ⎞ ⎛ mπx ⎞ ⎛ nπx ⎞ ⎛ mπx ⎞ ⎟ and sin ⎜ ⎟ and evaluate ∫ sin ⎜ ⎟ sin ⎜ ⎟dx : L L ⎝ L ⎠ ⎝ L ⎠ ⎝ ⎠ ⎝ ⎠ 0 L

Substitute for sin ⎜

πx ⎤ ⎡ sin ⎢(n − m ) ⎥ 1⎧ ⎡ L πx ⎤ πx ⎤ ⎫ L⎦ ⎡ ⎣ ∫−∞ 2 ⎨⎩cos⎢⎣(n − m) L ⎥⎦ − cos⎢⎣(n + m ) L ⎥⎦ ⎬⎭dx = π n−m ∞

πx ⎤ ⎡ sin ⎢(n + m ) ⎥ L L⎦ ⎣ − n+m π Because n and m are integers and n ≠ m, the sine functions vanish at the two limits x = 0 L

and x = L. Therefore,

⎛ nπx ⎞ ⎛ mπx ⎞ ⎟ sin ⎜ ⎟dx = 0 for n ≠ m. L ⎠ ⎝ L ⎠

∫ sin⎜⎝ 0

General Problems 32 •• Picture the Problem We can use the wave functions ψ1(x) and ψ2(x) and the definitions of <x> and <x2> to evaluate these quantities and the wave functions at x = 0. (a) The wave functions ψ1(x) and ψ2(x) are:

ψm =

2 ⎛ mπ ⎞ sin ⎜ x ⎟, m = 2n L ⎝ L ⎠

and

ψm =

2 ⎛ mπ ⎞ x ⎟, m = 2n + 1 cos⎜ L ⎝ L ⎠

Applications of the Schrödinger Equation 1161 where n = 0, 1, 2, … Evaluate these functions at x = 0 to obtain:

2 ⎡π ⎤ sin ⎢ (0)⎥ = 0 L ⎣L ⎦

ψ 1 (0) = and

2 ⎡π ⎤ cos ⎢ (0)⎥ = L ⎣L ⎦

ψ 2 (0) = (b) Because ψ m ( x ) is an even

L2

2

function of x in all cases, xψ m2 ( x ) is

x =

∫ xψ (x ) dx = 2 m

0

−L 2

an odd function of x and: L2

(c) For n = 1:

x

2

2 π = x 2 sin 2 xdx ∫ L −L 2 L

From integral tables: 2 2 ∫ x sin (ax ) dx =

1 ⎞ x cos(2ax ) x3 ⎛ x3 − ⎜⎜ − 3 ⎟⎟ sin (2ax ) − 4a 2 6 ⎝ 4 a 8a ⎠

Use this integral with a = π/L to obtain:

x

2

L2 ⎛ 6 ⎞ = ⎜1 + 2 ⎟ 12 ⎝ π ⎠ L2

For n = 2:

x2 =

2 2π x 2 cos 2 xdx ∫ L −L 2 L

From integral tables:

1 ⎞ x cos(2ax ) x3 ⎛ x3 ∫ x cos (ax ) dx = 6 − ⎜⎜⎝ 4a − 8a 3 ⎟⎟⎠ sin (2ax ) + 4a 2 2

2

Use this integral with a = 2π/L to obtain:

x2 =

Remarks: Note that for any value of m, x 2 =

L2 ⎛ 3 ⎞ ⎜1 + ⎟ 12 ⎝ 2π 2 ⎠

L2 ⎛ 6 ⎞ ⎜ 1 + 2 2 ⎟. m π ⎠ 12 ⎝

2 L

1162 Chapter 35 *33 •• Picture the Problem We can determine the energies of the state by identifying the four lowest quantum states that are occupied in the ground state and computing their combined energies. We can then find the energy difference between the ground state and the first excited state and use this information to find the energy of the excited state. Each n, m state can accommodate only 2 particles. Therefore, in the ground state of the system of 8 fermions, the four lowest quantum states are occupied. These are:

(1,1), (1,2), (2,1) and (2,2)

The energies are quantized to the values given by:

⎛ h2 ⎞ 2 ⎟ n + n22 En1 ,n2 = 2⎜⎜ 2 ⎟ 1 8 mL ⎝ ⎠

Note that the states (1,2) and (2,1) are distinctly different states because the x and y directions are distinguishable.

(

)

The energy of the ground state is the sum of the energies of the four lowest quantum states:

E0 = E1,1 + E1, 2 + E2,1 + E2, 2 ⎛ h2 ⎞ 2 2 ⎛ h2 ⎞ 2 ⎛ h2 ⎞ 2 2 ⎛ h2 ⎞ 2 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟ 2 + 22 = 2⎜⎜ 1 + 1 + 2 1 + 2 + 2 2 + 1 + 2 2 ⎟ 2 ⎟ ⎜ 8mL2 ⎟ ⎜ 8mL2 ⎟ 8 mL 8 mL ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(

)

(

)

(

)

(

⎛ h2 ⎞ ⎟ (2 + 5 + 5 + 8) = 2⎜⎜ 2 ⎟ 8 mL ⎝ ⎠ 2 5h = mL2 The next higher state is achieved by taking one fermion from the (2,2) state and raising it to the next higher unoccupied state. That state is the (1,3) state. The energy difference between the ground state and this state is: Hence, the energies of the degenerate states (1,3) and (3,1) are: The three lowest energy levels are therefore:

∆E = E1,3 − E2, 2

(

)

(

h2 h2 2 2 = 1 +3 − 22 + 22 2 2 8mL 8mL 2 h h2 ( ) = − = 10 8 8mL2 4mL2

E1,3 = E3,1 = E0 + ∆E =

5h 2 h2 21h 2 + = mL2 4mL2 4mL2

5h 2 E0 = mL2 and two states of energy

)

)

Applications of the Schrödinger Equation 1163 21h 2 E1 = E2 = 4mL2 34 •• Picture the Problem The energy levels are the same as for a two-dimensional box of widths L and 3L. (a) The energies of the bound states are given by:

E n ,m = =

The three lowest energy states are:

h2 ⎛ n2 m2 ⎞ ⎜ + ⎟ 8m ⎜⎝ L2 9 L2 ⎟⎠

(

h2 9n 2 + m 2 2 72mL

)

E1,1 =

h2 5h 2 ( ) 9 + 1 = 72mL2 36mL2

E1, 2 =

13h 2 h2 ( ) 9 4 + = 72mL2 72mL2

and

E1,3

h2 h2 (9 + 9) = = 72mL2 4mL2

None of these states are degenerate. (b) Express the condition that must be satisfied for two states to be degenerate:

(

)

9 n12 − n22 = m22 − m12

This condition is first satisfied for:

n1 = 2, m1 = 3, and n2 = 1 and m2 = 6

Find the energy of this doubly degenerate state:

E2,3 =

h2 5h 2 ( ) 36 + 9 = 72mL2 8mL2

35 ••• Picture the Problem We can use the definition of the classical expectation value (average value) to show that the classical expectation value of x2 for a particle in a onedimensional box of length L centered at the origin is L2/12. In (b) we’ll proceed as in (a) using the definition of the quantum expectation value of x2.

1164 Chapter 35 (a) The classical expectation value is given by:

(x )

L2

2

av

=

1 x 2 dx ∫ 1 1 2 L − (− 2 L ) − L 2 L2

1 ⎡ x3 ⎤ 1 ⎛ L3 ⎞ = ⎢ ⎥ = ⎜⎜ ⎟⎟ L ⎣ 3 ⎦ − L 2 L ⎝ 12 ⎠ = (b) For a particle in the nth state in a one-dimensional box:

L2 12 L2

x

2

2 nπ = x 2 sin 2 xdx ∫ L −L 2 L

From integral tables: 2 2 ∫ x sin (ax ) dx =

In the limit n >> 1:

x3 ⎛ x3 1 ⎞ x cos(2ax ) − ⎜⎜ − 3 ⎟⎟ sin (2ax ) − 6 ⎝ 4 a 8a ⎠ 4a 2

x2 =

L2 12

36 •• Picture the Problem We can solve Equation 35-28 for T and substitute for R using Equation 35-27. Letting r = k2/k1 and simplifying will lead to the given result. Equation 35-28 is:

T + R =1 ⇒ T =1− R 2

From Equation 35-27:

R=

(k1 − k 2 )2 (k1 + k 2 )2

⎛ k2 ⎞ ⎜⎜1 − ⎟⎟ k1 ⎠ (1 − r )2 =⎝ = 2 (1 + r )2 ⎛ k2 ⎞ ⎜⎜1 + ⎟⎟ k1 ⎠ ⎝

where r = k2/k1 Substitute for R to obtain:

T = 1− =

Substitute for k2/k1 for r and simplify to obtain:

T=

(1 − r )2 = (1 + r )2 − (1 − r )2 (1 + r )2 (1 + r )2

4r (1 + r )2 4k1k 2 (k1 + k 2 )2

Applications of the Schrödinger Equation 1165 37 •• Picture the Problem We can use the energies in the regions U = 0 and U = U0 to express the ratio of the wave numbers k1 and k2 in these regions in terms of E and U0 and the definition of the reflection coefficient R to show that R =

In the region U0 = 0:

In the region U = U0:

E=

(1 − r )2 . (1 + r )2

h 2 k12 ⇒ k1 = 2m

E −U0 =

2 m( E − U 0 ) h2 = 2mE h2

k r= 2 = k1

Factor k1 from the numerator and denominator to obtain:

Substitute for k2/k1 to obtain:

R=

2m(E − U 0 ) h2

h 2 k 22 ⇒ k2 = 2m

Let r equal the ratio of k2 to k1:

The reflection coefficient R is given by:

2mE h2

1−

(k1 − k2 )2 (k1 + k2 )2 2

⎛ k2 ⎞ ⎜⎜1 − ⎟⎟ k1 ⎠ R=⎝ 2 ⎛ k2 ⎞ ⎜⎜1 + ⎟⎟ k1 ⎠ ⎝ R=

(1 − r )2 (1 + r )2

38 •• Picture the Problem (a) From Problem 37 we have:

Because E = αU0, R can be written: From Problem 36 we have:

2 ( 1− r) R= , where r = (1 + r )2

r = 1−

1

α 4r T= (1 + r )2

1−

U0 E

U0 E

1166 Chapter 35 A spreadsheet program to plot R and T as functions of α is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A2 B2

Content/Formula 1.0 SQRT(1−1/A2)

C2

(1−B2)^2/(1+B2)^2

D2

4*B2//(1+B2)^2

Algebraic Form

α

1−

1

α (1 − r )2 (1 + r )2 4r (1 + r )2

1 2 3 4 5

A alpha 1.0 1.5 2.0 2.5

B r 0.000 0.577 0.707 0.775

C R 1.000 0.072 0.029 0.016

D T 0.000 0.928 0.971 0.984

16 17 18 19 20

8.0 8.5 9.0 9.5 10.0

0.935 0.939 0.943 0.946 0.949

0.001 0.001 0.001 0.001 0.001

0.999 0.999 0.999 0.999 0.999

The following graph of R and T as functions of α was plotted using the data in the table:

1.0

0.8

0.6

R T

0.4

0.2

0.0 1

2

3

4

5

6 alpha

7

8

9

10

Applications of the Schrödinger Equation 1167 (b)

From the graph, we note that, as α → ∞, T → 1 and R → 0. The graph also shows that, as α → 1, T → 0 and R → 1.

39 ••• Picture the Problem We require that ∞

A22

∫ (2ax

∞

)

(

)

− 12 e −2 ax dx = 2 A22 ∫ 2ax 2 − 12 e −2 ax dx = 1. 2

2

−∞

2

2

2

0

Expand the integrand to obtain:

(2ax

2

)

2

(

)

− 12 e −2 ax = 4a 2 x 4 − 2ax 2 + 14 e −2 ax = 4a 2 x 4 e −2 ax − 2ax 2 e −2 ax + 14 e −2 ax 2

2

2

2

Substitute in the integral expression:

∫ (4a

∞

2A

2 2

2

)

x 4 e −2 ax − 2ax 2 e −2 ax + 14 e −2 ax dx = 1 2

2

2

0

or ∞

∞

∞

8a 2 A22 ∫ x 4 e −2 ax dx − 4aA22 ∫ x 2 e −2 ax dx + 12 A22 ∫ e −2 ax dx = 1 2

2

0

0

∞

Use the definite integrals

−bx ∫ e dx = 2

0

∞

∫x

2 n −bx 2

e

dx =

0

2

(1)

0

1 π and 2 b

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ (2n − 1) π , n ≥ 1 (see Table D-5) to integrate equation 2 n+1 b n b

(1) term by term:

⎡ 3 8a 2 A22 ⎢ 3 2 ⎣ 2 (2a )

π ⎤

⎡ 1 π ⎤ 1 2 ⎡1 π ⎤ 2 ⎥ − 4aA2 ⎢ 2 ⎥ + 2 A2 ⎢ ⎥ =1 2a ⎦ ⎣ 2 (2a ) 2a ⎦ ⎣ 2 2a ⎦

or

⎡3 π 1 π 1 π ⎤ A22 ⎢ − + ⎥ =1 ⎣ 4 2a 2 2a 4 2a ⎦ or

⎡1 π ⎤ A22 ⎢ ⎥ =1 a 2 2 ⎣ ⎦ Solve for A2:

A2 = 2

2a

π

2

1168 Chapter 35 Because a =

mω0 mω0π = 2h h

A2 =

4

8mω0 h

40 ••• (a) Let x = −x. The second derivative is an even operator, that is,

d 2ψ (− x ) d (− x ) = d 2ψ (x ) dx 2 . Therefore, if U (− x ) = U ( x ), the Schrödinger 2

equation for ψ (−x) = ψ (x) and must give the same values for the energy E. If ψ (−x) differs from ψ (x), the ratio ψ (−x)/ψ (x) cannot be a function of x and must be a constant. Hence, ψ (x) = Cψ (−x). (b) The previous result means that replacing the argument of the wave function by its negative is equivalent to multiplication by C. Thus, if Cψ (−x) is a good wave function and we replace its argument by its negative, that is, by x, we must multiply by C again. Thus, ψ (x) = C 2ψ (x), C 2 = 1, and C = ±1. *41 ••• Picture the Problem We can follow the step-by-step procedure outlined in the problem statement to show that (Eav)min = + 12 ħ ω. 1. The total classical energy is:

Eav = U av + K av

( ) + ( p2m)

= 12 mω 2 x 2 2. Express the standard deviation of ∆p:

Because pav = 0: 3. Express the standard deviation of ∆x:

Because xav = 0:

av

av

(∆p )2 = [( p − pav )2 ]av

[

= p 2 − 2 ppav − p av2

]

(∆p )2 = ( p 2 )av

(∆x )2 = [(x − xav )2 ]av

[

= x 2 − 2 xxav − x av2

(∆x )2 = (x 2 )av

(1)

2

]

av

av

Applications of the Schrödinger Equation 1169 4. Use the uncertainty principle ∆p = h/2∆x to eliminate (p2)av from the average energy in equation (1):

( ) + (∆2pm )

Eav = 12 mω 2 x 2

2

av

( )

1 ⎡ h2 ⎤ ⎢ ⎥ 2m ⎣ 4(∆x )2 ⎦

( )

h2 8m x 2

= 12 mω 2 x 2 = 12 mω 2 x 2

av +

av +

( )

av

Let Z = (x2)av to obtain:

Eav = 12 mω 2 Z +

5. Differentiate Eav with respect to Z and set this derivative equal to zero:

dEav d ⎡1 h2 ⎤ 2 = ω + m Z 2 8mZ ⎥⎦ dZ dZ ⎢⎣ h2 = 12 mω 2 − = 0 for extrema 8mZ 2

Solve for Z to find the value of Z that minimizes Eav (see the remark below):

Z=

6. Evaluate Eav when Z = h/2mω:

(Eav )min = 12 mω 2 ⎛⎜

h2 8mZ

h 2mω

h ⎞ h 2 ⎛ 2mω ⎞ ⎟+ ⎜ ⎟ ⎝ 2mω ⎠ 8m ⎝ h ⎠

=

1 2

hω

Remarks: All we’ve shown is that Z = h/2mω is an extreme value, i.e., either a maximum or a minimum. To show that Z = h/2mω minimizes Eav, we must either 1) show that the second derivative of Eav with respect to Z evaluated at Z = h/2mω is positive, or 2) confirm that the graph of Eav as a function of Z opens upward at Z = h/2mω. 42 ••• Picture the Problem The classically allowed region is for E ≥ U(z). In the figure below, this region extends from z = 0 to z = zmax.

1170 Chapter 35 The kinetic energy is E − U(z). In this case, K(z) is a straight line extending from E at z = 0 to 0 at z = zmax.

A sketch of the wave functions for the lowest three energy states is shown to the right:

43 •• Picture the Problem If f(x) = 0 everywhere on the interval 1 < x < 2, then the slope of f(x) is zero everywhere on the interval; and if the slope remains zero everywhere on the interval, then the rate of change of the slope (with respect to x) also remains zero everywhere on the interval; the rate of change of slope remains zero everywhere on the interval, then the rate of change of the rate of change of the slope also remains zero everywhere on the interval; and so on. More concisely, if f(x) = 0 everywhere on the interval 1 < x < 2, then derivatives of f(x) with respect to x of order 1, 2, 3, … are each equal to zero everywhere on the interval. Calculating the first three derivatives of f we obtain:

df = 3 Ax 2 + 2 Bx + Cx dx d2 f = 6 Ax + 2 B dx 2 and

d3 f = 6A dx 3 Using d3f/df 3 = 0 and solving for A one obtains:

A=0

Substituting 0 for A in the expression for d2f/dx2 gives:

d2 f = 0 + 2B = 2B dx 2

Using d2f/dx2 =0 and solving for B yields:

B=0

Substituting 0 for both A and B in the expression for df/dx yields:

df = 0 + 0 + Cx = Cx dx

Using df/dx =0 and solving for C

C=0

Applications of the Schrödinger Equation 1171 one obtains: Substituting 0 for A, B, and C in the expression for f gives:

f=0+0+0+D=D

Using f = 0 and solving for D gives:

D=0

Thus, we’ve shown that if f(x) = Ax3 + Bx2 + Cx + D = 0 everywhere on the interval 1 < x < 2, it follows that A = B = C = D = 0.

1172 Chapter 35

Chapter 36 Atoms Conceptual Problems *1 • Determine the Concept Examination of Figure 35-4 indicates that as n increases, the spacing of adjacent energy levels decreases. 2 • Picture the Problem The energy of an atom of atomic number Z, with exactly one electron in its nth energy state is given by En = − Z 2 Express the energy of an atom of atomic number Z, with exactly one electron, in its nth energy state: For lithium (Z = 3) in its first excited state (n = 1) this expression becomes:

E02 , n = 1, 2, 3, ... . n2 E02 , n = 1, 2, 3, ... n2

En = − Z 2

where E0 is the atom’s ground state energy.

E0 = −9 E0 12 and (a ) is correct. E2 = −(3)

2

3 • Determine the Concept Bohr’s postulates are 1) the electron in the hydrogen atom can move only in certain non-radiating, circular orbits called stationary states, 2) if Ei and Ef are the initial and final energies of the atom, the frequency f of the emitted radiation during a transition is given by f = [Ei – Ef]/h, and 3) the angular momentum of a circular orbit is constrained by mvr = nh. (a ) is correct. 4 •• Picture the Problem We can express the kinetic energy of the orbiting electron as well as its total energy as functions of its radius r. Express the total energy of an orbiting electron:

E = K +U

Express the orbital kinetic energy of an electron:

kZe 2 K= 2r

Express the potential energy of an orbiting electron:

U =−

Substitute and simplify to obtain:

(1)

kZe 2 r

kZe 2 kZe 2 kZe 2 2kZe 2 E= − = − 2r r 2r 2r 2 kZe =− 2r

1173

1174 Chapter 36 Thus, as r increases, E becomes less negative and therefore increases. Examination of the expression for K makes it clear that if r increases, K decreases. 5 • Picture the Problem We can relate the kinetic energy of the electron in the n = 2 state to its total energy using E2 = K 2 + U 2 . Express the total energy of the hydrogen atom in its n = 2 state:

Express the energy of hydrogen in its nth energy state:

E2 = K 2 + U 2 = K 2 − 2 K 2 = − K 2 or

K 2 = − E2 2 E02 E02 2 E0 En = − Z 2 = −(1) 2 = − 2 n n n 2

where E0 is hydrogen’s ground state energy and Z = 1. Substitute to obtain:

E E E02 and K 2 = 20 = 0 2 n 2 4 (d ) is correct.

Kn =

6 • Picture the Problem The orbital radius r depends on the n = 1 orbital radius a0, the atomic number Z, and the orbital quantum number n according to r = n2a0/Z. The radius of the n = 5 orbit is:

r5 = 52

a0 = 25a0 1

because Z = 1 for hydrogen.

(b ) is correct.

*7

•

Determine the Concept We can find the possible values of l by using the constraints on the quantum numbers n and l . The allowed values for the orbital quantum number l for n = 1, 2, 3, and 4 are summarized in table shown to the right: From the table it is clear that l can have 4 values.

l n 1 0 2 0, 1 3 0, 1, 2 4 0, 1, 2, 3 (a ) is correct.

Atoms

1175

8 • Picture the Problem We can find the number of different values ml can have by enumerating the possibilities when the principal quantum number n = 4.

l = 0,1, 2, 3

The allowed values for the orbital quantum number l and the magnetic quantum number ml for n = 4 are summarized to the right:

and

ml = −3, − 2, − 1, 0,1, 2, 3

From this enumeration we can see that m can have 7 values.

(c) is correct.

9 • Picture the Problem We can visualize the relationship between the quantum number l and the electronic configuration as shown in the table below. s

p d

f

g h

l value 0 1 2 3 4 5 Because the p state corresponds to l = 1, (c ) is correct. *10 •• Determine the Concept The s state, with l = 0, is a ″penetrating″ state in which the probability density near the nucleus is significant. Consequently, the 3s electron in sodium is in a region of low potential energy for a significant portion of the time. In the state l = 1, the probability density at the nucleus is zero, so the 2p electron of sodium is shielded from the nuclear charge by the 1s electrons. In hydrogen, the 3s and 2p electrons experience the same nuclear potential. 11 •• Determine the Concept In conformity with the exclusion principle, the total number of electrons that can be accommodated in states of quantum number n is n2 (see Problem 48). The fact that closed shells correspond to 2n2 electrons indicates that there is another quantum number that can have two possible values. 12 •• Picture the Problem We can group these elements by using Table 35-1 to look for a common outer electronic configuration in the ground states. The following elements have an outer 4s2 configuration in the ground state:

titanium, manganese, and calcium

1176 Chapter 36 The following elements have an outer 4s configuration in the ground state:

potassium, chromium, and copper.

Remarks: It is to be expected that atoms of the first group will have similar properties, and, likewise, that atoms of the second group will have similar properties. 13 • Picture the Problem We can use the fact that the sum of the exponents in the electronic configuration representation is the atomic number to identify these two elements. (a) Adding the exponents yields a sum of 15. Because this sum is the atomic number, Z, the element must be phosphorus. (b) Adding the exponents yields a sum of 24. Because this sum is the atomic number, Z, the element must be chromium. Remarks: Checking the electronic configurations in Table 35-1 further confirms these conclusions. *14 • Picture the Problem We can apply the constraints on the quantum numbers l and ml to find the possible values for each when n = 3. Express the constraints on the quantum numbers n, l , and ml:

n = 1, 2, 3, ... , l = 0,1, 2, ..., n − 1 , and

ml = −l, − l + 1, ..., l So, for n = 3, the constraints on l limit it to the values:

l = 0,1, and 2.

ml can take on the values:

ml = − 2, − 1, 0,1, 2

15 • Determine the Concept The correspondence between the letter designations K, L, M, N, O, and P for the shells and the principal quantum number n is summarized in the table below.

Atoms

Shell designation n

K 1 0

l

L 2 0 1

M 3 0 1 2

N 4 0 1 2 3

O 5 0 1 2 3 4

P 6 0 1 2 3 4 5

1177

Q 7 0 1 2 3 4 5 6

While n = 2 for the L shell, l can be either 0 or 1. (d ) is correct. 16 •• Picture the Problem The strengths and weaknesses of each model are summarized in the following table. Bohr Theory

Schrödinger Theory

Ease of application

Easy

Difficult

Prediction of stationary state energies

Correct predictions

Correct predictions

Prediction of angular momenta

Predicts incorrect results

Predicts correct results

Spatial distribution of electrons

Predicts incorrect results

Predicts correct probabilistic distribution

17 •• Determine the Concept The optical spectrum of any atom is due to the configuration of its outer-shell electrons. Ionizing the next atom in the periodic table gives you an ion with the same number of outer-shell electrons, and almost the same nuclear charge. Hence, the spectra should be very similar. *18 •• Determine the Concept The Ritz combination principle is due to the quantization of energy levels in the atom. We can use the relationship between the wavelength of the emitted photon and the difference in energy levels within the atom that results in the emission of the photon to express each of the wavelengths and then the sum of the reciprocals of the first and second wavelengths and the sum of the reciprocals of the third and fourth wavelengths.

1178 Chapter 36 Express the wavelengths of the spectral lines λ1, λ2, λ3, and λ4 in terms of the corresponding energy transitions:

hc E3 − E2 hc λ2 = E2 − E0 hc λ3 = E3 − E1

λ1 =

and

Add the reciprocals of λ1 and λ2 to obtain:

λ4 =

hc E1 − E0

1

1

λ1

+

λ2

=

E3 − E2 E2 − E0 + hc hc

E − E0 = 3 hc Add the reciprocals of λ3 and λ4 to obtain:

1

λ3

+

1

λ4

=

E3 − E1 E1 − E0 + hc hc

E − E0 = 3 hc Because the right-hand sides of equations (1) and (2) are equal:

1

λ1

+

1

λ2

=

1

λ3

+

1

λ4

One possible set of energy levels is shown to the right:

19 • Determine the Concept An allowed transition must satisfy the selection rules ∆ ml = 0 or ±1 and ∆l = ±1. (a) ∆l = −1 and ∆ ml = 0:

The transition is allowed.

(b) (3,0,1) does not exist.

The transition is not allowed.

(c) ∆l = −1 and ∆ ml = 2:

The transition is not allowed.

(d) ∆l = +1 and ∆ ml = 1:

The transition is allowed.

(1)

(2)

Atoms (e) ∆l = −1 and ∆ ml = 0:

1179

The transition is allowed.

Estimation and Approximation *20 •• Picture the Problem The number of photons need to stop a 85Rb atom traveling at 300 m/s is the ratio of its momentum to that of a typical photon. (a) The number N of photon-atom collisions needed to bring an atom to rest is the ratio of the change in the momentum of the atom as it stops to the momentum brought to the collision by each photon: The kinetic energy of an atom whose temperature is T is: Substitute for v to obtain:

N=

∆patom mv mvc = = E pphoton E c

where m is the mass of the atom.

1 2

mv 2 = 32 kT ⇒ v =

N=

3kT m

mc 3kT c = 3mkT E m E

For an atom use mass is 50 u:

N=

⎛ 3 × 108 m/s 1.66 × 10 −27 kg ⎞ ⎜ ⎟⎟ 1.38 × 10 −23 J/K (500 K ) ≈ 105 × 3 50 u −19 ⎜ 1.6 × 10 J ⎝ u ⎠ 1 eV × eV

(b) The number N of ping-pong ballbowling ball collisions needed to bring the bowling ball to rest is the ratio of the change in the momentum of the bowling ball as it stops to the momentum brought to the collision by each ping-pong ball: Provided the speeds of the approaching bowling ball and pingpong ball are approximately the same: (c) The number of photons N needed to stop a 85Rb atom is the ratio of the change in the momentum of the atom to the momentum brought to the collision by each photon: Substitute numerical values and evaluate N:

(

N=

N=

N=

)

∆pbowling ball pping-pong ball

∆pbowling ball pping-pong ball

=

mbb vbb mppb vppb

≈

mbb 6 kg ≈ ≈ 103 mppb 4g

∆patom mv mvλ = = h pphoton h

λ

1180 Chapter 36 85(1.66 × 10 −27 kg )(300 m/s )(780.24 nm ) N= = 4.98 × 10 4 −34 6.63 × 10 J ⋅ s

21 •• Picture the Problem We can use the relationship between the kinetic energy of an atom and its momentum, together with the de Broglie equation, to derive the expression for the thermal de Broglie wavelength. In Part (b), we can use the definition of the number density of atoms and the result from Part (a), with the interatomic spacing set equal to the thermal de Broglie wavelength, to estimate the temperature needed to create a Bose condensate. (a) Express the kinetic energy of an atom in terms of its momentum: Use the de Broglie relationship to express the atom’s momentum in terms of its de Broglie wavelength: Substitute for p to obtain:

p2 2m

K= p=

h

λT

where λT is the thermal de Broglie wavelength.

K=

h2 2mλT2

The kinetic energy of an atom is also a function of its temperature T:

K=

3 kT 2

Equate these expressions for K to obtain:

3 h2 kT = 2 2mλT2

Solve for λT:

(b) The number density of atoms ρ is given by:

λT = ρ=

h2 3mkT

N V

where N is the number of atoms and V is the volume they occupy. Assume that the atoms are arrayed on a cubic lattice of lattice spacing d to obtain:

V = Nd 3 and ρ =

Solve for d to obtain:

d = ρ −1 3

Setting d = λT yields:

ρ −1 3 =

h2 3mkT

N 1 = 3 3 Nd d

Atoms Solve for T to obtain:

1181

h2 ρ 2 3 T= 3mk

Substitute numerical values and evaluate T: 23

106 cm 3 ⎞ ⎟ (6.63 × 10 J ⋅ s) ⎛⎜⎜1012 atoms × cm 3 m 3 ⎟⎠ ⎝ T= = 75.2 nK ⎛ − 27 kg ⎞ ⎛ − 23 J ⎞ 3(85 u ) ⎜1.66 × 10 ⎟ ⎜1.38 × 10 ⎟ K⎠ u ⎠⎝ ⎝ 2

−34

The Bohr Model of the Hydrogen Atom 22

•

h2 Picture the Problem The radius of the first Bohr orbit is given by a0 = . mke2 Equation 36-12 is:

a0 =

h2 mke2

Substitute numerical values and evaluate a0:

a0 =

(9.11× 10

−31

)(

(1.05 × 10

−34

J ⋅s

)

2

kg 8.99 × 10 N ⋅ m /kg 9

2

2

)(1.60 × 10

−19

C

)

2

= 5.26 × 10 −11 m

= 0.0526 nm 23 • Picture the Problem We can use the equation relating the wavelength of the radiation emitted during a transition between two energy states to find the wavelengths for the transitions specified in the problem statement. Express the wavelength of the radiation emitted during an energy transformation from one energy state to another:

Because Ef = −13.6 eV:

λ=

1240 eV ⋅ nm 1240 eV ⋅ nm = ∆E Ei − Ef

provided the energies are expressed in eV. Note that this relationship tells us that the longest wavelength corresponds to the smallest energy difference.

λ=

1240 eV ⋅ nm Ei + 13.6 eV

1182 Chapter 36 Express the energy of the nth energy state of the atom:

En = −

Substitute to obtain:

λ=

E0 13.6 eV =− 2 n n2

1240 eV ⋅ nm 13.6 eV − + 13.6 eV n2 1240 eV ⋅ nm = 1⎞ ⎛ 13.6 eV⎜1 − 2 ⎟ ⎝ n ⎠

(a) Evaluate equation (1) for n = n1 = 3:

λ=

(b) Evaluate equation (1) for n = n1 = 4:

λ=

(1)

1240 eV ⋅ nm = 103 nm 1⎞ ⎛ 13.6 eV⎜1 − 2 ⎟ ⎝ 3 ⎠ 1240 eV ⋅ nm = 97.3 nm 1⎞ ⎛ 13.6 eV⎜1 − 2 ⎟ ⎝ 4 ⎠

24 • Picture the Problem For the Balmer series, Ef = E2 = −3.40 eV. The wavelength associated with each transition is related to the difference in energy between the states by λ =

1240 eV ⋅ nm . Ei − Ef

Express the wavelength of the radiation emitted during an energy transformation from one energy state to another:

λ=

1240 eV ⋅ nm ∆E

(1)

provided the energies are expressed in eV. Note that this relationship tells us that the longest wavelength corresponds to the smallest energy difference.

E0 − E2 n2

Evaluate ∆E for the transition n = 3 to n = 2:

∆En→2 = En − E2 =

Because Ef = E2 = −3.40 eV and E0 = −13.6 eV:

∆En→2 = −

13.6 eV + 3.40 eV n2

Evaluate equation (2) for n = 3:

∆E3→2 = −

13.6 eV + 3.40 eV 32

= 1.89 eV

(2)

Atoms

1183

1240 eV ⋅ nm = 656 nm 1.89 eV

Substitute in equation (1) to obtain:

λ=

Evaluate equation (2) for n = 4:

∆E4→2 = −

13.6 eV + 3.40 eV 42

= 2.55 eV 1240 eV ⋅ nm = 486 nm 2.55 eV

Substitute in equation (1) to obtain:

λ=

Evaluate equation (2) for n = 5:

∆E5→2 = −

13.6 eV + 3.40 eV 52

= 2.86 eV Substitute in equation (1) to obtain:

λ=

1240 eV ⋅ nm = 434 nm 2.86 eV

25 •• Picture the Problem We can use Bohr’s second postulate to relate the photon energy to its frequency and use λ =

1240 eV ⋅ nm to find the wavelengths of the three longest Ei − Ef

wavelengths in the Paschen series. (a) Use Bohr’s second postulate to express the energy of the photons in the Paschen series: For the series limit:

hf = ∆E = Ei − Ef

n = ∞ and Ei = 0

Substitute to obtain:

⎛ E ⎞ E ∆E = − Ef = −⎜⎜ − 20 ⎟⎟ = 20 ⎝ n2 ⎠ n2

Evaluate the photon energy for n2 = 3:

hf =

Express the wavelength of the radiation resulting from an energy transition ∆E = hf:

λ=

(1)

13.6 eV = 1.51eV 32

1240 eV ⋅ nm ∆E

(2)

provided the energies are expressed in eV.

1184 Chapter 36 Evaluate λmin for the transition n = ∞ to n2 = 3: (b) For the three longest wavelengths:

λmin =

1240 eV ⋅ nm = 821 nm 1.51eV

ni = 4, 5, and 6

E0 ⎛ E0 ⎞ ⎟ − ⎜− ni2 ⎜⎝ n22 ⎟⎠ (3) ⎛ 1 ⎛1 1 ⎞ 1 ⎞ = E0 ⎜⎜ 2 − 2 ⎟⎟ = E0 ⎜⎜ − 2 ⎟⎟ ⎝ n2 ni ⎠ ⎝ 9 ni ⎠

Equation (1) becomes:

hf = Ei − Ef = −

Evaluate equation (3) for n = 4:

⎛1 1 ⎞ ∆E4→3 = (13.6 eV ) ⎜ − ⎟ ⎝ 9 16 ⎠ = 0.661 eV 1240 eV ⋅ nm = 1876 nm 0.661eV

Evaluate equation (2) for ∆E = 0.661 eV:

λ4→3 =

Evaluate equation (3) for n = 5:

⎛1 1 ⎞ ∆E5→3 = (13.6 eV ) ⎜ − ⎟ ⎝ 9 25 ⎠ = 0.967 eV 1240 eV ⋅ nm = 1282 nm 0.967 eV

Evaluate equation (2) for ∆E = 0.967 eV:

λ5→3 =

Evaluate equation (3) for n = 6:

⎛1 1 ⎞ ∆E6→3 = (13.6 eV ) ⎜ − ⎟ ⎝ 9 36 ⎠ = 1.13 eV

Evaluate equation (2) for ∆E = 1.13 eV:

λ6→3 =

1240 eV ⋅ nm = 1097 nm 1.13 eV

The positions of these lines on a horizontal linear scale are shown below with the wavelengths and transitions indicated. 6→3 5→3 4→3 ---|--------------|---------------------------------------------|----1097 nm 1282 nm 1876 nm

Atoms

1185

*26 •• Picture the Problem We can use Bohr’s second postulate to relate the photon energy to its frequency and use λ =

1240 eV ⋅ nm to find the wavelengths of the three longest Ei − Ef

wavelengths in the Brackett series. (a) Use Bohr’s second postulate to express the energy of the photons in the Paschen series: For the series limit:

hf = ∆E = Ei − Ef

n = ∞ and Ei = 0

Substitute to obtain:

⎛ E ⎞ E ∆E = − Ef = −⎜⎜ − 20 ⎟⎟ = 20 ⎝ n2 ⎠ n2

Evaluate the photon energy for n2 = 4:

hf =

Express the wavelength of the radiation resulting from an energy transition ∆E = hf:

λ=

Evaluate λmin for the transition n = ∞ to n2 = 4: (b) For the three longest wavelengths:

(1)

13.6 eV = 0.850 eV 42

1240 eV ⋅ nm ∆E

(2)

provided the energies are expressed in eV.

λmin =

1240 eV ⋅ nm = 1459 nm 0.850 eV

ni = 5, 6, and 7

E0 ⎛ E0 ⎞ −⎜− ⎟ ni2 ⎜⎝ n22 ⎟⎠ (3) ⎛ 1 1⎞ ⎛1 1⎞ = E0 ⎜⎜ 2 − 2 ⎟⎟ = E0 ⎜⎜ − 2 ⎟⎟ ⎝ n2 ni ⎠ ⎝ 16 ni ⎠

Equation (1) becomes:

∆E = Ei − Ef = −

Evaluate equation (3) for n = 5:

1 ⎞ ⎛1 ∆E5→4 = (13.6 eV ) ⎜ − ⎟ ⎝ 16 25 ⎠ = 0.306 eV

Evaluate equation (2) for ∆E = 0.306 eV:

λ5→4 =

1240 eV ⋅ nm = 4052 nm 0.306 eV

1186 Chapter 36 Evaluate equation (3) for n = 6:

1 ⎞ ⎛1 ∆E6→4 = (13.6 eV ) ⎜ − ⎟ ⎝ 16 36 ⎠ = 0.472 eV 1240 eV ⋅ nm = 2627 nm 0.472 eV

Evaluate equation (2) for ∆E = 0.472 eV:

λ6→4 =

Evaluate equation (3) for n = 7:

1 ⎞ ⎛1 ∆E7→4 = (13.6 eV ) ⎜ − ⎟ ⎝ 16 49 ⎠ = 0.572 eV

Evaluate equation (2) for ∆E = 0.572 eV:

λ7→4 =

1240 eV ⋅ nm = 2168 nm 0.572 eV

The positions of these lines on a horizontal linear scale are shown below with the wavelengths and transitions indicated.

7→4 6→4 5→4 ---|---------|---------------------------------------------|----2168 nm 2627 nm 4052 nm

27 •• Picture the Problem We can use the grating equation to find the wavelength of the given spectral line and the Rydberg-Ritz formula to evaluate R. (a) The grating equation is:

mλ = d sin θ where m = 1, 2, 3, …

Solve for λ:

Substitute numerical values and evaluate λ for m = 1:

λ= λ=

d sin θ m

(3.377 µm )sin 11.233° 1

= 657.8 nm (b) The Rydberg-Ritz formula is:

⎛ 1 1 ⎞ = R⎜⎜ 2 − 2 ⎟⎟ λ ⎝ n2 n1 ⎠

1

Atoms Solve for R to obtain:

⎛ ⎞ ⎜ ⎟ 1⎜ 1 ⎟ R= ⎜ ⎟ 1 1 λ ⎜ n2 − n2 ⎟ 1 ⎠ ⎝ 2

Substitute numerical values and evaluate R:

⎛ ⎞ ⎜ 1 ⎟ 1 ⎜ ⎟ R= 0.6578 µm ⎜ 1 − 1 ⎟ ⎜ 2 ⎟ 32 ⎠ ⎝2

1187

= 10.946 µm −1 = 1.0946 × 10 7 m −1 Remarks: The data used here came from a real experiment. The value for R differs by approximately 0.2% from the commonly accepted value. 28 ••• Picture the Problem This is an extreme value problem in which we need to identify the relationship between E and r, differentiate it with respect to r, and set that derivative equal to zero. Solving the latter expression for r will give us rm.

ke 2 h2 − 2mr 2 r

Express the total energy of the electron:

E=

Differentiate this expression with respect to r to obtain:

dE d ⎡ h 2 ke 2 ⎤ = − ⎢ ⎥ dr dr ⎣ 2mr 2 r ⎦ d ⎡ h 2 ⎤ d ⎡ ke 2 ⎤ ⎢ ⎥− ⎢ ⎥ dr ⎣ 2mr 2 ⎦ dr ⎣ r ⎦ ke 2 h2 =− 3 + 2 mr r = 0 for extreme values

=

Solve for r to obtain:

r=

h2 ke 2 m

Differentiate E a second time to obtain:

d 2 E 3h 2 2ke 2 = − 3 dr 2 mr 4 r

Evaluate d2E/dr2 at r to obtain:

d 2E dr 2

= r=

h2 ke2m

k 4 e8 m 3 >0 h6

Therefore, our extreme value is a minimum and the value for r that minimizes the

1188 Chapter 36 energy is r =

Note that this is just the Bohr radius a0. Consequently, the energy is the ground state energy of the hydrogen atom and:

h2 ke 2 m

Emin = 13.6 eV

*29 ••• Picture the Problem We can express the total kinetic energy of the electron-nucleus system as the sum of the kinetic energies of the electron and the nucleus. Rewriting these kinetic energies in terms of the momenta of the electron and nucleus will lead to K = p2/2mr. (a) Express the total kinetic energy of the electron-nucleus system:

K = Ke + Kn

Express the kinetic energies of the electron and the nucleus in terms of their momenta:

Ke =

Substitute to obtain:

K= =

p2 p2 and K n = 2me 2M p2 p2 p2 ⎛ 1 1 ⎜⎜ + = + 2me 2M 2 ⎝ me M

⎞ ⎟⎟ ⎠

p 2 ⎛ M + me ⎞ p2 ⎜⎜ ⎟⎟ = 2 ⎝ me M ⎠ ⎛ mM ⎞ ⎟⎟ 2⎜⎜ e + M m e ⎠ ⎝ p2 2mr

=

provided we define µ = meM/(M + me). (b) From Equation 36-14 we have:

⎛ ⎜ m mr k 2e 4 e ⎜ R= = C 4πch 3 ⎜ 1 + me ⎜ M ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

where

C=

Use the Table of Physical Constants at the end of the text to obtain:

k 2e 4 4πch 3

C = 1.204663 × 1037 m −1 / kg

(1)

Atoms ⎛ ⎜ me RH = C ⎜⎜ me ⎜1+ ⎝ mp

For H:

1189

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Substitute numerical values and evaluate RH:

⎛ ⎞ ⎜ ⎟ −31 9.11 × 10 kg ⎟ ⎜ 37 −1 RH = 1.204663 × 10 m / kg ⎜ = 1.096850 × 10 7 m −1 9.11 × 10 −31 kg ⎟ ⎜1+ ⎟ − 27 ⎝ 1.67 × 10 kg ⎠

(

)

RH,approx = Cme

Let M → ∞ in equation (1) to obtain RH,approx:

Substitute numerical values and evaluate RH,approx:

(

)(

)

RH,approx = 1.204663 × 1037 m −1 / kg 9.11 × 10 −31 kg = 1.097448 × 10 7 m −1

RH and RH,approx agree to three significant figures. (c) Express the ratio of the kinetic energy K of the electron in its orbit about a stationary nucleus to the kinetic energy of the reduced-mass system K′:

p2 2m 1 ⎛⎜ mp me ⎞⎟ µ K = 2e = = p me me ⎜⎝ mp + me ⎟⎠ K' 2 mr =

Substitute numerical values and evaluate the ratio of the kinetic energies:

mp mp + me

=

1 m 1+ e mp

K = K'

1 9.11 × 10 −31 kg 1+ 1.67 × 10 −27 kg = 0.999455

or

K = 0.999455 K' and the correction factor is the ratio of the masses or 0.0545%

1190 Chapter 36 Remarks: The correct energy is slightly less than that calculated neglecting the motion of the nucleus. *30 •• Picture the Problem We can use Equation 36-15 with Z = 2 to explain how it is that every other line of the Pickering series is very close to a line in the Balmer series. We can use the relationship between the energy difference between two quantum states and the wavelength of the photon emitted during a transition from the higher state to the lower state to find the wavelength of the photon corresponding to a transition from the n = 6 to the n = 4 level of He+. (a) From Equation 36-15, the energy levels of an atom are given by:

En = − Z 2

E0 n2

where E0 is the Rydberg constant (13.6 eV). For He+, Z = 2 and:

En = −4

E0 n2

Because of this, an energy level with even principal quantum number n in He+ will have the same energy as a level with quantum number n/2 in H. Therefore, a transition between levels with principal quantum numbers 2m and 2p in He+ will have almost the same energy as a transition between level m and p in H. In particular, transitions from 2m to 2p = 4 in He+ will have the same energy as transitions from m to n = 2 in H (the Balmer series).

hc E6 − E 4

(b) Transitions between these energy levels result in the emission or absorption of a photon whose wavelength is given by:

λ=

Evaluate E6 and E4:

⎛ 13.6 eV ⎞ E6 = −4⎜ ⎟ = −1.51 eV 2 ⎝ 6 ⎠

(1)

and

⎛ 13.6 eV ⎞ E4 = −4⎜ ⎟ = −3.40 eV 2 ⎝ 4 ⎠ Substitute for E6 and E4 in equation (1) and evaluate λ:

λ=

1240 eV ⋅ nm = 656 nm − 1.51 eV − (− 3.40 eV )

which is the same as the n = 3 to n = 2 transition in H.

Quantum Numbers in Spherical Coordinates 31 • Picture the Problem We can use the expression relating L to l to find the magnitude of the angular momentum and the constraints on the quantum numbers to determine the

Atoms

1191

allowed values for m. (a) Express the angular momentum as a function of l : Substitute numerical values and evaluate L:

L = l(l + 1)h

L = 1(1 + 1)h = 2h

(

= 2 1.055 × 10 −34 J ⋅ s

)

= 1.49 × 10 −34 J ⋅ s (b) Because ml = − l , ... , 0, ... , l the allowed values for l = 1 are:

ml = − 1, 0, + 1

(c) The vector diagram is shown on the right. Note that because Lz = mlħ and L = 2h , the vectors for ml = −1 and ml= 1 must make angles of 45° with the z axis.

32 • Picture the Problem We can use the expression relating L to l to find the magnitude of the angular momentum and the constraints on the quantum numbers to determine the allowed values for m. (a) Express the angular momentum as a function of l : Substitute numerical values and evaluate L:

L = l(l + 1)h L = 3(3 + 1)h = 2 3h

(

= 2 3 1.055 × 10 −34 J ⋅ s

)

= 3.65 × 10 −34 J ⋅ s (b) Because ml = − l , ... , 0, ... , l the allowed values for l = 1 are:

ml = − 3, − 2, − 1, 0, + 1, + 2, + 3

1192 Chapter 36 (c) The vector diagram is shown on the right. Note that because Lz = mlħ and L = 2 3h , the angles between the vectors and the z axis are determined by cos θ m = ml 2 3 . Thus θ3 = 30°, θ2 = 54.7°, and θ1 = 73.2°. The spacing between the allowed values of Lz is constant and equal to h. 33 • Picture the Problem We can find the possible values of l by using the constraints on the quantum numbers. A more analytical solution is to first derive the number of electron states for an arbitrary value of n and then substitute the specific value of n. (a) When n = 3:

l = 0,1, 2

(b) For l = 0:

ml = 0

For l = 1:

ml = − 1, 0,+1

For l = 2:

ml = − 2, − 1, 0,+1, + 2 n 3 3 3 3 3 3 3 3 3

(c) We can find the total number of electron states by enumerating the possibilities as shown in the table.

l

ml 0 −1 0 1 −2 −1 0 1 2

0 1 1 1 2 2 2 2 2

Note that there are 9 m states. Because N , the number of electron states is twice the number of ml states, the number of electron states is 18. Alternatively, we can derive an expression for the number of electron states for an arbitrary value of n and then substitute specific values of n: The number of ml states for a given n is given by:

n −1

n −1

n −1

l =0

l =0

l =0

N m = ∑ (2l + 1) = 2∑ l + ∑ (1)

Atoms Express the sum of all integers from 0 to p: n−1

Use this result to evaluate 2

∑l : l =0

Evaluate the second term to obtain:

p

∑l = l =0

1 2

1193

p( p + 1)

n−1

2∑ l = 2[ 12 (n − 1)n] = n 2 − n l =0

n−1

∑ (1) = n l =0

Nm = n2 − n + n = n2

Substitute to obtain:

and, because N, the number of electron states is twice the number of m states, the number of electron states is N = 2n2. Hence, for n = 3, the number of electron states is:

N = 2n 2 = 2(3) = 18 2

34 • Picture the Problem While we could find the number of electron states by finding the possible values of l from the constraints on the quantum numbers and then enumerating the states, we’ll take a more analytical approach by deriving an expression for the number of electron states for an arbitrary value of n and then substitute specific values of n. The number of m states for a given n is given by: Express the sum of all integers from 0 to p: n−1

Use this result to evaluate 2

∑l : l =0

Evaluate the second term to obtain:

n −1

n −1

n −1

l =0

l =0

l =0

N m = ∑ (2l + 1) = 2∑ l + ∑ (1) p

∑l = l =0

1 2

p( p + 1)

n−1

2∑ l = 2[ 12 (n − 1)n] = n 2 − n l =0

n−1

∑ (1) = n l =0

Substitute to obtain:

Nm = n2 − n + n = n2 and, because N, the number of electron states is twice the number of m states, the number of electron states is N = 2n2.

1194 Chapter 36 (a) For n = 4, the number of electron states is:

N = 2n 2 = 2(4 ) = 32

(b) For n = 2, the number of electron states is:

N = 2n 2 = 2(2 ) = 8

2

2

*35 •• r Picture the Problem The minimum angle between the z axis and L is the angle between

r

the L vector for m = l and the z axis. Express the angle θ as a function of Lz and L:

r

⎛ Lz ⎞ ⎟ ⎝ L ⎠

θ = cos −1 ⎜

Relate the z component of L to ml and l :

Lz = ml h = lh

Express the angular momentum L:

L = l(l + 1)h

Substitute to obtain:

θ = cos −1 ⎜⎜

(a) For l = 1:

θ = cos −1 ⎜⎜

(b) For l = 4:

θ = cos −1 ⎜⎜

(c) For l = 50:

θ = cos −1 ⎜⎜

⎞ ⎛ ⎞ lh ⎟ = cos −1 ⎜ l ⎟ ⎜ l +1 ⎟ ⎟ ⎝ ⎠ ⎝ l(l + 1)h ⎠ ⎛

⎛

1 ⎞ ⎟ = 45.0° ⎟ 1 1 + ⎝ ⎠ ⎛

4 ⎞ ⎟ = 26.6° ⎟ ⎝ 4 +1 ⎠

⎛

50 ⎞ ⎟ = 8.05° ⎟ ⎝ 50 + 1 ⎠

36 •• Picture the Problem We can use constraints on the quantum numbers in spherical coordinates to find the possible values of n and ml for each of the values of l . The constraints on n, ml, and l are:

n = 1, 2, 3, ... l = 0, 1, 2, ..., n − 1 ml = − l , (− l + 1), ... 0, 1, 2, ... l

Atoms

1195

(a) For l = 3:

n ≥ 4 and ml = −3, − 2, − 1, 0, 1, 2, 3

(b) For l = 4:

n ≥ 5 and ml = −4,−3,−2,−1,0,1,2,3,4

(c) For l = 0:

n ≥ 1 and ml = 0

37 •• Picture the Problem The magnitude of the orbital angular momentum L of an electron is

l(l + 1)h and the z component of the angular momentum of the electron is given by Lz = mh.

related to orbital quantum number l by L =

(a) For the l = 2 state, the square magnitude of the angular momentum is:

L2 = 2(2 + 1)h 2 = 6h 2

(b) For the l = 2 state, the maximum 2 value of Lz is:

L2z = 2 2 h 2 = 4h 2

(c) The smallest value of Lx + L y is given by:

L2x + L2y = L2 − L2z = 6h 2 − 4h 2 = 2h 2

2

2

Quantum Theory of the Hydrogen Atom 38

• 32

1 ⎛ Z ⎞ −Zr a0 ⎜ ⎟ e to evaluate the normalized Picture the Problem We can use ψ (r ) = π ⎜⎝ a0 ⎟⎠

ground-state wave function and its square at r = a0 and P (r ) = 4πr 2 ψ

2

to find the radial

probability density at the same location. (a) Noting that Z = 1 for hydrogen, evaluate ψ (a0 ) to obtain:

32

1 ⎛ 1 ⎞ −a0 ⎜ ⎟ e ψ (a0 ) = π ⎜⎝ a0 ⎟⎠ =

(b) Square ψ (a0 ) to obtain:

a0

1 ea0 πa0

⎛ 1 ψ 2 (a0 ) = ⎜ ⎜ ea πa 0 ⎝ 0

2

⎞ 1 ⎟ = 2 ⎟ e a03π ⎠

1196 Chapter 36 (c) Use the result from part (b) to evaluate P(a0):

⎛ 1 2 P(a0 ) = 4πa02 ψ = 4πa02 ⎜⎜ 2 3 ⎝ e a0π 4 = 2 e a0

⎞ ⎟⎟ ⎠

*39 • Picture the Problem We can use the constraints on n, l, and m to determine the number of different wave functions, excluding spin, corresponding to the first excited energy state of hydrogen. For n = 2:

l = 0 or 1

(a) For l = 0, ml= 0 and we have:

1 state

For l = 1, ml = −1, 0, +1 and we have:

3 states

Hence, for n = 2 we have:

4 states n 2 2 2 2

(b) The four wave functions are summarized to the right.

40

ml (n, l , ml) 0 (2,0,0) −1 (2,1,−1) 0 (2,1,0) 1 (2,1,1)

l 0 1 1 1

•• 32

1 ⎛ Z ⎞ −Zr a0 ⎜ ⎟ e to evaluate the normalized Picture the Problem We can use ψ (r ) = π ⎜⎝ a0 ⎟⎠

ground-state wave function and its square and P (r ) = 4πr 2 ψ

2

to find the radial

probability density at the same location. Because the range ∆r is so small, the variation in the radial probability density P(r) can be neglected. The probability of finding the electron in some small range ∆r is then P(r) ∆r. Express the probability of finding the electron in the range ∆r:

The radial probability density function is:

Probability = ∫ P(r )dr

(1)

where P(r) is the radial probability density function.

P(r ) = 4πr 2 ψ

2

(2)

Atoms 32

Express the normalized ground-state wave function:

1 ⎛ Z ⎞ −Zr a0 ⎜ ⎟ e ψ (r ) = π ⎜⎝ a0 ⎟⎠

Evaluate the normalized groundstate wave function evaluated at r = a0 to obtain:

1 ⎛ 1 ⎞ −a0 ⎜ ⎟ e ψ (a0 ) = π ⎜⎝ a0 ⎟⎠

Square ψ (a0 ) to obtain:

32

⎛ 1 ψ (a0 ) = ⎜ ⎜ ea πa 0 ⎝ 0 2

a0

=

1 ea0 πa0

2

⎞ ⎟ = 1 ⎟ e 2 a03π ⎠

Substitute in equation (2) to obtain:

⎛ 1 2 P (a0 ) = 4πa02 ψ = 4πa02 ⎜⎜ 2 3 ⎝ e a0 π

(a) Substitute in equation (1) to find the probability of finding the electron in the small range ∆r = 0.03a0:

Probability = ∫ P(a0 )dr ≈ P(a0 )∆r

(b) Evaluate the normalized groundstate wave function at r = 2a0 to obtain:

1 ⎛ 1 ⎞ −2 a0 ⎜ ⎟ e ψ (2a0 ) = π ⎜⎝ a0 ⎟⎠

Square ψ (2a0 ) to obtain:

1197

=

⎞ 4 ⎟⎟ = 2 ⎠ e a0

4 (0.03a0 ) = 0.0162 e a0 2

32

⎛ 1 ψ (2a0 ) = ⎜ 2 ⎜ e a πa 0 0 ⎝ 2

a0

=

1 e a0 πa0 2

2

⎞ ⎟ = 1 ⎟ e 4 a03π ⎠

Substitute in equation (2) to obtain:

⎛ 1 2 2 P(2a0 ) = 4π (2a0 ) ψ = 16πa02 ⎜⎜ 4 3 ⎝ e a0π 16 = 4 e a0

Substitute in equation (1) to find the probability of finding the electron in some small range ∆r = 0.03a0:

Probability = ∫ P(2a0 )dr ≈ P(2a0 )∆r =

⎞ ⎟⎟ ⎠

16 (0.03a0 ) = 0.00879 e 4 a0

Remarks: There is about a 2% chance of finding the electron in this range at r = a0, but at r = 2a0, the chance is only about 0.9%.

1198 Chapter 36 41 •• Picture the Problem We can use Equation 36-36 and the given expression for C2,0,0 to evaluate the spherically symmetric wave function ψ for n = 2, l = 0, ml = 0, and Z = 1 and then use this result to evaluate ψ2 and P(r) for r = a0.

⎛

(a) Express the spherically symmetric wave function for n = 2, l = 0, ml = 0, and Z = 1 (Equation 35-36):

ψ 2,0,0 = C2,0, 0 ⎜⎜ 2 −

Evaluate this expression for r = a0:

1 ψ 2, 0,0 (a0 ) = 4 2π

⎝

1 = 4 2π

[ψ (a )] 2, 0,0

0

⎛1⎞ ⎜⎜ ⎟⎟ ⎝ a0 ⎠

32

⎛ r ⎞ ⎜⎜ 2 − ⎟⎟e −r 2 a0 a0 ⎠ ⎝

⎛1⎞ ⎜⎜ ⎟⎟ ⎝ a0 ⎠

1 4 2eπ a03 2

= (b) Square ψ 2, 0, 0 (a0 ) to obtain:

r ⎞ −r 2 a0 ⎟e a0 ⎟⎠

32

⎛ a ⎞ ⎜⎜ 2 − 0 ⎟⎟e −a0 a0 ⎠ ⎝ 0.0605 = a03 2 2

2

⎛ 0.0605 ⎞ 0.00366 = ⎜⎜ 3 2 ⎟⎟ = a03 ⎝ a0 ⎠

(c) Express the radial probability density:

P(r ) = 4πr 2ψ 2 (r )

Substitute to obtain:

P(a0 ) = 4πa02

0.00366 0.0460 = 3 a0 a0

42 ••• Picture the Problem We can use the definition of the radial probability density and the wave function (Equation 35-37) for the state (2, 1, 0) to obtain the result given in the problem statement. Using Equation 35-37, express the wave function for the state (2, 1, 0): Square ψ 2,1, 0 (r ) to obtain:

ψ 2,1, 0 (r ) = C2,1, 0

[ψ (r )] 2 ,1, 0

2

Zr −Zr 2 a0 e cosθ a0

⎛ ⎞ Zr −Zr 2 a0 = ⎜⎜ C2,1, 0 cosθ ⎟⎟ e a0 ⎝ ⎠ 2 2 Z r = C22,1, 0 2 e −Zr a0 cos 2 θ a0

2

2 a0

Atoms

Express the radial probability density: Substitute and simplify to obtain:

1199

P(r ) = 4πr 2ψ 2 (r )

P(r ) = 4πr 2C22,1,0

Z 2 r 2 −Zr a0 e cos 2 θ a02

= Ar 4 cos 2 θ e −Zr a0 where

4πC22,1,0 Z 2 A == a02 43 ••• Picture the Problem In this instance,

∫ P(r )dr extends over a sufficiently narrow

interval ∆r << 0.02a0 that we can neglect the dependence of P(r) on r. Hence, we can set P(r )dr = P(r )∆r and use the wave function (Equation 36-36) for the state

∫

(2, 0, 0) and the expression for C2,0,0 from Problem 41 to find ψ2,0,0, ψ 22, 0, 0 , P(r), and the probability of finding the electron in the range specified at r = a0 and r = 2a0. (a) Express the probability of finding the electron in the range ∆r:

The radial probability density function is: The normalized wave function for the (2, 0, 0) state of hydrogen is given by Equation 36-36: From Problem 41 we have, for hydrogen:

Substitute to obtain:

Probability = ∫ P(r )dr

(1)

where P(r) is the radial probability density function.

P(r ) = 4πr 2 ψ

2

(2)

⎛

ψ 2, 0,0 (r ) = C2, 0,0 ⎜⎜ 2 − ⎝

C2 , 0 , 0

1 = 4 2π

⎛1⎞ ⎜⎜ ⎟⎟ ⎝ a0 ⎠ ⎛

32

ψ 2, 0,0 (r ) = C2,0,0 ⎜⎜ 2 − ⎝

1 = 4 2π

r ⎞ −r 2 a0 ⎟e a0 ⎟⎠

r ⎞ −r 2 a0 ⎟e a0 ⎟⎠

⎛1⎞ ⎜⎜ ⎟⎟ ⎝ a0 ⎠

32

⎛ r ⎞ ⎜⎜ 2 − ⎟⎟e −r 2 a0 a0 ⎠ ⎝

1200 Chapter 36 32

⎛ 1 ⎞ −a0 2 a0 ⎜⎜ ⎟⎟ e ⎝ a0 ⎠ 1 0.0605 = = 32 a03 2 4 2πea0

Evaluate the normalized groundstate wave function at r = a0 to obtain:

1 ψ 2, 0,0 (a0 ) = 4 2π

Square ψ 2, 0, 0 (a0 ) to obtain:

2

[ψ (a )]

2

2, 0,0

0

⎛ 0.0605 ⎞ 0.00366 = ⎜⎜ 3 2 ⎟⎟ = a03 ⎝ a0 ⎠

Substitute in equation (2) to obtain:

⎛ 0.00366 ⎞ ⎟⎟ P(a0 ) = 4πa02ψ 2 (a0 ) = 4πa02 ⎜⎜ 3 a 0 ⎠ ⎝ 4π (0.00366) 0.0460 = = a0 a0

Substitute in equation (1) to find the probability of finding the electron in some small range ∆r = 0.02 a0:

Probability = ∫ P(a0 )dr ≈ P(a0 )∆r =

0.0460 (0.02a0 ) a0

= 9.20 × 10 −4 (b) Evaluate the normalized ground-state wave function at r = 2a0 to obtain:

1 ψ 2,0,0 (2a0 ) = 4 2π

⎛1⎞ ⎜⎜ ⎟⎟ ⎝ a0 ⎠

32

⎛ 2a ⎞ ⎜⎜ 2 − 0 ⎟⎟e −2 a0 a0 ⎠ ⎝

Square ψ 2, 0, 0 (2a0 ) to obtain:

[ψ (a )]

Substitute in equation (2) to obtain:

P(2a0 ) = 0

Substitute in equation (1) to find the probability of finding the electron in some small range ∆r = 0.02 a0:

2, 0,0

0

2

2 a0

=0

=0

Probability = ∫ P(2a0 )dr ≈ P(2a0 )∆r = 0

*44 •• 32

1 ⎛ Z ⎞ −Zr a0 ⎜ ⎟ e = Ce −Zr a0 is a Picture the Problem We wish to show that ψ 1, 0, 0 = π ⎜⎝ a0 ⎟⎠ kZe 2 h 2 ∂ ⎛ 2 ∂ψ ⎞ ( ) ( ) r + U r ψ = E ψ , where . Because the solution to − U r = − ⎜ ⎟ 2mr 2 ∂r ⎝ ∂r ⎠ r

Atoms

1201

ground state is spherically symmetric, we do not need to consider the angular partial derivatives in Equation 36-21. The normalized ground-state wave function is:

32

ψ 1,0, 0

1 ⎛ Z ⎞ −Zr a0 ⎜⎜ ⎟⎟ e = = Ce −Zr a0 π ⎝ a0 ⎠

Differentiate this expression with respect to r to obtain:

∂ψ 1,0, 0 ∂ −Zr a0 Z =C e = −C e −Zr a0 ∂r ∂r a0

Multiply both sides of this equation by r2:

r2

[

]

∂ψ 1,0, 0 Z = −C r 2 e − Zr a0 ∂r a0

Differentiate this expression with respect to r to obtain: 2 ⎤ ∂ ⎛ 2 ∂ψ 1,0, 0 ⎞ Z ∂ 2 −Zr a0 ⎡ 2 Zr 2⎛ Z ⎞ r e = ⎢− + r ⎜⎜ ⎟⎟ ⎥Ce −Zr a0 ⎜⎜ r ⎟⎟ = −C ∂r ⎝ ∂r ⎠ a0 ∂r ⎢⎣ a0 ⎝ a0 ⎠ ⎥⎦

(

)

Substitute in Schrödinger’s equation to obtain:

h2 − 2mr 2

2 ⎡ 2 Zr ⎤ kZe 2 −Zr a0 − Zr a0 2⎛ Z ⎞ ⎜ ⎟ ⎢− ⎥ − Ce = ECe −Zr a0 + r ⎜ ⎟ Ce r ⎢⎣ a0 ⎝ a0 ⎠ ⎥⎦

Solve for E:

2 ⎤ kZe 2 h 2 ⎡ 2 Zr 2⎛ Z ⎞ ⎢− E=− + r ⎜⎜ ⎟⎟ ⎥ − r 2mr 2 ⎢ a0 ⎝ a0 ⎠ ⎥⎦ ⎣

Because a0 =

h2 : mke2

2 2⎤ h 2 ⎡ 2mke 2 Zr kZe 2 kZe 2 Z 2 k 2e 4 m kZe 2 2 ⎛ Zmke ⎞ ⎟ ⎜ ⎢ E=− − + r ⎥ − = − − ⎜ h2 ⎟ h2 r r r 2mr 2 ⎢ 2h 2 ⎥ ⎠ ⎝ ⎣ ⎦

= −

Z 2 k 2e 4 m 2h 2

Because this is the correct ground state energy, we have shown that Equation 36-33, is a solution to Schrödinger’s Equation 36-21 with the potential energy function Equation 36-26. 45 •• Picture the Problem We can substitute the dimensions of the physical quantities for the physical quantities in Equation 36-28 and simplify the resulting expression to show that it has dimensions of energy.

1202 Chapter 36 Equation 36-28 is:

mk 2e 4 E0 = 2h 2

The units of this equation are:

2 [kg ] ⎡⎢ N ⋅ m2 ⎤⎥ [C]4 2 [ kg ] N ⋅ m 2 ⎣ C ⎦ = [N ⋅ m ⋅ s] 2 [J ⋅ s] 2

2

[

]

2 [ kg ][m ] = [s] 2 = [J ]

46

••

h2 . We can substitute the dimensions Picture the Problem The Bohr radius is a0 = mke 2 of the physical quantities for the physical quantities in this equation and simplify the resulting expression to show that it has dimensions of length. Because the SI units of h are J⋅s, its dimensions are:

M ⋅ L2 M ⋅ L2 T ⋅ = T2 T

Because the SI units of k are N⋅m2/C2, its dimensions are:

M⋅L 2 1 M ⋅ L3 ⋅ L ⋅ = T2 Q2 T 2 ⋅ Q2 where Q is the dimension of charge.

Substitute the dimensions in a0 =

2

h to obtain: mke 2

2

⎡ M ⋅ L2 ⎤ M 2 ⋅ L4 ⎢ T ⎥ ⎦ ⎣ T2 = L = ⎡ M ⋅ L3 ⎤ 2 M 2 ⋅ L3 M⋅⎢ 2 2 ⎥ ⋅Q T2 ⎣T ⋅ Q ⎦

47 •• Picture the Problem This is an extreme value problem. We'll begin its solution with the radial probability distribution function, differentiate it with respect to its independent variable r, set this derivative equal to zero, and solve for the value for an extreme value for r. We can show that this value corresponds to a maximum by evaluating the second derivative of P(r) at the location found from the first derivative.

Atoms Differentiate the radial probability distribution function with respect to r to obtain:

1203

dP(r ) d 2 −2 Zr a0 r e =C dr dr ⎡ 2Zr 2 −2 Zr a0 ⎤ e = C ⎢2re −2 Zr a0 − ⎥ a0 ⎣ ⎦ 2CZr −2 Zr a0 ⎛ a0 ⎞ e = ⎜ −r⎟ a0 ⎝Z ⎠

[

]

= 0 for extrema Solve for r to obtain:

To show that this value for r corresponds to a maximum, differentiate dP(r)/dr to obtain:

Evaluate this derivative at r = a0/Z:

r=

a0 Z

d 2 P(r ) 2CZr −2 Zr a0 ⎛ a0 ⎞ e =− +⎜ −r⎟ 2 dr a0 ⎝Z ⎠ ⎛ 4CZ 2 2CZ ⎞ −2 Zr a0 ⎟e × ⎜⎜ − 2 + a0 a0 ⎟⎠ ⎝ d 2 P(r ) = −2Ce −2 < 0 2 dr r = a0 Z

because C is a positive constant. Hence, P(r) has its maximum value at r =

a0 Z

48 ••• Picture the Problem We can double the sum of the number of m states for a given n to show that the number of states in the hydrogen atom for a given n is 2n2. The number of ml states for a given n is: The sum of all integers from 0 to p is:

n −1

n −1

n −1

l =0

l =0

l =0

N ml = ∑ (2l + 1) = 2∑ l + ∑ (1) p

∑l = l =0

1 2

p( p + 1)

and n−1

2∑ l = 2( 12 )(n − 1)(n ) = n 2 − n l =0

The second term is:

n−1

∑ (1) = n l =0

1204 Chapter 36 Substitute to obtain:

N ml = n 2 − n + n = n 2

Because N, the number of electron states, is twice the number of ml states, the number of electron states is:

N = 2 N ml = 2n 2

49 ••• Picture the Problem The ground state of a hydrogen atom is the state described by n = 1, l = 0, ml = 0. We can calculate the probability that the electron in the ground state of the a0

hydrogen atom is in the region 0 < r < a0 by evaluating the integral 4πr 2ψ 12, 0, 0 (r )dr .

∫ 0

a0

Express the probability that the electron in the ground state of a hydrogen atom is in the region 0 < r < a0:

Probability = ∫ 4πr 2ψ 12,0, 0 (r )dr

Express the ground-state wave function for hydrogen:

1 ⎛ 1 ⎞ −r a0 ⎜ ⎟ e ψ 1, 0,0 (r ) = π ⎜⎝ a0 ⎟⎠

Square the wave function to obtain:

ψ 12, 0,0 (r ) =

Substitute to obtain:

0

32

1 −2r a0 e πa03 a0

⎛ 1 ⎞ Probability = ∫ 4πr 2 ⎜⎜ 3 e −2r a0 ⎟⎟dr ⎝ πa0 ⎠ 0 a

=

Use a table of integrals to find:

Use this integral to show that:

2 bx ∫ x e dx =

4 0 2 −2r a0 r e dr a03 ∫0

(

ebx 2 2 b x − 2bx + 2 b3

) a0

Probability = − e

−2 r a0

⎛ 2 r 2 2r ⎞ ⎜⎜ 2 + + 1⎟⎟ ⎝ a0 a0 ⎠0

= 1 − 5e −2 = 0.323

Atoms

1205

The Spin-Orbit Effect and Fine Structure *50 • Picture the Problem The energy difference between the two possible orientations of an electron in a magnetic field is 2µB and the wavelength of the photons required to induce a spin-flip transition can be found from hc/∆E. The magnetic moment µB associated with the spin of an electron is 5.79×10−5 eV/T. (a) Relate the difference in energy between the two spin orientations in terms of the difference in the potential energies of the two states:

∆E = 2µB

(

)

= 2 5.79 × 10−5 eV/T (0.6 T ) = 6.95 × 10 −5 eV

(b) Relate the wavelength of the photon needed to induce such a transition to the energy required:

λ=

hc ∆E

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 1.78 ×107 nm −5 6.95 × 10 eV

= 1.78 cm 51 • Determine the Concept j and l are constrained according to j = l ± 12 . For j = ½, l = 12 ± 12 or l = 0 or 1 . 52 • Determine the Concept j and l are constrained according to j = l ± 12 . For

l = 2, j = 2 ± 12 , or j =

3 2

or

5 2

53 • Picture the Problem The total angular r momentum vector J is the sum of the

r

orbital momentum vector L and the spin

r

orbital angular momentum vector S .The quantum number j can be either l + 12 or

l − 12 , where l ≠ 0. Hence, j can take on

the values 3 + 1/2 = 7/2 and 3 − 1/2 = 5/2. The scaled vector diagrams are shown to the right.

1206 Chapter 36

The Periodic Table 54 • Determine the Concept The total number of quantum states of hydrogen with quantum number n is 2n2. For n =4, we have 2(4)2 = 32. (c) is correct. 55 • Determine the Concept From Table 36-1, oxygen’s electronic configuration is 1s22s22p4. Because there are 4 electrons in the p state, (c) is correct. *56 • Determine the Concept We can use the atomic numbers of carbon and oxygen to determine the sum of the exponents in their electronic configurations and then use the rules for the filling of the shells to find their electronic configurations. (a) The atomic number Z of carbon is 6. So we must fill the subshells of the electronic configuration until we have placed its 6 electrons. This is accomplished by writing:

1s 2 2s 2 2p 2

(b) The atomic number Z of oxygen is 8. So we must fill the subshells of the electronic configuration until we have placed its 8 electrons. This is accomplished by writing:

1s 2 2s 2 2p 4

57 • Determine the Concept We can find the z component of the orbital angular momentum using Lz = mħ and the relationship between the quantum numbers l (which we know from the state of the electrons) and ml (which is related to l through ml = −l, (−l + 1), …, 0, 1, 2, … l). (a) For a p electron l = 1. For l = 1, ml = −1, 0, or 1. Because Lz = −mħ, …, mħ:

Lz = − 2h, − h, 0, h, 2h

(b) For an f electron, l = 4. For l = 4, ml = −4, −3, −2, −1, 0, 1, 2, 3, 4. Because Lz = −mħ, …, mħ:

Lz = − 4h, − 3h, − 2h, − h, 0, h, 2h, 3h, 4h

Atoms

1207

Optical Spectra and X-Ray Spectra 58 • Determine the Concept Lithium, sodium, potassium, chromium, and cesium have one outer s electron and hence belong in the same group. Beryllium, magnesium, calcium, nickel, and barium have two outer s electrons and, hence, belong in the same group. 59 • Determine the Concept We can use Table 35-1 to find the electronic configurations for the first excited states of these elements. (a) For H, E depends only on n and the lowest excited state is:

2s or 2p

(b) For Na, the 3p state is higher energy than the 3s state and the lowest excited state is:

1s 2 2s 2 2p 6 3p

(c) For He, the lowest excited state has one electron in the 2s state and the lowest excited state is:

1s2s

60 • Determine the Concept Atoms with one outer electron have spectra similar to H: Li, Rb, Ag, Fr. Atoms with two outer electrons have spectra similar to He: Ca, Ti, Hg, Cd, Ba, Ra. Therefore, the table should be completed as shown below: Optical Spectra Similar to Hydrogen

Optical Spectra Similar to Helium

Li, Rb, Ag, Fr

Ca, Ti, Hg, Cd, Ba, Ra

*61 • Picture the Problem When an electron from state n drops into a vacated state in the n = 1 shell, a photon of energy ∆E = En − E1 is emitted. We can find the wavelength of this photon using λ = hc ∆E . The second and third longest wavelengths in the K series correspond to transitions from n = 3 to n = 1 and n = 4 to n = 1 and the shortest wavelength to the transition from n = ∞ to n = 1. Express the wavelength of the emitted photon in terms of the energy transition within the atom:

λ=

hc 1240 eV ⋅ nm = En − E1 En − E1

1208 Chapter 36 Express the energy of the nth energy state:

En = −(Z − 1)

2

E0 n2

where n = 1, 2, … Substitute to obtain:

λ=

hc En − E1

1240 eV ⋅ nm ⎛ 2 E 2 E ⎞ − (Z − 1) 20 − ⎜ − (Z − 1) 20 ⎟ 1 ⎠ n ⎝ 1240 eV ⋅ nm = (Z − 1)2 E0 ⎛⎜1 − 12 ⎞⎟ ⎝ n ⎠

=

(a) Evaluate this expression with n = 3 and Z = 42 to obtain:

λ3 =

1240 eV ⋅ nm (42 − 1)2 (13.6 eV )⎛⎜1 − 12 ⎞⎟ ⎝ 3 ⎠

= 0.0610 nm Use n = 4 and Z = 42 to obtain:

λ4 =

1240 eV ⋅ nm (42 − 1)2 (13.6 eV )⎛⎜1 − 12 ⎞⎟ ⎝ 4 ⎠

= 0.0578 nm (b) The shortest wavelength in the series corresponds to the largest energy difference between the initial and final states. Repeat the calculation in part (a) with n = ∞ to obtain:

λ∞ =

1240 eV ⋅ nm (42 − 1)2 (13.6 eV )(1 − 0)

= 0.0542 nm

62 • Picture the Problem When an electron from state n drops into the vacated state in the n = 1 shell, a photon of energy En − E1 is emitted. The wavelength of this photon is λ =

hc

(Z − 1) (13.6 eV )⎛⎜1 − 12 ⎞⎟ ⎝ n ⎠

. Hence, if we know the wavelength of the Kα line we

2

can solve for the atomic number of the element and use its value to identify the element.

Atoms Express the wavelength of the Kα line as a function of the atomic number of the element: Solve for Z:

λ=

hc

(Z − 1)2 (13.6 eV )⎛⎜1 − ⎝

Z = 1+

Substitute numerical values and evaluate Z:

1209

Z = 1+

1⎞ ⎟ n2 ⎠

hc ⎛ ⎝

λ (13.6 eV )⎜1 −

1 ⎞ ⎟ n2 ⎠

1240 eV ⋅ nm

(0.3368 nm )(13.6 eV )⎛⎜1 − ⎝

1⎞ ⎟ 22 ⎠

= 20 The element whose atomic number is 20 is calcium. 63 • Picture the Problem The Kα corresponds to a transition from n = 2 to n = 1. Equation 36-16 relates the atomic number Z to the wavelength of the emitted photon When an electron from state n drops into a vacated state in the n = 1 shell, a photon of energy En − E1 is emitted. We can find the wavelength of this photon using λ = hc (En − E1 )

(

)

and En from En = − Z 2 E0 n 2 .

hc 1240 eV ⋅ nm = En − E1 En − E1

Express the wavelength of the Kα line:

λ=

Express the energy of the atom’s nth energy state:

En = − Z 2

Substitute and simplify to obtain:

λ=

(a) Substitute n = 2, Z = 12, and E0 = 13.6 eV to obtain:

λ=

E0 n2

1240 eV ⋅ nm 1240 eV ⋅ nm = E 1 − Z 2 20 + Z 2 E0 Z 2 E0 ⎛⎜1 − 2 ⎞⎟ n ⎝ n ⎠ 1240 eV ⋅ nm = 1.00 nm 1⎞ ⎛ 2 11 (13.6 eV )⎜1 − 2 ⎟ ⎝ 2 ⎠

1210 Chapter 36 (b) Substitute n = 2, Z = 29, and E0 = 13.6 eV to obtain:

λ=

1240 eV ⋅ nm 1⎞ ⎛ 282 (13.6 eV )⎜1 − 2 ⎟ ⎝ 2 ⎠

= 0.155 nm

General Problems 64 • Picture the Problem The energy associated with a transition from an initial state to some final state is given by ∆E = Ei − Ef and the wavelength λ of a photon emitted in such a transition is given by λ = hc/∆E. Hence, the shortest wavelength corresponds to the largest energy difference. Express the wavelength of the emitted photon in terms of the energy difference ∆E between the atom’s initial and final states: For λmin, ∆E will be the energy required to ionize a hydrogen atom:

λ=

hc hc or ∆E = ∆E λ

∆Emax =

hc

λmin

= 13.6 eV

65 • Picture the Problem This spectral line is due to a transition from some initial state ni to a final state nf (we’re given that the final state is the ground state). The wavelength of the spectral line is related to the difference in energy ∆E between these states according to λ = 1240 eV ⋅ nm ∆E and the energy of the nth state is given (for hydrogen, Z = 1) by En = (12)(−13.6 eV)/n2.

1240 eV ⋅ nm ∆E

Relate the wavelength of a spectral line to the energy transition within the atom:

λ=

Express the energy difference ∆E in a transition:

Z 2 E0 Z 2 E0 ∆E = Ei − Ef = − 2 + 2 ni nf ⎛ 1 1 ⎞ = Z 2 E0 ⎜⎜ 2 − 2 ⎟⎟ ⎝ nf ni ⎠

For Z = 1 and E0 = 13.6 eV:

⎛ 1 1⎞ ∆E = (13.6 eV )⎜⎜ 2 − 2 ⎟⎟ ⎝ nf ni ⎠

(1)

Atoms Substitute in equation (1) to obtain:

λ=

=

1211

1240 eV ⋅ nm ⎞ ⎛ (13.6 eV )⎜⎜ 12 − 12 ⎟⎟ ⎝ nf ni ⎠ 91.2 nm 1 1 − nf2 ni2

or

1 1 91.2 nm − = λ nf2 ni2 For λ = 97.254 nm and nf = 1 this expression simplifies to:

1−

Solve for ni to obtain:

ni = 4

1 91.2 nm = = 0.938 2 ni 97.254 nm

The transition that produced the given wavelength was from ni = 4 to nf = 1. 66 • Picture the Problem This spectral line is due to a transition from some initial state ni to a final state nf (we’re given that the final state is the ground state). The wavelength of the spectral line is related to the difference in energy ∆E between these states according to λ = 1240 eV ⋅ nm ∆E and the energy of the nth state is given (for hydrogen, Z = 1) by En = (12)(−13.6 eV)/n2.

1240 eV ⋅ nm ∆E

Relate the wavelength of a spectral line to the energy transition within the atom:

λ=

Express the energy difference ∆E in a transition:

∆E = Ei − Ef =−

Z 2 E0 Z 2 E0 + 2 ni2 nf

⎛ 1 1 ⎞ = Z 2 E0 ⎜⎜ 2 − 2 ⎟⎟ ⎝ nf ni ⎠ For Z = 1 and E0 = 13.6 eV:

⎛ 1 1⎞ ∆E = (13.6 eV )⎜⎜ 2 − 2 ⎟⎟ ⎝ nf ni ⎠

(1)

1212 Chapter 36 Substitute in equation (1) to obtain:

λ=

=

1240 eV ⋅ nm ⎞ ⎛ (13.6 eV )⎜⎜ 12 − 12 ⎟⎟ ⎝ nf ni ⎠ 91.2 nm 1 1 − nf2 ni2

or

1 1 91.2 nm − = λ nf2 ni2 For λ = 1093.8 nm this expression simplifies to:

1 1 91.2 nm − 2 = = 0.0834 2 nf ni 1093.8 nm

Because the only constraints on nf and ni are that they be integers, we can solve this equation by trial and error. One way to do this is to plot a graph of ni as a function of nf and look for integer solutions visually or with a trace of the trajectory of the curve. The following graph was plotted using a spreadsheet program. Note that a solution to our equation is ni = 6 and nf = 3.

3.5

3.0

2.5

nf 2.0

1.5

1.0 0

1

2

3

4

5

6

7

8

9

10

ni

Thus, the transition that produces the given wavelength is from ni = 6 to nf = 3.

Atoms

1213

67 • Picture the Problem These spectral lines are due to transitions in singly ionized helium from some initial state ni to a final state nf. The wavelengths of the spectral lines are related to the difference in energy ∆E between these states according to λ = 1240 eV ⋅ nm ∆E and the energy of the nth state is given (for helium, Z = 2) by En = (22)(−13.6 eV)/n2.

1240 eV ⋅ nm ∆E

Relate the wavelength of a spectral line to the energy transition within the atom:

λ=

Express the energy difference ∆E in a transition:

∆E = Ei − Ef =−

(1)

Z 2 E0 Z 2 E0 + 2 ni2 nf

⎛ 1 1 ⎞ = Z 2 E0 ⎜⎜ 2 − 2 ⎟⎟ ⎝ nf ni ⎠ For Z = 2 and E0 = 13.6 eV:

⎛ 1 1⎞ ∆E = 22 (13.6 eV )⎜⎜ 2 − 2 ⎟⎟ ⎝ nf ni ⎠ ⎛ 1 1⎞ = (54.4 eV )⎜⎜ 2 − 2 ⎟⎟ ⎝ nf ni ⎠

Substitute in equation (1) to obtain:

λ=

22.8 nm 1240 eV ⋅ nm = 1 1 ⎛ ⎞ (54.4 eV ) ⎜⎜ 12 − 12 ⎟⎟ n 2 − n 2 f i ⎝ nf ni ⎠

or

1 1 22.8 nm − = nf2 ni2 λ For λ = 164 nm this expression becomes:

1 1 22.8 nm − = = 0.139 nf2 ni2 164 nm

Because the only constraints on nf and ni are that they be integers, we can solve this equation by trial and error. One way to do this is to plot a graph of ni as a function of nf and look for integer solutions visually or with a trace of the trajectory of the curve. The following graph was plotted using a spreadsheet program. Note that a solution to our equation is ni = 3 and nf = 2.

1214 Chapter 36 2.6 2.4 2.2 2.0

nf 1.8 1.6 1.4 1.2 1.0 0

1

2

3

4

5

6

7

8

9

10

ni

Thus, the transition that produces the given wavelength is from ni = 3 to nf = 2. Similarly, for λ = 230.6 nm:

The transition that produces the given wavelength is from ni = 9 to nf = 3.

For λ = 541 nm:

The transition that produces the given wavelength is from ni = 7 to nf = 4.

*68 •• Picture the Problem We can show that ke2 = 1.44 eV⋅nm by solving the equation for the ground state energy of an atom for ke2.

ke 2 2a0

Express the ground state energy of an atom as a function of k, e, and a0:

E0 =

Solve for ke2:

ke 2 = 2 E0 a0

Substitute for E0 and a0 to obtain:

ke 2 = 2(13.6 eV )(0.0529 nm ) = 1.44 eV ⋅ nm

Atoms

1215

69 •• Picture the Problem Because the energies of the photons emitted by potassium during these transitions are related to their wavelengths through hf = (1240 eV⋅ nm/λ) eV where λ is in nm, we can use this relationship to find the energies of the given photons. The difference in energy between these states can be found using its definition and is related to the magnetic field through ∆E = 2 µ B B . (a) For λ = 766.41 nm:

For λ = 769.90 nm:

(b) Using its definition, express the difference in energy between these two states: (c) Relate the energy difference between these states ∆E to the magnetic field B and the quantum unit of magnetic moment (a Bohr magneton) µB:

hf =

1240 eV ⋅ nm = 1.6179 eV 766.41 nm

hf =

1240 eV ⋅ nm = 1.6106 eV 769.90 nm

∆E = 1.6179 eV − 1.6106 eV = 0.00730 eV ∆E = 2µ B B

Solve for B:

B=

∆E 2µ B

Substitute numerical values and evaluate B:

B=

0.0073 eV = 63.0 T 2 5.79 × 10-5 eV/T

(

)

Remarks: This magnetic field is about 42 times that of commercial magnetic resonance imagers. 70 •• Picture the Problem One 1s electron must be released from the atom. It is shielded from the nuclear charge Z by one other 1s electron. Thus, the effective charge is Z − 1, and the ionization energy for that 1s electron is Emin = (Z − 1)2E0. (a) For tungsten, Z = 74, and:

Emin = 732 (13.6 eV ) = 72.5 keV

(b) For molybdenum, Z = 42, and:

Emin = 412 (13.6 eV ) = 22.9 keV

1216 Chapter 36 (c) For copper, Z = 29, and:

Emin = 282 (13.6 eV ) = 10.7 keV

*71 •• Picture the Problem We can show that α is dimensionless by showing that it has no units. In part (b) we can use Bohr’s 3rd postulate and the expression for the radii of the Bohr orbits, together with the definition of α, to show that the speed of the electron in a stationary state of quantum number n is related to α according to vn = cα /n. (a) Express the units of α:

⎛ N ⋅ m2 ⎞ ⎟⎟ C 2 ⎜⎜ 2 2 ⎝ C ⎠ = N⋅m =1 J⋅m (J ⋅ s ) m s

Because α is unitless, it is also dimensionless.

nh mrn

(b) Apply the quantization of angular momentum postulate to obtain:

vn =

The radii of the Bohr orbits are given by:

rn = n 2

h2 mkZe 2

or, because Z = 1 for hydrogen,

rn = n 2

h2 mke 2

nh ke 2 = 2 nh 2 h mn 2 mke

Substitute and simplify to obtain:

vn =

Divide this expression by the definition of α to obtain:

ke 2 c vn = n2h = α ek n hc

Solve for vn:

vn =

αc n

72 •• Picture the Problem We can use Problem 29 to express the energy levels of positronium in terms of the reduced mass of the electron-positron system. In part (b) we can find the

Atoms

1217

energies corresponding to 400 nm and 700 nm to decide whether the transitions between any of the levels found in (a) fall in the visible range of wavelengths Express the energy of positronium as a function of the quantum number n:

En = −

From Problem 29 we have:

mr =

mr k 2e 4 Z 2 2h 2 n 2 m e m pos

m e + m pos me me m = e 2 me + me

Because me = mpos:

mr =

Substitute and simplify to obtain:

me k 2e 4 1 E En = − = − 02 2 2 4h n 2n

(a) Evaluate En for n = 1, 2, 3, 4, and 5 to obtain:

n 1 2 3 4 5

hc 1240 eV ⋅ nm = ∆E ∆E

Relate the wavelength of the emitted photons to the energy-level differences:

λ=

Solve for ∆E:

∆E =

Evaluate ∆E for λ = 400 nm and λ = 700 nm:

En (eV) −6.80 −1.70 −0.756 −0.425 −0.272

1240 eV ⋅ nm

∆E400 nm = and

∆E700 nm =

λ 1240 eV ⋅ nm = 3.10 eV 400 nm 1240 eV ⋅ nm = 1.77 eV 700 nm

Because none of the energies in the table shown above are in the interval 1.77 eV to 3.10 eV, no transitions are in the visible range of wavelengths.

1218 Chapter 36 73 • Picture the Problem We can use E = hf to find the frequency of the photon and λ = hc/E to find its wavelength. (a) The energy of the photon whose energy is equal to the Lamb shift energy is given by: Solve for f to obtain:

E = hf

f =

E h

Substitute numerical values and evaluate f:

f =

4.372 × 10 −6 eV = 1.06 GHz 4.14 × 10 −15 eV ⋅ s

(b) The wavelength of this photon is given by:

λ=

hc E

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 28.4 cm 4.372 × 10−6 eV

This wavelength is in the microwave portion of the electromagnetic spectrum. *74 • Picture the Problem The ionization energy of the electron is the magnitude of the energy of the atom in the given state. We can use E = −E0/n2, where E0 is the ground-state energy, to find the energy levels in the 44th and 45th states and, hence, the energy level separation between the states. The wavelength of a photon resonant with this transition can be found from λ = hc/∆E. We’ll approximate the size of the atom in the n = 45 state by finding the radius of the outer-shell electron. (a) The energy of the atom in its nth state is:

En = −

The energy of the atom in the n = 45 state is:

E45 = −

The ionization energy is the negative of the energy in the state n = 45:

Eionizing = − E45 = 6.72 meV

(b) The energy level separation between the n = 45 and n = 44 state is:

⎛ 13.6 eV 13.6 eV ⎞ ⎟ E45→44 = −⎜⎜ − 2 (44) 2 ⎟⎠ ⎝ (45)

E0 n2 13.6 eV = −6.72 meV (45)2

= 3.09 × 10 −4 eV (c) The photon wavelength is:

λ=

hc ∆E

Atoms

1219

1240 eV ⋅ nm = 4.01× 106 nm 3.09 × 10−4 eV

Substitute numerical values and evaluate λ:

λ=

(d) The radii of the Bohr orbits are given by:

r = n2

Substitute numerical values and evaluate the radius of the 45th Bohr orbit:

r = (45)

a0 Z 2

0.0529 nm = 107 nm 1

75 •• Picture the Problem We can use the definition of the Rydberg constant and the equation for the reduced mass from Problem 29 to calculate the Rydberg constant for hydrogen and for deuterium. We can find the wavelength difference between the longest wavelength Balmer lines of hydrogen and deuterium by finding the longest wavelengths from the Rydberg-Ritz equation, using the appropriate value for R, and taking their difference. (a) From Equation 36-14 we have:

⎛ ⎜ m mr k 2e 4 e ⎜ R= = C 4πch 3 ⎜ 1 + me ⎜ M ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

where

C=

k 2e 4 = 1.204662 × 1037 m −1 / kg 4πch 3

For H:

⎛ ⎜ me RH = C ⎜⎜ me ⎜1+ ⎝ mp

Substitute numerical values and evaluate RH:

RH = 1.204662 × 1037 m −1 / kg

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

(

)

⎛ ⎞ ⎜ ⎟ −31 9.109390 × 10 kg ⎟ ⎜ ×⎜ 9.109390 × 10 −31 kg ⎟ ⎜1+ ⎟ − 27 ⎝ 1.672623 × 10 kg ⎠ = 1.096776 × 10 7 m −1

1220 Chapter 36 For deuterium:

⎛ ⎜ me RD = C ⎜⎜ me ⎜1+ ⎝ 2mp

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Substitute numerical values and evaluate RD:

RD = 1.204662 × 1037 m −1 / kg

(

)

⎛ ⎜ 9.109390 × 10 −31 kg × ⎜⎜ 9.109390 × 10 −31 kg 1 + ⎜ 2 1.672623 × 10 −27 kg ⎝

(

)

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

= 1.097075 × 10 7 m −1 (b) Express the wavelength difference between the longest wavelength Balmer lines of hydrogen and deuterium: Use the Rydberg-Ritz formula to express the reciprocal wavelength:

∆λ = λlongest, H − λlongest, D

⎛ 1 1⎞ = R⎜⎜ 2 − 2 ⎟⎟ λ ⎝ n2 n1 ⎠ 1

where n1 and n2 are integers and n1 > n2. Solve for λ to obtain:

The longest wavelength in the Balmer series corresponds to a transition from n1 = 3 to n2 = 2. Use R = RH to evaluate λlongest, H: Find λlongest,D using R = RD:

Substitute to obtain:

λ=

n12 n22 R n12 − n22

(

) ( )

32 2 2 1.096776 × 107 m −1 32 − 2 2 = 656.470 nm

λlongest, H =

λlongest, D

(

( )

)(

32 2 2 = 1.097075 × 107 m −1 32 − 22 = 656.291nm

(

)(

∆λ = 656.470 nm − 656.291 nm = 0.179 nm

)

)

Atoms

1221

76 •• Picture the Problem We can use Problem 29 to express the energy levels of muonium in terms of the reduced mass of the muonium-proton system. In Part (b) we can find the energies corresponding to 400 nm and 700 nm to decide whether the transitions between any of the levels found in (a) fall in the visible range of wavelengths Express the energy of muonium as a function of the quantum number n:

En = −

From Equation 35-47 in Problem 17 we have:

mr =

Because mµ −1 = 207 me :

mr =

mr k 2e 4 Z 2 2h 2 n 2

(1)

mp mµ -1 mp + mµ -1 207mp me mp + 207me

=

207me m 1 + 207 e mp

207me = 186me 207 1+ 1836

Because mp = 1836me:

mr =

Substitute in equation (1) and simplify to obtain:

186me k 2e 4 1 186 E0 En = − =− 2 2 2h n n2

(a) Evaluate En for n = 1, 2, 3, 4, and 5 to obtain:

n 1 2 3 4 5

hc 1240 eV ⋅ nm = ∆E ∆E

Relate the wavelength of the emitted photons to the energy-level differences:

λ=

Solve for ∆E:

∆E =

Evaluate ∆E for λ = 400 nm and λ = 700 nm:

1240 eV ⋅ nm

∆E400 nm = and

En (keV) −2.53 −0.633 −0.281 −0.158 −0.101

λ 1240 eV ⋅ nm = 3.10 eV 400 nm

1222 Chapter 36 ∆E700 nm =

1240 eV ⋅ nm = 1.77 eV 700 nm

Because none of the energies in the table shown above are in the interval 1.77 eV to 3.10 eV, no transitions are in the visible range of wavelengths. 77 •• Picture the Problem We can use the definition of the Rydberg constant and the equation for the reduced mass from Problem 29 to calculate the Rydberg constant for hydrogen, tritium, and deuterium. We can find the wavelength difference between the longest wavelength Balmer lines of tritium and deuterium and tritium and hydrogen by finding the longest wavelengths from the Rydberg-Ritz equation, using the appropriate value for R, and taking their difference. (a) From Problem 29 we have:

⎛ ⎜ m mr k 2e 4 e ⎜ R= = C 4πch 3 ⎜ 1 + me ⎜ M ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

where

C=

k 2e 4 = 1.204662 × 1037 m −1 / kg 4πch 3

For tritium:

⎛ ⎜ me RT = C ⎜⎜ me ⎜1+ ⎝ mp + 2mn

Evaluate the expression in parentheses to obtain:

mr = 9.107738 × 10 −31 kg

Substitute numerical values and evaluate RT:

RT = 1.204662 × 1037 m −1 / kg

(

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

(

× 9.107738 × 10 −31 kg = 1.097175 × 10 7 m −1

For deuterium:

⎛ ⎜ me RD = C ⎜⎜ me ⎜1+ ⎝ 2mp

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

)

)

Atoms Substitute numerical values and evaluate RD:

(

RD = 1.204662 × 1037 m −1 / kg

)

⎛ ⎜ 9.109390 × 10 −31 kg ×⎜ ⎜ 9.109390 × 10 −31 kg ⎜1+ 2 1.672623 × 10 −27 kg ⎝ = 1.097075 × 10 7 m −1

(

(b) Express the wavelength difference between the longest wavelength Balmer lines of hydrogen and deuterium: Use the Rydberg-Ritz formula to express the reciprocal wavelength:

1223

)

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

∆λ = λlongest, D − λlongest, T

⎛ 1 1⎞ = R⎜⎜ 2 − 2 ⎟⎟ λ ⎝ n2 n1 ⎠ 1

where n1 and n2 are integers and n1 > n2. Solve for λ to obtain:

The longest wavelength in the Balmer series corresponds to a transition from n1 = 3 to n2 = 2. Use R = RT to evaluate λlongest, T: Find λlongest,D using R = RD:

Substitute to obtain:

n12 n22 λ= R n12 − n22

(

λlongest, T

) ( )

32 2 2 = 1.097175 × 107 m −1 32 − 22 = 656.231nm

(

( )

)(

32 2 2 1.097075 × 107 m -1 32 − 2 2 = 656.291 nm

λlongest, D =

(

)(

∆λ = 656.291nm − 656.231 nm = 0.0600 nm

Proceed similarly to show that for hydrogen and hydrogen:

∆λ = 656.4695 nm − 656.2314 nm = 0.238 nm

)

)

1224 Chapter 36

Chapter 37 Molecules Conceptual Problems *1 • Determine the Concept Yes. Because the center of charge of the positive Na ion does not coincide with the center of charge for the negative Cl ion, the NaCl molecule has a permanent dipole moment. Hence, it is a polar molecule. 2 • Determine the Concept Because a N2 molecule has no permanent dipole moment, it is a non-polar molecule. 3 • Determine the Concept No. Neon occurs naturally as Ne, not Ne2. Neon is a rare gas atom with a closed shell electron configuration. 4 • Determine the Concept (a) Because an electron is transferred from the H atom to the F atom, the bonding mechanism is ionic. (b) Because an electron is transferred from the K atom to the Br atom, the bonding mechanism is ionic. (c) Because the atoms share two electrons, the bonding mechanism is covalent. (d) Because each valence electron is shared by many atoms, the bonding mechanism is metallic bonding. *5 •• Determine the Concept The diagram would consist of a non-bonding ground state with no vibrational or rotational states for ArF (similar to the upper curve in Figure 37-4) but for ArF* there should be a bonding excited state with a definite minimum with respect to inter-nuclear separation and several vibrational states as in the excited state curve of Figure 37-13. 6 • Determine the Concept Elements similar to carbon in outer shell configurations are silicon, germanium, tin, and lead. We would expect the same hybridization for these as

1225

1226

Chapter 37

for carbon, and this is indeed the case for silicon and germanium whose crystal structure is the diamond structure. Tin and lead, however, are metallic and here the metallic bond is dominant. 7 • Determine the Concept The effective force constant from Example 37-4 is 1.85×103 N/m. This value is about 25% larger than the given value of the force constant of the suspension springs on a typical automobile. 8 • Determine the Concept As the angular momentum increases, the separation between the nuclei also increases (the effective force between the nuclei is similar to that of a stiff spring). Consequently, the moment of inertia also increases. 9 • Determine the Concept For H2, the concentration of negative charge between the two protons holds the protons together. In the H +2 ion, there is only one electron that is shared by the two positive charges such that most of the electronic charge is again between the two protons. However, the negative charge in the H +2 ion is not as effective as the larger charge in the H2 molecule, and the protons should be farther apart. The experimental values support this argument. For H2, r0 = 0.074 nm, while for H +2 , r0 = 0.106 nm. 10 • Determine the Concept The energy of the first excited state of an atom is orders of magnitude greater than kT at ordinary temperatures. Consequently, practically all atoms are in the ground state. By contrast, the energy separation between the ground rotational state and nearby higher rotational states is less than or roughly equal to kT at ordinary temperatures, and so these higher states are thermally excited and occupied. 11 •• Determine the Concept With more than two atoms in the molecule there will be more than just one frequency of vibration because there are more possible relative motions. In advanced mechanics these are known as normal modes of vibration.

Estimation and Approximation 12 •• Picture the Problem We can estimate the value of the quantum number ν for which the improved formula corrects the original formula by 10 percent by setting the ratio of the correction term to the first term equal to 10 percent and solving for ν.

Molecules Express the ratio of the correction term to the first term of the expression for Eν and simplify to obtain: For a correction of 10 percent: Solve for ν to obtain:

Substitute numerical values and evaluate ν : 13

1227

(ν + 12 )2 hfα = (ν + 1 )α 2 (ν + 12 )hf (ν + 12 )α = 0.1 ν=

1 1 − 10α 2

ν=

1 1 − = 12.7 ≈ 13 −3 10 7.6 × 10 2

(

)

••

Picture the Problem We can solve Equation 37-12 for l and substitute for the moment of inertia and rotational kinetic energy of the baseball to estimate the quantum number l and spacing between adjacent energy levels for a baseball spinning about its own axis. The rotational energy levels are given by Equation 37-12:

l(l + 1) h 2 E= 2I where l = 0, 1, 2, … is the rotational quantum number and I is the moment of inertia of the ball.

Solve for l(l+1):

l(l + 1) =

Factor l from the parentheses to obtain:

⎛ 1 ⎞ 2 IE l 2 ⎜1 + ⎟ = 2 ⎝ l⎠ h

The result of our calculation of l will show that l >> 1. Assuming for the moment that this is the case:

l2 ≈

Because the energy of the ball is rotational kinetic energy:

E = K rot = 12 Iω 2

Substitute for E in the expression for l to obtain:

l≈

The moment of inertia of a ball about an axis through its diameter is (see Table 9-1): Substitute for I to obtain:

2 IE h2

2 IE and l ≈ h2

2I

(

1 2

Iω 2

h

I = 25 mr 2

l≈

2mr 2ω 5h

2 IE h

) = (I ω ) = Iω 2

h

2

h

1228 Chapter 37 Substitute numerical values and evaluate l:

2π rad 1 min ⎞ 2 ⎛ 20 rev ⎟ × × 2(0.3 kg )(0.03 m ) ⎜⎜ min rev 60 s ⎟⎠ ⎝ l≈ = 2.15 ×1030 −34 5 1.05 ×10 J ⋅ s

(

)

Set l = 0 to express the spacing between adjacent energy levels:

h2 5h 2 E0 r = = 2 I 4mr 2

Substitute numerical values and evaluate E0r:

5 1.05 × 10 −34 J ⋅ s E0 r = 2 4(0.3 kg )(0.03 m )

(

)

2

= 5.10 ×10 −65 J Remarks: Note that our value for l justifies our assumption that l >> 1. *14

••

Picture the Problem We can solve Equation 37-18 for ν and substitute for the frequency of the mass-and-spring oscillator to estimate the quantum number ν and spacing between adjacent energy levels for this system. The vibrational energy levels are given by Equation 37-18: Solve for ν :

The vibrational energy of the object attached to the spring is: Substitute for Eν in the expression for ν to obtain: The frequency of oscillation f of the mass-and-spring oscillator is given by:

Eν = (ν + 12 )hf where ν = 0, 1, 2, … Eν 1 − hf 2 or, because ν >> 1, E ν≈ ν hf

ν=

Eν = 12 kA2 where A is the amplitude of its motion.

ν=

kA2 2hf

f =

1 2π

ν=

πkA2 h

k m m πA2 = mk k h

Molecules 1229 Substitute numerical values and evaluate ν :

ν=

π (0.02 m )2 6.63 ×10

−34

(5 kg )(1500 N/m ) =

J ⋅s

Set ν = 0 in Equation 37-18 to express the spacing between adjacent energy levels: Substitute numerical values and evaluate E0ν:

1.64 ×1032

E0ν =

1 h hf = 2 4π

k m

E0ν =

6.63 × 10 −34 J ⋅ s 1500 N/m 4π 5 kg

= 9.14 × 10 −34 J Remarks: Note that our value for ν justifies our assumption that ν >> 1.

Molecular Bonding 15 • Picture the Problem The electrostatic potential energy with U at infinity is given by

U = − ke2 r. Relate the electrostatic potential energy of the ions to their separation:

Ue = −

Solve for r:

r=−

ke 2 r

ke 2 Ue

Substitute numerical values and evaluate r:

(8.99 ×10 r=−

)(

N ⋅ m 2 / C 2 1.6 × 10 −19 C (− 1.52 eV ) 1.6 ×10−19 J/eV 9

(

)

)

2

= 0.946 nm

16 • Picture the Problem We can find the energy absorbed or released per molecule by computing the difference between dissociation energy of Cl and the binding energy of NaCl.

1230 Chapter 37 Noting that the dissociation energy per Cl atom is 1.24 eV, express the net energy change per molecule ∆E:

∆E = 1.24 eV − Ebinding, NaCl

The binding energy of NaCl is (see page 1210):

4.27 eV

Substitute and evaluate ∆E:

∆E = 1.24 eV − 4.27 eV = − 3.03 eV

Because ∆E < 0, energy is released. The reaction is exothermic. 17 • Picture the Problem We can use conversion factors to convert eV/molecule into kcal/mol.

1 (a)

eV eV 1kcal 6.02 ×10 23 molecules 1.60 × 10−19 J =1 × × × molecule molecule 4184 J mole eV

= 23.0 kcal/mol (b) The dissociation energy of NaCl, in eV/molecule, is (see page 1205):

4.27 eV/molecule

Using the conversion factor in (a), express this energy in kcal/mol:

4.27 eV 23.0 kcal × = 98.2 kcal/mol molecule mol

*18 • Picture the Problem The percentage of the bonding that is ionic is given by

⎛p ⎞ 100⎜⎜ meas ⎟⎟. ⎝ p100 ⎠ Express the percentage of the bonding that is ionic: Express the dipole moment for 100% ionic bonding:

⎛p ⎞ Percent ionic bonding = 100⎜⎜ meas ⎟⎟ ⎝ p100 ⎠

p100 = er

Molecules 1231 ⎛p ⎞ Percent ionic bonding = 100⎜ meas ⎟ ⎝ er ⎠

Substitute to obtain:

Substitute numerical values and evaluate the percent ionic bonding:

⎡ ⎤ 6.40 × 10 −30 C ⋅ m Percent ionic bonding = 100 ⎢ ⎥ = 43.6% −19 ⎣ 1.60 × 10 C (0.0917 nm )⎦

(

)

19 •• Picture the Problem If we choose the potential energy at infinity to be ∆E, the total potential energy is Utot = Ue + ∆E + Urep, where Urep is the energy of repulsion, which is found by setting the dissociation energy equal to −Utot.

U tot = U e + ∆E + U rep

Express the total potential energy of the molecule:

U rep = −(∆E + U e + Ed )

The core-repulsive energy is : Calculate the energy ∆E needed to form Rb+ and F− ions from neutral rubidium and fluorine atoms:

∆E = 4.18 eV − 3.40 eV = 0.78 eV

Express the electrostatic potential energy is:

Ue = −

ke 2 r

Substitute numerical values and evaluate Ue:

Ue = −

(8.99 ×10

)(

N ⋅ m 2 / C 2 1.60 × 10 −19 C (0.227 nm ) 1.60 ×10−19 J/eV

Substitute numerical values and evaluate Urep:

9

(

)

)

2

= −6.34 eV

U rep = −(0.78 eV − 6.34 eV + 5.12 eV ) = 0.44 eV

20 •• Picture the Problem The potential energy of attraction of the ions is U e = − ke 2 r . We can find the dissociation energy from the negative of the sum of the potential energy of attraction and the difference between the ionization energy of potassium and the electron

1232 Chapter 37 affinity of chlorine. (a) The potential energy of attraction of the ions is given by:

Ue = −

ke 2 r

where ke2 = 1.44 eV⋅nm Substitute numerical values and evaluate Ue: (b) Express the total potential energy of the molecule:

Ue = −

1.44 eV ⋅ nm = − 5.39 eV 0.267 nm

U tot = U e + ∆E + U rep or, neglecting any energy of repulsion,

U tot = U e + ∆E The dissociation energy is the negative of the total potential energy: ∆E is the difference between the ionization energy of potassium and the electron affinity of Cl: Substitute numerical values and evaluate Ed,calc:

The energy due to repulsion of the ions at equilibrium separation is given by: Substitute numerical values and evaluate Urep:

Ed,calc = −U tot = −(U e + ∆E )

∆E = 4.34 eV − 3.62 eV = 0.72 eV

Ed,calc = −(− 5.39 eV + 0.72 eV ) = 4.67 eV U rep = Ed,calc − Ed,meas

U rep = 4.67 eV − 4.49 eV = 0.18 eV

21 •• Picture the Problem Assume that U(r) is of the form given in Problem 24 with n = 6.

Molecules 1233 The potential energy curve is shown in the figure. The turning points for vibrations of energy E1 and E2 are at the values of r, where the energies equal U(r). It is apparent that the average value of r depends on the energy and that r2,av is greater than r1,av.

22 •• Picture the Problem We can use U e = − ke 2 r0 to calculate the potential energy of attraction between the Na+ and Cl− ions at the equilibrium separation r0 = 0.236 nm. We can find the energy due to repulsion of the ions at the equilibrium separation from U rep = −(U e + Ed + ∆E ) . The potential energy of attraction between the Na+ and Cl− ions at the equilibrium separation r0 is given by:

Ue = −

ke 2 r0

where ke2 = 1.44 eV⋅nm.

1.44 eV ⋅ nm = − 6.10 eV 0.236 nm

Substitute numerical values and evaluate Ue:

Ue = −

From Figure 37-1:

Ed = 4.27 eV

The ratio of the magnitude of the potential energy of attraction to the dissociation energy is: Urep is related to Ue, Ed, and ∆E according to: From Figure 37-1: Substitute numerical values and evaluate Urep:

U e 6.10 eV = = 1.43 Ed 4.27 eV U rep = −(U e + Ed + ∆E )

∆E = 1.52 eV U rep = −(− 6.10 eV + 4.27 eV + 1.52 eV ) = 0.310 eV

23 •• Picture the Problem The potential energy of attraction of the ions is U e = − ke 2 r . We can find the dissociation energy from the negative of the sum of the potential energy of

1234 Chapter 37 attraction and the difference between the ionization energy of potassium and the electron affinity of fluorine. (a) The potential energy of attraction between the K+ and F− ions at the equilibrium separation r0 is given by: Substitute numerical values and evaluate Ue: (b) Express the total potential energy of the molecule:

Ue = −

ke 2 r0

where ke2 = 1.44 eV⋅nm.

Ue = −

1.44 eV ⋅ nm = − 6.64 eV 0.217 nm

U tot = U e + ∆E + U rep or, neglecting any energy of repulsion,

U tot = U e + ∆E The dissociation energy is the negative of the total potential energy: ∆E is the difference between the ionization energy of potassium and the electron affinity of fluorine: Substitute numerical values and evaluate Ed,calc:

The energy due to repulsion of the ions at equilibrium separation is given by: Substitute numerical values and evaluate Urep:

Ed,calc = −U tot = −(U e + ∆E )

∆E = 4.34 eV − 3.40 eV = 0.94 eV

Ed,calc = −(− 6.64 eV + 0.94 eV ) = 5.70 eV U rep = Ed,calc − Ed,meas

U rep = 5.70 eV − 5.07 eV = 0.63 eV

*24 ••• Picture the Problem U(r) is the potential energy of the two ions as a function of separation distance r. U(r) is chosen so U(∞) = −∆E, where ∆E is the negative of the energy required to form two ions at infinite separation from two neutral atoms also at infinite separation. Urep(r) is the potential energy of the two ions due to the repulsion of the two closed-shell cores. Ed is the disassociation energy, the energy required to separate the two ions plus the energy ∆E required to form two neutral atoms from the two ions at

Molecules 1235 infinite separation. The net force acting on the ions is the sum of Frep and Fe. We can find Frep from Urep and Fe from Coulomb’s law and then use dU/dr = Fnet = 0 at r = r0 to solve for n. Express the net force acting on the ions: Find Frep from Urep:

Fnet = Frep + Fe

Frep =

dU rep

dr nC = − n+1 r

The electrostatic potential energy of the two ions as a function of separation distance is given by:

ke 2 Ue = − r

Find the electrostatic force of attraction Fe from Ue:

Fe =

=

(1)

[

]

d Cr −n = −nCr −n−1 dr

dU e d ⎡ ke 2 ⎤ ke 2 = − = dr dr ⎢⎣ r ⎥⎦ r 2

nC ke 2 = − n+1 + 2 r r

Substitute for Frep and Fe in equation (1) to obtain:

Fnet

Because dU/dr = Fnet = 0 at r = r0:

0=−

nC ke 2 + r0n+1 r02

Multiply both sides of this equation by r0 to obtain:

0=−

nC ke 2 + = −nU rep (r0 ) + U e (r0 ) r0n r0

Solve for n to obtain:

n=

U e (r0 ) U rep (r0 )

25 ••• Picture the Problem Urep at r = r0 is related to Ue, Ed, and ∆E through U rep = −(U e + Ed + ∆E ) . The net force is the sum of Frep and Fe. We can find Frep from Urep and Fe from Coulomb’s law. Because Fnet = 0 at r = r0, we can obtain simultaneous equations in C and n that we can solve for each of these quantities. (a) Urep is related to Ue, Ed, and ∆E according to:

U rep = −(U e + Ed + ∆E ) where ∆ENaCl = 1.52 eV

1236 Chapter 37 Ue(r0) is given by:

ke 2 Ue = − r0 where ke2 = 1.44 eV⋅nm

Substitute numerical values and evaluate Ue: Substitute numerical values and evaluate Urep:

(b) Express the net force acting on the Na+ and Cl− ions: Find Frep from Urep:

Ue = −

1.44 eV ⋅ nm = −6.10 eV 0.236 nm

U rep = −(− 6.10 eV + 4.27 eV + 1.52 eV ) = 0.31eV Fnet = Frep + Fe

Frep =

dU rep

dr nC = − n+1 r

The electrostatic force of attraction is:

ke 2 Fe = 2 r

Substitute for Frep and Fe in equation (1) to obtain:

Fnet = −

Because Fnet = 0 at r = r0:

0=−

=

(1)

[

]

d Cr −n = −nCr −n−1 dr

nC ke 2 + r n+1 r 2

nC ke 2 + r0n+1 r02

or

U e (r0 ) = n

Solve for n and C to obtain:

n=

C = nU rep (r0 ) r0n

U e (r0 ) U rep (r0 )

and

C = U rep (r0 )r0n

From (a):

U e (r0 ) = −6.10 eV U rep (r0 ) = 0.31eV and, from Figure 37-1,

Molecules 1237 r0 = 0.236 nm Substitute for Ue(r0) and Urep(r0) and evaluate n and C:

n=

6.10 eV = 19.7 0.31eV

and

C = (0.31eV )(0.236 nm )

19.7

= 1.37 × 10 −13 eV ⋅ nm19.7

Energy Levels of Spectra of Diatomic Molecules 26 • Picture the Problem We can relate the characteristic rotational energy E0r to the moment of inertia of the molecule and model the moment of inertia of the N2 molecule as two point objects separated by a distance r. The characteristic rotational energy of a molecule is given by:

h2 E0 r = 2I

Express the moment of inertia of the molecule:

⎛r⎞ I = 2M N ⎜ ⎟ = 12 M N r 2 ⎝2⎠

Substitute for I to obtain:

E0 r =

h2 h2 h2 = = 2 12 M N r 2 M N r 2 14mp r 2

Solve for r:

r=h

1 14 E0 r mp

2

(

)

Substitute numerical values and evaluate r:

(

r = 1.055 × 10 −34 J ⋅ s

) 14(2.48 ×10

−4

1 eV 1.602 × 10 −19 J/eV 1.673 × 10 −27 kg

)(

)(

)

= 0.109 nm *27 • Picture the Problem We can relate the characteristic rotational energy E0r to the moment of inertia of the molecule and model the moment of inertia of the O2 molecule as two point objects separated by a distance r.

1238 Chapter 37 The characteristic rotational energy of a molecule is given by:

h2 E0 r = 2I

Express the moment of inertia of the molecule:

⎛r⎞ I = 2M O ⎜ ⎟ = 12 M O r 2 ⎝2⎠

Substitute for I to obtain:

E0 r =

Solve for r:

r=

2

h2 h2 h2 = = 2 12 M O r 2 M O r 2 16mp r 2

(

)

h 1 4 E0 r mp

Substitute numerical values and evaluate r:

⎛ 1.055 × 10 −34 J ⋅ s ⎞ 1 ⎟⎟ r = ⎜⎜ −4 −19 −27 4 ⎝ ⎠ 1.78 × 10 eV 1.6 × 10 J/eV 1.67 × 10 kg

(

)(

)(

)

= 0.121 nm 28 •• Picture the Problem We can use the definition of the reduced mass to show that the reduced mass is smaller than either mass in a diatomic molecule. Express the reduced mass of a twobody system:

µ=

m1m2 m1 + m2

Divide the numerator and denominator of this expression by m2 to obtain:

µ=

m1 m 1+ 1 m2

Divide the numerator and denominator of this expression by m1 to obtain:

µ=

m2 m 1+ 2 m1

Because the denominator is greater than 1 in equations (1) and (2): (a) For H2, m1 = m2 = 1 u:

(1)

(2)

µ < m1 and µ < m2

µH = 2

(1u )(1u ) = 1u + 1u

0.500 u

Molecules 1239

µN =

(14 u )(14 u ) =

7.00 u

(c) For CO, m1 = 12 u and m2 = 16 u:

µCO =

(12 u )(16 u ) =

6.86 u

(d) For HCl, m1 = 1 u and m2 = 35.5 u:

µ HCl =

(1u )(35.5 u ) =

(b) For N2, m1 = m2 = 14 u:

2

14 u + 14 u

12 u + 16 u

1 u + 35.5 u

0.973 u

29 •• Picture the Problem We can solve Equation 37-18 for ν and substitute for the frequency of the CO molecule (see Example 37-4) and its binding energy to estimate the quantum number ν . The vibrational energy levels are given by Equation 37-18:

Eν = (ν + 12 )hf where ν = 0, 1, 2, …

Solve for ν :

ν=

Substitute numerical values and evaluate ν :

1.60 × 10−19 J 1 eV ν= − −34 13 6.63 × 10 J ⋅ s 6.42 × 10 Hz 2

Eν 1 − hf 2 11eV ×

(

)(

)

= 40.8 ≈ 41 *30 •• Picture the Problem We can use the expression for the rotational energy levels of the diatomic molecule to express the energy separation ∆E between the l = 3 and l = 2 rotational levels and model the moment of inertia of the LiH molecule as two point objects separated by a distance r0. The energy separation between the l = 3 and l = 2 rotational levels of this diatomic molecule is given by:

∆E = El=3 − El=2

El=3 = 3(3 + 1)E0 r = 12 E0 r

Express the rotational energy levels El = 3 and El = 2 in terms of E0r:

and

Substitute for El = 3 and El = 2 to

∆E = 12 E0 r − 6 E0 r = 6 E0 r

El=2 = 2(2 + 1)E0 r = 6 E0 r

1240 Chapter 37 obtain:

or

E0 r = 16 ∆E h2 1 3h 2 = 6 ∆E ⇒ ∆E = 2I I

The characteristic rotational energy of a molecule is given by:

E0 r =

Express the moment of inertia of the molecule:

I = µ r02 where µ is the reduced mass of the molecule.

Substitute for I to obtain:

∆E =

3h 2

µ r02

3h 2 m Li m H

=

m Li + m H

=

r02

3h 2 (m Li + m H ) m Li m H r02

Substitute numerical values and evaluate ∆E:

(

)

3 1.055 × 10 −34 J ⋅ s (6.94 u + 1 u ) ∆E = = 5.61 meV (6.94 u )(1u )(0.16 nm )2 1.602 ×10−19 J/eV 1.660 ×10−27 kg/u 2

(

)(

)

*31 •• Picture the Problem Let the origin of coordinates be at the point mass m1 and point mass m2 be at a distance r0 from the origin. We can express the moment of inertia of a diatomic molecule with respect to its center of mass using the definitions of the center of mass and the moment of inertia of point particles. Express the moment of inertia of a diatomic molecule:

I = m1r12 + m2 r22

The r coordinate of the center of mass is:

rCM =

The distances of m1 and m2 from the center of mass are:

(1)

m2 r0 m1 + m2

r1 = rCM and

r2 = r0 − rCM = r0 − =

m1 r0 m1 + m2

m2 r0 m1 + m2

Molecules 1241 2

Substitute for r1 and r2 in equation (1) to obtain:

⎛ m2 ⎞ ⎛ m1 ⎞ I = m1 ⎜⎜ r0 ⎟⎟ + m2 ⎜⎜ r0 ⎟⎟ ⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠

Simplifying this expression leads to:

I=

2

m1m2 2 r0 m1 + m2

or

I = µ r02

36-14

where

µ=

m1m2 m1 + m2

36-15

32 •• Picture the Problem We can relate the characteristic rotational energy E0r to the moment of inertia of the molecule and model the moment of inertia of the KCl molecule as two point objects of reduced mass µ separated by a distance r.

h2 2I

The characteristic rotational energy of a molecule is given by:

E0 r =

Express the moment of inertia of the molecule:

I = µ r02 where µ =

mK mCl mK + mCl

h2 h 2 (mK + mCl ) = E0 r = 2µ r02 2mk mCl r02

Substitute for I to obtain:

Substitute numerical values and evaluate E0r:

(1.055 ×10

)

J ⋅ s (39.1 u + 35.5 u ) E0 r = 2 2(39.1 u )(35.5 u )(0.267 nm ) 1.660 ×10 −27 kg/u 1 = 2.53 × 10−24 J × = 0.0158 meV 1.602 ×10 −19 J/eV −34

2

(

)

33 •• Picture the Problem We can use the expression for the vibrational energies of a molecule to find the lowest vibrational energy. Because the difference in the vibrational energy levels depends on both ∆f and the moment of inertia I of the molecule, we can relate these quantities and solve for I. Finally, we can use I = µ r 2 , with µ representing

1242 Chapter 37 the reduced mass of the molecule, to find the equilibrium separation of the atoms.

Eν = (ν + 12 )hf, ν = 0,1, 2, ...

(a) The vibrational energy levels are given by:

E0 = 12 hf

The lowest vibrational energy corresponds to ν = 0: Substitute numerical values and evaluate E0:

E0 =

1 2

(6.63 ×10

−34

)(

)

J ⋅ s 8.66 ×1013 Hz = 2.87 × 10 −20 J ×

1 1.6 × 10 −19 J/eV

= 0.179 eV (b) For ∆l = ±1:

∆El =

lh 2 = lh∆f I

h2 h2 h = = 2 h∆f 4π h∆f 4π 2 ∆f

Solve for I:

I=

Substitute numerical values and evaluate I:

6.63 × 10−34 J ⋅ s I= 4π 2 6 × 1011 Hz

(

)

= 2.80 × 10 −47 kg ⋅ m 2 I = µ r2

(c) The moment of inertia of a HCL molecule is given by: Replace µ by the reduced mass of a HCl molecule and r by r0 to obtain:

I=

Solve for r0:

r0 =

mH mCl 2 r0 mH + mCl mH + mCl I mH mCl

Substitute numerical values and evaluate r0:

(

)

⎡ 1 u + 34.453 u ⎤ 2.80 × 10 −47 kg ⋅ m 2 r0 = ⎢ = 0.132 nm ⎥ − 27 ⎣ (1 u )(34.453 u ) ⎦ 1.66 × 10 kg/u

Molecules

1243

34 •• Picture the Problem Let the numeral 1 refer to the H+ and the numeral 2 to the Cl− ion. For a two-mass and spring system on which no external forces are acting, the center of mass must remain fixed. We can use this condition to express the net force acting on either the H+ or Cl− ion. Because this force is a linear restoring force, we can conclude that the motion of the object whose mass is m1 will be simple harmonic with an angular frequency given by ω = K eff m1 . Substitution for Keff will lead us to the result given in (b). If the particle whose mass is m1 moves a distance r1 from (or toward) the center of mass, then the particle whose mass is m2 must move a distance: Express the force exerted by the spring:

∆r2 =

m1 ∆r1 from (or toward) the center m2

of mass.

F = − K∆r = − K (∆r1 + ∆r2 )

Substitute for ∆r2 to obtain:

⎛ ⎞ m F = − K ⎜⎜ ∆r1 + 1 ∆r1 ⎟⎟ m2 ⎝ ⎠ ⎛ m + m2 ⎞ ⎟⎟∆r1 = − K ⎜⎜ 1 ⎝ m2 ⎠

A displacement ∆r1 of m1 results in a restoring force:

⎛ m + m2 ⎞ ⎟⎟∆r1 = − K eff ∆r1 F = − K ⎜⎜ 1 ⎝ m2 ⎠ ⎛ m + m2 ⎞ ⎟⎟ where K eff = K ⎜⎜ 1 m 2 ⎠ ⎝

Because this is a linear restoring force, we know that the motion will be simple harmonic with:

ω= or

f =

Substitute for Keff and simplify to obtain:

K eff m1

1 ω = 2π 2π

K eff m1

⎛ m + m2 ⎞ ⎟⎟ K ⎜⎜ 1 ⎝ m1m2 ⎠ m1 m2 or, because µ = is the reduced m1 + m2 f =

ω 1 = 2π 2π

1244 Chapter 37 mass of the two-particle system,

f =

Solve for K:

1 2π

K

µ

K = 4π 2 f 2 µ = 4π 2 f 2

mH mCl mH + mCl

Substitute numerical values and evaluate K:

(

)

(

)

4π 2 8.66 × 1013 Hz (1 u )(35.453 u ) 1.66 × 10 −27 kg/u K= = 478 N/m (1u + 35.453 u ) 2

35 •• Picture the Problem We’re given the population of rotational states function:

f (l ) = (2l + 1)e − El / kT where

El = l(l + 1) E0 r and E0 r = The moment of inertia I of an oxygen molecule is given by:

We’ll assume, as in Example 37-3, that:

h2 2I

I = 12 mr02 where m is the reduced mass and r0 is the separation of the atoms in a molecule.

r0 = 0.1 nm

A spreadsheet program to plot f(l) is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 A13

Formula/Content 1.00E−10 16 2.66E−26 1.05E−34 1.38E−23 4.15E−23 100 200 300 500 0

Algebraic Form r0 m (u) m (kg) h k E0r T (K) T (K) T (K) T (K) l

Molecules B13

l(l + 1)E0r

A13*(A13+1)*$B$6

C13 (2*A13 + 1)*EXP(−B13/($B$5*$B$7)) D13 (2*A13 + 1)*EXP(−B13/($B$5*$B$8)) E13 (2*A13 + 1)*EXP(−B13/($B$5*$B$9)) F13 (2*A13 + 1)*EXP(−B13/($B$5*$B$10)) A 1 r_0= 2 m= 3 m= 4 h_bar= 5 k= 6 E_0r= 7 T= 8 T= 9 T= 10 T= 11 12 l 13 0.0 14 0.5 15 1.0 16 1.5 17 2.0

B 1.00E−10 16 2.656E−26 1.05E−34 1.38E−23 4.15E−23 100 200 300 500

m u kg J.s J/K eV K K K K

E_l 0.00E+00 3.11E−23 8.30E−23 1.56E−22 2.49E−22

E_100 K E_200 K E_300 K E_500 K 1.00 1.00 1.00 1.00 1.96 1.98 1.99 1.99 2.82 2.91 2.94 2.96 3.57 3.78 3.85 3.91 4.17 4.57 4.71 4.82

29 30 31 32 33

2.99E−21 3.35E−21 3.74E−21 4.14E−21 4.57E−21

8.0 8.5 9.0 9.5 10.0

C

1.95 1.59 1.27 1.00 0.77

D

f(l, T = 100 K) f(l, T = 200 K) f(l, T = 300 K) f(l, T = 500 K)

5.76 5.34 4.91 4.46 4.02

E

8.26 8.01 7.71 7.36 6.97

F

11.02 11.07 11.06 10.98 10.83

1245

1246 Chapter 37

The following graph shows f(l) as a function of temperature. 12

T = 100 K T = 200 K

10

T = 300 K T = 500 K

f (l )

8

6

4

2

0 0

2

4

6

8

10

rotational quantum number l

*36 •• Picture the Problem For a two-mass and spring system on which no external forces are acting, the center of mass must remain fixed. We can use this condition to express the net force acting on either object. Because this force is a linear restoring force, we can conclude that the motion of the object whose mass is m1 will be simple harmonic with an angular frequency given by ω =

keff . Substitution for keff will lead us to the result m1

given in (b). (a) If the particle whose mass is m1 moves a distance ∆r1 from (or toward) the center of mass, then the particle whose mass is m2 must move a distance: Express the force exerted by the spring: Substitute for ∆r2 to obtain:

∆r2 =

m1 ∆r1 from (or toward) the center m2

of mass.

F = −k∆r = −k (∆r1 + ∆r2 )

⎞ ⎛ m F = −k ⎜⎜ ∆r1 + 1 ∆r1 ⎟⎟ m2 ⎠ ⎝ ⎛ m + m2 ⎞ ⎟⎟∆r1 = − k ⎜⎜ 1 ⎝ m2 ⎠

Molecules (b) A displacement ∆r1 of m1 results in a restoring force:

⎛ m + m2 ⎞ ⎟⎟∆r1 = −keff ∆r1 F = −k ⎜⎜ 1 ⎝ m2 ⎠ ⎛ m + m2 ⎞ ⎟⎟ where keff = k ⎜⎜ 1 ⎝ m2 ⎠

Because this is a linear restoring force, we know that the motion will be simple harmonic with:

ω=

Substitute for keff and simplify to obtain:

1247

keff m1

or

f =

ω 1 = 2π 2π

f =

⎛ m + m2 ⎞ ω 1 ⎟⎟ = k ⎜⎜ 1 2π 2π ⎝ m1m2 ⎠

keff m1

m1 m2 is the reduced m1 + m2

or, because µ =

mass of the two-particle system,

f =

1 2π

k

µ

.

37 ••• Picture the Problem We can use the definition of the reduced mass to find the reduced mass for the H35Cl and H37Cl molecules and the fractional difference ∆µ µ . Because the rotational frequency is proportional to 1/I, where I is the moment of inertia of the system, and I is proportional to µ, we can obtain an expression for f as a function of µ that we differentiate implicitly to show that ∆f f = − ∆µ µ . For H35Cl:

For H37Cl:

µ=

(35 u )(1u ) = 35 u =

0.9722 u

µ=

(37 u )(1u ) = 37 u =

0.9737 u

35 u + 1 u

37 u + 1 u

36

38

The fractional difference is:

37 35 36 × 37 − 35 × 38 u− u u ∆µ 36 36 × 38 = 38 = = 0.00150 1 ⎛ 35 37 ⎞ 35 × 38 + 36 × 37 µ u ⎜ u + u⎟ 2(36 )(38) 2 ⎝ 36 38 ⎠

1248 Chapter 37 The rotational frequency is proportional to 1/I, where I is the moment of inertia of the system. Because I is proportional to µ:

f =

C

µ

and

df = −Cµ −2 dµ

Divide df by f to obtain:

df dµ ∆f ∆µ =− and ≈− f µ f µ

From Figure 36-17:

∆f ≈ 0.01× 1013 Hz = 1011 Hz

For f = 8.40×1013 Hz:

∆f 1011 Hz ≈ = 0.00119 f 8.40 × 1013 Hz

This result is in fair agreement (about 21% difference) with the calculated result. Note that ∆f is difficult to determine precisely from Figure 36 - 17.

General Problems 38 • Picture the Problem We can use the definition of the reduced mass to show that when one atom in a diatomic molecule is much more massive than the other the reduced mass is approximately equal to the mass of the lighter atom. Express the reduced mass of a twobody system:

µ=

m1m2 m1 + m2

Divide the numerator and denominator of this expression by m2 to obtain:

µ=

m1 m 1+ 1 m2

If m2 >> m1, then:

m1 << 1 and µ ≈ m1 m2

39 •• Picture the Problem The rotational energy levels are given by

E=

l(l + 1)h 2 , l = 0,1, 2, .... 2I

Express the energy difference between these rotational energy

∆E1, 2 = E2 − E1

Molecules levels: Express E2 and E1:

E2 = and

E1 =

2(2 + 1)h 2 3h 2 = 2I I 1(1 + 1)h 2 h 2 = 2I I 3h 2 h 2 2h 2 = − = I I I

Substitute to obtain:

∆E1, 2

The moment of inertia of the molecule is:

I = µ r02 where µ is the reduced mass of the molecule.

Substitute for I to obtain:

∆E1, 2 =

=

2h 2 2h 2 = mC mO 2 µ r02 r0 mC + mO

2h 2 (mC + mO ) mC mO r02

Substitute numerical values and evaluate ∆E1,2:

∆E1, 2

(

) (

2 1.055 × 10 −34 J ⋅ s (12 u + 16 u ) 1.53 × 10 −22 J = = (16 u )(12 u )(0.113 nm )2 1.66 ×10−27 kg/u 1.6 ×10−19 J/eV 2

)

= 0.955 meV *40 •• Picture the Problem We can use the result of Problem 36 to find the frequency of vibration of the HF molecule. In Problem 36 it was established that:

f =

1 2π

The reduced mass is:

µ=

mH mF mH + mF

k

µ

1249

1250 Chapter 37 Substitute for µ to obtain:

f =

1 2π

k 1 = mH mF 2π mH + mF

k (mH + mF ) mH mF

Substitute numerical values and evaluate f:

f =

1 2π

(970 N/m )(1 u + 19 u ) = (1 u )(19 u )(1.66 ×10−27 kg/u )

1.25 × 1014 Hz

41 •• Picture the Problem We can use the result of Problem 36 to find the effective force constant for NO. In Problem 36 it was established that:

f =

Solve for k:

k = 4π 2 f 2 µ

The reduced mass is:

µ=

Substitute for µ to obtain:

4π 2 f 2 mN mO k= mN + mO

1 2π

k

µ

mN mO mN + mO

Substitute numerical values and evaluate k:

(

)

(

)

4π 2 5.63 × 1013 s −1 (14 u )(16 u ) 1.66 × 10 −27 kg/u k= = 1.55 kN/m 14 u + 16 u 2

42 •• Picture the Problem We can use the expression for the vibrational energy levels of a molecule and the expression for the frequency of oscillation from Problem 36 to find the four lowest vibrational levels of the given molecules. The vibrational energy levels are given by: In Problem 36 we showed that the frequency of oscillation is:

Eν = (ν + 12 )hf,ν = 0,1, 2, ...

f =

1 2π

k

µ

Molecules

1251

where

µ= Substitute for f and µ to obtain:

m1m2 m1 + m2

Eν = =

(ν + 12 )h (m1 + m2 )k 2π

m1m2

(ν + 12 )h 2π

k

(m1 + m2 ) m1m2

Substitute numerical values to obtain:

Eν

( ν + 12 ) (4.136 × 10−15 eV ⋅ s ) = 2π

= (ν + 12 )(0.389 eV ⋅ u )

580 N/m 1.661× 10−27 kg/u

(m1 + m2 ) m1m2

(m1 + m2 ) m1m2

Substitute for m1 and m2 and evaluate E0 for H2:

E0 =

1 2

(0.389 eV ⋅ u ) (1u + 1u ) = 0.275 eV (1u )(1u )

Proceed similarly to complete the table to the right:

H2 (eV)

HD (eV)

D2 (eV)

0

0.275

0.238

0.195

1

0.825

0.715

0.584

2

1.375

1.191

0.973

3

1.925

1.667

1.362

The energies of the photons resulting from transitions between adjacent vibrational levels of these molecules are given by:

∆E = hf =

Solve for λ:

λ=

hc ∆E

hc

λ

1252 Chapter 37 1240 eV ⋅ nm = 2.25 µm 0.550 eV

Evaluate λ(H2):

λ (H 2 ) =

Evaluate λ(HD):

λ (HD ) =

Evaluate λ(D2):

λ (D 2 ) =

1240 eV ⋅ nm = 2.60 µm 0.477 eV

1240 eV ⋅ nm = 3.19 µm 0.389 eV

43 •• Picture the Problem We can set the derivative of the potential energy function equal to zero to find the value of r for which it is either a maximum or a minimum. Examination of the second derivative of this function at the value for r obtained from setting the first derivative equal to zero will establish whether the function is a relative maximum or relative minimum at this point. Differentiate the potential energy function with respect to r:

12 6 dU d ⎧⎪ ⎡⎛ a ⎞ ⎛ a ⎞ ⎤ ⎫⎪ = ⎨U 0 ⎢⎜ ⎟ − 2⎜ ⎟ ⎥ ⎬ dr dr ⎪⎩ ⎣⎢⎝ r ⎠ ⎝ r ⎠ ⎦⎥ ⎪⎭ 11 5 U0 ⎡ ⎛ a ⎞ ⎛a⎞ ⎤ =− ⎢12⎜ ⎟ − 12⎜ ⎟ ⎥ r ⎣⎢ ⎝ r ⎠ ⎝ r ⎠ ⎦⎥

Set the derivative equal to zero:

Solve for r0 to obtain, as our candidate for r that minimizes the Lenard-Jones potential: To show that r0 = a corresponds to a minimum, differentiate U a second time to obtain:

11 5 ⎛a⎞ ⎤ dU U0 ⎡ ⎛ a ⎞ ⎢12⎜ ⎟ − 12⎜⎜ ⎟⎟ ⎥ =− dr r0 ⎢ ⎜⎝ r0 ⎟⎠ ⎝ r0 ⎠ ⎥⎦ ⎣ = 0 for extrema

r0 = a

11 5 d 2U d ⎧⎪ U 0 ⎡ ⎛ a ⎞ ⎛ a ⎞ ⎤ ⎫⎪ = ⎢12⎜ ⎟ − 12⎜ ⎟ ⎥ ⎬ ⎨− dr 2 dr ⎩⎪ r ⎣⎢ ⎝ r ⎠ ⎝ r ⎠ ⎦⎥ ⎪⎭

=

Evaluate this second derivative of the potential at r0 = a:

d 2U dr 2

U0 r2

= r =a

4 ⎡ ⎛ a ⎞10 ⎛a⎞ ⎤ − 132 60 ⎜ ⎟ ⎥ ⎢ ⎜ ⎟ ⎝ r ⎠ ⎥⎦ ⎢⎣ ⎝ r ⎠

U0 [132 − 60] = 72U2 0 > 0 2 a a

Therefore, we can conclude that r0 = a

Molecules

1253

minimizes the potential function. Evaluate Umin:

U min

⎡⎛ a ⎞12 ⎛ a ⎞ 6 ⎤ = U (a ) = U 0 ⎢⎜ ⎟ − 2⎜ ⎟ ⎥ ⎝ a ⎠ ⎥⎦ ⎢⎣⎝ a ⎠ = −U0

From Figure 37-4:

r0 = 0.074 nm and

U 0 = 4.52 eV 44 •• Picture the Problem We can use Equation 21-10 to establish the dependence of E on x and the dependence of an induced dipole on the field that induces it to establish the dependence of p and U on x. (a) In terms of the dipole moment, the electric field on the axis of the dipole at a point a great distance x away has the magnitude (see Equation 21-10):

(b) Because the induced dipole moment is proportional to the field that induces it:

E=

2kp x

3

or

1

E∝

x p∝

3

1 x3

and

r r U = − p⋅ E ∝

(c) Differentiate U with respect to x to obtain:

Fx = −

1 x6

1 dU ∝ 7 dx x

45 •• Picture the Problem the case of two polar molecules, p does not depend on the field E. Because p does not depend on the electric field in which the polar molecules find themselves:

U∝

1 x3

1254 Chapter 37 Differentiate U with respect to x to obtain:

Fx = −

dU 1 ∝ 4 dx x

46 •• Picture the Problem We can use the expression for the vibrational and rotational energies of a molecule, in conjunction with Figure 37-17 to find E0r, f, and hf. (a) Except for a gap of 4E0r/h at the vibrational frequency f, the absorption spectrum contains frequencies equally spaced at:

f =

2 E0 r h

Solve for E0r:

E0 r = 12 hf

From Figure 37-17:

f = 8.66 × 1013 Hz

Substitute numerical values and evaluate E0r:

E0 r =

1 2

(6.63 ×10

−34

)(

)

J ⋅ s 8.66 × 1013 Hz = 2.87 × 10 −20 J ×

1 1.6 × 10−19 J/eV

= 0.179 eV (b) The vibrational energy levels are given by:

Eν = (ν + 12 )hf,

ν = 0,1, 2, ...

E0 = 12 hf

The lowest vibrational energy corresponds to ν = 0:

and

hf = 2E0

Determine f from Figure 37-17:

(1)

f = 8.66 × 1013 Hz

Substitute for f and h and evaluate E0:

E0 =

1 2

(6.63 ×10

−34

)(

)

J ⋅ s 8.66 × 1013 Hz = 2.87 × 10 −20 J ×

1 1.6 × 10 −19 J/eV

= 0.179 eV Substitute in equation (1) and evaluate hf:

hf = 2(0.179 eV ) = 0.358 eV

Molecules 1255 *47 •• Picture the Problem We can find the reduced mass of CO and the moment of inertia of a CO molecule from their definitions. The energy level diagram for the rotational levels for l = 0 to l = 5 can be found using ∆El ,l−1 = 2lE0 r . Finally, we can find the wavelength of the photons emitted for each transition using λl ,l−1 =

hc hc . = ∆El ,l−1 2l∆E0 r

I = µ r02

(a) Express the moment of inertia of CO:

where µ is the reduced mass of the CO molecule.

Find µ:

µ=

(12 u )(16 u ) = 6.86 u mC mO = mC + mO 12 u + 16 u

In Problem 39 it was established that r0 = 0.113 nm. Use this result to evaluate I:

(

)

I = (6.86 u ) 1.66 × 10 −27 kg/u (0.113 nm ) = 1.45 × 10 −46 kg ⋅ m 2 The characteristic rotational energy E0r is given by:

2

E0 r =

h2 2I

Substitute numerical values and evaluate E0r:

(6.58 ×10 eV ⋅ s) (1.6 ×10 = 2(1.45 × 10 kg ⋅ m ) −16

E0 r

2

−46

−19

2

J/eV

)=

0.239 meV

1256 Chapter 37 (b) The energy level diagram is shown to the right. Note that ∆El,l−1, the energy difference between adjacent levels for ∆l = −1, is

∆El ,l−1 = 2lE0 r .

∆El ,l−1 = hf l ,l−1

(c) Express the energy difference ∆El.l−1 between energy levels in terms of the frequency of the emitted radiation: Because c = f l ,l−1λl ,l−1 :

λl ,l−1 =

hc hc = ∆El ,l−1 2l∆E0 r

Substitute numerical values to obtain:

λl ,l−1 =

(4.136 ×10

)(

)

eV ⋅ s 3 × 108 m/s 2596 µm = l 2l(0.239 meV) −15

For l = 1:

λ1,0 =

2596 µm = 2596 µm 1

For l = 2:

λ2,1 =

2596 µm = 1298 µm 2

For l = 3:

λ3, 2 =

2596 µm = 865 µm 3

Molecules 1257 For l = 4:

λ4,3 =

2596 µm = 649 µm 4

For l = 5:

λ5, 4 =

2596 µm = 519 µm 5

These wavelengths fall in the microwave region of the spectrum. *48 ••• Picture the Problem The wavelength resulting from transitions between adjacent harmonic oscillator levels of a LiCl molecule is given by λ =

2π c

ω

. We can find an

expression for ω by following the procedure outlined in the problem statement.

hc hc 2π c = = ∆E hω ω

The wavelength resulting from transitions between adjacent harmonic oscillator levels of this molecule is given by:

λ=

From Problem 24 we have:

U (r ) = −

The Taylor expansion of U(r) about r = r0 is:

⎛ dU ⎞ U (r ) = U (r0 ) + ⎜ ⎟ (r − r0 ) ⎝ dr ⎠ r0

(1)

ke 2 C + n , where ∆E is constant. r r

1 ⎛ d 2U ⎞ 2 + ⎜⎜ 2 ⎟⎟ (r − r0 ) + ... 2 ⎝ dr ⎠ r 0 Because U(r0) is a constant, it can be dropped without affecting the physical results and because

1 ⎛ d 2U ⎞ 2 U (r ) ≈ ⎜⎜ 2 ⎟⎟ (r − r0 ) 2 ⎝ dr ⎠ r 0

⎛ dU ⎞ ⎜ ⎟ = 0: ⎝ dr ⎠ r0 Differentiate U(r) twice to obtain: Because dU/dr = Fnet = 0 at r = r0:

d 2U ke 2 C = − 2 + n(n − 1) n+2 2 3 dr r r 2 nC ke 0 = − n+1 + 2 r0 r0

(2)

1258 Chapter 37 Solving for C yields:

Substitute for C and evaluate

⎛ d 2U ⎞ ⎜⎜ 2 ⎟⎟ to obtain: ⎝ dr ⎠ r0

⎛ d 2U ⎞ Substitute for ⎜⎜ 2 ⎟⎟ in ⎝ dr ⎠ r0

ke 2 r0n+1 ke 2 r0n−1 C= = nr02 n ⎛ d 2U ⎞ ke 2 n(n − 1) ke 2 r0n−1 ⎜⎜ 2 ⎟⎟ = −2 3 (n − 1) + n+2 r0 r0 n ⎝ dr ⎠ r0 =

ke 2 (n − 1) r03

⎤ 1 ⎡ ke 2 2 U (r ) ≈ ⎢ 3 (n − 1)⎥ (r − r0 ) 2 ⎣ r0 ⎦

equation (2): Because the potential energy of a simple harmonic oscillator is given 2 by U SHO = 12 mω 2 (r − r0 ) : Solve for ω to obtain:

Substitute µLiCl for m to obtain:

⎤ 1 1 ⎡ ke 2 2 2 mω 2 (r − r0 ) ≈ ⎢ 3 (n − 1)⎥ (r − r0 ) 2 2 ⎣ r0 ⎦

ω≈

ω≈

=

From Problem 24:

Urep is related to Ue, Ed, and ∆E according to:

n=

(n − 1)ke2 mr03

(n − 1)ke2 mLi mCl 3 r0 mLi + mCl

(n − 1)(mLi + mCl )ke

(3) 2

mLi mCl r03 U e (r0 ) U rep (r0 )

U rep = −(U e + Ed + ∆E )

The energy needed to form Li+ and Cl− from neutral lithium and chlorine atoms is:

∆E = Eionization − Eelectron affinity

Ue(r0) is given by:

Ue = −

(4)

(5)

= 5.39 eV − 3.62 eV = 1.77 eV

ke 2 1.44 eV ⋅ nm =− r0 r0

Molecules

Substitute r0 and evaluate Ue:

Substitute numerical values in equation (5) and evaluate Urep: Substitute for Urep(r0) and Ue(r0) in equation (4) and evaluate n:

Ue = −

1.44 eV ⋅ nm = −7.13 eV 0.202 nm

U rep = −(− 7.13 eV + 4.86 eV + 1.77 eV ) = 0.500 eV n=

− 7.13 eV = 14.3 0.500 eV

Substitute numerical values in equation (3) and evaluate ω:

ω≈

(14.3 − 1)(6.941u + 35.453 u )(1.44 eV ⋅ nm )(1.60 × 10 −19 J/eV ) (6.941u )(35.453 u )(1.66 ×10 −27 kg/u )(0.202 nm )3

= 1.96 × 1014 s −1 Substitute numerical values in equation (1) and evaluate λ:

1259

λ=

(

)

2π 3 × 108 m/s = 9.62 µm 1.96 × 1014 s −1

1260 Chapter 37

Chapter 38 Solids and the Theory of Conduction Conceptual Problems 1 • Determine the Concept The energy lost by the electrons in collision with the ions of the crystal lattice appears as Joule heat (I2R). *2 • Determine the Concept The resistivity of brass at 4 K is almost entirely due to the ″residual resistance, ″ the resistance due to impurities and other imperfections of the crystal lattice. In brass, the zinc ions act as impurities in copper. In pure copper, the resistivity at 4 K is due to its residual resistance, which is very low if the copper is very pure. 3

•

Picture the Problem The contact potential is given by Vcontact =

φ1 − φ2 e

, where φ1 and φ2

are the work functions of the two different metals in contact with each other. (a) Express the contact potential in terms of the work functions of the metals:

Vcontact =

φ1 − φ2 e

Examining Table 38 - 2, we see that the greatest difference between the work functions will occur when potassium and nickel are joined. (b) Substitute numerical values and evaluate Vcontact:

Vcontact =

4

(5.2 eV − 2.1eV )(1.60 ×10−19 J/eV ) = 1.60 × 10 −19 C

•

Picture the Problem The contact potential is given by Vcontact =

3.10 V

φ1 − φ2 e

, where φ1 and φ2

are the work functions of the two different metals in contact with each other. (a) Express the contact potential in terms of the work functions of the metals:

Vcontact =

1261

φ1 − φ2 e

1262

Chapter 38 Examining Table 38 - 2, we see that the least difference between the work functions will occur when silver and gold are joined.

(b) Substitute numerical values and evaluate Vcontact:

Vcontact

( 4.8 eV − 4.7 eV ) (1.60 × 10−19 J/eV ) = = 1.60 × 10−19 C

0.100 V

5 • Determine the Concept If the valence band is only partially full, there are many available empty energy states in the band, and the electrons in the band can easily be raised to a higher energy state by an electric field.

(c ) is correct.

6 • Determine the Concept Insulators are poor conductors of electricity because there is a large energy gap between the full valence band and the next higher band where electrons can exist.

(b ) is correct.

7 • (a) True (b) False. The classical free-electron theory predicts heat capacities for metals that are not observed experimentally. (c) True (d) False. The Fermi energy is the energy of the last filled (or half-filled) level at T = 0. (e) True (f) True (g) False. Because semiconductors conduct current by electrons and holes, their conduction is in both directions. *8 • Determine the Concept The resistivity of copper increases with increasing temperature; the resistivity of (pure) silicon decreases with increasing temperature because the number of charge carriers increases. 9 • Determine the Concept Because a gallium atom can accept electrons from the valence band of germanium to complete its four covalent bonds, (b ) is correct.

Solids and the Theory of Conduction 1263 10 • Determine the Concept Because phosphorus has 3 electrons that it can donate to the conduction band of germanium without leaving holes in the valence band,

(d ) is correct.

11 • Determine the Concept The excited electron is in the conduction band and can conduct electricity. A hole is left in the valence band allowing the positive hole to move through the band also contributing to the current. 12 • Determine the Concept (a) Phosphorus and antimony will make n-type semiconductors since each has one more valence electron than silicon. (b) Boron and thallium will make p-type semiconductors since each has one less valence electron than silicon.

13 • Determine the Concept Because a pn junction solar cell has donor impurities on one side and acceptor impurities on the other, both electrons and holes are created.

(c ) is correct.

Estimation and Approximation 14 • Picture the Problem We can use the list of tables on the inside back covers of volumes 1 and 2 to find tables of material properties. A representative sample is included in the following table in which all the units are SI: Table

Material property

13-1

Mass density

20-3

Thermal conductivity

20-1

Thermal expansion

12-8

Tensile strength

Largest value 22.5×103 (Osmium)

Smallest value 0.08994 (Hydrogen)

Ratio (order of magnitude)

429 (Ag)

0.026 (air)

104

51×10−6 (ice)

10−6 (invar)

102

520 (steel)

2 (concrete)

102

105

1264

Chapter 38

12-8

Young’s modulus

200 (steel)

9 (bone)

10

18-1

Heat capacity

4.18 (water)

0.900 (Al)

1

15 • Picture the Problem Figure 38-21 is reproduced below. We can draw tangent lines at each of the voltages and estimate the slope. The differential resistance is the reciprocal of the slope.

V (V)

1/slope (Ω)

–20

∞

+0.2

40

+0.4

20

+0.6

10

+0.8

5

Remarks: Note that, because of the difficulty in determining the slopes, these results are only approximations.

Solids and the Theory of Conduction 1265

The Structure of Solids 16 • Picture the Problem We can use the geometry of the ion to relate the volume per mole to the length of its side r0 and the definition of density to express the volume per mole in terms of its molar mass and density.

Vmol = 2 N A r03

Because the cube length/ion is r0, the volume/mole is given by: Solve for r0:

The volume/mole is given by:

r0 = 3

Vmol 2NA

Vmol =

M

ρ

where M is the molar mass of KCl. Substitute for Vmol in the expression for r0:

r0 = 3

M 2 ρN A

Substitute numerical values and evaluate r0:

r0 = 3

74.55 g/mol = 0.315 nm 2 1.984 g/cm 6.02 × 10 23 particles/mol

(

3

)(

)

17 • Picture the Problem We can use the definition of density and the geometry of the ions to compute the density of LiCl. The density of LiCl is given by:

ρ=

Express the volume of the unit cell:

Vunit cell = (2r0 )

Because the unit cell has four molecules, its mass is given by:

M unit cell =

Substitute for Vunit cell and Munit cell to obtain:

4M 4M N ρ = A3 = (2r0 ) N A (2r0 )3

Substitute numerical values and evaluate ρ:

M unit cell Vunit cell 3

4M NA

1266 Chapter 38

ρ=

(6.02 ×10

4(42.4 g/mol) 23

)[ (

particles/mol 2 0.257 × 10 −9 m

)]

⎛ 1m ⎞ ⎟⎟ = 2.07 × 10 g/m × ⎜⎜ 2 ⎝ 10 cm ⎠ 6

3

3

= 2.07 g/cm3 *18 • Picture the Problem We can solve Equation 38-6 for n.

ke 2 ⎛ 1 ⎞ U (r0 ) = −α ⎜1 − ⎟ r0 ⎝ n ⎠

Equation 38-6 is:

and

U (r0 ) = α Solve for n to obtain:

n=

ke 2 ⎛ 1 ⎞ ⎜1 − ⎟ r0 ⎝ n ⎠

1 U (r0 ) r0 1− αke 2

Substitute numerical values and evaluate n:

n=

1 ⎛ 1 eV/ion pair ⎞ ⎟(0.257 nm ) (741 kJ/mol)⎜⎜ ⎟ 96 . 47 kJ/mol ⎝ ⎠ 1− (1.7476)(1.44 eV ⋅ nm )

= 4.64

19 •• Picture the Problem We can substitute numerical values in Equation 38-6 to evaluate U(ro) for n = 8 and n = 10. (a) Equation 38-6 is:

U (r0 ) = −α

ke 2 ⎛ 1 ⎞ ⎜1 − ⎟ r0 ⎝ n ⎠

Substitute numerical values and evaluate U(r0):

U (ro ) = −

(1.7476)(8.99 ×109 N ⋅ m 2 /C 2 )(1.60 ×10−19 J )2 ⎛1 − 1 ⎞ ⎛ −9

0.208 × 10 m

⎜ ⎝

1eV ⎞ ⎟⎜ −19 ⎟ 8 ⎠ ⎝ 1.60 × 10 J ⎠

= − 10.6 eV (b) The fractional change is given by:

∆U (r0 ) U n=10 − U n=8 U n=10 = = −1 U (r0 ) U n=8 U n=8

3

Solids and the Theory of Conduction 1267 Substitute numerical values and evaluate U(r0) for n = 10:

U n=10 (ro ) = −

(1.7476)(8.99 ×109 N ⋅ m 2 /C 2 )(1.60 ×10−19 J )2 ⎛1 − ⎜ ⎝

−9

0.208 × 10 m

1 ⎞⎛ 1eV ⎞ ⎟⎜ −19 ⎟ 10 ⎠ ⎝ 1.60 × 10 J ⎠

= −10.9 eV Substitute numerical values and evaluate the fractional change in U(r0):

∆U (r0 ) − 10.9 eV = − 1 = − 2.83% U (r0 ) − 10.6 eV

A Microscopic Picture of Conduction 20 • Picture the Problem We can use the expression for the volume occupied by one electron 13

⎛ 3 ⎞ ⎟⎟ to show that rs = ⎜⎜ ⎝ 4π n ⎠

.

(a) The volume occupied by one electron is:

1 4 3 = π rs n 3

Solve for rs:

⎛ 3 ⎞ ⎟⎟ rs = ⎜⎜ ⎝ 4π n ⎠

(b) From Table 38-1:

nCu = 8.47 × 10 28 m −3

Substitute numerical values and evaluate rs for copper:

rs = 3

13

3 = 0.141 nm 4π 8.47 × 10 28 m −3

(

)

21 • Picture the Problem We can use the expression for the resistivity of the copper in terms of vav and λ to find the classical value for the resistivity ρ of copper. In (b) we can use

vav =

3kT to relate the average speed to the temperature. me

(a) In terms of the mean free path and the mean speed, the resistivity is:

ρ=

me vav ne e 2λ

1268 Chapter 38 Substitute numerical values and evaluate ρ (see Table 38-1 for the free-electron number density of copper):

ρ=

(9.11×10

(8.47 ×10

28

−31

)(

kg 1.17 × 105 m/s

)(

)

)

= 0.123 µΩ ⋅ m

m −3 1.60 × 10 −19 C (0.4 nm ) 2

(b) Relate the average speed of the electrons to the temperature:

vav =

Substitute for vav in the expression for ρ to obtain:

ρ=

3kT me

me ne e 2λ

3kT 1 = 3me kT me ne e 2λ

Substitute numerical values and evaluate ρ:

ρ=

(

)( )(1.60 ×10

)

3 9.11 × 10 −31 kg 1.38 × 10 −23 J/K (100 K )

(8.47 ×10

28

m −3

−19

)

C (0.4 nm ) 2

= 0.0708 µΩ ⋅ m

*22 •• Picture the Problem We can use Equation 38-14 to estimate the resistivity of silicon. (a) From Equation 38-14:

The speed of the electrons is given by:

Substitute numerical values and evaluate vav:

ρ=

me vav ne e 2λ

vav = vF =

vav =

(1)

2 EF me

2(4.88 eV ) 1.60 × 10 −19 J 9.11× 10 −31 kg 1 eV

(

)

= 1.31× 106 m/s The electron density of Si is given by:

ne = MN A N atom where Natom is the number of electrons per atom.

Substitute numerical values and evaluate ne:

kg ⎞⎛ 6.02 × 10 23 atoms ⎞⎛ 2 e ⎞ ⎛ 29 3 ⎟⎟⎜ ne = ⎜ 2.41× 103 3 ⎟⎜⎜ ⎟ = 1.03 × 10 e/m m ⎠⎝ 0.02809 kg ⎠⎝ atom ⎠ ⎝ Substitute numerical values in equation (1) and evaluate ρ:

Solids and the Theory of Conduction 1269

(9.11×10 kg )(1.31×10 m/s) ρ= = (1.60 ×10 C) (1.03 ×10 e/m )(27.0 ×10 m ) −31

−19

6

2

29

−9

3

1.66 × 10 −8 Ω ⋅ m

(b) The accepted resistivity of 640 Ω⋅m is much greater than the calculated value. We assume that valence electrons will produce conduction in the material. Silicon is a semiconductor and a gap between the valence band and conduction band exists. Only electrons with sufficient energies will be found in the conduction band.

The Fermi Electron Gas 23 • Picture the Problem The number density of free electrons is given by n = ρN A M , where NA is Avogadro’s number, ρ is the density of the element, and M is its molar mass. Relate the number density of free electrons to the density ρ and molar mass M of the element:

n ρ ρN A = ⇒ n= NA M M

(a) For Ag:

(10.5 g/cm )(6.02 ×10 = 3

nAg

23

)

23

)

electrons/mol = 5.86 × 1022 electrons/cm3 107.87 g/mol

(b) For Au:

nAg =

(19.3 g/cm )(6.02 ×10 3

electrons/mol = 5.90 × 1022 electrons/cm3 196.97 g/mol

Both these results agree with the values in Table 38 - 1. 24 • Picture the Problem The number of free electrons per atom ne is given by ne = nM ρN A , where NA is Avogadro’s number, ρ is the density of the element, M is its molar mass, and n is the free electron number density for the element. The number of free electrons per atom is given by: Substitute numerical values and evaluate ne:

ne =

nM ρN A

1270 Chapter 38

(18.1 × 10 electrons/cm )(26.98 g/mol) = = (2.7 g/cm )(6.02 × 10 electrons/mol) 22

ne

3

3

23

3.00

25 • Picture the Problem The number of free electrons per atom ne is given by ne = nM ρN A , where NA is Avogadro’s number, ρ is the density of the element, M is its molar mass, and n is the free electron number density for the element. The number of free electrons per atom is given by:

ne =

nM ρN A

Substitute numerical values and evaluate ne:

ne =

(14.8 ×10 electrons/cm )(118.69 g/mol) = (7.3 g/cm )(6.022 ×10 electrons/mol) 22

3

3

23

4.00

*26 • Picture the Problem The Fermi temperature TF is defined by kTF = EF, where EF is the Fermi energy. The Fermi temperature is given by:

TF =

EF k

(a) For Al:

TF =

11.7 eV = 1.36 × 105 K −5 8.62 × 10 eV/K

(b) For K:

TF =

2.11eV = 2.45 × 104 K 8.62 × 10−5 eV/K

(c) For Sn:

TF =

10.2 eV = 1.18 × 105 K −5 8.62 × 10 eV/K

27 • Picture the Problem We can solve the expression for the Fermi energy for the speed of a conduction electron. Express the Fermi energy in terms of the Fermi speed a conduction electron:

EF = 12 meuF2

Solids and the Theory of Conduction 1271 Solve for uF:

(a) Substitute numerical values (see Table 38-1 for EF) and evaluate uF for Na:

(b) Substitute numerical values and evaluate uF for Au:

uF =

2 EF me

uF =

2(3.24 eV ) 1.60 × 10 −19 J/eV 9.11×10 −31 kg

(

)

= 1.07 × 106 m/s

(

2(5.53 eV ) 1.60 × 10−19 J/eV 9.11× 10 −31 kg

uF =

)

= 1.39 × 106 m/s (c) Substitute numerical values and evaluate uF for Sn:

(

2(10.2 eV ) 1.60 ×10 −19 J/eV 9.11×10 −31 kg

uF =

)

= 1.89 ×106 m/s 28 • Picture the Problem The Fermi energy at T = 0 depends on the number of electrons per 23 unit volume (the number density) according to EF = 0.365 eV ⋅ nm 2 (N V ) .

(

The Fermi energy at T = 0 is given by:

)

(

)

⎛N⎞ EF = 0.365 eV ⋅ nm 2 ⎜ ⎟ ⎝V ⎠

23

(a) For Al, N/V = 18.1×1022 electrons/cm3 (see Table 38-1) and:

(

)(

)

= 11.7 eV

)

= 2.12 eV

EF = 0.365 eV ⋅ nm 2 18.1× 10 22 electrons/cm3

23

(b) For K, N/V = 1.4×1022 electrons/cm3 and:

(

)(

EF = 0.365 eV ⋅ nm 2 1.4 × 10 22 electrons/cm3

23

(c) For Sn, N/V = 14.8×1022 electrons/cm3 and:

(

)(

EF = 0.365 eV ⋅ nm 2 14.8 × 1022 electrons/cm3

)

23

= 10.2 eV

1272 Chapter 38 29 • Picture the Problem The average energy of electrons in a Fermi gas at T = 0 is threefifths of the Fermi energy. The average energy at T = 0 is given by:

Eav = 53 EF

Eav = (0.6 )(7.04 eV ) = 4.22 eV

(a) For copper, EF = 7.04 eV (see Table 38-1) and: (b) For lithium, EF = 4.75 eV and:

Eav = (0.6 )(4.75 eV ) = 2.85 eV

30 • Picture the Problem The Fermi energy at T = 0 is given by

EF = (0.365 eV ⋅ nm 2 )( N V ) , where N/V is the free-electron number density and the Fermi temperature is related to the Fermi energy according to kTF = EF . 32

(a) The Fermi temperature for iron is given by:

TF =

EF k

Substitute numerical values (see Table 38-1) and evaluate TF:

TF =

11.2 eV = 1.30 × 105 K −5 8.62 × 10 eV/K

(b) The Fermi energy at T = 0 is given by:

⎛N⎞ EF = 0.365 eV ⋅ nm 2 ⎜ ⎟ ⎝V ⎠

(

)

32

Substitute numerical values (see Table 38-1) and evaluate EF:

(

)(

EF = 0.365 eV ⋅ nm 2 17.0 × 1022 electrons/cm3 *31 •• Picture the Problem We can use ne = ρV =

ρN A N atom m

)

23

= 11.2 eV

, where Natom is the number of

electrons per atom, to calculate the electron density of gold. The Fermi energy is given by EF = 12 me vF2 . (a) The electron density of gold is given by: Substitute numerical values and evaluate ne:

ne = ρV =

ρN A N atom m

Solids and the Theory of Conduction 1273

(

)

⎛ 1e ⎞ ⎛ 3 kg ⎞ 23 ⎟ ⎜19.3 × 10 3 ⎟ 6.02 × 10 atoms ⎜ m ⎠ ⎝ 1 atom ⎠ = 5.90 × 10 28 e/m3 ne = ⎝ 0.197 kg

EF = 12 me vF2

(b) The Fermi energy is given by: Substitute numerical values and evaluate EF:

EF =

(

)(

)

2⎛ 1 1 eV ⎞ 9.11× 10−31 kg 1.39 × 106 m/s ⎜ = 5.50 eV −19 ⎟ 2 ⎝ 1.60 × 10 J ⎠

(c) The factor by which the Fermi energy is higher than the kT energy at room temperature is:

f =

EF kT

At room temperature kT = 0.026 eV. Substitute numerical values and evaluate f:

f =

5.50 eV = 212 0.026 eV

(d) EF is 212 times kT at room temperature. There are so many free electrons present that most of them are crowded, as described by the Pauli exclusion principle, up to energies far higher than they would be according to the classical model. *32 •• Picture the Problem We can solve PV = 23 NEav for P and substitute for Eav in order to express P in terms of N/V and EF. Solve PV = 23 NEav for P:

Because Eav = 35 EF :

P=

2⎛ N ⎞ ⎜ ⎟ Eav 3⎝V ⎠

P=

2 ⎛ N ⎞⎛ 3 ⎞ 2 ⎛ N ⎞ ⎜ ⎟⎜ EF ⎟ = ⎜ ⎟ EF 3 ⎝ V ⎠⎝ 5 ⎠ 5 ⎝ V ⎠

Substitute numerical values (see Table 38-1) and evaluate P:

(

)

(

)

2 8.47 × 10 22 electrons/cm3 (7.04 eV ) 1.60 × 10−19 J/eV 5 1atm = 3.82 × 1010 N/m 2 = 3.82 × 1010 N/m 2 × 101.325 × 103 N/m 2

P=

= 3.77 × 105 atm

1274 Chapter 38 33 •• Picture the Problem We can follow the procedure given in the problem statement to show that P =

2 NEF 5 2 NEF . = CV −5 3 and B = P = 5V 3 3V

PV = 23 NEav

(a) From Problem 32 we have: Because Eav = 35 EF :

P=

2 ⎛ N ⎞⎛ 3 ⎞ 2 ⎛ N ⎞ ⎜ ⎟⎜ EF ⎟ = ⎜ ⎟ EF 3 ⎝ V ⎠⎝ 5 ⎠ 5 ⎝ V ⎠ h 2 ⎛ 3N ⎞ ⎜ ⎟ 8me ⎝ πV ⎠

23

(1)

23

h 2 ⎛ 3N ⎞ −2 3 ⎜ ⎟ V 8me ⎝ π ⎠

The Fermi energy is given by:

EF =

Substitute to obtain:

2 ⎛ N ⎞ h 2 ⎛ 3N ⎞ −2 3 P= ⎜ ⎟ ⎜ ⎟ V 5 ⎝ V ⎠ 8me ⎝ π ⎠

=

23

23

N 5 3h 2 ⎛ 3 ⎞ −5 3 = = CV −5 3 ⎜ ⎟ V 20me ⎝ π ⎠ 23

N 5 3h 2 ⎛ 3 ⎞ where C = ⎜ ⎟ is a constant. 20me ⎝ π ⎠

[

]

dP d = −V CV −5 3 dV dV ⎛ 5 ⎞ 5 = −CV ⎜ − V −8 3 ⎟ = CV −5 3 ⎝ 3 ⎠ 3 5 = P 3

(b) The bulk modulus is given by:

B = −V

Substitute for P from equation (1) to obtain:

B=

5 ⎡2 ⎛ N ⎞ ⎤ 2⎛ N ⎞ ⎜ ⎟ EF ⎜ ⎟ EF ⎥ = ⎢ 3 ⎣5 ⎝ V ⎠ ⎦ 3⎝V ⎠

(c) Substitute numerical values and evaluate B for copper:

B=

(

)

(

)

2 8.47 × 1022 electrons/cm3 (7.04 eV ) 1.60 × 10 −19 J/eV = 63.6 × 109 N/m 2 3

From Table 13-2:

BCu = 140 GN/m2

Solids and the Theory of Conduction 1275 Divide the calculated value for B by the value from Table 13-2 to obtain:

B 63.6 GN/m 2 = = 0.455 BCu 140 GN/m 2 or

B = 0.454 BCu 34

•

Picture the Problem The contact potential is given by Vcontact =

φ1 − φ2 e

, where φ1 and φ2

are the work functions of the two different metals in contact with each other. The contact potential is given by:

Vcontact =

φ1 − φ2 e

(a) For Ag and Cu (see Table 38-1):

(4.7 eV − 4.1eV )(1.60 ×10−19 J/eV ) =

0.6 V

( 5.2 eV − 4.7 eV ) (1.60 × 10−19 J/eV ) = =

0.5 V

Vcontact =

1.60 × 10−19 C

(b) For Ag and Ni:

Vcontact

1.60 × 10−19 C

(c) For Ca and Cu (see Table 38-1):

Vcontact =

(4.1eV − 3.2 eV )(1.60 ×10−19 J/eV ) = 1.60 × 10 −19 C

0.9 V

Heat Capacity Due to Electrons in a Metal *35 •• Picture the Problem We can use Equation 38-29 to find the molar specific heat of gold at constant volume and room temperature. The molar specific heat is given by Equation 38-29: The Fermi energy is given by:

c'V =

π 2 RT 2TF

EF = kTF ⇒ TF =

EF k

1276 Chapter 38 Substitute for TF to obtain:

c'V =

π 2 RkT 2 EF

Substitute numerical values and evaluate c'V :

⎞ 1eV ⎟(300 K ) −19 ⎟ ⎝ 1.60 × 10 J ⎠ ⎛

c'V =

π 2 (8.31 J/mol K ) (1.38 ×10 −23 J/mol)⎜⎜ 2(5.53 eV )

= 0.192 J/mol ⋅ K Remarks: The value 0.192 J/mol K is for a mole of gold atoms. Since each gold atom contributes one electron to the metal, a mole of gold corresponds to a mole of electrons.

Quantum Theory of Electrical Conduction 36 • Picture the Problem We can solve the equation giving the resistivity of a conductor in terms of the mean free path and the mean speed for the mean free path and use the Fermi speeds from Problem 27 as the mean speeds. In terms of the mean free path and the mean speed, the resistivity is:

ρ=

me vav ne 2λ

Solve for λ to obtain:

λ=

me vav ne 2 ρ

From Problem 27 we have:

µ F,Na = 1.07 × 106 m/s µ F,Au = 1.39 × 106 m/s and

µ F,Sn = 1.89 × 106 m/s Using the Fermi speeds as the average speeds, substitute numerical values and evaluate the mean free path of Na:

λNa =

(2.65 ×10

(9.11×10

22

)( electrons/cm )(1.60 × 10 −31

kg 1.07 × 106 m/s 3

−19

C

)

) (4.2 µΩ ⋅ cm) 2

= 34.2 nm

Solids and the Theory of Conduction 1277

Proceed as above for Au:

λAu =

(5.90 ×10

(9.11×10 22

−31

)(

)

= 41.1 nm

)

= 4.29 nm

kg 1.39 × 106 m/s

)(

electrons/cm3 1.60 × 10 −19 C

) (2.04 µΩ ⋅ cm) 2

Proceed as above for Sn:

λSn =

(14.8 ×10

(9.11×10

22

−31

)(

kg 1.89 × 106 m/s

)(

)

electrons/cm3 1.60 × 10 −19 C (10.6 µΩ ⋅ cm ) 2

*37 •• Picture the Problem We can solve the resistivity equation for the mean free path and then substitute the Fermi speed for the average speed to express the mean free path as a function of the Fermi energy. (a) In terms of the mean free path and the mean speed, the resistivity is:

ρi =

me vav meuF = ne 2λi ne 2λi

Solve for λ to obtain:

λi =

meuF ne 2 ρi

Express the Fermi speed uF in terms of the Fermi energy EF:

uF =

2 EF me

Substitute to obtain:

λi =

2me EF ne 2 ρ i

Substitute numerical values (see Table 38-1) and evaluate λi:

λi =

(

)

(

2 9.11 × 10 −31 kg (7.04 eV ) 1.60 × 10 −19 J/eV

(8.47 × 10

28

)(

electrons/m 3 1.60 × 10 −19 C

) (10 2

1 nπr 2

(b) From Equation 38-16 we have:

λ=

Solve for π r2:

π r2 =

1 nλ

−8

)

Ω⋅m

)

= 66.1 nm

1278 Chapter 38 Substitute numerical values and evaluate πr2:

π r2 =

1 8.47 × 10 m −3 (66.1 nm )

(

)

28

= 1.79 × 10−22 m 2 = 1.79 × 10−4 nm 2

Band Theory of Solids 38 • Picture the Problem We can use Eg = hc λ to calculate the energy gap for this semiconductor. The lowest photon energy to increase conductivity is given by: Substitute numerical values and evaluate Eg :

Eg =

hc

λ

(6.63 ×10

−34

)(

)

J ⋅ s 3 × 108 m/s Eg = 380 × 10−9 m 1.60 × 10−19 J/eV

(

)(

)

= 3.27 eV *39 • Picture the Problem We can relate the maximum photon wavelength to the energy gag using ∆E = hf = hc λ .

hc

Express the energy gap as a function of the wavelength of the photon:

∆E = hf =

Solve for λ:

λ=

hc ∆E

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 1.09 µm 1.14 eV

λ

40 • Picture the Problem We can relate the maximum photon wavelength to the energy gag using ∆E = hf = hc λ . Express the energy gap as a function of the wavelength of the photon:

∆E = hf =

Solve for λ:

λ=

hc ∆E

hc

λ

Solids and the Theory of Conduction 1279 Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 1.68 µm 0.74 eV

41 • Picture the Problem We can relate the maximum photon wavelength to the energy gag using ∆E = hf = hc λ .

hc

Express the energy gap as a function of the wavelength of the photon:

∆E = hf =

Solve for λ:

λ=

hc ∆E

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 177 nm 7.0 eV

λ

42 •• Picture the Problem We can use ∆E = hf = hc λ to find the energy gap between these bands and T = Eg/k to find the temperature for which kT equals this energy gap. (a) Express the energy gap as a function of the wavelength of the photon:

∆E =

hc

Substitute numerical values and evaluate ∆E:

∆E =

1240 eV ⋅ nm = 0.370 eV 3.35 µm

(b) The temperature is related to the energy gap Eg according to:

T=

Eg

Substitute numerical values and evaluate T:

T=

0.370 eV = 4.29 × 103 K −5 8.617 × 10 eV/K

λ

k

Semiconductors 43 • Picture the Problem We can use ∆E = kT to find the temperature for which kT = 0.01 eV Express the temperature T in terms of the energy gap ∆E:

T=

∆E k

1280 Chapter 38 Substitute numerical values and evaluate T:

T=

0.01eV 1.38 × 10 −23 J/K ×

1eV 1.60 × 10 −19 J

= 116 K *44 •• Picture the Problem We can use E = hf to find the energy gap of this semiconductor. The energy gap of the semiconductor is given by:

Eg = hf =

hc

λ

where hc = 1240 eV⋅nm Substitute numerical values and evaluate Eg:

Eg =

1240 eV ⋅ nm = 0.670 eV 1.85 µm

45 •• Picture the Problem We can make the indicated substitutions in the expression for a0 (= 0.0529 nm) to obtain an expression the Bohr radii for the outer electron as it orbits the impurity arsenic atom in silicon and germanium. Make the indicated substitutions in the expression for a0 to obtain:

aB =

κ ∈0 h 2 κ ∈0 me h 2 = π meff e 2 π me meff e 2

κme ∈0 h 2 κme = = a0 meff π me e 2 meff For silicon:

aB =

12me (0.0529 nm ) = 3.17 nm 0.2me

For germanium:

aB =

16me (0.0529 nm ) = 8.46 nm 0.1me

*46 •• Picture the Problem We can make the same substitutions we made in Problem 45 in the expression for E1 (= 13.6 eV) to obtain an expression that we can use to estimate the binding energy of the extra electron of an impurity arsenic atom in silicon and germanium.

Solids and the Theory of Conduction 1281 Make the indicated substitutions in the expression for E1 to obtain:

E1 = −

e 2 meff e 2 me meff = − 2 8meκ 2 ∈02 h 2 8(κ ∈0 ) h 2

=−

meff e 2 me meκ 2 ∈02 8 ∈02 h 2

=−

meff E1 meκ 2 ∈02

(a) For silicon:

E1 = −

0.2me (13.6 eV ) = 18.9 meV 2 me (12 )

(b) For germanium:

E1 = −

0.1me (13.6 eV ) = 5.31meV 2 me (16 )

47 •• Picture the Problem We can use the expression for the resistivity ρ of the sample as a function of the mean free path λ of the conduction electrons in conjunction with the expression for the average speed vav of the electrons to derive an expression that we can use to calculate the mean free path of the electrons. Express the resistivity ρ of the sample as a function of the mean free path λ of the conduction electrons: Solve for λ to obtain:

ρ=

me vav ne e 2λ

λ=

me vav ne e 2 ρ

The average speed vav of the electrons is given by:

vav ≈ vrms =

Substitute for vav in the expression for λ to obtain:

λ=

me ne e 2 ρ

(1)

3kT me

3kmeT 3kT = me ne e 2 ρ

Substitute numerical values and evaluate λ:

λ=

(

)

(

)

3 1.38 ×10 −23 J/K (0.2 ) 9.11×10 −31 kg (300 K )

(10

16

)(

)( 2

cm −3 1.6 × 10 −19 C 5 ×10 −3 Ω ⋅ m

The number density of electrons ne is related to the mass density ρm, Avogadro’s number NA, and the molar mass M.:

ne =

)

= 37.2 nm

ρm N A M

Substitute numerical values (For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol.) and

1282 Chapter 38 evaluate ne:

ne =

(8.93 g/cm )(6.02 ×10 3

)

23

electrons/mol = 8.47 × 1028 electrons/m3 63.5 g/mol

Using equation (1), evaluate λCu (see Table 25-1 for the resistivity of copper and Example 38-4 for uF):

λCu =

(8.47 ×10

(9.11×10 28

−31

)(

kg 1.57 × 106 m/s

)(

electrons/m 1.6 × 10 3

−19

)(

)

2

C 1.7 × 10 −8 Ω ⋅ m

)

= 38.8 nm

The mean free paths agree to within 4.02%. 48 •• Picture the Problem We can use the expression for the Hall coefficient to determine the type of impurity and the concentration of these impurities. (a) and (b) The Hall coefficient is given by:

1 nq

R=

(1)

Because R > 0, q > 0 and conduction is by holes and the sample contains acceptor impurities. Solve equation (1) for n:

n=

1 Rq

Substitute numerical values and evaluate n:

n=

1 (0.04 V ⋅ m/A ⋅ T ) 1.6 ×10−19 C

(

= 1.56 × 1020 m −3

Semiconductor Junctions and Devices 49 •• Picture the Problem The following graph of I/I0 versus Vb was plotted using a spreadsheet program.

)

Solids and the Theory of Conduction 1283

50 45 40 35

I /I 0

30 25 20 15 10 5 0 -5 -0.10

-0.05

0.00

0.05

0.10

V b (V)

50 • Picture the Problem The base current is the difference between the emitter current and the plate current. The base current IB is given by:

IB = I − IC

We’re given that:

I C = 0.88I = 25.0 mA

Solve for I to obtain:

I=

Substitute numerical values for I and IC and evaluate IB:

I B = 28.4 mA − 25.0 mA = 3.4 mA

25.0 mA = 28.4 mA 0.88

*51 •• Picture the Problem We can use its definition to compute the voltage gain of the amplifier. The voltage gain of the amplifier is given by:

Voltage gain =

Substitute numerical values and evaluate the voltage gain:

Voltage gain =

I c RL I b Rb

(0.5 mA)(10 kΩ ) (10 µA )(2 kΩ )

= 250 52 •• Picture the Problem The number of electron-hole pairs N is related to the energy E of the incident beam and the energy gap Eg.

1284 Chapter 38 (a) The number of electron-hole pairs N is given by:

N=

E Eg

Substitute numerical values and evaluate N:

N=

660 keV = 9.17 × 10 5 0.72 eV

(b) The energy resolution of the detector is given by:

∆E ∆N = E N

For ∆N = 1 and N = 9.17 × 10 5 :

1 ∆E = = 1.04 × 10−3 5 E 9.17 × 10 = 0.104%

53 •• Picture the Problem The nearly full valence band is shown shaded. The Fermi level is shown by the dashed line.

(a)

(b)

*54 •• Picture the Problem We can use Ohm’s law and the expression for the current from Problem 49 to find the resistance for small reverse-and-forward bias voltages. (a) Use Ohm’s law to express the resistance:

R=

From Problem 47, the current across a pn junction is given by:

I = I 0 e eVb

For eVb << kT:

Substitute to obtain:

Vb I

(1)

(

e eVb

kT

I = I0

kT

−1 ≈ 1+

eVb kT

)

−1

(2)

eVb eV −1 = b kT kT

Solids and the Theory of Conduction 1285 Substitute for I in equation (1) and simplify:

R=

Substitute numerical values and evaluate R:

R=

Vb kT = eV I 0 b eI 0 kT

(0.025 eV )(1.60 × 10−19 J / eV )

(1.60 × 10

−19

C )(10− 9 A )

= 25.0 MΩ (b) Substitute equation (2) in equation (1) to obtain: Evaluate

eVb for Vb = − 0.5 V: kT

Evaluate equation (3) for Vb = − 0.5 V:

(c) Evaluate

eVb for kT

R=

eVb kT R=

V

(3) ( − 1) (1.60 ×10 C)(− 0.5 V ) = −19.8 = (1.38 ×10 J/K )(293 K )

I0 e

b eVb kT

−19

−23

− 0.5 V = 500 MΩ 10 A e −19.8 − 1

(

−9

)(

)

(

) )

eVb 1.60 ×10 −19 C (0.5 V ) = = 19.8 kT 1.38 ×10 −23 J/K (293 K )

(

Vb = + 0.5 V: Evaluate equation (3) for Vb = +0.5 V: (d) Evaluate Rac = dV/dI to obtain:

R=

0.5 V = 1.26 Ω 10 A e19.8 − 1

(

Rac =

−9

)(

)

dV ⎛ dI ⎞ =⎜ ⎟ dI ⎝ dV ⎠

[ (

⎧ d =⎨ I 0 e eVb ⎩ dV ⎧ eI = ⎨ 0 e eVb ⎩ kT Substitute numerical values and evaluate Rac:

−1

kT

)]

⎫ −1 ⎬ ⎭

−1

−1

kT

kT −eVb ⎫ e ⎬ = eI 0 ⎭

kT

Rac = (25 MΩ )e −19.8 = 0.0629 Ω

55 •• Picture the Problem We can use the Hall-effect equation to find the concentration of charge carriers in the slab of silicon. We can determine the semiconductor type by determining the directions of the magnetic and electric fields. Use the expression for the Halleffect voltage to relate the concentration of charge carriers n to

VH =

IB nte

1286 Chapter 38 the Hall voltage VH: Solve for n to obtain:

n=

IB teVH

Substitute numerical values and evaluate n:

n=

(0.2 A )(0.4 T ) (1.0 mm) (1.60 × 10 −19 C)(5 mV )

= 1.00 × 10 23 m −3 r r Referring to Figure 26 - 28, note that B points out of the page. E is in the y direction. Therefore, the charge carriers are holes and the semiconductor is p - type.

The BCS Theory 56 • Picture the Problem We can calculate Eg using Eg = 3.5kTc and find the wavelength of a photon having sufficient energy to break up Cooper pairs in tin at T = 0 using

λ = hc Eg .

(a) From Equation 38-24 we have: Substitute numerical values and evaluate Eg:

Eg = 3.5kTc

(

)

Eg = 3.5 8.617 × 10 −5 eV/K (3.72 K ) = 1.12 meV

Eg

Express the ratio of Eg to Eg,measured:

Eg,measured

=

1.12 meV = 1.87 6 × 10−4 eV

or

Eg ≈ 2 Eg,measured (b) The wavelength of a photon having sufficient energy to break up Cooper pairs in tin at T = 0 is given by:

λ=

hc Eg

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 2.07 × 106 nm 6 × 10 −4 eV

= 2.07 mm

Solids and the Theory of Conduction 1287 *57 • Picture the Problem We can calculate Eg using Eg = 3.5kTc and find the wavelength of a photon having sufficient energy to break up Cooper pairs in tin at T = 0 using

λ = hc Eg .

(a) From Equation 38-24 we have: Substitute numerical values and evaluate Eg:

Eg = 3.5kTc

(

)

Eg = 3.5 8.62 × 10 −5 eV/K (7.19 K ) = 2.17 meV

Eg

Express the ratio of Eg to Eg,measured:

Eg,measured

=

2.17 meV = 0.795 2.73 × 10−3 eV

or

Eg ≈ 0.8Eg,measured (b) The wavelength of a photon having sufficient energy to break up Cooper pairs in tin at T = 0 is given by:

λ=

hc Eg

Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 4.54 × 105 nm 2.73 × 10 −3 eV

= 0.454 mm

The Fermi-Dirac Distribution 58 •• Picture the Problem We can evaluate the Fermi factor at the bottom of the conduction band for T near room temperature to show that this factor is given by exp(−Eg/2kT). (a) At the bottom of the conduction band: We can neglect the 1 in the denominator of the Fermi function to obtain:

e ( E − EF ) kT = e

Eg 2 kT

>>> 1 for T near room

temperature.

f (12 Eg ) =

1 e

Eg 2 kT

= e

− Eg 2 kT

1288 Chapter 38

Substitute numerical values and evaluate f

(

1 2

Eg ) for T = 300 K:

⎡ ⎤ − 1eV f (12 Eg ) = exp ⎢ = 4.01× 10 −9 ⎥ −5 ⎣ 2 8.62 × 10 eV/K (300 K )⎦

(

)

Given that low a probability of finding an electron in a state near the bottom of the conduction band, the exclusion principle has no significant impact on the distribution function. With 10 22 valence electrons per cubic centimeter, the number of electrons in the conduction band will be about 4 × 1013 per cm3 . (b) Evaluate f

(

1 2

Eg ) for T = 300 K and Eg = 6 eV:

⎡ ⎤ − 6 eV f (12 Eg ) = exp ⎢ = 4.15 × 10 −51 ⎥ −5 ⎣ 2 8.62 × 10 eV/K (300 K )⎦

(

)

The probability of finding even one electron in the conduction band is negligibly small (approximately 4 × 10−51 ). 59 •• Picture the Problem The number of energy states per unit volume per unit energy interval N is given by N ≈ g ( E )∆E , where N is only approximate, because ∆E is not infinitesimal and g (E ) =

8 2πme3/2V 1/2 E is the density of states. h3

The number of states N is the product of the density of states and the energy interval:

N ≈ g ( E )∆E

The density of states is given by:

g (E ) =

(1)

8 2πme3/2V 1/2 E h3

Substitute numerical values and evaluate g(E):

g (E ) =

(

8 2π 9.11 × 10 −31 kg

(6.63 × 10

= 6.15 × 10 37 J −1

) (1 × 10 J ⋅ s)

− 34

32

3

−3

)

m ⎛ 1.60 × 10 −19 ⎜⎜ 2.1 eV × eV ⎝ 3

J⎞ ⎟⎟ ⎠

1/2

Solids and the Theory of Conduction 1289 Substitute numerical values in equation (1) and evaluate N:

(

)

N ≈ 6.15 × 1037 J −1 (2.20 eV − 2.00 eV )×

1.60 × 10 −19 J = 1.97 × 1018 eV

*60 •• Picture the Problem Equation 38-22a expresses the dependence of the Fermi energy EF on the number density of free electrons. Once we’ve determined the Fermi energy for silver, we can find the average electron energy from the Fermi energy for silver and then use the average electron energy to find the Fermi speed for silver. (a) From Equation 38-22a we have:

h 2 ⎛ 3N ⎞ EF = ⎜ ⎟ 8me ⎝ πV ⎠

Use Table 27-1 to find the freeelectron number density N/V for silver:

N electrons = 5.86x10 22 V cm 3 electrons = 5.86x10 28 m3

3/2

Substitute numerical values and evaluate EF:

(6.63 × 10 = 8(9.11 × 10

EF

) ( )

)

J ⋅ s ⎡ 3 5.86 × 10 28 electrons/m 3 ⎤ ⎢ ⎥ −31 π kg ⎣ ⎦

−34

2

23

⎛ ⎞ 1eV ⎜⎜ ⎟ −19 ⎟ ⎝ 1.60 × 10 J ⎠

= 5.51eV (b) The average electron energy is given by:

E av =

3 EF 5

Substitute numerical values and evaluate Eav:

E av =

3 (5.51 eV ) = 3.31eV 5

(c) Express the Fermi energy in terms of the Fermi speed of the electrons:

E F = 12 me v F2

Solve for vF:

Substitute numerical values and evaluate vF:

vF =

2EF me

vF =

2(3.31eV ) ⎛ 1.60 × 10 −19 J ⎞ ⎜ ⎟⎟ 1eV 9.11 × 10 −31 kg ⎜⎝ ⎠

= 1.08 × 10 6 m/s

1290 Chapter 38 61 •• Picture the Problem We can evaluate the f(EF) at E = EF to show that F = 0.5.

f (E ) =

The Fermi factor is:

1

e

f ( EF ) =

Evaluate f(EF):

( E − EF ) kT

+1

1

e

( EF − EF ) kT

+1

=

1 = 0.5 1+1

62 •• Picture the Problem We can find the difference between the energies at which the Fermi factor has the given values by solving the expression for Fermi factor for E and then deriving an expression for ∆E.

f (E ) =

(a) The Fermi factor is:

1

e

( E − EF ) kT

+1

⎛ 1 ⎞ E = EF + kT ln⎜⎜ − 1⎟⎟ ⎝ f (E ) ⎠

Solve for E:

The difference between the energies is given by:

∆E = E (0.1) − E (0.9) = EF

(1.38 ×10 +

)

J/K (300 K ) ⎛ 1 ⎞ ln⎜ − 1⎟ −19 1.60 × 10 J/eV ⎝ 0.1 ⎠ −23

(

)

⎧ 1.38 × 10−23 J/K (300 K ) ⎛ 1 ⎞⎫ ln⎜ − ⎨ EF + − 1⎟⎬ −19 1.60 × 10 J/eV ⎝ 0.9 ⎠⎭ ⎩ 1.38 × 10−23 J/K (300 K ) ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞⎤ = ln⎜ − 1⎟ − ln⎜ − 1⎟⎥ −19 ⎢ 1.60 × 10 J/eV ⎝ 0.9 ⎠⎦ ⎣ ⎝ 0.1 ⎠

(

)

= 0.114 eV (b) and (c) Because ∆E is independent of EF , ∆E is the same as in (a ).

Solids and the Theory of Conduction 1291 *63 •• Picture the Problem The probability that a conduction electron will have a given kinetic energy is given by the Fermi factor. The Fermi factor is:

Because EF – 4.9 eV >> 300k:

f (E ) =

1

e

( E − EF ) kT

f (4.9 eV ) =

+1

1 = 1 0 +1

64 •• Picture the Problem We can solve Equation 38-22a for V and substitute in Equation 3841 to show that g (E ) = (3 N 2)EF-3 2 E 1 2 . 23

From Equation 38-22a we have:

h 2 ⎛ 3N ⎞ EF = ⎜ ⎟ 8me ⎝ πV ⎠

Solve for V to obtain:

3N ⎛ h 2 ⎞ ⎜ ⎟ V = π ⎜⎝ 8me EF ⎟⎠

The density g(E) is given by Equation 38-41:

g (E ) =

Substitute for V and simplify to obtain:

8π 2me3 2 g (E ) = h3 =

32

8π 2me3 2V 1 2 E h3 ⎡ 3 N ⎛ h 2 ⎞3 2 ⎤ ⎟⎟ ⎥ E1 2 ⎜⎜ ⎢ ⎢⎣ π ⎝ 8me EF ⎠ ⎥⎦

3 N −3 2 1 2 EF E 2

65 •• Picture the Problem We can use the expression for g(E) from Problem 64 to show that the average energy at T = 0 is 53 EF . From Problem 64 we have:

g (E ) =

3 N −3 2 1 2 EF E 2

1292 Chapter 38 Substitute in the expression for Eav and simplify to obtain:

1 Eav = N

EF

1 N

EF

=

∫ Eg (E )dE 0

⎛ 3 N −3 2 1 2 ⎞ EF E ⎟dE 2 ⎠

∫ E⎜⎝ 0

E

=

Integrate the expression for Eav:

3 −3 2 F 3 2 EF ∫ E dE 2 0 E

E av = =

3 −3 2 F 3 2 E F ∫ E dE 2 0 3 −3 2 2 5 2 3 EF EF = EF 2 5 5

66 •• Picture the Problem We can integrate g(E) from 0 to EF to show that the total number of states is 23 AEF3 2 . (a) Integrate g(E) from 0 to EF:

N=

EF

∫ AE

12

dE =

2 3

AEF3 2

0

(b) Express the fraction of N within kT of EF: (c) Substitute numerical values and evaluate the expression obtained in (b) for copper:

kTg (EF ) kTAEF1 2 3kT = 2 = 32 N 2 EF 3 AEF

(

)

3kT 3 8.62 × 10 −5 eV/K (300 K ) = 2 EF 2(7.04 eV ) = 5.51× 10−3

67 •• Picture the Problem The probability that a conduction electron in metal is the Fermi factor. Express the Fermi factor:

Calculate the dimensionless exponent in the Fermi factor:

f (E ) =

1

e

( E − EF ) kT

+1

E − EF 5.49 eV − 5.50 eV = kT 8.62 × 10 −5 eV/K (300 K )

(

= −0.387

)

Solids and the Theory of Conduction 1293 Use this result to calculate the Fermi factor:

f (5.49 eV ) =

1 e

−0.387

+1

= 0.596

68 •• Picture the Problem We can integrate the density-of-states function, Equation 38-41, to find the number of occupied states N. The fraction of these states that are within kT of EF can then be found from the ratio of kTg(EF) to N. The density of states function is:

g (E ) = AE1 2 where

A=

8π 2me3 2V h3 EF

Integrate g(E) from 0 to EF to find the total number of occupied states:

N=

Express the fraction of N within kT of EF:

kTg (EF ) kTAEF1 2 3kT = 2 = 32 N 2 EF 3 AEF

(a) Substitute numerical values and evaluate the expression obtained above for copper at T = 77 K: (b) At T = 300 K:

∫ AE

12

dE = 23 AEE3F2

0

(

)

3kT 3 8.62 × 10−5 eV/K (77 K ) = 2 EF 2(7.04 eV ) = 1.41× 10−3

(

)

3kT 3 8.62 × 10 −5 eV/K (300 K ) = 2 EF 2(7.04 eV ) = 5.51× 10−3

69 •• Picture the Problem The distribution function of electrons in the conduction band is given by n(E ) = g (E ) f (E ) where f(E) is the Fermi factor and g(E) is the density of states in terms of EF. Express the number of electrons n with energy E:

n(E ) = g (E ) f (E ) where

g (E ) = and

3 N −3 2 1 2 EEF E 2

(1)

1294 Chapter 38 f (E ) = The dimensionless exponent in the Fermi factor is:

1

e

( E − EF ) kT

+1

E − EF E − 12 Eg = >> 1 kT kT and

⎛ E − 12 Eg ⎞ ⎟⎟ >> 1 exp⎜⎜ ⎝ kT ⎠ Hence:

Substitute in equation (1) and simplify to obtain:

1 E f (E ) = (E − 1 E ) kT ≈ e g 2 g e

(

n(E ) =

3 2

NEF−3 2e

Eg 2 kT

2 kT

)E

e − E kT

1 2 − E kT

e

There is an additional temperature dependence that arises from the fact that EF depends on T. At room temperature, exp[(E − Eg/2)/kT] ≥ exp(0.35 eV/0.0259 eV) = 7.4×105, so the approximation leading to the Boltzmann distribution is justified. *70 ••• Picture the Problem We can follow the step-by-step procedure outlined in the problem statement to obtain the indicated results. (a) The Fermi factor is:

f (E ) =

1

e

( E − EF ) kT

+1

=

e

− EF kT

1 e E kT + 1

1

=

Ce +1 − EF kT provided C = e (b) If C >> e−E/kT:

f (E ) =

E kT

1 Ce

E kT

+1

≈

1 = Ae − E kT E kT Ce

where A = 1/C (c) The energy distribution function is:

n(E ) dE = g (E ) dE f (E ) where

g (E ) = Substitute for g(E)dE and f(E) in the expression for N to obtain:

8π 2me3 2V 1 2 E h3 ∞

N=A

8π 2me3 2V E1 2e − E kT dE 3 ∫ h 0

Solids and the Theory of Conduction 1295 ∞

The definite integral has the value:

∫E

1 2 − E kT

e

32 ( kT ) dE =

π

2

0

8π 2me3 2V (kT ) 2 h3

32

Substitute to obtain:

N=A

Solve for A:

A=

π

2h 3 ⎛ N ⎞ 1 ⎜ ⎟ 8π 3 2 me3 2 ⎝ V ⎠ (kT )3 2

(d) Evaluate A at T = 300 K:

A=

8π

32

(9.11×10

(

)

3

2 6.63 × 10 −34 J ⋅ s n −31

kg

) [(1.38 ×10 3 2

−23

]

)

J/K (300 K )

3 2

≈ 4 × 10−26 n

where the units are SI.

The valence electron concentration is typically about 1039 m −3 . To satisfy the condition that A << 1 at room temperature, n should be less than 10 23 m −3 , or about one millionth of the valence electron concentration. Because A depends on T −3 2 , the electron concentration may be greater the higher the temperature.

(e)

1017 cm −3 = 10 23 m −3 . So, according to the criterion in (d ), the classical approximation is applicable.

71 ••• Picture the Problem We can approximate the separation of electrons in the gas by (V/N)1/3 and use the for A from Problem 70 and de Broglie’s equation to express the separation d of electrons in terms of the de Broglie wavelength λ and the constant A. 13

The separation d of electrons is approximately:

⎛V ⎞ d =⎜ ⎟ ⎝N⎠

From Problem 70:

⎛V ⎞ ⎜ ⎟ ⎝N⎠

Substitute to obtain:

d=

13

=

=

21 6 h 1 12 13 13 12 12 8 π me (kT ) A

21 6 h 1 12 13 13 12 12 8 π me (kT ) A 21 6 h

π

12

1 2me kT A1 3

1296 Chapter 38 Express the momentum of an electron in the gas in terms of its de Broglie wavelength λ: Substitute for

2me kT in the

p=

d=

expression for d to obtain:

h

λ

= 2mK = 2me kT

21 6 h λ 21 6 λ = π 1 2 hA1 3 π 1 2 A1 3

= 0.633

λ

A1 3 Thus, if A << 1, d >> λ 72 ••• Picture the Problem We can follow the procedure outlined in the problem statement to determine the rms energy of a Fermi distribution. Express the Erms in terms of g(E):

Erms

The density of states g(E) is given by:

g (E ) =

Substitute to obtain:

Erms

Evaluate the integral and simplify:

Erms

⎛1 = ⎜⎜ ⎝N

12

⎞ ∫0 g (E )E dE ⎟⎟ ⎠

EF

2

3 N −3 2 1 2 EF E 2

⎛ 1 = ⎜⎜ 32 ⎝ 2 NEF

12

⎞ ∫0 E dE ⎟⎟ ⎠

EF

32

3 EF = 0.655 EF 7 > Eav because the process of averaging the square of the energy weighs Erms =

larger energies more heavily.

General Problems 73 • Picture the Problem The number of free electrons per atom ne is given by ne = nM ρN A where NA is Avogadro’s number, ρ is the density of the element, M is its molar mass, and n is the free electron number density for the element. The number of electrons per atom is given by:

ne =

nM ρN A

Solids and the Theory of Conduction 1297 Substitute numerical values and evaluate ne:

ne =

(1.4 ×10 electrons/cm )(39.098 g / mol) = (0.851g/cm )(6.02 ×10 electrons/mol) 22

3

3

23

1.07

74 • Picture the Problem The number of free electrons per atom ne is related to the number density of free electrons n by ne = nM ρN A , where NA is Avogadro’s number, ρ is the density of the element, and M is its molar mass. The number of electrons per atom is given by:

ne =

nM ρN A

Solve for n to obtain:

n=

ne ρN A M

(a) Substitute numerical values and evaluate n for Mg:

n=

(2)(1.74 g/cm3 )(6.02 ×1023 electrons/mol) 24.31g / mol

= 8.62 × 1022 electrons/cm3

(b) Substitute numerical values and evaluate n for Zn:

( 2) (7.1g/cm3 )(6.02 × 10 23 electrons/mol) n= 65.38 g / mol

= 13.1× 1022 electrons/cm3

Both results agree with the values in Table 38 - 1 to within 1%. 75 •• Picture the Problem We can integrate g(E) from 0 to EF to show that the total number of states is 23 AEF3 2 and then use this result to find the fraction of the free electrons that are above the Fermi energy at the given temperatures. Integrate g(E) from 0 to EF:

N=

EF

∫ AE

12

dE = 23 AEE3F2

0

Express the fraction of N within kT of EF:

kTg (EF ) kTAEF1 2 3kT = 2 = 32 N 2 EF 3 AEF

1298 Chapter 38

(

)

3kT 3 8.62 × 10−5 eV/K (300 K ) = 2 EF 2(7.04 eV )

(a) Substitute numerical values and evaluate this fraction for copper at 300 K:

= 5.51× 10 −3

(

)

3kT 3 8.62 × 10 −5 eV/K (1000 K ) = 2 EF 2(7.04 eV )

(b) Evaluate the same fraction at 1000 K:

= 1.84 × 10−2 *76 •• Picture the Problem The Fermi factor gives the probability of an energy state being occupied as a function of the energy of the state E, the Fermi energy EF for the particular material, and the temperature T. The Fermi factor is:

f (E ) =

For 10 percent probability:

0.1 =

1 e

( E − E F ) kT

+1

1 e

( E − E F ) kT

+1

or

e ( E − EF ) kT = 9

Take the natural logarithm of both sides of the equation to obtain:

E − EF = ln 9 kT

Solve for E to obtain:

E = E F + kT ln 9

From Table 37-1, EF(Mn) = 11.0 eV. Substitute numerical values and evaluate E:

(

)

1 eV ⎛ ⎞ E = 11.0 eV + 1.38 × 10 − 23 J/K (1300 K ) ⎜ ⎟ ln 9 = 11.2 eV −19 ⎝ 1.60 × 10 J ⎠ 77 •• Picture the Problem The energy gap for the semiconductor is related to the wavelength of the emitted light according to E g = hc λ . Express the energy gap Eg in terms of the wavelength λ of the emitted light: Solve for λ:

Eg =

λ=

hc

λ

hc Eg

Solids and the Theory of Conduction 1299 Substitute numerical values and evaluate λ:

λ=

1240 eV ⋅ nm = 689 nm 1.8 eV

Remarks: This wavelength is in the red portion of the visible spectrum. *78 ••• Picture the Problem The rate of production of electron-hole pairs is the ratio of the incident energy to the energy required to produce an electron-hole pair. (a)The number of electron-hole pairs N produced in one second is:

N=

IA IAλ = hc hc

λ

Substitute numerical values and evaluate N:

N=

(4.0 W/m )(2 ×10 m )(775 nm) = (1240 eV ⋅ nm )(1.60 ×10 J/eV ) 2

(b) In the steady state, the rate of recombination equals the rate of generation. Therefore: (c) The power radiated equals the power absorbed: Substitute numerical values and evaluate Prad:

−4

2

−19

3.12 × 1015 s −1

N = 3.12 × 1015 s −1

Prad = IA Prad = (4.0 W/m 2 )(2 × 10 −4 m 2 ) = 0.800 mJ/s

1300 Chapter 38

Chapter 39 Relativity Conceptual Problems *1 • Picture the Problem The total relativistic energy E of a particle is defined to be the sum of its kinetic and rest energies. The total relativistic energy of a particle is given by:

E = K + mc 2 = 12 mu 2 + mc 2 and (a ) is correct.

*2 • Determine the Concept The gravitational field of the earth is slightly greater in the basement of the office building than it is at the top floor. Because clocks run more slowly in regions of low gravitational potential, clocks in the basement will run more slowly than clocks on the top floor. Hence, the twin who works on the top floor will age more quickly. (b) is correct. 3 • (a) True (b) True (c) False. The shortening of the length of an object in the direction in which it is moving is independent of the velocity of the frame of reference from which it is observed. (d) True (e) False. Consider two explosions equidistant, but in opposite directions, from an observer in the observer’s frame of reference. (f) False. Whether events appear to be simultaneous depends on the motion of the observer. (g) True 4 • Determine the Concept Because the clock is moving with respect to the first observer, a time interval will be longer for this observer than for the observer moving with the spring-and-mass oscillator. Hence, the observer moving with the system will measure a

1301

1302

Chapter 39

period that is less than T. (b) is correct. 5 • Determine the Concept Although ∆y = ∆y′, ∆t ≠∆t′. Consequently, uy = ∆y/∆t′ ≠ ∆y′/∆t′ = uy′.

Estimation and Approximation 6 •• Picture the Problem We can calculate the sun’s loss of mass per day from the number of reactions per second and the loss of mass per reaction.

∆M = N∆m ∆t

Express the rate at which the sun loses mass:

where N is the number of reactions per second and ∆m is the loss of mass per reaction. Solve for ∆M:

∆M = N∆m∆t

Find the number of reactions per second, N:

N=

The loss of mass per reaction ∆m is:

∆m =

(1)

P E / reaction 4 × 10 26 J/s = MeV J × 1.60 × 10−19 25 reaction eV 38 −1 = 10 s E / reaction c2 MeV J 25 × 1.60 × 10 −19 reaction eV = 2 8 3 × 10 m/s

(

= 4.44 × 10

)

−29

kg

Substitute numerical values in equation (1) and evaluate ∆M:

(

)(

)

∆M = 1038 s −1 4.44 × 10 −29 kg (1d )(86.4 ks/d ) = 3.84 × 1014 kg

Relativity 1303 *7 •• Picture the Problem We can use the result from Problem 30, for light that is Doppler-

⎛ u 2 −1 ⎞ ⎟⎟ , where u = z + 1 and z is the red-shift 2 ⎝ u +1⎠

shifted with respect to an observer, v = c ⎜⎜

parameter, to find the ratio of v to c. In (b) we can solve Hubble’s law for x and substitute our result from (a) to estimate the distance to the galaxy. (a) Use the result of Problem 30 to express v/c as a function of z:

v (z + 1) − 1 = c (z + 1)2 + 1 2

Substitute for z and evaluate v/c:

v (5 + 1) − 1 = = 0.946 c (5 + 1)2 + 1

(b) Solve Hubble’s law for x:

x=

Substitute numerical values and evaluate x:

2

x =

v H 0.946c H

=

0.946(3 × 10 5 km/s ) 75

= 3.78 × 10 Mpc × 3

km/s Mpc

3.26 × 10 6 c ⋅ y Mpc

= 12.3 Gc ⋅ y

Time Dilation and Length Contraction 8 • Picture the Problem We can find the mean lifetime of a muon as measured in the laboratory using t' = γ t where γ = 1 1 − (v c ) and t is the proper mean lifetime of the muon. The distance L that the muon travels is the product of its speed and its mean lifetime in the laboratory. 2

(a) The mean lifetime of the muon, as measured in the laboratory, is given by:

t' =

Substitute numerical values and evaluate t′:

t' =

t ⎛v⎞ 1− ⎜ ⎟ ⎝c⎠

2

2 µs ⎛ 0.95c ⎞ 1− ⎜ ⎟ ⎝ c ⎠

2

= 6.41 µs

1304 Chapter 39 (b) The distance L that the muon travels is related to its mean lifetime in the laboratory: Substitute numerical values and evaluate L:

L = vt'

L = 0.95ct'

(

)

= 0.95 3 × 108 m/s (6.41 µs ) = 1.83 km

9 •• Picture the Problem The proper length Lp of the beam is its length as measured in a reference frame in which it is not moving. The proper length is related to its length in the frame in which it is measured by Lp = γL . (a) Relate the proper length Lp of the beam to its length L in the laboratory frame of reference:

Lp = γL

The energy of the beam also depends on γ :

E = γ mc 2

Solve for and evaluate γ :

γ =

Substitute numerical values and evaluate Lp:

E 50 GeV = = 9.785 × 10 4 2 mc 0.511 MeV

(

)

Lp = 9.785 × 10 4 (1cm ) = 978.5 m and

The width w of the beam is unchanged. (b) Express the length of the accelerator in the electron beam’s frame of reference:

Lacc =

Set Lacc = Lp:

Lp =

Solve for Lacc,p:

Lacc,p

γ

Lacc,p

γ

Lacc,p = γLp

Relativity 1305 Substitute numerical values and evaluate Lp:

(c) The length of the positron bundle in the electron’s frame of reference is: Substitute numerical values and evaluate Lpos:

(

)

Lacc,p = 9.785 × 10 4 (978.5 m ) = 9.57 × 107 m

Lpos = Lpos =

L

γ 1cm = 0.102 µm 9.785 × 104

*10 •• Picture the Problem The time required for the particles to reach the detector, as measured in the laboratory frame of reference is the ratio of the distance they must travel to their speed. The half life of the particles is the trip time as measured in a frame traveling with the particles. We can find the speed at which the particles must move if they are to reach the more distant detector by equating their half life to the ratio of the distance to the detector in the particle’s frame of reference to their speed. (a) The time required to reach the detector is the ratio of the distance to the detector and the speed with which the particles are traveling:

∆t =

∆x ∆x = v 0.866c

Substitute numerical values and evaluate ∆t:

∆t =

1000 m = 3.85 µs 0.866(3 × 108 m/s ) ∆t

2

(b) The half life is the trip time as measured in a frame traveling with the particles:

⎛v⎞ ∆t' = = ∆t 1 − ⎜ ⎟ γ ⎝c⎠

Substitute numerical values and evaluate ∆t′:

⎛ 0.866c ⎞ ∆t' = 3.85 µs 1 − ⎜ ⎟ = 1.93 µs ⎝ c ⎠

(c) In order for half the particles to reach the detector:

⎛v⎞ ∆x' 1 − ⎜ ⎟ ∆x' ⎝c⎠ ∆t' = = γv v

2

2

where ∆x′ is the distance to the new detector.

1306 Chapter 39 v

Rewrite this expression to obtain:

⎛v⎞ 1− ⎜ ⎟ ⎝c⎠

2

=

∆x' ∆t'

Squaring both sides of the equation yields:

⎛ ∆x' ⎞ =⎜ ⎟ 2 ⎝ ∆t' ⎠ ⎛v⎞ 1− ⎜ ⎟ ⎝c⎠

Substitute numerical values for ∆x′ and ∆t′ and simplify to obtain:

⎛ 10 4 m ⎞ ⎟⎟ = (17.3c )2 = ⎜⎜ 2 ⎛v⎞ ⎝ 1.93 µs ⎠ 1− ⎜ ⎟ ⎝c⎠

Divide both sides of the equation by c2 to obtain:

v2 2 c2 = (17.3) 2 ⎛v⎞ 1− ⎜ ⎟ ⎝c⎠

Solve this equation for v2/c2:

(17.3) = 0.9967 v2 = 2 2 c 1 + (17.3)

Finally, solving for v yields:

v = 0.998c

v2

2

2

v2

2

11 •• Picture the Problem We can use the time-dilation relationship to find the speed of the spacecraft. The distance to the second star is the product of the new gamma factor, the speed of the spacecraft, and the elapsed time. Finally, the time that has elapsed on earth (your age) is the sum of the elapsed times for the three legs of the journey. (a) From the point of view of an observer on earth, the time for the trip will be:

∆t =

From the point of view of an observer on the spaceship, the time for the trip will be:

∆t' =

Substitute for γ to obtain:

∆t' =

L v

∆t

γ

=

L γv

L v2 1− 2 v c

Relativity 1307 Solve for v:

Substitute numerical values and evaluate v:

Lc

v=

L2 + c 2 (∆t' )

2

(27 c ⋅ y )c (27 c ⋅ y )2 + c 2 (12 y )2

v=

= 0.914c

Note that from the point of view of an earth observer, this part of the trip has taken 27 c⋅ y/0.914c = 29.5 y. (b) The distance the ship travels, from the point of view of an earth observer, in 5 y is:

∆L' = 2γ∆L = 2γv∆t where γ is the gamma factor for the first

The gamma factor in Part (a) is:

γ =

part of the trip.

1

⎛v⎞ 1− ⎜ ⎟ ⎝c⎠ = 2.46

Substitute numerical values and evaluate ∆L′:

2

=

1 ⎛ 0.914c ⎞ 1− ⎜ ⎟ ⎝ c ⎠

2

∆L' = 2(2.46)(0.914c )(5 y ) = 22.5 c ⋅ y

(c) The elapsed time ∆t on earth (your age) is the sum of the times for the spacecraft to travel to the star 27 c⋅ y away, l to the second star, and to return home from the second star:

∆t = 29.5 y + 22.5 y + ∆treturning home

The elapsed time on earth while the spacecraft is returning to earth is:

∆t returning home = 2γ∆tship's time

Substitute for ∆treturning home and evaluate ∆t:

∆t = 29.5 y + 22.5 y + 49.2 y

= 2(2.46)(10 y ) = 49.2 y

= 101 y

12 • Picture the Problem We can use ∆t = L/v, where L is the distance to the star and v is the speed of the spaceship to find the time ∆t for the trip as measured on earth. The travel time as measured by a passenger on the spaceship can be found using ∆t′ = ∆t/γ. (a) The travel time as measured on earth is the ratio of the distance

∆t =

L v

1308 Chapter 39 traveled L to speed of the spaceship: Substitute numerical values and evaluate ∆t:

(a) The travel time as measured by a passenger on the spaceship is given by: Substitute numerical values and evaluate ∆t′:

35 c ⋅ y 35 y = 8 2.7 × 10 m/s 2.7 × 108 m/s c 35 y = = 38.9 y 0.9

∆t =

∆t' =

∆t

⎛v⎞ = ∆t 1 − ⎜ ⎟ γ ⎝c⎠

2

∆t' = (38.9 y ) 1 − (0.9 ) = 17.0 y 2

13 • Picture the Problem We can use the definition of γ and the binomial expansion of (1 + x)n to show that each of these relationships holds provided v << c. (a) Express the gamma factor:

⎛ v2 ⎞ γ = = ⎜⎜1 − 2 ⎟⎟ v2 ⎝ c ⎠ 1− 2 c

Expand the radical factor binomially to obtain:

⎛ v2 ⎞ γ = ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠

1

−1 2

−1 2

2 ⎛ 1 ⎞⎛ v ⎞ = 1 + ⎜ − ⎟⎜⎜ − 2 ⎟⎟ + higher order terms ⎝ 2 ⎠⎝ c ⎠

For v << c:

(b) Express the reciprocal of γ :

γ ≈ 1+

1

γ Expand the radical binomially to obtain:

1 v2 2 c2

= 1−

v2 c2 12

⎛ v2 ⎞ = ⎜⎜1 − 2 ⎟⎟ γ ⎝ c ⎠ 2 ⎛ 1 ⎞⎛ v ⎞ = 1 + ⎜ ⎟⎜⎜ − 2 ⎟⎟ + higher order terms ⎝ 2 ⎠⎝ c ⎠

1

Relativity 1309 For v << c:

1

γ

1 v2 2 c2

≈ 1−

(c) Express the gamma factor:

⎛ v2 ⎞ = ⎜⎜1 − 2 ⎟⎟ γ = v2 ⎝ c ⎠ 1− 2 c

Subtract one from both sides of the equation to obtain:

⎛ v2 ⎞ γ − 1 = ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠

Expand the radical binomially to obtain:

γ − 1 = 1 + ⎜ − ⎟⎜⎜ −

For v << c:

1

−1 2

−1 2

−1

2 ⎛ 1 ⎞⎛ v ⎞ ⎟ −1 2 ⎟ ⎝ 2 ⎠⎝ c ⎠ + higher order terms

γ −1 ≈

1 v2 2 c2

14 •• Picture the Problem We can express the fractional difference in your time-interval measurements as a function of γ and solve the resulting equation for the relative speed of the two spaceships. Express the fractional difference in the time-interval measurements of the two observers:

∆t − ∆t' ∆t' = 1− = 0.01 ∆t ∆t

Since ∆t′/∆t = 1/γ:

1 ∆t − ∆t' = 1 − = 0.01 γ ∆t

From Problem 13(b) we have:

1 v2 ≈ 1− γ 2 c2

Substitute to obtain:

1−

1

⎛ 1 v2 ⎞ ⎟ = 0.01 ≈ 1 − ⎜⎜1 − 2 ⎟ γ 2 c ⎝ ⎠ 1

or

1 v2 = 0.01 2 c2

1310 Chapter 39 Solve for v to obtain:

v = 0.02c = 0.141c = 4.23 × 107 m/s

15 •• Picture the Problem We can use the time dilation equation to relate the time lost by the clock to the speed of the plane and the time it must fly. Express the time δ t lost by the clock:

δ t = ∆t − ∆tp = ∆t −

Because V << c, we can use part (b) of Problem 13:

1V2 ≈ 1− γ 2 c2

Substitute to obtain:

δ t = ∆t ⎢1 − ⎜⎜1 −

Solve for ∆t:

Substitute numerical values and evaluate ∆t:

⎛ 1⎞ = ∆t ⎜⎜1 − ⎟⎟ γ ⎝ γ⎠

∆t

1

∆t =

⎡

⎛

⎣

⎝

1 V 2 ⎞⎤ 1 V 2 ⎟⎥ = ∆t 2 c 2 ⎟⎠⎦ 2 c 2

2δt c 2 V2

(

)

2(1s ) 3 × 108 m/s ∆t = (2000 km/h ×1h/3600 s )2 1y = 5.83 × 1011 s × 31.56 Ms 2

= 1.85 × 10 4 y

The Lorentz Transformation, Clock Synchronization, and Simultaneity 16 •• Picture the Problem We can use the inverse Lorentz transformations and the result of Problem 13(c) to show that when u << c the transformation equations for x, t, and u reduce to the Galilean equations. The inverse transformation for x is: From Problem 13(c):

x' = γ (x − vt )

1 v2 γ = 1+ 2 2c

Relativity 1311 Substitute for γ and expand to obtain:

When v << c: The inverse transformation for t is: Substitute for γ and expand to obtain:

⎛ 1 v2 ⎞ ⎟ (x − vt ) x' = ⎜⎜1 + 2 ⎟ 2 c ⎝ ⎠ 1 v2 1 v3 = x − vt + x − 2 c2 2 c2 x' ≈ x − vt

⎛ vx ⎞ t' = γ ⎜ t − 2 ⎟ ⎝ c ⎠ ⎛ 1 v 2 ⎞⎛ vx ⎞ ⎟ t− 2⎟ t' = ⎜⎜1 + 2 ⎟⎜ ⎝ 2 c ⎠⎝ c ⎠ 1 v3 vx 1 v 2 =t− 2 + t− x 2 c2 2 c4 c

When v << c: The inverse velocity transformation for motion in the x direction is: When v << c:

t' ≈ t

u x' =

ux − v vu 1 − 2x c

u x' ≈ u x − v

The inverse velocity transformation for motion in the y direction is:

u y' =

Substitute for γ and expand to obtain:

u y' =

uy ⎛ vu ⎞ γ ⎜1 − 2x ⎟ c ⎠ ⎝ uy

⎛ 1 v 2 ⎞⎛ vu x ⎞ ⎜⎜1 + ⎟ 1− 2 ⎟ 2 ⎟⎜ c c ⎠ 2 ⎝ ⎠⎝ uy = vu 1 v 2 1 v 3u x 1 − 2x + − c 2 c2 2 c4

When v << c:

u y' ≈ u y

Proceed similarly to show that:

u z' ≈ u z

*17 •• Picture the Problem Let S be the reference frame of the spaceship and S′ be that of the earth (transmitter station). Let event A be the emission of the light pulse and event B the reception of the light pulse at the nose of the spaceship. In (a) and (c) we can use the

1312 Chapter 39 classical distance, rate, and time relationship and in (b) and (d) we can apply the inverse Lorentz transformations. (a) In both S and S′ the pulse travels at the speed c. Thus:

tA =

Lp v

=

400 m = 1.76 µs 0.76c

t B = tpulse to travel length of ship + tA

(c) The elapsed time, according to the clock on the ship is:

400 m 2.998 × 108 m/s = 1.33 µs

Find the time of travel of the pulse to the nose of the ship:

tpulse to travel length of ship =

Substitute numerical values and evaluate tB:

t B = 1.33 µs + 1.76 µs = 3.09 µs

(b) The inverse time transformation is:

⎛ vx ⎞ t B' = γ ⎜ t − 2 ⎟ ⎝ c ⎠ where

1

γ =

v 1− 2 c 2

=

1 2 ( 0.76c ) 1−

= 1.54

c2

(− 0.76c )(400 m ) ⎞ ⎛ t B' = (1.54)⎜ 3.09 µs − ⎟ c2 ⎝ ⎠

Substitute numerical values and evaluate t 'B :

⎛ (− 0.76)(400 m ) ⎞⎟ = (1.54)⎜⎜ 3.09 µs − 3 × 108 m/s ⎟⎠ ⎝ = 6.32 µs

x' = γ (x − vt )

(d) The inverse transformation for x is: Substitute numerical values and evaluate x′:

[

(

)(

)]

x' = (1.54 ) 400 m − (− 0.76 ) 3 × 108 m/s 3.09 × 10 −6 s = 1.70 km 18 •• Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to find the required speed of the observer.

Relativity 1313

Use Equation 39-12 to obtain:

v ⎡ ⎤ t B' − t A' = γ ⎢(t B − t A ) − 2 ( xB − x A )⎥ c ⎣ ⎦ v∆x ⎞ ⎛ = γ ⎜ ∆t − 2 ⎟ c ⎠ ⎝ where ∆t = tB − tA and ∆x = xB − xA.

Events A and B are simultaneous if:

Solve for v:

Substitute numerical values and evaluate v:

∆t −

v=

v∆x =0 c2

c 2 ∆t ∆x

(3 ×10 v=

)

m/s (2 µs ) 1.5 km 8

2

= 1.20 × 108 m/s = 0.4c Yes, t B' will be less than t A' if V > 0.4c. 19 •• Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to express the separation in time between the two explosions as measured in S′ as a function of the speed of the observer and Equation 39-11, the inverse position transformation equation, to find the speed of the observer. Use Equation 39-12 to obtain:

From Equation 39-11: Because the explosions occur at the same point in space, ∆x′ = 0: Solve for v:

v ⎡ ⎤ ∆t' = γ ⎢∆t − 2 ∆x ⎥ c ⎣ ⎦ v ∆t − 2 ∆x c = 2 ⎛v⎞ 1− ⎜ ⎟ ⎝c⎠ ∆x' = γ (∆x − v∆t ) 0 = γ (∆x − v∆t )

v=

∆x ∆t

(1)

1314 Chapter 39 Substitute numerical values and evaluate v:

v=

1200 m − 480 m = 1.44 × 108 m/s 5 µs

Substitute numerical values in equation (1) and evaluate ∆t′:

5 µs − ∆t' =

1.44 × 108 m/s

(3 ×10

8

m/s

)

2

(1200 m − 480 m )

⎛ 1.44 × 10 m/s ⎞ ⎟⎟ 1 − ⎜⎜ 8 ⎝ 3 × 10 m/s ⎠ 8

= 4.39 µs

2

20 ••• Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to establish the results called for in this problem. (a) Use Equation 39-12 to obtain:

v ⎡ ⎤ t2' − t1' = γ ⎢(t 2 − t1 ) − 2 ( x2 − x1 )⎥ c ⎣ ⎦ vD ⎞ ⎛ = γ ⎜T − 2 ⎟ c ⎠ ⎝ where T = t2 − t1 and D = x2 − x1.

(b) Events 1 and 2 are simultaneous in S′ if:

t2' = t1' or

T− Because v ≤ c:

vD c 2T ⇒ = D = 0 v c2

D ≥ cT

(c) If D < cT , then t2' > t1' and the events are not simultaneous in S' . (d) If D = c′T > cT, then:

T−

vD ⎡ v c' ⎤ = T ⎢1 − = t 2' − t1' 2 c ⎣ c c ⎥⎦

In this case, t2' − t1' could be negative; i.e., t2' could be less than t1' , or the effect could precede the cause. 21 ••• Picture the Problem Let S be the ground reference frame, S′ the reference frame of the rocket, and v = 0.9c be the speed of the rocket relative to S. Denote the tail and nose of

Relativity 1315 the rocket by T and N, respectively. The initial conditions in S′ are t N ' = 0 , xN ' = 0 ,

xT '= 0 , and xT ' = − L' = −700 m . (a) The reading of the tail clock is given by:

vx ⎞ γvx ⎛ tT' = γ ⎜ tT − 2T ⎟ = − 2T c ⎠ c ⎝ because tT = 0

L'

We can find xT using the length contraction equation:

xT = −

Substitute to obtain:

tT' =

Substitute numerical values and evaluate tT ':

tT' =

(b) The time for the rocket to move a distance L′ is given by:

tT' =

L' L' = v 0.9c

Substitute numerical values and evaluate tT ':

tT' =

700 m = 2.59 µs 0.9 3 × 108 m/s

(c) As seen by an observer on the ground:

t N = ∆t' = 2.59 µs − 2.10 µs = 0.49 µs

(d) Because the clocks are synchronized in S′:

t N' = tT' = 2.59 µs

(e) The time the signal is received on the ground is the sum of the time when the signal is sent and the time for it to travel to the ground: Find ∆t, the time the signal is sent:

γ

vL' c2

(0.9)(700 m ) = 3 × 108 m/s

(

2.10 µs

)

trec = ∆t + ∆t travel

∆t = γ∆tp =

1h 2 ( 0.9c ) 1−

= 2.294 h

c2

Find ∆ttravel, the time for the signal to travel to the ground:

∆x (2.294 h )(0.9c ) = c c = 2.065 h

∆t travel =

1316 Chapter 39 Substitute for ∆t and ∆ttravel and evaluate trec: (f) Find ∆x when the signal is sent: In S, the signal arrives at 0.1c relative to the rocket. The time required for the signal to travel to the rocket is:

trec = 2.294 h + 2.065 h = 4.36 h ∆x = (4.36 h )(0.9c ) = 3.924 c ⋅ h

∆t =

∆x 3.924 c ⋅ h = = 39.24 h 0.1c 0.1c

t = 39.24 h + 3.924 h = 43.16 h

Find the time when the signal reaches the rocket: Finally, use the time dilation equation to find tN′:

t N' =

t

γ

2 ( 0.9c ) = (43.16 h ) 1 − 2

c

= 18.8 h *22 ••• Picture the Problem We can use the inverse time dilation equation to derive an expression for the elapsed time between the flashes in S′ in terms of the elapsed time between the flashes in S, their separation in space, and the speed v with which S′ is moving.

v ⎡ ⎤ ∆t' = γ ⎢∆t − 2 ∆x ⎥ c ⎣ ⎦

From the inverse time transformation we have:

where ∆t′ is the time between the flashes in S′ and ∆t and ∆x are the elapsed time between the flashes and their separation in S.

− ∆t

Set ∆t′ = −∆t to obtain:

γ

= ∆t −

v ∆x c2

or

− ∆t 1 −

v2 v = ∆t − 2 ∆x 2 c c

Square both sides of the equation:

(∆t )2 − v 2 (∆t )2 = (∆t )2 − 2 v2 ∆x∆t + v 4 (∆x )2 2

c

2

c

c

Relativity 1317 Simplify to obtain:

− v(∆t ) = −2∆x∆t +

Solve for v:

∆x ∆t v= 2 ⎛ 1 ∆x ⎞ 1+ ⎜ ⎟ ⎝ c ∆t ⎠

2

v (∆x )2 2 c

2

Substitute numerical values and evaluate v:

v=

⎛ 2400 m ⎞ ⎟⎟ 2⎜⎜ ⎝ 5 µs ⎠ ⎡ ⎛ 2400 m ⎞⎤ 1 ⎜⎜ ⎟⎟⎥ 1+ ⎢ 8 ⎣ 3 × 10 m/s ⎝ 5 µs ⎠⎦

2

= 2.697 × 108 m/s = 0.899c Because v is positive, S ' is moving in the positive x direction.

The Velocity Transformation 23 •• Picture the Problem We can make the substitutions given in the hint in Equation 39-18a and simplify the resulting expression to show that ux < c. Equation 39-18a gives the x direction relativistic velocity transformation: Make the substitutions given in the hint to obtain:

ux =

(1 − ε1 )c + (1 − ε 2 )c ux = c c + (1 − ε 2 ) c (1 − ε1 )c c 2 − (ε1 + ε 2 ) = 1 + (1 − ε 2 )(1 − ε1 ) =

Because ε1 and ε2 are small positive numbers that are less than 1:

u x '+v u u '+v or x = x vu ' c c + vu x ' 1 + 2x c c

2 − (ε1 + ε 2 ) 2 − (ε1 + ε 2 ) + ε1ε 2

ux < 1⇒ u x < c c

1318 Chapter 39 *24 •• Picture the Problem We’ll let the velocity (in S) of the spaceship after the ith boost be vi and derive an expression for the ratio of v to c after the spaceship’s (i + 1)th boost as a function of N. We can use the definition of γ, in terms of v/c to plot γ as a function of N. (a) and (b) The velocity of the spaceship after the (i + 1)th boost is given by relativistic velocity addition equation: Factor c from both the numerator and denominator to obtain:

vi+1 =

vi+1

γi is given by:

vi + 0.5c (0.5c )vi 1+ c2

vi + 0.5 c = v 1 + 0.5 i c

γi =

1 ⎛v ⎞ 1− ⎜ i ⎟ ⎝c⎠

A spreadsheet program to calculate v/c and γ as functions of the number of boosts N is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell A3 B2 B3 C1

Content/Formula 0 0 (B2+0.5)/(1+0.5*B2) 1/(1−B2^2)^0.5

1 2 3 4 5 6 7 8 9 10 11 12

A boost 0 1 2 3 4 5 6 7 8 9 10

Algebraic Form N v0 vi+1

γ

B v/c 0.000 0.500 0.800 0.929 0.976 0.992 0.997 0.999 1.000 1.000 1.000

C gamma 1.00 1.15 1.67 2.69 4.56 7.83 13.52 23.39 40.51 70.15 121.50

Relativity 1319 A graph of v/c as a function of N is shown below: 1.0

v /c

0.8 0.6 0.4 0.2 0.0 0

2

4

6

8

10

6

8

10

N

A graph of γ as a function of N is shown below: 120 100

gamma

80 60 40 20 0 0

2

4 N

Examination of the spreadsheet or of the graph of v/c as a function of (c) N indicates that, after 8 boosts, the velocity of the spaceship is greater than 0.999c. (d) After 5 boosts, the spaceship has traveled a distance ∆x, measured in the earth frame of reference (S), given by:

∆x = ∆x1→2 + ∆x2→3 + ∆x3→4 + ∆x4→5

= (0.5c )(10 s )γ 1→2 + (0.8c )(10 s )γ 2→3 + (0.929c )(10 s )γ 3→4 + (0.976c )(10 s )γ 4→5 + (0.992c )(10 s )γ 5→6

= (0.5c )(10 s )(1.15) + (0.8c )(10 s )(1.67 ) + (0.929c )(10 s )(2.69) + (0.976c )(10 s )(4.56) + (0.992c )(10 s )(7.83) = 166 c ⋅ s

1320 Chapter 39 The average speed of the spaceship, between boost 1 and boost 5, as measured in S is given by:

vav =

∆x ∆t

where ∆t is the travel time as measured in the earth frame of reference.

Express ∆t as the sum of the times the spaceship travels during each 10-s interval following a boost in its speed:

∆t = ∆t1→2 + ∆t2→3 + ∆t3→4 + ∆t4→5

= (10 s )γ 1→2 + (10 s )γ 2→3 + (10 s )γ 3→4 + (10 s )γ 4→5 + (10 s )γ 5→6 = (10 s )(γ 1→2 + γ 2→3 + γ 3→4 + γ 4→5 + γ 5→6 )

Substitute numerical values and evaluate ∆t:

∆t = (10 s )(1.15 + 1.67 + 2.69 + 4.56 + 7.83) = 179 s Substitute for ∆x and ∆t and evaluate vav:

vav =

166 c ⋅ s = 0.927c 179 s

Remarks: This result seems to be reasonable. Relativistic time dilation implies that the spacecraft will be spending larger amounts of time at high speed (as seen in reference frame S).

The Relativistic Doppler Shift 25 • Picture the Problem We can substitute, using v = fλ, in the relativistic Doppler shift equation and solve for the speed of the source. Using the expression for the relativistic Doppler shift, express f ′ as a function of v: Substitute using v = fλ and simplify to obtain:

f' = f

1+ v c 1− v c

v v 1+ v c = λ' λ 1 − v c

λ 1+ v c = λ' 1− v c or

⎛ λ ⎞ 1+ v c ⎜ ⎟ = ⎝ λ' ⎠ 1 − v c 2

Relativity 1321 Solve for v to obtain:

⎡ ⎛ λ ⎞2 ⎤ ⎢ ⎜ ⎟ − 1⎥ λ' v = ⎢ ⎝ ⎠2 ⎥c ⎢⎛ λ ⎞ ⎥ ⎢ ⎜ ⎟ + 1⎥ ⎣ ⎝ λ' ⎠ ⎦

Substitute numerical values and evaluate v:

⎡ ⎛ 589 nm ⎞ 2 ⎤ ⎟ − 1⎥ ⎢ ⎜⎜ 547 nm ⎟⎠ ⎥ 3 × 108 m/s ⎢ ⎝ v=⎢ 2 ⎥ ⎢ ⎛⎜ 589 nm ⎞⎟ + 1⎥ ⎥⎦ ⎢⎣ ⎜⎝ 547 nm ⎟⎠

(

)

= 2.22 × 107 m/s 26 • Picture the Problem We can use the relativistic Doppler shift, when the source and the receiver are receding, to relate the frequencies of the two wavelengths and c = fλ to express the ratio of the wavelengths as a function of the speed of the galaxy. When the source and receiver are moving away from each other, the relativistic Doppler shift is given by: Use the relationship between the wavelength and frequency to obtain:

Solve for λ′/λ0:

v c f f '= v 0 1+ c v v 1− 1− c c c ⇒ λ0 = c = v v λ' λ' 1+ 1 + λ0 c c 1−

v

1+ λ' c = v λ0 1− c

Express the fractional redshift:

v

1+ λ' − λ0 λ' c −1 = −1 = v λ0 λ0 1− c

1322 Chapter 39 Substitute numerical values and evaluate (λ′−λ0)/ λ0:

1.85 × 107 m/s λ' − λ0 2.998 × 108 m/s = −1 1.85 × 107 m/s λ0 1− 2.998 × 108 m/s 1+

= 0.0637 27 •• Picture the Problem We can begin the derivation by expressing the number of waves encountered by the observer, in the rest frame of the source, in a time interval ∆t. We can then relate this time interval to the time interval in the rest frame of the observer to complete the derivation of Equation 39-16a.

(c + v )∆ts

Express the number of waves n encountered by the observer, in the rest frame of the source, in a time interval ∆ts:

n=

This time interval in the rest frame of the observer is given by:

∆to =

λ

=

(c + v ) f o ∆ts c

⎛ v⎞ = f o ⎜1 + ⎟∆ts ⎝ c⎠ ∆ts

γ

Express the frequency heard by the observer and simplify to obtain:

v v 1+ 1+ n ⎛ v⎞ c f = cf = = γ ⎜1 + ⎟ f o = fo = o o 2 v ∆to ⎝ c⎠ v − 1 1− 2 c c

v2 1− 2 c f v o 1− c

28 • Picture the Problem We can use the expression for the relativistic Doppler shift to show that, to a good approximation, ∆f/f ≈±v/c. Express the fractional Doppler shift in terms of f and f0: When the source and receiver are approaching each other, the relativistic Doppler shift is given by:

f − f0 f ∆f = = −1 f0 f0 f0 v v 1+ f c c f ⇒ = f = v v 0 f0 1− 1− c c 1+

Relativity 1323 Substitute in the expression for ∆f/f0 to obtain:

v 1+ ∆f c −1 = v f0 1− c 12

⎛ v⎞ ⎛ v⎞ = ⎜1 + ⎟ ⎜ 1 − ⎟ ⎝ c⎠ ⎝ c⎠ Keeping just the lowest order terms in v/c, expand binomially to obtain:

−1 2

−1

∆f ⎛ 1 v ⎞ ⎛ 1 v ⎞ = ⎜1 + ⎟ ⎜1 + ⎟ −1 f0 ⎝ 2 c ⎠ ⎝ 2 c ⎠ v v ≈ 1+ −1 = c c

The sign depends on whether the source and receiver are approaching or receding. Here we have assumed that they are approaching. *29 •• Picture the Problem Due to its motion, the orbiting clock will run more slowly than the earth-bound clock. We can use Kepler’s third law to find the radius of the satellite’s orbit in terms of its period, the definition of speed to find the orbital speed of the satellite from the radius of its orbit, and the time dilation equation to find the difference δ in the readings of the two clocks. Express the time δ lost by the clock:

δ = ∆t − ∆t p = ∆t −

Because v << c, we can use part (b) of Problem 13:

1 v2 ≈ 1− γ 2 c2

Substitute to obtain:

δ = ∆t ⎢1 − ⎜⎜1 −

Express the square of the speed of the satellite in its orbit:

⎛ 1⎞ = ∆t ⎜⎜1 − ⎟⎟ γ ⎝ γ⎠

∆t

1

⎡

⎛

⎣

⎝

1 v 2 ⎞⎤ 1 v 2 ⎟⎥ = ∆t 2 c 2 ⎟⎠⎦ 2 c 2

4π 2 r 2 ⎛ 2π r ⎞ v =⎜ ⎟ = T2 ⎝ T ⎠

(1)

2

2

(2)

where T is its period and r is the radius of its (assumed) circular orbit. Use Kepler’s third law to relate the period of the satellite to the radius of its orbit about the earth:

T2 =

4π 2 3 4π 2 3 r = r GM e gRe2

1324 Chapter 39 Solve for r:

r=3

gRe2T 2 4π 2

Substitute numerical values and evaluate r:

r=3

(9.81m/s )(6370 km) (90 min × 60 s/min ) 2

2

4π

2

2

Substitute numerical values in equation (2) and evaluate v2:

v2 =

= 6.65 × 106 m

(

)

4π 2 6.65 × 106 m (90 min × 60 s/min )2 2

= 5.99 × 107 m 2 / s 2 Finally, substitute for v2 in equation (1) and evaluate δ:

(

)

1 5.99 × 107 m 2 / s 2 (1 y × 31.56 Ms/y ) δ= = 10.5 ms 2 2 3 × 108 m/s

(

)

30 •• Picture the Problem We can use the definition of the redshift parameter and the

⎛ u 2 −1 ⎞ ⎟⎟ , where u = z + 1. 2 ⎝ u +1⎠

relativistic Doppler shift equation to show that v = c ⎜⎜

f − f' f'

The red-shift parameter is defined to be:

z=

The relativistic Doppler shift is given by:

f' = f

Substitute to obtain:

f−f z=

= Letting u = z + 1:

1+ v c 1− v c

1+ v c 1− v c

1+ v c f 1− v c 1+ v c −1 1− v c

u = z +1 =

1+ v c 1− v c

1− =

1+ v c 1− v c 1+ v c 1− v c

Relativity 1325 Solve for v to obtain:

⎛ u2 −1 ⎞ v = c⎜⎜ 2 ⎟⎟ ⎝ u +1⎠

31 • Picture the Problem We can use the velocity transformation equations for the x and y directions to express the x and y components of the velocity of the light beam in frame S. (a) The x and y components of the velocity of the light beam in frame S are:

ux =

u x' + v vu ' 1 + 2x c

and

uy =

Because ux′ = 0:

(b) The magnitude of the velocity of the light beam in S is given by:

u y' ⎛ vu ' ⎞ γ ⎜1 + 2x ⎟ c ⎠ ⎝

c

u x = v and u y =

γ

u = u +u = v + 2 x

2 y

2

c2

γ2

⎛ v2 ⎞ 2 = v + ⎜⎜1 − 2 ⎟⎟c = c ⎝ c ⎠ 2

32 • Picture the Problem Let S be the earth reference frame and S′ be that of the ship traveling east (positive x direction). Then in the reference frame S′, the velocity of S is directed west, i.e., v = −ux. We can apply the inverse velocity transformation equation to express u' x in terms of ux. Apply the inverse velocity transformation equation to obtain: Substitute for v:

ux − v vu 1 − 2x c 2u x u +u u x' = x 2x = u u2 1 + x2 1 + x2 c c u x' =

1326 Chapter 39 Because ux = 0.90c:

u x' =

2(0.90c ) = 0.994c 2 ( 0.90c ) 1+ c2

Picture the Problem We can apply the inverse velocity transformation equation to express the speed of the particle relative to both frames of reference. (a) Express u x' in terms of u x'' :

u x' =

u x'' + v vu '' 1 + 2x c

where V of S ′, relative to S″, is 0.8c. Substitute numerical values and evaluate u x' :

(b) Express ux in terms of u x' :

u x' =

ux =

0.8c + 0.8c 1.6c = = 0.976c 2 1.64 ( 0.8c ) 1+ c2 u x' + v vu ' 1 + 2x c

where v of S, relative to S ′, is 0.8c.

Substitute numerical values and evaluate ux:

ux =

0.976c + 0.8c 1.776c = (0.8c )(0.976c ) 1.781 1+ c2

= 0.997c

Relativistic Momentum and Relativistic Energy *34 • Picture the Problem We can use the relation for the total energy, momentum, and rest energy to find the momentum of the proton and Equation 39-26 to relate the speed of the proton to its energy and momentum. Relate the energy of the proton to its momentum:

( )

E 2 = p 2c 2 + mc 2

(b) Solve for p to obtain:

E 2 − (mc 2 ) c2

2

p=

2

Relativity 1327 Substitute numerical values and evaluate p:

p=

(2200 MeV )2 − (938 MeV )2 c

= 1.99 (a) From Equation 39-26 we have:

Solve for v to obtain:

Substitute numerical values and evaluate v:

35

•

GeV c

v pc = c E v=

pc c E

v=

1.99 GeV c = 0.905c 2200 MeV

( )

2

Picture the Problem We can use E 2 = p 2 c 2 + mc 2 (from Problem R-37) to find the relativistic momentum of the particle in terms of γ and the fact that the kinetic energy of the particle equals twice its rest energy to find the error made in using mv for the momentum of the particle. Express the error e in using p′ = mv for the momentum of the particle: From Problem R-37, the relationship between the total energy E, momentum p, and rest energy mc2 of the particle is: Solve for p to obtain:

e=

p − p' p' = 1− p p

E 2 = p 2c 2 + (mc 2 )

2

p=

2 ⎞ E2 2 2 2 2⎛ E ⎜ m c m c − = − 1⎟⎟ 2 2 4 ⎜ c ⎝m c ⎠

Because E = γmc2:

p = mc γ 2 − 1

Substitute for p and p′ in equation (1) to obtain:

e = 1−

From the definition of γ:

(1)

mu mc γ − 1

1 u = 1− 2 γ c

2

= 1−

u c γ 2 −1

(2)

1328 Chapter 39 Eliminate v/c in equation (2) to obtain:

The kinetic energy of the particle is related to its rest energy: Solve for γ to obtain:

1

e = 1−

γ −1 2

1−

1

γ

2

= 1−

1

γ

(3)

K = (γ − 1)mc 2

γ = 1+

K mc 2

Because the kinetic energy of the particle is twice its rest energy:

γ = 1+

2mc 2 =3 mc 2

Substitute for γ in equation (3) and evaluate e:

e = 1−

1 = 0.667 = 66.7% 3

36 •• Picture the Problem We can use the result of Problem R-37 to find the energy of the particle and its energy in a reference frame in which its momentum is 4 MeV/c. We can apply the inverse velocity transformation equation to find the relative velocities of the two reference frames. (a) From Problem R-37 we have:

E 2 = p 2 c 2 + m02 c 4 = p 2 c 2 + E02

Solve for E0:

E0 =

E 2 − p 2c 2

Substitute numerical values and evaluate E0:

E0 =

(8 MeV )2 − (6 MeV/c )2 c 2

=

(8 MeV )2 − (6 MeV )2

= 5.29 MeV (b) Because E0 is independent of the reference frame:

E=

p 2 c 2 + E02

Substitute numerical values and evaluate E:

E=

(4 MeV/c )2 c 2 + (5.29 MeV )2

(c) The inverse velocity transformation is:

ub =

= 6.63 MeV ua − v vu 1 − 2a c

where the subscripts refer to the velocities

Relativity 1329 in parts (a) and (b) of the problem. Solve for v to obtain:

v=

Relate the relativistic energy of the particle in (a) to its velocity:

u a − ub uu 1− a 2 b c E0

E=

u a2 1− 2 c

(1)

⎛E ⎞ ⇒ ua = c 1 − ⎜ 0 ⎟ ⎝ E ⎠

2

Substitute numerical values and evaluate ua:

⎛ 5.29 MeV ⎞ ⎟⎟ = 0.750c u a = c 1 − ⎜⎜ ⎝ 8 MeV ⎠

Relate the relativistic momentum of the particle in (b) to its velocity:

p=

Substitute numerical values and evaluate ub:

2

m0 u b

1−

ub =

u c

2 b 2

⇒ ub =

pc p + m0c 2 2

(4 MeV/c )c (4 MeV/c )2 + (5.29 MeV/c )2

= 0.603c Substitute in equation (1) and evaluate V:

V=

0.750c − 0.603c = 0.268c ( 0.750c )(0.603c ) 1− c2

37 •• Picture the Problem We can use the rule for the derivative of a quotient to establish the result given in the problem statement. Use the expression for the derivative of a quotient to obtain:

1− ⎞ d ⎛⎜ mu ⎟= 2 2 ⎜ du 1 − u c ⎟ ⎝ ⎠

u2 mu 2 m + c2 c2 1−

1 u2 1− 2 c

u2 c2

Multiply the numerator and denominator of the right-hand side of this expression by

1330 Chapter 39 1−

u2 and simplify to obtain: c2

1− ⎞ d ⎛⎜ mu ⎟= du ⎜ 1 − u 2 c 2 ⎟ ⎝ ⎠

u u mu 1− 2 m + 2 2 c c c 2

2

2

⎛ u2 ⎞ u2 ⎜⎜1 − 2 ⎟⎟ 1 − 2 c ⎝ c ⎠

⎛ u2 ⎞ = m⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠

u2 c2 u2 1− 2 c = 1−

⎛ u2 ⎞ mu 2 ⎜⎜1 − 2 ⎟⎟m + 2 c ⎝ c ⎠ 3 2 ⎛ u2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠

−3 2

and 2 −3 2 ⎛ ⎞ ⎛ u mu ⎟ = m⎜1 − ⎞⎟ du d⎜ ⎜ c2 ⎟ ⎜ 1 − u 2 c2 ⎟ ⎠ ⎝ ⎝ ⎠

38 •• Picture the Problem We will first consider the decay process in the center of mass reference frame and then transform to the laboratory reference frame in which one of the pions is at rest. Apply energy conservation in the center of mass frame of reference to obtain: Solve for γ :

Substitute numerical values and evaluate γ : Because one of the pions is at rest in the laboratory frame, γ = 1.78 for the transformation to the laboratory frame. The kinetic energy of the K0 particle is: The total initial energy in the laboratory frame is:

mK 0 c 2 = 2mπ 0 γ c 2

γ =

γ=

mK 0 2mπ 0

497.7 MeV/c 2 = 1.78 2 139.6 MeV/c 2

(

)

K K 0 = (γ − 1)E

= (1.78 − 1)(497.7 MeV ) = 388.2 MeV

E = 497.7 MeV + 388.2 MeV = 885.9 MeV

Relativity 1331 Express the energy of the other pion: Substitute numerical values and evaluate Eπ:

Eπ = E − 2m0π c 2 Eπ = 885.9 MeV − 2(139.6 MeV ) = 607 MeV

*39 •• Picture the Problem The total kinetic energy of the two protons in part (a) is the sum of their kinetic energies and is given by K = 2(γ − 1)E0 . Part (b) differs from part (a) in that we need to find the speed of the moving proton relative to frame S. (a) The total kinetic energy of the protons in frame S′ is given by: Substitute for γ and E0 and evaluate K:

K = 2(γ − 1)E0

⎛ ⎞ ⎜ ⎟ 1 ⎜ ⎟ − 1⎟ (938.28 MeV ) K = 2⎜ 2 ⎜ 1 − (0.5c ) ⎟ ⎜ ⎟ 2 c ⎝ ⎠ = 290 MeV

(b) The kinetic energy of the moving proton in frame S is given by:

K = (γ − 1)E0

(1)

where

γ =

1 1−

Express the speed u of the proton in frame S:

u=

Substitute numerical values and evaluate u:

u=

Evaluate γ:

γ =

uv c2

u x' + v vu ' 1 + 2x c 0.5c + 0.5c = 0.800c ( 0.5c )(0.5c ) 1+ c2 1 = 1.67 ( 0.8c )(0.8c ) 1− c2

1332 Chapter 39 Substitute numerical values in equation (1) and evaluate K:

K = (1.67 − 1)(938.28 MeV ) = 629 MeV

40 •• Picture the Problem We can find the speed of each proton by equating their total relativistic kinetic energy to 2mc2. In (b) we can use the inverse velocity transformation with V = u and ux = −u to find ux′. In part (c) we’ll need to evaluate γ ′ for the kinetic energy transformation K L = (γ' − 1)E0 . (a) Set the relativistic kinetic energy of the protons equal to 2mc2 to obtain:

2(γ − 1)E0 = 2mc 2 ⇒ γ = 2

Substitute for γ :

1 u2 1− 2 c

Solve for u to obtain:

(b) Use the inverse velocity transformation with v = u and ux = −u to find ux′:

(c) The kinetic energy of the moving proton in the laboratory’s frame is given by:

u=

3 c = 0.866c 2

3 3 c− c ux − v 2 2 u x' = = vu ⎛ 3 ⎞⎛ 3 ⎞ 1 − 2x ⎜− ⎟⎜ ⎟ ⎜ 2 c⎟⎜ 2 c⎟ c ⎠⎝ ⎠ 1− ⎝ c2 − 4 3c = = − 0.990c 7 −

K L = (γ' − 1)E0 where

1

γ' =

(u x' )

2

1−

Substitute for γ ′ and E0 and evaluate KL:

=2

c

2

=

1

⎛ 4 3 ⎞ ⎜− c ⎟⎟ ⎜ 7 ⎠ 1− ⎝ c2

K L = (7 − 1) mc 2 = 6mc 2

2

=7

Relativity 1333 41 ••• Picture the Problem (a) and (b) The initial speed of the particle can be found from its total energy and its total energy found using E = K + E0 = γE0 . (c) We can solve

( )

2

E 2 = p 2c 2 + mc 2 for the initial momentum of the system. In (d) and (e) we can use conservation of energy and conservation of momentum to find the total kinetic energy after the collision and the mass of the system after the collision. (a) Express the total energy of the particle: Because the kinetic energy of the particle is twice its energy: Solve the factor γ for u:

Substitute for γ and evaluate u:

(b) The total energy of the particle is:

E = K + E0 = γE0 2 E0 + E0 = γE0 and γ = 3

u = c 1−

u = c 1−

1

γ2 1 8 =c = 0.943c 2 3 9

E = K + E0 = γE0 = 3E0

Substitute for E0 and evaluate E:

E = 3(1 MeV ) = 3 MeV

(c) The initial momentum of the incoming particle is related to its energy and mass according to:

E 2 = p 2c 2 + mc 2

Solve for p:

p=

1 E 2 − mc 2 c

Substitute for E and mc2 and simplify to obtain:

p=

1 c

Substitute for E0 and evaluate p:

(d) and (e) From conservation of energy we have:

( )

p=

2

( )

2

(3E0 )2 − (E0 )2

=

8 E0 c

8 (1 MeV ) = 2.83 MeV/c c

Ef = Ei = 5 MeV

1334 Chapter 39 From conservation of momentum we have:

pf = pi

The final momentum of the system is related to its energy and mass according to:

Ef2 = pf2c 2 + Ef02

Solve for Ef0:

Ef0 = Ef2 − pf2c 2

Substitute numerical values and evaluate Ef0:

Ef0 =

Because Ef0 = mf0c 2 :

The total kinetic energy after the collision is given by:

(5 MeV )2 − (2.83 MeV/c )2 c 2

= 4.122 MeV mf0 =

Ef0 = 4.12 MeV/c 2 2 c

K f = Ef − Ef0 = 5 MeV − 4.122 MeV = 0.878 MeV

General Relativity *42 •• Picture the Problem Let m represent the mass equivalent of a photon. We can equate the change in the gravitational potential energy of a photon as it rises a distance L in the gravitational field to h∆f and then express the wavelength shift in terms of the frequency shift.

c

The speed of the photons in the light beam are related to their frequency and wavelength:

c = fλ ⇒ f =

Differentiate this expression with respect to λ to obtain:

df c = −cλ−2 = − 2 λ dλ

Approximate df/dλ by ∆f/∆λ and solve for ∆f:

∆f = −

Divide both sides of this equation by f to obtain:

∆f = f

c

∆λ

λ2

−

λ

c

λ2

c

λ

∆λ

=−

∆λ

λ

Relativity 1335 Solve for ∆λ:

The change in the energy of the photon as it rises a distance L in a gravitational field is given by:

∆λ = −λ

∆f f

(1)

∆E = ∆U = mgL

Because ∆E = h∆f :

h∆f = mgL

(2)

Letting m represent the mass equivalent of the photon:

E = mc 2 = hf

(3)

Divide equation (2) by equation (3) to obtain:

h∆f mgL gL ∆f = = 2 ⇒ 2 hf mc f c

Substitute for ∆f/f in equation (1):

∆λ = −

Substitute numerical values and evaluate ∆λ:

∆λ = −

gLλ c2 9.81 m/s 2 (100 m )(632.8 nm )

(

)

(3 ×10

8

m/s

)

2

= − 6.90 × 10 −12 nm 43 •• Picture the Problem In a freely falling reference frame, both cannonballs travel along straight lines, so they must hit each other, as they were pointed at each other when they were fired. 44 ••• Picture the Problem Consider the turntable to be a giant hollow cylinder in space that is spinning about its axis. Someone on the inside surface of the cylinder would experience a centripetal acceleration caused by the normal force of the surface pushing them toward the rotation axis. Alternatively, they can consider that they are not accelerating but a r 2 gravitational field g = ω rrˆ is pushing them away from the axis ( rˆ is away from the axis). This is the principle of equivalence. From this perspective, up is toward the axis and the points closer to the axis are at the higher gravitational potential. (Just like the electric field points in the direction of decreasing electric potential, the gravitational field points in the direction of decreasing gravitational potential.) (a) From the time dilation equation we have:

∆tr = and

∆t0

γ

1336 Chapter 39 ∆tr − ∆t0 1 = −1 γ ∆t0 Because rω/c << 1 (see Problem 14):

1

γ

≈ 1−

1 u2 r 2ω 2 = − 1 2 c2 2c 2

∆t r − ∆t0 r 2ω 2 r 2ω 2 1 = 1− − = − ∆t0 2c 2 2c 2

Substitute to obtain:

FP = −ma

(b) The pseudoforce is given by:

where a is the acceleration of the noninertial reference frame.

a = −rω 2 ⇒ FP = Fr = mrω 2

In this case a is the centripetal acceleration:

To relate this problem to Equation 39-31, point 2 is a distance r from the axis and point 1 is on the axis. The term in parentheses on the right hand side of Equation 2 is φ2 − φ1, which translates to φr − φ0. Because φr is at a lower potential than φ0, this term is negative. Hence: r

r

r

r

r

r

φr − φ0 = − ∫ g ⋅ d l = − ∫ ω 2 r rˆ ⋅ d l = − ∫ ω 2 r dr = − 12 r 2ω 2 0

From Equation 39-31:

0

0

1 ∆t r − ∆t0 = 2 (φr − φ0 ) ∆t0 c =

(

1 1 2 2 −2r ω c2

= −

)

r 2ω 2 2c 2

General Problems 45 • Picture the Problem We can use the definition of γ and the time dilation equation to find the speed of the muon.

Relativity 1337 (a) From the definition of γ we have:

γ=

Solve for u/c:

1 u = 1− 2 c γ

Relate the mean lifetime of the muon to its proper lifetime:

∆t = γ∆t p ⇒ γ =

Substitute in the expression for u/c to obtain:

⎛ ∆t p ⎞ u ⎟⎟ = 1 − ⎜⎜ c ∆ t ⎠ ⎝

Substitute numerical values and evaluate u/c:

⎛ 2 µs ⎞ u ⎟⎟ = 0.999 = 1 − ⎜⎜ c ⎝ 46 µs ⎠

1 ⎛u⎞ 1− ⎜ ⎟ ⎝c⎠

2

∆t ∆t p

2

2

or

u = 0.999c 46 • Picture the Problem We can use the relativistic Doppler shift, when the source and the receiver are receding, to relate the frequencies of the two wavelengths and c = fλ to express the ratio of the wavelengths as a function of the speed of the galaxy. When the source and receiver are moving away from each other, the relativistic Doppler shift is given by: Use the relationship between the wavelength and frequency to obtain:

Because λ′ = 2λ0:

v c f f' = v 0 1+ c v v 1− 1− c λ c c c ⇒ 0 = = v v λ0 λ' λ ' 1+ 1+ c c 1−

v λ0 1 c = = v 2λ0 2 1+ c 1−

1338 Chapter 39 Solve for v/c:

v 3 = ⇒ v = 0.600c c 5

*47 •• Picture the Problem We can use Equation 39-12, the inverse time transformation equation, to relate the elapsed times and separations of the events in the two systems to the velocity of S′ relative to S. We can use this same relationship in (b) to find the time at which these events occur as measured in S′. (a) Use Equation 39-12 to obtain:

v ⎡ ⎤ ∆t' = t2' − t1' = γ ⎢(t2 − t1 ) − 2 (x2 − x1 )⎥ c ⎣ ⎦ v ⎡ ⎤ = γ ⎢∆t − 2 ∆x ⎥ c ⎣ ⎦

Because the events occur simultaneously in frame S′, ∆t′ = 0 and:

0 = ∆t −

Solve for v to obtain:

Substitute for ∆t and ∆x and evaluate V: Because ∆t = t2 − t1 = −0.5 y :

(b) Use the inverse time transformation to obtain:

Substitute numerical values and evaluate t2′ and t1′:

v=

v ∆x c2

c 2 ∆t ∆x

c 2 (0.5 y − 1 y ) v= = − 0.5c 2.0 c ⋅ y − 1.0 c ⋅ y S ' moves in the negative x direction.

vx ⎞ ⎛ t2' = γ ⎜ t2 − 22 ⎟ = c ⎠ ⎝

t2' = t1' =

0.5 y −

= 1.73 y

vx2 c2 v2 1− 2 c

t2 −

(− 0.5c )(2.0 c ⋅ y )

c2 (− 0.5 c )2 1− c2

Relativity 1339 48 •• Picture the Problem We can use the relationship between distance, speed, and time and the length contraction relationship to find the speed of the ship relative to the earth. The elapsed time between the departure of the spaceship and the receipt of the signal at earth is the sum of the travel time to the distant star system and the time it takes the signal to return to earth.

L' L = u γu

(a) Express the travel time as measured on the spaceship:

∆t' =

Solve for γu:

γu =

L ∆t'

Substitute numerical values and evaluate γu:

γu =

12 c ⋅ y = 0.8c 15 y

or

1 1−

u2 c2

=

0.8c u

Solve for u to obtain:

u = 0.625c

(b) The elapsed time T before earth receives the signal is the sum of the travel time to the distant star system and the time it takes the signal to return:

T=

L L + u c

Substitute numerical values and evaluate T:

T=

12 c ⋅ y 12 c ⋅ y + = 31.2 y 0.625c c

49 •• Picture the Problem We can use conservation of energy to find γ in the CM frame of reference and then use the definition of γ to find the speed u of the projectile proton. We can then use the velocity transformation equation to find the speed and kinetic energy of this proton in the laboratory frame of reference.

Ef Ei

Use conservation of energy to find γ in the CM frame of reference:

γEi = Ef ⇒ γ =

Ei and Ef are:

Ei = 938 Mev + 938 MeV = 1876 MeV

1340 Chapter 39 and

Ef = 938 Mev + 938 MeV + 135 MeV = 2011 MeV Substitute Ei and Ef and evaluate γ :

γ =

Express γ as a function of the speed u of the projectile proton:

γ =

2011 MeV = 1.072 1876 MeV 1 1−

Solve for u to obtain:

u2 c2 1

u = c 1−

Substitute for γ and evaluate u:

γ2 1 = 0.360c (1.072)2

u = c 1−

u −v 0.360c − (− 0.360c ) = (0.360c )(− 0.360c ) vu 1− 2 1− c c2 = 0.637c

Transform to the laboratory frame and find u′:

u' =

K L = (γ L − 1)E0

The kinetic energy of the moving proton in the laboratory’s frame is given by:

where

γL =

1 1−

=

(u' )2 c2 1

(0.637c )2 1−

= 1.30

c2

K L = (1.30 − 1)(938 MeV ) = 281 MeV

Substitute for γL and E0 and evaluate KL:

Remarks: In Problem 55 we show that the threshold kinetic energy of the projectile is given by K th =

(∑ m + ∑ m )(∑ m in

fin

2m target

fin

− ∑ min )c 2

.

Relativity 1341 50 •• Picture the Problem We can use ∆t p = Lp u , where u is the speed of the bullet relative to the rocket, to find the elapsed time in the frame of the rocket. In (b) and (c) we can proceed similarly, finding the speed of the bullet relative to the rocket as seen from the ground frame in (b) and, in (c), using the speed of the bullet relative to the rocket. (a) In the rocket frame:

∆t = ∆tp =

Lp u

=

Lp 0.8c

1000 m = 4.17 µs 0.8 2.998 × 108 m/s

Substitute numerical values and evaluate ∆t:

∆t =

(b) In the ground frame of reference, the elapsed time is given by:

∆tground =

(

)

L u'

(1)

where u′ is the speed of the bullet relative to the rocket as seen from the ground.

urocket + V Vu 1 + ground c2

The speed of the ground is given by:

uground =

Substitute for urocket and V and evaluate uground:

uground =

The speed of the bullet relative to the rocket as seen from the ground is:

u ' = 0.946c − 0.6c = 0.346c

Relate Lground to Lp:

Substitute for Lp and uground and evaluate Lground:

0.6c + 0.8c = 0.946c ( 0.8c )(0.6c ) 1+ c2

Lground =

Lp

γ

= Lp 1 −

2 uground

c2

2 ( 0.6c ) Lground = (1000 m ) 1 − 2

c

= 800 m Substitute numerical values in equation (1) and evaluate ∆tground:

∆tground =

800 m 0.346 2.998 × 108 m/s

(

= 7.71 µs

)

1342 Chapter 39 (c) In the bullet’s frame of reference, the elapsed time is given by: The length L′ of the rocket in the bullet’s frame is given by: Substitute in equation (2) to obtain:

∆tbullet =

Lp

L' =

L'

(2)

ubullet

= Lp 1 −

γ

∆tbullet =

Lp ubullet

2 ubullet c2

1−

2 ubullet c2

Substitute numerical values and evaluate ∆tbullet:

∆tbullet =

1000 m 0.8 2.998 × 108 m/s

(

)

1−

(0.8c )2 c2

= 2.50 µs

*51 ••• Picture the Problem We can use conservation of energy to express the recoil velocity of the box and the relationship between distance, speed, and time to find the distance traveled by the box in time ∆t = L/c. Equating the initial and final locations of the center of mass will allow us to show that the radiation must carry mass m = E/c2. (a) Apply conservation of momentum to obtain: Solve for v:

(b) The distance traveled by the box in time ∆t = L/c is: Substitute for v from (a):

(c) Let x = 0 be at the center of the box and let the mass of the photon be m. Then initially the center of mass is at:

E + Mv = pi = 0 c v= −

E Mc

d = v∆t =

d=

vL c

L⎛ E ⎞ LE ⎜− ⎟= − c ⎝ Mc ⎠ Mc 2

xCM =

− 12 mL M +m

Relativity 1343 ⎡ − MEL EL ⎞⎤ ⎛1 ⎢ Mc 2 + m⎜ 2 L − Mc 2 ⎟⎥ ⎝ ⎠⎦ =⎣ M +m

When the photon is absorbed at the other end of the box, the center of mass is at:

xCM

Because no external forces act on the system, these expressions for xCM must be equal:

⎡ − MEL EL ⎞⎤ ⎛1 + − m L ⎜ ⎟ 2 2 ⎢ Mc 2 ⎠⎥⎦ − 12 mL ⎣ Mc ⎝ = M +m M +m

Solve for m to obtain:

m=

Because Mc2 is of the order of 1016 J and E = hf is of the order of 1 J for reasonable values of f, E/Mc2 << 1 and:

m=

E E ⎞ ⎛ c 2 ⎜1 − 2 ⎟ ⎝ Mc ⎠ E c2

52 ••• Picture the Problem We can apply a velocity transformation equation to find the speed of the particle and use the distance and time transformation equations to find the distance and direction the particle traveled from t′1 to t′2 and the time the particle traveled as observed in frame S.

u x' + v vu ' 1 + 2x c

(a) The velocity transformation equation for motion at speed v along the x axis is:

ux =

Evaluate ux for ux′ = −c/3 and v = 0.6c:

1 3 c+ c 3 5 = ux = ⎛ 3 ⎞⎛ 1 ⎞ ⎜ c⎟⎜− c⎟ 5 3 ⎠ 1 + ⎝ ⎠ ⎝2 c

(b) The distance traveled by the particle from t′1 to t′2 is given by: To find x2, we must first find x2′ and ∆t′:

−

∆x = x2 − x1

1 3

c

(1)

⎛c⎞ x2' = 10 m − (60 m/c )⎜ ⎟ = −10 m ⎝ 3⎠ and

∆t ' = t2' − t1' = 60 m/c = 200 ns

1344 Chapter 39 x2 is related to x2′ through the relativistic transformation:

x2 = γ (x2' + v∆t' ) =

Substitute numerical values and evaluate x2:

x2 =

x2' + v∆t' 1−

v2 c2

− 10 m + (0.6c )(200 ns ) 2 ( 0.6c ) 1−

= 32.5 m

c2

x1 is given by:

x1'

x1 = γx1' =

v2 c2

1−

Substitute numerical values and evaluate x1:

x1 =

10 m

(0.6c )2 1−

= 12.5 m

c2

Substitute numerical values in equation (1) and evaluate ∆x: (c) The time the particle traveled is given by: Express and evaluate t1:

∆x = 32.5 m − 12.5 m = 20.0 m ∆t = t2 − t1

t1 = γt1' =

(2)

t1' 1−

=

v2 c2

6 m/c

(0.6c )

2

1−

= 7.50 m/c = 25.0 ns

c2

Express and evaluate t2:

vt' vt' ⎞ ⎛ c2 t2 = γ ⎜ t' + 2 ⎟ = c ⎠ ⎝ v2 1− 2 c 60 m/c + (− 6 m/c ) = = 225 ns 2 ( 0.6c ) 1− c2

Substitute in equation (2) and evaluate ∆t:

∆t = 225 ns − 25 ns = 200 ns

t' +

Relativity 1345 53 ••• Picture the Problem We can evaluate the differentials of Equations 39-19a, b, and c and 39-10 and express their ratio to obtain expressions for ax', ay', and az'. From Equation 39-19a we have:

⎛ ⎜ u −v du x' = d ⎜ x ⎜ 1 − vu x ⎜ c2 ⎝

⎞ ⎟ ⎟= ⎟ ⎟ ⎠

⎛ v2 ⎞ ⎛ vu x ⎞ ⎛v ⎞ ⎜⎜1 − 2 ⎟⎟ ( ) du u v du 1 − + − ⎜ ⎟ x ⎜ 2⎟ x x c ⎠ c2 ⎠ ⎝ ⎝c ⎠ du x = ⎝ 2 2 ⎛ vu x ⎞ ⎛ vu x ⎞ ⎜1 − 2 ⎟ ⎜1 − 2 ⎟ c ⎠ c ⎠ ⎝ ⎝

From Equation 39-10:

γv γv dx ⎛ vx ⎞ ⎛ vu ⎞ dt' = γd ⎜ t − 2 ⎟ = γdt − 2 dx = 2 dt = γ ⎜1 − 2x ⎟dt c c dt c ⎠ ⎝ ⎝ c ⎠ Divide dux′ by dt′ to obtain:

⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎝ c ⎠ du x 2 ⎛ v2 ⎞ ⎛ vu x ⎞ ⎜⎜1 − 2 ⎟⎟ 1 − ⎜ ⎟ c ⎠ du x 1 du x' ⎝ c2 ⎠ = 3 3 ax = = ⎝ a x' = 3 γ δ dt' ⎛ vu ⎞ ⎛ vu ⎞ dt γ ⎜1 − 2x ⎟dt γ ⎜1 − 2x ⎟ c ⎠ ⎝ c ⎠ ⎝ vu where δ = 1 − 2x c Proceeding in exactly the same manner, one obtains:

a y' =

1

γ 2δ 2

ay +

vu y

γ 3δ 3c 2

ax

and an identical expression for az′ with z replacing y. 54 ••• Picture the Problem Without loss of generality, we’ll consider the absorption case. We’ll assume that the electron is initially at rest and that it travels with a speed v after it absorbs the photon. Applying the conservation of energy and the conservation of momentum will lead us to an absurd conclusion that, in turn, will force us to abandon our initial assumption that an electron can absorb a photon. Such an argument is known as a reductio ad absurdum argument.

1346 Chapter 39 When the electron absorbs a photon, the conservation of relativistic momentum requires that its momentum become: From the conservation of energy:

p = γmv

mc 2 + pc = γmc 2 ⇒ p = (γ − 1)mc

Equate these expression for p to obtain:

γmv = (γ − 1)mc

Solving for v yields:

⎛ γ −1⎞ v = ⎜⎜ ⎟⎟c ⎝ γ ⎠

Square both sides of the equation to obtain: From the definition of γ :

Solve for v2 to obtain:

Substitute for v2 in equation (1) and simplify to obtain:

(1)

2

⎛ γ − 1 ⎞ 2 γ 2 − 2γ + 1 2 v = ⎜⎜ c ⎟⎟ c = γ2 ⎝ γ ⎠ 2

γ2 =

v2 =

(2)

1 c2 = v2 c2 − v2 1− 2 c

γ 2 −1 2 c γ2

γ 2 − 1 2 γ 2 − 2γ + 1 2 c = c γ2 γ2 or

− 1 = −2γ + 1 ⇒ γ = 1

Substitute for γ in equation (1) and evaluate v:

⎛ 1−1⎞ v=⎜ ⎟c = 0 ⎝ 1 ⎠

Our assumption that an electron can absorb a photon has led to the contradictory conclusion that its speed after the absorption is zero. Hence, we must conclude that the electron cannot absorb a photon. *55 ••• Picture the Problem Let mi denote the mass of the incident (projectile) particle. Then ∑min = mi + mtarget and we can use this expression to determine the threshold kinetic energy of protons incident on a stationary proton target for the production of a proton– antiproton pair.

Relativity 1347 E 2 − p 2c 2 = ∑ mfin c 2

Consider the situation in the center of mass reference frame. At threshold we have:

Note that this is a relativistically invariant expression.

Etarget = Et = Et,0

In the laboratory frame, the target is at rest so:

− pi2c 2 = (∑ mfin c 2 )

We can, therefore, write:

(E + E )

For the incident particle:

Ei2 − pi2c 2 = Ei,02

i

2

t,0

2

and

Ei = Ei,0 + K th where Kth is the threshold kinetic energy of th incident particle in the laboratory frame.

(E

Express Kth in terms of the rest energies:

+ Ei.0 ) + 2 K th Et,0 = 2

t,0

(∑ m c )

2 2

fin

where

Et,0 + Ei.0 = ∑ mfin c 2

and

Et,0 = mtarget c 2

(∑ m

Substitute to obtain:

fin

c2

)

2

+ 2 K th mtarget c 2 =

(∑ m

Solve for Kth to obtain:

K th =

(∑ m + ∑ m )(∑ m in

For the creation of a proton - antiproton pair in a proton proton collision:

fin

fin

− ∑ min )c 2

2mtarget

∑m ∑m

in

= 2mp

fin

= 4mp

and

mtarget = mp Substitute to obtain:

K th

(2m =

p

+ 4mp )(4mp − 2mp )c 2 2mp

(6m )(2m )c = p

p

2mp

2

= 6mp c 2

fin

c2

)

2

1348 Chapter 39 in agreement with Problem 40. 56 ••• Picture the Problem We’ll solve the problem for the general case of a particle of rest mass M decaying into two identical particles each of rest mass m. In the center of mass reference frame:

Mc 2 = 2mc 2 = 2γmc 2

Solve for u/c to obtain:

u ⎛ 2m ⎞ = 1− ⎜ ⎟ c ⎝M ⎠

2

where u is the speed of each of the decay particles in the CM frame. Next we determine the speed v of the laboratory frame relative to the CM frame. The energy of the particle of rest mass M is:

γ CM Mc 2 where

γ CM =

1 1−

v2 c2

and

1 v = β CM = 1 − 2 γ CM c Use Equation 39-18a to express ulab, the speeds of the decay products in the laboratory reference frame:

ulab

u c c = u 1 ± β CM c

β CM ±

where ± refers to the fact that one of the decay particles will travel in the direction of M, and the other in the direction opposite to that of M. In this problem we have:

γ CM = 4 , β CM = 0.968 , and

Substitute to obtain:

u = 0.8 c

ulab = and

2m0 = 0.6 , M0

0.968 + 0.8 c = 0.996c 1 + (0.968)(0.8)

Relativity 1349 ulab =

0.968 − 0.8 c = 0.745c 1 − (0.968)(0.8)

57 ••• Picture the Problem We can write the components of the stick in its reference frame and then apply the Lorentz length contraction equation to obtain the given result. In its reference frame, the stick has x and y components:

Lpx = Lp cosθ and

Lpy = Lp sin θ

Only Lpx is Lorentz contracted to:

Hence, the length in the reference frame S′ is:

The angle that L′ makes with the x′ axis is given by:

Lpx

Lx' =

γ

[

L' = (Lx' ) + (Ly' )

]

2 12

2

12

⎞ ⎛ cos 2 θ = Lp ⎜⎜ + sin 2 θ ⎟⎟ 2 ⎠ ⎝ γ tan θ' =

Ly' Lx'

=

sin θ = γ tan θ cos θ

γ

58 ••• Picture the Problem We can express the tangent of the angle u′ makes with the x′ axis and then use the velocity transformation equations to obtain the given result.

u y'

Express the tangent of the angle u′ makes with the x′ axis:

tan θ' =

Substitute for uy′ and ux′:

uy ⎛ vu ⎞ γ ⎜1 − 2x ⎟ uy c ⎠ = tan θ' = ⎝ ux − v γ (u x − v ) vu x 1− 2 c

Substitute for uy and ux and simplify to obtain:

tan θ' =

u x'

u sin θ sin θ = γ (u cosθ − v ) γ (cosθ − v u )

1350 Chapter 39 *59 ••• r Picture the Problem We can use the expressions for p and E in S together with the relations we wish to verify and the inverse velocity transformation equations to establish

( ) + (u ' )

the condition u' 2 = (u x' ) + u y' 2

2

z

2

= v2 +

u2

γ2

and then use this result to verify

the given expressions for px′, py′, pz′ and E′/c. In any inertial frame the momentum and energy are given by:

r p=

r mu

and E =

mc 2

u2 u2 1− 2 1− 2 c c r where u is the velocity of the particle and u is its speed.

r

The components of p in S are:

px =

pz =

mu x u2 1− 2 c mu z 1−

Because ux = uz = 0 and uy = u:

, py =

mu y u2 1− 2 c

, and

u2 c2

px = pz = 0 and

py =

mu 1−

Substituting zeros for px and pz in the relations we are trying to show yields:

In S′ the momentum components are:

u2 c2

vE ⎞ vE ⎛ p x' = γ ⎜ 0 − 2 ⎟ = −γ 2 , p y' = p y , c ⎠ c ⎝ p z' = 0 , and E' E ⎛E ⎞ = γ ⎜ − 0⎟ = γ c c ⎝c ⎠ p x' =

p z' =

mu x' u' 1− 2 c mu z'

2

1−

u' 2 c2

, p y' =

mu y' u' 2 1− 2 c

, and

Relativity 1351 uy ux − v , u y' = , and vu y vu x 1− 2 1− 2 c c uz u z' = vu 1 − 2z c

The inverse velocity transformations are:

u x' =

Substitute ux = uz = 0 and uy = u to obtain:

u x' = −v , u y' = γu , and u z' = 0 u' 2 = (u x' ) + (u y' ) + (u z' ) 2

Thus:

= v2 + First we verify that pz′ = pz = 0:

p z' =

2

2

u2

γ2

m(0) u' 2 1− 2 c

= pz = 0

Next we verify that py′ = py:

p y' =

mu y' 1−

=

u' 2 c2

mu 1−

u2 c2

mu

=

γ 1−

v2 u2 − c 2 γ 2c 2

=

mu 1−

1−

u2 c2

u2 v2 u2 1 − − γ c2 c 2 γ 2c 2

⎛ u 2 ⎞ ⎛ v2 ⎞ v2 u 2 ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎜⎜1 − 2 ⎟⎟ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ c c ⎝ c ⎠ ⎝ c ⎠⎝ c ⎠ = p y v2 u 2 ⎛ v2 ⎞ v2 u 2 ⎛ v2 ⎞ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ c c ⎝ c ⎠ c c ⎝ c ⎠

= py ⎛ ⎝

Next we verify that p x' = γ ⎜ p x −

vE ⎞ ⎟: c2 ⎠

1352

Chapter 39

p x' =

mu x'

=

u' 2 1− 2 c

− mv v2 u2 γ 1− 2 − 2 2 c γ c

=−

u2 c2 v2 u2 1− 2 − 2 2 c γ c

γ −1 1 −

γv

mc 2

c2

u2 1− 2 c

⎛ u2 ⎞ ⎛ v2 ⎞ v2 u 2 ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎜⎜1 − 2 ⎟⎟ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ c c ⎠ c c ⎝ c ⎠ γv γv = − 2 E ⎝ 2 ⎠ 2⎝ =− 2 E 2 2 c c v u ⎛ v ⎞ v u 2 ⎛ v2 ⎞ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ c c ⎝ c ⎠ c c ⎝ c ⎠ = −

γv c2

E

Finally, we verify that

E' =

E' E ⎛ E vp ⎞ = γ ⎜ − x ⎟ = γ , or E' = γE : c c ⎠ c ⎝c

mc 1−

2

u' c2

2

=

γmc 1−

2

γ −1 1 −

u2 c2

1−

u2 c 2 = γE

u' 2 c2

u2 c2 v2 u2 1− 2 − 2 2 c γ c

γ −1 1 −

⎛ u 2 ⎞ ⎛ v2 ⎞ v2 u 2 ⎛ v2 ⎞ ⎜⎜1 − 2 ⎟⎟ ⎜⎜1 − 2 ⎟⎟ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ c c c c ⎝ c ⎠ ⎠ = γE = γE ⎝ 2 ⎠ 2⎝ 2 2 v u ⎛ v ⎞ v u 2 ⎛ v2 ⎞ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ 1 − 2 − 2 ⎜⎜1 − 2 ⎟⎟ c c ⎝ c ⎠ c c ⎝ c ⎠ = γE The x, y, z, and t transformation equations are:

x' = γ ( x − vt ) y' = y z' = z and

⎛ vx ⎞ t' = γ ⎜ t − 2 ⎟ ⎝ c ⎠ The x, y, z, and ct transformation equations are:

v ⎞ ⎛ x' = γ ⎜ x − ct ⎟ c ⎠ ⎝ y' = y z' = z and

Relativity 1353 v ⎛ ct' = γ ⎜ ct − c ⎝ The px, py, pz, and E/c transformation equations are:

⎞ x⎟ ⎠

v E⎞ ⎛ p x' = γ ⎜ p x − ⎟ c c⎠ ⎝ p y' = p y p z' = p z and

E' ⎛E v ⎞ = γ ⎜ − px ⎟ c ⎝c c ⎠ Note that the transformation equations for x, y, z, and ct and the transformation equations for px, py, pz, and E/c are identical. 60 ••• Picture the Problem The Lorentz transformation was derived on the basis of the postulate that the speed of light is c in any inertial reference frame. Thus, if the clocks in S and S′ are synchronized at t = t′ = 0, then it follows from the Einstein postulate that r2 = c2t2 and r′2 = c2t′2 or r2 − c2t2 = 0 = r′2 − c2t′2. In other words, the quantity s2 = r2 − c2t2 = 0 is a relativistic invariant, which can also be written as x2 + y2 + z2 − c2t2 = 0. Using the Lorentz transformation equations for x, y, z, and t we have: x′2 + y′2 + z′2 − (ct′)2 = γ 2(x2 − 2vxt + v 2t2) + y2 + z2 − γ 2(c2t2 − 2vxt + v 2x2/c2) The terms linear in x cancel and the terms in x2 combine to give:

γ 2x2(1 − v 2/c2) =

The coefficients of the terms in (ct)2 give:

γ 2(v 2/c2 − 1) = −1

x2

Thus, r2 − c2t2 = r′2 − c2t′2 as required by the Einstein postulate. 61 ••• Picture the Problem We’ll use Equation 39-27 to show that this quantity has the value −mc2 in both the S and S′ reference frames. From Equation 39-27, the relationship between total energy E, momentum p, and rest energy mc2 is:

( )

E 2 = p 2c 2 + mc 2 or

2

( )

p 2c 2 − E 2 = − mc 2

2

1354

Chapter 39 2

Divide both sides of this equation by c2 to obtain:

⎛E⎞ 2 p − ⎜ ⎟ = −(mc ) ⎝c⎠

We can relate p to px, py, and pz:

p 2 = p x2 + p y2 + p z2

Substitute for p2 in equation (1) to obtain:

⎛E⎞ p + p + p − ⎜ ⎟ = −m 2c 2 ⎝c⎠

Because m is the mass of the particle in its rest frame, it is constant. Hence:

⎛E⎞ p 2 − ⎜ ⎟ must be a relativistic invariant. ⎝c⎠

2

(1)

2

2 x

2 y

2 z

2

Also, in Problem 59 we saw that the components of p and the quantity E/c transform like the components of r and the quantity ct. In Problem 60 we demonstrated that r2 − (ct)2 is a relativistic invariant. Consequently, p2 − (E/c)2 must also be relativistically invariant. *62 ••• Picture the Problem We can use the inverse Lorentz transformation for time to show that the observer will conclude that the rod is bent into a parabolic shape. In frame S where the rod is not moving along the x axis, the height of the rod at time t is:

y (t ) = − 12 gt 2

The inverse Lorentz time transformation is:

vx ⎞ ⎛ t = γ ⎜ t' + 2 ⎟ c ⎠ ⎝

Express y′(t) in the moving frame of reference:

vx ⎞ ⎛ y' (t ) = − 12 gγ ⎜ t' + 2 ⎟ c ⎠ ⎝

Evaluate y′(t) at t′ = 0 to obtain:

y' (t ) = −

gγv 2 2 x 2c 2

2

(1)

Because equation (1) is the equation of a parabola, we' ve shown that the moving observer will conclude that the rod is bent into a parabolic shape. Because the coefficient of x 2 is negative, the parabola is concave downward.

Chapter 40 Nuclear Physics Conceptual Problems 1 • Determine the Concept Two or more nuclides with the same atomic number Z but different N and A numbers are called isotopes. (a) Two other isotopes of 14N are:

15

N, 16N

(b) Two other isotopes of 56Fe are:

54

Fe, 55Fe

(c) Two other isotopes of 118Sn are:

54

Fe, 55Fe

2 • Determine the Concept The parent of that series, 237Np, has a half-life of 2×106 y that is much shorter than the age of the earth. There is no naturally occurring Np remaining on earth. 3 • Determine the Concept Generally, β-decay leaves the daughter nucleus neutron rich, i.e., above the line of stability. The daughter nucleus therefore tends to decay via β – emission which converts a nuclear neutron to a proton. *4 • Determine the Concept 14C is found on earth because it is constantly being formed by cosmic rays in the upper atmosphere in the reaction 14N + n → 14C + 1H. 5 • Determine the Concept It would make the dating unreliable because the current concentration of 14C is not equal to that at some earlier time. 6 • Determine the Concept An element with such a high Z value would either fission spontaneously or decay almost immediately by α emission (see Figure 40-3). 7 • Determine the Concept The probability for neutron capture by the fissionable nucleus is large only for slow (thermal) neutrons. The neutrons emitted in the fission process are fast (high energy) neutrons and must be slowed to thermal neutrons before they are likely to be captured by another fissionable nucleus.

1355

1356 Chapter 40 8 • Determine the Concept The process of ″slowing down″ involves the sharing of energy of a fast neutron and another nucleus in an elastic collision. The fast particle will lose maximum energy in such a collision if the target particle is of the same mass as the incident particle. Hence, neutron-proton collisions are most effective in slowing down neutrons. However, ordinary water cannot be used as a moderator because protons will capture the slow neutrons and form deuterons. 9 • Determine the Concept Beta decay occurs in nuclei that have too many or too few neutrons for stability. In β decay, A remains the same while Z either increases by 1 (β − decay) or decreases by 1 (β + decay). (a) The reaction is:

22 11

0 β + decay Na →22 10 Ne+ +1 β ⇒

(b) The reaction is:

24 11

0 Na →24 β − decay 12 Mg + −1 β ⇒

10

• Advantages

The reactor uses 238U, which, by neutron capture and subsequent decays, produces 239Pu. Thus plutonium isotope fissions by fast neutron capture. Thus, the breeder reactor uses the plentiful uranium isotope and does not need a moderator to slow the neutrons needed for fission.

Disadvantages The fraction of delayed neutrons emitted in the fission of 239Pu is very small. Consequently, control of the fission reaction is very difficult, and the safety hazards are more severe than for the ordinary reactor that uses 235U as fuel.

11 • (a) False. The nucleus does not contain electrons. (b) True. (c) False. After two half-lives, three-fourths of the radioactive nuclei in a given sample have decayed. (d) True (given an unlimited supply of 238U).

Nuclear Physics 1357 12 • Determine the Concept Pressure and temperature changes have no effect on the internal structure of the nucleus. They do have an effect on the electronic configuration; consequently, they can influence K-capture processes. *13 • Determine the Concept Knowing the parent nucleus and one of the decay products, we can use the conservation of charge, the conservation of energy, and the conservation of the number of nucleons to identify the participants in the decay. (a) beta decay of 16N

16 7

(b) alpha decay of 248Fm (c) positron decay of 12N (d) beta decay of 81Se (e) positron decay of 61Cu (f) alpha decay of 228Th

N →168 O + −10β + 00ν + Q

248 100

4 Fm→244 98 Cf + 2 He + Q

12 7

N→126 C+ +10β + 00ν + Q

81 34

81 Se→35 Br + −10β + 00ν + Q

61 29

61 Cu →28 Ni + +10β + 00ν + Q 4 Th →224 88 Ra + 2 He + Q

228 90

*14 • Determine the Concept We can use the information regarding the daughter nuclei to write and balance equations for each of the reactions. 147 Pu →301 n+90 38 Sr + 56 Ba

(a)

240 94

(b)

72 32

(c)

127 53

I+ 21H→01 n +128 54 Xe

(d)

235 92

121 U + 01n →201 n +113 47 Ag + 45 Rh

(e)

55 25

75 Ge+ 42 He→01 n + 34 Se

59 Mn + 73 Li→31 H + 27 Co

1 239 (f) 238 92 U + 0 n → 92 U;

239 92

U→−10β + 00ν + 239 93 Np;

239 93

Np→ −10β + 00ν +

239 94

Pu

1358 Chapter 40

Estimation and Approximation 15 • Picture the Problem There is no table of half lives in the text although the information is mentioned in the alpha particle discussion for alpha decay (about 15 orders of magnitude). Mass density in an atom ranges roughly as the cube of the radius of an atom to that of the nucleus, also about 15 orders of magnitude. Nuclear masses only range 2 orders of magnitude. Material property

Ratio (order of magnitude)

Mass density

1015

Half life

1015

Nuclear masses 16

2

••

Picture the Problem The mass of 235U required is given by m235 =

N M 235 , where NA

M235 is the molecular mass of 235U and N is the number of fissions required to produce 1020 J. The mass of deuterium and tritium required can be found similarly. (a) Relate the mass of 235U required to the number of fissions N required:

m235 =

N M 235 NA

(1)

where M235 is the molecular mass of 235U.

Eannual Eper fission

Determine N:

N=

Substitute numerical values and evaluate N:

10 20 J N= 1.60 × 10 −19 J 200 MeV × eV 30 = 3.13 × 10

Substitute numerical values in equation (1) and evaluate m235:

m235 =

3.13 × 10 30 (235 g/mol) = 5.20 ×10 6 kg 6.02 ×10 23 nuclei/mol

Nuclear Physics 1359 (b) Relate the mass of 2H and 3H required to the number of fusions N required:

m 2 H +3 H =

N M 2 H+3 H NA

(2)

where M 2 H + 3 H is the molecular mass of 2

H + 3H.

Determine N:

N=

Substitute numerical values and evaluate N:

N=

Eannual Eper fission

10 20 J 1.60 × 10 −19 J 18 MeV × eV 31 = 3.47 × 10

Substitute numerical values in equation (2) and evaluate m 2 H + 3 H :

m235 =

3.47 × 10 31 (5 g/mol) = 2.88 ×10 6 kg 23 6.02 ×10 nuclei/mol

Properties of Nuclei *17 • Picture the Problem To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2. To convert to MeV we multiply this result by 931.5 MeV/u. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus. (a) For 12C, Z = 6 and N = 6. Add the mass of the neutrons to that of the protons:

6mp + 6mn = 6 × 1.007825 u + 6 × 1.008665 u = 12.098940 u Subtract the mass of 12C from this result:

(6m

p

+ 6mn ) − m12 C = 12.098940 u − 12 u = 0.098940 u

Multiply the mass difference by c2 and convert to MeV:

Eb = (∆m )c 2 = (0.098940 u ) c 2 ×

931.5 MeV/c 2 = 92.2 MeV 1u

1360 Chapter 40 and the binding energy per nucleon is

Eb 92.2 MeV = = 7.68 MeV A 12

(b) For 56Fe, Z = 26 and N = 30. Add the mass of the neutrons to that of the protons:

26mp + 30mn = 26 × 1.007825 u + 30 × 1.008665 u = 56.463400 u Subtract the mass of 56Fe from this result:

(26m

p

+ 30mn ) − m12 C = 56.463400 u − 55.934942 u = 0.528458 u

Multiply the mass difference by c2 and convert to MeV:

931.5 MeV/c 2 = 492 MeV 1u E 492 MeV and the binding energy per nucleon is b = = 8.79 MeV A 56 Eb = (∆m ) c 2 = (0.528458 u ) c 2 ×

(c) For 238U, Z = 92 and N = 146. Add the mass of the neutrons to that of the protons:

92mp + 146mn = 92 × 1.007825 u + 146 × 1.008665 u = 239.984990 u Subtract the mass of 238U from this result:

(92m

p

+ 146mn ) − m 238 U = 239.984990 u − 238.050783 u = 1.934207 u

Multiply the mass difference by c2 and convert to MeV:

931.5 MeV/c 2 = 1802 MeV 1u E 1802 MeV = 7.57 MeV and the binding energy per nucleon is b = A 238 Eb = (∆m ) c 2 = (1.934207 u )c 2 ×

Nuclear Physics 1361 18 • Picture the Problem To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2. To convert to MeV we multiply this result by 931.5 MeV/u. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus. (a) For 6Li, Z = 3 and N = 3. Add the mass of the neutrons to that of the protons:

3mp + 3mn = 3 × 1.007825 u + 3 × 1.008665 u = 6.049470 u Subtract the mass of 6Li from this result:

(3m

p

+ 3mn ) − m6 Li = 6.049470 u − 6.015122 u = 0.034348 u

Multiply the mass difference by c2 and convert to MeV:

931.5 MeV/c 2 = 32.0 MeV 1u E 32.0 MeV = 5.33 MeV and the binding energy per nucleon is b = A 6 Eb = (∆m )c 2 = (0.034348 u )c 2 ×

(b) For 39K, Z = 19 and N = 20. Add the mass of the neutrons to that of the protons:

19mp + 20mn = 19 × 1.007825 u + 20 × 1.008665 u = 39.321975 u Subtract the mass of 39K from this result:

(19m

p

+ 20mn ) − m39 K = 39.321975 u − 38.963707 u = 0.358268 u

Multiply the mass difference by c2 and convert to MeV:

931.5 MeV/c 2 Eb = (∆m )c = (0.358268 u )c × = 334 MeV 1u E 334 MeV and the binding energy per nucleon is b = = 8.56 MeV A 39 2

2

(c) For 208Pb, Z = 82 and N = 126. Add the mass of the neutrons to that of the protons:

1362 Chapter 40 82mp + 126mn = 82 × 1.007825 u + 126 × 1.008665 u = 209.733440 u Subtract the mass of 208Pb from this result:

(82m

p

+ 126mn ) − m 208 Pb = 209.733440 u − 207.976636 u = 1.756804 u

Multiply the mass difference by c2 and convert to MeV:

931.5 MeV/c 2 = 1636 MeV 1u E 1636 MeV = 7.87 MeV and the binding energy per nucleon is b = A 208 Eb = (∆m ) c 2 = (1.756804 u ) c 2 ×

19 • Picture the Problem The nuclear radius is given by R = R0 A1 3 where R0 = 1.2 fm. (a) The radius of 16O is:

R16 O = (1.2 fm )(16 ) = 3.02 fm

(b) The radius of 56Fe is:

R56 Fe = (1.2 fm )(56 ) = 4.59 fm

(c) The radius of 197Au is:

R197 Au = (1.2 fm )(197 ) = 6.98 fm

13

13

13

20 • Picture the Problem The nuclear radius is given by R = R0 A1 3 where R0 = 1.2 fm. The radii of the daughter nuclei are given by:

R = R0 A1 3

Because the ratio of the mass numbers of the daughter nuclei is 3 to 1:

1 3 A1 = (239) and A2 = (239) 4 4

Substitute in equation (1) to obtain:

⎛ 3 × 239 ⎞ R1 = (1.2 fm )⎜ ⎟ ⎝ 4 ⎠

(1)

1/3

= 6.77 fm

and 1/3

⎛ 239 ⎞ R2 = (1.2 fm )⎜ ⎟ ⎝ 4 ⎠

= 4.69 fm

*21 •• Picture the Problem The speed of the neutrons can be found from their thermal energy. The time taken to reduce the intensity of the beam by one-half, from I to I/2, is the half-

Nuclear Physics 1363 life of the neutron. Because the beam is monoenergetic, the neutrons all travel at the same speed. (a) The thermal energy of the neutron is:

Ethermal = kT

(

)

= 1.38 × 10−23 J/K (25 + 273)K = 4.11× 10 −21 J = 4.11× 10 −21 J ×

1eV 1.60 × 10 −19 J

= 25.7 meV (b) Equate Ethermal and the kinetic energy of the neutron to obtain: Solve for v to obtain:

Ethermal = 12 mn v 2 v=

2 Ethermal mn

Substitute numerical values and evaluate v:

v=

2 4.11 × 10 − 21 J = 2.22 km/s 1.67 × 10− 27 kg

(c) Relate the half-life, t1 2 , to the

t1 2 =

x v

t1/2 =

1350 km 1min = 608 s × 2.22 km/s 60 s

speed of the neutrons in the beam: Substitute numerical values and evaluate t1 2 :

(

)

= 10.1min 22 • Picture the Problem We can use the definition of density, the equation for the volume of a sphere, and the given approximation to calculate the density of nuclear matter in grams per cubic centimeter.

m V

Express the density of a spherical nucleus:

ρ=

The approximate mass is:

m = 1.66 × 10−27 kg A

Express the volume of the nucleus:

V = 43 π R0 A1 3 = 43 πR03 A

(

)

(

)

3

1364 Chapter 40 Substitute for m and V to obtain:

ρ=

Substitute numerical values and evaluate ρ:

(1.66 × 10 4 3

(

−27

)

kg A

πR A 3 0

=

3 1.66 × 10− 27 kg 4πR03

ρ=

3 1.66 × 10 −27 kg 3 4π (1.2 fm )

(

) )

= 2.29 × 1017 kg/m 3 = 2.29 × 1014 g/cm3 23 •• Picture the Problem The separation of the nuclei when they are just touching is the sum of their radii, which is given by R = R0 A1 3 . The electrostatic potential energy of the system is given by:

U =k

(Z e)(Z La e ) q1q2 = k Mo R RMo + RLa

where R is the distance from the center of the 95Mo nucleus to the center of the 139La nucleus. The radii of the nuclei are:

1/3 1/3 RMo = Ro AMo and RLa = Ro ALa

Substitute for RMo and RLa and simplify to obtain:

U = ke 2 =

(Z Mo )(Z La ) 1/3 1/3 Ro AMo + Ro ALa

ke 2 (Z Mo )(Z La ) 1/3 1/3 Ro AMo + ALa

(

)

Substitute numerical values and evaluate U:

(8.99 ×10 Nm /C )(1.60 ×10 U= (1.2 ×10 m) 9

2

2

−19

C

)

−15

= 4.71× 10−11 J ×

2

(42)(57 )

(95

1/3

+ 1391/3

)

1eV = 295 MeV 1.60 × 10−19 J

*24 •• Picture the Problem The Heisenberg uncertainty principle relates the uncertainty in position, ∆x, to the uncertainty in momentum, ∆p, by ∆x∆p ≥ Solve the Heisenberg equation for ∆p:

∆p ≈

h 2∆x

1 h. 2

Nuclear Physics 1365 Substitute numerical values and evaluate ∆p:

∆p ≈

1.05 × 10−34 J ⋅ s 2 10 × 10−15 m

(

= 5.25 × 10 The kinetic energy of the electron is given by: Substitute numerical values and evaluate K:

)

− 21

kg ⋅ m/s

K = pc K = (5.25 × 10 −21 kg ⋅ m/s )(3 × 108 m/s ) 1eV = 1.58 × 10 −12 J × 1.60 × 10−19 J = 9.88 MeV

This result contradicts experimental observations that show that the energy of electrons in unstable atoms is of the order of 1 to 1000 eV.

Radioactivity 25 • Picture the Problem The counting rate, as a function of the number of half-lives n, is n given by R = ( 12 ) R0 . (a) The counting rate after n halflives is:

R = ( 12 ) R0 n

ln (R R0 ) ln ( 12 )

Solve for n to obtain:

n=

Substitute numerical values and evaluate n:

⎛ 1000 counts/s ⎞ ⎟ ln⎜⎜ 4000 counts/s ⎟⎠ ⎝ n= =2 ln ( 12 )

Because there are two half-lives in 10 min:

t1 2 = 5 min

(b) At the end of 4 half-lives:

R = ( 12 ) (4000 counts/s ) = 250 Bq 4

26 • Picture the Problem The counting rate, as a function of the number of half-lives n, is given by R = ( 12 ) R0 . n

(a) When t = 4 min, two half-lives will have passed and n = 2:

R = ( 12 ) (2000 counts/s ) = 500 Bq 2

1366 Chapter 40 (b) When t = 6 min, three half-lives will have passed and n = 3: (c) When t = 8 min, four half-lives will have passed and n = 4:

R = ( 12 ) (2000 counts/s ) = 250 Bq 3

R = ( 12 ) (2000 counts/s ) = 125 Bq 4

27 • Picture the Problem The counting rate, as a function of the number of half-lives n, is n given by R = ( 12 ) R0 and the decay constant λ is related to the half-life by t1 2 = ln 2 λ . (a) Relate the counting rate at time t = 10 min to the counting rate at t = 0:

R10 min = ( 12 ) R0 n

ln (R10 min R0 ) ln ( 12 )

Solve for n:

n=

Substitute numerical values and evaluate n:

⎛ 1000 counts/s ⎞ ⎟ ln⎜⎜ 8000 counts/s ⎟⎠ ⎝ n= =3 ln ( 12 )

Therefore, 3 half-lives have passed in 10 min:

3t1 2 = 10 min ⇒ t1 2 = 200 s

(b) The decay constant λ is related to the half-life by:

t1 2 =

Substitute for t1/2 and evaluate λ:

λ=

(c) Six half-lives will have passed in 20 min:

ln 2

λ

⇒λ =

ln 2 t1 2

ln 2 = 3.47 × 10−3 s −1 200 s

R20 min = ( 12 ) (8000 Bq ) = 125 Bq 6

28 • Picture the Problem We can use R = λN 0e − λ t to show that the disintegration rate is approximately 1 Ci. The decay rate is given by:

R = λN 0 e − λ t where N0 is the number of nuclei at t = 0.

Nuclear Physics 1367 The decay constant λ is related to the half-life by:

t1 2 =

Substitute for t1/2 and evaluate λ:

λ=

ln 2

λ

⇒λ =

ln 2 t1 2

ln 2 y = 4.28 × 10−4 y −1 × 1620 y 31.56 Ms

= 1.356 × 10−11 s −1 The number of nuclei at t = 0 is given by:

N0 = NA/M where M is the atomic mass of radium and NA is Avogadro’s number.

Substitute numerical values and evaluate N0:

N0 =

6.02 × 1023 = 2.664 × 1021 226

Substitute numerical values for λ and N0 and evaluate R:

(

)(

)

R = 1.356 × 10 −11 s −1 2.664 × 10 21 e − (1.356×10

−11 −1

s

)(1200 s ) = 3.61× 1010 s −1

≈ 3.7 × 1010 s −1 = 1Ci 29 • n Picture the Problem We can use R = ( 12 ) R0 to relate the counting rate R to the number of half-lives n that have passed since t = 0. The detection efficiency depends on the probability that a radioactive decay particle will enter the detector and the probability that upon entering the detector it will produce a count. If the efficiency is 20 percent, the decay rate must be 5 times the counting rate. (a) When t = 2.4 min, n = 1 and:

R2.4 min = ( 12 ) (1000 counts/s ) = 500 Bq

When t = 4.8 min, n = 2 and:

R4.8 min = ( 12 ) (1000 counts/s ) = 250 Bq

1

2

(b) The number of radioactive nuclei is related to the decay rate R, and the decay constant λ:

R = λN ⇒ N =

The decay constant is related to the half-life:

λ=

R

λ

(1)

0.693 0.693 0.693 = = t1 2 2.4 min 144 s

= 4.813 × 10 −3 s −1 Calculate the decay rate at t = 0

R0 = 5 × 1000 counts/s = 5000 s −1

1368 Chapter 40 from the counting rate:

R0

5000 s −1 = 1.04 × 106 −3 −1 4.813 × 10 s

Substitute in equation (1) and evaluate N0 at t = 0:

N0 =

Calculate the decay rate at t = 2.4 min from the counting rate:

R2.4 min = 5 × 500 counts/s = 2500 s −1

Substitute in equation (1) and evaluate N2.4 min at t = 0:

N 2.4 min =

λ

=

R2.4 min

λ

=

2500 s −1 4.813 × 10 −3 s −1

= 5.19 × 105 (c) The time at which the counting rate will be about 30 counts/s is the product of the number of half-lives that will have passed and the halflife: The counting rate R after n halflives is related to the counting rate at t = 0 by:

t = nt1 2

(2)

R = ( 12 ) R0 n

Solve for n:

n=

ln (R R0 ) ln ( 12 )

Substitute numerical values and evaluate n:

n=

ln (30 counts/s 1000 counts/s ) ln ( 12 )

= 5.059 Substitute numerical values for n and t1/2 in equation (2) and evaluate t:

t = (5.059)(2.4 min ) = 12.1 min

30 • Picture the Problem Knowing each of these reactions, we can use Table 40-1 to find the differences in the masses of the nuclei and then convert this difference into the energy released in each reaction. (a) Write the reaction:

226

Ra →222 Rn + 4 He

Nuclear Physics 1369 Use Table 40-1 to find ∆E:

∆E =

931.5 MeV/c 2 (226.025403 u − 222.017571u − 4.002603 u )c 2 = 4.87 MeV 1u

(b) Write the reaction:

242

Pu →238 U + 4 He

Use Table 40-1 to find ∆E:

∆E =

931.5 MeV/c 2 (242.058737 u − 238.050783 u − 4.002603 u )c 2 = 4.98 MeV 1u

*31 •• Picture the Problem Each 239Pu nucleus emits an alpha particle whose activity, A, depends on the decay constant of 239Pu and on the number N of nuclei present in the ingested 239Pu. We can find the decay constant from the half-life and the number of nuclei present from the mass ingested and the atomic mass of 239Pu. Finally, we can use the dependence of the activity on time to find the time at which the activity be 1000 alpha particles per second. (a) The activity of the nuclei present in the ingested 239Pu is given by:

A = λN

Find the constant for the decay of 239 Pu:

λ=

(1)

ln 2 0.693 = t1 / 2 (24360 y )(31.56 Ms/y )

= 9.02 × 10 −13 s −1 Express the number of nuclei present in the quantity of 239Pu ingested: Substitute numerical values and evaluate N:

Substitute numerical values in equation (1) and evaluate A:

(b) The activity varies with time according to:

N = mPu

NA M Pu

where MPu is the atomic mass of 239Pu.

⎛ 6.02 × 1023 nuclei/mol ⎞ ⎟⎟ N = (2.0 µg )⎜⎜ 239 g/mol ⎝ ⎠ 15 = 5.04 × 10 nuclei A = (9.02 × 10 −13 s −1 )(5.04 × 1015 α ) = 4.55 × 103 α/s

A = Aoe− λt

1370 Chapter 40 Solve for t to obtain:

⎛ A⎞ ln⎜⎜ ⎟⎟ A t= ⎝ o⎠ −λ

Substitute numerical values and evaluate t:

⎛ 1× 103α / s ⎞ ⎟ ln⎜⎜ 4.55 × 103α / s ⎟⎠ ⎝ t= ⎛ 31.56 Ms ⎞ ⎟⎟ − 9.02 × 10−13 s −1 ⎜⎜ ⎝ 1y ⎠

(

)

= 5.32 × 104 y 32 •• Picture the Problem We can use conservation of energy and conservation of linear momentum to relate the momenta and kinetic energies of the nuclei to the decay’s Q value (a) Express the kinetic energies of the alpha particle and daughter nucleus:

pα2 Kα = mα vα = 2mα 1 2

2

(1)

and

pY2 KY = m v = 2mY 1 2

Solve equations (1) and (2) for pα2 and pD2 :

From the conservation of linear momentum we have:

2 Y Y

(2)

pα2 = 2mα Kα and

pY2 = 2mY K Y r r pi = pf or, because the parent is initially at rest, 0 = pα − pY and pα = pY

Because the momenta are equal:

2mY K Y = 2mα Kα

Solve for KY:

KY =

Because the daughter nucleus and the alpha particle share the Q-value:

Q = K Y + Kα

4 mα Kα = Kα mY A−4

4 ⎛ 4 ⎞ + 1⎟ Kα Kα + Kα = ⎜ A−4 ⎝ A−4 ⎠ ⎛ A ⎞ =⎜ ⎟ Kα ⎝ A−4⎠ =

Nuclear Physics 1371 Solve for Kα:

⎛ A−4⎞ Kα = ⎜ ⎟Q ⎝ A ⎠

(b) Substitute for Kα in the expression for Q to obtain:

⎛ A−4⎞ Q = KY + ⎜ ⎟Q ⎝ A ⎠

Solve for KY:

4Q ⎛ A−4⎞ KY = Q − ⎜ ⎟Q = A ⎝ A ⎠

*33 •• Picture the Problem We can write the equation of the decay process by using the fact that the post-decay sum of the Z and A numbers must equal the pre-decay values of the parent nucleus. The Q value in the equations from Problem 32 is given by Q = −(∆m)c2. 239

Pu undergoes alpha decay according to: The Q value for the decay is given by:

239 94

4 Pu → 235 92 U + 2 α + Q

⎛ 931.5 MeV ⎞ Q = [(mPu ) − (mU + mα )]⎜ ⎟ 1u ⎝ ⎠

Substitute numerical values and evaluate Q:

⎛ 931.5MeV ⎞ Q = [(239.052156 u ) − (235.043923 u + 4.002603 u )]⎜ ⎟ = 5.24 MeV 1u ⎝ ⎠ From Problem 32, the kinetic energy of the alpha particle is given by:

⎛ A−4⎞ Kα = ⎜ ⎟Q ⎝ A ⎠

Substitute numerical values and evaluate Kα:

⎛ 239 − 4 ⎞ Kα = ⎜ ⎟(5.24 MeV ) ⎝ 239 ⎠ = 5.15 MeV

From Problem 32, the kinetic energy of the 239U is given by:

KU =

4Q A

Substitute numerical values and evaluate KU:

KU =

4(5.24 MeV ) = 87.7 keV 239

34 • n Picture the Problem We can find the age of the sample using Rn = ( 12 ) R0 to find n and then applying t = nt1 2 . Express the age of the bone in terms of the half-life of 14C and the

t = nt1 2

(1)

1372 Chapter 40 number n of half-lives that have elapsed: The decay rate Rn after n half-lives is related to the counting rate R0 at t = 0 by:

Rn = ( 12 ) R0 n

ln (R R0 ) ln ( 12 )

Solve for n:

n=

Because there are 15.0 decays per minute per gram of carbon in a living organism:

R0 = 15.0

Substitute numerical values for R and R0 and evaluate n:

⎛ 8.1 Bq ⎞ ⎟ ln⎜⎜ 43.75 Bq ⎟⎠ ⎝ n= = 2.433 ln ( 12 )

Substitute numerical values in equation (1) and evaluate t:

decays 1 min × × 175 g min ⋅ g 60 s = 43.75 Bq

t = (2.433)(5730 y ) = 13,940 y

35 • Picture the Problem We can solve R = R0 e − λt for λ to find the decay constant of the sample and use t1 2 =

ln 2

λ

to find its half-life. The number of radioactive nuclei in the

sample initially can be found from R0 = λN 0 . (a) The decay rate is given by:

R = R0e − λt

Solve for λ to obtain:

⎛R⎞ ln⎜⎜ ⎟⎟ R λ= ⎝ 0⎠ −t

Substitute numerical values and evaluate λ:

⎛ 85.2 Bq ⎞ ⎟ ln⎜⎜ 115 Bq ⎟⎠ ⎝ λ= = 0.133 h −1 − 2.25 h

Nuclear Physics 1373 ln 2

ln 2 = 5.20 h 0.133 h −1

The half-life is related to the decay constant:

t1 2 =

(b) The number N0 of radioactive nuclei in the sample initially is related to the decay constant λ and the initial decay rate R0:

R0 = λN 0 ⇒ N 0 =

Substitute numerical values and evaluate N0:

N0 =

=

λ

R0

λ

115 Bq 1h 0.133 h −1 × 3600 s

= 3.11× 106

*36 •• Picture the Problem We can use R0 = λN to find the initial activity of the sample and

R = Ro e − λ t to find the activity of the sample after 1.75 y. (a) The initial activity of the sample is the product of the decay constant λ for 60Co and the number of atoms N of 60Co initially present in the sample:

R0 = λN

Express N in terms of the mass m of the sample, the molar mass M of 60 Co, and Avogadro’s number NA:

N=

(1)

m NA M

Substitute numerical values and evaluate N:

⎛ 1.00 × 10 −6 g ⎞ ⎟⎟ 6.02 × 10 23 nuclei/mol = 1.00 × 1016 nuclei N = ⎜⎜ 60 g/mol ⎠ ⎝

(

)

The decay constant is given by:

λ=

ln2 t1/2

Substitute numerical values and evaluate λ:

λ=

0.693 (5.27 y )(31.56 Ms/y )

= 4.17 × 10 −9 s −1 Substitute numerical values in equation (1) and evaluate A0:

(

)(

R0 = 4.17 × 10 −9 s −1 1.00 × 1016 nuclei = 4.17 × 10 7 s −1 × = 1.13 mCi

1 Ci 3.7 × 1010 s −1

)

1374 Chapter 40 (b) The activity varies with time according to:

R = Ro e

Evaluate R at t = 1.75 y:

−λ t

= Ro e

R = (1.13 mCi ) e

⎛ 0.693t ⎞ −⎜⎜ ⎟⎟ ⎝ 5.27y ⎠

⎛ 0.693×1.75y ⎞ −⎜⎜ ⎟⎟ 5.27y ⎝ ⎠

= 0.898 mCi 37 •• Picture the Problem The following graph was plotted using a spreadsheet program. Excel’s ″Add Trendline″ feature was used to determine the equation of the line. 9

8

ln(R )

7

6

5 ln(R ) = -0.0771t + 8.3395 4

3 0

10

20

30

40

50

60

t (min)

The linearity and negative slope of this graph tell us that it represents an exponential decay. The decay rate equation is:

R = R0e − λt

Take the natural logarithm of both sides of the equation to obtain:

ln R = ln e − λt + ln R0

This equation is of the form:

= −λt + ln R0

y = mx + b where y = ln R, x = t, m = −λ, and b = lnR0.

The decay constant is the negative of the slope of the graph:

λ = 0.0771 min −1

Nuclear Physics 1375 The half-life of the radioisotope is:

t1 2 =

ln 2

λ

=

ln 2 = 8.99 min 0.0771 min −1

38 •• Picture the Problem We can solve Equation 40-7 for λ to show that λ = t1−1ln(R0/R1).

ln 2

(a) Express the half-life as a function of the decay constant λ:

t1 2 =

From Equation 40-7 it follows that:

R0 = eλ t R1

Solve for λ:

⎛R ⎞ ln⎜⎜ 0 ⎟⎟ R ⎛R ⎞ λ = ⎝ 1 ⎠ = t −1 ln⎜⎜ 0 ⎟⎟ t ⎝ R1 ⎠

(b) Substitute numerical values for t, R1, and R0 and evaluate λ:

λ=

Use the decay constant to find the half-life:

t1 2 =

(1)

λ

1 ⎛ 1200 Bq ⎞ ⎟ = 0.00676 s −1 ln⎜ 60 s ⎜⎝ 800 Bq ⎟⎠

ln 2

λ

=

ln 2 = 103 s 0.00676 s −1

39 •• Picture the Problem The required mass is given by M = (5 counts/min)/R, where R is the current counting rate per gram of carbon. We can use the assumed age of the casket to find the number of half-lives that have elapsed and R = ( 12 ) R0 to find the current n

counting rate per gram of 14C. The mass of carbon required is:

Because there were about 15.0 decays per minute per gram of the living wood, the counting rate per gram is: We can find n from the assumed age of the casket and the half-life of 14C:

M =

5 counts/min R

(1)

R = ( 12 ) R0 = ( 12 ) (15 counts/min ⋅ g ) n

n=

n

18,000 y = 3.141 5730 y

1376 Chapter 40 Substitute for n and evaluate R:

R = ( 12 )

3.141

(15 counts/min ⋅ g )

= 1.70 counts/min ⋅ g Substitute for R in equation (1) and evaluate M:

M =

5 counts/min = 2.94 g 1.70 counts/min ⋅ g

40 •• Picture the Problem The decay constant λ can be found from the decay rate R and the number of radioactive nuclei N at the moment of interest and the half-life, in turn, can be found from the decay constant. The decay rate R is related to the decay constant λ and the number of radioactive nuclei N at the moment of interest:

R = λN ⇒ λ =

The number of radioactive nuclei N at the moment of interest can be found from Avogadro’s number, the mass m of the sample, and the molar mass M of the sample:

N = NA

R N

(1)

m M

Substitute numerical values and evaluate N:

N = (6.02 × 1023 nuclei/mol)

Substitute numerical values in equation (1) and evaluate λ:

10 −3 g = 1.004 × 1019 59.934 g / mol

λ=

3.7 × 1010 Bq Ci 1.004 × 1019

1.131Ci ×

= 4.17 × 10−9 s −1 Find the half-life from the decay constant:

ln 2 λ 4.17 × 10−9 s −1 1y = 1.67 × 108 s × 31.56 Ms

t1 2 =

ln 2

=

= 5.27 y

Nuclear Physics 1377 *41 •• Picture the Problem The following graph was plotted using a spreadsheet program. Excel’s ″Add Trendline″ feature was used to determine the equation of the line.

7.0 6.8 6.6 6.4 ln( R )

6.2 6.0 5.8 ln(R ) = -0.198t + 6.9076

5.6 5.4 5.2 5.0 0

1

2

3

4

5

6

7

t (min)

The linearity and negative slope of this graph tells us that it represents an exponential decay. The decay rate equation is:

R = R0e − λt

Take the natural logarithm of both sides of the equation to obtain:

ln R = ln e − λt + ln R0

This equation is of the form:

= −λt + ln R0

y = mx + b where y = ln R, x = t, m = −λ, and b = lnR0.

The half-life of the radioisotope is:

t1 2 =

ln 2

λ

=

ln 2 = 3.50 min 0.198 min −1

42 •• Picture the Problem We can use the decay rate equation R = R0e − λt and the expression relating the half-life of a source to its decay constant to find the half-life of the sample. Solving the decay-rate equation for t will yield the time at which the activity level drops to any given value.

1378 Chapter 40 ln 2

(a) The half-life of the material is given by:

t1 2 =

The decay rate is given by:

R = R0e − λt

Solve for λ:

⎛R ⎞ ln⎜ 0 ⎟ R λ= ⎝ ⎠ t

λ

(1)

(2)

Substitute for λ in equation (1) to obtain:

t1 2 =

ln 2 ln 2 = t ⎛ R0 ⎞ ⎛ R0 ⎞ ln⎜ ⎟ ln⎜ ⎟ ⎝ R⎠ ⎝R⎠ t

Substitute numerical values and evaluate t1/2:

t1 2 =

ln 2 (101h ) ⎛ 115 decays/min ⎞ ⎟⎟ ln⎜⎜ ⎝ 73.5 decays/min ⎠

= 156 h (b) Solve equation (2) for t:

⎛R⎞ ln⎜⎜ ⎟⎟ R t= ⎝ 0⎠ −λ

Express λ in terms of t1/2:

λ=

Substitute for λ in the expression for t to obtain:

⎛R⎞ ln⎜⎜ ⎟⎟ R t = ⎝ 0 ⎠ t1 2 − ln 2

Substitute numerical values and evaluate t:

⎛ 10 decays/min ⎞ ⎟ ln⎜⎜ 115 decays/min ⎟⎠ ⎝ (156 h ) t= − ln 2 1d = 550 h × = 22.9 d 24 h

ln 2 t1 2

43 •• Picture the Problem We can use the decay rate equation R = R0e − λt and the expression relating the half-life of a source to its decay constant to find the age of the fossils.

Nuclear Physics 1379 The decay rate is given by:

R = R0e − λt

Solve for t to obtain:

⎛R⎞ ln⎜⎜ ⎟⎟ R t= ⎝ 0⎠ −λ

Express λ in terms of t1/2:

λ=

Substitute for λ in the expression for t to obtain:

⎛R⎞ ln⎜⎜ ⎟⎟ R t = ⎝ 0 ⎠ t1 2 − ln 2

ln 2 t1 2

or, because the activity at any time is proportional to the number of radioactive nuclei present,

⎛ N ⎞ ln⎜⎜ Rb ⎟⎟ N 0,Rb ⎠ t= ⎝ t1 2 − ln 2

(1)

The number of 87Sr nuclei present in the rocks is given by:

NSr = N 0,Rb − N Rb ⇒ N 0,Rb = N Sr + N Rb

We’re given that:

NSr = 0.01N Rb ⇒

Express the ratio of N0,Rb to NRb:

N 0,Rb N Sr + N Rb N Sr = = +1 N Rb N Rb N Rb

N Sr = 0.01 N Rb

= 0.01 + 1 = 1.01 Substitute numerical values in equation (1) and evaluate the age of the fossils:

⎛ 1 ⎞ ln⎜ ⎟ 1.01 ⎠ ⎝ (4.9 ×1010 y) t= − ln 2 = 7.03 × 108 y

44 ••• Picture the Problem We can evaluate this integral by changing variables to obtain a form that we can find in a table of integrals. Change variables by letting:

x = λt

1380 Chapter 40 Then:

Substitute to obtain:

From integral tables:

dx = λdt , dt = ∞

dx

λ ∞

, and t =

x

τ = ∫ tλe dt = ∫ λe λ 0 0 ∞

∫ xe

− λt

−x

−x

dx

λ

x

λ =

1

∞

xe λ∫

−x

dx

0

dx = 1

0

Substitute in the expression for τ to obtain:

τ=

1

λ

Nuclear Reactions 45 • Picture the Problem We can use Q = −(∆m )c 2 to find the Q values for these reactions. (a) Find the mass of each atom from Table 40-1:

m1 H = 1.007825 u m 3 H = 3.016049 u m 3 He = 3.016029 u mn = 1.008665 u

Calculate the initial mass mi of the incoming particles: Calculate the final mass mf:

mi = 1.007825 u + 3.016049 u = 4.023874 u mf = 3.016029 u + 1.008665 u = 4.024694 u

Calculate the increase in mass:

∆m = mf − mi = 4.024694 u − 4.023874 u = 0.000820 u

Calculate the Q value:

Q = −(∆m ) c 2 ⎛ MeV/c 2 ⎞ ⎟⎟ = −(0.000820 u ) c 2 ⎜⎜ 931.5 u ⎝ ⎠ = − 0.764 MeV

Nuclear Physics 1381 (b) Proceed as in (a) to obtain:

⎛ MeV/c 2 ⎞ 2 ⎟⎟c Q = (0.003510 u )⎜⎜ 931.5 u ⎠ ⎝ = 3.27 MeV

Remarks: Because Q < 0 for the first reaction, it is endothermic. Because Q > 0 for the second reaction, it is exothermic. 46 • Picture the Problem We can use Q = −(∆m )c 2 to find the Q values for these reactions. (a) Find the mass of each atom from Table 40-1:

m 2 H = 2.014102 u m 3 H = 3.016049 u m1 H = 1.007825 u

Calculate the initial mass mi of the incoming particles: Calculate the final mass mf:

mi = 2(2.014102 u ) = 4.028204 u mf = 3.016049 u + 1.007825 u = 4.023874 u

Calculate the increase in mass:

∆m = mf − mi = 4.023874 u − 4.028204 u = −0.004330 u

Calculate the Q value:

Q = −(∆m ) c 2 ⎛ MeV/c 2 ⎞ ⎟⎟ = −(− 0.004330 u ) c ⎜⎜ 931.5 u ⎝ ⎠ 2

= 4.03 MeV (b) Proceed as in (a) to obtain:

Q = −(∆m ) c 2 ⎛ MeV/c 2 ⎞ 2 ⎟⎟c = −(− 0.019703 u )⎜⎜ 931.5 u ⎝ ⎠ = 18.4 MeV

1382 Chapter 40 (c) Proceed as in (a) to obtain:

Q = −(∆m ) c 2 ⎛ MeV/c 2 ⎞ 2 ⎟⎟c = −(− 0.005135 u )⎜⎜ 931.5 u ⎝ ⎠ = 4.78 MeV

*47 •• Picture the Problem We can use Q = −(∆m )c 2 to find the Q values for this reaction. (a) The masses of the atoms are:

m 14 C = 14.003 242 u m14 N = 14.003 074 u

Calculate the increase in mass:

∆m = mf − mi = 14.003 074 u − 14.003 242 u = −0.000168 u

Calculate the Q value:

Q = −(∆m )c 2 ⎛ MeV/c 2 ⎞ ⎜ ⎟⎟ = −(− 0.000168 u ) c ⎜ 931.5 u ⎝ ⎠ 2

= 0.156 MeV (b)

The masses given are for atoms, not nuclei, so for nuclear masses the masses are too large by the atomic number times the mass of an electron. For the given nuclear reaction, the mass of the carbon atom is too large by 6me and the mass of the nitrogen atom is too large by 7 me . Subtracting 6me from both sides of the reaction equation leaves an extra electron mass on the right. Not including the mass of the beta particle (electron) is mathematically equivalent to explicitly subtracting 1me from the right side of the equation. 48 •• Picture the Problem We can use Q = −(∆m ) c 2 to find the Q values for this reaction. (a) The masses of the atoms are:

m13 N = 13.005 738 u m13 C = 13.003 354 u

Nuclear Physics 1383 For β+ decay:

Q = (mi − mf − 2me ) c 2

= (mi − mf )c 2 − 2me c 2

Calculate mi − mf:

mi − mf = 13.005 738 u − 13.003 354 u

= 0.002 384 u Calculate the Q value:

⎛ MeV/c 2 ⎞ ⎟⎟ − 2(0.511 MeV ) = 1.20 MeV Q = (0.002 384 u )c 2 ⎜⎜ 931.5 u ⎝ ⎠

The atomic masses include the masses of the electrons of the neutral atoms. In this reaction the initial atom has 7 electrons and the final atom only has (b)

6 electrons. Moreover, in addition to one electron not included in the atomic masses, a positron of mass equal to that of an electron is created. Consequently, one must add the rest energies of two electrons to the rest energy of the daughter atomic mass when calculating Q.

Fission and Fusion *49 • Picture the Problem The power output of the reactor is the product of the number of fissions per second and energy liberated per fission. Express the required number N of fissions per second in terms of the power output P and the energy released per fission Eper fission:

N=

Substitute numerical values and evaluate N:

N=

P Eper fission

500 MW 200 MeV

J 1 eV 5 × 108 × s 1.60 × 10 −19 J = 200 MeV = 1.56 × 1019 s −1

1384 Chapter 40 50 • Picture the Problem If k = 1.1, the reaction rate after N generations is 1.1N. We can find the number of generations by setting 1.1N equal, in turn, to 2, 10, and 100 and solving for N. The time to increase by a given factor is the number of generations N needed to increase by that factor times the generation time. (a) Set 1.1N equal to 2 and solve for N:

(1.1)N

(b) Set 1.1N equal to 10 and solve for N:

(1.1)N

(c) Set 1.1N equal to 100 and solve for N:

(1.1)N

(d) Multiply the number of generations by the generation time:

=2 N ln 1.1 = ln 2 ln 2 N= = 7.27 ln 1.1 = 10 N ln 1.1 = ln 10 ln 10 N= = 24.2 ln 1.1 = 100 N ln 1.1 = ln 100 ln 100 N= = 48.3 ln 1.1

t2 = Nt1 = (7.27 )(1 ms ) = 7.27 ms t10 = Nt1 = (24.2 )(1 ms ) = 24.2 ms t100 = Nt1 = (48.3)(1 ms ) = 48.3 ms

(e) Multiply the number of generations by the generation time:

t2 = Nt1 = (7.27 )(100 ms ) = 0.727 s t10 = Nt1 = (24.2 )(100 ms ) = 2.42 s t100 = Nt1 = (48.3)(100 ms ) = 4.83 s

*51 •• Picture the Problem We can use Q = −(∆m)c2, where ∆m = mf − mi, value. The Q value is given by:

Calculate the change in mass ∆m:

Q = −(∆m ) c 2 ×

to calculate the Q

931.5 MeV/c 2 1u

Nuclear Physics 1385 ∆m = mf − mi

= 94.905842 u + 138.906348 u + 2(1.008665 u ) − (235.043923 u + 1.008665 u ) = −0.223068 u

Substitute for ∆m and evaluate Q:

Q = −(− 0.223068 u ) ×

931.5 MeV 1u

= 208 MeV The ratio of Q to U found in Problem 23 is:

Q 208 MeV = = 88.1% U 236 MeV

52 •• Picture the Problem We can find the number of neutrons per second in the generation of 4 W of power from the number of reactions per second. The number of neutrons emitted per second is:

N n = 12 N where N is the number of reactions per second.

The number of reactions per second is:

J 1eV ⎛ ⎞ ⎜ 4 × ⎟ −19 s 1.60 × 10 J ⎟ N = 2⎜ ⎜ 3.27 MeV + 4.03 MeV ⎟ ⎟ ⎜ ⎝ ⎠ 12 −1 = 6.85 × 10 s

Substitute for N and evaluate Nn:

Nn =

1 2

(6.85 ×10

12

s −1

)

= 3.43 × 1012 neutrons/s 53 •• Picture the Problem We can use the energy released in the reactions of Problem 50, together with the 17.6 MeV released in the reaction described in this problem, to find the energy released using 5 2H nuclei. Finding the number of D atoms in 4 L of H2O, we can then find the energy produced if all of the 2H nuclei undergo fusion. Find the energy released using 5 2H nuclei:

Q = 3.27 MeV + 4.03 MeV + 17.6 MeV = 24.9 MeV

1386 Chapter 40 ⎛ mH 2O ⎞ ⎟⎟ N A N H = 2⎜⎜ ⎝ 18 g/mol ⎠

The number of H atoms in 4 L of H2O is: Substitute numerical values and evaluate NH:

⎛ 4 kg ⎞ ⎟⎟ 6.02 × 10 23 atoms/mol = 2.676 × 10 26 N H = 2⎜⎜ ⎝ 18 g/mol ⎠

(

The number of D atoms in 4 L of H2O is:

)

( ) = (1.5 × 10 )(2.676 × 10 )

N D = 1.5 × 10−4 N H −4

26

= 4.01× 10 22 ND Q 5

The energy produced is given by:

E=

Substitute numerical values and evaluate E:

4.01 × 10 22 (24.9 MeV ) E= 5 1.60 × 10 −19 J = 1.997 × 10 23 MeV × eV

= 3.20 × 1010 J *54 ••• Picture the Problem We can use the conservation of momentum and the given Q value to find the final energies of both the 4He nucleus and the neutron, assuming that the initial momentum of the system is zero. Apply conservation of energy to obtain: Apply conservation of momentum to obtain: Solve equation (2) for vHe:

2 in equation (1): Substitute for vHe

2 18.6 MeV = 12 mHe vHe + 12 mn vn2

(1)

= K He + K n

mHe vHe + mn vn = 0

(2)

2

vHe

⎛m ⎞ mv 2 = − n n ⇒ vHe = ⎜⎜ n ⎟⎟ vn2 mHe ⎝ mHe ⎠ 2

⎛m ⎞ 18.6 MeV = 12 mHe ⎜⎜ n ⎟⎟ vn2 + 12 mn vn2 ⎝ mHe ⎠ or

Nuclear Physics 1387 ⎛ m ⎞ 18.6 MeV = 12 mn vn2 ⎜⎜1 + n ⎟⎟ ⎝ mHe ⎠ ⎛ m ⎞ = K n ⎜⎜1 + n ⎟⎟ ⎝ mHe ⎠ Solve for Kn:

Kn =

18.6 MeV m 1+ n mHe

Substitute numerical values for mn and mHe and evaluate Kn:

Kn =

18.6 MeV = 14.86 MeV 1.008665 u 1+ 4.002603 u

Use equation (1) to find KHe:

K He = 18.6 MeV − K n = 18.6 MeV − 14.86 MeV = 3.74 MeV

55 ••• Picture the Problem Adding the three reactions will yield their net effect. We can use (∆m)c2 to find the rest energy released in the cycle and find the rate of proton consumption from the ratio of the sun’s power output to the released per proton in fusion. (a) Add the three reactions to obtain: 1

H + 1H + 1H + 2H + 1H + 3He → 2H + β + + νe + 3He + γ + 4He + β

Simplify to obtain:

(b) Express the rest energy released in this cycle:

+

+ νe

41 H→ 4 He + 2 β + + 2ν e + γ

(∆m )c 2 = (4mp − mα − 4me )c 2

Use Table 40-1 to find the masses of the participants in the reaction and evaluate (∆m)c2:

(∆m )c 2 = [4(1.007825 u ) − 4.002603 u ]c 2 × 931.5 MeV/c u

= 24.7 MeV

2

− 4(0.511 MeV )

1388 Chapter 40 (c) Express the rate R of proton consumption:

R=

P E

(1)

where E is the energy released per proton in fusion. Find N, the number of protons in the sun:

N=

1 2

(

msun 12 1.99 × 1030 kg = mp 1.67 × 10−27 kg

)

= 5.96 × 1056 where we have assumed that protons constitute about half of the total mass of the sun.

E=

The energy released per proton in fusion is:

1 4

(26.7 MeV ) = 6.675 MeV

1.60 × 10 −19 J = 6.675 MeV × eV −12 = 1.07 × 10 J

Substitute numerical values in equation (1) and evaluate R:

R=

The time T for the consumption of all protons is:

T=

4 × 10 26 W = 3.74 × 1038 s −1 −12 1.07 × 10 J

N 5.96 × 1056 = R 3.74 × 1038 s −1 1y = 1.59 × 1018 s × 31.56 Ms = 5.04 × 1010 y

General Problems 56 • Picture the Problem We can use the values of k, e, h, and c and the appropriate conversion factors to show that ke2 = 1.44 MeV⋅fm and hc = 1240 MeV⋅fm (a) Evaluate ke2 to obtain:

(

)(

)

2

ke 2 = 8.99 × 109 N ⋅ m 2 / C 2 1.60 × 10 −19 C = 2.307 × 10 −28 N ⋅ m 2 1 eV = 2.307 × 10 −28 J ⋅ m × = 1.44 × 10 −9 eV ⋅ m 1.60 × 10 −19 J 1 fm 1 MeV = 1.44 × 10 −9 eV ⋅ m × −15 × 6 = 1.44 MeV ⋅ fm 10 m 10 eV

Nuclear Physics 1389 (b) Evaluate hc to obtain:

(

)(

)

hc = 6.63 × 10 −34 J ⋅ s 3 × 108 m/s = 1.99 × 10−25 J ⋅ m = 1.99 × 10−25 J ⋅ m ×

1eV = 1240 × 10 −9 eV ⋅ m −19 1.60 × 10 J

= 1240 × 10−9 eV ⋅ m ×

1fm 1 MeV × 6 = 1240 MeV ⋅ fm −15 10 m 10 eV

*57 • Picture the Problem We can use the given information regarding the half-life of the source to find its decay constant. We can then plot a graph of the counting rate as a function of time. The decay constant is related to the half-life of the source:

λ=

ln 2 ln 2 = = 0.0693 s −1 t1 2 10 s

−1 R = R0 e − λt = (6400 Bq ) e − (0.0693 s )t

The activity of the source is given by:

−1 The following graph of R = (6400 Bq ) e − (0.0693 s )t was plotted using a spreadsheet

program.

7000 6000

R (Bq)

5000 4000 3000 2000 1000 0 0

10

20

30

40

50

60

t (s)

58 • Picture the Problem The energy needed to remove a neutron is given by Q = (∆m ) c 2 where ∆m is the difference between the sum of the masses of the reaction products and the mass of the target nucleus.

1390 Chapter 40 (a) The reaction is: The masses are (see Table 40-1):

4

He + Q= 3 He + n

m 4 He = 4.002603 u m 3 He = 3.016029 u mn = 1.008665 u

Calculate the final mass:

mf = 3.016029 u + 1.008665 u = 4.024694 u

Calculate the increase in mass:

∆m = mf − mi = 4.024694 u − 4.002603 u = 0.022091 u

Calculate the energy Q needed to remove a neutron from 4He:

Q = (∆m )c 2 ⎛ MeV/c 2 ⎞ ⎟⎟ = (0.022091u )c 2 ⎜⎜ 931.5 u ⎝ ⎠ = 20.6 MeV

(b) The reaction is: The masses are (see Table 40-1):

7

Li + Q= 6 Li + n

m 7 Li = 7.016004 u m 6 Li = 6.015122 u mn = 1.008665 u

Calculate the final mass:

mf = 6.015122 u + 1.008665 u = 7.023787 u

Calculate the increase in mass:

∆m = mf − mi = 7.023787 u − 7.016004 u = 0.007783 u

Calculate the energy Q needed to remove a neutron from 4He:

Q = (∆m )c 2 ⎛ 931.5 MeV/c 2 ⎞ ⎟⎟ = (0.007783 u )c ⎜⎜ 1u ⎝ ⎠ 2

= 7.25 MeV

Nuclear Physics 1391 59 • Picture the Problem The maximum kinetic energy of the electron is given by K max = Q = m14 C − m14 N c 2 .

(

)

The maximum kinetic energy of the electron is the Q value for the reaction: Find the mass of each atom from Table 40-1: Calculate ∆m = m14 C − m14 N :

Calculate the maximum kinetic energy of the electron:

(

)

K max = Q = m14 C − m14 N c 2

m14 C = 14.003242 u m14 N = 14.003074 u ∆m = 14.003242 u − 14.003074 u = 0.000168 u Q = (∆m ) c 2 ⎛ MeV/c 2 ⎞ ⎟⎟ = (0.000168 u ) c 2 ⎜⎜ 931.5 u ⎝ ⎠ = 156 keV

60 • Picture the Problem We can use the definition density to find the radius of the neutron star. Relate the mass of the neutron star to the mass of the sun M, the volume V of the star and the nuclear density ρ:

M = ρV = 43 πρR 3 where R is the radius of the star.

Solve for R:

R=3

In Problem 20 it was established that:

ρ = 1.174 × 1017 kg/m 3

Substitute numerical values and evaluate R:

R=3

3M 4πρ

(

)

3 1.99 × 10 30 kg 4π 1.174 × 1017 kg/m 3

(

= 15.9 km

)

1392 Chapter 40 *61 •• Picture the Problem We can show that 109Ag is stable against alpha decay by demonstrating that its Q value is negative. The Q value for this reaction is:

⎛ MeV/c 2 ⎞ ⎟⎟ Q = − (mRh + mα ) − mAg c 2 ⎜⎜ 931.5 u ⎝ ⎠

[

]

Substitute numerical values and evaluate Q:

Q = −[(4.002603 u + 104.905250 u ) − 108.904756 u ](931.5MeV/u ) = − 2.88 MeV Remarks: Alpha decay occurs spontaneously and the Q value will equal the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus, Q = K α + K D . Kinetic energy cannot be negative; hence, alpha decay cannot occur unless the mass of the parent nucleus is greater than the sum of the masses of the alpha particle and daughter nucleus, m P > mα + m D . Alpha decay cannot take place unless the total rest mass decreases. 62 •• Picture the Problem We can use E threshold = hf threshold = hc λthreshold , where Ethreshold is the binding energy of the deuteron, to find the threshold wavelength for the given nuclear reaction. Express the threshold energy of the photon:

Ethreshold = hf threshold =

Solve for the threshold wavelength:

λthreshold =

hc

λthreshold

hc Ethreshold

(1)

The threshold energy equals the binding energy of the deuteron:

[

]

Ethreshold = EB = mD − (mp + mn ) c 2 ×

931.5 MeV/c 2 u

Substitute numerical values and evaluate Eth, the energy that must be added to the deuteron that will cause it to fission:

Ethreshold = [2.014102 u − (1.007825 u + 1.008665 u )](931.5 MeV/u ) = −2.22 MeV ×

1.60 × 10 −19 J = −3.55 × 10 −13 J eV

Nuclear Physics 1393 Substitute numerical values in equation (1) and evaluate λthreshold:

λthreshold =

(6.63 ×10

−34

)(

J ⋅ s 3.00 × 108 m/s 3.55 × 10 −13 J

)

= 5.60 × 10 −13 m = 0.560 pm 63 • Picture the Problem The activity of a radioactive source is the product of the number of radioactive nuclei present and their decay constant. The activity of the isotope 40K in the student is:

R = N 40 λ =

N 40 ln 2 t1 2

Find N, the number of K nuclei in the student:

N = 0.0036

mN A M

(1)

where m is the mass of the student and M is the atomic mass of K. Substitute numerical values and evaluate N:

N = 0.0036

(60 kg )(6.02 ×1023 nuclei/mol) = 3.326 ×1024 39.098 g/mol

The number N40 of 40K nuclei in the student is the product of the relative abundance and the number of K nuclei in the student:

N 40 = Relative abundance × N

Substitute numerical values in equation (1) and evaluate R:

R=

= (1.2 × 10 −4 )(3.326 × 10 24 ) = 3.991× 10 20

(3.991×10 )ln 2 20

1.3 × 109 y ×

31.56 Ms y

= 6.74 × 103 Bq 64 •• Picture the Problem We can find the energy released in the reaction β + + β − → Q by recognizing that a total of 2 electron masses are converted into energy in this annihilation. The energy released when a positron-electron pair annihilate is given by: Substitute numerical values and evaluate Q:

Q = E = 2me c 2

1394 Chapter 40

(

)(

)

2

Q = 2 9.11× 10−31 kg 3 × 108 m/s = 1.64 × 10 −13 J ×

1eV = 1.02 MeV 1.60 × 10−19 J

65 •• Picture the Problem We can use the fact that, after n half-lives, the decay rate of the 24Na n isotope is R = ( 12 ) R0 , where R0 is its decay rate at t = 0. The counting rate after n half-lives is related to the initial counting rate:

R = ( 12 ) R0 n

Divide both sides of the equation by the volume V of blood in the patient:

R 1 n R0 = (2 ) V V

We’re given that n = 2/3, R0 = 600 kBq, and, after n half-lives, the decay rate per unit volume is 60 Bq/mL:

60 Bq/mL = ( 12 )

Solve for and evaluate V:

V = ( 12 )

23

23

600 kBq V

600 kBq = 6.30 × 103 mL 60 Bq/mL

= 6.30 L *66 •• Picture the Problem We can solve this problem in the center of mass reference frame for the general case of an α particle in a head-on collision with a nucleus of atomic mass M u and then substitute data for a nucleus of 197Au and a nucleus of 10B. In the CM frame, the kinetic energy is:

K CM =

At the point of closest approach:

K CM =

K lab K lab = 4u m 1+ α 1+ M M kq1q2 k (2e )(Ze ) ke 2 2Z = = Rmin Rmin Rmin

or, because ke2 = 1.44 MeV⋅fm,

K CM =

Solve for Rmin to obtain:

Rmin =

(1.44 MeV ⋅ fm )(2Z ) Rmin

(1.44 MeV ⋅ fm )(2Z ) K CM

(1)

Nuclear Physics 1395 (a) Neglecting the recoil of the target nucleus is equivalent to replacing KCM by Klab. Evaluate equation (1) for 197Au:

Rmin =

Evaluate equation (1) for 10B:

Rmin =

(1.44 MeV ⋅ fm )(2 × 79) 8 MeV

= 28.4 fm

(1.44 MeV ⋅ fm )(2 × 5) 8 MeV

= 1.80 fm (b) Find KCM for the 197Au nucleus:

K CM =

Substitute numerical values in equation (1) and evaluate Rmin:

Rmin =

8 MeV = 7.841 MeV 4u 1+ 197 u

(1.44 MeV ⋅ fm )(2 × 79) 7.841 MeV

= 29.0 fm Note that this result is about 2% greater that Rmin calculated ignoring recoil. Find KCM for the 10B nucleus:

K CM =

Substitute numerical values in equation (1) and evaluate Rmin:

Rmin =

8 MeV = 5.714 MeV 4u 1+ 10 u

(1.44 MeV ⋅ fm )(2 × 5) 5.714 MeV

= 2.52 fm Note that this result is about 40% greater that Rmin calculated ignoring recoil. 67 •• Picture the Problem The allowed energy levels in a one-dimensional infinite square well

⎛ h2 ⎞ ⎟. 2 ⎟ ⎝ 8mL ⎠

are given by Equation 35-13: En = n 2 ⎜⎜

1396 Chapter 40 (a) The lowest energy of a nucleon of mass 1 u in the well corresponds to n = 1:

(6.63 ×10

) )

2

−34

J ⋅s E1 = 2 −27 8(1 u ) 1.66 × 10 kg/u (3 fm ) 1eV = 3.678 × 10 −12 J × 1.60 × 10 −19 J

(

= 23.0 MeV (b) Because neutrons are fermions, there can be only two per state:

(

E = 2(E1 + E2 + E3 + E4 + E5 + E6 ) = 2 E1 + 22 E1 + 32 E1 + 4 2 E1 + 52 E1 + 6 2 E1

)

= 182 E1 = 182(23.0 MeV ) = 4.19 GeV (c) Find E for 4 protons and 4 neutrons:

(

E = 4(E1 + E2 + E3 ) = 4 E1 + 22 E1 + 32 E1

)

= 56 E1 = 56(23.0 MeV ) = 1.29 GeV 68 •• Picture the Problem We can apply BE = (∆m)c2 to the model to find the binding energies and the binding energies/bond. (a) Find the binding energy BE for this model:

BE = (4mα − m16 O )c 2

= [4(4.002603 u ) − 15.994915 u ]c 2 = (0.015497 u ) c 2

There are 6 bonds for the regular tetrahedron:

BE = 16 BE = bond =

1 6

1 6

(0.015497 u )c 2

(0.015497 u )c 2 × 931.5 MeV/u 2 c

= 2.406 MeV (b) 12C has 3 pairwise α particle bonds. Find the total BE for 12C with this model: Calculate BE(4He):

BE

( C) = 3 × BE( He)+ 3(2.406 MeV ) 12

4

Nuclear Physics 1397

(

) [

]

BE 4 He = 2(mp + mn ) − m 4 He c 2 = [2(1.007825 u + 1.008665 u ) − 4.002603 u ]c 2 ×

931.5 MeV/c 2 u

= 28.30 MeV Substitute numerical values and evaluate BE

BE

( C): 12

( C) = 3(28.30 MeV ) + 3(2.406 MeV ) = 12

Use Table 40-1 to find BE

92.1 MeV

( C): 12

[

]

BE(12 C ) = 6(mp + mn ) − m12 C c 2 = [6(1.007825 u + 1.008665 u ) − 12.000000 u ]c 2 ×

931.5 MeV/c 2 u

= 92.2 MeV Note that this result is good agreement with the model. 69 •• Picture the Problem We can separate the variables in the differential equation dN/dt = Rp – λN and integrate to express N as a function of t. When dN/dt ≈ 0, Rp – λN∞ = 0, a condition we can use to find N∞. (a) Separate the variables in the differential equation to obtain:

dN = dt Rp − λN

Integrate the left side of the equation from 0 to N and the right side from 0 to t to obtain:

N

Let u = Rp − λN′. Then:

du = −λdN '

t

dN ' ∫0 Rp − λN ' = ∫0 dt '

and l

l

N

2 dN' 1 2 du 1 = − = − u ln ∫0 Rp − λN' λ l∫ u λ l 1 1

=−

1

λ

ln (Rp − λN ')

N

0

1 1 = − ln (Rp − λN ) + ln (Rp )

λ

=

⎛ Rp ⎞ ⎟ ln⎜ λ ⎜⎝ Rp − λN ⎟⎠ 1

λ

1398 Chapter 40 ⎛ Rp ⎞ ⎟=t ln⎜ λ ⎜⎝ Rp − λN ⎟⎠

t

Because

1

∫ dt ' = t : 0

Solve for N to obtain:

N=

(1 − e ) λ

Rp

−λ t

The following graph of N(t) = (Rp/λ)(1 −e−λt) was plotted using a spreadsheet program. Note that N(t) approaches Rp/λ in the same manner that the charge on a capacitor approaches the value CV.

1.2

1.0

N

0.8

0.6

0.4

0.2

0.0 0

1

2

3

4

λt

(b) When dN/dt = 0:

The decay constant is:

Substitute for λ to obtain:

Substitute numerical values and evaluate N∞:

Rp − λN ∞ = 0 ⇒ N ∞ =

λ=

Rp

λ

ln 2 t1 2

N∞ =

N∞ =

Rp ln 2

t1 2

100 s −1 ⎛ 60 s ⎞ ⎜10 min × ⎟ ln 2 ⎝ min ⎠

= 8.66 × 104

Nuclear Physics 1399 *70 •• Picture the Problem The mass of 235U required is given by m235 =

N M 235 , where NA

M235 is the molecular mass of 235U and N is the number of fissions required to produce 7.0×1019 J. Relate the mass of 235U required to the number of fissions N required:

m235 =

N M 235 NA

(1)

where M235 is the molecular mass of 235U. Determine N:

N=

Substitute numerical values and evaluate N:

N=

Eannual Eper fission

7.0 × 1019 J 1.60 × 10−19 J 200 MeV × eV 30 = 2.18 × 10

Substitute numerical values in equation (1) and evaluate m235:

m235

2.18 × 1030 (235 g/mol) = 8.51×105 kg = 23 6.02 × 10 nuclei/mol

71 •• Picture the Problem In the ground state of a one-dimensional infinite square well of length L the wavelength of a particle is 2L. We can use de Broglie’s equation to find p for the particle and the relationship E 2 = E02 + p 2 c 2 with E0 << pc to show that E ≈ pc. (a) In the ground state of a onedimensional infinite square well of length L:

λ = 2 L = 2(2 fm ) = 4.00 fm

h

hc λc

(b) Use de Broglie’s relation to obtain:

p=

Substitute numerical values and evaluate p:

p=

(c) Relate the total energy of the electron to its rest energy and

⎛ E2 ⎞ E 2 = E02 + p 2c 2 = p 2c 2 ⎜⎜1 + 2 0 2 ⎟⎟ pc ⎠ ⎝

λ

=

1240 eV ⋅ nm = 310 MeV/c (4 fm )c

1400 Chapter 40 momentum: Because E0 << pc:

(d) The kinetic energy of an electron in the ground state of this well is given by:

E 2 ≈ p 2c 2 ⇒ E ≈ pc K = E − E0 ≈ E = pc = (310 MeV/c )c = 310 MeV

72 •• Picture the Problem When a single proton is removed from a 12C nucleus, a 11B nucleus remains and we can use Q = ∆mc 2 to determine the minimum energy required to remove a proton.

C + Q→115 B+11 H

The nuclear reaction is:

12 6

The minimum energy Q required is:

Q = m11 B + m1 H − m12 C c 2

(

)

Substitute numerical values and evaluate Q:

⎛ 931.5 MeV ⎞ Q = [(11.009306 u + 1.007825 u ) − 12.000000 u ]⎜ ⎟ = 16.0 MeV u ⎝ ⎠ *73 ••• Picture the Problem The momentum of the electron is related to its total energy through E 2 = p 2 c 2 + E02 and its total relativistic energy E is the sum of its kinetic and rest energies. (a) Relate the total energy of the electron to its momentum and rest energy: The total relativistic energy E of the electron is the sum of its kinetic energy and its rest energy: Substitute for E in equation (1) to obtain: Solve for p:

Substitute numerical values and evaluate p:

E 2 = p 2 c 2 + E02

E = K + E0

(K + E0 )2 = p 2c 2 + E02 p=

K ( K + 2 E0 ) c

(1)

Nuclear Physics 1401 p=

(0.782 MeV )(0.782 MeV + 2 × 0.511MeV )

(b) Because pp = −pe:

c Kp =

Substitute numerical values (see Table 7-1 for the rest energy of a proton) and evaluate Kp:

Kp =

(c) The percent correction is:

Kp K

=

= 1.188 MeV/c

p p2 2mp

(1.188 MeV/c )2

(

2 938.28 MeV/c 2

)=

752 eV

752 eV = 0.0962% 0.782 MeV

74 ••• Picture the Problem Conservation of momentum and conservation of energy allow us to find the final velocities. Because the initial kinetic energy of the nucleus is zero, its final kinetic energy equals the energy lost by the neutron. (a) Apply conservation of momentum to the collision to obtain:

(m + M )V

mvL m+M

Solve for V:

V =

(b) In the CM frame, VMi = V and so:

VMi = V

In the CM frame, Vf = −Vi and so:

VMf = − V

(c) Use conservation of momentum to obtain one relation for the final velocities: The equality of the initial and final kinetic energies provides a second equation relating the two final velocities. This is implemented by equating the speeds of recession and approach:

= mvL

mvL = mvf + MVMf

(1)

VMf − vf = −(VMi − vL ) = 0 + vL and so

vf = VMf − vL

1402 Chapter 40 mvL = m(VMf − vL ) + MVMf

To eliminate vf, substitute in equation (1) :

2m vL M +m

Solve for VMf:

V Mf =

(d) The kinetic energy of the nucleus after the collision in the laboratory frame is:

K M = 12 MVM2f

2

Substitute for VMf and simplify to obtain:

K Mf

⎛ 2m ⎞ = M⎜ vL ⎟ ⎝M +m ⎠ 4mM 1 2 mv = (M + m )2 2 L 1 2

(

)

(e) The fraction of the energy lost by the neutron in the elastic collision is given by:

(

)

4mM 1 2 2 mvL 4mM ∆E − K Mf (M + m )2 = 1 2 = = = 2 1 E (M + m )2 2 mvL 2 mvL

4m M = 2 2 m⎞ m⎞ ⎛ ⎛ 1 + M 2 ⎜1 + ⎟ ⎜ ⎟ ⎝ M⎠ ⎝ M⎠ 4mM

75 ••• Picture the Problem We can use the result of Problem 74, part (e), to find the fraction f = Ef E0 of its initial energy lost per collision and then use this result to show that, after N collisions, E = (0.714)NE0. (a) Determine f = Ef E0 per collision: From Problem 74, part (e):

Substitute for ∆E/E0 in the expression for f to obtain:

f =

E0 − ∆E ∆E = 1− E0 E0

∆E = E0

4m

m⎞ ⎛ M ⎜1 + ⎟ ⎝ M⎠

f = 1−

2

4m

m⎞ ⎛ M ⎜1 + ⎟ ⎝ M⎠

2

Nuclear Physics 1403 4(1.008665 u )

Substitute numerical values and evaluate f:

f = 1−

After N collisions:

E fN = f N E0 =

(b) Solve equation (1) for N:

⎛ E fN ⎞ ⎟ ln⎜⎜ E 0 ⎟⎠ ⎝ N= ln(0.714)

Substitute numerical values and evaluate N:

⎛ 0.02 eV ⎞ ⎟⎟ ln⎜⎜ 2 MeV ⎠ = 54.7 N= ⎝ ln (0.714 )

⎛ 1.008665 u ⎞ ⎟⎟ 12.000000 u ⎜⎜1 + ⎝ 12.000000 u ⎠ = 0.714

(0.714)N E0

2

(1)

55 head - on collisions are required to reduce the energy of the neutron from 2 MeV to 0.02 eV. 76 ••• Picture the Problem We can use the result of Problem 74, part (e), to find the fraction f = Ef E0 of its initial energy lost per collision. Note the difference between the energy loss per collision specified here and that of the preceding problem. In the preceding problem it was assumed that all collisions are head-on collisions. (a) Determine f = Ef E0 per collision: In a collision with a hydrogen atom:

f =

∆E E0 − ∆E = 1− E0 E0

∆E = 0.63 E0 and so

f = 0.37

After N collisions: Solve equation (1) for N:

E fN = f N E0 = (0.37 ) E0 N

⎛E ⎞ ln⎜⎜ fN ⎟⎟ E N= ⎝ 0 ⎠ ln (0.37 )

(1)

(2)

1404 Chapter 40 Substitute numerical values and evaluate N:

⎛ 0.02 eV ⎞ ⎟ ln⎜⎜ 2 MeV ⎟⎠ ⎝ N= = 18.5 ln (0.37 )

19 head - on collisions with an atom of hydrogen are required to reduce the energy of the neutron from 2 MeV to 0.02 eV. (b) In a collision with a carbon atom:

∆E = 0.11 E0 and so

f = 0.89

Equation (2) becomes:

⎛E ⎞ ln⎜⎜ fN ⎟⎟ E N= ⎝ 0 ⎠ ln (0.89 )

Substitute numerical values and evaluate N:

⎛ 0.02 eV ⎞ ⎟ ln⎜⎜ 2 MeV ⎟⎠ ⎝ N= = 158 ln (0.89 )

158 head - on collisions with an atom of carbon are required to reduce the energy of the neutron from 2 MeV to 0.02 eV. *77

•••

Picture the Problem We can differentiate N B (t ) =

N A0λA −λAt −λBt e −e with respect λB − λA

(

)

to t to show that it is the solution to the differential equation dNB/dt = λANA − λBNB. (a) The rate of change of NB is the rate of generation of B nuclei minus the rate of decay of B nuclei. The generation rate is equal to the decay rate of A nuclei, which equals λANA. The decay rate of B nuclei is λBNB. (b) We’re given that:

dN B = λA N A − λB N B dt N λ N B (t ) = A0 A e − λ A t − e − λ B t λB − λA

(

N A = N A0e − λAt

(1)

)

(2) (3)

Nuclear Physics 1405 Differentiate equation (2) with respect to t to obtain:

[(

)]

[

d [N B (t )] = N A0λA d e−λAt − e−λBt = N A0λA − λA e−λAt + λBe−λBt λB − λA λB − λA dt dt

]

Substitute this derivative in equation (1) to get:

[

]

⎡N λ ⎤ N A0λA − λA e −λAt + λBe −λBt = λA N A0e −λAt − λB ⎢ A0 A e −λAt − e −λBt ⎥ λB − λA ⎣ λB − λA ⎦ Multiply both sides by

N A0

λB

[− λ e A

−λAt

(

)

λB − λA and simplify to obtain: λBλA

]

λB − λA N A0e −λ t − N A0 (e −λ t − e −λ t ) λB N λ = N A0e −λ t − A0 A e −λ t − N A0e −λ t + N A0e −λ t λB N λ = − A0 A e −λ t + N A0e −λ t λB

+ λBe −λBt =

A

A

A

A

=

N A0

λB

A

[− λ e A

−λAt

B

A

B

B

+ λBe −λBt

]

which is an identity and confirms that equation (2) is the solution to equation (1).

If λA > λB the denominator and the expression in the parentheses are both (c) negative for t > 0. If λA < λB the denominator and the expression in the

parentheses are both positive for t > 0.

1406 Chapter 40

(d) The following graph was plotted using a spreadsheet program. 1.0 0.9 0.8

Number of parent (A) nuclei Number of daughter ( B) nuclei

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0

5

10 Time , arbitrary units

15

20

78 ••• Picture the Problem We can express the time at which the number of isotope B nuclei will be a maximum by setting dNB/dt equal to zero and solving for t.

dN B = λA N A − λB N B = 0 for extrema dt

From Problem 77 we have:

Replace λANA by λA N A0e

− λAt

N A0λA −λAt −λBt NB by e −e : λB − λA

(

)

Simplify to obtain:

and

λA N A0e −λ t − λB A

N A0λA −λAt −λBt e −e =0 λB − λA

Solve for t:

)

λB (e −λ t − e −λ t ) = 0 λB − λA (λB − λA )e − λ t − λB (e − λ t − e − λ t ) = 0

e −λAt −

A

A

Remove the parentheses and combine like terms to obtain:

(

A

λA e − λ t = λBe − λ t A

t=

B

B

ln (λB λA ) λB − λA

B

Nuclear Physics 1407 Remarks: Note that all we’ve shown is that an extreme value exists at

t=

ln( λB λA ) . To show that this value for t maximizes NB, we need to either λB − λA

1) examine the second derivative at this value for t, or 2) plot a graph of NB as a function of time (see Problem 77) . 79 ••• Picture the Problem We can show that, provided τ A >> τ B , e − λAt − e − λBt ≈ 1 and

λA λ ≈ A and, hence, that NB = (λA/λB)NA. λB − λA λB

N A λA −λAt −λBt e −e λB − λA

We have, from Problem 77 (b):

N B (t ) =

Because τ A >> τ B :

λA << λB

When several years have passed, because λA t << 1 :

(

e − λAt − e − λBt ≈ 1

Also, when λA << λB :

λA λ ≈ A λB − λA λB

Substitute (2) and (3) in (1) to obtain:

N B (t ) =

λA N λB A

)

(1)

(2)

(3)

1408 Chapter 40

Chapter 41 Elementary Particles and the Beginning of the Universe Conceptual Problems 1

• Similarities

Differences

Baryons and mesons are hadrons, i.e., they participate in the strong interaction. Both are composed of quarks.

Baryons consist of three quarks and are fermions. Mesons consist of two quarks and are bosons. Baryons have baryon number +1 or −1. Mesons have baryon number 0.

2 • Determine the Concept The muon is a lepton. It is a spin-½ particle and is a fermion. It does not participate in strong interactions. It appears to be an elementary particle like the electron. The pion is a meson. Its spin is 0 and it is a boson. It does participate in strong interactions and is composed of quarks. *3 • Determine the Concept A decay process involving the strong interaction has a very short lifetime (∼10−23 s), whereas decay processes that proceed via the weak interaction have lifetimes of order 10−10 s. 4 • (a) True (b) False. There are two kinds of hadrons-baryons, which have spin

1 2

(or

3 2

, 52 , and so

on), and mesons, which have zero or integral spin. 5 • False. Mesons have zero or integral spins. 6 • Determine the Concept A meson has 2 quarks, a baryon has 3 quarks. 7 • Determine the Concept No; from Table 41-2 it is evident that any quark-antiquark combination always results in an integral or zero charge.

1409

1410

Chapter 41

8 • (a) False. Leptons are not made up of quarks. (b) True (c) False. Electrons are leptons and leptons interact via the weak interaction. (d) True (e) True *9 • Determine the Concept No. Such a reaction is impossible. A proton requires three quarks. Three quarks are not available because a pion is made of a quark and an antiquark and the antiproton consists of three antiquarks.

Estimation and Approximation 10 •• Picture the Problem Assuming that the lifetime of a proton is 1032 y, one proton out of every 1032 protons should decay every year on average. Hence, we can estimate the expected time between proton-decays that occur in the water of a filled Olympic-size swimming pool by determining the number of protons N in the pool and dividing 1032 y by this number. The mean time between disintegrations is the ratio of the lifetime of the protons to the number of protons N in the pool:

∆tmean =

1032 y N

The number of protons N in the pool is related to the mass of water in the pool Mwater, the molar mass of water mmolar, water, and the number of protons per molecule n:

N nN A = M water mmolar, water

Solve for N to obtain:

N=

Because the mass of the water is the product of its density and the volume of the pool:

N=

nN A M water mmolar, water

nN A ρ waterVpool mmolar, water

(1)

Elementary Particles and the Beginning of the Universe 1411 Substituting for N in equation (1) yields:

∆tmean =

1032 y nN A ρ waterVpool mmolar, water

(10 y )m 32

=

Because each molecule of water has 10 protons:

n = 10

molar, water

nN A ρ waterVpool

protons molecule

Substitute numerical values and evaluate ∆tmean:

g 1 kg ⎞ ⎟ (10 y )⎛⎜⎜18 mol × 10 g ⎟ 32

3

⎝ ⎠ protons ⎞ ⎛ ⎛ 23 molecules ⎞ ⎛ 3 kg ⎞ ⎜10 ⎟ ⎜ 6.02 × 10 ⎟ ⎜10 3 ⎟ (100 m )(25 m )(2 m ) mol ⎠ ⎝ m ⎠ ⎝ molecule ⎠ ⎝ 365.24 d = 0.0598 y = 0.0598 y × = 21.8 d y

∆tmean =

11 • 2 2 2 2 Picture the Problem We can use Fem = kqproton rnucleus and Fgrav = Gmproton rnucleus to estimate the ratio of the electromagnetic and gravitational forces between two protons located in a nucleus. The electromagnetic force between two protons located in a nucleus is given by: The gravitational force between these same protons is given by:

Fem =

Fgrav =

2 kqproton 2 rnucleus

2 Gmproton 2 rnucleus 2 kqproton

Divide Fem by Fgrav to obtain:

2 kqproton Fem r2 = nucleus = 2 2 Fgrav Gmproton Gmproton 2 rnucleus

Substitute numerical values and evaluate Fem/Fgrav:

1412 Chapter 41

Fem Fgrav

⎛ 2 N ⋅ m2 ⎞ ⎜⎜ 8.99 × 109 ⎟⎟ (1.60 × 10 −19 C ) 2 C ⎠ = ⎝ = 1.24 × 1036 2 ⎛ 2 N⋅m ⎞ ⎜⎜ 6.67 × 10 −11 ⎟ (1.67 × 10 −27 kg ) 2 ⎟ kg ⎠ ⎝

Spin and Antiparticles *12 • Picture the Problem We can use both conservation of energy and momentum to explain why the energies of the two γ-rays must be equal. We can find the energy of each γ-ray in Table 41-1 and find their wavelengths using λ = hc/E.

The initial momentum is zero; therefore, the final momentum must be (a)

zero. The momentum of a photon is E/c. To conserve both momentum and energy the two photons must have the same momentum magnitude. Hence, they have the same energy.

(b) From Table 41-1:

Eγ = 139.6 MeV

(c) The wavelength of each γ ray is given by:

λ=

hc 1240 MeV ⋅ fm = E E

Substitute numerical values and evaluate λ:

λ=

1240 MeV ⋅ fm = 8.88 fm 139.6 MeV

13 • Picture the Problem In each case, the required energy is given by E = 2mc 2 where m is mass of each particle produced in the pair-production reaction. These masses can be found in Tables 41-1 and 41-3. (a) For γ → π + + π –:

(

)

E = 2mπ c 2 = 2 139.6 MeV/c 2 c 2

= 279.2 MeV (b) For γ → p + p –:

(

)

E = 2mp c 2 = 2 938.3 MeV/c 2 c 2 = 1877 MeV

(c) For γ → µ – + µ +:

(

)

E = 2mµ c 2 = 2 105.659 MeV/c 2 c 2 = 211.3 MeV

Elementary Particles and the Beginning of the Universe 1413

The Conservation Laws 14 • Picture the Problem We need to check for conservation of energy, charge, baryon number, and lepton number. (a) Energy conservation:

Because mp < mn , energy conservation is violated.

Charge conservation: +e → 0 + e + 0 = +e

Because the net charge is + e before and after the decay, charge is conserved.

Baryon number: +1 → +1 + 0 + 0 = +1

Because B is + 1 before and after the

Lepton number; electrons: 0→0+0+0=0

decay, baryon number is conserved.

Because Le = 0 before and after the decay, the lepton number for electrons is conserved.

The process is not allowed because it violates conservation of energy. (b) Energy conservation:

Because mn < mp + mπ − , energy conservation is violated.

Charge conservation: 0 → +e + (−e) = 0

Because the net charge is 0 before and after the decay, charge is conserved.

Baryon number: 1→1+0=1

Because B = 0 before and after the decay, baryon number is conserved.

Lepton number; electrons: 0→0+0=0

Because L = 0 before and after the decay, lepton number is conserved.

Because energy is not conserved, this decay is not allowed.

1414 Chapter 41 (c)

Momentum conservation is violated; two (or more) γ rays must be emitted to conserve momentum.

(d) Energy conservation:

Energy is conserved.

Charge conservation: +1 + (−1) → 0 + 0 = 0

Because the net charge is zero before and after the decay, charge is conserved.

Baryon number: +1 + (−1) → 0 + 0 = 0

Because B = 0 before and after the

Lepton number; electrons: 0→0+0+0=0

decay, baryon number is conserved.

Because Le = 0 before and after the decay, the lepton number for electrons is conserved.

Because none of the conservation laws are violated, this is an allowed process. (e) Energy conservation:

Because mp > mn + me+ , energy is conserved.

Charge conservation: 0+1→0+1 =1

Because the net charge is one before and after the decay, charge is conserved.

Baryon number: 0+1→1+0 =0

Because B = 1 before and after the

Lepton number; electrons: −1 + 0 → 0 + (−1) = −1

decay, baryon number is conserved.

Because Le = − 1 before and after the decay, the lepton number for electrons is conserved.

Because none of the conservation laws are violated, this is an allowed process.

Elementary Particles and the Beginning of the Universe 1415 15 • Picture the Problem The decay will occur via the strong interaction if strangeness is conserved. If ∆S = ±1, it will occur via the weak interaction. If S changes by more than 1, the decay will not occur. (a) List the strangeness of Ω– , Ξ0, and π –:

Ω–: S = −3 Ξ0: S = −2 π –: S = 0

Determine ∆S:

∆S = −2 − (− 3) = + 1

Because ∆S = + 1, the reaction can proceed via the weak interaction. (b) List the strangeness of Ξ0, p, π –, and π 0:

Ξ0: S = −2 p: S = 0 π –: S = 0 π 0: S = 0

Determine ∆S:

∆S = 0 − (− 2 ) = + 2 Because ∆S = + 2, the reaction is not allowed.

(c) List the strangeness of Λ0, p + , and π –:

Λ0: S = −1 p +: S = 0 π –: S = 0

Determine ∆S:

∆S = 0 − (− 1) = + 1

Because ∆S = + 1, the reaction can proceed via the weak interaction. 16 • Picture the Problem The decay will occur via the strong interaction if strangeness is conserved. If ∆S = ±1, it will occur via the weak interaction. If S changes by more than 1, the decay will not occur. (a) List the strangeness of Ω– , Λ0, and K–:

Ω–: S = −3 Λ0: S = −1 K–: S = −1

Determine ∆S:

∆S = −1 − 1 − (− 3) = + 1

1416 Chapter 41 Because ∆S = + 1, the reaction can proceed via the weak interaction. (b) List the strangeness of Ξ0, p, and π –:

Ξ0: S = −2 p: S = 0 π –: S = 0

Determine ∆S:

∆S = 0 − (− 2 ) = + 2 Because ∆S = + 2, the reaction is not allowed.

17 • Picture the Problem The decay will occur via the strong interaction if strangeness is conserved. If ∆S = ±1, it will occur via the weak interaction. If S changes by more than 1, the decay will not occur. (a) List the strangeness of Ω– , Λ0, ν e , and e –:

Ω–: S = −3 Λ0: S = −1 νe : S = 0 e –: S = 0

∆S = −1 − (− 3) = + 2

Determine ∆S:

Because ∆S = + 2, the reaction is not allowed. (b) List the strangeness of Σ+, p, and π0:

Σ+: S = −1 p: S = 0 π0: S = 0

Determine ∆S:

∆S = 0 − (− 1) = + 1

Because ∆S = + 1, the reaction can proceed via the weak interaction. 18 • Picture the Problem We can decide whether the given decays of the τ particle are possible by determining whether energy conservation is satisfied and whether conservation of both the τ and µ lepton numbers is satisfied.

(a)

The first decay is allowed. It satisfies energy conservation and conservation of both the τ and µ lepton numbers.

Elementary Particles and the Beginning of the Universe 1417 (b)

The second decay scheme is not allowed because it does not conserve τ and µ lepton numbers.

(c) The total kinetic for the decay in (a) is:

K tot = mτ c 2 − mµ c 2

From Table 41-3 we have:

mτ = 1784 MeV/c 2 and

mµ = 105.659 MeV/c 2 Substitute numerical values and evaluate Ktot:

(

)

(

)

K tot = 1784 MeV/c 2 c 2 − 106 MeV/c 2 c 2 = 1678 MeV

Remarks: Note that the kinetic energy of the individual decay products cannot be determined from the decay scheme alone. 19 •• Picture the Problem Examination of the decay products will reveal whether all the final products are stable. A decay process is allowed if energy, charge, baryon number, and lepton number are conserved. (a) No; the neutron is not stable:

n → p + + e − +ν e

(b) Add the reactions to obtain:

Ω − → p + + e + + 3e − +ν e + 3ν e + 2ν µ + 2ν µ (c) Charge conservation:

Because Q = − 1 before and after

− 1 → 1 + 1 − 3 + 0 + 0 + 0 + 0 = −1

the decay, charge is conserved.

Baryon number: 1→1+0+0+0+0+0+0=1

Because B = 1 before and after

Lepton number: 0 → 0 −1 + 3 + 1 − 3 − 2 + 2 = 0

Because Le = 0 before and after

the decay, baryon number is conserved.

the decay, the lepton number for electrons is conserved.

1418 Chapter 41 Strangeness: −3 → 0

Strangeness is not conserved. However, in each baryon decay ∆S = +1, and each decay is allowed via the weak interaction.

*20 •• Picture the Problem A decay process is allowed if energy, charge, baryon number, and lepton number are conserved. (a) Energy conservation:

Because mn > 2mπ + 2mµ , energy conservation is not violated.

Charge conservation: 0 → +1 + (−1) + 0 + 0 = 0

Because the total charge is 0 before and after the decay, charge is conserved.

Baryon number: 1→0+0+0+0=0

Because baryon number changes from + 1 to 0, conservation of baryon number is violated.

Lepton number: 0 → 0 + 0 + 1 + (−1) = 0

Because Lµ = 0 before and after the decay, the lepton number for muons is conserved.

The process is not allowed because it violates conservation of baryon number. (b) Energy conservation:

Because mπ > 2me , energy conservation is not violated.

Charge conservation: 0 → +1 + (−1) + 0 + 0 = 0

Because the total charge is 0 before and after the decay, charge is conserved.

Baryon number: 0→0+0+0+0=0

Because B = 0 before and after the decay, the baryon number is conserved.

Elementary Particles and the Beginning of the Universe 1419 Because Le = 0 before and after the

Lepton number: 0→0+0+0+0=0

decay, the lepton number is conserved.

The decay satisfies all conservation laws and is allowed.

Quarks 21 • Picture the Problem For each quark combination we can determine the baryon number B, the charge Q, and the strangeness S and then use Table 41-1 to find a match and complete the following table.

(a) (b) (c) (d) (e) (f)

Combination uud udd uus dds uss dss

B 1 1 1 1 1 1

Q +1 0 +1 −1 0 −1

S 0 0 −1 −1 −2 −2

hadron p+ n Σ+ Σ− Ξ0 Ξ−

22 • Picture the Problem For each quark combination we can determine the baryon number B, the charge Q, and the strangeness S and then use Table 41-1 to find a match and complete the following table. Combination (a) (b) (c) (d)

ud ud us us

B 0

Q +1

S 0

0 0 0

−1 +1 −1

0 +1 −1

hadron

π+ π− K+ K−

23 • Determine the Concept From Table 41-2 we see that to satisfy the conditions of charge = +2 and zero strangeness, charm, topness, and bottomness, the quark combination must be uuu. 24 • Picture the Problem Because K + and K0 are mesons, they consist of a quark and an antiquark. We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles.

1420 Chapter 41 (a) For K + we need:

Q = +1 B=0 S = +1

A combination of quarks with these properties is us . (b) For K0 we need:

Q=0 B=0 S = +1

A combination of quarks with these properties is ds . 25 • Determine the Concept Because D + and D− are mesons, they consist of a quark and an antiquark. (a) B = 0, so we must look for a combination of quark and antiquark. Because it has charm of +1, one of the quarks must be c. Because the charge is +e, the antiquark must be

d . The possible combination for D+ is cd . . (b) Because D− is the antiparticle of D+, the quark combination is c d . 26 • Picture the Problem We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles. Because K – and K 0 are mesons, they consist of a quark and an antiquark. (a) For K – we need:

Q = −1 B=0 S = −1

A combination of quarks with these properties is u s. (b) For K 0 we need:

Q=0 B=0 S = −1

A combination of quarks with these properties is d s.

Elementary Particles and the Beginning of the Universe 1421 Remarks: An alternative solution could take advantage of our results in Problem 20 for the antiparticles K + and K 0. *27 •• Picture the Problem Because Λ0, p – , and Σ– are baryons, they are made up of three quarks. We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles. (a) For Λ0 we need:

Q=0 B = +1 S = −1

The quark combination that satisfies these conditions is uds. (b) For p – we need:

Q = −1 B = −1 S = +1

The quark combination that satisfies these conditions is u u d . (c) For Σ– we need:

Q = −1 B = +1 S = −1

The quark combination that satisfies these conditions is dds. 28 •• Picture the Problem Because n , Ξ0 , and Σ+ are baryons, they are made up of three quarks. We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles. (a) For n we need:

Q=0 B = −1 S=0

The quark combination that satisfies these conditions is u d d . (b) For Ξ0 we need:

Q=0 B = +1 S = −2

1422 Chapter 41 The quark combination that satisfies these conditions is uss. (c) For Σ+ we need:

Q = +1 B = +1 S = −1

The quark combination that satisfies these conditions is uus. 29 •• Picture the Problem Because Ω– and Ξ– are baryons, they are made up of three quarks. We can use Tables 41-1 and 41-2 to find combinations of quarks with the correct values for electric charge, baryon number, and strangeness for these particles. (a) For Ω– we need:

Q = −1 B = +1 S = −3

The quark combination that satisfies these conditions is sss. (b) For Ξ– we need:

Q = −1 B = +1 S = −2

The quark combination that satisfies these conditions is ssd . 30 •• Picture the Problem We can use Table 41-2 to identify the properties of the particles made up of the given quarks. (a) For ddd:

Q = −1 B = +1 S= 0

(b) For uc :

Q= 0 B= 0 S= 0 charm = − 1

Elementary Particles and the Beginning of the Universe 1423 Q = +1

(c) For ub :

B= 0 S= 0 bottomness = − 1 (d) For s s s :

Q = +1 B = −1 S = +3

The Evolution of the Universe *31 • Picture the Problem We can use Hubble’s law to find the distance from the earth to this galaxy.

v = Hr

The recessional velocity of galaxy is related to its distance by Hubble’s law: Solve for r:

r=

v H

( 0.025) c (0.025) (3 × 105 km/s ) = r=

Substitute numerical values and evaluate r:

23 km/s 106 c ⋅ y

23 km/s 106 c ⋅ y

= 3.26 × 108 c ⋅ y 32 • Picture the Problem We can use Hubble’s law to find the speed of the galaxy.

v = Hr

The recessional velocity of galaxy is related to its distance by Hubble’s law: Substitute numerical values and evaluate v:

⎛ 23 km/s ⎞ c ⎛ ⎞ ⎟⎟ 12 × 109 c ⋅ y ⎜ v = ⎜⎜ 6 ⎟ = 0.920c 5 ⎝ 3.00 × 10 km/s ⎠ ⎝ 10 c ⋅ y ⎠

(

)

1424 Chapter 41 33 •• Picture the Problem We can substitute for f ′ and f0, using v = fλ, in Equation 39-16b to

1 + v/c . 1 − v/c

show that the relativistic wavelength shift is λ' = λ0 From Equation 39-16b:

f' = f 0

Express f ′ and f0 in terms of λ′ and λ0:

f' =

Substitute for f ′ and f0 to obtain:

c

λ' Solve for λ′:

=

1 − v/c 1 + v/c

c c and f 0 = λ' λ0 c

λ0

λ' = λ0

1 − v/c 1 + v/c 1+ v c 1− v c

*34 •• Picture the Problem Using Hubble’s law, we can rewrite the equation from Problem 31 as λ' = λ0

1 + Hr/c . 1 − Hr/c

From Problem 33 we have:

Use Hubble’s law to relate v to r: Substitute for v to obtain:

λ' = λ0

1 + v/c 1 − v/c

v = Hr

λ' = λ0

1 + Hr/c 1 − Hr/c

(a) For r = 5×106 c⋅y:

⎛ 23 km/s ⎞ (5 × 10 6 c ⋅ y ) ⎟ 1 + ⎜⎜ 6 10 c ⋅ y ⎟⎠ (3 × 10 5 km/s ) ⎝ λ' = 656.3 nm = 656.6 nm ⎛ 23 km/s ⎞ (5 × 10 6 c ⋅ y ) ⎟⎟ 1 − ⎜⎜ 6 5 ⎝ 10 c ⋅ y ⎠ (3 × 10 km/s ) (b) For r = 50 ×106 c⋅y:

Elementary Particles and the Beginning of the Universe 1425 ⎛ 23 km/s ⎞ (50 × 106 c ⋅ y ) ⎟ 1 + ⎜⎜ 6 10 c ⋅ y ⎟⎠ (3 × 105 km/s ) ⎝ λ' = 656.3 nm = 658.8 nm ⎛ 23 km/s ⎞ (50 × 106 c ⋅ y ) ⎟⎟ 1 − ⎜⎜ 6 5 ⎝ 10 c ⋅ y ⎠ (3 × 10 km/s ) (c) For r = 500 ×106 c⋅y:

⎛ 23 km/s ⎞ ⎟ 1 + ⎜⎜ 6 10 c ⋅ y ⎟⎠ ⎝ λ' = 656.3 nm ⎛ 23 km/s ⎞ ⎟⎟ 1 − ⎜⎜ 6 ⎝ 10 c ⋅ y ⎠

(500 ×10 c ⋅ y ) (3 ×10 km/s) = (500 ×10 c ⋅ y) (3 ×10 km/s)

682.0 nm

(5 ×10 c ⋅ y) (3 ×10 km/s) = (5 ×10 c ⋅ y ) (3 ×10 km/s)

983.0 nm

6

5

6

5

(d) For r = 5×109 c⋅y:

⎛ 23 km/s ⎞ ⎟ 1 + ⎜⎜ 6 10 c ⋅ y ⎟⎠ ⎝ λ' = 656.3 nm ⎛ 23 km/s ⎞ ⎟⎟ 1 − ⎜⎜ 6 ⎝ 10 c ⋅ y ⎠

9

5

9

5

General Problems 35 • Determine the Concept (a) It must be a meson, and it must consist of a quark and its antiquark. (b) The π0 is its own antiparticle. (c) The Ξ 0 is a baryon; it cannot be its own antiparticle; the antiparticle is the Ξ0 = u s s . 36 •• Picture the Problem Examination of the decay products will reveal whether all the final products are stable. A decay process is allowed if energy, charge, baryon number, and lepton number are conserved. (a) Yes, the final products are stable. (b) Add the reactions to obtain:

1426 Chapter 41 Ξ 0 → p + + e − + ν e +ν µ + ν µ + 2γ (c) Charge conservation:

0 → e+ + e− = 0

Charge is conserved.

Baryon number: 1→1+0=1

Baryon number is conserved.

Lepton number: 0→0+1−1+1−1=0

Lepton number is conserved.

Strangeness: −2 → 0

Strangeness is not conserved. However, the reaction is allowed via the weak interaction because in the first two decays ∆S = +1.

(d)

No; the rest masses of the decay products would be greater than the rest mass of the Ξ 0 , violating energy conservation.

*37 •• Picture the Problem The π o particle is composed of two quarks, uu . Hence, the reaction π o → γ + γ is equivalent to uu → γ + γ . (a) The u and u annihilate resulting in the photons. (b) Two or more photons are required to conserve linear momentum. 38 •• Picture the Problem A decay process is allowed if energy, charge, baryon number, and lepton number are conserved. (a) Energy conservation:

Because mΛ > mp + mπ , energy conservation is not violated.

Charge conservation: 0 → 1 −1 = 0

Because the total charge is 0 before and after the decay, charge conservation is not violated.

Elementary Particles and the Beginning of the Universe 1427 Baryon number: 1→1+0=1

Because there is no change in baryon number, baryon number is conserved.

Lepton number: 0→0+0=0

Because lepton number is 0 on both sides, lepton number is conserved.

The decay satisfied all conservation laws and is allowed. (b) Energy conservation:

Because mΣ < mn + m p , energy is not conserved.

Charge conservation: −1 → 0 − 1 = −1

Because the total charge does not change, charge is conserved.

Baryon number: +1 → 1 − 1 = 0

Because B changes from + 1 to 0,

Lepton number: 0→0+0 =0

Because L is 0 on both sides,

baryon number is not conserved.

lepton number is conserved.

Because the decay violates both energy conservation and baryon number, it is not allowed. (c) Energy conservation:

Energy is conserved.

Charge conservation: −1 → − 1 + 0 + 0 = −1

Because the total charge does not

Baryon number: 0→0+0+0 =0 Lepton number: 1→1−1+1 =1

change, charge is conserved. Because B does not change, baryon number is conserved. Because L does not change, lepton number is conserved.

The decay satisfied all conservation laws and is allowed.

1428 Chapter 41 Remarks: The decay in Part (c) is the decay process for the muon µ− (see Example 41-2). *39 •• Picture the Problem We can systematically determine Q, B, S, and s for each reaction and then use these values to identify the unknown particles. (a) For the strong reaction:

p + π − → Σo + ?

Charge number:

+1 − 1 = 0 + Q ⇒ Q = 0

Baryon number:

+1 + 0 = +1 + B ⇒ B = 0

Strangeness:

0 + 0 = −1 + S ⇒ S = +1

Spin:

+ 12 + 0 = + 12 + s ⇒ s = 0

These properties indicate that the particle is the kaon, K 0 . (b) For the strong reaction:

p + p → π + + n + K+ + ?

Charge number:

+1 + 1 = +1 + 0 + 1 + Q ⇒ Q = 0

Baryon number:

+1 + 1 = 0 + 1 + 0 + B ⇒ B = +1

Strangeness:

0 + 0 = 0 + 0 + 1 + S ⇒ S = −1

Spin:

+ 12 +

1 2

= 0 + 12 + 0 + s ⇒ s = + 12

These properties indicate that the particle is either the Σ 0 or the Λ0 baryon. (c) For the strong reaction:

p + K − → Ξ− + ?

Charge number:

+1 − 1 = −1 + Q ⇒ Q = +1

Baryon number:

+1 + 0 = +1 + B ⇒ B = 0

Strangeness:

0 − 1 = −2 + S ⇒ S = −1

Spin:

+ 12 + 0 = + 12 + s ⇒ s = 0

These properties indicate that the particle is the kaon, K + . 40 •• Picture the Problem We can systematically determine Q, B, S, and s for the reaction and then use these values to identify the unknown particle. The Q value for the reaction is given by Q = −(∆m ) c 2 and the expression for the threshold energy for the reaction is given in the problem statement.

Elementary Particles and the Beginning of the Universe 1429 (a) For the strong reaction:

p + p → Λo + K o + p + ?

Charge number:

+1 + 1 = 0 + 0 + 1 + Q ⇒ Q = +1

baryon number:

+1 + 1 =+1 + 0 + 1 + B ⇒ B = 0

strangeness:

0 + 0 = −1 + 1 + 0 + S ⇒ S = 0

spin:

+ 12 +

1 2

= + 12 + 0 +

1 2

+s ⇒s=0

These properties indicate that the unknown particle is a pion, π + .

p + p → Λo + K o + p + π +

(b) The reaction is: The Q-value for the reaction is:

[

]

Q = (mp + mp ) − (M Λo + M Ko + M p + M π + ) c 2 Use Table 41-1 to find the mass-energy values:

Q = [(938.3 + 938.3) − (1116 + 497.7 + 938.3 + 139.6 )]MeV = − 815 MeV Because Q < 0, the reaction is endothermic. (c) The threshold energy for this reaction is:

K th = −

Q (mp + mp + M Λo + M Ko + M p + M π + ) 2mp

Using Table 41-1 to find the mass-energy values, substitute numerical values and evaluate Kth:

K th = −

− 815 MeV (938.3 + 938.3 + 1116 + 497.7 + 938.3 + 139.6)MeV 2(938.3 MeV )

= 1984 MeV = 1.984 GeV 41 •• Picture the Problem We can solve the equation derived in Problem 31 for the recessional velocity of the galaxy and then use Hubble’s equation to estimate the distance to the galaxy. (a) From Problem 31 we have:

λ' = λ0

1 + v/c 1 − v/c

1430 Chapter 41 Solve for the radicand:

1 + v/c ⎛ λ' ⎞ =⎜ ⎟ 1 − v/c ⎜⎝ λ0 ⎟⎠

Substitute numerical values for λ′ and λ0:

1 + v/c ⎛ 1458 nm ⎞ =⎜ ⎟ = 4.935 1 − v/c ⎝ 656.3 nm ⎠

Simplify to obtain:

v ⎛ v⎞ 4.953⎜1 − ⎟ = 1 + c ⎝ c⎠

2

2

and

v 5.953 = 3.953 c Solve for v:

v = 0.664c = 0.664(3 ×108 m/s ) = 1.99 ×108 m/s = 1.99 ×10 5 km/s

(b) From the Hubble equation we have:

r=

Substitute numerical values and evaluate r:

r=

v H 1.99 × 105 km/s = 8.65 × 109 c ⋅ y 23 km/s 106 c ⋅ y

42 ••• Picture the Problem We can use the masses of the parent and daughters to find the total kinetic energy of the decay products under the assumption that the Λ0 is initially at rest. Application of conservation of energy and the definition of kinetic energy will yield the ratio of the kinetic energy of the pion to the kinetic energy of the proton. Finally, we can use our results in (a) and (b) to find the kinetic energies of the proton and the pion for this decay. (a) The total kinetic energy of the decay products is given by:

K tot = (mΛ − mp − mπ )c 2

Substitute numerical values (see Table 41-1) and evaluate Ktot:

MeV MeV MeV ⎞ ⎛ K tot = ⎜1116 2 − 938.3 2 − 139.6 2 ⎟c 2 = 38.1 MeV c c c ⎠ ⎝ (b) The ratio of the kinetic energies is given by:

Kπ 12 mπ vπ2 mπ vπ2 = 1 = 2 Kp mp vp2 2 mp vp

Elementary Particles and the Beginning of the Universe 1431

vπ mp = vp mπ

Use conservation of momentum (nonrelativistic) to obtain:

mπ vπ = mp v p ⇒

Substitute for the ratio of the speeds to obtain:

Kπ mπ ⎛ mp ⎜ = K p mp ⎜⎝ mπ

Substitute numerical values and evaluate the ratio of the kinetic energies:

MeV 938.3 2 Kπ c = 6.72 = K p 139.6 MeV c2

(c) Express the total kinetic energy in terms of Kπ and Kp:

K p + Kπ = K tot

Use our results in (a) and (b) to obtain:

K p + 6.72 K p = 38.1 MeV

Solve for Kp:

K p = 4.94 MeV

Substitute in equation (1) to obtain:

Kπ = K tot − K p = 33.2 MeV

2

m ⎞ ⎟⎟ = p mπ ⎠

(1)

*43 ••• Picture the Problem The total kinetic energy of the decay products is the rest energy of the Σ0 particle. We can find the momentum of the photon from its energy and use the conservation of momentum to calculate the kinetic energy of the Λ0. (a) The total kinetic energy of the decay products is given by:

K tot = (mΣ ) c 2

Substitute numerical values (see Table 41-1) and evaluate Ktot:

MeV ⎞ ⎛ K tot = ⎜1193 2 ⎟ c 2 = 1193 MeV c ⎠ ⎝

(b) The momentum of the photon is given by:

pγ =

Substitute numerical values and evaluate pγ:

Eγ E − mΛ c 2 = c c

MeV ⎞ ⎛ 1193 MeV − ⎜1116 2 ⎟ c 2 c ⎠ ⎝ pγ = c MeV = 77.0 c

1432 Chapter 41 (c) The kinetic energy of the Λ0 is given by:

pΛ2 KΛ = 2mΛ or, because pΛ = pγ ,

KΛ =

pγ2 2mΛ 2

Substitute numerical values and evaluate KΛ:

MeV ⎞ ⎛ ⎜ 77.0 ⎟ c ⎠ ⎝ = 2.66 MeV KΛ = MeV ⎞ ⎛ 2⎜1116 2 ⎟ c ⎠ ⎝

(d) A better estimate of the energy of the photon is:

Eγ = E − mΛ c 2 − K Λ

Substitute numerical values and evaluate Eγ:

MeV ⎞ ⎛ Eγ = 1193 MeV − ⎜1116 2 ⎟c 2 − 2.66 MeV = 74.3 MeV c ⎠ ⎝ The improved estimate of the momentum of the photon is:

pγ =

Eλ 74.3 MeV MeV = = 74.3 c c c

44 ••• Picture the Problem The solution strategy is outlined in the problem statement. (a) Express ∆t = t2 − t1 in terms of u2 and u1: Noting that u1u2 ≈ c2, we have:

∆t = t2 − t1 =

∆t ≈

x x x(u1 − u2 ) − = u2 u1 u1u2

x∆u c2

(1)

where ∆u = u1 − u2 (b) From Equation 39-25 we have:

2 ⎛ ⎛ mc 2 ⎞ 2 ⎞ ⎛ mc 2 ⎞ u ⎟⎟ = ⎜1 − ⎜⎜ ⎟⎟ ⎟ = 1 − ⎜⎜ ⎜ c E E ⎝ ⎠ ⎠ ⎟⎠ ⎝ ⎝

Expand binomially to obtain:

u 1 ⎛ mc 2 ⎞ ⎟ = 1 − ⎜⎜ c 2 ⎝ E ⎟⎠

2

12

Elementary Particles and the Beginning of the Universe 1433 (c) Express u1 − u2 in terms of E1, E2, and mc2:

u1 − u2 = =

1 2

(mc ) ⎛⎜⎜ E1

−

( ) (E

− E22

2 2

⎝

c mc

2 2

2 2

2 1 2 2 1 2

1 ⎞ ⎟ E12 ⎟⎠

)

2E E

Substitute numerical values and evaluate ∆u: 2

[

]

⎛ eV ⎞ 2 2 c⎜ 20 2 c 2 ⎟ (20 MeV ) − (5 MeV ) c ⎠ ∆u = ⎝ = 7.50 × 10 −12 c 2 2 2(20 MeV ) (5 MeV )

(1.7 ×10

)(

c ⋅ y 7.5 × 10−12 c c2 31.56 Ms = 1.275 × 10−6 y × y

Use equation (1) to evaluate ∆t:

∆t ≈

5

)

= 40.2 s (d) Using mc2 = 40 eV for the rest energy of a neutrino: 2

[

]

⎛ eV ⎞ 2 2 c⎜ 40 2 c 2 ⎟ (20 MeV ) − (5 MeV ) c ⎠ ∆u = ⎝ = 3.00 × 10 −11 c 2 2 2(20 MeV ) (5 MeV ) Use equation (1) to evaluate ∆t:

(1.7 ×10

)(

c ⋅ y 3 × 10−11 c c2 31.56 Ms = 5.1× 10−6 y × y

∆t ≈

5

)

= 161s Remarks: Note that the spread in the arrival time for neutrinos from a supernova can be used to estimate the mass of a neutrino.