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 �