The Classical Mechanics

when(~~)

=IV
is basically the maximum rate of change of
normal direction to the surface
A=

Q

~ o

and since,

d

o

d

• In spherical polar coordinates, x = r sin 0 cos q>, y = r sin 0 sin q>, z = r cos 0 where i = j =

and,

a ax -a Oy

a az

(r sinO cosq> + ecosO cosq> - q, sinq»

(r sinO sinq> + ecosO sinq> + q, cosq», k = r cosS a a a -+ sinO cosq> or rcosO cosq> ae r sinO sinq> oq> a a+ a -+ sin 0 sin q> or r cosO sinq> ae r sinO cosq> oq> a a ------

= cosO or

esinS

rsinSoS· •• This gradient operation gives the directional derivatives along a direction (given) which is given by Directional Directive == Vq> . whena is unit vector along given direction w.r.t. the surface q> = o.

a

16

The Classical Mechanics

. dqJ dqJ Here, SInce, --;;; is rate of change of qJ along PQ and dn is the maximum rate or change of qJ along nonnal direction PR then dn and

= dr cos 8 = dr . n

dqJ d(~ -cos8<-'I'dn dn

dqJ dr

~: (n.dr) = ~: dn = dqJ

and so,

(When VqJ ==

The divergence operation on a vector quantity, say, quantity which is mathematically given by _ _ aA x aAy aA z S=V.A = fu+ay+8;""' This physically gives the amount of

A,

~: n)

gives a scalar

y

outward flux of vector A per unit volume. aA x aAy aA z Actually, fu' ay' 8;"" are the rates of

change of these components in respective x, y, and z axes. Taking a cuboidal box of volume dv = dx, dy, dz the net out flow or the flux of vector A in the x-direction is

" a~ x dx ) dy dz ( Ax + __ ax

z

A x dy dz

aA dx dy dz ax

= __ x

aAy Similarly, the flux of same vector.A.. in y and z directions are a; dy dx dz aAz and az dz dx dy resp. so the net outflow per unit volume is aA x + aAy + aAz) __ _ ( ax ay az - V.A.

Let us now discuss about curl of a vector Awhich is mathematically given by another vector field

as

when

B

Vector

17

( VxA)

= y

(OA oz

X

_

OAz)and(vxA)

ax

= z

(OA y _ OAx) ax oy

Physically every component of curl of any vector is obtained from line integra1* of that vector per unit area of a surface. From figure, we see that

(v x A) y

=

(!

J'o abc 0

A.dr)!dx dz

OAt) OAz) dz-A x dx ] - dx1dz [ z dz+ (A x +--" oz dz dx- (A z +--dx ax

_ -- A

=

(OA x _ OAz) oz ax' z a

~---------------y

x The basic formulae involving V for two differentiable vector function A and :i3 and for two scalar functions (
(iv) V.(
*

See line integral of a field vector.

18

The Classical Mechanics

VX(AX13) = (13.V)A-13(V.A)-(A.V)13+A(V.13). V(A.13) = (13.V)A+(A.V)13+13x(VxA)+Ax(Vx13). V.(V
•(vii) (viii) (ix) (.r:)

Vx(V
(xi)

V.(VxA)=O.

(xii)

Vx(VXA) = V(V.A)-V 2,A..

1.17. Laplacian and D' Alambertian Operator:

In vector analysis, another two important operators are: (i) Laplacian operator

(V2)

(ii) D' Alambertian operator (IT) which are very important in several branch of Physics. This operators are mathematically given by,

(cylinderical coordinate)

1 0 ( r 2 -0) = -

and where

*

IT =

r2 or

or

V2

0

__ 1 c2

2

+

1 - 0 (. 0)+ 1 sm8-0 r2 sin8 08 09 r2 sin 2 9 0
2

0(2

C = velocity of light or e.m. wave in free space.

Here this property of curl

(VX)

operation gives a most popular expression of

pseudo vector was,

Vxv = Vx(wxr) = (r.V)w-r(V.w)+ w(V.r)-(w.V)r. See, get,

w is constant vector, V opertion on

Vxv =w(V.r)-(w.V)r =3w-w = 2w.

w gives zero. So we

Vector

19

1.18. Vector Integration: Vector Integeration is basically identical with usual procedures of integral calculus. There are three types of vector integration can be mentioned which are (i) Line integral (ii) Surface integral (iii) Volume integral (I) Line Integral: In physics we usually have to go through several type of integration like

L q> dr, LA. dr, L A x dr and all of these are line integral. Here q> and A are both position dependent or not and they are respective scalar and vector fields. This integration are also known as contour integrations and c is a contour, either close or not A

such

A

A

xi + yj + zk

r

A

dr

A

A

dx i + dy j + dz k . ~

hI

i f q> dx + J f q> dy + k f q> dz

and

hI

LA. dr and

LAxdr

b:J

=

=

~

b:J

fAx dx + fAy dy + f Az dz

i f(Aydz-Azdy)+J f(Azdx-Axdz)+k f(AxdY-Aydx).

Here all the limits of integrations are obtained from the nature of contour, 'C'. (ii) Surface Integral: This is commonly known as double integration which is taken over a surface area (given). The integrals like

Lq> as, LA. as, LA x as

are common surface integrals and here

surface element which can be written as, (n is unit normal to

d'S

is small

d'S = ds n

ds).

To evaluate surface integrals, it is convenient to express them as double integrals taken over the projected area of the surface'S' on one of the coordinate planes. Mathematically, the integral over projected region is given by,

20

The Classical Mechanics

I I A.nlit d!1 R ln.k A

=

I I A.nd:, ~Zl ln.l

R

IIA.nlitIA ~Zl R n.]

or where,

or

n is mathematically given by

n=

V'cp IVcpl

for given surface,

cp =

cost.

(iii) Volume Integral: This is a tripple integral of a vector point function over a certain volume element, say V. This is simple given by,

Iv Adv

=

iI Ax dv + JI Ay dv + kI A z dv.

1.19. Gauss's Divergence Theorem This theorem states that if

A is a continuously differentiable vector point

function and '8' is a closed surface enclosing a volume 'v', then

Is A.nds Iv V.Adv

.... (1 )

=

n

where, is the unit outward drawn normal vector. Briefly, this theorem states that the normal surface integral of a function

A over the boundary of a closed region is equal to the volume integral of V.A taken throughout the region. We will now suppose that the region V is such that it is possible to choose rectangular cartesian co-ordinate system such that each line parallel to any coordinate axis which has internal points in common with the region meets the boundary 8 in two points. In terms of cartesian co-ordinates, the equation (1) stated as follows,

r (A x dy dz + A y dz dx + A_• dx dy)

Js

Consider now the volume integral

=

Jv (a axAx + aayAy + aazAz) dxdydz.

Iv( a~z

)

litdydz

let R3 denote the

projection of the region V, on the OXY plane. Every line through a point (x, y, 0) of R3 meets the boundary 8 in two points. Let the z co-ordinates of these points be, z = cp (x, y), z = \If (x, y) and cp (x, y) ~ \If (x, y).

Vector

21

Now, f oA z dxdydz = f OZ

IR) [l oAOZz dZ] dxdy \jI

f f[Az (x,y,
=

R)

Let us now denote the parts of the surface S corresponding to z =

f f A z (x,y,
=

R3

f fAz(x,y,\jf)dxdy

=

R3

1s, A (it. k) ds -1s, Az.(it.k)ds z

.

for the outward drawn normal at any point of SI makes an acute angle with positive direction of z axis and then at any point of S2 makes an obtuse angle with z axis. From above equation we now have

fv oAzoz dxdydz

=

r A z it. k ds + Jr

Js,

S2

A z it. k ds

fs A z (ii ,k) ds Similarly, it may be proved that

oAy

f -Oy- dxdydz

=

and

r

=

oAx dxdydz

Jv ox

J!r Ay n.j~ds A

V

s

r Ax ii.t ds

Js

So we get,

(OAx fVox

oAy --+- +OAz) - - dxdydz Oy

oz

=

f( A iA+A j'A+A kA) .nds x y z

This is Gauss divergence theorem.

1.20. Green's Theorem This theorem states that if

22

The Classical Mechanics

point functions such that

V
and

V\jf

fJ
=

are also continuously differentiable, then

fJ
where V is the volume which is involved by S. Now,

V .(
div(
V

=

fvV
Now, D
=

fJ
and this result of this theorem can also be written as,

fv(
2 )

=

) s\.

f' (O\jf

directional derivatives of \jf and any point of S.

1::

uS

and

o\jf o

are called

1.21. Stoke's Theorem: This theorem states that if A be any continuously differentiable vector point function and S be a surface bounded by a curve 'C' then,

LA.

dr

=

DV A) .n

ds

x

n

where the unit normal vector at any point of S is drawn in the sense in which a right handed screw would move when rotated in the sense of description of C. We can now easily prove this Stoke's theorem for plane which is also often referred to as Green's theorem. Take a system of cartesian rectangular co-ordinate axes such that the plane of a given surface S is the OXY plane and the z-axis lies along the direction of the normal vector ii. A A A Let Ax i + Ay j + A z k A We have,

l A.

dr

lA .

nds

A A) (A oX Aoy Aoz) +kA +j - +kf (AAx + jAy os os os A ox + A ay ) ds ( f os os c

c

i

z . i -

x

y

x ds

23

Vector

dz for dx

=

0, the tangent at any point lying in the OXY plane.

fJv

Also,

x

fJv

A). nds =

x

A). kds

f'(OA y _ OAx)dxdY ~ ax Oy .

=

Thus for the case of plane surface the theorem is equivalent to showing A dx + A dy)

fC (

y

X

=

f

[a Ay _ aOyA x ) dxdy .

cox

x=a x=b We shall now prove it. Firstly suppose y that the region'S' is such that any line Y= \jI(x) parallel to either axis meets the boundary 'c' in at the most two points. Let the region be included between the lines x = a and x = b and any line parallel to y axis with abscissa x meet c in the points given by y = \jf (x) y =

fs

OA x dxdy

oy

fb[fljl (x) aoyAx dY] dx a

cp(x)

b

f[Ax (x, \jf(x)) - Ax (x,

-fAx (x,y)dxC2

J

Ax (x,y)dx

CI

-LAx (x,y)dx. Similarly, we may show that

JOAax s

y

dxdy

=

JAy(x,y)dy C

24

The Classical Mechanics Finally,

f( A

dx+A dy) == Y

eX

f~ ox

OAy_OA) _ _ x

oy

dxdy

1.23. Reciprocal Vectors: For every set of vectors, say, such that

(a, b, c) , there is a reciprocal set (a', b', c')

a.a' == b.b' == c.i? == 1. And for existence of this reciprocal set, the restriction is that the vectors mustnotbecoplanner, i.e., a.(bxc) :;t:0. Now for this case, the vectors of reciprocal set· for the given set of vectors

(a,b,c)

(a, b, c)

are given by,

a'

==

bx c c xa a. (b x c)' b' == a.(b x c)'

_,

c ==

ax b

a.(b x c)"

1.23. Scalar and Vector Field: Basically, any scalar or vector field means a set or collection of a no. of scalars or vectors. For example, a vector field V can be expressed as V ==

(a),a2' ....)

where, a, eV i.e., a;'s are component vectors along several direction in field. Similarly, a scalar field, say F is a collection of a no. of scalars i.e.,F=(u]>U2'U3' ... ·) and uieF. This scalar or vector fields are very significant to construct vector space, having finite dimensionality. 1.25. Elementary Idea about Vector Space: A space G over the vector field V and scalar field F is said to be a vector space for binary operations EB among vectors and 0 among scalars and vectors, when the following conditions are satisfied. (i)

(ii) (iii)

aEB h eV

for all

a, h eV.

aEB h == hEB a for all a, h e V . a EB (h EB c) = (a $ h) $ c for all a, h, c eV.

(iv) There exists an element

*

e in V such that aEB e= a for all ae V .

For reciprocal set of vectors, we should apply the property of vectors tripple product a .(b x == b .(C x a) == (a x b)to satisfy the condition a.a' = b .b' = .c'.

c)

=

c

I

For this set of reciprocal vector, one should remember that

a'.b == a'.c = h'.c = b'.a = c'.a == c'.b =0.

c

Vector

25

(v) For each {i in V there exist an element (i' in V such that (vi) a 0 (vii)

(xi)

a EV, o.a EV for all a EF,a EV.

o.0(P0a)

=

(o.p)0a for all o.,pEF,all aEV.

(a EEl b)

=

(a 0 a) EEl (P 0 a) for all a, p E F, a E V .

(viii) a 0

(o.+p)0a

(x) 10

aEEl a' = e.

=

(o.0a) +(P0a) for all o.,p EF,a EV.

a = a, 1 being the identity element in F.

For these characteristics, any vector space*, 'G' is denoted by (V, F, +,., EEl, 0) . Two important parameters of a vector space are its dimensionality and basis. The dimensionality of the space is given by the maximum possible number of linearly independent vectors in that space. For example, in a three dimensional space, the maximum possible number of linearly independent vectors can be three. Suppose, in three dimensional space, al> a2' a3 are three linearly independents vectors. Now any vector

h

=

a, a, +

b in the space can be expressed as,

0.2

a2 + 0.3 a3'

when, o. p 0.2 , 0.3 E F and at least one of them is non-zero. On the other hand, the basis of a vector space are those linearly independent vectors which are essential to construct any vector of the space. So more clearly, a set of n linearly independent vector is known as a basis for the space. One should remember that for a given space, basis is not unique. It can be selected in an infinite no. of ways. As for example, for a 3-dimensional space, one of the basis is well known set of three unit vectors

i, J and k along x, y

and z axes of cartesian co-ordinate system, such that for any vector

P=

A

A

p,

A

Px i + Py j + pz k

1.25. Linear Operator in Vector Space: A linear mapping T : V ~ V is called a linear mapping on V. T is also called linear operator on the vector field V of the given space. A set of all linear operators on a vector space (V, F, +,., $, 0) over V and F, in its own right, a linear space over F.

*

For vector space (V, F, +,., $, 0), the elements of V are vectors, elements of Fare scalars, and F is called ground field of the vector space. Four symbols. +, ., $, 0 are four different compositions +: F x F ~ F,.: F x F ~ F, EEl: V x V ~ V, o : F x V ~ V when: stands for mapping.

26

The Classical Mechanics

Summary 1. Except algebraical laws, the scalar quantities also obeys commutative, associative and distributive laws. 2. Unit vector is essential to represent a vector and mathematically, A = A Ii . 3. Opposite vector has same magnitude, but opposite direction to the original vector. i.e.,

1- ·~I = IAI = A, -A = A (-Ii) .

4. For collinear vectors, all vectors are parallel to each other.

5. For linearly independent vectors A/s

La

i Ai = 0 when all a, = O. But for at least one a., O. They are dependent vectors.

'*

6. Magnitude of resultant of two vectors A and B is IA + BI = (N + B2 + 2AB cos a)112 Bsina and if this resultant makes angle 9 with A then, tan 9 = A + B cos a .

7. Magnitude of subtraction of two vectors A and B , is IA - BI = (A' + B2 2AB cos a)1/2 and if this subtraction vector makes angle 9 with vector A, then B sin a tan9=A - B cosa .

8. For resultant of any no. of vectors by perpendicular resolution method, if

it =

" " Ai L.J

,

R= (R; + R;)

2

I

and when,

9 = tan- 1 (RylRx).

Rx = LAi cos 9 i , Ry = LAi sin 9 i i

i

~

9. Position vector at any point P is r = OP when, '0' is arbitrarily choosen origin. This position vector has following representations,

r

r

=

xi + yJ

r = r and

r

=

(2 dim cartesian coordinates)

r (2 dim polar coordinates)

xi + yJ + zk

r

(3 dim cartesian coordinates)

r = r (3 dim spherical polar coordinates) r = pp + z k (3 dim cylinderical polar coordinates).

Vector

27

A,B,C, we have C = cx.A + PB, which will 13 .

10. For three co-planner vectors

be co-planner with

A and

A, IAI will remain unchanged for

11. For rotational invarience of any vector

rotation of the frame.

-

-

12. For dot product of two vectors A and B, A.B = A B cos ex

·A.B (ii) A.B

When

(i)

o for A..L 13

B.A

~ Commutative property.

J.] = k.k i.] = J.k = k.i = i .i

(iii)

=

1,

O.

X and 13 X x 13 = IX x 131 n = (AB sin a)n (i) Xx13 = 6 for XII 13 . (ii) Xx13 = -13 xX ~ Anti commutative property (iii) i xi = J x J = k x k = 0

13. For cross product of two vectors

when

and

i x J = k, Jx Ie = t., Ie x t = J. Jx i = -k, k x J = -i, i x k = - J.

14. For scalar tripple product of

A.(B xC)

=

B.(C xX)

=

(X, 13, C),

C.(X xB)

15. For vector tripple product of three vectors

A, B, C

Xx(13 xC) = (A.C) 13 - (A.B) C. and Ax(13 xC) + 13 x(C x A) + Cx(X x B) = 6 16. Pseudo vectors does not change sign under co-ordinate inversion or mirror reflection. 17. Pseudo scalars changes sign under co-ordinate inversion or mirror reflection. 18. For ordinary derivative of a vector

X,

~dAx -:dAy AdA z z--+}--+k--. da da da dcx. and it follows the rule of ordinary differentiation. dX

-

=

19. For partial derivatives of vector

X = X(Xi)

28

The Classical Mechanics

dA

=

~:~ dx V

20. For gradiant operator _

i

-

dp

A

1-

IA

A = VP = dn n = VP n when, dp

=

V p.d;

V.

21. For divergenee operator

V.A

=

"oAa LJ OU . a

Vx

22. For Curl operator

- -

VxA =

OA y)

~(OAz

~(OAx

OAz) +kA(OAy OAx) ----

1 - - - - - +} - - - - -

oy

oz

23. Laplacian operator V 2

oz

=

D' Alambartian operator

02 OU 2

La

V. V =

IT =

ox

ox

Oy

•

1 02

V2

-""""2 C

!:l

ut

2 .

24. Line integral of a vector is given by 2

fA.d;

=

C

f[Axdx+AydY+ Azdzl. I

25. Surface integral of a vector - f A.ds

.f =

S

r-

dxdy JRA.nAI IA n.k

f f A.n =

R

A

dIY~ZI

fA dv = f f f

= f fAn dxdz

n.i

26. Volume integral of a vector

=

A is

R·

In·JI

A is given by

A dx dy dz

i JJJAx dxdydz + JJJJAxdxdydz + k JJJAzdxdydz.

Vector

29

27. Gauss's Divergence theorem states that

1A.n

ds =

Iv div(A) dv

28. According to Green's theorem,

fJ~V2\V - \VV2~)dv

=

1(~~~ - :!)as \V

29. By Stoke's theorem,

(a', b', c') of (a, b, c)

30. For reciprocal set

a'

bXc a.(b x c) , b'

a.a'

=

cxa a.(b x c) , c'

=

axb a.(b x c)

c.c'

when, = b.b' = = 1. 31. A vector space is represented as (V, F, +, ., EB, 0)

32. For basis vector

aj

(inn dimension), any vector

b = Lujaj

(i= 1,2,., n)

Worked Out Examples Example 1. If

A=

J - 4k, 8 = -2f + 4J - 3k, 3A - 28 + 4(: . 3A - 28 + 4(:

3/ -

Find a unit vector parallel to

Ans. Let the vector

or or,

c = / + 2J -

a= a = 3( 3f - J - 4k) - 2(-2f + 4J - 3k) + 4(f + 2J - k) a = (9 + 4 + 4)/ + (- 3 - 8 + 8)J + (- 12 + 6 - 4)k a = 17 f - 3J - 10k a is

so, the unit vector along _

A

a

n = lal

=

A

,..

17i - 3j - 10k ± .J398 .

Example 2. The position vectors of points P and Q are given by

Fr

=

2f + 3J - k,

'2

=

4/ - 3J + 2k.

k,

The Classical Mechanics

30 ~

Determine PQ and its magnitude.

or = ~ = 21 + 3J - k.

Ans. Here,

00 =- r2 = 41 - 3J+ 2k PO

=

OO-OP

r2 -~

=

IpQI = 7 unit. a and b are

=

21 -6J+3k.

So, its magnitude is Example 3. If

(x + 4y)a + (2x + y + I)b and andy

13

=

non-collinear vectors and A (y - 2x + 2)a + (2x - 3y - I)b, find x

3A = 213 Ans. Here 3A = 213 such that

or, 3(x + 4y)a + 3(2x + y + I)b = 2(y - 2x + 2)a + 2(2x - 3y -I)b

a

Equating coefficient of and b from both side, 3x + 12y = 2y - 4x + 4 => 7x + lOy = 4 6x + 3y + 3 = 4x - 6y - 2 => 2x + 9x = -S Solving this two equations, x = 2, y = -I. Example 4. Examine whether the vectors Sa + 6b + 7(5 , 7a - 8b + 9(5 and 3a + 20b + S(5 , (a , b ' (5 being non-coplanar vectors) are linearly independent or dependent. Ans. If possible, let the given vectors are linearly dependent. Then there exist scalars ai' a 2 , a 3 not all zero, such that al(Sa + 6b + 7c) + a2(7a - 8b + 9c) + a3(3a + 20b + or, (5a l + 7a 2 + 3( 3 )a + (6a l

-

8a 2 + 20( 3 )b

+ (7a, + 9a 2 + 5( 3 )c As,

=

Sc) = 0 ... 1

O.

a, b,

(5 non-eo-planner vectors, 5a l + 7a 2 + 3a.3 = 0 6a l - 8a 2 + 20('.(.3 = 0 7a l + 9a 2 + 5a 3 = O. From 1st two equations, we get al

-

a,2

= -

2 -1 -k, a 3 = -k

a3 = =k(say).

-1

.

:. a l = 2k, a 2 = These values also satisfy the third equation. Hence, there exist scalars ai' a 2, a 3 such that equation (1) holds. Hence, given vectors are linearly dependent.

Vector

31

Example 5. Consider three vectors

A,

Band

C has respective magnitude

1, 2 and 3 units and these are directed along arms of a equilataral triangle, taken in the same sense. Find magnitude of resultant of these three vectors by resolution method. Ans. By perpendicular resolution method, let us take X component, Rx = A cos 0 + B cos 120 + C cos 240 = -105 unit. and take Y component, Ry = A sin 0 + B sin 120 + C sin 240 = - 0.866 unit.

A is taken along +ve x-axis.

where the vector

R= .[(R; + Rn = J3 unit.

So magnitude of resultant

A= 41 - J+ 3k perpendicular to both A and E. Example 6. If

and

E = 21 + J- 2k,

Ans. Here, we know that for any two vectors

find a unit vector

A and E, (A x E)

is perpen-

dicular to the plane of A and B so, the unit vector perpendicular to both •

B

(AxE)

A

IS

n

=

1- -I

+ AxB'

AxE

Now A

=

A and

A

i

j

4

-1

-2

1

k 3 =

-2

1(2-3)+ J(-6+8)+k(4-2)

A

-i + 2j + 2k.

(-7 + 2J + 2k)

:. unit vector

iz

or,

iz=+

=

±"""'---------"3

(7- 2J - 2k)

.

3

Example 7. Find the magnitude projection of the vector the vector

7+ 2J + 2k.

Ans. The projection is R when and

=

(A.iz)

A = 2f - 3J + 6k

n is unit vector along (1 + 2J + 2k) A

and

A

A

i + 2j + 2k

n=

----"'--

R=

~ (2 -

3 6 + 12)

=

%unit.

27 - 3J + 6k

on

32

The Classical Mechanics

(A +i3).(i3 +c) x (C +A) (A +i3).(i3 +c) x (c + A)

Example 8. Simplify: Ans.

= (A+B).[BxC+BxA+CxA] [':CxC=O] = A.(i3 x c) +A.(i3 x A) + A.(C x A) + i3.(i3 x c) +i3.(i3 x A) + i3.(C x A) = A. (i3 x C) +0 + 0 +0 +0 +A. (i3 xC) = 2A.(i3 x C).

-

dA Example 9 . Solve -dt dA Ans. dt

_

dB-

= B 'dt -- = A-'

_ d2 A - dB = B => - =dt 2 dt

_

= -A.

2

d A --+A =0 2

dt

•

A = C1 sin t - C2 cos t B = C1 cos t - C2 sin t . where, C1 and C2

and solution is similarly,

are two constant vec-

tors.

A is a constant vector, prove V(r.A) A = Axi + Ay} + A/•.

Example 10. If Ans. Now let

when (Ax , Ay , A) all are constants for constant vector z

A.

r- • A- =xAx + yAy + zAz

Now

V(r.A)

= -

a (r.A-)~i +a (r.A-)j~ +-a (r.A-)-k

rx

ry

- ( 2 ~ 2 ~ 2~) Ans. V. 2x z i - xy z j + 3y z k

=

~(2x2z) + ~(-xiz) + ~(3lz)

=

4xz - 2xyz + 3y2.

rx

=

ry

Example 12. Evaluate

rz'

V.(r 3r)

rz

A.

Vector

33

-(3-) __ + 3r 3 r r = V-(3)r .r + r3-V.r = 3rr.r

Ans. V. =

3,-3 + 3,-3 = 6,-3.

(V2 rn)

Example 13. Prove, Ans. We know that

V(rn)

=

==

n(n + l)~ - 2.

nr n- 2r.

V2(rn) = V.V(rn) = V.(nr n- 2r) =

=

n V.(r n- 2 r) = n[V(r n- 2 ).r + rn- 2 V.r] n[(n - 2)r n- 4r.r + 3r n- 2]

= n(n - 2) ~-2 + 3~-2.n = (n2 - 2n + 3n) ~-2 = (n2 = n(n + 1) ~-2.

Example 14. Prove that

fnx (a x r) ds

+ n) ~-2

= 2aV.

s

a

Where is constant vector ADS. By Gauss's divergence theorem

fnx (a x r) ds fvx (a x r)dv =

s

v

V x (a x r) = _

Now

~A L..,.i

°(_ _) axr

x-

ox

~~

=L..,.IX

(oa-xr+ax__ or) ox

=

Li

=

3ii - ii

=

f2a dv = 2a v .

ox

(0 + ax i) = L[(i.i)a - (i.a)i] = L(a -a.i) = La- Lai

fn (a x

s

x r)ds

x

=

2ii

r

yJ).

Example 15. Verify Stoke's theorem for the function F = x(xi + integrated round the square in the plane z = whose sides are along the lines x = 0, y = 0, x = a, y = a.

°

The Classical Mechanics

34 Ans. We have, V x [x(xi + yi)]

=

yk

=

V x F.

s a a

=

a a

ff kyo Xdx dy ffy dx dy =

00

00

a

fa

o

0

= f ydy. dx

f(V

x

F).as

a2

-a

=

1

2"a 3 .

2

1 3 2

-a

=

s

t F.dr f F.dr + f F.dr + f F.dr + f F.dr =

Again,

C

t F.dr

Now

=

OA

AB

Be

co

fx 2 dX+ f(a 2 i + ay)).)dy+ f(x 2 i + a)x).(-i)dx + fO(-))dY

=

coo

0

0

1 3 1 3- -a 1 3 =-a 13 + -a = -a 3

:. tF.dr

2

3

=

c

2

f(v x F).ds s

Hence, Stoke's theorem is verified.

EXERCISES 1. Simplify

2A+13+3e-{A-213-2(2A-313-e)}

2. Show that

[ADs.SA-313+e]

IA +13 +el ~ IAI +1131 +lei·

3. A quadrilateral ABeD has masses of 1, 2, 3 and 4 units located respectively at its vertices A(-I, -2,2), B (3, 2, -1), C(1, -2,4) and D(3, 1,2). Find the [ADS. (2, 0, 2)] co-ordinates of the centroid. 4. In each case determine whether the vectors are linearly independent or linearly dependent.

(a) A = 2i + J- 3k, 13 (b) A = i - 3J+ 2k, 13

i - 4k, e = 4i + 3) - k = 2i - 4J- k, C = 3i + 2) - k =

[ADS. (a) Linearly dependent. (b) Linearly independent]

Vector

35

5. For two vectors

A = iF - 3) + k, B = -i +2) - k.

Find the projection of

[Ans.-~] ..[6.

A on B.

6. If 111 and 112 be the unit vectors and e is the angle between them, show that

I

. -ellA A. sm = - nl - n2 2 2 7. For three vectors find

A=i-

2) -

3k. B = 2i + J- k, (: = i + 3) - 2k

(A x B) (B. C) . (A x 13).(C x D) + (13 x C).(A x D) + (c x A).(13 x D)

8. Prove that

= 0.

9. If (:1 and (:2 are constant ve\;Lors and I.. is a constant scalar, show that e-Ax

(C 1 sin I..y + (:2 cos I..y)

(iB a2B ax2 + 8y2

11. Evaluate V2(2n 12. Evaluate

satisfies the partial differential equation

=0.

V.( cpA)

10. Show that

=

Vcp.A + cpV.A

r)

[ADS.

v.[r vL; )1

A = ~,find V(V.A) Prove V2(cp\jf) = cpV2\jf + 2Vcp. V\jf + \jfV2cp .

15. Prove that

F=

[ADS.-2r-3 F]

r2F is conservative and find the scalar potential. [ADS. cp =

16. Evaluate

r;]

[ADs.3r~]

13. If 14.

B=

fA.dF

along a curve x 2 + y = 1,

Z

=

r: +

constant]

1 in the positive direction

C

from (0,1, I) to (I, 0, I) if 17. Evaluate

x 2 + y2

ff ~ x

2

A = (yz + 2x)i +xz)+(xy+2z)k

[ADS. 1]

+ l dxdy over the region R in the xy plane bounded by

R

=

36.

[ADS. 144n]

36

The Classical Mechanics

18. Evaluate

fr.ds over the surface ofasphere ofradius

a with centres at (0,0,0).

s

[Ans.41ta 3 ]

19. Evaluate

fF

dv where V is the region bounded by the surfaces x

= 0, y = 0,

v

y

= 6, Z = X2, Z = 4. and

F=

2xzt - xJ + ifc

[Ans.128t - 24J + 384 fc]

20. A fluid of density p(x, y, z) moves with velocity vex, y, z, t). If there are no

- - op

sources or sinks, prove that V.J + rt

21. Verify Stoke's theorem for

A=

=

°

where, j

= pv.

(2x - y)t - yz2J- iz fc, where S is the

upper half surface of the sphere x 2 + y2 + Z2 = 1 and c is its boundary. 22. Prove

f ?dV

fr.n -;Ids.

=

v

s

fnds °for any close surface S. 24. Prove fr x as 6 for any close surface S. 25'. Prove ~cp dr fas x Vcp .

23. Prove

=

s

=

s

=

c

s

-:0:-

Chapter-2

Linear Motion

2.1. Introduction: Classical mechanics basically deals with motion and cause of motion. At the very beginning, it is important to have a clear conception about linear motion and its characteristics to study whole mechanics when a particle is displaced in a particular direction then it is said to be in linear motion. Such change in position along a given direction may occur under external force or not, but the fact is that characteristics of linear motion in several situation is very significant at several situation of classical mechanics. Linear motion is thus, a starting branch of mechanics and we will actually go through this chapter with kinematic view of study.

2.2. Kinematics: A part of mechanics which deals with forces causing the motion is called Dynamics. Kinematics is that branch of dynamics which deals with motion alone without considering its cause. In this branch, we shall discuss the path described by a particle and determine the velocity and acceleration at any point along its path. At every instant of motion, it is required to find the instantaneous position of particle about an arbitrarily choosen origin and then taking variation of it w.r.t. time, one can deal with the nature of particles motions. We will now deal with the kinematical treatment for the particle, moving linearly.

2.3. Basic Definitions of Required Parameters: It is very urgent to define some basic parameters required in dealing any mechanical problem. These parameters are given below: (i) Displacement: For motion of particle in a given direction, its change of position in that direction is called displacement. It is a vector quantity and is denoted by

S. 37

38

The Classical Mechanics

So mathematically, S = t.;i =

r2 - r)

= change in position.

where rj and Y2 are instanteneous positions at initial and final position ofthe moving particle. (ii) Velocity: For a moving particle, its rate of change of position or rate of displacement is its velocity. Here, this velocity is a vector quantity and is denoted by v. So, But we should always remember the instantaneous velocity of the moving particle at any instant of its motion is fly dy Lt - = t1t~O M dt which is now estimated by the slope* of a tangent, drawn at any instantaneous point of "position-time" graph. (iii) Acceleration: It is another significant parameter for linear motion which

v=

is defined by the rate of change of velocity and is mathematically given by, flv

J=

v- u

v and ii are respective final and initial velocities of this moving particle. iv - iii f= =Now

where,

I~

iv-iii

and since,

where,

I

= (v 2 +u 2 -2uv cos

eis the angle between ii

and in case,

vII ii

t

and

in same direction,

and

e)2

v. So,/= ~(v2 + u

I

2

- 2uv cos ~)2

e= 0

f= v-u

f= - - (where, v> or < u) t

*

Practically, this slope of position-time graph at any instantaneous point gives the magnitude of velocity at that instant and is precisely given by,

I

Ivl = t1~;O ~I point.

=

I~I

= tan

e= Slope of "position-time" curve at instantaneous

Linear Motion

39

Also for particles motion, the instantaneous acceleration* is given by,

1

= Lt i\t -+ 0

.ilv .ilt

= dV dt

L I.ilVI 1 = IJiiI = L\/-! 0 dt = IdVl dt I

when,

which is estimated by the slope of tangent drawn at instantaneous point of velocity-time graph. (iv) Average velocity and acceleration: For motion of particle, average velocity is estimated by taking ratio of total displacement and total time taken for the motion of particle, provided that the particle traverse several displacement in several interval of time. So mathematically Total displacement Average velocity = Total time taken SI + S2 + '" .iltl + Llt2 + ...

v

:. < >

n

n

where, SI'

82 ,

....•

are respective displacement occurs in respective time

intervals .ilt I' .ilt2 , ••••• etc. We can also have, for average velocity,

< V> =

I il2 v-dt -t--t2 -

1 1\

when we study the variation of

v(t)

in a particular

interval of time from t\ to t 2 • Similarly, the average acceleration is the ratio of total change in velocity to total time taken and it is then given by



.ilvI + .ilv2 + '" Llt)

+ .ilt2 + '"

and also,

I

= -t2 - t)

112 1- dt. 1\

Momentum: It is such a parameter which is the sample product of velocity with mass of particle and is given by, Momentum,

p=

mv.

v and thus for the instantaneous momentum, the instantaneous velocity v is now, v = v(t) where

*

p is directed along

The direction of acceleration is along the direction for the change of velocity, occured in a certain time interval. The retardation or negative acceleration is taken in reversed direction of acceleration.

40

The Classical Mechanics

Again, the average value of

p is given by,

- = --J, m rl2 vdt

= -1- il2 pdt t2 - tl

t2 - tl

II

II

m· (vi) Relative velocity and acceleration : For two moving particle, the velocity of one w.r.t. another is the relative velocity of 1st particle w.r.t. 2nd particle. If VI and V2 be the instantaneous velocity of particle -1 and 2 respectively then,

v2

Relative velocity of particle -1 w.r.t. particle 2 is vl2 = VI -Similarly, the relative acceleration is the acceleration of one particle w.r.t. another particle which is given by,

--

-

fi2 = fi - 12' we should now remember that

!VI2!

=

!V2I! and V12

=

-V21

also, I

when

V 12

=

V 21

= (vf

+ vi - 2vlv2 cos

a)2 I

and

J..2 =1;1 = (fi2 + Il- 2fih cos a)2

2.4. Velocity and Acceleration in Several Co-ordinate System : We have already discussed in our earlier section that the instantaneous velocity and acceleration are given by, di _ dV d2 ; V = dt ' 1 = dt = dt 2 So we see that, these parameters entirely depends on position vector ; and its variation W.r.t. time, thus it is important to express them in several coordinate system. In certesian co-ordinate system (3 dim)

di

v

or,

v

and

1

-

dt

dx

A

= -i

dt

dy

A

dz

A

+-j+-k dt

dt

Linear Motion

41

xi + yJ +zk. In cy1inderical polar co-ordinates (p,
v = vpP + v + v) I

=

PP + pq, + ik

Ii) + l + I) (15 - pq, 2)p + (PiP + 2pq,) + z"k

In spherical polar coordinates (r, e,
v=

We have,

vrr + vee + v

r ee

v

or = f +r + r sin 9 q, . and similarly for acceleration,

J= (r - r 82- r sin 29 q, 2) r + (r e+ 2f 8 - r sin 9 cos 9q, 2) e+ (r sine

V

=

8r Lt cSt~O 8t

dr dt

=-

8r

fu' dr ds ds dt

= -.-

A

when, in limiting case, as P approaches pi, dr ds i and ds

x

A

dr

=

_ v

:- ds dt

I

A

= I

~

= :-

= A

vi

~2

t

Here, v = dt and i is the unit vector along the tangent at P. So we see that the velocity of the moving particle has a direction, tangent to the path and a ds magnitude dt' Let i + 8i be the unit vector along the tangent at pl. If

42

The Classical Mechanics

inclinations of the vectors £ and £ + 8£ to OX and to

J is unit vector at P normal

£ then, from figure, we have . 8m

Slll-'t'

2

= ~~

1 for, 8

2

8i I .. 18

-

=

1 and I:8i

= J.. A

u". m

di dj Similarly, d

A

-i .

Hence we get the acceleration,

_

f

dV

=

- =

f

dt i

=

d(-:dS) dt I dt

=

.d2s ds d£ I dt2 + dt dt

d2s + ds J d

=

2 -:d s ds ii d

2 £d s + Jd
=

;(:7":fit;; :7v~ at is

~u~ili~ ;;:~s :~

P

ds p = dp .

dv So the tangential component of acceleration is dt and the normal

v2

component of acceleration is - . p

2.6. Radial and Transverse Component of Velocity and Acceleration: Consider instantaneous position of a moving particle is given by per, 8) by using plane polar co-ordinates. If then,

v=

vxi + vyj

v be instantaneous velocity at that position,

... 1

Now the component of this instantaneous velocity in radial direction is called radial component of velocity whereas the component, perpendicular to the radial direction is called cross-radial or transverse component of velocity. These two components are denoted by vr and va by which, we may write

v= Now from fig. i =

r cos 8 -

Vr

r + va e

esin 8

...2

Linear Motion

J = r sin S + ecos S.

and then

43

So, if (x, y) be the instantaneous position in Cartesian Co-ordinate system, -:J x = r cos S r y = r sin S and we now have, A

dx . vx = dt =rcosS-rsinS.S.

dy

.

vy = dt = rsinS+rcosS.8.

Using these equations in expression 1, we get

v

= (r

cos S - r sinS. 8)(r cos S -

=> v =

2

esin S) + (r sin S + r cosS. 8)(r sin 8 + ecos 8)

r-; cos 9 sin 9 e- r sin 9 cos 9 8 r 9 e+ r sin 9 r + r sin9 cos9 e+ r cos 9 sin 9 8 r + r cos

; cos 9

+ 8 r sin 2

2

v=

2

9.8. ... 3

;r+r8e

Comparing this equation with equation (2), we get

v

r

fa

=

dr . dS v = r 9 = rdt' e dt

. = -

r

which are respective radial and transverse components of velocity. Similarly, for acceleration having radial and cross radial component f,. and respectively, we have,

J=f,.r+/ee

.. .4

and,

J = f) + fyJ

... 5

When,

!, = -dv x dt

x

=

and

J:

= -d (.r

r cosS -

= dvl... =

y

dt =

dt

cos S - r S·· sm 9)

2; sin9 8 - r e sin9 - r e 2 cose

~(;sinS+r8cosS) dt

r sin S + 2; 8 cos S + r e cos S - r 8 2 sin 9

Thus from equation 5, we get,

J = (r cosS - 2; 8 sinS - resinS - r 8 2 cosS)(r cosS - esinS) + (r sin S + 2; 8 cos S + r ecos S - r 8 2 sin S) (r sin 8 + ecos 9 )

44

The Classical Mechanics On solving this equation, we finally get,

j

=

(r - dF)P + (r 8+ 2f e)8 .

. .. 6

Comparing this equation with equation (4) We get, 2

Ie

and

which are respective

. . _ r d e + 2(dr) (de) -_.. r e + 2 r e - dt 2 dt dt r~dial

and transverse components of acceleration.

2.7. Newton's Laws of Motion: In kinematics, Newton gave his three famous laws which are essential for any mechanical problem. These laws are, (i) Law of Inertia: This law states that ifno force be applied from outside, then rest particle will always remain at rest and moving particle will always remain in motion along a straight line. That is, every particle will try to retain their self inertia. (ii) Law of Force : This law states that if forces be applied on a moving system then its change in momentum will occur in the direction of resultant force and the resultant force will be directly proportional to the rate of change of momentum of that system as a whole. By this law, mathematically, we have,

_ F where,

P=

L Pi

=

"

_

= L..J Fi =

.J = djJdt'

d (" _ dt ~ pl

momentum of the system as a whole.

i

(iii) Law of Interaction: This 3rd law of Newton states that to every action, there is an equal and opposite reaction, when these two are acting on two separate

bodies. So if

F and R be the respective action and reaction then F = - R when, F = R.

One should now remember that application of Newton's laws has some limitations. (i) These must be valid in inertial frame (ii) This should be applied on a moving particle having velocity much less than that of light in free space. This is because of the fact that for non-inertial frame, a pseudo force* will act on system which can not be detected from outside and for particle velocity tend to that of

* See

the chapter 'Reference Frame'.

45

Linear Motion

light in free space, we should pass through relativistic zone* in which particles mass will inverse with increase of velocity. But Newton's laws are only valid for fixed mass only. 2.S. Accelerated Linear Motion : When a force or a no. of force be applied on a moving particle, then the motion will be accelerated and that unbalance force (resultant) creates acceleration of the particle or system of particles. The acceleration is then given by

LFi

J = _i_

by Newton's law and for such accelerated motion, the equation

M 2

d ; of motion will be M dt 2

=" F. =F --

.L..J

effective.

I

i

2.9. Graphical Treatment of Linear Motion : When a particle or system of particle is in linear motion, either accelerated or not, the motion can also be studied graphically. Since, during its motion, position is given by, ;

=

;(t)

which changes with time, the position-time and t velocity-time graph gives r (i) Magnitude of instantaneous velocity which is estimated by the slope of tangent drawn at any instantaneous position of positive-time graph. i.e.,

Ivl =

/).r t.t~ 0 /).t

=

dr dt

* -.;>/

1M 1M I I I,

I I II

I

I

I I

o

I I

t~

= tan O.

(ii) Magnitude of displacement which is estimated by the area of velocityt2

time graph and it is given by,

lsi = It, v dt .

= Area of velocity time graph. (iii) Average velocity (magnitude only) can also be estimated from the area of velocity-time graph divided by the interval of time. It is now given by

*

In relativistic region, mass varies with velocity as m = where,

p = vic, m =

moving mass, mo = rest mass.

46

The Classical Mechanics



rt2

1

--J, vdt. t2 - t) II

=

i

(iv) Magnitude of instantaneous acceleration can

u

be estimated by the slope of tangent drawn at instantaneous point of velocity-time graph,

~ -->\

1M I~U I I I I I I I I I I I

I I

I I I I I I I I I

L'1v dv o t~ = - = tan 8. M~ 0 L'1t dt (v) Nature of motion can also be studied from position-time and velocitytime graph. i.e.,

I~

Lt -

2.10. Conservation of Linear Motion: If no force be applied on a moving particle or system or if the resultant of all applied force be two, then the momentum of that particle or system will be constant. This is principle of conservation* of linear momentum. Now from Ne~~'s law O(f fOr~e) d L.li = · Smce, . dt "p. ~ I i

I

So for

IFi = 0, i

or,

I

Pi

=

!!'-(Ipi) =0 dt i

constant

This is conservation of Linear Motion.

2.11. Time Integral of Force (Impulse) : The time integral of force is basically the impulse of force and it is denoted by j. Mathematically j = If

or,

*

jdJ

=

r II

FCt) dt

F, the force be time independent, then

J =

FM

=

Force x Time of action

This conservation of linear momentum also can be obtained from Newton's 3rd law and so for collision of two particles, mutually colliding or mutually interacting without any applied force, the conservation of linear momentum holds. ~ See the chapter, 'Collision Theory'.

47

Linear Motion But from Newton's law,

m(v - u)

mv-mu

M

!!.t

F

J =

FM = mv - mu = change of momentum.

Similarly, also for time dependent force,

J

r r

=

II

l2

J =

or,

r J~

Fdt

=

II

dj; dt dt

dj; = (p\ 2 - (p) I I =

P2 - PI

= !J.p = change in momentum Thus the time integral* of force or impulse of force is estimated by the change in momentum. Also for time varying force, the average of force in given time interval can also be obtained from this time integral of force, and is given by,

Fav

1

= -t2 - tl

112 Fdt II

.

2.12. Work: The work done by the force F causing displacement Si of a particle is given by,

!J.W = F.!J.i = F!J.r cos 8 = F S cos 8 where, 8 is the angle between the vectors

F and

SF and S is the magnitude of !J.i which also denotes an element

of length along the path of the particle and F cos 8 denotes the projection of

F

on the tangent to the path. Generally, mathematical representation of work done is represented as W=

f

dW =

f

F dr cos8 =

fF.di.

where, F may not be constant and may depend on the co-ordinates of the particles. The work done has particular sign convention that it is taken when work is done against force, and is considered -ve when work is done in favour of force.

* Time

integral of force or impulse is also estimated by the area of "Force-time" graph.

48

The Classical Mechanics

2.13. Power: The power* of force F or its activity is the time rate at which work is done by the force and is given by

~

p =

In case,

:t (F.r).

=

F is time independent, - dr

p

F'd(

=

=

F.v

Thus the power is equal to the scalar product of the force and the velocity of the point of application of the force.

2.14. Energy: Kinetic and Potential: The energy of a particle or system of particles is the ability of doing work. When work is done on the system, its energy will increase and also if work is done by the system, then the energy will decrease. So, normally, the work done is related with change of energy, and is mathematically given by,

=>

f-

W

=

W

= ~E =

F.dr =

fE2 E dE = E2 - El . J

E2 - E\

or, work done = change in energy. Now for any system, the energy or total energy is the sum·· of kinetic and potential energy and thus it is given by,

E = EK + Ep

=

K.E + P.E.

Now for system of particle or motion, its kinetic energy is

,,1 2,,1_E = K.E = L..J -2 mivi = L..J -2 mivi' vi K

.

I

.

I

On the other hand, the potential energy is the energy which a body possesses by virtue of its position or configuration and is measured by the work done by the body in passing from its present position or configuration to some standard position or configuration. The potential energy in the standard position is taken to be zero. • It can be shown that for particle in linear motion, if the power P remain constant,

then the distance'S' traversed by the particle along straight path will be 3 proportional to the power of time taken. i.e., for P = constant, S a(tj312 .

'2

.. This sum of kinetic and potential energy is taken in Newtonian Mechanics. But in relativistic zone, the total energy F0 is the sum of Kinetic -energy and rest energy of particle. --+ See any book on Relativity.

Linear Motion

49

Thus the potential energy is now given by

where, F denotes the force acting on the particle whose position vector is '1;

r2

denotes the position vector of the fixed point.

2.15. Conservative Force: A force is said to be conservative if the work done by the force along a close path be zero. So for this conservative force, if energy loss occurs in one part of the path, then it will overcome in another remaining part of the path and total work done along close path will be zero, and ultimately no change in energy will occur. So for conservative* force, F

f F.dr

=

o.

c

Now by Stokes theorem,

fF.dr

f(\7 x F).as

c

s

and then for conservative

force.

\7 x F = Since, for any scalar
F = \7
for

O.

\7 x F = 0

But conventionally, conservative force, F is mathematically represented as F

= -\7

2.16. Conservation of Energy: For conservative force

F = -\7<j) , where

easily show the conservation energy. Since,

*

F.dr = - \7
Examples of conservative force are gravitational force, coulomb force, magneto static force. etc, all of which are negative gravient of gravitational potential energy, coulomb potential, magnetostatic potential energy ... respectively. ** Here, -ve gradient is taken for requirement of energy conservation.

50

The Classical Mechanics

we have,

Also,

m

=

f

V

2

vdv

I 2 - mv2

or,

2

+
1

2

= - mVI

2

2)

m (') vi - VI 2

= -

VI

+
This shows that in a conservative field, the sum of the kinetic and potential energy of a particle at any point remains constant. This is called conservation of energy.

2.17. Center of Mass and Its Motion: A fixed point with respect to a body at which the total mass of the body is supposed to be concentrated, is called center of mass of the body. If r is the position of center of mass of the body w.r.t. an arbitrarily choosen origin, '0'. then, mathematically,

r

when, M

=

L

mi =

M

Total mass of the body and

mj

is mass of ith constituent

of the body.

L

mj~ = Mr is the total moment of mass* of that rigid body. Here, Now in actual case, for every constituent particle of a rigid body, two types of force will act on every particle. One is external force and another is internal force, one to mutual interaction with one constituent by other remaining constituent particles. So for ith constituent particle,

* For rotation of body about an axis passing through its center of mass, total moment n

of mass will be two. i.e.,

L i= I

mi

r; = 0 for this rotation.

51

Linear Motion n

F + IFij j

j=l

(j,. j)

when, body contains total 'n' constituents and here, Fj is external force on ith particle. We can now write down, n

n

11

IF + I

IFij

j

j=

1

j=

1

=

Im;

dZ dt;j

i= 1

(J" i)

But here, 2nd term of L.H.S. is zero since the sum of pair of forces Fij and Fji is zero.

ov, or,

F

This is equation of motion for center of mass. Thus the motion of center of mass of a system of particles relative to an inertial frame is the same as that of a single particle of mass equal to the total mass of the system under the action or some interval force acting on the system.

F which is equal to the vector sum of the external forces

2.1S. The Two Body Problem:

We now consider two particles of mass m 1 and m 2 separated by a distance and acted on by external forces FI> Fz and internal forces F1Z and FZl respectively.

r

Now, by Newton's 3rd law,

F1Z

=

FZl .

y

So the equation of motion for the two particles are

d zrl ml dt Z

where,

il

and

respectively and

r2

=

F12 + Fl

x are instantaneous position vectors of two masses

r = il - rz .

52

The Classical Mechanics

Now,

-

2 d i).

= FI2

dt 2

ml

-

+!L ml

and

If the motion of two particles is only due to internal forces arising from the interaction between the two particles, then

FI

= F2 =

6

and the above equation is reduced to

d 2(i).- r2) dt 2

where, 11

=

_ = FI2

mlm2 ml +m2

is reduced mass of the system such

relative position vector of mass

ml

W.r.t. mass

nl2

d (-

and dt

r = rl -)

rl -r2

=

r2

is the

di). dr2 Tt-Tt

VI - v2 = v12 = relative velocity of m l w.r.t. m2, this two body motion is thus possible to replace by n single particle motion for the particle of mass 11, having velocity, equal to the relative velocity of one particle w.r.t. another. Here one should remember, that, for two body problem, six to internal force only, center of mass or the system remains stationary, only for this case, such replacement of two body motion by one body motion is possible otherwise, two body motion can be considered as motion of reduced mass '11' and center of mass* (nil + m 2)

=

2.19. Application of the Principle of Linear Motion: Principle of linear motion means principle of linear momentum conservation. This principle can be applied in several dynamical case, such as

*

For two body motion with center or mass Chanj'ng its position, it can energetically

1 2 1 2 be shown that -mini + -m2n2 2 2 where, U =

(mlvl (ml

+ m2 v2) + m2)

= -1 (

ml m2

2 ml + m2

(

VI - v2

)2

+ -1 ( ml + m2 ) xU 2 2

Linear Motion

53

collision theory, Firing of gun, ... etc. In many cases another type of motion may be replaced by equivalant linear motion and then in that case, all axioms will hold for such kind of motion. A study of linear motion thus helps to have basic knowledge of mechanics, and when required principle of linear motion can be applied to deal that problem. 2.20. Mechanics of Variable Mass : Principle of Rocket Motion: Uptill now we have considered the equation of motion and the laws of conservation in such cases when the mass of the system was constant during the motion of the system. We will now consider the motion of a system when the mass varies with time. There are so many examples of such system in nature and also in technology. As for example, when a drop of rain falls through a cloud, it will gain in mass as it descends. Rocket motion is another example of such system when mass varies with time for linkage of exhaust material or fuel after burning from it. A rocket fired from the earth will always be affected by the gravitational pull of the earth. We will at first consider rocket motion in horizontal direction and then will modify it for its vertical motion against gravitational pulling. M-~M

M

......------...>

-4

~D I

>~

~M

M-~M

Let M be the instantaneous mass of a rocket with its fuel, while moving along horizontal direction. V is its instantaneous velocity and we also consider that due to burning of fuel, the exhaust material or gases leaves from rocket with constant velocity Il W.r.t. rocket in backward direction. Hence, if at instant, (t + Ot), the mass of rocket with its remaining fuel be (M - ~M) and velocity be (v - Ov) (obeying momentum conservation) then in that said interval M, ~M mass of exhaust gas leaves from the rocket. Hence, from momentum conservation, Mv = (M - ~M) (v - ~v) -~M(u - v) ~ M ~v = u ~M (.,' ~v.~M = small) Dividing throughout by ~t and taking the limit as M ~ 0, we get

dv

dM

M- = - u dt dt

...

(1)

Here -ve sign is added on the right hand side to indicate that velocity v increases as mass M decreases. So integrating equation (1). w.r.t. time, we get v

JVodv

=

1M -dM

-u

Mo M

54

The Classical Mechanics

=>

v = Vo -

=>

v = Vo

U

lOge( ~: )

-UIOgc(~)

... (2)

when Vo and Mo are respective initial velocity and mass at instant t = O. This equation 2, thus gives an expression for instantaneous velocity of rocket motion. dM and it lasts Let us suppose that the fuel is burnt at constant rate K

=Tt

for time T. If the mass of the vehicle be Mv and that of the fuel at t = 0 be

MFo then Mo = Mv + MFo . The mass of the vehicle-fuel system at any instant t can be written as

M

=

Mv +MFo (1-;)

=

M v +M1\0 -M1\0 -T

t

t

=>

M = Mo - M1\o -T for 0 < t
and M = Mv for t ~ T. ' Then for equation 2, _ dx _ Vo -ulog. (1- MFo .!....) dt I Mo T

v -

... (3)

Again, integrating this equation with respect to time we get.

x =

Xo

+ vot -

t

S

U 0 loge

(

MFO

t)

1 - Mo T dt.

But we have,

Thus,

x

= Xo +

vol -{(I - ~:~ }Og. (1 -'::: ~) -I]

... (4)

This is the distance covered by the rocket in time t. Since the rocket attains maximum velocity at t = T when all its fuel is burnt out. The maximum velocity calculated from equation (3) is given by,

Linear Motion

55

v = Vo -ulog (1- MFO)* max e Mo V = Vo + u loge (Mo ) max Mv

Vmax=VO+U10g{1+~O)

=>

... (5)

If the rocket is moving vertically upward and if the gravitational pull of the earth on it is assumed to be constant, then the equation of motion of the rocket can be written as dv dM M- = -u--Mg dt dt ... (6)

Integrating this equation (1) w.r.t. time, we get, v=

U10g{~: )-gt (taking, Vo = 0 at t = 0)

and similarly, x = ut-'!gt

2

2

-(t-

.!.-). (taking 11

MoT)lOge(l_ MFo MFo Mo T

= 0

o att = 0) This is all about rocket motion.

SUMMARY 1. The displacement of a moving particle is change in position in a given direction. i. e. S = & = r2 - rj . 2. The velocity of a moving particle is the rate of change of position. i.e. -= v

!1r

dr

Lt-=t.t~O M dt .

3. Acceleration of a moving particle is the rate of change of velocity. i.e.

* Here

- = Lt !1v = dV M~O!1t

f

MF the ratio M

0

dt .

has a practical limit and it can not be increased indefinitely.

v

A single rocket therefore will not attain high velocity that is required.

56

The Classical Mechanics

4. Average velocity is the ratio of total displacement and total time taken by the moving particle.

i.e.

LSi

r/L

1 == - - J, v dt . L../l t2 - tl /,

v

< > == "

5. Average acceleration is the ratio of total change in velocity and total time taken by the accelerated particle.

i.e.

_

L~vi 1 < f > == , , - == - L.ji

t2 - tl

1/2-fdt /,

.

6. Relative velocity is the velocity of one moving particle w.r.t. another particle in motion or at rest. i.e.

Vru = V2 -

vI

Similarly, relative acceleration is the acceleration of one accelerated particle w.r.t. another accelerated particle. i.e.

-

fru ==

i2- - fi- .

7. For particle moving in three dimension, the velocity ==

or,

or,

xi + yJ + iu

A

A

A

vxi + vyi + vzk

(cartesian)

vee + vee == i-; + ree + r sin Oe

v ==

v=

vr ; +

(spherical polar)

v == vpP + v
== PP + pci>cP + ik (cylinderical polar). 8. For any moving particle in three dimension, the acceleration,

J == f) + fyJ + fJ == xi + yJ + zk (cartesian) or,

or,

J == fr.p+fe·e+/z.z 2 == (r - re ); + (re + 08) e+ zz . (cylinderical polar) J == 1,.; + fee + flp == (i- - re 2 - rsin 2 94>2); + (re + 2;.8 - r sin 9 cos 94>2) e + (r sin 9 + 2i-4> sin 9 + 20r4> cos9)cP (spherical polar).

9. Tangential component of velocity and acceleration

VI

ds == dt

dv d 2s ,1; == dt == dt2

Linear Motion

57

Nonnal component of velocity and acceleration v2

vN=O,jN= - . P 10. Radial and Transverse component of velocity are vr =

r, va = ,e

11. Radial and Transverse component of acceleration are

J,. =

r - rB. 2 ,fa =

..

.

rB + 2Ye. 12. For accelerated motion of particle, equation of motion is

d2M_r = "F. = dt 2 ~ I Feffective I

13. For conservation of linear momentum,

L Pi

= constant when Fapplied = O.

14. Time integral of force or impulse, is t2

j

=

Jrtl Fdt

P2 - PI

=

=

/).P

= change in momentum.

15. Work done by the force is W=

fF.dr

W = -ve for work done by the force = +ve for workdone against the force. 16. Power of a system is mathematically given by, dw

P=

dt

F.v·

=

17. Energy of a system is given by E = K.E + P.E

K.E =

L ~ mi(Vj'Vj ) j

P.E

f

= W = F.dr .

18 For conservative force and

Vx F = 0

F,

fl dr

when,

=

0

F = -V

58

The Classical Mechanics

when,

;

Imi~ =" = M

"m. L.-J I

L.-Jmi i 20. For two body motion, with stationary center of motion, d 2; /..1.-- dt2 - F when, F12 = F21 = /..I. =

IFI

= mutual interaction

mlm2 = Reduced mass. ml +m2

and dv dM 21. For Rocket motion M- = - U - . 'dt dt But for motion in vertical direction the equation of motion is given by, dv M dt =

dM

-uTt - Mg . Worked Out Examples

1. A car covers is quarter of its journey at a speed 30 kmIhr, second quarter at 40 kmlhr and the rest at 60 kmIhr. What is the average speed of the car during its journey? Ans. Let the total distance covered by the car is S. So if v be the average velocity then Total distance travelled v= Total time taken S - S - - - S - - - S - = 43.63 kmIhr

--+--+--

4 x 30 4 x 40 2 x 60 2. A motor car moves 40 km due east and then 30 km due north in 6 hr. Calculate its average speed. Ans. Here, the total distance covered 1

S = (40 2 + 30 2 )"2 = 50 km. so the average speed, v =

50

6'

= 8.34 kJnlhr.

3. A car moves due north at 50 kmlhr, while a wind blows from the north east at 30 kmIhr. Find the speed and direction of the wind as they appear to a man in the car. Let, VI = velocity of man, v 2 = velocity of wind Here, if Vo be the relative velocity as appear to a man, then

Linear Motion

59 N

I

Vo = [1-l\12 + IV212 + 21-vlllv21 cos 45)2

= (50

2

~

2

+ 30 + 2 x 50 x 30 x

I

y

--------~-------E

= 74 kmlhr. If it makes angle 8, with south of west,

30 sin 45 then, tan 8 = 50 + 30 cos 45

4. To a cyclist riding towards north at 10 kmlhr the wind appears to be blowing from the north west at 12 kmIhr. Find the true velocity of the wind. Let,

vI

= velocity of cyclist,

vI

= 10 kmlhr

N

True velocity of wind, which makes angle 8 with east or north. Vo = Relative velocity of wind w.r.t. -------r~-L~~---E the cyclist. and Vo = 12 kmIhr Now from fig, 122 = 102 + + 20 2 cos(90 + 8) => 44 = 20 2 sin 8. V2 =

v/

v/- v

Now tan 45

=1=

V2

cos 45 =

sin (90 + 8)

10 + vI cos(90 + 8)

1

But

v

Ji

10 2 + 122 _ v 2

_ _ _ _ _=-z

=> 10 =>

V22

2 x 10 x 12

vI

88 = v2 cos 8. 244 -

120Ji , v2 =

8.38

kmlhr and Sin 9

=

(vi - 44)/20 v2 , 9

~

10° 20'.

5. A man is walking on a horizontal road at 3 k.rnItr. The rain appears to him to come down vertically at the rate of 3 kmIhr. Calculate the actual velocity and the direction of the rainfall. Here,

VI

= velocity of man,

VI

= 3 kmIhr

Vo = Relative velocity of rain w.r.t. man, Vo = 3 kmIhr

True or actual velocity of rain fall, which makes angle 8 with vertical. 32 = 32 + + 2 x 3 x 2 cos(90 + 9) 6 sin 8 = 0 :. 6 sin 8 = v 2 v2 =

.. =>v2 -

v/

v

v2 sin(90 + 9) Again, tan 90

= 00 = 3 + v2 cos(90 + 8)

-vt

v+ 2

60

The Classical Mechanics :. v2 sin 8 = 3 sin2 8

= 112 :::::> 8 = 45°, v2 = 6 sin 45 =

3.J2 kmIhr. 6. A projectile shot from the ground has range R and the maximum height it reaches is it. Find the magnitude and the direction of its initial velocity. Ans. Let v be the initial velocity for projection and 8 is the angle of projection. I I For maximum height reached H, I :.

H

=

I

IH II

v2 sin 2 8 - - - . If T be the total 2g

time of flight, v 08 = g T/2. .. T = 2 v sin 8/g.

I

R

(

So for range, R, R = v cos 8.T = v cos 8.

2v sin8

------+)

v sin 28

2

g

g

e~

cos-

I

(

~~::

2gH

Again, v2

This is angle ofprojection.

).

2gH 1- (R 2g/8H) .. v

= -1--"'::co'--s~2-e =

16gH2

=

8H _ R 2 g ~ velocity

of projection 7. A particle has total energy E and the force on it is due to potential field v(x). Show that the time taken by the particle to go from x\ to x 2 is t2 - t\ =

/2

[2E - v(x)t dx.

Ans. We have, total energy E =

v=

..

1

"2 mv2 + v(x).

~ ~ [E - v(x)]

= \

:

5: [E -

=

2

:::::>

I

dx \ = v(x)]2

{~ [E - V(X)]Y {2 rl2 dt V-;;; JI1

g

[E -

v(x)]~

Linear Motion

61

8. A particle is projected vertically with velocity v. What height it will reach from ground? If it reaches upto height 'h' then 1 2 "2 mu = W = 1 2 "2mv

=

rR+h

JR

aMmh R(R+h)

aMm --;:z-dr where, R is the radius of earth. 2aMh

~v2= R(R+h)'

aM 2gRh Brg= R2 ~GM=gR2.:. v2= (R+h)'

~ _ R _ (2g R _ 1)

R + h _ 2gR h - v2

- h=

-

h -

1(2~~

v2

.

- 1) This is the height reached

9. A particle is allowed to fall under gravity from height h. If the air damping on it be proportional to instantaneous velocity of the particle, find the velocity attend at time t. Here, for instanteneous velocity 'v', this air damping F a. v ~ F = -Rv. Here, R = resistive constant. dv Since, the equator of motion of the falling particle is m dt = mg - Rv.

~ f g--v d~ fdt ~ s:g ~i =

=

m

m

g--v loge

(

v = lotdt

R gm

J

= _

~

R y - - v = -Rt/m mg e

This is instantaneous velocity of falling particle. 10. Given force

F=

xy

i - i J' find the work done in moving a particle

from (0. 0) to (2, 1). For displacement from (0, 0) to (2, 1), ut, Y = t, x = 2t So, t runs from 0 to 1

62

The Classical Mechanics and

dy = dt, dx = 2dt. workdone

J

J

= JF.dr = xydx - idy

= Jrl 2t 2 .2dt -

f

r)

2

2

t dt = J 3t dt = 1 unit o o 11. Two particles of mass m) and m2 are moving respective with velocity v) and v 2 • Show that the total kinetic energy of the system is

+ m2v2 (m) + m2)

m)v)

when, v =

Here, total energy of the system is 1 2 1 2 v2 //) E = -2 m)v) + -m2 2 ···1' m)v)

But let,

U

= v) - v 2 '

V

=

m)

+ m2v2 + m2

m1v1 + m2v 2 = (m l + m2)v m)(u + v2 ) + m2v 2 = (m l + m2)v. m)u

v = v---:...2 m) + m2 similarly,v1 = v +

m2 u m) +m2

So, from equation (l), the total energy is

E

1(

=-

) 2

In)

1

+ m2 v + - (

m)m2

22m)

+ m2

)u

2

12. Two particles of mass m and 2m approaches to each other due to their mutual attractive force. Find velocity of their center of mass. Ans. Since, no external force acts on the system, are from momentum conservation principle, mV I = 2mv2. where, VI and v2 are their resp. velocities.

Linear Motion

63

and for, center of mass position,

mixi - m2 x 2 Xc

=

ml +m2

the velocity of center of mass, dx ml vI - m2 v2 v = - c = c dt ml + m2

mVI - 2mv2 ml + 1n2

=0

13. Find acceleration and tension of string for At wood's machine. Ans. For tension of string T and acceleration of the system 'f, m 2g - T m2f T-m2g= mJ

=>

f=

(m2 - md2 (ml +m2) T=m 2g- m 2f

T

2m l m 2g

=

=

~f (m l

T

+ m 2)

Now, we can hence have,

[(ml + m2)2

J

- 4mlm2 g 2

(ml + m2)

2

This is the relation between acceleration f and tension T. Here, are all that for At wood machine,! increases for T decreases. 14. A rocket rises from Launching pad with an exhaust speed of 2kg sec-I. For what value of the ratio MJM y can the rocket escape the gravitational pull of the earth? Here Mo is the initial total mars of the rocket and My is the mars of the vehicle. Ans. For rocket velocity, v Now,

= Vo

+ u loge (

~: ) .

Vo = 0 (say).

v u

12

= escape velocity = 12 km/sec. =

2 km/sec.

= 210 ge (

~~). where, Mo = initial mars

M = My final mars. Mo M = (e)6 = 403.43. y

This is the ratio of (

~~ ) .

64

The Classical Mechanics

EXERCISES 1. Define the following parameters. Displacement, Speed, Velocity, Average velocity, Instantaneous velocity, Instantaneous acceleration, Average acceleration, Momentum. 2. What is relative velocity? Find its limit of magnitude. 3. Two bodies are allowed to fall freely from same height in 5 sec interval. Find their average velocity and acceleration after further 10 sec. 4. Draw velocity-time and acceleration-time curve in the following cases: (i) Particle is moving with uniform velocity. (ii) Particle is moving with uniform acceleration. (iii) Acceleration of particle increases uniformly with time. (iv) Acceleration of particle decreases uniformly with time. S. Give the significance of velocity time graph. 6. Rain is falling vertically in downward direction and water is accumulating in a container. Now, if wind flows horizontally, then do you think that water will accumulate in that container at the same rate. 7. A train starts from a station at rest and then moves with uniform acceleration./; along time tl. After that it moves with uniform velocity along time t2 and after that it fmally moves with retardation/2 along time t3 and comes at rest in another station. Draw velocity-time and acceleration-time graph and also find out the distance between that two stations. 8. A man is walking on a horizontal road at 3kglhr. The rain appears to him to come down vertically at the rate of 3 kmIhr. Calculate the actual velocity and the direction of the rain fall. 9. Three masses 2 gm, 3 gm and 5 gm are situated at the corners of an equilateral triangle with sides 5 cm long. Find the position of center of mars. 10. A rocket has a mars of 1100 kg and contains 900 kg of fuel. The maximum exhaust speed of gases equated from it is 104 km/sec. If the rocket rises vertically, then what should be the minimum rate of fuel consumption for just lifting it off the launching pad.

-:0:-

Chapter-3

Rotational Motion

Rigid Body Rotation

3.1 Introduction For particle motion along any curved path, the instantaneous position vector of particle will change its direction and magnitude at every instant of motion. Now for circular motion the magnitude of position will remain invarient and due to rotation of particle along any circular path, the instantaneous velocity will be directed tangentially at every position of particle and for any curved motion, except circular motion, this instantaneous velocity will be tangential. * In this chapter, we will deal the motion through several circular or angular parameters, like, angular velocity, angular momentum... etc and then we will extend the whole mechanism to the rotation of system of particle, i.e., to rigid body rotation. So our main objectivity is that in this chapter we will try to study the whole rotational theory in general view and try to draw the several conclusions to make the chapter easy. 3.2 Angular velocity and Angular Momentum: For point particle rotation along any circular path, the angular velocity is defined by the rate of change of angular position, i.e. the rate of angular displacement. This is basically a vector quantity and is denoted by w Mathematically, it is given by, w

*

Lt

~e

tl.t - t 0

!1.t

=

de dt .

For any curved motion, the instantaneous velocity has two components. One is radial component vr =r and other is transverse or cross radial component va = r

e.

So,

v=

v/ + veS

=

rr + res. When radial means along r and cross radial r, i.e along transverse direction. Now for circular

direction means normal to motion. r = constant,

r = 0, and velocity is only tangential, v = va = r e = r w. 65

66

The Classical Mechanics

This is actually, an axial vector* and is directed along the axis of rotation. For point

w

particle rotation, this angular velocity is normal to the plane of circle, i.e. the plane of and r when, is instanteneous linear velocity and r is instantaneous position vector.

v

v

It can be shown that v = because, the instantaneous displacement for circular motion is s = r.

wxr; small

ex

Now, Hence

v =

r xv

wx r

when,

w

de dt·

rx(wxF) = (r.r)w-(r.w)r = r2 w (·:r1-w,;:.w=O). =

1 (~ ~) w=zrxv. r

r

w

v

So, is normal to the plane of and and it is along the axis of rotation. On the other hand, for rotation of point mass. 'm', the angular momentum is the moment oflinear momentum. This is also an axial vector, which is directed along the axis of rotation, and it is now given by,

[, = r x (mv) = m(r x v) = m[r x ( wx r)] Now we have [, =

m[(r.r)w - (r. w)F]

=

mr2 w

r

So, [, = Iw. (some, r.w = 0 for 1- w) where, for a point mass rotating along a circle of radius r, I = mr2 = moment of inertia** of that point mass about axis of rotation. 3.3 Angular Acceleration: For rotational motion of the particle, the angular acceleration is defined by the rate of change of angular velocity. This is a vector quantity and is denoted

*

In our previous chapter, we have discussed about axial vector which is actually pseudo vector and is obtained by the cross product of two polar vectors. Basically, this axial vector does not change sign under coordinate inversion (i.e. mirror reflection). ** This moment of inertia gives inertia of rotation. For point particle rotation, it is treated as scalar quantity, but we will see very soon that for rigid body rotation it will become a 2nd rank tensor having 9 components.

Rotational Motion : Rigid Body Rotation

67

by ,-; which is mathematically given by, ,-; VI.

VI.

~w

=

Lt ~ !'J.t ~ 0 ut -

dw dt .

When magnitude of velocity of rotating particle along tangential direction changes with time, then body will have a linear acceleration which for circular

j

motion is given by and

j

ii x r .

=

has magnitude f

=

ar

for ii .1 rand ii is directed in perpendicular direction to the plane of circle, i.e. parallel to the axis of rotation.

_ de Since we have,

dt

W =

2

_

d e d (de) we can write, a = dt 2 = dt dt

which is another mathematical representation for angular acceleration.

3.4 Moment of Inertia and Torque: Radius of Gyration : When a particle rotates along a circular path, its rotational inertia is represented by a rotational parameter, called, moment of inertia. This is very similar with 'mass' for linear motion and gives rotational inertia of the rotating body. It is a scalar quantity and is in general denoted by I. When particle of mass m (point mass) rotates along circle of radius r, this moment of inertia is estimated by the product of mass and square of radius of the circular path, w M i.e. I = mr2. r, But for a rigid body rotation or for the rotation of a system of particles about an axis of rotation, the moment of inertia will be the sum of all moment of inertias for all point particles, existing within the system. Mathematically, N

1=

Im;r/ ;=1

N

If M =

I

mi i= 1

be the total mass of the whole system then for rigid body

rotation, this moment of inertia can be written as

I

=

Imir/ i

=

M.!c2.

Where, k is another rotational parameter, called 'swing radius' or 'radius

68

The Classical Mechanics

of gyration'. *

k=~ =

Here,

This radius of gyration for rigid body rotation is defined by such normal distance of an imaginary point from axis of rotation, when the total mass of the system is supposed to be concentrated at that imaginary point. With respect to that radius of gyration, we have the moment of inertia.

Since for continuous mass distribution. I

=

Lm/i2 = tr2dm, i

a

When limits of integration depends on the geometric structure of body and mode of rotation, the radius of gyration is now given by

k = [~

f: dm y. 1

r

2

Now for point particle or rigid body rotation about some axis of rotation, the 'torque' is such parameter, which is rotational homolog of 'force'. This is defined by moment of force and is that parameter, which, on application on the body, gives angular acceleration mathematically, it is given by

1

= =

r x F = r x (mJ) r x m(a x r) = m[r x (a x r)]

For point particle rotation,

m[(r.r)a - (r.a)r] = mr 2a

1

=

t

= lei ** (when,

r.l..
-

2

=

0) and it is also given by,

e

1dt 2 .

't

* One

should remember that the imaginary point taken to define radius of gyration is no longer the center of mass of the body. It is purely an imaging point whose position depends on the axis of rotation choosen. ** This expression for torque corresponds to the conservation of angular momentum, since, dw

i = lei = I dt

d

tiL

_

= dt (Iw)

=

dt

dL So for i = 0, - = 0, L = constant. dt

Rotational Motion : Rigid Body Rotation

69

Here, since, angular momentum is given L =

for

r x p . are can have,

L =rxp =

r x (mv)

m(r x v)

the torque is

-'t

= r x

F = r- xdP dt

dr _ _ dP dL - xdP =-xp+rx- = r dt dt dt dt _ _ dr _ (because, since, lip, v x p = - x p = 0). dt

Hence,

-* 't

v

dL

i

=

dt

dL dt

=

0,

and for i

L=

=

0,

constant.

This is conservation of angular momentum. Similarly, for system of particle, L

dL

";:: x

~

dt

I

dPi (_ dt

_ d~ = dt x Pi =

.: vi x Pi

I

which also leads to conservation of angular momentum for

° )

i

= 0, L = constant.

dL is taken with respect to a fixed frame or body frame dt of reference. If we take the time derivatives in a rotating frame of reference then from the chapter 'Reference Frame' we will see that

* Actually the relation t' =

d1 (-I dt J

.. t

=

- (d) -

FIX -

dl") (dt

dt

Fix =

_

+wX Rot

(tiL) dt + _Rot

wx L

For rigid body rotation we will see later that

L= I w= j w:. t = j .t + wx L

(where, I is inertia tensor) This is Euler's equation for rigid body rotation.

70

The Classical Mechanics

3.5 Centrifugal force: It is basically a pseudo force* which appears within a rotating body and acts in radially outward direction. It originates in a rotating frame and is balanced by centripetal force, required for rotation. Since we know that for rotation of a particle along any circular path, the direction of its instantaneous velocity changes continuously, so, the centripetal force can now be given by F

=m

dv = m~(vii) dt

dt'

dii

mVdi . (ii since,

Ivl

=

x

=

unit vector along velocity direction)

v remains stationary for uniform circular motion.

J

Now from figure, ii = cose - i sine where, e is instantaneous angular position. So,

dii di

=

(A A ) de - j sin e - i cos e di

) -ro (Ai cos e + jA sin e

=

F = -m u ro(i cose + Jsine). If Fe be the centrifugal force developed, then

and

Fe

=

Fe

=

-F'

m u ro(i cose + Jsine)

So, this centrifugal force has magnitude

u ro =

=

m

=

mro2r.

u

mvr

r

(.: u = ror)

If we want to write down the expression of this centrifugal force in vector notation, then we can write it as

Fe *

=

m if>

X

(r X (0)

= -

mif>

X

(if>

X

r)

This centrifugal force only appears in rotating forms, i.e. the frame on rotating body. It has no essence in inertial frame. It will not appear for real interaction and thus it is a pseudo for::e, We will also discuss the appearance of it in next chapter 'Reference Frame',

Rotational Motion .' Rigid Body Rotation =

-mro x v

=

71

m(v x ro)

Because, in that case, since,

v

i

v cos(90 + e) + v cose A

J

x

A

= -vsinei+vcosej We get,

vx

Since,

x

* = -v sin e, Vy = v cos e.

ro = rok' we can write down

vxro

i

j

k

-v sinO 0

v cose

0

0

ro

A

= i

A

rov cose + j rov sinO

and finally,

Fe

=

m(v x ro)

J)

ro v( cosO i + sinO which is identical with equation (1). So, in vector notation, the centrifugal force, appeared in rotating particle, is expressed as = m

Fe =

m(v x ro)

=

mro X (F x ro)

and for point particle rotation, since, F 1. ffi

Fe

=

m(ffi.ro)F

=

.

mro 2F. which is directed in radially outward direction.

3.6 Rotational Kinetic Energy For rotation of a particle along any circular path, the rotational kinetic energy is the energy necessary for rotation of particle. For single particle rotation, this energy is given by, 1 1-1 2 1 __ =-mv.v KE . . =-mv 2 2

or,

* Here,

K.E

=

21 m(_roxr_) .(_ro xr_)

=

1 ro. _ {_ -m r x (ro x r_)}

2

the components of linear velocity along x and y directions are (90 + 8) = -v sin 8 and, Vy = v cos 8.

Vx

=

v cos

72

The Classical Mechanics

=

__ )_} "21 mro_ .{(r.r w

(·.or l.ro)

__ ) = -1 mr 2 ro 2 = -1 m (-ro.ro-) (r.r 2 2 2 K .E. = !Iro 2

where, I is the moment of inertia for particle's rotation about said axis. Similarly, for rotation of a system of particles, this kinetic energy is the sum of kinetic energies for all constituent particles. and is given by,

K.E.=T=

ITi=I~mdro2 i

( •.o

ro

i ...

has some magnitude for all constituent particle)

K.E.

when, I =

I

~ H~mir? )m2 ~ ~Im2. ? = moment of inertia for system of particles.

mir

3.7 Angular Momentum for Rigid Body Rotation: Rigid body means collection of a number of point particles, interacting* mutually and rigidly such that their mutual separation

Ir)

=

I~

- r)

=

constant throughout the rotation about any axis.

For rigid body rotation, the angular velocity for rotation will remain constant for all constituent particles and hence, the angular momentum is given by,

*

This condition ~ -

01

= constant

for rigid body rotation is called constraint and

thus it is constrained motion. We will discuss it in other chapter.

Rotational Motion : Rigid Body Rotation

73

(x;i' + Yi +Zi u)} i~:>'li{(r/

-xl)())x +

(-XiY,)())y + (-XiZ,)())Z}

so we get, 1= i(lxx())x +Ixy ())y + Ixz())z) + J(lyx())x +Iyy ())y +Iyz ())z)+ k(lzx()) x + Izy()) y + I

L=

Lx i + Ly

J+ LJ

zz ()) z)

... 1

Lx = Ixx ())x + IXY ())y + Ixz ())z ... 2 Ly = Iyx ())x + Iyy ())y + Iyz ())z ... 3 Lz = Izx ())x + Izy ())y + Izz ())z .. .4 These are the relations among components of angular momentum and components of angular velocities for rigid body rotation. When,

Here, Ixx

=

Imi(r/ - x;) = Imi (Y; + z;) i

I

>y

=

Imi(r/ - Yl) = Imi(x; +zl) ,,

,,

and similarly for Izz' which are called principal moment of inertia about x, Y and z axis respectively. On the other hand, I.xy

="L.J i

-m,x,y' " " yIz

="L.J

-m,y,z, 111

i

and similarly for Izx' which are called product of inertia. * Here, equations (2), (3) and (4) can be arranged together as,

*

We will see in the next section of this chapter, how can this be obtained in several symmetrical case.

74

The Classical Mechanics L

or,

I

ro .

This is angular momentum for rigid body rotation. Here, I is called moment of inertia tensor* of rank-2. This has 9 components, 3 of which are principal moment of inertia and remaining 6 are product of inertias. This 2nd rank tensor

is given by,

I

Ixx = Iyx ( Izx

Ixy Iw Izy

Ixz) Iyz I zz

3.8 Kinetic Energy for Rigid Body Rotation: When rigid body rotation about a given axis is taken, its kinetic energy is the sum of kinetic energies of all point particles. This kinetic energy is given by

=

L ~ mj{ro.[~ x(ro x~)]} I

=

L ~ mj {(r/) (ro; + ro; + w;) - (Xjro

x

2 + yjro y + Zjro z) }

I

{2 (2 ro x + ro 2 y + ro 2) z - (2 Xj ro 2 x + Yj2ro 2 y + Zj2ro 2 z + Xj Yj ro x ro y +

= "" L.. -1 mj ri

. 2 I

+ YjXjro yro x + YjZjro yro z + ZjYjro zro y + XjZjro x ro z + ZjXjro zro x)} So, we get, the kinetic energy for rotating body T

*

(2 - Xj2) ro x2+ "" "" -1 mj (2 L.. -1 mj (2 rj - Yj2) ro y2+ L.. rj - Zj2) ro z2+

= "" L.. -1 mj ri

. 2 I

. 2

I

. 2

I

The characteristics of this tensor will be discussed in next section.

75

Rotational Motion : Rigid Body Rotation

where, lxx' of inertias.

Iyy'

Izz are principal moment of inertias and

3.9 Axes theorem for Moment

~f

I xy '

Iyz' ... are product

Inertia:

The moment of inertia for rotation of rigid body about any axis parallel or perpendicular to the axis passing through center of gravity, can be obtained by two axes theorem, which are (i) Perpendicular Axis theorem (ii) Parallel Axis theorem Let us discuss this two theorems. z (i) Perpendicular Axis theorem : (a) For two dimensional body: For a two dimensional laminated body, if lx, Iy and Iz be the respective moment of inertia for rotation about x, y and z axis respectively, then by the perpendicular axis theorem. Iz = Ix + Iy when the body initially exist y inxy plane Ix = Iy + Iz when the body initially exist inyz plane Iy = Ix + Iz when the body initially exist is xz plane Because, Let the body exist in xy plane. Here, Ix = Emy2 (Em = total mass) Iy = Emx2 and 1z = Em? when the point P has co-ordinates (x, y) and the mass is supposed to be concentrated at P. Now since,? = x2 + y2 I z = Em(x2 + y2) = Emx2 + Emy 1=1+1 =1+1. z y x x y This is perpendicular axis theorem.

The Classical Mechanics

76

(b) For three dimensional body: In this case, by perpendicular axis theorem,

if Ix , Iy , I.• be principal moment of inertia of a three dimensional body, when rotating about x, y and z axis respectively, then z p Ix + Iv + Iz = 2 Lm(Op)2. wheri P is imaginary point at which total mass is supposed to be concentrated, and '0' is arbitrarily y choosen origin. Here, by definition, It = Lm(y2 + z2). I = Lm(x 2 + z2). Y I, = Lm(x2 + y2). 0 x Because, Ix == ~m(PC2) = Lm(AC2 + PA2) == Lm(y2 + z2) and similarly for

Iv' I z• .

Since, OP2 == x 2 + y2 + z2

r

2Lm(x2 + + z2) = 2Lm(OP)2. (ii) Parallel Axis theorem : This theorem states that if AB and CD be two parallel axes such that the former axis passes through the center of gravity of the body then for moment of inertia lAB and ICD about AB and CD axis respectively, A ICD = lAB + Ma 2 • (M = mass of the body). C Where, 'a' is the normal separation of two a parallel axis AB and CD. Here, for point P on the body, if it has normal distance x from the axis AB, then I AB = Lmx2 and ICD = Lm(x + a)2 Now, ICD = Lm(x + a)2 = Lm(x2 + 2ax + a2) = Lm(x 2) + 2aLmx + a2 Lm 2 ICD == lAB + 2a Lmx + Ma . But in rotational equilibrium*, Lmx = O. D 2 8 So, ICD = lAB + Ma . This is parallel axis theorem. ..

Ix

+ Iy + Iz

=

3.10 Calculation of Moment of Inertia in different cases: (a) Circular Ring: Consider a circular ring of radius 'a' and mass M. It is rotating about an axis passing through its center and perpendicular to the plane of circle. If A be mass per unit length of the ring, then

* Here, we have the position of center of mass through the center of mass then x = O. For that case, 'L.nlX = O. i.e. the moment of mass will be zero.

x ==

'L.nlX 'L.m .; Now if the axis passes

Rotational Motion : Rigid Body Rotation

77

w

M

A =21ta' Now, for a small elementary segment 'dl', having mass dm = Adl, the moment of inertia is elI = dm.a 2 . So, for whole ring, the moment of inertia* I =

f

dI =

f

dm.a

2

=

fA dl.a

2

= 'A.a

M

2

f

dl

2

--.a .2n:a = Ma 2• 21ta I = Ma 2 • (b) Circular Disc: Here, we take a circular disc of mass 'M' and radius 'a'. It is rotating about an axis passing though center of it and normal to the plane of the disc. If eJ be the mass per unit area of the disc, then eJ = Ml1ta 2• Here the disc can be divided into a large no. of co-axial circular rings, one of which has radius x and width dx. If dm be the mass of this elementary disc, then dm = 21tXdx. eJ. So the moment of inertia of that elementary ring about the said axis, is elI = dm.x 2 = 21tXdx. eJ. x 2 = 21teJ .x3dx. So, the moment of inertia for the whole disc about the axis passing though its center and normal to its plane is =

I

=

f 1:

=

a1 21teJ.-

elI

=

4

3

21teJ.x dx

1 4

M 1ta

4

1 2

= 2·-1t·--2 ·a = - Ma

2

1 2 1= -Ma 2

This is the moment of inertia of circular disc, rotating about the given axis. (c) Hollow Sphere: Here we take the rotation of a hollow sphere of mass

*

Here, if we want to find out the moment of inertia of the ring about its diameter then taking the ring on xy plane, we have from perpendicular axis theorem, Iz = Ix

+~.

But for symmetry of rotation about x or y axis

1=1 . 1x =.!..I x y'" 2 z

2 =.!..Ma 2

78

The Classical Mechanics

M and radius 'a', about its any diametric axis. The mass per unit area of the sphere M

is

= -4 2· If this sphere be divided 7ta into a large no. of co-axial circular rings. One of which has radius a sin e and width a d8, the mass of this elementary ring is dm = 27t a sin 8. and 8.a = 27ta 2a sin 8 d8. Now, the moment of inertia of this elementary ring is dI = dm.(a sin 8)2 = 27ta 2.a=sin8 dv. a 2 sn 28 dI = 27ta4a.sin38 d8. So, the moment of inertia of the whole sphere is now given by a 0"

I = fdI = 27ta 4 0". = 27ta 4 a. Jolt

rer sin 2 8 d8 J9 =0

(1- cos 2 8) sine de

4 f+l( -I 1- z 2) dz

= 27ta a

(when, z = cos 9)

M (1)

M4

-_ 27ta 4 .--2.2 1- - = 27ta 4 . - - 2 .-. 47ta 3 47ta 3

2

2

1= -Ma 3 This is the required result. (d) Solid Sphere : For a solid sphere of

mass M, radius 'a', and density p =

1i

7ta3 .

If we take the rotation about its any diametric axis, then dividing the whole sphere into a large no. of co-axial circular disc, we get an elementary disc of mass dm = 7tYdx.p. when, y = radius of elementary disc, and dx is its width. Now from the figure, y2 = (a 2 - x 2 ) where, x is normal position of the elementary disc w.r.t. the center of sphere. So, we get, dm = 7t(a2 - x 2 ) dx.p. Now, the moment of inertia of this disc about the said axis is

Rotational Motion : Rigid/Body Rotation

dI

79

1 1 (2 y 2 = 2,dm = 2,.dm. a -x 2) .

(2

1 =2,.7ta -x

2)2 p.dx.

So the moment of inertia of the whole sphere, is I =

f

2 2)2 dx

1 f+a( dI = 2,7t P -a a - x 3

5

+a

1 4 2 X x I - -7tp. a x-2a . - + - 2 3 5 ] [

-a

I = (

15 -10 + 3) 5 3M 15 tta . 47ta 3

2

2

1= -Ma 5 This is the required result. (e) Rectangular Lamina: We take a Y rectangular laminated disc of length 'a' and breadth 'b'. So ifM be its mass, then mass M per unit length cr = db. b x If we now want to calculate its moment of inertia about any of its two adjacent sides, then placing it in xy plane with x and x y axis along its two adjacent sides, ( a ) Let us now find out its moment of inertia about y-axis. Taking a rectangular narrow strip of mass dm = bdxcr, its moment of inertia dl = dm.x 2 = bx2dxcr. So the moment of inertia of the whole foil about y-axis, is

1

2

Iy = fdI = f;bx .dx.cr. = bcr·i a3 . I y

1 M 3 1 2 = -b.-.a = -Ma 3

ab

3

80

The Classical Mechanics

Similarly, the moment of inertia* of the same foil about x-axis is I

= x

1Mb 2

3

(j) Elliptic Foil : Consider an elliptic foil, having semimajor axis 'a' and semimirror axis 'b' and mass M. We can simply find out the moment of inertia of this foil about any of its axes. If cr be the mass per unit area of the foil, then

P(x, y)

x

M nab· Let us now find out moment of inertia of this foil about its minor axis, along which y-axis is taken. If we now take a rectangular narrow strip, at a distance x from y axis corresponding to the terminal point P(x, y) on the ellipse, then mass of this strip is dm = (2y.dx).cr.

cr

=

g

2 bl2 2 But for ellipse, y = b 1 - 2" = - '\j a - x

a

a

2bcr ~ 2 2 dm = - a - x .dx a

Now the moment of inertia of this linear strip about minor axis, is dl

= dm.x 2 =

2bcr x a

2

~a 2 -

2

x .dx

So the moment of inertia of the whole foil about that said axis is 1=

Let, for

f

f+a

2bcr 2 I 2 2 dl=--· x'\ja -x dx a -a

dx = a cose de. x =a, sine = l,e=nI2. x = -a sine = -1, e = -nI2. X

= a sin 8. :.

2bcr J+1t/2 2 I = --. a sin 2 e.acose.acose de a -1t/2

*

Here, if we take the rotation of foil about z-axis, then I, + Iy =

~ M( a2 + b2 ) •

(By perpendicular axis theorem) Here, if we want to find moment of inertia of the same foil about axis passing though center of foil and parallel to z-axis, we get, 2

a => I=-M 1 (2 1 2 I = I =I+Ma +b2) --Ma z Z 4 3 4

Rotational Motion : Rigid Body Rotation =

I

81

2ber.a 3 • -.!.. f+1t/2(4 sin 2 8 cos 2 8)d8 4 -1t12

1 1 f+1t/2 bera 3 . (1 - cos 8 48) d8 2 2 -1t/2

= -

(I). = -.!...nb.~.a3 mmor 4 nab

=

-.!..Ma 2 4

This is moment of inertia of this elliptic foil about its minor axis. Similarly, the moment of inertia of this foil about its major axis*, is 2 (I)major = -.!..Mb 4

(g) Right Circular Cylinder: (A) About its own axis: Here, if we divide the whole cylinder into a large no. of circular disc, one of which has mass 'm' when M = Lm = mass of whole cylinder, then, the moment of inertia of the whole cylinder about its own axis is

I

,,1 =

LJ 2. rna

2 =

2.I Ma 2 .

(B) About an axis perpendicular to its own axis and passing through center o/gravity o/the cylinder:

Here, we take an axis AB which passes though the center of gravity of the cylinder. Ifwe now consider a small elementary circular disc at a distance x from the center of gravity and parallel to the axis AB, then for another parallel axis CD, passing through the center of gravity of this elementary disc,

c

A \

1\ J.l___ _ j)ta__

I

I

1

1 I I

1 I I

x B

2

the moment of inertia of this disc about axis CD, is dIeD = ±dm.a

.

**

When, dm is the mass of the elementary disc. and 'a' is radius of the cylinder.

*

Similarly, from perpendicular axis theorem, if z axis be taken perpendicular to the foil, passing though the center of foil, then I 2 2 I - I major + I mmor - -4 M(a + b ) ** It is followed from perpendicular axis theorem.

82

The Classical Mechanics M Now if P

be the density of the cylinder (where, L is total Length 7ta L of cylinder) then, dm = 7ta 2.dx. p. = --2-

dl co

=

1 2 2 "41ta pdx.a

=

1.1ta 4 pdx

4

.

Now by parallel axis theorem, the moment of inertia of this elementary disc about AB axis is dI AB = dIco + dm.x 2 • dI

AB

=

1 4 2 2 -1ta .pdx + 1ta pdx.x

4

So, the moment of inertia of the whole cylinder about the said axis is I

= AB

I

JdI

= AB

1

AB

AB

I

AB

4

L 2 L 2 2 2 J+-L!2 / dx +1ta p J+ / x dx -L!2

2

1[L3) 4 .

= -1tpa .L + 1ta p.-. -

4

=

I

1.1tpa 4 4

3

3 M 1 M 4 1 -7t--a L + -7ta 2 L . -22 4 7ta L 12 1ta L

1 4

2

1 12

2

= -Ma +-ML

=1.4 M (a 2 +.!.3 L2)·

This is the moment of inertia of a right circular cylinder about an axis passing through its center of gravity and normal to the axis of cylinder. (h) For a Right Circular Cone about its own axis: We now consider a right circular cone of mass M, base radius a and height p=

M

3M .!. 7ta 2 L - 1ta 2 L·

3 Here we take an elementary circular disc of radius 'y', at a distance x from the vertex of the cone, then moment of inertia of that elementary disc about axis of cone is

83

Rotational Motion " Rigid Body Rotation 1 z dI = -dm.y 2 where, dm is mass of the elementary disc and dm = (nYdx).p. 1 z z dI = -7tPY dx.y 2 =

But

4

L

x

Y

1

27tPY dx.

=-::::}

a 4

4 1 a dI = -np.-.x dx 2 L4 .

So the moment of inertia of the whole cone about its own axis is

I

=

=

f

4

1 a 4 L5 -np.-.2

lL

1 a dl = -np.-. x 4 dx 2 L4 0

L4

5

=

143M

-na L.-10 na zL

3 z 1= -.Ma 10 This is the required result. 3.11 Momental Ellipsoid or Ellipsoid of Inertia: Let us now consider the rotation of any rigid body about any axis 'OA' having direction cosines (t, m, n). If P be a point within the body at which the mass of the body is supposed to be concentrated, then for perpendicular PM, drawn from P on the axis OA, we get, Op2 = x2 + y2 + z2. and, OM = Ix ~ my + nz. where, (x, y, z) are coordinates ofP. Now from the fig, the moment of inertia x of this rotating body for rotation about axis OA, is lOA = I = :Lm'(PM)2. where, :Lm' = Total mass 1= :Lm' (OP2 - OM2) = :Lm' [(xl + Y + z2) - (Ix + my + nzf]

z

y

84

The Classical Mechanics

1= Lm'[x2(1 - Z2) + y(1- m2) + z2(1 - n2) -21m xy - 2mnyz - 2/nxz] = Lm'[x2(m 2 + n2) + y(n2 + [2) + z2([2 + m2) -2 Imxy - 2mnyz2nlxz] = Lm'(I + z2) [2 + Lm'(x2 + z2)m 2 + Lm'(x2' + y)n 2 + 2 1m L(-mxy) + 2mn L(-myz) + 2 nl'ir--m'zx I = Ixx z2 + Iyy m2 + Izz'n 2 + 21m Ixy + 2 mn Iyz + 2nl Izx . Since for several axis, (t, m, n) are variables, so let such axis for which, body has constant moment of inertia for rotation, I = X, m = Y, n = Z We now get I = Ixx X2 + Iyy y2 + Izz Z2 + 21xy XY + 21yzYZ + 21zx ZX :::::> Ax2 + By + Cz2 + Dxy + Eyz + Fzx = 1 Ixx Iyy When, A= -1-' B = -1- ... 1 I zz 2Ixy 2Iyz 2Izx C=-I ,D=-I- E=-I-F=-I-'

and all of these are constants. Now the equations (1) is an equation of an ellipsoid, * which is an ellipsoid of inertia or momental ellipsoid. More clearly, we can say that if several axes of rotation are possible to rigid body rotation, for which moment of inertia of the body remains invariant, then in this condition, the locus of all direction cosines for several axes, will be ellipsoid. This elliposid is known as ellipsoid of inertia.

3.13 Moment and product of Inertia and Ellipsoid of inertia of some symmetrical bodies : We will now find out z principal moment of inertia and product of inertia and also ellipsoid of inertia for some symmetrical bodies. (i) A Thin Rectangular Foil: Here we take a rectangular foil or lamina in xy plane having length a, breadth b and mass per M

unit axis cr = ab'

*

~(----

a

----~)

We have the general equation of the ellipse,

i

x2 2xy -+-+-cosS = sin2 8 => A'x2 + BY + C'xy a2 b2 ab Similarly, for ellipsoid, the equation should be, A'x2 + BY + C'z2 + D'xy + Eyz + F'zx + a'. => Ax2 + By + Cz2 + Dxy + Eyz + Fzx = 1. This is equation of an ellipsoid.

= D'

Rotational Motion : Rigid Body Rotation

85

Assuming uniform mass distribution, We have, the principal moment of inertia Ixx

=

I

mdyl + zl)

=

Sf (l + z2) dm

i

=

Sf (l + z2) crdxdy = Sf lcrdxdy (.,' z = 0) +a12

I xx

2

Similarly,

Ma 2

J+b/2

2

= cr. J y dy dx = - -a12 -b/2 12

(2

Mb 1=- . 1 =M - a +b 12

Y.Y

12

zz

2)

(By perpendicular axis theorem). Also the product of inertia

I

IX)! = IyX =

-mjXjYi = -

JJ cr xy dxdy

i

=

and some z

-cr. f

+b/2 f+al2 xdx ydy -b/2 -a12

=

°

= 0, Iyz = Izy = 0,

Izx = Ixz = 0. So for this thin rectangular foil with '0' at the centre, the ellipsoid of inertia is given by, Ixx x2+IY.Y y2+Izz z2= 1

~Ma2x2 +~Mb2l +~M(a2 +b 2)z2

=> 12 => =

12

12

~ [a 2x 2 + b2l

2

=

1

2

+ (a + b )z2] = 1

which is the equation of a spheroid. (ii) Circular disc : Here, we consider a uniform circular disc of radius' a', mass per unit area cr = M! 1ta2.

Assuming uniform mass distribution and using cylinderical co-ordinates x = r cos e, y = r sin e, z = z and taking any elementary area dA = dx dy = dr.rde which has mass dm = rdrde.cr We may write I zz =

Sf (x 2 + l) rdrdecr = a4

Ma 2

4

2

= cr.-.21t = - -

cr f: r3 dr fo;;e

86

The Classical Mechanics which is our expected result Also,

=

a

4

(j-'lt

4

Ma 2 = - - =1

4

y.y

We can also show that, in this case of circular disc, Ixy = Iyz = Izx = O. So ellipsoid of inertia is Ixx x 2 + Iyy y2 + Izz z2 = 1

2

2

Ma- (2 Ma 2= 1 =>x2+y2+zz2=4 => x +y2) +--z a 2 which is equation of an oblate spheroid. (iii) Thin spherical shell : We now consider a thin spherical shell of radius 'r', thickness dr, mass per unit volume M

p=

4'ltr 2dr· Assuming uniform mass distribution and using spherical polar coordinates, x = r sin S cos S y = r sin S sin S z = r cos S. We now have, I

=

zz

~~~----~~~"--- -

',---------- ,-' ".,,-,-

--------- r-~-\ \ \

, .... __

0

\ ~-\ \ \ I

.....,

I:".' I 1.--_ .....

-------------j--, , I ,

I

I

I

I

I

/"./

V

f(x2 + i)pdv = H(r sinS COS
prZ sinS drdSd
I

7t

zz

= pr 4 dr r sin 3 SdS

Jo

r~S

Jo

2'lt = ~Mr2 .i. 3 3

= pr 4 dr .

Similarly, we can show that

Iy'y

=

J(z 2+ x 2) pdv = "32 Mr 2.

=

Ixx.

All of these are moment of inertia of the shell about any of its diameter. Again,

Ixy

= -

I

=

xy

and

Iyz =

J

xypdv

= -

J

r sin S cos

-pr 4 dr rx sin 3 SdS r2~in

=

0

Rotational Motion : Rigid Body Rotation

87

Hence the ellipsoid of inertia will be

%Mr2(x 2+ i +z2) = or,

1

x2+ y2 + z2 = %(M~2 J.

This is equation of a sphere. And, similarly, we can get all of these for other symmetric bodies.

3.12 Moment of Inertia Tensor: From our previous discussion we have seen that due to rigid body rotation about a fixed axis, the angular momentum is given by,

L = i ro

and kinetic energy T =

±~~lijQ)iQ) ,

j

J

Here, I is a 2nd rank tensor, called inertia tensor. It has 9 components and

Ixz] Iyz = I..

has the form. I

I

1J

zz

where,

I xx

="L...."m· (,} - x~),

and

I.\Y

= -

Izx = -

"m.xy. I = L.... 111,yz

L mizixi

- L.... " m,Yi z ,·

... etc.

In tensor notation we can now write down I ij

="L.... m(r o.. - r.r.) 2

1J

IJ'

where, oij = 1 for i = j = 0 for i:t; j Here, I.. = I .. which is called symmetric tensor, and it has rank-2, with three 1J Jl diagonal elements, principal moment of inertia where as, the off diagonal elements are product of inertia.

3.14 Routh's Rule: By this rule, we can also find out the moment of inertia of symmetrical

The Classical Mechanics

88

bodies. According to this rule, the moment of inertia of a body about any of the three mutually perpendicular axes of symmetry passing through the center of mass ofthe body is given by, mass of body x sum of the squares of the other two perpendicular semi axes n where, n = 3, 4 or 5 according as the body is rectangle (or parallelopiped), a circle (or ellipse) or a sphere (or ellipsoid). As for example, the moment of inertia of a rectangular lamina about the required axis is given by,

I

~ ,(~)' + (%)'] 3 =

~ (a 2 + b2 )

where, M = mass of lamina and a, b are respectively the length and breadth of the lamina For uniform circular disc,

2

1.

I = M(r2 + r ) = Mr2 4 2

z

M(r2 + r2) For solid sphere, 1= -"'-----'-

y,

5

... etc.

y

3.15 Euler's Angles

Uptill now, we have taken the motion of a rigid body to a body frame of reference, the axes z

x

x,

89

Rotational Motion: Rigid Body Rotation

of which are coincident with the principal axes of the body. These axes rotates along with the body. We will now introduce axes fixed in space which is actually the inertial frame of reference. Now the orientation of a rotating body can be completely specified by giving three angles called Euler's angles. Let us first take OXYZ frame on the body and the body is at first rotated through angle

SUMMARY 1. For point particle rotation, v = w x r So angular velocity

1 (_ _) w=-Zrxv

and angular momentum

1=

r

m(rxv)=m{rx(wxv)}

2. For point particle rotation angular acceleration ii =

i-(r x 1) r

3. Moment of inertia

I I

= mr (for point particle rotation) = 'Lmr =Mk2 (for rigid body rotation)

4. Radius of gyration

k

= ~ =~'L;~2

5. For point particle rotation,

t

=Torque = Moment of force = r x F= Iii

where, t

= dldt

(-I

= r- x p- = mr-x-v)

6. For rotation of particle, Centrifugal force developed

Fc = mwx (w x r)

7. Rotational kinetic energy for point particle rotation K.E

= "21 1

0)

2 ,

where, I

='Lmr

8. Angular Momentum for Rigid Body Rotation is given by

90

The Classical Mechanics

9. Kinetic energy for Rigid Body Rotation

!

K.E

2

LLlij.coi'CO j i

k

10. For Rigid Body Rotation. Parallel Axis Theorem : I = (I)CM + Ma 2 Perpendicular Axis Theorem (2 dim) = Iz = Ix Iy ..... etc. Perpendicular Axis Theorem (3 dim) : Ix + Iy + Iz = 2 L.m (OP)2. 11. Moment of Inertia for rotation about symmetric axis in several cases (i) Circular Ring: I = mil (ii) Circular Disc: I =

1

mr2

2 (iii) Circular Disc (about diameter) : 1= ±mr (iv) Spherical Shell : I = (v) Solid Sphere : I =

2

32 mr 2

"'5 mr

2

1

(vi) Right Circular Cylinder (about axis) : 1= 2mr2

2 (vii) Right Circular Cylinder: 1= ±m(r

+~b2)

(about axis .1 to the

axis of Cylinder) (viii) Rectangular lamina (about axis passing through comer and .1 to the

3m (2 a + b2)

plane on lamina) : I = 1 (ix) Elliptic lamina: I

=

±m(af + bf)

(x) Circular cone (about own axis) = I =

1~ mr2

Rotational Motion.' Rigid Body Rotation

(xi) Uniform rod : 1=

91

121 mL2

12. Equation for ellipsoid of inertia: Ax2

+ By2 + Cz 2 + Dxy + Eyz + Fzx

=1

13. Moment of Inertia Tensor: I = l.m (ro - r r) J IJ

I)

I

mass of body x Sum of the squares of the other two perpendicular semi axes 14. Routh's Rule: 1 = - - - - - - " - - ' ' - - - - - - - - n n 3 (rectangle) = 4 (Circle or ellipse) = 5 (Sphere or ellipsoid) 15. Euler's angles : (
=

Worked Out Examples 1. Find the moment of inertia about the diagonal of a rectangle of side 'a' and 'b' respectively. Ans : Here from the equation for ellipsoid of inertia, the moment of inertia about the diagonal is, I = Ixx COS2 a + Iyy Cos2 (90 - a) + Izz Cos2 90 = I.u Cos 2 a + Iyy Sin2 a . But

and,

I

xx

=

1 2 1 2 -mb I =-ma 3 'yy 3

2 2

= -2 ma2-b- -2

3 a +b 2. A body of mass hangs from one end of a light rope which is wound on the surface of the horizontal cylinder of radius 'a', which is free to rotate about its axis. Prove that the body descends with an acceleration

Ans : Here, from energy conservation,

92

The Classical Mechanics

!mv 2 +!IW2 2 2 1

2

1 v

=

mgh

=

mgh

=

1 "2m(2gh-i)

2

-mv +-12 2 a2 1 v2

"21:;

(: v = wa)

m(2gh - i)a

I

2

m(2gh - 2fh)a 2

v2

2fh

m(n- f)a 2 f (': v2 = 2fh,J= linear acceleration)

3. For a thin spherical shell of radius or', thickness 'dr' and mass per unit volume p, Show that the equation of ellipsoid of inertia is the equation of a sphere, Ans : Here, x = r sin a cos cp, y = r sin a sin cp, z = r cos S for spherical polar co-ordinates (r, a, cp), So, we have, I zz

J(x 2 + i)pdv J[(r sin S COScp)2 + (r sinS sin cp)2 ]p,r 2 sin a drdSdcp 4 pr dr sin ada dcp = pr dr'3.27t o 0

J

21<

J

1t

4

3

4

(': M = 47trdrp)

Simiiarly, Iyy

2

= I zz = -Mr 3

2

But the produce or inertia

IXY

-

Jxypdv= - JrsinScoscprsinasincp.p.r sinSdrdadcp 2

1t

4

f

21t

3

f

- pr dr sin adS sincp coscp dcp = 0 = lyx

o

0

Rotational Motion " Rigid Body Rotation

93

Similarly, Iyz = Izy = 0, Ixz = 0 But, Ixx x 2 + Iff,,2 + Izz Z2 = 1 =

jMr2(x2 + i +z2)

= 1

3 1 2 (say) 2Mr2 This is equation of a sphere. 4. Find the moment of inertia of a uniform vertical rod of mass m and length

=> x 2 + y2 + Z2

= --- = a

L about a horizontal axis passing through a point

1

'3 L

from one end.

Ans : From parallel axis theorem,

I = (I)

eM

1 1 . L L)2 +M( - - - = -ML2 +-ML2 2 3 12 36

1_ 5. For rigid body rotation, show that T = 2"(J).L where the symbols has usual significance.

II I··w-w·

Ans : We have, T = -2

lJ'

I

)

i,j

But

L= =

~

~

~

Lxi + Lyj + Lzk (Ixxwx +Ixywy +Ixzwz)~ + (Iyxwx + Iy'ywy + Iyzwz )] + (Izxwx + Izywy + Izzwz)i

(i

=

x, y,

Z,

j = x, y, z)

EXERCISES 1. State and prove the theorem of parallel axis for moment of inertia. 2. Find an expression for the moment of inertia about any axis of a rigid body. 3. Considering expression for kinetic energy of a rigid body, find its angular momentum and also show that

aT aw = L

94

The Classical Mechanics

4. Find the principal moment of inertia of a cylindrical tube of length L, mass M and inner and outer radii R J and ~ respectively.

5. A rigid body is rotating with an angular velocity w about an axis lying in the y-z plane and making an angle of 45° with the y-axis. Obtain the K.E of the body in terms of the moments and products of inertia. Also obtain the components of angular momentum of the rigid body. 6. A rectangular thin plate of dimension a spins about an axis along the diagonal at a constant angular velocity w. Given Iu ==

1~ ma 2 , Iyv == lma 2 (i) What

is the direction and magnitude of the angular momentum? (ii) Calculate the torque about the axis of rotation. 7. A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the velocity of the other and when it hits the floor, assuming that the end on the floor does not slip. 8. Find out a relation between the linear velocity and the distance travelled, for a sphere of radius a and mass m rolling down an inclined plane having inclination 9. Consider two masses m J and m2 separated by a distance ro when connected with a massless rigid rod. Show that the moment of inertia of the two masses about an axis through their center of mass in a plane perpendicular to the

e.

"d d' rlgl ro lS

m)m2,.,2 0 . m) +m2

10. The products of inertia in the principal axis are zero - how is it realised in a symmetrical body?

-:0:-

Chapter-4

Reference Frame

4.1 Introduction When we deal with any physical system either in rest or in motion, we should take everything with respect to a certain region, called frame of reference. This is very important and significant to consider such a reference frame with respect to which the physical system and its interaction can be viewed. The frame or the region of interest can be classified into two types in respect of its nature. This two are inertial and non inertial frame. The frame which is at absolute rest or in uniform motion, is called "inertial frame." But the frame which has some acceleration or retardation is called "non inertial frame". Basically, earth is itself a non inertial frame and any frame associated with earth is non inertial. So the idea about inertial frame is absolute and it is physically impossible to obtain inertial frame in reality. On the other hand, the frame can be classified into 3 classes in respect of its geometrical construction, when considered in three dimension. These are cartesian, cylinderical polar and spherical polar system. Similarly, in 2 dimension, the frame will have two classes which are cartesian and plane polar system. The other side about the nature of a frame which is essential to keep in mind that the law of physics will not alter in any frame but its form of representation will change from frame to frame. More precisely, if we move our interest from inertial frame to non inertial frame, we will see that the nature of the action of force effective on particle motion will change because of the appearance of some 'so called' pseudo force. So it is a fact that every interaction with a physical system when considered in a frame, will occur in an organised way with the frame by obeying the law of physics in every respect.

4.2 Non Inertial Frame and Pseudo Force When we consider particle motion or the motion of a system in an accelerated frame or non inertial frame, the particle or the system as a whole, 95

96

The Classical Mechanics

will experience a force in opposite to the direction of acceleration of the frame. This force appeared in non inertial frame is called 'pseudo force'. For the presence of such 'pseudo force', the effect of force acting on the particle will change than that in inertial frame. It can easily be shown that if F be the force applied on the particle in non inertial frame, the effective force acting on it will be Feff

= F+(-Fp) = F+Fo

Where, Fo = - Fp is the pseudo force appearing in non inertial frame. The nature of this pseudo force entirely depends on the mode of acceleration of the frame. We can clear this point in the following p cases(i) Case - I : Pseudo Force in non inertial frames having translational accelerated motion: Consider two frames S and S' when S is an inertial frame and S' is non inertial or accelerated frame having acceleration in its translational

J

motion with respect to S. R is the instantaneous position vector of the point 0' of S' with respect to the origin '0' ofS frame. So if rand r' be the respective position vector of the instantaneous position P of a moving particle then from fundamental idea of vector algebra. r R+r'

r' d 2r' dt 2

r-R d 2r

d 2R

dt

dt

---2 2

d', (d'R 1

=:>

d 2r' m-dt 2

=:>

Feff

F' = F + m( -

Feff

- (-F=m ddt r) F+(-Fp). 2

m--+m --dt 2 dt 2

J) = F + (-m J) 2

actual force applied from outside. where, - Fp = inertial frame.

mJ is the pseudo force acting on that particle considered in non-

Reference Frame

97

This pseudo force sharply acts in direction opposite to the acceleration of the frame. (ii) Case-II : Pseudo Force in Z non inertial rotating frame: , Here we consider that'S' (oxyz) is Z \ an inertial frame where as'S" (0' X' Y' P(x, y, z) Z') is non inertial rotating frame having (x', y', z') origin coinciding with that of'S' frame. It is obvious that the basis unit vectors

k®

(i,},k)

of S-frame are fixed where as

the other set of unit vectors

y'

(i'),k')

changes their orientation w.r.t. time t during rotation of S'-frame. So if rand r' be the respective positions of same instanteneous* position 'P' of the moving particle relative to Sand S' frame then A

r

r'

y

A

j

A

A

xi +yj+zk A

A

A

x'i' + y'j' + z'k' Also for this two frames Sand S', (

dr) dt fixed

dr') and also ( dt rot =

dx dt

~

dy,: dt

dz dt

A

- 1 + - } +-k

dx'~, -1

dt

dy' ':, dt

in S frame or fixed frame.

dz' k dt A

+-} + -

,

When (t',}',k') are itself fixed w.r.t. rotational frame S' and these are rotating unit vectors when viewed in S-frame.

dx'~,

dy' ':,

dz' A, dt' d}' die' +-k +x'-+y'-+z'-

(

dr') dt fixed

-1 +-}

(

dr') dt fixed

dr') dt' i' z'die' ( - dt rot +x'-+y,JL+ dt dt dt

~

~

~

~

~

~

* Here, in Newtonian machines, the position when measured w.r.t. the frame Sand S' will be the same in t' = t. So, the position which is instantaneous in one frame, will be also instantaneous in another frame. But it is not possible in relativistic case.

98

The Classical Mechanics But we have'"

dr __

v

-=wxr dt

di' dt so,

dj' -. ':, dk' - k , dt A

A

,

wxi"-dt =WX)

-=WX

(dr') dt Fixed

+X'(w xj,) + Y'(w xl') +z'(w xk') ( dr') dt rot

(dr') dt Fixed

dr') (-dt rot +wx(x'i'+Y'J'+z'k') dr') __, (-dt rot +wxr A

-

and we usually have,"''''

(!!..) dt

Fixed

=(!!..)

dt

rot

A

A

+ wx .

This is the relation between time variation operators in fixed frame S(inertial) and rotating frame S' (non-inertial). Since, in our case,

r' =r

~

= OP

~

= O'P

so, we should have,

(d2;] dt

= Fixed

(!!..) (!!..) dt dt Fixed

=

=

[(!!..) dt

rot

(r) Fixed

x][( dr) dt

+w

+ wX r]

_ (dr) _ (dr) +wx(wxr) _ __ d2r] +wx +wx ( dt rot dt rot dr rot -2-

... from fundamental rotational dynamics, v

s = ex F;

rot

= wr and

v= Wx F.

d(e- x r-) = dt de x r-+ e- x dt' dF But dt dF = 0 :.

:. ii = dt

Because, ii = ~

- - v = w x r when

_ de W=-

dt

__ + w x w; But Wx W= 0 . (dW) Fixed = (dW) dt rot

...... Here, dt

dW) :. (dt

= FIxed

(dW) dt

i.e. angular acceleration will be the same in any frame. rot

,

99

Reference Frame

Where W, the rotational velocity is time

indepen~ent.

:. m(d r) = m(d2~) +2mwx(dr) +mwx(wr) dt 2 Fixed dt rot dt rot 2

dr) (dt rot

But if V =

r) then, m dt

=

velocity of moving particle in rotating frame

r)

d2 m -2(

2

d

(

2

dt

Fixed

r)

dt

2

d m -2(

Feff

dt

=

V

+

X

X

rot

m(d2~)

2

m(d r) dt 2 rot

where

_ _ mw_ (w_ r)_

+ 2mw X

+(-2mwx v)+[-mwx(wxr)] Fixed

. Ie·m rotatmg . non-mertla .. I effi· ectlve fiorce on partlc

rot

frame.

r)

2

F=

d m( -2-

FI = Fc

dt

=

Actual force acting on particle in ftxed inertial frame.

Fixed

wx V) = Pseudo force which is commonly known as and F2 = Fc = -mw x (w x r) = Pseudo force which is

= -2m(

'coriolis force'. '" popularly known as 'centrifugal force'. So we see that in a rotating non inertial frame, two pseudo forces will appear, one of which is called corio lis force and other is centrifugal force. The action of 1st one is conditional and it is not effective for a rest particle w.r.t. rotating frame, where as the 2nd force always acts in rotating frame.

4.3 Effect of rotation of earth on acceleration due to gravity Neglecting earth's translational motion around sun, if we only consider earth's rotation then we can show that the acceleration due to gravity at any place on earth surface except at pole, will change effectively due to earth's rotation. '" This force -2m( WX v) will act on a moving particle in non-inertial frame. As for example, since earth is it self non inertial frame, this coriolis force will act on any moving particle on earth surface. But this force will be zero for any rest particle.

100

The Classical Mechanics

Let us now consider two frames Sand S' when the frame S is inertial and is taken at the center of earth and the non-inertial frame S' at any point on earth surface. Now if for any particle motion with velocity

v=(dr) dt

measured from that frame S' on earth surface, we take its weight

mg

as Tot

directed

towards the center of earth and any real force (actual) F acting on it, then the effective force on it as seen from the surface of earth due to earth's rotation, will become,

m(~~~J

Tot

m(d2~) dt

+ [-mw x (w x r)] + [-2mw x

v]

Fixed

(F + mg) + [-mw (w r)] + [-2m(w v)]

Feff

X

X

X

F+ m[g - wX ( WX r)] - 2m(wX v)

F+ mgeff - 2m( wX v) where geff =

g- wX (w X r)

is the effective gravitational acceleration of

the place on earth surface. This is the resultant of actual gravitational acceleration

g

and centrifugal acceleration due to earth's rotation

-w X (w X r)

and as a

result we have the nature of this effective gravitational acceleration as, geff g at pole =

g-wx(wxr)

at any place on earth surface other than at equator (not directed towards the center of earth). =

g - wX (w X Ii)

at equator (directed

towards the center of earth.) If we now be interested to calculate the maximum effect of this centrifugal acceleration on actual gravitational acceleration, then we will see that for earth radius R = 6400 km and angular velocity

21t 86400 7.29 x 10-5 rod/sec, the maximum value of centrifugal acceleration at equator is for earth rotation

W

= - - rad/ sec =

Reference Frame

101

a2 = w2.R = 3.4 x 10-2 mlsec2 which is about 0.2% of earths gravitational acceleration. So at equator the reduction of actual gravitational acceleration due to centrifugal acceleration will be maximum and the effective value of the gravitational acceleration at equator will be. gefT= g - w2R = 9.81 - 0.034 = 9.776 rnlsec 2 It is a fact that for rotation of frame, the centripetal acceleration will act in opposite to the centrifugal acceleration towards the center at which inertial rest frame exist. This fact is also true for any particle rotation above a fixed point.

o

4.4 Effect of Coriolis Force on a particle moving on the surface of earth Here we are now interested about the effect of corio lis force on a particle moving on earth surface. We have from our earlier discussion that due to earth rotation, the coriolis force acting on particle moving with velocity v w.r.t. earth surface is

Fe = -

2m( Wx v)

v

w.

which will be at right angle to the plane of and Now we consider particle motion at P on earth surface having a latitude e, with velocity along y-axis. Here for the system OXYZ ofS' frame taken at P, the plane XY is horizontal plane w.r.t. earth surface at P and Z is vertical axis to the same point P, the effective component of earth's angular velocity along Z axis is

v

Wz

= w cos(

~ - e) = wsin e.

The Classical Mechanics

102

v

z

So for particles velocity along y-axis the coriolis acceleration will be -2{ Wz x

v)

and its magnitude is 2wzv

A

k

2w sin 0 v. This is a fact that for such acceleration in transverse direction to the particle motion, the particle will deviate from its original path (along y-axis) in transverse direction (in XY plane) and will finally move along the deflected path in XY plane as shown in figure. The maximum effect of this x transverse corio lis acceleration will occur at any pole and it is =

---

ac =

2 w sin

1t

"2.v = 2wv = 2 x

7.29

x

...................... Deflected'" , path ",

1O-5v

ac = 1.458 x 10-4v. When, v is the velocity in horizontal plane. If we now take a typical example of a particle moving horizontally on earth surface with velocity lkm!sec or 3600 kmIhr, the magnitude of corio lis acceleration will become ac = 1.458 x 10-4 x 10 5 = 14.58 cmlhr.2 or, ac Ri 0.15 mlsec 2 which is about 0.015 g. We can now conclude that although the magnitude of coriolis acceleration is small, it plays an important role in many phenomenon in the earth. It is important to take into consideration the effects of the corio lis acceleration in the flight of missiles, the velocity and the time of flight which are considerably large.

v

in the horizontal plane on earth Here for particle moving with velocity surface, we can immediately find out the angle of deflection for particle motion along the duration t as

a=

distance travelled in time t in the deflected direction (for sufficiently small deflection) distance travelled in time t in the direction of projection 1 (2wsinO. v)t

2

.

-'------'- = w sm O. t

2 vt Thus, on the north pole (9 = 90°), a = wt which is maximum and for any t = 180 sec. a = 7 x 10-3 radian = 0.04° which is found to be quite small, but it assumes considerably important in guided missiles.

Reference Frame

103

4.5 Effect of Coriolis force on a particle falling freely under gravity We consider a particle which is falling freely (neglecting air resistance) from height 'h' on earth surface and in this case, due to corioUs acceleration only the acceleration of particle will become

a=g-2wx v a v

When w is angular velocity for earth rotation, and both are measured in rotating frame on earth surface. Now from our previous topic of discussion, we have, at latitude 8 on earth surface, where we have considered the falling of particle in -k direction (along

-ve z direction), Wx = v, Wy cos 8, W z = W sin 8 Also, for particle motion along -ve z direction,

0, y ~ 0, i = - gt Again, for quite small deflection produced by coriolis force the components of the acceleration will be

i

~

x = 1-2(wx v)1 = 1-2(-0) cos8 gt)1

ax

2w gt cos 8 ay

ji = 0

az

z= - g

Here the acceleration ax along X direction is due to corioUs force and we thus get

d 2x -2dt

z when, t

=

1 3 = 2wgt cos8 => x = -wgt cos8

3

2 d z 1 2 -=-g => z=zo --gt dt 2 2

= 0, z = zO' x = 0, y

=

0 is the initial position, with initial velocities

i( 0) = 0 = i( 0) . Since, the time of fall from height 'h' is t=

f¥ '

the deflection of a particle towards the east when it is dropped

from rest is 8h 3 )1/2 8 = "3wcos82 (

1

This is called 'eastward effect' due to the effect of coriolis force on a freely falling body. As for example, if the particle be dropped from a height of 100

104

The Classical Mechanics

m from rest at latitude e = 45°, it will be deflected by about 1.55 x 1O-2m toward east in northen hemisphere.

4.6 Principle of Foucault's Pendulum The French scientist Leon Foucault showed that the small effect of coriolis force may be remarkable for the oscillation of a pendulum. He noticed that the right angled coriolis deflection on one swing of the pendulum could not be undone in the return swing. The effect of corio lis force of terrestrial origin moved from the domain of theory and outdoor observations to that of observation in a laboratory experiment. The equation of motion for a pendulum* including the coriolis term is 2-

d r 2-2 + k r

dt

where, k =

~ g~ff

=

(

-2 Wx v)

= stiffness constant for pendulum and' l' is the effective

length of pendulum. Now taking cartesian components of this equation, we get x+k2x

2(Ywz -iwy)~2Ywz

For i and z negligible in compare to X, y and X, y . Thus the above equations are the coupled equations of motion in x and y. To solve the above coupled equations, let us choose a complex variable

.J=l) .

u = x + iy (i = Then, after some algebraic manipulation, we finally get,

eu

0 or, it + 2iwz u+ The general solution of this equation is now given by,

*

For pendulum motion in simple harmonic manner under restoring force only, the equation of motion becomes, d2-r k_ _ or d 2rm/=-k r => m-=-kor => +-=0 2

o

m 2 ko g d ;: But -;;; = I for pendulum oscillation:. dt 2

other force act

dt

dP

+~r= 0 => I

2 d ;: 2dt 2 + k r = 0 when no

Reference Frame

105

y

,,

",;r"-

/ /

\

/

\

I

\

I

\

I

\

I

\

J

J

I

I

I

I

I

I

I

J J

I

\ \ \

I \

x

,,

I \ \

,

\

"

/ /

"

~

rotation of plane of oscillation In clock wise sense for Foucaults pendulum.

where,

u

exp(-iwzt)[c\e ik 't

k'

~X2 +/.

u

=

+C2 e-

ik t ' ]

. )e-iw t . = (Xo + lyO

X + ly

Z

where, Xo and Yo are the solutions when Corio lis force is absent and u' :::; k for k> > wz' So from this solution, we can conclude that the plane of oscillation of the pendulum rotates with an angular velocity

-w),

in opposite to the sense of

rotation of the earth. The period for a complete rotation of the plane of oscillation is

21t T So at the poles, T

21t

w = wsinEl at latitude El. z

24 hr, and at the equator

106

The Classical Mechanics

So that no rotation of the plane of oscillation is observed at the equator. Foucault had demonstrated the truth of the above facts through his historic pendulum experiment in 1851. By measuring the period of oscillation, T, Foucault measured the period of earth rotation. This was the first experimental proof that the earth is in fact rotating w.r.t. the inertial frame in which Newton's Laws are valid, with an angular velocity which is precisely the same as that inferred frame the apparent durinal rotation of the sun, moon and the star sphere.

4.7 Flow of River on Earth Surface Basically, rivers flow approximately in a horizontal plane for which a slight downward slope in the direction of flow is required to maintain the speed of the flow. So the component of gravitational force along the direction of river flow is primarily important depending on the magnitude of the slope of the down stream. But there is absolutely no component of geff (including the effect of centrifugal acceleration) acting along the breadth of a river. Hence, the component of corio lis acceleration, however small, can act freely on the moving water across the direction of the stream. To find the effect of such transverse coriolis acceleration on the flow of river, let us choose the direction of river flow along x-axis. y-axis is taken in transverse direction to x-axis on horizontal plane and z -axis in vertical direction. The flow is now considered at latitude 'e' of earth surface, making angle e (in the anti cloth wise sense) w.r.t. the geographical north direction. Therefore, the earth's angular velocity vector by,

w w.r.t. the said reference frame is now given ~

w .. w

v= vi

=

=

~

~

wsinek + wcose coscpi - wcose sincp}

w(sinek+cos9coscpt -cos9sincp}). and the velocity of flow is

along x-direction. So the coriolis acceleration is

lie = -2(wxv) = -2vw(sin9}+cos9sincpk) Thus, a corio lis force will be experienced by the water in the rivers flowing in any direction, causing a direction towards the right of the flow direction in the northen hemisphere and to the left of the flow direction in the southern hemisphere. As a result, the corresponding banks of the river will be denuded, which is actually observed.

SUMMARY 1. Frame is a certain region w.r.t. rest or motion of particle or system of particles is considered. 2. Frame has two types which are inertial and noninertial frame. Inertial frame is at rest or in uniform motion where as non inertial frame is accelerated frame.

Reference Frame

107

3. Pseudo force will appear in non inertial frame and in that case, the effective force in non inertial S '-frame is

F' = F + (- Fp )., F = Real or actual force. - Fp =Pseudo force appeared. 4. For rotation of the frame with angular velocity W, (;). FIxed

=(;)

+wX rot

5. Centrifugal force developed in rotating frame is

-w x (w x r). For earth's rotation, it reduces the effective gravitational acceleration resulting its effective value as geff

=g-wx(wxr)

6. Coriolis force developed in a moving body w.r.t. the rotating frame on earth surface is -2m( wx

v).

7. The maximum value of centrifugal acceleration at equator is a~ 3.4 x 10-2 mls2 •

= w2R =

=0.3% of g.

8. The maximum effect of transverse coriolis acceleration at any pole of earth is

ac = 2 w sin 90 v = 2 w v = 1.458 X lO-4v. 9. Angular deflection of particle moving in time t at latitude e of earth surface due to transverse coriolis acceleration is a = w sin e. t

10. If any latitude e on northen hemisphere of earth, the components of angular speed 'w' of earth rotation are, Wx = Wy = W cos e, W z = w sin e. 11. The deflection of particle when dropped from certain height 'h' above earth surface, due to coriolis acceleration is 8

1

8h 3

= "3wcose ( g

)1/2

and this

deflection occurs towards east.

12. For Foucault's pendulum, the time period for each oscillation at any latitude e on earth surface is T =

~. wsine

\Vorked Out Examples Example 1 : An object is thrown downward with initial speed vo. Prove that after time t, it is deflected east of the vertical by an amount 2 wvot2 cose +-wgt 3 cose 3

108

The Classical Mechanics

Ans : Here, the effective component of angular velocity along x, )', z axis of the frame S' taken on earth surface, are Wx = 0, Wy = W cos 9, W = W sin 9 z Now, for velocity in time t, if X, y,

wfor earth rotation

z be the components then x = 0 , y = 0,

z = Vo + gt. Again the instantaneous transverse coriolis acceleration

x = d2~ = 12(w x v)1 = 2wcos9( Vo + gt)

dt 2wvo cos 9 + 2wg cos 9 t.

x

=

wvot 2 cose + ~wgt3 cose

3 This is the deflection towards east of vertical. Example 2 : Show that if the particle moves near the earth's surface, then the equation of motion are given by

x

2wsin9y

ji

-2(wsin9 x + wcos9z)

z

-

g + 2w cos 9 y

for latitude '9' on earth surface. Ans : We have for any frame S' (xyz) on the earth surface, A

A

A

W =

wxi +wy}+wzk

=

w) +wy}+wzk

=

(-wsin9y)i +(wsinex+wcos9z)}-(wcoseY)k

A

wx v

A

A

But the equation of motion is given by,

d 2r

-

dt 2

=

g-2(wx v)

=

-gu + 2wsine yi - 2(w sine x + xcose z)} + 2wcosey k

x=

2wsin9 y

ji = -2(wsinex+wcosi)

i

= -

g + 2w cos 9 y

Example 3 : An object is thrown vertically upward at latitude 9 with speed va- Prove that when it returns, it will be at a distance west ward from its starting 4

point equal to

3

cos9

3 WVo --2g

Reference Frame

109

Ans : Here, by following Ex. 1. the deflection is

x = WVo cose t 2 +.!.wgt 3 cose.

3 But t is the time of ascending which is actually the time of fall.

so, Vo

=

gt ... x

Also, t =

Vo

-,

g

=

"39 wgt 3 cose

and then,

4 v3 the total deflection x = -. wg-i} cos e 3 g

x=

4 3 cose 3WVO-2-'

g

Example 4 : An object at the equator is thrown vertically upward with a speed of 60 mph. How far from its orginal position will it land? Ans : Here the deflection is given by

x=

4 3 cose 3 WVO - 2-

g But w = 7.29 x 10-5 rad/sec Vo = 60 mph = 2.682 x 103 cm/sec q = 0° at equator g = 980 cm/sec x = l.952 Ri 2 cm. Example 5 : How long would it take for the plane of oscillation of a Foucault pendulum to make one complete revolution if the pendulum is located at a plane of colatitude 45°. Ans : Here latitude e = 90 - 45 = 45°. So for Foucault's pendulum, the time taken for one complete revolution of the plane of oscillation . T

=~ = w sine

2n = 33.84 hr. 7.29 x 10-5 x sin45

Example 6 : Prove that if an object is thrown downward from height h with initial velocity Vo above the earth surface then it will hit the earth at a point east of the vertical at a distance. (at latitude e) w;;;e (

~V5 + 2gh _ Vo )2( ~v5 + 2gh + 2vo).

Ans : Here by following the similar problem of Ex. 1. we have the east

110

The Classical Mechanics

ward deflection x = WVo COSe(2

+ .!wg cose (3

v=

3 the downward instanteneous velocity Vo + gt

(=

-(v- vo)

But for instant

T,

1

But

v2 = 1=

and

X

=

g

v5 + 2gh

~[~vJ + 2gh - Va] WCOSe 12 ( 3) gl + Vo 3 w

r(

;;~ e (~ v5 + 2gh - Vo ~ vJ + 2gh + 2vo )

This is eastward linear deflection of that falling body. Example 7 : A train having a maximum speed equal moving Vo is moving round a curve with radius of curvature 'p'. Prove that if there is no lateral deflection through out the outer tract, then the track should be at a height above the inner track is given by

aV5/ ~v6 + p2g2

where

'a' is the distance between

tracks. ADS :

Here, if x be the hight difference between outer and inner track,

then

x

v

2

tane= _=_0_ a pletT But

fetT = effective accn. =

IN = normal accn. =

~IJ + /f

Vo2

-

p

IT = Tangential accn;. = g

EXERCISES 1. What do you mean by a frame? Give its significance.

111

Reference Frame

2. Discuss the appearance of Pseudo force in a non inertial accelerated frame in translational motion. 3. Discuss the appearance of Pseudo force in a non inertial rotating frame. 4. Discuss the effect of transverse coriolis force on a particle when falling from a certain height above earth surface. S. Discuss the effect of centrifugal acceleration on the gravitational acceleration of earth surface. 6. Discuss the appeamce of cyclone. 7. Discuss the effect of Coriolis force on riverflow. 8. Di8cuss the priaciple of foucaults pendulum. 9. Show that the angular deviation of a plumb-line frame through vertical at a point on the earth surface at a latitude e is

ro w 2 sin ecos e 2

g-row cos

2

e

where, r0 is the radius of the earth.

10. Suppose that the mass m of a conical pendulum of length I moves in a

I

horizontal circle of radius a. Prove that (a) Tht speed is ali and (b) The tension in the string is

I

mg

(P - a 2 ) 1/4

l

vP _a 2

11. If an object is dropped to the earth's surface prove that its path is a semicubical parabola. 12. How long would it take the plane of oscillation of a Foucault pendulum to make one complete revolution if the pendulum is located at 45° Latitude or at the north pole. 13. Explain physically why the true vertical and apparent vertical would coincide at the equator and also the north and south poles.

-:0:-

Chapter-S

Central Force

5.1 Introduction The discussion of 'Central force' is very richest discussion in classical mechanics, since the knowledge about central force is important for dealing with several inverse square force, specially, in case of planetory motion. Several forces, like Newton's force of gravitation, Van' dar Waal' s force between' atoms and molecules in a gas, or even the Yukawa force between the nucleous in the nucleons of atoms are examples of central forces. The idea about it is in general, applicable in discussion on tides, dynamical manifulations of the orbits of spaceships in the space age, the geometry of the orbits of plannets ... etc. In this chapter, we will restrict ourself in discussion of central force characteristics and in several applications of it. Basically, the chapter will go through the depth of central force phenomena. 5.2 Definition and Characteristics of Central force The force field, either attractive or repulsive which is always in radial direction towards or away from the center (origin) of the force, is called central force. Mathematically, it is represented as FCentral

=

F

= ± f(r).c. = ± f(r)r . r

Where, '+' sign indicates repulsive force and '-' sign gives the attractive nature of force, and the force is always directed along

(±r) radial direction.

The basic characteristics of the force are, (i) The force may be repulsive or attractive, which is directed towards or away from the centre of force. (ii) This force obeys inverse square law of force, and this kind of force field is derivable from a scalar potential field, cp(r) such that

112

Central Force

113

f(r)r = - v
conservative in nature. (iv) Energy and momentum conservation will always hold under action of central force. (v) For conservation of central force, angular momentum m(r x v) will be the constant of motion. (vi) The magnitude of central force f(r) could be uniquely represented by a power in r, i.e. +00

f(r)

"~ knr-n

=

n=-oo

where,

k~s

are either constants including zeros or at most functions

of time. (vii) Only radial dependence in f(r) implies that the isotropy of space is preserved by f(r) about the origin.

5.3 Conservation of Angular Momentum under Central Force: We now consider particle motion in a plane under central force. So, the torque acting on it is

rxF

't

But

F = ±f(r)r

rX[±f(r)r] But

r x r = 0 :.

i

=

=

±f(r)(rxr)

0

Since we have for effective angular momentum L _ 't

=

dL

-

dL

dt :. dt

-

= 0 =:> L = constant.

So for particle motion in a plane under conservative central force, its angular momentum will remain conserved.

5.4 Conservation of energy under central force Consider the motion of a particle under central force

F(r)

=

f(r) r

along radial direction.

Now if Cr, 9) be the instanteneous plane polar co-ordinates of the particle then the energy of the particle will be

E = Ek +Ep

=~m(f2 +r 2e2)+Ep

114

The Classical Mechanics r

Ep = - f f(r) dr = Potential energy.

where

So the expression for total energy will be E =

~m(rZ+rZeZ)+{-ff(r)dr}

Now for our radial central force having no transverse component,

IF(r)1

=

mIr

o = mfe

where, f,. and fa are respectively radial and cross radial components of the acceleration.

re Z)

=

f(r)

m(re + 2iB)

=

0

m(r -

But from equation (2) we have

; (rZe) rZe

... (1) ... (2)

. re.+.2re

Z"·

= 0 => r 8 + 2rre = 0

o constant = h (say).

This is an outcome for conservative central force. So, we have from equation (1)

;: _re. 2 =

(.2r + ?h

2

"21 m

)

=

f(r)

--

m

ff(r )dr +

constant

115

Central Force

=>

~ m(,:2 + r 2( 2) + {-

f

j(r) dr} = constant

=> E = constant. This gives energy conservation under central force.

5.5 Equation of motion under attractive central force For attractive central force, we have the radial form of central force F(r) = - j(r)r which gives the equations of motion for particles motion under

such central force as

m(r _re 2 )

m(,B+ ue)

- f(r)

... (1)

o

... (2)

Then equation (2) gives the constant of motion r2

e=

Constant = h But to simplify the form of 1sl equation, let us now make a change in 1

variable as, r = - . U

,: Also, But

Also,

dr dt

1 .

--u

u2

dr du du . de .

e

1 = --;;

edu de

~=hU2 r2

d (dr) _ d ( h dU) dt dt dt de

So we have from equation (1),

-~~) m

116

The Classical Mechanics

This is equation of motion under attractive central force.

5.6 Application of central force theory to gravitation: Deduction of Keplar's law Actually, the gravitational force between two mass points m] and m2 obeys inverse square law of force and in that case the force is attractive central force in nature. So the gravitational central force is

F{r) = where,

f(r)

Gm]m2 =

2

r Now for planetory motion under such attractive central force for a planet of mass m arround the sun of mass M, the equation of motion will be

1~)

d 2u --+u = +--de 2 mh 2u2

1

where, u =- ,(r, e) r

=

position of planet

2• 'j 2 re=eu=h

and

So, for planetory motion under gravitational force, the equation of motion should be,

d 2u --+u 2

+

de

GM

where,

..

GMmu 2 GM =+--=+f.! 2 2 mh u h2

f.!

d 2u' --+u' de 2

-2- =

Constant

h 0

... (1)

for u' u - f.!. We can now apply this equation of motion (1) to deduce Keplar's law of planetory motion. For planetory motion, Keplar gave three basic laws which are

Central Force

117

(i) Law-I : Every planet will rotate arround sun in elliptic orbit with

sun at any of its focie. (ii) Law-II : For orbital motion of planet, its areal velocity will remain

constant. (iii) Law-III : For revolution of planet around sun, the square of time

period of revolution will be proportional to the cube if semi major axis of that elliptic orbit. We now obtain the solution of equation (1), u' A cos e => u - 11 = A cos e. (A = Const.) u 11 + A cos e

=>

GM -2-+Acose

h

r

1+( ~~)cose

r

I = 1+

E

cos e

... (2)

r

and

This is polar equation of the planet orbit where 'l' is semi latus rectum E is eccentricity of the orbit. It is now obvious that the eccentricity

Ah 2 GM GM ~ 10+ 19 SI unit for So the orbit is elliptic in nature and hence 1st law of Keplar is now established. But for planetory motion, the area swept out by the planet w.r.t. sun'S' E

= --<1

P,

1

in time is dA = -r.{rde) 2

s

1 2 o dA=-r de 2 So the areal velocity for such planetory motion is now given by

dA 1 2 de 1 2· 1 -=-r -=-r e=-h=Constant dt 2 dt 2 2 nd Thus 2 law of Keplar is also established. Since, the orbit of planet is elliptic, if its semi major axis and semi minor axis be 'a' and 'b' respectively then, v

=

A

b2 h2 Semi latus rectum, I = -;; = GM .

[From equation (2)]

The Classical Mechanics

118

b2 =ah 2 /GM So the time period of revolution of the planet along elliptic orbit is T

=

area of ellipse area of velocity

nab h/2

2

T2

4n a2 b2 h2

T2

4n 2 ah 4n 2 3 --a . = --a h2 GM GM

2

=>

2

2

T2

4n = __ .a 3 => T2 ex. a3 GM

i.e. the square of time period will be proportional to cube of semi major axis of the elliptic orbit. Thus Keplar' s 3 rd law is established.

5.7 Energy conservation for planetory motion For planetory motion, the total energy is now given by E Ek + Ep

(m = mass of planet)

But

Now in presence of gravitational force field offered by sun to the planet, the potential energy is

W = + S' GMm dr = _ GMm r2 r 00

1 (.2 +r 2e· 2)-GMm -mr -2 r

E

e e

1

Now considering

r = - and r 2 = - 2 = h u u

we have,

U

=

lIE

- =-

I

r

+ - cos e I

Where, I is semi latus rectum of the planetory orbit and

..

1 --r r2

--sinee

=>

r

-~sine.h

E

E

is eccentricity.

.

I

I

(.: r

2

e= h)

119

Central Force

2

2)

{ 22

(

1 E 2. 2 h GMm E - -m - h SIn 8 + --- 2 P r2 r

(

E

=

1 E'"h.2 21 E -m - - S I n 8+h -+-cos8 2 P I I

E

=

l -m

2 {

E'"22 h

. 28

--SIn

P

1 E )2} -GMm(-+-cos8 ) I I

h 2 2 Eh 2 cos8 2 h2 E 2 8} +-+ +--cos

P

P

P

GMm ---(1+ E cos8)

=

=

I 2 2 h h GMm 1 {E2 -m --+-(1+2 Ecos8) } ---(1+ Ecos8) 2 P P I

2)

E }

1 {h2 2h2 -m - ( 1+ E +--cos8 -GMm - - -GMm - - Ecos8 2 P P I I

But taking our previous notations GMm

~ = --2-

~

h

h2

and

I

GM = ~h2

1

= GM =;- , we can write down

GMmE I

and then we have, the total energy of planet is 2 2 mh ( 2) mh E ~2m mh2 E 2 E = - - 1+ E'" +--cos8------cos 8

2P

E=

But also,

2 mh (

2P

PIP

2)

~h2m

l+E --1-

1

~ =

I

E=

-(1+~)--

E

mh 2

mh 2

2P

P

mh 2

= ---(1-~) = constant (E < 1)

2P

120

The Classical Mechanics

Thus energy conservation holds for planetory motion. This energy is also -ve which implies that the planetory system is a bound system.

5.8 Stability of Orbits The orbit is the path of an object moving under central force. This orbit is taken as stable orbit if after giving small disturbance to the radial co-ordinate keeping energy or angular momentum unchanged, the orbit will be disturbed small. For effective potential energy for central force, ~(r) = vefr
o2veff(r) 2

or

>0

and for particular r = ro' [

02 Vefr (r)

1= '='0

or2

0

For central force, potential energy v( r)

=a T'+ 1, a =const.

and centrifugal potential energy verI r)

= ~, b =const. > 0 r

· . ~ ov(r) Slllce at eqm'l'b 1 num, lorce

Tr =Centn'fugaI ~lorce

..

=>

(n + 1) aT' n + 1) a

a v(r)

Now,

OVeff ] [or2 r=ro

= 12b r-31 = + 2b rn - 3 2b -(n+3) = -n+1 -r 2b_2 = -n+v-1r -(n+3) r n+1 = -n+ -r 1

-4-4(

= 6bro

= 2bro

Now for any circular orbit with r o2Veff( v) or

2

3 + n)

=ro the stability condition is

>0

n + 3 > 0 => n > -3. and for circular orbit n = -3.

SUMMARY 1. Central force has the standard form F(r) = ± j(r); .

Central Force

121

2. This force can be derivable from a scal:!r potential energy

~l(r)

= vef~r),

i.e., F= - \7lj)(r) 3. Magnitude of central force f( r) could be uniquely represented as a power in r is, +C()

f(r)

=

I

knr-n n=-C()

4. Angular momentum will be conserved for planner motion of particle under central force. 5. Only radial dependence of force preserves the isotropy of space. 6. Conservative nature of force restricts the time independence of the force. 7. Energy conservation holds for conservative central force. S. The equation of motion of particle under attractive central force is

d2u --+u 2

de

1 h =r where, u = -, r

f(~)

= --2 2 mh u

2"j e = e u2 .

9. The equation of planetory orbit under central force is

I

-r = 1 + e

cos

e

2

h2 Ah where, I = --, e = - - ; A GM GM and e < 1

=constant

10. The areal velocity of planet under gravitational central force is V A --

I 2' 2 r = h/2 = constant .

-

e

11. For planetory motion, T2 a a 3, T planetory elliptic orbit 12. Total energy of planet E

= Time period. a = Semi major axis of

mh 2

=- -2- (1- e 2 ) = const (- ve) .

21 13. For stability of orbit under effective central force potential a 1"'+1, n > -3 and for circular orbit, n =-3.

Worked out Examples 1. A particle describes the parabola r = asec

2

e '2

such that the cross radial

The Classical Mechanics

122

component of the velocity is constant. Show that

d 2r dt

Ans : Here for the cross radial component, r 2 r

But

=

is constant.

-2

e = constant = k (say)

o

asec 2 2

o

0 2

0 . 1 2 2

a.2 sec- sec - tan -0.2

asec 2 ~tan~e = 2 2

i-

k tan~ 2

F

d r = dt 2

2

d 2r

r8 tan~ 2

k~sec2~e =~k~e 2

2

2 a

k2 = constant 2a dt 2 2. A particle describes a circular orbit given by r = 2a cos 0 under the influence of an attractive central force directed towards a point on the circle. Show that the force varies as the inverse fifth power of the distance. ADS : Here the equation of the circle r = 2a cos 0

..

-

1 secO Putting r = - we have u = - u 2a du secO tanO dO 2a d u

2

sec 2 0+secOtan 2 0

d0 2

2a

Central Force

123

f(r)

'

1

:, f (r)a s

=

r i. e., the force varies as the inverse fifth power of the distance. 3. A particle of mass m moves in a central force field by

F:;,; - ~ r

r4 Show that if E be the total energy supplied to the particle, then its speed is given by 1

2E)2

v = _k + ( mr2 m ADs:

F

Here,

IFI

F=~

v(r)

-fFdr=-~

So the energy will be

r3

2r2

E

Ek + vCr)

E

.!. m(r2 + r2{)2) _ ~ 2r2

2

12k -my - 2 2r2 I

v =

e:+ m: Y 2

So the central forc.e is a conservative force. ADs: We know that if a force

fF.dr

or,

=

F be conservative, then V x F= 0

0

c

Now for central force,

V F~

F

f(r)!'" r

x vx(J(r)~)~ Vx[~) (xi + yJ +Zk)]

124

The Classical Mechanics i 0

ox

k

j

0

0

By

oz

-

x --j(r) l.. f(r) :.. f(r) r r r =

But

of ox

-

=

i(~ of _l.. of) + j(~ of _:.. 8f)+k(l.. of _~ 81)

rBy roz of or x of ---=-or ox r or

of _ l.. of of By - ror' oz

VXF

roz

rox

rox

=:.. of ror

=

VX(f(r)~)

=

i 1. of (zy - yz) + j 1. of (xz - zx) + k1. of (yx r or

rBy

r or

r or

xy)

=0 :. The central force is conservative

EXERCISES 1. What is central force? Give its characteristics. 2. Consider the motion of a particle under central force

_.!.. Show that its 2

areal velocity is constant. r 3. Find the differential equation of orbit for motion of particle under central force. 4. Find the equivalent one dimensional problem for the motion of particle under central force. S. Discuss Keplar's problem under inverse square law of force. 6. Discuss the condition for stability of orbit. 7. A particle of mass m moves under the action of central force whose potential v(r) = kmrZ (r> 0), then, (i) For what energy and angular momentum, will the orbit become circle of radius 'a' about the origin. (ii) Calculate the period for circular motion. (iii) If the problem be slightly disturbed from circular. motion what will be the period for small radial oscillation about r = a. 8. Show that the gravitational force is conservative. 9. A particle moves in a plane under the force F(r) = ar, where r is the distance from the center of force. Find the path of the particle. -:0:-

Chapter-6

Theory of Collision

6.1 Introduction Collision is a process through which two moving particle or one moving and one rest particle comes in contact and also transfer their momentum mutually keeping total momentum conserve. For two elastic bodies, this collision is commonly known as 'impact'. Since, du~ing collision between two bodies, no external force acts on the system from outside, the total momentum of the system will remain conserve through the impact. Basically, during collision or impact between two elastic bodies, the property of elasticity plays its role and thus the time* during which the deformation (slight) in shape takes place through collision, is called 'resistitution time' and within this time the shape is restored. During collision between two objects, say two spherical bodies, they exerts mutual forces to each other along the common normal to their point of contact and this common normal is called 'line of impact'. This line is the line joining the centers of two impinging spheres. For such line of impact, we can classify wheather the collision is direct or oblique collision because, collision takes place such that the direction of motion of both colliding objects before and after collision be along the line of impact, then it is direct collision. Otherwise, for oblique collision, the direction of motion is not along the line of impact. But for any collision, particle comes in closer and they exchange their momentum with each other keeping total momentum constant. 6.2 Characteristics of Collision Newton's Collision Rule Any type of collision has the following basic characters which are(j) Momentum Conservation takes place during collision. (ii) Kinetic energy (total) may be conserved or energy loss may occur during collision.

125

126

The Classical Mechanics (iii) Momentum transfer takes place during collision and through such

transfer impulsive action takes place during collision. (iv) Collision may be perfectly elastic, or perfectly inelastic energy conservation takes place along with momentum conservation. For perfectly semielastic collision, kinetic energy loss occurs and the particle moves together after collison, and for semi elastic collision, although energy loss takes place, both particles moves separately after collision. During any collision, either direct or oblique collision, Newton's collision rule is that the magnitude of relative velocity of two colliding object after collision will be proportional to that before collision. This constant of proportionality is called coefficient of resistitution* (e) and mathematically it is given by

where, UI and U2 are respective velocities of two particles before collision and

VI and V2 are that after collision.

°

Here, by conservation principle, ~ e ~ 1 and, for e = 1, collision is said to be elastic, for e = 0, collision is inelastic. and, for < e < 1, collision is semielastic.

°

So by this rule, !VI - V2! = e!ul -U2!' and this is in general true for oblique collision where as for direct collision. If particles moves in the same direction before and after collision, (VI - v2) = e (u 2 - u I ) and if they move opposite to each other before and after collision, then (VI + v2) = e (u l + u2). ~

6.3 Lab Frame and Center of Mass Frame : In general, the collision between two particles is oblique collision where both moving particles approaches to each other in a non linear manner with respect to their line of impact. After collision, they move away from each other in

v,

~

m,

u,

m,

"GENERAL CASE"

* For iii' u2' Vl> lI2 in the same direction. Because,

ml(u l -

v 2)

= m 2(v2 --v2)

U) - VI2) ( 2 2) ",""ml (2

ml UI - V)

=

m hour for elastic collision

(Conservation of mometum)

(2 2)

m2 U2 - U2

.

(Conservation of energy)

e = VI

- V2 =

u2 -ul

1

Theory of Collision

127

the same fashion. So in general case, collision takes place in a plane and obviously, the direct or linear collision is a special case of oblique collision in which particles moves along their line m, m, U, ----+------..--------------of impact before and after collision. ~ U,=o Experiments are often carried out in which one of the particles is at rest in the laboratory and the other approaches it and collision takes place. Such a set up in which a particle "LABORATORY FRAME" collides with another particle at rest is called the laboratory frame of reference. But the discussion of collision between two particles becomes very simple and symmetric for motions of particles before _ .....m, u,_ _ _ _ _ ' U, and after collision which are taken such that x / m, the center of mass of the system be at rest throughout the collision. Such setup is thus called center of mass frame, and obviously "CENTER OF MASS FRAME" in this system the collision between two particles is treated as they have equal and opposite momenta initially. ~

~

_~.

-

v:~ /-;t

..

ym,

6.4 Direct or Linear Collision Consider the collision between two m, t: m, particles when they are in motion along --------.-~-------.-~-----their line of impact before and after ~ R collision. So it is direct or linear m, m, F collision. ---------~-----~----)----------If m\ and m 2 be the respective masses of two impinging particles m, v, m, v, having initial velocities u\ and U2 --------------------.-~----.-~

u:

~

~

~

before collision (in the same direction) and final velocities 1\ and v2 after collision (in the same direction along the same line of impact) then for such direct or head on collision, we have from momentum conservation, m\u\

+ m 2u 2

= m\v\

+ m2v 2

...

(1)

...

(2)

But since for u\ > u2' v2 > vi' we have from Newton's collision rule, v 2 - v\ = e(u\ - u 2)

Multiplying equation (2) by m2 and then substracting from (1) we get,

=>

m\v\ + m 2 v\ m\u\ + m 2u 2 - em 2u\ + em 2u 2• v\(m\ + m 2) = u\(m\ - em 2) + m 2u 2 (1 + e)

The Classical Mechanics

128

vI

=

1

[(ml- em 2)ul

ml +m2

+(1 + e)m2u2]

... (3)

Similarly, multiplying equation (2) m l and then adding with (1), we get, (m l + m2) v2 m l u l (1 + e) + (m2 - em l )u2

v2

=

1

nil +m2

d

[(m2 - em u2 + (1 + e)mlud .

... (4)

This two equations (3) and (4) respectively gives the velocities VI and v2 of two impinging particles after direct impact. For this general direct impact, if we want to find out the loss of kinetic energy due to impact, then We have ~E =

loss in K.E

~E

-[(mIVI

+ m2 v2)2 + m\m2(vl -

2

V2)2J} 2 2J

1( 1 ) [ mjm2(uIU2) -m\m2(UI-u2) e 2 m\ +m2

[Useing equation (2) and eauation (1)1

1 mlm2 ( l-e 2)(U\ -U2 )2 ' ... (5) 2 ml +m2 This is the general expression for loss in K.E for direct collision and we should have, for perfectly in elastic collision, e = 0

=

V

I

nllUI nil

+ f1'12 U 2 + m2

= V2, and loss,

~E

1 mlm2 ( = --- - Uj 2

nil

+ nil

-

u2

)2

and this two

Theory of Collision

129

particles will move together after collision also, for perfectly elastic collision, e=l then,

VI

v2

(

=

(

1 )[(ml-m2)ul+2m2u2] m1 +m2 1

m) +mz

) [2m)u) + (m2 - md u2]

and then the loss, ilE = O. i.e. for elastic collision, particles will move separately along the same line of impact after collision without any loss in kinetic energy.

6.5 Characteristic of Direct Collision We have from our previous discussion that for direct collision between two bodies moving in the same direction before and after collision along their line of impact, the velocities after collision. I

=

V2

=

V

Where symbols has their usual meaning and also, the loss in K.E for such collision. ilE =..!.. m)mZ (l-eZ)(u) -uZ)2. (O:S; e:S; 1) 2 m) +mz With this results, we can have the following characteristics for direct impact, which are, (i) For m) = m 2 , VI = u 2' v2 = u l for e = 1 i.e. for elastic collision between two bodies having same mass, the velocity exchange takes place between them. (ii) For elastic collision, e = 1, ilE = 0 i.e. no loss in K.E occurs for elastic collision between two particles. (iii) For elastic collision, e = 1, if u2 = 0 i.e. if 2nd particle is at rest initially, then

V

I

=

m) -mz m) +mz

1- mz m)

u) = - - - u ) and Vz

1+ mz m)

2m)u)

2u)

m) +mz

l+_m_z

=---'-...!....- =- - m)

(a) for m) = m2 VI = 0, v2 = u) i.e. the 1st body is stopped and the 2nd one takes off with the velocity which the 1st one originally had. Both the momentum and the kinetic

130

The Classical Mechanics energy of the first are completely transferred to the 2nd. (b) for m 2 » mp VI ::::; - u l and v 2 ::::; o. i.e. when a light body collides with a much heavier body at rest, the velocity of light body is approximately reversed and the heavier body remains approximately at rest. viz, a ball dropped on earth rebounds with reverse velocity when the collision is elastic. (c) for m 2 « mp vI ::::; u l and u2 ::::; 2u I i.e. when heavy body collide elastically which remains practically unchanged, but the light body rebounds approximately with twice the velocity of the heavy body. By these considerations, we can explain the 'slow down' of the fast nutrons in a nuclear reactor, taking proton of hydrogeneous material (paraffin) has nearly the same mass as that of a neutron in a good moderator.

6.6 Maximum Energy transfer due to head on elastic collision Consider a particle of mass m l which collides with another mass m2, initially at rest. For such head on collision, if the initial velocity u l of mass m l changes to the velocity VI then, for u2 = o. We have, and the fractional loss in kinetic energy of the mass m l for such collision IS

L\E

E

Now let 4x

Here the transfer in energy will be maximum if

Ef =

I

2

"2mlVI

4x = I =0 is L\E = I E => (l+x)2

Theory of Collision

131

This is the condition for maximum energy transfer for head on collision.

6.7 Oblique Collision Here we consider the oblique collision of two masses m l and m 2 with initial velocities UI and U2 before collision at resp. angles (II and (I2 with line of impact and the velocities after collision are vI and v2 with the same line of impact. Now from momentum conservation along the horizontal line of impact, we get, mlu l cos (II + m 2u 2 cos (I2 = miv i cos PI + m 2v 2 cos P2 ... (1) But since the transverse components of momenta remains unchanged, we have, mlu l sin (II = miv i sin PI; m 2u 2 sin (I2 = m 2v 2 sin P2 Also, from Newton's collision rule, ... (2) ... (3) (v2 cos P2 - vI cos PI) = e(u l cos (II - u 2 cos (I2) By solving these equations, we now get, (ml - em2)uI COS(II + m2(1 + e)u2 COS(I2 vI cos PI = (ml + m2) (ml - em2)uI COS(I2 + m2(1 + e)u2 COS(I2 or,

vI

and,

v 2

... (4)

(ml + +m2)cosPI

=

(m2 - emdu2

COS(I2 + ml(l+e)uI COS(II ... (5)

(ml + m2)cosP2

So for such oblique impact, if we want to obtain the loss of kinetic energy, we get

=

2"1 [{ mlul2( cos 2 (II + sm. 2(II) + mlu22( cos 2 (I2 + sm. 2 (I2 )} {ml

~E

=

V?( cos 2 PI + sin 2 PI) + m2V~( cos 2 P2 + sin 2 P2)}]

2"1 [mlul2 cos2 (II + m2 u22 cos 2 (I2 -

A A )] ( mivi2 cos 2 1-' 1 + m2 v 22 cos 2 1-' 2

(Using equation (2))

132

The Classical Mechanics

=

(

1

2 ml +m2

[22

2

22

2

) mlul cos UI +m2u2 cos u2

2 +mlm2(ufcoS UI +uiCOS2(2)

+mlm2(vf cos

2

-{mfvfcos2~1 +mivicos2~2

~I + vi cos 2 ~2)}J

=

=

E=

(m l m2

)(1-e2)(UICOSUI-U2COSU2)2

2 ml +m2

... (6)

This is the energy loss for oblique impact. So for perfect inelastic oblique collision, this loss will be ilE=

1 mlm2 ( )2 (fore =0) UICOSUI-U2COSU2 2 ml +m2

SUMMARY 1. For any collision, momentum conservation takes place. 2. The ratio of magnitude of relative velocity after collision to that before collision is called coefficient of resistitution (e) which is 1 or 0 or 0 < e < 1 according as the collision is elastic, inelastic or sernielastic.

3. For any collision between two particles having initial velocities iii and ii2 before collision and

vI and v2

after collision,

liil -ii21 = IV2 - vII· 4. For direct collision, we have from momentum conservation

= ml VI + m2 v2 . 5. The velocities of particles after direct collision; are mlii l

+ m2 ii2

133

Theory of Collision

6. The loss in kinetic energy for direct collision, llE

=

1 mjm2 ( 1- e2)( Uj - U2 )2 .

2 mj +m2

7. For u 2 = 0, Vj = 0, v2 = u j for m l = m2

°

vI ~ -u I, v 2 = for m 2 » mj vI ~ up Vz ~ 2u I for m2 « m l 8. For Uz = 0, the maximum energy transfer takes place for m l 9. For oblique collision, mlu l cos a l + m2uZ cos a z =miv i cos ~I + m2 v2 cos ~z (u 2 cos ~2 - vI cos ~I) = e (u l cos a l - Uz cos a z) when,

vI cos ~I

=

cos ~2

=

= m2

(mj - em2)Uj cosaj - U2 cosa2 (mj + m2)

+ (1 + e)mjUj cosaj (mj + m2)

(m2 - emdU2 cosa2 V2

Energy loss for oblique collision llE

=

M

mj m 2/(mj

+ m2)](1- e

2

)(u2 cosa2 - Uj cosaj)2

Worked Out Examples 1. Two masses 5 gm and 10 gm approaches to each other with respectively velocity 5 rnIs and 10 rnIsec. They makes head on collision and after collision, the 1st particle comes at rest. What will be the velocity of 2 nd particle after collision. ADS: Here, given that m l = 5 gm, m2 = 10 gm u l = 5 rnIsec u2 = -10 rnIsec After collision, Vj = so from momentum conservation mju l + m 2v2 =miv i + m2v2 => 5 x 5 - 10 x 10 = 5 x + 10 v2

°

°

..

vI

75

= -10 =-7.5 rnIsec.

So, after collision the 2 nd particle moves with velocity 7.5 rnIsec in its original direction. 2. A particle collides elastically with another particle of same mass initially at rest. So that after collision, they moves at right angles to each other.

134

The Classical Mechanics Ans : Let both particles has the same mass 'm'. Here, u l = U, u2 = 0, e = 1 So from momentum conservation, mu

mVI +mv2

vI + V2

U

=:}

But, if after collision, this two particles be scattered at an angle a with each other, then I

_

e -

1=

IV2+ vll (v?+v~-2Vlv2cosa)2

= -'-----------'--

U

U

2 vI2+22 v2 - vlv2 cosa --u

... (1)

Also, for elastic collision, we have from energy conServation

I 2

2

v?

+ v~

-mu

I

2

I

2

-mvl +- mv2

2

2

u2

... (2) 2

= u2

So from this two equations, u - 2vI v 2 cos a 2v l v 2 cos a = or, cos a = 0, a = '!t/2 i.e. they will move at right angle to each other after collision. 3. Show that if two equal masses collides elastically, they will exchange velocities. Ans : Let, both has same mass m and they has respective velocities u l and u2 before collision and has velocities vI and v 2 after collision. so from momentum conservation,

°

mU I + mU 2 = mV I + mV2 u l + u2 = vI + v2

... (1)

Again for elastic collision, e = I ..

v 2 - vI = u l - v 2

... (2)

By solving this two equations (1) and (2), we get vI = u 2' v2 = u l

So they exchange velocity through collision. 4. Show that for perfectly elastic collision of particles, the total kinetic energy will remain conserved. Ans : Let m l and m 2 are two masses and u l and u 2 are the respective velocities before collision and VI and v2 are that after collision. So for elastic collision, (taken to be direct or head on collision) v2 - vI = u l - u2... (2)(·,' e = 1) and from momentum conservation, ... (2)

135

Theory of Collision

(ml - m2 )UI + 2m2u2

we can have,

ml +m2

v2

=

ml +m2

so the total kinetic energy after collision, 1 2 1 2 E = -mivi +-m2 v2 'f 2 2

2 1 2 1 2 2 = Ej = Total kinetic energy before collision. 5. Consider the direct collision of two particles of mass m l and m 2 having initial velocities U I and u2 before collision. If e be the coefficient of resistitution, then show that the momentum which is transferred from the 1st particle to the 2 nd particle is

= -mlul +-m2u2

mlm2 (l+e)(ul -U2) ml +m2 Ans : For this direct collision, mlu l + m2u2 = mlv} + m 2v2 ... (1) and v2 - VI = e(u l - u2) ... (2) Where VI and v2 are their respective velocities after collision. So the change in momentum due to such impact is I'!.p = m 2v2 - m 2u2 = m 2(v 2 - u2)

[Because by solving equation (1) and (2), we get

m}(l+e) v = 2

(m\ + m 2)

u\

+

(m2 -em l ) U (m l + m2) 2

m2(1+e) (m\ -em 2) ~+ U2 (m\ + m2) (ml + m 2) So, this momentum transfer v = I

The Classical Mechanics

136

:. f'o..p =

mlm2 ml +m2

(1 + e)(ul -

U2)

6. A ball which is dropped from a height H onto a floor rebounds to a height h < H. Determine the coefficient of resistitution Ans : If the particle after falling from height H, collides with the floor with velocity u and rebounds with instantaneous velocity v after collision then v the cofficient of resistitution, e = -

u

Here

u2 =

2gH, y2 = 2gh

e=~ 7. A billiard ball strikes another identical billiard ball obliquely at an angle 45° with their line of centers at the time of impact. If the coefficient of resistitution is

1

"2' find the angle at which the 1sl ball will 'bounce off'.

Ans : For such oblique impact, initially, u l = U, a l = a = 45°. u2 = 0 If vI and v2 be their respective velocities after impact then for their angles 13 1 and 13 2 with the line of impact, mu cos a = mV I cos 13 1 + mV2 cos 13 2 U

vI cos 13 1 + v2 cos 13 2 = Again,

v 2 cos

13 2 - vI cos 13 1 =

U

cos 45

e(u

=

..fi

cos 45°) = .!.u. ~ 2 v2

u

=2J2 2vI cos 13 1 =

~(1-~) = 2:n

Again for normal momentum about the line of impact, mu sin a = mV I sin 13 1

... (1)

Theory of Collision

137

vI sin PI = u sin 45 =

U

fi

... (2)

tan PI = 2. => PI = tan- I(2) This is the angle at which the 1sl ball will 'bounce off after collision.

EXERCISES 1. What are the basic characteristics of collision? 2. State and explain Newton's collision rule. 3. Show that for inelastic direct collision between two particles, the loss in 1 -mlm2 kinetic energy is ~E = 2 (VI - V2)2 . ml +m2 4. Find the velocities of two particles after oblique collision between this two particles. Also find loss in kinetic energy in this regard. 5. Discuss collision between two particles in center of mass frame. 6. A gun fires a bullet of mass m with horizontal velocity v into a block of mass m which rests on a horizontal frictionless plane. If the bullet becomes embedded in the block, find the loss in kinetic energy in this impact. 7. Repeat the previous problem (6) if the block is moving away from the gun with velocity vO' 8. A mass m l travelling with speed v on a horizontal plane hits another mass m2 which is at rest. If the coefficient of resistitution is E, prove that there

mlm2(1- ~)v2 is a loss of kinetic energy equal to

2(

)

+m2 9. A ball is dropped from a height' h' above a horizontal plane on to an inclined plane or angle a which is resting on the horizontal plane. Prove that if the cofficient of resistitution is e, then the ball will next hit the inclined at a point which is at a distance 4e(1 + e) sin a below the original point of impact.

-:0:-

ml

Chapter-7

Conservation Principle and Constrained Motion

7.1 Characteristics of Conservation Principle In classical mechanics, we deal several problems with motion of a single particle or a system of particles. For such motion, we have to face a term 'conservation' many times. The conservation of a parameter, involve with the motion, is basically the constantness of that parameter with respect to time. Clearly, the parameter which is a conserved quantity, will not change with time throughout. We also observe that the parameter conservation demands one or more than one conditions which are basically constrained of motion and these conditions restricts the conservation of that parameter. With the dealing with several conserved parameter, we can now say that, (i) Conservation principles are independent on the trajectories of the particles or system, but it depends on some restrictions. (ii) Conservation of any parameter means its constantness and all conserved quantities are constant of motion. (iii) Conservation laws have an intimate relation with invariance. Their failure in certain cases may result in the discovery of new and not yet understood phenomenon. (iv) From conservation law we can find equation of motion even when the effective force action on the system is not totally known. (v) Conservation laws assure us many times that some aspects of motion are impossible and must be left out. (vi) Conservation phenomenon tells us a great deal about the motion even if forces affecting motion are not known in advance. In our physical world, there exist a no. of conservation principles or laws, some exact and some approximate. These principles are related with energy, linear momentum, angular momentum change ... etc. and various other quantities. We will discuss all such conservations in our present chapter which are very

138

Conservation Principle and Constrained Motion

139

powerful tools in our present chapter which are very powerful tools in solving several mechanical problems.

7.2 Mechanics of a single particle and system of particles The motion of a single particle or a system of particles can primarily be studied by applying Newton's 2 nd law of motion which estimate the effective force acting on that particle or system. To study Newtonian mechanics for the motion of particle, it is mandatory to realize the nature and interaction of force on it. To get a sharp knowledge about this for the motion of particle or for the motion of system of particles, we should have the following discussionFor a single particle motion, we have from Newton's 2 nd law, the applied . . - djJ d _ dV d 2; force or the effective force 1S F = - = -(my) = m- = m---:x dt dt dt dt Where, p is instantaneous momentum of the particle 'm' is its mass and

v is instantaneous velocity which has magnitude much less than that of light wave in vacuum. So motion is restricted in the region of Newtonian mechanics which is far apart from relativistic region. Here, ; , is instantaneous position of particle during its motion w.r.t some stationary point. But if we now consider the system ofN particles, then from Newton's law, the effective force on ith particle, F j

Where,

F}e)

~

-+ =

= mj d

2

;,

-(e) + "~ Fij - . F j

dt j*j is instantaneous position of ith particle having mass

is external force acting on ith particle and

L Fij

'm;'.

Here,

is the sum of internal

forces on ith particle by all Jth particles (j #- i). So for any constituent of the system, the effective force is the resultant of external force and all possible internal forces given by the neighbours particles. But here considering all particles together for the whole system, the net force, F

= LFj = LFj(e) + LLFij j*i

j

F = LFj(e) + L LFij = Lmj d2~ =

and

j

j

- But we must have, Fij = -Fji

""p.. ~~ j

j

-

lj

=

1 -2

(j*j)

i

dt

d: [Lmj~)

dt

j

L(-P.·+F·· -) =0 ij

I}

Jl

i. e., the net internal force for motion of all particles in the system is zero, and as a result, for motion of a system of particles,

140

The Classical Mechanics 2

IF/e) = dtd lIm,r;].

F

2

j

j

If we now take the average of all radius vectors for all constituents particles, then we get the instantaneous position of center of mass of the system,

Mrc where, M

=

I mj

=

Total mass,

~F/e) =M( :~]

F

Which is a single particle equation. So the motion of a system of particles can in general be replaced by the motion of the center of mass of that system under the net force which is the sum of all external forces, applied on that system.

7.3 Conservation of linear momentum The principle of linear momentum conservation states that if the net effective force on a single particle or system of particles be zero then the momentum of that particle or the vector sum of momentum of all particles of the system will be constant or will remain invariant throughout the motion of that particle or that system. This conservation can easily be understood from Newton's law and for single particle motion,

_

F

So for

F= 0,

dV

d ( _)

dt

dt

= m- = - mv

mv = constant or conserved.

Also, for system of particles, 2

F

=

"F-(e) _ " L.J j - L.J I

so for,

I

dV -!!...-" mj ddt2r; -_" L.J mj dt - dt L.J m, j

Vj

I

I

F

:, (

~ miVi) ~ 0 => L mi V' ~ Constant

This gives the conservation of linear momentum. So for effective force to

Conservation Principle and Constrained Motion

141

be zero for the motion, the net momentum will become a conserved quantity or constant of motion.

7.4 Conservation of Angular Momentum For a single particle rotation or for rigid body rotation if net torque acting on the particle or the system be equal to zero then the angular momentum will be conserved. This is angular momentum conservation. We can now discuss such conservation principle for single particle motion or for motion of the system of particles, in the following wayFor single particle rotation, the angular momentum is the moment of linear momentum which is given by

L = F x Ii =F x (mv) = m(F x v) So the torque acting on it, is t

-t

-

= r x

- djJ d (- -) = r x dt = dt r x p

dL

dF fi dF -) ( . :dt- x p= 0, or-lip dt

dt

0,

and for

F-

dL - = 0· L = Constant

dt

'

This is angular momentum conservation for single particle rotation. Also for system of particles, the torque on ith particle. .r.::. r..-I x "p. I

=r.-

I

d( m·v·-) =r..- x F· x -dt I I I I

But,

So, But we have, for all constituent particles in the system

L~ x Fij I#i

~~(~ x Fij +Fj x Fji) J"#l

~ - Fj = ~j and this IFijl gives the separation between ith and jth particle. Since, the internal force between ith and jth particle Fij acts along Here,

142

The Classical Mechanics

:. (~ - 0) x Fij

=

0 and the total internal torque I~ x Fij = 0 I~j

So the effective torque on ith particle is effectively the torque for external acting on that particle, force F(e) I ~ x

Ie.

-

Fj

=

~ x

-(e)

Fi

and the net torque on the whole system 1:

=

"i:. ~

I

x

I

dPj dt

=~("i:. dt ~ x P')= dL dt I

I

I

d"l; _ . d"i_) I Ilpi ( for, -dt x P = 0, SInce, _ dt dL L = I("l; x Pj) = Constant - =0 , dt This is conservation of angular momentum for system of particles by which if net torque for all external forces acting on the system be zero, the total angular momentum for the whole system will become constant. Now, for

:r =0,

7.5 Conservation of Energy For a single particle motion or for motion of a system of particles, if conservative force acts on that particle or on that system, then its total energy is the sum of kinetic and potential energy will be constant.

F is said to be conservative if the close line along a close path 'C' becomes zero. i.e., if F be conservative, Now a force

integral of it

fF.dr = 0 c

and in that case we have from Stoke's theorem

V x F= 0

the conservative

nature of force demands that this force can be expressed as a -ve gradient of some scalar function, v(r), called scalar potential. I.e.

F

=

-Vv(r)

Under action of such conservative force, we have for a single particle system,

Conservation Principle and Constrained Motion

Work done,

W

2

2

f

f

1

1

143

d-

= F.dr = m d~ .di

When particle goes from state 1 to state 2. 2

W

2

2

f mv.av f~md(v. v) fd(~mv2) =

1

=

1

1

1 2 1 2 - mv2 --mvl 2 2 But also, with respect to conservative force F , the work done in this case, 2

W

2

f F.dr fVY.dr =-

I

W

I

Y I - Y 2 and in this case, we finally have,

1 2 1 2 - mv2 --mvl 2 2

= =

1 2 -m~+~ 2

1 2 -m~+~ 2

(KE)I + (PE)I) (KE)2 + (PE)2 i.e. the sum of kinetic and potential energy will be constant in every state of the particle. This is energy conservation. Similarly, for N-particle system the total potential energy

when,

Fint

_Vjyint

I

Here, one thing is very clear that if the mutual interacting forces be Newtonian, yIn! must be a function of the relative distance between two interacting particles, i. e., for ith particle.

y~nt )

yint

Iyr

=

~I(Yrt + yr) j

144

The Classical Mechanics

=

~LLVAI~ -rjl) i

Here clearly, the factor

1

"2

/F-i

comes because of the fact that while summing

the mutual potential energies, a pair of particle (i, j) appears twice, once for ith due to Jth and other for jth due ith particle, where both are the same and should appear only once. So, for ith particle of the system,

p. = p(e) + "pi.nt ~lJ I

I

fF-i

"F-ij' d-1";

-(e) dF i . ri + ~ joF-i

But

"F-y' d-r

-(e) dF, . ri + ~

i

j~i

d(l )dt dt

_

_ m .f..2 2 I I

=

-(e)d"F-dF i . ri + ~ ij' 1"; j~i

Thus for whole system,

L ~(!..mjvl)dt = LPi(e).d~ + LLPij.d~ dt 2 i i j~i j

But also,

L~Pij.d~ I

=

~LL(Pij.dr, +Pji .d0) i

rF-I

=

j~,

~L~Pij.d(~ -0) ,

J~'

Conservation Principle and Constrained Motion

145

Also, We then finally have,

d(""l

.2] dt -_- "V(V(e)) ~ d'i.. -2l",,~ ~~ 'VI} ( v,; ).dr~

dt L2mi~

j

i

j

y

. •

, 1*'

'

"1

1""

'

,,(e) --~~Vij+Constant ~-mi~.2 = -~Vj . 2 , 2 , ., I , , J*'

"

..!.m.f,2

~2 ,

I'

+"v(e) +..!.2 ""v.1,nt ~' ~~ y ,

,

T+V where,

T

J*'

Constant

=

Constant

,,1

.2

total K.E = ~ 2miri I

and

"V(e) l " " V int V = total potential energy = ~ j + 2~ ~ ij ,

,

J

This is energy conservation for system of particles under conservative forces acting on it.

7.6 Constrained Motion Degree of Motion, Constraints: The motion of a particle or system of particles under some conditions or restrictions is called constrained motion. For much motion, we must have some knowledge about the degree of freedom and constraints. The minimum no. of independent variables which are required to specify the motion of particle or system of particles as a whole, is called degree of freedom for that particle or system. It can be shown frum mathematical version, that if motion of a system ofN-particles are involve with 'm' no. of restrictions, then degree of freedom of that system is d"" 3N - m. We will analyse the reason behind it later on, but at this junction, what we want to say that if we take all example of gas molecules, then for mono atomic molecule, N ~-= 1. m = 0, d= 3. For diatomic molecule, N = 2, m == I, d = 5 ... etc. Now the mathematical restrictions which can be expressed through the equations or inequations, are called constraints. So the constrained motion is restricted by the 'so called' constraints. We can now consider some example of constrained motion to clear these points.

The Classical Mechanics

146

Example 1 : Consider motion of a simple pendulum. If (x, y) be the cartesian coordinates of the instantaneous position of bob and (r, 8) be the polar coordinates of it, then the constraint is x 2 + y2 = r2 = 12 where, r = I = effective length. So here the degree of freedom is 1 for independent variable x or y in caltesian system or 8 in plane polar system. Example 2 : Consider the motion of a particle which is confined on the surface ofa sphere of radius 'r'. So here, the constraint is x 2 + y2 + z2 = r2 where, the degree of freedom is (3 -- 1) = 2 and independent variables are (x, y) or (y, z) or (z, x) in cartesian system and (r, 8) in spherical polar system. Example 3 : Consider the motion of a monatomic gas molecule within a box of dimension (a x b x c). So here if (x, y, z) be the instantaneous position of that molecule within box, then the constraints are, o : ; x ::;; a, 0 ::;; y ::;; b, 0 :5 z ::;; c. and here, the degree of freedom = (3 - 0) = 3 where, the independent coordinates are (x, y, z), in cartesian coordinate system, or (r, 8,
-I 1- -I IrIJ.. I = r: - r'lJ = Constant I

This is the restriction for rigid body motion for which the separation between any two constituent particles within rigid body will be constant during rigid body motion. Clearly, such constraint is holonomic and scleronomous constraint. 7.7 Generalised Co-ordinates and other Generalised Parameters The independent coordinates which are involve with the motion of a single

* Constraints are bilateral which can be expressed in equations and non holonomic constramts are unilateral whIch can not be expressed in equations.

Conservation Principle and Constrained Motion

147

particle or system of particles are commonly known as generalised coordinates. So it is a fact that for a given system, the no. of generalised cordinates gives the degree of freedom (d). These generalized coordinates are in general represented by where,j = 1,2, ... d Basically, for motion ofN-particle system having no 'so called' restrictions over the motion, d = 3N and for q; 's, j = 1, 2, 3 ... 3N. As for examples, for pendulum motion, ql = 8, degree of freedom = 1 for motion of mono atomic molecule within cuboidal box, ql = X, q2 = y, q3 = z, degree of freedom = 3 for motion of particle confined on the surface of a sphere of radius r, ql = 8, q2 = a, a = radius of sphere) for motion of a N particle system,

q/s

ql

= xi'

q2

= Yi'

q3

= zi'

... , q3N

= zN

and degree of freedom = 3N. It is a fact that for motion of any system, under any no. of constraints, the generalised coordinate can be expressed in terms of general coordinates and the converse is also true. As for example for motion of a simple pendulum, x2 + y2

=f2

..

x =

ql

= ~P - i

or,

y

ql

=~12

also,

ql =

8=

= ./i(xlyd

_X2 =J;(x,y)

tan-l(~) = j(XI y)

Thus, we can now conclude that for generalised cordinates have,

... etc.

a/s we usually

qAI~1) = qjh) and,

~ (qj)

where, for i = 1, 2, ... N j = 1, 2, ... d. (d = 3N) for a N-particle system. We will now discuss about some other relevent generalised parameters which are significant to study with the motion of a single particle or the system of particles. These parameters are given below. (i) Generalised Velocity: The time rate of change of generalized coordinate i.e.

iI, is called generalised velocity.

148

The Classical Mechanics

Since, we have, F; 3N ",-

oF;

=

.

"~o ~a q}

]=\

qj

Here, or, are arbitrary virtual displacement which is totally independent of time and ot = o. So here, oq/ s are called generalised displacements.

F;(qj,t)

But for,

dF; dt Here, the terms qj are generalised velocity and so we can conclude that the ordinary velocity ~ are seen to be the linear functions of the generalised velocity components, no matter how the generalised coordinates are defined. (iiJ Generalised Acceleration We have,

F;

=

F;(qj,t)

Conservation Principle and Constrained Motion

rj Here, the term

aaqj

"-.!lq

L:

=

'!

iij

7"-:

+""

J

149

a2 ': qqk + aqj.aqk J

a a q +-~ 2"L: _---.!L oqj.Dt ot 2-

2-

J

2

is called generalised acceleration and here, we see that

the ordinary acceleration is not linear functions of components of generalised acceleration alone but depends quadratically and linearly on the generalised velocity

iI j

as well.

(iii) Generalised Momentum We originally have the kinetic energy ofN-free particles, as, N

T

=

"~2mjrj 1 .2

(-'--'-) = ,,1 ~2mj rj.ri

i=1

i

N {3N ~3N ~- } uri ~- 1 "N (-) "uri. "uri. ari .(-ari-) ~-qj + ~--qk . - +- ~mi -

+-1 "~mi 2

i=1

j=1 aqj

k=1 aqk

at

2

i=1

at

at

- ) qj+-~mi - .(aF;) "~~mi " (--. aF; ) (aF; 1" (aF;) " J aqj' at 2, at at I

Now if

F;

a~ = 0 at

where,

I

be not function of time explicitly then and T =

LLajkiJAk j

ajk =

~2

k

Imi( aqjaF; ).( aqkaF; ) i

and in that case, T is homogeneous function of generalised velocities in degree 2. Now the generalised momentum Pj corresponding to generalised velocity iI j is defined as

The Classical Mechanics

150

So, for stationary generalised system (

oF. a:-

)

= 0 ,this generalised momentum

is given by p. = }

L L a,i! I

k

k

which is again linear function of generalised velocities.

(iv) Generalised Force

L Fi

The amount of workdone by the force arbitrary small displacement

where,

oy;

on the system during an

i

of the system, is

,,- oF.

' Q.= L..Fi·oqj

i

J

This parameter Q. is called generalised force associated with the co-ordinates qj' Here we should follow that whatever dimension a generalised coordinate has, the product of the generalised force and generalised displacement must have the dimension of work. For that reason the generalised force need not have always the dimension of force. (v) Generalised Potential If V be the position dependent potential energy of the system then dw

= oV = - "L..aq. oV oq.} ="Q .oq . L.. } } j

}

j

Since,

= -

,,- oF. L..Fi·' = -Qj i

oqj

So, the definition of ~ as generalised force is a natural one.

Conservation Principle and Constrained Motion

151

But when the system is not conservative for which the potential energy V is also function of generalised velocities

qj, we can define the generalised force

Q]

as,

Qj

oV

d

(avo 1

= - oqj + dt oqj

How, V 0 may be called a 'velocity dependent potential' or simply 'generalised potential', since it gives rise to generalised force Qj"

7.8 Limitation of Newton's Law The limitations for applicability of Newton's law are (i) These laws are only valid in inertial frame. For non inertial frame, the transformed version of Newton's Laws are made because of the presence of pseudo force in that frame. (ii) The velocity of particle on which Newton's law is required to apply, must be much less than that of light in vacuum. Because, of the particle velocity be close to that of light in vacuum (v ~ c), then by Einstein's relativity, mass will change with velocity. But Newton's law only is applicable for stationary mass.

SUMMARY 1. Conservation principle gives the constantness of parameter. 2. Conservation law gives the informations about motion even if forces affecting motion are not known. 3. Force for a single particle motion,

F= m~

4. Effective force on ith particle for the system of particles is -

Fi

-(e)

,,-int

:.;

= Fi + L.'pij = mj'f ft-i

5. Effective force on system of particles

where

6. For momentum conservation, LmiVj = Constant for effective force,

F= LFi(e) =0 i

7. For angular momentum conservation,

152

The Classical Mechanics

L)::'i == L>li(~ x vi) ~ Constant i

for effective torque

,,rj i

L

8. For conservative force,

-

~(e)

,,-

x Fi :::: L. ri x Fi

fF.dr

= 0

i

== 0 and

F== - v( v)

c

9. For energy conservation,

1 2 -mvi + vI 2

I 2 mv2 + V2 2 /

(Single particle)

= -

~ ~ m/? + ~ vi(e) + ~ ~ ~ Vbnt I

I

I

= Constant (System of N particles)

)

10. For constrained motion: No. of constraints = No of restrictions involbe = m No. of particles in the system = N No. of degree of freedom = d = 3N - m No. of generalised coordinates = d i.e. for q/s,j = 1,2,3, ... d Constraints: Holonornic ~ Velocity dependent Non holonomic ~ Velocity independent Bilateral ~ Can be expressed in equations Unilateral ~ Can not be expressed in equations Scleronomic ~ Not explicit function of time Rheonomic ~ Depends on time explicitly Generalised parameters: Generalised coordinates: qj ; qj = qj(r;) Generalised displacement: oq).;

or..

1

=

I

J

Generalised velocity :

Generalised acceleration:

iij

Generalised momentum:

Pj

fJi _I

oq. J

oqj

(for 01

=

0)

Conservation Principle and Constrained Motion

153

Generalised potential :

Worked Out Examples 1. Find the generalised coordinates for the following system, fly-wheel, a particle moving inside the surface of a cone, Bead in abacus, Hydrogen mole<.:Ule. Ans : (i) Fly wheel: 8, the angle between a finite radius of the fly wheel and fixed line perpendicular to the axis. (ii) A particle moving inside the surface of a cone: r, 8; r =The radius vector drawn from the vertex as the origin to the position of the particle 8 = the angle of the radius vector with a fixed slant edge of the cone. (iii) Beads in abacus: x, the cartesian co-ordinate along the horizontal wire. (v) Hydrogen molecule: x, y, Z,
Kinitic energy,

T = ~ L '21 m/ .2i i

" .. =" L..JL..Jajkqjqk j

k

I ~mi( a~ ).( aqk a~ ) 2 aqj

where,

i

so, this kinetic energy is a homogeneous function of generalised velocity in degree 2. Hence, from Euler's theorem

T

. = -2II p.q. J J j

154

The Classical Mechanics

Example 4 :Write down Jacobian (1) for the transformation from a set of coordinates q} (j = 1, 2, ... , d) to the set ri(i = 1, 2, ... N). (d = 3N) Ans : J(ql'q2 ... ,q3N} == (O(QbQ2 ... ,q3N )) xby)···,zN

O(XbYb···ZN)

aQ)

aQ2

ax) aQ)

ax) aQ2

aX2

aX2

aX2

aQ)

aQ2 aZN

aQ3N aZN

aZN

aQ3N ax) aQ3N

EXERCISES 1. Prove the laws of conservation of linear momentum, angular momentum and energy for a system of interacting particles. 2. What are the advantages of using generalised coordinates. 3. Discuss the statement -"~ 8Qj must have the dimension of work". 4. What are constrains? How do they affect motion of a mechanical system. S. Show that of the equations of transformations do not involve time, the kinetic energy can be expressed as a homogeneous quadratic function of velocities.

-:0:-

Chapter-8

Variational Principle and Lagrangian Mechanics

8.1 Introduction In our present discussion, we will try to develope a new field of mechanics, popularly known as Lagrangian Mechanics which is some thing different from Newtonian mechanics. In this new field, we can deal the several mechanical problems by considering the energy of the system, both kinetic and potential rather than the force or the momentum. The development of Lagrangian formalism can be made through the zone of Newtonian mechanics (by the consideration of D'Alambert's principle) or from the basis of calculus of variation. It is now important and also significant to realize the calculus of variation at first and then its role to develop the new field, the Lagrangian field in classical mechanics. In our provinces chapter we have discussed about the limitations of applicability of Newtonian mechanics. Keeping such restriction in mind, we will discuss how to develop the range if the applicability of that Newtonian mechanics in a new version through energetic consideration. This approach is at first made by Lagrange and in that sense, Lagrangian mechanics was mostly acceptable to several mechanical systems. This mechanics, or more clearly, the laws of such mechanics, were at first developed on the basis of Hamilton's integral principle which was actually an outcome of the calculus of variation. After that, this laws are shown to be equivalent with Newtonian mechanics through the developement ofD' Alambert's principle. We will try to cover more or less all such discussions in our present chapter and also we will try to make interest on such discussions as far as possible.

8.2 Forces of Constraint The mathematical restrictions which is the constraints of the constrained motion, not only interfere with the solution of the problems, but they are always associated with the forces by virtue of which they restrict the motion of the 155

156

The Classical Mechanics

system. Such forces are known as "forces of constraints". We generally formulate the laws of mechanics in such a way that the work done by the forces of constraints is zero when the system is in motion. We basically require our formulation to side track the effect of forces of constraints without violating them. So for the force of constraint F , the workdone for small time independent is displacement

Or ,

OW= F.or = 0 By definition, the forces of constraint are workless forces. As for example, if we take the motion of simple pendulum the tension of the string will be the force of constraint which is perpendicular to every instantaneous displacement of the bob of pendulum. Similarly, if we take the sliding of a body down a friction less inclined plane, the normal reaction given by the plane on the body will be the force of constraint which is also workless since it is perpendicular to every instantaneous displacement of the body along the plane. So we see that the role of the forces of constraints are very significant for constrained motion although the forces are themselves workless.

8.3 Virtual Displacement A virtual displacement is an arbitrary, instantaneous, infinitesimal displacement of a dynamical system independent of time and consistent with the constraints of the system. Since we have ~ =~(qi,t) for N particle system we have, virtual displacement, d or. o~ = I-J oq oqj j=l

(Since, it is independent of time)

j

So, the virtual displacement can be expressed as the linear superposition of infinitesimal generalised displacements, all taken as independent of time.

8.4 Principle of Virtual Work The work due to virtual displacement is known as virtual work. By the principle of virtual work, A system with workless constraints is in equilibrium under applied forces if and only if zero virtual work is done by the applied force in an arbitrary infinitesimal displacement satisfying the constraints. Let us now consider a dynamical system which is in equilibrium. Then from the total force one each particle of the system vanishes, i.e.,

Fi = 0 .

Now let us give a virtual displacement to the system, then the virtual work of the ith particle is,

Fi .o~

where o~ is the virtual displacement if i-th particle.

So the total virtual work for all the particle is

Variational Principle and Lagrangian Mechanics

IFj.8~

0

pta) , + p(c) ,

Fj

But

p.(a)

where,

157

externally applied force

I

Ftc) ,

force of constraint

But Fj(c),s are all workless forces

"F(e) ~L..J' .ur,

=

0

This is principle of virtual work.

8.S D' Alembert's Principle We have from Newton's 2nd law, if

Pi

be the force on with particle of a

dynamical system,

o Now for infinitesimal, time independent virtual displacement 8F; we have from the principle of virtual work.

I[Pi +(-mi~)]

0

i

But externally applied force,

where, p(c) , " { - (a)

L..J F;

But,

force of constraints

- (e) ( + F; + -m,r;:;)}.8r;

o

=

0

158

The Classical Mechanics because,

F(e) I

is workless.

:. I{F/a)+(-mj~)}.DF;

=0

j

This is D' Alembert's law. Here, the force (-mj~) is called reverse effective force and so by this law, the total workdone for all particles of the system by the resultant of externally applied force and reverse effective force for infinitesimal, time independent displacement of all particles of the system will become zero. Here, we should note that we can equate the coefficients of

D~

to zero since oF;' s are not independent to each other. There it is necessary to transform oF; into the changes of generalised co-ordinates, dqp which are independent to each other. The coefficient of every dqj will then be equated to zero. We should also note that, as this principle does not involve forces of constraints in any way, it is sufficient to specify all the applied forces only. Also, this principle is valid for all rheonomic and scleronomic systems that are either holonomic or nonholonomic.

8.6 Lagrange's equations for a holonomic System For a holonomic dynamical system, we have, the instantaneous position vector of ith particle,

F;

=

F;(qj,t)

i = 1,2, ... N

j

=

1,2, ... d (d

=

3N)

we get,

... (1)

Also the kinetic energy for the whole system, T

II mjrj = I .!.mj(~'~) 2 . 2 .2

!

or Oqk and

or oqk

I

I

~ o~' m·r.·.-11

... (2)

I mj'i'-- o~

... (3)

j

j

0qk

oqk

Variational Principle and Lagrangian Mechanics

~ a~

159

~

a~

" d ( - .) L..Jnljrj.-. + "L..Jnljrj..

aqk

I

But,

.

I

dt aqk

(From basic rule of calculus)

So from equation (2),

[from equation (3)]

where,

Qk =

,,- a;.

1

L..JFj. j

aqk

= Generalised force.

So, we finally get,

L[~( a: )-~-Qkl8qk k dt aqk aqk

=

0

Since, the system is holonomic, all 8q k 's are independent and arbitrary. So it follow from the above equation,

d(aT) aT

dt aq k - aq k - Qk = 0 for all k. This equation is called Lagranges equation of 1st kind for the holonomic system. Now if the system be conservative, then there exist a potential energy V = V(q) (j = 1, ... , d) and

(k = 1, ... , d)

The Classical Mechanics

160 :. We have from above equation,

d(OT) oT

dt oq k - oq k

OV =

- oq k

oV Since, V is independent of qj, oq. = 0 _

J

We now get,

~(OT dt oqk

d

=>

_oqk oV )-~(T- V) oqj

(OL) - oqk oL

dt oqk

= 0

= 0

This is Lagrange's equation of 2nd kind for a conservative holonomic system and here the term L T - V is called 'Lagrangian' of the system Here,

L(qk' qk' t) (in general).

L

8.7 Lagrange's equation for a conservative, non-bolonomic system Basically, for nonholonomic system, dqj's are n()t independent and they are related by 'm' nonintegrable relations. (m < d), such that

~)aikdqk + aikdt)

=

0,

(l = 1,2, .... m)

k

where, afk = afk(qk' t). Now these above equations are the equations of the constraints for a non holonomic system. Hence, for time independent virtual displacements C5q k' we have from above equation. . .. (1) Now from Hamilton's integral principle (which will be developed latter on) we have for a conservative system

... (2) We now have, by multiplying equation (1) by some undetermined multipliers

Variational Principle and Lagrangian Mechanics

161

(Lagrange's) 1...[ (l = 1, 2, ... , m) and then integrating from tl to t2 and finally adding with equation (2), we get, ... (3)

Since, all the 8q k 's are not independent for non holonomic system, we now assume that some of them, 8q i' 8q 2' ... , 8qd-m are independent and the last m are fixed. We now choose A;U = 1,2, ... , m) such that oL- d (OL) -. + ~ 4...Alalk oqk

dt oqk

=

1=1

0

... (4)

(k = d - m + I, ... , d) With these A/ 's determined by equation (4), we can write equation (3) as,

tr{d-m[ oL d ( OL) m J L ---. + LAlalk ]} 8qkdt t\

dt oqk

k=1 oqk

Therefore, 8a\, 8a 2

...

8a d_m are independent which follows

oL- d(OL) ~ - . + 4...Alqlk oqk

dt oqk

... (5) .

=0

1=1

1=1

= 0

... (6)

(k=I,2, .. d-m) Combining equation (4) with equation (6), we have fmally the complete set of Lagrange's equations for non holonomic system,

~

oL- = 4...Alalk -d (OL) -- -dt oqk

oqk

... (7)

1=1

(k = 1, 2, ... , d)

S.S Introduction to Calculus of variations For the solution of a dynamical Y ,, ~ problem, we usually are interested about the ,, , , position or location of the system or the ,, ,,' ~ particle at a particular instant of time. We ,,are also interested about the path adoped by ,, the system and its information, wheather it is maximum or minimum or extremum. This whole study can be made easily by the technique of the 'calculus of variation'. ol----------x--

,

162

The Classical Mechanics

Here, consider the motion of the particle from the pointds ,I I I (xi' Y2) to the point-II (x 2, Y2) along the path' A'. This I I path may be the shortest path 'Ao' which is a straight line :, dy path connecting this two said positions of the particle. I I We can find this shortest straight line path for motion _____________ III of the particle by the method of calculus of variation in the dx following way If ds be the length of infinitesimal portion of the path of particle then total length of the path

~d)2

fds=

f~(cbl+(dy)2 = fV1+li)

I

I

2

S

2

x2

dx

Xl

X2

J

f(y,y', x) dx

S

fly, y', x) = ~1 + y,2

where,

... (1)

( where, y' = : )

The equation (1) is now called the action integral in the calculus of variation and here, since, 'S' is minimum or shortest, we must have, in general, oS 0 X2 ... (2) f(y,y', x)dx = 0 or,

oJ Xl

This is the formulation of the problem of calculus of variations. Here we should note that the o-variation is defined as the variation in the quantity to which it is applied at the fixed value of the independent parameter i.e.

Oy = Y2 - y\ = (y)X=X2 -(y)X=Xl' Here since we consider the path for motion of particle, the shortest path is taken as the extremum path and the other path with which we compare it (in our present discussion of finding the extremum path) are called comparison path. Now, But 0

o [f(y .

"

y' x)] = of oy+ of oy' + of ox By By' ax

~ (o:)oa, where, ais some parameter.

Since x is not a function of a

Ox

=

ax -oa=O oa

(for,

:= 0) =

Variational Principle and Lagrangian Mechanics

X2

163

XJ2[Of oy + ~f Oy,ldx oy OY' Xl ...

oJf(y,y',x)dx

X2[of 7.f d ] dx -oy+~-(Sy) oy oy'dx

J

Xl

XJ2[8f d(Of) d(Of -oy+--oy - --oy)] dx oy dx By' dx oy' Xl

XJ2[Of d (Of)] d (Of By - dx By' oydx + XJ2 dx By,oy ) dx Xl

=

Xl

!

X2[Of d (Of)] [Of ]X2 By - dx By' oydx+ By,oy Xl

But there is no o-variation of y at x = x\ and also at x = x2. i.e.,

(OY)X=XI = (OY)X=X2 = 0 X2

So we get, 0

Jf(y,

y', x) dx

... (3)

Xl

So fmally, we have from equation (2)

XJ2[of _~( Of)]OYdx=O. By dx oy' Xl

But oy represents some arbitrary variation of y(x) with respect to the arbitrary parameter a. about its extremum value (a. = 0). Since oy is arbitrary, We get,

of = 0 oy

of = 0 By

These are the conditions that should be satisfied by the functionj(y, the integral'S' of equation (1) is to be extremum.*

*

'" (4)

y1 if

Here we should note that if this calculus of variation be applied in formulating Lagrange's equation, the path of a particle in configuration space of Lagrangian formulation is uncertain and consequently a discussion of extremum path is an important consideration for the system in motion. Also, Readers would note that Lagrange's equation for conservative system (holonomic) bear a great similarity with equation (4) which will be derived from this.

164

The Classical Mechanics

Now in our case of extremum path,j{y, y', x) = ~1 + y,2 . of =0 of = oy 'oy'

y'

~1+y'2

J

y' So from equation (4), -d ( r:--:? = 0 dx V1+ y,2 y'

Const.

=

A (Say).

y' = c.

y = cx+ b. (b = another const.) This is an equation representing a straight line. So the shortest or extremum path must be a straight line path which is usually obtained from the technique of this calculus of variation. In another example, we can also apply this ~=-----------y method of the calculus of variation to the 'brachistocrone problem'. In this problem, we find a curve joining two points along which a particle falling from rest under the influence of gravity travels from higher to the lower point in the minimum time. Here, the time spent in traversing ds portion 2 of the curve with speed v along the curve is x 2

112 =

ds

f-;· I

But for vertical distance to fall x upto the point-2 we have from energy conservation.

mgx :. v = ~2gx

Variational Principle and Lagrangian Mechanics

165

2

f

f(y,y' x)dx

Where, in this case,j(y, y' x)

~(aj)dx

oy'

aj

of

But

ay

= 0,

=

'" (6)

R

~2gx

ajay ° y'

ay' =- ~2gx ~1 + y,2

So, we get, -d [ y, dx ~2gx ~1 + y,2

x( 1+ y' 2)

J ° =

= Constant = C (Say).

y'2(~ -x) =x=>y'2(~_x2) y' =

On integration, y

=

=x2.

x

a cos -I (1 - xl a) - ( 2ax - x 2 )

1/2

1

+ ci for 2a = ~

Here, c 1 is the new constant of integration. Obviously, if c 1 be zero, then y will be zero for x to be zero. 1

In such case, Y = a cos -I

(1 - ~)-(2ax - x )"z 2

Which represents an inverted cycloid with its base along y-axis and cusp at the origin. This is an well known result for brachistochrone problem which is not obtained from the calculus of variation. 8.9 Variational Technique for many independent variables: EulerLagrange's differential equation: Let us consider that the function 'f is a function of many independent variables y/x) i.e. f= j(yj' yj', x). (j = 1, 2, ... , d) In that case from the calculus of variation.

166

The Classical Mechanics 2

8 ff(Yj,Yj,x)dx = 0 )

Now taking, 8 4

~8u OU

for an arbitrary parameter u, we get,

fI{-Of au + of o~ }8U dx = 0 00. Oy", 00. oY} 1 },

JI{ of oYj ) j oYj

+

OU

ox 8u = 0 as ox ( for 8x = OU 00.

=>

I{J, oY)'of oYj )

)

OU

,

But at end pomts,

o~ OYj} 8u dx = 0

oYj

= 0 because, x is independent of

80. dx

)

U)

+[ oY)'o~ OYj]2 - J~( o~ 1OYj} dt oY)' 00.

)

OU

)

80. dx

=0

(Oyj) (Oyj) =0 OU 2 = OU )

:. I{J[ Oy)Of, _.!!..-Ifdx oY)o~.ll(OYj)8o..dx} j

OU

00.

= 0

But 8u.dx is arbitrary and independent. So, we must have each coefficient of the previous equation zero. i.e.,

of _.!!..-( Of Oyj dx Oyj

1

= 0 for all}.

This mathematical equation is known, as 'Euler-Lagrange's equation' for all independent variables Yj 'so

8.10 Hamilton's Variational Principle: Following the method of the calculus of variation, Hamilton stated that the motion of a dynamical system from time t) to time t2 will be such that

167

Variational Principle and Lagrangian Mechanics ~

~

I = f L dt will be extremum for the path of the motion. i. e., of L dt = 0 . This is Hamilton's variational principle. where,

L=L(qj,qj)

=

T(qj,qj)-V(qj)

is Lagrangian for that conservative system. To study with this principle, we should at first have some knowledge about the motion of the system. For motion of a single particle we should require 3-independent co-ordinate. Similarly, for two body system 6 independent coordinates are required and in that case, that two body system can be replaced by a single particle in 6 dimensional space. Similarly, any system ofN particles can be replaced by the motion of a single particle in 3N dimensional space and that particle (single) is called 'system point' and the required 3N dimensional space is called 'configuration space'. So the motion ofa system in realistic space can be replaced by the motion of a system point in configuration space.

8.11 Derivation of Hamilton's principle from Lagrange's equation d

From Lagrange's equation, we have dt where, L = conservative.

L(qj' qj)

(aL) aL aqj aqj

= Lagrangian of the system which is holonomic,

t2

Now,

ofLdt t1

where ,

(oq j ) t2 =

(oq j )

t1

=0

for no o-variation of qj at the end points. t2

of Ldt = o·

=

168

The Classical Mechanics

This is Hamilton's principle which follows Lagrange's equation for conservative, holonomic system. 8.12 Derivation of Lagrange's equations from Hamilton's principle: t2

From Hamilton's principle, of Ldt =

o.

tl

and ~ = 0) ( Since, 0 ~ ~oa aa aa

",[t --oa 1 -oa 1 o. 2

~

L..J f aL aqj . aqj aa }

~

dt + -aL. aqj aqj aa

dt

=

~

qj ",[t aL aqj aL 1 d ( aL) aqj ] L..J f --oadt+-. -d (a - ) oadt- t J-dt -aqj. -Badt . aqj aa aqj dt aa aa 2

~

}

But,

tl

tl

aq}. oa ]

[ aa

tl

So we finally get,

tl

Variational Principle and Lagrangian Mechanics But since, &ij is arbitrary, each [for,

~q

j

= ';:: ~a]

169

coefficient of above

equation must be zero. So in that case,

aL

8qj

or,

d [aL) dt 8qj

!!.-( aL ) _ aL dt 8qj

=

0

=

0

aqj

This is Lagrange's equation of 2nd kind: Also for conservative system. L = L{qj,qj) = T(qj,qj) - v(qj)

h T = k·· -aL = -aT were, me tIc energy 8qj 8qj

so,

:. ~(8:) _~(T ~ ~(o:) dt 8qj

aqj

dt aqj

aT aqj

- V)

=-

=

o.

av = Q. = ~ This is Lagrange's equation of 1st

aqj

J

kind. 8.13 Derivation of Lagrange's equation from D' Alambert's principle: We have from D'Alambert's principle,

l:[F, +(-mi~)].8F;

=

0

i

But for,

... (1)

170

The Classical Mechanics

where,

Q.

=

J

,,- aF

~ Fi' aqjI ~ Generalised force. I

But

"1 ~ i

. af::) mi~a. . a (.)} m;i1'a-~ ~ q} q}

-

d( dt

. a~=af; -) [ smce , aqj aqj

,,[!!...~(~ m.F })] 7 dt aq j 2 m.F';) - ~(~ aq j 2 I I

I

I I

I

!!...~(L ~m/i2)-~(L~m/i2) dt aq j 2 aq j 2 i

i

But for whole system, total kinetic energy T =

,,1

.2 ~2m,r,. i

draT) aT

---dt aqj

aqj

We thus have from equation (1) and (2) as,

=>

L[l!!...( ~ )- aT} -Qj]8 j .

}

dt aqj

q

aqj

= 0

Since, all 8qj's are arbitrary, we have,

av av

Q. = - - , - . =0 J

aqj

.. !!...~(T -V) -~(T -V) dt aqj

or,

!!.. ( a~ ) dt aqj

aqj

=

0

aqj

aL

aqj

= O. (for L =

T-

V)

... (2)

Variational Principle and Lagrangian Mechanics

171

This is Lagrange's equation of 2nd kind for the conservative holonomic system.

8.14 Derivation of Hamilton's Principle from D' Alambert's Principle From D' Alambert's principle, we have,

L(Fi-mi~).Or; =~ i

when,

L(Fi-Pi).Or; i

Pi

=

~

mj~.

LPi.Or;

... (l)

But

d( dr; s:-) -m·dr; s:(dr;) dt m·dt· dt· dt

-

uf,·

1

u -

1

1

!!"(m. dt dr;dt .0;:;) _o["!'m.(dr; 2 dt ).(dr;)] dt I

1

I

d;:; -) -0 (1 2) -dtd ( m·-'.od· -m·v· dt 2 1

I

1

d;:;) (forv. dt =_1

I

1

!!..["(m. ..!.m.v.2 ) dt ~ dr;dt .0;:;)]-0(" ~2 I

I

I

I

1

I

So from equation (1)

:,[~(m, ~.or;)] But

miv/ Li i

... (2) = T = Total kinetic energy of the system,

and, So from equation (2)

:. :, [~( m, ~ .or, )] ~ o(T - V) ~ oL (where, L

=

T - V)

172

The Classical Mechanics

[

dr; "m.-.or,. ~ 'dt

12

']

,

o

II

(for

or;' s

zero at the end points)

(0 L)dt + U(dt) = (oL)dt for o(dt) = 0 This is Hamilton's principle

where,

o(L dt)

=

8.15 Cyclic or Ignorable Co-ordinates Among all generalised co-ordinates qj (j = I, 2, ... , d), if q k 's are some generalised coordinates (k = I, 2, ... p, p < d) such that the Lagrangian of the system (L) does not depend on these co-ordinates, then all these generalised co-ordinates qk 's are called cyclic or ignorable co-ordinates. So, for qk's are to be cyclic, L:;:. L (qk)' oL

:. -;-- = 0 for all such

uqk

k = 1,2, ,,, , p.

Now we have from Lagrange's equation. d(OL) oL dt oq k - oq k = O. oL d(OL) So for oqk = 0, dt oqk = 0 oL or -;:uqk

OT = -;:- =

uqk

constant (since for conservative system, V :;:.V(qk)'

Pk

=

Constant

So we see that generalised momentum corresponding to cyclic co-ordinate will be conserved.

8.16 Conservation Theorems: (a) Conservation of Linear momentum in Lagrangian formulation :

For linear motion (uniform) of any dynamic system, K.E = T = is independent of some generalised co-ordinates, say qj OT .. -=0. oqj

L ~ m, i

Vi

2

173

Variational Principle and Lagrangian Mechanics

Also, we have the generalised force,

8V ,,- 8i} Q. = - - = ~Fi'8qj

J

i

8qj

where, .~ represents the component of the total force along the direction of translatIOn of q .. Thus, if th6 component of the total applied force along a given direction vanishes, then ~ = O.

8V

oand qj will be also cyclic in V. 8

-(T-V) =0 8qj

and

d( 8LJ

dt 8qj

=

d(8TJ -d -8' t qj

=

d -(Pj) dt

=

O.

.. Pi = Constant. i.e., in that case, the corresponding component of the linear momentum will be conserved. This is linear momentum conservation. (b) Conservation of Angular momentum in Lagrangian formulation: N ow consider that dq. represents a rotation of the system along the 1\ generalised co-ordinate q. J J n then Idi}1 = ri sin e dqj dr; dqj and

8i} 8qj

ri sin e

n xri

Also, since q. cannot appear in total kinetic energy T (because trans(ational velocities are not affected by rotation)

Also the generalised force (which has not always the dimension of force)

174

The Classical Mechanics

"n ~".. (r. x F.) '= n A

;

(L-

F-)

r,·x· I I

•

;

where

i

total torque =

Li

j

Also, the generalised momentum '

P}.=

OT=~( ~'. ~,;. L2..!..mv2) VLfJ j

uq}

j

j

(

Where,

ot.' = ot.' =_, or. ) Because 'oqj

oqj

oqj

L = I L; = Total angular momentum.

If now the effective torque along ii for rotation be zero then

Q.)

oV

=

0- - oqj

So qj is also cyclic in V.

oL

:,[ ~)= :,(Pj)=O

..

oqj

..

Pj = const. ii.L = const.

=> This gives conservation of angular momentum, by which, when component of torque along ii vanishes, the component of angular momentum along ii will be conserved.

Variational Principle and Lagrangian Mechanics

175

(C) Conservation of Energy in Lagrangian Formulation

From Lagrange's equation, we have,

! (:~ ) :~ =

.

Now if we take that the constraints are all time independent then the Lagrangian will not be the function of time explicity.

L

L(qj,qj)

dL dt

and

dL dt

7

"'d[.8L) dt qj 8qj

~ ~(LIqj 8~) =0 dt j 8qj But in that case, the total kinetic energy T will be a homogeneous quadratic function of

qj •

Thus

2T

..

2T

..

d -(L-2T) dt

I'qj-;;8T j qj I'qj-;;8L j

(By Euler's theorem)

qj

(for V

= V(q) only)

0

d -(2T- L) 0 dt .. 2T- L Constant. .. 2T - (T - V) Constant T+V E = Total energy = Constant. .. This is conservation of energy. ~

8.17 Gauge Function for Lagrangian From the calculus of variation, we know that Euler's Lagrangian is given by

equatio~

176

The Classical Mechanics

:,(~)-:~ ~O. This equation does not change its form if we add the time derivative of any arbitrary function F(qj' t) with the Lagrangian L{qj,qj' t) of the system. This function F( q j' t) is then called Gauge function. Because, let the Lagrangian will now become,

But,

aq; aq; aqj = 0, aqj = By = I for i = j only

and

aF aqj = 0 for F

* F(q;)

Thus taking i = j, we get

:'[a:):)]-a:):) ~ :,(:;)- ~;qr a:):)

Variational Principle and Lagrangian Mechanics

a Fq. +~(aF)_ a Fq. _~(aFI aq; ) aqj at aq; } aqj at) 2

2

=

177

=

0

So we get from equation (1)

d [aLI) au aqj aqj

dt

d =

dt

(aL) aL aqj aqj

Thus Lagrange's equation will remain invarient even if

dF

dt

added with Lagrangian of the system. This arbitary function

for

F(q"

F(q"

t) be

t) is called

Gauge function for the Lagrangian L( q}, q}, t)

8.18 Invarience of Lagrange's equations under Generalised Co-ordinate transformations For generalised co-ordinate transformation, let co-ordinates and the corresponding Lagrangian as

qj and

Qj are two sets of

L(q}, qj,t), L' = L/(Q},Oj' t) qj = qi~, t), Qj = ~(qj' t) aq}. aq} aQ j Qj+Tt L

where

=

where

and

Basically, such transformations are called point transformation for which we usually consider

aq} aQ}

= 1

Now, to show invarience of the form of Lagrange's equation we have, d

dt

(au)

aL'

ao} - aQ}

=

178

The Classical Mechanics

=

d ( 8L) 8L]( 8qj ) [ dt 8qj - 8q} 8Q j

= -d (8L -

dt 8qj

1

8qi - -8L for - - 1 for point transformation. 8qj

8Q j -

Thus the form of Lagrange's equation remains unchanged under generalised co-ordinate transformation. 8.19 Concept of Symmetry: Homogeneity and Isotropy If a function does not change its property under some operation, then that function is said to be symmetric about that operation. As for example, if a cylinder rotates along the surface about its axis, the cylinder's apparent shape will not change and in that case, the cylinder is said to have rotational symmetry about its axis. In our case, the Lagragian of the system may have similar kind of symmetry under some operations, which are, (i) Homogeneity of time : For such symmetry operation, the Lagrangian of the close system will not be function of time explicity and in that case, 8L

at = 0

and there is no external force.

From our earlier discussion, this leads to the conservation of energy i.e. E = T + V = Constant. (ii) Homogeneity of Space: For this symmetry, Lagrangian of a closed system, should not change due to any arbitrary small uniform translation of all particles. As for closed system, external force is zero, we have generalised force ~ =

O. Thus for such symmetry 8L = 0 and we have from our previous discussion oqj

the total linear momentum of that closed system will be conserved due to this homogeneity of space. (iii) Isotropy of Space: If for any arbitrary rotation about the origin of any reference frame, the physical property of any closed system remains unaffected, we say that space is isotropic.

Variational Principle and Lagrangian Mechanics For such rotation about

8F;

8F;

8qj

89

iz,

179

we have from our earlier discussion,

A_

--=-=nxr: I

and for such isotropy, we come to conclusion that if rotation co-ordinate is cyclic, the angular momentum is conserved.

8.20 Invarience of Lagrange's equation under Galilean Transformation For two inertial frames Sand S', when S' is moving with velocity Vo w.r.t. S frame, we have from Galilean transformation, F'

v'

=

r - vot

=

v- Vo

Now in S frame, the Lagrangian of the system 1

2'

L = -mv -V 2

1 ,2 - V' L' = -mv 2 But since, the potential energy V only depends on interparticle separation

and in S' frame,

IF; - F21 ' we must have,

V(li) -F2 1)

=

V'(Ii)'-Fil)

i.e., the potential energy V will remain unchanged under Galilean transformation. L But

L'

1 (_

_) (_

_)

- m v - Vo . v - Vo - V

2

1 2 +-mvo 1 2 --mv.vo-V -mv

2

1

2

2

(1

2

__

I

-mv - V + -mvo - mv.vo I 2 2 )

=> were, h F

d( - -

2)

L'

1 L+- -mvo.r +-mvot

L'

L+

dt

2

dF (F,t) dt

- ) = - mvo.r - - + -mvot 1 2 ·1S th e G auge lunctlon c. . h = F(r,t an d we h ave sown 2

180

The Classical Mechanics

in our earlier discussion that for additional term

dF

with the original dt Lagrangian, the Lagrange's equation will remain also unchanged in S' frame. So Lagrange's equation will remain unchanged under Galilean transformation.

8.21 Application's of Lagrange's equation of motion in several mechanical systems (i) Linear Harmonic Oscillator : For linear harmonic oscillator in one dimension,

m

the kinetic energy of the system is T =

.!.'nX 2 2

and the potential energy x

V =

-fFdx

=

Hence, the Lagrangian of the system

L

So, here,

=

T- V=

-f(-/a;)dx=-l/a;2 o

.!.lni 2 2

.!./a;2

2

aL _ Ini aL - _ /a; ax - 'ax-

we now have from Lagrange's equation of motion, d dt

(aL) aL ax ax x + w5x

O~mX+/a;=O

= 0 for Wo =

g

= Angular frequency

This is equation of motion for one dimensional linear harmonic oscillator. (ii) Simple Pendulum 0 At angular position of the pendulum e(L4°) the kinetic energy

2 ='!'m(/8)2 T = .!.mv 2 2

=.!.m1 282 2

.

Also, taking horizontal reference plane through the equilibrium position of the bob, the potential energy V = mg(1 - I cos e) = mg/(1 - cos e) Thus, Lagrangian of this pendulum is given by, L = T- V=

.!.mP8 2 - mg/(1- cose) 2

m

Variational Principle and Lagrangian Mechanics

aL

so that,

ae

ml2e,

aL as

181

= -mgl sin S

Putting these in Lagrange's equation, d

~

(aL) aL

ae - as

o

ml29 + mglsinS

0

dt

But for small S (L4°), sin 8 ::: 8 9+(7)S

=

o~

9+w58

=0

This is equation of motion of a simple pendulum where, Wo =

Jf

is the

circular or angular frequency of the pendulum with time period of oscillation

T

=

21t Wo

=

21t

fI.

Vg

(iii) Spherical Pendulum : For such pendulum, bob is constrained to move on a sphere rather than in a circle. The position of bob is then located by independent spherical co-ordinates (8, cp), such that for length of pendulum r, the kinetic energy of the bob is

T=~mr2(e2 +sin2S~2) Also the potential energy of the bob due to gravity relative to the horizontal plane, = mgr cos 8. The Lagrangian for this system is thus given by L So here,

T- V

=

~mr2(e2 + sin 2 Scj>2) -

aL

ae

aL as

aL acj>

mr2 sinS cosScj>2 + mgrsinS

aL

.2S·cp, -acp = 0 mr 2 sm

:. cp is cyclic Here, Lagrange's equations can be written as,

mgrcosS.

182

The Classical Mechanics d (aL) aL dt a8 - as

o

... (1)

:t(~~)

o

'" (2)

From equation (2),

~ (mr2 sin 2 S<j» or, mr2

:t

0

2

(sin S<j»

0

'" (3)

Also from equation (l) mr 2e-mr 2 sin9cosS<j>2 -mgrsine = 0 '" (4) These equations (3) and (4) are the equations of motion of a spherical pendulum. (iv) Compound Pendulum: When a rigid body is capable for oscillating in a vertical plane about a fixed horizontal axis then that body is called a compound pendulum. For such pendulum of mass m and moment of inertia I about the axis of rotation, if I be the distance of its center of mass from the point of suspension, then for angle of deflection e of that body the kinetic energy T = -.!. 18 2 . 2

Also the potential energy relative to a horizontal plane through the point of suspension, is v = - mgt cos S So the Lagrangian of the system, L

T- V

aL

a8

.

d (aL) aL dt a8 - as

o

Ie + mglsinS

o

2

Ie + mgl cos e

aL

1e; as

so from Lagrange's equation,

1 .2

= -

= -

mgl sin S

Variational Principle and Lagrangian Mechanics

183

8+( mt)e e+W2 S

0

This is equation of motion for compound pendulum with time period

T = 27t = w

27t~

I

mgl'

(v) Isotropic Oscillator in three dimension: For such oscillator, the vibration is acted upon by a force directed always towards or away from the position of equilibrium and the magnitude of such force varies linearly with the distance from the position of equilibrium. This force is given by F = - kr when k is stiffness constant.

1

Jo dr ="2 r

Now the potential energy, V = -

F

2

kr .

Taking polar co-ordinates, the kinetic energy is then given by T=

k

m(f2 + r 2e 2 + r2 sin 2 Sej>2)

we have, Lagrangian of the system 2 2 L = T - V = -km(r2 +r 2e +r2 sin eej>2) - kkr 2 . Here, we have,

aL af

mr

aL ar

m,{)2 + mrsin 2 S<j>2 - kr

aL ae cL as

mr

2.

e

mr2 sin 2 S cosSej>2

aL

mr2 sin 2 Sej>

aej> aL

a
=

0 where,