CHAPTER SEVEN

SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

7.1 INTRODUCTION

In the earlier chapters we primarily considered the motion

of a single particle. (A particle is ideally represented as a

point mass having no size.) We applied the results of our

study even to the motion of bodies of finite size, assuming

that motion of such bodies can be described in terms of the

motion of a particle.

Any real body which we encounter in daily life has a

finite size. In dealing with the motion of extended bodies

(bodies of finite size) often the idealised model of a particle is

inadequate. In this chapter we shall try to go beyond this

inadequacy. We shall attempt to build an understanding of

the motion of extended bodies. An extended body, in the

first place, is a system of particles. We shall begin with the

consideration of motion of the system as a whole. The centre

of mass of a system of particles will be a key concept here.

We shall discuss the motion of the centre of mass of a system

of particles and usefulness of this concept in understanding

the motion of extended bodies.

A large class of problems with extended bodies can be

solved by considering them to be rigid bodies. Ideally a

rigid body is a body with a perfectly definite and

unchanging shape. The distances between all pairs of

particles of such a body do not change. It is evident from

this definition of a rigid body that no real body is truly rigid,

since real bodies deform under the influence of forces. But in

many situations the deformations are negligible. In a number

of situations involving bodies such as wheels, tops, steel

beams, molecules and planets on the other hand, we can ignore

that they warp (twist out of shape), bend or vibrate and treat

them as rigid.

7.1.1 What kind of motion can a rigid body have?

Let us try to explore this question by taking some examples

of the motion of rigid bodies. Let us begin with a rectangular

7.1 Introduction

7.2 Centre of mass

7.3 Motion of centre of mass

7.4 Linear momentum of a

system of particles

7.5 Vector product of two

vectors

7.6 Angular velocity and its

relation with linear velocity

7.7 Torque and angular

momentum

7.8 Equilibrium of a rigid body

7.9 Moment of inertia

7.10 Theorems of perpendicular

and parallel axes

7.11 Kinematics of rotational

motion about a fixed axis

7.12 Dynamics of rotational

motion about a fixed axis

7.13 Angular momentum in case

of rotation about a fixed axis

7.14 Rolling motion

Summary

Points to Ponder

Exercises

Additional exercises

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142 PHYSICS

block sliding down an inclined plane without any

sidewise movement. The block is taken as a rigid

body. Its motion down the plane is such that all

the particles of the body are moving together,

i.e. they have the same velocity at any instant

of time. The rigid body here is in pure

translational motion (Fig. 7.1).

In pure translational motion at any

instant of time, all particles of the body have

the same velocity.

Consider now the rolling motion of a solid

metallic or wooden cylinder down the same

inclined plane (Fig. 7.2). The rigid body in this

problem, namely the cylinder, shifts from the

top to the bottom of the inclined plane, and thus,

seems to have translational motion. But as Fig.

7.2 shows, all its particles are not moving with

the same velocity at any instant. The body,

therefore, is not in pure translational motion.

Its motion is translational plus ‘something else.’

In order to understand what this ‘something

else’ is, let us take a rigid body so constrained

that it cannot have translational motion. The

most common way to constrain a rigid body so

that it does not have translational motion is to

fix it along a straight line. The only possible

motion of such a rigid body is rotation. The line

or fixed axis about which the body is rotating is

its axis of rotation. If you look around, you will

come across many examples of rotation about

an axis, a ceiling fan, a potter’s wheel, a giant

wheel in a fair, a merry-go-round and so on (Fig

7.3(a) and (b)).

(a)

(b)

Fig. 7.3 Rotation about a fixed axis

(a) A ceiling fan

(b) A potter’s wheel.

Let us try to understand what rotation is,

what characterises rotation. You may notice that

in rotation of a rigid body about a fixed axis,

Fig 7.1 Translational (sliding) motion of a block down

an inclined plane.

(Any point like P

or P

of the block moves

with the same velocity at any instant of time.)

Fig. 7.2 Rolling motion of a cylinder. It is not pure

translational motion. Points P

, P

and P

have different velocities (shown by arrows)

at any instant of time. In fact, the velocity of

the point of contact P

is zero at any instant,

if the cylinder rolls without slipping.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 143

every particle of the body moves in a circle,

which lies in a plane perpendicular to the axis

and has its centre on the axis. Fig. 7.4 shows

the rotational motion of a rigid body about a fixed

axis (the z-axis of the frame of reference). Let P

be a particle of the rigid body, arbitrarily chosen

and at a distance r

from fixed axis. The particle

describes a circle of radius r

with its centre

on the fixed axis. The circle lies in a plane

perpendicular to the axis. The figure also shows

another particle P

of the rigid body, P

is at a

distance r

from the fixed axis. The particle P

moves in a circle of radius r

and with centre C

on the axis. This circle, too, lies in a plane

perpendicular to the axis. Note that the circles

described by P

and P

may lie in different planes;

both these planes, however, are perpendicular

to the fixed axis. For any particle on the axis

like P

, r = 0. Any such particle remains

stationary while the body rotates. This is

expected since the axis of rotation is fixed.

Fig. 7.5 (a) A spinning top

(The point of contact of the top with the

ground, its tip O, is fixed.)

Fig. 7.5 (b) An oscillating table fan with rotating

blades. The pivot of the fan, point O, is

fixed. The blades of the fan are under

rotational motion, whereas, the axis of

rotation of the fan blades is oscillating.

Fig. 7.4 A rigid body rotation about the z-axis (Each

point of the body such as P

describes a circle with its centre (C

or C

) on the axis of rotation. The radius of

the circle (r

or r

) is the perpendicular

distance of the point (P

or P

) from the

axis. A point on the axis like P

remains

stationary).

Axis of oscillation

Axis of

rotation

from blades

In some examples of rotation, however, the

axis may not be fixed. A prominent example of

this kind of rotation is a top spinning in place

[Fig. 7.5(a)]. (We assume that the top does not

slip from place to place and so does not have

translational motion.) We know from experience

that the axis of such a spinning top moves

around the vertical through its point of contact

with the ground, sweeping out a cone as shown

in Fig. 7.5(a). (This movement of the axis of the

top around the vertical is termed precession.)

Note, the point of contact of the top with

ground is fixed. The axis of rotation of the top

at any instant passes through the point of

contact. Another simple example of this kind of

rotation is the oscillating table fan or a pedestal

fan [Fig.7.5(b)]. You may have observed that the

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144 PHYSICS

axis of rotation of such a fan has an oscillating

(sidewise) movement in a horizontal plane about

the vertical through the point at which the axis

is pivoted (point O in Fig. 7.5(b)).

While the fan rotates and its axis moves

sidewise, this point is fixed. Thus, in more

general cases of rotation, such as the rotation

of a top or a pedestal fan, one point and not

one line, of the rigid body is fixed. In this case

the axis is not fixed, though it always passes

through the fixed point. In our study, however,

we mostly deal with the simpler and special case

of rotation in which one line (i.e. the axis) is fixed.

Thus, for us rotation will be about a fixed axis

only unless stated otherwise.

The rolling motion of a cylinder down an

inclined plane is a combination of rotation about

a fixed axis and translation. Thus, the

‘something else’ in the case of rolling motion

which we referred to earlier is rotational motion.

You will find Fig. 7.6(a) and (b) instructive from

this point of view. Both these figures show

motion of the same body along identical

translational trajectory. In one case, Fig. 7.6(a),

the motion is a pure translation; in the other

case [Fig. 7.6(b)] it is a combination of

translation and rotation. (You may try to

reproduce the two types of motion shown, using

a rigid object like a heavy book.)

We now recapitulate the most important

observations of the present section: The motion

of a rigid body which is not pivoted or fixed in

some way is either a pure translation or a

combination of translation and rotation. The

motion of a rigid body which is pivoted or fixed

in some way is rotation. The rotation may be

about an axis that is fixed (e.g. a ceiling fan) or

moving (e.g. an oscillating table fan [Fig.7.5(b)]).

We shall, in the present chapter, consider

rotational motion about a fixed axis only.

7.2 CENTRE OF MASS

We shall first see what the centre of mass of a

system of particles is and then discuss its

significance. For simplicity we shall start with

a two particle system. We shall take the line

joining the two particles to be the x- axis.

Fig. 7.7

Let the distances of the two particles be x

and x

respectively from some origin O. Let m

and m

be respectively the masses of the two

Fig. 7.6(a) Motion of a rigid body which is pure

translation.

Fig. 7.6(b) Motion of a rigid body which is a

combination of translation and

rotation.

Fig 7.6 (a) and 7.6 (b) illustrate different motions of

the same body. Note P is an arbitrary point of the

body; O is the centre of mass of the body, which is

defined in the next section. Suffice to say here that

the trajectories of O are the translational trajectories

and Tr

of the body. The positions O and P at

three different instants of time are shown by O

, O

and O

, and P

, P

and P

, respectively, in both

Figs. 7.6 (a) and (b) . As seen from Fig. 7.6(a), at any

instant the velocities of any particles like O and P of

the body are the same in pure translation. Notice, in

this case the orientation of OP, i.e. the angle OP makes

with a fixed direction, say the horizontal, remains

the same, i.e.

. Fig. 7.6 (b) illustrates a

case of combination of translation and rotation. In

this case, at any instants the velocities of O and P

differ. Also,

and

may all be different.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 145

particles. The centre of mass of the system is

that point C which is at a distance X from O,

where X is given by

1 1 2 2

1 2

m x m x

m m

(7.1)

In Eq. (7.1), X can be regarded as the mass-

weighted mean of x

and x

. If the two particles

have the same mass m

= m

then

1 2 1 2

2 2

mx mx x x

+ +

= =

Thus, for two particles of equal mass the

centre of mass lies exactly midway between

them.

If we have n particles of masses m

, m

...m

respectively, along a straight line taken as

the x- axis, then by definition the position of the

centre of the mass of the system of particles is

given by.

m x m x m x

m m m

m x

n n

i i

= = =

∑

1 1 2 2

1 2

+ + ... +

+ +... +

(7.2)

where x

, x

,...x

are the distances of the

particles from the origin; X is also measured from

the same origin. The symbol

∑

(the Greek letter

sigma) denotes summation, in this case over n

particles. The sum

m M

∑

is the total mass of the system.

Suppose that we have three particles, not

lying in a straight line. We may define x– and y–

axes in the plane in which the particles lie and

represent the positions of the three particles by

coordinates (x

), (x

) and (x

) respectively.

Let the masses of the three particles be m

, m

and m

respectively. The centre of mass C of

the system of the three particles is defined and

located by the coordinates (X, Y) given by

1 1 2 2 3 3

1 2 3

m x m x m x

m m m

+ +

(7.3a)

1 1 2 2 3 3

1 2 3

m y m y m y

m m m

+ +

(7.3b)

For the particles of equal mass m = m

= m

1 2 3 1 2 3

( )

3 3

m x x x x x x

+ + + +

= =

1 2 3 1 2 3

( )

3 3

m y y y y y y

+ + + +

= =

Thus, for three particles of equal mass, the

centre of mass coincides with the centroid of the

triangle formed by the particles.

Results of Eqs. (7.3a) and (7.3b) are

generalised easily to a system of n particles, not

necessarily lying in a plane, but distributed in

space. The centre of mass of such a system is

at (X, Y, Z ), where

i i

m x

∑

(7.4a)

i i

m y

∑

(7.4b)

and

i i

m z

∑

(7.4c)

Here M =

∑

is the total mass of the

system. The index i runs from 1 to n; m

is the

mass of the i

particle and the position of the

particle is given by (x

, y

, z

Eqs. (7.4a), (7.4b) and (7.4c) can be

combined into one equation using the notation

of position vectors. Let

be the position vector

of the i

particle and R be the position vector of

the centre of mass:



i i i i

x y z

= + +

r i j k

 

and



X Y Z

= + +

R i j k

 

Then

i i

∑

(7.4d)

The sum on the right hand side is a vector

sum.

Note the economy of expressions we achieve

by use of vectors. If the origin of the frame of

reference (the coordinate system) is chosen to

be the centre of mass then

i i

∑

for the

given system of particles.

A rigid body, such as a metre stick or a

flywheel, is a system of closely packed particles;

Eqs. (7.4a), (7.4b), (7.4c) and (7.4d) are

therefore, applicable to a rigid body. The number

of particles (atoms or molecules) in such a body

is so large that it is impossible to carry out the

summations over individual particles in these

equations. Since the spacing of the particles is

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146 PHYSICS

small, we can treat the body as a continuous

distribution of mass. We subdivide the body into

n small elements of mass; ∆m

, ∆m

... ∆m

; the

element ∆m

is taken to be located about the

point (x

, y

, z

). The coordinates of the centre of

mass are then approximately given by

( ) ( ) ( )

, ,

i i i i i i

i i i

m x m y m z

X Y Z

m m m

∆ ∆ ∆

= = =

∆ ∆ ∆

∑ ∑ ∑

As we make n bigger and bigger and each

∆m

smaller and smaller, these expressions

become exact. In that case, we denote the sums

over i by integrals. Thus,

∆m m M

→ =

∫

∑

d ,

( ) ,∆m x x m

i i

→

∫

∑

( ) ,∆m y y m

i i

→

∫

∑

and

(∆m z z m

i i

) →

∫

∑

Here M is the total mass of the body. The

coordinates of the centre of mass now are

x m Y

y m Z

z m= = =

∫∫ ∫

1 1 1

d d d, and

(7.5a)

The vector expression equivalent to these

three scalar expressions is

R r=

∫

(7.5b)

If we choose, the centre of mass as the origin

of our coordinate system,

R 0

i.e.,

r 0dm =

∫

x m y m z md d d= = =

∫ ∫∫

(7.6)

Often we have to calculate the centre of mass

of homogeneous bodies of regular shapes like

rings, discs, spheres, rods etc. (By a

homogeneous body we mean a body with

uniformly distributed mass.) By using symmetry

consideration, we can easily show that the

centres of mass of these bodies lie at their

geometric centres.

Let us consider a thin rod, whose width and

breath (in case the cross section of the rod is

rectangular) or radius (in case the cross section

of the rod is cylindrical) is much smaller than

its length. Taking the origin to be at the

geometric centre of the rod and x-axis to be

along the length of the rod, we can say that on

account of reflection symmetry, for every

element dm of the rod at x, there is an element

of the same mass dm located at –x (Fig. 7.8).

The net contribution of every such pair to

the integral and hence the integral

itself

is zero. From Eq. (7.6), the point for which the

integral itself is zero, is the centre of mass.

Thus, the centre of mass of a homogenous thin

rod coincides with its geometric centre. This can

be understood on the basis of reflection symmetry.

The same symmetry argument will apply to

homogeneous rings, discs, spheres, or even

thick rods of circular or rectangular cross

section. For all such bodies you will realise that

for every element dm at a point (x, y, z) one can

always take an element of the same mass at

the point (–x, –y, –z). (In other words, the origin

is a point of reflection symmetry for these

bodies.) As a result, the integrals in Eq. (7.5 a)

all are zero. This means that for all the above

bodies, their centre of mass coincides with their

geometric centre.

Example 7.1 Find the centre of mass of

three particles at the vertices of an

equilateral triangle. The masses of the

particles are 100g, 150g, and 200g

respectively. Each side of the equilateral

triangle is 0.5m long.

Answer

Fig. 7.9

Fig. 7.8 Determining the CM of a thin rod.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 147

With the x–and y–axes chosen as shown in Fig.

7.9, the coordinates of points O, A and B forming

the equilateral triangle are respectively (0,0),

(0.5,0), (0.25,0.25

). Let the masses 100 g,

150g and 200g be located at O, A and B be

respectively. Then,

1 1 2 2 3 3

1 2 3

m x m x m x

m m m

+ +

(

)

100 0 150(0.5) 200(0.25) g m

(100 150 200) g

+ +

75 50 125 5

m m m

450 450 18

= = =

100(0) 150(0) 200(0.25 3) g m

450 g

+ +

50 3 3 1

m m m

450 9

3 3

= = =

The centre of mass C is shown in the figure.

Note that it is not the geometric centre of the

triangle OAB. Why? t

Example 7.2 Find the centre of mass of a

triangular lamina.

Answer The lamina (∆LMN) may be subdivided

into narrow strips each parallel to the base (MN)

as shown in Fig. 7.10

Fig. 7.10

By symmetry each strip has its centre of

mass at its midpoint. If we join the midpoint of

all the strips we get the median LP. The centre

of mass of the triangle as a whole therefore, has

to lie on the median LP. Similarly, we can argue

that it lies on the median MQ and NR. This

means the centre of mass lies on the point of

concurrence of the medians, i.e. on the centroid

G of the triangle. t

Example 7.3 Find the centre of mass of a

uniform L-shaped lamina (a thin flat plate)

with dimensions as shown. The mass of

the lamina is 3 kg.

Answer Choosing the X and Y axes as shown

in Fig. 7.11 we have the coordinates of the

vertices of the L-shaped lamina as given in the

figure. We can think of the

L-shape to consist of 3 squares each of length

1m. The mass of each square is 1kg, since the

lamina is uniform. The centres of mass C

, C

and C

of the squares are, by symmetry, their

geometric centres and have coordinates (1/2,1/2),

(3/2,1/2), (1/2,3/2) respectively. We take the

masses of the squares to be concentrated at

these points. The centre of mass of the whole

L shape (X, Y) is the centre of mass of these

mass points.

Fig. 7.11

Hence

[

]

( )

1(1/2) 1(3/2) 1(1/2) kg m

1 1 1 kg

+ +

[

]

( )

1(1/2) 1(1/2) 1(3/2) kg m

1 1 1 kg 6

+ +

= =

+ +

The centre of mass of the L-shape lies on

the line OD. We could have guessed this without

calculations. Can you tell why? Suppose, the

three squares that make up the L shaped lamina

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148 PHYSICS

of Fig. 7.11 had different masses. How will you

then determine the centre of mass of the lamina?

7.3 MOTION OF CENTRE OF MASS

Equipped with the definition of the centre of

mass, we are now in a position to discuss its

physical importance for a system of n particles.

We may rewrite Eq.(7.4d) as

1 1 2 2

...

i i n n

M m m m m= = + + +

∑

R r r r r

(7.7)

Differentiating the two sides of the equation

with respect to time we get

1 2

d d

...

d d d

M m m m

t t t dt

= + + +

r r

1 1 2 2

...

n n

M m m m

= + + +

V v v v

(7.8)

where

(

)

1 1

d /d

=v r

is the velocity of the first

particle

(

)

2 2

d dt

=v r

is the velocity of the

second particle etc. and

d /d

V R

is the

velocity of the centre of mass. Note that we

assumed the masses m

, m

, ... etc. do not

change in time. We have therefore, treated them

as constants in differentiating the equations

with respect to time.

Differentiating Eq.(7.8) with respect to time,

we obtain

1 2

dd d

...

d d d d

M m m m

t t t t

= + + +

v v

1 1 2 2

...

n n

M m m m= + + +

A a a a

(7.9)

where

(

)

1 1

d /d

=a v

is the acceleration of the

first particle,

(

)

2 2

d /d

=a v

is the acceleration

of the second particle etc. and

(

)

d / d

=A V

the acceleration of the centre of mass of the

system of particles.

Now, from Newton’s second law, the force

acting on the first particle is given by

1 1 1

F a

The force acting on the second particle is given

2 2 2

F a

and so on. Eq. (7.9) may be written

1 2

...

M = + + +

A F F F

(7.10)

Thus, the total mass of a system of particles

times the acceleration of its centre of mass is

the vector sum of all the forces acting on the

system of particles.

Note when we talk of the force

on the first

particle, it is not a single force, but the vector

sum of all the forces on the first particle; likewise

for the second particle etc. Among these forces

on each particle there will be external forces

exerted by bodies outside the system and also

internal forces exerted by the particles on one

another. We know from Newton’s third law that

these internal forces occur in equal and opposite

pairs and in the sum of forces of Eq. (7.10),

their contribution is zero. Only the external

forces contribute to the equation. We can then

rewrite Eq. (7.10) as

ext

M =

A F

(7.11)

where

ext

represents the sum of all external

forces acting on the particles of the system.

Eq. (7.11) states that the centre of mass

of a system of particles moves as if all the

mass of the system was concentrated at the

centre of mass and all the external forces

were applied at that point.

Notice, to determine the motion of the centre

of mass no knowledge of internal forces of the

system of particles is required; for this purpose

we need to know only the external forces.

To obtain Eq. (7.11) we did not need to

specify the nature of the system of particles.

The system may be a collection of particles in

which there may be all kinds of internal

motions, or it may be a rigid body which has

either pure translational motion or a

combination of translational and rotational

motion. Whatever is the system and the motion

of its individual particles, the centre of mass

moves according to Eq. (7.11).

Instead of treating extended bodies as single

particles as we have done in earlier chapters,

we can now treat them as systems of particles.

We can obtain the translational component of

their motion, i.e. the motion of the centre of mass

of the system, by taking the mass of the whole

system to be concentrated at the centre of mass

and all the external forces on the system to be

acting at the centre of mass.

This is the procedure that we followed earlier

in analysing forces on bodies and solving

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 149

problems without explicitly outlining and

justifying the procedure. We now realise that in

earlier studies we assumed, without saying so,

that rotational motion and/or internal motion

of the particles were either absent or negligible.

We no longer need to do this. We have not only

found the justification of the procedure we

followed earlier; but we also have found how to

describe and separate the translational motion

of (1) a rigid body which may be rotating as

well, or (2) a system of particles with all kinds

of internal motion.

Fig. 7.12 The centre of mass of the fragments

of the projectile continues along the

same parabolic path which it would

have followed if there were no

explosion.

Figure 7.12 is a good illustration of Eq.

(7.11). A projectile, following the usual parabolic

trajectory, explodes into fragments midway in

air. The forces leading to the explosion are

internal forces. They contribute nothing to the

motion of the centre of mass. The total external

force, namely, the force of gravity acting on the

body, is the same before and after the explosion.

The centre of mass under the influence of the

external force continues, therefore, along the

same parabolic trajectory as it would have

followed if there were no explosion.

7.4 LINEAR MOMENTUM OF A SYSTEM OF

PARTICLES

Let us recall that the linear momentum of a

particle is defined as

p v

(7.12)

Let us also recall that Newton’s second law

written in symbolic form for a single particle is

(7.13)

where F is the force on the particle. Let us

consider a system of n particles with masses m

,...m

respectively and velocities

1 2

, ,.......

v v v

respectively. The particles may be interacting

and have external forces acting on them. The

linear momentum of the first particle is

1 1

of the second particle is

2 2

and so on.

For the system of n particles, the linear

momentum of the system is defined to be the

vector sum of all individual particles of the

system,

1 2

...

= + + +

P p p p

1 1 2 2

...

n n

m m m= + + +

v v v

(7.14)

Comparing this with Eq. (7.8)

P V

(7.15)

Thus, the total momentum of a system

of particles is equal to the product of the

total mass of the system and the velocity of

its centre of mass. Differentiating Eq. (7.15)

with respect to time,

d d

M M

t t

= =

P V

(7.16)

Comparing Eq.(7.16) and Eq. (7.11),

ext

(7.17)

This is the statement of Newton’s second law

of motion extended to a system of particles.

Suppose now, that the sum of external

forces acting on a system of particles is zero.

Then from Eq.(7.17)

= Constant (7.18a)

Thus, when the total external force acting

on a system of particles is zero, the total linear

momentum of the system is constant. This is

the law of conservation of the total linear

momentum of a system of particles. Because of

Eq. (7.15), this also means that when the

total external force on the system is zero

the velocity of the centre of mass remains

constant. (We assume throughout the

discussion on systems of particles in this

chapter that the total mass of the system

remains constant.)

Note that on account of the internal forces,

i.e. the forces exerted by the particles on one

another, the individual particles may have

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150 PHYSICS

complicated trajectories. Yet, if the total external

force acting on the system is zero, the centre of

mass moves with a constant velocity, i.e., moves

uniformly in a straight line like a free particle.

The vector Eq. (7.18a) is equivalent to three

scalar equations,

= c

, P

= c

and P

= c

(7.18 b)

Here P

, P

and P

are the components of the

total linear momentum vector P along the x–, y–

and z–axes respectively; c

, c

and c

are

constants.

(a) (b)

Fig. 7.13 (a) A heavy nucleus radium (Ra) splits into

a lighter nucleus radon (Rn) and an alpha

particle (nucleus of helium atom). The CM

of the system is in uniform motion.

(b) The same spliting of the heavy nucleus

radium (Ra) with the centre of mass at

rest. The two product particles fly back

to back.

As an example, let us consider the

radioactive decay of a moving unstable particle,

like the nucleus of radium. A radium nucleus

disintegrates into a nucleus of radon and an

alpha particle. The forces leading to the decay

are internal to the system and the external

forces on the system are negligible. So the total

linear momentum of the system is the same

before and after decay. The two particles

produced in the decay, the radon nucleus and

the alpha particle, move in different directions

in such a way that their centre of mass moves

along the same path along which the original

decaying radium nucleus was moving

[Fig. 7.13(a)].

If we observe the decay from the frame of

reference in which the centre of mass is at rest,

the motion of the particles involved in the decay

looks particularly simple; the product particles

move back to back with their centre of mass

remaining at rest as shown in Fig.7.13 (b).

In many problems on the system of

particles, as in the above radioactive decay

problem, it is convenient to work in the centre

of mass frame rather than in the laboratory

frame of reference.

In astronomy, binary (double) stars is a

common occurrence. If there are no external

forces, the centre of mass of a double star

moves like a free particle, as shown in Fig.7.14

(a). The trajectories of the two stars of equal

mass are also shown in the figure; they look

complicated. If we go to the centre of mass

frame, then we find that there the two stars

are moving in a circle, about the centre of

mass, which is at rest. Note that the position

of the stars have to be diametrically opposite

to each other [Fig. 7.14(b)]. Thus in our frame

of reference, the trajectories of the stars are a

combination of (i) uniform motion in a straight

line of the centre of mass and (ii) circular

orbits of the stars about the centre of mass.

As can be seen from the two examples,

separating the motion of different parts of a

system into motion of the centre of mass and

motion about the centre of mass is a very

useful technique that helps in understanding

the motion of the system.

7.5 VECTOR PRODUCT OF TWO VECTORS

We are already familiar with vectors and their

use in physics. In chapter 6 (Work, Energy,

Power) we defined the scalar product of two

vectors. An important physical quantity, work,

is defined as a scalar product of two vector

quantities, force and displacement.

(a) (b)

Fig. 7.14 (a) Trajectories of two stars, S

(dotted

line) and S

(solid line) forming a

binary system with their centre of

mass C in uniform motion.

(b) The same binary system, with the

centre of mass C at rest.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 151

We shall now define another product of two

vectors. This product is a vector. Two important

quantities in the study of rotational motion,

namely, moment of a force and angular

momentum, are defined as vector products.

Definition of Vector Product

A vector product of two vectors a and b is a

vector c such that

(i) magnitude of c = c

sin

where a and b

are magnitudes of a and b and

is the

angle between the two vectors.

(ii) c is perpendicular to the plane containing

a and b.

(iii) if we take a right handed screw with its head

lying in the plane of a and b and the screw

perpendicular to this plane, and if we turn

the head in the direction from a to b, then

the tip of the screw advances in the direction

of c. This right handed screw rule is

illustrated in Fig. 7.15a.

Alternately, if one curls up the fingers of

right hand around a line perpendicular to the

plane of the vectors a and b and if the fingers

are curled up in the direction from a to b, then

the stretched thumb points in the direction of

c, as shown in Fig. 7.15b.

(a) (b)

Fig. 7.15 (a) Rule of the right handed screw for

defining the direction of the vector

product of two vectors.

(b) Rule of the right hand for defining the

direction of the vector product.

A simpler version of the right hand rule is

the following : Open up your right hand palm

and curl the fingers pointing from a to b. Your

stretched thumb points in the direction of c.

It should be remembered that there are two

angles between any two vectors a and b . In

Fig. 7.15 (a) or (b) they correspond to

(as

shown) and (360

–

). While applying either of

the above rules, the rotation should be taken

through the smaller angle (<180

) between a

and b. It is

here.

Because of the cross (×) used to denote the

vector product, it is also referred to as cross product.

• Note that scalar product of two vectors is

commutative as said earlier, a.b = b.a

The vector product, however, is not

commutative, i.e. a × b ≠ b × a

The magnitude of both a × b and b × a is the

same (

sin

); also, both of them are

perpendicular to the plane of a and b. But the

rotation of the right-handed screw in case of

a × b is from a to b, whereas in case of b × a it

is from b to a. This means the two vectors are

in opposite directions. We have

× = − ×

a b b a

• Another interesting property of a vector

product is its behaviour under reflection.

Under reflection (i.e. on taking the plane

mirror image) we have

and ,

x x y y z z

→ − → − → −

. As a result all

the components of a vector change sign and

thus

a a

→ −

b b

→ −

. What happens to

a × b under reflection?

a × b

( ) ( )

→ − × − = ×

a b a b

Thus, a × b does not change sign under

reflection.

• Both scalar and vector products are

distributive with respect to vector addition.

Thus,

.( ) . .

+ = +

a b c a b a c

( )

× + = × + ×

a b c a b a c

• We may write c = a × b in the component

form. For this we first need to obtain some

elementary cross products:

(i) a × a = 0 (0 is a null vector, i.e. a vector

with zero magnitude)

This follows since magnitude of a × a is

sin0 0

° =

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152 PHYSICS

From this follow the results

(i)

ˆ ˆ ˆ ˆ ˆ ˆ

, ,

× = × = × =

i i 0 j j 0 k k 0

(ii)

ˆ ˆ ˆ

× =

i j k

Note that the magnitude of

ˆ ˆ

i j

is sin90

or 1, since

and

both have unit

magnitude and the angle between them is 90

Thus,

ˆ ˆ

i j

is a unit vector. A unit vector

perpendicular to the plane of

and

related to them by the right hand screw rule is

. Hence, the above result. You may verify

similarly,

ˆ ˆ ˆ ˆ ˆ ˆ

and

× = × =

j k i k i j

From the rule for commutation of the cross

product, it follows:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

, ,

× = − × = − × = −

j i k k j i i k j

Note if

ˆ ˆ ˆ

, ,

i j k

occur cyclically in the above

vector product relation, the vector product is

positive. If

ˆ ˆ ˆ

, ,

i j k

do not occur in cyclic order,

the vector product is negative.

Now,

ˆ ˆ ˆ ˆ ˆ ˆ

( ) ( )

x y z x y z

a a a b b b× = + + × + +

a b i j k i j k

ˆ ˆ ˆ ˆ ˆ ˆ

x y x z y x y z z x z y

a b a b a b a b a b a b

= − − + + −

k j k i j i

= + +( )



( )



( )



a b a b a b a b a b a b

y z z y z x x z x y y x

− − −i j k

We have used the elementary cross products

in obtaining the above relation. The expression

for a × b can be put in a determinant form

which is easy to remember.

ˆ ˆ ˆ

x y z

a a a

b b b

× =

i j k

a b

Example 7.4 Find the scalar and vector

products of two vectors. a = (3i

– 4j

+ 5k

)

and b = (– 2i

+ j

– 3k

)

Answer

ˆ ˆ ˆ ˆ ˆ ˆ

(3 4 5 ) ( 2 3 )

6 4 15

= − + − + −

= − − −

= −

a b i j k i j k

i i

ˆ ˆ ˆ

3 4 5 7 5

2 1 3

× = − = − −

− −

i j k

a b i j k

Note

ˆ ˆ ˆ

7 5

× = − + +

b a i j k

7.6 ANGULAR VELOCITY AND ITS

RELATION WITH LINEAR VELOCITY

In this section we shall study what is angular

velocity and its role in rotational motion. We

have seen that every particle of a rotating body

moves in a circle. The linear velocity of the

particle is related to the angular velocity. The

relation between these two quantities involves

a vector product which we learnt about in the

last section.

Let us go back to Fig. 7.4. As said above, in

rotational motion of a rigid body about a fixed

axis, every particle of the body moves in a circle,

Fig. 7.16 Rotation about a fixed axis. (A particle (P)

of the rigid body rotating about the fixed

(z-) axis moves in a circle with centre (C)

on the axis.)

which lies in a plane perpendicular to the axis

and has its centre on the axis. In Fig. 7.16 we

redraw Fig. 7.4, showing a typical particle (at a

point P) of the rigid body rotating about a fixed

axis (taken as the z-axis). The particle describes

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 153

a circle with a centre C on the axis. The radius

of the circle is r, the perpendicular distance of

the point P from the axis. We also show the

linear velocity vector v of the particle at P. It is

along the tangent at P to the circle.

Let P′ be the position of the particle after an

interval of time ∆t (Fig. 7.16). The angle PCP′

describes the angular displacement ∆

of the

particle in time ∆t. The average angular velocity

of the particle over the interval ∆t is ∆

/∆t. As

∆t tends to zero (i.e. takes smaller and smaller

values), the ratio ∆

/∆t approaches a limit which

is the instantaneous angular velocity d

/dt of

the particle at the position P. We denote the

instantaneous angular velocity by

(the

Greek letter omega). We know from our study

of circular motion that the magnitude of linear

velocity v of a particle moving in a circle is

related to the angular velocity of the particle

by the simple relation

υ ω

, where r is the

radius of the circle.

We observe that at any given instant the

relation

v r

applies to all particles of the

rigid body. Thus for a particle at a perpendicular

distance r

from the fixed axis, the linear velocity

at a given instant v

is given by

i i

v r

(7.19)

The index i runs from 1 to n, where n is the

total number of particles of the body.

For particles on the axis,

, and hence

v =

r = 0. Thus, particles on the axis are

stationary. This verifies that the axis is fixed.

Note that we use the same angular velocity

for all the particles. We therefore, refer to

ωω

as the angular velocity of the whole body.

We have characterised pure translation of

a body by all parts of the body having the same

velocity at any instant of time. Similarly, we

may characterise pure rotation by all parts of

the body having the same angular velocity at

any instant of time. Note that this

characterisation of the rotation of a rigid body

about a fixed axis is just another way of saying

as in Sec. 7.1 that each particle of the body moves

in a circle, which lies in a plane perpendicular

to the axis and has the centre on the axis.

In our discussion so far the angular velocity

appears to be a scalar. In fact, it is a vector. We

shall not justify this fact, but we shall accept

it. For rotation about a fixed axis, the angular

velocity vector lies along the axis of rotation,

and points out in the direction in which a right

handed screw would advance, if the head of the

screw is rotated with the body. (See Fig. 7.17a).

The magnitude of this vector is

d dt

ω θ

referred as above.

Fig. 7.17 (a) If the head of a right handed screw

rotates with the body, the screw

advances in the direction of the angular

velocity

ωω

. If the sense (clockwise or

anticlockwise) of rotation of the body

changes, so does the direction of

ωω

Fig. 7.17 (b) The angular velocity vector

ωω

ω is

directed along the fixed axis as shown.

The linear velocity of the particle at P

is v =

ωω

ω ×

r. It is perpendicular to both

ω ω

and r and is directed along the

tangent to the circle described by the

particle.

We shall now look at what the vector product

ωω

r corresponds to. Refer to Fig. 7.17(b) which

is a part of Fig. 7.16 reproduced to show the

path of the particle P. The figure shows the

vector

ωω

ω directed along the fixed (z–) axis and

also the position vector r =

of the particle

at P of the rigid body with respect to the origin

O. Note that the origin is chosen to be on the

axis of rotation.

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154 PHYSICS

Now

ωω

ω × r =

ωω

ω × OP =

ωω

ω × (OC + CP)

But

ωω

ω × OC =

0 0

0 as

ω ω

ω is along OC

Hence

ωω

ω × r =

ωω

ω × CP

The vector

ωω

CP is perpendicular to

ωω

ω, i.e.

to the z-axis and also to CP, the radius of the

circle described by the particle at P. It is

therefore, along the tangent to the circle at P.

Also, the magnitude of

ωω

CP is ω (CP) since

ωω

ω and CP are perpendicular to each other. We

shall denote CP by

⊥

and not by r, as we did

earlier.

Thus,

ωω

r is a vector of magnitude

⊥

and is along the tangent to the circle described

by the particle at P. The linear velocity vector v

at P has the same magnitude and direction.

Thus,

v =

ω ω

r (7.20)

In fact, the relation, Eq. (7.20), holds good

even for rotation of a rigid body with one point

fixed, such as the rotation of the top [Fig. 7.6(a)].

In this case r represents the position vector of

the particle with respect to the fixed point taken

as the origin.

We note that for rotation about a fixed

axis, the direction of the vector

ωω

ω does not

change with time. Its magnitude may,

however, change from instant to instant. For

the more general rotation, both the

magnitude and the direction of

ω ω

ω may change

from instant to instant.

7.6.1 Angular acceleration

You may have noticed that we are developing

the study of rotational motion along the lines

of the study of translational motion with which

we are already familiar. Analogous to the kinetic

variables of linear displacement (s) and velocity

(v) in translational motion, we have angular

displacement (

θθ

θ) and angular velocity (

ωω

ω) in

rotational motion. It is then natural to define

in rotational motion the concept of angular

acceleration in analogy with linear acceleration

defined as the time rate of change of velocity in

translational motion. We define angular

acceleration

αα

α as the time rate of change of

angular velocity; Thus,

ωω

αα

(7.21)

If the axis of rotation is fixed, the direction

ω ω

ω and hence, that of

αα

α is fixed. In this case

the vector equation reduces to a scalar equation

(7.22)

7.7 TORQUE AND ANGULAR MOMENTUM

In this section, we shall acquaint ourselves with

two physical quantities (torque and angular

momentum) which are defined as vector products

of two vectors. These as we shall see, are

especially important in the discussion of motion

of systems of particles, particularly rigid bodies.

7.7.1 Moment of force (Torque)

We have learnt that the motion of a rigid body,

in general, is a combination of rotation and

translation. If the body is fixed at a point or along

a line, it has only rotational motion. We know

that force is needed to change the translational

state of a body, i.e. to produce linear

acceleration. We may then ask, what is the

analogue of force in the case of rotational

motion? To look into the question in a concrete

situation let us take the example of opening or

closing of a door. A door is a rigid body which

can rotate about a fixed vertical axis passing

through the hinges. What makes the door

rotate? It is clear that unless a force is applied

the door does not rotate. But any force does not

do the job. A force applied to the hinge line

cannot produce any rotation at all, whereas a

force of given magnitude applied at right angles

to the door at its outer edge is most effective in

producing rotation. It is not the force alone, but

how and where the force is applied is important

in rotational motion.

The rotational analogue of force in linear

motion is moment of force. It is also referred to

as torque or couple. (We shall use the words

moment of force and torque interchangeably.)

We shall first define the moment of force for the

special case of a single particle. Later on we

shall extend the concept to systems of particles

including rigid bodies. We shall also relate it to

a change in the state of rotational motion, i.e. is

angular acceleration of a rigid body.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 155

Fig. 7.18

τ = τ =

τ = r

× ×

× F,

τ τ

is perpendicular to the plane

containing r and F, and its direction is

given by the right handed screw rule.

If a force acts on a single particle at a point

P whose position with respect to the origin O is

given by the position vector r (Fig. 7.18), the

moment of the force acting on the particle with

respect to the origin O is defined as the vector

product

ττ

τ = r

× F (7.23)

The moment of force (or torque) is a vector

quantity. The symbol

τ τ

τ stands for the Greek

letter tau. The magnitude of

τ τ

τ is

= r F sin

(7.24a)

where r is the magnitude of the position vector

r, i.e. the length OP, F is the magnitude of force

F and

is the angle between r and F as

shown.

Moment of force has dimensions M L

-2

Its dimensions are the same as those of work

or energy. It is, however, a very different physical

quantity than work. Moment of a force is a

vector, while work is a scalar. The SI unit of

moment of force is newton metre (N m). The

magnitude of the moment of force may be

written

( sin )

r F r F

τ θ

⊥

= =

(7.24b)

sin

r F rF

τ θ

⊥

= =

(7.24c)

where

⊥

= r sin

is the perpendicular distance

of the line of action of F from the origin and

( sin )

F F

⊥

is the component of F in the

direction perpendicular to r. Note that

= 0 if

r = 0, F = 0 or

= 0

or 180

. Thus, the moment

of a force vanishes if either the magnitude of

the force is zero, or if the line of action of the

force passes through the origin.

One may note that since r × F is a vector

product, properties of a vector product of two

vectors apply to it. If the direction of F is

reversed, the direction of the moment of force

is reversed. If directions of both r and F are

reversed, the direction of the moment of force

remains the same.

7.7.2 Angular momentum of a particle

Just as the moment of a force is the rotational

analogue of force in linear motion, the quantity

angular momentum is the rotational analogue

of linear momentum. We shall first define

angular momentum for the special case of a

single particle and look at its usefulness in the

context of single particle motion. We shall then

extend the definition of angular momentum to

systems of particles including rigid bodies.

Like moment of a force, angular momentum

is also a vector product. It could also be referred

to as moment of (linear) momentum. From this

term one could guess how angular momentum

is defined.

Consider a particle of mass m and linear

momentum p at a position r relative to the origin

O. The angular momentum l of the particle with

respect to the origin O is defined to be

l = r

× ×

× p (7.25a)

The magnitude of the angular momentum

vector is

sin

l r p

(7.26a)

where p is the magnitude of p and

is the angle

between r and p. We may write

l r p

⊥

r p

⊥

(7.26b)

where

⊥

(= r sinθ) is the perpendicular distance

of the directional line of p from the origin and

( sin )

p p

⊥

is the component of p in a direction

perpendicular to r. We expect the angular

momentum to be zero (l = 0), if the linear

momentum vanishes (p = 0), if the particle is at

the origin (r = 0), or if the directional line of p

passes through the origin

= 0

or 180

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156 PHYSICS

The physical quantities, moment of a force

and angular momentum, have an important

relation between them. It is the rotational

analogue of the relation between force and linear

momentum. For deriving the relation in the

context of a single particle, we differentiate

= r × p with respect to time,

d d

( )

d d

= ×

r p

t t

Applying the product rule for differentiation

to the right hand side,

d d d

( )

d d d

t t t

× = × + ×

r p

r p p r

Now, the velocity of the particle is v = dr/dt

and p = m v

Because of this

× = × =

p v v

as the vector product of two parallel vectors

vanishes. Further, since dp / dt = F,

r F× = × =

ττ

Hence

( )r p× =

ττ

or (7.27)

Thus, the time rate of change of the angular

momentum of a particle is equal to the torque

acting on it. This is the rotational analogue of

the equation F = dp/dt, which expresses

Newton’s second law for the translational motion

of a single particle.

Torque and angular momentum for a system

of particles

To get the total angular momentum of a system

of particles about a given point we need to add

vectorially the angular momenta of individual

particles. Thus, for a system of n particles,

The angular momentum of the i

particle

is given by

= r

× p

where r

is the position vector of the i

particle

with respect to a given origin and p = (m

) is

the linear momentum of the particle. (The

particle has mass m

and velocity v

) We may

write the total angular momentum of a system

of particles as

(7.25b)

This is a generalisation of the definition of

angular momentum (Eq. 7.25a) for a single

particle to a system of particles.

Using Eqs. (7.23) and (7.25b), we get

t t t

i i

( )

= =

∑∑

(7.28a)

An experiment with the bicycle rim

Take a

bicycle rim

and extend

its axle on

both sides.

Tie two

strings

at both ends

A and B,

as shown

in the

adjoining

figure. Hold

both the

strings

together in

one hand such that the rim is vertical. If you

leave one string, the rim will tilt. Now keeping

the rim in vertical position with both the strings

in one hand, put the wheel in fast rotation

around the axle with the other hand. Then leave

one string, say B, from your hand, and observe

what happens.

The rim keeps rotating in a vertical plane

and the plane of rotation turns around the

string A which you are holding. We say that the

axis of rotation of the rim or equivalently

its angular momentum precesses about the

string A.

The rotating rim gives rise to an angular

momentum. Determine the direction of this

angular momentum. When you are holding the

rotating rim with string A, a torque is generated.

(We leave it to you to find out how the torque is

generated and what its direction is.) The effect

of the torque on the angular momentum is to

make it precess around an axis perpendicular

to both the angular momentum and the torque.

Verify all these statements.

Initially After

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 157

where

ττ

is the torque acting on the i

particle;

ττ

i i i

= ×r F

The force F

on the i

particle is the vector

sum of external forces

ext

acting on the particle

and the internal forces

int

exerted on it by the

other particles of the system. We may therefore

separate the contribution of the external and

the internal forces to the total torque

ττ ττ

= = ×

∑ ∑

r F

ττ ττ ττ

= +

ext int

where

ττ

ext i i

ext

= ×

∑

r F

and

ττ

int

= ×

∑

r F

i i

We shall assume not only Newton’s third law

of motion, i.e. the forces between any two particles

of the system are equal and opposite, but also that

these forces are directed along the line joining the

two particles. In this case the contribution of the

internal forces to the total torque on the system is

zero, since the torque resulting from each action-

reaction pair of forces is zero. We thus have,

ττ

int

0 and therefore

ττ

τ τ

ext

Since

ττ ττ

∑

, it follows from Eq. (7.28a)

that

ext

ττ

(7.28 b)

Thus, the time rate of the total angular

momentum of a system of particles about a

point (taken as the origin of our frame of

reference) is equal to the sum of the external

torques (i.e. the torques due to external forces)

acting on the system taken about the same

point. Eq. (7.28 b) is the generalisation of the

single particle case of Eq. (7.23) to a system of

particles. Note that when we have only one

particle, there are no internal forces or torques.

Eq.(7.28 b) is the rotational analogue of

ext

(7.17)

Note that like Eq.(7.17), Eq.(7.28b) holds

good for any system of particles, whether it is a

rigid body or its individual particles have all

kinds of internal motion.

Conservation of angular momentum

ext

= 0, Eq. (7.28b) reduces to

or L = constant. (7.29a)

Thus, if the total external torque on a system

of particles is zero, then the total angular

momentum of the system is conserved, i.e.

remains constant. Eq. (7.29a) is equivalent to

three scalar equations,

= K

, L

= K

and L

= K

(7.29 b)

Here K

, K

and K

are constants; L

, L

and

are the components of the total angular

momentum vector L along the x,y and z axes

respectively. The statement that the total

angular momentum is conserved means that

each of these three components is conserved.

Eq. (7.29a) is the rotational analogue of

Eq. (7.18a), i.e. the conservation law of the total

linear momentum for a system of particles.

Like Eq. (7.18a), it has applications in many

practical situations. We shall look at a few of

the interesting applications later on in this

chapter.

Example 7.5 Find the torque of a force

+ 3j

– 5k

about the origin. The force

acts on a particle whose position vector is

– j

+ k

Answer Here

ˆ ˆ ˆ

= − +

r i j k

and

ˆ ˆ ˆ

7 3 5

= + −

F i j k

We shall use the determinant rule to find the

torque

ττ

= ×r F

ττ

= + +2 12 10

ˆ ˆ ˆ

i j k

Example 7.6 Show that the angular

momentum about any point of a single

particle moving with constant velocity

remains constant throughout the motion.

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158 PHYSICS

Answer Let the particle with velocity v be at

point P at some instant t. We want to calculate

the angular momentum of the particle about an

arbitrary point O.

Fig 7.19

The angular momentum is l = r × mv. Its

magnitude is mvr sin

, where

is the angle

between r and v as shown in Fig. 7.19. Although

the particle changes position with time, the line

of direction of v remains the same and hence

OM = r sin

. is a constant.

Further, the direction of l is perpendicular

to the plane of r and v. It is into the page of the

figure.This direction does not change with time.

Thus, l remains the same in magnitude and

direction and is therefore conserved. Is there

any external torque on the particle? t

7.8 EQUILIBRIUM OF A RIGID BODY

We are now going to concentrate on the motion

of rigid bodies rather than on the motion of

general systems of particles.

We shall recapitulate what effect the

external forces have on a rigid body. (Henceforth

we shall omit the adjective ‘external’ because

unless stated otherwise, we shall deal with only

external forces and torques.) The forces change

the translational state of the motion of the rigid

body, i.e. they change its total linear momentum

in accordance with Eq. (7.17). But this is not

the only effect the forces have. The total torque

on the body may not vanish. Such a torque

changes the rotational state of motion of the

rigid body, i.e. it changes the total angular

momentum of the body in accordance with Eq.

(7.28 b).

A rigid body is said to be in mechanical

equilibrium, if both its linear momentum and

angular momentum are not changing with time,

or equivalently, the body has neither linear

acceleration nor angular acceleration. This

means

(1) the total force, i.e. the vector sum of the

forces, on the rigid body is zero;

F F F F

1 2

+ + + = =

∑

...

n i

(7.30a)

If the total force on the body is zero, then

the total linear momentum of the body does

not change with time. Eq. (7.30a) gives the

condition for the translational equilibrium

of the body.

(2) The total torque, i.e. the vector sum of the

torques on the rigid body is zero,

ττ ττ ττ ττ

1 2

+ + + = =

∑

...

n i

(7.30b)

If the total torque on the rigid body is zero,

the total angular momentum of the body does

not change with time. Eq. (7.30 b) gives the

condition for the rotational equilibrium of the

body.

One may raise a question, whether the

rotational equilibrium condition [Eq. 7.30(b)]

remains valid, if the origin with respect to which

the torques are taken is shifted. One can show

that if the translational equilibrium condition

[Eq. 7.30(a)] holds for a rigid body, then such a

shift of origin does not matter, i.e. the rotational

equilibrium condition is independent of the

location of the origin about which the torques

are taken. Example 7.7 gives a proof of this

result in a special case of a couple, i.e. two forces

acting on a rigid body in translational

equilibrium. The generalisation of this result to

n forces is left as an exercise.

Eq. (7.30a) and Eq. (7.30b), both, are vector

equations. They are equivalent to three scalar

equations each. Eq. (7.30a) corresponds to

∑

and

∑

(7.31a)

where F

, F

and F

are respectively the x, y

and z components of the forces F

. Similarly,

Eq. (7.30b) is equivalent to three scalar

equations

∑

and

∑

(7.31b)

where

and

are respectively the x, y and

z components of the torque

ττ

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 159

Eq. (7.31a) and (7.31b) give six independent

conditions to be satisfied for mechanical

equilibrium of a rigid body. In a number of

problems all the forces acting on the body are

coplanar. Then we need only three conditions

to be satisfied for mechanical equilibrium. Two

of these conditions correspond to translational

equilibrium; the sum of the components of the

forces along any two perpendicular axes in the

plane must be zero. The third condition

corresponds to rotational equilibrium. The sum

of the components of the torques along any axis

perpendicular to the plane of the forces must

be zero.

The conditions of equilibrium of a rigid body

may be compared with those for a particle,

which we considered in earlier chapters. Since

consideration of rotational motion does not

apply to a particle, only the conditions for

translational equilibrium (Eq. 7.30 a) apply to

a particle. Thus, for equilibrium of a particle

the vector sum of all the forces on it must be

zero. Since all these forces act on the single

particle, they must be concurrent. Equilibrium

under concurrent forces was discussed in the

earlier chapters.

A body may be in partial equilibrium, i.e., it

may be in translational equilibrium and not in

rotational equilibrium, or it may be in rotational

equilibrium and not in translational

equilibrium.

Consider a light (i.e. of negligible mass) rod

(AB) as shown in Fig. 7.20(a). At the two ends (A

and B) of which two parallel forces, both equal

in magnitude and acting along same direction

are applied perpendicular to the rod.

Fig. 7.20 (a)

Let C be the midpoint of AB, CA = CB = a.

the moment of the forces at A and B will both

be equal in magnitude (aF), but opposite in

sense as shown. The net moment on the rod will

be zero. The system will be in rotational

equilibrium, but it will not be in translational

equilibrium;

F 0≠

∑

Fig. 7.20 (b)

The force at B in Fig. 7.20(a) is reversed in

Fig. 7.20(b). Thus, we have the same rod with

two forces of equal magnitude but acting in

opposite diretions applied perpendicular to the

rod, one at end A and the other at end B. Here

the moments of both the forces are equal, but

they are not opposite; they act in the same sense

and cause anticlockwise rotation of the rod. The

total force on the body is zero; so the body is in

translational equilibrium; but it is not in

rotational equilibrium. Although the rod is not

fixed in any way, it undergoes pure rotation (i.e.

rotation without translation).

A pair of forces of equal magnitude but acting

in opposite directions with different lines of

action is known as a couple or torque. A couple

produces rotation without translation.

When we open the lid of a bottle by turning

it, our fingers are applying a couple to the lid

[Fig. 7.21(a)]. Another known example is a

compass needle in the earth’s magnetic field as

shown in the Fig. 7.21(b). The earth’s magnetic

field exerts equal forces on the north and south

poles. The force on the North Pole is towards

the north, and the force on the South Pole is

toward the south. Except when the needle points

in the north-south direction; the two forces do

not have the same line of action. Thus there is

a couple acting on the needle due to the earth’s

magnetic field.

Fig. 7.21(a) Our fingers apply a couple to turn

the lid.

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160 PHYSICS

Fig. 7.21(b) The Earth’s magnetic field exerts equal

and opposite forces on the poles of a

compass needle. These two forces form

a couple.

Example 7.7 Show that moment of a

couple does not depend on the point about

which you take the moments.

Answer

Fig. 7.22

Consider a couple as shown in Fig. 7.22

acting on a rigid body. The forces F and -F act

respectively at points B and A. These points have

position vectors r

and r

with respect to origin

O. Let us take the moments of the forces about

the origin.

The moment of the couple = sum of the

moments of the two forces making the couple

= r

× (–F) + r

× F

= r

× F – r

× F

= (r

–r

)

× F

But r

+ AB = r

, and hence AB = r

– r

The moment of the couple, therefore, is

× F.

Clearly this is independent of the origin, the

point about which we took the moments of the

forces. t

7.8.1 Principle of moments

An ideal lever is essentially a light (i.e. of

negligible mass) rod pivoted at a point along its

length. This point is called the fulcrum. A see-

saw on the children’s playground is a typical

example of a lever. Two forces F

and F

, parallel

to each other and usually perpendicular to the

lever, as shown here, act on the lever at

distances d

and d

respectively from the

fulcrum as shown in Fig. 7.23.

Fig. 7.23

The lever is a system in mechanical

equilibrium. Let R be the reaction of the support

at the fulcrum; R is directed opposite to the

forces F

and F

. For translational equilibrium,

R – F

– F

= 0 (i)

For considering rotational equilibrium we

take the moments about the fulcrum; the sum

of moments must be zero,

– d

= 0 (ii)

Normally the anticlockwise (clockwise)

moments are taken to be positive (negative). Note

R acts at the fulcrum itself and has zero moment

about the fulcrum.

In the case of the lever force F

is usually

some weight to be lifted. It is called the load

and its distance from the fulcrum d

is called

the load arm. Force F

is the effort applied to lift

the load; distance d

of the effort from the

fulcrum is the effort arm.

Eq. (ii) can be written as

(7.32a)

or load arm

× load = effort arm

× effort

The above equation expresses the principle

of moments for a lever. Incidentally the ratio

is called the Mechanical Advantage (M.A.);

M.A. =

1 2

2 1

F d

(7.32b)

If the effort arm d

is larger than the load

arm, the mechanical advantage is greater than

one. Mechanical advantage greater than one

means that a small effort can be used to lift a

large load. There are several examples of a lever

around you besides the see-saw. The beam of a

balance is a lever. Try to find more such

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 161

examples and identify the fulcrum, the effort and

effort arm, and the load and the load arm of the

lever in each case.

You may easily show that the principle of

moment holds even when the parallel forces F

and F

are not perpendicular, but act at some

angle, to the lever.

7.8.2 Centre of gravity

Many of you may have the experience of

balancing your notebook on the tip of a finger.

Figure 7.24 illustrates a similar experiment that

you can easily perform. Take an irregular-

shaped cardboard having mass M and a narrow

tipped object like a pencil. You can locate by trial

and error a point G on the cardboard where it

can be balanced on the tip of the pencil. (The

cardboard remains horizontal in this position.)

This point of balance is the centre of gravity (CG)

of the cardboard. The tip of the pencil provides

a vertically upward force due to which the

cardboard is in mechanical equilibrium. As

shown in the Fig. 7.24, the reaction of the tip is

equal and opposite to Mg and hence the

cardboard is in translational equilibrium. It is

also in rotational equilibrium; if it were not so,

due to the unbalanced torque it would tilt and

fall. There are torques on the card board due to

the forces of gravity like m

g, m

g …. etc, acting

on the individual particles that make up the

cardboard.

Fig. 7.24 Balancing a cardboard on the tip of a

pencil. The point of support, G, is the

centre of gravity.

The CG of the cardboard is so located that

the total torque on it due to the forces m

g, m

…. etc. is zero.

If r

is the position vector of the ith particle

of an extended body with respect to its CG, then

the torque about the CG, due to the force of

gravity on the particle is

ττ

= r

× m

g. The total

gravitational torque about the CG is zero, i.e.

ττ ττ

i i

m= = × =

∑ ∑

r g 0

(7.33)

We may therefore, define the CG of a body

as that point where the total gravitational torque

on the body is zero.

We notice that in Eq. (7.33), g is the same

for all particles, and hence it comes out of the

summation. This gives, since g is non-zero,

∑

i i

= 0. Remember that the position vectors

) are taken with respect to the CG. Now, in

accordance with the reasoning given below

Eq. (7.4a) in Sec. 7.2, if the sum is zero, the origin

must be the centre of mass of the body. Thus,

the centre of gravity of the body coincides with

the centre of mass in uniform gravity or gravity-

free space. We note that this is true because

the body being small, g does not

Fig. 7.25 Determining the centre of gravity of a body

of irregular shape. The centre of gravity G

lies on the vertical AA

through the point

of suspension of the body A.

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162 PHYSICS

vary from one point of the body to the other. If

the body is so extended that g varies from part

to part of the body, then the centre of gravity

and centre of mass will not coincide. Basically,

the two are different concepts. The centre of

mass has nothing to do with gravity. It depends

only on the distribution of mass of the body.

In Sec. 7.2 we found out the position of the

centre of mass of several regular, homogeneous

objects. Obviously the method used there gives

us also the centre of gravity of these bodies, if

they are small enough.

Figure 7.25 illustrates another way of

determining the CG of an irregular shaped body

like a cardboard. If you suspend the body from

some point like A, the vertical line through A

passes through the CG. We mark the vertical

. We then suspend the body through other

points like B and C. The intersection of the

verticals gives the CG. Explain why the method

works. Since the body is small enough, the

method allows us to determine also its centre

of mass.

Example 7.8 A metal bar 70 cm long

and 4.00 kg in mass supported on two

knife-edges placed 10 cm from each end.

A 6.00 kg load is suspended at 30 cm from

one end. Find the reactions at the knife-

edges. (Assume the bar to be of uniform

cross section and homogeneous.)

Answer

Fig. 7.26

Figure 7.26 shows the rod AB, the positions

of the knife edges K

and K

, the centre of

gravity of the rod at G and the suspended load

at P.

Note the weight of the rod W acts at its

centre of gravity G. The rod is uniform in cross

section and homogeneous; hence G is at the

centre of the rod; AB = 70 cm. AG = 35 cm, AP

= 30 cm, PG = 5 cm, AK

= BK

= 10 cm and K

= K

G = 25 cm. Also, W= weight of the rod =

4.00 kg and W

= suspended load = 6.00 kg;

and R

are the normal reactions of the

support at the knife edges.

For translational equilibrium of the rod,

–W

–W = 0 (i)

Note W

and W act vertically down and R

and R

act vertically up.

For considering rotational equilibrium, we

take moments of the forces. A convenient point

to take moments about is G. The moments of

and W

are anticlockwise (+ve), whereas the

moment of R

is clockwise (-ve).

For rotational equilibrium,

–R

G) + W

(PG) + R

G) = 0 (ii)

It is given that W = 4.00g N and W

= 6.00g

N, where g = acceleration due to gravity. We

take g = 9.8 m/s

With numerical values inserted, from (i)

+ R

– 4.00g – 6.00g = 0

or R

+ R

= 10.00g N (iii)

= 98.00 N

From (ii), – 0.25 R

+ 0.05 W

+ 0.25 R

= 0

or R

– R

= 1.2g N = 11.76 N (iv)

From (iii) and (iv), R

= 54.88 N,

= 43.12 N

Thus the reactions of the support are about

55 N at K

and 43 N at K

. t

Example 7.9 A 3m long ladder weighing

20 kg leans on a frictionless wall. Its feet

rest on the floor 1 m from the wall as shown

in Fig.7.27. Find the reaction forces of the

wall and the floor.

Answer

Fig. 7.27

The ladder AB is 3 m long, its foot A is at

distance AC = 1 m from the wall. From

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 163

Pythagoras theorem, BC =

2 2

m. The forces

on the ladder are its weight W acting at its centre

of gravity D, reaction forces F

and F

of the wall

and the floor respectively. Force F

perpendicular to the wall, since the wall is

frictionless. Force F

is resolved into two

components, the normal reaction N and the

force of friction F. Note that F prevents the ladder

from sliding away from the wall and is therefore

directed toward the wall.

For translational equilibrium, taking the

forces in the vertical direction,

N – W = 0 (i)

Taking the forces in the horizontal direction,

F – F

= 0 (ii)

For rotational equilibrium, taking the

moments of the forces about A,

2 2 F

−

(1/2) W = 0 (iii)

Now W = 20 g = 20 × 9.8 N = 196.0 N

From (i) N = 196.0 N

From (iii)

4 2 196.0/4 2 34.6 N

F W= = =

From (ii)

34.6 N

F F= =

2 2

199.0

F F N= + =

The force F

makes an angle

with the

horizontal,

tan 4 2 , tan (4 2) 80

N F

α α

−

= = = ≈



7.9 MOMENT OF INERTIA

We have already mentioned that we are

developing the study of rotational motion

parallel to the study of translational motion with

which we are familiar. We have yet to answer

one major question in this connection. What is

the analogue of mass in rotational motion?

We shall attempt to answer this question in the

present section. To keep the discussion simple,

we shall consider rotation about a fixed axis

only. Let us try to get an expression for the

kinetic energy of a rotating body. We know

that for a body rotating about a fixed axis, each

particle of the body moves in a circle with linear

velocity given by Eq. (7.19). (Refer to Fig. 7.16).

For a particle at a distance from the axis, the

linear velocity is

i i

υ ω

. The kinetic energy of

motion of this particle is

2 2 2

1 1

2 2

i i i i i

k m m r

υ ω

= =

where m

is the mass of the particle. The total

kinetic energy K of the body is then given by

the sum of the kinetic energies of individual

particles,

2 2

1 1

( )

n n

i i i

i i

K k m r

= =

∑ ∑

Here n is the number of particles in the body.

Note

is the same for all particles. Hence, taking

out of the sum,

2 2

( )

i i

K m r

∑

We define a new parameter characterising

the rigid body, called the moment of inertia I

given by

i i

I m r

∑

(7.34)

With this definition,

K I

(7.35)

Note that the parameter I is independent of

the magnitude of the angular velocity. It is a

characteristic of the rigid body and the axis

about which it rotates.

Compare Eq. (7.35) for the kinetic energy of

a rotating body with the expression for the

kinetic energy of a body in linear (translational)

motion,

K m

Here, m is the mass of the body and v is its

velocity. We have already noted the analogy

between angular velocity

(in respect of rotational

motion about a fixed axis) and linear velocity v (in

respect of linear motion). It is then evident that

the parameter, moment of inertia I, is the desired

rotational analogue of mass in linear motion. In

rotation (about a fixed axis), the moment of inertia

plays a similar role as mass does in linear motion.

We now apply the definition Eq. (7.34), to

calculate the moment of inertia in two simple

cases.

(a) Consider a thin ring of radius R and mass

M, rotating in its own plane around its centre

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164 PHYSICS

with angular velocity

. Each mass element

of the ring is at a distance R from the axis,

and moves with a speed R

. The kinetic

energy is therefore,

2 2 2

1 1

2 2

K M MR

υ ω

= =

Comparing with Eq. (7.35) we get I = MR

for the ring.

Fig. 7.28 A light rod of length l with a pair of masses

rotating about an axis through the centre of

mass of the system and perpendicular to

the rod. The total mass of the system is M.

(b) Next, take a rigid rod of negligible mass of

length of length l with a pair of small masses,

rotating about an axis through the centre of

mass perpendicular to the rod (Fig. 7.28).

Each mass M/2 is at a distance l/2 from the

axis. The moment of inertia of the masses is

therefore given by

(M/2) (l/2)

+ (M/2)(l/2)

Thus, for the pair of masses, rotating about

the axis through the centre of mass

perpendicular to the rod

I = Ml

/ 4

Table 7.1 simply gives the moment of inertia of

various familiar regular shaped bodies about

specific axes. (The derivations of these

expressions are beyond the scope of this textbook

and you will study them in higher classes.)

As the mass of a body resists a change in its

state of linear motion, it is a measure of its inertia

in linear motion. Similarly, as the moment of

inertia about a given axis of rotation resists a

change in its rotational motion, it can be

regarded as a measure of rotational inertia of

the body; it is a measure of the way in which

different parts of the body are distributed at

different distances from the axis. Unlike the

mass of a body, the moment of inertia is not a

fixed quantity but depends on distribution of

mass about the axis of rotation, and the

orientation and position of the axis of rotation

with respect to the body as a whole. As a

measure of the way in which the mass of a

rotating rigid body is distributed with respect to

the axis of rotation, we can define a new

parameter, the radius of gyration. It is related

to the moment of inertia and the total mass of

the body.

Notice from the Table 7.1 that in all

cases, we can write I = Mk

, where k has

the dimension of length. For a rod, about

the perpendicular axis at its midpoint,

i.e.

2 2

12,

k L=

k L

. Similarly, k = R/2

for the circular disc about its diameter. The

length k is a geometric property of the body and

axis of rotation. It is called the radius of

gyration. The radius of gyration of a body

about an axis may be defined as the distance

from the axis of a mass point whose mass is

equal to the mass of the whole body and whose

moment of inertia is equal to the moment of

inertia of the body about the axis.

Thus, the moment of inertia of a rigid body

depends on the mass of the body, its shape and

size; distribution of mass about the axis of

rotation, and the position and orientation of the

axis of rotation.

From the definition, Eq. (7.34), we can infer

that the dimensions of moments of inertia are

and its SI units are kg m

The property of this extremely important

quantity I, as a measure of rotational inertia of

the body, has been put to a great practical use.

The machines, such as steam engine and the

automobile engine, etc., that produce rotational

motion have a disc with a large moment of

inertia, called a flywheel. Because of its large

moment of inertia, the flywheel resists the

sudden increase or decrease of the speed of the

vehicle. It allows a gradual change in the speed

and prevents jerky motions, thereby ensuring

a smooth ride for the passengers on the vehicle.

7.10 THEOREMS OF PERPENDICULAR AND

PARALLEL AXES

These are two useful theorems relating to

moment of inertia. We shall first discuss the

theorem of perpendicular axes and its simple

yet instructive application in working out the

moments of inertia of some regular-shaped

bodies.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 165

Table 7.1 Moments of inertia of some regular shaped bodies about specific axes

Z Body Axis Figure I

(1) Thin circular Perpendicular to M R

ring, radius R plane, at centre

(2) Thin circular Diameter M R

ring, radius R

(3) Thin rod, Perpendicular to M L

/12

length L rod, at mid point

(4) Circular disc, Perpendicular to M R

radius R disc at centre

(5) Circular disc, Diameter M R

radius R

(6) Hollow cylinder, Axis of cylinder M R

radius R

(7) Solid cylinder, Axis of cylinder M R

radius R

(8) Solid sphere, Diameter 2 M R

radius R

Theorem of perpendicular axes

This theorem is applicable to bodies which are

planar. In practice this means the theorem

applies to flat bodies whose thickness is very

small compared to their other dimensions (e.g.

length, breadth or radius). Fig. 7.29 illustrates

the theorem. It states that the moment of

inertia of a planar body (lamina) about an axis

perpendicular to its plane is equal to the sum

of its moments of inertia about two

perpendicular axes concurrent with

perpendicular axis and lying in the plane of

the body.

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166 PHYSICS

Fig. 7.29 Theorem of perpendicular axes

applicable to a planar body; x and y axes

are two perpendicular axes in the plane

and the z-axis is perpendicular to the

plane.

The figure shows a planar body. An axis

perpendicular to the body through a point O is

taken as the z-axis. Two mutually perpendicular

axes lying in the plane of the body and

concurrent with z-axis, i.e., passing through O,

are taken as the x and y-axes. The theorem

states that

z x y

I I I

= +

(7.36)

Let us look at the usefulness of the theorem

through an example.

Example 7.10 What is the moment of

inertia of a disc about one of its diameters?

Fig. 7.30 Moment of inertia of a disc about a

diameter, given its moment of inertia about

the perpendicular axis through its centre.

Answer We assume the moment of inertia of

the disc about an axis perpendicular to it and

through its centre to be known; it is MR

/2,

where M is the mass of the disc and R is its

radius (Table 7.1)

The disc can be considered to be a planar

body. Hence the theorem of perpendicular axes

is applicable to it. As shown in Fig. 7.30, we

take three concurrent axes through the centre

of the disc, O, as the x–, y– and z–axes; x– and

y–axes lie in the plane of the disc and z–axis is

perpendicular to it. By the theorem of

perpendicular axes,

z x y

I I I

= +

Now, x and y axes are along two diameters

of the disc, and by symmetry the moment of

inertia of the disc is the same about any

diameter. Hence

and I

= 2I

But I

= MR

So finally, I

= I

/2 = MR

Thus the moment of inertia of a disc about

any of its diameter is MR

/4 . t

Find similarly the moment of inertia of a

ring about any of its diameters. Will the theorem

be applicable to a solid cylinder?

Fig.7.31 The theorem of parallel axes The z and z

′

axes are two parallel axes separated by a

distance a; O is the centre of mass of the

body, OO

’

= a.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 167

7.10.1 Theorem of parallel axes

This theorem is applicable to a body of any

shape. It allows to find the moment of inertia of

a body about any axis, given the moment of

inertia of the body about a parallel axis through

the centre of mass of the body. We shall only

state this theorem and not give its proof. We

shall, however, apply it to a few simple situations

which will be enough to convince us about the

usefulness of the theorem. The theorem may

be stated as follows:

The moment of inertia of a body about any

axis is equal to the sum of the moment of

inertia of the body about a parallel axis passing

through its centre of mass and the product of

its mass and the square of the distance

between the two parallel axes. As shown in

the Fig. 7.31, z and z′ are two parallel axes,

separated by a distance a. The z-axis passes

through the centre of mass O of the rigid body.

Then according to the theorem of parallel axes

′

(7.37)

where I

and

′

are the moments of inertia of the

body about the z and z

′

axes respectively, M is the

total mass of the body and a is the perpendicular

distance between the two parallel axes.

Example 7.11 What is the moment of

inertia of a rod of mass M, length l about

an axis perpendicular to it through one

end?

Answer For the rod of mass M and length l,

I = Ml

/12. Using the parallel axes theorem,

′

= I + Ma

with a = l/2 we get,

2 2

12 2 3

l l Ml

I M M

 

′

= + =

 

 

We can check this independently since I is

half the moment of inertia of a rod of mass 2M

and length 2l about its midpoint,

2 2

4 1

2 .

12 2 3

l Ml

I M

′

= × =

Example 7.12 What is the moment of

inertia of a ring about a tangent to the

circle of the ring?

Answer

The tangent to the ring in the plane of the ring

is parallel to one of the diameters of the ring.

The distance between these two parallel axes is

R, the radius of the ring. Using the parallel axes

theorem,

Fig. 7.32

I I MR

MR MR

diatangent

= + = + =

2 2

7.11 KINEMATICS OF ROTATIONAL MOTION

ABOUT A FIXED AXIS

We have already indicated the analogy between

rotational motion and translational motion. For

example, the angular velocity

ωω

ω plays the same

role in rotation as the linear velocity v in

translation. We wish to take this analogy

further. In doing so we shall restrict the

discussion only to rotation about fixed axis. This

case of motion involves only one degree of

freedom, i.e., needs only one independent

variable to describe the motion. This in

translation corresponds to linear motion. This

section is limited only to kinematics. We shall

turn to dynamics in later sections.

We recall that for specifying the angular

displacement of the rotating body we take any

particle like P (Fig.7.33) of the body. Its angular

displacement

in the plane it moves is the

angular displacement of the whole body;

measured from a fixed direction in the plane of

motion of P, which we take to be the x

′

-axis,

chosen parallel to the x-axis. Note, as shown,

the axis of rotation is the z – axis and the plane

of the motion of the particle is the x - y plane.

Fig. 7.33 also shows

, the angular

displacement at t = 0.

We also recall that the angular velocity is

the time rate of change of angular displacement,

= d

/dt. Note since the axis of rotation is fixed,

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168 PHYSICS

there is no need to treat angular velocity as a

vector. Further, the angular acceleration,

/dt.

The kinematical quantities in rotational

motion, angular displacement (

), angular

velocity (

) and angular acceleration (

)

respectively are analogous to kinematic

quantities in linear motion, displacement (x),

velocity (v) and acceleration (a). We know the

kinematical equations of linear motion with

uniform (i.e. constant) acceleration:

v = v

+ at (a)

0 0

x x t at

= + +

(b)

2 2

υ υ

= +

(c)

where x

= initial displacement and v

= initial

velocity. The word ‘initial’ refers to values of the

quantities at t = 0

The corresponding kinematic equations for

rotational motion with uniform angular

acceleration are:

= +

ω ω α

(7.38)

0 0

t t

= + +

θ θ ω α

(7.39)

and

2 2

0 0

2 ( – )

= +

ω ω α θ θ

(7.40)

where

= initial angular displacement of the

rotating body, and

= initial angular velocity

of the body.

Fig.7.33 Specifying the angular position of a rigid

body.

Example 7.13 Obtain Eq. (7.38) from first

principles.

Answer The angular acceleration is uniform,

hence

constant

= =

(i)

Integrating this equation,

ω α

= +

∫

dt c

(as is constant)

t c

α α

= +

At t = 0,

(given)

From (i) we get at t = 0,

= c =

Thus,

t +

as required.

With the definition of

= d

/dt we may

integrate Eq. (7.38) to get Eq. (7.39). This

derivation and the derivation of Eq. (7.40) is

left as an exercise.

Example 7.14 The angular speed of a

motor wheel is increased from 1200 rpm

to 3120 rpm in 16 seconds. (i) What is its

angular acceleration, assuming the

acceleration to be uniform? (ii) How many

revolutions does the engine make during

this time?

Answer

(i) We shall use

= initial angular speed in rad/s

= 2

× angular speed in rev/s

2 angular speed in rev/min

60 s/min

2 1200

rad/s

= 40

rad/s

Similarly

= final angular speed in rad/s

2 3120

rad/s

= 2

× 52 rad/s

= 104

rad/s

∴ Angular acceleration

ω ω

−

= 4

rad/s

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 169

The angular acceleration of the engine

= 4π rad/s

(ii) The angular displacement in time t is

given by

t t

θ ω α

= +

(40 16 4 16 )

π π

= × + × ×

rad

(640 512 )

π π

= +

rad

= 1152π rad

Number of revolutions =

1152

576

7.12 DYNAMICS OF ROTATIONAL MOTION

ABOUT A FIXED AXIS

Table 7.2 lists quantities associated with linear

motion and their analogues in rotational motion.

We have already compared kinematics of the

two motions. Also, we know that in rotational

motion moment of inertia and torque play the

same role as mass and force respectively in

linear motion. Given this we should be able to

guess what the other analogues indicated in the

table are. For example, we know that in linear

motion, work done is given by F dx, in rotational

motion about a fixed axis it should be

τ θ

since we already know the correspondence

d d

→

and

→

. It is, however, necessary

that these correspondences are established on

sound dynamical considerations. This is what

we now turn to.

Before we begin, we note a simplification

that arises in the case of rotational motion

about a fixed axis. Since the axis is fixed, only

those components of torques, which are along

the direction of the fixed axis need to be

considered in our discussion. Only these

components can cause the body to rotate about

the axis. A component of the torque

perpendicular to the axis of rotation will tend

to turn the axis from its position. We specifically

assume that there will arise necessary forces of

constraint to cancel the effect of the

perpendicular components of the (external)

torques, so that the fixed position of the axis

will be maintained. The perpendicular

components of the torques, therefore need not

be taken into account. This means that for our

calculation of torques on a rigid body:

(1) We need to consider only those forces that

lie in planes perpendicular to the axis.

Forces which are parallel to the axis will

give torques perpendicular to the axis and

need not be taken into account.

(2) We need to consider only those components

of the position vectors which are

perpendicular to the axis. Components of

position vectors along the axis will result in

torques perpendicular to the axis and need

not be taken into account.

Work done by a torque

Fig. 7.34 Work done by a force F

acting on a particle

of a body rotating about a fixed axis; the

particle describes a circular path with

centre C on the axis; arc P

P′

(ds

) gives

the displacement of the particle.

Figure 7.34 shows a cross-section of a rigid

body rotating about a fixed axis, which is taken

as the z-axis (perpendicular to the plane of the

page; see Fig. 7.33). As said above we need to

consider only those forces which lie in planes

perpendicular to the axis. Let F

be one such

typical force acting as shown on a particle of

the body at point P

with its line of action in a

plane perpendicular to the axis. For convenience

we call this to be the x

′

–y

′

plane (coincident

with the plane of the page). The particle at P

describes a circular path of radius r

with centre

C on the axis; CP

= r

In time ∆t, the point moves to the position

′. The displacement of the particle ds

therefore, has magnitude ds

= r

and

direction tangential at P

to the circular path

as shown. Here d

is the angular displacement

of the particle, d

1 1

P CP

∠

′

.The work done by

the force on the particle is

= F

. ds

= F

cos

= F

)sin

where

is the angle between F

and the tangent

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170 PHYSICS

at P

and

is the angle between F

and the

radius vector OP

;

= 90°.

The torque due to F

about the origin is

× F

. Now OP

= OC + OP

. [Refer to

Fig. 7.17(b).] Since OC is along the axis, the

torque resulting from it is excluded from our

consideration. The effective torque due to F

ττ

= CP × F

; it is directed along the axis of rotation

and has a magnitude

= r

sinα , Therefore,

If there are more than one forces acting on

the body, the work done by all of them can be

added to give the total work done on the body.

Denoting the magnitudes of the torques due to

the different forces as

, … etc,

1 2

d ( ...)d

τ τ θ

= + +

Remember, the forces giving rise to the

torques act on different particles, but the

angular displacement d

is the same for all

particles. Since all the torques considered are

parallel to the fixed axis, the magnitude

of the

total torque is just the algebraic sum of the

magnitudes of the torques, i.e.,

+ .....

We, therefore, have

d d

τ θ

(7.41)

This expression gives the work done by the

total (external) torque τ which acts on the body

rotating about a fixed axis. Its similarity with

the corresponding expression

dW= F ds

for linear (translational) motion is obvious.

Dividing both sides of Eq. (7.41) by dt gives

d d

t t

τ τω

= = =

τω

(7.42)

This is the instantaneous power. Compare

this expression for power in the case of

rotational motion about a fixed axis with that of

power in the case of linear motion,

P = Fv

In a perfectly rigid body there is no internal

motion. The work done by external torques is

therefore, not dissipated and goes on to increase

the kinetic energy of the body. The rate at which

work is done on the body is given by Eq. (7.42).

This is to be equated to the rate at which kinetic

energy increases. The rate of increase of kinetic

energy is

ω ω ω













( )

We assume that the moment of inertia does

not change with time. This means that the mass

of the body does not change, the body remains

rigid and also the axis does not change its

position with respect to the body.

Since

d /d ,

α ω

we get

ω α













Equating rates of work done and of increase

in kinetic energy,

τω ω α

Table 7.2 Comparison of Translational and Rotational Motion

Linear Motion Rotational Motion about a Fixed Axis

1 Displacement x Angular displacement

2 Velocity v = dx/dt Angular velocity

= d

/dt

3 Acceleration a = dv/dt Angular acceleration

= d

/dt

4 Mass M Moment of inertia I

5 Force F = Ma Torque

= I

6 Work dW = F ds Work W =

7 Kinetic energy K = Mv

/2 Kinetic energy K = I

8 Power P = F v Power P =

τω

9 Linear momentum p = Mv Angular momentum L = I

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 171

τ α

(7.43)

Eq. (7.43) is similar to Newton’s second law

for linear motion expressed symbolically as

F = ma

Just as force produces acceleration, torque

produces angular acceleration in a body. The

angular acceleration is directly proportional to

the applied torque and is inversely proportional

to the moment of inertia of the body. In this

respect, Eq.(7.43) can be called Newton’s second

law for rotational motion about a fixed axis.

Example 7.15 A cord of negligible mass

is wound round the rim of a fly wheel of

mass 20 kg and radius 20 cm. A steady

pull of 25 N is applied on the cord as shown

in Fig. 7.35. The flywheel is mounted on a

horizontal axle with frictionless bearings.

(a) Compute the angular acceleration of

the wheel.

(b) Find the work done by the pull, when

2m of the cord is unwound.

wheel at this point. Assume that the

wheel starts from rest.

(d) Compare answers to parts (b) and (c).

Answer

Fig. 7.35

(a) We use I

the torque

= F R

= 25 × 0.20 Nm (as R = 0.20m)

= 5.0 Nm

I = Moment of inertia of flywheel about its

axis

20.0 (0.2)

= 0.4 kg m

= angular acceleration

= 5.0 N m/0.4 kg m

= 12.5 s

–2

(b) Work done by the pull unwinding 2m of the

cord

= 25 N × 2m = 50 J

be the final angular velocity. The

kinetic energy gained =

since the wheel starts from rest. Now,

2 2

0 0

2 , 0

ω ω αθ ω

= + =

The angular displacement θ = length of

unwound string / radius of wheel

= 2m/0.2 m = 10 rad

2 2

2 12 5 10 0 250= × × =. . )(rad/s

∴

(d) The answers are the same, i.e. the kinetic energy

gained by the wheel = work done by the force.

There is no loss of energy due to friction. t

7.13 ANGULAR MOMENTUM IN CASE OF

ROTATION ABOUT A FIXED AXIS

We have studied in section 7.7, the angular

momentum of a system of particles. We already

know from there that the time rate of total

angular momentum of a system of particles

about a point is equal to the total external torque

on the system taken about the same point. When

the total external torque is zero, the total angular

momentum of the system is conserved.

We now wish to study the angular momentum

in the special case of rotation about a fixed axis.

The general expression for the total angular

momentum of the system of n particles is

L r p= ×

∑

i i

(7.25b)

We first consider the angular momentum of

a typical particle of the rotating rigid body. We

then sum up the contributions of individual

particles to get L of the whole body.

For a typical particle l = r × p. As seen in the

last section r = OP = OC + CP [Fig. 7.17(b)]. With

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172 PHYSICS

p = m v ,

(

)

(

)

= × + ×

OC v CP v

m ml

The magnitude of the linear velocity v of the

particle at P is given by v =

⊥

where r

⊥

is the

length of CP or the perpendicular distance of P

from the axis of rotation. Further, v is tangential

at P to the circle which the particle describes.

Using the right-hand rule one can check that

CP × v is parallel to the fixed axis. The unit

vector along the fixed axis (chosen as the z-axis)

. Hence

(

)

⊥

× =

CP v k

m r mv

⊥

(since

⊥

)

Similarly, we can check that OC × v is

perpendicular to the fixed axis. Let us denote

the part of l along the fixed axis (i.e. the z-axis)

by l

, then

= ×

CP v

⊥

and

= + ×

OC v

l l

We note that l

is parallel to the fixed axis,

but l is not. In general, for a particle, the angular

momentum l is not along the axis of rotation,

i.e. for a particle, l and

ωω

ω are not necessarily

parallel. Compare this with the corresponding

fact in translation. For a particle, p and v are

always parallel to each other.

For computing the total angular momentum

of the whole rigid body, we add up the

contribution of each particle of the body.

Thus

We denote by

⊥

and

the components of

respectively perpendicular to the z-axis and

along the z-axis;

L OC v

⊥

= ×

∑

i i i

(7.44a)

where m

and v

are respectively the mass and

the velocity of the i

particle and C

is the centre

of the circle described by the particle;

and

L k

(7.44b)

The last step follows since the perpendicular

distance of the i

particle from the axis is r

;

and by definition the moment of inertia of the

body about the axis of rotation is

I m r

i i

∑

Note

⊥

= +

L L L

(7.44c)

The rigid bodies which we have mainly

considered in this chapter are symmetric about

the axis of rotation, i.e. the axis of rotation is

one of their symmetry axes. For such bodies,

for a given OC

, for every particle which has a

velocity v

, there is another particle of velocity

–v

located diametrically opposite on the circle

with centre C

described by the particle. Together

such pairs will contribute zero to

⊥

and as a

result for symmetric bodies

⊥

is zero, and

hence

= =

L L k

(7.44d)

For bodies, which are not symmetric about

the axis of rotation, L is not equal to L

and

hence L does not lie along the axis of rotation.

Referring to Table 7.1, can you tell in which

cases L = L

will not apply?

Let us differentiate Eq. (7.44b). Since

is a

fixed (constant) vector, we get

t t

IL k

( )













Now, Eq. (7.28b) states

ττ

As we have seen in the last section, only

those components of the external torques which

are along the axis of rotation, need to be taken

into account, when we discuss rotation about a

fixed axis. This means we can take

ττ

Since

⊥

= +

L L L

and the direction of L

(vector

) is fixed, it follows that for rotation about a

fixed axis,

(7.45a)

and

⊥

(7.45b)

Thus, for rotation about a fixed axis, the

component of angular momentum

perpendicular to the fixed axis is constant. As

L k

, we get from Eq. (7.45a),

(

)

ω τ

(7.45c)

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 173

If the moment of inertia I does not change with

time,

(

)

d d

I I I

t t

ω α

= =

and we get from Eq. (7.45c),

τ α

(7.43)

We have already derived this equation using

the work - kinetic energy route.

7.13.1 Conservation of angular momentum

We are now in a position to revisit the principle

of conservation of angular momentum in the

context of rotation about a fixed axis. From Eq.

(7.45c), if the external torque is zero,

= I

= constant (7.46)

For symmetric bodies, from Eq. (7.44d), L

may be replaced by L .(L and L

are respectively

the magnitudes of L and L

This then is the required form, for fixed axis

rotation, of Eq. (7.29a), which expresses the

general law of conservation of angular

momentum of a system of particles. Eq. (7.46)

applies to many situations that we come across

in daily life. You may do this experiment with

your friend. Sit on a swivel chair (a chair with a

seat, free to rotate about a pivot) with your arms

folded and feet not resting on, i.e., away from,

the ground. Ask your friend to rotate the chair

rapidly. While the chair is rotating with

considerable angular speed stretch your arms

horizontally. What happens? Your angular

speed is reduced. If you bring back your arms

closer to your body, the angular speed increases

again. This is a situation where the principle of

conservation of angular momentum is

applicable. If friction in the rotational

mechanism is neglected, there is no external

torque about the axis of rotation of the chair

and hence I

is constant. Stretching the arms

increases I about the axis of rotation, resulting

in decreasing the angular speed

. Bringing

the arms closer to the body has the opposite

effect.

A circus acrobat and a diver take advantage

of this principle. Also, skaters and classical,

Indian or western, dancers performing a

pirouette (a spinning about a tip–top) on the toes

of one foot display ‘mastery’ over this principle.

Can you explain?

7.14 ROLLING MOTION

One of the most common motions observed in

daily life is the rolling motion. All wheels used

in transportation have rolling motion. For

specificness we shall begin with the case of a

disc, but the result will apply to any rolling body

rolling on a level surface. We shall assume that

the disc rolls without slipping. This means that

at any instant of time the bottom of the disc

Fig 7.36 (a) A demonstration of conservation of

angular momentum. A girl sits on a

swivel chair and stretches her arms/

brings her arms closer to the body.

Fig 7.36 (b) An acrobat employing the principle of

conservation of angular momentum in

her performance.

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174 PHYSICS

which is in contact with the surface is at rest

on the surface.

We have remarked earlier that rolling motion

is a combination of rotation and translation. We

know that the translational motion of a system

of particles is the motion of its centre of mass.

Fig. 7.37 The rolling motion (without slipping) of a

disc on a level surface. Note at any instant,

the point of contact P

of the disc with the

surface is at rest; the centre of mass of

the disc moves with velocity, v

. The disc

rotates with angular velocity

about its

axis which passes through C; v

where R is the radius of the disc.

Let v

be the velocity of the centre of mass

and therefore the translational velocity of the

disc. Since the centre of mass of the rolling disc

is at its geometric centre C (Fig. 7. 37), v

the velocity of C. It is parallel to the level

surface. The rotational motion of the disc is

about its symmetry axis, which passes through

C. Thus, the velocity of any point of the disc,

like P

, P

or P

, consists of two parts, one is the

translational velocity v

and the other is the

linear velocity v

on account of rotation. The

magnitude of v

is v

= r

, where

is the angular

velocity of the rotation of the disc about the axis

and r is the distance of the point from the axis

(i.e. from C). The velocity v

is directed

perpendicular to the radius vector of the given

point with respect to C. In Fig. 7.37, the velocity

of the point P

) and its components

and

are shown; v

here is perpendicular to CP

It is easy to show that v

is perpendicular to the

line P

. Therefore the line passing through P

and parallel to

ωω

ω is called the instantaneous axis

of rotation.

At P

, the linear velocity, v

, due to rotation

is directed exactly opposite to the translational

velocity v

Further the magnitude of v

here is

, where R is the radius of the disc. The

condition that P

is instantaneously at rest

requires v

= R

. Thus for the disc the condition

for rolling without slipping is

υ ω

(7.47)

Incidentally, this means that the velocity of

point P

at the top of the disc (v

) has a

magnitude v

+ R

or 2 v

and is directed

parallel to the level surface. The condition (7.47)

applies to all rolling bodies.

7.14.1 Kinetic Energy of Rolling Motion

Our next task will be to obtain an expression

for the kinetic energy of a rolling body. The

kinetic energy of a rolling body can be separated

into kinetic energy of translation and kinetic

energy of rotation. This is a special case of a

general result for a system of particles,

according to which the kinetic energy of a

system of particles (K) can be separated into

the kinetic energy of translational motion of the

centre of mass (MV

/2) and kinetic energy of

rotational motion about the centre of mass of

the system of particles (K

′

). Thus,

(7.48)

We assume this general result (see Exercise

7.31), and apply it to the case of rolling motion.

In our notation, the kinetic energy of the centre

of mass, i.e., the kinetic energy of translation,

of the rolling body is mv

/2, where m is the

mass of the body and v

is the centre of the

mass velocity. Since the motion of the rolling

body about the centre of mass is rotation, K

′

represents the kinetic energy of rotation of the

body; , where I is the moment of

inertia about the appropriate axis, which is the

symmetry axis of the rolling body. The kinetic

energy of a rolling body, therefore, is given by

(7.49a)

Substituting I = mk

where k = the

corresponding radius of gyration of the body

and v

= R ω, we get

or (7.49b)

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 175

Equation (7.49b) applies to any rolling body:

a disc, a cylinder, a ring or a sphere.

Example 7.16 Three bodies, a ring, a solid

cylinder and a solid sphere roll down the

same inclined plane without slipping. They

start from rest. The radii of the bodies are

identical. Which of the bodies reaches the

ground with maximum velocity?

Answer We assume conservation of energy of

the rolling body, i.e. there is no loss of energy

due to friction etc. The potential energy lost by

the body in rolling down the inclined plane

(= mgh) must, therefore, be equal to kinetic

energy gained. (See Fig.7.38) Since the bodies

start from rest the kinetic energy gained is equal

to the final kinetic energy of the bodies. From

Eq. (7.49b),

K m

 

= +

 

 

, where v is the

final velocity of (the centre of mass of) the body.

Equating K and mgh,

Fig.7.38

mgh m

 

= +

 

 

2 2

k R

 

 

 

Note is independent of the mass of the

rolling body;

For a ring, k

= R

1 1

ring

For a solid cylinder k

= R

1 1 2

disc

For a solid sphere k

= 2R

1 2 5

sphere

From the results obtained it is clear that among

the three bodies the sphere has the greatest and

the ring has the least velocity of the centre of mass

at the bottom of the inclined plane.

Suppose the bodies have the same mass. Which

body has the greatest rotational kinetic energy while

reaching the bottom of the inclined plane? t

SUMMARY

1. Ideally, a rigid body is one for which the distances between different particles of the

body do not change, even though there are forces on them.

2. A rigid body fixed at one point or along a line can have only rotational motion. A rigid

body not fixed in some way can have either pure translational motion or a combination

of translational and rotational motions.

3. In rotation about a fixed axis, every particle of the rigid body moves in a circle which

lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in

the rotating rigid body has the same angular velocity at any instant of time.

4. In pure translation, every particle of the body moves with the same velocity at any

instant of time.

5. Angular velocity is a vector. Its magnitude is

= d

/dt and it is directed along the axis

of rotation. For rotation about a fixed axis, this vector

ωω

ω has a fixed direction.

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176 PHYSICS

6. The vector or cross product of two vector a and b is a vector written as a×b. The

magnitude of this vector is absinθ and its direction is given by the right handed screw

or the right hand rule.

7. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by

v =

ωω

ω × r, where r is the position vector of the particle with respect to an origin along

the fixed axis. The relation applies even to more general rotation of a rigid body with

one point fixed. In that case r is the position vector of the particle with respect to the

fixed point taken as the origin.

8. The centre of mass of a system of n particles is defined as the point whose position

vector is

∑

9. Velocity of the centre of mass of a system of particles is given by V = P/M, where P is the

linear momentum of the system. The centre of mass moves as if all the mass of the

system is concentrated at this point and all the external forces act at it. If the total

external force on the system is zero, then the total linear momentum of the system is

constant.

10. The angular momentum of a system of n particles about the origin is

L r p

= ×

∑

The torque or moment of force on a system of n particles about the origin is

ττ

= ×

∑

r F

i i

The force F

acting on the i

particle includes the external as well as internal forces.

Assuming Newton’s third law of motion and that forces between any two particles act

along the line joining the particles, we can show

ττ

int

= 0 and

ext

ττ

11. A rigid body is in mechanical equilibrium if

(1) it is in translational equilibrium, i.e., the total external force on it is zero :

F 0

∑

and

(2) it is in rotational equilibrium, i.e. the total external torque on it is zero :

ττ

i i i

= × =

∑ ∑

r F 0

12. The centre of gravity of an extended body is that point where the total gravitational

torque on the body is zero.

13. The moment of intertia of a rigid body about an axis is defined by the formula

I m r

i i

∑

where r

is the perpendicular distance of the ith point of the body from the axis. The

kinetic energy of rotation is

K I

14. The theorem of parallel axes:

z z

I I Ma

′

= +

, allows us to determine the moment of

intertia of a rigid body about an axis as the sum of the moment of inertia of the body

about a parallel axis through its centre of mass and the product of mass and square of

the perpendicular distance between these two axes.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 177

15. Rotation about a fixed axis is directly analogous to linear motion in respect of kinematics

and dynamics.

16. For a rigid body rotating about a fixed axis (say, z-axis) of rotation, L

= I

, where I is

the moment of inertia about z-axis. In general, the angular momentum L for such a

body is not along the axis of rotation. Only if the body is symmetric about the axis of

rotation, L is along the axis of rotation. In that case,

L I

= =L

. The angular

acceleration of a rigid body rotating about a fixed axis is given by I

. If the external

torque τ acting on the body is zero, the component of angular momentum about the

fixed axis (say, z-axis), L

(=I

) of such a rotating body is constant.

17. For rolling motion without slipping v

= R

, where v

is the velocity of translation (i.e.

of the centre of mass), R is the radius and m is the mass of the body. The kinetic energy

of such a rolling body is the sum of kinetic energies of translation and rotation:

2 2

1 1

2 2

K m v I

= +

POINTS TO PONDER

1. To determine the motion of the centre of mass of a system no knowledge of internal

forces of the system is required. For this purpose we need to know only the external

forces on the body.

2. Separating the motion of a system of particles as the motion of the centre of mass, (i.e.,

the translational motion of the system) and motion about (i.e. relative to) the centre of

mass of the system is a useful technique in dynamics of a system of particles. One

example of this technique is separating the kinetic energy of a system of particles K as

the kinetic energy of the system about its centre of mass K′ and the kinetic energy of

the centre of mass MV

/2,

K = K

′

+ MV

3. Newton’s Second Law for finite sized bodies (or systems of particles) is based in Newton’s

Second Law and also Newton’s Third Law for particles.

4. To establish that the time rate of change of the total angular momentum of a system of

particles is the total external torque in the system, we need not only Newton’s second

law for particles, but also Newton’s third law with the provision that the forces between

any two particles act along the line joining the particles.

5. The vanishing of the total external force and the vanishing of the total external torque

are independent conditions. We can have one without the other. In a couple, total

external force is zero, but total torque is non-zero.

6. The total torque on a system is independent of the origin if the total external force is

zero.

7. The centre of gravity of a body coincides with its centre of mass only if the gravitational

field does not vary from one part of the body to the other.

8. The angular momentum L and the angular velocity

ωω

ω are not necessarily parallel vectors.

However, for the simpler situations discussed in this chapter when rotation is about a

fixed axis which is an axis of symmetry of the rigid body, the relation L = I

ωω

ω holds good,

where I is the moment of the inertia of the body about the rotation axis.

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178 PHYSICS

EXERCISES

7.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv)

cube, each of uniform mass density. Does the centre of mass of a body necessarily

lie inside the body ?

7.2 In the HCl molecule, the separation between the nuclei of the two atoms is about

1.27 Å (1 Å = 10

-10

m). Find the approximate location of the CM of the molecule,

given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and

nearly all the mass of an atom is concentrated in its nucleus.

7.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V

on a smooth horizontal floor. If the child gets up and runs about on the trolley in any

manner, what is the speed of the CM of the (trolley + child) system ?

7.4 Show that the area of the triangle contained between the vectors a and b is one half

of the magnitude of a × b.

7.5 Show that a

(b × c) is equal in magnitude to the volume of the parallelepiped formed

on the three vectors , a, b and c.

7.6 Find the components along the x, y, z axes of the angular momentum l of a particle,

whose position vector is r with components x, y, z and momentum is p with

components p

, p

and p

. Show that if the particle moves only in the x-y plane the

angular momentum has only a z-component.

7.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel

lines separated by a distance d. Show that the angular momentum vector of the two

particle system is the same whatever be the point about which the angular momentum

is taken.

7.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible

weight as shown in Fig.7.39. The angles made by the strings with the vertical are

36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the

centre of gravity of the bar from its left end.

Fig. 7.39

7.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its

centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the

level ground on each front wheel and each back wheel.

7.10 (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the

moment of inertia of the sphere about any of its diameters to be 2MR

/5, where

M is the mass of the sphere and R is the radius of the sphere.

(b) Given the moment of inertia of a disc of mass M and radius R about any of its

diameters to be MR

/4, find its moment of inertia about an axis normal to the

disc and passing through a point on its edge.

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 179

7.11 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both

having the same mass and radius. The cylinder is free to rotate about its standard

axis of symmetry, and the sphere is free to rotate about an axis passing through its

centre. Which of the two will acquire a greater angular speed after a given time.

7.12 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s

-1

The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the

rotation of the cylinder? What is the magnitude of angular momentum of the cylinder

about its axis?

7.13 (a) A child stands at the centre of a turntable with his two arms outstretched. The

turntable is set rotating with an angular speed of 40 rev/min. How much is the

angular speed of the child if he folds his hands back and thereby reduces his

moment of inertia to 2/5 times the initial value ? Assume that the turntable

rotates without friction.

(b) Show that the child’s new kinetic energy of rotation is more than the initial

kinetic energy of rotation. How do you account for this increase in kinetic energy?

7.14 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius

40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a

force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no

slipping.

7.15 To maintain a rotor at a uniform angular speed of 200 rad s

-1

, an engine needs to

transmit a torque of 180 N m. What is the power required by the engine ?

(Note: uniform angular velocity in the absence of friction implies zero torque. In

practice, applied torque is needed to counter frictional torque). Assume that the

engine is 100% efficient.

7.16 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre

of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity

of the resulting flat body.

7.17 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass

5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be

balanced at 45.0 cm. What is the mass of the metre stick?

7.18 A solid sphere rolls down two different inclined planes of the same heights but

different angles of inclination. (a) Will it reach the bottom with the same speed in

each case? (b) Will it take longer to roll down one plane than the other? (c) If so,

which one and why?

7.19 A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre

of mass has a speed of 20 cm/s. How much work has to be done to stop it?

7.20 The oxygen molecule has a mass of 5.30 × 10

-26

kg and a moment of inertia of

1.94×10

-46

kg m

about an axis through its centre perpendicular to the lines joining

the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and

that its kinetic energy of rotation is two thirds of its kinetic energy of translation.

Find the average angular velocity of the molecule.

7.21 A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom

of the inclined plane the centre of mass of the cylinder has a speed of 5 m/s.

(a) How far will the cylinder go up the plane?

(b) How long will it take to return to the bottom?

Additional Exercises

7.22 As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and

hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from

a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless

and neglecting the weight of the ladder, find the tension in the rope and forces

exerted by the floor on the ladder. (Take g = 9.8 m/s

)

(Hint: Consider the equilibrium of each side of the ladder separately.)

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180 PHYSICS

Fig.7.40

7.23 A man stands on a rotating platform, with his arms stretched horizontally holding a

5 kg weight in each hand. The angular speed of the platform is 30 revolutions per

minute. The man then brings his arms close to his body with the distance of each

weight from the axis changing from 90cm to 20cm. The moment of inertia of the

man together with the platform may be taken to be constant and equal to 7.6 kg m

(a) What is his new angular speed? (Neglect friction.)

(b) Is kinetic energy conserved in the process? If not, from where does the change

come about?

7.24 A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded

exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is

hinged at one end and rotates about a vertical axis practically without friction. Find

the angular speed of the door just after the bullet embeds into it.

(Hint: The moment of inertia of the door about the vertical axis at one end is ML

/3.)

7.25 Two discs of moments of inertia I

and I

about their respective axes (normal to the

disc and passing through the centre), and rotating with angular speeds

and

are brought into contact face to face with their axes of rotation coincident. (a) What

is the angular speed of the two-disc system? (b) Show that the kinetic energy of the

combined system is less than the sum of the initial kinetic energies of the two discs.

How do you account for this loss in energy? Take

≠

7.26 (a) Prove the theorem of perpendicular axes.

(Hint : Square of the distance of a point (x, y) in the x–y plane from an axis through

the origin and perpendicular to the plane is x

(b) Prove the theorem of parallel axes.

(Hint : If the centre of mass of a system of n particles is chosen to be the origin

i i

r =

∑

7.27 Prove the result that the velocity v of translation of a rolling body (like a ring, disc,

cylinder or sphere) at the bottom of an inclined plane of a height h is given by

( )

2 2

1 /

k R

using dynamical consideration (i.e. by consideration of forces and torques). Note k is

the radius of gyration of the body about its symmetry axis, and R is the radius of the

body. The body starts from rest at the top of the plane.

7.28 A disc rotating about its axis with angular speed

is placed lightly (without any

translational push) on a perfectly frictionless table. The radius of the disc is R. What

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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 181

are the linear velocities of the points A, B and C on the disc shown in Fig. 7.41? Will the disc roll in

the direction indicated ?

Fig. 7.41

7.29 Explain why friction is necessary to make the disc in Fig. 7.41 roll in the direction indicated.

(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling

begins.

(b) What is the force of friction after perfect rolling begins ?

7.30 A solid disc and a ring, both of radius 10 cm are placed on a horizontal table

simultaneously, with initial angular speed equal to 10 π rad s

-1

. Which of the two will start to roll

earlier ? The co-efficient of kinetic friction is

= 0.2.

7.31 A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30

. The co-

efficient of static friction

= 0.25.

(a) How much is the force of friction acting on the cylinder ?

(b) What is the work done against friction during rolling ?

of the plane is increased, at what value of

does the cylinder begin to skid,

and not roll perfectly ?

7.32 Read each statement below carefully, and state, with reasons, if it is true or false;

(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM

of the body.

(b) The instantaneous speed of the point of contact during rolling is zero.

(d) For perfect rolling motion, work done against friction is zero.

(e) A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling)

motion.

7.33 Separation of Motion of a system of particles into motion of the centre of mass and motion about the

centre of mass :

(a) Show

p p V= ′ +

i i

where p

is the momentum of the ith particle (of mass m

) and p

′′

′

= m

′′

′

. Note v

′′

′

is the velocity

of the ith particle relative to the centre of mass.

Also, prove using the definition of the centre of mass

∑

′ =p

(b) Show

½K K MV

′

= +

where K is the total kinetic energy of the system of particles,

′

is the total kinetic energy of the

system when the particle velocities are taken with respect to the centre of mass and MV

/2 is the

kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the

system). The result has been used in Sec. 7.14.

L L R V= ′ + × M

where

′L

′ × ′

∑

r p

is the angular momentum of the system about the centre of mass with

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182 PHYSICS

velocities taken relative to the centre of mass. Remember

–

i i

′

r r R

; rest of the

notation is the standard notation used in the chapter. Note

′

and

R V

can

be said to be angular momenta, respectively, about and of the centre of mass of

the system of particles.

(d) Show

′

= ′ ×

′

∑

Further, show that

ext

′

= ′

ττ

where

ext

′

ττ

is the sum of all external torques acting on the system about the

centre of mass.

(Hint : Use the definition of centre of mass and third law of motion. Assume the

internal forces between any two particles act along the line joining the particles.)

Pluto - A Dwarf Planet

The International Astronomical Union (IAU) at the IAU 2006 General Assembly

held on August 24, 2006, in Prague in Czech Republic, adopted a new

definition of planets in our Solar System. According to the new definition,

Pluto is no longer a planet. This means that the Solar System consists of

eight planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and

Neptune. According to the IAU usage, the ‘planet’ and ‘other bodies’ in our

Solar System, except satellites, are to be defined into three distinct categories

of celestial objects in the following way:

1. A ‘planet’ is a celestial body that (a) is in orbit around the Sun, (b) has

sufficient mass for its self-gravity to overcome rigid body forces so that it

assumes a hydrostatic equilibrium (nearly round) shape, and (c) has

cleared the neighbourhood around its orbit.

2. A ‘dwarf planet’ is a celestial body that (a) is in orbit around the Sun,

(b) has sufficient mass for its self-gravity to overcome rigid body forces so

that it assumes a hydrostatic equilibrium (nearly round) shape, (c) has

not cleared the neighbourhood around its orbit, and (d) is not a satellite.

3. All ‘other objects’, except satellites, orbiting the Sun, shall be referred to

collectively as ‘Small Solar-System Bodies’.

Unlike other eight planets in the Solar System, Pluto’s orbital path overlaps

with ‘other objects’ and the planet Neptune. The ‘other objects’ currently

include most of the Solar System asteroids, most of the Trans-Neptunian

Objects (TNOs), comets, and other small bodies.

Pluto is a ‘dwarf planet’ by the above definition and is recognised as the

prototype of a new category of Trans-Neptunian Objects.

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