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
1
or P
2
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
1
, P
2
,
P
3
and P
4
have different velocities (shown by arrows)
at any instant of time. In fact, the velocity of
the point of contact P
3
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
1
be a particle of the rigid body, arbitrarily chosen
and at a distance r
1
from fixed axis. The particle
P
1
describes a circle of radius r
1
with its centre
C
1
on the fixed axis. The circle lies in a plane
perpendicular to the axis. The figure also shows
another particle P
2
of the rigid body, P
2
is at a
distance r
2
from the fixed axis. The particle P
2
moves in a circle of radius r
2
and with centre C
2
on the axis. This circle, too, lies in a plane
perpendicular to the axis. Note that the circles
described by P
1
and P
2
may lie in different planes;
both these planes, however, are perpendicular
to the fixed axis. For any particle on the axis
like P
3
, 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
1
or
P
2
describes a circle with its centre (C
1
or C
2
) on the axis of rotation. The radius of
the circle (r
1
or r
2
) is the perpendicular
distance of the point (P
1
or P
2
) from the
axis. A point on the axis like P
3
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
1
and x
2
respectively from some origin O. Let m
1
and m
2
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
Tr
1
and Tr
2
of the body. The positions O and P at
three different instants of time are shown by O
1
, O
2
,
and O
3
, and P
1
, P
2
and P
3
, 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.
α
1
=
α
2
=
α
3
. 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,
α
1
,
α
2
and
α
3
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
X
m m
+
=
+
(7.1)
In Eq. (7.1), X can be regarded as the mass-
weighted mean of x
1
and x
2
. If the two particles
have the same mass m
1
= m
2
= m
,
then
1 2 1 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
1
, m
2
,
...m
n
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.
X
m x m x m x
m m m
m x
m
m x
m
n n
n
i i
i
n
i
i
n
i i
i
= = =
=
=
1 1 2 2
1 2
1
1
+ + ... +
+ +... +
(7.2)
where x
1
, x
2
,...x
n
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
i
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
1
,y
1
), (x
2
,y
2
) and (x
3
,y
3
) respectively.
Let the masses of the three particles be m
1
, m
2
and m
3
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
m x m x m x
(7.3a)
1 1 2 2 3 3
1 2 3
m y m y m y
Y
m m m
+ +
=
+ +
(7.3b)
For the particles of equal mass m = m
1
= m
2
= m
3
,
1 2 3 1 2 3
( )
3 3
m x x x x x x
X
m
+ + + +
= =
1 2 3 1 2 3
( )
3 3
m y y y y y y
Y
m
+ + + +
= =
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
X
M
=
(7.4a)
i i
m y
Y
M
=
(7.4b)
and
i i
m z
Z
M
=
(7.4c)
Here M =
i
m
is the total mass of the
system. The index i runs from 1 to n; m
i
is the
mass of the i
th
particle and the position of the
i
th
particle is given by (x
i
, y
i
, z
i
).
Eqs. (7.4a), (7.4b) and (7.4c) can be
combined into one equation using the notation
of position vectors. Let
i
r
be the position vector
of the i
th
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
m
M
=
r
R
(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
0
i i
m
=
r
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
1
,m
2
... m
n
; the
i
th
element m
i
is taken to be located about the
point (x
i
, y
i
, z
i
). 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
i
smaller and smaller, these expressions
become exact. In that case, we denote the sums
over i by integrals. Thus,
m m M
i
=
d ,
( ) ,m x x m
i i
d
( ) ,m y y m
i i
d
and
(m z z m
i i
)
d
Here M is the total mass of the body. The
coordinates of the centre of mass now are
X
M
x m Y
M
y m Z
M
z m= = =
1 1 1
d d d, and
(7.5a)
The vector expression equivalent to these
three scalar expressions is
R r=
1
M
md
(7.5b)
If we choose, the centre of mass as the origin
of our coordinate system,
=
R 0
i.e.,
r 0dm =
or
x m y m z md d d= = =
0
(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.
u
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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 147
u
u
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
3
). Let the masses 100 g,
150g and 200g be located at O, A and B be
respectively. Then,
1 1 2 2 3 3
m x m x m x
(
)
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
Y
+ +
=
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
1
, C
2
and C
3
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
X
+ +
=
+ +
5
m
6
=
[
]
( )
1(1/2) 1(1/2) 1(3/2) kg m
5
m
1 1 1 kg 6
Y
+ +
= =
+ +
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?
t
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
1 2
d
d d
d
...
d d d
n
n
M m m m
t t t dt
= + + +
r
r r
R
or
1 1 2 2
...
n n
M m m m
= + + +
V v v v
(7.8)
where
(
)
1 1
d /d
t
=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
t
=
V R
is the
velocity of the centre of mass. Note that we
assumed the masses m
1
, m
2
, ... 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
1 2
dd d
d
...
d d d d
n
n
M m m m
t t t t
= + + +
v
v v
V
or
1 1 2 2
...
n n
M m m m= + + +
A a a a
(7.9)
where
(
)
1 1
d /d
t
=a v
is the acceleration of the
first particle,
(
)
2 2
d /d
t
=a v
is the acceleration
of the second particle etc. and
(
)
d / d
t
=A V
is
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
m=
F a
.
The force acting on the second particle is given
by
2 2 2
m=
F a
and so on. Eq. (7.9) may be written
as
1 2
...
n
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
1
F
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
F
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
m
=
p v
(7.12)
Let us also recall that Newton’s second law
written in symbolic form for a single particle is
d
d
t
=
p
F
(7.13)
where F is the force on the particle. Let us
consider a system of n particles with masses m
1
,
m
2
,...m
n
respectively and velocities
1 2
, ,.......
n
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
m
v
,
of the second particle is
2 2
m
v
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
...
n
= + + +
P p p p
1 1 2 2
...
n n
m m m= + + +
v v v
(7.14)
Comparing this with Eq. (7.8)
M
=
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
d d
M M
t t
= =
P V
A
(7.16)
Comparing Eq.(7.16) and Eq. (7.11),
d
d
ext
t
=
P
F
(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)
or
d
0
d
t
=
P
P
= 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,
P
x
= c
1
, P
y
= c
2
and P
z
= c
3
(7.18 b)
Here P
x
, P
y
and P
z
are the components of the
total linear momentum vector P along the x–, y–
and z–axes respectively; c
1
, c
2
and c
3
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
1
(dotted
line) and S
2
(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
ab
θ
=
where a and b
are magnitudes of a and b and
θ
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
θ
(as
shown) and (360
0
θ
). While applying either of
the above rules, the rotation should be taken
through the smaller angle (<180
0
) between a
and b. It is
θ
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
ab
θ
); 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
2
sin0 0
a
° =
.
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152 PHYSICS
u
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
0
or 1, since
ˆ
i
and
ˆ
j
both have unit
magnitude and the angle between them is 90
0
.
Thus,
ˆ ˆ
×
i j
is a unit vector. A unit vector
perpendicular to the plane of
ˆ
i
and
ˆ
j
and
related to them by the right hand screw rule is
ˆ
k
. 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
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
ˆ
– 4j
ˆ
+ 5k
ˆ
)
and b = (– 2i
ˆ
+ j
ˆ
– 3k
ˆ
)
Answer
ˆ ˆ ˆ ˆ ˆ ˆ
(3 4 5 ) ( 2 3 )
6 4 15
25
= + +
=
=
a b i j k i j k
i i
ˆ ˆ ˆ
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
t
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
θ
of the
particle in time t. The average angular velocity
of the particle over the interval t is
θ
/t. As
t tends to zero (i.e. takes smaller and smaller
values), the ratio
θ
/t approaches a limit which
is the instantaneous angular velocity d
θ
/dt of
the particle at the position P. We denote the
instantaneous angular velocity by
ω
(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
ω
by the simple relation
r
υ ω
=
, where r is the
radius of the circle.
We observe that at any given instant the
relation
v r
ω
=
applies to all particles of the
rigid body. Thus for a particle at a perpendicular
distance r
i
from the fixed axis, the linear velocity
at a given instant v
i
is given by
i i
v r
ω
=
(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,
0
=
r
, and hence
v =
ω
r = 0. Thus, particles on the axis are
stationary. This verifies that the axis is fixed.
Note that we use the same angular velocity
ω
for all the particles. We therefore, refer to
ωω
ωω
ω
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
ωω
ω
. If the sense (clockwise or
anticlockwise) of rotation of the body
changes, so does the direction of
ωω
ωω
ω
.
Fig. 7.17 (b) The angular velocity vector
ωω
ωω
ω is
directed along the fixed axis as shown.
The linear velocity of the particle at P
is v =
ωω
ωω
ω ×
r. It is perpendicular to both
ω ω
ω 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
ωω
ω
×
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
ωω
ω directed along the fixed (z–) axis and
also the position vector r =
OP
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
ωω
ω × r =
ωω
ωω
ω × OP =
ωω
ωω
ω × (OC + CP)
But
ωω
ωω
ω × OC =
0 0
0 0
0 as
ω ω
ω ω
ω is along OC
Hence
ωω
ωω
ω × r =
ωω
ωω
ω × CP
The vector
ωω
ωω
ω
×
CP is perpendicular to
ωω
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
ωω
ω
×
CP is ω (CP) since
ωω
ωω
ω and CP are perpendicular to each other. We
shall denote CP by
r
and not by r, as we did
earlier.
Thus,
ωω
ωω
ω
×
r is a vector of magnitude
ω
r
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
ω 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
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 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
θθ
θ) and angular velocity (
ωω
ω
ω
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
αα
α as the time rate of change of
angular velocity; Thus,
d
d
t
=
ω
ωω
ω
α
αα
α
(7.21)
If the axis of rotation is fixed, the direction
of
ω ω
ω ω
ω and hence, that of
αα
αα
α is fixed. In this case
the vector equation reduces to a scalar equation
d
d
t
ω
α
=
(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 = τ =
τ = r
×
×
× ×
× F,
τ τ
τ 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
ττ
τ = r
× F (7.23)
The moment of force (or torque) is a vector
quantity. The symbol
τ τ
τ τ
τ stands for the Greek
letter tau. The magnitude of
τ τ
τ τ
τ is
τ
= r F sin
θ
(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
θ
is the angle between r and F as
shown.
Moment of force has dimensions M L
2
T
-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)
or
sin
r F rF
τ θ
= =
(7.24c)
where
r
= r sin
θ
is the perpendicular distance
of the line of action of F from the origin and
( sin )
F F
θ
=
is the component of F in the
direction perpendicular to r. Note that
τ
= 0 if
r = 0, F = 0 or
θ
= 0
0
or 180
0
. 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
=
θ
(7.26a)
where p is the magnitude of p and
θ
is the angle
between r and p. We may write
l r p
=
or
r p
(7.26b)
where
r
(= r sinθ) is the perpendicular distance
of the directional line of p from the origin and
( sin )
p p
θ
=
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
θ
= 0
0
or 180
0
.
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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
l
= r × p with respect to time,
d d
( )
d d
= ×
r p
t t
l
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
d
0,
d
m
t
× = × =
r
p v v
as the vector product of two parallel vectors
vanishes. Further, since dp / dt = F,
r
p
r F× = × =
d
dt
ττ
Hence
d
dt
( )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
th
particle
is given by
l
i
= r
i
× p
i
where r
i
is the position vector of the i
th
particle
with respect to a given origin and p = (m
i
v
i
) is
the linear momentum of the particle. (The
particle has mass m
i
and velocity v
i
) 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
d
d
d
d
d
d
L
t t t
i i
ii
=
( )
= =
l
l
τ
(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
2020-21
SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 157
u
u
where
ττ
τ
τ
τ
i
is the torque acting on the i
th
particle;
ττ
i i i
= ×r F
The force F
i
on the i
th
particle is the vector
sum of external forces
F
i
ext
acting on the particle
and the internal forces
int
i
F
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
= = ×
i
i
i
i
i
r F
as
ττ ττ ττ
= +
ext int
,
where
ττ
ext i i
ext
i
= ×
r F
and
ττ
int
int
= ×
r F
i 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
ττ
τ
int
=
0 and therefore
ττ
ττ
τ
=
τ τ
τ τ
τ
ext
.
Since
ττ ττ
=
i
, it follows from Eq. (7.28a)
that
d
d
L
t
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
d
d
P
F
t
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
If
τ
τ
τ
τ
τ
ext
= 0, Eq. (7.28b) reduces to
d
0
d
t
=
L
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,
L
x
= K
1
, L
y
= K
2
and L
z
= K
3
(7.29 b)
Here K
1
, K
2
and K
3
are constants; L
x
, L
y
and
L
z
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
7i
ˆ
+ 3j
ˆ
– 5k
ˆ
about the origin. The force
acts on a particle whose position vector is
i
ˆ
j
ˆ
+ k
ˆ
.
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
or
ττ
= + +2 12 10
ˆ ˆ ˆ
i j k
t
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
θ
, where
θ
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
θ
. 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
1
+ + + = =
=
...
n i
i
n
0
(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
1
+ + + = =
=
...
n i
i
n
0
(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
F
ix
i
n
=
=
0
1
,
F
iy
i
n
=
=
0
1
and
F
iz
i
n
=
=
0
1
(7.31a)
where F
ix
, F
iy
and F
iz
are respectively the x, y
and z components of the forces F
i
. Similarly,
Eq. (7.30b) is equivalent to three scalar
equations
τ
ix
i
n
=
=
0
1
,
τ
iy
i
n
=
=
0
1
and
τ
iz
i
n
=
=
0
1
(7.31b)
where
τ
ix
,
τ
iy
and
τ
iz
are respectively the x, y and
z components of the torque
ττ
ττ
τ
i
.
2020-21
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
u
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
1
and r
2
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
1
× (–F) + r
2
× F
= r
2
× F r
1
× F
= (r
2
r
1
)
× F
But r
1
+ AB = r
2
, and hence AB = r
2
r
1
.
The moment of the couple, therefore, is
AB
× 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
1
and F
2
, parallel
to each other and usually perpendicular to the
lever, as shown here, act on the lever at
distances d
1
and d
2
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
1
and F
2
. For translational equilibrium,
RF
1
F
2
= 0 (i)
For considering rotational equilibrium we
take the moments about the fulcrum; the sum
of moments must be zero,
d
1
F
1
d
2
F
2
= 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
1
is usually
some weight to be lifted. It is called the load
and its distance from the fulcrum d
1
is called
the load arm. Force F
2
is the effort applied to lift
the load; distance d
2
of the effort from the
fulcrum is the effort arm.
Eq. (ii) can be written as
d
1
F
1
=
d
2
F
2
(7.32a)
or load arm
× load = effort arm
× effort
The above equation expresses the principle
of moments for a lever. Incidentally the ratio
F
1
/F
2
is called the Mechanical Advantage (M.A.);
M.A. =
1 2
2 1
F d
F d
=
(7.32b)
If the effort arm d
2
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
2020-21
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
1
and F
2
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
1
g, m
2
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
1
g, m
2
g
…. etc. is zero.
If r
i
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
ττ
τ
i
= r
i
× m
i
g. The total
gravitational torque about the CG is zero, i.e.
ττ ττ
g
i
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,
m
i i
r
= 0. Remember that the position vectors
(r
i
) 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
1
through the point
of suspension of the body A.
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162 PHYSICS
u
u
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
AA
1
. 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
1
and K
2
, 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
1
= BK
2
= 10 cm and K
1
G
= K
2
G = 25 cm. Also, W= weight of the rod =
4.00 kg and W
1
= suspended load = 6.00 kg;
R
1
and R
2
are the normal reactions of the
support at the knife edges.
For translational equilibrium of the rod,
R
1
+R
2
–W
1
–W = 0 (i)
Note W
1
and W act vertically down and R
1
and R
2
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
R
2
and W
1
are anticlockwise (+ve), whereas the
moment of R
1
is clockwise (-ve).
For rotational equilibrium,
–R
1
(K
1
G) + W
1
(PG) + R
2
(K
2
G) = 0 (ii)
It is given that W = 4.00g N and W
1
= 6.00g
N, where g = acceleration due to gravity. We
take g = 9.8 m/s
2
.
With numerical values inserted, from (i)
R
1
+ R
2
– 4.00g – 6.00g = 0
or R
1
+ R
2
= 10.00g N (iii)
= 98.00 N
From (ii), – 0.25 R
1
+ 0.05 W
1
+ 0.25 R
2
= 0
or R
1
R
2
= 1.2g N = 11.76 N (iv)
From (iii) and (iv), R
1
= 54.88 N,
R
2
= 43.12 N
Thus the reactions of the support are about
55 N at K
1
and 43 N at K
2
. 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
2020-21
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
1
and F
2
of the wall
and the floor respectively. Force F
1
is
perpendicular to the wall, since the wall is
frictionless. Force F
2
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,
NW = 0 (i)
Taking the forces in the horizontal direction,
F – F
1
= 0 (ii)
For rotational equilibrium, taking the
moments of the forces about A,
1
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)
1
4 2 196.0/4 2 34.6 N
F W= = =
From (ii)
1
34.6 N
F F= =
2 2
2
199.0
F F N= + =
N
The force F
2
makes an angle
α
with the
horizontal,
1
tan 4 2 , tan (4 2) 80
N F
α α
= = =
t
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
r
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
i
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
1
( )
2
n n
i i i
i i
K k m r
ω
= =
= =
Here n is the number of particles in the body.
Note
ω
is the same for all particles. Hence, taking
ω
out of the sum,
2 2
1
1
( )
2
n
i i
i
K m r
ω
=
=
We define a new parameter characterising
the rigid body, called the moment of inertia I
,
given by
2
1
n
i i
i
I m r
=
=
(7.34)
With this definition,
2
1
2
K I
ω
=
(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,
2
1
2
K m
υ
=
Here, m is the mass of the body and v is its
velocity. We have already noted the analogy
between angular velocity
ω
(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
ω
. Each mass element
of the ring is at a distance R from the axis,
and moves with a speed R
ω
. 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
2
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)
2
+ (M/2)(l/2)
2
Thus, for the pair of masses, rotating about
the axis through the centre of mass
perpendicular to the rod
I = Ml
2
/ 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
2
, where k has
the dimension of length. For a rod, about
the perpendicular axis at its midpoint,
i.e.
2 2
12,
k L=
=
12
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
ML
2
and its SI units are kg m
2
.
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
2
ring, radius R plane, at centre
(2) Thin circular Diameter M R
2
/2
ring, radius R
(3) Thin rod, Perpendicular to M L
2
/12
length L rod, at mid point
(4) Circular disc, Perpendicular to M R
2
/2
radius R disc at centre
(5) Circular disc, Diameter M R
2
/4
radius R
(6) Hollow cylinder, Axis of cylinder M R
2
radius R
(7) Solid cylinder, Axis of cylinder M R
2
/2
radius R
(8) Solid sphere, Diameter 2 M R
2
/5
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
u
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
/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
I
x
=
I
y
and I
z
= 2I
x
But I
z
= MR
2
/2
So finally, I
x
= I
z
/2 = MR
2
/4
Thus the moment of inertia of a disc about
any of its diameter is MR
2
/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
u
u
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
I
z
=
I
z
+
Ma
2
(7.37)
where I
z
and
I
z
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
2
/12. Using the parallel axes theorem,
I
= I + Ma
2
with a = l/2 we get,
2
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
= × =
t
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 MR
diatangent
= + = + =
2
2
2 2
2
3
2
.
t
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
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
θ
in the plane it moves is the
angular displacement of the whole body;
θ
is
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
θ
0
, the angular
displacement at t = 0.
We also recall that the angular velocity is
the time rate of change of angular displacement,
ω
= d
θ
/dt. Note since the axis of rotation is fixed,
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168 PHYSICS
u
u
there is no need to treat angular velocity as a
vector. Further, the angular acceleration,
α
=
d
ω
/dt.
The kinematical quantities in rotational
motion, angular displacement (
θ
), angular
velocity (
ω
) and angular acceleration (
α
)
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
0
+ at (a)
2
0 0
1
2
x x t at
υ
= + +
(b)
2 2
0
2
ax
υ υ
= +
(c)
where x
0
= initial displacement and v
0
= 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:
0
t
= +
ω ω α
(7.38)
2
0 0
1
2
t t
= + +
θ θ ω α
(7.39)
and
2 2
0 0
2 ( )
= +
ω ω α θ θ
(7.40)
where
θ
0
= initial angular displacement of the
rotating body, and
ω
0
= 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
d
d
constant
t
ω
α
= =
(i)
Integrating this equation,
ω α
= +
dt c
(as is constant)
t c
α α
= +
At t = 0,
ω
=
ω
0
(given)
From (i) we get at t = 0,
ω
= c =
ω
0
Thus,
ω
=
α
t +
ω
0
as required.
With the definition of
ω
= d
θ
/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
ω
=
ω
0
+
α
t
ω
0
= initial angular speed in rad/s
= 2
π
× angular speed in rev/s
=
2 angular speed in rev/min
60 s/min
π
×
=
2 1200
rad/s
60
π
×
= 40
π
rad/s
Similarly
ω
= final angular speed in rad/s
=
2 3120
rad/s
60
π
×
= 2
π
× 52 rad/s
= 104
π
rad/s
Angular acceleration
0
t
ω ω
α
=
= 4
π
rad/s
2
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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 169
The angular acceleration of the engine
= 4π rad/s
2
(ii) The angular displacement in time t is
given by
2
0
1
2
t t
θ ω α
= +
2
1
(40 16 4 16 )
2
π π
= × + × ×
rad
(640 512 )
π π
= +
rad
= 1152π rad
Number of revolutions =
1152
576
2
π
π
=
t
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
d
τ θ
,
since we already know the correspondence
d d
x
θ
and
F
τ
. 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
1
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
1
P
1
(ds
1
) 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
1
be one such
typical force acting as shown on a particle of
the body at point P
1
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
1
describes a circular path of radius r
1
with centre
C on the axis; CP
1
= r
1
.
In time t, the point moves to the position
P
1
. The displacement of the particle ds
1
,
therefore, has magnitude ds
1
= r
1
d
θ
and
direction tangential at P
1
to the circular path
as shown. Here d
θ
is the angular displacement
of the particle, d
θ
=
1 1
P CP
.The work done by
the force on the particle is
dW
1
= F
1
. ds
1
= F
1
ds
1
cos
φ
1
= F
1
(r
1
d
θ
)sin
α
1
where
φ
1
is the angle between F
1
and the tangent
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170 PHYSICS
at P
1,
and
α
1
is the angle between F
1
and the
radius vector OP
1
;
φ
1
+
α
1
= 90°.
The torque due to F
1
about the origin is
OP
1
× F
1
. Now OP
1
= OC + OP
1
. [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
1
is
ττ
ττ
τ
1
= CP × F
1
; it is directed along the axis of rotation
and has a magnitude
τ
1
= r
1
F
1
sinα , Therefore,
dW
1
=
τ
1
d
θ
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
τ
1
,
τ
2
, … etc,
1 2
d ( ...)d
W
τ τ θ
= + +
Remember, the forces giving rise to the
torques act on different particles, but the
angular displacement d
θ
is the same for all
particles. Since all the torques considered are
parallel to the fixed axis, the magnitude
τ
of the
total torque is just the algebraic sum of the
magnitudes of the torques, i.e.,
τ
=
τ
1
+
τ
2
+ .....
We, therefore, have
d d
W
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
d d
W
P
t t
θ
τ τω
= = =
or
P
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
d
d
d
dt
I
I
t
ω ω ω
2
2
2
2
=
( )
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 ,
t
α ω
=
we get
d
dt
I
I
ω
ω α
2
2
=
Equating rates of work done and of increase
in kinetic energy,
I
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
θ
2 Velocity v = dx/dt Angular velocity
ω
= d
θ
/dt
3 Acceleration a = dv/dt Angular acceleration
α
= d
ω
/dt
4 Mass M Moment of inertia I
5 Force F = Ma Torque
τ
= I
α
6 Work dW = F ds Work W =
τ
d
θ
7 Kinetic energy K = Mv
2
/2 Kinetic energy K = I
ω
2
/2
8 Power P = F v Power P =
τω
9 Linear momentum p = Mv Angular momentum L = I
ω
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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 171
u
I
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.
(c) Find also the kinetic energy of the
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
α
=
τ
the torque
τ
= F R
= 25 × 0.20 Nm (as R = 0.20m)
= 5.0 Nm
I = Moment of inertia of flywheel about its
axis
2
2
MR
=
=
2
20.0 (0.2)
2
×
= 0.4 kg m
2
α
= angular acceleration
= 5.0 N m/0.4 kg m
2
= 12.5 s
–2
(b) Work done by the pull unwinding 2m of the
cord
= 25 N × 2m = 50 J
(c) Let
ω
be the final angular velocity. The
kinetic energy gained =
2
1
2
I
ω
,
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
ω
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
i
N
1
(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 =
ω
r
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)
is
ˆ
k
. Hence
(
)
ˆ
× =
CP v k
m r mv
=
2
ˆ
ω
k
mr
(since
υ
=
ω
r
)
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
z
, then
z
= ×
CP v
ml
=
2
ˆ
ω
k
mr
and
= + ×
OC v
z
m
l l
We note that l
z
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
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
L
and
z
L
the components of
L
respectively perpendicular to the z-axis and
along the z-axis;
L OC v
= ×
i i i
m
(7.44a)
where m
i
and v
i
are respectively the mass and
the velocity of the i
th
particle and C
i
is the centre
of the circle described by the particle;
and
or
ˆ
ω
=
L k
z
I
(7.44b)
The last step follows since the perpendicular
distance of the i
th
particle from the axis is r
i
;
and by definition the moment of inertia of the
body about the axis of rotation is
I m r
i i
=
2
.
Note
z
= +
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
i
, for every particle which has a
velocity v
i
, there is another particle of velocity
v
i
located diametrically opposite on the circle
with centre C
i
described by the particle. Together
such pairs will contribute zero to
L
and as a
result for symmetric bodies
L
is zero, and
hence
ˆ
ω
= =
L L k
z
I
(7.44d)
For bodies, which are not symmetric about
the axis of rotation, L is not equal to L
z
and
hence L does not lie along the axis of rotation.
Referring to Table 7.1, can you tell in which
cases L = L
z
will not apply?
Let us differentiate Eq. (7.44b). Since
ˆ
k
is a
fixed (constant) vector, we get
d
d
d
d
z
t t
IL k
( )
=
( )
ω
ˆ
Now, Eq. (7.28b) states
d
d
t
=
L
τ
ττ
τ
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
ˆ
τ
=
k
τ
ττ
τ
.
Since
z
= +
L L L
and the direction of L
z
(vector
ˆ
k
) is fixed, it follows that for rotation about a
fixed axis,
d
ˆ
d
τ
=
L
k
z
t
(7.45a)
and
d
0
d
t
=
L
(7.45b)
Thus, for rotation about a fixed axis, the
component of angular momentum
perpendicular to the fixed axis is constant. As
ˆ
ω
=
L k
z
I
, we get from Eq. (7.45a),
(
)
d
d
I
t
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
d d
I I I
t t
ω
ω α
= =
and we get from Eq. (7.45c),
I
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,
L
z
= I
ω
= constant (7.46)
For symmetric bodies, from Eq. (7.44d), L
z
may be replaced by L .(L and L
z
are respectively
the magnitudes of L and L
z
.)
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
ω
is constant. Stretching the arms
increases I about the axis of rotation, resulting
in decreasing the angular speed
ω
. 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
0
of the disc with the
surface is at rest; the centre of mass of
the disc moves with velocity, v
cm
. The disc
rotates with angular velocity
ω
about its
axis which passes through C; v
cm
=R
ω
,
where R is the radius of the disc.
Let v
cm
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
cm
is
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
0
, P
1
or P
2
, consists of two parts, one is the
translational velocity v
cm
and the other is the
linear velocity v
r
on account of rotation. The
magnitude of v
r
is v
r
= r
ω
, where
ω
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
r
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
2
(v
2
) and its components
v
r
and
v
cm
are shown; v
r
here is perpendicular to CP
2
.
It is easy to show that v
z
is perpendicular to the
line P
O
P
2
. Therefore the line passing through P
O
and parallel to
ωω
ωω
ω is called the instantaneous axis
of rotation.
At P
o
, the linear velocity, v
r
, due to rotation
is directed exactly opposite to the translational
velocity v
cm
.
Further the magnitude of v
r
here is
R
ω
, where R is the radius of the disc. The
condition that P
o
is instantaneously at rest
requires v
cm
= R
ω
. Thus for the disc the condition
for rolling without slipping is
υ ω
cm
R=
(7.47)
Incidentally, this means that the velocity of
point P
1
at the top of the disc (v
1
) has a
magnitude v
cm
+ R
ω
or 2 v
cm
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
/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
cm
/2, where m is the
mass of the body and v
cm
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
2
where k = the
corresponding radius of gyration of the body
and v
cm
= R ω, we get
or (7.49b)
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SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 175
u
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),
2
2
2
1
1
2
k
K m
R
υ
= +
, where v is the
final velocity of (the centre of mass of) the body.
Equating K and mgh,
Fig.7.38
2
2
2
1
1
2
k
mgh m
R
υ
= +
or
2
2 2
2
1
gh
k R
υ
=
+
Note is independent of the mass of the
rolling body;
For a ring, k
2
= R
2
2
1 1
ring
gh
υ
=
+
,
=
gh
For a solid cylinder k
2
= R
2
/2
2
1 1 2
disc
gh
υ
=
+
=
4
3
gh
For a solid sphere k
2
= 2R
2
/5
2
1 2 5
sphere
gh
υ
=
+
=
10
7
gh
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
ω
= d
θ
/dt and it is directed along the axis
of rotation. For rotation about a fixed axis, this vector
ωω
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
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
R
r
=
m
M
i
i
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
= ×
=
i
i
n
i
1
The torque or moment of force on a system of n particles about the origin is
ττ
= ×
r F
i i
1
The force F
i
acting on the i
th
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
ττ
τ
int
= 0 and
d
dt
ext
L
=
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
i
=
,
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
=
2
where r
i
is the perpendicular distance of the ith point of the body from the axis. The
kinetic energy of rotation is
2
1
2
K I
ω
=
.
14. The theorem of parallel axes:
2
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.
2020-21
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
z
= I
ω
, 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,
z
L I
= =L
ω
. The angular
acceleration of a rigid body rotating about a fixed axis is given by I
α
=
τ
. If the external
torque τ acting on the body is zero, the component of angular momentum about the
fixed axis (say, z-axis), L
z
(=I
ω
) of such a rotating body is constant.
17. For rolling motion without slipping v
cm
= R
ω
, where v
cm
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
cm
K m v I
= +
ω
.
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
/2,
K = K
+ MV
2
/2
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
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
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
x
, p
y
and p
z
. 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
2
/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
2
/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
2
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
2
)
(Hint: Consider the equilibrium of each side of the ladder separately.)
2020-21
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
2
.
(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
2
/3.)
7.25 Two discs of moments of inertia I
1
and I
2
about their respective axes (normal to the
disc and passing through the centre), and rotating with angular speeds
ω
1
and
ω
2
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
ω
1
ω
2
.
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
2
+y
2
).
(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
m
i i
r =
0
).
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 2
2
1 /
gh
v
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
ω
o
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
µ
k
= 0.2.
7.31 A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30
o
. The co-
efficient of static friction
µ
s
= 0.25.
(a) How much is the force of friction acting on the cylinder ?
(b) What is the work done against friction during rolling ?
(c) If the inclination
θ
of the plane is increased, at what value of
θ
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.
(c) The instantaneous acceleration 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
m
where p
i
is the momentum of the ith particle (of mass m
i
) and p
i
= m
i
v
i
. Note v
i
is the velocity
of the ith particle relative to the centre of mass.
Also, prove using the definition of the centre of mass
=p
i
0
(b) Show
2
½K K MV
= +
where K is the total kinetic energy of the system of particles,
K
is the total kinetic energy of the
system when the particle velocities are taken with respect to the centre of mass and MV
2
/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.
(c) Show
L L R V= ′ + × M
where
L
=
×
r p
i
i
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
L
and
M
×
R V
can
be said to be angular momenta, respectively, about and of the centre of mass of
the system of particles.
(d) Show
d
dt
d
dt
i
= ×
L
r
p
Further, show that
d
dt
ext
=
L
ττ
where
ext
τ
ττ
τ
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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