CONTENTS
7.1
Introduction
7.2
Centre of mass
7.3
Angular displacement
7.4
Angular velocity
7.5
Angular acceleration
7.6
Equations of linear motion and rotational motion
7.7
Moment of inertia
7.8
Radius of gyration
7.9
Theorem of parallel axes
7.10
Theorem of perpendicular axes
7.11
Moment of inertia of two point masses about their centre of mass
7.12
Analogy between translatory motion and rotational motion
7.13
Moment of inertia of some standard bodies about different axes
7.14
Torque
7.15
Couple
7.16
Translatory and rotatory equilibrium
7.17
Angular momentum
7.18
Law of conservation of angular momentum
7.19
Work, Energy and Power for rotating body
7.20
Slipping, Spinning and Rolling
7.21
Rolling without slipping
7.22
Rolling on an inclined plane
7.23
Rolling sliding and falling of a body
7.24
Velocity, acceleration and time for different bodies
7.25
Motion of connected mass
7.26
Time period of compound pendulum
Sample Problems
Practice Problems (Basic and Advance Level)
Answer Sheet of Practice Problems
Chapter
7
In 1988, Greg Louganis of the United States,
arguably the greatest Olympic diver in history,
cracked his head on the springboard while
attempting a reverse 2.5 pike. After receiving
stitches, Louganis won gold medal in diving.
Greg Louganis jumping from the spring
board exhibits somersaults in air before touching
the water surface. After leaving the spring board,
he curls his body by rolling the arms and the legs
inwards. Due to this, his moment of inertia
decreases and he spins in mid air with large
angular speed. As he is about to touch the water
surface, he stretches out his limbs. This increases
his moment of inertia and he enters the water at a
gentle speed.
108 Rotational Motion
7.1 Introduction.
Translation is motion along a straight line but rotation is the motion of wheels, gears,
motors, planets, the hands of a clock, the rotor of jet
engines and the blades of helicopters. First figure
shows a skater gliding across the ice in a straight line
with constant speed. Her motion is called translation
but second figure shows her spinning at a constant
rate about a vertical axis. Here motion is called
rotation.
Up to now we have studied translatory motion of
a point mass. In this chapter we will study the rotatory motion of rigid body about a fixed axis.
(1) Rigid body : A rigid body is a body that can rotate with all the parts locked together and
without any change in its shape.
(2) System : A collection of any number of particles interacting with one another and are
under consideration during analysis of a situation are said to form a system.
(3) Internal forces : All the forces exerted by various particles of the system on one another
are called internal forces. These forces are alone enable the particles to form a well defined
system. Internal forces between two particles are mutual (equal and opposite).
(4) External forces : To move or stop an object of finite size, we have to apply a force on
the object from outside. This force exerted on a given system is called an external force.
7.2 Centre of Mass.
Centre of mass of a system (body) is a point that moves as though all the mass were
concentrated there and all external forces were applied there.
(1) Position vector of centre of mass for n particle system : If a system consists of n
particles of masses
, whose positions vectors are
n
rrrr ........,, 321
respectively then position vector of centre of mass
n
nn
mmmm
rmrmrmrm
r............
............
321
332211
m1
m2
m3
y
x
3
r
2
r
1
r
r
C.M.
Rotational Motion 109
Hence the centre of mass of n particles is a weighted average of the position vectors of n
particles making up the system.
(2) Position vector of centre of mass for two particle system :
21
2211 mm
rmrm
r
and the centre of mass lies between the particles on the line joining them.
If two masses are equal i.e.
21 mm
, then position vector of centre of mass
221
rr
r
(3) Important points about centre of mass
(i) The position of centre of mass is independent of the co-ordinate system chosen.
(ii) The position of centre of mass depends upon the shape of the body and distribution of
mass.
Example : The centre of mass of a circular disc is within the material of the body while that
of a circular ring is outside the material of the body.
(iii) In symmetrical bodies in which the distribution of mass is homogenous, the centre of
mass coincides with the geometrical centre or centre of symmetry of the body.
(iv) Position of centre of mass for different bodies
S. No.
Body
Position of centre of mass
(a)
Uniform hollow sphere
Centre of sphere
(b)
Uniform solid sphere
Centre of sphere
(c)
Uniform circular ring
Centre of ring
(d)
Uniform circular disc
Centre of disc
(e)
Uniform rod
Centre of rod
(f)
A plane lamina (Square, Rectangle,
Parallelogram)
Point of inter section of diagonals
(g)
Triangular plane lamina
Point of inter section of medians
(h)
Rectangular or cubical block
Points of inter section of diagonals
(i)
Hollow cylinder
Middle point of the axis of cylinder
(j)
Solid cylinder
Middle point of the axis of cylinder
(k)
Cone or pyramid
On the axis of the cone at point distance
4
3h
from the vertex where h is the height
of cone
110 Rotational Motion
(v) The centre of mass changes its position only under the translatory motion. There is no
effect of rotatory motion on centre of mass of the body.
(vi) If the origin is at the centre of mass, then the sum of the moments of the masses of the
system about the centre of mass is zero i.e.
0
ii rm
.
(vii) If a system of particles of masses
,......,, 321 mmm
move with velocities
,......,, 321 vvv
then the velocity of centre of mass
i
ii
cm m
vm
v
.
(viii) If a system of particles of masses
,......,, 321 mmm
move with accelerations
,......,, 321 aaa
then the acceleration of centre of mass
i
ii
cm m
am
A
(ix) If
r
is a position vector of centre of mass of a system
then velocity of centre of mass
......
......
321
3
32211 mmm
rmrmrm
dt
d
dt
rd
vcm
(x) Acceleration of centre of mass
.......
.......
321
2211
2
2
2
2
mmm
rmrm
dt
d
dt
rd
dt
vd
Acm
cm
(xi) Force on a rigid body
2
2
dt
rd
MAMF cm
(xii) For an isolated system external force on the body is zero
0
cm
v
dt
d
MF
constant
cm
v
.
i.e., centre of mass of an isolated system moves with uniform velocity along a straight-line
path.
Sample problems based on centre of mass
Problem 1. The distance between the carbon atom and the oxygen atom in a carbon monoxide molecule
is 1.1 Å. Given, mass of carbon atom is 12 a.m.u. and mass of oxygen atom is 16 a.m.u.,
calculate the position of the center of mass of the carbon monoxide molecule [Kerala (Engg.) 2001]
(a) 6.3 Å from the carbon atom (b) 1 Å from the oxygen atom
(c) 0.63 Å from the carbon atom (d) 0.12 Å from the oxygen atom
Solution : (c) Let carbon atom is at the origin and the oxygen atom is placed at x-axis
12
1m
,
16
2m
,
jirjir ˆ
0
ˆ
1.1and
ˆ
0
ˆ
021
y
m1
m2
x
O
C
Rotational Motion 111
21
2211 mm
rmrm
r
i
ˆ
28
1.116
ir ˆ
63.0
i.e. 0.63 Å from carbon atom.
Problem 2. The velocities of three particles of masses 20g, 30g and 50 g are
kji
10and,10,10
respectively. The velocity of the centre of mass of the three particles is [EAMCET 2001]
(a)
532 kji
(b)
)(10 kji
(c)
kji
53020
(d)
kji
50302
Solution : (a) Velocity of centre of mass
321
332211 mmm
vmvmvm
vcm
100
ˆ
1050
ˆ
1030
ˆ
1020 kji
kji ˆ
5
ˆ
3
ˆ
2
.
Problem 3. Masses
24, 2,8, kg
are placed at the corners
A, B, C, D
respectively of a square
ABCD
of
diagonal
cm80
. The distance of centre of mass from
A
will be [MP PMT 1999]
(a)
cm20
(b)
cm30
(c)
cm40
(d)
cm60
Solution : (b) Let corner A of square ABCD is at the origin and the mass 8 kg is placed at this corner
(given in problem) Diagonal of square
cmad 802
cma240
,8
1kgm
,2
2kgm
,4
3kgm
kgm2
4
Let
4321 ,,, rrrr
are the position vectors of respective masses
jir ˆ
0
ˆ
0
1
,
jiar ˆ
0
ˆ
2
,
jaiar ˆˆ
3
,
jair ˆˆ
0
4
From the formula of centre of mass
4321
44332211 mmmm
rmrmrmrm
r
ji ˆ
215215
co-ordinates of centre of mass
)215,215(
and co-ordination of the corner
)0,0(
From the formula of distance between two points
),( 11 yx
and
),( 22 yx
distance
2
12
2
12 )()( yyxx
=
22 )0215()0215(
=
900
=
cm30
Problem 4. The coordinates of the positions of particles of mass
gm10 and 4,7
are
)7,5(2,3),5,(1,
and
cm1) 3, (3,
respectively. The position of the centre of mass of the system would be
(a)
cm
7
1
,
17
85
,
7
15
(b)
cm
7
1
,
17
85
,
7
15
(c)
cm
7
1
,
21
85
,
7
15
(d)
cm
3
7
,
21
85
,
7
15
Solution: (c)
gmm7
1
,
gmm4
2
,
gmm10
3
and
),752(),
ˆ
3
ˆ
5
ˆ
(21 kjirkjir
)
ˆ
ˆ
3
ˆ
3(
3kjir
Position vector of center mass
1047
)
ˆ
ˆ
3
ˆ
3(10)
ˆ
7
ˆ
5
ˆ
2(4)
ˆ
3
ˆ
5
ˆ
(7
kjikjikji
r
21
)
ˆ
3
ˆ
85
ˆ
45(kji
kjir ˆ
7
1
ˆ
21
85
ˆ
7
15
. So coordinates of centre of mass
7
1
,
21
85
,
7
15
.
7.3 Angular Displacement.
It is the angle described by the position vector
r
about the axis of rotation.
Q
P
S
r
y
B
8kg
x
2kg
4kg
2kg
C
D
A
(0,
0)
(a,
0)
(a,
a)
(0,
a)
402
C.M
.
112 Rotational Motion
Angular displacement
)( Radius
)( ntdisplacemeLinear
)( r
s
(1) Unit : radian
(2) Dimension :
][ 000 TLM
(3) Vector form
rS
i.e., angular displacement is a vector quantity whose direction is given by right hand rule. It
is also known as axial vector. For anti-clockwise sense of rotation direction of
is
perpendicular to the plane, outward and along the axis of rotation and vice-versa.
(4)
.revolution 1360radian 2
(5) If a body rotates about a fixed axis then all the particles will have same angular
displacement (although linear displacement will differ from particle to particle in accordance
with the distance of particles from the axis of rotation).
7.4 Angular Velocity.
The angular displacement per unit time is defined as angular velocity.
If a particle moves from P to Q in time
t
,
t
where
is the angular displacement.
(1) Instantaneous angular velocity
dt
d
t
t
0
lim
(2) Average angular velocity
total time
ntdisplacemeangular total
av
12
12 tt
(3) Unit : Radian/sec
(4) Dimension :
][ 100
TLM
which is same as that of frequency.
(5) Vector form
rv
[where
v
= linear velocity,
r
= radius vector]
is a axial vector, whose direction is normal to the rotational plane and its direction is
given by right hand screw rule.
(6)
n
T
2
2
[where T = time period, n = frequency]
(7) The magnitude of an angular velocity is called the angular speed which is also
represented by
.
7.5 Angular Acceleration.
The rate of change of angular velocity is defined as angular acceleration.
If particle has angular velocity
1
at time
1
t
and angular velocity
2
at time
2
t
then,
Angular acceleration
12
12
tt
Q
P
Rotational Motion 113
(1) Instantaneous angular acceleration
2
2
0
lim dt
d
dt
d
t
t
.
(2) Unit :
2
/secrad
(3) Dimension :
][ 200
TLM
.
(4) If
0
, circular or rotational motion is said to be uniform.
(5) Average angular acceleration
12
12 tt
av
.
(6) Relation between angular acceleration and linear acceleration
ra
.
(7) It is an axial vector whose direction is along the change in direction of angular velocity
i.e. normal to the rotational plane, outward or inward along the axis of rotation (depends upon
the sense of rotation).
7.6 Equations of Linear Motion and Rotational Motion.
Linear Motion
Rotational Motion
(1)
If linear acceleration is 0, u = constant and s =
u t.
If angular acceleration is 0,
= constant and
t
(2)
If linear acceleration a = constant,
If angular acceleration
= constant then
(i)
t
vu
s2
)(
(i)
t
2
)( 21
(ii)
t
uv
a
(ii)
t12
(iii)
atuv
(iii)
t
12
(iv)
2
2
1atuts
(iv)
2
12
1tt
(v)
asuv 2
22
(v)

2
2
1
2
2
(vi)
)12(
2
1 nausnth
(vi)
2
)12(
1
n
nth
(3)
If acceleration is not constant, the above
equation will not be applicable. In this case
If acceleration is not constant, the above
equation will not be applicable. In this case
(i)
dt
dx
v
(ii)
2
2
dt
xd
dt
dv
a
(iii)
dsavdv
(i)
dt
d
(ii)
2
2
dt
d
dt
d
(iii)
dd
114 Rotational Motion
Sample problems based on angular displacement, velocity and acceleration
Problem 5. The angular velocity of seconds hand of a watch will be [CPMT 2003]
(a)
secrad /
60
(b)
secrad /
30
(c)
secrad /60
(d)
secrad /30
Solution : (b) We know that second's hand completes its revolution (2
) in 60 sec
secrad
t/
3060
2
Problem 6. The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the
accelerator for 10 sec it starts rotating at 4500 revolutions per minute. The angular
acceleration of the wheel is [MP PET 2001]
(a) 30 radians/sec2 (b) 1880 degrees/sec2 (c) 40 radians/sec2 (d) 1980 degrees/sec2
Solution: (d) Angular acceleration (
) = rate of change of angular speed
t
nn )(2 12
10
60
12004500
2
2
2
360
10
60
3300
2
sec
degree
2
/1980 secdegree
.
Problem 7. Angular displacement
)(
of a flywheel varies with time as
32 ctbtat
then angular
acceleration is given by [BHU 2000]
(a)
2
32 ctbta
(b)
tb 62
(c)
tba 62
(d)
ctb62
Solution: (d) Angular acceleration
)( 32
2
2
2
2ctbtat
dt
d
dt
d
ctb62
Problem 8. A wheel completes 2000 rotations in covering a distance of
km5.9
. The diameter of the
wheel is [RPMT 1999]
(a)
m5.1
(b)
cm5.1
(c)
m5.7
(d)
cm5.7
Solution: (a) Distance covered by wheel in 1 rotation =
Dr
2
(Where D= 2r = diameter of wheel)
Distance covered in 2000 rotation = 2000
D
=
m
3
105.9
(given)
meterD 5.1
Problem 9. A wheel is at rest. Its angular velocity increases uniformly and becomes 60 rad/sec after 5
sec. The total angular displacement is
(a)
rad600
(b)
rad75
(c)
rad300
(d)
rad150
Solution: (d) Angular acceleration
2
12 /12
5
060 secrad
t
Now from
2
12
1tt
=
.150)5)(12(
2
1
02rad
Problem 10. A wheel initially at rest, is rotated with a uniform angular acceleration. The wheel rotates
through an angle
1
in first one second and through an additional angle
2
in the next one
second. The ratio
1
2
is
(a) 4 (b) 2 (c) 3 (d) 1
Rotational Motion 115
Solution: (c) Angular displacement in first one second
2
)1(
2
12
1
......(i) [From
2
12
1tt
]
Now again we will consider motion from the rest and angular displacement in total two
seconds
2)2(
2
12
21
......(ii)
Solving (i) and (ii) we get
2
1
and
2
3
2
3
1
2
.
Problem 11. As a part of a maintenance inspection the compressor of a jet engine is made to spin
according to the graph as shown. The number of revolutions made by the compressor
during the test is
(a) 9000 (b) 16570 (c) 12750 (d) 11250
Solution: (d) Number of revolution = Area between the graph and time axis = Area of trapezium
=
3000)55.2(
2
1
= 11250 revolution.
Problem 12. Figure shows a small wheel fixed coaxially on a bigger one of double the radius. The system
rotates about the common axis. The strings supporting A and B do not slip on the wheels. If
x and y be the distances travelled by A and B in the same time interval, then
(a)
yx 2
(b)
yx
(c)
xy 2
(d) None of these
Solution: (c) Linear displacement (S) = Radius (r) × Angular displacement (
)
rS
(if
constant)
)( mass by travelledDistance
)( mass by travelledDistance
yB
xA
)2( mass withconcerned of pulley Radius
)( mass withconcerned of pulley Radius
rB
rA
2
1
xy 2
.
Problem 13. If the position vector of a particle is
)
ˆ
4
ˆ
3( jir
meter and its angular velocity is
)
ˆ
2
ˆ
(kj
rad/sec then its linear velocity is (in m/s)
(a)
)
ˆ
3
ˆ
6
ˆ
8( kji
(b)
)
ˆ
8
ˆ
6
ˆ
3( kji
(c)
)
ˆ
6
ˆ
6
ˆ
3( kji
(d)
)
ˆ
3
ˆ
8
ˆ
6( kji
0
50
0
100
0
1500
200
0
250
0
300
0
(in rev per
min)
1
2
3
4
5
t (in
min)
A
B
116 Rotational Motion
Solution: (a)
rv
=
)
ˆ
2
ˆˆ
0()
ˆ
0
ˆ
4
ˆ
3( kjikji
kji
kji
ˆ
3
ˆ
6
ˆ
8
210
043
ˆ
ˆˆ
7.7 Moment of Inertia.
Moment of inertia plays the same role in rotational motion as mass plays in linear motion.
It is the property of a body due to which it opposes any change in its state of rest or of uniform
rotation.
(1) Moment of inertia of a particle
2
mrI
; where r is the perpendicular distance of particle
from rotational axis.
(2) Moment of inertia of a body made up of number of particles (discrete distribution)
.......
2
33
2
22
2
11 rmrmrmI
(3) Moment of inertia of a continuous distribution of mass, treating the element of mass
dm
at
position
r
as particle
2
rdmdI
i.e.,
dmrI 2
(4) Dimension :
][ 02 TML
(5) S.I. unit : kgm2.
(6) Moment of inertia depends on mass, distribution of mass and on the position of axis of
rotation.
(7) Moment of inertia does not depend on angular velocity, angular acceleration, torque,
angular momentum and rotational kinetic energy.
(8) It is not a vector as direction (clockwise or anti-clockwise) is not to be specified and
also not a scalar as it has different values in different directions. Actually it is a tensor quantity.
(9) In case of a hollow and solid body of same mass, radius and shape for a given axis,
moment of inertia of hollow body is greater than that for the solid body because it depends
upon the mass distribution.
7.8 Radius of Gyration.
Radius of gyration of a body about a given axis is the perpendicular distance of a point from
the axis, where if whole mass of the body were concentrated, the body shall have the same
moment of inertia as it has with the actual distribution of mass.
r
m
r1
m1
r2
r3
m2
m3
r
dm
Rotational Motion 117
When square of radius of gyration is multiplied with the mass of the body gives the
moment of inertia of the body about the given axis.
2
MkI
or
M
I
k
.
Here k is called radius of gyration.
From the formula of discrete distribution
2
2
3
2
2
2
1....... n
mrmrmrmrI
If m1 = m2 = m3 = ....... = m then
)..........( 2
2
3
2
2
2
1n
rrrrmI
........(i)
From the definition of Radius of gyration,
2
MkI
........(ii)
By equating (i) and (ii)
)............( 2
2
3
2
2
2
1
2n
rrrrmMk
)..........( 2
2
3
2
2
2
1
2n
rrrrmnmk
[As
nmM
]
n
rrrr
kn
2
2
3
2
2
2
1...........
Hence radius of gyration of a body about a given axis is equal to root mean square distance
of the constituent particles of the body from the given axis.
(1) Radius of gyration
)(k
depends on shape and size of the body, position and configuration
of the axis of rotation, distribution of mass of the body w.r.t. the axis of rotation.
(2) Radius of gyration
)(k
does not depends on the mass of body.
(3) Dimension
][ 010 TLM
.
(4) S.I. unit : Meter.
(5) Significance of radius of gyration : Through this concept a real body (particularly
irregular) is replaced by a point mass for dealing its rotational motion.
Example : In case of a disc rotating about an axis through its centre of mass and
perpendicular to its plane
2
)21( 2R
M
MR
M
I
k
So instead of disc we can assume a point mass
M
at a distance
)2/(R
from the axis of
rotation for dealing the rotational motion of the disc.
Note
: For a given body inertia is constant whereas moment of inertia is variable.
r4
m
m
r2
m
m
m
r3
r5
r1
k
M
118 Rotational Motion
7.9 Theorem of Parallel Axes.
Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of
the body about an axis parallel to given axis and passing through centre of mass of the body Ig
and
2
Ma
where M is the mass of the body and a is the perpendicular
distance between the two axes.
2
MaII g
Example: Moment of inertia of a disc about an axis through its
centre and perpendicular to the plane is
2
2
1MR
, so moment of inertia
about an axis through its tangent and perpendicular to the plane will be
2
MaII g
22
2
1MRMRI
2
2
3MRI
7.10 Theorem of Perpendicular Axes.
According to this theorem the sum of moment of inertia of a plane lamina about two
mutually perpendicular axes lying in its plane is equal to its moment of inertia about an axis
perpendicular to the plane of lamina and passing through the point of intersection of first two
axes.
yxz III
Example : Moment of inertia of a disc about an axis through its centre of mass and
perpendicular to its plane is
2
2
1MR
, so if the disc is in x–y plane then by theorem of
perpendicular axes
i.e.
y
I
x
I
z
I
D
IMR 2
2
12
[As ring is symmetrical body so
D
I
y
I
x
I
]
2
4
1MRID
I
IG
R
G
a
I
G
IG
X
Z
Y
X
Z
Y
O
ID
ID
Rotational Motion 119
Note
: In case of symmetrical two-dimensional bodies as moment of inertia for all axes
passing through the centre of mass and in the plane of body will be same so the
two axes in the plane of body need not be perpendicular to each other.
7.11 Moment of Inertia of Two Point Masses About Their Centre of Mass.
Let
1
m
and
2
m
be two masses distant
r
from each-other and
1
r
and
2
r
be the distances of
their centre of mass from
1
m
and
2
m
respectively, then
(1)
rrr 21
(2)
2211 rmrm
(3)
r
mm
m
rr
mm
m
r
21
1
2
21
2
1and
(4)
2
22
2
11 rmrmI
(5)
22
21
21 rr
mm
mm
I
[where
21
21 mm
mm
is known as reduced mass
1
m
and
2
m
.]
(6) In diatomic molecules like
HClH ,
2
etc. moment of inertia about their centre of mass is
derived from above formula.
7.12 Analogy Between Tranlatory Motion and Rotational Motion.
Translatory motion
Rotatory motion
Mass
)(m
Moment of
Inertia
)(I
Linear
momentum
mvP
mEP2
Angular
Momentum
IL
IEL2
Force
maF
Torque
I
Kinetic energy
2
2
1mvE
m
P
E2
2
2
2
1
IE
I
L
E2
2
7.13 Moment of Inertia of Some Standard Bodies About Different Axes.
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
m1
Centre of
mass
r1
r2
m2
120 Rotational Motion
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
Ring
About an axis
passing
through C.G.
and
perpendicular
to its plane
2
MR
R
1
Ring
About its
diameter
2
2
1MR
2
R
2
1
Ring
About a
tangential axis
in its own
plane
2
2
3MR
R
2
3
2
3
Ring
About a
tangential axis
perpendicular
to its own
plane
2
2MR
R2
2
Disc
About an axis
passing
through C.G.
and
perpendicular
to its plane
2
2
1MR
2
R
2
1
Disc
About its
Diameter
2
4
1MR
2
R
4
1
Rotational Motion 121
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
Disc
About a
tangential axis
in its own
plane
2
4
5MR
R
2
5
4
5
Disc
About a
tangential axis
perpendicular
to its own
plane
2
2
3MR
R
2
3
2
3
Annular disc
inner radius = R1
and outer radius
= R2
Passing
through the
centre and
perpendicular
to the plane
][
22
2
2
1RR
M
–
–
Annular disc
Diameter
][
42
2
2
1RR
M
–
–
Annular disc
Tangential and
Parallel to the
diameter
]5[
42
2
2
1RR
M
–
–
Annular disc
Tangential and
perpendicular
to the plane
]3[
22
2
2
1RR
M
–
–
R2
R1
122 Rotational Motion
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
Solid cylinder
About its own
axis
2
2
1MR
2
R
2
1
Solid cylinder
Tangential
(Generator)
2
2
3MR
R
2
3
2
3
Solid cylinder
About an axis
passing through
its C.G. and
perpendicular
to its own axis
412
22 RL
M
412
22 RL
Solid cylinder
About the
diameter of one
of faces of the
cylinder
43
22 RL
M
43
22 RL
Cylindrical shell
About its own
axis
MR2
R
1
L
R
Rotational Motion 123
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
Cylindrical shell
Tangential
(Generator)
2MR2
R2
2
Cylindrical shell
About an axis
passing through
its C.G. and
perpendicular
to its own axis
212
22 RL
M
212
22 RL
Cylindrical shell
About the
diameter of one
of faces of the
cylinder
23
22 RL
M
23
22 RL
Hollow cylinder
with inner radius =
R1 and outer radius
= R2
Axis of cylinder
)(
22
2
2
1RR
M
Hollow cylinder
with inner radius =
R1 and outer radius
= R2
Tangential
)3(
22
2
2
1RR
M
R2
R1
124 Rotational Motion
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
Solid Sphere
About its
diametric axis
2
5
2MR
R
5
2
5
2
Solid sphere
About a
tangential axis
2
5
7MR
R
5
7
5
7
Spherical shell
About its
diametric axis
2
3
2MR
R
3
2
3
2
Spherical shell
About a
tangential
axis
2
3
5MR
R
3
5
3
5
Hollow sphere of
inner radius R1
and outer radius
R2
About its
diametric axis
3
1
3
2
5
1
5
2
5
2
RR
RR
M
Hollow sphere
Tangential
2
2
3
1
3
2
5
1
5
2
)(5
][2 MR
RR
RRM
L
L
Rotational Motion 125
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
Long thin rod
About on axis
passing through
its centre of
mass and
perpendicular
to the rod.
12
2
ML
12
L
Long thin rod
About an axis
passing through
its edge and
perpendicular
to the rod
3
2
ML
3
L
Rectangular
lamina of length l
and breadth b
Passing through
the centre of
mass and
perpendicular
to the plane
][
12 22 bl
M
Rectangular
lamina
Tangential
perpendicular
to the plane and
at the mid-point
of breadth
]4[
12 22 bl
M
Rectangular
lamina
Tangential
perpendicular
to the plane and
at the mid-point
of length
]4[
12 22 bl
M
Rectangular
parallelopiped
length l, breadth b,
thickness t
Passing through
centre of mass
and parallel to
(i) Length (x)
(ii) breadth (z)
(iii) thickness
(y)
(i)
12
][ 22 tbM
(ii)
12
][ 22 tlM
(iii)
12
][ 22 lbM
b
ii
iii
l
i
t
ii
iii
i
126 Rotational Motion
Body
Axis of
Rotation
Figure
Moment of
inertia
k
k2/R2
Rectangular
parallelepiped
length l, breath b,
thickness t
Tangential and
parallel to
(i) length (x)
(ii) breadth (y)
(iii)
thickness(z)
(i)
]3[
12 222 tbl
M
(ii)
]3[
12 222 tbl
M
(iii)
]3[
12 222 tbl
M
Elliptical disc of
semimajor axis = a
and semiminor axis
= b
Passing through
CM and
perpendicular
to the plane
][
422 ba
M
Solid cone of
radius R and
height h
Axis joining the
vertex and
centre of the
base
2
10
3MR
Equilateral
triangular lamina
with side a
Passing through
CM and
perpendicular
to the plane
6
2
Ma
Right angled
triangular lamina
of sides a, b, c
Along the edges
(1)
6
2
Mb
(2)
6
2
Ma
(3)
22
22
6ba
baM
Rotational Motion 125
Sample problem based on moment of inertia
Problem 14. Five particles of mass = 2 kg are attached to the rim of a circular disc of radius 0.1 m and
negligible mass. Moment of inertia of the system about the axis passing through the centre
of the disc and perpendicular to its plane is [BHU 2003]
(a) 1 kg m2 (b) 0.1 kg m2 (c) 2 kg m2 (d) 0.2 kg m2
Solution: (b) We will not consider the moment of inertia of disc because it doesn't have any mass so
moment of inertia of five particle system
22 )1.0(255 mrI
2
-1.0 mkg
.
Problem 15. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of
radius 4R is made from an iron plate of thickness
4
t
. Then the relation between the
moment of inertia IX and IY is [AIEEE 2003]
(a) IY = 64IX (b) IY = 32IX (c) IY = 16IX (d) IY = IX
Solution: (a) Moment of Inertia of disc I =
2
2
1MR
22 )(
2
1RtR
4
2
1Rt
[As
VM
=
tR2
where
t
thickness,
=
density]
4
x
y
x
y
x
y
R
R
t
t
I
I
[If
= constant]
64)4(
4
14
x
y
I
I
[Given
xy RR 4
,
4
x
yt
t
]
xy II 64
Problem 16. Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia
about an axis perpendicular to its plane and passing through a point on its rim will be [UPSEAT 2002]
(a) 5 I (b) 6 I (c) 3 I (d) 4 I
Solution: (b) Moment of inertia of disc about a diameter =
IMR
2
4
1
(given)
IMR 4
2
Now moment of inertia of disc about an axis perpendicular to its plane and passing through
a point on its rim
=
IIMR 6)4(
2
3
2
32
.
Problem 17. Four thin rods of same mass M and same length l, form a square as shown in figure.
Moment of inertia of this system about an axis through centre O and perpendicular to its
plane is [MP PMT 2002]
(a)
2
3
4Ml
(b)
3
2
Ml
(c)
6
2
Ml
(d)
2
3
2Ml
Solution: (a) Moment of inertia of rod AB about point
2
12
1MlP
l
l
O
l
l
P
B
D
C
A
126 Rotational Motion
M.I. of rod AB about point O
2
2
212
l
M
Ml
2
3
1Ml
[by the theorem of parallel axis]
and the system consists of 4 rods of similar type so by the symmetry
2
3
4MlISystem
.
Problem 18. Three rings each of mass M and radius R are arranged as shown in the figure. The moment
of inertia of the system about YY will be [MP PET 2000]
(a) 3 MR2
(b)
2
2
3MR
(c) 5 MR2
(d)
2
2
7MR
Solution: (d) M.I of system about
'YY
321 IIII
where I1 = moment of inertia of ring about diameter, I2 = I3 = M.I. of inertia of ring about a
tangent in a plane
222 2
3
2
3
2
1mRmRmRI
2
2
7mR
Problem 19. Let
l
be the moment of inertia of an uniform square plate about an axis
AB
that passes
through its centre and is parallel to two of its sides.
CD
is a line in the plane of the plate
that passes through the centre of the plate and makes an angle
with
AB
. The moment of
inertia of the plate about the axis
CD
is then equal to
[IIT-JEE 1998]
(a)
l
(b)
2
sinl
(c)
2
cosl
(d)
2
cos2
l
Solution: (a) Let
Z
I
is the moment of inertia of square plate about the axis which is passing through the
centre and perpendicular to the plane.
'''' DCCDBAABZIIIII
[By the theorem of perpendicular axis]
'''' 2222 DCCDBAABZIIIII
[As AB, A' B' and CD, C' D' are symmetric axis]
Hence
lII ABCD
Problem 20. Three rods each of length L and mass M are placed along X, Y and Z-axes in such a way that
one end of each of the rod is at the origin. The moment of inertia of this system about Z
axis is [AMU 1995]
(a)
3
22
ML
(b)
3
42
ML
(c)
3
52
ML
(d)
3
2
ML
Solution: (a) Moment of inertia of the system about z-axis can be find out by calculating the moment of
inertia of individual rod about z-axis
3
2
21 ML
II
because z-axis is the edge of rod 1 and 2
Y
Y
1
2
3
A
B
A
B
C
D
C
D
x
z
y
3
1
2
Rotational Motion 127
and
0
3I
because rod in lying on z-axis
321system IIII
0
33
22 MLML
3
22
ML
.
Problem 21. Three point masses each of mass
m
are placed at the corners of an equilateral triangle of
side a. Then the moment of inertia of this system about an axis passing along one side of
the triangle is [AIIMS 1995]
(a)
2
ma
(b)
2
3ma
(c)
2
4
3ma
(d)
2
3
2ma
Solution: (c) The moment of inertia of system about AB side of triangle
CBA IIII
2
00 mx
2
2
3
a
m
2
4
3ma
Problem 22. Two identical rods each of mass M. and length l are joined in crossed position as shown in
figure. The moment of inertia of this system about a bisector would be
(a)
6
2
Ml
(b)
12
2
Ml
(c)
3
2
Ml
(d)
4
2
Ml
Solution: (b) Moment of inertia of system about an axes which is perpendicular to plane of rods and
passing through the common centre of rods
61212
222 MlMlMl
Iz
Again from perpendicular axes theorem
21 BBz III
6
22 2
21
Ml
II BB
[As
21 BB II
]
12
2
21
Ml
II BB
.
Problem 23. The moment of inertia of a rod of length l about an axis passing through its centre of mass
and perpendicular to rod is I. The moment of inertia of hexagonal shape formed by six such
rods, about an axis passing through its centre of mass and perpendicular to its plane will be
(a) 16I (b) 40 I (c) 60 I (d) 80 I
Solution: (c) Moment of inertia of rod AB about its centre and perpendicular to the length =
12
2
ml
= I
Iml 12
2
B1
B2
B
C
m
A
a
a
x
a
m
m
128 Rotational Motion
Now moment of inertia of the rod about the axis which is passing through O and
perpendicular to the plane of hexagon Irod=
2
2
12 mx
ml
[From the theorem of parallel
axes]
6
5
2
3
12
2
2
2ml
lm
ml
Now the moment of inertia of system Isystem =
6
5
66 2
rod ml
I
2
5ml
Isystem = 5 (12 I) = 60 I [As
Iml 12
2
]
Problem 24. The moment of inertia of HCl molecule about an axis passing through its centre of mass and
perpendicular to the line joining the
H
and
Cl
ions will be, if the interatomic distance is
1 Å
(a)
247 .1061.0 mkg
(b)
247 .1061.1 mkg
(c)
247 .10061.0 mkg
(d) 0
Solution: (b) If
1
r
and
2
r
are the respective distances of particles
1
m
and
2
m
from the centre of mass
then
2211 rmrm
)(5.351xLx
)1(5.35 xx
Åx 973.0
and
ÅxL 027.0
Moment of inertia of the system about centre of mass
2
2
2
1)( xLmxmI
22 )027.0(5.35)973.0(1 ÅamuÅamuI
Substituting 1 a.m.u. = 1.67 10–27 kg and 1 Å = 10–10 m
I
247
1062.1 mkg
Problem 25. Four masses are joined to a light circular frame as shown in the figure. The radius of
gyration of this system about an axis passing through the centre of the circular frame and
perpendicular to its plane would be
(a)
2/a
(b)
2/a
(c) a
(d) 2a
Solution: (c) Since the circular frame is massless so we will consider moment of inertia of four masses
only.
22222 8232 mamamamamaI
.....(i)
Now from the definition of radius of gyration
2
8mkI
.....(ii)
2m
m
2m
3m
A
A
O
a
H
m1
Cl
m2
C.M.
x
L – x
A
B
O
x
l
l
l
Rotational Motion 129
comparing (i) and (ii) radius of gyration
ak
.
Problem 26. Four spheres, each of mass
M
and radius
r
are situated at the four corners of square of
side
R
. The moment of inertia of the system about an axis perpendicular to the plane of
square and passing through its centre will be
(a)
)54(
2
522 RrM
(b)
)54(
5
222 RrM
(c)
)54(
5
222 rrM
(d)
)54(
2
522 rrM
Solution: (b) M. I. of sphere A about its diameter
2
'5
2MrIO
Now M.I. of sphere A about an axis perpendicular to the plane of square and passing
through its centre will be
2
'2
R
MII OO
25
22
2MR
Mr
[by the theorem of parallel axis]
Moment of inertia of system (i.e. four sphere)=
O
I4
25
2
42
2MR
Mr
22 54
5
2RrM
Problem 27. The moment of inertia of a solid sphere of density
and radius R about its diameter is
(a)
5
176
105 R
(b)
2
176
105 R
(c)
5
105
176 R
(d)
2
105
176 R
Solution: (c) Moment of inertia of sphere about it diameter
2
5
2MRI
23
3
4
5
2RR
[As
VM
=
3
3
4R
]
I =
5
15
8R
55 105
176
715
228RR
Problem 28. Two circular discs
A
and
B
are of equal masses and thickness but made of metals with
densities
A
d
and
B
d
)( BA dd
. If their moments of inertia about an axis passing through
centres and normal to the circular faces be
A
I
and
B
I
, then
(a)
BA II
(b)
BA II
(c)
BA II
(d)
BA II
Solution : (c) Moment of inertia of circular disc about an axis passing through centre and normal to the
circular face
O
R
r
O
O
2/R
B
A
C
D
130 Rotational Motion
2
2
1MRI
t
M
M
2
1
[As
VM
tR2
t
M
R
2
]
t
M
I2
2
or
1
I
If mass and thickness are constant.
So, in the problem
A
B
B
Ad
d
I
I
BA II
[As
BA dd
]
7.14 Torque.
If a pivoted, hinged or suspended body tends to rotate under the action of a force, it is said to
be acted upon by a torque. or The turning effect of a force about the axis of rotation is called
moment of force or torque due to the force.
If the particle rotating in
xy
plane about the origin under the
effect of force
F
and at any instant the position vector of the particle
is
r
then,
Torque
=
Fr
sinFr
[where
is the angle between the direction of
r
and
F
]
(1) Torque is an axial vector. i.e., its direction is always perpendicular to the plane
containing vector
r
and
F
in accordance with right hand screw rule. For a given figure the
sense of rotation is anti-clockwise so the direction of torque is perpendicular to the plane,
outward through the axis of rotation.
(2) Rectangular components of force
forceof componentradial cos
FFr
,
forceof component transversesin
FF
As
sinFr
or
Fr
= (position vector) (transverse component of force)
F
F cos
F sin
X
r
P
d
90o
Y
P
F
Rotatio
n
(A)
O
r
(B)
r
F
O
Rotatio
n
Rotational Motion 131
Thus the magnitude of torque is given by the product of transverse component of force and
its perpendicular distance from the axis of rotation i.e., Torque is due to transverse component
of force only.
(3) As
sinFr
or
)sin(
rF
Fd
[As
figure thefrom sin
rd
]
i.e. Torque = Force Perpendicular distance of line of action of force from the axis of
rotation.
Torque is also called as moment of force and d is called moment or lever arm.
(4) Maximum and minimum torque : As
sinor FrFr
rF
m
aximum
1maxsin When
i.e.,
90
F
is perpendicular to
r
0
minimum
180or 0 . 0minsin When
i.e
F
is collinear to
r
(5) For a given force and angle, magnitude of torque depends on r. The more is the value of
r, the more will be the torque and easier to rotate the body.
Example : (i) Handles are provided near the free edge of the Planck of the door.
(ii) The handle of screw driver is taken thick.
(iii) In villages handle of flourmill is placed near the circumference.
(iv) The handle of hand-pump is kept long.
(v) The arm of wrench used for opening the tap, is kept long.
(6) Unit : Newton-metre (M.K.S.) and Dyne-cm (C.G.S.)
(7) Dimension :
][ 22
TML
.
(8) If a body is acted upon by more than one force, the total torque is the vector sum of
each torque.
........
321
(9) A body is said to be in rotational equilibrium if resultant torque acting on it is zero i.e.
0
.
(10) In case of beam balance or see-saw the system will
be in rotational equilibrium if,
0
21
or
0
2211 lFlF
2211 lFlF
However if,
2
1
, L.H.S. will move downwards and if
2
1
. R.H.S. will move downward. and the system will not
be in rotational equilibrium.
(11) On tilting, a body will restore its initial position due to torque of weight about the
point O till the line of action of weight passes through its base on tilting, a body will topple due
to torque of weight about O, if the line of action of weight does not pass through the base.
G
R
Torque
Tilt
O
Torqu
e
Tilt
O
R
G
l1
R
F1
F2
l2
132 Rotational Motion
(12) Torque is the cause of rotatory motion and in rotational motion it plays same role as
force plays in translatory motion i.e., torque is rotational analogue of force. This all is evident
from the following correspondences between rotatory and translatory motion.
Rotatory Motion
Translatory
Motion
I
amF
dW
dsFW
P
vFP
dt
dL
dt
dP
F
7.15 Couple .
A special combination of forces even when the entire body is free to move can rotate it. This
combination of forces is called a couple.
(1) A couple is defined as combination of two equal but oppositely directed force not acting
along the same line. The effect of couple is known by its moment of couple or torque by a couple
Frτ
.
(2) Generally both couple and torque carry equal meaning. The basic difference between
torque and couple is the fact that in case of couple both the forces are externally applied while
in case of torque one force is externally applied and the other is reactionary.
(3) Work done by torque in twisting the wire
2
2
1
CW
.
Where
C
; C is known as twisting coefficient or couple per unit twist.
7.16 Translatory and Rotatory Equilibrium .
F
F
r
Rotational Motion 133
Forces are equal and
act along the same
line.
0 F
and
0
Body will remain stationary
if initially it was at rest.
Forces are equal and
does not act along the
same line.
0 F
and
0
Rotation i.e. spinning.
Forces are unequal and
act along the same
line.
0 F
and
0
Translation i.e. slipping or
skidding.
Forces are unequal and
does not act along the
same line.
0 F
and
0
Rotation and translation
both i.e. rolling.
Sample problems based on torque and couple
Problem 29. A force of
)
ˆ
2
ˆ
4
ˆ
2( kji
N acts at a point
)
ˆ
4
ˆ
2
ˆ
3( kji
metre from the origin. The magnitude of
torque is
(a) Zero (b) 24.4 N-m (c) 0.244 N-m (d) 2.444 N-m
Solution: (b)
NkjiF )
ˆ
2
ˆ
4
ˆ
2(
and
)
ˆ
423( kir
meter
Torque
Fr
242
423
ˆ
ˆˆ
kji
kji ˆ
16
ˆ
14
ˆ
12
and
222 )16()14()12(||
= 24.4
N-m
Problem 30. The resultant of the system in the figure is a force of
N8
parallel to the given force through
R
. The value of
PR
equals to
(a)
RQ41
(b)
RQ83
(c)
RQ53
(d)
RQ52
Solution: (c) By taking moment of forces about point R,
035 RQPR
RQPR 5
3
.
Problem 31. A horizontal heavy uniform bar of weight W is supported at its ends by two men. At the
instant, one of the men lets go off his end of the rod, the other feels the force on his hand
changed to
(a) W (b)
2
W
(c)
4
3W
(d)
4
W
F
F
F
F
F2
F1
F2
F1
R
P
5 N
3 N
Q
134 Rotational Motion
Solution: (d) Let the mass of the rod is M Weight (W) = Mg
Initially for the equilibrium
MgFF
2/MgF
When one man withdraws, the torque on the rod
2
l
MgI
23
2l
Mg
Ml
[As I = Ml 2/ 3]
Angular acceleration
l
g
2
3
and linear acceleration
4
3
2
gl
a
Now if the new normal force at A is
'F
then
MaFMg '
4
3
'Mg
MgMaMgF
4
Mg
4
W
.
7.17 Angular Momentum .
The turning momentum of particle about the axis of rotation is called the angular
momentum of the particle.
or
The moment of linear momentum of a body with respect to any axis of rotation is known as
angular momentum. If
P
is the linear momentum of particle and
r
its
position vector from the point of rotation then angular momentum.
PrL
nPrL ˆ
sin
Angular momentum is an axial vector i.e. always directed perpendicular to the plane of
rotation and along the axis of rotation.
(1) S.I. Unit : kg-m2-s–1 or J-sec.
(2) Dimension :
][ -12TML
and it is similar to Planck’s constant
)(h
.
(3) In cartesian co-ordinates if
kPjPiPPkzjyixr zyx ˆ
ˆˆ
and
ˆ
ˆˆ
Then
zyx PPP
zyx
kji
PrL
ˆ
ˆˆ
=
kyPxPjzPxPiPzP xyxzyz ˆ
)(
ˆ
)(
ˆ
)(y
(4) As it is clear from the figure radial component of momentum
cosPPr
P
P cos
P sin
X
Y
r
P
d
L
P
r
Mg
A
F
F
B
Mg
A
F
B
B
Rotational Motion 135
Transverse component of momentum
sinPP
So magnitude of angular momentum
sinPrL
PrL
Angular momentum = Position vector × Transverse component of angular momentum
i.e., The radial component of linear momentum has no role to play in angular momentum.
(5) Magnitude of angular momentum
PdLrPL )sin(
[As
sinrd
from the figure.]
Angular momentum = (Linear momentum) (Perpendicular distance of line of action of force
from the axis of rotation)
(6) Maximum and minimum angular momentum : We know
PrL
sinsin][ rPrvmvrmL
[As
vmP
]
mvrLm
aximum
1maxsin When
i.e.,
90
v
is perpendicular to
r
0
minimum L
180or 0 . 0minsin When
i.e
v
is parallel or anti-parallel
to
r
(7) A particle in translatory motion always have an angular momentum unless it is a point
on the line of motion because
sinmvrL
and
1L
if
o
0
or
o
180
(8) In case of circular motion,
)( vrmPrL
=
sinmvr
mvrL
2
mr
[As
vr
and
rv
]
or
IL
[As mr2 = I]
In vector form
IL
(9) From
IL
dt
d
I
dt
Ld
=
I
[As
dt
d
and
I
]
i.e. the rate of change of angular momentum is equal to the net torque acting on the
particle. [Rotational analogue of Newton's second law]
(10) If a large torque acts on a particle for a small time then 'angular impulse' of torque is
given by
136 Rotational Motion
2
1
t
t
va dtdtJ
or Angular impulse
LtJ av
Angular impulse = Change in angular momentum
(11) The angular momentum of a system of particles is equal to the vector sum of angular
momentum of each particle i.e.,
n
LLLLL .......
321
.
(12) According to Bohr theory angular momentum of an electron in nth orbit of atom can be
taken as,
2
h
nL
[where n is an integer used for number of
orbit]
7.18 Law of Conservation of Angular Momentum.
Newton’s second law for rotational motion
dt
Ld
So if the net external torque on a particle (or system) is zero then
0
dt
Ld
i.e.
.......
321 LLLL
= constant.
Angular momentum of a system (may be particle or body) remains constant if resultant torque
acting on it zero.
As
IL
so if
0
then
constant
I
1
I
Since angular momentum
I
remains constant so when
I
decreases, angular velocity
increases and vice-versa.
Examples of law of conservation of angular momentum :
(1) The angular velocity of revolution of a planet around the sun in an elliptical orbit
increases when the planet come closer to the sun and vice-versa because when planet comes
closer to the sun, it's moment of inertia I decreases there fore
increases.
(2) A circus acrobat performs feats involving spin by bringing his arms and legs closer to
his body or vice-versa. On bringing the arms and legs closer to body, his moment of inertia I
decreases. Hence
increases.
Rotational Motion 137
(3) A person-carrying heavy weight in his hands and standing on a rotating platform can
change the speed of platform. When the person suddenly folds his arms. Its moment of inertia
decreases and in accordance the angular speed increases.
(4) A diver performs somersaults by Jumping from a high diving board keeping his legs and
arms out stretched first and then curling his body.
(5) Effect of change in radius of earth on its time period
Angular momentum of the earth
constant
IL
constant
2
5
22 T
MRL
2
RT
[if M remains constant]
If R becomes half then time period will become one-fourth i.e.
.6
4
24 hrs
Sample problems based on angular momentum
Problem 32. Consider a body, shown in figure, consisting of two identical balls, each of mass M
connected by a light rigid rod. If an impulse J = Mv is imparted to the body at one of its
ends, what would be its angular velocity
[IIT-JEE (Screening) 2003]
(a) v/L
(b) 2v/L
(c) v/3L
(d) v/4L
Solution: (a) Initial angular momentum of the system about point O
= Linear momentum × Perpendicular distance of linear momentum from the axis of rotation
2
L
Mv
....(i)
Final angular momentum of the system about point O
21 II
)( 21 II
22
22
L
M
L
M
....(ii)
Applying the law of conservation of angular momentum
M
M
L
J = Mv
M
M
O
L/2
L/2
1
2
138 Rotational Motion
2
2
2
2
L
M
L
Mv
L
v
Problem 33. A thin circular ring of mass M and radius R is rotating about its axis with a constant
angular velocity
. Four objects each of mass m, are kept gently to the opposite ends of two
perpendicular diameters of the ring. The angular velocity of the ring will be [CBSE PMT 2003]
(a)
mM
M
4
(b)
M
mM
)4(
(c)
mM
mM
4
)4(
(d)
m
M
4
Solution: (a) Initial angular momentum of ring
2
MRI
If four object each of mass m, and kept gently to the opposite ends of two perpendicular
diameters of the ring then final angular momentum =
')4( 22
mRMR
By the conservation of angular momentum
Initial angular momentum = Final angular momentum
')4( 222
mRMRMR
mM
M
4
'
.
Rotational Motion 137
Problem 34. A circular platform is free to rotate in a horizontal plane about a vertical axis passing
through its center. A tortoise is sitting at the edge of the platform. Now, the platform is
given an angular velocity
0. When the tortoise moves along a chord of the platform with a
constant velocity (with respect to the platform), the angular velocity of the platform
(t)
will vary with time t as [IIT-JEE (Screening) 2002]
(a) (b) (c) (d)
Solution: (b) The angular momentum (L) of the system is conserved i.e. L = I
= constant
When the tortoise walks along a chord, it first moves closer to the centre and then away
from the centre. Hence, M.I. first decreases and then increases. As a result,
will first
increase and then decrease. Also the change in
will be non-linear function of time.
Problem 35. The position of a particle is given by :
)
ˆ
ˆ
2
ˆ
(kjir
and momentum
)
ˆ
2
ˆ
4
ˆ
3( kjiP
. The
angular momentum is perpendicular to [CPMT 2000]
(a) X-axis
(b) Y-axis
(c) Z-axis
(d) Line at equal angles to all the three axes
Solution: (a)
prL
243
121
ˆ
ˆˆ
kji
kjkji ˆ
2
ˆ
ˆ
2
ˆˆ
0
and the X- axis is given by
kji ˆ
0
ˆ
0
Dot product of these two vectors is zero i.e. angular momentum is perpendicular to X-axis.
Problem 36. Two discs of moment of inertia I1 and I2 and angular speeds
21 and
are rotating along
collinear axes passing through their centre of mass and perpendicular to their plane. If the
two are made to rotate together along the same axis the rotational KE of system will be [RPMT 2000]
(a)
)(2 21
2211 II
II
(b)
2
)()( 2
2121
II
(c)
)(2
)(
21
2
2211 II
II
(d) None of these
Solution: (c) By the law of conservation of angular momentum
212211 IIII
Angular velocity of system
21
2211 II
II
(t)
O
t
(t)
O
t
(t)
O
t
(t)
O
t
138 Rotational Motion
Rotational kinetic energy
2
21
2
1
II
2
21
2211
21
2
1
II
II
II
)(2
)(
21
2
2211 II
II
.
Problem 37. A smooth uniform rod of length L and mass M has two identical beads of negligible size,
each of mass
m
, which can slide freely along the rod. Initially the two beads are at the
centre of the rod and the system is rotating with angular velocity
0
about an axis
perpendicular to the rod and passing through the mid point of the rod (see figure). There
are no external forces. When the beads reach the ends of the rod, the angular velocity of
the system is [IIT-JEE 1998]
(a)
0
(b)
mM
M
12
0
(c)
mM
M
2
0
(d)
mM
M
6
0
Solution: (d) Since there are no external forces therefore the angular momentum of the system remains
constant.
Initially when the beads are at the centre of the rod angular momentum
0
2
112
ML
L
.....(i)
When the beads reach the ends of the rod then angular momentum
'
1222
2
22
MLL
m
L
m
.....(ii)
Equating (i) and (ii)
'
12212
22
0
2
MLmLML
mM
Mo
6
'
.
Problem 38. Moment of inertia of uniform rod of mass M and length L about an axis through its centre
and perpendicular to its length is given by
12
2
ML
. Now consider one such rod pivoted at its
centre, free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet
of mass
M
moving horizontally at a speed v strikes and embedded in one end of the rod.
The angular velocity of the rod just after the collision will be
(a)
Lv
(b)
Lv2
(c)
Lv 23
(d)
Lv6
Solution: (c) Initial angular momentum of the system = Angular momentum of bullet before collision
2
L
Mv
.....(i)
L/2
L/2
Rotational Motion 139
let the rod rotates with angular velocity
.
Final angular momentum of the system
2
2
212
L
M
ML
......(ii)
By equation (i) and (ii)
4122
22 MLMLL
Mv
or
Lv 2/3
Problem 39. A solid cylinder of mass 2 kg and radius
m2.0
is rotating about its own axis without friction
with angular velocity
srad /3
. A particle of mass 0.5 kg and moving with a velocity 5 m/s
strikes the cylinder and sticks to it as shown in figure. The angular momentum of the
cylinder before collision will be
(a) 0.12 J-s
(b) 12 J-s
(c) 1.2 J-s
(d) 1.12 J-s
Solution: (a) Angular momentum of the cylinder before collision
2
2
1MRIL
3)2.0()2(
2
12
= 0.12 J-s.
Problem 40. In the above problem the angular velocity of the system after the particle sticks to it will be
(a) 0.3 rad/s (b) 5.3 rad/s (c) 10.3 rad/s (d) 89.3 rad/s
Solution: (c) Initial angular momentum of bullet + initial angular momentum of cylinder
= Final angular momentum of (bullet + cylinder) system
')( 211
IIImvr
1
Imvr
=
'
2
122
mrMr
12.02.055.0
')2.0()5.0()2.0(2
2
122
3.10'
rad/sec.
7.19 Work, Energy and Power for Rotating Body.
(1) Work : If the body is initially at rest and angular displacement is
d
due to torque then work
done on the body.
dW
[Analogue to work in translatory motion
dxFW
]
3
rad/s
M
v
140 Rotational Motion
(2) Kinetic energy : The energy, which a body has by virtue of its rotational motion is
called rotational kinetic energy. A body rotating about a fixed axis possesses kinetic energy
because its constituent particles are in motion, even though the body as a whole remains in
place.
Rotational kinetic energy
Analogue to translatory kinetic energy
2
2
1
IKR
2
2
1mvKT
LKR2
1
PvKT2
1
I
L
KR2
2
m
P
KT2
2
(3) Power : Rate of change of kinetic energy is defined as power
dt
d
II
dt
d
K
dt
d
PR
2
2
1
)(

II
In vector form Power
[Analogue to power in translatory motion
vFP
]
7.20 Slipping, Spinning and Rolling.
(1) Slipping : When the body slides on a surface without rotation then its motion is called
slipping motion.
In this condition friction between the body and surface
0F
.
Body possess only translatory kinetic energy
2
2
1mvKT
.
Example : Motion of a ball on a frictionless surface.
(2) Spinning : When the body rotates in such a manner that its axis of rotation does not
move then its motion is called spinning motion.
In this condition axis of rotation of a body is fixed.
Example : Motion of blades of a fan.
In spinning, body possess only rotatory kinetic energy
2
2
1
IKR
.
or
2
2
2
2
2
22
1
2
1
R
K
mv
R
v
mKKR
i.e., Rotatory kinetic energy =
2
2
R
K
times translatory kinetic energy.
= 0
v
Rotational Motion 141
Here
2
2
R
K
is a constant for different bodies. Value of
1
2
2
R
K
(ring),
2
1
2
2
R
K
(disc) and
2
1
2
2
R
K
(solid sphere)
(3) Rolling : If in case of rotational motion of a body about a fixed axis, the axis of rotation
also moves, the motion is called combined translatory and rotatory.
Example : (i) Motion of a wheel of cycle on a road.
(ii) Motion of football rolling on a surface.
In this condition friction between the body and surface
0F
.
Body possesses both translational and rotational kinetic energy.
Net kinetic energy = (Translatory + Rotatory) kinetic energy.
RTN KKK
2
2
22 2
1
2
1
R
K
mvmv
2
2
21
2
1
R
K
mvKN
7.21 Rolling Without Slipping.
In case of combined translatory and rotatory motion if the object rolls across a surface in
such a way that there is no relative motion of object and surface at the point of contact, the
motion is called rolling without slipping.
Friction is responsible for this type of motion but work done or dissipation of energy
against friction is zero as there is no relative motion between body and surface at the point of
contact.
Rolling motion of a body may be treated as a pure rotation about an axis through point of
contact with same angular velocity
.
By the law of conservation of energy
22 2
1
2
1
ImvKN
[
As
Rv
]
222 2
1
2
1
ImR
=
][
2
122 ImR
=
222
2
1
][
2
1
P
ImRI
[As
2
mRIIP
]
By theorem of parallel axis, where I = moment of inertia of rolling body about its centre ‘O’
and IP = moment of inertia of rolling body about point of contact ‘P’.
(1) Linear velocity of different points in rolling : In case of rolling, all points of a rigid
body have same angular speed but different linear speed.
Let A, B, C and D are four points then their velocities are shown in the following figure.
v
v
O
P
v
v
v
v
A
B
C
D
v
v
v
v = 0
A
B
C
D
2v
v = 0
B
C
D
2 v
142 Rotational Motion
(2) Energy distribution table for different rolling bodies :
Body
2
2
R
K
Translatory
(KT)
2
2
1mv
Rotatory
(KR)
2
2
2
2
1
R
K
mv
Total (KN)
2
2
1
2
2
1
R
K
mv
N
T
K
K
(%)
N
R
K
K
(%)
Ring
Cylindrical
shell
1
2
2
1mv
2
2
1mv
2
mv
2
1
(50%)
2
1
(50%)
Disc
solid cylinder
2
1
2
2
1mv
2
4
1mv
2
4
3mv
3
2
(66.6%)
3
1
(33.3%)
Solid sphere
5
2
2
2
1mv
2
5
1mv
2
10
7mv
7
5
(71.5%)
7
2
(28.5%)
Hollow sphere
3
2
2
2
1mv
2
3
1mv
2
6
5mv
5
3
(60%)
%)40(
5
2
Sample problems based on kinetic energy, work and power
Problem 41. A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with an angular
velocity of 20 rad/s. Its kinetic energy is [MP PET 2003]
(a) 10 J (b) 100 J (c) 500 J (d) 250 J
Solution: (d) Rotational kinetic energy
222 2
1
2
1
2
1
MRI
J25020)5.0(10
2
1
2
12
2
Problem 42. An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min. What
torque does it deliver [CBSE PMT 2000]
(a) 350 N-m (b) 440 N-m (c) 531 N-m (d) 628 N-m
Solution: (c)

P
mN -531
60
1800
2
10100 3
Problem 43. A body of moment of inertia of 3 kg-m2 rotating with an angular velocity of 2 rad/sec has
the same kinetic energy as a mass of 12 kg moving with a velocity of [MH CET (Med.) 1999]
(a) 8 m/s (b) 0.5 m/s (c) 2 m/s (d) 1 m/s
Solution: (d) Rotational kinetic energy of the body =
2
2
1
I
and translatory kinetic energy
2
2
1mv
According to problem
22 2
1
2
1mvI
22 12
2
1
)2(3
2
1v
smv /1
.
+
=
Rotational Motion 143
Problem 44. A disc and a ring of same mass are rolling and if their kinetic energies are equal, then the
ratio of their velocities will be
(a)
3:4
(b)
4:3
(c)
2:3
(d)
3:2
Solution: (a)
2
2
21
2
1
R
k
mvKddisc
2
4
3d
mv
discfor
2
1
2
2
R
k
As
2
2
1
2
1
R
k
mvKrring
2
r
mv
ringfor 1
2
2
R
k
As
According to problem
ringdisc KK
22
4
3rd mvmv
3
4
r
d
v
v
.
Problem 45. A wheel is rotating with an angular speed of
secrad /20
. It is stopped to rest by applying a
constant torque in
s4
. If the moment of inertia of the wheel about its axis is 0.20 kg-m2,
then the work done by the torque in two seconds will be
(a) 10 J (b) 20 J (c) 30 J (d) 40 J
Solution: (c)
20
1
rad/sec,
.4,0
2sect
So angular retardation
2
21 /5
4
20 secrad
t
Now angular speed after 2 sec
t
12
2520
=
10
rad/sec
Work done by torque in 2 sec = loss in kinetic energy =
2
2
2
1
2
1
I
))10()20(()20.0(
2
122
3002.0
2
1
= 30 J.
Problem 46. If the angular momentum of a rotating body is increased by 200%, then its kinetic energy
of rotation will be increased by
(a) 400% (b) 800% (c) 200% (d) 100%
Solution: (b) As
I
L
E2
2
2
1
2
1
2
L
L
E
E
2
1
1
3
L
L
[As
112 .%200 LLL
= 3L1]
12 9EE
%800
1 E
of
1
E
Problem 47. A ring, a solid sphere and a thin disc of different masses rotate with the same kinetic
energy. Equal torques are applied to stop them. Which will make the least number of
rotations before coming to rest
(a) Disc (b) Ring
(c) Solid sphere (d) All will make same number of rotations
Solution: (d) As

W
= Energy
n2
Energy
So, if energy and torque are same then all the bodies will make same number of rotation.
Problem 48. The angular velocity of a body is
kji ˆ
4
ˆ
3
ˆ
2
and a torque
kji ˆ
3
ˆ
2
ˆ
acts on it. The
rotational power will be
(a) 20 W (b) 15 W (c)
17
W (d)
14
W
Solution: (a) Power
.)( P
)
ˆ
4
ˆ
3
ˆ
2(.)
ˆ
3
ˆ
2( kjikji
1262
= 20 W
144 Rotational Motion
Problem 49. A flywheel of moment of inertia 0.32 kg-m2 is rotated steadily at
sec/120 rad
by a
W50
electric motor. The kinetic energy of the flywheel is
(a)
J4608
(b)
J1152
(c)
J2304
(d)
J6912
Solution: (c) Kinetic energy
2
2
1
IKR
2
)120()32.0(
2
1
= 2304 J.
7.22 Rolling on an Inclined Plane.
When a body of mass m and radius R rolls down on inclined plane of height ‘h’ and angle of
inclination
, it loses potential energy. However it acquires both linear and angular speeds and
hence, gain kinetic energy of translation and that of rotation.
By conservation of mechanical energy
2
2
21
2
1
R
k
mvmgh
(1) Velocity at the lowest point :
2
2
1
2
R
k
gh
v
(2) Acceleration in motion : From equation
aSuv 2
22
By substituting
2
2
1
2
and
sin
,0
R
k
gh
v
h
Su
we get
2
2
1
sin
R
k
g
a
(3) Time of descent : From equation
atuv
By substituting u = 0 and value of v and a from above expressions
2
2
1
2
sin
1
R
k
g
h
t
From the above expressions it is clear that,
2
2
2
2
2
21;
1
1
;
1
1
R
k
t
R
k
a
R
k
v
Note
: Here factor
2
2
R
k
is a measure of moment of inertia of a body and its value is
constant for given shape of the body and it does not depend on the mass and radius
of a body.
Velocity, acceleration and time of descent (for a given inclined plane) all depends
on
2
2
R
k
. Lesser the moment of inertia of the rolling body lesser will be the value of
h
B
S
Translatio
n
C
Rotation
Rotational Motion 145
2
2
R
k
. So greater will be its velocity and acceleration and lesser will be the time of
descent.
If a solid and hollow body of same shape are allowed to roll down on inclined plane
then as
,
2
2
2
2
HS R
k
R
k
solid body will reach the bottom first with greater velocity.
If a ring, cylinder, disc and sphere runs a race by rolling on an inclined plane then
as
minimum
sphere
2
2
R
k
while
maximum
Ring
2
2
R
k
, the sphere will reach the bottom
first with greatest velocity while ring at last with least velocity.
Angle of inclination has no effect on velocity, but time of descent and acceleration
depends on it.
velocity
, time of decent
1
and acceleration
.
7.23 Rolling Sliding and Falling of a Body.
Figure
Velocity
Acceleration
Time
Rolling
0
2
2
R
k
22
1
2
Rk
gh
22
1
sin
RK
g
2
2
1
2
sin
1
R
k
g
h
Sliding
0
2
2
R
k
gh2
sing
g
h2
sin
1
Falling
0
2
2
R
k
= 90o
gh2
g
g
h2
7.24 Velocity, Acceleration and Time for Different Bodies.
Body
2
2
R
k
Velocity
2
2
1
2
R
k
gh
v
Acceleration
2
2
1R
k
gsin
a
θ
Time of descent
2
2
1
21
R
k
g
h
sin
tθ
Ring or
Hollow cylinder
1
gh
sin
2
1g
g
h4
sin
1
Disc or solid
cylinder
5.0or
2
1
3
4gh
sin
3
2g
g
h3
sin
1
h
146 Rotational Motion
Solid sphere
4.0or
5
2
gh
7
10
sin
7
5g
g
h
5
14
sin
1
Hollow sphere
66.0or
3
2
gh
5
6
sin
5
3g
g
h
3
10
sin
1
Sample problems based on rolling on an inclined plane
Problem 50. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of
length L and height h. What is the speed of its centre of mass when the cylinder reaches its
bottom [CBSE PMT 2003]
(a)
gh
4
3
(b)
gh
3
4
(c)
gh4
(d)
gh2
Solution: (b) Velocity at the bottom (v)
gh
gh
R
K
gh
3
4
2
1
1
2
1
2
2
2
.
Problem 51. A sphere rolls down on an inclined plane of inclination
. What is the acceleration as the
sphere reaches bottom [Orissa JEE 2003]
(a)
sin
7
5g
(b)
sin
5
3g
(c)
sin
7
2g
(d)
sin
5
2g
Solution: (a) Acceleration (a)
sin
7
5
5
2
1
sin
1
sin
2
2g
g
R
K
g
.
Problem 52. A ring solid sphere and a disc are rolling down from the top of the same height, then the
sequence to reach on surface is [RPMT 1999]
(a) Ring, disc, sphere (b) Sphere, disc, ring (c) Disc, ring, sphere (d) Sphere, ring, disc
Solution: (b) Time of descent
moment of inertia
2
2
R
k
4.0
2
2
sphere
R
k
,
5.0
2
2
disc
R
k
,
1
2
2
ring
R
k
ringdiscsphere ttt
.
Problem 53. A thin uniform circular ring is rolling down an inclined plane of inclination 30° without
slipping. Its linear acceleration along the inclined plane will be [CBSE PMT 1992; BHU 1998]
(a)
2g
(b)
3g
(c)
4g
(d)
32g
Solution: (c)
2
2
1
sin
R
k
g
a
11
30sin
o
g
4
g
[As
1
2
2
R
k
and
o
30
]
Problem 54. A solid sphere and a disc of same mass and radius starts rolling down a rough inclined
plane, from the same height the ratio of the time taken in the two cases is
(a) 15 : 14 (b)
14:15
(c) 14 : 15 (d)
15:14
Rotational Motion 147
Solution: (d) Time of descent
2
2
1
2
sin
1
R
k
g
h
t
disc
shpere
t
t
=
disc
2
2
sphere
2
2
1
1
R
k
R
k
2
1
1
5
2
1
15
14
3
2
5
7
Problem 55. A solid sphere of mass
kg1.0
and radius
cm2
rolls down an inclined plane
m4.1
in length
(slope 1 in 10). Starting from rest its final velocity will be
(a)
sec/4.1 m
(b)
sec/14.0 m
(c)
sec/14 m
(d)
sec/7.0 m
Solution: (a)
2
2
1
2
R
k
gh
v
5
2
1
sin8.92
l
[As
5
2
2
2
R
k
,
sin
h
l
and
10
1
sin
given]
5/7 10
1
4.18.92
v
./4.1 sm
Problem 56. A solid sphere rolls down an inclined plane and its velocity at the bottom is v1. Then same
sphere slides down the plane (without friction) and let its velocity at the bottom be v2.
Which of the following relation is correct
(a) v1 = v2 (b)
21 7
5vv
(c)
21 5
7vv
(d) None of these
Solution: (d) When solid sphere rolls down an inclined plane the velocity at bottom
ghv7
10
1
but, if there is no friction then it slides on inclined plane and the velocity at bottom
ghv2
2
7
5
2
1 v
v
.
7.25 Motion of Connected Mass.
A point mass is tied to one end of a string which is wound round the solid body [cylinder,
pulley, disc]. When the mass is released, it falls vertically downwards and the solid body rotates
unwinding the string
m
= mass of point-mass, M = mass of a rigid body
R
= radius of a rigid body,
I
= moment of inertia of rotating body
(1) Downwards acceleration of point mass
2
1mR
I
g
a
(2) Tension in string
2
mRI
I
mgT
(3) Velocity of point mass
2
1
2
mR
I
gh
v
(4) Angular velocity of rigid body
2
2
mRI
mgh
Sample problems based on motion of connected mass
m
T
h
148 Rotational Motion
Problem 57. A cord is wound round the circumference of wheel of radius r. The axis of the wheel is
horizontal and moment of inertia about it is I. A weight mg is attached to the end of the
cord and falls from rest. After falling through a distance h, the angular velocity of the
wheel will be [MP PMT 1994; DPMT 2001]
(a)
mrI
gh
2
(b)
2
2
mrI
mgh
(c)
2
2
2
mrI
mgh
(d)
gh2
Solution : (b) According to law of conservation of energy
22)(
2
1
mrImgh
2
2
mrI
mgh
.
Problem 58. In the following figure, a body of mass m is tied at one end of a light string and this string
is wrapped around the solid cylinder of mass M and radius R. At the moment t = 0 the
system starts moving. If the friction is negligible, angular velocity at time
t
would be
(a)
)( mM
mgRt
(b)
)2(
2
mM
Mgt
(c)
)2(
2
mMR
mgt
(d)
)2(
2
mMR
mgt
Solution : (d) We know the tangential acceleration
2
1mR
I
g
a
Mm
mg
mR
MR
g
2
2
2/1
12
2
[As
2
2
1MRI
for
cylinder]
After time t, linear velocity of mass m,
atuv
Mm
mgt
2
2
0
So angular velocity of the cylinder
R
v
)2(
2
mMR
mgt
.
Problem 59. A block of mass
kg2
hangs from the rim of a wheel of radius
m5.0
. On releasing from rest
the block falls through
m5
height in
s2
. The moment of inertia of the wheel will be
(a) 1 kg-m2
(b) 3.2 kg-m2
(c) 2.5 kg-m2
(d) 1.5 kg-m2
Solution : (d) On releasing from rest the block falls through 5m height in 2 sec.
2
)2(
2
1
05 a
[As
2
2
1atutS
]
2
/5.2 sma
m
M
2kg
Rotational Motion 149
Substituting the value of a in the formula
2
1mR
I
g
a
and by solving we get
2
)5.0(2
1
10
5.2
I
2
5.1 mkgI
7.26 Time Period of Compound Pendulum.
Time period of compound pendulum is given by,
g
L
T
2
where
l
kl
L22
Here l = distance of centre of mass from point of suspension
k = radius of gyration about the parallel axis passing through centre of mass.
Body
Axis of rotation
Figure
l
k2
l
kl
L22
g
L
πT2
Ring
Tangent passing
through the rim and
perpendicular to the
plane
R
2
R
R2
g
R
T2
2
Tangent parallel to
the plane
R
2
2
R
R
2
3
g
R
T2
3
2
Disc
Tangent,
Perpendicular to
plane
R
2
2
R
R
2
3
g
R
T2
3
2
Tangent parallel to
the plane
R
4
2
R
R
4
5
g
R
T4
5
2
Spherical
shell
Tangent
R
2
3
2R
R
3
5
g
R
T3
5
2
Solid
sphere
Tangent
R
2
5
2R
R
5
7
g
R
T5
7
2
150 Rotational Motion
Sample problems based on time period of compound pendulum
Problem 60. A ring whose diameter is 1 meter, oscillates simple harmonically in a vertical plane about a
nail fixed at its circumference. The time period will be
(a)
sec4/1
(b)
sec2/1
(c)
sec1
(d)
sec2
Solution: (d)
g
R
T2
2
g
1
2
= 2 sec [As diameter
12 R
meter given]
Problem 61. A number of holes are drilled along a diameter of a disc of radius
R
. To get minimum time
period of oscillations the disc should be suspended from a horizontal axis passing through a
hole whose distance from the centre should be
(a)
2
R
(b)
2
R
(c)
22
R
(d) Zero
Solution: (b)
g
L
T
2
where
l
kl
L22
Here
2
2
2R
k
l
R
l
L2
2
2
l
R
l2
2
For minimum time period
L
should be minimum
0
dl
dL
0
2
2
l
R
l
dl
d
2
21
2
1l
R
0
2
12
2 l
R
2
R
l
.