Chapter 12 Rotational Motion Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes
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.
 Rigid body : A rigid body is a body that can rotate with all the parts locked together and without any change in its
 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
 Internal forces : All the forces exerted by various particles of the system on one another are called internal These forces are alone enable the particles to form a well defined system. Internal forces between two particles are mutual (equal and opposite).
 External forces : To move or stop an object of finite size, we have to apply a force on the object from This force exerted on a given system is called an external force.
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
m1, m2, m3 ……mn , whose positions vectors are
® ® ® ®
r1, r2, r3………… rn
respectively then position vector of centre of mass
® ® ® ®
® m1 r1 + m2 r2 + m3 r3 +…………….. mn rn
r = m + m + m +…………. m


3 n
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 :
® = m1 r1 + m2 r2
r
and the centre of mass lies between the particles on the line joining them.
m1 + m2
® ®
® r + r
If two masses are equal i.e. m = m , then position vector of centre of mass = ^{1} ^{2}
1 2 r 2
(3) Important points about centre of mass
 The position of centre of mass is independent of the coordinate system
 The position of centre of mass depends upon the shape of the body and distribution of
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.
 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
 Position of centre of mass for different bodies
 The centre of mass changes its position only under the translatory There is no effect of rotatory motion on centre of mass of the body.
 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. S mi ri = 0 .
 If a system of particles of masses m1 , m2 , m3 ,…… move with velocities v1 , v2 , v3 ,……
then the velocity of centre of mass vcm
= å mivi .
å m
i
 If a system of particles of masses m1, m2, m3,…… move with accelerations a1 , a2 , a3 ,……
then the acceleration of centre of mass
Acm
= å miai
å m
i
®
 If r is a position vector of centre of mass of a system
® ® æ ® ® ® ö
then velocity of centre of mass
vcm = d r
= d ç m_{1} r_{1} + m_{2} r_{2} + m_{3} r 3 +…. ÷

dt dt ç
è
m1 + m2 + m3 +……….. ÷
® ® 2 ®
2 æ ® ® ö
 Acceleration of centre of mass
Acm = d vcm = d r = d ç m_{1} r_{1} + m_{2} r_{2} +………… ÷
dt dt ^{2} dt ^{2} ç m_{1} + m_{2} + m_{3} +…… ÷
è ø
® ® 2 ®
 Force on a rigid body F =M Acm = M d r
dt ^{2}
 For an isolated system external force on the body is zero
® = d æ®
ö = 0 Þ ®
= constant .
F M dt ç v cm ÷
v cm
è ø
i.e., centre of mass of an isolated system moves with uniform velocity along a straightline path.
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]
 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 xaxis
^{®} ˆ ˆ ^{®} ˆ ˆ
m_{1} = 12 , m_{2} = 16 ,
r 1 = 0i + 0 j and r 2 = 1.1i + 0 j
® ®
® m r + m r
16 ´ 1.1 ˆ
r = 1 1 2 2 =
m1 + m2
28 i

®
r = 0.63 i
i.e. 0.63 Å from carbon atom.
Problem 2. The velocities of three particles of masses 20g, 30g and 50 g are r r
r
respectively. The velocity
10i ,10 j, and 10k
of the centre of mass of the three particles is [EAMCET 2001]
r r r
r r r
r r r
r r r
(a)
2i + 3 j + 5k
(b)
10(i + j + k)
(c)
20i + 30 j + 5k
(d)
2i + 30 j + 50k
Solution : (a) Velocity of centre of mass vcm
= m1v1 + m2v2 + m3v3 m1 + m2 + m3
= 20 ´ 10ˆi + 30 ´ 10ˆj + 50 ´ 10kˆ = 2ˆi + 3ˆj + 5kˆ.
100
Problem 3. Masses
8, 2, 4, 2 kg
are placed at the corners A, B, C, D respectively of a square ABCD of diagonal
80 cm . The distance of centre of mass from A will be [MP PMT 1999]
(a)
20 cm
(b)
30 cm
(c)
40 cm
(d)
60 cm
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 d = a
= 80 cm Þ a = 40
2cm
m_{1} = 8kg,
m_{2} = 2kg,
m_{3} = 4kg,
m4 = 2kg
Let
r 1 , r 2 , r 3 , r 4
are the position vectors of respective masses
r 1 = 0ˆi + 0ˆj ,
r2 = aˆi + 0ˆj , r3 = aˆi + aˆj , r4
= 0ˆi + aˆj
From the formula of centre of mass
r = m1 r1 + m2 r2 + m3 r3 + m4 r4 m1 + m2 + m3 + m4
= 15
2i + 15
2ˆj
\ coordinates of centre of mass = (15
of the corner = (0, 0)
2,15
2) and coordination
From the formula of distance between two points (x1, y1) and (x2, y2)
distance = =
Problem 4. The coordinates of the positions of particles of mass 7, 4
=
and 10 gm
= 30cm
are (1, 5, – 3), (2, 5,7)
and (3, 3, – 1) cm




respectively. The position of the centre of mass of the system would be

(a)
æ 15 , 85 , 1 ö cm
(b)
æ 15 , 85 , 1 ö cm
(c)
æ 15 , 85 , 1 ö cm
(d)
æ 15 , 85 , 7 ö cm
ç 17 ÷
ç 17 ÷
ç 21 ÷
ç 21 ÷











Solution: (c)
m1 = 7gm , m2 = 4 gm , m3 = 10gm and
r1 = (ˆi + 5ˆj – 3kˆ), r2 = (2i + 5 j + 7k),
r3 = (3ˆi + 3ˆj – kˆ)
Position vector of center mass
r = 7(ˆi + 5ˆj – 3kˆ) + 4(2ˆi + 5ˆj + 7kˆ) + 10(3ˆi + 3ˆj – kˆ) = (45ˆi + 85ˆj – 3kˆ)
7 + 4 + 10 21
Þ r = 15 ˆi + 85 ˆj – 1 kˆ . So coordinates of centre of mass é15 , 85 , – 1ù .
7 21 7
Angular Displacement.
êë 7
21 7 úû
It is the angle described by the position vector r about the axis of rotation.
Angular displacement (q ) = Linear displacement (s)
Radius (r)
 Unit : radian
 Dimension : [M ^{0} L^{0} T ^{0} ]
® ® ®
 Vector form S = q ´ r
i.e., angular displacement is a vector quantity whose direction is given by right hand rule. It is also known as axial vector. For anticlockwise sense of rotation direction of q is perpendicular to the plane, outward and along the axis of rotation and viceversa.
 2p radian = 360° = 1
 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).
Angular Velocity.
The angular displacement per unit time is defined as angular velocity.
If a particle moves from P to Q in time Dt , w = Dq
Dt
where Dq
is the angular displacement.
 Instantaneous angular velocity w = lim Dq
Dt ®0 Dt
= dq
dt
 Average angular velocity w _{av}
= total angular displacement
total time
= q _{2} – q_{1}
t – t
2 1
 Unit : Radian/sec
 Dimension : [M ^{0} L^{0}T ^{1} ] which is same as that of
® ® ® ® ®
 Vector form
v = w ´ r [where v = linear velocity, r = radius vector]
®
w is a axial vector, whose direction is normal to the rotational plane and its direction is given by right hand
screw rule.
(6) w = 2p
T
= 2pn
[where T = time period, n = frequency]
(7) The magnitude of an angular velocity is called the angular speed which is also represented by w .
Angular Acceleration.
The rate of change of angular velocity is defined as angular acceleration.
If particle has angular velocity w_{1}
at time t1 and angular velocity w _{2}
at time t _{2}
then,
® ®

Angular acceleration ® = w 2 – w 1
t 2 – t1
®
®
Dw d
d 2 ®
 Instantaneous angular acceleration a = lim
Dt ®0 Dt
= w =
dt
q .
dt ^{2}
 Unit : rad/sec ^{2}
 Dimension : [M ^{0} L^{0} T ^{2} ] .
 If a = 0 , circular or rotational motion is said to be
 Average angular acceleration a
= w _{2} – w_{1} .

av
2
 t1
® ® ®
 Relation between angular acceleration and linear acceleration a = a ´ r .
 It is an axial vector whose direction is along the change in direction of angular velocity e. normal to the rotational plane, outward or inward along the axis of rotation (depends upon the sense of rotation).
Equations of Linear Motion and Rotational Motion.
Problem 5. The angular velocity of seconds hand of a watch will be [CPMT 2003]
(a)
p rad / sec
60
(b)
p rad / sec
30
(c)
60 p
rad / sec
(d)
30 p
rad / sec
Solution : (b) We know that second’s hand completes its revolution (2p) in 60 sec \ w = q
t
= 2p 60
= p rad/sec
30
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/sec^{2} (b) 1880 degrees/sec^{2} (c) 40 radians/sec^{2} (d) 1980 degrees/sec^{2} Solution: (d) Angular acceleration (a) = rate of change of angular speed
2p æ 4500 – 1200 ö
3300
2p (n – n )
ç 60 ÷ 2p
360 degree
= 2 1
t
= è ø
10
= 60 ´
10 2p
sec ^{2}
= 1980 degree / sec ^{2} .
Problem 7. Angular displacement (q ) of a flywheel varies with time as q = at + bt ^{2} + ct ^{3}
then angular acceleration is given
by [BHU 2000]
(a)
a + 2bt – 3ct ^{2}
(b)
2b – 6t
(c)
a + 2b – 6t
(d)
2b + 6ct
Solution: (d) Angular acceleration a = d2q = d2 (at + bt ^{2} + ct^{3} ) = 2b + 6ct
dt^{2} dt^{2}
Problem 8. A wheel completes 2000 rotations in covering a distance of 9.5 km . The diameter of the wheel is [RPMT 1999]
(a)
1.5 m
(b)
1.5 cm
(c)
7.5 m
(d)
7.5 cm
Solution: (a) Distance covered by wheel in 1 rotation = 2pr = pD (Where D= 2r = diameter of wheel)
\ Distance covered in 2000 rotation = 2000 pD = 9.5 ´ 10^{3}m (given)
\ D = 1.5 meter
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)
600 rad
(b)
75 rad
(c)
300 rad
(d)
150 rad
Solution: (d) Angular acceleration a = w 2 – w1
t
= 60 – 0 = 12rad / sec ^{2}
5
Now from q = w_{1}t + 1 a t ^{2} = 0 + 1 (12)(5)^{2} = 150 rad.
2 2
Problem 10. A wheel initially at rest, is rotated with a uniform angular acceleration. The wheel rotates through an angle q _{1}
in first one second and through an additional angle q _{2} in the next one second. The ratio q 2 is
q_{1}
(a) 4 (b) 2 (c) 3 (d) 1
Solution: (c) Angular displacement in first one second q
= 1 a (1)^{2} = a ……(i) [From q = w t + 1 a t ^{2} ]
^{1} 2 2 ^{1} 2
Now again we will consider motion from the rest and angular displacement in total two seconds
q_{1} + q _{2} = 1 a (2)^{2} = 2a
2
Solving (i) and (ii) we get q = a
and q = 3a
……(ii)
\ q 2 = 3 .
^{1} 2 ^{2} 2 q_{1}
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
= 1 ´(2.5 + 5)´ 3000 = 11250 revolution.
2
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
 x = 2y
 x = y
 y = 2x
 None of these
Solution: (c) Linear displacement (S) = Radius (r) × Angular displacement (q)
\ S µ r
(if q =
constant)
Distance travelled by mass A (x) = Radius of pulley concerned with mass A (r) = 1
Þ y = 2x .
Distance travelled by mass B (y) Radius of pulley concerned with mass B (2r) 2
Problem 13. If the position vector of a particle is ® = (3ˆi + 4ˆj) meter and its angular velocity is ® = (ˆj + 2kˆ) rad/sec then its
r w
linear velocity is (in m/s)
(a)
(8ˆi – 6ˆj + 3kˆ)
(b)
(3ˆi + 6ˆj + 8kˆ)
ˆi ˆj
(c)
kˆ
– (3ˆi + 6ˆj + 6kˆ)
(d)
(6ˆi + 8ˆj + 3kˆ)
Solution: (a)
v = w ´ r
=(3ˆi + 4ˆj + 0kˆ) ´(0ˆi + ˆj + 2kˆ) = 3 4
0 1
0 = 8ˆi – 6ˆj + 3kˆ
2
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.
 Moment of inertia of a particle I = mr ^{2} ; where r is the perpendicular distance of particle from rotational
 Moment of inertia of a body made up of number of particles (discrete distribution)
I = m r ^{2} + m r ^{2} + m r ^{2} + …….
1 1 2 2 3 3
 Moment of inertia of a continuous distribution of mass, treating the element of mass dm at position r as particle
dI = dmr ^{2}
i.e.,
I = ò r ^{2} dm
 Dimension : [ML^{2}T ^{0} ]
 I. unit : kgm^{2}.
 Moment of inertia depends on mass, distribution of mass and on the position of axis of
 Moment of inertia does not depend on angular velocity, angular acceleration, torque, angular momentum and rotational kinetic
 It is not a vector as direction (clockwise or anticlockwise) is not to be specified and also not a scalar as it has different values in different Actually it is a tensor quantity.
 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
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.
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.
I = Mk ^{2} or k = .
Here k is called radius of gyration.
From the formula of discrete distribution

I = mr12 + mr2 2 + mr3 2 +……….. + mr ^{2}
If m_{1} = m_{2} = m_{3} =….. = m then

I = m(r12 + r2 2 + r3 2 +………….. r ^{2} )
From the definition of Radius of gyration,
I = Mk ^{2}
By equating (i) and (ii)

Mk ^{2} = m(r12 + r2 2 + r3 2 +…………….. + r ^{2} )
……..(i)
……..(ii)

nmk ^{2} = m(r12 + r2 2 + r3 2 +…………… + r ^{2} )
[As
M = nm ]
\ k =
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.
 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.
 Radius of gyration (k) does not depends on the mass of
 Dimension [M ^{0} L^{1}T ^{0} ].
 I. unit : Meter.
 Significance of radius of gyration : Through this concept a real body (particularly irregular) is replaced by a point mass for dealing its rotational
Example : In case of a disc rotating about an axis through its centre of mass and perpendicular to its plane
k = = = R
So instead of disc we can assume a point mass M at a distance (R / the rotational motion of the disc.
2) from the axis of rotation for dealing
Note : @ For a given body inertia is constant whereas moment of inertia is variable.
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 I_{g} and body and a is the perpendicular distance between the two axes.
I = I g + Ma ^{2}
Ma^{2}
where M is the mass of the
Example: Moment of inertia of a disc about an axis through its centre and
perpendicular to the plane is
1 MR ^{2} , so moment of inertia about an axis through its
2
tangent and perpendicular to the plane will be
I = Ig + Ma^{2}
I = 1 MR^{2} + MR^{2}
2
\ I = 3 MR^{2}
2
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.
I z = I x + I y
Example : Moment of inertia of a disc about an axis through its centre of mass and perpendicular to its plane is
1 MR^{2} , so if the disc is in x–y plane then by theorem of perpendicular axes
2
i.e.
Þ
Þ
I z = I x + Iy

MR^{2} = 2ID
2
I = 1 MR^{2}
D 4
[As ring is symmetrical body so
I x = Iy = I D ]
Note : @In case of symmetrical twodimensional 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.
Moment of Inertia of Two Point Masses About Their Centre of Mass.
Let m1
and m2
be two masses distant r from eachother and
r1 and r2
be the distances of their centre of
mass from m_{1} and m_{2}
(1) r_{1} + r_{2} = r
(2) m1r1 = m2r2
respectively, then
 r1
= m2
m + m
r and
r2 =
m1 r

m + m
1 2 1 2
 I = m r ^{2} + m r ^{2}
I = é
1 1
m1m2
2 2
ù r ^{2} = m r ^{2}
[where m =
m1m2
is known as reduced mass
m < m
and m < m .]
êm + m ú
m + m ^{1} ^{2}
ë 1 2 û 1 2
 In diatomic molecules like above
H 2 , HCl
etc. moment of inertia about their centre of mass is derived from
Analogy Between Tranlatory Motion and Rotational Motion.
Translatory motion Rotatory motion  
Mass  (m)  Moment of Inertia  (I)  
Linear momentum  P = mv  Angular Momentum  L =  Iw 
P = 2mE  L =  2IE  
Force  F = ma  Torque  t =  Ia 
Kinetic energy  E = 1 mv^{2}
2 
E =  1 Iw ^{2}
2 

E = P 2  L2  
2m 
E = 
2I 
Moment of Inertia of Some Standard Bodies About Different Axes.
Body Axis of
Rotation
Ring About an axis passing through
C.G. and perpendicular to its plane
Ring About its diameter
Figure Moment of inertia
MR ^{2}
k k^{2}/R^{2}
R 1

1
2

Ring About a tangential axis in its own plane
3 3
2 R 2
Ring About a tangential axis perpendicular to its own plane
2MR ^{2} 2
Disc About an axis passing through
C.G. and perpendicular to its plane
Disc About its Diameter
1


2
R 1
2 4
Disc About a tangential axis in its own plane
5 R 5
2 4
Disc About a tangential axis perpendicular to its own plane
3 3
2 R 2
Annular disc inner radius = R_{1} and outer radius = R_{2}
Passing through – –
the centre and
R2
perpendicular to
R1
the plane
Annular disc Diameter – –
Annular disc Tangential and – –
Parallel to the diameter
Annular disc Tangential and – –
perpendicular to the plane

Solid cylinder About its own axis 1

R 2
L
Solid cylinder Tangential (Generator)
3 3
2 R 2
Solid cylinder About an axis
L2 R 2
passing through its
C.G. and perpendicular to its own axis
+
12 4
Solid cylinder About the diameter
L2 R 2
of one of faces of 3 + 4
the cylinder
Cylindrical shell

About its own axis
MR^{2} R 1
Cylindrical shell Tangential
(Generator)
2MR^{2} 2
Cylindrical shell About an axis
L2 R 2
passing through its
C.G. and perpendicular to its own axis
+
12 2
Cylindrical shell About the diameter
L2 R 2
of one of faces of 3 + 2
the cylinder
Hollow cylinder with inner radius = R_{1} and outer radius = R_{2}
Axis of cylinder
R2 R1
Hollow cylinder with inner radius = R_{1} and outer radius = R_{2}
Tangential
Solid Sphere About its diametric
axis
2 2
5 R 5
Solid sphere About a tangential
axis
7 7
5 R 5
Spherical shell About its diametric
axis
2 2
3 R 3
Spherical shell About a tangential
axis
5 5
3 R 3
Hollow sphere of About its diametric
inner radius R_{1}
and
axis
outer radius R_{2}
Hollow sphere Tangential
Long thin rod About on axis
passing through its centre of mass and
L
perpendicular to
the rod.
Long thin rod About an axis
ML^{2}


12


ML^{2}
Rectangular lamina of length l and breadth b
passing through its 3
edge and L
perpendicular to the rod
Passing through the
centre of mass and b
perpendicular to
the plane l
Rectangular lamina Tangential
perpendicular to the plane and at the midpoint of breadth
Rectangular lamina Tangential
perpendicular to the plane and at the midpoint of length
ii iii

b
Rectangular
Passing through
i (i)
parallelopiped length l, breadth b, thickness t
centre of mass and t
parallel to l
 Length (x)
Rectangular parallelepiped length l, breath b, thickness t
Elliptical disc of semimajor axis = a and semiminor axis = b
 breadth (z)
 thickness (y)
Tangential and parallel to
 length (x)
 breadth (y)
 thickness(z)
Passing through CM and perpendicular to the plane


i ii
iii
(ii)
(iii)
(i)
(ii)
(iii)
M [l^{2} + b^{2} + 3t^{2}] 12
Solid cone of radius R and height h
Axis joining the vertex and centre of the base
Equilateral triangular lamina with side a
Passing through CM and perpendicular to the plane
Ma ^{2}
6
Right angled triangular lamina of sides a, b, c
Along the edges
 Mb ^{2}
6
b c
 Ma ^{2}
a 1 6
2 3