Motion of system of particles and rigid

bodies - I

Topics: Centre of mass of two particle

system, motion of CM numerical examples
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION
 RIGID BODY: In it, on application of forces the relative
positions of particles of the body do not change.

CENTRE OF MASS OF A RIGID BODY

It is the point where all the mass of a body can be
assumed to be concentrated.
When the net force acting on the body is applied at this
point, it produces the same effect as with actual positions
on application of forces.
𝑭𝟏
𝑭𝟐

𝒎𝟏 𝑭 c.m.
𝒎𝟐 M
𝑭𝟑
𝒎𝟑

𝑭 = 𝑭𝟏 + 𝑭𝟐 + 𝑭𝟑 𝑴 = 𝒎𝟏 + 𝒎𝟐 + 𝒎𝟑

C.M. may or may not lie on the body.

 CENTRE OF MASS OF TWO PARTICLE SYSTEM
For a system consisting mass
particles 𝒎𝟏 and 𝒎𝟐 at positions 𝒓𝟏 Y 𝑭𝟏
𝒎𝟏
and 𝒓𝟐 respectively, 𝒇𝟏 𝑭𝟐
The internal forces are 𝒇𝟏 and 𝒇𝟐 . M
𝒓𝟏 𝒇𝟐
External forces acting on the 𝒓
particles are 𝑭𝟏 and 𝑭𝟐 . 𝒎𝟐
𝒅𝑷𝟏 𝒅𝒗 𝒓𝟐
Net force on 𝒎𝟏 , = 𝑭𝟏 + 𝒇𝟏 = = 𝒎𝟏 𝟏
𝒅𝒕 𝒅𝒕 X
𝒅𝑷𝟐 𝒅𝒗
O
Net force on 𝒎𝟐 , = 𝑭𝟐 + 𝒇𝟐 = = 𝒎𝟐 𝟐
𝒅𝒕 𝒅𝒕 internal forces,
𝒅𝒗𝟏 𝒅𝒗𝟐
Net force on the system, 𝑭 = 𝑭𝟏 + 𝑭𝟐 = 𝒎𝟏 + 𝒎𝟐 𝒇𝟏 + 𝒇𝟐 = 𝟎
𝒅𝒕 𝒅𝒕

𝒅 𝒅𝒓𝟏 𝒅 𝒅𝒓𝟐 𝒅𝟐 𝒓𝟏 𝒅𝟐 𝒓𝟐 𝒅𝟐 𝒎𝟏 𝒓𝟏 𝒅𝟐 𝒎𝟐 𝒓𝟐
So, 𝑭 = 𝒎𝟏 + 𝒎𝟐 = 𝒎𝟏 𝟐 + 𝒎𝟐 𝟐 = +
𝒅𝒕 𝒅𝒕 𝒅𝒕 𝒅𝒕 𝒅𝒕 𝒅𝒕 𝒅𝒕𝟐 𝒅𝒕𝟐
𝒅𝟐 𝒎𝟏 𝒓𝟏 +𝒎𝟐 𝒓𝟐 𝒅𝟐 𝒓𝒄𝒎
So, 𝑭 = 𝒎𝟏 + 𝒎𝟐 𝟐
𝒅𝒕 𝒎𝟏 +𝒎𝟐
= 𝑴 𝟐
𝒅𝒕
𝒅𝟐 𝒓
𝑭=𝒎 𝟐
𝒅𝒕
𝒎𝟏 𝒓𝟏 + 𝒎𝟐 𝒓𝟐
𝒓𝒄𝒎 = C.M. lies on the line joining the masses.
𝒎𝟏 + 𝒎𝟐
 C.M. in Cartesian co-ordinates

𝒎𝟏 𝒓𝟏 + 𝒎𝟐 𝒓𝟐
𝒓𝒄𝒎 = If c.m. lies at (x, y, z)
𝒎𝟏 + 𝒎𝟐

𝒎𝟏 𝒙𝟏 + 𝒎𝟐 𝒙𝟐 𝒎𝟏 𝒚𝟏 + 𝒎𝟐 𝒚𝟐 𝒎𝟏 𝒛 𝟏 + 𝒎𝟐 𝒛𝟐
𝒙= 𝒚= 𝒛=
𝒎 𝟏 + 𝒎𝟐 𝒎𝟏 + 𝒎𝟐 𝒎 𝟏 + 𝒎𝟐
𝒎𝟏 𝒓𝟏 +𝒎𝟐 𝒓𝟐
If C.M. lies at the origin, 𝒓𝒄𝒎 = =𝟎
𝒎𝟏 +𝒎𝟐
𝒎𝟏 𝒓𝟏 + 𝒎𝟐 𝒓𝟐 = 𝟎 𝒎𝟏 𝒓𝟏 = −𝒎𝟐 𝒓𝟐 , If 𝒎𝟏 > 𝒎𝟐 ,then 𝒓𝟏 < 𝒓𝟐
C.M. lies closer to the heavier mass.
𝒓𝟏 𝒎𝟐 𝑚1
Also = reverse ratio. 𝒓𝟏
𝒓𝟐 𝒎𝟏

For n-particle system,

𝒎𝟏 𝒓𝟏 +⋯……+𝒎𝒏 𝒓𝒏
𝒓𝒄𝒎 = 𝑚2
𝒓𝟐
𝒎𝟏 +⋯…..+𝒎𝒏
Centre of mass of a body of uniform mass distribution
Their centre of mass coincides with their geometric centre.

CM
CM h/4
CM

CM CM CM

CM
 Motion of centre of mass of a body

𝒅𝒓𝒄𝒎 𝒅 𝒎𝟏 𝒓𝟏 +⋯……+𝒎𝒏 𝒓𝒏
Velocity of c.m. 𝒗𝒄𝒎 = = ( )
𝒅𝒕 𝒅𝒕 𝒎𝟏 +⋯…..+𝒎𝒏
𝒅 𝒅
𝒎𝟏 𝒅𝒕𝒓𝟏 +⋯……+𝒎𝒏 𝒅𝒕𝒓𝒏 𝒎𝟏 𝒗𝟏 +⋯……+𝒎𝒏 𝒗𝒏
𝒗𝒄𝒎 = =
𝒎𝟏 +⋯…..+𝒎𝒏 𝒎𝟏 +⋯…..+𝒎𝒏

𝒎 𝟏 𝒗𝟏 + ⋯ … … + 𝒎 𝒏 𝒗𝒏
𝒗𝒄𝒎 =
𝒎𝟏 + ⋯ … . . +𝒎𝒏

Conservation and c.m. motion

(𝒎𝟏 + ⋯ … . . +𝒎𝒏 )𝒗𝒄𝒎 = 𝒎𝟏 𝒗𝟏 + ⋯ … … + 𝒎𝒏 𝒗𝒏

𝑴𝒗𝒄𝒎 = 𝒎𝟏 𝒗𝟏 + ⋯ … … + 𝒎𝒏 𝒗𝒏
Momentum of c.m. 𝑷𝒄𝒎 = 𝑷𝟏 + ⋯ … … + 𝑷𝒏
 Examples of motion of centre of mass
When a stationary radium (Ra) nucleus breaks into
alpha particle and radon (Rn), the products fly off in
such a way that the c.m. of the two always remain at
rest.

𝑴𝒗 = 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐 = 𝟎
𝜶 Ra Rn 𝒎𝟏 𝒗𝟏 = −𝒎𝟐 𝒗𝟐

When Ra is moving CM continues to

move with same
velocity
Ra
CM

Rn
𝑴𝒗 = 𝒎𝟏 𝒗𝟏 + 𝒎𝟐 𝒗𝟐
Example for motion of centre of mass independent
of internal forces
Y

CM

O
CM X
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 centre of mass under the influence of the external
force continues along the same parabolic trajectory as it
would have followed if there were no explosion.
Q. Two bodies of masses 1kg and 2kg are
located at (1,2) and (-1,3) respectively. Find the
location of their centre of mass.

Co-ordinates of CM,

𝒎𝟏 𝒙𝟏 +𝒎𝟐 𝒙𝟐 𝒎𝟏 𝒚𝟏 +𝒎𝟐 𝒚𝟐
𝒙= ,𝒚=
𝒎𝟏 +𝒎𝟐 𝒎𝟏 +𝒎𝟐

𝟏𝑿𝟏 + 𝟐𝑿 − 𝟏 −𝟏
𝒙= =
𝟏+𝟐 𝟑

𝟏𝑿𝟐 + 𝟐𝑿𝟑 𝟖
𝒚= =
𝟏+𝟐 𝟑
−𝟏 𝟖
So the CM is located at ( , )
𝟑 𝟑
−𝟏 𝟖
Position vector of CM 𝒓 = 𝒊Ƹ + 𝒋Ƹ
𝟑 𝟑
 Three identical spheres of radius r and uniform
mass distribution are placed on a table touching
each other. Find the location of their CM.
Co-ordinates of CM of the
spheres can be taken as(0,0),
(2r,0), (𝐫, 𝟑𝒓).So the CM of Y
system,
𝒎 𝒙 +𝒎 𝒙 +𝒎 𝒙
𝒙 = 𝟏 𝟏 𝟐 𝟐 𝟑 𝟑, m (𝐫, 𝟑𝒓)
𝒎𝟏 +𝒎𝟐 +𝒎𝟑
𝒎 𝟏 𝒚𝟏 + 𝒎 𝟐 𝒚𝟐 + 𝒎 𝟑 𝒚𝟑
𝒚=
𝒎𝟏 + 𝒎𝟐 + 𝒎𝟑
m CM m
O X
𝒎𝑿𝟎 + 𝒎𝑿𝟐𝒓 + 𝒎𝒓 𝟑𝒎𝒓 (0,0) (2r,0)
𝒙= =
𝒎+𝒎+𝒎 𝟑𝒎
=𝒓
𝒓
𝒎𝑿𝟎 + 𝒎𝑿𝟎 + 𝒎 𝟑𝒓 So the CM is located at (𝐫, )
𝒚= 𝟑
𝒎+𝒎+𝒎 Position vector of CM
𝒎𝒓 𝟑 𝒓 𝒓
= = 𝒓 = 𝒓𝒊Ƹ + 𝒋Ƹ
𝟑𝒎 𝟑
𝟑 It lies at the centroid of the triangle.
 From uniform disc of radius R, a circular hole of
radius R/2 has been removed. The centre of the
hole is at R/2 distance from the centre of the disc.
Locate the centre of gravity of the remaining part.
𝑹 𝟐 𝝅𝑹𝟐
The area of the hole 𝑨 = 𝝅( ) =
𝟐 𝟒
If the mass of the whole disc was M,
−𝑴
The mass of the removed disc 𝒎𝟏 =
𝟒
𝟑𝑴 3M/4
The mass of the remaining disc 𝒎𝟐 = OM −𝑴/𝟒
𝟒 O’
Co-ordinates of CM of remaining part, CM
−𝑹 (0,0) (R/2, 0)
𝒎 𝒙 +𝒎 𝒙 𝒎 𝒚 +𝒎 𝒚 ( , 𝟎)
𝒙= 𝟏 𝟏 𝟐 𝟐 ,𝒚= 𝟏 𝟏 𝟐 𝟐 𝟔
𝒎𝟏 +𝒎𝟐 𝒎𝟏 +𝒎𝟐

−𝑴 𝑹
𝑴𝑿𝟎+ 𝑿 −𝑴𝑹/𝟖 𝑹
𝟒 𝟐
𝒙= = =− ,
𝑴−𝑴/𝟒 𝟑𝑴/𝟒 𝟔
−𝑹
So the CM is located at ( , 𝟎)
𝟔
−𝑴 −𝑹
𝑴𝑿𝟎 + 𝑿𝟎 Position vector of CM 𝒓 = 𝒊Ƹ
𝒚= 𝟒 =𝟎 𝟔
𝑴 − 𝑴/𝟒
 Motion of system of particles and rigid
bodies - 2

Topics: Rotational motion ,Torque, angular

momentum, moment of inertia and relations
between them.
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION
 RIGID BODY
Ideally a rigid body is a body with a perfectly definite
shape. The distances between particles of such a
body do not change.
Note: No real body is truly rigid, since real bodies
deform under the influence of forces.

Types of motion a rigid body can have:

Pure translation
P2
P1 Ex-1: Motion of a block on an
inclined plane
In pure translation all particles
of the body have the same
velocity at any instant of time.

P2

P1
MOTION OF A VASE ON A PARABOLIC PATH (PROJECTILE)

In this example, the body is restricted to move only with

translation.
All particles of the body have the same velocity at any instant
of time.
The direction of velocities of the various particles will be
parallel (tangential) to the trajectory at any instant.
 ROTATION
In rotation of a rigid body about a fixed axis, 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.

Axis of rotation
 TRANSLATION AS WELL AS ROTATION OF A VASE AS A
PROJECTILE

In this example, the body is moving with translation as well as

rotation.
The particles of the body have different velocities at any given
instant of time.
The direction of velocities of the various particles will be
different w.r.t. the trajectory at any given instant.
 Examples

Z A table fan rotating about its axis (┴ to screen)

A spin rotating about its axis (vertical) A pot rotating about its vertical axis
A rigid body rotation about Y-axis: The particles describe
circular motion with the planes of the circles lying in XZ
Y perpendicular to Y-axis.
plane. i.e.

A rigid body rotation about Z-axis: Z

The particles describe circular motion with the planes of

the circles lying in XY plane. i.e. perpendicular to Z-axis.
 Translational Motion

All the particles cover

equal distance on
parallel paths

Particles follow circular paths about

the axis of rotation. Their linear
displacements are different.
Angular displacements of all
O the particles are equal
 Torque or moment of force
The torque or moment of
Axis
force is the turning effect
O
of the force about the axis 𝒓
of rotation. 𝒓 𝐬𝐢𝐧 𝜽
It is measured as the ∟ 𝑭
product of the force and
its ⊥distance from the axis
of rotation.
Torque is an axial vector.
Magnitude of torque, It is directed along the
𝝉 = 𝒓 𝐬𝐢𝐧 𝜽 𝑿𝑭 = 𝒓𝑭 𝐬𝐢𝐧 𝜽 axis of rotation given by
In vector form, Right hand screw rule.
𝝉 = 𝒓𝑿𝑭
Torque for anticlockwise rotation is positive
Torque for clockwise rotation is negative

The SI unit of torque is N m Dimensional formula 𝑴𝑳𝟐 𝑻−𝟐

 Special cases

(1) When 𝜽 = 𝟎° 𝒐𝒓 𝟏𝟖𝟎°, 𝒔𝒊𝒏𝜽 = 𝟎,

𝝉 = 𝒓𝑭 𝐬𝐢𝐧 𝜽 = 𝟎 ,
r
When line of action of force passes F
though axis of rotation, it cannot
produce turning effect.
(2) When 𝜽 = 𝟗𝟎° , 𝒔𝒊𝒏𝜽 = 𝟏(𝒎𝒂𝒙), F
𝝉 = 𝒓𝑭(𝒎𝒂𝒙) ,
Maximum turning effect.

(3) When 𝜽 = 𝟗𝟎° , 𝒔𝒊𝒏𝜽 = 𝟏(𝒎𝒂𝒙),

𝝉 = 𝒓𝑭(𝒎𝒂𝒙) ,
When r is increased, large torque
can be produced by using smaller
force.
A wrench with long arm is preferred.
 Angular momentum
Axis
It is moment of linear O
momentum of a body 𝒓
about an axis. 𝒓 𝐬𝐢𝐧 𝜽
It is measured as the
∟ 𝑷
m
product of the linear
momentum and its
⊥distance from the axis of
rotation.
Angular momentum is an
Magnitude of angular axial vector. It is directed
momentum, along the axis of rotation
𝑳 = 𝒓 𝐬𝐢𝐧 𝜽 𝑿𝒑 = 𝒓𝑷 𝐬𝐢𝐧 𝜽 given by Right hand
In vector form, screw rule.
𝑳 = 𝒓𝑿𝑷
The SI unit of torque is 𝒌𝒈 𝒎𝟐 𝒔−𝟏
Dimensional formula 𝑴𝑳𝟐 𝑻−𝟏
 Relation between linear and rotational motion
Y Anticlockwise rotation

Linear displacement,
𝝎 𝒗 𝚫𝒓 = 𝚫𝜽𝑿𝒓
𝚫𝜽 is angular
𝚫𝜽
displacement along the
X
axis of rotation (+Y).
𝒓
Linear velocity,
𝒗 = 𝝎𝑿𝒓
𝝎 is angular velocity along
the axis of rotation (+Y).

Linear acceleration,
𝒂 = 𝜶𝑿𝒓 , 𝜶 is angular acceleration.
 Moment of inertia & rotational KE
M.I. of a body about an axis of Y
rotation resists any change in state
of uniform rotational motion.
Consider a rigid body of N 𝒗𝟏
particles is rotating about Y axis 𝒎𝟐
𝒓𝟐
with angular velocity 𝝎. The mass 𝒓𝟏
particles 𝒎𝟏 ,………, 𝒎𝑵 are placed 𝒗𝟐 𝒎𝟏
at distances 𝒓𝟏 ,……, 𝒓𝑵 from axis. 𝒓𝟑
𝒎𝟑
So velocities of the particles are,
𝒗𝟏 = 𝒓𝟏 𝝎 ,………, 𝒗𝑵 = 𝒓𝑵 𝝎 𝒗𝟑
𝟏 𝟏
Total Kinetic energy of body= 𝒎𝟏 𝒗𝟏 𝟐 + ⋯ … + 𝒎𝑵 𝒗𝑵 𝟐
𝟐 𝟐
𝟏 𝟏 𝟏
𝑲𝑬 = 𝒎𝟏 (𝒓𝟏 𝝎) + ⋯ + 𝒎𝑵 (𝒓𝑵 𝝎) = (𝒎𝟏 𝒓𝟏 𝟐 … + 𝒎𝑵 𝒓𝑵 𝟐 )𝝎𝟐
𝟐 𝟐
𝟐 𝟐 𝟐
𝟏
𝑲𝑬 = 𝑰𝝎𝟐 , Moment of inertia 𝑰 = (𝒎𝟏 𝒓𝟏 𝟐 … + 𝒎𝑵 𝒓𝑵 𝟐 )
𝟐
𝟏 𝟏
Rotational 𝑲𝑬 = 𝑰𝝎𝟐 Compare it with linear 𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝟐
M.I., 𝑰 = 𝒎𝟏 𝒓𝟏 𝟐 … + 𝒎𝑵 𝒓𝑵 𝟐 = 𝜮𝒎𝒓𝟐
 Radius of gyration
Y
Radius of gyration of a body
about an axis of rotation is
the distance from the axis at
which, if all the mass of the
body is concentrated, its K M
moment of inertia would
remain same as with actual
distribution of mass.

𝑴. 𝑰. = 𝜮𝒎𝒓𝟐 = 𝑴𝑲𝟐
 Relation between torque and angular momentum

Angular momentum, In vector form,

𝑳 = 𝒓𝑿𝑷
𝒅𝑳 𝒅𝒓 𝒅𝑷
= 𝑿𝑷 + 𝒓𝑿
𝒅𝒕 𝒅𝒕 𝒅𝒕
𝒅𝑳 𝒅𝒓 𝒅𝑷
= 𝒗𝑿𝒎𝒗 + 𝒓𝑿𝑭 ( = 𝒗, 𝑷 = 𝒎𝒗 and = 𝑭)
𝒅𝒕 𝒅𝒕 𝒅𝒕
𝒅𝑳
=𝟎+𝝉 (𝒗𝑿𝒎𝒗 = 𝟎, 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒗𝒆𝒄𝒕𝒐𝒓𝒔)
𝒅𝒕

𝒅𝑳
𝝉=
𝒅𝒕
So, torque is given by rate of change of angular momentum.

𝒅𝑳 𝒅𝑷
𝝉= can be compared with 𝑭 =
𝒅𝒕 𝒅𝒕
Relation between angular momentum and angular velocity
Consider a rigid body of N Y
particles is rotating about Y axis
with angular velocity 𝝎. The mass
particles 𝒎𝟏 ,………, 𝒎𝑵 are placed
𝒗𝟏
at distances 𝒓𝟏 ,……, 𝒓𝑵 from axis. 𝒎𝟐
So velocities of the particles are, 𝒓𝟐
𝒓𝟏
𝒗𝟏 = 𝒓𝟏 𝝎 ,………, 𝒗𝑵 = 𝒓𝑵 𝝎 𝒗𝟐 𝒎𝟏
And 𝑳𝟏 = 𝒓𝟏 𝑷𝟏 ,………, 𝑳𝑵 = 𝒓𝑵 𝑷𝑵 𝒎𝟑 𝒓𝟑

Total angular momentum 𝒗𝟑

𝐋 = 𝑳𝟏 + ⋯ … + 𝑳𝑵
𝐋 = 𝒓𝟏 𝑷𝟏 +…… + 𝒓𝑵 𝑷𝑵
𝐋 = 𝒓𝟏 𝒎𝟏 𝒗𝟏 +…… + 𝒓𝑵 𝒎𝑵 𝒗𝑵
𝐋 = 𝒓𝟏 𝒎𝟏 𝒓𝟏 𝝎+…… + 𝒓𝑵 𝒎𝑵 𝒓𝑵 𝝎
𝐋 = 𝒎𝟏 𝒓𝟏 𝟐 … + 𝒎𝑵 𝒓𝑵 𝟐 𝝎 = 𝑰𝝎
Where Moment of inertia 𝑰 = (𝒎𝟏 𝒓𝟏 𝟐 … + 𝒎𝑵 𝒓𝑵 𝟐 )
𝑳 = 𝑰𝝎 or 𝑳 = 𝑰𝝎 Compare it with 𝐏 = 𝒎𝒗 or 𝑷 = 𝒎𝒗
Relation between torque and angular acceleration
Consider a rigid body of N Y
particles is rotating about Y axis
with angular acceleration 𝜶. The
mass particles 𝒎𝟏 ,………, 𝒎𝑵 are
𝒂𝟏
placed at distances 𝒓𝟏 ,……, 𝒓𝑵 from 𝒎𝟐
axis. So linear acc of the particles 𝒓𝟐
𝒓𝟏
are, 𝒂𝟏 = 𝒓𝟏 𝜶 ,………, 𝒂𝑵 = 𝒓𝑵 𝜶 𝒂𝟐 𝒎𝟏
And 𝝉𝟏 = 𝒓𝟏 𝑭𝟏 ,………, 𝝉𝑵 = 𝒓𝑵 𝑭𝑵 𝒎𝟑 𝒓𝟑

Total torque,
𝒂𝟑
𝛕 = 𝝉𝟏 + ⋯ … + 𝝉𝑵
𝝉 = 𝒓𝟏 𝑭𝟏 +…… + 𝒓𝑵 𝑭𝑵
𝝉 = 𝒓𝟏 𝒎𝟏 𝒂𝟏 +…… + 𝒓𝑵 𝒎𝑵 𝒂𝑵
𝝉 = 𝒓𝟏 𝒎𝟏 𝒓𝟏 𝛂+…… + 𝒓𝑵 𝒎𝑵 𝒓𝑵 𝜶
𝛕 = 𝒎𝟏 𝒓𝟏 𝟐 … + 𝒎𝑵 𝒓𝑵 𝟐 𝜶 = 𝑰𝜶
Where Moment of inertia 𝑰 = (𝒎𝟏 𝒓𝟏 𝟐 … + 𝒎𝑵 𝒓𝑵 𝟐 )
𝝉 = 𝑰𝜶 or 𝝉 = 𝑰𝜶 Compare it with 𝐅 = 𝒎𝒂 or 𝑭 = 𝒎𝒂
 Motion of system of particles and rigid
bodies - 3

Topics: Law of conservation of angular

momentum, moment of inertia of regular
shaped bodies.
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION
 Law of conservation of angular momentum
When no external torque acts on a system of bodies in
rotational in motion, its total angular momentum remains
conserved.
𝒅𝑳
When external torque on a system is zero, 𝝉 = =𝟎 ⇒
𝒅𝒕
𝑳 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, 𝑳 = 𝑰𝝎 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
Example: Ice skating,
When the girl closes her body,
her Moment of Inertia
decreases(𝑴. 𝑰. = 𝜮𝒎𝒓𝟐 ). As no
external torque is acting on
the system, the angular
momentum,
𝐋 = 𝑰𝝎 = 𝑰′ 𝝎′ = 𝒄𝒐𝒏𝒕𝒂𝒏𝒕
When 𝑰 𝒅𝒆𝒄𝒓𝒂𝒔𝒆𝒔, 𝝎 𝒊𝒏𝒄𝒓𝒆𝒂𝒔𝒆.
So the girl rotates faster.
For somersault from a spring
board in a swimming pool or
for jumping from high bar, the
person closes his body to
decrease (𝑴. 𝑰. = 𝜮𝒎𝒓𝟐 ) his
moment of inertia, as a result
his angular velocity
increases(𝐋 = 𝑰𝝎 = 𝒄𝒐𝒏𝒕𝒂𝒏𝒕)
. So he rotates in air. Just
before entering water or
reaching ground, he extends
his limbs to increase his
Moment of Inertia so his
angular velocity deceases and
rotation stops.
𝐋 = 𝑰𝝎 = 𝑰′ 𝝎′ = 𝒄𝒐𝒏𝒕𝒂𝒏𝒕
 Parallel axes theorem
Moment of inertia of a body about Y
axis which is parallel to an axis
passing through its centre of mass is
given by the sum of its MI about the
axis through the CM and product of
its mass and square of its distance d
C.M
from CM.
𝑴. 𝑰. = 𝑰𝑪𝑴 + 𝑴𝒅𝟐 = 𝑴𝑲𝟐
Perpendicular axes theorem
Moment of inertia of a lamina about Y
an axis which is normal to it, is
given by the sum of its MI about two
perpendicular axes in the plane of
lamina which meet the given axis at O
X
the same point.
𝑴. 𝑰. , 𝑰𝒀 = 𝑰𝑿 + 𝑰𝒁 Z
Applications of perpendicular and parallel axes theorems
M.I. of a ring about a diameter Z
Due to symmetry, the MI of ring about
diameters along X and Y axes will be
equal. 𝑰𝑿 = 𝑰𝒀 = 𝑰𝑫
MI of ring about Z axis (normal to its O
plane and passing through its C.M.),
applying perpendicular axes theorem,
𝟏 𝑴𝑹𝟐
𝑰𝒁 = 𝑰𝑿 + 𝑰𝒀 = 𝟐𝑰𝑫 , 𝑰𝑫 = 𝑰 = Y X
𝟐 𝒁 𝟐
𝑴𝑹𝟐
MI of ring about any diameter,𝑰𝑫 =
𝟐
𝑴𝑹 𝟐 𝑹
𝑴𝑲𝟐 = , radius of gyration 𝑲 =
𝟐 𝟐 R
M.I. of a ring about a tangent in plane of ring CM

applying parallel axes theorem,

𝑰𝑻 = 𝑰𝑪𝑴 + 𝑴𝑹𝟐
𝑴𝑹𝟐 𝟑𝑴𝑹 𝟐
= + 𝑴𝑹𝟐 = 𝟐
𝑴𝑲 =
𝟑𝑴𝑹𝟐
, radius of gyration 𝑲 = 𝑹
𝟑
𝟐 𝟐 𝟐 𝟐
 COUPLE
A couple consists of two equal and opposite forces
with different line of action, acting on a body produce
turning effect.

Examples1. opening or
closing of a tap

Examples2. Turning
steering of a car.

Examples3. rotating wheels

of a bicycle
 Torque of a couple
𝑭
The torque or moment of
couple is the turning effect 𝒓
𝒓 𝐬𝐢𝐧 𝜽
of the forces about the ∟ 𝑭
axis of rotation.
It is measured as the
product of the force and Magnitude of torque,
⊥distance between the 𝝉 = 𝒓 𝐬𝐢𝐧 𝜽 𝑿𝑭 = 𝒓𝑭 𝐬𝐢𝐧 𝜽
lines of action of forces. In vector form,
𝝉 = 𝒓𝑿𝑭
 Equilibrium in rotational motion

R The rod is balanced on the

tip of a wedge by using
𝒍𝟏 𝒍𝟐 forces 𝑭𝟏 𝒂𝒏𝒅 𝑭𝟐 .
R is the reaction in upward
direction. 𝒍𝟏 𝒂𝒏𝒅 𝒍𝟐 are
fulcrum distances from the centre
𝑭𝟏 𝑭𝟐
of balance.
For translational equilibrium,
Net force F=0
𝑹 − 𝑭𝟏 + 𝑭𝟐 = 𝟎
𝑹 = 𝑭𝟏 + 𝑭𝟐
For Rotational equilibrium,
Net torque 𝝉 = 𝟎
𝑭𝟏 𝒍 𝟏 − 𝑭𝟐 𝒍 𝟐 = 𝟎
𝑭𝟏 𝒍𝟏 = 𝑭𝟐 𝒍𝟐
loadXload arm = effortXeffort arm
Q. A wooden meter scale of mass 100g is supported at 40cm
mark. A 100g wt is suspended at 100cm mark. Where should
a200g wt be placed so that the scale remains horizontal?

The C.G. of the left side of rod R

will be at 20 cm mark, So its wt
40gf will act at this point. 0 x 40 70 100
Similarly the C.G. of the right 20 30 30
side of rod will be at 70 cm 40gf 60gf
(mid point) mark, So its wt 200gf 100gf
60gf will act at this point.
200gf is placed at x cm mark.

Load X load arm = effort X effort arm

𝟐𝟎𝟎𝑿 𝟒𝟎 − 𝒙 + 𝟒𝟎𝑿𝟐𝟎 = 𝟔𝟎𝑿𝟑𝟎 + 𝟏𝟎𝟎𝑿𝟔𝟎
𝟖𝟎𝟎𝟎 − 𝟐𝟎𝟎𝒙 + 𝟖𝟎𝟎 = 𝟕𝟖𝟎𝟎
𝟐𝟎𝟎𝒙 = 𝟖𝟎𝟎𝟎 − 𝟕𝟎𝟎𝟎 = 𝟏𝟎𝟎𝟎
𝒙 = 𝟓 𝒄𝒎
 Motion of system of particles and rigid bodies - 4

Topics: Equations of rotational motional, work

done and power in rotational motion with
numerical problems
PREPARED BY,
NIRUPAMA,
P.G.T. PHYSICS,
KENDRIYA VIDYALAYA,
NEW TEHRI TOWN,
DEHRADUN REGION
 Equations of rotational motion for uniform acceleration
Symbols used,
Initial angular velocity= 𝝎𝟎 , Final angular velocity= 𝝎
Constant angular acceleration= 𝜶, Angular displacement= 𝜽
Time taken= 𝒕
1st Equation
𝒅𝝎
𝜶=
𝒅𝒕
, 𝒅𝝎 = 𝜶 𝒅𝒕 , 𝝎 = 𝝎𝟎 + 𝜶 𝒕

𝝎 𝒕
‫𝒕𝒅 𝜶 𝟎׬ = 𝝎𝒅 𝝎׬‬
𝟎 Compare it with
𝒗 = 𝒖 + 𝒂𝒕
𝒕
(𝝎 )𝝎
𝝎𝟎 = 𝜶 (𝒕)𝟎

𝝎 − 𝝎𝟎 = 𝜶 𝒕
 2nd equation

𝝎=
𝒅𝜽
, 𝒅𝜽 = 𝝎 𝒅𝒕 𝟏 𝟐
𝒅𝒕 𝜽 = 𝝎𝟎 𝒕 + 𝜶𝒕
𝟐
𝜽 𝒕
න 𝒅𝜽 = න 𝝎 𝒅𝒕
𝟎 𝟎
Compare it with
𝒕 𝟏 𝟐
( 𝜽 )𝜽𝟎 = න (𝝎𝟎 +𝜶 𝒕) 𝒅𝒕 𝒔 = 𝒖𝒕 + 𝒂𝒕
𝟎 𝟐
𝒕 𝒕
𝜽 − 𝟎 = න 𝝎𝟎 𝒅𝒕 + න 𝜶 𝒕 𝒅𝒕
𝟎 𝟎

𝒕
𝜽 = 𝝎𝟎 𝒕 + 𝜶 ‫𝟎׬‬ 𝒕 𝒅𝒕

𝒕𝟐
𝜽 = 𝝎𝟎 𝒕 + 𝜶
𝟐
 3rd equation
𝒅𝜽 𝒅𝜽 𝒅𝝎 𝒅𝜽
𝝎= = = 𝜶
𝒅𝒕 𝒅𝝎 𝒅𝒕 𝒅𝝎 𝝎𝟐 = 𝝎𝟎 𝟐 + 𝟐𝜶 𝜽
𝝎 𝜽
න 𝝎 𝒅𝝎 = න 𝜶 𝒅𝜽 Compare it with
𝝎𝟎 𝟎
𝒗𝟐 = 𝒖𝟐 + 𝟐𝒂 𝒔
𝝎𝟐 𝝎
( ) =𝜶𝜽
𝟐 𝝎𝟎

𝝎𝟐 𝝎𝟎 𝟐
( − )=𝜶𝜽
𝟐 𝟐

𝝎𝟐 − 𝝎𝟎 𝟐 = 𝟐𝜶 𝜽

𝝎𝟐 = 𝝎𝟎 𝟐 + 𝟐𝜶 𝜽
 Work done in rotational motion 𝑭 𝒔𝒊𝒏𝜶
𝑭
When force is applied at the mass
particle, it gets displaced through angle 𝑭 𝒄𝒐𝒔𝜶
𝐫𝐝𝜽
𝐝𝜽. So its linear displacement ,
𝒅𝒔 = 𝒓𝒅𝜽
The component of force responsible 𝒓
Axis
for doing work is 𝑭 𝒔𝒊𝒏𝜶. So the work O
done,
𝒅𝑾 = 𝑭 𝒔𝒊𝒏𝜶 𝒅𝒔 = 𝑭 𝒔𝒊𝒏𝜶 𝒓𝒅𝜽

𝒅𝑾 = (𝒓𝑭 𝒔𝒊𝒏𝜶)𝒅𝜽 𝒅𝑾
𝒑𝒐𝒘𝒆𝒓 𝑷 =
𝒅𝒕
𝒅𝑾 = 𝝉𝒅𝜽 𝝉. 𝒅𝜽 𝒅𝜽
Or 𝑷= = 𝝉.
𝒅𝒕 𝒅𝒕
𝒅𝑾 = 𝝉. 𝒅𝜽 𝑷 = 𝝉. 𝝎

Compare with 𝒅𝑾 = 𝑭. 𝒅𝒙 and 𝑷 = 𝑭. 𝒗

A solid cylinder, a ring and a hollow sphere
having same radii are rolling down an inclined
plane from same height. What will be their
velocities when they reach the bottom?
At the highest point the
potential energy =mgh, 𝝎
h
Kinetic energy=0 𝒗
So the Mechanical
energy 𝑬 = 𝒎𝒈𝒉
At the bottom the body has both linear and rotational motion ,
𝟏 𝟏 𝟏 𝒗 𝟐
so total KE= 𝒎𝒗𝟐 + 𝑰𝝎𝟐 = 𝒎𝒗𝟐 + 𝒎𝑲𝟐 , where K
𝟐 𝟐 𝟐 𝑹
is radius of gyration.
As mechanical energy is conserved in gravitational field,
𝟏 𝒗 𝟐
𝑬 = 𝒎𝒗𝟐 + 𝒎𝑲𝟐 = 𝒎𝒈𝒉
𝟐 𝑹
𝟐
𝑲
𝒗𝟐 𝟏 + 𝟐 = 𝟐𝒈𝒉
𝑹
 𝟐 𝟐𝒈𝒉 𝟐𝒈𝒉
𝒗 = 𝑲𝟐
or 𝒗 = 𝑲𝟐
𝟏+ 𝟐 𝟏+ 𝟐
𝑹 𝑹

𝟐𝒈𝒉
For ring, 𝑲 = 𝑹 , 𝒗 = = 𝒈𝒉
𝟏+𝟏

𝑹 𝒈𝒉
For solid cylinder, 𝑲 = ,𝒗 = 𝟐
𝟐 𝟑

𝟐 𝟔𝒈𝒉
For hollow sphere, 𝑲 = 𝑹 ,𝒗 =
𝟑 𝟓
 O R
F

(a) Torque acting on the flywheel 𝝉 = 𝒓𝑭 = 𝟎. 𝟐𝟎 𝒎𝑿𝟐𝟓 𝑵 = 𝟓 𝑵𝒎

𝟏
Moment of inertia of the wheel about the axis of rotation 𝑰 = 𝑴𝑹𝟐
𝟐
𝟏
𝑰 = 𝑿𝟐𝟎𝑿𝟎. 𝟐𝟎𝑿𝟎. 𝟐𝟎 = 𝟎. 𝟒 𝒌𝒈 𝒎𝟐
𝟐
𝝉 𝟓
Also, 𝝉 = 𝑰𝜶 , ⇒ 𝜶 = = = 𝟏𝟐. 𝟓 𝒓𝒂𝒅 𝒔−𝟐
𝑰 𝟎.𝟒
 (b) Work done 𝑾 = 𝑭𝑿𝑺 = 𝟐𝟓𝑿𝟐 = 𝟓𝟎 𝒋𝒐𝒖𝒍𝒆

(c ) When the linear displacement is 2m, the angular displacement,

𝒔 𝟐
𝜽= = = 𝟏𝟎 𝒓𝒂𝒅
𝑹 𝟎.𝟐
Initial angular velocity 𝝎𝟎 = 𝟎(𝒓𝒆𝒔𝒕)
final angular velocity will be given by 𝝎𝟐 = 𝝎𝟎 𝟐 + 𝟐𝜶 𝜽

𝝎𝟐 = 𝟎 + 𝟐𝑿𝟏𝟐. 𝟓𝑿𝟏𝟎 = 𝟐𝟓𝟎 , 𝝎 = 𝟐𝟓𝟎 = 𝟓 𝟏𝟎 𝒓𝒂𝒅 𝒔−𝟏

𝟏 𝟐 𝟏
Rotational KE = 𝑰𝝎 = 𝑿𝟎. 𝟒𝑿𝟐𝟓𝟎 = 𝟓𝟎 𝒋𝒐𝒖𝒍𝒆
𝟐 𝟐

Parts (b) and (c ) give same results.

Work done by the force is converted to rotational Kinetic Energy.