1. For a uniform rod the moment of inertia about an axis through the centre
is
Therefore the moment of inertia about an axis through one end can be obtained
From
Moment of inertia of a sphere of radius R and mass M about an axis through the point on its circumference can be obtained.
 Figure 9 IG 5 = 2MR2 From parallel axes theorem I = IG + 2Mh h = R
1. 3 . Moment of inertia of a disc of radius r and mass m about an axis through a point on the circumference can be obtained.
2
IG = 2MR
5
From parallel axes theorem
2
I = IG + Mh
2 2
I = 1MR + MR But h = R
2
###### 2 2
I = MR + 2MR
2
2
I = 3MR
2
PERPENDICULAR AXES THEOREM
The perpe
ndicular axis theorem states that “the moment of inertia of plane body about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the plane body about any two mutually perpendicular axes in the plane of the body which intersect the first axis”
COMBINED ROTATIONAL AND TRANSLATIONAL MOTION
Consider a wheel rolling on a flat surface without slipping. Each particle of the wheel is undergoing two types of motion at the same time.
The centre of the wheel, which is wheel‘s centre of mass, is moving horizontally with speed VG.
Figure 10 (a)
At the same time, the wheel is rotating about its centre of mass with angular speed ω.

Figure 10 (b)
Thus rolling is the superposition of two motions
1. Translation of objects centre of mass at velocity VG along a straight line
2. Rotation about the center of mass at angular velocity ω.
Therefore the wheel possesses both translational and rotational K.E.
M = Mass of the object
VG = Velocity of object’s centre of mass
I = Moment of inertia of the object about an axis through the centre of mass.
The total K.E of an object undergoing both translational and rotational motion about its centre of mass axis is equal to the sum of translation K.E of the centre of mass and the rotational K.E about the centre of mass.
ROLLING OBJECTS
Consider a rigid body rolling without slipping down the plane inclined at an angle θ with the horizontal.
Required to determine the velocity, acceleration and minimum required coefficient of static friction (μs) for rolling.
Figure. 11
By the conservation of energy we have
Total energy at A = total energy at B E .
EA = EB
K.EA +
P.EA =
1. Rolling a Hollow Sphere
I = MR2
From
2 2 2
But v = R ω
Then
2 2 2 2
MgH = 1 MR ω + 1 MR ω
2 2
2 2
MgH = MR ω
gH = MR2
2
v = gH
Is the velocity of a hollow sphere
From the figure
For Acceleration
rd
From the 3 equation of motion
2 2
v = u + 2aL
Assuming the body started from rest
u = 0
Then
2 v = 2aL
Also
2
v = gH
2aL = gH
Is the acceleration of hollow sphere
Then from
2.
ROLLING A SOLID SPHERE From
2 2 2
v = R ω
2
7V = 10gH
2
V = 10gH
7
Then
Is the velocity of solid sphere
For acceleration
From,
2 v = u2 + 2aL
But u = 0
2
v = 2aL
gH = 2aL
a = 5 . gH
7 L
3. ROLLING OF A SOLID CYLINDER
From
2
I = 1 MR
2
2 2 2
v = R ω
Is the velocity of solid cylinder.
For acceleration
From,
2 2
v = u + 2aL
u = 0
2 v = 2aL
But,
Is a acceleration of solid cylinder.
ALTERNATIVELY
Consider a solid cylinder of radius R and mass M rolling down a plane inclined at an angle θ to the horizontal.
For rotational motion about the centre of mass, the only force that produce the torque is. Torque on cylinder
If I is the moment of inertia and α is the angular acceleration about the axis of rotation, then torque on the cylinder is FS
Equating equation (ii) and (iii)
Since the cylinder rolls without slipping
Then
Putting the value of FS into equation (i)
For a solid cylinder, the moment of inertia about its symmetry axis is
All uniform rolling solid cylinders have the same linear acceleration down the incline irrespective of their masses and radii.
The linear acceleration is just 2/3 as large as it would be if the cylinder could slide without friction down the slope
Minimum Required for Rolling
From
1.The frictional force is static, the cylinder rolls without slipping and there is no relative motion between the cylinder and the inclined plane at the point of contact. 2. If there were no friction between the cylinder and the incline, the cylinder would have slipped instead of rolled.
MOTION OF A MASS TIED TO THE STRING WOUND ON A CYLINDER
Consider a solid cylinder of radius R and mass M capable of rotating freely without friction about its symmetry axis.
Suppose a string of negligible mass is wrapped around the cylinder and a heavy point mass m is suspended from the free end.
Figure 12 (a)
When the point mass m is released, the cylinder rotates about its axis and at the same time the mass falls down due to gravity and unwinds the string wound on the cylinder
Figure 12 (b)
Linear Acceleration of Point Mass (m)
We first examine the forces of the point mass m as shown below
Looking at the cylinder end; we see that tension in the string exerts a torque on the cylinder
If I is the moment of inertia of the cylinder about the axis of rotation and α is the angular acceleration produced in the cylinder.
……………………..(iii)
By equating equations (ii) and (iii)

Since the string unwinds without slipping the linear acceleration a of the point mass m and the angular acceleration α of the cylinder are related as
Putting the value of T into equation (i)
1. The tension T in the string is always less than mg.
2. As moment of inertia increases T approaches mg.
NEWTON’S LAWS OF ROTATIONAL MOTION
1. First law
States that “everybody continues in its state of rest or of uniform rotational motion about an axis until it is compelled by some external torque to change that state”.
1. Second law
States that “the rate of change of angular momentum about an axis is directly proportional to the impressed external torque and the change angular momentum takes place in the direction of the applied torque”.
where k = 1.
1. Third law
To every external torque applied, there is an equal and opposite restoring torque.
SIMPLE PENDULUM
Figure. 13
It consists of a string of length l suspended from a fixed point O and carrying a bob of mass m at its free end.
When the bob is slightly displaced from the equilibrium position and then released, it executes rotational motion about an axis through the suspension point O.
Suppose at anytime t, the bob is at point P making an angle θ with the vertical let be the angular velocity and α be the angular acceleration at that instant.
The only force which exerts a torque on the bob is that due to the weight of the bob.
The negative sign is used since this is a restoring torque i.e. a torque which force the bob toward θ = 0°.
The angular momentum of the bob about the axis of rotation at the considered instant (at point P)
L = mvr v = ωr
L = mωr.r
2
L = mωr
But r = lsinθ where θ = 90º

Therefore L = mωl2
From equation (i) and (ii) we have
= α
dt
Now , the angular acceleration of the bob at time t
This governs the oscillation of simple pendulum,in the vertical plane.
1. When θ = 0º
Therefore, the bob has zero angular acceleration at the equilibrium position.
1. The negative sign shows that the tangential acceleration is directed toward the mean position.
1. If θ is small, Sin θ = θ
1. Time period
Also
From
For S.H.M
From
Parallel axis theorem
Then
Then

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