Consider a mass m attached to the end of a massless rod as shown in Fig. 5.8.let us assume that the bearing at the pivot point O is frictionless. Let the system be in a horizontal plane. A force F is acting on the mass perpendicular to the rod and hence, this will accelerate the mass according to
Fig. 5.8: The force F causes a torque about the axis O and gives the mass man angular acceleration about the pivot point.
F = ma
In doing the force will cause the mass to rotate about O. Since tangential acceleration at is related to angular acceleration a by the equation. 
So, F = mra
As turning effect is produced by torque τ, it would, therefore, be better to write the equation for rotation in terms of torque. This can be done by multiplying both sides of the above equation by r. Thus 
Which is rotational analogue of the Newton’s second law of motion, F = ma.
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Here F is replaced by τ, a by a and m by mr2. The quantity mr2 is known as the moment of inertia and is represented by I. the moment of inertia plays the same role in angular motion as the mass in linear motion. It may be note that moment of inertia depends not only on mass m but also on r2.most rigid bodies have different mass concentration at different distances from the axis of rotation, which means the mass distribution is not uniform. As shown in Fig. 5.9(a), the rigid body is made up of n small pieces of masses
Fig. 5.9: Each small piece of mass within a large, rigid body undergoes the same angular acceleration about the pivot point.
m1,m2,………mn at distances r1,r2,……..rn from the axis of rotation O. let the body be rotating with the angular acceleration a,so the magnitude of the torque acting on m1 is 
And so on.
Since the body is rigid, so all the masses are rotating with the same angular acceleration a,
Total torque τ total is then given by 
Or τ = I a …………… (5.16)
Where I is the moment of inertia of the body and is expressed as
| For your information |
 Moments of inertia of various bodies about AA’. |