

universe.

(The axis is outside the body).









Moment of inertia of a rigid body about an axis is defined as the sum of the product and the square of the distance of the particle from the axis of rotation.
That is, I =

Here, m is mass of particle and r is distance of particle from axis of rotation.
Note : Moment of inertia plays the same role in rotational motion as mass plays in linear motion. That is, it is a measure of inertia in rotational motion and is also
called rotational inertia.

Suppose a body of mass M has moment of inertia I about an axis. The radius of gyration , k, of the body about that axis is defined as
I = Mk2
That is, k is the distance of the point mass M from the axis of rotation such that this point mass has the same moment of inertia about that axis as the given body.

According to this theorem, the moment of inertia of body about moment of inertia of body about any axis is equal to the moment of inertia of body about a parallel
axis through the centre of mass plus the product of the mass of the body and the square of the distance between the two axes.
I = ICM + Md2

The moment of inertia of a plane body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes
in its own plane intersecting at the point through which the perpendicular axis passes.
i.e.,



Where I is the moment of inertia of the body. From this expression it is clear that moment of inertia is the rotational analogous of mass.


Its unit is Newton meter (Nm)

Angular momentum of particle about a given axis is the moment of linear momentum of the particle about axis. It is equal to the product of linear distance of the
line of action of linear momentum from axis of rotation and linear momentum.
i.e.,

The magnitude of angular momentum is


if



If the resultant external torque on a system is zero, its total angular momentum remains constant.
I

This is the rotational analogue of the law of conservation of linear momentum.

It can now be easily shown that if a torque



and, therefore, the total work done in rotating the body from



For a constant torque this becomes W = t





The rate at which work is done by a torque, i.e., power in rotational motion is

p =


It can be easily shown using the equations




We have as I is constant for a rigid body. Comparing this with


we get,

Which is the rotational analogue of p = mv.

(a) Translational Equilibrium. For a body to be in translational equilibrium, the vector sum of all the external forces on the body must be zero.
(b) Rotational Equilibrium. For a body to be in rotational equilibrium, the vector sum of all the external torques on the body about any axis must be zero.