Obtain the relation between torque and moment of inertia.

(N/A) The rotational kinetic energy of a rigid body is given by $K = \frac{1}{2} I \omega^{2}$.
The rate at which work is done on the body is equal to the rate at which its kinetic energy increases. The rate of increase of kinetic energy is:
$\frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} I \omega^{2} \right)$
$P = \frac{1}{2} I \frac{d}{dt} (\omega^{2})$
Since $I$ is constant for a rigid body:
$P = \frac{1}{2} I \times 2 \omega \frac{d\omega}{dt}$
$P = I \omega \alpha$,where $\alpha = \frac{d\omega}{dt}$ is the angular acceleration.
We also know that the power delivered by a torque is $P = \tau \omega$.
Equating the two expressions for power:
$\tau \omega = I \omega \alpha$
$\tau = I \alpha$
This equation is the rotational analogue of Newton's second law for linear motion,$F = ma$. Thus,the angular acceleration $\alpha$ is directly proportional to the applied torque $\tau$ and inversely proportional to the moment of inertia $I$ of the body.