We consider a particle of mass \(m\) which moves in a circle of radius \(r\) under the action of a constant tangential force \(\vec F_\text{tan}\). The displacement of the particle is \(d\theta\) measured in radian in a very small time interval \(dt\). If \(ds\) is the arc length subtended by the angle \(d\theta\), we have \(ds = rd\theta \). So the small work done by the tangential force for the time interval \(dt\) is
\[dW = {F_{\tan }}Rd\theta = \tau d\theta \tag{1} \label{1}\]
If the tangential force is constant, the torque is constant, so the work done by the constant torque from initial angular displacement \(\theta_1\) to final angular displacement \(\theta_2\) is obtained by integrating the above expression from \(\theta_1\) to \(\theta_2\).
\[W = \tau \int\limits_{{\theta _1}}^{{\theta _2}} {d\theta } = \tau ({\theta _2} - {\theta _1}) = \tau \Delta \theta \tag{2} \label{2}\]
The Eq. \eqref{2} is analogous to \(W = F\Delta s\) for translational motion. We also know the constant acceleration equation of rotational motion \({\omega _2^2 = \omega _1^2 + 2\alpha {\kern 1pt} \Delta \theta }\) for initial angular velocity \({\omega _1}\), final angular velocity \({\omega _2}\) and angular displacement \(\Delta \theta\). Substituting the value of \(\tau = I\alpha \) and \(\Delta \theta = \frac{{\omega _2^2 - \omega _1^2}}{{2\alpha }}\) in Eq. \eqref{2}, you'll get,
\[W = \frac{1}{2}I\omega _2^2 - \frac{1}{2}I\omega _1^2 \tag{3} \label{3}\]
This expression is analogous to the work-energy theorem for the translational motion. The quantity \(\frac{1}{2}I\omega^2\) is the kinetic energy of a particle in rotational motion. You saw that the work done by the torque on the particle in rotational motion is equal to the change in rotational kinetic energy of the particle.
What happens when the tangential force \(\vec F_\text{tan}\) is not constant? The Eq. \eqref{3} is valid even if the tangential force varies. From Eq. \eqref{1} the small work done for small time interval \(dt\) is
\[\begin{align*} dW &= \tau d\theta \\ &= I\alpha d\theta = I\frac{{d\omega }}{{dt}}d\theta \\ &= I\frac{{d\theta }}{{dt}}d\omega = I\omega {\kern 1pt} {\kern 1pt} d\omega \end{align*}\]
Now the total work done from initial angular velocity \(\omega_1\) to final angular velocity \(\omega_2\) is determined by integrating the above expression from \(\omega_1\) to \(\omega_2\).
\[W = I\int\limits_{{\omega _1}}^{{\omega _2}} {\omega {\kern 1pt} {\kern 1pt} d\omega } = \frac{1}{2}I\omega _2^2 - \frac{1}{2}I\omega _1^2\]
This is the same as Eq. \eqref{3} for the constant tangential force. So you can conclude that Eq. \eqref{3} is valid even if the tangential force varies.
Let us consider a rolling body as shown in Figure 1. The body is moving with both rotational and translational motion.
If the body has moment of inertia \(I\) about its axis of rotation, mass \(m\), angular speed \(\omega\) and linear speed \(v\) , the total kinetic energy of the body at any instant is the sum of the rotational kinetic energy and the translational kinetic energy at that instant. Therefore the total kinetic energy \(K_\text{total}\) is
\[K_\text{total} = \frac{1}{2}I \omega ^2 + \frac{1}{2}mv^2\]
Power is the rate of change of work. It means how much work is done per unit time. From Eq. \eqref{1} the small work done on the particle in time \(dt\) is \(dW = \tau d\theta \) and therefore the power \(P\) is
\[P = \frac{{dW}}{{dt}} = \frac{{\tau d\theta }}{{dt}} = \tau \omega \tag{4} \label{4}\]
