Explain the work done by torque.

(N/A) Consider a rigid body rotating about a fixed axis,which is taken as the $Z$-axis. This axis is perpendicular to the plane $X^{\prime} Y^{\prime}$.
Let a force $\overrightarrow{F}_{1}$ act on a particle of the body at point $P_{1}$. The particle rotates on a circle of radius $r_{1}$ with center $C$ on the axis,such that $CP_{1} = r_{1}$.
In a small time interval $\Delta t$,the point moves from $P_{1}$ to $P_{1}^{\prime}$. The displacement of the particle is $\Delta S_{1} = r_{1} \Delta \theta$,and it is in the tangential direction at $P_{1}$.
Here,$\Delta \theta = \angle P_{1} C P_{1}^{\prime}$ is the angular displacement of the particle. The work done by the force $\overrightarrow{F}_{1}$ on the particle is given by:
$dW_{1} = \overrightarrow{F}_{1} \cdot d\overrightarrow{S}_{1}$
$dW_{1} = F_{1} \Delta S_{1} \cos \phi_{1}$
Since $\Delta S_{1} = r_{1} \Delta \theta$ and $\phi_{1} = 90^{\circ} - \alpha_{1}$,where $\alpha_{1}$ is the angle between the force $\overrightarrow{F}_{1}$ and the radius vector $\overrightarrow{r}_{1}$,we have:
$dW_{1} = F_{1} (r_{1} \Delta \theta) \cos(90^{\circ} - \alpha_{1})$
$dW_{1} = F_{1} r_{1} \sin \alpha_{1} \Delta \theta$
Since the torque $\tau_{1} = r_{1} F_{1} \sin \alpha_{1}$,the work done is:
$dW_{1} = \tau_{1} \Delta \theta$
For the entire body,the total work done is the sum of the work done on all particles:
$dW = \sum dW_{i} = \sum \tau_{i} \Delta \theta = \tau \Delta \theta$
where $\tau$ is the total torque acting on the body.