Explain work done by torque.

Suppose a rigid body rotating about a fixed axis, which is taken as the $Z$-axis. This axis is perpendicular to plane $X^{\prime} Y^{\prime}$.

Let force $\overrightarrow{\mathrm{F}}_{1}$ acting on a particle of the body at point $\mathrm{P}_{1}$ and it rotates on a circle of radius $r_{1}$ with centre $\mathrm{C}$ on the axis $\mathrm{CP}_{1}=r_{1}$.

In time $\Delta t$, the point moves to the position $P_{1}^{\prime}$ from $P_{1} .$ The displacement of the particle is $\Delta \mathrm{S}_{1}=r_{1} \Delta \theta$ and it is in the direction tangential at $\mathrm{P}_{1}$

$\therefore \Delta \theta=\angle \mathrm{P}_{1} \mathrm{OP}_{1}^{\prime}$ is the angular displacement of the particle. The work done by the force $\overrightarrow{\mathrm{F}_{1}}$ on the particle is

$d \mathrm{~W}_{1} =\overrightarrow{\mathrm{F}}_{1} \cdot d \overrightarrow{\mathrm{S}_{1}} $

$=\mathrm{F}_{1} \Delta \mathrm{S}_{1} \cos \phi_{1}$

$=\mathrm{F}_{1}\left(r_{1} \Delta \theta \cos \left(90^{\circ}-\alpha_{1}\right)\right.$

where $\Delta \mathrm{S}_{1}=r_{1} \Delta \theta$ and $\phi_{1}+\alpha_{1}=90^{\circ}$

where $\phi_{1}$ is the angle between $\vec{F}_{1}$ and the tangent at $P_{1}$ and $\alpha_{1}$ is the angle between $\vec{F}_{1}$ and the

radius vector $\overrightarrow{\mathrm{OP}}=\overrightarrow{r_{1}}, \phi$