A torque can transfer mechanical energy when a rigid body rotates. For rotation about a fixed axis, let $\tau$ be the signed torque component about that axis and let $d\theta$ be the corresponding angular displacement.

The small amount of rotational work is

$$\boxed{dW=\tau,d\theta}.$$

If the torque varies with angle, the work from $\theta_i$ to $\theta_f$ is

$$\boxed{W=\int_{\theta_i}^{\theta_f}\tau(\theta),d\theta}.$$

For constant torque,

$$W=\tau,\Delta\theta.$$

The sign is determined by whether torque and angular displacement have the same rotational sense.

Rotational power

Instantaneous power is the rate of doing work:

$$P=\frac{dW}{dt}.$$

Using $dW=\tau,d\theta$,

$$P=\tau\frac{d\theta}{dt}.$$

Since

$$\omega=\frac{d\theta}{dt},$$

we obtain

$$\boxed{P=\tau\omega}$$

for fixed-axis rotation.

This is the rotational counterpart of

$$P=\mathbf F\cdot\mathbf v.$$

Worked example

A motor applies constant torque

$$\tau=50,\mathrm{N,m}$$

to a shaft through an angular displacement

$$\Delta\theta=10,\mathrm{rad}.$$

The work transferred is

$$W=(50)(10)=500,\mathrm J.$$

If the shaft is rotating at

$$\omega=20,\mathrm{rad/s}$$

at some instant while the same torque acts, the mechanical power is

$$P=(50)(20)=1000,\mathrm W=1.0,\mathrm{kW}.$$

Rotational work-energy theorem

For a rigid body with constant moment of inertia $I$ rotating about a fixed axis,

$$\tau_{\rm net}=I\alpha.$$

Using

$$\alpha=\frac{d\omega}{dt} =\frac{d\omega}{d\theta}\frac{d\theta}{dt} =\omega\frac{d\omega}{d\theta},$$

we have

$$\tau_{\rm net},d\theta =I\omega,d\omega.$$

Integrating gives

$$W_{\rm net} =\frac12I\omega_f^2-\frac12I\omega_i^2.$$

Therefore

$$\boxed{W_{\rm net}=\Delta K_{\rm rot}}.$$

The translational and rotational work-energy relations are not separate principles; they are the same energy-transfer idea expressed in different coordinates.