Work Done by Torque

The formula

For a constant torque $\tau$ acting through angular displacement $\theta$:

$W = \tau\theta$

For variable torque:

$W = \int_{\theta_1}^{\theta_2} \tau \, d\theta$

Derivation

A tangential force $F_t$ acts at radius $r$ on a rotating body. As the body rotates through angle $d\theta$, the point moves through arc length $ds = r \, d\theta$.

Work done:

$dW = F_t \cdot ds = F_t \cdot r \, d\theta = \tau \, d\theta$

Integrating:

$W = \int_{\theta_1}^{\theta_2} \tau \, d\theta$

For constant $\tau$: $W = \tau(\theta_2 - \theta_1) = \tau\theta$

Work-energy theorem for rotation

$W = \Delta KE_{rot} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

The analogy

Force $\leftrightarrow$ Torque. Displacement $\leftrightarrow$ Angular displacement.

Example

A motor applies a constant torque of 50 N·m to a flywheel, rotating it through 100 revolutions:

$\theta = 100 \times 2\pi = 200\pi \text{ rad}$

$W = \tau\theta = 50 \times 200\pi = 10000\pi \approx 31416 \text{ J}$