Rotational Work & Power

Engineering Physics | AQA A-Level Physics

Work done by a rotating object

Work has to be done on a rigid body when a torque turns it through an angle about an axis. Think of rotating cranes and fairground rides.
In linear systems, work $W$ is the product of force and distance moved. For a rotating object, the work done is: $$W = \tau\theta$$
Where: $W$ = work done (J), $\tau$ = torque (N m), $\theta$ = angular displacement (rad).
Work can also be found by calculating the area under a torque-angular displacement graph. This is directly analogous to the area under a force-displacement graph.

Power output of a rotating object

Power is the rate of doing work: $$P = \frac{\Delta W}{\Delta t} = \frac{\Delta\tau\theta}{\Delta t} = \tau\frac{\Delta\theta}{\Delta t}$$
Since $\frac{\Delta\theta}{\Delta t} = \omega$, this simplifies to: $$P = \tau\omega$$
Where: $P$ = power (W), $\omega$ = angular velocity (rad s$^{-1}$).
This is the angular version of the linear equation $P = Fv$.

Frictional torque

Frictional torqueThe difference between the applied torque and the resulting net (observed) torque. It is the torque caused by friction between surfaces in contact during rotation. is the torque caused by the frictional force when two objects in contact move past each other during rotation.
The net torque is the sum of the applied and frictional torques: $$\text{Net torque} = \text{applied torque} + \text{frictional torque}$$
In rotating machinery, power must be expended to overcome frictional torque. In most cases, frictional torque is minimised to reduce kinetic energy losses transferred to heat and sound.
Frictional torque is calculated using the same equation: $\tau = Fr = I\alpha$, but $F$ is the frictional force instead of an externally applied force.

Worked example: circular saw

A circular saw rotates at 3100 rev min$^{-1}$. A horizontal force of 45 N acts at an effective radius of 0.22 m. A constant frictional torque of 2.7 N m acts at the bearings.
Torque on the saw blade: $\tau = Fr = 45 \times 0.22 = 9.9$ N m.
Total torque: $9.9 + 2.7 = 12.6$ N m.
Convert angular velocity: $\omega = 3100 \times \frac{2\pi}{60} = 324.63$ rad s$^{-1}$.
Output power: $P = \tau\omega = 12.6 \times 324.63 = 4100$ W.

Common Mistake

Always make sure $\theta$ is in radians when you are doing conversions from rev s$^{-1}$ or rev min$^{-1}$. Forgetting to multiply by $2\pi$ is one of the most common errors in rotational work and power calculations.