Rotational Kinetic Energy and Work by Torque said a torque turning something through $\Delta\theta$ does work $\tau\Delta\theta$. Work by a Variable Force and Angular Velocity and Acceleration with Calculus extend that to torques that change, and to the rate of work.
Key idea
$$W = \int_{\theta_1}^{\theta_2} \tau\,d\theta, \qquad P = \dfrac{dW}{dt} = \tau\omega, \qquad W_{\text{net}} = \Delta\left(\tfrac{1}{2}I\omega^2\right),$$ with angles in radians and $\omega$ in rad/s.
Why it works
A force $F$ applied tangentially at radius $r$ moves its point of application along an arc $ds = r\,d\theta$. The work is $F\,ds = (Fr)\,d\theta = \tau\,d\theta$. Divide by $dt$ for the power: $\tau\,d\theta/dt = \tau\omega$. The work-energy theorem becomes $\int\tau_{\text{net}}\,d\theta = \Delta K_{\text{rot}}$, the turning version of The Work-Energy Theorem from Newton's Second Law.
Worked example: Engine power
An engine delivers $300\ \text{N·m}$ of torque at $3000\ \text{rpm}$. What is its power?
- $\omega = 3000 \times \dfrac{2\pi}{60} = 314\ \text{rad/s}$.
- $P = \tau\omega = 300(314) = 94\,200\ \text{W} \approx 94.2\ \text{kW}$.
Answer: About $94.2\ \text{kW}$.
Worked example: Spinning up a flywheel
A steady $5\ \text{N·m}$ torque turns a flywheel ($I = 2\ \text{kg·m}^2$), starting from rest, through 10 revolutions. How fast does it spin?
- Angle in radians: $10(2\pi) = 62.8\ \text{rad}$; work $W = 5(62.8) = 314\ \text{J}$.
- $\tfrac{1}{2}(2)\omega^2 = 314$, so $\omega = 17.7\ \text{rad/s}$.
Answer: $17.7\ \text{rad/s}$.
Worked example: A torque that fades
A spring motor's torque falls as it unwinds, $\tau(\theta) = 20 - 4\theta$ (N·m, rad). How much work does it do from $\theta = 0$ to $3\ \text{rad}$?
- $W = \int_0^3 (20 - 4\theta)\,d\theta = \left[20\theta - 2\theta^2\right]_0^3 = 60 - 18$.
- $W = 42\ \text{J}$, the shaded area in the picture. Using the starting torque for the whole turn would give $60\ \text{J}$, too much.
Answer: $42\ \text{J}$.
Common mistake
Multiplying the torque by an angle in degrees, or by an arc length. Turning $5\ \text{N·m}$ through $90^\circ$ does $5 \times \pi/2 = 7.85\ \text{J}$, not $450\ \text{J}$. And torque times arc length has units of N·m², not joules. Check: in $W = \tau\theta$, the radian is what makes N·m of torque become N·m of energy.
Friction torques work the same way with a minus sign. A grinding wheel ($I = 0.5\ \text{kg·m}^2$) spinning at $40\ \text{rad/s}$ stores $\tfrac{1}{2}(0.5)(40)^2 = 400\ \text{J}$; if a $2\ \text{N·m}$ friction torque stops it, it turns through $400/2 = 200\ \text{rad}$, about 32 revolutions, before stopping.
Gears and bicycle gears trade torque for speed at nearly constant power: $P = \tau\omega$, so a gear that turns the wheel half as fast can deliver about twice the torque. That is why a low gear helps on a hill.