Topic 6.2 Torque and Work: compute the work done by a torque as the integral of torque over angle, apply the rotational work-energy theorem, and define rotational power as $P = \tau\omega$.

What this topic is asking

The College Board (Topic 6.2) wants you to compute the work done by a torque as the integral of torque over angular displacement, to apply the rotational work-energy theorem, and to define rotational power as $P = \tau\omega$. Each is the rotational counterpart of a translational result from Unit 3, so the calculus and the reasoning transfer directly, with angles replacing distances.

Work done by a torque

This parallels translational work exactly: where linear work integrates force over distance, rotational work integrates torque over angle. A constant torque turning a body through an angle does work $\tau\Delta\theta$; a varying torque must be integrated. The torque does positive work when it acts in the direction of rotation (speeding the body up) and negative work when it opposes the rotation (a braking torque). Like its translational cousin, this integral is the area under a torque-versus-angle graph.

The rotational work-energy theorem

Integrating the rotational form of Newton's second law over angle gives the rotational work-energy theorem: the net work done by torques equals the change in rotational kinetic energy.

This is the rotational mirror of $W_{net} = \Delta K$ from Unit 3. Its value is the same: you can find the final angular velocity from the net work without solving the detailed angular motion. It is especially handy when the torque varies with angle, so the angular acceleration is not constant and the kinematic equations do not apply; the energy method sidesteps that entirely.

Rotational power

The rate at which a torque does work is the rotational power:

This is the rotational analogue of $P = Fv$. A motor delivering torque $\tau$ to a shaft spinning at angular velocity $\omega$ outputs power $\tau\omega$; for a fixed power, $\tau = P/\omega$ falls as the shaft spins faster, the rotational version of the constant-power vehicle. Engine and motor specifications are exactly this relationship: power equals torque times rotational speed, which is why peak power and peak torque occur at different speeds.

Try this

Q1. A constant torque of $3.0$ N m turns a wheel through $6.0$ rad. Calculate the work done. [2 points]

• Cue. $W = \tau\Delta\theta = (3.0)(6.0) = 18$ J.

Q2. A motor delivers $12$ N m of torque to a shaft spinning at $50$ rad/s. Calculate the power output. [2 points]

• Cue. $P = \tau\omega = (12)(50) = 600$ W.