Topic 6.2 Torque and Work: calculate the work done by a torque through an angular displacement and apply the work-energy theorem to rotation.

What this topic is asking

The College Board (Topic 6.2) wants you to calculate the work done by a torque as it turns an object through an angular displacement, and to apply the work-energy theorem in rotational form. The pattern mirrors translation exactly: where linear work is $W = Fd$, rotational work is $W = \tau\,\Delta\theta$, and the work done equals the change in rotational kinetic energy.

The work done by a torque

The substitution that builds the whole topic is force becomes torque and linear displacement becomes angular displacement. A constant torque turning a wheel through an angle $\Delta\theta$ does work $\tau\,\Delta\theta$, exactly as a constant force pushing through a distance $d$ does work $Fd$. Because work is a transfer of energy, positive work from a torque speeds the rotation up and negative work (a torque opposing the motion, like friction in a bearing) slows it down.

The rotational work-energy theorem

This theorem is powerful because it lets you bypass the angular acceleration entirely. If you know the net torque and the angle turned, you know the work done, and therefore the change in rotational kinetic energy, and therefore the final angular velocity, without ever solving $\tau_{net} = I\alpha$ for $\alpha$ and integrating. It is the rotational equivalent of using $W = \Delta K$ to find a final speed without kinematics.

Rotational power

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

the analogue of $P = Fv$ from Topic 3.5. A motor delivering a torque $\tau$ to a shaft spinning at angular velocity $\omega$ delivers power $\tau\omega$. This relation is why a car engine quotes both torque and the rpm at which it is produced: the useful power output depends on both. The deeper point of this topic is that the entire energy framework of Unit 3 carries over to rotation by analogy. Wherever you used force, distance, $\tfrac{1}{2}mv^2$, and $Fv$ for translation, you use torque, angular displacement, $\tfrac{1}{2}I\omega^2$, and $\tau\omega$ for rotation. Recognizing this analogy lets you choose energy methods for rotational problems, often the fastest route, and connects torque (Topic 5.3) to the energy of rotating systems (Topic 6.1).

Try this

Q1. A torque of $2.0$ N$\cdot$m turns a wheel through $6.0$ rad. Calculate the work done. [2 points]

• Cue. $W = \tau\,\Delta\theta = (2.0)(6.0) = 12$ J.

Q2. A torque does $30$ J of work on a wheel initially at rest with rotational inertia $0.60$ kg$\cdot$m squared. Calculate the final angular velocity. [2 points]

• Cue. $30 = \tfrac{1}{2}(0.60)\omega^2 = 0.30\,\omega^2$, so $\omega^2 = 100$ and $\omega = 10$ rad/s.