Unit 6 · Topic 6.2

6.2 Torque and Work

A torque that acts while an object turns does work on it: $W = \int \tau\,d\theta$ , which is the area under a torque–angle graph. The net work done by all the torques equals the change in the object's rotational kinetic energy.

Key terms

work done by a torque
angular displacement
work–energy theorem
torque–angle graph
rotational kinetic energy

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of the Rotation unit (old Unit 5). It's now in Unit 6, Energy and Momentum of Rotating Systems, so older videos may file it under rotation.

Work done by a torque

In Unit 3, a force did work when its point of application moved: $W = \int F\,dx$ . For rotation, the twin is $W = \int_{\theta_1}^{\theta_2} \tau\,d\theta$ For a constant torque this is just $W = \tau\,\Delta\theta$ , with Δθ in radians. The unit is the joule, the same as any work.

This isn't a new kind of work. A tangential force F at radius r moves its point of application an arc length ds = r dθ, so F ds = (Fr) dθ = τ dθ. Writing it with torque is just easier when an object turns.

Positive and negative work

A torque in the same direction as the rotation does positive work. It transfers energy into the object, the way a motor does. A torque opposite the rotation does negative work and takes energy out, the way brakes or friction in an axle do.

A torque on an object that isn't turning does no work. Pushing hard on a stuck bolt transfers no energy to it until it starts to turn.

Graphs of torque against angle

On a graph of torque (vertical) against angular position (horizontal), the work done is the area between the curve and the axis. Area above the axis is positive work; area below is negative.

Use triangles and rectangles for straight-line graphs. If the torque is given as a function, like τ(θ) = cθ, integrate.

The work–energy theorem for rotation

The net work done by all the torques on a rigid object equals the change in its rotational kinetic energy: $W_{\text{net}} = \Delta K_{\text{rot}} = \tfrac{1}{2}I\omega_f^2 - \tfrac{1}{2}I\omega_i^2$

Use the net work, not the work by just one torque. If a motor turns a wheel while friction resists, the wheel's kinetic energy grows by the motor's work minus the energy friction takes out.

This is often faster than finding α and using kinematics, especially when the torque changes with angle.

Two ways to see the same work

Picture pulling a string off a spool with a steady force F. The string comes off at radius R. When the spool turns through Δθ, a length RΔθ of string unwinds, so your hand moves that far.

Counted as a force, your work is F times the distance your hand moves: F(RΔθ). Counted as a torque, it's τΔθ = (FR)Δθ. These are the same number. Use whichever description matches the information you have, but don't count the same work twice.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Constant torque spinning up a wheel
A constant 12 N·m torque acts on a wheel with I = 0.60 kg·m², starting from rest, while the wheel turns 5.0 revolutions. Ignore friction. How much work is done, and what is the wheel's final angular velocity?
Show the solutionHide the solution
1. Step 1: Convert the angle: 5.0 rev × 2π = 10π rad ≈ 31.4 rad.
2. Step 2: $W = \tau\,\Delta\theta = (12)(10\pi) \approx 377$ J.
3. Step 3: All of it becomes rotational kinetic energy: $377 = \tfrac{1}{2}(0.60)\omega^2$ , so $\omega = \sqrt{\dfrac{2(377)}{0.60}} \approx 35$ rad/s.
Answer: W ≈ 377 J; ω ≈ 35 rad/s
Example 2
Torque that changes with angle
A wound-up spring in a toy exerts a torque τ = (4.0 N·m/rad)θ on a wheel, where θ is how far the spring is wound. How much work does the spring do as it unwinds from θ = 3.0 rad to θ = 0?
Show the solutionHide the solution
1. Step 1: The work is the area under the τ–θ graph between 0 and 3.0 rad: $W = \displaystyle\int_0^{3.0} 4.0\,\theta\,d\theta = 2.0\,\theta^2\Big|_0^{3.0} = 18$ J.
2. Step 2: Graph check: the torque grows in a straight line from 0 to 12 N·m, so the area is a triangle: ½(3.0)(12) = 18 J.
3. Step 3: The spring's torque points the way the wheel turns as it unwinds, so the work is positive.
Answer: 18 J
Example 3Calculator allowed
Motor against friction (classic trap)
A motor applies 8.0 N·m to a wheel (I = 0.60 kg·m²) that starts at rest. A friction torque of 2.0 N·m opposes the motion. Find the wheel's angular velocity after it turns 20 rad.
Show the solutionHide the solution
1. Step 1: The trap is using only the motor's work, 160 J. Some of that energy goes to friction.
2. Step 2: Motor work: (8.0)(20) = +160 J. Friction work: −(2.0)(20) = −40 J. Net work: 120 J.
3. Step 3: $120 = \tfrac{1}{2}(0.60)\omega^2$ , so $\omega = 20$ rad/s.
Answer: 20 rad/s

Common mistakes

Using degrees or revolutions in W = τΔθ. The angle must be in radians.
Setting the work by one torque equal to ΔK when other torques act too. Only the net work equals ΔK.
Giving work a sign from the torque alone. The sign depends on whether the torque points with or against the rotation.
Multiplying a changing torque by the angle. If τ varies, use the area or the integral.

On the exam

Expect a torque–angle graph and a question about the energy transferred or the final ω. Find the area, then use the work–energy theorem.
In free response, say which torques do positive and negative work and why, before you add them.

Connected topics

Videos

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Check yourself

4 questions on 6.2 Torque and Work. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A motor exerts a constant torque of 8.0 N·m on a wheel while the wheel turns through 3.0 revolutions. How much work does the motor do?

A wheel with rotational inertia 0.60 kg·m² starts from rest on a frictionless axle. A single torque acts on it. The torque grows steadily from 0 at θ = 0 to 12 N·m at θ = 4.0 rad. It then stays at 12 N·m until θ = 6.0 rad.

Described situation

Question 2 of 4Calculator allowed

How much work does the torque do from θ = 0 to θ = 6.0 rad?

Question 3 of 4Calculator allowed

What is the wheel's angular speed at θ = 6.0 rad?

Question 4 of 4Calculator allowed

The torque on a rotor depends on its angle as $\tau(\theta) = 3\theta^2$ , with τ in N·m and θ in radians. How much work does this torque do as the rotor turns from θ = 0 to θ = 2.0 rad?

0 of 4 answered