AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/6/6-2)
Unit 6 · Topic 6.2
6.2 Torque and Work
A torque that acts while an object turns does work on it, moving energy into or out of the object. For a constant torque, W = τΔθ with Δθ in radians; for a changing torque, the work is the area under a graph of torque against angular position.
Key terms
- work done by a torque
- angular displacement
- torque versus angle graph
- energy transfer
- rotational kinetic energy
Work done by a torque
In Unit 3 a force did work when it acted over a distance: W = F∥d. The rotational version: a torque does work when it acts while the object turns through an angle. For a constant torque, W = τΔθ, with Δθ in radians. The units are N·m × rad, which is just joules.
The two ideas are the same thing. A tangential force F at radius r acting while the object turns through Δθ pushes along an arc of length s = rΔθ. The work is Fs = F(rΔθ) = (rF)Δθ = τΔθ.
Positive, negative or zero
- Positive work: the torque is in the same sense as the rotation. A motor spinning up a wheel adds energy.
- Negative work: the torque opposes the rotation. Friction at an axle takes energy out of a spinning wheel, usually turning it into thermal energy.
- Zero work: if nothing rotates (Δθ = 0), the torque does no work, however large it is. Straining on a stuck bolt transfers no energy to the bolt.
Work and rotational kinetic energy
The total work done by all the torques on a rigid object equals the change in its rotational kinetic energy: ΣτΔθ = ΔK_rot = ½Iω_f² − ½Iω_i². This is the work–energy theorem from 3.2, in rotational form.
It's the fastest way to connect torque to speed when you know an angle but not a time. To find how far a wheel turns before friction stops it, set the friction's negative work equal to the wheel's loss of kinetic energy.
When the torque changes: use the area
If the torque changes as the object turns, W = τΔθ won't work directly. Instead, plot torque against angular position. The area between the graph and the angle axis is the work, just as the area under a force–position graph was work in Unit 3. Area below the axis counts as negative work.
Watch the horizontal axis. Area under a torque–angle graph is work. Area under a torque–time graph is angular impulse, which changes angular momentum (6.3), not energy.
| Graph | Area means | Unit |
|---|---|---|
| torque vs angular position | work done by the torque | J |
| torque vs time | angular impulse | N·m·s |
| force vs position | work done by the force | J |
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A motor spins up a wheel
A motor applies a constant 12 N·m torque to a wheel (I = 2.0 kg·m²) that starts from rest. Ignore friction. Find the work done in the first 5.0 revolutions and the wheel's angular speed afterward.
Show the solutionHide the solution
- Step 1: Convert: Δθ = 5.0 rev × 2π rad/rev = 10π ≈ 31.4 rad.
- Step 2: W = τΔθ = (12)(31.4) ≈ 377 J.
- Step 3: All of the work becomes rotational kinetic energy: 377 = ½(2.0)ω², so ω² ≈ 377 and ω ≈ 19.4 rad/s.
Answer: W ≈ 380 J; ω ≈ 19 rad/s
- Example 2Calculator allowed
Friction stops a wheel
A wheel with I = 0.40 kg·m² spins at 25 rad/s. Friction at the axle exerts a constant 0.50 N·m torque. Through what angle does the wheel turn before it stops, and how much work does friction do?
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- Step 1: Initial kinetic energy: ½(0.40)(25)² = 125 J. Final kinetic energy: 0.
- Step 2: Friction's work equals the change in kinetic energy: W = 0 − 125 = −125 J. It's negative because the torque opposes the spin.
- Step 3: Size of the work = τΔθ: 125 = (0.50)Δθ, so Δθ = 250 rad.
- Step 4: In revolutions: 250 ÷ 2π ≈ 40 rev.
Answer: 250 rad (about 40 revolutions); friction does −125 J of work
- Example 3Calculator allowed
Work from a torque–angle graph (classic trap)
A torque on a wheel increases steadily from 0 to 8.0 N·m as the wheel turns from θ = 0 to θ = 2.0 rad, then stays at 8.0 N·m until θ = 5.0 rad. The wheel has I = 0.50 kg·m² and starts from rest. Ignore friction. Find the work done and the final angular speed.
Show the solutionHide the solution
- Step 1: Area of the triangle from 0 to 2.0 rad: ½(2.0)(8.0) = 8.0 J.
- Step 2: Area of the rectangle from 2.0 to 5.0 rad: (3.0)(8.0) = 24 J.
- Step 3: Total work = 32 J. The trap is using the maximum torque for the whole turn: (8.0)(5.0) = 40 J is too big.
- Step 4: 32 = ½(0.50)ω², so ω² = 128 and ω ≈ 11.3 rad/s.
Answer: 32 J; ω ≈ 11 rad/s
Common mistakes
- Using degrees or revolutions in W = τΔθ. The angle must be in radians for the answer to be in joules.
- Saying a torque always does work. With no rotation there is no work.
- Forgetting the sign: a torque opposing the rotation does negative work and removes kinetic energy.
- Using area under a torque–time graph as work. That area is angular impulse.
On the exam
- Questions that give an angle (or number of turns) and ask for a final speed are energy questions; use ΣτΔθ = ΔK_rot rather than kinematics.
- Graph questions may give torque against angle and ask for work or final angular speed. Show the area calculation.
Connected topics
Videos
Check yourself
4 questions on 6.2 Torque and Work. Pick an answer to see if you got it, and why.
A constant torque of 15 N·m turns a wheel through 2.0 revolutions. How much work does the torque do?
A brake exerts a constant 20 N·m torque on a flywheel with rotational inertia 2.0 kg·m², which is spinning at 10 rad/s. Through what angle does the flywheel turn before it stops?
A spinning wheel slows to a stop because friction at its axle exerts a torque opposite to its rotation. Which statement best describes the energy?
Wheel P has rotational inertia I and wheel Q has rotational inertia 4I. Both start from rest, and each is turned through the same angle by the same constant torque. Which statement is correct afterward?
0 of 4 answered