Read work from a torque–angle graph
You will be able to: Calculate signed graph areas and distinguish torque–angle work from torque–time impulse.
How do you find work when torque changes during a turn?
A motor’s torque increases from 0 to 6 N·m during the first 2 radians of a turn. Using the final torque for the whole turn overestimates the energy transferred. A triangle under the torque–angle line gives the correct work: 6 J.
A useful starting point: A torque does work through an angle →
Words and symbols before equations
- Torque–angle graph
- Torque in N·m vertically, angular position in rad horizontally.
- Signed area
- Area above zero torque adds work for increasing angle; area below subtracts.
- Trapezoid area
- Average of the two heights multiplied by the horizontal width.
What this picture assumes
Initial K=10 J. Net torque rises linearly from 0 to 6 N·m over 0–2 rad, then is −2 N·m over 2–4 rad. The torque jump adds no area by itself. Motion remains forward.
Connect the picture to the physics
During a small angular step, work is approximately torque times the angle step. Adding the thin strips across a torque–angle graph gives total work. For straight segments, rectangles, triangles and trapezoids are enough; no calculus is needed here.
Separate positive and negative contributions. A 6 J positive area followed by a −4 J area gives +2 J net work. The body must have enough kinetic energy to reach each plotted angle; drawing a torque function alone does not guarantee that it gets there.
Always inspect the horizontal axis. Area under torque versus angle gives work (J). Area under torque versus time gives angular impulse (N·m·s). These are distinct transfers, even if the two graphs have similar shapes.
| Graph | Signed area | Unit |
|---|---|---|
| Torque vs angle | Work / change in kinetic energy | J |
| Torque vs time | Angular impulse / change in angular momentum | kg·m²/s |
A worked example, step by step
Torque rises linearly from 0 to 6 N·m over θ=0 to 2 rad, then is −2 N·m from θ=2 to 4 rad. The wheel initially has 10 J. Find work and final K.
- First area is a triangle: ½(2 rad)(6 N·m)=6 J.
- Second area is a negative rectangle: (2 rad)(−2 N·m)=−4 J.
- Net work=6−4=2 J.
- Final K=10+2=12 J. It rises to 16 J at 2 rad, then falls to 12 J, so the assumed forward motion remains possible.
The height of a torque–angle graph is torque, not work. Its area gives work.
Is a negative torque–angle area a negative kinetic energy?
Compare with an explanation
No. It represents a negative change in kinetic energy, not the energy itself.
Predict. Change one thing. Explain.
Move the endpoint through the increasing-angle path. Use the triangle first, then subtract the growing negative rectangle. Compare accumulated work with the energy readout.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At θ=3 rad, accumulated work=4 J and K=14 J. Above-axis area adds energy; below-axis area removes it.
Initial K=10 J. Net torque rises linearly from 0 to 6 N·m over 0–2 rad, then is −2 N·m over 2–4 rad. The torque jump adds no area by itself. Motion remains forward.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant physical relationship or contact condition to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionTorque rises linearly from 2 to 6 N·m over a positive 3 rad turn. Initial K=5 J. (a) Sketch labeled axes and endpoints. (b) Find work by a trapezoid. (c) Find final K. (d) Explain what would change if the horizontal axis were time in seconds.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: θ in rad horizontally, τ in N·m vertically; endpoints (0,2) and (3,6).
- 1 point: W=[(2+6)/2]×3=12 J.
- 1 point: K_f=17 J.
- 1 point: Area would give angular impulse 12 N·m·s, not work; initial angular momentum would be needed to obtain final L.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Torque–angle area measures what?
Work.
RECALL 2Torque–time area measures what?
Angular impulse.
RECALL 3How do below-axis areas enter the sum?
As negative contributions for an increasing horizontal coordinate.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Read work from a torque–angle graph
- W = signed area under τ versus θ.
- For a straight segment: W=½(τ_start+τ_end)Δθ.
Remember: The height of a torque–angle graph is torque, not work. Its area gives work.
Conditions: Initial K=10 J. Net torque rises linearly from 0 to 6 N·m over 0–2 rad, then is −2 N·m over 2–4 rad. The torque jump adds no area by itself. Motion remains forward.
Refresh Kid · Unit 6 · Objectives 6.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.2, objectives 6.2.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
Want to work through this with a tutor?
Bring your question about Read work from a torque–angle graph. Your explanation and answers remain free to access.