A pulley connects translation and rotation
You will be able to: Combine a mass’s force equation with a pulley’s torque equation and a no-slip constraint.
Why does a hanging mass fall more slowly when it must spin a pulley?
A hanging mass pulls a cord wrapped around a pulley. Some of the gravitational pull must create the tension that turns the pulley. A pulley with appreciable rotational inertia therefore changes the mass’s acceleration compared with free fall.
A useful starting point: Rotational second law →
Words and symbols before equations
- Tension T
- Pull exerted by the light cord, in N.
- No slipping
- Cord and pulley rim have matching tangential motion, so a=Rα.
- Pulley radius R
- Distance from axle to the cord’s point of tangency.
- Separate systems
- Use a force diagram for the mass and a torque diagram for the pulley.
What this picture assumes
1 kg hanging mass, pulley radius 0.5 m, g=10 m/s². Fixed frictionless axle; light nonstretching cord unwinds without slipping. I=0 is the ideal massless-pulley limit. Downward a and unwinding α are positive.
Connect the picture to the physics
Take the hanging mass’s downward direction positive. Its forces give mg−T=ma. The cord’s tension acts tangentially on the pulley, so TR=Iα about its fixed axle, neglecting axle friction.
No slip gives a=Rα, hence T=Ia/R². Substitute into the mass equation: mg=(m+I/R²)a, so a=mg/(m+I/R²). A larger I lowers a at fixed m and R.
For nonzero I, tension is less than mg because the mass accelerates downward, but is not zero. In the ideal I=0 limit of this unwinding setup, T=0 and a=g. This is not the same geometry as every two-mass pulley problem; draw the actual arrangement first.
A worked example, step by step
A 1 kg mass unwinds a cord from a pulley of radius 0.5 m and I=0.5 kg·m². Take g=10 m/s². Find a, T and α.
- I/R²=0.5/(0.5²)=2 kg.
- a=mg/(m+I/R²)=10/(1+2)=10/3≈3.33 m/s² downward.
- T=mg−ma=10−10/3=20/3≈6.67 N.
- α=a/R=20/3≈6.67 rad/s². Check: TR=(20/3)(0.5)=10/3 N·m and Iα=(0.5)(20/3)=10/3 N·m.
Do not set T=mg for an accelerating hanging mass or assume a massive pulley has zero net torque.
Why do we need the no-slip condition?
Compare with an explanation
It connects the mass’s linear acceleration to the rim’s tangential acceleration; without it, a=Rα need not hold.
Predict. Change one thing. Explain.
Increase pulley inertia while keeping m=1 kg and R=0.5 m fixed. Track downward acceleration, tension and angular acceleration. Check the I=0 ideal limit and the torque equation at every setting.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
a=3.33 m/s² downward; T=6.67 N; α=6.67 rad/s² unwinding. TR=3.33 N·m and Iα=3.33 N·m.
1 kg hanging mass, pulley radius 0.5 m, g=10 m/s². Fixed frictionless axle; light nonstretching cord unwinds without slipping. I=0 is the ideal massless-pulley limit. Downward a and unwinding α are positive.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use an angular-motion or torque relationship 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 questionFor this setup use m=2 kg, R=0.5 m, I=0.5 kg·m² and g=10 m/s². (a) State all three governing equations. (b) Find a. (c) Find T. (d) Find α and verify TR=Iα.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: mg−T=ma, TR=Iα, and a=Rα.
- 1 point: a=20/(2+2)=5 m/s² downward.
- 1 point: T=20−2(5)=10 N.
- 1 point: α=5/0.5=10 rad/s²; TR=5 N·m=Iα.
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 1Mass force balance?
mg−T=ma with downward positive.
RECALL 2Pulley torque balance?
TR=Iα for the ideal fixed axle.
RECALL 3When does a=Rα hold?
For a taut, nonstretching cord that does not slip at the rim.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A pulley connects translation and rotation
- mg−T=ma; TR=Iα; a=Rα when cord does not slip.
- a=mg/(m+I/R²) for this single-mass unwinding setup.
Remember: Do not set T=mg for an accelerating hanging mass or assume a massive pulley has zero net torque.
Conditions: 1 kg hanging mass, pulley radius 0.5 m, g=10 m/s². Fixed frictionless axle; light nonstretching cord unwinds without slipping. I=0 is the ideal massless-pulley limit. Downward a and unwinding α are positive.
Refresh Kid · Unit 5 · Objectives 5.6.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.6, objectives 5.6.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.
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