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Torque and rotational dynamics practice problems AP Physics 1, Unit 5

Written by STEM-Tutor's tutors · Updated October 2026 · 10 questions and 1 free-response question

Rotational kinematics, torque, rotational inertia, equilibrium, and Newton's second law for rotation. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Free-response practice
  5. Common mistakes
  6. Keep going

Key equations for torque and rotational dynamics

τ = rF sinθtorque about an axis
α = Στ / INewton's second law for rotation
I = Σmr2rotational inertia of point masses
ω = ω0 + αtrotational kinematics, like v = v0 + at
v = rωlinking linear and angular speed

All questions use g = 9.8 m/s², as on the AP Physics 1 equation sheet. A calculator is allowed, as in the exam. The full sheet is in our AP Physics 1 equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
A 30 N force is applied at right angles to a door, 0.40 m from the hinge. What torque does it produce about the hinge?
N·m

Hint

Torque = distance from the axis × perpendicular force.

Worked solution
  1. τ = rF sinθ, with θ = 90° so sinθ = 1.
  2. τ = 0.40 × 30 = 12 N·m.
Q2Warm-up
A wheel starts from rest with a constant angular acceleration of 3.0 rad/s². What is its angular velocity after 4.0 s?
rad/s

Hint

Rotational kinematics works exactly like linear kinematics.

Worked solution
  1. ω = ω0 + αt = 0 + 3.0 × 4.0.
  2. ω = 12 rad/s.
Q3Warm-up
Why is it harder to open a door by pushing near the hinge than at the handle, using the same force?

Hint

Torque depends on where the force is applied, not just how big it is.

Worked solution
  1. τ = rF sinθ: the same force at a smaller distance r from the axis gives a smaller torque.
  2. Less torque means less angular acceleration, so the door opens more slowly.
  3. The answer is B. The rotational inertia about the hinge does not depend on where you push.
Q4Warm-up
Two 2.0 kg point masses are fixed to a light rod, each 0.50 m from the axis of rotation. What is the rotational inertia of the system?
kg·m²

Hint

Add mr2 for each mass.

Worked solution
  1. I = Σmr2 = 2.0 × 0.502 + 2.0 × 0.502.
  2. I = 0.50 + 0.50 = 1.0 kg·m².

Exam-style questions

Questions at the level of the AP Physics 1 exam. The ones marked Challenge combine more than one idea.

Q5Exam-style
A 30 kg child sits 2.0 m from the pivot of a seesaw. How far from the pivot, on the other side, must a 40 kg child sit to balance it?
m

Hint

For balance, the clockwise and anticlockwise torques about the pivot are equal.

Worked solution
  1. Torque balance: 30g × 2.0 = 40g × d, so g cancels.
  2. d = 60 / 40 = 1.5 m.
Q6Exam-style
A uniform 4.0 m beam of mass 20 kg rests on supports at each end. A 50 kg person stands 1.0 m from the left end. What force does the left support exert?
N

Hint

Take torques about the right-hand support, so its force has no torque.

Worked solution
  1. About the right support: the beam's weight acts 2.0 m away and the person 3.0 m away.
  2. Balance: FL × 4.0 = (20 × 9.8)(2.0) + (50 × 9.8)(3.0) = 392 + 1470.
  3. FL = 1862 / 4.0 = 466 N. Check: the right support gives 686 − 466 = 220 N, and the person is nearer the left, so the left force is larger, as expected.
Q7Exam-style
A net torque of 6.0 N·m acts on a disc with rotational inertia 0.50 kg·m². What is its angular acceleration?
rad/s²

Hint

Use Newton's second law for rotation.

Worked solution
  1. α = Στ / I = 6.0 / 0.50.
  2. α = 12 rad/s².
Q8Exam-style
A string is wrapped round a pulley that is a uniform disc of mass 2.0 kg and radius 0.10 m (I = ½MR2). The string is pulled with a steady 4.0 N. What is the pulley's angular acceleration?
rad/s²

Hint

Find the rotational inertia and the torque separately.

Worked solution
  1. I = ½ × 2.0 × 0.102 = 0.010 kg·m².
  2. The string pulls tangentially at the rim: τ = 0.10 × 4.0 = 0.40 N·m.
  3. α = 0.40 / 0.010 = 40 rad/s².
Q9Exam-style
A wheel spins at a constant rate. How does the linear speed of a point on the rim compare with that of a point halfway between the axle and the rim?

Hint

Every point on the wheel has the same angular velocity ω.

Worked solution
  1. v = rω, and ω is the same for every point on the rigid wheel.
  2. The rim point is twice as far from the axle, so its linear speed is twice as large.
  3. The answer is B.
Q10Challenge
A 1.0 kg block hangs from a light string wrapped round a pulley: a uniform disc of mass 2.0 kg and radius 0.10 m on a frictionless axle. The block is released. What is its acceleration?
m/s²

Hint

Write Newton's second law for the block and for the pulley, and link them with a = Rα.

Worked solution
  1. Block: mg − T = ma.
  2. Pulley: TR = Iα = (½MR2)(a/R), so T = ½Ma = 1.0a.
  3. Add the equations: mg = (m + ½M)a, so 9.8 = 2.0a.
  4. a = 4.9 m/s², less than g because some of the energy goes into turning the pulley.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Mathematical routines style · 6 points · about 20 minutes

A uniform beam of mass M and length L is attached to a wall by a hinge at one end. It is held horizontal by a light cable fixed to its other end, making an angle of 30° with the beam.

hingecable30°uniform beam, mass M, length L
A uniform beam hinged to a wall and held horizontal by a cable from its far end.
  1. (a) Derive an expression for the tension T in the cable, in terms of M and g.
  2. (b) Calculate the tension when M = 10 kg.
  3. (c) The cable is moved so that it attaches to the middle of the beam, still at 30°. Does the tension increase, decrease or stay the same? Justify your answer.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Take torques about the hinge, so the hinge force has no torque. The weight Mg acts at L/2, and the perpendicular component of the tension, T sin 30°, acts at L. Balance: T sin 30° × L = Mg × L/2, so T = Mg / (2 sin 30°) = Mg.

Part (b)

T = Mg = 10 × 9.8 = 98 N.

Part (c)

It increases (it doubles). The weight's torque about the hinge is unchanged, but the cable now acts at L/2 instead of L, so it needs twice the force to give the same torque.

Your mark: 0 of 6

Common mistakes in torque and rotational dynamics

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics 1 past papers, with official scoring guidelines. · All AP Physics 1 units