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Energy and momentum of rotating systems practice problems AP Physics 1, Unit 6

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

Rotational kinetic energy, rolling without slipping, angular momentum and its conservation. 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 energy and momentum of rotating systems

K = ½Iω2rotational kinetic energy
L = Iωangular momentum of a rigid object
L = mvr sinθangular momentum of a point mass
ΔL = τΔtangular impulse
vcm = rωrolling without slipping

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 flywheel with rotational inertia 0.20 kg·m² spins at 10 rad/s. What is its rotational kinetic energy?
J

Hint

Use the rotational version of ½mv2.

Worked solution
  1. K = ½Iω2 = ½ × 0.20 × 102.
  2. K = 10 J.
Q2Warm-up
A spinning ice skater pulls her arms in. What happens to her angular velocity and her angular momentum?

Hint

Is there an external torque on the skater?

Worked solution
  1. With negligible external torque, angular momentum L = Iω is conserved.
  2. Pulling the arms in reduces I, so ω must increase to keep L constant.
  3. The answer is B.
Q3Warm-up
A disc with rotational inertia 0.50 kg·m² spins at 8.0 rad/s. What is its angular momentum?
kg·m²/s

Hint

L = Iω.

Worked solution
  1. L = 0.50 × 8.0 = 4.0 kg·m²/s.
Q4Warm-up
A solid sphere and a hoop of the same mass and radius are released together from rest at the top of a ramp, and roll without slipping. Which reaches the bottom first?

Hint

Both lose the same potential energy per kilogram. Which one has to put more of it into spinning?

Worked solution
  1. Each object's energy at the bottom is shared between moving along and spinning.
  2. The hoop has all its mass at the rim, so a larger I (MR2) than the sphere (⅖MR2).
  3. The hoop puts more of its energy into rotation, leaving less for forward motion.
  4. The sphere is faster all the way down: B.

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 skater spins at 2.0 revolutions per second with rotational inertia 3.0 kg·m². She pulls her arms in, reducing it to 1.2 kg·m². What is her new rate of spin, in revolutions per second?
rev/s

Hint

Angular momentum is conserved. You can keep the rates in revolutions per second, since the units cancel.

Worked solution
  1. I1ω1 = I2ω2.
  2. 3.0 × 2.0 = 1.2 × ω2.
  3. ω2 = 5.0 rev/s.
Q6Exam-style
A solid cylinder (I = ½MR2) rolls without slipping from rest down a slope, dropping a height of 1.5 m. What is its speed at the bottom?
m/s

Hint

Share the potential energy between translational and rotational kinetic energy, and use v = Rω.

Worked solution
  1. Energy: Mgh = ½Mv2 + ½Iω2.
  2. With I = ½MR2 and ω = v/R, the rotational term is ¼Mv2.
  3. So gh = ¾v2, giving v = √(4gh/3) = √(4 × 9.8 × 1.5 / 3) = √19.6.
  4. v = 4.43 m/s.
Q7Exam-style
A 3.0 kg hoop (I = MR2) rolls without slipping at 2.0 m/s. What is its total kinetic energy?
J

Hint

Add the translational and rotational kinetic energies.

Worked solution
  1. Translational: ½Mv2 = ½ × 3.0 × 4.0 = 6.0 J.
  2. Rotational: ½(MR2)(v/R)2 = ½Mv2 = 6.0 J.
  3. Total: 12 J. For a hoop, the two parts are always equal.
Q8Challenge
A 0.10 kg lump of clay moving at 5.0 m/s hits the rim of a disc at rest, tangentially, and sticks. The disc has radius 0.50 m and rotational inertia 0.40 kg·m², and turns freely about its centre. What is the angular velocity afterwards?
rad/s

Hint

Angular momentum about the axle is conserved. Find the clay's angular momentum before it hits.

Worked solution
  1. Clay's angular momentum about the centre: L = mvr = 0.10 × 5.0 × 0.50 = 0.25 kg·m²/s.
  2. Afterwards the clay is part of the system: I = 0.40 + 0.10 × 0.502 = 0.425 kg·m².
  3. ω = L / I = 0.25 / 0.425 = 0.59 rad/s.
Q9Exam-style
A constant torque of 5.0 N·m turns a wheel (I = 0.25 kg·m²) from rest through 10 rad. What is its final angular velocity?
rad/s

Hint

The work done by a torque is τΔθ.

Worked solution
  1. Work done: W = τΔθ = 5.0 × 10 = 50 J.
  2. This becomes rotational kinetic energy: 50 = ½ × 0.25 × ω2.
  3. ω2 = 400, so ω = 20 rad/s.
Q10Exam-style
Sand is sprinkled onto a turntable spinning freely with no external torque, and sticks to it. What happens to the turntable's angular momentum and rotational kinetic energy?

Hint

Write K in terms of L: K = L2 / 2I.

Worked solution
  1. With no external torque, L stays constant.
  2. The sand increases I, so ω falls.
  3. K = L2/2I: with L fixed and I larger, K decreases. The lost energy becomes internal energy as the sand sticks.
  4. The answer is B.

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.

Qualitative/quantitative translation style · 6 points · about 20 minutes

Disc A spins at angular velocity ω0 on a frictionless axle. An identical disc B, not spinning, is dropped onto it on the same axle. Friction between the discs makes them spin together.

  1. (a) Derive an expression for the final angular velocity of the two discs, in terms of ω0.
  2. (b) Find the fraction of the original kinetic energy that remains after the discs spin together.
  3. (c) Explain why angular momentum is conserved in this process but kinetic energy is not.

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)

No external torque about the axle, so angular momentum is conserved: Iω0 = (2I)ω, so ω = ω0/2.

Part (b)

Before: ½Iω02. After: ½(2I)(ω0/2)2 = ¼Iω02. So half the kinetic energy remains.

Part (c)

The friction torques between the discs are internal: they are equal and opposite, so the total angular momentum is unchanged. But the discs slide over each other while friction acts, and that converts kinetic energy into internal (thermal) energy.

Your mark: 0 of 6

Common mistakes in energy and momentum of rotating systems

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