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Energy and momentum of rotating systems practice problems AP Physics C: Mechanics, Unit 6

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 free-response question

Rotational kinetic energy, rolling, angular momentum and its conservation, including collisions with rotating objects. 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
K = ½Mv2 + ½Iω2rolling
L = Iωangular momentum of a rigid body
L = r × pangular momentum of a particle
τ = dL/dttorque changes angular momentum

Use g = 9.8 m/s², as on the AP reference sheet. You can type answers like 4.9, 1/3 or 3.6e2. The full sheet is in our AP Physics C 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 wheel with I = 0.50 kg·m² spins at 4.0 rad/s. Find its rotational kinetic energy.
J

Hint

K = ½Iω2.

Worked solution
  1. K = ½ × 0.50 × 16 = 4 J.
Q2Warm-up
Find the angular momentum of a body with I = 2.0 kg·m² spinning at 3.0 rad/s.
kg·m²/s

Hint

L = Iω.

Worked solution
  1. L = 2.0 × 3.0 = 6 kg·m²/s.
Q3Warm-up
A spinning ice skater pulls her arms in. What happens?

Hint

There is no external torque about her axis.

Worked solution
  1. L = Iω is conserved, and I decreases, so ω increases.
  2. The answer is B. Her kinetic energy increases, from the work her arms do.

Exam-style questions

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

Q4Exam-style
A skater spins at 2.0 rev/s with rotational inertia 4.0 kg·m², then pulls in her arms to 1.6 kg·m². Find her new rate of spin in rev/s.
rev/s

Hint

Conserve angular momentum. Units of rev/s work on both sides.

Worked solution
  1. I1ω1 = I2ω2: 4.0 × 2.0 = 1.6ω2.
  2. ω2 = 5 rev/s.
Q5Exam-style
A solid cylinder rolls without slipping from rest down a ramp, dropping 1.5 m in height. Find its speed at the bottom.
m/s

Hint

Share the energy between translation and rotation, with I = ½MR2 and v = Rω.

Worked solution
  1. Mgh = ½Mv2 + ½(½MR2)(v/R)2 = ¾Mv2.
  2. v = √(4gh/3) = √(4 × 9.8 × 1.5/3) = √19.6.
  3. v = 4.43 m/s.
Q6Exam-style
A uniform rod of length 1.2 m is pivoted at one end and released from rest in a horizontal position. Find its angular speed as it swings through the vertical.
rad/s

Hint

The centre of mass falls L/2. Use I = ML2/3 about the pivot.

Worked solution
  1. Mg(L/2) = ½(ML2/3)ω2, so ω = √(3g/L).
  2. ω = √(3 × 9.8/1.2) = √24.5.
  3. ω = 4.95 rad/s.
Q7Exam-style
A particle moves in a straight line at constant velocity, not through point P. Its angular momentum about P:

Hint

τ = dL/dt. Is there a torque about P?

Worked solution
  1. No net force acts, so there is no torque about P.
  2. So L = mvr⊥ is constant, with r⊥ the fixed perpendicular distance from P to the line: C.
Q8Challenge
A uniform rod of mass 1.2 kg and length 1.0 m can rotate freely about its centre on a horizontal frictionless surface. A 0.20 kg ball of clay moving at 5.0 m/s, perpendicular to the rod, hits one end and sticks. Find the angular speed afterwards.
rad/s

Hint

Conserve angular momentum about the centre; the clay adds m(L/2)2 to I.

Worked solution
  1. Before: L = mv(L/2) = 0.20 × 5.0 × 0.50 = 0.50 kg·m²/s.
  2. After: I = ML2/12 + m(L/2)2 = 0.10 + 0.05 = 0.15 kg·m².
  3. ω = 0.50/0.15 = 3.33 rad/s.

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 · 7 points · about 15 minutes

A turntable with rotational inertia 0.40 kg·m² spins freely at 3.0 rad/s. A 0.50 kg lump of putty is dropped vertically onto it and sticks 0.40 m from the axis.

  1. (a) Find the new angular speed.
  2. (b) Find the kinetic energy lost.
  3. (c) Explain why angular momentum is conserved 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)

Inew = 0.40 + 0.50(0.40)2 = 0.48 kg·m2. ω = 0.40 × 3.0/0.48 = 2.5 rad/s.

Part (b)

Before: ½(0.40)(9.0) = 1.8 J. After: ½(0.48)(6.25) = 1.5 J. Lost: 0.30 J.

Part (c)

No external torque acts about the axis. The putty sticks, so the collision is inelastic and friction between putty and turntable turns kinetic energy into internal energy.

Your mark: 0 of 7

Common mistakes in energy and momentum of rotating systems

Keep going

Revise the ideas behind these questions:

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