AP Physics 1 Energy and Momentum of Rotating Systems — Worked Answer Explanations

Unit 6 · 7% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Physics 1 Energy and Momentum of Rotating Systems practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Energy and Momentum of Rotating Systems practice test and come back here to review, or head back to the Energy and Momentum of Rotating Systems unit overview.

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Question 1 · Easy
A solid disk of moment of inertia $I = 0.20 kg\cdotm^{2}$ spins at angular speed $ω = 10 rad/s$ . What is its rotational kinetic energy?
- A
  $1.0 J$
  Why not A: Missed the factor of $\frac{1}{2}$ and forgot to square.
- B
  $2.0 J$
  Why not B: Used $I ω$ .
- C
  $10 J$ Correct
- D
  $20 J$
  Why not D: Forgot the $\frac{1}{2}$ factor.
Explanation
$K E_{r o t} = \frac{1}{2} I ω^{2} = \frac{1}{2} \times 0.20 \times 100 = 10 J$ .

Key takeaway
Rotational kinetic energy: $\tfrac{1}{2} I \omega^2$ — analog of $\tfrac{1}{2} m v^2$.
Question 2 · Easy
A wheel of moment of inertia $I = 0.50 kg\cdotm^{2}$ has a constant net torque of $2.0 N\cdotm$ applied for $4.0 s$ starting from rest. What is its angular momentum at $t = 4.0 s$ ?
- A
  $2.0 kg\cdotm^{2} / s$
  Why not A: Used torque only.
- B
  $4.0 kg\cdotm^{2} / s$
  Why not B: Off by a factor of 2.
- C
  $8.0 kg\cdotm^{2} / s$ Correct
- D
  $16 kg\cdotm^{2} / s$
  Why not D: Doubled the answer.
Explanation
Angular impulse: $Δ L = τ \cdot Δ t = 2.0 \times 4.0 = 8.0 kg\cdotm^{2} / s$ .

Key takeaway
Angular impulse–momentum theorem: $\Delta L = \tau \Delta t$.
Question 3 · Medium
A figure skater spinning with arms extended pulls her arms in. As a result, her angular velocity increases by a factor of 3. By what factor does her rotational kinetic energy change?
- A
  Decreases by $1/3$ .
  Why not A: Confused with conservation of energy (which doesn't apply when she does work).
- B
  Stays the same.
  Why not B: Angular momentum is conserved, but KE is not.
- C
  Increases by a factor of 3.Correct
- D
  Increases by a factor of 9.
  Why not D: Used $ω^{2}$ scaling without considering the change in $I$ .
Explanation
Conservation of angular momentum: $I_{1} ω_{1} = I_{2} ω_{2}$ . With $ω_{2} = 3 ω_{1}$ , we get $I_{2} = I_{1} /3$ . KE scales: $K E_{2} / K E_{1} = (I_{2} / I_{1}) (ω_{2} / ω_{1})^{2} = (1/3) (9) = 3$ . The skater does positive work pulling her arms in, supplying the extra KE.

Key takeaway
When $I$ shrinks at constant $L$, KE *increases* — the rotator does work pulling mass inward.
Question 4 · Medium
Two identical disks (same mass $M$ and radius $R$ ) start from rest at the top of an incline. Disk A slides down without rotating; Disk B rolls without slipping. Which reaches the bottom with greater speed?
- A
  Disk A (sliding).Correct
- B
  Disk B (rolling).
  Why not B: Rolling diverts energy into rotational KE.
- C
  Same speed (energy is conserved in both).
  Why not C: Sliding has all KE in translation; rolling splits it.
- D
  Cannot be determined without knowing the angle.
  Why not D: Independent of angle.
Explanation
Sliding: $m g h = \frac{1}{2} m v^{2} \Rightarrow v_{A} = 2 g h$ . Rolling: $m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$ . With $I = \frac{1}{2} M R^{2}$ and $v = ω R$ : $v_{B} = 4 g h /3 < v_{A}$ .

Key takeaway
Rolling shares energy between translation and rotation — slower than sliding from the same height.
Question 5 · Medium
A spinning ice skater has angular momentum $L$ and rotational kinetic energy $K E$ . By pulling in his arms, he reduces his moment of inertia by half. What are the new $L$ and $K E$ ?
- A
  $L$ unchanged; $K E$ doubled.Correct
- B
  $L$ doubled; $K E$ unchanged.
  Why not B: $L$ is conserved when no external torque acts.
- C
  Both $L$ and $K E$ doubled.
  Why not C: $L$ doesn't change.
- D
  $L$ halved; $K E$ unchanged.
  Why not D: $L$ doesn't depend on internal redistribution.
Explanation
$L = I ω$ is conserved (no external torque). With $I \to I /2$ , $ω \to 2 ω$ . $K E = \frac{1}{2} I ω^{2} \to \frac{1}{2} (I /2) (2 ω)^{2} = 2 \cdot \frac{1}{2} I ω^{2} = 2 K E$ .

Key takeaway
$L$ is conserved internally; KE is not — the skater does work pulling arms in.
Question 6 · Medium
A solid sphere ( $I = \frac{2}{5} M R^{2}$ ), a solid disk ( $I = \frac{1}{2} M R^{2}$ ), and a hoop ( $I = M R^{2}$ ) all roll without slipping down the same incline starting from rest. In what order do they reach the bottom?
- A
  Sphere, disk, hoop.Correct
- B
  Hoop, disk, sphere.
  Why not B: Reversed the order — larger $I$ means more energy in rotation, so slower.
- C
  All tie.
  Why not C: Different $I$ leads to different translational speeds.
- D
  Disk, sphere, hoop.
  Why not D: Sphere is faster than disk.
Explanation
Smaller $I / (M R^{2})$ ratio means less energy goes into rotation, so faster translational speed. Sphere ( $0.4$ ) < disk ( $0.5$ ) < hoop ( $1.0$ ), so the sphere wins, then disk, then hoop.

Key takeaway
Smaller $I/(MR^2)$ ratio rolls faster from the same height.
Question 7 · Hard
A sphere rolls without slipping down a frictionless incline. The friction at the contact point does:
- A
  Positive work, increasing rotational KE.
  Why not A: Static friction does no work since contact point is instantaneously at rest.
- B
  Negative work, decreasing translational KE.
  Why not B: Same reason — no work.
- C
  No work, but provides the torque to roll the sphere.Correct
- D
  Work cannot be calculated without knowing $μ$ .
  Why not D: $μ$ doesn't matter — static friction at a rolling contact does no work regardless.
Explanation
Rolling without slipping means the contact point is instantaneously at rest, so static friction acts over zero displacement and does no work. It still produces a torque about the center of mass.

Key takeaway
Static friction in rolling-without-slipping does no work but provides torque.
Question 8 · Hard
A child of mass $m$ jumps onto the edge of a stationary merry-go-round (a disk of mass $M$ and radius $R$ on a frictionless bearing) with tangential speed $v$ . What is the angular speed of the system after the child lands?
- A
  $\frac{m v}{( M + m ) R}$
  Why not A: Used point-mass moment of inertia for the disk.
- B
  $\frac{m v}{( M /2 + m ) R}$ Correct
- C
  $\frac{v}{R}$
  Why not C: Ignored the disk's moment of inertia entirely.
- D
  $\frac{m v}{M R}$
  Why not D: Forgot the child's contribution to moment of inertia after landing.
Explanation
Conserve angular momentum about the disk's axis: $L_{i} = m v R$ . After: $I_{f} = \frac{1}{2} M R^{2} + m R^{2} = (M /2 + m) R^{2}$ . So $ω = L_{i} / I_{f} = m v / [(M /2 + m) R]$ .

Key takeaway
When something attaches to a rotor, total $I$ becomes $I_{disk} + m R^2$ for the new mass at radius $R$.