AP Physics C: Mechanics, AP exams

Unit 6: Energy and Momentum of Rotating Systems

Unit 6 applies conservation laws to rotation. Rotational kinetic energy, work done by a torque, angular momentum, and angular impulse are defined, and conservation of angular momentum handles problems that force analysis cannot reach easily. Rolling motion and orbital mechanics are the standard applications.

Position: 6 of 7 units
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Physics C: Mechanics. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Rotational kinetic energy: Energy of rotation, and combining rotational with translational kinetic energy for rolling bodies.
Work and power in rotation: Work as the integral of torque over angular displacement, and rotational power.
Angular momentum and angular impulse: Angular momentum for particles and rigid bodies, and its change under an applied torque.
Conservation of angular momentum: Identifying when external torques vanish, and analyzing systems whose rotational inertia changes.
Rolling motion: Rolling without slipping, the role of static friction, and energy partition between translation and rotation.
Orbits and gravitational systems: Applying energy and angular momentum conservation to circular and elliptical orbits.

Study tips

How to practice Rotating Systems.

Section-specific guidance that follows from how this material is usually tested.

Work the classic rolling-race problem and be able to explain the ordering from rotational inertia alone.
Check for external torques explicitly before invoking conservation of angular momentum.
Solve orbit problems with energy and angular momentum together rather than with force alone.

Where this fits

Unit 6: Energy and Momentum of Rotating Systems is 6 of 7 units in the AP Physics C: Mechanics guide. Use the full guide to balance it against everything else you have to review.

Back to the AP Physics C: Mechanics guide Official exam information

Sample Rotating Systems questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
A uniform solid disk of mass 2.0 kg and radius 0.40 m spins about its central axis at 10 rad/s, with rotational inertia (1/2)MR^2. What is its rotational kinetic energy?
1. A4.0 J
2. B8.0 J
3. C16 J
4. D80 J
Show answer
Answer: B. 8.0 J
The rotational inertia is (1/2)(2.0)(0.40)^2 = 0.16 kg m^2, so the kinetic energy is (1/2)(0.16)(10)^2 = 8.0 J. The 16 J choice omits the factor of one half in the kinetic energy expression.
Key insight: Rotational kinetic energy uses the same one-half factor and squared rate as its translational counterpart.
Question 2Foundational
A 0.20 kg particle moves in a circle of radius 0.50 m at a constant speed of 4.0 m/s. What is the magnitude of its angular momentum about the center of the circle?
1. A0.10 kg m^2/s
2. B0.40 kg m^2/s
3. C0.80 kg m^2/s
4. D1.6 kg m^2/s
Show answer
Answer: B. 0.40 kg m^2/s
For a particle moving perpendicular to the radius, L = mvr = (0.20)(4.0)(0.50) = 0.40 kg m^2/s. The 0.80 kg m^2/s choice leaves the radius out, and 1.6 kg m^2/s divides by it instead of multiplying.
Key insight: Angular momentum of a particle is defined relative to a chosen point and depends on the perpendicular distance to it.
Question 3Intermediate
A uniform solid cylinder, whose rotational inertia about its axis is (1/2)MR^2, is released from rest and rolls without slipping down an incline through a vertical drop of 1.5 m. Using g = 10 m/s^2, what is the speed of its center of mass at the bottom?
1. A2.7 m/s
2. B3.9 m/s
3. C4.5 m/s
4. D5.5 m/s
Show answer
Answer: C. 4.5 m/s
Energy conservation with the rolling constraint v = (omega)R gives Mgh = (3/4)Mv^2, so v = sqrt(4gh/3) = sqrt(20) = 4.5 m/s. The 5.5 m/s choice is sqrt(2gh), the frictionless sliding result, which ignores the energy stored in rotation.
Key insight: A rolling body arrives more slowly than a sliding one because part of the released energy goes into spin.
Question 4Intermediate
A horizontal turntable of rotational inertia 0.20 kg m^2 and radius 0.40 m rotates freely at 6.0 rad/s. A 0.50 kg lump of clay is dropped straight down onto the rim and sticks there. What is the new angular speed of the turntable?
1. A3.0 rad/s
2. B4.3 rad/s
3. C6.0 rad/s
4. D8.4 rad/s
Show answer
Answer: B. 4.3 rad/s
The clay adds mR^2 = (0.50)(0.40)^2 = 0.080 kg m^2, so angular momentum conservation gives (0.20)(6.0) = (0.28)(omega) and omega = 4.3 rad/s. The 8.4 rad/s choice multiplies by the inertia ratio instead of dividing, which would speed the turntable up.
Key insight: Dropping mass onto a rotating body adds rotational inertia with no external torque, so the spin must slow.
Question 5Advanced
A satellite travels in a circular orbit around a planet. If the satellite is moved to a new circular orbit whose radius is four times as large, what happens to its orbital period?
1. AIt becomes 2 times as long
2. BIt becomes 4 times as long
3. CIt becomes 8 times as long
4. DIt becomes 16 times as long
Show answer
Answer: C. It becomes 8 times as long
Setting the gravitational force equal to the centripetal requirement gives T^2 proportional to r^3, so multiplying r by 4 multiplies T by 4^(3/2) = 8. The 4 times choice assumes period scales directly with radius, which would ignore that the orbital speed also drops.
Key insight: Kepler's third law follows directly from Newtonian gravitation plus uniform circular motion.

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Unit 6: Energy and Momentum of Rotating Systems