AP Physics C: Mechanics, AP exams

Unit 5: Torque and Rotational Dynamics

Unit 5 rebuilds dynamics for rotating rigid bodies. Rotational inertia is computed by integration for continuous bodies, the parallel-axis idea extends those results, and torque plays the role of force in the rotational form of Newton's second law. Equilibrium problems with extended bodies also live here.

Position: 5 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 kinematics: Angular position, velocity, and acceleration, related by differentiation and integration.
Connecting linear and rotational motion: Relating tangential and angular quantities, and the rolling constraint between them.
Torque: Torque from force and lever arm, net torque on an extended body, and the sign convention for rotation.
Rotational inertia: Computing rotational inertia by integration for rods, disks, and spheres, and combining bodies.
Rotational equilibrium: Balanced forces and torques for beams and supported bars, including choosing a convenient pivot.
Newton's second law for rotation: Net torque equal to rotational inertia times angular acceleration, applied to rigid-body problems.

Study tips

How to practice Torque and Rotation.

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

Derive the rotational inertia of a uniform rod about its end and its center; those two derivations cover most of the technique.
Maintain a translational-to-rotational analogy table and add to it as the unit progresses.
Pick the pivot to eliminate an unknown force in equilibrium problems, and say why you chose it.

Where this fits

Unit 5: Torque and Rotational Dynamics is 5 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 Torque and Rotation 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 thin rod of mass M and length L rotates about an axis through one end and perpendicular to the rod. What is its rotational inertia about that axis?
1. AML^2/12
2. BML^2/3
3. CML^2/2
4. DML^2
Show answer
Answer: B. ML^2/3
Integrating r^2 dm with dm = (M/L)dr from 0 to L gives (M/L)(L^3/3) = ML^2/3. The ML^2/12 choice is the result about the rod's center, which is smaller because the mass sits closer to the axis.
Key insight: The parallel-axis theorem connects the two rod results: ML^2/12 plus M(L/2)^2 equals ML^2/3.
Question 2Foundational
A 50 N force is applied at the end of a 0.80 m wrench handle, directed at 30 degrees to the handle. What is the magnitude of the torque about the bolt at the other end of the handle?
1. A10 N m
2. B20 N m
3. C35 N m
4. D40 N m
Show answer
Answer: B. 20 N m
Torque is rF sin(theta) = (0.80 m)(50 N)sin(30 degrees) = (0.80)(50)(0.50) = 20 N m. The 40 N m choice ignores the angle, and 35 N m uses the cosine, which picks out the component along the handle that produces no turning.
Key insight: Only the force component perpendicular to the lever arm produces torque.
Question 3Intermediate
A uniform solid disk of mass 4.0 kg and radius 0.50 m turns on a frictionless axle through its center, where its rotational inertia is (1/2)MR^2. A constant 6.0 N force is applied tangentially at the rim. What is the angular acceleration of the disk?
1. A1.5 rad/s^2
2. B3.0 rad/s^2
3. C6.0 rad/s^2
4. D12 rad/s^2
Show answer
Answer: C. 6.0 rad/s^2
The rotational inertia is (1/2)(4.0)(0.50)^2 = 0.50 kg m^2 and the torque is (6.0 N)(0.50 m) = 3.0 N m, so alpha = 3.0/0.50 = 6.0 rad/s^2. The 3.0 rad/s^2 choice uses MR^2 for the disk, the value that belongs to a hoop.
Key insight: Rotational inertia depends on how mass is distributed, so the shape factor must be applied before dividing the torque.
Question 4Intermediate
A uniform rod of mass 2.0 kg and length 1.2 m has rotational inertia ML^2/12 about a perpendicular axis through its center. What is its rotational inertia about a parallel axis 0.30 m from the center?
1. A0.24 kg m^2
2. B0.30 kg m^2
3. C0.42 kg m^2
4. D0.60 kg m^2
Show answer
Answer: C. 0.42 kg m^2
The parallel-axis theorem gives I = I_cm + Md^2 = (2.0)(1.2)^2/12 + (2.0)(0.30)^2 = 0.24 + 0.18 = 0.42 kg m^2. The 0.24 kg m^2 choice reports only the center-of-mass term and forgets the shift.
Key insight: Moving the axis away from the center of mass always increases the rotational inertia, by exactly Md^2.
Question 5Advanced
A uniform rod of mass M and length L is pivoted at one end on a frictionless hinge, held horizontally, and then released from rest. In terms of g and L, what is the magnitude of its angular acceleration immediately after release?
1. Ag/L
2. B2g/(3L)
3. C3g/(2L)
4. D3g/L
Show answer
Answer: C. 3g/(2L)
Gravity acts at the center of mass, giving a torque Mg(L/2), and the rotational inertia about the end is ML^2/3, so alpha = [Mg(L/2)]/(ML^2/3) = 3g/(2L). Note that the far end therefore begins with a linear acceleration of 3g/2, which is larger than g.
Key insight: For rigid-body rotation, weight can be treated as acting entirely at the center of mass.

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Unit 5: Torque and Rotational Dynamics