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AP Physics C: Mechanics Practice Quiz: Rotational Kinematics

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

Which of the following physical quantities is defined as the rate at which angular velocity changes with respect to time?

All Questions (10)

Which of the following physical quantities is defined as the rate at which angular velocity changes with respect to time?

A) Angular displacement

B) Angular velocity

C) Angular acceleration

D) Rotational inertia

Correct Answer: C

According to the provided content, 'Angular acceleration is the rate at which angular velocity changes with respect to time.' The relevant equation is α = dω/dt.

A rigid body rotates with a constant, non-zero angular velocity. Which of the following statements about its motion must be true?

A) The angular acceleration is constant and non-zero.

B) The angular acceleration is zero.

C) The angular displacement is zero.

D) The angular acceleration is increasing.

Correct Answer: B

Angular acceleration is the rate of change of angular velocity (α = dω/dt). If the angular velocity (ω) is constant, its rate of change is zero. Therefore, the angular acceleration (α) must be zero.

A flywheel, initially at rest, is subjected to a constant angular acceleration of 2.0 rad/s². What is its angular velocity after 5.0 seconds?

A) 2.5 rad/s

B) 5.0 rad/s

C) 10.0 rad/s

D) 25.0 rad/s

Correct Answer: C

Using the kinematic equation ω = ω₀ + αt. Given that the flywheel is initially at rest, ω₀ = 0. With α = 2.0 rad/s² and t = 5.0 s, the final angular velocity is ω = 0 + (2.0 rad/s²)(5.0 s) = 10.0 rad/s.

The angular position of a spinning disk is described by the equation θ(t) = 4t² + 2t - 1, where t is in seconds and θ is in radians. What is the angular velocity, ω, of the disk at t = 3 s?

A) 14 rad/s

B) 26 rad/s

C) 35 rad/s

D) 41 rad/s

Correct Answer: B

Angular velocity is the time derivative of angular position: ω = dθ/dt. Taking the derivative of θ(t) = 4t² + 2t - 1 gives ω(t) = 8t + 2. At t = 3 s, ω(3) = 8(3) + 2 = 24 + 2 = 26 rad/s.

A wheel starts from rest and accelerates uniformly. After 4.0 seconds, it has rotated through an angular displacement of 48 radians. What is the angular acceleration of the wheel?

A) 3.0 rad/s²

B) 6.0 rad/s²

C) 12.0 rad/s²

D) 24.0 rad/s²

Correct Answer: B

Use the kinematic equation θ = θ₀ + ω₀t + (1/2)αt². Since the wheel starts from rest, ω₀ = 0. We can set the initial position θ₀ = 0. The equation simplifies to θ = (1/2)αt². Rearranging to solve for α gives α = 2θ / t². Plugging in the values: α = 2(48 rad) / (4.0 s)² = 96 / 16 = 6.0 rad/s².

The angular position of a rotating object is given by θ(t) = t³ - 6t². At what time t > 0 is the angular acceleration of the object equal to zero?

A) 1 s

B) 2 s

C) 3 s

D) 4 s

Correct Answer: B

First, find the angular velocity by taking the first derivative: ω(t) = dθ/dt = 3t² - 12t. Then, find the angular acceleration by taking the second derivative: α(t) = dω/dt = 6t - 12. To find when the angular acceleration is zero, set α(t) = 0, which gives 6t - 12 = 0. Solving for t gives 6t = 12, so t = 2 s.

The motion of a rotating object is described by the equation θ = 10 + 4t - 2t², where θ is in radians and t is in seconds. Which of the following describes the motion of the object?

A) Initial angular position of 10 rad, initial angular velocity of 4 rad/s, and constant angular acceleration of -2 rad/s².

B) Initial angular position of 10 rad, initial angular velocity of 4 rad/s, and constant angular acceleration of -4 rad/s².

C) Initial angular position of 4 rad, initial angular velocity of 10 rad/s, and constant angular acceleration of 2 rad/s².

D) Initial angular position of 10 rad, initial angular velocity of -2 rad/s, and constant angular acceleration of 4 rad/s².

Correct Answer: B

The given equation θ = 10 + 4t - 2t² is compared to the standard kinematic equation θ = θ₀ + ω₀t + (1/2)αt². By matching the terms, we find θ₀ = 10 rad, ω₀ = 4 rad/s, and (1/2)α = -2 rad/s². Solving for α gives α = -4 rad/s².

A centrifuge accelerates from rest with a constant angular acceleration of 5.0 rad/s². What is its angular displacement after it has reached an angular velocity of 20 rad/s?

A) 20 rad

B) 40 rad

C) 80 rad

D) 100 rad

Correct Answer: B

This is a two-step problem. First, find the time it takes to reach the final velocity using ω = ω₀ + αt. Given ω = 20 rad/s, ω₀ = 0, and α = 5.0 rad/s², we have 20 = 0 + 5.0t, which gives t = 4.0 s. Next, use this time to find the angular displacement with θ = θ₀ + ω₀t + (1/2)αt². The displacement is Δθ = θ - θ₀ = ω₀t + (1/2)αt². Plugging in the values: Δθ = (0)(4.0) + (1/2)(5.0)(4.0)² = 0.5 * 5.0 * 16 = 40 rad.

The angular velocity of a propeller is given by ω(t) = 15 - 3t². Which statement correctly describes the angular acceleration α of the propeller for t > 0?

A) The angular acceleration is constant.

B) The angular acceleration is zero at all times.

C) The magnitude of the angular acceleration is increasing with time.

D) The magnitude of the angular acceleration is decreasing with time.

Correct Answer: C

Angular acceleration is the time derivative of angular velocity: α = dω/dt. Taking the derivative of ω(t) = 15 - 3t² gives α(t) = -6t. The magnitude of the acceleration is |α(t)| = 6t. As time t increases for t > 0, the magnitude of the angular acceleration increases linearly.

A merry-go-round is spinning with a constant positive angular velocity ω₀. At time t=0, the operator applies a brake that provides a constant negative angular acceleration, which has a magnitude of α. Which equation correctly describes the angular displacement Δθ of the merry-go-round for t>0 until it stops, assuming Δθ is measured from its position at t=0?

A) Δθ = ω₀t + (1/2)αt²

B) Δθ = ω₀t - (1/2)αt²

C) Δθ = -(1/2)αt²

D) Δθ = ω₀ + αt

Correct Answer: B

The general equation for angular displacement under constant acceleration is θ = θ₀ + ω₀t + (1/2)α_actual*t². The displacement is Δθ = θ - θ₀ = ω₀t + (1/2)α_actual*t². The problem states the initial angular velocity is ω₀ and the acceleration is negative with magnitude α, so α_actual = -α. Substituting this into the equation gives Δθ = ω₀t + (1/2)(-α)t², which simplifies to Δθ = ω₀t - (1/2)αt².