AP Physics 1 Quiz: Rotational Kinematics
20 questions · exam conditions
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Rotational KinematicsQuestion 1 of 20

A turntable has constant angular acceleration α=1.5 rad/s2\alpha=1.5\,\text{rad/s}^2 from rest. What is ω\omega at t=4 st=4\,\text{s}?

6.0 rad6.0\,\text{rad}
3.0 rad/s3.0\,\text{rad/s}
6.0 rad/s6.0\,\text{rad/s}
24 rad/s224\,\text{rad/s}^2
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AP Physics 1 Quiz

AP Physics 1 Quiz: Rotational Kinematics

Practice Rotational Kinematics in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rotational Kinematics, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A turntable has constant angular acceleration α=1.5 rad/s2\alpha=1.5\,\text{rad/s}^2 from rest. What is ω\omega at t=4 st=4\,\text{s}?

  1. 6.0 rad6.0\,\text{rad}
  2. 3.0 rad/s3.0\,\text{rad/s}
  3. 6.0 rad/s6.0\,\text{rad/s} (correct answer)
  4. 24 rad/s224\,\text{rad/s}^2
Explanation: This problem tests applying constant angular acceleration from rest. For rotational motion starting from rest (ω0=0\omega_0 = 0) with constant angular acceleration, the angular velocity at time t is ω=ω0+αt=0+αt\omega = \omega_0 + \alpha t = 0 + \alpha t. With α=1.5 rad/s2\alpha = 1.5 \, \text{rad/s}^2 and t=4 st = 4 \, \text{s}, we get ω=1.5×4=6.0 rad/s\omega = 1.5 \times 4 = 6.0 \, \text{rad/s}. Choice A (6.0 rad6.0 \, \text{rad}) has incorrect units for velocity, while choice D (24 rad/s224 \, \text{rad/s}^2) might come from incorrectly multiplying all values together. For constant acceleration from rest, final velocity equals acceleration times time.

Question 2

A fan has ωi=8 rad/s\omega_i=8\ \text{rad/s} and constant α=1 rad/s2\alpha=1\ \text{rad/s}^2. How long until ωf=11 rad/s\omega_f=11\ \text{rad/s}?

  1. 3 s3\ \text{s} (correct answer)
  2. 19 s19\ \text{s}
  3. 88 s88\ \text{s}
  4. 311 s\tfrac{3}{11}\ \text{s}
Explanation: This problem requires finding the time needed to reach a specific angular velocity with constant acceleration. The fan has ω₀ = 8 rad/s, α = 1 rad/s², and we need to find when ω = 11 rad/s. Using ω = ω₀ + αt, we get 11 = 8 + 1(t), which gives t = 3 s. The time is simply the change in angular velocity divided by the angular acceleration. Choice B (19 s) might result from incorrectly multiplying values instead of solving for time. When finding time in kinematics problems, rearrange the velocity equation to isolate time as (ω - ω₀)/α.

Question 3

A turntable rotates at constant ω=5 rad/s\omega=5\ \text{rad/s} for 4 s4\ \text{s}. What is the angular displacement Δθ\Delta\theta?

  1. 1.25 rad1.25\ \text{rad}
  2. 9 rad9\ \text{rad}
  3. 20 rad20\ \text{rad} (correct answer)
  4. 5 rad/s5\ \text{rad/s}
Explanation: This problem involves calculating angular displacement for constant angular velocity motion. The turntable rotates at constant ω = 5 rad/s for time t = 4 s. For constant angular velocity, angular displacement is Δθ = ωt = 5 × 4 = 20 rad. Since there's no angular acceleration, the motion follows the simplest rotational kinematics relationship. Choice D (5 rad/s) confuses angular displacement with angular velocity by giving the wrong units. When solving constant velocity problems, remember that displacement equals velocity multiplied by time, whether for linear or rotational motion.

Question 4

A record's angular velocity changes from +8+8 to +2 rad/s+2\,\text{rad/s} in 3 s3\,\text{s}. What is α\alpha?

  1. +2 rad/s2+2\,\text{rad/s}^2
  2. −2 rad/s2-2\,\text{rad/s}^2 (correct answer)
  3. +6 rad/s+6\,\text{rad/s}
  4. −6 rad-6\,\text{rad}
Explanation: This problem requires finding angular acceleration from a change in angular velocity. Angular acceleration is defined as α = Δω/Δt = (ωf - ωi)/Δt. The angular velocity changes from +8 rad/s to +2 rad/s over 3 seconds, so α = (2 - 8)/3 = -6/3 = -2 rad/s². The negative sign indicates the object is slowing down (decelerating) even though it continues rotating in the positive direction. Choice C (+6 rad/s) has incorrect units for acceleration, while choice A (+2 rad/s²) has the wrong sign. When velocity decreases (even if still positive), acceleration is negative.

Question 5

A motor's angular position is θ(t)=5−3t\theta(t)=5-3t (rad). What is its angular velocity?

  1. 5 rad/s5\,\text{rad/s}
  2. −3 rad/s-3\,\text{rad/s} (correct answer)
  3. −3 rad/s2-3\,\text{rad/s}^2
  4. 2 rad2\,\text{rad}
Explanation: This problem requires finding angular velocity from a position function. Angular velocity is the first derivative of angular position: ω = dθ/dt. Given θ(t) = 5 - 3t rad, taking the derivative gives ω = d(5 - 3t)/dt = -3 rad/s. The negative sign indicates clockwise rotation, and the constant value means there's no angular acceleration. Choice A (5 rad/s) incorrectly uses the initial position value, while choice C (-3 rad/s²) has incorrect units for velocity. When position is linear in time, velocity is the constant coefficient of t.

Question 6

A wheel's angular velocity changes from −2-2 to −6 rad/s-6\,\text{rad/s} in 2 s2\,\text{s}. What is α\alpha?

  1. −4 rad/s2-4\,\text{rad/s}^2
  2. −2 rad/s2-2\,\text{rad/s}^2 (correct answer)
  3. −8 rad-8\,\text{rad}
  4. 2 rad/s22\,\text{rad/s}^2
Explanation: This problem involves calculating angular acceleration from changing angular velocity in rotational kinematics. Angular acceleration is α = Δω/Δt, where Δω is the change in angular velocity. The velocity changes from -2 rad/s to -6 rad/s, so Δω = -6 - (-2) = -4 rad/s over Δt = 2 s. Therefore, α = (-4 rad/s)/(2 s) = -2 rad/s². Choice A (-4 rad/s²) gives the change in velocity without dividing by time. When both initial and final velocities are negative, carefully compute the change by subtracting initial from final.

Question 7

A wheel's angular velocity changes from −6-6 to −2 rad/s-2\,\text{rad/s} in 2 s2\,\text{s}. What is the angular acceleration?

  1. −2 rad/s2-2\,\text{rad/s}^2
  2. +2 rad/s2+2\,\text{rad/s}^2 (correct answer)
  3. −4 rad/s2-4\,\text{rad/s}^2
  4. +4 rad+4\,\text{rad}
Explanation: This question evaluates rotational kinematics through the calculation of angular acceleration from changing angular velocity. Angular velocity ω indicates the speed and direction of rotation, while angular acceleration α represents the rate at which ω changes, positive if speeding up in the positive direction or slowing in the negative. Here, ω shifts from -6 to -2 rad/s, a change of +4 rad/s over 2 s, resulting in α = +2 rad/s², showing acceleration opposes the initial direction. This relationship parallels linear motion where acceleration is Δv/Δt, emphasizing direction matters in vector quantities. Choice A (-2 rad/s²) might tempt those who average velocities without considering the sign of change. For transferable strategy, compute changes in velocity carefully, including signs, and divide by time to find acceleration in kinematics problems.

Question 8

A wheel's angular velocity is given by ω(t)=4−2t\omega(t)=4-2t (rad/s) for 0≤t≤2 s0\le t\le2\,\text{s}. What is Δθ\Delta\theta?

  1. 0 rad0\,\text{rad}
  2. 4 rad4\,\text{rad} (correct answer)
  3. 8 rad8\,\text{rad}
  4. −4 rad/s-4\,\text{rad/s}
Explanation: This question evaluates rotational kinematics by integrating angular velocity to find displacement. Given ω(t) = 4 - 2t, Δθ is the integral from 0 to 2 s, yielding [4t - t²] evaluated as 4 rad, representing area under ω-t curve. Angular displacement accumulates from varying velocity, decreasing here from positive to zero. This is like finding displacement from varying linear velocity via integration. Choice C (8 rad) might come from ignoring the -2t term's effect. For variable motion, integrate velocity over time to compute displacement accurately in kinematics.

Question 9

A wheel's angular velocity is constant and positive. Which statement about angular acceleration is correct?

  1. α\alpha is positive because ω\omega is positive
  2. α=0\alpha=0 because ω\omega is constant (correct answer)
  3. α\alpha must be increasing over time
  4. α\alpha equals the angular position divided by time
Explanation: This question tests recognizing angular acceleration for constant angular velocity in rotational kinematics. Constant velocity means no change in rotational speed, implying zero acceleration. Acceleration only exists if velocity is varying, either in magnitude or direction. Qualitatively, this distinguishes uniform from accelerated rotational motion, akin to linear cases. Choice A is a distractor linking acceleration sign to velocity sign, but acceleration is independent when velocity is constant. A transferable strategy is to check if velocity changes over time to determine if acceleration is zero or nonzero.

Question 10

A spinner's angular position is constant at θ=3\theta=3 rad for 55 s. What are ω\omega and α\alpha during this interval?

  1. ω=3 rad/s\omega=3\ \text{rad/s}, α=0\alpha=0
  2. ω=0\omega=0, α=3 rad/s2\alpha=3\ \text{rad/s}^2
  3. ω=0\omega=0, α=0\alpha=0 (correct answer)
  4. ω=35 rad/s\omega=\tfrac{3}{5}\ \text{rad/s}, α=0\alpha=0
Explanation: This question assesses identifying angular velocity and acceleration from constant angular position in rotational kinematics. Constant position means no rotation is occurring, so velocity is zero. With no change in velocity, acceleration is also zero. Qualitatively, this represents a state of rest in rotational terms, with no motion or change. Choice A is a distractor that misinterprets the constant position value as velocity, ignoring the lack of change. A transferable strategy is to examine changes in position and velocity to infer velocity and acceleration values.

Question 11

A wheel's angular velocity is constant at −5 rad/s-5\,\text{rad/s} for 2 s2\,\text{s}. What is its angular acceleration?

  1. −10 rad-10\,\text{rad}
  2. 0 rad/s20\,\text{rad/s}^2 (correct answer)
  3. −5 rad/s2-5\,\text{rad/s}^2
  4. −2.5 rad/s-2.5\,\text{rad/s}
Explanation: This problem involves understanding constant angular velocity motion. When angular velocity is constant, the angular acceleration must be zero by definition: α = dω/dt = 0. The fact that ω = -5 rad/s (negative, indicating clockwise rotation) doesn't change this - if velocity is constant, acceleration is zero regardless of the velocity's sign or magnitude. Choice C (-5 rad/s²) incorrectly assumes the acceleration equals the velocity value, while choice A (-10 rad) confuses displacement with acceleration. Remember: constant velocity always means zero acceleration.

Question 12

A disk's angular velocity changes from +6+6 to +2 rad/s+2\ \text{rad/s} in 2 s2\ \text{s}. What is its angular acceleration?

  1. α=+4 rad/s2\alpha=+4\ \text{rad/s}^2
  2. α=−2 rad/s2\alpha=-2\ \text{rad/s}^2 (correct answer)
  3. α=+2 rad/s2\alpha=+2\ \text{rad/s}^2
  4. α=−8 rad/s2\alpha=-8\ \text{rad/s}^2
Explanation: This question tests determining angular acceleration from changes in angular velocity in rotational kinematics. Angular acceleration measures how quickly angular velocity changes over time, positive if speeding up in the positive direction and negative if slowing down. For constant acceleration, it's the change in velocity divided by time, capturing the average rate of change. This is qualitatively like linear acceleration describing speed changes. Choice A is a distractor that ignores the sign, treating the magnitude incorrectly as positive despite deceleration. A transferable strategy is to always include signs in calculations to account for direction in rotational motion.

Question 13

A fan's angular velocity is constant and negative: ω=−5 rad/s\omega=-5\,\text{rad/s} for 3 s3\,\text{s}. What is Δθ\Delta\theta?

  1. 15 rad15\,\text{rad}
  2. −15 rad-15\,\text{rad} (correct answer)
  3. −5 rad/s2-5\,\text{rad/s}^2
  4. −53 rad-\tfrac{5}{3}\,\text{rad}
Explanation: This question assesses rotational kinematics by calculating angular displacement with constant negative velocity. Constant ω = -5 rad/s over 3 s gives Δθ = ωΔt = -15 rad, where the negative sign indicates direction opposite to positive reference. Angular displacement accumulates based on velocity's magnitude and direction, akin to linear displacement depending on velocity sign. With no acceleration mentioned, motion is uniform in the negative direction. Choice A (15 rad) ignores the negative sign, a frequent oversight in directional quantities. Remember to include signs in calculations for vector quantities like displacement and velocity in kinematics problems.

Question 14

A fan's angular velocity increases uniformly from 44 to 10 rad/s10\,\text{rad/s} over 2 s2\,\text{s}. What is its angular acceleration?

  1. 14 rad/s14\,\text{rad/s}
  2. 3 rad/s23\,\text{rad/s}^2 (correct answer)
  3. 6 rad6\,\text{rad}
  4. 12 rad/s212\,\text{rad/s}^2
Explanation: This problem tests understanding of angular acceleration in rotational kinematics. Angular acceleration is the rate of change of angular velocity: α = Δω/Δt. The angular velocity changes from 4 rad/s to 10 rad/s, giving Δω = 10 - 4 = 6 rad/s over Δt = 2 s. Therefore, α = (6 rad/s)/(2 s) = 3 rad/s². Choice D (12 rad/s²) incorrectly multiplies the change in velocity by time instead of dividing. Remember that acceleration is always the change in velocity divided by the time interval.

Question 15

A wheel's angular position increases linearly from θ=1\theta=1 to θ=9 rad\theta=9\,\text{rad} in 4 s4\,\text{s}. What is ω\omega?

  1. 2 rad/s2\,\text{rad/s} (correct answer)
  2. 8 rad/s8\,\text{rad/s}
  3. 4 rad/s24\,\text{rad/s}^2
  4. 10 rad10\,\text{rad}
Explanation: This question probes rotational kinematics with linear change in angular position implying constant velocity. When θ increases linearly from 1 to 9 rad in 4 s, the change Δθ = 8 rad over time gives ω = Δθ/Δt = 2 rad/s, as constant slope in θ vs. t graph indicates constant ω. In angular motion, linear θ(t) means no acceleration, paralleling constant velocity in linear kinematics. This contrasts with quadratic θ(t) which would indicate acceleration. Choice B (8 rad/s) could stem from using Δθ without dividing by time. For strategy, plot or visualize the position-time relationship to infer velocity and acceleration in kinematics.

Question 16

A disk's angular position changes from 2 rad2\,\text{rad} to −4 rad-4\,\text{rad} in 3 s3\,\text{s}. What is its average angular velocity?

  1. −2 rad/s-2\,\text{rad/s} (correct answer)
  2. −6 rad-6\,\text{rad}
  3. 2 rad/s2\,\text{rad/s}
  4. −23 rad/s-\tfrac{2}{3}\,\text{rad/s}
Explanation: This problem requires calculating average angular velocity from rotational kinematics. Average angular velocity is defined as ω_avg = Δθ/Δt, where Δθ is the change in angular position. The angular displacement is Δθ = θ_f - θ_i = (-4 rad) - (2 rad) = -6 rad over Δt = 3 s. Therefore, ω_avg = (-6 rad)/(3 s) = -2 rad/s. Choice B (-6 rad) gives the displacement instead of velocity, missing the division by time. When finding average angular velocity, always calculate the change in position and divide by the time interval.

Question 17

A disk's angular position changes from 1.01.0 to 7.0 rad7.0\,\text{rad} in 2.0 s2.0\,\text{s}. What is average ω\omega?

  1. 3.0 rad/s3.0\,\text{rad/s} (correct answer)
  2. 6.0 rad6.0\,\text{rad}
  3. 4.0 rad/s24.0\,\text{rad/s}^2
  4. 0.33 rad/s0.33\,\text{rad/s}
Explanation: This problem requires calculating average angular velocity from angular displacement. Average angular velocity is defined as ωavg=Δθ/Δt=(θf−θi)/Δtω_{\text{avg}} = \Deltaθ/\Delta t = (θ_f - θ_i)/\Delta t. The angular position changes from 1.0 rad1.0 \, \text{rad} to 7.0 rad7.0 \, \text{rad}, giving Δθ=7.0−1.0=6.0 rad\Deltaθ = 7.0 - 1.0 = 6.0 \, \text{rad} over Δt=2.0 s\Delta t = 2.0 \, \text{s}. Therefore, ωavg=6.0/2.0=3.0 rad/sω_{\text{avg}} = 6.0/2.0 = 3.0 \, \text{rad/s}. Choice B (6.0 rad6.0 \, \text{rad}) gives the displacement rather than velocity, while choice C (4.0 rad/s24.0 \, \text{rad/s}^2) has incorrect units for velocity. For average velocity problems, always divide total displacement by total time.

Question 18

A wheel starts with ω0=3 rad/s\omega_0=3\,\text{rad/s} and has constant α=2 rad/s2\alpha=2\,\text{rad/s}^2 for 2 s2\,\text{s}. Find Δθ\Delta\theta.

  1. 10 rad10\,\text{rad} (correct answer)
  2. 7 rad/s7\,\text{rad/s}
  3. 14 rad14\,\text{rad}
  4. 4 rad/s24\,\text{rad/s}^2
Explanation: This problem involves calculating angular displacement with constant acceleration. Using the kinematic equation Δθ = ω₀t + ½αt², with ω₀ = 3 rad/s, α = 2 rad/s², and t = 2 s, we get Δθ = 3(2) + ½(2)(2²) = 6 + ½(2)(4) = 6 + 4 = 10 rad. The first term (6 rad) represents displacement due to initial velocity, while the second term (4 rad) is the additional displacement from acceleration. Choice B (7 rad/s) confuses final velocity with displacement, while choice C (14 rad) might come from incorrect calculation. For constant acceleration problems, use the complete kinematic equation including both velocity and acceleration terms.

Question 19

A fan starts at θ=0\theta=0 and rotates with constant ω=6 rad/s\omega=6\,\text{rad/s} for 5 s5\,\text{s}. What is Δθ\Delta\theta?

  1. 30 rad30\,\text{rad} (correct answer)
  2. 1.2 rad/s21.2\,\text{rad/s}^2
  3. 6 rad6\,\text{rad}
  4. 0 rad/s0\,\text{rad/s}
Explanation: This problem involves calculating angular displacement for constant angular velocity motion. For rotational motion with constant angular velocity, the angular displacement is Δθ = ωt, where ω is the angular velocity and t is the time interval. With ω = 6 rad/s and t = 5 s, we get Δθ = 6 × 5 = 30 rad. The motion starts at θ = 0, so the final position is θ = 30 rad, making Δθ = 30 - 0 = 30 rad. Choice B (1.2 rad/s²) incorrectly suggests an acceleration value, while choice D (0 rad/s) confuses angular velocity with displacement. For constant angular velocity problems, remember that displacement equals velocity times time.

Question 20

A wheel's angular position is θ(t)=2.0t2\theta(t)=2.0t^2 (rad) for 0≤t≤3 s0\le t\le3\,\text{s}. What is its angular acceleration?

  1. 2.0 rad/s22.0\,\text{rad/s}^2
  2. 4.0 rad/s24.0\,\text{rad/s}^2 (correct answer)
  3. 12 rad12\,\text{rad}
  4. 4.0 rad/s4.0\,\text{rad/s}
Explanation: This problem tests understanding of rotational kinematics relationships between angular position, velocity, and acceleration. Angular acceleration αα is the second derivative of angular position with respect to time: α=d2θdt2α = \frac{d^2θ}{dt^2}. Given θ(t)=2.0t2θ(t) = 2.0t^2, we first find angular velocity by taking the first derivative: ω=dθdt=4.0t rad/sω = \frac{dθ}{dt} = 4.0t \, \text{rad/s}. Then, taking the derivative of ωω gives us α=dωdt=4.0 rad/s2α = \frac{dω}{dt} = 4.0 \, \text{rad/s}^2. Choice C (12 rad12 \, \text{rad}) has incorrect units for acceleration, while choice D (4.0 rad/s4.0 \, \text{rad/s}) would be the angular velocity at t = 1 s, not the acceleration. When given position as a function of time, always differentiate twice to find acceleration.