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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.

Question 1 / 20

0 of 20 answered

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

Select an answer to continue

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

  1. 30 rad30\,\text{rad}30rad (correct answer)
  2. 1.2 rad/s21.2\,\text{rad/s}^21.2rad/s2
  3. 6 rad6\,\text{rad}6rad
  4. 0 rad/s0\,\text{rad/s}0rad/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 2

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

  1. +2 rad/s2+2\,\text{rad/s}^2+2rad/s2
  2. −2 rad/s2-2\,\text{rad/s}^2−2rad/s2 (correct answer)
  3. +6 rad/s+6\,\text{rad/s}+6rad/s
  4. −6 rad-6\,\text{rad}−6rad

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 3

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

  1. 6.0 rad6.0\,\text{rad}6.0rad
  2. 3.0 rad/s3.0\,\text{rad/s}3.0rad/s
  3. 6.0 rad/s6.0\,\text{rad/s}6.0rad/s (correct answer)
  4. 24 rad/s224\,\text{rad/s}^224rad/s2

Explanation: This problem tests applying constant angular acceleration from rest. For rotational motion starting from rest (ω0=0\omega_0 = 0ω0​=0) with constant angular acceleration, the angular velocity at time t is ω=ω0+αt=0+αt\omega = \omega_0 + \alpha t = 0 + \alpha tω=ω0​+αt=0+αt. With α=1.5 rad/s2\alpha = 1.5 \, \text{rad/s}^2α=1.5rad/s2 and t=4 st = 4 \, \text{s}t=4s, we get ω=1.5×4=6.0 rad/s\omega = 1.5 \times 4 = 6.0 \, \text{rad/s}ω=1.5×4=6.0rad/s. Choice A (6.0 rad6.0 \, \text{rad}6.0rad) has incorrect units for velocity, while choice D (24 rad/s224 \, \text{rad/s}^224rad/s2) might come from incorrectly multiplying all values together. For constant acceleration from rest, final velocity equals acceleration times time.

Question 4

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

  1. 2.0 rad/s22.0\,\text{rad/s}^22.0rad/s2
  2. 4.0 rad/s24.0\,\text{rad/s}^24.0rad/s2 (correct answer)
  3. 12 rad12\,\text{rad}12rad
  4. 4.0 rad/s4.0\,\text{rad/s}4.0rad/s

Explanation: This problem tests understanding of rotational kinematics relationships between angular position, velocity, and acceleration. Angular acceleration ααα is the second derivative of angular position with respect to time: α=d2θdt2α = \frac{d^2θ}{dt^2}α=dt2d2θ​. Given θ(t)=2.0t2θ(t) = 2.0t^2θ(t)=2.0t2, we first find angular velocity by taking the first derivative: ω=dθdt=4.0t rad/sω = \frac{dθ}{dt} = 4.0t \, \text{rad/s}ω=dtdθ​=4.0trad/s. Then, taking the derivative of ωωω gives us α=dωdt=4.0 rad/s2α = \frac{dω}{dt} = 4.0 \, \text{rad/s}^2α=dtdω​=4.0rad/s2. Choice C (12 rad12 \, \text{rad}12rad) has incorrect units for acceleration, while choice D (4.0 rad/s4.0 \, \text{rad/s}4.0rad/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.

Question 5

A disk rotates with constant ω=−4 rad/s\omega=-4\,\text{rad/s}ω=−4rad/s for 5 s5\,\text{s}5s. What is its angular displacement?

  1. 20 rad20\,\text{rad}20rad
  2. −0.8 rad/s-0.8\,\text{rad/s}−0.8rad/s
  3. −20 rad-20\,\text{rad}−20rad (correct answer)
  4. 4 rad/s4\,\text{rad/s}4rad/s

Explanation: This problem involves angular displacement with constant angular velocity in rotational kinematics. For constant angular velocity, the angular displacement is Δθ = ωt. With ω = -4 rad/s (negative indicating clockwise rotation) and t = 5 s, we get Δθ = (-4 rad/s)(5 s) = -20 rad. The negative sign indicates the displacement is in the clockwise direction. Choice A (20 rad) has the correct magnitude but wrong sign, ignoring the direction of rotation. Always maintain the sign of angular velocity to correctly determine the direction of angular displacement.

Question 6

A rotor has ω(t)=10−2t\omega(t)=10-2tω(t)=10−2t (rad/s) for 0≤t≤4 s0\le t\le4\,\text{s}0≤t≤4s. What is its angular acceleration?

  1. 8 rad/s8\,\text{rad/s}8rad/s
  2. −2 rad/s2-2\,\text{rad/s}^2−2rad/s2 (correct answer)
  3. −2 rad-2\,\text{rad}−2rad
  4. 2 rad/s22\,\text{rad/s}^22rad/s2

Explanation: This problem tests finding angular acceleration from a velocity function. Given ω(t)=10−2t rad/s\omega(t) = 10 - 2t \, \text{rad/s}ω(t)=10−2trad/s, angular acceleration is the derivative: α=dωdt\alpha = \frac{d\omega}{dt}α=dtdω​. Taking the derivative of ω(t)=10−2t\omega(t) = 10 - 2tω(t)=10−2t gives α=−2 rad/s2\alpha = -2 \, \text{rad/s}^2α=−2rad/s2. The negative acceleration indicates the rotor is slowing down from its initial velocity of 10 rad/s10 \, \text{rad/s}10rad/s. Choice A (8 rad/s8 \, \text{rad/s}8rad/s) would be the velocity at t = 1 s, not the acceleration, while choice D (+2 rad/s22 \, \text{rad/s}^22rad/s2) has the wrong sign. When given velocity as a function of time, differentiate once to find acceleration.

Question 7

A wheel starts from rest and reaches 12 rad/s12\ \text{rad/s}12 rad/s after 3 s3\ \text{s}3 s with constant angular acceleration. What is α\alphaα?

  1. 36 rad/s236\ \text{rad/s}^236 rad/s2
  2. 4 rad/s24\ \text{rad/s}^24 rad/s2 (correct answer)
  3. 9 rad/s29\ \text{rad/s}^29 rad/s2
  4. 0.25 rad/s20.25\ \text{rad/s}^20.25 rad/s2

Explanation: This problem tests understanding of rotational kinematics with constant angular acceleration. The wheel starts from rest (ω₀ = 0 rad/s) and reaches ω = 12 rad/s in time t = 3 s. For constant angular acceleration, we use ω = ω₀ + αt, which gives us 12 = 0 + α(3), so α = 12/3 = 4 rad/s². Choice A (36 rad/s²) incorrectly multiplies 12 × 3 instead of dividing. To solve rotational kinematics problems, identify given quantities, select the appropriate equation, and check that units match throughout your calculation.

Question 8

A disk has ωi=10 rad/s\omega_i=10\ \text{rad/s}ωi​=10 rad/s and α=−2 rad/s2\alpha=-2\ \text{rad/s}^2α=−2 rad/s2 for 3 s3\ \text{s}3 s. What is ωf\omega_fωf​?

  1. 16 rad/s16\ \text{rad/s}16 rad/s
  2. 4 rad/s4\ \text{rad/s}4 rad/s (correct answer)
  3. 8 rad8\ \text{rad}8 rad
  4. −6 rad/s-6\ \text{rad/s}−6 rad/s

Explanation: This problem requires applying rotational kinematics with constant negative angular acceleration. The disk starts with ω₀ = 10 rad/s and experiences α = -2 rad/s² for t = 3 s. Using ω = ω₀ + αt, we get ω = 10 + (-2)(3) = 10 - 6 = 4 rad/s. The negative acceleration causes the disk to slow down from its initial angular velocity. Choice A (16 rad/s) incorrectly adds the acceleration term instead of subtracting it. To handle problems with deceleration, pay careful attention to the sign of acceleration and ensure it reduces the velocity magnitude.

Question 9

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

  1. 3 s3\ \text{s}3 s (correct answer)
  2. 19 s19\ \text{s}19 s
  3. 88 s88\ \text{s}88 s
  4. 311 s\tfrac{3}{11}\ \text{s}113​ 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 10

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

  1. 1.25 rad1.25\ \text{rad}1.25 rad
  2. 9 rad9\ \text{rad}9 rad
  3. 20 rad20\ \text{rad}20 rad (correct answer)
  4. 5 rad/s5\ \text{rad/s}5 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 11

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

  1. −2 rad/s2-2\,\text{rad/s}^2−2rad/s2
  2. +2 rad/s2+2\,\text{rad/s}^2+2rad/s2 (correct answer)
  3. −4 rad/s2-4\,\text{rad/s}^2−4rad/s2
  4. +4 rad+4\,\text{rad}+4rad

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 12

A turntable’s angular position is recorded as θ(t)=2t2\theta(t)=2t^2θ(t)=2t2 (radians) from t=0t=0t=0 to 2 s2\,\text{s}2s. What is its angular acceleration?

  1. 2 rad/s22\,\text{rad/s}^22rad/s2
  2. 4 rad/s24\,\text{rad/s}^24rad/s2 (correct answer)
  3. 8 rad/s28\,\text{rad/s}^28rad/s2
  4. 2 rad/s2\,\text{rad/s}2rad/s

Explanation: This question assesses understanding of rotational kinematics by relating angular position to angular acceleration. In angular motion, the angular position θ describes the rotational displacement, angular velocity ω is the rate of change of θ over time, and angular acceleration α measures how ω changes. For a position function like θ(t) = 2t², ω is found by differentiating θ with respect to time, yielding ω(t) = 4t, which shows velocity increasing linearly. Differentiating again gives α = 4 rad/s², constant throughout the motion, analogous to linear kinematics where acceleration is the second derivative of position. A common distractor, like choice A (2 rad/s²), might result from taking the coefficient of t² directly without differentiating twice. To solve similar problems, always differentiate the given position function step-by-step to find velocity and acceleration, verifying units match expectations.

Question 13

A wheel rotates with constant ω=−5 rad/s\omega=-5\ \text{rad/s}ω=−5 rad/s for 2 s2\ \text{s}2 s. Which describes its angular displacement?

  1. +10 rad+10\ \text{rad}+10 rad, because speed is 5 rad/s5\ \text{rad/s}5 rad/s
  2. −10 rad-10\ \text{rad}−10 rad, because the rotation is in the negative direction (correct answer)
  3. −2.5 rad-2.5\ \text{rad}−2.5 rad, because Δθ=ω/t\Delta\theta=\omega/tΔθ=ω/t
  4. 0 rad0\ \text{rad}0 rad, because ω\omegaω is constant

Explanation: This question probes understanding angular displacement with negative angular velocity in rotational kinematics. Negative velocity indicates rotation in the opposite direction, affecting the sign of displacement. Displacement is velocity times time, preserving the direction information. Qualitatively, this shows how direction influences total angle covered, unlike speed which ignores sign. Choice A is a distractor that uses the magnitude but assigns a positive sign, ignoring the negative direction. A transferable strategy is to consistently track signs in rotational variables to determine direction-dependent quantities.

Question 14

A wheel starts from rest and rotates with constant α=2 rad/s2\alpha=2\ \text{rad/s}^2α=2 rad/s2 for 5 s5\ \text{s}5 s. What is ωf\omega_fωf​?

  1. 25 rad25\ \text{rad}25 rad
  2. 10 rad/s10\ \text{rad/s}10 rad/s (correct answer)
  3. 5 rad/s25\ \text{rad/s}^25 rad/s2
  4. 2.5 rad/s2.5\ \text{rad/s}2.5 rad/s

Explanation: This problem tests finding final angular velocity starting from rest with constant acceleration. The wheel starts from rest (ω₀ = 0 rad/s) with α = 2 rad/s² for t = 5 s. Using ω = ω₀ + αt, we get ω = 0 + 2(5) = 10 rad/s. The final angular velocity is directly proportional to both acceleration and time when starting from rest. Choice A (25 rad) incorrectly calculates angular displacement instead of velocity using ½αt². To find final velocity from rest, multiply acceleration by time; to find displacement from rest, use ½αt².

Question 15

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

  1. 2 rad/s2\,\text{rad/s}2rad/s (correct answer)
  2. 8 rad/s8\,\text{rad/s}8rad/s
  3. 4 rad/s24\,\text{rad/s}^24rad/s2
  4. 10 rad10\,\text{rad}10rad

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

  1. −10 rad-10\,\text{rad}−10rad
  2. 0 rad/s20\,\text{rad/s}^20rad/s2 (correct answer)
  3. −5 rad/s2-5\,\text{rad/s}^2−5rad/s2
  4. −2.5 rad/s-2.5\,\text{rad/s}−2.5rad/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 17

A wheel has ω(t)=8−2t\omega(t)=8-2tω(t)=8−2t (rad/s). At what time does it momentarily stop rotating?

  1. t=2 st=2\ \text{s}t=2 s
  2. t=4 st=4\ \text{s}t=4 s (correct answer)
  3. t=6 st=6\ \text{s}t=6 s
  4. It never stops because ω\omegaω is changing

Explanation: This question evaluates determining when angular velocity reaches zero from a time-dependent function in rotational kinematics. The velocity function decreasing linearly suggests constant negative acceleration, eventually crossing zero. Setting the function to zero solves for the time of momentary stop. Qualitatively, this illustrates how changing velocity can reverse direction or halt rotation. Choice D is a distractor assuming continuous change prevents stopping, ignoring the mathematical zero crossing. A transferable strategy is to set kinematic functions to target values and solve for unknowns like time.

Question 18

A spinner’s angular position is constant at θ=3\theta=3θ=3 rad for 555 s. What are ω\omegaω and α\alphaα during this interval?

  1. ω=3 rad/s\omega=3\ \text{rad/s}ω=3 rad/s, α=0\alpha=0α=0
  2. ω=0\omega=0ω=0, α=3 rad/s2\alpha=3\ \text{rad/s}^2α=3 rad/s2
  3. ω=0\omega=0ω=0, α=0\alpha=0α=0 (correct answer)
  4. ω=35 rad/s\omega=\tfrac{3}{5}\ \text{rad/s}ω=53​ rad/s, α=0\alpha=0α=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 19

A wheel rotates with constant ω=−4 rad/s\omega=-4\ \text{rad/s}ω=−4 rad/s for 3 s3\ \text{s}3 s. What is Δθ\Delta\thetaΔθ?

  1. 12 rad12\ \text{rad}12 rad
  2. −12 rad-12\ \text{rad}−12 rad (correct answer)
  3. −1.33 rad/s-1.33\ \text{rad/s}−1.33 rad/s
  4. 0 rad/s20\ \text{rad/s}^20 rad/s2

Explanation: This problem tests understanding of angular displacement with constant negative angular velocity. The wheel rotates with ω = -4 rad/s for t = 3 s. Using Δθ = ωt, we get Δθ = (-4)(3) = -12 rad. The negative angular displacement indicates rotation in the negative direction (typically clockwise when viewed from above). Choice A (12 rad) has the correct magnitude but wrong sign, missing that negative velocity produces negative displacement. When working with signed quantities in rotational motion, maintain consistency in sign conventions throughout the problem.

Question 20

A wheel has constant angular acceleration α=−2 rad/s2\alpha=-2\,\text{rad/s}^2α=−2rad/s2 while ω\omegaω goes from 666 to 2 rad/s2\,\text{rad/s}2rad/s. How long does this take?

  1. −2 s-2\,\text{s}−2s
  2. 1 s1\,\text{s}1s
  3. 2 s2\,\text{s}2s (correct answer)
  4. 4 s4\,\text{s}4s

Explanation: This question explores rotational kinematics under constant angular acceleration to find time. With α = -2 rad/s² and ω changing from 6 to 2 rad/s, Δω = -4 rad/s, so t = Δω/α = 2 s, using the equation ω_f = ω_i + αt rearranged. Negative α slows positive ω, showing deceleration qualitatively. This mirrors linear kinematics where time is found from velocity change and acceleration. Choice D (4 s) might result from using magnitudes without signs. Always rearrange kinematic equations carefully, preserving signs, to solve for time or other variables.