AP Physics 1 Quiz: Rotational Kinematics

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.

Question 1

1. $6.0\,\text{rad}$
2. $3.0\,\text{rad/s}$
3. $6.0\,\text{rad/s}$ (correct answer)
4. $24\,\text{rad/s}^2$

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

Question 2

1. $3\ \text{s}$ (correct answer)
2. $19\ \text{s}$
3. $88\ \text{s}$
4. $\tfrac{3}{11}\ \text{s}$

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

1. $1.25\ \text{rad}$
2. $9\ \text{rad}$
3. $20\ \text{rad}$ (correct answer)
4. $5\ \text{rad/s}$

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

1. $3.0\,\text{rad/s}$ (correct answer)
2. $6.0\,\text{rad}$
3. $4.0\,\text{rad/s}^2$
4. $0.33\,\text{rad/s}$

Explanation: This problem requires calculating average angular velocity from angular displacement. Average angular velocity is defined as $ω_{\text{avg}} = \Deltaθ/\Delta t = (θ_f - θ_i)/\Delta t$. The angular position changes from $1.0 \, \text{rad}$ to $7.0 \, \text{rad}$, giving $\Deltaθ = 7.0 - 1.0 = 6.0 \, \text{rad}$ over $\Delta t = 2.0 \, \text{s}$. Therefore, $ω_{\text{avg}} = 6.0/2.0 = 3.0 \, \text{rad/s}$. Choice B ($6.0 \, \text{rad}$) gives the displacement rather than velocity, while choice C ($4.0 \, \text{rad/s}^2$) has incorrect units for velocity. For average velocity problems, always divide total displacement by total time.

Question 5

1. $10\,\text{rad}$ (correct answer)
2. $7\,\text{rad/s}$
3. $14\,\text{rad}$
4. $4\,\text{rad/s}^2$

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 6

1. $30\,\text{rad}$ (correct answer)
2. $1.2\,\text{rad/s}^2$
3. $6\,\text{rad}$
4. $0\,\text{rad/s}$

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 7

1. $2.0\,\text{rad/s}^2$
2. $4.0\,\text{rad/s}^2$ (correct answer)
3. $12\,\text{rad}$
4. $4.0\,\text{rad/s}$

Explanation: This problem tests understanding of rotational kinematics relationships between angular position, velocity, and acceleration. Angular acceleration $α$ is the second derivative of angular position with respect to time: $α = \frac{d^2θ}{dt^2}$. Given $θ(t) = 2.0t^2$, we first find angular velocity by taking the first derivative: $ω = \frac{dθ}{dt} = 4.0t \, \text{rad/s}$. Then, taking the derivative of $ω$ gives us $α = \frac{dω}{dt} = 4.0 \, \text{rad/s}^2$. Choice C ($12 \, \text{rad}$) has incorrect units for acceleration, while choice D ($4.0 \, \text{rad/s}$) 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.

Question 8

1. $36\ \text{rad/s}^2$
2. $4\ \text{rad/s}^2$ (correct answer)
3. $9\ \text{rad/s}^2$
4. $0.25\ \text{rad/s}^2$

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 9

1. $8\,\text{rad/s}$
2. $-2\,\text{rad/s}^2$ (correct answer)
3. $-2\,\text{rad}$
4. $2\,\text{rad/s}^2$

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

Question 10

1. $+10\ \text{rad}$, because speed is $5\ \text{rad/s}$
2. $-10\ \text{rad}$, because the rotation is in the negative direction (correct answer)
3. $-2.5\ \text{rad}$, because $\Delta\theta=\omega/t$
4. $0\ \text{rad}$, because $\omega$ 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 11

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

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 12

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

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 13

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

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 14

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

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 15

1. $2\,\text{rad/s}^2$
2. $4\,\text{rad/s}^2$ (correct answer)
3. $8\,\text{rad/s}^2$
4. $2\,\text{rad/s}$

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 16

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

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 17

1. $2\,\text{rad/s}$ (correct answer)
2. $8\,\text{rad/s}$
3. $4\,\text{rad/s}^2$
4. $10\,\text{rad}$

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 18

1. $-10\,\text{rad}$
2. $0\,\text{rad/s}^2$ (correct answer)
3. $-5\,\text{rad/s}^2$
4. $-2.5\,\text{rad/s}$

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 19

1. $t=2\ \text{s}$
2. $t=4\ \text{s}$ (correct answer)
3. $t=6\ \text{s}$
4. It never stops because $\omega$ 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 20

1. $\omega=3\ \text{rad/s}$, $\alpha=0$
2. $\omega=0$, $\alpha=3\ \text{rad/s}^2$
3. $\omega=0$, $\alpha=0$ (correct answer)
4. $\omega=\tfrac{3}{5}\ \text{rad/s}$, $\alpha=0$

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.