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

A ceiling fan blade rotating at 10 rad/s10 \, \text{rad/s} is switched off and slows down with a constant angular acceleration of magnitude 2.0 rad/s22.0 \, \text{rad/s}^2. Through what angle does the blade rotate before coming to rest?

5.0 rad5.0 \, \text{rad}
12.5 rad12.5 \, \text{rad}
25 rad25 \, \text{rad}
50 rad50 \, \text{rad}
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Rotational Kinematics

Practice Rotational Kinematics in AP Physics C Mechanics 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 C Mechanics.

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.

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Question 1

A ceiling fan blade rotating at 10 rad/s10 \, \text{rad/s} is switched off and slows down with a constant angular acceleration of magnitude 2.0 rad/s22.0 \, \text{rad/s}^2. Through what angle does the blade rotate before coming to rest?

  1. 5.0 rad5.0 \, \text{rad}
  2. 12.5 rad12.5 \, \text{rad}
  3. 25 rad25 \, \text{rad} (correct answer)
  4. 50 rad50 \, \text{rad}
Explanation: Use the rotational kinematic equation ω2=ω02+2αΔθ\omega^2 = \omega_0^2 + 2\alpha\Delta\theta. The final angular velocity is ω=0\omega = 0, the initial angular velocity is ω0=10 rad/s\omega_0 = 10 \, \text{rad/s}, and the angular acceleration is α=−2.0 rad/s2\alpha = -2.0 \, \text{rad/s}^2 (negative because it's slowing down). Solving for Δθ\Delta\theta: 02=(10)2+2(−2.0)Δθ0^2 = (10)^2 + 2(-2.0)\Delta\theta. This gives 0=100−4.0Δθ0 = 100 - 4.0\Delta\theta, so 4.0Δθ=1004.0\Delta\theta = 100, and Δθ=25 rad\Delta\theta = 25 \, \text{rad}.

Question 2

The angular position of a spinning wheel is given by θ(t)=13t3−52t2+6t\theta(t) = \frac{1}{3}t^3 - \frac{5}{2}t^2 + 6t, where tt is in seconds and θ\theta is in radians. For t>0t>0, at what time does the wheel first momentarily reverse its direction of rotation?

  1. 1.0 s1.0 \, \text{s}
  2. 2.0 s2.0 \, \text{s} (correct answer)
  3. 2.5 s2.5 \, \text{s}
  4. 3.0 s3.0 \, \text{s}
Explanation: The wheel reverses its direction of rotation when its angular velocity ω\omega changes sign, which occurs when ω(t)=0\omega(t)=0. To find ω(t)\omega(t), we differentiate θ(t)\theta(t) with respect to time: ω(t)=dθdt=t2−5t+6\omega(t) = \frac{d\theta}{dt} = t^2 - 5t + 6. Setting ω(t)=0\omega(t) = 0 gives t2−5t+6=0t^2 - 5t + 6 = 0, which factors to (t−2)(t−3)=0(t-2)(t-3) = 0. The solutions are t=2 st=2 \, \text{s} and t=3 st=3 \, \text{s}. The first time this occurs for t>0t>0 is at t=2.0 st=2.0 \, \text{s}.

Question 3

A spinning top has an angular velocity given by ω(t)=ω0e−kt\omega(t) = \omega_0 e^{-kt}, where ω0\omega_0 and kk are positive constants. If the top's angular position at t=0t=0 is θ=0\theta=0, what is its angular position θ(t)\theta(t) for t>0t>0?

  1. θ(t)=−ω0ke−kt\theta(t) = -\omega_0 k e^{-kt}
  2. θ(t)=ω0ke−kt\theta(t) = \frac{\omega_0}{k} e^{-kt}
  3. θ(t)=ω0k(1−e−kt)\theta(t) = \frac{\omega_0}{k}(1-e^{-kt}) (correct answer)
  4. θ(t)=ω0(1−e−kt)\theta(t) = \omega_0(1-e^{-kt})
Explanation: Angular position θ\theta is the integral of angular velocity ω\omega with respect to time. θ(t)=∫0tω(t′)dt′+θ(0)\theta(t) = \int_0^t \omega(t') dt' + \theta(0). Given θ(0)=0\theta(0)=0, we have θ(t)=∫0tω0e−kt′dt′=ω0[−1ke−kt′]0t=−ω0k(e−kt−e0)=−ω0k(e−kt−1)=ω0k(1−e−kt)\theta(t) = \int_0^t \omega_0 e^{-kt'} dt' = \omega_0 \left[ -\frac{1}{k}e^{-kt'} \right]_0^t = -\frac{\omega_0}{k}(e^{-kt} - e^0) = -\frac{\omega_0}{k}(e^{-kt} - 1) = \frac{\omega_0}{k}(1-e^{-kt}).

Question 4

The angular acceleration of a wheel is given by α(t)=6t−12\alpha(t) = 6t - 12, in units of rad/s2^2. The wheel starts from rest at t=0t=0. At what time t>0t>0 does the wheel's direction of rotation reverse?

  1. t=2 st=2 \, \text{s}
  2. t=3 st=3 \, \text{s}
  3. t=4 st=4 \, \text{s} (correct answer)
  4. The wheel's direction of rotation never reverses.
Explanation: To find when the direction reverses, we must find when the angular velocity ω\omega is zero (and changing sign). First, find ω(t)\omega(t) by integrating α(t)\alpha(t): ω(t)=∫(6t−12)dt=3t2−12t+C\omega(t) = \int (6t-12) dt = 3t^2 - 12t + C. Since the wheel starts from rest, ω(0)=0\omega(0)=0, which implies C=0C=0. So, ω(t)=3t2−12t=3t(t−4)\omega(t) = 3t^2 - 12t = 3t(t-4). The angular velocity is zero at t=0t=0 and t=4 st=4 \, \text{s}. For 0<t<40 < t < 4, ω(t)\omega(t) is negative. For t>4t > 4, ω(t)\omega(t) is positive. Therefore, the direction of rotation reverses at t=4 st=4 \, \text{s}.

Question 5

A rigid body, initially rotating with angular velocity ω0\omega_0, is subjected to a time-varying angular acceleration α(t)\alpha(t). The change in the body's angular velocity from t=0t=0 to a final time t=tft=t_f is equal to which of the following?

  1. The value of the α\alpha versus tt graph at t=tft=t_f
  2. The slope of the α\alpha versus tt graph at t=tft=t_f
  3. The area under the ω\omega versus tt graph from t=0t=0 to t=tft=t_f
  4. The area under the α\alpha versus tt graph from t=0t=0 to t=tft=t_f (correct answer)
Explanation: By definition, α=dωdt\alpha = \frac{d\omega}{dt}. Rearranging gives dω=αdtd\omega = \alpha dt. To find the total change in angular velocity, Δω\Delta\omega, we integrate this expression over the time interval: Δω=∫0tfα(t)dt\Delta\omega = \int_{0}^{t_f} \alpha(t) dt. The definite integral of a function represents the area under the curve of that function. Thus, the change in angular velocity is the area under the α\alpha versus tt graph.

Question 6

A rotating object starts from rest at t=0t=0 and is subjected to a time-dependent angular acceleration given by α(t)=6t rad/s2\alpha(t) = 6t \, \text{rad/s}^2. What is its total angular displacement after the first 2.0 s2.0 \, \text{s}?

  1. 6.0 rad6.0 \, \text{rad}
  2. 8.0 rad8.0 \, \text{rad} (correct answer)
  3. 12 rad12 \, \text{rad}
  4. 24 rad24 \, \text{rad}
Explanation: To find the angular displacement, we must integrate the angular acceleration twice. First, find the angular velocity ω(t)\omega(t) by integrating α(t)\alpha(t): ω(t)=∫α(t)dt=∫6tdt=3t2+C\omega(t) = \int \alpha(t) dt = \int 6t dt = 3t^2 + C. Since the object starts from rest, ω(0)=0\omega(0) = 0, so C=0C=0. Thus, ω(t)=3t2\omega(t) = 3t^2. Next, find the angular displacement Δθ\Delta\theta by integrating ω(t)\omega(t) from t=0t=0 to t=2.0t=2.0 s: Δθ=∫023t2dt=[t3]02=23−03=8.0 rad\Delta\theta = \int_0^2 3t^2 dt = \left[ t^3 \right]_0^2 = 2^3 - 0^3 = 8.0 \, \text{rad}.

Question 7

A wheel with moment of inertia I=1.2 kg\cdotpm2I=1.2\ \text{kg·m}^2 must speed up from ω0=5.0 rad/s\omega_0=5.0\ \text{rad/s} to ωf=17 rad/s\omega_f=17\ \text{rad/s} in t=4.0 st=4.0\ \text{s} under a constant net torque (friction included). Key equations: α=(ωf−ω0)/t\alpha=(\omega_f-\omega_0)/t and τ=Iα\tau=I\alpha. Using the given conditions, what is the torque required to achieve this change?

  1. τ=1.8 N\cdotpm\tau=1.8\ \text{N·m}
  2. τ=3.6 N\cdotpm\tau=3.6\ \text{N·m} (correct answer)
  3. τ=14 N\cdotpm\tau=14\ \text{N·m}
  4. τ=0.30 N\cdotpm\tau=0.30\ \text{N·m}
Explanation: This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating required torque for a desired angular acceleration. The problem requires working backwards from kinematic information to find the necessary torque. For a wheel with I = 1.2 kg·m² accelerating from ω0 = 5.0 rad/s to ωf = 17 rad/s in t = 4.0 s, we first find the angular acceleration. Choice B is correct because α = (ωf - ω0)/t = (17 - 5.0)/4.0 = 12/4.0 = 3.0 rad/s², and then τ = Iα = 1.2 × 3.0 = 3.6 N·m. Choice C at 14 N·m might result from calculation errors or using wrong values. To help students: emphasize the two-step process of finding acceleration first, then torque; practice problems that work backwards from desired motion to required forces/torques; and always verify that calculated values make physical sense.

Question 8

A disk with moment of inertia I=0.30 kg\cdotpm2I=0.30\ \text{kg·m}^2 is acted on by a constant torque τ=1.5 N\cdotpm\tau=1.5\ \text{N·m} for t=2.0 st=2.0\ \text{s}, starting from rest; friction is negligible. Key equations: τ=Iα\tau=I\alpha and θ=12αt2\theta=\tfrac12\alpha t^2. Using the given conditions, calculate the total angular displacement during the 2.0 s interval.

  1. θ=10 rad\theta=10\ \text{rad} (correct answer)
  2. θ=2.5 rad\theta=2.5\ \text{rad}
  3. θ=20 rad\theta=20\ \text{rad}
  4. θ=5.0 rad\theta=5.0\ \text{rad}
Explanation: This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating angular displacement under constant angular acceleration. This parallels the translational kinematic equation for displacement under constant acceleration. For a disk with I = 0.30 kg·m² experiencing torque τ = 1.5 N·m for t = 2.0 s starting from rest, we need to find total angular displacement. Choice A is correct because first we find α = τ/I = 1.5/0.30 = 5.0 rad/s², then use θ = ½αt² = ½(5.0)(2.0)² = ½(5.0)(4.0) = 10 rad. Choice C at 20 rad might result from forgetting the factor of ½ in the kinematic equation. To help students: draw parallels between rotational and translational kinematic equations; emphasize that the ½ factor comes from integration just as in linear motion; and practice problems starting from rest to simplify calculations.

Question 9

A gyroscope has spin angular momentum magnitude L=0.40 kg\cdotpm2/sL=0.40\ \text{kg·m}^2/\text{s}. Gravity exerts a torque of magnitude τ=0.080 N\cdotpm\tau=0.080\ \text{N·m} about the pivot, producing steady precession; assume the spin magnitude stays constant. Key equation: Ω=τ/L\Omega=\tau/L. Using the given conditions, determine the precession rate Ω\Omega.

  1. Ω=0.20 rad/s\Omega=0.20\ \text{rad/s} (correct answer)
  2. Ω=5.0 rad/s\Omega=5.0\ \text{rad/s}
  3. Ω=0.032 rad/s\Omega=0.032\ \text{rad/s}
  4. Ω=2.0 rad/s\Omega=2.0\ \text{rad/s}
Explanation: This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically gyroscopic precession. Precession occurs when a torque acts perpendicular to a spinning object's angular momentum, causing the spin axis to rotate. For a gyroscope with angular momentum L = 0.40 kg·m²/s experiencing torque τ = 0.080 N·m, the precession rate is found using the gyroscopic equation. Choice A is correct because Ω = τ/L = 0.080/0.40 = 0.20 rad/s. Choice B at 5.0 rad/s would result from inverting the fraction or other calculation errors. To help students: explain that precession is different from ordinary rotation; emphasize that the precession equation Ω = τ/L applies when spin angular momentum is much larger than precession angular momentum; and use demonstrations or videos to visualize gyroscopic motion.

Question 10

A rigid disk rotates about a fixed axis. Its angular position is given by the equation θ(t)=At3−Bt\theta(t) = At^3 - Bt, where AA and BB are positive constants. What is the angular velocity ω(t)\omega(t) of the disk as a function of time?

  1. ω(t)=3At2−B\omega(t) = 3At^2 - B (correct answer)
  2. ω(t)=At3−B\omega(t) = At^3 - B
  3. ω(t)=14At4−12Bt2\omega(t) = \frac{1}{4}At^4 - \frac{1}{2}Bt^2
  4. ω(t)=6At\omega(t) = 6At
Explanation: Angular velocity ω\omega is the time derivative of angular position θ\theta. Taking the derivative of the given function θ(t)=At3−Bt\theta(t) = At^3 - Bt with respect to time tt gives ω(t)=dθdt=ddt(At3−Bt)=3At2−B\omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(At^3 - Bt) = 3At^2 - B.

Question 11

A flywheel's angular velocity is described by the function ω(t)=C−Dt2\omega(t) = C - Dt^2, where CC and DD are positive constants and tt is time. What expression represents the angular acceleration α(t)\alpha(t) of the flywheel?

  1. α(t)=Ct−13Dt3\alpha(t) = Ct - \frac{1}{3}Dt^3
  2. α(t)=C−Dt2\alpha(t) = C - Dt^2
  3. α(t)=−2D\alpha(t) = -2D
  4. α(t)=−2Dt\alpha(t) = -2Dt (correct answer)
Explanation: Angular acceleration α\alpha is the time derivative of angular velocity ω\omega. Taking the derivative of the given function ω(t)=C−Dt2\omega(t) = C - Dt^2 with respect to time tt gives α(t)=dωdt=ddt(C−Dt2)=−2Dt\alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt}(C - Dt^2) = -2Dt.

Question 12

The angular position of a rotating object is given by θ(t)=2t3\theta(t) = 2t^3, where θ\theta is in radians and tt is in seconds. What is the average angular velocity of the object over the time interval from t=1 st=1 \, \text{s} to t=3 st=3 \, \text{s}?

  1. 24 rad/s24 \, \text{rad/s}
  2. 26 rad/s26 \, \text{rad/s} (correct answer)
  3. 30 rad/s30 \, \text{rad/s}
  4. 52 rad/s52 \, \text{rad/s}
Explanation: Average angular velocity is defined as the total angular displacement divided by the time interval, ωavg=ΔθΔt\omega_{avg} = \frac{\Delta\theta}{\Delta t}. First, find the angular positions at the endpoints of the interval: θ(3)=2(3)3=54 rad\theta(3) = 2(3)^3 = 54 \, \text{rad} and θ(1)=2(1)3=2 rad\theta(1) = 2(1)^3 = 2 \, \text{rad}. The angular displacement is Δθ=θ(3)−θ(1)=54−2=52 rad\Delta\theta = \theta(3) - \theta(1) = 54 - 2 = 52 \, \text{rad}. The time interval is Δt=3−1=2 s\Delta t = 3 - 1 = 2 \, \text{s}. Therefore, ωavg=52 rad2 s=26 rad/s\omega_{avg} = \frac{52 \, \text{rad}}{2 \, \text{s}} = 26 \, \text{rad/s}.

Question 13

The graph of angular velocity ω\omega versus time tt for a rotating rigid body is a straight line with a positive slope, passing through the origin. What does this indicate about the body's motion?

  1. The body has a constant positive angular velocity.
  2. The body is undergoing a constant positive angular acceleration, starting from rest. (correct answer)
  3. The body is undergoing an angular acceleration that increases linearly with time.
  4. The body has a constant positive angular displacement from its starting point.
Explanation: The angular acceleration α\alpha is the slope of the ω\omega versus tt graph. A straight line with a positive slope indicates a constant positive angular acceleration. Since the line passes through the origin, the initial angular velocity at t=0t=0 is zero, meaning the body started from rest.

Question 14

A merry-go-round is rotating counter-clockwise and its rate of rotation is decreasing. If the counter-clockwise direction is defined as positive, which of the following correctly describes the signs of its angular velocity ω\omega and angular acceleration α\alpha?

  1. ω\omega is positive, α\alpha is negative. (correct answer)
  2. ω\omega is positive, α\alpha is positive.
  3. ω\omega is negative, α\alpha is positive.
  4. ω\omega is negative, α\alpha is negative.
Explanation: Since the merry-go-round is rotating counter-clockwise and this direction is positive, its angular velocity ω\omega is positive. Since it is slowing down, its angular acceleration α\alpha must be in the opposite direction to its angular velocity. Therefore, the angular acceleration α\alpha is negative.

Question 15

A carousel starts from rest and accelerates at a constant rate α1\alpha_1 for a time t1t_1. It then rotates at a constant angular velocity for an additional time t2t_2. What is the total angular displacement Δθ\Delta\theta of the carousel during the entire time interval t1+t2t_1 + t_2?

  1. 12α1(t1+t2)2\frac{1}{2}\alpha_1 (t_1+t_2)^2
  2. α1t12+α1t1t2\alpha_1 t_1^2 + \alpha_1 t_1 t_2
  3. 12α1t12+α1t1t2\frac{1}{2}\alpha_1 t_1^2 + \alpha_1 t_1 t_2 (correct answer)
  4. 12α1t12+12α1t22\frac{1}{2}\alpha_1 t_1^2 + \frac{1}{2}\alpha_1 t_2^2
Explanation: The motion has two parts. Part 1 (acceleration): Δθ1=ω0t1+12α1t12=0+12α1t12\Delta\theta_1 = \omega_0 t_1 + \frac{1}{2}\alpha_1 t_1^2 = 0 + \frac{1}{2}\alpha_1 t_1^2. The angular velocity at the end of this part is ω1=ω0+α1t1=α1t1\omega_1 = \omega_0 + \alpha_1 t_1 = \alpha_1 t_1. Part 2 (constant velocity): Δθ2=ω1t2=(α1t1)t2\Delta\theta_2 = \omega_1 t_2 = (\alpha_1 t_1) t_2. The total displacement is the sum: Δθ=Δθ1+Δθ2=12α1t12+α1t1t2\Delta\theta = \Delta\theta_1 + \Delta\theta_2 = \frac{1}{2}\alpha_1 t_1^2 + \alpha_1 t_1 t_2.

Question 16

A computer hard drive platter accelerates from rest to its operating angular velocity of 72007200 revolutions per minute (rpm) over an angular displacement of 5050 revolutions. What is the magnitude of its average angular acceleration, assuming it is constant?

  1. 72π rad/s272\pi \, \text{rad/s}^2
  2. 144π rad/s2144\pi \, \text{rad/s}^2
  3. 288π rad/s2288\pi \, \text{rad/s}^2 (correct answer)
  4. 576π rad/s2576\pi \, \text{rad/s}^2
Explanation: First, convert all quantities to SI units. Final angular velocity: ωf=7200 rpm×2π rad1 rev×1 min60 s=240π rad/s\omega_f = 7200 \, \text{rpm} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = 240\pi \, \text{rad/s}. Angular displacement: Δθ=50 rev×2π rad1 rev=100π rad\Delta\theta = 50 \, \text{rev} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} = 100\pi \, \text{rad}. Use the kinematic equation ωf2=ω02+2αΔθ\omega_f^2 = \omega_0^2 + 2\alpha\Delta\theta. Since it starts from rest, ω0=0\omega_0=0. (240π)2=2α(100π)(240\pi)^2 = 2\alpha(100\pi). 57600π2=200πα57600\pi^2 = 200\pi\alpha. Solving for α\alpha gives α=57600π2200π=288π rad/s2\alpha = \frac{57600\pi^2}{200\pi} = 288\pi \, \text{rad/s}^2.

Question 17

The angular velocity of a motor is given by ω(t)=At3+B\omega(t) = At^3 + B, where A=1.0 rad/s4A=1.0 \, \text{rad/s}^4 and B=2.0 rad/sB=2.0 \, \text{rad/s}. What is the average angular acceleration over the interval from t=0 st=0 \, \text{s} to t=2.0 st=2.0 \, \text{s}?

  1. 4.0 rad/s24.0 \, \text{rad/s}^2 (correct answer)
  2. 6.0 rad/s26.0 \, \text{rad/s}^2
  3. 8.0 rad/s28.0 \, \text{rad/s}^2
  4. 12.0 rad/s212.0 \, \text{rad/s}^2
Explanation: Average angular acceleration is the change in angular velocity divided by the time interval, αavg=ΔωΔt\alpha_{avg} = \frac{\Delta\omega}{\Delta t}. First, find the angular velocities: ω(2)=(1.0)(2.0)3+2.0=8.0+2.0=10.0 rad/s\omega(2) = (1.0)(2.0)^3 + 2.0 = 8.0 + 2.0 = 10.0 \, \text{rad/s}. ω(0)=(1.0)(0)3+2.0=2.0 rad/s\omega(0) = (1.0)(0)^3 + 2.0 = 2.0 \, \text{rad/s}. The change in angular velocity is Δω=10.0−2.0=8.0 rad/s\Delta\omega = 10.0 - 2.0 = 8.0 \, \text{rad/s}. The time interval is Δt=2.0 s\Delta t = 2.0 \, \text{s}. So, αavg=8.0 rad/s2.0 s=4.0 rad/s2\alpha_{avg} = \frac{8.0 \, \text{rad/s}}{2.0 \, \text{s}} = 4.0 \, \text{rad/s}^2.

Question 18

A rigid body rotates with a constant positive angular velocity ω\omega about a fixed axis. Which of the following statements correctly describes its angular acceleration α\alpha and its angular position θ\theta, assuming θ(0)=0\theta(0)=0?

  1. α=0\alpha=0 and θ\theta is constant.
  2. α\alpha is constant and positive, and θ\theta increases quadratically with time.
  3. α=0\alpha=0 and θ\theta increases linearly with time. (correct answer)
  4. α\alpha is not constant and θ\theta increases linearly with time.
Explanation: If the angular velocity ω\omega is constant, the angular acceleration α=dωdt\alpha = \frac{d\omega}{dt} must be zero. The angular position is the integral of the constant angular velocity, θ(t)=∫ωdt=ωt+C\theta(t) = \int \omega dt = \omega t + C. Since θ(0)=0\theta(0)=0, the constant of integration is zero, so θ(t)=ωt\theta(t) = \omega t. This shows that the angular position increases linearly with time.

Question 19

The angular position θ\theta of an oscillating rigid body is plotted as a function of time tt. The resulting curve is a cosine function. At which of the following points in time is the magnitude of the angular velocity at a minimum and the magnitude of the angular acceleration at a maximum?

  1. At the times when the angular displacement is zero.
  2. At the times corresponding to the maximum and minimum angular displacements. (correct answer)
  3. At the times when the slope of the θ\theta vs. tt graph is steepest.
  4. At all points in time, because the motion is periodic and these conditions always hold.
Explanation: The angular velocity ω\omega is the slope of the θ\theta vs. tt graph. For a cosine curve, the slope is zero at the peaks and troughs (maximum and minimum displacements), so the magnitude of angular velocity is at a minimum (zero) at these points. Angular acceleration α\alpha is the second derivative of θ(t)\theta(t), which is proportional to −θ(t)-\theta(t) for simple harmonic motion. Therefore, the magnitude of α\alpha is maximum when the magnitude of θ\theta is maximum.

Question 20

A wheel starts from rest and rotates with constant angular acceleration α\alpha. After rotating through an angle θ1\theta_1, its angular velocity is ω1\omega_1. After rotating through an additional angle θ2\theta_2, its angular velocity becomes ω2\omega_2. Which expression correctly relates these quantities?

  1. ω22=ω12+2αθ2\omega_2^2 = \omega_1^2 + 2\alpha\theta_2 (correct answer)
  2. ω22=ω12+2α(θ1+θ2)\omega_2^2 = \omega_1^2 + 2\alpha(\theta_1 + \theta_2)
  3. ω2=ω1+α2θ2\omega_2 = \omega_1 + \alpha\sqrt{2\theta_2}
  4. ω22=2α(θ1+θ2)\omega_2^2 = 2\alpha(\theta_1 + \theta_2)
Explanation: This problem requires careful application of rotational kinematic equations. The wheel starts from rest and after angle θ1\theta_1, has angular velocity ω1\omega_1. Then after an additional angle θ2\theta_2, it has angular velocity ω2\omega_2. For the second phase of motion (from ω1\omega_1 to ω2\omega_2 through angle θ2\theta_2), we use ωf2=ωi2+2αθ\omega_f^2 = \omega_i^2 + 2\alpha\theta, which gives ω22=ω12+2αθ2\omega_2^2 = \omega_1^2 + 2\alpha\theta_2. Choice B incorrectly uses the total angle from the start. Choice C incorrectly takes the square root of the angle term. Choice D ignores the initial angular velocity ω1\omega_1 for the second phase.