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AP Physics C: Mechanics Unit 5 practice questions

Try 10 multiple-choice questions from Unit 5 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. Question 1

    A rigid body rotates with an angular position given by θ(t)=2t3−4t\theta(t) = 2t^3 - 4t, where θ\theta is in radians and tt is in seconds. How does the angular acceleration α\alpha at time t=4t = 4 s compare to the angular acceleration at time t=2t = 2 s?

    The angular acceleration at t=4t = 4 s is two times the angular acceleration at t=2t = 2 s

    The angular acceleration at t=4t = 4 s is four times the angular acceleration at t=2t = 2 s

    The angular acceleration at t=4t = 4 s is half the angular acceleration at t=2t = 2 s

    The angular acceleration at t=4t = 4 s is equal to the angular acceleration at t=2t = 2 s

    Show answer

    Answer: (A)

    Differentiating the position function twice yields α(t)=12t\alpha(t) = 12t. Comparing values shows α(4)=48\alpha(4) = 48 rad/s² and α(2)=24\alpha(2) = 24 rad/s², resulting in a ratio of 2:1.

  2. Question 2

    Two flywheels, A and B, start from rest and rotate with constant angular accelerations αA\alpha_A and αB\alpha_B, respectively. If αB=4αA\alpha_B = 4\alpha_A, how does the time tAt_A required for flywheel A to rotate through a displacement Δθ\Delta\theta compare to the time tBt_B required for flywheel B to rotate through the same displacement?

    The time tAt_A is two times the time tBt_B

    The time tAt_A is four times the time tBt_B

    The time tAt_A is half the time tBt_B

    The time tAt_A is equal to the time tBt_B

    Show answer

    Answer: (A)

    Using Δθ=12αt2\Delta\theta = \frac{1}{2}\alpha t^2, time is inversely proportional to the square root of acceleration (t∝1/αt \propto 1/\sqrt{\alpha}). Since αB=4αA\alpha_B = 4\alpha_A, tB=tA/2t_B = t_A/2.

  3. Question 3

    A turbine blade rotates with an angular velocity given by ω(t)=kt\omega(t) = k\sqrt{t}, where kk is a positive constant. How does the angular acceleration α\alpha at time t=Tt = T compare to the angular acceleration at time t=4Tt = 4T?

    The acceleration at TT is eight times the acceleration at 4T4T

    The acceleration at TT is four times the acceleration at 4T4T

    The acceleration at TT is two times the acceleration at 4T4T

    The acceleration at TT is half the acceleration at 4T4T

    Show answer

    Answer: (C)

    Differentiating ω(t)\omega(t) gives α(t)=k2t\alpha(t) = \frac{k}{2\sqrt{t}}. Comparing values, α(T)∝1\alpha(T) \propto 1 and α(4T)∝1/2\alpha(4T) \propto 1/2, so the ratio is 2:1.

  4. Question 4

    A disk rotates with constant angular acceleration α\alpha. It takes time t1t_1 to accelerate from rest to angular velocity ωf\omega_f, and time t2t_2 to accelerate from ωf\omega_f to 2ωf2\omega_f. How does t2t_2 compare to t1t_1?

    The interval t2t_2 is equal to the interval t1t_1

    The interval t2t_2 is twice the interval t1t_1

    The interval t2t_2 is half the interval t1t_1

    The interval t2t_2 is four times the interval t1t_1

    Show answer

    Answer: (A)

    Using Δω=αΔt\Delta\omega = \alpha \Delta t, the change in velocity for both intervals is ωf\omega_f (ωf−0\omega_f - 0 and 2ωf−ωf2\omega_f - \omega_f). Since α\alpha is constant, the time intervals must be identical.

  5. Question 5

    A system rotates with angular acceleration α(t)=3t2\alpha(t) = 3t^2. If the system starts from rest at t=0t=0, how does the angular velocity ω\omega at t=2t=2 s compare to the angular velocity at t=1t=1 s?

    The velocity at t=2t=2 s is eight times the velocity at t=1t=1 s

    The velocity at t=2t=2 s is four times the velocity at t=1t=1 s

    The velocity at t=2t=2 s is two times the velocity at t=1t=1 s

    The velocity at t=2t=2 s is sixteen times the velocity at t=1t=1 s

    Show answer

    Answer: (A)

    Integrating acceleration α(t)=3t2\alpha(t) = 3t^2 gives ω(t)=t3\omega(t) = t^3. Comparing values, ω(2)=8\omega(2) = 8 and ω(1)=1\omega(1) = 1, resulting in a ratio of 8.

  6. Question 6

    Two points, P and Q, are located on a rigid rotating disk. Point P is at distance RR from the axis, and Point Q is at distance 2R2R. How does the angular acceleration αP\alpha_P of Point P compare to the angular acceleration αQ\alpha_Q of Point Q?

    The angular acceleration αP\alpha_P is equal to αQ\alpha_Q

    The angular acceleration αP\alpha_P is half of αQ\alpha_Q

    The angular acceleration αP\alpha_P is twice αQ\alpha_Q

    The angular acceleration αP\alpha_P is four times αQ\alpha_Q

    Show answer

    Answer: (A)

    For a rigid body, all points share the same angular quantities (θ\theta, ω\omega, α\alpha) regardless of their distance from the axis of rotation.

  7. Question 7

    A wheel rotates with an angular velocity ω\omega related to its angular position θ\theta by the equation ω=cθ\omega = c\theta, where cc is a constant. How does the angular acceleration α\alpha at position θ2=2\theta_2 = 2 rad compare to the angular acceleration at θ1=1\theta_1 = 1 rad?

    The acceleration at θ2\theta_2 is two times the acceleration at θ1\theta_1

    The acceleration at θ2\theta_2 is four times the acceleration at θ1\theta_1

    The acceleration at θ2\theta_2 is equal to the acceleration at θ1\theta_1

    The acceleration at θ2\theta_2 is half the acceleration at θ1\theta_1

    Show answer

    Answer: (A)

    Using the chain rule, α=dωdt=dωdθω\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta}\omega. With ω=cθ\omega = c\theta, α=(c)(cθ)=c2θ\alpha = (c)(c\theta) = c^2\theta. Thus, acceleration is directly proportional to θ\theta, so doubling θ\theta doubles α\alpha.

  8. Question 8

    A fan blade slows down with an angular acceleration given by α(t)=−kt\alpha(t) = -k t. If the blade stops at time TT, how does the magnitude of the angular displacement Δθ\Delta\theta in the first half of the time interval (00 to T/2T/2) compare to the displacement in the second half (T/2T/2 to TT)?

    The displacement in the first half is greater than in the second half

    The displacement in the first half is less than in the second half

    The displacement in the first half is equal to that in the second half

    The displacement in the first half is zero compared to the second half

    Show answer

    Answer: (A)

    Since the blade is slowing down, its angular velocity is higher during the first interval than the second. Therefore, it covers more angular displacement in the first half of the time.

  9. Question 9

    Two gears, A and B, are meshed. Gear A has radius RR and Gear B has radius 3R3R. If Gear A accelerates with angular acceleration αA\alpha_A, how does the angular acceleration αB\alpha_B of Gear B compare to αA\alpha_A?

    The acceleration αB\alpha_B is one-third of αA\alpha_A

    The acceleration αB\alpha_B is three times αA\alpha_A

    The acceleration αB\alpha_B is nine times αA\alpha_A

    The acceleration αB\alpha_B is equal to αA\alpha_A

    Show answer

    Answer: (A)

    The tangential acceleration at the contact point must be equal: at=RαA=3RαBa_t = R\alpha_A = 3R\alpha_B. Solving for αB\alpha_B yields αB=αA/3\alpha_B = \alpha_A / 3.

  10. Question 10

    An object rotates with angular position θ(t)=Acos⁡(bt)\theta(t) = A \cos(bt). How does the magnitude of the angular acceleration α\alpha at t=0t = 0 compare to the magnitude of the angular acceleration at t=π/(2b)t = \pi/(2b)?

    The magnitude at t=0t=0 is greater than at t=π/(2b)t=\pi/(2b)

    The magnitude at t=0t=0 is less than at t=π/(2b)t=\pi/(2b)

    The magnitude at t=0t=0 is equal to that at t=π/(2b)t=\pi/(2b)

    The magnitude at t=0t=0 is half that at t=π/(2b)t=\pi/(2b)

    Show answer

    Answer: (A)

    Differentiating twice gives α(t)=−Ab2cos⁡(bt)\alpha(t) = -Ab^2 \cos(bt). At t=0t=0, ∣α∣=Ab2|\alpha| = Ab^2. At t=π/(2b)t=\pi/(2b), the cosine term is 0, so ∣α∣=0|\alpha| = 0.

What's on the AP Physics C: Mechanics Unit 5 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

  • 20 Multiple Choice Questions (40 minutes)
  • 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
  • 1 FRQ – FRQ 3 – Experimental Design (24 minutes)

Questions are drawn from 1022 multiple-choice questions and 19 FRQs written for Unit 5, Torque and Rotational Dynamics.

Practice AP Physics C: Mechanics Unit 5 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 5 before you start

Other AP Physics C: Mechanics unit tests