AP Physics 1 Torque and Rotational Dynamics — Worked Answer Explanations

Unit 5 · 12% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Physics 1 Torque and Rotational Dynamics practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Torque and Rotational Dynamics practice test and come back here to review, or head back to the Torque and Rotational Dynamics unit overview.

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Question 1 · Easy
A force of $20 N$ is applied perpendicular to the end of a $0.50 m$ wrench. What torque does it produce about the bolt?
- A
  $10 N\cdotm$ Correct
- B
  $20 N\cdotm$
  Why not B: Forgot the lever arm.
- C
  $40 N\cdotm$
  Why not C: Doubled the answer.
- D
  $0.50 N\cdotm$
  Why not D: Reported just the lever arm.
Explanation
$τ = r F sin θ = 0.50 \times 20 \times 1 = 10 N\cdotm$ .

Key takeaway
Torque is force times perpendicular lever arm.
Question 2 · Easy
A solid disk of mass $M$ and radius $R$ rotates about an axis through its center. What is its moment of inertia?
- A
  $M R^{2}$
  Why not A: Hoop value, not disk.
- B
  $\frac{1}{2} M R^{2}$ Correct
- C
  $\frac{2}{5} M R^{2}$
  Why not C: Solid sphere.
- D
  $\frac{1}{12} M L^{2}$
  Why not D: Thin rod about its center.
Explanation
For a solid disk (or solid cylinder) about its central axis, $I = \frac{1}{2} M R^{2}$ .

Key takeaway
Memorize standard moments of inertia: hoop $MR^2$, disk $\tfrac{1}{2}MR^2$, solid sphere $\tfrac{2}{5}MR^2$, rod (center) $\tfrac{1}{12}ML^2$.
Question 3 · Easy
A net torque of $4.0 N\cdotm$ is applied to a wheel with moment of inertia $I = 2.0 kg\cdotm^{2}$ . What is the angular acceleration?
- A
  $0.5 rad/s^{2}$
  Why not A: Inverted Newton's law for rotation.
- B
  $2.0 rad/s^{2}$ Correct
- C
  $6.0 rad/s^{2}$
  Why not C: Added instead of dividing.
- D
  $8.0 rad/s^{2}$
  Why not D: Multiplied instead of dividing.
Explanation
$τ = I α$ , so $α = τ / I = 4.0/2.0 = 2.0 rad/s^{2}$ .

Key takeaway
Rotational analog of $F=ma$: $\tau = I\alpha$.
Question 4 · Easy
An object moves in uniform circular motion. Which statement about the net force is true?
- A
  Net force is zero.
  Why not A: Velocity direction changes — non-zero acceleration.
- B
  Net force points tangent to the circle.
  Why not B: Tangential force would change speed; uniform means constant speed.
- C
  Net force points toward the center.Correct
- D
  Net force points outward (centrifugal).
  Why not D: Centrifugal is a fictitious force in the rotating frame.
Explanation
In uniform circular motion, the net force is centripetal — always pointing toward the center of the circle.

Key takeaway
Uniform circular motion: net force is centripetal (toward center).
Question 5 · Easy
A heavy door is harder to push open if you push close to the hinge. The reason is that:
- A
  The mass of the door is larger near the hinge.
  Why not A: Mass distribution doesn't change.
- B
  The lever arm is shorter, requiring more force for the same torque.Correct
- C
  Friction is greater near the hinge.
  Why not C: Hinge friction is small and unrelated to push location.
- D
  Air resistance is higher near the hinge.
  Why not D: Negligible at low speeds.
Explanation
Torque $= r F$ . At small $r$ , you need a larger $F$ to produce enough torque to open the door against any resistive torque (hinge friction).

Key takeaway
Forces applied far from a pivot produce more torque per newton of effort.
Question 6 · Medium
A uniform meterstick (mass $0.20 kg$ ) is balanced at its center. A $0.10 kg$ mass is hung $30 cm$ from the pivot on one side. Where on the other side should a $0.30 kg$ mass be hung to balance the stick?
- A
  $10 cm$ Correct
- B
  $15 cm$
  Why not B: Used wrong mass ratio.
- C
  $30 cm$
  Why not C: Same distance, ignored mass difference.
- D
  $90 cm$
  Why not D: Inverted the mass ratio.
Explanation
Balance: $m_{1} d_{1} = m_{2} d_{2}$ . So $0.10 \times 30 = 0.30 \times d$ , giving $d = 10 cm$ . (The stick is uniform and pivoted at its center, so its weight contributes no torque about the pivot.)

Key takeaway
For a balanced beam: sum of torques on each side equals zero — $m_1 d_1 = m_2 d_2$ for two point masses.
Question 7 · Medium
A $2.0 kg$ object on a string moves in a horizontal circle of radius $0.50 m$ at $4.0 m/s$ . What is the tension in the string?
- A
  $8.0 N$
  Why not A: Used $v$ rather than $v^{2}$ .
- B
  $16 N$
  Why not B: Off by a factor of 4.
- C
  $32 N$
  Why not C: Doubled the radius.
- D
  $64 N$ Correct
Explanation
Centripetal: $T = m v^{2} / r = 2.0 \times 16/0.50 = 64 N$ .

Key takeaway
Centripetal force: $F_c = m v^2 / r$ — quadratic in speed.
Question 8 · Medium
A wheel rotating at $10 rad/s$ slows uniformly to rest in $5.0 s$ . Through what angle does it rotate during this time?
- A
  $10 rad$
  Why not A: Used initial speed alone.
- B
  $25 rad$ Correct
- C
  $50 rad$
  Why not C: Forgot to apply average angular velocity.
- D
  $100 rad$
  Why not D: Multiplied by extra factor of 2.
Explanation
Average angular velocity is $(10 + 0) /2 = 5.0 rad/s$ . $θ = ω_{a v g} t = 5.0 \times 5.0 = 25 rad$ .

Key takeaway
Use rotational kinematics analog: $\theta = \omega_{avg} t$ for uniform $\alpha$.