Unit 5 practice test: Torque and Rotational Dynamics
12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value; data sets share one table, graph or scene diagram.
Covers: Rotational Kinematics; Linking Linear and Rotational Motion; Torque; Rotational Inertia; Rotational Equilibrium; Newton's Second Law for Rotation.
Data table
Spinning up a wheel
A student applies a steady torque to a wheel on a low-friction axle and measures its angular acceleration with a rotation sensor. She repeats this for four torques.
| Torque (N·m) | Angular acceleration (rad/s²) |
|---|---|
| 0.50 | 1.0 |
| 1.00 | 2.1 |
| 1.50 | 2.9 |
| 2.00 | 4.0 |
1. Which plot gives a straight line whose slope tells you the wheel’s rotational inertia?
- α against τ; the slope is 1/I
- α against τ; the slope is I
- α against 1/τ; the slope is I
- τ against α²; the slope is I
Show the answer
Newton’s second law in rotational form, τ = Iα, gives α = (1/I)τ: a line through the origin with slope 1/I (or plot τ against α for slope I).
- Correct: α against τ; the slope is 1/I: Right: α = τ/I is a line through the origin with slope 1/I.
- α against τ; the slope is I: Rearranged, α = (1/I)τ, so the slope is the reciprocal of I.
- α against 1/τ; the slope is I: α grows with τ, so against 1/τ the points curve.
- τ against α²; the slope is I: τ is proportional to α, not α².
2. A metal ring is clamped to the wheel, doubling its rotational inertia. With a torque of 1.00 N·m, its angular acceleration would be about
- 1.0 rad/s²
- 4.0 rad/s²
- 2.0 rad/s²
- 0.50 rad/s²
Show the answer
α = τ/I. At 1.00 N·m the wheel alone had α ≈ 2.0 rad/s²; doubling I halves it to about 1.0 rad/s².
- Correct: 1.0 rad/s²: Right: α = τ/I, so doubling I halves α from 2.0 rad/s².
- 4.0 rad/s²: A larger rotational inertia makes the wheel harder to speed up, not easier.
- 2.0 rad/s²: The torque is the same but the inertia doubled, so α changes.
- 0.50 rad/s²: α ∝ 1/I, not 1/I²: doubling I halves α.
3. A second student repeats the experiment with a rusty axle. Her line of α against τ is parallel to the first but crosses the τ-axis at 0.30 N·m. What best explains this?
- Friction at the axle exerts a torque of about 0.30 N·m that opposes the applied torque
- Rust on the axle gives the wheel a larger rotational inertia than before
- Her torque sensor reads each torque 0.30 N·m lower than the real value
- Below 0.30 N·m of applied torque, the wheel turns in the opposite direction
Show the answer
The net torque is τ_applied − τ_friction = Iα. The same slope means the same I; the intercept at 0.30 N·m means α stays zero until the applied torque exceeds a friction torque of about 0.30 N·m.
- Correct: Friction at the axle exerts a torque of about 0.30 N·m that opposes the applied torque: Right: the wheel speeds up only once the applied torque beats the friction torque.
- Rust on the axle gives the wheel a larger rotational inertia than before: A larger I would change the slope, but the lines are parallel.
- Her torque sensor reads each torque 0.30 N·m lower than the real value: A sensor reading too low would shift the line the other way.
- Below 0.30 N·m of applied torque, the wheel turns in the opposite direction: Below 0.30 N·m the wheel does not turn at all.
Diagram
A potter’s wheel speeding up
4. About how many revolutions does the wheel make in the 3.0 s?
- 1.4 rev
- 9.0 rev
- 2.9 rev
- 0.72 rev
Show the answer
One revolution is 2π rad, so 9.0 rad ÷ 2π = 1.43, about 1.4 rev.
- Correct: 1.4 rev: Right: 9.0 rad ÷ 2π rad per revolution = 1.4 rev.
- 9.0 rev: This treats each radian as a whole turn. One revolution is 2π rad, about 6.28 rad.
- 2.9 rev: This divides by π instead of 2π. A full turn is 2π rad.
- 0.72 rev: This halves the correct value. A revolution is 2π rad, not 4π rad.
5. A second chalk mark is drawn halfway between the first mark and the axle. During the same 3.0 s, its angular displacement is
- the same as the first mark’s
- half the first mark’s
- twice the first mark’s
- one quarter of the first mark’s
Show the answer
The wheel is a rigid system: every point on it turns through the same angle in the same time, so both marks have the same angular displacement (9.0 rad), the same ω and the same α.
- Correct: the same as the first mark’s: Right: a rigid wheel turns as one piece, so every mark turns through the same angle.
- half the first mark’s: Halving the distance from the axle changes how far the mark travels, not the angle it turns through.
- twice the first mark’s: Being closer to the axle does not make a mark turn through more angle; the wheel turns as one piece.
- one quarter of the first mark’s: No square law applies here: every mark on a rigid wheel turns through the same angle.
6. The wheel keeps the same angular acceleration but runs for 6.0 s instead of 3.0 s, starting from rest. Its angular displacement is
- 4 times as large
- 2 times as large
- the same
- 8 times as large
Show the answer
From rest, θ = ½αt². With α fixed, θ ∝ t², so twice the time gives 2² = 4 times the angle (36 rad).
- Correct: 4 times as large: Right: from rest θ = ½αt², so doubling t multiplies θ by 2² = 4.
- 2 times as large: That would be true at constant ω. With constant α from rest, θ grows with t².
- the same: A longer time at the same α gives more turning, not the same.
- 8 times as large: θ depends on t², not t³.
7. A Ferris wheel of radius 15 m makes one revolution every 60 s. What is the speed of a rider?
Type a number in m/s.
Show the answer
ω = 2π rad / 60 s = 0.105 rad/s, so v = rω = (15 m)(0.105 rad/s) = 1.57 m/s, about 1.6 m/s.
- Answer: 1.6 m/s
8. A mechanic pulls on the end of a 0.30 m wrench with an 80 N force at 60° to the handle. What is the size of the torque about the bolt?
Type a number in N·m.
Show the answer
τ = rF sin θ = (0.30 m)(80 N)(sin 60°) = 20.8 N·m, about 21 N·m.
- Answer: 21 N·m
9. A solid disk has mass 2.0 kg and radius 0.30 m. Its rotational inertia about its center is I = ½MR² (given). What is I?
Type a number in kg·m^2.
Show the answer
I = ½MR² = ½(2.0 kg)(0.30 m)² = 0.090 kg·m².
- Answer: 0.090 kg·m^2
10. A 25 kg child sits 1.8 m from the pivot of a seesaw. How far from the pivot, on the other side, must a 30 kg child sit to balance it? Ignore the plank’s mass.
Type a number in m.
Show the answer
Balanced torques about the pivot: (25 kg)g(1.8 m) = (30 kg)g r, so r = 45/30 = 1.5 m. g cancels.
- Answer: 1.5 m
11. A record turntable spins at 45 revolutions per minute. What is its angular velocity in rad/s?
Type a number in rad/s.
Show the answer
ω = (45 rev/min)(2π rad/rev)(1 min/60 s) = 4.71 rad/s, about 4.7 rad/s.
- Answer: 4.7 rad/s
12. A point on the rim of a grinding wheel of radius 0.40 m has a tangential acceleration of 2.0 m/s². What is the wheel’s angular acceleration?
Type a number in rad/s^2.
Show the answer
α = a_T/r = (2.0 m/s²)/(0.40 m) = 5.0 rad/s².
- Answer: 5.0 rad/s^2
Keep going
Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.