This diagnostic is for the algebra-based Physics 1 course, Unit 5 (Torque and Rotational Dynamics). It shows you which topics to revisit. Allow about 30 minutes, then use the table at the end.
These are original Marlbridge practice questions, not past exam questions. They have not been calibrated against real exam results, so your total does not predict an exam score. Use a scientific calculator, g = 9.8 m/s² and 1 rev = 2π rad. Rotation senses are clockwise or counterclockwise about a stated axis. Treat strings as light and axles as frictionless unless told otherwise.
Question 1 (multiple choice · 5.1)
Take counterclockwise as positive. At t = 0 an exercise-bike flywheel turns clockwise at 8.0 rad/s. A motor gives it a constant angular acceleration of +2.0 rad/s². What is its angular displacement from t = 0 to t = 6.0 s?
- (A) −12 rad
- (B) −84 rad
- (C) +12 rad
- (D) +20 rad
Answer and explanation
Answer: (A). With ω₀ = −8.0 rad/s, Δθ = ω₀t + ½αt² = −48 + 36 = −12 rad. Check by stages: the wheel stops at 4.0 s after turning 16 rad clockwise, then turns 4.0 rad counterclockwise, so the net is −12 rad.
- (B) subtracts the ½αt² term. Substitute the signed α and let the signs do the work.
- (C) has the right size but loses the sign of ω₀.
- (D) is the total angle turned (16 + 4.0). Displacement counts the reverse turn as negative.
If you missed this: read “Speeding up or slowing down?” and Worked example 2 in the Topic 5.1 study guide.
Question 2 (multiple choice · 5.2)
A grinding wheel has radius 0.10 m. For safety, points on its rim must not move faster than 30 m/s. What is the greatest rotation rate allowed?
- (A) 3.0 × 10² rpm
- (B) 48 rpm
- (C) 2.9 × 10³ rpm
- (D) 1.8 × 10⁴ rpm
Answer and explanation
Answer: (C). ω = v ÷ r = 30 ÷ 0.10 = 300 rad/s. In rpm: 300 × 60 ÷ 2π = 2865 rpm, about 2.9 × 10³ rpm.
- (A) is the value in rad/s with the wrong unit.
- (B) is 47.7 rev/s, labelled as revolutions per minute.
- (D) multiplies by 60 but does not divide by 2π rad per revolution.
If you missed this: read “Linear speed and tangential acceleration” in the Topic 5.2 study guide.
Question 3 (multiple choice · 5.3)
A vertical mast stands on a hinge at ground level, which is the axis. A rope is tied to the mast 3.0 m above the hinge and runs down to a stake in the ground. The rope makes 30° with the mast and pulls with a tension of 200 N. What is the size of the rope’s torque about the hinge?
- (A) 600 N·m
- (B) 520 N·m
- (C) 300 N·m
- (D) 0 N·m
Answer and explanation
Answer: (C). τ = rF sin θ = 3.0 × 200 × sin 30° = 300 N·m. Equivalently, the lever arm from the hinge to the rope’s line of action is 3.0 sin 30° = 1.5 m.
- (A) leaves out sin θ, as if the rope pulled at right angles to the mast.
- (B) uses cos 30°. The angle is already between r and F, so use sin.
- (D) would be right only if the rope’s line of action passed through the hinge. It passes 1.5 m away.
If you missed this: read “Two other ways to read the same equation” in the Topic 5.3 study guide.
Question 4 (multiple choice · 5.3)
A uniform rod hangs from a pivot at its top end. It is pulled aside and held at 30° from the vertical, then at 60° from the vertical. By what factor does the torque of gravity about the pivot change?
- (A) 2.0
- (B) 1.7
- (C) 0.58
- (D) 1.0
Answer and explanation
Answer: (B). The weight acts at the centre, L/2 from the pivot, and the angle between r and the downward weight is the angle from the vertical. So τ = Mg(L/2) sin φ, and the factor is sin 60° ÷ sin 30° = 1.7.
- (A) treats the torque as proportional to the angle itself.
- (C) uses cos φ, which measures the vertical distance, not the lever arm.
- (D) notices that r and F are unchanged but forgets the angle.
If you missed this: read “Predicting changes: functional dependence” and Worked example 3 in the Topic 5.3 study guide.
Question 5 (multiple choice · 5.4)
Three small masses are fixed to a light flat frame: 2.0 kg at (0, 0), 1.0 kg at (0.30 m, 0) and 1.0 kg at (0, 0.40 m). What is the rotational inertia about an axis perpendicular to the frame through the 1.0 kg mass at (0.30 m, 0)?
- (A) 0.25 kg·m²
- (B) 0.43 kg·m²
- (C) 0.27 kg·m²
- (D) 1.1 kg·m²
Answer and explanation
Answer: (B). Measure every r from the axis. The 2.0 kg mass is 0.30 m away. The other 1.0 kg mass is √(0.30² + 0.40²) = 0.50 m away. The mass on the axis adds nothing. I = 2.0 × 0.30² + 1.0 × 0.50² = 0.18 + 0.25 = 0.43 kg·m².
- (A) measures from the origin, a different axis.
- (C) uses only the horizontal distance (0.30 m) for the top mass.
- (D) forgets to square the distances.
If you missed this: read “Several objects: add them up” in the Topic 5.4 study guide.
Question 6 (multiple choice · 5.5)
A hanging mobile is a uniform rod of mass 0.30 kg and length 0.90 m. A 0.20 kg ornament hangs from the left end and a 0.40 kg ornament from the right end. How far from the left end must the supporting string be tied for the rod to hang level?
- (A) 0.45 m
- (B) 0.60 m
- (C) 0.35 m
- (D) 0.55 m
Answer and explanation
Answer: (D). Take torques about the string, distance x from the left end. Counterclockwise: 0.20g·x + 0.30g(x − 0.45). Clockwise: 0.40g(0.90 − x). Setting them equal gives 0.90x = 0.495, so x = 0.55 m.
- (A) is the midpoint. It ignores the unequal ornaments.
- (B) forgets the rod’s own weight, acting at its centre.
- (C) is the right distance measured from the wrong end.
If you missed this: read “Choosing the pivot” and Worked example 1 in the Topic 5.5 study guide.
Question 7 (multiple choice · 5.6)
A stepped wheel turns on a fixed axle; I = 0.50 kg·m². A rope wound on the outer step (radius 0.30 m) pulls with 20 N, giving a clockwise torque. A rope wound on the inner step (radius 0.10 m) pulls with 30 N, giving a counterclockwise torque. What is the wheel’s angular acceleration?
- (A) 6.0 rad/s² clockwise
- (B) 18 rad/s² clockwise
- (C) 6.0 rad/s² counterclockwise
- (D) Zero, because the axle force balances the rope forces
Answer and explanation
Answer: (A). Clockwise torque 0.30 × 20 = 6.0 N·m; counterclockwise 0.10 × 30 = 3.0 N·m. Net 3.0 N·m clockwise, so α = 3.0 ÷ 0.50 = 6.0 rad/s² clockwise.
- (B) adds the torques instead of letting them oppose.
- (C) lets the larger force decide the sense. Torque, not force, decides.
- (D) mixes the linear and rotational analyses. The axle force acts at the axis, so it exerts no torque.
If you missed this: read “Linear and rotational analyses are separate” in the Topic 5.6 study guide.
Question 8 (short answer · 5.5)
A wheelbarrow is held at rest with its legs just off the ground. Measure all distances horizontally from the wheel’s axle. The 12 kg barrow has its centre of mass 0.50 m behind the axle. It carries a 60 kg load whose centre of mass is 0.40 m behind the axle. The gardener lifts straight up on the handles, 1.4 m behind the axle.
(a) Find the lifting force. (b) Find the upward force of the ground on the wheel. (c) The gardener says: “If I move the load to 0.20 m behind the axle, half as far, I will need half the lifting force.” Evaluate this claim.
Worked answer
(a) Take torques about the axle, so the ground’s force drops out. Clockwise (weights): 12 × 9.8 × 0.50 + 60 × 9.8 × 0.40 = 58.8 + 235.2 = 294 N·m. Counterclockwise (lift): F × 1.4. So F = 294 ÷ 1.4 = 210 N.
(b) ΣF_y = 0: N + 210 = 72 × 9.8 = 705.6 N, so N = 5.0 × 10² N (496 N).
(c) Incorrect. Only the load’s torque halves, to 117.6 N·m. The barrow’s 58.8 N·m stays. F = 176.4 ÷ 1.4 = 126 N, which is 0.60 of 210 N, not half (105 N).
Check yourself: 1 point each for choosing the axle as pivot, 210 N, 496 N from the force balance, and rejecting the claim because the barrow’s own torque does not change.
If you missed this: read “Choosing the pivot” and Worked example 1 in the Topic 5.5 study guide.
Question 9 (short answer · 5.6)
A playground roundabout has rotational inertia 180 kg·m² (given) and radius 1.5 m. Its bearing exerts a constant friction torque of 30 N·m. Starting from rest, a parent pushes tangentially on the rim with a constant 80 N.
(a) Find the net torque and the angular acceleration. (b) Find how long it takes to reach 2.0 rad/s. (c) A 20 kg child now sits on the rim (treat the child as a small mass). With the same push and friction, find the new time to reach 2.0 rad/s from rest.
Worked answer
(a) Push: 1.5 × 80 = 120 N·m. Friction opposes it, so τ_net = 120 − 30 = 90 N·m. α = 90 ÷ 180 = 0.50 rad/s².
(b) From rest, ω = αt, so t = 2.0 ÷ 0.50 = 4.0 s.
(c) I = 180 + 20 × 1.5² = 225 kg·m². α = 90 ÷ 225 = 0.40 rad/s², so t = 5.0 s.
Check yourself: 1 point each for subtracting the friction torque, 0.50 rad/s², 4.0 s, adding mr² for the child, and 5.0 s.
If you missed this: read “Using functional dependence” and Worked example 1 in the Topic 5.6 study guide.
Your next step
Mark each question right or wrong. Count Questions 8 and 9 as right only if you earned every point.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 5.1 Rotational kinematics | 1 | Topic 5.1 study guide |
| 5.2 Connecting linear and rotational motion | 2 | Topic 5.2 study guide |
| 5.3 Torque | 3, 4 | Topic 5.3 study guide |
| 5.4 Rotational inertia | 5 | Topic 5.4 study guide |
| 5.5 Rotational equilibrium | 6, 8 | Topic 5.5 study guide |
| 5.6 Newton’s second law in rotational form | 7, 9 | Topic 5.6 study guide |
How to use your result
- Missed one question in a topic? Re-read the section named in that answer, then try the topic’s practice set.
- Missed two or more questions in a topic? Start again with that study guide, including its worked examples, before doing any mixed practice.
- Guessed? Treat the question as missed.
- Look for patterns. Errors in Questions 1 and 7 usually mean a sign slip: state clockwise or counterclockwise as positive and use it for every quantity. Errors in Questions 3, 4 and 8 often mean the lever arm, not r, was the problem. Question 9(c) failing alone points back to Topic 5.4.
- All correct? Move straight to the mixed unit review, which combines the topics in longer exam-style questions.