What this is for. This diagnostic shows which Unit 5 (Torque and Rotational Dynamics) topics to revisit. These are original Marlbridge practice questions, not past exam questions. It is not calibrated against exam results, so it gives no predicted score.
How to sit it. Allow about 30 minutes. Close your notes. Use a calculator for arithmetic and trigonometry, but do the calculus by hand. Use g = 9.8 m/s². Each question states its positive sense of rotation. θ is in rad, ω in rad/s, α in rad/s² and t in s; coefficients carry the units that make each term correct. Strings do not slip or stretch.
Question 1 (multiple choice · 5.1)
Take counterclockwise as positive. A rotor has ω₀ = −9.0 rad/s (clockwise) at t = 0. From then on its angular acceleration is α(t) = 2.0t. At what time is the rotor momentarily at rest?
- (A) 2.1 s
- (B) 3.0 s
- (C) 4.5 s
- (D) Never: α is positive, so the rotor keeps spinning faster.
Answer and explanation
Answer: (B). ω = −9.0 + ∫₀ᵗ 2.0t dt = −9.0 + 1.0t², which is zero at t = 3.0 s.
- (A) writes ω = ω₀ + α(t)·t = −9.0 + 2.0t². That formula needs a constant α.
- (C) treats α as a constant 2.0 rad/s². It grows with time.
- (D) ignores the signs: ω and α are opposite, so the rotor slows.
If you missed this: read “Angular velocity and angular acceleration as derivatives” and Worked example 2 in the Topic 5.1 study guide.
Question 2 (multiple choice · 5.2)
A motor turns a small pulley of radius 0.10 m at 60 rad/s. A belt that does not slip runs round it and round a large pulley of radius 0.30 m. What is the angular velocity of the large pulley?
- (A) 6.0 rad/s
- (B) 20 rad/s
- (C) 60 rad/s
- (D) 180 rad/s
Answer and explanation
Answer: (B). Both rims move with the belt: v = 0.10 × 60 = 6.0 m/s. For the large pulley, ω = v/R = 6.0 ÷ 0.30 = 20 rad/s.
- (A) is the belt speed in m/s, labelled as an angular velocity.
- (C) assumes the pulleys share ω. They share the belt speed, not ω.
- (D) multiplies by the radius ratio instead of dividing.
If you missed this: read “Strings, hoses and belts that do not slip” in the Topic 5.2 study guide.
Question 3 (multiple choice · 5.3)
A 50 N force acts at the end of a bar, 0.80 m from the axis. The force’s line of action passes 0.60 m from the axis. What is the size of the torque about the axis?
- (A) 23 N·m
- (B) 26 N·m
- (C) 30 N·m
- (D) 40 N·m
Answer and explanation
Answer: (C). The lever arm is the perpendicular distance from the axis to the line of action: τ = r⊥F = 0.60 × 50 = 30 N·m. (Check: sin θ = 0.75, so rF sin θ = 30 N·m.)
- (A) uses the sine twice. Use the lever arm or the perpendicular component, not both.
- (B) uses cos θ = 0.66, the component along the bar, which gives no torque.
- (D) is rF, correct only when the force is perpendicular to the bar.
If you missed this: read “Only the perpendicular component turns” in the Topic 5.3 study guide.
Question 4 (multiple choice · 5.3)
Take +x to the right, +y up the page and +z out of the page. A force F = (−15î + 10ĵ) N acts at r = (0.60î + 0.20ĵ) m from an axle along the z-axis. What is the torque τ = r × F?
- (A) 9.0 N·m out of the page (counterclockwise)
- (B) 9.0 N·m into the page (clockwise)
- (C) 3.0 N·m out of the page (counterclockwise)
- (D) 11 N·m out of the page (counterclockwise)
Answer and explanation
Answer: (A). τ_z = xF_y − yF_x = (0.60)(10) − (0.20)(−15) = 6.0 + 3.0 = +9.0 N·m. +z (out of the page) means counterclockwise.
- (B) is F × r. Reversing the order reverses the direction.
- (C) drops the minus sign in the second term: 6.0 + (0.20)(−15) = 3.0.
- (D) is |r||F| = 11.4 N·m, as if F were perpendicular to r. The angle between them is about 128°.
If you missed this: read “Torque as a vector: the cross product” in the Topic 5.3 study guide.
Question 5 (multiple choice · 5.4)
A uniform rod of mass 0.60 kg and length 0.90 m turns about a perpendicular axis through one end. A small 0.20 kg ball is fixed to the far end. What is the rotational inertia of the rod and ball about the axis?
- (A) 0.20 kg·m²
- (B) 0.22 kg·m²
- (C) 0.32 kg·m²
- (D) 0.65 kg·m²
Answer and explanation
Answer: (C). I = ML²/3 + mL² = 0.162 + 0.162 = 0.324 kg·m², both about the same axis.
- (A) uses ML²/12, the rod’s value about its centre.
- (B) treats the ball as part of the rod, (M + m)L²/3. The ball is all at distance L.
- (D) puts all the mass at the far end, (M + m)L². Most of the rod is closer to the axis.
If you missed this: read “Point masses: I = mr²” and “Deriving the standard results” in the Topic 5.4 study guide.
Question 6 (multiple choice · 5.5)
A uniform metre stick of mass 0.15 kg rests on a pivot at the 40 cm mark. Where must a 0.10 kg mass hang for the stick to balance horizontally?
- (A) At the 25 cm mark
- (B) At the 33 cm mark
- (C) At the 55 cm mark
- (D) Nowhere: the stick can balance only on its centre of mass.
Answer and explanation
Answer: (A). The stick’s weight acts at its centre of mass, the 50 cm mark, 0.10 m right of the pivot. Torques about the pivot: (0.10 kg)g d = (0.15 kg)g(0.10 m), so d = 0.15 m left of the pivot, at the 25 cm mark.
- (B) inverts the mass ratio: d = 0.10 × 0.10 ÷ 0.15 = 0.067 m.
- (C) is the right distance on the wrong side, where both torques are clockwise.
- (D) forgets the hanging mass: the combined centre of mass can sit over the pivot.
If you missed this: read “Extended force diagrams” and “Choosing the point for torques” in the Topic 5.5 study guide.
Question 7 (multiple choice · 5.6)
A 2.0 kg block hangs from a light string wound round a uniform disk pulley of mass 4.0 kg and radius 0.10 m, which turns on a frictionless axle (I = ½MR²). The block is released. What is the angular acceleration of the pulley?
- (A) 0.49 rad/s²
- (B) 4.9 rad/s²
- (C) 33 rad/s²
- (D) 49 rad/s²
Answer and explanation
Answer: (D). I = ½(4.0)(0.10)² = 0.020 kg·m². Block: mg − T = ma. Pulley: TR = Iα, with a = Rα. Combining, a = mg ÷ (m + I/R²) = 19.6 ÷ (2.0 + 2.0) = 4.9 m/s², so α = a/R = 49 rad/s².
- (A) multiplies a by R instead of dividing: α = a/R, not aR.
- (B) is the block’s linear acceleration, in m/s², not α.
- (C) treats the pulley as a hoop, a = mg ÷ (m + M).
If you missed this: read “Linear and rotational analyses are separate” in the Topic 5.6 study guide.
Question 8 (short answer · 5.2)
Take counterclockwise as positive for the drum. A winch drum of radius 0.15 m lifts a bucket on a rope wound round it. Starting from rest, the drum’s angular velocity is ω(t) = 0.80t².
(a) Find the bucket’s speed and acceleration at t = 3.0 s. (b) Find how far the bucket rises in the first 3.0 s. (c) Find the size of a rim point’s acceleration at t = 3.0 s, and explain why it exceeds the bucket’s.
Answer and explanation
(a) ω(3.0) = 7.2 rad/s, so v = Rω = 0.15 × 7.2 = 1.1 m/s (1.08 m/s). α = dω/dt = 1.6t = 4.8 rad/s², so a = Rα = 0.72 m/s².
(b) θ = ∫₀³ 0.80t² dt = 0.80 × 27 ÷ 3 = 7.2 rad. The bucket rises s = Rθ = 0.15 × 7.2 = 1.1 m (1.08 m).
(c) a_T = 0.72 m/s² and a_c = ω²R = 7.2² × 0.15 = 7.8 m/s², so |a| = √(0.72² + 7.78²) = 7.8 m/s². The rim point moves in a circle, so it also has centripetal acceleration; the bucket does not.
If you missed this: read “Deriving v = rω and a_T = rα” and Worked example 2 in the Topic 5.2 study guide.
Question 9 (short answer · 5.4)
A thin rod lies along x from 0 to L. Its mass per unit length is λ = cx², where c is a constant, and its mass is M.
(a) Find c in terms of M and L. (b) Derive I about a perpendicular axis through x = 0. (c) Find x_cm and I_cm, then use the parallel axis theorem to find I about x = L. (d) Evaluate the three rotational inertias for M = 0.40 kg and L = 1.0 m.
Answer and explanation
(a) M = ∫₀ᴸ cx² dx = cL³/3, so c = 3M/L³.
(b) I₀ = ∫₀ᴸ x² (cx²) dx = cL⁵/5 = 3ML²/5.
(c) x_cm = (1/M)∫₀ᴸ x(cx²) dx = cL⁴/(4M) = 3L/4. I_cm = I₀ − M(3L/4)² = 3ML²/5 − 9ML²/16 = 3ML²/80. The end x = L is L/4 from the centre of mass: I_L = 3ML²/80 + M(L/4)² = ML²/10.
(d) I₀ = 0.24 kg·m², I_cm = 0.015 kg·m², I_L = 0.040 kg·m². I_L < I₀ because most of the mass sits near x = L.
If you missed this: read “Continuous objects: I = ∫r² dm”, “The parallel axis theorem” and Worked example 2 in the Topic 5.4 study guide.
Question 10 (short answer · 5.6)
Take counterclockwise as positive. A grinding wheel is a uniform disk of mass 2.0 kg and radius 0.10 m. It spins counterclockwise at 150 rad/s with the motor off. A tool presses on the rim straight towards the axle with a normal force of 20 N. The coefficient of kinetic friction is 0.50. Ignore bearing friction.
(a) Find the angular acceleration of the wheel. (b) Find the time to stop and the number of revolutions it makes while stopping. (c) Explain why the 20 N normal force exerts no torque about the axle.
Answer and explanation
(a) I = ½MR² = 0.010 kg·m². Friction f = μN = 10 N acts along the rim against the rotation, with lever arm R: τ = −(10)(0.10) = −1.0 N·m. α = τ/I = −100 rad/s² (clockwise, so the wheel slows).
(b) α is constant. Time: 150 ÷ 100 = 1.5 s. Angle: Δθ = ω₀² ÷ (2|α|) = 150² ÷ 200 = 112.5 rad, which is 18 revolutions (17.9).
(c) The normal force points along the radius, so its line of action passes through the axis and its lever arm is zero. Only the tangential friction turns the wheel.
If you missed this: read “Newton’s second law in rotational form” in the Topic 5.6 study guide. For (c), see “Only the perpendicular component turns” in Topic 5.3.
Your next step
Mark each question. Count a short answer as missed if any part went wrong.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 5.1 Rotational Kinematics | 1, 10(b) | Topic 5.1 study guide |
| 5.2 Connecting Linear and Rotational Motion | 2, 8 | Topic 5.2 study guide |
| 5.3 Torque | 3, 4, 10(c) | Topic 5.3 study guide |
| 5.4 Rotational Inertia | 5, 9 | Topic 5.4 study guide |
| 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form | 6 | Topic 5.5 study guide |
| 5.6 Newton’s Second Law in Rotational Form | 7, 10 | Topic 5.6 study guide |
How to use your result
- Read every explanation, even for questions you got right. A right answer for a wrong reason is still a gap.
- Missed one question in a topic? Read the named section, then try its practice set.
- Missed two or more in a topic? Work through its whole study guide, practice set and checklist.
- Missed questions across several topics? Start with Topic 5.3. Torque feeds Topics 5.5 and 5.6, which also use I from Topic 5.4.
- All right? Go straight to the mixed unit review.
Your result here is a guide to what to study next. It does not predict an exam score.