OxfordAQA International AS and A-level Physics: Rotational motion and angular momentum (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.13.1 (Rotational motion) of OxfordAQA International AS and A-level Physics (9630), Version 4.4. The specification labels section 3.13 International A-level only. Read the study guide first, or use the revision notes. Use g = 9.81 m s⁻². Give answers to a sensible number of significant figures. Not sure what to practise? Take a free diagnostic.

Questions

1. A light rod of length 0.90 m carries a 0.40 kg mass at one end and a 0.60 kg mass at the other. It turns about an axis perpendicular to the rod, 0.30 m from the 0.40 kg mass. (a) Calculate the moment of inertia of the system about this axis. [2] (b) The axis is moved to the midpoint of the rod. Calculate the new moment of inertia and state whether it is larger or smaller. [2]

2. Two cylinders have the same mass and the same radius. One is solid. The other is a thin-walled hollow tube. Each is free to turn about its central axis, and the same constant torque is applied to each from rest. State which gains angular speed more quickly and explain why. [3]

3. A turntable speeds up uniformly from rest to 33⅓ revolutions per minute in 1.8 s. (a) Show that the final angular speed is 3.49 rad s⁻¹. [1] (b) Calculate the angular acceleration. [2] (c) Calculate the angle turned through, in radians and in revolutions. [2]

4. A fairground ride turns at 5.0 rad s⁻¹ and is then slowed uniformly with an angular acceleration of −0.40 rad s⁻² until it stops. (a) Calculate the time taken to stop. [2] (b) Calculate the number of revolutions made while slowing. [2]

5. A wheel of moment of inertia 0.50 kg m² has a steady tangential force of 12 N applied at a radius of 0.25 m. A constant frictional torque of 0.50 N m acts at the axle. The wheel starts from rest. (a) Calculate the resultant torque and the angular acceleration. [3] (b) Calculate the angular speed after 4.0 s. [1] (c) Calculate the angle turned through in this time. [1]

6. A horizontal turntable of moment of inertia 0.045 kg m² spins freely at 4.0 rad s⁻¹. A lump of clay of mass 0.080 kg is dropped onto it, lands 0.12 m from the axis and sticks. (a) Calculate the angular momentum of the turntable before the clay lands. [1] (b) Calculate the angular speed afterwards. [3] (c) Show that kinetic energy is not conserved. [1]

7. A uniform rod of mass 0.90 kg and length 1.0 m is hinged at one end and released from rest in a horizontal position. About the hinge it has I = ML²/3. (a) Calculate I about the hinge. [1] (b) Calculate the loss of gravitational potential energy as the rod swings to the vertical. [2] (c) Calculate the angular speed of the rod as it passes the vertical. [2] (d) Calculate the speed of the free end at that moment. [1] (e) Explain why the fall of the centre of mass is not 1.0 m. [1]

8. A motor applies a steady torque of 18 N m to a shaft turning at 2400 revolutions per minute. (a) Calculate the angular speed in rad s⁻¹. [1] (b) Calculate the power delivered. [1] (c) Calculate the work done during 50 revolutions. [2]

9. State the rotational quantity that corresponds to each of the following. (a) Mass. [1] (b) Force. [1] (c) Linear momentum. [1] (d) Write the rotational equation that corresponds to P = Fv. [1]

10. A student sits on a stool that turns freely. Each hand holds a 2.5 kg mass, treated as a point mass, 0.70 m from the axis. The student and stool together have I = 1.8 kg m² and the system turns at 1.2 rad s⁻¹. The student pulls both masses in to 0.20 m from the axis. (a) Calculate the moment of inertia before and after. [2] (b) Calculate the new angular speed. [2] (c) Calculate the change in kinetic energy and explain where the energy comes from. [3]

11. A string is wound round a drum of radius 0.20 m. The drum has moment of inertia 0.10 kg m² about its axle and turns without friction. A 1.5 kg mass hangs from the free end and is released from rest. (a) Write an equation for the motion of the mass, using F for the tension. [1] (b) Write an equation for the rotation of the drum. [1] (c) State the relationship between the acceleration a of the mass and the angular acceleration α. [1] (d) Calculate a. [2] (e) Calculate the tension F. [1] (f) The mass falls 2.0 m. Show that the total kinetic energy gained equals the loss of gravitational potential energy. [2] (g) Explain why a is less than 9.81 m s⁻². [1]

Answers

1. (a) Distances from axis: 0.30 m and 0.60 m [1]. I = 0.40 × 0.30² + 0.60 × 0.60² = 0.036 + 0.216 = 0.25 kg m² [1] (b) I = (0.40 + 0.60) × 0.45² = 0.20 kg m² [1]. It is smaller, because the heavier mass is now nearer the axis [1]. [4] Examiner insight: Show each mr² term separately; a single wrong distance is easier to spot, and to credit, when the working is laid out.

2. The hollow tube has its mass at the full radius, so it has the larger moment of inertia [1]. α = T/I, so for the same torque the solid cylinder has the larger angular acceleration [1]. The solid cylinder gains angular speed more quickly [1]. [3] Examiner insight: Name the quantity that differs (moment of inertia) and link it to T = Iα; “it is heavier” is not the point, since the masses are equal.

3. (a) ω = 33⅓ × 2π / 60 = 3.49 rad s⁻¹ [1] (b) α = (3.49 − 0) / 1.8 [1] = 1.9 rad s⁻² [1] (c) θ = (0 + 3.49) × 1.8 / 2 = 3.1 rad [1], which is 3.14 / 2π = 0.50 revolutions [1] [5] Examiner insight: A “show that” answer should carry an unrounded or more precise value before the quoted one, and the conversion factor 2π/60 must be visible.

4. (a) 0 = 5.0 + (−0.40)t [1], so t = 12.5 s [1] (b) 0 = 5.0² + 2(−0.40)θ gives θ = 31.25 rad [1], and 31.25 / 2π = 5.0 revolutions [1] [4] Examiner insight: Keep the minus sign on α. Dropping it gives a negative time, which is a sign something is wrong.

5. (a) Applied torque = 12 × 0.25 = 3.0 N m [1]. Resultant = 3.0 − 0.50 = 2.5 N m [1]. α = 2.5 / 0.50 = 5.0 rad s⁻² [1] (b) ω = 5.0 × 4.0 = 20 rad s⁻¹ [1] (c) θ = ½ × 5.0 × 4.0² = 40 rad [1] [5] Examiner insight: Using 3.0 N m in T = Iα gives 6.0 rad s⁻²; say that friction opposes the drive.

6. (a) L = Iω = 0.045 × 4.0 = 0.18 kg m² s⁻¹ [1] (b) I of clay = 0.080 × 0.12² = 1.15 × 10⁻³ kg m² [1]. Total I = 0.0462 kg m² [1]. ω = 0.18 / 0.0462 = 3.9 rad s⁻¹ [1] (c) Ek before = ½ × 0.045 × 4.0² = 0.36 J; after = ½ × 0.0462 × 3.90² = 0.35 J, so it falls [1] [5] Examiner insight: Say angular momentum is conserved because no external torque acts; compare both energies.

7. (a) I = 0.90 × 1.0² / 3 = 0.30 kg m² [1] (b) The centre of mass falls 0.50 m [1]. ΔEp = 0.90 × 9.81 × 0.50 = 4.4 J [1] (c) ½Iω² = 4.41 [1], so ω² = 2 × 4.41 / 0.30 = 29.4 and ω = 5.4 rad s⁻¹ [1] (d) v = rω = 1.0 × 5.4 = 5.4 m s⁻¹ [1] (e) The weight acts at the centre of the rod, which starts 0.50 m from the hinge and ends 0.50 m below it; the whole rod does not fall 1.0 m [1] [7] Examiner insight: State that the potential energy lost becomes rotational kinetic energy before equating them.

8. (a) ω = 2400 × 2π / 60 = 251 rad s⁻¹ [1] (b) P = Tω = 18 × 251 = 4.5 kW [1] (c) θ = 50 × 2π = 314 rad [1]. W = Tθ = 18 × 314 = 5.7 kJ [1] [4] Examiner insight: Convert to rad s⁻¹ once and carry the full value into parts (b) and (c).

9. (a) Moment of inertia [1] (b) Torque [1] (c) Angular momentum [1] (d) P = Tω [1] [4] Examiner insight: For (d), write the equation rather than describing it in words.

10. (a) Before: I = 1.8 + 2 × 2.5 × 0.70² = 4.25 kg m² [1]. After: I = 1.8 + 2 × 2.5 × 0.20² = 2.0 kg m² [1] (b) 4.25 × 1.2 = 2.0 × ω [1], so ω = 2.6 rad s⁻¹ (2.55) [1] (c) Ek before = ½ × 4.25 × 1.2² = 3.06 J; after = ½ × 2.0 × 2.55² = 6.50 J [1]. Increase = 3.4 J [1]. The student does work pulling the masses inwards against the need for a centripetal force, and this work is the source of the extra energy [1] [7] Examiner insight: Link the rise in energy to work done by the student, not to conservation of angular momentum.

11. (a) mg − F = ma, that is 14.7 − F = 1.5a [1] (b) Fr = Iα, that is 0.20F = 0.10α [1] (c) a = rα, because the string does not slip on the drum [1] (d) F = Iα/r = Ia/r² = 2.5a [1]. 14.7 = 1.5a + 2.5a, so a = 14.7/4.0 = 3.7 m s⁻² [1] (e) F = 2.5 × 3.68 = 9.2 N [1] (f) v² = 2 × 3.68 × 2.0, so v = 3.84 m s⁻¹ and ω = v/r = 19.2 rad s⁻¹. Ek(trans) = ½ × 1.5 × 3.84² = 11.0 J and Ek(rot) = ½ × 0.10 × 19.2² = 18.4 J [1]. The total is 29.4 J, and mgh = 1.5 × 9.81 × 2.0 = 29.4 J [1] (g) The tension is less than the weight, because part of the weight’s effect goes into the drum’s angular acceleration [1] [9] Examiner insight: Eliminate F to combine the equations, and keep a unrounded for part (f).

Where marks are usually lost

• Using rev min⁻¹ in a formula instead of converting to rad s⁻¹.
• Taking the distance in mr² to the wrong point, or forgetting to square it.
• Using one value of r for several masses instead of adding one mr² term for each.
• Using the applied torque, rather than the resultant torque, in T = Iα.
• Saying angular momentum is conserved without giving the condition: no external resultant torque.
• Assuming kinetic energy is conserved when two rotating objects join.
• In drum and pulley problems, taking the string tension to be mg.
• Rounding early, then failing to reproduce a “show that” value.

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Visit the course hub and the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Published by OxfordAQA: International AS and A-level Physics (9630), specification Version 4.4, which applies to exams from May/June 2020. Source for this page: section 3.13.1, Rotational motion.