IB DP Physics: Rotational dynamics: torque, moment of inertia and angular momentum (DP Physics) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise sub-topic A.4, Rigid body mechanics, from the IB Diploma Programme Physics guide, first assessment 2025. A.4 is additional higher level (HL only), so every question here is HL only. Use g = 9.81 N kg⁻¹. Where a question needs a moment of inertia for a particular shape, the formula is given.

Read the study guide and the revision notes first if you need to. See also the IB Physics hub and the printable checklist.

Questions

1. A force of 45 N acts on a spanner at 0.30 m from the nut, at 30° to the line of the spanner. Calculate the magnitude of the torque about the nut. [2]

2. A uniform horizontal plank of length 2.4 m and mass 12 kg rests on two vertical supports, one at its left end and one 0.60 m from its right end. State the sense of the torque of the plank’s weight about the left support, and calculate the force exerted by the other support. [4]

3. A wheel turning at 50 rad s⁻¹ is brought uniformly to rest in 20 s.

(a) Calculate the magnitude of its angular acceleration. [2] (b) Calculate the number of complete turns it makes while stopping. [2]

4. Three small masses are fixed to a light rigid frame: 0.25 kg at 0.60 m, 0.40 kg at 0.30 m and 0.15 kg at 0.90 m from an axis.

(a) Calculate the moment of inertia of the system about the axis. [2] (b) A resultant torque of 0.50 N m acts. Calculate the angular acceleration. [2]

5. A flywheel with I = 0.18 kg m² turns at 30 rad s⁻¹. A constant frictional torque of 0.36 N m acts at its bearing and no other torque acts. Calculate the angular acceleration, the time to stop, and the angle turned through in revolutions. [4]

6. A playground roundabout has I = 240 kg m² and turns at 1.2 rad s⁻¹. A child of mass 30 kg, initially at rest, jumps onto its rim, 1.5 m from the axis, and stays there. Calculate the new angular speed, and state the condition that makes your method valid. [4]

7. A diver in a straight position has I = 4.2 kg m² and turns at 1.8 rad s⁻¹. She tucks, reducing I to 1.5 kg m².

(a) Calculate her new angular speed. [2] (b) Calculate her rotational kinetic energy before and after. [2] (c) Explain the source of the increase. [1]

8. A torque of 1.2 N m acts on a body of I = 0.80 kg m², initially at rest, for 4.0 s. Determine the angular impulse, the final angular speed and the final rotational kinetic energy, and show that two expressions for the energy agree. [4]

9. A uniform solid cylinder of mass 2.0 kg and radius 0.050 m has I = ½MR². It rolls without slipping at 1.8 m s⁻¹.

(a) Calculate its angular speed. [1] (b) Calculate its translational, rotational and total kinetic energy. [3] (c) State the fraction of the total energy that is rotational. [1]

10. A mass of 1.5 kg hangs from a light string wound around a uniform disc pulley of mass 4.0 kg and radius 0.20 m, with I = ½MR². The string does not slip. The mass is released from rest.

(a) Show that I = 0.080 kg m². [1] (b) Calculate the acceleration of the mass. [3] (c) Calculate the tension in the string. [1] (d) Calculate the speed of the mass after it falls 1.2 m. [2] (e) Use energy to check your answer to (d). [1]

11. Disc A (I = 0.45 kg m²) spins freely at 24 rad s⁻¹. Disc B (I = 0.15 kg m²) on the same axis is at rest. They are pressed together by a clutch and turn as one.

(a) Calculate their common angular speed. [3] (b) Calculate the kinetic energy transferred to other forms. [3] (c) The joined discs are brought to rest in 6.0 s by a constant torque. Calculate its magnitude. [2]

Answers

1. τ = Fr sin θ [1] = 45 × 0.30 × sin 30° = 6.75 N m [1] [2] Examiner insight: Show the substitution with the angle in the sine. A bare number gives no evidence that you used the angle between the force and the line from the axis.

2. The weight acts at the midpoint, 1.2 m from the left support, and its torque there is clockwise (the plank extends to the right of that support) [1]. Equilibrium: counter-clockwise torque of the right support equals clockwise torque of the weight [1]: N × 1.8 = 12 × 9.81 × 1.2 [1]. N = 141.3/1.8 = 78.5 N [1] [4] Examiner insight: Choosing the left support as the axis avoids the unknown force there. Answers that use the full rod length for the weight, instead of the centre of mass, lose the main step.

3. (a) α = (0 − 50)/20 [1] = −2.5, so magnitude 2.5 rad s⁻² (a slowing) [1] (b) Δθ = ((50 + 0)/2) × 20 = 500 rad [1]; 500/2π = 79.6, so 79 complete turns [1] [4] Examiner insight: “Complete turns” asks for a whole number, so 79 is safer than rounding up to 80. Give the angle in radians before converting.

4. (a) I = Σmr² = 0.25 × 0.60² + 0.40 × 0.30² + 0.15 × 0.90² [1] = 0.09 + 0.036 + 0.1215 = 0.248 kg m² [1] (b) τ = Iα [1], so α = 0.50/0.2475 = 2.02 rad s⁻² [1] [4] Examiner insight: Square each distance before multiplying by its mass. Carrying the unrounded I into part (b) avoids drifting the final digit.

5. α = τ/I = 0.36/0.18 = 2.0 rad s⁻² (a deceleration) [1]. t = ω/α = 30/2.0 = 15 s [1]. Δθ = ω²/2α = 900/4.0 = 225 rad [1] = 225/2π = 35.8 revolutions [1] [4] Examiner insight: The frictional torque is the resultant torque here, so it is the one that goes in τ = Iα. Say that it opposes the motion rather than leaving a sign unexplained.

6. Initial L = 240 × 1.2 = 288 kg m² s⁻¹ (the child has none) [1]. I after = 240 + 30 × 1.5² = 307.5 kg m² [1]. 288 = 307.5 ω, so ω = 0.937 rad s⁻¹ [1]. Valid because no external resultant torque acts on the roundabout and child system (bearing friction negligible) [1] [4] Examiner insight: A conservation answer needs the condition stated, not only the calculation. Treat the child as a point mass at the rim when adding to I.

7. (a) L conserved: 4.2 × 1.8 = 1.5 ω [1], so ω = 5.04 rad s⁻¹ (5.0 to 2 s.f.) [1] (b) Ek before = ½ × 4.2 × 1.8² = 6.80 J [1]; after = ½ × 1.5 × 5.04² = 19.1 J [1] (c) The diver does work with her muscles in pulling her mass closer to the axis, which transfers chemical energy to rotational kinetic energy [1] [5] Examiner insight: “Energy increases while L is conserved” is only a good explanation if you name where the energy comes from. Do not say that energy is created.

8. ΔL = τΔt = 1.2 × 4.0 = 4.8 kg m² s⁻¹ [1]. ω = 4.8/0.80 = 6.0 rad s⁻¹ [1]. Ek = L²/2I = 23.04/1.6 = 14.4 J [1]. Also ½Iω² = ½ × 0.80 × 6.0² = 14.4 J, the same [1] [4] Examiner insight: When a question asks you to show two expressions agree, compute both separately. Initial rest means initial L is zero, so ΔL equals the final L.

9. (a) ω = v/R = 1.8/0.050 = 36 rad s⁻¹ [1] (b) Translational ½Mv² = ½ × 2.0 × 1.8² = 3.24 J [1]. I = ½ × 2.0 × 0.050² = 0.0025 kg m², so rotational ½Iω² = ½ × 0.0025 × 36² = 1.62 J [1]. Total = 4.86 J [1] (c) 1.62/4.86 = one third [1] [5] Examiner insight: Rolling without slipping is the condition that lets you use v = ωR. Keep the translational and rotational energies as separate lines so the total is easy to follow.

10. (a) I = ½ × 4.0 × 0.20² = 0.080 kg m² [1] (b) Mass: mg − T = ma [1]. Pulley: TR = Iα with α = a/R, so T = (I/R²) a = 2.0a [1]. a = 1.5 × 9.81/(1.5 + 2.0) = 4.20 m s⁻² [1] (c) T = 2.0 × 4.2043 = 8.41 N [1] (d) v² = 2as = 2 × 4.2043 × 1.2 [1] = 10.09, so v = 3.18 m s⁻¹ [1] (e) mgh = 1.5 × 9.81 × 1.2 = 17.7 J, and ½mv² + ½Iω² = 7.57 + 10.08 = 17.7 J (ω = v/R = 15.9 rad s⁻¹), so they agree [1] [8] Examiner insight: The tension acts on both the mass and the pulley, so write separate equations for each and link them with a = αR. In part (e), do not leave out the pulley’s rotational energy.

11. (a) Initial L = 0.45 × 24 = 10.8 kg m² s⁻¹ [1]. Final I = 0.45 + 0.15 = 0.60 kg m² [1]. ω = 10.8/0.60 = 18.0 rad s⁻¹ [1] (b) Ek before = ½ × 0.45 × 24² = 129.6 J [1]. Ek after = ½ × 0.60 × 18.0² = 97.2 J [1]. Energy transferred = 32.4 J [1] (c) τ = ΔL/Δt = 10.8/6.0 = 1.8 N m [1] (a retarding torque, opposite to the rotation) [1] [8] Examiner insight: L is conserved in the clutch but Ek is not, and part (b) tests whether you know that. Giving the sense of the braking torque shows you have linked it to the direction of rotation.

Where marks are usually lost

• Leaving out sin θ, or using the wrong angle, in a torque calculation.
• Forgetting that the weight of a uniform body acts at its centre of mass.
• Taking torques about a point and then missing a force from the equation.
• Using the wrong axis or the wrong distance in Σmr².
• Not converting between revolutions, degrees and radians.
• Stating that L is conserved without saying that no external resultant torque acts.
• Assuming kinetic energy is conserved when bodies join.
• Leaving out the pulley’s rotational energy in an energy check on a coupled system.
• Rounding intermediate values too early.

Next steps

• Review the revision notes and the study guide.
• Return to the IB Physics hub and the printable checklist.
• Try the free 10-minute diagnostics to find your next weak spot.
• Book a free trial class.

Official syllabus

These questions follow sub-topic A.4 (Rigid body mechanics) of the IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. A.4 is additional higher level content only.