A.4 Rigid Body Mechanics Questions

1A-1A-05

Rolling without slipping·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A bicycle moves at a constant speed of 6.0 m s⁻¹ along a level road. Its wheels have a radius of 0.35 m and roll without slipping.

Which row gives the angular speed of a wheel and the speed, relative to the road, of the point at the top of a tyre?

Angular speed / rad s⁻¹Speed of the top of the tyre / m s⁻¹

Show mark scheme

Marking point	Mark	Notes
Step 1Rolling without slipping: the point of the tyre in contact with the road is momentarily at rest, so the axle moves at v = ωR: ω = 6.0/0.35 = 17 rad s⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Every point of the tyre has the forward velocity v of the axle plus a velocity ωR = v due to the rotation, directed along the tangent.	—
Step 3At the top the two are in the same direction: speed = v + ωR = 2v = 12 m s⁻¹ (at the bottom they cancel, giving zero).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis uses the diameter, 0.70 m, in ω = v/r, and gives the top of the tyre only the speed of the axle, ignoring the rotation.
BThe speed of the top is right, but ω = 6.0/0.70 uses the diameter instead of the radius.
CThe angular speed is right, but only the axle moves at 6.0 m s⁻¹: the top of the tyre also has the rotational velocity ωR in the forward direction.
DCorrect: ω = v/R = 17 rad s⁻¹ and the top moves at v + ωR = 12 m s⁻¹.

Syllabus understandingA.4 — that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration (guidance: simultaneous rotational and translational motion will be restricted to rolling without slipping); A.2 — the angular velocity ω related to the linear speed v by v = ωr Command term: Determine

2A-1A-06

Angular momentum·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

An ice skater spins freely with angular speed ω. She pulls her arms in so that her moment of inertia halves.

What happens to her rotational kinetic energy?

Show mark scheme

Marking point	Mark	Notes
Step 1No external torque acts about the axis, so angular momentum L = Iω is conserved: halving I doubles ω.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Rotational KE = ½Iω² becomes ½(I/2)(2ω)² = 2 × ½Iω².	—
Step 3The energy doubles; the extra energy is the work her muscles do pulling her arms inwards against the centripetal requirement.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis assumes the energy is proportional to I with ω fixed. But ω doubles when I halves.
BIt is angular momentum that is conserved, not kinetic energy — her muscles do work.
CCorrect: E_k = L²/(2I), so halving I at constant L doubles the energy.
DThis doubles ω but forgets that I has halved.

Syllabus understandingA.4 — the conservation of angular momentum; rotational kinetic energy as given by E_k = ½Iω² = L²/2I Command term: Deduce

3A-1A-07

Torque and angular acceleration·A.4 Rigid body mechanics (HL)

Paper 1AHard1 mark

Multiple choice · 1 mark4 steps to full marksDetermine

A uniform rod of length L and mass M is pivoted freely at one end and released from rest in a horizontal position. The moment of inertia of the rod about the pivot is ⅓ML².

What is the initial angular acceleration of the rod?

Show mark scheme

Marking point	Mark	Notes
Step 1The weight Mg acts at the centre of mass, a distance L/2 from the pivot, perpendicular to the rod.	—	All 4 steps must be completed — there is no mark for a part-answer.
Step 2Torque about the pivot: τ = Mg × L/2 = ½MgL.	—
Step 3Newton's second law for rotation: τ = Iα, so ½MgL = ⅓ML² × α.	—
Step 4α = 3g/(2L). Note the free end then accelerates at αL = 1.5g, faster than free fall.	✓ 1	Answer D

Answer: D · 4 stages of work, one mark

Every option, and why

AThis divides the torque ½MgL by ML², treating the rod as a point mass at its far end.
BThe fractions ½ and ⅓ have been combined the wrong way round.
CThis takes the weight as acting at the far end (torque MgL) and uses the point-mass moment of inertia ML².
DCorrect: α = (½MgL)/(⅓ML²) = 3g/(2L).

Syllabus understandingA.4 — that torque τ of a force about an axis is given by τ = Fr sin θ; Newton's second law for rotation as given by τ = Iα Command term: Determine

4A-1A-12

Conservation of angular momentum·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A turntable of moment of inertia 0.060 kg m² rotates freely about a vertical axis at 4.0 rad s⁻¹. A thin ring of mass 0.50 kg and radius 0.20 m, not rotating, is dropped gently onto the turntable so that it lands centred on the axis and then turns with it. The moment of inertia of a thin ring about its axis is MR².

What is the angular speed of the turntable and ring after the ring has stopped slipping?

Show mark scheme

Marking point	Mark	Notes
Step 1Moment of inertia of the ring: MR² = 0.50 × 0.20² = 0.020 kg m².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The frictional forces between ring and turntable are internal, and no resultant external torque acts about the axis, so the angular momentum L = Iω of the system is constant; the ring initially has none.	—
Step 30.060 × 4.0 = (0.060 + 0.020)ω, so ω = 0.24/0.080 = 3.0 rad s⁻¹.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: 0.060 × 4.0 = 0.080ω gives ω = 3.0 rad s⁻¹.
BThis conserves kinetic energy instead: ω = 4.0 × √(0.060/0.080) = 3.5 rad s⁻¹. Kinetic energy is not conserved — some is transferred to thermal energy while the ring slips (0.48 J falls to 0.36 J).
CThis forgets to add the moment of inertia of the ring, so the angular speed appears unchanged.
DThis inverts the ratio of the moments of inertia: 4.0 × 0.080/0.060 = 5.3 rad s⁻¹. Adding mass away from the axis must slow the turntable, not speed it up.

5A-1A-13

Rotational kinematics·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark4 steps to full marksDetermine

A wheel rotating at 20 rad s⁻¹ is brought uniformly to rest in 40 complete revolutions.

What is the magnitude of the angular deceleration?

Show mark scheme

Marking point	Mark	Notes
Step 1Convert the angular displacement to radians: θ = 40 × 2π = 251 rad.	—	All 4 steps must be completed — there is no mark for a part-answer.
Step 2The rotational analogue of v² = u² + 2as is ω_f² = ω_i² + 2αθ, with ω_f = 0 and ω_i = 20 rad s⁻¹.	—
Step 3α = ω_i²/(2θ) = 400/(2 × 251).	—
Step 4α = 0.80 rad s⁻².	✓ 1	Answer C

Answer: C · 4 stages of work, one mark

Every option, and why

AThis divides by 4θ instead of 2θ — a factor-of-two slip in the equation.
BThis is 20/40 — the angular speed divided by a number of revolutions, which mixes units and uses no valid equation.
CCorrect: α = 20²/(2 × 251) = 0.80 rad s⁻².
DThis uses ω²/θ, dropping the 2 in ω² = 2αθ.

6A-1A-17

Rotational equilibrium·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A uniform beam of length 4.0 m and weight 200 N is hinged at one end and held horizontal by a vertical cable attached 3.0 m from the hinge.

What is the tension in the cable?

Show mark scheme

Marking point	Mark	Notes
Step 1Take moments about the hinge so that the unknown hinge force does not appear.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Clockwise moment of the weight (acting at the centre, 2.0 m from the hinge) = 200 × 2.0 = 400 N m; anticlockwise moment of the tension = T × 3.0.	—
Step 3Equilibrium: 3.0T = 400, so T = 133 N (the hinge supplies the remaining 67 N upwards).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis would be the tension if the weight acted at 1.5 m from the hinge.
BCorrect: T = 200 × 2.0/3.0 = 133 N.
CThis assumes the cable carries the whole weight, ignoring the upward force at the hinge.
DThis has the moment arms the wrong way round (200 × 4.0/3.0).

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ; that bodies in rotational equilibrium have a resultant torque of zero; A.2 — translational equilibrium (Newton’s first law) Command term: Determine

7A-1A-21

Rolling and rotational kinetic energy·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A solid sphere, a solid cylinder and a thin hoop, all of the same mass m and radius r, are released together from rest at the top of the same incline and roll without slipping. Their moments of inertia about their axes of rotation are ⅖mr², ½mr² and mr² respectively.

In what order do they reach the bottom?

Show mark scheme

Marking point	Mark	Notes
Step 1For rolling from rest, mgh = ½mv²(1 + k) where I = kmr², so v² = 2gh/(1 + k).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2k = 2/5 (sphere), 1/2 (cylinder), 1 (hoop): the larger k, the larger the share of energy locked in rotation and the smaller the linear speed.	—
Step 3So the sphere is fastest and the hoop slowest — independent of mass and radius.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the acceleration down the slope is g sin θ/(1 + k), largest for the smallest k.
BThis assumes that the body with the largest moment of inertia gains the most speed. In fact a larger k in I = kmr² puts a larger share of the energy into rotation, leaving less as translational kinetic energy, so the hoop is slowest.
CThey would arrive together only if sliding without friction; rolling shares the energy differently for each shape.
DThis ranks the sphere’s ⅖mr² as larger than the cylinder’s ½mr² (comparing the denominators 5 and 2 instead of the fractions 0.4 and 0.5). The sphere has the smaller k, so it beats the cylinder.

Syllabus understandingA.4 — that the moment of inertia I depends on the distribution of mass of an extended body about an axis of rotation; the kinetic energy of rotational motion as given by E_k = ½Iω² (guidance: rolling without slipping); A.3 — the principle of the conservation of energy Command term: Deduce

8A-1A-26

Moment of inertia·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A uniform rod of mass M and length L has a moment of inertia of ML²/12 about an axis through its centre and ML²/3 about a parallel axis through one end, both axes perpendicular to the rod.

The rod is made to rotate with the same angular speed ω first about the central axis and then about the end axis. What is the ratio (rotational kinetic energy about the end axis)/(rotational kinetic energy about the central axis)?

Show mark scheme

Marking point	Mark	Notes
Step 1Rotational kinetic energy is E_k = ½Iω², so at the same ω it is proportional to I.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = (ML²/3)/(ML²/12) = 12/3 = 4.	—
Step 3About the end axis the mass is, on average, further from the axis, so the same angular speed needs four times the energy.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis inverts the ratio of the moments of inertia (central/end instead of end/central).
BThis assumes the rotational kinetic energy depends only on the mass and the angular speed, which are unchanged; it also depends on how the mass is distributed about the axis.
CThis is the ratio of the greatest distances of any part of the rod from the axis (L and L/2); I depends on the square of the distance, summed over all the mass.
DCorrect: at the same ω, E_k ∝ I, and (ML²/3)/(ML²/12) = 4.

9A-1A-30

Principle of moments·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A uniform plank of length L and weight W rests on a pivot a distance L/3 from its left end. A child stands on the left end of the plank, as shown, and the plank is horizontal and in equilibrium.

What is the weight of the child?

The plank is uniform; distances are measured along the plank.

Show mark scheme

Marking point	Mark	Notes
Step 1The weight of a uniform plank acts at its centre, L/2 from the left end, i.e. L/2 − L/3 = L/6 to the right of the pivot.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The child is L/3 to the left of the pivot. Moments about the pivot: (weight of child) × L/3 = W × L/6.	—
Step 3Weight of child = W/2. The pivot force has no moment about the pivot.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the plank's weight has a moment arm of only L/6, half that of the child.
BThis assumes that a balanced plank carries equal weights on either side, ignoring the different moment arms.
CThis measures the moment arm of the plank's weight from the end of the plank (L/2) instead of from the pivot: W × (L/2)/(L/3).
DThis inverts the ratio of the moment arms: W × (L/3)/(L/6).

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ; that bodies in rotational equilibrium have a resultant torque of zero (guidance: the mass of an extended body may be taken as concentrated at the centre of mass) Command term: Determine

10A-1A-47

Torque and its sense·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A cyclist pushes vertically downward with a force of 400 N on a pedal. The crank is 0.17 m long and, at this instant, points forward at 30° above the horizontal, as shown.

Which row gives the magnitude of the torque of this force about the axle and its sense as seen in the diagram?

Side view of the crank (not to scale). The force on the pedal is vertically downward.

Torque / N mSense

Show mark scheme

Marking point	Mark	Notes
Step 1τ = Fr sin θ, where θ is the angle between the force and the line from the axle to the pedal.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The crank is 30° above the horizontal and the force is vertical, so the angle between them is 120° (or 60°); sin 120° = sin 60° = 0.866.	—
Step 3τ = 400 × 0.17 × 0.866 = 59 N m. A downward push on a pedal that is to the right of the axle turns the crank clockwise in the diagram.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses sin 30°, the angle between the crank and the horizontal rather than between the crank and the force: 400 × 0.17 × 0.5 = 34 N m.
BWrong angle (34 N m) and wrong sense: the pedal is to the right of the axle and is pushed down, which is a clockwise turning effect.
CCorrect: 400 × 0.17 × sin 60° = 59 N m, clockwise.
DThe magnitude is right, but a downward force applied to the right of the pivot turns the crank clockwise, not anticlockwise.

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ; the sense (clockwise or counter-clockwise) of a torque Command term: Determine

11A-1A-48

Moment of inertia·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Three small spheres, each of mass 0.20 kg, are fixed at the corners of a light, rigid frame in the shape of an equilateral triangle of side 0.30 m. The frame rotates about an axis perpendicular to its plane that passes through one of the spheres.

What is the moment of inertia of the system about this axis? Treat the spheres as point masses.

Show mark scheme

Marking point	Mark	Notes
Step 1For point masses I = Σmr², with r the perpendicular distance from the axis.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The sphere on the axis has r = 0 and contributes nothing; the other two are each 0.30 m from the axis.	—
Step 3I = 2 × 0.20 × 0.30² = 0.036 kg m².	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis takes the axis through the centre of the triangle, where each sphere is 0.30/√3 = 0.173 m away: 3 × 0.20 × 0.173² = 0.018 kg m².
BCorrect: 2 × 0.20 × 0.30² = 0.036 kg m².
CThis counts all three spheres at 0.30 m: 3 × 0.20 × 0.30² = 0.054 kg m². The sphere on the axis has r = 0.
DThis forgets to square the distance: 2 × 0.20 × 0.30 = 0.12 (and the unit would be kg m, not kg m²).

Syllabus understandingA.4 — the moment of inertia for a system of point masses as given by I = Σmr²; that the moment of inertia depends on the distribution of mass about an axis of rotation Command term: Determine

12A-1A-49

Rotational kinematics·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksCalculate

The rotor of a laboratory centrifuge accelerates uniformly from rest to 6000 revolutions per minute in 25 s.

How many revolutions does the rotor make during this time?

Show mark scheme

Marking point	Mark	Notes
Step 1Final angular speed: 6000 rev min⁻¹ = 100 rev s⁻¹ (= 628 rad s⁻¹).	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2For uniform angular acceleration Δθ = ½(ω_i + ω_f)t = ½ × (0 + 100) × 25 = 1250 revolutions (7850 rad).	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: the average angular speed is 50 rev s⁻¹, and 50 × 25 = 1250.
BThis multiplies the final angular speed by the time (100 × 25), as if the rotor had been at full speed throughout.
CThis is the angular displacement in radians (½ × 628 × 25 = 7850 rad), not the number of revolutions; divide by 2π.
DThis treats 6000 rev min⁻¹ as 6000 rev s⁻¹: ½ × 6000 × 25 = 75 000.

Syllabus understandingA.4 — that equations of motion for uniform angular acceleration can be used to predict the angular displacement Δθ = ½(ω_f + ω_i)t; that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration Command term: Calculate

13A-1A-50

Angular impulse·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

A satellite contains two reaction wheels, P and Q, both initially at rest. The moment of inertia of Q is three times that of P. The motor of each wheel then applies the same constant torque for the same time.

Which row gives (angular momentum of Q)/(angular momentum of P) and (rotational kinetic energy of Q)/(rotational kinetic energy of P)?

L_Q/L_PE_k_Q/E_k_P

Show mark scheme

Marking point	Mark	Notes
Step 1The angular impulse is τΔt = ΔL. The same torque for the same time gives both wheels the same angular momentum: L_Q/L_P = 1.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2E_k = L²/2I. With equal L, the kinetic energy is inversely proportional to I: E_k_Q/E_k_P = 1/3.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: equal angular impulse gives equal L; E_k = L²/2I gives 1/3.
BThis uses E_k = ½Iω² as if both wheels had the same angular speed. Q actually spins at one third of the angular speed of P.
CThis assumes L ∝ ω alone: Q spins three times more slowly, but its moment of inertia is three times larger, so L is the same.
DThis assumes both wheels reach the same angular speed; the same torque gives Q only one third of the angular acceleration of P.

Syllabus understandingA.4 — that the action of a resultant torque constitutes an angular impulse ΔL as given by ΔL = τΔt = Δ(Iω); the kinetic energy of rotational motion as given by E_k = ½Iω² = L²/2I Command term: Deduce

14A-1A-51

Conservation of angular momentum·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Seen from above, disc P, of moment of inertia I, rotates clockwise with angular speed ω on a frictionless vertical axle. Disc Q, of moment of inertia 2I, rotates anticlockwise with angular speed ω on the same axle. Q is lowered onto P and friction between the discs makes them rotate together.

What is the final angular velocity of the discs?

Show mark scheme

Marking point	Mark	Notes
Step 1There is no external torque about the axle, so the total angular momentum is conserved. Take anticlockwise as positive.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Initial angular momentum = 2Iω − Iω = +Iω (anticlockwise).	—
Step 3Final: 3Iω_f = Iω, so ω_f = ω/3 anticlockwise, in the original sense of the disc with the larger angular momentum.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThe size is right but the sense is that of P; Q has twice the angular momentum of P, so the combination turns the way Q did.
BThis assumes opposite rotations at equal angular speeds cancel. It is the angular momenta (Iω and 2Iω) that must be compared, not the angular speeds.
CThis adds the magnitudes of the angular momenta (Iω + 2Iω = 3Iω), ignoring that they are in opposite senses.
DCorrect: net angular momentum Iω anticlockwise shared by a total moment of inertia 3I.

15A-1A-60

Rotational kinematics·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A wheel of moment of inertia 0.25 kg m² rotates about a fixed axis. The graph shows how its angular velocity ω varies with time t.

What is the greatest magnitude of the resultant torque on the wheel during the 10 s?

Angular velocity ω of the wheel against time t (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The resultant torque is τ = Iα, and α is the gradient of the ω–t graph; the torque is greatest where the graph is steepest.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2From 7.0 s to 9.0 s: \|α\| = 12/2.0 = 6.0 rad s⁻² (from 0 to 3.0 s it is only 12/3.0 = 4.0 rad s⁻²).	—
Step 3τ = 0.25 × 6.0 = 1.5 N m.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis divides 12 rad s⁻¹ by the time coordinate 9.0 s instead of the duration of the slowing, 2.0 s: 0.25 × 12/9.0.
BThis uses the speeding-up section (0.25 × 4.0); the slowing-down section is steeper.
CCorrect: 0.25 kg m² × 6.0 rad s⁻² = 1.5 N m.
DThis is the angular acceleration (6.0 rad s⁻²); the moment of inertia has been omitted.

Syllabus understandingA.4 — that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration; Newton's second law for rotation as given by τ = Iα Command term: Determine

16A-1A-71

Torque and its sense·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A square plate of side 0.40 m can rotate freely about a fixed axis through its centre O, perpendicular to the plate. Three forces act in the plane of the plate, as shown: 6.0 N along the top edge, 4.0 N along the right-hand edge and 5.0 N at the bottom-left corner, directed towards O.

Which row gives the magnitude of the resultant torque on the plate about O and its sense, as seen in the diagram?

Forces on the square plate (drawn to scale).

resultant torque / N msense

Show mark scheme

Marking point	Mark	Notes
Step 1Each force acts along an edge 0.20 m from O, so the perpendicular distance from O to its line of action is 0.20 m. The 5.0 N force acts along a line through O, so its torque is zero.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 26.0 N to the right along the top edge: 6.0 × 0.20 = 1.2 N m clockwise. 4.0 N upwards along the right-hand edge: 4.0 × 0.20 = 0.80 N m anticlockwise.	—
Step 3Resultant torque = 1.2 − 0.80 = 0.40 N m, clockwise.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis adds the magnitudes of the two torques, (6.0 + 4.0) × 0.20 = 2.0 N m, ignoring that they act in opposite senses.
BThis gives the 5.0 N force a torque equal to 5.0 × 0.28 m (the distance from O to the corner) and adds it to 0.40 N m. Its line of action passes through O, so its torque is zero.
CThe magnitude is right but the sense is reversed: the larger torque, from the 6.0 N force above O pushing to the right, is clockwise.
DCorrect: 1.2 N m clockwise − 0.80 N m anticlockwise = 0.40 N m clockwise; the 5.0 N force has no torque.

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ; the sense (clockwise or counter-clockwise) of a torque; that an unbalanced torque applied to an extended, rigid body will cause angular acceleration Command term: Determine

17A-1A-112

Equations of uniform angular acceleration·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A turntable starts from rest and rotates with uniform angular acceleration. During the first 2.0 s it turns through an angle θ.

What angle does it turn through during the next 2.0 s?

Show mark scheme

Marking point	Mark	Notes
Step 1Starting from rest with uniform angular acceleration, Δθ = ω_it + ½αt² = ½αt², so the angle turned from the start is proportional to t².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2After 4.0 s the total angle is (4.0/2.0)² × θ = 4θ.	—
Step 3The angle turned during the next 2.0 s is 4θ − θ = 3θ.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis assumes the turntable turns at a constant angular speed, so that equal times give equal angles. The angular speed is increasing.
BThis scales the angle with the angular speed reached at the end of each interval, which doubles from 2.0 s to 4.0 s. The angle in an interval depends on the average angular speed during it, which is three times larger in the second interval (the angular speed at 3.0 s against that at 1.0 s).
CCorrect: the total angle after 4.0 s is 4θ, so the angle turned in the second interval is 3θ.
DThis is the total angle turned in the first 4.0 s, not the angle turned during the second 2.0 s.

18A-1A-113

Angular impulse·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A wheel of moment of inertia 0.40 kg m² is at rest on a frictionless axle. A motor then applies a torque τ to the wheel. The graph shows how τ varies with time t.

What is the angular speed of the wheel at t = 7.0 s?

Torque τ on the wheel against time t (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1The area under a torque–time graph is the angular impulse τΔt, which equals the change in angular momentum ΔL = Δ(Iω).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Area = ½ × 3.0 × 2.0 + 3.0 × 3.0 + ½ × 3.0 × 1.0 = 3.0 + 9.0 + 1.5 = 13.5 N m s.	—
Step 3The wheel starts from rest, so ω = 13.5/0.40 = 34 rad s⁻¹. After 6.0 s there is no torque and the angular speed stays constant.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis multiplies the angular impulse by the moment of inertia (13.5 × 0.40 = 5.4) instead of dividing by it.
BThis uses only the constant-torque section, 3.0 × 3.0 = 9.0 N m s, and leaves out the two sloping sections: 9.0/0.40 = 23 rad s⁻¹.
CCorrect: angular impulse 13.5 N m s divided by 0.40 kg m² gives 34 rad s⁻¹.
DThis assumes the peak torque of 3.0 N m acts for the whole 6.0 s: 18/0.40 = 45 rad s⁻¹. The torque is smaller while it rises and falls.

19A-1A-114

Describing rotation·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A rigid wheel rotates about a fixed axis through its centre. Its angular speed is increasing.

Which quantities are the same for every point on the wheel that is not on the axis?
I. angular velocity
II. angular acceleration
III. linear speed

Show mark scheme

Marking point	Mark	Notes
Step 1A rigid body turns as a whole: in any time interval every point turns through the same angle, so every point has the same angular velocity ω (I).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Every point therefore also has the same rate of change of angular velocity, the angular acceleration α (II).	—
Step 3The linear speed v = ωr depends on the distance r from the axis, so it is different for points at different radii (III is false).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: ω and α describe the rotation of the whole rigid body; the linear speed ωr grows with r.
BThis treats the linear speed as common to every point and the angular acceleration as varying with radius. It is the linear quantities (ωr, αr) that depend on r, not the angular ones.
CThis forgets that the angular velocity is shared by every point of a rigid body, and wrongly includes the linear speed, which is ωr.
DThis assumes every point of a rigid body moves with the same speed. Points further from the axis travel round larger circles in the same time, so they move faster.

Syllabus understandingA.4 — that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration; A.2 — the angular velocity ω related to the linear speed v by v = ωr Command term: Deduce

20A-1A-115

Couples and resultant torque·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A driver turns a steering wheel of radius 0.18 m by pushing with forces of 25 N tangentially at opposite ends of a horizontal diameter, one upwards and one downwards, as shown.

Which row gives the magnitude of the resultant force on the steering wheel and the magnitude of the resultant torque about its centre O?

The steering wheel seen by the driver (not to scale).

resultant force / Nresultant torque / N m

Show mark scheme

Marking point	Mark	Notes
Step 1The two forces are equal in size and opposite in direction, so the resultant force is zero: the wheel is not pushed sideways.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each force is tangential, so it is perpendicular to the radius and τ = Fr sin 90° = 25 × 0.18 = 4.5 N m.	—
Step 3Upwards on the left and downwards on the right both turn the wheel clockwise, so the torques add: 2 × 4.5 = 9.0 N m.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThe resultant force is right, but this counts the torque of one hand only. Both forces turn the wheel in the same sense, so their torques add.
BCorrect: the forces cancel (resultant force zero) but their torques add, 2 × 25 × 0.18 = 9.0 N m.
CThis adds the forces as if they were in the same direction and lets the torques cancel; it is the other way round for this pair of forces.
DThe torque is right, but the forces are in opposite directions and cancel; a resultant force of 50 N would accelerate the wheel sideways.

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ; the sense (clockwise or anticlockwise) of a torque; A.2 — the resultant force on a body (free-body diagrams) Command term: Determine

21A-1A-116

Rotational kinetic energy and angular momentum·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

A flywheel of moment of inertia 0.80 kg m² has a rotational kinetic energy of 360 J.

What is the angular momentum of the flywheel?

Show mark scheme

Marking point	Mark	Notes
Step 1E_k = L²/2I, so L = √(2IE_k).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2L = √(2 × 0.80 × 360) = √576 = 24 kg m² s⁻¹.	—
Step 3Check: ω = L/I = 30 rad s⁻¹ and ½ × 0.80 × 30² = 360 J.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis drops the factor 2: √(0.80 × 360) = 17.
BCorrect: L = √(2IE_k) = 24 kg m² s⁻¹.
CThis is the angular speed, √(2E_k/I) = 30 rad s⁻¹; it must still be multiplied by I to give L.
DThis is L² = 2IE_k = 576; the square root has not been taken.

Syllabus understandingA.4 — the kinetic energy of rotational motion as given by E_k = ½Iω² = L²/2I; that an extended body rotating with an angular speed has an angular momentum L as given by L = Iω Command term: Calculate

22A-1A-117

Angular displacement–time graph·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A wheel starts from rest and turns with uniform angular acceleration. The graph shows how its angular displacement θ varies with time t.

What is the angular acceleration of the wheel?

Angular displacement θ of the wheel against time t (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1From rest with uniform angular acceleration, θ = ½αt².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2From the graph, θ = 20 rad at t = 4.0 s (or 5.0 rad at 2.0 s).	—
Step 3α = 2θ/t² = 2 × 20/4.0² = 2.5 rad s⁻².	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses θ = αt², leaving out the ½: 20/16 = 1.25 rad s⁻².
BCorrect: α = 2θ/t² = 2.5 rad s⁻².
CThis is θ/t = 20/4.0, the average angular velocity over the 4.0 s (the gradient of the chord), not the angular acceleration.
DThis is 2θ/t = 10 rad s⁻¹, the angular velocity at 4.0 s (the gradient of the tangent there), not the angular acceleration.

23A-1A-118

Coupled rotating bodies·A.4 Rigid body mechanics (HL)

Paper 1AHard1 mark

Multiple choice · 1 mark4 steps to full marksDeduce

Disc P, of moment of inertia I and radius R, rotates freely with angular speed ω₀ on a fixed axle. Disc Q, of moment of inertia 16I and radius 2R, is at rest on a second fixed axle parallel to the first. The axles are moved together until the rims of the discs press against each other. Friction at the contact acts until the rims stop slipping. The axles are frictionless.

What is the final angular speed of disc P?

Show mark scheme

Marking point	Mark	Notes
Step 1The frictional forces on the two rims are equal and opposite (Newton’s third law), but they act at different radii, R and 2R. The axles exert forces too, so the angular momentum of the pair is not conserved.	—	All 4 steps must be completed — there is no mark for a part-answer.
Step 2For each disc, angular impulse = change in angular momentum. If J is the impulse of the friction force: I(ω₀ − ω_P) = JR and 16Iω_Q = J(2R).	—
Step 3No slipping means equal rim speeds: ω_PR = ω_Q(2R), so ω_Q = ω_P/2.	—
Step 4Then JR = 8Iω_Q = 4Iω_P, so ω₀ − ω_P = 4ω_P and ω_P = ω₀/5.	✓ 1	Answer C

Answer: C · 4 stages of work, one mark

Every option, and why

AThis gives both discs the same angular speed and conserves Iω: Iω₀ = 17Iω. The discs are on separate axles; once slipping stops their rims, not their angular speeds, are equal.
BThis conserves the total angular momentum, Iω₀ = Iω_P + 16I(ω_P/2). That would need equal and opposite torques on the discs, but the equal frictional forces act at radii R and 2R, and the axles exert external forces.
CCorrect: angular impulse on each disc from the same friction force, with equal rim speeds at the end, gives ω₀/5.
DThis conserves kinetic energy, ω₀² = ω_P² + 16(ω_P/2)². Energy is dissipated while the rims slip, so kinetic energy is not conserved.

Syllabus understandingA.4 — that the action of a resultant torque constitutes an angular impulse ΔL as given by ΔL = τΔt = Δ(Iω); that an extended body rotating with an angular speed has an angular momentum L as given by L = Iω; that angular momentum remains constant unless the body is acted upon by a resultant torque (guidance: situations include coupled pairs of bodies); A.2 — Newton’s third law Command term: Deduce

24A-1A-119

Rolling without slipping·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A thin-walled pipe of mass m and radius r, with moment of inertia mr² about its axis, rolls without slipping along level ground with speed v. It then rolls, without slipping, up a slope.

What is the greatest height that the pipe reaches above the level ground? Rolling resistance and air resistance are negligible.

Show mark scheme

Marking point	Mark	Notes
Step 1Rolling without slipping: ω = v/r. Total kinetic energy = ½mv² + ½(mr²)(v/r)² = mv².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The static friction at the contact does no work (the contact point is momentarily at rest), so all the kinetic energy becomes gravitational potential energy at the highest point.	—
Step 3mgh = mv², so h = v²/g.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis ignores the rotational kinetic energy, as for a block sliding without friction. A rolling pipe has as much rotational as translational kinetic energy.
BThis uses the moment of inertia of a solid sphere, ⅖mr², instead of mr².
CThis uses the moment of inertia of a solid cylinder, ½mr², instead of mr² for a thin-walled pipe.
DCorrect: the total kinetic energy mv² is converted to mgh.

Syllabus understandingA.4 — the kinetic energy of rotational motion as given by E_k = ½Iω² = L²/2I (guidance: simultaneous rotational and translational motion is restricted to rolling without slipping); that the moment of inertia I depends on the distribution of mass of an extended body about an axis of rotation; A.3 — the principle of the conservation of energy Command term: Deduce

25A-1A-120

Sense of torque and angular impulse·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A wheel rotates clockwise, as seen from the front, about a fixed axis. Its angular speed is decreasing.

Which row gives the sense, as seen from the front, of the resultant torque on the wheel and the sense of the change in its angular momentum?

sense of resultant torquesense of change in angular momentum

Show mark scheme

Marking point	Mark	Notes
Step 1The wheel is slowing down, so its angular acceleration, and therefore the resultant torque (τ = Iα), is in the opposite sense to its rotation: anticlockwise.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Angular impulse: ΔL = τΔt, so the change in angular momentum has the same sense as the resultant torque: anticlockwise.	—
Step 3The angular momentum itself is still clockwise; it is getting smaller because an anticlockwise change is being added to it.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the resultant torque opposes the rotation, and ΔL = τΔt has the sense of the torque.
BThe torque is right, but this gives the sense of the angular momentum itself (clockwise) instead of the sense of its change.
CThis assumes the resultant torque acts in the sense of rotation, as if a torque were needed to keep the wheel turning. A clockwise torque would make it speed up.
DThis assumes that everything acts in the sense in which the wheel turns. A clockwise resultant torque would increase the clockwise angular speed.

Syllabus understandingA.4 — the sense (clockwise or anticlockwise) of a torque; that an unbalanced torque applied to an extended, rigid body will cause angular acceleration; that the action of a resultant torque constitutes an angular impulse ΔL as given by ΔL = τΔt = Δ(Iω) Command term: Deduce

26A-1A-121

Torque of a force at an angle·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A light bar is hinged at H. Any one of four forces, P, Q, R or S, can be applied to it in the plane of the diagram. The diagram shows the size of each force, the distance from H of the point where it acts and the angle between the force and the bar.

Which force, acting alone, produces the largest torque about H?

The bar hinged at H and the four forces (not to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1τ = Fr sin θ, where θ is the angle between the force and the bar (the line from H to the point of application).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2P: 100 × 0.20 × sin 90° = 20.0 N m. Q: 60 × 1.00 × sin 20° = 20.5 N m.	—
Step 3R: 30 × 0.80 × sin 90° = 24.0 N m. S: 50 × 0.60 × sin 70° = 28.2 N m, the largest.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis chooses the largest force. Its point of application is only 0.20 m from H, so its torque is only 20.0 N m.
BThis multiplies the force by the distance from H (60 N m) without the sin θ factor, or uses cos 20° (56 N m). Q acts almost along the bar, so its torque is only 20.5 N m.
CThis assumes that a force perpendicular to the bar at the largest distance always wins. R does act perpendicular to the bar, but 30 × 0.80 = 24.0 N m is less than the torque of S.
DCorrect: 50 × 0.60 × sin 70° = 28.2 N m, more than R (24.0 N m), Q (20.5 N m) and P (20.0 N m).

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ Command term: Deduce

27A-1A-122

Torque and angular acceleration·A.4 Rigid body mechanics (HL)

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A toy consists of a uniform disc of mass m and radius r, with moment of inertia ½mr². A light string is wound around the rim of the disc and its free end is held fixed. The disc is released from rest and falls vertically, rotating as the string unwinds without slipping. The string stays vertical.

What is the tension in the string while the disc falls?

Show mark scheme

Marking point	Mark	Notes
Step 1Translation of the disc: mg − T = ma.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Rotation about the centre: only the tension has a torque, Tr = Iα = ½mr² × (a/r), using a = αr because the string unwinds without slipping. So T = ½ma.	—
Step 3Then mg = ma + ½ma, so a = 2g/3 and T = mg/3.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the two equations give a = 2g/3 and T = ½ma = mg/3.
BThis puts a = g into T = ½ma, as if the disc were in free fall. The tension reduces the acceleration below g.
CThis is ma, the resultant force on the disc (m × 2g/3), not the tension.
DThis assumes the string supports the whole weight, as it would if the disc were in equilibrium. The disc accelerates downwards, so T < mg.

Syllabus understandingA.4 — Newton’s second law for rotation as given by τ = Iα where τ is the average torque; that an unbalanced torque applied to an extended, rigid body will cause angular acceleration (guidance: simultaneous rotational and translational motion is restricted to rolling without slipping); A.2 — Newton’s second law Command term: Deduce

28A-1A-123

Moment of inertia and angular momentum·A.4 Rigid body mechanics (HL)

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two uniform discs X and Y are cut from the same steel sheet, so they have the same thickness. The radius of Y is twice the radius of X. The moment of inertia of a uniform disc of mass M and radius R about its axis is ½MR². Both discs rotate about their axes with the same angular speed.

What is (angular momentum of Y)/(angular momentum of X)?

Show mark scheme

Marking point	Mark	Notes
Step 1Same thickness and density: mass ∝ area ∝ R², so Y has 4 times the mass of X.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2I = ½MR² ∝ R² × R² = R⁴: Y has 2⁴ = 16 times the moment of inertia of X.	—
Step 3At the same angular speed L = Iω ∝ I, so the ratio is 16.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis takes L as proportional to the radius, as for the linear speed of the rim at the same ω.
BThis uses I ∝ R² with the mass unchanged. The larger disc cut from the same sheet also has 4 times the mass.
CThis takes the mass as proportional to the radius (2 times) instead of the area (4 times): 2 × 2² = 8.
DCorrect: mass × 4 and R² × 4, so I and L are 16 times larger.

Syllabus understandingA.4 — that the moment of inertia I depends on the distribution of mass of an extended body about an axis of rotation; that an extended body rotating with an angular speed has an angular momentum L as given by L = Iω Command term: Deduce

29A-1A-124

Angular speed and linear speed·A.4 Rigid body mechanics (HL)

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksCalculate

The disc in a computer hard drive rotates at 7200 revolutions per minute. A read head is positioned 40 mm from the axis of rotation.

What is the speed of the part of the disc that is passing under the read head?

Show mark scheme

Marking point	Mark	Notes
Step 1Angular speed: ω = 7200 × 2π/60 = 754 rad s⁻¹ (120 revolutions per second).	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2v = ωr = 754 × 0.040 = 30 m s⁻¹.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThis uses 120 revolutions per second as if it were the angular speed in rad s⁻¹ (120 × 0.040); the factor 2π is missing.
BCorrect: ω = 754 rad s⁻¹ and v = ωr = 30 m s⁻¹.
CThis multiplies 7200 revolutions per minute directly by 0.040 m, without converting to rad s⁻¹.
DThis converts revolutions to radians (× 2π) but not minutes to seconds: 7200 × 2π × 0.040 = 1810.

30A-1B-01

Rolling and rotational kinetic energy·A.4 Rigid body mechanics (HL)

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A uniform solid cylinder is released from rest at the top of a straight ramp inclined at θ = 10.0° to the horizontal and rolls without slipping. Light gates record the time t for the cylinder to roll a distance s from rest. Each time is the mean of three runs and has an uncertainty of ±0.02 s; each distance has an uncertainty of ±0.005 m.

For a body of moment of inertia I = kMR² rolling without slipping down the ramp, the acceleration is predicted to be a = g sin θ/(1 + k). For a uniform solid cylinder k = 0.5. The graph shows s against t² with the line of best fit.

s / m	t / s	t² / s²
0.40	0.86	0.74
0.60	1.06	1.12
0.80	1.22	1.49
1.00	1.36	1.85
1.20	1.49

Graph drawn to scale. The point for s = 1.20 m has not been plotted.

(a)

(i)

Calculate t² for s = 1.20 m together with its absolute uncertainty.

(2)

(b)

(i)

Determine the acceleration of the cylinder, using the line of best fit and your point from (a)(i).

(2)

(ii)

The uncertainty in the gradient is ±3 %. Deduce whether your value of a is consistent with the prediction for a uniform solid cylinder.

(2)

(iii)

Suggest one reason for the difference between the measured and predicted values in (b)(ii).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
t² = 1.49² = 2.22 s²	✓ 1
Percentage uncertainty doubles on squaring: Δ(t²) = 2 × (0.02/1.49) × 2.22 = ±0.06 s²	✓ 1	Absolute uncertainty to 1 s.f. Accept ±0.059. ±0.03 s² (uncertainty not doubled) scores 0 for this mark.
Part (b)(i)
Gradient of the line through the origin, e.g. using the full range up to (2.22, 1.20): gradient = 1.20/2.22 = 0.54 m s⁻²	✓ 1	Accept 0.52–0.56 m s⁻². Allow ECF from (a)(i).
s = ½at², so a = 2 × gradient = 1.08 m s⁻²	✓ 1	Accept 1.04–1.12 m s⁻². Award [0] for a = gradient.
Part (b)(ii)
Predicted a = 9.81 × sin 10.0°/1.5 = 1.14 m s⁻²	✓ 1
Experimental range 1.05 to 1.11 m s⁻² (±3 %) does not include 1.14 m s⁻², so the result is not consistent: the measured acceleration is about 5 % too small	✓ 1	Allow ECF from (b)(i): the conclusion must agree with the candidate's own range.
Part (b)(iii)
Energy is dissipated, so less gravitational potential energy becomes kinetic energy: rolling resistance (deformation of the ramp or cylinder surface) or air resistance; OR the ramp angle is in fact slightly less than 10.0° (a systematic error in θ)	✓ 1	Allow ECF from (b)(ii) (a candidate who found agreement may suggest why the uncertainty is large). Do not accept "friction" alone: static friction at the contact does no work in rolling without slipping.

Answers: (a)(i) 2.22 ± 0.06 s² · (b)(i) a = 1.08 m s⁻² · (b)(ii) predicted 1.14 m s⁻²; not consistent (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the moment of inertia I depends on the distribution of mass; rolling without slipping; the rotational kinetic energy E_k = ½Iω²; Tools 3 — propagation of uncertainty for a power; gradient using the full data range; comparing a result with a prediction Command term: Determine

31A-1B-07

Torque and angular acceleration·A.4 Rigid body mechanics (HL)

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A steel flywheel of moment of inertia 0.20 kg m² is mounted on an axle supported by a bearing assembly of mass 0.30 kg. The flywheel is spun up by a motor, which is then disconnected. An optical sensor reads the angular speed ω every 10 s while the flywheel slows down under a constant frictional torque in the bearing. Air resistance is negligible. The specific heat capacity of steel is 450 J kg⁻¹ K⁻¹.

The graph shows ω against t with the line of best fit.

t / s	0	10	20	30	40
ω / rad s⁻¹	50.2	43.7	37.9	31.5	25.7

Graph drawn to scale

(a)

(i)

Determine, using the full range of the data, the angular acceleration of the flywheel and the frictional torque.

(2)

(ii)

Determine, by extrapolating the graph, the number of revolutions the flywheel makes from t = 0 until it stops.

(2)

(iii)

All the initial rotational kinetic energy of the flywheel is transferred to internal energy of the bearing assembly. Estimate the temperature rise of the bearing assembly.

(2)

(iv)

Show that the angular impulse of the frictional torque, from t = 0 until the flywheel stops, is consistent with the initial angular momentum of the flywheel.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
α = gradient = (25.7 − 50.2)/40 = −0.61 rad s⁻²	✓ 1	Accept −0.59 to −0.63 rad s⁻². Using only two adjacent readings scores 0 for this mark.
τ = Iα = 0.20 × 0.61 = 0.12 N m	✓ 1	Accept 0.118–0.126 N m. Sign not required.
Part (a)(ii)
Line reaches ω = 0 at t = 50.0/0.612 = 81.8 s	✓ 1	Allow ECF from (a)(i). Accept 80–84 s.
Angle = area under graph = ½ × 50.0 × 81.8 = 2046 rad; 2046/2π = 326 revolutions	✓ 1	Accept 315–335 revolutions. Allow ECF from the stopping time.
Part (a)(iii)
E_k = ½Iω² = ½ × 0.20 × 50.2² = 252 J	✓ 1	Accept 250–252 J (using the intercept 50.0 rad s⁻¹ gives 250 J).
ΔT = E_k/(mc) = 252/(0.30 × 450) = 1.9 K	✓ 1	Accept 1.8–1.9 K. Allow ECF from the energy.
Part (a)(iv)
Angular impulse τΔt = 0.122 × 81.8 = 10.0 N m s, equal to Iω₀ = 0.20 × 50.2 = 10.0 kg m² s⁻¹ within the precision of the data	✓ 1	Allow ECF from (a)(i) and (a)(ii) (stopping time).

Answers: (a)(i) α = −0.61 rad s⁻²; τ = 0.12 N m · (a)(ii) 326 revolutions · (a)(iii) 1.9 K · (a)(iv) 10.0 N m s (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — Newton's second law for rotation τ = Iα; the equations of motion for uniform angular acceleration; angular impulse ΔL = τΔt = Δ(Iω); the kinetic energy of rotational motion E_k = ½Iω²; B.1 — specific heat capacity; Tools 3 — gradient using the full data range, extrapolation Command term: Determine

32A-1B-09

Conservation of angular momentum·A.4 Rigid body mechanics (HL)

Paper 1BHard7 marks

Data-based question7 steps to full marksDetermine

A student sits on a stool that rotates freely about a vertical axis, holding a 2.0 kg mass in each hand. The system is set rotating and the student then slowly pulls both masses inwards. A sensor on the stool records the angular speed ω when the masses are at several distances r from the axis. Friction in the stool bearing is negligible and the masses may be treated as point masses.

If the angular momentum L of the system is constant, 1/ω = I₀/L + (2m/L)r², where I₀ is the moment of inertia of the student and stool and m = 2.0 kg.

r / m	ω / rad s⁻¹	r² / m²	(1/ω) / s
0.60	2.42	0.36	0.413
0.50	2.99	0.25	0.334
0.40	3.62	0.16	0.276
0.30	4.42	0.09	0.226
0.20	5.18	0.04	0.193

Graph drawn to scale

(a)

(i)

Explain why the angular momentum of the system is constant as the masses are pulled in.

(1)

(b)

(i)

Determine L, using the gradient of the graph.

(2)

(ii)

Determine I₀, by extrapolating the graph.

(2)

(iii)

Determine the work done by the student in pulling the masses in from 0.60 m to 0.20 m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
No external torque acts about the axis (the bearing is frictionless; the forces pulling the masses in are internal), so L is conserved	✓ 1	Must refer to zero external torque.
Part (b)(i)
Gradient = (0.413 − 0.193)/(0.36 − 0.04) = 0.69 s m⁻²	✓ 1	Accept 0.66–0.71 s m⁻².
2m/L = gradient, so L = 4.0/0.69 = 5.8 kg m² s⁻¹	✓ 1	Accept 5.6–6.1 kg m² s⁻¹. Award [0] for using m = 2.0 kg instead of 2m.
Part (b)(ii)
Intercept at r² = 0: 0.165 s	✓ 1	Accept 0.160–0.170 s.
I₀ = intercept × L = 0.165 × 5.8 = 0.96 kg m²	✓ 1	Allow ECF from (b)(i). Accept 0.90–1.03 kg m².
Part (b)(iii)
E_k = ½Lω: ½ × 5.8 × 2.42 = 7.0 J and ½ × 5.8 × 5.18 = 15.0 J	✓ 1	Allow ECF from (b)(i). Accept ½Iω² with I = I₀ + 2mr² using (b)(ii).
Work done = increase in rotational kinetic energy = 8.0 J	✓ 1	Accept 7.5–8.5 J.

Answers: (b)(i) 5.8 kg m² s⁻¹ · (b)(ii) 0.96 kg m² · (b)(iii) 8.0 J (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — angular momentum L = Iω and its conservation when the resultant external torque is zero; the moment of inertia of point masses I = Σmr²; rotational kinetic energy E_k = L²/2I; Tools 3 — linearising a relationship, gradient and extrapolated intercept Command term: Determine

33A-1B-18

Torque and angular acceleration·A.4 Rigid body mechanics (HL)

Paper 1BHard7 marks

Data-based question7 steps to full marksDetermine

A light string is wound round a drum of radius 2.50 cm fixed to the axle of a turntable, passes over a light frictionless pulley and supports a hanging mass m. The mass is released from rest and the time t for it to fall 0.600 m is measured; each time is ±0.05 s. Friction at the axle is assumed negligible, and the acceleration of the mass is a = 2 × 0.600/t².

Applying Newton's second law to the mass and to the turntable gives 1/a = 1/g + (I/gr²)(1/m), where I is the moment of inertia of the turntable and r the drum radius.

m / kg	t / s	a / m s⁻²	(1/m) / kg⁻¹	(1/a) / s² m⁻¹
0.050	4.41		20.0
0.075	3.60	0.0926	13.3	10.80
0.100	3.16	0.1202	10.0	8.32
0.150	2.54	0.1860	6.67	5.38
0.200	2.24	0.2392	5.00	4.18
0.250	1.98	0.3061	4.00	3.27

Graph drawn to scale. The point for m = 0.050 kg has not been plotted.

(a)

(i)

For m = 0.050 kg, calculate a, its absolute uncertainty, and 1/a.

(3)

(ii)

Determine I, using the full range of the data.

(2)

(iii)

There is in fact a small constant frictional torque at the axle. State and explain the effect on the value of I found in (a)(ii).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
a = 1.20/4.41² = 0.0617 m s⁻²	✓ 1
Fractional uncertainty = 2 × 0.05/4.41 = 2.3 %, so Δa = ±0.001 m s⁻²	✓ 1	Absolute uncertainty to 1 s.f. (0.0014).
1/a = 16.2 s² m⁻¹	✓ 1
Part (a)(ii)
Gradient from (4.00, 3.27) to (20.0, 16.2): (16.2 − 3.27)/(20.0 − 4.00) = 0.81 kg s² m⁻¹	✓ 1	Allow ECF from (a)(i). Accept 0.78–0.84.
Gradient = I/gr², so I = 0.81 × 9.81 × 0.0250² = 5.0 × 10⁻³ kg m²	✓ 1	Accept 4.8–5.2 × 10⁻³ kg m². Radius must be in m.
Part (a)(iii)
Friction reduces a for every mass, so every 1/a is too large; the effect is proportionally greatest for the smallest masses (largest 1/m), where the driving torque is smallest	✓ 1
So the gradient is too large and I is overestimated	✓ 1	Allow ECF from (a)(ii).

Answers: (a)(i) 0.062 ± 0.001 m s⁻²; 16.2 s² m⁻¹ · (a)(ii) 5.0 × 10⁻³ kg m² (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — an unbalanced torque causes angular acceleration; Newton's second law for rotation τ = Iα; the torque τ = Fr sin θ; Tools 3 — propagation of uncertainty for a power, linearising a relationship, gradient using the full data range, systematic errors Command term: Determine

34A-1B-19

Rotational equilibrium·A.4 Rigid body mechanics (HL)

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

A uniform plank of length 2.40 m and unknown mass M rests horizontally on a knife-edge pivot at one end P and on a digital scale at the other end S. A block of unknown mass m is placed with its centre a distance x from P and the scale reading F is recorded; each reading is ±0.5 N. With nothing on it, the scale reads 1.5 N.

Taking torques about P gives F_S = Mg/2 + (mg/2.40)x for the force exerted by the scale. The graph shows F against x with the line of best fit extended to x = 0.

x / m	F / N
0.40	62.8
0.80	82.9
1.20	101.7
1.60	122.0
2.00	141.2

The plank rests on a pivot at P and on a scale at S (not to scale).

Graph drawn to scale

(a)

(i)

Determine M, using the intercept of the graph.

(3)

(ii)

Determine m, using your intercept and the reading at x = 2.00 m.

(2)

(iii)

Determine the force exerted on the plank by the pivot when x = 2.00 m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Intercept on the F axis = 43.4 N	✓ 1	Accept 42.5–44.0 N.
Subtract the zero offset: Mg/2 = 43.4 − 1.5 = 41.9 N	✓ 1
M = 2 × 41.9/9.81 = 8.5 kg	✓ 1	Accept 8.3–8.7 kg. Ignoring the offset (8.8 kg) scores [2 max].
Part (a)(ii)
Gradient = (141.2 − 43.4)/2.00 = 49 N m⁻¹	✓ 1	Allow ECF from (a)(i). Accept 47–51 N m⁻¹.
m = gradient × 2.40/g = 49 × 2.40/9.81 = 12 kg	✓ 1	Accept 11.5–12.5 kg. The zero offset does not affect the gradient.
Part (a)(iii)
Translational equilibrium: pivot force = (M + m)g − true scale force = (8.5 + 12.0) × 9.81 − (141.2 − 1.5)	✓ 1	Allow ECF from (a)(i) and (a)(ii). The zero offset must be removed from the scale reading.
= 61 N upwards	✓ 1	Accept 59–63 N. Direction required.

Answers: (a)(i) 8.5 kg · (a)(ii) 12 kg · (a)(iii) 61 N upwards (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — bodies in rotational equilibrium have a resultant torque of zero; the torque τ = Fr sin θ; A.2 — translational equilibrium; Tools 3 — gradient and extrapolated intercept, zero (systematic) error Command term: Determine

35A-1B-36

Torque and rotational equilibrium·A.4 Rigid body mechanics (HL)

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

A student tests the relationship τ = Fr sin θ. A disc can turn about a horizontal axle through O with negligible friction. A load of unknown mass M hangs from a string wound on a drum of radius 0.0400 m fixed to the disc. A force sensor pulls on a peg P, 0.150 m from O, with a force F at an angle θ to the line OP, as shown. For each angle the student adjusts F until the disc is at rest with OP horizontal. Each angle is ±1° and each force is ±0.05 N.

In equilibrium F × 0.150 × sin θ = τ₀, where τ₀ is the torque of the load about O, so F = (τ₀/0.150) × (1/sin θ). The graph shows F against 1/sin θ.

θ / °	F / N	1/sin θ
20	3.81
30	2.63	2.000
40	2.02	1.556
50	1.72	1.305
65	1.43	1.103
90	1.31	1.000

The disc held at rest by the force sensor (not to scale).

Graph drawn to scale. The point for θ = 20° has not been plotted.

(a)

(i)

Calculate 1/sin θ for θ = 20°, together with its absolute uncertainty.

(2)

(b)

(i)

Determine the gradient of the line. Give its unit.

(2)

(ii)

Determine τ₀ and hence M.

(2)

(c)

(i)

After the experiment the student finds that P is actually 0.155 m from O. State and explain the effect of this on the value of M found in (b)(ii).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1/sin 20° = 2.92	✓ 1	Accept 2.924.
1/sin 19° = 3.07 and 1/sin 21° = 2.79, so the uncertainty is ±0.14	✓ 1	Accept ±0.13 to ±0.15 (or ±0.1). ±1 or ±0.017 (uncertainty in θ not converted) scores 0 for this mark.
Part (b)(i)
Gradient using the full range of the line, e.g. (3.81 − 1.31)/(2.924 − 1.000) = 1.30	✓ 1	Allow ECF from (a)(i). Accept 1.27–1.34.
Unit: N	✓ 1	Do not accept N m or N rad. 1/sin θ has no unit.
Part (b)(ii)
τ₀ = gradient × 0.150 = 1.30 × 0.150 = 0.195 N m	✓ 1	Allow ECF from (b)(i). Accept 0.19–0.20 N m.
M = τ₀/(g × 0.0400) = 0.195/(9.81 × 0.0400) = 0.50 kg	✓ 1	Allow ECF from the torque. Accept 0.49–0.51 kg. Using the drum diameter scores 0 for this mark.
Part (c)(i)
The forces measured are unchanged, so the gradient is unchanged, but τ₀ = gradient × (distance OP) is larger than calculated: M was underestimated (by about 3 %, true value 0.51 kg)	✓ 1	Allow ECF from (b)(ii). The direction of the effect with a reason is needed; the numerical estimate is not required.

Answers: (a)(i) 2.92 ± 0.14 · (b)(i) 1.30 N · (b)(ii) 0.195 N m; 0.50 kg · (c)(i) M underestimated (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ; that bodies in rotational equilibrium have a resultant torque of zero; Tools 3 — propagating an uncertainty through a function (sin); linearising a relationship; gradient with units; systematic error Command term: Determine

36A-1B-37

Equations of uniform angular acceleration·A.4 Rigid body mechanics (HL)

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

A laboratory turntable is rotating freely when, at time t = 0, a motor is engaged that exerts a constant torque of 0.46 N m on it. A constant frictional torque of 0.060 N m acts on the turntable throughout. A rotary sensor on the axle records the angular displacement θ of the turntable from its position at t = 0. The uncertainty in each value of θ is ±0.02 rad, and the uncertainty in t is negligible.

For uniform angular acceleration α from an initial angular speed ω_i, θ = ω_it + ½αt², so θ/t = ω_i + ½αt. The graph shows θ/t against t.

t / s	θ / rad	(θ/t) / rad s⁻¹
1.0	2.83	2.83
2.0	6.36	3.18
3.0	10.86	3.62
4.0	15.95
5.0	22.06	4.41
6.0	28.80	4.80

Graph drawn to scale. The line of best fit has been extended to t = 0. The point for t = 4.0 s has not been plotted.

(a)

(i)

Calculate θ/t for t = 4.0 s.

(1)

(b)

(i)

Determine α, using the gradient of the graph.

(2)

(ii)

Determine ω_i.

(1)

(c)

(i)

Determine the moment of inertia of the turntable.

(2)

(ii)

The sensor actually started recording 0.30 s after the motor was engaged, and θ and t were measured from that instant. State and explain the effect on the values of α and ω_i determined from the graph.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
15.95/4.0 = 3.99 rad s⁻¹	✓ 1	Accept 3.99 or 3.988.
Part (b)(i)
Gradient using the full range of the line, e.g. (4.80 − 2.83)/(6.0 − 1.0) = 0.394 rad s⁻²	✓ 1	Accept 0.38–0.42 rad s⁻². A unit is not required for this mark.
Gradient = ½α, so α = 2 × 0.394 = 0.79 rad s⁻²	✓ 1	Allow ECF from the gradient. Accept 0.76–0.84 rad s⁻². α = gradient scores 0 for this mark.
Part (b)(ii)
Intercept on the θ/t axis: ω_i = 2.4 rad s⁻¹	✓ 1	Accept 2.3–2.5 rad s⁻¹. Accept a value calculated from the candidate’s gradient and a point on the line.
Part (c)(i)
Resultant torque = 0.46 − 0.060 = 0.40 N m	✓ 1	Using 0.46 N m (friction ignored) scores 0 for this mark.
I = τ/α = 0.40/0.79 = 0.51 kg m²	✓ 1	Allow ECF from (b)(i). Accept 0.48–0.53 kg m².
Part (c)(ii)
α is unaffected because the angular acceleration is uniform, so the gradient is the same; the intercept gives the angular speed at the start of the recording, which is larger than the angular speed when the motor was engaged, so ω_i is overestimated (by α × 0.30 ≈ 0.24 rad s⁻¹)	✓ 1	Allow ECF from (b)(i). Both effects are needed for the mark. The numerical estimate is not required.

Answers: (a)(i) 3.99 rad s⁻¹ · (b)(i) 0.79 rad s⁻² · (b)(ii) 2.4 rad s⁻¹ · (c)(i) 0.51 kg m² (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that equations of motion for uniform angular acceleration can be used to predict the body’s angular position θ, angular displacement Δθ, angular speed ω and angular acceleration α, as given by Δθ = ω_it + ½αt²; Newton’s second law for rotation as given by τ = Iα where τ is the average torque; Tools 3 — linearising a relationship; gradient and intercept with units; systematic error Command term: Determine

37A-1B-38

Rotational equilibrium·A.4 Rigid body mechanics (HL)

Paper 1BEasy7 marks

Data-based question6 steps to full marksDetermine

A uniform metre rule balances on a pivot at its centre. A metal sinker of mass m and volume V hangs from a light thread 0.300 m to the left of the pivot and is fully immersed in a liquid of density ρ. A counterweight of mass 0.100 kg is moved along the right-hand side of the rule until the rule balances horizontally, at a distance x from the pivot, as shown. The measurement is repeated with liquids of different known density. Each value of x is ±0.002 m.

The tension in the thread is (m − ρV)g, so x = (0.300m/0.100) − (0.300V/0.100)ρ. The graph shows x against ρ.

ρ / kg m⁻³	790	880	1000	1100	1260
x / m	0.287	0.272	0.256	0.239	0.217

The balance with the sinker immersed (not to scale).

Graph drawn to scale. The line of best fit has been extended to ρ = 0.

(a)

(i)

Determine the gradient of the line. Give its unit in fundamental SI units.

(2)

(ii)

Determine the volume V of the sinker.

(2)

(b)

(i)

Determine the mass m of the sinker, using the graph.

(1)

(ii)

With the sinker immersed in an unknown liquid, the rule balances when x = 0.248 m. Determine the density of this liquid.

(1)

(iii)

The pivot is in fact 2 mm to the left of the centre of mass of the rule. State and explain whether this affects the value of V found in (a)(ii).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient using the full range of the line, e.g. (0.217 − 0.287)/(1260 − 790) = −1.5 × 10⁻⁴	✓ 1	Accept −1.45 × 10⁻⁴ to −1.53 × 10⁻⁴. The negative sign is required.
Unit: m⁴ kg⁻¹	✓ 1	Accept m/(kg m⁻³) only if simplified to m⁴ kg⁻¹.
Part (a)(ii)
Gradient = −0.300V/0.100 = −3.00V	✓ 1	Sign errors are not penalised here.
V = 1.5 × 10⁻⁴/3.00 = 5.0 × 10⁻⁵ m³	✓ 1	Allow ECF from (a)(i). Accept 4.8–5.1 × 10⁻⁵ m³.
Part (b)(i)
Intercept at ρ = 0: 0.404 m (accept 0.398–0.410 m); m = 0.404 × 0.100/0.300 = 0.135 kg	✓ 1	Accept 0.133–0.137 kg. The intercept corresponds to the sinker in a vacuum (no buoyancy force).
Part (b)(ii)
ρ = (0.404 − 0.248)/1.49 × 10⁻⁴ = 1050 kg m⁻³	✓ 1	Allow ECF from (a)(i) and (b)(i). Accept 1020–1080 kg m⁻³, including a value read from the graph.
Part (b)(iii)
No: the weight of the rule adds the same constant torque for every liquid, so every value of x is shifted by the same amount; the gradient, and hence V, is unchanged (only the intercept, and so m, is affected)	✓ 1	Allow ECF from (a)(ii). Do not accept "no" without a reason.

Answers: (a)(i) −1.5 × 10⁻⁴ m⁴ kg⁻¹ · (a)(ii) 5.0 × 10⁻⁵ m³ · (b)(i) 0.135 kg · (b)(ii) 1050 kg m⁻³ · (b)(iii) no effect on V (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that bodies in rotational equilibrium have a resultant torque of zero; the torque τ of a force about an axis as given by τ = Fr sin θ; A.2 — the buoyancy force F_b = ρVg; Tools 3 — gradient and intercept with units in fundamental SI units; extrapolation; systematic error Command term: Determine

38A-2-02

Rotational kinetic energy and flywheels·A.4 Rigid body mechanics (HL)

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksDetermine

A hybrid bus stores energy in a flywheel. The flywheel is a uniform disc of mass 60 kg and radius 0.25 m that spins in a vacuum about a fixed axis through its centre. The moment of inertia of a uniform disc is ½MR². The maximum safe rotation rate is 16 000 revolutions per minute.

When the bus, of mass 1.2 × 10⁴ kg, brakes, 60 % of its kinetic energy is transferred to the flywheel. The graph shows the speed of the bus during one stop.

Speed v of the bus against time t during one stop (drawn to scale).

(a)

(i)

Show that the moment of inertia of the flywheel is about 1.9 kg m².

(1)

(ii)

Calculate the maximum energy that the flywheel can store.

(2)

(iii)

Show that the maximum energy stored per unit mass of any uniform disc depends only on the speed v of its rim, and hence suggest why flywheel designers aim for a high rim speed rather than a large mass.

(2)

(b)

(i)

Use the graph to determine the kinetic energy that the bus loses during the stop.

(1)

(ii)

Before the stop the flywheel rotates at 8000 revolutions per minute. Determine its angular speed at the end of the stop.

(2)

(iii)

Determine the average torque exerted on the flywheel during the stop.

(2)

(c)

(i)

Later the flywheel drives a generator that supplies the electric motor of the bus at 600 V. The motor draws a current of 250 A. 85 % of the energy captured in (b) reaches the motor. Determine how long the captured energy can run the motor.

(3)

(ii)

Suggest why the flywheel is made to spin in a vacuum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
I = ½ × 60 × 0.25² = 1.875 kg m²	✓ 1	Full substitution or 1.88 needed.
Part (a)(ii)
ω = 16 000 × 2π/60 = 1676 rad s⁻¹	✓ 1	Using 16 000 as ω scores 0 for this mark.
E_k = ½ × 1.875 × 1676² = 2.6 × 10⁶ J	✓ 1	Allow ECF from (a)(i). Accept 2.6–2.7 × 10⁶ J.
Part (a)(iii)
E_k/M = ½(½MR²)ω²/M = ¼R²ω² = ¼v², since v = ωR	✓ 1	Symbolic working required; no numbers.
The energy per kilogram rises as v², so a lighter, stronger rotor that can survive a higher rim speed stores more energy per unit mass than adding mass	✓ 1	Accept: the limit is the strength of the material at the rim, not the mass.
Part (b)(i)
E_k = ½ × 1.2 × 10⁴ × 15.0² = 1.35 × 10⁶ J	✓ 1	Speed read from the graph: 15 m s⁻¹.
Part (b)(ii)
Initial E_k = ½ × 1.875 × 838² = 6.6 × 10⁵ J; energy added 0.60 × 1.35 × 10⁶ = 8.1 × 10⁵ J	✓ 1	Allow ECF from (a)(i) and (b)(i).
ω = √(2 × 1.47 × 10⁶/1.875) = 1251 rad s⁻¹	✓ 1	Accept 1240–1260 rad s⁻¹. Adding the angular speeds linearly is a named error: [1 max].
Part (b)(iii)
τ = ΔL/Δt = IΔω/Δt with Δt = 6.0 s read from the graph	✓ 1
τ = 1.875 × (1251 − 838)/6.0 = 129 N m	✓ 1	Allow ECF from (b)(ii). Accept 125–135 N m.
Part (c)(i)
Electrical power P = VI = 600 × 250 = 1.5 × 10⁵ W	✓ 1
Energy delivered = 0.85 × 8.1 × 10⁵ = 6.9 × 10⁵ J	✓ 1	Allow ECF from (b)(i).
t = 6.9 × 10⁵/1.5 × 10⁵ = 4.6 s	✓ 1	Accept 4.5–4.7 s.
Part (c)(ii)
There is no air resistance (drag) on the spinning disc, so the stored rotational kinetic energy is not dissipated as thermal energy while the flywheel is waiting to be used	✓ 1	Do not accept "to reduce friction" without reference to the air.

Answers: (a)(ii) 2.6 × 10⁶ J · (b)(i) 1.35 × 10⁶ J · (b)(ii) 1251 rad s⁻¹ · (b)(iii) 129 N m · (c)(i) 4.6 s (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the kinetic energy of rotational motion E_k = ½Iω²; the moment of inertia depends on the distribution of mass; angular impulse ΔL = τΔt = Δ(Iω); A.3 — energy transfers, dissipation and efficiency; B.5 — electrical power P = VI Command term: Determine

39A-2-05

Rolling and rotational kinetic energy·A.4 Rigid body mechanics (HL)

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksDetermine

A uniform solid ball of mass 0.10 kg and radius 2.0 cm is released from rest and rolls without slipping down a ramp inclined at 30° to the horizontal. It descends a vertical height of 0.50 m. The moment of inertia of a uniform solid sphere about a diameter is ⅖mr².

(a)

(i)

Show that the speed of the ball after descending 0.50 m is about 2.6 m s⁻¹.

(2)

(ii)

Show that the fraction of the kinetic energy of the ball that is rotational is 2/7, whatever the mass and radius of the ball.

(2)

(b)

(i)

Determine the magnitude of the frictional force on the ball as it rolls down the ramp.

(3)

(ii)

Calculate the minimum coefficient of static friction between the ball and the ramp.

(1)

(c)

(i)

The ball is placed in a spherical bowl of radius R = 0.40 m and released from a small displacement. It rolls without slipping and performs simple harmonic motion of period T = 2π√(7(R − r)/5g). Calculate T.

(1)

(ii)

A small block sliding without friction in the same bowl has a shorter period, 2π√(R/g) = 1.27 s. Explain, with reference to (a)(ii), why the rolling ball oscillates more slowly.

(2)

(iii)

The ball is released from a point 5.0 mm above the lowest point of the bowl. Determine the maximum speed of the ball and the amplitude of its motion, measured along the path of its centre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
mgh = ½mv² + ½(⅖mr²)(v/r)² = (7/10)mv²	✓ 1
v = √(10 × 9.81 × 0.50/7) = 2.65 m s⁻¹	✓ 1	Must see 2.65 or full substitution.
Part (a)(ii)
Rotational E_k = ½(⅖mr²)(v/r)² = ⅕mv²; total = (7/10)mv²	✓ 1	Symbolic working required.
Fraction = (1/5)/(7/10) = 2/7: m and r cancel	✓ 1
Part (b)(i)
Linear: mg sin θ − f = ma; rotational: fr = Iα = ⅖mr²(a/r), so f = ⅖ma	✓ 1
a = (5/7)g sin 30° = 3.50 m s⁻²	✓ 1
f = ⅖ × 0.10 × 3.50 = 0.14 N	✓ 1	Accept 0.14 N. Equivalent: f = (2/7)mg sin θ.
Part (b)(ii)
μ ≥ f/F_N = 0.140/(0.10 × 9.81 × cos 30°) = 0.16	✓ 1	Allow ECF from (b)(i). Accept 0.16–0.17.
Part (c)(i)
T = 2π√(7 × 0.38/(5 × 9.81)) = 1.46 s	✓ 1	Using 0.40 m for R − r gives 1.50 s: [0].
Part (c)(ii)
Part of the gravitational potential energy (2/7 of the kinetic energy) goes into rotation, so the ball moves more slowly at each position	✓ 1	Allow ECF from (a)(ii).
Equivalently, part of the weight component is used to give angular acceleration (friction opposes the motion), so the restoring acceleration at a given displacement is only 5/7 as large and T is longer	✓ 1	Accept either line of reasoning for the second mark if linked to the period.
Part (c)(iii)
v_max = √(10gh/7) = √(10 × 9.81 × 0.0050/7) = 0.265 m s⁻¹	✓ 1	Allow ECF from (a)(i) method.
x₀ = v_max/ω = 0.265 × 1.46/2π = 0.062 m	✓ 1	Allow ECF from (c)(i). Accept 0.061–0.063 m.

Answers: (b)(i) 0.14 N · (b)(ii) 0.16 · (c)(i) 1.46 s · (c)(iii) 0.265 m s⁻¹, 0.062 m (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the kinetic energy of rotational motion; Newton's second law for rotation τ = Iα; rolling without slipping; the moment of inertia depends on the distribution of mass; A.2 — static friction F_f ≤ μ_sF_N; C.1 — the conditions for simple harmonic motion; v_max = ωx₀ Command term: Determine

40A-2-08

Torque and angular acceleration·A.4 Rigid body mechanics (HL)

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksDetermine

Two blocks of mass m₁ = 3.0 kg and m₂ = 2.0 kg hang from a light string that passes over a pulley. The pulley is a uniform aluminium disc of mass M = 2.0 kg and radius 0.10 m, with moment of inertia ½MR², turning on a frictionless axle. The string does not slip on the pulley. The system is released from rest.

(a)

(i)

Show that the acceleration of the blocks is a = (m₁ − m₂)g/(m₁ + m₂ + M/2).

(2)

(ii)

Calculate a.

(1)

(iii)

Determine the tension in the string on each side of the pulley.

(2)

(b)

(i)

The 3.0 kg block falls 1.0 m. Determine the rotational kinetic energy of the pulley at this instant and show that the total energy of the system is conserved.

(3)

(c)

(i)

A strong magnet is now fixed close to the rim of the aluminium pulley, so that its field passes through part of the disc. Explain why a torque now opposes the rotation of the pulley.

(2)

(ii)

With the magnet in place the blocks reach a steady speed of 0.40 m s⁻¹. Calculate the rate at which thermal energy is generated in the pulley.

(1)

(iii)

The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. Determine the rise in temperature of the pulley in 30 s, assuming no energy is lost to the surroundings.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
m₁g − T₁ = m₁a and T₂ − m₂g = m₂a	✓ 1	Symbolic working required.
Pulley: (T₁ − T₂)R = ½MR²(a/R), so T₁ − T₂ = ½Ma; adding the three equations gives the result	✓ 1	The no-slip condition α = a/R must be used.
Part (a)(ii)
a = 1.0 × 9.81/6.0 = 1.64 m s⁻²	✓ 1	Allow ECF from (a)(i) expression.
Part (a)(iii)
T₁ = 3.0 × (9.81 − 1.64) = 24.5 N	✓ 1	Allow ECF from (a)(ii).
T₂ = 2.0 × (9.81 + 1.64) = 22.9 N	✓ 1	Equal tensions on both sides scores [0] for this part.
Part (b)(i)
v = √(2 × 1.635 × 1.0) = 1.81 m s⁻¹, ω = v/R = 18.1 rad s⁻¹	✓ 1	Allow ECF from (a)(ii).
Rotational E_k = ½ × 0.010 × 18.1² = 1.64 J	✓ 1
Translational E_k = ½ × 5.0 × 1.81² = 8.18 J; total 9.81 J = loss of E_p, (3.0 − 2.0) × 9.81 × 1.0 = 9.81 J	✓ 1
Part (c)(i)
Parts of the rotating disc move through the magnetic field, so emfs are induced in the disc (Faraday's law) and currents flow in the conducting aluminium	✓ 1	Do not accept "the magnet attracts aluminium".
By Lenz's law the induced currents produce forces that oppose the relative motion, so the force on the disc acts against its rotation at a distance from the axle, giving an opposing torque	✓ 1
Part (c)(ii)
P = (m₁ − m₂)gv = 1.0 × 9.81 × 0.40 = 3.9 W	✓ 1	At steady speed all the gravitational potential energy lost is dissipated.
Part (c)(iii)
Energy = 3.92 × 30 = 118 J	✓ 1	Allow ECF from (c)(ii).
ΔT = 118/(2.0 × 900) = 0.065 K	✓ 1	Accept 0.065 K.

Answers: (a)(ii) 1.64 m s⁻² · (a)(iii) 24.5 N and 22.9 N · (b)(i) 1.64 J · (c)(ii) 3.9 W · (c)(iii) 0.065 K (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — Newton's second law for rotation τ = Iα; an unbalanced torque causes angular acceleration; the kinetic energy of rotational motion; A.3 — conservation of energy; D.4 — Faraday's law of induction and Lenz's law; B.1 — specific heat capacity Command term: Determine

41A-2-13

Angular impulse and braking torque·A.4 Rigid body mechanics (HL)

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksDetermine

A bicycle is held upside down on a stand so that its front wheel can spin freely about its axle. The moment of inertia of the wheel about the axle is 0.18 kg m² and the radius of the rim is 0.34 m. The wheel is spun to an angular speed of 25.0 rad s⁻¹. Friction at the axle and air resistance are negligible.

(a)

The spinning wheel.

(i)

Calculate the angular momentum of the wheel.

(1)

(ii)

Calculate the rotational kinetic energy of the wheel.

(1)

(b)

The rim brake is applied: a brake pad presses on the rim with a constant normal force of 40 N. The coefficient of dynamic friction between pad and rim is 0.60.

(i)

Show that the frictional torque about the axle is about 8.2 N m.

(2)

(ii)

Use the idea of angular impulse to determine the time taken for the wheel to stop.

(2)

(iii)

Determine the average rate at which thermal energy is generated at the pad and rim while the wheel stops.

(2)

(c)

In a second trial the wheel is again spun to 25.0 rad s⁻¹, but the pad is applied gradually: the normal force rises uniformly from zero to 40 N in 0.40 s and then remains constant.

(i)

Determine the total time taken for the wheel to stop.

(3)

(ii)

Sketch or describe the variation of the angular momentum of the wheel with time during this second trial, and state what the gradient of the graph represents.

(2)

(d)

Wheel design.

(i)

A second wheel has the same mass and rim radius but more of its mass close to the hub. It is spun to 25.0 rad s⁻¹ and braked as in (b). State and explain how its stopping time compares with your answer to (b)(ii).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
L = Iω = 0.18 × 25.0 = 4.5 kg m² s⁻¹	✓ 1	Accept N m s.
Part (a)(ii)
E_k = ½Iω² = ½ × 0.18 × 25.0² = 56 J	✓ 1	Accept 56.3 J.
Part (b)(i)
Friction force F_f = μ_dF_N = 0.60 × 40 = 24 N	✓ 1
τ = F_fr = 24 × 0.34 = 8.16 ≈ 8.2 N m	✓ 1	The friction force is tangential to the rim, so sin θ = 1.
Part (b)(ii)
The angular impulse of the braking torque equals the change in angular momentum: τΔt = ΔL = 4.5 N m s	✓ 1
Δt = 4.5/8.16 = 0.55 s	✓ 1	Or α = τ/I = 45.3 rad s⁻² and t = 25.0/45.3.
Part (b)(iii)
All 56 J of rotational kinetic energy is transferred to thermal energy by the friction at the pad	✓ 1
Average power = 56.25/0.551 = 1.0 × 10² W	✓ 1	Allow ECF from (a)(ii) and (b)(ii).
Part (c)(i)
The torque rises uniformly from 0 to 8.16 N m, so the angular impulse in the first 0.40 s is the area of a triangle: ½ × 8.16 × 0.40 = 1.63 N m s	✓ 1
Angular momentum remaining at 0.40 s = 4.5 − 1.63 = 2.87 kg m² s⁻¹, removed at 8.16 N m in 2.87/8.16 = 0.35 s	✓ 1
Total time = 0.40 + 0.35 = 0.75 s	✓ 1
Part (c)(ii)
L falls from 4.5 kg m² s⁻¹, starting with zero gradient and curving increasingly steeply until 0.40 s (reaching 2.87 kg m² s⁻¹), then decreasing linearly to zero at 0.75 s	✓ 1	Shape and key values both required.
The gradient ΔL/Δt equals (minus) the resultant torque on the wheel	✓ 1
Part (d)(i)
Shorter: its moment of inertia is smaller, so at the same ω its angular momentum L = Iω is smaller, and the same braking torque removes it in less time (Δt = Iω/τ)	✓ 1

Answers: (a)(i) 4.5 kg m² s⁻¹ · (a)(ii) 56 J · (b)(ii) 0.55 s · (b)(iii) 1.0 × 10² W · (c)(i) 0.75 s (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that the action of a resultant torque constitutes an angular impulse ΔL as given by ΔL = τΔt = Δ(Iω); that an extended body rotating with an angular speed has an angular momentum L = Iω; the kinetic energy of rotational motion; A.2 — dynamic friction F_f = μ_dF_N Command term: Determine

42A-2-15

Equilibrium of a ladder·A.4 Rigid body mechanics (HL)

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksDetermine

A uniform ladder of length 5.00 m and mass 20.0 kg rests against a smooth vertical wall, making an angle of 60.0° with rough horizontal ground. A person of mass 60.0 kg stands on the ladder 3.50 m from its foot. The ladder is in equilibrium.

A 5.0 m ladder resting against a smooth wall with a person 3.5 m from its foot.

(a)

(i)

Explain why the force exerted by the wall on the ladder is horizontal.

(1)

(ii)

By taking torques about the foot of the ladder, show that the force exerted by the wall is about 295 N.

(2)

(iii)

Determine the minimum coefficient of static friction between the ladder and the ground.

(2)

(b)

(i)

The coefficient of static friction is in fact 0.40. Determine how far up the ladder the person can stand before the ladder slips.

(3)

(ii)

Show that a uniform ladder alone, leaning against a smooth wall at an angle θ to the ground, needs a coefficient of friction of at least 1/(2 tan θ).

(2)

(iii)

Hence deduce whether the ladder alone, at 60.0°, would stand on a floor where the coefficient of static friction is 0.25.

(1)

(c)

(i)

A second person stands on the bottom rung, at the foot of the ladder, while the first person climbs. Explain why this makes slipping less likely.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The wall is smooth, so it exerts no frictional (parallel) force, only a normal force perpendicular to the wall, which is horizontal	✓ 1
Part (a)(ii)
N × 5.00 sin 60.0° = 20.0 × 9.81 × 2.50 cos 60.0° + 60.0 × 9.81 × 3.50 cos 60.0°	✓ 1	Correct perpendicular distances are required.
N = (245 + 1030)/4.33 = 295 N	✓ 1	Must see 294.5 or full substitution.
Part (a)(iii)
Vertical: R = 80.0 × 9.81 = 785 N; horizontal: friction = N = 295 N	✓ 1	Allow ECF from (a)(ii).
μ ≥ 295/785 = 0.38	✓ 1	Accept 0.37–0.38.
Part (b)(i)
Maximum friction = 0.40 × 785 = 314 N = N	✓ 1	Allow ECF from (a)(iii) for R.
Torques about the foot: 314 × 4.33 = 245 + 60.0 × 9.81 × d × cos 60.0°	✓ 1
d = 3.8 m	✓ 1	Accept 3.7–3.9 m.
Part (b)(ii)
Torques about the foot: NL sin θ = mg(L/2) cos θ, so N = mg/(2 tan θ)	✓ 1	Symbolic working required.
Friction = N and R = mg, so μ ≥ N/R = 1/(2 tan θ)	✓ 1
Part (b)(iii)
1/(2 tan 60.0°) = 0.29 > 0.25, so the ladder slips	✓ 1	Allow ECF from (b)(ii).
Part (c)(i)
The second person's weight acts (almost) through the foot, so it adds no torque about the foot and the wall force N (the friction needed) is unchanged	✓ 1
but the normal force R from the ground increases, so the maximum friction μR increases	✓ 1	Both ideas needed for 2 marks. "Holds the ladder down" alone scores 0.

Answers: (a)(iii) 0.38 · (b)(i) 3.8 m · (b)(iii) it slips (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the torque τ of a force about an axis, τ = Fr sin θ; bodies in rotational equilibrium have a resultant torque of zero; A.2 — translational equilibrium; static friction F_f ≤ μ_sF_N Command term: Determine

43A-2-17

Conservation of angular momentum·A.4 Rigid body mechanics (HL)

Paper 2Hard13 marks

Short answer & extended response8 steps to full marksDetermine

The core of a massive star has a mass of 2.8 × 10³⁰ kg and a radius of 6.0 × 10⁶ m, and rotates with a period of 3.0 × 10³ s. It collapses to form a neutron star of radius 12 km. Model the core before and after the collapse as a uniform sphere, with moment of inertia ⅖MR², and assume that no mass is lost.

(a)

(i)

Show that the period of rotation after the collapse is T₂ = T₁(R₂/R₁)², where 1 and 2 refer to before and after.

(2)

(ii)

Calculate T₂.

(1)

(b)

(i)

Determine the rotational kinetic energy of the core before and after the collapse.

(3)

(ii)

Explain how the rotational kinetic energy can increase although angular momentum is conserved.

(2)

(c)

(i)

The gravitational potential energy released in the collapse is (3GM²/5)(1/R₂ − 1/R₁). Determine the fraction of this energy that becomes rotational kinetic energy.

(2)

(ii)

Determine the speed of a point on the equator of the neutron star, and deduce whether a relativistic treatment of its rotation is needed.

(2)

(iii)

State one feature of the model that makes your answer to (a)(ii) unreliable.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
No external torque acts, so angular momentum is conserved: ⅖MR₁²(2π/T₁) = ⅖MR₂²(2π/T₂)	✓ 1	Symbolic working required.
M and 2π cancel: T₂ = T₁R₂²/R₁²	✓ 1
Part (a)(ii)
T₂ = 3.0 × 10³ × (1.2 × 10⁴/6.0 × 10⁶)² = 0.012 s	✓ 1	Allow ECF from (a)(i). Accept 12 ms.
Part (b)(i)
I₁ = ⅖ × 2.8 × 10³⁰ × (6.0 × 10⁶)² = 4.0 × 10⁴³ kg m²; I₂ = 1.6 × 10³⁸ kg m²	✓ 1
Before: ½ × 4.0 × 10⁴³ × (2π/3.0 × 10³)² = 8.8 × 10³⁷ J	✓ 1
After: ½ × 1.6 × 10³⁸ × (2π/0.012)² = 2.2 × 10⁴³ J	✓ 1	Allow ECF from (a)(ii). Accept 2.2 × 10⁴³ J.
Part (b)(ii)
The gravitational forces on the core matter are internal and act towards the centre, so they exert no torque and L is constant	✓ 1
These forces do work on the matter as it falls inwards; gravitational potential energy is transferred to (rotational) kinetic energy, and E_k = L²/2I rises as I falls	✓ 1	Allow ECF from (b)(i). Reference to work done by the gravitational force needed.
Part (c)(i)
Energy released = 0.6 × 6.67 × 10⁻¹¹ × (2.8 × 10³⁰)² × (1/1.2 × 10⁴ − 1/6.0 × 10⁶) = 2.6 × 10⁴⁶ J	✓ 1
Fraction = 2.2 × 10⁴³/2.6 × 10⁴⁶ = 8.5 × 10⁻⁴	✓ 1	Allow ECF from (b)(i). Accept 8–9 × 10⁻⁴.
Part (c)(ii)
v = 2πR/T = 2π × 1.2 × 10⁴/0.012 = 6.3 × 10⁶ m s⁻¹ (0.021c)	✓ 1	Allow ECF from (a)(ii).
γ = 1.0002, so relativistic corrections are about 0.02 %: not needed	✓ 1	Accept any valid comparison of v with c leading to "not needed".
Part (c)(iii)
Any one: a real core is much denser at the centre (not uniform), so I ≠ ⅖MR²; mass (and angular momentum) is carried away in the explosion; magnetic fields exert torques	✓ 1

Answers: (a)(ii) 0.012 s · (b)(i) 8.8 × 10³⁷ J; 2.2 × 10⁴³ J · (c)(i) 8.5 × 10⁻⁴ · (c)(ii) 6.3 × 10⁶ m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that angular momentum remains constant unless the body is acted upon by a resultant torque; that the moment of inertia I depends on the distribution of mass of an extended body; the kinetic energy of rotational motion E_k = ½Iω² = L²/2I; D.1 — the gravitational potential energy of a system; A.5 — the Lorentz factor; E.5 — the effect of stellar mass on the evolution of a star Command term: Determine

44A-2-27

Rotational equilibrium·A.4 Rigid body mechanics (HL)

Paper 2Medium14 marks

Short answer & extended response8 steps to full marksDetermine

A small lifting bridge over a canal has a uniform deck of length 8.0 m and mass 2400 kg, hinged at one end H. A cable attached to the free end of the deck runs to the top of a tower directly above the hinge. When the deck is horizontal the cable makes an angle of 30° with the deck, as shown.

The bridge deck held horizontal by the cable (drawn to scale).

(a)

(i)

Show that the tension in the cable is about 2.4 × 10⁴ N.

(2)

(ii)

Determine the magnitude of the force exerted by the hinge on the deck.

(3)

(iii)

Show that, for a cable at an angle φ to the horizontal deck, the tension is Mg/(2 sin φ), and hence explain why bridge designers use a tall tower.

(2)

(iv)

State the sense, as seen in the diagram, of the torques about H of the weight of the deck and of the tension in the cable.

(1)

(b)

(i)

A winch raises the deck slowly from the horizontal to 60° above the horizontal in 90 s. Determine the average power that the winch delivers to the deck.

(2)

(ii)

Determine the torque of the weight of the deck about H when the deck has been raised to 60° above the horizontal, and compare it with the torque when the deck is horizontal.

(2)

(iii)

Many lifting bridges carry a counterweight on an arm on the other side of the hinge. Explain how this reduces the work that the winch must do.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Torques about H: T × 8.0 × sin 30° = 2400 × 9.81 × 4.0	✓ 1	The weight acts at the centre of the deck.
T = 23544 N	✓ 1	Must see 2.35 × 10⁴ or full substitution.
Part (a)(ii)
Horizontal component = T cos 30° = 2.04 × 10⁴ N	✓ 1	Allow ECF from (a)(i).
Vertical component = 2400 × 9.81 − T sin 30° = 1.18 × 10⁴ N upwards	✓ 1
Magnitude = √((2.04 × 10⁴)² + (1.18 × 10⁴)²) = 2.4 × 10⁴ N	✓ 1	Accept 2.3–2.4 × 10⁴ N.
Part (a)(iii)
Torques about H: TL sin φ = MgL/2, so T = Mg/(2 sin φ)	✓ 1	Symbolic working required.
A taller tower gives a larger φ, so sin φ is larger and the tension (and hinge force) is smaller	✓ 1
Part (a)(iv)
Weight: clockwise; tension: anticlockwise	✓ 1	Both needed.
Part (b)(i)
The centre of mass rises by 4.0 sin 60° = 3.46 m, so the work done = 2400 × 9.81 × 3.46 = 8.2 × 10⁴ J	✓ 1	Raised slowly, so no kinetic energy is gained.
Average power = 8.2 × 10⁴/90 = 9.1 × 10² W	✓ 1	Allow ECF from the work done. Accept 0.90–0.91 kW.
Part (b)(ii)
Perpendicular distance from H to the line of action of the weight = 4.0 cos 60° = 2.0 m	✓ 1	Or τ = Fr sin θ with the angle between the deck and the vertical weight equal to 30°.
τ = 2400 × 9.81 × 2.0 = 4.7 × 10⁴ N m, half of the value when horizontal (9.4 × 10⁴ N m)	✓ 1	Both the value and the comparison are needed.
Part (b)(iii)
As the deck rises the counterweight descends, and its loss of gravitational potential energy supplies most of the gain of the deck	✓ 1
Its torque about the hinge opposes that of the deck's weight, so the resultant torque the winch must overcome is small (mainly friction)	✓ 1

Answers: (a)(ii) 2.4 × 10⁴ N · (b)(i) 9.1 × 10² W · (b)(ii) 4.7 × 10⁴ N m (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the torque τ = Fr sin θ and its sense; bodies in rotational equilibrium have a resultant torque of zero; A.2 — translational equilibrium; A.3 — work done, gravitational potential energy and power as the rate of work done Command term: Determine

45A-2-28

Torque and angular acceleration·A.4 Rigid body mechanics (HL)

Paper 2Hard13 marks

Short answer & extended response9 steps to full marksDetermine

A small d.c. motor has a rectangular coil of 80 turns. Each of the two sides of the coil parallel to the axle has length 0.050 m and is 0.015 m from the axle. These sides move in a radial magnetic field of flux density 0.40 T, so the magnetic forces on them are always perpendicular to the plane containing the side and the axle. The coil has resistance 6.0 Ω and is connected to a 15.0 V supply. The moment of inertia of the rotor is 4.0 × 10⁻⁵ kg m².

The graph shows how the angular speed of the rotor varies with time after the motor is switched on.

Angular speed ω of the rotor against time t after the motor is switched on (drawn to scale).

(a)

(i)

Show that the torque on the coil at the instant the motor is switched on is about 0.12 N m.

(2)

(ii)

Calculate the initial angular acceleration of the rotor, ignoring friction.

(1)

(b)

(i)

Use the graph to determine the average resultant torque on the rotor during the first 0.20 s.

(2)

(ii)

Calculate the rotational kinetic energy of the rotor at its steady speed.

(1)

(c)

(i)

Show that the emf induced in the coil when it rotates at angular speed ω is NBLwω, where w is the width of the coil.

(2)

(ii)

Explain why the angular acceleration of the rotor decreases as its speed increases.

(2)

(iii)

At the steady speed shown on the graph, determine the current in the coil and the frictional torque on the rotor.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Current = 15.0/6.0 = 2.5 A; force on each side = NBIL = 80 × 0.40 × 2.5 × 0.050 = 4.0 N	✓ 1
τ = 2 × 4.0 × 0.015 = 0.12 N m	✓ 1	Both sides must be included.
Part (a)(ii)
α = τ/I = 0.12/4.0 × 10⁻⁵ = 3.0 × 10³ rad s⁻²	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
ω at 0.20 s ≈ 256 rad s⁻¹; ΔL = 4.0 × 10⁻⁵ × 256 = 1.0 × 10⁻² kg m² s⁻¹	✓ 1	Accept 250–262 rad s⁻¹.
Average torque = ΔL/Δt = 0.051 N m	✓ 1	Accept 0.050–0.052 N m.
Part (b)(ii)
E_k = ½ × 4.0 × 10⁻⁵ × 300² = 1.8 J	✓ 1	Steady speed 300 rad s⁻¹ read from the graph.
Part (c)(i)
Each side moves at speed v = ωw/2 perpendicular to the field, so the emf in one side of one turn is BvL = BLωw/2	✓ 1	Symbolic working required.
The two sides add in series for each turn and there are N turns: ε = 2NBLωw/2 = NBLwω	✓ 1
Part (c)(ii)
By Lenz's law the induced emf opposes the supply p.d., so the current falls as ω increases (I = (V − ε)/R)	✓ 1	Allow ECF from (c)(i).
The magnetic torque is proportional to the current, so the resultant torque, and hence α, decreases	✓ 1
Part (c)(iii)
ε = 80 × 0.40 × 0.050 × 0.030 × 300 = 14.4 V	✓ 1	Allow ECF from (c)(i) and graph value.
I = (15.0 − 14.4)/6.0 = 0.10 A	✓ 1
At steady speed the magnetic torque balances friction: τ = 0.12 × 0.10/2.5 = 4.8 × 10⁻³ N m	✓ 1	Allow ECF from (a)(i): torque ∝ current.

Answers: (a)(ii) 3.0 × 10³ rad s⁻² · (b)(i) 0.051 N m · (b)(ii) 1.8 J · (c)(iii) 0.10 A; 4.8 × 10⁻³ N m (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the torque τ = Fr sin θ; Newton's second law for rotation τ = Iα; angular impulse τΔt = ΔL; the kinetic energy of rotational motion; D.3 — the magnetic force on a current-carrying conductor F = BIL sin θ; D.4 — the emf induced in a conductor moving in a magnetic field ε = BvL; Lenz's law; B.5 — resistance and emf Command term: Determine

46A-2-29

Conservation of angular momentum·A.4 Rigid body mechanics (HL)

Paper 2Medium14 marks

Short answer & extended response9 steps to full marksDetermine

A diver of mass 60 kg performs a dive from a springboard. About the axis through her centre of mass for somersaults, her moment of inertia is 14.0 kg m² when her body is straight. She leaves the board with an upward velocity of 4.0 m s⁻¹, with her centre of mass 4.0 m above the water. Air resistance is negligible. The graph shows her angular speed during the dive.

Angular speed ω of the diver about her centre of mass against time t after take-off (drawn to scale).

(a)

(i)

At the end of her take-off the board pushes her vertically upwards, from rest to 4.0 m s⁻¹, in 0.25 s. Determine the average force exerted on her by the board.

(2)

(b)

(i)

Calculate her angular momentum as she leaves the board.

(1)

(ii)

Explain why her angular speed increases when she pulls into the tucked position.

(2)

(iii)

Use the graph to determine her moment of inertia in the tucked position and the increase in her rotational kinetic energy.

(2)

(iv)

Use the graph to determine the average rate at which her muscles do work as she pulls into the tucked position.

(1)

(c)

(i)

Show that her centre of mass reaches the water about 1.4 s after take-off.

(2)

(ii)

Use the graph to determine the number of somersaults she completes before entering the water.

(2)

(iii)

Explain why her centre of mass follows the same path whether or not she tucks.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Average resultant force = Δp/Δt = 60 × 4.0/0.25 = 960 N	✓ 1
Force from the board = 960 + 60 × 9.81 = 1.5 × 10³ N (upwards)	✓ 1	Accept 1.5–1.6 × 10³ N. 960 N (weight omitted) scores [1 max].
Part (b)(i)
L = 14.0 × 2.0 = 28 kg m² s⁻¹	✓ 1	Angular speed read from the graph.
Part (b)(ii)
Her weight acts through her centre of mass, so there is no external torque about it and her angular momentum is constant	✓ 1
Tucking brings mass closer to the axis, reducing I; since Iω is constant, ω increases	✓ 1
Part (b)(iii)
I = 28/8.0 = 3.5 kg m²	✓ 1	Allow ECF from (b)(i).
ΔE_k = L²/2I − L²/2I₀ = 28²/7.0 − 28²/28 = 112 − 28 = 84 J	✓ 1
Part (b)(iv)
The tuck takes 0.10 s (0.10 s to 0.20 s on the graph); power = 84/0.10 = 8.4 × 10² W	✓ 1	Allow ECF from (b)(iii). The increase in rotational kinetic energy is the work done by her muscles.
Part (c)(i)
Taking upwards as positive: −4.0 = 4.0t − ½ × 9.81 × t²	✓ 1
t = (4.0 + √(16 + 78.5))/9.81 = 1.40 s	✓ 1	Must see 1.40 or better.
Part (c)(ii)
Angle turned = area under the ω–t graph = 0.2 + 0.5 + 8.0 + 0.5 + 0.2 = 9.4 rad	✓ 1	Accept 9.2–9.6 rad.
Number of somersaults = 9.4/2π = 1.5	✓ 1	Allow ECF from the area.
Part (c)(iii)
The only external force is her weight, which does not change, so the acceleration of her centre of mass is g throughout	✓ 1
The forces between parts of her body when she tucks are internal and cancel in pairs: they change I and ω but not the motion of the centre of mass	✓ 1

Answers: (a)(i) 1.5 × 10³ N · (b)(i) 28 kg m² s⁻¹ · (b)(iii) 3.5 kg m²; 84 J · (b)(iv) 8.4 × 10² W · (c)(ii) 1.5 (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that angular momentum remains constant unless the body is acted upon by a resultant torque; that the moment of inertia depends on the distribution of mass; that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration; E_k = L²/2I; A.2 — that the applied external impulse equals the change in momentum of the system; A.3 — power as the rate of work done; A.1 — the equations of motion for uniformly accelerated motion; Tool 3 — areas under a graph Command term: Determine

47A-2-33

Angular momentum in a collision·A.4 Rigid body mechanics (HL)

Paper 2Hard13 marks

Short answer & extended response10 steps to full marksDetermine

A uniform door of mass 12.0 kg and width 0.800 m is free to rotate about a vertical hinge along one edge. Its moment of inertia about the hinge is ⅓ML², where L is its width. The door is at rest.

A beanbag of mass 0.600 kg moving horizontally at 8.00 m s⁻¹, perpendicular to the door, strikes it 0.700 m from the hinge and sticks to it.

(a)

The collision.

(i)

Explain why the angular momentum of the door and beanbag about the hinge is conserved in the collision, but their linear momentum is not.

(2)

(ii)

Show that the moment of inertia of the door about the hinge is 2.56 kg m².

(1)

(iii)

Determine the angular speed of the door immediately after the collision.

(3)

(iv)

Calculate the kinetic energy transferred to other forms in the collision.

(2)

(b)

After the collision.

(i)

The door turns through 90° before coming to rest, under a constant frictional torque at the hinge. Determine this torque.

(2)

(ii)

Calculate the time taken for the door to stop.

(1)

(c)

The hinge.

(i)

Determine the magnitude of the impulse exerted by the hinge on the door during the collision and state its direction.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The hinge force acts through the axis, so it exerts no torque about the hinge: no external torque, so angular momentum about the hinge is conserved	✓ 1
The hinge exerts an external (impulsive) force on the door during the collision, so linear momentum is not conserved	✓ 1
Part (a)(ii)
I = ⅓ × 12.0 × 0.800² = 2.56 kg m²	✓ 1
Part (a)(iii)
Initial angular momentum = mvr = 0.600 × 8.00 × 0.700 = 3.36 kg m² s⁻¹	✓ 1
Total moment of inertia after = 2.56 + 0.600 × 0.700² = 2.85 kg m²	✓ 1
ω = 3.36/2.854 = 1.18 rad s⁻¹	✓ 1	Allow ECF.
Part (a)(iv)
Initial kinetic energy = ½ × 0.600 × 8.00² = 19.2 J; final = ½ × 2.854 × 1.177² = 1.98 J	✓ 1
Energy transferred = 17.2 J	✓ 1	Accept 17.1–17.3 J; allow ECF.
Part (b)(i)
ω² = 2αθ with θ = π/2: α = 1.177²/π = 0.441 rad s⁻²	✓ 1
τ = Iα = 2.854 × 0.441 = 1.26 N m	✓ 1	Accept 1.2–1.3 N m; allow ECF.
Part (b)(ii)
t = ω/α = 1.177/0.441 = 2.67 s	✓ 1	Or t = 2θ/ω; accept 2.6–2.7 s.
Part (c)(i)
Linear momentum after = ω(0.600 × 0.700 + 12.0 × 0.400) = 1.177 × 5.22 = 6.15 kg m s⁻¹	✓ 1	Centre of mass of the door is 0.400 m from the hinge.
Impulse = 6.15 − 4.80 = 1.3 N s (1.35 N s) in the direction of the beanbag’s original motion	✓ 1	Accept 1.3–1.4 N s; direction required.

Answers: (a)(iii) 1.18 rad s⁻¹ · (a)(iv) 17.2 J · (b)(i) 1.26 N m · (b)(ii) 2.67 s · (c)(i) 1.35 N s (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — angular momentum L = Iω and its conservation when no external torque acts; coupled bodies; Newton’s second law for rotation τ = Iα and the equations for uniform angular acceleration; rotational kinetic energy; A.2 — impulse and the conservation of linear momentum only in the absence of external forces; A.3 — kinetic energy Command term: Determine

48A-2-39

Conservation of angular momentum·A.4 Rigid body mechanics (HL)

Paper 2Hard14 marks

Short answer & extended response10 steps to full marksDetermine

Tides raised by the Moon exert a resultant torque on the rotating Earth, so that the length of the day increases by 2.3 ms every century. Treat the Earth and the Moon as an isolated system in which the Moon, a point mass, moves in a circular orbit about the centre of the Earth.

Data: mass of the Earth 5.97 × 10²⁴ kg; radius of the Earth 6.37 × 10⁶ m; moment of inertia of the Earth about its axis 0.33MR²; period of rotation of the Earth 8.62 × 10⁴ s; mass of the Moon 7.35 × 10²² kg; orbital radius of the Moon 3.84 × 10⁸ m.

(a)

(i)

Show that the angular momentum of the Earth due to its rotation is about 6 × 10³³ kg m² s⁻¹.

(2)

(ii)

Determine the average torque that the tides exert on the Earth.

(3)

(iii)

Determine the average rate at which the rotational kinetic energy of the Earth decreases.

(2)

(b)

(i)

Show that the orbital angular momentum of the Moon is L_M = m_M√(GM_Er), where m_M is the mass of the Moon, M_E the mass of the Earth and r the orbital radius.

(2)

(ii)

The angular momentum lost by the Earth is gained by the Moon. Determine the increase in the orbital radius of the Moon in one year.

(3)

(iii)

State and explain whether the orbital speed of the Moon increases or decreases as it gains angular momentum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
I = 0.33 × 5.97 × 10²⁴ × (6.37 × 10⁶)² = 8.0 × 10³⁷ kg m²	✓ 1
L = Iω = 8.0 × 10³⁷ × 2π/8.62 × 10⁴ = 5.8 × 10³³ kg m² s⁻¹	✓ 1	An answer to at least 2 s.f. is needed.
Part (a)(ii)
Fractional change of the period = fractional change of ω: Δω = 7.29 × 10⁻⁵ × 2.3 × 10⁻³/8.62 × 10⁴ = 1.95 × 10⁻¹² rad s⁻¹ per century	✓ 1	Accept a direct calculation of 2π/8.62 × 10⁴ − 2π/(8.62 × 10⁴ + 2.3 × 10⁻³).
Angular impulse ΔL = IΔω = 8.0 × 10³⁷ × 1.95 × 10⁻¹² = 1.6 × 10²⁶ kg m² s⁻¹	✓ 1	Allow ECF from (a)(i).
τ = ΔL/Δt = 1.56 × 10²⁶/(100 × 3.16 × 10⁷) = 4.9 × 10¹⁶ N m	✓ 1	Accept 4.8–5.0 × 10¹⁶ N m. Using one year for Δt scores [2 max].
Part (a)(iii)
ΔE_k = ½I(ω₁² − ω₂²) = ½I(ω₁ + ω₂)(ω₁ − ω₂) ≈ IωΔω = 5.83 × 10³³ × 1.95 × 10⁻¹² = 1.1 × 10²² J per century	✓ 1	Allow ECF from (a)(i) and (a)(ii).
Rate = 1.13 × 10²²/3.16 × 10⁹ = 3.6 × 10¹² W	✓ 1	Accept 3.5–3.7 × 10¹² W. Calculating ½Iω² before and after needs at least 9 s.f.; accept a correct result.
Part (b)(i)
The gravitational force provides the centripetal force: GM_Em_M/r² = m_Mv²/r, so v = √(GM_E/r)	✓ 1
L_M = Iω = (m_Mr²)(v/r) = m_Mvr = m_M√(GM_Er)	✓ 1	The point-mass moment of inertia, or mvr, must be seen.
Part (b)(ii)
L_M = 7.35 × 10²² × √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 3.84 × 10⁸) = 2.9 × 10³⁴ kg m² s⁻¹	✓ 1
L_M ∝ √r, so Δr/r = 2ΔL/L_M, with ΔL = τ × 1 year = 4.93 × 10¹⁶ × 3.16 × 10⁷ = 1.6 × 10²⁴ kg m² s⁻¹	✓ 1	Allow ECF from (a)(ii) and (b)(i). Accept a method that finds r from L_M + ΔL directly.
Δr = 2 × 3.84 × 10⁸ × 1.56 × 10²⁴/2.87 × 10³⁴ = 4.2 × 10⁻² m (about 4 cm)	✓ 1	Accept 4.0–4.4 cm. Omitting the factor 2 gives 2.1 cm and scores [2 max].
Part (b)(iii)
It decreases	✓ 1	Award this mark only with a valid reason.
v = √(GM_E/r) falls as r increases; L_M = m_Mvr ∝ √r still increases because r increases faster than v decreases	✓ 1	Allow ECF from (b)(i).

Answers: (a)(i) 5.8 × 10³³ kg m² s⁻¹ · (a)(ii) 4.9 × 10¹⁶ N m · (a)(iii) 3.6 × 10¹² W · (b)(ii) 4.2 cm · (b)(iii) decreases (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that an extended body rotating with an angular speed has an angular momentum L as given by L = Iω; that angular momentum remains constant unless the body is acted upon by a resultant torque; that the action of a resultant torque constitutes an angular impulse ΔL = τΔt = Δ(Iω); the kinetic energy of rotational motion E_k = ½Iω² = L²/2I; the moment of inertia for a system of point masses I = Σmr²; D.1 — the gravitational force as the centripetal force for a circular orbit, v = √(GM/r) Command term: Determine

49A-2-55

Rotational dynamics of a washing-machine drum·A.4 Rigid body mechanics (HL)

Paper 2Easy13 marks

Short answer & extended response10 steps to full marksDetermine

The drum of a washing machine rotates about a horizontal axis. The moment of inertia of the empty drum about this axis is 0.34 kg m². During the spin cycle, 4.0 kg of wet clothes are pressed against the inside wall of the drum, 0.24 m from the axis; treat the clothes as a thin layer at this radius. The drum accelerates uniformly from rest to 1200 revolutions per minute in 15 s.

(a)

(i)

Show that the final angular speed of the drum is about 126 rad s⁻¹.

(1)

(ii)

Calculate the angular acceleration of the drum.

(1)

(iii)

Determine the number of revolutions the drum makes while it accelerates.

(2)

(b)

(i)

Show that the moment of inertia of the drum and clothes is about 0.57 kg m².

(2)

(ii)

Calculate the average resultant torque on the drum and clothes while they accelerate.

(1)

(iii)

Determine the rotational kinetic energy of the drum and clothes at 1200 revolutions per minute and the average useful power output of the motor during the spin-up.

(2)

(c)

(i)

A sock of mass 0.050 kg is pressed against the drum wall at full speed. Determine the force that the drum wall exerts on the sock and compare it with the weight of the sock.

(2)

(ii)

The drum wall has many small holes. Explain how the spin cycle removes water from the clothes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
ω = 1200 × 2π/60 = 125.7 rad s⁻¹	✓ 1	Must see the conversion with 2π and 60, or 125.7.
Part (a)(ii)
α = 125.7/15 = 8.4 rad s⁻²	✓ 1	Allow ECF from (a)(i). Accept 8.38 rad s⁻².
Part (a)(iii)
Δθ = ½(ω_i + ω_f)t = ½ × 125.7 × 15 = 943 rad	✓ 1	Allow ECF from (a)(i) or (a)(ii). Accept ½αt².
Number of revolutions = 943/2π = 150	✓ 1	Accept 150 directly from ½ × 20 rev s⁻¹ × 15 s.
Part (b)(i)
Clothes: I = Σmr² = 4.0 × 0.24² = 0.230 kg m²	✓ 1	The clothes are all at the same distance from the axis.
Total = 0.34 + 0.230 = 0.570 kg m²	✓ 1	Must see 0.570 or the full substitution.
Part (b)(ii)
τ = Iα = 0.570 × 8.38 = 4.8 N m	✓ 1	Allow ECF from (a)(ii) and (b)(i).
Part (b)(iii)
E_k = ½ × 0.570 × 125.7² = 4.5 × 10³ J	✓ 1	Allow ECF from (a)(i) and (b)(i).
Average power = 4.5 × 10³/15 = 3.0 × 10² W	✓ 1	Allow ECF from the energy. Accept 300 W.
Part (c)(i)
Centripetal force = mω²r = 0.050 × 125.7² × 0.24 = 1.9 × 10² N	✓ 1	Allow ECF from (a)(i). The force from the wall provides the centripetal force (the weight is negligible).
Weight = 0.050 × 9.81 = 0.49 N, so the force is about 400 times the weight	✓ 1	Accept any valid comparison (ratio 380–400).
Part (c)(ii)
Water at a hole is not in contact with the drum wall (or the clothes no longer push on it), so no centripetal force acts on it	✓ 1	Do not accept "centrifugal force throws the water out".
With no resultant force it moves in a straight line, along the tangent to its circular path (Newton’s first law), so it leaves through the holes	✓ 1	Reference to straight-line (tangential) motion needed.

Answers: (a)(ii) 8.4 rad s⁻² · (a)(iii) 150 · (b)(ii) 4.8 N m · (b)(iii) 4.5 × 10³ J; 3.0 × 10² W · (c)(i) 1.9 × 10² N, about 400 times its weight (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration; that equations of motion for uniform angular acceleration can be used to predict the body’s angular position θ, angular displacement Δθ, angular speed ω and angular acceleration α; the moment of inertia for a system of point masses as given by I = Σmr²; Newton’s second law for rotation as given by τ = Iα where τ is the average torque; the kinetic energy of rotational motion as given by E_k = ½Iω² = L²/2I; A.2 — centripetal force F = mω²r; Newton’s first law; A.3 — power as the rate of work done Command term: Determine

50A-2-56

Rotational equilibrium: tipping or sliding·A.4 Rigid body mechanics (HL)

Paper 2Medium12 marks

Short answer & extended response10 steps to full marksDetermine

A uniform cabinet of mass 60.0 kg is 1.80 m tall and 0.50 m deep. It stands on a level floor. A horizontal force F, perpendicular to its back face, is applied at a height h above the floor, as shown. E is the lower front edge of the cabinet. The coefficient of static friction between the cabinet and the floor is 0.40 and the coefficient of dynamic friction is 0.30.

Side view of the cabinet (not to scale).

(a)

(i)

Show that the maximum frictional force that the floor can exert on the cabinet is about 235 N.

(1)

(ii)

The force is applied at h = 1.50 m and slowly increased from zero. Determine the value of F at which the cabinet is about to tip about E.

(2)

(iii)

Deduce whether, as F increases from zero, the cabinet first tips or first slides.

(1)

(b)

(i)

Show that, for a cabinet of depth w, the cabinet slides before it tips if h < w/(2μ_s).

(2)

(ii)

Calculate the greatest height at which the force can be applied if the cabinet is to slide without tipping.

(1)

(c)

(i)

The cabinet is now pushed at h = 0.40 m and slides across the floor at a constant speed of 0.60 m s⁻¹. Determine the rate at which the person pushing does work on the cabinet.

(2)

(ii)

While the cabinet slides as described in (c)(i), the resultant normal force from the floor does not act through the centre of the base. Determine the distance from E of the line of action of the resultant normal force.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
F_f = μ_sF_N = 0.40 × 60.0 × 9.81 = 235.4 N	✓ 1	Must see the full substitution or 235.4. The normal force equals the weight because F is horizontal.
Part (a)(ii)
When the cabinet is about to tip, the normal force and the friction act at E, so they have no torque about E; then Fh = mg × (0.50/2)	✓ 1	Recognition that the back of the cabinet loses contact, so the floor forces act at E, is needed for this mark.
F = 60.0 × 9.81 × 0.25/1.50 = 98 N	✓ 1	Accept 98.1 N.
Part (a)(iii)
It tips: 98 N is less than the 235 N needed to make it slide, so the tipping condition is reached first	✓ 1	Allow ECF from (a)(i) and (a)(ii).
Part (b)(i)
The force needed to tip it is mgw/(2h); the force needed to make it slide is μ_smg	✓ 1	Symbolic working required; Allow ECF from the method of (a)(ii).
It slides first if μ_smg < mgw/(2h), which gives h < w/(2μ_s)	✓ 1	The inequality must be correctly reversed when rearranged.
Part (b)(ii)
h = 0.50/(2 × 0.40) = 0.625 m	✓ 1	Allow ECF from (b)(i). Accept 0.63 m.
Part (c)(i)
At constant speed F = dynamic friction = 0.30 × 60.0 × 9.81 = 177 N	✓ 1	Using the coefficient of static friction (235 N, giving 141 W) scores 0 for this mark.
P = Fv = 177 × 0.60 = 1.1 × 10² W	✓ 1	Allow ECF from the force. Accept 106 W.
Part (c)(ii)
The cabinet does not rotate, so the resultant torque about E is zero; the friction acts along the floor through E and the normal force equals the weight, 589 N	✓ 1	Allow torques about any point, e.g. the centre of mass, if all four forces are included.
Fh + F_Nx = mg × 0.25: 177 × 0.40 + 589x = 589 × 0.25	✓ 1	Allow ECF from (c)(i). Correct perpendicular distances are needed.
x = (147 − 70.6)/589 = 0.13 m from E (0.12 m in front of the centre of the base)	✓ 1	Accept 0.12–0.14 m. Award [2 max] for 0.12 m given as the distance from E.

Answers: (a)(ii) 98 N · (a)(iii) it tips · (b)(ii) 0.625 m · (c)(i) 1.1 × 10² W · (c)(ii) 0.13 m (the remaining parts are explanations — see the table above)

51A-2-57

Coupled rotating bodies: a bicycle drive·A.4 Rigid body mechanics (HL)

Paper 2Medium12 marks

Short answer & extended response10 steps to full marksDetermine

The diagram shows the drive of a bicycle. A rider pushes on a pedal with a force of 300 N perpendicular to the crank, which is 0.170 m long. The crank turns a chainring of radius 0.085 m. A chain transmits the motion to a sprocket of radius 0.034 m that is fixed to the rear wheel, of radius 0.340 m. The chain does not slip, and friction in the bearings and chain is negligible.

The bicycle drive (not to scale).

(a)

(i)

Calculate the torque of the pedal force about the axle of the crank.

(1)

(ii)

Assume that the chainring is in rotational equilibrium and that the rotational inertia of the rear wheel is negligible. Determine the tension in the upper part of the chain, the torque that the chain exerts on the rear wheel and the horizontal force that the road exerts on the rear tyre.

(3)

(b)

(i)

The rider turns the pedals at 90 revolutions per minute. Show that the speed of the bicycle is about 8.0 m s⁻¹.

(2)

(ii)

Show that, at this speed, the rate at which the rider does work on the pedal is equal to the rate at which the road force in (a)(ii) does work on the bicycle.

(2)

(c)

(i)

Explain why the rotation of the wheels reduces the acceleration of the bicycle when it starts from rest, compared with a body of the same total mass that does not rotate.

(2)

(ii)

Each wheel has a moment of inertia of 0.12 kg m² and the total mass of the bicycle and rider is 80 kg. Determine the fraction of the total kinetic energy that is rotational when the bicycle moves at the speed in (b)(i).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
τ = Fr sin 90° = 300 × 0.170 = 51 N m	✓ 1
Part (a)(ii)
Only the upper part of the chain is under tension: T × 0.085 = 51, so T = 6.0 × 10² N	✓ 1	Allow ECF from (a)(i).
Torque on the rear wheel = 600 × 0.034 = 20 N m	✓ 1	Allow ECF from the tension. Accept 20.4 N m.
F = 20.4/0.340 = 60 N (forwards)	✓ 1	Allow ECF from the torque. Using the radius of the chainring or of the sprocket instead of the wheel scores 0 for this mark.
Part (b)(i)
Crank: ω = 90 × 2π/60 = 9.42 rad s⁻¹; chain speed = 9.42 × 0.085 = 0.801 m s⁻¹, which is also the rim speed of the sprocket	✓ 1	The chain links the two rims, so their speeds (not their angular speeds) are equal.
Rear wheel: ω = 0.801/0.034 = 23.6 rad s⁻¹; v = 23.6 × 0.340 = 8.01 m s⁻¹	✓ 1	Must see 8.01 or the full substitution.
Part (b)(ii)
Speed of the pedal = 9.42 × 0.170 = 1.60 m s⁻¹, so the rider’s power = 300 × 1.60 = 481 W	✓ 1	Allow ECF from (b)(i).
Road force: P = Fv = 60 × 8.01 = 481 W, the same (no energy is dissipated in the drive)	✓ 1	Allow ECF from (a)(ii) and (b)(i).
Part (c)(i)
Part of the work done by the rider is transferred to the rotational kinetic energy ½Iω² of the wheels	✓ 1	Or: part of the driving torque is needed to give the wheels angular acceleration (τ = Iα).
So for the same work done less becomes translational kinetic energy: the bicycle reaches a smaller speed in the same distance, so its acceleration is smaller	✓ 1	Or: so the forward force on the bicycle is less than 60 N. Award only if linked to the first point.
Part (c)(ii)
Rotational: 2 × ½ × 0.12 × (8.01/0.340)² = 67 J; translational: ½ × 80 × 8.01² = 2.57 × 10³ J	✓ 1	Allow ECF from (b)(i). Both wheels must be included; ω = v/R for rolling without slipping.
Fraction = 67/(67 + 2567) = 0.025 (2.5 %)	✓ 1	Accept 0.025–0.026. Accept 67/2567 = 0.026.

Answers: (a)(i) 51 N m · (a)(ii) 600 N; 20 N m; 60 N · (b)(ii) 481 W · (c)(ii) 0.025 (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — the torque τ of a force about an axis as given by τ = Fr sin θ; that bodies in rotational equilibrium have a resultant torque of zero (guidance: situations include coupled pairs of bodies); that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration; the kinetic energy of rotational motion as given by E_k = ½Iω² = L²/2I (guidance: simultaneous rotational and translational motion is restricted to rolling without slipping); A.2 — v = ωr; A.3 — power P = Fv; conservation of energy Command term: Determine

52A-2-58

Rolling without slipping: a cable reel·A.4 Rigid body mechanics (HL)

Paper 2Hard12 marks

Short answer & extended response11 steps to full marksDetermine

A cable reel of mass 120 kg stands on level ground. Its two end flanges, of radius R = 0.60 m, roll on the ground without slipping. The cable is wound on a central core of radius r = 0.25 m. The moment of inertia of the reel and cable about the axle is 26 kg m². A worker pulls the free end of the cable horizontally with a constant force F = 150 N. Rolling resistance and air resistance are negligible, and the mass of the unwound cable can be ignored.

In case 1 the cable leaves the top of the core; in case 2 it leaves the bottom of the core, as shown.

Side view of the cable reel (not to scale).

(a)

(i)

For case 1, show that the acceleration of the axle of the reel is a = F(1 + r/R)/(M + I/R²), where M is the mass of the reel and I its moment of inertia. Take the frictional force f from the ground to act in the direction of F.

(2)

(ii)

Calculate a for case 1.

(1)

(iii)

Determine the magnitude and direction of the frictional force that the ground exerts on the reel in case 1.

(2)

(b)

(i)

Determine the acceleration of the axle, including its direction, in case 2.

(2)

(ii)

In case 2 the worker walks backwards so that the cable stays taut. Deduce whether the cable winds onto the core or unwinds from it.

(2)

(c)

(i)

In case 1 the reel starts from rest. Show that, when the axle has moved 2.0 m, the work done by the worker equals the total kinetic energy of the reel.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Translation: F + f = Ma; rotation about the axle: Fr − fR = Iα with α = a/R (rolling without slipping)	✓ 1	Both equations and the rolling condition are needed. Symbolic working required.
Eliminating f: f = Fr/R − Ia/R², so F(1 + r/R) = a(M + I/R²), which gives the result	✓ 1
Part (a)(ii)
a = 150 × (1 + 0.25/0.60)/(120 + 26/0.60²) = 212.5/192.2 = 1.1 m s⁻²	✓ 1	Allow ECF from (a)(i). Accept 1.10–1.11 m s⁻².
Part (a)(iii)
f = Ma − F = 120 × 1.105 − 150 = −17 N	✓ 1	Allow ECF from (a)(ii). Accept 16–18 N in magnitude.
The negative sign means the friction acts opposite to F (away from the worker), magnitude 17 N	✓ 1	Direction must be consistent with the sign of the candidate’s value.
Part (b)(i)
The force on the bottom of the core now has a torque about the axle in the opposite sense: replace r by −r, so a = F(1 − r/R)/(M + I/R²)	✓ 1	Allow ECF from the method of (a)(i). Equations written afresh for case 2 are equally acceptable.
a = 150 × 0.583/192.2 = 0.46 m s⁻², positive: the reel still rolls towards the worker	✓ 1	Accept 0.45–0.46 m s⁻². The direction is required.
Part (b)(ii)
Rolling without slipping: the bottom of the core moves forwards at v − ωr = v(1 − r/R), less than the speed v of the axle; the cable (and the worker’s hand) moves at this lower speed	✓ 1	Allow an argument based on the instantaneous rotation about the contact point P: points of the core below the axle move more slowly than the axle.
So the axle gains on the worker, the free length of cable decreases and the cable winds onto the core	✓ 1	Allow ECF from (b)(i) for the direction of motion of the reel. Award only with a valid speed comparison. "It unwinds because the cable is pulled" scores 0.
Part (c)(i)
The top of the core moves at v(1 + r/R), so the worker’s hand moves 2.0 × 1.417 = 2.83 m and does work 150 × 2.83 = 425 J	✓ 1	Using 2.0 m for the distance moved by the force (300 J) scores 0 for this mark.
v = √(2 × 1.105 × 2.0) = 2.10 m s⁻¹; E_k = ½ × 120 × 2.10² + ½ × 26 × (2.10/0.60)² = 265 + 160 = 425 J	✓ 1	Allow ECF from (a)(ii).
The two are equal: the friction does no work because the point of contact P is momentarily at rest	✓ 1	Award for the comparison together with the reason.

Answers: (a)(ii) 1.1 m s⁻² · (a)(iii) 17 N, opposite to F · (b)(i) 0.46 m s⁻² towards the worker · (b)(ii) it winds onto the core (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — Newton’s second law for rotation as given by τ = Iα where τ is the average torque; that an unbalanced torque applied to an extended, rigid body will cause angular acceleration; the torque τ of a force about an axis as given by τ = Fr sin θ; the kinetic energy of rotational motion as given by E_k = ½Iω² = L²/2I (guidance: simultaneous rotational and translational motion is restricted to rolling without slipping); A.2 — Newton’s second law; static friction; A.3 — work done by a force; conservation of energy Command term: Determine

53A-2-59

Angular momentum of coupled bodies: a reaction wheel·A.4 Rigid body mechanics (HL)

Paper 2Hard14 marks

Short answer & extended response12 steps to full marksDetermine

A satellite turns to point its telescope in a new direction by using a reaction wheel. The wheel, of moment of inertia 0.050 kg m², is driven by an electric motor fixed to the body of the satellite. The wheel’s axle passes through the centre of mass of the satellite, and the moment of inertia of the satellite body (without the wheel) about this axis is 150 kg m². Initially the wheel and the satellite are both at rest, and no external torque acts on the satellite.

The graph shows how the angular velocity ω of the wheel, measured relative to the distant stars, varies with time t during one manoeuvre.

Angular velocity ω of the reaction wheel against time t (drawn to scale).

(a)

(i)

Explain why the satellite body starts to rotate, in the opposite sense to the wheel, when the motor speeds up the wheel.

(2)

(ii)

Show that the greatest angular speed of the satellite body is about 0.04 rad s⁻¹.

(2)

(iii)

Determine the torque exerted by the motor on the wheel during the first 10 s.

(2)

(b)

(i)

Determine the total angle, in degrees, through which the satellite body turns during the manoeuvre.

(3)

(ii)

State the angular speed of the satellite body at t = 45 s, and explain your answer.

(1)

(c)

(i)

Show that, while the wheel spins at 120 rad s⁻¹, (rotational kinetic energy of the body)/(rotational kinetic energy of the wheel) = (moment of inertia of the wheel)/(moment of inertia of the body), and evaluate this ratio.

(2)

(ii)

The motor is supplied from the 28 V electrical system of the satellite and is 75 % efficient. Determine the average current in the motor during the first 10 s.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The motor exerts a torque on the wheel and, by Newton’s third law, the wheel exerts an equal and opposite torque on the satellite body (through the motor mounting)	✓ 1	Accept: the torques between wheel and body are internal to the system.
No external torque acts, so the total angular momentum stays zero: the angular momentum gained by the wheel in one sense is matched by equal angular momentum of the body in the opposite sense	✓ 1	Must refer to the conservation of (zero) total angular momentum.
Part (a)(ii)
Greatest angular momentum of the wheel = 0.050 × 120 = 6.0 kg m² s⁻¹ (read from the graph)	✓ 1
ω_body = 6.0/150 = 0.040 rad s⁻¹	✓ 1	Allow ECF from the angular momentum in the first marking point. Must see 0.040 or the full substitution.
Part (a)(iii)
α = gradient of the graph = 120/10 = 12 rad s⁻²	✓ 1
τ = Iα = 0.050 × 12 = 0.60 N m	✓ 1	Accept ΔL/Δt = 6.0/10.
Part (b)(i)
At every instant the angular momenta are equal and opposite, so ω_body = (0.050/150)ω_wheel = ω_wheel/3000	✓ 1	Allow ECF from (a)(ii) for the ratio of moments of inertia.
Angle turned by the wheel = area under the graph = ½ × 10 × 120 + 20 × 120 + ½ × 10 × 120 = 3600 rad	✓ 1	Accept 3.6 × 10³ rad.
Angle turned by the body = 3600/3000 = 1.2 rad = 69°	✓ 1	Allow ECF from the area. Accept 68–70°.
Part (b)(ii)
Zero: the total angular momentum is still zero and the wheel is at rest, so the body must also be at rest (although it has turned through a new angle)	✓ 1	Allow ECF from (a)(i).
Part (c)(i)
The body and the wheel have angular momenta of the same magnitude L, and E_k = L²/2I, so E_k ∝ 1/I and the ratio is I_wheel/I_body	✓ 1	Symbolic working required.
Ratio = 0.050/150 = 3.3 × 10⁻⁴ (wheel 360 J, body 0.12 J)	✓ 1	Allow ECF from (a)(ii).
Part (c)(ii)
Energy gained in 10 s ≈ ½ × 0.050 × 120² = 360 J, so the input power = 360/(10 × 0.75) = 48 W	✓ 1	Allow ECF from (c)(i). The kinetic energy of the body (0.12 J) is negligible.
I = P/V = 48/28 = 1.7 A	✓ 1	Accept 1.71 A. Omitting the efficiency (1.3 A) scores [1 max].

Answers: (a)(iii) 0.60 N m · (b)(i) 69° · (b)(ii) zero · (c)(i) 3.3 × 10⁻⁴ · (c)(ii) 1.7 A (the remaining parts are explanations — see the table above)

Syllabus understandingA.4 — that an extended body rotating with an angular speed has an angular momentum L as given by L = Iω; that angular momentum remains constant unless the body is acted upon by a resultant torque (guidance: situations include coupled pairs of bodies); Newton’s second law for rotation as given by τ = Iα where τ is the average torque; the kinetic energy of rotational motion as given by E_k = ½Iω² = L²/2I; that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration; A.2 — Newton’s third law; A.3 — efficiency; B.5 — electrical power P = IV; Tools 3 — area under a graph Command term: Determine

54A-2-66

Rotating space habitat·A.4 Rigid body mechanics (HL)

Paper 2Hard20 marks

Short answer & extended response14 steps to full marksDetermine

A space habitat is a ring of radius 112 m joined to a central hub by spokes. It is in a circular orbit around the Earth. The habitat is made to rotate about its axis so that astronauts standing on the inside of the rim experience the same apparent weight as on the Earth's surface. The moment of inertia of the habitat about its axis is 5.0 × 10⁹ kg m². Mass of the Earth = 5.97 × 10²⁴ kg; radius of the Earth = 6.37 × 10⁶ m.

The habitat seen along its axis of rotation (not to scale). The four thrusters act tangentially at the rim in two opposite pairs.

(a)

(i)

Show that the angular speed of the habitat is about 0.30 rad s⁻¹.

(2)

(ii)

Calculate the rate of rotation of the habitat in revolutions per minute.

(1)

(iii)

An astronaut of height h stands on the rim. Show that the ratio of the centripetal acceleration of her head to that of her feet is 1 − h/R, and hence suggest why habitats are designed with a large radius.

(2)

(iv)

An astronaut standing on the rim releases a spanner from rest relative to her hand, 2.0 m above the floor. By considering the motion of the spanner in a non-rotating frame moving with the centre of the habitat, determine the time the spanner takes to reach the floor.

(3)

(b)

(i)

The habitat is set rotating from rest by four thrusters on the rim, each giving a tangential force of 450 N, as shown. Determine the time taken to reach the angular speed in (a)(i).

(2)

(ii)

Outline why the thrusters are fired in opposite pairs.

(1)

(c)

(i)

With the habitat rotating at the angular speed in (a)(i), all 100 crew members, each of mass 75 kg, move from the rim to the hub for a meeting. Determine the new angular speed of the habitat.

(2)

(ii)

Calculate the apparent value of g now experienced by a person remaining on the rim.

(1)

(d)

(i)

The habitat completes one orbit of the Earth every 92.6 minutes. Determine its height above the Earth's surface.

(3)

(ii)

Explain why the orbital period of the habitat does not depend on the mass of the habitat.

(1)

(iii)

Estimate by how much a clock on the habitat falls behind a clock on the Earth in one year because of the orbital speed of the habitat, and state one assumption you make.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The floor provides the centripetal force: ω²R = g	✓ 1
ω = √(9.81/112) = 0.296 rad s⁻¹	✓ 1	Must see 0.296 or full substitution.
Part (a)(ii)
60 × 0.296/2π = 2.8 rev min⁻¹	✓ 1	Allow ECF from (a)(i). Unit rev min⁻¹ required.
Part (a)(iii)
Same ω for every point: a_head/a_feet = ω²(R − h)/ω²R = 1 − h/R	✓ 1	Symbolic working required.
A large R makes the ratio close to 1, so the variation of apparent weight over the body is small (and the rotation rate needed is lower, reducing dizziness)	✓ 1
Part (a)(iv)
After release no resultant force acts on the spanner (the habitat is in free fall), so it moves in a straight line at its speed when released: ω × (112 − 2.0) = 0.296 × 110 = 32.6 m s⁻¹	✓ 1	The hidden step: Newton’s first law in a non-rotating frame. Allow ECF from (a)(i).
The straight line meets the rim after a distance √(112² − 110²) = 21.1 m	✓ 1	Award for the geometry, e.g. a sketch with the right angle at the release point.
t = 21.1/32.6 = 0.65 s (close to √(2 × 2.0/9.81) = 0.64 s for the same drop on the Earth)	✓ 1	Accept 0.63–0.66 s.
Part (b)(i)
Torque = 4 × 450 × 112 = 2.02 × 10⁵ N m	✓ 1
t = Iω/τ = 5.0 × 10⁹ × 0.296/2.02 × 10⁵ = 7.3 × 10³ s (about 2 hours)	✓ 1	Allow ECF from (a)(i). Accept 7.2–7.5 × 10³ s.
Part (b)(ii)
The forces of each pair are equal and opposite, so the resultant force is zero (the habitat's orbit is not changed) while the torques add	✓ 1
Part (c)(i)
Moment of inertia falls by 100 × 75 × 112² = 9.4 × 10⁷ kg m² (the hub radius is negligible)	✓ 1
ω′ = 5.0 × 10⁹ × 0.296/(5.0 × 10⁹ − 9.4 × 10⁷) = 0.302 rad s⁻¹	✓ 1	Allow ECF from (a)(i). Angular momentum is conserved: no external torque.
Part (c)(ii)
ω′²R = 0.3016² × 112 = 10.2 m s⁻² (about 4 % more than before)	✓ 1	Allow ECF from (c)(i).
Part (d)(i)
Gravitational force provides the centripetal force: r³ = GMT²/4π²	✓ 1
r³ = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (92.6 × 60)²/4π², r = 6.78 × 10⁶ m	✓ 1
Height = 6.78 × 10⁶ − 6.37 × 10⁶ = 4.1 × 10⁵ m (about 410 km)	✓ 1	Accept 4.0–4.2 × 10⁵ m. Forgetting to subtract the radius of the Earth scores [2 max].
Part (d)(ii)
The gravitational force GMm/r² provides the centripetal force m(4π²r/T²); the habitat mass m appears on both sides and cancels, so T depends only on r (and the mass M of the Earth)	✓ 1	A statement that the mass cancels without the force equation scores 0.
Part (d)(iii)
Orbital speed = 2πr/T = 7.66 × 10³ m s⁻¹; γ − 1 ≈ ½(v/c)² = 3.3 × 10⁻¹⁰, so in 3.2 × 10⁷ s the lag is ≈ 1 × 10⁻² s (about 10 ms)	✓ 1	Allow ECF from (d)(i). A calculator value of γ that rounds to 1 scores 0 unless γ − 1 is found.
Assumption: the Earth clock is treated as at rest in an inertial frame (the Earth's rotation and orbit are ignored), or effects of the gravitational field on the clocks are ignored	✓ 1	Any one valid assumption.

Answers: (a)(ii) 2.8 rev min⁻¹ · (a)(iv) 0.65 s · (b)(i) 7.3 × 10³ s · (c)(i) 0.302 rad s⁻¹ · (c)(ii) 10.2 m s⁻² · (d)(i) 4.1 × 10⁵ m · (d)(iii) ≈ 1 × 10⁻² s (the remaining parts are explanations — see the table above)

Syllabus understandingA.2 — circular motion and centripetal acceleration a = ω²r; Newton's first law; that bodies moving along a circular trajectory at a constant speed have a speed v = ωr; A.4 — torque; Newton's second law for rotation; conservation of angular momentum L = Iω; A.5 — time dilation; D.1 — Newton's universal law of gravitation and orbital motion; Kepler's third law Command term: Determine

A.4 Rigid body mechanics: IB Physics HL exam-style questions

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