105 free, high-quality past-paper-style questions on Rigid body mechanics (HL) for IB Physics HL, each with a full mark scheme. Covers: Conservation of angular momentum; Rotational kinematics; Moment of inertia; and more. For when you have finished the past papers, or want to practise one topic at a time.Read the A.4 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.520 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1A-1A-12Conservation of angular momentumPaper 1AEasy1 mark
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?
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Step 1Moment of inertia of the ring: MR² = 0.50 × 0.20² = 0.020 kg m².
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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.
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Step 30.060 × 4.0 = (0.060 + 0.020)ω, so ω = 0.24/0.080 = 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.
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)?
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Step 1Rotational kinetic energy is Ek = ½Iω², so at the same ω it is proportional to I.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = (ML²/3)/(ML²/12) = 12/3 = 4.
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Step 3About the end axis the mass is, on average, further from the axis, so the same angular speed needs four times the energy.
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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, Ek ∝ I, and (ML²/3)/(ML²/12) = 4.
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
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Step 1τ = Fr sin θ, where θ is the angle between the force and the line from the axle to the pedal.
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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.
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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.
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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.
Two uniform solid discs, A and B, have the same mass. Disc B has twice the radius of disc A. The same torque acts on each disc about its central axis. What is \(\frac{\alpha_B}{\alpha_A}\), the ratio of the angular acceleration of B to that of A?
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For a uniform disc, I = ½MR2. Doubling R with M unchanged makes I four times larger. Since α = τ/I and τ is the same, αB is one quarter of αA. Option C assumes I ∝ R. Options A and B assume that a larger disc accelerates more.
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Answer D
Answer: D
Common error: Moment of inertia depends on R², not R. Doubling the radius at constant mass quadruples I.
A reaction wheel in a small satellite is initially at rest. A motor exerts a constant torque of 0.40 N m on the wheel for 3.0 s. What is the change in the angular momentum of the wheel?
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Angular impulse equals the change in angular momentum: ΔL = τΔt = 0.40 × 3.0 = 1.2 kg m2 s−1. Option A divides τ by t. Option C wrongly uses ½τt2, which has the form of an angular displacement. Option D divides t by τ.
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Answer B
Answer: B
Exam tip: Angular impulse τΔt is the rotational analogue of FΔt. You do not need the moment of inertia to find ΔL.
8IBP2H-007-00-01Torque and its sensePaper 1AEasy1 mark
A technician tightens a bolt with a spanner. She applies a force of 40 N at a point 0.25 m from the centre of the bolt. The force acts at 30° to the line of the spanner handle. What is the magnitude of the torque about the bolt?
Spanner on a bolt, seen from above (not to scale)
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Torque = rF sin θ, where θ is the angle between the force and the line from the axis to the point where the force acts. This gives τ = 0.25 × 40 × sin 30° = 5.0 N m. Only the component of the force perpendicular to the handle (40 sin 30° = 20 N) produces a turning effect. Using cos 30° gives 8.7 N m, ignoring the angle gives 10 N m, and dividing F by r gives 160 N m.
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Answer A
Answer: A
Common error: The angle in τ = rF sin θ is measured between the force and the handle. If the force is given at an angle to the handle, use sin, not cos.
9IBP2H-007-00-06Torque and its sensePaper 1AEasy1 mark
Three forces act in the plane of a wheel that can turn about a fixed axle. Force 1: 10 N tangential to the rim at radius 0.50 m, turning the wheel clockwise. Force 2: 30 N tangential at radius 0.20 m, turning the wheel anticlockwise. Force 3: 20 N applied at the rim, directed radially towards the axle. What is the resultant torque on the wheel about the axle?
Forces on the wheel, seen from the front (not to scale)
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Force 1 gives a clockwise torque of 10 × 0.50 = 5.0 N m. Force 2 gives an anticlockwise torque of 30 × 0.20 = 6.0 N m. Force 3 acts along a line through the axle, so its perpendicular distance from the axle is zero and its torque is zero. The resultant is 6.0 − 5.0 = 1.0 N m anticlockwise. Option C wrongly gives the radial force a torque of 10 N m. Option D adds torques that act in opposite senses.
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Answer A
Answer: A
Exam tip: Give every torque a sign for its sense before adding. A force whose line of action passes through the axis produces no torque.
10IBP2H-007-02-01Angular momentumPaper 1AEasy1 mark
A student sits on a freely rotating office chair holding a heavy dumbbell in each outstretched hand. While rotating, the student pulls the dumbbells in close to the chest. Friction at the chair bearing is negligible. What happens to the angular momentum L and the rotational kinetic energy Ek of the student–chair system?
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No external torque acts about the axis, so L = Iω is conserved. Pulling the masses in reduces I, so ω rises. Since Ek = L2/2I with L constant, a smaller I gives a larger Ek. The extra energy comes from work done by the student's arms. Option A wrongly assumes kinetic energy is conserved alongside L; option C assumes faster spin means more angular momentum; option D confuses the effect of a smaller I.
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Answer B
Answer: B
Common error: Angular momentum conservation does not imply kinetic energy conservation. Use Ek = L²/2I to see how Ek changes when I changes at constant L.
11IBP2H-007-02-02Moment of inertiaPaper 1AEasy1 mark
A wheelchair-racing engineer compares four objects, each of mass M and outer radius R: a solid disc, a thin ring, a solid sphere and a thin-walled hollow sphere. The disc and the ring each rotate about an axis through their centre, perpendicular to their plane. Each sphere rotates about an axis along a diameter. Which object has the largest moment of inertia about its stated axis?
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Moment of inertia depends on how far the mass is from the axis. In a thin ring rotating about its perpendicular central axis, all the mass is at distance R. This gives I = MR2, the largest possible value for this mass and radius. The hollow sphere (2/3 MR2) has some of its mass close to the axis, near the poles. The disc (1/2 MR2) and the solid sphere (2/5 MR2) have their mass spread from the axis outwards. Choosing the hollow sphere comes from thinking that 'hollow' always means all the mass is at the full radius from the axis.
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Answer B
Answer: B
Exam tip: For equal mass and radius, rank shapes by how far their mass sits from the rotation axis, not by whether they are hollow. Always check which axis is stated.
12IBP3H-A4-20r5-01Torque and its sensePaper 1AEasy1 mark
A force of 20 N is applied to a uniform rod at a point 0.30 m from a fixed pivot. The force acts at 30° to the rod. What is the magnitude of the torque of the force about the pivot?
Rod and pivot, seen from above (not to scale)
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Torque = Fr sin θ = 20 × 0.30 × sin 30° = 3.0 N m, so C is correct. A (6.0 N m) ignores the angle. B (5.2 N m) uses cos 30° instead of sin 30°, taking the wrong component. D (1.5 N m) applies sin 30° twice.
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Answer C
Answer: C
Common error: Torque uses the force component perpendicular to the line from the pivot: with the angle measured from the rod, that is F sin θ.
13IBP3H-A4-20r5-02Angular momentumPaper 1AEasy1 mark
A skater spins on ice with arms outstretched and then pulls both arms in close to the body. Friction with the ice is negligible. Which row describes the skater's angular velocity ω and rotational kinetic energy Ek after the arms are pulled in?
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No external torque acts, so L = Iω is constant and ω rises as I falls. Since Ek = L2/2I, a smaller I gives a larger Ek because the skater's muscles do work, so B is correct. A wrongly treats energy as conserved. C has the energy falling, which contradicts L2/2I. D is wrong because ω must change when I changes at constant L.
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Answer B
Answer: B
Common error: Angular momentum being conserved does not mean rotational kinetic energy is conserved; check the energy separately.
14IBP3H-A4-20r24-01Rotational kinematicsPaper 1AEasy1 mark
The rotor of a wind turbine has moment of inertia 2.4 × 106 kg m2 and rotates at a steady 15 revolutions per minute. What is the angular momentum of the rotor?
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Angular speed ω = 15 × 2π / 60 = 1.57 rad s−1, so L = Iω = 2.4 × 106 × 1.57 = 3.77 × 106 kg m2 s−1. Using 15 as if it were already in rad s−1 gives 3.6 × 107. Converting only minutes to seconds (15/60 = 0.25) and omitting 2π gives 6.0 × 105. Multiplying by 2π as well as 15 without dividing by 60 gives 2.26 × 108.
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Answer B
Answer: B
Common error: Angular momentum formulae need ω in rad s⁻¹; convert revolutions per minute by multiplying by 2π and dividing by 60.
15IBP3H-A4-20r24-02Rotational equilibriumPaper 1AEasy1 mark
A library door is 0.90 m wide and hinged along one vertical edge. A person pushes on the handle at the outer edge with a force of 40 N. The force makes an angle of 30° with the plane of the door. What is the torque of the force about the hinge?
Plan view of the door; the force makes 30° with the plane of the door (not to scale)
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Only the component of force perpendicular to the door produces torque: τ = Fr sin θ = 40 × 0.90 × sin 30° = 18 N m. Using the full force gives 36 N m. Using the component along the door (cos 30°) gives 31 N m. Dividing by the lever arm instead of multiplying (40 / 0.90) gives 44 N m, which also has the wrong units.
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Answer C
Answer: C
Exam tip: Check which angle is given: sin θ applies when θ is between the force and the line from the pivot to the point of application.
16IBP3H-A4-20r24-08Rolling and rotational kinetic energyPaper 1AEasy1 mark
A library trolley of mass 40 kg moves at a constant 1.5 m s−1 along a straight aisle. Point P is 3.0 m from the line of the aisle, measured perpendicular to it. What is the magnitude of the trolley's angular momentum about P?
Plan view of the aisle (not to scale)
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For a particle, L = mvd where d is the perpendicular distance from P to the line of motion: L = 40 × 1.5 × 3.0 = 180 kg m² s−1. The value 60 is the linear momentum with no distance. The value 270 comes from squaring the speed (mv²d). The value 80 comes from dividing by the speed (md/v).
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Answer B
Answer: B
Exam tip: A body moving in a straight line has angular momentum about a point even though it is not rotating; use the perpendicular distance from the point to the line of motion.
17A-1A-05Rolling without slippingPaper 1AMedium1 mark
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
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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.
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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.
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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).
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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.
19A-1A-17Rotational equilibriumPaper 1AMedium1 mark
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?
Uniform beam held horizontal (not to scale)
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Step 1Take moments about the hinge so that the unknown hinge force does not appear.
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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.
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Step 3Equilibrium: 3.0T = 400, so T = 133 N (the hinge supplies the remaining 67 N upwards).
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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).
20A-1A-21Rolling and rotational kinetic energyPaper 1AMedium1 mark
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?
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Step 1For rolling from rest, mgh = ½mv²(1 + k) where I = kmr², so v² = 2gh/(1 + k).
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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.
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Step 3So the sphere is fastest and the hoop slowest — independent of mass and radius.
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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.
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?
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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.
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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.
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Step 3Weight of child = W/2. The pivot force has no moment about the pivot.
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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).
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.
Light rigid frame with three point masses (plan view)
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Step 1For point masses I = Σmr², with r the perpendicular distance from the axis.
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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.
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Step 3I = 2 × 0.20 × 0.30² = 0.036 kg m².
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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).
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)?
LQ/LPEkQ/EkP
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Step 1The angular impulse is τΔt = ΔL. The same torque for the same time gives both wheels the same angular momentum: LQ/LP = 1.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ek = L²/2I. With equal L, the kinetic energy is inversely proportional to I: EkQ/EkP = 1/3.
24A-1A-51Conservation of angular momentumPaper 1AMedium1 mark
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?
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Step 1There is no external torque about the axle, so the total angular momentum is conserved. Take anticlockwise as positive.
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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
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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.
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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.
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Step 3Resultant torque = 1.2 − 0.80 = 0.40 N m, clockwise.
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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.
27IBP2H-007-01-03Rolling and rotational kinetic energyPaper 1AMedium1 mark
A table-tennis ball can be modelled as a thin-walled hollow sphere, with moment of inertia \(\frac{2}{3}MR^2\). The ball rolls without slipping across a table. What fraction of its total kinetic energy is rotational?
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With ω = v/R, Erot = ½(⅔MR2)(v/R)2 = ⅓Mv2 and Etrans = ½Mv2. The fraction is (1/3)/(1/2 + 1/3) = 2/5. Option D is the ratio Erot/Etrans, not a fraction of the total. Option B is the translational fraction. Option A comes from taking ½ × ⅔ and ignoring the translational term.
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Answer C
Answer: C
Exam tip: Write both energies in terms of Mv² by substituting ω = v/R. Then read off the ratio required.
28IBP2H-007-01-04Rolling and rotational kinetic energyPaper 1AMedium1 mark
Four objects are released together from rest at the top of the same slope. Each rolls without slipping. Which object reaches the bottom first?
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Energy conservation gives v2 = 2gh/(1 + k), where I = kMR2. Mass and radius cancel, so only k matters. The values are: solid sphere 0.4, solid cylinder 0.5, hollow sphere 0.67, hoop 1. The solid sphere has the smallest k, so it puts the largest share of its energy into translation and wins. Options B and D tempt those who think heavier or larger objects roll faster. Option C tempts those who think a cylinder rolls better than a sphere.
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Answer A
Answer: A
Common error: Rolling races are decided by the shape factor I/(MR²) alone. Mass and size are deliberate distractions.
29IBP2H-007-01-05Angular impulsePaper 1AMedium1 mark
The chuck of a power drill has a moment of inertia of 0.015 kg m2 and starts from rest. The motor torque on it increases uniformly from zero to 6.0 N m over 0.50 s. What is the angular speed of the chuck at 0.50 s? Ignore friction.
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The angular impulse is the area under the torque–time relation: ½ × 6.0 × 0.50 = 1.5 N m s. Then ω = ΔL/I = 1.5/0.015 = 100 rad s−1. Option B uses the maximum torque as if it were constant. Option A gives τmax/I, which is an angular acceleration, not an angular speed. Option D halves the answer a second time.
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Answer C
Answer: C
Exam tip: When the torque varies, use the area under the τ–t graph (here, the average torque × time) for the angular impulse.
30IBP2H-007-01-06Rolling and rotational kinetic energyPaper 1AMedium1 mark
A solid ball rolls without slipping down a rough slope. Which statement about the frictional force from the slope on the ball is correct?
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Friction is static and acts up the slope. It reduces the linear acceleration and provides the torque that increases ω. Its point of application, the contact point, is instantaneously at rest, so it does no work and no energy is dissipated. Option C confuses zero work with zero force. Option B treats static friction as if it were kinetic friction. Option A gives the wrong direction.
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Answer D
Answer: D
Common error: Static friction in rolling without slipping is not zero. It does zero work, because the contact point does not slide.
31IBP2H-007-00-02Angular momentumPaper 1AMedium1 mark
An ice skater spins on frictionless ice with her arms outstretched. She pulls her arms in, and her moment of inertia about the spin axis falls to one third of its original value. What happens to her rotational kinetic energy?
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No external torque acts, so L = Iω is constant. Reducing I by a factor of 3 increases ω by a factor of 3. Rotational kinetic energy can be written as Ek = L2/2I. With L fixed and I divided by 3, Ek triples. The skater supplies the extra energy by doing work with her muscles as she pulls her arms in. 'Nine times' comes from squaring the new ω but forgetting that I has also fallen.
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Answer B
Answer: B
Exam tip: When L is conserved, use Ek = L²/2I. This avoids having to track ω and I separately.
32IBP2H-007-00-03Conservation of angular momentumPaper 1AMedium1 mark
The core of a massive star rotates with period T. It collapses to a neutron star with no loss of mass, and its radius falls by a factor of 100. Model the core as a uniform sphere both before and after the collapse, and assume no external torque acts. What is the new rotation period?
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For a uniform sphere, I = (2/5)MR2, so I ∝ R2. Dividing R by 100 divides I by 104. Angular momentum Iω is conserved, so ω increases by a factor of 104. Since T = 2π/ω, the period becomes T/104. Option B assumes I ∝ R. Option D confuses period with angular speed.
✓ 1
Answer C
Answer: C
Common error: Moment of inertia depends on the square of the distance from the axis. A change in size therefore affects I by the square of the scale factor.
33IBP2H-007-00-04Torque and its sensePaper 1AMedium1 mark
A patient holds a 20 N weight in her hand with her forearm horizontal. The forearm pivots at the elbow. The biceps muscle pulls vertically upward on the forearm at a point 4.0 cm from the elbow, and the weight is 32 cm from the elbow. Ignore the weight of the forearm. What force does the elbow joint exert on the forearm?
Forearm held horizontal (not to scale)
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Take moments about the elbow: Fbiceps × 0.040 = 20 × 0.32, which gives Fbiceps = 160 N upward. For translational equilibrium the upward and downward forces must balance: 160 = 20 + Fjoint, so Fjoint = 140 N downward. Option C gives the biceps force, not the joint force. Option D adds the forces instead of balancing them. Option B has the wrong direction.
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Answer A
Answer: A
Exam tip: A full equilibrium analysis needs two steps. Use moments to find one unknown force, then use the force balance to find the other.
34IBP2H-007-00-08Conservation of angular momentumPaper 1AMedium1 mark
A rigid body is in translational and rotational equilibrium. Consider these statements. I. The resultant force on the body is zero. II. The resultant torque on the body about any axis is zero. III. The body must be at rest. Which statements are correct?
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Translational equilibrium requires zero resultant force. Rotational equilibrium requires zero resultant torque. When the resultant force is zero, a resultant torque of zero about one axis means it is zero about every axis. So I and II are correct. The body need not be at rest: it can move at constant velocity or rotate at constant angular velocity, so III is wrong. The answer is I and II only.
✓ 1
Answer B
Answer: B
Exam tip: Because the torque is zero about every axis, you can take moments about whichever point removes the most unknown forces.
35IBP2H-007-02-03Rolling and rotational kinetic energyPaper 1AMedium1 mark
In a product test, a solid sphere, a solid cylinder and a thin hoop are released together from rest at the top of the same incline. Each rolls without slipping. The objects have different masses and radii. Which reaches the bottom first?
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For rolling from height h, Mgh = ½Mv2(1 + k), where I = kMR2. So v2 = 2gh/(1 + k), which does not depend on mass or radius. The smallest k gives the greatest speed at every point. The values of k are: sphere 2/5, cylinder 1/2, hoop 1. The sphere wins. 'All together' ignores the energy taken up by rotation. The hoop is a common wrong choice by those who think a larger moment of inertia helps.
✓ 1
Answer A
Answer: A
Exam tip: Write I = kMR². Then mass and radius cancel, and only k decides the rolling race.
36IBP2H-007-02-04Rotational kinematicsPaper 1AMedium1 mark
A flywheel in a vehicle test rig is rotating with angular speed ω0. It decelerates uniformly and comes to rest in time t. How many revolutions does it make while stopping?
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For uniform deceleration, the angular displacement is θ = ½(ω0 + 0)t = ω0t/2 radians. One revolution is 2π rad, so the number of revolutions is ω0t/(4π). Option A gives radians, not revolutions. Option B uses the initial speed instead of the average speed. Option D makes both errors in opposite directions.
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Answer C
Answer: C
Common error: Check two things: use the average angular speed for uniform acceleration, and convert radians to revolutions by dividing by 2π.
37IBP2H-007-02-05Angular momentumPaper 1AMedium1 mark
A space probe spins freely about its axis of symmetry. It then extends two long booms, and its moment of inertia about the spin axis triples. No external torque acts. What is the ratio (final rotational kinetic energy) / (initial rotational kinetic energy)?
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L is conserved. Since Ek = L2/2I, tripling I divides Ek by 3. As a check: ω falls to ω/3, and ½(3I)(ω/3)2 = (1/3)(½Iω2). Option A forgets that I also changes. Option C assumes energy is conserved. Option D uses Ek ∝ I at constant ω.
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Answer B
Answer: B
Exam tip: When L is fixed, Ek = L²/2I is the quickest route: Ek is inversely proportional to I.
38IBP2H-007-02-06Angular impulse and braking torquePaper 1AMedium1 mark
When a cordless drill is switched on, the angular momentum of its rotating parts increases uniformly from zero to 0.60 kg m2 s−1 in 0.30 s. What is the resultant torque on the rotating parts?
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Resultant torque equals the rate of change of angular momentum: τ = ΔL/Δt = 0.60/0.30 = 2.0 N m. Option A multiplies instead of divides. Option B inverts the ratio. Option D doubles the answer, as if an average had to be taken, although the change is already uniform.
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Answer C
Answer: C
Exam tip: τ = ΔL/Δt is the rotational form of Newton's second law, F = Δp/Δt.
39IBP3H-A4-20r5-03Angular impulsePaper 1AMedium1 mark
In a rail workshop test rig, a flywheel of moment of inertia 0.40 kg m2 spins at 6.0 rad s−1 on a frictionless axle. A second flywheel of moment of inertia 0.20 kg m2, initially at rest on the same axle, is clutched to it so that both rotate together. What is the final angular velocity?
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Angular momentum is conserved: 0.40 × 6.0 = (0.40 + 0.20)ω, so ω = 2.4/0.60 = 4.0 rad s−1 (D). A (12) divides the initial angular momentum by only the added flywheel's I. B (3.0) averages 6.0 and 0 as though the flywheels were equal. C (2.0) uses 0.20 × 6.0 as the initial angular momentum.
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Answer D
Answer: D
Exam tip: Write I₁ω₁ = (I₁ + I₂)ωf before substituting, and be sure each I is paired with the right ω.
40IBP3H-A4-20r5-04Rotational equilibriumPaper 1AMedium1 mark
A uniform horizontal signal arm of length 4.0 m and weight 200 N is hinged to a post at one end. A vertical cable attached to the free end holds the arm horizontal. A lamp of weight 100 N hangs from the arm 1.0 m from the hinge. What is the tension in the cable?
Uniform arm held horizontal (not to scale)
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Taking torques about the hinge: T × 4.0 = 200 × 2.0 + 100 × 1.0 = 500, so T = 125 N (A). B (25 N) ignores the arm's own weight (100 × 1.0 / 4.0). C (150 N) assumes the cable carries half the total weight. D (300 N) assumes the cable carries the whole weight, which ignores the hinge force.
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Answer A
Answer: A
Common error: The weight of a uniform arm acts at its midpoint, so its torque arm is half the length, not the full length.
41IBP3H-A4-20r5-05Torque and its sensePaper 1AMedium1 mark
A driver turns a steering wheel of diameter 0.40 m by pushing one hand tangentially downwards with a 15 N force on one side while pulling the other hand tangentially upwards with a 15 N force on the opposite side. What are the net force and the net torque about the wheel's axle on the wheel?
Steering wheel seen from the driver's seat (not to scale)
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The two forces are equal and opposite, so the net force is zero. Each force has a torque of 15 × 0.20 = 3.0 N m in the same sense, so the net torque is 6.0 N m (B), a couple. A treats equal and opposite forces as cancelling their turning effect. C adds the force magnitudes. D counts the torque of only one force.
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Answer B
Answer: B
Common error: A pair of equal, opposite, non-collinear forces gives zero resultant force but a non-zero resultant torque.
42IBP3H-A4-20r5-06Angular impulse and braking torquePaper 1AMedium1 mark
In a reaction-time training device, a turntable of moment of inertia 0.55 kg m2 about its axle is initially at rest. A lump of clay of mass 0.20 kg moving at 5.0 m s−1 tangentially to the rim, at a radius of 0.50 m, sticks to the rim. What is the angular velocity of the turntable and clay immediately afterwards?
Plan view of the turntable (not to scale)
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Initial angular momentum = 0.20 × 5.0 × 0.50 = 0.50 kg m2 s−1. Final I = 0.55 + 0.20 × 0.502 = 0.60 kg m2, so ω = 0.50/0.60 = 0.83 rad s−1 (D). A (0.91) leaves the clay out of the final moment of inertia. B (1.7) omits the radius from mvr. C (0.77) uses mr rather than mr2 for the clay.
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Answer D
Answer: D
Exam tip: For a point mass at radius r, use L = mvr before the collision and I = mr² after it.
43IBP3H-A4-20r24-03Angular momentumPaper 1AMedium1 mark
A swimmer performing a tumble turn rotates freely in the water with an extended body, then pulls into a tuck so that the moment of inertia falls to 0.40 of its original value. Assume the net external torque about the swimmer's axis is zero. Which row describes the change in angular speed and in rotational kinetic energy?
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With no external torque, L = Iω is constant, so ω increases by 1/0.40 = 2.5. Kinetic energy is L² / 2I, which also increases by a factor 2.5 because I falls while L stays fixed; the extra energy comes from the swimmer's muscles. Kinetic energy is not conserved here, so 'unchanged' is wrong. A factor of 6.3 comes from squaring the speed factor while ignoring that I has changed. A decrease would require I to increase.
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Answer A
Answer: A
Common error: Conservation of angular momentum does not mean conservation of rotational kinetic energy when the body changes shape.
44IBP3H-A4-20r24-04Rolling and rotational kinetic energyPaper 1AMedium1 mark
A uniform solid sphere of mass m and radius r rolls without slipping across a library floor. Its moment of inertia about its centre is 2⁄5mr2. What fraction of the total kinetic energy is translational?
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Translational KE = ½mv². Rotational KE = ½(2⁄5mr²)(v/r)² = 1⁄5mv². The total is 0.7mv², so the translational fraction is 0.5 / 0.7 = 0.714. The value 0.286 is the rotational fraction. An even 0.500 split assumes the two energies are equal. The value 0.600 comes from subtracting the factor 0.4 from 1 instead of comparing the energies.
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Answer D
Answer: D
Exam tip: Write both kinetic energy terms using v and cancel m and r before taking the ratio.
45IBP3H-A4-20r24-05Rotational kinematicsPaper 1AMedium1 mark
During a gust, the net torque on a wind-turbine rotor is 1.5 × 104 N m. The rotor has moment of inertia 3.0 × 105 kg m2 and starts from rest. How long does it take to reach an angular speed of 2.0 rad s−1?
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α = τ / I = 1.5 × 104 / 3.0 × 105 = 0.050 rad s−2, so t = ω / α = 2.0 / 0.050 = 40 s. Dividing α by ω instead of ω by α gives 0.025 s. Taking the mean angular speed as ω/2 in ω = αt gives 20 s. Using τ = ½Iα (copying the ½ from kinetic energy) gives 80 s.
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Answer D
Answer: D
Common error: Newton's second law for rotation is τ = Iα with no factor ½; the ½ belongs to energy and displacement formulae.
46IBP3H-A4-20r24-06Conservation of angular momentumPaper 1AMedium1 mark
A circular library book carousel has moment of inertia 4.0 kg m2 about its vertical axis and rotates freely at 2.0 rad s−1. A 2.0 kg book is placed gently on the carousel 0.60 m from the axis and does not slip. What is the angular speed afterwards?
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No external torque acts, so Iω is conserved: 4.0 × 2.0 = (4.0 + 2.0 × 0.60²) ωf = 4.72 ωf, giving ωf = 1.69 rad s−1. The value 1.84 comes from conserving kinetic energy, which is not conserved in this inelastic process. The value 1.54 comes from adding mr instead of mr2 to the moment of inertia. The value 2.00 ignores the book's contribution to the moment of inertia.
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Answer A
Answer: A
Exam tip: Treat the placed object as a point mass adding mr² to the total moment of inertia, then conserve angular momentum, not energy.
47A-1A-07Torque and angular accelerationPaper 1AHard1 mark
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?
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Step 1The weight Mg acts at the centre of mass, a distance L/2 from the pivot, perpendicular to the rod.
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2Torque about the pivot: τ = Mg × L/2 = ½MgL.
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Step 3Newton's second law for rotation: τ = Iα, so ½MgL = ⅓ML² × α.
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Step 4α = 3g/(2L). Note the free end then accelerates at αL = 1.5g, faster than free fall.
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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.
48IBP2H-007-01-07Rolling and rotational kinetic energyPaper 1AHard1 mark
A bowling ball, modelled as a uniform solid sphere, is released onto a lane. It slides at 7.0 m s−1 with no initial rotation. Kinetic friction acts until the ball rolls without slipping. What is its speed when rolling without slipping begins?
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Friction acts along the lane surface, so it has no torque about a fixed point on the lane. The weight and normal force are vertical, equal and opposite, so their torques also cancel. Angular momentum about that point is therefore conserved: Mv0R = MvR + ⅖MR2(v/R) = (7/5)MvR. This gives v = (5/7)v0 = 5.0 m s−1. Option C wrongly conserves kinetic energy, but kinetic friction dissipates energy. Option D assumes friction has no effect. Option A is (2/7)v0.
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Answer B
Answer: B
Exam tip: For skid-to-roll problems, take angular momentum about a point on the surface. Friction then has zero moment and drops out.
49IBP2H-007-00-05Angular impulsePaper 1AHard1 mark
In a clutch, plate X has moment of inertia I and rotates freely with angular speed ω. Plate Y, on the same axis, has moment of inertia 3I and is initially at rest. The plates are pressed together and soon rotate at a common angular speed. No external torque acts. What fraction of the original rotational kinetic energy is transferred to internal energy?
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Angular momentum is conserved: Iω = 4Iω', so ω' = ω/4. The final kinetic energy is ½(4I)(ω/4)2 = Iω2/8. This is one quarter of the initial value ½Iω2, so 3/4 of the energy is dissipated by friction between the plates. Option B is the fraction retained, not the fraction lost. Option D uses ω' = ω/4 but forgets that the moment of inertia has increased. Option A wrongly assumes that conserving L also conserves kinetic energy.
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Answer C
Answer: C
Common error: Angular momentum is conserved in a rotational 'sticking' collision, but kinetic energy is not. It behaves like a perfectly inelastic linear collision.
50IBP2H-007-00-07Torque and its sensePaper 1AHard1 mark
A space probe drifts in deep space at constant velocity along a straight line. The line passes a small asteroid P at a closest distance d. Gravitational effects are negligible. Which statement describes the angular momentum of the probe about P?
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The angular momentum of a particle about a point is L = mvr sin θ. Here r sin θ is the perpendicular distance from P to the line of motion, which is always d. So L = mvd, which is non-zero and the same at every point. This is consistent with conservation: no force acts on the probe, so there is no torque about P. Options C and D wrongly use the instantaneous distance r instead of the perpendicular distance. Option A wrongly assumes that only rotating objects can have angular momentum.
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Answer B
Answer: B
Common error: A particle moving in a straight line has angular momentum about any point that is not on its line of motion.
51IBP2H-007-02-07Angular impulsePaper 1AHard1 mark
A potter's wheel spins freely with angular speed ω. Its moment of inertia about its axis is 3mr2. A lump of clay of mass m is dropped vertically onto the wheel, lands at distance r from the axis and sticks. What fraction of the initial rotational kinetic energy is lost?
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Angular momentum is conserved: 3mr2ω = 4mr2ω′, so ω′ = 3ω/4. The final energy divided by the initial energy is (4mr2)(3ω/4)2 / (3mr2ω2) = 3/4, so 1/4 is lost. Option A treats the collision as elastic. Option C gives the fraction retained, not the fraction lost. Option D squares the speed ratio but forgets that I has increased.
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Answer B
Answer: B
Common error: In a 'sticking' rotational collision, both I and ω change. Include both when you compare kinetic energies.
52IBP2H-007-02-08Rolling and rotational kinetic energyPaper 1AHard1 mark
A solid steel ball rolls without slipping along a horizontal track in a marble-run toy. The moment of inertia of a solid sphere about a diameter is (2/5)MR2. What fraction of the ball's total kinetic energy is rotational?
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Rolling without slipping means v = ωR. The rotational energy is ½(2/5 MR2)(v/R)2 = (1/5)Mv2, and the translational energy is (1/2)Mv2. The fraction is (1/5)/(1/2 + 1/5) = 2/7. Option A confuses the moment-of-inertia coefficient with the energy fraction. Option C assumes the energy is shared equally. Option D is the translational fraction.
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Answer B
Answer: B
Exam tip: Express both kinetic energies in terms of Mv² using ω = v/R, then take the ratio.
53IBP3H-A4-20r5-07Angular impulse and braking torquePaper 1AHard1 mark
A spinning satellite module has moment of inertia 0.30 kg m2 about its axis. It docks along that axis with a stationary module of moment of inertia 0.10 kg m2 about the same axis, and the two rotate together. No external torque acts. What fraction of the initial rotational kinetic energy is transformed to other forms?
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L is conserved, so ωf = 0.30ωi/0.40 = 0.75ωi. Ef/Ei = 0.40(0.75)2/0.30 = 0.75, so 25% is transformed (A). B confuses conservation of L with conservation of energy. C (50%) averages the two states. D (75%) is the fraction that remains rather than the fraction lost.
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Answer A
Answer: A
Common error: A perfectly sticking rotational collision conserves angular momentum but not kinetic energy, even when no outside torque acts.
54IBP3H-A4-20r24-07Rolling and rotational kinetic energyPaper 1AHard1 mark
A solid cylinder (moment of inertia ½mr2) and a thin hollow cylinder (moment of inertia mr2) have the same mass and radius. Both roll without slipping from rest down the same incline through the same vertical drop. What is the ratio (speed of solid cylinder) / (speed of hollow cylinder) at the bottom?
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Energy conservation gives mgh = ½mv²(1 + I/mr²), so v² = 2gh/(1 + k) with k = ½ for the solid cylinder and 1 for the hollow one. The ratio of speeds is √(2 / 1.5) = 1.15. The value 1.33 is the ratio of the squares of the speeds. The value 1.41 comes from treating the solid cylinder as if it had no rotational energy, giving √(2 / 1). The value 0.866 is the inverse, which would mean the hollow cylinder wins.
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Answer C
Answer: C
Common error: The object with the smaller moment of inertia per unit mass and radius has less energy tied up in rotation, so it reaches a greater speed.
55A-1B-19Rotational equilibriumPaper 1BEasy7 marks
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 FS = 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)
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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.
56A-1B-01Rolling and rotational kinetic energyPaper 1BMedium7 marks
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).
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)
57A-1B-07Torque and angular accelerationPaper 1BMedium7 marks
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)
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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)
Ek = 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 = Ek/(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)
58A-1B-09Conservation of angular momentumPaper 1BHard7 marks
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)
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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
59A-1B-18Torque and angular accelerationPaper 1BHard7 marks
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)
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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)
60IBP2H-007-01-09Rotational kinematicsPaper 2Easy5 marks
The rotor of a wind turbine has a moment of inertia of 3.0×106 kg m2 about its axle. In high winds, the blades are turned so the wind exerts no net torque. A mechanical brake then applies a constant torque of 1.5×105 N m. At the moment the brake is applied, the rotor is turning at 1.8 rad s−1.
(a)
Outline why the angular acceleration produced by a given torque depends on how the mass of the rotor is distributed.
(2)
(b)
Calculate the magnitude of the angular deceleration of the rotor.
(1)
(c)
Calculate the number of revolutions the rotor makes before it stops.
(2)
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Part (a)
Moment of inertia (I = Σmr2) depends on the distribution of mass about the axis: mass further from the axle gives a larger I . Since α = τ/I, a larger I gives a smaller angular acceleration for the same torque .
✓ 2
Part (b)
α = τ/I = 1.5×105/3.0×106 = 0.050 rad s−2 .
✓ 1
Part (c)
Allow follow-through from the previous part. θ = ω2/(2α) = 1.82/(2 × 0.050) = 32.4 rad . Number of revolutions = 32.4/(2π) ≈ 5.2 .
✓ 2
Exam tip: The angular equations of motion mirror the linear ones. Convert radians to revolutions only at the final step.
61IBP2H-007-00-16Torque and its sensePaper 2Easy7 marks
Each crank of a bicycle is 0.17 m long, measured from the axle to the pedal. The two cranks point in opposite directions. A cyclist pushes vertically downward on the forward pedal with a force of 450 N.
(a)
State what is meant by the torque of a force about an axis.
(1)
(b)
Calculate the torque about the axle when the crank is horizontal.
(1)
(c)
Calculate the torque when the crank is 40° below the horizontal, with the force still vertical.
(2)
(d)
Outline why the cyclist's downward push produces no torque when the crank points vertically downward.
(1)
(e)
With the cranks horizontal, the cyclist's other foot rests on the rising pedal and pushes vertically downward on it with 120 N. Determine the resultant torque on the crank assembly.
(2)
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Part (a)
Torque is the product of the force and the perpendicular distance from the axis to the line of action of the force .
✓ 1
Part (b)
τ = 450 × 0.17 = 77 N m (76.5 N m) .
✓ 1
Part (c)
Perpendicular distance = 0.17 × cos 40° = 0.130 m . τ = 450 × 0.130 = 59 N m .
✓ 2
Part (d)
The line of action of the vertical force then passes through the axle, so its perpendicular distance from the axle is zero .
✓ 1
Part (e)
The 120 N force on the opposite crank produces a torque in the opposite sense to the 450 N force . Resultant torque = (450 − 120) × 0.17 = 56 N m, in the sense of the 450 N force .
✓ 2
Common error: The two pedals are on opposite sides of the axle. Downward forces on them produce torques of opposite sense, so subtract them rather than adding.
62IBP3H-A4-20r5-09Principle of momentsPaper 2Easy5 marks
A level-crossing barrier is a uniform pole of length 6.0 m and mass 40 kg. It is pivoted at a point 1.0 m from end A, and a counterweight of mass M is fixed at end A. The pole is to stay horizontal in equilibrium. When viewed from the front, the long arm of the pole points to the right of the pivot.
Front view of the barrier (not to scale)
(a)
State the sense of the torque of the weight of the pole about the pivot.
(1)
(b)
Calculate the magnitude of this torque.
(2)
(c)
Determine the mass M needed.
(2)
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Part (a)
clockwise
✓ 1
Part (b)
weight of pole = 40 × 9.8 = 392 N acts at its centre, 2.0 m from the pivot
✓ 1
torque = 392 × 2.0 = 784 N m
✓ 1
Part (c)
torques balance: M × 9.8 × 1.0 = 784
✓ 1
M = 80 kg
✓ 1
Exam tip: Find where the centre of mass of a uniform object lies relative to the pivot before writing the torque equation.
63IBP3H-A4-20r5-10Conservation of angular momentumPaper 2Easy6 marks
A platform diver leaves the board and tucks into a ball with moment of inertia 1.6 kg m2 about her axis of rotation, rotating at 7.5 rad s−1. Before entering the water she stretches out to a straight body shape with moment of inertia 4.0 kg m2 about the same axis.
(a)
Explain why her angular momentum is constant during the dive.
(2)
(b)
Calculate her angular velocity when she is straight.
(2)
(c)
Calculate the decrease in her rotational kinetic energy.
(2)
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Part (a)
the only significant external force is her weight, acting at the centre of mass
✓ 1
so there is no external torque about the centre of mass, hence angular momentum is conserved
✓ 1
Part (b)
L = 1.6 × 7.5 = 12 kg m2 s−1
✓ 1
ω = 12/4.0 = 3.0 rad s−1
✓ 1
Part (c)
initial energy ½ × 1.6 × 7.52 = 45 J and final energy ½ × 4.0 × 3.02 = 18 J
✓ 1
decrease = 27 J
✓ 1
Common error: Using ½Iω² with a mismatched I and ω gives a wrong energy; pair each moment of inertia with its own angular velocity.
64IBP3H-A4-20r24-09Rotational kinematicsPaper 2Easy6 marks
The rotor of a wind turbine has moment of inertia 1.8 × 106 kg m2 and rotates at 12 revolutions per minute. Use unrounded values in later parts and give answers to 3 significant figures.
(a)
Calculate the angular momentum of the rotor.
(2)
(b)
A brake applies a constant torque of 3.0 × 104 N m. Determine the time taken for the rotor to stop.
(2)
(c)
Calculate the number of revolutions made by the rotor while it stops.
(2)
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Part (a)
M1 ω = 12 × 2π / 60 = 1.257 rad s−1
✓ 1
A1 L = Iω = 2.26 × 106 kg m2 s−1
✓ 1
Part (b)
M1 torque = rate of change of angular momentum, so t = L / τ
✓ 1
A1 75.4 s (ECF from (a), using the unrounded L)
✓ 1
Part (c)
M1 θ = ½ωt (constant deceleration) = 47.4 rad
✓ 1
A1 7.54 revolutions
✓ 1
Exam tip: Torque equals the rate of change of angular momentum, so a constant torque gives a time of ΔL / τ without needing the angular acceleration separately.
65IBP3H-A4-20r24-10Rolling and rotational kinetic energyPaper 2Easy6 marks
A lifeguard's lane-rope reel is a uniform solid cylinder of mass 24 kg and radius 0.30 m that turns freely about its axis. The moment of inertia of a uniform solid cylinder about its axis is ½MR2. A rope is wound round the rim and pulled with a steady tangential force of 36 N, starting from rest.
(a)
Calculate the torque exerted by the rope on the reel.
(1)
(b)
Determine the angular acceleration of the reel.
(2)
(c)
Calculate the angular speed of the reel after 3.0 s.
(1)
(d)
Calculate the rotational kinetic energy of the reel after 3.0 s.
(2)
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Part (a)
A1 τ = 36 × 0.30 = 10.8 N m
✓ 1
Part (b)
M1 α = τ / I with I = ½ × 24 × 0.30² = 1.08 kg m2
✓ 1
A1 α = 10 rad s−2
✓ 1
Part (c)
A1 ω = αt = 30 rad s−1 (ECF from (b))
✓ 1
Part (d)
M1 E = ½Iω²
✓ 1
A1 E = ½ × 1.08 × 30² = 486 J (ECF from (b) and (c))
✓ 1
Common error: Use the moment of inertia of the whole cylinder about its own axis; do not mix up the radius of the reel with the length of the rope.
66A-2-27Rotational equilibriumPaper 2Medium14 marks
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)
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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
67A-2-29Conservation of angular momentumPaper 2Medium14 marks
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)
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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
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)
68IBP2H-007-01-10Rolling and rotational kinetic energyPaper 2Medium7 marks
A student releases objects from rest at a vertical height of 0.50 m above the bottom of a ramp. Each object rolls without slipping. A light gate measures the speed v of each object at the bottom. Each object has moment of inertia I = kMR2.
Object — shape — v / m s−1 A — solid sphere — 2.65 B — solid cylinder — 2.56 C — thin hoop — 2.21 X — unknown — 2.38
(a)
Show that \(v=\sqrt{\frac{2gh}{1+k}}\), where h is the height of release.
(3)
(b)
Determine the value of k for object X.
(2)
(c)
Explain why the student did not need to measure the mass or radius of the objects.
(2)
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Part (a)
Energy conservation: Mgh = ½Mv2 + ½Iω2 . Substitute I = kMR2 and ω = v/R: Mgh = ½Mv2(1 + k) . Rearrange to v2 = 2gh/(1 + k), giving the required result .
✓ 3
Part (b)
1 + k = 2gh/v2 = 9.8/2.382 = 1.73 . k ≈ 0.73 .
✓ 2
Part (c)
Mass cancels, because every energy term is proportional to M . Radius cancels, because ω = v/R makes the rotational energy ½kMv2, independent of R .
✓ 2
Exam tip: In 'show that' questions, state the energy principle explicitly and show the substitution ω = v/R as a clear step.
69IBP2H-007-01-11Angular impulsePaper 2Medium7 marks
A spacecraft has a moment of inertia of 1.2×103 kg m2 about its central axis and is initially not rotating. Two small thrusters are mounted 1.5 m from the axis, on opposite sides. Each produces a force of 22 N, tangential to a circle about the axis, in a direction that produces torque in the same sense. Both fire for 4.0 s.
Looking along the central axis (not to scale)
(a)
Calculate the angular impulse delivered to the spacecraft.
(2)
(b)
Determine the final angular speed of the spacecraft.
(1)
(c)
The spacecraft contains a reaction wheel that is initially at rest relative to the spacecraft. Explain how the wheel can bring the spacecraft's rotation to a stop without using any propellant.
(2)
(d)
The reaction wheel has a moment of inertia of 0.80 kg m2. Calculate its final rotation rate in revolutions per minute.
(2)
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Part (a)
Total torque = 2 × 22 × 1.5 = 66 N m . Angular impulse = τΔt = 66 × 4.0 = 264 ≈ 2.6×102 N m s .
✓ 2
Part (b)
Allow error carried forward from (a). ω = ΔL/I = 264/(1.2×103) = 0.22 rad s−1 .
✓ 1
Part (c)
A motor spins the wheel. The wheel and the spacecraft body exert equal and opposite internal torques on each other, so the total angular momentum of the system stays constant . The wheel spins up in the original sense of rotation until it carries all 264 N m s, leaving the spacecraft body with zero angular momentum, so the body stops rotating .
✓ 2
Part (d)
Allow error carried forward from (a). ω = 264/0.80 = 330 rad s−1 . Rate = 330 × 60/(2π) ≈ 3.2×103 rev min−1 . Using the rounded value 260 N m s gives 3.1×103 rev min−1, which also earns full credit.
✓ 2
Common error: Using the torque of only one thruster halves every later answer. Count every force contributing to the torque, and carry unrounded values through to the final step.
70IBP2H-007-01-16Rolling without slippingPaper 2Medium8 marks
A car of total mass 1200 kg, including its four wheels, travels at 20 m s−1 on a level road. Each wheel has a mass of 15 kg, a radius of 0.30 m and a moment of inertia of 0.60MR2 about its axle, where M is the mass of the wheel and R its radius. The wheels roll without slipping.
(a)
State what is meant by rolling without slipping, and state the relationship between the speed v of the car and the angular speed ω of a wheel.
(2)
(b)
Calculate the angular speed of each wheel.
(1)
(c)
Calculate the total kinetic energy of the car.
(3)
(d)
The engine is disengaged at the foot of a hill. Ignoring resistive forces, determine the maximum vertical height the car can coast up.
(2)
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Part (a)
The point of the tyre in contact with the road is instantaneously at rest relative to the road . v = ωR, where v is the speed of the axle (wheel centre) .
✓ 2
Part (b)
ω = v/R = 20/0.30 ≈ 67 rad s−1 .
✓ 1
Part (c)
Translational kinetic energy = ½ × 1200 × 202 = 2.40×105 J . Rotational kinetic energy per wheel = ½(0.60MR2)(v/R)2 = ½ × 0.60 × 15 × 202 = 1800 J, so 7200 J for four wheels . Total ≈ 2.47×105 J .
✓ 3
Part (d)
Allow error carried forward from the total kinetic energy calculated above. All kinetic energy (translational and rotational) becomes gravitational potential energy: Mgh = 2.47×105 J . h = 2.47×105/(1200 × 9.81) ≈ 21 m .
✓ 2
Exam tip: The wheel mass is already included in the 1200 kg for the translational term. Add only the wheels' rotational energy separately.
71IBP2H-007-00-09Rotational equilibriumPaper 2Medium9 marks
A tower crane has a horizontal jib that rests on the top of the tower, which acts as a pivot. • The jib has mass 8.0×103 kg. Its centre of mass is 15 m from the tower, on the load side. • A counterweight is fixed 13 m from the tower, on the opposite side. • A load of mass 2.0×103 kg hangs from a trolley 40 m from the tower. Ignore the masses of the trolley and cable. Take g = 9.81 m s−2.
Side view of the crane (heights not to scale)
(a)
State the two conditions for a rigid body to be in equilibrium.
(2)
(b)
Calculate the mass of the counterweight needed for the jib to be in rotational equilibrium about the tower.
(3)
(c)
Calculate the force that the tower exerts on the jib.
(2)
(d)
The trolley moves the load out to 45 m from the tower without the counterweight being moved. Determine the magnitude and sense of the resultant torque on the jib about the tower.
(2)
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Part (a)
The resultant force on the body is zero . The resultant torque about any point is zero .
✓ 2
Part (b)
Write the moment balance about the tower: anticlockwise moment = clockwise moment, so Mg × 13 = (8.0×103 × 15 + 2.0×103 × 40)g . Cancelling g gives M = 2.0×105 / 13 . M = 1.5×104 kg (1.54×104 kg) .
✓ 3
Part (c)
The force equals the total weight supported: (8.0×103 + 2.0×103 + 1.54×104) × 9.81 . Force = 2.5×105 N, directed vertically upward . Allow error carried forward from the counterweight mass calculated earlier.
✓ 2
Part (d)
The jib was balanced before the move, so the resultant torque is the extra torque from the load: 2.0×103 × 9.81 × 5.0 . Resultant torque = 9.8×104 N m, acting in the sense that turns the load end of the jib downward .
✓ 2
Exam tip: When every force is a weight, g cancels from the moment equation. You can then balance mass × distance directly, which saves time and reduces rounding errors. Carry the unrounded mass into later parts.
72IBP2H-007-00-11Angular impulsePaper 2Medium8 marks
A small observation satellite changes its orientation using an internal reaction wheel driven by an electric motor. • The satellite body has moment of inertia 450 kg m2 about the wheel's axis. • The wheel has moment of inertia 0.060 kg m2 about the same axis. • Initially the satellite and the wheel are both at rest. • The motor spins the wheel up to 314 rad s−1, measured relative to the distant stars.
(a)
Explain why the satellite body starts to rotate in the opposite sense to the wheel.
(2)
(b)
Calculate the angular speed of the satellite body while the wheel spins at 314 rad s−1.
(2)
(c)
The motor brings the wheel up to speed in 4.0 s, applying a constant torque. Calculate the torque exerted on the satellite body during this time.
(2)
(d)
The satellite must turn through 30° to point its camera at a new target. Treat the spin-up and spin-down of the wheel as instantaneous. Calculate the time for which the wheel must run at 314 rad s−1.
(2)
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Part (a)
No external torque acts on the satellite–wheel system, so its total angular momentum stays at zero . The motor's torque on the wheel is paired with an equal and opposite torque on the body, so the body gains angular momentum equal in size and opposite in sense to the wheel's .
✓ 2
Part (b)
Use conservation of angular momentum: 450ω = 0.060 × 314 = 18.8 kg m2 s−1 . ω = 0.042 rad s−1 (0.0419 rad s−1) .
✓ 2
Part (c)
τ = ΔL/Δt = 18.8/4.0 . τ = 4.7 N m . Allow error carried forward from the angular momentum calculated earlier.
✓ 2
Part (d)
Convert the angle: 30° = π/6 = 0.524 rad . t = 0.524/0.0419 = 12.5 s; accept any answer from 12 s to 13 s, e.g. 12 s from using ω = 0.042 rad s−1 . Allow error carried forward from the angular speed of the body.
✓ 2
Exam tip: A system that starts at rest with no external torque always has zero total angular momentum. Any rotation of one part must be balanced by rotation of another part in the opposite sense.
73IBP2H-007-00-13Torque and angular accelerationPaper 2Medium8 marks
A motor applies a constant torque τ to a small flywheel used in a regenerative-braking demonstrator. The angular acceleration α is measured for several values of τ. A constant frictional torque τf acts in the bearings.
τ / N m: 0.50, 1.00, 1.50, 2.00, 2.50 α / rad s−2: 1.0, 3.1, 5.0, 7.1, 9.0
(a)
Show that τ = Iα + τf, where I is the moment of inertia of the flywheel.
(2)
(b)
Use the data to determine I.
(2)
(c)
Determine the frictional torque τf.
(2)
(d)
The motor is switched off when the flywheel rotates at 40 rad s−1. After this, the constant frictional torque is the only torque acting. Calculate the time taken for the flywheel to stop.
(2)
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Notes
Part (a)
The resultant torque is the applied torque minus the opposing frictional torque, τ − τf . Newton's second law for rotation gives τ − τf = Iα, which rearranges to τ = Iα + τf .
✓ 2
Part (b)
I is the gradient of τ against α, found from a best-fit line or from widely separated data points, e.g. (2.50 − 0.50)/(9.0 − 1.0) . I = 0.25 kg m2; accept 0.24–0.26 kg m2 .
✓ 2
Part (c)
τf is the intercept on the τ axis at α = 0, e.g. 0.50 − 0.25 × 1.0 . τf = 0.25 N m; accept 0.23–0.27 N m. Allow error carried forward from the value of I .
✓ 2
Part (d)
Angular deceleration = τf/I = 0.25/0.25 = 1.0 rad s−2 . t = 40/1.0 = 40 s. Allow error carried forward from the values of I and τf .
✓ 2
Exam tip: In linear-graph questions, match your equation to y = mx + c first. Here the gradient gives I and the intercept gives τf.
74IBP2H-007-00-15Rotational kinetic energy and flywheelsPaper 2Medium7 marks
The rotor of a hospital blood centrifuge has moment of inertia 0.012 kg m2 and spins at 1500 rad s−1. When the run ends, a brake applies a constant torque that brings the rotor to rest in 25 s.
(a)
Calculate the initial angular momentum of the rotor.
(1)
(b)
Calculate the magnitude of the braking torque.
(2)
(c)
Determine the number of revolutions the rotor makes while braking.
(2)
(d)
Calculate the mean power dissipated by the brake.
(2)
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Part (a)
L = Iω = 0.012 × 1500 = 18 kg m2 s−1 .
✓ 1
Part (b)
τ = ΔL/Δt = 18/25, or τ = Iα with α = 1500/25 = 60 rad s−2 . τ = 0.72 N m . Allow ECF from (a).
✓ 2
Part (c)
θ = ½(ωi + ωf)t = ½ × 1500 × 25 = 1.875×104 rad, or an equivalent method such as ωi2 = 2αθ . Number of revolutions = 1.875×104/2π ≈ 3.0×103 . Accept 2980 or 2984.
✓ 2
Part (d)
Energy dissipated = ½Iω2 = ½ × 0.012 × 15002 = 1.35×104 J, or mean angular speed = 750 rad s−1 . Mean power = 1.35×104/25, or P = τωmean = 0.72 × 750, giving 540 W . Allow ECF from (b).
✓ 2
Exam tip: For a constant torque you can find τ directly from τ = ΔL/Δt without working out α first. Keep unrounded values until the final answer.
75IBP2H-007-02-10Rotational kinetic energy and flywheelsPaper 2Medium9 marks
A hybrid city bus stores braking energy in a flywheel. The flywheel is a uniform solid steel cylinder of mass 120 kg and radius 0.25 m, rotating about its axis. The bus has mass 12 000 kg. When it brakes from 15 m s−1 to rest, 60% of its translational kinetic energy is transferred to the flywheel. The flywheel is initially rotating at 800 rad s−1.
(a)
Show that the moment of inertia of the flywheel is about 3.8 kg m2.
(2)
(b)
Calculate the energy transferred to the flywheel during the stop.
(2)
(c)
Determine the angular speed of the flywheel after the bus has stopped.
(3)
(d)
Some flywheels are designed with most of their mass concentrated in a heavy rim. Outline the advantage of this design.
(2)
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Part (a)
Uses I = ½MR2 for a solid cylinder . I = 0.5 × 120 × 0.252 = 3.75 kg m2 ≈ 3.8 kg m2
✓ 2
Part (b)
Kinetic energy of the bus = ½ × 12 000 × 152 = 1.35 × 106 J . Energy transferred = 0.60 × 1.35 × 106 = 8.1 × 105 J
✓ 2
Part (c)
Initial flywheel energy = ½ × 3.75 × 8002 = 1.2 × 106 J . Final energy = 1.2 × 106 + 8.1 × 105 = 2.01 × 106 J . ω = √(2E/I) = √(2 × 2.01 × 106 / 3.75) = 1.04 × 103 rad s−1 . ECF from (a) and (b).
✓ 3
Part (d)
For the same mass, moving mass further from the axis increases the moment of inertia . More energy is therefore stored at a given ω, or a lower ω (and so lower stresses in the rim) is needed to store the same energy .
✓ 2
Exam tip: When energy is added to a flywheel that is already spinning, add it to the existing rotational energy before finding the new ω. Do not use the added energy alone.
76IBP2H-007-02-11Torque and angular accelerationPaper 2Medium8 marks
Engineers test a reaction wheel for a small satellite. A pair of identical sliding masses of combined mass m can be clamped at distance r from the axis of the wheel. The total moment of inertia I of the assembly is measured for several values of r. The masses may be treated as point masses, so I = I0 + mr2, where I0 is the moment of inertia of the wheel alone.
r / m : 0.10, 0.20, 0.30, 0.40, 0.50 I / kg m2 : 0.0060, 0.0084, 0.0124, 0.0180, 0.0252
(a)
I is plotted on the vertical axis. State the quantity that should be plotted on the horizontal axis to obtain a straight-line graph, and identify what the gradient and the vertical intercept represent.
(2)
(b)
Determine m using the data.
(2)
(c)
Determine I0.
(1)
(d)
The masses are clamped at r = 0.35 m. A motor applies a constant resultant torque of 0.050 N m to the assembly, starting from rest. Calculate the time taken to reach an angular speed of 20 rad s−1.
(3)
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Part (a)
Plot r2 on the horizontal axis, i.e. I against r2 . The gradient is m and the intercept on the I axis is I0 .
✓ 2
Part (b)
Gradient = (0.0252 − 0.0060)/(0.250 − 0.010) = 0.0192/0.240 . m = 0.080 kg . The data lie exactly on a straight line, so any pair of points or a best-fit line gives the same result.
✓ 2
Part (c)
I0 = 0.0060 − 0.080 × 0.010 = 0.0052 kg m2 . ECF from (b).
✓ 1
Part (d)
I = 0.0052 + 0.080 × 0.352 = 0.0150 kg m2 . α = τ/I = 0.050/0.0150 = 3.3 rad s−2 . t = ω/α = 20/3.33 = 6.0 s . ECF from (b) and (c).
✓ 3
Exam tip: To linearise data, match the relationship to y = mx + c. Here y = I and x = r².
77IBP2H-007-02-12Conservation of angular momentumPaper 2Medium6 marks
At the end of its life, the core of a massive star has a radius of 7.0 × 106 m and rotates with a period of 30 days. The core collapses to form a neutron star of radius 12 km. Assume that the core remains a uniform sphere of constant mass throughout the collapse.
(a)
State the condition for the angular momentum of the core to be conserved during the collapse.
(1)
(b)
Determine the rotation period of the neutron star.
(3)
(c)
Explain, without further calculation, why the rotational kinetic energy of the core increases during the collapse, and identify the source of this energy.
(2)
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Part (a)
No external (resultant) torque acts on the core .
✓ 1
Part (b)
I = (2/5)MR2, so I ∝ R2, and Iω = constant gives T ∝ R2 . (R1/R2)2 = (7.0 × 106 / 1.2 × 104)2 = 3.4 × 105 . T2 = 30 × 86 400 / 3.4 × 105 = 7.6 s .
✓ 3
Part (c)
Ek = L2/2I with L constant, and I decreases, so Ek increases . The energy comes from the gravitational potential energy released as the core contracts, i.e. from work done by gravity .
✓ 2
Common error: Period and angular speed vary in opposite ways. If ω increases by a factor k, T decreases by the same factor k.
78IBP2H-007-02-15Torque and angular accelerationPaper 2Medium7 marks
During an emergency stop, the rotor of a wind turbine is brought to rest by a mechanical brake. The rotor has a moment of inertia of 3.2 × 106 kg m2 about its axis and is initially rotating at 1.8 rad s−1. It stops after 12 revolutions, with uniform angular deceleration. The brake disc is steel of mass 450 kg and specific heat capacity 460 J kg−1 K−1.
(a)
Calculate the initial rotational kinetic energy of the rotor.
(2)
(b)
Determine the angular deceleration of the rotor.
(2)
(c)
Calculate the braking torque.
(1)
(d)
Estimate the temperature rise of the brake disc, assuming that all the rotational kinetic energy is transferred to it as internal energy.
(2)
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Part (a)
Ek = ½Iω2 = ½ × 3.2 × 106 × 1.82 . Ek = 5.2 × 106 J (5.18 × 106 J) .
✓ 2
Part (b)
θ = 12 × 2π = 75.4 rad . From ω2 = ω02 − 2αθ with ω = 0: α = 1.82/(2 × 75.4) = 0.0215 rad s−2 (accept 0.021 rad s−2) .
✓ 2
Part (c)
τ = Iα = 3.2 × 106 × 0.0215 = 6.9 × 104 N m (accept 6.7 × 104 N m from α = 0.021 rad s−2; allow error carried forward from the angular deceleration) .
✓ 1
Part (d)
ΔT = Ek/(mc) . ΔT = 5.18 × 106 / (450 × 460) = 25 K (allow error carried forward from the kinetic energy) .
✓ 2
Exam tip: Convert an angular displacement given in revolutions to radians before you use the rotational equations of motion. Carry unrounded values into later parts so that rounding does not change your final answer.
79IBP3H-A4-20r5-11Angular momentum in a collisionPaper 2Medium7 marks
A boxing reaction trainer is a uniform rod of mass 1.2 kg and length 0.80 m hanging vertically from a frictionless pivot at its top end. The moment of inertia of a uniform rod about one end is ⅓ML2. A ball of clay of mass 0.15 kg travelling horizontally at 6.0 m s−1 hits the bottom end of the rod and sticks to it.
Not to scale
(a)
Explain why angular momentum about the pivot, and not linear momentum, is conserved in the collision.
(2)
(b)
Calculate the angular velocity of the rod and clay just after the collision.
(3)
(c)
Determine the percentage of the initial kinetic energy of the clay that is transformed to other forms in the collision. Use unrounded values.
(2)
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Part (a)
the pivot exerts an unknown external force on the rod, so linear momentum of the system is not conserved
✓ 1
this force has zero torque about the pivot so angular momentum about the pivot is conserved
✓ 1
Part (b)
angular momentum before = 0.15 × 6.0 × 0.80 = 0.72 kg m2 s−1
✓ 1
I after = ⅓ × 1.2 × 0.802 + 0.15 × 0.802 = 0.352 kg m2
✓ 1
ω = 0.72/0.352 = 2.05 rad s−1
✓ 1
Part (c)
initial kinetic energy ½ × 0.15 × 6.02 = 2.7 J, final energy ½ × 0.352 × 2.04552 = 0.736 J
✓ 1
percentage transformed = 72.7%
✓ 1
Common error: In a collision at a pivot, linear momentum is not conserved because the pivot supplies an external impulse; use angular momentum about the pivot.
80IBP3H-A4-20r5-12Equilibrium of a ladderPaper 2Medium8 marks
A uniform ladder of length 4.5 m and mass 15 kg leans against the smooth vertical side of a rail wagon, making an angle of 62° with the horizontal ground. The ladder is at rest.
Uniform ladder against a smooth wall (not to scale)
(a)
State the two conditions for the ladder to be in equilibrium.
(2)
(b)
State the vertical force exerted on the ladder by the ground.
(1)
(c)
Calculate the horizontal force exerted on the ladder by the wall.
(3)
(d)
Determine the minimum coefficient of static friction between the ladder and the ground. Use unrounded values.
(2)
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Part (a)
the resultant force on the ladder is zero
✓ 1
the resultant torque about any point is zero
✓ 1
Part (b)
147 N upward, equal to the weight, since the smooth wall provides no vertical force
✓ 1
Part (c)
torques about the base: Nw × 4.5 sin 62° = 147 × 2.25 cos 62°
✓ 1
Nw = 147 × 2.25 cos 62° / (4.5 sin 62°) = 147/(2 tan 62°)
✓ 1
Nw = 39.1 N
✓ 1
Part (d)
friction force at the base equals 39.1 N (horizontal balance)
✓ 1
μ = 39.08/147 = 0.266
✓ 1
Exam tip: Choosing the base of the ladder as the pivot removes the two unknown ground forces from the torque equation.
81IBP3H-A4-20r5-13Conservation of angular momentumPaper 2Medium8 marks
Astronomers model a star as a uniform sphere of mass 2.0 × 1030 kg and radius 7.0 × 108 m, rotating once every 25 days. After a supernova its core collapses to a uniform sphere of the same mass and radius 1.4 × 104 m. The moment of inertia of a uniform sphere is (2/5)MR2. Assume angular momentum is conserved.
(a)
Explain why the angular velocity of the core increases during the collapse.
(2)
(b)
Calculate the final angular velocity.
(3)
(c)
Suggest one reason why a real collapsing star may not conserve its angular momentum.
(1)
(d)
Calculate the rotation period of the collapsed core in milliseconds.
(2)
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Part (a)
angular momentum Iω is constant with no external torque
✓ 1
I decreases as the radius decreases, so ω must increase
✓ 1
Part (b)
initial ω = 2π/(25 × 86400) = 2.91 × 10−6 rad s−1 and initial I = 3.92 × 1047 kg m2
✓ 1
final I = 1.57 × 1038 kg m2
✓ 1
ω = 7.27 × 103 rad s−1
✓ 1
Part (c)
matter may be ejected, carrying angular momentum away (or an external torque from magnetic fields)
✓ 1
Part (d)
period = 2π/ω = 2π/7272
✓ 1
= 8.64 × 10−4 s = 0.864 ms
✓ 1
Exam tip: Mass cancels in ωf = ωi(Ri/Rf)² for equal-mass uniform spheres: use this as a quick check.
82IBP3H-A4-20r5-14Torque and its sensePaper 2Medium6 marks
A weightlifter holds his forearm horizontal with a dumbbell in his hand. The biceps muscle pulls vertically upwards on the forearm at a point 0.045 m from the elbow joint. The weight of the forearm is 15 N and acts 0.14 m from the elbow. The dumbbell weighs 60 N and is 0.32 m from the elbow. The forearm points to the right when viewed from the front.
Forearm held horizontal (not to scale)
(a)
State the sense of the torque of the dumbbell's weight about the elbow.
(1)
(b)
Calculate the tension in the biceps.
(3)
(c)
Determine the magnitude and direction of the force exerted by the upper arm on the forearm at the elbow.
(2)
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Part (a)
clockwise
✓ 1
Part (b)
anticlockwise torque of muscle equals clockwise torques of the weights
✓ 1
0.045T = 15 × 0.14 + 60 × 0.32 = 21.3 N m
✓ 1
T = 473 N
✓ 1
Part (c)
vertical forces balance: T = 15 + 60 + F so F = 473.3 − 75 (ECF from (b))
✓ 1
398 N directed downwards
✓ 1
Common error: The muscle attaches very close to the joint, so its force must be several times the weight it supports; do not expect the joint force to be small.
83IBP3H-A4-20r24-11Angular impulse and braking torquePaper 2Medium8 marks
A circular swimming-pool deck platform of moment of inertia 180 kg m2 turns freely about a vertical axis and is initially at rest. A swimmer of mass 55 kg runs along a tangent at 3.2 m s−1 and jumps on to the rim, which is 2.5 m from the axis. The swimmer stays at the rim. Treat the swimmer as a point mass.
Plan view of the platform (not to scale)
(a)
Determine the angular speed of the platform immediately after the swimmer lands.
(4)
(b)
Explain why the total kinetic energy is not conserved in this process.
(1)
(c)
Calculate the kinetic energy lost. Use unrounded values.
(3)
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Part (a)
M1 angular momentum conserved: mvr = (I + mr2)ω
✓ 1
A1 mvr = 55 × 3.2 × 2.5 = 440 kg m2 s−1
✓ 1
A1 total moment of inertia = 180 + 55 × 2.5² = 524 kg m2 (523.75)
✓ 1
A1 ω = 0.840 rad s−1
✓ 1
Part (b)
R1 the landing is inelastic: kinetic energy is transferred to thermal energy and sound, through friction and impact between swimmer and platform
✓ 1
Part (c)
M1 uses ½mv² before and ½Iω² after
✓ 1
A1 initial 282 J (281.6 J) and final 185 J (184.8 J)
✓ 1
A1 loss = 96.8 J
✓ 1
Common error: In a collision where two bodies end up moving together, angular momentum is conserved but kinetic energy is not; do not equate the two kinetic energies.
84IBP3H-A4-20r24-14Conservation of angular momentumPaper 2Medium7 marks
A diver leaves a springboard with her body straight, rotating about her centre of mass with moment of inertia 14 kg m2 at 1.6 rad s−1. She then pulls into a tuck, which reduces her moment of inertia to 4.2 kg m2.
(a)
Calculate her angular speed in the tuck position.
(2)
(b)
Explain why her angular momentum about her centre of mass stays constant even though gravity acts on her.
(1)
(c)
State the origin of the extra rotational kinetic energy gained in the tuck.
(1)
(d)
Calculate the increase in her rotational kinetic energy.
(3)
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Part (a)
M1 I1ω1 = I2ω2
✓ 1
A1 ω2 = 14 × 1.6 / 4.2 = 5.33 rad s−1
✓ 1
Part (b)
R1 her weight acts through the centre of mass, so it exerts no torque about that point
✓ 1
Part (c)
A1 work done by the diver's muscles (internal chemical energy of her body)
✓ 1
Part (d)
M1 uses ½Iω² for the initial and final states
✓ 1
A1 initial 17.9 J and final 59.7 J
✓ 1
A1 increase = 41.8 J
✓ 1
Exam tip: When the moment of inertia changes with no external torque, use L = Iω to find the new angular speed first, then compare the energies ½Iω² before and after.
85A-2-02Rotational kinetic energy and flywheelsPaper 2Hard14 marks
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)
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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.
Ek = ½ × 1.875 × 1676² = 2.6 × 106 J
✓ 1
Allow ECF from (a)(i). Accept 2.6–2.7 × 106 J.
Part (a)(iii)
Ek/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)
Ek = ½ × 1.2 × 104 × 15.0² = 1.35 × 106 J
✓ 1
Speed read from the graph: 15 m s−1.
Part (b)(ii)
Initial Ek = 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)
86A-2-05Rolling and rotational kinetic energyPaper 2Hard13 marks
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.
Ball rolling down the ramp (not to scale)
(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)
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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 Ek = 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/FN = 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.
87A-2-08Torque and angular accelerationPaper 2Hard13 marks
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.
Light string; no slipping on the pulley
(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 Ek = ½ × 0.010 × 18.1² = 1.64 J
✓ 1
Translational Ek = ½ × 5.0 × 1.81² = 8.18 J; total 9.81 J = loss of Ep, (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)
88A-2-13Angular impulse and braking torquePaper 2Hard14 marks
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)
Ek = ½Iω² = ½ × 0.18 × 25.0² = 56 J
✓ 1
Accept 56.3 J.
Part (b)(i)
Friction force Ff = μdFN = 0.60 × 40 = 24 N
✓ 1
τ = Ffr = 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)
Model sketch
Model sketch: angular momentum of the wheel against time
89A-2-15Equilibrium of a ladderPaper 2Hard13 marks
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)
90A-2-17Conservation of angular momentumPaper 2Hard13 marks
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
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 Ek = L²/2I rises as I falls
✓ 1
Allow ECF from (b)(i). Reference to work done by the gravitational force needed.
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)
91A-2-28Torque and angular accelerationPaper 2Hard13 marks
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)
Ek = 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)
92A-2-33Angular momentum in a collisionPaper 2Hard13 marks
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.
Plan view of the door (not to scale)
(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)
93A-2-39Conservation of angular momentumPaper 2Hard14 marks
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 LM = mM√(GMEr), where mM is the mass of the Moon, ME 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].
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 = √(GME/r) falls as r increases; LM = mMvr ∝ √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)
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: ahead/afeet = 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)
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)
95IBP2H-007-01-12Rolling and rotational kinetic energyPaper 2Hard9 marks
A cylindrical gas canister, modelled as a uniform solid cylinder of mass 12 kg and radius 0.15 m, rolls without slipping down a loading ramp. The ramp is inclined at 25° to the horizontal.
Solid cylinder rolling down a ramp (not to scale)
(a)
Explain why a frictional force is needed for the canister to roll without slipping.
(2)
(b)
Show that the acceleration of the canister is \(\frac{2}{3}g\sin\theta\).
(3)
(c)
Calculate the frictional force on the canister.
(2)
(d)
Determine the minimum coefficient of static friction between the canister and the ramp.
(2)
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Part (a)
The weight and the normal force act through, or along a line through, the centre of mass, so neither produces a torque about the axis . Friction at the contact point is the only force with a torque about the centre, and this torque provides the angular acceleration needed to keep ω = v/R as v increases .
✓ 2
Part (b)
Along the slope: Mg sinθ − f = Ma . Rotation about the centre: fR = ½MR2α with α = a/R, so f = ½Ma . Substituting gives Mg sinθ = (3/2)Ma, so a = ⅔g sinθ .
✓ 3
Part (c)
f = ½Ma = ⅓Mg sinθ . f = ⅓ × 12 × 9.8 × sin25° = 16.6 N ≈ 17 N .
✓ 2
Part (d)
Allow error carried forward from the frictional force calculated in the previous part. Normal force N = Mg cos25° = 12 × 9.8 × cos25° ≈ 107 N . μmin = f/N = 16.6/107 ≈ 0.155 ≈ 0.16; this is equivalent to ⅓tan25°. Accept 0.155–0.16 .
✓ 2
Exam tip: Static friction only needs to satisfy f ≤ μN. Find the minimum μ by setting the required friction equal to μN. Keep unrounded values until the final step: the radius cancels, so μmin = ⅓tanθ for any solid cylinder.
96IBP2H-007-01-13Rolling and rotational kinetic energyPaper 2Hard9 marks
A thin hoop of mass 0.50 kg and radius 0.40 m spins at 15 rad s−1 about a horizontal axis through its centre, with its centre at rest. It is placed gently upright on a horizontal floor. Its lowest point initially moves backwards relative to the floor. The coefficient of kinetic friction between the hoop and the floor is 0.30. Treat the hoop as a ring with I = MR2. Take g = 9.8 m s−2.
(a)
Explain why the centre of the hoop begins to move forwards while its angular speed decreases.
(2)
(b)
Determine the speed of the hoop when it starts to roll without slipping.
(3)
(c)
Calculate the time taken for rolling without slipping to begin.
(2)
(d)
Calculate the energy dissipated during the slipping phase.
(2)
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Part (a)
The contact point slides backwards, so kinetic friction on the hoop acts forwards. This gives the centre a forward linear acceleration . The torque of this friction about the centre opposes the spin, so ω decreases until v = ωR .
✓ 2
Part (b)
Angular momentum about a fixed point on the floor is conserved. Friction acts along the floor, through that point, and the weight and normal force are equal, opposite and collinear, so their torques cancel . MR2ω0 = MR2(v/R) + MvR = 2MvR . v = ω0R/2 = 15 × 0.40/2 = 3.0 m s−1 .
✓ 3
Part (c)
Linear acceleration a = μg = 0.30 × 9.8 = 2.94 m s−2 . t = v/a = 3.0/2.94 = 1.02 s ≈ 1.0 s. Allow error carried forward from the speed found earlier .
✓ 2
Part (d)
Initial kinetic energy = ½MR2ω02 = ½ × 0.50 × 0.402 × 152 = 9.0 J . Final kinetic energy = ½Mv2 + ½MR2(v/R)2 = Mv2 = 4.5 J, so the energy dissipated is 4.5 J. Allow error carried forward from the speed found earlier .
✓ 2
Common error: Mechanical energy is not conserved while the hoop slips, because kinetic friction does work. Use angular momentum about a fixed point on the floor instead.
97IBP2H-007-00-10Conservation of angular momentumPaper 2Hard10 marks
A diver leaves a springboard rotating at 1.8 rad s−1 in a straight (layout) position. In this position her moment of inertia about a horizontal axis through her centre of mass is 15.0 kg m2. She then pulls into a tuck, which reduces her moment of inertia to 3.6 kg m2. Air resistance is negligible.
(a)
Explain why the diver's angular momentum about her centre of mass is constant during the flight.
(2)
(b)
Calculate her angular speed in the tuck position.
(2)
(c)
Calculate the change in her rotational kinetic energy when she moves into the tuck, and identify where this energy comes from.
(3)
(d)
During the dive she spends a total of 0.40 s in the layout position and 1.0 s in the tuck. Assume the changes of position are instantaneous. Determine the number of revolutions she completes.
(3)
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Part (a)
The only external force is her weight, which acts through the centre of mass . This weight therefore produces no torque about the centre of mass, so angular momentum is conserved .
✓ 2
Part (b)
15.0 × 1.8 = 3.6 × ω . ω = 7.5 rad s−1 .
✓ 2
Part (c)
Initial Ek = ½ × 15.0 × 1.82 = 24 J, and final Ek = ½ × 3.6 × 7.52 = 101 J . The increase is 77 J . ECF from (b). The energy comes from work done by her muscles as she pulls her limbs towards the axis .
✓ 3
Part (d)
Angle turned in layout = 1.8 × 0.40 = 0.72 rad, and angle turned in tuck = 7.5 × 1.0 = 7.5 rad . Total angle = 8.2 rad . Number of revolutions = 8.2/2π = 1.3 . ECF from (b).
✓ 3
Common error: Do not assume rotational kinetic energy is conserved just because angular momentum is. Internal forces can do work without changing L.
98IBP2H-007-00-12Equilibrium of a ladderPaper 2Hard9 marks
A uniform ladder of length 5.0 m and mass 12 kg rests against a smooth vertical wall. It makes an angle of 65° with the rough horizontal ground. A window cleaner of mass 70 kg climbs the ladder. The coefficient of static friction between the ladder and the ground is 0.30.
Uniform ladder against a smooth wall; rough ground (not to scale)
(a)
State why the force exerted by the wall on the ladder is horizontal.
(1)
(b)
Show that the normal reaction force from the ground on the ladder is about 800 N.
(2)
(c)
Outline why the friction force at the foot of the ladder increases as the window cleaner climbs.
(2)
(d)
Determine the maximum distance along the ladder, measured from its foot, that the window cleaner can climb before the ladder starts to slip.
(4)
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Part (a)
The wall is smooth, so there is no friction and it exerts only a normal force, perpendicular to the vertical wall, which is therefore horizontal .
✓ 1
Part (b)
Vertical equilibrium: the normal reaction equals the total weight, since the wall force is horizontal . N = (12 + 70) × 9.8 = 804 N ≈ 800 N .
✓ 2
Part (c)
As he climbs, the moment arm of his weight about the foot increases, so its torque increases and the wall force must increase to balance it . For horizontal equilibrium the friction force equals the wall force, so friction increases too .
✓ 2
Part (d)
Maximum friction = 0.30 × 804 = 241 N, which equals the wall force R . Moments about the foot: R × 5.0 sin 65° = 118 × 2.5 cos 65° + 686 × x cos 65° . This gives 1092 = 124 + 290x . x = 3.3 m from the foot . Allow error carried forward from the value of the normal reaction found earlier; g = 9.81 N kg−1 is equally acceptable.
✓ 4
Exam tip: Take moments about the foot of the ladder. Both the friction force and the ground's normal reaction act there, so they drop out of the moment equation.
99IBP2H-007-00-14Angular momentum in a collisionPaper 2Hard10 marks
In a sports-science demonstration, a 0.50 kg lump of putty is thrown horizontally at 6.0 m s−1. It strikes a uniform door at right angles and sticks to it. • The door is at rest before the impact and can rotate freely about its vertical hinge. • The door has mass 20 kg and width 0.90 m. • The putty strikes the door 0.80 m from the hinge. The moment of inertia of a uniform door about its hinge is \(\frac{1}{3}ML^2\), where L is its width.
Plan view of the door (not to scale)
(a)
Show that the moment of inertia of the door about the hinge is about 5.4 kg m2.
(2)
(b)
Calculate the angular speed of the door immediately after the impact.
(3)
(c)
Determine the fraction of the initial kinetic energy that is dissipated in the collision.
(3)
(d)
Explain why the linear momentum of the putty–door system is not conserved in the collision, but its angular momentum about the hinge is.
(2)
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Part (a)
I = ⅓ × 20 × 0.902 . I = 5.4 kg m2 .
✓ 2
Part (b)
Angular momentum of the putty about the hinge = 0.50 × 6.0 × 0.80 = 2.4 kg m2 s−1 . Total moment of inertia afterwards = 5.4 + 0.50 × 0.802 = 5.72 kg m2 . ω = 2.4/5.72 = 0.42 rad s−1 .
✓ 3
Part (c)
Initial kinetic energy = ½ × 0.50 × 6.02 = 9.0 J . Final kinetic energy = ½ × 5.72 × 0.422 = 0.50 J . Fraction dissipated = (9.0 − 0.50)/9.0 = 0.94 . ECF from (b).
✓ 3
Part (d)
The hinge exerts an external force on the door during the impact, so linear momentum changes . This force acts through the axis of rotation, so it produces no torque about the hinge, and angular momentum about the hinge is conserved .
✓ 2
Common error: The putty adds mr² to the moment of inertia after the impact. Forgetting this term gives an angular speed that is too large.
100IBP2H-007-02-13Rolling and rotational kinetic energyPaper 2Hard9 marks
A bowler releases a bowling ball onto a horizontal lane. At release the ball slides with speed v0 = 8.0 m s−1 and has no rotation. The coefficient of kinetic friction between the ball and the lane is 0.20. The ball is a uniform solid sphere of mass M and radius R, with I = (2/5)MR2. Air resistance is negligible.
(a)
Explain why the linear speed of the ball decreases and its angular speed increases while it slides.
(2)
(b)
Show that the ball begins to roll without slipping when its speed is (5/7)v0.
(3)
(c)
Calculate the time for which the ball slides.
(2)
(d)
Determine the distance travelled by the ball before it rolls without slipping.
(2)
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Part (a)
The ball has no spin at first, so its contact point slides forward over the lane. Kinetic friction therefore acts backwards on the ball and decelerates its centre of mass . The friction force acts at a distance R from the centre, so it exerts a torque about the centre that increases ω from zero .
✓ 2
Part (b)
Linear motion: a = μg, so v = v0 − μgt . Rotation: α = μMgR / ((2/5)MR2) = 5μg/(2R), so ω = 5μgt/(2R) . Rolling begins when v = ωR: v0 − μgt = 5μgt/2, so t = 2v0/(7μg) and v = v0 − 2v0/7 = 5v0/7 .
✓ 3
Part (c)
t = 2v0/(7μg) = 2 × 8.0 / (7 × 0.20 × 9.8) . t = 1.2 s (1.17 s) .
✓ 2
Part (d)
Any one valid method :
✓ 1
s = v0t − ½μgt2 = 8.0 × 1.17 − 0.5 × 1.96 × 1.172, or
—
s = ½(8.0 + 5.71) × 1.17, or
—
s = (v02 − v2)/(2μg).
—
s = 8.0 m .
✓ 1
Allow error carried forward from the time calculated in the previous part; using t = 1.2 s gives 8.2 m, which earns full credit.
—
Exam tip: In skid-to-roll problems, write separate equations for v(t) and ω(t). Then apply the rolling condition v = ωR to find when slipping stops. Keep unrounded values until the final step.
101IBP2H-007-02-14Conservation of angular momentumPaper 2Hard8 marks
A space station rotates freely about its central axis to provide artificial gravity at its rim, 25 m from the axis. The station's rigid structure, excluding a cargo pod, has a constant moment of inertia of 4.5 × 105 kg m2 about this axis. A 400 kg cargo pod, treated as a point mass, is initially fixed at the rim, and the station rotates at 0.20 rad s−1. An internal winch then pulls the pod in slowly to 5.0 m from the axis, where it is fixed again.
Looking along the axis of rotation (not to scale)
(a)
Determine the angular speed of the station when the pod is at 5.0 m from the axis.
(3)
(b)
By calculating the centripetal acceleration at the rim before and after the change, compare how heavy crew members standing at the rim feel.
(2)
(c)
Calculate the work done by the winch.
(3)
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Part (a)
I1 = 4.5 × 105 + 400 × 252 = 7.0 × 105 kg m2 . I2 = 4.5 × 105 + 400 × 5.02 = 4.6 × 105 kg m2 . Only internal torques act, so angular momentum is conserved: I1ω1 = I2ω2, giving ω2 = (7.0 × 105 × 0.20) / (4.6 × 105) = 0.304 rad s−1; accept 0.30 rad s−1 .
✓ 3
Part (b)
At the rim a = ω2r. Before the change, a = 0.202 × 25 = 1.0 m s−2. After the change, a = 0.3042 × 25 = 2.3 m s−2; accept 2.25 m s−2 from ω = 0.30 rad s−1 . Their apparent weight (the normal force from the floor) increases by a factor of about 2.3, so they feel more than twice as heavy . Allow ECF from (a).
✓ 2
Part (c)
Initial rotational kinetic energy = ½ × 7.0 × 105 × 0.202 = 1.4 × 104 J . Final rotational kinetic energy = ½ × 4.6 × 105 × 0.3042 (or L2/2I2 = (1.4 × 105)2 / (2 × 4.6 × 105)) = 2.13 × 104 J . Work done = ΔEk = 2.13 × 104 − 1.4 × 104 = 7.3 × 103 J . Accept 6.7 × 103 J if ω2 = 0.30 rad s−1 is used; allow ECF from (a).
✓ 3
Exam tip: Keep unrounded values, or use Ek = L²/2I, when subtracting energies. A small rounding error in ω becomes a large error in the difference between two kinetic energies. Include the pod's mr² in both moments of inertia.
102IBP2H-007-02-16Torque and angular accelerationPaper 2Hard8 marks
A car-park barrier arm is a uniform rod of mass 15 kg and length 2.4 m, pivoted at one end. Its moment of inertia about the pivot is (1/3)ML2. The motor fails while the arm is vertical, and the arm falls from rest about the pivot. Friction at the pivot and air resistance are negligible.
Barrier arm, released from the vertical (not to scale)
(a)
Show that the angular speed of the arm is 3.5 rad s−1 when it reaches the horizontal.
(3)
(b)
Calculate the speed of the free end of the arm when it is horizontal. Compare it with the speed of an object dropped from rest through 2.4 m.
(2)
(c)
Determine the tangential acceleration of the free end at the instant the arm is horizontal, and compare it with g.
(3)
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Part (a)
The centre of mass falls L/2 = 1.2 m . Mg(L/2) = ½ × (1/3)ML2 × ω2 . ω = √(3g/L) = √(3 × 9.8/2.4) = √12.25 = 3.5 rad s−1 .
✓ 3
Part (b)
v = ωL = 3.5 × 2.4 = 8.4 m s−1 . A dropped object reaches √(2 × 9.8 × 2.4) = 6.9 m s−1, so the free end moves faster. This is because the rigid rod forces all points to share the same ω, so the tip gains more kinetic energy than its own loss of potential energy .
✓ 2
Part (c)
τ = Mg(L/2) = 15 × 9.8 × 1.2 = 176.4 N m . α = τ/I = 176.4 / ((1/3) × 15 × 2.42) = 176.4/28.8 = 6.125 rad s−2 . a = αL = 6.125 × 2.4 = 14.7 ≈ 15 m s−2, which is greater than g (about 1.5g) .
✓ 3
Common error: For a pivoted rod, the gravitational potential energy lost depends on the fall of the centre of mass (L/2), not the fall of the tip (L).
103IBP3H-A4-20r5-15Conservation of angular momentumPaper 2Hard8 marks
A railway workshop turntable is a horizontal disc free to rotate about a vertical axle. The disc has moment of inertia 1500 kg m2. A worker of mass 75 kg stands 4.0 m from the axle while the turntable rotates at 0.30 rad s−1. The worker then walks slowly to a point 1.0 m from the axle. Treat the worker as a point mass and ignore friction.
Plan view of the turntable (not to scale)
(a)
Calculate the initial moment of inertia of the turntable and worker together.
(2)
(b)
Calculate the final angular velocity of the turntable.
(2)
(c)
Explain why the rotational kinetic energy of the system changes even though angular momentum is conserved.
(1)
(d)
Calculate the increase in rotational kinetic energy. Use unrounded values.
(3)
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Part (a)
worker's I = 75 × 4.02 = 1200 kg m2
✓ 1
total = 2700 kg m2
✓ 1
Part (b)
L = 2700 × 0.30 = 810 kg m2 s−1 is conserved
✓ 1
final I = 1500 + 75 × 1.02 = 1575 kg m2 so ω = 810/1575 = 0.514 rad s−1
✓ 1
Part (c)
the worker does work against the (centripetal) force needed as moving inward, increasing the kinetic energy at the expense of chemical energy
✓ 1
Part (d)
initial energy ½ × 2700 × 0.302 = 121.5 J
✓ 1
final energy 8102/(2 × 1575) = 208.3 J
✓ 1
increase = 86.8 J
✓ 1
Exam tip: Use Ek = L²/2I when L is known: it avoids rounding the intermediate angular velocity.
104IBP3H-A4-20r5-16Rotational equilibriumPaper 2Hard11 marks
A breakdown crane on a railway has a uniform boom of length 8.0 m and mass 600 kg, hinged at its lower end to the crane body. The boom is held at 50° above the horizontal by a horizontal cable attached to its upper end. A load of mass 1400 kg hangs from the upper end of the boom. The boom is in equilibrium.
(a)
State two conditions for the boom to be in rotational and translational equilibrium.
(2)
(b)
Calculate the tension in the horizontal cable.
(4)
(c)
Calculate the magnitude of the force exerted by the hinge on the boom.
(3)
(d)
Determine the angle this hinge force makes with the horizontal, giving the answer to 3 significant figures.
(2)
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Part (a)
the resultant force on the boom is zero
✓ 1
the resultant torque about any point is zero
✓ 1
Part (b)
torque about the hinge of the cable: T × 8.0 sin 50°
✓ 1
torques of weights: 600 × 9.8 × 4.0 cos 50° + 1400 × 9.8 × 8.0 cos 50°
✓ 1
equate torques about the hinge
✓ 1
T = 1.40 × 104 N
✓ 1
Part (c)
vertical component = (600 + 1400) × 9.8 = 19600 N
✓ 1
horizontal component equals the cable tension 13979 N
✓ 1
magnitude = 2.41 × 104 N
✓ 1
Part (d)
tan θ = 19600/13979 (ECF from (b))
✓ 1
θ = 54.5°
✓ 1
Exam tip: Taking torques about the hinge removes the unknown hinge force from the equation: then find the hinge force by resolving horizontally and vertically.
105IBP3H-A4-20r24-16Rolling and rotational kinetic energyPaper 2Hard9 marks
A thin hollow plastic ball of mass m and radius r (moment of inertia 2⁄3mr2) rolls without slipping up a ramp at a swimming pool. The ramp is inclined at 15° to the horizontal. The ball has a speed of 4.0 m s−1 at the foot of the ramp. Use g = 9.8 m s−2.
Hollow ball rolling up a ramp (not to scale)
(a)
Show that the total kinetic energy of the ball at the foot of the ramp is 5⁄6mv2.
(2)
(b)
Calculate the distance the ball travels up the ramp before it stops momentarily.
(3)
(c)
Determine the minimum coefficient of static friction between the ball and the ramp that allows it to roll without slipping while moving up the ramp.
(4)
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Part (a)
M1 rotational energy = ½ × 2⁄3mr² × (v/r)² = 1⁄3mv²
✓ 1
A1 total = ½mv² + 1⁄3mv² = 5⁄6mv²
✓ 1
Part (b)
M1 all kinetic energy converted to potential energy: 5⁄6mv² = mgd sin 15°
✓ 1
A1 d = 5 × 4.0² / (6 × 9.8 × sin 15°)
✓ 1
A1 d = 5.26 m
✓ 1
Part (c)
M1 along the ramp: mg sin θ − f = ma and about the centre: fr = I(a/r)
✓ 1
A1 f = 2⁄3ma, so a = 3⁄5g sin θ
✓ 1
M1 f ≤ μmg cos θ with f = 0.4mg sin θ
✓ 1
A1 μmin = 0.4 tan 15° = 0.107
✓ 1
Exam tip: Friction acts up the slope here and provides the torque that slows the spin; its size is fixed by the rolling condition, so compare it with the maximum static friction μR.
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