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IB Physics A.4: Rigid body mechanics worked solutions

Worked solutions to every practice question in A.4: Rigid body mechanics of Photon Physics: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

Type A1 / torque-arm / Warm-up / P1A[1 mark]

A mechanic pulls vertically downwards with a force of 90\,\mathrm{N} on the end of a spanner 0.25\,\mathrm{m} long. The spanner makes an angle of 30^\circ with the horizontal. What is the torque about the nut?

A. 0 B. 11.3\,\mathrm{N\,m}

C. 19.5\,\mathrm{N\,m} D. 22.5\,\mathrm{N\,m}

Show solutionHide solution

The angle between the spanner and the vertical force is 60^\circ, so \tau=Fr\sin\theta=90(0.25)\sin60^\circ=19.5\,\mathrm{N\,m}. Equivalently, the perpendicular distance from the nut to the line of action is the horizontal distance 0.25\cos30^\circ=0.217\,\mathrm{m}. B uses the angle with the horizontal, D the whole length as the arm, A assumes a force that is not perpendicular gives no torque.

Final answer

Question A2

Type A2 / signed-torques / Warm-up / P1A[1 mark]

A pulley wheel on a fixed axle has a rim of radius 0.20\,\mathrm{m} and a hub of radius 0.050\,\mathrm{m}. Its moment of inertia is 0.040\,\mathrm{kg\,m^2}. A 12\,\mathrm{N} force acts tangentially on the rim, turning the wheel clockwise, and a 30\,\mathrm{N} force acts tangentially on the hub, turning it anticlockwise. What is the angular acceleration of the wheel?

A. 22.5\,\mathrm{rad\,s^{-2}} B. 90\,\mathrm{rad\,s^{-2}}

C. 97.5\,\mathrm{rad\,s^{-2}} D. 450\,\mathrm{rad\,s^{-2}}

Show solutionHide solution

Clockwise positive: \sum\tau=12(0.20)-30(0.050)=2.4-1.5=0.90\,\mathrm{N\,m}, so \alpha=0.90/0.040=22.5\,\mathrm{rad\,s^{-2}} clockwise. The larger force gives the smaller torque because its arm is a quarter as long. B subtracts the forces and uses the rim radius, C adds the two torques, D divides the difference of the forces by I and ignores the radii.

Final answer

Question B1

Type B1 / period-angular-speed / Warm-up / P1A[1 mark]

The blades of a wind turbine complete one revolution every 4.0\,\mathrm{s}. What is their angular velocity?

A. 0.25\,\mathrm{rad\,s^{-1}} B. 1.6\,\mathrm{rad\,s^{-1}}

C. 25\,\mathrm{rad\,s^{-1}} D. 90\,\mathrm{rad\,s^{-1}}

Show solutionHide solution

\omega=2\pi/T=2\pi/4.0=1.57\,\mathrm{rad\,s^{-1}}. A is the frequency (no 2\pi), C multiplies 2\pi by the period, D is 90^\circ per second written as radians.

Final answer

Question B2

Type B2 / rolling-link / Warm-up / P1A[1 mark]

A trundle wheel of diameter 0.30\,\mathrm{m} rolls without slipping along a corridor 12\,\mathrm{m} long. How many complete turns does it make?

A. 6 B. 12 C. 25 D. 40

Show solutionHide solution

Rolling without slipping: each turn moves the wheel one circumference, \pi d=0.942\,\mathrm{m}. Turns =12/0.942=12.7, so 12 complete turns. C uses the radius in place of the diameter, A takes the circumference as 2\pi d, D divides by the diameter.

Final answer

Question D1

Type D1 / energy-scaling / Warm-up / P1A[1 mark]

Flywheels P and Q are uniform discs of the same material and thickness. Q has twice the radius of P and spins at half the angular velocity of P. What is \dfrac{\text{rotational kinetic energy of Q}}{\text{rotational kinetic energy of P}}?

A. 1 B. 2 C. 4 D. 16

Show solutionHide solution

Same thickness and material: mass \propto R^2, so M_Q=4M_P and I\propto MR^2 gives I_Q=16I_P. E_k=\tfrac12I\omega^2: ratio =16\times(1/2)^2=4. A keeps the mass the same, B also uses \omega without squaring it, D ignores the change in \omega.

Final answer

Question C1

Type C1 / point-inertia / Warm-up / P1A[1 mark]

Four small balls, each of mass 0.20\,\mathrm{kg}, are fixed at the corners of a light square frame of side 0.40\,\mathrm{m}. The frame rotates about an axis through its centre, perpendicular to the frame. What is its moment of inertia?

A. 0.032\,\mathrm{kg\,m^2} B. 0.064\,\mathrm{kg\,m^2}

C. 0.128\,\mathrm{kg\,m^2} D. 0.226\,\mathrm{kg\,m^2}

Show solutionHide solution

Each ball is half a diagonal from the axis: r=0.40/\sqrt2=0.283\,\mathrm{m}, r^2=0.080\,\mathrm{m^2}. I=\sum mr^2=4(0.20)(0.080)=0.064\,\mathrm{kg\,m^2}. A uses half the side as the distance, C the whole side, D forgets to square r.

Final answer

Question C2

Type C2 / inertia-scaling / Warm-up / P1A[1 mark]

For a thin uniform rod of mass M and length L, turning about a perpendicular axis through its middle, I=\tfrac1{12}ML^2. Call this value I for a whole rod. A piece one third as long is sawn off. What is the moment of inertia of the piece about a perpendicular axis through its own middle?

A. I/3 B. I/9 C. I/27 D. I/81

Show solutionHide solution

The piece has mass M/3 and length L/3, so I'=\tfrac1{12}(M/3)(L/3)^2=I/27. A scales the mass only, B scales the length but keeps the whole mass, D squares the mass ratio as well.

Final answer

Question B3

Type B3 / linear-time / Warm-up / P1A[1 mark]

A disc starts from rest and a constant resultant torque acts on it. Which quantity increases in proportion to the time?

A. the torque B. the rotational kinetic energy

C. the angle turned D. the angular velocity

Show solutionHide solution

A constant torque gives a constant \alpha=\tau/I, so \omega=\alpha t is proportional to t (and so is L=I\omega). The torque itself is constant (A); E_k=\tfrac12I\omega^2 grows as t^2 (B); \Delta\theta=\tfrac12\alpha t^2 also grows as t^2 (C).

Final answer

Question E1

Type E1 / impulse-unit / Warm-up / P1A[1 mark]

A torque acts on a wheel for a short time. Which unit is correct for the change in the wheel's angular momentum?

A. \mathrm{kg\,m\,s^{-1}} B. \mathrm{kg\,m^2\,s^{-2}}

C. \mathrm{N\,s} D. \mathrm{kg\,m^2\,s^{-1}}

Show solutionHide solution

\Delta L=\tau\Delta t has unit \mathrm{N\,m\,s}=\mathrm{kg\,m^2\,s^{-1}}, the unit of I\omega. A and C are units of linear momentum and linear impulse; B is the unit of torque, missing the time.

Final answer

Question E2

Type E2 / equal-energy-momentum / Warm-up / P1A[1 mark]

Disc X has moment of inertia 0.10\,\mathrm{kg\,m^2} and disc Y has 0.40\,\mathrm{kg\,m^2}. They spin with equal rotational kinetic energies. What is the ratio L_X/L_Y of their angular momenta?

A. 0.25 B. 0.50 C. 2.0 D. 4.0

Show solutionHide solution

E_k=L^2/2I, so L=\sqrt{2IE_k} and at equal energy L\propto\sqrt I: \sqrt{0.10/0.40}=0.50. A forgets the square root, C inverts the ratio, D does both.

Final answer

Tutorial practice

Question A3

Type A3 / equilibrium-moments / Exam / P2[5 marks]

A shop sign hangs from a uniform horizontal beam of length 1.6\,\mathrm{m} and weight 60\,\mathrm{N}. The beam is hinged to a wall at one end. A cable joins the outer end of the beam to a point higher up the wall and makes 35^\circ with the beam. The sign, of weight 90\,\mathrm{N}, hangs from the beam 1.2\,\mathrm{m} from the wall.

(a) Calculate the tension in the cable. [2]

(b) Determine the magnitude and direction of the force exerted by the hinge on the beam. [3]

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(a) Take moments about the hinge, so the unknown hinge force has no torque (M1): T\sin35^\circ(1.6)=60(0.80)+90(1.2)=156\,\mathrm{N\,m}, so T=156/(1.6\sin35^\circ)=170\,\mathrm{N} (A1). (b) Horizontal: the hinge pushes the beam away from the wall with T\cos35^\circ=139\,\mathrm{N} (M1). Vertical: V=150-T\sin35^\circ=150-97.5=52.5\,\mathrm{N} upwards (M1). Size \sqrt{139.2^2+52.5^2}=149\,\mathrm{N} at \tan^{-1}(52.5/139.2)=20.7^\circ above the horizontal (A1). The cable tension exceeds the total load of 150\,\mathrm{N} because only its vertical component, 97.5\,\mathrm{N}, holds the beam up.

Final answer

(a) 170\,\mathrm{N}. (b) 149\,\mathrm{N} at 20.7^\circ above the horizontal, directed away from the wall.

Question B4

Type B4 / angular-suvat; Type B2 / rolling-link / Exam / P2[5 marks]

The wheels of a car have radius 0.32\,\mathrm{m} and roll without slipping. The car accelerates uniformly from 8.0\,\mathrm{m\,s^{-1}} to 20\,\mathrm{m\,s^{-1}} in 6.0\,\mathrm{s}.

(a) Calculate the initial and the final angular velocity of a wheel. [2]

(b) Determine the angular acceleration of a wheel. [1]

Show solutionHide solution

(a) No slipping, so \omega=v/r (M1): 8.0/0.32=25\,\mathrm{rad\,s^{-1}} and 20/0.32=62.5\,\mathrm{rad\,s^{-1}} (A1). (b) \alpha=(62.5-25)/6.0=6.25\,\mathrm{rad\,s^{-2}} (A1). (c) \Delta\theta=\tfrac12(\omega_i+\omega_f)t=\tfrac12(87.5)(6.0)=262.5\,\mathrm{rad} (M1); 262.5/2\pi=41.8 revolutions (A1). Check: the car travels \tfrac12(8.0+20)(6.0)=84\,\mathrm{m}=0.32\times262.5\,\mathrm{m}.

Final answer

(a) 25\,\mathrm{rad\,s^{-1}}; 62.5\,\mathrm{rad\,s^{-1}}. (b) 6.25\,\mathrm{rad\,s^{-2}}. (c) 41.8 revolutions.

Question C3

Type C3 / inertia-gradient / Exam / P1A[1 mark]

Different torques \tau are applied to a wheel and its angular acceleration \alpha is measured. Plotting \tau (vertical axis) against \alpha (horizontal axis) gives a line that passes through the origin and through the point (8.0\,\mathrm{rad\,s^{-2}},\ 2.0\,\mathrm{N\,m}). What is the moment of inertia of the wheel?

A. 0.25\,\mathrm{kg\,m^2} B. 2.0\,\mathrm{kg\,m^2}

C. 4.0\,\mathrm{kg\,m^2} D. 16\,\mathrm{kg\,m^2}

Show solutionHide solution

\tau=I\alpha, so with \tau on the vertical axis the gradient is I: 2.0/8.0=0.25\,\mathrm{kg\,m^2}. C inverts the gradient (correct only if \alpha were plotted against \tau), B reads the torque, D multiplies the two coordinates.

Final answer

Question C4

Type C4 / pulley-dynamics / Exam / P2[6 marks]

A 4.0\,\mathrm{kg} block on a smooth horizontal table is joined by a light string to a 1.0\,\mathrm{kg} mass that hangs over the edge of the table. The string passes over a pulley that is a uniform disc of mass 2.0\,\mathrm{kg} and radius 0.050\,\mathrm{m}, with moment of inertia \tfrac12MR^2. The string does not slip on the pulley and the axle is frictionless. The system is released from rest.

(a) Explain why the tensions in the string on the two sides of the pulley are different. [1]

(b) Show that the acceleration of the block is about 1.6\,\mathrm{m\,s^{-2}}. [3]

Show solutionHide solution

(a) The pulley has moment of inertia, so it needs a resultant torque (T_1-T_2)R=I\alpha to speed up; equal tensions would give zero torque (A1). (b) Hanging mass: 9.8-T_1=1.0a; block: T_2=4.0a (M1). Pulley: (T_1-T_2)R=\tfrac12MR^2(a/R), so T_1-T_2=\tfrac12Ma=1.0a (M1). Adding: 9.8=6.0a, a=1.63\,\mathrm{m\,s^{-2}} (A1). (c) T_2=4.0(1.63)=6.53\,\mathrm{N}; T_1=1.0(9.8-1.63)=8.17\,\mathrm{N} (A1 each). Check: T_1-T_2=1.63\,\mathrm{N}=1.0a. With a light pulley the answer would be 9.8/5.0=1.96\,\mathrm{m\,s^{-2}}.

Final answer

(a) A resultant torque is needed to give the pulley angular acceleration. (b) 1.63\,\mathrm{m\,s^{-2}}. (c) Hanging side 8.17\,\mathrm{N}; table side 6.53\,\mathrm{N}.

Question D2

Type D2 / energy-share / Exam / P1A[1 mark]

A trolley of total mass 12\,\mathrm{kg} runs on four wheels. Each wheel is a uniform disc of mass 0.50\,\mathrm{kg} and radius 0.050\,\mathrm{m}, with moment of inertia \tfrac12mr^2, and rolls without slipping. What fraction of the trolley's total kinetic energy is rotational?

A. 7.7\% B. 8.3\% C. 14\% D. 17\%

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For one wheel \tfrac12(\tfrac12mr^2)(v/r)^2=\tfrac14mv^2; four wheels give mv^2=0.50v^2. Translational KE =\tfrac12(12)v^2=6.0v^2. Fraction of the total =0.50/6.50=7.7\%; the radius cancels. B divides by the translational KE only, C treats each wheel as a ring, I=mr^2, D makes both slips.

Final answer

Question D3

Type D3 / flywheel-power / Exam / P1A[1 mark]

A flywheel energy store of moment of inertia 12\,\mathrm{kg\,m^2} slows uniformly from 600\,\mathrm{rad\,s^{-1}} to 400\,\mathrm{rad\,s^{-1}} in 50\,\mathrm{s} while it drives a generator. What is the average power delivered by the flywheel?

A. 4.8\,\mathrm{kW} B. 24\,\mathrm{kW}

C. 43\,\mathrm{kW} D. 48\,\mathrm{kW}

Show solutionHide solution

\Delta E_k=\tfrac12(12)(600^2-400^2)=1.2\times10^6\,\mathrm{J}; P=1.2\times10^6/50=24\,\mathrm{kW}. A squares the difference in speeds, \tfrac12(12)(200)^2, C uses all the energy at 600\,\mathrm{rad\,s^{-1}}, D drops the \tfrac12.

Final answer

Question D4

Type D4 / rolling-energy / Exam / P2[5 marks]

A thin-walled hollow cylinder of mass 1.2\,\mathrm{kg} and radius 0.050\,\mathrm{m}, with moment of inertia MR^2, is released from rest on a slope. It rolls without slipping and descends a vertical height of 0.40\,\mathrm{m}.

(a) Show that its speed at the bottom is about 2\,\mathrm{m\,s^{-1}}. [2]

(b) Calculate its angular velocity at the bottom. [1]

(c) A solid cylinder, I=\tfrac12MR^2, of the same mass and radius is released beside it at the same moment. Explain which cylinder reaches the bottom first. [2]

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(a) Mgh=\tfrac12Mv^2+\tfrac12(MR^2)(v/R)^2=Mv^2 (M1), so v=\sqrt{gh}=\sqrt{9.8(0.40)}=1.98\,\mathrm{m\,s^{-1}} (A1). (b) \omega=v/R=1.98/0.050=39.6\,\mathrm{rad\,s^{-1}} (A1). (c) Solid: Mgh=\tfrac12Mv^2+\tfrac14Mv^2, v=\sqrt{4gh/3}=2.29\,\mathrm{m\,s^{-1}} (M1). At every height its speed is larger by the same factor, so it arrives first; mass and radius play no part, only the ratio I/MR^2 (A1).

Final answer

(a) 1.98\,\mathrm{m\,s^{-1}}. (b) 39.6\,\mathrm{rad\,s^{-1}}. (c) The solid cylinder: a smaller share of its energy goes into rotation, so it is faster at every height (2.29\,\mathrm{m\,s^{-1}} at the bottom).

Question E3

Type E3 / redistribute-mass / Exam / P1A[1 mark]

A skater spins on ice with negligible friction. She pulls her arms in and her moment of inertia falls from 4.0\,\mathrm{kg\,m^2} to 1.6\,\mathrm{kg\,m^2}. Her rotational kinetic energy was 50\,\mathrm{J}. What is it now?

A. 8.0\,\mathrm{J} B. 20\,\mathrm{J} C. 50\,\mathrm{J} D. 125\,\mathrm{J}

Show solutionHide solution

No external torque, so L is constant and E_k=L^2/2I\propto1/I: 50\times4.0/1.6=125\,\mathrm{J}. Her arms do work pulling the mass inwards, which is where the extra 75\,\mathrm{J} comes from. C assumes KE is conserved, B scales the energy with I, A with I^2.

Final answer

Question E5

Type E5 / impulse-graph / Exam / P1A[1 mark]

A wheel of moment of inertia 0.50\,\mathrm{kg\,m^2} starts from rest. The torque on it rises steadily from 0 to 3.0\,\mathrm{N\,m} in 4.0\,\mathrm{s} and then stays at 3.0\,\mathrm{N\,m} for a further 2.0\,\mathrm{s}. What is the angular velocity at the end of the 6.0\,\mathrm{s}?

A. 6.0\,\mathrm{rad\,s^{-1}} B. 12\,\mathrm{rad\,s^{-1}}

C. 24\,\mathrm{rad\,s^{-1}} D. 36\,\mathrm{rad\,s^{-1}}

Show solutionHide solution

Angular impulse = area under the \tau–t graph =\tfrac12(4.0)(3.0)+2.0(3.0)=12\,\mathrm{N\,m\,s}=\Delta L, so \omega=12/0.50=24\,\mathrm{rad\,s^{-1}}. B stops at the angular impulse, A multiplies by I, D takes the torque as 3.0\,\mathrm{N\,m} for the whole 6.0\,\mathrm{s}.

Final answer

Question E7

Type E7 / attach-rotor / Exam / P2[6 marks]

A record turntable of moment of inertia 0.090\,\mathrm{kg\,m^2} spins freely at 3.0\,\mathrm{rad\,s^{-1}}. A metal ring of moment of inertia 0.060\,\mathrm{kg\,m^2}, not rotating, is dropped centrally onto it. After a short time the ring and turntable rotate together.

(a) State the condition for the angular momentum of the turntable and ring to be conserved. [1]

(b) Calculate their common angular velocity. [2]

(d) Suggest what happens to the lost energy. [1]

Show solutionHide solution

(a) The resultant external torque about the axis must be zero; the friction between ring and turntable is internal to the system (A1). (b) 0.090(3.0)=(0.090+0.060)\omega (M1), \omega=0.27/0.15=1.8\,\mathrm{rad\,s^{-1}} (A1). (c) Before \tfrac12(0.090)(3.0)^2=0.405\,\mathrm{J}; after \tfrac12(0.15)(1.8)^2=0.243\,\mathrm{J} (M1); fraction lost 0.162/0.405=0.40 (A1). It equals I_{\text{ring}}/(I_{\text{ring}}+I_{\text{table}}), the same pattern as a sticking collision. (d) While the surfaces slip, friction converts kinetic energy into internal energy (A1).

Final answer

(a) No resultant external torque about the axis. (b) 1.8\,\mathrm{rad\,s^{-1}}. (c) 0.40. (d) Internal energy of ring and turntable (heating by friction while they slip), with a little sound.

Advanced practice

Question A4

Type A4 / tipping-threshold / Stretch / P2[5 marks]

A uniform filing cabinet of mass 60\,\mathrm{kg} is 0.50\,\mathrm{m} wide and 1.4\,\mathrm{m} tall and stands on a level floor. The coefficient of static friction between cabinet and floor is 0.36. A worker pushes horizontally on the side of the cabinet, at a height h above the floor, with a slowly increasing force.

(a) Show that a push of about 210\,\mathrm{N} is needed to make the cabinet slide. [1]

(b) Determine the smallest height h for which the cabinet tips before it slides. [3]

Show solutionHide solution

(a) Maximum static friction =0.36(60)(9.8)=211.7\approx210\,\mathrm{N} (A1). (b) At the point of tipping the floor's force acts only at the front bottom edge, so take moments about that edge (M1). The weight's arm is half the width, 0.25\,\mathrm{m}: tipping needs Ph\ge588(0.25)=147\,\mathrm{N\,m} (M1). With the largest push before sliding, P=211.7\,\mathrm{N}, h=147/211.7=0.694\,\mathrm{m} (A1). (c) Push lower than 0.69\,\mathrm{m}: the moment Ph of a 212\,\mathrm{N} push is then less than the weight's 147\,\mathrm{N\,m}, so the cabinet slides before it can tip (A1).

Final answer

(a) \mu_smg=211.7\,\mathrm{N}. (b) 0.694\,\mathrm{m}. (c) Below 0.69\,\mathrm{m}, near the bottom: the push then has a smaller moment about the front edge.

Question B5

Type B5 / recover-angular-motion / Predicted type / Stretch / P2[5 marks]

A sensor records the angle turned by a rotor that has a constant angular acceleration. The angle is zero at t=0. At t=2.0\,\mathrm{s} the angle is 14.0\,\mathrm{rad} and at t=5.0\,\mathrm{s} it is 20.0\,\mathrm{rad}.

(a) Show that the initial angular velocity is 9.0\,\mathrm{rad\,s^{-1}} and the angular acceleration is -2.0\,\mathrm{rad\,s^{-2}}. [3]

(b) Determine the time at which the rotor is momentarily at rest and the largest angle it reaches. [2]

Show solutionHide solution

(a) Use \Delta\theta=\omega_it+\tfrac12\alpha t^2 for both readings (M1): 14.0=2.0\omega_i+2.0\alpha and 20.0=5.0\omega_i+12.5\alpha (M1). From the first, \omega_i=7.0-\alpha; substituting, 35-5\alpha+12.5\alpha=20, so \alpha=-2.0\,\mathrm{rad\,s^{-2}} and \omega_i=9.0\,\mathrm{rad\,s^{-1}} (A1). The rotor was not released from rest. (b) \omega_f=0: t=9.0/2.0=4.5\,\mathrm{s} (A1). \Delta\theta=9.0(4.5)-(4.5)^2=20.25\,\mathrm{rad} (A1). The 5.0\,\mathrm{s} reading, 20.0\,\mathrm{rad}, comes after the reversal, so it is not the maximum.

Final answer

(a) \omega_i=9.0\,\mathrm{rad\,s^{-1}}, \alpha=-2.0\,\mathrm{rad\,s^{-2}}. (b) 4.5\,\mathrm{s}; 20.25\,\mathrm{rad}.

Question D5

Type D5 / rolling-forces / Stretch / P2[5 marks]

A solid ball of mass 0.50\,\mathrm{kg}, moment of inertia \tfrac25mr^2, rolls without slipping down a slope at 15^\circ to the horizontal.

(a) Explain why the friction force on the ball acts up the slope. [1]

(b) Find the acceleration of the ball down the slope and the size of the friction force. [3]

Show solutionHide solution

(a) Weight and normal force pass through the centre and have no torque about it. Without friction the ball would slide without turning; static friction up the slope opposes the slip of the contact point and supplies the torque (A1). (b) Along the slope: mg\sin15^\circ-F_f=ma; about the centre: F_fr=\tfrac25mr^2(a/r), so F_f=\tfrac25ma (M1). Then a=\tfrac57g\sin15^\circ=1.81\,\mathrm{m\,s^{-2}} (A1) and F_f=\tfrac25(0.50)(1.81)=0.362\,\mathrm{N} (A1). (c) At 30^\circ rolling needs F_f/F_N=\tfrac27\tan30^\circ=0.165>0.10, so the ball slips (A1).

Final answer

(a) Friction is the only force with a torque about the centre; it must give the ball its angular acceleration and stop the contact point sliding down. (b) 1.81\,\mathrm{m\,s^{-2}}; 0.362\,\mathrm{N}. (c) No: rolling needs F_f/F_N=0.165, more than 0.10, so the ball slips.

Question E4

Type E4 / loss-radius / Exam / P1A[1 mark]

A horizontal disc of radius 0.50\,\mathrm{m} and moment of inertia 0.0080\,\mathrm{kg\,m^2} spins freely about a vertical axis. A 0.050\,\mathrm{kg} lump of clay falls onto it and sticks at a distance r from the axis. For which r is half of the rotational kinetic energy lost?

A. 0.16\,\mathrm{m} B. 0.28\,\mathrm{m} C. 0.40\,\mathrm{m} D. 0.50\,\mathrm{m}

Show solutionHide solution

L is conserved, so E_f/E_i=I/(I+mr^2) and the fraction lost is mr^2/(I+mr^2). Half is lost when mr^2=I: r=\sqrt{0.0080/0.050}=0.40\,\mathrm{m}. The loss is zero at the axis and grows as r^2 close to it, then levels off. A forgets the square root, B equates the loss to the clay's own KE \tfrac12mr^2\omega_i^2, D assumes the rim.

Final answer

Question E6

Type E6 / momentum-gradient / Predicted type / Exam / P1A[1 mark]

A graph of the angular momentum L of a wheel against its angular velocity \omega is a straight line through the origin and the point (40\,\mathrm{rad\,s^{-1}},\ 2.4\,\mathrm{kg\,m^2\,s^{-1}}). What is the rotational kinetic energy of the wheel at 40\,\mathrm{rad\,s^{-1}}?

A. 48\,\mathrm{J} B. 96\,\mathrm{J}

C. 1.9\times10^3\,\mathrm{J} D. 1.3\times10^4\,\mathrm{J}

Show solutionHide solution

L=I\omega, so the gradient is I=2.4/40=0.060\,\mathrm{kg\,m^2}. E_k=\tfrac12I\omega^2=\tfrac12(0.060)(40)^2=48\,\mathrm{J}, the same as \tfrac12L\omega. B drops the \tfrac12, C takes the L reading, 2.4, as the moment of inertia, D inverts the gradient.

Final answer

Question E8

Type E8 / particle-stick / Stretch / P2[6 marks]

A uniform board of mass 1.5\,\mathrm{kg} and length 0.80\,\mathrm{m} lies on a frictionless horizontal table and can turn about a vertical axle through one end. Its moment of inertia about the axle is \tfrac13ML^2. A 0.060\,\mathrm{kg} ball of clay slides across the table at 5.0\,\mathrm{m\,s^{-1}}, perpendicular to the board, and sticks to it 0.60\,\mathrm{m} from the axle.

(a) Calculate the angular momentum of the clay about the axle just before impact. [1]

(b) Determine the angular velocity of the board just after impact. [2]

(d) Explain why the linear momentum of the board and clay is not conserved in the impact, but their angular momentum about the axle is. [1]

Show solutionHide solution

(a) L=mvr_\perp=0.060(5.0)(0.60)=0.18\,\mathrm{kg\,m^2\,s^{-1}} (A1). (b) I_f=\tfrac13(1.5)(0.80)^2+0.060(0.60)^2=0.320+0.0216=0.3416\,\mathrm{kg\,m^2} (M1); \omega=0.18/0.3416=0.527\,\mathrm{rad\,s^{-1}} (A1). (c) Before \tfrac12(0.060)(5.0)^2=0.75\,\mathrm{J}; after L^2/2I_f=0.0474\,\mathrm{J} (M1); fraction lost 0.703/0.75=0.94 (A1). (d) During the impact the axle pushes on the board, an external force that changes the linear momentum; its line of action passes through the axis, so its torque about the axle is zero and L is conserved (A1).

Final answer

(a) 0.18\,\mathrm{kg\,m^2\,s^{-1}}. (b) 0.527\,\mathrm{rad\,s^{-1}}. (c) 0.94. (d) The axle exerts an external force, but it acts through the axis, so its torque is zero.

Connected questions

Question X1

A.4 + A.1 + A.3 / Connected / P2[8 marks]

A bucket of water of total mass 8.0\,\mathrm{kg} hangs from a rope wound round the drum of a well windlass. The drum is a solid cylinder of mass 20\,\mathrm{kg} and radius 0.10\,\mathrm{m}, with moment of inertia \tfrac12MR^2, on a frictionless axle. The handle is released and the bucket falls from rest down the 12\,\mathrm{m} shaft; the rope does not slip on the drum. Ignore air resistance and the mass of the rope.

(a) Explain why the tension in the rope is less than the weight of the bucket. [1]

(b) Show that the acceleration of the bucket is about 4.4\,\mathrm{m\,s^{-2}}. [3]

(d) Determine the speed of the bucket at the bottom and the fraction of the released energy that is stored in the drum. [2]

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(a) The bucket accelerates downwards, so mg-T=ma>0 (A1). (b) Bucket: 8.0(9.8)-T=8.0a (M1). Drum: TR=I\alpha=\tfrac12MR^2(a/R), so T=\tfrac12Ma=10a (M1). Adding: 78.4=18a, a=4.36\,\mathrm{m\,s^{-2}} (A1). (c) A.1: s=\tfrac12at^2 from rest (M1), t=\sqrt{2(12)/4.36}=2.35\,\mathrm{s} (A1). (d) v=at=10.2\,\mathrm{m\,s^{-1}} (A1). A.3: released energy mgh=8.0(9.8)(12)=941\,\mathrm{J}; drum \tfrac12I\omega^2=\tfrac12(0.10)(102)^2=523\,\mathrm{J}, a fraction 10/18=0.56 (A1). The bucket keeps only 4/9 of the energy, which is why its speed is well below \sqrt{2gh}=15.3\,\mathrm{m\,s^{-1}}.

Final answer

(a) The bucket accelerates downwards, so the resultant force on it is downwards. (b) 4.36\,\mathrm{m\,s^{-2}}. (c) 2.35\,\mathrm{s}. (d) 10.2\,\mathrm{m\,s^{-1}}; 0.56 (5/9).

Question X2

A.4 + D.1 / Connected / P2[8 marks]

Model the Earth as a sphere of mass 5.97\times10^{24}\,\mathrm{kg} and radius 6.37\times10^{6}\,\mathrm{m} with moment of inertia 0.33MR^2, spinning once every 86\,400\,\mathrm{s}. Tides raised by the Moon exert a small torque on the spinning Earth that opposes its rotation.

(a) Calculate the angular velocity of the Earth's spin. [1]

(b) Calculate the rotational kinetic energy of the Earth. [2]

(c) The day becomes 2.0\,\mathrm{ms} longer every century (3.16\times10^{9}\,\mathrm{s}). Determine the average torque that the tides exert on the Earth. [3]

(d) The Earth–Moon system has no significant external torque. Explain why the Moon's orbital radius must increase. You may use the orbital speed v=\sqrt{GM/r}. [2]

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(a) \omega=2\pi/86\,400=7.27\times10^{-5}\,\mathrm{rad\,s^{-1}} (A1). (b) I=0.33(5.97\times10^{24})(6.37\times10^{6})^2=7.99\times10^{37}\,\mathrm{kg\,m^2} (M1); E_k=\tfrac12I\omega^2=2.11\times10^{29}\,\mathrm{J} (A1). (c) \omega=2\pi/T, so a small change gives \Delta\omega/\omega=-\Delta T/T (M1): |\Delta\omega|=7.27\times10^{-5}(0.0020/86\,400)=1.68\times10^{-12}\,\mathrm{rad\,s^{-1}}. \tau=I\Delta\omega/\Delta t (M1) =7.99\times10^{37}(1.68\times10^{-12})/3.16\times10^{9}=4.3\times10^{16}\,\mathrm{N\,m} (A1). (d) The Earth's spin angular momentum falls, so the Moon's orbital angular momentum must rise by the same amount (M1). D.1: L=mvr=m\sqrt{GMr} increases with r, so the orbit widens (A1).

Final answer

(a) 7.27\times10^{-5}\,\mathrm{rad\,s^{-1}}. (b) 2.11\times10^{29}\,\mathrm{J}. (c) 4.3\times10^{16}\,\mathrm{N\,m}. (d) The Moon's orbital angular momentum increases; L=mvr=m\sqrt{GMr} grows with r.

Question X3

A.4 + E.1 + D.2 / Connected / P2[8 marks]

In the Bohr model of hydrogen the electron moves in a circle of radius 5.29\times10^{-11}\,\mathrm{m} about the proton in its ground state (n=1). Its angular momentum is quantised: mvr=nh/2\pi.

(a) Use the quantisation condition to find how fast the electron moves when n=1. [2]

(b) Treat the electron as a point mass. Calculate its angular velocity and its moment of inertia about the proton, and show that I\omega equals h/2\pi. [2]

(d) For n=2 the radius is four times larger. Determine \omega_1/\omega_2. [2]

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(a) v=h/(2\pi mr)=6.63\times10^{-34}/(2\pi\times9.11\times10^{-31}\times5.29\times10^{-11}) (M1) =2.19\times10^{6}\,\mathrm{m\,s^{-1}} (A1). (b) \omega=v/r=4.14\times10^{16}\,\mathrm{rad\,s^{-1}}; I=mr^2=2.55\times10^{-51}\,\mathrm{kg\,m^2} (M1); I\omega=1.06\times10^{-34}\,\mathrm{J\,s}=h/2\pi (A1): for a point mass L=I\omega and mvr are the same quantity. (c) D.2: Coulomb's law F=ke^2/r^2 with the booklet constants gives 8.22\times10^{-8}\,\mathrm{N} (M1); mv^2/r=8.26\times10^{-8}\,\mathrm{N}, agreeing within 1\% with the rounded constants used here (A1). (d) v=nh/(2\pi mr): doubling n and multiplying r by 4 halves v (M1). \omega=v/r falls by 2\times4, so \omega_1/\omega_2=8 (A1).

Final answer

(a) 2.19\times10^{6}\,\mathrm{m\,s^{-1}}. (b) 4.14\times10^{16}\,\mathrm{rad\,s^{-1}}; 2.55\times10^{-51}\,\mathrm{kg\,m^2}; I\omega=1.06\times10^{-34}\,\mathrm{J\,s}. (c) Electric: 8.22\times10^{-8}\,\mathrm{N}; centripetal: 8.26\times10^{-8}\,\mathrm{N}, agreeing within 1\%. (d) 8.

Question X4

A.4 + A.2 + A.3 / Connected / P2[8 marks]

A small solid ball, moment of inertia \tfrac25mr^2, rolls without slipping down a curved track and round a vertical loop of radius R=0.30\,\mathrm{m}. The radius of the ball is very small compared with R. It starts from rest, a height h above the lowest point of the loop.

(a) Show that the smallest speed at the top of the loop for the ball to stay on the track is about 1.7\,\mathrm{m\,s^{-1}}. [2]

(b) Show that the total kinetic energy of the ball is 0.7mv^2. [2]

(d) A small block that slides without friction needs a smaller release height. Explain why. [1]

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(a) A.2: at the top the track force is zero at the limit, so weight alone is the centripetal force: mg=mv^2/R (M1), v=\sqrt{9.8(0.30)}=1.71\,\mathrm{m\,s^{-1}} (A1). (b) \omega=v/r (M1); \tfrac12(\tfrac25mr^2)(v/r)^2=\tfrac15mv^2, total \tfrac12mv^2+\tfrac15mv^2=0.7mv^2 (A1). (c) A.3: mgh=mg(2R)+0.7mv^2 (M1) with v^2=gR (M1): h=2R+0.7R=2.7R=0.81\,\mathrm{m} (A1). (d) The block has no rotational KE, so h=2R+0.5R=0.75\,\mathrm{m}: the ball must carry an extra 0.2mv^2 as rotation (A1).

Final answer

(a) \sqrt{gR}=1.71\,\mathrm{m\,s^{-1}}. (b) \tfrac12mv^2+\tfrac15mv^2. (c) 0.81\,\mathrm{m} (2.7R). (d) None of its energy goes into rotation, so the same top speed needs less height (2.5R=0.75\,\mathrm{m}).

From the book

Photon Physics: Study Guide for the IB Diploma (SL & HL)

Rigid body mechanics from the core ideas to full exam questions, with question types, worked examples and graded practice.

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