Rigid body mechanics HL Paper 1&2

A.4 Rigid Body Mechanics

IB-style HL Paper 1 and Paper 2 question bank

This original practice bank contains 15 multiple-choice questions, 15 extended-response questions and a lab-style investigation. All worked answers expand below their questions. These are practice questions, not official IB examination questions.

Instructions and conventions

A.4 is an HL-only topic. This bank follows the rotational mechanics formulas on the supplied IB sheet.
Show working, include units and give numerical answers to 2–3 significant figures unless otherwise stated.
Use $g=9.81\,\mathrm{m\,s^{-2}}$ . Angular quantities in the motion equations must be in radians.
Take counterclockwise as positive unless another convention is stated. Torques and angular momenta about the same axis must be added with their signs.
Use the angular motion equations only for constant angular acceleration. Use $\tau_{\mathrm{net}}=I\alpha$ for a constant moment of inertia about the stated fixed axis.
Moments of inertia refer to the stated axis. Treat small masses as point masses and neglect the mass of connecting rods unless stated otherwise.

IB formula reference

$\tau=Fr\sin\theta$

$\Delta\theta=\frac{\omega_f+\omega_i}{2}t$

$\omega_f=\omega_i+\alpha t$

$\Delta\theta=\omega_i t+\frac12\alpha t^2$

$\omega_f^2=\omega_i^2+2\alpha\Delta\theta$

$I=\sum mr^2$

$\tau_{\mathrm{net}}=I\alpha$

$L=I\omega$

$\Delta L=\tau_{\mathrm{net,avg}}\Delta t$

$\Delta L=\Delta(I\omega)$

$E_k=\frac12I\omega^2=\frac{L^2}{2I}$

In the torque equation, $\theta$ is the angle between the force and the radius from the axis to the point of application. In angular motion, $\Delta\theta$ is angular displacement. The torque in the angular-impulse equation is the average resultant external torque about the chosen axis.

Useful derived connections: $v=\omega r$ , $a_t=\alpha r$ , $W=\tau\Delta\theta$ for constant torque, and $P=\tau\omega$ for instantaneous rotational power. These follow from linear motion and work–energy relationships.

Common misconceptions

“Zero resultant force guarantees equilibrium.” Static equilibrium requires zero resultant force and zero resultant torque.
“Torque depends only on force.” It also depends on the perpendicular distance from the axis to the force’s line of action.
“Moment of inertia depends only on mass.” Mass distribution and the chosen axis matter.
“Conserved angular momentum means conserved kinetic energy.” Internal work or an inelastic interaction can change kinetic energy while total angular momentum remains constant.
“An object at rest has zero angular acceleration.” Angular velocity can be zero at an instant while a non-zero resultant torque produces angular acceleration.

Part 1 — Multiple-choice questions

Multiple choice 1 — Torque at an angle [1 mark]

A 15 N force acts 0.40 m from an axis. The angle between the force and the radius is 30°. What is the torque magnitude?

6.0 N m
3.0 N m
12 N m
0 N m

Show worked answer

Correct option: B.

$\tau=(15)(0.40)\sin30^\circ=3.0\,\mathrm{N\,m}$

Multiple choice 2 — A force through the axis [1 mark]

A force’s line of action passes through the rotation axis. What is its torque about that axis?

Zero
Force multiplied by the body’s radius
Force multiplied by the body’s mass
It cannot be determined without angular velocity

Show worked answer

Correct option: A.

The perpendicular distance from the axis to the line of action is zero, so the torque is zero.

Multiple choice 3 — A couple [1 mark]

Two equal, opposite, parallel forces act on a body along different lines of action. Which statement is correct?

Both resultant force and resultant torque are zero.
The resultant force is non-zero and the resultant torque is zero.
The forces cannot cause rotation.
The resultant force is zero and the resultant torque is non-zero.

Show worked answer

Correct option: D.

The forces cancel translationally, but their torques form a couple. The torque magnitude is one force multiplied by the perpendicular separation of the lines of action.

Multiple choice 4 — Angular displacement [1 mark]

A wheel starts from rest and has constant angular acceleration 2.0 rad/s² for 3.0 s. What is its angular displacement?

6.0 rad
18 rad
9.0 rad
3.0 rad

Show worked answer

Correct option: C.

$\Delta\theta=\frac12\alpha t^2=\frac12(2.0)(3.0)^2=9.0\,\mathrm{rad}$

Multiple choice 5 — Angular acceleration [1 mark]

A rotor’s angular velocity increases uniformly from 4.0 rad/s to 10 rad/s in 3.0 s. What is its angular acceleration?

2.0 rad/s²
4.7 rad/s²
18 rad/s²
0.50 rad/s²

Show worked answer

Correct option: A.

$\alpha=\frac{10-4.0}{3.0}=2.0\,\mathrm{rad\,s^{-2}}$

Multiple choice 6 — Point masses [1 mark]

Two point masses, each of mass 0.20 kg, are each 0.30 m from a rotation axis. The connecting rod has negligible mass. What is the moment of inertia?

0.012 kg m²
0.018 kg m²
0.12 kg m²
0.036 kg m²

Show worked answer

Correct option: D.

$I=2mr^2=2(0.20)(0.30)^2=0.036\,\mathrm{kg\,m^2}$

Multiple choice 7 — Changing the mass distribution [1 mark]

All point masses on a rotor are moved to twice their original distances from the axis. The same resultant torque is then applied. What is the new angular acceleration compared with the original?

Twice as large
Four times as large
One-quarter as large
One-half as large

Show worked answer

Correct option: C.

$I_2=4I_1,\qquad\alpha_2=\frac{\tau}{4I_1}=\frac{\alpha_1}{4}$

Multiple choice 8 — Resultant torque [1 mark]

A wheel has moment of inertia 0.50 kg m². A 3.0 N m driving torque and an opposing 1.0 N m friction torque act. What is its angular acceleration in the driving direction?

8.0 rad/s²
4.0 rad/s²
6.0 rad/s²
2.0 rad/s²

Show worked answer

Correct option: B.

$\alpha=\frac{3.0-1.0}{0.50}=4.0\,\mathrm{rad\,s^{-2}}$

Multiple choice 9 — Rotational energy scaling [1 mark]

The angular speed of a rigid body about a fixed axis is tripled. Its moment of inertia is unchanged. By what factor does its rotational kinetic energy increase?

Show worked answer

Correct option: D.

$E_k\propto\omega^2\quad\Rightarrow\quad\frac{E_{k,2}}{E_{k,1}}=3^2=9$

Multiple choice 10 — Angular momentum [1 mark]

A rotor has moment of inertia 0.40 kg m² and angular velocity +5.0 rad/s. What is its angular momentum about that axis?

+0.080 kg m²/s
+12.5 kg m²/s
+2.0 kg m²/s
+10 kg m²/s

Show worked answer

Correct option: C.

$L=I\omega=(0.40)(5.0)=+2.0\,\mathrm{kg\,m^2\,s^{-1}}$

Multiple choice 11 — Conservation of angular momentum [1 mark]

A rotating system’s moment of inertia is halved. Resultant external angular impulse is negligible. What happens to its angular speed?

It doubles.
It halves.
It is unchanged.
It increases by a factor of four.

Show worked answer

Correct option: A.

$I_i\omega_i=I_f\omega_f,\qquad\omega_f=\frac{I_i}{I_i/2}\omega_i=2\omega_i$

Multiple choice 12 — Energy at constant angular momentum [1 mark]

A rotating system’s moment of inertia is halved while its angular momentum remains constant. What happens to its rotational kinetic energy?

It halves.
It doubles.
It is unchanged.
It increases by a factor of four.

Show worked answer

Correct option: B.

$E_k=\frac{L^2}{2I}\quad\Rightarrow\quad E_{k,f}=2E_{k,i}$

The increase can be supplied by internal work while the mass distribution changes.

Multiple choice 13 — Angular impulse [1 mark]

A constant resultant torque of +6.0 N m acts for 0.50 s. What is the change in angular momentum?

+12 kg m²/s
+6.0 kg m²/s
+3.0 kg m²/s
+0.083 kg m²/s

Show worked answer

Correct option: C.

$\Delta L=\tau\Delta t=(6.0)(0.50)=+3.0\,\mathrm{kg\,m^2\,s^{-1}}$

Multiple choice 14 — Signed braking torque [1 mark]

A wheel rotates counterclockwise and is slowing down under friction. Counterclockwise is positive. Which pair correctly gives the signs of friction torque and angular acceleration?

Positive, positive
Positive, negative
Negative, positive
Negative, negative

Show worked answer

Correct option: D.

The friction torque is clockwise, so it is negative. With positive constant I, angular acceleration has the same sign as resultant torque. The positive angular velocity is decreasing.

Multiple choice 15 — Two rotating discs [1 mark]

Two discs with equal moments of inertia rotate about the same axis with angular velocities +8.0 rad/s and −4.0 rad/s. They couple and rotate together. External angular impulse is negligible. What is their final angular velocity?

+6.0 rad/s
+2.0 rad/s
+4.0 rad/s
−2.0 rad/s

Show worked answer

Correct option: B.

$I(8.0)+I(-4.0)=(2I)\omega_f$

$\omega_f=+2.0\,\mathrm{rad\,s^{-1}}$

Part 2 — Extended-response questions

Extended response 1 — An angled force on a wrench [6 marks]

A force of 80 N is applied 0.250 m from a bolt’s axis. The angle between the force and the wrench’s radius is 40°.

Calculate the torque magnitude about the bolt. [2]
Determine the torque if the same force acts perpendicular to the wrench. [2]
Find the smallest force that can produce the original torque at the same point, and state its direction. [2]

Show worked answer

(a) [2]

$\tau=Fr\sin\theta=(80)(0.250)\sin40^\circ=12.9\,\mathrm{N\,m}$

(b) [2] A perpendicular force gives the maximum torque for this F and r.

$\tau_{\mathrm{max}}=(80)(0.250)=20.0\,\mathrm{N\,m}$

(c) [2] Minimise F by making the perpendicular lever arm equal to r. Use the unrounded original torque.

$F_{\mathrm{min}}=\frac{12.855752}{0.250}=51.4\,\mathrm{N}$

Answer: Original torque = 12.9 N m; perpendicular-force torque = 20.0 N m; minimum force = 51.4 N, perpendicular to the wrench.

Extended response 2 — A loaded beam in static equilibrium [8 marks]

A uniform horizontal beam is 4.0 m long and has mass 12 kg. It is supported by upward vertical reactions at both ends. A 20 kg load rests 3.0 m from the left end.

State the two conditions for static equilibrium. [2]
Use torques about the left end to calculate the right support reaction. [3]
Calculate the left support reaction. [2]
Explain why the left reaction is absent from the torque equation about the left end. [1]

Show worked answer

(a) [2] Resultant force is zero and resultant torque about any axis is zero.

(b) [3] The beam’s weight acts at its centre, 2.0 m from the left end. Taking counterclockwise torque as positive:

$R_R(4.0)-(12g)(2.0)-(20g)(3.0)=0$

$R_R=\frac{(12)(9.81)(2.0)+(20)(9.81)(3.0)}{4.0}=206\,\mathrm{N}$

(c) [2]

$R_L+R_R=(12+20)g$

$R_L=313.92-206.01=108\,\mathrm{N}$

(d) [1] Its line of action passes through the chosen axis, so its perpendicular lever arm is zero.

Answer: Right reaction = 206 N; left reaction = 108 N.

Extended response 3 — Balancing a seesaw [6 marks]

A uniform seesaw of mass 10 kg is pivoted at its centre. A 30 kg child sits 1.80 m to the left of the pivot. A 45 kg child sits to the right. The seesaw is horizontal and in static equilibrium.

Determine the distance of the 45 kg child from the pivot. [2]
Calculate the upward pivot reaction. [2]
Explain why the seesaw’s own weight affects the pivot reaction but not the torque balance about the pivot. [2]

Show worked answer

(a) [2] The children’s opposing torques balance.

$(30g)(1.80)=(45g)r$

$r=\frac{30(1.80)}{45}=1.20\,\mathrm{m}$

(b) [2] The reaction balances all three weights.

$R=(30+45+10)g=(85)(9.81)=834\,\mathrm{N}$

(c) [2] The uniform seesaw’s centre of mass is at the pivot, so its weight has zero lever arm. Its weight is nevertheless a downward force and must be included in vertical force balance.

Answer: The heavier child sits 1.20 m to the right; pivot reaction = 834 N.

Extended response 4 — Uniform angular acceleration [6 marks]

A wheel initially rotates at 3.0 rad/s. Its constant angular acceleration is 1.20 rad/s² for 5.0 s. The wheel’s radius is 0.200 m.

Calculate the final angular velocity. [2]
Calculate the angular displacement. [2]
Express this displacement in revolutions. [1]
Calculate the final tangential speed of a point on the rim. [1]

Show worked answer

(a) [2]

$\omega_f=\omega_i+\alpha t=3.0+(1.20)(5.0)=9.0\,\mathrm{rad\,s^{-1}}$

(b) [2]

$\Delta\theta=(3.0)(5.0)+\frac12(1.20)(5.0)^2=30.0\,\mathrm{rad}$

(c) [1] One revolution is $2\pi$ radians.

$n=\frac{30.0}{2\pi}=4.77\ \text{revolutions}$

(d) [1]

$v=\omega_f r=(9.0)(0.200)=1.80\,\mathrm{m\,s^{-1}}$

Answer: Final angular velocity = 9.0 rad/s; displacement = 30.0 rad = 4.77 revolutions; rim speed = 1.80 m/s.

Extended response 5 — A flywheel braking uniformly [8 marks]

A flywheel of moment of inertia 0.150 kg m² rotates initially at +24.0 rad/s. A constant braking torque brings it to rest in 6.0 s.

Calculate the angular acceleration. [2]
Calculate the angular displacement during braking. [2]
Determine the resultant braking torque. [2]
Calculate the decrease in rotational kinetic energy and identify where this energy goes if friction provides the braking torque. [2]

Show worked answer

(a) [2]

$\alpha=\frac{0-24.0}{6.0}=-4.0\,\mathrm{rad\,s^{-2}}$

(b) [2]

$\Delta\theta=\frac{24.0+0}{2}(6.0)=72.0\,\mathrm{rad}$

(c) [2]

$\tau_{\mathrm{net}}=I\alpha=(0.150)(-4.0)=-0.600\,\mathrm{N\,m}$

(d) [2]

$E_{k,i}=\frac12(0.150)(24.0)^2=43.2\,\mathrm{J}$

The rotational kinetic energy decreases by 43.2 J. This becomes internal energy of the brake, flywheel and surroundings.

Answer: Angular acceleration = −4.0 rad/s²; displacement = +72.0 rad; braking torque = −0.600 N m; energy decrease = 43.2 J.

Extended response 6 — Moment of inertia of point masses [6 marks]

Three point masses are attached to a rigid frame of negligible mass. Their masses and distances from the rotation axis are: 0.20 kg at 0.30 m; 0.40 kg at 0.20 m; 0.10 kg at 0.50 m.

Calculate the moment of inertia about the stated axis. [2]
A resultant torque of +0.118 N m acts. Calculate the angular acceleration. [2]
Determine the angular momentum when the angular velocity is +8.0 rad/s. [2]

Show worked answer

(a) [2] Add each mass’s contribution, using its own radius.

$I=(0.20)(0.30)^2+(0.40)(0.20)^2+(0.10)(0.50)^2$

$I=0.018+0.016+0.025=0.0590\,\mathrm{kg\,m^2}$

(b) [2]

$\alpha=\frac{\tau_{\mathrm{net}}}{I}=\frac{0.118}{0.0590}=+2.00\,\mathrm{rad\,s^{-2}}$

(c) [2]

$L=I\omega=(0.0590)(8.0)=+0.472\,\mathrm{kg\,m^2\,s^{-1}}$

Answer: Moment of inertia = 0.0590 kg m²; angular acceleration = +2.00 rad/s²; angular momentum = +0.472 kg m²/s.

Extended response 7 — Mass distribution: a ring and a disc [8 marks]

A thin ring and a uniform solid disc have the same mass M and radius R. Both rotate about their central symmetry axes. The disc’s moment of inertia is given as $I_{\mathrm{disc}}=MR^2/2$ .

Use the point-mass expression to show that the thin ring has moment of inertia MR². [3]
The same resultant torque acts on each. Compare their angular accelerations. [2]
Compare their rotational kinetic energies first at the same angular speed, then at the same angular momentum. [3]

Show worked answer

(a) [3] Divide the thin ring into small masses. Every mass is the same distance R from the axis.

$I_{\mathrm{ring}}=\sum m_jR^2=R^2\sum m_j=MR^2$

(b) [2] The ring has twice the disc’s moment of inertia, so its angular acceleration is half as large under the same torque.

$\frac{\alpha_{\mathrm{ring}}}{\alpha_{\mathrm{disc}}}=\frac{I_{\mathrm{disc}}}{I_{\mathrm{ring}}}=\frac12$

(c) [3] At equal angular speed, use $E_k=I\omega^2/2$ : the ring has twice the disc’s energy. At equal angular momentum, use $E_k=L^2/(2I)$ : the ring has half the disc’s energy.

Answer: Ring moment of inertia = MR²; ring angular acceleration = half the disc’s; energy ratios are 2 at equal angular speed and 1/2 at equal angular momentum.

Extended response 8 — Motor torque, friction and energy [10 marks]

A rotor of moment of inertia 0.800 kg m² starts from rest. A constant +5.0 N m motor torque and a constant −1.4 N m friction torque act for 6.0 s. The rotor turns counterclockwise throughout this interval.

Calculate the resultant torque and angular acceleration. [2]
Determine the final angular velocity and angular displacement. [3]
Calculate the final rotational kinetic energy. [2]
Calculate the motor’s work and the energy dissipated by friction. Check the energy balance. [3]

Show worked answer

(a) [2]

$\tau_{\mathrm{net}}=5.0-1.4=3.6\,\mathrm{N\,m}$

$\alpha=\frac{3.6}{0.800}=4.50\,\mathrm{rad\,s^{-2}}$

(b) [3]

$\omega_f=(4.50)(6.0)=27.0\,\mathrm{rad\,s^{-1}}$

$\Delta\theta=\frac12(4.50)(6.0)^2=81.0\,\mathrm{rad}$

(c) [2]

$E_{k,f}=\frac12(0.800)(27.0)^2=291.6\,\mathrm{J}$

(d) [3] For constant torque, work is torque multiplied by signed angular displacement.

$W_{\mathrm{motor}}=(5.0)(81.0)=405\,\mathrm{J}$

$W_{\mathrm{friction}}=(-1.4)(81.0)=-113.4\,\mathrm{J}$

$405-113.4=291.6\,\mathrm{J}=\Delta E_k$

Answer: Angular acceleration = 4.50 rad/s²; final angular velocity = 27.0 rad/s; displacement = 81.0 rad; motor work = 405 J; dissipated energy = 113 J; final rotational energy ≈ 292 J.

Extended response 9 — Angular impulse from a torque–time graph [10 marks]

A rotor has moment of inertia 2.0 kg m² and initial angular velocity −3.0 rad/s. The resultant external torque increases linearly from 0 to +4.0 N m between t = 0 and 2.0 s, remains +4.0 N m from 2.0 to 5.0 s, then decreases linearly to zero from 5.0 to 7.0 s.

Calculate the total angular impulse from the torque–time graph. [3]
Determine the initial and final angular momenta. [3]
Calculate the final angular velocity, including its sign. [2]
Calculate the change in rotational kinetic energy. [2]

Show worked answer

(a) [3] Add the two triangular areas and the rectangular area. All are above the time axis.

$\Delta L=\frac12(2.0)(4.0)+(3.0)(4.0)+\frac12(2.0)(4.0)$

$\Delta L=4.0+12.0+4.0=+20.0\,\mathrm{N\,m\,s}$

(b) [3]

$L_i=(2.0)(-3.0)=-6.0\,\mathrm{kg\,m^2\,s^{-1}}$

$L_f=L_i+\Delta L=-6.0+20.0=+14.0\,\mathrm{kg\,m^2\,s^{-1}}$

(c) [2]

$\omega_f=\frac{14.0}{2.0}=+7.0\,\mathrm{rad\,s^{-1}}$

(d) [2]

$\Delta E_k=\frac12(2.0)\left[7.0^2-(-3.0)^2\right]=+40.0\,\mathrm{J}$

Answer: Angular impulse = +20.0 N m s; final angular momentum = +14.0 kg m²/s; final angular velocity = +7.0 rad/s; rotational-energy change = +40.0 J.

Extended response 10 — A skater draws in their arms [7 marks]

A skater initially has moment of inertia 3.00 kg m² and angular velocity +2.00 rad/s. Drawing in their arms reduces the moment of inertia to 1.20 kg m². Resultant external angular impulse is negligible.

Calculate the initial angular momentum. [1]
Determine the final angular velocity. [2]
Calculate the change in rotational kinetic energy. [2]
Explain why the energy changes although angular momentum is conserved. [2]

Show worked answer

(a) [1]

$L_i=(3.00)(2.00)=+6.00\,\mathrm{kg\,m^2\,s^{-1}}$

(b) [2]

$I_i\omega_i=I_f\omega_f$

$\omega_f=\frac{6.00}{1.20}=+5.00\,\mathrm{rad\,s^{-1}}$

(c) [2]

$E_{k,i}=\frac12(3.00)(2.00)^2=6.00\,\mathrm{J}$

$E_{k,f}=\frac12(1.20)(5.00)^2=15.0\,\mathrm{J}$

$\Delta E_k=15.0-6.00=+9.00\,\mathrm{J}$

(d) [2] The skater’s muscles do work while changing the mass distribution. Chemical energy is transferred into rotational kinetic energy. Negligible external angular impulse conserves angular momentum, not necessarily kinetic energy.

Answer: Angular momentum = +6.00 kg m²/s; final angular velocity = +5.00 rad/s; rotational-energy increase = 9.00 J.

Extended response 11 — Two discs couple [8 marks]

Two coaxial discs rotate independently. Disc 1 has moment of inertia 1.20 kg m² and angular velocity +8.0 rad/s. Disc 2 has moment of inertia 0.800 kg m² and angular velocity −3.0 rad/s. They are brought into contact and eventually rotate together. External angular impulse is negligible.

Calculate the total initial angular momentum. [2]
Calculate their common final angular velocity. [2]
Calculate the decrease in total rotational kinetic energy. [3]
Explain where the lost rotational kinetic energy goes. [1]

Show worked answer

(a) [2] Add angular momenta with signs.

$L_i=(1.20)(8.0)+(0.800)(-3.0)=+7.20\,\mathrm{kg\,m^2\,s^{-1}}$

(b) [2] The final moment of inertia is the sum about the common axis.

$\omega_f=\frac{7.20}{1.20+0.800}=+3.60\,\mathrm{rad\,s^{-1}}$

(c) [3]

$E_{k,i}=\frac12(1.20)(8.0)^2+\frac12(0.800)(3.0)^2=42.0\,\mathrm{J}$

$E_{k,f}=\frac12(2.00)(3.60)^2=12.96\,\mathrm{J}$

$E_{\mathrm{decrease}}=42.0-12.96=29.0\,\mathrm{J}$

(d) [1] Friction between the discs transfers rotational energy into internal energy, with possible sound. Internal torques exchange angular momentum between the discs but do not change the system’s total angular momentum.

Answer: Total initial angular momentum = +7.20 kg m²/s; final angular velocity = +3.60 rad/s; rotational-energy decrease = 29.0 J.

Extended response 12 — A mass lands on a rotating platform [7 marks]

A horizontal platform has moment of inertia 120 kg m² about its vertical axis and rotates at +0.500 rad/s. A small 50.0 kg mass is lowered vertically onto the platform at radius 2.00 m and sticks to it. The mass initially has zero angular momentum about the vertical axis. External angular impulse about that axis is negligible.

Calculate the final moment of inertia of the platform and mass. [2]
Calculate their final angular velocity. [2]
Determine the decrease in rotational kinetic energy. [2]
State why angular momentum conservation is appropriate. [1]

Show worked answer

(a) [2] Treat the added mass as a point mass.

$I_f=120+(50.0)(2.00)^2=320\,\mathrm{kg\,m^2}$

(b) [2]

$L_i=(120)(0.500)=60.0\,\mathrm{kg\,m^2\,s^{-1}}$

$\omega_f=\frac{60.0}{320}=+0.1875\,\mathrm{rad\,s^{-1}}$

(c) [2]

$E_{k,i}=\frac12(120)(0.500)^2=15.0\,\mathrm{J}$

$E_{k,f}=\frac12(320)(0.1875)^2=5.625\,\mathrm{J}$

$E_{\mathrm{decrease}}=15.0-5.625=9.38\,\mathrm{J}$

(d) [1] The resultant external angular impulse about the vertical axis is negligible. Vertical forces have no torque component about that axis. The calculation concerns rotational energy only; any vertical landing energy is separate.

Answer: Final moment of inertia = 320 kg m²; final angular velocity ≈ +0.188 rad/s; rotational-energy decrease = 9.38 J.

Extended response 13 — Challenge: A falling load drives a pulley [11 marks]

A 2.0 kg load hangs from a light string wound around a fixed-axis pulley of radius 0.100 m and moment of inertia 0.0400 kg m². The system starts from rest and the load descends 1.50 m. The string does not slip. Neglect bearing friction and air resistance.

Combine the linear and rotational equations to determine the load’s acceleration. [4]
Calculate the string tension. [2]
Calculate the load’s final speed and the pulley’s final angular speed. [2]
Calculate the final translational and rotational kinetic energies, and check energy conservation. [3]

Show worked answer

(a) [4] Use downward as positive for the load and the unwinding sense as positive for the pulley. Denote tension by T and radius by R.

$mg-T=ma$

$TR=I\alpha,\qquad a=\alpha R$

$T=\frac{Ia}{R^2},\qquad a=\frac{mg}{m+I/R^2}$

$a=\frac{(2.0)(9.81)}{2.0+0.0400/(0.100)^2}=3.27\,\mathrm{m\,s^{-2}}$

(b) [2]

$T=m(g-a)=(2.0)(9.81-3.27)=13.1\,\mathrm{N}$

(c) [2]

$v=\sqrt{2ah}=\sqrt{2(3.27)(1.50)}=3.13\,\mathrm{m\,s^{-1}}$

$\omega_f=\frac{v}{R}=31.3\,\mathrm{rad\,s^{-1}}$

(d) [3] Using unrounded speeds:

$E_{k,\mathrm{load}}=\frac12mv^2=9.81\,\mathrm{J}$

$E_{k,\mathrm{pulley}}=\frac12I\omega_f^2=19.62\,\mathrm{J}$

$9.81+19.62=29.43\,\mathrm{J}=mgh$

Answer: Load acceleration = 3.27 m/s²; tension = 13.1 N; final speed = 3.13 m/s; angular speed = 31.3 rad/s; the 29.4 J gravitational-energy decrease supplies both kinetic energies.

Extended response 14 — Challenge: A rotor reverses direction [10 marks]

A rotor has moment of inertia 0.500 kg m² and initial angular velocity +6.0 rad/s. A constant resultant torque of −1.50 N m acts for 4.0 s. The torque is supplied by an active device and remains clockwise even after the rotor reverses direction.

Calculate the angular impulse over the interval. [2]
Calculate the final angular velocity. [2]
Determine when the rotor is momentarily at rest. [2]
Calculate the signed angular displacement over the whole interval. [2]
Calculate the change in rotational kinetic energy and explain the result in terms of work. [2]

Show worked answer

(a) [2]

$\Delta L=(-1.50)(4.0)=-6.00\,\mathrm{kg\,m^2\,s^{-1}}$

(b) [2] Initial angular momentum is $(0.500)(6.0)=+3.00\,\mathrm{kg\,m^2\,s^{-1}}$ .

$L_f=3.00-6.00=-3.00\,\mathrm{kg\,m^2\,s^{-1}}$

$\omega_f=\frac{-3.00}{0.500}=-6.0\,\mathrm{rad\,s^{-1}}$

(c) [2]

$\alpha=\frac{-1.50}{0.500}=-3.00\,\mathrm{rad\,s^{-2}}$

$0=6.0-3.00t\quad\Rightarrow\quad t=2.0\,\mathrm{s}$

(d) [2]

$\Delta\theta=(6.0)(4.0)+\frac12(-3.00)(4.0)^2=0\,\mathrm{rad}$

It turns +6.0 rad in the first half, then −6.0 rad in the second half. Its total angular distance is 12.0 rad, but its signed displacement is zero.

(e) [2] Initial and final angular speeds are equal, so rotational-energy change is zero. The device does negative work before reversal and equal positive work afterward.

$\Delta E_k=\frac12(0.500)\left[(-6.0)^2-(6.0)^2\right]=0\,\mathrm{J}$

Answer: Angular impulse = −6.00 kg m²/s; final angular velocity = −6.0 rad/s; reversal at 2.0 s; net displacement = 0 rad; rotational-energy change = 0 J.

Extended response 15 — Challenge: Energy supplied by a flywheel [10 marks]

A flywheel has moment of inertia 2.00 kg m². Its angular speed falls from 100 rad/s to 60.0 rad/s while supplying a device. The device’s efficiency is 80.0% and its useful output power remains constant at 800 W. Neglect other losses.

Calculate the decrease in the flywheel’s rotational kinetic energy. [2]
Calculate the time for which the useful output is supplied. [2]
Taking the original rotation as positive, calculate the average resultant torque on the flywheel over this interval. [2]
Find the magnitudes of the instantaneous energy-extraction torques at the initial and final speeds. [3]
State whether angular acceleration is constant. [1]

Show worked answer

(a) [2]

$E_{\mathrm{decrease}}=\frac12(2.00)(100^2-60.0^2)=6400\,\mathrm{J}$

(b) [2] Useful output is 80.0% of extracted energy.

$t=\frac{(0.800)(6400)}{800}=6.40\,\mathrm{s}$

(c) [2] Use the average torque in the angular-impulse relationship; do not assume uniform angular acceleration.

$\Delta L=I(\omega_f-\omega_i)=(2.00)(60.0-100)=-80.0\,\mathrm{kg\,m^2\,s^{-1}}$

$\tau_{\mathrm{net,avg}}=\frac{-80.0}{6.40}=-12.5\,\mathrm{N\,m}$

(d) [3] The rate of rotational-energy extraction is $800/0.800=1000\,\mathrm{W}$ . Since $P=|\tau|\omega$ :

$|\tau_i|=\frac{1000}{100}=10.0\,\mathrm{N\,m}$

$|\tau_f|=\frac{1000}{60.0}=16.7\,\mathrm{N\,m}$

(e) [1] No. The torque magnitude increases as angular speed falls, so the angular deceleration magnitude increases.

Answer: Rotational-energy decrease = 6.40 kJ; useful operating time = 6.40 s; average torque = −12.5 N m; extraction torque magnitudes = 10.0 N m initially and 16.7 N m finally.

Part 3 — Lab-style investigation

Investigating mass distribution and angular acceleration [20 marks]

A flywheel rotates about a fixed horizontal axle. Two small identical masses, each of mass $M=0.0500\,\mathrm{kg}$ , are attached at opposite positions, both at adjustable radius r. Treat them as point masses. Their opposing positions balance their gravitational torques.

A light string is wound around an axle of constant radius $R=0.0200\,\mathrm{m}$ . A hanging mass $m=0.0500\,\mathrm{kg}$ descends as the flywheel accelerates from rest. An optical encoder measures the time t for the first complete revolution. The string does not slip. Model bearing friction as a constant opposing torque $\tau_f$ . Let $I_0$ be the moment of inertia of the flywheel and all fixed rotating hardware, excluding the two adjustable masses.

The data below are simulated practice data. The time uncertainty is ±0.020 s. For the timing uncertainty calculation, treat one revolution as exactly $2\pi$ radians and neglect other uncertainties. Treat radians as dimensionless when expressing graph gradient and intercept units.

r / m	t / s	r² / m²	α / rad s⁻²	1/α / s²
0.050	1.296	Calculate	Calculate	Calculate
0.075	1.474	Calculate	Calculate	Calculate
0.100	1.691	Calculate	Calculate	Calculate
0.125	1.935	Calculate	Calculate	Calculate
0.150	2.196	Calculate	Calculate	Calculate

Identify the independent variable and directly measured dependent variable. State two control variables. [2]
Combine the hanging mass’s linear equation with the flywheel’s rotational equation to show the relationship below. Explain why the string tension is not simply mg. [3]
Use the first-revolution time to determine angular acceleration, then complete the calculated columns. Show one example calculation. [3]
Describe a suitable straight-line graph. Estimate its gradient b and intercept c, with units. [3]
Use b and c to estimate the bearing friction torque and I₀. [3]
Calculate the absolute uncertainty in angular acceleration for the largest radius. Explain why timing uncertainty alone cannot establish the model’s validity. [3]
Suggest two improvements and identify one limitation of the constant-friction model. [3]

Relationship to show in part (b):

$\frac{1}{\alpha}=\frac{I_0+mR^2}{mgR-\tau_f}+\frac{2M}{mgR-\tau_f}r^2$

Show worked answer

(a) [2] Independent variable: radius r of the two adjustable masses. Directly measured dependent variable: time t for the first revolution. Keep the hanging mass m and axle radius R fixed; also keep M, the release method and the rotating hardware unchanged.

(b) [3] Let T be string tension. Downward is positive for the hanging mass; the unwinding sense is positive for the flywheel.

$mg-T=ma,\qquad a=\alpha R$

$TR-\tau_f=(I_0+2Mr^2)\alpha$

Substitute $T=m(g-\alpha R)$ into the torque equation.

$mgR-\tau_f=(I_0+2Mr^2+mR^2)\alpha$

$\frac{1}{\alpha}=\frac{I_0+mR^2}{mgR-\tau_f}+\frac{2M}{mgR-\tau_f}r^2$

The hanging mass accelerates downward, so tension is less than its weight. Using mg as the string force would overestimate the driving torque.

(c) [3] With constant angular acceleration from rest, one full revolution gives:

$2\pi=\frac12\alpha t^2\quad\Rightarrow\quad\alpha=\frac{4\pi}{t^2}$

First row: $r^2=0.002500\,\mathrm{m^2}$ , $\alpha=4\pi/(1.296)^2=7.48\,\mathrm{rad\,s^{-2}}$ , and $1/\alpha=0.13366\,\mathrm{s^2}$ .

r / m	r² / m²	α / rad s⁻²	1/α / s²
0.050	0.002500	7.482	0.13366
0.075	0.005625	5.784	0.17290
0.100	0.010000	4.395	0.22755
0.125	0.015625	3.356	0.29796
0.150	0.022500	2.606	0.38376

(d) [3] Plot $1/\alpha$ vertically against $r^2$ horizontally. Draw a best-fit straight line with error bars. Using the end points to estimate the gradient:

$b\approx\frac{0.383756-0.133660}{0.022500-0.002500}=12.5\,\mathrm{s^2\,m^{-2}}$

$c\approx0.133660-(12.5)(0.002500)=0.1024\,\mathrm{s^2}$

The intercept is non-zero because the fixed hardware and accelerating hanging mass contribute to the effective inertia. A full best-fit calculation gives similar values.

(e) [3] Define $D=mgR-\tau_f$ . The gradient gives $b=2M/D$ .

$D=\frac{2M}{b}=\frac{2(0.0500)}{12.5}=0.00800\,\mathrm{N\,m}$

$\tau_f=mgR-D=(0.0500)(9.81)(0.0200)-0.00800$

$\tau_f\approx1.81\times10^{-3}\,\mathrm{N\,m}$

The intercept gives $c=(I_0+mR^2)/D$ .

$I_0=cD-mR^2=(0.1024)(0.00800)-(0.0500)(0.0200)^2$

$I_0\approx7.99\times10^{-4}\,\mathrm{kg\,m^2}$

(f) [3] Since $\alpha\propto t^{-2}$ , conservative first-order uncertainty propagation gives:

$\frac{\delta\alpha}{\alpha}=2\frac{\delta t}{t}=\frac{2(0.020)}{2.196}=0.0182$

$\delta\alpha=(2.60582)(0.0182)=0.0475\,\mathrm{rad\,s^{-2}}$

Thus $\alpha=(2.61\pm0.05)\,\mathrm{rad\,s^{-2}}$ for the largest radius. Timing uncertainty alone omits uncertainties in radius, masses and axle radius, and systematic effects such as slip or changing friction. Agreement of points with a line must be assessed using all relevant uncertainties.

(g) [3] Repeat trials at each radius and use their scatter to assess timing uncertainty. Measure both adjustable radii carefully and verify axle radius, string alignment and no slip. A limitation is that bearing friction may depend on angular speed or load, so it may not be a constant torque. Accept two specific improvements and one justified model limitation. Secure the adjustable masses and use a catcher for the falling load.

Answer: Gradient ≈ 12.5 s²/m²; intercept ≈ 0.102 s²; friction torque ≈ 1.81 × 10⁻³ N m; fixed moment of inertia ≈ 8.0 × 10⁻⁴ kg m². Moving the masses outward reduces angular acceleration.

Exam readiness checklist

Identify the axis and calculate perpendicular lever arms.
Use both force balance and torque balance for static equilibrium.
Apply angular motion equations with radians and consistent signs.
Calculate moment of inertia from the mass distribution.
Distinguish driving torque from resultant torque.
Use signed angular momentum and torque–time areas.
Separate angular momentum conservation from kinetic-energy conservation.
Analyse experimental gradients, intercepts, uncertainties and assumptions.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Rigid body mechanics HL Paper 1&2

Instructions and conventions

IB formula reference

Common misconceptions

Part 1 — Multiple-choice questions

Multiple choice 1 — Torque at an angle [1 mark]

Multiple choice 2 — A force through the axis [1 mark]

Multiple choice 3 — A couple [1 mark]

Multiple choice 4 — Angular displacement [1 mark]

Multiple choice 5 — Angular acceleration [1 mark]

Multiple choice 6 — Point masses [1 mark]

Multiple choice 7 — Changing the mass distribution [1 mark]

Multiple choice 8 — Resultant torque [1 mark]

Multiple choice 9 — Rotational energy scaling [1 mark]

Multiple choice 10 — Angular momentum [1 mark]

Multiple choice 11 — Conservation of angular momentum [1 mark]

Multiple choice 12 — Energy at constant angular momentum [1 mark]

Multiple choice 13 — Angular impulse [1 mark]

Multiple choice 14 — Signed braking torque [1 mark]

Multiple choice 15 — Two rotating discs [1 mark]

Part 2 — Extended-response questions

Extended response 1 — An angled force on a wrench [6 marks]

Extended response 2 — A loaded beam in static equilibrium [8 marks]

Extended response 3 — Balancing a seesaw [6 marks]

Extended response 4 — Uniform angular acceleration [6 marks]

Extended response 5 — A flywheel braking uniformly [8 marks]

Extended response 6 — Moment of inertia of point masses [6 marks]

Extended response 7 — Mass distribution: a ring and a disc [8 marks]

Extended response 8 — Motor torque, friction and energy [10 marks]

Extended response 9 — Angular impulse from a torque–time graph [10 marks]

Extended response 10 — A skater draws in their arms [7 marks]

Extended response 11 — Two discs couple [8 marks]

Extended response 12 — A mass lands on a rotating platform [7 marks]

Extended response 13 — Challenge: A falling load drives a pulley [11 marks]

Extended response 14 — Challenge: A rotor reverses direction [10 marks]

Extended response 15 — Challenge: Energy supplied by a flywheel [10 marks]

Part 3 — Lab-style investigation

Investigating mass distribution and angular acceleration [20 marks]

Exam readiness checklist

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