A.4 Rigid body mechanics (HL) – IB Physics exam questions with mark scheme

1StandardHL only6 marks

A flywheel of moment of inertia $0.45\ \mathrm{kg\,m^2}$ is initially at rest. A constant torque of $3.0\ \mathrm{N\,m}$ is applied for 5.0 s. Friction is negligible.

(a) Calculate the angular acceleration of the flywheel. [1]

(b) Calculate its angular speed after 5.0 s. [1]

(d) The work done by a constant torque is $W=\tau\theta$, where $\theta$ is the angle turned through. Show that the work done by the torque during the 5.0 s is consistent with your answer to (c). [2]

$\alpha=\tau/I = 3.0/0.45 = 6.7\ \mathrm{rad\,s^{-2}}$ (allow 6.67)(1)
$\omega = \omega_i+\alpha t = 0+6.67\times5.0 = 33\ \mathrm{rad\,s^{-1}}$ (allow 33.3 or 33.5 from rounded $\alpha=6.7$; allow ECF from (a))(1)
$E_K=\tfrac12 I\omega^2$ with correct substitution(1)
$E_K = 250\ \mathrm{J}$ (allow 245–252 J from rounded values; allow ECF from (b))(1)
Angle turned $\theta=\omega_i t+\tfrac12\alpha t^2 = \tfrac12\times6.67\times5.0^2 = 83$ rad (allow $\theta=\tfrac12(\omega_i+\omega_f)t$ or $\omega_f^2=2\alpha\theta$; allow 83–84 rad)(1)
Work $=\tau\theta = 3.0\times83.3 = 250$ J, equal to $E_K$ in (c) (allow 249–252 J from rounded values) and statement that they agree because, with negligible friction, all the work done by the torque becomes rotational KE (OWTTE)(1)
Answer: (a) 6.7 rad s$^{-2}$ (b) 33 rad s$^{-1}$ (c) 250 J (d) $W=\tau\theta=250$ J
Examiner tip: Use unrounded values from earlier parts in (c) and (d) so the two energies agree.

2CoreHL only3 marks

A mechanic uses a spanner to loosen a wheel nut. A force of 60 N is applied at the end of the spanner, 0.25 m from the axis of the nut, at an angle of 70° to the spanner as shown.

(a) Calculate the torque about the axis of the nut. [2]

(b) State how the mechanic could produce a larger torque with the same size of force. [1]

use of $\tau = Fr\sin\theta$ with $\theta = 70°$: $60\times0.25\times\sin70°$(1)
$\tau = 14\ \mathrm{N\,m}$ (allow 14.1); 15 N m (no sine) scores 0 for this mark; $\cos70°$ giving 5.1 N m scores 0(1)
apply the force perpendicular to the spanner / at 90° (to make $\sin\theta = 1$) OR apply it further from the axis / use a longer spanner; accept either(1)
Answer: (a) $14\ \mathrm{N\,m}$ (b) push at 90° to the spanner, or further from the nut
Examiner tip: $\theta$ in $\tau = Fr\sin\theta$ is the angle between the force and the line from the axis to the point of application, so check which angle the diagram gives before using sin.

3StandardHL only5 marks

A uniform plank of length 4.0 m and mass 25 kg rests horizontally on two supports, A and B. A is 0.50 m from the left-hand end and B is 0.50 m from the right-hand end. A child of mass 35 kg stands on the plank 0.70 m to the right of A.

(a) State the two conditions for the plank to be in equilibrium. [1]

(b) Calculate the upward force exerted on the plank by support B. [2]

(d) The child walks to the right-hand end of the plank. Show that the plank does not tip about B. [1]

resultant force is zero and resultant torque (about any point) is zero; both needed(1)
moments about A: $F_B\times3.0 = 25g\times1.5 + 35g\times0.70$ (= 62g)(1)
$F_B = 203\ \mathrm{N}$ (allow 200–203; 207 with $g = 10$)(1)
$F_A = 60g - F_B = 386\ \mathrm{N}$ (allow 390; ecf from (b))(1)
torque about B: plank's weight $25g\times1.5 = 368\ \mathrm{N\,m}$ (anticlockwise) is greater than child's $35g\times0.50 = 172\ \mathrm{N\,m}$ (clockwise), so $F_A$ stays positive / plank does not tip; accept tipping position 4.57 m from left end, beyond the plank(1)
Answer: (b) $203\ \mathrm{N}$ (c) $386\ \mathrm{N}$ (d) $368\ \mathrm{N\,m} > 172\ \mathrm{N\,m}$ about B
Examiner tip: Take moments about a support whose force you do not yet know, so that force has zero torque and drops out.

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4CoreHL only4 marks

Two small spheres, each of mass 0.40 kg, are fixed to the ends of a light rod of length 0.60 m. The spheres can be treated as point masses.

(a) Calculate the moment of inertia of the system about an axis perpendicular to the rod through its centre. [1]

(b) Calculate the moment of inertia about a parallel axis through one of the spheres. [1]

(c) The system is given the same angular momentum about each axis in turn. Determine the ratio $\dfrac{\text{kinetic energy about the centre}}{\text{kinetic energy about one end}}$. [2]

$I = 2\times0.40\times0.30^2 = 0.072\ \mathrm{kg\,m^2}$(1)
$I = 0.40\times0.60^2 = 0.144\ \mathrm{kg\,m^2}$ (sphere on the axis contributes zero); allow 0.14(1)
use of $E_k = \dfrac{L^2}{2I}$ so $E_k \propto 1/I$ for fixed $L$(1)
ratio $= \dfrac{0.144}{0.072} = 2$ (ecf from (a) and (b))(1)
Answer: (a) $0.072\ \mathrm{kg\,m^2}$ (b) $0.144\ \mathrm{kg\,m^2}$ (c) 2
Examiner tip: In $I=\Sigma mr^2$, $r$ is the distance from the axis, not the length of the rod; a mass on the axis adds nothing.

5StandardHL only5 marks

The drum of a washing machine is spinning at 1200 revolutions per minute when the motor is switched off. The drum slows down with uniform angular deceleration and stops after 8.0 s. The radius of the drum is 0.25 m.

(a) Show that the initial angular velocity of the drum is about $126\ \mathrm{rad\,s^{-1}}$. [1]

(b) Calculate the angular acceleration of the drum. [1]

(d) Calculate the linear speed of a point on the rim of the drum 4.0 s after the motor is switched off. [1]

$\omega = \dfrac{1200\times2\pi}{60} = 125.7\ \mathrm{rad\,s^{-1}}$; must see the conversion, not just the answer(1)
$\alpha = \dfrac{0-125.7}{8.0} = -15.7\ \mathrm{rad\,s^{-2}}$; allow magnitude 15.7 (16) if described as a deceleration(1)
$\Delta\theta = \dfrac{(\omega_i+\omega_f)t}{2} = \dfrac{125.7\times8.0}{2} = 503\ \mathrm{rad}$(1)
$\dfrac{503}{2\pi} = 80$ revolutions(1)
$\omega = 62.8\ \mathrm{rad\,s^{-1}}$ so $v = \omega r = 62.8\times0.25 = 15.7\ \mathrm{m\,s^{-1}}$ (allow 16)(1)
Answer: (b) $-15.7\ \mathrm{rad\,s^{-2}}$ (c) 80 revolutions (d) $16\ \mathrm{m\,s^{-1}}$
Examiner tip: Convert rev/min to rad/s ($\times 2\pi/60$) before using the angular equations, and convert radians back to revolutions at the end.

6HardHL only6 marks

A playground roundabout is rotating freely about a vertical axis with angular velocity $1.2\ \mathrm{rad\,s^{-1}}$. Its moment of inertia about the axis is $240\ \mathrm{kg\,m^2}$ and its radius is 1.5 m. A child of mass 32 kg, initially at rest, steps onto the edge of the roundabout. Friction at the axle is negligible and the child can be treated as a point mass.

(a) Calculate the moment of inertia of the child about the axis when on the edge. [1]

(b) Calculate the new angular velocity of the roundabout. [2]

(c) Determine the change in rotational kinetic energy of the system, and explain why the kinetic energy changes even though angular momentum is conserved. [3]

$I_c = mr^2 = 32\times1.5^2 = 72\ \mathrm{kg\,m^2}$(1)
angular momentum conserved: $240\times1.2 = (240+72)\,\omega$(1)
$\omega = 0.92\ \mathrm{rad\,s^{-1}}$ (allow 0.923; ecf from (a))(1)
$E_{k,\text{before}} = \tfrac12\times240\times1.2^2 = 173\ \mathrm{J}$ and $E_{k,\text{after}} = \tfrac12\times312\times0.923^2 = 133\ \mathrm{J}$(1)
loss of $40\ \mathrm{J}$ (allow 39–40; ecf)(1)
the interaction is like an inelastic collision: friction between the child's feet and the roundabout (internal forces) does work / transfers energy to thermal energy, but exerts no external torque on the system, so $L$ is unchanged; OWTTE(1)
Answer: (a) $72\ \mathrm{kg\,m^2}$ (b) $0.92\ \mathrm{rad\,s^{-1}}$ (c) decrease of about 40 J
Examiner tip: Angular momentum is conserved whenever there is no external torque, but kinetic energy is only conserved if no internal work is done to thermal energy.

7HardHL only5 marks

A solid cylinder is released from rest at the top of a ramp and rolls without slipping to the bottom. The top of the ramp is 0.80 m higher than the bottom. The moment of inertia of a solid cylinder of mass $M$ and radius $R$ about its axis is $\tfrac12 MR^2$.

(a) State the condition for rolling without slipping, in terms of the speed $v$ of the centre of the cylinder. [1]

(b) Show that the speed of the cylinder at the bottom of the ramp is about $3.2\ \mathrm{m\,s^{-1}}$. [2]

(c) A thin hoop of the same mass and radius ($I = MR^2$) is released from the same point. Explain, without further calculation of speed, why the hoop reaches the bottom more slowly. [2]

$v = \omega R$ (point of contact instantaneously at rest)(1)
$Mgh = \tfrac12 Mv^2 + \tfrac12\left(\tfrac12 MR^2\right)\left(\dfrac{v}{R}\right)^2 = \tfrac34 Mv^2$(1)
$v = \sqrt{\tfrac43 gh} = \sqrt{\tfrac43\times9.81\times0.80} = 3.23\ \mathrm{m\,s^{-1}}$; answer to at least 3 s.f. (3.23) or full substitution needed for a 'show that'(1)
the hoop has a larger moment of inertia (mass further from the axis) so a larger fraction of the same $Mgh$ goes into rotational KE(1)
so less goes into translational KE, giving a lower speed at each point / smaller (linear) acceleration down the ramp; allow hoop's KE is half rotational vs one third for the cylinder(1)
Answer: (a) $v = \omega R$ (b) $3.23\ \mathrm{m\,s^{-1}}$ (c) larger $I$, more of the energy is rotational, less translational
Examiner tip: In rolling, the loss of GPE is shared between translational and rotational KE; using only $\tfrac12 Mv^2$ gives the sliding answer of $4.0\ \mathrm{m\,s^{-1}}$.

8StandardHL only5 marks

The graph shows how the angular velocity $\omega$ of a turntable varies with time $t$. The moment of inertia of the turntable about its axis is $0.050\ \mathrm{kg\,m^2}$.

(a) Determine the angular acceleration of the turntable between $t = 0$ and $t = 2.0$ s. [1]

(b) Determine the total angle turned through between $t = 0$ and $t = 8.0$ s. [2]

(d) Calculate the angular impulse on the turntable between $t = 5.0$ s and $t = 8.0$ s. [1]

gradient: $\alpha = \dfrac{6.0}{2.0} = 3.0\ \mathrm{rad\,s^{-2}}$(1)
angle = area under graph, e.g. $\tfrac12(2)(6) + (3)(6) + \tfrac12(3)(6)$; allow trapezium $\tfrac12(8+3)\times6$(1)
$\Delta\theta = 33\ \mathrm{rad}$(1)
$\alpha = 6.0/3.0 = 2.0\ \mathrm{rad\,s^{-2}}$ so $\tau = I\alpha = 0.050\times2.0 = 0.10\ \mathrm{N\,m}$(1)
$\Delta L = I\Delta\omega = 0.050\times(0-6.0) = -0.30\ \mathrm{N\,m\,s}$ (allow $\mathrm{kg\,m^2\,s^{-1}}$; allow magnitude 0.30 or $\tau\Delta t = 0.10\times3.0$; ecf from (c))(1)
Answer: (a) $3.0\ \mathrm{rad\,s^{-2}}$ (b) $33\ \mathrm{rad}$ (c) $0.10\ \mathrm{N\,m}$ (d) $-0.30\ \mathrm{N\,m\,s}$
Examiner tip: On an $\omega$–$t$ graph, the gradient is $\alpha$ and the area is angle turned in radians, just like $v$–$t$ graphs in linear motion.

9HardHL only6 marks

A diver leaves a springboard rotating slowly with her body straight. In the air she pulls her arms and legs in tightly (a 'tuck') and rotates much faster. Just before entering the water she straightens out again.

Explain, in terms of torque, moment of inertia, angular momentum and energy, the changes in the diver's rotation during the dive. Air resistance is negligible. [6]

once in the air, the only external force is her weight, which acts through her centre of mass, so it has no torque about the centre of mass(1)
so her angular momentum $L = I\omega$ about the centre of mass is constant during the flight(1)
the springboard (reaction force not through the centre of mass) exerted a torque / angular impulse at take-off, giving her the initial angular momentum(1)
tucking moves mass closer to the axis of rotation, so $I = \Sigma mr^2$ decreases(1)
so $\omega$ increases in proportion (e.g. halving $I$ doubles $\omega$); straightening increases $I$ and reduces $\omega$, so she enters the water slowly rotating(1)
rotational KE $= L^2/2I$ increases when she tucks; the extra energy comes from work done by her muscles pulling limbs inwards (against the tendency of the limbs to move in straight lines)(1)
when she straightens, rotational KE decreases (energy transferred into her muscles / thermal energy)(1)
her centre of mass follows a parabolic path regardless of the rotation (projectile motion)(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent explanation linking all four ideas: no external torque in flight so angular momentum is conserved, tucking reduces the moment of inertia so angular velocity rises (and falls again on straightening), and rotational kinetic energy changes with the work done by the diver's muscles. Correct terms used throughout.
Level 2	3–4	Explains that angular momentum is conserved and that a smaller moment of inertia gives a larger angular velocity, but the reason for conservation (no torque from the weight) or the energy change is missing or unclear. Mostly correct terms.
Level 1	1–2	Isolated relevant statements, e.g. tucking makes her spin faster or mass is closer to the axis, with little or no linking reasoning. May state that angular momentum increases.
Level 0	0	No relevant physics.

Answer: Weight has no torque about the centre of mass, so $L$ is constant; tucking reduces $I$, raising $\omega$; rotational KE $=L^2/2I$ rises, supplied by muscle work.
Examiner tip: Many answers say angular momentum increases when the diver tucks; it is constant, and it is $\omega$ and kinetic energy that increase.

10StretchHL only6 marks

A student determines the moment of inertia $I$ of a wheel on a fixed horizontal axle. A light string is wound around the rim of the wheel, radius 0.10 m, and a 0.50 kg mass is attached to the free end. The mass is released from rest and falls 1.20 m to the floor in a measured time of 1.11 s. The string does not slip.

(a) Calculate the linear acceleration of the falling mass. [1]

(b) Calculate the tension in the string. [2]

(d) In fact there is a small frictional torque at the axle. State and explain whether the value of $I$ found in (c) is greater or smaller than the true value. [1]

$a = \dfrac{2h}{t^2} = \dfrac{2\times1.20}{1.11^2} = 1.95\ \mathrm{m\,s^{-2}}$(1)
$mg - T = ma$ applied to the mass(1)
$T = 0.50\times(9.81-1.95) = 3.93\ \mathrm{N}$ (ecf from (a))(1)
$\alpha = a/r = 19.5\ \mathrm{rad\,s^{-2}}$ and $\tau = Tr = 0.393\ \mathrm{N\,m}$(1)
$I = \tau/\alpha = 0.020\ \mathrm{kg\,m^2}$; allow 0.0202; ecf(1)
greater: friction reduces the acceleration, and the calculation assumes all of $Tr$ accelerates the wheel, so the smaller $\alpha$ is interpreted as a larger $I$; answer 'greater' with no reason scores 0(1)
Answer: (a) $1.95\ \mathrm{m\,s^{-2}}$ (b) $3.93\ \mathrm{N}$ (c) $0.020\ \mathrm{kg\,m^2}$ (d) greater than the true value
Examiner tip: The tension is not $mg$: the mass is accelerating, so apply $mg - T = ma$ before finding the torque on the wheel.

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