IB Physics HL A.4 Rigid body mechanics Questions

SLSL HLHL

Practise IB Physics HL A.4 by balancing torques, linking angular acceleration to moment of inertia, and using rotational equations, angular momentum and energy to analyse motion.

51 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Find torque using the perpendicular distance from the chosen axis; opposite forces on opposite sides can produce moments in the same sense.
For rotational equilibrium, set net torque to zero and select a pivot that removes unknown support forces; locate a uniform rod’s weight at its midpoint.
For a ladder at a smooth wall, balance weight and wall-force moments with their different perpendicular arms, then apply horizontal force balance.
Explain angular acceleration through unbalanced torque: opposed top and bottom friction forces on a rolling rod need not cancel their moments.
Define inertia as resistance to changes in rotation, and relate it to mass distribution about the axis; “resistance to rotation” alone is insufficient.
Sum every point-mass mr² term about the actual axis, adding any extended body’s given inertia rather than treating that body as a point at its centre.
In a supplied uniform-cylinder formula, distinguish combined mass and length from each component’s values when joining cylinders or modelling propeller blades.
Relate linear motion to angular motion at the correct contact radius: the string axle radius can differ from the radius used in a flywheel’s inertia.
For no-slip rolling, use centre travel in θ = s/R; a rod rolling between stationary ground and a moving boat travels half as far as the boat.
For first-arrival rotation time, read the direction, convert the remaining angle to radians and use t = Δθ/ω.
Use average angular acceleration Δω/Δt for stated speed changes; use constant-α equations only where that condition applies.
Convert complete turns to 2πN radians before using ω² = ω₀² + 2αθ; combine the resulting angular speed with I when angular momentum is requested.
In braking calculations, use the angular speed at the start of that stage and its own deceleration; convert stopping radians into revolutions at the end.
For uniform angular-speed change, find angular distance from the mean of initial and final speeds times time, retaining nonzero final speed.
Distinguish graph operations: angular acceleration is an ω–time gradient, while constant acceleration from rest gives a curved θ–time graph.
Use net torque and same-axis inertia in τ = Iα; for a timed driving stage, follow with ω = ω₀ + αt.
Infer I from τ/α; when α is plotted vertically against torque, take the reciprocal of the graph gradient.
For coupled falling-mass and cylinder motion, combine the mass’s force equation, the cylinder’s torque equation and a = αR; derive results in order.
For a massive pulley, keep its two string tensions different and include both blocks and pulley rotation in a force or energy derivation.
For a rolling hoop, combine linear force and frictional torque equations with a = αr; mass distribution also explains why a ring trails a solid disk.
For post-collision braking, use the inertia of the full attached system when converting the braking torque to angular deceleration.
For uniform flywheel slowing at fixed inertia, find torque from IΔω/Δt or the transferred average power divided by the matching mean angular speed.
Calculate rotating-body angular momentum as Iω, carrying the correct earlier speed, inertia and time from the linked motion stage.
For a tangential incoming particle at a fixed pivot, use initial L = mvr and conserve it into the complete final Iω after sticking.
When bodies link about one axis, add their final inertias and conserve total angular momentum if external torque about that axis is negligible.
For inward or outward mass redistribution with no external torque, use Iω = constant and explain both the inertia change and the speed response.
When L is conserved but I changes, use Ek = ½Lω for energy ratios; using the same I in both ½Iω² expressions can invalidate the whole calculation.
Check torque before claiming conservation: the unequal hanging weights in the pulley system create a net external torque about its axle.
At an orbit’s closest and furthest points use L = mrv; changing speed changes momentum magnitude, while radial gravity gives zero torque about the star.
Use angular impulse as the angular-momentum change; its unit is N m s, and from rest a constant torque acting for Δt supplies L = τΔt.
For rotational KE loss after bodies link, calculate initial and final ½Iω² with their respective inertias and subtract final energy from initial.
For a rod falling about an end pivot, use the centre-of-mass drop L/2 in energy conservation; keep a driven cylinder’s KE separate from its falling block’s KE.
For rolling total energy, include both translation and rotation; in wheel-to-car ratios use each wheel’s mass in I and the total car mass in translation.
At equal rotational KE, use L²/(2I) to obtain angular-momentum ratios proportional to √I rather than I.
For flywheel power, divide ½I(ωi² − ωf²) by time; capacity increases through greater inertia or greater maximum angular speed.

Covered syllabus points

A.4.1 (HL)—Torque A.4.2 (HL)—Rotational equilibrium A.4.3 (HL)—Unbalanced torque A.4.4 (HL)—Angular motion quantities A.4.5 (HL)—Angular motion equations A.4.6 (HL)—Moment of inertia A.4.7 (HL)—Point-mass moment of inertia A.4.8 (HL)—Rotational Newton’s second law A.4.9 (HL)—Angular momentum A.4.10 (HL)—Angular momentum conservation A.4.11 (HL)—Angular impulse A.4.12 (HL)—Rotational kinetic energy

Exam points

Use the perpendicular lever arm from the axis to the force’s line of action; halve a given cylinder diameter before calculating each rocket’s torque.
Assign torque signs by rotational sense: opposite force directions on opposite sides of an axis can produce torques that add.

[Maximum number: 1]

A student models a rotating dancer using a system that consists of a vertical cylinder, a horizontal rod and two spheres.

The cylinder rotates from rest about the central vertical axis. A rod passes through the cylinder with a sphere on each side of the cylinder. Each sphere can move along the rod. Initially the spheres are close to the cylinder.

A horizontal force of 50 N is applied perpendicular to the rod at a distance of 0.50 m from the central axis. Another horizontal force of 40 N is applied in the opposite direction at a distance of 0.20 m from the central axis. Air resistance is negligible.

Show that the net torque on the system about the central axis is approximately 30 Nm .

$\Sigma \Gamma=50 \times 0.5+40 \times 0.2$

OR

33 «Nm»

Marking guidance:

Accept opposite rotational sign convention

Exam points

State rotational equilibrium as zero resultant torque or moment; a zero resultant force alone is not this condition.
For a uniform rod, take its weight at the midpoint and balance moments about the pivot using the actual support distance.
For an inclined ladder, use separate perpendicular arms for weight and wall force, then equate wall reaction to ground friction by horizontal balance.

[Maximum number: 1]

A uniform cylinder, of mass M and length L, has a moment of inertia of $\frac{1}{12} M L^{2}$ when rotated about an axis through its centre.

State the condition for rotational equilibrium.

Exam points

Explain a rolling rod’s acceleration: opposite friction forces at its top and bottom produce a net torque in the same rotational sense.

[Maximum number: 2]

A boat is moved from land to water by rolling it across a set of cylindrical airbags.

Another way to move a boat down a ramp is to place it on rigid rods. Assume that the same boat is now moved on rods of diameter 1.80 m with the same boat acceleration as before.

Outline how two frictional forces lead to the angular acceleration of the rod.

Exam points

For average angular acceleration, use Δω/Δt; with string-speed data, first calculate ω = v/r using the axle radius where the string unwinds.
For rolling without slipping at fixed radius, relate the centre’s linear travel to angular displacement using θ = s/R.
For a rod rolling without slipping between ground and a moving boat, use rod-centre travel, half the boat’s travel, in θ = s/r.
For rotation about a fixed central axis, calculate a surface point’s angular velocity from ω = v/r, using radius rather than diameter.
Read the rotation direction before using t = Δθ/ω; convert the remaining angle to radians when timing the first arrival at the top.

[Maximum number: 3]

A boat is moved from land to water by rolling it across a set of cylindrical airbags.

(a)

Another way to move a boat down a ramp is to place it on rigid rods. Assume that the same boat is now moved on rods of diameter 1.80 m with the same boat acceleration as before.

[ 1 ]

(i)

Show that the angular displacement of a rod during the first 8.0 s of the motion is about 8 rad.

[ 1 ]

$\theta=2 \pi$ revolutions $=2 \pi \frac{s}{2 \pi r}=\frac{7}{0.9}=7.78 \mathrm{rad}$

(b)

During the testing of the rod, it is rotated about its central axis. Point P is on the surface of the rod. At time t=0, P is at a point $45^{\circ}$ above the horizontal. The linear speed of P is $2.0 \mathrm{~ms}^{-1}$ .

The vertical displacement x of P can be modelled with an equation $x=x_{0} \sin (\omega t+\phi)$ .

[ 2 ]

(i)

Calculate the angular velocity $\omega$ of P.

[ 1 ]

$\omega=\frac{v}{r}=\frac{2}{0.9}=2.22 \mathrm{rads}^{-1}$

(ii)

Calculate the time it will take for P to reach the top of the circle.

[ 1 ]

Alternative 1
$t=\frac{\left(\frac{\mathrm{C}}{8}\right)}{v}=\frac{\left(\frac{2 \pi \mathrm{r}}{8}\right)}{2.0}=0.35 \mathrm{~s}$

Alternative 2
$t=\frac{\left(\frac{\pi}{4}\right)}{\omega}=0.35 \mathrm{~s}$

Allow 0.35-0.36 s.

Exam points

Before using rotational SUVAT, check that angular acceleration is constant; a falling hinged rod has changing gravitational torque as its angle changes.
For speed after a stated number of turns, convert the angle to 2πN radians before using ω² = ω₀² + 2αθ.
For stopping revolutions, use the angular speed at the start of braking with ωf = 0, then divide the angular stopping distance by 2π.
Treat acceleration and braking as separate stages: find the braking deceleration from its own angular distance before calculating stopping time.
With uniform angular acceleration or deceleration, use θ = ½(ω₀ + ω)t; retain any nonzero final speed and convert radians to revolutions.
Sketch θ against time for constant angular acceleration from rest as a curve from the origin with increasing gradient, rather than a straight line.
For constant torque and fixed inertia, calculate α = τ/I, then use t = (ω − ω₀)/α to find the time to the target angular speed.
For a disk rolling from rest without slipping, combine θ = L/R with θ = ½αt², or find linear acceleration first and divide by R.
Find angular acceleration from the gradient of the rising ω–time segment; do not use the later constant-speed plateau in that gradient.

[Maximum number: 3]

A horizontal rigid bar of length 2 R is pivoted at its centre. The bar is free to rotate in a horizontal plane about a vertical axis through the pivot. A point particle of mass M is attached to one end of the bar and a container is attached to the other end of the bar.

A point particle of mass $\frac{M}{3}$ moving with speed v at right angles to the rod collides with the container and gets stuck in the container. The system then starts to rotate about the vertical axis.

The mass of the rod and the container can be neglected.

A torque of 0.010 Nm brings the system to rest after a number of revolutions. For this case $R=0.50 \mathrm{~m}, M=0.70 \mathrm{~kg}$ and $v=2.1 \mathrm{~ms}^{-1}$ .

Calculate the number of revolutions made by the system before it comes to rest.

$\theta=\frac{\omega_i^{2}}{2\alpha}$ , from $\omega_f^{2}=\omega_i^{2}+2\alpha\theta$ $\theta=\frac{v^{2}}{32R^{2}\alpha}=\frac{2.1^{2}}{32\times0.5^{2}\times0.043}=12.8\text{ or }12.9\ \mathrm{rad}$
number of rotations $=\frac{12.9}{2\pi}=2.0$ revolutions

Exam points

Define moment of inertia as resistance to a change in rotation, or use I = Σmr² with r measured from the axis; “resistance to rotation” alone is insufficient.
For two identical cylinders joined end to end, use combined mass 2M and length 2L in the supplied central-axis formula, giving I = ⅔ML².
In the two-blade propeller model, distinguish each blade’s mass and length from the totals; show full substitution or extra precision in “show that”.
For no-slip rolling at equal mass and radius, a ring’s larger I puts more energy into rotation; explain why the disk reaches the bottom first.

[Maximum number: 3]

A uniform cylinder, of mass M and length L, has a moment of inertia of $\frac{1}{12} M L^{2}$ when rotated about an axis through its centre.

(a)

Outline what is meant by moment of inertia.

[ 1 ]

quotes $\mathrm{I}=\Sigma \mathrm{mr}^{2}$ with r= «perpendicular» distance to axis

OR

resistance to change in rotation

OR

ratio of torque «applied» to angular acceleration

OR

analog to mass in rotational mechanics

In MP1, accept r= radius.

In MP2, do not accept resistance to rotation.

In MP3 accept the expression as a formula if both symbols identified.

(b)

Two identical cylinders, each of mass M and length L, are connected end to end. Show that the moment of inertia when these cylinders are rotated about their combined centre is $\frac{2}{3} M L^{2}$ .

[ 1 ]

$\frac{1}{12} 2 M(2 L)^{2}=<\frac{8}{12} M L^{2}=\frac{2}{3} M L^{2}>\vee$

(c)

A two-blade propeller can be modelled using the two-cylinder arrangement in (a)(iii).

The following data for the two-blade propeller are available:
Length of each blade: 0.60 m
Mass of each blade: 2.2 kg
Show that the moment of inertia of the two-blade propeller is about $0.5 \mathrm{~kg} \mathrm{~m}^{2}$ .

[ 1 ]

$\frac{2}{3} \times 2.2 \times 0.6^{2}$

OR
$\frac{1}{12} \times 4.4 \times 1.2^{2}$

OR

0.53 «kg $\mathrm{m}^{2}$ »

Answer of $0.5 \mathrm{~kg} \mathrm{~m}^{2}$ given, so award the
mark if candidates show a correct full substitution OR the value with an extra significant figure

Exam points

After a particle sticks to a pivoted system, include every final point-mass mr² term before writing the system’s angular momentum as Iω.
For a particle stuck to an extended rod, add mr² to the rod’s given inertia; use the particle’s distance from the pivot, half the rod length here.
For equal small spheres joined by a massless rod and rotating about its midpoint, add both mr² terms using half the separation for each radius.
In the given orbital I = mr² model, r is the distance between the planet’s centre and the star’s centre, rather than the planet’s own radius.

[Maximum number: 1]

Just after the collision the system begins to rotate about the vertical axis with angular velocity $\omega$ . Show that the angular momentum of the system is equal to $\frac{4}{3} M R^{2} \omega$ .

evidence of use of $L=I\omega=\left(MR^{2}+\frac{M}{3}R^{2}\right)\omega=\frac{4}{3}MR^{2}\omega$

evidence of use of conservation of angular momentum, $\frac{MvR}{3}=\frac{4}{3}MR^{2}\omega$ , rearranging to get $\omega=\frac{v}{4R}$

Exam points

Calculate the resultant torque about the stated axis before dividing by I; use the actual radius or centre-of-mass lever arm in Fr.
If a constant torque acts for a stated time from rest, find α = τ/I first, then ω = αt using the same rotational stage.
For a flywheel with a separate axle, use wheel radius in the supplied inertia formula but axle radius in the string’s v = rω relation.
After a particle sticks to a rod, use the combined system inertia in braking α = τ/I; the bare rod’s inertia omits the attached mass.
To infer inertia from measured torque and acceleration, rearrange I = τ/α and keep the net torque distinct from either applied force.
For an α-versus-τ graph, the gradient equals 1/I; take its reciprocal to obtain the moment of inertia.
For a falling mass driving a cylinder, combine the mass’s F = ma with TR = Iα and a = αR; tension is less than the falling weight.
Once the cylinder’s angular acceleration is established, substitute it into T = Iα/R to show the string tension, including the final cancellation.
For a massive pulley, use two different string tensions: combine each block’s F = ma with (T₁ − T₂)r = Iα and the no-slip condition.
For a no-slip rolling hoop, combine ma = mg sinθ − Ff with Ffr = Iα and a = αr before evaluating its reduced downhill acceleration.
For uniform flywheel slowing, find torque from IΔω/Δt, or divide the transferred average power by the matching average angular speed.
From an ω–time graph, use its accelerating gradient in τ = Iα, or use ΔL/Δt over that same interval; include the supplied disk inertia.
For direct torque calculations, multiply the stated inertia by angular acceleration and give torque in N m, retaining small powers of ten.

[Maximum number: 2]

Determine the moment of inertia of the system about the central axis.

$\begin{aligned} I=\frac{\Gamma}{\alpha} \text { OR } 33=I \times 4 I=8.25<\mathrm{kg} \mathrm{~m}^{2}> \end{aligned}$

Allow ECF from (a) and (b)

Award [2] for a BCA

Exam points

Calculate a rod’s or merry-go-round’s angular momentum using L = Iω and the preceding angular speed.
Convert complete turns to Δθ = 2πN, use ω² = 2αΔθ from rest, then calculate L = Iω.
For a cylinder starting from rest, combine I = ½MR² and ω = αt to find L at a given time.

[Maximum number: 1]

Write down an expression, in terms of M, v and R, for the angular momentum of the system about the vertical axis just before the collision.

$\frac{M}{3} v R$

Exam points

When bodies join about a common axis, conserve total angular momentum and add their final inertias, including an added rim mass mr².
For a tangential particle sticking to a pivoted rod, equate its initial mvr to the full rod–particle Iω and show the required precision.
With zero resultant external torque, use Iω = constant: moving mass inward raises angular speed, while moving it outward lowers angular speed.
At constant angular momentum, use Ek = ½Lω for rotational-energy ratios; do not keep I fixed when the mass distribution changes.
Before claiming angular-momentum conservation, check external torque about the chosen axis; unequal hanging weights can give a nonzero net torque.
At the closest and furthest points of an elliptical orbit, use L = mrv with velocity perpendicular to radius; smaller r requires larger speed.
In an elliptical orbit, explain changing momentum magnitude through changing speed, while radial gravity gives zero torque about the star.

[Maximum number: 4]

(a)

When the system has reached its maximum angular speed, the two forces are removed. The spheres now move outward, away from the central axis.

[ 4 ]

(i)

Outline why the angular speed $\omega$ decreases when the spheres move outward.

[ 2 ]

moment of inertia increases

Angular momentum is conserved

Marking guidance:

Allow algebraic expressions e.g. $\omega=\frac{L}{I}$ so $\omega$ decreases for MP2

(ii)

When the spheres move outward, the angular speed decreases from $20 \mathrm{rad} \mathrm{s}^{-1}$ to $12 \mathrm{rads}^{-1}$ . Calculate the percentage change in rotational kinetic energy that occurs when the spheres move outward.

[ 2 ]

$\begin{aligned} & \boldsymbol{«} \mathrm{E}_{\mathrm{k}}=» \frac{1}{2} L \omega_{1}=1 / 2 L \omega_{2} & \boldsymbol{O} \boldsymbol{R} & \frac{E_{k 1}}{E_{k 2}}=\frac{\omega_{1}}{\omega_{2}} \end{aligned}$

OR

« L is constant so» $E_{\mathrm{k}}$ is proportional to $\omega$

40 \% «energy loss»

MP1 is for understanding that angular momentum is constant so change in rotational kinetic energy is proportional to change in angular velocity

Award [0] if $E=0.5 \mathrm{I} \omega^{2}$ is used with the same I value for both values of E

Award [2] for BCA

Exam points

Use N m s for angular impulse, from torque × time; distinguish it from N s for linear impulse and N m for torque.

[Maximum number: 1]

What is the unit of angular impulse?

$\mathrm{Nms}^{-1}$

Nms

Exam points

For rotational-energy loss when coaxial bodies join, calculate each state’s ½Iω² with its own inertia, then subtract final energy from initial.
For a uniform rod falling from horizontal to vertical about one end, equate its weight × L/2 to ½Iω² using the supplied pivot-axis inertia.
When asked for the cylinder’s KE in a falling-block system, use its own ½Iω² with ω from the preceding motion; keep the block’s KE separate.
Derive Ek = ½Lω by writing ½Iω² = ½(Iω)ω and substituting L = Iω; the identity does not require angular momentum to be constant.
For a no-slip solid sphere, add translational and rotational KE and use the supplied I = ⅖mr² with ω = v/r to show total Ek = ⁷⁄₁₀mv².
At equal rotational KE, use L²/(2I) = constant to obtain LX/LY = √(IX/IY); angular momentum scales with the square root of inertia.
For four rolling wheels, sum four rotational energies using each wheel’s mass, then compare with ½Mv² using the car’s total mass.
For average power released by a slowing flywheel, use ½I(ωi² − ωf²)/Δt; squaring the angular-speed difference loses the energy-change method.
Increase flywheel energy capacity through greater inertia, using more mass or mass farther from the axis, and a greater maximum angular speed.

[Maximum number: 1]

When the system has reached its maximum angular speed, the two forces are removed. The spheres now move outward, away from the central axis.

Show that the rotational kinetic energy is $\frac{1}{2} L \omega$ where L is the angular momentum
of the system.

$E_{\kappa} \ll=\frac{1}{2} I \omega^{2}=>\frac{1}{2}(I \omega) \omega=\frac{1}{2} L \omega \vee$

Marking guidance:

Accept equivalent methods

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