This whole subtopic is higher level. Nothing in it is on an SL paper.
Educerie · IB Diploma · Physics
Theme A Space, time and motion · A.4 Rigid body mechanics
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Calculate the torque of a force about an axis from τ = Fr sin θ, and state its sense | HL only | "Calculate the torque about the hinge" (2 marks) |
| Use zero resultant torque (and zero resultant force) to solve a body in equilibrium | HL only | "Determine the force exerted by support B" (3 marks) |
| Explain that an unbalanced torque gives an angular acceleration | HL only | Paper 1A, or one line of an "explain" |
| Describe rotation with angular displacement, angular velocity and angular acceleration | HL only | Converting rev min⁻¹ to rad s⁻¹ inside a longer question |
| Use the four equations of uniform angular acceleration | HL only | "Calculate the number of revolutions before the wheel stops" (3 marks) |
| Explain how the moment of inertia depends on the distribution of mass, and calculate I = Σmr² | HL only | "Explain why the moment of inertia increases…" (2 marks), or a calculation for point masses |
| Apply τ = Iα, including to a wheel coupled to a falling mass | HL only | "Show that the acceleration of the mass is…" (4 marks) |
| Calculate angular momentum L = Iω and use its conservation | HL only | A skater, a stool or two discs: "Calculate the final angular velocity" (2 marks) |
| Find an angular impulse from ΔL = τΔt or from the area under a torque–time graph | HL only | Paper 1B or Paper 2 graph question (2 to 3 marks) |
| Calculate rotational kinetic energy from ½Iω² or L²/2I, including rolling without slipping | HL only | "Explain why the kinetic energy increases although L is constant" (3 marks) |
Before you start
You need A.1 (the suvat equations and what each symbol means), A.2 (Newton's second law, momentum and impulse, and circular motion with v = ωr and ω = 2π/T) and A.3 (kinetic energy and conservation of energy). Every equation in this subtopic is one of those with a new symbol in place of an old one, so if the linear version is shaky, fix that first. Angles are in radians throughout: one revolution is 2π rad.
1The idea in one paragraph
So far every object has been a point: push it and it moves in a straight line. Real objects are extended bodies, and where you push them matters. Push a door at the handle and it swings; push it at the hinge and nothing happens. A rigid body is an extended body that keeps its shape, so every part of it turns through the same angle in the same time. Its rotation is described by three quantities that copy displacement, velocity and acceleration; it is changed by torque, which copies force; and it resists change through its moment of inertia, which copies mass. Once the dictionary is in place, Newton's second law, momentum, impulse and kinetic energy all have rotational twins, and the same conservation laws hold.
2Torque: the turning effect of a force
The torque τ of a force about an axis is the force multiplied by the perpendicular distance from the axis to the force's line of action.
τ = Fr sin θ, where r is the distance from the axis to the point where the force acts and θ is the angle between F and the line r
Figure 1 shows why the sin θ is there. Only the part of r that is at right angles to the force does any turning. That part is r sin θ, the perpendicular distance from the axis to the line of action of the force (the line along which the force points, extended both ways). You can read the equation the other way too: F sin θ is the component of the force at right angles to r, multiplied by the full r. Both give the same number.
A spanner is 0.25 m long. You pull its end with a force of 80 N at 60° to the handle.
Pull at 90° instead and the torque is 80 × 0.25 = 20 N m, the most this force can give. Pull along the handle (θ = 0) and the torque is zero, however hard you pull, because the line of action passes through the axis. This is why you push a door at right angles and far from the hinge.
Three details carry marks.
The unit is N m. It is not written as J, although the base units are the same, because torque is not energy.
Torque has a sense. The guide asks you to state it as clockwise or anticlockwise (you do not need the vector treatment). When several torques act, pick one sense as positive, write it down, and add with signs.
Torque depends on the axis you choose. The same force has a different torque about a different point. Always say "about the hinge" or "about A".
Two equal and opposite forces that do not act along the same line give a turning effect with no resultant force at all. This pair is called a couple, and it is how you turn a steering wheel or a tap: the body turns without being shoved sideways.
3Rotational equilibrium
A body is in rotational equilibrium when the resultant torque on it is zero: the clockwise torques about any axis add to the same size as the anticlockwise torques. It is then either not rotating or rotating at a constant rate. For a body to be in full equilibrium, at rest and staying at rest, two conditions hold together:
- resultant force = 0 (translational equilibrium, from A.2), and
- resultant torque about any axis = 0 (rotational equilibrium).
To use them you need to know where the weight acts. The centre of mass of an extended body is the point where its whole mass can be treated as concentrated when you look at its linear motion, and its weight acts there. For a uniform beam, that is the middle. You will not be asked to calculate a centre of mass; you will be told it or it will be obvious from symmetry.
Figure 2 is a 3.0 m uniform beam of mass 20 kg resting on supports at its two ends, A and B. A painter of mass 70 kg stands 1.0 m from A. Find the two support forces. Take g = 9.81 m s⁻².
The trick is the choice of axis. Take torques about A: the force at A passes through A, so it has no torque there and drops out of the equation, leaving one unknown.
Sense-check: the painter stands nearer A, so A should carry more. It does. Take torques about the support whose force you do not want, and finish with the force balance.
4Describing rotation, and the equations of uniform angular acceleration
An unbalanced torque does to rotation what an unbalanced force does to linear motion: it causes an angular acceleration. To say that precisely you need three quantities, each the twin of one you know.
- Angular displacement Δθ: the angle a body turns through, in radians.
- Angular velocity ω: the rate of change of angular displacement, in rad s⁻¹. The guide lets you treat it as a speed, without vectors.
- Angular acceleration α: the rate of change of angular velocity, in rad s⁻².
Every point on a rigid body shares the same θ, ω and α. A point at distance r from the axis moves along an arc s = rθ, at speed v = ωr, with a tangential acceleration a = αr. Figure 3 lays out the whole dictionary; keep it open for the rest of the page.
Because α copies a and ω copies v, the suvat equations copy straight across. They hold only when α is constant.
Here ωi is the initial and ωf the final angular velocity. Choose the equation the same way as before: find the one quantity you neither know nor want, and use the equation without it.
A washing machine drum is spinning at 1200 revolutions per minute when the cycle ends. It slows uniformly and stops in 15 s. Find its angular acceleration and the number of turns it makes while stopping.
The minus sign says the drum is slowing. If the drum's radius is 0.25 m, a sock on the rim was moving at v = ωr = 125.7 × 0.25 = 31.4 m s⁻¹ at the start. The conversion in the first line is where most marks are dropped: rev min⁻¹ is not rad s⁻¹.
5Moment of inertia: where the mass is matters
Mass measures how hard it is to change a body's linear motion. The rotational twin is the moment of inertia I: how hard it is to change a body's rotation about a particular axis. It depends on how much mass there is and on how far from the axis that mass sits. For a set of point masses:
I = Σmr², where each mass m is at a perpendicular distance r from the axis. Unit: kg m²
The r² is the whole story. Mass twice as far from the axis counts four times as much. Figure 4 takes two 0.50 kg masses on a light rod 0.80 m long and puts the axis in three places.
Same mass, same rod, and I changes by a factor of eight. So "the moment of inertia of this object" means nothing until the axis is named.
For solid bodies the sum becomes a formula that depends on the shape, such as MR² for a thin hoop about its centre or ½MR² for a uniform disc. You do not have to remember or derive these: the guide says the formula will be given when you need it. What you must be able to do is explain the difference. A hoop has all its mass at distance R; a disc of the same mass and radius has most of its mass closer in, so its I is smaller.
6Newton's second law for rotation
τ = Iα, where τ is the resultant (average) torque about the axis
This is F = ma with the dictionary applied. A larger resultant torque gives a larger angular acceleration; a larger moment of inertia gives a smaller one for the same torque. If the torques balance, α = 0 and the rotation carries on unchanged, which is Newton's first law for rotation.
The exam's favourite use is a coupled system: a wheel on a fixed axle, with a string wrapped round it and a mass hanging from the string, as in Figure 5. The wheel is a uniform disc, M = 4.0 kg and R = 0.15 m, so I = ½MR² = 0.045 kg m². The hanging mass is 0.50 kg. Find the acceleration of the mass and the tension in the string.
Write one equation for each body, then one that ties them together.
Notice what the answer says. The mass falls at 1.96 m s⁻², far less than g, because the falling mass has to spin up the wheel as well as itself, and the wheel "feels like" an extra I/R² = 2.0 kg on the end of the string. Notice also that T (3.92 N) is less than mg (4.91 N): if it were equal, the mass would not accelerate at all.
7Angular momentum and angular impulse
A rotating body has angular momentum:
L = Iω, unit kg m² s⁻¹
For a single point mass m moving at speed v in a circle of radius r, I = mr² and ω = v/r, so L = mvr. You will need that form when a person jumps onto a roundabout.
In A.2, a resultant force applied for a time changes momentum: FΔt = Δp. The twin here is the angular impulse:
ΔL = τΔt = Δ(Iω)
When the torque changes with time, the angular impulse is the area under a torque–time graph, exactly as impulse was the area under a force–time graph. Figure 6 shows a motor's torque on a wheel of I = 0.30 kg m² that starts from rest.
Now the law that makes this subtopic worth studying.
The angular momentum of a body, or of a system of bodies, stays constant unless a resultant external torque acts on it.
The consequence is that I and ω trade off. If I falls, ω must rise to keep Iω the same. Figure 7 is a student on a freely rotating stool holding two dumbbells at arm's length, spinning at 2.0 rad s⁻¹, with a total moment of inertia of 3.6 kg m². They pull the dumbbells in to their chest and the moment of inertia drops to 1.2 kg m².
The pull on the dumbbells is inward, towards the axis, so it has no torque about the axis. The friction in the stool's bearing is the external torque you ignore; it is why the student slowly stops in real life. Figure skaters, divers tucking into a somersault, and a collapsing star that spins up into a pulsar are all this one line.
Coupled pairs. The guide also asks for two bodies that join. A turntable (I = 0.080 kg m²) spins freely at 3.0 rad s⁻¹. A second disc (I = 0.040 kg m²), not rotating, is dropped onto it on the same axis, and friction between them makes them turn together. The friction torques on the two discs are internal to the system (equal in size and opposite in sense), so the total angular momentum is conserved.
This is the rotational version of two trolleys sticking together in A.2, and it has the same energy story, which is the next section.
8Rotational kinetic energy
A rotating body has kinetic energy even if its centre is not moving, because every piece of it is moving. Adding up ½mv² for every piece, with v = ωr, gives:
Ek = ½Iω² = L²/2I
The second form comes from substituting ω = L/I, and it is the one to reach for whenever L is conserved, because then Ek depends only on I.
Go back to the student on the stool. L is fixed at 7.2 kg m² s⁻¹.
The kinetic energy has tripled while the angular momentum stayed the same. That is not a contradiction. Angular momentum is conserved because there is no external torque; energy is conserved because the extra 14.4 J came from somewhere, namely the chemical energy in the student's muscles, which did work pulling the dumbbells inwards against their tendency to fly outwards. If the student lets the dumbbells drift back out, the energy goes back the other way.
Now the coupled discs:
Angular momentum survived; kinetic energy did not. Exactly as in an inelastic collision.
9Rolling without slipping
A wheel rolling along the ground both translates (its centre moves) and rotates. The guide limits combined motion to rolling without slipping, which means the contact point is momentarily at rest on the ground, and that fixes the link between the two motions:
v = ωr for the centre of a body rolling without slipping
Its total kinetic energy is the sum of both kinds:
Ek = ½mv² + ½Iω²
Release three objects of the same mass from rest at the top of a slope with a vertical drop of 1.2 m: a block that slides without friction, a solid cylinder (I = ½mr²) and a thin hoop (I = mr²). All three start with the same gravitational potential energy mgh. The block puts all of it into ½mv². The rollers must also spin, so part of mgh goes into ½Iω² and less is left for ½mv².
Figure 8 shows the split. The mass and the radius cancel every time, so the result depends only on the shape: how the mass is distributed.
The hoop comes last because all of its mass sits at the rim, so a larger share of the energy goes into rotation. That sentence is the whole answer to "explain why the hoop is slower", and it is a favourite Paper 1A question.
These ideas reach well beyond wheels. Conservation of angular momentum keeps a satellite's orbit in one plane and sets the radius of the electron's orbit in the Bohr model of hydrogen (Theme E). A torque that grows with angular displacement produces simple harmonic motion (Theme C). A coil rotating in a magnetic field is how a generator makes a current (Theme D).
10Where marks are lost
- Leaving ω in rev min⁻¹ or rev s⁻¹. Every equation here needs rad s⁻¹. Multiply revolutions by 2π and divide minutes by 60, and write the conversion as its own line so the method mark is visible.
- Using the distance along the body instead of the perpendicular distance. τ = Fr only when the force is at 90° to r. Otherwise include sin θ, and make sure θ is the angle between F and r, not the angle with the vertical.
- Taking moments about an axis and then forgetting forces that act there. A force through the axis has zero torque about it, which is why you choose that axis, but it still appears in the force balance.
- Saying the moment of inertia depends only on mass. It depends on mass and on its distribution about the named axis. Without "distance from the axis" in the answer, the mark goes.
- Treating kinetic energy as conserved when angular momentum is. When a skater pulls in, L is constant and Ek rises; when two discs couple, L is constant and Ek falls. Check each quantity separately and name the reason for each.
- Setting the tension equal to mg in a pulley problem. If T = mg the mass would not accelerate. Write Newton's second law for the mass and τ = Iα for the wheel separately.
- Missing the rotational share of energy in rolling. A rolling ball down a slope does not reach √(2gh). Add ½Iω² and use v = ωr.
- Calling an inward pull a torque. A force directed along the radius, towards or away from the axis, has θ = 0 and no torque. That is why angular momentum is conserved when a skater pulls in.
11Draw it right
- Mark the axis of rotation clearly, as a dot for an axle seen end on or a dashed line for an axis in the plane of the page.
- Draw each force from the point where it acts, along its true line of action, with an arrowhead and a label. Weights go from the centre of mass.
- When you use τ = Fr sin θ, mark θ between the force and the line r, and draw the perpendicular distance if it helps.
- Show the sense of rotation or torque with a curved arrow, and label it clockwise or anticlockwise.
- In a coupled system (Figure 5), draw separate forces on each body: T acts down on the wheel and up on the mass.
- On a torque–time graph, label τ / N m and t / s, and shade the area you are using for ΔL.
- In a before-and-after sketch (Figure 7), write I and ω under each state, so the conservation line can be read off the page.
12Try it
Marks in brackets. Take g = 9.81 m s⁻². Answers and marker's notes follow.
Q1. A door is 0.90 m wide. A force of 40 N is applied at the handle, 0.85 m from the hinge line, at an angle of 30° to the plane of the door. Calculate the torque about the hinges, and the smallest force that would give the same torque from the handle. 3 marks
Q2. A uniform see-saw 4.0 m long has a mass of 15 kg and is pivoted at its centre. A child of mass 25 kg sits 1.8 m from the pivot. A second child of mass 30 kg sits on the other side so that the see-saw balances horizontally. 4 marks
(a) Calculate the distance of the second child from the pivot. 2 marks
(b) Calculate the force the pivot exerts on the see-saw. 2 marks
Q3. A potter's wheel has a moment of inertia of 0.60 kg m² about its axle. It starts from rest and reaches 90 revolutions per minute in 4.0 s with uniform angular acceleration. 6 marks
(a) Calculate the angular acceleration. 2 marks
(b) Calculate the average resultant torque on the wheel. 2 marks
(c) Determine the number of revolutions it makes in the 4.0 s. 2 marks
Q4. (Data-based, Paper 1B style. Invented data.) A student applies a known torque τ to a turntable by a string wound round its hub and measures the angular acceleration α each time.
| τ / N m | 0.020 | 0.040 | 0.060 | 0.080 | 0.100 |
|---|---|---|---|---|---|
| α / rad s⁻² | 0.22 | 0.58 | 1.02 | 1.38 | 1.81 |
(a) Explain why a graph of α against τ is expected to be a straight line. 2 marks
(b) Determine the moment of inertia of the turntable from the gradient. 2 marks
(c) The best-fit line crosses the τ axis at about 0.010 N m instead of passing through the origin. Suggest what this intercept represents. 2 marks
Q5. A roundabout in a playground has a moment of inertia of 240 kg m² and a radius of 1.5 m, and is at rest. A child of mass 30 kg runs at 4.0 m s⁻¹ along a line tangent to its edge and jumps on at the rim. Treat the child as a point mass. 6 marks
(a) Calculate the angular momentum of the child about the axis just before landing. 1 mark
(b) Calculate the angular velocity of the roundabout and child just after landing. 2 marks
(c) Calculate the change in kinetic energy, and explain why kinetic energy is not conserved although angular momentum is. 3 marks
Q6. A solid cylinder and a thin hoop have the same mass and the same radius. They are released together from rest at the top of the same slope and roll without slipping. Explain which one reaches the bottom first. 3 marks
13In one breath
A rigid body turns about an axis, and its rotation copies linear motion symbol for symbol: angular displacement θ in radians, angular velocity ω, angular acceleration α, with s = rθ, v = ωr and a = αr, and four suvat-style equations for constant α. Torque τ = Fr sin θ is the turning effect of a force, measured in N m, with a clockwise or anticlockwise sense; a body in equilibrium has zero resultant force and zero resultant torque about any axis. An unbalanced torque gives angular acceleration through τ = Iα, where the moment of inertia I = Σmr² depends on how far the mass sits from the named axis. Angular momentum L = Iω changes only through an angular impulse τΔt, the area under a torque–time graph, so with no external torque Iω stays constant and ω rises as I falls. Rotational kinetic energy is ½Iω² = L²/2I, which can rise (a skater pulling in) or fall (two discs coupling) while L holds steady. A body rolling without slipping has v = ωr and kinetic energy ½mv² + ½Iω², so the more its mass sits at the rim, the slower it rolls.
Answers
Q1. τ = Fr sin θ = 40 × 0.85 × sin 30° = 17 N m. The smallest force is the one applied at 90°: F = τ ÷ r = 17 ÷ 0.85 = 20 N. M1 for substituting into τ = Fr sin θ with sin 30°, A1 for 17 N m, A1 for 20 N. Using cos 30° gives 29.4 N m and scores M0 A0 for the first part. The 0.90 m door width is a distractor; using it scores M0.
Q2. (a) Torques about the pivot: the see-saw's own weight acts at the pivot and has no torque. 25 × 9.81 × 1.8 = 30 × 9.81 × d, so d = (25 × 1.8) ÷ 30 = 1.5 m. (b) The resultant force is zero, so the pivot pushes up with the total weight: (25 + 30 + 15) × 9.81 = 687 N upwards. (a) M1 for equating clockwise and anticlockwise torques about the pivot, A1 for 1.5 m. (b) M1 for including all three weights, A1 for 687 N with direction. Leaving out the see-saw's weight gives 540 N and scores M0 A0.
Q3. (a) ωf = 90 × 2π ÷ 60 = 9.42 rad s⁻¹. α = (9.42 − 0) ÷ 4.0 = 2.36 rad s⁻². (b) τ = Iα = 0.60 × 2.36 = 1.41 N m. (c) Δθ = ((ωf + ωi) ÷ 2) × t = (9.42 ÷ 2) × 4.0 = 18.8 rad; 18.8 ÷ 2π = 3.0 revolutions. (a) M1 for the conversion to rad s⁻¹, A1 for 2.36 rad s⁻². (b) M1 for τ = Iα, A1 for 1.4 N m (error carried forward from (a) allowed). (c) M1 for a correct equation for Δθ, A1 for 3.0 revolutions. Using 90 directly as ω gives α = 22.5 and loses the first M1 but can earn the later marks on carried-forward values.
Q4. (a) τ = Iα, so α = (1/I) × τ. I is a constant for the turntable, so α is directly proportional to τ, and the graph is a straight line through the origin with gradient 1/I. (b) Gradient of the best-fit line = (1.81 − 0.22) ÷ (0.100 − 0.020) ≈ 19.9 rad s⁻² per N m. I = 1 ÷ gradient = 1 ÷ 19.9 = 0.050 kg m². (c) The intercept is the torque needed before the turntable starts to accelerate at all: a friction torque in the bearing of about 0.010 N m that opposes the applied torque, so the resultant torque is τ − 0.010 N m. (a) R1 for quoting τ = Iα, R1 for I constant so α ∝ τ with gradient 1/I. (b) M1 for a gradient from the line using points far apart (19 to 21 accepted), A1 for I = 0.048 to 0.053 kg m². (c) R1 for friction, R1 for the idea that it must be overcome before any angular acceleration, or that the resultant torque is less than the applied one. "Human error" scores 0.
Q5. (a) L = mvr = 30 × 4.0 × 1.5 = 180 kg m² s⁻¹. (b) After landing, I = 240 + 30 × 1.5² = 307.5 kg m². Angular momentum is conserved, so ω = 180 ÷ 307.5 = 0.585 rad s⁻¹. (c) Before: Ek = ½ × 30 × 4.0² = 240 J. After: Ek = L² ÷ 2I = 180² ÷ (2 × 307.5) = 52.7 J. Change = −187 J, a loss of about 78%. The child and roundabout exert equal and opposite frictional forces on each other while the child's speed matches the rim's, so there is no external torque and L is conserved; but those frictional forces do work that transfers kinetic energy to internal energy (and sound), so kinetic energy is not conserved. It is a rotational inelastic collision. (a) A1. (b) M1 for adding the child's mr² to I, A1 for 0.585 rad s⁻¹. (c) A1 for both energies and the change (loss of 187 J), R1 for no external torque so L conserved, R1 for work done by friction or internal forces transferring energy away. Leaving the child out of the final I gives 0.75 rad s⁻¹ and scores M0 A0 in (b).
Q6. The solid cylinder arrives first. Both start with the same gravitational potential energy mgh, and at every point they share it between ½mv² and ½Iω² with ω = v/r. The hoop has all its mass at the rim, so it has the larger moment of inertia (mr² against ½mr²). A larger share of its energy goes into rotation, leaving less for translation, so at every height it is moving more slowly than the cylinder. R1 for the energy split between translational and rotational kinetic energy, R1 for the hoop's larger I because its mass is further from the axis, A1 for the conclusion that the cylinder has greater v at every point and arrives first. "The hoop is heavier" scores 0: the masses are equal.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section A.4 Rigid body mechanics. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.