OxfordAQA International AS and A-level Physics: Rotational motion and angular momentum (9630) – Revision Notes

These notes condense section 3.13.1 (Rotational motion) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4. Section 3.13 is labelled International A-level only. For the reasoning behind each result, see the study guide. Then try the practice questions. Tick off what you know on the printable checklist and see the course hub. Not sure where your gaps are? Take a free diagnostic.

Key formulas

Quantity	Equation	Unit
Moment of inertia, point mass	I = mr²	kg m²
Moment of inertia, extended object	I = Σmr²	kg m²
Rate of turning (angular speed)	ω = Δθ/Δt	rad s⁻¹
Rate of change of ω (angular acceleration)	α = Δω/Δt	rad s⁻²
Torque	T = Fr = Iα	N m
Angular momentum	Iω	kg m² s⁻¹
Rotational kinetic energy	Ek(rot) = ½Iω²	J
Work done by a torque	W = Tθ	J
Power of a torque	P = Tω	W

The specification says that formulae for the moment of inertia of particular solids are supplied when needed. Learn I = mr² and I = Σmr², and practise using a given expression.

Equations for uniform angular acceleration:

ω = ω₀ + αt
θ = (ω₀ + ω)t / 2
θ = ω₀t + ½αt²
ω² = ω₀² + 2αθ

Definitions to know

• Moment of inertia: a measure of how hard it is to change the rotation of an object about a given axis. It depends on the mass, how the mass is distributed about the axis, and the position of the axis.
• Angular displacement: the angle turned through, in radians.
• Angular speed: the rate of change of angle. Angular velocity is angular speed with a sense of rotation.
• Angular acceleration: the rate of change of angular velocity.
• Torque: the turning effect of a force, T = Fr, with r measured at right angles from the axis to the line of the force.
• Conservation of angular momentum: the total angular momentum of a system is constant if no external resultant torque acts on it.

Factors that change I (qualitative)

• More mass gives a larger I.
• Mass further from the axis gives a much larger I, because r is squared.
• A different axis gives a different I for the same object.

For the same mass and radius, a hollow ring has a larger I than a solid disc. It is harder to start and harder to stop, so for the same torque it gains angular speed more slowly.

Method boxes

Uniform angular acceleration.

1. Convert to radians and rad s⁻¹ (multiply rev min⁻¹ by 2π/60).
2. Write the symbols you know: ω₀, ω, α, θ, t.
3. Name the symbol you need and the one you are not given.
4. Pick the equation that leaves out the missing one.
5. Give the sign of α: negative if the body is slowing.

Torque and acceleration.

1. Find each torque as force times perpendicular distance.
2. Subtract opposing torques to get the resultant.
3. Find I for the correct axis.
4. α = resultant torque / I.
5. Use the uniform-acceleration equations for ω or θ.

Conservation of angular momentum.

1. Check that no external torque acts about the axis.
2. Write Iω before = Iω after.
3. Add the moments of inertia of everything that ends up rotating together.
4. Solve for ω. Compare the kinetic energies if asked about energy.

Small worked reminders

Net torque. A wheel with I = 0.080 kg m² is driven by a torque of 0.75 N m against an opposing torque of 0.15 N m.

resultant = 0.75 − 0.15 = 0.60 N m
α = 0.60 / 0.080 = 7.5 rad s⁻²

Changing I. A diver in the open position has I = 3.6 kg m² and turns at 1.5 rad s⁻¹. In the tucked position I = 1.2 kg m².

3.6 × 1.5 = 1.2 × ω
ω = 4.5 rad s⁻¹
Ek: 4.05 J → 12.15 J

The kinetic energy rises threefold, because ω rises threefold while I falls by the same factor (Ek = L²/2I with L fixed). The extra energy comes from the work done by the diver’s muscles in pulling the body in.

Ring and disc. A ring and a disc each have mass 0.40 kg and radius 0.25 m. The ring has I = mr² = 0.025 kg m². The disc is given as ½MR² = 0.0125 kg m². The ring has twice the moment of inertia because all its mass is at the rim.

Work and energy. A constant torque of 0.80 N m acts through 15 rad on a rotor of I = 0.060 kg m², starting from rest.

W = Tθ = 0.80 × 15 = 12 J
½Iω² = 12  →  ω² = 2 × 12 / 0.060 = 400
ω = 20 rad s⁻¹

Power. A torque of 50 N m turning at 60 rad s⁻¹ gives P = 50 × 60 = 3000 W.

Must-know distinctions

Idea	Be clear about
Angular speed and angular velocity	The first has size only; the second also has a sense of rotation
Torque and moment of inertia	Torque is the cause; I is the resistance to change
Applied torque and resultant torque	T = Iα needs the resultant
Mass and moment of inertia	Mass is fixed; I changes with the axis and the distribution
Conservation of angular momentum and of energy	The first holds with no external torque; kinetic energy may change
W = Tθ and Ek(rot)	The first is energy transferred; the second is energy stored

Analogy table

Translational	Rotational
s, v, a	θ, ω, α
m	I
F	T
F = ma	T = Iα
mv	Iω
½mv²	½Iω²
Fs, Fv	Tθ, Tω

Quick self-test

1. Find I for a 0.25 kg mass fixed 0.40 m from an axis on a light arm.
2. A point mass of 0.30 kg and a point mass of 0.10 kg are 0.20 m and 0.50 m from an axis. Find I for the pair.
3. Convert 450 revolutions per minute to rad s⁻¹.
4. A wheel accelerates uniformly from 4.0 rad s⁻¹ to 10.0 rad s⁻¹ in 3.0 s. Find α and θ.
5. A turntable starts from rest and has α = 2.5 rad s⁻². Find its angular speed after it has turned through 20 rad.
6. A force of 15 N acts at right angles to a spanner 0.28 m long. Find the torque.
7. A resultant torque of 4.5 N m acts on a body with I = 0.60 kg m². Find α.
8. A rotor with I = 2.0 kg m² turns at 5.0 rad s⁻¹. Find its angular momentum and its kinetic energy.
9. A spinning object has I = 0.40 kg m² and ω = 6.0 rad s⁻¹. Its I is then halved with no external torque. Find the new ω.
10. A shaft transmits 3.6 kW at 120 rad s⁻¹. Find the torque.
11. State the rotational equivalent of mass, force and momentum.
12. Give two reasons why the same rod has different I about different axes.

Answers

1. I = mr² = 0.25 × 0.40² = 0.040 kg m².
2. I = 0.30 × 0.20² + 0.10 × 0.50² = 0.012 + 0.025 = 0.037 kg m².
3. 450 × 2π / 60 = 47 rad s⁻¹.
4. α = (10.0 − 4.0)/3.0 = 2.0 rad s⁻². θ = (4.0 + 10.0) × 3.0 / 2 = 21 rad.
5. ω² = 0 + 2 × 2.5 × 20 = 100, so ω = 10 rad s⁻¹.
6. T = 15 × 0.28 = 4.2 N m.
7. α = 4.5 / 0.60 = 7.5 rad s⁻².
8. Iω = 2.0 × 5.0 = 10 kg m² s⁻¹. Ek(rot) = ½ × 2.0 × 5.0² = 25 J.
9. Iω is constant: 0.40 × 6.0 = 0.20 × ω, so ω = 12 rad s⁻¹.
10. T = P/ω = 3600 / 120 = 30 N m.
11. Moment of inertia, torque, angular momentum (Iω is the equivalent of mv).
12. The distances of the mass elements from the axis change, and I depends on r², so the sum Σmr² changes. The total mass stays the same.

Where marks are usually lost

• Using rev min⁻¹ or degrees directly, instead of converting to rad s⁻¹ and radians.
• Writing I = mr² with r measured to the wrong point, or forgetting to square r.
• Adding the masses and using one r for an extended object, instead of summing each mr².
• Using the applied torque in T = Iα when a frictional or opposing torque is present.
• Saying angular momentum is conserved without stating that no external resultant torque acts.
• Claiming kinetic energy is conserved when two spinning bodies lock together, or forgetting that the energy has gone to thermal energy.
• Giving a numerical answer to a “describe the effect” question without saying which way the quantity changes and why.
• Omitting units, or mixing N m with J as if they were interchangeable. Torque is in N m; work and energy are in J.
• Using a value of I for one axis when the object turns about another.

Official syllabus

Published by OxfordAQA: International AS and A-level Physics (9630), specification Version 4.4, which applies to exams from May/June 2020. Source for this page: section 3.13.1, Rotational motion.