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

This guide teaches section 3.13.1 (Rotational motion) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, in exams from May/June 2020. The topic sits under 3.13 (Energy sources), which the specification labels International A-level only, so International AS candidates can leave it out. Use it with the course hub and tick points off on the printable checklist.

Other pages for this topic: revision notes and practice questions. You should already be comfortable with circular motion and angular speed and with vectors and moments. To find which topics need work first, take a free diagnostic.

What the specification asks for

Outcome	What you must be able to do
I = mr² and I = Σmr²	Find the moment of inertia of a point mass and of a set of point masses
Factors affecting I	Say qualitatively how mass, its distribution and the axis change I
Angular quantities	Describe and use displacement, speed, velocity and acceleration for turning motion
Uniform angular acceleration	Use the four equations for constant α
Torque = Fr = Iα	Find torque and use it to find angular acceleration
Angular momentum Iω	Calculate it, and use conservation of angular momentum
Ek(rot) = ½Iω²	Calculate rotational kinetic energy
W = Tθ and P = Tω	Find work done and power for a torque
Analogy	Link each rotational quantity to its translational partner

The specification says that formulae for the moment of inertia of particular solids are supplied when needed. So you do not need to memorise the formula for a particular solid. You do need to use a given expression correctly, and you must know I = mr² and I = Σmr².

Moment of inertia

Moment of inertia, I, measures how hard it is to change the rotation of an object about a chosen axis. It plays the part that mass plays in straight-line motion. Its unit is kg m².

For one point mass at perpendicular distance r from the axis:

I = mr²

For an extended object, imagine it split into small pieces. Add up the value for each piece:

I = Σmr²

Factors that affect I

• Total mass. More mass gives a larger I, if the distribution is similar.
• How the mass is distributed. Mass far from the axis counts for much more than mass near it, because r is squared.
• The position of the axis. The same object has a different I about a different axis.

Shape matters because of the second factor. A ring and a solid disc can have the same mass and radius, yet the ring has the larger I because all its mass sits at the full radius.

Worked example 1: adding point masses. Three small beads are fixed to a light rod. The rod turns about an axis through one end. The beads are 0.20 kg at 0.10 m, 0.50 kg at 0.30 m and 0.10 kg at 0.60 m from the axis. Calculate I, and the rotational kinetic energy when the rod turns at 6.0 rad s⁻¹.

I = 0.20 × 0.10² + 0.50 × 0.30² + 0.10 × 0.60²
  = 0.0020 + 0.045 + 0.036
  = 0.083 kg m²
Ek(rot) = ½ × 0.083 × 6.0² = 1.5 J

The 0.10 kg bead is the lightest, yet it contributes almost as much as the 0.50 kg bead, because it is twice as far out.

Worked example 2: a given expression. A uniform rod of mass 3.0 kg and length 1.2 m is given as having I = ML²/12 about an axis through its centre and I = ML²/3 about an axis through one end.

centre: I = 3.0 × 1.2² / 12 = 0.36 kg m²
end:    I = 3.0 × 1.2² / 3  = 1.44 kg m²

The same rod is four times harder to spin about its end. Half the rod is further from the axis, and r is squared.

Angular displacement, speed, velocity and acceleration

These match the linear quantities one for one.

• Angular displacement, θ. The angle turned through, in radians. One revolution is 2π rad.
• Angular speed, ω. The rate of change of angle: ω = Δθ/Δt, in rad s⁻¹. It has size only.
• Angular velocity. Angular speed with a direction, which is the sense of rotation. In a problem with a single axis, choose one sense as positive and give the opposite sense a negative sign.
• Angular acceleration, α. The rate of change of angular velocity: α = Δω/Δt, in rad s⁻².

Convert before you calculate. For a rotation rate in revolutions per minute, multiply by 2π and divide by 60. A point at distance r from the axis has linear speed v = rω and tangential acceleration a = rα.

Equations for uniform angular acceleration

For constant α these have the same form as the equations for constant linear acceleration:

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

Here ω₀ is the initial angular speed and ω the final. List the known symbols, name the unknown, and choose the equation that links them without needing a fifth quantity.

Worked example 3: a carousel. A carousel speeds up uniformly from rest to 1.5 rad s⁻¹ in 12 s. Find α and the angle turned through, in revolutions.

α = (1.5 − 0) / 12 = 0.125 rad s⁻²
θ = ½αt² = ½ × 0.125 × 12² = 9.0 rad
revolutions = 9.0 / 2π = 1.4

Worked example 4: slowing to rest. A grinding wheel turns at 3000 revolutions per minute. It is brought to rest uniformly in 40 revolutions. Find the angular acceleration and the time taken.

ω₀ = 3000 × 2π / 60 = 314 rad s⁻¹
θ  = 40 × 2π = 251 rad
ω² = ω₀² + 2αθ  →  0 = 314² + 2α(251)
α  = −314² / (2 × 251) = −196 rad s⁻²
t  = 2θ / ω₀ = 2 × 251 / 314 = 1.6 s

The negative sign means the angular acceleration opposes the rotation. A graph of ω against t is a straight line, its gradient is α, and the area under it is θ.

Torque and Fr = Iα

The torque of a force about an axis is

T = Fr

where r is measured at right angles from the axis to the line along which the force acts. The unit is N m. Torque is the cause of angular acceleration, as force is for linear acceleration:

T = Iα

Here T is the resultant torque. If a driving torque and an opposing torque act, subtract them first. A larger I means a smaller α for the same torque.

Worked example 5: a pulley wheel. A string is wrapped round a uniform disc of mass 2.0 kg and radius 0.15 m. The disc is given as having I = ½MR² about its axis. A steady tangential pull of 2.4 N acts at the rim, and the disc starts from rest. Find α, the angular speed after 2.5 s, and the work done by the torque in that time.

I = ½ × 2.0 × 0.15² = 0.0225 kg m²
T = Fr = 2.4 × 0.15 = 0.36 N m
α = T / I = 0.36 / 0.0225 = 16 rad s⁻²
ω = αt = 16 × 2.5 = 40 rad s⁻¹
θ = ½αt² = ½ × 16 × 2.5² = 50 rad
W = Tθ = 0.36 × 50 = 18 J

Check: Ek(rot) = ½ × 0.0225 × 40² = 18 J. The work done equals the energy gained.

Angular momentum and its conservation

The angular momentum of a body turning about a fixed axis is

L = Iω

with unit kg m² s⁻¹ (equivalently N m s). Like linear momentum, it is conserved: the total angular momentum of a system stays constant when no external resultant torque acts on it.

Two common cases:

1. I changes, same object. If I falls, ω rises so that Iω stays constant.
2. Two bodies join. When one rotating body locks onto another about the same axis, (I₁ + I₂)ω_f = I₁ω₁ + I₂ω₂.

Worked example 6: a disc dropped onto a disc. Disc A (I = 0.50 kg m²) spins freely at 8.0 rad s⁻¹. Disc B (I = 0.30 kg m²), at rest on the same axis, is lowered onto it and the two turn together.

0.50 × 8.0 = (0.50 + 0.30) × ω_f
ω_f = 4.0 / 0.80 = 5.0 rad s⁻¹
Ek before = ½ × 0.50 × 8.0² = 16 J
Ek after  = ½ × 0.80 × 5.0² = 10 J

Angular momentum is conserved but 6 J of kinetic energy is not. It becomes thermal energy as the surfaces slip against each other before they lock, as in an inelastic collision.

Rotational kinetic energy, work and power

Rotational kinetic energy is Ek(rot) = ½Iω². Each small mass in the object has kinetic energy ½m(rω)², and adding them gives ½Iω².

A constant torque T turning a body through θ does work W = Tθ. Dividing by time gives the power P = Tω.

Worked example 7: power from a motor. A motor turns a shaft at 1800 revolutions per minute against a steady torque of 12 N m.

ω = 1800 × 2π / 60 = 188 rad s⁻¹
P = Tω = 12 × 188 = 2.3 × 10³ W

Use W = Tθ for energy transfer over an angle, and Ek(rot) = ½Iω² for energy stored in the rotation. When a resultant torque speeds up a body, W = Tθ equals the gain in Ek(rot), as in example 5.

The analogy with translational dynamics

Translational	Rotational
displacement s	angular displacement θ
velocity v	angular velocity ω
acceleration a	angular acceleration α
mass m	moment of inertia I
force F	torque T
F = ma	T = Iα
momentum mv	angular momentum Iω
Ek = ½mv²	Ek(rot) = ½Iω²
W = Fs	W = Tθ
P = Fv	P = Tω

If you forget a rotational equation, write the translational one and swap each symbol. The one thing that does not carry over is the role of distance: I depends on where the mass is, but mass m does not.

Common errors

• Using revolutions per minute, or degrees, in a formula. Convert to rad s⁻¹ and radians first.
• Measuring r to the wrong place. In I = mr² the distance is to the axis, not to the centre of the object.
• Using the applied torque in T = Iα when an opposing torque also acts. Use the resultant.
• Using a value of I for the wrong axis, for example the centre value when the object turns about its end.
• Assuming kinetic energy is conserved because angular momentum is.
• Leaving out the sign on α when the body slows down.

Next steps

Test yourself with the practice questions, and condense the topic with the revision notes.

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.