The big idea: Spin on an office chair and pull your arms in: you suddenly whirl faster. That stored spin is angular momentum (L), rotation's version of momentum.
Straight-line momentum
- p = mv
- mass × velocity
Spinning momentum
- L = Iω
- moment of inertia × angular velocity
A fast, heavy, spread-out spin has lots of it.
- angular momentum (kg m² s⁻¹)
- moment of inertia (kg m²)
- angular velocity (rad s⁻¹)
A disc has a moment of inertia of 0.20 kg m² and spins at 15 rad s⁻¹. Find its angular momentum.
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Angular impulse: Angular momentum stays the same unless a resultant torque acts. A torque τ acting for a time Δt changes it by the angular impulse:
ΔL = τΔt = Δ(Iω), unit N m s (the same as kg m² s⁻¹).
A 0.50 N m torque for 4.0 s on a wheel of I = 0.20 kg m² from rest: ΔL = 2.0 kg m² s⁻¹, so ω = 2.0 ÷ 0.20 = 10 rad s⁻¹.
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Spin is conserved: If no external torque acts, angular momentum stays constant: L = Iω is the same before and after.
Why an ice skater speeds up
Pull mass inward
The body pulls its mass inward, so I gets smaller.
L stays the same
With no external torque, L = Iω cannot change.
Spin gets faster
So ω gets faster: this is why a skater speeds up when they pull their arms in.
Smaller I, faster ω.
A skater spins at 2.0 rad s⁻¹ with a moment of inertia of 4.0 kg m². They pull their arms in, lowering it to 1.5 kg m². Find their new angular velocity.
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A spinning body that shrinks: For a sphere or disc, I ∝ MR². If it shrinks to one tenth of its radius with no external torque, I falls to 1/100, so ω rises 100 times.
Its rotational kinetic energy L²/2I rises 100 times too: the work comes from the forces pulling it in.
Is angular momentum conserved?: Look for an external torque on the system. A skater on smooth ice: none, so L is conserved. Two blocks on a pulley: their weights give a resultant torque about the axle, so L of the system changes as they speed up.
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A spinning body also stores kinetic energy — flywheels use this to store energy. It mirrors ½mv², with I in place of m and ω in place of v.
- rotational kinetic energy (J)
- moment of inertia (kg m²)
- angular velocity (rad s⁻¹)
- angular momentum (kg m² s⁻¹)
A rolling object has both: Something that rolls (a wheel, a ball) is moving and spinning, so its total KE = ½mv² + ½Iω² — translational plus rotational.
The disc above (I = 0.20 kg m²) spins at 15 rad s⁻¹. Find its rotational kinetic energy.
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Same energy, different spin: Two wheels with the same rotational kinetic energy: L = √(2IEk), so the one with four times the moment of inertia has twice the angular momentum.
Rolling down a slope
A solid ball (I = ⅖MR²) rolls without slipping from rest down a slope, dropping 0.90 m. Find its speed at the bottom (g = 9.8 m s⁻²).
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How this is tested — angular momentum and rotational energy are HL only (A.4):
Paper 1A
- A quick L = Iω, ½Iω², or 'what happens to ω when I changes?'.
Paper 2
- A conservation problem (something lands on a spinning disc) or comparing the rotational and translational KE of a rolling object.
The classic trap: Using energy conservation for a sticking collision. When objects stick, angular momentum is conserved but kinetic energy is not — always use I₁ω₁ = I₂ω₂.
Three easy marks: (1) Spot the magic words 'no external torque' → use I₁ω₁ = I₂ω₂. (2) Energy is not conserved when objects stick (some is lost), but angular momentum is. (3) For rolling, remember the two KE terms.
A turntable (moment of inertia 0.50 kg m²) spins freely at 12 rad s⁻¹. A lump of clay is dropped onto it, adding 0.10 kg m² to the moment of inertia. Determine the new angular velocity.
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