Rotational Dynamics

Every linear quantity has a rotational partner: mass → moment of inertia $I=\sum mr^2$ , force → torque $T$ , momentum → angular momentum $L=I\omega$ . Newton's second law becomes $T=I\alpha$ ; kinetic energy is $\frac12I\omega^2$ ; power is $T\omega$ . With no external torque, angular momentum is conserved.

AQA 7408 Paper 3B≈ 45 minRevision 1 · updated 2026-10-27

Section 1 of 14

Learning objectives

Calculate moments of inertia and rotational kinetic energy.
Use $T=I\alpha$ , work $T\theta$ and power $T\omega$ .
Use angular equations of motion.
Apply conservation of angular momentum.

Pause and think

A spinning ice skater pulls her arms in and spins faster. Where does the extra speed come from?

Section 2 of 14

Core idea

Start hereSpinning faster for free

An ice skater starts a slow spin with arms out, then pulls them in and whirls. No one pushed her. Rotation has its own versions of mass, force and momentum, and the last one is conserved.

The big idea

Rotational analogues: $I$ for $m$ , $T$ for $F$ , $\omega$ for $v$ . $T=I\alpha$ , $E_k=\frac12I\omega^2$ , $L=I\omega$ conserved without external torque.

1. Moment of inertia

Rotational kinetic energy

$E_k=\sum\tfrac12m_iv_i^2$
Add up every particle.
$v_i=r_i\omega$
All share the same $\omega$ .
$E_k=\tfrac12\left(\sum m_ir_i^2\right)\omega^2=\tfrac12I\omega^2$
Define $I=\sum mr^2$ .

Mass far from the axis counts for much more.

Worked exampleEngine power

Torque $50$ N m at $600$ rpm.

Goal: Power.
Strategy: Convert to rad s⁻¹; $P=T\omega$ .

$\omega=\frac{600\times2\pi}{60}=62.8\text{ rad s}^{-1}$
Revolutions to radians.
$P=50\times62.8=3100\text{ W}$
$P=T\omega$ .

Answer

3.1

kW.

✓ CheckLike

P=Fv

✓.

2. Angular momentum

Worked exampleDisc dropped on a disc

$I=0.20$ kg m² at $10$ rad s⁻¹; an identical still disc lands on it.

Goal: KE lost.
Strategy: Conserve $L$ , compare KE.

$0.20\times10=0.40\,\omega\Rightarrow\omega=5.0$
No external torque.
$E_1=\tfrac12(0.20)(10^2)=10\text{ J}$
Before.
$E_2=\tfrac12(0.40)(5^2)=5.0\text{ J}$
After.

Answer

5.0

J lost (to friction/heat).

✓ CheckLike a perfectly inelastic collision ✓.

Linear and rotational

Drawing diagram…

Swap each linear quantity for its rotational partner.

Common misconceptionrpm Convert rpm to rad s⁻¹ before any formula.

Key takeawayFlywheels A large

I

stores energy and resists changes in speed, smoothing an engine's power strokes.

Pause and think

Why do flywheels have most of their mass in the rim?

Summary

$I=\sum mr^2$ .
$T=I\alpha$ ; $P=T\omega$ .
$L=I\omega$ conserved.

Moment of inertia: Resistance to angular acceleration.
Torque: Turning effect of a force.
Angular momentum: $I\omega$ .

HistorySpinning planets. Kepler's second law, equal areas in equal times, is really conservation of angular momentum: planets speed up as they approach the Sun.

Fun factFlywheel buses. In the 1950s, Swiss "gyrobuses" ran on a 1.5-tonne flywheel spun up at each stop, travelling a few kilometres per charge.

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