Angular Momentum: Definition and Conservation Revision Notes | A Level AQA Physics

1. Why Angular Momentum Matters

A figure skater pulls their arms inward and suddenly spins faster — no engine, no push, no extra force. How? The answer lies in angular momentum, one of the most powerful conservation laws in physics.

Angular momentum is the rotational counterpart of linear momentum. You already know that linear momentum $p = mv$ tells you how hard it is to stop a moving object. Angular momentum does exactly the same job for spinning objects — it tells you how hard it is to stop or change a rotation.

This topic sits within AQA's Engineering Physics option and builds directly on your earlier work on moment of inertia and torque. The two specification requirements here are tightly linked: first, the definition $L = I\omega$; second, the conservation principle that follows from it.

2. Defining Angular Momentum: $L = I\omega$

Think back to linear momentum: $p = mv$. Mass $m$ measures resistance to linear acceleration; velocity $v$ describes how fast the object moves. Now swap to rotation: moment of inertia $I$ replaces mass, and angular velocity $\omega$ replaces linear velocity.

The equation $L = I\omega$ is deceptively compact. The moment of inertia $I$ depends not only on the total mass of the object but on how that mass is distributed around the axis. A hollow cylinder has a larger $I$ than a solid cylinder of the same mass and radius, because more of its mass sits far from the axis.

Angular momentum is a vector quantity — it has both magnitude and direction. The direction is along the axis of rotation, determined by the right-hand rule. For AQA A Level, most questions treat rotation about a fixed axis, so you work with the magnitude $L = I\omega$ and track sign by specifying a positive direction of rotation.

Worked example: Calculating Angular Momentum of a Rotating Disc

A solid disc of moment of inertia $I = 0.40$ kg m² rotates at $\omega = 12$ rad s⁻¹. Calculate its angular momentum.

1. $L = I\omega$ — This is the defining equation for angular momentum. [M1]
2. $I = 0.40$ kg m², $\omega = 12$ rad s⁻¹ — Both values are already in SI units, so no conversion is needed.
3. $L = 0.40 \times 12$ — Substituting directly. [M1]
4. $L = 4.8$ kg m² s⁻¹ — Units: kg m² × rad s⁻¹ = kg m² s⁻¹ (radians are dimensionless). [A1]

Answer: $L = 4.8$ kg m² s⁻¹

3. Conservation of Angular Momentum

Now that you can calculate $L$, the next question is: what happens to it over time? The answer is the conservation principle, and it is one of the most elegant results in classical mechanics.

The condition is crucial: no resultant external torque. Internal forces within the system — such as a skater's muscles pulling their arms inward — do not count as external torques. They redistribute angular momentum within the system but cannot change the total.

This mirrors conservation of linear momentum exactly. Linear momentum is conserved when no resultant external force acts; angular momentum is conserved when no resultant external torque acts. The parallel is not a coincidence — both arise from Newton's laws applied to their respective domains.

The physical insight here is powerful: if $I$ decreases (mass moves closer to the axis), then $\omega$ must increase to keep $L$ constant. If $I$ increases (mass moves further from the axis), $\omega$ decreases. This is exactly what the figure skater exploits.

4. Worked Examples: Applying Conservation

The conservation equation $I_1\omega_1 = I_2\omega_2$ is straightforward to apply, but exam questions often disguise the change in moment of inertia. The key skill is identifying what has changed about the mass distribution and calculating the new $I$ before substituting.

Worked example: Conservation of Angular Momentum: Skater Pulling Arms In

A skater spins with arms outstretched, giving a moment of inertia of $I_1 = 3.6$ kg m² and angular velocity $\omega_1 = 2.0$ rad s⁻¹. She pulls her arms in, reducing her moment of inertia to $I_2 = 1.2$ kg m². Calculate her new angular velocity $\omega_2$.

1. $I_1\omega_1 = I_2\omega_2$ (no external torque acts) — Angular momentum is conserved because the ice exerts negligible torque on the skater about the vertical axis. [M1]
2. $I_1 = 3.6$ kg m², $\omega_1 = 2.0$ rad s⁻¹, $I_2 = 1.2$ kg m² — All values are in SI units. The unknown is $\omega_2$.
3. $\omega_2 = \frac{I_1\omega_1}{I_2}$ — Divide both sides by $I_2$. [M1]
4. $\omega_2 = \frac{3.6 \times 2.0}{1.2}$ — Numerator: $3.6 \times 2.0 = 7.2$. Denominator: $1.2$.
5. $\omega_2 = \frac{7.2}{1.2} = 6.0$ rad s⁻¹ — Units: (kg m² × rad s⁻¹) ÷ kg m² = rad s⁻¹. Sanity check: $I$ decreased by a factor of 3, so $\omega$ should increase by a factor of 3. $2.0 \times 3 = 6.0$ ✓ [A1]

Answer: $\omega_2 = 6.0$ rad s⁻¹

Notice the sanity check in step 5. When $I$ reduces by a factor of 3, $\omega$ must increase by a factor of 3 — their product $L$ stays fixed. Building this proportional reasoning into your checking routine will catch arithmetic errors before they cost marks.

Worked example: Conservation of Angular Momentum: Two Discs Coupling

A disc of moment of inertia $I_A = 0.80$ kg m² rotates at $\omega_A = 15$ rad s⁻¹. A second stationary disc of moment of inertia $I_B = 0.40$ kg m² is dropped onto it and the two discs couple together (rotate as one). Calculate the final angular velocity of the combined system.

1. $L_i = I_A\omega_A + I_B\omega_B = (0.80 \times 15) + (0.40 \times 0)$ — Disc B is stationary, so $\omega_B = 0$ and contributes no angular momentum. [M1]
2. $L_i = 12 + 0 = 12$ kg m² s⁻¹ — $0.80 \times 15 = 12$. The initial angular momentum of the system is 12 kg m² s⁻¹.
3. $L_f = (I_A + I_B)\omega_f = (0.80 + 0.40)\omega_f = 1.20\,\omega_f$ — After coupling, both discs rotate together with the same $\omega_f$. [M1]
4. $12 = 1.20\,\omega_f$ — No external torque acts during the coupling, so total angular momentum is conserved.
5. $\omega_f = \frac{12}{1.20} = 10$ rad s⁻¹ — Units: kg m² s⁻¹ ÷ kg m² = rad s⁻¹. Sanity check: total $I$ increased from 0.80 to 1.20 kg m², so $\omega$ should decrease. $15 \to 10$ rad s⁻¹ ✓ [A1]

Answer: $\omega_f = 10$ rad s⁻¹

5. Physical Examples and Conceptual Analysis

Conservation of angular momentum appears across a wide range of physical systems. Being able to describe and explain these examples is a core exam skill — 'describe' questions ask you to state what happens, while 'explain' questions require you to link the observation to the principle $I_1\omega_1 = I_2\omega_2$.

• Figure skater: pulls arms inward, reducing $I$, so $\omega$ increases — spins faster without any external torque.
• Diver tucking: reduces $I$ by curling into a ball, increasing $\omega$ to complete more rotations before entering the water.
• Neutron star formation: a collapsing stellar core dramatically reduces its radius (and hence $I$), causing it to spin at hundreds of rotations per second.
• Two discs coupling: a stationary disc dropped onto a spinning disc increases the total $I$, so the combined system rotates more slowly.
• Gyroscope: the large angular momentum of a spinning gyroscope resists changes in orientation — the direction of $L$ is conserved unless an external torque acts.

The neutron star example is worth dwelling on. A star's core might have a radius of around $10^5$ km before collapse and shrink to roughly 10 km — a reduction in radius by a factor of $10^4$. Since $I \propto r^2$ for a uniform sphere, $I$ decreases by a factor of roughly $10^8$. To conserve $L$, the angular velocity must increase by the same factor. This is why neutron stars (pulsars) can rotate hundreds of times per second.

6. Conditions for Conservation: When Does It Apply?

Angular momentum is always conserved in an isolated system — one with no resultant external torque. The word 'resultant' matters: individual external torques may act, but if they cancel, the total angular momentum still does not change.

Compare this with the condition for conservation of linear momentum: no resultant external force. The two principles are structurally identical. In both cases, internal interactions (forces or torques) within the system cannot change the total conserved quantity — they only transfer it between parts of the system.

A common scenario where conservation does NOT apply is a braking flywheel. When a brake pad presses against a spinning flywheel, it exerts an external torque on the flywheel system. Angular momentum is not conserved — it decreases as the flywheel slows. The angular momentum is transferred to the brake and surroundings as heat.

7. Exam Technique and Common Mistakes

AQA questions on angular momentum tend to follow predictable patterns. Knowing the traps in advance is worth marks. The guidance below consolidates the most important exam technique points for this topic.

For extended-response questions asking you to 'evaluate' or 'explain fully', structure your answer as: state the conservation principle and its condition → identify what changes in the system (which quantity, and in which direction) → apply $I_1\omega_1 = I_2\omega_2$ → state the consequence for $\omega$ → link back to the physical scenario described in the question.

That final worked example captures the highest-level demand in this topic: recognising that angular momentum conservation and energy conservation are separate principles, and that one can hold while the other appears to be violated — until you account for all energy inputs.