Physics · Engineering Physics · A2 Level

Conservation of Angular Momentum

Watch an ice skater pull their arms in mid-spin and they visibly speed up, with no extra push from anywhere. Nothing external is driving that increase — it's a direct consequence of one of the cleanest conservation laws in physics, and it turns out to matter just as much inside a piece of rotating machinery as it does on an ice rink.

8 min read

The principle

The previous lesson showed that a system's angular momentum can only change if an external torque acts on it: $T\Delta t = \Delta(I\omega)$. It follows immediately that if the net external torque acting on a system is zero, its angular momentum cannot change at all — it stays exactly constant. This is the principle of conservation of angular momentum:

$$I_1\omega_1 = I_2\omega_2$$

for any isolated system (no net external torque), where the subscripts label the system's state before and after some internal change. The subtlety, and the reason this principle is so useful, is that the system's own moment of inertia $I$ is allowed to change internally — for instance, by the system redistributing its own mass closer to or further from the rotation axis — even while no external torque acts on it at all. If $I$ changes while $L = I\omega$ must stay fixed, then $\omega$ is forced to change to compensate. Pull mass inward (decrease $I$) and $\omega$ must increase; push mass outward (increase $I$) and $\omega$ must decrease.

Worked exampleApplying I₁ω₁ = I₂ω₂

A rotating system initially has moment of inertia $I_1 = 4.8\text{ kg m}^2$ and angular velocity $\omega_1 = 2.5\text{ rad s}^{-1}$. Internal rearrangement pulls mass inward, reducing the moment of inertia to $I_2 = 3.0\text{ kg m}^2$, with no external torque acting throughout.

Find the new angular velocity $\omega_2$.

Apply conservation of angular momentum.

$I_1\omega_1 = I_2\omega_2$

Rearrange to make $\omega_2$ the subject.

$\omega_2 = \dfrac{I_1\omega_1}{I_2}$

Substitute and evaluate.

$\omega_2 = \dfrac{4.8 \times 2.5}{3.0} = 4.0\text{ rad s}^{-1}$

ω₂ = 4.0 rad s⁻¹

The moment of inertia dropped by a factor of $1.6$, and the angular velocity rose by exactly that same factor — the product $I\omega$ is unchanged throughout, as it must be with no external torque acting.

Sporting examples — the same physics, a different context

The redistribution of mass driving a change in moment of inertia doesn't have to happen inside a machine — it happens inside a rotating human body just as readily, which is why this principle shows up so often in sporting contexts.

The spinning skater. Arms and legs extended outward place mass further from the body's spin axis, giving a larger moment of inertia. Pulling the arms in close to the body moves that same mass much closer to the axis, sharply reducing $I$. Since no external torque acts on the skater once they're already spinning freely on the ice, angular momentum must be conserved, and $\omega$ increases dramatically — the skater visibly speeds up.
The tucking diver or gymnast. A diver launching off a board with some initial rotation, or a gymnast mid-tumble, can pull into a tight tuck — knees drawn to chest, arms wrapped in — reducing their moment of inertia and speeding up their rotation, allowing more somersaults in the same time in the air. Extending back out into a straight, open body position before entering the water or landing increases $I$ again, slowing the rotation back down to a controlled speed for a clean entry or landing.

In every one of these cases, the underlying mechanism is identical: the athlete does internal work to redistribute their own mass, changing their own moment of inertia, while no external torque acts on them once they're already in free rotation — so $I\omega$ is conserved and $\omega$ must adjust to compensate.

The core idea, stated precisely

Conservation of angular momentum does not require that a system's moment of inertia stays fixed — only that no external torque acts on it. A system is entirely free to change its own internal mass distribution, and therefore its own $I$, and $\omega$ will change automatically to keep $I\omega$ constant throughout.

Why this matters for rotating machinery

The same reasoning applies directly inside real engineering systems. Any rotating machine component whose effective mass distribution shifts during operation — for instance, if part of it moves radially inward or outward relative to the rotation axis while the machine is spinning freely with no driving or braking torque applied — will experience a change in rotational speed as a direct consequence of conservation of angular momentum, exactly as the skater does. Recognising this is essential for understanding controlled energy storage devices such as flywheels, where the deliberate concentration of mass at a large, fixed radius is what makes their rotational behaviour predictable and useful — the subject of the next and final lesson in this topic.

Key takeaways

If no net external torque acts on a system, its total angular momentum $L = I\omega$ remains constant — the principle of conservation of angular momentum.
A system's own moment of inertia can still change internally, through redistribution of its own mass relative to the rotation axis.
If $I$ decreases, $\omega$ must increase to keep $I\omega$ constant, and vice versa: $I_1\omega_1 = I_2\omega_2$.
A spinning skater pulling their arms in reduces $I$ and increases $\omega$, speeding up their spin with no external push.
A diver or gymnast tucking into a tighter shape spins faster for the same reason, then extends back out to slow their rotation before landing or entry.
The same principle applies to rotating machinery: a shift in a component's mass distribution during free rotation changes its rotational speed.
This principle underlies the predictable behaviour of flywheels, covered next.

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Angular Momentum

Flywheels