Conservation of Angular Momentum

A figure skater pulling in her arms spins faster. A planet sweeping closer to its sun speeds up. A gyroscope refuses to topple. All three involve the same powerful idea: angular momentum is conserved when there is no external torque.

Concept

Conservation law: If the net external torque on a system about a point is zero, the total angular momentum about that point is constant:

$\vec{\tau}_{\text{ext}} = 0 \quad \Longrightarrow \quad \vec{L} = \text{constant}$

For a rigid body rotating about a fixed axis,

$L = I\omega = \text{constant}$

So if $I$ decreases (mass pulled in), $\omega$ increases — and vice versa.

Applications:

• Figure skater. Arms outstretched: large $I$, slow $\omega$. Arms pulled in: small $I$, fast $\omega$. KE actually increases because the skater does work pulling her arms in against the centrifugal effect.
• Planetary motion (Kepler's second law). Gravity acts along $\vec{r}$ (toward Sun), so $\vec{\tau} = \vec{r}\times\vec{F} = 0$ about the Sun. Angular momentum of the planet about the Sun is conserved → equal areas in equal times.
• Gyroscope and spinning top. Large $L$ resists changes in direction. External torque (gravity) causes precession rather than toppling.

Derivation

From $\vec{\tau}_{\text{ext}} = d\vec{L}/dt$, if $\vec{\tau}_{\text{ext}} = 0$:

$\frac{d\vec{L}}{dt} = 0 \quad \Longrightarrow \quad \vec{L} = \text{constant}$

For a rigid body where $\vec{L} = I\vec{\omega}$ along a fixed axis,

$I_1 \omega_1 = I_2 \omega_2$

Kepler's second law from conservation. In a small time $dt$, the radius vector from Sun to planet sweeps a triangle of area

$dA = \tfrac{1}{2}|\vec{r}\times\vec{v}|\,dt = \frac{L}{2m}\,dt$

So $dA/dt = L/(2m) = \text{constant}$. Equal areas in equal times.

Worked Example

A figure skater spins at 1.5 rev/s with arms extended. Her moment of inertia is 4 kg·m². When she pulls her arms in, the moment of inertia drops to 1.6 kg·m². Find her new angular speed.

Solution:

By conservation, $I_1 \omega_1 = I_2 \omega_2$:

$\omega_2 = \frac{I_1 \omega_1}{I_2} = \frac{4 \times 1.5}{1.6} = 3.75 \, \text{rev/s}$

She spins 2.5× faster.

Energy check. KE = $\tfrac{1}{2}I\omega^2$.

KE$_1 = \tfrac{1}{2}(4)(1.5 \cdot 2\pi)^2 = \tfrac{1}{2}(4)(88.83) \approx 178 \, \text{J}$.

KE$_2 = \tfrac{1}{2}(1.6)(3.75 \cdot 2\pi)^2 = \tfrac{1}{2}(1.6)(555.2) \approx 444 \, \text{J}$.

KE increases by ~266 J — the skater did this work by pulling her arms in against the rotation.

Common Confusions

• Conservation needs zero net torque, not zero force. Earth's gravity on a planet has no torque about the Sun because force is along the radius.
• KE is not conserved when angular momentum is. In the skater example, KE rises because internal forces do work.
• Mass falling vertically near a rotating Earth. Angular momentum changes? No — gravity acts toward Earth's centre; torque about Earth's centre is zero. The mass's angular speed about Earth's centre changes, but $r$ changes too, so $L = mvr$ remains constant.
• Skater example reversibility. If she extends her arms again, $\omega$ slows and KE decreases — the body does negative work, getting it back.

Key Takeaways

• $\vec{L}$ is conserved when $\vec{\tau}_{\text{ext}} = 0$.
• For fixed axis: $I_1 \omega_1 = I_2 \omega_2$ when $I$ changes (no external torque).
• Kepler's 2nd law follows from conservation of planetary angular momentum.
• Gyroscopes: large $L$ resists direction changes; gravity causes precession.
• KE may change even when $L$ is conserved (internal work done).