Iω = constant. Why does the skater speed up when arms are pulled in?

Angular Momentum Conservation — Ice Skater Spinning

Question

An ice skater is spinning with angular velocity $\omega_1$ with arms stretched out. She pulls her arms close to her body, reducing her moment of inertia from $I_1$ to $I_2$ where $I_2 < I_1$ . What happens to her angular velocity, and why?

More precisely: if $I_1 = 2\, \text{kg·m}^2$ , $\omega_1 = 3\, \text{rad/s}$ , and she pulls in to $I_2 = 0.5\, \text{kg·m}^2$ , find $\omega_2$ .

Solution — Step by Step

There’s no external torque acting on the skater — the ice gives negligible friction and her internal muscle forces are internal to the system. When net external torque is zero, angular momentum $L = I\omega$ is conserved.

L_i = L_f

I_1 \omega_1 = I_2 \omega_2

This is the entire physics. Everything else is just substitution.

2 \times 3 = 0.5 \times \omega_2

6 = 0.5 \, \omega_2

\omega_2 = 12 \, \text{rad/s}

We can verify by the ratio form: $\omega_2 = \omega_1 \times \dfrac{I_1}{I_2} = 3 \times \dfrac{2}{0.5} = 3 \times 4 = 12\, \text{rad/s}$ . ✓

Moment of inertia dropped by a factor of 4, so angular velocity increased by a factor of 4.

The final answer: $\omega_2 = 12\, \text{rad/s}$

Why This Works

Angular momentum $L = I\omega$ is the rotational analogue of linear momentum $p = mv$ . Just as linear momentum is conserved when no external force acts, angular momentum is conserved when no external torque acts.

When the skater pulls her arms in, she is redistributing mass closer to the rotation axis. Moment of inertia $I = \sum mr^2$ — the $r^2$ dependence means even a small reduction in radius causes a large drop in $I$ . To keep the product $I\omega$ constant, $\omega$ must shoot up.

The energy angle is interesting too. The skater’s kinetic energy increases — she does positive work with her muscles pulling her arms in. Angular momentum is conserved; energy is not (she added energy from her muscles). This trips up students who try to apply energy conservation here.

Alternative Method — Ratio Approach

For MCQs, you rarely need to calculate absolute values. Use the ratio directly:

\frac{\omega_2}{\omega_1} = \frac{I_1}{I_2}

If the problem says “moment of inertia becomes one-fourth”, you immediately write $\omega_2 = 4\omega_1$ . No algebra needed. This saves 30 seconds in NEET, which matters.

Whenever a spinning object changes shape with no external torque, write $I_1\omega_1 = I_2\omega_2$ first, then identify whether you need the absolute answer or just the ratio. For most MCQs, the ratio gets you there in one line.

Common Mistake

Applying energy conservation instead of angular momentum conservation.

Students write $\frac{1}{2}I_1\omega_1^2 = \frac{1}{2}I_2\omega_2^2$ — this is wrong. The skater does muscular work while pulling her arms in, so kinetic energy is NOT conserved. Angular momentum is conserved because there is no external torque, but the skater adds energy to the system through internal forces.

This mistake appeared as a trap option in NEET 2024 — the energy-conservation answer was listed as an option specifically to catch students who confuse the two conservation laws.

The check: after getting $\omega_2 = 12\, \text{rad/s}$ , calculate KE before and after.

KE_1 = \frac{1}{2}(2)(3^2) = 9 \, \text{J}

KE_2 = \frac{1}{2}(0.5)(12^2) = 36 \, \text{J}

KE increased fourfold. That extra $27\, \text{J}$ came from the skater’s muscles — completely consistent with physics, and a useful cross-check that your answer is in the right direction (angular velocity should increase, KE should increase).

Question

Solution — Step by Step

Why This Works

Alternative Method — Ratio Approach

Common Mistake

Try These Next