Conservation of angular momentum — figure skater spinning faster

Question

A figure skater spins with arms outstretched at 2 rev/s. Her moment of inertia with arms extended is 4 kg m². When she pulls her arms in, her moment of inertia decreases to 1.6 kg m². Find her new angular velocity. Why does she spin faster?

(NCERT Class 11, Chapter 7 — System of Particles and Rotational Motion)

Solution — Step by Step

When the skater pulls her arms in, there is no external torque acting about her vertical axis of rotation (the ice exerts negligible friction torque). So angular momentum is conserved:

L_i = L_f

I_i \omega_i = I_f \omega_f

$I_i = 4$ kg m², $\omega_i = 2$ rev/s, $I_f = 1.6$ kg m²

4 \times 2 = 1.6 \times \omega_f

\omega_f = \frac{8}{1.6}

\boxed{\omega_f = 5 \text{ rev/s}}

She spins 2.5 times faster after pulling her arms in.

Initial KE: $\frac{1}{2}I_i\omega_i^2 = \frac{1}{2} \times 4 \times (2 \times 2\pi)^2 = \frac{1}{2} \times 4 \times 16\pi^2 = 32\pi^2$ J

Final KE: $\frac{1}{2}I_f\omega_f^2 = \frac{1}{2} \times 1.6 \times (5 \times 2\pi)^2 = \frac{1}{2} \times 1.6 \times 100\pi^2 = 80\pi^2$ J

KE increases from $32\pi^2$ to $80\pi^2$ J. The extra energy comes from the internal muscular work the skater does while pulling her arms inward.

Why This Works

Angular momentum $L = I\omega$ is conserved when net external torque is zero. When the skater pulls her arms in, she reduces her moment of inertia $I$ (mass moves closer to the rotation axis). Since $L$ must stay the same, $\omega$ must increase proportionally.

Think of it as a trade-off: less spread-out mass → faster spin. The product $I\omega$ stays constant, but the individual values change inversely.

This is the same principle behind a diver tucking in during a somersault (spins faster) and opening up before entering the water (slows the spin for a clean entry).

Alternative Method — Using angular momentum in SI units

Convert to rad/s first: $\omega_i = 2 \times 2\pi = 4\pi$ rad/s

L = I_i \omega_i = 4 \times 4\pi = 16\pi \text{ kg m}^2\text{/s}

\omega_f = \frac{L}{I_f} = \frac{16\pi}{1.6} = 10\pi \text{ rad/s} = 5 \text{ rev/s}

You can work in rev/s throughout — no need to convert to rad/s — as long as you’re consistent. The $2\pi$ factor cancels on both sides of $I_i\omega_i = I_f\omega_f$ . This saves time in NEET where every second counts.

Common Mistake

Students often assume that kinetic energy is also conserved here. It is NOT. Angular momentum is conserved (no external torque), but kinetic energy increases because the skater does internal work with her muscles. Conservation of energy and conservation of angular momentum are different laws — one does not imply the other. If a question asks “is KE conserved?”, the answer is no.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using angular momentum in SI units

Common Mistake

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