Problem: An ice skater has $I_i = 6$ kg·$m^{2}$ and $\omega_i = 2$ rev/s. She pulls her arms in to $I_f = 2$ kg·$m^{2}$. Find: (a) $\omega_f$, (b) $KE_f/KE_i$.

Step 1: Convert units. $\omega_i = 2 \text{ rev/s} = 2 \times 2\pi = 4\pi$ rad/s.

Step 2: Apply conservation of angular momentum ($\tau_{net} = 0$ — no friction between skate and ice in the spin direction).

$I_i\omega_i = I_f\omega_f$ $(6\,\text{kg·m}^2)(4\pi\,\text{rad/s}) = (2\,\text{kg·m}^2)\omega_f$ $\omega_f = \frac{6 \times 4\pi}{2} = 12\pi\,\text{rad/s} = 6\,\text{rev/s}$

Step 3: Calculate kinetic energies.

$KE_i = \frac{1}{2}I_i\omega_i^2 = \frac{1}{2}(6)(4\pi)^2 = 3 \times 16\pi^2 = 48\pi^2\,\text{J}$

$KE_f = \frac{1}{2}I_f\omega_f^2 = \frac{1}{2}(2)(12\pi)^2 = 1 \times 144\pi^2 = 144\pi^2\,\text{J}$

$\frac{KE_f}{KE_i} = \frac{144\pi^2}{48\pi^2} = 3$

Interpretation: The angular momentum is conserved (L = const), but kinetic energy triples. The extra energy comes from muscular work done by the skater in pulling her arms in. Angular momentum conservation does NOT imply kinetic energy conservation.

Alternative formula: $KE = \frac{L^2}{2I}$. Since $L$ is constant: $KE \propto 1/I$. Ratio = $I_i/I_f = 6/2 = 3$. ✓