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An ice-skater spinning with arms extended ($I_1 = 4$ kg·m², $\omega_1 = 3$ rad/s) pulls her arms in to $I_2 = 1$ kg·m². Her new $\omega_2$ and kinetic-energy change are:
A12 rad/s; KE quadruples (18→72 J)
B12 rad/s; KE halved
C1.5 rad/s; KE doubled
D6 rad/s; KE constant
Answer & Solution
Correct answer: A. 12 rad/s; KE quadruples (18→72 J)
Conservation of angular momentum: $I_1\omega_1 = I_2\omega_2$ ⇒ $\omega_2 = 12$ rad/s. KE$_1 = \tfrac{1}{2}(4)(9) = 18$ J. KE$_2 = \tfrac{1}{2}(1)(144) = 72$ J. Extra 54 J came from the skater's muscles doing work against centrifugal forces while pulling arms in.
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