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An ice skater spinning with arms outstretched at $\omega_1$ pulls arms in, halving her moment of inertia. Her new angular speed becomes:
A$2\omega_1$, by conservation $I\omega$ = constant
B$\omega_1/2$, since the skater slows when compact
C$\omega_1$, unchanged since mass is conserved on the rink
D$4\omega_1$, the moment of inertia squared in the law
Answer & Solution
Correct answer: A. $2\omega_1$, by conservation $I\omega$ = constant
No external torque: $I\omega$ conserved; halving $I$ doubles $\omega$.
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