A skater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular velocity of one revolution per second. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg⋅m², and the distance of the masses from the axis changes from 1 m to 0.1 m? A) 6 B) 3 C) 9 D) 4 E) 7

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Quizplus · Accepted Answer

The skater is initially rotating with angular velocity $\omega_i = 2\pi$ rad/s. When she brings her arms in, her moment of inertia decreases from $I_i = 2(5	ext{ kg})(1	ext{ m})^2 = 10	ext{ kg}\cdot	ext{m}^2$ to $I_f = 2(5	ext{ kg})(0.1	ext{ m})^2 = 0.1	ext{ kg}\cdot	ext{m}^2$, since the mass is now 0.1 m from the axis of rotation. By conservation of angular momentum, $I_i\omega_i = I_f\omega_f$. Solving for $\omega_f$, we find $\omega_f = \frac{I_i}{I_f}\omega_i = \frac{10	ext{ kg}\cdot	ext{m}^2}{0.1	ext{ kg}\cdot	ext{m}^2}(2\pi	ext{ rad/s}) = 200\pi	ext{ rad/s}$. Converting to revolutions per second, we have $\omega_f = \frac{200\pi}{2\pi} = \boxed{	extbf{(B) }3}$ rev/s.

A Skater Extends Her Arms Horizontally, Holding a 5-Kg Mass