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Conservation of Angular Momentum

AP Physics 1· difficulty 4/5

A skater spins at angular speed ω0\omega_0 with arms extended, having moment of inertia I0I_0. She pulls her arms in until her moment of inertia becomes I0/3I_0/3. What is her new angular speed and the ratio of final to initial rotational kinetic energy?

  • A

    ω=9ω0\omega = 9\omega_0, KEf/KEi=9KE_f/KE_i = 9

  • B

    ω=3ω0\omega = 3\omega_0, KEf/KEi=3KE_f/KE_i = 3

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  • C

    ω=ω0/3\omega = \omega_0/3, KEf/KEi=1/3KE_f/KE_i = 1/3

  • D

    ω=3ω0\omega = 3\omega_0, KEf/KEi=1KE_f/KE_i = 1

Explanation

Conservation of angular momentum: I0ω0=(I0/3)ωI_0\omega_0 = (I_0/3)\omega, so ω=3ω0\omega = 3\omega_0. Rotational KE =12Iω2= \tfrac{1}{2}I\omega^2, ratio =(1/3)(3)2=3= (1/3)(3)^2 = 3. The skater does work pulling her arms in.

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