school
Acemy

Conservation of Angular Momentum

AP Physics 1· difficulty 4/5

arms out: I, ω arms in: I/3

A skater spins with moment of inertia II and angular speed ω\omega. She pulls her arms in, reducing her moment of inertia to I/3I/3. What is her new angular speed and the ratio of new to old rotational kinetic energy?

  • A

    ω′=3ω\omega' = 3\omega, K′/K=9K'/K = 9

  • B

    ω′=ω/3\omega' = \omega/3, K′/K=1/3K'/K = 1/3

  • C

    ω′=3ω\omega' = 3\omega, K′/K=1K'/K = 1

  • D

    ω′=3ω\omega' = 3\omega, K′/K=3K'/K = 3

    check_circle

Explanation

Angular momentum is conserved: Iω=(I/3)ω′I\omega = (I/3)\omega' so ω′=3ω\omega' = 3\omega. Then K′=12(I/3)(3ω)2=32Iω2=3KK' = \tfrac{1}{2}(I/3)(3\omega)^2 = \tfrac{3}{2}I\omega^2 = 3K.

Want 10 more like this — adaptive to your weak spots?

Practice 10 like thisarrow_forwardPredict my AP score

Related questions

auto_awesomeAll Toolsquestion_answerHomework Q&Amenu_bookAP & SAT PreptimelineLearn Mode