Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 65

Answer

$1.2 kg \cdot m^2$ By bringing her arms close to her body.

Work Step by Step

There is no net external torque on the skater, so her net angular momentum is conserved. $$L_{i}=L_{i}$$ $$I_i \omega_i = I_f \omega_f$$ $$I_f = \frac{\omega_i }{\omega_f}I_i $$ $$I_f = \frac{1.0rev/1.5s }{2.5rev/s}(4.6 kg \cdot m^2)=1.2 kg \cdot m^2$$ Physically, she pulls her arms closer to her body (the rotation axis) to reduce her moment of inertia.
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