Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 15 - Oscillations - Exercises and Problems: 59

Answer

$I = 0.087~kg~m^2$

Work Step by Step

We can find the moment of inertia of the lower leg about the knee joint as: $f = \frac{1}{2\pi}~\sqrt{\frac{Mgd}{I}}$ $2\pi~f = \sqrt{\frac{Mgd}{I}}$ $(2\pi~f)^2 = \frac{Mgd}{I}$ $I = \frac{Mgd}{(2\pi~f)^2}$ $I = \frac{(5.0~kg)(9.80~m/s^2)(0.18~m)}{(2\pi)^2~(1.6~Hz)^2}$ $I = 0.087~kg~m^2$
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