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: 31

Answer

$L = 0.54~m$

Work Step by Step

We can use the following equation for the period of a physical pendulum: $T = 2\pi~\sqrt{\frac{I}{mgd}}$ Here, $I$ is the moment of inertia of the object $d$ is the distance from the center of mass to the pivot Therefore; $T = 2\pi~\sqrt{\frac{I}{mgd}}$ $T = 2\pi~\sqrt{\frac{\frac{1}{3}mL^2}{mg~(\frac{L}{2})}}$ $T = 2\pi~\sqrt{\frac{2L}{3g}}$ $L = \frac{T^2~(3g)}{2~(2\pi)^2}$ $L = \frac{(1.2~s)^2~(3)(9.80~m/s^2)}{2~(2\pi)^2}$ $L = 0.54~m$
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