Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems: 68

Answer

$1.532 \ s$

Work Step by Step

We know the following equation for period: $T= 2\pi \sqrt{\frac{I}{mgcL}}$, where k is the percent as a decimal down the initial rod that you are. Thus, we see that $c=.25$. Using the equations for the moments of inertias, we find: $T = 2\pi \sqrt{\frac{\frac{1}{12}mL^2+ \frac{1}{16}mL^2}{mgcL}}$ $T = 2\pi \sqrt{\frac{\frac{1}{12}L+ \frac{1}{16}L}{.25g}}$ Since L=1, we find: $T = 2\pi \sqrt{\frac{\frac{1}{12}+ \frac{1}{16}}{.25g}}=1.532 \ s$
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