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 - Page 240: 70


$m=.303\ kg$

Work Step by Step

We know that the moment of intertia of a rod about this point is $\frac{1}{3}ML^2$, and we know that the moment of inertia of a point mass a distance R away from the axis of rotation is $mR^2$. Thus, we find: $\frac{T^2K}{4\pi^2}=I$ $\frac{T^2K}{4\pi^2}=\frac{1}{3}ML^2+ mR^2+mR^2$ $\frac{T^2K}{4\pi^2}=\frac{1}{3}ML^2+ 2mR^2$ $\frac{T^2K}{4\pi^2}-\frac{1}{3}ML^2= 2mR^2$ $m= \frac{ \frac{T^2K}{4\pi^2}-\frac{1}{3}ML^2}{2R^2}$ $m= \frac{ \frac{5.6^2(.63)}{4\pi^2}-\frac{1}{3}(.85)(.75)^2}{2(.75)^2}$ $m=.303\ kg$
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