Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 13 - Exercises and Problems - Page 248: 81


$m=.06488 kg$

Work Step by Step

Simplifying the equation for frequency, we know: $I=\frac{K}{4\pi^2f^2}$ Using the values of moments of inertias, we find: $\frac{1}{2}Mr^2 + mr^2+mr^2 = \frac{K}{4\pi^2f^2}$ $\frac{1}{2}Mr^2 + 2mr^2 = \frac{K}{4\pi^2f^2}$ $ 2mr^2 = \frac{K}{4\pi^2f^2}-\frac{1}{2}Mr^2$ $m = \frac{\frac{K}{4\pi^2f^2}-\frac{1}{2}Mr^2}{2r^2 }$ $m = \frac{\frac{5}{4\pi^2(2.6)^2}-\frac{1}{2}(.34)(.25)^2}{2(.25)^2 }$ $m=.06488 kg$
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