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 245: 22


$ .824\ Nm$

Work Step by Step

We first must find the moment of inertia: $ I = \frac{2}{3}mr^2 = \frac{2}{3}(.622)(.12^2)=.006$ Thus, we now can find the value of k by simplifying the equation for the angular velocity of a wire and by recalling that omega is equal to $2\pi f$. Doing this, we find: $k = 4\pi^2f^2I = 4\pi^2(1.87^2)( .006) = .824\ Nm$
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