Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 10 - Exercises and Problems - Page 194: 70

Answer

$\omega_{min}=\sqrt{\omega_{max}^2-1.24\times \frac{1}{R}}$

Work Step by Step

When the disk is at the top of its path, the center of mass is shifted up by $\frac{1}{30}R$. The mass of the disk is $\frac{15}{16}M$, since $M-(\frac{1}{4})^2M =\frac{15}{16}M$. Thus, using conservation of energy, it follows: $ I\omega_{max}^2=I\omega_{min}^2+\frac{15}{16}M(\frac{1}{30}R)g$ Thus, using the result from problem 65 and simplifying, we find: $\omega_{min}=\sqrt{\omega_{max}^2-1.24\times \frac{1}{R}}$
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