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 11 - Exercises and Problems - Page 209: 55

Answer

a) $\omega d(\frac{1}{2}-\frac{I}{2md^2})$ b) $\omega d$ c) $\omega d(2+\frac{I}{md^2})$

Work Step by Step

a) We find using conservation of angular momentum that: $mvd+I\omega=(md^2+I)(\frac{1}{2}\omega)$ $mvd=\frac{1}{2}md^2\omega-\frac{1}{2}I\omega$ $v=\frac{1}{2}\omega d-\frac{I\omega}{2md}$ $v=\omega d(\frac{1}{2}-\frac{I}{2md^2})$ b) $mvd+I\omega=(md^2+I)(\omega)$ $mvd=md^2\omega$ $v= \omega d$ c) We use a similar process to find: $mvd+I\omega=(md^2+I)(2\omega)$ $mvd=2md^2\omega +I\omega$ $v=2\omega d+\frac{I\omega}{md}$ $v= \omega d(2+\frac{I}{md^2})$
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