Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 334: 74


(a) $\omega = \sqrt{\frac{3g}{L}}$ (b) $v = \sqrt{3gL}$

Work Step by Step

(a) The final rotational kinetic energy of the rod will be equal to the initial potential energy of the rod. Note that the center of mass of the rod is initially at a height of $\frac{L}{2}$. $KE_{rot} = PE$ $\frac{1}{2}I~\omega^2 = \frac{MgL}{2}$ $\frac{1}{2}(\frac{1}{3}ML^2)~\omega^2 = \frac{MgL}{2}$ $\omega^2 = \frac{3g}{L}$ $\omega = \sqrt{\frac{3g}{L}}$ (b) We can find the speed of the tip of the rod. $v = \omega~L$ $v = \sqrt{\frac{3g}{L}}~L$ $v = \sqrt{3gL}$
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