Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 57

Answer

The speed of the ball at the lowest point of the track will be $v = (3.74~\sqrt{R-r}~)~m/s$

Work Step by Step

The kinetic energy at the bottom will be equal to the magnitude of the change in potential energy. Note that the change in height is $R-r$. So, $PE = K$ $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I_{CM}\omega^2$ $Mg(R-r) = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}Mr^2)(\frac{v^2}{r^2})$ $Mg(R-r) = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2$ $Mg(R-r) = \frac{7}{10}Mv^2$ $v = \sqrt{\frac{10}{7}(9.80~m/s^2)(R-r)}$ $v = (3.74~\sqrt{R-r})~m/s$ Therefore, the speed of the ball at the lowest point of the track will be $v = (3.74~\sqrt{R-r}~)~m/s$
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