Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 15 - Exercises and Problems: 72


a) $P_t = \frac{1}{2}\rho \omega v^2 r^2+P_a + \rho h_0 g$ b) $h_t = h_0+\frac{\omega^2 r^2}{2g}$

Work Step by Step

a) In this problem, there are three sources of pressure at the bottom of the pan. For starters, there is the atmospheric pressure, which we will call $P_a$. In addition, there is the pressure due to the water above the bottom of the pan. We know that pressure is force per unit area, so we find the gravitational force per unit area to be: $P=\rho h_0 g$ (For there is $h_0$ meters of water above the point.) Finally, there is pressure due to centripetal acceleration, which we find is: $P = \frac{1}{2}\rho v^2$ $P = \frac{1}{2}\rho \omega^2 r^2$ Adding all of these gives: $P_t = \frac{1}{2}\rho \omega^2 r^2+P_a + \rho h_0 g$ b) We know that the height will occur where the centripetal force and gravitational force cancel. Thus, we find: $\rho h g=\frac{1}{2}\rho \omega^2 r^2$ $h =\frac{\omega^2 r^2}{2g}$ We add this to the initial height to find: $h_t = h_0+\frac{\omega^2 r^2}{2g}$
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