Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 10 - Exercises and Problems - Page 186: 68



Work Step by Step

We use conservation of energy to find: $mg(h-2R)=\frac{1}{2}mv^2 + \frac{1}{2} I \omega^2$ $mg(h-2R)=\frac{1}{2}mv^2 + \frac{1}{2} \frac{2}{5}mr^2\frac{v^2}{r^2}$ $g(h-2R)=\frac{1}{2}v^2 + \frac{1}{5} v^2$ $g(h-2R) = .7v^2 $ We know the following: $F_n+mg=\frac{mv^2}{R}$ At the minimum velocity, $F_n$ will equal 0. Thus, we find: $v=\sqrt{gR}$ We plug this into the above equation to obtain: $g(h-2R) = .7gR $ $h=2.7R$
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