College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 234: 113


$h = \frac{2R}{3}$

Work Step by Step

We can use conservation of energy to find an expression for $v^2$, where $v$ is the speed when the skier leaves the surface: $\frac{1}{2}mv^2+mgh = mgR$ $\frac{1}{2}mv^2 = mg~(R-h)$ $v^2 = 2g~(R-h)$ Let $\theta$ be the angle above the horizontal, from the center of the half dome, when the skier leaves the surface. Note that $sin~\theta = \frac{h}{R}$. At this moment, the component of the skier's weight directed into the surface of the slope is equal in magnitude to the required centripetal force: $mg~sin~\theta = \frac{mv^2}{R}$ $v^2 = gR~sin~\theta$ $v^2 = (gR)~(\frac{h}{R})$ $v^2 = gh$ We can equate the two expressions for $v^2$: $gh = 2g~(R-h)$ $h = 2~(R-h)$ $3h = 2R$ $h = \frac{2R}{3}$
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