Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 6 - Work and Energy - Problems - Page 168: 50

Answer

$\theta=48.2^o$

Work Step by Step

The person starts to leave the surface when there is no longer normal force $F_N$, leaving gravitational force to provide the only centripetal force there. From the figure below, the component of $mg$ that provides for $F_c$ is $mg\cos\theta$ $$mg\cos\theta=F_c=\frac{mv^2}{r}$$ $$g\cos\theta=\frac{v^2}{r}$$ $$v=\sqrt {gr\cos\theta}$$ The person slides down, so his KE increases while his PE decreases by the same amount according to the principle of energy conservation. We have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m(v_f^2-v_0^2)=-mg\Delta h$$ $$(v_f^2-v_0^2)=-2g\Delta h$$ We know $v_0=0$. We take the finish point to be the point the person leaves the surface, so $v_f=\sqrt {gr\cos\theta}$. From the figure below, we also have $\Delta h=h_f-h_0=r\cos\theta-r$ $$gr\cos\theta=-2g(r\cos\theta-r)$$ $$gr\cos\theta=-2gr(\cos\theta-1)$$ $$\cos\theta=-2\cos\theta+2$$ $$3\cos\theta=2$$ $$\cos\theta=\frac{2}{3}$$ $$\theta=48.2^o$$
Small 1581338517
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.