Physics (10th Edition)

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

Chapter 9 - Rotational Dynamics - Problems - Page 247: 55


a) The height of the hill is $3.7m$ b) The translation speed at the bottom is $6.95m/s$

Work Step by Step

a) The basketball has $v_0=0$, $\omega_0=0$ and $h_0=0$. Therefore, its initial energy $E_0=0$ According to the principle of energy conservation: $$E_f=E_0=0$$ $$\frac{1}{2}mv_f^2+\frac{1}{2}I\omega_f^2+mg(-h_f)=0$$ The basketball is a thin-walled spherical shell rotating around its center, so $I=\frac{2}{3}mr^2$. Also, because the basketball has a rolling motion, $\omega_f=\frac{v_f}{r}$ $$\frac{1}{2}mv_f^2+\frac{1}{3}mr^2\Big(\frac{v_f^2}{r^2}\Big)+mg(-h_f)=0$$ $$\frac{1}{2}mv_f^2+\frac{1}{3}mv_f^2=mgh_f$$ $$\frac{5}{6}v_f^2=gh_f$$ $$h_f=\frac{5v_f^2}{6g}$$ We have $v_f=6.6m/s$ and $g=9.8m/s^2$, so $h_f=3.7m$, which is the height of the hill. b) The frozen juice can is a solid cylinder, so $I=\frac{1}{2}mr^2$. Similarly, we have $$\frac{1}{2}mv_f^2+\frac{1}{2}I\omega_f^2+mg(-h_f)=0$$ $$\frac{1}{2}mv_f^2+\frac{1}{4}mv_f^2=mgh_f$$ $$\frac{3}{4}v_f^2=gh_f$$ $$v_f=\sqrt{\frac{4gh_f}{3}}=6.95m/s$$
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