# Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 297: 9.43

(a) h = 0.673 m (b) The pulley has 45.5% of the total kinetic energy.

#### Work Step by Step

Let $v$ be the speed of the stone after it drops a height $h$. We can calculate the required speed $v$. $KE = \frac{1}{2}I\omega^2 = 4.50~J$ $\frac{1}{2}(\frac{1}{2}m_p~r^2)(\frac{v}{r})^2 = 4.50~J$ $v^2 = \frac{(4)(4.50~J)}{m_p}$ $v^2 = \frac{(4)(4.50~J)}{2.50~kg}$ $v^2 = 7.20~m^2/s^2$ We can find the required height $h$ for the stone to fall. $m_sgh = \frac{1}{2}m_sv^2+\frac{1}{2}I\omega^2$ $m_sgh = \frac{1}{2}m_sv^2+\frac{1}{2}(\frac{1}{2}m_p~r^2)(\frac{v}{r})^2$ $m_sgh = \frac{1}{2}m_sv^2+\frac{1}{4}m_p~v^2$ $h = \frac{v^2~(\frac{1}{2}m_s+\frac{1}{4}m_p)}{m_s~g}$ $h = \frac{(7.20~m^2/s^2)~[\frac{1}{2}(1.50~kg)+\frac{1}{4}(2.50~kg)]}{(1.50~kg)(9.80~m/s^2)}$ $h = 0.673~m$ (b) We can find the kinetic energy of the stone. $KE = \frac{1}{2}m_sv^2$ $KE = \frac{1}{2}(1.50~kg)(7.20~m^2/s^2)$ $KE = 5.40~J$ We can find the percent of kinetic energy in the pulley. $\frac{4.50~J}{4.50~J+5.40~J}\times 100\% = 45.5\%$ The pulley has 45.5% of the total kinetic energy.

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