Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - General Problems - Page 227: 78

Answer

a) $1.06\frac{m}{s}$ b) $93\%$

Work Step by Step

$m=2\times0.050kg+0.0050kg=0.105kg$ $d=0.075m$ $I=\frac{2M_cR_c^2}{2}+\frac{M_hR_h^2}{2}=\frac{(0.050kg)(0.038m)^2}{2}+\frac{(0.005kg)(0.0065m)^2}{2}$ $=2\times3.61\times10^{-5}kg m^2+1.06\times10^{-7}kg m^2$$=7.23\times10^{-5}kgm^2$ $mgh=(2m_c+m_h)gh=(0.105kg)(9.8\frac{m}{s^2})(1.0m)=1.03J$ a) $E_P=E_K+E_R$ $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ $mgh=\frac{mv^2}{2}+\frac{Iv^2}{2r^2}$ $v=\sqrt{\frac{2r^2mgh}{r^2m+I}}$ Here, we can use the radius of the hub because the torque is acting on it. $v=\sqrt{\frac{2(0.0065m)^2(1.03J)}{(0.0065m)^2(0.105kg)+7.23\times10^{-5}kgm^2}}=1.06\frac{m}{s}$ b) $\frac{E_R}{E_P}=\frac{Iv^2}{2r^2mgh}=\frac{(7.23\times10^{-5}kgm^2)(1.06\frac{m}{s})^2}{2(0.0065m)^2(0.105kg)(9.8\frac{m}{s^2})(1.0m)}\times100\%=93\%$
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