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) $\frac{E_R}{E_P}=\frac{Iv^2}{2r^2mgh}=\frac{(7.23\times10^{-5}kgm^2)(1.06\frac{m}{s})^2}{2(0.0065m)^2(0.105kg)(9.8\frac{m}{s^2})(1.0m)}\times100\%=93\%$