Answer
(a) $\omega = 33.9~rad/s$
(b) $v = 2.71~m/s$
Work Step by Step
(a) We can set up an equation with the torque from the tension in the string.
$\tau = I\alpha$
$TR = (mR^2)(\frac{a}{R})$
$T = ma$
We can use this expression in the equation for the center of mass.
$\sum F = ma$
$mg-T = ma$
$mg - ma = ma$
$a = \frac{g}{2}$
We can find the angular acceleration.
$\alpha = \frac{a}{R}$
$\alpha = \frac{g}{2R}$
$\alpha = \frac{9.80~m/s^2}{(2)(0.0800~m)}$
$\alpha = 61.25~rad/s^2$
We can find the angular position $\theta$ after the hoop has dropped a distance of 75.0 cm.
$\theta = \frac{y}{R}$
$\theta = \frac{0.750~m}{0.0800~m}$
$\theta = 9.375~rad$
We can find the angular speed.
$\omega^2 = \omega_0+2\alpha ~\theta$
$\omega = \sqrt{2\alpha ~\theta}$
$\omega = \sqrt{(2)(61.25~rad/s^2)(9.375~rad)}$
$\omega = 33.9~rad/s$
(b) We can find the speed of the center of mass.
$v = \omega ~R$
$v = (33.9~rad/s)(0.0800~m)$
$v = 2.71~m/s$
