Answer
$\alpha = 13.9~rad/s^2$
Work Step by Step
We can find the required tangential speed of the marble when it leaves the cup. We can use conservation of energy to find this speed.
$KE = PE$
$\frac{1}{2}mv^2 = mgh$
$v = \sqrt{2gh} = \sqrt{(2)(9.80~m/s^2)(12.0~m)}$
$v = 15.34~m/s$
We can find the required angular speed when the marble leaves the cup.
$\omega = \frac{v}{r} = \frac{15.34~m/s}{0.260~m}$
$\omega = 59.0~rad/s$
We can find the angular acceleration.
$2~\theta ~\alpha = \omega^2$
$\alpha = \frac{\omega^2}{2~\theta}$
$\alpha = \frac{(59.0~rad/s)^2}{(2)(20.0~rev)(2\pi~rad/rev)}$
$\alpha = 13.9~rad/s^2$
