Answer
$\sqrt{\frac{6}{5}gdsin\theta}$
Work Step by Step
We use conversation of energy to find:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$
$mgdsin\theta = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2) \frac{v^2}{R^2}$
$gdsin\theta = \frac{1}{2}v^2 + \frac{1}{3}v^2$
$v=\sqrt{\frac{6}{5}gdsin\theta}$
