## Essential University Physics: Volume 1 (4th Edition) Clone

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