## University Physics with Modern Physics (14th Edition)

$I = \frac{1}{2} mR^2$
Let $v$ be the speed of the bucket. $KE_{pulley} = \frac{1}{2}~KE_{bucket}$ $\frac{1}{2}I\omega^2 = \frac{1}{2}\times \frac{1}{2}mv^2$ $I(\frac{v}{R})^2 = \frac{1}{2}\times mv^2$ $I = \frac{1}{2} mR^2$