## Essential University Physics: Volume 1 (3rd Edition)

We first find the moment of inertia: $I=mr^2=(380)(1.1)^2=459.8 \ kgm^2$ 'We now use conservation of energy to obtain: $mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \\ mgh=\frac{1}{2}mr^2\omega^2+\frac{1}{2}I\omega^2\\ \omega = \sqrt{\frac{2Mgh}{I+Mr^2}} \\\omega = \sqrt{\frac{2(509)(9.81)(16)}{459.8+(509)(1.1)^2}} =\fbox{12.2 radians per second}$