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

We substitute the equation for the moment of inertia of a ring and the unit conversion for angular velocity into the equation for rotational kinetic energy to find: $k_{rotational}=\frac{1}{2}I\omega^2$ $k_{rotational}=\frac{1}{2}mr^2\omega^2$ $k_{rotational}=\frac{1}{2}(1.8252)(.195)^2(30,000\times .104)^2$ $k_{rotational}=\fbox{9 MJ}$