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

The translational Kinetic energy is: $K_t=\frac{1}{2}(.15)(33^2)=81.675 \ J$ The rotational Kinetic energy is: $K_r=\frac{1}{2}I\omega^2=\frac{1}{2}\frac{2}{5}(.15)(.037^2)(42^2)=.0724$ Thus, we see that .0086 percent of the energy is rotational.