Answer
.0086 percent
Work Step by Step
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.
