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

We know that the translational kinetic energy is given as $K_{t}=(\frac{1}{2})mv^2$ The rotational kinetic energy is given as $K_r=(\frac{1}{2})I\omega^2=(\frac{1}{2})(\frac{1}{2}mr^2)(\frac{v}{r^2})=\frac{1}{4}mv^2$ Now the total kinetic energy is given as $K_{total}=K_t+K_r=(\frac{1}{2})mv^2+(\frac{1}{4})mv^2=(\frac{3}{4})mv^2$ Now we can find the required ratio as $\frac{K_r}{K_{total}}=\frac{(\frac{1}{4})mv^2}{(\frac{3}{4})mv^2}=\frac{1}{3}$