A solid uniform cylinder rolls forward on a flat surface without slipping. What percentage of its to

Question

A solid uniform cylinder rolls forward on a flat surface without slipping. What percentage of its total kinetic energy is strictly translational?

Professor PC Thomas Classes · Answer

(b) : For a solid cylinder rolling without slipping, the translational kinetic energy is $KE_{trans} = \frac{1}{2}mv^2$. Its moment of inertia is $I = \frac{1}{2}mR^2$, so rotational kinetic energy is $KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{4}mv^2$. Total kinetic energy is $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$. The required fraction is $\frac{KE_{trans}}{KE_{total}} = \frac{1/2}{3/4} = \frac{2}{3}$. Converting to a percentage gives approximately $66.67\%$.