rolling motion

Question

A solid cylinder produces only Rolling Motion on rough horizontal surface with initial velocity v. The ratio of translational kinetic energy to total kinetic energy is:

RankGuru · Accepted Answer

1. **Translational Kinetic Energy**: $K_t = \frac{1}{2}mv^2$.
2. **Rotational Kinetic Energy**: $K_r = \frac{1}{2}I\omega^2$. For a solid cylinder, $I = \frac{1}{2}mr^2$. In pure rolling, $v = r\omega \implies \omega = v/r$.
   $$K_r = \frac{1}{2}\left(\frac{1}{2}mr^2ight)\left(\frac{v}{r}ight)^2 = \frac{1}{4}mv^2$$
3. **Total Kinetic Energy**: $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
4. **Ratio calculation**: $	ext{Ratio} = \frac{K_t}{K_{total}} = \frac{\frac{1}{2}mv^2}{\frac{3}{4}mv^2} = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}$.

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