rolling motion

Question

A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping down an incline. What is the ratio of its rotational kinetic energy to translational kinetic energy?

RankGuru · Accepted Answer

1. **Translational KE**: $K_t = \frac{1}{2} M v^2$.
2. **Rotational KE**: $K_r = \frac{1}{2} I \omega^2$.
3. **Rolling Condition**: For no slipping, $\omega = v/R$.
4. **Sphere Inertia**: $I = \frac{2}{5} M R^2$.
5. **Substitution**:
   - $K_r = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{v}{R})^2 = \frac{1}{5} M v^2$.
6. **Ratio**: $\frac{K_r}{K_t} = \frac{(1/5) M v^2}{(1/2) M v^2} = \frac{2}{5}$.

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