rolling energy partition

Question

A solid sphere (I = 2/5 MR²) rolls without slipπng at speed v. Total kinetic energy?

RankGuru · Accepted Answer

1. **Energy Components**: Total kinetic energy $K_{total} = K_{trans} + K_{rot}$.
2. **Translation**: $K_{trans} = \frac{1}{2} M v^2$.
3. **Rotation**: $K_{rot} = \frac{1}{2} I \omega^2$. Under no-slip condition, $\omega = v/R$.
   - $K_{rot} = \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{5} M v^2$.
4. **Summation**:
   - $K_{total} = (\frac{1}{2} + \frac{1}{5}) M v^2$
   - $K_{total} = (\frac{5}{10} + \frac{2}{10}) M v^2 = \frac{7}{10} M v^2$.
5. **Partition**: $71.4\%$ is translational and $28.6\%$ is rotational.

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