Combined Translation And Rotation

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A cylinder of mass 10 kg and radius 0.2 m rolls without slipping with a linear velocity of 5 m/s. What is its total kinetic energy?

RankGuru · Accepted Answer

1. **Total Energy $K$**: $K = K_{translational} + K_{rotational}$.
2. **Translational**: $K_t = \frac{1}{2} M v^2 = \frac{1}{2}(10)(5^2) = 125$ J.
3. **Rotational**: $K_r = \frac{1}{2} I \omega^2$.
   - For cylinder, $I = \frac{1}{2}MR^2$.
   - For rolling, $\omega = v/R$.
   - $K_r = \frac{1}{2}(\frac{1}{2}MR^2)(v/R)^2 = \frac{1}{4} M v^2$.
   - $K_r = \frac{1}{4} (10)(25) = 62.5$ J.
4. **Total**: $125 + 62.5 = 187.5$ Joules.

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Combined Translation And Rotation

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