JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A sphere of radius \('a'\) and mass \('m'\) rolls along a horizontal plane with constant speed \(v_{0}\). It encounters an inclined plane at angle \(\theta\) and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

- A \(\frac{10 v_{0}^{2}}{7 g \sin \theta}\)
- B \(\frac{v_{0}^{2}}{5 g \sin \theta}\)
- C \(\frac{2}{5} \frac{v_{0}^{2}}{ g \sin \theta}\)
- D \(\frac{v_{0}^{2}}{2 g \sin \theta}\)
Answer & Solution
Correct Answer
(A) \(\frac{10 v_{0}^{2}}{7 g \sin \theta}\)
Step-by-step Solution
Detailed explanation
Angular momentum conservation about A \(mv _{0} a \cos \theta+\frac{2}{5} ma ^{2} \omega\) \(= mva +\frac{2}{5} ma ^{2} \omega^{1}\) \(mv _{0} a \left[\frac{2}{5}+\cos \theta\right]=\frac{7}{5} mva\) \(v =\frac{5}{7}= v _{0}\left[\frac{2}{5}+\cos \theta\right]\)…
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