A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

(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}

JEE Mains · Physics · STD 11 - 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?

Pick an option to check your answer.

Verified Solution

Correct Answer

(A) \(\frac{10 v_{0}^{2}}{7 g \sin \theta}\)

Detailed explanation

Angular momentum conservation about A

\(mv _{0} a \cos \theta+\frac{2}{5} ma ^{2} \omega\)

\(= mva +\frac{2}{5} ma ^{2} \omega^{1}\)

\(mv _{0} a \left[\frac{2}{5}+\cos \theta\right]=\frac{7}{5} mva\)

\(v =\frac{5}{7}= v _{0}\left[\frac{2}{5}+\cos \theta\right]\)

\(\frac{1}{2} mv ^{2}+\frac{1}{2} I \omega^{2}=\frac{7}{10} mv ^{2}= mgh\)

\(d=\frac{10 v_{0}^{2}}{7 g \sin \theta}\)

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