Question

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. None of these

Answer 1

Answer: (D)

Angular momentum conservation about A
$\operatorname{mv}_{0}$ a $\cos \theta+\frac{2}{5} \operatorname{ma}^{2} \omega$
$=m v a+\frac{2}{5} m a^{2} \omega^{1}$
$mv_0a \left[\frac{2}{5}+\cos \theta\right]=\frac{7}{5} mva$
$v =\frac{5}{7}= v _{0}\left[\frac{2}{5}+\cos \theta\right]$
$\frac{1}{2} mv ^{2}+\frac{1}{2} I \omega^{2}=\frac{7}{10} mv ^{2}= mgh$