Question 215 of 280

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$\dfrac{2}{5}\dfrac{{v}_{0}^{2}}{g\sin \theta }$B$\dfrac{7{v}_{0}^{2}}{10g\sin \theta }$C$\dfrac{{v}_{0}^{2}}{5g\sin \theta }$D$\dfrac{{v}_{0}^{2}}{2g\sin \theta }$

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Answer: (B) $\dfrac{7{v}_{0}^{2}}{10g\sin \theta }$

From energy conservation
$mgh=\dfrac{1}{2}{mv}_{0}^{2}+\dfrac{1}{2}{I\omega }^{2}$
$\Rightarrow mgh=\dfrac{1}{2}{mv}_{0}^{2}+\dfrac{1}{2}\times \dfrac{2}{5}{ma}^{2}\times \dfrac{{v}_{0}^{2}}{{a}^{2}}$
$\Rightarrow gh=\dfrac{1}{2}{v}_{0}^{2}+\dfrac{1}{5}{v}_{0}^{2}$
$\Rightarrow gh=\dfrac{7}{10}{v}_{0}^{2}$
From triangle, $\sin \theta =\dfrac{h}{l}$
Then $h=l\sin \theta$
$\Rightarrow l\sin \theta =\dfrac{7}{10}\dfrac{{v}_{0}^{2}}{g}$
Thus, the sphere will travel distance, $l=\dfrac{7}{10}\dfrac{{{v}_{0}}^{2}}{g\sin \theta }$.