A sphere of radius ' a ' and mass ' m ' rolls along


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 ?

  1. $\frac{10 v_{0}^{2}}{7 g \sin \theta}$

  2.  $\frac{v_{0}^{2}}{5 g \sin \theta}$

  3. $\frac{2}{5} \frac{v_{0}^{2}}{g \sin \theta}$

  4. $\frac{v_{0}^{2}}{2 g \sin \theta}$

Correct Option: 1


Angular momentum conservation about A

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

$=m v a+\frac{2}{5} m a^{2} \omega^{1}$

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

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

$\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\frac{7}{10} \mathrm{mv}^{2}=\mathrm{mgh}$

No option Maching

Leave a comment


May 26, 2022, 2:56 p.m.
I don't know why is everyone doing mistakes. At half of the places, they have directly conserved mechanical energy which is wrong since impulsive normal and friction force acts on colliding with the incline which reduces mechanical energy. The methodology shown in this is correct but there is some unknown random error unreasonable just to match the answer. And everyone is copying the random error

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