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

Question:

A sphere of radius 'a' and mass 'm' rolls along horizontal plane with constant speed $u_{-} 0$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping how far up the sphere will travel?

Note: NTA has dropped this question in the final official answer key.

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

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

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

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


Correct Option: , 2

Solution:

From energy conservation

$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}_{0}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$

$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}_{0}^{2}+\frac{1}{2} \times \frac{2}{5} \mathrm{ma}^{2} \times \frac{\mathrm{v}_{0}{ }^{2}}{\mathrm{~s}^{2}}$

$g h=\frac{1}{2} v_{0}^{2}+\frac{1}{5} v_{0}^{2}$

$g h=\frac{7}{10} v_{0}^{2}$

$h=\frac{7}{10} \frac{v_{0}^{2}}{g}$

from triangle, $\sin \theta=\frac{\mathrm{h}}{\ell}$

then $\mathrm{h}=\ell \sin \theta$

$\ell \sin \theta=\frac{7}{10} \frac{v_{0}^{2}}{9}$

$\ell=\frac{7}{10} \frac{v_{0}^{2}}{g \sin \theta}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now