Question:
A thin spherical shell lying on a rough horizontal surface is hit by cue in such a way that the line of action passes through the centre of the shell. As a result the shell starts moving with the linear speed v without any initial angular velocity. Find the linear speed of the shell after it starts pure rolling on the surface.
Solution:
By angular momentum conservation at bottom most point
$L_{i}=L_{f}$
$m v R=m v^{\prime} R+I \omega^{\prime}$
$m v R=\left(\frac{2}{3} m R^{2}\right)\left(\frac{v^{\prime}}{R}\right)+m v^{\prime} R$
$v^{\prime}=\frac{3 v}{5}$
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.