question_answer

A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. Where should it be hit so that the shell does not slip on the surface?

A) \[\frac{2R}{3}\]

B) \[\frac{3R}{2}\]

C) \[\frac{5R}{3}\]                        

D) \[\frac{3R}{5}\]

Correct Answer: A

Solution :

Method: let v is the velocity attained by its center of mass and \[\omega \]be the angular velocity about Centre of mass by the impact of the cue.
Suppose \[F\]is the force exerted by cue for small duration\[\Delta t\], then
\[F\Delta t=m\left( v-0 \right)\]	..(i)
and \[Fh\Delta t=I(\omega -0)\]	...(ii)

From equations (i) and (ii),

\[mv\times h=I\omega \]

For pure rolling \[\omega =\frac{v}{R}\]

\[\therefore mv\times h=\frac{2}{3}m{{R}^{2}}\times \frac{v}{R}\Rightarrow h=\frac{2R}{3}\]

Method II: due to the impact of the cue, friction starts acting on the shell at the point of the contact, which constitutes a torque about Centre of the shell. But the process of impact is of very short duration\[\left( \Delta t\to 0 \right)\]and so there is negligible change in angular momentum due to this torque. So applying conservation of angular momentum about the Centre of the shell. If v is the velocity of the Centre of mass of the shell, then we can write, \[mvh={{I}_{cm}}\omega \]

=\[\left( \frac{2}{3}m{{R}^{2}} \right)\times \frac{v}{R}or\,h=\frac{2R}{3}.\]