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, \[mvh={{I}_{cm}}\omega \] |
=\[\left( \frac{2}{3}m{{R}^{2}} \right)\times \frac{v}{R}or\,h=\frac{2R}{3}.\] |
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