question_answer

Consider a uniform rod of mass M = 4 m and length l pivoted about its centre. A mass moving with velocity v making angle\[\theta =\frac{\pi }{4}\]to the rod?s long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is                                              [JEE MAIN Held On 08-01-2020 Morning]

A) \[\frac{4}{7}\frac{V}{l}\]                      

B) \[\frac{3}{7}\frac{V}{l}\]

C) \[\frac{3}{7\sqrt{2}}\frac{V}{l}\]

D) \[\frac{3\sqrt{2}}{7}\frac{V}{l}\]

Correct Answer: D

Solution :

[d] About point P, angular momentum (L) of the System \[{{L}_{\operatorname{initial}}}=\frac{mv}{\sqrt{2}}\times \frac{1}{2}\] \[{{L}_{final}}=\left[ \frac{(4m){{l}^{2}}}{12}+\frac{m{{l}^{2}}}{4} \right]\omega \] As \[{{L}_{initial}}={{L}_{final}}\] \[\omega =\frac{3\sqrt{2}v}{7l}\]