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Question

Moment of inertia of uniform rod of mass 'M' and length 'L' about an axis through its centre and perpendicular to its length is given by ML212. Now consider one such rod pivoted at its centre, free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass 'M' moving horizontally at a speed 'v' strikes and embedded in one end of the rod. The angular velocity of the rod just after the collision will be

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Solution

The correct option is **C** 3v2L

Initial angular momentum of the system = Angular momentum of bullet before collision =Mv(L2)

.....(i) let the rod rotates with angular velocity ω.

Final angular momentum of the system =(ML212)ω+M(L2)2ω ....(ii)

By equation (i) and (ii) MvL2=(ML212+ML24)ω or ω=3v2L

Initial angular momentum of the system = Angular momentum of bullet before collision =Mv(L2)

.....(i) let the rod rotates with angular velocity ω.

Final angular momentum of the system =(ML212)ω+M(L2)2ω ....(ii)

By equation (i) and (ii) MvL2=(ML212+ML24)ω or ω=3v2L

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