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Question

A uniform rod of length l & mass M is free to rotate about frictionless pin through one end. The rod is released from rest in a horizontal position. Find momentum of system (rod) when rod is at its lowest position.

A
M5gl
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B
M3gl
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C
M2gl
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D
M23gl
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Solution

The correct option is D M23gl
COM is at l2.
By conservation of energy:
loss in PE = gain in KEMgl2=12Mll3ω2
ω2=6gl2l2=3gl
ω=3gl
vCM=l2ω=l23gl=3gl4,p=MvCM=M3gl4

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