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Question

A light rod of length L can revolve in a vertical circle around point O. The rod carries two equal masses of mass m each such that one mass is connected at the end of the rod and the second mass is fixed at the middle of the rod. u is the velocity imparted to the end P to deflect the rod to the horizontal position. Again mass m in the middle of the rod is removed and mass at end P is doubled. Now v is the velocity imparted to end P to deflect it to the horizontal position. Then (uv)2 is

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Solution

In first case,
As velocity of mass attached to the middle of rod is u2
Using conservation of energy
12mu2+12m(u2)2=mgL+mgL2
u=(125gL)
In second case,
Using conservation of energy
12(2m)v2=2mgLv=(2gL)12
Now,
(uv)2=(125gL)[2gL]=[65]=1.2

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