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Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length l and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position on the rod through which the axis should pass in order that the work required to set the rod rotating with angular velocity ω0 be minimum is

a
M2LM1+M2
b
M1+M2M2L
c
M1LM1-M2
d
M1-M2M2L

detailed solution

Correct option is A

In order  From work-energy theorem W=ΔKE=12m1x2+m2(l-x)2ω02For minimum work, dWdx=0or  2m1x+2m2(l-x)=0⇒ x=M2LM1+M2

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