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Point masses m(1) and m(2) are placed at...

Point masses `m_(1) and m_(2)` 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 perpendicualr to it. The position of point `P` on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity `omega_(0)` is minimum, is given by :

Text Solution

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`(KE)_(t)=0, (KE)_(F)=1/2(I_(1)+I_(2))omega_(0)^(2)`
So by work energy theorem,
`W=(KE)_(F)-(KE)_(I)`
`=1/2(I_(1)+I_(2))omega_(0)^(2)=1/2[M_(1)x^(2)+M_(2)(L-x)^(2)]omega_(0)^(2)`

For `W` to be minimum
`(dW)/(dx)=0`, i.e. `2M_(1)x-2M_(2)(L-x)=0`
so, `x=(M_(2)L)/((M_(1)+M_(2)))`
This the position of cenre of mass of the rod from `M_(1)`. So the required work is minimum when the rod is rotating about an axis passing through its centre of mass and perpendicular to the length of the rod.
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