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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 :
image
A. `x = (m_(2)L)/(m_(1) + m_(2))`
B. `x = (m_(1)L)/(m_(1) + m_(2))`
C. `x = (m_(1))/(m_(2))L`
D. `x = (m_(2))/(m_(1))L`

1 Answer

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Best answer
Correct Answer - A
As `K.E.` of rotation `= (1)/(2)I omega^(2)`
`I` is min. about the centre of mass
`:. M_(1)x = m_(2) (L - x) or x = (m_(2)L)/(m_(1) + m_(2))`

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