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 :

Options

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

Answer

C

Explanation

`K.E. = (1)/(2)I omega_(0)^(2)`
I is min. about the centre of mass
So, `(m_(1))(x)=(m_(2))(L-x)implies x=(m_(2)L)/(m_(1)+m_(2))`