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_(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`

Answer

A

Explanation

`I=m_(1)x^(2)+m_(2)(L-x)^(2)`
`I=m_(1)x^(2)+m_(2)L^(2)+m_(2)x^(2)-2m_(2)Lx `
`(dI)/(dx)=2m_(1)x+0+2xm_(2)-2m_(2)L=0`
`x(2m_(1)+2m_(2))=2m_(2)L`
`x=(m_(2)L)/(m_(1)+m_(2))`
When I is minimum work done `W=(1)/(2)Iomega_(0)^(2)`
will also be minimum.