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 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 ω0 is minimum, is given by

Moment of inertia of the system about the axis of rotation (through point P ) isI=m1x2+m2(L-x)2 By work energy theorem,  Work done to set the rod rotating with angular velocity ω0= Increase in rotational kinetic energy W=12Iω02=12m1x2+m2(L-x)2ω02 For W to be minimum, dWdx=0 i.e.   122m1x+2m2(L-x)(-1)ω02=0 or   m1x-m2(L-x)=0  ∵ω0≠0 or m1+m2x=m2L or x=m2Lm1+m2