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 angu…

Question

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

Accepted Answer

(2) x=(m₂L)/(m₁+m₂). Given: m₁ and m₂ at the ends of a massless rod of length L, axis at a distance x from m₁. Idea: W=1/2Iω₀² with ω₀ fixed, so minimising the work means minimising I. I(x)=m₁x²+m₂(L-x)² (dI)/(dx)=2m₁x-2m₂(L-x)=0 m₁x=m₂L-m₂x⇒ x(m₁+m₂)=m₂L x=(m₂L)/(m₁+m₂) (d²I)/(dx²)=2(m₁+m₂)>0, so this is the minimum. It is the centre of mass of the pair.

Step-by-step solution

Why the other options are wrong