Two point masses of 0.3 kg and 0.7kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of

Options

A. `0.42 m` from the mass of `0.3 kg`
B. `0.70 m` from the mass of `0.7 kg`
C. `0.98 m` from the mass of `0.3 kg`
D. `0.98 m` from the mass of `0.7 kg`

Answer

C

Explanation

The moment of inertia of the system about axis of rotation `O` is
`I=I_(1)+I_(2)`
`=0.3x^(2)+0.7(1.4-x)^(2)`
`=0.3x^(2)+0.7(1.96+x^(2)-2.8pi)`
`= x^(2)+1.372-1.96x`

The final done in rotating the rod is converted into its rotational kinetic energy.
`:.W=1/2Iomega^(2)=1/2[x^(2)+1.372-1.96]omega^(2)`
For work done to be minimum
`(domega)/(dx)=0`
`implies 2x-1.96=0`
`implies x=1.96/2=0.98m`