Two point masses of `0.3 kg` and `0.7` kg 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)`
`=0.3 x^(2) + 0.7(1.4 -x)^(2)`
`=0.3x^(2) + 0.7 (1.96 +x^(2) -2.8x)`
`=x^(2) + 1.372 -1.96x`

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