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 in from mass of 0.3 kg
B. 0.70 m from mass of 0.7 kg
C. 0.98 m from mass of 0.3 kg
D. 0.98 m from mass of 0.7 kg.

Answer

C

Explanation

Let the axis of rotation pass through O.
`I=mr^2` for point mass.
`therefore I=I_1+I_2`
`=0.3x^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 for rotation of the rod is stored as rotational kinetic energy, `1/2Iomega^2`, of rod
`W=(Iomega^2)/2=1/2 (x^2-1.372-1.96x)omega^2`
For work done to be minimum , `(dW)/(dx)=0`
`therefore d/(dx)[(x^2+1.372-1.96x)]omega^2/2=0`
2x+0-1.96=0
2x=1.96
x=0.98 m