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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 |
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Answer» 0.42 in from mass of 0.3 kg `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 
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