Let the axis of rotationPass through O.I=mr2 for point mass∴I=I1+I20.3x2+0.7(1.4−x2)=0.3x2+0.7(1.96+x2−2.8x)x2+1.372−1.96xThe work done for rotation ofThe rod is stored as rotational kinetic energy,31Iω2, of rodOr
W=2Iω2=21(x2+1.372−1.96x)ω2
For work done to be minimum,dxdW=0∴dxd(x2+1.372−1.96x)2ω2=0Or 2x+0−1.96=0Or 2x=1.96 or x=0.98m