A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod passing through the centre of mass will be what?

Dear Student ,
Here in this case ,
centre of mass :Xcm=m1x1+m2x2m1+m2=m1×0+m2×lm1+m2=m2lm1+m2Determination of the moment of inertia :r1=Xcm=m2lm1+m2And r2=l-Xcm=l-m2lm1+m2=m1lm1+m2Therefore , I=m1r12+m2r22=m1m22l2(m1+m2)2+m12m2l2(m1+m2)2=m1m2l2(m1+m2)2m1+m2=m1m2l2(m1+m2)

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