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Two particles of masses m1 and m2 are connected by a light rod of length r to constitute a dumb-bell.The moment of inertia of the dumb=bell about an axis perpendicular to the rod passing through the center of mass of the two particles is
consider a rod , two particles of mass m1 , m2 are at its two ends then
distance of center of mass from m1 is r1= m2r/(m1+m2) ..........1 (using formula)
distance of center of mass from m2 is r2 = m1r/(m1+m2) .........2
now moment of inertia of two particles is
I1 = m1r12 & I2 = m2r22
I = I1+I2 = m1r12 + m2r22 ................3
from 2 & 3 above expression becomes
I = m1m2r2/(m1+m2)
approve my ans if u like ....
also see question new13 , i have posted the ans
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