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Question

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 and passing through the centre of mass is:

A
(m1+m2)l2
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B
m1m2l2
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C
m1m2m1+m2l2
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D
m1+m2m1m2l2
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Solution

The correct option is C m1m2m1+m2l2Given: A light rod of length l has two masses m1 and m2 attached to its two ends. To find the moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of massSolution:Moment of inertia for system of particle -I=m1r21+...+mnr2nApplied when masses are placed discretely.Here the center of mass isXcm=m1x1+m2x2m1+m2⟹Xcm=m1×0+m2×lm1+m2⟹Xcm=m2lm1+m2And so, r1=Xcm=m2lm1+m2and r2=l−Xcm=l−m2lm1+m2⟹r2=m1lm1+m2So the moment of inertia of the present system isI=m1r21+m2r22⟹I=m1(m2lm1+m2)2+m2(m1lm1+m2)2⟹I=(m1m2)(m2+m1)l2(m1+m2)2⟹I=m1m2l2m1+m2is the moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass

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