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
√m1m2l2
(m1+m2)l2
m1m2m1+m2l2
m1+m+2m1m2l2
A
√m1m2l2
B
m1+m+2m1m2l2
C
(m1+m2)l2
D
m1m2m1+m2l2
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Solution
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The distance of center of mass from mass m1 is x1=m2lm1+m2
The distance of center of mass from mass m2 is x2=m1lm1+m2
Hence moment of inertia of masses about the center of mass=m1x21+m2x22
=m1m2m1+m2l2
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