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```        Twp Point Masses m1 and m2 are joined by a weightless rod of length r. Calculate the moment of inertia of the system about an axis passing through the center of mass and perpendicular to the rod.
Ans: (m1m2r^2)/(m1+m2)```
7 years ago

147 Points
```										Dear Nehal Wani

Center of mass of the system from mass m1  =  (m1*0 + m2*r)/(m1+m2)
= m2 r/(m1+m2)
distance of COM from the mas m2  = [r- m2 r/(m1+m2)]
=m1 r/(m1+m2)

moment of inertia I = m1 [m2 r/(m1+m2)]2 + m1[m1 r/(m1+m2)]2
=m1 m2 r2/(m1+m2)

Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
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