Q.

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                 [2016]
 

1 m1m2m1+m2l2  
2 m1+m2m1m2l2  
3 (m1+m2)l2  
4 m1m2l2  

Ans.

(1)

Here, l1+l2=l

Centre of mass of the system,

l1=m1×0+m2×lm1+m2=m2lm1+m2

          l2=l-l1=m1lm1+m2

Required moment of inertia of the system,

     I=m1l12+m2l22

       =(m1m22+m2m12)l2(m1+m2)2

       =m1m2(m1+m2)l2(m1+m2)2=m1m2m1+m2l2