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
- m1m2m1+m2l2
- m1+m2m1m2l2
- (m1+m2)l2
- m1m2 l2
Answer
(1) m1m2m1+m2l2
Solution
Distances of m1 and m2 from the centre of mass:
r1 = m2lm1+m2, r2 = m1lm1+m2
I = m1r12 + m2r22
I = m1m22l2 + m2m12l2(m1+m2)2
I = m1m2(m1+m2)l2(m1+m2)2
I = m1m2m1+m2l2
Answer: (1)