Two masses `m_(1) and m_(2)` are placed at a distance r from each other. Find out the moment of inertia of system about an axis passing through their centre of mass.

Explanation

`m_(1)r_(1) = m_(2)r_(2) and r_(1)+r_(2) = r `
` rArr r_(1) (m_(2)r)/(m_(1)+m_(2)) , r_(2) = (m_(1)r)/(m_(1)+m_(2))`
Moment of inertia
` I = m_(1)r_(1)^(2) + m_(2)r_(2)^(2) = m_(1) ((m_(2)r)/(m_(1)+m_(2)))^(2) + m_(2) ((m_(1)r)/(m_(1)+m_(2)))^(2)`
`= ((m_(1)m_(2))/(m_(1)+m_(2))) r^(2)`
Note Here `I = mur^(2)` where `mu = ` reduced mass ` = (m_(1)m_(2))/(m_(1)m_(2))`