Two particles of masses `m_(1)` and `m_(2)` are separated by a distance 'd'. Then moment of inertia of the system about an axis passing through centre of mass and perpendicular the line joining them is

Explanation

`m_1 r_1 = m_2 r_` and `r_1 + r_2 = 4 Phi r_1 = (m_2 r)/(m_1 + m_2)`
`r_2 = (m_1 r)/(m_1 + m_2)`

Moment or inertia `I = m_1 r_1^2 + m_2 r_2^2 `
`=m_1= ((m_2r)/(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 = mu r^2` where `mu` ( reduced mass)
` = (m_1 m_2)/(m_2 + m_2)`