Calculate the moment of inertia of a diatomic molecule about an axis passing through its centre of mass and perpendicular to the line joining the two atoms .
Given : The two masses are `m_(1)` and `m_(2)` separated by a distance r .

Explanation

`m_(1) r_(1) = m_(2) (r_(2)) = m_(2) (r- r_(1))`
`therefore m_(1) r_(1) = m_(2) r - m_(2) r_(1)`
or `r_(1) = (m_(2) r)/((m_(1) + m_(2)))`
Now `r_(2) = r- r_(1) = r - (m_(2) r)/((m_(1) + m_(2)))`
`r_(2) = (m_(1) r)/((m_(1) + m_(2))) therefore I = m_(1) r_(1)^(2) + m_(2) r_(2)^(2)`
or `I = (m_1 xx (m_(2) r)^(2))/((m_(1) + m_(2))) + (m_(2) xx (m_(1) r)^(2))/((m_(1) + m_(2))^(2))`
or `I = (m_(1) m_(2)r^(2) (m_(2) + m_(1)))/((m_(1) + m_(2))^(2)) or I = (m_(1) m_(2) r^(2))/((m_(1) + m_(2)))`
or `I = mu r^(2)` , where `mu = (m_(1) m_(2))/((m_(1) + m_(2)))` = reduced mass of system .