Two particles of masses `m_1 and m_2` are joined by a light rigid rod of length r. The system rotates at an angular speed `omega` about an axis through the centre of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is `L=mur^2omega` where `mu` is the reduced mass of the system defined as `mu=(m_1m_2)/(m_1+m_2)`

Explanation

`L = Iomega, I = m_(1)r_(1)^(2) + m_(2)r_(2)^(2), l=r=r_(1)+r_(2)`
`m_(1)r_(1) = m_(2)r_(2)=m_(2)(r-r_(1)), r_(1)=m_(2)r//m_(1) + m_(2)`
`r_(2) = m_(1)r//m_(1) + m_(2),I =m_(1)m_(2)r^(2)//m_(1) + m_(2) = mur^(2)`, Here `r=l, L = Iomega = mul^(2)omega`