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)`

Answer

A

Explanation

Angular momentum due to the mass `m_1 ` at the centre of system
`=m_1((m_2r)/(m_1+m_2))^2omega`
=(m_1m_2^2r^2)/((m_1+m_2)^2)` ………1
similarly, the angular momentum due the mass `m_2` at the centre of system
`=m_2=((m_1r)/(m_1m_2)^2)` omega`
`=(m_2m_1^2r_2)/((m_1+m_2)^2)omega`..........2
Therefore net angular momentum
`=(m_1m_2^2r^2omega)/((m_1+m_2)^2)+(m_2m_1^2r^2omega)/((m_1+m_2)^2)`
`=(m_1m_2(m_1+m_2)r^2omega)/((m_1+m_2)^2)`
`=(m_1m_2)/((m_1+m_2))r^2omega
`=mur^2omega` Proved