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)

Two particles of masses m_1 and m_2 are joined by a light rigid rod of

1.	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» SOLUTION :`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`