Two particles of masses m1 and m2 are joined by a light rigid rod of length r. The system rotates at an angular speed ω 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 \[L = \mu r^2 \omega\] where \[\mu\] is the reduced mass of the system defined as \[\mu = \frac{m_1 + m_2}{m_1 + m_2}\]
Solution
Distance of centre of mass from mass m1,
\[d_1 = \frac{m_2 r}{m_1 + m_2}\]
Angular momentum due to the mass m1 at the centre of system,
\[L_1 = I_1 \omega = m_1 {d_1}^2 \omega = m_1 \left( \frac{m_2 r}{m_1 + m_2} \right)^2 \omega\]
\[\Rightarrow L_1 = \frac{m_1 m_2^2 r^2}{\left( m_1 + m_2 \right)^2}\omega...........(1)\]
Similarly, the angular momentum due to the mass m2 at the centre of system,
\[L_2 = m_2 \left( \frac{m_1 r}{m_1 m_2} \right)^2 \omega\]
\[ = \frac{m_2 m_1^2 r^2}{\left( m_1 + m_2 \right)^2}\omega .........(2)\]
Therefore, net angular momentum,
\[L = L_1 + L_2 \]
\[ = \frac{m_1 m_2^2 r^2 \omega}{\left( m_1 + m_2 \right)^2} + \frac{m_2 m_1^2 r^2 \omega}{\left( m_1 + m_2 \right)^2}\]
\[ = \frac{m_1 m_2 \left( m_1 + m_2 \right) r^2 \omega}{\left( m_1 + m_2 \right)^2}\]
\[ = \frac{m_1 m_2}{\left( m_1 + m_2 \right)^2} r^2 \omega\]
\[ = \mu r^2 \omega\]
