Question

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 =$\mu r^2\omega$ where $\mu$ is the reduced mass of the system defined as $\mu =\frac{m_1+m_2}{m_1+m_2}$

Answer 1

Angular momentum due to the mass $m_1$ at the centre of system

$=m_1{\left(\frac{m_{2}r}{m_1+m_2}\right)}^2\omega$

$=\frac{m_1{m}_2^2{r}^2}{(m_1+m_2{)}^2}\omega$ ... (1)

Similarly, the angular momentum due to the mass $m_2$ at the centre of system

$=m_2={\left(\frac{m_{1}r}{m_1{m}_2}\right)}^2\omega$

$=\frac{m_2{m}_1^2{r}^2}{(m_1+m_2{)}^2}\omega$ ...(2)

Therefore, Net angular momentum

$=\frac{m_1{m}_2^2{r}^2\omega }{(m_1+m_2{)}^2}+\frac{m_2{m}_1^2{r}^2\omega }{(m_1+m_2{)}^2}-$

$=\frac{m_1{m}_2(m_1+m_2)r^2\omega }{(m_1+m_2{)}^2}$

$=\frac{m_1{m}_2}{(m_1+m_2)}{r}^2\omega$

$=\mu r^2\omega \left(\text{Proved}\right)$