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 $L = μ r^{2} ω$ where $μ$ is the reduced mass of the system defined as $μ = \frac{m_{1} + m_{2}}{m_{1} + m_{2}}$ .

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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,
$L_{1} = I_{1} ω = m_{1} {d_{1}}^{2} ω = m_{1} {(\frac{m_{2} r}{m_{1} + m_{2}})}^{2} ω$

$\Rightarrow L_{1} = \frac{m_{1} m_{2}^{2} r^{2}}{{(m_{1} + m_{2})}^{2}} ω . . . (1)$

Similarly, the angular momentum due to the mass m₂ at the centre of system,
$L_{2} = m_{2} {(\frac{m_{1} r}{m_{1} m_{2}})}^{2} ω = \frac{m_{2} m_{1}^{2} r^{2}}{{(m_{1} + m_{2})}^{2}} ω . . . (2)$

Therefore, net angular momentum,

$L = L_{1} + L_{2} = \frac{m_{1} m_{2}^{2} r^{2} ω}{{(m_{1} + m_{2})}^{2}} + \frac{m_{2} m_{1}^{2} r^{2} ω}{{(m_{1} + m_{2})}^{2}} = \frac{m_{1} m_{2} (m_{1} + m_{2}) r^{2} ω}{{(m_{1} + m_{2})}^{2}} = \frac{m_{1} m_{2}}{{(m_{1} + m_{2})}^{2}} r^{2} ω = μ r^{2} ω$

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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

$ω$ 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 = $μ r^{2} ω$ where $μ$ is the reduced mass of the system defined as $μ = \frac{m_{1} + m_{2}}{m_{1} + m_{2}}$