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# Two Particles of Masses M1 and M2 Are Joined by a Light Rigid Rod of Length R. - Physics

Sum

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 $\mu = \frac{m_1 + m_2}{m_1 + m_2}$

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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,

$L_1 = I_1 \omega = m_1 {d_1}^2 \omega = m_1 \left( \frac{m_2 r}{m_1 + m_2} \right)^2 \omega$

$\Rightarrow L_1 = \frac{m_1 m_2^2 r^2}{\left( m_1 + m_2 \right)^2}\omega...........(1)$

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,

$L = L_1 + L_2$

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

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

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

$= \mu r^2 \omega$

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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
Q 49 | Page 198
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