A light rod of length l has two masses m_1 an | vedclass

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(C) Let the distances of masses $m_1$ and $m_2$ from the center of mass be $l_1$ and $l_2$ respectively.Given that the total length of the rod is $l$,

Vedclass · Accepted Answer

$\frac{m_1 m_2}{m_1 + m_2} l^2$

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(C) Let the distances of masses $m_1$ and $m_2$ from the center of mass be $l_1$ and $l_2$ respectively.Given that the total length of the rod is $l$,so $l_1 + l_2 = l$.By the definition of the center of mass,$m_1 l_1 = m_2 l_2$.Substituting $l_2 = l - l_1$,we get $m_1 l_1 = m_2 (l - l_1)$,which simplifies to $l_1 = \frac{m_2 l}{m_1 + m_2}$.Similarly,$l_2 = \frac{m_1 l}{m_1 + m_2}$.The moment of inertia $I$ about an axis passing through the center of mass is given by $I = m_1 l_1^2 + m_2 l_2^2$.Substituting the values of $l_1$ and $l_2$:$I = m_1 \left( \frac{m_2 l}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 l}{m_1 + m_2} \right)^2$$I = \frac{m_1 m_2^2 l^2 + m_2 m_1^2 l^2}{(m_1 + m_2)^2}$$I = \frac{m_1 m_2 l^2 (m_2 + m_1)}{(m_1 + m_2)^2}$$I = \frac{m_1 m_2}{m_1 + m_2} l^2$.

$A$ light rod of length $l$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

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