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(A) Let $r_1$ and $r_2$ be the distances of masses $m_1$ and $m_2$ from the center of mass $(CM)$ respectively.By the definition of the center of mass

Vedclass · Accepted Answer

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

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(A) Let $r_1$ and $r_2$ be the distances of masses $m_1$ and $m_2$ from the center of mass $(CM)$ respectively.By the definition of the center of mass,$m_1 r_1 = m_2 r_2$ and $r_1 + r_2 = r$.Solving these equations,we get $r_1 = \frac{m_2 r}{m_1 + m_2}$ and $r_2 = \frac{m_1 r}{m_1 + m_2}$.The moment of inertia $I$ about an axis passing through the center of mass is given by $I = m_1 r_1^2 + m_2 r_2^2$.Substituting the values of $r_1$ and $r_2$:$I = m_1 \left( \frac{m_2 r}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 r}{m_1 + m_2} \right)^2$$I = m_1 \frac{m_2^2 r^2}{(m_1 + m_2)^2} + m_2 \frac{m_1^2 r^2}{(m_1 + m_2)^2}$$I = \frac{m_1 m_2 r^2 (m_2 + m_1)}{(m_1 + m_2)^2}$$I = \left( \frac{m_1 m_2}{m_1 + m_2} \right) r^2$.

Two masses $m_1$ and $m_2$ are placed at a distance $r$ from each other. Find the moment of inertia of the system about an axis passing through the center of mass and perpendicular to the line joining the masses.

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