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Two particles of masses $m_1$ and $m_2$ are connected to a rigid massless rod of length r to constitute a dumb bell which is free to move in the plane. The moment of inertia of the dumb bell about an axis perpendicular to the plane passing through the centre of mass

\[\begin {array} {1 1} (a)\;\frac{m_1m_2}{(m_1+m_2)}r^2 & \quad (b)\;(m_1+m_2)r^2 \\ (c)\;\frac{m_1m_2}{(m_1-m_2)} r^2 & \quad  (d)\;(m_1-m_2)r^2 \end {array}\]

1 Answer

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$m_1r_1=m_2r_2$
$r_1=\large\frac{m_2r}{(m_1+m_2)}$
$r_2=\large\frac{m_1r}{(m_1+m_2)}$
$I= m_1r_1^2+m_2r_2^2$
$\quad= \large\frac{m_1m_2^2r^2}{(m_1+m_2)^2}+\frac{m_2m_1^2r^2}{(m_1+m_2)^2}$
$\quad= \large\frac{m_1m_2 r^2}{(m_1+m_2)}$
answered Nov 27, 2013 by meena.p
edited Jun 19, 2014 by lmohan717
 

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