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Point masses $m_1$ and $m_2$ are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity $\omega_0$ is minimum, is given by:
A
$x = \frac{m_2L}{m_1 + m_2}$
B
$x = \frac{m_1L}{m_1 + m_2}$
C
$x = \frac{m_1}{m_2}L$
D
$x = \frac{m_2}{m_1}L$
Detailed Solution
Work required $= \frac{1}{2}I\omega_0^2$, which is minimum when I is minimum, i.e. when the axis passes through the centre of mass.
$m_1x = m_2(L - x)$
$x = \frac{m_2L}{m_1 + m_2}$
$m_1x = m_2(L - x)$
$x = \frac{m_2L}{m_1 + m_2}$