Point masses m1 and m2 are placed at the opposite ends of a rigid rod… | NEET 2015 (AIPMT)

Q177NEET 2015 · AIPMTMedium

Moment of Inertia & Radius of Gyration

Point masses m1 and m2 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

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Answer: (b)

Work $= \frac{1}{2}I\omega_0^2$ , so $I$ must be minimum.
$I = m_1x^2 + m_2(L-x)^2$ ; $\frac{dI}{dx} = 2m_1x - 2m_2(L-x) = 0$ .
$x = \frac{m_2L}{m_1+m_2}$ (the centre of mass), so the answer is (b).

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