The work required to set the rod rotating with angular velocity $\omega_0$ is equal to its rotational kinetic energy, $W = \Delta K = \frac{1}{2}I\omega_0^2$. For the work to be minimum, the moment of inertia $I$ must be minimum. Let the axis of rotation pass through a point at a distance $x$ from mass $m_1$. Then its distance from mass $m_2$ is $(L - x)$. The moment of inertia of the system about this axis is: $I = m_1x^2 + m_2(L - x)^2$ To find the condition for minimum moment of inertia, we differentiate $I$ with respect to $x$ and set it to zero: $\frac{dI}{dx} = \frac{d}{dx}[m_1x^2 + m_2(L - x)^2] = 0$ $2m_1x + 2m_2(L - x)(-1) = 0$ $m_1x - m_2(L - x) = 0$ $m_1x = m_2L - m_2x$ $(m_1 + m_2)x = m_2L$ $x = \frac{m_2L}{m_1 + m_2}$ This implies that the axis must pass through the centre of mass of the two-particle system for the moment of inertia, and thus the work done, to be minimum.