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 in rotation about an axis perpendicular to it. Find the position on this rod through which the axis should pass in order that the work required to set the rod in rotation with angular velocity $\omega_0$ should be minimum.
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Let the axis of rotation be at a distance x from mass $m_1$. Therefore, the distance of the axis of rotation from mass $m_2$ is (L - x).
When the rod is set into rotation, the increase in the rotational K.E. is given by
$\text{K.E.}=\frac{1}{2}\text{I}_1\omega^2_0+\frac{1}{2}\text{I}_2\omega_0^2$
$=\frac{1}{2}\text{m}_1\text{x}^2\omega^2_0+\frac{1}{2}\text{m}_2(\text{L}-\text{x})^2\omega^2_0$
According to work energy theorem
Work done, SW= increase in rotational K.E.
$\text{i.e.,}\text{ W}=\frac{1}{2}\text{m}_1\text{x}^2\omega_0^2+\frac{1}{2}\text{m}_2(\text{L}-\text{x})^2\omega_0^2$
Work will be minimum if $\frac{\text{dW}}{\text{dx}}=0$
$\text{or }\frac{\text{d}}{\text{dx}}\Big[\frac{1}{2}\text{m}_1\text{x}^2\omega^2_0+\frac{1}{2}\text{m}_2(\text{L}-\text{x})^2\omega^2_0\Big]=0$
$\text{m}_1\text{x}\omega^2_0-\text{m}_2(\text{L}-\text{x})\omega^2_0=0$
$\omega^2_0[\text{m}_1\text{x}-\text{m}_2(\text{L}-\text{x})]=0$
Since $\omega^2_0\neq0$
$\therefore\text{m}_1\text{x}-\text{m}_2(\text{L}-\text{x})=0$
$\text{or }(\text{m}_1+\text{m}_2)\text{x}=\text{m}_2\text{L}$
$\therefore\text{x}=\frac{\text{m}_2\text{L}}{(\text{m}_1+\text{m}_2)}$
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