Pure rotational motion and its kinetic energy

Question

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

Difficult

Solution

Moment of inertia of the system about the axis of rotation (through point P ) is

$I={m}_{1}{x}^{2}+{m}_{2}(L-x{)}^{2}$

$\text{By work energy theorem,}$

$\text{Work done to set the rod rotating with angular velocity}{\omega}_{0}$

$=\text{Increase in rotational kinetic energy}$

$W=\frac{1}{2}I{\omega}_{0}^{2}=\frac{1}{2}\left[{m}_{1}{x}^{2}+{m}_{2}(L-x{)}^{2}\right]{\omega}_{0}^{2}$

$\text{For}W\text{to be minimum,}\frac{dW}{dx}=0$

$\text{i.e.}\frac{1}{2}\left[2{m}_{1}x+2{m}_{2}(L-x)(-1)\right]{\omega}_{0}^{2}=0$

$\text{or}{m}_{1}x-{m}_{2}(L-x)=0\left(\because {\omega}_{0}\ne 0\right)$

$\text{or}\left({m}_{1}+{m}_{2}\right)x={m}_{2}L\text{or}x=\frac{{m}_{2}L}{{m}_{1}+{m}_{2}}$

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