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 ω$$0$$ is minimum, is given by

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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 ω$$0$$ is minimum, is given by

Infinity Learn · Accepted Answer

Moment of inertia of the system about the axis of rotation (through$$ point P $$) $$isI=m1x2+m2(L-x)2$$ By work energy theorem,  Work done to set the rod rotating with angular velocity ω$$0=$$ Increase in rotational kinetic energy $$W=12I$$ω$$02=12m1x2+m2(L-x)2$$ω$$02$$ For W to be minimum, $$dWdx=0$$ i.e.   $$122m1x+2m2(L-x)(-1)$$ω$$02=0$$ or   $$m1x-m2(L-x)=0$$  ∵ω$$0$$≠$$0$$ or $$m1+m2x=m2L$$ or $$x=m2Lm1+m2$$

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