InterviewSolution
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InterviewSolution
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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 perpendicualr 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 : A. `x=(m_(2)L)/(m_(1)+m_(2))`B. `x=(m_(1)L)/(m_(1)+m_(2))`C. `x=(m_(1))/(m_(2))L`D. `x=(m_(2))/(m_(1))L` |
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Answer» Correct Answer - A `I=m_(1)x^(2)+m_(2)(L-x)^(2)` `I=m_(1)x^(2)+m_(2)L^(2)+m_(2)x^(2)-2m_(2)Lx ` `(dI)/(dx)=2m_(1)x+0+2xm_(2)-2m_(2)L=0` `x(2m_(1)+2m_(2))=2m_(2)L` `x=(m_(2)L)/(m_(1)+m_(2))` When I is minimum work done `W=(1)/(2)Iomega_(0)^(2)` will also be minimum. |
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