In a rectangle ABCD where BC = 2AB,the moment | vedclass

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(D) The moment of inertia $I$ of a body about an axis is given by $I = \int r^2 dm$,where $r$ is the perpendicular distance of the mass element $dm$ f

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$EG$

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(D) The moment of inertia $I$ of a body about an axis is given by $I = \int r^2 dm$,where $r$ is the perpendicular distance of the mass element $dm$ from the axis. To minimize the moment of inertia,the mass distribution should be as close to the axis as possible.For a rectangle of sides $a$ and $b$ (where $b = 2a$),the axes $HF$ and $EG$ pass through the center. Axis $HF$ is parallel to the sides $AB$ and $CD$ (length $a$),and axis $EG$ is parallel to the sides $AD$ and $BC$ (length $2a$).The moment of inertia about an axis passing through the center and parallel to a side of length $L$ is given by $I = \frac{1}{12} M L^2$ (where $L$ is the side perpendicular to the axis). For axis $HF$ (parallel to $AB$),the perpendicular side is $BC = 2a$,so $I_{HF} = \frac{1}{12} M (2a)^2 = \frac{4}{12} M a^2 = \frac{1}{3} M a^2$.For axis $EG$ (parallel to $BC$),the perpendicular side is $AB = a$,so $I_{EG} = \frac{1}{12} M a^2$.Comparing the two,$I_{EG} < I_{HF}$. Thus,the moment of inertia is minimum along the axis $EG$.

In a rectangle $ABCD$ where $BC = 2AB$,the moment of inertia will be minimum along which of the following axes?

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