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

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(D) Let the side $AB = B$ and $BC = L = 2B$. The mass of the rectangle is $M$.For an axis passing through the center of mass and parallel to the sides

Vedclass · Accepted Answer

$EG$

Vedclass · Answer

(D) Let the side $AB = B$ and $BC = L = 2B$. The mass of the rectangle is $M$.For an axis passing through the center of mass and parallel to the sides:$I_{EG} = \frac{MB^2}{12}$$I_{HF} = \frac{ML^2}{12} = \frac{M(2B)^2}{12} = \frac{4MB^2}{12} = \frac{MB^2}{3}$For the diagonal $BD$,the moment of inertia is given by $I_{BD} = \frac{M B^2 L^2}{6(B^2 + L^2)}$.Substituting $L = 2B$:$I_{BD} = \frac{M B^2 (2B)^2}{6(B^2 + (2B)^2)} = \frac{4MB^4}{6(5B^2)} = \frac{4MB^2}{30} = \frac{2MB^2}{15} \approx 0.133 MB^2$.Comparing the values:$I_{EG} = 0.0833 MB^2$$I_{HF} = 0.333 MB^2$$I_{BD} = 0.133 MB^2$Thus,the moment of inertia is minimum about the axis $EG$.

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

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