The moment of inertia of a uniform square pla | vedclass

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(B) Let $I_0$ be the moment of inertia about the axis passing through the center $O$ and perpendicular to the plane,given by $I_0 = \frac{Ma^2}{6}$.Th

Vedclass · Accepted Answer

$\frac{2 Ma^2}{3}$

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(B) Let $I_0$ be the moment of inertia about the axis passing through the center $O$ and perpendicular to the plane,given by $I_0 = \frac{Ma^2}{6}$.The distance $h$ from the center $O$ to any corner $A$ is half the length of the diagonal of the square.The diagonal of the square is $\sqrt{a^2 + a^2} = \sqrt{2}a$.Therefore,the distance $h = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.Using the parallel axis theorem,the moment of inertia $I_A$ about an axis passing through the corner $A$ and parallel to the axis through the center is:$I_A = I_0 + Mh^2$$I_A = \frac{Ma^2}{6} + M\left(\frac{a}{\sqrt{2}}\right)^2$$I_A = \frac{Ma^2}{6} + \frac{Ma^2}{2}$$I_A = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$.

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