A thin uniform square lamina of side a is pla | vedclass

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(B) Let the mass of the square lamina be $m$. The moment of inertia of the square lamina about an axis passing through its center of mass and parallel

Vedclass · Accepted Answer

$\frac{\sqrt{47}}{12} a$

Vedclass · Answer

(B) Let the mass of the square lamina be $m$. The moment of inertia of the square lamina about an axis passing through its center of mass and parallel to the $X$-axis is $I_{cm,x} = \frac{ma^2}{12}$.Using the parallel axis theorem,the moment of inertia about an axis parallel to the $X$-axis passing through $(0, 2a)$ is:$I_1 = I_{cm,x} + m(2a)^2 = \frac{ma^2}{12} + 4ma^2 = \frac{49ma^2}{12}$.The moment of inertia of the square lamina about an axis passing through its center of mass and perpendicular to the $xy$-plane is $I_{cm,z} = I_{cm,x} + I_{cm,y} = \frac{ma^2}{12} + \frac{ma^2}{12} = \frac{ma^2}{6}$.Using the parallel axis theorem,the moment of inertia about an axis passing through $(d, 0)$ and perpendicular to the $xy$-plane is:$I_2 = I_{cm,z} + md^2 = \frac{ma^2}{6} + md^2$.Equating $I_1$ and $I_2$:$\frac{49ma^2}{12} = \frac{ma^2}{6} + md^2$.$\frac{49a^2}{12} - \frac{2a^2}{12} = d^2$.$d^2 = \frac{47a^2}{12}$.$d = \sqrt{\frac{47}{12}} a$.

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