For the given uniform square lamina ABCD,whos | vedclass

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Question

(D) Let the square lamina have mass $M$ and side length $a$. The moment of inertia of a square lamina about an axis passing through its centre $O$ and

Vedclass · Accepted Answer

$I_{AC} = I_{EF}$

Vedclass · Answer

(D) Let the square lamina have mass $M$ and side length $a$. The moment of inertia of a square lamina about an axis passing through its centre $O$ and perpendicular to its plane $(I_z)$ is given by $I_z = I_x + I_y$,where $I_x$ and $I_y$ are moments of inertia about the axes passing through the centre and parallel to the sides. By symmetry,$I_x = I_y = I_{EF} = I_{GH}$ (where $EF$ and $GH$ are axes passing through the midpoints of opposite sides). Thus,$I_z = 2 I_{EF}$.Similarly,the moment of inertia about the diagonal axes $AC$ and $BD$ are equal by symmetry,so $I_{AC} = I_{BD}$. According to the perpendicular axes theorem,$I_z = I_{AC} + I_{BD} = 2 I_{AC}$.Equating the two expressions for $I_z$,we get $2 I_{EF} = 2 I_{AC}$,which simplifies to $I_{AC} = I_{EF}$.

For the given uniform square lamina $ABCD$,whose centre is $O$,which of the following relations is correct?

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