For the given square frame ABCD with center O | vedclass

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(C) Let the side length of the square be $a$ and its mass be $M$.The moment of inertia of a square frame about an axis passing through its center and

Vedclass · Accepted Answer

$I_{AC} = I_{EF}$

Vedclass · Answer

(C) Let the side length of the square be $a$ and its mass be $M$.The moment of inertia of a square frame about an axis passing through its center and perpendicular to its plane is $I_z = Ma^2$.By the perpendicular axis theorem,$I_z = I_x + I_y$. Since the square is symmetric,$I_x = I_y$,so $I_x = I_y = \frac{1}{2} I_z = \frac{1}{2} Ma^2$.Here,$I_{EF}$ is the moment of inertia about an axis passing through the midpoints of opposite sides,which is equivalent to $I_x$ or $I_y$. Thus,$I_{EF} = \frac{1}{2} Ma^2$.$I_{AC}$ is the moment of inertia about the diagonal $AC$. For a square,the moment of inertia about any axis in its plane passing through the center is the same. Therefore,$I_{AC} = I_{EF} = \frac{1}{2} Ma^2$.

For the given square frame $ABCD$ with center $O$,which of the following relations is correct?

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