Determine the moment of inertia of a uniform square plate having...

Let the square plate lie in the $xy$-plane with its center at the origin and its edges parallel to the $x$ and $y$ axes.

The moment of inertia of the square plate about the $x$-axis passing through its center is $I_x = \dfrac{ma^2}{12}$.

The moment of inertia of the square plate about the $y$-axis passing through its center is $I_y = \dfrac{ma^2}{12}$.

By the perpendicular axis theorem, the moment of inertia about the $z$-axis passing through the center and perpendicular to the plane is $I_z = I_x + I_y = \dfrac{ma^2}{12} + \dfrac{ma^2}{12} = \dfrac{ma^2}{6}$.

Let the two diagonals of the square plate be the $x'$ and $y'$ axes. Since the diagonals are perpendicular to each other and lie in the plane of the plate, we can apply the perpendicular axis theorem again.

By symmetry, the moment of inertia about both diagonals is the same, so $I_{x'} = I_{y'}$.

Using the perpendicular axis theorem, $I_z = I_{x'} + I_{y'} = 2I_{x'}$.

Substituting the value of $I_z$, we get $2I_{x'} = \dfrac{ma^2}{6}$.

$I_{x'} = \dfrac{ma^2}{12}$.

Thus, the moment of inertia of the square plate about one of its diagonals is $\dfrac{ma^2}{12}$.