Square Wire Moment

From the symmetry of the figure we can say that the wire frame has equal moment of inertia about its two diagonals $AC$ and $BD$ .

Also, these diagonals are orthogonal therefore, according to the Perpendicular axis theorem. $I_O = I_{AC} + I_{BD}.$ Here, $I_O$ is the moment of inertial of the wire frame about an axis passing through the center $O$ and perpendicular to the plane of the wire frame. Also, $I_{AC} = I_{BD}.$ Therefore, we can write $I_O = 2 I_{AC}.$ The four wires will have equal moment of inertia about the axis passing through $O$ and perpendicular to the plane of the figure. Therefore, the moment of inertia of the complete wire frame about this axis can be calculated by finding the moment of inertia of one of the wire and multiplying it by 4. The moment of inertia of the wire $AB$ about the axis passing though its center of mass and perpendicular to plane of the figure is

$I_{\text{cmAB}} = \frac{M_{\text{AB}}L^2}{12} = \frac{ML^2}{48}.$ The moment of inertia of $AB$ about the axis passing though $O$ can be calculated using the parallel axis theorem. $I_{OAB} = I_{cmAB} + M_{AB}d^2.$ Here, $d$ is the distance between these two axis. Hence,

$\begin{aligned} I_{OAB} &= \frac{ML^2}{48} + \frac{M}{4} \left( \frac{L}{2} \right)^2 \\ I_{OAB} &= \frac{ML^2}{12} \\ I_O &= 4 \times I_{OAB} = \frac{ML^2}{3} \\ I_{AC} &= \frac{1}{2} \times I_O \\ I_{AC} &= \frac{ML^2}{6} \\ I_{AC} &= \frac{ML^2}{\alpha} \\ \alpha &= \boxed{6}. \end{aligned}$

Let the diagonal be the $x$ -axis with 0 at the lower end and the mass per unit length of the square be $\sigma$ , the moment of inertia of the square is as below:

$\begin{aligned} I & = 4 \int_0^L x^2 \sigma dL & \small \color{#3D99F6} \text{Note that } dL = \sqrt 2 \ dx \\ & = 4 \int_0^\frac L{\sqrt 2} \sqrt 2 \sigma x^2 dx & \small \color{#3D99F6} \text{Note that } \sigma = \frac M{4L} \\ & = 4 \int_0^\frac L{\sqrt 2} \frac {\sqrt 2 Mx^2}{4L} dx \\ & = \int_0^\frac L{\sqrt 2} \frac {\sqrt 2 Mx^2}L dx \\ & = \frac {\sqrt 2 M x^3}{3L} \\ & = \frac {ML^2}6 \end{aligned}$

$\implies \alpha = \boxed{6}$ .