Physics - Rotation Question with Solution | TestHub

The moment of inertia of one rod about an axis passing through its centre and perpendicular to its length is $I_{CM}=\frac{M{L}^2}{12}$.

The distance of the centre of the square from the centre of each rod is $d=\frac{L}{2}$.

Using the parallel axis theorem, the moment of inertia of one rod about the axis passing through the centre of the square and perpendicular to its plane is:

$I_1=I_{CM}+M{d}^2$

$I_1=\frac{M{L}^2}{12}+M{\left(\frac{L}{2}\right)}^2$

$I_1=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}=\frac{M{L}^2}{3}$

Since the square is made of four identical rods, the total moment of inertia of the system is:

$I=4\times I_1=4\times \frac{M{L}^2}{3}=\frac{4M{L}^2}{3}$

Answer: $\frac{4{\mathrm{ML}}^2}{3}$