Physics - Rotation Question with Solution | TestHub

The distance of each particle from the centre of the square is $r=\frac{L}{\sqrt{2}}$.

The moment of inertia of the system about an axis perpendicular to the square and passing through its centre is

$I=4\times M{r}^2=4M{\left(\frac{L}{\sqrt{2}}\right)}^2=2M{L}^2$

The total mass of the system is $M_{total}=4M$.

The radius of gyration $K$ is given by $I=M_{total}{K}^2$.

$2M{L}^2=4M{K}^2$

$K^2=\frac{L^2}{2}$

$K=\frac{L}{\sqrt{2}}$

Answer: $\frac{\mathrm{L}}{\sqrt{2}}$