Question

Four particles each of mass $m$ are placed at the corners of a square of side length $l$. The radius of gyration of the system about an axis perpendicular to the square and passing through its center is:

• A. $\frac{l}{\sqrt{2}}$
• B. $\frac{l}{2}$
• C. $l$
• D. $\left(\sqrt{2}\right)l$

Answer 1

Answer: (A)

$\frac{l}{\sqrt{2}}$

The distance of mass from the center of square is $r=\frac{l}{\sqrt{2}}$

So the total moment of inertia of system $I=4m{r}^2=2m{l}^2$

Let the radius of gyration be k

$I=2m{l}^2=4m{k}^2\Rightarrow k=\frac{l}{\sqrt{2}}$