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 centre is

• A. $\frac{l}{\sqrt{2}}$
• B. $\frac{l}{2}$
• C. $l$
• D. $\sqrt{2}l$

Answer 1

The correct option is A $\frac{l}{\sqrt{2}}$


For this system, $I_{total}=I_A+I_B+I_C+I_D$
or $I_{total}=4\times I_A$

calculation of moment of inertia for point $A$,
$I_A=M(\frac{\sqrt{2}l}{2}{)}^2{I}_A=\frac{M{l}^2}{2}$

So, $I_{total}=4\times \frac{M{l}^2}{2}=2M{l}^2$ ....(1)

When converts into a closed ring,
where, $I_{ring}=4M{k}^2$ ....(2)

Now, (1)=(2)
$2m{l}^2=4M{k}^{2}k=\frac{l}{\sqrt{2}}$

hence, option A is correct.