Question

A wire of mass $M$ and length $L$ is bent in form of a square. Calculate the MI of the square about the central axis perpendicular to the plane.

• A. $\frac{M{L}^2}{2}$
• B. $\frac{M{L}^2}{24}$
• C. $\frac{M{L}^2}{36}$
• D. $\frac{M{L}^2}{48}$

Answer 1

The correct option is D $\frac{M{L}^2}{48}$

Each side of the square can be considered as rod of length $m=\frac{M}{4}$ and mass $m=\frac{M}{4}$
Moment of inertia of rod about center of square is
$I=[\frac{m{l}^2}{12}+m{\left(\frac{l}{2}\right)}^2]$

Moment of inertia due four sides of square
$4I=[\frac{m{l}^2}{12}+m{\left(\frac{l}{2}\right)}^2]\times 4$
Here, $l=\frac{L}{4}$ and $m=\frac{M}{4}$
$=4\frac{m{l}^2}{3}=\frac{4}{3}\times \frac{M{L}^2}{64}=\frac{M{L}^2}{48}$