Question

Four identical thin rods each of mass $M$ and length $l$ form a square frame. Moment of inertia of this frame about an axis through the center of the square and perpendicular to its plane is

• A. $\frac{4}{3}M{l}^2$
• B. $\frac{1}{3}M{l}^2$
• C. $\frac{M{l}^2}{6}$
• D. $\frac{2}{3}M{l}^2$

Answer 1

The correct option is A $\frac{4}{3}M{l}^2$


$I_{total}=I_{AB}+I_{BC}+I_{CD}+I_{DA}$
Since all rods are symmetric,
therefore, $I_{total}=4\times I_{AB}$ ...(1)

For rod AB:
Moment of inertia of rod about transverse axis passing through its centre,
$I_{y{y}^{\prime }}=\frac{M{l}^2}{12}$
Using parallel axis theorem,
$I_{o{o}^{\prime }}=\frac{M{l}^2}{12}+\frac{M{l}^2}{4}{I}_{o{o}^{\prime }}=\frac{M{l}^2}{3}$
From equation (1),
$I_{total}=4\times \frac{M{l}^2}{3}=\frac{4M{l}^2}{3}$