Question

Four thin rods of same mass $M$ and same length $l$, form a square as shown in the figure. Moment of inertia of this system about an axis passing through centre $O$ and perpendicular to plane is :

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

Answer 1

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

Moment of inertia of rod $AB$ about

$P=\frac{1}{12}M{I}^2$

$MI$ of rod $AB$ about

$O=\frac{M{I}^2}{12}+M{\left(\frac{I}{2}\right)}^2=\frac{1}{3}M{I}^2$ (from the theorem of parallel axis)

and the system consists of $4$ rods of similar type so by the symmetry

$I_{system}=\frac{4}{3}M{I}^2$