Question

Four thin rods of same mass $M$ and same length $I$, form a square as shown in figure. Moment of inertia of this system about an axis through centre $O$ and perpendicular to its 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

Answer: (A)

Moment of inertia of rod $AB$ about point
$P=\frac{1}{12}M{I}^2$
MI of rod $AB$ about point
$O=\frac{M{I}^2}{12}+M{\left(\frac{1}{2}\right)}^2$
$=\frac{1}{3}M{I}^2$ [from the theorem of parallel axis]
and the system consists of $4$ rods of similar type so by the symmetry
$I_{\text{system }}=\frac{4}{3}M{I}^2$