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. $\frac{4}{3}MI^2$
• B. $\frac{MI^2}{3}$
• C. $\frac{MI^2}{6}$
• D. $\frac{2}{3}MI^2$

Answer 1

Answer: (A)

Moment of inertia of rod $A B$ about point
$P=\frac{1}{12} M I^{2}$
MI of rod $A B$ 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
$I_{\text {system }}=\frac{4}{3} M I^{2}$