Q152 / 200 · Manipal (MET) 2014 · Physics
Four thin rods of same mass $M$ and same length $l$, 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
- Correct answerA$\frac{4}{3} M l^2$
- B$\frac{M l^2}{3}$
- C$\frac{M l^2}{6}$
- D$\frac{M l^2}{6}$
Solution
Moment of inertia of rod $A B$ about point
$P=\frac{1}{12} M I^2$
Ml of rod $A B$ about point $O=\frac{M l^2}{12}+M\left(\frac{I}{2}\right)^2=\frac{1}{3} M l^2$ [from the theorem of parallel axis] and the system consists of 4 rods of similar type so by the symmetry $I_{\text {system }}=\frac{4}{3} M l^2$
Ml of rod $A B$ about point $O=\frac{M l^2}{12}+M\left(\frac{I}{2}\right)^2=\frac{1}{3} M l^2$ [from the theorem of parallel axis] and the system consists of 4 rods of similar type so by the symmetry $I_{\text {system }}=\frac{4}{3} M l^2$
Hence, the correct answer is option: 'A'.