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 centre of the square and perpendicular to its plane is?

• A. $\frac{1}{3}M{l}^2$
• B. $\frac{8}{3}M{l}^2$
• C. $\frac{2}{3}M{l}^2$
• D. $\frac{13}{3}M{l}^2$

Answer 1

The correct option is B $\frac{8}{3}M{l}^2$

Moment of inertia about centre of rod is $I_0=\frac{m{l}^2}{12}$

By parallel axis theorem for rod $AB,MI$ about $0$,

$I=I_0+m{d}^2(d=l/2)$

$=\frac{m{l}^2}{12}+\frac{m{l}^2}{4}=\frac{m{l}^2}{3}$

So $MI$ due to all four rods,

$I=4I=\frac{4m{l}^2}{3}=\frac{8}{3}{I}_0$