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. $\frac{2}{3}M{l}^2$
• B. $\frac{13}{3}M{l}^2$
• C. $\frac{1}{3}M{l}^2$
• D. $\frac{4}{3}M{l}^2$

Answer 1

Answer: (D)

Given, mass $=M$, length $=L$
So the moment of inertia of each rod through the center and about an axis perpendicular to the ends.
$I=\frac{M{L}^2}{12}$
Thus the moment of inertia of each rod about axis is:
From parallel axis theorem,
$I^{\prime }=4\times (I+M){\left(\frac{L}{2}\right)}^2$
$\Rightarrow \left[\frac{M{L}^2}{12}\right]+\left[\frac{M{L}^2}{4}\right]=\left[\frac{4M{L}^2}{3}\right]$