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

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.
$I =\frac{ ML ^{2}}{12}$
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} $
$\Rightarrow\left[\frac{ ML ^{2}}{12}\right]+\left[\frac{ ML ^{2}}{4}\right]=\left[\frac{4 ML ^{2}}{3}\right]$