Question

Three identical uniform thin rods each of length $l$ and mass $m$ are joined together to form a letter $H$. The moment of inertia of the system about one of the sides of $H$ is

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

Answer 1

The correct option is D $\frac{4M{l}^2}{3}$

Divide the three rods of H as

$\Rightarrow H=I_1+I_2+I_3$

We assume that the MOI is about $I_1$

Now, MOI of $I_1=0$

By applying theorem of perpendicular axis, the momentum of $I_2=\frac{m{l}^2}{3}$

By applying theorem of parallel axis, MOI of $I_3=0$ ( for $I_1$ ) $+m{l}^2$ ( for $I_3$ at length L ) $=m{l}^2$

$I$ ( Moment of inertia of H ) $=0+\frac{1}{3}m{l}^2+m{l}^2$

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