Question

An arrangement of rods, each of mass $m$ and length $l$ are welded (wherever required) to form a shape as shown. The moment of inertia about an axis passing through point $C$ and perpendicular to the plane of figure is,

• A. $m{l}^2$
• B. $\frac{3}{2}m{l}^2$
• C. $3m{l}^2$
• D. $\frac{9}{2}m{l}^2$

Answer 1

Answer: (C)

Moment of inertia of a rod of mass $m$ And length $l$ about its end is $\frac{m{l}^2}{3}$
Moment of inertia of a rod of mass $m$ And length $l$ about its centre is $\frac{m{l}^2}{12}$
Moment of Inertia of the given system about Point $C$ and perpendicular to plane can be calculated as

$I=I_{AC}+I_{BC}+I_{CD}+I_{DE}+I_{AB}+I_{DE}I=\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+I_{AB}+I_{DE}..\left(1\right)$
Let's calculate Moment of inertia of $AB\&DE$ Using parallel axis theorem
$I_{AB}=I_f+m{\left(cf\right)}^2{I}_{AB}=\frac{m{l}^2}{12}+m\frac{3l^2}{4}$
$I_{DE}=I_g+m{\left(cg\right)}^2{I}_{DE}=\frac{m{l}^2}{12}+m\frac{3l^2}{4}$
Substituting the values of $I_{AB}\&I_{DE}$ in equation first we have
$I=\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+\frac{m{l}^2}{3}+I_{AB}+I_{DE}I=\frac{4m{l}^2}{3}+2\left[\frac{m{l}^2}{12}+\frac{3m{l}^2}{4}\right]=3m{l}^2$