Question

Four identical rods are joined end to end to form a square. The mass of each rod is $M$. The moment of inertia of the square about the central line $y{y}^{\prime }$ as shown in figure is

• A. $\frac{M{l}^2}{3}$
• B. $\frac{M{l}^2}{4}$
• C. $\frac{M{l}^2}{6}$
• D. none of these

Answer 1

The correct option is D none of these

From given,
Moment of inertia about $Y{Y}^{\prime }$
$I_{Y{Y}^{\prime }}=I_{AB}+I_{BC}+I_{CD}+I_{DA}$ ....(1)
here, road $AB$ and $CD$ are symmectric.

so, moment of inertia about $Y{Y}^{\prime }$
$I_{Y{Y}^{\prime }}=\frac{M{l}^2}{12}$
$I_{\left(Y{Y}^{\prime }forABandCD\right)}=\frac{M{l}^2}{12}\times 2=\frac{M{l}^2}{6}$

Moment of inertia for rods $AD$ and $BC$

$dI=\int dm(\frac{l}{2}{)}^2$
$dI=\frac{l^2}{4}\int dm$
$I_{Y{Y}^{\prime }}=\frac{M{L}^2}{4}$
for rod $AD$ and $BC$
$I_{Y{Y}^{\prime }\left(ADandBC\right)}=\frac{M{L}^2}{4}\times 2=\frac{M{L}^2}{2}$

by using eq. (1)
$I_{Y{Y}^{\prime }}=I_{AB}+I_{BC}+I_{CD}+I_{DA}$
$I_{Y{Y}^{\prime }}=\frac{M{l}^2}{6}+\frac{M{L}^2}{2}=\frac{2M{L}^2}{3}$