Question

A wire of length $16\text{m}$ and mass $8\text{kg}$ is bent in the form of a rectangle $\text{ABCD}$ with $\frac{\text{AB}}{\text{BC}}=2$. The moment of inertia of this wire frame about side $\text{BC}$ is

• A. $88.49{\text{kg-m}}^2$
• B. $108.5{\text{kg-m}}^2$
• C. $50.5{\text{kg-m}}^2$
• D. $38{\text{kg-m}}^2$

Answer 1

The correct option is A $88.49{\text{kg-m}}^2$

Given,
Length of the wire $\left(l\right)=16\text{m}$
Mass of the wire $\left(m\right)=8\text{kg}$
and, $\frac{AB}{BC}=2$
Perimeter of rectangle $=2(AB+BC)$
$\Rightarrow AB+BC=8$
Since, $AB=2BC$, we get $BC=\frac{8}{3}\text{m}\&AB=\frac{16}{3}\text{m}$
$\therefore AD=BC=\frac{8}{3}\text{m}$ and $AB=CD=\frac{16}{3}\text{m}$

As the wire is uniform, linear mass density $\lambda =\frac{m}{l}=\frac{8}{16}=\frac{1}{2}\text{kg/m}$
Using this,
$m_{AB}=\lambda \left(AB\right)=\frac{8}{3}\text{kg};m_{BC}=\frac{4}{3}\text{kg}$
By symmetry, $m_{AB}=m_{CD}=\frac{8}{3}\text{kg}$ and $m_{BC}=m_{AD}=\frac{4}{3}\text{kg}$

Moment of inertia of rectangle $ABCD$ about $BC$
$I=I_{AB}+I_{BC}+I_{CD}+I_{DA}$
$=2\left(I_{AB}\right)+I_{DA}$
$(\because I_{AB}=I_{BC};I_{BC}=0)$

$I_{AB}=\frac{m_{AB}A{B}^2}{3}=[\frac{8}{3}{\left(\frac{16}{3}\right)}^2.\frac{1}{3}]=\frac{2048}{81}$
$I_{DA}=m_{DA}A{B}^2=\frac{4}{3}\times {\left(\frac{16}{3}\right)}^2=\frac{1024}{27}$

$\therefore I=2\times \frac{2048}{81}+\frac{1024}{27}$
$=\frac{7168}{81}=88.49{\text{kg-m}}^2$