Question

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

• A. $\frac{11}{252}m{l}^2$
• B. $\frac{8}{203}m{l}^2$
• C. $\frac{5}{136}m{l}^2$
• D. $\frac{7}{162}m{l}^2$

Answer 1

Answer: (D)

$\frac{AB}{BC}=2\therefore AB=DC=\frac{l}{3}$
and $BC=AD=\frac{l}{6}$
Similarly, $m_{AB}=m_{Dc}=\frac{m}{3}$
and $m_{BC}=m_{AD}=\frac{m}{6}$
$\therefore$ Moment of inertia of the wire frame about the given axis is
$I=I_{AB}+I_{AD}+I_{Dc}+I_{BC}$
$=\frac{1}{3}\left(\frac{m}{3}\right){\left(\frac{l}{3}\right)}^2+\left(\frac{m}{6}\right){\left(\frac{l}{3}\right)}^2+\frac{1}{3}\left(\frac{m}{3}\right){\left(\frac{l}{3}\right)}^2+0$
$=\frac{m{l}^2}{81}+\frac{m{l}^2}{54}+\frac{m{l}^2}{81}=\frac{7}{162}m{l}^2$