Question

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

$\frac{7}{162}m{l}^2$

The perimeter of the rectangle is given as,

$l=2\left(AB+BC\right)$

$l=2\left(2BC+BC\right)$

$BC=AD=\frac{l}{6}$

$AB=DC=\frac{l}{3}$

The mass of the length $AB$ is given as,

$m_{AB}=\frac{m}{3}$

The mass of the length $DC$ is given as,

$m_{DC}=\frac{m}{3}$

The mass of the length $AD$ is given as,

$m_{AD}=\frac{m}{6}$

The total moment of inertia about the side $BC$ is given as,

$M.I.=M.I{.}_{AB}+M.I{.}_{DC}+M.I{.}_{AD}$

$M.I.=\frac{m{l}^2}{81}+\frac{m{l}^2}{81}+\frac{m{l}^2}{27}$

$M.I.=\frac{7m{l}^2}{162}$