Question

A wire of length $l$ and mass $m$ is bent in the form of a rectangle $ABCD$ with $\dfrac{{AB}}{{BC}} = 2$. The moment of inertia of this wire frame about the side BC is
A. $\dfrac{{11}}{{252}}m{l^2}$
B. $\dfrac{8}{{203}}m{l^2}$
C. $\dfrac{5}{{136}}m{l^2}$
D. $\dfrac{7}{{162}}m{l^2}$

Answer 1

Hint From the relation $\dfrac{{AB}}{{BC}} = 2$ we can establish the length of each side of the rectangle using the perimeter formula. Then using this value we can calculate the moment of inertia of each side of the rectangle with respect to the side $BC$
Formula used
Perimeter of a rectangle $ = 2\left( {l + b} \right)$ where $l$ is the length and $b$ is the breadth of the rectangle.
$I = m{l^2}$where $m$ is the mass and $l$ is the length of the body.

Complete step by step answer
The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis.
We are given
$
  \dfrac{{AB}}{{BC}} = 2 \\
   \Rightarrow AB = 2BC \\
$
Now $ABCD$ forms a rectangle of sides $AB$ and $BC$
Therefore, the perimeter of the rectangle is $2\left( {AB + BC} \right)$ which is equal to the total length of the wire.
Therefore,
 $
  2\left( {AB + BC} \right) = l \\
   \Rightarrow 2\left( {2BC + BC} \right) = l \\
   \Rightarrow 6BC = l \\
   \Rightarrow BC = \dfrac{l}{6} \\
$
From this expression we can also derive $AB = \dfrac{l}{3}$
As $m$ is the mass of the whole rectangular wire, so the mass of the length $AB = \dfrac{m}{3}$ and the mass of the length $BC = \dfrac{l}{6}$
Now, we know that the moment of inertia of a thin uniform rod passing through one of its ends and is perpendicular to its length is $\dfrac{1}{3}m{l^2}$
So, moment of inertia of $AB$ about $BC$${I_1}$ is $\dfrac{1}{{3 \times 3 \times {3^2}}}m{l^2} = \dfrac{1}{{81}}m{l^2}$
Similarly, moment of inertia of $CD$about $BC$${I_2}$ is $\dfrac{1}{{3 \times 3 \times {3^2}}}m{l^2} = \dfrac{1}{{81}}m{l^2}$
The moment of inertia of $DA$about $BC$${I_3}$ is $\dfrac{{m{l^2}}}{{6 \times {3^2}}} = \dfrac{1}{{54}}m{l^2}$ where $DA$ is parallel to $BC$
So resultant moment of inertia about side $BC$ is
$I = {I_1} + {I_2} + {I_3}$
$ \Rightarrow I = \dfrac{1}{{81}}m{l^2} + \dfrac{1}{{81}}m{l^2} + \dfrac{1}{{54}}m{l^2}$
$ \Rightarrow I = \dfrac{7}{{162}}m{l^2}$

Therefore, the correct option is D.

Note We know that the kinetic energy of a body in translational motion is $\dfrac{1}{2}m{v^2}$, but the kinetic energy in rotational motion is $\dfrac{1}{2}I{\omega ^2}$. Comparing these two expressions for kinetic energy, we see that since $\omega $ is the rotational analogue of $v$, $I$ corresponds to $m$ and represents the effect of the mass in rotational motion.