Question

A square frame ABCD is formed by four identical rods each of mass $m$ and length $l$. This frame is in the x-y plane such that side AB coincides with x- axis and side AD along y -axis. The moment of inertia of the frame about x axis is
A. $\dfrac{5}{3}m{l^2}$
B. $\dfrac{2}{3}m{l^2}$
C. $\dfrac{4}{3}m{l^2}$
D. $\dfrac{1}{{12}}m{l^2}$

Answer 1

Hint: We can calculate the moment of inertia of the square frame with respect to x-axis by adding the individual moment of inertia of all the four sides with respect to this axis. Since the side AB lies along the axis the moment of inertia of this side will be zero. The moment of inertia of the sides AD and BC will be equal to the value of moment of inertia of a rod about an axis passing through one of its ends. For the side CD all the particles in the rod are at equal distance from the axis. Using this information, we can find the total moment of inertia of the system.

Complete step by step answer:
It is given that a square frame is formed by four identical rods having mass m and length l.
The frame is lying in the x-y plane and the axis of rotation is along the side AB which coincides with the x axis.
Side AD coincides with the y-axis.

We need to find the moment of inertia of the frame about the x-axis.
We know that the moment of inertia of rod about an axis passing through one of its ends can be calculated as
$I = \int {{r^2}dm} $
Where, $dm$ is the mass per unit length of the rod. That is, $dm = \dfrac{m}{l}$ and let dx be the small distance then we will get the total length by integrating from 0 to l.
$ \Rightarrow I = \int\limits_0^l {\dfrac{m}{l}{x^2}dx} $
$ \Rightarrow I = \dfrac{m}{l}\int\limits_0^l {{x^2}dx} $
$ \Rightarrow I = \dfrac{m}{l}\left[ {\dfrac{{{x^3}}}{3}} \right]_0^l$
$ \Rightarrow I = \dfrac{m}{{3l}}{l^3}$
$I = \dfrac{1}{3}m{l^2}$ ……………..(1)
There are four rods. Let us calculate the moment of inertia of each rod.
Moment of inertia of rod AB about x axis will be zero because AB lies along the axis.
So, $r = 0$
$ \Rightarrow {I_{AB}} = m{r^2} = 0$
The moment of inertia of BC can be written using equation (1) since the rotation axis is at one end of BC.
$ \Rightarrow {I_{BC}} = \dfrac{1}{3}m{l^2}$
Similarly, moment of inertia of DA is also
$I = \dfrac{{m{l^2}}}{3}$
Since the axis is at one end of DA.
The moment of inertia of CD is
 ${I_{CD}} = m{l^2}$
Since all the particles in this rod are at a distance l from the axis.
Now we can calculate the total moment of inertia as the sum of all the moments of inertia of each rod forming the square frame.
Total moment of inertia is
$ \Rightarrow I = {I_{AB}} + {I_{BC}} + {I_{CD}} + {I_{DA}}$
$ \Rightarrow I = 0 + \dfrac{{m{l^2}}}{3} + m{l^2} + \dfrac{{m{l^2}}}{3}$
$ \Rightarrow I = m{l^2}\left( {\dfrac{1}{3} + 1 + \dfrac{1}{3}} \right)$
$ \Rightarrow I = \dfrac{5}{3}m{l^2}$
This is the moment of inertia of the square frame when it rotates about an axis passing through the side AB.

Hence the correct answer is option A.

Note:
Remember that moment of inertia of a body depends upon the axis about which it rotates. Though all the four rods in the square frame are of equal length and mass, the moment of inertia of each rod will vary depending upon its orientation with respect to the axis of rotation. For the sides BC and AD, the axis of rotation passes through one of its ends. For the side AB axis of rotation and length of the side coincides. For side CD all the particles in it lie at a distance l from the axis of rotation. Hence the moment of inertia of each side should be calculated accordingly.