Question

Find the moment of inertia of a rectangular bar magnet about an axis passing through its center and parallel to its thickness. Mass of the magnet is 100g, length is 12cm, breadth is 3cm, and thickness is 2cm.

Answer 1

Moment of inertia is equivalent to mass in rotational theory. It is a measure of rotational inertia of a body and we know that mass itself is inertia. The dimensional analysis of moment of inertia is $[M]{{[L]}^{2}}$.Now for a body with mass m and placed at a distance r from the axis, the moment of inertia will be $I=m{{r}^{2}}$, where I is moment of inertia.

Complete step by step answer:
In our case, we will take a small mass dm and calculate its moment of inertia and after that we will integrate it over the full body of the magnet.
The volume of the magnet is V=72cube centimeter. The mass of the magnet in unit volume is $\rho$=100/72 gm/cubic centimeter. We are taking length as x, breadth as y and thickness as z.
Now let's take a part of the magnet, whose mass is dm=$\rho$dv=$\rho$dxdydz. It is placed at r from the axis of the rotation. Now, $r={{x}^{2}}+{{y}^{2}}+{{z}^{2}}$.
Now moment of inertia of that part will be $dI=\rho ({{x}^{2}}+{{y}^{2}}+{{z}^{2}})dxdydz$. Now, the total moment of inertia of the body will be, $I=\int{dI}$.
Since the axis of rotation is in the center the limit of the integration will run from –x/2 to x/2, -y/2 to y/2 and –z/2 to z/2, since we are taking that center of the body as origin.
So, total moment of Inertia of the bar magnet will be $I=\int{dI}=\int{\rho ({{x}^{2}}+{{y}^{2}}+{{z}^{2}})dxdydz}$, where $\rho$is constant.
$\begin{aligned} & I=\int{dI}=\int{\rho ({{x}^{2}}+{{y}^{2}}+{{z}^{2}})dxdydz} \\\ & \Rightarrow I=\int\limits_{-5}^{5}{\int\limits_{-3/2}^{3/2}{\int\limits_{-1}^{1}{\rho ({{x}^{2}}+{{y}^{2}}+{{z}^{2}})dxdydz}}} \\\ \end{aligned}$

Note:
Now by doing the whole integration and putting the value of $\rho$as stated above, we get the total moment of inertia of the bar magnet as 11300/6 gm.cm square.