Moment of Inertia of a Cube

Moment of inertia plays the same role in rotational motion as the mass does in translational motion. In other words, the moment of inertia is the measurement of the resistance of the body to a change in its rotational motion. For example, if the body is at rest. The larger the moment of inertia of the body, the more difficult it is to put the body into rotational motion. Similarly larger the moment of inertia of the body, the more difficult is to stop its rotational motion.

The moment of inertia of the cube is calculated using different equations depending on the location of its axis. When the axis of rotation is at the center:

• \[I = (\frac{1}{6}) ma^{2} = \frac{ma^{2}}{6}\] when the axis of rotation passes through the center.

• \[I = \frac{2mb^{2}}{3}\] when the axis of rotation passes through its edge.

Concept of Area Density (\[\sigma\])

Area Density can be found by selecting and defining a tiny strip of mass with differential width. Now write an expression for the area density for the whole cube and then the tiny strips of differential widths. Add all of the individual strips using integral Calculus.

From the concept of area density, which is mass divided by area, Area density (\[ \sigma \]) is an intensive property, meaning that it doesn't depend on the amount of the material and also as long as the mass is uniform, its area density is the same whether you have chosen the entire or small strip of differential width.

The macro-scale area density is given in this equation: \[\frac{dm}{da} = \sigma \]

Derivation of the Moment of Inertia of Cube

To derive the moment of inertia of a cube when its axis is passing through the center, we will assume the solid cube has mass m, height h, width w and depth d. Now the moment of inertia of the cube is similar to that of a square laminar with a side about an axis through the center. Also, we will be assuming the area density of the lamina to be ρ. We will then take the element of the lamina with cartesian coordinates x, y in the plane to be dx -dy. Thus, its mass is = \[\rho~dxdy.\]

To find the Moment of inertia of the body, we will use \[ \rho (x^{2} + y^{2}) dx dy\].

Now the next step involves integration where we integrate over the entire lamina. We obtain;

\[\int_{-a/2} ^{a/2} \int_{-a/2} ^{a/2} \rho (x^{2} + y^{2}) dx dy = \frac{\rho a^{4}}{6} \].

We will have to then substitute the values for the mass of the lamina which is \[\rho~=~ma^{2}\].

And we obtain, \[ I = \frac{ma^{2}}{6}\]

Now the next case is when the axis is passing through the edge, we will understand how the derivation is carried out below.

The equation for moment of inertia is written as:

\[I = \int  r^{2}dm\]

Now we need to find the MOI about an axis through the edge, we will take the z-axis.

Further moving forward, we will have to consider the cube to be broken down into infinitesimally small masses. We can thus assume their sizes to be dy, dx, and dz. With this, we get;

dm = \[ \rho \] dxdydz

Here, \[ \rho \] = density

If we look at the moment of inertia formula given above, we have r as well. It is the distance from the z-axis to mass dm. Let the coordinates of the mass ‘dm’ be x,y, and z). Now the distance ‘r’ will be;

\[r= \sqrt{( x^{2} + y^{2})}\]

\[r^{2} =  x^{2} + y^{2}\]

Meanwhile, the value of x,y, and z will range from O to b according to the length of the edges.

We will now have to substitute the values that we have obtained so far in the moment of the inertia equation and finally carry out the integration.

\[I = \int  r^{2}dm\]

\[I = \int_{o}^{b} \int_{o}^{b} \int_{o}^{b} (x^{2} + y^{2}) \rho dxdydz\]

\[I = \frac{\rho 2b^{5}}{3}\]

Mass of cube, \[m = \rho b^{3}\].

Finally substituting it in the equation, we get;

\[ \frac{\rho 2mb^{2}}{3}\]

Finding the Moment of Inertia through Face Diagonal of the Cube

It is clear from the figure that the Moment of Inertia of a square plate can be given as

I=2×(Moment of Inertia of the triangular plate)

Thus, the Moment of Inertia of the triangular plate is given as 

\[I_{tri}~=~\frac{Mr^{2}}{6}\]

\[\therefore I = 2~\times I_{tri}\]

\[I~=2~\times~\frac{Mr^{2}}{6}\]

And from the above figure \[r~=~\frac{a}{\sqrt{2}}\]

Now, we can consider the cube to be made up of square plates of dm mass stacked upon each other till height a. 

So, the total moment of inertia of the cube about the diagonal is

\[\int_{0}^{I} dI = \frac{1}{12}(\int_{0}^{M}a^{2}dm)\]

\[I~=~\frac{Ma^{2}}{12}\]​

Useful Tips 

• Know the Syllabus Well and Buy the Required Material: Make sure to have all the study material at the earliest. Specific books are important as they help in identifying what topics are important for the exam and what isn't worth the trouble, because time is a crucial factor during preparation. Popular JEE books are carefully engineered to help students understand the pattern and extent of the testing areas. While those reference books are essential, a common misconception of students is that NCERT isn't that important, ignoring the fact that it is the book that helps in forming a strong base. Solving all examples and exercises from NCERT becomes the foundation of higher-order problems.

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