# Moment of Inertia of a Cube

## Cube’s Moment of Inertia

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

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

$I~=~\frac{Ma^{2}}{12}$​

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## FAQs on Moment of Inertia of a Cube

1. What is the Moment of Inertia of a Cube?

The moment of inertia is the property of the mass of the rigid body that defines the total net torque needed for a desired or required angular acceleration about an axis of rotation. The mass moment of inertia of a solid cube (axis of rotation at the center of a face) is related to the length of its side.

2. How can we increase the Moment of Inertia of the body?

Moment of inertia is mainly the calculation of the required force to rotate an object. This value can be increased or decreased by the corresponding increase in the radius from the axis of rotation, the moment of inertia increases thus decreasing the speed of rotation.

3. On what factors does the moment of inertia of any given body depend?

The formula for the moment of Inertia depends upon m = mass. r = Distance from the axis of the rotation. Also, the dimensional formula of the moment of inertia can be given by,$M^{1}~L^{2}~T^{0}$.

4. Why is the importance of Moment of Inertia important?

Rotational inertia is important in Physics as this involves the mass in rotational motion. It is used to calculate the angular momentum which also allows us to explain (via conservation of angular momentum) how rotational motion changes when the distribution of mass changes.

5. What shape has the lowest Moment of Inertia?

For any given shape, the moment of inertia through the center of mass of that body will be the minimum, since any moment of inertia through another axis would add mr2 to the result.

6. What is the significance or importance of the radius of Gyration?

Since the mass of any rotating rigid body is considered to be distributed with respect to the axis of rotation, we have defined a new parameter known as the radius of gyration. It is that distance whose square when multiplied with the mass of that body gives us the moment of inertia of the body about that given axis.

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