Moment of Inertia of an Ellipse

Moment of inertia is an important topic in Physics and is used in solving problems involving mass in rotational motion and angular momentum. It is a property exhibited by objects in rotational motion and is interpreted about a rotational axis. In this article, students will learn about the definition of the moment of inertia and its derivation on Vedantu.

The moment of inertia (also known as the second moment) is a physical quantity that quantifies the rotational inertia of an object. The moment of inertia of an ellipse can be considered as the rotational analogue of mass in linear motion. The moment of inertia of a body is always interpreted about a rotation axis. This is a geometrical property of an area that speaks of how its points are distributed in context to an arbitrary axis. The unit of the dimension of the moment of inertia of area is length to 4th power, \[L^{4}\]. This must not be confused with the mass moment of inertia. If the piece is thin, nonetheless, the mass moment of inertia is equivalent to the area density times the area moment of inertia.

Step Involved in Moment of Inertia of Ellipse Derivation

It is not very usual that we find the derivation for the moment of inertia of an ellipse or elliptical object. This makes it even more intriguing to know how to derive the equation for calculating the moment of inertia of an elliptical disc. Remember that there is more than one way of calculating the moment of inertia of an ellipse. Here we are describing the more generalized method that can also be used for a variety of different shapes. Let’s first check the steps involved in the derivation of the moment of inertia of an ellipse.

The steps are as follows:

1. Identifying the parametric equation of the shape.

2. Using the equation of parametric as a transformation.

3. Determining the Jacobian.

4. Computing the Moment of Inertia.

Moment of Inertia of Ellipse Derivation

Let us illustrate the process of derivation via the example of an ellipse.

We will begin with the parametric equation of an ellipse which is given by,

\[(a cos t)(b sin t)\]

To obtain the equations of transformation, we are required to modify the parametric equation as,

\[ = (r cos \theta,  \lambda r sin \theta) \] \[(\lambda cos \theta,   r \lambda  sin \theta) \]

In which,

λ = ba λ =ba.

= b / a

Remember that ‘a’ and ‘b’ are the semi-major and semi-minor axis of an ellipse respectively. Here we take ‘a’ as the semi-major axis and ‘b’ as the semi-minor. However, it's only a matter of choice and you can choose however you want. In this case, the positioning of the elliptical plate needs to be rotated to again have the semi-major axis on the x-axis.

Now, the transformation equation shall be,

x= r cosθ 

y=λ r sinθ

Next, we require the Jacobian to obtain the area element here. Jacobian will be given as:

\[J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta } \\ \frac{\partial y}{\partial r}  & \frac{\partial y}{\partial \theta} \end{vmatrix}\]

\[J = \begin{vmatrix} cos \theta & -r sin \theta\\ \lambda sin \theta  &  r \lambda cos \theta \end{vmatrix}\]

\[= r\lambda\]

With this, we are done with the setup of transformation. Now, all we need is the area and then the moment of inertia.

Calculating Area Moment of Inertia

One thing to mention is that r will be integrated from 0 to the largest value it will acquire, which is the semi-major axis i.e. ‘a’.

\[ A = _{0}\int ^{a}  _{0}\int ^{2\pi } \lambda r d\theta  A\]

\[ A = \lambda a^{2} \pi \]

\[  = \pi  ab\]

Calculating Moment of Inertia

The Formula for a moment of inertia, in this case, will be given by:

\[ I = \rho  \int (x^{2} +y^{2})dA \]

\[ = \rho  \int_{0}^{a}  \int_{0}^{2\pi } r^{3}(cos ^{2}\theta + \lambda ^{2} sin ^{2}\theta  ) dr d\theta  \]

\[ = \rho \lambda \frac{a^{4}}{4} \int_{0}^{2\pi } (cos^{2}\theta + \lambda^{2} sin^{2}\theta )d\theta\]

Taking into account the symmetry we can express this as,

\[ I = \rho \lambda a^{4} \frac{{1}}{2} 2 \int_{0}^{2\pi } (cos^{2}\theta + \lambda^{2} sin^{2}\theta )d\theta\]

\[ = \rho \lambda a^{4} \frac{{1}}{2}  (B (\frac{1}{2}, \frac{3}{}) ) \lambda^{2} b\]

\[(\frac{3}{2}, \frac{1}{2})\]

\[(\frac{3}{2}, \frac{1}{2})\]

Where ‘B’ denotes the beta function.

Using the beta-gamma relation we get,

\[(I=\frac{\rho \lambda ab }{4(a^{2}+b^{2})}\]

\[(I=\frac{M(a^{2}+b^{2})}{4}\]

This brings us to the answer with complete derivation

Moment of Inertia of Ellipse Derivation Using Routh’s Rule

For symmetrical solid like an elliptical cylinder rotated about the symmetry axis or an elliptical or circular disk about any of the axis of symmetry, Routh’s rule implies that the second moment (moment of inertia) “I” of a body of mass M about an axis is mathematically given by:

I = M Sum of the Squares of the Perpendicular Axis/3, 4 or 5

Where,

M = The mass of the body

3 = A rectangular body

4 = An elliptical body

5 = An ellipsoidal body.

Since it is quite simple to express this rule, let's work out for one of the body types taking the example of an elliptical disc.

As you might know, area moments of inertia are common in engineering disciplines such as mechanical and civil. However, the mass moment of inertia is much more common in physics. Therefore, working out the moment of inertia shall provide us with an intuitive idea of the working of this object.

We shall work out the analytical treatment of the elliptical body to resolve any doubts you might have with respect to the moment of inertia of an ellipse. Once you are clear with finding the moment of inertia of an ellipse, calculating the moment of inertia for other common object types shall be easy.

MOI of Ellipse By Routh’s Rule

(Image Will be Uploaded Soon)

In the above diagram, we have two axes as ‘a’ and ‘b’. Thus, in the denominator, we have:

\[(a^{2}+b^{2})\]

In the denominator, we are having 4 for an elliptical disc. Therefore by Routh’s rule, we have the moment of inertia as:

\[(I_{z}-\frac{M(a^{2}+b^{2})}{4}\]

Derivation of Moment of Inertia in Ellipse

First, there are some steps to be followed to start with the derivation of the moment of inertia of an ellipse which is given as follows:

• Identification of the parametric equation of the ellipse.

• Using the parametric equation of an ellipse, taking its transformation and then determining the Jacobian.

• Calculating the moment of inertia.

Let's Elaborate on the Above Process of Derivation:

The parametric equation of an ellipse can be given by (a cos t)(b sin t), where a is the semi-major axis and b is the semi-minor axis of the ellipse. Then the equation can be modified and given as:

\[(r cos \theta, \lambda r sin\theta) (r cos \theta, \lambda r sin\theta)\]

where λ = b/a.

Now, the transformation can be written as:

x= r cosθ, y=λ r sinθ, then we can get the value of the Jacobian as rλ. From this value, the moment of inertia of the area can be calculated on 0 to 'a' as:

\[A=\int_{0}^{a}\int_{0}^{2\pi}\lambda r d\theta\ A = \pi ab\]

Next, to find out the moment of inertia, the formula will be given by:

\[I = \rho \int(X^{2}+Y^{2})dA\]

From which we can get the following:

\[I = \frac{\rho \lambda ab}{4(a^{2}+b^{2})}\]

\[ = \frac{M(a^{2}+b^{2})}{4}\]

This is the value of the moment of inertia for an ellipse.

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