 # Moment of Inertia of a Circle

Moment of Inertia of a Circle

First of all, let us discuss the basic concept of moment of inertia, in simple terms.

It can be inferred that inertia is related to the mass of a body. When a body starts to move in rotational motion about a constant axis, every element in the body travels in a loop with linear velocity, which signifies, every particle travels with angular acceleration.

Yes, the proper definition of the moment of inertia is that a body tends to fight the angular acceleration. Mathematically, it is the sum of the product of the mass of each particle in the body with the square of its length from the axis of rotation.

The moment of Inertia formula can be coined as:

I = Moment of inertia = Σ miri2

Here, m = mass of the body

r = radius of the circular path

• The moment of a circle area or the moment of inertia of a circle is frequently governed by applying the given equation:

The moment of inertia = I = πR4/4

• Similarly, the moment of inertia of a circle about an axis tangent to the perimeter(circumference) is denoted as:

The moment of inertia = I = 5πR4/4

• In the case of a circle, the polar moment of inertia is given as:

The moment of inertia = I = 5πR4/2

• In case of a semi-circle the formula is expressed as:

The moment of inertia = I = πR4/8

• In the case of a quarter circle the expression is given as:

The moment of inertia = I = πR4/16

Moment of Inertia of a Circle about its Diameter

If we consider the diameter of a circle D, then we must also take ‘r’ the radius as D/2.

The axis passes through the center.

This expression for the moment of inertia of a circle about its diameter can be given as

I = π D4/64.

Moment of Inertia of a Circle Derivation

Do you know how to find the moment of inertia of a circle? To learn about this, we need to understand the derivation of the moment of inertia of a circle, as explained below. It will be easier to understand how to find moment of inertia of a circle with this derivation.

The moment of inertia formula of a circle, as per the derivation, the circular cross-section will be calculated with the radius and an axis going exactly through the center. This explanation will follow certain steps, such as:

1. Describe the coordinate system.

2. Discover the differential area.

3. Finally, Integrate it.

Step: 1

Let us just begin by remembering the equation for the second-moment area. That equation is expressed below:

A∫ z2 dy dz = Iy

∫ z2 da = Iy

Iy is the component of MOI in the y-axis.

Presently, let us just describe the coordinates by applying the polar system.

After applying, we acquire;

r sin θ = z

r cos θ = y

Step: 2

After the completion of step-1, we need to calculate the differential area, which can be achieved by declaring the area of the sector.

The picture is drawn below:

From the image, the area of the sector can be expressed as;

A = ½r2

Now after the differentiation, the expression will be obtained as;

dA = rdrdθ

Step: 3

Now, let us just transcribe the integral for the moment of inertia of a circle.

The expression will be;

oor(r sin θ)2rdrdθ = Iy

Let us just simplify this integration with proper steps:

oor r3 sin2θ dr dθ = Iy

osin2θ [or r3dr ] dθ = Iy

o sin2θ [r4/4]ordθ = Iy

r4/4osin2θ = Iy

As we know about the trigonometric identity i.e.

sin2θ = 1 - cos2θ/2 -------eq(1)

Let us just put the value in the integral. The expression will be:

r4/4 o1 - cos 2θ/2 x dθ = Iy

r4/4 o½ – ½cos(2θ)dθ = Iy

r4/4 [½θ – ½sin(2θ)]o = Iy

r4/4 [½(2π) – ¼sin(4π) – ½(0) + ¼sin(0)]o = Iy

r4/4 (π – 0 – 0 + 0) = Iy

πr4/4 = Iy

In some cases, there is a possibility to calculate the M.O.I of the circle regarding its axis tangent to the perimeter, then we will practice the parallel axis theorem.

According to the theorem;

Well, we need to consider some of the factors such as,

d = r

I = πR4/4

A = πR2

Let us just put the value in eq (1):

πR4/4 + πR2/4= II

5πR4/4 = II

Moment of Inertia of a Circle Formula

1. I = πR4/4 - MOI of a circle

2. I = πD4/64 - MOI of a circular section about an axis perpendicular to the section

3. I = 5πR4/4 - MOI of a circle about an axis tangent to the perimeter(circumference)

4. I = 5πR4/2 - The polar moment of inertia

5. I = πR4/8 - The case of a semi-circle

6. I = πR4/16 - The case of a quarter circle

Q1. Explain about IXX, Iyy, and Izz.

Ans: The very nature of a material that calculates the resistance of the body to the rotational acceleration is known as the mass moment of inertia of the body about a fixed axis. IXX, Iyy, and Izz are the codes used in the 3-D structure to determine MOI.

Here,

Ixx represents 3D codes components in the x-axis

Iyy represents 3D codes components in the y-axis

Izz represents 3D codes components in the z-axis

Q2: Describe Some Examples of Inertia in Everyday Life.

Ans: Let us talk about the law of inertia. The object will possess its stationary state in, or state of motion, except an external force, makes a change.

The perfect illustration is, if the ball is rolling, it will keep rolling except when friction or external factors halt it by force.

Q3: Describe the Calculation of Rotational Inertia.

Ans: For the knowledge, you must know that rotational inertia is a scalar quantity.

It is reliant on upon the radius of rotation as per the formula:

Rotational inertia = mass * radius2

The perfect explanation for the rotational inertia is the measurement of the resistance of an object to alter its rotation.

Q4: Calculate the Moment of Inertia of a Steel Disk having the Diameter D = 60cm.

Ans: The answer is very simple.

As we describe the principles and formulas regarding MOI of a circular disc,

It will be π D4 / 64

We can write D = 60 cm = 0.6 m

So answer is, I = π (0.6)4 / 64 = 0.636 m4