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:

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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:

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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 the 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 the 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 by DZ = Iy 

∫ z2 da = Iy 

It 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. 

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

A = ½r2dθ

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;

o∫2πo∫r(r sin θ)2rdrdθ = Iy

Let us just simplify this integration with proper steps:

o∫2πo∫r r3 sin2θ dr dθ = Iy  

o∫2πsin2θ [o∫r r3dr ] dθ = Iy 

o∫2π sin2θ [r4/4]ordθ = Iy  

r4/4o∫2πsin2θ = 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 o∫2π1 - cos 2θ/2 x dθ = Iy

r4/4 o∫2π½ – ½cos(2θ)dθ = Iy 

r4/4

½θ–½sin(2θ)

½θ–½sin(2θ)o2π = Iy 

r4/4

½(2π)–¼sin(4π)–½(0)+¼sin(0)

½(2π)–¼sin(4π)–½(0)+¼sin(0)o2π = 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;

I + Ad2 = II

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

Law of Inertia-Application in Mechanics.

According to Newton's first  law of motion nobody by itself can change its state of rest or of uniform motion in a straight line. This inertness or   in ability of the body is called inertia. Obviously an external force is required to change the state of rest or of uniform motion in a  straight line of the body.Larger the force required to change the state of the body, greater  is its inertia.  therefore to produce or for destroy a given  even acceleration force is required  is directly proportional to the mass of the body .clearly heavier  the body greater is the force  required to change it state and hence greater is it's inertia.The reverse is also true.mass of the body is therefore a measure of its inertia in translatory motion.the inertia of body depends only upon its mass, the moment of inertia of a body about a given axis depends upon its mass , position and direction of the axis of rotation.It also depend on the shape of the body. It is due to  the reason that  two bodies having the same mass, but different shapes will have different moments of inertia about a given axis of rotation.

Table of Contents:

The following contents are present in the moment of inertia in Physics of Class 11

1. What is the Law of Inertia? 

2. Moment of inertia of a rigid body

3. Moment  of inertia of a straight rod about an Axis through its center

4. Moment of inertia of a square lamina about its diagonals

Types of Inertia

Following are the three types of inertia:

1. Inertia of Rest

2. Inertia of Direction

3. Inertia of Motion