Moment Of Inertia Of A Circle

Moment of inertia of a circle or the second-moment area of a circle is usually determined using the following expression;

I = π R4 / 4

Here, r is the radius and the axis is passing through the centre. This equation is equivalent to I = π D2 / 64 when we express it taking the diameter (D) of the circle.

Similarly, a circle’s moment of inertia about an axis tangent to the circumference is given by;

I = 5 π R4 / 4

The polar moment of inertia of a circle is expressed as;

I = 5 π R4 / 2

Semi Circle

For a semi-circle the formula is given as;

I = π R4 / 8

Quarter Circle

For a quarter circle the equation is expressed as;

I = π R4 / 16

For the full derivation of the equations, students can visit the given links.

Moment Of Inertia Of Circle Derivation

For the derivation of the moment of inertia formula of a circle, we will consider the circular cross-section with the radius and an axis passing through the centre. In this derivation, we have to follow certain steps.

  • Define the coordinate system.
  • Find the differential area
  • Integration

Moment Of Inertia Of Circle

1. We will first begin with recalling the expression for the second-moment area. It is given as;

Iy = ∫A∫ z2dy dz

Iy = ∫ z2 da

Now we define the coordinates using the polar system. We get;

z = r sin θ

y = r cos θ

2. After this, we have to determine the differential area which is obtained by stating the area of the sector. It is given as;

Moment Of Inertia Of Circle

Area of the sector, A = ½ r2

We now carry out the differentiation and obtain;

dA =rdrd θ

3. We then write the integral for the moment of inertia of a circle. It is stated as;

Iy = oor (r sin θ)2 rdrdr θ

We have to solve this.

Iy = oor r3 sin2θ drd θ

Iy = o sin2θ [ or r3dr ] d θ

Iy = o sin2θ [ r4 / 4]or d θ

Iy = r4 / 4 o sin2θ

Now we apply the trigonometric identity: sin2θ = 1-cos 2 θ / 2 and calculate the integral. It will be;

Iy = r4 / 4 o 1-cos 2 θ / 2 X d θ

Iy = r4 / 4 o ½ – ½ cos (2 θ) d θ

Iy = r4 / 4 [ ½ θ – ½ sin(2 θ)]o

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

Iy = r4 / 4 (π – 0 – 0 + 0)

Iy = πr4 / 4

This equation also stands valid for the x-axis and similar derivation can be followed. In addition, the concept that we have discussed above can be further extended to find or determine the moment of inertia of a semicircle and a quarter circle.

Meanwhile, if we want to determine the M.O.I of the circle with respect to its axis tangent to the circumference we use the parallel axis theorem.

The theorem states;

II = I + Ad2

In this case, we have to consider,

d=r, A = πR2, I = πR4 / 4

If we apply the theorem we get;

II = πR4 / 4 + πR2 R2

II = 5 π R4 / 4