# Moment Of Inertia Of Annular Disc

Moment of inertia of an annular disc that is uniform and having mass (m), thickness (t), inner radius (R1) and outer radius (R2) is expressed as;

 I = ½ M (R22 + R12 )

Here, we have to consider that the disc is rotating about an axis passing through the centre.

## Derivation Of Moment Of Inertia Of Annular Disc

1. We will start by recalling the moment of inertia expression which is given as;

dI = r2 dm

In this, we can consider dm as the mass of volume dV. We will then first consider the moment of inertia about the z-axis and we get the expression;

Izz = OR r2dm

Like the relation in a disc, the mass dm of the volume element dV is related to volume and density by;

dm = ρ dV

We then have to calculate dV. Here we will assume that the disc will have a uniform density. Here we will also consider a ring to be at radius r with width dr and thickness t. We will get;

dV = (2 π r dr)t

If we substitute the values for dm we get;

dm = (2 πpt ) r dr

2. We can then write a new expression for Izz. It will be;

Izz = R1R2 r2dm

Izz = 2 π pt R1R2 r3dm

Now we integrate between R1 and R2. We will now have;

Izz = 2 π pt [R24 – R14 / 4]

Izz = ½ πpt [ R24 – R14] ……………(Equation 1)

However, we can also express it as [ R22 + R12][ R22 – R12]. Hence, the equation now becomes;

Izz = ½ πpt [ R22 + R12][ R22 – R12] ……………..(Equation 2)

3. Now we need to find the total mass of the annular disk. We will consider the total mass (M2) of a disc of radius R2 and subtract it from the mass (M1) of a disc of radius R1.

Mass = Density X Area X Thickness

We will get,

M2 = π pt r22

M1 = πpt r12

So the mass of the annular disc is given as;

Ma = M2 – M1

Ma = π pt r22 – πpt r12

Ma = π pt (r22 – r12)

4. In the next step, we will substitute the above values in equation 2, which is.

Izz = ½ π pt [ R22 + R12][ R22 – R12]

Once the values are added and after integration, we can now write the expression for moment of inertia of an annular disc as;

Izz = ½ Ma (R22 + R12)