Moment of Inertia of Annular Disc

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What is the Moment of Inertia?

The moment of inertia is also known as the mass moment of inertia, rotational inertia or angular mass of a rigid body. It is a quantity determining the torque needed for a desired angular acceleration about a rotational axis. It is similar to knowing how mass determines the force needed for a desired acceleration. Thus, the amount of torque needed to cause any given angular acceleration or the rate of alteration in angular velocity is proportional to the moment of inertia of the body. 

The dimensional formula of moment of inertia is M L2  

SI unit of moment of inertia is kg m2.


Calculating Moment of Inertia of an Annular Disc

Let us know how to calculate the moment of inertia of an annular disc which is uniform with mass (m), thickness (t), inner radius (R1) and outer radius (R2). The Moment of inertia of a uniform angular with mass m, thickness t and inner and outer radius as R1 and R2 respectively can be expressed as:

I = ½ M (R22 + R12 )

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


Derivation of Moment of Inertia of Annular Disc

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

dI = r2 dm

Here, dm is the mass of volume dV. When we consider the moment of inertia about the z-axis, we get the expression;

Izz = O∫R r2dm

The mass dm of the volume element dV, related to volume and density, is given as:

dm = ρ dV

For the calculation of dV, we will assume that the disc will have a uniform density. We will also consider a ring with a radius r with width dr and thickness t. We will represent it as:

dV = (2πr dr)t

If we substitute the values for dm here, we get,

dm = (2 πpt ) r dr

2. Now, a new expression for Izz will be obtained which is as follows:

Izz = R1∫R2 r²dm

Izz = 2 π pt R1∫R2 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)

Also, it can be expressed as [ R2²  + R1²][ R2²  – R1²]. 

Now, therefore, the equation becomes:

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

3. Next step is to find the total mass of the annular disk.

Suppose the total mass of a disc is (M2) and radius is R2, on subtracting it from the mass (M1) of a disc of radius R1, we get equations as mentioned below:

Mass = Density X Area X Thickness

We will get,

M2 = π pt r2²

M1 = πpt r1²

Therefore, the mass of the annular disc is given as;

Ma = M2 – M1

Ma = π pt r2²– πpt r12

Ma = π pt (r2² – r1²)

4. Further, on substituting the above values in equation 2, we get:

Izz = ½ π pt [ R2²+ R1²][ R2² – R1²]

After the addition of values and on integrating, we can represent the moment of inertia of an annular disc as;

Izz = ½ Ma (R2² + R1² )