Moment of Inertia of a ring can be determined using various expressions. It includes;
- When the axis is passing through the centre (z-axis) and is perpendicular to it. For this we use;
|I = mR2|
- Along the axis which passes through the diameter. For this we have;
|Ix = Iy = mR2 / 2|
We will further look at how the equations are derived.
We will derive the moment of inertia of a ring for both instances below.
1. First, we will look at a ring about its axis passing through the centre.
We will assume the mass of the ring to be M and radius be R.
Now we need to cut an elemental ring (dx) at the circumference of the ring. Hence, the mass (dm) of the elemental ring will be;
dm = (m / 2πR) dx
Next, we calculate dI = (dm) R2
dI = [(m / 2πR) dx]R2
Substituting the values, we get;
dI = (m / 2πR) R2 dx
I = (m R/ 2π) o∫2πR dx
I = (m / 2π) [x] 02πR
I = (mR/2π) [2πR – 0]
I = mR2
2. For the second expression, we will be dealing with the moment of inertia of a ring about an axis passing through its diameter.
Here we will recall the formula;
I = ∫ r′2dm
We need to find dm. We will get;
dm = M / 2π dθ
Meanwhile, r′ = r cos θ
The next step is to carry out the integration. We will have;
I = o∫2π r2 cos2 θ (M / 2π) dθ
I = Mr2 / 2π o∫2π cos2 θ dθ
I = Mr2 / 2π [θ / 2 + sin 2 θ / 4] |o2π
I = Mr2 / 2π [ ( π + 0) – (0 + 0)]
I = Mr2 / 2