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# Moment Of Inertia Of A Ring

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 / 2Ï€R) dx

Next, we calculate dI = (dm) R2

dI =Â Â [(m / 2Ï€R) dx]R2

Substituting the values, we get;

dI = (m / 2Ï€R) R2Â dx

Using integration;

I = (m R/ 2Ï€)Â oâˆ«2Ï€R dx

I = (m / 2Ï€) [x] 02Ï€R

I = (mR/2Ï€) [2Ï€R – 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 = âˆ« râ€²2dm

We need to find dm. We will get;

dm = M / 2Ï€ dÎ¸

Meanwhile, râ€² = r cos Î¸

The next step is to carry out the integration. We will have;

I = oâˆ«2Ï€ r2 cos2 Î¸ (M / 2Ï€) dÎ¸

I = Mr2 / 2Ï€ oâˆ«2Ï€ cos2 Î¸ dÎ¸

I = Mr2 / 2Ï€ [Î¸ / 2 + sin 2 Î¸ / 4] |o2Ï€

I = Mr2 / 2Ï€ [ ( Ï€ + 0) – (0 + 0)]

I = Mr2 / 2