In this article, we will learn the moment of inertia of a ring, along with examples, calculation, etc.

Let’s explore!

## What is Moment of Inertia of a Ring?

Let’s try to understand the moment of inertia of rod basics.

### Moment of Inertia of Ring Basics

Let’s try to understand the basics of moment of inertia of ring. The rotational inertia or the moment of inertia can be defined as the force preventing a body from moving in a circular motion.

Now, one should know what factors or forces make a body move in rotational motion. The main force that makes a circular body rotate about its axis is torque. Torque is the twisting force. The process of action of torque requires two forces in opposite directions on parallel sides of a circular object. These two forces act in opposite directions but induce a coupling effect. Due to that coupling effect, the body rotates in a circular motion about a fixed point known as the pivot point. It is how the torque makes a body rotate in a circular motion.

Now talking about the moment of inertia, this is the resistance of a rotating body to rotate in a circular motion about a fixed point. Through this thing, we can assume that the moment of inertia of a circular body is the oppressing force of that body against rotation.

Some of the basics related to the moment of inertia of a ring are stated below:

- Any entity’s moment of inertia having a shape, including a ring, can be described using a mathematical equation and is usually calculated using integral calculus. While for bodies with indescribable shapes, the moment of inertia is possibly calculated using experiments.
- The moment of inertia of the ring, like other bodies, will be specified to the axis. The axis is either external or internal while it can be fixed as well as not fixed.

### Moment of Inertia of Ring Definition

The moment of inertia of a ring is the opposing force by the ring’s mass towards the torque that tends to rotate it.

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## Moment of Inertia of a Ring Formula or Equation

The formula or equation of the moment of inertia of a ring can be provided based on the two following conditions:

### Ring Axis Passing Through its Centre

The moment of inertia of a ring can be stated as below when the ring axis is passing through its center.

I = mR^{2}

### Ring Axis Passing Through its Diameter

The moment of inertia of a ring can be stated as below when the ring axis is passing through its diameter.

I = mR^{2}/2

Let’s try to derive the formula of the moment of inertia of a ring.

## How to Derive Moment of Inertia of a Ring?

Let’s derive MI of a ring in both the conditions:

### Ring Axis Passing Through its Centre & Perpendicular to the Ring’s Plane

As we have to derive the moment of inertia of a ring, let’s consider,

- m = the total mass of that ring
- R = the radius of that ring of mass m

Like the moment of inertia of sphere, we have to cut the ring into an infinitely small part and calculate the MI of the small part. Once, MI is known for the small part, it will be very easy to find out the entire ring MI with the help of integration.

The ring has mass “m,” and the length of the ring is 2πR. Moreover, the mass per unit length is given by:

- λ= m/ 2 πR

Now, suppose that a small ring element is at a polar angle “θ” from a specific reference radius. This element at the center subtends a specific angle “dθ.”

Length of this element is given below:

- L= Rdθ

Mass of this element is given below:

- dm= mass per unit length x length of element
- dm = λ x Rdθ
- dm = λRdθ

Now, let’s begin with the calculation. We use the following equation in the case of continuous mass distribution:

- I= ʃ (dm) R
^{2}

So, the moment of inertia of the small element,

- dI = (λrdθ) R
^{2}

Now, integrating, we have:

- ʃ dI= ʃ
_{0}^{2π }(λRdθ) R^{2} - I= ʃ
_{0}^{2π }(λRdθ) R^{2} - I= λ R
^{3 }ʃ_{0}^{2π}dθ - I= (m/ 2πR) R
^{3 }2π - I= mR
^{2}

### Ring Axis Passing Through its Diameter

Let’s write the basic MI formula,

I = ∫ r_{1}^{2}dm

The ring has mass “m,” and the length of the ring is 2πR. Moreover, the mass per unit length is given by:

- λ= m/ 2 πR

However, in this case, the axis is passing through the diameter, so, the mass per unit length is given by:

- λ= m/ 2π

Length of this element is given below:

- L= dθ [ as, it is through the diameter]

Mass of this element is given below:

- dm= mass per unit length x length of the element
- dm = λ x dθ
- dm = λdθ
- dm = (m/ 2π) dθ

Now, we can get the relationship between r1 & R from the triangle,

r_{1} = R cos θ

The following step is to complete the integration. We will have;

- I = ∫ r
_{1}^{2}dm - I = ∫
_{0}^{2π}R^{2}cos^{2}θ (m/ 2π) Rdθ - I = mR
^{2}/2π ∫_{0}^{2π}cos^{2}θ dθ - I = mR
^{2}/2π ∫_{0}^{2π}1/2(1 + cos2θ) dθ [as, cos2θ =2cos^{2}θ -1] - I = mR
^{2}/ 2π [θ / 2 + sin 2 θ / 4] |o2π - I = mR
^{2}/ 2π [ ( π + 0) – (0 + 0)] - I = mR
^{2}/ 2

## Radius of gyration

In case we study the rotational motion of a body, the radius of gyration is important. It is defined as the distance from the axis of rotation where a particle of equal mass is placed to possess the same moment of inertia as the given mass distribution. The radius of gyration for an entity of mass “m” and moment of inertia “I,” the expression is hereunder:

K= √ I/M

## Moment of Inertia of Ring Calculation

### Moment of Inertia of Ring Calculation#1

#### Problem

Calculating moment of inertia of a ring having mass “m” and radius “R” about an axis which passes through its center and is perpendicular to the ring’s plane

#### Solution

Now, let’s begin with the calculation. We use the following equation in the case of continuous mass distribution:

I= ʃ (dm) r^{2}

The ring has mass “m,” and the length of the ring is 2 π r. Moreover, the mass per unit length is given by:

λ= M/ 2 πr

Suppose that a small ring element is at a polar angle “θ” from a specific reference radius. This element at the center subtends a specific angle “dθ.” Mass, length, and moment of inertia of this element are given below:

L= Rdθ

m= λRdθ

moment of inertia= (λRdθ) R^{2}

for the moment of inertia, we have:

I= ʃ _{0} ^{2π }(λRdθ) R^{2}

I= λ R^{3 }ʃ _{0} ^{2π }dθ

I= (M/ 2 π R) R^{3 }2 π

I= MR^{2}

**Radius of gyration**

In case we study the rotational motion of a body, the radius of gyration is important. It is defined as the distance from the axis of rotation where a particle of equal mass is placed to possess the same moment of inertia as the given mass distribution. The radius of gyration for an entity of mass “m” and moment of inertia “I,” the expression is hereunder:

K= √ I/M

### Moment of Inertia of Ring Calculation#2

#### Problem

Determine the moment of inertia of a ring about a tangent parallel to the diameter when the moment of inertia of the ring about one of its diameter is I.

#### Solution

The ring’s moment of inertia about its diameter is given by the expression hereunder:

I_{d}= I= ½ MR^{2}

where R means radius.

R is the distance between tangent and diameter.

The moment of inertia about the tangent can be calculated using the parallel axis theorem:

I_{T} = I_{d} + MR^{2 }= I+ 2I =3I (Since MR^{2 }= 2I)

### Calculating moment of inertia of a ring of mass 0.5 kg

#### Problem

What is the moment of inertia of a ring having mass 0.5 kg and radius 1m about an axis passing through its center and is perpendicular to the plane? B) also calculate the moment of inertia about the diameter as the axis.

#### Solution

Mass, m= 0.5 kg

Radius, R= 1m

The expression for moment of inertia of ring about an axis passing through its center and is perpendicular to its plane is given by:

I = MR^{2}

I= 0.5 . (1)^{2}

I = 0.5 kg m ^{2}

Moment of inertia about the diameter as the axis= MR^{2}/2= 0.5/2 = 0.25 kg/m^{2}

## Applications of MI of Ring

The shape of a ring is not only associated with the accessories; rather, it has many different functions. There are many nuts having ring shapes used in mechanical physics. Many objects around us have the geometry of a ring. So talking about the applications of the MI of the Ring, they are mostly its benefits. By using ring-shaped pieces in a piece of machinery, engineers learn about the strength of bending of different rings by comparing their moment of inertia.

# Conclusion

Hence, we can say that finding the moment of inertia of a rotating body, especially a ring is not a difficult job. The formula is the same, but the dimensions change each time. Moreover, people should also have basic information about inertia and how this physical property works.

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