Moment of inertia also known as the angular mass or rotational inertia can be defined w.r.t. rotation axis, as a quantity that decides the amount of torque required for a desired angular acceleration or a property of a body due to which it resists angular acceleration. The formula for the moment of inertia is the “sum of the product of mass” of each particle with the “square of its distance from the axis of the rotation”. The formula of Moment of Inertia is expressed as I = Σ m_{i}r_{i}^{2}.

### What is Moment of Inertia

The moment of inertia of an object is a determined measurement for a rigid body rotating around a fixed axis. The axis might be internal or external, and it can be fixed or not. However, the moment of inertia (I) is always described in relation to that axis.

The moment of Inertia depends on the distribution of the mass around its axis of rotation. MOI varies depending upon the position of the axis that is chosen. That is, depending on the location and direction of the axis of rotation, the same item might have various moment of inertia values.

Angular mass or rotational inertia are other names for the moment of inertia.

The SI unit of moment of inertia is kg m^{2}.

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## Moment of Inertia Example

Imagine you are on a bus right now. You find a seat and sit down. The bus starts moving forward. After a few minutes, you arrive at a bus stop and the bus stops. What did you experience at this point? Yes. When the bus stopped, your upper body moved forward whereas your lower body did not move.

Why is that? It is because of Inertia**. **Your lower body is in contact with the bus but your upper body is not in contact with the bus directly. Therefore, when the bus stopped, your lower body stopped with the bus but your upper body kept moving forward, that is, it resisted change in its state.

Similarly, when you board a moving train, you experience a force that pushes you backwards. That is because before boarding the train you were at rest. As soon as you board the moving train, your lower body comes in contact with the train but your upper body is still at rest. Therefore, it gets pushed backwards, that is, it resists change in its state.

*Understand the Theorem of Parallel and Perpendicular Axis here in detail.*

**Browse more Topics Under System Of Particles And Rotational Dynamics**

- Introduction to Rotational Dynamics
- Vector Product of Two Vectors
- Centre of Mass
- Motion of Centre of Mass
- Moment of Inertia
- Theorems of Parallel and Perpendicular Axis
- Rolling Motion
- Angular Velocity and Angular Acceleration
- Linear Momentum of System of Particles
- Torque and Angular Momentum
- Equilibrium of a Rigid Body
- Angular Momentum in Case of Rotation About a Fixed Axis
- Dynamics of Rotational Motion About a Fixed Axis
- Kinematics of Rotation Motion about a Fixed Axis

## What is Inertia?

What is Inertia? It is the property of a body by virtue of which it resists change in its state of rest or motion. But what causes inertia in a body? Let’s find out.

Inertia in a body is due to its mass. More the mass of a body more is the inertia. For instance, it is easier to throw a small stone farther than a heavier one. Because the heavier one has more mass, it resists change more, that is, it has more inertia.

## Moment of Inertia Definition

So we have studied that inertia is basically mass. In rotational motion, a body rotates about a fixed axis. Each particle in the body moves in a circle with linear velocity, that is, each particle moves with an angular acceleration. Moment of inertia is the property of the body due to which it resists angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

### Moment of Inertia Formula

*Moment of Inertia (I) = Σ m _{i}r_{i}^{2}*

where,

m = Sum of the product of the mass.

r = Distance from the axis of the rotation.

And the Integral form of MOI is as follows:

*I = ∫ d I = ∫ _{0}^{M} r^{2} dm*

where,

dm = The mass of an infinitesimally small component of the body

r = (perpendicular) distance between the point mass and the axis of rotation

## Moment of Inertia of a System of Particles

For a system of point particles revolving about a fixed axis, the moment of inertia is:

*Moment of Inertia (I) = Σ m _{i}r_{i}^{2}*

where r_{i} is the perpendicular distance from the axis to the i^{th} particle which has mass m_{i}.

**Example**

A system of point particles is shown in the following figure. Each particle has a mass of 0.3 kg and they all lie in the same plane. What is the moment of inertia of the system about the given axis?

**Answer:**

The mass of all the particles is equal. Hence, m = 0.3 kg

I = Σ m_{i}r_{i}^{2} = m Σ r_{i}^{2} = 0.3 [0.6^{2} + 0.4^{2} + 0.2^{2}] …..(Converting the distance of the particles to metre)

I = 0.168 kg m^{2}

## Moments of Inertia for Different Objects

**For Rod**

**For Sphere**

**For Hoop**

**For Cylinder**

**For Rectangular objects**

**For Ring/Disc**

The moment of inertia, as we all know, is affected by the axis of rotation. After selecting two distinct axes, you will notice that the object resists the rotational change differently. As a result, the following theorems can be used to calculate the moment of inertia along any given axis:

- Parallel Axis Theorem
- Perpendicular Axis Theorem

## Parallel Axis Theorem

The Parallel Axis Theorem states,

The moment of inertia of a body about an axis parallel to the body passing through its centre is equal to the sum of moment of inertia of the body about the axis passing through the centre and product of the mass of the body times the square of the distance between the two axes.

Its formula is,

*I = I _{c} + Mh^{2}*

where,

I = moment of inertia of the body

I_{c} = moment of inertia about the center

M = mass of the body

h^{2} = square of the distance between 2 axes

Let us see how the Parallel Axis Theorem helps us to determine the moment of inertia of a rod whose axis is parallel to the axis of the rod and it passes through the center of the rod.

Moment of inertia of rod is given as:

I = (1/3) ML^{2}

The distance between the end of the rod and its centre is:

h = L/2

Therefore, the parallel axis theorem of the rod is:

I_{c} = (1/3) ML^{2} – M(L/2)^{2}

I_{c} = (1/3) ML^{2} – ML^{2}/4

I_{c} = (1/12) ML^{2}

## Kinetic Energy in Rotational Motion

What is the analogue of mass in rotational motion? To answer this question, we have to derive the equation of kinetic energy in rotational motion.

Consider a particle of mass *m *at a distance from the axis with linear velocity =* v _{i} = r_{i}ω. *Therefore, the kinetic energy of this particle is,

k_{i} = 1/2 *m _{i}v_{i}^{2} *= 1/2

*m*

_{i}(r_{i })^{2}ω^{2}

where *m _{i}* is the mass of the particle. The total kinetic energy K of the body is thus the sum of the kinetic energies of individual particles. ∴ K = Σ k

_{i }= 1/2 ( Σ

*m*

_{i}(r_{i })^{2}ω^{2}*)*

**where**

*n*is the number of particles in the body. We know that angular acceleration

*ω*is the same for all particles. Therefore, taking

**out of the sum, we get,**

*ω*K = 1/2*ω ^{2}* (Σ

*m*………………….(I)

_{i}(r_{i })^{2 })Let Σ *m _{i}(r_{i })*

^{2}**be I which is a new parameter characterising the rigid body known as the Moment of Inertia**

^{ }**.**Therefore, with this definition,

K = 1/2I*ω ^{2 }*

*……………..*(II)

The parameter I is independent of the magnitude of the angular velocity. It is the characteristic of the rigid body and the axis about which it rotates. We already know that linear velocity in linear motion is analogous to angular acceleration in rotational motion.

On comparing equation II with the formula of the kinetic energy of the whole rotating body in linear motion, it is evident that mass in linear motion is analogous to the moment of inertia in rotational motion. Hence, the question is answered. In simple words, the moment of inertia is the measure of the way in which different parts of the body are distributed at different distances from the axis.

## Radius of Gyration

As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we define a new parameter known as the radius of gyration**.** It is related to the moment of inertia and the total mass of the body. Notice that we can write I = M*k*^{2 }where *k* has the dimension of length.

Therefore, the radius of gyration is the distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis. Therefore, the moment of inertia depends not only on the mass, shape, and size of the body but also on the distribution of mass in the body about the axis of rotation.

Learn more about Torque and Angular Momentum here.

## Here are a few Solved Questions for You

**Question 1.** The moment of inertia depends on:

a) mass of the body b) shape and size of the body

c) distribution of mass about the axis of rotation d) all of the above

**Answer: **d) all of the above. All the factors together determine the moment of inertia of a body.

**Question 2:** Find the moment of inertia of a disc of mass 5 kg and radius 30 cm about the following axes.

- Axis passing through the center and perpendicular to the plane of the disc.
- Axis touching the edge and perpendicular to the plane of the disc.

**Answer.**

Given: Mass (M) = 5 kg, Radius (R) = 30 cm = 0.3 m

i. I = ½ MR^{2} = ½ x 5 x (0.3)^{2} = 0.225 kg m^{2}

ii. I = I_{c} + Md^{2} = ½ MR^{2} + MR^{2} = 3/2 MR^{2} = (3/2) x 5 x (0.3)^{2} = 0.675 kg m^{2}

**Question 3:** Determine the moment of inertia about the geometric centre of the given construction, which consists of a single thin rod connecting two identical solid spheres as shown in the figure.

**Answer.**

The above image consists of 3 structures: 1 thin rod and 2 solid spheres.

The mass of the rod (M) = 3 kg, total length of the rod (l) = 80 cm = 0.8 m

The moment of inertia of the rod about its center of mass is,

**I _{rod} = (1/12) x Ml^{2}** = (1/12) x 3 x (0.8)

^{2}=

**0.16 kg m**

^{2}The mass of the sphere (M) = 5 kg, radius of the sphere (R) = 10 cm = 0.1 m

The moment of inertia of the sphere about its center of mass is,

**I _{c} = (2/5) MR^{2}**

But the moment of inertia of the sphere about the geometric center of the structure is,

**I _{sph} = I_{c} + Md^{2}**

where, d = 40 cm + 10 cm = 50 cm = 0.5 m

**I _{sph} = (2/5) MR^{2} + Md^{2}**

I_{sph} = (2/5) x 5 x (0.1)^{2} + 5 x (0.5)^{2} = **1.27 kg m ^{2}**

As there is one thin rod and 2 similar solid spheres,

Total Moment of Inertia **(I _{total})= I_{rod} + (2 x I_{sph})**

I_{total} = (0.16) x (2 x 1.27) = **2.7 kg m ^{2}**

Hence, the moment of inertia of the given construction figure is 2.7 kg m^{2}.

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