Moment of inertia is an essential concept when it comes to physics problems involving mass in rotational motion, as it is commonly asked in such questions. It is primarily used to calculate angular momentum. In the following paragraphs, we will dive deeper into this topic.

What is Moment of Inertia?

Moment of Inertia (also known as angular mass or rotational inertia) is a quantity that expresses the body’s resistance to angular acceleration. It is calculated by summing the product of the mass of each particle with the square of its distance from the axis of rotation. In simpler terms, it is the amount of torque needed for a specific angular acceleration in a rotational axis. The SI unit of moment of inertia is kg m².

Moment of inertia (MOI) is typically determined relative to a selected axis of rotation. It largely depends on the arrangement of mass around the axis of rotation, and the MOI can differ depending on the chosen axis.

Table of Contents

• Moment of Inertia Formula
• What are the Factors on which Moment of Inertia Depends?
• Moment of Inertia of a System of Particles
• Moment of Inertia of Rigid Bodies
• Calculation of Moment of Inertia
• Moment of Inertia for Different Objects
• Parallel Axes Theorem
• Radius of Gyration
• Solved Examples

Moment of Inertia Formula

In general form, the Moment of Inertia is expressed as I = m × r²

Where,

$$m = \sum m \times v$$

r = Distance from the axis of rotation.

Integral form: $$I = \int_{0}^{M} r^2\ dm$$

⇒ The dimensional formula of the moment of inertia is given by, M¹L²T⁰.

The role of moment of inertia is analogous to that of mass in linear motion. It is a measure of the resistance of a body to a change in its rotational motion and is constant for a particular rigid frame and a specific axis of rotation.

Moment of inertia, $$I = \sum m_i r_i^2$$ (1)

$K=½\ast I\ast {\omega }^2$ (2)

What Factors Affect Moment of Inertia?

The factors which affect the moment of inertia are:

• The Density of the Material
• Shape and Size of the Body
• Axis of Rotation (Distribution of Mass Relative to the Axis)

We can further categorize rotating body systems as follows:

1. Discrete (System of Particles)

2. Continuous (Rigid body)

Moment of Inertia of a System of Particles

The moment of inertia of a system of particles is denoted by

$$I = \sum m_i r_i^2$$ [from equation (1)]

The perpendicular distance from the axis to the $i^{th}$ particle, which has mass m_i, is denoted by r_i.

Example:

Moment of Inertia of Rigid Bodies

The moment of inertia of a continuous mass distribution can be found by using integration. If we divide the system into an infinitesimal element of mass dm and denote the distance from the mass element to the axis of rotation as x, then the moment of inertia is:

I = ∫r²dm (3)

Calculation of Moment of Inertia

A Step-by-Step Guide for Calculating the Moment of Inertia is Given Below:

Moment of Inertia of a Uniform Rod about a Perpendicular Bisector

Consider a uniform rod of mass M and length L, and the moment of inertia should be calculated about the bisector AB with origin at 0.

The mass element ‘dm’ considered is between x and x + dx from the origin.

As the rod is uniform, the linear mass density remains constant.

M/L = dm/dx

$$\frac{dM}{dx} = \frac{M}{L}$$

Moment of inertia of dm

dI = dm x 2

$$dI = \frac{M}{L} x^2\cdot dx$$

$$I = \frac{-L}{2}\int_{-L/2}^{L/2} dI = \frac{M}{L} \times \frac{-L}{2}\int_{-L/2}^{L/2} x^2dx$$

The element covers the entire rod, as x changes from -L/2 to +L/2, where x = -L/2 is the left end of the rod.

$$I = \frac{M}{L} \times \left[\frac{x^3}{3} + \frac{L}{2} - \frac{L}{2}\right]$$

$$I = \frac{ML^2}{12}$$

Therefore, the moment of inertia of a uniform rod about a perpendicular bisector $$(I) = \frac{ML^2}{12}$$.

Moment of Inertia of a Circular Ring about its Axis

A line perpendicular to the plane of the ring, with its centre as the origin, is considered. The radius of the ring is denoted as $R$ and its mass as $M$. All elements are located at a distance of $R$ from the axis of rotation.

Linear mass density remains constant.

M/(2π) = (dm)/(dθ)

$$dm = \frac{M}{2\pi} \times d\theta$$

$$I = \int{R^2} dm = R^2 \int_0^{2\pi} \frac{M}{2\pi} d\theta$$

Limits: θ = 0 to 2π (inclusive)

$I=[R^2\ast M/(2\pi )]\ast (\theta /2\pi {)}_0$

Therefore, the moment of inertia of a circular ring about its axis (I) = M R².

Note that in one-dimensional bodies, if it is uniform, their linear mass density (M/L) remains constant. Similarly, for two-dimensional and three-dimensional bodies, surface density (M/A) and volume density (M/V) remain constant, respectively.

Moment of Inertia of a Rectangular Plate about a Line Parallel to an Edge and Passing Through the Centre

The mass element can be taken between x and x + dx from the axis AB.

As the plate is uniform, the mass-to-area ratio (M/A) is constant.

M/A = dm/da

M/(lxb) = dm/dx * b

dm = (M/lb) × b × dx = (M/l) dx

$$I = \int x^2 , dm = \frac{M}{l} \times -\frac{l}{2}\int_{-\frac{l}{2}}^{\frac{l}{2}} x^2 , dx$$

Limits: The left end of the rectangular plate is at x = -l/2, and the plate is covered from x = -l/2 to x = +l/2.

$$ I = \frac{M}{l} \left[ \frac{x^3}{3} - \frac{l}{2} + \frac{l}{2} \right] = \frac{Ml^2}{12}$$

Therefore, the moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre $$ (I) =  \frac{Ml^2}{12}$$.

Note: If the mass element is chosen parallel to the length of the plate, then the moment of inertia would be, $$I = \frac{Mb^2}{12}$$.

Moment of Inertia of a Uniform Circular Plate about its Axis of Rotation

Let the mass of the plate be M and the radius be R. The centre is at O and the axis is perpendicular to the plane of the plate. The mass element considered is a thin ring between x and x+dx with thickness dx and mass dm.

The surface mass density of the plane is constant as it is uniform.

M/A = dm/da

M/(πR^2) = dm/(2πx∙dx)

$$ I = \int x^2 , dm = \frac{2M}{R^2} \times \int_0^R x^3 dx$$

Limits: Taking the area of all mass elements from x=0 to x=R covers the whole plate.

I = MR²/2

Therefore, the moment of inertia of a uniform circular plate about its axis $$ (I) = \frac{MR^2}{2}$$.

Moment of Inertia of a Thin Spherical Shell or Uniform Hollow Sphere

Let M and R be the mass and the radius of the sphere, O at its centre and OY be the given axis. The mass is spread over the surface of the sphere and the inside is hollow.

Let us consider the radii of a sphere at an angle θ and at an angle θ + dθ with the axis OY, and take an element (thin ring) of mass dm with radius $R\sin \theta$ as we rotate these radii about OY. The width of this ring is $Rd\theta$ and its periphery is $2\pi R\sin \theta$.

The surface mass density (M/A) of the hollow sphere is constant, due to its uniformity.

$$M/A = \frac{dm}{da}$$

$$\frac{M}{4\pi R^2} = \frac{dm}{2\pi R \sin{\theta} \cdot R \cdot d\theta}$$

$[M/2]\times sin\theta d\theta =[M/4\pi R^2]\times 2\pi R^2.sin\theta d\theta$

$$ I = \int x^2 , dm = \int_0^{\pi} \left(R \sin \theta \right)^2 \times \left[ \frac{M}{2} \right] \sin \theta , d\theta$$

As θ increases from 0 to π, the elemental rings completely cover the spherical surface.

$$ I = \frac{MR^2}{2} \times 0 \int_{0}^{\pi} \sin^3 \theta \ d\theta = \frac{MR^2}{2} \times 0 \int_{0}^{\pi} [\sin^2 \theta \times \sin \theta] \ d\theta$$

$$\int_0^{\pi} (1 - \cos^2 \theta) \sin\theta d\theta$$

Now, by integrating the above equation using the substitution method, we get,

u = -sin θ dθ

Then, $$du = -\sin\theta\ d\theta$$

When θ = 0, changing the limits results in u = 1

When, $$\theta = \pi$$, $$u = -1$$

$I=\frac{M{R}^2}{2}\times {\int }_{-1}^1(1-u^2)du$

$=\frac{M{R}^2}{2}\times {\int }_{-1}^1(u^2-1)du$

$[M{R}^2/2]\times [u^3/3-u{]}^1-1$

$[M{R}^2/2]\times [4/3]=[2M{R}^2/3]$

Therefore, the moment of inertia of a thin spherical shell and a uniform hollow sphere (I) is equal to 2MR²/3.

Moment of Inertia of a Uniform Solid Sphere

Let us consider a sphere of radius R and mass M. A thin spherical shell of radius x, mass dm and thickness dx is taken as a mass element. Volume density (M/V) remains constant as the solid sphere is uniform.

M/V = dm/dV

$[4/3\times \pi R^3]=[4\pi x^2.dx]$

$M/(4/3\pi R^3)\times 4\pi x^{2}dx=[3M/R^3]\times 2dx$

$I=\int dI=\frac{2}{3}\times \int dm\cdot x^2$

$2/3\times \int [3M/R^{3}dx]x^4$

$$\frac{2M}{R^3} \times 0 \int_R^4 x^4 dx$$

As x increases from 0 to R, the elemental shell covers the entire spherical surface.

$$I = \frac{2M}{R^3}\left[\frac{x^5}{5}\right]R^0$$

(2M/3) × (5R/5)

Therefore, the moment of inertia of a uniform solid sphere $$(I) = \frac{2MR^2}{5}$$.

Calculating Moment of Inertia for Various Objects

As we can see in the table above, the moment of inertia is dependent on the axis of rotation. So far, we have calculated the moment of inertia of the objects when the axis is passing through their centre of masses (Icm). If we choose two different axes, we will notice that the object resists the rotational change differently. To find the moment of inertia through any given axis, the following theorems are useful.

Parallel Axis Theorem

Perpendicular Axis Theorem

Parallel Axis Theorem

The moment of inertia of an object about an axis through its centre of mass is the minimum moment of inertia for an axis in that direction in space. The moment of inertia about an axis parallel to that axis through the centre of mass is given by,

$I=Icm+M\cdot d^2$

Where d is the distance between the two axes.

Radius of Gyration

If the moment of inertia (I) of a body of mass m about an axis can be written in the form:

$$I = \mathrm{Mk}2$$

The radius of gyration, denoted by k, is the distance from the given axis of rotation at which the entire mass of the body can be assumed to be concentrated, thus keeping its rotational inertia unchanged.

The radius of gyration for a solid sphere about its axis is:

$k=\surd [2/5]\times R$

Solved Example

1. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc can be calculated by using the formula $$I = \frac{1}{2}MR^2$$, where M is the mass of the disc and R is the radius of the disc.

Solution:

The moment of inertia of the part removed about the axis passing through the centre of mass and perpendicular to the plane of the disc = $I_{cm}+m{d}^2$

$[m\times (R/3{)}^2]/2+m\times [4R^2/9]=m{R}^2/2$

Therefore, the moment of inertia of the remaining portion = the moment of inertia of the complete disc minus the moment of inertia of the removed portion

$9m{R}^2/2-m{R}^2/2=8m{R}^2/2$

Therefore, the moment of inertia of the remaining portion (I_remaining) = 4mR².

2. What is the moment of inertia of the system with two balls of masses 700 grams and 500 grams connected by a rod about AB?

Given

Given

The rotation axis is AB

m_X = 700 grams = 0.7 kg

m_Y = 500 grams = 0.5 kg

r_X = 10cm = 0.1m

r_Y = 40cm = 0.4m

Solution:

I = mX rX² + mY rY²

I = (0.7)× (0.1)² + (0.5)× (0.4)²

I = (0.7) x (0.01) + (0.5) x (0.16)

I = 0.007 + o.08

I = 0.087 kg m²

Moment of inertia of the system is 0.087 kg m²

3. What is the moment of inertia of the two balls connected by a rod, as shown in the figure below (Ignore rod’s mass)?

Given:

m_X = 0.3 kg = 300 grams

m_Y = 0.5 kg = 500 grams

r_X = 0 m

r_y = 0.3m

Solution:

I = mXr² + mYr²

I = 0.09 + 0.045

I = 0 + 0.045

I = 0.045 kg m²

The moment of inertia of the system is 0.045 kg m²

4. What is the moment of inertia of the two 200 gram balls about the axis of rotation (Ignoring the mass of the cord)?

Given

Mass of ball = m1 = m2 = m3 = m4 = 200 grams = 0.2 kg

Distance between ball and the axis of rotation (r1) = 40cm = 0.4 m

Distance between ball 2 and the axis of rotation (r2) = 40 cm = 0.4 m

Distance between ball 3 and the axis of rotation (r₂) = 40 cm = 0.4 m

Distance between ball 4 and the axis of rotation (r₂) = 40 cm = 0.4 m

Solu:

I = m1r²₁ + m2r²₂ + m3r²₃ + m4r²₄

I = 0.2 × (0.4)² + 0.2 × (0.4)² + 0.2 × (0.4)² + 0.2 × (0.4)²

I = 0.128

I = 0.128 kg·m²

Moment of inertia of the balls about the axis: 0.128 kg m²

Frequently Asked Questions on Moment of Inertia

Does the Moment of Inertia of a Rigid Body Change with the Speed of Rotation?

The moment of inertia of a rigid body depends only on the distribution of mass of the body about the axis of rotation and is independent of the speed of rotation. Hence, the moment of inertia of a rigid body does not change with the speed of rotation.

Which shape will have a larger moment of inertia: a disc or a hollow and thin cylinder of the same radius?

The moment of inertia of a hollow cylinder will be larger than that of a disc due to its mass being further away from the axis of rotation.

The moment of inertia is a scalar quantity.

The moment of inertia is a scalar quantity.

What is the Moment of Inertia of a Solid Sphere About its Axis of Rotation?

$I=(\mathrm{⅖})M{R}^2$