 # Moment of Inertia

## What is Moment of Inertia?

The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation. The SI unit of moment of inertia is kg m2. It is also known as the angular mass or rotational inertia.

### Moment of Inertia Formula

In General form Moment of Inertia is expressed as: I = mr

and, Integral form: I = ∫dI = ∫[0→M] r2 dm

⇒ The dimensional formula of the moment of inertia is given by, MLT0.

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

Moment of inertia, I = ∑mi ri2. . . . . . . (1)

Kinetic Energy, K = ½ I ω2 . . . . . . . . . (2)

### What are the Factors on which Moment of Inertia Depends?

The moment of inertia depends on the following factors,

• The density of the material
• Shape and size of the body
• Axis of rotation (distribution of mass relative to the axis)

We can 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 given by,

I = ∑ mi ri2 [from equation (1)]

where ri is the perpendicular distance from the axis to the ith particle which has mass mi. Example: ### Moment of Inertia of Rigid Bodies The moment of inertia of continuous mass distribution is found by using the integration technique. If the system is divided into an infinitesimal element of mass ‘dm’ and if ‘x’ is the distance from the mass element to the axis of rotation, the moment of inertia is:

I = ∫ r2 dm . . . . . . (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. Origin is at 0.

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

As the rod is uniform, mass per unit length (linear mass density) remains constant.

∴  M/L = dm/dx

dm = (M/L)dx

Moment of inertia of dm = dI = dm x2

dI = (M/L) x2.dx

I = ∫+L/2-L/2 dI = M/L × -L/2+L/2 xdx

Here, x = -L/2 is the left end of the rod and ‘x’ changing from –L/2 to +L/2, the element covers the entire rod.

I = M/L × [x3/3]+L/2-L/2

I = R2 × M/2π × [θ]0

Therefore, the moment of inertia of a uniform rod about a perpendicular bisector (I) = ML2/12.

### Moment of Inertia of a Circular Ring about its Axis The line perpendicular to the plane of the ring through its centre – Consider the radius of the ring as R and its mass as M. All the elements are at the same distance from the axis of rotation, R.

Linear mass density is constant.

∴ M/2πR = dm/dθ

dm = M/2π × dθ

I = ∫ R2 dm = R2 0 [M/2π] dθ

⇒ Check: Limits and Continuity

Limits: θ = 0 to 2π includes the whole mass of the ring

∴ I = R2 [M/2π] × [θ]0

Therefore, the moment of inertia of a circular ring about its axis (I) = MR2.

⇒ Note that in one-dimensional bodies, if it’s uniform, their linear mass density (M/L) remains constant. Similarly, for 2D and 3D, M/A (surface density) and M/V (volume density) 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, M/A is constant.

M/A = dm/da

M/[l×b] = dm/dx.b

dm = M/lb × b × dx = M/Ll dx

I = ∫ x2 dm = M/l × -l/2+l/2 [x2] dx

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

I = M/l [x3/3]-l/2+l/2 = Ml2/12

Therefore, the moment of inertia of a rectangular plate about a line parallel to an edge and passing through the centre (I) =  Ml2/12.

⇒ Note: If the mass element is chosen parallel to the length of the plate, then the moment of inertia would be, I = Mb2/12.

### Moment of Inertia of a Uniform Circular Plate about its Axis 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.

As the plane is uniform, the surface mass density is constant.

M/A = dm/da

M/πR2 = dm/[2π.x.dx]

I = ∫ x2 dm = 2M/R2 × 0R x3 . dx

Limits: As we take the area of all mass elements from x=0 to x=R, we cover the whole plate.

I = 2M/R2 [x4/4]R0 = 2M/R2 × R4/4 = MR2/2

Therefore, the moment of inertia of a uniform circular plate about its axis (I) = MR2/2.

### Moment of Inertia of thin Spherical Shell or Uniform Hollow Sphere Let M and R be the mass and the radius of the sphere, O its centre and OX the given axis. The mass is spread over the surface of the sphere and the inside is hollow.

Let us consider radii of the sphere at an angle θ and at an angle θ+dθ with the axis OY and the element (thin ring) of mass dm with radius Rsinθ is taken as we rotate these radii about OY. The width of this ring is Rdθ and its periphery is 2πRsinθ.

As the hollow sphere is uniform, the surface mass density (M/A) is constant.

M/A = dm/da

M/4πR2 = dm/[2π × Rsinθ . Rdθ]

dm = [M/4πR2] × 2πR. sinθ dθ = M/2 × sinθ dθ

I = ∫ x2 dm = 0π (R sin θ)2 × M/2 sinθ dθ

Limits: As θ increases from 0 to π, the elemental rings cover the whole spherical surface.

I = MR2/2 × 0π sin3 θ dθ = MR2/2 × 0π [sin2 θ . sin θ] × dθ

= MR2/2 0π (1 – cos2 θ) sin θ × dθ

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

Take u = cos θ dθ

Then, du = – sin θ dθ

Changing limits, when θ = 0, u = 1

When, θ = π, u = -1

I = MR2/2 × 1-1 (1 – u2) (-du)

= MR2/2 × 1-1 (u2 – 1) (du)

= MR2/2 × [u3/3 – u]1-1

= MR2/2 [-2/3 + 2] = MR2/2 × [4/3] = 2MR2/5

Therefore, the moment of inertia of thin spherical shell and uniform hollow sphere (I) = 2MR2/5.

### 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

M/[4/3 × πR3] = dm/[4πx2.dx]

dm = M/[4/3 × πR3] × 4πx2 dx = [3M/R3] x2 dx

I = ∫ dI = 2/3 × ∫ dm . x2

= 2/3 × ∫ [3M/R3 dx] x4

= 2M/R3 × 0R x4 dx

Limits: As x increases from 0 to R, the elemental shell covers the whole spherical surface.

I = 2M/R3 [x5/5]R0

= 2M/R3 × R5/5

Therefore, the moment of inertia of a uniform solid sphere (I) = 2MR2/5.

### Moment of Inertia for different Objects Moment of Inertia

As we note in the table above, the moment of inertia depends upon the axis of rotation. Whatever we have calculated so far are the moment of inertia of those objects when the axis is passing through their centre of masses (Icm). Having chosen, two different axis you will observe that the object resists the rotational change differently. Therefore, 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 + Md2

Where d is the distance between the two axis.

Read more on Parallel and Perpendicular axis theorem If the moment of inertia (I) of a body of mass m about an axis be written in the form:

I = Mk2

Here, k is called the radius of gyration of the body about the given axis. It represents the radial distance from the given axis of rotation where the entire mass of the body can be assumed to be concentrated so that its rotational inertia remains unchanged.

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

Mk2 = 2MR2/5, or k = √[2/5] × R.

## Solved Example

From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as shown in the figure. Calculate 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. Solution:

The moment of inertia of removed part abut the axis passing through the centre of mass and perpendicular to the plane of the disc = Icm + md2

= [m × (R/3)2]/2 + m × [4R2/9] = mR2/2

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

= 9mR2/2 – mR2/2 = 8mR2/2

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