Moment of Inertia

What is the Moment of Inertia?

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. SI unit of moment of inertia is kg.m2. It is also known as the angular mass or rotational inertia.

Moment of inertia formula:

General form: I = mr2

Integral form: I = ∫dI = ∫[0→M] r2 dm

The role of the moment of inertia is the same as mass (m) 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 body and a particular rotation axis.

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

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

Moment of Inertia depends on

  • 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)

Also Read: Rotational Motion

Moment of Inertia of a System of Particles

It is given by, from equation (1)

I = ∑mi ri2

where ri is the perpendicular distance from the axis to the ith particle, which has mass mi.

Moment of Inertia of a System of Particles

Moment of Inertia of a System of Particles

Moment of Inertia of Rigid Bodies

Moment of Inertia of Rigid Bodies

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 on how to calculate the moment of inertia is given below.

Moment of Inertia of a uniform rod about a perpendicular bisector

moment of inertia rod centre Calculation of Moment of Inertia

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

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

I=ML2/12

Moment of inertia of a circular ring about its axis

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.

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.

Rectangular plate about a line parallel to an edge and passing through the centre

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.

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.

If the mass element is chosen parallel to the length of the plate, then the result would be,

I = Mb2 /12

Moment of Inertia of a uniform circular plate about its axis

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.

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

Moment of Inertia of a thin spherical shell or a uniform hollow sphere

Moment of Inertia of a thin spherical shell or a 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.

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

Integration by substitution method,

Take u = cos θ dθ

Then, du = – sin θ dθ

Changing limits, when θ = 0, u = 1

When, θ = π, u = -1

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.

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

Moment of Inertia for different objects

Moment of Inertia for different objects

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 axes, 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 Axes 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

Parallel Axes Theorem

I = Icm + Md2

Where d is the distance between the two axes.

Parallel Axes Theorem

Perpendicular Axes Theorem

This theorem is applicable only to the plane bodies. Let X and Y-axes be chosen in the plane of the body and Z-axis perpendicular to this plane, three axes being mutually perpendicular. Then mathematically,

Perpendicular Axes Theorem

Example:

Perpendicular Axes Theorem

Radius of Gyration

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

I = Mk2

then 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.

Example: Radius of gyration for a solid sphere about its axis is

 

Type 1: Moment of Inertia of Compound bodies

Type 2: Moment of Inertia of rigid bodies with the cavity

Perpendicular Axes Theorem

Perpendicular Axes Theorem

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. 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 is :

Perpendicular Axes Theorem

Perpendicular Axes Theorem

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