Parallel Axis Theorem

Mathematically, the parallel axis theorem is written as

I_O = I_C + Md²

I_C is the moment of inertia about an axis passing through the center of mass.

I_O is the moment of inertia about an arbitrary axis parallel to the axis passing through the center of mass.

M is the object’s mass.

d is the distance between the two parallel axes.

In the image above, AB represents the axis passing through the center of mass C of the rigid body. A’B’ represents the axis passing through any arbitrary point D. A’B’ is at a distance d from AB. Consider an infinitesimal mass dm at point E. 

We have the following quantities.

M: Mass of the body

dm: Infinitesimal mass of point D

I_C: Moment of inertia about AB

I_O: Moment of inertia about A’B’

h: Distance between the two parallel axes AB and A’B’

rₒ: Distance between points C and D

r: Distance between points O and D

x: Distance between points C and C’

The expressions for the moment of inertia are as follows: 

Consider triangle OC’D.

(OD)² = (OC’)² + (C’D)²

=> r² = (d + x)² + (C’D)²

=> r² = (d + x)² + r_o² – x²

=> r² = d² + x² + 2dx + r_o² – x²

=> r² = r_o² + d² + 2dx

Multiplying both sides by dm and integrating

The term \( \int x dm \) is zero since the integral of the moments of infinitesimal masses about the center of mass is always zero when the body is in equilibrium. Therefore, \( \int x dm = 0 \), and we get, 

I_O = I_C + Md²

Thus, we have derived the parallel axis theorem.

The parallel axis theorem can determine the moment of inertia of a given rigid body about any axis. Let us look at a few examples.

1. Uniform Rod

The moment of inertia I_C of a thin rod of mass M and length L about an axis passing through its center is

Let us apply the parallel axis theorem to find the moment of inertia about its edge I_O. We know that the edge is at a distance L/2 from the center; d = L/2.

2. Ring

The moment of inertia I_C of a ring of mass M and radius R about an axis passing through its center and perpendicular to the surface is

I_C = MR²

The edge is at a distance R from the center. Therefore, the moment of inertia about an axis tangential to the edge and perpendicular to the surface is

I_O = MR² + MR²

=> I_O = 2MR²

3. Disc

The moment of inertia I_C of a disc of mass M and radius R about an axis passing through its center and perpendicular to the surface is

The edge is at a distance R from the center. Therefore, the moment of inertia about an axis tangential to the edge and perpendicular to the surface is

4. Sphere

The moment of inertia I_C of a solid sphere of mass M and radius R about its central axis is

The surface is at a distance R from the center of the sphere. Therefore, the moment of inertia about an axis tangential to the surface is

Problem 1. A body of mass 55 kg has a moment of inertia of 35 kg m² along an axis perpendicular to its center of gravity. What is its moment of inertia along a different axis parallel to and 40 cm from the center of mass axis?

Solution

Given

I_C = 35 kg m², M = 55 kg, and d = 40 cm = 0.4 m

From the parallel axis theorem,

I_O = I_C + Md²

=> I_O = 35 kg m² + 55 kg x (0.4 m)²

=> I_O = 43.8 kg m²

Problem 2. Find the moment of inertia of a uniform rod with I_C = 0.06 kg m², L = 0.3 m, and mass = 1.50 kg about an axis perpendicular to the rod and passing through a point at 1/8 of the length of the rod.

Solution

Given I_C = 0.06 kg m², L = 0.3 m, and mass = 1.50 kg

From the parallel axis theorem

I_O = I_C + Md²

Since the point is at a distance 1/8 of the length of the rod, we have

d = L/2 – L/8 = (3/8) L

=> d = (3/8) x 0.3 m = 0.1125 m

Therefore, the moment of inertia about an axis perpendicular to the rod and passing through the point in question is

I_O = 0.06 kg m² + 1.50 kg x (0.1125 m)²

=> I_O = 0.079 kg m²