Parallel Axes Theorem

Theorem of Parallel axes gives the moment of inertia of a body about any axis in terms moment of inertia of a body about a axis parallel to first axis.

• We know that, the weight of the any body is due to its inertia. The moment of inertia of a body about any axis is nothing but the sum of product of mass of that each particle and square of the distance of each particle from that axis of rotation.
• That means, I = ∑ m_i r_i²

 

• The moment of inertia of the body depends on the mass, shape and size of the body.
• It also depends on the mass distribution of particles about the axis of rotation of the body.
• And also, on the position of the axis of rotation of the body.
• The SI unit of moment of inertia of the body is kg m².

Statement:

Theorem of Parallel axes states that the moment of inertia of a body about any axis is equal to the sum of moment of inertia about a axis parallel to it and passing through centre of mass of the body and the product of the mass of the body & square of the perpendicular distance between those two parallel axes.

The following is neat labelled diagram which explains the parallel axes theorem.

• Thus, from figure we can write mathematical statement of parallel axes theorem as,

I_o = I_c + Mh²

• Where, I_o is the moment of inertia of a body about the axis passing through point O
• I_c is the moment of inertia of a body about the parallel axis passing through centre of mass C of the body.
• M is the total mass of the body
• And h is the perpendicular distance between the two parallel axis.

Applications:

• Parallel axis theorem can be used to find the moment of inertia of about an axis in case of any body.
• Parallel axes theorem can be used to find moment of inertia of a thin rod.
• It is used to find moment of inertia of a ring about a tangent in its plane.
• It is also used to find the moment of inertia of ring about a tangent perpendicular to its plane.

Example:

What is the moment of inertia of a ring about a tangent in its plane?

Solution:

Let I_c be the moment of inertia about the diameter of the ring as shown in figure.

And I_o be the moment of inertia of a ring about the tangent which is parallel to diameter and the distance  between two parallel axis is h = R.

Then by parallel axes theorem,

I_o = I_c + Mh²

We know that, the moment of inertia of a ring about its diameter is I_c = 1/2MR²

Thus, I_o = 1/2MR² + MR²

I_o = 3/2 MR²

Thus, the moment of inertia of a ring about tangent in its plane is 3/2MR².