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_{i}^{2}

- 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 m
^{2}.

**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} + Mh^{2}**

- 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} + Mh^{2}**

We know that, the moment of inertia of a ring about its diameter is I_{c} = 1/2MR^{2}

Thus, I_{o }= 1/2MR^{2 }+ MR^{2}

**I _{o} = 3/2 MR^{2}**

Thus, the moment of inertia of a ring about tangent in its plane is 3/2MR^{2}.