Question

State and prove the theorem of the parallel axis about the moment of inertia.

Answer 1

A body's moment of inertia along an axis is equal to the product of two things: Its moment of inertia about a parallel axis through its center of mass; and the product of the body's mass and the square of the distance between the two axes. This is known as the parallel axis theorem.

Proof: Let I_CM represent a body of mass M moment of inertia (MI) about an axis passing through its centre of mass C and let I stand for that body's MI about a parallel axis passing through any point O. Let h represent the separation of the two axis.

Think about the body's minuscule mass element dm at point P. It is perpendicular to the rotation axis through point C and to the parallel axis through point O, with a corresponding perpendicular distance of OP. CP2 dm is the MI of the element about the axis through C. As a result, `I_(CM) = int CP^2 dm` is the body's MI about the axis through the CM. In a similar vein, `I = int OP^2 dm` is the body's MI about the parallel axis through O.

Theorem of parallel axis

Draw PQ perpendicular to OC produced, as shown in the figure. Then, from the figure,

`I = int OP^2 dm`

= `int (OQ^2 + PQ^2) dm`

= `int [(OC + CQ)^2 + PQ^2] dm`

= `int (OC^2 + 2OC.CQ + CQ^2 + PQ^2) dm`

= `int (OC^2 + 2OC.CQ + CP^2)dm`   ...(∵ CQ² + PQ² = CP²)

= `int OC^2 dm + int 2OC.CQ dm + int CP^2 dm`

= `OC^2 int dm + 2OC int CQ dm + int CP^2 dm`

Since OC = h is constant and `int dm = M` is the mass of the body,

`I = Mh^2 + 2h int CQ  dm + I_(CM)`

The integral `int CQ  dm` now yields mass M times a coordinate of the CM with respect to the origin C, based on the concept of the centre of mass. This position and the integral are both zero because C is the CM in and of itself.

∴ I = I_CM +Mh²

This proves the theorem of the parallel axis.