The Moment of Inertia of a Regular Polygon

The Polygon can be divided into n congruent triangles. Let us find the Moment of Inertia of one such triangle about the polygon’s center.

And since Moment of Inertia is a scalar quantity, multiplying it by n, will give us the Moment of Inertia of the entire polygon.

The triangle can be imagined to be made up of infinite number of rods. Let us take such a rod, of length 2x(figure), and width dy.

Let the Mass per unit area of the polygon (or, the triangle) be λ.

 

 Area of rod        =             2.x.dy

Mass of rod        =             λ.2.x.dy

By the parallel axis theorem,

Moment of Inertia of the rod about O    =             Moment of Inertia of rod about C + m.OC².

Where C is the Center of Mass of the rod and m is the mass of the rod (=2λ.x.dy).

Therefore,                                         

Which gives,

Now, since                                         

 We have

Let the total mass of the polygon   =  M

Therefore                          

Therefore,

Further,

Therefore,

Multiplying the above value of I by n, we have the Moment of Inertia for the entire polygon

Here L represents the length of one side of the polygon.

If the circum-radius R of the polygon is given, then putting

We have

Clearly for a circular disc we have n→ ∞,

the value of I becomes