 # Moment Of Inertia Of A Semicircle

Moment of inertia of a semicircle is generally expressed with the following equation;

 I = π R4 / 8

## How To Find The Moment Of Inertia Of A Semicircle

In order to find the moment of inertia of a semicircle, we need to recall the concept of deriving the moment of inertia of a circle. The concept can be used to easily determine the moment of inertia.

1. We will first begin with recalling the expression for a full circle.

I = πr4 / 4

In order to find the moment of inertia, we have to take the results of a full circle and basically divide it by two to get the result for a semicircle.

Now, in a full circle because of complete symmetry and area distribution, the moment of inertia relative to the x-axis is the same as the y-axis.

Ix = Iy = ¼ πr4

M.O.I relative to the origin, Jo = Ix + Iy = ¼ πr4 + ¼ πr4 = ½ πr4

Now we need to pull out the area of a circle which gives us;

Jo = ½ (πr2) R2

Similarly, for a semicircle, the moment of inertia of the x-axis is equal to the y-axis. Here, the semi-circle rotating about an axis is symmetric and therefore we consider the values equal. Here the M.O.I will be half the moment of inertia of a full circle. Now this gives us;

Ix = Iy = ⅛ πr4 = ⅛ (Ao) R2 = ⅛ πr2) R2

Now to determine the semicircle’s moment of inertia we will take the sum of both the x and y-axis.

M.O.I relative to the origin, Jo = Ix + Iy = ⅛ πr4 + ⅛ πr4 = ¼ πr4

## Derivation

We will basically follow the polar coordinate method. Now we define the coordinates using the polar system. We get;

z = r sin θ

y = r cos θ

2. Next, we determine the differential area by stating the area of the element. It is given as;

Area of the sector, ABCD = (r⋅d θ) ⋅ dr = r ⋅ drd θ

The centroid of this elemental area from x-axis = y sin θ

3. We will now determine the first moment of inertia about the x-axis. We get;

Ix = oπoR y ⋅ dA

= oπoR r sin θ ⋅ rdrd θ

= oπ sin θ ( oR r2dr) d θ

= oπ R3 / 3 sin θ d θ

= R3 / 3 [cos θ]oπ

= 2R3 / 3

Meanwhile, Ix = A = π / 2 R2 y

We get,

π / 2 R2 y = 2R3 / 3

y = 4R / 3π

Now to find the second moment of inertia.

IXX = oπoR y2 dA

Ixx = oπ sin2θ [ oR r3dr ] d θ

Ixx = oπ sin2θ [ r4 / 4]or d θ

Ixx = r4 / 4 oπ sin2

Applying trigonometric identity: sin2θ = 1-cos 2 θ / 2 we calculate the integral.

Ixx = r4 / 8 oπ (1 – cos 2 θ / 2) d θ

Ixx = r4 / 8 [ θ – sin 2 θ / 2)]oπ

Ixx = πr4 / 8