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
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 = o∫πo∫R y ⋅ dA
= o∫πo∫R r sin θ ⋅ rdrd θ
= o∫π sin θ ( o∫R r2dr) d θ
= o∫π R3 / 3 sin θ d θ
= R3 / 3 [cos θ]oπ
= 2R3 / 3
Meanwhile, Ix = A = π / 2 R2 y
π / 2 R2 y = 2R3 / 3
y = 4R / 3π
Now to find the second moment of inertia.
IXX = o∫πo∫R y2 dA
Ixx = o∫π sin2θ [ o∫R r3dr ] d θ
Ixx = o∫π sin2θ [ r4 / 4]or d θ
Ixx = r4 / 4 o∫π sin2 dθ
Applying trigonometric identity: sin2θ = 1-cos 2 θ / 2 we calculate the integral.
Ixx = r4 / 8 o∫π (1 – cos 2 θ / 2) d θ
Ixx = r4 / 8 [ θ – sin 2 θ / 2)]oπ
Ixx = πr4 / 8