 # 3-How to find the area and Cg for a semi-circle?

## Area and Cg for a semi-circle.

### Reference handbook 10.00 value for the area and Cg for a semi-circle.

There is a list of the common round shapes area and CG value. Our third case is the case of a semi-circle.

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### Area and Cg for a semi-circle- select an area dA.

We have a semi-circle with radius a, for which it is required to get the area and CG. We can find that there’s an internal axis Y that divides the semi-circle into two similar parts. Due to that symmetry, we expect that the Cg or the center of gravity will be located along the y axis at a certain distance y from the external axis x. We have used the radius of the semi-circle as equal to r.

### Area and Cg for a semi-circle-first moment of area about the x-axis.

For the area and Cg for a semi-circle about X-axis. We have two intersecting axes X&Y, we will select a small area dA that has a radius of ρ from the intersection of the two axes X and Y.

The first moment of area for the small area dA about the X-axis is the product of that area by the vertical distance to the X-axis. The angle θ is the angle enclosed between the Cg of the strip dA with the X-axis. The angle dθ is the enclosed angle by the strip dA.

The vertical distance is y which is equal to ρ*sin θ. The moment dMx=dA*(y)=(ρ*dρ*d θ)*(ρ*sin θ). It will be simplified to (ρ^2*dρ*sin θ*dθ).

For the first moment of area for the whole semi-circle, we will use double integration since we have to integrate for ρ from ρ=0 to ρ=r. The second integration os for the angle dθ from θ equals to zero to dθ equal to π or 180 degrees. dA*y=∬(ρ^2*dρ*sin θ*dθ).

This is the value of the first integration, ∫ρ^2*dρ from zero to r will be equal to 1/3*ρ^3 after substitution will lead to 1/3*(r^3-0)=r^3/3. While for the second integration ∫(sin θ*dθ)=-cos(θ), after substituting from zero to π. The value will be (- (cos(π)- cos(0))=(-(-1)-(1)=+2. The total value of the first moment of area for the semi-circle will be equal to =2*(r^3/3).

But the semi-circle area can be found from the integration of dA= ∬(ρ*dρ*dθ) from ρ=0 to ρ=a and for dθ from θ equals zero to dθ=π, the final expression for the area is dA=ρ^2*0.5*(θ), substitute to get A=0.50*(r^2-0)*(π-0)=0.50*π*r^2.

Divide A*y/A to get Y bar will lead to ybar Cy=2*(r^3/3)/0.5*π*a^2=4a/3*π. This includes that the Cg lies on the X-axis.

### Area and Cg for a semi-circle-first moment of area about the Y-axis.

For the area and Cg for a semi-circle about the y-axis. The first moment of area dA about the y’-axis is dMy=dA*(x)=(ρ*dρ*d θ)*(ρ*cos θ). it will be simplified to (ρ^2*dρ*cos θ*dθ).

For the first moment of area for the whole semi-circle, we will use double integration since we have to integrate for ρ from ρ=0 to ρ=r. The second integration os for the angle dθ from θ equals to zero to dθ equal zero to π or 180 degrees. dA*y=∬(ρ^2*dρ*cos θ*dθ).

The value of the first integration is equal to ∫ρ^2*dρ from zero to r=1/3*ρ^3 after substitution will lead to 1/3*(a^3-0)=a^3/3, while the ∫(cos θ*dθ)=-cos(θ), after substitute from zero to π. The value will be (+ (sin(π)-sin(0))=zero.

We have completed the subject of the area and Cg for a semi-circle.

The next post will be on how to estimate the area and Cg for a circular sector.

This is a link to a very useful site: Engineering statics open and interactive. Scroll to Top
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