# 2nd moment of area + centroid

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## Homework Statement

I need to get the centroid of this shape

http://img713.imageshack.us/img713/5607/section.jpg [Broken]

it's a semi-circular section basically. On the left most side, the distance from the origin to to where it cuts the y-axis is 'R' and the distance from the x-axis to the end of the upper arc is 'r'.

## Homework Equations

$$I_x = \int y^2 dA$$

$$I_y= \int x^2 dA$$

$$A \bar{x}= \int xy dA$$

$$A \bar{y} = \int y dA$$

## The Attempt at a Solution

I just need to know if my integrals are set up properly

The equation of the entire circle would have been x2+y2=R2.

$$A= \int y dx = \int_{0} ^{r} \sqrt{R^2-x^2}$$

Assuming my 'A' is correct

dA=y dx

$$A \bar{x} = \int xy dA = \int xy^2 dx = \int_{0} ^{r} x(R^2-x^2) dx$$

my object is symmetrical about the x-axis so [itex]\bar{y} = 0[/tex]

$$I_x = \int y^2 dA = \int y^3 dx = \int_{0} ^{r} (R^2-x^2)^{\frac{3}{2}}dx$$

$$I_y = \int x^2 dA = \int x^2 y dx = \int_{0} ^{r} x^2\sqrt{R^2-x^2)dx$$

I think my limits may be wrong, but the integrands should be correct.

Also, if someone can just link me to a site with the centroid and second moment of area of this shape that would be helpful as I don't need to actually derive it in my report, I just need to use the result.

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## Answers and Replies

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LCKurtz
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## Homework Statement

I need to get the centroid of this shape

http://img713.imageshack.us/img713/5607/section.jpg [Broken]

I just need to know if my integrals are set up properly

The equation of the entire circle would have been x2+y2=R2.

$$A= \int y dx = \int_{0} ^{r} \sqrt{R^2-x^2}$$
Typo you forgot the dx.
Assuming my 'A' is correct

dA=y dx

$$A \bar{x} = \int xy dA = \int xy^2 dx = \int_{0} ^{r} x(R^2-x^2) dx$$

my object is symmetrical about the x-axis so [itex]\bar{y} = 0[/tex]
These look OK.

$$I_x = \int y^2 dA = \int y^3 dx = \int_{0} ^{r} (R^2-x^2)^{\frac{3}{2}}dx$$
This one is wrong. If you set it up as this double integral

$$\int_0^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} y^2\ dydx = 2\int_0^R\int_{0}^{\sqrt{R^2-x^2}} y^2\ dydx$$

and work the inner integral, you will see why.
$$I_y = \int x^2 dA = \int x^2 y dx = \int_{0} ^{r} x^2\sqrt{R^2-x^2)dx$$
and the last one is OK.

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Homework Helper
thanks very much, I will have to recheck the double integral one.

Homework Helper
I need to also find the mass moment of inertia of that section if it were rotated about the x-axis.

I assumed an elemental section such that x2+y2=R2

dI= y2 dm =y2 ρ dV = ρ y2 (πy2 dx) =ρπy4dx.

$$I = \rho \pi \int_{0} ^{r} (R^2 -x^2)dx$$

is this correct? This somehow gives me answers like 2400 kgm2 (ρ=7850 kg/m3) with the total integrand being like 0.9. The section is very small in fact and is hollow, I don't think it should be that much.