Finding Centroid and Second Moment of Area for Semi-Circular Section

[rock.freak667]

Homework Statement

I need to get the centroid of this shape

http://img713.imageshack.us/img713/5607/section.jpg

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]<br /> <br /> $$I_x = \int y^2 dA = \int y^3 dx = \int_{0} ^{r} (R^2-x^2)^{\frac{3}{2}}dx$$<br /> <br /> $$I_y = \int x^2 dA = \int x^2 y dx = \int_{0} ^{r} x^2\sqrt{R^2-x^2)dx$$<br /> <br /> I think my limits may be wrong, but the integrands should be correct.<br /> <br /> 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.[/itex]

 

[LCKurtz]

Homework Statement

I need to get the centroid of this shape

http://img713.imageshack.us/img713/5607/section.jpg

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][/itex]

[itex] <br /> These look OK.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $$I_x = \int y^2 dA = \int y^3 dx = \int_{0} ^{r} (R^2-x^2)^{\frac{3}{2}}dx$$ </div> </div> </blockquote><br /> This one is wrong. If you set it up as this double integral<br /> <br /> $$\int_0^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} y^2\ dydx = 2\int_0^R\int_{0}^{\sqrt{R^2-x^2}} y^2\ dydx$$<br /> <br /> and work the inner integral, you will see why.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $$I_y = \int x^2 dA = \int x^2 y dx = \int_{0} ^{r} x^2\sqrt{R^2-x^2)dx$$ </div> </div> </blockquote><br /> and the last one is OK.[/itex]

 

[rock.freak667]

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

 

[rock.freak667]

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.