Centroid of a shape with a half circle hole

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SUMMARY

The discussion focuses on calculating the centroid of a shape with a half-circle hole at the top. The area of the semicircle is established as (r^2*pi)/2, with a radius of 0.9 inches and a distance from the edge of the cross-section to the center of the semicircle being 1.5 inches. The x-coordinate of the centroid is determined to be 0 due to symmetry, while the y-coordinate can be calculated using the integral y_{cm} = \frac{\int_0^R{yw(y)dy}}{\frac{\pi R^2}{2}}, where w(y) = 2*\sqrt{R^2-y^2}. This method provides a definitive approach to finding the centroid of the semicircle.

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cathliccat
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I'm not sure if this is physics, but I thought I'd ask my question in case someone knows. In my effort to find the centroid of a cross section, I have a half-circle hole at the top of the shape. I know that the area is (r^2*pi)/2 and from the edge of cross-section to the center of the half-circle is 1.5 in. (The radius is .9 in.) How do I figure out the x and the y for the half-circle. Someone told me the x would be the distance 1.5 in. I'm not so sure about that. Thanks in advance if anyone knows what it is.
 
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It's difficult to understand the problem without a picture. Could you please post one?
 
picture

Here's the picture, I hope this works.
 

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I'm assuming that you're just asking about the semicircle here and that once you know that, you'll know how to find the centroid of the given cross section. Here's one way to go about it:

Consider a semicircle that is the top half of a circle (radius R) centered at the origin. The semicircle is clearly symmetric about the y-axis, so xcm = 0. To find ycm, compute the following integral:

y_{cm} = \frac{\int_0^R{yw(y)dy}}{\frac{\pi R^2}{2}}

where w(y) is the width of the element of area, given by

w(y) = 2*\sqrt{R^2-y^2}

If you calculate that, you have the distance from the "base" of the semicircle (the flat side) to its centroid. Keep that in mind when you flip it upside down.
 

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