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I am doing a project which involves finding the center of mass/gravity of half of a thin-walled, hollow, cylinder capped with ends of the same material and thickness. Basically like half of an empty soup can, as if the can were standing on a counter top and was sliced in half from top to bottom, leaving 2 "boat hulls".

I'm surprised at how much trouble I'm having finding out how to do this. Can someone please tell me how to do it?

I was using this for the body, but I wasn't (at all) sure it is the right formula:

$y_{cg} =\frac{1}{2r} \int^r_{-r} (r^2 -x^2)^.5 dr $

where r is the radius (a constant) and integrating from -x to x because x=r

Thanks!

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Two of the "coordinates" of the centre of mass are obvious by symmetry. Put the thing flat (open) side down. So the cross-section is a semi-circle, say of radius $r$, though I would prefer $1$.

There are two things to worry about: (i) the two ends and (ii) the rest. We find the height of the centroid of each part, say $h_1$ and $h_2$. Then the height of the centroid of the combination is the weighted average of the centroids of the two parts. The combined weight of the two ends is $w_1=\pi r^2$, and the weight of the rest is $w_2=\pi r l$, where $l$ is the length. The centroid of the combination is at height $$\frac{w_1h_1+w_2h_2}{w_1+w_2}.$$

Now we need $h_1$ and $h_2$. I will leave $h_1$ to you, standard formula.

Calculating $h_2$ is a bit more interesting. As usual we divide the moment about the $x$-axis by the circumference of the rounded part of the semicircle, assuming unit linear density.

To find the moment, look at the piece that is above the little interval from $x$ to $x+dx$. This is at height $\sqrt{r^2-x^2}$. Its length is roughly $\sqrt{1+\left(\frac{dy}{dx} \right)^2\,dx}$. So the moment is equal to $$\int_{-r}^r \sqrt{r^2-x^2}\sqrt{1+\left(\frac{dy}{dx} \right)^2}\,dx.$$ When you calculate $\sqrt{1+\left(\frac{dy}{dx} \right)^2}$, you will have an extremely pleasant surprise!

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Forgive me, but you lost me when you began calculating h2. Here is where my mind is: Why is it necessary to calculate a moment? What does dividing "the moment about the x-axis by the circumference of the rounded part of the semicircle" do? Maybe you can tell from these questions where/why I'm not understanding the big picture. Thanks! –  ChrisC70 Oct 3 '12 at 0:45
    
I am starting from a standard definition of height of centroid. It is the height $h_2$ at which if you placed the whole "mass" (in this case $\pi r$, half the circumference), the resulting moment about the $x$-axis would be the same as the actual moment $m$. that is, $(h_2)(\pi r)=m$. So we find $m$, divide by $\pi r$. Note that we think of the half-circumference as being made of wire of unit density. –  André Nicolas Oct 3 '12 at 0:51
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