Expert: Paul Klarreich - 12/5/2006

Question
It was here:  http://experts.about.com/q/Advanced-Math-1363/Findig-location-Centroid.htm
and the subject line was "Findig the location of a Centroid!" and yes, the word "findiNg" was misspelled (in the subject and the web address).  The date was 10/12/05.  Thanks again.
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The text above is a follow-up to ...

-----Question-----
I read your answer to another student's question about finding the centroid of a half annulus with radius2 = 2*radius1 and it was very helpful.  However, it would have been nice if you had shown exactly what the written out integral was for each of the three separate sections (-2b to -b, -b to b, and b to 2b).  Also, is there a formula for finding the centroid of a half annulus where the outer radius is not necessarily equal to twice the outer radius?  That would be useful information to know.  This is for my calculus class (Vector Calculus).  Thanks a lot.
-----Answer-----
Hi, Ken,

You wrote:

I read your answer to another student's question about finding the centroid of a half annulus with radius2 = 2*radius1 and it was very helpful.  

>> Thank you for the kind comment.

However, it would have been nice if you had shown exactly what the written out integral was for each of the three separate sections (-2b to -b, -b to b, and b to 2b).  Also, is there a formula for finding the centroid of a half annulus where the outer radius is not necessarily equal to twice the outer radius?  That would be useful information to know.  This is for my calculus class (Vector Calculus).  Thanks a lot.

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I must be getting old.  I have a vague recollection of this but I can't find it online or in my own notes.  Can you tell me when I answered it, and, perhaps, what the subject line was?

In the meantime, I'll take a look at the problem again, so expect a followup in some reasonable time.  It's very possible that when the ratio is not 2/1, the intersection point becomes a messy affair.
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Answer
Hi, Ken,

Okay, I am satisfied that I can do it, now.  I have fastened upon the following two methods:

A. Start with finding the centroid for a semicircle with radius r.  Sort of a thin wire, if you like, formed as a semicircle.

Obviously for this and all the other stuff, the center is at the origin, and so the x-coordinate (xC, I shall write) of the centroid is always zero.

Centroid = moment / total mass.
In this case, it's moment / total length.

In doing this integration, like any other application, we look at a sample piece and integrate that.  The sample piece here is a bit of wire from the semicircle, and it is not a dx, or a dy, but a ds.  Seeing that, it becomes easiest to do the integration in polar coordinates:

Moment =
{
| y ds
}

Now remember that  ds = r dt    [I can't make a theta on this computer, so when you see a 't', read 'theta'.]
and that  y = r sin t

{
| r sin t r dt
}

{
| r^2 sin t dt
}

= - r^2 cos t,  from  0 to pi.

= -r^2[(-1) - (1)]
= -r^2(-2)
= 2r^2
Okay, that's the moment, now divide by pi r, the length of a semicircle.
    2r^2   2r
yC = ---- = --
    pi r   pi

Now that we have the semicircle, we can integrate these semicircles outward from r=a to r=b.  Each 'sample' semicircle has 'mass' equal to pi r, its length, and a centroid as we just found.  So we can compute the moment for the semi-annulus:

Moment =

{  2r
| ----- pi r dr
}  pi

{   
| 2r^2  dr
}  

2 r^3
-----,  from  r = a to b:
 3
2
---(b^3 - a^3)
3

Now divide by the 'mass' of the annulus:  (the area, of course)
    pi
A = ---(b^2 - a^2)
    2
    2/3(b^3 - a^3)
yC = ----------------
    pi/2(b^2 - a^2)

    4(b^3 - a^3)
yC = --------------
    3pi(b^2 - a^2)

AND THAT IS THE FORMULA FOR THE CENTROID OF THE ANNULUS.

...........................................................
B. I promised you a second method.  Here it is:

What is the centroid of a complete semicircle disk of radius r?

The answer is 4r/3pi.  It's an easy integration, and I leave that to you.

Now the annulus is the 'difference' between two semicircles.  The basic idea of 'centroid' is that we can view a massive object as if it were a 'point mass' located at its centroid.  So we have:

For the big semi:

mb = pi b^2/2  (mass     of semi with radius b)
yb = 4b/3pi    (centroid of  ..............  b)

For the small semi:

ma = pi a^2/2
yb = 4a/3pi

For the Annulus: (Big A is not the same as small a.)

mA = pi(b^2 - a^2)/2
yA is to be found.

Now the moments of the objects follow the rule:

Moment of small semi + Moment of Annulus = Moment of big semi.

Or:

Moment of Annulus = Moment of big semi - Moment of small semi

And for a point mass, the moment is just the mass times the centroid.  We're in business:

Moment of Annulus = Moment of big semi - Moment of small semi  (again)

mA yA = mb yb - ma ya
    mb yb - ma ya
yA = --------------
         mA

    (pi b^2/2)(4b/3pi) - (pi a^2/2)(4a/3pi)
yA = ---------------------------------------
         pi(b^2 - a^2)/2


    (b^2/2)(4b/3) - (a^2/2)(4a/3)
yA = ------------------------------
         pi(b^2 - a^2)/2

    (b^2)(2b/3) - (a^2)(2a/3)
yA = ---------------------------
         pi(b^2 - a^2)/2

    (2b^3/3) - (2a^3/3)
yA = ----------------------
       pi(b^2 - a^2)/2

    2/3(b^3 - a^3)
yA = -----------------
    pi(b^2 - a^2)/2


    4(b^3 - a^3)
yA = ---------------   [HOORAY!]   
    3pi(b^2 - a^2)

...........................................
And, finally, suppose that we have the case where  b = 2a.

    4(8a^3 - a^3)
yA = ---------------
    3pi(4a^2 - a^2)

    4(7a^3)
yA = ----------
    3pi(3a^2)

    28a
yA = ----
    9pi

Basically, the same answer I got a year ago.  So I did it right.
(THE OLD GUY CAN STILL HACK IT!)