Expert: Paul Klarreich - 10/12/2005

Question
Hi, I am trying to answer this question but keep getting it wrong. I have a semi-circular Laminar that looks like half of a polo mint (so it has an inside and outside diameter). The question asks: Find the location of the centroid of the lamina from first principles, the radius of the outer circle is "a" and for the inner circle 0.5a. If you could shed any light on how the question should be done it would be much appreciated.

Answer
Hi, Adam,
I am not sure what a polo mint looks like (I was always partial to Milky Way bars.) but I am assuming it is half of an 'annulus' -- the area between two concentric circles.  

[In the discussion below, I have changed your conditions a little -- the outer circle has radius 2b and the inner circle has radius b.  Obviously, 2b = a for your purposes.]

I am also not sure what the 'first principles' are.  Normally you find a centroid by integration.  For each 'piece' of the lamina you obtain a 'moment' by doing these integrations:

x-coordinate of the lamina = x-moment / area,  and
x-moment = Int ( x (f1(x) - f2(x)) dx

y-coordinate of the lamina = y-moment / area,  and
y-moment = Int ( y (f1(x) - f2(x)) dx

Where f1 is the upper function (boundary) and f2 is the lower function.

(yes, you integrate with respect to x both times, but multiply by the appropriate coordinate.)

If you draw your 'mint' so that the centers are the origin and it lies in the upper half-plane, then it is clear that the x-coordinate is zero, so you just have to find the y-coordinate.

To do the integration, you put together the following ideas:

1. The y in the integral is the y-coordinate of the center of mass of each 'slice' of the integration.  Obviously that is (f1(x) + f2(x))/2

2. The integration goes from x = -2b to  x = +2b.

3. The upper function, f1, is the outer circle, which is sqrt(4b^2 - x^2).

4A. The lower function, f2, is the inner circle, which is
sqrt(b^2 - x^2) only for x going from  -b to +b

4B. The lower function, f2, is the x-axis, or zero, for
-2b to -b  and for  b to 2b.
{remember, your a = 2b)

5. You thus have THREE separate integrations to do.  

However, when you put in both the y and the f1-f2, you will find that the square roots disappear very nicely.

So you will get the moment after adding them all up.

The moment comes out to 14b^3/3.

Now you divide by the area, which you get from 'zeroth' principles:  Annulus = pi(r1^ - r2^2) = pi(4b^2 - b^2)
= 3pi b^2.

And your 'mint' is half of that.

Now divide: (14b^3/3) / (3pi b^2/2) = 28b/9pi

Your answer is  14a/9pi.  It comes out to just under a/2, so the centroid is just 'under the arch.'

=============================================
The integrals are actually easy ones, but the fact that there are three of them makes this messy.

Have you learned integrals in polar coordinates?  If so, that's much easier.