# Finding the center of mass of laminae

1. May 3, 2012

### dlivingston

Calculus 3 project – any and all help is appreciated.
We've just gone over center of mass with double integrals, so it's a bit peculiar to see this project feature only one integral. I went over that in calc 2 – and as a result, know how to calculate it that way.

However (as you will see in the formula below), I don't think he wants me to calculate it that way. I have no idea how to proceed and the book is of absolutely no help.

Anything you could do to help me out would be fantastic.

And here's the problem:

1. The problem statement, all variables and given/known data
There are three type of laminae we are trying to find:
1) A triangle with sides 3", 4", and 5" respectively.
2) A semicircle
3) A "horseshoe", i.e. a half-ring, whose inner and outer edges are composed of circular arcs of radii 3" and 5" respectively.

Definition 1:
The first moment of a planar mass distribution about a line l is the integral over the region of the (area) density σ(P) times the distance of P from the line. For example, the first moment about the line x=k is
$M_k=∫_R\left(x-k\right) σ\left(P\right) dA$

Definition 2:
In order for a line to be a balancing line, the first moment about this line must be zero.

2. Relevant equations
See the M_k above
I found, I believe, the equations for the laminae:
For the triangle,
$y=4-\frac{4}{3}x$
For the semicircle,
$x^2+y^2=1$
And the horseshoe,
$x^2+y^2=25$
$x^2+y^2=9$

3. The attempt at a solution
Welp, I'm lost. As I said, I can figure it out easily using the calculus 2 method of $x_{cm}=\frac{M_y}{M}$, but I'm not sure the way he wants it done.

What's the density formula? Is it just the equations I listed above? What's this (x-k) nonsense?

2. May 3, 2012

*cricket*
Anyone?

3. May 3, 2012

### gopher_p

The $M_k$ gives, roughly speaking, the potential torque on the lamina if one were to try to balance it on the line $x=k$. In order for the lamina to be balanced on that line, this "tourque" would need to be zero.

If you set $M_k=0$ and solve for $k$, I think you'll see what the "$x-k$ nonsense" is for.

4. May 3, 2012

### HallsofIvy

Staff Emeritus
Wow, you waited a whole hour before bumping? Sorry, I was eating supper!

5. May 3, 2012

### HallsofIvy

Staff Emeritus
You are not given any "density" function so this is a purely mathematical problem, not physics. You are really looking for the "centroid", not the "center of mass". That is the same as assuming the "density" is a constant.

This is a single lamina. "laminae" is the plural.

Well, you mean $x_{cm}=\frac{M_x}{M}$, and $y_{cm}=\frac{M_y}{M}$

x- k is, just your quote above says, the horizontal distance from point (x, y) to the line x= k.
So this is a right triangle. Set up a coordinate system so the right angle is at the origin, one vertex is to (3, 0), and another at (0, 4). The equation of the hypotenuse is y= -(4/3)x+ 4. The area of the triangle is, of course (1/2)(3)(4)= 6 so the x-coordinate of the centroid is $(1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx$ and the y coordinate is $(1/6)\int_{x=0}^3\int_{y= 0}^{-(4/3)x+ 4} x dydx$.

The area of a semicircle of radius R (you don't give a radius) is $\pi R^2/2$. It's probably best to use polar coordinates to do the integrals:
$$(2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} x d\thetadr= (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} rcos(\theta) d\thetadr$$
(It might occur to you that, by symmetry, the x coordinate of the centroid is 0)
and
$$(2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} y d\thetadr= (2/\pi R^2)\int_{r= 0}^R\int_{\theta= 0}^{\pi} rsin(\theta) d\thetadr$$

So this is the area between two semi-circles. Its area is $(pi(5)^2/2- \pi(3)^2/2= 8\pi$.The coordinates of the centroid are
$$(1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r cos(\theta)d\theta dr$$
(Again, by symmetry, this should be 0.)
and
$$(1/8\pi)\int_{r=3}^5\int_{\theta= 0}^\pi r sin(\theta)d\theta dr$$

Last edited: May 3, 2012
6. May 3, 2012

### dlivingston

Ah, you both are absolutely fantastic! Thank you.