Center of mass of a triangle (uniform density)

[Kenny Lee]

The triangle has uniform surface mass density of. The method that I've been taught involves some basic integration and substitution of dm for (Density X dA). And then substituting dA with one expression in terms of dy, and another in terms of dx.

I don't really like the method I was taught... 'feels' inconsistent. I don't like having to look at the graph to determine an expression for dA... I keep thinking that if the problem gets tougher, like with some complex graph, then I'm screwed.

What I was wondering however, is if we could use double integration to solve it? I haven't really learned double integration, but I've sort of used it for moments of areas.
So for example, maybe we could make the substitution dm = (density) dx dy?
Then do some fancy math? It just seems more logical. I hope I'm making sense. Any advise?

 

[Tide]

The answer is "Yes, you can use double integration to find the center of mass." E.g., finding the x coordinate of the CM would look like

[tex]\bar x = \int x dx dy[/itex]<br /> <br /> and involves somewhat complicated boundary values in x and y for arbitrary orientations of the triangle. However, if you are interested in a shortcut, notice that the integrated function eventually gets evaluated at the vertices of the triangle and, in the case of uniformly distributed mass, the coordinates of the CM is just the average position of the 3 vertices![/tex]

 

[Kenny Lee]

Thx for rply. So for the right angled triangle, what would the boundary values be? Would that be 'complicated' as well? Cause if it is, then I guess I'll just forget about it... at least now I know its possible to do it by double integration.

 

[daveed]

it might also be good to use Pappus' theorem here, wouldn't it?

 

[vanesch]

Thx for rply. So for the right angled triangle, what would the boundary values be?

Let's put the right angled triangle with its right angle in the x-y crossing.
Now let x run from 0 to a (the length of the x-side). Within this x-integral, x has a definite value, and for that value, you have to find the y-range.
Clearly for x=0, the y-range is the entire y-side (call it b): so y runs from 0 to b.
Also, for x = a, the y-range is reduced to 0. So the limits on the y-range of the y-integral will be dependent on x. The lower boundary of y will always be 0, it is the upper boundary that is changing. And it will clearly be a linear function of x.
So y_up(x) = A x + B and we know that y_up(x=0) = b and y_up(x=a) = 0, which allow you to solve for A and B
It will result in:
y_up(x) = b/a(a-x) = b - b/a x
So we have now everything:
$$\int _{x=0}^a dx \int_{y=0}^{b-\frac{b}{a}x} dy x$$

 

[Kenny Lee]

Huh? Yea, it'll take a bit to digest everything. Thx for additional info. I'll also have a look at Pappus' theorem... saw it on google. Wasn't sure if it applied tho.