Is there actually such thing as center of mass in polar coordinates?

flyingpig
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Homework Statement

Or any coordinates really. In the normal Cartesian plane, the center of mass is defined from the x, y , and z distance as follows

\bar{x} = \frac{1}{Area(R)}\iint_R x dA

\bar{y} = \frac{1}{Area(R)}\iint_R y dA

\bar{z} = \frac{1}{Area(R)}\iint_R z dA

Now is there one for polar coordinates where you find the center of mass from the average "r" and "theta"?

\bar{r} = \frac{1}{Area(R)}\iint_R r dA

\bar{\theta} = \frac{1}{Area(R)}\iint_R \theta dA

The Attempt at a Solution



I feel the derivation is going to be a bit lengthy and this really isn't as much of "HW", but just ponder.

In my Calculus book, although we do double integrals in polar coord and find the center of mass, we still stick to the cartesian coordinates.
 
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Sure. Coordinates are just a way of identifying a point in the plane or space, or wherever. The center of mass could be at (1, 1) in rectangular (or Cartesian) coordinates, or at (sqrt(2), pi/4) in polar coordinates.
 
No, I mean as a formula.
 
Like if I give you f(r,\theta), without converting, is there a way to find r and theta's center of mass?
 
flyingpig said:
Like if I give you f(r,\theta), without converting, is there a way to find r and theta's center of mass?
As stated, this doesn't make any sense, but I think I understand what you're asking, which is, "Are there integrals that give r-bar and theta-bar?"

There might be, but I don't recall ever seeing any.
 
If you are given r = f(θ) then the answer is yes and no.:rolleyes:

You can certainly calculate the center of mass:

\overline x = \frac{\iint_R x\ dydx}{\iint_R 1\ dydx}= \frac{\iint_R r\cos\theta\ r dr d\theta}{\iint_R \ rdrd\theta}

and similary for \overline y. And you will get the r coordinate of the center of mass:

\overline r = \sqrt{\overline x^2+\overline y^2}

But that doesn't give you the same as if you tried to use the first r moment to define the r value:

\overline r = \frac{\iint_r r\ rdrd\theta}{\iint_R rdrd\theta}
 
flyingpig said:
No, I mean as a formula.

You can find the formulas yourself by suitable substitution in integral for x-bar and y-bar. and using from both of them try to get separate integrals for r and theta
 
Mark44 said:
As stated, this doesn't make any sense, but I think I understand what you're asking, which is, "Are there integrals that give r-bar and theta-bar?"

There might be, but I don't recall ever seeing any.

I am also interested in these integrals. Some functions are just easier to work with in polar coordinates. If I remember correctly, the integral used to find area is derived my summing up areas of infinitesimal sectors of a circle. The formula for area is

A=\frac{1}{2} \int f(θ)^{2} dθ
 
iknowless said:
I am also interested in these integrals. Some functions are just easier to work with in polar coordinates. If I remember correctly, the integral used to find area is derived my summing up areas of infinitesimal sectors of a circle. The formula for area is

\displaystyle \text{A}=\frac{1}{2}\int (f(\theta))^{2}\, d\theta
That's true for a planar object whose boundary is defined by r = f(θ) .
 
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