Polar coordinates of the centroid of a uniform sector

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Homework Statement:

Find the centre of mass of a uniform sector of radius a and angle θ.

Relevant Equations:

##A=\frac{1}{2}a^2\theta##
Average value of value of ##f## over ##A=\frac{1}{A}\iint f\, dA##
If I use cartesian co-ordinates, I get:
##\bar{x}=\frac{1}{A}\iint x\, dA=\frac{1}{A} \iint r^2\cos\theta\, dr\, d\theta= \frac{2a\sin\theta}{3\theta}##
##\bar{y}=\frac{1}{A}\iint y\, dA=\frac{1}{A}\iint r^2\sin\theta\, dr\, d\theta= \frac{2a(1-\cos\theta)}{3\theta}##

But if I use polar co-ordinates, I get:
##\bar{r}=\frac{1}{A}\iint r\,dA=\frac{1}{A}\iint r^2\, dr\, d\theta=\frac{2a}{3}##
##\bar{\theta}=\frac{1}{A}\iint \theta\,dA=\frac{1}{A}\iint r\theta\, dr\, d\theta=\frac{\theta}{2}##

I believe the cartesian co-ordinates are correct and obviously the value of ##\bar{\theta}## is correct, but the polar co-ordinates of the centroid do not correspond with the cartesian co-ordinates. What is wrong with my calculation of ##\bar{r}##?
 
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Answers and Replies

  • #2
pasmith
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Centroids in non-cartesian coordinates don't work that way.

We are looking for [tex]\bar{\mathbf{r}} = \frac{\int_A \mathbf{r}\,dA }{\int_A\,dA}.[/tex] Now in cartesian coordinates you can do that by components easily: [itex]\mathbf{r} = x\mathbf{e}_x + y\mathbf{e}_y[/itex] and the basis vectors are constant, so [tex]
\bar{\mathbf{r}} = \frac{\int_A x\,dA \mathbf{e}_x + \int_A y\,dA \mathbf{e}_y}{\int_A\,dA}.[/tex]
But in plane polars we have [itex]\mathbf{r} = r\mathbf{e}_r(\theta)[/itex], and in this case [tex]
\bar{\mathbf{r}} = \frac{2}{a^2 \Theta} \int_0^a r^2\,dr \int_0^{\Theta} \mathbf{e}_r(\theta)\,d\theta[/tex]. To integrate [itex]\mathbf{e}_r(\theta)[/itex] you must first express it in terms of the constant cartesian basis: [tex]
\mathbf{e}_r(\theta) = \cos \theta\,\mathbf{e}_x + \sin\theta\,\mathbf{e}_y
[/tex] Once you have done that, you can find the polar coordinates of the centroid from [tex]
\bar{r}^2 = \bar{x}^2 + \bar{y}^2 \qquad \tan{\bar\theta} = \frac{\bar y }{\bar x}.[/tex]

I agree with your results for [itex]\bar x[/itex] and [itex]\bar y[/itex], but as you've discovered [tex]
\bar x \neq \frac{2a}{3} \cos (\Theta/2)\quad\mbox{and}\quad
\bar y \neq \frac{2a}{3} \sin(\Theta/2).[/tex]
 
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  • #3
pasmith
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As it happens, [tex]
\frac{\bar y}{\bar x} = \frac{1 - \cos \Theta}{\sin\Theta} = \frac{2 \sin^2 (\frac12\Theta)}{2\sin(\frac12\Theta)\cos(\frac12\Theta)} = \tan(\tfrac12\Theta)[/tex] but this is purely a coincidence.
 
  • #4
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Thank you so much for your answer! So I gather that my error was thinking of r as a scalar instead of a vector.

So does that mean that ##\frac{\iint r^2\,dr\,d\theta}{\iint_A dA}## is the average distance of a point in the sector from the origin, but the average displacement of a point in the sector from the origin (which is the displacement of the centrodi) is given by ##\frac{\iint r^{2}(\cos\theta \hat{i} + \sin\theta \hat {j})\,dr\,d\theta}{\iint_A dA}##?
 
  • #5
haruspex
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Thank you so much for your answer! So I gather that my error was thinking of r as a scalar instead of a vector.

So does that mean that ##\frac{\iint r^2\,dr\,d\theta}{\iint_A dA}## is the average distance of a point in the sector from the origin, but the average displacement of a point in the sector from the origin (which is the displacement of the centrodi) is given by ##\frac{\iint r^{2}(\cos\theta \hat{i} + \sin\theta \hat {j})\,dr\,d\theta}{\iint_A dA}##?
Yes.
 
  • #6
haruspex
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but this is purely a coincidence.
Not sure what you mean by that. Isn't it evident that the centroid will lie on the angle bisector?
 

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