# Polar coordinates of the centroid of a uniform sector

## 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

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pasmith
Homework Helper
Centroids in non-cartesian coordinates don't work that way.

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

I agree with your results for $\bar x$ and $\bar y$, but as you've discovered $$\bar x \neq \frac{2a}{3} \cos (\Theta/2)\quad\mbox{and}\quad \bar y \neq \frac{2a}{3} \sin(\Theta/2).$$

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parsesnip
pasmith
Homework Helper
As it happens, $$\frac{\bar y}{\bar x} = \frac{1 - \cos \Theta}{\sin\Theta} = \frac{2 \sin^2 (\frac12\Theta)}{2\sin(\frac12\Theta)\cos(\frac12\Theta)} = \tan(\tfrac12\Theta)$$ but this is purely a coincidence.

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}##?

haruspex
Homework Helper
Gold Member
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.

haruspex