Polar coordinates of the centroid of a uniform sector

[parsesnip]

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 $\bar{r}$?

 

[pasmith]

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 $$\bar x \neq \frac{2a}{3} \cos (\Theta/2)\quad\mbox{and}\quad \bar y \neq \frac{2a}{3} \sin(\Theta/2).$$

 

[pasmith]

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.

 

[parsesnip]

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]

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]

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?