# Polar coordinates of the centroid of a uniform sector

• parsesnip
In summary: Yes, it is evident. But what you are saying is that the centroid will lie on the angle bisector, but the angle bisector is not necessarily the origin.
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}##?

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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
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}##?

parsesnip said:
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.

pasmith said:
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?

## 1. What are polar coordinates?

Polar coordinates are a way of representing points in a two-dimensional plane using a distance from the origin and an angle from a reference line.

## 2. What is the centroid of a uniform sector?

The centroid of a uniform sector is the point at which the two lines connecting the center of the sector to its two ends intersect. It is also the center of mass of the sector, meaning it is the point where the sector would balance perfectly if it were cut out of a uniform material.

## 3. How do you find the polar coordinates of the centroid of a uniform sector?

To find the polar coordinates of the centroid of a uniform sector, you need to know the radius of the sector and the angle of the sector. Then, you can use the formula r = (2R/3)(sinθ/θ) to calculate the distance from the origin, and the formula θ = 3sinθ/2 to calculate the angle from the reference line.

## 4. Can the polar coordinates of the centroid of a uniform sector be negative?

No, the polar coordinates of the centroid of a uniform sector cannot be negative. The distance from the origin (r) is always positive, and the angle from the reference line (θ) is measured counterclockwise from the positive x-axis, so it will always be between 0 and 2π.

## 5. How are polar coordinates used in real life?

Polar coordinates are used in many real-life applications, such as navigation systems, radar systems, and astronomy. They are also commonly used in mathematics and physics to describe the position and movement of objects in a two-dimensional plane.

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