- #1

parsesnip

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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}##?

##\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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