# Homework Help: Calculating Center of Mass in Cylindrical Coordinates

1. Mar 25, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
Determine the center of mass in cylindrical coordinates of a cone with constant density $\rho(\vec{r})$. (The cone is inverted, i.e. it's thinnest point is at $z=0$.)

2. Relevant equations
$m=\int\int\int_C \rho r \, drdzd\theta$
$\overline{r}=\int\int\int_C r\cdot r\, drdzd\theta$
$r_{CM}=\frac{\overline{r}}{m}$

3. The attempt at a solution
I'm quite confused about calculating the center of mass in cylindrical coordinates since my results don't make any sense intuitively. Since the integrations are pretty trivial I will just post how I calculated $m$ and then the results of the other integrals. $m=\int_{0}^{2\pi}\int_{0}^{h}\int_{0}^{\frac{Rz}{h}}\rho r\, drdzd\theta=\frac{\pi \rho R^2h}{3}$, $\overline{r}=\frac{\pi R^3h}{6}$, $\overline{z}=\frac{\pi R^2h^2}{4}$, $\overline{\theta}=\frac{\pi^2 R^2h}{3}$, $r_{CM}=\frac{R}{2\rho}$, $z_{CM}=\frac{3h}{4\rho}$, $\theta_{CM}=\frac{\pi}{\rho}$, now clearly these three values won't give the position of the center of mass, i.e. $f_{CM}=(r_{CM}, \theta_{CM}, z_{CM})$. To me it seems like this will just give the center of mass with respect to each axis and therefore it doesn't make sense to have have a point defined by the three center of mass locations for each axis. Also the value for $\theta_{CM}$ does not make any sense, it doesn't even make sense how there can be a $\theta_{CM}$ when the cone has constant density.

2. Mar 25, 2016

### andrewkirk

You didn't say on which axis the cone is centred. I'll assume it's the z axis, as otherwise the cylindrical coordinates will be inappropriate for the calculations.
Then symmetry considerations dictate that the COM is on the z axis. So your result $r_{CM}=\frac{R}{2\rho}$ looks like it can't be correct. Also, $\theta_{CM}$ will be undefined, because the COM is on the z axis. THe only value that needs to be calculated is $z_{CM}$.

Can you set out your calculations behind the results you quote?

3. Mar 25, 2016

### Potatochip911

Yep it is along the z axis.

Yea my interpretation of this result is that it made sense when considering a cross section (Right Triangle) of the various radii.

Ok... with $m=\frac{\pi\rho R^2h}{3}$.
$$\overline{r}=\int_{0}^{2\pi}\int_{0}^{h}\int_{0}^{\frac{Rz}{h}}r^2\,drdzd\theta=2\pi\int_{0}^{h}\frac{r^3}{3}\Big |_{0}^{\frac{Rz}{h}}dz=\frac{2\pi R^3}{3h^3}\int_0^{h}z^3\,dz=\frac{2\pi R^3}{3h^3}\frac{h^4}{4}=\frac{\pi R^3h}{6} \\ \overline{z}=\int_0^{2\pi}\int_0^h\int_0^{\frac{Rz}{h}}rz\,drdzd\theta=2\pi\int_0^h \frac{R^2z^2}{2h^2}z\,{dz}=\frac{\pi R^2}{h^2}\frac{z^4}{4}\Big |_0^h=\frac{\pi R^2h^2}{4}\\ \overline{\theta}=\int_0^h\int_0^{\frac{Rz}{h}}\int_0^{2\pi}r\theta\,d\theta drdz=2\pi^2\int_0^h\int_0^\frac{Rz}{h}r\,drdz=\pi^2\int_0^h\frac{R^2z^2}{h^2}dz=\frac{\pi^2 R^2}{h^2}\frac{z^3}{3}\Big|_0^h\\ \overline{\theta}=\frac{\pi^2R^2h}{3}$$
Now to calculate the COMs:

$$r_{CM}=\frac{\overline{r}}{m}=\frac{R}{2\rho}\\ z_{CM}=\frac{\overline{z}}{m}=\frac{3h}{4\rho} \\ \theta_{CM}=\frac{\overline{\theta}}{m}=\frac{\pi}{\rho}$$

Edit: It appears as though my mistake is calculating things such as $r_{CM}$ and $\theta_{CM}$, to obtain the proper results all the integrals should start in cartesian and then be transformed into cylindrical which will give our answer in cartesian. By inspection this makes both the $\overline{x}$ and $\overline{y}$ integrals 0 since they will have a $\cos\theta$ and $\sin\theta$ term respectively and this will be integrated from 0 to 2$\pi$. The $\overline{z}$ integral remains the same and thus the COM becomes $(0,0,\frac{3}{4}h)$

Last edited: Mar 25, 2016