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Calculating Center of Mass in Cylindrical Coordinates

  1. Mar 25, 2016 #1
    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. jcsd
  3. Mar 25, 2016 #2

    andrewkirk

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    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?
     
  4. Mar 25, 2016 #3
    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
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