Another CM problem with var. density

[scorpius1782]

Homework Statement

This is similar to another problem I've posted but is supposedly more difficult.
Cylinder with dimensions radius=R and height=2L has density of $\rho=\rho_o(1+cos(\phi))$

Find the position of the cylinders center of mass.

Homework Equations

$r_{cm}=\frac{\int\vec{r}dm}{\int dm}$

The Attempt at a Solution

I have Mass$=\int_0^R\int_0^2L\int_0^{2\pi}\rho_o(1+cos(\phi))s ds d \phi dz$
So, $2L\frac{R^2}{2}\int_0^{2\pi}\rho_o(1+cos(\phi))d \phi$
=$2L\frac{R^2}{2}(\rho_o2\pi+\int_0^{2\pi}\rho_ocos(\phi))d \phi$
where $\int_0^{2\pi}\rho_ocos(\phi)=0$
so M=$2L\frac{R^2}{2}(\rho_o2\pi)=2LR^2\rho_o\pi$For the positions then:

$\int_0^R\int_0^{2L}\int_0^{2\pi}\rho_o(1+cos(\phi))s(s) ds d \phi dz$ for $\hat{s}$
$\int_0^R\int_0^{2L}\int_0^{2\pi}\rho_o(1+cos(\phi))s(\phi) ds d \phi dz$ for $\hat{\phi}$
$\int_0^R\int_0^{2L}\int_0^{2\pi}\rho_o(1+cos(\phi))s(z) ds d \phi dz$ for $\hat{z}$

Then I should just divide each integral by the mass from before. Is this the correct method? It seems a little too straightforward for being "more difficult."Thanks for the help.

 

[LCKurtz]

Homework Statement

This is similar to another problem I've posted but is supposedly more difficult.
Cylinder with dimensions radius=R and height=2L has density of $\rho=\rho_o(1+cos(\phi))$

Find the position of the cylinders center of mass.

Homework Equations

$r_{cm}=\frac{\int\vec{r}dm}{\int dm}$

The Attempt at a Solution

I have Mass$=\int_0^R\int_0^2L\int_0^{2\pi}\rho_o(1+cos(\phi))s ds d \phi dz$

Are $ds,d\phi,dz$ what are usually called $dr,d\theta,dz$ in cylindrical coordinates? If so, why does $s$ go from $0$ to $2\pi$? If not, what are they?

So, $2L\frac{R^2}{2}\int_0^{2\pi}\rho_o(1+cos(\phi))d \phi$
=$2L\frac{R^2}{2}(\rho_o2\pi+\int_0^{2\pi}\rho_ocos(\phi))d \phi$
where $\int_0^{2\pi}\rho_ocos(\phi)=0$
so M=$2L\frac{R^2}{2}(\rho_o2\pi)=2LR^2\rho_o\pi$For the positions then:

$\int_0^R\int_0^2L\int_0^{2\pi}\rho_o(1+cos(\phi))s(s) ds d \phi dz$ for $\hat{s}$
$\int_0^R\int_0^2L\int_0^{2\pi}\rho_o(1+cos(\phi))s(\phi) ds d \phi dz$ for $\hat{\phi}$
$\int_0^R\int_0^2L\int_0^{2\pi}\rho_o(1+cos(\phi))s(z) ds d \phi dz$ for $\hat{z}$

What is this $s()$ function that just showed up in the integrand? And where did these formulas come from?

 

[scorpius1782]

Sorry, it's a little annoying to have physics do cylindrical one way and math classes another.
cylindrical here is $(s,\phi,z)$

I used the volume $d\tau$ for cylindrical which is $d\tau=(s)dsd \phi dz$

I may not have the integrals in the right order? I want s to go from 0→R, $\phi$: 0→2$\pi$ and z: 0→2L

Perhaps the s variable isn't required for the position integrals but only the volume? I assumed that it would be.

 

[LCKurtz]

I think the more serious problem is those last 3 formulas in the first place. You need the first moments in the x,y,z directions, requiring corresponding moment arms in the integrals, expressed in cylindrical coordinates. And what you should be calculating, if I understand your notation, is $\hat x,\hat y, \hat z$. Once you have the centers of mass in the three directions $(\bar x,\bar y, \bar z)$ you could convert that to cylindrical coordinates.

 

[scorpius1782]

I'm sorry, I don't follow what you're saying.

 

[LCKurtz]

Your first formula in your original post has the formula$$
r_{cm} = \frac {\int \vec r~dm}{\int dm}$$Do you understand that formula? There, $\vec r$ is the position vector $\langle x,y,z\rangle$. That numerator is a vector$$
\langle \int x~dm,\int y~dm, \int z~dm\rangle$$You need to calculate those three integrals for your numerators. They will give your three moments you need for your numerators.

 

[scorpius1782]

I don't understand why I cannot leave it in cylindrical coordinates. It just seem that changing to cartesian makes this really complicated. for instance $s=\sqrt{x^2+y^2}$ and $\phi=tan^{-1}(\frac{y}{x})$.

Or could I simply do $x=scos(\phi)$ and $y=sin(\phi)$? The only problem with these two I have is trying to figure out what the limits should be. Should I further substitute for $\phi$? Or can I just evaluate the integral as:

$\int_0^R\int_0^2L\int_0^{2\pi}\rho_o(1+cos(\phi))s(scos(\phi)) ds d \phi dz$

Where $x=scos(\phi)$ replaced the (s) from before.

 

[LCKurtz]

I don't understand why I cannot leave it in cylindrical coordinates. It just seem that changing to cartesian makes this really complicated. for instance $s=\sqrt{x^2+y^2}$ and $\phi=tan^{-1}(\frac{y}{x})$.

You don't have to change to rectangular. Just include the moment arms.

Or could I simply do $x=scos(\phi)$ and $y=ssin(\phi)$?

Yes.

The only problem with these two I have is trying to figure out what the limits should be. Should I further substitute for $\phi$? Or can I just evaluate the integral as:

$\int_0^R\int_0^2L\int_0^{2\pi}\rho_o(1+cos(\phi))s(scos(\phi)) ds d \phi dz$

Where $x=scos(\phi)$ replaced the (s) from before.

Yes, that's the idea. But put your limits in the right order for the order of your variables. I have to leave for now, but you should be able to take it from here.

 

[scorpius1782]

Wonderful, thank you for the help!