# How to calculate this tricky integral? Can it be done?

1. Sep 10, 2015

### will eastman

I am trying to calculate the force acting on the end cap of a rotating cylinder containing water, this is part of a university project and any help would be much appreciated.
Due to the rotation the water in the container has a pressure defined by
P=((row)*(omega)^2*r^2)/2.
As the pressure is different at each radii I believe you must us and integral in order to calculate the force F=P*A. But I am really struggling to put together the integral force F without getting a dr^2 function. Can anybody help me out with this, it would be much appreciated.

2. Sep 10, 2015

### BvU

Hi Will, welcome to PF !

Do you realize that there are many ways to rotate a cylinder, even when it contains water ? Can you be more specific ? And show what integral you have set up in what coordinates, etc ?

3. Sep 10, 2015

### Staff: Mentor

A couple of nits:
The Greek letter $\rho$ is spelled "rho" not "row".

4. Sep 10, 2015

### will eastman

Yes I do, sorry I will try to be a little clearer.

The cylinder is rotating about an axis running through its centre connecting the centre of the circular faces on each end, (sorry not sure how to explain this better)

I am unsure about the coordinates, maybe this is where I am going wrong, I though I may be able to just integrate it using the radius as the only variable.

The equation I have put together is:

dP=(rho)*(omega)^2*r^2)/2 and dA={r^2-(r-dr)^2}*(pi)

dF={(rho)*(omega)^2*r^2)/2}*{r^2-(r-dr)^2}*(pi)

this then expands to dF={(rho)*(omega)^2*(pi/2)}*r^2*{2*r*dr+dr^2}

as rho and omega are constant in this case and when b={(rho)*(omega)^2*(pi/2)} we get the simplified formula of...

dF=b*{(2*(r^3)*dr)+(r^2*dr^2)}

is this correct and if so can it be integrated

5. Sep 11, 2015

### will eastman

Sorry, am now thinking that is wrong, this (I think) is correct:

dP=(rho)*(omega)^2*r^2)/2 and dA={r+dr)^2}*(pi)

dF={(rho)*(omega)^2*r^2)/2}*{r+dr)^2}*(pi)

this then expands to dF={(rho)*(omega)^2*(pi/2)}*r^2*{r^2*2*r*dr*dr^2}

as rho and omega are constant in this case and when b={(rho)*(omega)^2*(pi/2)} we get the simplified formula of...

dF=b*{(2*(r^3)*dr)+(r^2*dr^2)+(r^4)}

is this correct and if so can it be integrated

6. Sep 11, 2015

### HallsofIvy

Staff Emeritus
Then English transliteration of the Greek letter $\rho$ is "rho".

7. Sep 11, 2015

### BvU

Good. That narrows it down a little. I take it the cylinder is 'full' -- otherwise you get situations like they are pictured here ( I don't want to steal their illustrations without permission).

Then
or rather $dp = \rho\; \omega^2r^2/2$ and $dA = \pi (r+dr)^2$ both have the problem that on the left there is a differential and on the right there is not. And the first one doesn't have matching dimensions. Can't be.

If you clarify what things stand for we could try to make an inroad. Start from simple things like $$p = A\ F\quad \Rightarrow \quad dp =A dF + F dA$$ with p = pressure (N/m2), F = force (N), A = area (m2), then explain what dF and dA are in the configuration etc.

With $\vec F = m \vec a \Rightarrow d|\vec F| = |\vec a| \, dm$ you make a next step.

Likewise, if $A = \pi r^2 h$ then $dA = d( \pi r^2 h ) = 2\pi r h dr$ if h is constant.
(By the way, I think it should be $A = 2 \pi\, r\, h$ for a cylinder, but I still don't know what you are calculating)

Gradually work towards an integral, probably a double one (unless you can ignore hydrostatic pressure difference due to gravity).

8. Sep 11, 2015

### BvU

You want to be reported for nit2picking or for nitpicking2 ? Will is a new member and he wants some useful assistance, not two (now three) Waldorf and Statler types haggling among themselves !

Don't worry Will, you'll get used to this kind of thing in PF. After a while it's even kind of fun. Admittedly depending on your sense of humour. And on whether you are on the asking side or on the answering side .
--

9. Sep 11, 2015

### Nidum

Diagram would help .

Is this what you have in mind :

A rotating axle has a radial arm attached . On the outer end of the arm is mounted a tube full of water . The axis of the tube is on the axis of the radial arm .

This is basically a centrifuge arrangement and you want to establish the pressure seen by the outer cap of the tube for a given geometry and rotational speed .

The derivation of pressure can be simple or more complex depending on the construction details .

If the tube is relatively small in diameter compared to the length of the radial arm plus tube then a one dimensional analysis is possible .

If the tube is relatively large then much more complex two or three dimensional analysis is needed .

May help :

Last edited: Sep 11, 2015
10. Sep 11, 2015

### will eastman

Yes the cylinder is full.

Right to tidy somethings up:

P = pressure

\rho\; = water density

\omega = rotational speed