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

  1. Sep 10, 2015 #1
    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
    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. jcsd
  3. Sep 10, 2015 #2


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    Hi Will, welcome to PF :smile: !

    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 ?
  4. Sep 10, 2015 #3


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    A couple of nits:
    The Greek letter ##\rho## is spelled "rho" not "row".
    should be "at each radius" - radii is plural.
  5. Sep 10, 2015 #4
    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)


    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...


    is this correct and if so can it be integrated
  6. Sep 11, 2015 #5
    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)


    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...


    is this correct and if so can it be integrated
  7. Sep 11, 2015 #6


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    Then English transliteration of the Greek letter ##\rho## is "rho".

  8. Sep 11, 2015 #7


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    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).

    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).
  9. Sep 11, 2015 #8


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    You want to be reported for nit2picking or for nitpicking2 :smile: ? 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 :wink:.
  10. Sep 11, 2015 #9


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    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
  11. Sep 11, 2015 #10
    Yes the cylinder is full.

    Right to tidy somethings up:

    P = pressure

    \rho\; = water density

    \omega = rotational speed

    r = radius

    F = force acting on end of cylinder (due to the water pressure)

    A = area of end of cylinder

    The basic equations I am working from are:

    ##P = \rho\; \omega^2r^2/2## ##A = \pi (r+dr)^2## ##F = P * A

    Now I think the issue I am having is that I don't understand how to use the equations to form an integral. I have calculated this in excel using a multiple slice method but would like to do this correctly.

    I am fairly sure that the equation of the left does have matching units.
  12. Sep 11, 2015 #11
    Also I have no done some small sketches to explain this better.

  13. Sep 11, 2015 #12


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  14. Sep 11, 2015 #13


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  15. Sep 11, 2015 #14


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    Well, let's start with your (Will's) last claim... in SI units: [P] = N/m2 = kg/(ms2). And that's the same as on the righthand side, so I'm the one here with o:) egg on his face.

    Now the other issue: ##A = \pi (r+dr)^2##

    I can agree with ##A(r) = \pi r^2## and with ##A(r + dr) = \pi (r+dr)^2##
    The way to make a differential out of this is via ##dA = A(r) - A(r + dr) = \pi r^2 - \pi (r+dr)^2 = 2\pi r \, dr + \pi (dr)^2 = 2\pi r \, dr##
    (or with ##{dA\over dr} = 2\pi r##)

    Forgive me my physicists way of dealing with differentials (we write (dr)2 = 0 when we mean that in the limit for ##dr\rightarrow 0## this term disappears next to the term with dr).
    (seen the picture in the mean time. My intuition says the final answer/experimental result is probably zero or negative :rolleyes: . But let's not be over-hasty.)

    OK, now the assembly phase: You want F and realize it depends on r only since P depends on r only. So you write $$
    F = \int_0^R dF(r) \ \ = \ \ \int_0^R {dF\over dr} dr \ \ = \ \ \int_0^R {dPA(r)\over dr} dr \ \ = \\ \qquad \ \ \int_0^R {d \over dr} \left ( {\rho \omega^2 r^2 \over 2} 2\pi r^2 \right ) dr
    = \ \ \rho\pi \omega^2\, \int_0^R {d \over dr} \left ( r^4 \right ) dr $$
    and that's not a difficult integral anymore.

    Bringing in the pressure from gravity is a detail/complication that may be unnecessary for reasonable ##\omega##.

    (Dear Nidum: no offence intended, really. Why don't you think along with us, because my intuition is worrying me)

    Reason I'm wavering: this would make the pressure on the axis of rotation zero. In fact, since we are dealing with an incompressible liquid, I can well imagine that the pressure on the side walls -- that realistically have some elasticity -- will in fact make the liquid pull in the end plugs with a huge force (depending on ##\omega##, the elasticity and the length/radius ratio).

    Unless you have a little hole on the axis to neutralize that effect ! :cool:

    Interesting project !
    But you posted with what is basically a math problem, so let's first make sure that is all clear and understood. Will ?

  16. Sep 11, 2015 #15
    It is not clear to me why, when the cylinder is rotating about its own axis, any additional pressure than the one due to gravity should act on the plug. The axis of rotation is perpendicular to the plug. I see why more pressure would act on the walls of the cylinder, though.

    edit: Upon further thought, pressure doesn't work like my statement implies. Clearly, the lower part of the walls of a filled, upright, stationary cylinder have more pressure placed on them than the higher part of the walls, yet the force of gravity pulls downward.
    Last edited: Sep 11, 2015
  17. Sep 17, 2015 #16
    OK I have used this approach:

    Instead of integrating the area over the pressure which seems to be the suggestion of the OP (and why it would lead him to an awkward ##dr^2## expression), I integrated the pressure over the area:
    $$\int P~dA$$
    Consider the circular area of the plug in question. We will integrate the pressure over that area by adding up the pressure exerted on each concentric ring of area ##dA=2\pi r~dr##.
    Each ring will have pressure ##P=\frac{\rho~\omega^2~r^2}{2}## applied to it. All that's left is to integrate from ##r=0## to ##r=R##:
    F_{plug} &= \int_{r=0}^{r=R} P~dA\\
    F_{plug} &= \int_{r=0}^{r=R} \frac{\rho~\omega^2~r^2}{2} 2\pi r~dr\\
    F_{plug} &= \frac{\rho~\omega^2 \pi R^4}{4}

    edit: my answer is off by a factor of ##\frac{1}{4}## from BvU but I am not sure where he got that ##2\pi r^2## term inside his integrand.
    Last edited: Sep 17, 2015
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