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Bernoulli's spinning-top-cylinder problem

  1. Nov 2, 2007 #1
    Lets assume that vacuum empty cylinder has a spinning top. Rotational speed at circumference is V. Cylinder axis is oriented verticaly. Cylinder is surranded by air at athmospheric pressure and weights slightly more then displaced air.

    According to Bernoulli principle there should be pressure differential dp between top and bottom becuase of fluid speed difference. Bottom is stationary Vb=0, Average speed of top is Vt = 2/3*V (if I calculated correctly). This means that dp = <air density> * Vt * Vt / 2.

    Question: will cylinder fly if top spins fast enough ?

  2. jcsd
  3. Nov 8, 2007 #2

    Shooting Star

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    Hi, I'm sure it's my shortcoming, but could you write it up again, in more detail? Where exactly is the top in relationship to the cylinder?
  4. Nov 8, 2007 #3
    OK. Cylinder has a top (shape round) and a bottom (shape round) and the 2 are connected with a help of rectangular piece that is rolled up to create a tube. In other words, top and bottom are of the shape of both ends of a tube. This is all I can say about cylinder.

    As for the rest, I'm relaying on an imagination of the reader to allow round top spinning around it's central axis (like bike wheel, clock, roulette) while letting all the remaining parts of the cylinder rest. Cylinder is evacuated and placed in normal environment.

    Hint: it's close to a Magnus effect, but I expect it works quite different.

  5. Nov 8, 2007 #4
    Since you're having so much difficulty explaining a question in a way that other people can understand, you might like to try attaching a clearly labelled diagram next time.

    I think though that the answer is no, it doesn't work at all. You're assuming that by spinning just the top surface of an object (around its axis of symmetry), you can (by Bernoulli's principle) lessen the air pressure on just that top surface (causing the object as a whole to fly upward). This sounds like a misapplication of Bernoulli's principle (if you consider an air particle reflecting from the moving surface, the movement will have zero effect due to the symmetry with respect to every angle of rotation).
  6. Nov 8, 2007 #5
    Believe me, you don't want to see my drawing, they are worst then explanation.

    I'm not sure what do you mean (perhaps a little drawing... :tongue2:). Bernoulli effect describes pressure difference caused by fluid in motion. The direction of motion is immaterial, e.g. pressure in a stright pipe will change as much as in a precel bend pippe (as long as we are talking same speeds). If there is a pressure difference between arbitrary small element on top versus corresponding element at the bottom (and this is a big IF) then overall resulting force should be a stright sum of accross elements since pressure acts normal to the surface. The forces acting in other then axial direction sum up to null, due to the symmetry, but this is trivial.

    I guess, the part I'm missing is, why Magnus effect works and spinning-top does not, even though I can transform former into later by a series of small steps, each of them not affecting underlaying phenomenon.

  7. Nov 8, 2007 #6


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    Still not understanding your cylinder setup, but Magnus effect requires both an object to have both spin and linear speed in a gas or liquid.
  8. Nov 8, 2007 #7
    I'm not saying your contraption woudn't curve if tossed it sideways, but the Magnus effect is perpendicular to what I think you intend.
  9. Nov 8, 2007 #8
    Lets try to transform Magnus into spinning-top (actually spinning-front). This perhaps will clarify:

    Magnus> have a clockwise spinning cylinder enveloped with horizontal (right to left) linear air flow normal to the axis . At the bottom air is dragged by rotation and airflow speed increase causes pressure decrease. At the top air flow decrease is less because rotation opposes airflow. (see Wiki). The overall force is pointing down.

    Step 1> Let's stop the air flow and conclude that pressure drop from ambient anywhere on the side walls is equal. The overall force is 0.

    Step 2> Instead of considering cylinder side wall lets change attention to the front and back side sections close to perimeter and conclude that pressure drop there is almost same as at the perimeter. The overall force is 0.

    Spinning-front> Let stop back side and wall from turning and leave only front side spinning. Since pressure at the back equals to ambient and at the front is somehow less. The overall force is toward front.

    And where did I stray ?
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