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Surface of Water in a Rotating Bucket via Calc. of Variations

  1. Jun 6, 2014 #1
    I am referring to "Newton's bucket" -- essentially a hollow cylinder filled with water inside which rotates about the cylinder's axis. The result is that the surface of the water is a parabola. The argument for this is that the surface normal of the water must be directly opposing the other forces at the surface. This is fine.

    However, I am wondering if such a problem can be approached with calculus of variations to solve for a function describing the surface.

    The problem I ran into, when naively trying to do it this way, is that the energy will of course be minimum when the height of the water is minimum, leading to obtaining no stationary solution, and the useless conclusion that the energy is reduced when the total amount of water is reduced. So, I want only want to consider variations of the height function that satisfy and additional constraint that the total amount of water is the same.
     
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  3. Jun 7, 2014 #2

    BruceW

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    hmm. I think you'll have more success in extremizing the action of the system, i.e. the kinetic energy minus the potential energy. But yes, you'll still need to use the constraint for the total amount of water. I'm pretty sure that trying to minimise the total energy (even while using the constraint) will not work. If you think about it for a bit, we know that what happens is that the water surface takes a parabola shape, with the water curving upwards at the edge of the bucket. But also, the water is moving faster at the edge of the bucket. In other words, the curve is such that there is more fast moving water than there would have been with a flat horizontal curve. So energy is not minimised by the choice of curve. Also, we know that energy is not maximised because that would simply be a curve that goes up very sharply near the edge of the bucket. So I think you need to extremise the action, not the energy.
     
  4. Jun 7, 2014 #3
    I was working in the rotating frame, where the water was assumed to be stationary. Sorry if that was not clear. Does it make sense then?
     
  5. Jun 7, 2014 #4

    BruceW

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    oh, ok. yeah, that would work too. But it is a bit less intuitive, in my opinion. You would need to introduce a fictitious force, and from that, you can calculate a 'fictitious potential' (I'm not even sure if that is the proper terminology, which is why I've put scare quotes around it). And from there, you can add this to the potential due to gravitational field. Also, since there is zero kinetic energy in the rotating frame, this means Extremizing the action is essentially the same as extremizing the total energy (since there is only potential energy). So anyway, if you extremize the potential energy, you should get the correct answer for the shape of the surface of the water.

    edit: also, you still need to include the constraint of the volume of water, which you can do by introducing a Lagrange multiplier. so you should have something like:
    [tex]\int \left( \text{P.E. per volume} + \lambda \right) \ dV[/tex]
    and you should then extremize this expression with respect to the function ##z(r)##.
     
    Last edited: Jun 7, 2014
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