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The stable beaker

  1. Aug 6, 2014 #1
    1. The problem statement, all variables and given/known data
    An empty cylindrical beaker of mass 100g, radius 30mm and negligible wall thickness, has its centre of gravity 100mm above its base. To what depth should it be filled with water so as to make it as stable as possible?

    55.9mm

    2. Relevant equations
    Center of mass equations

    3. The attempt at a solution
    I tried to find the point which the beaker's center of mass gets furthest from the vertical by rotating the base, while in equilibrium, but the problem is to find the distance of the water's center of gravity to the vertical. How should I get it?
    Besides, I've found a solution which says: "Clearly, for maximum stability the overall centre of gravity must lie in the water surface". It's correct, but how can I prove it?
     
  2. jcsd
  3. Aug 7, 2014 #2

    ehild

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    To maximum stability, the position of the overall centre of gravity has to be as low as possible.

    At what height is the centre of gravity when the height of water is h? Find the minimum of the expression with respect h.

    ehild
     
  4. Aug 7, 2014 #3

    haruspex

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    The given solution is wrong.
    In general, stability can be defined in various ways. E.g., resistance (restorative force or torque) to a small perturbation. With that definition, maximum stability will be for a full beaker.
    The 'lowest centre of gravity' answer typically corresponds to a definition based on the degree of perturbation (specifically, tilt angle) required to make it unstable. However, that only works for solid objects. When the beaker is tilted, the contents will shift. At the point of instability, the water will occupy a more complicated region. (Particularly complicated if it no longer covers the base of the beaker.)
     
  5. Aug 7, 2014 #4

    AlephZero

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    The question seems to assume the "most stable" position is with the lowest center of mass, though the objections to that definition in the other posts are valid ones.

    Think how the center of mass will move, if you add a small amount of water
    (1) below the current center of mass position
    (2) above the current center of mass position.
     
  6. Aug 8, 2014 #5

    ehild

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    Yes it is the question, how the most stable state is defined. It is usual to define stability with the potential energy, and a system is in a stable steady state if it is at the bottom of a potential well - with respect to some variable.

    Here the variable is the height of the water, and as the mass changes, you can define the most stable configuration of lowest potential energy per unit mass. With that definition, the given solution is correct.

    The OP thanked me for my hint, so I think he has found the expected solution.

    ehild
     
  7. Aug 8, 2014 #6

    haruspex

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    Two objections:
    1. The lowest PE state of a given assemblage is not the same as the lowest PE per unit mass of an assemblage that is allowed to vary.
    2. Your first definition of stability does not involve degrees of stability, and defines locally stable states. In mechanical terms, when extending this to degrees of stability it makes more sense to consider the strength of the response to perturbation than whether a stable state is also the globally lowest energy.
    Oh, I'm sure it's what the questioner wanted, but that doesn't make it right. I think it very important to point out to the student that the answer is not really valid.
     
  8. Aug 8, 2014 #7

    ehild

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    How would you define the degree of stability in this case? How would you solve the problem?


    ehild
     
  9. Aug 8, 2014 #8

    haruspex

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    It depends why you care about stability.

    If your concern is the beaker tipping over, what matters is the extent of perturbation required to produce instability. In your PE terms, that's the depth of the well. So that would be maximising the tilt angle at which the total centre of mass is above the lowest point of the beaker.

    If you care about the beaker not moving much from the vertical when buffeted by sideways forces, you would consider the energy required to produce a given small tilt. In PE terms, that's the shape of the bottom of the well. Not certain, but I think the answer that would give is to fill the beaker to the brim.
     
  10. Aug 8, 2014 #9

    ehild

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    If you fill the beaker to the brim the water will spill out at the slightest tilt.

    You said "it depends why you care about stability". You can ask maximum tilt when the beaker is filled to height h, and assign maximum stability to maximum tilt angle, but you also can say "I mean maximum stability of the beaker if the overall CM is at lowest position". At high-school level I would mean so. But I would like to see your solution.


    ehild
     
    Last edited: Aug 9, 2014
  11. Aug 9, 2014 #10

    haruspex

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    It's still the theoretical limit. If you stop anywhere short of the brim then (with that definition of stability) you can make it more stable by adding just a little more.
    Of course you can define it that way, but it does not correspond to any actual physical exemplification of stability. It only seems reasonable because for rigid bodies the two turn out the same.
    EDIT: To expand on that...
    Any reasonable definition of stability should take the form "the extent of a stimulus of a specified type that produces a specified response". In the present context, the stimulus could be a force, a torque or work done (and yet other possibilities). The response could be an infinitesimal rotation, a specific angle of rotation, or the rotation to the tipping point. I cannot think of any definition within this form which will lead to the answer "lowest centre of gravity of water+beaker".
     
    Last edited: Aug 9, 2014
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