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A bucket spins

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A bucket spins at angular velocity ω. When the water stabilizes, what shape will the water take? Express your answer as a function of the radius.

    2. Relevant equations
    a = ω^2r
    a = v^2/r

    3. The attempt at a solution
    I honestly have no idea how to do this problem. Any help would be awesome
     
  2. jcsd
  3. Sep 21, 2015 #2

    billy_joule

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    Is that the entire problem statement word for word?
    Can you draw a diagram? Do you know what axis the bucket is spinning about?
     
  4. Sep 21, 2015 #3
    Here, sadly this is the only diagram that I have to go off of and yes, it is word for word.
     

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  5. Sep 22, 2015 #4

    haruspex

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    Consider a drop of water somewhere on the surface. You can think of the rest of the water as a solid, as far as that drop is concerned. If that drop is to stay at a given radius, what is the slope of the surface there?
     
  6. Sep 22, 2015 #5
    So would I just take the derivative of the a=w^2r equation with respect to w? and get 2wr?
     
  7. Sep 22, 2015 #6

    billy_joule

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    Nope, you are told theta is constant so you're not interested in the case where theta changes.

    Quickly stir a cup of water then remove the spoon and look at the shape of the water surface, hopefully this will give you an idea of the curve type you are looking for..

    The water has two forces acting on it, gravity and centrifugal. The water surface is perpendicular to the net force vector acting on it. At r=0 only gravity acts so the net force is downward and the surface is horizontal. As r increases centrifugal force increases and the direction of the net force is no longer vertical.
    Like Haruspex said, consider a drop somewhere on the surface. Draw a well labelled diagram.
     
  8. Sep 22, 2015 #7
    So what you are saying is that this water molecule will have the centrifugal force acting in the -r hat direction as well as the force of gravity acting vertically downward. This will cause newton's second law to be written as Fc = mv^2/r as the water moves in the circular horizontal motion and then Fg = may --> mg = may --> g = ay where y is the negative direction. So to get an image of water moving in a circular pattern, I would assume it looks like a spiral and thus would simply look at its horizontal movement and take into account the Fc = mv^2/r force. Would I then twice integrate this equation with respect to r in order to determine a position function while maintaining r?
     
  9. Sep 22, 2015 #8

    haruspex

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    Having determined those two forces, you do not need to worry any more about the circular motion. Just think of it as a droplet on an incline. What is the slope there?
     
  10. Sep 23, 2015 #9
    Oh! I think I figure it out thank you very much for your help. I'm assuming that the droplet can be represented as a mass, I find the motion along the slope and use this to equate it then take the integral.
     
  11. Sep 23, 2015 #10

    haruspex

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    That doesn't sound quite right. There is no motion along the slope, and that allows you to find the slope, which you then integrate.
     
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