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Forces of a basin surrounding a drain

  1. Mar 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A basin surrounding a drain has the shape of a circular cone opening upward, having everywhere an angle of 34.5° with the horizontal. A 28.5-g ice cube is set sliding around the cone without friction in a horizontal circle of radius R.

    (a) Find the speed the ice cube must have as a function of R.

    (b) Is any piece of data unnecessary for the solution? [YES]

    (c) Suppose R is made two times larger. Will the required speed increase, decrease, or stay constant?

    (d) Will the time required for each revolution increase, decrease, or stay constant?

    If it changes, by what factor? (If it does not change, enter CONSTANT.)

    (e) Do the answer to parts (c) and (d) seem contradictory? Explain.


    2. Relevant equations
    Sum of the forces = mac = mv2/R

    3. The attempt at a solution
    Attempted to draw a FBD and I still don't get it. From what I'm thinking, there's the force of gravity at an angle, and I guess that minus some sort of "normal" force would equal the centripetal force, but I don't think that's correct and I wouldn't even be able to get a normal force from that.
     
  2. jcsd
  3. Mar 1, 2016 #2

    haruspex

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    Do you mean, only a component of gravity? Which angle? The normal and the centripetal are not in the same direction either.
    There are two approaches you can take.
    1. Just go ahead and consider the force balance equation ΣF=ma in each of the horizontal and vertical directions.
    2. Since you don't care to know the normal force, consider only the force balance equation tangentially to the surface.
     
  4. Mar 1, 2016 #3
    I'm kind of confused now. I don't know if what I said made sense. I meant that if you did a force balance equation along the surface of the water, you would get the component of mg that goes along the surface as well as some opposite (which is what I meant by "normal" because I didn't know a better term), and I claimed the difference between these would be the centripetal force but now that I think about it I don't think I could do that and I don't think it'd work.

    Horizontally, I can only imagine there being the inward centripetal force, and vertically I can only imagine the force of gravity. Is there something I'm missing?
     
  5. Mar 2, 2016 #4

    haruspex

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    The centripetal force is a resultant, not an applied force. I.e. it is the ma part of the equation.
    You seem to have forgotten the normal force now. That has horizontal and vertical components.
    You mentioned water in your last post.... what water?
     
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