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I Circular motion of a bucket filled with water

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  1. Aug 23, 2016 #1
    Consider a bucket filled with water in vertical circular motion.
    Why does there exist a contact force between water and the bucket when the bucket is at the top of center of rotation?
    What will happen if I remove the bottom part of the bucket when it is on the top?
     
  2. jcsd
  3. Aug 23, 2016 #2

    A.T.

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    Because the bucket accelerates downwards at more than 1g.
    What do you think?
     
  4. Aug 23, 2016 #3

    sophiecentaur

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    The water will leave through the bottom and travel (as the theory always tells us) on a tangential path. This is exactly the same as if any mass were being swung on a string in vertical plane and going fast enough to be kept in a circle. Naturally, there are practicalities involved with actually getting all the water to leave at once and, in reality, there would be a 'fan' of water out of the bottom, each bit would be tangential to the path of the bucket.
    There exists a contact force between bucket and water because the bucket is providing a centripetal force to constrain the water to move in a circular path. It has to be going fast enough to have a centripetal acceleration of at least g, for the bucket to 'overtake' the Earth's gravitational acceleration of g downwards and still be 'pressing inwards' on the water. At the bottom, the contact force will be at least twice the weight of the water, if the rotation is uniform.
     
  5. Aug 23, 2016 #4
    How do you make up the assumption that acceleration is greater than g. Isn't it possible that the reaction force cancels some of the weight leaving the centripetal acceleration smaller than g?
     
  6. Aug 23, 2016 #5

    A.T.

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    Well, try to rotate it slower and report back.
     
  7. Aug 23, 2016 #6
    What do you get if you do a force balance on the water in the bucket?
     
  8. Aug 24, 2016 #7
    R+mg = mv^2/r
     
  9. Aug 24, 2016 #8
    Excellent. So this is the equation that applies at the moment that the bucket is directly overhead. What can you conclude from this equation?
     
  10. Aug 24, 2016 #9
    Many things can be concluded. What are we emphasizing on?
     
  11. Aug 24, 2016 #10
    Conclusions related to the questions you asked.
     
  12. Aug 24, 2016 #11
    Well if you're hinting about the reaction force then we can conclude that
    R = mv^2/r - mg
     
  13. Aug 24, 2016 #12
    So what does that tell you?
     
  14. Aug 24, 2016 #13
    Well it tells me that as long as v^2/r > g, there will be a reaction force. If v^2/r = g then reaction force will be 0
     
  15. Aug 24, 2016 #14
    Excellent!!!!

    Now, if there were a small hole in the bottom, would water spray out if the inequality applied?
     
  16. Aug 24, 2016 #15
    No because centripetal acceleration is moving the water towards itself more as compared to gravity.
     
  17. Aug 24, 2016 #16
    But the reaction force tells you that there is fluid pressure at the bottom of the bucket. The pressure is equal to the reaction force divided by the cross sectional area of the bucket.
     
  18. Aug 24, 2016 #17
    How about when v^2/r = g. What will happen to reaction force then?
     
  19. Aug 24, 2016 #18
    It would be zero(as you said), and the fluid pressure at the bottom would be zero(gauge). So the water would be in free fall, and none would spray out the hole.
     
  20. Aug 24, 2016 #19
    Okay one more thing, if the tangential velocity is towards the right at top. The water must be pushed from the left side of the bucket. Wouldn't that give rise to a reaction from the left side?
     
  21. Aug 24, 2016 #20
    Is the tangential velocity constant?
     
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