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Rings of slide

  1. Mar 6, 2007 #1
    Suppose two cooper rings,each having the same surface smoothness and of same size(geometry).The only difference is that interior of the first ring is hollow while of the second isn't.m1<<m2.
    If we send two rings,with same inital velocities,to glide accross a frozen long path of ice,which one goes further if we disregard effect of air resistance?
    Explain your answer.
     
  2. jcsd
  3. Mar 6, 2007 #2

    cristo

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    Is this a homework question? Please show your work. What factors determine how far an object will slide on the ice?
     
  4. Mar 7, 2007 #3
    Force of the friction if F=K*mg,where K is the coefficient of the friction.
    From F=ma ,deceleration is:
    a=K*g

    But I don't think for the two rings ice sliding K is the same!
    I feel that the ring with m=10 kg will slide somewhat further than ring with m=0.1 kg.
    Will you help me to analyze the problem cristo?
     
  5. Mar 7, 2007 #4

    Dick

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    You've just correctly cancelled the mass out of your expression for the deceleration. The problem says to neglect air friction. If you include air friction would the results be more like you feel they should be?
     
  6. Mar 8, 2007 #5
    I feel that without air friction two Ks aren't the same.
    Think .The material is an ice.
     
  7. Mar 8, 2007 #6

    HallsofIvy

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    You might try thinking yourself. If the ice were completely frictionless, neither ring would ever stop. If there is any coefficient of friction at all, the friction will be proportional to mass- the lighter ring will go farther.
     
  8. Mar 8, 2007 #7

    Dick

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    But as the poster has computed, the acceleration is independent of the mass. Both rings go the same distance. He also seems to doubt that kinetic friction is a accurate physical model. That could only be shown by experiment.
     
  9. Mar 8, 2007 #8
    I think that the state of ice changes under higher pressure of the heavier ring .It kinda melts more.
    You watched curling (the sport) ?
     
  10. Mar 8, 2007 #9

    Dick

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    Good point. In fact, my wife has played. Then the two rings have different values of kinetic friction. That is a more complete answer than I was thinking of.
     
    Last edited: Mar 8, 2007
  11. Mar 8, 2007 #10
    But which one do you expect to go further?
    I would place my bet on the heavier one ,but can't argumented it.
     
  12. Mar 8, 2007 #11

    Dick

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    I think you've argued it about as far as you can argue it. You've shown IF they have the same value for kinetic friction (which they would - ignoring effects like ice pressure melting), they will slide the same distance. I would suspect that this is what the expected answer is. You've also argued the heavier one will, in the real world, cause more melting and slide farther. What else can you say?
     
  13. Mar 8, 2007 #12
    I want to know what the reality says.Not an inaccurate theoretical model of reality.
    Curling and Ice Hockey experiments for example.
     
  14. Mar 8, 2007 #13

    Dick

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    Then you will need to look up or perform these experiments. Calculating the effect from first principles is a seriously difficult problem.
     
  15. Mar 8, 2007 #14
    Can you ask your wife for help?What does she think guided by her experience?
    I don't want to wait for a year on freezing cold days to perform experiments.
     
  16. Mar 8, 2007 #15

    Dick

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    Curling stones are a standardized weight. I think your intuitive reasoning is quite correct. A sufficiently lighter stone of the same geometry would not go as far.
     
  17. Mar 9, 2007 #16
    If you don't mind an add on, this past summer at a water park, we were watching a half pipe ride while on the stairs for another ride. People were not allowed to push off for obvious reasons.

    Some of the riders got very close to the ascent end while most got maybe 3/4 up. But the most successful riders varied in size and weight, which got me to thinking about the best strategy to maximize altitude. It seemed to me that the frictional retarding force would be independent of mass. And that the drag forces would favor a heavier payload cuz that force is not dependent on mass, only velocity,etc.

    So too really fat guys would be best. Sort of like bob sledding, except there power to weight ratio is critical in obtaining the best V0 in the shortest distance. To my surprise out of 30 rides, a couple of small maybe 10 ten year old girls came closest to going over the edge.
     
  18. Mar 9, 2007 #17
    Problem is that different people have not just different mass but also different shapes of the body so the air resistance is different and significant factor too.
    Problem is that I want to exclude air resistance and consider only ice friction factors.
     
  19. Nov 6, 2010 #18
    I think the hollow ring will travel a much longer distance when compared to the other ring. There is a suggestion that the ice is frictionless and therefore the rings will never stop. I don’t think that this is correct. Ice has got some minimal amount of friction. Even if we admit that the ice is frictionless then also we should remember that acceleration depends on the mass. In this case the hollow ring will travel more because it has got less mass.
     
  20. Nov 6, 2010 #19
    Am I Correct.
     
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