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Analysis on the circular motion

  1. Dec 14, 2004 #1
  2. jcsd
  3. Dec 14, 2004 #2

    Doc Al

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    Do you really think the child loses contact with the slide before even reaching the rounded section? :rolleyes: Don't just guess.

    Consider that in maintaining contact over the rounded section, the child must undergo circular motion. And that requires a centripetal force. (What force holds the child to the slide?) At some point, the child will be going too fast for the force to maintain the circular motion---off he goes.
     
  4. Dec 14, 2004 #3

    Tide

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    At some point on the circular section the child's speed will be such that mv^2/r exceeds the normal force holding him on the slide.
     
  5. Dec 14, 2004 #4

    Pyrrhus

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    When the normal force = 0 the boy loses contact with the surface.

    So on your analysis on the circular motion

    [tex] n - mg \cos \theta = -m \frac{v^2}{R} [/tex]
     
    Last edited: Dec 14, 2004
  6. Dec 14, 2004 #5
    but how does that get me the height?
     
  7. Dec 14, 2004 #6

    Tide

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    You should be able to determine the speed of the child in terms of her height.

    HINT: Energy is conserved!
     
  8. Dec 14, 2004 #7

    Pyrrhus

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    What i said above means when [itex] v^2 = Rg \cos \theta [/itex] it will be at the point it leaves the surface.

    Hint: Use this fact and Tide's hints.
     
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