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Slippery Problem please help!

  1. Sep 26, 2007 #1
    A boy is initially seated on the top of a hemispherical ice mound of radius R = 2.6 m. He begins to slide down the ice, with a negligible initial speed (Fig. 8-47). Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?
     
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  3. Sep 26, 2007 #2

    Avodyne

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    Ah, a classic problem!

    What are your ideas to solve it? Can you say what is special about the position or velocity or acceleration at the point that contact is lost?
     
  4. Sep 26, 2007 #3
    i really have no idea where to start. the velocity at that point is very small and the acceleration would be positive, the position would be a height less than 2.6 but thats all i can figure out.
     
  5. Sep 26, 2007 #4

    Avodyne

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    Will the velocity really be "very small"? What would "very small" mean?

    Before he leaves the ice, how is his acceleration in the radial direction related to his speed? What forces produce this acceleration? (Try drawing a free-body diagram.)

    Are all those forces still present at the moment he loses contact?
     
  6. Sep 26, 2007 #5

    Avodyne

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    After he leaves the ice, yes, but before he leaves there is another force.
     
  7. Sep 26, 2007 #6
    oh centripetal force is also acting on him before he looses contact
     
  8. Sep 26, 2007 #7
    v=sqrt(2g(r-h))
     
  9. Sep 26, 2007 #8

    Avodyne

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    Well, I would call it the normal force of the ice on the boy.
     
  10. Sep 26, 2007 #9

    Avodyne

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    Yes. (Can you explain why this equation is valid?)

    So, you have one relation between v and h.

    At the moment he loses contact, the normal force of the ice on the boy is zero. So, his centripetal acceleration must be produced by the force of ???? And the component of that force in the radial direction is ??? (Why do we want the component in the radial direction?)
     
  11. Sep 26, 2007 #10

    Avodyne

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    That's a sloppy way of saying it. I would say the centripetal acceleration is produced by the force of gravity. It's best not to confuse cause (force) with effect (acceleration).
     
  12. Sep 26, 2007 #11
    so, centripetal force =0, so gravity must produce his centripetal acceleration. how do we find the radial component with no angle?
     
  13. Sep 26, 2007 #12
    i set the centripetal force equal to gravity to solve for v, then plugged v into my first equation to get height but that was not right.
     
  14. Sep 26, 2007 #13

    Avodyne

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    There is an angle; the boy has moved off the top, and now a line from him to the center of the sphere makes an angle theta with respect to vertical. How is this angle related to h and R?
     
  15. Sep 26, 2007 #14

    Avodyne

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    Show your work!
     
  16. Sep 26, 2007 #15
    sin(theta)=h/r
     
  17. Sep 26, 2007 #16
    so, so far i have F=mv^2/R, v=sqrt(2g(r-h)), and sin(x)=h/r
     
  18. Sep 26, 2007 #17

    Avodyne

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    What is F?
     
  19. Sep 26, 2007 #18
    the magnitude of the centripetal force
     
  20. Sep 26, 2007 #19

    Avodyne

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    And that should be equal to ???
     
  21. Sep 26, 2007 #20

    Avodyne

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    No, sorry. I think you need to find someone locally who can help you with some basics. (A force is not going to be equal to a length ...)
     
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