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Skier on a Circular Hill

  1. Apr 29, 2009 #1
    A skier weighing 0.80 kN comes down a frictionless ski run that is circular (R = 30 m) at the bottom, as shown. If her speed is 12 m/s at point A, what is her speed at the bottom of the hill (point B)?

    http://www.physics.gatech.edu/people/faculty/larry/images/16212.gif

    I keep getting mixed results.

    I did 30(cos(40))*-1 to get the distance from A to B... but now what?
     
  2. jcsd
  3. Apr 29, 2009 #2

    Doc Al

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    Staff: Mentor

    I assume you mean 30*(1 - cos40). What's conserved?
     
  4. Apr 29, 2009 #3

    Energy is conserved. But I honestly do not have a clue where to proceed. I am obviously missing the portion of this in the book. The answer is in front of me, but I am being blind.
     
  5. Apr 29, 2009 #4

    Doc Al

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    Yes, mechanical energy (KE + gravitational PE) is conserved. So how would you express that mathematically?
     
  6. Apr 29, 2009 #5

    PE = mgy
    KE = .5mv^2

    so... 800N / 9.8 = 81.63

    (.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29/1000.. still wrong. where did I mess up?
     
  7. Apr 29, 2009 #6

    Doc Al

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    The left side is perfectly OK, but I'm not sure what you're doing on the right. Set the left side equal to .5mv^2 and solve for v.
     
  8. Apr 29, 2009 #7
    I was converting to kN... it did not matter, I get it.

    Bingo!


    (.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29

    11492.29/.5/81.63 = 281.57 ; 281.57^1/2 = 16.78m/s

    Excellent, thank you for your help!
     
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