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Virtical Circles

  1. Aug 5, 2006 #1
    Here's the question.

    A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius r. A passenger feels the seat of the car pushing upward on her with a force equal to twice her weight as she goes through the dip. If r=20.0 m, how fast is the roller coaster traveling at the bottom of the dip?

    I'm having some real troubles answering this problem. Thus far, here's
    what I've come up with.

    Fc = ( mv^2 ) / r
    FN - 2mg = ( mv^2 ) / r
    2mg - mg = ( mv^2 ) / r
    (2mg - mg )(r) = ( mv^2 )
    (2mg - mg )(r) = ( v^2 )
    sqrt( (2mg - mg )(r) ) = v

    sqrt( ( 19.6 - 9.8 m/s^2 )( 20.0m ) ) = v
    14 m/s = v

    I've been informed that 14 m/s is not the correct answer and that to one digit the correct answer is 10 m/s.

    It seems to me that I have all the right pieces, but I may be missing something. Will someone help, please? Big thanks!
     
  2. jcsd
  3. Aug 5, 2006 #2
    Create a relationship at the bottom of the coaster. Draw a picture of all of the forces and find the resultant force.
     
  4. Aug 5, 2006 #3
    These are the only forces I see:

    UP: F_c, F_N
    Down: W

    Am I missing any?
     
  5. Aug 5, 2006 #4
    CORRECTION:

    UP: F_N
    Down: W (mg)
     
  6. Aug 5, 2006 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What do you mean by F_N? The forces felt by the person at the bottom will be the centrifugal force (your F_c?) and the person weight, both downward.
     
  7. Aug 5, 2006 #6

    Doc Al

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    Staff: Mentor

    Correct. Those are the only forces acting on the person.

    Your analysis--and answer--in post #1 seem correct to me. (It's also true that to one significant digit, 10 m/s is the answer--14 rounds off to 10. But why in the world would you settle for one digit?)
     
  8. Aug 6, 2006 #7
    Thanks for the help.

    There was something lost in translation when the professor told me 'to one digit.'

    Thanks again.
     
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