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Circular motion-roller coaster

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    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 3.0 times her weight as she goes through the dip.If r=25m, how fast is the roller coaster traveling at the bottom of the dip?

    3. The attempt at a solution

    F=mv*2/r
    3Mg=mv*2/r
    3g=v*2/25
    V=27.12ms-1
    Is this the correct solution?
     
    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 6, 2012 #2

    PhanthomJay

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    No, you must look at the net force acting on the passenger before applying Newton's laws. You only looked at one of the forces. What is the other?
     
  4. Oct 6, 2012 #3
    But, isnt the magnitude of the force is equal to 3 times of the weight...where w is equal to mg
    ( base on the question)
     
  5. Oct 6, 2012 #4

    BruceW

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    the question does say that the force due to the seat is 3 times her weight. But she does have another force on her (as Phanthomjay said).
     
  6. Oct 6, 2012 #5
    So, the net force is N-mg...where N(upward force) is equal to 3 times of mg(weight)...???
     
  7. Oct 6, 2012 #6

    PhanthomJay

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    Yes....
     
  8. Oct 6, 2012 #7
    Thanks...
     
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