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Roller Coaster Loop Problem

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    The diagram below shows a roller coaster ride which contains a circular loop of radius r. A car of mass m begins at rest from point A and moves down the frictionless track from A to B where it then enters the vertical loop (also frictionless), travelling once around the circle from B to C to D to E and back to B, after which it travels along the flat portion of the track from B to F (which is not frictionless).

    1. Find the centripetal acceleration of the car at points B and C.
    2. Find the normal force at point B.

    2. Relevant equations
    Fnet = mv^2/r
    ac = v^2/r


    3. The attempt at a solution
    1. I got Ac = Fn/m - g for point B. For point C i got Ac = -Fn/m + g
    2. I got Fn = mv^2/r + g


    I got -2 marks off for this problem.. where did I go wrong?

    By the way, the loop goes counter clockwise (B, C, D, E)
     
  2. jcsd
  3. Feb 21, 2009 #2

    LowlyPion

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    If those are your answers, I'd say you didn't answer the question.

    They asked for centripetal acceleration. Just curious why you involved g?
     
  4. Feb 21, 2009 #3
    Fn - mg = mAc
     
  5. Feb 21, 2009 #4

    LowlyPion

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    And what is Fn?
     
  6. Feb 21, 2009 #5
    Fn is.. normal force? What do you mean. There is no value given.
     
  7. Feb 21, 2009 #6

    LowlyPion

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    Well you are expected to come up with a value for that. Your answer is no answer at all.

    For instance what is it at the point B?

    I can't see your diagram but I presume it is at the base of the loop where the horizontal angle is 0°. So what is the Fn at that point?
     
  8. Feb 21, 2009 #7
    Fn at point B is Fg ??
     
  9. Feb 21, 2009 #8

    LowlyPion

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    Which means what as far as the centripetal acceleration?
     
  10. Feb 21, 2009 #9
    it equals 0? =S
     
  11. Feb 21, 2009 #10

    LowlyPion

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    That might have made a better answer then. Now if your point B is just a hair past the 0° point its acceleration will be v2/r. In fact it may be open to debate whether right at B there is centripetal acceleration depending on how you want to interpret the drawing, (which I am disadvantaged in not being able to see).

    What I can tell you is that answering Fn - mg as the centripetal acceleration is not showing how much you should know.
     
    Last edited: Feb 21, 2009
  12. Feb 21, 2009 #11
    Right understandable.

    So at point C, what would be the appropriate answer if the tangent to the circle at this point is perpendicular to the x axis?
     
  13. Feb 21, 2009 #12

    LowlyPion

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    I will guess though that it is at 3 o'clock.

    In which case the centripetal acceleration will be determined by its speed after losing 1 radius worth of potential energy.
     
  14. Feb 21, 2009 #13
    I don't understand. We didn't learn potential energy yet.. How can I calculate the acceleration without it?

    And also for part B, There is no way to get value for Fn because there is no mass??
     
  15. Feb 21, 2009 #14

    LowlyPion

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    That's surprising, because these problems are typically addressed in terms of kinetic and potential energy.

    Specifically

    1/2*mVC2 = 1/2mVB2 - m*g*(HC-HB)

    The m cancels out and you are left with

    Vc2 = VB2 - 2*g*(HC-HB)

    ac = 1/R*(VB2 - 2*g*(HC-HB))

    And what is Hc - HB ?

    That's simply R, so

    ac = 1/R*(VB2 - 2*g*R)
     
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