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Phys Unit 4 - Swings

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Question 6:
    http://www.xtremepapers.com/Edexcel/Advanced%20Level/Physics/2008%20Jan/6754_01_que_20080121.pdf [Broken]

    "On one type of theme park ride, a...."


    2. Relevant equations



    3. The attempt at a solution
    So at the bottom of the loop the ride is at its fastest
    GPE = KE
    mgh = 0.5*m*v2
    max speed = 18.8

    The net force is mv2/r = Tension - mg

    How do I continue???
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 29, 2012 #2

    Doc Al

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    Consider the forces on the rider. (Not tension, but what?)
     
  4. Jan 29, 2012 #3
    reaction force from seat?
     
  5. Jan 29, 2012 #4

    Doc Al

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    Right. The reaction or normal force from the seat.
     
  6. Jan 29, 2012 #5
    ok I think I am working on it...
     
  7. Jan 29, 2012 #6
    so
    R = mv^2/r + mg
    R/mg = g-force
    g-force = 3
    right?
     
  8. Jan 29, 2012 #7
    but the next question 6b) is

    Two students in the queue are having a discussion.
    A says: “If they made a new ride twice as big the g-force at the bottom would be
    amazing!”
    B says: “I think the g-force wouldn’t be any different.”
    With reference to your calculation, explain which student is correct.

    If we look at our equation r (radius) is clearly a factor
    but the answer says B is correct>
    how?
     
  9. Jan 29, 2012 #8
    or would this cancel out due to a higher max speed?
     
  10. Jan 29, 2012 #9

    Doc Al

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    Right.
     
  11. Jan 29, 2012 #10

    Doc Al

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    mv2 also depends on the radius. Express that term in terms of the radius and see what happens.
     
  12. Jan 29, 2012 #11
    yes they then cancel out
    so (2g+g)/g = g-force
    3 = g-force
     
  13. Jan 29, 2012 #12

    Doc Al

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    Good.
     
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