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Centripetal acceleration of fairgrounds ride

  1. Oct 1, 2014 #1
    A fairgrounds ride spins its occupants inside a flying-saucer-shaped container. If the horizontal circular path the riders follow has a 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration 2.60 times that of gravity?
     
  2. jcsd
  3. Oct 1, 2014 #2

    TumblingDice

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    Gold Member

    Welcome to PF plowboy! How far have you gotten on your end? Do you understand the physics and forces involved? Have you formulated a mathematical approach?
     
  4. Oct 1, 2014 #3
    I get you have to use ac=Vt/r and that r will equal 8 and ac= 2.60. I think you will change the formlua to ac*r=Vt
     
  5. Oct 1, 2014 #4

    TumblingDice

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    I think you're close and on the right track. Take a look at the wiki on Uniform Circular Motion here:
    http://en.wikipedia.org/wiki/Circular_motion#Uniform

    The equations mentioned after these texts can be simplified/adapted:
    and
    If we assume a mass of '1', can you see how the second equation for centripetal force simplifies to the first equation for acceleration? You mentioned an acceleration 2.60 times that of gravity. Do you know how to compute 2.6g?

    I hope we're both on the right track! :w
     
  6. Oct 1, 2014 #5

    Nugatory

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    Staff: Mentor

    This thread really belongs in the homework forum, so I'm moving it there.
     
  7. Oct 2, 2014 #6
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