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Centripetal acceleration and force

  1. Mar 14, 2008 #1
    The cars on a theme park ride each have a mass of 500 kg and travel around a vertical loop of diameter 20m.
    What is the minimum speed at which the cars must enter the loop in order to remain in contact with the track and What will then be the maximum reaction of the track?



    M=mass
    g=gravity
    r=radius of loop

    i dont know how to work out the mimimum speed required becuase there is no time value or angular acceleration given.
    to work out the maximum reaction
    Square root of (MgXr)
    500X9.81=4905
    4905X10=49050
    square root of 49050 = 221.472 newtons (not sure on the forumla though)
     
  2. jcsd
  3. Mar 14, 2008 #2

    tiny-tim

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    Hi Batman121! Welcome to PF! :smile:

    Hint: what equation must you apply at the top of the loop for the car to remain in contact with the track? :smile:
     
  4. Mar 14, 2008 #3
    the only one i can think of is
    centripetal force + gravity has to equal the centrifugal force acting on the cars
     
  5. Mar 14, 2008 #4

    tiny-tim

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    That's the one! :smile:

    So … putting the numbers in … the minimum velocity, at the top, for a car to remain in contact with the track is … ? :smile:
     
  6. Mar 14, 2008 #5
    so it would be mass of 500 X acceleration to get centrigufal force. but i dont know acceleration or how to work it out.
     
  7. Mar 14, 2008 #6

    tiny-tim

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    Acceleration of something moving in a circle of radius r with speed v is [tex]\frac{v^2}{r}[/tex] towards the centre of the circle.

    That also equals [tex]\omega^2r[/tex], where [tex]\omega[/tex] is the angular velocity, v/r.
     
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