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Amusement Park Centripetal Force Around A Loop

  1. Oct 28, 2008 #1
    In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane. Suppose the ring rotates once every 4.90 s. and the rider's mass is 58.0 kg.

    A.) With how much force does the ring push on her at the top of the ride?

    B.) With how much force does the ring push on her at the bottom of the ride?

    C.) What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

    If anyone could help me with these questions that would be great. Thanks



    I tried this using the equations to find v = (2pi(r)) / T

    but then i realized that the velocity is different at the top than at the bottom. I am stumped as to how to find the velocity then find the force.

    I think the equation for force at the top is (m(vtop)2) / R

    Can anyone help me out with this though please
     
  2. jcsd
  3. Oct 28, 2008 #2

    LowlyPion

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    No. That's not what's going on.

    Draw a force diagram. You have 2 forces acting on the 58 kg person.

    There's weight - which always acts down and there is the radial acceleration from the rotation of the ride. So there are two forces. What is the difference between looking at the forces between the top and the bottom then?
     
  4. Oct 28, 2008 #3
    The force at the top would be adding to each other while at the bottom they would be opposing. But how do u find the normal force?
     
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