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Circular Motion: On a Ferris Wheel

  1. Oct 6, 2009 #1
    On a Ferris Wheel

    A chemistry student with a mass of 75 kg is riding a steadily moving Ferris wheel. When she is at the top of the Ferris wheel, the normal force from the seat on to her body has a magnitude of 550 N.

    a) What is the magnitude of the normal force on her body when she is at the bottom of the Ferris wheel's arc?
    b.) What would the normal force be on the student at the top of the wheel if the wheel's velocity were doubled? FN =-7.25


    F = (mv^2)/R

    Well I managed to solve for b by figuring out FN = 4(mg - 550) - 4(mg) = -7.25
    But I'm really stuck on a. I don't understand why it just wouldn't be 550.
     
  2. jcsd
  3. Oct 7, 2009 #2

    Delphi51

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    Homework Helper

    Use F = ma for the person at the top point. I want to take "down" as positive because I know the centripetal force ma must be down:
    mg - 550 = ma (assuming the seat pushes UP on the person).
    Here the a is the circular acceleration v^2/r. Another way to look at it is that gravity provides the centripetal force Fc plus 550 extra:
    mg = Fc + 550. You may prefer to change all the signs so up is positive - and acceleration is negative.
     
  4. Oct 7, 2009 #3
    Ok so I'm a little more confused. Are you saying that at the bottom of the wheel, Fc = Normal force?
     
  5. Oct 7, 2009 #4
    Never mind I figured it out.

    Fc = mg - N so Fc = 735.75 -550 = 185
    Mv^2 / r = 185 so 185 / M = v^2 / r = 2.47

    so again Fc = mg - N so N = 75(2.47) + 735.75 = 921
     
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