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Ferris Wheel Rotation physics

  1. Oct 1, 2007 #1
    A Ferris wheel that rotates three times each minute and has a diameter of 16.0 m

    What force [magnitude and direction (measured inward from the vertical)] does the seat exert on a 43.0 kg rider when the rider is halfway between top and bottom, going up?

    Ok, the magnitude is easy...

    Fnet = m(centripetal acceleration^2 + acceleration due to gravity^2)^1/2 = 422.77 N

    What I can't figure out to save my life is the direction "measured inward from the vertical"..

    My thinking is to basically make a triangle (similarly to the method used to find Fnet) with 33.95 N being the opposite side (pushing in), and the adjacent side being 421.4 N (pushing up). This gives 0.08 degrees in from vertical, which is incorrect.

    What am I doing wrong here? :confused:

    Thanks!
     
  2. jcsd
  3. Oct 2, 2007 #2

    learningphysics

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    opposite is 33.95. adjacent is 421.4. What is arctan(33.95/421.4) ?
     
  4. Oct 2, 2007 #3

    andrevdh

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    The rider is experiencing two forces, the seat (diagonally up and inwards) and his weight (down) these two together produce the (horizontally inwards) centripetal force.

    Fnet (the diagonal I get a bit different from you 427 N). The angle that is required is therefore between the weight and Fnet vector.

    Also I do not get a vector of 33.95 N in my calculations.
     
  5. Oct 2, 2007 #4

    learningphysics

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    I'm getting 33.95N. v = 3.14(16)*3/60 = 2.513m/s. mv^2/r = 43*(2.513)^2/8 = 33.95
     
  6. Oct 2, 2007 #5
    OK... This is the second time this semester I've made myself look like a fool because I was in radian mode...

    Sorry... the answer is indeed arctan(33.95/421.4).

    :frown:
     
  7. Oct 2, 2007 #6

    learningphysics

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    Don't feel bad. I've made that same mistake a couple of times right here on the forum!
     
  8. Oct 3, 2007 #7

    andrevdh

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    Ok, I took the diameter for the radius in my calcs.
     
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