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Angular velocity of a ferris wheel

  1. Oct 21, 2004 #1
    This is a hard problem!! :cry: Anyone know what my mistakes are??

    -----
    A passenger on the ferris wheel normally
    weighs 367 N. The ferris wheel has a 14 m radius and is
    equipped with a powerful motor. The operator revs it up so that the customers at the top of the wheel feel zero g's (they momentarily
    lift slightly of their seats). The acceleration of gravity is 9.8 m/s^2

    1) At what angular velocity will this occur?
    Answer in units of 1=s.

    N = 367 N
    R = 14m
    g = 9.8 m/s^2
    N = 0

    Fc = mg = m(V^2/R)
    gR = V^2

    (9.8)(14) = v^2 = 137.2 = 11.71

    (Angular Velocity) W= V/R

    11.71 / 14 = .835


    2) Assume: The rotating angular velocity is
    same as in Part 1.
    What weight does the customer feel at the
    bottom of the wheel? Answer in units of N.

    Fc = N - mg = m(V^2/R)

    N = mg + m(V^2/R)

    V = WR

    N = mg + m [(W^2 x R^2)/(R)]

    = mg + mW^2R

    = (367)(9.8) + (367) (.84)^2 (14)

    = 7221.9728
     
  2. jcsd
  3. Oct 21, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks OK to me. What are the units of your answer?
    One problem: 367 N is the passenger's weight, not mass!
     
  4. Oct 21, 2004 #3

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    Another hint for part 2 is that the magnitude of the centrifugal force is always the same. If you understand that, you should be able to immediately write down the correct answer.
     
  5. Oct 23, 2004 #4

    So for part 2...

    it should be uhm

    (11.71) (9.8) + (11.71) (.84)^2 (14)

    =230.43 N?
     
  6. Oct 23, 2004 #5
    How did you get a mass of 11.71 kg? The passenger's normal weight is 367 newtons....and

    weight = mass * g.
     
  7. Oct 23, 2004 #6

    So would it be...

    (37.4) (9.8) + (37.4) (.84)^2 (14)

    = 735.97 ?
     
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