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Ferris wheel problem - vertical circles and centripetal acceleration

  1. Dec 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Given a Ferris wheel that rotates 5 times each minute and has a diameter of 19 m, with the acceleration of gravity as 9.8 m/s^2, what is the centripetal acceleration of a rider? Answer in units of m/s^2.

    (There's a diagram that shows a ferris wheel with radius 9.5 m spinning at ω radians per second.)

    2. Relevant equations
    centripetal acceleration = v2 / r = ω2r


    3. The attempt at a solution
    one rotation was equal to 2pi * r, so
    5 rotations = 5(2pi * r) radians in one minute
    or 10pi * r / 60 radians/second = ω

    I used centripetal acceleration = ω2r = (10pi * r / 60)2 * 9.5. I got 4.97 for centripetal acceleration. But this was wrong! I don't know where I made a mistake.

    Oh, also, how does the force due to gravity figure into this problem? Shouldn't it be ignored since the ferris wheel is being powered by a motor?

    ----------
    While I'm at it, I might as well ask about the other three parts of the problem.
    Part b
    1. The problem statement, all variables and given/known data
    What force does the seat exert on a 72 kg rider at the lowest point of the ride? Answer in units of N.
    2. Relevant equations
    Fnet = Fcentripetal, so since net force must be upwards to maintain a centripetal force, normal force will be stronger than net force.
    So: Fnormal - Fg = Fnet = Fc = m * v2/r
    Fnormal = Fg + Fc = m(g + ac)
    3. The attempt at a solution
    Isn't the force exerted by the seat the normal force?
    I think it would just be the mass times the sum of acceleration due to gravity and the centripetal acceleration, or m(g + ac). I can't do this though as I have not found centripetal acceleration yet. That was part one.

    Part c
    1. The problem statement, all variables and given/known data
    What force does the seat exert on a 72 kg rider at the highest point of the ride? Answer in units of N.
    2. Relevant equations
    At the highest point, Fnet = Fc = Fg - Fnormal
    Fnormal = mg - mac
    3. The attempt at a solution
    Operation seems wasy with the information from part a.

    Part d
    1. The problem statement, all variables and given/known data
    What force (magnitude) does the seat exert on a rider when the rider is halfway between top and bottom? Answer in units of N.
    2. Relevant equations
    Fg = mg
    3. The attempt at a solution
    Wouldn't this just be the weight of the rider since there would be no centripetal force pulling him up or down? Or would it be the equilibrant to the force due to gravity nd the centripetal force?

    Thanks for help in advance!
    Bonus information: I have only two hours to submit the answer to this problem electronically. Because I am a procrastinator. :(
     
    Last edited: Dec 15, 2009
  2. jcsd
  3. Dec 15, 2009 #2
    The way I solve it, I consider that the Ferris wheel seat is a part of the wheel, and since the seat is at the rim of the wheel, we can say that the centripetal acceleration of the seat is (w^2)r, where w=(2pi x no. of rotations per min)/ 60.

    we get a = 2.6018 m/s^2.
    This gives us part (a).

    Now for (b) and (c) :
    We consider, the person to be in a non inertial frame that is accelerating at 2.6018 m/s^2 towards the centre. As a result, i consider, he is facing a pseudo force of the same, in a dir'n opposite to the dir'n of the centripetal force.

    At the highest point, we use the equation of equilibrium.
    We know Normal force, N is in the upward direction, and the centrifugal force F is in the upper direction too.
    The weight of the person, W, is in the downward direction now.

    This gives us an equation : N=W-F.

    Now substitute values, and the answer is N=518.2676 N.

    (d) halfway between top and bottom: the seat is again vertical, but the centripetal force is in the direction of the centre. Hence the centrifugal force (non inertial frame again) is in the horizontal outward direction ! As a result the net force experienced by the man is N=W. This solves it.

    Now, we are considering the man sits still, and does not cause the seat to oscillate. :)

    part (b) again : see, we consider the centripetal force as the net force.
    hence : the net force is inward.
    this means that the weight is greater than the normal reaction by the seat.
    as a result : centripetal force=W-N.

    therefore : N=W-ma ( considering that we subsequently note centripetal acc- centrifugal acc, only dir'ns are opp. as the frame of references are not the same.)
    Also, centripetal force is real, while centrifugal force is pseudo, a concept :)

    Somak.
     
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