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The Ferris Wheel

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data


    The figure below shows a Ferris wheel that rotates three times each minute. It carries each car around a circle of diameter 19.0 m.

    a) What is the centripetal acceleration of a rider?
    Answer in m/s^2

    (b) What force does the seat exert on a 40.0 kg rider at the lowest point of the ride?
    Answer in N

    (c) What force does the seat exert on the rider at the highest point of the ride?
    Answer in N

    (d) What force (magnitude and direction) does the seat exert on a rider when the rider is halfway between top and bottom, going up?
    Answer in N
    Answer in ° (measured inward from the vertical)

    2. Relevant equations



    3. The attempt at a solution

    a) Angular velocity w = 3 rev/min = 3 * 2pi/60 rad/s
    = pi/10 rad/s
    Or w = 0.314 rad/s
    Radius r = 19.0 m/2 = 9.5 m
    Centripetal acceleration a = w^2 * r
    = 0.314^2 * 9.5
    = 0.937 m/s^2

    b) Mass M = 40.0 kg
    Let force = F
    F is upward and weight Mg is downward. Centripetal force is upward.
    F - Mg = Ma
    Or F = M(a+g) = 40 * (9.8 + 0.937) = 429.5 N

    c) Mass M = 40.0 kg
    Let force = F
    F is upward and Mg is downward. Centripetal force is downward.
    Mg - F = Ma
    Or F = M(g - a) = 40 * (9.8 - 0.937) = 354 N

    d) Let force = F at angle theta below vertical.
    Vertical component of F = F cos(theta) upward.
    Horizontal component of F = F sin(theta)

    There is no acceleration in vertical direction.
    Therefore F cos(theta) = Mg--------------------------(1)
    Centripetal force is horizontal.
    Therefore F sin(theta) = Ma---------------------------(2)
    Dividing (2) by (1),
    tan(theta) = a/g
    Or theta = atan(a/g) = atan(0.937/9.8) = 5.46 deg
    From (1),
    F = Mg/cos(theta) = 40 * 9.8/cos(5.46 deg) = 394 N
    Therefore, force by the seat is 394 N at 5.46 deg below vertical.
    (Note: I do not know what exactly is meant by inward from vertical. I have calculated the angle from vertically upward. If you want the angle from vertically downward, then it is 180 - 5.46 = 174.54 deg



    I only have one chance to enter this, please check to see if the way i did it is correct.
     
  2. jcsd
  3. Nov 2, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    All calcs look good. For (d) it would be the 5.44 degree answer because the Fg is much larger than the Fc, so only a little off vertical.
     
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