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Ferris wheel circular motion (dealing with apparent and true weight)

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data

    While at a country fair, you decide to ride the Ferris Wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 15m, and you use your watch to find that each loop around takes 25s.
    a) What are your speed and magnitude of your acceleration?
    b) What is the ratio of your weight at the top of the ride to your weight while standing on the ground?
    c) What is the ratio of your weight at the bottom of the ride to your weight while standing on the ground?



    2. Relevant equations

    where T= period, r=radius

    v = 2(pie)r/T
    a = v^2/r


    3. The attempt at a solution

    For part a) I got v= 3.7699 m/s and a= 0.947 m/s^2

    For part b) since we are dealing with a ferris wheel, the FBD has normal force and weight pointing in the downward direction at the top of the ferris wheel. Would the apparent weight, which is the normal force, be (note: I made the positive y coordinate pointing down so n and w are positive):
    n+w=ma
    n=ma-mg
    Wapp=m(a-g)

    and because Wtrue=mg

    Wapp/Wtrue= (m(a-g))/mg

    m's cancel so,

    (a-g)/g

    (0.947-9.8)/9.8 = -0.9

    However, that number should be positive. So, I am assuming that it should not have been n=ma-mg BUT instead n=mg-ma so n=m(g-a) so that Wapp/Wtrue= (g-a)/g so the ratio is 0.9. IF SO, WHY would n= mg-ma???

    for part c) I got n-mg=ma so n=mg + ma so Wapp/Wtrue= (g+a)/g so ratio= 1.1


    PLEASE HELP this question is due tomorrow morning, so help asap would be GREATLY appreciated!
     
  2. jcsd
  3. Feb 18, 2009 #2
    at the top of the wheel, if your y coordiate is pointing down, then the acceleration (a) will be negative as it acts away from the wheels centre, ie. Normal force is up at the top of the wheel and down at the bottom of the wheel.

    does that help?
     
  4. Feb 18, 2009 #3
    So at the top of the wheel, just like at the botttom of the wheel, the normal force is in the opposite direction of the gravitational force, except for the normal force at top it is negative and for the normal force at the bottom it is positive?
     
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