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Ferris Wheel Problem

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data
    A Ferris wheel of radius 100 feet is rotating at a constant angular speed ω rad/sec counterclockwise. Using a stopwatch, the rider finds it takes 5 seconds to go from the lowest point on the ride to a point Q, which is level with the top of a 44 ft pole. Assume the lowest point of the ride is 3 feet above ground level.

    Let Q(t)=(x(t),y(t)) be the coordinates of the rider at time t seconds; i.e., the parametric equations. Assuming the rider begins at the lowest point on the wheel, then the parametric equations will have the form: [tex]x(t)=rcos(ωt-π/2)[/tex] and [tex]y(t)=rsin(ωt -π/2)[/tex], where r,ω can be determined from the information given. Provide answers below accurate to 3 decimal places. (Note: We have imposed a coordinate system so that the center of the ferris wheel is the origin. There are other ways to impose coordinates, leading to different parametric equations.)

    Find r, find ω. During the first revolution find the times when the riders height is 80ft.


    2. Relevant equations
    [tex]x(t)=rcos(ωt-π/2)[/tex]
    [tex]y(t)=rsin(ωt -π/2)[/tex]

    3. The attempt at a solution
    The radius is obviously already given to me. To find ω I drew a triangle with vertices at the origin, Q, and at (0, 44). The hypotenuse was 100 ft, and the adjacent side to the angle I was trying to find was (103-44) or 59 ft. To find the angle I need I found the inverse cosine of (59/100) and divided it by 5. The angular speed rounded to 3 decimal places is .188

    To find the times when the rider's height is 80ft is where I'm having problems. According to the problem the origin is the center of the wheel does that mean I set y=(80-103)?
     
  2. jcsd
  3. Jan 10, 2013 #2

    haruspex

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    Looks right to me.
     
  4. Jan 14, 2013 #3
    Thank you, turns out I was on the right track. Here's what I did.

    [tex]-23=100sin(.1879474972t-π/2)[/tex]
    [tex]-.23=sin.1879474972t-π/2[/tex]
    [tex]-.2320776829=.1879474972t-π/2[/tex]
    [tex]t=7.123[/tex]

    Now the problem is I'm not sure how to find the second time.
     
  5. Jan 14, 2013 #4

    haruspex

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    When you invert the sine function you get, in principle, infinitely many solutions. What are all the values of arcsin(sin(x))?
     
  6. Jan 14, 2013 #5
    All I knew is this.
    [tex]arcsin(sin.1879474972t-π/2)=1879474972t-π/2[/tex]

    Did I do something wrong or what should I do?
     
  7. Jan 14, 2013 #6

    haruspex

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    The sine function is periodic, i.e. as x increases y=sin(x) keeps taking the same set of values again and again. Therefore 'the angle whose sine is y' has many possible values. The arcsin function is defined, by convention, to take the value between -pi/2 and +pi/2, but there is an infinity of other x values.
    Draw a circle. Starting at the bottom, go around it clockwise until you have gone through some angle theta < pi. Draw a horizontal line through that point. Now continue around until you hit that line again on the right hand half of the circle. What angle have you gone through now (in total)?
     
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