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Ferris wheel rotation

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    The Singapore Flyer is te world's largest Ferris wheel. It's diameter is 150m and it rotates once every 30 min.
    a) Find the Magnitude of the average velocity
    b) Find the average acceleration at the wheel's rim.

    3. The attempt at a solution

    a) |v→| = Δs/Δt=75m.2π/30mins = 15.7m min^-1 (Could it be zero? p1 = (0,75) p2= (0,75); (p2-p1)/30mins = 0ms^-1)
    b)
     
    Last edited: Jan 18, 2014
  2. jcsd
  3. Jan 18, 2014 #2

    vela

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    What's the definition of average velocity? Is it change in displacement over time or distance over time?
     
  4. Jan 18, 2014 #3
    Yes, it's change in displacement/ change in time.
     
  5. Jan 18, 2014 #4

    vela

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    Right. So you should know which of your answers for part (a) is correct.
     
  6. Jan 18, 2014 #5
    The answer to part(a) is 0ms^-1

    b) a→ = v^2/r = [r^2Θ^2/t^2]/r = rΘ^2/t^2 = 75m(2π)^2/5^2 = 118.43m min^-1
     
  7. Jan 18, 2014 #6

    vela

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    You found the instantaneous acceleration, but the problem asked for the average acceleration. (You didn't state it explicitly, but I assume the problem asks you for the average velocity and acceleration for one complete turn.)

    Your answer for part (a) is correct.
     
  8. Jan 18, 2014 #7
    I made the mistake of leaving out a vital piece of information which was "over an interval of 5 mins"

    b) Find the average acceleration at the wheel's rim over an interval of 5 mins.
     
  9. Jan 18, 2014 #8

    vela

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    Does that apply to part (a) too? If so, you need to redo that part too.
     
  10. Jan 18, 2014 #9
    No, it applies only to part (b)

    average acceleration = delta v/ delta t?

    I'm having a little confusion here.
     
  11. Jan 18, 2014 #10
    Here's another go:

    a = Δv/Δt = (vf - vi )/Δt

    vi = rω = 75m(0π/ 5) = 0m min^-1

    vf = rω = 75m (2π/5) = 94.2m min^-1

    94.2/5 = 18.8m min^-1
     
  12. Jan 18, 2014 #11

    vela

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    This result would say the ferris wheel is starting from rest (v=0).

    And this one says that it's rotating at a rate of once every 5 minutes (##2\pi## per 5 minutes).

    ##v=\omega r## relates the linear speed ##v## of a point a distance ##r## away from the axis of rotation with the angular speed ##\omega##. If the ferris wheel is rotating at a constant rate, the speed will be constant. You need velocity, however, which is a vector. The speed gives you the magnitude of the vector, but there's still the direction of the vectors to worry about.
     
  13. Jan 18, 2014 #12
    Have I manage to obtain the correct answer then?
    I would have thought the average acceleration is the change in velocity/change in time
     
  14. Jan 18, 2014 #13

    vela

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    No, you haven't done the problem correctly. You didn't determine the velocities correctly yet.

    Draw a picture of the Ferris wheel. Label the point at the bottom as t=0. Which way does the velocity point and what is its magnitude? Now consider where that point is when t=5 min. The Ferris wheel takes 30 minutes to do a complete revolution, so after 5 minutes, by how many degrees will it have rotated? Which way does the velocity now point and what is its magnitude?
     
  15. Jan 18, 2014 #14
    Should the point at the top be labeled as t = 0?

    Untitled.jpg

    aaverage→ = Δv/Δt


    change in Θ = 2π/60 . 5 = 0.523 radians

    ratio of change in radial displacement to change in velocity
    Δr/r = Δv/v

    Δr: Δr^2 = r1^2 + r2^ - 2.r1.r2 cosΘ
    Δr = 38.78m

    velocity = Δp/Δt
    p(2) = (75 sin 0.523 , 75 cos 0.523) = (37, 65)
    p(1) = (75 sin 0 , 75 cos 0) = (0 , 75 )
    Δt = 5

    ∴[p(2) - p(1)] / 5 = | [(37 , 65 ) - ( 0 , 75) ] | /5 = | (37, -10) |/5 = 7.7ms^-1

    Δv = (Δr/r). v = 4ms^-1

    aaverage→ = Δv/Δt
    ∴4ms^-1 / 5 = 0.8ms^-2
     
    Last edited: Jan 18, 2014
  16. Jan 18, 2014 #15

    vela

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    That's not correct. First, what is ##\omega##, the angular speed of the Ferris wheel?
     
  17. Jan 19, 2014 #16
    ΔΘ/Δt

    Why is it not?
     
  18. Jan 19, 2014 #17

    vela

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    Give me a number. You use that quantity in various places, but every time you do, it's a different number. It seems like you plug in numbers at random.
     
  19. Jan 19, 2014 #18
    what is this number you want? Which variable is it for?
     
  20. Jan 19, 2014 #19

    vela

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    ##\omega##
     
  21. Jan 19, 2014 #20
    I thought about it and had another go with the workings. It does make complete sense to me. Why is it flawed?
     
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