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Fairground wheel Kinematics Question

  1. Oct 4, 2004 #1
    A fairground wheel is rotating in a vertical plane and carriages travel round a circle of diameter 40m at a constant speed. The carriages complete one revolution in 3.5min
    I need to calculate the speed of the carriage from the top of the wheel to the bottom:

    Circumference = 2 pi r
    = 125.7m

    top of wheel to bottom is therefore half this: 62.8m

    Convert 3.5mins to seconds = 210s
    half revolution = 105s

    speed = distance/time = 62.8/105

    speed = 0.6ms-1

    I think this is right! :-s i'm a beginner!!

    The problem i'm having is that i have to find the velocity aswell.....but am unsure how the distance affects the result in circular motion!

    Please help me!!
  2. jcsd
  3. Oct 4, 2004 #2
    Edit that might be too complicated for what you're looking for. Velocity is defined as the change in distance over the change in time. It doesn't matter HOW you get to a place just what that place is at the end.

    Edit #2 I didn't plug your numbers into a calculator but the method you used looks right. Velocity is simple displacement/time though not total distance traveled over time.
    Last edited by a moderator: Oct 4, 2004
  4. Oct 4, 2004 #3
    Does this mean that i don't have to use the circumference as the displacement? Basicly i just take the distance between the top of the wheel to the bottom of the wheel (the diameter 40m)?


    velocity = displacement/time
    = 40m/105s

    v = 0.38ms-1
  5. Oct 4, 2004 #4
    It is my belief that that is correct.
  6. Oct 4, 2004 #5
    Woohoo! thanks for your help!
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