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Ferris Wheel physics problem

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A passenger on the ferris wheel described in problem 18 (Problem 18: Fairgoers ride a Ferris wheel with a radius of 5.00m The wheel completes one revolution every 32.0s) drops his keys when he is on he way up and at the 10 o'clock position. Where do the keys land relative to the base of the ride?

    Also: the diagram reveals the ferris wheel is 1.75m above the ground, and the center of the wheel is = base of the ride.
    2. Relevant equations
    all projectile motion equations

    3. The attempt at a solution

    I found the solution to be 58.0cm but im pretty sure i didn't do it right. my steps were
    1)Setup a diagram using the ten o'clock position to be 30degrees with a hypotenuse of 5m.
    2)Found dx
    3)t=(1/12)<rotation of wheel*(32s)<How long for the whole wheel to rotate
    4) Found Vox
    5) Found Voy
    6) Used Voy in y(t) equation, where y(t)=0, gave me a quadratic i solved for t=3.44s (Supposing 0 = ground)
    7) found x(3.44)
    8) Subtracted the distance from center of the wheel from the radius

    =58.0cm to the right of the base.

    My intuition tells me i did something wrong at the start. My teacher told me today, the radius vector is perpinduclar to the Vo vector, so i think i can use that with pythag to find Vo somehow, any insight?
  2. jcsd
  3. Sep 25, 2007 #2
    How did you obtain your current v0? I don't see it in your procedure.

    I don't think you can obtain v0x and v0y without that.
  4. Sep 25, 2007 #3
    i used dx and dy (part 2) divided by t(part 3) to find Vo, which i'm pretty sure is wrong
  5. Sep 25, 2007 #4
    There's a simple way for getting v0

    You know the time it takes for a revolution.


    Can you figure out the distance it travels in one revolution?
  6. Sep 25, 2007 #5

    C=50(pie) m.


    .... this gives me velocity, a ferris wheel would have constant velocity but changing acceleration, due to the change in direction. Therefore if it has constant velocity, unless the wheel was at rest, this V value we just found is Vo?
  7. Sep 25, 2007 #6
    Check your circumference formula first. ;P

    About your other question, though the velocity vector of the person in the ferris wheel is constantly changing direction, at the instant the coin is released it is no longer moving in a circle and will behave as a projectile.

    In short, this V you obtained (once you correct your formula) is V0 for the purpose of projectile motion.
  8. Sep 25, 2007 #7
    Ok that makes sense, one more thing, can we assume that a clock at 10 o'clock, has an angle of 150degrees from standard position? Seeing as how from 9-12 forms a right angle with three angles in between
  9. Sep 25, 2007 #8
    basically now i just want to make sure i have the right angle for Vo, i think it's 60 degrees, using basic geometric properties but can anyone double check this?
  10. Sep 25, 2007 #9
    bump, anyone know the initial angle for Vo?
  11. Sep 26, 2007 #10
    60 sounds right.
  12. May 11, 2008 #11
    new question about this question

    I am currently working on this exact same question. I have my Vo but am really unsure what needs to be done to get Vox and Voy.
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