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Vertical Circular Motion problem (Carnival)

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data

    You've taken your neighbor's young child to the carnival to ride the rides. She wants to ride The Rocket. Eight rocket shaped cars hang by chains from the outside edge of a large steel disk. A vertical axle through the center of the ride turns the disk, causing the cars to revolve in a circle. You've just finished taking physics, so you decide to figure out the speed of the cars while you wait. You estimate that the disk is 4.19 m in diameter and the chains are 7.55 m long. The ride takes 13.9 s to reach full speed, then the cars swing out until the chains are 18.4° from vertical. What is the car's speed?

    2. Relevant equations
    Fr= mv^2/r

    Known-> (theta) = 18.4
    (radius)= (4.19/2)+7.55 = 9.645
    (time) = 13.90s

    3. The attempt at a solution
    I drew a circle and decomposed the components, got the radial axis and the tangential axis equations as shown below:
    Fr=T + mgsin(theta)=mv^2/R
    Ft=mgcos(theta) = ma(tangential)

    For Ft I subbed in all the known values, got a"t" = 9.299m/s^2, then i'm kind of lost, don't really know what's next, but i did gave some thought on the "Angular velocity equation" (Wf=Wi + at/R*(∆t)) figure "Wi" then i should be able to use V=Wr to figure out its velocity...

    Thankyou for taking the time to read this,
    it would be great if you guys could guide me through this problem,
    i would love to learn how to solve it.
  2. jcsd
  3. Jan 11, 2009 #2
    I lied. Force is prob the easier way
    Last edited: Jan 11, 2009
  4. Jan 11, 2009 #3
    Hi first of all thankyou so much for replying this post,
    since i'm actually not very bright in physics,
    would you mind further clarifying your method of approach? (ie how would you get acceleration in the first place without using forces?)

  5. Jan 11, 2009 #4
    PS: and the angular velocity,
    i dont know where to start with all the given variables, i'm truly lost..
  6. Jan 11, 2009 #5
    Let me go edit someone's philosophy paper first. ._. I will be back.

    I just reread the question, wonder if my approach is the right one. Hmm. I shall be back
  7. Jan 11, 2009 #6
  8. Jan 11, 2009 #7
    It turn out the essay is 3 pages long..so I guess this would be quicker.

    We know that the mass at the end of the chain is in equilibrium

    so Tcos([tex]\theta[/tex])=mg [formula 1]
    Tsin([tex]\theta[/tex]) = mar=mv^2/r [formula 2]

    r = sin (Length of chain sin [tex]\theta[/tex] + radius of the disk)

    and then you sub r into mv^2/r ==> a

    A trick with these problem is to division of the Tcos [tex]\theta[/tex] function by the Tsin [tex]\theta[/tex] function (formula 1/formula 3). So you will get

    tan [tex]\theta[/tex] = a/g

    you know a and g, and theta is given...so you can solve for v
  9. Jan 11, 2009 #8
    Why are you adding the radius of the disk to the angle all within the sin function?
  10. Jan 11, 2009 #9
    because the effective radius is the radius of the disk and the horizontal part of the chain. Since the chain is lifted up (due to the rotation), the sin theta portion also contribute to the radius.
  11. Jan 11, 2009 #10
    i still dont get it .... damn.. =(.....
    im really a physics noob

    *shouldn;t it be Tsin(theta) =mg and TCos(theta) =mv^2/R?
    *How would u get the acceleration even u have these 2 equations? since you would need the acceleration to figure out the theta, then to figure out the velocity ...
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