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Centripetal Force of park ride

  1. Oct 8, 2005 #1
    I have a question on what it is I might be doing wrong here... I've tried this answer in the grader and it is wrong...

    At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s and an 83 kg person feels a 495 N force pressing against his back. Radius of the chamber?

    Here is what I have done..

    v = 3.2 m/s
    m = 83 kg
    F = 495 N
    r = ?

    Fc = mv2/r
    r = Fc(1/mv2)

    r = 495 N (1/(83kg * 3.2 m/s2))
    r = .582 m

    Even just looking at it it looks wrong, heh...

    Any help would be great!

  2. jcsd
  3. Oct 8, 2005 #2


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    Notice this:

    [tex] F = m \frac{v^2}{r} [/tex]

    [tex] \frac{r}{F} F = \frac{r}{F} m \frac{v^2}{r} [/tex]

    [tex] r = m \frac{v^2}{F} [/tex]
  4. Oct 8, 2005 #3


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    I was just going to say that it looks like you didn't square the velocity.
  5. Oct 8, 2005 #4
    Ahhh thanks much - so I guess it was just a matter of arranging my equation? The mass needs to multiply out to those other variables for it to work...
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