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Centripital force - did i do it correctly?

  1. Oct 17, 2007 #1

    klm

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    centripetal force - did i do it correctly?

    In an amusement park ride called The Roundup, passengers stand inside a 19.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure View Figure .

    [​IMG]

    Suppose the ring rotates once every 4.40 s. If a rider's mass is 59.0 kg, with how much force does the ring push on her at the top of the ride?

    Suppose the ring rotates once every 4.40 s. If a rider's mass is 59.0 kg, with how much force does the ring push on her at the bottom of the ride?

    okay so I am not sure if i did this correctly, but this is what I have done so far:

    Fc= m(V^2) / r

    and V=(2 x pi x r) / t

    so i got v = 2pi8/ 4.4 = 11.4
    and then
    Fc= 59(11.4^2) / 8 = 958

    is that correct? and what is the diff. b/w the force at the top and bottom

    THANKS!
     
    Last edited: Oct 17, 2007
  2. jcsd
  3. Oct 17, 2007 #2

    Doc Al

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    Staff: Mentor

    Start by identifying the forces acting on the person at the top of the ride and apply Newton's 2nd law. Draw yourself a diagram. The acceleration is centripetal, so you'll need to use that formula at some point.

    Also: If the diameter is 19 m, what's the radius?
     
  4. Oct 17, 2007 #3

    klm

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    oh crap its 9.5 not 8. sorry i dont know what i was thinking!
     
  5. Oct 17, 2007 #4

    klm

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    aren't the only forces acting on them gravity(weight) and normal force that is pushing on them from the ring?
     
  6. Oct 17, 2007 #5

    Doc Al

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    Right! So set up the equation for net force.
     
  7. Oct 17, 2007 #6

    klm

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    okay this might be incorrect, because I am kind of confused about how centripetal acceleration and all of this really works...

    so basically all i have for my free body diagraphm is Nforce pointing straight up on the y axis and Fg pointing straight down..would that be right?

    so Fnety= N-Fg= m(ay) ... N-mg= m (ay) and we know m= 59 and g=9.8
    but i dont know what to do now.

    does that (ay) mean (ar)= V^2 / r ?
    i hope that isnt too confusing!
     
  8. Oct 17, 2007 #7

    klm

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    n - mg = m ( V^2)/ r

    m=59
    v = 2 pi r / t
    so 2 pi 9.5 / 4.4 = 13.56
    r= 9.5
    g=9.8

    so n = 1721.157 ??
     
  9. Oct 17, 2007 #8

    klm

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    would that be the correct way to use the formula?
     
  10. Oct 17, 2007 #9

    Doc Al

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    Almost: Which way does the normal force act at the top? (And what about at the bottom? That's what makes the difference.)
     
  11. Oct 17, 2007 #10

    klm

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    umm i am not really sure. i know normal has to push back against the person. so like the ring is pushing back against the person. so maybe normal force goes horizontally?
     
  12. Oct 17, 2007 #11

    klm

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    i dont understand how normal force is going to change from the top and bottom.
     
  13. Oct 17, 2007 #12

    Doc Al

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    The normal force is always perpendicular to the surface. When you're at the top, where's the surface with respect to you? Which way must it be pushing?
     
  14. Oct 17, 2007 #13

    klm

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    so is the ring considered to be the surface, so the ring is pushing horizontally. so in the positive x direction at the top?
     
  15. Oct 17, 2007 #14

    klm

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    no wait, would it be in the neg x direction at top, and positive x direction at the bottom
     
  16. Oct 17, 2007 #15

    klm

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    i have to go now, but i hope that maybe tomorrow you can help me some more and help me understand about this normal force!
     
  17. Oct 18, 2007 #16

    klm

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    hi doc al, do you think you can help me finish this problem please!
     
  18. Oct 18, 2007 #17

    Doc Al

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    Exactly right.
     
  19. Oct 18, 2007 #18

    klm

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    ahh thank you! i think i understand.
    so does that mean i have to use Fnetx = -N = m ( V^2)/ r
    and then Fnety = -mg = m ( V^2)/ r

    but shouldnt one of the Forces = 0?
     
  20. Oct 18, 2007 #19

    Doc Al

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    Not exactly.

    At the top, the net force is: -N -mg
    So Newton's 2nd law gives:
    -N -mg = -mv^2/r (note that the acceleration points down so is also negative)

    At the bottom, the net force is: +N -mg
     
  21. Oct 18, 2007 #20

    klm

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    ooh. but can you maybe explain a little on why acc is neg. i know you wrote that it is pointing down, but isn't that already considered when you draw the free body diagram, so you wouldnt have to make it neg in the equation?
     
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