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Gravity and its Components?

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data
    AmusementParkRide.jpg

    An amusement park ride consists of a rotating circular platform 8.07m in diameter from which 10kg seats are suspended at the end of 2.87m massless chains. When the system rotate, it makes an angle of 37.3[tex]\circ[/tex] with the vertical. The acceleration of gravity is 9.8m/s^2.

    What is the speed of each seat?

    2. Relevant equations
    Fc=[tex]\frac{Vt^2m}{r}[/tex]


    3. The attempt at a solution
    I already have a good idea of how to start, with the tension in the rope in the x direction, but what I was wondering is...can gravity have a horizontal component? Gravity pulls straight down, but to pull down on the seat as far as possible, it has to move in closer to the center (i.e. there has to be an x force). I'm pretty sure that's tension, but in this case, aren't tension and gravity an action-reaction pair? Which would imply that gravity is active to some extent in the x direction?
     
  2. jcsd
  3. Nov 25, 2009 #2

    mgb_phys

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    The outward force is provided by centrifugal force.

    Were you given the equation or did you derive it, it seems a pointless question if you just have to stick numbers in
     
  4. Nov 25, 2009 #3
    I thought centrifugal force wasn't real, just the inertial tendency to stay on the same path?
     
  5. Nov 25, 2009 #4

    Doc Al

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    Staff: Mentor

    Gravity acts down, thus it has no horizontal component. Tension and gravity are not 'action-reaction' pairs. If the rope pulls on the seat, the seat pulls on the rope. That's the correct 3rd law pair. (Gravity is the earth pulling down on the seat, thus the 'action-reaction' pair would be the seat exerting a gravitational force on the earth.)
     
  6. Nov 25, 2009 #5

    Doc Al

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    Centrifugal force is a fictitious force that is introduced when analyzing things from a rotating frame. No need for that here--just stick to the usual inertial frame and apply Newton's 2nd law. (Derive any needed equation from first principles--Newton's laws.)
     
  7. Nov 26, 2009 #6
    So...then the only force in the x direction is the tension, right? So I would use Fc=Tx.

    [tex]\frac{Vt^2m}{r}[/tex]=Tsin([tex]\Theta[/tex])

    And to find tension, I would have to use the y direction equation.

    [tex]\sum[/tex]F=Tcos[tex]\Theta[/tex]-mg
    [tex]mg/cos(\Theta[/tex])=T

    Then put that back into the first equation and solve for Vt, right?

    [tex]\frac{Vt^2m}{r}[/tex]=[tex]\frac{mgsin\Theta}{cos\Theta}[/tex]


    Vt=[tex]\sqrt{grtan\Theta}[/tex] , keeping in mind that the radius is the horizontal distance from the center of the ride.

    And I get 6.56565m/s.
    Right?
     
  8. Nov 26, 2009 #7

    Doc Al

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    I don't quite understand this equation. The right hand side doesn't seem to have the dimensions of a force.

    Edit: LOL... I just got what you mean. Fc = mVt2/r, not mVt2/r.

    All good! :approve:
     
  9. Nov 26, 2009 #8
    Yeah, sorry. I was a little lazy with the subscript at that part. And thanks a bunch.
     
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