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Six Flags rotation problem

  1. Jan 11, 2004 #1
    Passengers riding in the Great Six Flags Air Racer are spun around a tall steel tower. At top speed the planes fly at a 56 degree bank approximately 46m from the tower. In this position the support chains make an angle of 56 degrees with the vertical. Calculate the speed of the planes.
     
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  3. Jan 12, 2004 #2

    NateTG

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    Can you determine the tension in the chains?
     
  4. Jan 12, 2004 #3
    How? and what do I do then?
     
  5. Jan 12, 2004 #4

    Doc Al

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    Consider the forces acting on the plane and apply F=ma.
     
  6. Jan 12, 2004 #5
    I am so lost.

    So there is gravity, and there is also centripetal force right?
     
  7. Jan 12, 2004 #6

    Doc Al

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    The forces on the plane are gravity and the tension in the chain. The plane is centripetally accelerated. Apply F=ma to the vertical and horizontal components of the forces.
     
  8. Jan 12, 2004 #7

    HallsofIvy

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    Draw a picture showing the chain out to the seats at a 56 degree angle. The tension force, T, is along the hypotenuse of that right triangle. The vertical component, T sin 56 degrees;, must offset the weight so it must be mg. The horizontal component, T cos 56 degrees, is Rω2.
     
  9. Jan 12, 2004 #8
    Okay, I think I got it all figured out, and I managed to completely bypass figuring the tension in the chain. Can someone please check my work?

    I drew a right triangle with the top angle as 56 degrees, the bottom leg centripetal force and the side leg force of gravity. Then I set up [itex] \tan 56 = \frac{m\omega^2 r}{m g} [/itex]. The m's cancel and when you solve for [itex] \omega [/itex], you get [itex] \omega = \sqrt{\frac{g\tan 56}{r}} [/itex] = 0.562 rad/sec. Does that make sense?
     
  10. Jan 13, 2004 #9

    Doc Al

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    I'm not sure I understand your reasoning with the triangles, but [itex] \omega = \sqrt{\frac{g\tan 56}{r}} [/itex] is correct. The problem asks for speed, which I presume means linear speed not angular; but [itex]v=r\omega[/itex].

    I would solve it like so:

    (vertical forces) Tcos(56)=mg
    (horizontal forces) Tsin(56)=mv2/r

    Dividing gives you: tan(56) = v2/rg, etc.
     
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