What is the speed of the planes on Great Six Flags Air Racer?

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In summary, the Great Six Flags Air Racer spins passengers around a tall steel tower at a speed of 0.562 rad/sec, with a bank of 56 degrees and a distance of 46m from the tower. To calculate the speed of the planes, one must consider the forces acting on the plane and apply F=ma to the vertical and horizontal components of the forces. The tension in the chains must also be taken into account. To determine the tension, one can draw a right triangle with the top angle as 56 degrees and solve for the angular velocity. The linear speed can then be found using v=r\omega. Another method is to use the equations Tcos(56)=mg and Tsin(56)=mv
  • #1
tandoorichicken
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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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  • #2
Can you determine the tension in the chains?
 
  • #3
How? and what do I do then?
 
  • #4
Originally posted by tandoorichicken
How? and what do I do then?
Consider the forces acting on the plane and apply F=ma.
 
  • #5
I am so lost.

So there is gravity, and there is also centripetal force right?
 
  • #6
Originally posted by tandoorichicken
So there is gravity, and there is also centripetal force right?
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.
 
  • #7
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.
 
  • #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?
 
  • #9
Originally posted by tandoorichicken
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'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.
 

1. What is the "Six Flags rotation problem"?

The "Six Flags rotation problem" refers to a mathematical puzzle in which six flags must be arranged on a flagpole in a specific order, while following a set of rules, in order to spell out a specific word or phrase.

2. What are the rules of the "Six Flags rotation problem"?

The rules of the "Six Flags rotation problem" are as follows: 1) Each flag must be rotated to a specific angle before being placed on the flagpole, starting from the bottom and moving up. 2) Flags cannot be placed upside down. 3) The order of the flags on the flagpole must spell out a specific word or phrase. 4) Each flag must touch the one above and below it.

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4. How is the "Six Flags rotation problem" related to other mathematical concepts?

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