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

[tandoorichicken]

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.

 

[NateTG]

Can you determine the tension in the chains?

 

[tandoorichicken]

How? and what do I do then?

 

[Doc Al]

Originally posted by tandoorichicken
How? and what do I do then?

Consider the forces acting on the plane and apply F=ma.

 

[tandoorichicken]

I am so lost.

So there is gravity, and there is also centripetal force right?

 

[Doc Al]

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.

 

[HallsofIvy]

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.

 

[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 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 \tan 56 = \frac{m\omega^2 r}{m g}. The m's cancel and when you solve for \omega, you get \omega = \sqrt{\frac{g\tan 56}{r}} = 0.562 rad/sec. Does that make sense?

 

[Doc Al]

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 \omega = \sqrt{\frac{g\tan 56}{r}} is correct. The problem asks for speed, which I presume means linear speed not angular; but v=r\omega.

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.