How can I calculate the elongation of a steel rod in an amusement park ride?

[superdave]

An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 14.5 m and a cross-sectional area of 8.25 {\rm cm}^{2}.

When operating, the ride has a maximum angular speed of 8.50 rev/min. How much is the rod stretched then?
Take the Young's modulus for the rod to be Y = 2.00×1011 Pa and the free fall acceleration to be g = 9.80 m/s^2

Assume that each car plus two people seated in it has a total weight of 1930 N.

Now, I somehow need to find theta or r using only omega. I'm not really sure how to do this.

 

[OlderDan]

Hint: The only two forces acting on a plane with its cargo are the tension in the rod and gravity. The vector sum of these two forces is horizontal and that force is causing the plane to go in a circle.

 

[PhanthomJay]

An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 14.5 m and a cross-sectional area of 8.25 {\rm cm}^{2}.

When operating, the ride has a maximum angular speed of 8.50 rev/min. How much is the rod stretched then?
Take the Young's modulus for the rod to be Y = 2.00×1011 Pa and the free fall acceleration to be g = 9.80 m/s^2

Assume that each car plus two people seated in it has a total weight of 1930 N.

Now, I somehow need to find theta or r using only omega. I'm not really sure how to do this.

you should take a FBD of the car ,and examine the forces acting on it. In the y direction, there is no movement. In the x direction, the horizontal component of the tension force provides the centripetal acceleration. Write the centripetal force equation in terms of omega (where omega =v/r), do a little trig, and solve for T. Show your work, please. I then presume you are familiar with the formula for length expansion under load?