# Simple rotation/elasticity problem

1. Jun 28, 2005

### ninjagowoowoo

An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 14.0 m and a cross-sectional area of 0.000780m^2. When operating, the ride has a maximum angular speed of 8.00 rev/min. How much is the rod stretched?

Take the Young's modulus for the rod to be Y = 2×10^11 Pa and the free fall acceleration to be g = 9.80 m/s^2

Umm... Im really just stumped on this one. Really its just the rotation part thats messing me up. What outward force would be caused by the rotation? Would the rotation even change the amount stretched from a still position? Haha, I would really appreciate a push in the right direction.

2. Jun 28, 2005

### OlderDan

The cars are going around in circles. Each car is subject to a centripetal force provided by the rods. The rods are being stretched by the force of the car and attachment at the axis of rotation.

Was the rotation axis specified? Is the motion horizontal?

3. Jun 28, 2005

### ninjagowoowoo

ok so to find the tensile force, I will use the centripetal force. so F=(mv^2)/R. And v=R(omega). So F=R(omega)^2(m). But to find this force wouldn't I need to know the radius of the rotation?

4. Jun 28, 2005

### FredGarvin

You'll have to decide if you want to assume that the rods are horizontal at the time you are interested in. I would simply because it would make things easier.

5. Jun 28, 2005

### ninjagowoowoo

Nope still not working. using the formula for F that I gave earlier, and that
changeinlength = F(originallength)/(crosssectionalarea)(modulus)
I get that:
change in length = r(omega^2)m(orig length) / AY where Y is the modulus.
so I use the following values:
R = 14m
omega = 3015.92895rad/s (converted from rev/min)
m = 203.0612245 (1990N/9.8m/s/s)
orig length = 14m
A = 0.00078m^2
Y=2*10^11
and when i plug in the numbers I get a rediculous number (2320.602127m). Any other suggestions?

6. Jun 28, 2005

### OlderDan

See if you can fix that ridiculous value for omega
I did not check the rest of your calculation

7. Jun 28, 2005

### ninjagowoowoo

well if it was in rev/min wouldnt i have to convert it to rad/s?

*(2pi*60)???

$$\omega = \frac{8\ \ rev}{min} \cdot \frac{2\pi\ \ rad}{rev} \cdot \frac{1\ \ min}{60\ \ sec}$$