# Artificial Gravity Problem

1. Oct 17, 2006

### wyiyn

I've been staring at this problem for quite a while now, and I don't really know how to approach it. Any guidance would be greatly appreciated.

What we have is a rotating cylinder to create artificial gravity (equivalent gravitation acceleration of ag where a is a constant). We are asked to determine the limit on the radius in terms of the elastic limit, E, the density of the material, rho, and the desired fractional gravity, ag.

There was a hint given to assume the cylinder wall is "thin" and start by considering forces on a small segment of a ring rotating with angular velocity omega. Considering limit of delta (theta) goes to 0 would be useful.

The only thing i've got out of this is that the force has to equal the centipetal force, and so ag = mr(omega)^2. I don't know how to deal with the elastic limit, or get the omega out of the equation.

Any help would be great!

2. Oct 17, 2006

### OlderDan

Where does the force come from that is providing the centripetal acceleration? Think about dividing the ring into two halves. What holds the halves together?

3. Oct 20, 2006

### wyiyn

Ok, so just like a rope, this is held together with tension. But I'm just confused about how to relate tension to this elastic limit stuff.

4. Oct 20, 2006

### OlderDan

The elastic limit is in terms of stress, which is force per unit area. If you can calculate the force required to keep the two halves together and divide by the contact area between the two halves you have the stress. The stress cannot exceed the elastic limit.

5. Oct 22, 2006

### wyiyn

Yeah, I guess the problem is determining the force. I know that the total force (centripetal) has to equal the ag, but I can't quite seem to get a handle on how to define this other force.

I have a sneaking suspicion that an integral over theta will be involved, setting that equal to ag, and solving for r. But the problem is finding what's all involved in the restoring force. I've been sitting playing with elastic bands, and all I've noticed is that the tension will depend on the radius, the angular speed and the thickness (but we have a 'thin' cylinder)... so I guess what I'm saying is that I'm not sure what to do with restoring force.

6. Oct 22, 2006

### OlderDan

There is an integral involved. Every bit of mass in the ring is accelerating toward the center of the circle, so it experiences a force in that direction. The net force acting on half the ring is perpendicular to the diameter that cuts the ring in half. You need to integrate over the force components that are additive and ignore the ones that cancel. If you take the right half of the ring using a vertical diameter, the additive components are horizontal to the left. The force that keeps that half of the ring connected to the other half is the tension in ring at the top and bottom.