Consider a tire rolling without slip and a tire spinning in place. When a point on the rotating tire is at the top of its arc it is moving with velocity r*w+v where r is the radius, w is the angular acceleration, and v is the velocity of the center of the wheel (or the car). At the bottom of its arc the velocity is v-r*w. So the change in speed is 2*r*w. On the tire spinning in place, the velocity at the top is r*w and the velocity at the bottom is -r*w. Thus the change in velocity is also 2*r*w. Because acceleration is a change in velocity, and the change in velocity is the same in the two wheels, we come to the conclusion that the acceleration (and therefore the force) is the same.

The only thing that would be different in a wheel on a car driving with constant velocity and a wheel on a dyno is the normal force on the tire of the car. This would be a constant force perpendicular to the ground.

I think I have to agree with your explanation. Only the way I thought was on a moving wheel we were to assume twice the speed a similar stationary wheel makes for a point on top of the wheel that is in linear motion. This is because the OP did not ask about stress on the whole wheel per se but on the edge. But since he never said the wheel is in contact with the ground i don't see why it has to be any different from one just spinning fixed in space relative to a stationary observer.

Are we considering a wheel that is rotating in space or rolling along the road? A wheel spinning freely only has to overcome the centrifugal force of its own weight, whereas one with a load on it is subject to different force equations because the force acting against it isn't uniform and homogenous, there will be more stress at certain points in the wheel, less stress on the ground where the weight of the car is holding against centrifugal force. However this also would put slightly more stress on the rest of the wheel with the most force being at the left and right edges.

Also the greater force would be on the inner diameter of the wheel which has less area, these cracks from stress related to rotation always go from the inside out when caused by only the rotation.

If you would like an example I can supply one, I have lots of equations relating to stress and tensil strength, shear strength and so on. I have used these on my own dynamos.

Thanks alot dhcracker. Just supply a formula for calculating stress at the edge of said wheel. We can start with stress at 9 and 3 o'clock.

I'll just spell it out for you, I need a math thing so I can actually type equations out I haven't done that yet normally I just perform calculations, so you'll have to just take the formula from plain english.

First you use the formula posted above and get the hoop stress, then you just add the force from the weight of the car that is loaded on this wheel and divide it by the area of the wheel.. you add this to the hoop stress (though its not loaded evenly over the rim this is close enough to get a rough estimate unless you are planning on reaching the sound barrier or something).. howver there is also a force placed on a rim by the tire itself the inflated PSI is pressing against the inside of the rim, on the flange its pressing that out parallel to the axis however over the rest of the rim it actually presses against the hoop stress, so over the area of the inner facing rim you could subtract the tire pressure from the stress and add some stress over the area of the flange.

Talk about details, but normally there is no need to do all this unless you plan on running your car over 200 mph, then simply get a carbon fiber rim or aircraft alloy. There is no way you could shear that stuff, it would help if I knew exactly what your calculating this for?. Also what kind of rim? Are you meaning a flywheel rim or cylinder? The two are very very different figures and formulas.

Let's put the wheel aside for a moment. Let me show where my reasoning is heading (a bit off what the OP wanted, it's him that wanted the formula remember?)
This is my scenario: A merry-go-round is on board a Cruise ship. The tangential speed of the merry-go-round equals the ship's velocity. For illustration purpose let's say
a] Ship heading North
b]Merry-go-round is rotating clockwise viewed from above.
This thought experiments tells me that:
*The seats arriving at position 9 O'clock will hang perpendicularly down.
*The seats arriving at position 3 O'clock will be thrown up higher than usual
*The seats arriving at 6 & 12 O'clock will be thrown no higher than usual
NB: Usual is when ship is stationary.

If I understand Flipmastacash correctly, there will not be additional stress at 3 O'clock. The additional amplitude is accounted for by the ship's motion. Why I agreed with him is cause a vehicle moving at say a constant speed of 100 m.p.h is not experiencing any more force than one traveling at 50 mph (assuming no air resistance). In this respect the tug on the chain holding suspending the seat is not double or magnitude of force increased. This is cause it is not centrifugal force per se causing this. It is the linear speed of the sheep which can be subtracted from the equation. The end result is there is no additional tension (centripetal pull) on the chain.
Did I understand you correctly in view of what Ive said?

A simpler explanation why the wouldn't be additional stress as long as the vehicle is not accelerating is: assume you're on a train doing 50 mph. You throw a rock dead ahead at 66 mph. You don't feel more stress on your shoulder as you would if you threw the rock at 116 mph at rest. Now back to my merry-go-sound. Maybe the highest swing will be at 12 o'clock and lowest at 6 o'clock. Also i should have said the rotation is counter-clockwise. Also I wrote sheep instead of ship. Too late to edit the earlier post.