J.C.MacSwell

I agree...except how is that 2 events?

No. It is not just the rate. Observers in various rest frames can not agree on a common starting point in time for different locations.

That's correct, mathematically. What would have to happen for a material to be 100% inflexible? It would have to respond at infinite speed. Mathematically pretty straight forward but physically impossible with our current understanding of physical law.

But unless you are vitamin A deficient, it won't make any difference. Is it too early to mention the Nazis in this thread? http://www.unsolvedmysteries.com/usm434311.html

When you are ready, you will welcome it.

I'll start. A slowly decaying orbit will actually increase in speed as the orbit decays.

For an idealized ring rolling on an idealized surface, 1 and only 1 point is at rest. So no element is ever at rest, if n is a finite number.

So why would the elements take the paths shown (or described) after the break?

I agree. At least in principle.

It is not true. If a chain or rod is being pulled along it's length unconstrained it will accelerate. There will be a force gradient along that length, greatest at the pulled end and zero at the other depending on the distribution of mass. The force at any point will simply be proportional to the amount of mass behind that point. If the links are all the same, the force will be proportional to the number of links remaining behind the point in question. The force is reduced between each successive link, due to the acceleration/reaction of the preceding link. If you drew a free body diagram of any link, or set of links, the acceleration would be proportional to the net force, and inversely proportional to the mass of the link or links. So if, say, you had a 10 lb force pulling a chain of 10 links of mass 1 slug each the acceleration would be 1 foot per second per second. The force on the last 1 slug link would be 1 lb and it would nicely accelerate at 1 foot per second per second along with the rest of the chain. Or if, say, you had a 10 newton force pulling a chain of 10 links of mass 1 kilogram each the acceleration would be 1 meter per second per second. The force on the last 1 kilogram link would be 1 newton and it would nicely accelerate at 1 meter per second per second along with the rest of the chain.

Why would you even suggest that it might?

He'd be a little embarrassed trying to explain his "lorentz contraction": "But...It was cold... and I was approaching light speed"

There is probably some survival advantage in being more aware of a foreign smelling fart, putting us on high alert, and also for wanting others to smell it first, in a less diluted state. If they and we survive unscathed, we can allow ourselves to become more "accustomed to it" for next time.

Good find. That might be most likely to succeed on another planet, after space travel, as well. Might "we" possibly have gone through a stage of that?

How about with a Science theme: Max Swell Kneels Bore

It will take more than a thousand years regardless of what the rope is made of.

Based on?

The human force is backwards trying to decelerate but puts no torque on the object (as you stated). It acts through the center of mass. However: The friction force of the ground is opposite that of the human's. It pushes it forwards or better stated constrains it not to slip backwards at the contact point. It applies a torque that counters the rotation, but does no work. The acceleration is dependent on the moment of inertia. The stopping time and stopping distance are not the same for both objects. The linear momentum transferred to the surface is, however, the same in both cases, it just takes longer in the higher moment of inertia case. It may seem counterintuitive, but that is how it works. If however the human applied more force, to make up for the force of friction in each case, then the stopping time and distance would be the same for each object.

Without slipping Why gain? Why not lose? The body gains or loses momentum from the surface from the static friction only as the surface loses it or gains it respectively.

There will be the same net transfer of linear momentum, assuming they have the same final velocity as well.

For the same force? For the same time? Transferred to the rolling body, or to the ground, or both?

I would say that is somewhat misleading with regard to the angular momentum, to say it is simply transferred to what it is rolling on. It is often not obvious how it is conserved IMO. For linear momentum it is relatively straight forward. In, say, a 2 body system, one must gain what the other loses. For angular momentum one must look at the relationship between the two (or more) bodies to see that angular momentum is conserved.

Different mass distribution.

It is different, but exactly compensated for by the traction force. So effectively no. What do you mean by velocity frame? Frame of reference?

For translational momentum: It is different, but exactly compensated for by the traction force. For Energy: It is different. They have different kinetic energies and they expend different energies. For angular momentum: Less straight forward than the other two above. Just don't think of this as a source of energy, or translational momentum. It is neither.