How fast does a ring need to spin for is to be indistinguishable from a sphere?

Question

1) Say I had a ring with the formula x^2 + y^2 = r^2 and mounted it to a motor that would spin the ring about the y-axis, how fast would the motor need to spin to make the ring appear indistinguishable from a sphere?
2) Do the limits of this question approach the Heizenburg principle?

Answer 1

Interesting question! Is this from a test or something, or is it just out of curiousity?

At any rate, I guess it depends on what you mean by 'indistinguishable'. If you mean merely from a human's visual perspective... not that fast in the grand scheme of things. Apparently, the human eye can interpet somewhere around 200 frames per second, using cinematography terminology. So, the point where a ring would begin to 'look' like a sphere would be somewhere on the order of a few hundred revolutions per second.

If you're talking in the physics sense, I guess there are 2 different ways of looking at it. (I'm making this stuff up as I go, so bear with me). You could look at it from a relativistic sense; as when it starts rotating at a speed where the equator of the ring approaches the speed of light, then even to light itself the ring would start to behave like a hollow spherical shell. You'd have to include things like length contraction for a proper analysis... but I think that a simple approach gets the point across.

You could also look at it in terms of a quantum mechanical sense. Let's say you're 'observing' the ring by scattering particles of energy E off the ring. Quantum mechanics says that these particles have a DeBroglie frequency of E/hbar. I imagine that when the angular frequency of the ring approaches this value, it would begin to appear like a sphere.

I guess that the uncertainty principle figures into this because according to the nonrelativistic quantum mechanics, no matter how fast you spin the ring, particles of sufficient energy could distinguish it as a ring. The criteria for the energy is something like E > hbar w. Rearranging this, we can find h < E t, where t is the period of the ring's rotation. This is a form of the uncertainty principle, known as the time/ energy uncertainty principle.

Of course... this all has been rough order of magnitude stuff. I hope it helps. Again... cool, thought provoking question.

Answer 2

1) It depends on the visual sampling rate of the observer. Let us define the term "visual sampling rate" as the inverse of the smallest time slice in which an observer can detect change in his/her visual field. The motor would need to spin the ring faster than 1/2 this number of revolutions per time interval for the ring to appear as a sphere to that observer, since in 1/2 of a rotation, the ring would traverse the entire surface of a sphere.

For example, let us say, hypothetically, that you can detect change in your visual field in a time slice no smaller than 1/30 of a second. In other words, any movement that occured within a time slice less than 1/30 of second would appear as a blur to you. That would make your "visual sampling rate" equal to 1/(30 seconds) or 30 Hz. For you, the motor would need to spin faster than 15 revolutions per second for you to observer the ring as a sphere.

2) My answer is based on relatively low speeds. At very high rates of speed, I have no idea what would happen...

Answer 3

1) The frame hold time of an eyeball is about 1/60th of a second. Human beings dont see events shorter than that.

2) Heisenberg says that momentum and position or observation time and kinetic energy can only be resolved to a certain limit. Super-colliders don't work at a Planck scale (the order of mass, volume, velocity, or observation time for Heisenberg's principle to come into play. You are more likely to create a wormhole with your ring than to get to a Planck scale, its dialated mass would have to be literally astronomical in scale.

Answer 4

It will never appear to be a soild sphere, so there goes that proposed quantum interpretation.

Answer 5

i have no clue, but i do that with coins too!