What happens if you put a sphere (ball) on the top of a pyramid?

[Baluncore]

The Vickers hardness test indents a surface with a diamond pyramid.
The Brinell hardness test indents a surface with a hard sphere.

The point of the pyramid will offer an infinite pressure to the sphere and form a pyramidal socket in the sphere. Or, the sphere will crush the point of the pyramid, forming a concave platform close to the top of the pyramid. One or both of those two mechanisms will effectively lock the sphere in a stable position at the top of the pyramid.

The sphere therefore, will not fall from the apex of the pyramid.

 

[Vanadium 50]

A classical pyramid (assuming made out of a continuous material and not atoms) will have an infinitely sharp point. When the sphere is placed on it, the pyramid point will penetrate it (because the pressure is infinite) to a depth determined by Young's modulus. The sphere will not roll without a push.

If your response is "this doesn't count", I would respond that the question needs to be better defined then.

 

[DrStupid]

yes and it is well described in terms of deterministic ODE

It is described in terms of ODEs that may have different solutions for the same starting conditions. We discussed an example above. The ODEs alone don't guarantee that the system is full predictable - not even in theory.

 

[Halc]

Concerning reality instead of mathematical idealizations:

A classical pyramid (assuming made out of a continuous material and not atoms) will have an infinitely sharp point. When the sphere is placed on it, the pyramid point will penetrate it (because the pressure is infinite) to a depth determined by Young's modulus. The sphere will not roll without a push.

OK, but we've already discarded actual physics if we're talking about some fictional continuous material and not a collection of particles held in a vague shape by mutual forces. Young's modulus need not apply here since there is no reason to assume limited hardness for the purposes of this topic.

Yes, a physical (atoms this time) ball-ish shape perched perfectly on a symmetrical (ish) location like that will deform enough to form a finite surface area which can reasonably be expected to make the equilibrium stable. We're not unaware of this.

The Vickers hardness test indents a surface with a diamond pyramid.
The Brinell hardness test indents a surface with a hard sphere.

The point of the pyramid will offer an infinite pressure to the sphere and form a pyramidal socket in the sphere.

This set of comments makes a sort of implication that a mathematical sphere resting on a flat or other non-sharp surface will not offer infinite pressure. If we're talking physical and not mathematical, then the deformation (strain) will occur well before pressure becomes infinite, so this singularity is never reached. Colliding billiard balls do not generate infinite pressure.

 

[Vanadium 50]

OK, but we've already discarded actual physics

Yes we have.

Classical physics has a well-defined answer. The ball is impaled and doesn't move.
Quantum mechanics has a well-defined answer: the state cannot be prepared as specified.

What this whole argument about involves some sort of part-classical part-quantum set of rules that have never been clearly specified.

 

[samalkhaiat]

I cannot find the original article about the pencil, which computed that a perfect balanced pencil shaped object has a 50% chance of falling before 30 seconds at Earth gravity.

Actually, the pencil will fall after maximum time of 3-4 sec. Try to do the experiment.
If you solve Newton’s equation for the CM of the pencil and assume that the best initial conditions (i.e., the angle from the vertical and the associated angular velocity) are determined by the uncertainty principle, $\dot{\theta}_{0} = \frac{\hbar}{mr^{2}\theta_{0}}$, you find $t_{max} \approx 3.6 \ \mbox{sec}$. The example in the OP is no different from the pencil.

 

[Baluncore]

This set of comments makes a sort of implication that a mathematical sphere resting on a flat or other non-sharp surface will not offer infinite pressure.

A “sort of implication” is your subjective response to the pressure of conflating the mathematics of perfect geometry with the physics of real materials. If you insist on cross-coupling those different domains you will continue to trip over the idiosyncratic contradictions.

Only for perfect geometry can the area of contact be zero. The sphere would require mass, and acceleration to generate an infinite mathematical contact pressure. In the OP, the mathematical sphere has zero specified mass, so gravity is irrelevant, and nothing can fall. Indeed, a perfect mathematical sphere will have lower density than our real atmosphere, so it cannot fall but will rise to escape the perfect-Platonic-solids model of the Solar system. As is obvious, the two anti-domains must be disentangled, or all hell breaks loose.

For the real material world there is deformation at the point of contact. That real deformation may be elastic, plastic, or a melding of the materials to form a welded connection. As Hertz identified, the applied mathematics of real material contacts are tractable when the geometry can be simplified. That is why the well defined geometry of a sphere, or the point of a pyramid, is used as an indenter for material hardness testing.

 

[Halc]

Back to the pencil thing. I certainly take back my off-the-cuff claim about the ideal pencil claim when initially brought up. It seems that an ideal pencil is far more stable than a ball on Norton's dome. A pencil moving upward would never actually become vertical, taking infinite time to do so. This was illustrated by the link in post 59 here:

Here is the approach in the tilting pencil, mentioned a ways back...

Tilting Pencils

To quote the bottom line of that: "This idealized pencil would take 1 year ... to tip over (theoretically) if its initial stationary position is θ0 = 10^–117498396 degrees from vertical."
So they managed to have it take a year to fall, and only by cheating by giving it an initial displacement from vertical. Norton's ball will fall at any time spontaneously, and take a very finite (short) time to do so once it starts.

As for the physical pencil, of course it will fall quickly.

If you solve Newton’s equation for the CM of the pencil and assume that the best initial conditions (i.e., the angle from the vertical and the associated angular velocity) are determined by the uncertainty principle, $\dot{\theta}_{0} = \frac{\hbar}{mr^{2}\theta_{0}}$, you find $t_{max} \approx 3.6 \ \mbox{sec}$. The example in the OP is no different from the pencil.

Yes, you forgot that the real pencil is made of atoms and is not balanced on a point, but on a limited area, and random forces like moon tides and the bug in the next room will surely throw the thing. I'm not disputing any of that.

Actually, the pencil will fall after maximum time of 3-4 sec.

OK, I dispute this statement. There is a finite probability that the random forces and whatnot actually correct any imbalance that arise, and the pencil stays up for minutes at least (indefinitely actually). You can say there some very high probability of it falling within 3-4 seconds, but that probability for an object placed within tolerances of vertical can never be 100%.

 

[jbriggs444]

In the OP, the mathematical sphere has zero specified mass, so gravity is irrelevant, and nothing can fall.

That depends on how one models gravity acting on an entity with zero mass. If we are going to extend Newtonian gravity to such objects, I prefer a model where objects with and without mass accelerate identically.

 

[wrobel]

It is described in terms of ODEs that may have different solutions for the same starting conditions. We discussed an example above. The ODEs alone don't guarantee that the system is full predictable - not even in theory.

ODE alone does not even guarantee existence but ODE+ well-known regularity assumption do guarantee it all

We discussed an example above.

I thought you discussed physics because such examples of ODE are discussed in textbooks

 

[samalkhaiat]

As for the physical pencil, of course it will fall quickly.
you forgot that the real pencil is made of atoms and is not balanced on a point, but on a limited area,

No, I did not "forget" that. The thermal motions of atoms in the contact “area” prevent the pencil from being in stable equilibrium state for more than 4 sec. To account for this experimental result (the < 4sec.), one has to assume that the best initial conditions are restricted by the uncertainty principle which accounts for the fact that “the real pencil is made of atoms”.

OK, I dispute this statement.

Sir, I talk physics not gibberish. Unless you do the experiment and manage to balance the pencil for more than 4 seconds, you cannot dispute physical statement which is based on a very realistic mathematical model that predicts the experimental (< 4 seconds) result.

There is a finite probability that the random forces and whatnot actually correct any imbalance that arise, and the pencil stays up for minutes at least (indefinitely actually).

Again, I ask you to make a realistic mathematical model and calculate for us this “finite probability” of yours. If you cannot do that, then you should listen to the experts.

 

[jbriggs444]

Unless you do the experiment and manage to balance the pencil for more than 4 seconds, you cannot dispute physical statement which is based on a very realistic mathematical model that predicts the experimental (< 4 seconds) result.

The mathematical model in question has flaws which have been pointed out. To whit, it assumes that the only relevant uncertain inputs are present in the initial setup.

Direct experimental testing [balance a pencil and time the fall] is clearly infeasible or impractical for the sort of quibble that is being suggested. Indirect experimental evidence has been known for nearly 200 years [Brownian motion].

 

[.Scott]

I don't see any point in getting too attached to any specific real-world model for this experiment. Perfect spheres, perfect pyramids, and perfect balance are inconsistent with Physics. Spheres and pyramids are made of subatomic particles and those objects can only be positioned to within the limits of the HUP. That may seem to be a unimportant limit, but the falling pencil link by @bahamagree in post #59 shows that astronomically small offsets from "perfect" will quickly compound.

In the most vanilla worlds were "perfect" is possible, the sphere will remain in place.

 

[Baluncore]

In the most vanilla worlds were "perfect" is possible, the sphere will remain in place.

That is also consistent with Brownian motion being a martingale.
Any rotation of the pencil or sphere, will gyroscopically overcome the minor θ₀ = 10^–117498396 degrees from vertical.

 

[.Scott]

Any rotation of the pencil or sphere, will gyroscopically overcome the minor θ₀ = 10^–117498396 degrees from vertical.

Since that would result in the pencil toppling in about a year, the rotation would have to be on the order of once a year at a minimum.