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Yeah, it's a simple question.
Of course, they didn't add humans; Mент is right.
Uranium235
Posts from Quiz
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Uranium235
Драккошка
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I was going to reply in a funny way, but I changed my mind.
А, вот почему.
🏹🎪+⛓️♿=💕
🏹🎪+⛓️♿=💕
Ment
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There's a slightly harder and a slightly easier option (but neither is too difficult). Both seem like interesting tasks. Which one should I give?
Uranium235
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And it's a good thing you changed your mind; I wouldn't call a complex question a simple one.
Menti, come on, start with the first one – let's do the first one, which is easier – so we can get to the second one faster.
Ment
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Okay, let's consider this:
Imagine we are constructing polyhedra and implementing them, endowing them with weight. Obviously, we can construct a polyhedron such that, when placed on a flat horizontal plane (on one of its faces, or even on all its faces in turn), it remains in a state of rest, in equilibrium. At the same time, such a polyhedron can be irregular.
Question: is it possible to create such a polyhedron so that it would not be in a state of equilibrium (it would tip over) when placed on any of its faces? In other words, would there be no face from which it would not tip over?
Of course, this can be proven.
Imagine we are constructing polyhedra and implementing them, endowing them with weight. Obviously, we can construct a polyhedron such that, when placed on a flat horizontal plane (on one of its faces, or even on all its faces in turn), it remains in a state of rest, in equilibrium. At the same time, such a polyhedron can be irregular.
Question: is it possible to create such a polyhedron so that it would not be in a state of equilibrium (it would tip over) when placed on any of its faces? In other words, would there be no face from which it would not tip over?
Of course, this can be proven.
Uranium235
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I have a feeling that it doesn't exist; there should be a state of equilibrium, otherwise such a constantly rotating polyhedron would already be exploited as a perpetual motion machine.
A regular sphere can essentially be represented as a polyhedron if the faces are made 2-3 atoms long, but I think we know that if the plane is truly flat, then gravity + friction will prevent the sphere from rolling anywhere. And the less the polyhedron resembles a sphere, the less likely it is to be able to roll anywhere at all – and essentially, that's what's needed: it shouldn't be in a state of equilibrium = it should roll.
Without the application of an external force, the sphere will not roll. In some way, for example, fill the sphere (okay, the polyhedron) with some volatile substance like iodine and heat the horizontal surface. Then, iodine vapor from the bottom evaporates and settles on top – the center of mass shifts, the ball rolls, and the iodine crystals end up near the horizontal surface and re-sublimate... In short, here, heating is the external force.
But I don't have any proper proof.
Hermit
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A concave polyhedron can rest on its edges.
Added 3 minutes later
Yes, it can. It will be a concave polyhedron that will roll from faces to edges.
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Let's take, for example, a polyhedron formed by gluing two regular tetrahedra together along one face. Its center of gravity will be in the middle. This point will be the midpoint of the faces by which the tetrahedra are glued. Let's cut off almost at the very edge of the common face after gluing the tetrahedra together at both ends with a small pyramid - a regular tetrahedron. And let's hollow out three faces symmetrically inward from each side. The position of the center of gravity of the polyhedron formed in this way will not change, and if it is placed on any face after this, then the projection of the center of mass onto the plane of this face will be outside it - which means the polyhedron will roll onto its edges.
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Yes, it can. It will be a concave polyhedron that will roll from faces to edges.
Added 17 minutes later
Let's take, for example, a polyhedron formed by gluing two regular tetrahedra together along one face. Its center of gravity will be in the middle. This point will be the midpoint of the faces by which the tetrahedra are glued. Let's cut off almost at the very edge of the common face after gluing the tetrahedra together at both ends with a small pyramid - a regular tetrahedron. And let's hollow out three faces symmetrically inward from each side. The position of the center of gravity of the polyhedron formed in this way will not change, and if it is placed on any face after this, then the projection of the center of mass onto the plane of this face will be outside it - which means the polyhedron will roll onto its edges.
Всё не так плохо как Вы думаете. Всё намного хуже!
Злобный Ящур
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Sure. It will be a concave polyhedron that will rotate from faces to edges.
Added 42 minutes later
If the polyhedron is placed on an edge (or on a vertex), such a balance will not be stable (because the center of gravity will be above the support), and the polyhedron will fall onto one of the faces. If, at the same time, the projection of the center of gravity turns out to be within the area of the table (or wherever the polyhedron is placed) on which this face lies, then the polyhedron, placed on this face, will not fall. Such a face will always be found (the polyhedron will "fall" until it "finds" such a face), so the answer is no.
P.S. Of course, if the polyhedron is convex, but it seems that the question implies that it is convex. Otherwise, it is generally impossible to place it on "any face."
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if the plane is really flat, then gravity + friction - and the ball will not roll anywhere.
Ничто не возникает из ничего и ничто не пропадает бесследно.
Если где-то чего-то убудет, то в другом месте добавится.
(Закон сохранения).
Ment
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Let's take, for example, a polyhedron formed by joining two regular tetrahedra along one of their faces. Its center of gravity will be in the middle. This point will be the midpoint of the faces by which the tetrahedra are joined. Let's cut off almost at the very edge of the common face after joining the tetrahedra, removing a pyramid from both ends – a regular tetrahedron. And let's hollow out three faces symmetrically inward on each side. The position of the center of gravity of the polyhedron formed in this way will not change, but if we place it on any face after this, the projection of the center of mass onto the plane of this face will be outside it – which means the polyhedron will tip over onto its edges.
Such a face will always be found (the polyhedron will "fall" until it "finds" such a face), so the answer is no.
Hermit
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A polyhedron can lie on its face if its center of gravity is projected onto that face. (Here, the center of gravity must be projected orthogonally onto the plane in which the face lies. :) ) Since it turns out that the polyhedron must be convex, otherwise it will not be possible for it to lie on any of its faces, the center of gravity is located inside the polyhedron, which means that if perpendiculars are dropped from it onto all the faces, then at least one of them must intersect the face onto which it is dropped.
Всё не так плохо как Вы думаете. Всё намного хуже!
Uranium235
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This sphere will roll very well. Except in cases where its center of gravity lies on a diameter perpendicular to the tabletop.
Hermit
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Ideally, if there is no friction or resistance to movement, it will not come to a standstill. The same applies even to a polyhedron. In principle, if you release a cylinder whose center of gravity is shifted and not located on the axis, then in the absence of friction and resistance to movement, it will continue to roll back and forth. The same applies to, for example, a polyhedron that is "almost a cylinder." Of course, energy will be expended when lifting the almost cylinder onto an edge, but it will also be released when the almost cylinder is lowered onto a face.
Всё не так плохо как Вы думаете. Всё намного хуже!
Ment
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A polyhedron can lie on its face if its center of gravity is projected onto it. (Here, the center of gravity must be projected perpendicularly onto the plane in which the face it is lying on is located.) Since it turns out that the polyhedron must be convex, otherwise it will not be possible for it to lie on any of its faces, the center of gravity is located inside the polyhedron, which means that if perpendiculars are dropped from it onto all faces, then at least one of them must intersect the face onto which it is dropped.
Guys, this is not a dynamics problem. We are interested in equilibrium states, not how long a particular figure can rotate. An equipotential sphere is in a state of equilibrium by default, albeit an unstable one.
Ment
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What did Kosha write? What if it's true? You never know...
Драккошка
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I wrote that the problem can be reduced to a 2D plane and it can be proven that for any point inside a convex polygon, it is possible to draw a perpendicular to one of the sides, such that the end of the perpendicular falls on the side itself, and not its extension.
Well, there you basically divide this thing into triangles with angles vertex-vertex-center and calculate the heights, maybe it's also worth considering the 360 degrees and so on, I don't feel like solving it further.
А, вот почему.
🏹🎪+⛓️♿=💕
🏹🎪+⛓️♿=💕