- #1
User69
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Good day all.
First of all, I apologize if this has been asked a million times before. I have not been able to find a straight-forward answer to my wonderings; maybe because they do not yield a straight-forward question.
As part of a large exercise of thinking for the joy of thinking*, I am trying to model the behavior of a simple isolated system consisting of two identical spherical particles with basic ideal properties. For simplicity, I am initially considering them to be bidimensional discs capable of moving in two perpendicular axes on the same plane, but I would eventually like to understand the behavior of the system in three dimensions.
Both particles have an equal radius and the same amount of matter distributed homogeneously within their enclosed volume; they are solid, rigid, indivisible and identical. Their surface is not "sticky" -they will not become attached upon collision- but it cannot slide either: when both particles are in contact, for the surface of one to be able to move laterally, the surface of the other must move laterally by the same amount.
If one particle is still and the other moves directly towards its center, having both no rotational motion, their collision will be head-on, with no tangential velocity and perfectly elastic: both particles will exchange all their kinetic energy, since no energy can be dissipated as deformation or loss of internal structure, nor lost due to friction (the system is isolated and therefore in absolute vacuum, and the only interaction has no tangential velocity).
However, when the collision is oblique, the inability of the surfaces to slide against each other will surely introduce a force parallel to the tangential velocity (and therefore to both surfaces) but in opposite direction, causing some of that tangential velocity to be transformed into angular motion of the particles involved. After the collision, the sum of the kinetic energies of both particles will be smaller than the kinetic energy the moving particle had before the collision (i.e. it will be an inelastic collision), and the difference will be equal to the rotational energy gained by both particles.
The process should work in reverse. If a collision can be inelastic due to an increment in rotational energy, it can also be superelastic due to a decrement in rotational energy, as can be seen when two tops spinning in the same direction collide laterally: part of their rotational motion is transformed into linear motion and each moves away from the other at a speed greater than that with which they approached.
Up to this point, I may have made many conceptual mistakes and I would very much appreciate anyone's effort to correct them. If not, my question is how to model this transformation between linear and angular motion when two surfaces cannot slide at all. I'm afraid I lack the mathematical support needed to derive the equations myself :-(
If anyone understands the described scenario and would like to help me solve it in mathematical terms, I would be very grateful. Nevertheless, I appreciate any time anyone may have taken to read this.
Thank you very much.
First of all, I apologize if this has been asked a million times before. I have not been able to find a straight-forward answer to my wonderings; maybe because they do not yield a straight-forward question.
As part of a large exercise of thinking for the joy of thinking*, I am trying to model the behavior of a simple isolated system consisting of two identical spherical particles with basic ideal properties. For simplicity, I am initially considering them to be bidimensional discs capable of moving in two perpendicular axes on the same plane, but I would eventually like to understand the behavior of the system in three dimensions.
Both particles have an equal radius and the same amount of matter distributed homogeneously within their enclosed volume; they are solid, rigid, indivisible and identical. Their surface is not "sticky" -they will not become attached upon collision- but it cannot slide either: when both particles are in contact, for the surface of one to be able to move laterally, the surface of the other must move laterally by the same amount.
If one particle is still and the other moves directly towards its center, having both no rotational motion, their collision will be head-on, with no tangential velocity and perfectly elastic: both particles will exchange all their kinetic energy, since no energy can be dissipated as deformation or loss of internal structure, nor lost due to friction (the system is isolated and therefore in absolute vacuum, and the only interaction has no tangential velocity).
However, when the collision is oblique, the inability of the surfaces to slide against each other will surely introduce a force parallel to the tangential velocity (and therefore to both surfaces) but in opposite direction, causing some of that tangential velocity to be transformed into angular motion of the particles involved. After the collision, the sum of the kinetic energies of both particles will be smaller than the kinetic energy the moving particle had before the collision (i.e. it will be an inelastic collision), and the difference will be equal to the rotational energy gained by both particles.
The process should work in reverse. If a collision can be inelastic due to an increment in rotational energy, it can also be superelastic due to a decrement in rotational energy, as can be seen when two tops spinning in the same direction collide laterally: part of their rotational motion is transformed into linear motion and each moves away from the other at a speed greater than that with which they approached.
Up to this point, I may have made many conceptual mistakes and I would very much appreciate anyone's effort to correct them. If not, my question is how to model this transformation between linear and angular motion when two surfaces cannot slide at all. I'm afraid I lack the mathematical support needed to derive the equations myself :-(
If anyone understands the described scenario and would like to help me solve it in mathematical terms, I would be very grateful. Nevertheless, I appreciate any time anyone may have taken to read this.
Thank you very much.
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