[Fluffy]

Hi, I am trying to compute a 3d model, I am love all this vectors and stuff but I am really stuck with collisions.


I have a lot of questions about physics, but lets start with a simple one.

I am trying to distinguish between 2 movements translation and rotation, when a solid is coliided a part of the energy from the collision goes to translate the object, and another goes to rotation.

Say there is a solid sfere in the space subject to no force and somethign collides with it on a superficial point on a trayectory totally tangencial to its surface. I think this is the case that maximices the rotation/traslation ratio. Will theoretically be any traslation at all?

[helene_t]

Fluffy, on 2010-October-23, 04:22, said:

Hi, I am trying to compute a 3d model, I am love all this vectors and stuff but I am really stuck with collisions.


I have a lot of questions about physics, but lets start with a simple one.

I am trying to distinguish between 2 movements translation and rotation, when a solid is coliided a part of the energy from the collision goes to translate the object, and another goes to rotation.

Say there is a solid sfere in the space subject to no force and somethign collides with it on a superficial point on a trayectory totally tangencial to its surface. I think this is the case that maximices the rotation/traslation ratio. Will theoretically be any traslation at all?

Yes there will be translation, too. Conversation of momentum requires that any momentum lost by the object is compensated by a momentum of the sphere, and since rotation does not contribute to momentum, there must be translation as well.

You have three unknowns after the collision: The speed of the object, the translational speed of the sphere, and the rotational speed. So you need to write down three equations to solve the system. These could be coversation of energy, momentum, and angular momentum relative to the center of the sphere.

[gwnn]

You have a nice equation for the kinetic energy of a rotating object, it is

E_krot=I*O²/2.

O should be the Greek letter omega and is the angular velocity (rad/s).

I is the moment of inertia and it's related to the shape and size and mass of your object, as well as the position of the rotation axis. For a sphere (of uniform density, rotating around its centre) it is
2/5Mr², where M is the mass of the sphere and r is its radius.

Then you have two equations:
the conservation of momentum (in which rotation is not a part)
and
the conservation of energy

[kenberg]

My physics was a long time ago, but here are my thoughts:

The meaning of "Tangential strike" needs a little filling out. Start by thinking of an ice cube sliding past a glass, barely striking it. The friction is low and hardly anything will happen. I think there are a couple of ways of drawing this out:

1. Assume the glass and the cube are covered in sandpaper. This will cause frictional force on both objects.

2. Ignore friction, but assume that the strike is not quite tangential but analyze what happens as the blow becomes more and more tangential. Take it to the limit.

My guess about the second option is that in the limiting case is the same as the actual tangential but frictionless blow: Nothing happens. Another way of saying this is that if the blow is very close to tangential, very little happens. Equivalently, the object that does the striking will barely be slowed. This seems to agree with my physical experience. Think of a pool ball making a very glancing strike against another ball.

So I think some friction needs to be involved. There has to be some sort of "sticky contact", and I think this will complicate the analysis.

Those more involved with physics on a daily basis will now explain my error.

[Fluffy]

kenberg, on 2010-October-23, 06:31, said:

My physics was a long time ago, but here are my thoughts:

The meaning of "Tangential strike" needs a little filling out. Start by thinking of an ice cube sliding past a glass, barely striking it. The friction is low and hardly anything will happen. I think there are a couple of ways of drawing this out:

1. Assume the glass and the cube are covered in sandpaper. This will cause frictional force on both objects.

2. Ignore friction, but assume that the strike is not quite tangential but analyze what happens as the blow becomes more and more tangential. Take it to the limit.

My guess about the second option is that in the limiting case is the same as the actual tangential but frictionless blow: Nothing happens. Another way of saying this is that if the blow is very close to tangential, very little happens. Equivalently, the object that does the striking will barely be slowed. This seems to agree with my physical experience. Think of a pool ball making a very glancing strike against another ball.

So I think some friction needs to be involved. There has to be some sort of "sticky contact", and I think this will complicate the analysis.

Those more involved with physics on a daily basis will now explain my error.

Maybe thinking of spheres works this way, but then I switch to think of a pencil being barelly hit in its extreme by a big mass at high speed and my intuition is that there is some traslation, but there is a big rotation, the question is how much.

The equations don't help me much at this stage, maybe I need to study more physincs to solve the problem :/.


Ok, new question, what if the sphere is hit tangencial, but outside its surface?, think of it as the earth being spheric and having uniform density, but there is a tall ultra rigid-tower but with no (relevant) mass, that goes outside the atmosphere, and you hit the tower on its extreme with a big enough mass, to get zero traslation and 100% rotation, would the tower need to be of infinite lenght?

It think this is equivalent of saying a big sphere with all its mass condensed in a little core

Not sure why I ask this because the answer ain't gonna help me I think .

[kenberg]

A sphere with a little target sticking up would work fine for a model. Some other stipulations are necessary, for example 'Assume the collision is elastic". Or don't.

"Working fine" means that with some thinking I could do it, those more into physics day to day could do it much quicker.

Here is a question: I'm fine with doing fun problems for the fun of it, and I can give my thoughts, on a fwiw basis, on a real problem. I am not sure which category this one is in.
Oops, I didn't get a ? mark in to that, but you get it I think.

K

[gwnn]

No matter what scenario, you will always have a little bit of translation. You are right that the further you hit an object from its centre, the smaller the ratio of translation:rotation will be, but it will never be 0.

The reason is this:

• momentum must be conserved (despite friction)
  
• the object that hit our sphere has lost some of its velocity, so it lost some of the momentum
  
• rotation carries zero momentum
  
• therefore there must be some translation
  
• QED

[Fluffy]

yes thanks, I don't fully understand what momentum is even when I studied it not tha tlong ago, but I understand the logic that says that some must be lost and it must go into traslation. I will try to solve what is the ratio for the optimum hit when the sphere has uniform density.

[gwnn]

momentum is just mass*velocity

but it's a vectorial quantity (it has a direction) so when something is rotating around an axis that passes through its centre of mass, half of it is going ahead and half of it backwards and it all compensates so the sum of all the little momenta is zero.

now there is also angular momentum but it is not needed at the moment (sorry for the stupid pun) to talk about translation (because the angular momentum is sort of zero for translation)

[kenberg]

Csaba,

How much of the following is correct? Assume the case of a sphere with a small target hit by a pellet. Say that the sphere is uniformly solid of mass M but that changes only the details.

Imagine all following statements have an "Is that right" appended.

We choose our system so the sphere is initially stationary, the pellet of mass m moving with velocity v striking the target. The direction of v is tangential to the sphere, perpendicular to the small target.

Imagine all following statements have an "Is that right" appended.

The initial linear momentum of the system is m|v| where |v| is the speed (velocity stripped of its direction).

The initial angular momentum, for the entire system, about the center of the sphere comes entirely from the tangential linear velocity of the pellet and is given by by rm|v| where r is the radius of the sphere.

The initial energy is m|v|^2/2

What happens after the collision depends on the nature of the collision.

Assume an elastic collision if that is the right word. What I mean is that we assume no mechanical energy is transferred to heat, or other forms of energy. If we do not assume this, we have to say what we do assume about the collision. The elastic assumption is more warranted in some cases, such as a golf ball, than in others, such as a snow ball.

With this elastic assumption, the mechanical energy and the momentum, linear and angular, are preserved.

The new velocities, say u for the pellet and w for center of the sphere. will still be along the same( or parallel) line as it was for v before the collision.

Although the sphere is 3D, this is really a planar problem. The velocity vectors all lie in the same plane and the angular velocities are with respect to an axis perpendicular to the plane.

If we believe all this stuff so far, then it all boils down to formulas for inertia (here is where solid or hollow matters for the sphImagine all following statements have an "Is that right" appended.
ere) and then doing some algebra.

So far, so good?

[helene_t]

Yeah this is all right but as you said before, if v is tangential to the sphere and the collision is elastic then nothing happens.

[kenberg]

helene_t, on 2010-October-24, 11:56, said:

Yeah this is all right but as you said before, if v is tangential to the sphere and the collision is elastic then nothing happens.

No, there is (in my current attempt) a little upright target. We have a sphere with a small, flat straight up protrusion that the pellet hits and bounces off of, keeping its motion along the same line.

I hope I am not taking this in some direction different from the intended one.

[gwnn]

Yes Ken that's all correct. And it's all maths from that point. I think we need all three equations to get the right answers because we have no relationship between the angular velocity and linear velocity of the sphere (as opposed to the case where it is rolling on a surface)

[kenberg]

A further point that was troubling me for a bit. I was worried about the calculation of angular momentum after the collision. But it all works out I think. We have to be able to calculate the angular momentum of the system after the collision, and at least at first thought we have to calculate it with reference to the same point as we did earlier. The original computation was done with respect to the center of the stationary sphere and now it is in motion. But I think there is no difficulty here. Say that the sphere began at the origin and suppose E is a unit vector pointing in the direction of v. After the collision the center of the sphere moves at speed |w| in the E direction and so is at position t|w|E, with t being time since impact.

Angular momentum involves the cross-product of the momentum vector with the position vector, and the cross-product of parallel vectors is zero. If I am thinking right, this means that for a sphere that is rotating about an axis and moving at a constant velocity, the angular momentum of the sphere with respect to a point on the path of the center is independent of which point you choose. The upshot being that the sphere's contribution to the angular momentum, about the original point of reference, is the same as its angular momentum about its axis, despite the fact that it is in (uniform) motion.


The contribution to angular momentum from the pellet can also be (easily) referred to the original point of reference.


This may not be expressed very clearly. But I think that for the sphere with target, and with an elastic collision, it's sort of down to some calculations.


All of my physics was pre-quark, almost pre-meson, but of course this part hasn't changed and I sort of remember it. I helped a college girl (my wife's daughter) with her basic physics a while back. No cross-products, but it was something like this.

[gwnn]

yes there's no need for any cross products I think because I think both the bullet and the sphere will have a translation along the same axis as the bullet had initially. this means that when you're calculating the "torque" of the momentum of the bullet at a later time, it will always be m*|u|*r

[kenberg]

This time I am straying from the OP original intent.

As mentioned, the sphere problem is, in essence, two dimensional. But it reminded me of something in Scientific American about forty years ago. As I recall, the author was an Olympic diver and a physicist. The subject was the physics behind the twists of divers and, in particular, some counter-intuitive matters regarding angular momentum. First imagine someone running along the diving board and leaping forward. His center of mass moves along the familiar parabolic arc and there is nothing to be done about it. Next suppose he springs from the board imparting rotation for a somersault. Say he does a one-and-a-half turn and opens with his head downward towards the water. He has angular momentum about a line roughly running from side to side through his hips. By making suitable moves contracting and extending his limbs, the axis of rotation can be altered so that he now rotates about a vertical axis.

We see them do this. When I would attempt such things in my youth, I just assumed that somehow I had to get this extra twist into my exit from the board. After a few disastrous attempts, I gave up. The idea that the axis of rotation could be altered in flight, without pushing off of anything (pushing off the air is not remotely enough), violated my sense of physics. I was wrong.


Mathematically, I think that the issues is that the two-dimensional rotation group is Abelian, the three-dimensional one isn't. This means that in two dimensions if you rotate twice, say by 30 degrees and 45 degrees, it doesn't matter which rotation you do first. In three dimensions there is a standard example to the contrary. Take a block of wood, hold it in front of you, and rotate it 90 degrees "forward" (the top rotates away from your body) and then rotate it 90 degrees clockwise. Now repeat, only do the clockwise rotation first. The results will be different.


Anyway, I recommend this article if it could ever be found.

[Fluffy]

regarding bounces, if I give a ball a elasticity variable that goes form 0 to 1 and determines the percentage of speed that it will have when bouncing into a solid non elastic ground, meaning that 0 elasticity means the ball just his the ground and stays there, while 1 would mean it hits and bounces off at the same speed.

Does this make sense?

now regarding 2 elastic objects, if a non-elastic small stone hits a big ball will it bounce with the ball's elasticity? what about 2 elastic objects colliding?

[phil_20686]

The first question you posed is actually very difficult (conceptually). You need to note that in order to apply Conservation of angular momentum you need to specific a point about which !every part! of the problem rotates. This means that the angular momentum will come in 3 parts, rotation of pellet, rotation of the sphere, and rotation of the whole system about some point.

Hard to show this without diagrams, but here goes:

Suppose we think of the ball as having a target sticking out of it off which the pellet is reflected. The whole problem is elastic for simplicity of calculation. Initially I choose the target as the point about which the system "rotates", thus I initially have no angular momentum. After the pellet bounces off at v return for a total change in momentum of (v initial + v return)*mass. where I have chosen my sign convention such that all velocities are positive in my equations.

The sphere now has momentum v sphere* mass sphere = change in momentum of pellet.

Thus the linear momentum of the sphere is seen to be known, and linear momentum is conserved.

The angular momentum gained by the sphere is formed by its back reaction of its inertia vs the force. Its inertia acts through the centre of gravity, so this will translate part of the force from the pellet into rotation of the sphere. In order to work out how much it is easiest to see that around the initial point of my target the angular momentum of the system remains zero, so the "internal angular momentum" of the spheres rotation, must be exactly balanced by the "external" momentum of the sphere moving around this point. So linear momentum of the sphere*radius = Inertia about its centre*angular momentum of the sphere about its centre.

If you were actually going to calculate this rather than demonstrate how momentum is conserved it would be best to start with a observer at the cetnre of the sphere. Then you get automatically that change in momentum of the pellet*radius of the sphere = intertia of sphere about its centre * rotation of sphere about its centre. {In this frame the linear momentum of the sphere contributes no angluar momentum as it has no impact parameter. - you can see that this is equivalent to the frame above if you note that the conservation of the linear momentum means that the equation here is exactly equivalent to above}

The reason the second method is better is that in teh first method I have assumed that I can compute the angular momentum of a finite sized body as its linear momentum through its centre of mass - which is not easy to prove with any rigour, although the equivalence of teh two expression less you it must be true in this case and a little thought will convince you that it is "intuitavely clear" that it should be true in all cases.


The key point is to remember that angular momentum must be evaluate as rotaions about one point for every body in teh problem. Thus a particle flying in a straight line pas this point contributes angular momentum - that is why an asteroid can form an orbit and things do not just fall towards the centre of the solar system.

[phil_20686]

Fluffy, on 2010-October-25, 07:12, said:

regarding bounces, if I give a ball a elasticity variable that goes form 0 to 1 and determines the percentage of speed that it will have when bouncing into a solid non elastic ground, meaning that 0 elasticity means the ball just his the ground and stays there, while 1 would mean it hits and bounces off at the same speed.

Does this make sense?

now regarding 2 elastic objects, if a non-elastic small stone hits a big ball will it bounce with the ball's elasticity? what about 2 elastic objects colliding?

Not really, elasticity should really refer to the % of energy lost in a collision. However, for various historical reasons it has been defined in a variety of ways depending on what set up one is talking about. Momentum is always conserved, elasticity refers to the ability of macroscopic entities to dissipate energy into microscopic degrees of freedom. The energy is still conserved, it is just "lost" at the level of a classical physics problem. Most likely in real world applications it comes out as heat, or fractures.

Mostly you should choose a definition such that elasticity will have nice additive properties for repeat collision, if you are only thing one thing it doesn't really matter. For an elastic and a non elastic entity colliding there total elasticity will depend on a variety of factors. except in a thought experiment one cannot generally abstract from an objects elasticity in one collision to its elasticity of another. E.g. If I dive into water at 10 m/s it will be very different from if I dive in at 1000 m/s which is a way of saying its elasticity is a function of velocity. Even the angle of collision can make a large difference - skipping stones is pretty close to elastic collions, if i through the stone straight down it would sink though.

[phil_20686]

kenberg, on 2010-October-25, 06:22, said:

Mathematically, I think that the issues is that the two-dimensional rotation group is Abelian, the three-dimensional one isn't. This means that in two dimensions if you rotate twice, say by 30 degrees and 45 degrees, it doesn't matter which rotation you do first. In three dimensions there is a standard example to the contrary. Take a block of wood, hold it in front of you, and rotate it 90 degrees "forward" (the top rotates away from your body) and then rotate it 90 degrees clockwise. Now repeat, only do the clockwise rotation first. The results will be different.

The fact that it is non abelien doesnt mean you must be able to change your axis, only that if you do the way in which you change it matters.

I suspect that the answers as to how a diver does it may be quite tough. Probably they are large part illusion: we have all seen the ice skaters who's angular rotation accelerates when they draw in their arms due to their decreased moment of inertia. Its easy to imagine that you can impart only a small angular velocity in one direction initially and then accelerate it by changing body position. If you suppressed inertia in one direction and increased it in another you could change your axis of rotation. I remember an old finals problem about a disk made of nano bots that is rotation around a diameter. The bots reposition themselves into a hemisphere, what is the new axis of rotation.

It is also possible that precession plays a part as in the rotation of the Earth's axis. It would be a very tough problem to work that out.