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On September 30 2008 08:44 Juicyfruit wrote:
It's basically the opposite of an explosion.
You would expect the net momentum of the system (the system being the can + the air) to be 0 before and after the puncture (conservation of momentum).
Once the can is punctured, the air particles gains net momentum. That means the can needs to move in the opposite direction to conserve the momentum.
The can won't stop moving until the net momentum of the air becomes 0 again, which occurs at equilibrium.
Arguing that the can doesn't move at all is saying air got into the can without net motion.
Right?
I'm interested in Klockan3's answer....
I don't think anyone is arguing that there is no motion at all.
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On September 30 2008 06:48 betaben wrote:
If so, how do you say again that air is propelled in that direction?
I think you're saying that in the process where the gas on the unpunctured side pushes against the can, there will be an equal amount of gas moving in the opposite direction within the gas (as it's overall momentum is beforehand zero). this does not transfer it's momentum to anything, (unlike the gas moving against the can) so continues travelling away from the can.
Is this true?
If we isolate the two surfaces, one were we puncture it and the other is the one exactly on the opposite side.
Each reflects molecules proportional to its surface all of the time. When we puncture one of the sides, that side will now reflect less molecules until the can is filled, as such the excess molecules reflected by the far side will then not be weighted up by the predicted reflection of molecules in the other side.
So basically the can takes its momentum from the air it reflects on the far side and it can do this due to a temporary lack of reflection on the near side. The reason the amount of momentum transfered to it should be exactly proportional to the amount of air required to fill the can is because that is exactly how much air it fails to reflect in total no matter hole size/can size/air speed etc. (Of course if we neglect air resistance and other wasted energy/transferal of momentum to the outside air)
If you want a physical system with similar forces but less abstract ones how about this:
Imagine a box. It is getting hit in even intervals by one ball from north and one ball from the south at exactly the same time, due to this the forces from the impulses always takes out each other.
Now we alter the box, we remove one of the walls, the south one. The box is made in such a way that the time it takes for one ball to traverse to the far end of the box is equal for the time it takes for a new ball to arrive at the near end. The box is much heavier than the balls and as such any movement gained by it should be negligible compared to the movement of the balls.
What would happen to the box in this system? Were did it gain its momentum from? How large is that momentum?
Edit: If you can answer that, just multiply the number of balls, make them minuscule and put them at random hit intervals and you got the old example. Just note that the smaller the hole the longer it takes for the box to fill and the less the removed force area will be so it still applies there.
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On September 30 2008 16:27 Klockan3 wrote:Show nested quote +On September 30 2008 06:48 betaben wrote:
If so, how do you say again that air is propelled in that direction?
I think you're saying that in the process where the gas on the unpunctured side pushes against the can, there will be an equal amount of gas moving in the opposite direction within the gas (as it's overall momentum is beforehand zero). this does not transfer it's momentum to anything, (unlike the gas moving against the can) so continues travelling away from the can.
Is this true?
I'm guessing by the fact that you've quoted this, you're confirming the momentum of the can is equalised by a body of air moving away from the can on the side opposite to the hole.
On September 30 2008 16:27 Klockan3 wrote:
If we isolate the two surfaces, one were we puncture it and the other is the one exactly on the opposite side.
Each reflects molecules proportional to its surface all of the time. When we puncture one of the sides, that side will now reflect less molecules until the can is filled, as such the excess molecules reflected by the far side will then not be weighted up by the predicted reflection of molecules in the other side.
So basically the can takes its momentum from the air it reflects on the far side and it can do this due to a temporary lack of reflection on the near side.
so far, this is describing a differences in pressure on either side which moves the can. this process is agreed upon - it is not the query; the query is:
the air that moves in order that can momentum is equalised in the end state - how does it get moving?
On September 30 2008 16:27 Klockan3 wrote:
The reason the amount of momentum transfered to it should be exactly proportional to the amount of air required to fill the can is because that is exactly how much air it fails to reflect in total no matter hole size/can size/air speed etc. (Of course if we neglect air resistance and other wasted energy/transferal of momentum to the outside air)
This is agreed (I'm guessing by airspeed, you are not referring to the speed of the body of air filling the can, nor the body of air on the opposite side of the can, as a small amount of air at high speed has the same momentum as a large amount of air at low speed.)
If you want a physical system with similar forces but less abstract ones how about this:
Imagine a box. It is getting hit in even intervals by one ball from north and one ball from the south at exactly the same time, due to this the forces from the impulses always takes out each other.
Now we alter the box, we remove one of the walls, the south one.
so far again, this is agreed upon.
The box is made in such a way that the time it takes for one ball to traverse to the far end of the box is equal for the time it takes for a new ball to arrive at the near end. The box is much heavier than the balls and as such any movement gained by it should be negligible compared to the movement of the balls.
What would happen to the box in this system? Were did it gain its momentum from? How large is that momentum?
It gains it's momentum from the air on the far side having more total momentum transfer than those on the hole-side (due to the hole). This is where I'm trying to interpret your view:
the air on the the far side has air moving in all directions; as one fraction of air transfers it's momentum to the can, there is a residual amount of air which is moving away from the can, (which, due to the dispersive nature of the air, would probably travel away as a shock wave from the can in the form of sound.)
This is the only way I can think of for your air to be moving away from the can on the far side.
is this your interpretation?
However, a similar process occurs on the hole side, as the air which enters the can has a component which moves away from the can, in an equal and opposite shock wave. the over all momentum of the outside air would be 0.
unless you have a different notion.
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On September 30 2008 17:21 betaben wrote:
the air on the the far side has air moving in all directions; as one fraction of air transfers it's momentum to the can, there is a residual amount of air which is moving away from the can, (which, due to the dispersive nature of the air, would probably travel away as a shock wave from the can in the form of sound.)
Wrong, the air on the far side bounces of the surface just as if it were a normal surface. The shock waves only comes from the temporary lack of reflection on the side with the hole, it creates a large negative shock wave. The right side looks just as normal all the time, the left side (With the hole) creates a negative shock wave and thus gets a negative momentum compared to normal. Now since the right side has normal momentum and the left side have less momentum than normal, the momentum balancing up the cans momentum is the fact that the right air have more momentum than the left air. This momentum difference is as I described before the same as the momentum of the air which entered the can.
But yeah a really minor shock wave will also be created due to the can accelerating but if we ignore air resistance we should also ignore that since they are a part of the same phenomenon.
Edit: If you still do not understand ask again and I can try to describe it in a new way.
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you are continually using terms which are vague. "normal momentum" -what is normal? net momentum 0? "more momentum" - without a direction, this is unclear as net momentum is not scalar (unless you mean the air is heated up). please define a net momentum direction for the air. "negative shockwave" does this mean to the left? have you defined an axis, with negative = left?
Also I don't understand:
Wrong, the air on the far side bounces of the surface just as if it were a normal surface.
and
The right side looks just as normal all the time
if the air on the right size never changes overall momentum, it never transfers net momentum to the can, and the can won't move. It is not the same as 'bouncing off a normal surface' because it makes the surface move, transferring momentum. the can's leftwards momentum can only come from this side.
I think when you say:
Now since the right side has normal momentum and the left side have less momentum than normal, the momentum balancing up the cans momentum is the fact that the right air have more momentum
again, you must define the right hand sides airs' 'more momentum'. if it balances the cans momentum it must be momentum away from the can.
you mean precisely what I wrote when you said "wrong":
the air on the the far side has air moving in all directions; as one fraction of air transfers it's momentum to the can, there is a residual amount of air which is moving away from the can
as a minor point:
+ Show Spoiler +let's forget about shock waves, we'll just talk about air momentum. but still...
But yeah a really minor shock wave will also be created due to the can accelerating but if we ignore air resistance we should also ignore that since they are a part of the same phenomenon.
this I think is incorrect - shock waves are a pressure phenomenon, nothing to do with air resistance which involves an object colliding with the air and dissipating the objects' energy within the air as heat. shock waves should lose energy slowly, just have the energy dispersed as it spreads out. air resistance would dissipate the energy as random isotropic movement in the air - i.e heat.
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On October 01 2008 05:22 betaben wrote:
if the air on the right size never changes overall momentum, it never transfers net momentum to the can
...
Air pressure transfers a huge amount of momentum against everything all of the time, it is just that it usually gets that back exactly due to an exactly balanced other side also transferring momentum. The momentum in this case going to the left comes from the balance being broken by lessening the momentum transferring on the left side, and thus the right sides momentum transfer wont be neutralized and thus the can will receive a net momentum from the air on the right side but the only sides air which changed is the left side.
I repeat: Normal circumstances is that air transfers 10 tonne*meter momentum per second and square meter. Normal circumstances is also that this process gets exactly equaled out by other similar forces. However if we removed the amount transfered on one side, the other side would then be transferring all of this unhindered and thus the can can get a momentum from the right side without changing the air on the right side at all since having the air transferring momentum to the can is the normal situation, the left side stopping transferring momentum to the can is the altered situation and as such any change to the air momentum occurs on the left side and it takes its form as a negative shock wave which carriers negative momentum compared to normal air.
It is not really negative momentum, it is just that normally in air it have a net momentum of 0, but if you remove all molecules going one way we would thus get a lower than normal momentum in that direction and as such we have a negative momentum compared to normal air and if this shock wave hit anything it would propel it towards the direction the shock wave came from.
On October 01 2008 05:22 betaben wrote:
this I think is incorrect - shock waves are a pressure phenomenon, nothing to do with air resistance which involves an object colliding with the air and dissipating the objects' energy within the air as heat. shock waves should lose energy slowly, just have the energy dispersed as it spreads out. air resistance would dissipate the energy as random isotropic movement in the air - i.e heat.
Air resistance comes from really small shock waves being created which gets larger the faster you move compared to the air. Shock waves are just relatively compressed/thinned out air, you get that the instant you move in any form of gas.
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Dunn dunn dunn! Come back tomorrow for more monstrous PHYSICISTS slugging it out!
WHO WILL WIN?
(oh come on you know i'm just poking fun... this is what i do when i'm lost)
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+ Show Spoiler +
1c
just a feeling, once the car is moving the speed should not change with changing mass. as long as we dont look at friction at least
2a
p = mv , the velocity stays the same but the mass increases, thus the momentum increases
3a
E = 1/2 mv² - same as above
1c
2b
3b
Conservation of Energy, everything should go back to when the rain started. Maybe thats a bit easy though..
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I think I'm coming round to your view - please tell me if I'm incorrect:
you view the air as having infinite momentum - capable of transferring momentum in one direction without losing it itself. thus, you believe there is no net momentum change in the air on the right side, and the can's momentum is not balanced by air on this side. (here i was confused because I thought by quoting my option with this in it, I thought you were advocating it.) The can's momentum is equaled by a wave of low pressure traveling in the same direction as the can, which you justify as having momentum opposite to it's direction of movement by the fact that it will move objects towards itself as it hits.
please directly say which parts of this are correct/incorrect
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EPT
