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United States24665 Posts
On September 28 2008 16:34 15vs1 wrote:
Maybe it is my bad english but is it possible to ROLL without a friction.
It's fairly typical for the word friction to be used interchangeably in two ways when talking about wheels.
1) The friction between the wheel surface and the ground. As long as this is large enough, the wheel will 'roll without slipping'.
2) The friction acting on the axle etc... provides a negative torque which decreases angular velocity. Does not apply in simplified cases where friction is ignored.
The problem is just saying not to worry about 2.
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On September 28 2008 12:43 micronesia wrote:+ Show Spoiler +For those who didn't read the introductory entry, head over to http://www.teamliquid.net/blogs/viewblog.php?topic_id=79275This is entry 3. Last week the question of the week was: Question 2
Newton's Laws of Motion
If a can of compressed air is punctured and the escaping air blows to the right, the can will move to the left in a rocket-like fashion. Now consider a vacuum can that is punctured. The air blows in the left as it enters the can. After the vacuum is filled the can will:
a) be moving to the left
b) be moving to the right
c) not be moving The correct answer is c. First of all, for those who were confused about which way the can was punctured, I agree the wording was a bit tricky... but it didn't matter! It's very easy to get confused by this problem because there is going to be a momentarily slight oscillation about the center of mass. Just as the can is punctured, the pressure that was being exerted by the atmosphere on the now-absent piece of metal doesn't exist anymore... and the force acting on that surface is weaker than that acting on the opposite side of the can. However, the air rushing into the can will eventually hit the opposite wall, which will provide that force (plus a bit extra until everything equals out). These two drawings I made should clear it up: I assign this problem a difficulty level of 3 out of 5 because it's easy to confuse the physics principles. Congratulations to Xeofreestyler for being the first to provide the correct answer and provide a reasonable justification. Dr.Dragoon named the problem. There were a lot of explanations that were either wrong (even among people who chose C) or not backed up. For example: saying 'C because conservation of momentum' isn't necessarily wrong, but is not sufficient in my opinion. There were a few explanations that I am not sure if they are correct or not... although nobody quite used the one that I provided. Feel free to provide further analysis.
I am sorry but your answer is wrong.
Assuming that all of the air moves at the same speed(Really makes no difference of the answer but makes it easier to reason).
Now there are two phases you have to consider which are constantly interchanging:
Phase 1:
In this phase the air have not reached the far end of the can yet, at this time there is a net force towards the hole since the hole is a net lack of force.
Phase 2:
In this phase the air have reached the far end for the first time, there is no net force except for air resistance which should be negligible in these small time frames. During this phase the amount of air in the bottle is area of hole/area of a bottle side.
Phase 1:
Now the air which hit the far end have came back to the near end again. Since we were at an equilibrium just before this moment we now get an extra force towards the hole direction. Note however that this is smaller than the first phase 1 since we lost a bit of air going out the hole, but at the same time that slows down the filling of the bottle by the same amount and at the end the total work on the bottle is directly proportional to the volume of it and the force never stops until it is completely full.
In the end we will thus receive a net force towards the hole of the bottle as long as the bottle is not full. The bottle will move, case closed.
And just to note, the fact that the air moves faster in one direction than the other and thus creates a force on the bottle as long as it is moving is air resistance, it will not be fast enough to slow it down ever and unless we count other things such as friction it will for sure be moving.
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Klockan, what happens when the particles that you are talking about reach the other side again?
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klockan3, since the can and air starts in rest (relative each other) AND ends up in rest (the air in the can has to have the same velocity as the can itself macroscopically) the can cannot possibly move AFTER the equillibrium has been reached again (i.e no net velocity of the system)
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On September 28 2008 18:21 fight_or_flight wrote:
Klockan, what happens when the particles that you are talking about reach the other side again?
Phase 2, aka a new equilibrium of forces, until the stream hits the near end again etc.
Now in reality we would not have phases since the velocity of air is a distribution, however we can look at every velocity in isolation and then add them all up to a constant net force relative to how full the bottle is.
On September 28 2008 18:31 glassmazarin wrote:
klockan3, since the can and air starts in rest (relative each other) AND ends up in rest (the air in the can has to have the same velocity as the can itself macroscopically) the can cannot possibly move AFTER the equillibrium has been reached again (i.e no net velocity of the system)
The air/outside system do not end at rest though, I explained this in my post at the relevant topic.
On September 28 2008 10:25 Klockan3 wrote:Much wrong use of the laws of Newton in this thread. If someone finds anything wrong with this please poke it, but for now I believe that most in this thread are wrong: + Show Spoiler +Actually, I am pretty sure that we would have a net force on the can during the fillup process, and once it is full we would have an equilibrium again and there is no way air resistance would stop it in an instant.
This do not break any of newtons laws since when it is absorbing air it will reflect less molecules and thus while it will be moving towards the hole the closed system would start moving in the other direction due to the air pressure getting slightly lower on one end. (Or the average of the gas will be moving in the direction the can is not moving)
The easiest way to picture this is to imagine a square. Remove one of the walls and imagine that its interior gets filled by gas molecules, until the molecules who entered fills the whole square up the air pressure on the other side of it will create a net force and thus speed it up, and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval). The net force on the square gets exactly evened out by the net loss of reflected air from the removal of one of the walls.
If you want an open system just imagine that we have an artificial air around it, all with a speed directed towards the can and only in a volume to have exactly enough air to hit the can untill it has reached the critical point were we stop observing it. In such a system we would get a gas sphere around the can except were the hole is, there we would also get an air hole in the sphere, and that is were the momentum gained by the can is taken from.
In the end, the answer is B: It is moving to the right.
I can not believe that so many missed this...
Also yes even if the hole is extremely small, the fact that there is no air resistance unless the object is moving, and that until the last of the air have moved in there is always a net force working on the can from the higher pressure from the wall wo a hole it got to be moving at the time when equilibrium is reached.
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ok, lets say the air molecules moves with speed v_air microscopically.
then as the can starts moving to the right with speed v_can, the air molecules hitting the can on the left outside wall will hit with the relative speed v_can - v_air.
when the air that enters the can starts hitting the left inside wall, the molecules will hit with speed v_can + v_air and create a higher pressure than that on the left outside wall.
the effect is that as long as the can is moving to the right, air will create a force on the can towards the left..
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On September 28 2008 18:44 glassmazarin wrote:
ok, lets say the air molecules moves with speed v_air microscopically.
then as the can starts moving to the right with speed v_can, the air molecules hitting the can on the left outside wall will hit with the relative speed v_can - v_air.
when the air that enters the can starts hitting the left inside wall, the molecules will hit with speed v_can + v_air and create a higher pressure than that on the left outside wall.
the effect is that as long as the can is moving to the right, air will create a force on the can towards the left..
Which is the air resistance, I have already covered that.
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On September 28 2008 18:31 Klockan3 wrote:
The easiest way to picture this is to imagine a square. Remove one of the walls and imagine that its interior gets filled by gas molecules, until the molecules who entered fills the whole square up the air pressure on the other side of it will create a net force and thus speed it up, and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval). The net force on the square gets exactly evened out by the net loss of reflected air from the removal of one of the walls.
i assume this is where you covered air resistance, please correct me if im wrong 
"and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval)."
this is wrong imo. as soon as air hits the left inner wall, the air will create a force on the can in the opposite direction of the cans movement, lasting until the can has stopped.
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On September 28 2008 19:13 glassmazarin wrote:
this is wrong imo. as soon as air hits the left inner wall, the air will create a force on the can in the opposite direction of the cans movement, lasting until the can has stopped.
I covered air resistance in more than one place, anyhow:
The can gets a force trying to slow it down by air resistance as long as it moves, this force is proportional to the square of the velocity.
As long as the can is filling it will get a force trying to speed it up.
Now the force of air resistance is proportional to the movement speed of the object, as such even if it took 100 million years to fill the bottle and as such the air resistance got almost infinite time to equalize with the accelerating force, the fact that there is a force until the very end of the process to fill it means that air resistance would need to slow it down to a stop in 0 time or it will be moving at the exact moment it gets full.
And that is just the extreme case of air resistance > all, normally air resistance would not have much of a say the 1/10 of a second the can is filling up.
But you are right though that the whole system converges against everything being at a relative rest, but that were not the question.
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aha.. i was assuming that the "full" condition = everything is in equillibrium again, but just as the can gets filled with the last air molecule, it will move to the right.
seems that you have the right answer there : )
EDIT: Crap, im wasting too much time on this right now >.<
If one would define the hole as being "small enough" (i.e infinitesimal) we can argue that the molecules enter the can one by one.
when the last molecule enters (filling the can) the can will be moving to the right, but as soon as the molecule has hit the opposite wall, the can will stop.
air molecules moves with speeds ~500 m/s, i.e the time between "filling" and "stopping" will be of order ~1/10000 s for a can 5 cm wide in this case. compared to the time it takes to fill the can from scratch this can be argued to be negligable and thus the can stops "at the same time" as the can is filled.
however the question did not say anything about the size of the hole, so the correct answer would still be B, but i believe this is what the question was supposed to say
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Germany2896 Posts
Assume: the car moves horizontally.
Assume: The rain which is currently in the car counts as part of the car. If that assumption is wrong please clearify the task.
+ Show Spoiler +1b 2c 3b 4c 5b 6b Reason for 2: the impact of the rain can be regarded a inelastic impact. The (horizontal component of) the momentum of the rain is 0. Conservation of momentum applies to inelastic impact. for 1: v=p/m p2=p1, m2>m1 =>v2<v1 for 3: E=0.5*p^2/, p2=p1, m2>m1 =>E2<E1 Reason for 4: In the moved system the emmited water is evenly distributed in all(horizontal) directions. Thus no force applies to the car and its speed is constant. 5: Decrease, p=v*m v2=v1 m2<m1 =>p2<p1 The total momentum is conserved because the emmited rain carries part of the momentum 6: Decrease, E=0.5*v^2*m v2=v1 m2<m1 =>E2<E1
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United States24665 Posts
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points.
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On September 29 2008 01:22 micronesia wrote:
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points.
I am also a physicist...
Anyway:
1: In an open system it is the air which it did not reflect where the hole is. It might be easier to imagine if you see it as a closed system though, imagine that the can is inside a box. As soon as the can gets punctured it reflects less air molecules against one side than the other which means that while it accelerates itself towards one direction it would at the same time accelerate the box it is inside towards the other.
2: Yes, the leftwards pressure always comes before the rightwards pressure equalizes it out. However since a system with no acting forces does not change in velocity we can sum together all of these gaps into the total mechanical work done on the can.
About the center of mass, think like this: Remove everything else from the model except just the gas molecules which are hitting the can. You can see that if it is a regular can the reflected air will create a spherical pattern around the can. However if we open a hole with vacuum we also create a hole in this sphere and the movement of the can is only to make up for this lack of molecules moving in that direction.
And no, as I said, air resistance should not even be considered in the time frame it takes to fill an ordinary sized can with gas under ordinary temperature and pressure and even if we make ridiculous assumptions which would mean it would be a factor it is still debatable what the answer would be.
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United States24665 Posts
On September 29 2008 03:55 Klockan3 wrote:Show nested quote +On September 29 2008 01:22 micronesia wrote:
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points. I am also a physicist...
This was not obvious to me.
How do you define physicist?
Anyway:
1: In an open system it is the air which it did not reflect where the hole is. It might be easier to imagine if you see it as a closed system though, imagine that the can is inside a box. As soon as the can gets punctured it reflects less air molecules against one side than the other which means that while it accelerates itself towards one direction it would at the same time accelerate the box it is inside towards the other.
2: Yes, the leftwards pressure always comes before the rightwards pressure equalizes it out. However since a system with no acting forces does not change in velocity we can sum together all of these gaps into the total mechanical work done on the can.
About the center of mass, think like this: Remove everything else from the model except just the gas molecules which are hitting the can. You can see that if it is a regular can the reflected air will create a spherical pattern around the can. However if we open a hole with vacuum we also create a hole in this sphere and the movement of the can is only to make up for this lack of molecules moving in that direction.
And no, as I said, air resistance should not even be considered in the time frame it takes to fill an ordinary sized can with gas under ordinary temperature and pressure and even if we make ridiculous assumptions which would mean it would be a factor it is still debatable what the answer would be.
The example you gave is very similar to the one provided in a book which analyzes this problem. I'll let you see the actual drawings rather than my own creations:
http://img141.imageshack.us/my.php?image=poofandfoopgx4.jpg
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On September 29 2008 05:23 micronesia wrote:Show nested quote +On September 29 2008 03:55 Klockan3 wrote:On September 29 2008 01:22 micronesia wrote:
Klockan3, glassmazarin:
I don't think you guys arrived at a correct conclusion, but I don't fully understand your explanation so I'd like to clarify.
1) When the air had just finished entering the can, if the can (and its contents) were moving to the right, then what was moving to the left with equal and opposite momentum?
2) As the can fills, there will be pressure on the inside of the wall that has the small puncture, but that should be balanced by the pressure acting on the inside wall on the opposite side of the can that isn't directly across from the puncture. The difference as I see it is that there is a leftward pressure sooner than a rightward pressure. Unless the question writer (who is a physicist) is wrong, the motion of the can will be a motion about the center of mass which ends when the can fills up.
Please respond specifically to those two points. I am also a physicist... This was not obvious to me. How do you define physicist?
Someone who have worked in the field of physics? I have done work on how to utilize the inhomogenous effects of acoustic radiation pressure to guide objects through fluids on a micrometer scale.
That book is wrong though, if you consider the watercart picture 2 and instead of having a fixed inner containers you would put wheels on them too, then the outer cart and the inner cart would both accelerate in different directions which is obvious. At leasts with the small cart being filled, the other would not as much since fluids are much slower got and a much higher resistance to movement than gases and thus most of the work would just become heat.
And I already explained what he puts in the last picture, that the effects of the pressure during the time it takes for the air to bridge of the volume is certainly not negligible. As I explained in my last post the bouncing nature of the air means that we get the same total work on the objects as if we had just totally removed one of the walls, and if we just removed one of the walls and assume that the can have a volume of 1 cubic decimeter we would get 1000 newtons of force over 1/5000 seconds and a can weighting 10g would then get accelerated to 20 m/s during that fill up process.(Notice that the cans speed is still negligible compared to the speed of the particles of air)
If we just had 1 square centimeter of a hole the process would take ~200-300 times longer but the net work would be exactly the same and the air resistance can not slow such a speed at these time intervals.
Edit: And that would hardly be the first time a science book is wrong.
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The can problem turned out to be tricky but klockan's arguments looks stronger so far.
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klockan, I'm still not sure of your position.
if:
*there is no air resistance (the can moves very slow), but
*air pressure is not negligible (the air moves fast and can still exert pressure)
*open system
(no constraints on hole size and outer pressure, because these are not independent when considering the outcome.)
does the can move once a steady state has been reached? (all acceleration has stopped)
if yes, what was the total momentum before and after the puncture?
if no, what slows the can down?
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On September 30 2008 04:17 betaben wrote:
does the can move once a steady state has been reached? (all acceleration has stopped)
If we totally neglect air resistance, yes. (Even though air resistance would slow down anything to a relative near stop no matter how slow it moves if we let time go on)
On September 30 2008 04:17 betaben wrote:
if yes, what was the total momentum before and after the puncture?
Total momentum of what? The can would logically gain momentum exactly equal to the momentum of the air required to fill the can as long as the can were in relative rest compared to the air from the start and that air resistance is totally negligible, because thats how much momentum the can robbed from the outside system by refusing to reflect that much air in a specific direction.
There could be some waste due to shock waves or something else like that but they are also negligible under normal circumstances.
Also I just realized in what way the book author were thinking wrong. He were assuming that you would think that the instream of gas would create a force on the wall and as such it would accelerate due to that for some reason, which of course is a ridiculous thought. However when he made the problem he probably missed that the momental lack of force in the other direction would also make he himself wrong.
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ok, so ..
Total momentum of what?
of the entire system
The can would logically gain momentum exactly equal to the momentum of the air required to fill the can.
so, your end state is:
1). the can moving in one direction with a body of air (of the same momentum) outside of the can, moving in the opposite direction?
or,
2).with the body of air inside the can?
your wording, "required to fill the can" would imply the latter case, the air inside the can ("to fill the can") will collide with the can (as it must be moving int he opposite direction to the can to balance), nullifying the momentum of both, and the can will stop, giving the over all end state as no can movement.
so this cannot be what you mean (or is it? - do you consider the steady state to be before the collision? if so, then you are just having a disagreement with when 'end state' is called.)
So I'd guess your end state is the first of the options, which has a body of air on the other side of the can, moving away from it.
you say:
... that's how much momentum the can robbed from the outside system by refusing to reflect that much air in a specific direction.
which would support the above, with the air outside moving away from the can.
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?
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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?
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EPT![[image loading]](http://img81.imageshack.us/img81/3089/watercartkx1.png)
![[image loading]](http://img81.imageshack.us/img81/3171/puncturedcansa1.png)
