Michael C
Graduate Poster
But, once more: that's not an inertial frame.
Now please explain what you said here:
How can the direction of movement of the ball change without its velocity or momentum changing?
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Merged DDWFTTW in a “Real” Wind
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But, once more: that's not an inertial frame.
Now please explain what you said here:
How can the direction of movement of the ball change without its velocity or momentum changing?
Only for your kiddies!
Sorry Dude,
"Just because I say so." Only works with your kids, not in physics.
If you believe this, go ahead and declare "The cart doesn't work-just because I say so." and then you don't need to state anything else!
Since you consider this a "perfectly legitimate and ofetn (sic) used method of analysis", then why don't I just play God and say everything that I say is correct and beyond question? I guess you'll have to agree with this and say you are wrong!
At this point in the discussion it has become easy to see why its not worth wasting anymore time on you, like Humber!
WindGrins
Sorry Dude,
"Just because I say so." Only works with your kids, not in physics.
If you believe this, go ahead and declare "The cart doesn't work-just because I say so." and then you don't need to state anything else!
Since you consider this a "perfectly legitimate and ofetn (sic) used method of analysis", then why don't I just play God and say everything that I say is correct and beyond question? I guess you'll have to agree with this and say you are wrong!
At this point in the discussion it has become easy to see why its not worth wasting anymore time on you, like Humber!
WindGrins
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profpangloss
Critical Thinker
- Joined
- May 1, 2011
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- 391
You have not said why it isn't, and please don't mention the earth's orbit around the sun. Maybe I should say the "ground" instead of the earth so you don't try to bring in orbital mechanics. It is perfectly reasonable to define an inertial frame of reference where the ground is at rest and the air and cart are moving. State your reasons for objecting to this.
I'm only interested in the absolute value of momentum so it does not matter that it changed direction. The ball bounces off the earth with the same velocity it had when it hit it. But I should remind you that this is not even the problem set that we were discussing! The problem set was about the wind and the cart moving on the stationary ground and in that frame both momentum and energy must be conserved. I said nothing about vertical motion. If you want the earth to move vertically with impact of the ball to conserve the vector quantity of momentum, please be my guest and make yourself happy as it has nothing at all to do with the original problem.
profpangloss
Critical Thinker
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- 391
Sorry Dude,
"Just because I say so." Only works with your kids, not in physics.
If you believe this, go ahead and declare "The cart doesn't work-just because I say so." and then you don't need to state anything else!
Since you consider this a "perfectly legitimate and ofetn (sic) used method of analysis", then why don't I just play God and say everything that I say is correct and beyond question? I guess you'll have to agree with this and say you are wrong!
At this point in the discussion it has become easy to see why its not worth wasting anymore time on you, like Humber!
WindGrins
That's fine, as I have not noticed anything of value from you in this discussion either.
It couldn't be more obvious that harchrispangloss actually knows he's completely wrong. Why else would he refuse to answer sub's short list of simple questions?
How can this not be in someone's sig?
How can this not be in someone's sig?
Let's play your game!
How about you explain to us the following:
A car with mass of 1000KG is sitting motionless on the ground.
According to you the momentum is mass*velocity so the momentum of the car and ground is 0 since velocity is 0 for the car.
The driver accelerates to 100 kph.
Now the momentum is 1000*100=100,000 in these units. The momentum of the earth is still 0.
Since 100,000 does not equal 0. Momentum is not conserved in this closed system. Note this is PPG physics-not WG physics.
Please explain why momentum is not conserved since these are all your assumptions and rules.
Yes, I'll just drop off if you answer this one sensibly!
WindGrins
How about you explain to us the following:
A car with mass of 1000KG is sitting motionless on the ground.
According to you the momentum is mass*velocity so the momentum of the car and ground is 0 since velocity is 0 for the car.
The driver accelerates to 100 kph.
Now the momentum is 1000*100=100,000 in these units. The momentum of the earth is still 0.
Since 100,000 does not equal 0. Momentum is not conserved in this closed system. Note this is PPG physics-not WG physics.
Please explain why momentum is not conserved since these are all your assumptions and rules.
Yes, I'll just drop off if you answer this one sensibly!
WindGrins
Are you kidding? I had the 8 runs isolated and graphed the night I got it. It's one of the things I do for a living. The only one I missed at that time was the aborted run.
for one thing, did you notice what I said at the top of the post? I'll repeat:
For the hell of it.
and did you notice run 6?
LMAO The graphs and charts are of the raw data. I can present whatever average of whatever section of that raw data I want. And unlike you, I do know what I'm doing.
profpangloss
Critical Thinker
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How about you explain to us the following:
A car with mass of 1000KG is sitting motionless on the ground.
According to you the momentum is mass*velocity so the momentum of the car and ground is 0 since velocity is 0 for the car.
The driver accelerates to 100 kph.
Now the momentum is 1000*100=100,000 in these units. The momentum of the earth is still 0.
Since 100,000 does not equal 0. Momentum is not conserved in this closed system. Note this is PPG physics-not WG physics.
Please explain why momentum is not conserved since these are all your assumptions and rules.
Yes, I'll just drop off if you answer this one sensibly!
WindGrins
This is one game you cannot hope to win.
The law of conservation of linear momentum is a fundamental law of nature, and it states that if no external force acts on a closed system of objects the momentum of the closed system remains constant.
the original problem concerns a two-body collision, not a single body being accelerated by an external force. OBJECTS, not a single object! Would you like now to discuss one hand clapping?
Moreover, since the car is accelerating, it is acting under the external force of the engine, this is not a closed system as fuel is being burned.
You lost. Now keep your end of the bargain!
Ummm, he said if you answer sensibly, not laughably.
Sorry, try again.
Since I stated "if you answer sensibly..." I feel no compulsion to acquiesce. If you don't think the law of COM applies to a single object, but only applies to multiple objects, I'll let the other posters explain it to you. I'm sorry to inform you that COM does still apply even when fuel is being burned. Your physics and your explanation is what makes no sense in this example. It is no surprise that you can't analyze the real world cart if you can't analyze the trivial.
WindGrins
Since I stated "if you answer sensibly..." I feel no compulsion to acquiesce. If you don't think the law of COM applies to a single object, but only applies to multiple objects, I'll let the other posters explain it to you. I'm sorry to inform you that COM does still apply even when fuel is being burned. Your physics and your explanation is what makes no sense in this example. It is no surprise that you can't analyze the real world cart if you can't analyze the trivial.
WindGrins
Michael C
Graduate Poster
If you only count the values you're interested in, you can't hope to get correct results. Can you answer the question Windgrins just posted about a car that accelerates from 0 km/h to 100 km/h?
ETA: I see you've answered with more nonsense. Remember that momentum is not energy. In the closed system of objects that contains both the Earth and the car, momentum must be conserved. If one part of this system changes momentum, there must be a corresponding equal and opposite change in another part of the system.
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Bouncing Ball problem solved!
Michael C-At least this explains your bouncing ball issue:
PPG:"the original problem concerns a two-body collision, not a single body being accelerated by an external force. OBJECTS, not a single object!"
Apparently, COM does not apply because you only had one ball bouncing! I wonder what happens if you have two balls bouncing?
Regards,
WindGrins
Michael C-At least this explains your bouncing ball issue:
PPG:"the original problem concerns a two-body collision, not a single body being accelerated by an external force. OBJECTS, not a single object!"
Apparently, COM does not apply because you only had one ball bouncing! I wonder what happens if you have two balls bouncing?
Regards,
WindGrins
tobermoryphil
Thinker
- Joined
- Apr 23, 2011
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Of course they ran where it suited them- they were trying their darndest to go DDW! The cads!
Perhaps in the Humberpropfloss universe, you can arrange for the wind to blow from a constant direction at a constant speed for hours on end.
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An imaginative solution!
Actually, its quite simple in our universe-its called a treadmill in the case of DDW situations.
Regards,
WindGrins
Of course they ran where it suited them- they were trying their darndest to go DDW! The cads!
Perhaps in the Humberpropfloss universe, you can arrange for the wind to blow from a constant direction at a constant speed for hours on end.
Actually, its quite simple in our universe-its called a treadmill in the case of DDW situations.
Regards,
WindGrins
Wrong
The wind did not vary by 100 degrees, that is the difference between the errors generated when the CV first accelerated through wind speed, and at the end when it decelerated through wind speed. The period with reliable results indicates a total variation magnitude of 40 degrees. And that period represents the relevant portion of the run.
No, it was being measured by an anemometer on the Chase Vehicle. With the fixed station used as verification.
Unfortunately, since this all based on an incorrect assumption, this can be ignored.
profpangloss, I do not entirely understand your position. Do you believe that the law of conservation of momentum holds in all reference frames, including non-inertial ones? Or do you believe that a reference frame in which the Earth remains stationary, even while subject to forces from the cart, is an inertial reference frame?
Anyway, here's a relevant quote by Einstein. It's from his book on relativity, but this particular quote deals with classical mechanics. (What he calls a "Galileian system of co-ordinates" is what people in this thread are calling an "inertial reference frame".)
And here's a quote from pp. 15-16 of the book Classical Mechanics, by John R. Taylor, which I happen to have in front of me:
Anyway, here's a relevant quote by Einstein. It's from his book on relativity, but this particular quote deals with classical mechanics. (What he calls a "Galileian system of co-ordinates" is what people in this thread are calling an "inertial reference frame".)
And here's a quote from pp. 15-16 of the book Classical Mechanics, by John R. Taylor, which I happen to have in front of me:
Newton's laws cannot be true in all conceivable reference frames. To see this, consider just the first law and imagine a reference frame - we'll call it S - in which the first law is true. For example, if the frame S has its origin and axes fixed relative to the earth's surface, then, to an excellent approximation, the first law (the law of inertia) holds with respect to the frame S: A frictionless puck placed on a smooth horizontal surface is subject to zero force and, in accordance with the first law, it moves with constant velocity. Because the law of inertia holds, we call S an inertial frame. If we consider a second frame S' which is moving relative to S with constant velocity and is not rotating, then the same puck will also be observed to move with constant velocity relative to S'. That is, the frame S' is also inertial.
If, however, we consider a third frame S'' that is accelerating relative to S, then, as viewed from S'', the puck will be seen to be accelerating (in the opposite direction). Relative to the accelerating frame S'' the law of inertia does not hold, and we say that S'' is noninertial. I should emphasize that there is nothing mysterious about this result. Indeed it is a matter of experience. The frame S' could be a frame attached to a high-speed train traveling smoothly at constant speed along a straight track, and the frictionless puck, an ice cube place on the floor of the train, as in Figure 1.4. As seen from the train (frame S'), the ice cube is at rest and remains at rest, in accord with the first law. As seen from the ground (frame S), the ice cube is moving with the same velocity as the train and continues to do so, again in obedience to the first law. But now consider conducting the same experiment on a second train (frame S'') that is accelerating forward. As this train accelerates forward, the ice cube is left behind, and, relative to S'', the ice cube accelerates backward, even though subject to no net force. Clearly, the frame S'' is noninertial, and neither of the first two laws can hold in S''. A similar conclusion would hold if the frame S'' had been attached to a rotating merry-go-round. A frictionless puck, subject to zero net force, would not move in a straight line as seen in S'', and Newton's laws would not hold.
Evidently, Newton's two laws hold only in the special, inertial (nonaccelerating and nonrotating) reference frames.
If, however, we consider a third frame S'' that is accelerating relative to S, then, as viewed from S'', the puck will be seen to be accelerating (in the opposite direction). Relative to the accelerating frame S'' the law of inertia does not hold, and we say that S'' is noninertial. I should emphasize that there is nothing mysterious about this result. Indeed it is a matter of experience. The frame S' could be a frame attached to a high-speed train traveling smoothly at constant speed along a straight track, and the frictionless puck, an ice cube place on the floor of the train, as in Figure 1.4. As seen from the train (frame S'), the ice cube is at rest and remains at rest, in accord with the first law. As seen from the ground (frame S), the ice cube is moving with the same velocity as the train and continues to do so, again in obedience to the first law. But now consider conducting the same experiment on a second train (frame S'') that is accelerating forward. As this train accelerates forward, the ice cube is left behind, and, relative to S'', the ice cube accelerates backward, even though subject to no net force. Clearly, the frame S'' is noninertial, and neither of the first two laws can hold in S''. A similar conclusion would hold if the frame S'' had been attached to a rotating merry-go-round. A frictionless puck, subject to zero net force, would not move in a straight line as seen in S'', and Newton's laws would not hold.
Evidently, Newton's two laws hold only in the special, inertial (nonaccelerating and nonrotating) reference frames.
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It is fundamental, but it holds only in inertial reference frames.
The law of conservation of momentum is a consequence of Newton's three laws of motion, and those laws apply to a single object as well as to multiple objects.
The engine is part of the car. In what sense is it external to the car?
I agree that the car is not a closed system, but the reason is that the external earth exerts a frictional force on the tires, allowing the engine to move the car forward. A car, even with its engine burning fuel, will not go anywhere on very slippery ice.
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