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NILL
2005-Oct-06, 03:54 AM
I class today my teacher was talking about gravitational acceleration. He said that if a ball is lunched into the air and it falls back down to a point level with the origin that the time it took to go from start to the highest point was about the same as the time it took to go from the highest point back down. He went on to say that because of the air it would go from start to hidgest faster then it would go from highest back down. I thought about it and it didn't make sense. When I asked him in class he didn't have a answer for me other than the air caused the slow down on the return part.

He also said that friction with air causes matter to heat up when falling to earth. I approched him and said it was air compressing in front of the object. He said no it was friction and I brought in a print out I got from Bad Astronomy. He retracted what he said and said we're both right and couldn't understand way the paper said friction didn't have a factor. He asked me why I belived some web site over him and I told him he doesn't have nearly as many people scrutinizing him as Bad Astronomy does.

Does anybody have a good argument that he will believe.

gopher65
2005-Oct-06, 04:07 AM
Ok, it is late here, so I might be way out to lunch:).

Here's what I'm thinking: What causes the air to compress in front of the object?

That's as far as I can get without sleep. Nightynight.

Bob B.
2005-Oct-06, 05:00 AM
I class today my teacher was talking about gravitational acceleration. He said that if a ball is lunched into the air and it falls back down to a point level with the origin that the time it took to go from start to the highest point was about the same as the time it took to go from the highest point back down. He went on to say that because of the air it would go from start to hidgest faster then it would go from highest back down. I thought about it and it didn't make sense. When I asked him in class he didn't have a answer for me other than the air caused the slow down on the return part.
He is right about the ball; the descent will take longer than the ascent. This is because when you throw the ball upward its high initial velocity will create a large amount of drag. The ball will decelerate greatly within the first fraction of a second, but in will also travel a large distance during that first fraction of a second. After reaching its apex the ball will fall toward the ground, however drag will keep it from reaching the same velocity it was given initially. Since the ball is moving more slowly during the descent, it will take longer to cover the same distance.

In a vacuum the ascent and descent times will be the same.

Van Rijn
2005-Oct-06, 05:54 AM
He also said that friction with air causes matter to heat up when falling to earth. I approched him and said it was air compressing in front of the object. He said no it was friction and I brought in a print out I got from Bad Astronomy. He retracted what he said and said we're both right and couldn't understand way the paper said friction didn't have a factor. He asked me why I belived some web site over him and I told him he doesn't have nearly as many people scrutinizing him as Bad Astronomy does.

Does anybody have a good argument that he will believe.


Think of it this way:

What happens when you use a hand air pump? The air in the cylinder gets hot and there is resistance to compression.

At the velocity a spacecraft or meteor is entering the atmosphere, the air literally doesn't have time to get out of the way. The object is acting like a piston to compress a cylinder of air, the air becomes very hot, and the compressing air resists the object. Skin friction does affect air flow about and heat transfer to the object but air compression is the primary component in reentry drag.

hhEb09'1
2005-Oct-06, 10:35 AM
At the velocity a spacecraft or meteor is entering the atmosphere, the air literally doesn't have time to get out of the way. The object is acting like a piston to compress a cylinder of air, the air becomes very hot, and the compressing air resists the object. Skin friction does affect air flow about and heat transfer to the object but air compression is the primary component in reentry drag.Part of the problem is the distinction between friction and compression. Both involve interaction of the atmosphere with the surface of the spacecraft or meteor.

Can the atmosphere cause significant friction at all? I guess it depends upon what you mean by friction :)

Welcome to the BAUT, NILL

Sock puppet
2005-Oct-06, 01:09 PM
I think it is perfectly understandable to be skeptical of a website making claims contrary to what you believe, particularly if you've never heard of the website. He might be more likely to believe you if you can show him references to this effect from a physics textbook (try your local library). Even better if you can find the appropriate calculations done step by step. (you're not likely to find that in your library, however)

BTW, go easy on your teacher. If he get something wrong or if you don't understand something he says, bring it up, but remember he's only human. (written as someone who is seriously considering being a physics teacher if he ever grows up, and also someone who gave his physics teachers way too hard a time when he was in school.)

Grey
2005-Oct-06, 01:19 PM
Moreover, I'm not certain that he was necessarily wrong about friction being the source of drag for a ball thrown up in the air. For high velocity spacecraft during reentry, it's certainly compression that's the primary factor. But is that true for a low velocity projectile? Anyone done the math to know?

Sock puppet
2005-Oct-06, 01:40 PM
Moreover, I'm not certain that he was necessarily wrong about friction being the source of drag for a ball thrown up in the air. For high velocity spacecraft during reentry, it's certainly compression that's the primary factor. But is that true for a low velocity projectile? Anyone done the math to know?

I'll do it in a few hours and post it then (gotta go to labs soon). However, there is a problem with distinguishing friction and compression. That is to say, you can do it, but it seems rather artificial. I mean, air dragging over the sides of your craft, that's friction. But when that air gets a bit compressed by your motion through the air and the drag increases, that's compression. So really you have to declare 'friction' to be the loss that would arise if the air was not compressing at all (i.e. had sufficient time to spread out) and 'compression' to be everything else.

IIRC, friction (as I used the term above) is the major source of drag for objects where (velocity<<speed of sound) - the air has plenty of time to get away so you only have a small sheath of compressed air. I'll post again with (back of the envelope) calculations in about 5 hours.

Sorry for any typos. I tried to catch them all but I was making about three per sentence, so I'm sure some slipped through.

worzel
2005-Oct-06, 01:54 PM
I can only imagine how hard it is for a physics teacher to describe things in a simple enough fashion to be understood by the majority of the kids and yet be correct in every detail. Simply writing instructions for our staff on how to use our computer systems is hard enough in that respect.

novaderrik
2005-Oct-06, 06:25 PM
the physics teacher might have made AN error, but you made a grammatical error..

we all make mistakes.

Sock puppet
2005-Oct-06, 06:45 PM
Ok, I know I said I'd do them calculations and post them by now, but I'm gonna have to do it tomorrow instead (sorry). Something wierd happened in the lab and I can't find any reference to it happening anywhere in the relevent literature, so I had to stay a coupla' hours more and still not figure what's going on. Now I'm going home to bed.

NILL
2005-Oct-07, 12:09 AM
I'll do it in a few hours and post it then (gotta go to labs soon). However, there is a problem with distinguishing friction and compression. That is to say, you can do it, but it seems rather artificial. I mean, air dragging over the sides of your craft, that's friction. But when that air gets a bit compressed by your motion through the air and the drag increases, that's compression. So really you have to declare 'friction' to be the loss that would arise if the air was not compressing at all (i.e. had sufficient time to spread out) and 'compression' to be everything else.

IIRC, friction (as I used the term above) is the major source of drag for objects where (velocity<<speed of sound) - the air has plenty of time to get away so you only have a small sheath of compressed air. I'll post again with (back of the envelope) calculations in about 5 hours.

Sorry for any typos. I tried to catch them all but I was making about three per sentence, so I'm sure some slipped through.

I was thinking about it and thought that the air would have a greater cooling effect than heating that would be caused by friction. I mean if you stick your arm out the window of a car it feels cooler not hoter.



He is right about the ball; the descent will take longer than the ascent. This is because when you throw the ball upward its high initial velocity will create a large amount of drag. The ball will decelerate greatly within the first fraction of a second, but in will also travel a large distance during that first fraction of a second. After reaching its apex the ball will fall toward the ground, however drag will keep it from reaching the same velocity it was given initially. Since the ball is moving more slowly during the descent, it will take longer to cover the same distance.

In a vacuum the ascent and descent times will be the same.

I'm sorry, thats still doesn't make sense to me. If the air is there the hole time then wouldn't it affect it equalty the entiar time? I understand that the air is limiting the speed it would reach if it were in a vacuum, but I thought it would also have the exact same effect on the accent.

Laser Jock
2005-Oct-07, 12:55 AM
I'm sorry, thats still doesn't make sense to me. If the air is there the hole time then wouldn't it affect it equalty the entiar time? I understand that the air is limiting the speed it would reach if it were in a vacuum, but I thought it would also have the exact same effect on the accent.

The air resistance depends on velocity. When the ball is first released, it is traveling faster up than it will ever travel down. The air resistance saps energy from the ball so that it won't go as high as it would in a vacuum, and it won't be going as fast when it comes back down and arrives at the same height it was released. Therefore, it takes more time for the ball to come down than it took to go up.

Hope that helps. If not, I'll try explaining it a different way.

Bob B.
2005-Oct-07, 01:18 AM
I'm sorry, thats still doesn't make sense to me. If the air is there the hole time then wouldn't it affect it equalty the entiar time? I understand that the air is limiting the speed it would reach if it were in a vacuum, but I thought it would also have the exact same effect on the accent.
Let's say I throw a softball straight upward at an initial velocity of 25 m/s. At that speed and direction, the deceleration on the ball as soon as it leaves my hand is about 7 g (1 g gravity + 6 g drag). This causes the ball to rapidly slow down. After just 0.4 seconds it has slowed to about 10 m/s and has risen to a height of about 6.5 m. The ball will continue to a maximum height of about 10 m after 1.2 seconds of flight.

The ball will now begin to fall toward the ground, but its maximum velocity is limited to its terminal velocity. This is the velocity at which the acceleration caused by drag is exactly equal to, but opposite in direction of, the acceleration of gravity. For a softball the terminal velocity is about 10 m/s. The falling ball can not possibly reach the 25 m/s I gave it initially, therefore it will take longer to reach the ground. The ball will take 1.7 seconds to fall from its maximum height back to its starting point.

Below is a plot of height versus time for a softball thrown upward with an initial velocity of 25 m/s. Data points are separated by 0.1 second. As you can see, the ball rises more quickly than it falls.

http://www.braeunig.us/pics/softball.gif

ZaphodBeeblebrox
2005-Oct-07, 01:45 AM
Ok, I know I said I'd do them calculations and post them by now, but I'm gonna have to do it tomorrow instead (sorry). Something wierd happened in the lab and I can't find any reference to it happening anywhere in the relevent literature, so I had to stay a coupla' hours more and still not figure what's going on. Now I'm going home to bed.
WOW ....

The Sock puppet Effect?

;)

worzel
2005-Oct-07, 01:59 AM
For something launched over its terminal velocity just imagine throwing a feather really hard and watching it float back down. I'm having difficulty generalising that though - the best way I can figure it is that while going up, air resistance is working with gravity to deccelerate it, while on the way down air resistance is working against gravity, so it will deccelearate upwards faster than it will accelerate downwards.

pghnative
2005-Oct-07, 02:25 AM
Another (hopefully simpler) way to look at the time question is to realize that the ball is losing energy to friction the whole time. So no matter what, it'll hit the ground at a slower speed than it took off at.

From there, it is simple to realize that this is true at any given height. Going upward, the ball at 30 ft is travelling faster than the ball (at 30 ft) going downward.

Finally, since it has to travel the same distance going up than it came down, and time = distance / speed, clearly it spent less time going up than down.

NILL
2005-Oct-07, 03:11 AM
Let's say I throw a softball straight upward at an initial velocity of 25 m/s. At that speed and direction, the deceleration on the ball as soon as it leaves my hand is about 7 g (1 g gravity + 6 g drag). This causes the ball to rapidly slow down. After just 0.4 seconds it has slowed to about 10 m/s and has risen to a height of about 6.5 m. The ball will continue to a maximum height of about 10 m after 1.2 seconds of flight.

The ball will now begin to fall toward the ground, but its maximum velocity is limited to its terminal velocity. This is the velocity at which the acceleration caused by drag is exactly equal to, but opposite in direction of, the acceleration of gravity. For a softball the terminal velocity is about 10 m/s. The falling ball can not possibly reach the 25 m/s I gave it initially, therefore it will take longer to reach the ground. The ball will take 1.7 seconds to fall from its maximum height back to its starting point.

Below is a plot of height versus time for a softball thrown upward with an initial velocity of 25 m/s. Data points are separated by 0.1 second. As you can see, the ball rises more quickly than it falls.

http://www.braeunig.us/pics/softball.gif

OK now I get it. I forgot about terminal velocity. Thanks.

Gillianren
2005-Oct-07, 03:36 AM
and that, boys and girls, is why Gillian sticks to literature, history, and music. whew!

Enzp
2005-Oct-07, 08:02 AM
Well, Gillian, even if you get in over your head, it seems you come down slower than you went up.

That was a mixed metaphor, wasn't it? Sorry.

Sock puppet
2005-Oct-07, 01:46 PM
Apologies everyone. I thought I could remember how to do the relevent calculations, but apparently my goldfish-like mind has betrayed me.

papageno
2005-Oct-08, 01:10 PM
I'll do it in a few hours and post it then (gotta go to labs soon). However, there is a problem with distinguishing friction and compression. That is to say, you can do it, but it seems rather artificial. I mean, air dragging over the sides of your craft, that's friction. But when that air gets a bit compressed by your motion through the air and the drag increases, that's compression. So really you have to declare 'friction' to be the loss that would arise if the air was not compressing at all (i.e. had sufficient time to spread out) and 'compression' to be everything else.

IIRC, friction (as I used the term above) is the major source of drag for objects where (velocity<<speed of sound) - the air has plenty of time to get away so you only have a small sheath of compressed air. I'll post again with (back of the envelope) calculations in about 5 hours.

A quick Google search gave this (http://www.aerodyn.org/aero.html):
From a physical point of view, drag is the resultant of forces acting normally and tangentially to a surface, the former ones being pressure terms, and the latter ones viscous terms.
where the forces are due to the atmosphere acting on the object.

Fortunate
2005-Oct-13, 12:30 PM
BTW, go easy on your teacher. If he get something wrong or if you don't understand something he says, bring it up, but remember he's only human. (written as someone who is seriously considering being a physics teacher if he ever grows up, and also someone who gave his physics teachers way too hard a time when he was in school.)

NILL,
The atmosphere in school deliberately creates an artificial heirarchical separation between students and teachers. Remember that your teacher is a human being, subject to the same frailties as you and your best friends. You can see from the discussion here that even a knowledgeable group had to struggle to come up with the best answers to your questions.

Please be respectful of him, not only because he is your teacher, but also because he is a fellow human being. Don't "call him out" in front of the class. Showing him the printout was a good idea. Sometimes a person needs time to digest, maybe get a pencil and jot through a few calculations.

Avoid the issue of who is right. That's not the point.

Ken G
2005-Oct-14, 06:20 AM
I don't think the energy loss should be characterized either as compression or any normal sense of the word friction, interestingly! Here's a way to attack the question of where the drag comes from. If it is due to compression, the lost energy will show up as heat in the air (Van Rijn's bicycle pump). But that's also what any normal concept of friction will do (like rubbing your hands together). But for v << speed of sound, the place the lost energy shows up is in flowing air patterns that cascade into turbulence. Like the wake of a boat in water. Would you call that friction or compression? Neither, it's viscosity. It happens because air tends to "stick" to the sides of the ball! Note that for meteors, it is compression because they are supersonic, and the energy shows up as heat, which is obvious when you look at them crossing the sky!

grant hutchison
2005-Oct-14, 12:18 PM
Would you call that friction or compression? Neither, it's viscosity. It happens because air tends to "stick" to the sides of the ball! But surely "sticking to the sides of the ball" is friction by another name?
I agree that turbulence dissipates energy, and that turbulence is significant around most objects falling in air: so I agree with your main point. But that seems to me to be different from the "stuff sticking together" phenomenon of viscosity (objects falling in treacle ...), which I've always thought of under the heading of "friction", mediated at a smaller scale than the friction between two solid surfaces. The Reynolds Number is a way of deciding whether viscosity or turbulence is going to be the dominant form of energy dissipation in a particular situation.

Grant Hutchison

Ken G
2005-Oct-14, 01:49 PM
The Reynold number, as I understand it, determines the dissipation mechanism of any flows you have. If the Reynolds number is low, the flows dissipate directly, whereas if it is high, there is a turbulent cascade to small scales where you get some sort of anomalous resistivity or some such thing that maybe you would not call viscosity. But an ideal fluid can have a ball fly through it without "sticking" at all, and would suffer no resistance at all. To be honest, I'm not sure what the best semantics are here, so we are in agreement on the main point and also on the fact that the right teminology is not so clear. I think the main physics has two pieces. One is that the air around the ball tends to move with the ball (but that's not friction, because it's not sliding along the ball, but moving with it). And the second piece is that air sliding through air gets disrupted into turbulent motions that will cascade and dissipate. The key to having small air resistance is having only a small amount of air "stuck" to the ball so the turbulent rolls are harder to create (this is what is accomplished by the dimples on a golf ball or laces on a baseball). So the answer might depend on whether you are talking about a baseball or a smooth ball!

I don't think the disruption is due to pressure gradients (a fluid concept), because you can have those even in ideal fluids (like in the Bernoulli equation), and an ideal fluid won't yield air resistance. But maybe you're right and it's not viscosity either. It's whatever creates the tendency for the air to stick to the ball, since once you have that the cascade instability will do the rest. What is the proper word for that which allows a spoon to stir tea?

grant hutchison
2005-Oct-14, 02:55 PM
I think the main physics has two pieces. One is that the air around the ball tends to move with the ball (but that's not friction, because it's not sliding along the ball, but moving with it). And the second piece is that air sliding through air gets disrupted into turbulent motions that will cascade and dissipate.But at low Reynolds numbers you see laminar flow, without turbulence, which is what I was thinking of when I mentioned an object falling through treacle. Here the viscous forces between fluid laminae travelling at different speeds are what dissipate energy. So I guess I'm thinking that under laminar flow there's no friction at the surface of the ball (as you say, because there's a boundary layer), or very distant from the ball (where the fluid is at rest); but there's a zone of transition between those two regions where "friction" is occurring.

Grant Hutchison

Edited to add the text in bold.

Ken G
2005-Oct-14, 04:44 PM
Yes I agree, at low Reynolds number friction would dissipate the flows without turbulence, so a ball in treacle would certainly be friction over the volume of the treacle (not at the ball itself). But air has a high Reynolds number, so the dissipation is by turbulent cascase, and that's why I don't think friction is the right term here. But I'm also not sure that viscosity quite makes it either. But that's all semantics, we agree on the physics, the issue is just what to tell the teacher! I certainly agree with Fortunate, we're past who is right and into what is the best "sound byte" for describing it. Maybe you really have to describe the whole process and avoid catchy terms like friction and viscosity, which both suggest the treacle-like situation. How would a hydrodynamicist describe it?

grant hutchison
2005-Oct-15, 11:41 AM
But I'm also not sure that viscosity quite makes it either.I agree: once you're established in the turbulent regime the viscosity is pretty much irrelevant - the terminal velocity of our falling ball depends on the air density, and not on its viscosity.

No idea what to tell the teacher: "You're both right, and both wrong, depending," doesn't seem entirely satisfactory.

Grant Hutchison

Ken G
2005-Oct-15, 01:03 PM
Yeah, the "sticking" process is an interaction between the air and the ball, whereas viscosity is an interaction between air and air, and there's not much of it anyway. Friction sounds like an interaction between air and the ball, but requires slippage. What are we left with? Need a new word, I think! Or maybe static friction? Tell the teacher, you have a tough job!

pghnative
2005-Oct-17, 01:52 PM
But air has a high Reynolds number, so the dissipation is by turbulent cascase[/nitpick]Reynold number is a function of flow, so it isn't technically correct to say that "air" has a high Reynolds number. At low flow rates, the Reynolds number is lower than at higher flow rates.[/nitpick]

Ken G
2005-Oct-17, 04:33 PM
Good point, not a nitpick, it's important to get the physics as right as possible.:exclaim: Any opinion on the best terminology for air sticking to a ball?

pghnative
2005-Oct-17, 04:45 PM
I'm not sure. My experience is not in gas flow, but rather in liquid flow. (Of course, in this thread we aren't talking about gas flow, such as in a pipe, but rather an object travelling through the gas. But I would guess that many of the same principles apply.)

Reynolds number is proportional to density and inversely proportional to viscosity. Perhaps it is accurate to say that the low viscosity of air outweighs its low density, such that for most flow rates the Reynolds number is in the turbulent regime.

Ken G
2005-Oct-17, 05:00 PM
OK, it seems the best thing for the physics teacher to say would simply be that the ball stirs the air as it moves through it, and that causes the ball to lose energy and the air to follow after the ball and become turbulent.

Bob
2005-Oct-17, 07:07 PM
I think Ken G is giving the clearest explanation since he is the first to bring up separation of flow from the ball and Reynolds number. The separation of flow creates a pressure differential between front and rear of the ball which slows the ball down. That's why golf balls are dimpled: to minimize the area from which the flow has separated.
http://www.aerospaceweb.org/question/aerodynamics/q0215.shtml

Ken G
2005-Oct-18, 08:39 AM
That link seems to give a very complete explanation (although they didn't mention the no-slip boundary layer). I didn't follow all of it, but they are certainly attributing a place for both friction and pressure gradients, although not compression-- these pressures are of the Bernoulli type and are found even in incompressible gas (like a ball moving at a low, subsonic speed). So although flow separation (as pointed out by Bob) and turbulence play key roles, it would appear that the physics teacher was more correct than our thread-starter, at least for subsonic balls, since apparently friction (as grant reasoned) is necessary (though I couldn't figure out why from the link).

seohtu
2005-Oct-19, 03:39 AM
I was thinking about it and thought that the air would have a greater cooling effect than heating that would be caused by friction. I mean if you stick your arm out the window of a car it feels cooler not hoter.
The cooling effect you feel with your arm out the window is from rapid evaporation of the ever-present moisture on your skin caused by the airflow. It's the same principle by which swamp-coolers work. Water molecules absorb heat when changing state from liquid to vapor, and one of the most efficient means of causing that change of state is to apply forced air, "dragging" the molecules out of their liquid state. "Dragging" is not a terribly accurate term there, but it gets the idea across.

On a very humid day the physics chance somewhat, because, while you're still forcing evaporation by wind, your arm will also be forcing condensation as it impacts water vapor. Just as heat must be absorbed in evaporation, heat must be rejected by the molecule during condensation, thus returning some heat back to your arm. This returned heat probably wouldn't make up for the additional evaporation caused by the added moisture, however.

Sincerely,
Derrick Baumer