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  • Today we're going to continue with playing with liquids.

  • If I have an object that floats, a simple cylinder that floats in some liquid, the area

  • is A here, the mass of the cylinder is M. The density of the cylinder is rho and its

  • length is l and the surface area is A. So this is l.

  • And let the liquid line be here, and the fluid has a density rho fluid.

  • I call this level y1, this level y2. The separation is h, and right on top here,

  • there is the atmospheric pressure P2, which is the same as it is here on the liquid.

  • And here we have a pressure P1 in the liquid. For this object to float we need equilibrium

  • between, on the one hand, the force Mg and the buoyant force.

  • There is a force up here which I call F1, and there is a force down here which I call

  • F2-- barometric pressure.

  • The force is always perpendicular to the surface. There couldn't be any tangential component

  • because then the air starts to flow, and it's static.

  • And here we have F1, which contains the hydrostatic pressure.

  • So P1 minus P2-- as we learned last time from Pascal--

  • equals rho of the fluid g to the minus y2 minus y1, which is h.

  • So that's the difference between the pressure P1 and P2.

  • For this to be in equilibrium, F1 minus F2 minus Mg has to be zero, and this we call

  • the buoyant force. And "buoyant" is spelt in a very strange way:

  • b-u-o-y-a-n-t. I always have to think about that.

  • It's the buoyant force. F1 equals the area times P1 and F2 is the

  • area times P2, so it is the area times P1 minus P2, and that is rho fluids times g times

  • h. And when you look at this, this is exactly

  • the weight of the displaced fluid. The area times h is the volume of the fluid

  • which is displaced by this cylinder, and you multiply it by its density, that gives it

  • mass. Multiply it by g, that gives it weight.

  • So this is the weight of the displaced fluids. And this is a very special case of a general

  • principle which is called Archimedes' principle. Archimedes' principle is as follows: The buoyant

  • force on an immersed body has the same magnitude as the weight of the fluid which is displaced

  • by the body. According to legend Archimedes thought about

  • this while he was taking a bath, and I have a picture of that here--

  • I don't know from when that dates-- but you see him there in his bath, but what

  • you also see are there are two crowns. And there is a reason why those crowns are

  • there. Archimedes lived in the third century B.C.

  • Archimedes had been given the task to determine whether a crown that was made for King Hieron

  • II was pure gold. The problem for him was to determine the density

  • of this crown-- which is a very irregular-shaped object--

  • without destroying it. And the legend has it that as Archimedes was

  • taking a bath, he found the solution. He rushed naked through the streets of Syracuse

  • and he shouted, "Eureka! Eureka! Eureka!" which means, "I found it! I found it!" What

  • did he find? What did he think of? He had the great vision to do the following: You

  • take the crown and you weigh it in a normal way.

  • So the weight of the crown-- I call it W1--

  • is the volume of the crown times the density of which it is made.

  • If it is gold, it should be 19.3, I believe, and so this is the mass of the crown and this

  • is the weight of the crown. Now he takes the crown and he immerses it

  • in water. And he has a spring balance, and he weighs

  • it again. And he finds that the weight is less and so

  • now we have the weight immersed in water. So what you get is the weight of the crown

  • minus the buoyant force, which is the weight of the displaced fluid.

  • And the weight of the displaced fluid is the volume of the crown--

  • because the crown is where... the water has been removed where the crown

  • is-- times the density of the fluid--

  • which is water, which he knew very well-- times g.

  • And so this part here is weight loss. That's the loss of weight.

  • You can see that, you can measure that with a spring.

  • It's lost weight, because of the buoyant force. And so now what he does, he takes W1 and divides

  • that by the weight loss and that gives you this term divided by this term, which immediately

  • gives you rho of the crown divided by rho of the water.

  • And he knows rho of the water, so he can find rho of the crown.

  • It's an amazing idea; he was a genius. I don't know how the story ended, whether

  • it was gold or not. It probably was, because chances are that

  • if it hadn't been gold that the king would have killed him--

  • for no good reason, but that's the way these things worked in those days.

  • This method is also used to measure the percentage of fat in persons' bodies, so they immerse

  • them in water and then they weigh them and they compare that with their regular weight.

  • Let's look at an iceberg. Here is an iceberg.

  • Here is the water-- it's floating in water.

  • It has mass M, it has a total volume V total, and the density of the ice is rho ice, which

  • is 0.92 in grams per cubic centimeter. It's less than water.

  • This is floating, and so there's equilibrium between Mg and the buoyant force.

  • So Mg must be equal to the buoyant force. Now, Mg is the total volume times rho ice

  • times g, just like the crown. The buoyant force is the volume underwater,

  • which is this part, times the density of water, rho water, times g.

  • You lose your g, and so you find that the volume underwater divided by the total volume

  • equals rho ice divided by rho of water, which is 0.92.

  • That means 92% of the iceberg is underwater, and this explains something about the tragedy

  • on April 15, 1912, when the Titanic hit an iceberg.

  • When you encounter an iceberg, you literally only see the tip of the iceberg.

  • That's where the expression comes from. 92% is underwater.

  • I want to return now to my cylinder, and I want to ask myself the question, when does

  • that cylinder float? What is the condition for floating? Well, clearly, for that cylinder

  • to float the buoyant force must be Mg, and the buoyant force is the area times h--

  • that's the volume underwater-- multiplied by the density of the fluid times

  • g must be the total volume of the cylinder, which is the area times l, because that was

  • the length of the cylinder, times the density of the object itself times g.

  • I lose my A, I lose my g, but I know that h must be less than l; otherwise it wouldn't

  • be floating, right? The part below the water has to be smaller than the length of the cylinder.

  • And if h is less than l, that means that the density of the fluid must be larger than the

  • density of the object, and this is a necessary condition for floating.

  • And therefore, if an object sinks then the density of the object is larger than the density

  • of the fluid. And the amazing thing is that this is completely

  • independent of the dimensions of the object. The only thing that matters is the density.

  • If you take a pebble and you throw it in the water, it sinks, because the density of a

  • pebble is higher than water. If you take a piece of wood, which has a density

  • lower than water, and you throw it on water, it floats independent of its shape.

  • Whether it sinks or whether it floats, the buoyant force is always identical to the weight

  • of the displaced fluid. And this brings up one of my favorite questions

  • that I have for you that I want you to think about.

  • And if you have a full understanding now of Archimedes' principle, you will be able to

  • answer it, so concentrate on what I am going to present you with.

  • I am in a swimming pool, and I'm in a boat. Here is the swimming pool and here is the

  • boat, and I am sitting in the boat and I have a rock here in my boat.

  • I'm sitting in the swimming pool, nice rock in my boat.

  • I mark the waterline of the swimming pool very carefully.

  • I take the rock and I throw it overboard. Will the waterline go up, or will the waterline

  • go down, or maybe the waterline will stay the same? Now, use your intuition--

  • don't mind being wrong. At home you have some time to think about

  • it, and I am sure you will come up with the right answer.

  • Who thinks that the waterline will go up the swimming pool? Who thinks that the waterline

  • will go down? Who thinks that it will make no difference, that the waterline stays the

  • same? Amazing--

  • okay. Well, the waterline will change, but you figure

  • it out. Okay, you apply Archimedes' principle and

  • you'll get the answer. I want to talk about stability, particularly

  • stability of ships, which is a very important thing--

  • they float. Suppose I have an object here which is floating

  • in water. Here is the waterline, and let here be the

  • center of mass of that object. Could be way off center.

  • It could be an iceberg, it could be boulders, it could be rocks in there, right? It doesn't

  • have to be uniform density. The center of mass could be off the center...

  • of the geometric center. So if this object has a certain mass, then

  • this is the gravitational force. But now look at the center of mass fluid that

  • is displaced. That's clearly more here, somewhere here,

  • the displaced fluid. That is where the buoyant force acts.

  • And so now what you have... You have a torque on this object relative

  • to any point that you choose. It doesn't matter where you pick a point,

  • you have a torque. And so what's going to happen, this object

  • is clearly going to rotate in this direction. And the torque will only be zero when the

  • buoyant force and the gravitational force are on one line.

  • Then the torque becomes zero, and then it is completely happy.

  • Now, there are two ways that you can get them on one line.

  • We discussed that earlier in a different context. You can either have the center of mass of

  • the object below the center of mass of the displaced fluid or above.

  • In both cases would they be on one line. However, in one case, there would be stable

  • equilibrium. In the other, there would not be a stable

  • equilibrium. I have here an object which has its center

  • of mass very low. You can't tell that--

  • no way of knowing. All you know is that the weight of the displaced

  • fluid that you see here is the same as the weight of the object.

  • That's all you know. If I took this object and I tilt it a little

  • with the center of mass very low-- so here is Mg and here is somewhere the waterline--

  • so the center of mass of the displaced fluid is somewhere here, so Fb is here, the buoyant

  • force, you can see what's going to happen. It's going to rotate towards the right--

  • it's a restoring torque, and so it's completely stable.

  • I can wobble it back and forth and it is stable. If I would turn it over, then it's not stable,

  • because now I would have the center of mass somewhere here, high up, so now I have Mg.

  • And the center of the buoyant force, the displaced water, is about here, so now I have the buoyant

  • force up, and now you see what's going to happen.

  • I tilt it to the side, and it will rotate even further.

  • This torque will drive it away from the vertical. And that's very important, therefore, with

  • ships, that you always build the ship such that the center of mass of the ship is as

  • low as you can get it. That gives you the most stable configuration.

  • If you bring the center of mass of ships very high--

  • in the 17th century, they had these very massive cannons which were very high on the deck--

  • then the ship can capsize, and it has happened many times because the center of mass was

  • just too high. So here... the center of mass is somewhere

  • here. Very heavy, this part.

  • And so now, if I lower it in the water notice it goes into the water to the same depth,

  • because the buoyant force is, of course, the same, so the amount of displaced water is

  • the same in both cases. But now the center of mass is high and this

  • is very unstable. When I let it go, it flips over.

  • So the center of mass of the object was higher than the center of mass of the displaced fluid.

  • And so with ships, you have to be very careful about that.

  • Let's talk a little bit about balloons. If I have a balloon, the situation is not

  • too dissimilar from having an object floating in a liquid.

  • Let the balloon have a mass M. That is the mass of the gas in the balloon

  • plus all the rest, and what I mean by "all the rest"...

  • That is the material of the balloon and the string--

  • everything else that makes up the mass. It has a certain volume v, and so there is

  • a certain rho of the gas inside and there is rho of air outside.

  • And I want to evaluate what the criterion is for this balloon to rise.

  • Well, for it to rise, the buoyant force will have to be larger than Mg.

  • What is the buoyant force? That is the weight of the displaced fluid.

  • The fluid, in this case, is air. So the weight of the displaced fluid is the

  • volume times the density of the air-- that's the fluid in which it is now--

  • times g, that is the buoyant force. That's... the weight of the displaced fluid

  • has to be larger than Mg. Now, Mg is the mass of the gas, which is the

  • volume of the gas times the density of the gas.

  • That's the mass times g-- because we have to convert it to a force--

  • plus all the rest, times g. I lose my g, and what you see...

  • that this, of course, is always larger than zero.

  • There's always some mass associated with the skin and in this case with the string.

  • But you see, the only way that this balloon can rise is that the density of the gas must

  • be smaller than the density of air. Density of the gas must be less than the density

  • of the air. This is a necessary condition for this to

  • hold. It is not a sufficient condition, because

  • I can take a balloon, put a little bit of helium in there--

  • so the density of the gas is lower than the density of air--

  • but it may not rise, and that's because of this term.

  • But it is a necessary condition but not a sufficient condition.

  • Now I'm going to make you see a demonstration which is extremely nonintuitive, and I will

  • try, step by step, to explain to you why you see what you see.

  • What you're going to see, very nonintuitive, so try to follow closely why you see what

  • you will see. I have here a pendulum with an apple, and

  • here I have a balloon filled with helium. I cut this string and I cut this string.

  • Gravity is in this direction. The apple will fall, the balloon will rise.

  • The balloon goes in the opposite direction than the gravitational acceleration.

  • If there were no gravity, this balloon would not rise and the apple would not fall.

  • Do we agree so far? Without gravity, apple would not fall, balloon would not rise.

  • Now we go in outer space. Here is a compartment and here is an apple.

  • I'm here as well. None of us have weight, there's no gravity,

  • and here is a helium-filled object, a balloon, and there's air inside.

  • We're in outer space, there's no gravity. Nothing has any weight.

  • We're all floating. Now I'm going to accelerate.

  • I have a rocket-- I'm going to accelerate it in this direction

  • with acceleration a. We all perceive, now, a perceived gravity

  • in this direction. I call it g.

  • So the apple will fall. I'm standing there, I see this apple fall.

  • I'm in this compartment, closed compartment. I see the apple go down.

  • A little later, the apple will be here. I myself fall; a little later, I'm there.

  • I can put a bathroom scale here and weigh myself on the bathroom scale.

  • My weight will be M times this a, M being my mass, a being this acceleration.

  • I really think that it is gravity in this direction.

  • The air wants to fall, but the balloon wants to go against gravity.

  • The balloon will rise. The air wants to fall, so inside here you

  • create a differential pressure between the bottom, P1, and the top of the air, P2, inside

  • here. Just like the atmosphere on earth--

  • the atmosphere is pushing down on us-- the pressure is here higher than there.

  • So you get P1 is higher than P2. So you create yourself an atmosphere, and

  • the balloon will rise. The balloon goes in the opposite direction

  • of gravity. If there were no air in there, then clearly

  • all of us would fall: The apple would fall, I would fall, and the helium balloon would

  • fall. The only reason why the helium balloon rises

  • is because the air is there and because you build up this differential pressure.

  • Now comes my question to you: Instead of accelerating it upwards and creating perceived gravity

  • down, I'm now going to accelerate it in this direction, something that I'm going to do

  • shortly in the classroom. I'm going to accelerate all of us in this

  • direction a. In which direction will the apple go? In which

  • direction will the balloon go? What do you think? The apple will go in the direction

  • that it perceives gravity. The apple will go like this.

  • I will go like this. The air wants to go like this.

  • But helium-- balloon--

  • goes in the opposite direction of gravity, so helium goes in this direction.

  • In fact, what you're doing, you're building here an atmosphere where pressure P1 here

  • will be higher than the pressure P2 there. The air wants to go in this direction.

  • The pressure here is higher than the pressure there--

  • larger than zero. If there's no air in there, we would all fall.

  • Helium would fall... helium balloon would fall, apple would fall,

  • and I would fall. I have here an apple on a string in a closed

  • compartment, not unlike what we have there except I can't take you out to an area where

  • we have no gravity. So here is that closed compartment, and here

  • is the apple. There is gravity in this direction.

  • It wants to fall in that direction of gravity if I cut the wire.

  • Now I'm going to accelerate it in this direction, and when I do that, I add a perceived component

  • of gravity in the opposite direction. So I add a perceived component of gravity

  • in this direction. So this apple wants to fall down because of

  • the gravity that I cannot avoid, and it wants to fall in this direction.

  • So what will the string do? It's very clear, very intuitive, no one has any problem with

  • that-- the string will do this.

  • Now I have a balloon here. Helium.

  • There is gravity in this direction. That's why the balloon wants to go up.

  • It opposes gravity. I'm going to accelerate the car in this direction.

  • I introduce perceived gravity in this direction. What does the balloon want to do? It wants

  • to go against gravity. I build up in here, and it must be a closed

  • compartment... I must build up there a pressure differential.

  • The air wants to fall in this direction. I build up a pressure here which is larger

  • than the pressure there. That's why it has to be a closed compartment.

  • What will the helium balloon do? It will go like that.

  • That is very nonintuitive. So I accelerate this car.

  • As I will do, the apple will go back, which is completely consistent with all our intuition,

  • but the helium balloon will go forward. Let's first do it with the apple, which is

  • totally consistent with anyone's intuition. I'm going to make sure that the apple is not

  • swinging too much. Now, it only happens during the acceleration,

  • so it's only during the very short portion that I accelerate that you see the apple go

  • back, and then of course it starts to swing-- forget that part.

  • So watch closely-- only the moment that I accelerate the apple

  • will come this way. It goes in the direction of the extra component

  • of perceived gravity. Ready?

  • Boy, it almost hit this glass here. Everyone could see that, right? Okay.

  • Now we're going to do it with the balloon. We're going to take this one off.

  • And now let's take one of our beautiful balloons. We're going to put a balloon in here.

  • Has to be a closed compartment so that the air can build up the pressure differential.

  • There's always problems with static charges on these systems.

  • Okay. Only as long as I accelerate will the balloon

  • go in a forward direction, so I accelerate in this direction, and what you're going to

  • see is really very nonintuitive. Every time I see it, I say to myself, "I can

  • reason it, but do I understand it?" I don't know, what is the difference between reasoning

  • and understanding? There we go. The balloon went this way.

  • You can do this in your car with your parents. It's really fun to do it.

  • Have a string with an apple or something else and have a helium balloon.

  • Close the windows. They don't have to be totally closed, but

  • more or less, and ask your dad or your mom to slam the brakes.

  • If you slam the brakes, what will happen? The apple will go...

  • what do you think? If you slam the brakes, the apple will go forwards, balloon will go

  • backward. If you accelerate the car all of a sudden,

  • the apple will go backwards and the balloon will go forward.

  • You can do that at home. You can enjoy... entertain your parents at

  • Thanksgiving. They'll get some of their $25,000 tuition

  • back. [class laughs]

  • When fluids are moving, situations are way more complicated than when they are static.

  • And this leads to, again, very nonintuitive behavior of fluids.

  • I will derive in a short-cut way a very famous equation which is called Bernoulli's equation,

  • which relates kinetic energy with potential energy and pressure.

  • Suppose I have a fluid, noncompressible, like so.

  • This cross-sectional area is A2 and the pressure here is P2.

  • And I have a velocity of that liquid which is v2 and this level is y2.

  • Here I have a cross-sectional area A1. I have a pressure P1.

  • My level is y1; this is increasing y. And I have a much larger velocity because

  • the cross-section is substantially smaller there.

  • Now, if this fluid were completely static, if it were not moving--

  • so forget about the v1 and forget about the v2; it's just sitting still--

  • then P1 minus P2 would be rho g times y2 minus y1 if rho is the density of the fluid.

  • That's Pascal's Law. So it would just be sitting still, and we

  • know that the pressure here would be lower than the pressure there.

  • This is also, if you want to, rho gh if you call this distance h.

  • Rho gh-- that reminds me of mgh, and mgh is gravitational

  • potential energy. When I divide m by volume, I get density.

  • So this is really a term which is gravitational potential energy per unit volume.

  • That makes the m divided by volume become density.

  • Therefore, pressure itself must also have the dimension of energy per unit volume.

  • And if we now set this whole machine in motion, then there are three players: There is, on

  • the one hand, kinetic energy-- of motion--

  • kinetic energy... I take it, per unit volume.

  • There is gravitational potential energy... I will take it, per unit volume.

  • And then there is pressure. They're equal partners.

  • And if I apply the conservation of energy, the sum of these three should remain constant.

  • That's the idea behind Bernoulli's law, Bernoulli's equation.

  • When I take a fluid element and I move it from one position in the tube to another position,

  • it trades speed for either height or for pressure. What is the kinetic energy per unit volume?

  • Well, the kinetic energy is one-half mv squared. I divide by volume, I get one-half rho v squared.

  • What is gravitational potential energy? That is mgy.

  • I divide by volume, and so I get rho gy plus the pressure at that location y, and that

  • must be a constant. And this, now, is Bernoulli's equation.

  • It is a conservation of energy equation. And as I will show you, it has very remarkable

  • consequences. First I will show you an example whereby I

  • keep y constant. So I have a tube which changes diameter, but

  • the tube is not changing with level y, as I do there.

  • So I come in here, cross-sectional area A1. I widen it, cross-sectional area A2.

  • This is y-- it's the same for both.

  • I have here inside pressure P1 and here inside I have pressure P2 and this is the density

  • of the fluid. There is here a velocity v2, and there is

  • here a velocity v1. And clearly v1 is way larger than v2 because

  • A1 times v1 must be A2 times v2 because the fluid is incompressible.

  • So the same amount of matter that flows through here in one second must flow through here

  • in one second. And so these have to be the same, and since

  • A1 is much smaller than A2, this velocity is much larger than v2.

  • Now I'm going to apply Bernoulli's equation. So the first term tells me that one-half rho

  • v1 squared... I can forget the second term because I get

  • the same term here as I get there because I measure the pressure here and I measure

  • the pressure there. They have the same level of y.

  • So I can ignore the second term. Plus P1 must be one-half rho v2 squared plus

  • P2. That's what Bernoulli's equation tells me.

  • Now, v1 is larger than v2. The only way that this can be correct, then,

  • is that P1 must be less than P2. So you will say, "Big deal." Well, it's a

  • big deal, because I would have guessed exactly the other way around, and so would you, because

  • here is where the highest velocity is, and all our instincts would say, "Oh, if the velocity

  • is high, there's a lot of pressure." It's exactly the other way around.

  • Here is the low pressure, and here is the high pressure, which is one quite bizarre

  • consequence of Bernoulli's equation. You must all have encountered in your life

  • what we call a siphon. They were used in the medieval and they're

  • still used today. You have here...

  • A bucket in general is used with water-- lakes.

  • We have water here, but it could be any liquid. And I stick in here a tube which is small

  • in diameter, substantially smaller than this area here.

  • And there will be water in here up to this level--

  • this level P2, y2. This is y1, increasing value of y.

  • This height difference is h. P2 is one atmosphere.

  • I put a one there-- it's atmosphere.

  • And here, if it's open, then P1 is also one atmosphere.

  • So there's air in here and there's liquid in here.

  • I take this open end in my mouth and I suck the water in so that it's filled with this

  • water, full with this water. And strange as it may be, it's like making

  • a hole in this tank. If I take my finger off here, the water will

  • start to run out, and I will show you that. And you have here a velocity v1.

  • The water will stream down into this here and the velocity here is approximately zero,

  • because this area is so much larger than this cross-sectional area that to a good approximation

  • this water is going down extremely slowly. Let's call this height difference d.

  • I apply Bernoulli's law. So now we have a situation where the y's are

  • different but the pressure is the same, because right here at this point of the liquid I have

  • one atmosphere, which is barometric pressure, and since this is open with the outside world,

  • P1 is also one atmosphere. So now I lose my P term.

  • There I lost my y term; now I lose my P term. So now I have that one-half rho... rho--

  • this is rho of the liquid-- v1 squared plus rho g times y1 must be one-half

  • rho v2 squared, but we agreed that that was zero, so I don't have that term.

  • So I only have rho gy2. I lose my g's...

  • no, I don't lose my g's. One-half rho v squared--

  • no, that's fine. And so... I lose my rho.

  • This is one-half. I lose my rho.

  • And so you get that one-half v1 squared equals g times y2 minus y1, which is h.

  • And so what do you find? That the speed with which this water is running out here, v1,

  • is the square root of 2gh. And you've seen that before.

  • If you take a pebble and you release a pebble from this level and you let it fall, it will

  • reach this point here, this level with the speed the square root of 2gh.

  • We've seen that many times. So what is happening here--

  • since the pressure terms are the same here and there, now there's only a conversion.

  • Gravitational potential energy-- which is higher here than there--

  • is now converted to kinetic energy. This siphon would only work if d is less than

  • ten meters. Because of the barometric pressure you can

  • never suck up this water-- no one can; a vacuum pump can't either--

  • to a level that is higher than ten meters. When I did the experiment there with the cranberry

  • juice, I was able to get it up to five meters, but ten meters would have been the theoretical

  • maximum. So this has to be less than ten meters that

  • you go up. If I would have made a hole in this tank here,

  • just like this, down to exactly this level, and I would have asked you to calculate with

  • what speed the water is running out, you would have found exactly the same if you had applied

  • Bernoulli's equation. This is a way that people...

  • I've seen people steal other people's gasoline in the time that gasoline was very scarce

  • and that there were no locks yet on the gasoline caps.

  • You would put a hose in the gasoline tank and you would have to suck on it a little--

  • you have to sacrifice a little bit-- you get a little bit of gasoline in your mouth,

  • and then you can just empty someone's gasoline tank by having a canister or by having a jerrican

  • and fill it with gasoline. And I'm going to show that now to you by emptying...

  • That's still cranberry juice, by the way, from our last lecture.

  • So let's put this up on a stool. So there is the hose--

  • it's that thing-- and I'm going to transfer this liquid from

  • here to here. So first I have to fill it with cranberry

  • juice. And there it goes.

  • And as long as this level is below that level, it keeps running.

  • Not so intuitive. I remember, I was at a summer camp when I

  • was maybe six or seven years old. I couldn't believe it when I saw this for

  • the first time. We had these outdoor sinks where we washed

  • ourselves and brushed our teeth, and the sink was clogged, it was full with water.

  • And one of the camp leaders took a hose, sucked up and it emptied itself.

  • And I really thought, you know, you'd have to take spoonfuls of water or maybe buckets

  • and scoop it out. This is the way you do it.

  • Very nonintuitive. The nonintuitive part is that it runs up against

  • gravity there. So we can let it sit there and we have a transfer,

  • mass transfer of cranberry juice. Last time I was testing my lungs to see how

  • strong I was. I wasn't very good, right? I could only blow

  • up one meter of water and only suck one meter water.

  • Differential pressure only one-tenth of an atmosphere.

  • Today I would like to test one of the students who, no doubt, is more powerful than I am.

  • And I have here a funnel... with a Ping-Pong ball here, very lightweight,

  • and we're going to have a contest to see who can blow it the highest.

  • I have two funnels, so it's very hygienic. I will try it with this one.

  • They're clean, they just... We just got them from the chemistry department.

  • And so I would like to see a volunteer-- woman or man, it doesn't matter.

  • You want to try it, see whether you can reach the ceiling? You don't want to try it? Come

  • on! You want to try it? You're shy? You don't want to? Can I persuade you? I can.

  • Okay, come along. Come right here.

  • You think you can make it to the ceiling? It's only a very light Ping-Pong ball.

  • So, you go like this, blow as hard as you can.

  • STUDENT: Okay. LEWIN: Try it, don't be nervous.

  • STUDENT: All right. LEWIN: Straight up.

  • STUDENT: No... LEWIN: Blow as hard as you can--

  • get it out. Amazing! Do it again.

  • Come on, there must have been something wrong. [class laughs]

  • LEWIN: You're not sick today, are you? Blow. Harder! STUDENT: Is this a trick? LEWIN: No,

  • there's nothing, there's no trick in here. I mean, my goodness--

  • this is a Ping-Pong ball, I'm not a magician. [class laughs]

  • LEWIN: Come on, blow it up! Hey, it doesn't work.

  • It's amazing. Why don't you sit down?

  • [class laughs] LEWIN: Why doesn't it work? Why doesn't it

  • work? The harder you blow, the least it will work.

  • Air is flowing here... and right here, where there is very little

  • room, the air will have very high speed, way higher than it has where it has lots of room.

  • And so at the highest speed, you get the lowest pressure.

  • And so the Ping-Pong ball is sucked in while you're blowing it.

  • And to give you the conclusive proof of that I will do it this way.

  • I will put the Ping-Pong ball like so, and I'm going to blow like this, and if I blow

  • hard enough, the Ping-Pong ball will stay in there because I generate a lower pressure

  • right here where the passage is the smallest, but I have to blow quite hard.

  • [inhales deeply, blowing hard] You see that? Isn't that amazing? That's the

  • reason why she couldn't get it up. [inhales deeply, blowing hard]

  • That's what Bernoulli does for you. Not so intuitive, is it?

  • I have here an air flow, a hose with air coming out, and I can show you there something that

  • is equally nonintuitive. Let's start the air flow.

  • [air hissing] It's coming out.

  • I take a Ping-Pong ball. It stays there.

  • Is that due to Mr. Bernoulli? No. No, that's more complicated physics, because

  • it has to do with turbulence. It has to do with vortices, which is very

  • difficult. What is happening here is that as the air

  • flows, you get turbulence above here and the turbulence creates a lower pressure.

  • So the vortices, which are the turbulence, are keeping this up, because there's a lower

  • pressure here and there. But why is it so stable? I can see that I

  • have... because of this turbulence, that it's held

  • up. Why is it so stable? If I give it a little

  • push it doesn't... it's sucked back in again.

  • It's very stable-- that is Bernoulli.

  • Because if I blow air, like so... then the velocity here is the highest, because

  • it's diverging the air as it's coming out, but in the center, it is the highest, and

  • so when this Ping-Pong ball goes to this side, it clearly has a lower pressure here than

  • there and so it's being sucked back in again. So the stability is due to Bernoulli, but

  • the fact that it is held up is more difficult physics.

  • It is so stable that I can even tilt this... and it will still stay there.

  • Now I have something that I want you to show your parents on Thanksgiving.

  • It's a little present for them, and that is something that you can very easily do at home.

  • You take a glass and you fill it with cranberry juice--

  • not all the way, up to here. Take a thin piece of cardboard, the kind of

  • stuff that you have on the back of pads. You put it on top.

  • The table is beautifully set-- turkey, everything is there--

  • and you suggest to your parents that you turn this over.

  • Your mother will scream bloody murder, because she would think that the cranberry juice will

  • fall out. In fact, it may actually fall out.

  • I can't guarantee you that it won't. [class laughs]

  • LEWIN: But it may not, in which case you now have all the tools to explain that.

  • Please do invite me to your Thanksgiving dinner and I'll show it to your parents.

Today we're going to continue with playing with liquids.

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B1 中級

Lecture 28 Hydrostatics Archimedes' Principle Fluid Dynamics What Makes Your Boat Float Bernoulli's Equation

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    fisher 發佈於 2013 年 04 月 12 日
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