Got it right, but barely so. My thinking is that the combined system wants to minimize the total surface area. And this is accomplished best by a single sphere.
It might have to do with the fact that in rubber, entropy is the major contributor to the restoring force. And a minimum surface area minimises the number of chains being stretched (ordered).
This is closer to how I'm thinking about it. I think this is a different solution than the one Steve gave, but maybe I'm wrong. I was thinking that the amount of stretch per (change in) volume of air is less in the bigger balloon, so the bigger balloon pushes less on a given amount of air than the littler balloon does. It gives the same predictions in this case, but this way of thinking seems to give a better explanation. (or does it?)
You also have to take into account that the SAME TENSION in the surface of a balloon gives a normal force different depending on the radius of the ballon. If the radius is small (and has a high curvature) it gives a higher normal force than if the radius is big (and has a low curvature). This normal force that is distributed along its surface and points to its center must equilibrate the pressure of the air that is inside.
Due to how difficult it is to start blowing a balloon compared to when it is already blown somewhat, I'm going to go out on a limb and say the big one gets bigger...
Wait, did you just reply to yourself?? But yeah that was what I thought of too. The fact that them equalizing wouldn't have been a very interesting result also tipped it off :P
i'd guessed option one because i was considering not the forces acting on the balloons, but on the air inside them, which i thought should be equal and consequently the air would have no reason to move, i completely forgot that one balloon might have more potential than the other
consider this too, the balloon is getting stretched by outward force from the pressure. Force is equal to pressure times surface. Since the surface of the bigger balloon is, well, bigger, it experiences a net larger force than the smaller balloon does since the pressure is equal when the clamp is released.
My prediction is that the air will balance between them so they end up the same size, but that would make for a boring video so I also predict that I'm wrong about that.
Huntracony you're also more honest than the people saying they reasoned it was C on their own. I knew it would be C, but only because I had seen the demonstration before, and had gotten it wrong that time.
that prediction of "air balancing" is the same as option b, cuz once the air from one gets pushed to the other and they become the same size, there's nowhere for it to go, or in other words, the other balloon becomes the larger one and they endlessly swap between them
The surface tension formula for a thin film soap bubble can also be used to describe this analogous relationship. The difference is external and internal pressure is given by 4*gamma/R; i.e inversely proportional to R. This leads to the smaller balloon having a larger pressure inside wrt the larger balloon.
Can this be pushed so far that, although initially, both balloons are capable of holding in the air, the bigger balloon pops before they reach equilibrium?
I'd say yes, but it might be hard to set up, because there can be quite a bit of variance in balloons' popping points. This would be a cool thing to do for a follow up video!
+Kram1032 Something like that happens to Helium balloons! They fly higher and higher and get stretched more and more and actually lose resistance to being stretched. So eventually they pop. This is why you have to use nonelastic material for weather-balloons.
+CaesarsSalad Except that happens because the external pressure gets lower. The idea here would be to have the stronger elastic forces of the less inflated balloon pop the larger balloon by increasing the internal pressure.
+CaesarsSalad The weather balloons I've heard of were all elastic, but maybe some types use non-elastic materials? The elastic ones that I've heard were generally made of latex and they carry a radiosonde up into the atmosphere broadcasting measurements down. On the ground you fill them, usually with hydrogen, but only part way. As the altitude increases and the pressure decreases, the hydrogen expands to fill the balloon, and eventually it pops and then the payload drops (on a parachute normally, I believe). Since you don't really want these balloons floating around indefinitely, possibly causing hazards, it makes sense to design them like this. If some weather balloons have a non-elastic envelope, I wonder if they still burst? Or if the non-elastic envelope is for balloons designed to stay at altitude for an extended period?
I go with c. If you've ever let an inflated balloon go it flies around making a comical sound and in the last second it accelerates I guess this is the rubber contracting faster towards its relaxed state.
Chris Harney I know I'm late to the party with this comment, but I would imagine that the decrease in area, and so air resistance, as well as the lower total combined mass the balloon and air inside at the end would make more of a difference than the increased force.
But just "acceleration" wouldn't be enough, you'd have to measure it and confirm that it accelerates differently than what you'd expect if you don't consider the rubber contracting.
My first instinct is that the big one will get smaller, but I feel like that's too obvious a result. Still gonna guess it but I'm probably wrong or you wouldn't be making a video about it.
My intuition said B, but then I reasoned that you wouldn't be making a TH-cam video to demonstrate something that everyone would assume to be the obvious answer, so I wasn't surprised when the opposite of what we all thought would happen actually happened.
Justin Nanu That's what I thought too. Figured I better watch to see why I was wrong. Makes sense after thinking about it more, as it's much harder to blow a balloon up in the beginning.
well...almost. it is hard to blow up a balloon at first and it gets much easier and then gets harder again as capacity gets reached... I could give a clear argument for "B" in the demonstration, but I don't know the capacity of the balloons. "A" is also possible, if the smaller balloon has never been inflated at all.
Additionally: if you inflate a balloon, it gets plastic deformed a bit. You'll notice this, when you inflate one balloon and let it deflate and compare it to one, which was never before inflated. That was my thought, when deciding for option C. In this process of inflating, you permanently weakening the structure of the bigger inflated balloon. Also, if you ever inflated a loon with you lungs, you noticed that the first bit is the most hardest part to inflate.
I felt it was quite intutive. I thought about how hard it is to make the first blow, and how each blow thereafter becomes easier. So clearly, everyday experience with balloons should tell us this exact relationship.
I bet option C on the basis of the fact that it's harder to inflate a balloon at the start when the rubber is thick and hard to stretch. It has more potential to compress the air.
It might be worth noting that this effect is observable when you try to inflate a balloon by mouth. The hardest part is always getting the balloon to begin stretching. Also notable: I'm under the impression that the rubber used for balloons is not linear elastic, but is rather an elastomer. Which means that it isn't bound by the special case rule of hooke's law, but in this instance seems to demonstrate it well enough
I bet it's also due to the energy reduced by transferring a small amount of air. The small balloon gets deflated more quickly since you're talking about air volume vs. rubber surface area. At the end, it reduces the total energy to have one more inflated balloon than two partly inflated balloons.
+ElectroLlama This is a great insight, thanks ElectroLlama. With these sorts of problems you can usually tackle it in two different ways: Either look at the forces involved or try to find the lowest energy. Like if I put a ball on a slope at the side of a valley, I could look at the forces to figure out where the ball will go or I could just look for the spot at the bottom of the valley where the energy is lowest.
This might also explain why a balloon doesn't blow up evenly, but usually develops a bulging thinned-out section that gets bigger until you worry it might burst, and later the rest fills, spreading out from there. Unfortunately the rest does eventually stretch out too, so there has to be something wrong with the explanation. And the explanation was going so well! And if the explanation were completely right, the rest of the balloon would never fill up. Instead one tiny spot where the balloon was slightly weaker than average would aways be where it inflates, getting ever thinner, while the rest stayed permanently small and thick. Most things that need to be inflated to high pressure, like tires and basketballs, in fact have a tough exterior that hardly stretches, with embedded non-stretching reinforcement, and a very stretchy something inside that presses against the outside.
Great thinking. Yes, the full explanation is that Hooke's law is only linear up to a point. When you stretch the rubber beyond that point it very quickly gets much harder to stretch. That's when the rest of the balloon fills out. Rubber is made up of a jumble of long molecules. When you stretch it, these long molecules are pulled out straight. Once they are all stretched out there's nowhere else to go a Hooke's law breaks down.
As soon as you mentioned the final option, I thought it was going to be that for a different reason; inconsistencies in the balloon manufacturing process.
Haha I guessed correctly. My reasoning was that when you inflate a balloon its typically pretty difficult to get the first bit of air in, but after that it just gets easier and easier. I also figured the total area to volume ratio would be minimized, and I'm fairly sure that one large balloon + one small balloon has a lower total surface area than two medium ones.
Love your work. I really appreciate it an enjoy them a lot. Would you consider including a second chapter of the videos where you explain the math and physics behind? You are very clear concise so imho viewers really would benefit for this. Anyway thanks for your work
Pre-video prediction: "How could it be anything but b)? Perhaps if it actually takes more pressure to inflate a balloon the smaller it is. In that case it would probably be c). But I take b)." Edit: Dammit!
Fabrizio Lungo in fact, you could argue that it follows, as fast food is much cheaper than healthy food, especially the premium people spend on "organic, GMO free, and gluten free" food (I understand some people can't process gluten but not eating gluten is not any healthier for the average person)
I guessed it right: the C, but with the help of a second thinking, because of course the first intuition was that both ballons would go to equilibrium at half the total inflation for each, but the very presence of this video suggest that the obvious won't happen. So the small ballon becomes smaller, because we know by experience that it's always harder to blow a ballon in the beginning, and it becomes easier and easier as it inflates. Therefore they rightly go to equilibrium, but not the half/half equilibrium of the first guess.
Another example is that the first blow of air to a balloon is much harder than the second, even though there's no pressure differences from the outside.
What he described about rubber causes the graph of Stress vs Strain for rubber to be such that Stress/Strain is not constant and the slope of the graph becomes smaller and smaller with increase in stress, that is, you get larger strain for a smaller change in stress as you increase the stress.
This totally explains why it's is harder to inflate a balloon just when you start (when the rubber is thicker) vs. when you have already some air inside (the rubber is thinner). That is a really interesting concept. Great video.
I know this one. The less surface area the more surface tension, meaning the smaller balloon will have more pressure on forcing the air out than the larger balloon.
this problem is what happens on premature babies due to their alveoli not being fully developed, they died if not for the invention of a machine that helps to avoid this collapsing/overexpanding of their alveoli until they get to the age where surfactant is naturally released and the lungs stabilize
My guess is the tension in the smaller balloon is stronger and will inflate the larger balloon. That’s why balloons are hardest to blow up at the start.
C, I learnt about bubbles in my physics class. The smaller bubbles have higher pressure than the bigger ones. Mathematically, P = (4S/R), P= pressure, S =surface tension, R = radius
I got it right, but not for Hooke's Law. My hypothesis was that although it looks it this is not a closed system, there's actually a very much bigger system acting against both, atmospheric pressure against which they are both miniscule. The larger balloon is infalated to a greater pressure than the smaller by a tiny tiny miniscule amount and resists air pressure slightly more than the little one, the little one is slightly weaker than the larger one and since air pressure is the same against both the air in it is forced into the larger one.
I may be wrong, but I believe the reduction in thickness by the increase in radius is due to the balloon having a 2 dimensional surface, and like all 2 dimensional surfaces, area is Length * Width. Radius would be just either L or W. This is something I frequently need to explain to customers when printing photos. If you want a photo printed twice the length and width, you'll need 4x the amount of pixels, since the area is now 4x larger.
I don't think that you need to "beat" Hooke's law to explain this result. Hooke's law tells us that the energy used to stretch the balloon is proportional to the square of the radius. In other words, it is proportional to the surface area of the balloon. The total surface area of the balloon is maximal if the air is equally partitioned between both balloons (assuming constant pressure) and minimal if the air is in one balloon, the other being empty. Therefore, Hooke's law predicts that, as the system tends to a state of minimum energy, one balloon will be filled, and the other one will be empty.
For a moment i thought 'B' but then considered the tension or force per square centimeter of rubber on the small balloon and how inflated it was comparatively and remembered that if takes more effort to force air into a deflated balloon than one that already has a substantial amount of air in it and concluded that 'C' was most likely. Nailed it.
what about volume vs. surface area? same volume of air added to small balloon vs. same volume of air added to large balloon -> large area increase(large strech) vs. small area increase(small strech) -> large force change vs. small force change I don't think thickness of the rubber is the major factor but i'd love to be wrong as long as i hear about it.
C Just think of inflating a very strong balloon (like a water balloon). The start is always the hardest. You have to create a huge amount of pressure to get it to start. Every following breath gets easier. So, in a small balloon there is a higher pressure than in a larger (identical) balloon. Thus, a bit of air will go from the smaller one to the larger one. But then it only becomes more unequal and some more air will continue to flow in the same direction...
This can be explained another way: Potential energy stored in the balloon's shape is directly proportional to it's surface area. The system is trying to reach a point of minimal energy, hence it will try to minimise the surface area. It's well known, that if you fix a volume V, the smallest possible surface area of any shape with volume V is the sphere. Hence, the system will try to reach a spherical shape.
I originally thought A was the correct answer. The idea was that because both balloons were stationary, they must had been in equilibrium with the outside pressure, and because of that they both had 1 atm of pressure inside. Clearly this is not the case because I did not factor in the rubber thing. But I have one question, if there's a different amount of pressure inside the 2 ballons, why are they both in equilibrium with the same outside constant pressure? I'm starting to think that in order to expand they need to also win the resistance of the rubber, but I really don't know. How many equilibrium states are there?
If you've ever tried to squeeze a balloon into two equal lobes, you will automagically understand that option C is the correct answer. Squeezing a ballon into two equal lobes was just one of those things that you could do, but not for very long, so I played a game with my friends where you had to keep the lobes in equilibrium for as long as possible.
Yeah! This is also the exact reason why I knew it was option C, because I remembered as a kid playing with balloons, squeezing it into two parts, and I remembered how the smaller bulge would always flow into the larger one, rather than them equaling out.
That is a nice demonstration of an unstable system, like the inverted pendulum. If we started with two balloon inflated the same, then induce random pressure on either side, eventually, one win and get all the air (maybe explaining the many comments about rich becoming richer). It also explain why it is so hard to start inflating a balloon.
Alternative rationale: On the larger baloon the curve is less sharp, whereas the smaller radius baloon the curve is obviously more sharp. We can look at the problem of hanging a mass on a horzontal cable. If we put a 1kilo mass on a cable that spans a 1 meter gap, and give the wire more or less slack we observe that more "flat" the cable is, the more tension is required to suspend the mass (the vertical component of tension is greater with a sharper angle). Intuitively: A horizontal cable has no ability to pull up, whereas the extreme of sag is a verticle cable, where tension is all going to suspending the mass. So the sharper angle on the smaller balloon means that the restorative force is GREATER for a given tension (and due to Bernoulli we know that air pressure is the same between the two).
I actually knew this, and guessed correctly. Saw it once before in baby lungs. The oxygen collecting parts of the lung works similarly, and in some cases babies are born with some of these not being in equilibrium, this collapses once and enlarges the other. The procedure to alleviate this won a Nobel prize I think. To get it right I also imagine blowing up a balloon, it is much harder at the start. So smaller ones push back harder.
Hey there Steve. Could the phenomenon be due to a difference in Laplace pressures between the two balloons? When disconnected and at equilibrium, the internal pressure in the smaller balloon is greater than that of the larger balloon. p_inside = p_atm +2("surface tension")/r (Assuming a sphere) An analogy can be drawn to foam coarsening..
Then why did the exchange stop? The thinning effect was counteracted by the stretching of rubber having a limiting effect. Otherwise the thinning effect would have led to the complete emptying of the smaller balloon
My reasoning was the following: balloons and bubbles naturally go to a spherical shape, in bubbles because of the surface tension of the liquid, and in balloons because of the elastic tension of the rubber. The balloons are linked by the more rigid hose, but still, they should naturally tend towards a single sphere, not an 8 or infinity like shape. The large balloon inflating further is closer to a sphere than both balloons reaching equilibrium.
the amount that the rubber has to stretch to fit extra air may be less as the the balloon gets bigger meaning that the smaller balloon will squeze the air more.
This will be an excellent demo for students of thermodynamics, most of whom will go for Option B, as if this system is equivalent to two rigid containers separated by a piston.
What about atmospheric pressure? The larger balloon has more surface area for the outside atmosphere pressure to push against so you would think there would be more pressure in the larger balloon than the smaller balloon?
Hey, I actually guessed it right! I figured the same thing you did after explaining. The less-inflated balloon has more rubber contracting on the air inside forcing it into the already inflated-balloon.
i figured it was c. similar to how you can warp balloons by squashing them while blowing them up. if the answer was a or b the the balloons would not be able to stay warped, but i know they can stay warped. also when you blow up the long balloons they fill up from one side and work their war to the other as they inflate more, again if the answer was a or b this would not happen the entire balloon would inflate at the same rate.
I chose C because of how a balloon is harder to blow up in the beginning than after a couple of breaths. I assumed that the smaller balloon wanted to contract more than the big one did.
Excess pressure inside a balloon is inversely proportional to its radius. This means that less inflated balloon has more pressure. When the clip will be removed, the gas will always flow from high pressure to low pressure. Means the gas will flow from less inflated balloon to more inflated balloon. This makes less inflated balloon even lesser inflated and more inflated balloon even more inflated.
there is also the fact that the volume of a sphere rises faster than the surface, so one big sphere should energetically preferable to two spheres, even if the tension increased with the surface (to a degree)
Apart from that, you had to be sure that before doing this experiment you first stretched *both* balloons to the same full blown size, because new balloons that are partially inflated still have very strong rubber. You should have noticed that a new balloon is always very hard to inflate, but once you fully stretch it, the next time that you inflate it, it has become softer. Anyway, the effect would be the same if you didn't do that, because the new partially inflated balloon would be stronger than the stretched one.
Force from surface of the balloon is proportional to the curvature of the surface, so the smaller balloon is more curved and will apply more force to the gas than the bigger balloon which is less curved. So I think the bigger balloon will get bigger and the smaller smaller.
I got this correct not from knowing Hooke's law or the math behind it, but from experience. Anyone who's ever blown up a balloon with their mouth has surely noticed how much more pressure it takes initially to get it started and felt the resistance go down as inflation gets started and proceeds. Similarly, while letting the air out of a balloon, you can hear the pitch of the valve flapping increase as it gets smaller, suggesting higher pressure. Having the small one get smaller was intuitive to me for these reasons.
My fist instinct was the correct one and it was based on the fact that when one fills a balloon by 'lung power', it is hardest at the start and then becomes easier as the balloon gets bigger. I think the real explanation for the effect is that when the balloon is bigger, its curvature is smaller and this makes it weaker at resisting internal pressure.
That is because. P(Pressure)=F(Force)/A(Area) Or we can say that Pressure is inversely proportional Area . Means If Area is greater than Pressure is lower that is in case of large ballon. But opposite in smaller i.e., Area is lower but Pressure is more. And we know Air travels from High Pressure area to Low Pressure area. Thus all this happens.
I chose option C because hooks laws is linear but the increase in surface area is quadratic. Therefore the rubber will get stretched at a much faster rate than its ability to resist stretching. I remember when I was a kid balloons were always hard to get started but got easy very quickly once they filled so C matched my intuition.
This video ends rather too
+Chris James I don't get what you
+Chris James You get the idea, can't be bothered to edit it more, let's just upload it to
+Chris James That's so annoying! The file I uploaded has the whole ending. It's just "Hooke's Law" though so we're not missing too much!
+Steve Mould Haha no worries, its a great video regardless! I was just trying to be funny ;)
Steve Mould was there a part where you push back air into the smaller balloon enough so that it grows itself?
"[...]beats Hooke's Law. Thanks for watching."
brianpso haha yeah I was wondering why or how it ended on a
@@Hockeyluke1042 Indeed, quite sur
Got it right, but barely so. My thinking is that the combined system wants to minimize the total surface area. And this is accomplished best by a single sphere.
I like that way of thinking!
+Ben Hambrecht Nice!
It might have to do with the fact that in rubber, entropy is the major contributor to the restoring force. And a minimum surface area minimises the number of chains being stretched (ordered).
I'm pretty sure your explanation is correct.
This is closer to how I'm thinking about it. I think this is a different solution than the one Steve gave, but maybe I'm wrong. I was thinking that the amount of stretch per (change in) volume of air is less in the bigger balloon, so the bigger balloon pushes less on a given amount of air than the littler balloon does. It gives the same predictions in this case, but this way of thinking seems to give a better explanation. (or does it?)
You also have to take into account that the SAME TENSION in the surface of a balloon gives a normal force different depending on the radius of the ballon. If the radius is small (and has a high curvature) it gives a higher normal force than if the radius is big (and has a low curvature). This normal force that is distributed along its surface and points to its center must equilibrate the pressure of the air that is inside.
Due to how difficult it is to start blowing a balloon compared to when it is already blown somewhat, I'm going to go out on a limb and say the big one gets bigger...
+Matthew Mitchell Yep, anyone who has blown up a balloon ought to get this really.
Wait, did you just reply to yourself??
But yeah that was what I thought of too. The fact that them equalizing wouldn't have been a very interesting result also tipped it off :P
I guess I did...
i'd guessed option one because i was considering not the forces acting on the balloons, but on the air inside them, which i thought should be equal and consequently the air would have no reason to move, i completely forgot that one balloon might have more potential than the other
MinuteEarth/physics already did this
consider this too, the balloon is getting stretched by outward force from the pressure. Force is equal to pressure times surface. Since the surface of the bigger balloon is, well, bigger, it experiences a net larger force than the smaller balloon does since the pressure is equal when the clamp is released.
My prediction is that the air will balance between them so they end up the same size, but that would make for a boring video so I also predict that I'm wrong about that.
+Huntracony And I was correct in being wrong. Interesting though.
+Huntracony Nailed it!
Huntracony Wow, I keep running into you on TH-cam.
Huntracony you're also more honest than the people saying they reasoned it was C on their own. I knew it would be C, but only because I had seen the demonstration before, and had gotten it wrong that time.
that prediction of "air balancing" is the same as option b, cuz once the air from one gets pushed to the other and they become the same size, there's nowhere for it to go, or in other words, the other balloon becomes the larger one and they endlessly swap between them
The surface tension formula for a thin film soap bubble can also be used to describe this analogous relationship.
The difference is external and internal pressure is given by 4*gamma/R; i.e inversely proportional to R. This leads to the smaller balloon having a larger pressure inside wrt the larger balloon.
Can this be pushed so far that, although initially, both balloons are capable of holding in the air, the bigger balloon pops before they reach equilibrium?
I'd say yes, but it might be hard to set up, because there can be quite a bit of variance in balloons' popping points. This would be a cool thing to do for a follow up video!
+Kram1032 Something like that happens to Helium balloons! They fly higher and higher and get stretched more and more and actually lose resistance to being stretched. So eventually they pop. This is why you have to use nonelastic material for weather-balloons.
+CaesarsSalad Except that happens because the external pressure gets lower. The idea here would be to have the stronger elastic forces of the less inflated balloon pop the larger balloon by increasing the internal pressure.
+BlameItOnGreg yep, maybe filling a baloon from a gas bottle would be closer although thats not exactly elastic forces
+CaesarsSalad The weather balloons I've heard of were all elastic, but maybe some types use non-elastic materials?
The elastic ones that I've heard were generally made of latex and they carry a radiosonde up into the atmosphere broadcasting measurements down. On the ground you fill them, usually with hydrogen, but only part way. As the altitude increases and the pressure decreases, the hydrogen expands to fill the balloon, and eventually it pops and then the payload drops (on a parachute normally, I believe). Since you don't really want these balloons floating around indefinitely, possibly causing hazards, it makes sense to design them like this.
If some weather balloons have a non-elastic envelope, I wonder if they still burst? Or if the non-elastic envelope is for balloons designed to stay at altitude for an extended period?
Balloons are harder to inflate when starting out, and get easier and easier as they get more inflated, so C.
My thoughts exactly.
BlameItOnGreg i thought exactly the same
I voted c because this experiment was in my physic text book
Exactly!
I go with c. If you've ever let an inflated balloon go it flies around making a comical sound and in the last second it accelerates I guess this is the rubber contracting faster towards its relaxed state.
Love this line of reasoning. Not heard it before.
+Chris Harney yes cool, was going to put B until i read this haha and you were right fantastic!
Chris Harney I know I'm late to the party with this comment, but I would imagine that the decrease in area, and so air resistance, as well as the lower total combined mass the balloon and air inside at the end would make more of a difference than the increased force.
But just "acceleration" wouldn't be enough, you'd have to measure it and confirm that it accelerates differently than what you'd expect if you don't consider the rubber contracting.
Chris Harney eeeeerrrrrrr💩💩💩💩
There's also the effect of the Laplace equation, which shows that the pressure is inversely proportional to the radius.
My first instinct is that the big one will get smaller, but I feel like that's too obvious a result. Still gonna guess it but I'm probably wrong or you wouldn't be making a video about it.
+12tone Huh. Interesting. I guess that explains why the hardest part of blowing up a balloon is getting it started.
wow strange to see you here. love your videos
12tone why are you saying that🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔🤔
My intuition said B, but then I reasoned that you wouldn't be making a TH-cam video to demonstrate something that everyone would assume to be the obvious answer, so I wasn't surprised when the opposite of what we all thought would happen actually happened.
Justin Nanu That's what I thought too. Figured I better watch to see why I was wrong. Makes sense after thinking about it more, as it's much harder to blow a balloon up in the beginning.
C, it's harder to blow into a balloon when it's smaller. Once it gets big, it gets really easy to blow into
jpsousa4 that was my reasoning for picking C too, knowing that the rubber thins as it gets inflated and becomes even easier to inflate.
well...almost. it is hard to blow up a balloon at first and it gets much easier and then gets harder again as capacity gets reached... I could give a clear argument for "B" in the demonstration, but I don't know the capacity of the balloons. "A" is also possible, if the smaller balloon has never been inflated at all.
Additionally: if you inflate a balloon, it gets plastic deformed a bit. You'll notice this, when you inflate one balloon and let it deflate and compare it to one, which was never before inflated.
That was my thought, when deciding for option C.
In this process of inflating, you permanently weakening the structure of the bigger inflated balloon. Also, if you ever inflated a loon with you lungs, you noticed that the first bit is the most hardest part to inflate.
+Illu07 very good point about plastic deformation. I hadn't thought of that. Thanks!
But to inflate a loon, you'll have to catch him first!
This also allows you to control the answer to Steve's puzzle. The action lab covered this: th-cam.com/video/GiG0e1s6nV4/w-d-xo.htmlm
@@bennylofgren3208 well loons are getting inflated up here jn Canada
It ends a bit abruptly
I felt it was quite intutive. I thought about how hard it is to make the first blow, and how each blow thereafter becomes easier. So clearly, everyday experience with balloons should tell us this exact relationship.
I bet option C on the basis of the fact that it's harder to inflate a balloon at the start when the rubber is thick and hard to stretch. It has more potential to compress the air.
It might be worth noting that this effect is observable when you try to inflate a balloon by mouth. The hardest part is always getting the balloon to begin stretching.
Also notable: I'm under the impression that the rubber used for balloons is not linear elastic, but is rather an elastomer. Which means that it isn't bound by the special case rule of hooke's law, but in this instance seems to demonstrate it well enough
I bet it's also due to the energy reduced by transferring a small amount of air. The small balloon gets deflated more quickly since you're talking about air volume vs. rubber surface area. At the end, it reduces the total energy to have one more inflated balloon than two partly inflated balloons.
+ElectroLlama This is a great insight, thanks ElectroLlama. With these sorts of problems you can usually tackle it in two different ways:
Either look at the forces involved or try to find the lowest energy.
Like if I put a ball on a slope at the side of a valley, I could look at the forces to figure out where the ball will go or I could just look for the spot at the bottom of the valley where the energy is lowest.
(ballon surface ~r^2 ) (volume~r^3)
one bigger ballon means less stretching per volume
This might also explain why a balloon doesn't blow up evenly, but usually develops a bulging thinned-out section that gets bigger until you worry it might burst, and later the rest fills, spreading out from there. Unfortunately the rest does eventually stretch out too, so there has to be something wrong with the explanation. And the explanation was going so well!
And if the explanation were completely right, the rest of the balloon would never fill up. Instead one tiny spot where the balloon was slightly weaker than average would aways be where it inflates, getting ever thinner, while the rest stayed permanently small and thick.
Most things that need to be inflated to high pressure, like tires and basketballs, in fact have a tough exterior that hardly stretches, with embedded non-stretching reinforcement, and a very stretchy something inside that presses against the outside.
Great thinking. Yes, the full explanation is that Hooke's law is only linear up to a point. When you stretch the rubber beyond that point it very quickly gets much harder to stretch. That's when the rest of the balloon fills out. Rubber is made up of a jumble of long molecules. When you stretch it, these long molecules are pulled out straight. Once they are all stretched out there's nowhere else to go a Hooke's law breaks down.
+Steve Mould,
This more-complete explanation should definitely be in the video. If you get time, please add it. Cheers.
+Tom Hartley,
PS: Very interesting concept! Many thanks to Steve.
As soon as you mentioned the final option, I thought it was going to be that for a different reason; inconsistencies in the balloon manufacturing process.
i was sure about B, but turns out i was wrong. this experiment shows that you should always be a bit sceptical even if the result seems to be obvious.
"E" the experiment is intentionally flawed.
Forgot to make any prediction, watched the video, and still think option B is gonna happen.
Haha I guessed correctly. My reasoning was that when you inflate a balloon its typically pretty difficult to get the first bit of air in, but after that it just gets easier and easier. I also figured the total area to volume ratio would be minimized, and I'm fairly sure that one large balloon + one small balloon has a lower total surface area than two medium ones.
Same, I actually used the same reasoning lol
This explains why that initial little part in the beginning of blowing up a balloon is so hard.
I predict option C because it's the one that sounds the craziest.
Love your work. I really appreciate it an enjoy them a lot.
Would you consider including a second chapter of the videos where you explain the math and physics behind?
You are very clear concise so imho viewers really would benefit for this. Anyway thanks for your work
Pre-video prediction:
"How could it be anything but b)? Perhaps if it actually takes more pressure to inflate a balloon the smaller it is. In that case it would probably be c). But I take b)."
Edit: Dammit!
Can we explain it using laplaces law
P= 2T / R
Where p = pressure
R= radius
T = surface tension
And the rich get richer.
Following this analogy, it must be because the poor are thicker.
Fabrizio Lungo Which they are due to depression eating and bad nutrition.
Fabrizio Lungo in fact, you could argue that it follows, as fast food is much cheaper than healthy food, especially the premium people spend on "organic, GMO free, and gluten free" food (I understand some people can't process gluten but not eating gluten is not any healthier for the average person)
I was going to say that in the comments too! XD
jesam11 it's actually bad to not eat gluten from what I've read
I guessed it right: the C, but with the help of a second thinking, because of course the first intuition was that both ballons would go to equilibrium at half the total inflation for each, but the very presence of this video suggest that the obvious won't happen. So the small ballon becomes smaller, because we know by experience that it's always harder to blow a ballon in the beginning, and it becomes easier and easier as it inflates. Therefore they rightly go to equilibrium, but not the half/half equilibrium of the first guess.
ha, nailed it. I called C.
also, the video ends too soon
Another example is that the first blow of air to a balloon is much harder than the second, even though there's no pressure differences from the outside.
B
edit: lol Am I the only one?
+tggt00 yep
perhaps also because the volume goes up cubicly and now the total area is minimized?
Shouts to all those people in the comments that said C and definitely didn't watch the video to get the answer.
What he described about rubber causes the graph of Stress vs Strain for rubber to be such that Stress/Strain is not constant and the slope of the graph becomes smaller and smaller with increase in stress, that is, you get larger strain for a smaller change in stress as you increase the stress.
This totally explains why it's is harder to inflate a balloon just when you start (when the rubber is thicker) vs. when you have already some air inside (the rubber is thinner). That is a really interesting concept. Great video.
It's kind of apparent if you've ever blown up a balloon -- it's hard at first, gets easier, and then harder again
This is why the first puff in blowing up a balloon is alyways the jardest and then gets easier.
I know this one. The less surface area the more surface tension, meaning the smaller balloon will have more pressure on forcing the air out than the larger balloon.
this problem is what happens on premature babies due to their alveoli not being fully developed, they died if not for the invention of a machine that helps to avoid this collapsing/overexpanding of their alveoli until they get to the age where surfactant is naturally released and the lungs stabilize
My guess is the tension in the smaller balloon is stronger and will inflate the larger balloon. That’s why balloons are hardest to blow up at the start.
I figured that would happen since when you blow into a balloon, it is hardest at the beginning and gets progressively easier.
that explains why it's hard to blow up balloons at first, and it gets easier. and why when you let the air out, it moves faster at the last moment.
C. Balloons get easier to inflate as they get more inflated, so the smaller one is exerting more pressure and will force it's air into the larger one.
Also, as the balloon size increases, it takes less air volume to stretch each area of rubber. It's the volume vs surface area thing.
C, I learnt about bubbles in my physics class. The smaller bubbles have higher pressure than the bigger ones. Mathematically, P = (4S/R), P= pressure, S =surface tension, R = radius
I got it right, but not for Hooke's Law. My hypothesis was that although it looks it this is not a closed system, there's actually a very much bigger system acting against both, atmospheric pressure against which they are both miniscule. The larger balloon is infalated to a greater pressure than the smaller by a tiny tiny miniscule amount and resists air pressure slightly more than the little one, the little one is slightly weaker than the larger one and since air pressure is the same against both the air in it is forced into the larger one.
That must be why it's hard to begin blowing up a balloon and easier to continue.
I may be wrong, but I believe the reduction in thickness by the increase in radius is due to the balloon having a 2 dimensional surface, and like all 2 dimensional surfaces, area is Length * Width. Radius would be just either L or W.
This is something I frequently need to explain to customers when printing photos. If you want a photo printed twice the length and width, you'll need 4x the amount of pixels, since the area is now 4x larger.
I don't think that you need to "beat" Hooke's law to explain this result. Hooke's law tells us that the energy used to stretch the balloon is proportional to the square of the radius. In other words, it is proportional to the surface area of the balloon. The total surface area of the balloon is maximal if the air is equally partitioned between both balloons (assuming constant pressure) and minimal if the air is in one balloon, the other being empty. Therefore, Hooke's law predicts that, as the system tends to a state of minimum energy, one balloon will be filled, and the other one will be empty.
+Quozz yes, I think you're right. Eloquently put.
For a moment i thought 'B' but then considered the tension or force per square centimeter of rubber on the small balloon and how inflated it was comparatively and remembered that if takes more effort to force air into a deflated balloon than one that already has a substantial amount of air in it and concluded that 'C' was most likely. Nailed it.
I knew this would happen because it is harder to inflate a baloon than it is to keep inflating it
what about volume vs. surface area?
same volume of air added to small balloon vs. same volume of air added to large balloon
-> large area increase(large strech) vs. small area increase(small strech)
-> large force change vs. small force change
I don't think thickness of the rubber is the major factor but i'd love to be wrong as long as i hear about it.
C
Just think of inflating a very strong balloon (like a water balloon).
The start is always the hardest. You have to create a huge amount of pressure to get it to start. Every following breath gets easier. So, in a small balloon there is a higher pressure than in a larger (identical) balloon. Thus, a bit of air will go from the smaller one to the larger one. But then it only becomes more unequal and some more air will continue to flow in the same direction...
Is this why it can sometimes be hard to blow up a balloon at first but then once you get it going you can easily pop it?
Nothing is identical, just something with itself in the same time.
You'll have an intuitive understanding of this from blowing up balloons. It's way harder to blow in the beginning when it's small than when it's big
This can be explained another way:
Potential energy stored in the balloon's shape is directly proportional to it's surface area. The system is trying to reach a point of minimal energy, hence it will try to minimise the surface area. It's well known, that if you fix a volume V, the smallest possible surface area of any shape with volume V is the sphere. Hence, the system will try to reach a spherical shape.
I originally thought A was the correct answer. The idea was that because both balloons were stationary, they must had been in equilibrium with the outside pressure, and because of that they both had 1 atm of pressure inside. Clearly this is not the case because I did not factor in the rubber thing. But I have one question, if there's a different amount of pressure inside the 2 ballons, why are they both in equilibrium with the same outside constant pressure? I'm starting to think that in order to expand they need to also win the resistance of the rubber, but I really don't know. How many equilibrium states are there?
If you've ever tried to squeeze a balloon into two equal lobes, you will automagically understand that option C is the correct answer. Squeezing a ballon into two equal lobes was just one of those things that you could do, but not for very long, so I played a game with my friends where you had to keep the lobes in equilibrium for as long as possible.
Yeah! This is also the exact reason why I knew it was option C, because I remembered as a kid playing with balloons, squeezing it into two parts, and I remembered how the smaller bulge would always flow into the larger one, rather than them equaling out.
+MrGoat you guys are my kind of people.
Unfortunately there's very few of us in the world :(
That is a nice demonstration of an unstable system, like the inverted pendulum. If we started with two balloon inflated the same, then induce random pressure on either side, eventually, one win and get all the air (maybe explaining the many comments about rich becoming richer).
It also explain why it is so hard to start inflating a balloon.
That explains why it's so difficult to start a balloon.
Alternative rationale:
On the larger baloon the curve is less sharp, whereas the smaller radius baloon the curve is obviously more sharp.
We can look at the problem of hanging a mass on a horzontal cable. If we put a 1kilo mass on a cable that spans a 1 meter gap, and give the wire more or less slack we observe that more "flat" the cable is, the more tension is required to suspend the mass (the vertical component of tension is greater with a sharper angle).
Intuitively: A horizontal cable has no ability to pull up, whereas the extreme of sag is a verticle cable, where tension is all going to suspending the mass.
So the sharper angle on the smaller balloon means that the restorative force is GREATER for a given tension (and due to Bernoulli we know that air pressure is the same between the two).
I actually knew this, and guessed correctly. Saw it once before in baby lungs. The oxygen collecting parts of the lung works similarly, and in some cases babies are born with some of these not being in equilibrium, this collapses once and enlarges the other. The procedure to alleviate this won a Nobel prize I think.
To get it right I also imagine blowing up a balloon, it is much harder at the start. So smaller ones push back harder.
Hey there Steve. Could the phenomenon be due to a difference in Laplace pressures between the two balloons? When disconnected and at equilibrium, the internal pressure in the smaller balloon is greater than that of the larger balloon.
p_inside = p_atm +2("surface tension")/r (Assuming a sphere)
An analogy can be drawn to foam coarsening..
Then why did the exchange stop? The thinning effect was counteracted by the stretching of rubber having a limiting effect.
Otherwise the thinning effect would have led to the complete emptying of the smaller balloon
My reasoning was the following: balloons and bubbles naturally go to a spherical shape, in bubbles because of the surface tension of the liquid, and in balloons because of the elastic tension of the rubber. The balloons are linked by the more rigid hose, but still, they should naturally tend towards a single sphere, not an 8 or infinity like shape. The large balloon inflating further is closer to a sphere than both balloons reaching equilibrium.
Also, the smaller balloon is more curved, so a greater proportion of the tension is directed inwards.
the amount that the rubber has to stretch to fit extra air may be less as the the balloon gets bigger meaning that the smaller balloon will squeze the air more.
This is something we all have experienced if you've blown up a balloon. The first breath is much harder than the second.
This will be an excellent demo for students of thermodynamics, most of whom will go for Option B, as if this system is equivalent to two rigid containers separated by a piston.
That would be why it's harder to start blowing up a balloon but easy to finish.
What about atmospheric pressure? The larger balloon has more surface area for the outside atmosphere pressure to push against so you would think there would be more pressure in the larger balloon than the smaller balloon?
Hey, I actually guessed it right! I figured the same thing you did after explaining. The less-inflated balloon has more rubber contracting on the air inside forcing it into the already inflated-balloon.
I had somewhere same kind of question. The question was: which balloon would deflate faster in outer space (moon)
Like how short these sort of videos are, please don't change the format :)
I knew one of the balloons was going to completely deflate but I couldn't decide which one it was gonna be
Thank you for for not including stupid stuff in the end and ending your video right where you said
i figured it was c. similar to how you can warp balloons by squashing them while blowing them up. if the answer was a or b the the balloons would not be able to stay warped, but i know they can stay warped. also when you blow up the long balloons they fill up from one side and work their war to the other as they inflate more, again if the answer was a or b this would not happen the entire balloon would inflate at the same rate.
I chose C because of how a balloon is harder to blow up in the beginning than after a couple of breaths. I assumed that the smaller balloon wanted to contract more than the big one did.
Makes intuitive sense afterwards. Ballons are really hard to let air from your lungs in at first, but then it gets much easier.
I thought it was option B. good video! thanks for the knowledge
Excess pressure inside a balloon is inversely proportional to its radius. This means that less inflated balloon has more pressure. When the clip will be removed, the gas will always flow from high pressure to low pressure. Means the gas will flow from less inflated balloon to more inflated balloon. This makes less inflated balloon even lesser inflated and more inflated balloon even more inflated.
there is also the fact that the volume of a sphere rises faster than the surface, so one big sphere should energetically preferable to two spheres, even if the tension increased with the surface (to a degree)
+theholk yes! That's the "energy way of looking at it"
Apart from that, you had to be sure that before doing this experiment you first stretched *both* balloons to the same full blown size, because new balloons that are partially inflated still have very strong rubber. You should have noticed that a new balloon is always very hard to inflate, but once you fully stretch it, the next time that you inflate it, it has become softer. Anyway, the effect would be the same if you didn't do that, because the new partially inflated balloon would be stronger than the stretched one.
Force from surface of the balloon is proportional to the curvature of the surface, so the smaller balloon is more curved and will apply more force to the gas than the bigger balloon which is less curved. So I think the bigger balloon will get bigger and the smaller smaller.
I got this correct not from knowing Hooke's law or the math behind it, but from experience. Anyone who's ever blown up a balloon with their mouth has surely noticed how much more pressure it takes initially to get it started and felt the resistance go down as inflation gets started and proceeds. Similarly, while letting the air out of a balloon, you can hear the pitch of the valve flapping increase as it gets smaller, suggesting higher pressure. Having the small one get smaller was intuitive to me for these reasons.
makes sense if you think about trying to blow up a balloon. It's really hard to get it started
It's C because of the surface tension that is lower on the big baloon so it needs less force to get inflated
My fist instinct was the correct one and it was based on the fact that when one fills a balloon by 'lung power', it is hardest at the start and then becomes easier as the balloon gets bigger. I think the real explanation for the effect is that when the balloon is bigger, its curvature is smaller and this makes it weaker at resisting internal pressure.
I knew this one! As a kid I used to play with balloons a lot and this phenomenon really puzzled me. Now I know the physics behind it.
That is because.
P(Pressure)=F(Force)/A(Area)
Or we can say that Pressure is inversely proportional Area .
Means
If Area is greater than Pressure is lower that is in case of large ballon.
But opposite in smaller i.e., Area is lower but Pressure is more.
And we know Air travels from High Pressure area to Low Pressure area.
Thus all this happens.
See my explanation above...
I'd say C because it gets significantly easier to blow it up when you overcome the beginning
your Predict Observe Explain ( POE ) teaching method was awesome
I chose option C because hooks laws is linear but the increase in surface area is quadratic. Therefore the rubber will get stretched at a much faster rate than its ability to resist stretching. I remember when I was a kid balloons were always hard to get started but got easy very quickly once they filled so C matched my intuition.
+bmenrigh And damn, you explained the linear versus quadratic at the end so nothing new here.