Helium Balloon: Newsletter Challenge (07/06/10)

...If the car were accelerating at a rate of one G, the balloon would move forward at roughly the same rate ...

I just noticed that. Sorry, for me being so picky, but the froward acceleration of the balloon won't be even close to the car's acceleration.

The balloon acelerates faster than the car accelerates, if you use the same reference frame for each. (I.e., from the perspective of an observer on the street, the balloon moves forward more quickly than the car. You can see this effect by sliding a bubble level horizontally.)

Air resistance affects both a balloon floating upward in still air, and one floating forward in an accelerating car. At one G of forward acceleration the buoyant force (directed forward) on the mass of the balloon (which obviously includes the mass of the helium inside -- there is no need to separate the two) is the same as the buoyant force on the balloon under the one G of acceleration supplied by gravity when the car is still.

There is nothing in your cacluations that shows that air resistance in the vertical direction is different than that in the horizontal direction. Therefore MBs statement that the floating speeds would be comparable, seems correct, given his stipulation to ignore friction, etc.

The balloon acelerates faster than the car accelerates, if you use the same reference frame for each.

I didn't use the same frame reference for both. I think it was obvious since I calculated the inertial force that the balloon experiences. I cannot see how one could solve the problem if he would need to find out the forces applied to the balloon, being in the car's framework. (In such a case, one would also need to take into account the acceleration of the air in the car, etc.)

In #191, I explained why I treated the balloon and helium mass as separate: It was an engineering trick to bring down the calculation to the knowledge three factors that can be easily acquired: the ratio of the fluid densities (about 7 in our case), the weight of the balloon material and the volume of the balloon. You can twist the formula to match your own needs of course. But I think it's too difficult to get the total mass of the balloon+helium, as this would require a balance in vacuum. Or filling the balloon yourself and have a dial showing the mass passed to the balloon (I doubt gases are measured this way, though.)

Finally, you are right that the balloon would accelerate towards the front at the same amount as when rising up through the air, given that the car's acceleration equals g. But how do you know that the balloon rises with acceleration g in the first place? It's certainly wrong! After all, it is obvious that the acceleration needs to be a function of the density of the fluid in the balloon, isn't it? Would a balloon containing a vacuum behave the same? (in fact, in this case indeed the acceleration would be equal to g). What if it was filled with air?

I'm curious to know if I made a mistake in my maths (I saw too much criticism for the math, but no alternative formula so far), but what I know for sure, just out of mere intuition, is that the acceleration of the balloon won't be even close to the acceleration of the car.

Crystal clear or is that KrisDel^TM

I have to agree with Del, Newton rules the roost on this.The balloon will roll to the rear of the vehicle, at least initially, then when enough velocity has transferred from the roof of the vehicle to overcome the balloon's inertia it will roll back to the highest point which I must assume is the center of the roof, otherwise the balloon wouldn't have been there to begin with. this is assuming there are no air currents from the AC or open wndows to interfere with that behavior. if they exist, then all bets are off.

I think you cheated & have actually had balloons in your car before [JE has]

I am enjoying the various explanations of the differences in density, which are highly entertaining.

a car is not a sealed container, usually having vents in the back & an intake in the front

what is the difference in barometric pressure between inside & outside of a car

is there really a measurable difference from the from to the back seat caused by acceleration, that is greater than the difference caused by the car moving through the air?

I think everyone who is thinking along the lines of buoyancy and density gradients are getting lost in the weeds here. Yes all of those vectors are in play, however by the time they begin acting on the balloon (the vectors are very small in relation to the initial acceleration vector as well as the inertia of the balloon AND the air resistance which is quite large so all of these vectors take time to act.) the balloon and surrounding air will have reached what amounts to a steady state scenario again. They are failing to factor in the time vector in this.

Guys, remember this is a trick question.

You must not have kids who have traveled to birthday parties with helium balloons.

The balloons move to the inside of curves, and forward when the car accelerates, just as buoyancy effects would predict.

The balloon is snatched up and inhaled by my friend Ken in the back seat because he simply cannot resist asking "Are we there yet?" repeatedly in a falsetto voice just to annoy me.

wasn't this posted already a few years ago. Or a variant of while taking a corner.

CR4 reruns and remakes

Yes. Or it was in the general discussion forum.

Anonymous Hero is right. I've had helium balloons in a car before, and they move forward with forward acceleration. It's due to pressure/mass gradient of the air. Years of mechanical engineering say so.

There is no mention of the shape of said balloon.

If for example the balloon has a humanoid shape with certain "embellishments" it should be constrained by a seat belt.

Such as.

And hopefully the ballon does not deflate where you have to manual inflate otto the auto pilot like elaine did..

Best movie ever.

Luckiest 18 oz, of polystyrene on the planet.

Stupid Otto.

I have done this on a number of occasions, the balloons can be annoying flying onto the windscreen as you accelerate.

Air is heavier than the balloon so it wins over the balloon and takes its place in the back seat.

Acceleration is equivalent to gravity. As the mass of air in the car moves backwards in the car (down towards the gravity equivalent), the helium filled balloon, being lighter than the air, will float forward in the car and roll down the windshield to the dash board, as long as the acceleration is adequate.

would not everything (not bolted down) in the car move to the rear during acceleration. The driver is attached to the car and is pressed back in the seat. Is not the balloon "attached" to the car at the ceiling? (how much does the balloon itself weigh? and how much upward force is there before acceleration ?) and how does " friction " play a part during acceleration? At the "moment of acceleration", would not the balloon try to stay attached to the ceiling, but the weight of the balloon would move to the rear, just like the driver?. The helium in the balloon has some mass too.

Another way to look at it is to say that an un inflated balloon on the dash of the car is being held there by gravity. The acceleration is allot because my 65' Malibu SS has 533 gears, and the balloon would slide off the dash.

Depending on the mass of the balloon, mass of helium in the balloon and the air pressure inside the cabin, the balloon will move opposite to the direction of acceleration. This is due to inertial resistance of a free standing body. Not knowing the total mass of the combined cabin air pressure, balloon and helium or the rate of acceleration there is no way to determine the distance it will move given the known parameters of the question. The occupants of the vehicle will also move in the opposite (positive g direction relative to forward momentum) as the balloon. In reality, the balloon and the vehicles occupants will be attempting to remain stationary with respect to the accelerating force of the vehicle until equilibrium is achieved. So it is an illusion. The vehicle will move first, not the balloon or the vehicles occupants.

Being lighter than air inertial forces will make it move forward as the heavier air mass tries to remain static.

Bioramani

A couple of thought experiments:

1) A "large" centrifuge with balls of different sizes and densities. Add a big bag of smaller, lighter balls.

2) A tractor trailer with the same collection of balls. Throw in a couple of bowling balls to really see the effect.

In both cases the smaller, lighter balls, including the bag of balls (think density), move in the opposite direction of the acceleration. This is just a simple application of Newton's First Law or Inertia. My students love this one!

You describe a static state. --- stationary --- You also change the confinement of the balloon. --- The car being on slope allows the balloon to travel up at an angle farther than where the balloon already is. --- Your perspective does not make it clear, considering these two points.

Welcome one2playwitt,

[p]

You must be on a mac or using a browser other than IE or firefox

[p] equals a hard return

enjoy

Yes, I am using a helium filled browser. I tried to get rid of it, but when I ran home away from it, it beat me to my house. Perhaps I should run backwards away from it ... I will keep you posted.

It stays in place.. for the most part.

Just like the cabin full of non moving air.

You don't drive around with the windows open..

I used to deliver balloons in H.S.

sometimes by the thousands..

It seems to me that there are multiple forces acting on the balloon from different effects and the final motion will be determined by which of these effects is greatest.

First, the inertial aspect of the balloon would cause it to want to remain still relative to the earth, thus causing the balloon to move backwards relative to the car frame of reference.

Second, there is the drag force of air on the balloon resisting any motion relative to the air. This force is likely negligible for this experiment as I don't believe the air in the car will have significant distance or velocity of motion.

Third, there is the friction force of the ceiling and the balloon surface. This will probably cause the ball to roll instead of sliding in any motion it takes but will likely have no other effect.

Fourth, there is the force on the balloon due to any density difference in the fluid in which it is immersed. This is the crux of most of the discussion above. If the air inside the vehicle compresses enough due to acceleration so that the air in the back of the car is more dense than the air in the front of the car, then there will be a forward force on the balloon due to the pressure gradient. My intuition says that the change in density in the air in the vehicle is going to be very small (as I don't usually feel a breeze rushing backwards during acceleration, though I do drive a Geo and don't experience much acceleration) so I think this force will be small as well.

My net thought is that the inertial effect is going to outweigh the "buoyant" effect due to a pressure gradient, so I think the balloon will roll backwards in the car.

Another point with some of the discussion of buoyancy and shifting the direction of the vector for the buoyant force is that gravity is applied to all objects equally (as long as they have mass). However, the acceleration of the car is applied only through the tires of the car and through the structure of the car. So at the initial moment that acceleration starts, there is force exerted on the car but NOT on the balloon. The balloon is still seeing a fully downward gravitational pull, so that its buoyant force is still straight upwards. Once acceleration has started, the air in the car reacts to the moving structure by moving along with the car, exerting force on the balloon (due to basic drag air friction) and taking the balloon along with it. At this point the balloon starts seeing acceleration both forward and upward and thus the buoyant force direction could change. However, in the very initial moment of acceleration the inertial effect of the balloon will move it backwards before the bouyancy force changes directions.

You've got it exactly right...in the very initial moment of acceleration the inertial effect of the balloon will move it backwards...relative to the car. It's actually close to standing still with respect to the road at that initial moment..

Yes. I think psoderman commented on this earlier. It takes about 3 milliseconds for the balloon to respond to the pressure gradient fully.

The balloon moves forward as the denser air in the cabin moves rearward. The denser air resists motion.

This same effect can be felt with extreme levels of cabin heat or cooling. Temperature gradients can be felt as the cooler dense air sloshes around the cabin.

Assuming the windows are closed, and the A/C fan is not circulating air, and that the balloon is free to roll around on the ceiling, it will, because of its mass, try to stay in one place in space; therefore as the car accelerates forward, the balloon will be left behind, and as perceved from inside the car, move towards the rear of the car.

Its motion will be retarded by the friction against the ceiling of the car, which will cause the balloon to roll. It is also retarded by the air in the car, which essentially accelerates with the car, so the motion of the balloon relative to the car is pretty slight.

Its motion (relative to the inside of the car) will also be retarded by the curvature of the ceiling, as the balloon wants to rise to the ceiling's highest point.

__________________________________________________________

You don't have to be a rocket scientist

to be a rocket scientist.

The balloon will roll forward on the ceiling towards the windshield, because the heavier air in the car will be pushed toward the rear of the car by the car's acceleration from a dead stop. Although the balloon is subject to the same rearward acceleration as the surrounding air, the greater mass density of the air forward of the ballon will enable that air to flow around and behind the balloon displacing the ballon towards the front of the car. In other words while the car is accelerating, its as if the car were parked on a hill with the front of the car higher than the back and the ballloon simply rises to the highest point inside the car - at the front edge of the ceiling.

All the discussions about pressure/density gradients are really addressing the fundamental question of why buoyancy exists, i.e. why things like helium balloons appear to defy gravity. Given that buoyancy does exist, the balloon moves forward when the car accelerates for the same reason that it moves upward if you let go of it outside; it's defying the prevailing "gravitational" force.

I should have put "defy" in quotes. I just meant that the buoyancy force on an object acts in the opposite direction to its weight, and that's all you need to know to answer the challenge question.

But to add my voice to the general pressure/density debate:-

The pessure at a given depth in a fluid is the wieght of the column of fluid above that depth divided by the cross-sectional area of the column. (It doesn't matter how wide the column is; the cross-sectional area cancels out in the calculation and you get the same answer for any width.) Hence the deeper you go, the higher the pressure, so there's a pressure gradient whether or not the fluid is compressible. If you integrate the pressure over the surface of an object - allowing for the gradient - you get an upward force equal to the weight of displaced fluid (Archimedes principle). This is easiest to see for a cylinder with its axis vertical, but it works for any shape. There may or may not be a density gradient too, but it's the pressure gradient that produces the buoyancy effect.

In summary: Archimedes principle and pressure gradients are two sides of the same coin. For the purposes of the challenge question it doesn't matter which view you take.

The balloon will do nothing. The scenario was such that the balloon was "pressed up against the center of the ceiling". It is not floating free.

Presuming we are not looking for molecular-level explanations and that friction exists, the balloon (a pear-shaped object) will remain in contact with the roof of the car while the bottom swings toward the rear (effectively tipping or rolling backward) until that motion is stopped by the upper side surface of the balloon contacting the roof liner. This motion will be more pronounced if there is a string or ribbon attached to the bottom of the balloon. Think of an inflatable punching clown with sand or water used for ballast.

If the car was very fast, both friction and upward force could be temporarily overcome, skidding or tumbling the balloon to the rear where it would take up a new position; from which it may make inch its way back to the high point in the car as various forces act on it.

"An object at rest tends to stay at rest..." The balloon is at rest inside the car and upon acceleration it will appear as though the balloon moves to the rear of the car, but it is the car changing position around the balloon as the balloon sits still. At least that is the way it appears every time I have balloons in my car...

The balloon moves forward because the air, which is heavier and trying to remain at rest moves back relative to the vehicle which displaces the balloon forward.

The most fascinating part of this for me is that people can so confidently state they've experienced what they hypothesize to be true, when only one or the other can be.

The balloon responds to local air pressure, which is uniform horizontally while the car is stopped. The car's acceleration will create a pressure gradient with higher pressure at the rear of the passenger compartment. The balloon will move forward in the direction of the car's acceleration staying pressed to the ceiling. Another way to think of it is that gravity and acceleration are interchangeable. A pressure gradient is induced in the car by acceleration just as a pressure gradient in created in the Earth's atmosphere due to gravity.

The balloon should move to the front of the car; toward the windscreen as your car moves forward. The reason is that the Helium filled balloon which is lighter than air and less dense floats in a air medium that is denser than it. Air being almost incompressible within the car cabin causes the balloon to move forward as the movement of the car forward increases the density and pressure at the back of the car. The 'pressure' in the front lowers simultaneously causing the balloon to gravitates towards it.

Eric10

I'm sorry, but air being incompressible means it will suffer NO change in density when subject to pressure ... apart from this little issue I agree with what you say.

> I'm sorry, but air being incompressible means it will suffer NO change in density when subject to pressure.

Oh, c'mon, now -- everyone KNOWS air is incompressible!

"This is Wolf Blitzer, reporting live from Miami, Florida, where, this morning, thousands of dead swimmers mysteriously washed up on shore, wearing peculiar steel bottle-like contraptions on their backs. Several survivors of what appears to be a strange new cult -- which I'm told involves someone or something called "Scuba" -- claimed to have actually achieved the compression of air [shakes head, smiling], which they claim will enable them to breathe underwater for 10 minutes or even longer, from a tank clearly large enough to hold at most 6 or 7 lungfuls of air...." [gazes out at washed up bodies] "Poor, deluded fools....[sighs] Back to you, Katie."

>;-)

The balloon at rest, will rotate (due to skin friction of the ceiling of the car unto the balloon) and roll back due to the inertia effect (object at rest tend to stay at rest).

Depending on how quick the acceleration,

for instance a racing start acceleration will cause the balloon to roll cack and wedge itself between the rear window and the parcel tray of a sedan shape car and stay there. The friction of the window with the balloon is high enough to keep it there.

for a normal acceleration the balloon will roll back due to skin friction and hit the angled rear window and bounce down unto the rear seat and then fly up retaining its position once the car stabilises its speed. The lighter density balloon tends to find the highest point in the car.

Regards

MT

Have a look at THIS.

Its an extremely dry day and the static electricity build up of the rubber balloon sticks to the ceiling of the car due to the attrctive nature of the material lining the car ceiling so in effect the balloon stays where it is.

If the balloon occupies 25% or less than the general volume of the car cabin above the general line of sight of the seats it will move forward due to reasoning as given in post #60 and the like

If the balloon occupies more that 25% of the above parameter, then it will still move forward due to reasoning given in post #1 and the like

However not purely so in either case. It will move only a percentage as much as it would have for the above arguments, as the car is also traveling during - (note the original question ) - the acceleration phase. Then when the car stops accelerating at whatever /sec/sec gathering moment,,, and holds a steady speed, then the balloon will continue to move very slowly forward until touching the front windscreen for the same reasons as set out in #60 and the like

Well that's onlywhat i think,,
I must admit i have no engineering or mathematical qualification for what i say, it only comes from a lifetime of familiarising fluids and solids properties for using in 3D modelling animation.

sorry i was too late in editing my mistook.
Posts #1 and #60 are essentially the same. a smaller balloon would be subject to those ( pressure gradient ) influences, a larger balloon due to bouyancy as called up by other posts

---

I take it "Good answers" are what participants give to one another, but does the poser of the original question identify which answer is closest to the the final revelation?
Ive been following a few now and sometimes i wonder if some of the discussions have a sounder argument than the supposed real answer, like the recent Funny Space Drag one. How do we know the offical answer is right?

looking at past "questions", i notice there's more than one that don't have the official answer posted back, as promised by the date claimer? where is the Slick Roads answer, for example? wasn't an answer supposed to be up for this one by now?

no takers?

Hello, anybody home?

nah just a room full of vacant people

<sigh>

I would have to say the balloon will move forward due to the fact that it is lighter than the air which will momentarily compress backwards as the car accelerates. This is a take off from the trick question about holding a helium balloon on a string in a car as it makes a hard left turn. Of course one needs to consider as it is against the ceiling, the friction will only allow it to rock or roll slightly.

Easy, intuitive answer without resorting to pressure and density gradients (which are in fact related through energy functions)...

The helium balloons will be forced by inertia toward the back of the car. The heavier air will be forced back harder, displacing the balloons toward the front. End result...the balloons move to the front of the car. If you don't believe it, try it.

The balloon will move in the opposite direction of vector sum of two vectors -- Vector 1 being 1 g vertically (down), and vector 2 being (some number) horizontally rearward. With constant (short lived) acceleration, the balloon will move towards the spot where an extension of the tail of the vector sum intersects the roof line. That is, forward of its starting of its starting position.

No-one seems to have thought about potential energy (PE). You can ignore compressibility, pressure gradients, etc. and explain many flotation phenomena by PE.

A "system" in an enclosure will seek a minimum-PE configuration. When the system consists of a fluid in a container under acceleration, the fluid wants to be as close to "down" (relative to the acceleration) as possible. Thus a liquid in a glass will have its top surface horizontal. Any change from this requires kinetic energy to displace some of the liquid "up". Minimum PE for a liquid and a submerged object (a steel ball, another liquid, a balloon, etc) has the denser stuff on the bottom and the less-dense stuff on top. Again, changing that situation would require application of kinetic energy.

The buoyant force on a submerged object of volume V and density D is F = a*V*(D0-D), where a is total acceleration (g plus whatever other acceleration there is), and D0 is the density of the liquid medium. If both the medium and the object are incompressible this force is constant, regardless of depth. The PE is the product of F and depth.

Helium balloons, like flames, don't "rise" in the space shuttle because there's no up or down in zero-g conditions. On Earth, He balloons rise to whatever is "up" as defined by the total acceleration (e.g., forward and up in an accelerating car). As long as they are not as high as allowed by constraints, they are displacing denser air that wants to fall, so the system is not in a minimum-PE state. In moving as far as they can in the acceleration direction they minimize the PE of that particular accelerating system. The tiny change in air density from top-front to bottom-rear is almost totally irrelevant to the behavior of the balloon, as long as the density is greater than that of helium. But that density gradient is itself in a minimum-PE configuration with the density highest at the bottom-rear, and if there were water in the car it too would find its way to the bottom-rear.

so...

what does the balloon do then, according to your spiel??, ( that ends up relying on density anyway )

Question:

You're in your car at a stoplight, with a helium-filled balloon pressed up against the center of the ceiling. When the light turns and you accelerate, what does the balloon do and why?

Answer

The balloon will go forward towards the windshield because it is lighter than the air surrounding it which will compress towards the rear.

Leonard Neumann

An interesting discussion. I think most of us are agreed that the balloon will move forward in the car during acceleration, but the explanations have been all over the map. Perhaps that is because there are several factors to consider.

As I explained in note #65, the car's acceleration, which is analogous to a gravity field, induces a pressure gradient in the cabin air with higher pressure toward the rear of the car. We could work with density since density and pressure are related along with temperature, but the density gradient is very small and the temperature gradient in unknown. That is why, to the first order, we compute the pressure in a column of water at depth h as rho*g*h, where rho is the nominal water density, g is gravity, and the pressure is taken relative to the pressure at h = 0. The pressure gradient is caused by the weight of water above the point of interest. At great depths we would have to integrate to account for the change in density. In the helium balloon problem that is not necessary, we can use a nominal density. Besides, pressure is force per unit area which is what we want so as to use Newton's laws, which requires that we find all the forces acting on the balloon to compute its motion.

The difference in pressure from one side of the balloon to the other causes a buoyancy force that tends to push the balloon toward the car's acceleration direction. Incidentally, the buoyancy force does not depend on the density of the gas in the balloon; a solid ball of the same volume would experience the same force.

There is also a friction force at the top of the balloon as the car ceiling tries to slide over the balloon. That force is weak because of the weak normal force, is smaller than the buoyancy force, and is in the direction of the car's acceleration.

The buoyancy force acting through the balloon center and the friction force at the top create a couple which, if the car is moving to the left, causes a clockwise rotation of the balloon. It rolls toward the front of the car (neglecting effects of strings and odd balloon geometries). If the ceiling friction was zero, the balloon would slide forward.

Someone asked what would happen to a steel ball on the floor during acceleration. Assuming the same volume as the balloon, the buoyancy force would be the same as the balloon, but the friction force would be large because of the large weight of the steel ball. The two forces would create a couple that rotates the ball clockwise, and it would roll toward the rear of the car.

The comments above are for a coordinate system fixed to the car. The problem is slightly more complicated with a coordinate system fixed to earth because the air in the car moves also, but we would get the same answers.

The balloon will appear inclined with the balloon top facing the front and the balloon bottom towards the rear of the car. As the Car accelerates, the Helium balloon will tend to remain in the same spot with a backward incline caused by the raising of balloon bottom which will try to touch the car ceiling. There are two reasons for this. The Balloon top will be kissing the ceiling due to Helium bouyancy causing surface friction between the balloon surface and the ceiling surface. This will not allow the balloon to rub the ceiling surface or to cause the ballon to move backwards as the car is accelerating. The lower part of the balloon will be pushed towards the rear end of the car due to the slight air pressure differential between the front and rear end of the cabin. As this force is not strong enough to over come the surface friction between the balloon top and the car celing, the balloon will tend to remain in the same position even as the lower end of the balloon will get pushed towards the rear by the air dynamics.

The balloon will move forward, because it has less inertia (per unit volume) than the air.

#102 = most concise correct answer so far

As I see it, the pressure upward is greater than the pressure downward on the bottom of the balloon, so it must accelerate upward.

Rotate the diagram and substitute the pressure due to acceleration to find the horizontal component.

http://www.epa.gov/chp/

.

http://www.google.com/search?hl=&q=combined+heat+and+power+efficiency&sourceid=navclient-ff&rlz=1B7GGLL_enUS387US387&ie=UTF-8&aq=3&oq=Combined+Heat+and+Power

I missed

my reply was meant for this thread

http://cr4.globalspec.com/thread/57026/Sterling-Engine-to-Utilize-The-Wasted-Heat-Energy-in-Thermal-Power-Plant

Carry on

Car will not have the same acceleration, proportionate to the size of balloon, as the wheel pressure to ground reduce, the wheel may spin, instead of moving the car forward.

Gopalakrishnan.H

2-18, TVS nagar,

Hosur, India.

+91 9449566102

hgtechnical@gmail.com

This depends on the mass of the balloon and the friction force on the ceiling!

The helium filled balloon displaces some air - if the mass of the balloon and the included helium is greater than the mass of the displaced air the balloon will slip to the back of the car, if the mass of balloon and helium is less than the mass of the displaced air the balloon drifts to the front of the air; if the masses are equal there will be nothing happend.

if the friction force of the pressed balloon on the ceiling is greater than or equal to the acceleration force the balloon stays between the ceiling and the pressing hand.

otherwise the balloon moves to the direction described before.

if there's no pressing hand the balloon is touching the ceiling but is not pressed and so the balloon is beginning to rotate if the car is accelerated!

correction - if there's no pressing hand it depends on the mass of the balloon and included helium; if theses masses are greater than the mass of displaced air the balloon can not touch the ceiling and so moves to the ground.

but this has no effect to the result, in this case the balloon slips to the back, if the balloon floats (same mass of balloon with helium and displaced air) the balloon will stay somewhere in the car and if the mass of filled blloon is lower than the balloon moves to the front.

the accelleration of the balloon should be the difference between the mass of displaced air and the mass of balloon (surface and helium) as part of the mass of displaced air multiplied with the accelertion of the car; if the total mass of the balloon is greater than the mass of displaced air the mballoon moves backward, otherwise the balloon stands still or moves forward:

aballoon = acar*(malloon + mhelium -mdisplacedair)/mdisplacedair

What a bunch of variables!! The amount of helium in the balloon, the coefficients of friction of the headliner and the balloon, the rate of acceleration. So, ignoring all that.. the balloon has so much holding it in place, (the surrounding atmosphere, it's tiny mass versus friction) that it won't move significantly in any direction.

The pressure gradient is insignificant, and only applies if the conditions in the car are static. Newton's laws governing inertia affects the air in the cabin as well (the air has mass, too.) If the pressure gradient were significant, it had to result from the air "rushing" from the front of the cabin to the rear, which would have "blown" the balloon to the rear, anyway. The correct answer is - the inertia of the balloon will cause it to stay at rest while the car moves forward. With respect to the car, the balloon will move rearward. With respect to the road, the balloon will move forward, based on the amount of friction between the balloon and the roof of the car.

Since the air in the car is heavier than the helium filled balloon the balloon will move forward when the car accelerates.

The balloon will go to the front of the vehicle. The air is heavier than the helium so when you accelerate from the stop light the heavier air will rush to the back of the car pushing the balloon to the front. Just like the coffee does in your cup.

Where does the coffee go when you hit the gas???

FORWARD

THE AIR IN THE CAR IS DENSER THAN THE BALLOON. THE AIR GOES TO THE BACK OF THE CAR, FORCING THE BALOON FORWARD

The latex that the balloon is made of is denser than the helium inside it AND the air it displaces. See my post #132.

The balloon is lighter than the air in the car. The air does not want to stay in one place when the car "pushes" the air forward, therefore the air pushes to the back of the car, and the lighter balloon is therefore displaced toward the front of the car.

Jeez Louise!

I just bought a nice "Happy Birthday" mylar balloon at the Dollar Store. I have a cloth headliner and it has a sizeable ridge at one point. I let the balloon rise against the headliner, not at the ridge, with about 6" of slack.

A good acceleration (Old fogey "good", not teenager "good") rolled the bottom forward. The contact point did not move. A sharp braking did the reverse.

I then took up all the slack so the balloon was just below the headliner. Same effects except the whole balloon now tilted withe top free to move. Looked about the same angle, though.

Turned on the dash blowers to low and got the same effect roughly toward the back.

Put the balloon back on the headliner, turned on the blower to low and got pretty much the same effect, except the top slowly jiggled backwards. At higher blower speeds, the top skipped back quite well, even "hopping" the ridge.

Don't nobody do no spermentin' no more?

Well, this seems consistent with what many of us have said. During the car's acceleration a horizontal pressure gradient will be established that tends to push the balloon toward the front of the car via buoyancy. That motion is resisted by friction at the ceiling. If the acceleration is not strong enough to overcome the static friction between the balloon and ceiling, the balloon will just incline with the bottom leaning forward. With the balloon held below the headliner by a string, it will incline with the top moving forward. The blower part of the experiment doesn't have much relevance to the question at hand because the establishment of the pressure gradient does not require a flow of air - the gradient is established at the molecular level just as it is in a gravity field. That is, the balloon isn't blown forward by the air, it floats forward just as a cork would float upward in water.

I disagree. Head is head. And, there is a movement of air relative to the car when I accelerate, else I'd get no density gradient. It's just not continuous.

OK, you are talking about the transient events, which I hate to get into because this group is having enough problem with the steady-state solution. Besides, the transient events occur in a very small fraction of time, but here goes. Consider a minute cube of air against the back window at time = 0 with the car about to accelerate right to left. At time t = 0+ the rear window accelerates and pushes the cube of air. That pressure disturbance radiates throughout the cabin at the speed of sound. The cube of air feels the force from the window and the resistance of all the cubes of air in a row between it and the front windshield along the line of action. This is analogous to a cube of air at the earth's surface that feels the weight of all the air in a column above it. Anyway, the cube is squeezed, the molecules move slightly closer together and the pressure and density increase. The next cube to the left feels the first cube push on it and the resistance of the cubes to its left. The pressure is slightly less than the first cube because there is less air to its left. It squeezes...and so on as the force propagates to the front window. This process does not require a convection of air, it only requires a propagation of forces at the molecular level, which travel at the speed of sound. If the balloon is moved by this disturbance, it would travel so little and happen so quickly that I doubt it could be seen by a human eye. But who knows, maybe is does shake - transient events are tricky. The original problem, I believe, was addressed to the steady-state problem of balloon motion under steady acceleration. For that problem, the balloon moves forward to the front of the car unless stopped by friction.

I agree with you. We are splitting hairs. I am not arguing convection. I am arguing simple experimentation.

The balloon rolled forward on the bottom because the friction with the ceiling dragged the top backwards.

Actually, the friction kept the top stationary (in the reference frame of the car).

If the windows are rolled up when you accelerate then it would seem that the balloon would accelerate with the car, because you are accelerating the volume of air inside the car that the balloon is floating in. The balloon may drift towards the rear of the vehicle as you move but probably not a whole lot, as any drift would depend upon a certain amount of mass, of which the balloon would have very little.

The balloon will roll forward on the roof of the car. The cause is that the colder air that sits lower down near the floor of the car, flows back, due to inertia, toward the rear of the car as the car accelerates. This causes the lighter warm air in the upper area of the cab to move toward the windscreen, pushing the helium filled balloon with it.

The reason it rolls forward is because the balloon is round with the heaviest part hanging directly under the apex of it's curvature like a plumb bob or a pendulum. When the car accelerates forward the mass at the bottom of the balloon being heavier then the top shifts the center of gravity and the bottom of the balloon is now being "drug" by the upper half. The helium in the balloon now starts to act on the upper front side of the balloon which is an arched fulcrum that counter acts the G force and causes the ballast to shift forward as the G force drops off and sends the whole thing tumbling forward head over heals until it finds equilibrium again. Simple really. This is ignoring the effects of static though.

This is just like liquid in a tanker truck without ballasts. As the liquid is compressed against the rear of the tank until the Gs drop off. Then the liquid goes flooding forward. In this case the back wall of the tank is replaced with the helium acting on the curvature of the front upper inside of the round ballon. It isn't untill the G force drops off that the ballon starts to tumble forward.

post #147 is mine as well

Oh yea one other thing. The only reason that the balloon is effected by the G force is because it is in contact with the roof of the car.

And yet another after thought. If there is a string attached to the balloon it will act to dampen the motion and in this case the balloon will not likely tumble but will instead jitter bug across the ceiling. Depending on how taught the balloon is there will all so be a bouncing effect as well. Of course the variables in this scenario are not stated. Such as weight to gas ratios and the percentage of the max volume of said balloon displaced by the gas. Nor does it state the amount of weight that the balloon will be out of balance by. ie the knot. The presence of a string as I have mentioned changes the dynamics quite a bit too. So how the balloon reacts will vary by degree but the driving mechanism is the same.