# A balloon in a car...

• September 14th, 2006, 07:32 PM
william
A balloon in a car...
Here is a little question;
Let's say you are driving along in your car on a cold-autumn day. You have a passenger in the center of the back seat that is holding a helium-filled balloon in front of them by the string. You make a sharp left turn around a bend in the road.

Which way does the balloon move relative to the car?

Cheers,
william
• September 14th, 2006, 07:39 PM
jeheron
moving balloon
Until such time as a force is applied to the balloon its motion will continue in a straight line. Therefore, until the string tied to the balloon begins to drag the balloon around the left bend it will continue travelling in the same direction. Note that if there is air present in the car, as the car turns, and the air in the car follows, it to will exert a force on the balloon.

There are probably an incredible number of other effects on the balloon i have missed.
• September 14th, 2006, 08:11 PM
william
I should have made this a 'poll' question (oops). The answer is either right, left, forward, or backward. And remember this is relative to the car.
• September 14th, 2006, 08:30 PM
L.E.A.P.
Forward. The baloon will crash on a window. The car moes but not the air inside of it.
• September 15th, 2006, 02:20 AM
anandsatya
According to inertia of motion a body will continue to move in a straight line with uniform velocity unless an external force is acted on it.

So if the car turns left the ballon will move to right relative to the car.
• September 15th, 2006, 01:00 PM
william

Billco PM'd me the correct answer. I would like to see someone else do the same. It seems billco is very good at seeing the subtleties.

Cheers,
william
• September 15th, 2006, 02:13 PM
william
'Its All Relative' also PM'd me the correct answer.

Good work!

Cheers,
wm
• September 15th, 2006, 02:20 PM
Quote:

Originally Posted by william

Billco PM'd me the correct answer. I would like to see someone else do the same. It seems billco is very good at seeing the subtleties.

Cheers,
william

Subtelties? - no I have forgotten what they look like, aren't they the writing on foreign films? LOL.
• September 15th, 2006, 02:32 PM
Jon
Quote:

Originally Posted by L.E.A.P.
Forward. The baloon will crash on a window. The car moes but not the air inside of it.

So, if the air inside doesn't move, then a fly, flying inside the car, is flying at up to 70 mph?

Relative to the car, the air outside doesn't move. If the air inside did the same, we would have the same WOOSH feeling as we drove, and bad hair-days would be more prevelent. Not that they aren't already for some of us :( .

J0N
• September 15th, 2006, 03:07 PM
L.E.A.P.
I dont get it... The answer? Air does not move inside a car, no? I think it doesnt...
• September 18th, 2006, 12:04 PM
william
Hi everyone,
Well... the correct answer still hasn't been posted. I guess I'll give it a few more days, then I'll post the answer and let this thread sink to the bottom of the 'stale-thread sea'.

Cheers
• September 18th, 2006, 12:54 PM
river_rat
Hehe - I think i got the answer

Spoiler : The balloon moves towards the left if the car turns left since helium is lighter then the air in the car so will be displaced.
• September 18th, 2006, 12:58 PM
william
Thank you!!!

Cheers,
william
• September 18th, 2006, 04:25 PM
L.E.A.P.
I don't get it... When you move inside a car that is going at 100mp/h, if the air inside the car is relative to the air outside the car, we would feel a draft; would'nt we? The baloon stays where it is! It doesn't move.
• September 18th, 2006, 04:55 PM
billiards
From my experience of being in cars that make sharp left turns I'd say that it would move to the right relative to the car.
• September 18th, 2006, 04:58 PM
L.E.A.P.
But I most certanly think the baloon does not move!
• September 18th, 2006, 05:01 PM
Bettina
I don't exactly remember why but if you had a load of different objects inside the car, people, beach balls, marbles, and whatever... and only one helium balloon, all the objects would move right on a sharp left turn.... except the helium filled balloon.

Bettina
• September 18th, 2006, 05:11 PM
L.E.A.P.
Hi Bettina, welcome to the forum!

I hope you will enjoy yourself here!

As I was stating previously, The baloon does not move relatively to the air outside the car, wich would come to your answer...

:x Should've sticked to my first thought.
• September 18th, 2006, 05:12 PM
william
Maybe this will help a bit...
If the car is sitting still, gravity is pulling everything down - except the balloon.

Think buoyancy....

cheers
• September 18th, 2006, 05:17 PM
L.E.A.P.
Quote:

Originally Posted by Wikipedia
The buoyant force is equal to the weight of the displaced fluid.

I have problems understanding this... Let's say the air is a fluid... Can anyone explain it to me?

Quote:

Originally Posted by Wikipedia
Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum. Suppose that when the rock is lowered by the string into water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs will be 10 newtons minus the 3 newtons of buoyant force: 10 − 3 = 7 newtons.

Could we do the same but with air?
• September 18th, 2006, 05:25 PM
william
Quote:

Originally Posted by L.E.A.P.
Quote:

Originally Posted by Wikipedia
The buoyant force is equal to the weight of the displaced fluid.

I have problems understanding this... Let's say the air is a fluid... Can anyone explain it to me?

Well, actually, the air is a fluid. Yes, yes, I know someone will say it's a gas, but fluid dynamics still apply.

Seems like someone said everything in the car will move to the right (via Newton). This includes the air inside the car. We'll then have a density gradient provided the windows are shut (which they should be on a cold-autumn day :wink: ). The balloon, being a buoyant object, will move towards the less dense region (i.e., left).

The next time you have a helium-filled balloon handy, take it for a ride in a car and see what happens.

Cheers,
william
• September 18th, 2006, 05:31 PM
L.E.A.P.
Thanks, that clarified it, because the air to the right of tha car is compressed by the right side of the car's enterior, that makes sense.
• September 18th, 2006, 06:43 PM
Bettina
Quote:

Originally Posted by william
The next time you have a helium-filled balloon handy, take it for a ride in a car and see what happens.

Cheers,
william

I'm sure you can explain it to the police.... :wink:

Bee
• September 18th, 2006, 09:01 PM
billiards
Quote:

Originally Posted by william
Quote:

Originally Posted by L.E.A.P.
Quote:

Originally Posted by Wikipedia
The buoyant force is equal to the weight of the displaced fluid.

I have problems understanding this... Let's say the air is a fluid... Can anyone explain it to me?

Well, actually, the air is a fluid. Yes, yes, I know someone will say it's a gas, but fluid dynamics still apply.

Seems like someone said everything in the car will move to the right (via Newton). This includes the air inside the car. We'll then have a density gradient provided the windows are shut (which they should be on a cold-autumn day :wink: ). The balloon, being a buoyant object, will move towards the less dense region (i.e., left).

The next time you have a helium-filled balloon handy, take it for a ride in a car and see what happens.

Cheers,
william

I can see how that would work in theory, the ballon would still move to the right in the initial "gust", but would be pushed back if a sufficient air pressure gradient were established in the car. I think the car would have to be travelling really fast and take a really sharp turn for a sufficient air pressure gradient to develop.
• September 18th, 2006, 10:06 PM
MacGyver1968
Quote:

Originally Posted by billiards
I think the car would have to be travelling really fast and take a really sharp turn for a sufficient air pressure gradient to develop.

Quote:

Originally Posted by Bettina
I'm sure you can explain it to the police...

(Welcome to the board Bee)

Oh yeah....I can just imagine me trying to explain this to a Texas State trooper..It would probably go something like this:

Trooper: (tapping on the window with his mag-lite) Licence and registration please............Mr. MacGyver, can you please tell me WHY you felt the need to drive over 3 TIMES the posted speed limit?

Mac: um....er....

Trooper: ...and Mr. Macgyver, can you please explain why you decided to do a 4-wheel powerslide through the drivethru of this here Dairy Queen, narrowly missing my patrol car, and that group of mentally challenged children over there?

Mac: um...er...

Trooper: ...and Mr. Macgyver, could you please explain to me WHY you have a helium balloon tied to you beltbuckle, clearly impeding your vision?

Mac: Well..sir...It was for a science experiment.

Trooper: A science experiment ehh?? I suppose all these beers in the floorboard are for your "science experiment" too?

Mac: Well..actually they were..I was dropping them off a cliff to see which way the bubbles went.

Trooper: Please step out of the car, sir. (pulling out the handcuffs)
• September 18th, 2006, 10:12 PM
billiards
Quote:

Originally Posted by MacGyver1968
Quote:

Originally Posted by billiards
I think the car would have to be travelling really fast and take a really sharp turn for a sufficient air pressure gradient to develop.

Quote:

Originally Posted by Bettina
I'm sure you can explain it to the police...

(Welcome to the board Bee)

Oh yeah....I can just imagine me trying to explain this to a Texas State trooper..It would probably go something like this:

Trooper: (tapping on the window with his mag-lite) Licence and registration please............Mr. MacGyver, can you please tell me WHY you felt the need to drive over 3 TIMES the posted speed limit?

Mac: um....er....

Trooper: ...and Mr. Macgyver, can you please explain why you decided to do a 4-wheel powerslide through the drivethru of this here Dairy Queen, narrowly missing my patrol car, and that group of mental challenged children over there?

Mac: um...er...

Trooper: ...and Mr. Macgyver, could you please explain to me WHY you have a helium balloon tied to you beltbuckle, clearly impeding your vision?

Mac: Well..sir...It was for a science experiment.

Trooper: A science experiment ehh?? I suppose all these beers in the floorboard are for your "science experiment" too?

Mac: Well..actually they were..I was dropping them off a cliff to see which way the bubble went.

Trooper: Please step out of the car, sir. (pulling out the handcuffs)

Hilarious :lol:
• September 18th, 2006, 10:17 PM
william
Trooper: ... and Mr. MacGyver1968, just what do you plan to do with this hammer and feather?
• September 19th, 2006, 08:22 AM
captaincaveman
not seen it yet but it must have something to do with temperature as the statement mentioned a cold autumn day, so has it something to do with density of the helium in the balloon and therefore the balloon wouldnt be floating or something like that?
• September 19th, 2006, 09:01 AM
Quote:

Originally Posted by william
Trooper: ... and Mr. MacGyver1968, just what do you plan to do with this hammer and feather?

I was just on my way to Billco's to shove them.. I mean give them back.
• September 19th, 2006, 10:10 AM
william
Quote:

Originally Posted by captaincaveman
not seen it yet but it must have something to do with temperature as the statement mentioned a cold autumn day, so has it something to do with density of the helium in the balloon and therefore the balloon wouldnt be floating or something like that?

Hi CCMan,
The 'cold-autumn day' was a hint - but only a hint that the windows should be shut.

Cheers
• September 20th, 2006, 09:35 AM
Scifor Refugee
Quote:

Originally Posted by william
Seems like someone said everything in the car will move to the right (via Newton). This includes the air inside the car. We'll then have a density gradient provided the windows are shut (which they should be on a cold-autumn day :wink: ). The balloon, being a buoyant object, will move towards the less dense region (i.e., left).

It's not at all clear to me that the buoyancy force would necessarily win out over inertia. It seems like it would depend on the density and compressibility of the air in the car, the density difference between the balloon and the outside air, the mass of the balloon, etc. There would be some threshold for the compressibility of the air in the car below which the pressure gradient wouldn’t be enough to cause the balloon to float to the left. Perhaps air is well above that limit, but who knows that off the top of their head? That “compressibility threshold” would also depend on the balloon’s mass, since you will need a larger gradient for a more massive balloon. The momentum of the balloon is directly proportional to the velocity of the car. Is the pressure gradient created during the turn also directly proportional to the velocity of the car? And so on… Without having those numbers and running the calculations, I don’t see any way of knowing (other than just trying it, of course).

I normally enjoy your posts william, but this seems like something that a person couldn’t really figure out off the top of their head without having a lot more information. It’s kind of like asking “if a book is resting on a table and the table suddenly accelerates, does the book fall off?” The answer is “it depends on the exact values of a lot of numbers that weren’t given in the question.”
• September 20th, 2006, 09:47 AM
My considered opinion is that the ballon will move to the left while the hammer will move to the right..... 8)

May I propose a non-fatal experiment? 1 Large box, top removed helium balloon in side, four strings attached [1 to each upper corner] top replaced with transparent film, swing it around, IF balloon moves away from you scifor wins if balloon moves toward you William wins.

N.B you may need to consider electrostatics in this experiment if so use a deeper box with a cotton grid woven behind the film.

Wear a crash helmet though in case you get dizzy!

Maybe a polystyrene ball tied via string to a lid - bottle filled with water turned USD [or laid flat] swing it around? think about it.
• September 20th, 2006, 11:27 AM
Scifor Refugee
Quote:

Originally Posted by billco
Maybe a polystyrene ball tied via string to a lid - bottle filled with water turned USD [or laid flat] swing it around? think about it.

See, that's the sort of thing I was talking about. Water is basically non-compressible, so there would be no significant density gradient in the water as it accelerated around a curve. It all depends on how much of a density gradient is created, which depends on the compressibility of the fluid that surrounds the buoyant object. Air is pretty compressible, so I imagine you could get a fairly good density gradient during a turn - but will it be enough to overcome the balloon's inertia? Unless you actually get in a car and try it, there's no way to know without doing a lot of really complicated fluid dynamics calculations.
• September 20th, 2006, 11:44 AM
I just thought I'd found a use for all that old packaging.... The gradient density will affect how the heavier accelerates through the lighter, your point is valid, it affects how much the balloon will move, along with any air-surface friction, I do not know what the effect will be other than any friction free motion in a vacuum would continue straight on. I suspect it will 'motion toward' rather than violently swing.
• September 20th, 2006, 11:47 AM
william
Quote:

Originally Posted by Scifor Refugee
Quote:

Originally Posted by william
Seems like someone said everything in the car will move to the right (via Newton). This includes the air inside the car. We'll then have a density gradient provided the windows are shut (which they should be on a cold-autumn day :wink: ). The balloon, being a buoyant object, will move towards the less dense region (i.e., left).

It's not at all clear to me that the buoyancy force would necessarily win out over inertia. It seems like it would depend on the density and compressibility of the air in the car, the density difference between the balloon and the outside air, the mass of the balloon, etc. There would be some threshold for the compressibility of the air in the car below which the pressure gradient wouldn’t be enough to cause the balloon to float to the left. Perhaps air is well above that limit, but who knows that off the top of their head? That “compressibility threshold” would also depend on the balloon’s mass, since you will need a larger gradient for a more massive balloon. The momentum of the balloon is directly proportional to the velocity of the car. Is the pressure gradient created during the turn also directly proportional to the velocity of the car? And so on… Without having those numbers and running the calculations, I don’t see any way of knowing (other than just trying it, of course).

I normally enjoy your posts william, but this seems like something that a person couldn’t really figure out off the top of their head without having a lot more information. It’s kind of like asking “if a book is resting on a table and the table suddenly accelerates, does the book fall off?” The answer is “it depends on the exact values of a lot of numbers that weren’t given in the question.”

Thank you scifor.
Well, we need a volunteer (besides MacGyver1968!).
Okay, I think we can all agree that the balloon shouldn't move forward or backward. So it remains to be determined if it will move right, left, or not move at all. And the windows must be closed otherwise it won't work.

I propose doing this in an empty parking lot and I think a simple circular path will suffice. I also recommend starting at small speeds. And an "observer" (passenger) would be a good idea so the driver can concentrate on staying alive.

I think the balloon will move to the left with only a small speed of the vehicle. (This is a hunch.)

And keep in mind, we are not concerned with how far the balloon moves - only which direction. If I were to attempt this experiment, I first would simply have the passenger hold the string and see if there is a noticable direction in which the balloon moves. Another technique may be to somehow attach the balloon so that it just rests against the right side of the vehicle's window. If the balloon leaves the surface during a smooth (left) turn (or circular path), then we know buoyancy wins out - especially if the balloon maintains some distance from the window during a turn at constant speed.

If I were to attempt some sort of a calculation, it might go as follows:
I would start with centripetal (or centrifugal if you prefer) acceleration using the radius of the turn (circular path). For the mass, I would either measure the mass of a balloon, look up the number, or simply take a best guess (afterall, we're not sending a rocket to Mars). I suppose I would need a number for the volume of the balloon and also I might see how the mass of the Helium inside the balloon compares with the mass of the balloon itself to see if that must be taken into account (I'm guessing not). From this, we should be able to get the centripetal force on the balloon and the centripetal acceleration felt by all the objects in the car. Finally, I would need to calculate the buoyant force and see how that compares to the centripetal force. The largest number wins.

Let's do it:
density of air at STP: rho = 1.25 kg/m^3
mass of balloon: mb = 0.00185 kg (googled "mass of balloon")
volume of balloon: Vb = 0.00625 m^3 (via the above google)
mass of Helium inside balloon: mHe = 0.00104 kg

Okay, the first thing I notice is that I should include the mass of the Helium since it is comparable to the balloon's mass. Good thing I checked....

Now we just compare the centripetal force and the buoyant force on the balloon. (Careful thought should lead you to the conclusion that they act in opposite directions - and yes... I'm using "centripetal" rather loosely - in this case maybe "centrifugal" is more appropriate.)

Equations:
centripetal force: Fc = (mHe + mb)*ac
buoyant force: B = rho*Vb*ac

"ac" is the centripetal acceleration. The astute eye would have seen something funny in the "B" equation. Normally it is B = rho*V*g where g = 9.8 m/s^2. But here we want the acceleration in the horizontal direction --> "ac". I also notice that I need not assume a turning radius as you'll see why below.

Now we just compare the two (B and Fc):
(A note on notation: I will use "?" until the end. At the end of the calculation, "?" will become either "<", ">", or "=".)

Fc ? B
=> (mHe + mb)*ac ? rho*Vb*ac
=> (mHe + mb) ? rho*Vb
=> 0.00289 kg < 0.00781 kg

Therefore the buoyant force wins.
=> the balloon moves left.

It is interesting to note that this result is independent of the acceleration! Notice that it cancels from both sides of the equation. What this means is that any counterclockwise nonzero angular velocity (providing the centripetal acceleration) should move the balloon to the left!

To directly compare the forces, we just reinsert it;

Fc = ac*0.00289 kg < ac*0.00781 kg = B

This was fun!

Cheers,
william
• September 20th, 2006, 12:15 PM
Why don't you just weigh the balloon with the helium in it? Using a sensitive scale add a weight then, tie the balloon. The reduction will be the 'negative weight!" your way of doing it does not seem to take into account the compression of the helium.

Also just hang an ordinary balloon from the ceiling, it will have approoximately the same negative bouancy as the helium balloon has 'positive' - If the air filled swings to the right the helium will swing left.
• September 23rd, 2006, 01:05 PM
billiards
If you wanted to do this experiment you wouldn't even need to turn, you would just need to accelerate in a straight line, if the ballon moves forward then William is right. I think that William is wrong, his maths doesn't consider convection currents which would be significant in this problem.
• September 25th, 2006, 10:25 AM
william
Quote:

Originally Posted by billiards
If you wanted to do this experiment you wouldn't even need to turn, you would just need to accelerate in a straight line, if the ballon moves forward then William is right. I think that William is wrong, his maths doesn't consider convection currents which would be significant in this problem.

Hi Billiards,
If we stick with the turning form of this problem, any currents would reduce in a matter of a short time - especially once we maintained a constant angular velocity. For the linear problem, practically, I think it would be harder to do the experiment due to the inability to maintain a linear acceleration for very long. But in principle, it seems correct... I have to give it more thought....

Convection currents - formed from the generation of thermal energy - I think would be insignificant. How significant do you think they would be?

Cheers,
william
• September 25th, 2006, 10:55 AM
river_rat
Its easier to do this experiment by slamming on breaks instead of taking a corner - the balloon moves backward instead of forward like anything else not "tied down" in the car.
• September 25th, 2006, 11:05 AM
william
Quote:

Originally Posted by river_rat
Its easier to do this experiment by slamming on breaks instead of taking a corner - the balloon moves backward instead of forward like anything else not "tied down" in the car.

Hi double-r,
Do you mean "the balloon moves backward instead of forward like everything else not "tied down" in the car." ???
• September 25th, 2006, 11:13 AM
river_rat
anything, everything - potato, poetaatoe ;)
• September 25th, 2006, 11:16 AM
william
Oh good!
So do I understand correctly that you agree with me??? :)
• September 25th, 2006, 11:20 AM
river_rat
With the fact that the balloon moves in the opposite direction you would expect (ie. towards the inside of the turn) then yes i agree with you.
• September 25th, 2006, 11:39 AM
Trust a yank to get something going off in the wrong direction.. :wink:
• September 25th, 2006, 05:35 PM
william
Quote:

Originally Posted by billco
Trust a yank to get something going off in the wrong direction.. :wink:

That's funny! :lol:
• September 25th, 2006, 05:43 PM
Nothing personal old chap, just the British sense of humour!
• September 25th, 2006, 06:00 PM
MacGyver1968
Quote:

Originally Posted by billco
Trust a yank to get something going off in the wrong direction.. :wink:

Just don't ever call a Texan a "yank". 'Cause them's fightn' words. :)

This quote also makes me think of a masterbation joke...but I'll leave that one alone.

Like mama always said..If you can't say something relevent..at least say something funny.
• September 25th, 2006, 06:07 PM
Shit! I guess '13' is unlucky for you guys as well then :wink: :wink:
• September 25th, 2006, 07:01 PM
wallaby
if the car windows were rolled down would the balloon still move in the same manner?
• September 25th, 2006, 07:07 PM
Quote:

Originally Posted by wallaby
if the car windows were rolled down would the balloon still move in the same manner?

No, the air would tend to be compressed on the left side of the car and be forced through the windows, draughting the ballon to the right.
• September 25th, 2006, 11:07 PM
billiards
Look William,

what's the principle behind your maths. I figured it was just Newton's third law - every reaction has an equal and opposite right? So your air pushes against the window, the window pushes the air back. Then you assumed that the force that a balloon's worth of air pushes against the window is equal to the force that the air pushes against the balloon. Am I right?
You then compared this force to the inertial acceleration "force" or whatever you wanna call it, of the balloon. You found the buoyancy force was always greater, but this is only because air is denser than hydrogen. That's all you have shown, because the acceleration cancels from both side the denser material will always win, thus by your theory anything less dense than air will go the wrong way.
I figured that some of the air would go around the balloon, seeing as the balloon is not an impenetrable plane that covers all the gaps. I think you might have overestimated the buoyancy force, and as such the balloon might still actually go "the right way".
• September 26th, 2006, 04:17 AM
Quote:

Originally Posted by billiards
Look William,

what's the principle behind your maths. I figured it was just Newton's third law - every reaction has an equal and opposite right? So your air pushes against the window, the window pushes the air back. Then you assumed that the force that a balloon's worth of air pushes against the window is equal to the force that the air pushes against the balloon. Am I right?
You then compared this force to the inertial acceleration "force" or whatever you wanna call it, of the balloon. You found the buoyancy force was always greater, but this is only because air is denser than hydrogen. That's all you have shown, because the acceleration cancels from both side the denser material will always win, thus by your theory anything less dense than air will go the wrong way.
I figured that some of the air would go around the balloon, seeing as the balloon is not an impenetrable plane that covers all the gaps. I think you might have overestimated the buoyancy force, and as such the balloon might still actually go "the right way".

With the windows closed the balloon will, if free to move, go left. relative density is the only thing you need consider.

Just imagine if the car were half full of water and the rest was a balloon.
• September 26th, 2006, 04:24 AM
Its All Relative
A quick google search reveals someone else that was not willing to believe what would happen so decided to try it for themselves:

Clicky
• September 26th, 2006, 04:53 AM
Just love that last sentences don't you?
• September 26th, 2006, 11:36 AM
william
Quote:

Originally Posted by Its All Relative
A quick google search reveals someone else that was not willing to believe what would happen so decided to try it for themselves:

Clicky

Thank you very much relative!
Theory must be supported by experiment/observations.

Now all we need is someone on this forum - the next time you have a helium balloon - to corroborate the experiment. (Unless we want to take the above link as sufficient experimental evidence, which I'm willing to do.) :)

Cheers,
william
• September 26th, 2006, 12:29 PM
A's theory, agrees with 'B's observations, and my instinct, so good enough for me, unless it was a lead balloon - but that's another thread!.