Balloons, Cars and Newton's First Law

Is this supposed to be a pure thought experiment? This would be a very easy experiment to do.

It would certainly be an experiment that they could perform, but I would like the students to think through the problem using their knowledge of inertial and buoyancy.

My fifth grade son made an interesting observation this morning at breakfast. His comment was that it would be a function of the acceleration's magnitude. I am now wondering if a consideration of static vs. dynamic drag must taken into account.

That would depend on the level of certainty that you want in your answer. Are you interested in the main players or all of the players in the experiment?

If the effect of secondary components are small enough, they can be ignored in most cases.

The way I saw it, the question asks what is the significant effect and how does it work.

I believe it will move, but it would depend on where it is located in the cabin at the moment that the car accelerates.

My rational is that the surrounding air in the cabin of the car will form a density gradient when the car is under forward acceleration with the denser air in the aft of the cabin.

A neutrally buoyant object in a steady state environment (no acceleration), will remain in a fixed location. If the object is predominantly in the rear of the cabin, it will try to reach a homeostatic state by moving forward to where the air density is equal to its initial state.

If the object is biased forward at the start of the experiment, it will drift rearward.

There are other factors that come into play like the reaction mass of the balloon, which will tend to exert a force towards the rear of the car when the car accelerates and the aerodynamic drag induced by the ballon surface, which will impede its motion.

Finally, if the car accelerates too quickly, the screaming from the passenger will induce eddie currents in the air and foil the whole experiment.

Per one of my students:

If your idea about moving forward is correct, would you then assume the balloon would move down since it is now in less dense air or would the internal pressure of the balloon now increase its volume such that the ratio of densities of the balloon and the air in its new position in the car remains the same as before and therefore the balloon maintains the same "altitude"?

Good question. Here is my thought. The balloon is in a state of equilibrium at T0, which is just prior to acceleration.

At Tn the air density gradient appears as acceleration is present. If the balloon is biased aft at T0, then to reach equilibrium the balloon must drift forward to a point in the cabin that represents the same ambient air pressure at T0.

Since the system also has inertia to deal with the balloon will lag somewhat as it moves forward. Since the ambient air pressure is somewhat greater at Tn around the balloon, the volume in the balloon will shrink proportional to the ambient pressure.

That should cause the balloon to drop altitude slightly. Then, as the balloon moves forward to its equilibrium point it should return to its original altitude at T0, but slightly forward of its original position at T0.

Inertia will cause the balloon to overshoot its mark slightly.

All of these changes will probably be very minute, particularly in altitude.

That's my educated guess. ;-)

Thanks for the continued responses. It gets very interesting the more you look at the various qualities of the system. While not very significant, it does seem that the system should have some inertial dampening at either end of the acceleration.

Thanks again and have a great day! The GTA Gang

I like your answer, but I think we'll have to wait until this afternoon to get graded. I believe there will be be a slight oscillation backwards and forward, however the motion will be downward. (I posted that answer anonymously just in case it's wrong,).

You've already taken into consideration that the point of equal pressure will vary with acceleration rate, which I had previously failed to take into consideration. However I do question one part of your answer.

The final altitude should be determined by acceleration rate and time under acceleration, with the only downward limit been the floor of the car.

There is one other variable, a balloon is not a spherical object, there would be constant acceleration with a balloon that is not neutrally buoyant, (until it's restricted by part of the vehicle) in this case the balloon will be seeking a point of equilibrium pressure wise in the vehicle. I'm speculating there will be a slight rocking motion when it reaches this point, what do you think?

Double checking your answer is warranted. A physics class will be looking at us. Don't let us down.

Anon, are you saying that one can fractionate air into its component gases simply by mild acceleration of a closed volume? What's this about "denser air" moving to the rear during mild acceleration of something like a car? The air density should remain essentially constant in an accelerating vehicle, an elevator, etc. What about a 747 taking off, with much greater G forces than a car in acceleration? Are we to believe that "dense" air moves to the back of the aircraft? Air density is a function of air temperature and composition. Do you think that a boyant balloon in a 747 on takeoff will rocket to the back of the aircraft? Since a balloon filled with a low density gas (H2 or He) will tend to displace denser air and the balloon will rise, would you expect the balloon to move forward (towards lower density air) in an 747 taking off, similar to rising phenomenon? Wake up the theoretical physicists. We need to examine this closer!

Hi, Cardio!

Well, I didn't think that the air would stratify under acceleration, but I suppose that would happen to some extent.

However, I would expect that upon acceleration the air molecules would exhibit the same behavior a marble on the back deck of a car would. That is, it would tend to remain at rest and roll/move towards the rear of the car when the car accelerates forward.

I was thinking that the air outside the car does not move forward with the car (ignoring the aerodynamic effects of the car). The air inside will try to do the same thing, but since it is contained within the cabin of the vehicle, the rear window will exert a force in the forward direction on the air.

Since air is not a solid, there will be a slight tendency for the air to compress slightly more at the rear of the car than at the front inside of the car. This is the gradient I spoke of.

Under these conditions the balloon will tend to move towards a point inside the car's atmospheric gradient that represents the same temperature/pressure that the balloon was in prior to acceleration.

Overall, the effect will probably be subtle since the acceleration is not that great. As said before there is inertia of the balloon, aerodynamic friction/drag, and a few other factors that are players in this band, too.

Bonus points: What color is the car?

Asked by Mssrs Click & Clack.

The air density at the rear of the passenger compartment will increase. This will tend to push the balloon forward into less dense air. The balloon will no longer have neutral buoyancy and will begin to go down and toward the rear of the car. Net result is the balloon going stay in the same place in relation the front and rear of the vehicle, its motion will be downward.

PS. Should my answer be wrong I intend to invoke the uncertainty principle.

Should my answer be right, I'll tell you who posted it.

When I was much much younger, I conducted an experiment whilst standing in the corridor of a fast moving train. I jumped up and landed in exactly the same spot!

Tony

The simple answer is that it would stay where it is. A more complicated answer: if you make it neutrally bouyant by attaching a small weight to counter the lift of the helium, it will stay where it is but will rotate so that the weight is opposite the direction of acceleration - to the outside of a turn, forward if the car is stopping, or rearward if the car is accelerating forward.

Great question,

.........Before Anonymous Hero brought up the density gradient created in the car, I thought it was trivial: i.e. the balloon being neutrally buoyant acts like any other packet of air in the car, and stays exactly where it was (relative to the car). However now we have a change in pressure we need to determine whether the balloon gets more or less dense relative to the surrounding air.

First of all: some assumptions:

1.) That we can ignore the neck and knot.

2.) That the rubber is not at its elastic limit (i.e. following Hook's law). This is reasonable if the balloon has reached its state of neutral buoyancy by becoming slightly deflated, but, not if it has got to this state by being over inflated.

3.) This is a big one (assumption that is). Because I have no idea how to relate the tension in the rubber to the differential pressure between the inside and the outside of the balloon: I'm going to assume that the slight difference in size does not affect the tension in the rubber significantly. This assumption has now overridden assumption 2 (apart from the last sentence of 2).

Having pointed out that other changes can be ignored, I'm going to extrapolate to a ridiculous scenario. Most of the mass of the balloon is contained in the rubber and this (the density at least) changes as the square of linear dimensions whereas the density of the gas, and, surrounding air changes as the cube of the linear dimensions. So suppose the balloon is at the back of the car and the pressure of the air increases enough to halve the linear dimensions of the balloon; then the density of the balloon will have quadrupled whereas the density of the surrounding air will have octupled (got eight times as great). The opposite argument applies to one at the front.

Conclusion: under acceleration a balloon at the back will float up and towards the middle (front to back middle), and a balloon at the front will sink down and towards the middle.

Where's Fyz when you need him.

Confused, that's where. Seriously, though, we do need to stratify the problem (not the air), and I think STL and you have pretty-much canned the basics.

So I'd first consider the movement and pressure of the air inside a rigid accelerating sealed box. Assuming that the acceleration changes sufficiently slowly (the reference would be the decay time of sound), the air would create a pressure gradient just as it would under gravity. So all the would move slightly backwards, creating a pressure gradient that would support its accelerating mass.

If the balloon remains (or maybe becomes) neutral under these conditions, it will simply move with the air. If it was initially neutral, a real balloon will change density less rapidly than the surrounding air. So it will remain neutral at the point where the pressure was unchanged (a nominal centre), and move towards the nominal centre if it is either in front of or behind that point (and, as Randall realised, if it is forward of the nominal centre it will take a downward path to get there, and if behind it will take a downward path; but it should end up very near both to the centre and to its original height).

In practise you will find it almost impossible to achieve the sort of conditions described above, as a car is not a rigid sealed box that is devoid of internal variable influences.

For example, it is at least vented in near the rear, and in a position that becomes de-pressurised when the car is moving. As the car is accelerating, either the pressure in the cabin will reduce with the increasing velocity or the air will flow (more-or-less continuously) from front to back of the cabin. This will be a much larger effect than the buoyancy changes described above, so the balloon will move towards the rear of the car. Also, with most vehicles, the pressure in the cabin tends to fall with increasing velocity, so the density of the air will fall more rapidly than the density of the balloon - so the balloon will also fall under gravity - unless of course it's a very cold day and the air in the car was previously warm, or the particular car was designed with a large "ram" effect into the cabin, or...

Have fun

Fyz

It will move to the front since the air in the vehicle is pushed to the rear creating a higher pressure in he back and lower pressure in the front of the cab thus causing the balloon to head for the low pressure area.

As a physics teacher, I am trained to not give students answers to their questions. The students should be encouraged to find the answers for themselves. This is a simple problem that can be resolved without a huge amount of testing. A van works better than a car. The engineer in me permits me to answer the question without performing the experiment. Under forward acceleration, the air density increases toward the rear of the vehicle, driving the balloon forward in the van.

I just brought balloons home for a birthday party in my hotrod Lincoln. The balloons in the back seat move to the front under acceleration and retreat to the back when braking. It was a very annoying lesson in physics.

Regardless of buoyancy, the string, balloon, and helium still have mass and the laws of motion will continue to apply. One might better devise an experiment to measure the changes in pressure of the air trapped within the closed car from the front to rear in the direction of acceleration. Air, too, have mass and the laws of motion still apply.

Here is still another experiment one my devise: What is the difference in gravitational acceleration of a mass it the equator verses at the polls? Note that at the equator, one must consider the acceleration contributed by centrifugal force of the earth's rotation which is not present at the polls. This can, further, be set up for a calculated difference to be proved with actual experiments. Then, bring into the study the velocity required to produce zero gravity, weightlessness, and condition for geocentricism. This is directly related to both the space program and to Physics I as well and does point directly to the relationship math and physics.

The baloon moves forward because initially the air mass in the vehicle is at rest and remains at rest (briefly) when the vehicle accelerates. Remaining at rest (global reference) moves to the back of the vehicle (local reference). This then creates a void, or low pressure at the front of the vehicle which exerts an unbalanced force on the baloon. The baloon moves forward.

Travis

1. You won't be able to get the balloon neutrally bouyant. Have your class try in the classroom. In time, gas will diffuse, etc. , so it will not be neutrally bouyant for more than an instant.

2. You assume a significant density gradient, front to back, when the car accelerates. Unless the car and tires are very strange, it will not accelerate at more than one G. You already have a one G acceleration when the car is still, with the air supposedly more dense near the floor. Does that mean the balloon will sink or rise until the density matches and the balloon hovers neutally bouyant? I think not. Again, have the class try to make a balloon that will fall from the ceiling and stop before it gets to the floor, given the air density gradient with altitude.

3. Play your thought experiment games, but you will not be able to conduct the actual experiment with any sort of reliable results because the relevant forces are small relative to the chaos of the real world.

"In which direction will a neutrally buoyant balloon move when the car accelerates?"

We have a neutrally buoyant balloon, i.e. one filled with air which is subjected to the vertical accelearation of gravity. For simplicity we assume the material of the balloon is of the same density as air.

Accelerating the car in a horizontal direction will have NO effect on the balloon since it is the same as a volume of air.

Now fill the ballodn with helium and tie the string so that the balloon is at the mid height of the cabin. Only horizontal acceleration of the car will affect the ballon which will tend to move in a horizontal direction opposite to that of the car. Since it is less dense it has less force to move it rearward than the cabin air. Capice?

As always, I and now my students are enjoying and gaining from the comments shared on this blog. Thank you to all who have taken the time to respond and share their thoughts and ideas.

From the varied and details responses, we have concluded the following.

1) While conducting an experiment to examine the gross effects of this situation, the more subtle aspects of the forces at work will create a great degree of complexity when trying to establish controls on the various variables.

2) There is no "obvious" answer with regard to the question of a neutrally buoyant balloon in an accelerating frame of reference.

3) There is a great deal to be learned by way of a thought experiment and the discussions that ensue. Our class has had a lot of fun exploring the various ideas presented in this blog and directions in which they have taken us.

Thank you again to all who have shared.

Have a great day.

The Physics Class

Th Grand Traverse Academy

Thank you for a great little thought experiment. A lot more complicated than it looked on first sight. The possibility even existed that the pressure change on the neutral buoyancy balloon would've allowed the balloon to expand and thus retaining its neutral buoyancy, in a less dense surrounding atmosphere.

However I have one comment to make. I think you should collect some empirical data. The suspense is killing me.

Very good.

I just wanted to add one more of my favorites to this story. The problem is one of those problems that the effect we are looking for is very tiny and difficult to measure due to a host of subtle variables that are hard to control or measure. The result can lead to some bad conclusions by the experimenter.

This can be summed up by a great Nobel prize winner and researcher named Irving Langmuir.

Langmuir offered six characteristics of pathological science:

1. The magnitude of the effect is substantially independent of the intensity of the causative agent.

2. The effect is of a magnitude that remains close to the limits of detectability; or, many measurements are necessary because of the very low statistical significance of the results.

3. It makes claims of great accuracy.

4. It puts forth fantastic theories contrary to experience.

5. Criticisms are met by ad hoc excuses.

6. The ratio of supporters to critics rises up to somewhere near 50 percent and then falls gradually to oblivion.

These are good points to always keep in mind when looking at problems. ;-)

Hi AH,

"In 1953 Langmuir coined the term "pathological science", describing research conducted with accordance to the scientific method, but tainted by unconscious bias or subjective effects."

The points you described are good info for all of us.

-John

I was in process of preparing a comprehensive answer to the size and timing of the effects. Unfortunately, there was an internet crash, and I lost what I had written. I haven't time now to repeat the detail, and there could be factors of 2 errors in the numbers, but the conclusions are as follows:

1) If you are to be able to get meaningful measurements of the direct effects of the acceleration, the balloon will need to be inside a rigid sealed container (rigidity also applies to the contents, of course). The container must be sufficiently well insulated and the system will need to be in static thermal equilibrium so that convection is not an issue. Even Schrodinger's cat can't be in there to observe (it's OK if its cold, dead, and tied down).

2) It is perfectly possible to make a neutral balloon that will retain its fill. You need to use the same sort of material that is used for children's helium balloons. For maximum stability, it should not be overfilled (to allow the internal pressure is very close to the external), and the content should be predominantly N₂, with content of lighter gases (including water and ammonium vapour) that match the surrounding atmosphere. Density may be corrected (increased) by including first Ar (maximum amount being that to match the surrounding atmosphere) and then O₂. In the unlikely event that there is noticeable inward diffusion of heavier species, this may be offset by adjusting the internal concentration of the lighter gases (and then the heavier ones to recover neutrality).

3) The movements are very small. For a 10' long cacity accelerationg at 1gn, the equilibrium movement of the air at the centre of the cavity will be about 0.004". The resonance frequency for this direction of movement would be about 5-ms. The cavity will need to include lossy obstructions if this is to be damped in a reasonable time (seconds); but I think that seats and carpets would be adequate. Initially, the balloon would move with the air. For this same 1-gn acceleration, the balloon would over time move half-way towards the centre, and the same distance up (at the back) or down (at the front). The greater the acceleration the smaller would be the vertical proportion of the movement (I haven't checked, but I imagine that the trajectory of the final position as you increase the acceleration is a circle whose diameter runs horizontally and front-to-back from the original position of the balloon towards the centre of the cavity). However, because this is in a car, the time that you can sustain 1-gn is rather limited - perhaps 8-seconds for an extreme road car and 10-seconds for a dragster. In 8-seconds, the furthest that the balloon might move (from either the very front or back and at 1-gn) is about 0.013". [If the gas was effectively contained rigidly, this would increase to 0.4"].

So, contrary to some comments, the experiment is possible, but contrary to others it would be very difficult.

This question (in the idealised version with a rigid, insulating, and sealed cabin) continued to throw up surprises as I thought further about it.

For example, if the acceleration time is short compared with the time to reach the accelerating-equilibrium position, the horizontal distance moved during a period of constant acceleration appears to be proportional to [final velocity times initial distance from centre of the vehicle]. More startling perhaps is that, for a given final velocity, the vertical distance moved is proportional to [final velocity times time spent accelerating times initial distance from centre of vehicle].
That means that an acceleration of gn/50, the vertical movement could be a perfectly respectable 0.4" even with a flexible balloon as described in #33. (A more rigid balloon would virtually reach its equilibrium** position in this time - and could of course move a great deal further at moderately higher accelerations).

**The equilibrium position under gravity may be visualised as follows:
Model the effect of the acceleration inside the cavity as equivalent to a tilt and a change in gravity. We can see that (relative to the modified gravity) the final position will be vertically above the original position. If we take a point at the middle of the vehicle that was level with the balloon relative to unmodified gravity, the final position will be level with that point in the new gravity.

Thank you again for posing this very challenging thought-experiment.

Fyz