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It's well known that a helium balloon inside of a car moves forward when the car accelerates, and backward when it slows down. What would happen, though, if a lower density gas was used instead of helium? Would the movement change? Would it stay the same?

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One more question: How does the air pressure vary inside the car during acceleration?

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    $\begingroup$ What are your thoughts on this so far? $\endgroup$ Commented Oct 24, 2018 at 0:25
  • $\begingroup$ What do you mean here by the "deviation angle?" As far as I know, the balloon always moves in the direction of the acceleration. So when the car accelerates forward, the balloon moves forward within the car. If the car turns left (accelerates inward/to the left), the balloon should move that way too... I don't think there's any other complicated angle in play. And while the effect would be more pronounced the more different the balloon pressure is from the air pressure, there should otherwise be no change in this behavior. $\endgroup$
    – Bunji
    Commented Oct 24, 2018 at 3:21
  • $\begingroup$ I should have said that the balloon is tied to the ground, and the deviation angle is the angle between the thread and the vertical line. I've added an image to illustrate. My way to solve it was to put it in the non-inertial reference of the car, where the gravity acceleration and the fictitious acceleration sum into a resulting acceleration that should have the same direction of the buouyancy force the air makes in the balloon. This way, considering that the balloon stabilizes in an angle, it should depend just on the acceleration of the gravity and of the car, but not on the gas density. $\endgroup$
    – Danilo
    Commented Oct 25, 2018 at 0:09

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The air pressure will be greater in the direction opposite the car's acceleration. Intuitively it is because almost all of the air in the car is not attached to the engine (whereas the frame of the car is). Maybe some oxygen or nitrogen molecules are adsorbed on the inside of the back window, taking a free ride from your engine... but the air closer to the front of the car has no reason to do anything other than stay where it is (of course each air molecule is moving very fast between collisions with neighboring air molecules but never mind: A fast moving molecule in the middle of the car would go almost nowhere before colliding with a neighboring air molecule).

As the car accelerates, the colliding molecules more or less stay in the neighborhood they are in, bumping into one another. But the rear window, attached to the frame of the car, flies forward. This causes the rear window to "catch up" with the air molecules in the middle of the car. Well, not quite. Even though there is a net drift of the air toward the back window [due only to the window moving forward], the electric repulsion between neighboring air molecules prevents them from compressing too much (this situation is very much like the atmosphere around our planet: gravity "tries" to pull the atmosphere to the ground but it can only do so much before it no longer has the force to compress the air any further).

So as the density of the air increases toward the rear of the car, a point is reached where the air density will increase no more. But by then enough has happened, a pressure gradient has already built up in the car, there is in fact more air in the back of the car than in the front. The helium ballon feels more pressure on the side of it that is closer to the rear window, and less pressure on the side of it that is closer to the front window. So the collisions between the the rear of the balloon and the the "rear air" combine to exert a larger force than the collisions with the front of the ballon and the "front air". This puts a net force on the balloon toward the front of the car, making the balloon move in that direction.

And yes, if you had two balloons of the same volume, both less dense than the air in the car, the lighter of the two balloons would accelerate toward the front of the car more rapidly under identical conditions. This is because there would be the same force on the new balloon but now it weighs less... so the acceleration will be larger.

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  • $\begingroup$ Thank you for the reply! Excellent explanation for the phenomenon. Indeed, a lighter balloon will accelerate more rapidly. I should have specified before that the balloon is tied to the ground, and I wanted to relate the deviation angle to the density of the gas. $\endgroup$
    – Danilo
    Commented Oct 25, 2018 at 0:32
  • $\begingroup$ I certainly didn't address that. I suppose that the balloon will somehow end up on the sphere whose radius is the string the balloon is tied to. And then (as you mentioned in your comment above) you will have to use the combined effect of the earth's gravitational pull on the air in the car with the "acceleration gravity" to find the direction of the density gradient. I would say that the string would point in the direction of the density gradient. Interesting problem. $\endgroup$
    – okcapp
    Commented Oct 25, 2018 at 1:31

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