Balloon bamboozle

As the balloon is lighter than air I would guess that the air would move the balloon toward the centre of the curve?

John.

The car will roll over to the outside. The balloon should then move to the inside . This roll is usually countered by banking the road around the corner.

The slight mass of the balloon and gas will however want too continue in its original direction and would therefore move too the outside.

The downward slope of the offramp is ignored.

The eventual movement will therefore depend on the balance of the forces but moving to the inside should be predominant.

After Mum gets bopped in the head several times, it falls promptly to the floor in little latex particles. Sharpened claws (like Del's) work wonders on annoying balloons.

Away from the center of the curve.

The balloon is in motion it is being pulled along by the string tied to the seat. As the car turns the balloons will want to continue in a straight line. Centripetal force will be applied to the string forcing balloon to swing away from the center of the curve. As the balloon tries to continue its straight forward path the string will hold it to its out side path.

The mass of air within the car remains stationary relative to the inside of the car, whilst the car undergoes acceleration towards the centre of the curve resulting in pressure from the outside wall of the car on the contained air, and decreased preasure on the side moving towards the centre of curvature.

The helium is therefore likely to float towards the least density of gas...towards the inside of the curve.

the offramp is level....the linear velocity is constant.....

Newtons first law applys to all. Apply to the balloon as well as the air around it. since the air around it is denser it overcomes the balloon.

Never though of it that way.

but then....

"The helium is therefore likely to float towards the least density of gas...towards the inside of the curve."

since no absolutes are stated.....I'll go with comment #3 with pieces all over

..........soooo all this applies unless the cetrifugal force exerted by the mass of the rubber and string exceed the difference in gas density resulting in the "floatation" force exerted towards the cetre of the curve in the direction of the least air density.....ans maybe this also depends on how rapidly the pressure differrential is corrected within the volume of the car. Could it initially lean into the curve until the centrifugal forces exceed the normalised air pressure gradient, then lean out of the curve?

possible,

I tried it but the balloon keeps floating away.............should have put the top on

Away from the centre of the curve.

The lift from the balloon causes the car to experience a slight reduction in the normal frictional force against the tires which causes the car to slide off the road and overturn. Therefore, the balloon tends to find force equilibrium somewhere around the rear floorboard, or, if Jane wasn't wearing a seatbelt, somewhere in Jane's seat.

Try it and figure out why by how much.

interesting food for thought

I think the air currents (like breathing, AC or heater) in the car will out power the rest.

Sorry no level semicircular offramp to be found here, ice snow or rain would lessen the centripetal force by becoming a tangential rotating mass at the the point friction of the tires was exceeded. Followed by a energy absorbing sudden stop.

Let's change the density of the two fluids to enable a clearer visualisation as to what is going on.

• Change the air in the car for water. Yep. Fill it up to the sun-roof.
• Change the helium in the balloon for air.

The balloon is still buoyant in the water in the car.

Now what happens? Is it the same or is it different?

oil/water

helium/air

same problem....same physics except that liquids at the macroscopic level are incompressible.....I concede that it removes one of the only variables that make the question a curly one

I think there are 2 forces in concert. 1) The Helium is trying to move towards lower air pressure. 2) centrifugal force is trying to throw the mass of the balloon towards to outside of the curve.

I don't think the air pressure is any less towards the inside of the curve so the helium is still drawn upward as normal through the curve. The only lateral force that has affect on the ballon is the centrifugal force. The balloon will go towards the outside of the curve.

Why wouldn't the balloon always point "straight" up, essentially a straight line from the center of the Earth, at an angle to whatever the car is doing? There should be no air flow in the closed car and the balloon should always point "up".

So does anybody dispute that the car is accellerating towards the center of the curve.....ie a constant change in velocity......as the car continues around the curve?

.... that the air in the car should continue to move at the same velocity as it was while travelling stationary within the confines of the car moving in a straight line?

....that the wall of the car pushing against the air will increase the pressure relative to the wall moving away from the air on the opposite wall? (rather like the piston of a speaker cone but on a grand scale)

.... that the constant change in velocity of the vehicle as it moves around the circle would maintain a constant pressure gradient?

.....that the centrifugal force would act equally on the unrestrained masses of gas in the car the balloon and even the car itself (only resisted by the coeeficient of friction in the case of the car tyres, and the tethering string in the case of the balloon)?

or that the balloon would "float" towards a region of equal pressure, much like a weather ballon released heading to the stratosphere?