Balloons in a Car: The Science Behind Why They Move Forward When You Take Off

  • Thread starter Thread starter yakabod
  • Start date Start date
  • Tags Tags
    Car
AI Thread Summary
When a car accelerates, the balloons inside move forward due to the pressure difference created by the car's motion. As the car speeds up, high pressure builds at the rear and low pressure at the front, causing lighter-than-air helium balloons to float toward the front. This phenomenon occurs because the air in the car moves backward, pushing the balloons forward. It is also noted that air-filled balloons can demonstrate this effect as long as there is a buoyant force from the pressure gradient. Understanding the buoyant force is key to grasping why the balloons behave this way during acceleration.
yakabod
You are in a stop light and you have plenty of balloons in the back of your car. The light turns green and you proceed all the balloons move forward. Why when you take off in a car the balloons move forward?





Thanks Alot to whoever answers :)
 
Physics news on Phys.org
The forward acceleration of the car causes a high pressure in the rear of the car and a low pressure in the front of the car, thus the He balloons float to the front of the car.
 
Another way of saying the same thing: The balloons are lighter than air (did you notice that FrankR assumed Helium balloons even though that was not specifically stated- if your balloons are blown up with air this won't happen) and so have less inertia than the air in the car. The air moves backward pushing the balloons forward.
 
You can do the same thing in a turn. In an American car if the helium balloon is floating above the front passenger seat and you take a left turn, the balloon slaps you in the face...provided the right window is not open!
:wink:

Creator
 
Originally posted by HallsofIvy
Another way of saying the same thing: The balloons are lighter than air (did you notice that FrankR assumed Helium balloons even though that was not specifically stated- if your balloons are blown up with air this won't happen) and so have less inertia than the air in the car. The air moves backward pushing the balloons forward.

Sure it will... (won't it?)

The innards of the balloon are completely separate from what's going on outside.

When the car accelerates, there is an uneven pressure distribution on the outside of the balloon, which pushes it to the front of the car.

How will the the forces sum to zero if you've got air in the balloon, when they won't if you've got helium in it?
 
You don't necessarily need helium filled balloons. It works with air filled balloons as well, as long as there is a buoyant force on the balloon caused by a pressure gradient in the car, since the pressure in the balloon remains constant. I've actually tested both helium and air filled balloons, I barrowed balloons from a kid while riding a car to test a physics principle. I found it actually works better with air balloons, the He balloons tend to stick to the roof of the car, with static electricity and friction because they remain stuck to the roof if you don’t grab them. Saying they're filled with He helps to understand the concept. Just understand where the buoyant force comes from and it's fairly easy to grasp.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top