Why does the Cartesian diver sink when the bottle is turned upside down?

  • Thread starter Thread starter frensel
  • Start date Start date
  • Tags Tags
    Cartesian
AI Thread Summary
The Cartesian diver experiment demonstrates buoyancy changes based on pressure and air volume. When the bottle is turned upside down, the diver sinks and does not float again, indicating a change in buoyancy. This may occur due to air escaping from the eyedropper, which alters its internal pressure and affects buoyancy. Despite the external air pressure being equal to atmospheric pressure when the bottle is opened, the diver remains submerged. Understanding these dynamics is crucial for explaining the behavior of the Cartesian diver in different orientations.
frensel
Messages
20
Reaction score
0
Hi,

I follow the instruction in wikipedia to make a Cartesian diver (Coca Cola bottle + eyedropper), and it works well. I press the bottle, the eyedropper sinks; I release, it floats.

However, if I turn the bottle upside down, the eyedropper sinks and it never floats again. That means the buoyancy of the eyedropper is somewhat changed but I cannot understand the mechanism. If the buoyancy of the eyedropper decreases, the air pressure inside the bottle should increase, but it seems that it is not the case. I can open the bottle and the air pressure inside the bottle is equal to the atmospheric pressure, the eyedropper still sinks.

Can anyone repeat the experiment and explain why it happens? Thank you!
 
Physics news on Phys.org
Perhaps some air came out of the dropper when you turned it around.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top