Cartesian Diver

1. Jan 5, 2009

liltinkerbell

We had to do a lab on the cartesian diver and there were a few questions that I was unsure on.

Here they are:
What are some possible modifications to the diver that would cause it to sink quicker or slower?
What would happen if the bottle was only half full of water or there was no cap?
If the diver was placed in a bottle containing a different liquid what would you predict to happen?

2. Jan 7, 2009

m.e.t.a.

Let's say that we're talking about a well-balanced Cartesian diver inside a plastic drinks bottle. By "well-balanced" I mean that the diver itself is floating, but only just.

As you know, to make the diver sink you squeeze the sides of the bottle. When you do this, you reduce the volume inside the bottle. The only things in the bottle are air, water, and the diver; and of these, only air is significantly compressible (squashable). So when you squeeze the bottle you know that you are compressing the air only, while the water (and the diver) stay the same size. Remember that this goes for all of the air in the bottle: the pocket at the top and the trapped bubble inside the diver. All the air is compressed proportionally (i.e. the volume of the pocket and the volume of the bubble are both scaled down by the same factor).

So you're squeezing the bottle; the pressure inside the bottle has increased as a result; and, importantly, this pressure increase has compressed the air bubble inside the diver, so the bubble is now smaller in volume than when you were not squeezing. This allows a little extra water to creep into the diver through its open base. The diver as a whole is now more dense, and therefore less buoyant, than it was before. The diver is denser because it has the same volume as before, but it now has more mass. (The water that crept in obviously has far more mass than the equal volume of air that it replaced. Density = mass/volume.) The diver is less buoyant because buoyancy is proportional to the volume of fluid displaced, but is inversely proportional to mass. If you increase the density of an object, that object's buoyancy clearly goes down. (This gives you a big hint for your question 3.)

We know that the volume of the diver's bubble shrinks by a certain percentage when you apply a certain pressure increase. The diver's buoyancy decreases as a result. But its buoyancy definitely does not decrease by the same percentage. Think of the bubble as representing an upward force inside the diver. The bigger the bubble, the greater this upward force. Consider what things you could change in order to cause this upward force to cause a bigger or smaller change in acceleration of the diver.

I think this question would have been clearer if it had read: "What are some possible modifications to the diver that would cause it to sink quicker or slower for a given increase in pressure inside the bottle?" Obviously you could just squeeze the sides of the bottle harder or softer to control the internal pressure and thus make the diver sink faster or slower, but I don't think that counts as a "modification".

These are two different questions. (1) If you increase the ratio of air to water in the bottle, ask yourself:
• How much force will now be required to cause the same increase in pressure inside the bottle as before? More force, less force, or the same?
• How far will you have to squeeze the sides of the bottle to cause the same increase in pressure inside the bottle as before? That is, by how much will you have to reduce the internal volume of the bottle in order to cause the same pressure increase as before? More than before, less than before, or the same?
(2) If you take off the cap, what will happen to the air inside the bottle when you squeeze the sides? Why won't the diver work with the cap off?

Sorry for dumbing this down so much. It was more for the benefit of my own understanding than yours! I have tried not to give too much away because this seems like a homework question.

- m.e.t.a.

Edit: Removed a mistake.

Last edited: Jan 8, 2009