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Underwater Kinematics
By Andrew Hodges on Tuesday, October 23, 2001 - 11:53 am:

Could anyone tell me a little, or direct me to a good website about the forces acting on a ball in water. I am trying to model the situation where if you drop a ball (which floats in water) from a particular height, how deep will it fall into the water before rising to the surface?

By James Oldfield on Tuesday, October 23, 2001 - 04:59 pm:

If you're prepared to be a little approximate about when the ball isn't fully in the water then all you need is Archimedes' Principle. This states that the force of buoyancy on an object is equal to the weight of the fluid (in this case of water) that it displaces. Do you know how to solve the problem from there?

If you are worried about when the ball is partially in the water, or about resistence due to the water, the modelling becomes a bit more difficult (though Archimedes' Principle still holds). If you want a bit more information about that please ask.

Jim

By Arun Iyer on Wednesday, October 24, 2001 - 07:52 pm:

refer to this link...
http://webug.physics.uiuc.edu/courses/phys113/fall97/lectures/Lect12/sld011.htm

love arun

By Andrew Hodges on Wednesday, October 24, 2001 - 09:12 pm:

Could you expand on resistance due to water please Jim, also would the force change as the ball went deeper under the water, due to increased pressure above it?

If the depth was kept the same, would the shape of the container make a difference to the calculations?

By James Oldfield on Thursday, October 25, 2001 - 05:59 pm:

When the object is deeper in the water it would indeed experience more pressure acting on all surfaces (if this is enough the object might even implode). However, this pressure acts on both the top and bottom of the object, so that the net force still obeys Archimedes' Principle (the force comes from the bottom of the body being slightly deeper than the top of the body, so the difference in pressure results in a net force).

OK, fluid resistance. You can never get an exact formula for this, but you can approximate it. The simplest way is that at a low speed v the force of resistence R is proportional to v. At higher speeds tubulence comes into effect (the ball disrupts the flow of liquid past it as it moves) so that R is proportional to v2. In practice R = av + bv2 for some constants a and b (so that the av term has the most effect when v is small and the bv2 has more effect for large v), and as far as I know the only way to find a and b are by experimentation. These constants depend mainly on the shape and size of the object, so someone may have found there approximate values for basic shapes of bodies in water (a web search would be useful for that). As you say, the shape of the container would have an effect on this (particularly the tubulence term), but again the only way to find out the numbers would be to perform an experiment. Such is the messiness of physics.

Jim