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Daniel Zetterberg July 3, 2000 14:42

Splash of Water
 
Hello

We're competing in IYPT (International Young Physicist Tournament) in Budapest on Saturday and we have a question about one of the problems:

"Measure the height reached by splashes of water when a spherical body is dropped into water. Find a relationship between the height of the splashes, the height from which the body is dropped and other relevant parameters."

By approximations, we have an expression for the "splash energy", but when we transform it into height a new unknown appears - the mass of the splash.

Have anyone any suggestions about this?

Any help is appreciated

//Daniel Zetterberg

John C. Chien July 3, 2000 16:40

Re: Splash of Water
 
(1). At the time the sphere touches the water surface, you can calculate the velocity, momentum, and kinetic energy of the sphere, the water and the combined energy and momentum. (2). Assuming that at the time the sphere is totally submerged into water, the velocity of the sphere is zero, then one can apply the mass, the momentum and the energy conservation to the system. (3). The initial state is the sphere moving at velocity V_sphere, the water velocity, V_water=0. The final state is V_sphere=0, and the water moving at some velocity. (4). Since the sphere at final state is submerged in the water, the water displaced must be equal to the volume of the sphere. This is the basic assumption of a sphere entering the water. If none of the water is displaced, then the sphere is still above the water. (5). We are assuming that the final velocity of the sphere is relatively small when compared with the initial velocity of the sphere. (6). If you change the shape of a sphere into a very thin needle, then the velocity of the needle will remain about the same, no splash will generated and no energy transfer will take place. (7). SO, a simple model is: the energy of the sphere is transferred to the water of the same volume at the end of the impact and the subsequent motion of the splash will determine its height. And the system should satisfy the conservation law before and after the impact. (8). It is a solution, but I don't think it is the solution. (the angle of the splash is also a parameter)

Z T Xie July 10, 2000 12:55

Re: Splash of Water
 
You may consider a cavity behind the spheric body if the velocity is not small. You could asume the presure Pc in the cavity is the atmosphic presure. IF the body enter the fluid not far, the presure of the fluid around the body is nearly the presure of fluid surface, so the cavitation number sigma is nearly zero. For this case, the cavity behand the body is alike a cylinder. So you can evaluate how much fluid is replaced by the body and cavity. You can simpily assume this is the mass of splash. But if the body goes too deep, it need some revision to the cavitation number, and the length of cavity could be found from some refereces(cavitation flows).

Good luck


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