Let me try this one more time. At 4:47 PM 3/7/5, Robin van Spaandonk wrote:
>The question is, where does the energy come from to increase the >velocity of the water? > >If the water is stationary to start with, then it comes from the >change in height of the water as it leaves the tank, and the >velocity at the edge automatically adjusts itself accordingly. Water in a tank, once spinning, does not stop for an amazingly long time. It is generally thought water goes down the drain a differing direction in the Southern Hemisphere than the Northern Hemisphere. However, if you check it a day or two after filling the tank it usually forms a vortex in the vortex direction that resulted when the tank was filled. I recall an article about this in the early 1960's - maybe in Scientific American or Popular Science. The rotation that results is a function of both gravity and initial angular momentum. Some of the gravity potential energy converts to angular velocity via the coreolis force. If the initial angular velocity of the water is zero, then your assertion that the energy comes entirely from the water "falling" to the drain is true, and the angular momentum of the vortex comes from the coreolis force. >However if the water is already rotating before the plug is >pulled, then it has to end up going faster than can be accounted >for by gravity. Yes, except when the coreolis force is in a direction opposing the initial rotation, and its effect exceeds that of the inital angular momentum. > >If the radius decreases by a factor of ten before the water >reaches the drain, then the velocity has to increase 10 fold, and >the energy per unit mass must increase 100 fold. > >So where does the energy come from, or for some reason, does it >simply not happen, and if not, then what does happen? Ignoring the coreolis force for a moment, the fallacy in the above statement is the assumption that the instantaneous speed v of some small chunk of the water changes as it approaches the drain. The speed of the chunk remains constant at all times, except for the speed added by converting gravitational potential energy PE to kinetic energy. Thus the instantaneous linear kinetic energy KE = 1/2 m v^2 of the chunck remains constant except for speed added by falling in the gravitational field. The angular velocity w increases however, because w = v/r. Now, you might say that for a rotational system KE = 1/2 I w^2, and w is increasing with reduction in radius, so where does the free energy come from? Well, the answer is that in a vortex the moment of inertia I of a chunk is not constant. We have I = m R^2, and w = v/R, so when we substitute these into KE = 1/2 I w^2 and we have: KE = 1/2 (m R^2) (v/R)^2 = 1/2 m v^2 which is constant, except for the PE converted to KE by falling down hill. Since the KE of every chunk remains constant the energy of what remains in the tank is the original gravitational potential energy PE plus KE less the KE of what went down the drain. I hope that is all a bit more correct. Regards, Horace Heffner
