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 hemiphere 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 Sci. Am. *If* the angular velocity of the water is zero, then the above is true - the energy comes form the water "falling" to the drain, and the angular momentum comes from the coreolis force. The rotation that results is a function of *both* gravity and initial angular momentum. The gravity potential energy is manifest as angular velocity via 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, unless the coreolis force exactly cancels the rotation. > >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 fallicy in the above statement is the assumption that the velocity v of (some small chunk) of the water changes as it approaches the drain. The velocity of the chunk remains constant at any time. Thus the instantaneous linear kinetic energy KE = 1/2 m v^2 remains constant. 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 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 substitue 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. Since the KE of every chunk remains constant the energy of what remains in the bowl is the the original PE + KE less the KE of what went down the drain. I hope I got all that right. 8^) Regards, Horace Heffner
