At 11:29 AM 3/12/5, Robin van Spaandonk wrote: >In reply to Horace Heffner's message of Fri, 11 Mar 2005 07:45:11 >-0900: >Hi, >[snip] >>direction the water goes down the drain. Any angular momentum exhibited by >>the vortex must be there initially, and some of that is lost by transfer to >>the tank bottom. >> >>In all cases the overall angular momentum is conserved. >[snip] >Correct, but not really the issue. I had already assumed that such >was the case. The issue is where does the *energy* come from?
The COAM statement above was just an afterthought. My main point earlier was *what energy?*. You don't have an accounting of the energy on a system basis. > >Put simply, 1 kg of water that changes it's radius by a factor of >10, undergoes roughly[1] a ten fold velocity increase. This >implies a 100 fold kinetic energy *increase*. There again is an example of treating a part of the system like it was the whole system! You just can't do that. The short simple answer is the 1 kg of water, if that's all you have, can't all go down the drain at a 10 fold increase in angular velocity. The kinetic energy of parts of the system remain either as (1) water which can't go down the drain (frictionless version, or rotating tank) due to centrifugal force or (2) water which depletes its kinetic energy before it goes down the drain and transfers its momentum in part to the earth via the tank and viscosity (normal verison). > >Where does that energy come from? What energy? You haven't intergated anything. You are treating the entirety of the water as a single lump at fixed radius. You just can't do that. You don't know from these particulars (1) what the initial velocities are (on a finite element basis), (2) the total system angular momentum, (3) the total initial starting kinetic energy of the system, (4) what angular momentum and kinetic energy is actually dumped down the drain (though it is clearly not nearly as much as you estimate unless the initial total momentum is very small), (5) how much kinetic energy and momentum remains in water which can not go down the drain due to centrifugal force or (6) what kinetic energy is dumped into heat and corresponding momentum dumped to the earth in order to allow all the water to go down the drain. > >It can't all come from gravity, because gravity can only supply a >fixed amount, equal to m x g x h, whereas the increase in energy >is a function of the initial velocity, and is therefore different >for each initial velocity, and consequently not always equal to m >x g x h. The m x g x h integrated throughout the mass may well be way too *much* energy. Almost all that energy could easily be converted to a combination of vertical velocity, water heat, and tank heat, provided the initial system angular velocity is very small. If m x g x h provides too little energy, then there is water which cannot go down the hole until its angular velocity is reduced by turning it into heat. The excess angular momentum in this case is indeed transfered to the earth as its corresponding energy is reduced to heat. >You pointed out that angular momentum goes down the drain with the >water. This is something I hadn't really considered, but it only >serves to sharpen the problem. It provides a perspective to solve the problem. All the energy ends up as heat, kinetic energy going down the drain, or (in a frictionless version or rotating tank version only) as kinetic energy of water remaining in the tank and which can not go down the drain. All the initial momentum of the system ends up either going down the drain or being transferred to the earth in a process where the corresponding kinetic energy is reduced to heat. >It is precisely the angular >momentum that goes down the drain, which is *not* passed out of >the system before the drain is reached. That angular momentum is still in the water if that water has not gone down the drain. If there is not enough kinetic energy available to make its radius reduce, then that radius remains fixed - it can not go down the drain, it can not reduce its radius of rotation unless some other energy M x g x h comes from some other M in the system. In a friction world, however, the angular momentum of water which can not go down the drain is eventually reduced due to viscosity and heat generation, so that water can all eventually go down the drain. It just doesn't carry all the angular momentum that you would estimate based on the simplistic model. > >[1] In fact it will be slightly less, due to friction as you >pointed out. Making the tank deeper will reduce this loss on >average, because friction with the bottom will apply to a smaller >fraction of the water. Or if you prefer, we can fill it with >liquid helium instead, so that there is no friction. > The case for a frictionless liquid is similar to the case for the rotating tank. The mass containing the angular momentum for which there is no energy available to put down the drain remains behind. Superfluids, though, would probably find a way to crawl over the outside of the tank! 8^) I hope all this makes sense. Regards, Horace Heffner
