At 4:16 PM 3/11/5, Robin van Spaandonk wrote: >In reply to Horace Heffner's message of Wed, 9 Mar 2005 00:31:22 >-0900: >Hi, >[snip] >>>mv1r1 = mv2r2, i.e. v2 = v1 x r1/r2. (m is the same before and >>>after, because we are dealing in both cases with the identical >>>chunk of water). >> >>Yes, you are certainly right about this. Momentum is conserved. The speed >>of a chunck is thus not constant in a vortex, as I had assumed. In fact, I >>think in your imaginary tank the tangential speed is proportional to 1/r, > >i.e. v2 = v1 x r1/r2 Where r2 is your r, and V1 x r1 is the >proportionality constant.
It is a highly idealized picture. This describes the motion of an idealized chunk unaffected by its surroundings other than the force required to reduce its radius. > > >>and the vertical speed to 1/r as well. This is shown in Feynman's *Lectures >>on Physics*, Vol II, 40-10 ff. Fig. 40-12 is a great drawing of your tank, >>showing the surface countour of a sample vortex. >> >>The reason kinetic energy is increased is that work is done on the chunck >>as it moves inward. In the case of a cylinder of water the work is just >>the work of falling, m*g*h. The work is completely analogous to the work > >This is what I also initially assumed, and it must be the case for >a tank in which the water is essentially initially motionless, and >is eventually brought into motion by the vortex action spreading >from the centre. Assuming that some random minor current in the >water near the drain starts the vortex rotating. This is almost never the case though. The vortex is not "started" by some random motion, but rather by residual angular momentum from the filling of the tank. Did you try an experiment? If the water *were* completely still, then it would be started rotating by the coreolis force, but this is a nominal force, and would not exist if the experiment were done on the equator. If the water were completely still, and located on the equator, then it would flow radially only down the drain. >In this case, the velocity at the rim of the tank will eventually >assume a value determined by the gravitational energy gain, call >this velocity V0, with matching AM m x V0 x R0 (where R0 is the >radius of the rim of the tank). The water at the exact rim of the tank always has tangential velocity zero. The water part way between the rim and the hole carries most of the initial angular momentum. >The gravitational component of the final velocity can't exceed >sqrt(2 x g x h), where h is the initial height of the chunk of >water, and g is the Earth's gravitational acceleration at the >surface. This implies a maximum rim velocity due to gravity of >V0 = sqrt(2 x g x h) x Rd/R0. This is not true in general I think. You are attempting to apply conservation to a single chunk. Conservation of angular momentum and energy only applies in the aggregate. The above relation might apply to water on the surface of the vortex which is "falling" along the surface into the drain, but not to water deeper down. It falls along the water-air surface having a cross section curve h = k/r^2 + h0, where h is the height, and r the radius. Water falling inside the surface falls from a lessor height and contracts from a lessor radius. The pressure and work required to accelerate the chunks is done by water falling which is closer to the rim and above the chunk. > >The problem arises, when one starts with a tank full of water that >is already rotating, especially where the initial velocity at the >rim > V0. >If we now follow this down the drain, we find that the energy gain >of the water is greater than would ensue purely from gravity. I think maybe you are ignoring the fact that the water to go down the drain last has less angular momentum and energy than the water going down the drain earlier. The energy of the rim water falling is translated into increased kinetic energy and angular momentum of the water nearer the drain. The water at the bottom and near the rim doesn't even move much toward the drain until the very end, and then its height is low and it has low initial angular momentum as well. > >The energy *gain* per unit mass is: > >1/2 x V0^2 x ((R0/Rd)^2 -1) which clearly is a function of V0! > >(R0 and Rd being constants). > >IOW by choosing a larger V0, we gain more energy, which implies >that either gravity is not the only source, or the water doesn't >flow, or part of the angular momentum is passed off, such that the >velocity can remain constant. > >This was the reasoning that led to my initial question. > >If a large part of the angular momentum is not passed to the >Earth, then where does the energy come from? It appears the problem is you don't have a good estimate of conditions throughout the tank, and thus you don't have valid integration of the energy and angular momentum that goes down the drain. A visualization that may be helpful is a tank that rotates, and is initially rotating. In that case the boundary conditions of the tank are very different, more like you imagine. If it dumps into a lower tank that is free to rotate frictionlessly, and contains a material that soaks up the angular momentum, then a measure could be taken of the angular momentum and energy transferred down the drain. One interesting thing is that some water would actually remain in the upper tank, unable to go down the drain, as you imagined, held there by centrifugal force. Regards, Horace Heffner
