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. >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. 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 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. 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. 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? The only other possibility I can see, is that we have a situation where the centrifugal force is so large at the drain, that the inner radius of the vortex is in fact larger than the drain, and no water goes down at all, which means that the water eventually slows down due to internal friction, until a point is reached that the weakening centrifugal force allows it flow. Regards, Robin van Spaandonk All SPAM goes in the trash unread.
