# A water drop in a falling lift

Consider a lift, which is at rest in an homogeneous gravity field. There is a thin layer(with thickness $h$) of water on the floor of the lift. At some moment a single cable, supporting the lift, breaks and the lift begins free falling(forever). It is easy to describe qualitatively what happens with water, i think: the formation of a drop begins. During this process the drop jumps up from the floor and once the total kinetic energy of the drop is dissipated into the heat, the center of the drop stands at some height $H$ from the floor. (All the kinetic energy of the drop is coming from the difference in surface tension energy of water between initial and end moments.)

Question: How to determine(approximately) $H$ ?

For simplicity let's assume that the cross-section of the lift is a disc-shaped with radius $R$. $(h<<R)$

Remark: A key point is to determine how the total kinetic energy distributes between internal kinetic energy and translational kinetic energy of the drop, i think. With this the problem will be solved.

So it is convenient at the early stage of formation of the drop ignore viscosity effects of water and air resistance.

Edit:

I would like to clarify that by internal kinetic energy i mean the kinetic energy of macroscopic motion of water inside the drop. At early stage there is no energy loss assumed, for simplicity.

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Downvoted because the question is much too vague. Even if a drop formed and started moving, why would it stop? The answer depends on knowing about the surface of the floor of the elevator, the air in the elevator, and requires taking into account thermodynamics effects. –  Mark Eichenlaub Mar 15 '11 at 17:18

The only reason the water would form into a drop is the surface tension. You can work out the reduction in the energy stored in the surface tension in being a drop rather than a flat surface using a simple calc of the surface area. There's no particular reason I can see that the drop would float off the floor at all.

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You said it looses energy, well this has to go somewhere, so most likely into translation kinetic energy. –  user1708 Mar 15 '11 at 10:29
@kakemonsteret: If the energy goes into translation kinetic energy then it will move and continue moving until it hits an obstacle. It will not float to a certain height and then stop. –  delete Mar 15 '11 at 10:53
@mast Its in contact with air, and there is a temperature difference due to that kinetic energy, so as its coming into thermal equilibrium with the air its slowing down. This could take a while and it may never come to a full stop. Infact it could be in thermal equilibirum and still ahve translation kinetic energy, like rivers. –  user1708 Mar 15 '11 at 10:56
@Master: The drop must stop because of air resistance. We neglect this resistance for simplicity during the formation of the drop only. –  Martin Gales Mar 15 '11 at 11:22

Unless there is some inital temperature difference, its total kinetic energy is just its translational kinetic energy, which is the energy gained from its change in shape due to surface tension, from which you can get its temperature and also its velocity $v_0$.

You have the surface area, temperatures, conductivities, so you can get the heat transfer per time from fourier's law.

From which you can get the energy E(t), which gives you v(t), a solution to v(t)=0, if it exists, is T, then integrate v(t) from 0 to T to get H.

Most likely it wont come to a full stop, it could be in thermal equilibrium and still have translation kinetic energy, and even more likely it approaches velocity 0 asymptotically.

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Temperature is irrelevant in the problem, i think. At the early stage we ignore viscosity and assume that energy does not dissipate. By internal kinetic energy i meant macroscopic motion of water inside the drop. The translational motion of the drop decelerates by Stoke's law and it stops certainly. –  Martin Gales Mar 15 '11 at 11:44