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Re: [Phys-L] [ext] Re: Ex: Re: Fluids problem

I thought that too, but if we imagine moving the cylindrically shaped volume of water so that its center of mass moves up to ground level, without assuming any shape-shifting, half of the water is above the CoM, half below. Is it not true that then the shape-shift itself has a zero-energy cost? For every slice at depth y below the CoM, there's a matching slice at height y above the CoM: no net change in potential energy. I would expect this to be -- ahem! -- a wash. <grin!> But it can't be, since we get two different answers using slices and CoM calculations.

Still musing,


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-----Original Message-----
From: Phys-l <> On Behalf Of Brian Whatcott
Sent: Tuesday, 8 December, 2020 10:38
Subject: [ext] Re: [Phys-L] Ex: Re: Fluids problem

Key concept: a cylinder of water shape-shifts to an infinitesimally thin lamina of water when pumped at lowest energy cost. On Tuesday, December 8, 2020, 09:27:40 AM CST, Albert J. Mallinckrodt <> wrote:

Yes. I’m guessing Peter had that “block“ frozen.

On Dec 8, 2020, at 6:44 AM, John Denker via Phys-l <> wrote:

Setting aside typos, the key idea is this:

The center of mass is given by:
        ∫ X dm / ∫ dm          [1]

pretty much by definition, where dm is an element of mass, and X is
Note X can be one dimensional in the simple introductory situation, or
higher-dimensional if you want.

Given the symmetry of the situation, you can find the CM by
inspection, based on physicist's intuition and experience, without
doing the calculus.  It's in the middle.

If you want to do the calculus, it's
        ∫ X dX / ∫ dX

since in this situation dm is proportional to dX.
Turn the crank and find that the CM is halfway between the limits of
integration ... in agreement with the aforementioned intuition and

This comes up All The Time.

Note that the same formula [1] is also the formula for weighted
average, where dm tells you how things get weighted.  In the case
where dm = dX this reduces to a simple unweighted average.

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