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

["Albert J. Mallinckrodt" <ajm@cpp.edu>]

?  What two different answers?  In Peter’s original post he got two answers, one of which was simply wrong. 

John Mallinckrodt
Cal Poly Pomona

> On Dec 8, 2020, at 7:52 AM, Ken Caviness <caviness@southern.edu> wrote:
>
> 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,
>
> Ken
>
> Kenneth E. Caviness, Ph.D.
> Chair & Professor, Physics & Engineering
> Southern Adventist University
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> -----Original Message-----
> From: Phys-l <phys-l-bounces@mail.phys-l.org> On Behalf Of Brian Whatcott
> Sent: Tuesday, 8 December, 2020 10:38
> To: Phys-L@phys-l.org
> 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 <ajm@cpp.edu> wrote:
>
> Yes. I’m guessing Peter had that “block“ frozen.
>
> > On Dec 8, 2020, at 6:44 AM, John Denker via Phys-l <phys-l@mail.phys-l.org> 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
> > position.
> > 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
> > experience.
> >
> > 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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