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- Newsgroups: sci.physics
- Path: sparky!uunet!elroy.jpl.nasa.gov!nntp-server.caltech.edu!jboyce
- From: jboyce@cco.caltech.edu (Jack K. Boyce)
- Subject: Re: Rotating Bucket of Water in a Gravitation Field
- Message-ID: <1992Sep1.004013.157@cco.caltech.edu>
- Sender: news@cco.caltech.edu
- Nntp-Posting-Host: scratchy
- Organization: California Institute of Technology, Pasadena
- References: <30AUG199222561108@zeus.tamu.edu> <ZOWIE.92Aug30125336@daedalus.stanford.edu>
- Date: Tue, 1 Sep 1992 00:40:13 GMT
- Lines: 31
-
- zowie@daedalus.stanford.edu (Craig "Powderkeg" DeForest) writes:
-
- >In an earlier article, wdb3926@zeus.tamu.edu (BRUTON, WILLIAM DANIEL) writes:
- > Anyone ever proved that the surface of water in a bucket rotating
- > in a gravitational field is a paraboloid?
-
- > I remember seeing this problem worked out is some text, but I can't seem to
- > relocated it. I seem to remember using Bernoulli's Equation....but
- > the form of Bernoulli's Equation that I have seen is not gereralized
- > to rotational motion.
-
- >A better way to do it is [am I giving this away?] to transform into a
- >better reference frame where there isn't any motion; since this will
- >be a non-inertial frame, there'll be fictitious forces in it that (in
- >general) vary with position.
- >
- >From there, you can use variational techniques (see _The_Feynman_Lectures_,
- >II-19, p. 3 for a description) to find the shape of the surface.
-
-
- You don't really need variational methods to solve this one. Just make
- sure that the potential energy everywhere on the surface is constant.
- In the rotating frame there is a centrifugal force (w^2)r acting outward,
- giving a "potential" of -0.5(w^2)(r^2). So mgh-0.5(w^2)(r^2) = constant,
- and you're done.
-
- Gee, I hope this wasn't someone's homework problem...
-
- Jack
- jboyce@physics.berkeley.edu
-
-