Re: A ball in a dish

[Robert A Cohen <bbq@ESU.EDU>]

On Mon, 2 Aug 1999, Ludwik Kowalski wrote:

> > ... the ball will remain stationary (relative to the surface)
> > no matter where the ball is placed as long as the ball is
> > initially stationary (relative to the surface).
>
> He was actually referring to a parabolic not to a spherical
> surface. Perhaps this is more significant than I can see. The
> cos(TET) formula was "derived (?)" for any rotationally
> symmetric surface. I still suspect that something was wrong
> in the derivation. What was it?

Yes, the shape is the key.  Let's see if I can derive all the relevant
stuff...

If one rotates a fluid in "solid-body" rotation, the radial component of
the pressure gradient will balance out the centrifugal force (for
simplicity the masses are not written):

    (1/rho)dP/dr = omega^2 r

Using P = rho*h*g, and assuming rho is constant...

     dh/dr = omega^2 r / g

This gives the slope of the surface at any radial point r.  Now, not only
will the horizontal pressure gradient equal the centrifugal force at any
radius, but so will the horizontal component of the normal force exerted
on a object on the surface.

For any sloping surface, the horizontal component of the normal force
is...

    mg tan(theta).

In this case, the slope of the surface is...

     theta(r) = inverse tangent of (dh/dr).

so that the horizontal component of the normal force is...

    mg tan(theta) = mg dh/dr = m omega^2 r

which, is exactly equal to the centrifugal force.  As the radius
increases, the centrifugal force increases but so does the slope (and thus
the centripetal force that is provided by the normal force).

Getting back to the spherical surface then, one can see that for a given
omega, there will be only one radial distance in which the normal force
balances the centrifugal force.  Beyond that distance, the slope is larger
than that for the parabola and the object accelerates centripetally
(toward the center).  Closer than that distance, the slope is less than
that for the parabola and the object accelerates centrifugally (away from
the center). For any given omega, there will always be a minimum radial
distance below which the object will be accelerated centrifugally.  Not so
for the h(r) surface described above.  For a particular omega, there is a
specific surface such that the object will remain stationary (relative to
the surface) no matter where it is placed, whether it is placed at axis or
any other place.  Of course, if the surface shape doesn't match the omega,
then this won't be true.

As far as I can tell, there is nothing wrong with your derivation - I will
look at it in more detail, though, and see what I can find.  I just wanted
to provide some of the mathematical rigor that I had left out of my
previous posts.

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| Robert Cohen               Department of Physics       |
|                            East Stroudsburg University |
| bbq@esu.edu                East Stroudsburg, PA  18301 |
| http://www.esu.edu/~bbq/   (570) 422-3428              |
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