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Re: A bawl in a rotating dish (was Hot air rising ...)



Referring to a ball in a rotating dish, such as a salad bawl on
a turn-table, Robert Cohen 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).

Two of Bob's massages are quoted at the end.

I am not sure the ball will remain stationary. What is the
roll of friction in a situation like this?

I tried to analyze this but was stock without going too far.
So I need some help.

To avoid singularity the initial ball position must NOT be at
the bottom. By increasing the rotating frequency of the dish, f,
I expect to see the ball rising higher and higher toward the rim.
To avoid an idealization (a rolling ball with a single contact point)
I was considering of small piece of wood or plastic whose bottom
surface matches the inner radius of curvature of the spinning dish.
I was planning to make the coefficients of friction very small and
very large.

Suppose the dish is a perfect sphere and there is no friction.
Then the restoring force acting on the ball is exactly the same
as in a pendulum, m*g*sin(TET). I will use TET as the position
coordinate of the piece; it is the angle between the axis of rotation
and the radius vector toward the piece. The origin of that radius
vector is at the center of the sphere; it is the same point at which
the corresponding pendulum would be suspended.

To produce equilibrium, in the rotating frame, the tangential
component of the centrifugal force and the restoring force must
have the same magnitude.

    m*v^2*cos(TET)/(R*sin(TET) = m*g*sin(TET).

Substituting 2*Pi*R*sin(TET)*f for v, leads to a relation
between the expected angle TET (position of the small piece
on the dish surface) and the rotating frequency of the dish.

                  cos(TET)=g/(R*omega^2)

where omega=2*Pi*f. Using g=9.8 m/s^2 and R=0.2 m gives

TET=30.46 degrees    when f=1.2 rot/s
          56.52           f=1.5
          71.92           f=2.0
          82.07           f=3.0
          87.00           f=5.0
          89.29           f=10
          89.99           f=100

Something is not right; my formula does not allow for
f<1.114 (which is the frequency of the corresponding
pendulum).  It also keeps the piece on the surface, no
matter how large is f. What is wrong?
Ludwik Kowalski
*********************************************
Here are two original messages from Bob:
----------------------------------------------------------------------
How about spinning a parabolic surface about its center?  At a particular
rotation rate, a ball placed on its surface and at rest relative to the
surface will stay stationary relative to the surface.  At this speed, the
normal force (acting centripetally) will balance the centrifugal force
(or, in other words, the normal force will be providing the centripetal
force necessary for the ball to follow the circular path prescribed by the
moving surface...whew!).  Thus, any motion relative to the surface will
appear to be influenced by the coriolis force only (not the centrifugal
force), as is the case with the geostrophic winds.

If we drill a hole in the surface and attach the ball (that is residing on
the surface) to an object that is free to hang down through the hole, the
hanging bob will provide a force on the ball that will attract it to the
hole, much like a low pressure center does on the atmosphere.  From the
point of view of a camera mounted to rotate with the surface, the ball
will appear to spiral around the hole, rather than go directly into it.
This should be sufficient to demonstrate what you are looking for.

By the way, the direction of the spiral will be the same direction
(clockwise or counter-clockwise) as the spin of the surface.  From the
"rest" frame, the stationary ball is spinning around the hole and so as it
approaches the hole it spins faster, conserving angular momentum.
------------------------------------------------------------------------
I'm not sure if I understand the problem, but here is my thinking.  If you
take a dish of liquid and rotate it then for "solid body" rotation (i.e.,
the fluid is rotating as a solid body rather than sloshing around) the
surface will be parabolic.  A cork can be placed on the surface (as
it rotates) and, if the cork is moving with the fluid, it will remain
stationary relative to the fluid.   Thus, from the cork's point of view,
there are no horizontal (relative to the fluid surface) forces on it, no
matter where it is placed on the fluid (close to axis, far from axis,
etc.).

If we replace the fluid with a solid surface and the replace the cork with
a ball, 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).  From the ball's point of view,
there are no "apparent" horizontal forces acting on it as long as it
doesn't move (relative to the surface).  As soon as it starts moving,
there will be a Coriolis force.

I haven't done the math - I'm just guessing that the surface exerts a
normal force that has a component directed equal but opposite to the
centrifugal force, no matter where the ball is placed.  I'm guessing that
this is what you want, since you want to observe the effect of the
coriolis force.  Centrifugal forces will complicate things for you.

P.S. A good physics question is to ask why the winds feel only a coriolis
force and not a centrifugal force.  Another good one is to ask how one
might roll a ball down a plane and have it *not* experience a coriolis
force (granted, the coriolis force would be small in any case, but there
is a way to get it to go to zero).

----------------------------------------------------------
| 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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