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

[brian whatcott <inet@INTELLISYS.NET>]

At 10:37 7/31/99 +0100, Ludwik wrote:

> Referring to a ball in a rotating dish, such as a salad bowl 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).
...
> 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)....
> 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)....

> 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

I took pleasure in playing with this puzzle for an hour or so.
Then after equating sin(theta) and cos(theta) and tan(theta)
to theta for small theta, and approximating cos theta to
1 - theta^2/2.... I sat back and contemplated the shakiness
of the edifice - so lacking the shortcut insight, I simply note
that (in answer to Ludwick's second question) no matter how fast
a spherical bowl is rotated, the internal ball goes nowhere
when it hits the vertical surface until the bowl shatters!

I would not be displeased to see one of those magnificently
clear developments of Ludwick's puzzle for which this list
has been noted.



brian whatcott <inet@intellisys.net>
Altus OK