Re: A ball in a rotating dish

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

brian whatcott wrote:

> I simply note that (in answer to Ludwik'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!

Yes, this is no longer a puzzle; the cos(TET) formula describes
the situation at TET=90 correctly. [90 degrees  corresponds
to the largest distance r from the axis of rotation and the ball
would naturally stay at r=R for a very large f. The effect of
the rim is a different issue. Thinking about a dish deeper than
R helps to avoid this issue.]

But I am still stuck by what happens near TET=30 degrees.
I did watch a conical pendulum (slightly elliptical trajectories)
at very small TET; the mathematical formalism for the conical
pendulum is the same as for the ball in a spherical dish. Thus
something must be wrong in the reasoning below. What is it?

> > ... 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?