Re: "simple" pendulum

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

Referring to:

> > ... For example, in the rotating frame of reference the bob
> > of a conical pendulum moves in a fixed plane (PHI=const).

Leigh Palmer wrote:

> That's incorrect. The bob of a conical pendulum is in
> equilibrium, at rest, in the only rotating frame that
> makes sense.

I am sitting on a horizontal platform which rotates below
the bob. The axis of rotation coincides with the line along
which an inertial observer A would see the bob hanging
in equilibrium. As far as I am concerned (observer B) the
bob is swinging in the vertical plane. To me it is a simple
pendulum to A it is a conical pendulum. The bob is not
at rest in my frame, that frame makes sense to me.

But let play this logical game, perhaps we will find
another paradox, or a misconception.

First, we agree that relativistic effects are negligible
at our speeds and we use the same clock on the wall.
Then we messier vertical accelerations; both are
9.8 m/s^2. And we understand why the simple
pendulum in the B frame becomes a conical pendulum
in the A frame.

At that point the observer B asks A to construct a
pendulum of the same L as that he is playing with.
She does this and her simple pendulum now
swings with the same amplitude as his. And she
discovers a paradoxical situation. The period she
measures is not the same as he measures.

How can it be? Two pendula are simple (in their
own frames) and observers agree on both L and g.
The only difference is that m*v^2/r is real to him
but fictitious to her. Same clock, same newtonian
physics.
                   Ludwik Kowalski