Re: apparent weight

[LUDWIK KOWALSKI <KOWALSKIL@alpha.montclair.edu>]

> Date: Mon, 16 Feb 1998 22:26:04 -0600 (CST)
> From: "A. R. Marlow" <marlow@loyno.edu>
> Subject: Re: Apparent weight
>
> ...  For example, in the case of the experiences in a carnival centrifuge, 
> if we simply reported what we experienced, we would report a force on our 
> backs exerted by the wall of the centrifuge pushing us toward the center 
> of the centrifuge, and we would be perfectly accurate.  This very real 
> centripetal force can leave bruises, do work and be perceived, three 
> things that fictitious "forces" cannot do.  
>
> Somehow, however, people are tempted to create the psychological fiction 
> of an outward force pressing them against the wall; there is no such 
> force, and no amount of coordinate redefinition can create one or 
> transform away the very real effects of the centripetal force that 
> actually exists. 
.........................................................................
Let me defend a naive position. It is a risky busines but what do I have
to loose? I am saying that a cetrifugal force is real, as far as a bead 
mounted on a rotating stick is concerned. The stick is horizontal and turns 
about a vertical axis. Suppose I am that bead and I experience acceleration 
with respect to the beam on which I am sitting. I am gaining speed while 
sliding away from the axis of rotation. What is wrong with saying that m*a 
is the centrifugal force? It is very real in my rotating frame, it can hurt 
by making me dizzy, etc. 

I know it is often simpler to solve problems in non-inertial frames but
why should this prevent me from calling 0.5*m*v^2 a real force? (I can 
measure v with resect to circular markings on the soil below me at any 
time because I happen to have a stopwatch. v<<c)

What is the cause of the net force acting on me? I do not have to know this; 
I am simply recognizing a force by an acceleration which I can measure. And 
I happen to have a bathroom scale attached to my back. What happens when I 
slide to the end of the beam where a wall is mounted to protect me? The 
scale is between me and the wall. What does it read? It reads exactly 
0.5*m*v^2, where v refers to the speed at my current radial distance.

The third law? My body acts on the wall, and the wall acts on my body,
with a force of equal magnitude. Can I predict the outcome of a collision
between two marble balls colliding along the radius (in a tube inside of my 
transparent beam)? Why do I have to answer this question? It has nothing 
to with calling m*a a real force. What is wrong with my reasoning?
..................................................................
> ... Newton's third law of motion does not [ALWAYS] apply in a noninertial 
> frame of reference. Momentum is not [ALWAYS] conserved within such a 
> frame, either.
...................................................................
I took the liberty of adding ALWAYS because Leigh later implied this. 
Forgive me if I am wrong. Sure, non-inertial frames are often easier to 
work with. But sometimes the idea of a real centrifugal force is usueful. 
The rapid sedimentaion inside of a centrifuge tube, for example, is easier 
to comprehand in the rotating frame than in laboratory frame.

                                      Ludwik Kowalski