Re: The old loop-de-loop

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

OK, only two forces as described by Hugh (ignoring
negligible forces, such as friction, Coriolis, etc.).
OK, the resultant of these two forces has a radial
component "associated with" the centripetal acceleration.
Also OK for the not very spring-like reaction forces of
JohnD; "a force of constraint does no work," he wrote.

But not yet OK for the abandoning of the concept of force
as a cause of what happens in mechanics. Why would I
need a free-body force diagram if the cause-and-effect
way of thinking were rejected? Everything would be
reduced to kinematics. Here is the simple question that
started our debate: Why does a constraint force appear
(in the laboratory frame) when an object slides inside a
vertical loop? Note that I did not ask how to calculate
the constraint force; the question was qualitative.

We know how this question used to be answered in old
physics textbooks. Millikan, for example, would say that
the constraining force is the normal reaction force to the
centrifugal force. But I am not allowed to say this because
centrifugal forces "do not exist" in inertial frames of
reference. What should I do? How should I explain the
origin of the second force on my free-body diagram?

Some suggested to lean on Newton's first law. An object
has a "tendency" to move along a straight line. The track
generates a constraint force to prevent rectilinear motion.
Why is the tendency to "move with constant v" acceptable
but the "tendency to stay away from the center of rotation"
is not? Yes, I know, modeling reality in terms of hard to
define terms, such as "tendency," should be avoided.
But that issue is more general than our problem. Is it not
true that Newton's law of inertia refers to a "tendency?"
Is not true that a "potential well" is a mathematical
description of a field "tendency?"

If we agree that the constraint force can be attributed to
the law of inertia then we are ready to ask how to
calculate the magnitude of this force, and how to explain
its centripetal orientation. JohnD wrote:

> This exercise suffers from a bit of a chicken-and-egg
> problem.  Let me say it in the most confusing possible
> terms, to illustrate the nature of the confusion: we
> wish to find the acceleration, but we cannot find the
> acceleration until we know C [the second force], but
> since C is a force of constraint we don't know C until
> we know what the acceleration "would have been" ...
> or something like that .....
>
> Now, to cut to the chase:  This is why algebra was
> invented.  I don't see any way to attack this
> problem in "calculator mode" where you punch in
> a value for mg and then punch in a value for C
> and subtract.  I think you need to write down
> one equation in one unknown and solve it.
>
> So here's what we know:
>   -- local tangent direction of track
>   -- local curvature of track.
>   -- direction mg
>   -- magnitude of mg
>   -- direction of C
>   -- constraint:  object stays on track
>   -- assumption:  two-dimensional problem,
>      confined to a single vertical plane.
>
> Here's what we don't know:
>   -- magnitude of C.
>
> In the absence of a boldface font, let C be the
> vector and define c := |C| to be its magnitude.
>
> Algorithm:
>
> 1) Find the net force on the object in terms of
> mg, known track parameters, and the unknown c.
>
> 2) Find the corresponding (note I didn't say
> resulting) acceleration of the object.  Again
> this will involve the unknown c as well as the
> various knowns.
>
> 3) Find the instantaneous radius of curvature of
> the said accelerated motion.
>
> 4) Set that equal to the known local curvature of
> the track.
>
> 5) Solve for c.

I do not know how to implement the algorithm.
Please share the solution of your one equation
with one unknown. Is it really c = m*v^2/r?
Ludwik Kowalski