From: "William Haloupek" <haloupek@voicenet.com>
Subject: Re: Circular Motion....
Date: Mon, 27 Nov 2000 02:17:59 GMT
Newsgroups: sci.math
Summary: [missing]

John Prussing wrote:

>But I'm interested in the solution Bill Haloupek has discovered and
>how it was determined.


Here's the solution.  Of course, any mistakes are mine and not
Schoenberg's!  Sorry about the clumsy ascii.  You'll probably
have you write this out for yourself if you want to make any
sense of it.  I find that most mathematicians are surprised
to learn that there is more than one way to move along a curve
with constant magnitude of acceleration.

To recap, we are looking for motions on a circle of radius R>0
for which the magnitude of acceleration is 1.

If  v  is speed, then the normal and tangential components of
acceleration are  (v^2)/R and  dv/dt, respectively.
Thus  |a| = 1  becomes

(dv/dt)^2 + (v^4)/(R^2) = 1.              (1)

This is equivalent to the equation (dw/dt)^2 + w^4 = 1  that
John Prussing posted.

There is an obvious class of solutions in which
dv/dt = 0.  But since (1) is a 2^nd order DE,
we would expect another one-parameter family
of solutions.  From here on let's assume dv/dt
is not identically zero.

Now, the standard trick is to use the identity

dv/dt = (dv/ds)*(ds/dt) = (dv/ds)*v

where  s = arc length.  Thus (1) becomes

(v^2)*(dv/ds)^2 = 1 - (v^4)/(R^2).            (2)

Multiplying both sides by 16v^4 gives

((4v^3)*(dv/ds))^2 = (16v^4)*(1-(v^4)/(R^2)).   (3)

Next, use the substitution  u = v^4.  You get

(du/ds)^2 = 16u*(1-u/R^2).                      (4)

This can actually be solved!  Integrating,

4s = +/-  int(du/sqrt(u*(1-u/R^2)))
     = +/-  2r*arcsin(sqrt(u)/R) +c

where  c  is constant.  Inverting,

u = (R*sin(2*(s-c)/R))^2,   or

v = sqrt(R*sin(2*(s-c)/R)).       (5)

This motion starts from rest (when s = c),
goes 90 degrees around the circle, and
comes to rest again.  At the rest points
you can patch together two solutions.
In between rest points, the speed reaches
a maximum of  1/sqrt(R), which is the speed
of the constant-speed motion with |a| = 1.
So at these points one can patch together
the two different types of solutions.

Attempting to solve (5) results in

t = (1/sqrt(R))*int(sqrt(csc(2s/R))ds)

which, I think, is nonelementary.

This little example can be found in

I.J. Schoenberg,  The Landau Problem, Parts I, II and III.

These articles appeared in an obscure Scandinavian journal,
but most people can probably find them most easily in

I.J. Shoenberg: Sellected Works, Carl de Boor, Ed,
Birkhauser, 1988.

Shoenberg generalized the qualitative approach to
an equivalent problem called Landau's Problem.
He actually made some progess on the general
approach started by  spamless@nil.nil  in this
thread.

And in the "sorry you asked" category,


>And I'm curious whether he pronounces his name Hal'oupek, Halou'pek,
>or Haloupek'.
>--
>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
>John E. Prussing
>Dept. of Aeronautical & Astronautical Engineering
>University of Illinois at Urbana-Champaign
>http://www.uiuc.edu/~prussing
>=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

The accent is on the first syllable.  According to
my grandfather, the original in Bohemian has
an un-pronounceable (by Americans!) diphthong
at the front.  It means "little cottage in the woods".
Grandpa said that, if you had a peck of peas,
and you hulled them, you would "hull a peck"
of peas, and that's how you pronounce Haloupek.
I think he probably was not being true to the
original Bohemian, but was just trying to give
Americans something they could pronounce.

Bill Haloupek