RE: Re: Ball Rolling down on Cosh[t] Path

• To: mathgroup at smc.vnet.net
• Subject: [mg36751] RE: [mg36738] Re: Ball Rolling down on Cosh[t] Path
• From: "DrBob" <drbob at bigfoot.com>
• Date: Sun, 22 Sep 2002 04:32:37 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```I've modified Selwyn's solution to make it more general.  In particular,
the height can be specified (up to about 35 meters).  The differential
equation is solved for t=0 to 5 first.  The quarter-period is computed
by finding a zero; then the differential equation is solved again for
t=0 to the quarter-period, and the solution is extended using reflection
and periodicity.  This yields a higher-precision solution.  Then I graph
the solution, with a stepsize equal to period/40, from t=0 to 'period',
labeling each frame with the values of t, x[t], and y[t].  Using a step
size that divides period/4 guarantees the lowest point is reached ON a
frame when appropriate.

Here's the solution as a notebook expression:

Notebook[{

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Cell[TextData[StyleBox["Having noticed your statement \" ... BEYOND \
MY MEANS \" I thing you aren't yet\nfamiliar with Lagrangian \
formalism. It's quite easy to derive a general\nequation of motion \
for a point mass, subjected to gravity and to moving on a\ncurve f = \
y(x) (i.e. f = Cosh[#]&).\n\n1) I'll leave re-deriving equation to \
you, here is what I've got (just copy\npaste it).:",
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Cell["\<\
3) Here follows animation code, specially for linear versus cosh \
case, apply my initial conditions (below)\
\>", "Text"],

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Bobby Treat

-----Original Message-----
From: Selwyn Hollis [mailto:slhollis at earthlink.net]
To: mathgroup at smc.vnet.net
Subject: [mg36751] [mg36738] Re: Ball Rolling down on Cosh[t] Path

Matthias,

The simplest way to get the equation of motion is to set up the
Lagrangian. Let's assume a 1 kg mass. Then the kinetic energy is

KE = Simplify[(1/2)*(x'[t]^2 + D[Cosh[x[t]],t]^2)]

and the potential energy is

PE = 9.8*Cosh[x[t]]

The Lagrangian is

L = KE - PE

and the equation of motion is

diffeq =
Simplify[ D[D[L, x'[t]], t] ] == Simplify[ D[L, x[t]] ]

Now solve and animate ...

xx[t_] = x[t]/. First[
NDSolve[{diffeq, x[0] == -1, x'[0] == 0}, x[t], {t, 0, 5}]]

curve = Plot[Cosh[x], {x, -1, 1}]

Do[
Show[curve,
Graphics[Disk[{xx[t], Cosh[xx[t]]}, 0.025]],
PlotRange -> {{-1.2, 1.2}, {0.9, 1.65}},
AspectRatio -> Automatic, Axes->None],
{t, 0, 5, 0.1}]

----
Selwyn Hollis

Matthias.Bode at oppenheim.de wrote:
> Dear Colleagues,
>
> I intend to make an animation in which
>
> ball A rolls down on an inclined plane from the left whilst
>
> ball B - starting from the same height - rolls down Cosh[t]'s path
from the
> right.
>
> x-axis is time t, y-axis is height h.
>
> Ball A is fine; ball B - which should arrive at h=0 before A - is
beyond my
> means.
>
> Thank you for your consideration,
>
> Matthias Bode
> Sal. Oppenheim jr. & Cie. KGaA
> Koenigsberger Strasse 29
> D-60487 Frankfurt am Main
> GERMANY
> Tel.: +49(0)69 71 34 53 80
> Mobile: +49(0)172 6 74 95 77
> Fax: +49(0)69 71 34 95 380
> E-mail: matthias.bode at oppenheim.de
> Internet: http://www.oppenheim.de
>
>

```

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