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Re: Ball Rolling down on Cosh[t] Path
*To*: mathgroup at smc.vnet.net
*Subject*: [mg36738] Re: Ball Rolling down on Cosh[t] Path
*From*: Selwyn Hollis <slhollis at earthlink.net>
*Date*: Sat, 21 Sep 2002 02:22:08 -0400 (EDT)
*References*: <ambv6f$r44$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
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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