Newsgroups: comp.theory.dynamic-sys
Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!zaphod.mps.ohio-state.edu!magnus.acs.ohio-state.edu!csn!boulder!boulder!weverka
From: weverka@boulder.colorado.edu (Robert T. Weverka)
Subject: Re: double pendulum - revisited.
Message-ID: <1991Apr24.154100.17128@colorado.edu>
Sender: news@colorado.edu (The Daily Planet)
Nntp-Posting-Host: espresso.colorado.edu
Reply-To: weverka@espresso.Colorado.EDU (Robert T. Weverka)
Organization: University of Colorado, Boulder
References: <puchm.672390379@cutmcvax>
Date: Wed, 24 Apr 1991 15:41:00 GMT
Lines: 38

mail bounced so this gets posted...
In article <puchm.672390379@cutmcvax> puchm@cutmcvax.cutmcvax.cs.curtin.edu.au (RichardPuchmayer) writes:
>
>                T1 = 1/2 * m1 * l1 * th1.
>                T2 = 1/2 * m2 * (l1 * th1. + l2 * th2.)
>
>                U1 = -m1 * g * l1 * cos(th1)
>                U2 = -m2 * g * l2 * cos(th2) * l1 * cos(th1)
>
>            Are the potentials correct ?
>            If YES, why ?
>            If NO, why and what are the correct ones?
>            The above questions should indicate a total lack of
>            understanding as to the formulation of the potentials.
>
U = m * g * h

so...

you should have
             U2 = -m2 * g * ( l2 * cos(th2) + l1 * cos(th1) )
since the quantity in parenthesis is the height.

For the kinetic   T= 1/2 m v^2

so ...

          v1 = th1. * l
          v2 = l * ~th1.  + l * ~th2.
where ~ denotes vector quantity.  Add these vectorally and compute the magnitude
squared.

Note: check your equations with dimensional analysis.
this would have shown you the error you have in T1 and T2.

have fun.  -Ted