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MathGroup Archive 2002

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36755] Re: [mg36730] Re: Ball Rolling down on Cosh[t] Path
  • From: Selwyn Hollis <slhollis at earthlink.net>
  • Date: Mon, 23 Sep 2002 03:32:52 -0400 (EDT)
  • References: <000001c261fc$b8901b60$0200a8c0@HolyCow>
  • Sender: owner-wri-mathgroup at wolfram.com

Bobby,

Your ODE is different from both mine and Borut's. You do however get the 
same as mine with your approach, if you differentiate the total energy 
rather than the Lagrangian.

   KE = Simplify[(x'[t]^2 + D[f[x[t]], t]^2)/2];
   PE = g*f[x[t]];
   totalE = KE + PE;
   treat = D[totalE, t] == 0

   x''[t] - (x''[t] /. First[Solve[treat, x''[t]]]//Apart) == 0

Borut's ODE differs only in his factor of 2 in the first-order term. I 
suspect that might have been a typo... Borut?

By the way, some nice animations of mechanical systems (some based on 
the same sort of approach) can be seen here: 
http://www.math.armstrong.edu/faculty/hollis/DEmovies/


--Selwyn



DrBob wrote:

> The two of you derived slightly different ODE's. I think Selwyn's is
> correct, but I only had one physics course, 30 years ago.
>
> Here's a notebook expression showing how Selwyn's approach can be used
> to derive the ODE for y[t] = f[x[t]], with arbitrary f. It shows that
> solution in a form that is easily compared with Borut's ODE, and then it
> shows my own solution (same as Borut's, by a slightly simpler method).
>
> Bobby Treat

<snip>

 > KE = Simplify[(x'[t]^2 + D[f[x[t]], t]^2)/2];
 > PE = g*f[x[t]];
 > L = KE - PE;
 > treat = D[L, t] == 0

<snip>

>




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