Re: Elastic potential energy (repeat)

[Brian Whatcott <inet@INTELLISYS.NET>]

Does Joel's expression for elastic potential energy cover the case where
the fully compressed length of the spring is "coil-bound" and needs
finite step force for any extension?  [ A case mentioned recently]

Brian W

At 10:35 AM 11/20/01, you wrote:
> > > ... Does the relaxed position of the spring define a unique
> > > origin for a spring or am I missing something here?
>
> John M responded:
>
> >
> > Not at all, but the correct equation for any choice of
> > the "reference stretch" xo is
> >
> >   elastic potential energy = (k/2)x(x + 2*xo)
> >
> > This reduces to the familiar (k/2)x^2 with the convenient (but
> > not mandatory) choice, xo = 0.
>
> If I'm not mistaken, there is a minus sign error,
> I get for (1dim motion along x axis, F=-k(x-xo);
>
> elastic potential energy = (k/2)x(x-2*xo)
>
> One can take advantage of the ability to add an arbitrary constant to put
> this into a more "elegant" form
>
> namely add (k/2)xo^2 to get
>
> elastic potential energy = (k/2)(x-xo)^2 which is in a familiar form
> compared to the usual
>
> (k/2)x^2  , which is to say in words
>
> "elastic potential energy = half k times (distance spring is stretched or
> compressed) squared" for any choice of coordinate origin.

Brian Whatcott
  Altus OK                      Eureka!