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-----Original Message-----
From: Brian Whatcott [mailto:inet@INTELLISYS.NET]
Sent: Tuesday, November 20, 2001 12:06 PM
To: PHYS-L@lists.nau.edu
Subject: Re: Elastic potential energy (repeat)
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:
constant to put... 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
this into a more "elegant" formstretched or
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
compressed) squared" for any choice of coordinate origin.
Brian Whatcott
Altus OK Eureka!