How to decide the sign of a potential

  • Thread starter laharl88
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  • #1
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Hi! I'm a new user of this forum, although i've been reading a few threads for a while...
Mi question is this: in rational mechanics, how do i decide the sign of a potential?
Let me explain better: in some exercises it may happen that a mass particle is subject to both gravity and, for example, the force of a spring. If by hypothesis, the two forces are always perpendicular, how should i write the potential?

V= mgy -1/2*k*x^2 or V=mgy + 1/2*k*x^2

Let me underline that this question is not trivial!!! Infact that mere sign can change everything, including the stability of equilibria points!!
I'd be really grateful to anyone who can answer this question.
If i wasn't clear in explaining my doubts, please fell free to tell me.
 

Answers and Replies

  • #2
Doc Al
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In the case of spring PE, x is displacement from the unstretched position and the PE is 1/2kx² not -1/2kx².
 
  • #3
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I know that, i just want to know if the two potentials must have the same sign, or the opposite sign. Suppose, instead of a spring, the plane in which the motion happens to rotate with a constant w. In this case there would be a potential, due to the centrifugal force, given by
V1= 1/2*m*w^2*x^2
In this case, do the two potentials have the same sign or the opposite?
By the way, thanks for answering :)
 
  • #4
Doc Al
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I know that, i just want to know if the two potentials must have the same sign, or the opposite sign.
I guess I don't understand your question then. Each term in the potential has its own sign, independent of the other terms.

Suppose, instead of a spring, the plane in which the motion happens to rotate with a constant w. In this case there would be a potential, due to the centrifugal force, given by
V1= 1/2*m*w^2*x^2
Since the centrifugal force is outward, that potential must be negative, assuming x is the distance from the center: V = -1/2mω²x²
 
  • #5
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Just think of the force that the potential represents. The potential must always increase in the direction opposite the force.
 
  • #6
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I guess I don't understand your question then. Each term in the potential has its own sign, independent of the other terms.


Since the centrifugal force is outward, that potential must be negative, assuming x is the distance from the center: V = -1/2mω²x²
Just think of the force that the potential represents. The potential must always increase in the direction opposite the force.
Ok,i think i understand now, thanks a lot guys :)
 

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