How to decide the sign of a potential

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    Potential Sign
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Discussion Overview

The discussion revolves around determining the sign of potential energy in mechanics, particularly when multiple forces, such as gravity and spring forces, are involved. Participants explore how to express potential energy mathematically and the implications of sign choices on stability and equilibrium points.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • A user questions how to decide the sign of potential energy when multiple forces are acting on a mass particle, specifically gravity and spring force.
  • One participant asserts that the potential energy of a spring is given by V = 1/2kx², indicating a positive sign.
  • Another participant seeks clarification on whether the potentials from different forces should have the same sign or opposite signs, introducing the concept of centrifugal potential energy.
  • There is a suggestion that the potential due to centrifugal force should be negative, depending on the direction of displacement.
  • A participant emphasizes that the potential must increase in the direction opposite to the force it represents.

Areas of Agreement / Disagreement

Participants express differing views on the signs of potential energy terms and whether they should be consistent or vary based on the forces involved. The discussion remains unresolved regarding the general rule for determining the signs of potentials in combined force scenarios.

Contextual Notes

Participants do not reach a consensus on the conditions under which potential energies should be considered positive or negative, highlighting the complexity of the topic.

laharl88
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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! In fact 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.
 
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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.
 
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 :)
 
laharl88 said:
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
 
Just think of the force that the potential represents. The potential must always increase in the direction opposite the force.
 
Doc Al said:
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

DaleSpam said:
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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