How to decide the sign of a potential

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SUMMARY

In rational mechanics, the sign of a potential is determined by the direction of the forces acting on a mass particle. When considering gravitational potential energy (V = mgy) and spring potential energy (V = 1/2 kx²), the signs of these potentials are independent of each other. If the centrifugal force is involved, as in a rotating plane, the potential due to this force is negative (V = -1/2 mω²x²). Understanding the relationship between force direction and potential sign is crucial for analyzing stability in equilibrium points.

PREREQUISITES
  • Understanding of gravitational potential energy (V = mgy)
  • Knowledge of spring potential energy (V = 1/2 kx²)
  • Familiarity with centrifugal force and its implications in mechanics
  • Basic principles of stability in equilibrium points
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  • Study the concept of potential energy in rotating systems
  • Learn about the stability of equilibrium points in mechanical systems
  • Explore the implications of force direction on potential energy calculations
  • Investigate the relationship between conservative forces and potential energy
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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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