How to prove that a force is conservative?

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  • #36
jbriggs444
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I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.
Yet i am also told that the spring force is conservative despite only knowing the force in 1-dimension and not anywhere else.
I fail to see the difference between the 2 cases
The difference that I see is whether the force falls within the same sub-space as the motion. How can you have an inner product space if you can't figure out what space you are talking about?
 
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Orodruin
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I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.
Yet i am also told that the spring force is conservative despite only knowing the force in 1-dimension and not anywhere else.
I fail to see the difference between the 2 cases
You have been told the difference repeatedly. However, you seem to be ignoring this.

In the first case you are looking at the entire force field in three dimensions and a single possible solution to the motion. There is nothing a priori telling you that the motion is one-dimensional. Any particle in three dimensions will by definition move along what is a one-dimensional path.

In the second case, all possible solutions are one-dimensional and typically constrained by other forces to move in one dimension only. (Note: In the constrained case, curl is not an applicable concept.)
 
  • #38
dyn
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The university exam paper that i took the initial case from regarding the motion in a circle says that the force is conservative. If i accept that then i can also understand why the spring force is conservative,
Thank you to everyone who replied
 
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Orodruin
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The university exam paper that i took the initial case from regarding the motion in a circle says that the force is conservative. If i accept that then i can also understand why the spring force is conservative,
Thank you to everyone who replied
Again, the original question is ill posed. You should not pay too much attention to it. Being in a university exam paper is no guarantee that it is correct.
 
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Orodruin
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The model solution writes the components of F such as Fx = -mω2x.
To be more specific, the solution quotes the force field ##\vec F_1 = - mω^2 \vec x##. However, all we can conclude from the problem statement is that this force field coincides with the actual force field along a circle of radius ##R##. The actual force field may have any other behaviour away from this circle. We just don’t know. An example would be
$$
\vec F_2 = - mω^2 \vec x + (R^2 - x^2 - y^2) \vec k \times \vec x
$$
where ##\vec k## is a constant vector. The curl of this field is non-zero. We now have two fields, ##\vec F_1## and ##\vec F_2## that would both have the given circular path as a possible solution for the particle motion. One of them is conservative and the other not. It is therefore clear that it cannot be concluded with the given information whether the field is conservative or not.
 
  • #41
Office_Shredder
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I have been told that force in #1 is not conservative because we only know the force on the circle which can be considered 1-dimensional and not anywhere else.

You don't know the force on the circle. You know the force on the circle conditional on the particle moving in the way it moves. That's what I mean by you only know one path. If another particle moves another way (e.g. at a different speed around the circle), then you don't know what the force is going to be.

If the force is being generated by gravity, then the particle will move off the path of the circle. If the force is being generated by normal force from walls that constrain you to be inside the circle, then the particle may continue to be in the circle.
 

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