Are these forces conservative?

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In summary, the conversation discusses three different types of forces: one that is dependent on time and not position, one that is not conservative, and one that is conservative but only for certain conditions. The group also mentions the criteria for determining if a force is conservative and gives examples of counter-examples to these criteria.
  • #1
geoffrey159
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Homework Statement



a - ##\vec F = \vec F_0 \sin(at) ##
b - ##F = A\theta \hat r##, A constant and ## 0 \le \theta < 2\pi ##. ##\vec F## is limited to the (x,y) plane
c - A force which depends on the velociity of a particle but which is always perpendicular to the velocity

Homework Equations

The Attempt at a Solution



I don't know what to say for (a)

a -
b - Not conservative because ##\vec \nabla \times \vec F = -\frac{A}{r} \hat k \neq \vec 0 ##
c - Conservative because instantaneous work is 0, so the work done on any closed path is 0 : ##\vec F.d\vec r = \vec F.\vec v\ dt = 0 ##
 
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  • #2
You could look at the criteria ... and get mixed results. So perhaps it's reasonable to check them (what comes out ?) and then think of a counter-example.
 
  • #3
The force depends on time and not on position, so the curl test is useless.
Remains the integral test:
##\oint \vec F.d\vec r = \int_0^T \vec F.\vec v\ dt = \int_0^T \vec F_0.\vec v \sin(at)\ dt##
But I don't know what to do with that.

I found something, with the assumption that F is a net force on a body of mass m, with initial velocity equal to v0=0.
Using Newton's 2nd law and integration, I can calculate the position x.
In order to find a closed path, one must find t>0 such that x(t) = x(0) <=> at = sin(at). It is not possible with that force, because the equation x = sin(x) has one solution: x = 0, so at = 0, and t = 0. Therefore, the force is not conservative.
 
  • #4
What happens if you let the force act on an object that takes the path ##\vec x(t) = \vec x_0 \cos(at)## over one full period?
 
  • #5
Hello,
If there is no mistakes, I find that the work on this path is ##W = -\pi \vec F_0.\vec x_0 \neq 0##, which proves that the force is non conservative. Thanks, that makes it clear !
 
  • #6
Yes, but, note that the other conditions such as ##\nabla\times \vec F = 0## are fulfilled. The point is that the conditions are equivalent only for force fields that do not depend explicitly on time. Even if a time dependent force field has a potential for every fixed time, the time dependence breaks time translation invariance, which is what leads to energy conservation through Noether's theorem.
 
  • #7
Thank you for your counter example !
( Last post is too advanced for me, sorry, I wish I could reflect on that )
 
  • #8
[edit] wrote this earlier, forgot to post...

The force depends on time and not on position, so the curl test is useless.
Curl is zero and F can be written as the gradient of a potential. But the equivalence of these two with the integral test is lost (at least for trajectories that take a finite time). Oro has a sophisticated way of saying that in #6.

His counter-example is one of the many possible.
The one I was thinking of is this force driving a harmonic oscillator (without damping and with the same natural frequency). The the oscillator amplitude keeps growing, meaning work is transferred to it. Work that the force field does. [edit] I realize it's almost the same example!
A closer-by example: pushing a swing. A little push every time at the right time gives a big amplitude after a while.
 
  • #9
Ahhh, ok ! I understand now. Thank you !
 

1. Is the gravitational force conservative?

Yes, the gravitational force is a conservative force. This means that the work done by the force is independent of the path taken and only depends on the initial and final positions of the object.

2. How can you determine if a force is conservative or not?

A force is considered conservative if the work done by the force on an object is the same for any path taken between two points. This can be mathematically represented by the integral of the force over a closed loop being equal to zero.

3. What is the difference between conservative and non-conservative forces?

Conservative forces are those in which the work done by the force is independent of the path taken. Non-conservative forces, on the other hand, are those in which the work done depends on the path taken.

4. Can a force be both conservative and non-conservative?

No, a force cannot be both conservative and non-conservative. It is either one or the other. However, some forces may have both conservative and non-conservative components.

5. Why is it important to know if a force is conservative or not?

Knowing whether a force is conservative or not can help in understanding the behavior of a system and predicting its motion. It can also help in simplifying mathematical calculations and determining the energy of a system.

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