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Are these forces conservative?

  • #1
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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 ##
 

Answers and Replies

  • #2
BvU
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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
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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
Orodruin
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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
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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
Orodruin
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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
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Thank you for your counter example !
( Last post is too advanced for me, sorry, I wish I could reflect on that )
 
  • #8
BvU
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[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
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Ahhh, ok ! I understand now. Thank you !
 

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