Unsuccessful DIY - Seeking Help

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In summary, the conversation discusses different methods of solving a problem involving a conservative force and a closed line integral. One individual tries to use substitution and Stokes' Theorem, while another suggests using the differential equation for motion and the concept of work. Ultimately, it is determined that the force is not conservative, as the line integral is time dependent and cannot be 0 unless the particle is at rest. Another individual suggests a simpler solution involving the direction of the velocity and the dot product between the force and displacement vectors.
  • #1
Brad Barker
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I tried to do this myself, but I was unsuccessful.

Thank you.
 
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  • #2
Ok, I was able to do this after all.

My first method was to substitute in -bv in for F and try to show that the closed line integral of F with dr is not equal to 0. This did not get me anywhere, although I imagine that some masters of vector calculus could get the result this way.

What was successful for me was exploiting Stokes' Theorem, and also solving the differential equation

m dv/dt = -bv.

I got v from this, plugged it back into my equation for force, and then computed the curl, which is not equal to zero so long as the velocity is not equal to zero.

Could someone verify that this is correct?

Thank you.
 
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  • #3
Brad,

No fair! You didn't really say that the force had to be in the same direction as the velocity! In that case it's obvious; the velocity of the mass is always increasing. So if it travels ina closed loop, it's going faster at the end than it was at the beginning. KE isn't conserved.
 
  • #4
Consider the work done on a particle in going from a to b.
Now consider going along the same path, but with twice the velocity.
 
  • #5
Brad Barker said:
Ok, I was able to do this after all.

My first method was to substitute in -bv in for F and try to show that the closed line integral of F with dr is not equal to 0. This did not get me anywhere, although I imagine that some masters of vector calculus could get the result this way.

What was successful for me was exploiting Stokes' Theorem, and also solving the differential equation

m dv/dt = -bv.

I got v from this, plugged it back into my equation for force, and then computed the curl, which is not equal to zero so long as the velocity is not equal to zero.
For the differential equation of motion:

[tex]\frac{dv}{dt} + \frac{b}{m}v = 0[/tex]

the integrating factor is [tex]e^{\frac{b}{m}t[/tex]

So:
[tex]e^{\frac{b}{m}t}\frac{dv}{dt} + e^{\frac{b}{m}t}\frac{b}{m}v = \frac{d}{dt}(ve^{\frac{b}{m}t}) = 0[/tex]

This means that:

[tex]v = v_0e^{{\frac{-b}{m}t}[/tex]

If the force is conservative, the work done is a function of position only, so the line integral of Fds along a closed path must always be 0:

[tex]\oint m\frac{dv}{dt}ds = - \oint bvds = W[/tex]

Edit: oops - integration corrected as per subsequent posts:

[tex]W = -b \oint vds = -b\oint v^2dt = -b\int_{t_0}^{t_1} v_0^2e^{\frac{-2b}{m}t}dt + -b\int_{t_1}^{t_2} v_0^2e^{\frac{-2b}{m}t}dt[/tex]

[tex]W = -b(\frac{-m}{2b}v_0^2(e^{\frac{-2b}{m}t_1}-e^{\frac{-2b}{m}t_0}) + (-b(\frac{-m}{2b}v_0^2(e^{\frac{-2b}{m}t_2}-e^{\frac{-2b}{m}t_1})) [/tex]

[tex]W = \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_1} - e^{\frac{-2b}{m}t_0}) + \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_2} - e^{\frac{-2b}{m}t_1}) [/tex]

[tex]W = \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_2} - e^{\frac{-2b}{m}t_0}) = 0[/tex] only if [itex]t_2 = t_0[/itex], which is impossible or if v_0 = 0.

In other words, the line integral along any path is time dependent, not position dependent and cannot be 0, so it is not a conservative force,

AM
 
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  • #6
Andrew Mason said:
[tex]W = -b \oint vds = -b\int_A^B v_0e^{\frac{-b}{m}t} + -b\int_B^A v_0e^{\frac{-b}{m}t}[/tex]

[tex]W = -b(\frac{-m}{b}v_0(e^{\frac{-b}{m}t_1}-e^{\frac{-b}{m}t_0}) + (-b(\frac{-m}{b}v_0(e^{\frac{-b}{m}t_2}-e^{\frac{-b}{m}t_1})) [/tex]

[tex]W = mv_0(e^{\frac{-b}{m}t_1} - e^{\frac{-b}{m}t_0}) + mv_0(e^{\frac{-b}{m}t_2} - e^{\frac{-b}{m}t_1}) [/tex]

[tex]W = mv_0(e^{\frac{-b}{m}t_2} - e^{\frac{-b}{m}t_0}) = 0[/tex] only if [itex]t_2 = t_0[/itex], which is impossible or if v_0 = 0.

In other words, the line integral along any path is time dependent, not position dependent and cannot be 0, so it is not a conservative force,

AM

Thanks, Andrew.

It looks like you integrated with respect to time, although the line above, you have a "ds." I think this might change the answer, although the salient feature is the same.
 
  • #7
Andrew,your analysis and computations are incorrect.

[tex] W_{1\rightarrow 2}=:\int_{1}^{2} \vec{F}\cdot d\vec{s}=-b\int_{1}^{2}\vec{v}\cdot d\vec{s} [/tex]

Since [itex] \vec{v}\uparrow\uparrow \vec{r} [/itex],then

[tex] W_{1\rightarrow 2}=-b\int_{1}^{2} v \ ds=-b\int_{t_{1}}^{t_{2}} \left(v\frac{ds}{dt}\right)dt=-b\int_{t_{1}}^{t_{2}} v^{2} dt [/tex]

Do everything again.

Daniel.
 
  • #8
dextercioby said:
Andrew,your analysis and computations are incorrect.

[tex] W_{1\rightarrow 2}=:\int_{1}^{2} \vec{F}\cdot d\vec{s}=-b\int_{1}^{2}\vec{v}\cdot d\vec{s} [/tex]

Since [itex] \vec{v}\uparrow\uparrow \vec{r} [/itex],then

[tex] W_{1\rightarrow 2}=-b\int_{1}^{2} v \ ds=-b\int_{t_{1}}^{t_{2}} \left(v\frac{ds}{dt}\right)dt=-b\int_{t_{1}}^{t_{2}} v^{2} dt [/tex]

Do everything again.

Daniel.
Thanks for pointing that out - of course E_0 = mv^2/2 and not mv. See above correction. The bottom line is that W is not position dependent so it is not conservative.

AM
 
  • #9
Another way of saying it, entirely equivalent to AM's analysis, is that [tex]\vec{F} \cdot \vec{ds}[/tex] has the same sign along the entire path of the particle. Further, this dot product is only zero where V is zero, the velocity has the same direction as [tex]\vec{ds}[/tex]. Thus the only way a closed loop integral could be zero is if it has zero length--the particle doesn't move.
 
  • #10
Galileo said:
Consider the work done on a particle in going from a to b.
Now consider going along the same path, but with twice the velocity.

Why isn't anyone considering this elegant solution?
 
  • #11
vinter said:
Why isn't anyone considering this elegant solution?
Thank you vinter. :smile:
 

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