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Brad Barker
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I tried to do this myself, but I was unsuccessful.
Thank you.
Thank you.
For the differential equation of motion: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.
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 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.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.
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.
Thank you vinter.vinter said:Why isn't anyone considering this elegant solution?
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