Astrophysics
  • Astrophysics
@empty hey hey
Mathematics
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Astrophysics
  • Astrophysics
@empty hey hey
Mathematics
schrodinger
  • schrodinger
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Astrophysics
  • Astrophysics
|dw:1436077425511:dw| leggo
Astrophysics
  • Astrophysics
I didn't have the chance to till now, so now I'll dooooo it! \[W = \int\limits \vec F \cdot dr\]
Empty
  • Empty
Good luck I'll be watching haha.

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Astrophysics
  • Astrophysics
Oh that dr should be a vector but w/e ok lol
Astrophysics
  • Astrophysics
\[\frac{ \vec dr }{ dt } = \vec r'(t) \implies \vec r'(t) dt\] \[\vec r'(t) = <1,-2(t-2)>\] \[W = \int\limits <0,-g> \cdot <1,-2(t-2)> dt\]
Astrophysics
  • Astrophysics
|dw:1436077905456:dw| just seeing the drawing again XD \[W = \int\limits_{0}^{4} (0) \vec i + (2g(t-2)) \vec j dt \]
Michele_Laino
  • Michele_Laino
is: \[\Large {\mathbf{F}} = \left( {0, - mg} \right)\]?
Astrophysics
  • Astrophysics
Got 0? It's a made up question haha
Astrophysics
  • Astrophysics
But yeah it would be -mg otherwise
Michele_Laino
  • Michele_Laino
since, I see: \[\Large {\mathbf{F}} = \left\langle {0, - g} \right\rangle \]
Empty
  • Empty
Yeah I just made this question up on the fly last night really late, so yeah throw an m in there haha.
Astrophysics
  • Astrophysics
|dw:1436078560687:dw| hey empty quick question, I can't really remember, but would I have just a constant in the i direction as I'm integrating respect to 0
Michele_Laino
  • Michele_Laino
I got: 8g
Astrophysics
  • Astrophysics
Mhm sec, maybe I made a mistake
Astrophysics
  • Astrophysics
Oh im dumb
Astrophysics
  • Astrophysics
I messed up at distributing XD
Michele_Laino
  • Michele_Laino
please note this: \[\Large W = \int_0^4 {dt\left( {2gt - 4g} \right)} \]
Astrophysics
  • Astrophysics
ok let me redo it |dw:1436078733805:dw|
Astrophysics
  • Astrophysics
Thanks for checking @Michele_Laino !!
Michele_Laino
  • Michele_Laino
so: \[\Large W = \int_0^4 {dt\left( {2gt - 4g} \right)} = 16g - 16g = 0\]
Michele_Laino
  • Michele_Laino
Empty
  • Empty
Ok so you have the calculation done correctly which is excellent, but what is the significance of what you've done? @Astrophysics That's what I want you to explain to us.:D
Michele_Laino
  • Michele_Laino
I think that a possible answer can be this: "the work done by the weight force, along that trajectory is 0"
Astrophysics
  • Astrophysics
The work done by the field is pulling us down via trajectory giving us 8g, but if we had a mass it would be W = mg.
Astrophysics
  • Astrophysics
Well Michele put it much nicer than me
Michele_Laino
  • Michele_Laino
we gave the same answer! :)
Astrophysics
  • Astrophysics
Haha, yeah well work problems are pretty fun
Michele_Laino
  • Michele_Laino
yes! they are related to vector calculus, and I like very much the vector calculus! as you know @Astrophysics eh eh :)
Astrophysics
  • Astrophysics
But if we had a circle lets say, that's the field and the circle is going counter clockwise, how would we define the work exactly? We'll just assume x = cost, y = sint, w/e |dw:1436079570813:dw|
Astrophysics
  • Astrophysics
|dw:1436079731168:dw|
Michele_Laino
  • Michele_Laino
I think that we have to use this parametrization: \[\Large \begin{gathered} x = R\cos t \hfill \\ \hfill \\ y = R\sin t \hfill \\ \end{gathered} \] |dw:1436079812068:dw|
Astrophysics
  • Astrophysics
Haha yeah, you're probably right, all of these questions in this post are just made up out of our heads and I guess that's not very useful...xD but to say the least it's fun, I just want to know what happens with the field and the path taken
Astrophysics
  • Astrophysics
If we give it an interval [0,2pi], we would get -2 pi, so let me just ask you a question to clarify, aside from the math what exactly does this mean, it's opposing the field, or just saying it's going in the opposite direction?
Michele_Laino
  • Michele_Laino
what is the vector field?
Astrophysics
  • Astrophysics
Oh yi-xj
Empty
  • Empty
This is because you have a nonconservative field in this case. In the previous case, we could have written that particular vector field as the gradient of a scalar field while this one has no corresponding scalar field. So think, what is the corresonding scalar field to the Force field from the first question? Just let yourself be stumped on this for a while so you will be satisfied when learning the answer since it is important and should be remembered, as it has a fairly well known associated equation. :)
Michele_Laino
  • Michele_Laino
I got this: \[\Large W = \int_0^{2\pi } {\left( {R\sin t,R\cos t} \right) \cdot \left( { - R\sin t,R\cos t} \right)} dt\]
Astrophysics
  • Astrophysics
Ah, I see thanks empty, I did not think of it as a nonconservative field, and no worries Michele, the math is fine, but I was just trying to figure out the theoretical part of the problem!
Michele_Laino
  • Michele_Laino
oops.. I have made an error of sigh, here is the right integral: \[\Large W = \int_0^{2\pi } {\left( {R\sin t, - R\cos t} \right) \cdot \left( { - R\sin t,R\cos t} \right)} dt\]
Michele_Laino
  • Michele_Laino
sign*
Astrophysics
  • Astrophysics
It's ok :)
Astrophysics
  • Astrophysics
Thanks @Michele_Laino and @Empty !! :)
Michele_Laino
  • Michele_Laino
we can note that your field can be rewritten as follows: \[\Large {\mathbf{F}} = \left( {y, - x,0} \right)\] now according to the observation of @Empty , we have: \[\Large \nabla \times {\mathbf{F}} = \left( {0,0, - 2} \right)\] namely our vector field is an irrotational field

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