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Astrophysics

  • one year ago

@empty hey hey

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  1. Astrophysics
    • one year ago
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    |dw:1436077425511:dw| leggo

  2. Astrophysics
    • one year ago
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    I didn't have the chance to till now, so now I'll dooooo it! \[W = \int\limits \vec F \cdot dr\]

  3. Empty
    • one year ago
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    Good luck I'll be watching haha.

  4. Astrophysics
    • one year ago
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    Oh that dr should be a vector but w/e ok lol

  5. Astrophysics
    • one year ago
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    \[\frac{ \vec dr }{ dt } = \vec r'(t) \implies \vec r'(t) dt\] \[\vec r'(t) = <1,-2(t-2)>\] \[W = \int\limits <0,-g> \cdot <1,-2(t-2)> dt\]

  6. Astrophysics
    • one year ago
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    |dw:1436077905456:dw| just seeing the drawing again XD \[W = \int\limits_{0}^{4} (0) \vec i + (2g(t-2)) \vec j dt \]

  7. Michele_Laino
    • one year ago
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    is: \[\Large {\mathbf{F}} = \left( {0, - mg} \right)\]?

  8. Astrophysics
    • one year ago
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    Got 0? It's a made up question haha

  9. Astrophysics
    • one year ago
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    But yeah it would be -mg otherwise

  10. Michele_Laino
    • one year ago
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    since, I see: \[\Large {\mathbf{F}} = \left\langle {0, - g} \right\rangle \]

  11. Empty
    • one year ago
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    Yeah I just made this question up on the fly last night really late, so yeah throw an m in there haha.

  12. Astrophysics
    • one year ago
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    |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

  13. Michele_Laino
    • one year ago
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    I got: 8g

  14. Astrophysics
    • one year ago
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    Mhm sec, maybe I made a mistake

  15. Astrophysics
    • one year ago
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    Oh im dumb

  16. Astrophysics
    • one year ago
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    I messed up at distributing XD

  17. Michele_Laino
    • one year ago
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    please note this: \[\Large W = \int_0^4 {dt\left( {2gt - 4g} \right)} \]

  18. Astrophysics
    • one year ago
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    ok let me redo it |dw:1436078733805:dw|

  19. Astrophysics
    • one year ago
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    Thanks for checking @Michele_Laino !!

  20. Michele_Laino
    • one year ago
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    so: \[\Large W = \int_0^4 {dt\left( {2gt - 4g} \right)} = 16g - 16g = 0\]

  21. Michele_Laino
    • one year ago
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    :) @Astrophysics

  22. Empty
    • one year ago
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    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

  23. Michele_Laino
    • one year ago
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    I think that a possible answer can be this: "the work done by the weight force, along that trajectory is 0"

  24. Astrophysics
    • one year ago
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    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.

  25. Astrophysics
    • one year ago
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    Well Michele put it much nicer than me

  26. Michele_Laino
    • one year ago
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    we gave the same answer! :)

  27. Astrophysics
    • one year ago
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    Haha, yeah well work problems are pretty fun

  28. Michele_Laino
    • one year ago
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    yes! they are related to vector calculus, and I like very much the vector calculus! as you know @Astrophysics eh eh :)

  29. Astrophysics
    • one year ago
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    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|

  30. Astrophysics
    • one year ago
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    |dw:1436079731168:dw|

  31. Michele_Laino
    • one year ago
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    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|

  32. Astrophysics
    • one year ago
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    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

  33. Astrophysics
    • one year ago
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    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?

  34. Michele_Laino
    • one year ago
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    what is the vector field?

  35. Astrophysics
    • one year ago
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    Oh yi-xj

  36. Empty
    • one year ago
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    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. :)

  37. Michele_Laino
    • one year ago
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    I got this: \[\Large W = \int_0^{2\pi } {\left( {R\sin t,R\cos t} \right) \cdot \left( { - R\sin t,R\cos t} \right)} dt\]

  38. Astrophysics
    • one year ago
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    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!

  39. Michele_Laino
    • one year ago
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    oops.. I have made an error of sigh, here is the right integral: \[\Large W = \int_0^{2\pi } {\left( {R\sin t, - R\cos t} \right) \cdot \left( { - R\sin t,R\cos t} \right)} dt\]

  40. Michele_Laino
    • one year ago
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    sign*

  41. Astrophysics
    • one year ago
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    It's ok :)

  42. Astrophysics
    • one year ago
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    Thanks @Michele_Laino and @Empty !! :)

  43. Michele_Laino
    • one year ago
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    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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