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Newton's second law integration

  1. Sep 19, 2007 #1
    I need to integrate

    [tex]\frac{mdv}{F(v)}[/tex] = dt

    to show that

    t = m [tex]\int[/tex][tex]^{v}_ {vo}[/tex] [tex]\frac{dv '}{F(v ')}[/tex]

    Now, I now that velocity is the derivative of position, and the mass is constant, but how does one integrate F(v)?

    [Integrating is a weak spot for me] Thanks in advance!
  2. jcsd
  3. Sep 19, 2007 #2
    in getting the first equation, since F(v) is the only force, F(v) = mv'.
    In rearranging, how does the dt come about?
  4. Sep 19, 2007 #3


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    Just integrate both sides of that first equation... the side with the dt becomes t... and the right side gives you the integral...

    integral of dt is t.
  5. Sep 19, 2007 #4
    I'm just wondering how Newtons second law: F(v) = mv' obtained the t...
  6. Sep 19, 2007 #5


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    oh... v' = dv/dt then rearrange...
  7. Sep 19, 2007 #6
    I don't understand how i miss things sometimes...

    doesn't adding the primes cancel out the integral?
  8. Sep 19, 2007 #7


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    Ah I see your confusion... the prime in the equation with the integral has nothing to do with derivative...

    They're just using another variable to do the integration...

    they could just as well write

    [tex]t = m \int_{v0}^v\frac{du}{F(u)}[/tex]

    they just want to keep one variable v for the limit of the integration... ie we're integrating from v0 to the final velocity v... so inside the integral they want a different variable (they used v' which was a bad choice because it can cause confusion with the derivative)
  9. Sep 19, 2007 #8
    Oohhh. haha

    What about the integral and dv?
  10. Sep 19, 2007 #9


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    Yeah, when they put in the integral... they just switched from dv to dv'... you can use any variable name you want...

    this would also be equally valid:

    [tex]t = m \int_{v0}^{v_{final}}\frac{dv}{F(v)}[/tex]

    The only thing is you need a different variable inside the integral than you have at your limits of integration... ie: this would technically be wrong:

    [tex]t = m \int_{v0}^v}\frac{dv}{F(v)}[/tex]

    people often use this type of notation... but you shouldn't really have v as the limit of integration, when the variable inside the integral is also v.
  11. Sep 19, 2007 #10
    Would that just become v' though since its dv' and in an integral?
  12. Sep 19, 2007 #11


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    Not sure what you mean.
  13. Sep 19, 2007 #12
    same question as before, but instead of prime i mean the d in front of the v.
    Derivative of v (dv) and the integral... = v?

    I think not because the dt is seperated though.
  14. Sep 19, 2007 #13


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    I'm sorry... I'm still not following... let's start with this:

    [tex]\frac{mdv}{F(v)} = dt[/tex]

    Then I can take the integral of both sides

    [tex]\int_{v_0}^{v_{final}}\frac{mdv}{F(v)} = \int_0^{t_{final}}dt[/tex] so that gives

    [tex]\int_{v_0}^{v_{final}}\frac{mdv}{F(v)} = t_{final}[/tex]

    I can rewrite t_final as t... and take out the m...

    [tex]m\int_{v_0}^{v_{final}}\frac{dv}{F(v)} = t[/tex]
  15. Sep 19, 2007 #14
    yeah, that is where I'm up to (correctly, compared to your example). I'm basically wondering, what CAN I do with dv (the numerator)?
  16. Sep 19, 2007 #15


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    why do you need to do anything? Isn't that the final result? If you want you can switch from v to some other variable... v', u or anything else... but v' is NOT the derivative of v. It's just another variable like u or x etc...

    [tex]\int_3^4{f(t)dt}[/tex] is exactly the same thing as [tex]\int_3^4{f(x)dx}[/tex]... just switched the name of the variable... makes no difference.
  17. Sep 19, 2007 #16
    I'm now supposed to solve for v, now that I have solved this, to get v as a function of t. I think I need to get rid of the integral to rearrange?

    (I was partly reading too far into the question again)
  18. Sep 19, 2007 #17


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    Can you show exactly you need to solve? I'm confused... is there a formula for F(v) ?
    Last edited: Sep 19, 2007
  19. Sep 19, 2007 #18
    K the whole question is this:

    There are certain simple one dimensional problems where the equatin of motion can always be solve,d or at least reduced to the problem of doing an integral. One of these is the motion of a done dimensional particle subject to a force that depends on only the velocity v, that is F=F(v). Write down N2L and seperate the variables by rewriting it as mdv/F(v) = dt. Now integrate both sides of this equation and show that

    t = m..... (we did this)

    Provided you can do the integral, this gives t as a function of v. You can also solves to give v as a function of t. Use this method to solve the special case that F(v) = Fo, a constant, and comment on your result.
  20. Sep 19, 2007 #19


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    Yes, just plug in Fo into the integral, instead of F(v)... and then solve the integral... since Fo is a constant, it just comes out of the integral... leaving dv inside
  21. Sep 19, 2007 #20
    so just solve for the integral of dv?
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