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Lagrangian mech.: Action for a particle under constant force

  1. Mar 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Find Scl for a particle under constant force f, that is:

    L = (m/2)v2 + fx


    2. Relevant equations

    S = ∫Ldt

    d(∂L/∂[itex]q^{.}[/itex])/dt = ∂L/∂q


    3. The attempt at a solution

    Apologies if this belongs in the Introductory Physics section. Apologies for terrible formatting.

    ∂L/∂v = mv
    d(mv)/dt = ma

    ∂L/∂x = f

    From the Euler eqn it follows that f = ma... surprise surprise!

    This makes the Lagrangian:

    L = (m/2)v2 + m.a.x

    So when we integrate Ldt over ta and tb the dependencies confuse me... the m.a.x part, which is also f.x., should have the m.a (or f) just taken out of the integral because it's constant over time. On the left hand side m/2 is constant so it can be taken out. This leaves me with

    S = m^2 * a/2 ∫ (v^2)xdt

    I know this is wrong, but I can't tell you why... even if it's not wrong I have no idea on how to continue because Lagrangian mechanics confuses me on implicit / explicit dependencies. I can't tell what is a function of what and which one can be taken as a constant over dt. I'd appreciate any help since I'm lost on the subject.
     
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  3. Mar 17, 2013 #2

    vela

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    It's certainly not true that
    $$\frac{1}{2}mv^2 + max = \frac{m^2a}{2}v^2x.$$ You can't just ignore the rules of algebra.

    The Euler-Lagrange equation told you ##m\ddot{x} = f##. From it, you can solve for x(t) and v(t). Then you can find L as a function of t, which you can then integrate.
     
  4. Mar 17, 2013 #3
    Whoops, that obviously wasn't correct... i made an algebraic mistake while writing the equations down here.

    What I meant was simply taking m/2 out of the first term and ma out of the second term since they should be constant:

    S = (m/2) ∫v^2dt + (ma) ∫ xdt.

    And can you please provide an example of which equation to use to solve for x(t) and v(t)... the dependencies totally throw me off, and I can't come up with what to use.
     
  5. Mar 17, 2013 #4

    vela

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    Solve the differential equation ##m\ddot{x} = f##.
     
    Last edited: Mar 17, 2013
  6. Mar 17, 2013 #5
    Okay... So that produces:

    x(t) = t2f/2m + t
    v(t) = f.t/m

    That makes the Lagrangian

    L = (m/2)f2t2/m2 + f (t2f/2m + t)

    So this is L as nothing but a function of t (and f is constant)

    If we integrate this wrt time over ta-tb that should be the answer?
     
  7. Mar 17, 2013 #6

    vela

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    Your expressions for x(t) and v(t) aren't quite correct. You forgot the constant of integration when finding v(t). It should be v(t) = v0 + (f/m)t. Hopefully, you recognize that equation from intro physics. You made the same mistake with x(t). Also, I'm not sure where that lone t came from.
     
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