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Free Fall with Viscous Friction

  1. May 10, 2013 #1
    Hello, First post hear so bear with me.

    I have a mass in free fall with a viscous friction which can be derived from the dissipative potential Kv2/2. I must find the Lagrangian and proove that the maximum speed is v=mg/K. I have the following Lagrangian:

    L=T-V=[itex]\frac{1}{2}[/itex]m[itex]\dot{y}[/itex][itex]^{2}[/itex]-mgy-[itex]\frac{1}{2}[/itex]k[itex]\dot{y}[/itex][itex]^{2}[/itex]=[itex]\frac{1}{2}[/itex](m-k)[itex]\dot{y}[/itex][itex]^{2}[/itex]-mgy

    When I do the Euler-Lagrange:

    [itex]\ddot{y}[/itex]=-mg/(m-k)

    However, from this equation I cant proove that maximum velocity.

    Any help will be appreciated.
     
    Last edited: May 10, 2013
  2. jcsd
  3. May 11, 2013 #2

    George Jones

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    I you sure "dissipative potential" was used in the question? Did you use the Euler-Lagrange equation that is appropriate for dissipative systems?
     
  4. May 11, 2013 #3
    Hello, the question does say "Dissipative Potential". From what I've seen in class, if the force can be derived from a potential, the normal Euler-Lagrange equations are still valid. Therefore I used the following:

    [itex]\mathcal{L}=\frac{1}{2}(m-k)\dot{y}^{2}-mgy[/itex]

    Euler-Lagrange:

    [itex]\frac{\partial \mathcal{L}}{\partial y}-\frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{y}})=0[/itex]

    And I get:

    [itex]-mg-\frac{d}{dt}((m-k)\dot{y})=-mg-(m-k)\ddot{y}=0[/itex]

    Finally:

    [itex]\ddot{y}=\frac{-mg}{(m-k)}[/itex]

    I think my problem is with the dissipative potential. How do I deal with it?
     
  5. May 11, 2013 #4

    Jano L.

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