How to Prove Maximum Speed in Free Fall with Viscous Friction?

[fuchini]

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=\frac{1}{2}m\dot{y}^{2}-mgy-\frac{1}{2}k\dot{y}^{2}=\frac{1}{2}(m-k)\dot{y}^{2}-mgy

When I do the Euler-Lagrange:

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

However, from this equation I can't proove that maximum velocity.

Any help will be appreciated.

 

[George Jones]

I you sure "dissipative potential" was used in the question? Did you use the Euler-Lagrange equation that is appropriate for dissipative systems?

 

[fuchini]

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:

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

Euler-Lagrange:

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

And I get:

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

Finally:

\ddot{y}=\frac{-mg}{(m-k)}

I think my problem is with the dissipative potential. How do I deal with it?

 

[Jano L.]

Try to read the section "Generalized forces" and "Dissipation function", in the middle of the article.

http://en.wikipedia.org/wiki/Lagrangian_mechanics

Does it help?