# Free Fall with Viscous Friction

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 cant proove that maximum velocity.

Any help will be appreciated.

Last edited:

## Answers and Replies

George Jones
Staff Emeritus
Science Advisor
Gold Member
I you sure "dissipative potential" was used in the question? Did you use the Euler-Lagrange equation that is appropriate for dissipative systems?

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.
Gold Member