Free Fall with Viscous Friction

  • Thread starter fuchini
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  • #1
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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.
 
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Answers and Replies

  • #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?
 
  • #3
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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?
 
  • #4
Jano L.
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