Find dissipative function for the non-linear force f=-bv^n

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The discussion focuses on determining the dissipative function for the non-linear force f = -bv^n in a Lagrangian framework. It is established that the dissipative function can be expressed as D = - (1/(n+1))bv^(1+n). There is clarification regarding the relationship between the force and the dissipative function, correcting a typo in the initial query. Participants confirm the correctness of the derived dissipative function. The conversation emphasizes the importance of accurately relating force and dissipative functions in nonconservative systems.
mcconnellmelany
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Homework Statement
For a nonconservative force,
What would be the dissipative function for a force f=-bvⁿ in Lagrangian
(Where v is the velocity)
[#qoute for a nonconservative force f=-bv
The dissipative function is D=-(1/2)bv² ]
Relevant Equations
##\frac{d}{dt}(\frac{\partial L}{\partial \dot x})=\frac{\partial L}{\partial x} - \frac{\partial F}{\partial \dot x}##
For a nonconservative force,
What would be the dissipative function for a force f=-bvⁿ in Lagrangian
(Where v is the velocity)
[#qoute for a nonconservative force f=-bv
The dissipative function is D=-(1/2)bv² ]

Since ##f=\frac{\partial D}{\partial \dot x}## so the dissipative function should be ##D=-\frac{1}{n+1}bv^{1+n}##, isn't it?
 
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mcconnellmelany said:
Since ##f=\frac{\partial D}{\partial \ddot x}##
Do you mean ##f=\frac{\partial D}{\partial \dot x}## ?
 
haruspex said:
Do you mean ##f=\frac{\partial D}{\partial \dot x}## ?
Hmm! It was a typo.
 
mcconnellmelany said:
Hmm! It was a typo.
Your answer looks right to me.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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