Free Fall with Viscous Friction

In summary, the conversation discusses finding the Lagrangian and proving the maximum velocity of a mass in free fall with viscous friction. The given Lagrangian is used to solve the Euler-Lagrange equation, but the presence of the dissipative potential complicates the process. The individual is directed to read about generalized forces and the dissipation function in order to better understand how to deal with the dissipative potential.
  • #1
fuchini
11
0
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 can't proove that maximum velocity.

Any help will be appreciated.
 
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  • #2
I you sure "dissipative potential" was used in the question? Did you use the Euler-Lagrange equation that is appropriate for dissipative systems?
 
  • #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


Hello, thank you for sharing your question and work on this problem. I would first like to commend you for your efforts in deriving the Lagrangian and using the Euler-Lagrange equation to solve for the acceleration. This shows a good understanding of classical mechanics principles.

To prove that the maximum speed is v=mg/K, we can use the concept of energy conservation. In this system, the total energy is given by the sum of kinetic and potential energies:

E = T + V = \frac{1}{2}m\dot{y}^{2}-mgy-\frac{1}{2}k\dot{y}^{2}

Since there is no external force acting on the system, the total energy must be conserved. This means that the energy at any point in time must be equal to the initial energy when the object is released.

At the maximum velocity, the kinetic energy is at its maximum while the potential energy is at its minimum. This occurs when the object reaches the bottom of its trajectory, where the height is zero. Therefore, we can set the potential energy to zero and solve for the maximum kinetic energy:

E_{max} = \frac{1}{2}m(v_{max})^2 = \frac{1}{2}m\left(\frac{mg}{k}\right)^2

Solving for v_{max}, we get:

v_{max} = \frac{mg}{k}

Which is the same as the maximum speed you were asked to prove.

In conclusion, using the concept of energy conservation, we can prove that the maximum speed in this system is v=mg/K. I hope this helps and good luck with your further studies!
 

Related to Free Fall with Viscous Friction

1. What is free fall with viscous friction?

Free fall with viscous friction refers to the motion of an object falling under the influence of gravity, while also experiencing resistance from a fluid, such as air or water. This resistance is known as viscous friction, and it can affect the speed and trajectory of the falling object.

2. How does viscous friction affect an object in free fall?

Viscous friction creates a force that opposes the motion of the falling object, slowing it down and changing its direction. This force increases as the speed of the object increases, eventually reaching a point where it balances out the force of gravity, resulting in a constant speed known as terminal velocity.

3. What factors influence the amount of viscous friction experienced by a falling object?

The amount of viscous friction experienced by a falling object depends on several factors, including the density and viscosity of the fluid, the surface area and shape of the object, and the speed of the object. Objects with larger surface areas, such as parachutes, experience more viscous friction than smaller, streamlined objects.

4. How is free fall with viscous friction different from regular free fall?

In regular free fall, an object is only influenced by the force of gravity, resulting in a constant acceleration towards the ground. However, in free fall with viscous friction, the resistance from the fluid creates a counter force, causing the object to slow down and reach a terminal velocity.

5. Can viscous friction be beneficial in any way during free fall?

Yes, viscous friction can be beneficial in certain situations. For example, it can help slow down skydivers or parachutes as they approach the ground, reducing the impact force and preventing injury. In some cases, it can also help stabilize the motion of falling objects, making them less susceptible to turbulence or unpredictable winds. However, in most cases, viscous friction is considered a hindrance and is often minimized or avoided in engineering and design processes.

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