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Spring pendulum with friction (Lagrange?)

  • Thread starter Nick89
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  • #1
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Homework Statement


Consider a pendulum with a spring as in the following diagram:
2ro6obb.jpg

Please note the 'rotated' coordinates.
The bob has a mass [itex]m[/itex]. The spring has a spring constant [itex]k[/itex] and an unextended length [itex]\ell[/itex]. We can not ignore air friction.

Assume the initial velocity and horizontal displacement are parallel, so that the motion of the pendulum is in the plane. It is not a 'spherical' pendulum.

I need to show that the equations of motion satisfy the following:
[tex]m\left( r'' - r (\theta ')^2 \right) + cr' + k(r-\ell) = mg \cos \theta[/tex]
[tex]m \left( r \theta '' + 2 r' \theta ' \right) + cr \theta ' = - mg \sin \theta[/tex]


Homework Equations


The forces on the bob are the force of gravity, the spring force and the air friction:
[tex]\mathbf{F}_g = mg \mathbf{\hat{i}}[/tex]
[tex]\mathbf{F}_s = -k \left(r - \ell \right) \mathbf{\hat{e_r}}[/tex]
[tex]\mathbf{F}_f = -c \mathbf{r}'[/tex]
(where c is a positive friction coefficient)


The Attempt at a Solution


I can use the Lagrangian to find the equations of motion, but I don't have a clue what to do with the friction force..??

Using the Lagrangian, the kinetic energy T is:
[tex]T = \frac{1}{2} m v^2 = \frac{1}{2} m (x' ^2 + y'^2) = \frac{1}{2} m \left[ (r')^2 + r^2 (\theta ')^2\right][/tex]

The potential V is the sum of the gravitational potential V_g, the spring potential V_s, and...??? Frictional potential? I don't think I can write the frictional force as a potential, can I?
Anyway:
[tex]V = -mgr \cos \theta + \frac{1}{2}k(r-\ell)^2 + ...?[/tex]

[tex]L = T - V = \frac{1}{2} m \left[ (r')^2 + r^2 (\theta ')^2\right] +mgr \cos \theta - \frac{1}{2}k(r-\ell)^2[/tex]

For r, we have:
[tex]\frac{\partial}{\partial t}\left( \frac{\partial L}{\partial r'} \right) = \frac{\partial L}{\partial r}[/tex]
From this I get:
[tex]m\left( r'' - r (\theta ')^2 \right) + k(r-\ell) = mg \cos \theta[/tex]

For theta, we have:
[tex]\frac{\partial}{\partial t}\left( \frac{\partial L}{\partial \theta '} \right) = \frac{\partial L}{\partial \theta}[/tex]
And I get:
[tex]m \left( r \theta '' + 2 r' \theta ' \right) = - mg \sin \theta[/tex]

So as expected, I get the correct equations, without the frictional terms...


How can I calculate the frictional terms? Can I use the Lagrangian, or do I have to use normal force analysis? It doesn't make sense in my mind to use both the Lagrangian and force analysis... ?

Thanks!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Hao
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In the case of dissipative forces, the Lagrangian is unchanged. However, the Euler-Lagrange equation changes.

The Euler-Lagrange equation is typically derived by extremizing the action:

[tex]\delta S = \delta\int{L(x, \dot x)} dt = 0[/tex]

However, another equivalent starting point is the "[URL [Broken] principle[/URL], which states that the virtual work done by a system is 0.

The derivation from then on is relatively involved (Goldstein 3Ed pg16 - pg24), but one finds that if we assume that the force is conservative, we get the usual Euler-Lagrange equation. However, if we assume that the force has conservative and dissipative terms, we find out that the Euler-Lagrange equation equation needs to be modified by adding a dissipation function.

The results are summarized in this "www.phys.uri.edu/~gerhard/PHY520/mln9.pdf"[/URL].

In the end, what all this really means is just

[tex]\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q_j}\right) - \frac{\partial L}{\partial q_j} = Q_j[/tex]

, where [tex]Q_j[/tex] is the dissipative force. Which is probably not surprising at all.
 
Last edited by a moderator:
  • #3
555
0
Thanks, got it!
 

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