1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring pendulum with friction (Lagrange?)

  1. May 28, 2009 #1
    1. The problem statement, all variables and given/known data
    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]


    2. Relevant 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)


    3. 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!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 28, 2009 #2

    Hao

    User Avatar

    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: May 4, 2017
  4. May 28, 2009 #3
    Thanks, got it!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spring pendulum with friction (Lagrange?)
  1. Spring pendulum (Replies: 38)

Loading...