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Lagrange Multipliers

  1. Feb 28, 2006 #1
    A long first post, but not too hard!

    dont worry about this i already solved it thanks anyway!

    The lagrangian of a particle of mass m moving under constant gravity is
    [tex] \mathcal{L} = \frac{1}{2} m (\dot{x}^2 + \dot{z}^2 - mgz = \frac{1}{2}m (\dot{\rho}^2 + \rho^2 \dot{\phi}^2) - mg \rho \cos \phi [/tex]
    If the particle is contrained to move on the surface rho= R where R>0 is a constant (such as a parrticle slifing off a sphere or a cylinder) thjen the constraint is rho =R can be taken into account, in teh Lagrangian Formalism by the Lagrange multiplier method i.e. by starting wiuth the lagrangian
    [tex] \mathcal{L}' = \mathcal{L} + \lambda (\rho - R) [/tex]

    a) Work out the two equations of motion [tex] \frac{d}{dt} \left(\frac{\partial L'}{\partial \dot{q_{i}}} \right) = \frac{\partial L'}{\partial q_{i}}, q_{i}= (\rho,phi) [/tex]

    Do i have to solve these? I mean i will be forming coupled non linear second order DEs...

    [tex] \frac{\partial L'}{\partial \phi} = mg \rho \sin \phi [/tex]
    [tex] \frac{\partial L'}{\partial \dot{\phi}} = m \rho^2 \dot{\phi} [/tex]
    [tex] \frac{d}{dt} \frac{\partial L'}{\partial \dot{\phi}} = 2 m \rho \dot{\rho} \dot{\phi} + m \rho^2 \ddot{\phi} [/tex]
    into the Euler Lagrange
    [tex] 2 m \rho \dot{\rho} \dot{\phi} + m \rho^2 \ddot{\phi} = mg \rho \sin \phi [/tex]

    [tex] \frac{\partial L'}{\partial \rho} = m \rho \dot{\phi}^2 - mg \cos \phi - \lambda [/tex]
    [tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{p}} = \frac{d}{dt} m \dot{\rho} = m \ddot{\rho} [/tex]
    Euler Lagrange
    [tex] m \ddot{\rho} = m \rho \dot{\phi}^2 - mg \cos \phi - \lambda [/tex]

    i dont think it is possibel to solve them...

    b) Incorporate the constraint rho = R into the equations obtained in a). Detremine the dimensions of the Lagrange multiplier lambda and explain teh physical meaning of lambda. (use a diagram)

    [tex] R \dot{\phi}^2 + g \cos \phi + \frac{\lambda}{m} = 0 [/tex]
    [tex] \ddot{\phi} - mR \sin \dot{\phi} - mg \sin \phi = 0 [/tex]
    rearrange the first one for lambda yields
    [tex] \lambda = -m(R \dot{\phi}^2 + g \cos \phi) [/tex]
    so lambda's units are
    [tex] \lambda = kg(m \frac{rad^2}{s^2} + \frac{m}{s^2}) [/tex]
    so lambda is N rad^2 + N ...? How si this siginificant >??

    c)Determine whether thi sytem is conservative. If so derive the expression for the cosnant energy. If the system is not conservativem explain why not.
    System is conservative if the force can be written as the derivative of some function dpeendant on one variable only (??)
    well isnt hte force simply mg?
    to find the constant energy expression do i find the Hamiltonian of the system? Two variables so there will be two of them?

    d)Use your results of b and c to obtai the expression for lambda at angle phi given that the particel starts at phi =0 (t hte top of the sphere or cylinder) and is given an intiial speed of v0 = R omega0

    e) At hwat angle would hte particle leave teh surface (of the sphere or cylinder)? What would this angle be if v0 = 0? What is tha largest vbalue that v0 could be for the particle to slide along the sphere (or cylinder) at least briefly? Evaluate this limiting value of v0 if R=1 meter.

    last two i will psot solutions for later on
    Last edited: Feb 28, 2006
  2. jcsd
  3. Mar 1, 2006 #2


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    [itex] \lambda [/itex]'s units are of force...So...

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