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Lagrangian Question

  1. Sep 22, 2004 #1
    A point mass is constrained to move on a massless hoop of radius a fixed in a vertical plane that rotates about its vertical symmetry axis with constant angular speed [tex] \omega [/tex].

    a. Obtain the Lagrange's equations of motion assuming that the only external forces arise from gravity.

    Should I have seperate KE components for the linear velocities as well as the angular velocity? I have this so far (with seperate x,y and z velocity components written in spherical coordinates) [tex]T=\frac{m}{2} v^2 + I\omega^2[/tex]. I'm pretty sure that is correct, but I don't know what to use for the moment of inertia? Can I just use the moment of inertia for a spherical shell, or would I use that of a ring or something else entirely?

    EDIT:
    Since [tex]\omega[/tex] is the rate of change of the angle [tex]\theta[/tex] in spherical coordinates, could I set that equal to [tex]\frac{d}{dt}\theta[/tex] in the kinetic energy term? Or can I not ignore the moment of inertia like that?
     
    Last edited: Sep 22, 2004
  2. jcsd
  3. Sep 23, 2004 #2

    cepheid

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    I just had a similar question in an assignment, and I'm curious if my approach was correct:

    First of all, note that the hoop is massless. So where is this term for kinetic energy due to the hoop's angular velocity coming from? My reasoning was that the only particle in the system with any mass is the one with mass m moving around the wire. So the total kinetic energy of the system comes entirely from this particle's kinetic energy. Correct?

    [tex] \sum{T} = \frac{1}{2}m|\vec{v}|^2 [/tex]

    You are also told (as were we) that external forces are limited to graviational forces. More specifically, it was spelled out for us that the particle's potential energy is [itex] V = mgz [/itex], where z is the height above some reference point (wherever you want). Of course, it would seem that we are done:

    [tex] \mathcal{L} = T - V [/tex]

    However, when introducing the Lagrangian formulation of dynamics, my prof stressed that any system of particles can be expressed completely using a certain number of generalized coordinates. I don't quite understand the fine aspects of this point, but in this case I used two generalized coordinates:

    [tex] \phi \ \ \text{and} \ \ \dot{\phi} [/tex] representing the angular position and angular velocity of the particle respectively.

    Note that the velocity, although in some crazy direction, is the resultant of two components. The first, [itex] \vec{v_1} [/itex], is the tangential component of the velocity due to it's revolution around the wire.. The second, [itex] \vec{v_2} [/itex], is the component of the velocity in a plane perpendicular to the first (in which the hoop lies). If you have trouble seeing this, draw a "top" view, which just ends up being a circle of radius [itex] a\cos\phi [/itex] on the verge of being swept out by the particle on the wire at that instant.

    [tex] |\vec{v_1}| = \dot\phi a [/tex]

    [tex] |\vec{v_2}| = \omega a\cos\phi [/tex]

    [tex] z = a \sin\phi [/tex]

    Now the Lagrangian can be expressed entirely in terms of the two generalized coordinates. Is this method essentially correct?
     
  4. Sep 23, 2004 #3
    I was thinking that the moment of inertia would come from the rotating mass, not the hoop itself, but then I realized that I had two terms that were representing the same angular velocity of the mass. I also completely missed the fact that since [tex]\omega[/tex] is constant, the height of the mass in the hoop is a constant and the number ov velocity components reduces to two. I'm pretty sure I can get the rest of the problem now.
     
  5. Oct 18, 2010 #4
    what will the solution look like for this problem? I would appreciate to see the entire solution to this problem
     
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