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Effective potential of a particle sliding on an inclined rod

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Hey guys,

    Here is the question:

    A pointlike mass [itex]m[/itex] can slide along a rigid rod of length [itex]l[/itex] and negligible mass. One extremity of the rod is fixed at the origin [itex]O[/itex] of an inertial system [itex](x,y,z)[/itex], and the rod forms a constant angle [itex]α[/itex] with the [itex]z[/itex]-axis. The rod rotates about the [itex]z[/itex]-axis with constant angular velocity [itex]ω[/itex]. Gravity acts in the negative [itex]z[/itex]-direction.

    (i) How many degrees of freedom does the system have?

    (ii) Write down the Lagrangian and the Lagrange equations.

    (ii) Recast the system as a 1-dimensional motion in an effective potential. Find an expression for the effective potential and determine the equilibrium positions as well as their stability

    2. Relevant equations
    [itex]T=1/2mv^{2}[/itex]
    Lagrangian: [itex]L = T - V[/itex]
    Lagrange Equation: [itex]\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}[/itex]

    3. The attempt at a solution
    Here are my solutions:

    (i) one degree of freedom - the distance from the origin to the point mass. call this distance [itex]r[/itex].

    (ii) I wont put the calculations here, I'll just give my results:

    Lagrangian [itex]L = T - V = \frac{1}{2}m[\dot{r}^{2}+r^{2}ω^{2}sin^{2}α] - mgrcosα[/itex]


    Lagrange equation: [itex]\ddot{r}-rω^{2}sin^{2}α + gcosα = 0 [/itex]

    (iii)

    Okay, so here is the issue. How do you get the effective potential from this? and do I just differentiate that to get the equilibrium positions and then differentiate again to determine stability?
     
  2. jcsd
  3. Apr 26, 2013 #2

    TSny

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    If you had a particle of mass m confined to 1-dimension of space (a straight line) and you let r be the distance from the origin, how would you express the kinetic energy T? How would you construct a potential energy function V(r) for this particle so that it would have the same Lagrangian L as your three-dimensional problem of the particle on the rotating wire?

    Sounds good.
     
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