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Setting Up Lagrangian, David Morin 6.25

  1. Jan 24, 2016 #1
    1. The problem statement, all variables and given/known data

    A rigid “T” consists of a long rod glued perpendicular to another rod of length l that is pivoted at the origin. The T rotates around in a horizontal plane with constant frequency ω. A mass m is free to slide along the long rod and is connected to the intersection of the rods by a spring with spring constant k and equilibrium length zero. Find r(t), where r is the position of the mass along the long rod. There is a special value of ω; what is it, and why is it special? Figure 6.27 in David Morin Classical Mechanics

    2. Relevant equations
    L = T - V

    3. The attempt at a solution

    I am having trouble setting up the Lagrangian for this system so that it is in appropriate coordinate system.

    So far T = ½mv^2 + ½ω^2 + ½kx^2

    V = -kx - mghsin(ωt)
     
  2. jcsd
  3. Jan 25, 2016 #2
    Your expression for the Lagrangian is wrong. I think that the kinetic energy term should be:
    (1/2)Iω2 + (1/2)mv2 and the potential energy should be
    -(1/2)kr2. I is the moment of inertia of the rod and v is the velocity of the mass in the horizontal plane. I suggest writing the velocity in terms of the rotational angle and the x and y components in the plane, i.e.
    x = rcos(ϑ), y = rsin(ϑ),
    dx/dt = -rdϑ/dt sin(ϑ) +dr/dtcos(ϑ),
    dy/dt =rdϑ/dt + dr/dtsin(ϑ),
    and v2=((dx/dt)2 + (dy/dt)2)
     
  4. Jan 31, 2016 #3
    okay thanks! Except l is changing with time as well based on where the mass is on the rod, so l = sqrt(l2 + r2)
     
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