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Lagrangian equation from this free body diagram

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Here's the free body diagram with variables.
    txbPQ.png
    I am looking for the lagrangian mechanics equation.
    [itex]M[/itex] is mass of the bottom wheel.
    [itex]m[/itex] is the mass of the top wheel.
    [itex]R[/itex] is the radius of the bottom wheel.
    [itex]r[/itex] is the radius of the top wheel.
    [itex]θ_{1}[/itex] is the angle from vertical of the bottom wheel.
    [itex]θ_{2}[/itex] is the angle from vertical of the top wheel.
    [itex]\dot{θ}_{1}[/itex] is the angular velocity of the bottom wheel.
    [itex]\dot{θ}_{2}[/itex] is the angular velocity of the top wheel.
    [itex]x[/itex] is the linear distance.
    [itex]\dot{θ}[/itex] is linear velocity of the whole contraption.

    2. Relevant equations
    Here are some relationships of these variables according to the free body diagram.
    [itex]l_{cm}=\frac{m(R+r)}{M+m}[/itex] is the distance to center of mass from center of the bottom wheel.
    [itex]\dot{θ}_{1}R=-\dot{θ}_{2}r[/itex] is just the relationship of the two wheel's angular velocity.
    [itex]I=\frac{2}{5}MR^{2}[/itex] is the moment of inertia of the bottom wheel.
    [itex]I=\frac{1}{4}MR^{2}[/itex] is the moment of inertia of the top wheel.
    [itex]\dot{θ}_{1}R=\dot{x}[/itex] just means that there is no slipping.

    I am looking for mechanical Lagrangian equation of [itex]L=T-V[/itex].
    While i know [itex]V=mgl_{cm}cosθ_{1}[/itex], I am not sure what T would look like, I know it would have to do with at least 2 terms, transitional kinetic energy and rotational energy terms, but I am not sure how the interaction of the two wheels would play out.

    3. The attempt at a solution
    [itex]L=T-mgl_{cm}cosθ_{1}[/itex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 18, 2012 #2
    Can the wheel on the bottom slide? Assuming that the small one sticks to the big one (is this OK?), then just write the x, y position of the big one as a function of theta sub one and distance from origin, and write down the position of the small one relative to the big one. Then just add them to get the position w.r.t to the ground, and use this to write the lagrangian.
     
  4. Jan 19, 2012 #3
    the wheel on the bottom does not slide. the small one is actually a motor that turns the big one. so they are stuck together.

    I don't get what you mean "and write down the position of the small one relative to the big one. Then just add them to get the position w.r.t to the ground, and use this to write the lagrangian."
     
  5. Jan 19, 2012 #4
    Ooh ok, then that makes the math simpler. :-) Forget about what I said.

    Set the origin to be the center of the wheel. Using basic trig, find the x,y coordinates of the small wheel in terms of theta 1 and theta 2. (The small one can slide, right?) Then plug it into
    [itex]T_{small}=\frac{1}{2}I \dot{\theta_{1}}+\frac{1}{2}mv^{2}=\frac{1}{2}I \dot{\theta_{1}}+\frac{1}{2}m \left ( \dot{x^{2}} + \dot{y^{2}} \right ) [/itex]
    The kinetic energy for the big wheel is easy. Plug those expressions for x and y for the small wheel into [itex]V=mgy[/itex], then get the Lagrangian.
     
  6. Jan 19, 2012 #5
    the small one would kinda be orbiting around the big one, but ideally it should stay pretty much on top of it. this is kind of like a inverted pendulum problem.
    so for the small wheel, the transitional motion would be
    [itex]\frac{1}{2}m((R+r)\dot{θ}cosθ)^2[/itex].

    you are missing the square on the [itex]\dot{θ}[/itex] on the rotational term right?
     
  7. Jan 19, 2012 #6
    ^Oops! Yeah, I forgot that...
     
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