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Lagrange mechanics

  1. Nov 1, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves over the cylindrical surface of radius R. The particle is connected to the origin O, located on the central point of the cylindrical surface axis, by a spring with spring constant k and lenght R. Ignore force of gravity.
    2447ayt.jpg


    a) State how many degrees of freedom the particle has and choose a set of generalized coordinates.

    b) Write the Lagrangian of the system and obtain the equations of motion

    c) Identify two constants of the motion.

    d) Obtain the equation of motion for the coordinate along the cylinder axis on the limit of small deviations of the plane that contais the origin. Calculate the first integral of the equations and say what's its physical meaning. (Note that: [itex]d^2z/dt^2=v dv/dz[/itex])




    2. Relevant equations
    [itex]L=T-V[/itex]

    [itex]\frac{\partial L}{\partial q} - \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}}) = 0 [/itex]


    3. The attempt at a solution

    a) The particle has 2 degrees of freedom, and the generalized coordinates I choose are the cylindrical coordinates: (R,θ,z) with R being constant

    b)

    First of all I define the distance from the origin to the mass [itex]r = \sqrt{R^2 + z^2}[/itex]

    [itex] L = T- V[/itex]

    [itex] V= \frac{1}{2} k (r - R)^2= \frac{1}{2} k (\sqrt{R^2 + z^2} - R)^2[/itex]

    [itex] T=\frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)[/itex]

    R is constant so :

    [itex] \dot{x}= -R sin(\theta) \dot{\theta}[/itex]
    [itex] \dot{y}= R cos(\theta) \dot{\theta}[/itex]
    [itex] \dot{z}=\dot{z}[/itex]

    [itex]T=\frac{1}{2} m (( R \dot{\theta})^2 + \dot{z}^2)[/itex]

    [itex] L = \frac{1}{2}m ( (R \dot{\theta})^2 + \dot{z}^2) - \frac{1}{2} k (\sqrt{R^2 + z^2} - R)^2[/itex]

    Using the formula:

    [itex]\frac{\partial L}{\partial q} - \frac{d}{dt}( \frac{\partial L}{\partial \dot{q}}) = 0 [/itex]

    For the equations of motion I got:

    [itex] k z- kz\frac{R}{\sqrt{R^2 + z^2 }} =m \ddot{z}[/itex]

    [itex]m R^2 \dot{\theta} = Constant [/itex]

    c) Identify two constants of motion

    One is the angular momentum which can be seen above [itex] L=m R^2 \dot{\theta}[/itex]

    To find the other I integrated the other equation and got :

    [itex]\frac{k}{2}z^2 - kR\sqrt{R^2+z^2}-\frac{m}{2} \dot{z}^2 = Constant[/itex] which I guess is the energy of the system.

    d) Still havn't put a lot of thinking into this question but for now I just need someone to confirm the above answers, if someone here has the time to do so I would be grateful. Thanks!
     
    Last edited: Nov 1, 2013
  2. jcsd
  3. Nov 1, 2013 #2

    TSny

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    Looks like you assumed that the spring has a natural length of R, but I didn't see that stated in the problem. [EDIT: OK, now I see it. Sorry.]

    Otherwise, your setup of the Lagrangian looks good to me.

    Is this equation dimensionally consistent?
     
    Last edited: Nov 1, 2013
  4. Nov 1, 2013 #3
    I just found a mistake on that equation, I'll fix it asap
     
  5. Nov 1, 2013 #4
    Ok, it's fixed, can you take a quick look at it?

    edit:
    Anyway, since the setup looks good to you I guess it's alright, I'll just look at it again later.

    Do you have any idea what d) is about? I'v never seen anything similar in class. Hints are appreciated, thanks!
     
    Last edited: Nov 1, 2013
  6. Nov 1, 2013 #5

    TSny

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    Rather than going back and making corrections in your original post, it is best to make the corrections in a new reply. That way, it will be less confusing for other readers of the thread.

    Your equation of motion for z looks good except for signs. Same for your integration of the z equation of motion to get part of the energy.

    I believe your comments regarding the angular momentum are correct.

    For (d), you want to approximate your equation of motion for small z (i.e., z << R).
     
  7. Nov 1, 2013 #6
    I see, sorry about that.

    This should be enough to get me started, many thanks!
     
  8. Nov 1, 2013 #7

    TSny

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    OK, good.
     
  9. Nov 2, 2013 #8
    Ok I just got d) done, I got this result

    [itex]Constant=1/2 mv^2[/itex], which is the kinetic energy ??


    I just did [itex]z=0[/itex] and [itex]d^2 z/dt^2=vdv/dz [/itex] on the equation of motion for z .
     
    Last edited: Nov 2, 2013
  10. Nov 3, 2013 #9

    TSny

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    You went too far in the approximation. You will need to approximate ##-kz+\frac{kRz}{\sqrt{R^2+z^2}}## to the lowest order non-zero approximation. Use $$\frac{1}{\sqrt{1+x}} \approx 1-x/2$$
     
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