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Lagrangian math problem

  1. Dec 5, 2006 #1
    1. The problem statement, all variables and given/known data
    A sphere of radius rho is constrained to roll without slipping on the lower half of a hollow cylinder of radius R. Determine the Lagrangian function, the equation of constraint, and equations of motion.


    2. Relevant equations
    U = mgh
    Lagrange's eqns of motion
    T= .5 mv^2 + .5 I w^2
    3. The attempt at a solution
    Let me work this in cylindrical coords. Let theta be the angle between where the sphere is at the bottom of the cylinder and where it might lie some distance up the side of the cylinder. With this, the height can be written as R cos(theta). This gives me the potential energy term in the L eqns. The kinetic energy is a sum of translational and rotational. The translational term looks like .5 m (r^2*theta_dot ^2 + z_dot^2). There is no r_dot term since it is constrained to lie on the surface. The rotational kinetic energy term should look like .5 (2/5 m rho^2)*w^2. (Sorry that I am using w for omega.) So is this okay so far? I am having trouble with the eqn of constraint. I can see that the arclength traversed by the sphere is equal to the arclength traced out on the cylinder. This seems to introduce another coordinate though- the angle rotated out on the sphere itself. Does that make sence? thanks
     
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  3. Dec 5, 2006 #2

    OlderDan

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    I don't think your potential energy term will work. If you take the top of the half cylinder as U = 0 then you could have U = -(R-ρ)cos(θ), or if you take U = 0 at the bottom you could have U = (R-ρ)[1-cos(θ)]. The reason I wrote (R-ρ) instead of R is to locate the CM of the ball taking its raduis into account. My guess is you need to do that everywhere unless you are given R>>ρ.

    For future use, here is ω :smile:

    In cylindrical coordinates you have r, θ, z for the CM with r constrained to be r = (R-ρ). That was the easy one. But you do need something for the angular displacement and velocity of the ball in terms of dz and dθ. Something along the lines of dφ = (∂φ/∂z)dz + (∂φ/∂θ)dθ comes to mind. The first partial should be easy since that is like rolling on a flat surface. The second one is not as easy if ρ is not negligably small compared to R, but at constant z it is quite tractable I think.
     
  4. Dec 6, 2006 #3
    Thank you my friend. The phi angle you refer to is the angle traced out on the sphere itself, right?
    I will see where I can get with this one now.
     
  5. Dec 6, 2006 #4

    OlderDan

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    Yes. φ is a rotation coordinate for the sphere. I think you only need to be concerned with infinitesimal rotations, dφ.
     
  6. Dec 11, 2006 #5
    Okay, I finally have some time to finish this one up, hopefully. From your hint, I looked at the angle phi. For the z direction, rho*phi = z. For the theta direction, can I say that rho*phi = R*theta, that seems too simple; what am I missing? If this is okay, or not, I have a differential equation that has d(phi) equal to a d(z) term and a d(theta) term. Can I just integrate both terms with respect to the differential variables to find the eqn. of constraint for phi? The eqns of motion should be straight forward after this.

    Also, super newb question: I can't find the forward slash on my keyboard so that I can start practicing Latex. Where is that? thanks a 10^9
     
  7. Dec 11, 2006 #6

    OlderDan

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    For the theta direction, I think you need to consider a circle of radius ρ rolling around the interior of a circle of radius R. The circumference of the large circle is 2πR and the circumference of the small circle is 2πρ. How much does φ change if you change θ by 2π? I don't think it is what you have because your rleationship suggests that as ρ approaches R, φ approaches θ and that cannot be. As ρ approaches R, φ hardly changes at all when θ changes by 2π. I think you will find that φ = θ when ρ = R/2. What will it be if ρ = R/4?

    Once you get this right you can write the differential relationship dφ = (∂φ/∂z)dz + (∂φ/∂θ)dθ. Do you need anything more than that for φ. Is it not true that the only place φ enters the problem is in the rotational kinetic energy term where you need dφ/dt? Am I oversimplifying? There is no way you can find φ(z,θ) because changes in φ are path dependent, but I don't think you have to.

    The / is in the last row of letters on the keyboard one position left of the right shift key. The \ is in the top row one position left of backspace.
     
  8. Dec 12, 2006 #7
    Ah, that makes a great deal of sense. I will see how much further I can get.
    Thanks for the keyboard help too.
     
  9. Dec 12, 2006 #8
    This is the eqn I found: [tex] \rho * \phi = (R - \rho) \theta [/tex]. Hmm my Latex isnt working out here? ANywho, does this constraint eqn look better now? I think that before I was looking at the arclength traced out by the edge of the disc and not by the center of mass. :tongue2:

    *edit* the LAtex did work, just not in the preview!
     
  10. Dec 12, 2006 #9

    OlderDan

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    I think that works for the angle ratio. From that you can get ∂φ/∂θ since that angle relationship is for constant z. Too bad about that LaTeX preview not working. It used to, but I guess it has not been working for some time now.
     
  11. Dec 14, 2006 #10
    I got it! Then I solved two more Lagrangian problems for fun.:biggrin:
     
  12. Dec 14, 2006 #11

    OlderDan

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    Great!!!!!
     
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