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Rolling hoop and lagrange multipliers

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A uniform hoop of mass m and radius r rolls without slipping on a fixed cylinder of radius R. The only external force is that of gravity. If the smaller cylinder starts rolling from rest on top of the bigger cylinder , use the method of lagrange multipliers to find the point at which the hoop falls off the cylinder.


    2. Relevant equations



    3. The attempt at a solution

    So, it will fall off when the radial force is 0, right? Should I use virtual forces and d'Alemberts principle to do this problem? I want to get the radial force as a function of the angle that the smaller hoop rotates along the cylinder, right?

    EDIT: the word smaller in the last sentence should not be there because there is only one hoop
     
    Last edited: Oct 18, 2007
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  3. Oct 18, 2007 #2

    Gokul43201

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    Is it a hoop or a cylinder(?)..seems to change back and forth.

    In any case, you've got 2 constraints. Write down their equations in the desired form, extract the coefficients of the derivatives of generalized co-ordinates, and use them in the full form of the Lagrange equations. It looks like you will end up with 5 equations (including the 2 constraint equations) in 3 generalized co-ordinates and 2 Lagrange multipliers. One of these Lagrange multipliers comes with the constraint that the hoop be one the surface of the cylinder. That's the one you care to solve for.
     
  4. Oct 18, 2007 #3
    So the 3 generalized coordinates would be the angle that the hoop has rotated, call it theta.

    Another would be the angle from the center of the cylinder to the center of mass of the hoop, call it phi.

    [tex] T = 1/2 m (r\dot{ \theta})^2 [/tex]

    [tex] V = -(R + r) m g sin \phi [/tex]

    Are those good generalized coordinates? I am not sure that they are even independent...
     
  5. Oct 18, 2007 #4

    Gokul43201

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    Those are two. You need a third, which it appears you have eliminated by how you've written the V term; and you've also eliminated one constraint equation in the process. But the third co-ordinate and this constraint equation are important.

    Also, there must be an additional term in T; you've covered rotation about the hoop's axis, but what about the "translational" term a.k.a. rotation about the cylinder's axis?

    PS: It looks like you are measuring \phi from the horizontal. It may be easier if you measured it from the vertical (starting position of hoop) instead.
     
    Last edited: Oct 18, 2007
  6. Oct 18, 2007 #5
    OK. I guess the third is the distance from the hoops center (of mass) to the center of the cylinder, call it d.

    Then

    [tex] T = 1/2 m (r\dot{ \theta})^2 + 1/2m ((R+d) \dot{ \phi})^2+ 1/2 m (\dot{d})^2 [/tex]

    [tex] V = -(R + d) m g \cos \phi [/tex]

    with the new definition of phi.

    So, I know how to plug that into the EL equation and get the three equations of motion. However, I am not sure how to get the other two equation that you referred for the restraints and I am not sure where the Lagrange multipliers will come in...
     
    Last edited: Oct 18, 2007
  7. Oct 19, 2007 #6

    Gokul43201

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    You will find this in any standard classical mechanics text, such as Goldstein.

    First, Identify the two constraints:
    1. Hoop is rolling without slipping
    2. Hoop is touching the surface of the cylinder

    Both these constraints can be written in the form : [itex]\sum_j a_{ij} \delta q_j = 0 [/itex]

    Example, for 2: [itex]\delta d = 0 [/itex]

    Now, for the virtual work done by the constraint forces (in this case friction, and the normal reaction) to be zero, we can write these forces as [itex]f_j = \sum _i \lambda_i a_{ij} [/itex]. Finally, all you need to do is use stick these forces in the (RHS of the) 3 EL EoMs that you have.
     
  8. Oct 19, 2007 #7
    So, the other constraint is [tex] \dot{\theta} = \dot{\phi} [/tex] (2), right?

    So, I guess if we write that in terms of virtual displacements, it would be

    [tex] \delta \theta = \delta \phi [/tex] (1)

    right?

    But, d'Alembert's principle is something I am struggling with, can you help me figure out rigorously how you go from (2) to (1)?

    I have the Goldstein book and indeed this is a question from it.
     
    Last edited: Oct 19, 2007
  9. Oct 19, 2007 #8

    Gokul43201

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    Actually, that should be [tex]d \dot{\phi} = r \dot{\theta}[/tex]. Make sure that's clear.

    Also, in post#5, both places where you have R+d, should only have d (since this is already the distance between centers).
     
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