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Lagrangian - rigid body problem

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    In a uniform gravitational field, there is a uniform solid disk of of mass [itex]M[/itex] and radius [itex]R[/itex]. A point mass [itex]m[/itex] is glued to the disk at a point that is at a distance [itex]a[/itex] from the center of the disk.
    The disk rolls without slipping. Find the frequency of small oscillations about the equilibrium point.
    UXmazpx.jpg
    I have been told to solve the problem in two ways:
    1. Considering two independent bodies, the disk and the point mass.
    2. Considering only one body by using their center of mass.


    2. Relevant equations
    Euler-Lagrange equations: [itex]\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial {\dot q}}=0[/itex] (1)
    Parallel axis theorem : [itex]I=I_{cm}+ MR^2[/itex] (For the second way) (2)

    3. The attempt at a solution
    First way:

    The position of the center of mass of the disk is:

    [itex]\begin{cases}
    x_{cm}=R\varphi \\
    y_{cm}=R
    \end{cases}[/itex]
    And the position of the point mass is:

    [itex]\begin{cases}
    x_{p}=x_{cm} - a\sin(\varphi)=R\varphi - a\sin(\varphi)\\
    y_{p}=y_{cm} - a\cos(\varphi) =R - a\cos(\varphi)
    \end{cases}[/itex]
    So I can compute the kinetic energy [itex]K[/itex] and the potential energy [itex]U[/itex] as:

    [itex]K=K_{cm}+K_{p}[/itex] and [itex]U=U_{cm}+U_{p}[/itex]

    where:

    [itex]K_{cm}=\frac{1}{2} M{v^2}_{cm} +\frac{1}{2}I_{disk}\dot{\varphi}^2 = \frac{3}{4}MR^2\dot{\varphi}^2 [/itex]

    (I have used [itex]I_{disk}=\frac{1}{2}MR^2[/itex])

    [itex]K_{p}=\frac{1}{2} m({\dot{x}^2}_{p}+{\dot{y}^2}_{p} )+\frac{1}{2}I_{p}\dot{\varphi}^2 = \frac{1}{2}mR^2 + \frac{1}{2}ma^2 -maR\cos(\varphi) + \frac{1}{2}ma^2\dot{\varphi}^2[/itex]

    (I have used [itex]I_{p}=ma^2[/itex])

    [itex]U_{disk}=MgR[/itex]

    [itex]U_{p}=mg(R-a\cos(\varphi))[/itex]

    Thus, the lagrangian is:

    [itex]L=\frac{1}{2}mR^2 + \frac{1}{2}ma^2 -maR\cos(\varphi) + \frac{1}{2}ma^2\dot{\varphi}^2 +\frac{3}{4}MR^2\varphi -MgR - mg(R-a\cos(\varphi)) [/itex]

    And using the Euler-Lagrange equation (1) and approximating [itex]sin(\varphi)=\varphi[/itex] I get

    [itex]\ddot{\varphi}+{\varphi}[\frac{mRa + mga}{\frac{3}{2}MR^2 +ma^2}]=0[/itex]

    Where [itex]\omega^2[/itex] is the term next to [itex]\varphi[/itex] wich is not the result I should get.

    Where is my error? I have spent hours looking and re-looking at this problem.

    Second way:

    My problem is that I don't know how to get the rolling condition for this system. The rest should be practically the same as I have done before, but computing the inertia moment by using the parallel axis theorem (2).

    Thank you very much, I really appreciate some help on this.
     
    Last edited: Nov 29, 2014
  2. jcsd
  3. Nov 29, 2014 #2

    OldEngr63

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    Gold Member

    Looks like you left off a dot and an exponent 2 in this expression .. just typos I presume.
    This is the MMOI for the point mass with respect to the center of the disk, but there is no reason to include this in the kinetic energy of the point mass. The point mass, by definition of the term point mass, has no MMOI about its own center of mass, and that would be the only term to include in addition to the point mass velocity^2 term.
    The rolling constraint is easy, and you already know it. It is simply
    s = R * phi
    This may be differentiated as needed.
     
  4. Nov 29, 2014 #3
    First of all thanks a lot for helping me with this problem.
    I have already corrected that typo :)
    You mean [itex]K_{p}=\frac{1}{2} m({\dot{x}^2}_{p}+{\dot{y}^2}_{p} )[/itex] with no additional rotating-motion terms right?
    I feel so foolish about considering the distance with respect to the center of the disk :H
    I thought this would change when considering only 1 body. Then, is there any consideration I should take appart from the MMOI? It looks too easy.
    Anyway, I'm going to solve it and edit #1
     
  5. Nov 29, 2014 #4

    OldEngr63

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    Gold Member

    As long as this is true, then one and the same constraint applies all the time.
     
  6. Nov 29, 2014 #5
    You're right, I have thought about it and seems totally logic.
     
  7. Nov 29, 2014 #6
    Now, the second part of the problem is solving it by using the center of mass of the system disk+point mass.
    I have a doubt when computing the Inertia Moment of my new system.
    QJHxdpf.jpg
    The system's center of mass position is:

    [itex]\begin{cases}
    x_{cm}=x_{0} - d\sin{\varphi}=R\varphi - d\sin{\varphi}\\
    y_{cm}=y_{0} - d\cos{\varphi} =R - d\cos{\varphi}
    \end{cases}[/itex]

    Where [itex]d[/itex] is the distance from the center of the disk to the center of mass:

    [itex]d=\frac{ma}{m+M}[/itex]

    Then, It is possible to write [itex]K[/itex] as:

    [itex]K=K_{trans}+K_{rot}[/itex]

    Where:

    [itex]K_{trans}=\frac{1}{2} (M+m)({\dot{x}^2}_{cm}+{\dot{y}^2}_{cm}) [/itex]

    And:

    [itex]K_{rot}=\frac{1}{2}I_{cm}\dot{\varphi}^2[/itex]

    Here is my doubt. I get stuck when computing [itex]I_{cm}[/itex].

    Should I use Steiner's theorem on each object and then sum them?
    [itex]\begin {cases}
    I^{disk}_{cm}=I^{disk}_0 -Md^2 \\
    I^{p}_{cm}=I^{p}_0 -m(d-a)^2
    \end {cases} [/itex]

    And then just do [itex]I_{cm}=I^{disk}_{cm} + I^{p}_{cm}[/itex]

    I don't see it clear.
     
  8. Nov 29, 2014 #7

    OldEngr63

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    Gold Member

    The MMOI for the combined body involves two applications of the parallel axis theorem:

    Jcombined = Jdisk + M*d^2 + Jpoint + m*(a-d)^2

    Then take the equations you have for xcm and ycm for the combined system, and differentiate them with respect to time to get the velocity components to write the kinetic energy. Don't make this a hard problem; it is not.
     
  9. Nov 29, 2014 #8

    OldEngr63

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    Gold Member

    Just as a matter of curiosity, where are you from?
     
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