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Equation of motion for a rigid rotating rod

  1. Nov 22, 2011 #1
    Hi all,

    1. The problem statement, all variables and given/known data
    My problem is a pretty basic one, in a exercise a rigid rod of mass M is rotating around a horizontal pivot point i one end. The rod has the length L. I now need to derive the equation of motion using the Lagrangian formalism.

    My question is:
    Can i view the rod as a pendulum with a bob mass equal to that of the rod?


    In this case the equation would simply be:
    [itex]\theta '' + \sqrt{\dfrac{g}{L}} sin(\theta)=0[/itex]

    Best Regards
    Cole.
     
  2. jcsd
  3. Nov 22, 2011 #2

    Doc Al

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    Staff: Mentor

    Not exactly. An extended mass such as a rod cannot be treated as a simple pendulum. Treat it as a physical (or compound) pendulum.
     
  4. Nov 22, 2011 #3
    Ach thanks your reference to compound pendulum got me on the right track.
    It's all about the center of mass:
    [itex]I \theta '' = -M_{cm} g L_{cm} sin \theta
    \Leftrightarrow \theta '' = -\dfrac{3 g}{2 L} sin (\theta)[/itex]
     
  5. Nov 22, 2011 #4

    Doc Al

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    Staff: Mentor

    Good. Now do it using the Lagrangian. :smile:
     
  6. Nov 22, 2011 #5
    Thanks for the quick reply, I am slightly confused as how to in corporate the moment of inertia for the rod. If I use the kinetic energy [itex]T=\dfrac{1}{2}M r'^{2}[/itex] and potential energy [itex]U=-mg r \mathrm{cos} \theta[/itex] there is no difference from the regular pendulum and i don't get the 3/2 factor!
    Could I use [itex]T=\dfrac{1}{2} I r'^{2}[/itex] where I is the inertia? A hint is appreciated :)

    Regards
    Cole.
     
  7. Nov 22, 2011 #6

    Doc Al

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    You must use the rotational KE = 1/2Iω2, not 1/2Mv^2. (You'll express in terms of θ', of course.)
     
  8. Nov 22, 2011 #7
    Great i figured it out!
    However I have run into a new problem. Suppose the rod has a bead of mass m attached at the pivot moving down is it rotates. I can't figure out what the kinetic energy of the system is.. Me guess is [itex]T_{tot}=T_{rod}+T_{bead}=\dfrac{1}{2}\dot{\theta}^{2}\left( I_{rod}+I_{bead}\right) + \dfrac{1}{2}m\left( \dot{r}+r\dot{\theta}\right) ^2[/itex] because the bead must contribute with kinetic energy from the radial speed. ([itex]\bar{v}=v_{r}+v_{\theta}=\dot{r}+r\dot{\theta}[/itex]) However my equations of motion don't seem right.. Any hints? You've been more than helpful so far!

    Regards
    Cole.
     
  9. Nov 23, 2011 #8

    Doc Al

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    The KE of the rod is the same. Careful how you write the KE of the bead: Don't count it twice and don't add perpendicular velocity components together before squaring them.
     
  10. Nov 23, 2011 #9
    Everything appears correct except for the velocity factor. When I expand it I get:
    [itex]\dfrac{1}{2}m\ddot{r}^{2}+m\dot{r}r\dot{\theta}+ \dfrac{1}{2} mr^{2}\dot{\theta}^{2}[/itex]
    and when I use the part of the Legrender equation relevant for the energy I get:
    [itex]\dfrac{\mathrm{d}}{\mathrm{t}}\dfrac{\mathrm{d}L}{\mathrm{d}\dot{\theta}}=m\dot{r}r\ddot{\theta} +m r^{2}\ddot{\theta}[/itex]
    I have deduced from another solution that the above result should simply be: [itex] 2mr \dot{r} \dot{\theta} [/itex]
    Where am I going wrong?
    Thanks a million for the help!

    Regards
    Cole.
     
  11. Nov 23, 2011 #10

    Doc Al

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    What did you expand to get this? If you are referring to the last term in the following, I'd say it was wrong:
    Reread my last post.
     
  12. Nov 23, 2011 #11
    Yes I am referring to the last term:
    [itex]\dfrac{1}{2}m( \dot{r}+r \dot{\theta} )^{2}=\dfrac{1}{2}m(\dot{r}^{2}+2r\dot{r} \dot{\theta}+r^{2}\theta ^{2})[/itex]
    What i think you mean in the above is to say:
    [itex]T_{b}=\dfrac{1}{2}I_{b}\dot{\theta}+\dfrac{1}{2}m \dot{r} ^{2}+\dfrac{1}{2}mr^{2} \dot{\theta}^{2}[/itex]
    but that doesn't either give me the expression I'm looking for..
     
  13. Nov 23, 2011 #12

    Doc Al

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    Yes, that's what I would put as the KE term. Except that Ib should be Irod in the first term. And the [itex]\dot{\theta}[/itex] should be squared.
    Why is that?
     
  14. Nov 23, 2011 #13
    Ahh but it does! Silly mistake from my part.
    Thank you so much it has been a real journey :)

    Regards
    Cole.
     
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