1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrangian Mechanics - coupled oscillator?

  1. Nov 14, 2005 #1
    Hello I'm having a bit of trouble with analysing some of the coupled oscillator questions in terms of the energy functions.

    Here is a coupled oscillator diagram:
    [​IMG]

    Now for this one my main problem is that I don't know how to come up with the kinetic energy functions because there are the three masses there. By the way the question just asks you to find the lagrangian and hence find the equation of motion for the system.

    What I know for this question is that it will obviously be moving about the point b/2 (the big block that is). I also don't really know how to find the moment of inertia for this object. I would have thought you would need the length of it, but it wasn't given in the question. And this is what I have for the kinetic energy function:

    KE = (1/2)*I*[tex]\theta' ^2[/tex] + (1/2)m(x'+y')^2

    Where x=2dsin[tex]\theta[/tex] and y = 2dcos[tex]\theta[/tex]

    The reason I have 2 there is because I thought that's what you would do since both rods are displaced in the same direction and all.

    For my potential energy function I have this:

    potential is taken to be 0 at the black support
    PE = -[tex]M_G[/tex]gdcos[tex]\theta[/tex] - 2mgdcos[tex]\theta[/tex]

    From all this I can solve for the equation of motion, but I'm really not sure about it. I'd be grateful for any advice here.

    Here is another diagram (not a coupled oscillator) that I'm having trouble with too.
    [​IMG]

    In this case I really don't know how to do it. I know that it'd be rotational motion and all, but is the moment of inertia just the centre of mass moment of inertia plus md^2 where d = [sqrt([tex]a^2+b^2[/tex])]/2

    This is what I have for the energies:

    KE = (1/2)*I*[tex]\theta'^2[/tex]
    PE = -[tex]M_c_m[/tex]gdcos[tex]\theta[/tex], where d is the distance of the centre of mass to the pivot point.

    Again any help on either of these two questions would be great.
    N.B. the prime (') means that it is the derivative with respect to time
     
    Last edited: Nov 14, 2005
  2. jcsd
  3. Nov 14, 2005 #2
    Second one looks okay to me. In equilibrium the CM of the lamina (or frame...sorry I don't remember what moments are) lies at the mid point of the diagonal which coincides with the vertical. When you displace it from this position, a restoring torque acts to bring it back to the equilibrium position. Of course in LM, you need just the two expressions that you have written. The obvious constraint is that the distance of the CM from the pivot is constant.
     
  4. Nov 14, 2005 #3
    Thanks for the response to the second one man.

    haha just need to find out if I'm on the right track for the first one.
     
  5. Nov 14, 2005 #4

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    I think you have the first one about right. I can't read your diagram very well so let me know if I have goofed up here, I see two rods and a block? If [tex] I [/tex] in your expression for the kinetic energy is the moment of inertia about the end of a rod, then you simply need to count the kinetic energy of both rods since it looks like you have only one contribution so far. Also, if [tex] x' [/tex] and [tex] y' [/tex] are the center of mass coordinates of the block, then the block's kinetic energy is as you have written. Here is a question for you, why don't you need to know the dimensions of the block?

    Edit: I don't think those two's are right. If d is the length of the rod, then when [tex] \theta = 0 [/tex], the center of mass of the block is at d (measured with positive down), not 2d.
     
    Last edited: Nov 14, 2005
  6. Nov 14, 2005 #5
    whoa sorry yeah I wasn't really paying attention for the problem. All the kinetic energy is rotational kinetic energy so ignore the (1/2)m(x'^2+y'^2).

    So for the rods you would just have 2 times the rotational kinetic energy of one rod.

    Then you would also have the rotational kinetic energy of the block right.
    This would just be (1/2)*[tex]I_b[/tex]*[tex]\theta' ^2[/tex], where [tex]I_b[/tex] means the moment of inertia of the block. Also I didn't say that you wouldn't need to know the dimensions of the block. You're given the one dimension, but I said I would have thought you would need the length as well. Right now you only have the dimension 'a'. The centre of mass of the block is just the point in the middle and is denoted G in the diagram.

    Also is the potential energy function right?
     
  7. Nov 14, 2005 #6

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Heh, well I did say you didn't need to know the dimensions of the block. :tongue2: Does the center of mass of the block translate? Is there any rotation about the center of mass?

    The potential energy is easy, it is just the sum of the contributions from each center of mass (of the block and the two rods). You have a slight mistake in the potential energy, where is the center of mass of each rod?
     
  8. Nov 14, 2005 #7
    Sorry man I just misread what you wrote.

    The centre of mass of each rod considering the mass that is attached to it would be very close to the pivot points on the block mass wouldn't they?
     
  9. Nov 14, 2005 #8

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Don't worry about it, I was just kidding around. Perhaps I misread the diagram, are the rods massive? If the rods are massive and uniform wouldn't the center of mass of each rod be at d/2 along the rod?
     
  10. Nov 14, 2005 #9
    Oh ok yup. Well yeah so you can treat them as separate then? I was just thinking that the attachment of the block would bring down the centre of mass, but yeah it makes sense that you wouldn't look at it that way since we are analysing the effect of gravity on the individual elements of the system.

    Thank you so much for your help.
     
  11. Nov 14, 2005 #10

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    You're welcome, and don't forget about the kinetic energy of the box. Hint: you don't need to know what its moment of inertia is (and you can't find it anyway).
     
  12. Nov 14, 2005 #11
    yup got it. cheers : )
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Lagrangian Mechanics - coupled oscillator?
  1. Lagrangian Mechanics (Replies: 2)

  2. Lagrangian mechanics (Replies: 3)

  3. Lagrangian Mechanics (Replies: 5)

  4. Lagrangian Mechanics (Replies: 1)

Loading...