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!

Mechanics - rod attached to cylinder

  1. Sep 23, 2006 #1
    A homogeneous cylinder of mass M and radius R can roll without
    slipping on a horisontal table. One end of the cylinder reaches a tiny
    bit out over the edge of the table. At a point of the pheriphery of the
    end surface of the cylinder, a homogeneous rod is attached to the
    cylinder by a frictionless joint. The rod has mass m and length l. Find
    Lagrange's equations for the generalised coordinates phi (angle in the cylinder) and theta (rod's angle)
    according to the figure, and determine the frequences of the principal
    modes of small oscillations. (Principal frequences = roots of
    characteristic equation.)


    I have a attached a figure.


    [tex]T = 1/4MR^2 \dot{\phi}^2 + 1/6ml^2 \dot{\theta}^2[/tex]



    I get the equation of motions



    which is the correct answer except a factor 1/3 in the equation for phi.
    But it feels like i have done some big misstakes computing T.
    I havent included that the rod is attached to the cylinder. And im not sure how to do this.
    And i havent included the the cylinder is rolling ?!?
    Any ideas?

    Attached Files:

    Last edited: Sep 23, 2006
  2. jcsd
  3. Sep 23, 2006 #2
    [tex] T_{cylinder}=\frac{1}{2}MR^2\dot{\phi}^2+\frac{1}{2}\frac{1}{2}R^2\dot{\phi}^2=\frac{3}{4}MR^2\dot{\phi}^2[/tex]

    since kinetic enefgy for a rolling cylinder is [tex] V=\frac{1}{2}MV^2+\frac{1}{2}I\dot{\phi}^2 [/tex]

    and rolling without gliding gives [tex] V=R\dot{\phi} [/tex]

    It also seems that you have only taken into account the kinetic energy part from the rotation of the rod and forgotten the kinetic energy from the movement of the center of mass of the rod?

    Start by writing the coordinates for the point in the cylinder the rod attaches to and then find the coordinates for the center of mass of the rod. From that you can find the speed of the center of mass of the rod.
  4. Sep 25, 2006 #3
    Thanks for answer!
    I wasnt even close :)
    But i did some reading over the weekend to fresh up on the basics and with your guidance i think i almost got it.
    I get a very complicated expression tho.
    Now im gonna check my calculations with matlab.
  5. Sep 25, 2006 #4
    Remember that you can get rid of ALOT of junk with the small angle aproximation :)

    I acctualy solved the exact same problem 2 weeks ago. Cant count how many hours I struggled with it before I got it right.
  6. Sep 26, 2006 #5
    When i finally got to lagrange equations i get terms, which i cant eliminate, including

    [tex] \dot{\phi}^2 , \dot{\psi}^2[/tex]

    Are you supposed to solve those equations? I must have done something wrong.

    I used small angle approximation already when i had the vector for the center of mass of the rod, to simplify the calculations, and then i used it for potential energy also.
  7. Sep 26, 2006 #6
    the [tex] \dot{\phi}^2 [/tex] and other squared angel derivates can be put equal to zero because of the small angle aproximation.

    Dont use the aproximation before you have the lagrange equations written out. Atleast its easier to not do any misstake doing it that way :)
  8. Sep 28, 2006 #7
    +20 hours on this problem now i think.
    Im starting to hate mechanics...:cry: :confused: :biggrin:
    If you have time maybe you could look at my attached word document which shows my calculations in Mathematica.
    Its probably a couple of easy misstakes but i cant seem to find them.
    Last edited: Sep 28, 2006
  9. Sep 28, 2006 #8
    I have found a couple of misstakes in my attachment. Im gonna change those and see if i get the right answer
  10. Sep 28, 2006 #9
    No it didnt help.
    x is phi and y is psi.
    When i used the small angle approximation
    I set sin x = x and all terms with squared derivatives I set equal to zero and cos x=1. Its possible ive done some mistakes here. I set cosx=1 and also expressions like cos(x-y)=1…im not sure if that is correct.
    See the attachment for all my calculations.

    Attached Files:

  11. Oct 1, 2006 #10
    Can somebody approve my attachment?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?