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Lagrangian Mechanics, bead on a hoop

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data

    This is the problem:
    http://i.imgur.com/OJyzfhz.png?1



    2. Relevant equations

    $$\frac{dL}{dq_i}-\frac{d}{dt}\frac{dL}{d\dot{q_i}}=0$$
    $$ L=\frac{1}{2}mv^2-U(potential energy)$$

    3. The attempt at a solution

    This is my attempt at question A):

    http://i.imgur.com/IeJVGm3.jpg


    Does this look right? If not, am I atleast in the right direction. If I am right, how do I tackle B? I have no clue how to even begin on B.

    Thanks
     
  2. jcsd
  3. Oct 6, 2013 #2
    Your solution of A is generally correct, but there is a sign error in it.

    To find the equilibrium positions, use the main property of equilibrium: everything stays constant. That means that all the dynamical variables are constant, and their derivatives are zero.
     
  4. Oct 6, 2013 #3
    Alright, thanks. I see I need a minus sign before the first term of my final answer. I'll try B with your tip!
     
  5. Oct 6, 2013 #4
    at C, do you think they want me to do a taylor expansion around theta_equilibrium?
     
  6. Oct 6, 2013 #5
    Yes, the idea is that you should end up with ## \ddot \theta + \omega ^2 \theta = 0 ##, where ##\omega## is to be determined from the expansion.
     
  7. Oct 6, 2013 #6
    I actually have no idea where/how to do the expansion. Shouldn't I do it on x(theta), y(theta) and z(theta) ?

    Alsol, this is how I did B):

    Eq. of motion is

    $$\frac{g}{R}\sin{\theta}+\ddot{\theta}-\omega^2\sin{\theta}\cos{\theta}=0$$

    the 2nd derivative is zero at equilibrium so i get

    $$\theta_{eq} = arccos(\frac{g}{R\omega^2})$$
     
    Last edited: Oct 6, 2013
  8. Oct 6, 2013 #7
    You do it in the equation of motion you obtained.

    Does it exist for all values of ##\omega##?

    What about ## \sin \theta = 0 ##?
     
  9. Oct 6, 2013 #8
    It exists for
    $$R\omega^2\geq g $$

    Expansion of sine around my theta_eq??

    $$\sin(\theta)=\sqrt{1-(\frac{g}{R\omega^2})^2}+\frac{g}{R\omega^2}\cdot (\theta -\arccos(\frac{g}{R\omega^2})-0.5\sqrt{1-(\frac{g}{R\omega^2})^2}\cdot (\theta -\arccos(\frac{g}{R\omega^2})^2+.. $$

    I'm not really sure about what you meant with ''what about sin(theta)=0 ?'' . Sorry
     
  10. Oct 6, 2013 #9
    My question was whether ## \sin \theta = 0 ## also gives equilibrium points.

    Your expansion is correct, but you do not need terms that are above linear in ##\theta##.
     
  11. Oct 6, 2013 #10
    Oh, ofcourse! I forgot sin(theta). theta = 0 is an equilibrium point too.(well and pi but I guess that's not very reasonable here)

    So I just have to expand sin(theta)cos(theta) now and then write it out?
     
  12. Oct 6, 2013 #11
    Also, how do you type in tex inside your normal sentences? The way I do it ([ tex] [/ tex]) , my text comes on a new line.
     
  13. Oct 6, 2013 #12

    BruceW

    User Avatar
    Homework Helper

    you can use itex instead. (sorry for jumping in on this thread). Don't be too quick to dismiss theta=Pi. What do questions b) and c) actually ask you for?
     
  14. Oct 6, 2013 #13
    Do not guess. Your task is to find out whether ##\theta = \pi ## is a stable equilibrium.

    Yes - about each equilibrium. Note that in all cases except ## \theta_{e} = 0 ##, you will have to change your variable to ## \theta^* = \theta - \theta_{e} ##, so that the resultant equation have the canonical form ##\ddot {\theta ^*} + k^2 \theta ^* = 0 ##.
     
  15. Oct 6, 2013 #14
    I think I have it now. I did the expansion around theta_eq=pi, and the differential equation was the one of an exponential function(so not stable).

    Around theta_eq=arccos(g/(Rω^2)), I found

    $$k^2=\frac{3g^2}{R^2\omega^2}-\omega^2$$
     
  16. Oct 6, 2013 #15
    At ## \theta = 0 ##, the linearized equation is ## \ddot \theta + (\frac g R - \omega^2) \theta = 0 ##. Clearly, the nature of its solution depends critically on ##(\frac g R - \omega^2)##, and three cases are possible. You should analyze all of them.

    Same for ## \theta = \pi ##.
     
  17. Oct 6, 2013 #16
    I do not think it is correct. I obtain a similar expression, yet a different one.
     
  18. Oct 6, 2013 #17

    Thanks again! forgot about that. Still get that with theta_eq=pi, it is unstable in every case. I also see I made a mistake with determining k in my previous post, I got a K now that makes it stable for the allowed \omega (minus the end point).
     
  19. Oct 6, 2013 #18
    I got this one now:

    $$k^2=\omega^2(1-(\frac{g}{R\omega})^2),$$
     
  20. Oct 6, 2013 #19
    This is not self-consistent dimensionally. ##(\frac{g}{R\omega})^2## has the dimension of frequency squared, while 1, from which it is subtracted, is dimensionless.
     
  21. Oct 6, 2013 #20
    By the way, how did you analyze the case ##\theta_e = 0## and ## R\omega^2 = g ##?
     
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