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Classical Mechanics, clyinder inside a cylinder.

  1. Nov 18, 2009 #1
    b]1. The problem statement, all variables and given/known data[/b]

    A cylinder (solid) of radius a rolls inside a fixed hollow cylinder of radius 4a; inside a homogenous gravity field.

    Find, langragian using the rolling angle of the little cylidner as a generalized co-ord.
    angular frequency of small oscilalations about equilibrium. is the the oscillation more or less rapid than that of a point particle sliding without friction inside the hollow cylinder?

    2. Relevant equations

    L= T - V
    [tex] T = 1/2 m \dot{r}^{2} + \frac{I}{2} \dot{\varphi}^{2} [/tex]
    [tex] V = mgh [/tex]
    [tex] chord length = 2Rsin(\frac{\theta}{2}) [/tex]


    3. The attempt at a solution

    i have the lagrangian;

    [tex] L = T-V = 1/2 M 12^{2} a^{2} \dot{\varphi}^{2} + 1/4 M a^{2} \dot{\varphi}^{2} - 3 \sqrt{2} a M g sin(2 \varphi) [/tex]

    now i have trouble as when i got the small angle approximation [tex] sin(2 \varphi) [\tex] becomes [tex] 2 \varphi [/tex] and when i put this int the euler lagrange equation i end up differentiating this with respect to [tex] \varphi [/tex] and so i get a constant, then im left with the [tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{\varphi}}[/tex] bit being equal to a constant, [tex] 3 \sqrt{2} a m g [/tex]
    which you cant solve with a function of cos and sine
     
    Last edited: Nov 18, 2009
  2. jcsd
  3. Nov 19, 2009 #2
    Isn't the point of making a small angle approximation to make the function linear, thus eliminating the need for cosine and sine terms?
     
  4. Nov 25, 2009 #3
    you need the final awnser in terms of q(double dot) = a*q where a is a cosntant.
    then you can get a trigonometric solution to the problem by making
    q = sin(sqrt(a)*t)
    its ok now ive done it :) the trick is to get the right hand side (the q bit) = cos(phi), then you can small angel aproximate to (phi^2)/2 (constants are thrown away was lagrangian makes constants dissapear) then when you differentiate the equation to get the euler lagrangian the phi^2 becomes a phi, lovely :)
     
  5. Nov 26, 2009 #4
    Correct me if I'm wrong, but isn't the small angle approximation of cosine [itex]\cos[x]=1-x^2/2[/itex]? And what constants were eliminated through the Lagrangian, because there shouldn't be any (maybe mass, but that depends on the system).
     
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