Classical Mechanics, clyinder inside a cylinder.

[Spoony]

b]1. Homework Statement [/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?

Homework Equations

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

The Attempt at a Solution

i have the lagrangian;

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

now i have trouble as when i got the small angle approximation [tex] sin(2 \varphi) [\tex] becomes $$2 \varphi$$ and when i put this int the euler lagrange equation i end up differentiating this with respect to $$\varphi$$ and so i get a constant, then I am left with the $$\frac{d}{dt} \frac{\partial L}{\partial \dot{\varphi}}$$ bit being equal to a constant, $$3 \sqrt{2} a m g$$
which you can't solve with a function of cos and sine

 

[jdwood983]

now i have trouble as when i got the small angle approximation $$sin(2 \varphi)$$ becomes $$2 \varphi$$ and when i put this int the euler lagrange equation i end up differentiating this with respect to $$\varphi$$ and so i get a constant, then I am left with the $$\frac{d}{dt} \frac{\partial L}{\partial \dot{\varphi}}$$ bit being equal to a constant, $$3 \sqrt{2} a m g$$
which you can't solve with a function of cos and sine

Isn't the point of making a small angle approximation to make the function linear, thus eliminating the need for cosine and sine terms?

 

[Spoony]

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 I've 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 :)

 

[jdwood983]

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 I've 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 :)

Correct me if I'm wrong, but isn't the small angle approximation of cosine $\cos[x]=1-x^2/2$? And what constants were eliminated through the Lagrangian, because there shouldn't be any (maybe mass, but that depends on the system).