Mechanics - rod attached to cylinder

[JohanL]

Problem:
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.

With

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

U=mg[R(1-cos\phi)+l/2(1-cos\theta)]

L=T-U

I get the equation of motions

\ddot{\phi}=\frac{2mg}{MR}\phi

\ddot{\theta}=\frac{3g}{2l}\theta

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 haven't included that the rod is attached to the cylinder. And I am not sure how to do this.
And i haven't included the the cylinder is rolling ?!?
Any ideas?

 

[Azael]

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

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

and rolling without gliding gives V=R\dot{\phi}

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.

 

[JohanL]

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 I am going to check my calculations with matlab.

 

[Azael]

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.

 

[JohanL]

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.

When i finally got to lagrange equations i get terms, which i can't eliminate, including

\dot{\phi}^2 , \dot{\psi}^2

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.
R(1-cos(phi))=2R[sin(phi/2)]^2=2R(phi/2)^2

 

[Azael]

the \dot{\phi}^2 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 :)

 

[JohanL]

+20 hours on this problem now i think.
Im starting to hate mechanics...
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 can't seem to find them.

 

[JohanL]

I have found a couple of misstakes in my attachment. I am going to change those and see if i get the right answer

 

[JohanL]

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 I've 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.

 

[JohanL]

Can somebody approve my attachment?