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Lagrangian problem (rigid body+particle)

1. The problem statement, all variables and given/known data
A hollow semi-cylinder of negligible thickness, radius a and mass M can rotate without slipping over the horizontal plane z=0, with its axis parallel to the x-axis. On its inside a particle of mass m slides without friction, constrained to move in the y-z plane. This system is set under a uniform gravitational field $-g \hat z$.
1)How many degrees of liberty does the system have? Write down the Lagrangian.
2)Find the configuration of stable equilibrium and write down the corresponding Lagrangian for small oscillations.
3)Find the frequencies of oscillations and their corresponding normal modes and describe these qualitatively.
4)If initially the system is at rest in its position of stable equilibrium and we apply a small impulse to the particle of mass m, write down the corresponding solution of the motion equations.

2. Relevant equations
$L=L_1+L_2$.

3. The attempt at a solution
1)1 degree of liberty, using intuition (i.e. you give me the position of either the hollow cylinder or the particle and I can tell you the position of the other body. Their motion isn't independent.).
Lagrangian $L_2$ of the hollow cylinder: $L_2=T_2=\frac{I \omega ^2}{2}=\frac{I \dot \phi ^2}{2}$. I'll come back to this angle phi later.
Lagrangian of the particle of mass m: $L_1=T_1-V_1$. I choose to use polar coordinates. The origin of my system of reference lies in the y-z plane at the height of the axis of rotation of the hollow cylinder. The kinetic energy of the particle is worth $T_1=\frac{ma^2\dot \theta }{2}$. The potential energy is worth $mga[1- \cos (\theta ) ]$.
However $\phi$ is related to $\theta$, they are dependent of each other. By intuition, $\phi I=m \theta$ so that $\dot \phi ^2 =\frac{m^2 \dot \theta ^2 }{I^2}$, I do not know how to prove this though and I know it should be possible. I'd like to know if this relation is correct.
All in all this makes the total Lagrangian worth $L=\frac{ma^2 \dot \theta }{2}+mga [\cos (\theta )-1 ]+\frac{m^2 \dot \theta ^2 }{I}$.
I know I need to calculate $I$. I tried very hard to do this but I'm really stuck on that (sigh). I considered the hollow semi-cylinder with a length L. I actually derived that the x component of the center of mass lies in the middle of the cylinder (L/2) which was obvious. The much less obvious position to find is the height over the ground of the center of mass (z component). I've drew a sketch. If I'm at a height dz over the ground, I need to find the arc length of a vertical transversal cut of the hollow semi cylinder. I still didn't find it. With intuition it should be proportional to the radius, a.
$z_{\text {center of mass}}=\frac{\int z \sigma dM}{\int dM}$ where $dM=SL$ and S is the arc length I'm looking for. I must express it in terms of $dz$. Then integrate from $z_0=0$ and $z_1=a$.
I'd appreciate if someone could tell me if what I've done so far is correct and how to get S in terms of dz. Thanks in advance!

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