i have managed to find the kinetic energy but it cant be right because i didnt do anything to do with the rotational inertia because i have no idea how to do part (a)
for part 3 the lagrangian is just $L=T-U$ where $T$ is the kinetic energy and $U$ is the potential energy?
part 4 seems straight forward but i need to calculate L first, but i'm struggling to find the potential energy and the correct kinetic energy, i've only been doing examples with one mass this time there is multiple so its confusing me.
Calling $I_O$ the hoop inertia moment regarding $O$ and
$p_M = a(\sin\phi,-\cos\phi),\ \ p_m = p_M + a(\sin\theta,-\cos\theta)$ we have
$$ T = \frac 12\left(I_O\dot\phi^2+m \|\dot p_m\|^2\right) $$
and
$$ V = g\left(M p_M + m p_m\right)\cdot(0,1) $$
then $L = T - V$ and the movement equations are
$$ \left\{ \begin{array}{rcl} \phi'' & = & -\frac{a \left(2 a m \sin (\phi-\theta ) \left(\phi '^2 \cos (\phi-\theta)+\theta '^2\right)+g (m+2 M) \sin (\phi)+g m \sin (\phi -2 \theta)\right)}{2 \left(a^2 (m+M)+a^2 (-m) \cos ^2(\phi -\theta )+I_O\right)} \\ \theta'' & = & -\frac{a^3 m \theta '^2 \sin (2 (\phi-\theta ))-g \sin (\theta) \left(a^2 (m+M)+2 I_O\right)+a^2 g (m+M) \sin (2 \phi-\theta)+2 a \phi '^2 \left(a^2 (m+M)+I_O\right) \sin (\phi-\theta )}{2 \left(a^3 m \cos ^2(\phi-\theta)-a \left(a^2 (m+M)+I_O\right)\right)} \\ \end{array} \right. $$
Considering now small variations $(|\alpha| << 1, \sin\alpha = \alpha,\ \cos\alpha = 1,\ \alpha^2=0)$ we have
$$ \left( \begin{array}{c} \phi''\\ \theta'' \end{array} \right)= \left( \begin{array}{cc} -\frac{a g (m+M)}{I_O} & \frac{a g m}{I_O} \\ \frac{a g (m+M)}{I_O} & -\frac{g \left(a^2 m+I_O\right)}{a I_O} \\ \end{array} \right)\left( \begin{array}{c} \phi\\ \theta \end{array} \right) $$