We're given a semicircle whose total mass $M$ is evenly distributed along its contour and a little particle of mass $m$ is dropped from its top left corner as in the image below all over an horizontal plane. There is no friction between the semicircle and the particle nor between the semicircle and the ground (so no rolling without slipping!).
$P_0$ represents the particle at the start of the movement and $P$ represents it on a random position after the release of it.
The center of the semicircle is $O$.
The angles that I used to describe this crazy movement were $\theta = \angle P_0OA$ and $\alpha = AOP$ and, as I'm not that great with rigid body dynamics, I actually solved this problem using Lagrange's equations and I couldn't help but to notice that one of the equations that describe this movement is fairly simple:
$$2\frac gR \cos (\alpha) = (\ddot \theta + \ddot \alpha) 2\cos (\theta) - \pi \ddot \theta \sin(\theta+\alpha)$$
it is pretty remarkable that this equation does not involve the masses $m$ nor $M$ and I've been wondering if there is an easy way to get to this equation by analysing the momentum of the semicircle. But I struggled a bit with the acceleration of the system in a non inertial frame and I wonder if someone could hint on how to tackle it without Lagrange's equations.
This is the Lagrangian I got and I'm 100% sure that it is correct:
$$ \mathcal L (\theta, \alpha)= -\frac{R^2}{2(M+m)}(\frac 2{\pi}M \dot \theta \cos(\theta) + m \sin(\theta+\alpha)(\dot \theta + \dot \alpha))^2 + \frac{MR^2 \dot \theta ^2}2 + \frac{mR^2(\dot \theta + \dot \alpha)^2}2 + mgR\sin(\theta +\alpha) + \frac 2{\pi} MR\cos(\theta)g$$
EDIT: @eyeballfrog managed to turn that equation in a pendulum like equation:
$$\frac gR \sin\theta +(\pi/2)\ddot{\theta} - \cos\theta \ddot{x} = 0$$
where $x$ is the size of $OO_0$ (with signal), looks like a pendulum in a non inertial frame