To what extent does this system behave as a pendulum?

[l4teLearner]

Homework Statement

This is exercise 9 from chapter 7 of Fasano, Marmi "Analytical Mechanics"

In a vertical plane, a homogeneous equilateral triangle $ABC$ with weight $p$ and side $l$ has the vertices $A$ and $B$ sliding without friction along a circular guide of radius $l$ (the point $C$ is located at the centre of the guide). A horizontal force of intensity $\frac{p}{\sqrt{3}}$ is applied at $C$. Find the equilibrium configurations and the corresponding constraint reactions. Study the stability of the (two) configurations and the small oscillations around the stable one. (Remark: the first part of the problem can be solved graphically.)

Relevant Equations

Second cardinal equation around $C$
As $C$ does not move, for equilibrium we need

$$0={dL}{dt}=M_C=\phi_A \times AC +\phi_B \times BC + F_p \times CP_0$$

where $P_0$ is the center of mass of the triangle and $F_p$ the weight of the triangle with magnitude $p$ and $\phi_A$ and $\phi_B$ are the costraint reactions at $A$ and $B$ respectively.

In the formula above I have that the mechanical momentum of the horizontal force with respect to $C$ is always $0$ because the point of application coincides with the pole. Also, the mechanical momentum of the costraint reactions is $0$ because the costraint is smooth so the reaction is always perpendicular to the circle, hence parallel to the radius vectors $AC$ and $BC$. This means that for the equilibrium the mechanical momentum of the weight needs to be $0$ as well, then I see that the two equilibrium positions are the one in which $AB$ is horizontal, either below $C$ (stable) or above $C$ (unstable).
So the horizontal force seems to play no role as for the equilibrium configurations.

Also, the system has one degree of freedom hence a single lagrangian coordinate which I can choose as $\theta$, the angle that the heigth of the triangle passing from $C$ does with the vertical direction.
The projection of the horizontal force $F$ on a tangent vector $\frac{\partial C}{\partial \theta}$ is always $0$ as well as $C$ does not move so $\frac{\partial C}{\partial \theta}=0$.

It seems then that the horizonal force has no role in the dynamic of the system as well and that the system behaves as a compound pendulum with moment of inertia $\frac{5}{12}\frac{p}{g}l^2$.

The only place where I can see the horizontal could play a role is in the calculation of the costraint reactions $\phi_A$ and $\phi_B$ which can be derived I think calculating the acceleration of the centre of mass $P$ of the triangle (e.g. solving lagrange equation for $\theta$) and then applying the first cardinal equation for the $x$ and the $y$ position (I have two equations and two unknowns, the magnitude of the costraint reactions as the direction is known, perpendicular to the circle).

Does the above make sense or am I loosing something?

thanks

 

[haruspex]

All seems reasonable to me.