- #1
l4teLearner
- 11
- 4
- 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
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