# Implicit definition of internal forces through choice of dynamical description of a rigid two mass pendulum

The central mass theorem states that the center of mass of a N-body system moves as if all external forces acting on the bodies of the system acted directly on the center of mass, it is a generic statement that can be made about N-body systems that obey Newton's laws. Furthermore, the motion of a system consisting of a pendulum composed of two masses ($$m_1$$,$$m_2$$ at $$l_1$$,$$l_2$$) on a (massless) rigid rod is typically solved using rotational dynamics $$(m_1 l_1^2 + m_2 l_2^2) \ddot{\phi} = -(m_1 l_1 + m_2 l_2) g \sin \phi$$. In principle, the central of mass theorem is also applicable, to solve for the motion of the pendulum. However, if one naively assumes the external forces for the system are the same as those when we described the system using rotational dynamics i.e. the gravitational forces acting on both masses, then we get (unsurprisingly) the equations of motion dictating a simple pendulum $$\ddot{\phi} = -\frac{g}{r} \sin \phi$$ (where $$r$$ is the center of mass). This is, of course, incorrect.

The mistake, that was made in applying the central of mass theorem is that we neglected to consider the tangential force applied by the rigid rod in order to keep the two masses on the rod. One can deduce this force from the discrepancy between the angular acceleration of the center of mass motion and the correct angular acceleration using rotational dynamics ($$M := m_1 + m_2$$, $$I := m_1 l_1^2 + m_2 l_2^2$$, $$I_r := Mr^2$$). $$$$-\frac{g}{r}\sin \phi + \frac{F_t r}{I_r} = -\frac{(m_1 l_1 + m_2 l_2)}{I} g \sin \phi$$$$ Rearranging and using the identity $$m_1 l_1 + m_2 l_2 = Mr$$ one finds: $$$$\begin{split} F_t &= -Mg \sin \phi \left(\frac{I_r}{I} - 1\right) = -F_{g_{\phi}}\frac{m_1 m_2 (l_2 - l_1)^2}{IM}\\ &= - F_{g_{\phi}} \frac{m_1}{M} \frac{m_2}{M}\frac{M(\delta l)^2}{I} = - F_{1g_{\phi}}\frac{m_2(\delta l)^2}{I} \end{split}$$$$ In other words, we see that the inertia of the second mass further along the rod drags on the center of mass because the period of the uncoupled simple pendulum systems is longer, the further along the rod we place the mass ($$\omega_{simple} = \sqrt{\frac{g}{l}}$$).

The question then remains, why is it that we have to consider the tension force ($$F_t$$) when considering the dynamics at the center of mass and not when considering the system using the rotational laws? The only answer that I can come up with is that, the central of mass theorem by definition can only consider the motion of objects with mass. Since the pivot of the pendulum has no mass we have to consider the tensional forces of the rod as external forces and explicitly handle them in our calculation. On the other hand, the rotational dynamics laws are derived from considering all point masses in the system and a pivot. Since, an equal and opposite force countering the tension force (acting on the center of mass), must be acting at the pivot, the tensional forces can be considered internal forces and cannot effect our rotational dynamics.

This seems reasonable enough, but also means that using the formalism built up with rotational dynamics implicitly expands what we consider internal forces of the system by a singular point i.e. the pivot (whether it does or does not have any mass). Is this correct, or am I over complicating things here?

You are using two different theorems as if they were equivalent, but they are not. Both theorems come from studying a system of particles. The theorem you use in rotational dynamics is that $$\tau=\dot{L}=I\ddot{ \alpha}$$. In this case you only consider the external forces that make a torque.
If instead, you want to use the theorem that states that $$F_{T}=Ma_{CM}$$, where M is the total mass and F the total external forces, then you need to use all the external forces. The external force you are missing is the force at the top of the rod, in the pivot that allows it to rotate (using this method is likely not advisable for this particular problem, but it is doable).