Why is it incorrect to find the period of a physical pendulum by considering forces on the centre of mass?

Question

Given a system of interacting particles subject to external forces, one can decompose the force on each particle into a net internal force and a net external force. Applying Newton's second law to each particle and summing over all the particles, one can then invoke Newton's third law to show that the derivative of the total momentum of the system with respect to time is equal to the resultant external force. In particular, the centre of mass of the system behaves like a point particle which has a mass equal to the total mass of the system and which is subject to the resultant external force on the system.

Suppose that a physical pendulum (rigid body) has a mass $M$ and is free to rotate about a point lying a distance $l$ from the centre of mass. As the pendulum oscillates, the centre of mass is constrained to move along an arc of a circle of radius length $l$ and is accelerated only by the tangential component of its weight. Why, then, does the centre of mass not oscillate like a simple pendulum of mass $M$ and length $l$? I understand that the correct procedure is to obtain the equation of motion using the rotational analogue of Newton's second law, but I do not understand why the problem cannot be solved simply by considering the centre of mass.

Answer 1

A simple pendulum is generally analyzed as a point mass. A physical pendulum has extended mass which must rotate about the center of mass. This requires the existence of an additional dedicated torque and consideration of a moment of inertia.

Answer 2

For a simple pendulum there is only one particle, because all the mass is concentrated in a point. For that particle it is possible to apply the Newton's second law. If we make the dot product with the elementary possible displacement, the tension force can be eliminated, because it is orthogonal to that displacement, resulting in: $$mgsin(\theta)L\omega dt = m\frac{dv}{dt}L\omega dt = mL\frac{d(L\omega)}{dt}\omega dt$$ Dividing by $\omega dt$ we get the expression for $\tau = I\alpha$ where $I = mL^2$: $$mgsin(\theta)L = mL^2\frac{d\omega}{dt}$$ The same expression for the torque can be used for any rigid body, pivoted at a distance $L$ from the center of mass. But the moment of inertia about the pivot point changes, it is a function of the geometry. The difference from the previous analysis is that applying the second law to a generic element of a rigid body, it is not possible to eliminate all forces except the tangential one.