When obtaining the equation of a pendulum following classical mechanics (Virtual Work) we state that:

The work is in equilibrium, therefore $\textbf{F} = 0$ and the Virtual Work is

$$\textbf{F} · \delta \textbf{r} = 0\tag{1}$$

But, is a pendulum in equilibrium? I mean, the velocity of the pendulum changes with the time, how can we say that the pendulum is in equilibrium?

Often is also used the expression

$$\textbf{F} – m \ddot{\textbf{r}} = 0\tag{2}$$

to express this equilibrium, but it isn't an equilibrium at all, since the only think we do is move to the left the inertial force from the Newton's second equation $\textbf{F} = m \ddot{\textbf{r}}$.

Goldstein sais in his book that equation (2) means: *that the particles in the system will be in equilibrium under a force equal to the actual force plus a "reversed effective force" $- m \ddot{\textbf{r}}$*.

What does it mean an how applies this to the pendulum?

## Best Answer

The equilibrium Goldstein is referring to is the equilibrium between the actual force $\vec F$ acting on the particle and the inertia force $-m\vec a$, i.e., $$\vec F-m\vec a=0.\tag 1$$ The idea, due to d'Alembert, is to extend the applicability of the virtual work principle from statics to dynamics and in some sense to transform the problem of motion to the problem of equilibrium. Note that this is consistent with the fact that we can always go to the reference frame where the particle is at rest (thus static) and in this frame we need to introduce a fictitious force (which is in equilibrium with the interaction force).

The above idea seems to be trivial however it is not. The whole point is that, since constraint forces do no virtual work, then Eq. (1) implies $$(\vec F_s-m\vec a)\cdot\delta \vec r=0,$$ where $\vec F_s$ are the specified or impressed forces and $\delta\vec r$ is a virtual displacement. This is actually the so-called d'Alembert Principle, and that is what Goldstein is intending to apply for the pendulum.