Question 1.8.1: Consider the inverted pendulum connected to two equal spring......
Consider the inverted pendulum connected to two equal springs, shown in Figure 1.40.
Assume that the springs are undeflected when in the vertical position and that the mass m of the ball at the end of the pendulum rod is substantially larger than the mass of the rod itself, so that the rod is considered to be massless. The total length of the rod is l and the springs are attached at the point l/2. Summing the moments around the pivot point (point O) yields
ml² \ddot{θ} = Σ M_{0}There are three forces acting. The spring force is the stiffness times the displacement (kx) where the displacement x is (l/2) sin θ. There are two such springs, so the total force acting on the pendulum by the springs is kl sin θ. This force acts through a moment arm of (l/2) cos θ. The gravitational force acting on the mass m is mg acting through a moment arm of l sin θ. Thus summing moments about point O yields
ml² \ddot{θ} = -\left(\frac{kl²}{2} \sin θ\right) \cos θ + m\mathtt{g}l \sin θand the equation of motion becomes
ml² \ddot{θ} + \left(\frac{kl²}{2} \sin θ\right) \cos θ – m\mathtt{g}l \sin θ = 0 (1.87)
For values of θ less than about π/20, sin θ and cos θ can be approximated by sin θ ≅ θ and cos θ ≅ 1. Applying this approximation to equation (1.87) yields
ml² \ddot{θ} + \frac{kl²}{2} θ – m\mathtt{g}l = 0
which upon rearranging becomes
2ml\ddot{θ}(t) + (kl – 2m\mathtt{g})θ(t) = 0
where θ is now restricted to be small (smaller than π/20). If k, l, and m are all such that the coefficient of θ, called the effective stiffness, is negative, that is, if
kl – 2m\mathtt{g} < 0
the pendulum motion will be unstable by divergence, as illustrated in Figure 1.38.
