Question 3.8.1: Consider the inverted pendulum of Example 1.8.1, illustrated......
Consider the inverted pendulum of Example 1.8.1, illustrated in Figure 3.21, and discuss its stability properties.
Summing the moments about the pivot point yields
\sum M_0=m l^2 \ddot{\theta}(t)=[-k l \sin \theta(t)][l \cos \theta(t)]+m g[l \sin \theta(t)]
Considering the small-angle approximation of the inverted pendulum equation results in the equation of motion
m l^2 \ddot{\theta}(t)+k l^2 \theta(t)=m g l \theta(t)
If m g l \theta is considered to be an applied force, the homogeneous solution is stable since m, l, and k are all positive. Writing the equation of motion as a homogeneous equation yields
m l^2 \ddot{\theta}(t)+\left[k l^2-m g l\right] \theta(t)=0
The forced response, however, is not bounded unless k l>m g and was shown in Example 1.8.1 to be divergent (unbounded) in this case. Hence the forced response of this system is Lagrange stable for F(t)=m g l \theta if k l>m g, and unbounded (unstable) if k l<m g.