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An inverted pendulum has its center of mass above its pivot. For a point mass $m$ on a massless rod of length $\ell$, let $\theta$ measure departure from the upward vertical. The gravitational potential energy is $mg\ell \cos \theta$, and the equation of motion is $\ddot{\theta }=(g/\ell )\sin \theta$.

For small $\theta$, linearization gives exponentially growing and decaying solutions with rate $\sqrt{g/\ell }$: the upright equilibrium is unstable. A moving support or an applied torque can stabilize it through feedback control. Balancing a rod on a hand makes this visible: the hand continually moves the pivot to counter the growing departure.

Mark Levi studies dynamical systems through geometry and analysis, with applications to mechanical and physical motion. His research description includes stability, geometric phases, averaging and charged-particle dynamics.

An example of the viewpoint is the inverted pendulum: a rapidly varying drive can stabilize a configuration that is unstable without it. Translating the motion into geometric or averaged descriptions can expose a simpler mechanism than direct inspection of the full trajectory.^[ref]

Static equilibrium is a state in which a mechanical system remains at rest in a specified reference frame. For a rigid body in an inertial frame, zero resultant external force and zero resultant external torque are the balance conditions:The body must also initially be at rest: zero acceleration alone does not exclude uniform translation or steady rotation.^[ref]

A free-body diagram identifies the body being analyzed and includes its external loads and support reactions. Internal action–reaction pairs cancel when both interacting parts belong to the chosen system. Choosing a different point $O$ for the torques changes their sum by a term proportional to the resultant force. Thus, when the resultant force is zero, a vanishing resultant torque about one point also vanishes about any other point.^[ref]

For an original example, take an ideal massless horizontal beam of length $4\mathrm{m}$, supported at its ends, with a downward $120\mathrm{N}$ load $1\mathrm{m}$ from the left support. Let the upward support reactions be $R_L$ and $R_R$. Vertical force balance and torque balance about the left end giveso $R_R=30\mathrm{N}$ and $R_L=90\mathrm{N}$. The nearer support carries more load because the farther reaction has the larger moment arm. Two equal and opposite forces alone would not guarantee equilibrium: a force couple can have zero resultant force and a nonzero torque.

A simple apparatus test suspends a beam from two spring balances and moves a known load along it. Account separately for the beam's own gravitational weight and its center of mass before comparing the measured reactions with the ideal calculation. The sum of the readings should track total downward load, while their difference changes with load position. Finally, static equilibrium does not imply dynamical stability: an exactly upright inverted pendulum balances at rest but departs from that state after a small disturbance.

Statics studies mechanical equilibrium: configurations in which the relevant forces and torques balance. For a rigid body at rest in an inertial frame, the conditions are $\sum \mathbf{F}=0$ and $\sum {\mathbf{\tau }}_O=0$, with all torques calculated about the same point $O$.

A horizontal beam carrying a downward load illustrates why both equations are necessary. Vertical force balance fixes the sum of the support reactions, while torque balance determines how the load is divided between supports. Equilibrium alone does not imply dynamical stability: a perfectly upright inverted pendulum satisfies force and torque balance but falls after a sufficiently small disturbance.