SHO - C H A P T E R 2 The Simple Pendulum 2.1 INTRODUCTION Our goals for this chapter are modest wed like to understand the dynamics of a pendulum Why a

SHO - C H A P T E R 2 The Simple Pendulum 2.1 INTRODUCTION...

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C H A P T E R 2 The Simple Pendulum 2.1 INTRODUCTION Our goals for this chapter are modest: we’d like to understand the dynamics of a pendulum. Why a pendulum? In part, because the dynamics of a majority of our multi-link robotics manipulators are simply the dynamics of a large number of coupled pendula. Also, the dynamics of a single pendulum are rich enough to introduce most of the concepts from nonlinear dynamics that we will use in this text, but tractable enough for us to (mostly) understand in the next few pages. g θ m l FIGURE 2.1 The Simple Pendulum The Lagrangian derivation (e.g, [35]) of the equations of motion of the simple pen- dulum yields: ¨ ( t ) + mgl sin θ ( t ) = Q, where I is the moment of inertia, and I = ml 2 for the simple pendulum. We’ll consider the case where the generalized force, Q , models a damping torque (from friction) plus a control torque input, u ( t ) : Q = ˙ ( t ) + u ( t ) . 2.2 NONLINEAR DYNAMICS W/ A CONSTANT TORQUE Let us first consider the dynamics of the pendulum if it is driven in a particular simple way: a torque which does not vary with time: ¨ + ˙ + mgl sin θ = u 0 . (2.1) These are relatively simple equations, so we should be able to integrate them to obtain θ ( t ) given θ (0), θ˙(0)... right? Although it is possible, integrating even the simplest case
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(0) . Let’s start by plotting x ˙ vs x for the case when u 0 = 0 : x ˙ x mgl b π - π The first thing to notice is that the system has a number of fixed points or steady states , which occur whenever x ˙ = 0 . In this simple example, the zero-crossings are x = { ..., π, 0 , π, 2 π, ... } . When the system is in one of these states, it will never leave that state. If the initial conditions are at a fixed point, we know that x ( ) will be at the same fixed point.
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