9780071086806, Chapter 1, Problem 5

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Chapter Problem 5 Problem Consider the pendulum discussed in Example Now add torsional spring at the pivot point such that moment MS applied the rod about the pivot, where with KT the torsional spring constant a. Show that the nonlinear equation motion now Hint: You can use whatever method you like but one way simply sum moments about the pivot and include the "inertial force" ma acting on the mass (with the acceleration appropriately b. Using small perturbation derive the Reference and Perturbation Sets equations associated with the pendulum's equation motion C. Find equilibrium conditions for this d. Assess the stability of each of the equilibrium conditions Step step solution Step Solution part sketch of the pendulum as shown where the acceleration the mass force the mass is then Step g m umming moments about the and noting that positive moment cou we have the equation motion given by or 1.SM.5) Note that the spring attempting to hold the pendulum at against gravity Solution to the reference and perturbation quantities through the following change variable the equation motion becomes By using the trig identity and making the small perturbation assumption we have Therefore (1 SM 6) under the perturbation ml or (1.SM.7) Noting that the first term parentheses is the original equation of which the reference displacement angle must we have the reference equation given by (1.SM.8) And after eliminating this equation from we are left with the following perturbation equation motion =0 (1.SM.9) Since there only one degree of freedom there only one reference and one perturbation Solution By observing the action of the which attempts to hold the pendulum would expect one equilibrium angle little less than depending on the spring stiffness But will be interesting to explore whether there are other equilibrium and upon what factors the results The equilibrium conditions are governed by where equilibrium requires that =0 the equilibrium conditions are all the solutions of or (1.SM.10) Since (mgl/?t) and the functions and are both symmetric about =0, we might expect only one solution the depending the value Equation (1 SM 10) plotted for (mgl/KT) and: 0 rad and there fact only one solution for And we see that the spring constant gets larger (or mgl/KT gets the single equilibrium condition approaches as one would Equilbrium Constraint Equation 3 mgl/Kt Constrain Equation 3 3 Reference Angle the spring constant gets sufficiently the solution should approach the case without and that case has an uncountable number solutions at 1.2. 3 And fact the case Shown plotted below the solution Eqn. (1. 10) for (mgl/?t) and rad the latter case we see that there are multiple solutions (here three for as the value continues to increase there will more equilibrium solutions But again, all equilibrium solutions are those that satisfy Eqn SM 10). Equilibrium Constraint Equation 6 mgl/Kt mgl/Kt -2 6 6 -4 8 Reference Solution The stability equilibrium conditions depends on the characteristic roots the linear perturbation equation motion or these characteristic roots have negative real part, the equilibrium condition stable Let's consider the case withy which corresponds to (mgl/?T) 0.6783 rad (We know from the above analysis that for this value rad the only equilibrium condition Now the perturbation equation becomes The characteristic equation for the above for which the two roots are Since we know that positive, the two characteristics roots are seen be imaginary So the equilibrium condition stable the sense that the pendulum is perturbed from the equilibrium condition, it will not diverge But will continuously oscillate at frequency of rad/sec This conditional sometimes called neutrally Other equilibrium conditions would be evaluated the same manner