the initial rest configuration theta1 =O will result in the system going back the same initial configuration, i.e. theta 2 = theta 1 = theta. However, ifN>Ner, a perturbation A theta <1 applied to the initial rest configuration theta 1 = 0 will result in a new,permanent, rest configuration 02 #0. The scope of this exercise is to find what is the value of Ner for a given generic length of the rigid bar l, stiffness k and mass m. In order to solve the problem and find Ner: you have to 1. apply the principle of work and energy, T1 + U1 =T2 + U2, and 2. consider an angle 0 very small so that you can assume the following Taylor series expansions trumeated to the first and second order \sin (\theta)=\theta+\ldots \ldots \cos (\theta)=1-\frac{\theta^{2}}{2}+\ldots
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