Classical Mechanics (Kibble & Berkshire) · Problem 12.22

The calculus of variations and Hamilton's principle — T.W.B. Kibble, F.H. Berkshire

Physics — Mechanics · lagrangian-mechanics-upside-down-pendulum

Problem

Upside-down pendulum A particle PP is attached to a support SS by a light rigid rod of length aa, which is freely pivoted at SS. PP moves in a vertical plane through SS and at the same time the support SS is made to oscillate vertically having upward displacement Z=ϵacos⁡ptZ = \epsilon a \cos pt at time tt. Take θ\theta, the angle between SPSP and the upward vertical, as generalised coordinate and show that Lagrange's equation is θ¨−(Ω2+ϵp2cos⁡pt)sin⁡θ=0,\ddot{\theta} - \left(\Omega^2 + \epsilon p^2 \cos pt\right)\sin\theta = 0, where Ω2=g/a\Omega^2 = g/a. The object is to show that, for suitable choices of the parameters, the pendulum is stable in the vertically upwards position! First write the equation in dimensionless form by introducing the dimensionless time τ=pt\tau = pt. Then θ(τ)\theta(\tau) satisfies d2θdτ2−(Ω2p2+ϵcos⁡τ)sin⁡θ=0.\frac{d^2\theta}{d\tau^2} - \left(\frac{\Omega^2}{p^2} + \epsilon\cos\tau\right)\sin\theta = 0. Solve this equation numerically with initial conditions in which the pendulum starts from rest near the upward vertical. Plot the solution θ(τ)\theta(\tau) as a function of τ\tau for about twenty oscillations of the support. Try ϵ=0.3\epsilon = 0.3 with increasing values of the parameter p/Ωp/\Omega in the range 1≤p/Ω≤101 \leq p/\Omega \leq 10. You will know that the upside-down pendulum is stable when θ\theta remains small in the subsequent motion.

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