410ahw2soln - HW#2—Solutions —Phys410—Fall 2011 Prof...

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Unformatted text preview: HW#2—Solutions —Phys410—Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ [email protected] [S2] ( motion about equilibria for bead on spinning hoop ) The Lagrangian (7.68) can be writ- ten in the form L = 1 2 mR 2 ˙ θ 2- U ( θ ), At an equilibrium point θ we have U ( θ ) = 0. (a) Make a Taylor expansion of U ( θ ) about an equilibrium point θ . Keep terms out to quadratic order in the displacement from equilbrium. (b) Stability is determined by the sign of the quadratic term in U ( θ ): if it’s positive the motion is a stable oscillation, if it’s negative the motion runs away exponentially. (i) Show that below the critical angular ve- locity, the equilibrium at θ = 0 is stable and that at θ = π is unstable. (ii) Show that above the critical angular velocity the equilibria at θ = 0 and π are both unstable, while the two new equilibrium points are stable. Solution : L = 1 2 mR 2 ( ˙ θ 2 + sin 2 θ Ω 2 )- mgR (1- cos θ ), which has the form L = 1 2 mR 2 ˙ θ 2- U ( θ ) with U ( θ ) =- 1 2 mR 2 Ω 2 sin 2 θ + mgR (1- cos θ ). Here Ω is the externally imposed angular frequency of the hoop. Notice that L is proportional to m , so m will not affect the equations of motion. This is because the problem involves only gravity and inertia, both of which are proportional to m . To clean up the formulas, let’s choose units with m = R = g = 1. At the end we can put these quantities back in using dimensional analysis. Hence we have U ( θ ) =- 1 2 Ω 2 sin 2 θ- cos θ + 1. This can be expanded about any θ as U ( θ ) = U ( θ ) + U ( θ )( θ- θ ) + 1 2 U 00 ( θ )( θ- θ ) 2 + O [( θ- θ ) 3 ] . If θ is an equilibrium point, U ( θ ) = 0, and the frequency of small oscillations around θ is given by ω 2 = U 00 ( θ ). If U 00 ( θ ) < 0 then actually there are no oscillations, but rather a runaway motion....
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410ahw2soln - HW#2—Solutions —Phys410—Fall 2011 Prof...

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