410ahw2soln - HW#2Solutions Phys410Fall 2011 Prof. Ted...

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Unformatted text preview: HW#2Solutions Phys410Fall 2011 Prof. Ted Jacobson Room 4115, (301)405-6020 www.physics.umd.edu/grt/taj/410a/ jacobson@umd.edu [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 2- U ( ), At an equilibrium point we have U ( ) = 0. (a) Make a Taylor expansion of U ( ) about an equilibrium point . Keep terms out to quadratic order in the displacement from equilbrium. (b) Stability is determined by the sign of the quadratic term in U ( ): if its positive the motion is a stable oscillation, if its negative the motion runs away exponentially. (i) Show that below the critical angular ve- locity, the equilibrium at = 0 is stable and that at = is unstable. (ii) Show that above the critical angular velocity the equilibria at = 0 and are both unstable, while the two new equilibrium points are stable. Solution : L = 1 2 mR 2 ( 2 + sin 2 2 )- mgR (1- cos ), which has the form L = 1 2 mR 2 2- U ( ) with U ( ) =- 1 2 mR 2 2 sin 2 + mgR (1- cos ). Here 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, lets choose units with m = R = g = 1. At the end we can put these quantities back in using dimensional analysis. Hence we have U ( ) =- 1 2 2 sin 2 - cos + 1. This can be expanded about any as U ( ) = U ( ) + U ( )( - ) + 1 2 U 00 ( )( - ) 2 + O [( - ) 3 ] . If is an equilibrium point, U ( ) = 0, and the frequency of small oscillations around is given by 2 = U 00 ( ). If U 00 ( ) < 0 then actually there are no oscillations, but rather a runaway motion....
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410ahw2soln - HW#2Solutions Phys410Fall 2011 Prof. Ted...

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