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Physics 325 Spring 2011 Problem Session 11 Solutions

Physics 325 Spring 2011 Problem Session 11 Solutions - 2...

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Physics 325 Spring 2011 Discussion 11 April 25, 2011 A bead of mass m is constrained to move on a parabola z = λtx 2 , where λ is a positive constant. There is a uniform gravity g pointing in the - z direction. (a) How many degrees of freedom are there? (b) What is the Lagrangian in terms of generalized coordinates (without constraint)? (c) Find the Hamiltonian. (d) Is the Hamiltonian conserved? Solutions : (a) There is one degree of freedom (bead can move on xz -plane, but constrained to a parabola). (b) Use x as generalized coordinate. z = λtx 2 ˙ z = λx 2 + 2 λtx ˙ x L = 1 2 m ( ˙ x 2 + ˙ z 2 ) - mgz L = 1 2 m ( ˙ x 2 + ( λx 2 + 2 λtx ˙ x ) 2 ) - mgλtx 2 L = 1 2 m (1 + 4 λ 2 t 2 x 2 ) ˙ x 2 + 2 2 t ˙ xx 3 - mgλtx 2 + 1 2 2 x 4 (c) The momentum is p = ∂L
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Unformatted text preview: λ 2 t 2 x 2 ) ˙ x + 2 mλ 2 tx 3 . Equivalently, ˙ x = p-2 mλ 2 tx 3 m (1 + 4 λ 2 t 2 x 2 ) . (1) 1 H = p ˙ x-L H = ± m (1 + 4 λ 2 t 2 x 2 ) ˙ x + 2 mλ 2 tx 3 ² ˙ x-³ 1 2 m (1 + 4 λ 2 t 2 x 2 ) ˙ x 2 + 2 mλ 2 t ˙ xx 3 ´ + mgλtx 2-1 2 mλ 2 x 4 H = 1 2 m (1 + 4 λ 2 t 2 x 2 ) ˙ x 2 + mgλtx 2-1 2 mλ 2 x 4 Use Eqn. (1) to write H in terms of only p , x , and t . H = ( p-2 mλ 2 tx 3 ) 2 2 m (1 + 4 λ 2 t 2 x 2 ) + mgλtx 2-1 2 mλ 2 x 4 (d) dH dt =-∂L ∂t Since L has an explicit time-dependence, the Hamiltonian is not conserved. 2...
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