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Unformatted text preview: Solution to Problem Set 4 Solution: a) Express the kinetic energy T in terms of the given coordinates: T cm = 1 2 Â· 2 m Â· ( Ë™ r 2 + r 2 Ë™ Ï† 2 ) , (1) T = T cm + 1 2 X i m i r 2 i Ï‰ 2 i = m ( Ë™ r 2 + r 2 Ë™ Ï† 2 ) + mL 2 Ë™ Î¸ + Ë™ Ï† 2 . (2) b) The Lagrangian of the system: L = T U, (3) U = GMm âˆš r 2 + L 2 + 2 rL cos Î¸ + GMm âˆš r 2 + L 2 2 rL cos Î¸ , (4) where M is the sunâ€™s mass. c) Since âˆ‚L/âˆ‚Ï† = 0, p Ï† is the constant of motion: p Ï† = âˆ‚L âˆ‚ Ë™ Ï† = 2 mr 2 Ë™ Ï† + 2 mL 2 Ë™ Ï† + Ë™ Î¸ = const. (5) d) Using EulerLagrange equation, itâ€™s straightforward to write down those equations of motion: for r : 2Â¨ r = 2 r Ë™ Ï† 2 GM r + L cos Î¸ ( r 2 + L 2 + 2 rL cos Î¸ ) 3 / 2 + r L cos Î¸ ( r 2 + L 2 2 rL cos Î¸ ) 3 / 2 ! , (6) for Ï† : 2 r Ë™ r Ë™ Ï† + ( r 2 + L 2 ) Â¨ Ï† = L 2 Â¨ Î¸, (7) for Î¸ : 2 L 2 Â¨ Î¸ = 2 L 2 Â¨ Ï† GMrL sin Î¸ 1 ( r 2 + L 2 + 2 rL cos Î¸ ) 3 / 2 1 ( r 2 + L 2 2 rL cos Î¸ ) 3 / 2 ! . (8) e) Find the Hamiltonian of the system and Hamiltonianâ€™s equations of motion: p r = âˆ‚L âˆ‚ Ë™ r = 2 m Ë™ r, (9) p Ï† = âˆ‚L âˆ‚ Ë™ Ï† = 2 mr 2 Ë™ Ï† + 2 mL 2 Ë™ Ï† + Ë™ Î¸ , (10) p Î¸ = âˆ‚L âˆ‚ Ë™ Î¸ = 2 mL 2 Ë™ Î¸ + Ë™ Ï† , (11) 1 and therefore, H = X i p i Ë™ q i L = p 2 r 4 m + ( p Ï† p Î¸ ) 2 4 mr 2 + p 2 Î¸ 4 mL 2 + GMm âˆš r 2 + L 2 + 2 rL cos Î¸ + GMm âˆš r 2 + L 2 2 rL cos Î¸ . (12) Hamiltonianâ€™s equations of motion: Ë™ p i = âˆ‚H âˆ‚q i , Ë™ q i = âˆ‚H âˆ‚p i , (13) thus: Ë™ r = p r 2 m , (14) Ë™ p r = ( p Î¸ p Ï† ) 2 2 mr 3 GMm r + L cos Î¸ ( r 2 + L 2 + 2 rL cos Î¸ ) 3 / 2 + r L cos Î¸ ( r 2 + L 2 2 rL cos Î¸ ) 3 / 2 ! ; (15) Ë™ Ï† = p Ï† p Î¸ 2 mr 2 , (16) Ë™ p Ï† = 0; (17) Ë™ Î¸ = p Ï† p Î¸ 2 mr 2 + p Î¸ 2 mL 2 , (18) Ë™ p Î¸ = GMmrL sin Î¸ 1 ( r 2 + L 2 + 2 rL cos Î¸ ) 3 / 2 1 ( r 2 + L 2 2 rL cos Î¸ ) 3 / 2 ! . (19) f) We need the initial conditions r,Î¸,Ï†, Ë™ r, Ë™ Î¸, Ë™ Ï† t =0 to solve the dynamical system r = R L, Î¸ = Ï† = 0 , (20) Ë™ r t =0 = 0 . (21) According to the previous discussions, p Ï† [Eq. (5)] is the total angular momentum of the system, which is conserved. We have p Ï† = 2 m ( R L ) 2 Ë™ Ï† t =0 + 2 mL 2 Ë™ Ï† t =0 + Ë™ Î¸ t =0 = m ( V e R + V v ( R 2 L )) . (22) Additionally, according to our coordinate system: V v V e = 2 L ( Ë™ Î¸ t =0 + Ë™ Ï† t =0 ) , (23) where V v = r GM R 2 L ,V e = r GM R . (24) Therefore using Eqs. (22) and (23), we find Ë™ Ï† t =0 = V e ( R + L ) + V e ( R 3 L ) 2 ( R L ) 2 , (?? â€”TW) (25) 2 Ë™ Ï† t =0 = ( V e + V v ) / (2 r ) (â€”TW) (26) Ë™ Î¸ t =0 = V v V e 2 L V e ( R + L ) + V e ( R 3 L ) 2 ( R L ) 2 . (?? â€”TW) (27) Ë™ Î¸ t =0 = Ë™ Ï† + 1 2 ( V v V e ) /L (â€”TW) (28) g) â€”see next page (â€”TW) 3 solution to problem set 4 part g The question asks you to discuss whether the motion is confined to small oscillations of Î¸ or whether the bridge rotates and Î¸ increases indefinitely. I wanted you to try to formulate a strategy like the one below.increases indefinitely....
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 Spring '10
 Pucker
 mechanics, Energy, Kinetic Energy, Constant of integration, Equilibrium point

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