This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 2 m ( r 2 + r 2 2 ) , (1) T = T cm + 1 2 X i m i r 2 i 2 i = m ( r 2 + r 2 2 ) + mL 2 + 2 . (2) b) The Lagrangian of the system: L = T U, (3) U = GMm r 2 + L 2 + 2 rL cos + GMm r 2 + L 2 2 rL cos , (4) where M is the suns mass. c) Since L/ = 0, p is the constant of motion: p = L = 2 mr 2 + 2 mL 2 + = const. (5) d) Using EulerLagrange equation, its straightforward to write down those equations of motion: for r : 2 r = 2 r 2 GM r + L cos ( r 2 + L 2 + 2 rL cos ) 3 / 2 + r L cos ( r 2 + L 2 2 rL cos ) 3 / 2 ! , (6) for : 2 r r + ( r 2 + L 2 ) = L 2 , (7) for : 2 L 2 = 2 L 2  GMrL sin 1 ( r 2 + L 2 + 2 rL cos ) 3 / 2 1 ( r 2 + L 2 2 rL cos ) 3 / 2 ! . (8) e) Find the Hamiltonian of the system and Hamiltonians equations of motion: p r = L r = 2 m r, (9) p = L = 2 mr 2 + 2 mL 2 + , (10) p = L = 2 mL 2 + , (11) 1 and therefore, H = X i p i q i L = p 2 r 4 m + ( p  p ) 2 4 mr 2 + p 2 4 mL 2 + GMm r 2 + L 2 + 2 rL cos + GMm r 2 + L 2 2 rL cos . (12) Hamiltonians equations of motion: p i = H q i , q i = H p i , (13) thus: r = p r 2 m , (14) p r = ( p  p ) 2 2 mr 3 GMm r + L cos ( r 2 + L 2 + 2 rL cos ) 3 / 2 + r L cos ( r 2 + L 2 2 rL cos ) 3 / 2 ! ; (15) = p  p 2 mr 2 , (16) p = 0; (17) = p  p 2 mr 2 + p 2 mL 2 , (18) p = GMmrL sin 1 ( r 2 + L 2 + 2 rL cos ) 3 / 2 1 ( r 2 + L 2 2 rL cos ) 3 / 2 ! . (19) f) We need the initial conditions r,,, r, , t =0 to solve the dynamical system r = R L, = = 0 , (20) r t =0 = 0 . (21) According to the previous discussions, p [Eq. (5)] is the total angular momentum of the system, which is conserved. We have p = 2 m ( R L ) 2 t =0 + 2 mL 2 t =0 + t =0 = m ( V e R + V v ( R 2 L )) . (22) Additionally, according to our coordinate system: V v V e = 2 L ( t =0 + 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 t =0 = V e ( R + L ) + V e ( R 3 L ) 2 ( R L ) 2 , (?? TW) (25) 2 t =0 = ( V e + V v ) / (2 r ) (TW) (26) t =0 = V v V e 2 L V e ( R + L ) + V e ( R 3 L ) 2 ( R L ) 2 . (?? TW) (27) t =0 = + 1 2 ( V v V e ) /L (TW) (28) g) see next page (TW) 3 solution to problem set 4 part g The question asks you to discuss whether the motion is confined to small oscillations of or whether the bridge rotates and increases indefinitely. I wanted you to try to formulate a strategy like the one below.increases indefinitely....
View
Full
Document
This note was uploaded on 03/02/2010 for the course PHYSICS 316 taught by Professor Pucker during the Spring '10 term at King's College London.
 Spring '10
 Pucker
 mechanics, Energy, Kinetic Energy

Click to edit the document details