solution4ttt - Solution to Problem Set 4 Solution a Express...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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 ˙ φ 2 ) , (1) T = T cm + 1 2 X i m i r 2 i ω 2 i = m ( ˙ r 2 + r 2 ˙ φ 2 ) + mL 2 ˙ θ + ˙ φ 2 . (2) b) The Lagrangian of the system: L = T- U, (3) U =- GMm √ r 2 + L 2 + 2 rL cos θ +- GMm √ r 2 + L 2- 2 rL cos θ , (4) where M is the sun’s mass. c) Since ∂L/∂φ = 0, p φ is the constant of motion: p φ = ∂L ∂ ˙ φ = 2 mr 2 ˙ φ + 2 mL 2 ˙ φ + ˙ θ = const. (5) d) Using Euler-Lagrange equation, it’s straightforward to write down those equations of motion: for r : 2¨ r = 2 r ˙ φ 2- GM r + L cos θ ( r 2 + L 2 + 2 rL cos θ ) 3 / 2 + r- L cos θ ( r 2 + L 2- 2 rL cos θ ) 3 / 2 ! , (6) for φ : 2 r ˙ r ˙ φ + ( r 2 + L 2 ) ¨ φ =- L 2 ¨ θ, (7) for θ : 2 L 2 ¨ θ =- 2 L 2 ¨ φ- GMrL sin θ 1 ( r 2 + L 2 + 2 rL cos θ ) 3 / 2- 1 ( r 2 + L 2- 2 rL cos θ ) 3 / 2 ! . (8) e) Find the Hamiltonian of the system and Hamiltonian’s equations of motion: p r = ∂L ∂ ˙ r = 2 m ˙ r, (9) p φ = ∂L ∂ ˙ φ = 2 mr 2 ˙ φ + 2 mL 2 ˙ φ + ˙ θ , (10) p θ = ∂L ∂ ˙ θ = 2 mL 2 ˙ θ + ˙ φ , (11) 1 and therefore, H = X i p i ˙ q i- L = p 2 r 4 m + ( p φ- p θ ) 2 4 mr 2 + p 2 θ 4 mL 2 +- GMm √ r 2 + L 2 + 2 rL cos θ +- GMm √ r 2 + L 2- 2 rL cos θ . (12) Hamiltonian’s equations of motion: ˙ p i = ∂H ∂q i , ˙ q i =- ∂H ∂p i , (13) thus: ˙ r = p r 2 m , (14) ˙ p r = ( p θ- p φ ) 2 2 mr 3- GMm r + L cos θ ( r 2 + L 2 + 2 rL cos θ ) 3 / 2 + r- L cos θ ( r 2 + L 2- 2 rL cos θ ) 3 / 2 ! ; (15) ˙ φ = p φ- p θ 2 mr 2 , (16) ˙ p φ = 0; (17) ˙ θ = p φ- p θ 2 mr 2 + p θ 2 mL 2 , (18) ˙ p θ = GMmrL sin θ 1 ( r 2 + L 2 + 2 rL cos θ ) 3 / 2- 1 ( r 2 + L 2- 2 rL cos θ ) 3 / 2 ! . (19) f) We need the initial conditions r,θ,φ, ˙ r, ˙ θ, ˙ φ t =0 to solve the dynamical system r = R- L, θ = φ = 0 , (20) ˙ r t =0 = 0 . (21) According to the previous discussions, p φ [Eq. (5)] is the total angular momentum of the system, which is conserved. We have p φ = 2 m ( R- L ) 2 ˙ φ t =0 + 2 mL 2 ˙ φ t =0 + ˙ θ t =0 = m ( V e R + V v ( R- 2 L )) . (22) Additionally, according to our coordinate system: V v- V e = 2 L ( ˙ θ t =0 + ˙ φ 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 ˙ φ t =0 = V e ( R + L ) + V e ( R- 3 L ) 2 ( R- L ) 2 , (?? —TW) (25) 2 ˙ φ t =0 = ( V e + V v ) / (2 r ) (—TW) (26) ˙ θ t =0 = V v- V e 2 L- V e ( R + L ) + V e ( R- 3 L ) 2 ( R- L ) 2 . (?? —TW) (27) ˙ θ t =0 =- ˙ φ + 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 θ or whether the bridge rotates and θ increases indefinitely. I wanted you to try to formulate a strategy like the one below.increases indefinitely....
View Full Document

{[ snackBarMessage ]}

Page1 / 9

solution4ttt - Solution to Problem Set 4 Solution a Express...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online