Solutions_hw2 - Theoretical Dynamics Homework 2 Instructor...

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Theoretical Dynamics September 16, 2010 Homework 2 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena 1 Goldstein 1.22 Taking the point of support as the origin and the axes as shown, the coordinates are ( x 1 , y 1 ) = ( l 1 sin θ 1 , - l 1 cos θ 1 ) (1) ( x 2 , y 2 ) = ( l 1 sin θ 1 - l 2 sin θ 2 , - l 1 cos θ 1 - l 2 cos θ 2 ) (2) The Lagrangian is L = T - V (3) where T = 1 2 m 1 ( ˙ x 2 1 + ˙ y 2 1 ) + 1 2 m 2 ( ˙ x 2 2 + ˙ y 2 2 ) = 1 2 m 1 l 2 1 ˙ θ 2 1 + 1 2 m 2 ( l 2 1 ˙ θ 2 1 + l 2 2 ˙ θ 2 2 - 2 l 1 l 2 ˙ θ 1 ˙ θ 2 cos( θ 1 + θ 2 )) (4) and V = - m 1 gl 1 cos θ 1 - m 2 g ( l 1 cos θ 1 + l 2 cos θ 2 ) (5) So, L = 1 2 ( m 1 + m 2 ) l 2 1 ˙ θ 2 1 + 1 2 m 2 l 2 2 ˙ θ 2 2 - m 2 l 1 l 2 ˙ θ 1 ˙ θ 2 cos( θ 1 + θ 2 )+ m 1 gl 1 cos θ 1 + m 2 g ( l 1 cos θ 1 + l 2 cos θ 2 ) (6) 2 - 1
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The derivatives are ∂L ˙ θ 1 = ( m 1 + m 2 ) l 2 1 ˙ θ 1 - m 2 l 1 l 2 ˙ θ 2 cos( θ 1 + θ 2 ) , ∂L ˙ θ 2 = m 2 l 2 2 ˙ θ 2 - m 2 l 1 l 2 ˙ θ 1 cos( θ 1 + θ 2 ) (7) ∂L ∂θ 1 = m 2 l 1 l 2 ˙ θ 1 ˙ θ 2 sin( θ 1 + θ 2 ) - m 1 gl 1 sin θ 1 - m 2 gl 1 sin θ 1 , ∂L ∂θ 2 = m 2 l 1 l 2 ˙ θ 1 ˙ θ 2 sin( θ 1 + θ 2 ) - m 2 gl 2 sin θ 2 (8) d dt ∂L ˙ θ 1 = ( m 1 + m 2 ) l 2 1 ¨ θ 1 - m 2 l 1 l 2 ¨ θ 2 cos( θ 1 + θ 2 ) + m 2 l 1 l 2 ˙ θ 2 ( ˙ θ 1 + ˙ θ 2 ) sin( θ 1 + θ 2 ) (9) d dt ∂L ˙ θ 2 = m 2 l 2 2 ¨ θ 2 - m 2 l 1 l 2 ¨ θ 1 cos( θ 1 + θ 2 ) + m 2 l 1 l 2 ˙ θ 1 ( ˙ θ 1 + ˙ θ 2 ) sin( θ 1 + θ 2 ) (10) The Euler-Lagrange equations are d dt ∂L ˙ θ 1 - ∂L ∂θ 1 = 0 d dt ∂L ˙ θ 2 - ∂L ∂θ 2 = 0 that is, ( m 1 + m 2 ) l 2 1 ¨ θ 1 - m 2 l 1 l 2 ¨ θ 2 cos( θ 1 + θ 2 ) + m 2 l 1 l 2 ˙ θ 2 2 sin( θ 1 + θ 2 ) + ( m 1 + m 2 ) gl 1 sin θ 1 = 0 (11) m 2 l 2 1 ¨ θ 2 - m 2 l 1 l 2 ¨ θ 1 cos( θ 1 + θ 2 ) + m 2 l 1 l 2 ˙ θ 2 1 sin( θ 1 + θ 2 ) + m 2 gl 2 sin θ 2 = 0 (12) 2 Goldstein 2.20 Kinetic Energy T = 1 2 M ˙ x 2 1 + 1 2 m ( ˙ x 2 2 + ˙ y 2 2 ) (13) Potential Energy V = mgy 2 (14) 2 - 2
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Constraint: G ( x 1 , x 2 , y 2 ) = y 2 - ( x 2 - x 1 ) tan α = 0 (15) Lagrangian: L = T - V = 1 2 M ˙ x 2 1 + 1 2 m ( ˙ x 2 2 + ˙ y 2 2 ) - mgy 2 (16) Constrained Lagrangian: L c = T - V - λG = 1 2 M ˙ x 2 1 + 1 2 m ( ˙ x 2 2 + ˙ y 2 2 ) - mgy 2 - λ [ y 2 - ( x 2 - x 1 ) tan α ] (17) The Euler-Lagrange equation, d dt ∂L ˙ x 1 - ∂L ∂x 1 = 0 (18) d dt ∂L ˙ x 2 - ∂L ∂x 2 = 0 (19) d dt ∂L ˙ y 2 - ∂L ∂y 2 = 0 (20) give M ¨ x 1 + λ tan α = 0 (21) m ¨ x 2 - λ tan α = 0 (22) m ¨ y 2 + mg + λ = 0 (23) Adding (21) and (22) we get M ¨ x 1 + m ¨ x 2 = 0 (24) which upon one integration wrt time, yields the expected result that the l inear momentum of the (block + wedge) system in the X -direction is constant . Multiplying (23) throughout by tan α , using (15) to write ¨ y 2 = (¨ x 2 - ¨ x 1 ) tan α and substituing λ tan α = - M ¨ x 1 from (21) we get m x 2 - ¨ x 1 ) tan α + mg + λ = 0 = ⇒ - ( M + m x 1 tan 2 α + mg tan α - M ¨ x 1 = 0 So, ¨ x 1 = m M g tan α ( 1 + m M ) tan 2 α + 1 (25) ¨ x 2 = - g tan α ( 1 + m M ) tan 2 α + 1 (26) ¨ y 2 = - 1 + m M g tan 2 α ( 1 + m M ) tan 2 α + 1 (27) λ = - mg ( 1 + m M ) tan 2 α + 1 (28)
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