Solutions_hw2 - Theoretical Dynamics September 16, 2010...

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Unformatted text preview: 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 1 ,- l 1 cos 1 ) (1) ( x 2 ,y 2 ) = ( l 1 sin 1- l 2 sin 2 ,- l 1 cos 1- l 2 cos 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 2 1 + 1 2 m 2 ( l 2 1 2 1 + l 2 2 2 2- 2 l 1 l 2 1 2 cos( 1 + 2 )) (4) and V =- m 1 gl 1 cos 1- m 2 g ( l 1 cos 1 + l 2 cos 2 ) (5) So, L = 1 2 ( m 1 + m 2 ) l 2 1 2 1 + 1 2 m 2 l 2 2 2 2- m 2 l 1 l 2 1 2 cos( 1 + 2 )+ m 1 gl 1 cos 1 + m 2 g ( l 1 cos 1 + l 2 cos 2 ) (6) 2- 1 The derivatives are L 1 = ( m 1 + m 2 ) l 2 1 1- m 2 l 1 l 2 2 cos( 1 + 2 ) , L 2 = m 2 l 2 2 2- m 2 l 1 l 2 1 cos( 1 + 2 ) (7) L 1 = m 2 l 1 l 2 1 2 sin( 1 + 2 )- m 1 gl 1 sin 1- m 2 gl 1 sin 1 , L 2 = m 2 l 1 l 2 1 2 sin( 1 + 2 )- m 2 gl 2 sin 2 (8) d dt L 1 = ( m 1 + m 2 ) l 2 1 1- m 2 l 1 l 2 2 cos( 1 + 2 ) + m 2 l 1 l 2 2 ( 1 + 2 )sin( 1 + 2 ) (9) d dt L 2 = m 2 l 2 2 2- m 2 l 1 l 2 1 cos( 1 + 2 ) + m 2 l 1 l 2 1 ( 1 + 2 )sin( 1 + 2 ) (10) The Euler-Lagrange equations are d dt L 1- L 1 = 0 d dt L 2- L 2 = 0 that is, ( m 1 + m 2 ) l 2 1 1- m 2 l 1 l 2 2 cos( 1 + 2 ) + m 2 l 1 l 2 2 2 sin( 1 + 2 ) + ( m 1 + m 2 ) gl 1 sin 1 = 0 (11) m 2 l 2 1 2- m 2 l 1 l 2 1 cos( 1 + 2 ) + m 2 l 1 l 2 2 1 sin( 1 + 2 ) + m 2 gl 2 sin 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 Constraint: G ( x 1 ,x 2 ,y 2 ) = y 2- ( x 2- x 1 )tan = 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- G = 1 2 M x 2 1 + 1 2 m ( x 2 2 + y 2 2 )- mgy 2- [ y 2- ( x 2- x 1 )tan ] (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 + tan = 0 (21) m x 2- tan = 0 (22) m y 2 + mg + = 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...
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Solutions_hw2 - Theoretical Dynamics September 16, 2010...

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