hw7_solution_to_1_and_2

# hw7_solution_to_1_and_2 - Problem 1 The position of M is q...

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Unformatted text preview: Problem 1 : The position of M is q M = ( x ( t ) ,x 2 ( t )), the position of M is q m = ( x ( t ) + l sin θ,x 2 ( t )- l cos θ ). So the velocity v M = ˙ q M = ( ˙ x, 2 x ˙ x ) , v m = ˙ q m = ( ˙ x + l cos θ ˙ θ, 2 x ˙ x + l sin θ ˙ θ ) . The kinetic energy is K E = 1 2 M k v M k 2 + 1 2 m k v m k 2 = 1 2 M ( ˙ x 2 + 4 x 2 ˙ x 2 ) + 1 2 m (( ˙ x + l cos θ ˙ θ ) 2 + (2 x ˙ x + l sin θ ˙ θ ) 2 ) = 1 2 M (4 x 2 + 1) ˙ x 2 + 1 2 m ( ˙ x 2 + l 2 cos 2 θ ˙ θ 2 + 2 ˙ xl cos θ ˙ θ + 4 x 2 ˙ x 2 + l 2 sin 2 θ ˙ θ 2 + 4 x ˙ xl sin θ ˙ θ ) = 1 2 ( M + m )(4 x 2 + 1) ˙ x 2 + 1 2 ml 2 ˙ θ 2 + ml ˙ x ˙ θ (cos θ + 2 x sin θ ) , and the potential energy is P E = Mgx 2 + mg ( x 2- l cos θ ) . So the Lagrangian for the system is L = K E- P E = 1 2 ( M + m )(4 x 2 + 1) ˙ x 2 + 1 2 ml 2 ˙ θ 2 + ml ˙ x ˙ θ (cos θ + 2 x sin θ )- Mgx 2- mg ( x 2- l cos θ ) ....
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hw7_solution_to_1_and_2 - Problem 1 The position of M is q...

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