hw7_solution_to_1_and_2

hw7_solution_to_1_and_2 - Problem 1 : The position of M is...

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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 ,x 2 ( t )- l cos ). So the velocity v M = q M = ( x, 2 x x ) , v m = q m = ( x + l cos , 2 x x + l sin ) . 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 ) 2 + (2 x x + l sin ) 2 ) = 1 2 M (4 x 2 + 1) x 2 + 1 2 m ( x 2 + l 2 cos 2 2 + 2 xl cos + 4 x 2 x 2 + l 2 sin 2 2 + 4 x xl sin ) = 1 2 ( M + m )(4 x 2 + 1) x 2 + 1 2 ml 2 2 + ml x (cos + 2 x sin ) , and the potential energy is P E = Mgx 2 + mg ( x 2- l cos ) . 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 2 + ml x (cos + 2 x sin )- Mgx 2- mg ( x 2- l cos ) ....
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hw7_solution_to_1_and_2 - Problem 1 : The position of M is...

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