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Unformatted text preview: Dynamics: NewtonEuler Equations of Motion with Matlab 2 Direct Dynamics NewtonEuler Equations of Motion The NewtonEuler equations of motion for a rigid body in plane motion are m r C = X F and I Czz = X M C , or using the cartesian components m x C = X F x , m y C = X F y , and I Czz = X M C . The forces and moments are known and the differential equations are solved for the motion of the rigid body (direct dynamics). 2.1 Double Pendulum A twolink planar chain (double pendulum) is considered, Fig. 1. The links 1 and 2 have the masses m 1 and m 2 and the lengths AB = L 1 and BD = L 2 . The system is free to move in a vertical plane. The local acceleration of grav ity is g . Numerical application: m 1 = m 2 = 1 kg, L 1 = 1 m, L 2 = 1 m, and g = 9 . 807 m/s 2 . Find and solve the equations of motion. Solution The plane of motion is xy plane with the yaxis vertical, with the positive sense directed downward. The origin of the reference frame is at A . The mass centers of the links are designated by C 1 ( x C 1 ,y C 1 , 0) and C 2 ( x C 2 ,y C 2 , 0). The number of degrees of freedom are computed using the relation M = 3 n 2 c 5 c 4 , where n is the number of moving links, c 5 is the number of one degree of freedom joints, and c 4 is the number of two degrees of freedom joints. For the double pendulum n = 2 , c 5 = 2 , c 4 = 0, and the system has two degrees of freedom, M = 2, and two generalized coordinates. The angles q 1 ( t ) and q 2 ( t ) are selected as the generalized coordinates as shown in Fig. 1. Kinematics The position vector of the center of the mass C 1 of the link 1 is r C 1 = x C 1 + y C 1 , Dynamics: NewtonEuler Equations of Motion with Matlab 1 where x C 1 and y C 1 are the coordinates of C 1 x C 1 = L 1 2 cos q 1 , y C 1 = L 1 2 sin q 1 . The position vector of the center of the mass C 2 of the link 2 is r C 2 = x C 2 + y C 2 , where x C 2 and y C 2 are the coordinates of C 2 x C 2 = L 1 cos q 1 + L 2 2 cos q 2 and y C 2 = L 1 sin q 1 + L 2 2 sin q 2 . The velocity vector of C 1 is the derivative with respect to time of the position vector of C 1 v C 1 = r C 1 = x C 1 + y C 1 , where x C 1 = L 1 2 q 1 sin q 1 and y C 1 = L 1 2 q 1 cos q 1 . The velocity vector of C 2 is the derivative with respect to time of the position vector of C 2 v C 2 = r C 2 = x C 2 + y C 2 , where x C 2 = L 1 q 1 sin q 1 L 2 2 q 2 sin q 2 , y C 2 = L 1 q 1 cos q 1 + L 2 2 q 2 cos q 2 ....
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 Fall '11
 Marghitu

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