This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Dynamics: Newton-Euler Equations of Motion with Matlab 2 Direct Dynamics Newton-Euler Equations of Motion The Newton-Euler 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 two-link 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 y-axis 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: Newton-Euler 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 ....
View Full Document
- Fall '11