3 Problem 3: Using the Lagrangian method, obtain the equations of motion in terms of the torque for the2-DOF robot arm shown in Figure 3, given the total kinetic energy K and total potential energy P asshown below. The center of mass for each link is at the center of the link. The moments of inertiaare I1₁ and 1₂.Yo4₁P=m₁g S₁ + m2g|l₁S₁ +41,4The total potential energy P of the system is+1/2/512)myCThe total kinetic energy K is1K = 0² ( _m₂l² + _m²l² + _m₂l² + 1/ m₂hhc₂ )12/1/2+ 0² ( 1 m₂l² ) + this ( m₂l² + / m₂/C₂)0₁У1BT₂Dm₂Figure 3 for Problem 30₂where, S1 = sin(01), C1 = cos(01), S12 = sin(0₁+02) and similarly others.

The total kinetic energy and the total potential energy are given for the system. The expression for the total kinetic energy is, K=θ˙1216m1l12+16m2l22+12m2l12+12m2l1l2C2+θ˙2216m2l22+θ˙1θ˙213m2l22+12m2l1l2C2 The expression for the total potential energy is, P=m1gl12S1+m2gl1S1+l22S12 It is required to find the equation of motion in terms of torque.Thus, the Lagrangian of the system is, L=K-P =θ˙1216m1l12+16m2l22+12m2l12+12m2l1l2C2+θ˙2216m2l22+θ˙1θ˙213m2l22+12m2l1l2C2-m1gl12S1+m2gl1S1+l22S12 The following two equations of motion are produced by taking the derivative of Lagrangian and inserting the terms into the Lagrangian dynamic equation. τ1=13m1l12+m2l12+13m2l22+m2l1l2C2θ¨1+13m2l22+12m2l1l2C2θ¨2 -m2l1l2S2θ˙1θ˙2-12m2l1l2S2θ˙22+12m1+m2gl1C1+12m2gl2C12 and, τ2=13m2l22+12m2l1l2C2θ¨1+13m2l22θ¨2+12m2l1l2S2θ˙12+12m2gl2C12

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Problem 3: Using the Lagrangian method, obtain the equations of motion in terms of the torque for the 2-DOF robot arm shown in Figure 3, given the total kinetic energy K and total potential energy P as shown below. The center of mass for each link is at the center of the link. The moments of inertia are 1₁ and 1₂. The total kinetic energy K is 4,4 K m²l²2 xo mi C 1 x = 0² ( @_m² ² + + m₂l² + 1/{m₂l² + = m₂ll₂ (₂) + 0² ( ½ m²l²³) + Ö‚Ò₂ (²m²l² + ½ m²ll2C₂) B T₂ * ។ D m₂ x1 0₂ Figure 3 for Problem 3 The total potential energy P of the system is h P=m₁g / S₁ + m28 (4₁S₁ + 1/25₁2) where, S1 = sin(01), C₁ = cos(01), S12 = sin(01+02) and similarly others.

Problem 3: Using the Lagrangian method, obtain the equations of motion in terms of the torque for the 2-DOF robot arm shown in Figure 3, given the total kinetic energy K and total potential energy P as shown below. The center of mass for each link is at the center of the link. The moments of inertia are 1₁ and 1₂. The total kinetic energy K is 4,4 K m²l²2 xo mi C 1 x = 0² ( @_m² ² + + m₂l² + 1/{m₂l² + = m₂ll₂ (₂) + 0² ( ½ m²l²³) + Ö‚Ò₂ (²m²l² + ½ m²ll2C₂) B T₂ * ។ D m₂ x1 0₂ Figure 3 for Problem 3 The total potential energy P of the system is h P=m₁g / S₁ + m28 (4₁S₁ + 1/25₁2) where, S1 = sin(01), C₁ = cos(01), S12 = sin(01+02) and similarly others.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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