Answer Happy

PROGRAMMING EXERCISE (PART 6) 1. Derive the dynamic equations of motion for the three-link manipulator (from Example 3.3). That is, expand Section 6.7 for the three-link case. The following numerical values describe the manipulator: - l1 = 12 = 0.5m, m1 = 4.6Kg, m2 = 2.3Kg, 1113 = 1.0Kg, 8 = 9.8m/s2. - For the first two links, we assume that the mass is all concentrated at the distal end of the link. For link 3, we assume that the center of mass is located at the origin of frame {3}—that is, at the proximal end of the link. The inertia tensor for link 3 is 3 Cs I = [ 0.05 0 0 0 0.1 0 0 0 0.1 Kg-m?.
VILI MODULI LIUL L CHUI VI was I luvattu at LIIT UITGIII OI frame {3}—that is, at the proximal end of the link. The inertia tensor for link 3 is Cs I 0.05 0 0 0 0.1 0 0 0 0.1 Kg-m 198 Chapter 6 Manipulator dynamics The vectors that locate each center of mass relative to the respective link frame are 1 Pai =111, = 122, 3 Pc =0. = 0 2 Рc. -