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RR Manipulator 써, Vi Wi V,(9)= M, gh, 'gh, V;Cq) = Maghz 1. Meghe } Transfer Vz= Vp+ 'Vz W2 = w, + %3D gla)=2VG> K(q, 4) = u = M(q)4 +q*r(q)q +g(q). -). aq be

RR Manipulator 써, Vi Wi V,(9)= M, gh, 'gh, V;Cq) = Maghz 1. Meghe } Transfer Vz= Vp+ 'Vz W2 = w, + %3D gla)=2VG> K(q, 4) = u = M(q)4 +q*r(q)q +g(q). -). aq be

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Obtain the dynamic equations of the RR manipulator in the figure using the Lagrangian method. write in standard form.

RR Manipulator I, w? V, hi M, Wi hz M2 V2 V,(9)= Mi Vz Cq) = Meghe Traas fer V2 = v. W, = W, t 91 Cosx ) = u = M(q)q+ġ*r(q)ġ +g(q). r,(4) = aq
Transcribed Image Text:RR Manipulator I, w? V, hi M, Wi hz M2 V2 V,(9)= Mi Vz Cq) = Meghe Traas fer V2 = v. W, = W, t 91 Cosx ) = u = M(q)q+ġ*r(q)ġ +g(q). r,(4) = aq
The kinetic energy of the first link can be expressed as where v2 and w2 are the linear velocity of the link center of mass and angular velocity of the link, respectively. We have S10.7) (10.7) - M(q)4 + C(q, à + g(q). where C(a.4 eR° is a vector of velocity product terms witl the lo X no matrix C(q. 4) linear in 4. ge gravitational forces, and Mq) is an no X no symmetric, positiv definite mass or inertia matrix.A matrix M is symmetric if My = Mj, where My is the entry in the th row and th column of M. F matrix Mis positive definite if v Mv > 0 holds for any nonzerc vector v, This is true if the determinant and trace sum of diagonal elements) of M are positive. 1 K1(q, 4) = E Ris a vector c + a = 4 yielding where v1 and w1 are the linear velocity of the link center of mass and angular velocity of the link, respectively. We have Ka(q. 4) = =m2(4; + (q>¢1>*) + h4i. (the %3D Equations (10.5) and (10.6) for the RP manipulator can be written in the standard form of equation (10.7), where vi = 41 The potential energy of the second link is Mg) -" V2(q) = m2a,42 sin q1. [4,(mn+) con n. yielding the expression The Lagrangian for the system is L = K1 + K2 – V1 - V2: The standard form (10,7) is compact, but the term C(q. 9 masks the fact that these velocity product terms can be derive from the inertia matrix Mg). If we consider the individual components of this vector, C(q. 4 = (c(q. )..... Cnglq. 4). we find that L-( +4+mr+ m)vi + m;) - a, sin q,(mr + g) K1(q, 4) = + Applying Lagrange's equations, we get (10.8) The potential energy of the first link is (10.5) (10.5) +ag(mirn + q2) cos q Vi(q) = m1agr1 sin 1. (10.6) (10.6) uz = mzä2 – m2qrq¡ +a,m2 sinq1. The kinetic energy of the second link can be expressed as (wr- (10.8) a(q. 4)== K2(q, 4) = j-l where (10.9) (10.9) le) =(2M(g) , aMalq) _ aMy(q) \
Transcribed Image Text:The kinetic energy of the first link can be expressed as where v2 and w2 are the linear velocity of the link center of mass and angular velocity of the link, respectively. We have S10.7) (10.7) - M(q)4 + C(q, à + g(q). where C(a.4 eR° is a vector of velocity product terms witl the lo X no matrix C(q. 4) linear in 4. ge gravitational forces, and Mq) is an no X no symmetric, positiv definite mass or inertia matrix.A matrix M is symmetric if My = Mj, where My is the entry in the th row and th column of M. F matrix Mis positive definite if v Mv > 0 holds for any nonzerc vector v, This is true if the determinant and trace sum of diagonal elements) of M are positive. 1 K1(q, 4) = E Ris a vector c + a = 4 yielding where v1 and w1 are the linear velocity of the link center of mass and angular velocity of the link, respectively. We have Ka(q. 4) = =m2(4; + (q>¢1>*) + h4i. (the %3D Equations (10.5) and (10.6) for the RP manipulator can be written in the standard form of equation (10.7), where vi = 41 The potential energy of the second link is Mg) -" V2(q) = m2a,42 sin q1. [4,(mn+) con n. yielding the expression The Lagrangian for the system is L = K1 + K2 – V1 - V2: The standard form (10,7) is compact, but the term C(q. 9 masks the fact that these velocity product terms can be derive from the inertia matrix Mg). If we consider the individual components of this vector, C(q. 4 = (c(q. )..... Cnglq. 4). we find that L-( +4+mr+ m)vi + m;) - a, sin q,(mr + g) K1(q, 4) = + Applying Lagrange's equations, we get (10.8) The potential energy of the first link is (10.5) (10.5) +ag(mirn + q2) cos q Vi(q) = m1agr1 sin 1. (10.6) (10.6) uz = mzä2 – m2qrq¡ +a,m2 sinq1. The kinetic energy of the second link can be expressed as (wr- (10.8) a(q. 4)== K2(q, 4) = j-l where (10.9) (10.9) le) =(2M(g) , aMalq) _ aMy(q) \
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