Yo m l. m, ,1
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Obtain the Lagrangian equations of the PR(prismatic+rotarary) manipulator in the figure. write in standard form.(by using formulas in photo)
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Transcribed Image Text:Yo
m, ,I,
m, ,I,
a1
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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