Kinematics - a motivating example
Basilio Bona
DAUIN – Politecnico di Torino
Semester 1, 2014-15
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Kinematics
Semester 1, 2014-15
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Motivating example
Figure: A sliding cart carrying an inverse pendulum. Consider the mass as a point
mass.
What is best the approach to model the system dynamics? One has three
choices: Newton-Euler approach, Lagrange approach, Bond-graph
approach
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Kinematics
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Motivating example
Newton-Euler: vector equations based on the dynamic equilibrium of
forces and moments.
Lagrange: scalar equations based on state functions related to kinetic
and potential energies.
Bond-graph: an approach based on the power flow between
interacting parts.
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Kinematics
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Motivating example
Lagrangian approach for mechanical systems
1
2
3
4
5
6
7
8
9
count the rigid bodies N.
associate to each one a reference frame and add an (pseudo-)inertial
one; total (N + 1) RF.
find a complete and independent set of n ≤ 6N generalized
coordinates that best describe the system motion.
compute the total kinetic (co-)energy and the potential energy.
compute the dissipative effects.
compute the external acting forces/torques.
compute the Lagrange function.
derive the n Lagrange equations, one for each coordinate.
if required transform them into state equations.
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Kinematics
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Motivating example
Our system includes two bodies, three reference frames, two coordinates
(q1 , q2 )
Figure: RFs and qi coordinates.
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Kinematics
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Motivating example
Some questions arise
How you define the independent complete set of coordinates?
How you define and include the constraints?
What is the effect of the constraints on the coordinates?
Each rigid body is defined by 6 coordinates (called d.o.f. or dof); 3 for
position x, 3 for orientation α, together called the pose p of the body.
Each constraint reduces the number of coordinates.
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Kinematics
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Motivating example
A rover moving in a plane has only 3 dof; indeed the third positional
coordinate is z = 0 and the two rotation angles around i and j are zero.
⎡%
⎡ ⎤
&T ⎤
! "
x
x
y
0
x
⎣
⎦
⎣
⎦
y
p=
= %
&T ⇒
α
0 0 θ
θ
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Kinematics
Semester 1, 2014-15
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Motivating example
Inverted pendulum (point mass M, gravity vector g)
⎡
⎤
q1 + L cos q2
x = ⎣ L + L sin q2 ⎦
0
⎡
⎤
q̇1 − L sin q2 q̇2
ẋ = v = ⎣ L cos q2 q̇2 ⎦
0
⎡ ⎤
0
α=⎣0⎦
q2
⎡ ⎤
0
⎣
α̇ = ω = 0 ⎦
q̇2
∥v∥2 = (q̇1 − L sin q2 q̇2 )2 + (L cos q2 q̇2 )2 = q̇12 + L2 q̇22 − 2L sin q2 q̇1 q̇2
Kinetic co-energy
1
1
C ∗ = M ∥v∥2 = M(q̇12 + L2 q̇22 − 2L sin q2 q̇1 q̇2 )
2
2
Potential energy
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P = −MgT x = MGL(1 + sin q2 )
Kinematics
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Motivating example
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Kinematics
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Motivating example
Lagrangian
L = C∗ − P
Equation 1
d
dt
)
∂L
∂ q̇1
*
−
∂L
=0
∂q1
d
dt
)
∂L
∂ q̇2
*
−
∂L
=0
∂q2
Equation 2
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Kinematics
Semester 1, 2014-15
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Motivating example
Equation 1
M
d
(q̇1 − L sin q2 q̇2 ) = 0
dt
M q̈1 − ML sin q2 q̈2 − ML cos q2 q̇22 = 0
Equation 2
M
d 2
(L q̇2 − L sin q2 q̇1 ) + ML cos q2 q̇1 q̇2 + MGL cos q2 = 0
dt
ML2 q̈2 − ML sin q2 q̈1 − ML cos q2 q̇1 q̇2 + ML cos q2 q̇1 q̇2 + MGL cos q2 = 0
ML2 q̈2 − ML sin q2 q̈1 + MGL cos q2 = 0
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Kinematics
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