Kinematics - Introductory Example Basilio Bona DAUIN Politecnico - - PowerPoint PPT Presentation

Kinematics - Introductory Example

Basilio Bona

DAUIN – Politecnico di Torino

Semester 1, 2015-16

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 1 / 16

The planar structure

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

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 2 / 16

Available modelling methods

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.

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 3 / 16

Lagrangian approach

Lagrangian approach for mechanical systems

1

count the rigid bodies N.

2

associate to each one a reference frame and add an (pseudo-)inertial

• ne; total (N + 1) RF.

3

find a complete and independent set of n ≤ 6N generalized coordinates that best describe the system motion.

4

compute the total kinetic (co-)energy and the potential energy.

5

compute the dissipative effects.

6

compute the external acting forces/torques.

7

compute the Lagrange function.

8

derive the n Lagrange equations, one for each coordinate.

9

if required transform them into state equations.

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 4 / 16

Analysis

Our system includes two bodies, three reference frames, two coordinates (q1, q2)

Figure: RFs and qi coordinates.

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Kinematics Semester 1, 2015-16 5 / 16

Some questions

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.

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 6 / 16

Kinematics: position and velocity

Inverted pendulum (point mass M, gravity vector g) x =   q1 + L cos q2 L + L sin q2   α =   q2   ˙ x = v =   ˙ q1 − L sin q2 ˙ q2 L cos q2 ˙ q2   ˙ α = ω =   ˙ q2   v2 = ( ˙ q1 − L sin q2 ˙ q2)2 + (L cos q2 ˙ q2)2 = ˙ q2

1 + L2 ˙

q2

2 − 2L sin q2 ˙

q1 ˙ q2 Kinetic co-energy K∗ = 1 2M v2 = 1 2M( ˙ q2

1 + L2 ˙

q2

2 − 2L sin q2 ˙

q1 ˙ q2) Potential energy P = −MgTx = MGL(1 + sin q2)

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 7 / 16

Coordinates and other parameters

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Kinematics Semester 1, 2015-16 8 / 16

Lagrangian and Lagrange equations

Lagrangian L = K∗ − P Equation 1 d dt ∂L ∂ ˙ q1

• − ∂L

∂q1 = 0 Equation 2 d dt ∂L ∂ ˙ q2

• − ∂L

∂q2 = 0

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Kinematics Semester 1, 2015-16 9 / 16

Lagrange equations

In classical mechanics, the kinetic co-energy is function of both positions and velocities, while potential energy is only function of positions K∗ = K∗(q, ˙ q) P = P(q) hence ∂L ∂ ˙ qi = ∂K∗ ∂ ˙ qi − ∂L ∂qi = −∂K∗ ∂qi + ∂L ∂qi therefore the i−th Lagrange equation can be written as d dt ∂K∗ ∂ ˙ qi

• − ∂K∗

∂qi + ∂P ∂qi = 0 Notice that no elastic elements, no dissipative elements and no external forces were included in this example. Usually these elements are present and the Lagrange equations must be modified accordingly.

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 10 / 16

Solution with point mass

Equation 1 M d dt ( ˙ q1 − L sin q2 ˙ q2) = 0 M ¨ q1 − ML sin q2¨ q2 − ML cos q2 ˙ q2

2 = 0

Equation 2 M d dt (L2 ˙ q2 − L sin q2 ˙ q1) + ML cos q2 ˙ q1 ˙ q2 + MGL cos q2 = 0 ML2¨ q2 − ML sin q2¨ q1 − ML cos q2 ˙ q1 ˙ q2 + ML cos q2 ˙ q1 ˙ q2 + MGL cos q2 = 0 ML2¨ q2 − ML sin q2¨ q1 + MGL cos q2 = 0

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 11 / 16

Solution with extended mass

Inverted pendulum (with extended mass M, gravity vector g) x =   q1 + L cos q2 L + L sin q2   α =   q2   ˙ x = v =   ˙ q1 − L sin q2 ˙ q2 L cos q2 ˙ q2   ˙ α = ω =   ˙ q2   v2 = ( ˙ q1 − L sin q2 ˙ q2)2 + (L cos q2 ˙ q2)2 = ˙ q2

1 + L2 ˙

q2

2 − 2L sin q2 ˙

q1 ˙ q2 One should consider also the rotational kinetic co-energy K∗ = 1 2M v2 + 1 2ωTΓω = 1 2M( ˙ q2

1 + L2 ˙

q2

2 − 2L sin q2 ˙

q1 ˙ q2) + 1 2 ˙ q2

2Γz

Potential energy does not change, since it depends only on the center-of-mass position P = −MgTx = MGL(1 + sin q2)

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 12 / 16

Simulation with Simulink

We rewrite the Lagrange equations of the point mass case as Equation 1 ¨ q1 = L(sin q2¨ q2 + cos q2 ˙ q2

2)

Equation 2 ¨ q2 = 1 L(sin q2¨ q1 − G cos q2)

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Kinematics Semester 1, 2015-16 13 / 16

Matrix representation

The two Lagrange equations can be rewritten in matrix form as H(q)¨ q(t) + C(q, ˙ q)˙ q(t) + G(q) = 0 where H(q) =

• M

−MLs2 −MLs2 ML2

• C(q, ˙

q) =

• MLc2 ˙

q2

• G(q) =
• MGLc2
• So the second order differential equation system is

¨ q(t) = −H(q)−1[C(q, ˙ q)˙ q(t) + G(q)] The determinant of H(q) is M2L2 − M2L2s2

2, that goes to zero when

s2 = ±1, i.e., q2 = 0, π, . . . This is the reason for the malfunction of the Simulink example presented at lesson.

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Kinematics Semester 1, 2015-16 14 / 16

Simulink model

From a physical point of view the singularity (or algebraic loop) encountered in the previous model is due to the fact that the rigid body associated with the cart has mass = 0. It is sufficient to add a mass M1 for the cart and a mass M2 for the pendulum, and everything works. Without giving much details, with this modification the system has the following H matrix H(q) = (M1 + M2) −M2Ls2 −M2Ls2 M2L2

• whose determinant is

det H(t) = M1M2L2 + M2

2L2 − M2L2s2 2

when s2 = ±1 the condition for a zero determinant is M1M2L2 = 0 that is never met, except when one of the masses is zero.

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 15 / 16

Simulink model

• B. Bona (DAUIN)

Kinematics Semester 1, 2015-16 16 / 16