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 count the rigid bodies N . 1 associate to each one a reference frame and add an (pseudo-)inertial 2 one; total ( N + 1) RF. find a complete and independent set of n ≤ 6 N generalized 3 coordinates that best describe the system motion. compute the total kinetic (co-)energy and the potential energy. 4 compute the dissipative effects. 5 compute the external acting forces/torques. 6 compute the Lagrange function. 7 derive the n Lagrange equations, one for each coordinate. 8 if required transform them into state equations. 9 B. Bona (DAUIN) Kinematics Semester 1, 2015-16 4 / 16
Analysis Our system includes two bodies, three reference frames, two coordinates ( q 1 , q 2 ) Figure: RFs and q i coordinates. B. Bona (DAUIN) 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 ) q 1 + L cos q 2 0 x = L + L sin q 2 α = 0 0 q 2 q 1 − L sin q 2 ˙ ˙ q 2 0 x = v = ˙ L cos q 2 ˙ q 2 α = ω = ˙ 0 0 q 2 ˙ � v � 2 = ( ˙ q 2 ) 2 + ( L cos q 2 ˙ q 2 ) 2 = ˙ 1 + L 2 ˙ q 2 q 2 q 1 − L sin q 2 ˙ 2 − 2 L sin q 2 ˙ q 1 ˙ q 2 Kinetic co-energy K ∗ = 1 2 M � v � 2 = 1 1 + L 2 ˙ q 2 q 2 2 M ( ˙ 2 − 2 L sin q 2 ˙ q 1 ˙ q 2 ) Potential energy P = − M g T x = MGL (1 + sin q 2 ) B. Bona (DAUIN) Kinematics Semester 1, 2015-16 7 / 16
Coordinates and other parameters B. Bona (DAUIN) Kinematics Semester 1, 2015-16 8 / 16
Lagrangian and Lagrange equations Lagrangian L = K ∗ − P Equation 1 � ∂ L � − ∂ L d = 0 d t ∂ ˙ q 1 ∂ q 1 Equation 2 � ∂ L d � − ∂ L = 0 d t ∂ ˙ q 2 ∂ q 2 B. Bona (DAUIN) 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 = ∂ K ∗ − ∂ L = − ∂ K ∗ + ∂ L ∂ ˙ q i ∂ ˙ q i ∂ q i ∂ q i ∂ q i therefore the i − th Lagrange equation can be written as � ∂ K ∗ � d − ∂ K ∗ + ∂ P = 0 d t ∂ ˙ q i ∂ q i ∂ q i 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 d t ( ˙ q 1 − L sin q 2 ˙ q 2 ) = 0 q 2 M ¨ q 1 − ML sin q 2 ¨ q 2 − ML cos q 2 ˙ 2 = 0 Equation 2 M d d t ( L 2 ˙ q 2 − L sin q 2 ˙ q 1 ) + ML cos q 2 ˙ q 1 ˙ q 2 + MGL cos q 2 = 0 ML 2 ¨ q 2 − ML sin q 2 ¨ q 1 − ML cos q 2 ˙ q 1 ˙ q 2 + ML cos q 2 ˙ q 1 ˙ q 2 + MGL cos q 2 = 0 ML 2 ¨ q 2 − ML sin q 2 ¨ q 1 + MGL cos q 2 = 0 B. Bona (DAUIN) Kinematics Semester 1, 2015-16 11 / 16
Solution with extended mass Inverted pendulum (with extended mass M , gravity vector g ) q 1 + L cos q 2 0 x = L + L sin q 2 α = 0 0 q 2 q 1 − L sin q 2 ˙ ˙ 0 q 2 L cos q 2 ˙ 0 x = v = ˙ q 2 α = ω = ˙ 0 q 2 ˙ � v � 2 = ( ˙ q 2 ) 2 + ( L cos q 2 ˙ q 2 ) 2 = ˙ 1 + L 2 ˙ q 2 q 2 q 1 − L sin q 2 ˙ 2 − 2 L sin q 2 ˙ q 1 ˙ q 2 One should consider also the rotational kinetic co-energy K ∗ = 1 2 M � v � 2 + 1 2 ω T Γ ω = 1 q 2 ) + 1 1 + L 2 ˙ q 2 q 2 q 2 2 M ( ˙ 2 − 2 L sin q 2 ˙ q 1 ˙ 2 ˙ 2 Γ z Potential energy does not change, since it depends only on the center-of-mass position P = − M g T x = MGL (1 + sin q 2 ) 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 q 2 q 1 = L (sin q 2 ¨ ¨ q 2 + cos q 2 ˙ 2 ) Equation 2 q 2 = 1 ¨ L (sin q 2 ¨ q 1 − G cos q 2 ) B. Bona (DAUIN) 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 � � � � � � − ML s 2 ML c 2 ˙ 0 0 M q 2 H ( q ) = C ( q , ˙ q ) = G ( q ) = ML 2 − ML s 2 0 0 MGL c 2 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 M 2 L 2 − M 2 L 2 s 2 2 , that goes to zero when s 2 = ± 1, i.e., q 2 = 0 , π, . . . This is the reason for the malfunction of the Simulink example presented at lesson. B. Bona (DAUIN) 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 M 1 for the cart and a mass M 2 for the pendulum, and everything works. Without giving much details, with this modification the system has the following H matrix � ( M 1 + M 2 ) � − M 2 L s 2 H ( q ) = M 2 L 2 − M 2 L s 2 whose determinant is det H ( t ) = M 1 M 2 L 2 + M 2 2 L 2 − M 2 L 2 s 2 2 when s 2 = ± 1 the condition for a zero determinant is M 1 M 2 L 2 = 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
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