Nonholonomic Constraints Examples
|
|
- Megan Strickland
- 5 years ago
- Views:
Transcription
1 Nonholonomic Constraints Examples Basilio Bona DAUIN Politecnico di Torino July 2009 B. Bona (DAUIN) Examples July / 34
2 Example 1 Given q T = [ x y ] T check that the constraint φ(q) = (2x + siny ye x ) dx dt + (x cos y + cos y + e x ) dy dt = 0 is holonomic. We have, by inspection a 1 (q) = 2x + siny ye x a 2 (q) = x cos y + cos y + e x B. Bona (DAUIN) Examples July / 34
3 We have to test the following conditions a 1 (q) y a 1(q) q 2? = a 2 (q) q 1 a 2(q) x but since a 1 (q) y a 2 (q) x = cos y e x = cos y e x the two terms are equal, and the constraint is holonomic φ(q) = x 2 + x sinx + ye x + siny = c i.e., x 2 + x sinx + ye x + siny c = 0 B. Bona (DAUIN) Examples July / 34
4 Example 2 Given q T = [ x y θ ] T, check that the constraint ẋ sinθ ẏ cos θ = 0 [ ] sinθ cos θ 0 dx dy = 0 dθ is nonholonomic. Computing the mixed partial derivative we obtain a 1 (q) y a 2 (q) x a 3 (q) x = 0 = 0 = 0 a 1 (q) θ a 2 (q) θ a 3 (q) y = cos θ = sinθ = 0 B. Bona (DAUIN) Examples July / 34
5 Hence the constraint is nonholonomic, since a 1 (q) y a 1 (q) θ a 2 (q) θ = a 2(q) x a 3(q) x a 3(q) y Using the integrating factor approach, we have to compute the terms in µa 1 q 2 = sinq 3 µ q 2 µa 2 q 1 = cos q 3 µ q 1 µa 1 q 3 = sinq 3 µ q 3 + µ cos q 3 µa 3 q 1 = 0 µa 2 q 3 = cos q 3 µ q 3 µ sinq 3 µa 3 q 2 = 0 B. Bona (DAUIN) Examples July / 34
6 and write the system: sinq 3 µ q 2 = cos q 3 µ q 1 sinq 3 µ q 3 = µ cos q 3 cos q 3 µ q 3 = µ sinq 3 Squaring and adding the two last equations yields µ θ in the system above gives = ±µ; substitution µ cos θ = µ sinθ µ sinθ = µ cos θ Since the only possible solution is µ = 0, the constraints are nonholonomic. B. Bona (DAUIN) Examples July / 34
7 Example 3 Figure: Example of nonholonomic constrained system: a disk rotating on a plane without sliding. B. Bona (DAUIN) Examples July / 34
8 The previous Figure represents a rigid wheel, with radius ρ and negligible thickness, rolling on a plane π c with a single contact point O c (t), that changes in time according to the wheel motion. The plane π r of the wheel is orthogonal to the axis of the wheel, described by the unit vector j r. This plane forms an angle α(t) wrt the normal of the plane π c, that, for simplicity is horizontal. The plane π r forms an angle β(t) wrt to a conventional direction on π c, e.g., the direction of the unit vector i 0 of the fixed reference frame R 0 (i 0,j 0,k 0 ). This angle may vary with time. The motion constraint implies in the first place that the contact point O c must in any circumstance belong to π c (no jumps allowed). B. Bona (DAUIN) Examples July / 34
9 In the second place, the non sliding motion implies two condition: (Transversal sliding) the impossibility for the wheel to move transversally; the wheel velocity has no components along the unit vector j c on π c. (Longitudinal sliding) the wheel displacement d(t) on the plane along i c must always be equal to the length ρθ(t) of the arc under θ. If not so, a sliding or slipping condition arise: the wheel moves, but does not rotate, or the wheel rotates but does not move. These conditions imply that the instantaneous velocity of O c or O r must be aligned with the unit vector i, by construction itself aligned with i c. Rotation around k r are allowed, since the rotation takes place around an ideal geometric point O c, that do not imply a sliding effect. B. Bona (DAUIN) Examples July / 34
10 In conclusion, the wheel translates along a local direction given by i = i c and rotates around the instantaneous rotation axis k r = k The homogeneous matrix T 0 r between the fixed reference frame R 0 and the local reference frame R r is from R 0 to R c : T 0 c = Trasl(d c (t)) Rot( k,β(t)), where d c = [ x c y c 0 ] T represents Oc in R 0 : from R c to R r : T c r = Rot(i,α(t)) Trasl(ρ) Rot(j,θ(t)), where ρ = [ 0 0 ρ ] T Without constraints, the generalized coordinates describing the system are the three components of d c (t) and the three angles α(t), β(t), e θ(t): q T = [ x c y c z c α β θ ] B. Bona (DAUIN) Examples July / 34
11 It seems convincing that the planar motion reduces the dof from six to five, considering the additional constraint d c k 0 = 0; q 3 (t) = 0, t Therefore the new generalized coordinates are q T c = [ x c y c α β θ ] The non sliding/slipping condition generates a relation between the contact point velocity and the wheel velocity; when we represent all the vectors in a common frame R 0, we have: ḋ c (t) = ρ θ(t)i c B. Bona (DAUIN) Examples July / 34
12 Using the quantities in Figure, we write ρ θ(t) c β s β d x c y c = 0 0 dt 0 or dx c dy c = ρ θ(t)c β dt = ρ θ(t)s β dt hence dx c dy c = 1 tanβ B. Bona (DAUIN) Examples July / 34
13 Replacing the infinitesimal displacements with the virtual ones, we have i.e., tanβ δx c + δy c = 0 or sinβ δx c + cos β δy c = 0 (1) δq 1 [ sinq4 cos q ] δq 2 δq 3 δq = 0 4 δq 5 This represent a constraint in Pfaffian form with a 1 = sinβ,a 2 = cos β,a 3 = a 4 = a 5 = 0. Computing the partial derivatives we have a 1 (q) q 4 = cos q 4 ; a 4 (q) q 1 = 0; a 2 (q) q 4 = sinq 4 ; a 4 (q) q 2 = 0 B. Bona (DAUIN) Examples July / 34
14 It appears immediately that equality relations among partial derivatives are not satisfied, so the constrain is non holonomic. We could ask how many degrees of freedom has the system; relation (1) introduces a constraint that links the virtual displacement δx c to δy c. Recalling the dof s definition as the number of virtual displacements independent and complete, the system has only four of such displacements, i.e., δx c, δα, δβ e δθ, while δy c is connected to δx c by (1) and δz c = 0 due to the plane motion. The accessibility space however has dimension 5, since, it is possible, with a suitable motion, to bring the state from any initial point q 0 R 5 to any final point q f R 5. B. Bona (DAUIN) Examples July / 34
15 Example: the unicycle The unicycle is illustrated in Figure 2. Figure: The unicycle scheme. B. Bona (DAUIN) Examples July / 34
16 The unicycle generalized coordinates are q = [ q 1 q 2 q 3 ] T = [ x y θ ] T As seen in a previous example, the non sliding motion on the plane determines the following single (m = 1) constraint ẋs θ ẏc θ = [ s θ c θ 0 ] q 1 q 2 = Aq = 0 }{{} q 3 A Now we compute the vector fields forming a basis of the null space of A; the null space has dimension n m = 2; moreover, since the null space conditions are expressed by the following equation [ sθ c θ 0 ] g 11 g 21 g 12 g 22 = [ s θ c θ 0 ][ ] g 1 g 2 g 13 g 23 we obtain g 1 = c θ s θ0 ; g 2 = B. Bona (DAUIN) Examples July / 34
17 Hence the kinematic model of the unicycle becomes ẋ ẏ = Gu = c θ s θ0 u u 2 θ 1 Physical considerations allows to define the following inputs: u 1 = v, where v is the driving velocity of the unicycle, obtained as the product of the wheel radius ρ and the wheel angular speed α, and u 2 = ω, where ω = θ is the steering velocity around the vertical axis. B. Bona (DAUIN) Examples July / 34
18 The Lie bracket of the two input vector fields is [g 1,g 2 ] = g 2 q g 1 g 1 q g 2 = g s θ 0 0 c θ g 2 = s θ c θ We can see that [g 1,g 2 ] g 3 is linearly independent from g 1 and g 2. We can also compute the next term [g 1,[g 1,g 2 ]] as [g 1,g 3 ] = g 3 q g 1 g 1 q g 3 = 0 0 c θ 0 0 s θ s θ 0 0 c θ s θ c θ = c θ s θ that is no more independent from the previous fields. B. Bona (DAUIN) Examples July / 34
19 If we consider the iterative procedure 1, 2,, κ = κ+1 to obtain the involutive closure, i.e, the accessibility distribution A, and recall that κ is called the degree of nonholonomy, we have 1 = = span {g 1,g 2 } 2 = span {g 1,g 2,[g 1,g 2 ]} 3 = span {g 1,g 2,[g 1,g 2 ],[g 1,[g 1,g 2 ]]} = 2 Hence 2 = 3, the accessibility distribution A is with dim A = 3 and κ = 2. A = 2 = span {g 1,g 2,[g 1,g 2 ]} B. Bona (DAUIN) Examples July / 34
20 Chained form: the unicycle Recalling the kinematic model of the unicycle ẋ ẏ = Gu = c θ s θ0 v ω θ 1 with the following coordinates change and the following input change we obtain the chained form z 1 = θ; z 2 = xc θ + ys θ ; z 3 = xs θ yc θ v u 1 = v 2 + z 3 v 1 ω u 2 = v 1 ż 1 = v 1 ż 2 = v 2 ż 3 = z 2 v 1 Notice that the new coordinates z 2,z 3 give the position of the unicycle in a reference frame that rotates with the robot body. B. Bona (DAUIN) Examples July / 34
21 Example: the bicycle The bicycle is illustrated in Figure 3, where C is the instantaneous rotation center, θ the body orientation, φ the steering wheel angle, and x,y the cartesian coordinates of the rear (passive) wheel contact point. Figure: The bicycle scheme. B. Bona (DAUIN) Examples July / 34
22 Possible bicycle generalized coordinates are q = [ q 1 q 2 q 3 q 4 ] T = [ x y θ φ ] T The vehicle motion is constrained by the two non-sliding constraints at the contact points of the posterior and anterior wheels: ẋ a s 34 ẏ a c 34 = 0 ẋs 3 ẏc 3 = 0 where x a,y a are the cartesian coordinates of the anterior wheel, s 3 = sinθ, c 3 = cos θ s 34 = sin(θ + φ), and c 34 = cos(θ + φ). The bicycle body fix the relation between the position of the two wheels x a = x + lc 3 ẋ a = ẋ ls 3 y a = y + ls 3 ẏ a = ẏ + lc 3 B. Bona (DAUIN) Examples July / 34
23 The constrains can be rewritten as ẋs 34 ẏc 34 l θc 4 = 0 ẋs 3 ẏc 3 = 0 The associated A(q) matrix is [ ] s3 c A(q) = s 34 c 34 lc 4 0 whose rank is ρ(a) = 2; the null space N(A) dimension is ν = n ρ = 2. A basis of the null space can be found setting A [ ] g 1 g 2 = 0. Consider g 11 g 21 g g 1 = 12 g g g 2 = g G = [ ] g 1 g 2 23 g 14 g 24 B. Bona (DAUIN) Examples July / 34
24 The condition AG = O gives the following constraints s 3 g 11 c 3 g 12 = 0 s 3 g 21 c 3 g 22 = 0 s 3 g 11 c 3 g 12 lc 4 g 13 = 0 s 3 g 21 c 3 g 22 lc 4 g 23 = 0 Since we have eight unknowns, but only four equations, we apply some heuristics: we set g 2 = [ ] T and g14 = 0, so we reduce to two equations in three unknowns. We can verify that the choice c 3 c 4 0 s g 1 (q) = 3 c 4 0 s 4 /l g 2 (q) = satisfies the constraints, yielding to the kinematic model of the bicycle ẋ c 3 c 4 0 ẏ θ = s 3 c 4 0 s 4 /l u u 2 φ 0 1 B. Bona (DAUIN) Examples July / 34
25 Consider the geometric relations among the command variables, if the bicycle is driven by the anterior wheel, we set u 2 as the steering velocity ω, while u 1 is the anterior driving velocity v. In this case ẋ c 3 c 4 0 ẏ θ = s 3 c 4 0 s 4 /l v + 0 ω φ 0 1 If, on the contrary, the driving velocity v is applied by the rear wheel, we have the geometric relation u 1 c 4 = v, and consequently ẋ c 3 c 4 0 ẋ c 3 0 ẏ θ = s 3 c 4 0 s 4 /l u ω ẏ θ = s 3 s 4 /c 4 l v ω φ 0 1 φ 0 1 B. Bona (DAUIN) Examples July / 34
26 Chained form: the bicycle Assume the rear wheel drive bicycle model ẋ c 3 0 ẏ θ = s 3 t 4 l v ω φ 0 1 where t 4 = tanq 4. Introducing the following coordinates change z 1 = x; z 2 = 1 l sec3 θ tanφ; z 3 = tanφ; z 4 = y B. Bona (DAUIN) Examples July / 34
27 and the following input transformation v = v 1 cos θ we obtain the chained form ω = 3 l v 1 sec θ sin 2 φ + 1 l v 2 cos 3 θ cos 2 φ ż 1 = v 1 ż 2 = v 2 ż 3 = z 2 v 1 ż 4 = z 3 v 1 B. Bona (DAUIN) Examples July / 34
28 Example: a planar space robot B. Bona (DAUIN) Figure: The planar Examples satellite scheme. July / 34 The planar satellite in Figure 4 consists of a free-flying main body, with two rotating arms; for simplicity their mass is concentrated at end of the arm.
29 Consider the following generalized coordinates: x,y the position of center-of-mass (com), θ its orientation wrt a fixed frame, φ 1,φ 2 the orientation of the two arms wrt to the body frame. Hence q = [ x y φ 1 φ 2 θ ] T Other parameters are: r the distance of the arm revolute joints from com and l the length of the two arms; M is the main body mass, Γ its inertia moment wrt to the com, and m are the equal masses of the two arms. With the usual angle convention (positive if anti-clockwise), the two masses have coordinates x 1 = x + r cos θ + lcos(θ + φ 1 ); y 1 = y + r sinθ + lsin(θ + φ 1 ) x 2 = x r cos θ lcos(θ + φ 1 ); y 2 = y r sinθ lsin(θ + φ 1 ) B. Bona (DAUIN) Examples July / 34
30 The Lagrangian function, assuming no gravity, is simply the kinetic energy T = 1 2 [M(ẋ2 + ẏ 2 ) + Γ θ 2 + m(ẋ1 2 + ẏ2 1 ) + m(ẋ2 2 + ẏ2 2 )] but it can also be written in simplified form as T = 1 2 (M + 2m) ẋ [ φ 1 φ1 φ2 θ] M φ ; M R 2 2 θ 3 3 where ẋ is the com velocity vector and M is diagonal, with the following elements m ij : m 11 = ml 2 m 12 = 0 m 13 = ml 2 + mr cos φ 1 m 22 = ml 2 m 23 = ml 2 + mr cos φ 2 m 33 = Γ + 2m(r 2 + l 2 ) + 2mlr(cos φ 1 + cos φ 2 ) We notice that L does not depend on θ, and we can reduce the generalized coordinates to the following q = [ φ 1 φ 2 θ ] T. B. Bona (DAUIN) Examples July / 34
31 If the satellite is free-floating and no other generalized force/torque is applied, and assuming that the satellite com remains in (0,0) for t t 0, then T reduces to T = 1 [ φ 2 1 φ 2 θ ] φ 1 M φ ; M R 2 θ 3 3 and the Lagrange equations reduce to d L L dt θ }{{} θ =0 = 0 since we know (from Hamilton equations) that L is the angular θ momentum p B. Bona (DAUIN) Examples July / 34
32 We have p = L θ = m 13 φ 1 + m 23 φ2 + m 33 θ If the initial angular momentum is zero, we have the constraint m 13 (q) q 1 + m 23 (q) q 2 + m 33 (q) q 3 = 0 Since the actuated coordinates are u 1 = φ 1 and u 2 = φ 2, we have q = g 1 (q)u 1 + g 2 (q)u 2 with q 1 q2 = q m 13 m 33 u m 23 m 33 u 2 B. Bona (DAUIN) Examples July / 34
33 It is sufficient to compute m 13 = 0 q 2 m 13 = 0 q 3 m 23 = 0 q 3 m 23 = 0 q 1 m 33 = 2mlr sinq 1 q 1 m 33 = 2mlr sinq 2 q 2 to verify that the constraints are nonholonomic. B. Bona (DAUIN) Examples July / 34
34 References [01] F. Bullo, A.D. Lewis, Geometric Control of Mechanical Systems, Springer, [02] H.K. Khalil, Nonlinear Systems, III Ed. Springer, [03] R.M. Murray, Z. Li, S. Sastry, A Mathematical Introduction to Robotic Manipulation, CRC Press, [04] S. Sastry, Nonlinear Systems: Analysis, Stability, and Control, Springer, [05] B. Siciliano, L. Sciavicco, L. Villani, G. Oriolo, Robotics: Modelling, Planning and Control, Springer, [06] Z. Qu, Cooperative Control of Dynamical Systems: Applications to Autonomous Vehicles, Springer, B. Bona (DAUIN) Examples July / 34
Non-holonomic constraint example A unicycle
Non-holonomic constraint example A unicycle A unicycle (in gray) moves on a plane; its motion is given by three coordinates: position x, y and orientation θ. The instantaneous velocity v = [ ẋ ẏ ] is along
More informationGeneralized coordinates and constraints
Generalized coordinates and constraints Basilio Bona DAUIN Politecnico di Torino Semester 1, 2014-15 B. Bona (DAUIN) Generalized coordinates and constraints Semester 1, 2014-15 1 / 25 Coordinates A rigid
More informationEN Nonlinear Control and Planning in Robotics Lecture 2: System Models January 28, 2015
EN53.678 Nonlinear Control and Planning in Robotics Lecture 2: System Models January 28, 25 Prof: Marin Kobilarov. Constraints The configuration space of a mechanical sysetm is denoted by Q and is assumed
More informationDifferential Kinematics
Differential Kinematics Relations between motion (velocity) in joint space and motion (linear/angular velocity) in task space (e.g., Cartesian space) Instantaneous velocity mappings can be obtained through
More informationMSMS Basilio Bona DAUIN PoliTo
MSMS 214-215 Basilio Bona DAUIN PoliTo Problem 2 The planar system illustrated in Figure 1 consists of a bar B and a wheel W moving (no friction, no sliding) along the bar; the bar can rotate around an
More informationPhysics 106a, Caltech 16 October, Lecture 5: Hamilton s Principle with Constraints. Examples
Physics 106a, Caltech 16 October, 2018 Lecture 5: Hamilton s Principle with Constraints We have been avoiding forces of constraint, because in many cases they are uninteresting, and the constraints can
More informationCase Study: The Pelican Prototype Robot
5 Case Study: The Pelican Prototype Robot The purpose of this chapter is twofold: first, to present in detail the model of the experimental robot arm of the Robotics lab. from the CICESE Research Center,
More informationLine following of a mobile robot
Line following of a mobile robot May 18, 004 1 In brief... The project is about controlling a differential steering mobile robot so that it follows a specified track. Steering is achieved by setting different
More informationChapter 10.A. Rotation of Rigid Bodies
Chapter 10.A Rotation of Rigid Bodies P. Lam 7_23_2018 Learning Goals for Chapter 10.1 Understand the equations govern rotational kinematics, and know how to apply them. Understand the physical meanings
More informationDynamics. Basilio Bona. Semester 1, DAUIN Politecnico di Torino. B. Bona (DAUIN) Dynamics Semester 1, / 18
Dynamics Basilio Bona DAUIN Politecnico di Torino Semester 1, 2016-17 B. Bona (DAUIN) Dynamics Semester 1, 2016-17 1 / 18 Dynamics Dynamics studies the relations between the 3D space generalized forces
More information5. Nonholonomic constraint Mechanics of Manipulation
5. Nonholonomic constraint Mechanics of Manipulation Matt Mason matt.mason@cs.cmu.edu http://www.cs.cmu.edu/~mason Carnegie Mellon Lecture 5. Mechanics of Manipulation p.1 Lecture 5. Nonholonomic constraint.
More informationChapter 3 Numerical Methods
Chapter 3 Numerical Methods Part 3 3.4 Differential Algebraic Systems 3.5 Integration of Differential Equations 1 Outline 3.4 Differential Algebraic Systems 3.4.1 Constrained Dynamics 3.4.2 First and Second
More informationMEAM 520. More Velocity Kinematics
MEAM 520 More Velocity Kinematics Katherine J. Kuchenbecker, Ph.D. General Robotics, Automation, Sensing, and Perception Lab (GRASP) MEAM Department, SEAS, University of Pennsylvania Lecture 12: October
More informationQuestion 1: A particle starts at rest and moves along a cycloid whose equation is. 2ay y a
Stephen Martin PHYS 10 Homework #1 Question 1: A particle starts at rest and moves along a cycloid whose equation is [ ( ) a y x = ± a cos 1 + ] ay y a There is a gravitational field of strength g in the
More informationNonholonomic Behavior in Robotic Systems
Chapter 7 Nonholonomic Behavior in Robotic Systems In this chapter, we study the effect of nonholonomic constraints on the behavior of robotic systems. These constraints arise in systems such as multifingered
More informationMultibody simulation
Multibody simulation Dynamics of a multibody system (Euler-Lagrange formulation) Dimitar Dimitrov Örebro University June 16, 2012 Main points covered Euler-Lagrange formulation manipulator inertia matrix
More informationPhysics 106a, Caltech 4 December, Lecture 18: Examples on Rigid Body Dynamics. Rotating rectangle. Heavy symmetric top
Physics 106a, Caltech 4 December, 2018 Lecture 18: Examples on Rigid Body Dynamics I go through a number of examples illustrating the methods of solving rigid body dynamics. In most cases, the problem
More informationSolving high order nonholonomic systems using Gibbs-Appell method
Solving high order nonholonomic systems using Gibbs-Appell method Mohsen Emami, Hassan Zohoor and Saeed Sohrabpour Abstract. In this paper we present a new formulation, based on Gibbs- Appell method, for
More informationCP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford TT 2017
CP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford TT 2017 1 OUTLINE : CP1 REVISION LECTURE 3 : INTRODUCTION TO CLASSICAL MECHANICS 1. Angular velocity and
More informationRobot Control Basics CS 685
Robot Control Basics CS 685 Control basics Use some concepts from control theory to understand and learn how to control robots Control Theory general field studies control and understanding of behavior
More informationAdvanced Dynamics. - Lecture 4 Lagrange Equations. Paolo Tiso Spring Semester 2017 ETH Zürich
Advanced Dynamics - Lecture 4 Lagrange Equations Paolo Tiso Spring Semester 2017 ETH Zürich LECTURE OBJECTIVES 1. Derive the Lagrange equations of a system of particles; 2. Show that the equation of motion
More informationN mg N Mg N Figure : Forces acting on particle m and inclined plane M. (b) The equations of motion are obtained by applying the momentum principles to
.004 MDEING DNMIS ND NTR I I Spring 00 Solutions for Problem Set 5 Problem. Particle slides down movable inclined plane. The inclined plane of mass M is constrained to move parallel to the -axis, and the
More information(W: 12:05-1:50, 50-N202)
2016 School of Information Technology and Electrical Engineering at the University of Queensland Schedule of Events Week Date Lecture (W: 12:05-1:50, 50-N202) 1 27-Jul Introduction 2 Representing Position
More informationCONTROL OF THE NONHOLONOMIC INTEGRATOR
June 6, 25 CONTROL OF THE NONHOLONOMIC INTEGRATOR R. N. Banavar (Work done with V. Sankaranarayanan) Systems & Control Engg. Indian Institute of Technology, Bombay Mumbai -INDIA. banavar@iitb.ac.in Outline
More informationThe Jacobian. Jesse van den Kieboom
The Jacobian Jesse van den Kieboom jesse.vandenkieboom@epfl.ch 1 Introduction 1 1 Introduction The Jacobian is an important concept in robotics. Although the general concept of the Jacobian in robotics
More informationLagrangian Dynamics: Derivations of Lagrange s Equations
Constraints and Degrees of Freedom 1.003J/1.053J Dynamics and Control I, Spring 007 Professor Thomas Peacock 4/9/007 Lecture 15 Lagrangian Dynamics: Derivations of Lagrange s Equations Constraints and
More informationROBOTICS 01PEEQW. Basilio Bona DAUIN Politecnico di Torino
ROBOTICS 01PEEQW Basilio Bona DAUIN Politecnico di Torino Kinematic Functions Kinematic functions Kinematics deals with the study of four functions(called kinematic functions or KFs) that mathematically
More informationConstrained motion and generalized coordinates
Constrained motion and generalized coordinates based on FW-13 Often, the motion of particles is restricted by constraints, and we want to: work only with independent degrees of freedom (coordinates) k
More informationRotational & Rigid-Body Mechanics. Lectures 3+4
Rotational & Rigid-Body Mechanics Lectures 3+4 Rotational Motion So far: point objects moving through a trajectory. Next: moving actual dimensional objects and rotating them. 2 Circular Motion - Definitions
More informationGeneralized Forces. Hamilton Principle. Lagrange s Equations
Chapter 5 Virtual Work and Lagrangian Dynamics Overview: Virtual work can be used to derive the dynamic and static equations without considering the constraint forces as was done in the Newtonian Mechanics,
More informationM2A2 Problem Sheet 3 - Hamiltonian Mechanics
MA Problem Sheet 3 - Hamiltonian Mechanics. The particle in a cone. A particle slides under gravity, inside a smooth circular cone with a vertical axis, z = k x + y. Write down its Lagrangian in a) Cartesian,
More information8 Velocity Kinematics
8 Velocity Kinematics Velocity analysis of a robot is divided into forward and inverse velocity kinematics. Having the time rate of joint variables and determination of the Cartesian velocity of end-effector
More informationRobot Dynamics Lecture Notes. Robotic Systems Lab, ETH Zurich
Robot Dynamics Lecture Notes Robotic Systems Lab, ETH Zurich HS 217 Contents 1 Introduction 1 1.1 Nomenclature.............................. 2 1.2 Operators................................ 3 2 Kinematics
More informationPosition and orientation of rigid bodies
Robotics 1 Position and orientation of rigid bodies Prof. Alessandro De Luca Robotics 1 1 Position and orientation right-handed orthogonal Reference Frames RF A A p AB B RF B rigid body position: A p AB
More informationPLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)
PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2) We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. As discussed in Chapter 16, when
More informationKinematics. Basilio Bona. Semester 1, DAUIN Politecnico di Torino. B. Bona (DAUIN) Kinematics Semester 1, / 15
Kinematics Basilio Bona DAUIN Politecnico di Torino Semester 1, 2016-17 B. Bona (DAUIN) Kinematics Semester 1, 2016-17 1 / 15 Introduction The kinematic quantities used to represent a body frame are: position
More informationLecture Schedule Week Date Lecture (M: 2:05p-3:50, 50-N202)
J = x θ τ = J T F 2018 School of Information Technology and Electrical Engineering at the University of Queensland Lecture Schedule Week Date Lecture (M: 2:05p-3:50, 50-N202) 1 23-Jul Introduction + Representing
More informationProblem Goldstein 2-12
Problem Goldstein -1 The Rolling Constraint: A small circular hoop of radius r and mass m hoop rolls without slipping on a stationary cylinder of radius R. The only external force is that of gravity. Let
More information06. Lagrangian Mechanics II
University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 2015 06. Lagrangian Mechanics II Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License
More informationRotational Kinematics
Rotational Kinematics Rotational Coordinates Ridged objects require six numbers to describe their position and orientation: 3 coordinates 3 axes of rotation Rotational Coordinates Use an angle θ to describe
More informationPHY6426/Fall 07: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT #1 due by 9:35 a.m. Wed 09/05 Instructor: D. L. Maslov Rm.
PHY646/Fall 07: CLASSICAL MECHANICS HOMEWORK ASSIGNMENT # due by 9:35 a.m. Wed 09/05 Instructor: D. L. Maslov maslov@phys.ufl.edu 39-053 Rm. 4 Please help your instructor by doing your work neatly.. Goldstein,
More informationRotational Motion and Torque
Rotational Motion and Torque Introduction to Angular Quantities Sections 8- to 8-2 Introduction Rotational motion deals with spinning objects, or objects rotating around some point. Rotational motion is
More informationVideo 3.1 Vijay Kumar and Ani Hsieh
Video 3.1 Vijay Kumar and Ani Hsieh Robo3x-1.3 1 Dynamics of Robot Arms Vijay Kumar and Ani Hsieh University of Pennsylvania Robo3x-1.3 2 Lagrange s Equation of Motion Lagrangian Kinetic Energy Potential
More informationChapter 4 Statics and dynamics of rigid bodies
Chapter 4 Statics and dynamics of rigid bodies Bachelor Program in AUTOMATION ENGINEERING Prof. Rong-yong Zhao (zhaorongyong@tongji.edu.cn) First Semester,2014-2015 Content of chapter 4 4.1 Static equilibrium
More informationProblem 1 Problem 2 Problem 3 Problem 4 Total
Name Section THE PENNSYLVANIA STATE UNIVERSITY Department of Engineering Science and Mechanics Engineering Mechanics 12 Final Exam May 5, 2003 8:00 9:50 am (110 minutes) Problem 1 Problem 2 Problem 3 Problem
More informationPhysical Dynamics (SPA5304) Lecture Plan 2018
Physical Dynamics (SPA5304) Lecture Plan 2018 The numbers on the left margin are approximate lecture numbers. Items in gray are not covered this year 1 Advanced Review of Newtonian Mechanics 1.1 One Particle
More informationLecture Outline Chapter 10. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.
Lecture Outline Chapter 10 Physics, 4 th Edition James S. Walker Chapter 10 Rotational Kinematics and Energy Units of Chapter 10 Angular Position, Velocity, and Acceleration Rotational Kinematics Connections
More informationKinematics. Basilio Bona. October DAUIN - Politecnico di Torino. Basilio Bona (DAUIN - Politecnico di Torino) Kinematics October / 15
Kinematics Basilio Bona DAUIN - Politecnico di Torino October 2013 Basilio Bona (DAUIN - Politecnico di Torino) Kinematics October 2013 1 / 15 Introduction The kinematic quantities used are: position r,
More informationForces of Constraint & Lagrange Multipliers
Lectures 30 April 21, 2006 Written or last updated: April 21, 2006 P442 Analytical Mechanics - II Forces of Constraint & Lagrange Multipliers c Alex R. Dzierba Generalized Coordinates Revisited Consider
More informationPosture regulation for unicycle-like robots with. prescribed performance guarantees
Posture regulation for unicycle-like robots with prescribed performance guarantees Martina Zambelli, Yiannis Karayiannidis 2 and Dimos V. Dimarogonas ACCESS Linnaeus Center and Centre for Autonomous Systems,
More informationControl of Mobile Robots Prof. Luca Bascetta
Control of Mobile Robots Prof. Luca Bascetta EXERCISE 1 1. Consider a wheel rolling without slipping on the horizontal plane, keeping the sagittal plane in the vertical direction. Write the expression
More informationCDS 205 Final Project: Incorporating Nonholonomic Constraints in Basic Geometric Mechanics Concepts
CDS 205 Final Project: Incorporating Nonholonomic Constraints in Basic Geometric Mechanics Concepts Michael Wolf wolf@caltech.edu 6 June 2005 1 Introduction 1.1 Motivation While most mechanics and dynamics
More informationEE Homework 3 Due Date: 03 / 30 / Spring 2015
EE 476 - Homework 3 Due Date: 03 / 30 / 2015 Spring 2015 Exercise 1 (10 points). Consider the problem of two pulleys and a mass discussed in class. We solved a version of the problem where the mass was
More informationPhys 7221 Homework # 8
Phys 71 Homework # 8 Gabriela González November 15, 6 Derivation 5-6: Torque free symmetric top In a torque free, symmetric top, with I x = I y = I, the angular velocity vector ω in body coordinates with
More informationPHY 5246: Theoretical Dynamics, Fall Assignment # 9, Solutions. y CM (θ = 0) = 2 ρ m
PHY 546: Theoretical Dnamics, Fall 5 Assignment # 9, Solutions Graded Problems Problem (.a) l l/ l/ CM θ x In order to find the equation of motion of the triangle, we need to write the Lagrangian, with
More informationAdvanced Robotic Manipulation
Lecture Notes (CS327A) Advanced Robotic Manipulation Oussama Khatib Stanford University Spring 2005 ii c 2005 by Oussama Khatib Contents 1 Spatial Descriptions 1 1.1 Rigid Body Configuration.................
More informationLecture 2: Controllability of nonlinear systems
DISC Systems and Control Theory of Nonlinear Systems 1 Lecture 2: Controllability of nonlinear systems Nonlinear Dynamical Control Systems, Chapter 3 See www.math.rug.nl/ arjan (under teaching) for info
More informationVideo 1.1 Vijay Kumar and Ani Hsieh
Video 1.1 Vijay Kumar and Ani Hsieh 1 Robotics: Dynamics and Control Vijay Kumar and Ani Hsieh University of Pennsylvania 2 Why? Robots live in a physical world The physical world is governed by the laws
More informationP321(b), Assignement 1
P31(b), Assignement 1 1 Exercise 3.1 (Fetter and Walecka) a) The problem is that of a point mass rotating along a circle of radius a, rotating with a constant angular velocity Ω. Generally, 3 coordinates
More information1.1. Rotational Kinematics Description Of Motion Of A Rotating Body
PHY 19- PHYSICS III 1. Moment Of Inertia 1.1. Rotational Kinematics Description Of Motion Of A Rotating Body 1.1.1. Linear Kinematics Consider the case of linear kinematics; it concerns the description
More informationRIGID BODY MOTION (Section 16.1)
RIGID BODY MOTION (Section 16.1) There are cases where an object cannot be treated as a particle. In these cases the size or shape of the body must be considered. Rotation of the body about its center
More informationKinematics. Basilio Bona. Semester 1, DAUIN Politecnico di Torino. B. Bona (DAUIN) Kinematics Semester 1, / 15
Kinematics Basilio Bona DAUIN Politecnico di Torino Semester 1, 2014-15 B. Bona (DAUIN) Kinematics Semester 1, 2014-15 1 / 15 Introduction The kinematic quantities used are: position r, linear velocity
More informationCOMPLETE ALL ROUGH WORKINGS IN THE ANSWER BOOK AND CROSS THROUGH ANY WORK WHICH IS NOT TO BE ASSESSED.
BSc/MSci EXAMINATION PHY-304 Time Allowed: Physical Dynamics 2 hours 30 minutes Date: 28 th May 2009 Time: 10:00 Instructions: Answer ALL questions in section A. Answer ONLY TWO questions from section
More informationAssignments VIII and IX, PHYS 301 (Classical Mechanics) Spring 2014 Due 3/21/14 at start of class
Assignments VIII and IX, PHYS 301 (Classical Mechanics) Spring 2014 Due 3/21/14 at start of class Homeworks VIII and IX both center on Lagrangian mechanics and involve many of the same skills. Therefore,
More informationRotation. Kinematics Rigid Bodies Kinetic Energy. Torque Rolling. featuring moments of Inertia
Rotation Kinematics Rigid Bodies Kinetic Energy featuring moments of Inertia Torque Rolling Angular Motion We think about rotation in the same basic way we do about linear motion How far does it go? How
More informationArtificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik. Robot Dynamics. Dr.-Ing. John Nassour J.
Artificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik Robot Dynamics Dr.-Ing. John Nassour 25.1.218 J.Nassour 1 Introduction Dynamics concerns the motion of bodies Includes Kinematics
More informationIn the presence of viscous damping, a more generalized form of the Lagrange s equation of motion can be written as
2 MODELING Once the control target is identified, which includes the state variable to be controlled (ex. speed, position, temperature, flow rate, etc), and once the system drives are identified (ex. force,
More informationPhysical Dynamics (PHY-304)
Physical Dynamics (PHY-304) Gabriele Travaglini March 31, 2012 1 Review of Newtonian Mechanics 1.1 One particle Lectures 1-2. Frame, velocity, acceleration, number of degrees of freedom, generalised coordinates.
More informationPhysics 351, Spring 2015, Final Exam.
Physics 351, Spring 2015, Final Exam. This closed-book exam has (only) 25% weight in your course grade. You can use one sheet of your own hand-written notes. Please show your work on these pages. The back
More informationChapter 10 Rotational Kinematics and Energy. Copyright 2010 Pearson Education, Inc.
Chapter 10 Rotational Kinematics and Energy Copyright 010 Pearson Education, Inc. 10-1 Angular Position, Velocity, and Acceleration Copyright 010 Pearson Education, Inc. 10-1 Angular Position, Velocity,
More informationMarion and Thornton. Tyler Shendruk October 1, Hamilton s Principle - Lagrangian and Hamiltonian dynamics.
Marion and Thornton Tyler Shendruk October 1, 2010 1 Marion and Thornton Chapter 7 Hamilton s Principle - Lagrangian and Hamiltonian dynamics. 1.1 Problem 6.4 s r z θ Figure 1: Geodesic on circular cylinder
More informationME 230: Kinematics and Dynamics Spring 2014 Section AD. Final Exam Review: Rigid Body Dynamics Practice Problem
ME 230: Kinematics and Dynamics Spring 2014 Section AD Final Exam Review: Rigid Body Dynamics Practice Problem 1. A rigid uniform flat disk of mass m, and radius R is moving in the plane towards a wall
More informationChapter 8 Lecture. Pearson Physics. Rotational Motion and Equilibrium. Prepared by Chris Chiaverina Pearson Education, Inc.
Chapter 8 Lecture Pearson Physics Rotational Motion and Equilibrium Prepared by Chris Chiaverina Chapter Contents Describing Angular Motion Rolling Motion and the Moment of Inertia Torque Static Equilibrium
More informationSOLUTIONS, PROBLEM SET 11
SOLUTIONS, PROBLEM SET 11 1 In this problem we investigate the Lagrangian formulation of dynamics in a rotating frame. Consider a frame of reference which we will consider to be inertial. Suppose that
More informationClassical Mechanics III (8.09) Fall 2014 Assignment 3
Classical Mechanics III (8.09) Fall 2014 Assignment 3 Massachusetts Institute of Technology Physics Department Due September 29, 2014 September 22, 2014 6:00pm Announcements This week we continue our discussion
More informationInstitute of Geometry, Graz, University of Technology Mobile Robots. Lecture notes of the kinematic part of the lecture
Institute of Geometry, Graz, University of Technology www.geometrie.tugraz.at Institute of Geometry Mobile Robots Lecture notes of the kinematic part of the lecture Anton Gfrerrer nd Edition 4 . Contents
More informationTorque and Rotation Lecture 7
Torque and Rotation Lecture 7 ˆ In this lecture we finally move beyond a simple particle in our mechanical analysis of motion. ˆ Now we consider the so-called rigid body. Essentially, a particle with extension
More information9 Kinetics of 3D rigid bodies - rotating frames
9 Kinetics of 3D rigid bodies - rotating frames 9. Consider the two gears depicted in the figure. The gear B of radius R B is fixed to the ground, while the gear A of mass m A and radius R A turns freely
More informationPhysics 141 Rotational Motion 2 Page 1. Rotational Motion 2
Physics 141 Rotational Motion 2 Page 1 Rotational Motion 2 Right handers, go over there, left handers over here. The rest of you, come with me.! Yogi Berra Torque Motion of a rigid body, like motion of
More informationRobotics & Automation. Lecture 25. Dynamics of Constrained Systems, Dynamic Control. John T. Wen. April 26, 2007
Robotics & Automation Lecture 25 Dynamics of Constrained Systems, Dynamic Control John T. Wen April 26, 2007 Last Time Order N Forward Dynamics (3-sweep algorithm) Factorization perspective: causal-anticausal
More informationSteering the Chaplygin Sleigh by a Moving Mass
Steering the Chaplygin Sleigh by a Moving Mass Jason M. Osborne Department of Mathematics North Carolina State University Raleigh, NC 27695 jmosborn@unity.ncsu.edu Dmitry V. Zenkov Department of Mathematics
More informationPhysics 235 Chapter 7. Chapter 7 Hamilton's Principle - Lagrangian and Hamiltonian Dynamics
Chapter 7 Hamilton's Principle - Lagrangian and Hamiltonian Dynamics Many interesting physics systems describe systems of particles on which many forces are acting. Some of these forces are immediately
More informationChapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics
Chapter 1: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational / / 1/ m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv / / 1/ I
More informationLecture Note 7: Velocity Kinematics and Jacobian
ECE5463: Introduction to Robotics Lecture Note 7: Velocity Kinematics and Jacobian Prof. Wei Zhang Department of Electrical and Computer Engineering Ohio State University Columbus, Ohio, USA Spring 2018
More informationRigid Body Kinetics :: Virtual Work
Rigid Body Kinetics :: Virtual Work Work-energy relation for an infinitesimal displacement: du = dt + dv (du :: total work done by all active forces) For interconnected systems, differential change in
More informationFinal Exam April 30, 2013
Final Exam Instructions: You have 120 minutes to complete this exam. This is a closed-book, closed-notes exam. You are allowed to use a calculator during the exam. Usage of mobile phones and other electronic
More informationLecture Notes Multibody Dynamics B, wb1413
Lecture Notes Multibody Dynamics B, wb1413 A. L. Schwab & Guido M.J. Delhaes Laboratory for Engineering Mechanics Mechanical Engineering Delft University of Technolgy The Netherlands June 9, 29 Contents
More informationLagrange s Equations of Motion with Constraint Forces
Lagrange s Equations of Motion with Constraint Forces Kane s equations do not incorporate constraint forces 1 Ax0 A is m n of rank r MEAM 535 Review: Linear Algebra The row space, Col (A T ), dimension
More informationPLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY Today s Objectives: Students will be able to: 1. Define the various ways a force and couple do work.
PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY Today s Objectives: Students will be able to: 1. Define the various ways a force and couple do work. In-Class Activities: 2. Apply the principle of work
More informationIn most robotic applications the goal is to find a multi-body dynamics description formulated
Chapter 3 Dynamics Mathematical models of a robot s dynamics provide a description of why things move when forces are generated in and applied on the system. They play an important role for both simulation
More informationTranslational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work
Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational
More informationHamilton-Jacobi theory on Lie algebroids: Applications to nonholonomic mechanics. Manuel de León Institute of Mathematical Sciences CSIC, Spain
Hamilton-Jacobi theory on Lie algebroids: Applications to nonholonomic mechanics Manuel de León Institute of Mathematical Sciences CSIC, Spain joint work with J.C. Marrero (University of La Laguna) D.
More informationGeneralized Coordinates, Lagrangians
Generalized Coordinates, Lagrangians Sourendu Gupta TIFR, Mumbai, India Classical Mechanics 2012 August 10, 2012 Generalized coordinates Consider again the motion of a simple pendulum. Since it is one
More informationPLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION
PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION I. Moment of Inertia: Since a body has a definite size and shape, an applied nonconcurrent force system may cause the body to both translate and rotate.
More informationRobot Dynamics II: Trajectories & Motion
Robot Dynamics II: Trajectories & Motion Are We There Yet? METR 4202: Advanced Control & Robotics Dr Surya Singh Lecture # 5 August 23, 2013 metr4202@itee.uq.edu.au http://itee.uq.edu.au/~metr4202/ 2013
More informationTrajectory-tracking control of a planar 3-RRR parallel manipulator
Trajectory-tracking control of a planar 3-RRR parallel manipulator Chaman Nasa and Sandipan Bandyopadhyay Department of Engineering Design Indian Institute of Technology Madras Chennai, India Abstract
More informationHW 6 Mathematics 503, Mathematical Modeling, CSUF, June 24, 2007
HW 6 Mathematics 503, Mathematical Modeling, CSUF, June 24, 2007 Nasser M. Abbasi June 15, 2014 Contents 1 Problem 1 (section 3.5,#9, page 197 1 2 Problem 1 (section 3.5,#9, page 197 7 1 Problem 1 (section
More informationRotation. Rotational Variables
Rotation Rigid Bodies Rotation variables Constant angular acceleration Rotational KE Rotational Inertia Rotational Variables Rotation of a rigid body About a fixed rotation axis. Rigid Body an object that
More informationkx m x B N 1 C L, M Mg θ
.004 MODELING DYNAMICS AND CONTROL II Spring 00 Solutions to Problem Set No. 7 Problem 1. Pendulum mounted on elastic support. This problem is an execise in the application of momentum principles. Two
More informationTwo-Dimensional Rotational Kinematics
Two-Dimensional Rotational Kinematics Rigid Bodies A rigid body is an extended object in which the distance between any two points in the object is constant in time. Springs or human bodies are non-rigid
More information