Rigid Body Dynamics Professor Sanjay Sarma November 16, 2007 1.0 Where are we in the course? Thus far we have completed Kinematics and Kinetics of single particles, systems of parti-cles and rigid bodies respectively. We are now well into the Lagrange portion of the class. System Particle System of particles Rigid Bodies Lagrangian formulation Oscillations Kinematics Kinetics & Constitutive Next 2.0 Generalized Coordinates The generalized coordinates of a mechanical system are the minimal group of parameters which can completely and unambiguously define the configuration of that system. Some generalized coordinates are more “natural” than others, but there might be many ways to define them for any one system. The number of generalized coordinates equals the number of degrees of freedom of the system as long as the system is holonomic. We only study holonomic systems in this class. Consider a system consisting of N rigid bodies in 2D space. Each rigid body has 3 degrees of freedom: two translational and one rotational. The N-body system has 3n degrees of freedom. Now let’s say that there are k kinematic constraints which can be expressed as algebraic equations. Then the system has d = 3Nk degrees of freedom.– The term “holonomic” refers to the fact that the kinematic constraints must be expressible as algebraic equalities. Some kinematic constraints can only be expressed as inequalities or differential equations. Such systems are called non-holonomic constraints. We will not consider non-holonomic systems in this class— if you are interested in such systems, you can talk to me about them outside class. 3.0 Why Lagrange? There are several reasons why the Lagrange Approach is important. 1. The Lagrange Approach automatically yields as many equations as there are degrees of freedom. It has the convenience of energy methods, but whereas energy conservation Rigid Body Dynamics November 16, 2007 Cite as: Sanjay Sarma, Nicholas Makris, Yahya Modarres-Sadeghi, and Peter So, course materials for 2.003J/1.053J Dynamics and Control I, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 1only yields just one equation, which isn’t enough for a multi-degree-of-freedom sys-tem, Lagrange yields as many equations as you need. 2. The Lagrange equations naturally use the generalized coordinates of the system. By contrast, Newton’s Equations are essentially Cartesian. You end up having to convert everything into Cartesian components of acceleration and Cartesian components of forces to use Newton’s Equation. Lagrange bypasses that conversion. 3. The Lagrange approach naturally eliminates non-contributing forces. You could do the same with the direct (Newtonian) approach, but your ability to minimize the number of variables depends very much on your skill; Lagrange takes care of it for you automati-cally because the generalized forces only include force components in directions of admissible motion. 4.0 The Lagrange Equations For a d-dof (degree-of-freedom) system with generalized coordinates qj ’s, it is possible to formulate the Lagrangian L = T – V where T is the kinetic energy and V is the poten-tial energy. The Lagrangian is a function of generalized coordinates qj ’s and generalized · velocities qj ’s: · · Lq1, , q1, …q · ) .1 (EQ 1) L = ( …qj…qd …qj d where d is the number of degrees of freedom. The Lagrange Equations are then: ddt ⎛⎝∂∂ Lq · j ⎞⎠ – ∂∂Lqj = Qj , (EQ 2) where Qj ’s are the external generalized forces. Since j goes from 1 to d, Lagrange gives us d equations of motion. But what are generalized forces? We derived them in class. Read on. 4.1 Generalized Forces The generalized force Qj is defined below: 1. There are some situations in which the Lagrangian is explicitly a function of time. Such systems are called rheonomic systems. We will not explore the implications in this course. Rigid Body Dynamics November 16, 2007 2 Cite as: Sanjay Sarma, Nicholas Makris, Yahya Modarres-Sadeghi, and Peter So, course materials for 2.003J/1.053J Dynamics and Control I, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].q x y δq Δy Δx FIGURE 1. A bead on a wire Qj = Fi i = 1 N ∑ • ∂qj ∂ri ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ (EQ 3) where Fi is the force at point i and ri is the position vector of point i. The index j corre-sponds to generalized coordinates. 4.2 The Intuition So why does the Lagrange formulation work? The insight is simple. The Lagrange formu-lation only considers admissible motions. 4.2.1 The Problem with the Newtonian Approach Consider a bead sliding without friction on a curved wire as shown in Figure 2. Clearly the bead can only move along the wire, which can be approximated locally as a direction tan-gential to the wire. Now, the Cartesian coordinates of the bead would be x and y. However, these coordinates are redundant. We can only eliminate the redundancy by introducing a geometric constrain between x and y of the form Constraint x y( , ) = 0 .1 For example, if the wire is in the form of a circle of radius R, the constraint will be x 2+ y 2– R2 = 0. No combination of Δx and Δy is legal if it does not satisfy x 2+ y 2– R2= 0. In the direct, or Newtonian approach, we waste a lot of time considering x and y motions as if the bead could get to any x and y (which it can’t), postulating reaction forces (which are actually irrelevant) and then solving for these reaction forces and motions such that the Δx and Δy satisfy the kinematic constraint (which is a waste of time). The problem, as 1. We will assume that this is an algebraic. If it is an inequality constraint or an unintegrable differential equation, we need more machinery which we will not cover in this course. Rigid Body Dynamics November 16, 2007 3 Cite as: Sanjay Sarma, Nicholas Makris, Yahya Modarres-Sadeghi, and Peter So, course materials for 2.003J/1.053J Dynamics and Control I, Fall 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].1 2 Let’s say you try to move in this direction. 5 A reaction force keeps the bead on the wire. 3 This reaction force is irrelevant because it adapts to counter any applied force, and it doesn’t do work. 4 So why even consider impossible motions and the forces we need to make