Kinematics1.0 What is “Dynamics?”2.0 How this Course is Laid Out3.0 Some Basics on Frames and Derivatives of Vectors4.0 Calculating Velocities and Accelerations of Given Points4.1 The really tedious but conceptually simple approach (RTBCSA)4.2 The Less Tedious Approach (LTA) Alternatively: Taking Derivatives with the Angular Velocity4.3 A Really Simple Formulaic Approach (RSFA)4.4 SummaryKinematics Professor Sanjay Sarma September 12, 2007 1.0 What is “Dynamics?” The goal of the field of dynamics is to understand how mechanical systems move under the effect of forces. There are 3 components to the study of dynamics: • Kinematics deals with the motions of bodies. Kinematics has to do with geometry and physical constraints. • Kinetics deals with the evolution of this motion under the effect of forces. Classical Dynamics invoke Newton’s laws. • Constitutive relationships are the relationships which capture the effects of springs, gravitation, electromagnetism, etc. By combining kinematic, kinetic and constitutive relationships, it is possible to generate a complete set of equations collectively called the equations of motion. Solving these equa-tions of motion gives us the motion of the system. 2.0 How this Course is Laid Out System Kinematics Kinetics & Constitutive Particle System of particles Rigid Bodies Lagrangian formulation Oscillations 3.0 Some Basics on Frames and Derivatives of Vectors Kinematics is all about reference frames, vectors, differentiation, constraints and coordi-nates. Later, we will use generalized coordinates and constraints, but not yet. Right now, we describe some of the basic terms. 1. A reference frame is a perspective from which a system is observed. The inertial frame of reference is a special frame which is important when we study kinetics, but has no relevance in kinematics per se. 2. It is customary to attach three mutually perpendicular unit vectors to frames. You can give them names like: i, j, k or er, eθ, et or a1, a2, a3. Kinematics September 12, 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]. 1• By the way, we underline vectors on the blackboard, but using a bold symbol is the equivalent in type-set form. I will do both for some time, but please interpret bold as a vector if you see it by itself in future material. • Consider a situation in which you have defined a number of frames with unit vec-tors in each. There is absolutely no reason why you couldn’t mix unit vectors. So, you could say: p = 3i + 7eθ You can differentiate vectors. However, differentiation of vectors has no meaning if you do not specify the frame with respect to which you are differentiating them. So it is important to say Ad (3a + 7b) instead of justd (3a + 7b) if you are dt dt differentiating with respect to frame A. Often, this frame is not stated explicitly, and must be inferred from the context in which the expression is written. I prefer to spell it out explicitly. Here’s the secret: if a1, a2, a3 are the unit vectors associated with a frame A, then Ada1 Ada2 Ada3 = 0; = 0; = 0. (EQ 1) dt dt dt In fact any vector which is merely translating in A, even if it is accelerating, has a zero derivative in A. A non-zero derivative occurs when the vector is stretching (in which case it is stretching in any frame) or rotating with respect to A. So in general we can say that: Adb ≠ 0 . (EQ 2) dt In other words, if you want to differentiate an expression with vectors in it, it is best to express all vector terms in terms of a1, a2, a3 and then use relationships (I) above. That way, the taking of the derivative becomes trivially easy. • Furthermore for any two vectors a and b, Ad AdAd(a b) = a • b + a • b• () ()dt dt dt and Ad AdAd× () a × b .(a b) = a × b + ()dt dt dt 4.0 Calculating Velocities and Accelerations of Given Points The last and the next homework are all about calculating velocities and accelerations of points defined in various frames of references. There are three approaches to this problem. Kinematics September 12, 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]. 2O P S garden frame A a1 a2 frisbee frame B The spider b2 b1 FIGURE 1. Spider on a Frisbee Problem θ l u v We will use the spider-on-the-frisbee problem used in class to illustrate the approaches. Spider photo courtesy of B. Smith. See Figure 1. 4.1 The really tedious but conceptually simple approach (RTBCSA) 1. Write down the position vector of the spider with respect to point O: rOS. = rOP+ rPQ. For convenience, you can write it in mixed form if you would like: rOS= ua1 + va2 + lb1 2. Now, say you want to calculate velocities and accelerations with respect to frame A. Then you are looking for: AS Ad rOS AdV = = (ua1 + va2 + lb1). (EQ 3) dt dt 3. Because of Equations 1, this simplifies to: ASV = a1 Ad ()u + a2 Ad v + Ad (lb1). (EQ 4) ()dt dt dt The first two terms are done (!) but how do we take care of the third term? Here we use the trick alluded to earlier. We simply rewrite b1 in terms of a1 and a2: b1 = a1 cosθ + a2 sinθ. (EQ 5) By the way, even though we don’t need it here, b2 = – a1 sin θ + a2 cosθ (EQ 6) Anyway, we can insert the expressions for b1 and b2 into Equation 4 above, take the derivative, and we are done! We get: Kinematics September 12, 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].AS · ·· V = a1u ·+ a2v ·– a1θ sinθ + a2θ co θ + l(a1 cosθ + a2 sin θ) . (EQ 7 ) 4. The acceleration, AaS is simply the derivative of AVS done exactly as before. The good news is that since we have already down-converted everything to a1 and a2, the task is simple. When everything is done, we get: AS AdAS ·· ··· · a = V = (u ··a1 + v ··a2) + (lcosθa1 + lsinθa2) + 2(– l · θ sinθa1 + l · θ cosθa2)dt ·· · · ·2·2(– lθ sinθa1 + lθ