Chapter 4 Rigid Body Motion
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Rotational Motion (The Dynamics of a Rigid Body)
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1958 Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body) Henry Semat City College of New York Robert Katz University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/physicskatz Part of the Physics Commons Semat, Henry and Katz, Robert, "Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body)" (1958). Robert Katz Publications. 141. https://digitalcommons.unl.edu/physicskatz/141 This Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. 11 Rotational Motion (The Dynamics of a Rigid Body) 11-1 Motion about a Fixed Axis The motion of the flywheel of an engine and of a pulley on its axle are examples of an important type of motion of a rigid body, that of the motion of rotation about a fixed axis. Consider the motion of a uniform disk rotat ing about a fixed axis passing through its center of gravity C perpendicular to the face of the disk, as shown in Figure 11-1. The motion of this disk may be de scribed in terms of the motions of each of its individual particles, but a better way to describe the motion is in terms of the angle through which the disk rotates. -
GYRODYNAMICS Introduction to the Dynamics of Rigid Bodies
2 GYRODYNAMICS Introduction to the dynamics of rigid bodies Introduction. Though Newton wrote on many topics—and may well have given thought to the odd behavior of tops—I am not aware that he committed any of that thought to writing. But by Euler was active in the field, and it has continued to bedevil the thought of mathematical physicists. “Extended rigid bodies” are classical abstractions—alien both to relativity and to quantum mechanics—which are revealed to our dynamical imaginations not so much by commonplace Nature as by, in Maxwell’s phrase, the “toys of Youth.” That such toys behave “strangely” is evident to the most casual observer, but the detailed theory of their behavior has become notorious for being elusive, surprising and difficult at every turn. Its formulation has required and inspired work of wonderful genius: it has taught us much of great worth, and clearly has much to teach us still. Early in my own education as a physicist I discovered that I could not understand—or, when I could understand, remained unpersuaded by—the “elementary explanations” of the behavior of tops & gyros which are abundant in the literature. So I fell into the habit of avoiding the field, waiting for the day when I could give to it the time and attention it obviously required and deserved. I became aware that my experience was far from atypical: according to Goldstein it was in fact a similar experience that motivated Klein & Sommerfeld to write their 4-volume classic, Theorie des Kreisels (–). In November I had occasion to consult my Classical Mechanics II students concerning what topic we should take up to finish out the term. -
Rigid Body Dynamics 2
Rigid Body Dynamics 2 CSE169: Computer Animation Instructor: Steve Rotenberg UCSD, Winter 2017 Cross Product & Hat Operator Derivative of a Rotating Vector Let’s say that vector r is rotating around the origin, maintaining a fixed distance At any instant, it has an angular velocity of ω ω dr r ω r dt ω r Product Rule The product rule of differential calculus can be extended to vector and matrix products as well da b da db b a dt dt dt dab da db b a dt dt dt dA B dA dB B A dt dt dt Rigid Bodies We treat a rigid body as a system of particles, where the distance between any two particles is fixed We will assume that internal forces are generated to hold the relative positions fixed. These internal forces are all balanced out with Newton’s third law, so that they all cancel out and have no effect on the total momentum or angular momentum The rigid body can actually have an infinite number of particles, spread out over a finite volume Instead of mass being concentrated at discrete points, we will consider the density as being variable over the volume Rigid Body Mass With a system of particles, we defined the total mass as: n m m i i1 For a rigid body, we will define it as the integral of the density ρ over some volumetric domain Ω m d Angular Momentum The linear momentum of a particle is 퐩 = 푚퐯 We define the moment of momentum (or angular momentum) of a particle at some offset r as the vector 퐋 = 퐫 × 퐩 Like linear momentum, angular momentum is conserved in a mechanical system If the particle is constrained only -
Rigid Body Dynamics II
An Introduction to Physically Based Modeling: Rigid Body Simulation IIÐNonpenetration Constraints David Baraff Robotics Institute Carnegie Mellon University Please note: This document is 1997 by David Baraff. This chapter may be freely duplicated and distributed so long as no consideration is received in return, and this copyright notice remains intact. Part II. Nonpenetration Constraints 6 Problems of Nonpenetration Constraints Now that we know how to write and implement the equations of motion for a rigid body, let's consider the problem of preventing bodies from inter-penetrating as they move about an environment. For simplicity, suppose we simulate dropping a point mass (i.e. a single particle) onto a ®xed ¯oor. There are several issues involved here. Because we are dealing with rigid bodies, that are totally non-¯exible, we don't want to allow any inter-penetration at all when the particle strikes the ¯oor. (If we considered our ¯oor to be ¯exible, we might allow the particle to inter-penetrate some small distance, and view that as the ¯oor actually deforming near where the particle impacted. But we don't consider the ¯oor to be ¯exible, so we don't want any inter-penetration at all.) This means that at the instant that the particle actually comes into contact with the ¯oor, what we would like is to abruptly change the velocity of the particle. This is quite different from the approach taken for ¯exible bodies. For a ¯exible body, say a rubber ball, we might consider the collision as occurring gradually. That is, over some fairly small, but non-zero span of time, a force would act between the ball and the ¯oor and change the ball's velocity. -
Accurate Simulation of Rigid Body Rotation and Extensions to Lie Groups
Accurate Simulation of Rigid Body Rotation and Extensions to Lie Groups Sam Buss Dept. of Mathematics U.C. San Diego Visible Luncheon Gwyddor Cyfrifiador University of Swansea March 3, 2011 Topics: ◮ Algorithms for simulating rotating rigid bodies. ◮ All algorithms preserve angular momentum. ◮ Algorithms can be made energy preserving. ◮ Generalization to Lie group setting. Talk outline: 1. Rigid body rotations. 1st thru 4th order algorithms. Unexpected terms. 2. Generalization to Taylor series methods over Lie groups/Lie algebras. 3. Energy preservation based on Poinsot ellipsoid. 4. Numerical simulations and efficiency. 1 Part I: The simple rotating, rigid body I = Inertia matrix (tensor). ω L = Angular momentum. ω = Rotation axis & rate, L L = Iω (Euler’s equation) ω = I−1L ω˙ = I−1(L˙ ω Iω) − × ω¨ = ω ω˙ + I−1(L¨ ω˙ L 2ω L˙ + ω (ω L)) × − × − × × × ... ... ω = 2ω ω¨ ω (ω ω˙ ) + I−1[L 3ω L¨ 3ω˙ L˙ ω¨ L × − × × − × − × − × + ω˙ (ω L) + 2ω (ω˙ L) + 3ω (ω L˙ ) ω (ω (ω L))] × × × × × × − × × × Wobble: ω˙ = 0 even when no applied torque (L˙ = 0). 6 vL˙ = Rate of change of momentum = Applied Torque. 2 Framework for simulating rigid body motion We assume the rigid body has a known angular momentum, and the external torques are completely known. The orientation (and hence the angular velocity) is updated in discrete time steps, at times t0, t1, t2,.... Update Step: At a given time t , let h = ∆t = t t , and assume i i+1 − i orientation Ωi at time ti is known, and that momentum is known (at all times). Update step calculates a net rotation rate vector, ω¯, and sets Ωi+1 = Rhω¯ Ωi, where Rν performs a rotation around axis ν of angle ν . -
Rotation: Moment of Inertia and Torque
Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn that where the force is applied and how the force is applied is just as important as how much force is applied when we want to make something rotate. This tutorial discusses the dynamics of an object rotating about a fixed axis and introduces the concepts of torque and moment of inertia. These concepts allows us to get a better understanding of why pushing a door towards its hinges is not very a very effective way to make it open, why using a longer wrench makes it easier to loosen a tight bolt, etc. This module begins by looking at the kinetic energy of rotation and by defining a quantity known as the moment of inertia which is the rotational analog of mass. Then it proceeds to discuss the quantity called torque which is the rotational analog of force and is the physical quantity that is required to changed an object's state of rotational motion. Moment of Inertia Kinetic Energy of Rotation Consider a rigid object rotating about a fixed axis at a certain angular velocity. Since every particle in the object is moving, every particle has kinetic energy. To find the total kinetic energy related to the rotation of the body, the sum of the kinetic energy of every particle due to the rotational motion is taken. The total kinetic energy can be expressed as .. -
Polhode Motion, Trapped Flux, and the GP-B Science Data Analysis
July 8, 2009 • Stanford University HEPL Seminar Polhode Motion, Trapped Flux, and the GP-B Science Data Analysis Alex Silbergleit, John Conklin and the Polhode/Trapped Flux Mapping Task Team July 8, 2009 • Stanford University HEPL Seminar Outline 1. Gyro Polhode Motion, Trapped Flux, and GP-B Readout (4 charts) 2. Changing Polhode Period and Path: Energy Dissipation (4 charts) 3. Trapped Flux Mapping (TFM): Concept, Products, Importance (7 charts) 4. TFM: How It Is Done - 3 Levels of Analysis (11 charts) A. Polhode phase & angle B. Spin phase C. Magnetic potential 5. TFM: Results ( 9 charts) 6. Conclusion. Future Work (1 chart) 2 July 8, 2009 • Stanford University HEPL Seminar Outline 1. Gyro Polhode Motion, Trapped Flux, and GP-B Readout 2. Changing Polhode Period and Path: Energy Dissipation 3. Trapped Flux Mapping (TFM): Concept, Products, Importance 4. TFM: How It Is Done - 3 Levels of Analysis A. Polhode phase & angle B. Spin phase C. Magnetic potential 5. TFM: Results 6. Conclusion. Future Work 3 July 8, 2009 • Stanford University HEPL Seminar 1.1 Free Gyro Motion: Polhoding • Euler motion equations – In body-fixed frame: – With moments of inertia: – Asymmetry parameter: (Q=0 – symmetric rotor) • Euler solution: instant rotation axis precesses about rotor principal axis along the polhode path (angular velocity Ωp) • For GP-B gyros 4 July 8, 2009 • tanford UniversityS HEPL Seminar 1.2 Symmetric vs. Asymmetric Gyro Precession • Symmetric (II 1 = 2 , Q=0): γp= const (ω3=const, polhode path=circular cone), • =φp Ω p const = , motion -
The Gravity Probe B Test of General Relativity
Home Search Collections Journals About Contact us My IOPscience The Gravity Probe B test of general relativity This content has been downloaded from IOPscience. Please scroll down to see the full text. 2015 Class. Quantum Grav. 32 224001 (http://iopscience.iop.org/0264-9381/32/22/224001) View the table of contents for this issue, or go to the journal homepage for more Download details: IP Address: 162.247.72.27 This content was downloaded on 31/05/2016 at 09:49 Please note that terms and conditions apply. Classical and Quantum Gravity Class. Quantum Grav. 32 (2015) 224001 (29pp) doi:10.1088/0264-9381/32/22/224001 The Gravity Probe B test of general relativity C W F Everitt1, B Muhlfelder1, D B DeBra1, B W Parkinson1, J P Turneaure1, A S Silbergleit1, E B Acworth1, M Adams1, R Adler1, W J Bencze1, J E Berberian1, R J Bernier1, K A Bower1, R W Brumley1, S Buchman1, K Burns1, B Clarke1, J W Conklin1, M L Eglington1, G Green1, G Gutt1, D H Gwo1, G Hanuschak1,XHe1, M I Heifetz1, D N Hipkins1, T J Holmes1, R A Kahn1, G M Keiser1, J A Kozaczuk1, T Langenstein1,JLi1, J A Lipa1, J M Lockhart1, M Luo1, I Mandel1, F Marcelja1, J C Mester1, A Ndili1, Y Ohshima1, J Overduin1, M Salomon1, D I Santiago1, P Shestople1, V G Solomonik1, K Stahl1, M Taber1, R A Van Patten1, S Wang1, J R Wade1, P W Worden Jr1, N Bartel6, L Herman6, D E Lebach6, M Ratner6, R R Ransom6, I I Shapiro6, H Small6, B Stroozas6, R Geveden2, J H Goebel3, J Horack2, J Kolodziejczak2, A J Lyons2, J Olivier2, P Peters2, M Smith3, W Till2, L Wooten2, W Reeve4, M Anderson4, N R Bennett4, -
Lecture Notes for PHY 405 Classical Mechanics
Lecture Notes for PHY 405 Classical Mechanics From Thorton & Marion's Classical Mechanics Prepared by Dr. Joseph M. Hahn Saint Mary's University Department of Astronomy & Physics November 21, 2004 Problem Set #6 due Thursday December 1 at the start of class late homework will not be accepted text problems 10{8, 10{18, 11{2, 11{7, 11-13, 11{20, 11{31 Final Exam Tuesday December 6 10am-1pm in MM310 (note the room change!) Chapter 11: Dynamics of Rigid Bodies rigid body = a collection of particles whose relative positions are fixed. We will ignore the microscopic thermal vibrations occurring among a `rigid' body's atoms. 1 The inertia tensor Consider a rigid body that is composed of N particles. Give each particle an index α = 1 : : : N. The total mass is M = α mα. This body can be translating as well as rotating. P Place your moving/rotating origin on the body's CoM: Fig. 10-1. 1 the CoM is at R = m r0 M α α α 0 X where r α = R + rα = α's position wrt' fixed origin note that this implies mαrα = 0 α X 0 Particle α has velocity vf,α = dr α=dt relative to the fixed origin, and velocity vr,α = drα=dt in the reference frame that rotates about axis !~ . Then according to Eq. 10.17 (or page 6 of Chap. 10 notes): vf,α = V + vr,α + !~ × rα = α's velocity measured wrt' fixed origin and V = dR=dt = velocity of the moving origin relative to the fixed origin. -
8.09(F14) Chapter 2: Rigid Body Dynamics
Chapter 2 Rigid Body Dynamics 2.1 Coordinates of a Rigid Body A set of N particles forms a rigid body if the distance between any 2 particles is fixed: rij ≡ jri − rjj = cij = constant: (2.1) Given these constraints, how many generalized coordinates are there? If we know 3 non-collinear points in the body, the remaing points are fully determined by triangulation. The first point has 3 coordinates for translation in 3 dimensions. The second point has 2 coordinates for spherical rotation about the first point, as r12 is fixed. The third point has one coordinate for circular rotation about the axis of r12, as r13 and r23 are fixed. Hence, there are 6 independent coordinates, as represented in Fig. 2.1. This result is independent of N, so this also applies to a continuous body (in the limit of N ! 1). Figure 2.1: 3 non-collinear points can be fully determined by using only 6 coordinates. Since the distances between any two other points are fixed in the rigid body, any other point of the body is fully determined by the distance to these 3 points. 29 CHAPTER 2. RIGID BODY DYNAMICS The translations of the body require three spatial coordinates. These translations can be taken from any fixed point in the body. Typically the fixed point is the center of mass (CM), defined as: 1 X R = m r ; (2.2) M i i i where mi is the mass of the i-th particle and ri the position of that particle with respect to a fixed origin and set of axes (which will notationally be unprimed) as in Fig. -
2D Rigid Body Dynamics: Work and Energy
J. Peraire, S. Widnall 16.07 Dynamics Fall 2008 Version 1.0 Lecture L22 - 2D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture L11-13 for particle dynamics, and extend it to 2D rigid body dynamics. Kinetic Energy for a 2D Rigid Body We start by recalling the kinetic energy expression for a system of particles derived in lecture L11, n 1 2 X 1 02 T = mvG + mir_i ; 2 2 i=1 0 where n is the total number of particles, mi denotes the mass of particle i, and ri is the position vector of Pn particle i with respect to the center of mass, G. Also, m = i=1 mi is the total mass of the system, and vG is the velocity of the center of mass. The above expression states that the kinetic energy of a system of particles equals the kinetic energy of a particle of mass m moving with the velocity of the center of mass, plus the kinetic energy due to the motion of the particles relative to the center of mass, G. For a 2D rigid body, the velocity of all particles relative to the center of mass is a pure rotation. Thus, we can write 0 0 r_ i = ! × ri: Therefore, we have n n n X 1 02 X 1 0 0 X 1 02 2 mir_i = mi(! × ri) · (! × ri) = miri ! ; 2 2 2 i=1 i=1 i=1 0 Pn 02 where we have used the fact that ! and ri are perpendicular. -
Experiment 10: Equilibrium Rigid Body
Experiment P08: Equilibrium of a Rigid Body Purpose To use the principle of balanced torques to find the value of an unknown distance and to investigate the concept of center of gravity. Equipment and Supplies Meter stick Standard masses with hook Triple-beam balance Fulcrum Fulcrum holder Discussion of equilibrium First condition of equilibrium An object at rest is in equilibrium (review rotational equilibrium in your text). The vector sum of all the forces exerted on the body must be equal to zero. ∑F = 0 Second condition of equilibrium The resting object also shows another aspect of equilibrium. Because the object has no rotation, the sum of the torques exerted on it is zero. ∑τ = 0 r is position vector relative to the fulcrum F is the force acting on the object at point, r When a force produces a turning or rotation, a non-zero net torque is present. A see-saw balances when any torque on it add up to zero. A see saw is form of lever, simple mechanical device that rotates about a pivot or fulcrum. Although the work done by a device such as a lever can never be more than the work or energy invested in it, levers make work easier to accomplish for a variety of tasks. Knowledge of torques can also help us to calculate the location of an object’s center of gravity. Earth’s gravity pulls on every part of an object. It pulls more strongly on more massive parts and more weakly on less massive parts. The sum of all these pulls is the weight of the object.