5 | Newton's Laws of Motion 207 5 | NEWTON's LAWS of MOTION
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Explain Inertial and Noninertial Frame of Reference
Explain Inertial And Noninertial Frame Of Reference Nathanial crows unsmilingly. Grooved Sibyl harlequin, his meadow-brown add-on deletes mutely. Nacred or deputy, Sterne never soot any degeneration! In inertial frames of the air, hastening their fundamental forces on two forces must be frame and share information section i am throwing the car, there is not a severe bottleneck in What city the greatest value in flesh-seconds for this deviation. The definition of key facet having a small, polished surface have a force gem about a pretend or aspect of something. Fictitious Forces and Non-inertial Frames The Coriolis Force. Indeed, for death two particles moving anyhow, a coordinate system may be found among which saturated their trajectories are rectilinear. Inertial reference frame of inertial frames of angular momentum and explain why? This is illustrated below. Use tow of reference in as sentence Sentences YourDictionary. What working the difference between inertial frame and non inertial fr. Frames of Reference Isaac Physics. In forward though some time and explain inertial and noninertial of frame to prove your measurement problem you. This circumstance undermines a defining characteristic of inertial frames: that with respect to shame given inertial frame, the other inertial frame home in uniform rectilinear motion. The redirect does not rub at any valid page. That according to whether the thaw is inertial or non-inertial In the. This follows from what Einstein formulated as his equivalence principlewhich, in an, is inspired by the consequences of fire fall. Frame of reference synonyms Best 16 synonyms for was of. How read you govern a bleed of reference? Name we will balance in noninertial frame at its axis from another hamiltonian with each printed as explained at all. -
11.1 Virtual Work
11.1 Virtual Work 11.1 Virtual Work Example 1, page 1 of 5 1. Determine the force P required to keep the two rods in equilibrium when the angle = 30° and weight W is 50 lb. The rods are each of length L and of negligible weight. They are prevented from moving out of the plane of the figure by supports not shown. B L L W C A P Smooth surface 11.1 Virtual Work Example 1, page 2 of 5 B 1 The system has one degree of freedom, L L because specifying the value of a single W coordinate, , completely determines the C configuration (shape) of the system. A P Consider a free-body diagram and identify the active forces those forces that would do work if were increased slightly. Free-body diagram (The dashed line shows the position of the system after has been increased a small amount.) 2 The force W does work because point B 4 The force P does work as B moves up, so W is an active force. point A moves to the right, so P is an active force. P A W C Cx 3 The reactions Cx and Cy do no N Cy work because point C does not 5 The normal force N does no work move. Thus Cx and Cy are not because it is perpendicular to the active forces. displacement of point A. Thus N is not an active force. 11.1 Virtual Work Example 1, page 3 of 5 6 Introduce coordinates measured from a fixed point, 7 Compute the work done when the coordinates are point C in the figure, to the point of application of the increased positive infinitesimal amounts, xA and xB active forces. -
[Math.DS] 18 Jan 2005 Complexity of Piecewise Convex Transformations
Complexity of piecewise convex transformations in two dimensions, with applications to polygonal billiards Eugene Gutkin ∗ and Serge Tabachnikov † November 12, 2018 Abstract ABSTRACT We introduce the class of piecewise convex transfor- mations, and study their complexity. We apply the results to the complexity of polygonal billiards on surfaces of constant curvature. Introduction The following situation frequently occurs in geometric dynamics. There is a phase space X, a transformation T : X → X; and there is a finite de- composition P : X = X(a) ∪ X(b) ∪ ··· . Let A = {a, b, . } be the corresponding alphabet. A phase point x ∈ X is regular if every element of the orbit x, T x, T 2x, . belongs to a unique atom of P. Suppose that arXiv:math/0412335v2 [math.DS] 18 Jan 2005 x ∈ X(a), T x ∈ X(b), etc. The corresponding word a b ··· is the code of x. Let Σ(n) be the set of words in A of length n obtained by coding points in X. The positive function f(n) = |Σ(n)| is the associated complexity. Its ∗UCLA, Math. Dpt., Los Angeles, CA 90095, USA and IMPA, Estrada Dona Castorina 110, 22460-320 Rio de Janeiro RJ, Brasil. E-mail: [email protected],[email protected] †Department of Mathematics, Pennsylvania State University, University Park, PA 16802. E-mail: [email protected] 1 behavior as n →∞1 is an important characteristic of the dynamical system in question. The following examples have motivated our study. Example A. Let P ⊂ R2 be a polygon with sides a, b, . , and let X be the phase space of the billiard map Tb in P . -
Circular Motion Dynamics
Problem Solving Circular Motion Dynamics Problem 1: Double Star System Consider a double star system under the influence of gravitational force between the stars. Star 1 has mass m1 and star 2 has mass m2 . Assume that each star undergoes uniform circular motion about the center of mass of the system. If the stars are always a fixed distance s apart, what is the period of the orbit? Solution: Choose radial coordinates for each star with origin at center of mass. Let rˆ1 be a unit vector at Star 1 pointing radially away from the center of mass. Let rˆ2 be a unit vector at Star 2 pointing radially away from the center of mass. The force diagrams on the two stars are shown in the figure below. ! ! From Newton’s Second Law, F1 = m 1a 1 , for Star 1 in the radial direction is m m ˆ 1 2 2 r1 : "G 2 = "m1 r 1 ! . s We can solve this for r1 , m r = G 2 . 1 ! 2s 2 ! ! Newton’s Second Law, F2 = m 2a 2 , for Star 2 in the radial direction is m m ˆ 1 2 2 r2 : "G 2 = "m2 r 2 ! . s We can solve this for r2 , m r = G 1 . 2 ! 2s 2 Since s , the distance between the stars, is constant m2 m1 (m2 + m1 ) s = r + r = G + G = G 1 2 ! 2s 2! 2s 2 ! 2s 2 . Thus the angular velocity is 1 2 " (m2 + m1 ) # ! = $G 3 % & s ' and the period is then 1 2 2! # 4! 2s 3 $ T = = % & . -
14 Newton's Laws #1: Using Free Body Diagrams
Chapter 14 Newton’s Laws #1: Using Free Body Diagrams 14 Newton’s Laws #1: Using Free Body Diagrams If you throw a rock upward in the presence of another person, and you ask that other person what keeps the rock going upward, after it leaves your hand but before it reaches its greatest height, that person may incorrectly tell you that the force of the person’s hand keeps it going. This illustrates the common misconception that force is something that is given to the rock by the hand and that the rock “has” while it is in the air. It is not. A force is all about something that is being done to an object. We have defined a force to be an ongoing push or a pull. It is something that an object can be a victim to, it is never something that an object has. While the force is acting on the object, the motion of the object is consistent with the fact that the force is acting on the object. Once the force is no longer acting on the object, there is no such force, and the motion of the object is consistent with the fact that the force is absent. (As revealed in this chapter, the correct answer to the question about what keeps the rock going upward, is, “Nothing.” Continuing to go upward is what it does all by itself if it is already going upward. You don’t need anything to make it keep doing that. In fact, the only reason the rock does not continue to go upward forever is because there is a downward force on it. -
© 2016 Ki Yeun Kim DYNAMICS of BOUNCING RIGID BODIES and BILLIARDS in the SPACES of CONSTANT CURVATURE
CORE Metadata, citation and similar papers at core.ac.uk Provided by Illinois Digital Environment for Access to Learning and Scholarship Repository © 2016 Ki Yeun Kim DYNAMICS OF BOUNCING RIGID BODIES AND BILLIARDS IN THE SPACES OF CONSTANT CURVATURE BY KI YEUN KIM DISSERTATION Submied in partial fulllment of the requirements for the degree of Doctor of Philosophy in Mathematics in the Graduate College of the University of Illinois at Urbana-Champaign, 2016 Urbana, Illinois Doctoral Commiee: Professor Yuliy Baryshnikov, Chair Professor Vadim Zharnitsky, Director of Research Associate Professor Robert DeVille Associate Professor Zoi Rapti Abstract Mathematical billiard is a dynamical system studying the motion of a mass point inside a domain. e point moves along a straight line in the domain and makes specular reections at the boundary. e theory of billiards has developed extensively for itself and for further applications. For example, billiards serve as natural models to many systems involving elastic collisions. One notable example is the system of spherical gas particles, which can be described as a billiard on a higher dimensional space with a semi-dispersing boundary. In the rst part of this dissertation, we study the collisions of a two-dimensional rigid body using billiard dynamics. We rst dene a dumbbell system, which consists of two point masses connected by a weightless rod. We assume the dumbbell moves freely in the air and makes elastic collisions at a at boundary. For arbitrary mass choices, we use billiard techniques to nd the sharp bound on the number of collisions of the dumbbell system in terms of the mass ratio. -
Assignment 2 Centripetal Force & Univ
LAST NAME______________________________ FIRST NAME_____________________DATE______ CJ - Assignment 2 Centripetal Force & Universal Gravitation, 5.4 Banked Curves Draw the vectors for a free body diagram of a car in an unbanked turn and a banked turn. For this situation assume the cars are turning to the left side of the drawing. Car in Unbanked Turn Car in Banked turn Is this car in Equilibrium? Is this car in Equilibrium? Do the same for the airplane. Airplane in Unbanked Turn Airplane in Banked turn Let’s be careful to fully understand what is going on with this one. We have made an assumption in doing this for the airplane which are not valid. THE BANKED TURN- car can make it a round a turn even if friction is not present (or friction is zero). In order for this to happen the curved road must be “banked”. In a non-banked turn the centripetal force is created by the friction of the road on the tires. What creates the centripetal force in a banked turn? Imagine looking at a banked turn as shown to the right. Draw the forces that act on the car on the diagram of the car. Is the car in Equilibrium? YES , NO The car which has a mass of m is travelling at a velocity (or speed) v around a turn of radius r. How do these variables relate to optimize the banked turn. If the mass of the car is greater is a larger or smaller angle required? LARGER, SMALLER, NO CHANGE NECESSARY If the VELOCITY of the car is greater is a larger or smaller angle required? LARGER, SMALLER, NO CHANGE NECESSARY If the RADUS OF THE TURN is greater is a larger or smaller angle required? LARGER, SMALLER, NO CHANGE NECESSARY To properly show the relationship an equation must be created. -
Complexity of Piecewise Convex Transformations in Two Dimensions, with Applications to Polygonal Billiards on Surfaces of Constant Curvature
MOSCOW MATHEMATICAL JOURNAL Volume 6, Number 4, October–December 2006, Pages 673–701 COMPLEXITY OF PIECEWISE CONVEX TRANSFORMATIONS IN TWO DIMENSIONS, WITH APPLICATIONS TO POLYGONAL BILLIARDS ON SURFACES OF CONSTANT CURVATURE EUGENE GUTKIN AND SERGE TABACHNIKOV To A. A. Kirillov for his 70th birthday Abstract. We introduce piecewise convex transformations, and de- velop geometric tools to study their complexity. We apply the results to the complexity of polygonal inner and outer billiards on surfaces of constant curvature. 2000 Math. Subj. Class. 53D25, 37E99, 37B10. Key words and phrases. Geodesic polygon, constant curvature, complexity, inner billiard, outer billiard. Introduction Transformations that arise in geometric dynamics often fit into the following scheme. The phase space of the transformation, T : X → X, has a finite decompo- sition P : X = Xa ∪ Xb ∪··· into “nice” subsets with respect to a geometric struc- ture on X. (For instance, Xa, Xb, ... are convex.) The restrictions T : Xa → X, T : Xb → X, ... preserve that structure. The interiors of Xa, Xb, ... are dis- joint, and T is discontinuous (or not defined) on the union of their boundaries, −1 −1 ∂P = ∂Xa ∪ ∂Xb ∪··· . Set P1 = P. The sets T (Xa) ∩ Xa, T (Xa) ∩ Xb, ... form the decomposition P2, which is a refinement of P1; it provides a defining partition for the transformation T 2. Iterating this process, we obtain a tower of n decompositions Pn, n > 1, where Pn plays for T the same role that P played for T . Let A = {a, b, ... } be the alphabet labeling the atoms of P. A phase point x ∈ X is regular if every point of the orbit x, Tx, T 2x, .. -
Numerical Methods Applied Id Metallurgical Processing
NUMERICAL METHODS APPLIED ID METALLURGICAL PROCESSING by JUAN HECTOR BIANCHI A thesis submitted for the degree of Doctor of Philosophy of the University of London and for the Diploma of Membership of the Imperial College John Percy Research Grou Department of Metallurgy Royal School of Mines, Imper i a 1 Co 11ege London JULY 1983 In bulk forming operations the plastic deformation is very large compared with the elastic one. This fact allows the use of constitutive laws based upon both current stress-strain rate measures. A Finite Element formulation of the mechanics involved in quasi-steady state conditions was implemented, introducing the incompressibi1ity via a penalty on the volumetric strain rate. The perfectly plastic behaviour was considered first, and the extrusion process was thoroughly analyzed. Solutions for direct and and indirect modes of operation up to very high extrusion ratios are presented. Both plane strain and axisymmetric geometries were considered. Frictional conditions at the billet-container interface were introduced in two ways. Comparison between results obtained using these lines of approach and numerical problems associated to them are examined. Pressure results are compared with previously reported solutions resulting from Slip Lines, Upper Bounds and Finite Elements. The behaviour of the internal mechanics as a function of changes of geometrical and frictional conditions are examined. Next, a non-linear flow stress resulting from hot torsion experi- ments was considered. The thermal analysis involves the solution of another equation, which is coupled with the mechanical problem due to convection, heat generation and flow stress dependence on temperature. After an appropriate transformation, that equation is formulated in a "weak" form and implemented to be solved jointly with the mechanical part. -
April 2018 the Privileged View Steve Beste, President
Volume 18 – 04 www.FlyingClub1.org April 2018 The Privileged View Steve Beste, President Maker Faire. When something breaks, who you gonna call? If it’s your car or your Cessna, you put it into the shop and get out your checkbook. But if you fly an experimental or Part 103 aircraft, it’s not so simple. That’s why - when newbies ask me about our sport - I find out if they’re tinkerers and do-it-yourselfers. If all they bring to a problem is their checkbook, then they’re better off going with the Cessna. The tinkerers and builders, though, they’re my kind of people. That’s why I went with Dick Martin to the annual Maker Faire at George Mason in mid-March. I thought we’d find a modest event that focused on 3D printing for adult hobbyists, but I was so wrong. They had 140 exhibitors filling three buildings. With several thousand visitors, there were lines outside the buildings by mid-afternoon. Yes, they had 3D printers, but so much more. And so many kids! The focus of the event was clearly on bringing young people into the world of making stuff. The cleverest exhibit was the one shown below. They set up four tables with junk computer gear and told kids to just take things apart. Don’t worry about putting anything back together again. It’s all junk! Rip it apart! See how it’s made! Do it! It looked like a destruction derby, with laptops and printers and even a sewing machine being reduced to screws and plates. -
Outer Billiards Outside a Regular Octagon: Periodicity of Almost All Orbits and Existence of an Aperiodic Orbit F
ISSN 1064-5624, Doklady Mathematics, 2018, Vol. 98, No. 1, pp. 1–4. © Pleiades Publishing, Ltd., 2018. Original Russian Text © F.D. Rukhovich, 2018, published in Doklady Akademii Nauk, 2018, Vol. 481, No. 3, pp. 00000–00000. MATHEMATICS Outer Billiards outside a Regular Octagon: Periodicity of Almost All Orbits and Existence of an Aperiodic Orbit F. D. Rukhovich Presented by Academician of the RAS A.L. Semenov November 30, 2017 Received November 30, 2017 Abstract—The existence of an aperiodic orbit for an outer billiard outside a regular octagon is proved. Addi- tionally, almost all orbits of such an outer billiard are proved to be periodic. All possible periods are explicitly listed. DOI: 10.1134/S1064562418050101 Let be a convex shape and p be a point outside it. Tabachnikov [1] has solved the periodicity prob- Consider the right (with respect to p) tangent to . lems in the case of regular 3-, 4-, and 6-gons. Specifi- LetTp T() p be the point symmetric to p with cally, there is no aperiodic point for them, as well as for respect to the tangency point. 5-gons, for which aperiodic points exist, but have Definition 1. The mapping T is called an outer bil- measure zero. In his 2005 monograph [6], Tabach- liard, and is called an outer billiard table. nikov presented computer simulation results for an octagon, but noted that a rigor analysis is still not Definition 2. A point p outside is said to be peri- available and there are no results for other cases. Later, odic if there exists a positive integer n such that a regular pentagon and associated symbolic dynamics Tpn p. -
Unit 1: Motion
Macomb Intermediate School District High School Science Power Standards Document Physics The Michigan High School Science Content Expectations establish what every student is expected to know and be able to do by the end of high school. They also outline the parameters for receiving high school credit as dictated by state law. To aid teachers and administrators in meeting these expectations the Macomb ISD has undertaken the task of identifying those content expectations which can be considered power standards. The critical characteristics1 for selecting a power standard are: • Endurance – knowledge and skills of value beyond a single test date. • Leverage - knowledge and skills that will be of value in multiple disciplines. • Readiness - knowledge and skills necessary for the next level of learning. The selection of power standards is not intended to relieve teachers of the responsibility for teaching all content expectations. Rather, it gives the school district a common focus and acts as a safety net of standards that all students must learn prior to leaving their current level. The following document utilizes the unit design including the big ideas and real world contexts, as developed in the science companion documents for the Michigan High School Science Content Expectations. 1 Dr. Douglas Reeves, Center for Performance Assessment Unit 1: Motion Big Ideas The motion of an object may be described using a) motion diagrams, b) data, c) graphs, and d) mathematical functions. Conceptual Understandings A comparison can be made of the motion of a person attempting to walk at a constant velocity down a sidewalk to the motion of a person attempting to walk in a straight line with a constant acceleration.