DOCSLIB.ORG

Vector Mechanics for Engineers: Statics

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Vector Mechanics for Engineers: Statics Ninth Edition CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. 9 Distributed Forces: Lecture Notes: Moments of Inertia J. Walt Oler Texas Tech University © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Edition Ninth Vector Mechanics for Engineers: Statics Contents Introduction Sample Problem 9.6 Moments of Inertia of an Area Sample Problem 9.7 Moment of Inertia of an Area by Mohr’s Circle for Moments and Integration Products of Inertia Polar Moment of Inertia Sample Problem 9.8 Radius of Gyration of an Area Moment of Inertia of a Mass Sample Problem 9.1 Parallel Axis Theorem Sample Problem 9.2 Moment of Inertia of Thin Plates Parallel Axis Theorem Moment of Inertia of a 3D Body by Moments of Inertia of Composite Integration Areas Moment of Inertia of Common Sample Problem 9.4 Geometric Shapes Sample Problem 9.5 Sample Problem 9.12 Product of Inertia Moment of Inertia With Respect to an Principal Axes and Principal Arbitrary Axis Moments of Inertia Ellipsoid of Inertia. Principle Axes of Axes of Inertia of a Mass © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 2 Edition Ninth Vector Mechanics for Engineers: Statics Introduction • Previously considered distributed forces which were proportional to the area or volume over which they act. - The resultant was obtained by summing or integrating over the areas or volumes. - The moment of the resultant about any axis was determined by computing the first moments of the areas or volumes about that axis. • Will now consider forces which are proportional to the area or volume over which they act but also vary linearly with distance from a given axis. - It will be shown that the magnitude of the resultant depends on the first moment of the force distribution with respect to the axis. - The point of application of the resultant depends on the second moment of the distribution with respect to the axis. • Current chapter will present methods for computing the moments and products of inertia for areas and masses. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 3 Edition Ninth Vector Mechanics for Engineers: Statics Moment of Inertia of an Area • Consider distributed forces F whose magnitudes are proportional to the elemental areas A on which they act and also vary linearly with the distance of from a given axis. • Example: Consider a beam subjected to pure bending. Internal forces vary linearly with distance from the neutral axis which passes through the section centroid. F kyA R k y dA 0 y dA Qx first moment M k y2dA y2dA second moment • Example: Consider the net hydrostatic force on a submerged circular gate. F pA yA R y dA 2 M x y dA © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 4 Edition Ninth Vector Mechanics for Engineers: Statics Moment of Inertia of an Area by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, 2 2 I x y dA I y x dA • Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. • For a rectangular area, h I y2dA y2bdy 1 bh3 x 3 0 • The formula for rectangular areas may also be applied to strips parallel to the axes, dI 1 y3dx dI x2dA x2 y dx x 3 y © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 5 Edition Ninth Vector Mechanics for Engineers: Statics Polar Moment of Inertia • The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs. 2 J0 r dA • The polar moment of inertia is related to the rectangular moments of inertia, 2 2 2 2 2 J0 r dA x y dA x dA y dA I y I x © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 6 Edition Ninth Vector Mechanics for Engineers: Statics Radius of Gyration of an Area • Consider area A with moment of inertia Ix. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix. I I k 2 A k x x x x A kx = radius of gyration with respect to the x axis • Similarly, I I k 2 A k y y y y A J J k 2 A k O O O O A 2 2 2 kO kx k y © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 7 Edition Ninth Vector Mechanics for Engineers: Statics Sample Problem 9.1 SOLUTION: • A differential strip parallel to the x axis is chosen for dA. 2 dIx y dA dA l dy • For similar triangles, l h y h y h y l b dA b dy b h h h Determine the moment of inertia of a triangle with respect • Integrating dI from y = 0 to y = h, to its base. x h h 2 2 h y b 2 3 Ix y dA y b dy hy y dy 0 h h 0 3 4 h b y y bh3 h h 3 4 I x 0 12 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 8 Edition Ninth Vector Mechanics for Engineers: Statics Sample Problem 9.2 SOLUTION: • An annular differential area element is chosen, 2 dJO u dA dA 2 u du r r 2 3 JO dJO u 2 u du 2 u du 0 0 J r4 O 2 a) Determine the centroidal polar • From symmetry, I = I , moment of inertia of a circular x y area by direct integration. J I I 2I r4 2I O x y x 2 x b) Using the result of part a, determine the moment of inertia I I r4 of a circular area with respect to a diameter x 4 diameter. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 9 Edition Ninth Vector Mechanics for Engineers: Statics Parallel Axis Theorem • Consider moment of inertia I of an area A with respect to the axis AA’ I y2dA • The axis BB’ passes through the area centroid and is called a centroidal axis. I y2dA y d 2 dA y2dA 2d ydA d 2 dA I I Ad 2 parallel axis theorem © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 10 Edition Ninth Vector Mechanics for Engineers: Statics Parallel Axis Theorem • Moment of inertia IT of a circular area with respect to a tangent to the circle, I I Ad 2 1 r 4 r 2 r 2 T 4 5 r 4 4 • Moment of inertia of a triangle with respect to a centroidal axis, 2 I AA I BB Ad 2 I I Ad 2 1 bh3 1 bh 1 h BB AA 12 2 3 1 bh3 36 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 11 Edition Ninth Vector Mechanics for Engineers: Statics Moments of Inertia of Composite Areas • The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 12 Edition Ninth Vector Mechanics for Engineers: Statics Moments of Inertia of Composite Areas © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 13 Edition Ninth Vector Mechanics for Engineers: Statics Sample Problem 9.4 SOLUTION: • Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section. • Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to The strength of a W14x38 rolled steel composite section centroidal axis. beam is increased by attaching a plate • Calculate the radius of gyration from the to its upper flange. moment of inertia of the composite Determine the moment of inertia and section. radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 14 Edition Ninth Vector Mechanics for Engineers: Statics Sample Problem 9.4 SOLUTION: • Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section. Section A, in 2 y, in. yA, in3 Plate 6.75 7.425 50.12 Beam Section 11.20 0 0 A 17.95 yA 50.12 yA 50.12 in3 Y A yA Y 2.792 in. A 17.95 in 2 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 15 Edition Ninth Vector Mechanics for Engineers: Statics Sample Problem 9.4 • Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis. 2 2 Ix,beam section Ix AY 385 11.202.792 472.3 in 4 3 I I Ad 2 1 9 3 6.75 7.425 2.792 2 x,plate x 12 4 145.2 in 4 Ix Ix,beam section Ix,plate 472.3 145.2 4 Ix 618 in • Calculate the radius of gyration from the moment of inertia of the composite section. 4 Ix 617.5 in kx kx 5.87 in. A 17.95 in 2 © 2010 The McGraw-Hill Companies, Inc. All rights reserved. 9 - 16 Edition Ninth Vector Mechanics for Engineers: Statics Sample Problem 9.5 SOLUTION: • Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis.
Load more

Recommended publications
  • Vector Mechanics: Statics
    PDHOnline Course G492 (4 PDH) Vector Mechanics: Statics Mark A. Strain, P.E. 2014 PDH Online | PDH Center 5272 Meadow Estates Drive Fairfax, VA 22030-6658 Phone & Fax: 703-988-0088 www.PDHonline.org www.PDHcenter.com An Approved Continuing Education Provider www.PDHcenter.com PDHonline Course G492 www.PDHonline.org Table of Contents Introduction ..................................................................................................................................... 1 Vectors ............................................................................................................................................ 1 Vector Decomposition ................................................................................................................ 2 Components of a Vector ............................................................................................................. 2 Force ............................................................................................................................................... 4 Equilibrium ..................................................................................................................................... 5 Equilibrium of a Particle ............................................................................................................. 6 Rigid Bodies.............................................................................................................................. 10 Pulleys ......................................................................................................................................
    [Show full text]
  • Knowledge Assessment in Statics: Concepts Versus Skills
    Session 1168 Knowledge Assessment in Statics: Concepts versus skills Scott Danielson Arizona State University Abstract Following the lead of the physics community, engineering faculty have recognized the value of good assessment instruments for evaluating the learning of their students. These assessment instruments can be used to both measure student learning and to evaluate changes in teaching, i.e., did student-learning increase due different ways of teaching. As a result, there are significant efforts underway to develop engineering subject assessment tools. For instance, the Foundation Coalition is supporting assessment tool development efforts in a number of engineering subjects. These efforts have focused on developing “concept” inventories. These concept inventories focus on determining student understanding of the subject’s fundamental concepts. Separately, a NSF-supported effort to develop an assessment tool for statics was begun in the last year by the authors. As a first step, the project team analyzed prior work aimed at delineating important knowledge areas in statics. They quickly recognized that these important knowledge areas contained both conceptual and “skill” components. Both knowledge areas are described and examples of each are provided. Also, a cognitive psychology-based taxonomy of declarative and procedural knowledge is discussed in relation to determining the difference between a concept and a skill. Subsequently, the team decided to focus on development of a concept-based statics assessment tool. The ongoing Delphi process to refine the inventory of important statics concepts and validate the concepts with a broader group of subject matter experts is described. However, the value and need for a skills-based assessment tool is also recognized.
    [Show full text]
  • Classical Mechanics
    Classical Mechanics Hyoungsoon Choi Spring, 2014 Contents 1 Introduction4 1.1 Kinematics and Kinetics . .5 1.2 Kinematics: Watching Wallace and Gromit ............6 1.3 Inertia and Inertial Frame . .8 2 Newton's Laws of Motion 10 2.1 The First Law: The Law of Inertia . 10 2.2 The Second Law: The Equation of Motion . 11 2.3 The Third Law: The Law of Action and Reaction . 12 3 Laws of Conservation 14 3.1 Conservation of Momentum . 14 3.2 Conservation of Angular Momentum . 15 3.3 Conservation of Energy . 17 3.3.1 Kinetic energy . 17 3.3.2 Potential energy . 18 3.3.3 Mechanical energy conservation . 19 4 Solving Equation of Motions 20 4.1 Force-Free Motion . 21 4.2 Constant Force Motion . 22 4.2.1 Constant force motion in one dimension . 22 4.2.2 Constant force motion in two dimensions . 23 4.3 Varying Force Motion . 25 4.3.1 Drag force . 25 4.3.2 Harmonic oscillator . 29 5 Lagrangian Mechanics 30 5.1 Configuration Space . 30 5.2 Lagrangian Equations of Motion . 32 5.3 Generalized Coordinates . 34 5.4 Lagrangian Mechanics . 36 5.5 D'Alembert's Principle . 37 5.6 Conjugate Variables . 39 1 CONTENTS 2 6 Hamiltonian Mechanics 40 6.1 Legendre Transformation: From Lagrangian to Hamiltonian . 40 6.2 Hamilton's Equations . 41 6.3 Configuration Space and Phase Space . 43 6.4 Hamiltonian and Energy . 45 7 Central Force Motion 47 7.1 Conservation Laws in Central Force Field . 47 7.2 The Path Equation .
    [Show full text]
  • Exercises in Classical Mechanics 1 Moments of Inertia 2 Half-Cylinder
    Exercises in Classical Mechanics CUNY GC, Prof. D. Garanin No.1 Solution ||||||||||||||||||||||||||||||||||||||||| 1 Moments of inertia (5 points) Calculate tensors of inertia with respect to the principal axes of the following bodies: a) Hollow sphere of mass M and radius R: b) Cone of the height h and radius of the base R; both with respect to the apex and to the center of mass. c) Body of a box shape with sides a; b; and c: Solution: a) By symmetry for the sphere I®¯ = I±®¯: (1) One can ¯nd I easily noticing that for the hollow sphere is 1 1 X ³ ´ I = (I + I + I ) = m y2 + z2 + z2 + x2 + x2 + y2 3 xx yy zz 3 i i i i i i i i 2 X 2 X 2 = m r2 = R2 m = MR2: (2) 3 i i 3 i 3 i i b) and c): Standard solutions 2 Half-cylinder ϕ (10 points) Consider a half-cylinder of mass M and radius R on a horizontal plane. a) Find the position of its center of mass (CM) and the moment of inertia with respect to CM. b) Write down the Lagrange function in terms of the angle ' (see Fig.) c) Find the frequency of cylinder's oscillations in the linear regime, ' ¿ 1. Solution: (a) First we ¯nd the distance a between the CM and the geometrical center of the cylinder. With σ being the density for the cross-sectional surface, so that ¼R2 M = σ ; (3) 2 1 one obtains Z Z 2 2σ R p σ R q a = dx x R2 ¡ x2 = dy R2 ¡ y M 0 M 0 ¯ 2 ³ ´3=2¯R σ 2 2 ¯ σ 2 3 4 = ¡ R ¡ y ¯ = R = R: (4) M 3 0 M 3 3¼ The moment of inertia of the half-cylinder with respect to the geometrical center is the same as that of the cylinder, 1 I0 = MR2: (5) 2 The moment of inertia with respect to the CM I can be found from the relation I0 = I + Ma2 (6) that yields " µ ¶ # " # 1 4 2 1 32 I = I0 ¡ Ma2 = MR2 ¡ = MR2 1 ¡ ' 0:3199MR2: (7) 2 3¼ 2 (3¼)2 (b) The Lagrange function L is given by L('; '_ ) = T ('; '_ ) ¡ U('): (8) The potential energy U of the half-cylinder is due to the elevation of its CM resulting from the deviation of ' from zero.
    [Show full text]
  • Post-Newtonian Approximation
    Post-Newtonian gravity and gravitational-wave astronomy Polarization waveforms in the SSB reference frame Relativistic binary systems Effective one-body formalism Post-Newtonian Approximation Piotr Jaranowski Faculty of Physcis, University of Bia lystok,Poland 01.07.2013 P. Jaranowski School of Gravitational Waves, 01{05.07.2013, Warsaw Post-Newtonian gravity and gravitational-wave astronomy Polarization waveforms in the SSB reference frame Relativistic binary systems Effective one-body formalism 1 Post-Newtonian gravity and gravitational-wave astronomy 2 Polarization waveforms in the SSB reference frame 3 Relativistic binary systems Leading-order waveforms (Newtonian binary dynamics) Leading-order waveforms without radiation-reaction effects Leading-order waveforms with radiation-reaction effects Post-Newtonian corrections Post-Newtonian spin-dependent effects 4 Effective one-body formalism EOB-improved 3PN-accurate Hamiltonian Usage of Pad´eapproximants EOB flexibility parameters P. Jaranowski School of Gravitational Waves, 01{05.07.2013, Warsaw Post-Newtonian gravity and gravitational-wave astronomy Polarization waveforms in the SSB reference frame Relativistic binary systems Effective one-body formalism 1 Post-Newtonian gravity and gravitational-wave astronomy 2 Polarization waveforms in the SSB reference frame 3 Relativistic binary systems Leading-order waveforms (Newtonian binary dynamics) Leading-order waveforms without radiation-reaction effects Leading-order waveforms with radiation-reaction effects Post-Newtonian corrections Post-Newtonian spin-dependent effects 4 Effective one-body formalism EOB-improved 3PN-accurate Hamiltonian Usage of Pad´eapproximants EOB flexibility parameters P. Jaranowski School of Gravitational Waves, 01{05.07.2013, Warsaw Relativistic binary systems exist in nature, they comprise compact objects: neutron stars or black holes. These systems emit gravitational waves, which experimenters try to detect within the LIGO/VIRGO/GEO600 projects.
    [Show full text]
  • Kinematics, Kinetics, Dynamics, Inertia Kinematics Is a Branch of Classical
    Course: PGPathshala-Biophysics Paper 1:Foundations of Bio-Physics Module 22: Kinematics, kinetics, dynamics, inertia Kinematics is a branch of classical mechanics in physics in which one learns about the properties of motion such as position, velocity, acceleration, etc. of bodies or particles without considering the causes or the driving forces behindthem. Since in kinematics, we are interested in study of pure motion only, we generally ignore the dimensions, shapes or sizes of objects under study and hence treat them like point objects. On the other hand, kinetics deals with physics of the states of the system, causes of motion or the driving forces and reactions of the system. In a particular state, namely, the equilibrium state of the system under study, the systems can either be at rest or moving with a time-dependent velocity. The kinetics of the prior, i.e., of a system at rest is studied in statics. Similarly, the kinetics of the later, i.e., of a body moving with a velocity is called dynamics. Introduction Classical mechanics describes the area of physics most familiar to us that of the motion of macroscopic objects, from football to planets and car race to falling from space. NASCAR engineers are great fanatics of this branch of physics because they have to determine performance of cars before races. Geologists use kinematics, a subarea of classical mechanics, to study ‘tectonic-plate’ motion. Even medical researchers require this physics to map the blood flow through a patient when diagnosing partially closed artery. And so on. These are just a few of the examples which are more than enough to tell us about the importance of the science of motion.
    [Show full text]
  • New Rotational Dynamics- Inertia-Torque Principle and the Force Moment the Character of Statics Guagsan Yu* Harbin Macro, Dynamics Institute, P
    Computa & tio d n ie a l l p M Yu, J Appl Computat Math 2015, 4:3 p a Journal of A t h f e o m l DOI: 10.4172/2168-9679.1000222 a a n t r ISSN: 2168-9679i c u s o J Applied & Computational Mathematics Research Article Open Access New Rotational Dynamics- Inertia-Torque Principle and the Force Moment the Character of Statics GuagSan Yu* Harbin Macro, Dynamics Institute, P. R. China Abstract Textual point of view, generate in a series of rotational dynamics experiment. Initial research, is wish find a method overcome the momentum conservation. But further study, again detected inside the classical mechanics, the error of the principle of force moment. Then a series of, momentous the error of inside classical theory, all discover come out. The momentum conservation law is wrong; the newton third law is wrong; the energy conservation law is also can surpass. After redress these error, the new theory namely perforce bring. This will involve the classical physics and mechanics the foundation fraction, textbooks of physics foundation part should proceed the grand modification. Keywords: Rigid body; Inertia torque; Centroid moment; Centroid occurrence change? The Inertia-torque, namely, such as Figure 3 the arm; Statics; Static force; Dynamics; Conservation law show, the particle m (the m is also its mass) is a r 1 to O the slewing radius, so it the Inertia-torque is: Introduction I = m.r1 (1.1) Textual argumentation is to bases on the several simple physics experiment, these experiments pass two videos the document to The Inertia-torque is similar with moment of force, to the same of proceed to demonstrate.
    [Show full text]
  • 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 25, 2003 revised November 29 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. The inertia tensor Consider a rigid body that is composed of N particles. Give each particle an index α = 1 : : : N. Particle α has a velocity vf,α measured in the fixed reference frame and velocity vr,α = drα=dt in the reference frame that rotates about axis !~ . Then according to Eq. 10.17: vf,α = V + vr,α + !~ × rα where V = velocity of the moving origin relative to the fixed origin. Note that the particles in a rigid body have zero relative velocity, ie, vr,α = 0. 1 Thus vα = V + !~ × rα after dropping the f subscript 1 1 and T = m v2 = m (V + !~ × r )2 is the particle's KE α 2 α α 2 α α N so T = Tα the system's total KE is α=1 X1 1 = MV2 + m V · (!~ × r ) + m (!~ × r )2 2 α α 2 α α α α X X 1 1 = MV2 + V · !~ × m r + m (!~ × r )2 2 α α 2 α α α ! α X X where M = α mα = the system's total mass. P Now choose the system's center of mass R as the origin of the moving frame.
    [Show full text]
  • A Short History of Physics (Pdf)
    A Short History of Physics Bernd A. Berg Florida State University PHY 1090 FSU August 28, 2012. References: Most of the following is copied from Wikepedia. Bernd Berg () History Physics FSU August 28, 2012. 1 / 25 Introduction Philosophy and Religion aim at Fundamental Truths. It is my believe that the secured part of this is in Physics. This happend by Trial and Error over more than 2,500 years and became systematic Theory and Observation only in the last 500 years. This talk collects important events of this time period and attaches them to the names of some people. I can only give an inadequate presentation of the complex process of scientific progress. The hope is that the flavor get over. Bernd Berg () History Physics FSU August 28, 2012. 2 / 25 Physics From Acient Greek: \Nature". Broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves. The universe is commonly defined as the totality of everything that exists or is known to exist. In many ways, physics stems from acient greek philosophy and was known as \natural philosophy" until the late 18th century. Bernd Berg () History Physics FSU August 28, 2012. 3 / 25 Ancient Physics: Remarkable people and ideas. Pythagoras (ca. 570{490 BC): a2 + b2 = c2 for rectangular triangle: Leucippus (early 5th century BC) opposed the idea of direct devine intervention in the universe. He and his student Democritus were the first to develop a theory of atomism. Plato (424/424{348/347) is said that to have disliked Democritus so much, that he wished his books burned.
    [Show full text]
  • NEWTON's FIRST LAW of MOTION the Law of INERTIA
    NEWTON’S FIRST LAW OF MOTION The law of INERTIA INERTIA - describes how much an object tries to resist a change in motion (resist a force) inertia increases with mass INERTIA EXAMPLES Ex- takes more force to stop a bowling ball than a basket ball going the same speed (bowling ball tries more to keep going at a constant speed- resists stopping) Ex- it takes more force to lift a full backpack than an empty one (full backpack tries to stay still more- resists moving ; it has more inertia) What the law of inertia says…. AKA Newton’s First Law of Motion An object at rest, stays at rest Because the net force is zero! Forces are balanced (equal & opposite) …an object in motion, stays in motion Motion means going at a constant speed & in a straight line The net force is zero on this object too! NO ACCELERATION! …unless acted upon by a net force. Only when this happens will you get acceleration! (speed or direction changes) Forces are unbalanced now- not all equal and opposite there is a net force> 0N! You have overcome the object’s inertia! How does a traveling object move once all the forces on it are balanced? at a constant speed in a straight line Why don’t things just move at a constant speed in a straight line forever then on Earth? FRICTION! Opposes the motion & slows things down or GRAVITY if motion is in up/down direction These make the forces unbalanced on the object & cause acceleration We can’t get away from those forces on Earth! (we can encounter some situations where they are negligible though!) HOW can you make a Bowling
    [Show full text]
  • Newtonian Mechanics Is Most Straightforward in Its Formulation and Is Based on Newton’S Second Law
    CLASSICAL MECHANICS D. A. Garanin September 30, 2015 1 Introduction Mechanics is part of physics studying motion of material bodies or conditions of their equilibrium. The latter is the subject of statics that is important in engineering. General properties of motion of bodies regardless of the source of motion (in particular, the role of constraints) belong to kinematics. Finally, motion caused by forces or interactions is the subject of dynamics, the biggest and most important part of mechanics. Concerning systems studied, mechanics can be divided into mechanics of material points, mechanics of rigid bodies, mechanics of elastic bodies, and mechanics of fluids: hydro- and aerodynamics. At the core of each of these areas of mechanics is the equation of motion, Newton's second law. Mechanics of material points is described by ordinary differential equations (ODE). One can distinguish between mechanics of one or few bodies and mechanics of many-body systems. Mechanics of rigid bodies is also described by ordinary differential equations, including positions and velocities of their centers and the angles defining their orientation. Mechanics of elastic bodies and fluids (that is, mechanics of continuum) is more compli- cated and described by partial differential equation. In many cases mechanics of continuum is coupled to thermodynamics, especially in aerodynamics. The subject of this course are systems described by ODE, including particles and rigid bodies. There are two limitations on classical mechanics. First, speeds of the objects should be much smaller than the speed of light, v c, otherwise it becomes relativistic mechanics. Second, the bodies should have a sufficiently large mass and/or kinetic energy.
    [Show full text]
  • RELATIVISTIC GRAVITY and the ORIGIN of INERTIA and INERTIAL MASS K Tsarouchas
    RELATIVISTIC GRAVITY AND THE ORIGIN OF INERTIA AND INERTIAL MASS K Tsarouchas To cite this version: K Tsarouchas. RELATIVISTIC GRAVITY AND THE ORIGIN OF INERTIA AND INERTIAL MASS. 2021. hal-01474982v5 HAL Id: hal-01474982 https://hal.archives-ouvertes.fr/hal-01474982v5 Preprint submitted on 3 Feb 2021 (v5), last revised 11 Jul 2021 (v6) HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Distributed under a Creative Commons Attribution| 4.0 International License Relativistic Gravity and the Origin of Inertia and Inertial Mass K. I. Tsarouchas School of Mechanical Engineering National Technical University of Athens, Greece E-mail-1: [email protected] - E-mail-2: [email protected] Abstract If equilibrium is to be a frame-independent condition, it is necessary the gravitational force to have precisely the same transformation law as that of the Lorentz-force. Therefore, gravity should be described by a gravitomagnetic theory with equations which have the same mathematical form as those of the electromagnetic theory, with the gravitational mass as a Lorentz invariant. Using this gravitomagnetic theory, in order to ensure the relativity of all kinds of translatory motion, we accept the principle of covariance and the equivalence principle and thus we prove that, 1.
    [Show full text]