DOCSLIB.ORG

Rotational Dynamics

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Rotational Dynamics ROTATIONAL DYNAMICS Newton's 3rd Law – A Closer Look ● Consider 2 particles with action/reaction pair of forces: – rd = − Newton's 3 Law → F 1 on 2 F 2 on 1 2 – Puts no restrictions on the direction of F 1 on 2 r 12 ● Symmetry Considerations 1 – Attempt to write a formula for the force F 1 on 2 – No “universal” xyz directions → What can F 1 on 2 depend on? – Relative position vector r 12 and relative velocity vector v 12 only ● If F 1 on 2 depends only on r 12 mathematically: – Direction of r 12 → only “defined” direction in space – F must point in direction of r (or opposite direction) 1 on 2 12 – Forces of this type are called “central” forces Central Forces and Torque 2 ● × = Mathematical definition of central force: r 12 F 1 on 2 0 r 12 – True in every reference frame! y 1 ● Calculating in a particular reference frame S: x × r × F = 0 The quantity r i F i is called the “torque” 12 1 on 2 th ( i) on the i particle − × = r 2 r 1 F 1 on 2 0 Internal central forces produce zero net torque × − × = r 2 F 1 on 2 r 1 F 1 on 2 0 on a system × × = r 2 F 1 on 2 r 1 F 2 on 1 0 External forces and non-central internal forces can exert non-zero net torque on a system ● Examples: – Central forces: gravity, electric – Non-central force: magnetic (depends on position and velocity) Static Equilibrium ● “Equilibrium” → Fnet = 0 → aCM = 0 ● “Static Equilibrium” → ai = 0 for every particle – Examples: buildings, bridges → (not perfectly static!) – = = ∑ × = Requires: F net , external 0 and net , external r i F i 0 i – Useful for calculating structural loads and stresses ● Examples: d – 3d/4 Shelf → calculate tension in cable m 2m – Calculate force of wall on plank d – Door of width w and height h → draw direction of each Fhinge Angular Momentum ● Consider the net torque on a system of particles d p d = ∑ × = ∑ × i = ∑ r × p net r i F i net r i net i i i i dt dt i – × th r i p i is referred to as “angular momentum”( L i ) of i particle – Internal, central forces exert zero net torque – Net torque must be provided by external forces: d L = total (Similar to Newton's 2nd Law ) net , external dt – If zero net external torque → angular momentum is conserved ● Both and L depend on choice of origin – Unlike force and momentum (only depend on xyz directions) – However, the equation above is true in all reference frames Examples ● Pendulum at some instant (angle θ, speed v) θ d – Using top of string as origin: – Calculate torque and angular momentum m – Plug in to Tnet = dL/dt Repeat, using mass's lowest point as origin d/4 ● Wooden board falls off table d – Mass m, starting from rest – Using edge of table as origin: – Calculate Tnet and aright edge of board at t=0 – (Assume board stays rigid → v proportional to r) ● Why does a helicopter need a tail rotor? CM Frame of a Physical System ● Empty space → homogeneous and isotropic – All points are physically the same → no point is “special” – All directions physically the same → no direction is “special” y y' y' x' y' vCM x' x' x ● System of particles → breaks this symmetry – A “special” reference frame can be defined for the system: – Origin = CM of system (called the “CM frame”) – x'y'z' axes → defined by particle positions (“principal axes”) – Angular velocity vector CM → defined by particle velocities Angular Momentum in CM Frame N = = 1 2 ∑ 1 2 psystem M system v CM KE system M system vCM mi v ' i 2 i = 1 2 ● For a system of particles measured in the CM frame: – Total momentum must be zero, but total KE can be non-zero! ● Angular Momentum (calculated in an inertial frame S): = ∑ [ × ] = ∑ [ × ] Lsystem r i pi rCM r i ' mi vCM vi ' i i = ∑ [ × × × × ] Lsystem mi r CM v CM r i ' v CM r CM v i ' r i ' vi ' i ∑ = ∑ = mi ri ' 0 mi vi ' 0 i i = ∑ × Lsystem LCM mi r i ' v i ' “spin” angular momentum i “orbital” angular momentum (zero in CM frame) Rotating CM Frame ● “Special” reference frame for a system of particles: – Must allow for rotating axes to account for angular momentum ● Let S be an inertial (non-rotating) CM frame – Let S' be a rotating CM frame with matrix R(t) [ ] – Recall: v ' = R v − × R r = ∑ × = Lsystem ' mi r i ' vi ' 0 For some “special” i = ∑ [ × − × ] = Ω Lsystem ' mi R r i R vi R r i 0 i = ∑ [ × ] − ∑ [ × × ] = Lsystem ' mi R r i R v i mi R r i R r i 0 i i = [ − ∑ [ × × ]] = Lsystem ' R Lspin , inertial mi r i r i 0 i = ∑ [ × × ] This can be used to calculate the “special” Lspin , inertial mi r i r i i for a given system (which makes L' = 0) Inertia Tensor = ∑ [ × × ] Lspin , inertial mi r i r i i = − ∑ [ × × ] Lspin , inertial mi r i r i i − − 0 z y 0 z y x i i i i = − ∑ − − L m [ z 0 x z 0 x ] spin , inertial i i i i i y i − − yi xi 0 yi xi 0 z 2 2 − − y z x y z x x i i i i i i = ∑ − 2 2 − ≡ L [ m ] I spin , inertial i xi yi xi zi yi zi y i −z x − y z x2 y 2 i i i i i i z “Inertia Tensor” – fully describes the distribution of Reference frame for a system of mass in a system particles is almost complete: Diagonal elements are called “moments of inertia” 1) origin → CM 2) angular velocity → using L and I Off-diagonal elements are called “products of inertia” 3) need to find “principal axes” Inertia Tensor for Mass Distributions y 2 z2 −x y −z x i i i i i i ● = ∑ − 2 2 − For a system of particles: I mi xi yi xi zi yi zi i − − 2 2 zi xi yi zi xi yi ● Inertia tensor extends naturally to mass distributions: 2 2 − − y z x y z x = ∫ − 2 2 − I x , y , z dxdydz x y x z y z −z x − y z x 2 y 2 ● Example: Calculate I for a flat disk in the xy-plane M 2 2 x y R 1 2 2 M R 0 0 x , y , z = z { R 4 = 1 2 z 0 elsewhere I 0 M R 0 4 1 2 0 0 M R 2 x y Principal Axes y 2 z2 −x y −z x I I I i i i i i i xx xy xz = ∑ − 2 2 − = I mi xi yi xi zi yi zi I yx I yy I yz i − − 2 2 I I I zi xi yi zi xi yi zx zy zz ● Elements of inertia tensor depend on choice of xyz axes – Is there a “special” set of xyz axes for a given system? – Yes! Possible (but difficult) to find “principal axes” such that L products of inertia are zero → in this reference frame: 2 2 = y z 0 0 L x I xx x i i = ∑ 2 2 = I mi 0 xi zi 0 L y I yy y i 0 0 x 2 y2 = i i Lz I zz z Note: L and can point in different ● For a system which is symmetric about an axis: directions! – The symmetry axis is one of the principal axes of the system Rigid Body Rotation ● Physics definition of “rigid body” – System of particles which maintains its shape (no deformation) – i.e. velocity of particles in CM frame comes from rotation only – Notation: ω = angular velocity of rigid body (inertial CM frame) ω Fcent ● View from rotating CM frame Tension – Every particle stands still in equilibrium (I is constant) – Centrifugal forces balanced by internal stresses ● Rigid body model for solid objects → reasonably good – In reality, solids can 1) deform and 2) dissipate energy as heat – Rigid body model → not applicable to fluids, orbits, stars, etc. Example: Rotating Skew Rod ω y ● Rigid massless rod (length 2d): – Rotates as shown with mass m at each end φ x – At the instant shown: – Calculate Lsystem, using 2 different methods: – × 1) r i p i , and 2) calculate the inertia tensor – Notice L and ω don't point same direction! ● Is an external torque needed to sustain this motion? – If so, calculate it ● Use symmetry to guess the principal axes – Verify guess by calculating inertia tensor in principal frame Fixed-Axis Rotation d L ● ≠ 0 Common practice → hold ω constant (not L) → dt – If L and ω not parallel → “axle” must exert an external torque! ● Example: Rotating ceiling fan – If mass is symmetrically distributed: – L and ω parallel → fan turns smoothly – If mass distribution has asymmetry (poor alignment, etc.): – L and ω not parallel → fan wobbles ● High rotation speeds → wheels, lathes, etc. – Mounting must be able to exert external torque – Enough to handle a “tolerable” amount of asymmetry Parallel-Axis Theorem ● Often: fixed axis does not pass through CM of system – ω Example: door hinge – how to calculate Ldoor? – = × Recall Lsystem M total vCM rCM I principal – = × For rigid body rotation: v CM r CM d = × × Plug in: Lsystem M total r CM r CM I principal = × × Lsystem M total r CM r CM I principal = Lsystem I orbital I principal ● Parallel-Axis Theorem: – Let d be the perpendicular distance from CM to fixed axis = 2 = 2 I orbital , axis M d I axis I principal M d = 1 2 2 = 1 2 I M 2 d M d M 2 d door about hinges 12 door door 3 door Rotational KE ● For a rotating rigid body (in inertial frame) : = ∑ 1 2 = ∑ 1 × ⋅ × KE system mi vi mi r i r i i 2 i 2 = 1 ⋅ ∑ × × KE system [ mi r i r i ] (Using vector identity) 2 i = 1 ⋅ = 1 ⋅ KE system I L (maximum angle between and L is 90º) 2 2 (For non-rigid bodies, must also include motion of particles toward/away from CM) = 1 2 1 2 1 2 ● KE I I I In principal axis frame: principal 2 xx x 2 yy y 2 zz z ● Example: “average” frisbee toss – Estimate KEtranslation and KErotation in Joules Rolling Without Slipping ● System's v CM and : independent of each other – No relation between translational / rotational motion in general – However, by using a force → the 2 can be coupled ω v ● Example: friction contact F – Drop a ball spinning at angular velocity ω on the floor – Relative velocity of ball's surface / floor causes kinetic friction – This force has 2 effects: – 1) pushes the ball to the right (affecting v CM ) – 2) exerts a CCW torque on the ball (affecting ) ● = Rolling without slipping → condition where vCM R – vcontact=0 → fixed-axis rotation about contact pt.
Load more

Recommended publications
  • The Experimental Determination of the Moment of Inertia of a Model Airplane Michael Koken [email protected]
    The University of Akron IdeaExchange@UAkron The Dr. Gary B. and Pamela S. Williams Honors Honors Research Projects College Fall 2017 The Experimental Determination of the Moment of Inertia of a Model Airplane Michael Koken [email protected] Please take a moment to share how this work helps you through this survey. Your feedback will be important as we plan further development of our repository. Follow this and additional works at: http://ideaexchange.uakron.edu/honors_research_projects Part of the Aerospace Engineering Commons, Aviation Commons, Civil and Environmental Engineering Commons, Mechanical Engineering Commons, and the Physics Commons Recommended Citation Koken, Michael, "The Experimental Determination of the Moment of Inertia of a Model Airplane" (2017). Honors Research Projects. 585. http://ideaexchange.uakron.edu/honors_research_projects/585 This Honors Research Project is brought to you for free and open access by The Dr. Gary B. and Pamela S. Williams Honors College at IdeaExchange@UAkron, the institutional repository of The nivU ersity of Akron in Akron, Ohio, USA. It has been accepted for inclusion in Honors Research Projects by an authorized administrator of IdeaExchange@UAkron. For more information, please contact [email protected], [email protected]. 2017 THE EXPERIMENTAL DETERMINATION OF A MODEL AIRPLANE KOKEN, MICHAEL THE UNIVERSITY OF AKRON Honors Project TABLE OF CONTENTS List of Tables ................................................................................................................................................
    [Show full text]
  • 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.
    [Show full text]
  • The Stress Tensor
    An Internet Book on Fluid Dynamics The Stress Tensor The general state of stress in any homogeneous continuum, whether fluid or solid, consists of a stress acting perpendicular to any plane and two orthogonal shear stresses acting tangential to that plane. Thus, Figure 1: Differential element indicating the nine stresses. as depicted in Figure 1, there will be a similar set of three stresses acting on each of the three perpendicular planes in a three-dimensional continuum for a total of nine stresses which are most conveniently denoted by the stress tensor, σij, defined as the stress in the j direction acting on a plane normal to the i direction. Equivalently we can think of σij as the 3 × 3 matrix ⎡ ⎤ σxx σxy σxz ⎣ ⎦ σyx σyy σyz (Bha1) σzx σzy σzz in which the diagonal terms, σxx, σyy and σzz, are the normal stresses which would be equal to −p in any fluid at rest (note the change in the sign convention in which tensile normal stresses are positive whereas a positive pressure is compressive). The off-digonal terms, σij with i = j, are all shear stresses which would, of course, be zero in a fluid at rest and are proportional to the viscosity in a fluid in motion. In fact, instead of nine independent stresses there are only six because, as we shall see, the stress tensor is symmetric, specifically σij = σji. In other words the shear stress acting in the i direction on a face perpendicular to the j direction is equal to the shear stress acting in the j direction on a face perpendicular to the i direction.
    [Show full text]
  • Basic Galactic Dynamics
    5 Basic galactic dynamics 5.1 Basic laws govern galaxy structure The basic force on galaxy scale is gravitational • Why? The other long-range force is electro-magnetic force due to electric • charges: positive and negative charges are difficult to separate on galaxy scales. 5.1.1 Gravitational forces Gravitational force on a point mass M1 located at x1 from a point mass M2 located at x2 is GM M (x x ) F = 1 2 1 − 2 . 1 − x x 3 | 1 − 2| Note that force F1 and positions, x1 and x2 are vectors. Note also F2 = F1 consistent with Newton’s Third Law. Consider a stellar− system of N stars. The gravitational force on ith star is the sum of the force due to all other stars N GM M (x x ) F i j i j i = −3 . − xi xj Xj=6 i | − | The equation of motion of the ith star under mutual gravitational force is given by Newton’s second law, dv F = m a = m i i i i i dt and so N dvi GMj(xi xj) = − 3 . dt xi xj Xj=6 i | − | 5.1.2 Energy conservation Assuming M2 is fixed at the origin, and M1 on the x-axis at a distance x1 from the origin, we have m1dv1 GM1M2 = 2 dt − x1 1 2 Using 2 dv1 dx1 dv1 dv1 1 dv1 = = v1 = dt dt dx1 dx1 2 dx1 we have 2 1 dv1 GM1M2 m1 = 2 , 2 dx1 − x1 which can be integrated to give 1 2 GM1M2 m1v1 = E , 2 − x1 The first term on the lhs is the kinetic energy • The second term on the lhs is the potential energy • E is the total energy and is conserved • 5.1.3 Gravitational potential The potential energy of M1 in the potential of M2 is M Φ , − 1 × 2 where GM2 Φ2 = − x1 is the potential of M2 at a distance x1 from it.
    [Show full text]
  • Moment of Inertia
    MOMENT OF INERTIA The moment of inertia, also known as the mass moment of inertia, angular mass or rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determines the force needed for a desired acceleration. It depends on the body's mass distribution and the axis chosen, with larger moments requiring more torque to change the body's rotation rate. It is an extensive (additive) property: for a point mass the moment of inertia is simply the mass times the square of the perpendicular distance to the rotation axis. The moment of inertia of a rigid composite system is the sum of the moments of inertia of its component subsystems (all taken about the same axis). Its simplest definition is the second moment of mass with respect to distance from an axis. For bodies constrained to rotate in a plane, only their moment of inertia about an axis perpendicular to the plane, a scalar value, and matters. For bodies free to rotate in three dimensions, their moments can be described by a symmetric 3 × 3 matrix, with a set of mutually perpendicular principal axes for which this matrix is diagonal and torques around the axes act independently of each other. When a body is free to rotate around an axis, torque must be applied to change its angular momentum. The amount of torque needed to cause any given angular acceleration (the rate of change in angular velocity) is proportional to the moment of inertia of the body.
    [Show full text]
  • Moment of Inertia – Rotating Disk
    ANGULAR ACCELERATION AND MOMENT OF INERTIA Introduction Rotating an object requires that one overcomes that object’s rotational inertia, better known as its moment of inertia. For highly symmetrical cases it is possible to develop formulas for calculating an object’s moment of inertia. The purpose of this lab is to compare two of these formulas with the moment of inertia measured experimentally for a disk and a ring. Equipment Computer with Logger Pro SW Bubble Level String PASCO Rotational Apparatus + Ring Vernier Caliper Mass Set Vernier Lab Pro Interface Triple Beam Balance Smart Pulley + Photogate Rubber Band Note: If you are also going to determine the moment of inertia of the ring then you need to perform the first portion of this lab with only a single disk. If you use both disks you will not be able to invert them and add the ring to the system. The second disk does not have mounting holes for the ring. Only the disk with the step pulley has these positioning holes for the ring. Theory If a force Ft is applied tangentially to the step pulley of radius r, mounted on top of the rotating disk, the disk will experience a torque τ τ = F t r = I α Equation 1. where I is the moment of inertia for the disk and α is the angular acceleration of the disk. We will solve for the moment of inertia I by using Equation 1 and then compare it 2 to the moment of inertia calculated from the following equation: Idisk = (1/2) MdR where R is the radius of the disk and Md is the mass of the disk.
    [Show full text]
  • Center of Mass Moment of Inertia
    Lecture 19 Physics I Chapter 12 Center of Mass Moment of Inertia Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics IN THIS CHAPTER, you will start discussing rotational dynamics Today we are going to discuss: Chapter 12: Rotation about the Center of Mass: Section 12.2 (skip “Finding the CM by Integration) Rotational Kinetic Energy: Section 12.3 Moment of Inertia: Section 12.4 PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Center of Mass (CM) PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Center of Mass (CM) idea We know how to address these problems: How to describe motions like these? It is a rigid object. Translational plus rotational motion We also know how to address this motion of a single particle - kinematic equations The general motion of an object can be considered as the sum of translational motion of a certain point, plus rotational motion about that point. That point is called the center of mass point. PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Center of Mass: Definition m1r1 m2r2 m3r3 Position vector of the CM: r M m1 m2 m3 CM total mass of the system m1 m2 m3 n 1 The center of mass is the rCM miri mass-weighted center of M i1 the object Component form: m2 1 n m1 xCM mi xi r2 M i1 rCM (xCM , yCM , zCM ) n r1 1 yCM mi yi M i1 r 1 n 3 m3 zCM mi zi M i1 PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Example Center
    [Show full text]
  • Center of Mass, Torque and Rotational Inertia
    Center of Mass, Torque and Rotational Inertia Objectives 1. Define center of mass, center of gravity, torque, rotational inertia 2. Give examples of center of gravity 3. Explain how center of mass affects rotation and toppling 4. Write the equations for torque and rotational inertia 5. Explain how to change torque and rotational inertia Center of Mass/Gravity • COM – point which is at the center of the objects mass – Middle of meter stick – Center of a ball – Middle of a donut • Donuts are yummy • COG – center of object’s weight distribution – Usually the same point as center of mass Rotation and Center of Mass • Objects rotate around their center of mass – Wobbling baseball bat tossed through the air • They also balance there – Balancing a broom on your finger • Things topple over when their center of gravity is not above the base – Leaning Tower of Pisa Torque • Force that causes rotation • Equation τ = Fr τ = torque (in Nm) F = force (in N) r = arm radius (in m) • How do you increase torque – More force – Longer radius Rotational Inertia • Resistance of an object to a change in its rotational motion • Controlled by mass distribution and location of axis • Equation I = mr2 I = rotational inertia (in kgm2) m = mass (in kg) r = radius (in m) • How do you increase rotational inertia? – Even mass distribution Angular Momentum • Measure how difficult it is to start or stop a rotating object • Equations L = Iω = mvtr • It is easier to balance on a moving bicycle • Precession – Torque applied to a spinning wheel changes the direction of its angular momentum – The wheel rotates about the axis instead of toppling over Satellite Motion John Glenn 6 min • Satellites “fall around” the object they orbit • The tangential speed must be exact to match the curvature of the surface of the planet – On earth things fall 4.9 m in one second – Earth “falls” 4.9 m every 8000 m horizontally – Therefore a satellite must have a tangential speed of 8000 m/s to keep from hitting the ground – Rockets are launched vertically and rolled to horizontal during their flights Sputnik-30 sec .
    [Show full text]
  • 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 ..
    [Show full text]
  • Lecture 18, Structure of Spiral Galaxies
    GALAXIES 626 Upcoming Schedule: Today: Structure of Disk Galaxies Thursday:Structure of Ellipticals Tuesday 10th: dark matter (Emily©s paper due) Thursday 12th :stability of disks/spiral arms Tuesday 17th: Emily©s presentation (intergalactic metals) GALAXIES 626 Lecture 17: The structure of spiral galaxies NGC 2997 - a typical spiral galaxy NGC 4622 yet another spiral note how different the spiral structure can be from galaxy to galaxy Elementary properties of spiral galaxies Milky Way is a ªtypicalº spiral radius of disk = 15 Kpc thickness of disk = 300 pc Regions of a Spiral Galaxy · Disk · younger generation of stars · contains gas and dust · location of the open clusters · Where spiral arms are located · Bulge · mixture of both young and old stars · Halo · older generation of stars · contains little gas and dust · location of the globular clusters Spiral Galaxies The disk is the defining stellar component of spiral galaxies. It is the end product of the dissipation of most of the baryons, and contains almost all of the baryonic angular momentum Understanding its formation is one of the most important goals of galaxy formation theory. Out of the galaxy formation process come galactic disks with a high level of regularity in their structure and scaling laws Galaxy formation models need to understand the reasons for this regularity Spiral galaxy components 1. Stars · 200 billion stars · Age: from >10 billion years to just formed · Many stars are located in star clusters 2. Interstellar Medium · Gas between stars · Nebulae, molecular clouds, and diffuse hot and cool gas in between 3. Galactic Center ± supermassive Black Hole 4.
    [Show full text]
  • Equation of Motion for Viscous Fluids
    1 2.25 Equation of Motion for Viscous Fluids Ain A. Sonin Department of Mechanical Engineering Massachusetts Institute of Technology Cambridge, Massachusetts 02139 2001 (8th edition) Contents 1. Surface Stress …………………………………………………………. 2 2. The Stress Tensor ……………………………………………………… 3 3. Symmetry of the Stress Tensor …………………………………………8 4. Equation of Motion in terms of the Stress Tensor ………………………11 5. Stress Tensor for Newtonian Fluids …………………………………… 13 The shear stresses and ordinary viscosity …………………………. 14 The normal stresses ……………………………………………….. 15 General form of the stress tensor; the second viscosity …………… 20 6. The Navier-Stokes Equation …………………………………………… 25 7. Boundary Conditions ………………………………………………….. 26 Appendix A: Viscous Flow Equations in Cylindrical Coordinates ………… 28 ã Ain A. Sonin 2001 2 1 Surface Stress So far we have been dealing with quantities like density and velocity, which at a given instant have specific values at every point in the fluid or other continuously distributed material. The density (rv ,t) is a scalar field in the sense that it has a scalar value at every point, while the velocity v (rv ,t) is a vector field, since it has a direction as well as a magnitude at every point. Fig. 1: A surface element at a point in a continuum. The surface stress is a more complicated type of quantity. The reason for this is that one cannot talk of the stress at a point without first defining the particular surface through v that point on which the stress acts. A small fluid surface element centered at the point r is defined by its area A (the prefix indicates an infinitesimal quantity) and by its outward v v unit normal vector n .
    [Show full text]
  • 2 Review of Stress, Linear Strain and Elastic Stress- Strain Relations
    2 Review of Stress, Linear Strain and Elastic Stress- Strain Relations 2.1 Introduction In metal forming and machining processes, the work piece is subjected to external forces in order to achieve a certain desired shape. Under the action of these forces, the work piece undergoes displacements and deformation and develops internal forces. A measure of deformation is defined as strain. The intensity of internal forces is called as stress. The displacements, strains and stresses in a deformable body are interlinked. Additionally, they all depend on the geometry and material of the work piece, external forces and supports. Therefore, to estimate the external forces required for achieving the desired shape, one needs to determine the displacements, strains and stresses in the work piece. This involves solving the following set of governing equations : (i) strain-displacement relations, (ii) stress- strain relations and (iii) equations of motion. In this chapter, we develop the governing equations for the case of small deformation of linearly elastic materials. While developing these equations, we disregard the molecular structure of the material and assume the body to be a continuum. This enables us to define the displacements, strains and stresses at every point of the body. We begin our discussion on governing equations with the concept of stress at a point. Then, we carry out the analysis of stress at a point to develop the ideas of stress invariants, principal stresses, maximum shear stress, octahedral stresses and the hydrostatic and deviatoric parts of stress. These ideas will be used in the next chapter to develop the theory of plasticity.
    [Show full text]