mechanics_chapter8
流体力学第八章(Fluid Mechanics)
2nd – Year Fluid Mechanics, Faculty of Engineering and Computing, Curtin University FLUID MECHANICS 230For Second-Year Chemical, Civil and Mechanical EngineeringFLUID MECHANICS LECTURE NOTESCHAPTER 8 FLOW OVER IMMERSED BODY8.1 IntroductionWe have discussed pipe flow in Chapter 6, which is internal flow because the fluid is confined within well-defined boundaries. This chapter deals with external flows, i.e. flows over bodies immersed in the fluid. Typical examples of external flows include the flow of water over a submarine (Figure 8-1a) or a fish (Figure 8-1b), the flow of air over an aircraft (Figure 8-1c).(a)(b) (c)Figure 8-1 Examples; a) submarine; b) fish; c) aircraft;The fluid forces (drag and lift) on the immersed bodies are of important considerations in practice. In this chapter, we will learn the fundamentals of drag and lift, as well as the methods for determining and optimising these forces in engineering applications.8.2 Drag and liftAs shown in Figure 8-2, the forces exerted on the surface of an aerofoil by the fluid include the pressure force (Figure 8-2a) and the viscous force (Figure 8-2b). The results of these forces are the net drag force F D and lift force F L (Figure 8-2c). Taking a small element dA on the aerofoil surface in Figure 8-2c, we can decompose the pressure force (PdA ) and viscous force (dA w τ) into x-direction force dF x and y-direction force dF y , as shown in Figure 8-2d,θτθsin )(cos )(dA PdA dF w x += θτθcos )(sin )(dA PdA dF w y +−=.Integrating dF x and dF y along the surface, we can calculate the drag and lift forces as, dA dA P dF F w x D ∫∫∫+==θτθsin cos (E8-1) ∫∫∫+−==dA dA P dF F w y L θτθcos sin . (E8-2)Equations E8-1 and E8-2 reveals that 1) both shear stress and pressure contribute to the drag and lift. In the case of drag, the former is called friction drag ; the later is called pressure drag ;2) to calculate drag and lift, sufficient knowledge are required on three aspects: the body shape as it determines the distribution of θ; the distribution of pressure P and the distribution of shear stress w τ along the surface.Practically, it is very difficult to obtain the distribution of pressure and shear stress. Therefore, E8-1 and E8-2 are generally not very useful. To make it easy for engineers, dimensionless coefficients are often used instead. These dimensionless numbers are called drag coefficient C D and lift coefficient C L , defined as A V F C D D 22ρ= (E8-3) A V F C L L 22ρ= (E8-4) where A is projected (or frontal) area, i.e. the project area of the body in the flow direction.Figure 8-2 Forces on an aerofoil [1]: a) pressure force; b) viscous force; c). resultant drag andlift; d) pressure and viscous forces on a small surface area dA in (c), plotted in an x-y coordination system.8.3 The boundary layer8.3.1 Hydraulician and hydrodynamicistFor a fundamental understanding on the cause of drag and lift, we must have a good understanding on the boundary layer concept. Let’s start with some historical background on two groups of fluid mechanicians who developed two different approaches in dealing with problems of fluid mechanics by late 19th century.One is the group of hydraulicians, focused on experiments and attempted to generalise useful design equations from experimental data. This group developed the filed of experimental hydraulics, delivering empirical solutions with little theoretical content. The other is the group of hydrodynamicists, who focused on differential equations describing flows and tried to apply them to practical problems. In order to solve these differential equations, the fluid was assumed to have zero viscosity and constant density. This group developed the filed of theoretical hydrodynamics, seeking pure theoretical solutions based on ideal-fluid flows.The ideal-fluid solutions of hydrodynamicists agreed well with the observations of flows did not involve solid surface, e.g. tides, however did not agree with observed behaviours in the problems that concerned the hydraulicians, e.g, flow over immersed bodies. The following are two typical examples.(1) Flow over a thin plateFigure 8-3 illustrates the differences between the ideal solutions by the hydrodynamicists and the experimental observations by the hydraulicians for flow over a thin plate. For an ideal-fluid flow (Figure 8-3a), the fluid is inviscid, there is no friction between the fluid and the surface of the thin plate. Therefore, the fluid will maintain its free stream velocity when it flows over the thin plate. However, for a real-fluid flow, the interaction between the viscous fluid with the surface leads to a velocity gradient near the surface of the plate although the fluid far away from the plate still maintains its free stream velocity.(2) Flow over a circular cylinderFigure 8-5 shows the patterns of ideal and real flow over a circular cylinder. In absence of viscous effects, the wall shear stress is zero therefore the streamlines are symmetrical (see Figure 8-6a). The fluid coasts from the front (point A), pasts the top (point C) and then reaches to the rear (point F) of the cylinder.However, the experimental observations by the hydraulicians are very different from the above analysis by the hydrodynamicists, as shown in Figure 8-4b. A real-fluid flow cannot coast along the cylinder surface down to point F. Instead, flow separation occurs from the surface at a point between point C and F, leading to the formation of turbulent wake in the downstream (see Figure 8-4b). The friction between the viscous fluid and the cylinder surface leads to inevitable energy loss so that the flow does not have sufficient kinetic energy to travel along the surface down to the rear of the cylinder.From the above examples, it seems that the hydrodynamicists calculated what can NOT be observed while the hydraulicians observed what can NOT be calculated [2]. By early 20th century, the hydraulicians and hydrodynamicists had completely gone to live in their own worlds. The hydraulicians continued to solve their problems by trial and error based on experiments while the hydrodynamicists kept publishing academic papers based on mathematics with little bearing on engineering problems.(a) Ideal flow over a thin plate(b) Real flow over a thin plateFigure 8-3 Ideal (a) and real (b) flow over a thin plate(a) Ideal flow a cylinder (b) Real turbulent flow over a cylinderFigure 8-4 Ideal (a) and real (b) flow over a cylinder8.3.2 The boundary layer conceptThe revolutionary thinking was finally due in 1904. The two fields of theoretical hydrodynamics and experimental hydraulics were united by a German professor, Ludwig Prandtl (1857-1953). Prandtl proposed the concept of the boundary layer , which has brought the hydrodynamics and hydraulics together and laid the foundation of modern fluid mechanics.F A CTurbulentwakeThe key concept of the boundary layer is to conceptually divide the flow into two regions: •The boundary layer, the region close to solid surfaces, the effects of viscosity are too large to be ignored. Within the boundary layer, ideal-fluid flow is unsatisfactory and a set of boundary layer equations should be used.•Free stream region, the region outside the boundary layer, the effects of viscosity is small and can be neglected. In this region, Ideal-fluid flow is satisfactory.At the edge of boundary layer, the pressures and velocities of the two regions should be matched. Prandtl arbitrarily suggested the boundary layer be considered that region in which the x component of the velocity, v, is less than 0.99 times of the free-stream velocity, V, as shown in Figure 8-5.Figure 8-5 Thickness of the boundary layerThe introduction of the boundary layer concept is revolutionary in seamlessly uniting the fields of theoretical hydrodynamics and experimental hydraulics. The boundary layer concept is commonly accepted as the foundation of modern fluid mechanics, because it has •clarified numerous unexplained phenomena;•provided a much better intellectual basis for discussing complicated flows;•become a standard idea in minds of fluid mechanicians;•brought analogous ideas in heat and mass transfer, generally with useful results. However, it should be noted that•the division of the flow filed by the boundary layer concept does NOT correspond any physically obvious boundary;•the edge of boundary layer does NOT correspond to any sudden change in the flow but rather corresponds to an arbitrary definition;•even with this simplification the calculation is still difficult, and in general only approximate mathematical solutions are possible.Let’s use the boundary layer concept to explain the patterns of flow over a thin plate and flow over a circular cylinder at various Reynolds’ numbers.(1). Flow over a thin plate (Figure 8-6) [1]At a very low Reynolds number, e.g. Re = 0.1 (Figure 8-6a), the viscous force is more important than the inertial force. The viscous effects are therefore strong and the plate affects the flow considerably in a wide range (see the grey area) in all directions. Consequently, there is an extensively wide range around the plate in which the streamlines deflected considerably. The boundary layer is very thick. Outside the boundary layer is the free stream region where viscous effects are no long important so that the ideal-flow solutions is applicable.At a moderate Reynolds number, e.g. Re = 10 (Figure 8-6b), the region in which the viscous effects are important becomes much smaller. The boundary layer is much thinner. The streamlines over the plate only deflected somewhat.At a large Reynolds number, e.g. Re = 107 (Figure 8-6c), the flow is dominated by inertial force. The viscous effects are negligible anywhere except in a thin boundary layer close to the surface of the plate and the wake region. As the boundary layer is very thin, the flow streamlines are largely unaffected except slightly deflected near and within the boundary layer.Figure 8-6 Patterns of flow over a thin plate [1] at (a) a low Reynolds number; (b) a moderate Reynolds number and (c) a large Reynolds numberFigure 8-7 Patterns of flows over a circular cylinder [1] at (a) a low Reynolds number; (b) a moderate Reynolds number and (c) a large Reynolds number2). Flow over a circular cylinder (Figure 8-7) [1]At a low Reynolds number, e.g. Re = 0.1 (Figure 8-7a), flow past a cylinder is dominated by the viscous force. The viscous effects influence a large portion of the flow field, stretching to several diameters in any direction of the cylinder. However, the flow can still coast along the surface of the cylinder slowly. Such flow is sometime called creeping flow, which has streamlines essentially symmetric about the centre of the cylinder.At a moderate Reynolds number, e.g. Re = 50 (Figure 8-7b), the inertial force becomes more important. The region ahead of the cylinder in which the viscous effects are important becomes much smaller. The viscous effects are convected downstream and the flow loses its symmetry. The flow inertia dominates so that it does not coast along the surface down to the rear of the body, resulting in the formation of flow separation bubbles behind the cylinder.At a large Reynolds number, e.g. Re = 105 (Figure 8-6c), the flow is dominated by the inertialforce. The region affected by the viscous forces is forced further downstream, leading to theformation of a very thin boundary layer on the front portion of the cylinder. The boundary lay can be laminar or turbulent, depending on Reynolds number. Due to the strong inertia force, the boundary layer separation occurs from the cylinder, leading to the formation of a turbulent wake region extending far downstream. In the free stream region outside the boundary layer and the wake region, the velocity gradient is zero and the fluid flows as if it were inviscid.8.4 Drag force and streamlined bodies8.4.1 Drag forceIn 1710, Isaac Newton (cited in [2]) dropped hollow spheres from the inside of the dome of St Paul’s Cathedral in London and measured their rate of fall. He calculated that the drag force F D on a sphere and concluded that the following equation holds 242222V D V A F D ρπρ== where A is called projected (or frontal) area – the projected area seen by a person lookingtoward the object from a direction parallel to the upstream (see Figure 8-8).Figure 8-8 Drag force on a free-fall sphereIn other words, Isaac Newton thought that 122=V A F D ρ. However, subsequent experimental investigations by many other researchers found that the above formula must be modified by a coefficient C D in the right-hand side, i.e. D D C V A F =22ρ This is essentially same as the definition of drag coefficient C D in E8-3. Normally, drag coefficient C D is not 1. We can calculate the drag force of the flow on an immersed body as (E8-5)Therefore, when calculate drag force using E8-5, we have accumulated the effects of all rest factors into a single coefficient, C D , i.e. drag coefficient. One can see that the drag coefficient C D in E8-5 plays a similar role of the friction factor f in E6-22. The key difference between E8-5 and E6-22 is that in the case of pipe flow, the geometry of pipes (with different length and diameter) varies while all spheres have the same shape.8.4.2 Drag coefficient chartAccording to E8-1 and the discussion in Section 8-2, the drag coefficient of an object will be a function of Reynolds number, the shape and the surface properties of the object. We haverties)face_prope ,shape,sur Φ(C D Re = (E8-6)Practically, finding the exact function of E8-6 is extremely difficult, if not impossible. Therefore, for engineering applications, what we need is a drag coefficient chart, similar to Moody chart used for determining friction factor in pipe flows. Figure 8-9 shows the dragcoefficient as a function of Reynolds number for a smooth sphere and a smooth cylinder.Figure 8-9 Drag coefficient of a smooth sphere and a smooth circular cylinder [1].EX8-1: An exampleWe caught two breams in two consecutive casts. If we pulled the strings at the same speed, explain why we felt it was much harder to pull the big bream?Solutions:Figure 8-10The projected area of the big bream, A bf , is a much bigger than that of the small bream A sf . As both fishes are bream, we can reasonable assume that the two fishes have similar shape and surface characteristics. As both fishes were caught in consecutive casts in the same water area and we pulled the string at the same speed V , the water flows over the two fishes have same Reynolds numbers. Therefore, the drag coefficients can be taken as same. According to E8-5, the water would have induced much higher drag force on the big bream when we pulled the string.8.4.3 Pressure dragEquation E8-1 in Section 8.2 clearly shows that the drag consists of two components, friction drag and pressure drag. For flow over an immersed circular cylinder, as shown in Figure 8-7c, the fluid friction within the boundary layer certainly leads to the friction drag. In this section, let’s have a detailed analysis on the cause of pressure drag.Let’s start with inviscid flows, as shown in Figure 8-11. In absence of viscosity, the flow will coast along the cylinder surface and the streamlines are symmetrical. Based on theoretical hydrodynamics (further details can be found in Chapter 6 of Reference [1]), the distribution of pressure and velocity along the surface of the cylinder are )sin 41(21220θρ−+=V P P (E8-7) θsin 2V V fs = (E8-8) which are plotted in Figures 8-11b and 8-11c.Figure 8-11c indicates that for an ideal fluid (0=μ and ρ is constant), the fluid velocity along the surface varies from 0=fs V at the very front and rear (stagnation point A and F) of the cylinder to the maximum of V V fs 2= at the bop (point C) and bottom of the cylinder.Figure 8-11 Ideal flow over a circular cylinder [1] (a) streamlines for the ideal flow; (b) pressure distribution of the ideal flow; (c) fluid velocity distribution on the cylinder surfaceFigure 8-12 Real flow over a circular cylinder [1] (a) boundary layer separation; (b)distribution of pressure coefficient.Similarly, as shown in Figure 8-11b, the pressure distribution is also symmetrical about thevertical midplane of the cylinder, varying from a minimum of 2023V P ρ−at the top orbottom of the cylinder to a maximum of 2021V P ρ+. The decrease in pressure in thedirection of flow along the front half of the cylinder is termed as favourable pressure gradientwhile the increase in pressure in the direction of flow along the rear half of the cylinder istermed adverse pressure gradient . In absence of viscous effects, the fluid travelling from thefront to the back of the cylinder coasts down the “pressure hill” (see Figure 8-5b) from°=0θat point A to °=90θat point C and then back up to the hill to °=180θ(from point C topoint F ). Therefore, there is only energy exchange between kinetic energy and pressureenergy without any energy loss.However, the experimental observations are very different from the above theoreticalanalysis, as shown in Figure 8-12. At a large Reynolds number, the flow forms the boundarylayer on the surface of the cylinder and cannot coast along the cylinder surface down to therear stagnation point F. The boundary layer separation occurs at point D on the rear surface, (a)(b)leading to the formation of turbulent wakes in the downstream (see Figure 8-12a). The separation of the boundary layer can be explained by the pressure distribution in Figure 12b. Due to the friction between the viscous fluid and the cylinder surface, energy loss is inevitable so that after pass through point C, the fluid does not have enough kinetic energy to climb the pressure hill up to point F which sits on the top of pressure hill therefore the boundary layer separates at point D in Figure 12a.Figure 8-12b also indicates that the location of separation therefore the width of the turbulent wake and the pressure distribution on the surface depend on the nature of the boundary layer. Compared with a laminar boundary layer, a turbulent boundary layer has more kinetic energy and momentum so that it can flow further around the cylinder, resulting in a narrower wake, less drag, corresponding to a decrease in drag coefficient from point D to E in Figure 8-10. Because of the boundary layer separation, the average pressure on the front half of the cylinder is significantly greater than that on the rear half. This leads to the development of a large pressure drag. Under turbulent conditions, the friction drag is insignificant compared with the pressure drag, as discussed blow.Figure 8-13 Two objects of significant different size that have the same drag force [1]: (a) a circular cylinder - a blunt body, C D = 1.2; (b) a streamlined strut C D = 0.12.Table 8-1 Drag coefficients of various bodies8.4.4 Streamlined bodiesIt is clear now that the drag force developed on an object immersed in turbulent fluid isdominantly pressure drag as a result of boundary layer separation. Therefore, we can optimisethe body shape of the object to minimise the boundary layer separation hence reduce pressuredrag. This requires us to design streamlined bodies .Figure 8-13 shows the significance of body streamlining. The streamlined strut has a sizemuch bigger than the circular cylinder. However, the two bodies have the same drag forces.The boundary layer separation on the streamlined strut has been postponed to the tail of thebody so that pressure drag is minimised compared with that on a circular cylinder, which is ablunt body. It is interesting to revisit the shapes of submarine, fish and aerofoil in Figures 8-1and 8-2, we would appreciate that these bodies are actually all streamlined.8.4.5 Drag coefficient for various objectsThe drag coefficient information for a wide range of objects is available in the literature [1,2].Some of this information is listed in Table 8-1.8.4.6 Drag coefficient at low Reynolds number (Re < 1)At Re < 1, the flow over an immersed body is called creeping flow , dominated by viscouseffects. The drag coefficient gives the 1/Re dependence. Table 8-2 shows the drag coefficientsfor various bodies at Re < 1.Table 8-2 Drag coefficient of various bodies at Re < 1, from Reference [1]8.6 Terminal velocityIf an object in a body of fluid is free to move subject only to gravity, or perhaps centrifugalforces in certain circumstances, then it accelerates to a particular velocity at which the tractiondeveloped and the other forces on the body balance. This velocity is called terminal velocity .Generally, the acceleration is very rapid so that the whole process can be treated as the objecttravelling at the terminal velocity.Figure 8-14a illustrates the concept of terminal velocity of a sphere settling in a fluid. Figure8-14b shows the force analysis of the sphere. When the sphere travels at the terminal velocityV t , the force is balanced. V V V V Re4.20Re6.13Re0.24Re2.22Therefore, as shown in Figure 8-14b, the gravity force F W , drag force F D and buoyancy forceF B are in balance,B D W F F F += (E8-9)We know thatobject object W gV F ρ= (E8-10)object fluid B gV F ρ= (E8-11) projected t fluid D D A V C F 22ρ= (E8-12) where6)2(3433D D V object ππ== (E8-13)42D A projected π=. (E8-14)Assuming Re < 1, for a sphere (see Table 8-2), the drag coefficient is DV C t fluid fluid D ρμ24Re 24==. (E8-15)Substituting E8-15 into E8-12, we have t fluid t fluid t fluid fluid projected t fluid D D DV D V D V A V C F πμπρρμρ342242222===. (E8-16)This is the famous Stock’s law, which is only valid when Re < 1.(a) Concept of terminal velocity (b) Force analysisFigure 8-14 Terminal velocity of a sphereSubstituting E8-11, E8-13 and E8-16 into E8-9, we have fluidfluid object t gD V μρρ18)(2−= (E8-17)Equation E8-17 is not generally given for calculations so that if not available we are requiredto go through the force analysis for its derivation.Once we have calculated the terminal velocity, one final critical step is to double check Re toensure that 0.1Re <=fluid t fluid D V μρ (E8-18)so that our assumption leading to E8-16 is indeed valid.Solutions:We can use a free-body diagram of a mineral particle, as shown in Figure 8-14b. The mineralparticle moves downward with a constant velocity V t (relative to the moving air flow)that isgoverned by a balance between weight of particle, F W , the buoyancy force of the surroundingair, F B , and the drag of air on the particle, F D . Please Note: it is acceptable that thebuoyancy force of the surrounding air, F B , is neglected.(a) We haveB D W F F F += (E8-19)63p p p p W D g gV F πρρ== (E8-20) 63p air p air B D g gV F πρρ== (E8-21)Assuming Re <1, we have the drag coefficient of an mineral particle pt air air D D V C ρμ24Re 24== (E8-22) Therefore the drag force p t air pt air air p t air D p t air D D V D V D V C D V F πμρμπρπρ3244214212222=== (E8-23) Substitute E8-20, E8-21 and E8-23 into E8-19, we have p air p air p p UD D g D g πμπρπρ36633+= Therefore, the terminal velocity of a mineral particle relative to the flowing air is air p air p t gD V μρρ18)(2−= or airp p t gD V μρ182= if buoyancy is neglected,Therefore, we haves m m Ns m s m m kg m kg gD V air p air p t /029.0)/1079.1(18)1020(/8.9)/29.1/2400(18)(25262332=×××××−=−=−−μρρCheck Reynolds number 0.10418.0/1079.11020/029.0/29.1Re 2563<<=××××==−−m Ns m s m m kg D V air p t air μρ Therefore, the assumption to use E8-22 is valid.(b) We can calculate the air travelling velocity in the vertical tube s m m s L m L D Q A Q V tube air /017.0)05.0(14.3)60min 110001min 0.2(44232=××××===π Therefore, the residence time of a mineral particle in the vertical tube is 3.16/)017.0029.0(75.0=+=+=sm m V V L t air t sReferences[1]. Munson BR, Young DF and Okiishi TH, Fundamentals of Fluid Mechanics, 4th Edition,John Wiley & Sons, Brisbane, 2002.[2]. Noel de Nevers, Fluid Mechanics for Chemical Engineers, 3rd Edition, McGraw-Hill’sChemical Engineering Series, Sydney 2005.。
机械设计基础 南理工
1.4 Purpose It is hoped that, after studying this textbook, the student will grasp(掌握 the basic theory and 掌握) 掌握 obtain the basic knowledge and skills(技能 技能) 技能 needed in mechanisms synthesis and kinematic and dynamic(动力学的 analysis of machinery. 动力学的) 动力学的 The knowledge to be obtained from this 基本的) textbook is therefore fundamental(基本的 in 基本的 analyzing existing(现有的 machines and 现有的) 现有的 designing new ones.
What is Machanism?
机构： 机构：具有机器的前两个特征
人为的实物组合（不是天然形成的）； 人为的实物组合（不是天然形成的）； 各运动单元具有确定的相对运动； 各运动单元具有确定的相对运动； Mechanism(机构 is a basic system of 机构) 机构 links(构件 which can transform(转换 or 构件) 转换) 构件 转换 transmit(传递 force and motion. 传递) 传递
大学物理实验的英语教材
大学物理实验的英语教材University Physics Laboratory: English TextbookIntroduction:The aim of this English textbook is to provide comprehensive guidance and instructions for university physics laboratory experiments. The textbook covers a wide range of topics, including fundamental laws and principles, experimental techniques, data analysis, and safety precautions. By following this textbook, students will enhance their laboratory skills, develop a deeper understanding of physics concepts, and improve their English proficiency.Chapter 1: Introduction to Laboratory Equipment1.1 Laboratory Safety1.2 Basic Laboratory Equipment1.2.1 Glassware and Containers1.2.2 Measuring Instruments1.2.3 Electrical Equipment1.2.4 Advanced EquipmentChapter 2: Measurement Techniques2.1 Units and Dimensions2.2 Uncertainty and Error Analysis2.3 Measurement Tools and Techniques2.3.1 Length Measurement2.3.2 Time Measurement2.3.3 Mass Measurement2.3.4 Temperature Measurement2.3.5 Other Important MeasurementsChapter 3: Experiments on Mechanics3.1 Introduction to Mechanics3.2 Experimental Procedures for Newton's Laws3.2.1 Experiment 1: Force and Motion3.2.2 Experiment 2: Frictional Forces3.3 Experiment on Gravitation3.3.1 Experiment 3: Gravitational Force and Acceleration due to Gravity 3.4 Experiment on Simple Harmonic Motion3.4.1 Experiment 4: Pendulum MotionChapter 4: Experiments on Optics4.1 Introduction to Optics4.2 Experiments on Geometrical Optics4.2.1 Experiment 5: Reflection4.2.2 Experiment 6: Refraction4.3 Experiments on Wave Optics4.3.1 Experiment 7: Interference of Light4.3.2 Experiment 8: Diffraction of LightChapter 5: Experiments on Electricity and Magnetism 5.1 Introduction to Electricity and Magnetism5.2 Experiments on DC Circuits5.2.1 Experiment 9: Ohm's Law and Resistors5.2.2 Experiment 10: Kirchhoff's Laws and DC Circuits 5.3 Experiments on Magnetism and Electromagnetism 5.3.1 Experiment 11: Magnetic Fields and Forces5.3.2 Experiment 12: Electromagnetic Induction Chapter 6: Experiments on Modern Physics6.1 Introduction to Modern Physics6.2 Experiments on Atomic and Nuclear Physics6.2.1 Experiment 13: Radioactivity and Half-Life6.2.2 Experiment 14: Atomic Spectra and Energy Levels 6.3 Experiments on Quantum Mechanics6.3.1 Experiment 15: Wave-Particle Duality6.3.2 Experiment 16: Photoelectric EffectChapter 7: Data Analysis and Error Propagation7.1 Data Collection and Recording7.2 Data Analysis Techniques7.3 Graphing and Curve Fitting7.4 Error Propagation and ReportingChapter 8: Laboratory Reports and Presentation8.1 Structure of a Laboratory Report8.2 Writing Style and Language8.3 Presenting Experimental Results8.4 Peer Review and FeedbackConclusion:This English textbook for university physics laboratory experiments offers a comprehensive guide for students to conduct practical experiments effectively. With a strong emphasis on safety, accurate measurements, data analysis, and clear reporting, the textbook equips students with the necessary skills to excel in the laboratory. By using this textbook, students will enhance their understanding of physics concepts, improve their English proficiency, and become adept researchers in the field of physics.。
Chapter 8 Mechanical Failure
Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.4. John Wiley and Sons, Inc., 1996.
Moderately Ductile vs. Brittle Failure
cup-and-cone fracture
brittle fracture
Adapted from Fig. 8.3, Callister & Rethwisch 8e.
Chapter 8 - 6
Brittle Failure
Adapted from Fig. 8.8(a), Callister & Rethwisch 8e. Chapter 8 - 10
Concentration of Stress at Crack Tip
Adapted from Fig. 8.8(b), Callister & Rethwisch 8e.
Chapter 8: Mechanical Failure
ISSUES TO ADDRESS...
• How do cracks that lead to failure form? • How is fracture resistance quantified? How do the fracture resistances of the different material classes compare? • How do we estimate the stress to fracture? • How do loading rate, loading history, and temperature affect the failure behavior of materials?
8_principal_stresses
MECHANICS OF MATERIALS
Sample Problem 8.2
SOLUTION:
Third
Beer • Johnston • DeWolf
• Determine reactions at A and D. • Determine maximum shear and bending moment from shear and bending moment diagrams. The overhanging beam supports a uniformly distributed load and a concentrated load. Knowing that for the grade of steel to used all = 24 ksi and all = 14.5 ksi, select the wideflange beam which should be used. • Calculate required section modulus and select appropriate beam section. • Find maximum normal stress. • Find maximum shearing stress.
M A 0 RD 59 kips M D 0 R A 41kips
物理专业英语名词解释
part one mechanics力学chapter 1 kinematics—uniformly accelerated motion运动学，匀加速运动frame of reference: 参照系position vector and displacement 位置向量和位移speedvelocity 速度instantaneous velocity瞬时速度velocity components速度分量acceleration 加速度graphical interpretation图像释义uniformly accelerated motion along a straight line 匀加速直线运动acceleration due to gravity (g) 重力加速度projectile problems落体问题relative motion相对运动chapter 2 dynamics-- newton's laws of motion动力学- 牛顿运动定律general properties of forces in mechanics:力的基本性质1 the law of universal gravitation万有引力定律2the weight 重量3 the tensile force 拉力4 the normal force 法向力5 the friction force 摩擦力6 dimensional analysis三维分析chapter 3 equilibrium平衡concurrent forces共点力are forces whose lines of action all pass through a common point. the forces acting on a point object are concurrent because they all pass through the same point, the point object.equilibrium 平衡rigid body刚体the torque (or moment) 转矩或力矩the two conditions for equilibrium 平衡的两个条件the center of gravity 重心axis轴chapter 4 work and energy功和能kinetic energy (ke) 动能conservative force保守力⑵gravitational (weight) potential energy (peg) 重力（重量）势能⑵elastic potential energy 弹性势能the efficiency 效率chapter 5 impulse and momentumthe linear momentum 冲量与动量an impulse 冲量collisions and explosions碰撞和爆炸a perfectly elastic collision 完全弹性碰撞coefficient of restitution恢复系数the center of mass重心chapter 6 rotation转动the angular speed 角速度the angular acceleration 角加速度tangential 【数】切线;正切centripetal acceleration ( )加速度the centripetal force 向心力chapter 7 rigid-body rotation刚体转动the moment of inertia 转动惯量parallel-axis theorem平行轴定理chapter 8 elasticity弹性elasticity弹性;弹力the stress 【物】应力[u][c]strain 应变the elastic limit弹性极限the shear modulus 切变模量standard atmospheric pressure标准大气压the hydrostatic pressure静水压力equation of continuity连续性方程the viscosity 粘度spring弹簧a restoring force 恢复力simple harmonic motion 简谐运动vibratory motion 振动运动the period ( ) 【数】循环节;周期the frequency ( ) 频率the elastic potential energy 弹性势能the simple pendulum 单摆chapter 11 wave motion波动a propagating wave 波传播wave terminology波术语in-phase vibrations同相振动standing waves驻波conditions for resonance共振的条件longitudinal (compressional) waves 纵向（挤压）波chapter 12 sound声音the intensey （i）强度loudness 响度beats节拍doppler effect 多普勒效应interference effects 干扰效应part two thermodynamics热力学chapter 1 the kinetic theory of gases第1章气体动力学理论avogadro's number ( ) 阿伏伽德罗数（）the root mean square speed根均方速度the absolute temperature绝对温度the mean free path (m.f.p.) 平均自由程（m.f.p.）the equipartition theorem of energy 能量均分定理ideal gas law理想气体定律heat 热the internal energy 内部能量an isobaric process is a process carried out at constant pressure. 等压过程是恒压进行的过程。
机械原理第七版西北工业大学课后习题答(8-11章)整本是的重点
第8章课后习题参考答案8-l 铰链四杆机构中，转动副成为周转副的条件是什么?在下图所示四杆机构ABCD 中哪些运动副为周转副?当其杆AB 与AD 重合时，该机构在运动上有何特点?并用作图法求出杆3上E 点的连杆曲线。
答：转动副成为周转副的条件是：（1）最短杆与最长杆的长度之和小于或等于其他两杆长度之和；（2）机构中最短杆上的两个转动副均为周转副。
图示ABCD 四杆机构中C 、D 为周转副。
当其杆AB 与AD 重合时，杆BE 与CD 也重合因此机构处于死点位置。
8-2曲柄摇杆机构中，当以曲柄为原动件时，机构是否一定存在急回运动，且一定无死点?为什么? 答：机构不一定存在急回运动，但一定无死点，因为：（1）当极位夹角等于零时，就不存在急回运动如图所示，（2）原动件能做连续回转运动，所以一定无死点。
8-3 四杆机构中的极位和死点有何异同?8-4图a 为偏心轮式容积泵；图b 为由四个四杆机构组成的转动翼板式容积泵。
试绘出两种泵的机构运动简图，并说明它们为何种四杆机构，为什么?解 机构运动简图如右图所示，ABCD 是双曲柄机构。
因为主动圆盘AB 绕固定轴A 作整周转动，而各翼板CD 绕固定轴D 转动，所以A 、D 为周转副，杆AB 、CD 都是曲柄。
8-5试画出图示两种机构的机构运动简图，并说明它们各为何种机构。
图a 曲柄摇杆机构图b 为导杆机构。
8-6如图所示，设己知四杆机构各构件的长度为240a mm =，600b =mm ,400,500c mm d mm ==。
试问:1)当取杆4为机架时，是否有曲柄存在?2)若各杆长度不变，能否以选不同杆为机架的办法获得双曲柄机构和双摇杆机构?如何获得?3)若a 、b ﹑c 三杆的长度不变，取杆4为机架，要获得曲柄摇杆机构,d 的取值范围为何值? :解 (1)因a+b=240+600=840≤900=400+500=c+d 且最短杆 1为连架轩．故当取杆4为机架时，有曲柄存在。
质心运动课件
一.质心动能定理 (科尼希定理)
一个质点组的质心在C，如图．
z S
ric C
mi
rc
对某参照系S， 定义：
O
ri
EC
1 2
MvC2
——质心动能
x
y
是否相等？
Ek
i
1 2
mi
vi2——质点组总动能
可以证明：
对 质某点参组照 总系 动， 能：Ek EC ErC
——质心动能定理 (科尼希定理)
质点组总动能 = 质心动能 + 质点组相对质心的动能
ErC
vrriiCC
i
1 2
mi
vi2C
是质点组相对质心的总动能
是第i个质点相对于质心C的位 速率矢
18
科尼希定理: Ek EC ErC 证明如下： z
r riC
是第i
个质点相对于质心C的位矢
如图：对某参照系S，
ri
v
2 i
rC
i
1 2
mi vi2C
i
1 2
mi
2vC
viC
19
Ek
i
1 2
mi vC2
i
1 2
mi vi2C
i
rr mivC viC
r r r mivC viC vC
r mi viC
vC
0
0
i
i
质心系中质点组总动量
＝质心系中的质心动量
Ek
i
1 2
dLrC dt
M rC
质点组对质心的 角动量变化定理
质点组的角动
质点组相对于质心的角动量的时间 量变化定理在
变化率 = 各外力对质心的总力矩
漆安慎力学
F1 p1 A, F2 p2 A m V A( y1 y2 )
p2 A p1 A Ag ( y1 y2 )
or p2 p1 g ( y1 y2 ) ★Pressure in a liquid
(10-7)
y1 0, p1 p0 and y2 h, p2 p
Pascal's Principle
d1 d2
Pistono : F2 , A2 , and The pressures on both sides are equal
Fi Fo Ai Ao
Ao Fo Fi Ai
(10-13)
Ao Ai ,
Fo Fi
The same volume of incompressible liquid is displaced at both pistons
mg The gravitational force of the
column
F2 acts at the bottem of the column;
10-4
F2 F1 mg
Fluids at Rest
(10-5) (10-6)
The column is in static equilibrium, these three forces balanced.
Eq. 10-7 :
level 1: surface; level 2: h below it
p p0 gh
(10-8)
10-4
p2 p1 g ( y1 y2 )
★Pressure
Fluids at Rest
(10-7) Level 2
y1 p1 , p1 p0 and y2 d , p2 p
《Mechanics 1历年考试真题分类汇编》
Section A Velocity and Acceleration（Chapter 1）1．A man runs in a straight line. He passes through a fixed point A with constantvelocity 7ms_1at time t = 0. At time t s his velocity is v ms_1. The diagram shows the graph of v against t for the period 0 ≤t ≤ 40. （02w）(i) Show that the man runs more than 154m in the first 24 s. [2](ii)Given that the man runs 20m in the interval 20 ≤t ≤ 24, find how far he is from A when t = 40.[2]2The diagram shows the velocity-time graphs for the motion of two cyclists P and Q, who travel in the same direction along a straight path. Both cyclists start from rest at the same point O and both accelerate at 2ms−2 up to a speed of 10ms−1. Both then continue at a constant speed of 10ms−1.(03s)Q starts his journey T seconds after P.(i) Show in a sketch of the diagram the region whose area represents thedisplacement of P, from O,at the instant when Q starts. [1]Given that P has travelled 16m at the instant when Q starts, find(ii) the value of T, [3](iii)the distance between P and Q when Q’s speed reaches 10ms−1. [2]3A boy runs from a point A to a point C. He pauses at C and then walks back towardsA until reaching the point B, where he stops. The diagram shows the graph ofv against t, where v ms−1 is the boy,s velocity at time t seconds after leavingA. The boy runs and walks in the same straight line throughout. (04s)(i) Find the distances AC and AB. [3](ii) Sketch the graph of x against t, where x metres is the boy,s displacement from A. Show clearly the values of t and x when the boy arrives at C, when he leaves C, and when he arrives at B. [3]4Particles P and Q start from points A and B respectively, at the same instant, and move towards each other in a horizontal straight line. The initial speeds of P and Q are 5ms−1 and 3ms−1 respectively. The accelerations of P and Q are constant and equal to 4ms−2 and 2ms−2 respectively (see diagram).（04w）(i) Find the speed of P at the instant when the speed of P is 1.8 times the speedof Q. [4](ii) Given that AB = 51 m, find the time taken from the start until P and Q meet.5The diagram shows the velocity-time graph for a lift moving between floors ina building. The graph consists of straight line segments. In the first stagethe lift travels downwards from the ground floor for 5 s, coming to rest atthe basement after travelling 10 m. (05s)(i) Find the greatest speed reached during this stage. [2]The second stage consists of a 10 s wait at the basement. In the third stage,the lift travels upwards until it comes to rest at a floor 34.5m above the basement,arriving 24.5 s after the start of the first stage. The lift accelerates at 2ms−2 for the first 3 s of the third stage, reaching a speed of V ms−1. Find(ii) the value of V, [2](iii) the time during the third stage for which the lift is moving at constant speed, [3](iv) the deceleration of the lift in the final part of the third stage. [2]6 A car travels in a straight line with constant acceleration a ms−2. It passesthe points A, B and C, in this order, with speeds 5ms−1, 7ms−1and 8ms−1respectively.The distances AB and BC are d1 m and d2 m respectively.(i) Write down an equation connecting (a) d1 and a,(b) d2 and a. [2](ii) Hence find d1 in terms of d2. [2]7、 The diagram shows the displacement-time graph for a car’s journey. The graphconsists of two curved parts AB and CD, and a straight line BC. The line BC isa tangent to the curve AB at B and a tangent to the curve CD at C. The gradientof the curves at t = 0 and t = 600 is zero, and the acceleration of the car is constant for 0 < t < 80 and for 560 < t < 600. The displacement of the car is 400m when t = 80. (05w)(i) Sketch the velocity-time graph for the journey. [3](ii) Find the velocity at t = 80. [2](iii) Find the total distance for the journey. [2](iv) Find the acceleration of the car for 0 < t < 80. [2]8The diagram shows the velocity-time graph for the motion of a small stone which falls vertically from rest at a point A above the surface of liquid in a container.The downward velocity of the stone t s after leaving A is v ms−1. The stone hits the surface of the liquid with velocity7ms−1when t = 0.7. It reaches the bottom of the container with velocity 5ms−1 whent = 1.2. （06s）(i) Find(a) the height of A above the surface of the liquid,(b) the depth of liquid in the container.[3](ii) Find the deceleration of the stone while it is moving in the liquid. [2](iii)Given that the resistance to motion of the stone while it is moving in the liquid has magnitude 0.7N, find the mass of the stone. [3]9 A train travels from A to B, a distance of 20 000m, taking 1000 s. The journeyhas three stages. In the first stage the train starts from rest at A andaccelerates uniformly until its speed is V ms−1. In the second stage the traintravels at constant speed V ms−1 for 600 s. During the third stage of the journeythe train decelerates uniformly, coming to rest at B. （08w）(i) Sketch the velocity-time graph for the train’s journey. [2](ii) Find the value of V. [3](iii) Given that the acceleration of the train during the first stage of thejourney is 0.15 ms−2, find the distance travelled by the train during thethird stage of the journey. [4]10A train starts from rest at a station and travels in a straight line until itcomes to rest again at the next station. The displacement-time graph above refersto the journey. (01w)(i) The speed of the train is constant from t = 120 to t = 440. Find this speed.[2](ii) Given that the acceleration of the train is constant from t = o tot= 120 and from t = 440 to t = 480, make a sketch of the velocity-timegraph for the journey, showing the maximum speed of the train. [3]。
施平 机械工程专业英语教程
施平机械工程专业英语教程Introduction to Mechanical EngineeringChapter 1: Introduction to Mechanical Engineering- Definition and scope of mechanical engineering- Historical background and evolution of the field- Overview of various disciplines within mechanical engineering Chapter 2: Mechanics- Principles of mechanics, including statics and dynamics- Laws of motion and their applications- Analysis of forces and moments in mechanical systems Chapter 3: Thermodynamics- Basic concepts and laws of thermodynamics- Energy and heat transfer in mechanical systems- Analysis of thermodynamic cycles and processesChapter 4: Materials Science and Engineering- Properties and behavior of materials used in mechanical engineering- Material testing and characterization methods- Selection of materials for specific applicationsChapter 5: Fluid Mechanics- Fundamentals of fluid mechanics- Analysis of fluid flow and pressure distribution- Applications of fluid mechanics in mechanical systems Chapter 6: Heat Transfer- Modes of heat transfer (conduction, convection, radiation)- Heat transfer analysis and calculations- Applications of heat transfer in mechanical systemsChapter 7: Energy Conversion and Power Systems- Energy conversion principles and devices- Analysis of power generation systems- Renewable energy sources and sustainabilityChapter 8: Machine Design and Control Systems- Design principles and methodologies for mechanical systems- Control systems and automation in mechanical engineering- Analysis and optimization of machine componentsChapter 9: Manufacturing Processes- Various manufacturing processes and methodologies- Machining, forming, casting, and joining processes- Quality control and inspection in manufacturingChapter 10: Engineering Ethics and Professionalism- Ethical considerations in engineering practice- Professional responsibility and accountability- Society and the engineer's role in sustainable development Chapter 11: Career Opportunities in Mechanical Engineering- Overview of career options and paths in mechanical engineering - Skills and qualities desired by employers- Professional organizations and resources for career advancement Chapter 12: Emerging Technologies in Mechanical Engineering- Trends and developments in the field of mechanical engineering - Introduction to advanced technologies like robotics, nanotechnology, and artificial intelligence- Potential impact of these technologies on the future of mechanical engineering.。
物理学类专业英语_基础课程
Introduction 1 参考译文：导论 3Chapter 1: Mechanics 61.1 Classical versus quantum 61.2 Einsteinian versus Newtonian 61.3 History 71.4 Types of mechanical bodies 81.5 Sub-disciplines in mechanics 81.5.1 Classical mechanics 81.5.2 Quantum mechanics 9参考译文：第一章力学 91.1 经典和量子 91.2 爱因斯坦和牛顿 101.3力学的历史 101.4力学中物体的种类 111.5力学的分支学科 111.5.1经典力学有如下学科构成： 111.5.2量子力学 12Chapter 2: Heat 132.1 Overview 132.2 Notation 142.3 Definitions 152.4 Thermodynamics 152.4.1 Internal energy 152.4.2 Heat capacity 162.4.3 Phase Changes 172.5 Heat transfer mechanisms 17 2.6 Heat dissipation 19参考译文：第二章热学 202.1 综述 202.2 符号 212.3 定义 212.4 热力学 212.4.1 内能 212.4.2 热容量 222.4.3 相变 232.5 热传递的机制 232.6 散热 24Chapter 3: Electromagnetism 25 3.1 History 253.2 Overview 273.3 Classical electrodynamics 27 3.4 The photoelectric effect 28 3.5 Maxwell's equations 293.6 Special relativity 30参考译文：第三章电磁学 343.1 发展历史 353.2 总论 363.3 经典电动力学 363.4 光电效应 373.5 麦克斯韦方程组 373.6 狭义相对论 37Chapter 4 Optics 414.1 History 414.2 Classical optics 434.2.1 Geometrical optics 44 4.2.2 Physical optics 46 4.3 Modern optics 514.3.1 Lasers 524.3.2 Nonlinear optics 59 参考译文：第四章光学 59 4.1 光学的历史 604.2 经典光学 614.2.1 几何光学 614.2.2 物理光学 634.3 现代光学 664.3.1激光 664.3.2 非线性光学 71Chapter 5 Atomic physics 725.1 Isolated atoms 725.2 Electronic configuration 725.3 History and developments 735.3.1 Introduction to Atomic Physics 74 5.3.2 Atomic Structure 745.3.3 Bohr atom structure model 755.3.4 Atomic Isotopes 765.3.5 Einstein's Equation 765.3.6 Radioactive Decay 77参考译文：第五章原子物理 795.1 孤立原子 795.2 电子图像 795.3 原子物理的历史和发展过程 805.3.1 原子物理引论 805.3.2 原子结构 805.3.3波尔的原子结构模型 815.3.4 原子的同位素 815.3.5 爱因斯坦方程 825.3.6 放射性衰变 82Chapter 6: Quantum mechanics 846.1 Overview 856.2 Quantum mechanics and classical physics 86 6.3 Theory 866.4 Mathematical formulation 89。
结构设计常用专业英语
结构设计常用专业英语词汇汇编Chapter 1 Loads and Action (1)第一章荷载与作用 (1)Chapter 2 Seismic Design (8)第二章抗震设计 (8)Chapter 3 Foundation (14)第三章地基基础 (14)Chapter 4 Reinforcement Concrete (22)第四章钢筋混凝土结构 (22)Chapter 5 Steel Structure (28)第五章钢结构 (28)Chapter 6 Composite Structure (37)第六章组合结构 (37)Chapter 7 Masonry Structure (40)第七章砌体结构 (40)Chapter 8 Others (42)第八章其它 (42)第一章荷载与作用 (43)Chapter 1 Loads and Action (43)第二章抗震设计 (50)Chapter 2 Seismic Design (50)第三章地基基础 (56)Chapter 3 Foundation (56)第四章钢筋混凝土结构 (65)Chapter 4 Reinforcement Concrete (65)第五章钢结构 (71)Chapter 5 Steel Structure (71)第六章组合结构 (80)Chapter 6 Composite Structure (80)第七章砌体结构 (83)Chapter 7 Masonry Structure (83)第八章其它 (85)Chapter 8 Others (85)上册Chapter 1 Loads and Action 第一章荷载与作用Chapter 2 Seismic Design第二章抗震设计Chapter 3 Foundation 第三章地基基础Chapter 4 Reinforcement Concrete第四章钢筋混凝土结构Chapter 5 Steel Structure第五章钢结构Chapter 6 Composite Structure第六章组合结构Chapter 7 Masonry Structure第七章砌体结构Chapter 8 Others 第八章其它下册第一章荷载与作用Chapter 1 Loads and Action。
牛津英语九年级Chapter8Twogeniuses全
Einstein's special and general theories of relativity reinterpreted gravity as a curvature of spacetime and explained the equivalence of mass and energy.
Formulation of calculus
Newton developed calculus, a branch of mathematics that deals with continuous change, independently of Gottfried Wilhelm Leibniz.
Greatest scientist of the 20th century
Einstein's theories of relativity and quantum mechanics reshaped our understanding of the universe, introducing concepts like curved spacetime and quantum entanglement.
Scientific revolution
The works of Newton and Darwin helped initiate the scientific revolution, which transformed the way humans view the natural world.
03
Newton's Main Achievements
The law of universal gravitation
Introduction_Fluid_Mechanics_8th_txtbk_Noted
8.3Air at40 Cflows in a pipe system in which diameter is decreased in two stages from25mm to15mm to10mm. Each section is2m long.As theflow rate is increased,which section will become turbulentfirst?Determine theflow rates at which one,two,and then all three sectionsfirst become turbulent.At each of theseflow rates,determine which sections,if any,attain fully developedflow.P8.38.4Forflow in circular tubes,transition to turbulence usuallyoccurs around Re%2300.Investigate the circumstancesunder which theflows of(a)standard air and(b)water at15 C become turbulent.On log-log graphs,plot:the averagevelocity,the volumeflow rate,and the massflow rate,atwhich turbulencefirst occurs,as functions of tube diameter.Laminar Flow between Parallel Plates8.5For the laminarflow in the section of pipe shown in Fig.8.1,sketch the expected wall shear stress,pressure,and centerlinevelocity as functions of distance along the pipe.Explain sig-nificant features of the plots,comparing them with fullydevelopedflow.Can the Bernoulli equation be applied any-where in theflowfield?If so,where?Explain briefly.8.6An incompressiblefluidflows between two infinite sta-tionary parallel plates.The velocity profile is given by u5u maxðAy21By1CÞ,where A,B,and C are constants and yis measured upward from the lower plate.The total gapwidth is h e appropriate boundary conditions toexpress the magnitude and units of the constants in termsof h.Develop an expression for volumeflow rate per unitdepth and evaluate the ratio V=u max.8.7The velocity profile for fully developedflow betweenstationary parallel plates is given by u5aðh2=42y2Þ,wherea is a constant,h is the total gap width between plates,and yis the distance measured from the center of the gap.Deter-mine the ratio V=u max.8.8Afluidflows steadily between two parallel plates.Theflow is fully developed and laminar.The distance betweenthe plates is h.(a)Derive an equation for the shear stress as a function of y.Sketch this function.(b)Forμ52:431025lbfÁs=ft2;@p=@x524:0lbf=ft2=ft,and h50:05in.,calculate the maximum shear stress,in lbf/ft2.8.9Oil is confined in a4-in.-diameter cylinder by a pistonhaving a radial clearance of0.001in.and a length of2in.Asteady force of4500lbf is applied to the piston.Assume theproperties of SAE30oil at120 F.Estimate the rate at whichoil leaks past the piston.8.10A viscous oilflows steadily between stationary parallelplates.Theflow is laminar and fully developed.The total gapwidth between the plates is h55mm.The oil viscosity is0.5NÁs/m2and the pressure gradient is21000N/m2/m.Findthe magnitude and direction of the shear stress on the upperplate and the volumeflow rate through the channel,permeter of width.8.11Viscous oilflows steadily between parallel plates.Theflow is fully developed and laminar.The pressure gradient is1.25kPa/m and the channel half-width is h51:5mm.Cal-culate the magnitude and direction of the wall shear stressat the upper plate surface.Find the volumeflow rate throughthe channel(μ50:50NÁs=m2).8.12A large mass is supported by a piston of diameterD54in.and length L54in.The piston sits in a cylinderclosed at the bottom,and the gap a50.001in.between thecylinder wall and piston isfilled with SAE10oil at68 F.The piston slowly sinks due to the mass,and oil is forced outat a rate of0.1gpm.What is themass(slugs)?P8.12,P8.168.13A high pressure in a system is created by a small piston-cylinder assembly.The piston diameter is6mm and itextends50mm into the cylinder.The radial clearancebetween the piston and cylinder is0.002mm.Neglect elasticdeformations of the piston and cylinder caused by pressure.Assume thefluid properties are those of SAE10W oil at35 C.When the pressure in the cylinder is600MPa,estimatethe leakage rate.8.14A hydraulic jack supports a load of9000kg.The fol-lowing data are given:Diameter of piston100mmRadial clearance between piston and cylinder0.05mmLength of piston120mmEstimate the rate of leakage of hydraulicfluid past the pis-ton,assuming thefluid is SAE30oil at30 C.8.15A hydrostatic bearing is to support a load of1000lbf/ftof length perpendicular to the diagram.The bearing is sup-plied with SAE10W-30oil at212 F and35psig through thecentral slit.Since the oil is viscous and the gap is small,theflow may be considered fully developed.Calculate(a)therequired width of the bearing pad,(b)the resulting pressuregradient,dp/dx,and(c)the gap height,if theflow rateisQ52.5gal/hr/ft.P8.158.16The basic component of a pressure gage tester consists of a piston-cylinder apparatus as shown.The piston,6mm in diameter,is loaded to develop a pressure of known mag-nitude.(The piston length is25mm.)Calculate the mass, M,required to produce 1.5MPa(gage)in the cylinder. Determine the leakageflow rate as a function of radial clearance,a,for this load if the liquid is SAE30oil at20 C. Specify the maximum allowable radial clearance so the vertical movement of the piston due to leakage will be less than1mm/min.8.17In Section8.2we derived the velocity profile between parallel plates(Eq.8.5)by using a differential control volume.Instead,following the procedure we used in Example5.9,derive Eq.8.5by starting with the NavierÀStokes equations(Eqs. 5.27).Be sure to state allassumptions.8.18Consider the simple power-law model for a non-Newtonianfluid given by Eq.2.16.Extend the analysis ofSection8.2to show that the velocity profile for fully devel-oped laminarflow of a power-lawfluid between stationaryparallel plates separated by distance2h may be writtenu5hkΔpL1=nnhn1112yhðn11Þ=nwhere y is the coordinate measured from the channel centerline.Plot the profiles u=U max versus y/h for n50:7,1.0,and1.3.8.19Viscous liquid,at volumeflow rate Q,is pumpedthrough the central opening into the narrow gap between theparallel disks shown.Theflow rate is low,so theflow islaminar,and the pressure gradient due to convective accel-eration in the gap is negligible compared with the gradientcaused by viscous forces(this is termed creepingflow).Obtain a general expression for the variation of averagevelocity in the gap between the disks.For creepingflow,the velocity profile at any cross section in the gap is the sameas for fully developedflow between stationary parallel plates.Evaluate the pressure gradient,dp/dr,as a function ofradius.Obtain an expression for p(r).Show that the net forcerequired to hold the upper plate in the position shown isF53μQR2h312R0R2"#hP8.198.20A sealed journal bearing is formed from concentriccylinders.The inner and outer radii are25and26mm,thejournal length is100mm,and it turns at2800rpm.The gap isfilled with oil in laminar motion.The velocity profile is linearacross the gap.The torque needed to turn the journal is0.2NÁm.Calculate the viscosity of the oil.Will the torqueincrease or decrease with time?Why?8.21Using the profile of Problem8.18,show that theflowrate for fully developed laminarflow of a power-lawfluidbetween stationary parallel plates may be written asQ5hkΔpL1=n2nwh22n11Here w is the plate width.In such an experimental setup thefollowing data on applied pressure differenceΔp andflowrate Q were obtained:Δp(kPa)102030405060708090100Q(L/min)0.4510.7591.011.151.411.571.661.852.052.25Determine if thefluid is pseudoplastic or dilatant,and obtainan experimental value for n.8.22Consider fully developed laminarflow between infiniteparallel plates separated by gap width d50.2in.The upperplate moves to the right with speed U255ft/s;the lower platemoves to the left with speed U152ft/s.The pressure gradientin the direction offlow is zero.Develop an expression for thevelocity distribution in the gap.Find the volumeflow rate perunit depth(gpm/ft)passing a given cross section.8.23Water at60 Cflows between two largeflat plates.Thelower plate moves to the left at a speed of0.3m/s;the upperplate is stationary.The plate spacing is3mm,and theflow islaminar.Determine the pressure gradient required to pro-duce zero netflow at a crosssection.8.24Two immisciblefluids are contained between infiniteparallel plates.The plates are separated by distance2h,andthe twofluid layers are of equal thickness h55mm.Thedynamic viscosity of the upperfluid is four times that ofthe lowerfluid,which isμlower50.1NÁs/m2.If the plates arestationary and the applied pressure gradient is250kPa/m,find the velocity at the interface.What is the maximumvelocity of theflow?Plot the velocity distribution.8.25Two immisciblefluids are contained between infiniteparallel plates.The plates are separated by distance2h,andthe twofluid layers are of equal thickness h;the dynamicviscosity of the upperfluid is three times that of the lowerfluid.If the lower plate is stationary and the upper platemoves at constant speed U520ft=s,what is the velocity atthe interface?Assume laminarflows,and that the pressuregradient in the direction offlow is zero.8.26The record-read head for a computer disk-drivememory storage system rides above the spinning disk on avery thinfilm of air(thefilm thickness is0.25μm).The headlocation is25mm from the disk centerline;the disk spins at8500rpm.The record-read head is5mm square.For stan-dard air in the gap between the head and disk,determine(a)the Reynolds number of theflow,(b)the viscous shearstress,and(c)the power required to overcome viscousshear.8.27The dimensionless velocity profile for fully developedlaminarflow between infinite parallel plates with the upperplate moving at constant speed U is shown in Fig.8.6.Find thepressure gradient@p/@x at which(a)the upper plate and(b)the lower plate experience zero shear stress,in terms of U,a,andμ.Plot the dimensionless velocity profiles for these cases.8.28Consider steady,fully developed laminarflow of a viscous liquid down an inclined surface.The liquid layer is of constant thickness,e a suitably chosen differential control volume to obtain the velocity profile.Develop anexpression for the volumeflowrate.8.29Consider steady,incompressible,and fully developed laminarflow of a viscous liquid down an incline with no pressure gradient.The velocity profile was derived in Example5.9.Plot the velocity profile.Calculate the kine-matic viscosity of the liquid if thefilm thickness on a30slope is0.8mm and the maximum velocity is15.7mm/s.8.30Two immisciblefluids of equal density areflowing downa surface inclined at a60 angle.The twofluid layers are of equal thickness h510mm;the kinematic viscosity of the upperfluid is1/5th that of the lowerfluid,which isνlower5 0.01m2/s.Find the velocity at the interface and the velocity at the free surface.Plot the velocity distribution.8.31The velocity distribution forflow of a thin viscousfilm down an inclined plane surface was developed in Example 5.9.Consider afilm7mm thick,of liquid with SG51.2and dynamic viscosity of1.60NÁs/m2.Derive an expression for the shear stress distribution within thefilm.Calculate the maximum shear stress within thefilm and indicate its direc-tion.Evaluate the volumeflow rate in thefilm,in mm3/s per millimeter of surface width.Calculate thefilm Reynoldsnumber based on averagevelocity.8.32Consider fully developedflow between parallel plates with the upper plate moving at U55ft/s;the spacing between the plates is a50.1in.Determine theflow rate per unit depth for the case of zero pressure gradient.If thefluid is air,evaluate the shear stress on the lower plate and plot the shear stress distribution across the channel for the zero pressure gradient case.Will theflow rate increase or decrease if the pressure gradient is adverse?Determine the pressure gradient that will give zero shear stress at y50.25a. Plot the shear stress distribution across the channel for thelattercase.8.33Glycerin at59 Fflows between parallel plates with gap width b50.1in.The upper plate moves with speed U52ft/s in the positive x direction.The pressure gradient is@p/@x5 250psi/ft.Locate the point of maximum velocity and determine its magnitude(let y50at the bottom plate). Determine the volume offlow(gal/ft)that passes a given cross section(x5constant)in10s.Plot the velocity andshear stressdistributions.8.34The velocity profile for fully developedflow of castor oil at20 C between parallel plates with the upper plate moving is given by Eq.8.8.Assume U51.5m/s and a55 mm.Find the pressure gradient for which there is no netflow in the x direction.Plot the expected velocity distribution and the expected shear stress distribution across the channel for thisflow.For the case where u51/2U at y/a50.5,plot the expected velocity distribution and shear stress distribution across the ment on features of theplots.8.35The velocity profile for fully developedflow of carbon tetrachloride at68 F between parallel plates(gap a5 0.05in.),with the upper plate moving,is given by Eq.8.8. Assuming a volumeflow rate per unit depth is1.5gpm/ft for zero pressure gradient,find U.Evaluate the shear stress on the lower plate.Would the volumeflow rate increase or decrease with a mild adverse pressure gradient?Calculate the pressure gradient that will give zero shear stress at y/a50.25.Plot the velocity distribution and the shear stress distribution for this case.8.36Free-surface waves begin to form on a laminar liquid filmflowing down an inclined surface whenever the Rey-nolds number,based on massflow per unit width offilm,is larger than about33.Estimate the maximum thickness of a laminarfilm of water that remains free from waves while flowing down a verticalsurface.8.37Microchips are supported on a thin airfilm on a smooth horizontal surface during one stage of the manufacturing process.The chips are11.7mm long and9.35mm wide and have a mass of0.325g.The airfilm is0.125mm thick.The initial speed of a chip is V051:75mm=s;the chip slows as the result of viscous shear in the airfilm.Analyze the chip motion during deceleration to develop a differential equa-tion for chip speed V versus time t.Calculate the time required for a chip to lose5percent of its initial speed.Plot the variation of chip speed versus time during deceleration. Explain why it looks as you have plotted it.8.38A viscous-shear pump is made from a stationary housing with a close-fitting rotating drum inside.The clear-ance is small compared with the diameter of the drum,so flow in the annular space may be treated asflow between parallel plates.Fluid is dragged around the annulus by vis-cous forces.Evaluate the performance characteristics of the shear pump(pressure differential,input power,and effi-ciency)as functions of volumeflow rate.Assume that the depth normal to the diagram is b.P8.38,P8.408.39The clamping force to hold a part in a metal-turning operation is provided by high-pressure oil supplied by a pump. Oil leaks axially through an annular gap with diameter D, length L,and radial clearance a.The inner member of the annulus rotates at angular speedω.Power is required both to pump the oil and to overcome viscous dissipation in the annular gap.Develop expressions in terms of the specified geometry for the pump power,3p,and the viscous dissipation power,3v. Show that the total power requirement is minimized when the radial clearance,a,is chosen such that3v533p.8.40The efficiency of the viscous-shear pump of Fig.P8.39is given byη56q ð122qÞð426qÞwhere q5Q=abRωis a dimensionlessflow rate(Q is the flow rate at pressure differentialΔp,and b is the depth normal to the diagram).Plot the efficiency versus dimen-sionlessflow rate,andfind theflow rate for maximum effi-ciency.Explain why the efficiency peaks,and why it is zero at certain values of q.8.41Automotive design is tending toward all-wheel drive to improve vehicle performance and safety when traction ispoor.An all-wheel drive vehicle must have an interaxledifferential to allow operation on dry roads.Numerousvehicles are being built using multiplate viscous drives forinteraxle differentials.Perform the analysis and designneeded to define the torque transmitted by the differentialfor a given speed difference,in terms of the design para-meters.Identify suitable dimensions for a viscous differentialto transmit a torque of150NÁm at a speed loss of125rpm,using lubricant with the properties of SAE30oil.Discusshow tofind the minimum material cost for the viscous dif-ferential,if the plate cost per square meter is constant. 8.42An inventor proposes to make a“viscous timer”by placing a weighted cylinder inside a slightly larger cylindercontaining viscous liquid,creating a narrow annular gapclose to the wall.Analyze theflowfield created when theapparatus is inverted and the mass begins to fall undergravity.Would this system make a satisfactory timer?If so,for what range of time intervals?What would be the effect ofa temperature change on measured time?8.43A journal bearing consists of a shaft of diameter D535 mm and length L550mm(moment of inertia I50.125 kgÁm2)installed symmetrically in a stationary housing such that the annular gap isδ51mm.Thefluid in the gap has viscosityμ50.1NÁs/m2.If the shaft is given an initial angular velocity ofω5500rpm,determine the time for the shaft to slow to100rpm.On another day,an unknownfluid is tested in the same way,but takes10minutes to slow from 500to100rpm.What is its viscosity?8.44In Example8.3we derived the velocity profile for laminarflow on a vertical wall by using a differential control volume.Instead,following the procedure we used in Example5.9,derive the velocity profile by starting with the NavierÀStokes equations(Eqs.5.27).Be sure to state all assumptions.8.45A continuous belt,passing upward through a chemical bath at speed U0,picks up a liquidfilm of thickness h,density ρ,and viscosityμ.Gravity tends to make the liquid drain down,but the movement of the belt keeps the liquid from running off completely.Assume that theflow is fully devel-oped and laminar with zero pressure gradient,and that the atmosphere produces no shear stress at the outer surface of thefilm.State clearly the boundary conditions to be satisfied by the velocity at y50and y5h.Obtain an expression for the velocity profile.8.46A wet paintfilm of uniform thickness,δ,is painted ona vertical wall.The wet paint can be approximated as a Binghamfluid with a yield stress,τy,and density,ρ.Derive an expression for the maximum value ofδthat can be sus-tained without having the paintflow down the wall.Calculate the maximum thickness for lithographic ink whose yield stressτy540Pa and density is approximately1000kg/m3.8.47When dealing with the lubrication of bearings,the governing equation describing pressure is the Reynolds equation,generally written in1D asddxh3μdpdxþ6U dhdx¼0where h is the step height and U is the velocity of the lower surface.Step bearings have a relatively simple design and are used with low-viscosityfluids such as water,gasoline,and solvents.The minimumfilm thickness in these applications is quite small.The step height must be small enough for good load capacity,yet large enough for the bearing to accom-modate some wear without losing its load capacity by becoming smooth andflat.Beginning with the1D equation forfluid motion in the x direction,show that the pressure distribution in the step bearing is as shown,wherep s¼6μðh2Àh1Þh31L1þh32L21ppP8.47Laminar Flow in a Pipe8.48Considerfirst water and then SAE10W lubricating oil flowing at40 C in a6-mm-diameter tube.Determine the maximumflow rate(and the corresponding pressure gradient, @p/@x)for eachfluid atwhich laminarflow would be expected.P8.458.49For fully developed laminarflow in a pipe,determine the radial distance from the pipe axis at which the velocityequals the averagevelocity.8.50Using Eq.A.3in Appendix A for the viscosity of water,find the viscosity at220 C and120 C.Plot the viscosity over this range.Find the maximum laminarflow rate(L/hr)in a 7.5-mm-diameter tube at these temperatures.Plot the max-imum laminarflow rate over this temperature range.8.51A hypodermic needle,with inside diameter d50:005in. and length L51in.,is used to inject saline solution with viscosityfive times that of water.The plunger diameter is D50:375in.;the maximum force that can be exerted by a thumb on the plunger is F57:5lbf.Estimate the volumeflow rate of saline that can be produced.8.52In engineering science,there are often analogies to be made between disparate phenomena.For example,the applied pressure difference,Δp,and corresponding volume flow rate,Q,in a tube can be compared to the applied DC voltage,V,across and current,I,through an electrical resistor,respectively.By analogy,find a formula for the “resistance”of laminarflow offluid of viscosity,μ,in a tube length of L and diameter D,corresponding to electrical resistance,R.For a tube250mm long with inside diameter 7.5mm,find the maximumflow rate and pressure difference for which this analogy will hold for(a)kerosene and(b)castor oil(both at40 C).When theflow exceeds this maximum,why does the analogy fail?8.53Consider fully developed laminarflow in the annulus between two concentric pipes.The outer pipe is stationary, and the inner pipe moves in the x direction with speed V. Assume the axial pressure gradient is zero(@p=@x50). Obtain a general expression for the shear stress,τ,as a function of the radius,r,in terms of a constant,C1.Obtain a general expression for the velocity profile,u(r),in terms of two constants,C1and C2.Obtain expressions for C1and C2.P8.538.54Consider fully developed laminarflow in a circular pipe. Use a cylindrical control volume as shown.Indicate the forces acting on the control ing the momentumequation,develop an expression for the velocitydistribution.P8.548.55Consider fully developed laminarflow in the annular space formed by the two concentric cylinders shown in the diagram for Problem8.53,but with pressure gradient,@p/@x, and the inner cylinder stationary.Let r05R and r i5kR. Show that the velocity profile is given byu52R24μ@p@x12rR2112k2lnð1=kÞlnrRObtain an expression for the location of the maximum veloc-ity as a function of k.Plot the location of maximum velocity (α5r=R)as a function of radius ratio pare the lim-iting case,k-0,with the corresponding expression forflow in a circular pipe.8.56For theflow of Problem8.55show that the volumeflow rate is given byQ52πR4@p@xð12k4Þ2ð12k2Þ2"#Find an expression for the average pare the limiting case,k-0,withthe corresponding expression for flow in a circular pipe.8.57It has been suggested in the design of an agricultural sprinkler that a structural member be held in place by a wire placed along the centerline of a pipe;it is surmised that a relatively small wire would have little effect on the pressure drop for a givenflow ing the result of Problem 8.56,derive an expression giving the percentage change in pressure drop as a function of the ratio of wire diameter to pipe diameter for laminarflow.Plot the percentage change in pressure drop as a function of radius ratio k for0.001# k#0.10.8.58Consider fully developed pressure-drivenflow in a cylindrical tube of radius,R,and length,L510mm,with flow generated by an applied pressure gradient,Δp.Tests are performed with room temperature water for various values of R,with afixedflow rate of Q510μL/min.The hydraulic resistance is defined as R hyd5Δp/Q(by analogy with the electrical resistance R elec5ΔV/I,whereΔV is the electrical potential drop and I is the electric current).Calculate the required pressure gradient and hydraulic resistance for the range of tube radii listed in the table.Based on the results,is it appropriate to use a pressure gradient to pump fluids in microchannels,or should some other driving mechanism be used?R(mm)Δp(Pa)R hyd(PaÁs/m3)110211022102310248.59Thefigure schematically depicts a conical diffuser, which is designed to increase pressure and reduce kinetic energy.We assume the angleαis small(α,10 )so that tanα%αand r e5r i+αl,where r i is the radius at the diffuser inlet,r e is the radius at the exit,and l is the length of the diffuser.Theflow in a diffuser is complex,but here we assume that each layer offluid in the diffuserflow is laminar, as in a cylindrical tube with constant cross-sectional area. Based on reasoning similar to that in Section8.3,the pres-sure differenceΔp between the ends of a cylindrical pipe is。
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方程的通解为
x(t) = Acos(ωt + ϕ0 )
振子作简谐振动
任何运动方程可以写成
d2 dt
x
2
+
ω
2x
=
0
的运动都是简谐振动。
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8
(1) 单摆的运动方程
定轴转动的角动量定理
I
d2θ
dt 2
= Mz
ml2
d2θ
dt 2
= −mgl sin θ
θ 很小时
sin θ ≈ θ
d2θ
dt 2
+
g l
θ
=
0
简谐振动
θ (t) = θ0 cos(ωt + ϕ0 ) ω = g
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l
9
(2) 复摆的运动方程
刚体的质心距转轴为l，当复摆稍离开 平衡位置θ角后，受到净力矩
M = −mgl sin θ
θ 很小时 sin θ ≈ θ
I d2θ = −mglθ
dt 2
d2θ
dt 2
+
mgl θ
ν =|ν2-ν1|
ω = 2πν
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x1 x2 x
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t t t
28
3. 垂直方向同频率简谐振动的合成
(a)分振动 (b) 合运动
x=A1cos(ω t+ϕ 1) y=A2cos(ω t+ϕ 2)
x2 A12
+
y2 A22
−
2
x A1
y A2
cos(ϕ 2
− ϕ1 )
如 A1=A2 , 则 A=0
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2. 同方向不同频率的简谐振动的合成
(a) 分振动
x1=Acosω1 t x2=Acosω2t
(b) 合振动 x = x1+ x2
x = 2 Acos( ω2 − ω1 ) t ⋅ cos( ω2 + ω1 ) t
2
2
合振动不是简谐振动
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y A1
轨迹称为李萨如图形
ω x : ω y=3:2 ϕ x=0, ϕ y=π/2
-A2
O
A2 x
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- A1
31
例：
半个周期
ωx = 2 ωy 1
一个周期
半个周期
沿Y方向的振动
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一个周期
沿X方向的振动
32
当0<Δϕ <π时，质点沿顺时针方向运动； 当π<Δϕ< 2π时，质点沿逆时针方向运动。
2012-3-9
19
6. 简谐振动的能量
(1)简谐振动系统的能量特点
(a) 动能
x(t)=Acos(ωt+ϕ)
v = dx = −ωAsin( ωt + ϕ)
dt
Ek
=
1 2
mυ2
=ห้องสมุดไป่ตู้
1 2
mω02 A2
sin 2 (ωt
+ϕ)
= 1 kA2 sin 2 (ωt + ϕ),
2
k = mω02
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2012-3-9
12
同相和反相
当 Δ ϕ =±2kπ ，
(k=0,1,2,…) ， 两
x A1
A2
x1 x2
同相
T
振动步调相同，
O - A2
t
称同相。
-A1
x
当Δϕ= ±(2k+1)π ,
A1
( k =0,1,2,…)，两
A2
振动步调相反，
O
称反相。
- A2
-A1
2012-3-9
x1 x2
反相
T t
tgϕ = A1 sinϕ1 + A2 sinϕ2
A1 cosϕ1 + A2 cosϕ 2
24
(c) 两种特殊情况
(1)若两分振动同相
ϕ2−ϕ1=±2kπ
(k=0,1,2,…)
则A=A1+A2 , 两分振动相互加强
(2)若两分振动反相
ϕ2 − ϕ1= ±(2k+1)π
(k=0,1,2,…)
则A=|A1-A2|, 两分振动相互减弱
23
二、简谐振动的合成
1. 同方向同频率的简谐振动的合成
(a) 分振动 : x1=A1cos(ω t+ϕ 1) x2=A2cos(ω t+ϕ 2)
(b) 合振动 : x = x1+ x2
x =A cos(ω t+ϕ )
合振动是简谐振动, 其频率仍为ω
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A = A12 + A22 + 2 A1 A2 cos(ϕ 2 − ϕ1 )
=
sin2 (ϕ 2
− ϕ1 )
Æ 合运动一般是在 2A1 ( x向 )、2A2 ( y向 ) 范围内的一个椭圆
Æ 椭圆的性质 (方位、长短轴、左右旋 )：
在 A1 、A2确定之后, 主要决定于Δϕ = ϕ 2-ϕ 1
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29
Δϕ = 0
Δϕ = π/4
Δϕ = π/2
Δϕ = 3π/4
• 加速度也是简谐振动
• a比x领先π
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15
ω 2A ωA
A
O
-A
-ω A -ω 2A
位移、速度和加速度
v
x
v> 0 a<0 减速
<0 <0 加速
<0
>0 减速
a
T t
>0 >0 加速
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例：设质点的位移为 x(t)=-3.0sin (πt-π/4)
试画出该简谐振动的位移、速度和加速度随时间变 化的曲线，并求出它们的频率、振幅和初相位。
I
=
0
简谐振动
θ (t) = θ0 cos(ωt + ϕ0 ) ω = mgl
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I
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4. 描述简谐振动的特征量
x(t)=Acos(ωt+ϕ)
(1) 振幅 A：物体离开平衡位置的最大位移 或角位移。(Amplitude)
(2) 周期T ：物体作一次完全振动所需的时间。
T = 2π ω
频率ν：在单位时间内物体所作的完全振动的
次数，它是周期的倒数。
ν=1= ω T 2π
ω = 2πν
角频率，圆频率
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(3) 相位ϕ =(ωt+ϕ0) ：确定振动系统的瞬时运动状
态(phase)。
(a) (ωt +ϕ0 )是 t 时刻的相位 (b) ϕ0 是 t =0时刻的相位 — 初相
[解] 为了进行比较，必须把简谐量改写成标准 形式，振幅A 必需取正值。
x(t) = −3.0sin(πt − π ) = 3.0 cos(πt + π )
4
4
v
=
dx
=
−3.0πsin(πt
+
π )
=
3.0π cos(πt
+
3π )
dt
4
4
a
=
d2x dt 2
=
−3.0π2sin(πt
+
3π )
4
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Ek max
=
1 kA2 2
, Ek min
=0
∫ Ek
=
1 T
t +T
Ek dt
t
=
1 kA2 4
(b) 势能
Ep
=
1 2
kx2
=
1 2
kA2
cos2 (ωt
+ϕ)
EP max
=
1 kA2 2
, EP min
=0
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∫ EP
=
1 T
t +T
EP dt
t
=
1 4
kA2
情况和动能相同。 21
Δϕ = π
Δϕ = 5π/4
Δϕ = 3π/2 Δϕ = 7π/4
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4. 垂直方向不同频率简谐振动的合成
• 两分振动频率相差很小
Δϕ = (ω 2-ω 1) t + (ϕ 2-ϕ 1)
可看作两频率相等而ϕ2-ϕ1随 t 缓慢变化
合运动轨迹将按上页图依次缓慢变化
• 两振动的频率成整数比
一般物理量“振动”：物理量在其某一 基准值附近往返变化。
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2. 简谐振动
简谐振动：用时间的正弦或余弦函数描述的 振动。（simple harmonic motion）
表达式
x(t)=Acos(ωt+ϕ)
特点 (1)等幅振动
(2)周期振动 x(t)=x(t+T )
(3)最简单、最基本的振动，复杂的振 动可以看作若干简谐振动的合成。
3. 求半径为5.00㎝的金属球与长尾25.0㎝的细棒组成的复摆的振动周期，设 金属球质量为m，细棒质量可忽略。.
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4. 如图，一细棒两端装有质量为m的小球A,B,可绕水平轴O自由转动，且 OA=l1，OB=l2，若细棒的质量可忽略不计，求细棒做角度很小的摆动时的 周期.
5. 图示两个弹簧系统，劲度系数为k1和k2， 分别求出两种情形的振动频率.
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当ω2≈ω1时 ω2 - ω1<<ω2 +ω1