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Chapter 2 KinematicsMechanics, the oldest of the physical science, is the study of the motion of objects. The calculation of the path of a baseball or of a space probe sent from Earth to Mars is among its problem. When we describe motion we are dealing with that part of mechanics called kinematics. When we relate motion to the force associated with it and to the properties of the moving objects, we are dealing with dynamics. We will discuss kinematics in this chapter and dynamics in the following chapters.Usually the motion of an object is very complicated. We will restrict to discuss the motion of a particle, or mass-point.2.1 Position, Displacement, Velocity, and Acceleration1.Position vector r is a vector thatextends from a reference point (usuallythe origin of a coordinate system) tothe object. We can write r as r =x i +y j+z k.Where x i, y j, and z k the vector components of r and the coefficients x, y, and z are its scalar components.The unit of position is meter in SI system.2. Displacement vector r ∆: As an object moves, its position vector changes. If the object has position vector r 1at time t 1 and position vector r 2 at later time t t ∆+1, as shown in above figure. The displacement vector is the difference of the two position vectors r 2 and r 1 during the time intervalt ∆, whichcan be written ask z z j y y i x x k z j y i x k z j y i x r r r )()()()()(12121211122212-+-+-=++-++=-=∆ See sample problem 4-1and checkpoint1 of P54 and P55.3. Average velocity vector : If a particle moves through a displacementr ∆ in a time interval t ∆, then its averagevelocity isk t z j t y i t x t k z z j y y i x x t r v ∆∆+∆∆+∆∆=∆-+-+-=∆∆=)()()(121212 Instantaneous velocity vector : is defined as the limit of the average velocity as the time intervalt ∆becomesinfinitesimally short. We can write it ask v j v i v k dt dz j dt dy i dt dx dt r d t r v z y x t ++=++==∆∆=→∆0lim The direction of the instantaneous velocity of a particle is always tangent to the path of the particle, as shown in the figure.The unit of velocity is 1-⋅s m .The magnitude of the velocity of a particle is its speed . Speedis a scalar quantity.dt r d dt r d v v ===Average speed is the total distance covered by the object over the time interval.t cedis total v ∆=tan4. Acceleration vector : When a partic le’s velocity changes from v 1 to v 2 in a time intervalt ∆, its average acceleration a during t ∆ ist v v t v ∆-=∆∆=12Instantaneous acceleration vector is defined as the limit of the average acceleration as the time intervalt ∆ becomes infinitesimally short. It can be express ask a j a i a k dtdv j dt dv i dt dv dt v d t v a z y x z y x t ++=++==∆∆=→∆0lim The unit of acceleration is2-⋅s m .2.2 One Dimensional Motion with Constant Acceleration.1. The rule of one-dimensional motion with constant acceleration can be derived as follow:200000000)(21)()]([)(sin 00t t a t t v s s dt t t a v ds t t a v v adt dv adtdv therefore a dt dv or c a ce t t v v -+-+=⇒-+=-+=⇒====⎰⎰ We also have:)(2])(21)([2)()(202020002020200202s s a v t t a t t v a v t t a t t a v v v -+=-+-+=-+-+= If we suppose t 0=0, Then we haveasv v at t v s s atv v 2212022000=-++=+= 2. Free-falling bodies: If you throw an object either up or down and could somehow eliminate the effects of air on its flight, you would find that the object accelerates downward at a certain rate. We call it free-fall acceleration g. The object is a free-falling body . The value of g change slightly with latitude and with height. At sea level in the mid-latitudes the value is9.8m/s 2, we will use it to solve the problem in this course. If we choose the upward direction along the y axis, then the free fall acceleration is negative. If we suppose t 0=0 and y 0=0, we will have:202200221v v ay gt t v y gtv v -=-=-=2.3 Projectile MotionIf a particle moves in a vertical plane during free fall, it undergoes projectile motion . The object can be called a projectile . Throughout we will assume that the air has no effect on the motion of the projectile.Projectile motion has the feature that the horizontal motion and the vertical motion are independent of each other. So we can deal with them respectively.If we choose the rectangular coordinate as the following figure, then000000sin cos θθv v and v v y x ==1. The horizontal motion is motion without acceleration. So we havet v t v x x x 0000cos θ==-2. The vertical motion is the motion of free fall. So we have)(2)sin ()sin (21)sin (210200200200200y y g v v gt t v v gt t v gt t v y y y y y --=-=-=-=-θθθ 3. The equation of the path : Let x 0=0 and y 0=0, eliminating the t between the equations above, we get20020)cos (2)(θθv gx x tg y -=It is the equation of a parabola , so the path is parabolic.4. The horizontal Range R is horizontal distance the projectile has covered when it returns to its initial height. We have 021)sin (21cos 200200000=-=-=-=-=gt t v gt t v y y tv x x R y θθ Eliminating the time t, we get02000202sin cos sin 2θθθgv g v R ==2.4 Uniform Circular Motion1. A particle is in uniform circular motion if it travels around a circle or circular arc at constant speed. Although the speed does not vary, it means the magnitude of the velocity is a constant quantity, but the direction of the velocity alters with the time. So the particle is accelerating.2. The feature of uniform circular motion: We have)()()(00t v v t r r t t v s s c v ==-=-⇒= 3. The acceleration of uniform circular motion:(1). Direction: (2). Magnitude: r v a 2=4. Another look at the acceleration of uniform circular motion: 0200000)()(n rv r r v v dt t d v dt t d v t dt dv t v dt d dt v d a =-==+=== 5. The figure shows the relation betweenthe velocity and acceleration vectorsat various stages during uniformcircular motion. Both vectors haveconstant magnitude as the motionprogress, but their directions change continuously. The velocity is always tangent to the circle in the direction of motion; the acceleration is always directed radially inward . Because of this, the acceleration associated with uniform circular motion is called a centripetal acceleration .2.5 Relative motion in two or three dimensions1. The figure shows reference frames A and B, now two dimensions, of course, it can be extended to three dimensions if you would like to. Two observers are watching a movingparticle P in the twodifferent frames. Weassume that the two framesare separating at a constantvelocity v BA (both of themare inertial frames ) and wefurther assume that their axes remain parallel to each other for simplicity.2. Position vectors: If two observers on frame A and B each measure the position of particle P at a certain instant. From the vector triangle in the figure, we have the vector equation BA PB PA r r r +=3. Velocity vectors : If we take the time derivative of above equation, we will get the velocity relation of the particle as measure by the two observersBA PB PA v v v +=4. Acceleration vectors : If we take the time derivative of above equation, we will have a connection between the two measured accelerationPB PA a a =。
Corresponding Relation Between SHM and UCMThe simple harmonic motion is the side view of circular motion.Draw x-t Diagram Using Circle of ReferenceExampleUsing the phasorExampleExample: π=+0v >ExampleExampleExampleExample:A wooden block floats in water. We press it until its upper surface2Example ——Vertical SHM:Suppose we hang a spring with force constant k and suspend from it a body with mass m. Oscillation will now be When the body hangs at rest, in equilibriumTake x=0 to be the equilibrium position, andThe body ’s motion is still SHMwith the angular frequency:kmω=ExampleWhen the body is at the position x, the total 0=)0l mg ∆−=0Example cont ’d§6 Damped OscillationsThe dissipative force causes the decrease in amplitude ——damping, the corresponding motion is called damped oscillation.Restoring force:Resistance force:Newton ’s second law:The solution:s F kx=−R bv=−F kx bv ma=−−=∑220d x b dx k x dt m dt m ++=(/2)cos()cos()b m t x Ae t A t ωφωφ−′=+=+22(/2)20, 22b m t k b b A Ae m m m ωω−⎛⎞⎛⎞′==−=−⎜⎟⎜⎟⎝⎠⎝⎠τ=2m /b is called damping time constant, or mean lift time.If the system no longer oscillates,and is called critically dampedDamped Oscillations Cont ’d§7 Forced OscillationsA forced oscillator is damped oscillator driven by an external force that varies periodically.A sinusoidally varying driving force:Newton’s Second Law:The solution:0()sin F t F t ω=202sin dxd xF t b kx m dt dt ω−−=202sin F d x b dx k x tdt m dt m m ω++=cos()x A t ωφ=+()022220/F mA b m ωωω=⎛⎞−+⎜⎟⎝⎠The forced oscillator in its “steady state”isoscillated with the frequency of driven force.The goblet breaks as it vibrates in the resonance In 1940, the Tacoma Narrows Bridge collapsed four months and six days after it was opened for traffic, due to gusty ResonanceUsing Circle of Reference12212cos()A A φφ+−1122122sin sin cos cos A A φφφφ++Chapter 12 Oscillatory Motion are in phase, the resultant amplitude take are out of phase, the resultant amplitude Chapter 12 Oscillatory MotionExample。
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运行这部分里的 .rm档需要RealOne™ 软件RealOne™ Player software is required to run the .rm files in this section.Walter Lewin 教授悬挂在单摆上证明单摆的周期与单摆的质量无关.这个展示在第十讲的录影里. (图片由Markos Hankin提供,物理系演讲演示组)Professor Walter Lewin demonstrates that the period of a pendulum is independent of the mass hanging from the pendulum. This demonstration can be viewed on the video of Lecture #10. (Image courtesy of Markos Hankin, Physics Department Lecture Demonstration Group).这个录影索引提供了一个关于本课程的细目录.它给出了每一课中相应段落的起始时间.进入某个特别的段落,只需要在RealPlayer视窗中打开该课并调整滑条到指定的起始时间The Video Index gives a breakdown of topics and the start time of the corresponding segment in each lecture. To access a particular segment, play the lecture in a RealPlayer window and adjust the slider bar to the designated start time.课课程单元影片大纲影片1 十的幂-单位- 尺度-测量- 不确定度-空间分析-缩放比例的讨论Powers of Ten - Units - Dimensions -Measurements - Uncertainties - DimensionalAnalysis - Scaling Arguments第1课录影索引Lecture 1VideoIndex(线上- 80K)(线上- 300K)(下载- 300K)2 一维运动学- 速率- 速度- 加速度1D Kinematics - Speed - Velocity - Acceleration第2课录影索引(线上- 80K)(线上- 300K)Forced Oscillations - Normal Modes - Resonance - Natural Frequencies - Musical Instruments 索引Lecture 31VideoIndex(线上- 300K)(下载- 300K)32 热- 热膨胀Heat - Thermal Expansion第32课录影索引Lecture 32VideoIndex(线上- 80K)(线上- 300K)(下载- 300K)33 气体动力论- 理想气体定理- 大气等温线- 相图- 相变Kinetic Gas Theory - Ideal Gas Law - IsothermalAtmosphere - Phase Diagrams - Phase Transitions第33课录影索引Lecture 33VideoIndex(线上- 80K)(线上- 300K)(下载- 300K)34 奇妙的量子世界- 经典力学的崩溃The Wonderful Quantum World - Breakdown ofClassical Mechanics第34课录影索引Lecture 34VideoIndex(线上- 80K)(线上- 300K)(下载- 300K)35 告别特别节目- 高能天体物理Farewell Special - High-energy Astrophysics第35课录影索引Lecture 35VideoIndex(线上- 80K)(线上- 300K)(下载- 300K)RealOne™ 是商标或RealNetworks公司的注册商标RealOne™ is a trademark or a registered trademark of RealNetworks, Inc.。
Chapter 12 The Kinetic Theory of GasesClassical thermodynamics has nothing to say about atoms or molecules. Its laws are concerned only with such macroscopic variables as pressure, volume, and temperature. However, we know that gas is made up of atoms or molecules (groups of atoms bound together). The pressure exerted by a gas must surely be related to steady drumbeat of its molecules on the walls of its container. The ability of a gas to take on the volume of its container must surely be due to the freedom of motion of its molecules. And the temperature and internal energy of a gas must surely be related to the kinetic energy of these molecules. Perhaps we can learn something about gases by approaching the subject from this direction. We call this molecular approach the kinetic theory of gases.12.1 Ideal Gases1、Our goal in this chapter is to explain the macroscopic properties of a gas, such as its pressure and its temperature, in terms of the behavior of the molecules that make it up.2、The experiments show that, at low enough densities, all real gases tend to obey the relation )pV , innRTgas(lawidealwhich p is the absolute pressure, A N N n /= is the number ofmoles of gas present, andR , the gas constant , has the same value for all gases, namely, K mol J R ⋅=/31.8. The temperature T must be expressed in kelvins. Above equation is called the ideal gas law . Provided the gas density is reasonably low, it holds for any type of gas, or a mixture of different types, with n being the total number of moles present.3、 Work done by an ideal gas at constant temperature :(1). Suppose that a sample of n moles of an ideal gas, confined to a piston-cylinder arrangement, is allowed to expand from an initial volume i V to a final volume f V .(2). Suppose further that the temperature T of the gas is held constant throughout the process. Such a process is called an isothermal expansion (and the reverse is called an isothermal compression ).4、Let us calculate the work done by an ideal gas during an isothermal expansion, we have⎰⎰==f i f i V V V V dV V nRT pdV W i fV V nRT ln =. Recall that the symbol ln specifies a naturallogarithm, that is, a logarithm to base e .12.2 Pressure, Temperature, RMS Speed, and Translational Kinetic Energy1、 Let n moles of an ideal gasbeing confined in a cubical box ofvolume V , as in the figure.2、 A typical gas molecule, of mass m and velocity v, is about to collide with the shaded wall. We assume that any collision of a molecule with a wall is elastic , so the change in the particle’s momentum is along the x axis and its magnitude is x x x x mv mv mv p 2)()(-=--=∆. Hence the momentum x p ∆ delivered to the wall by the molecule during the collision is x mv 2+.3、 The average rate at which momentum is delivered to the shaded wall by this single molecule is Lmv v L mv t p x x x x 2/22==∆∆. 4、 F rom Newton’s second law, the rate at which momentum is delivered to the wall is the force acting on that wall. To find the total force, we must add up the contributions of all molecules that strike the wall, allowing for the possibility that they all have different speeds. Dividing the total force x F by the area of the wall then gives the pressurep on that wall. Thus))((///2222132222212xN x x xN x x x v v v Lm L L mv L mv L mv L F p +++=+++== , where N is the number of molecules in the box.5、 Since A nN N = in which A N is Avogadro’s number ,above equation can be re-expressed as223x x A v V nM v L nmN p ==, where A mN M = is the molar mass of the gas , and V is thevolume of the box.6、 For any molecules, 2222z y x v v v v ++=. Because there aremany molecules and because they are all moving in random direction, the average values of the squares of their velocity components are equal, so that3/22v v x =. Thus V v nM p 32=. 7、 The square root of 2v is a kind of average speed, called the root-mean-square speed of the molecules and symbolized by rms v .8、 Using the relationnRT pV = for an ideal gas, it leads to M RT nM pV v v rms 332===.See the RMS speeds ofsome molecules in Table.9、 The averagetranslational kinetic energy of a single molecule of an ideal gas iskT T N R M RT m mv mv K A rms 23)(23)3)(21(212122=====, where the constantK J N R k A /1038.1/23-⨯== is the Boltzmann constant . So we come to the conclusion that at a given temperature T , all ideal gas molecules, no matter what their mass, have the same average translational kinetic energy . When we measure the temperature of a gas, we are also measuring the average translational kinetic energy of its molecules.12.3 Mean Free Path1、 Table gives us the RMS speed of some molecules, A question often arises: If molecules move so fast, why does it take as long as a minute or so before you can smell perfume if someone open a bottle across a room? This is because although the molecules move very fast between collisions, a given molecule will wander only very slowly away from its release point.2、 The right figure shows the path of atypical molecule as it moves through the gas,changing both speed and direction abruptlyas it collide elastically with other molecules.Between collisions, our typical moleculemoves in a straight line at constant speed.Although the figure shows all the other molecules as stationary,they too are moving in much the same way.3、 One useful parameter to describe this random motion is the mean free path λ. As its name implies, λ is the average distance traversed by a molecule between collisions.4、 The expression for the mean free path can be deduced from following steps:(1). First we assume that our molecule is traveling with a constant speed u and that all other molecules are at rest. We assume further that the molecules are spheres of diameter d. A collision will then take place if the centers of the molecules come within a distance d of each other. A help way to look at this situation is to consider our single molecule to have radius of d and all the other molecules to be points.(2). As our single molecule zigzags through the gas, it sweeps out a short cylinder of cross-sectional area 2d π between successive collision. If we watch this molecule for a time interval t ∆, it moves a distance t u ∆. So the volume of the cylinder is t u d ∆2π. The number of collisions that occur is then equal to the number of (point) molecules that lie within this cylinder. It is t u d V N ∆2)/(π.(3). The mean free path is the length of the path divided by the numberof collisions)/()/(22V N u d v V N t u d t v collisions of numbers path of length ππλ=∆∆==.(4). The u in the denominator is the mean speed of our single molecule relative to the other molecules, which are moving. A detailed calculation, taking into account the actual speed distribution of the molecules, gives thatv u 2=. So )/(212V N d πλ=.5、The mean free path of air molecules at sea level is about m μ1.0. At an altitude of km 100, the mean free path is about 16cm. At 300 km, the mean free path is about 20 km.12.4 The Distribution of Molecular Speeds1、 In 1852, Scottish physics James Clerk Maxwell first solved the problem of finding the speed distribution of gas molecules. His result, known as Maxwell’s speed distribution law , isRT v M e v RT M v P 2/22/32)2(4)(-=ππ. Here v is the molecular speed, Tis the gas temperature, M is the molar mass of the gas, and R is the gas constant. The quantity )(v P is a probabilitydistribution function , defined as follow: The product dv v P )( isthe fraction of molecules whose speeds lie in the range v to dv v +.2、 As figure (a) shows, thisfraction is equal to the area of astrip whose height is)(v P and whose width is dv . The totalarea under the distribution curvecorresponds to the fraction ofthe molecules whose speed liebetween zero and infinity. Allmolecules fall into this category,so the value of this total area isunit.3、 There are two otherspeeds. The most probable speed P v is the speed at which )(v P is a maximum. The average speed v is a simple average of the molecular speeds. We will find thatM RT v π8= and M RT v P 2=，so we have relationsrms P v v v <<.12.5 Degree of Freedom and Molar Specific Heats1、 The right figure showskinetic theory models ofhelium (a monatomic gas),oxygen (diatomic), and methane (polyatomic). On the basis of their structure, it seems reasonable to assume that monatomic molecules, which are essentially point-like and have only a very small rotational inertia about any axis, can store energy only in their translational motion. Diatomic and polyatomic molecules, however, should be able to store substantial additional amount of energy by rotating or oscillating.2、To take these possibilities into account quantitatively, we use the theorem of the equipartition of energy, introduced by James Clerk Maxwell: Every kind of molecules has a certain number f of degree of freedom, which are independent ways in which the molecule can store energy. Each such degree of freedom has associated with it, on average, an energy of 2/kT per molecule (or 2/RT per mole).3、For translational motion, there are three degrees of freedom. For rotational motion, a monatomic molecule has no degree of freedom. A diatomic molecule has two rotational degrees of freedom. A molecule with more than two atoms has six degrees of freedom, three rotational and three translational.4、Internal energy is the energy associated with random motion of atoms and molecules. So a sample of n moles of a gas containsnN atoms. The internal energy of the sample is then AnRT f kT f nN E A 2)2)((int ==. Thus, the internal energy int E of anideal gas is a function of the gas temperature only; it does not depend on any other variable .5、 Molar specific heat at constant volume : (1) According to the definition of molar specific heat, we have T nC Q v ∆=. (2) Substituting it into the first law of thermodynamics, we find T nC T nC W Q E V v ∆=-∆=-=∆0int . (3) So we will have the relation R f T n RT f n T n E C v 2)2(int =∆∆=∆∆=. 6、 Molar specific heat at constant pressure : (1) the heat is T nC Q p ∆=. (2) According to the first law of thermodynamics, wehave T nC T nR T nC V p T nC W Q E V p p ∆=∆-∆=∆-∆=-=∆int .(3) So we have relationR C C V p +=.12.6 The Adiabatic Expansion of an Ideal Gas1、 Using the first law of thermodynamics, we have pdV Q dE -=int pdV -=. It means pdV dT nC V -=.2、 From the ideal gas law ()nRT pV = we have nRdT Vdp pdV =+ 3、 Combining these two equations, we have 0)(=+V dV C C p dp V p . Replacing the ratio of the molar specific heats with γ and integrating yieldt cons a pV tan =γ.11 4、 Using the ideal gas law, we also have t cons a TV tan 1=-γ, and t cons a T p tan 1=--γγ.5、 Free expansion: Since a free expansion of a gas is an adiabatic process that involves no work done on or by the gas, and no change in the internal energy of the gas. A free expansion is thus quite different from the type of adiabatic process described above, in which work is done and the internal energy changes. Since the int ernal energy isn’t change for free expansion, we must havef i RT f n RT f n 22=. It means f i T T =, andso f f i i V p V p =.。
