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Date Created: 09/17/15
Potential Energy and Equilibrium in 1D Figures 627 628 and 629 of TiplerMosca dU Fx dag A particle is in equilibrium if the net force acting on it is zero dU F 0 x dx In stable equilibrium a small displacement results in a restoring force that accelerates the particle back toward its equilibirum position de d2U F1 dz de lt0 as 960 0 and Fm Fx1c0 l Ax Fx1c0 l Ax l OAx2 060 Example The rest position of a spring for which we have dFm d2U dx W In unstable equilibrium a small displacement results in a force that accelerates the particle away from its equilibrium position dF d2 U gt0 dx de Finally in neutral equilibrium a small displacement results in zero force and the particle remains in equilibrium Questions Assume the ball in the bowl has mass m and is positioned at a distance 5 8 of height h above the lowest point of the bowl Which of the following is true 1 The potential energy of the ball is given by mgs 2 The potential energy of the ball is given by mgh 3 The potential energy of the bowl is given by mgh 4 The potential energy of the bowl is given by mgs 5 None of the above statements holds Combine the correct statements to one of the following choices A 1121 d 0 holds for a conservative force where 81 and 82 denote the endpoints of the integration curve B f 13 d 0 holds for a conservative force where the integral is over a closed curve C Friction is a conservative force D The weight due to gravity here close to the surface ofthe earth is a conservative force E The force of friction allows to define a potential F A conservative force allows to define a potential 1ABDF 2BCD 3BCDEF 4ACE 5BDF 6ABCDEF 7ADF Example The potential energy function between two atoms in a diatomic molecule is defined for x gt 0 and given by Um 00 3112 36 U0 gt0 agt0 Find the equilibrium position dU a12 a6 0 U 12 12 dw 0 9653 x5 x0 12a12x5 120169653 gta6 968 gt 960 a Is the equilibrium stable or unstable d2U Q12 a6 U 12gtlt13 12gtlt7 de x0 0 JF x54 x3 U U 71 3 156 84 4261 lt 0 stable Energy Conservation Tipler Chapter 7 The law of conversation of mechanical energy For a system on which only conservative internal forces act the sum of kinetic and potential energy is constant E K U constant Proof Z AK AK Wtotal ZAU AU Therefore AK AU 0 If E K U is the intitial and Ef Kf l Uf is the final mechanical energy conservation of energy implies Example 72 of Tipler Mosca Pendulum without friction or drag forces 1 2 EKU mv mgh At the turning points EmghmaLX as 210 At the bottom 1 E Emvfnax as h0 Therefore 1 2 hmax 2 m Umax m g vmax 2 g hmax and hmax constant The conservation of mechanical energy is often much easier to use than Newton39s laws for solving certain problems of mechanics Example Find the speed at the bottom of the two blocks of figure 77 of Tipler Mosca Note the obtained results do not go beyond Newton39s laws because the conservation of mechanical energy was derived from them General Conservation of Energy Let ESys be the total energy of a given system Em be the energy that enters the system Eout be the energy that leaves the system The law of conservation of energy then states Ei Eout AEsys Alternatively The total energy of the universe is constant Energy can be converted from on form to another or transmitted from one region to another but energy can never be created or destroyed The WorkEnergy Theorem If work is the only form of energy transferred to a system the law of conservation of energy becomes Wext AEsys where Wext is the work done on the system by external forces WorkEnergy Theorem with Kinetic Friction The energy dissipated by friction is thermal energy heat f A8 AEtherm where f is the frictional force applied along the distance A8 The workenergy theorem reads then Wext AEmech l AEtherm Other Forms of Energy Chemical Energy It is due to differences in the molecular binding energies For instance food for people and gasoline for cars In these two cases chemical energy is released through the process of oxidation Nuclear Energy Due to the nuclear binding energy which is related to mass differences by Einstein39s famous equation Ezmc2 Announcement Dr Baba Jain will be holding a help session on Friday Oct 10 at 330 pm in UPL 110 Look up your mini scores on the CAPA system and compare with your records Dr Frawley is responsible for mini 3 Results were not good 79 average Mini 4 Next Thursday I will spend 15 20 minutes today on preparation Rotation Chapter 9 of TiplerMosca Figure 9 2 of Tipler Mosca shows a disk which rotates about a fixed axis perpendicular to the disk and through its center Rigid body As the disk turns the distance between any two particles that make up the disk remains fixed Consider a particle at distance 7 from the center and let 6 be the angle measures counterclockwise from a fixed reference line When the disk rotates through and angular displacement d6 measured in radians rad the particle moves through a circular arc of length d3 rd6 The time rate of change of the angle is the angular velocity d6 Question A CD ROM disk is rotating at 3000 revolutions per minute What is the angular speed in radians per second 1 284 rads 2 314 rads 3 334 rads Angular acceleration a dw e dt dt2 39 Tangential velocity ds d6 vt no Tangential acceleration dvt dw E TETOC at Do not confuse this with the centripetal acceleration 2 2 ac v t law MP 7 7 Rotational Kinetic Energy 1 1 1 Krot mz v 2 m w2 Ii W2 The quantity I 2 mi 7 Z where n is the distance of particle 239 from the rotation axis is called Moment of Inertia The kinetic energy becomes 1 Krot inZ Ir2dm For a continuous object Calculating the Moment of Inertia Discrete Systems of Particles Simply apply the definition I Emir 1 Examples PRS For figure 9 3 of Tipler Mosca the result is I nma2 where n is an integer Push this number For figure 9 4 of Tipler Mosca n is another or the same integer Push this number Continuous Objects The moment of inertia is calculated by integration Examples folow 1 Uniform stick about an axis perpendicular to the stick and through one end figure 9 5 of Tipler Mosca L L M 1 dem xZ dx 0 0 L pick one 1 ML3 2 ML2 3 ML2 2 Hoop of radius R about a perpendicular axis through its center figure 9 6 of Tipler Mosca Ir2dmR2dmMR2 3 Uniform disk of radius R about a perpendicular axis through its center figure 9 7 Tipler Mosca M dmZdAr Where A7rR2 and Ar7rr2 so that R R R dAr 27rrdr7rr2i07rR2A 0 0 Therefore dm R227TTdT R2 rd and 2M R Ir2dm 7 ng R2 0 pickone 1 1 1 2 1 MR4 2 MR4 3 MR2 4 MR2 4 2 2 3 3 Uniform cylinder of radius R about its axis figure 9 8 of Tipler Mosca Using the result for the uniform disk we have 1 1 I R2dmdisk MR2 Results for Uniform Bodies of Various Shapes are collected in Table 9 1 of Tipler Mosca Examples we had 1 hoop IMR2 full disk I MR2 Rolling Objects without Slipping The point of contact moves a distance 3 Rgb The velocity and acceleration of the CM are vcm Ru and acm Ra The kinetic energy of a rolling object is 1 1 K Mv m 1w2 Example Objects rolling down an inclined plane The terminal velocity follows from energy conservation 2 2 I 2 mghKmvm l I2UCRH12m l vcm 2 2mgh cm m where the moment of inertia I is about the CM axis A hoop a full disk and a body on tiny wheels 3 have equal masses Starting from rest they roll down the same inclined plane In which order do they arrive 1 123 2 132 3 213 4 231 5 312 6 321 Collisions In a collision two objects interact strongly for a very short time so that external forces can be neglected When the total kinetic energy of the two objects is the same after the collision as before the collision is called elastic Otherwise it is called inelastic In the perfectly inelastic coistion all the energy relative to the center of mass is converted to thermal or internal energy of the system and the two objects stick together Collisions in 1D Momentum conservation PM P2f m1 Ulf m2 U2f m1 7111 W2 7121 2911 P21 Perfectly Inelastic Collisions m1 7111 W2 7121 Ulf U2f vcm Wlth vcm m1 m2 Demonstration Airtrack Example Figure 830 of Tipler Mosca A bullet is fired into a hanging target which is at rest The target with the bullet embedded swings upward Find the speed of the bullet from the height reached Solution Energy conservation gives 1 5m1m2vf2 m1m2gh gt 39Uf v2gh Let m1 be the mass of the bullet and m2 be the mass of the target Now mlvim20 39Uf39Ucm m1i m2 and m1 m2 x29h m1 1 is the initial speed of the bullet Elastic Collisions The final kinetic energies are equal 1 2 1 2 1 2 1 2 m101f m2212f m1vh m2v2i Rearranging this equation gives 2 2 2 2 m2 02f U21 m1 U11 Ulf m2 02f U21 02f 7121 m1 U11 Ulf 7111 Ulf Momentum conservation may be rewritten as m2 U2f U21 m1 U11 Ulf Hence U2f U21 7111 Ulf U2f U1f U2i 7111 The righthandside of this equation is called speed of approach and the lefthand side speed of recession Example 1 Find Ulf and U2f for an elastic coission with m1m2m v1igt0 and 0210 Momentum conservation gives Ulf U2f U11 and the elastic equation becomes U2f Ulf 7111 Adding both equations we find 2112f 22111 18 Ugf U11 Example 2 Find Ulf and U2f for an elastic collission with m22m1 m1m U11gt0 and UgiZO Momentum conservation gives U1f 2 02f 7111 while he elastic equation stays the same U2f U1f 7111 Adding both equations we find 2 4 1 3U2fZ2U1i 18 Ung Uh gt 1fZU1 01 111 Kinetic Energy of a System Although the total momentum of a system is conserved when the net external force is zero the total mechanical energy changes because the internal forces may be nonconservative This is for instance the case in our example of the astronaut Her muscles use chemical energy to push the pannel away Before the push the kinetic energy of the system was zero after the push it is 2162 U 7 2 Kafter ma p 2 However it is possible to decompose the kinetic energy of a system into a CM energy which does not change when the net external force is zero and kinetic energies relative to the CM which may change due to internal forces The kinetic energy of a system of particles is TL 77 The velocity of each particle can be written as the sum of the velocity of the CM 27cm and the velocity of the particle relative to the CM reference frame TL s s s s mi 717 U7 vi vcm Wlth vcm 271 mi see figure 840 of Tipler Mosca for 239 1 2 Then and 7L1 7L1 1 n 1 a n a ngwgm Ucm 39 11 11 Now 39I L M cm 0 11 because the CM velocity is zero relative to the CM Therefore the kinetic energy of a system of particles is 1 n 1 K ingm Krel Where Krel 11 M is the total mass and Krel is the kinetic energy of the particles relative to the CM Impulse and Average Force Figure 824 of TiplerMosca The impulse of the force 13 on one of the particles is defined as tf 12 th t l where the integration is over the time interval At tf ti of the collision The magnitude of the impulse is the area under the Fversus t curve The impulse equals the total change in momentum during the time interval tfd r p I dt iA fti dt pr 29 p The average force for the time interval is a 1 in f Fav th Altti At Example Figure 826 of TiplerMosca A car with an 80 kg crash test dummy drives into a wall at 25 ms about 56 mih Estimate the force on the seat belt Steps 1 The dummy s initial momentum is mu 2000kg ms 2 The impulse is the change of the momentum I 2000N 8 3 Estimate the collision time With Ax 1m and vav 125m8 one gets At 008 s 4 Compute the average force FaV 25kN Rocket Propulsion mathematically ambitious We consider the simple model of a rocket where the fuel is burned at a constant rate dm constant dt and the speed of the exhaust gas relative to the rocket is another constant uex gt 0 The mass of the rocket at time t becomes mmtmO Rt where m0 is the rocket mass at the initial time t 0 The momentum of the rocket at time t is P mv We assume that the rocket moves against a constant gravitational acceleration In the instantaneous rest frame of the rocket momentum conservation reads dP mgdt mdv uexdm mdv uedet Note dP 0 if there is no external force ie g 0 We rewrite the momentum conservation equation as m0 Rtgdt m0 Rt do ueXRdt Dividing both sides by me Rt gives uexR dt do dt 9 m0 Rt Such an equation is solved by the method of separation of variables This simply means that the term with o is brought to one side of the equation and all terms with t to the other uexR dt m0 Rt do gdt Now both sides can be integrated 7 t 1 1 d dt eXR dt and the integration is elementary 1nmO Rt t t t exR 717 gu R O gt uex lnm0 Rt 1nm0 gt uex 1n me which is equation 843 of TiplerMosca we used m m0 Rt and the additionsubtraction rule for the logarithmic function in the last step Motion in Two and Three Dimension Displacement Vector A is defined by its magnitude and direction and graphically represented by an arrow The magnitude is written as or simply A Addition of displacement vectors A by putting the arrows together Tipler figure 31 Note this does not imply C A B Multiplication by a scalar number B3A Direction unchanged for 8 gt 0 inverted for 8 lt 0 magnitude B isi A Subtracting a vector 6 A 3 Add the negative vector direction inverted Components of vectors 2D Tipler figure 39 Ax A 3086 Ay A sin6 A A Ag Pythagoreas in3D A A3A Ag Addition in components Cm Ax Bx CyZAyBy Unit Vectors along the coordinate axis Tipler figure 3 12 Notations of the literature Sgt ll E3gt ll Cbgt x in m direction gtgt gt Qgt ll C13gt y in y direction k z ez in z direction Now every vector can be written like gt AAxEAy AzE Tipler table 3 1 Overview of Vector Properties Position Velocity and Acceleration Position vector Instantaneous velocity vector Tipler figure 3 14 77 mm A77 d7 At gtO Component by component dm dy dz 2 ZE dt Relative velocity 77pB T7pA 7714B Example Frame A a train and frame B ground Average acceleration vector A27 E Instantaneous acceleration vector 67 lim g d 77 At gt0 At dt dt2 aav Component notation Projectile Motion We neglect friction in the following Differential equations d2mdvx 0 WWa x d2ydvy WWa yg and 998177252 as approximately measured in the last lecture First integration 2775 Initial velocity 770 vac 710x voy gt 710 210 308090 voy v0 Sin90 Second integration solution 7775 mt 0 2101575 1 yo ont Possible initial positions 960 yo 0 Tipler Mosca figure 3 25 Path of a projectile with velocity components Illustration Air track Questions on Projectile Motion PRS Consider the 2D orbit given by 1 mtv0xt v0xgt0 ytv0yt gt2 voygt0 At which time is the highest point reached 2 1 tmaX tmaX voy g 9 At which time is the ground 3 0 reached again 2 2 1 751 00x 2 751 my 9 g Consider two canon balls and assume their initial conditions for 960 and 2103 are different while those for yo and my are identical Which of the following holds pick one Demonstration 1 The two canon balls hit the ground at distinct times 2 The two canon balls hit the ground at the same time 3 The provided information on the initial conditions is insufficient to decide between answer 1 and 2 Two point particles have the following ininital conditions 11010 11020 1095 2095 210 220 and ploy 0 Ugoy gt 0 Which of the following holds 1 At time zero the two particles are at different positions 2 At time zero the two particles are at the same position and they will remain to stay together 3 At time zero the two particles are at the same position and at all future times their positions are different 4 At time zero the two particles are at the same position and they will meet once again in the future Besides this their future positions will be different Tipler Mosca figure 3 31 Shooting a monkey Demonstration The gun is fired at the same instant when the monkey drops Will the dart hit PRS 1 Yes and 2No Waves Encountering Barriers Reflection a nd Refraction When a wave is incident on a boundary that separates two regions of different wave speed part of the wave is reflected and part is transmitted Figure 1517 of Tipler Mosca shows a A pulse on a light string that is attached to a heavier string c A pulse on a heavy string that is attached to a light string In the first case the replected pulse is inverted in the second not The same happens with a light and a heavy spring attached to one another In three dimensions a boundary between two regions of differing wave speed is a surface Figure 1519 of TiplerMosca shows a ray incident on such a surface The reflected ray makes an angle with the normal to the surface equal to that of the incident ray The transmitted ray is bent towards or away from the normal a process called refraction When the wave speed in the second medium is greater than that in the incident medium the ray is bent away from the normal As the angle of incidence is increased the angle of refraction increases unit it reaches 900 This defines a critical angle of incidence and for even greater incident angles there is no refracted ray a phenomenon called total internal reflection In total internal reflection the wave function drops to zero exponentially fast so that it becomes negligible within a few wavelength from the surface This can lead to barrier penetration or tunneling Diffraction A wave encountering a small obstacle tends to bend around the obstacle This bending of the wavefront is called diffraction When a wave encounters a barrier with an aperture which is much smaller than the wavelength the wave bends and spreads out as a spherical circular wave This distinguishes the waves from particles Figure 1522 of Tipler Mosca which can be regarded as waves with a wavelength so small that any aperture will be large in comparison In the ray approximation waves are traveling with no diffraction Sound waves with frequencies above 20000 Hz are called ultrasonic waves Because of their small wavelength they can be send out to be reflected from small objects Used by bats sonar sound navigation and ranging sonogram for diagnostic purposes The Doppler Effect When a wave source and a receiver are moving relative to each other the frequencey is not the same as that omitted This is called the Doppler effect after the Austrian physicist who predicted this phenomenon The change in the frequency is slightly different depending on whether the source or the receiver moves relative to the medium When the source moves the wavelength changes and the new frequency if found from the relation f v When the receiver moves the frequency changes while the wavelength is unchanged Consider that a source moves with speed us relative to the medium The frequency of the source is f0 The waves in front of the source are compressed whereas behind the source they are farther appart Figure 1524 of Tipler Mosca When 21 is the speed of the wave in the medium the new wavelength becomes Avius f0 In front of the source the minus sign applies behind the source the plus sign The receiver at rest gets then the frequency If3 f0 f0 vius liusU39 If the receiver moves with velocity u relative to the medium and the source is at rest the number of waves crests that pass by per second becomes Here the plus sign stand when the receiver moves towards the source otherwise the minus sign applies When receiver and source move one combines the two equations in the obvious way Example The radar used by police to catch speeders relies on the Doppler effect Electromagnetic waves emitted by the transmitter are reflected by the car and the Doppler shift is measured Schock Waves and Mach Number If a source moves with speed greater than the wave speed 21 there will be no waves in front of the source When a source accelerates to approach and pass wave speed the waves pile up as a shock wave In case of a sound wave this is heard as a sonic boom When the source travels at a constant speed u gt v the wave is confined to a cone of angle wt 21 sm6 Mach number ut u see Figure 1524 of TiplerMosca Superpositions of Waves Chapter 16 When two waves meet in space they add algebraically superposition The superposition of harmonic waves is called interference In 1801 Young observed the interference of light Davisson and Germer observed in 1927 the interference of electron waves The principle of superposition When two or more waves combine the resultant wave is the algebraic sum of the individual waves y3x7t y1x7t y2x7t Examples Figure 161 of Tipler Mosca Interference of Harmonic Waves Two wave sources that are in phase or have a constant phase difference are said to be coherent otherwise they are said to be incoherent We consider the superposition of two Figure 162 Tipler Mosca coherent waves 3J1 yo sinkx wt yg yo sinkx cut i 6 y3y1y2 2 yo Sink3x wty0 sink3x wt6 Using the trigonometric identity sin61 l sin62 2 cos61 622 sin61 l 622 yg 290 cos62 sinkx wt 62 The resulting wave has interesting properties If the two waves are in phase 6 0 the amplitude of y3 is 2y0 constructive interference Figure 163 of TiplerMosca If the two wave are 1800 out of phase 6 7r then yg 0 destructive interference Figure 164 of Tipler Mosca Beats This phenomenon is caused by the interference of sound waves with slightly different frequenciesWhat do we hear For equal amplitudes we have at a fixed point up to a phase constant the pressure fluctuation p 291 292 290 sinw1t 290 sian t p 2190 cosw1 C12 152 sinw1 02 152 2190 cosAw2 t si1r1cu8W t where Aw cal cog and wav cal l w22 The frequencies of the factors are 2A 2 av fbeat 2A1 and fav w 27r 27F The tone we hear has the average frequency fav whose amplitude 2190 00827r fbeat t is modulated by the beat frequency which is much smaller than the average frequency Figure 165 of TiplerMosca Beats can be used to tune a piano Phase Difference due to Path Difference The wave function from two coherent sources oscillating in phase can be written as Figure 166 of TiplerMosca 2991 192 2190 sinlltx1 wt 190 sink3x2 l wt An example is given in Figure 168 of Tipler Mosca The phase difference for these two wave function is A 6k32 ZB127TTx The amplitude is 2190 cos62 and Figure 169 of Tipler Mosca shows how the intensity varies with the path difference The DoubleSlit Experiment Interference of light is difficult to observe because a light beam is usually the result of millions of atoms radiating incoherently Coherence in optics is commonly achieved by splitting the light beam from a single source One method of achieving this is by diffraction of a light beam by two slits in a barrier Thomas Young 1801 Standing Waves Chapter 162 When waves are confined in space reflections at both ends cause the wave to travel in both directions For a string or pipe there are certain frequencies for which superposition results in a stationary pattern called standing wave The frequencies that produce these patterns are called resoncance frequencies Each such frequency with its accompanying wave function is called a mode of vibration The lowest frequency produces the fundamental mode or first harmonic For each frequency there are certain points on the string that do not move Such points are called nodes Midway between each pair of nodes is a point of maximum amplitude of vibration called an antinode String fixed at both ends Figure 1610 of TiplerMosca The standing wave condition is and with fn An v the resonance frequencies become f n U nf 71 1 2 3 n 1 7 7 where fl is the fundamental frequency Example pianos String fixed at one end Figure 1616 of Tipler Mosca The free end is an antinode The standing wave condition can thus be written n L n135 quot 4 and with fn An v the resonance frequencies become U 1 4L nflv 77 73757 fnn where fl is the fundamental frequency Example the air column in an organ pipe Wave Functions for Standing Waves Standing wave occur due to the superposition of the reflected waves When a sting vibrates in its nth mode a point on the string moves with simple harmonic motion Therefore the wave function is given by yx t Anx coswnt 6 where can is the angular frequency 6 the phase constant and Ax the amplitude which depends on the location on the string At an instant where the vibration is at its maximum amplitude the shape of the string is Anx An sinkn 16 where kin 27rn is the wave number The wave function for a standing wave in the nth harmonic can thus be written ynxt An sinkn x coswnt l 6 Superpositions of Standing Waves In general a vibrating system does not vibrate in a single harmonic mode Instead the motion consists of a mixture of the allowed harmonics and the wave function is a linear combination of the harmonic wave functions yxt 2A sinkn x coswnt 6 where k3 27rn can 27r fn and An 6 are constants which depend on the initial position and velocity of the string Interestingly each wave which fulfills the appropriate boundary conditions here y 0 at x 0 and x L can be expanded in this way Harmonic Analysis and Synthesis Waves can be analyzed in terms of harmonics Example Figure 1624 of Tipler Mosca shows the relative intensities for a tuning fork a clarinet and an oboe each playing a tone at a fundamental frequency of 440 Hz The inverse is harmonics synthesis the construction of a periodic wave from harmonic components Example Figure 1625 and 1626 of Tipler Mosca Wave Packets and Dispersion Pulses which are not periodic can also be expanded into sinusoidal waves of different frequencies However a continuous distributions of frequencies rather than a discrete set of harmonics is needed These are wave packets The characteristic feature of a wave pulse is that it has a beginning and an end If the duration of the pulse is At the range of frequencies Aw needed to describe the impulse is given by the relation Aw At N 1 Eg if At is very small Aw is very large and vice versa A wave pulse produced by a source of duration At has a width Ax vAt in space where v is the wave speed A range of frequencies Aw implies a range of wave numbers Ak Awv Therefore Aw At N 1 implies AkrAx N 1 If a wave packet is to maintain its shape as it travels all of the components must 15 travel at the same speed A medium where this happens is called nondispersive medium Air is a nondispersive medium for sound waves but solids and liquids are generally not A familiar example for the dispersion of light waves is the rainbow When the speed of the wave component depends only slightly on the their wavelength the wave packet changes shape only slowly as it travels However the speed of the wave packet called group velocity is not the same as the average speed of the components called phase velocity For example the group velocity of surface waves in deep water is half the phase velocity Kinematics Displacement of a point particle A f2 31 where E1 is the position at time t1 and E2 is the position at time t2 152 gt151 Instantaneous velocity dz E This is the slope of the tangent of the curve at at t and called derivative 17 The instantaneous acceleration is dt dt2 i Motion With Constant Acceleration d2 6 Javera e dt g Integration df 4 v d dt 0 Here 270 is the velocity at time zero the first initial condition l Second integration 1 IE 0U0t at2 Here ED is the second initial condition the position at time zero SI Units m and 5 1D 1 2 ZEZEOU0t CLt How long does an object take to fall from a 200 m heigh tower 1 2045 2 4075 3 6395 4 4525 What is the velocity of the object when it hits the ground 1 200 ms 2 100 ms 3 626 ms 4 1252 ms Newton39s Laws 1 Law of inertia An object continues to travel with constant velocity including zero unless acted on by an external force 2 The acceleration a of an object is given by m6 net 7 where m is the mass of the object and net the net external force 3 Action Reaction Forces always occur in equal and opposite pairs If object A exterts a force on object B an equal but opposite force is exterted by object B on A Friction An object may not move because the external force is balanced by the force f8 of static friction Its maximum value fsamm is obtained when any further increase of the external force will cause the object to slide To a good approximation fsamax is simply proportional to the normal force fsmar Ms Fn where us is called the coefficient of static friction Kinetic friction also called sliding friction Once the box slides a constant force is needed to keep it sliding at constant velocity The opposing force is the force of kinetic friction In a good approximation it is also simply proportional to the normal force fk LLan where uk is called the coefficient of kinetic friction Experimentally it is found that uk lt us Work and Energy Motion With Constant Force The work W done by a constant Force 13 whose point of application moves through a distance Azf is defined to be W F 3086 Ax where 6 is the angle between the vector 13 and the vector Af If Azf is along the xaxis then W Fm Ax holds Work is a scalar quantity that is positive if A96 and F96 have the same sign and negative otherwise The Si unit of work and energy is the joule J L 1N In 1kgm2s2 Work and Kinetic Energy There is and important theorem which relates the total work done on a particle to its initial and final speeds If 13 is the net force acting on a particle Newton39s second law gives 13 me The total work becomes 1 1 Wtot chAzf 5m m273 The kinetic energy of the particle is defined by 1 T2 K 277121 and the mechanical workkinetic energy theorem states The total work done on the particle is equal to the change in kinetic energy WtotKf K7L Potential Energy Often work done by external forces on a system does not increase the kinetic energy of the system but is instead stored as potential energy Conservative Forces A force is called conservative when its total work done on a particle along a closed path is zero figure 622 of Tipler Potential Energy Function For conservative forces a potential energy function U can be defined because the work done between two positions 1 and 2 does not depend on the path 82 AUU2 U1 Fd 31 dU 13 d for in nitesimal displacements Example Gravitational potential energy near the earth39s surface dU 13d mg 39 dxidy dzl mgdy y UdUmg dy mgy mgyo y 0 Work Energy Theorem with Kinetic Friction Non conservative Forces Not all forces are conservative Friction is an example of a nonconservative force The energy dissipated by friction is thermal energy heat f A8 AEther1r1r1 where f is the frictional force applied along the distance A8 The workenergy theorem reads then Wext AE1r1r1ech l AEther1r1r1 Momentum Conservation The Center of Mass CM The CM 77cm moves as if all the external foces acting on the system were acting on the total mass M of the system located at this point In particular it moves with constant velocity if the external forces acting on the system add to zero Definition 39I L 39I L MfcmZm F Where Mzzmi 71 71 Here the sum is over the particles of the system m7 are the masses and 7339 are the position vectors of the particles In case of a continuous object this becomes Mfcmde where dm is the position element of mass located at position 77 Momentum The mass of a particle times its velocity is called momentum 13 m Newton39s second law can be written as F d dm27 d27 m m net dt dt dt as the masses of our particles have been constant The total momentum 13 of a system is the sum of the momenta of the individual particles TL TL P 23 Zmi i M cm 21 21 Differentiating this equation with respect to time we obtain c113 dam a M M gem ne ex dt dt a t t The law of momentum conservation When the net external force is zero the total momentum is constant Fmth 0 gt P constant Example Inelastic scattering figure 830 of TiplerMosca A bullet of mass m1 is fired into a hanging target of mass m2 which is at rest The bullet gets stuck in the target Find the speed vi of the bullet from the joint velocity of of bullet and target after the collision PRS The result IS m1 m2 1 vi vf 2 vi m1 m2 m1m2 v Solution Momentum conservation gives pimlvim1 l m2vfpf Given the initial speed 217 of the bullet what is the speed of the combined systems after the inelastic scattering 1 U 7711717 2 vm1 l m2vi f m1m2 f m2 How high will the combined system swing 71f 1 1 h 2 h v29 Rotation The angular velocity LU Direction Righthandrule In accordance with the righthandrule the torque is defined as a vector 77x F Angular Momentum Definition E 77X 13 Like the torque angular momentum is defined with respect to the point in space where the position vector 77 originates For a rotation around a symmetry axis we find L LU magnitude L Iw Rotational kinetic energy Kmt fu Torque and Angular Momentum The net external torque acting on a system equals the rate of change of the angular momentum of the system a dL 7quot Ei zext Conservation of Angular Momentum If the net external torque acting on a system is zero the total angular momentum of the system is constant dt Tnet 0 gt L constant Gravitational foce Gravity m1 m2 F i 12i G T12 Fluids Chapter 13 Density mass m Densit p y volume V Liter L An often used unit for the volume of fluids 1L 103cm3 103m3 Because the gramm was originally defined as the mass of one cubic centimeter of water the weight of 1 L water at 40 C is 100 kg Pressure When a body is submerged in a fluid the fluid exerts a force perpendicular to the surface of the body at each point of the surface This force per unit area is called pressure P of the fluid P A The SI unit for pressure is Pascal Pa 1P3 1Nm2 Another common unit is the atmosphere atm which equals approximately the air pressure at sea level One atmosphere is now defined in kilopascals latm 101325 kPa 14701bin2 The weight of an incompressible liquid in a column of crosssectional area A and height Ah is w mg pVg pAAhg p constant If P0 is the pressure at the top and P is the pressure at the bottom we have PA POA pAAhg or P P0pAhg The pressure depends only on the depth of the water Pascal39s principle Blaire Pascal 1623 1662 c A pressure change applied to an enclosed liquid is transmitted undiminisched to every point in the liquid and to the walls of the container Archimedes Principle The force exerted by a fluid on a body wholly or partially submerged in it is called the buoyant force 0 A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid Fluids in Motion The general behavior of fluid in motion is very comples because of the phenomen of turbulence But there are some easy concepts governing the nonturbulent steadystate flow of an incompressible fluid Continuity equation Figure 1313 of TiplerMosca Let U the velocity of the flow and A be the crosssectional area the IU Av constant PRS Assume a velocity of 10 ms How many cubic meterssecond come out of a pipe of diameter 1cm 1 314x103m3s 2 785x104m3s 2O Bernoulli39s Equation Figures 1314 and 1315 of TiplerMosca 1 2 Ppgh pv constant Assume a water tower of height h At the bottom there is a hole At which approximate speed does the water leak out of the hole 1 v 2gh 2 v P where P is the pressure at the bottom 21 Static Equilibrium Chapter 12 1 The net external force acting on the body must remain zero 13 0 2 The next external torque about any point must remain zero 2 7 0 Conveniently the net torque due to gravity about any point 0 can be calculated as if the entire weight W were applied at a the center of gravity Tnet Tcg X W In a uniform gravitational field ch 77cm Balanced Bar Assume the is one pivot The condition is 20 for the torq ues Stability of Rotational Equilibrium Figures 1216 and 1217 of TiplerMosca Cone rod Sign hanging from a rod Figures 126 and 127 of TiplerMosca The weight of the sign and other numbers are given in the book Problem Find the tension T in the wire and the force 13 of the rod against the wall 1 Draw a free body diagram for the rod figure 127 of TiplerMosca 2 27 0 about point 0 L TLsiHQ MgL mg 0 3 Use trigonometry to solve for 6 1 1 o 6 tan 5 266 4 Solve for T M 9 m9 T 483N sine 2 sine 02FxFxTx and OZFyFyTy Mg mg Fx Tx TCOS6432N Fy TyMg i mg Tsin8Mgmg192N Ladder Figures 1210 and 1211 of TiplerMosca The weight of the ladder and other numbers are given in the figures Question What is the minimum coefficient of static friction us so that the ladder does not slip 1 Draw a free body diagram figure 1211 of Tipler Mosca 2 f8LSFTL 0Zszfs F1 and 02FyFn w 4 Solve for f8 and Fn fsF1 and Fnw60N 5 2739 0 about an axis directed out of the page and through the foot of the laddeh 4mF1 15mw 0 6 Solve for f8 15 fsF1 mw225N 4m 739 22 5N us E 39 20375 Stress and Strain When a solid object is subjected to forces that tend to stretch shear or compress it its shape changes If the object returns to its original shape when the forces are removed it is said to be elastic Most objects are elastic for forces up to a certain limit called the elastic limit A force streching an object is called a tensile force figure 1220 of TiplerMosca The fractional change in the length is called the strain AL Strain L The ratio of the force F to the crosssectional area A is called the tensile stress F St ress A Figure 1221 of TiplerMosca shows a graph of stress versus strain for a typical solid bar The ratio of stress to strain in the linear region Hooke39s law applies there of the graph is a constant called Young39s modulus T T stress FA strain ALL 39 The stress at which breakage occurs is called tensile strength If a bar is subjected to forces that compress it rather than stretch it the notation is compressive stress instead of stress and compressive strength for the breakoint force Similar treatment of shear forces Figure 1222 of TiplerMosca Fluids Chapter 13 Density mass m p Density volume V Liter L An often used unit for the volume of fluids 1L 103cm3 103m3 Because the gramm was originally defined as the mass of one cubic centimeter of water the weight of 1 L water at 40 C is 100 kg The ratio of the density of a substance to that of water is called its specific gravity Pressure When a body is submerged in a fluid the fluid exerts a force perpendicular to the surface of the body at each point of the surface This force per unit area is called pressure P of the fluid F P A The SI unit for pressure is Pascal Pa 1P3 1Nm2 In the US pressure is usually given in pounds per square inch lbin Another common unit is the atmosphere atm which equals approximately the air pressure at sea level One atmosphere is now defined in kilopascals latm 101325 kPa 14701101112 This is quite a large pressure on you Do you believe it Pressure tends to compress an object The ratio of the increase in the pressure AP to the fractional decrease in volume AVV is called bulk modulus AVV The compressibility is the reciprocal of the bulk modulus Liquids and solids are relatively incompressible They have large values of B relatively independent of temperature and pressure Gases are easily compressed and their values for B depend strongly on temperature and pressure The weight of an incompressible liquid in a column of crosssectional area A and height Ah is w mg pVg pAAhg p constant If P0 is the pressure at the top and P is the pressure at the bottom we have PA POApAAhg or P P0 pAhg The pressure depends only on the depth of the water Pascal39s principle Blaire Pascal 1623 1662 c A pressure change applied to an enclosed liquid is transmitted undiminisched to every point in the liquid and to the walls of the container Example Hydraulic lift Figure 134 of TiplerMosca F1 F2 AT A2 PRS Assume the large piston has a radius or 10cm and the small piston a radius of 1cm What force must be applied to the small piston to raise a car of15000 N weight 1 1500N 2 150N 3 15N Archimedes Principle The force exerted by a fluid on a body wholly or partially submerged in it is called the buoyant force 0 A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid Example Figure 139 amd 1310 of Tipler Mosca Application Density Measurements Circular Motion 02 am 73 The Period T The time T required for one complete revolution is called the period For constant speed 2 v 7 holds Example Circular Pendulum Figures 522 and 523 of Tipler Mosca Relation between angle and velocity U2 Tr T sin6 m Ty T 3086 mg sm6 21 2 3086 gr 2 v tan6 g T Example Forces on a car in a banked curve Figures 526 of Tipler Mosca The optimal angle 8 is the one for which the centrifugal force is balanced by the inward component of the normal force ie without friction Then Fn 3086 mg 0 U2 Fn sin6 m 0 r 2 tan6 U 97 Angular Velocity Definition w d6 dt For w constant and in radian we find 21 ma Namely for one period wT27rgtvT27rr Surface of Rotating Water dyr tan6 21 2 w2r2 W27 d7 g7 97 g Integration 2 w 0 m gt 2g m gt parabola For the mathematically ambitious only Another Derivation of the Acceleration Now 7715 r with r 3086 1 l sin6 13 6t wt Therefore 7 coswt g l sinwt 13 The velocity is 4 d7 dr A A v g rE rw smwty l rw coswtx v sinwt fl v cosw t 13 211 where A t sinwt 1 cosw t 56 is the tangential unit vector In the same way the acceleration follows L d2 dt 0A ty v vwcosw vwsmw x dt dt 9 2 2 2 v v v coswt A sinwti f T my T H T 2 A U arr Wlth aT 7 l Questions on Circular motion A particle of mass m moves with constant speed 21 on a circle of radius R The following holds pick one 1 The centripetal force is U2R towards the center 2 The centripetal force is mUZR towards the center 3 The centripetal force is WUZR away from the center 4 The centripetal force is U2R away from the center 5 The centripetal force is mUZR downward The acceleration of the particle is a constant vector The acceleration of the particle is a vector of constant magnitude The magnitude of the acceleration of the particle varies with time The acceleration of the particle is a vector which points up The acceleration of the particle is a vector which points down The acceleration of the particle is a vector which points towards the center of the circle Drag Forces When an object moves through a gas like air or a fluid like water it will be subject to a drag force or retarding force that reduces its speed For an object which falls in air under the influence of gravity one observes an acceleration like my be ma Where I is a constant and n is approximately one a low speed and two at high speeds The terminal speed vtwm is reached for a 0 1n buferm my gt vtwm For n 2 mg 21 term b Obviously the terminal speed for a free fall in air is highly material dependent Eg a feather versus an iron ball a man with our without a parachute Questions Uterm T Determine b for an 80 kg object n 2 and warm 200 kmh The result is pick one 1 b0254kgm 2 b0254kg8 3 b0021kg8 4 b 0021 kgm
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