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# COLLEGE PHYSICS A PHY 2053C

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This 139 page Class Notes was uploaded by Garett Kovacek on Thursday September 17, 2015. The Class Notes belongs to PHY 2053C at Florida State University taught by David Lind in Fall. Since its upload, it has received 7 views. For similar materials see /class/205521/phy-2053c-florida-state-university in Physics 2 at Florida State University.

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Date Created: 09/17/15

a Hand in your extra credit summaries for chapter 2 bring them down and place them on the front podium a Hand in your CAPA problem that you printed out you should also neatly write on the sheet of paper your name socsecno and the section you are registered in 9 Come down and pick up a PRS clicker any one numbered below 200 we will assign you each a specific number next week Ll Motion Forces Energy Heat Waves Dr David M Lind Dr Kun Yang Dr David Van Winkle Today 4 Kinematics in one dimension position velocity acceleration free fall xxx Labs CAPA assignments Bonus Points MiniExams Physical measurements require units We use the SI units 9 m s for mass length time 39l39l ch one of the following statements sounds right to you 1 Objects keep a constant speed indefinitely unless I do something to them 2 Objects tend to slow down by themselves and will stay at rest if left alone B Which statement is right 1 Atwo pound weight will hit the ground before a onepound weight when dropped from the same height 2 Atwo pound and a onepound weight will hit the ground simultaneously when dropped from the same height 00 Galileo observed three simple but in view of his peers radical facts 1 If you do nothing on an object and eliminate friction it keeps going indefinitely at a constant speed same rate independent of Galileo Galilei 15641 642 2 All objects fall at th their mass 3 The distance faen is proportional to the time squared I t2 Let39s see if he is right quot we want to describe motion mathematically we have to choose a frame of reference in which we measure the changing position of an object m providing its coordinates A typical choice for the zeropoint is a point fixed relative to the earth39s surface eg lm The bottom left corner of the UPL 101 projector screen All of you can now describe their position by giving three numbers in meters xyz l 3 a lm g D LACEMENT is the difference between final and original coordinates as defined by your Frame of Reference A x xf Xi The A stan s for change in Example m m Travel 70 m to the east quotquotquotquotquotquot y 39 3912 x then turn aroun sl 0 40 m 30 m East and travel 30 m to the west i Di splacemenl Displacement Here 40 m pos or neg vector Distance is the total length of the path always positive Here 100 m scalar displacement average veIOCIty time elapsed The symbol is A1283 T m E 70 n1 4T gt 39 t2 t1 A t 51 O 40 m 30 m Ea Unit is 1 meter 15quot 1 ms DiSP EEmCm Note velocity is displacement over time 5 ms vector speed is distance over time 125 ms scalar riv39ing on a straight street with A r constant velocity What is happening to the car There are Forces act on it but theyquot Force ogntires Yon39 Horizontal Air resistance and 39 J 15 eve Friction are balanced by i Motor39s Force Constant i velocity 39 or Force Sy we know that an object moves with constant velocity v Where will it be in the future X2X11Om 5m 2 2 2 1 1s Os XX0Vot X5m5mlst The policememan with the rada gun is not interested in the E average trip velocity but quot1 quot 50 your velocity in the instant FD 50 100 150 280 250 360 Time 15 We get the instantaneous 3m velocity by using very small time intervals and Ax 1 At Position I m Cfnge of position per time is velocity New Change of velocity per time is acceleration Change of velocity average acceleration time elapsed V2V1AV Unitz mSzm 0 f s 52 12 t1 At analogous definition for instantaneous acce make the time intervals very small or tangent to the vtgraph car gt Velocity Any change in speed or Slowing down acceleration is opposite the direction of motion driving on a straight street but is now accelerating What is the acceleration of the coffee liquid is tilted the other way direction is an acceleration Spee39 is in he direction of motion new going around a bend in the street accelerating The net force must be in the directionp heaccelerationi Any change in speed or Turning at a constant speed means that acceleration is perpendicular to the direction of motion directior is an acceleration Turning coffee liquid is tilted Rear View No object moves with constant acceleration starting from rest Where will it be in the future lme5 posmomm velocles o o 1 2 2 n Vz V16ms 2mls 2 8 e tz t1 15s o5s 3 18 1 2 14 a 4m 5 4 32 18 t t 5 50 22 V a39 1 6 72 Xt 5 at2 Position as a function oftime with initial position x0 and initial velocity v0 Initial means at t0 XtX0 vo39l lat2 2 t Velocity as a function of time V vt v0 at Velocity as a function of position v2t V0226X X0 t the Kinematic equations Galileo observed that all objects fall with the same constant acceleration 980 mISZW as long as airresistance is small Galileo Galilez39 15641642 Free Fall is oned mensional motion with a constant negative acceleration This means the acceleration is always pointing downward Position as afunction oftime different initial positions y0 3 39 39 39 39 yft IVfZPW 25 quot y025mv00 and initial velocities v0 20 1 gtlt y y0vo39tgt2 1 39 2 5 r o 39 39 39 39 Note that the acceleration WW asamncnonomme is constant independent 3 39 39 39 ixo9vofgws of initial conditions 2quot v025mro 10 s vtv0g39t 0 1o a 2o 39 430 0 05 15 25 3 35 time seconds we Will meet again Friday 1010 when we will talk only about the CAPA and homework problems Before then you should have looked at your CAPA assignment a Go to httpcaga1fsuedu b Your account is your garnetaccountname Most of you also use their garnetpassword Those who registered after Monday have ast eir initial password ilovephysics CHANGE lT If one does not work try the other If you have problems send Dr Yang email kunyangmagnetfsuedu Fall 2007 PHY 2053C College Physics A Motion Forces Energy Heat Waves Today zoFluids Fluids Liquids Pressure in Sink or Swim Bouyancy today Gases Fluids Fluid ow Bernoulli39s Lift amp the Introduction Fluids So far we have studied the motion kinematics dynamics and statics of solid rigid bodies In this chapter we look at propelties of fluids which include liquids and gases Fluids are phases of substance which flow without a fixed shape We will study both the static fluid not moving and dynamic moving fluid propelties of fluids Materials can be in solid liquid and gas phase depending on temperature Chapter 13 Characteristic of Fluids The shape can be changed without using forces but the volume usually is not Density The density of an object is defined as mass per vo ume units 1kgm3 eg Water has 1000 kgm3 density Density is characteristic of the material We say lead is heavier than wood but what we mean is lead has higher density than wood Estmate 15 mass of the air in this room pAIR 13 kgm3 Pressure in Fluids Pressure is defined as force per unit area units 1 Wm2 1 Pa 1 Pascal Pressure is the stress in fluids acts also on submerged bodies Forces act on all surfaces The forces on a or exerted by a surface act perpendicular to the surface Pressure can not pus1 sideways l l l L Pressure in Liquids The pressure in liquids varies with the depth The weight of the liquid inside r39 quot the cube is supported by the 7 A lower surface Fmwaterg I z I in 39A Note that this means that the pressure increases with depth Si Question Container shapes I For P depends only on depth P at equal depths is the same P does not depend on the shape or volume The water level is the same in all connected containers Q Why is P3P It seems like the bottom area of three supports a larger load A The walls of container 3 exert a Force perpendicular to their surface which has an upward vertical componen that component supports the additional weight example Faucet Pressure The ressure in domestic water supply is supplied by an elevated 2 tank 7 gt E i Wire k h P1000 Lagso m980 m 30m S P294 1O5Pa 29 atm l That is why there are water tanks mounted on towers and hilltops or high on top of towers constant pressure without big pumps only need a small pump to refill the amount of water used and maintain constant level Pascal39s Principle We have 39ust discussed P generated from within a fluid by its own weight Pressure can also be generated by external forces Pascal39s Principle pressure increase due to some external forces is Me same I I throughout a con ned uid l T gt y T I 1 w 2 ml l We mechaiiically multiplythe force available for a hydraulic lift In example quot i LF2 A1 A2 The previous example shows that one I I can increase forces with pistons of ll 1 J iquot different crosssectionalarea 7 Atmospheric and Gauge Pressure We have to accountjor air pressure The normal air pressure at sea level 1013 kPa 1013 bar 1atm 101300 NmZ E the weight of 10 tons on 1 m 4 V L Magdebem Sgheres famous demonstration from 1672where Lord Magdeberg med m pull lhe two halves of an evacuated sphere apartwilh mams of 16 horses Air pressure usually works on both sides of a hydraulic system so that it cancels out Atmospheric and Gauge Pressure y We have to accountigr air pressure I 7 Press The normal air pressure at sea level 39 7 1013 kPa 1013 bar 1atm 101300 N E 15 weight of 10 tons on 1 m Air pressure usually works on both sides of a hydraulic system LI Openlube nmuomelel39 so that it cancels out Scale The opentube manometer shows I g reading gauge pressure pressure 23 difference relative to surrounding air pressure Spring Barometric Pressure The air ressure itself must is measured relative to zero pressure vacuum The glass tube is completely lled with mercury and then inverted into a bowl of mercury The normal atmospheric pressure supports 760 mm of Hg 760 torr which is yet another unit of pressure Note that the pressure above the Hg in the tube is zero vacuum Question Atmospheric Pressure Buoyancy Archimedes39 Principle Since the ressuregin a liquid increases with depth a quot gt lt Force on Bottom gt Force on Top F3 F2 F1 l 1 01 g Ah2h1 z 1 Li ngAh l l weight of displaced liquid a a mgma e If you submerge objects the fluid exerts a buoyant force on the object which equals the weight of displaced uid Sink Swim 0139 Float Is the bou ant force weight of displaced water enough to suppont the weight of the object FszgpFV I szgpov FBFOgpFpO V The density decides 7 7 77 If pF gt pO 9 swim quotB 39r prFp09 oat 1 iT I prFlt p0 95ink 1 f t t example Floating Log The o is submerged only partially Howmuch Just enough so that the weight balances the bouyant force mg according to Archimedes The displaced volume of water equals the weight of the log A log with p500 kgm3 pF Volspgpo V09 Would be halfway submerged gt Vdm pa in water with p1000 kgm3 o Stay tuned Question L 7 Ch 5 Spring 2004 PHY 2053C College Physics A Motion Forces Energy Heat Waves Reminder Block on inclined plane A block is at rest on an inclined plane 1 quot7 ps042 pKO15 At What angle does the block start to slide What is the acceleration Free Body Diagram Solution We Will see that mass m does not matter Notice X and y aXis a Find the X and y components of weight F FGXImSin FGymmS FGX F39ff fffI e c Starting to slide FfrFGx usmgcosmgsin gt Ilsten n T 39 F m m b Find the unknown friction force FN FGymgCOS Ffr sFNIlsmgCOS Reminder I I I I Newton 5 Third Law of Motion whenever an object excerts a force on a second object the second excerts an equal 39 and opposite force on the rst r o X by sleLi EA Fm r t 1 i as i I Fst 39cht Fsc 17 E FN1 FAG Frigion Frce on Faro on Force on 39 ml ground ground awsmni sled exerted exeried exerted exened by dad hymsr slani by ground by ground Action and Reaction forces always act on different objects we swap the roles of object and world Question 1 Uniform Cirular Motion Ima ine an opject moving on a circular path with constant speed V1 v V V2 V2 V2 2 1 AV 7 i7 a 2 1 At 1 2 Although the speed length of v is constant the velocity is changing all the time The acceleration is pointing toward the center with constant magnitude V2 We call this centripetal acceleration 87 centeraiming acceleration E4 Q Notes on Circular Motionx L The centripetal acceleration points toward the center of the circle It is an automatic property of constant speed circular motion A number of common terms for circular motion ac Distance around the circle circumference 21tR 1 revolution Period T time to go once around circle Frequency f 11 number of revolutions per second ltgt T time T v Centri petal Force FCmaCmI3 From Newton s second law A centripetal acceleration must be created by a net centripetal force Some properties oz points to the center of motion z acts on the circling object oz It is not an independent force oz It is always delivered as a Tension E V Friction Normal or Gravitational Force See examples r V K i r r r Tm r I 3 iK l I Question 2 Centripetal Force Fcmacm From Newton s second law 39 39 A centripetal acceleration must be created by a net centripetal force u If not for the centripetal force provided by the string Some properties points to the center of motion acts on the circling object V It is not an independent force It is always delivered as a Tension K Friction Norma or Gravitational Force See examples There is no such thing as a centrifug I this is just a manifestation of inertia I 399 0 0 0 0 0 0 0 0 Example the UpSidE39Down A man holds a pail of water by the A r I handle and whirls i Faround a vertical circle r074m at constant speed m What is the minimum speed that the pail must have at the top of its circular motion if the water is not to spill out of the upsidedown pail 0 z I Solution the most critical point is the top A Here the centripetal force is supplied by the tension in the rope and the weight m F m g T R F What condition has to be fulfilled so that the water does not spill out Tension in the rope As the speed B FwsinH A is lowered lei FT needed to reach the required centripetal force Limit FT0 mg 39FW cos 9 FW 2 mg example Circular Motion Why do we bank racet and roadways r From Newton s second law the centripetal acceleration must be provided by a centripetal force For an unbanked roadway the entire centripetal acceleration must be provided by the frictional force between the car s tires and the roadway But when the roadway is banked a part of the normal force from the roadway pushes inward providing the centripetal force If banking is enough there doesn t need to be a radial frictional force Example 57 Newton s Law of Universal Gravitation All objects with mass feel an attraction to each other a clue to gravity 39 quotnthe surface of the earth F G my m1g78m2hm s EARTH MEI ulli Earquot Lilli GraVity cont j a Newton s Law of Universal Gravitation L Gravity is an attractive force acts along the line between two objects ozo Important BOTH objects feel this force and are attracted to each other The force is proportional to the The force is inversely proportional PFOdUCt 0f the tWO masses to the square of the distance between the masses G F I I F 9 m1 m2 l l l I l oz More mass either one means more force ozo Greater separation means less force Demo Cavendish balance delicate tool for measuring gravity lullll1li39539 Weightlessness in Orbit Astronauts in the Space Shuttle or International AL Space Station 155 feel and appear to be weightless Are they Q u e sti o n 3 L21 ChIIl5Ch 4 PHY 2053C College Physics A S rin 2004 Mo Ion orces Enerqy Heat Waves r David M Lind Dr Kun Yang Dr David Van Winkle Today 3 Wave Motion V Energy in waves V Wave speed V Resonances standing waves 0 v Sound amp Harmonics V Resonance harmonics V String and Wind instruments Announcements repeat Next week review for final Monday ch915 will include CAPA discussion rWedesy Mini Exam 7 Friday ch18 I T Q st The final exam will be held Wednesday December 8 at 10001200 and is scheduled for 275 Fisher Lecture Hall not in our current classroom st The final exam will be comprehensive and we next week st Labs will be held next week not this week Important Points from last L ture Simple Harmonic Oscillator is a consequence of k X The Period T is TZZWJE f k 3 Pendulum T27T l f VVaves i Waves are moving oscillations They carry energy they do NOT carry Think of water waves 1 i Cl ESIl pr peak i V 7 position Trough i ii i 39 Wavelength 1 length of one full cycle at fixed time I Period T time for one full cycle at fixed position Waves travel at speed v A VTI A V T Waves 2 Waves are moving oscillations iiasi 3T peak position 1Tough i 1 a The timeperiod T of the wave and t spaceperiod of the wave are connected by the wave speed v A A 1 Energy Transport in Waves We define Intensity of wave as the power transported through unit area the energy of the wave is proportional to Amplitudez Energy of simple harmonic oscillator E 12 kAZ for spherical eg sound wave If power output by source is consti MP gtIOC1 Area 47Tr2 r2 r1 I V2 ource Poweroc Amplitude2 S Area Distance2 Question Speed of Waves The speed of waves depends on the elasticity and mass of the medium transverse wave on a string FT tension in string Velocity of rope particle u quot R 7 7 I x fitwmty of wave L g J C Vs k V I 3VL JVVL 439 L x a x 7 x l t a r 35 a 2 Aquot A quot39 quoty I x A is 4 Q 4 a J V LN t 4 y x 1V longitudinal wave on a rod g E elastic modulus longitudinal compression wave in gas Question Re ection of Waves re m henever a medium changes A mm waves may be reflected v Reflection off a wall f reflected wave is negative km A of original w Reflection off a loose end Light Heavy reflected wave is positive MW of original 3 Transmltted pulse Reflection transmissmn EVE ew at change of medium Re eeeed b ight 2005 Pearson Prentice Hall Inc nterference of Waves Waves can pass through each other amp interfere destructively or constructively la iii 51 393 velocmes i 1 Pulses i nr apart zipproaci39lin g r 39quoti me 2 Pulses m eriap preciser Pulses i tir apart receding 1 l in 82 the waveenergy iS all ki EtiC motion but no amplitude 3 1 l in bZ the waveenergy iS all potential amplitude but no motion Standing Waves esonances on Strings A wave on a string is always being reflected back and forth on both ends with a negative sign which Node II J Antinode Node ifc Nodes stationary pointsquot and f2 f0 W waves moving in Antinode A O f f0 Antinodes maximum displacement pointsquot A Possible wavelengths frequencies 4 Node gimme Xx 2 A0 v QQ A f nE nfo aAn n n A i6 no a 1 Breaker at the Seashore r 39x l I gProperties of Surface Waves Surface Waves occur at the 4 1 boundaries of fluids Remember Waves do not transport matter only energy Surface waves may break when friction with ground suppresses Clrcular motion when they break matter is transported Sound Intensity Sound spans a wide ranqe of intensities Jet Plane at 30m 100 Wm2 Threshold of hearing minimum 1 X 103912 Wm2 The human perception of loudness follows a logarithmic scale measured in dB decibel 12W m2 Bin dB1OogL where 01O O Log is a measure of powers of ten so log counts the zeros log 1002 log001 2 An additive difference in dB corresponds to a multiplicative factor in Intensity Hearing Range of Frequencies The pitch of sound comes from the frequency udible range 20 20000 s1 Hertz he ear is not equally sensitive to all frequencies The curves descrIbe sounds TM M W emaci EL JEi IzII 1 of equal loudness to a s 39 J quot39 39 Jr SOUHCI at IIIII w 31 fay f III393 1 IS senSIthe 3r 231Hquot quot at f f 1 01 EffHa quot d J I low a nd h Ig h a m J I I 15 1 at quota h 339quot quotagarquot 3 19439 frequenCIes I I a H r I a ma 3quot u an L quotam E 40 quotH quotquot was If 939 III a II a a Ema I ffquot quotquot3 I 5 r If II I III 2 I I I I I I I I I I I I Tquotquot39i T39Ff I I 20 so 30 500 1000 some 10000 Fremmney Hal 1 Inl nsily Mimi oundWaves Sound travels through air as waves of pressure Sound waves are longitudinal as all waves In fluids quot quotquot quotquot quotquot quotquot displacement of molecules 29 lt 5 lt lgt lt 5 lt 5 4h gt i 18 VClOCity Of mOleCUICS l I i w i a veloo1ty of wave quot 00005 a X I k 0 8 a EaIquot I Mquoth 00005 g d1sp1acement x 3310002 395 density g 10000 x cm 09998 H DI The displacement of molecules stores potential energy which is continuously converted back and forth into kinetic energy motion of molecules 1 VQuestion 1 Speed of Sound i The speed of sound is given by I E I where B is bulk modulus of the V medium I II P I I f0l39 Air sea level arm pressure v331060 T Cms N343ms at20 C Example Distance from a lightning strike Light travels much faster c3x108 ms than sound Time between seeing lightning and hearing thunder is time it takes for sound to arrive 9 55 343 ms 1715 m 39 1 mile Speed of sound in Water 1440 ms String Instruments f1 You can raise the pitch of tone by effectively shortening the string 5 i The pitch is identified with the fundamental frequency f1 of the string g r 444quot l o g u Along with the fundamental 3 the string also vibrates at all harmonics nf1 along with the fundamental Z lllillzgl Illl The spectrum of harmonics 1 determine the sound 1 Illatl39net Resonances 1n a Tube E Most wind instruments use I l resonances in a tube I a Displacemenl H l L l Closed end displacement A i r node B E Open end displacement t39 cl A aquot 6 L m gt n L B 4 4 A 4 4 5 V even harmonics can not have an antinode LZZA5 gt f5 5Z L atthe open end Harmony in Music When several notes are played together they may sound consonant or dissonant Example 1 major chord composed from 3 tones at frequencies 1 54 64 fA so Hz fj3 60 Hz 1 I x n a m I I w I l I x r x v I r I Ir u i X I f 1 l y ix 94 r f I I I I I x I l l l I I l I I A r I I I J I W g I 5 I I N f X 2i i la Il will I amp V V J J V U 0 r20 055 r0ls 20155 7 AAAA VV WVV beat period 0010 s Copyright 2005 Pearson Prentice Hall Inc Example 2 this also shows up in the time sequence of waves as beat frequencies fbeat f1 f2 Dissonance in Music If notes are played together that do not have a imple harmonic relationship the sound is dissonant no simple period occurs Example tritonus major quart simple and complex harmonic motion IN llf39lll In fail fal JFI 39 H 05 Fall 39 m in If Js ismi gf l ll I I I II II 1 If II II l EI lhl Ix Lqtcf sk1gt1l 2 3lHXp ll Rf l39ll39 l39l39lll flquot 394 39 If I II I ll L I I I39 k I l l 39I U L In IV 395le LI lllj I I 39IU 39kvll Iva lgquot 11 PHY 2053C 7 College Physics B Fall 2UE7 Marian Fnrcai Energy Hair We a my Today Mo on Important Points from last SW 6 Harmoer OSOHamr S a cmsecueme of X a E a j e 7 a El w a we 1 Pendumm i Waves Waves are movln oscillations They carry energ mey do NOT carry ma um kaofwamwwu M e y pushun n y Wa elengthA engm ofone mu oydej Period T ume for one mx cyde at xed posmon Waves Uave at speed v Waves 2 I Waves are mo 39rm ur 3 i 39 Ammnb Yplmy position i r g oscillationsquot The timeperiod T of the wave and the spaceperiod A of the wave are connected by the wave spee W r 1 Properties of Surface Waves Breaker at the easlys Surface Waves occur at the boundaries of fluids Remember Waves do not transport 39 39 matter only energy Surface waves may break when friction with ground suppresses circular motion 4quot 39 when they break gt gt matter is transported oundWaves g Sound travels through air as waves of pressure Sound waves are longitudinal as all waves in uids g I displacement ofrnolecules I I I I I Iacement of molecules stores I energy which is continuously converted back and forth into kinetic energy motion of molecules 1 Speed of Waves The s eed of waves depends on the elasticity and mass of the medium transverse wave on a string F terswom n39smng 39 l lt Speed of Sound The 5 eed ofsound s gwen b w ere B medwum 5 UK modu us of the for Air sayEVE ampam l v331060 T39Cnis 343nis azo39c Detamce from a hghmmg smke nghtuave s M faster c3x1ox ms man s ound me be een seeing hghmmg and hearing mmder s ume t Bkes for sound m amve gt ES 343 mS 1715 m 1 mHe Speed ofsound W Water 1440 ms Que on Are the ve ocmes ofpamdes on a smng the same asths w A 1N l2YE Waves can pass ihrou destructivelv or constructive I Interference of Waves i gh each other amp interfere v iii llii DWWM velmfres l l l l 3 quot 1i139l39lquotquot ll I in aZJiheWaveenergv 39 u in b2 the waveenergy is all potential Wm Mmme quot1i l 1 Reflection of Waves henever a medium changes waves mav be ref ed39 Reflection off a wall reflected Wave is negative of original Reflection off a loose e1d reflected wave is positive of original Reflection transmission at change of medium Standing Waves Resonances 0n Strings A Wave on a string is alwavs being reflected back and forth on both ends With a negative sign which leads to standing waves Myerth an wavesriowwi re dunam T i ii i AA ffn l i i Nodes mammary penny and i 39mdesmmmum mapimmeni puinim A f2 f I Possible Wavelengths frequencies i in if i A i r3rn String Instruments Nodes at each end f stnng You can ta se tne pwtch of tone bv ettectwew shortening the stnng E l The pitch 5 dentw ed wnh t L the fundamental frequencv quot mg i i f of the stn m M A ong wnn tne fundamenta the stnng a so wbtates at aH harmonics nf a ong wnn the tundamentd The spectrum of harmonics deterrmne the sound Resonances in 2 Tube l M C osed end dwsp acement node dwsp acement Harmony n Music When severa notes are p aved together thev ma sound conso a tordissonant Example 1 mapr cnord composed from 3 tones at frequencwes 54 64 added together MKva M A A v3 V f I Example 2 this a so snows uo m the twme sequence 42 of Waves as beat frequencies tgeat u g 2 o 5 m a w 0 Sound Intensity Sound s ans a wrde range of intensities m 100 vvm2 reshold of hearing mrnrmum 1 x 10 2vvm2 The numan perception of loudness foHoWs a logarithmic scale measured m dB decrbeu measure of powers of ten so 09 comm Log rsa 00122 the zeros 09 100 2 09 addmve drrrerence m dB corresponds to a mumphcame factor m Intensity Hearing Range of Frequencies t of sound comes from the frequencw Iera e 20 20000 5quot Hertz Tne ear rs not equally sensitive to 5 frequencres The cuvesdesmbe sounds ofqual loudnessth A gg 7 r somd at IOOCHZ Jukhlm 7 39 Treearrslesssensmve atlow h andhig requenoes h a a A a 7 V k quotum g Que tion I Energy Transport in Waves We de ne Intensity of Wave as the power transported through unit area the energy of the Wave ts proporttonat to Amplitude2 Energy otstnpte Harmontc osoHator E v W Powe39 Area strum Power C AmEIitude2 Area Distance2 Question D sonance in Music If notes are ether that do not have a ple har etatronshrp the sound t5 dissonant no srrnpte penod occurs Example tritonus rnapr guar t t Ht ntt n Doppler effect g caused by a travelling source of Waves Approaching higher pltcn shorter Wavelength Receding lower pltcn longer Wavelength nlllllJIH Stay tuned Announcements repeat Announcements repem Damped Harmonic Oscillation Resonance example Resonance Bridge Collapes PHY 2053C College Physics A Moton Forces Energy Heat Waves WWWW inn thntlwuwsimnn Toda sLecture ur oseamp o Is nish Linear Momentum Collisions v Centerofmass motion ozo Rotational Momentum Kinematics 2 Angles amp Rotation v Dynamics 2 Torques Remember momenta add u example Collision of two We call the velocities before collision v1 v2 after collision V 39 V239 Pm1V1m2V2 Let spmve rat 1AtAp 2I392At 9 3 A p1 Newton39s3 SF 32 meansthatA 1 A 32 ltgtm1 7139m1 71m2 7239m2 72 ltgtm1 71mz 72m1 7139m2 7239 which is iust conservation of momentum quot72 V2 q Question reminder Elastic Collisio 1dimensional 1 momentum conservation 2 energy conservation rel velocity before Em1v1m2V2m1VW 2 v v V1V2V139V239 rel velocity after Elastic collisi iudimensional V1V2V139V239 relative velocity before relative velocity after This result is specific to elastic collisions where energy and momentum are conserved It is inde endent of the masses Question Simull Ene 1 Colli 39bm171mz i72m1 71 39m27239 So far two unknowns V391 V392 one equation 2 Elastic collision conservation of energy 1 1 1 1 KE m1vf m2V m1vf m2v KE Now we have two unknowns V 391 V392 two equations In the following we will use both properties to calculate the two unknowns A collision in which energy is conserved is called an elastic collision example I Newton 5 Cradle g W7 y does only one ball bounce back from Newton 5 cradle 7 all balls have the same mass m 1 Momentum conservation XmVout 2 ergy conervation 1 1 Evan gt therefore 7 ITIV1XITIV1 gt Two railroad cars collide and couple together ilnelastic collisions r 140 ml 13 7 n In 39039 mi rem m vNuu b After COIHMU Momentum is always conserved mv12mv39 gt v v1212ms Energy is not 1 V 2 in KE mvf out KEI 2m2 1KE2 the Missing KE goes into heat or vibrations reminder Important Points from Last Lecture In all collisions momentum is conserved In elastic collisions momentum and energy are conserved Elastic C96 21 a Totallv Inelastic a Balm collision gl Center of Mass So far we have considered motion of point objects not extended bodies We do not have to change much Extended objects move translational moton as 39 concentrated In tne quotcenter of mass p0nt In this picture the center of mass moves like a free projectile independent of the rotational component 7 b I Center of Mass Location The center of mass is a position the mass weighted average of the positions of objects that make up a system or body 2 objects m1X1m2 X2 XCM m X m X m X 3 objects XCM m1m2m3 The same is true for the ycoordinate of cm m1Y1m2 Y2m3 Ya yCM m1m2m3 m110 mm250 mm360 m X CM m1m2 m3 So if each mass equals m60 kg m10 m50 m60 m CM 3m 4m If m12mm2m3m m210 m150 m160 m325 m CM 4m Center of Gravity The effect of many small weight together is equal to the total weight acting on the center of gravity center of mass 3 In order to support a body support its center of gravity In order to nd the center of gravity suspend the body from two points CG will always be below both pivot points Extended Bodies amp Rotation Translational motion Objects move as if all of the extended object39s mass was concentrated in the centerofmass If T j is ir 39 r 7 V i 7 x e w 363 213141139 1 W i V J a i l Center of mass X m1x1mzxzmx y m1y1mzy2my CM m1mzmv CM m1mzm Extended objects can do one thing which point masses can39t they Rotate while they translate Why Rotational Kinematics What is the difference between the circular motion from chapter 5 and rotational motion Chapter 5 point objects moving in circular trajectories Here we are talking about extended rigid bodies Mim V1 Xx v m G quotu 1 l 2 Chapter 5 03221203111 0 Chapter 8 u Moon by Earth 3 Earth Gruvlulmnzil force exerted on hath 39 by he Mum Rotational Kinematics Rotational motion is described by angles of orientation the angular coordinate 7 is measured in radians P 39 r I 1 r this means that the distance traveled is given as I r0 NOTE that this formula is not true for angles measured in the customary degrees The translation radians 2 degrees is 21rrad2360 lt1 1radzg 513 Tr I Angular Veloaty i In describing angular motion the angle in radians serves as the coordinate displacement In analogy we define average angular velocity as 02 01 A 0 z 22 1 instantaneous angular velocity I Hum39canes I Van and Jeanne Enl hELPuvuusicnn a quot4 Iim A0 00 At OAt and angular acceleration Dz D1 Aw 2 2 2 1 F a Linear and Angular Kinematics summary iDisplacement xm angle erad Velocity vms angular vel oorads acceleration a msZangular acc a rad52 How do the angular kinematics translate to linear kinematics v 39 Sitting on a merrygoround What 39 is your tangential speed 39 vta quot rm quotgt 0 your tangential acceleration i abanro yourradial centripetal acceleration 2 2 arad v r rw Angular Kinematic Equations l Along with the angular coqrdinatee velocity w and acceleration a come the kinematic equations vvoat const a wwoot const o xxovot12 at2 6eowot 12cxt2 v2v02 2ax w2w022o6 r r C E G G CD 9 x l 1quot 0 057 17 157 2 7 25 7 l Note that if we use 9 as a coordinate it goes beyond the full circle or 211 Althou h a wheel goes back to its original position after one fu I turn the coordinate continues and measures full turns as multiples of2139r Question 2 example Rolling Motion What is the relation between angular velocity in and linear velocity v for a rolling wheel The point of contact between whe39 l and ground P is momentarily at rest The axle is moving at velocity v Now go into wheel39s frame of reference The point P is moving at velocity v Now we see vhnlw w is the same in both frames of ref Given rot freq frevsgtw21r frads nuind Tinran 3 f 1 ITrrnun39l n Owl Thad1 L15 Ch13Ch14 Spring 2004 PHY 2053C College Physics A Motion Forces Energy Heat Waves r David M Lind Dr Kun Yang Dr David Van Winkle lunar a r rem hemem In 39 39 Uh u y u 39i of the lengths F Thc 39ch Today st Kinematic Theory of Gas Ideal Gas Law PVnRT Maxwell distribution Internal energy genometram quot it i luncth lJI an on HQIL as shown in dimem in mm st Heat Heat definition Speci c Heat Thermal conductivity Ideal Gas Law g lPVnRTJ PressureVoume nomoles R Temp R8315 Jmol Kelvin universal gas constant 00821 L atmmo K absolute Temperature is used I absolute Pressure is used Wha is an ideal gas T well above liquefication point I We 58 PV P1V1 PZVZ constant if T and n constant oyle39s law PT P1T1 PZTZ constant if V and n constant GayLussac law Question Important points from last Lecture Temperature is related to un random motion of atoms Most solid and liquid materials ex39pa c contract with temp about lo39i t r a total length per 1 C J Absolute temperature is measWquot 39 O K is absolute zero Gases obey the ideal gas law PVn RT where pressure and temperature are absolute n is the number of moles and R is the gas constant 8315 Jmol39Kelvin 1 inetic Gas Theory PVnRT SIK AtomsMolecules bounce off the walls The pressure exerted by a gas I comes from the rate at which atoms molecules bounce per surface area times their average momentum Many many atoms are in the air around you 9 you feel no bouncing but constant pressure Temperature is related to average kin energy Kinetic Gas Theory Tem erature39 unordered random motion of atoms e inition average kinetic energy E of one atom moleculequot temperature I I5 1 2 KE 2m 2kBT x gm 23 are kB 138 x 10 JK j r f 73 The atoms molecules have a wide range of different speeds 39 called the Maxwell distribution Notice this is dependent both on 01 I lt g 3 atomicmolecular mass and on 0 300 400 700 800 1000 1200 temperature Molecular Speeds IllS 1 7 l I Percent of Molecules wrth gpecds mthm 1 ms of That Indicated Along the Horizontal Axis example peed of Molecules speed of airmolecules around you g 1 2 3 KE Emv EkBT xv a A forair O at24 C297K 39 2 I I f 3 J KE 1 3810 23 297K gr 2 K Ifquot lt 5 21 1 1 A 0 614391O J EmVZ Emvrzms 0 4 m02321671O 27kg531o 26kg 03 32 R 2 KE m 1 53 0 OE r m 2 S 01 273 K 53 0 1 r 139 MN BE I EIE l I I 39 I L l 39 E 0 300 Q00 00 800 1000 1200 0 500 I000 1500 2000 S Molecular S ecds ms Molecular Speed Ins Ideal Gases Temperature Heat and Internal Energy Here the kinetic energy is the only enfrgy an atom can have f 1 2 3 KE mV kBT in A where kI3 138 x 1023 JK quotX The total internal energy U in this case is the number of atoms times kinetie t energy per a om U2 NKBTZEnRT Heat transferred increases internal Energy 1 Questions example Heat capacity of Helium How much heat is required to raise the temxrature of 1 kg of Helium by 1 C is For the singleatom gas Helium i 1 QA U 2 n RA T 1 How many moles i5 1kg 2 x Mass number 4 49 He is one mole xi 250 moles K i i 1K3118J Q 250 mol 8315 2 K mol Internal Energy Molecular Gases Different materials have different forms of internal energy degrees of freedom z In addition to linear KE Molecules may have rotational KE Molecules may have elastic Potential E UN rotKE kBTkBT kBT Znumber of dimensions x 1 kB T energy forms The more degrees of freedom the higherx the specific heat capacity Real Gases Condensation Water and Vapor E It I l I quot51 l IllI ii if l quotI Solid 3 ll ix x1 3 39 Enrica lquot l quota lquot quottic n R T I point 39539 quotl a 39 ill KII39 1391quot I xxx P 1quot E 39 GIN P Liqmd a quot quot ii i x a 39 W in V31lif39quot L mhquot 39quot Thigh I H Dr 1 u h sl an I TWPlull g Liquid I Mini g rarer quotEa 5 000 0101 it39ll 374 region T 10W V Raise Pressure lower Tempgt condensation gas to liquid V gas and liquid coeXist vapor until all is converted Heat Oriqinallv heat was thought to be a separate quantiw connected to Temperature Heat is energy Mechanical energy can become heat through friction symbol Q unit 1 J 1 Joule m h v feign 1 unit 1 cal Heat required to 39 raise lg of Water Temp 1 C i 39C 1 kcal 1 Cal 39i E 2 6 l3 James Prescott Joule 18181889 showed that you can raise temperature by mechanical work 1 demo melting races Al Fe Pb Speci c Heat Different materials require d ferent amount of Q to change their temperature The difference is called Specific Heat c the amount of Heat required to raise the temperature of 1 kg of given material by 1 C Q ATm Q omAT positive Q heats up negative Q cools down Several examples C cwater 4186 JK kg cGlass 840 JK kg cIron 450 JK kg cProtein 1700 JK kg cAluminumn 900 JK kg clead 128 JK kg E li39lrgl il39i lkl Heat and Temperature gm Heat will flow from higher to lower temperature Heat is transfer of energy and energy is conserved so Q2 Q1 Q will raise T2 while Q will lower T1 gt until T1T2 and Heat stops flowing Heat is the change of the internal energy U Hi ln39r vi 3941 Heat and Temperature Thermal conductivity The rate of heat flow is proportional to the thermal conductivity of the materials Qt kA TlT2l several examples kcopper 380 Js m 0C km 2 Js m 0C kAluminum 200 Js m 0C kGlass 084 Js m 0C ksteel 40 Js m 0C kstyrofoam 0010 Js m C SQuestion Stay tuned Friday CAPA10 Recitation Monday Chapter 14 cont xMore Calorimetry Latent Heat Heat Exchange Wednesday MiniExam5 chapters 9 13 PHY 2053C College lesics A Motion Fumes Ene I ave r My 39 39 Rotational Motion Khemar Rotati nal motion amp is kinemat Tiesgt2 Tor u Linear and Angular nematics summary in angle slraal v ms angularvel mrads acceleration a rnsZ an ular acc alradsZ How do the angular kinematics Uansiate to I inearquot kl nemah cs 7 5m an a menygomlmd what rs yaw tmgmtla spaa yaw tmgenrlaiIEelavon 7 quot 39i39ot 39 our 1345 centnpetsa scmbrsnon Ang ular Kinematic Eq uations Alon With the angular coordinate a velocity wand acce erauon acome the kinematic equations vv at const a xx v t z at2 v2v 2 2ax Note that rtwe use a as a coordinate it goes beyond the M clrde or 2n Althou n awheel goes back to is ongrnal posuon atter one tul turn the coordinate conunues and rneasures tull turns as nuUpEs 02quot Torque and Force I example Striking a tangential component produces torque gt spin m otion radial component Force acts toward c m gt only linear motion Egtlttended bodies behave like mass points only when forces acting in the direction toward the center of mass If forces act at a radius tangentially they produce a turning force called torque which produces spin Here the torque is defined as 39 II 39 Torque A Turning Force A Force sets an object into linear motion A Torque sets an object into rotation Torque is de ned with rspect tn an axis of rotation 5 units 1 Newton meter 1Nm example To turn the door open you need a T h r larger force acting at the smaller radius To exert the same torque if r2ltrl gt F1ltF2 T F2r2Flrl remember Leverarm times force torque TF r Mm D r 1 a liculi 1quot F Lr F Ilullun allot example I Torque on the Elbow The e 5 muscle is attached 5 cm from the elbow What torque can the muscle erert on the lower arm in situations a and 7 below T F I Si Me a T700N005msin90 35Nm b T700N005msin12030Nm ll hamm a an angle llit Note how the sin8 can be used beyond the 90 angle in a triangle l g Quest39on 1 Rotational Dynamics g Torque and Moment of Inertla Unear Dyna as are based on Newton39s laws Rotational Dyna ics are similar 1 If there is no net torque acting on a system it remains at constant angular velocity w r iquot Inerti 2 sum of torques Moment of inertia times angular accelera lion F x ma 37mm 3 Action Reaclion 39 rtial mass Momentof Inertial Momentof inertia takes the role of We still have not define Moment of Inertla We have to find I in the formula use a case whEre we can look at the same situation in Inear an 39 nal We two ways d rota ho Let F act tangentially on m at a radius r That is equiv to 1Fr Tm 7 andil atxri In our equation ZTIIX the moment of inertia becomes which we define as the Moment of Inertla of a single mass Moment of Inertia2 i For extended bodies the moment of inertia is caicuiated bv averaging mr2 Within the bodv ii oi iiii i iii their are i The outside mass counts ihe mosti Moment of Inertia 3 The Moment ofil39ierna depends 0m the axis for thw it is de ned Usuaiiy We deine the rotational axis to go throng the center ofmass Cofa body 7 7 This leads m a separation of rotationat from linear women m i i i I Wiatdoe ng ww whee5 mean x 1 Putihe rotational axis middgn the ow 2 Make the rotational axis a symmetry axis ofihe mass Rotational Energy T e mple ofihe VorVo snows Ls matrotation contail39s energv inanaiogytolinearmolion Final velocity vb Mgh V vlvzhap V21 aim 2 Va is r Question 2 Hoop Racesquot 92 example Roll ng of different Objects I lump Depen on r QEmpw w moment ofln 39ulld u lmllcr UJAEcIIY ofme bodresdrfferenr H jrhcmlmumm O r mzmzn from me hnear Kmeuc energv any ne shppmg box eonens an of me poterma energv mm hnear Kmeuc energy a and therefore mm eed 1 2 2 WW M Vm 39 Icmwmh lmxMr i 2 7 2 H 2Mv gt V 7 Angular Momentum As no surpnse A er rotauona kmeuc energy comes LV angu ar momen nm Lana ogue to hnear momentum pIW Angular momentum nasanre 1kg mZs Mo 3 ever arm Nouce drfferentfrom hnear momenmm umt 1 kg ms Angu ar momentum rs a seperatew conserved quanu You can not enange angular momentum from me rnsrde Conservation of Angular Momentum it l5 far Easler ta ana your mamant of inertia 2 1 M R L 1 w Brll39vgll lg mass toward tine robauorl axis reduces I Conserve angilar omentum ll39lcrease Igt decrease Lu Rota I Motion analogue t ona s dsplacementAx velootyv AxAt acoeleranon a AvAt anguar displacementAs anguar velan a Asmt angular aoteleranmiz AvAt Vlzn W ahquotru anEVZr rm Kll iemaucal ecuauons vnal const a v t 2 atZ Forces Torque Moment oflnerua 21 In r W PM War Wauon Laws ZKE ZPE 2 Z Summary Rotational Mon39on follows veryslmllar laws as linear m l39on Torque tones from farce acmg 07 a leiEranquot MomentaHnema IrmIJs the 5 05 body that ressis EDgw39E saysanon Angular Momentum l a conserved quanuty like linear momenmrn Rotation carries kinetic energy Important points 2 1 Linear and Angular Kinematics displacement xm angle 6rad velocity v ms angular vel w rads wvlr acceleration a msz angular acc a rad52 Force F N Torque Nm 1r F sinl9 mass m kg moment of inertia I kg m2 I r2 momentum 1 kg ms ang mom L kg m2s L w L r 1 sin linear KE KE J rot KE J m v2 2 2 Vector Nature I of Angular Quantities So far we have treated rotational quantities as numbers They are vectors pointing along the axis of rotation The right hand rule 00 points 39m th39m h when the ngers point along rota When the rotation is ccw in to points in pos z direction An 00 in negative zdirectior means a clockwise rotation In x y The same is true for L 011 QUEStiOI I 3 Vector Nature of I Angular Momentum The vector nature of angular momentum as interesting consequences The person stands or sits on a rotatable platform at rest He holds a spinninf wheel in his hands He turns the axis upsidedown I What happens 7 g Stay tune Wednesday Chapter 9 gram Equmbmum amp MiniExam7 On chapter 2 g g Mnear momentum co servamm aws amp rotatwona monon n Friday mPASRecih hon Example39 Circular Platform 3 m 521kev 3 7mvmatesabcut A 79 3 kg smmtwak 5 qu ivtm me Hm me cmta The anmav vawmw m we Svstan s 3 3 rads Mm me smdmhs anhenm Fwd me away vdmtv mm svscenmmmesmm szsgmimmnecmca Anngar momentum conservation 3 Ly 390 vlvmv a meme 2 Bf me mame mm 1V gammaum 1 M7 MM Ru MMQMHWQ n rad 12532 rad 357W 68 Dr David M Lind Dr Kun Yang Dr David Van Winkle Today 399 Dynamics Newton s Laws of Motion Newton39s laws 3 Problem solving free body diagrams 0 Given V and 9 we can nd Vx and Vy vector form to component form Converting between the two forms of vectors Y Vx V c050 Vy V Sine X90 1 And vice versa 9 component form to vector form Q ggm 3 m v W 0 arc tanVy7Vx v 4 m X tan391 vyvx x 39239 move tail of V2 to end head of V1 39239 draw parallelogram using V1 V2 39239 draw resulting vector V from tail 39239 draw resulting vector V diagonally of V1 1 to head of V2 across parallelogram frorn tails of V1 gt V2 horizontal components le VZX vtx le V2x sz verticalcongponents Vly VTY VTY Viv VZy We have to add both x and y components Vtx segarately Pay attention to signs of every component Reminder wzo Newton s First Law All objects continue in their state of rest or of uniform speed in a straight line unless acted on by a net force Inertia oz Newton39s Second Law The net force acting on an object equals its mass times acceleration E F m l 4 Newtons Law Slr Issac Newt0n16421727 Whenever an object exerts a force on a second object the second object exerts an equal and opposite force on the rSt39 ActionReaction Newton s First Law of Motion All objects continue in their state of rest or of uniform speed in a straight line unless acted on by a net force ozo Constant velocity along a straight line ozo rest is zero velocity 2 called inertia satellite Newton s Second Law of Motion Net Force equals mass times acceleration world 2 F m 2 object a single object interact with the world st add all forces the world acting on the object st they cause acceleration of the object st mass determines the force per acceleration more force means more acceleration o more mass means less acceleration b on an object applying a causes the object to a The harder you something more the greater the l The more massive an object the less it in response to a given force and vice versa A chain of five links of equal mass is pulled by a constant force F at an acceleration of 25 msz What is the force excerted by link 3 on link 27 What is the total force F salmon 1 I Sepamte the object from the World a Link 12 are the object of choice for the rst question b What forces are acting on the object 7 The unknown force F1 and the object Weight Infrng point in opposite directions a g F2 FNET 7 F2 39 m1 m2 0 Newton39s nd FNETma F In m g m m2a 1111n12 g d s lve folr unzknown 1sz Solution 2 Now the object is the Whole chain A block is at rest on an inclined plane whose N elevation can be varie FN p5042 pK015 Fr At what angle does the block start to slide Bf g quot What is the acceleration 7 qty 39 X Free Body Diagram i 39 Solution We Will see that mass m does not matter 39ce x and yaxis l r 9 a Find the x and y components of Weight F FGXmgsin0 FGV mgoos6 b Find the unknown iction force c FNFeylmoos6 FnusFNusmgO S0 gt ustan9 r 5 Whenever an object excerts a force on a second object the second excen s an equal and opposite force on the rst Force on sled Force on cxcncd by assistant assistaul r lav 1 Force F0quot 0 n skater wa FM Friction Force on has an assistant sled exerted exerted exerted by gmund by sled by Istsislanl by ground Action and Reaction forces always act on different objects We swap the roles of object and World Acionreaction is the source of all propulsion cars use it your feet use it rockets use it Newton s Third Law of Motion Whenever an object excerts a force on a second object the second excerts an equal and opposite force on the rst Lets see what its like to ride a rocket sled I I K Warning I am a trained professional Don t try this at home AE DENTE HAPPEN This Friday Recitation CAPA 3 Next Monday we will talk about Circular Motion and Gravity Chap 5 May the force be with you Dr David M Lind Dr Kun Yang Dr David Van Winkle Today 4 Dynamics Newton s Laws of Motion Forces Friction normal force etc Newton39s laws 1 2 e wton s First Law All objects continue in their state of rest or of uniform speed in a straight line unless acted on by a net force oz Newton s Second Law The net force acting on an object equals its mass times acceleration E F m Sir Issac Newton 16421727 oz Newton s Third Law Whenever an object exerts a force on a second object the second object exerts an equal and opposite force on the rst gt We need to understand Forces osofyou intuitively know the effect of forces oto To set an object in motion or change its state of motion st To deform an object See also Chapter 9 ln Physics forces are vectors with both magnitude and direction Forces add up as vectors the sum of all forces acting on one object is called the net force ype of Forces include 39339 Gravity F mg 39339 Forces between collections A l of atoms or molecules I 139 I Friction opposes motion I Normal Force opposes penetration I Tension OppOSGS separation 01 compressio 9 v On subatomic level I Weak Force Strong Force I Electromagnetic Force e have used the fact that gravity pulls with constant acceleration of g 980 mls2 toward the center of the earth What force does that correspond to 8 gt gt is E F Fma1kg98ms2 L I 98 kgm52 m 98 Newtons g g The force of gravity is Fg mg pointing downward This force is called weight We strictly distinguish between mass b and aAstandardkilogram weighs about 98 N on earth b The same kilogram weighs Mam 16 N on timoon A is resting on a surface Nothing happens Why Newton The weight of the object is opposed by the normal force FN exerted by the surface to prevent penetration At the atomic level no two atoms can occupy the same location Normal forces Always perpendicular normal to surface r Always the same size opposite direction of acting force39s normal component Static friction force Ffr co oppose acting forces F A keeping the object at rest ma co Friction forces point along the surface so There is a maximum static friction force quot39quot Kinetic friction force Ffr so If more force is applied the object starts to slide and the kinetic friction force applies Fir i395 FN 5min Kinetic f c iun Friclinn 10 10 30 40 50 60 7 Applied force FA FfruKFN p5 r N lt sliding no motion Newton s First Law of Motion All objects continue in their state of rest or of uniform speed in a straight line unless acted on by a net force We understand this to mean 39zThe normal state for objects is constant velocity motion along a straight line 39z rest is motion with velocity zero ozThis property is called inertia ozNet force Sum of all forces acting on one object satellite z t 1 l 1 velocity Find the forces acting on the car The net force must be zero because a 0 Vertical Weight is balanced by Norman leorce driving on a straight street with constant 5K Horizontal Air resistance and frictionar quotquotquot H coffee liquid H wfr ialJIquotJ balanced by Motor39s Force lt w resistance wt quotas5 W Fo Rolling Frition l Rollin Friction Tlewton s Second Law of Motion Net Force equals mass times acceleration world How does a single object interact with the world a The important part is finding all the forces acting on one object in a given situation the world a The right side of the equation never changes for the same object a All the world can do is determine the acceleration of the object l a Mass determines theamount of force needed per acceleration object w r midi 1 r uw i i l l 1 l l l l v lrm r n k a Consta nt 39 velocity rce 1 2 A v 7 5 Whoa JI acceleration gt Velocity gt Slowing down Accel opposite motion Caused by Bra king A ans driving on a straight street but is now accelerating Find the additional forces on the car The net force must be in the direction of the Any ch direction is an acceleration and is caused by a Force Motor force Braking Motor force ange in speed or Speeding up accel same dir Yrion 1 more force means more acceleration oz more mass means less acceleration causes the object to a on an object applying 81 b The harder you something More the greater the l in response to a given force and vice versa The morean object the less it Wednesday Chapter 4 Forces and Newton s Laws This Friday Recitation CAPA 3 Next Monday we will talk about Circular Motion and Gravity Chap 5 May the force be with you Spring 2004 I I I0 Ch6Ch7 PHY 2053C 39 College Physics Motion Forces Energy Heat Waves Dr David M Lind i I Dr Kun Yang 39 i Dr David Van Winkle Today nish w r NonConservative Fc Linear Momentum I Momentum Conservatio Elastic and Inelastic Col Centerof mass motion i q review amp preview Conserved quantities energy amp momentum 1 The Total Energy is a conserved quantity You can not change energy from the inside of a system 1 EKEPE mv2mghmconstant You can change energy from the outside by performing work on an object AE Wchos9 The Total Momentum is also a conserved quantity You can not change momentum from the inside f E m 3 constant You can change the momentum of an object from the outside by applying a force AZ FAt Example Energy conservation 9 LooptheLoop n A small block of mass m10 kg H SI es WI ou nc f n along 9 the loopthe loop track shown I 7 5quot The block starts from the point P a distance h58m above the bottom of the loop of radius R20 m What is the kinetic energy of the mass at the point A on the loop Solution Point P E PEmgh10 kg98 m3258 m 5684J Point A5684JE PE KEmg2R KE gt KEA5684J mg2R5684J 1kg98m3240m 5684J 392J 1764J How fast is it at point A l2KEA KEAmi gt VA 188ms Conservative and Nonconservative Forces The conservation of mechanical energy applies only if no friction occurs v20 Friction is a nonconservative force meaning that Friction converts mechanical energy into heat another form of energy which cannot be directly recovered back into mech energy Ch 14 The work done by all no conservative quot forces such as friction I nge the total if mechanical energy 539 L Conservative and Nonconservative Forces 39 39 Friction is a nonconservative force The work done by all nonconservative forces I change the total mechanical energy NC EXAMPLE Let a block of ml kg slide down a path with 2m height difference and no friction Him Then it encounters a 4m long level surface which excerts 10 N offriction force on it What 39 is its energy at the enclof that path t i7 Ef k 9MZR4JL potential 39 quot 39i4mi quot E2E1784Jkinetic fl immn lehnf a l mmtimp3g4i Specialtreat I I Texts to evaluate We would like everyone in the class to evaluate the quality of a potential new textbook We will give everyone who turns in an evaluation sheet on Monday comparing their chapter on Work and Energy the same extra credit points as a chapter summary There will also be a raffle of all those who turn in an evaluation sheet for the chance at a free DVD player Momentum Momentum is defined as mass times velocity I mT Momentum is a vector Units kg ms unit does not have special name Newton wrote his second law as s A quot 2 F p At When mass is constant this is equivalent to A mAV At At ma Change of Momentum quotimpulsequot This definition of momentum is in accord with the momentum used in everyday life The more momentum an object has the harder it is to stop or to get it going a At gt A 73 I3 A t7 impulse The change of momentum equals quotforcequot times quottime appliedquot This is also called impulse Conservation of Momentum 39 Momentum is a conserved quantity just like energy if there is no external force acting 39bm171m272m373m474 constant That sounds like old news what is it good for Even for many objects total momentum does not change at any time if they are an isolated system 1 ffllllllwi N 2 You cannot change the momentum of a system from the inside Question 1 example Rocket Propulsion all The propulsion of a rocket is based on momentum conservation p E S PM Pril lit l 75x1 05 kg of propellant pushing 50x1 04 kg rocket In space there is no material to push against July 16 1969 Saturn V rocket Apollo 11 on the way to the moon I The momentum of hot gas expelled toward the back plus the forward momentum of the rocket remain ZEFO Remember momenta add up example Collision of two J We call the velocities before collision V1 V2 after collision 71 72 152771 71 l m272 W1 71 39 mz 72 39 i5 39 I x Let s prove thatgt A 1F1At A 2F2At Newton39s 3 F1 F 2 1 meansthatA is A 52 gt r gt gt gt m 1 1 v m2 V2 lt gt m1 V1 m1 V1 2 m2 V2 m2 V2 M 1 I ltgtm171m2 72m17139m272 which is just conservation of momentum Question 2 Si m u lta n eo u 5 Co n x f E 391 6 my a n cl M o m 1 1 Collisions conservation of gt 1 t 7 pm1V1m2V2f771V1llsz2IZ So far two unknowns V 1 V 2 one equation 2 Elastic collision conservation of energy KEm1 m2v m1v m2v39 KE39 Now we have two unknowns V 391 V392 two equations In the following we will use both properties to calculate the two unknowns A collision in which energy is conserved is called an elastic collision E Ia stic Col l isio39quot 1 s 1dimensional conservation lmV1V139l n72V2 V2 1 2 2 r 1 r 2 I 2 energy 5m 1 m2V2 m1v1 m2v2 conservation N 1cimensional I V1V2V1 V2 relative velocity before relative velocity after This result is specific to elastic collisions where energy and momentum are conserved It is independent of the masses ml m 1 I h Ii hi 1439 I 1quot I v1 v3 4 V V1 V2 I 1 momentum pm1 V1 szz m1 V1 a H 7 Question 3 4 kv x example Newton39s Cradle Why does only one ball bounce 1 back from Newton quot5 cradle all balls have the same mass m u 1 Momentum conservation X m vow 2 E ergy con ervation 1 1 V1Voutvout therefore xx a NM Fa mv1XgtIltmv1 gt X1 L20Ch10Ch11 Spring2004 PHY 2053C College Physics A Motion Forces Enerqy Heat Wave Dr David M Lind Dr Kun Yang Dr David Van Winkle Today 4o Fluids dynamics Fluid ow Continuity Equation Bernoulli39s Equation Lift amp the Airplane Wing 4o Vibrations and Simple Harmonic Announcements We don t have miniexam6 completely graded yet but will go over it on Wednesday The final exam will be held Wednesday December 8 at 10001200 and is scheduled for 275 Fisher Lecture Hall The place will not be in our current classroom The final exam will be comprehensive and we will spend most of the last week of class reviewing for it There will be labs held the last week of class this is different from the syllabus for the class review Important Points from Last Lecture Densit The density of an object We 1 is defined as mass per volume units 7 1 kg m3 Ip mVI 39 f The pressure in liquids varies a Pascal39s Principle pressure increase A due to some external forces is the same 9 I 2 I throughout a con ned uid 9 leads 2mm A1 1 to mechanical force multiplication in Hydra u l i C Lifts Bouyancy Since the pressure in a liquid quot increases with depth for submerged 9 objects the fluid exerts a buoyant force on the object which equals the weight of displaced fluid Continuity Equation In flowing liquids the mass going through a certain area equals the mass going through any other area Am1 p1AV1 0141AI1 A1 zij llq lv l At At At Am1 Am2 At At Continuity equation If the fluid is incompressible does not change so A1 V1 2 A2 V2 smaller area gt faster motion example 100d Flow All blood going through the Aorta will end up going through the capillaries Aorta ra 1 cm Flow 30 cms Capillary rc410394 cm Flow 510394 cms Question How many capillaries are there in the body from the Continuity Equation Aa Va N AC VC 1 1 l quot2 I39LLl th it Kt Vl 2 rms V Antler Head LLIITIi 1 a Jk ngeurl r Body39 39rgans Willis 5 A Timquot J 1 nKlElt l l Iquot J 4 1239 valves 139 capillaries vanriN vcnrigt NV a 4x109 0 c Bernoulli39s Equation g Look at fluid streaming through a pipe with decreasing diameter Bernoulli39s equation is pl conservation of energy divided by Volume All gt 1 g mv2mghFdoonst Energy pV2pgh 2 Ad g pvzpghPoonst doonstE V a Ml ihl Bernoulli39s Equation applied 1 2 l I H I 2 0 v Ip gnI IquotCOHSI Our previous hydrostatic pressure formula is contained in this for vO where d is depth under surface note d h Ppgd Important observation Where the flow is high the pres l example Airplane wing not pushed up but sucked up Tennis ball spin Bernoulli39s Equation applied 2 Leafblower Sailing against the Wind Carburator 4 lll l quotl p I Air w 39 Wind p I if ll quotI 39 HighP T T T p 39p 339 I i no flow p j Mil p i l L l 9 MW Jib I Atmospheric Vl l pressure ll ll 7 7 7 Low F p p pressure Gas 7 wind Mainsa lib Ub 11 y i curveball spin in baseball 1 example urricane and the roof Durin a hurricane the atmos heric ressure inside a house may blow off the roof because of the reduced pressure outside If air density 1293 kgm3 is blowing over the roof of a house at a speed of 133 kmhr whatis the pressure difference between the inside and the ouiside of the house A p212pair V2 1000m V 133kmh 133 S 3694ms A p121293kgm33694ms2882Pa What would be the lifting force on a roof of area 106 m2 FPA882Pa106mZ935347Nzweight of10t 1 XgBernoulli39s Equation applied 3 2 l J 1 2 tr r lz pv pghP const 322 A1 Combine the continuity equation with Bernoulli s equation A 2 1 1 1 A V1 A1 V2142gt szA 2 V1 2 P1P2 P Vf PA V1 2 1 2 A1 P P V 1 1 2 2 p 1 A2 We can measure the ow rate by measuring the pressure differential Vibrations and imple Harmonic Motion Vibration is a type of motion which repeats itself ack and forth after the timeperiod T Nonperiodic Noise Vibration Oscillation Periodic Motion If the oscillation follows a cosinewave we call it simple harmonic it AmplitudeA maximum displacement 39 CYCIe wave One a 3 complete path back and forth z Frequency f 1T Elia a The number of cycles per secondquot Harmonic motion f independent of amplitude Forces in the imple Harmonic Oscillator Force exerted by pendulum mg sine a Maximum displacement quot a mg sine b Equilibrium Position XO VV 30 c Maximum negative displ x L sine vO amg sine Forces in the imple Harmonic Oscillator Force exerted by spring F k x a Maximum displaceoment F x A V swing I all Li A 333235 m m m 1quot Li 1 A b Equilibrium Position 24 X O V2 V0 30 v illi ll gal my g 39 climatic c Maximum negative displ k m l39 x A vO aE F Jillilllllil m w 3t t ID Question Damped Harmonic Oscillation he simple harmonic oscillator would qo on forever c Attached to quot X car frame i 1 39 i z Plston a Viscous iquot ii uid Attached to Many technical applications such as shock absorbers try to reach critical dampingB where the oscillation goes away in one cycle while maintaining elasticiw Resonance When a harmonic oscillator is driven by an external oscillation External frequency f Natural osc frequency f0 f0 Response in resonance ff0 depends on damping A small amount of damping B strong damping Q X l plllll t U oscillating system 177 I T W39 f rWWWWU a quot I x Frequency example esonance Bridge Collapes When a harmonic oscillator is I 1 VW 0 2n a 39 a Tacoma Narrows bridge collapse oscillation driven by Wind driven by an external oscillation an garquot a 5 r V Oakld Freeway oscillation driven by earthquake ridge collapSe i lTlPlllLlUC Ul Oscillating system I Waves Waves are moving oscillations They carry energy they do NOT carry matter Creslor peak V V position Troiigli A p 39 Wavelength 3 length of one full cycle at fixed time 39 Period T time for one full cycle at fixed position Waves travel at speed v Stay tuned Announcements repeat Wednesday chapter 11 cent More b at s39 Energy Re ection Diffmction and Resonance r W otoWe will go over miniexam6 on Wednesday oonhe nal exam will be held Wednesday December 8 at 10001200 and is scheduled for 275 Fisher Lecture Hall not in our current classroom oonhe nal exam will be comprehensive and we next week Labs will be held next week not this week Fall zm7 352151 A yes Lift 52 the Amp Stay tuned Wednesday Einstein Centennial mm speudplznzhrimndmw wi mtbemizsis ewdidiscmsEmsmn mdhiswmkinmuchmme mum semesm IIYZEISAC Monday chapte quotsun scmamafVlannnsand Wives and My aim ughmu quotusury m Buoyancy39 Archimedes Principle 5 use mg mg Lena llquld increases Wlm depth Force on Bottom gt Force on Tap F2 F1 9 9 Ahzhl 9 9 AW lawman W Fg weight of displaced liquid If you submerge objects the uid exerls a buoyant fame on the object which equals the weiciht of displaced fluid Atmospheric and Gauge Pressure We have to acoocuyntfor air pressure The normal air pressure at sea level 1013 kPa 1013 bar 1atm 101300 Nm2 E hats he same as the weyhf 0f10 tons on 1111 Mageberg gheres famcus dmmstramn cm 1571 my Lord Magieberg mad m pull me WU halva at an mmmd sphere apart Wllh teams at 16 hnrsa 39 Air pressure usually works on both Sides ofa L J quot so that it cancels out Fluid Flow Continuity Equation In owing liquids the mass going 39 t roug a certain areaequals the j lJ w mass going through any other area M Ami plAl P1A1All A TTTW Vw A m1 A m2 At A t Continuity eguatio quot Canserval7bn anamenl39umJ If the uid is incompres 39ble p does not change so am smaller area 9 faster motion Bernoulli39s Equation Look at uid streaming through a pipe with decreasing diameter A 39Ar Bernoulli39s equation is divided by Volume 6 1Emu2mghFdconstEnergy I gtAIlt Al 7 i 1 F 21iZpghLdd constEV Bernoulli39s Equation applied 1 Our premous marosmuc pressure formu a S oomamed m Bernoum39s ecuanon for v0 Where d S depih under surface 0 39 gd important observanon vWere the ow 5 hwgh the pressure 5 ovw example Airpla in ne w 9 at not pushed Lp but sucked up Bernoulli39 Equation a plied 2 La bluw Saumg against the Wm Carbumlur gt mph Question 7 In the swtuatwom above W What dv echm S the ow quot 7 the bvrpass du ecte v 1 to the eft 2 no ow 3 to the nght example Hurricane and the roof 39 39 e the atmospheric pressure inside a house may blow off the roof because of the reduced pressure outside Ifair density 1293 kgm3 is blowing over the roof ofa house at a speed of 133 kmhr what395 re pramre d reice between re irisle and re altsbe of re house A p1l2par V2 1000m v133kmlh13336008 3694mls A p1l21293kglnf 94mls2882Pa What would be re lilh39lrg Iirrce on a mofofares 106 m2 FPA882P6166I772935347NEdeh10f lot I SI Bernoulli39s Equation applied 3 quot i r Combine the continuity equation with Bernoulli s equation A 1 1 2 l v1A1 v2A2gt vz v1 P1 P2 p ill EMT v1 A2 We can measure the ow rate by measuring the pressure differential l l Head example fA39 Blood Flow All blood oin through the Aorta gt 7 will end up going through Mm the capillaris A Aorta 1 cm Flow 30 cms it it lt i V l N K Capillary rE4quot1lr4 cm Flow 510 cms Qustion How marry cap749173 are rere in re body from the Continuity Equation Trunk J alleys L39 n valves AavaNAcvC rczipillmie 2 2 V r2 9 varrrachrrrcNV a 4x10 C re Stay tuned Wednesday Einstein Centennial 77 There wm beNO quiz um Wednesday mam wenal phnzlzriumshnw Wm nuibe un tests We will discuss Einstein and h5 vmrk in much mure dam nail semesta39Pi iY2S C humamimming Raa vily Iheury m Introduction Fluids Density Pressure in Fluids Pressure in Liquids Question Container shapes m 12 m Faucet Pressure 5 Question Hydraulic Lift 1 Atmospheric and Gauge Pressure Barometric Pressure Q Question Atmospheric Pressure Sink Swim or Float exam 2 Floating Log Question I20 I PHY 2053C College Physics A Fall 2007 Motion FarcesgtEneryy Heat Waves 9 Dynamics Newton s Three laws Newton s laws 3 Problem solving ryeehndy dizgxams 7 er dp 2dlmen5Ional PrOJectIle Motlon Mouon m me hor 01113 and verl ca independent There s no acceleration m me x d ec o There is constant acceleran39on m me dwrecnm downward aocelerauon ofgravilyl KNK llm pnur 5quot 939 Projectile Range 1 An omectrs lhrown ata speed of 25 ms Whalis he En765 L slance i an IravehorilanlaIy Debra hni WE VEyam Sumtan 1 WE lake the Vemcal motlon El nd the We uffhghttby requestmg y yn U assumngno ar resstance ytynvw39t lngtz 2 Thenwe use hammetftu deterrmrre tne rangex m the horlzomal quot quot quot zx vmz Exam PrOJectIle Range 2 An ObJSCtlS lhrowl39l ata speed of 25 ms ha the air7e dim7439s i an Iran haiIDHIaIy gramd assUmngno ar ress Enos ande l Flrstflnd 1f rerwnleny neeemeszen r WM El ylv 12 g2 lumen 2 Tnenrlnu EIUL new far x ne enleellravels nenzemallyln nalsamellme R xt Xtan39t E mu39 Projectile Range 3 An ObJSCtlS mown ata speed of 25 ms w 1 me i an IravehorizanlaIy assUmngno ar resstance 1 andeve 9741770 3 Now We have m relate vx al ld v 7 m me lnnal speed v and the angle a m 9 f R 5 g g Cm W 25ln8c05a ln29 r x a l ewton 5 Three Laws of Motion Newton39s First Law AH abjects commue lrl New State Elf rest Eler unlfoml speed ln 3 slralgnl llne unless acted an by a netrorce s m a Tne net force amng an an en 21 e uals lls masstlmes acceleration o Newton sx em mum 1727 Whenever n enleel exerts force an a seeenu ublecL lne seeenu enleel exerts an equal and opposlte force an me 395 gt We need In understand Forces Forces MW know the effect of forces ose ano Je motion or 39 motion v ut aiso to deform an ooiect see cnapter 9 Unit of force 1 1Ni We2 2 13 v t in Pnysics orces are vectors wtn ootn magnitude and direction Forces add as vectors tne sum ot aii torces acting on one object i5 caiied tne net force Fo rces Cummun T1225 DfFung inriude Gravity F mg Forces between i 39 39 collections of atoms or moiecuies Friction e opposes mutinn Norm orce Dppuses On subaatomic level Lndeiiies most ct the mm DSCEUiC fates shave Weak Force Strong Force Electromagretic Force 111a Force of Gravity We have used the tact tnat gravity puiis With constant acce ration ot g 30 mis2 toward tne center ot tne eartn What force does that correspond to v om mm 7 Fma The force of gravityis FEl mg pointing downward inwzdvhe u1huquotrru or m mm m ms 1mg This forceis Caiied weight mmn We strictiy distinguish between mass and we39 trey aemrerem llunws n f lCE memller zqilznllly m39sm i quotem h Normal Forces Ah eoleel ls resnhg an a surraee lhl 9 happens 39 Why on T e wagm erlhe eoleel ls Dle F pe h are alwaysthe s used by he normal force exerted by the surraee tn prevem slmace ame sue and h the opposlte dlrecnon er aenhg fume s normal Bumpunent elralluh Aline ammlclwd lhls ls because he lwu alum can awcuw lha same lacallah allhe same llmE Normal forces are always perpendlcularmomlal tn he Frictional Forces Static friction force F o o oses and exactiy balances me acnng force FA lhe Object at rest 9 Frlc ol l forces are ll l a dlrecuol39l along ne surface 9 There ls a maximum st 39 frlcnm forcel Kinetic friction force F 5 o Ifm mu objectsmr slaw mu me kln mourn pl l 2 3n on w m m LHllrunf quot luIN 7 alldlrlg 7 u We uhu o Thls prupeny ls ealle o Metrorce Sum at all rerees erslahu hrs In mean 9 The normal statequot rereoleels ls constant veloclty motlon alehg a slralghl llhe restquot ls melleh Wllh veluclty zere a menu actlng eh one eoleel 4 rm 4 mu bumpquot Constant Veloc ty SUEEL W th constant veloclm h The quota A caris drMng eh a straight Fwd the fumes actihgorit e car Rehemoei zen musmeZEm o ause iii Welght is warmed by Normal Force resistance aherrictiori are ba a ced by caffeellqmd yeast 3 force Vemce Horizonte Air How does a single object inerect With the Worldquot 0 The impunant pan is riheih aiithe forces acting eh one one in a given situation the wormquot o The right Slde er the e uatieh never hangesfurme same e iect 0 AM the worldquot can on is daermme the acceieratieh erme obiect 0 Mass determines he ameuht errerce heeeee peracceieratieh amp Acceleration amp Force A caris d vmg en a straight street but is NEW accelerating Find he additiuna fumes on me car The ne orce mus1 be h the dwectiun mt cceiernioni s Q39Q s M m Cmsm by Braking Motor force Braking Motor force EXquot 4quot9 Force Mass amp Acceleration H W m human 7 W quot mii E mmm m l g Question 1 Action and Reac o s always act a fferenl objecB We 5w ap the mes Df bbjetf and Wm7x Question 2 Newton39s third law Rocket sled 7 39 niiiiA 3 1mm r min I i am a trained prufess39und Don39t try this at humel Problem solving Free body diagrams amp forces Digpiuure Kedu ui pi or m carequy choose object to anaiyze Separate the world from the Draw a caretui diagram sno Object N object Wing wnat is nappenrng to he an Xycoordinale system for he WiH neip you zero some otme ihers cnoos ih origin and prooiem A good choice varraoies and simpth o List 3 the known and unknown u n u s Apin me reievant equations that indudes in he knowns and me paramiar unknown being sougwt m force eXcsrfedby nlt 3 m nlt 2 n 1 52pm me mm mm Lhevm man are me abject ufnh What fnrczs are a Pul g a chain g A chem of ve hnks ofequa mass 5 puHed upward bya oons znt force Fat an aoce erauon of 2 5 m 2 What9 0 75 Snlll b mug un me uh m 7 The unknu fume F m39 r mpmgg quot5sz 5 r quot I mi g a live furunknuwn WWW he Efa OHS F7 Snllnjn 2 n Nuw39he nhjecusthewhllz chain rld mce fur me rst quesuun wn 1 and me uh 31 weight m1mlg mm m sue duemuns Block on Inclined plane A Neck 5 atreston an mchned p ane I whose e evauon can be va ned 5m pK u 15 ac 539 17m Maraye doesmebocksrarrmym F i I What51779 25 a ammsrartssmfng i I Diagra g Stay tuned a Wednesday We W ta k abmtEimularMu un and Gravity Chap 5 May t e force 53 wit youJ p HY 2053C College Physics A Motion Fumes Energ y Heat Waves a mm m Energy Camembert uf Momenm Acc one s ordlng to Newton any object s whlch ls no se ve ll lcludll lg movll lg t subject to external net forces as the Poll39lt of Reference We call those Inertial Frames of Reference We can not choose borne ofreference whlch are accelerated and have thern be lnerual gtW Apparent Weightlessness HOW doe M4571 Emu fe 1135 m warm 71 werghtessnesez r a a WWWWW amtmmymy W mrbdamabouyamywlthwalg b If We use accelerated frames of reference obJe a t nge tnelr welgn eg a erson ln an devator Wlth constant veloily wlll feel ther norrrtal weght wm r meNIalrrlyrnul uu an evamr accelerating upward or downward WlH feel an Increased w s n mm m u lf the elevator cable breaks both the devatof and the person wlll ewerlmce free fall the person wlll be a parently weig less quotramquot 1m Iranquot wm 95 m Work Energy PrInCIple Work can become several forms of way ale lhe same land uf sniff i In 1 draws vi Mi Mk 5mm 39 39 energy lhe energy or mouon Gravit tio al potential energy E astic potential energy potenuai energy assoo ated With posmol l lat Halts 14 he WWIEVEI Heat is aiso erergy Energy is stored work which can be relrieved or changed iron one rorm to another units same as work quotJoule The net work done man Objedls ecuai to is change in kinetic energy K39ne 39c Ener y Assume a constant net force applied to a car oyer a distance d What5 me wak wne incrang the cars 5 Fmay WM a T 2 Wr FierdmadMVll 1 War2w 2m We define as the h a39islational kinetic energyquot of an obiect ote ihatihis is lhe net Work gnomect change in erergy N Retr39ev39ng Kinetic Energy The hammer s kinetic energy KEh is use up quot as if nail is sludi As itsinkes lhe nail139ie naii exam a roroe on me hammer slowing it down With a constant rorce F oz v2 Whthm 2d d rrorn Action rReaclion W F th1lt5h I The w done on the naii is equal m KEH Therefore aii KE is used Lp as work on the naii Potent39al Energy Gravitational g titt a brick ot mass m trorn v to v2 What5 the work 41076 on rhebrim 7 mgh cos039mgh W I But wm can be retrieved bake the hand away and the brch WlH retrieve th Work done as kinetic energy 1 2 m2 ghlmgh Work had been stored m raising the brldlt s V posluol l Thls torn ofenergv is Gravitational Potential Energy Potential Energy Elastic Springs I Hook39 La person stremhes or compresses a sp i With a torce F the sprirg exerB an opposing restoring force raaiiaiar whim F5 Fpr warring which is proportonai to the elonganm or compresle Ax m n 3 rm Dr W m 4 W m l r li a xn x restorll lg39 means the spring tries to go badlt o sition to x0 t Do The constant k is called sprll ig constantquot units Nm Potential Energy Elastic Springs How much wk 5 done DlL Dmpr sing me spring 7 F 46 increases while We are pusnirgi Butwe can use the average force to calcu ate the Work done l J AX AX lmiollllll llvlllk or porri pressirg it Work had e stored in the compressed sprll lgr r m PE WEAxf this is called the Elastic potential Energy Potent39al Energy Properties i potential Energ s an energy associated With the position 0 an Ject gravitational iThe gravitational PE depends O the height above a certain reterenee ievei Vou may thodee any paint as the y 0 art But be consistent H Therefore the vaiue ofPE is not unique even tor a gven posmm But APEis the physicaiiy meanngtui cuartuty and APEdoes NOT depend on the choice of the reterence ievei Conservation of Energy We ustiodred atan exampie Wrere tn it kinetic energy as it fai s F q nitiaiiy ianupi PEEymgh i i Finaiiy ratiiaoii KE 7mV2 For nan rrh energy is kinetic and part is potential but nstant ii the total energy is co A iQuestion 3 Pendulum Question 4 Water Slides Question 5 Water SIidesZ Momentum Momentum 5 de ned as mass umes ve ocm 1 I Momenmm S a vecl or UWB kgms umtdoes nothave speaa name Newton Wrote nts second law as wnen mass 5 constant tns S equwa ent to p mm Conserved quantities energy amp momentum Tine Total Energy is a oorserved quantity Vou can not dnange energy from the H iside ofa system EKEPE imv mgh conscam Vou canbcharge energy from the ouside by performing work 3 19 AE W Fd cos9 The Total Momentum is aiso a conserved quantity Vou can not Change rnornenturn from the H iSide by applying a force i2 constant Vou can change the rnornencurn ofan objectfrom tine outside A5F Example Energy conserva on J 39 r gh1 Dkgcg a WSZSENSEE 41 PE mg2RKE 4 J mg2 imgamlfvmrn sea 417 3921 176 43 mass mwngat poHM Conserva ve and Non conservative Forces in onservation of rnecnanicai energy appiies only if no friction ccurs Friction is a nonnconser a meaning that Friction comer mechanical energy into heat another torn or energy Wnrdn cannot be direcuy recovered badlt rnm necnanrcai energY m 14 Tne work done by aii nonnconservative forces sucn as friction nge the totai 39 rnecnanicai energy lt E mu 3 8 Conservative and 3 Non conservative Forces Frlctlorl rs a nonconservativequot force The work done by 3 nonnoonservanve forces mange the mar mednanrcar energy EXAMPLE Leta b0dlt orm1 Kg shde down a path wrtn 2m nergntdrnerenee and no fncuon Then rt encounters a rt Wrenss magyatme smoftEtpatn E was El AKQ QN Q EmSZ 7a 4 J pmmna e E2 754 J krrent B44N 10NSE4J 4m mm Momentum r Momentum is de ned as mass times verocrty I Momentum 5 a vec or on n r unrtdoes notnave speaar name I Newton Wrote n15 second law as When masS rs constant the rs equwa ent to 4 A6 Change of Momentum mpulse ans dennmon of momentum 5 n accord W th the momentum used n evervdav hfe we mane71m an m i cth s we harder smug 096115ng zF p At mpulsequot 391 re change of momentum equa s r rce nmes time aaplied an5 rs 5 50 2 r Hed mpwse g Conservation of Momentum omentum is a conserved quantity just like energy if there is no external force acting That sounds like old news what Is It good for Even for many objects total momentu d es not change at any time if they are an isolated system 1 quot393 N 2 You cannot change the momentum of a system from the inside Si Question 1 g Stay tuned Friday CAPA6Recitation Monday Chapter 7 Linear Momentum mg PHY2053C

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