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# Gen Physics IA Lab PHYS 110

Siena College

GPA 3.6

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This 39 page Class Notes was uploaded by Alyce Abshire on Tuesday October 20, 2015. The Class Notes belongs to PHYS 110 at Siena College taught by Rose Finn in Fall. Since its upload, it has received 41 views. For similar materials see /class/225359/phys-110-siena-college in Physics 2 at Siena College.

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108 Fluids in Motion Flow Rate and the Equation of Continuity If the flow of a fluid is smooth it is called streamline or laminar flow a Above a certain speed the flow becomes turbulent b Turbulent flow has eddies the viscosity of the fluid is much greater when eddies are present Ch 10 Problem 35 Using the data of Example 1011 calculate the average speed of blood flow in the major arteries of the body which have a total crosssectional area of about 20 cm2 109 Bernoulli s Equation l AI2 llt N l gt A fluid can also change its 13in height By looking at the 2 work done as it moves we find a P p U2 pgy constant This is Bernoulli s 7 equation One thing it tells quot 3 l us is that as the speed goes up the pressure 1 goes down 108 Fluids in Motion Flow Rate and the Equation of Continuity We will deal with laminar flow The mass flow rate is the mass that passes a given point per unit time The flow rates at any two points must be equal as long as no fluid is being added or taken away This gives us the equation of continuity P114191 PzAz Uz 1 43 108 Fluids in Motion Flow Rate and the Equation of Continuity If the density doesn t change typical for liquids this simplifies to A1 A292 Where the pipe is wider the flow is slower 1010 Applications of Bernoulli s Principle from Torricelli to Airplanes Baseballs and TIA Using Bernoulli s principle we find that the speed of fluid coming from a spigot on an open tank is 2ZSO 39 U1 V280 2 M 106 This is called 2 Vl Torricelli s theorem D1 1010 Applications of Bernoulli s Principle from Torricelli to Airplanes Baseballs and TIA Lift on an airplane wing is due to the different air speeds and pressures on the two surfaces of the wing Lower pressure 1010 Applications of Bernoulli s Principle from Torricelli to Airplanes Baseballs and TIA A ball s path will curve due to its Home plate spin which results in the air speeds on the two sides of the ball not being equal 1010 Applications of Bernoulli s Principle from Torricelli to Airplanes Baseballs and TIA m A person with constricted toebf in arteries will find that they w is may experience a Len l u 5Rigm temporary lack of blood to vertebral 1 vertebral the brain TIA as blood my K my speeds up to get past the swam constriction thereby artery reducing the pressure Corilriction VJ onrm 9 gt 1010 Applications of Bernoulli s Principle from Torricelli to Airplanes Baseballs and TIA A venturi meter can be used to measure fluid flow by measuring pressure differences Copyright l Home Fesrsan 1010 Applications of Bernoulli s Principle from Torricelli to Airplanes Baseballs and TIA Air flow across the top helps smoke go up a chimney and air flow over multiple openings can provide the needed circulation in underground burrows Ch 10 Problem 44 What is the lift in newtons due to Bernoulli s principle on a wing of area 78 m2 if the air passes over the top and bottom surfaces at speeds of 260 ms and 150 ms respectively Chapter 15 9 r The Laws of Thermodynamics 151 The First Law of Thermodynamics This is the law of conservation of energy written in a form useful to systems involving heat transfer AU L TQ W 151 Q energy added to the system W work done by the system ifwork is done on the system W lt 0 152 Thermodynamic Processes and the First Law An isothermal process is one where the temperature gtL EQ does not change v anuhlc piston Hi ghcr T B Lower T 152 Thermodynamic Processes and the First Law In order for an isothermal process to take place we assume the system is in contact with a heat reservoir In general we assume that the system remains in equilibrium throughout all processes 152 Thermodynamic Processes and the First Law An adiabatic process is one where there is no heat flow into or out of the system Isothermal B Adiabatic C 152 Thermodynamic Processes and the First Law An isobaric process a occurs at constant pressure an isovolumetric one b at constant volume A lsohm39ic lsnvolumciric a b Cnpyngmmuus Pearson Prentice Hall in 152 Thermodynamic Processes and the First Law If the pressure is constant the work done is the pressure multiplied by the change in volume W P AV 53 In an isometric process the volume does not change so the work done is zero 152 Thermodynamic Processes and the First Law For processes where the pressure varies the work done is the area under the PV curve Isovolumetric 7 PB 152 Thermodynamic Processes and the First Law TABLE 15 1 Simple Thermodynamic Processes and the First Law Process What is constant The rst law predicts Isothermal T constant AT 0 makes AU 0 so Q W lsobaric P constant Q AU W AU P AV Isovolumetric V constant AV mach W 0 so Q AU Adiabatic Q 0 AU W Copyrlglil 2005 Pearson Prentice Hall inc Ch 15 Problem 1 An ideal gas expands isothermally performing 340 X 103 J of work in the process Calculate a the change in internal energy of the gas and b the heat absorbed during this expansion Ch 15 Problem 3 One liter of air is cooled at constant pressure until its volume is halved and then it is allowed to expand isothermally back to its original volume Draw the process on a PV diagram Ch 15 Problem 6 The pressure in an ideal gas is cut in half slowly while being kept in a container with rigid walls In the process 265 kJ of heat left the gas a How much work was done during this process b What was the change in internal energy of the gas during this process 153 Human Metabolism and the First Law If we apply the first law of thermodynamics to the human body AU r W we know that the body can do work If the internal energy is not to drop there must be energy coming in It isn t in the form of heat the body loses heat rather than absorbing it Rather it is the chemical potential energy stored in foods 153 Human Metabolism and the First Law The metabolic rate is the rate at which internal energy is transformed in the body TABLE 152 Metabolic Rates askg human Metabolic Rate approximate Activity kculh walls Sleeping so 70 Sitting upright 00 115 Light activity 200 230 eatin drcssing household Chores Moduralc work 400 460 lcnnis walking Running S kmh 1000 1150 Bicycling Time 1100 1270 nmn saca t l w Ch 15 Problem 15 Calculate the average metabolic rate of a person who sleeps 80 h sits at a desk 80h engages in light activity 40h watches television 20h plays tennis 15h and runs 05h daily 146 Heat Transfer Conduction Heat conduction can be visualized as occurring through molecular collisions The heat flow per unit time is given by Q kA H 144 I l Hotter Cooler A X Heat 7 ow i gt I 1 T2 TABLE 144 Thermal Conductivities 1 4 6 Heat TranSfer39 Themml Conductivity k CO n d U Cth n W Substance Smca smCB 2 The constant k is called the sum 10 x 10 420 Copper 92 x 10 1 330 thermal conductiv1ty Aluminum 50 gtlt 10 2 200 Steel 11 x 10 2 40 Ice 5 x w 2 Materials with large k are GI 2M1 W called conductors those Brick 20 X 10 4 004 Comm 2 X 04 084 With small k are called Water 14 x 10quot4 050 insulators Human issue 05 X 10quot 02 Wood 03 x 10quot 01 Fiberglass 012 x 104 0048 Cork 01 x 10 4 0042 Wool 01 X 10 4 0040 Goose down 005 x 10 0025 Polyurethane 006 x 10 4 0024 Ai 0055 x 10 4 0023 ice Hail Inc 146 Heat Transfer Conduction Building materials are measured using R values rather than thermal conductivity 1 R7 Here I is the thickness of the material TABLE 14 5 Rvalues R Material Thickness fll lr WBlu G lass inch 1 Brick 3 3 inches 001 Plywood inch 06 Fiberglass insulation 4 inches 12 Ch 14 Problem 33 One end of a 33cmIong aluminum rod with a diameter of 20 cm is kept at 460 C and the other end is immersed in water at 22 C Calculate the heat conduction rate along the rod 147 Heat Transfer Convection Convection occurs when heat flows by the mass movement of molecules from one place to another It may be natural or forced both these examples are natural convection Cooler willcr 6 22 52 r2 Hotter w 147 Heat Transfer Convection Many home heating systems are forced hotair systems these have a fan that blows the air out of registers rather than relying completely on natural convection Our body temperature is regulated by the blood it runs close to the surface of the skin and transfers heat Once it reaches the surface of the skin the heat is released through convection evaporation and radiation Ch 14 Problem 37 Two rooms each a cube 40m per side share a 12cthick brick wall Because ofa number of 100W bulbs in one room the air is at 300C while in the other room it is at 100C How many 100W lightbulbs are needed to maintain the temperature difference across the wall 148 Heat Transfer Radiation The energy radiated has been found to be proportional to the fourth power of the temperature A Q errAT4 145 At The constant a is called the StefanBoltzmann constant 039 567 X 10 3W1 HZ39K4 The emissivity e is a number between zero and one characterizing the surface black objects have an emissivity near one while shiny ones have an emissivity near zero 148 Heat Transfer Radiation If you are in the sunlight the Sun s radiation will warm you In general you will not be perfectly perpendicular to the Sun s rays and will absorb energy at the rate AQ A 1000 Wm2eA cos a 46 148 Heat Transfer Radiation The most familiar example of radiation is our own Sun which radiates at a temperature of almost 6000 K Couyngm KL 2005 Pearson Pmuucu Hall Inc 148 Heat Transfer Radiation If you are sitting in a place that is too cold your body radiates more heat than it can produce You will start shivering and your metabolic rate will increase unless you put on warmer clothing 148 Heat Transfer Radiation quoti 39 Equator Earth uquot Earth June December a This cos 9 effect Is also responsible for the June Axis seasons x 7 lt7 7 A90 i FF Summer 4 Sun39s rays Juneli Va 7r quot 0090 B9509 Cold Winter 393 148 Heat Transfer Radiation Thermog raphy the detailed measurement of radiation from the body can be used in medical imaging Warmer areas may be a sign of tumors or infection cooler areas on the skin may be a sign of poor circulation 3 lb or 113 Frau54 m Ch 14 Problem 40 a Using the solar constant estimate the rate at which the whole Earth receives energy from the Sun b Assume the Earth radiates an equal amount back into space that is the Earth is in equilibrium Then assuming the Earth is a perfect emitter e 10 estimate its average surface temperature 1 Welcome to Phys 110 2 SE 4 SUI M Q Prof Rose Finn Office Roger Bacon 121125 Phone 7826764 email rfinnsienaedu Class MWF 11251215 Office Hours MWF 1302 or by appointment What is Physics 0 Divide into groups 5 7 students 0 Talk amongst yourselves 0 Formulate an answer in each group What do you expect to learn 0 Discuss in groups and present answers Teaching Philosophy 0 You learn best by doing oYou will be active learners 0 reading 0 working problems 0 discussing with neighbors 0 teaching peers 0 laboratory Group Cooperative Learning 0 ThinkPairShare 0 Group problems 0 Group discussions 0 Working together on homework is recommended 0 discuss problems 0 write your own solutions Course DeSIgn amp Grading Other 20 Reading Quizzes Homework lnclass problems Tests 2 Quizzes 5 each 2 Exams 15 each Laboratory 40 20 7U 8U 9U ioil You must pass lab to pass class Final Exam 20 Text book 0 Giancoli Physics Sixth Edition 0 Reading 0 Homework assignments 0 Some inclass problems 0 Additional resources 0 Student Study Guide Vol 1 Notes 0 My slides will be posted on the web 0 will tw to post BEFORE class 0 not eventhing is in handouts 0 You need paper for working out inclass problems as well as additional material not covered in slides 0 Calculator 0 Binder is a good option Attendance Policy 0 Students are expected to attend all classes and are responsible for all material covered in class even when absent 0 Students should understand that some material discussed in class is not covered in the textbook 0 Inclass problems and reading quizzes can not be made up Homework 0 Read Chapter 1 all sections 0 First reading quiz on Friday 0 sample reading quiz 0 Questions and problems that accompany each lecture are listed on the web 0 NOTE Test questions will closely re ect homework questions and inclass problems Chapter 13 Temperature and Kinetic Theory 131 Atomic Theory of Matter Atomic mass unit u This unit is defined so carbon12 has a mass of exactly 120000 u Expressed in kilograms 39I u 16605 x 10 27 kg Brownian motion is the result of collisions with individual water molecules 131 Atomic Theory of Matter oatomic nature of matter Democritus atom Brownian motion Robert Brown 1827 the result of collisions with individual water molecules quantified by Einstein 1905 131 Atomic Theory of Matter On a microscopic scale the arrangements of molecules in solids a liquids b and gases c are quite different Copyrighl 2005 Pearson Prentice Hall inc 132 Temperature and Thermometers Temperature is a measure of how hot or cold something is Most materials expand when heated 132 Temperature and Thermometers Thermometers are instruments designed to measure temperature In order to do this they take advantage of some property of matter that changes with temperature 7 3 z Early thermometers renlice Hail Inc a 132 Temperature and Thermometers Common thermometers used today include the liquidinglass type and the bimetallic strip Tube Bulb acts as a reservoir 132 Temperature and Thermometers Dioljtafc Most common temp scales Fahrenheit and Celsius Defined by phase change of water at atmospheric pressure ofreezing point of water 0 C or 32 F boiling point of water 100 C or 212 F J T 5T0 32 Copyright 2005 F Chapter 13 Problem 3a Room temperature is 68 F What is this in Celsius 133 Thermal Equilibrium and the Zeroth Law of Thermodynamics Two objects placed in thermal contact will eventually come to the same temperature When they do we say they are in thermal equilibrium zeroth law of thermodynamics if two objects are each in equilibrium with a third object they are also in thermal equilibrium with each other 134 Thermal Expansion il L0 at 70 Li J Linear expansion lALl occurs when an object MT 1 j is heated L L01 aAT 131b Here a is the coefficient of linear expansion Ch 13 Problem 7 A concrete highway is built of slabs 12 m long 20 C How wide should the expansion cracks between the slabs be at 20 C to prevent buckling ifthe range of temperature is 30 C and 50 C 134 Thermal Expansion Volume expansion is similar except that it is relevant for liquids and gases as well as solids AV 3V0 AT 132 Here 3 is the coefficient of volume expansion For uniform solids 3 3a 134 Thermal Expansion Water behaves differently from most other solids its minimum volume occurs when its temperature is 4 C As it cools further it expands oMaximum density at 4 C a h ll 5 in low l i i 1 i Iiinno 2 43 6 Tl C IOUquot I 100000 i 104343 H343 09099 7 rr wow 090 W 77 09998 quotE a B quot mum moon iinmmi llllll 1196 41 i ii i i ii i u i i N 0 5 10quot I000 0 20 4 6 low Tuinpumiurc 1 C Tcmpumlurc Cl 134 Thermal Expansion TABLE 131 Coef cients of Expansion near 2050 Coef cient of Linear Coef cient uqulIImu Muleriul Etpimsiun ltC39l Expmsinn C l Salim Aluminum 25 x 10 7S gtlt iu Ems w x m quot 56 x in quot Copper l7 gtlt IIIquot 5 x ilrquot Gold N x m39quot 42 x iir Irnnm slccl l2 x it 0 15 x H quot 29 X m39quot 37 x ll 1ltnquot uxiii39 Vxli39 27xm39 Ullilrll 4 1 ll quot 1 X ll quot Cnricrclu and brick 212 x 1039quot an x 1039quot Marble 1445 x m quot 47m x in quot Liqm iIi Gasoline 95 x iu39quot Mummy 1in x Urquot Elllyl nlcnlml Hun x in quot Glyctl ln sun x ilrquot Willcr 2m x M Gum Air 24nd must ulher gases illdlmusphe c prcssure 3400 x 10quot rerilice Hall inc 136 The Gas Laws and Absolute Temperature Equation of state relationship between the volume pressure temperature and mass of a gas Boyle s Law the volume of a given amount of gas is inversely proportional to the pressure as long as the temperature is constant V olt i P 136 The Gas Laws and Absolute Temperature The volume is linearly proportional to the temperature as long as the temperature is somewhat above the condensation point and the pressure is constant V x T Extrapolating the volume becomes zero at 27315 C this temperature is called absolute zero Vulnmc 727135 PC WPC anim UK milk ZUUK Milk Mix 500K 0 39l39rnmcmliu e T39l ii 39I39empsmium menu or Kl i inc 136 The Gas Laws and Absolute Temperature Kelvin scale 0 K at absolute zero molecules stop moving same temp per degree as the Celsius scale the freezing point of water is 27315 K the boiling point is 37315 K 137 The Ideal Gas Law A mole mol is defined as the number of grams of a substance that is numerically equal to the molecular mass of the substance 1 mol H2 has a mass of 2 g 1 mol Ne has a mass of 20 g 1 mol CO2 has a mass of 44 g The number of moles in a certain mass of material mass grams i n mo molecular mass gmol 137 The Ideal Gas Law ideal gas law PV nRT 133 where n is the number of moles and R is the universal gas constant R 8315111101 K 0082 L atmmol K 199 calories mol K II 138 Problem Solving with the Ideal Gas Law Useful facts and definitions Standard temperature and pressure STP T 273 K 0 C P 100 atm 1013 X 105NmZ 1013 kPa Volume of1 mol of an ideal gas is 224 L If the amount of gas does not change P1 V P2 V2 T T2 Always measure Tin kelvins P must be the absolute pressure Ch 13 Problem 29 If 300 m3 of a gas initially at STP is placed under a pressure of 320 atm the temperature of the gas rises to 38000 What is the volume 139 Ideal Gas Law in Terms of Molecules Avogadro s Number Avogadro s number the number of molecules in one mole NA 602 X 1023 The number of molecules in a gas is the number of moles times Avogadro s number NHNA 139 Ideal Gas Law in Terms of Molecules Avogadro s Number Therefore we can write PV NkT 134 where k is called Boltzmann s constant R 8315 JmollK k 138gtlt 10 23 K NA 602 X 1023m01 Ch13 Problem 42 How many moles of water are there is 1000 L How many molecules 1310 Kinetic Theory and the Molecular Interpretation of Temperature Assumptions of kinetic theory large number of molecules moving in random directions with a variety of speeds molecules are far apart on average molecules obey laws of classical mechanics and interact only when colliding collisions are perfectly elastic 1310 Kinetic Theory and the Molecular Interpretation of Temperature The average translational kinetic energy of the molecules in an ideal gas is directly proportional to the temperature of the gas nw2 ng 138 1310 Kinetic Theory and the Molecular Interpretation of Temperature We can invert this to find the average speed of molecules in a gas as a function of temperature 3kT vrms W I 139 I Ch 13 Problem 47 Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 6000 K 46 Weight the Force of Gravity and the Normal Force Weight is the force exerted on an object by gravity Close to the surface of the Earth where the gravitational force is nearly constant the weight is F mgquot 46 Weight the Force of Gravity and the Normal Force Normal Force perpendicuar to surface Not always parallel with gravity baances force on surface Where is F G 47 Solving Problems with Newton s Laws FreeBody Diagrams Lu 7 ma mm Em B When a cord or rope pulls on an object it is said to be under tension and the force it exerts is called a tension force mA um kg v 4mm Chapter 9 Static Equilibrium Units of Chapter 9 91 94 only The Conditions for Equilibrium Solving Statics Problems Applications to Muscles and Joints Stability and Balance 91 The Conditions for Equilibrium An object with forces acting on it but that is not moving is said to be in equilibrium Normal force i Gravity Y Comm m 7mm Pr atsm Fromm mu m 91 The Conditions for Equilibrium The first condition for equilibrium is that the forces along each coordinate axis add to zero 91 The Conditions for Equilibrium Conditions for E uilibrium The second condition of equilibrium is that q there be no torque around any axis the choice of axis is arbitrary 39 Ne t force 0 Net torque 0 92 Solving Statics Problems 1 Choose one object at a time and make a free Th m kPalrShare body diagram showing all the forces on it and where they act Problem 1 Three forces are applied to a 2 Choose a coordinate system and resolve tree Sapl39ng to Stab39l39ze 39t39 If FA 310 N forces into components and FB 425 N nd the magnitude and direction of FC 3 WrIte equmbrlum equatIons for the forces 4 Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation A clever choice here can simplify the problem enormously 5 Solve 92 Solving Statics Problems If there is a cable or cord in V the problem it can support L forces only along its length Forces perpendicular to that would cause it to bend Think Pair Share 0 Problem 3 Calculate the mass m needed in orderto suspend the leg below Assume the leg with cast has a mass of 150 kg and its CG is 350 cm from the hip joint This sling is 805 cm from the hipjoint ouuyugmo 2cu5 Peamuu qullue Hall m 93 Applications to Muscles and Joints These same principles can my be used to understand forces within the body Cupyrlglvltgz b Fuzusuu mm mu m 93 Applications to Muscles and Joints The angle at which this man s back is bent places an enormous force on the disks at the base of his spine as the lever arm for FVI is so wH 0071139 mm s mall sz mm Zarms m w Tmal weight E Lf EYV fFHUL WT 03946 of person Flml quot 3L L trunk lumbar 39 vcncbru 2 3 D a r quot W a F v 39quot Lever 31111 7 for FM Axis Toriquot calculation ThinkPair Share 0 Problem 34 The Achilles tendon is attached to the rear of the foot When a person elevates herselfjust barely off the floor on the ball of one foot estimate the tension in the Achilles tendon pulling upward and the downward force FB exerted by the lower leg bone on the foot Assume the person has a mass of 72 kg and D is twice as long as d W7 L Ballof foot FE gg 94 Stability and Balance If the forces on an object are such that they tend to return it to its equilibrium position it is said to be in stable equilibrium Net force a 94 Stability and Balance If however the forces tend to move it away from its equilibrium point it is said to be in unstable equilibrium 94 Stability and Balance An object in stable equilibrium may become unstable if it is tipped so that its center of gravity is outside the pivot point Of course it will be stable again once it lands 94 Stability and Balance People carrying heavy loads automatically adjust their posture so their center of mass is over their feet This can lead to injury if the contortion is too great Chapter 11 Vibrations and Waves 111 Simple Harmonic Motion If an object vibrates or oscillates back and forth over the same path each cycle taking the same amount of time the motion is called periodic The mass and spring system is a useful model for a lly y periodic system 1 w 111 Simple Harmonic Motion We assume that the surface is frictionless There is a point where the spring is neither stretched nor compressed this is the equilibrium position We measure displacement from that point x 0 on the previous gure The force exerted by the spring depends on the displacement F kx 111 111 Simple Harmonic Motion The minus sign on the force indicates that it is a restoring force it is directed to restore the mass to its equilibrium position k is the spring constant The force is not constant so the acceleration is not constant either 111 Simple Harmonic Motion 111 Simple Harmonic Motion m 7 Displacement is measured from the equilibrium point Amplitude is the maximum b 0 dIs lacement M P f A cycle is a full toandfro If the s n is hun l motion this gure shows halfa p g 9 x0 vertically the only change l cycle Is In the equrllbrlum mow quot quot Period is the time required to position which is at the measured Complete one CVCIe point where the spring from here Frequency is the number of g g rice cycles completed per second 9 a 39 111 Simple Harmonic Motion 112 Energy in the Simple Harmonic Oscillator We already know that the potential energy of a spring is given by Any vibrating system where the restoring force is proportional to the negative of PE kx2 the displacement is in simple harmonic motion SHM and is often caued a The total mechanical energy Is then simple harmonic oscillator E mv2 ka 113 The total mechanical energy will be conserved as we are assuming the system is frictionless l p 5 2 mux Y7A rel WA Hm Emr1k39l n Prentice Hall inc 112 Energy in the Simple Harmonic Oscillator If the mass is at the limits of its motion the energy is all potential If the mass is at the equilibrium point the energy is all kinetic We know what the potential energy is at the turning points 114a 112 Energy in the Simple Harmonic Oscillator The total energy is therefore kA2 And we can write mvZ kx2 kA2 Me This can be solved for the velocity as a function of position x2 v i vmax 1 F 395 where vfmx kmA2 M n Prentice Hall Inc 113 The Period and Sinusoidal Nature of SHM If we look at the projection onto the x axis of an object moving in a circle of radius A at a constant speed vmax we nd that the x component of its velocity varies as x2 E This is identical to SHM 391 vmx i 113 The Period and Sinusoidal Nature of SHM Therefore we can use the period and frequency of a particle moving in a circle to nd the period and frequency 27H Simplifying gives U T2w rA UziEA fli E m T 2w m 113 The Period and Sinusoidal Nature of SHM We can similarly nd the position as a function of time x A cos wt 118a A cos27rft 118b A cos277tT 1186 113 The Period and Sinusoidal Nature of SHM The top curve is a graph of the previous equation 4 Pupcr motion QMMOMMMMU The bottom curve is the same but shifted A period so that it is a sine function rather than a cosine 113 The Period and Sinusoidal Nature of SHM a The velocity and acceleration can be calculated as functions of time the results are below and are plotted at left I U1 meelliLul Velocity V v 39 max sin wt 119 k v A DJE AX a amX cos2mT 1110 amax kAl n Accelemlion u arson Prentice Hall inc ThinkPair Share Problem 9 A 060kg mass at the end of a spring vibrates 30 times per second with an amplitude of 013 m Determine a the velocity when it passes the equilibrium point b the velocity when it is 010 m from equilibrium c the total energy ofthe system and d the equation describing the motion of the mass assuming that x was a maximum at t0 114 The Simple Pendulum A simple pendulum consists of a mass at the end ofa lightweight cord We assume that the cord does not stretch and that its mass is negligible 114 The Simple Pendulum V In order to be in SHM the restoring force must be proportional to the negative of quot the displacement Here we haVei F mg sinf which is proportional to sin 9 My 9 and not to 9 itself TABLE 11 1 cw v Sin 9 at Small Angles 0 0 72 125225 Y di llv SiIIU Dill erence However If the 0 n a D l 01 745 00 I 745 0005 an 9 s I 5 008727 00571 0 W sin 9 z H7453 017365 05 026180 115882 11 034907 034232 20 30 052360 050000 47 N am 114 The Simple Pendulum Therefore for small angles we have 71 FP Tgx where x L0 The period and frequency are T 277 Z 1111a 8 1 5 1111b f L 114 The Simple Pendulum So as long as the cord can be considered massless and the amplitude is small the period does not depend on the mass CupyuglllIQQUDBFuMsonFHm m a nu ThinkPair Share Problem 29 How long must a simple pendulum be if it is to make exactly one swing persecond That is one complete vibration takes exactly 20 s Chapter 3 Kinematics in Two Dimensions Vectors 31 Vectors and Scalars A vector has magnitude as xquot well as direction 1 If 9 1 1 399 y x Some vector quantities displacement velocity force momentum A scalar has only a magnitude 39 Scale for velocity f l cm 90 kml v i a Some scalar quantities mass time temperature l k x all lnc Resullunl 14 km east r 6 km 32 Addition of Vectors Graphical Methods For vectors in one dimension simple addition and subtraction are all that is needed v km 8 km 1 km East a You do need to be careful about the signs as the Resultant 2 km east gure indicates llll Hkm 8km East b Capnghl x nus Pearson Prawn mu in 32 Addition of Vectors Graphical Methods If the motion is in two dimensions the situation is somewhat more complicated Here the actual travel paths are at right angles to one another we can nd the displacement by using the Pythagorean Theorem 0 we DRVDD 66 you 5quot 2as 9 X km East 32 Addition of Vectors Graphical Methods 32 Addition of Vectors Graphical Methods Adding the vectors in the opposite order gives the Even if the vectors are not at right same newquot V1 V2 V2 V1 angles they can be added graphically by y km using the tailtotip method North 6 1 4 2 A V 132 10 v V3 v 2 I 9 quotK VR WestlollllllllllJMkm V3 2 4 6 8 East Copyright 2005 Pearson Prennce Halllnc South V 33 Subtraction of Vectors and 33 Subtraction of Vectors and Multiplication of a Vector by a Scalar Multiplication of a Vector by a Scalar In order to subtract vectors we A vector V can be multiplied by a scalar c the g de ne the ne ative of a vec tor result is a vector N that has the same direction V V which has thg same ma nitud but but a magnitude cV If c is negative the resultant vector points in the opposite direction pomts In the oppOSIte dIrectIon Then we add the negative vector v v v v39 t a 39 gt V2 vw Copyright 2005 Pearson Prenuce Hall lnc 34 Adding Vectors by Components Any vector can be expressed as the sum of its components Components are perpendicular to each other N 011 h N 0th 9 30 9 30 4 0 I E3151 0 V East a I h mum Hall lmz 34 Adding Vectors by Components v g y 31quot 9 7 V If the components are L05 9 I perpendicular they can be V found using trigonometric tan 9 E functions Vquot W 34 Adding Vectors by Components The components are effectively one dimensional so they can be added arithmetically V Vb 34 Adding Vectors by Components Adding vectors mmth x Draw a diagram add the vectors graphically Choose x and y axes Resolve each vector into x and y components Calculate each component using sines and cosines Add the components in each direction To find the length and direction ofthe vector use A Vv tn0 39 a V x V V V 48 Applications Involving Friction Inclines 48 Static Friction On a microscopic scale most surfaces are rough The force can be modeled in Static friction is the frictional force between two a simple way surfaces that are not moving along each other gt A Static friction keeps objects on inclines from 1 I For kinetic sliding sliding and keeps objects from moving when a friction we write force is rst applied Ix Fir Mk FN l I 39 r I Mk is the coef cient F S SFN of kinetic friction and 1 is different for every pair of surfaces 48 Applications Involving Friction Inclines 48 Static versus Kinetic Friction The static frictional force increases as the applied force us gt k increases until it reaches its maximum Then the object starts to move and the kinetic frictional TABLE 4 2 Coefficients of Friction39 force takes over 50 Coe icienl of Cue icient of F F Surfaces Static Friction n Kinetic Frinian1k 40 fr H s N Wood on wood 04 02 Ice on ice 01 003 r 30 m Muial on metal Iuhricaicd 015 1117 Steel on sIeeI ulllubricaied 07 16 8 20 K I v 1116th RIIbhcr on dry concreie 10 08 g Stflnf f Rubber on wet con 07 15 a 0 fnCtlon mo Ruhhizr on nlhizr solid surfaces l4 1 10 20 30 40 50 60 70 Teflonm on Teflon in air 004 L114 uquot I I i 1 I I Tunnn on steel in air 004 1104 0 Applied force FA Lubricated ball bearing lt01 lt001 3 F N Synoviziljuinls in human limbs 001 101 Values are nppmximatc nmi imcndcd nnIy as a guide I motion Shdmg 48 Applications Involving Friction Inclines An object sliding down an incline has three forces acting on it the normal force gravity and the frictional force 0 The normal force is always perpendicular to the surface 0 The friction force is parallel to it 0 The gravitational force points down lfthe object is at rest the forces are the same except that we use the static frictional force and the sum ofthe forces is zero FG mg Components of Gravity on an Inclined Plane Chapter 8 81 Angular Quantities ln purely rotational motion all points on the object move in circles around the axis of rotation O Rotational Motion Al points on a straight line drawn through the axis move through the same angle in the same time The angle 9 in radians is defined 1 0 81a r quotI where l is the arc length Cupyngnmznoa Fuzvsmr Prunhue Hall in 81 Angular Quantities ThinkPaIrShare y Anguar displacement 1 Problem 1 Express the following angle in 1 A6 62 01 radians a 30 degrees b 57 degrees L Li Average angular velocity Give as numerical values and as fractions 3 I 39 A6 of i 74 p a At 7 quot 9 82 x 937 7 t The instantaneous angular a 1 a velocity 7 1 l X a 11m bi AI gt0 At 82b 81 Angular Quantities The angular acceleration is the rate at which the angular velocity changes with time oz ml Am a 84a At At The instantaneous acceleration l39 amp 83b a AIILTJO A ThinkPair Share Problem 4 The blades of a blender rotate at a rate of 6500 rpm When the motor is turned offduring operation the blades slow to rest in 30 3 What is the angular acceleration as the blades slow down 81 Angular Quantities Every point on a rotating body has an angular velocity to and a linear velocity v They are related 7 rw 84 V W 81 Angular Quantities Therefore objects farther from the axis of rotation will move faster 81 Angular Quantities If the angular velocity of a rotating object changes it a 2 has a tangential acceleration a quot P amll ra 85 81 Angular Quantities Here is the correspondence between linear and rotational quantities TABLE 8 1 Linear and Rotational Quantities 5R 1 Even if the anguar Linear Type Rotational Relation 1 velocity is constant each x displacement 0 x r0 point on the object has a v velocity w v rw centripetal acceleration am acceleration 01 am ra 2 2 Copynglu 2005 Pearson Prenllce Hall lnc 1 no 39 HR 1 wz 86 mmum qmrsm unwind m r f 81 An ular Quantities g ThinkPaIrShare The frequency is the number of complete revolutions per second w Problem 5 A child rolls a ball on a level E oor 35 m to another child Ifthe ball makes 150 revolutions what is its Frequencies are measured In hertz diameter 1 Hz 1 s 1 The period is the time one revolution takes T 1 88 82 Constant Angular Acceleration The equations of motion for constant angular acceleration are the same as those for linear motion with the substitution of the angular quantities for the linear ones ThinkPair Share Problem 15 A centrifuge accelerates uniformly from rest to 15000 rpm in 220 3 Through how many revolutions did it turn Angular Line in this time wwoat vvo t 0 cool TECH X vot at2 w2w52ae v2v62ax a no U M w 2 v T 84 Torque 83 Rolling Motion Without Slipping Rolling without slipping static friction whee contact point is a rest WRT ground The linear speed of the wheel is related to its angular speed To make an object start rotating a force is needed the position and direction of the force matter as well The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm 84 Torque 84 Torque T TFL VH1 a w UWF Alonger lever T TFsme arm is very Torque rotational analog of force helpful in rotating objects r distance from point of rotation to point where force is applied F force perpendicular to r FA applies maximum torque 85 Rotational Dynamics Torque and ThinkPairShare Rotational Inertia Knowing that F ma we see that 7 mr2a Problem 22 A 55kg person rIdIng a bike 811 puts all her weight on each pedal when This is fora single point climbing a hill The pedals rotate in a mass What lbout an circle of radius 17 cm a What is the tended Obie maximum torque she exerts s As the angular r F acceleration is the same 0 for the whole object we m can write 27 21ersz 812 85 Rotational Dynamics Torque and Rotational Inertia The quantity 1 Eer is called the rotational inertia of an object The distribution of mass matters here these two objects have the same mass but the one on the left has a greater rotational inertia as so much of its mass is far from the axis of rotation CupVHA llll w mus minim FirMu a mu lnu mm 85 Rotational Dynamics Torque and Rotational Inertia we The rotational inertia of an object depends not only on its mass distribution but also the location of the axis of rotation compare f and g for example mum w Copyright lt3 2005 Pearson Pranllce Hall lrv 85 Rotational Inertia I 277172 moment of inertia rotational equivalent of mass changes with axis of rotation Rotational equivalent of Newton s 2nd Law TIa ThinkPair Share Problem 27 Determine the moment of inertia of a 108 kg sphere of radius 0648 m when the axis of rotation is through its center 86 Solving Problems in Rotational Dynamics 1 Draw a diagram 2 Decide what the system comprises 3 Draw a freebody diagram for each object under consideration including all the forces acting on it and where they act 4 Find the axis of rotation calculate the torques around it 86 Solving Problems in Rotational Dynamics 5 Apply Newton s second law for rotation If the rotational inertia is not provided you need to find it before proceeding with this step 6 Apply Newton s second law for translation and other laws and principles as needed 7 Solve 8 Check your answer for units and correct order of magnitude

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