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# Gen Physics I PHYS 2010

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PHYS2010 General Physics I Course Lecture Notes Section XII Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn XII Thermal Physics A Thermal Equilibrium 1 If bodies A amp B are separately in thermal equilibrium with a 3rd body C then A amp B will be in thermal equilibrium with each other a This statement is often referred to as the 0th Law of Thermodynamics It simply means that if 2 objects in thermal equilibrium with each other are at the same temperature b Thermal equilibrium means that an object has the same temperature throughout its interior 2 Temperature is nothing more than a measure of how fast particles are moving due to the heat energy stored in the system a There are 3 different temperature scales i Fahrenheit scale archaic English system gt 32011 E water freezes 2121 E water boils at atmo spheric pressure ii Celsius scale metric system scale once called the Centigrade scale gt 0 C E water freezes 1000C E water boils at atmospheric pressure iii Kelvin scale Sl system gt the absolute tem perature scale gt 0 K E no atomic motions lowest possible tem perature gt Lowest temp recorded in lab is 10 6 K X11 1 Xll 2 PHYS 2010 General Physics b Converting between the temperature scales i F gt 0C 9 TF ETC 32 Xll l ii 0C gt K To T 27315 XII 2 iii Note that T without a subscript will always refer to the Kelvin scale in these notes except where noted Example X1171 Problem 108 Page 346 from the Serway amp Faughn textbook The temperature difference between the inside and the outside of an automobile engine is 4500C Express this temperature difference on the a Fahrenheit scale and b Kelvin scale Solution a We need to come up with a formula that changes ATC to ATF and AT where T is measured in Kelvin Using Eq Xll l we can write 191 Tc1 32 d an 9 TF2 5 TC2 32 Subtracting equation 1 from 2 yields 9 TF2 TF1 g lltTC2 TOM since the two 32 s cancel with each other We are given a temperature difference of ATC 4500C so from the equation above we get 9 450 C 810 F 9 Donald G Luttermoser ETSU Xll 3 Solution b From the same argument above Eq XII 2 gives T2 T1 Tc2 27315 Tc1 27315 T2 T1 ch Tch AT ATC 450K B Thermal Expansion 1 Heat energy added to matter causes the particles that make up the matter to speed up a Increased velocity of particles increases pressure b lncreased pressure causes matter to expand due to the increased internal force i Liquids and gases ll a larger volume ii Solids get longer c A loss of heat energy causes objects to shrink in size 0 For solids the change in heat as measured by temperature dic tates a change in linear size AL Via AL oz Lo AT XII 3 a LG E initial length b AT E change in temp usually measured in 0C in Eq XII 3 Xll 4 PHYS 2010 General Physics c oz E coef cient of linear expansion see Table 101 in text book 3 Surface areas also change in size AA via AA VAC AT XII4 2 b 7 E coef cient of area expansion 2a a AO E initial area m Volumes increase or decrease AV as well when heat is added AV VOAT a V0 E initial volume mg or taken away via XII5 b E coef cient of volume expansion 3a Example XIIEZ Problem 1011 Page 346 from the Serway amp Faughn textbook The New River Gorge bridge in West Virginia is a 518 m long steel arch How much will its length Change between temperature extremes of 20quotC and 350C Solution Use Eq XII 3 here AT 350C 200C 550C LO 518 m and oz 11 X 10 6 quotC 1 from Table 101 on page 313 in the textbook This gives a length change of AL ozLo AT 11 x 106 C 1518 m55 o 031 m Donald G Luttermoser E TSU C Heat and Internal Energy 1 Internal energy U is the energy associated with the micro scopic components of a system given by a b C d 2 Heat or thermal energy Q is a mechanism by which energy is transferred between a system and its environment because of U mtrans rm mVib Win01 39 mm E average translational linear kinetic en ergy of the atoms and molecules mm E average rotational kinetic energy gt the ro tation of a molecule about an axis Wm E average vibrational kinetic energy gt a con tinuous change of distance between the atoms that make up the molecule Wm E average intermolecular potential energy be tween the atoms that make up the molecules gt the bond energy a temperature difference between them a b c Heat is essentially related to kinetic energy gt energy due to the motion of particles in matter Heat is measured in calories in the cgs unit system gt one calorie of heat is required to raise 1 gram of water by a temperature of 10C Heat is measured in kilocalories in the SI system i One kilocalorie 103 calories of heat is required atoms and molecules It is XII6 X11 5 X11 6 PHYS 2010 General Physics to raise 1 kg of water by a temperature of 10C or 1 ii A kilocalorie is also called a Calorie capital C gt this is the energy unit that you see on food containers d Heat is measured in British Thermal Units BTU in the English system gt one BTU of heat is required to raise 1 lb of water by a temperature of 10F 3 In 1843 Joule showed that heat can be used to drive a machine gt mechanical energy and heat were 2 forms of the same thing a Heat is just another form of energy b 1 cal 4187 Joules c 4187 Jcal 4187 Jkcal is known as the mechanical equivalent of heat D Speci c Heat and Calorimetry 1 Calorimetry means the measurement of heat exchange 2 Heat capacity is the amount of heat required to change an entire object s temperature by 10C 3 The speci c heat c of a substance relates the amount of heat Q it takes to change the temperature T of an object of mass m by one unit degree Q C m Donald G Luttermoser ETSU XII 7 which is the heat capacity per unit mass so we can write the a In the cgs system c is measured in calg OC heat equation as b In the SI system c is measured in Jkg OC note that 0C is used instead of K here c Note that water has a speci c heat de ned as cW 1000 calgOC XII 9 1 Table 111 in your textbook lists speci c heats for various substances 4 Devices used to measure heat energy are called calorimeters 5 Heat ow from one system to another obeys the conservation of energy a Heat gained by one system positive number heat loss by another system negative number 0 b Or rephrasing Heat loss by one system Heat gained by another system anjn Q1055 m1C1T mQCQ T XII11 i 1 represents the system gaining heat ii 2 represents the system losing heat iii T represents the nal equilibrium temperature when everything is mixed Xll 8 PHYS 2010 General Physics Example XII73 Problem 119 Page 379 from the Serway amp Faughn textbook A 500 gm lead bullet traveling at 300 ms is stopped by a large tree If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree what is the increase in temperature of the bullet Solution The mechanical energy transformed into internal energy of the bullet is 1 1 1 2 1 2 Q i E K131 E Elmo 17ml 7 where m is the mass of the bullet and vi 300 ms is the initial speed of the bullet From Table 111 on page 335 in the textbook the speci c heat of lead is 128 JkgOC Using Eq XII 8 the change in temperature of the bullet is 1 2 2 3 2 AT Q ampEM 176 C me mc 40 4128 Jkg quot C Example XII74 A 040 kg iron horseshoe that is initially at 5000C is dropped into a bucket 20 kg of water at 220C What is the final equilibrium temperature Neglect any energy transfer to or from the surroundings Solution From Table 111 in the textbook the speci c heat of iron Fe is 448 JkgOC and water is 4186 JkgOC Using Eq XII 11 we get an equi librium temperature T of mwcw T TW mFecFe TFe T mWCWT mWCWTW mFecFeTFe mFeCFeT mWCWT T mFeCFeT mWCWTW T mFecFeTFe mwcw T mFecFe T mWCWTW T mFecFeTFe T i mWCWTW T mFecFeTFe mwcw T mFecFe Donald G Luttermoser ETSU Xll 9 with chWTW 20 kg4186 Jkg C22 C 184 x 106 J mFecFeTFe 040 kg448 Jkg C500 C 896 x 104 J and mch mpecpe 20 kg4186 Jkg C 040 kg448 Jkg C 839 x 104 J C Plugging this into out equilibrium temperature equation above we get 184 x 106 J 896 x 104 J T 839 x 104 J C 230C E Latent Heat amp Phase Changes 1 When a substance undergoes a physical alteration from one form to another e liquid to gas it is referred to as a phase change a Under such circumstances T does Lot change when heat is added or taken away from the system gt the energy goes into changing the phase XII12 i Q E heat gained or lost to the system b Mathematically ii m E mass of the system Xll 10 C d 8 PHYS 2010 General Physics iii L E latent heat iv You use the sign when energy is being added during a phase change e melting or boiling 7 v You use the sign when energy is being taken away during a phase change e freezing or con densation When the phase change is from gas to liquid or Vise Versa L is called the latent heat of vaporization LV i Changing from gas to liquid is called condensa tion ii Changing from liquid to gas is called boiling When the phase change is from liquid to solid or Vise Versa L is called the latent heat of fusion L i Changing from liquid to solid is called freezing ii Changing from solid to liquid is called melting Table 112 of the textbook shows latent heats of fusion and vaporization for some common substances We can deduce temperature changes in systems that undergo phase changes by using the conservation of energy for each state of the system Donald G Luttermoser E TSU 3 b c d Xll 11 Water has the following characteristics T C A E 100 D I STEAM 2 c WATER l 50 1 STEAM 0 B I WATER I ICE I I I ICE WATER 627 3967 8157 3076 7 Q J Heat ln states A C and E the heat equation follows Eq XII 8 ln states B and D the phase change states the heat equation follows Eq XII 12 Using the conservation of energy equation e Eq Xll 10 going from point 2 to point 1 on the above graph we can set up the following equation Qtot Qcof water at 2 QB QAof ice at 1 QC mwcwj 2 TB QB mg Lf positive since energy is being added QA miciTB T1 Here 01 and cW are the speci c heats of ice and water L is the latent heat of fusion of water mW is the initial mass of the water mi mW is the nal mass of the ice mi mi mW is the mass of the water freezing into ice Xll 12 PHYS 2010 General Physics T2 gt 0 the initial temperature of the water TB 0 the temperature of fusion in Celsius and T1 lt 0 the nal temperature of the ice so Qm chWT2 TB mil mg L miciTB T1 mwchg mWLf mwciTl Note that since T1 lt 0 mwciT1 gt 0 and hence the right hand side of this equation will always be positive e For those problems where a nal equilibrium temperature is being sought One chooses a Q as either a gain or a loss based upon whether the xed temperature of the Q equation is less than or greater than the equilibrium temperature i If T gt then Q Q1055 lt 0 ii If T lt then Q om gt 0 iii As a result note that we also could have written the conservation of heat energy e Eq XII 10 as Qcold Qhot Xll lB as given by the textbook Example X1175 A 300 gm lead bullet at 300 C is fired at a speed of 240 ms into a large block of ice at 00C in which it embeds itself What quantity of ice melts Solution Let s assume that the large block of ice does not completely melt as a result of the bullet embedding itself then the kinetic energy of the bullet Donald G Luttermoser ETSU Xll 13 will go towards heating the block of ice Also the temperature of the bullet will heat the ice block Let mm be the amount of mass in the ice block that melts mb 300 gm 300 X 10 3 kg be the mass of the bullet Tb 300 C be the initial temperature of the bullet vb 240 ms be the initial speed of the bullet and T 00C be the nal temperature of the bullet since it will cool to the same temperature of the block of ice Using the tables from the textbook the latent heat of fusion of water is L 333 X 105 Jkg and the speci c heat for lead is cb 128 JkgOC Using the conservation of energy and noting that J kg m2s2 we get anjn Q1055 Qmelt KEb Qbeloss 1 mmelth Embvg mbe W L 240 ms22 128 Jkg 0o300o 00C 333 x 105 J kg mmelt mb 300 gm 0294 gm F Transport of Heat Energy 1 Thermal energy ie heat can only ow by one of three dif ferent mechanisms conduction convection and radiation transport 2 Heat Transfer by Conduction a Conduction is the process by which heat is transferred Via collisions of internal particles that make up the object gt individual mass particle transport Xll 14 b C v PHYS 2010 General Physics i Heat causes the molecules and atoms to move faster in an object ii The hotter molecules those moving faster col lide with cooler molecules those moving slower which in turn speeds up the cooler molecules mak ing them warm iii This continues on down the line until the object reaches equilibrium The amount of heat transferred AQ from one location to another over a time interval At is i 73 E heat transfer rate XII14 ii 73 is measured in watts when Q is measured in Joules and At in seconds iii As such 73 is the same thing as power since they are both measured in the same units Heat will only flow if a temperature difference exists be tween 2 points in an object i For a slab of material of thickness L and surface area A the heat transfer rate for conduction is AQ 7 KA 73601111 lt Th Tc XII 15 gt lt gt Donald G Luttermoser ETSU Xll 15 Th on back side Heat Flow if Th gt T J u L ii Th is the temperature of the hotter side and TC is the temperature of the cooler side iii K E thermal conductivity of the material see Table 113 in textbook gt K JsmOC WmOC iv The larger K is the better the material is in conducting heat v The smaller K is the better the material is as a thermal insulator vi The effectiveness of thermal insulation is rated by another quantity gt thermal resistance R Where L g 7 Where in the US R ftghoFBTU h7 hour and elsewhere R m20CW see Table 114 in the textbook R E XII16 Xll 16 PHYS 2010 General Physics Example X1176 Problem 1135 Page 381 from the Ser way amp Faughn textbook A steam pipe is covered with 150 cm thick insulating material of thermal conductivity 0200 calcmquotCs How much energy is lost every second when the steam is at 2000C and the surrounding air is at 200 C The pipe has a circumference of 800 cm and a length of 500 m Neglect losses through the end of the pipe Solution Listing our given parameters we rst need to convert some of these input parameters to SI units cal 102 cm 4186 J J K 2 837 0 00cm Cslt1mgtlt1calgt msquotC Th 2000C TC 200C C 800 cm 800 m is the circumference of the pipe 6 500 m is the length of the pipe and the thickness of the insulation is L 150 cm 00150 In Since the problem gives us the circumference of the pipe the total surface area of the pipe is this circumference times the length of the pipe A or 800 m500 m 400 m2 Using Eq XII 15 then gives the heat loss rate via conduction as T Tc pcond J 2 20000 2000 83 7msocgt400mlt 00150m gt 402 x 108 Js 402 MW Donald G Luttermoser E TSU 3 Heat Transfer by Convection a b c When an ensemble of hot particles move in bulk to cooler regions of a gas or liquid the heat is said to ow via convection Boiling water and cumulus clouds are 2 examples of con vection Convection is complicated requiring graduate level physics and math to describe it i Convection will occur if an instability occurs in the gas or liquid ii The mixinglength theory is often used to de The heat of a gasliquid bub ble will glue up its heat energy to the cooler sur roundings after the bubble has traveled one missing length scribe convection 4 Heat Transfer by Radiation a 0 Of all the heat energy transport mechanisms only radia tion does not require a medium gt it can travel through a vacuum The rate at which an object emits radiant energy is given by the StefanBoltzmann Law 735m aAeT4 XII17 i 735m E power radiated emitted watts ii a E Stefan Boltzmann s constant 56696 x 10 8 WmZKA Xll 17 Xll 18 PHYS 2010 General Physics iii A E surface area of the object mg iv e E emissivity unitless e 1 for a perfect ab sorber or emitter v T E temperature c A body also can absorb radiation If a body absorbs a power of radiation 73135 it will change its temperature to To i The net power radiated by the system is then prad pnet pen abs or PM UA T4 aAeTj 73ml aAe T4 T04 XII19 ii In astronomy the total power radiated by an ob ject over its entire surface is called the luminosity L 73ml of the object Since the amount of energy falling on the surface of the Sun or any isolated star from interstellar space is negligible to that of the power radiated L 735m for isolated stars iii If an object is in equilibrium with its surround ings it radiates and absorbs energy at the same rate gt its temperature remains constant gt this ra diative equilibrium results in the object being in thermal equilibrium 73ml0 gt TTO where T0 is the temperature of the surroundings Donald G Luttermoser ETSU Xll 19 d An ideal absorber is de ned as an object that absorbs all of the energy incident upon it i In this case emissivity e 1 ii Such an object is called a blackbody iii Note that a blackbody radiator can be any color depending on its temperature gt red blackbod ies are cooler than blue blackbodies it does not appear black unless it is very cold iv The energy ux of such a radiator is be 7 0T4 v This radiative ux results from the condition that in order to be in thermal equilibrium the heat gained by absorbing radiation must be Virtually immediately radiated away by the object vi This is not the same thing as reflecting the radi ation off of the surface which does not happen in a blackbody The incident radiation does get ab sorbed by the atoms of the object and deposited in the thermal pool It is just that this radiation immediately gets re emitted just after absorption XH 20 PHYS 2010 General Physics Example X1177 Problem 1142 Page 382 from the Serway amp Faughn textbook The surface temperature of the Sun is about 5800 K Taking the Sun39s radius to be 696 X 108 m calculate the total energy radiated by the Sun each second Assume e 0965 Solution We will assume the Sun s shape to be a sphere where the surface area of a sphere is A 47ng 477696 x 108 m2 609 x 1018 m2 Making use of Eq XII 17 the total power energy per second that the Sun radiates to space is W 735m aAeT4 567 x 108 K4 609 x 1018 m209655800 K4 m2 377 x 1026 W This is the value of the Sun s luminosity LG 735m 377 x 1026 W PHYS2010 General Physics I Course Lecture Notes Section XIV Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn XIV Thermodynamics A The area of physics concerned with the relationships between heat and work is thermodynamics gt the study of the motion of heat B The First Law of Thermodynamics 1 ln words The change in internal energy ofa system equals the difference between the heat taken in by the system and the work done on the system a When an amount of heat Q is added to a system some of this added energy remains in the system increasing its internal energy by an amount AU b The rest of the added energy leaves the system as the system does work W 2 Mathematically note that AU Uf U1 and the First Law states 3 ln thermo there will always be two speci c regions in which we will be interested in a The system the region of interest where we wish to know the state parameters 69 P T and V Ther modynamic variables relating to the system will remain unsubscripted in these notes e W work on the sys tem XIV 1 XIV 2 PHYS 2010 General Physics the re gion that contains the system Note that some scientists b The environment also called the universe including me prefer to use the word environment over universe since in astronomy the word Universe means all of the cosmos and nQot just the immediate ie nearby surroundings which is what thermo means by universe Thermodynamic variables relating to the environment will be labeled with the env subscript in these notes ie WenV work on the environment c Note that work done Q the system W is the exact oppo site of work done Q the environment W Wenv 39 Note that we will often be interested in systems that are com pletely isolated from the environment Such a system is called a closed system Using the de nition of work Eq Vl l and pressure Eq Xl l we can show how work is related to the change in volume of gas a Work will increase in small increments AW as a piston in a cylinder moves by a length Ax then AW F Ax Ax A P AV XIV2 note that the negative sign in this relation results from the fact that when the environment does work Q the cylinder ie the system AV becomes smaller hence AV lt 0 due to the piston s motion As such the work done on the system is positive W gt 0 Since A06 is a very small length in this equation P will not change by much As such P is considered to be constant in this equation Donald G Luttermoser E TSU b The total work done W is then just the sum of all of the incremental increases in work WZAWZ ZPZAVZ XIV 3 However the shorthand notation of W PAV XIV 4 is often used for Eq XIV 3 and we will use this notation here c A gas lled cylinder with a piston is the simplest form of a heat engine Such heat engines are used to drive more complicated machines such as automobiles 6 For a monatomic ideal gas 16 the gas particles consist of a single atomic species recall Eq XIII 16 where the internal energy is given by U gnRT a The change in internal energy for such a gas is then given by AU SydEAT XIV5 assuming no gas escapes or is brought into the system b We will now introduce a new thermodynamic term call the molar speci c heat at constant volume for a monatomic ideal gas 0 E R XIV6 3 2 c The change in internal energy of an ideal gas can then be written as AU nCvAT XIV7 XIV 3 XIV 4 PHYS 2010 General Physics d For ideal gases this expression is always valid even when the volume isn t constant The value of the molar speci c heat however depends on the gas eg the numbers of degrees of freedom the gas can have and can vary under different conditions of temperature and pressure 7 There are four speci c types of processes involving energy in ther modynamics a Isobaric processes Pressure remains constant during the process The full form of the 1st law is used in these cases i Under these conditions the thermal energy trans ferred into the gas e system is given by rear ranging the terms in the rst law QAU WAUPAV where here we have made use of Eq XIV 4 in this relation ii Using Eq XIV 5 and Form 1 of the ideal gas law ie PAV nRAT in this relation gives 5 Q gnRAT nRAT EnRAT iii Another way to express this transfer of heat is Q nCpAT XIV 8 where 010 R is the molar speci c heat at constant pressure iv For ideal gases the molar heat capacity at con stant pressure is equal to the sum of the molar spe ci c heat at constant volume and the universal gas Donald G Luttermoser E TSU b constant 010 CU R XIV 9 Adiabatic processes Heat does not enter or leave the system only occurs for closed systems or e ectively closed systems A good example of an adiabatic process is anything that happens so rapidly that heat does not have time to ow in or out of the system during the process Then Q 0 and Eq XV l becomes AU W adiabatic processes i If the system does work on the environment W lt 0 negative and AU lt 0 negative gt the sys tem uses its internal energy to supply work to the outside environment note here that WenV gt 0 as a result of this ii If the system requires work from the environment W gt 0 positive and AU gt 0 positive gt the environment supplies work hence WenV lt 0 to the system and increases the systems internal energy as a result iii An ideal gas undergoing an adiabatic process obeys the following relation PVl constant XIV 10 with CP 7 XIV 11 7 CU where 7 is called the adiabatic index of the gas See Table 121 in the textbook for values of GP CU and 7 for various species XIV 5 XIV 6 C d PHYS 2010 General Physics Isothermal processes Temperature does not change during the process Processes that occur gradually can be considered isothermal since the temperature changes very slowly during process In isothermal processes AU 0 and Eq XV l becomes W c2 isothermal processes i Using methods in calculus it can be shown that the work done on the environment during an isother mal process is Vf WenV nRT ln XIV 12 where ln is the natural log function ii Since work on the system is just the negative of work on the environment Vf W i nRT ln i XIV13 Isochoric processes also called isovolumetric pro cesses Volume does not change during the process Since volume does not change no work can be done on the system As such Eq XV l then becomes AUQ isochoric processes i In an isochoric process the change in thermal en ergy of a system equals the energy transferred to the system by heat ii From Eq XIV 5 the energy transferred by heat in constant volume processes is given by Q 710va XIV14 Donald G Luttermoser E TSU 8 9 XIV 7 Enthalpy Heat energy associated with phase changes a b c Through the use of the 1st law of thermodynamics we can introduce a new state variable associated with phase changes Eq XII 12 showed the heat liberated or needed during a phase change is related the latent heat Usually there is also a volume change when matter changes from one state to another A volume change results in work either being performed on or by the system via Eq XIV 2 Using this equation along with Eq XII 12 in the 1st law e Eq XIV l we can write Ug U1mL PV2 V1 XIV 15 i Solving this equation for the latent heat term gives mLU2PV2 U1PV1H2 H1 XIV16 H U PV is called the enthalpy of the system where XlV 1 7 ii The enthalpy H is another state variable which is a function of the 3 other state variables T through U P and V As can be seen the latent heat of fusion and vaporization is just the difference of the enthalpy of the two different states of matter on either side of the fusion or vaporiza tion Up until now we have considered only processes in which the potential and macroscopic kinetic energies of a system remain XIV 8 PHYS 2010 General Physics constant We now relax this constraint to develop the general form of the 1st law of thermodynamics a The work energy theorem e Eq Vl 4 says that the work done on a system is equal to the change of kinetic energy AKE KEf KEW XIV18 b This kinetic energy change should be added to the in ternal energy change to represent the total kinetic energy gt both macroscopically e bulk kinetic energy microscopically e internal energy AU AKE Q W XIV19 c If conservative forces act on the system the work done by the conservative forces We is related to a change in the potential energy through Eq VI 6 WC APE XIV20 1 Note that the total work done on a system is the sum of the work due to conservative forces and non conservation forces labeled with Wm W WC Wm XIV21 e Using Eq XIV 21 along with Eq XIV 20 in Eq XIV 19 we can write AU AKE Q WC Wm Q We Wm AU AKE WC Q Wm AU AKE APE Q Wm XIV 22 f Now if we de ne the total energy E of the system as the sum of the internal energy the kinetic energy and the Donald G Luttermoser ETSU XIV 9 potential energy E U KE PE XIV 23 then we can write the general form of the rst law of thermodynamics as AE E E Q Wm XIV24 noting that if both the kinetic energy and potential energy are constant then AE AU and Vlm W which gives us the standard form of the 1st law as written in Eq XIV l Example XIVil Problem 128 Page 419 from the Serway amp Faughn textbook A movable piston having a mass of 800 kg and a cross sectional area of 500 cm2 traps 0200 mol of an ideal gas in a vertical cylinder If the piston slides without friction in the cylinder how much work is done on the gas when its temperature is increased from 200C to 3000C Solution For an ideal gas as the temperature increases the volume increases by nRTf nRTi AV Vf Vi 7 P 39 Assume the pressure is held constant B Pf P so we can write nR nRAT AVTf ET The change in temperature is AT 300 273 K 20 273 K 280 K The work done on the gas is W PAV nRAT 0200 mol831 Jmol K280 K 465 J XIV 10 PHYS 2010 General Physics Example XIV72 Problem 1220 Page 420 from the Serway amp Faughn textbook A thermodynamic system undergoes a process in which the internal energy decreases by 500 J If at the same time 220 J of work is done on the system find the energy transferred to or from it by heat Solution From the rst law as given in Eq XIV 1 QAUW 500J220J The negative sign in the result means that energy is transferred from the system by heat C Heat Engines 1 Many of the machines used in our current technology are driven by heat engines indeed your own body is a heat engine A heat engine is a device that converts internal energy into some other form of energy 69 electricity mechanical kinetic energy etc 2 Let s represent a typical heat engine as a cylinder with a piston in it The total work done by a heat engine is then the net heat ow into the cylinder via the 1st law of thermodynamics note that AU 0 for the complete cycle W IQHI chl a XIV25 where is the magnitude hence a positive number of the heat that owed into the system re engine and QC is the magnitude hence a positive number of the heat that owed out of the system Donald G Luttermoser ETSU XIV ll 3 Reversible versus non reversible processes a A process is said to be reversible if the nal state of a process can be returned to its initial state i No dissipative effects that convert mechanical en ergy to thermal energy e friction ii Such processes are nearly impossible to achieve in nature b A process is irreversible if the nal state of a process cannot be returned to its initial state gt nature behaves in this manner 4 The thermal ef ciency 6 of any system is the ratio of the work W a Using Eq XIV l4 we immediately see that done to the heat input 6 lQHl IQCI XIV27 IQHl b Kelvin showed through the ideal gas law that chl To 7 7 XIV 28 IQHI TH l where these temperatures are measured in K c Using Eq XIV 28 in Eq XIV 27 we see that the ther mal ef ciency of an ideal heat engine is T e l 0 ideal engine XIV 29 TH XIV 12 PHYS 2010 General Physics 1 As can be seen from this even with an ideal engine one with no internal friction it is impossible to make a 100 ef cient engine gt it is a physical impossibility to make a perpetual motion machine machines that require no energy to run and never run down 5 There are two things to always keep in mind when doing calcu lations with heat engines a W is positive and WenV is negative when the system contracts 239e AV lt 0 and negative and WenV posi tive when it expands 239e AV gt 0 b Q is positive for heat coming into the system and negative for heat leaving the system Example XIV73 Problem 1225 Page 420 from the Serway amp Faughn textbook The energy absorbed by an engine is three times greater than the work it performs a What is the thermal efficiency b What fraction of the energy absorbed is expelled by the cold reservoir Solution a Using Eq XIV 26 and the fact that 3W the thermal ef ciency IS W W 1 77 7 333 333 6QH3W 30 or Using Eq XIV 27 we can write lQHl chl chl iQHl IQHI l e Solution b iQHi Donald G Luttermoser E TSU XIV 13 D The Carnot Cycle 1 In the early 1800s Carnot pointed out the basic working of an ideal one without internal friction heat engine The Carnot cycle see Figure 1215 in your textbook can be described in 4 steps a b Step 1 The cycle starts with the piston positioned such that V is at a minimum At this point heat Q is added to the system through a heat reservoir at a high temperature TH The system absorbs the heat a constant temperature TH which causes the volume to expand doing work on the piston During this time the system s internal energy does not change AU 0 and since T is not changing it is an isothermal process From the 1st law of thermo the work done bl the system W lt 0 is equal to the negative of the heat input QH Step 2 The cylinder is moved off of the heat reservoir onto an insulator This isolates the heat in the system which makes this step adiabatic The load on the piston is reduced which allows the gas to expand in the cylinder hence the volume continues to increase This expansion of gas does work on the system which is the same thing as saying that the work is done bl the system W lt 0 caus ing the internal energy to go down since Q is constant here During this phase of the expansion the tempera ture decreases XIV l4 PHYS 2010 General Physics c Step 3 The cylinder is then moved to a heat sink at a cooler temperature To The internal heat flows out Q0 of the system at constant temperature hence an isothermal process The volume decreases which causes work to be done Q the system W gt 0 1 Step 4 The cylinder is moved back to the insulator The load on the piston is increased and the gas undergoes adiabatic compression Since volume is still decreasing work is still being done the system W gt 0 The internal energy returns to the value that it had at the start of Step 1 3 The Carnot engine is said to be a reversible engine E Refrigerators and Heat Pumps 1 Reversing the Carnot cycle we can put work into a system and transfer heat from a lower temperature to a higher one gt a refrigerator a The heat extracted from the cold reservoir to the work supplied similar to the thermal ef ciency is called the coef cient of performance 17 77mg XIV 30 b Using Eq XIV 25 and Eq XIV 28 we can write lQCl To 1 XIV31 mfg lQHl lQCl TH To gt 2 Air conditioners are refrigerators that move heat from within a house to outside a house Donald G Luttermoser ETSU XIV 15 3 The reverse of an air conditioner is a heat pump gt it cools the outdoors by delivering heat from the outdoors to the inside a The coef cient of performance for a heat pump is given by 171 XIV32 b Using Eq XIV 25 and Eq XIV 28 we can write IQH l TH XIV33 nhp IQHI IQCI TH To F The Second Law of Thermodynamics i The Classical Descrip tion 1 The second law of thermodynamics deals with how heat ows It is essentially a description of change a Change To make different the form nature and content of something b Change has over the course of time and throughout all space brought forth successively and successfully galax ies stars planets and life c Evidence for change is literally everywhere i In the macroscopic domain as viewed through tele scopes The Universe started out being fairly ho mogeneous ie the same everywhere until today it is very inhomogeneous Heavy elements ie C N 0 Fe etc have become more abundant over time Stars have formed from a few billion years af ter the Big Bang until the present epoch and will continue to form in the far distant future XlV 16 O PHYS 2010 General Physics ii In the mesoscopic domain as deduced from geo logic records and Via microscopes studying the bi ological world iii And in the microscopic domain as studied with high energy particle accelerators 1 Much of the change is subtle such as when the Sun fuses hydrogen into helium sedately over billions of years or when the Earth s tectonic plates drift sluggishly across the face of our planet over those same billions of years Indeed our perception of time is nothing more than our noticing changes on Earth and in the Universe as a whole 8 There are two classical formulations of this law both essentially mean the same thing a Clausius statement of the second law Heat cannot by itself pass from a colder to a warmer body b KelvinPlanck statement of the second law It is im possible for any system to undergo a cyclic process whose sole result is the absorption of heat from a single reser voir at a single temperature and the performance of an equivalent amount of work The 2nd law speci es the way in which available energy also called usable energy free energy or potential energy change occurs a This law s essence stipulates that a price is paid each time energy changes from one form to another Donald G Luttermoser E TSU b The price paid to Nature is a loss in the amount of available energy capable of performing work of some kind in the future c We de ne here a new term to describe this decrease of available energy entropy S It is derived from the Greek word tropae which means transformation 4 The second law of thermodynamics is different from the laws of mechanics It does not describe the interactions between in dividual particles but instead describes the overall behavior of collections of many particles a In classical mechanics an event is symmetric in time gt the laws are satis ed whether we run the experiment for ward or backward in time b The second law tells us about the sequence or order in which events naturally take place gt the second law of thermodynamics shows us the direction in which time progresses 5 The concept of entropy a Clausius introduced a new state variable called entropy b As de ned above Entropy is a measure of how much energy or heat is mavailable for conversion into work c When a system at temperature T in K undergoes a re versible process by absorbing an amount of heat Q its increase in entropy AS is AC2 i Q AS T i T XIV34 XIV 17 XlV 18 d 8 PHYS 2010 General Physics where often we simply write Q for the heat input or re moved instead of the more proper form of AQ i If Q gt 0 gt heat is absorbed by the system gt entropy increases ii If Q lt 0 gt heat is expelled by the system gt entropy decreases iii For reversible processes Q 0 and the entropy change is zero As can be seen by Eq XIV 34 entropy when multiplied by temperature is a measure of the amount of energy Q no longer capable of conversion in useful work We can use this concept of unavailable energy versus anall able energy to develop a different form for the rst law of thermodynamics We can express the total energy E see Eq XIV 23 as E F T3 XIV 35 where TS l e temperature times entropy represents the energy that is unavailable to be converted to work and F called the Helmholtz free energy is the energy avail able to be converted to work i The rst law can then be written by making this a difference equation AE AF TAS SAT XIV 36 where we have made use of the product rule from differential calculus for the TS term ii Eq XIV 36 shows that if the temperature change over time is small or zero which is typically the Donald G Luttermoser E TSU XIV 19 case for an isolated or closed system the conserva tion of energy dictates that as free energy decreases the energy unavailable for conversion to work in creases gt the entropy increases iii Given enough time all closed systems will run down until they are capable of performing no more work gt once again it is impossible to make a perpetual motion machinel G The Second Law of Thermodynamics i The Probabilistic De scription 1 In the late 1800s Boltzmann showed that an increase in entropy of a system corresponds to an increased degree of disorder in the atoms or molecules composing the substance a This realization lead to the creation of a new area of 0 physics known as statistical mechanics which rewrites the laws of thermodynamics in a probabilistic formalism Boltzmann rewrote the entropy equation de nition in terms of a probability equation S CBan XIV 37 where kg is Boltzmann s constant ln is the natural log arithm of base 6 and W is the number of different ar rangements of microscopic states e positions veloci ties compositions and any various arrangements of quan tum properties Note that W here does Lot mean work i Boltzmann actually has Eq XIV 37 carved on his gravestone in Vienna XIV 20 C d PHYS 2010 General Physics ii W is a measure of the inverse probability p of the occurrence of the possible microscopic states a system can have iii That is p lW 1 for a 100 chance gt Then it is completely certain that a process will occur when p 1 iv As a result we could also write the de nition of entropy e Eq XIV 37 as S kBln kBlnp XIV 38 With the help of Eq XIV 38 we can see the second law implies that any isolated system naturally tends towards an equilibrium state of minimum microscopic probability namely a uniformity of temperature pressure chemi cal composition and so on i Since ordered molecular states for example where molecules in one part of the system have one prop erty value but those in the remaining part have another are less probable than those of random or disordered states ii Boltzmann s law of entropy then signi es that or dered states tend to degenerate into disordered ones in a closed system As can be seen the concept of disorder is very di icult since it requires a detailed knowledge of probability and statistics i The best way to learn about probabilities is through example Let s say we have one die from a set of Donald G Luttermoser E TSU dice There are 6 sides with dots imprinted on the sides relating to the numbers 1 2 3 4 5 and 6 ii The probability of one number say 4 coming up is n a given state 7 XIV 39 p N total number of states 7 l and for this case n 1 and N 6 so p one side i 1 0167 7 total number of sides T 6 or a 167 chance that we would role the die with a 4 landing on top iii The probability of an even number 2 4 or 6 landing on top is n 3 and N 6 3 75 p 6 00 or a 50 chance to role such a number 2 Through this concept of entropy we can rewrite the second law of thermodynamics as any of the following statements a The entropy of the Universe as a whole increases in all natural processes b lsolated systems tend towards greater disorder and en tropy is a measure of that disorder c In a closed system entropy increases over time gt less and less energy can be converted into work 1 All of these statements are probabilistic in nature gt on average this is true XIV 21 XlV 22 PHYS 2010 General Physics Note that the second law written in this probabilistic way can be violated locally gt entropy can decrease locally Only over the whole isolated or closed system over a long enough period of time will necessitate an increase in entropy a Note that on the Earth entropy decreases all the time at the expense of an increase of entropy of the Sun b As such the second law of thermodynamics cannot be used as proof against the theory of biological evolution as some people have suggested The bottom line is that we no longer regard things as xed or being or even that they exist lnstead everything in the Universe is owing always in the act of becoming All enti are permanently changing ties living and non living alike Example XIV74 Problem 1237 Page 420 from the Serway amp Faughn textbook A 70 kg log falls from a height of 25 m into a lake If the log the lake and the air are all at 300 K find the Change of entropy of the Universe for this process Solution The potential energy lost by the log is carried away by heat that is the kinetic energy of the impact into the lake get converted into heat so AQ APE mgAy mgh 70 kg980 ms225 m 17x 104 J We will ignore air friction here and since the temperature remains con stant we just have one term in the entropy equation As such the change of entropy is AQ 17x104J AS T 300K PHYS2010 General Physics I Course Lecture Notes Section IV Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn IV Motion in Two Dimensions A Vector Arithmetic 1 We now will be working with both the x and y axes Vectors are represented on an x y graph as an arrow with a distinct direction indicated by the arrowhead y L 3 Vector A 6 39 X a Note that vector fl makes an angle 6 with the x axis b The x axis is always the reference line which vector direc tion angles are measured 2 Vector addition a Two or more vectors can be added graphically by plac ing the beginning of the 2nd vector at the tail of the Ist vector lVl PHYS 2010 General Physics Example IVil Graphically add vectors ii and g in the diagrams below y 2 a B l I l 2 X 5 x y a a a R A B 2 2 X Fl is 7 units long in the x direction and is 2 units long in the y direction Donald G Luttermoser E TSU b c Two or more vectors can be added algebraically by rewrit ing the vectors into 2 components 06 amp y then adding the respective components Example IVi2 Algebraically add vectors fl and g from the Example IV 1 1 zl where I I y and 13 z lx y Hylgy A 21mg 5A0A g Emmy 2A2A 132 zl AIBI AyByy 52 02g 7022Q Fl is 7 units long in the x direction and is 2 units long in the y direction The actual length of a vector is determined from the Pythagorean theorem IV 1 d PHYS 2010 General Physics gt note that the length of a vector is sometimes written as R 132 IV 2 and that the length is always taken to be the positive root of the square root in Equation IV l The angle that a vector makes with the x axis is deter mined by gt Donald G Luttermoser E TSU Example 1V7 What are the lengths and angles of the 3 vectors and in Example IV 2 0A 5393 0R X5 1415 Ay0 2922g 312 By2 1327922g RI7 Ry2 7 x5202 5 W m m 2 2828m3 v72 22 x494 v53 7280 m 73 tan 1 tan 10 0 2 tan 1 i tan 1l 45 2 tan 1 a m tan 102857 m 1509 m 16 though here I would also have accepted 200 as the answer for 03 Note that from this point forward we will use the standard equals sign even when the answer is not exact as indicated above with the m sign 3 Vector subtraction a Graphically Flip the 2nd vector so that it is pointing in the opposite direction then follow the rules for vector ad dition to 01 H H EN EN PHYS 2010 General Physics Example IVi4 Subtract g from if in Example IV l Y 2 gt I B I gt I A i B 39x AA U Y a A L gt I 395 7x B I U J 2 Donald G Luttermoser E TSU b Y gt gt gt RAB gt 2 A M 6 3 5 7x gt gt 2 R B Fl is 3 units in the x direction and is 2 units in the y direction Subtracting one vector from another algebraically is done by subtracting the respective components a a 12A BAz BzatAy Byg Example IV5 Algebraically subtract g from fl in EX ample VI 4 01 l 5 2920 2g 392 23 gt 216 3 units over in 06 direction 2 units down in y direction IV 8 PHYS 2010 General Physics remember to round off as per the signi cant digits of the input parameters B Velocity and Acceleration in 27D 1 Displacement AF 1 j IV 5 Donald G Luttermoser ETSU IV 9 2 Velocity a Average Velocity 7 IV 6 b Instantaneous Velocity E Velocity o AF d U Alimo g IV 7 3 Acceleration a Average Acceleration a AU 7 lV 8 a At l b lnstantaneous Acceleration E Acceleration a AU 117 a Alimo a IV 9 c Note that if d 5 then the acceleration remains constant during the time that this equation is valid C Projectile Motion 1 In this section we will make use of the following assumptions a The free fall acceleration g7 has a magnitude of 980 ms2 is constant over the range of motion and is directed down ward b The effect of air resistance is negligible hence the size of the object is relatively small and its surface is smooth c The rotation of the Earth does not affect the motion hence the distance traveled is small with respect to the Earth s radius IV 10 2 OD PHYS 2010 General Physics For such projectile motion the path is always a parabola The acceleration in the y direction is g just as in free fall a b The acceleration in the x direction is 0 air friction is neglected c 00 is called the projection angle Breaking motion into components The vectors in our equations of motion can be broken up a into 06 and y components vxvcos6 vyvsin6 b We can just use the l D equations of motion and set them up for each component Donald G Luttermoser ETSU IV 11 xdirection horizontal ydirection vertical xxovxot t0 yyovoyt to gt to2 a vz v voy gltt to b 2 i 2 vy i voy 2921 yo 0 IV10 c The speed 1 of the projectile can be computed at any 1 mg U IV 11 d The angle with respect to the ground at which the pro instant by jectile is moving can be found with 6 tan 1 IV 12 I 4 Problem solving strategy for 2 D motion a Select a coordinate system typically Cartesian b Resolve the initial velocity vector into 06 and y compo nents c Treat the horizontal motion and the vertical motion inde pendently 1 Horizontal motion has constant velocity e Vertical motion has constant acceleration IV 12 PHYS 2010 General Physics Example IV76 Problem 356 Page 77 from the Serway amp Faughn textbook A ball is thrown straight upward and returns to the thrower s hand after 300 sin the air A second ball is thrown at an angle 3000 with the horizontal At what speed must the sec ond ball be thrown so that it reaches the same height as the one thrown vertically y BALL 1 BALL 2 x Ball 1 00 Ball 2 7 to 0 tf 300 s Ax Step 1 Get ymax rst by realizing y ymax 7 that W 0 at that position yo 03570 vyvoy gt t 300 s 150 s my W gt 0 980 ms2 x 150 s 00y 147 rns Step 2 ymax now is determined with v vfy 29y 29y 02 v3 Donald G Luttermoser ETSU IV 13 y if 29 147 ms2 0 ms2 2980 ms2 ymax 110 m ymax Step 3 Now solve for v0 for the ball thrown at 300 up to ymax remember ymax 110 m and W 0 at the top of the parabola vi 7 vi 29y 2 2 voy vym 2911mm pig 0 2980 ms2110 m pig 2156 m2s 2 on 147 ms which is exactly the initial velocity for the ball thrown straight up as it should be Then to nd the total initial velocity 1 we just use a little trig my vo sin639 my 7 147 ms 7 147 ms sin6 sin30quot 05 v0 294 ms D Relative Velocity 1 Observers in different frames of reference may measure different displacements or velocities for an object in motion 2 For measuring motion in one frame of reference with respect to a different frame of reference we must add the motion of the other frame to our equations IV 14 a b PHYS 2010 General Physics Displacement F F t IV 13 F E displacement in our frame F l E displacement in the other frame 1 E velocity of the other reference frame wrt our frame If E time Velocity UUU IV l4 17 E velocity in our frame 17 l E velocity in the other frame Example IVE7 Problem 363 Page 78 from the Serway amp Faughn textbook A hunter wishes to cross a river that is 15 km wide and flows with a speed of 50 kmh parallel to its banks The hunter uses a small powerboat that moves at a maximum speed of 12 kmh with respect to the water What is the minimum time necessary for crossing Let 17 17b where br means boat wrt river and UTE where re vre means river wrt Earth then 501mmquot 17 17 17 UUbT12kmh02 UTe50 kmh v br X Donald G Luttermoser ETSU IV 15 In velocity space y A L 17y UTE V Vy Ur 17127 G X V Determine v amp 6 The actual velocity of the boat wrt the Earth is v Ube W 122 502 kmh 7 V 169 kmh 13 kmh and the angle that the boat s makes wrt the x axis is 50 kmh 0 t 1ltUigtt 1 an v1 an 12 kmh 50 tanil E 22 15 23quot The time it will take to cross the river is derived from v L260 setro0to0 7a or 157 t r t IV 16 PHYS 2010 General Physics 7 is now the path length of our trip 15 k 0086 m i 15 km i 0086 139 i 15 km 0032206 162 km A 15 km 7 162 km 15 7 7 0125h U 13 kmh 0125 h x 601 7477 min PHYS2010 General Physics I Course Lecture Notes Section X Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn X Rotational Equilibrium and Rotational Dynamics A Torque 1 The ability of a force to rotate a body about some axis is mea sured by a quantity called torque 739 Greek letter tau a The torque due to a force of 13 has a magnitude of i 739 E torque Nm Xl ii F E applied force iii d E lever arm also called the moment arm dis tance b The lever arm is the J distance from the axis of rotation to a line drawn along the direction of the force Note that d 7 sin b X Z where 7 is the magnitude of the displacement from the axis to the point of the applied force F and b is the angle between the direction of F and the direction of 13 c As a result we can rewrite the torque equation Eq X l as 77 F sinq X B X1 X 2 PHYS 2010 General Physics 2 The net torque is found by summing all torques e multiple forces acting on a rotating object N Tnet Z73 71Tg TN F1d1F2d2 FNdN X4 2391 a 739 is positive if the rotation is counterclockwise CCW b 739 is negative if the rotation is clockwise y 4 positive torque CCW negative torque CW Donald G Luttermoser E TSU X 3 Example X71 Problem 82 Page 253 from the Serway amp Faughn textbook A steel band exerts a horizontal force of 800 N on a tooth at point B in Figure P82 of the textbook What is the torque on the root of the tooth about point A y A I r 120 cm H a Solution The distance from point A to point B is 7 120 cm 120 X 10 2 In As can be seen from the gure above the lever arm is d rsinq 120 x 102 m 114200 803 x 103 In This gives a torque of 739 Fd 800 N803 gtlt 103m 0642 N n1 Since the torque is positive it is in the direction X 4 PHYS 2010 General Physics B Static Equilibrium 1 The rst condition of equilibrium is that the sum of all forces acting on an object is zero N E E 0 X 5 see Eq V 8 2 The second condition for equilibrium asserts that if an object is in rotational equilibrium the net torque acting on it about any N il 3 A body in static equilibrium must satisfy 2 conditions axis must be zero a Resultant external force must be zero Eq X 5 b Resultant external torque must be zero Eq X 6 4 If a body is in equilibrium the position of the axis of rotation is completely arbitrary when calculating net torque C Center of Gravity 1 The center of gravity of an object is the point where all of the weight of the object can be considered to be concentrated 2 This also is called the center of mass Sill mixi m1x1 mgxg meN X7 meg N 221mi m1m2mN EL miyz mm mm mNyN ycg N Elam m1m2mN Z 7 31 mizz i m1z1 mgzg mNzN X 9 cg i i ZZZLn m1m2mN Donald G Luttermoser E TSU 3 D ProblemSolving Strategy for Objects in Static Equilibrium 1 The center of gravity of a homogeneous symmetric body must lie on the axis of symmetry R center of gravity at A r0 yL2 I I I I I t I 9 L y L2 Y l quot center of r J y 0 gravrty at r0 Draw a diagram of the system Isolate the object that is being analyzed Draw a free body di agram showing all external forces acting on the system do this for all masses in the system Be careful to show the point where each force acts Chose coordinate axes and specify the positive sense of rotation Then resolve the forces into components along the axes Apply the rst condition of equilibrium ZFI0 amp ZFy0 X5 X6 PHYS 2010 General Physics 5 Chose an origin for calculating a net torque usually an axis of rotation Apply the second condition of equilibrium 270 6 Solve the set of simultaneous equations from steps 4 and 5 and solve for the unknowns Example X72 child with a mass of 42 kg sits 14 m from the center Where on the other Two children are playing on a balanced seesaw One side must the second child sit if her mass is 34 kg n n I x1 14 m x2 J 39l I 1 m2 positive negative torque torque CCW CW W1 w 5 2 Solution Since the seesaw is balanced the sum of all torques must be zero We will choose the axis at the point where the fulcrum is located With this location of the axis the torques from two of the forces the weight of the seesaw wS and the normal force on the fulcrum n are zero since the lever arm is zero for both forces Since the weight of mass l 101 will cause the seesaw to rotate in the counterclockwise CCW direction it will have a positive torque Since the weight of mass 2 wg will cause the seesaw to rotate in the clockwise CW direction it will have a negative torque Donald G Luttermoser E TSU X 7 Hence the torque equation becomes remembering that w mg 7 71 72TSTn0 w1x1 w2x2000 m1gx1 mggxg 0 Here 717273 and Tn are the torques of mass 1 mass 2 the mass of the seesaw and the normal force respectively Mass m1 42 kg and m2 34 kg The lever arm on mass 1 is 961 14 m We want to nd the lever arm on mass 2 which will be the distance that the second child is from the fulcrum 239e the balance point 062 since this distance is perpendicular to the force vector wg Since the equation above has only one unknown in it we do not need the rst condition of equilibrium for this problem 69 2F n wl wg wS 0 Solving the above torque equation gives m1 i 42 kg 061 714 m 17 m 062 i 34 kg m2 E Torque and Moment of Inertia 1 The tangential force on a rotating object is E mat a Multiplying this by the radius 7 of the rotation object gives E 7 mm b Since at 7 01 see Eq VIII 13 we can write E 7 mrgoz c Finally E 7 is nothing more than the torque so X8 PHYS 2010 General Physics where m is the mass of the rotating object 7 is its radius and oz its angular acceleration d The quantity mi is called the moment of inertia of mass m gt usually represented with an L The total torque on a rotating object is then the sum of all the torques on each mass of the rotating objects N N 7net Z a 7 il il where the N is the number of the indiVidual objects of mass m rotating about a common axis each at a distance 7 from that axis a The moment of inertia is then N 2 I 2 mm 2391 Im17 mgrgm1rr v X 12 01quot gt I kg m2 b We can then rewrite Eq X ll as c Note the similarity of this equation to Newton s 2nd Law of motion 2F ma gt Eq X lB is the rotational counterpart to Newton s 2nd law X 13 1 Force amp mass in linear motion correspond to torque amp moment of inertia respectively in rotational motion Donald G Luttermoser E TSU 3 Moments of inertia for common solids a Solid sphere with axis of rotation through center I 3MB2 M mass R radius of sphere b Hoop or cylindrical shell with axis of rotation through center parallel to the wall I IWR2 R outer radius Solid cylinder or disk with axis of rotation through cen C l ter parallel to wall I EIWR2 R radius of cylinder 1 1 Long thin rod rotation about center I EIWL2 L length of rod 1 Long thin rod rotation about one end I gMLZ e Example X73 Problem 835 Page 258 from the Serway amp Faughn textbook A 150 kg merry go round in the shape of a uniform solid horizontal disk of radius 150 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope What constant force must be exerted on the rope to bring the merry go round from rest to an angular speed of 0500 revs in 200 s Solution This problem is clear enough without drawing a diagram We rst need to calculate the moment of inertia for the merry go round Since we are told it has a solid disk shape rotating about its center 1 1 I EMR2 E150 kg150 m2 169 kg m2 We also can calculate the angular acceleration based on the information given 7 w wo a At 0500 revs 0 200 s 025 reVs2 27F radrev g rads2 X9 X 10 PHYS 2010 General Physics Using Eq X l in conjunction with Eq X43 we have 739 F d oz where the lever arm here is just the radius of the merry go round d 7 As such 2 2 F 7a 169 kg In 7T2 rads 177 N 7 150m F Rotational Kinetic Energy 1 A body rotating about some axis with an angular speed w has a rotational kinetic energy of KE E It X14 which can be shown through the following argument a Total kinetic energy of all rnass elements in a rotating body is N 1 2 2391 b However 1 M for an object in rotation thus N 1 2 2 1 2 N 2 KET KB Z gym73 in Z mlTi 1 2 1 where we have taken the angular speed out of the sum rnation since all of the mass elements of the rotating body have the same angular speed c Finally I Emrg so 1 KB Elna hence we have proven Eq X l4 Donald G Luttermoser ETSU X 11 2 The conservation of mechanical energy now becomes KB KB PE KB KB PEf X15 a KEt E translational straight line KE b KET E rotational KE c PE E potential energy Example X74 Problem 841 Page 258 from the Serway amp Faughn textbook A horizontal 800 N merry go round of radius 150 m is started from rest by a constant horizontal force of500 N applied tangentially to the merry go round Find the kinetic energy of the merry go round after 300 s Assume it is a solid cylinder Solution This problem is clear enough without drawing a diagram We rst need to calculate the moment of inertia for the merry go round Since we are told it is a solid cylinder rotating about its center 1 1 1 800 N I 7MP i 9122 a 2 150 m2 918 kg m2 2 2 g 2 980 ms We also can calculate the angular acceleration from the torque equations 739 F R 500 N150 m 2 E 7 817 a I I 918 kg m2 0 radS With this angular acceleration we can now calculate the angular speed after 300 s starting from rest with w w at 0 0817 rads2300 s 245 rads Finally we can calculate the rotational kinetic energy with Eq X l4 1 1 KB in E918 kg m2 245 lads2 276 J X 12 PHYS 2010 General Physics G Angular Momentum 1 We have the torque equation of Eq X43 739 oz a However 1 wo at or w wo aAt if tO y 0 thus w wo Aw a At E39 b Then we can rewrite torque as Ti1ltampgtin Iwo At 7 At 39 2 The angular momentum of a rotating object is the moment of inertia multiplied by the angular velocity 3 As a result the torque also can be de ned as the ratio of the change in angular momentum to the time interval At over which the change occurs T i change in ang mom L LO AL 7 X17 time interval At Kt a This is the rotational analogy to F ApAt see Eq VII 2 b If ET 0 then AllAt 0 and Li L or find Ifwf 4 Conservation of Angular Momentum a The angular momentum of a system is conserved when the net external torque acting on the system is zero If 270 thenLiLf Donald G Luttermoser ETSU X 13 b In an isolated or closed system the energy linear mo mentum and angular momentum of the system remains constant Example X75 a Calculate the angular momentum of Earth that arises from its spinning motion on its axis and b the angular momentum of Earth that arises from its orbital motion Solution a This problem is clear enough without drawing a diagram The angular momentum is L 01 If we treat the Earth as a solid sphere spinning about its center the moment of inertia is 2 2 2 1m gMR2 30469123 g598 x 1024 kg638 x 106 m2 974 x 1037 kg m2 The angular speed of the Earth s rotation is 27T 27T 1 hr 24hri3600s This then gives the angular momentum of the Earth s spin as Lg 1m cum 974 x 1037 kg m2727 x 105 rads 708x1033Js Considering the orbital motion we consider the Earth to be a cum 727 x 105 rads Solution b point mass a distance of re 1496 X 1011 m from the Sun in a circular orbit so orb Mgr 598x1024 kg1496gtlt1011m2 134x1047 kg mg The angular speed of the Earth in its orbit is 27T 27T 100 yr 100 day 100 hr Cquotjorb Tob 100 yr 36524 day 2400 hr 3600 s 7 199 x 107 rads This then gives the angular momentum of the Earth s orbital PHYS 2010 General Physics motion as L0b orbworb 134 x 1047 kg m2l99 x 107 rads 266gtlt 1040Js The Formation of the Solar System a b The solar system started out as a giant more or less spherical gas cloud a few light years in diameter and ro tating very slowly a few meters per second on the outer rim gt we currently see such clouds in star forming re gions of our Milky Way Galaxy An instability occurred in the cloud from either the pas sage of one of the spiral arms of the Milky Way or a super nova shock wave passing through the cloud aluminum iso tope ratios in the solar system suggest the latter of these two causes likely happened This set up a compression in the cloud which caused Fg gt P A the weight of the cloud exceeded the internal pressure that was keeping it stable gt the cloud began to collapse i lnitially these clouds called Giant Molecular Clouds GMC have thousands of solar mass worth of material in them ii Within the GMC turbulence is common just as in a thunderstorm here on Earth This turbulence causes localized spinning eddies to exist within the GMC Donald G Luttermoser ETSU X 15 iii During the collapse gravitational instabilities cause the GMC to fragment into smaller cloudlets that range in mass from approximately 01 MG up to approximately 100 M9 in mass each iv The eddies from the GMC induce an initial spin to each cloudlet as it begins to collapse see the diagram below we 1 J J39J quotCIoud1et v The more massive the cloudlet the faster it col lapses as shown in the following table Cloudlet Mass Contraction Time MG Years 01 108 10 107 30 106 10 105 100 104 C d 8 PHYS 2010 General Physics As the solar cloudlet often referred to as the solar neb ula collapsed it began to spin faster due to the conser vation of angular momentum Li Lf 1W1 Irwf 2 2 iMirigwi iMfrgwf 5 5 and since the mass is conserved during collapse Ml M f 2 riwi riwf 7 1 2 i 011 Tr The faster the cloud spins the atter it got gt centrifu CJf gal acceleration as shown in the following diagram Disk 2amp6 VJ L7 i A central bulge formed called protostars and in the case of the Sun the protosun ii A disk formed surrounding the central bulge P and T got so high in the central bulge that thermonu clear reactions began gt the Sun was born Donald G Luttermoser E TSU f Meanwhile in the cooler disk the planets initially called protoplanets formed through the process of condensa tion solids forrning from gas advection bigger particles forming from smaller ones and accretion large objects forming even larger objects through collisions PHYS2010 General Physics I Course Lecture Notes Section VIII Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn VIII Circular Motion A Angular Speed and Angular Acceleration 1 Arc length of a rotating or revolving object is analogous to linear displacement a Arc length 3 is swept out as an object rotates 6 degrees such that Reference i s is measured in the same units as 7 e a length ii 6 is measured in radians not degrees 1 revolution rev 27F radians rad 360O VIII 2 or 3600 27F 1 rad E 5703 VIII 1 Vlll 2 b C d PHYS 2010 General Physics Conversion of angles 639 E angular displacement 6 rad W F00 639 deg 1113 0 rad i745 x 1042 0 deg Sometimes you may run across angular measure dealing with very small angles There are 60 arominutes aromin per 10 and 60 aroseoonds aroseo per 1 ar ominute 1 60 3600 An object is said to be rotating if it is spinning about an aXis 69 the Earth rotates about its axis quotspinsquot Donald G Luttermoser ETSU VIII 3 e A body revolves around another object if the rst body is in orbit about the second object 69 the Earth re volves about the Sun quotorbitsquot Sun Eth at Sun Example VIIIil Problem 72 Page 218 from the Ser way amp Faughn textbook A wheel of radius 41 meters rotates at a constant velocity How far path length does a point on the Circumference travel if the wheel is rotated through angles of 30quot 30 rad and 30 rev Let 7 be the radius of the wheel 3 be the path length along the circumference and 6 be the angle that this point subtends from a reference point as the wheel rotates As such 7 41 In and 6 1 30 6392 30 rad and 03 30 rev We will use Eq VI l 36 r except that all of these angles must be expressed in radians so 7T rad 6391 300 X 1800 05236 rad 6392 30 rad which is what we want 27F rad 6 3 30 rev X 7 1885 rad 1 rev VIII 4 PHYS 2010 General Physics Plugging these values of 6 into Eq Vl l gives 01 05236 41 m admin 03 1885 41 m 81 21 m 82 83 2 The length 3 in Eq Vlll l is actually an arc length curved length However if 6 is small 216 6 ltlt 1 s gt D the actual linear size or diameter if the object is round a b c As such by measuring the angle that a distance object subtends we can calculate its actual size or diameter D by knowing the distance d to the object Since 6 is small here it is often more convenient to express this angle in terms of arcseconds instead of radians We can rewrite Eq Vlll l by setting 3 D 7 d and expressing 6 in arcseconds and relabeling it as oz then 7T rad 1O 0 d 7 Ia 1800 X 3600 arcsec aarcseconds X 7 oz arcsec i 206265 arcsecrad Using oz instead of 6 in Eq Vlll l gives 7 ad 206265 which is called the smallangle formula Here D is the VIII4 linear size ie diameter of the object at a distance d which subtends an angle oz measured in arcseconds the Donald G Luttermoser ETSU VIII 5 206265 is the conversion factor between arcseconds and radians The lengths d and D will be in the same units 69 if d is measured in km then D will be in km Note that this formula is only valid when 6 ltlt 1 typically one would want oz lt 1000 arcseconds in order to use this small angle approximation Example VIII72 The star Betelguese oz Ori has an angular size of 0040 arcsec 40 milliarcseconds 40 mas and it is at a distance of 200 pc What is the linear size of Betelgeuse How does this size compare to the planet39s distances from the Sun in our solar system d 200 pc x 309 x 1016 mpc 618 x 1018 m D 0040 x 618 x 1018 m 206265 120 x 1012 m The Sun is 139 X 109 m in diameter which means that Betelgeuse is i 120 x 1012 m 7 139 x 109 mDG where an Astronomical Unit AU is the size of the Earth s orbital semimajor axis note that 1 AU 1496 X 1011 m Betelgeuse is 860 times bigger than the Sunl This diameter gives a stellar 860 D9 80 AU radius of 40 AU for Betelgeuse If it was put in the place of the Sun then Mercury 039 AU Venus 072 AU Earth 10 AU and Mars 15 AU would be inside Betelgeusel The planet Jupiter 52 AU would be close 12 AU from the photosphere surface of Betelgeuse 3 Angular speed 01 is the change of angular displacement di vided by the time interval that the angular displacement took place VIII 6 PHYS 2010 General Physics a Average angular speed i 02 01 A0 7 VIII 5 w 152 t1 At gt b Instantaneous angular speed A6 1 Aliran Kt VIII 6 c Angular speed has units of radians second gt rad s or just s 1 since radian is a pseudo unit d w is positive when 6 is increasing counter clockwise mo tion and negative in the clockwise direction 4 Angular acceleration oz alpha is the change of angular speed over the time interval of interest a E change in angular speed wg 11 E VHLU time interval 152 t1 At which is the average angular acceleration a Whereas the instantaneous acceleration is Aw oz Aliran E VIII 8 b The units of angular acceleration are rads2 or just s 5 When a rigid object rotates about an axis every portion of the object has the same angular speed and same angular acceleration 6 Note that we have Lot written then angular variables as vectors In reality they are vectors but the unit vectors for the magni tudes we have written above point out of the plane in which the object is rotating Since this is a bit cornplicated we will just discuss the magnitudes of these angular motions at present Donald G Luttermoser ETSU VIII 7 B Rotation Motion under Constant Angular Acceleration 1 These are analogous to their counterparts in linear motion LINEAR ANGULAR 1 2 1 2 06 vat Eat 6 wot Eat VIII 10 Note that in the above equations the initial tiInes angles and displacements are all set to zero If these initial setting are not zero replace t with At 6 with A0 and x with A96 2 The tangential speed of a point on a rotating object equals the distance of that point from the axis of rotation 7 multiplied by the angular speed w A A0i 76 71 ef gy 1 A0 i 11 As AlIBOXt i AlIEOKt7 1 w iv 7 vt raj VIII 12 VIII 8 PHYS 2010 General Physics where v is the tangential velocity tangent to the curved path of motion the particle is taking 3 The tangential acceleration at of a point on a rotating object equals the distance of that point from the axis of rotation 7 multiplied by the angular acceleration oz Av rAw Am 7 7amp At 7 At Rf Rio 1 AU 7 1 Aw AltIBOTt i r AltIBOXt at roz VIII 13 C Centripetal Acceleration and Force 1 If an object travels at constant speed in a curved path it is accelerating since the velocity vector is continuously changing direction 2 Besides the tangential acceleration e acceleration tangent to the curved path there must be an acceleration perpendicular l to the tangent line pointing toward the center of the curved path Donald G Luttermoser E TSU 3 b C VIII 9 Otherwise the object would y off in a straight path in the direction of at as per Newton s 1st law of motion This center seeking acceleration is called the centripetal acceleration VIII14 i Gravity is a centripetal force hence acceleration for planets orbiting the Sun ii A string s tension is a centripetal force hence ac celeration for an object attached to that string be ing rotated in circular motion rock at 03 The total acceleration of an object in circular motion is d 1 in VIII15 Vlll 10 PHYS 2010 General Physics Example VIII73 Problem 718 Page 219 from the Ser way amp Faughn textbook A race car starts from rest on a circular track of radius 400 m The car s speed increases at a constant rate of 0500 ms2 At the point where the magnitudes of the centripetal and tangential accelerations are equal determine a the speed of the race car b the distance traveled and c the elapsed time Solution a From Eq VIII 14 the centripetal acceleration is aC Thus when aC at 0500 ms2 we have u M M400 m0500 ms2 Solution b The total distance traveled is just the total arc length traveled on the circular track 3 which can be found from the linear 1 D equation of motion 1 U 2 a s where v v 141 ms 1 0 starts from rest and a at 0500 ms2 Hence the total arc length re distance traveled is 2 2 2 141 0 5 U lt 18 2 200m 2at 20500 ms Solution c We can use another linear 1 D equation of motion to nd the elapse time u 1 at where v v 141 ms 1 0 starts from rest and a at 0500 ms2 Hence the time it take to reach the tangential Donald G Luttermoser ETSU Vlll ll velocity from part a is i 141 0 at 0500 ms 3 The centripetal force is just the mass times the centripetal accel eration 2 Fcmac quotWt mm VIII17 7 a Centripetal forces are center seeking forces i The gravitational force is a center seeking force gt planets are in orbit about the Sun ii The tension force is a centripetal force for an ob ject connected to a string being swung about some axis b The concept of centrifugal force the apparent force that forces an object outward while moving in a circu lar path is Lot really a force gt this tendency to follow a straight line path away from the curved path is nothing more than Newton s 1st law the law of inertia Example VIIIi4 Problem 748 Page 221 from the Serway amp Faughn textbook A 0400 kg pendulum bob passes through the lowest part of its path at a speed of 300 ms a What is the tension in the pendulum cable at this point if the pen dulum is 800 cm long b When the pendulum reaches its highest point what angle does the cable make with the vertical c What is the tension in the pendulum cable when the pendulum reaches its highest point Vlll 12 Solution a PHYS 2010 General Physics rL800cm The diagram above shows the system when the bob is at its low est point and shows the direction of the forces involved and the tangential velocity vector The mass of the bob is m 0400 kg and the length of the cable as L 800 cm 0800 In Since the bob follows a circular path about the pivot of the pendulum the radius of this path is just the length of the cable 7 0800 In The centripetal force is always directed toward the center of the orbit which we will de ne as the positive y direction as indi cated in the gure above As a result the sum of the forces in the y direction is just the total centripetal force Solving for the tension T gives T 842 N ZFyFcTFg FgFCmgmacmgac 2 m g 0400 kg 980 ms2 300 Ins2gt 0800 In Donald G Luttermoser ETSU Vlll 13 Solution b L39Yf vf0 quotYr Yf yi0 vi 300 ms The diagram above shows the bob at its maximum height To nd this height hence 0m we only need to use the conservation of mechanical energy We will set the lowest position of the bob at the ground level yi 0 At its highest position yf the velocity changes from the counter clockwise to the clockwise direction hence goes through zero at this point of 0 From the diagram above we can solve for yf in terms of 0m Lyf cos rm L L cos 6mm L yf yf L L cos 0m yf L 1 cos6 m We can now determine 0m from the conservation of mechanical energy by setting the lowest point as the initial position so 1 vt from part a and yi 0 and the highest point as the nal position so of 0 and yf is given in the equation above KE PEg KE PEgf i 1 2 1 2 Emvimgyi Emvfmgyf V111 14 PHYS 2010 General Physics 1 7m 0 0 mgL 1 cosem 2 1 2 mgL 1 cos 0m Emut 2 v 1 0m t cos 2g 2 cos m 1 Ut 29 2 0m cos 1 1 0081 1 300 ms2 gt 2980 ms20800 m cos 104260 Solution c The diagram above shows the forces acting on the bob at the highest point with respect to an arbitrary coordinate system cho sen such that the tension and total centripetal force points in the positive y direction From this we now construct a free body diagram Donald G Luttermoser ETSU Vlll 15 yn At T 5 6max Fg Since the tangential velocity at this highest point is zero as men tioned in part b the centripetal force must be zero 2 U E mac m it 0 7 Using this equilibrium equation in conjunction with the free body diagram above we get Fe Z Fy T Fm 0 T mgcos m T mg cos 0max 0400 kg980 n1s2 cos648o 167 N PHYS2010 General Physics I Course Lecture Notes Section XII Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn XII Thermal Physics A Thermal Equilibrium 1 If bodies A amp B are separately in thermal equilibrium with a 3rd body C then A amp B will be in thermal equilibrium with each other a This statement is often referred to as the 0th Law of Thermodynamics It simply means that if 2 objects in thermal equilibrium with each other are at the same temperature b Thermal equilibrium means that an object has the same temperature throughout its interior 2 Temperature is nothing more than a measure of how fast particles are moving due to the heat energy stored in the system a There are 3 different temperature scales i Fahrenheit scale archaic English system gt 32011 E water freezes 2121 E water boils at atmo spheric pressure ii Celsius scale metric system scale once called the Centigrade scale gt 0 C E water freezes 1000C E water boils at atmospheric pressure iii Kelvin scale Sl system gt the absolute tem perature scale gt 0 K E no atomic motions lowest possible tem perature gt Lowest temp recorded in lab is 10 6 K X11 1 Xll 2 PHYS 2010 General Physics b Converting between the temperature scales i F gt 0C 9 TF ETC 32 Xll l ii 0C gt K To T 27315 XII 2 iii Note that T without a subscript will always refer to the Kelvin scale in these notes except where noted Example X1171 Problem 108 Page 346 from the Serway amp Faughn textbook The temperature difference between the inside and the outside of an automobile engine is 4500C Express this temperature difference on the a Fahrenheit scale and b Kelvin scale Solution a We need to come up with a formula that changes ATC to ATF and AT where T is measured in Kelvin Using Eq Xll l we can write 191 Tc1 32 d an 9 TF2 5 TC2 32 Subtracting equation 1 from 2 yields 9 TF2 TF1 g lltTC2 TOM since the two 32 s cancel with each other We are given a temperature difference of ATC 4500C so from the equation above we get 9 450 C 810 F 9 Donald G Luttermoser ETSU Xll 3 Solution b From the same argument above Eq XII 2 gives T2 T1 Tc2 27315 Tc1 27315 T2 T1 ch Tch AT ATC 450K B Thermal Expansion 1 Heat energy added to matter causes the particles that make up the matter to speed up a Increased velocity of particles increases pressure b lncreased pressure causes matter to expand due to the increased internal force i Liquids and gases ll a larger volume ii Solids get longer c A loss of heat energy causes objects to shrink in size 0 For solids the change in heat as measured by temperature dic tates a change in linear size AL Via AL oz Lo AT XII 3 a LG E initial length b AT E change in temp usually measured in 0C in Eq XII 3 Xll 4 PHYS 2010 General Physics c oz E coef cient of linear expansion see Table 101 in text book 3 Surface areas also change in size AA via AA VAC AT XII4 2 b 7 E coef cient of area expansion 2a a AO E initial area m Volumes increase or decrease AV as well when heat is added AV VOAT a V0 E initial volume mg or taken away via XII5 b E coef cient of volume expansion 3a Example XIIEZ Problem 1011 Page 346 from the Serway amp Faughn textbook The New River Gorge bridge in West Virginia is a 518 m long steel arch How much will its length Change between temperature extremes of 20quotC and 350C Solution Use Eq XII 3 here AT 350C 200C 550C LO 518 m and oz 11 X 10 6 quotC 1 from Table 101 on page 313 in the textbook This gives a length change of AL ozLo AT 11 x 106 C 1518 m55 o 031 m Donald G Luttermoser E TSU C Heat and Internal Energy 1 Internal energy U is the energy associated with the micro scopic components of a system given by a b C d 2 Heat or thermal energy Q is a mechanism by which energy is transferred between a system and its environment because of U mtrans rm mVib Win01 39 mm E average translational linear kinetic en ergy of the atoms and molecules mm E average rotational kinetic energy gt the ro tation of a molecule about an axis Wm E average vibrational kinetic energy gt a con tinuous change of distance between the atoms that make up the molecule Wm E average intermolecular potential energy be tween the atoms that make up the molecules gt the bond energy a temperature difference between them a b c Heat is essentially related to kinetic energy gt energy due to the motion of particles in matter Heat is measured in calories in the cgs unit system gt one calorie of heat is required to raise 1 gram of water by a temperature of 10C Heat is measured in kilocalories in the SI system i One kilocalorie 103 calories of heat is required atoms and molecules It is XII6 X11 5 X11 6 PHYS 2010 General Physics to raise 1 kg of water by a temperature of 10C or 1 ii A kilocalorie is also called a Calorie capital C gt this is the energy unit that you see on food containers d Heat is measured in British Thermal Units BTU in the English system gt one BTU of heat is required to raise 1 lb of water by a temperature of 10F 3 In 1843 Joule showed that heat can be used to drive a machine gt mechanical energy and heat were 2 forms of the same thing a Heat is just another form of energy b 1 cal 4187 Joules c 4187 Jcal 4187 Jkcal is known as the mechanical equivalent of heat D Speci c Heat and Calorimetry 1 Calorimetry means the measurement of heat exchange 2 Heat capacity is the amount of heat required to change an entire object s temperature by 10C 3 The speci c heat c of a substance relates the amount of heat Q it takes to change the temperature T of an object of mass m by one unit degree Q C m Donald G Luttermoser ETSU XII 7 which is the heat capacity per unit mass so we can write the a In the cgs system c is measured in calg OC heat equation as b In the SI system c is measured in Jkg OC note that 0C is used instead of K here c Note that water has a speci c heat de ned as cW 1000 calgOC XII 9 1 Table 111 in your textbook lists speci c heats for various substances 4 Devices used to measure heat energy are called calorimeters 5 Heat ow from one system to another obeys the conservation of energy a Heat gained by one system positive number heat loss by another system negative number 0 b Or rephrasing Heat loss by one system Heat gained by another system anjn Q1055 m1C1T mQCQ T XII11 i 1 represents the system gaining heat ii 2 represents the system losing heat iii T represents the nal equilibrium temperature when everything is mixed Xll 8 PHYS 2010 General Physics Example XII73 Problem 119 Page 379 from the Serway amp Faughn textbook A 500 gm lead bullet traveling at 300 ms is stopped by a large tree If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree what is the increase in temperature of the bullet Solution The mechanical energy transformed into internal energy of the bullet is 1 1 1 2 1 2 Q i E K131 E Elmo 17ml 7 where m is the mass of the bullet and vi 300 ms is the initial speed of the bullet From Table 111 on page 335 in the textbook the speci c heat of lead is 128 JkgOC Using Eq XII 8 the change in temperature of the bullet is 1 2 2 3 2 AT Q ampEM 176 C me mc 40 4128 Jkg quot C Example XII74 A 040 kg iron horseshoe that is initially at 5000C is dropped into a bucket 20 kg of water at 220C What is the final equilibrium temperature Neglect any energy transfer to or from the surroundings Solution From Table 111 in the textbook the speci c heat of iron Fe is 448 JkgOC and water is 4186 JkgOC Using Eq XII 11 we get an equi librium temperature T of mwcw T TW mFecFe TFe T mWCWT mWCWTW mFecFeTFe mFeCFeT mWCWT T mFeCFeT mWCWTW T mFecFeTFe mwcw T mFecFe T mWCWTW T mFecFeTFe T i mWCWTW T mFecFeTFe mwcw T mFecFe Donald G Luttermoser ETSU Xll 9 with chWTW 20 kg4186 Jkg C22 C 184 x 106 J mFecFeTFe 040 kg448 Jkg C500 C 896 x 104 J and mch mpecpe 20 kg4186 Jkg C 040 kg448 Jkg C 839 x 104 J C Plugging this into out equilibrium temperature equation above we get 184 x 106 J 896 x 104 J T 839 x 104 J C 230C E Latent Heat amp Phase Changes 1 When a substance undergoes a physical alteration from one form to another e liquid to gas it is referred to as a phase change a Under such circumstances T does Lot change when heat is added or taken away from the system gt the energy goes into changing the phase XII12 i Q E heat gained or lost to the system b Mathematically ii m E mass of the system Xll 10 C d 8 PHYS 2010 General Physics iii L E latent heat iv You use the sign when energy is being added during a phase change e melting or boiling 7 v You use the sign when energy is being taken away during a phase change e freezing or con densation When the phase change is from gas to liquid or Vise Versa L is called the latent heat of vaporization LV i Changing from gas to liquid is called condensa tion ii Changing from liquid to gas is called boiling When the phase change is from liquid to solid or Vise Versa L is called the latent heat of fusion L i Changing from liquid to solid is called freezing ii Changing from solid to liquid is called melting Table 112 of the textbook shows latent heats of fusion and vaporization for some common substances We can deduce temperature changes in systems that undergo phase changes by using the conservation of energy for each state of the system Donald G Luttermoser E TSU 3 b c d Xll 11 Water has the following characteristics T C A E 100 D I STEAM 2 c WATER l 50 1 STEAM 0 B I WATER I ICE I I I ICE WATER 627 3967 8157 3076 7 Q J Heat ln states A C and E the heat equation follows Eq XII 8 ln states B and D the phase change states the heat equation follows Eq XII 12 Using the conservation of energy equation e Eq Xll 10 going from point 2 to point 1 on the above graph we can set up the following equation Qtot Qcof water at 2 QB QAof ice at 1 QC mwcwj 2 TB QB mg Lf positive since energy is being added QA miciTB T1 Here 01 and cW are the speci c heats of ice and water L is the latent heat of fusion of water mW is the initial mass of the water mi mW is the nal mass of the ice mi mi mW is the mass of the water freezing into ice Xll 12 PHYS 2010 General Physics T2 gt 0 the initial temperature of the water TB 0 the temperature of fusion in Celsius and T1 lt 0 the nal temperature of the ice so Qm chWT2 TB mil mg L miciTB T1 mwchg mWLf mwciTl Note that since T1 lt 0 mwciT1 gt 0 and hence the right hand side of this equation will always be positive e For those problems where a nal equilibrium temperature is being sought One chooses a Q as either a gain or a loss based upon whether the xed temperature of the Q equation is less than or greater than the equilibrium temperature i If T gt then Q Q1055 lt 0 ii If T lt then Q om gt 0 iii As a result note that we also could have written the conservation of heat energy e Eq XII 10 as Qcold Qhot Xll lB as given by the textbook Example X1175 A 300 gm lead bullet at 300 C is fired at a speed of 240 ms into a large block of ice at 00C in which it embeds itself What quantity of ice melts Solution Let s assume that the large block of ice does not completely melt as a result of the bullet embedding itself then the kinetic energy of the bullet Donald G Luttermoser ETSU Xll 13 will go towards heating the block of ice Also the temperature of the bullet will heat the ice block Let mm be the amount of mass in the ice block that melts mb 300 gm 300 X 10 3 kg be the mass of the bullet Tb 300 C be the initial temperature of the bullet vb 240 ms be the initial speed of the bullet and T 00C be the nal temperature of the bullet since it will cool to the same temperature of the block of ice Using the tables from the textbook the latent heat of fusion of water is L 333 X 105 Jkg and the speci c heat for lead is cb 128 JkgOC Using the conservation of energy and noting that J kg m2s2 we get anjn Q1055 Qmelt KEb Qbeloss 1 mmelth Embvg mbe W L 240 ms22 128 Jkg 0o300o 00C 333 x 105 J kg mmelt mb 300 gm 0294 gm F Transport of Heat Energy 1 Thermal energy ie heat can only ow by one of three dif ferent mechanisms conduction convection and radiation transport 2 Heat Transfer by Conduction a Conduction is the process by which heat is transferred Via collisions of internal particles that make up the object gt individual mass particle transport Xll 14 b C v PHYS 2010 General Physics i Heat causes the molecules and atoms to move faster in an object ii The hotter molecules those moving faster col lide with cooler molecules those moving slower which in turn speeds up the cooler molecules mak ing them warm iii This continues on down the line until the object reaches equilibrium The amount of heat transferred AQ from one location to another over a time interval At is i 73 E heat transfer rate XII14 ii 73 is measured in watts when Q is measured in Joules and At in seconds iii As such 73 is the same thing as power since they are both measured in the same units Heat will only flow if a temperature difference exists be tween 2 points in an object i For a slab of material of thickness L and surface area A the heat transfer rate for conduction is AQ 7 KA 73601111 lt Th Tc XII 15 gt lt gt Donald G Luttermoser ETSU Xll 15 Th on back side Heat Flow if Th gt T J u L ii Th is the temperature of the hotter side and TC is the temperature of the cooler side iii K E thermal conductivity of the material see Table 113 in textbook gt K JsmOC WmOC iv The larger K is the better the material is in conducting heat v The smaller K is the better the material is as a thermal insulator vi The effectiveness of thermal insulation is rated by another quantity gt thermal resistance R Where L g 7 Where in the US R ftghoFBTU h7 hour and elsewhere R m20CW see Table 114 in the textbook R E XII16 Xll 16 PHYS 2010 General Physics Example X1176 Problem 1135 Page 381 from the Ser way amp Faughn textbook A steam pipe is covered with 150 cm thick insulating material of thermal conductivity 0200 calcmquotCs How much energy is lost every second when the steam is at 2000C and the surrounding air is at 200 C The pipe has a circumference of 800 cm and a length of 500 m Neglect losses through the end of the pipe Solution Listing our given parameters we rst need to convert some of these input parameters to SI units cal 102 cm 4186 J J K 2 837 0 00cm Cslt1mgtlt1calgt msquotC Th 2000C TC 200C C 800 cm 800 m is the circumference of the pipe 6 500 m is the length of the pipe and the thickness of the insulation is L 150 cm 00150 In Since the problem gives us the circumference of the pipe the total surface area of the pipe is this circumference times the length of the pipe A or 800 m500 m 400 m2 Using Eq XII 15 then gives the heat loss rate via conduction as T Tc pcond J 2 20000 2000 83 7msocgt400mlt 00150m gt 402 x 108 Js 402 MW Donald G Luttermoser E TSU 3 Heat Transfer by Convection a b c When an ensemble of hot particles move in bulk to cooler regions of a gas or liquid the heat is said to ow via convection Boiling water and cumulus clouds are 2 examples of con vection Convection is complicated requiring graduate level physics and math to describe it i Convection will occur if an instability occurs in the gas or liquid ii The mixinglength theory is often used to de The heat of a gasliquid bub ble will glue up its heat energy to the cooler sur roundings after the bubble has traveled one missing length scribe convection 4 Heat Transfer by Radiation a 0 Of all the heat energy transport mechanisms only radia tion does not require a medium gt it can travel through a vacuum The rate at which an object emits radiant energy is given by the StefanBoltzmann Law 735m aAeT4 XII17 i 735m E power radiated emitted watts ii a E Stefan Boltzmann s constant 56696 x 10 8 WmZKA Xll 17 Xll 18 PHYS 2010 General Physics iii A E surface area of the object mg iv e E emissivity unitless e 1 for a perfect ab sorber or emitter v T E temperature c A body also can absorb radiation If a body absorbs a power of radiation 73135 it will change its temperature to To i The net power radiated by the system is then prad pnet pen abs or PM UA T4 aAeTj 73ml aAe T4 T04 XII19 ii In astronomy the total power radiated by an ob ject over its entire surface is called the luminosity L 73ml of the object Since the amount of energy falling on the surface of the Sun or any isolated star from interstellar space is negligible to that of the power radiated L 735m for isolated stars iii If an object is in equilibrium with its surround ings it radiates and absorbs energy at the same rate gt its temperature remains constant gt this ra diative equilibrium results in the object being in thermal equilibrium 73ml0 gt TTO where T0 is the temperature of the surroundings Donald G Luttermoser ETSU Xll 19 d An ideal absorber is de ned as an object that absorbs all of the energy incident upon it i In this case emissivity e 1 ii Such an object is called a blackbody iii Note that a blackbody radiator can be any color depending on its temperature gt red blackbod ies are cooler than blue blackbodies it does not appear black unless it is very cold iv The energy ux of such a radiator is be 7 0T4 v This radiative ux results from the condition that in order to be in thermal equilibrium the heat gained by absorbing radiation must be Virtually immediately radiated away by the object vi This is not the same thing as reflecting the radi ation off of the surface which does not happen in a blackbody The incident radiation does get ab sorbed by the atoms of the object and deposited in the thermal pool It is just that this radiation immediately gets re emitted just after absorption XH 20 PHYS 2010 General Physics Example X1177 Problem 1142 Page 382 from the Serway amp Faughn textbook The surface temperature of the Sun is about 5800 K Taking the Sun39s radius to be 696 X 108 m calculate the total energy radiated by the Sun each second Assume e 0965 Solution We will assume the Sun s shape to be a sphere where the surface area of a sphere is A 47ng 477696 x 108 m2 609 x 1018 m2 Making use of Eq XII 17 the total power energy per second that the Sun radiates to space is W 735m aAeT4 567 x 108 K4 609 x 1018 m209655800 K4 m2 377 x 1026 W This is the value of the Sun s luminosity LG 735m 377 x 1026 W PHYS2010 General Physics I Course Lecture Notes Section VIII Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn VIII Circular Motion A Angular Speed and Angular Acceleration 1 Arc length of a rotating or revolving object is analogous to linear displacement a Arc length 3 is swept out as an object rotates 6 degrees such that Reference i s is measured in the same units as 7 e a length ii 6 is measured in radians not degrees 1 revolution rev 27F radians rad 360O VIII 2 or 3600 27F 1 rad E 5703 VIII 1 Vlll 2 b C d PHYS 2010 General Physics Conversion of angles 639 E angular displacement 6 rad W F00 639 deg 1113 0 rad i745 x 1042 0 deg Sometimes you may run across angular measure dealing with very small angles There are 60 arominutes aromin per 10 and 60 aroseoonds aroseo per 1 ar ominute 1 60 3600 An object is said to be rotating if it is spinning about an aXis 69 the Earth rotates about its axis quotspinsquot Donald G Luttermoser ETSU VIII 3 e A body revolves around another object if the rst body is in orbit about the second object 69 the Earth re volves about the Sun quotorbitsquot Sun Eth at Sun Example VIIIil Problem 72 Page 218 from the Ser way amp Faughn textbook A wheel of radius 41 meters rotates at a constant velocity How far path length does a point on the Circumference travel if the wheel is rotated through angles of 30quot 30 rad and 30 rev Let 7 be the radius of the wheel 3 be the path length along the circumference and 6 be the angle that this point subtends from a reference point as the wheel rotates As such 7 41 In and 6 1 30 6392 30 rad and 03 30 rev We will use Eq VI l 36 r except that all of these angles must be expressed in radians so 7T rad 6391 300 X 1800 05236 rad 6392 30 rad which is what we want 27F rad 6 3 30 rev X 7 1885 rad 1 rev VIII 4 PHYS 2010 General Physics Plugging these values of 6 into Eq Vl l gives 01 05236 41 m admin 03 1885 41 m 81 21 m 82 83 2 The length 3 in Eq Vlll l is actually an arc length curved length However if 6 is small 216 6 ltlt 1 s gt D the actual linear size or diameter if the object is round a b c As such by measuring the angle that a distance object subtends we can calculate its actual size or diameter D by knowing the distance d to the object Since 6 is small here it is often more convenient to express this angle in terms of arcseconds instead of radians We can rewrite Eq Vlll l by setting 3 D 7 d and expressing 6 in arcseconds and relabeling it as oz then 7T rad 1O 0 d 7 Ia 1800 X 3600 arcsec aarcseconds X 7 oz arcsec i 206265 arcsecrad Using oz instead of 6 in Eq Vlll l gives 7 ad 206265 which is called the smallangle formula Here D is the VIII4 linear size ie diameter of the object at a distance d which subtends an angle oz measured in arcseconds the Donald G Luttermoser ETSU VIII 5 206265 is the conversion factor between arcseconds and radians The lengths d and D will be in the same units 69 if d is measured in km then D will be in km Note that this formula is only valid when 6 ltlt 1 typically one would want oz lt 1000 arcseconds in order to use this small angle approximation Example VIII72 The star Betelguese oz Ori has an angular size of 0040 arcsec 40 milliarcseconds 40 mas and it is at a distance of 200 pc What is the linear size of Betelgeuse How does this size compare to the planet39s distances from the Sun in our solar system d 200 pc x 309 x 1016 mpc 618 x 1018 m D 0040 x 618 x 1018 m 206265 120 x 1012 m The Sun is 139 X 109 m in diameter which means that Betelgeuse is i 120 x 1012 m 7 139 x 109 mDG where an Astronomical Unit AU is the size of the Earth s orbital semimajor axis note that 1 AU 1496 X 1011 m Betelgeuse is 860 times bigger than the Sunl This diameter gives a stellar 860 D9 80 AU radius of 40 AU for Betelgeuse If it was put in the place of the Sun then Mercury 039 AU Venus 072 AU Earth 10 AU and Mars 15 AU would be inside Betelgeusel The planet Jupiter 52 AU would be close 12 AU from the photosphere surface of Betelgeuse 3 Angular speed 01 is the change of angular displacement di vided by the time interval that the angular displacement took place VIII 6 PHYS 2010 General Physics a Average angular speed i 02 01 A0 7 VIII 5 w 152 t1 At gt b Instantaneous angular speed A6 1 Aliran Kt VIII 6 c Angular speed has units of radians second gt rad s or just s 1 since radian is a pseudo unit d w is positive when 6 is increasing counter clockwise mo tion and negative in the clockwise direction 4 Angular acceleration oz alpha is the change of angular speed over the time interval of interest a E change in angular speed wg 11 E VHLU time interval 152 t1 At which is the average angular acceleration a Whereas the instantaneous acceleration is Aw oz Aliran E VIII 8 b The units of angular acceleration are rads2 or just s 5 When a rigid object rotates about an axis every portion of the object has the same angular speed and same angular acceleration 6 Note that we have Lot written then angular variables as vectors In reality they are vectors but the unit vectors for the magni tudes we have written above point out of the plane in which the object is rotating Since this is a bit cornplicated we will just discuss the magnitudes of these angular motions at present Donald G Luttermoser ETSU VIII 7 B Rotation Motion under Constant Angular Acceleration 1 These are analogous to their counterparts in linear motion LINEAR ANGULAR 1 2 1 2 06 vat Eat 6 wot Eat VIII 10 Note that in the above equations the initial tiInes angles and displacements are all set to zero If these initial setting are not zero replace t with At 6 with A0 and x with A96 2 The tangential speed of a point on a rotating object equals the distance of that point from the axis of rotation 7 multiplied by the angular speed w A A0i 76 71 ef gy 1 A0 i 11 As AlIBOXt i AlIEOKt7 1 w iv 7 vt raj VIII 12 VIII 8 PHYS 2010 General Physics where v is the tangential velocity tangent to the curved path of motion the particle is taking 3 The tangential acceleration at of a point on a rotating object equals the distance of that point from the axis of rotation 7 multiplied by the angular acceleration oz Av rAw Am 7 7amp At 7 At Rf Rio 1 AU 7 1 Aw AltIBOTt i r AltIBOXt at roz VIII 13 C Centripetal Acceleration and Force 1 If an object travels at constant speed in a curved path it is accelerating since the velocity vector is continuously changing direction 2 Besides the tangential acceleration e acceleration tangent to the curved path there must be an acceleration perpendicular l to the tangent line pointing toward the center of the curved path Donald G Luttermoser E TSU 3 b C VIII 9 Otherwise the object would y off in a straight path in the direction of at as per Newton s 1st law of motion This center seeking acceleration is called the centripetal acceleration VIII14 i Gravity is a centripetal force hence acceleration for planets orbiting the Sun ii A string s tension is a centripetal force hence ac celeration for an object attached to that string be ing rotated in circular motion rock at 03 The total acceleration of an object in circular motion is d 1 in VIII15 Vlll 10 PHYS 2010 General Physics Example VIII73 Problem 718 Page 219 from the Ser way amp Faughn textbook A race car starts from rest on a circular track of radius 400 m The car s speed increases at a constant rate of 0500 ms2 At the point where the magnitudes of the centripetal and tangential accelerations are equal determine a the speed of the race car b the distance traveled and c the elapsed time Solution a From Eq VIII 14 the centripetal acceleration is aC Thus when aC at 0500 ms2 we have u M M400 m0500 ms2 Solution b The total distance traveled is just the total arc length traveled on the circular track 3 which can be found from the linear 1 D equation of motion 1 U 2 a s where v v 141 ms 1 0 starts from rest and a at 0500 ms2 Hence the total arc length re distance traveled is 2 2 2 141 0 5 U lt 18 2 200m 2at 20500 ms Solution c We can use another linear 1 D equation of motion to nd the elapse time u 1 at where v v 141 ms 1 0 starts from rest and a at 0500 ms2 Hence the time it take to reach the tangential Donald G Luttermoser ETSU Vlll ll velocity from part a is i 141 0 at 0500 ms 3 The centripetal force is just the mass times the centripetal accel eration 2 Fcmac quotWt mm VIII17 7 a Centripetal forces are center seeking forces i The gravitational force is a center seeking force gt planets are in orbit about the Sun ii The tension force is a centripetal force for an ob ject connected to a string being swung about some axis b The concept of centrifugal force the apparent force that forces an object outward while moving in a circu lar path is Lot really a force gt this tendency to follow a straight line path away from the curved path is nothing more than Newton s 1st law the law of inertia Example VIIIi4 Problem 748 Page 221 from the Serway amp Faughn textbook A 0400 kg pendulum bob passes through the lowest part of its path at a speed of 300 ms a What is the tension in the pendulum cable at this point if the pen dulum is 800 cm long b When the pendulum reaches its highest point what angle does the cable make with the vertical c What is the tension in the pendulum cable when the pendulum reaches its highest point Vlll 12 Solution a PHYS 2010 General Physics rL800cm The diagram above shows the system when the bob is at its low est point and shows the direction of the forces involved and the tangential velocity vector The mass of the bob is m 0400 kg and the length of the cable as L 800 cm 0800 In Since the bob follows a circular path about the pivot of the pendulum the radius of this path is just the length of the cable 7 0800 In The centripetal force is always directed toward the center of the orbit which we will de ne as the positive y direction as indi cated in the gure above As a result the sum of the forces in the y direction is just the total centripetal force Solving for the tension T gives T 842 N ZFyFcTFg FgFCmgmacmgac 2 m g 0400 kg 980 ms2 300 Ins2gt 0800 In Donald G Luttermoser ETSU Vlll 13 Solution b L39Yf vf0 quotYr Yf yi0 vi 300 ms The diagram above shows the bob at its maximum height To nd this height hence 0m we only need to use the conservation of mechanical energy We will set the lowest position of the bob at the ground level yi 0 At its highest position yf the velocity changes from the counter clockwise to the clockwise direction hence goes through zero at this point of 0 From the diagram above we can solve for yf in terms of 0m Lyf cos rm L L cos 6mm L yf yf L L cos 0m yf L 1 cos6 m We can now determine 0m from the conservation of mechanical energy by setting the lowest point as the initial position so 1 vt from part a and yi 0 and the highest point as the nal position so of 0 and yf is given in the equation above KE PEg KE PEgf i 1 2 1 2 Emvimgyi Emvfmgyf V111 14 PHYS 2010 General Physics 1 7m 0 0 mgL 1 cosem 2 1 2 mgL 1 cos 0m Emut 2 v 1 0m t cos 2g 2 cos m 1 Ut 29 2 0m cos 1 1 0081 1 300 ms2 gt 2980 ms20800 m cos 104260 Solution c The diagram above shows the forces acting on the bob at the highest point with respect to an arbitrary coordinate system cho sen such that the tension and total centripetal force points in the positive y direction From this we now construct a free body diagram Donald G Luttermoser ETSU Vlll 15 yn At T 5 6max Fg Since the tangential velocity at this highest point is zero as men tioned in part b the centripetal force must be zero 2 U E mac m it 0 7 Using this equilibrium equation in conjunction with the free body diagram above we get Fe Z Fy T Fm 0 T mgcos m T mg cos 0max 0400 kg980 n1s2 cos648o 167 N PHYS2010 General Physics I Course Lecture Notes Section XIV Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn XIV Thermodynamics A The area of physics concerned with the relationships between heat and work is thermodynamics gt the study of the motion of heat B The First Law of Thermodynamics 1 ln words The change in internal energy ofa system equals the difference between the heat taken in by the system and the work done on the system a When an amount of heat Q is added to a system some of this added energy remains in the system increasing its internal energy by an amount AU b The rest of the added energy leaves the system as the system does work W 2 Mathematically note that AU Uf U1 and the First Law states 3 ln thermo there will always be two speci c regions in which we will be interested in a The system the region of interest where we wish to know the state parameters 69 P T and V Ther modynamic variables relating to the system will remain unsubscripted in these notes e W work on the sys tem XIV 1 XIV 2 PHYS 2010 General Physics the re gion that contains the system Note that some scientists b The environment also called the universe including me prefer to use the word environment over universe since in astronomy the word Universe means all of the cosmos and nQot just the immediate ie nearby surroundings which is what thermo means by universe Thermodynamic variables relating to the environment will be labeled with the env subscript in these notes ie WenV work on the environment c Note that work done Q the system W is the exact oppo site of work done Q the environment W Wenv 39 Note that we will often be interested in systems that are com pletely isolated from the environment Such a system is called a closed system Using the de nition of work Eq Vl l and pressure Eq Xl l we can show how work is related to the change in volume of gas a Work will increase in small increments AW as a piston in a cylinder moves by a length Ax then AW F Ax Ax A P AV XIV2 note that the negative sign in this relation results from the fact that when the environment does work Q the cylinder ie the system AV becomes smaller hence AV lt 0 due to the piston s motion As such the work done on the system is positive W gt 0 Since A06 is a very small length in this equation P will not change by much As such P is considered to be constant in this equation Donald G Luttermoser E TSU b The total work done W is then just the sum of all of the incremental increases in work WZAWZ ZPZAVZ XIV 3 However the shorthand notation of W PAV XIV 4 is often used for Eq XIV 3 and we will use this notation here c A gas lled cylinder with a piston is the simplest form of a heat engine Such heat engines are used to drive more complicated machines such as automobiles 6 For a monatomic ideal gas 16 the gas particles consist of a single atomic species recall Eq XIII 16 where the internal energy is given by U gnRT a The change in internal energy for such a gas is then given by AU SydEAT XIV5 assuming no gas escapes or is brought into the system b We will now introduce a new thermodynamic term call the molar speci c heat at constant volume for a monatomic ideal gas 0 E R XIV6 3 2 c The change in internal energy of an ideal gas can then be written as AU nCvAT XIV7 XIV 3 XIV 4 PHYS 2010 General Physics d For ideal gases this expression is always valid even when the volume isn t constant The value of the molar speci c heat however depends on the gas eg the numbers of degrees of freedom the gas can have and can vary under different conditions of temperature and pressure 7 There are four speci c types of processes involving energy in ther modynamics a Isobaric processes Pressure remains constant during the process The full form of the 1st law is used in these cases i Under these conditions the thermal energy trans ferred into the gas e system is given by rear ranging the terms in the rst law QAU WAUPAV where here we have made use of Eq XIV 4 in this relation ii Using Eq XIV 5 and Form 1 of the ideal gas law ie PAV nRAT in this relation gives 5 Q gnRAT nRAT EnRAT iii Another way to express this transfer of heat is Q nCpAT XIV 8 where 010 R is the molar speci c heat at constant pressure iv For ideal gases the molar heat capacity at con stant pressure is equal to the sum of the molar spe ci c heat at constant volume and the universal gas Donald G Luttermoser E TSU b constant 010 CU R XIV 9 Adiabatic processes Heat does not enter or leave the system only occurs for closed systems or e ectively closed systems A good example of an adiabatic process is anything that happens so rapidly that heat does not have time to ow in or out of the system during the process Then Q 0 and Eq XV l becomes AU W adiabatic processes i If the system does work on the environment W lt 0 negative and AU lt 0 negative gt the sys tem uses its internal energy to supply work to the outside environment note here that WenV gt 0 as a result of this ii If the system requires work from the environment W gt 0 positive and AU gt 0 positive gt the environment supplies work hence WenV lt 0 to the system and increases the systems internal energy as a result iii An ideal gas undergoing an adiabatic process obeys the following relation PVl constant XIV 10 with CP 7 XIV 11 7 CU where 7 is called the adiabatic index of the gas See Table 121 in the textbook for values of GP CU and 7 for various species XIV 5 XIV 6 C d PHYS 2010 General Physics Isothermal processes Temperature does not change during the process Processes that occur gradually can be considered isothermal since the temperature changes very slowly during process In isothermal processes AU 0 and Eq XV l becomes W c2 isothermal processes i Using methods in calculus it can be shown that the work done on the environment during an isother mal process is Vf WenV nRT ln XIV 12 where ln is the natural log function ii Since work on the system is just the negative of work on the environment Vf W i nRT ln i XIV13 Isochoric processes also called isovolumetric pro cesses Volume does not change during the process Since volume does not change no work can be done on the system As such Eq XV l then becomes AUQ isochoric processes i In an isochoric process the change in thermal en ergy of a system equals the energy transferred to the system by heat ii From Eq XIV 5 the energy transferred by heat in constant volume processes is given by Q 710va XIV14 Donald G Luttermoser E TSU 8 9 XIV 7 Enthalpy Heat energy associated with phase changes a b c Through the use of the 1st law of thermodynamics we can introduce a new state variable associated with phase changes Eq XII 12 showed the heat liberated or needed during a phase change is related the latent heat Usually there is also a volume change when matter changes from one state to another A volume change results in work either being performed on or by the system via Eq XIV 2 Using this equation along with Eq XII 12 in the 1st law e Eq XIV l we can write Ug U1mL PV2 V1 XIV 15 i Solving this equation for the latent heat term gives mLU2PV2 U1PV1H2 H1 XIV16 H U PV is called the enthalpy of the system where XlV 1 7 ii The enthalpy H is another state variable which is a function of the 3 other state variables T through U P and V As can be seen the latent heat of fusion and vaporization is just the difference of the enthalpy of the two different states of matter on either side of the fusion or vaporiza tion Up until now we have considered only processes in which the potential and macroscopic kinetic energies of a system remain XIV 8 PHYS 2010 General Physics constant We now relax this constraint to develop the general form of the 1st law of thermodynamics a The work energy theorem e Eq Vl 4 says that the work done on a system is equal to the change of kinetic energy AKE KEf KEW XIV18 b This kinetic energy change should be added to the in ternal energy change to represent the total kinetic energy gt both macroscopically e bulk kinetic energy microscopically e internal energy AU AKE Q W XIV19 c If conservative forces act on the system the work done by the conservative forces We is related to a change in the potential energy through Eq VI 6 WC APE XIV20 1 Note that the total work done on a system is the sum of the work due to conservative forces and non conservation forces labeled with Wm W WC Wm XIV21 e Using Eq XIV 21 along with Eq XIV 20 in Eq XIV 19 we can write AU AKE Q WC Wm Q We Wm AU AKE WC Q Wm AU AKE APE Q Wm XIV 22 f Now if we de ne the total energy E of the system as the sum of the internal energy the kinetic energy and the Donald G Luttermoser ETSU XIV 9 potential energy E U KE PE XIV 23 then we can write the general form of the rst law of thermodynamics as AE E E Q Wm XIV24 noting that if both the kinetic energy and potential energy are constant then AE AU and Vlm W which gives us the standard form of the 1st law as written in Eq XIV l Example XIVil Problem 128 Page 419 from the Serway amp Faughn textbook A movable piston having a mass of 800 kg and a cross sectional area of 500 cm2 traps 0200 mol of an ideal gas in a vertical cylinder If the piston slides without friction in the cylinder how much work is done on the gas when its temperature is increased from 200C to 3000C Solution For an ideal gas as the temperature increases the volume increases by nRTf nRTi AV Vf Vi 7 P 39 Assume the pressure is held constant B Pf P so we can write nR nRAT AVTf ET The change in temperature is AT 300 273 K 20 273 K 280 K The work done on the gas is W PAV nRAT 0200 mol831 Jmol K280 K 465 J XIV 10 PHYS 2010 General Physics Example XIV72 Problem 1220 Page 420 from the Serway amp Faughn textbook A thermodynamic system undergoes a process in which the internal energy decreases by 500 J If at the same time 220 J of work is done on the system find the energy transferred to or from it by heat Solution From the rst law as given in Eq XIV 1 QAUW 500J220J The negative sign in the result means that energy is transferred from the system by heat C Heat Engines 1 Many of the machines used in our current technology are driven by heat engines indeed your own body is a heat engine A heat engine is a device that converts internal energy into some other form of energy 69 electricity mechanical kinetic energy etc 2 Let s represent a typical heat engine as a cylinder with a piston in it The total work done by a heat engine is then the net heat ow into the cylinder via the 1st law of thermodynamics note that AU 0 for the complete cycle W IQHI chl a XIV25 where is the magnitude hence a positive number of the heat that owed into the system re engine and QC is the magnitude hence a positive number of the heat that owed out of the system Donald G Luttermoser ETSU XIV ll 3 Reversible versus non reversible processes a A process is said to be reversible if the nal state of a process can be returned to its initial state i No dissipative effects that convert mechanical en ergy to thermal energy e friction ii Such processes are nearly impossible to achieve in nature b A process is irreversible if the nal state of a process cannot be returned to its initial state gt nature behaves in this manner 4 The thermal ef ciency 6 of any system is the ratio of the work W a Using Eq XIV l4 we immediately see that done to the heat input 6 lQHl IQCI XIV27 IQHl b Kelvin showed through the ideal gas law that chl To 7 7 XIV 28 IQHI TH l where these temperatures are measured in K c Using Eq XIV 28 in Eq XIV 27 we see that the ther mal ef ciency of an ideal heat engine is T e l 0 ideal engine XIV 29 TH XIV 12 PHYS 2010 General Physics 1 As can be seen from this even with an ideal engine one with no internal friction it is impossible to make a 100 ef cient engine gt it is a physical impossibility to make a perpetual motion machine machines that require no energy to run and never run down 5 There are two things to always keep in mind when doing calcu lations with heat engines a W is positive and WenV is negative when the system contracts 239e AV lt 0 and negative and WenV posi tive when it expands 239e AV gt 0 b Q is positive for heat coming into the system and negative for heat leaving the system Example XIV73 Problem 1225 Page 420 from the Serway amp Faughn textbook The energy absorbed by an engine is three times greater than the work it performs a What is the thermal efficiency b What fraction of the energy absorbed is expelled by the cold reservoir Solution a Using Eq XIV 26 and the fact that 3W the thermal ef ciency IS W W 1 77 7 333 333 6QH3W 30 or Using Eq XIV 27 we can write lQHl chl chl iQHl IQHI l e Solution b iQHi Donald G Luttermoser E TSU XIV 13 D The Carnot Cycle 1 In the early 1800s Carnot pointed out the basic working of an ideal one without internal friction heat engine The Carnot cycle see Figure 1215 in your textbook can be described in 4 steps a b Step 1 The cycle starts with the piston positioned such that V is at a minimum At this point heat Q is added to the system through a heat reservoir at a high temperature TH The system absorbs the heat a constant temperature TH which causes the volume to expand doing work on the piston During this time the system s internal energy does not change AU 0 and since T is not changing it is an isothermal process From the 1st law of thermo the work done bl the system W lt 0 is equal to the negative of the heat input QH Step 2 The cylinder is moved off of the heat reservoir onto an insulator This isolates the heat in the system which makes this step adiabatic The load on the piston is reduced which allows the gas to expand in the cylinder hence the volume continues to increase This expansion of gas does work on the system which is the same thing as saying that the work is done bl the system W lt 0 caus ing the internal energy to go down since Q is constant here During this phase of the expansion the tempera ture decreases XIV l4 PHYS 2010 General Physics c Step 3 The cylinder is then moved to a heat sink at a cooler temperature To The internal heat flows out Q0 of the system at constant temperature hence an isothermal process The volume decreases which causes work to be done Q the system W gt 0 1 Step 4 The cylinder is moved back to the insulator The load on the piston is increased and the gas undergoes adiabatic compression Since volume is still decreasing work is still being done the system W gt 0 The internal energy returns to the value that it had at the start of Step 1 3 The Carnot engine is said to be a reversible engine E Refrigerators and Heat Pumps 1 Reversing the Carnot cycle we can put work into a system and transfer heat from a lower temperature to a higher one gt a refrigerator a The heat extracted from the cold reservoir to the work supplied similar to the thermal ef ciency is called the coef cient of performance 17 77mg XIV 30 b Using Eq XIV 25 and Eq XIV 28 we can write lQCl To 1 XIV31 mfg lQHl lQCl TH To gt 2 Air conditioners are refrigerators that move heat from within a house to outside a house Donald G Luttermoser ETSU XIV 15 3 The reverse of an air conditioner is a heat pump gt it cools the outdoors by delivering heat from the outdoors to the inside a The coef cient of performance for a heat pump is given by 171 XIV32 b Using Eq XIV 25 and Eq XIV 28 we can write IQH l TH XIV33 nhp IQHI IQCI TH To F The Second Law of Thermodynamics i The Classical Descrip tion 1 The second law of thermodynamics deals with how heat ows It is essentially a description of change a Change To make different the form nature and content of something b Change has over the course of time and throughout all space brought forth successively and successfully galax ies stars planets and life c Evidence for change is literally everywhere i In the macroscopic domain as viewed through tele scopes The Universe started out being fairly ho mogeneous ie the same everywhere until today it is very inhomogeneous Heavy elements ie C N 0 Fe etc have become more abundant over time Stars have formed from a few billion years af ter the Big Bang until the present epoch and will continue to form in the far distant future XlV 16 O PHYS 2010 General Physics ii In the mesoscopic domain as deduced from geo logic records and Via microscopes studying the bi ological world iii And in the microscopic domain as studied with high energy particle accelerators 1 Much of the change is subtle such as when the Sun fuses hydrogen into helium sedately over billions of years or when the Earth s tectonic plates drift sluggishly across the face of our planet over those same billions of years Indeed our perception of time is nothing more than our noticing changes on Earth and in the Universe as a whole 8 There are two classical formulations of this law both essentially mean the same thing a Clausius statement of the second law Heat cannot by itself pass from a colder to a warmer body b KelvinPlanck statement of the second law It is im possible for any system to undergo a cyclic process whose sole result is the absorption of heat from a single reser voir at a single temperature and the performance of an equivalent amount of work The 2nd law speci es the way in which available energy also called usable energy free energy or potential energy change occurs a This law s essence stipulates that a price is paid each time energy changes from one form to another Donald G Luttermoser E TSU b The price paid to Nature is a loss in the amount of available energy capable of performing work of some kind in the future c We de ne here a new term to describe this decrease of available energy entropy S It is derived from the Greek word tropae which means transformation 4 The second law of thermodynamics is different from the laws of mechanics It does not describe the interactions between in dividual particles but instead describes the overall behavior of collections of many particles a In classical mechanics an event is symmetric in time gt the laws are satis ed whether we run the experiment for ward or backward in time b The second law tells us about the sequence or order in which events naturally take place gt the second law of thermodynamics shows us the direction in which time progresses 5 The concept of entropy a Clausius introduced a new state variable called entropy b As de ned above Entropy is a measure of how much energy or heat is mavailable for conversion into work c When a system at temperature T in K undergoes a re versible process by absorbing an amount of heat Q its increase in entropy AS is AC2 i Q AS T i T XIV34 XIV 17 XlV 18 d 8 PHYS 2010 General Physics where often we simply write Q for the heat input or re moved instead of the more proper form of AQ i If Q gt 0 gt heat is absorbed by the system gt entropy increases ii If Q lt 0 gt heat is expelled by the system gt entropy decreases iii For reversible processes Q 0 and the entropy change is zero As can be seen by Eq XIV 34 entropy when multiplied by temperature is a measure of the amount of energy Q no longer capable of conversion in useful work We can use this concept of unavailable energy versus anall able energy to develop a different form for the rst law of thermodynamics We can express the total energy E see Eq XIV 23 as E F T3 XIV 35 where TS l e temperature times entropy represents the energy that is unavailable to be converted to work and F called the Helmholtz free energy is the energy avail able to be converted to work i The rst law can then be written by making this a difference equation AE AF TAS SAT XIV 36 where we have made use of the product rule from differential calculus for the TS term ii Eq XIV 36 shows that if the temperature change over time is small or zero which is typically the Donald G Luttermoser E TSU XIV 19 case for an isolated or closed system the conserva tion of energy dictates that as free energy decreases the energy unavailable for conversion to work in creases gt the entropy increases iii Given enough time all closed systems will run down until they are capable of performing no more work gt once again it is impossible to make a perpetual motion machinel G The Second Law of Thermodynamics i The Probabilistic De scription 1 In the late 1800s Boltzmann showed that an increase in entropy of a system corresponds to an increased degree of disorder in the atoms or molecules composing the substance a This realization lead to the creation of a new area of 0 physics known as statistical mechanics which rewrites the laws of thermodynamics in a probabilistic formalism Boltzmann rewrote the entropy equation de nition in terms of a probability equation S CBan XIV 37 where kg is Boltzmann s constant ln is the natural log arithm of base 6 and W is the number of different ar rangements of microscopic states e positions veloci ties compositions and any various arrangements of quan tum properties Note that W here does Lot mean work i Boltzmann actually has Eq XIV 37 carved on his gravestone in Vienna XIV 20 C d PHYS 2010 General Physics ii W is a measure of the inverse probability p of the occurrence of the possible microscopic states a system can have iii That is p lW 1 for a 100 chance gt Then it is completely certain that a process will occur when p 1 iv As a result we could also write the de nition of entropy e Eq XIV 37 as S kBln kBlnp XIV 38 With the help of Eq XIV 38 we can see the second law implies that any isolated system naturally tends towards an equilibrium state of minimum microscopic probability namely a uniformity of temperature pressure chemi cal composition and so on i Since ordered molecular states for example where molecules in one part of the system have one prop erty value but those in the remaining part have another are less probable than those of random or disordered states ii Boltzmann s law of entropy then signi es that or dered states tend to degenerate into disordered ones in a closed system As can be seen the concept of disorder is very di icult since it requires a detailed knowledge of probability and statistics i The best way to learn about probabilities is through example Let s say we have one die from a set of Donald G Luttermoser E TSU dice There are 6 sides with dots imprinted on the sides relating to the numbers 1 2 3 4 5 and 6 ii The probability of one number say 4 coming up is n a given state 7 XIV 39 p N total number of states 7 l and for this case n 1 and N 6 so p one side i 1 0167 7 total number of sides T 6 or a 167 chance that we would role the die with a 4 landing on top iii The probability of an even number 2 4 or 6 landing on top is n 3 and N 6 3 75 p 6 00 or a 50 chance to role such a number 2 Through this concept of entropy we can rewrite the second law of thermodynamics as any of the following statements a The entropy of the Universe as a whole increases in all natural processes b lsolated systems tend towards greater disorder and en tropy is a measure of that disorder c In a closed system entropy increases over time gt less and less energy can be converted into work 1 All of these statements are probabilistic in nature gt on average this is true XIV 21 XlV 22 PHYS 2010 General Physics Note that the second law written in this probabilistic way can be violated locally gt entropy can decrease locally Only over the whole isolated or closed system over a long enough period of time will necessitate an increase in entropy a Note that on the Earth entropy decreases all the time at the expense of an increase of entropy of the Sun b As such the second law of thermodynamics cannot be used as proof against the theory of biological evolution as some people have suggested The bottom line is that we no longer regard things as xed or being or even that they exist lnstead everything in the Universe is owing always in the act of becoming All enti are permanently changing ties living and non living alike Example XIV74 Problem 1237 Page 420 from the Serway amp Faughn textbook A 70 kg log falls from a height of 25 m into a lake If the log the lake and the air are all at 300 K find the Change of entropy of the Universe for this process Solution The potential energy lost by the log is carried away by heat that is the kinetic energy of the impact into the lake get converted into heat so AQ APE mgAy mgh 70 kg980 ms225 m 17x 104 J We will ignore air friction here and since the temperature remains con stant we just have one term in the entropy equation As such the change of entropy is AQ 17x104J AS T 300K PHYS2010 General Physics I Course Lecture Notes Section XI Dr Donald G Luttermoser East Tennessee State University Edition 23 Abstract These Class notes are designed for use of the instructor and students of the course PHYS 2010 General Physics I taught by Dr Donald Lutternioser at East Tennessee State University These notes make reference to the College Physics 7th Edition 2005 textbook by Serway and Faughn XI Solids and Fluids A States of Matter 1 Matter exists in 3 different states a A solid is a rigid body gt takes a lot of energy to change its shape Solids can be classi ed into two types i Crystalline solids have atoms that are struc tured in an orderly fashion ii Amorphous solids have randomly arranged atoms b A liquid is uid in nature gt moderate energy required to change its shape c A gas is also uid in nature gt little energy required to change its shape 2 If a gas gets hot enough electrons circling the nucleus of the atoms in the gas are ripped away from the nucleus gt the gas becomes ionized gt ionized gas is called a plasma 3 Matter consists of a distribution of particles atoms and molecules a Atoms consist of a nucleus surrounded by electrons which are negatively charged The nucleus consists of pro tons positive H charge and neutrons no 0 charge b H He Li Be B C N 0 etc are the elements of the periodic table atoms The number of protons in the nucleus de nes each atom c For elements heavier than H typically the number of neu trons is equal to the number of protons Isotopes of X11 XI 2 PHYS 2010 General Physics atoms contain different numbers of neutrons in the nu cleus 69 12C 13C and 14C are isotopes of carbon 1 Neutral Atoms protons electrons i If electrons are taken away from the atom such that the number of protons exceeds the number of electrons the atom becomes a positive ion 69 H H H singly ionized hydrogen ii If the number of electrons exceeds the number of protons in the nucleus the atom becomes a neg ative ion 69 H negative hydrogen ion a hydrogen atom with two electrons instead of one iii The ionization stage of an atom can be labeled in a variety of different ways gtRoman numerals l neutral atom 69 He 1 neutral helium ll singly ionized 69 Fell lll doubly ionized 69 C III etc gt exponents positive ions if no expo nent appears then we have a neutral atom 69 He neutral helium singly ion ized 69 Fe 3 triply ionized 69 C 3 etc gt exponents negative ions one ex tra electron 69 H two extra elec trons 69 Cquot etc e Molecules are a collection of atoms that are bound to gether by molecular bonds i Salt NaCl 1 sodium atom 1 chlorine atom Donald G Luttermoser ETSU XI 3 ii Water H20 2 hydrogen atoms 1 oxygen atom iii Methane CH4 1 carbon atom 4 hydrogen atoms f Molecules can adhere to each other through chemical bonds making a structured lattice gt solids B The Deformation of Solids 1 The amount of shape change deformation for a given amount of energy supplied to a solid is called the elasticity of the solid a Stress is related to the force causing the deformation t force F s ress E E area A b Strain is a measure of the degree of deformation change in shape stra1n E 1n1t1al shape c For small stresses strain is directly proportional to stress stress elastic modulus stra1n 2 Young s Modulus Elasticity in Length a Tensile stress is the ratio of the magnitude of the ex ternal force applied F to the cross sectional area of the object A b Stress is the same thing as pressure P P E E Xl l A lt gt where pressure is measured in pascals Pa lPa ElNmg 3 c Young s Modulus tensile stress i FA i FLO tensile strain i ALLO i AAL gt force is J to the cross sectional area 1 PHYS 2010 General Physics XI 2 1 Shear Modulus Elasticity in Shape shear stress 7 FA Axshear strain i Ageh s bottom face is fixed XI 3 Donald G Luttermoser ETSU XI 5 where S E Pa gt force F is tangential to the cross sectional area A 4 Bulk Modulus Elasticity in Volume volume stress F A AP B E Xl 4 volume strain AV V AV V l l where B E Pa gt force exerted in all directions 16 squeez ing a The negative sign means that an increase in pressure AP results in a decrease in volume AV b is called the compressibility of the material 5 Both solids and liquids have bulk moduli 6 Only solids have shear and Young s moduli liquids can ow Example XIil A steel wire of diameter 1 mm can support tension 02 kN A cable to support tension 20 kN should have a diameter of what order of magnitude Solution The cross sectional area of a round cableWire is A 7TB2 7r4D2 where R is the radius and D the diameter of the cablewire Call cable 1 the 1 mm diameter Wire and cable 2 the one that is supporting the 20 kN tension Then the cross sectional area of cable 1 is A1 7r4D 7r4 mm2 08 mm2 Let s assume that both the Wire and the cable are both made of steel hence both have the same Young s modulus and have the same lengths Solving the Young s modulus formula for A gives i FLO YAL 39 X1 6 PHYS 2010 General Physics Now taking the ratio of the cross sectional areas of these cables gives A2 7 FgLoYAL i F2 A1 7 FlLoYAL i F1 OIquot Ang lt20kN E 1 02 kN Using our area formula from above gives 4A 32 2 D2 72 B 7T 7139 gt 08 mm2 80 mm2 C Density and Pressure 1 The density p of a substance of uniform composition is de ned as the mass per unit volume T p kgm3 in the 81 system a The density of a material changes as the pressure and temperature changes following the formula 27315 P p p R T 27315 Km where pO is the density when the material is at pressure PC the atmospheric pressure at 00C and p is the density at pressure P and temperature T measured in 0C see Xll of the notes b Often the density of material is given in terms of its value as compared to the density of water E 1000 kgmg When the density is described in this manner it is called the speci c gravity of the material 69 lead has a spe ci c gravity of 114 since it has a density of 11400 kgmg Donald G Luttermoser ETSU XI 7 2 The average pressure P of a uid is the ratio of the magnitude of the force exerted by the uid to the surface area A that the uid is pushing against F P27 Xl 7 A lt gt P Nm2 Pa pascal a Consider a column of uid of height h in a cylinder of cross sectional area A The force on th uid is just the gravitational force Fg mg pl9 WW9 where we have used Eq XI 5 and used the volume for mula for a cylinder V Ah b Using this force in Eq XI 7 we get why 13 A A pgh XI 8 c This formula is valid for all states of matter see the fol lowing gure Fluid in Cylinder Ground PHYS 2010 General Physics Example X172 pire State Building which is 1200 ft high What gauge pressure Water is to be pumped to the top of the Em is needed in the water line at the base of the building to raise the water to that height Solution We just need to use Eq XI 8 Converting the height of the building to SI units we get h 1200 X 03048 rnft 370 In Assuming we are using fresh water with density p 1000 kg m3 we get a pressure of Ppgh 103kgnn5980nyagx370n0313x106Pa 36 MPa If we wish to compare points between two different depths in a uid all we need to do is make use of Eq XI 8 and subtract the pressure at the these two different depths either h or y are acceptable here to describe depth APpgAhpgAy XI 9 D Pascal s Principle 1 ln words Pascal s Principle states Pressure applied to an en closed fluid is transmitted undirninished to every point of the fluid and t0 the walls of the containing vessel Since we often rneasure pressure in uids that are open to air at one end of a container we can use Pascal s principle to write PRWh mum Donald G Luttermoser ETSU X1 9 Where PO is the atmospheric pressure at the Earth s surface Po 101325 x 105 Pa XI11 a The atmospheric pressure in other units PO 101325 X 106 dynescm2 cgs system 147 lbsin2 English system 100 atm atmosphere 100 bars 1 bar m 107 Nmg 760 mm Hg mm of Mercury b Note that Eq XI 9 is essentially a description of Pascal s principle as well c Also note that the pressure as described by Eqs XI 9 and XI 10 is not affected by the shape of the vessel 3 A barometer is used to measure the atmospheric pressure and is based upon Pascal s Principle P2 P 0 evacuated y1h XI 10 For the device shown in the gure above Pascal s Principle Eqs XI 910 gives the following relationship for a mercury barometer using the exact value of surface gravity for sea level at 45quot north latitude P2 131 pgy2 y1 P Po pg0 h Po pgh Po pgh 13595 x 103 kgm3980665 ms207600 m 101325 x 105 Pa gt There s about 30000 lbs of air pushing in on our bodies Why don t we collapse 4 When taking pressure measurements P is called the absolute pressure and the difference P PO is called the gauge pressure Example X173 Problem 924 Page 313 from the Serway amp Faughn textbook Piston 1 in Figure P924 see textbook has a diameter of 025 in piston 2 has a diameter of 15 in In the absence of friction determine the force F necessary to support the 500 lb weight Solution Normally we would convert the English units to SI but since we will be taking ratios there is no need to do this The force is a product of pressure times area The downward force at the opening of the larger piston is F2 is the weight Fg w 500 lb of the mass which will produce a pressure P2 in the larger piston The pressure in the smaller piston is P1 which will produce an upward directed force F1 Pascal s law tells us that the pressures between the two connected pistons must be in equilibrium hence P1 P2 The cross sectional area of each piston is A 7TB2 Dg PHYS 2010 General Physics Donald G Luttermoser ETSU XI 11 hence the force that the small piston produces from the 500 lb mass via Eq XI 7 is P1 P2 F1 F2 w 141 X2 i 142 A1 ltD1gt2 w F1 w A won 792 39 2 0251 5001b 141b 15m Now to gure the force required on the handle F to produce an equilib rium condition we must use the second condition of equilibrium concern ing torques Z 739 0 For this torque problem F1 is pointing upward at a lever arm of d1 20 inches from the pivot point hence producing a pos itive torque since this force wants to push the rotation counter clockwise and the applied force F is pointing downward as shown in the textbook diagram with a lever arm of d 20 inch 10 inch 12 inches hence producing a negative torque due to the implied clockwise rotation These are the only forces acting on this balance As such the second condition of equilibrium gives 7 0 F1d1 Fd 0 Fldl Fd d1 20 in F i EFli 12in1411b 23 lb E Buoyancy 1 Archimedes Principle Any body completely or partially sub merged in a uid is buoyed up by a force equal to the weight of XI 12 PHYS 2010 General Physics the uid displaced by the body 2 Mathematically note that the f subscript means uid B AP A pfghA prQULA prQVr Jlfg wf XI 12 3 Sink or Float a The weight of a submerged object is we meg pol09 7 where the o subscript represents object b From Archimedes principle for an object to oat Bw0 gt wfwo SO pngf pogVo XI 13 Donald G Luttermoser ETSU XI 13 or gig mun c What if B y we B we pngf pegVe y 0 XI 15 i If an object is completely submerged Vf Ve and Eq XI 15 becomes BwomAV XL1 ii If pf gt pe B we gt 0 gt the object oats iii If pf lt pe B we lt 0 gt the object sinks 4 From Eq XI l5 we see that there are 2 competing forces in volved for submerged objects From this we can de ne an e ec twe weight of such an object as wee we B Fg Fe mg prfg XI 17 Since Ve Vf and Ve mpe we can rewrite this equation as m wee mg pfltigtgmglt1 gt Po P o 7 wmltd vampgt xrrg po 5 But what exactly do we mean by e ectwe weight Since weights are usually measured Via its effect on an equilibrium force e the normal force is the object rests on a scale or tension if hang ing from a scale the effective weight is nothing more than the tension on a cord that suspends the mass from a scale see the gures associated with Example 98 on page 285 in the textbook XI 14 PHYS 2010 General Physics As such we also could write Eq XI 17 as T we B XI19 Note that if we were to take the object out of the water we would get T w0 as we saw in V of the notes since the buoyancy force would be zero then Example X174 Problem 936 Page 314 from the Serway amp Faughn textbook An object weighing 300 N in air is immersed in water after being tied to a string connected to a balance The scale now reads 265 N Immersed in oil the object appears to weigh 275 N Find a the density of the object and b the density of the oil Solution a This is simple enough problem that we don t need to draw a gure This is nothing more than an effective weight type of problem Note that we have wajr 300 N wwater 265 N and won 275 N with wwater and won representing our effective weights We know that the density of water is pWater 1000 kgmg Using Eq XI l8 we can solve for p0 the density of the object wwater wajr w p0 wwater 1 wajr p0 pwater 1 M p0 wajr pwater N 7 1 71 08830117 p 300 N 1 p0 7 857 pwater p0 857pm 8571000 kgmg 8570 kgm3 Donald G Luttermoser ETSU XI 15 Solution b Now we use the same technique as in a but use the oil weight for the effective weight woil wajr p0 1 wajr p0 1 p0 wajr poil 275 N i 7 l 0917 00833 pg 300 N pen 0083390 00833 8570 kgmg 714 kgmg F Fluid Dynamics i Fluids in Motion 1 We will often make the assumption that a uid is ideal a uid is ideal when a The uid is nonviscous gt no internal friction b The uid is incompressible gt density is constant c The uid is steady gt no uid acceleration d The uid is nonturbulent gt zero angular velocity in uid XI 16 PHYS 2010 General Physics 2 Equation of Continuity 3 b c The continuity condition requires the mass ux of a uid through a pipe is constant Mass Flux pAv constant kgs p1A1U1 p2A2U2 If the uid is ideal p1 p2 so A1U1 AQ UQ i A E cross sectional area of pipe at two different points 1 and OIquot XI20 XI21 ii 1 E velocity of uid at two different points 1 and iii Av E ow rate iv Eq XI 21 is known as the equation of conti nuity gt the smaller you make the cross sectional area of a pipe the faster the uid ow Instead of the word trajectory the path that a uid fol lows is called a streamline i Smooth streamline ow the streamlines don t cross one another is referred as laminar ow ii Fluid ow where streamlines cross one another hence an angular velocity is present is referred as turbulent ow Donald G Luttermoser E TSU 3 Bernoulli s Equation Fluid ow with change in height a b C In uids the work done by a uid is given by W P1 AV P2 AV gpAVULg hl XI 22 According to the work energy theorem AKE W XI 23 where 1 1 KB Emu pAVv2 XI 24 Combining Eqs XI 222324 we get 1 1 EpAva EpAVi P1AV P2 AV gpAVh2 hl OIquot XI25 1 1 P1 5m pgh1 P2 5 We OIquot XI26 1 P Em2 pgh constant Y2h2 XI 18 PHYS 2010 General Physics d Bernoulli s equation is nothing more than the conserva tion of energy for a uid e This equation says that swiftly moving uids exert less pressure than do slowly moving uids Example X175 Problem 950 Page 315 from the Serway amp Faughn textbook An airplane is cruising at altitude 10 km The pressure outside the craft is 0287 atm within the passenger com partment the pressure is 100 atm and the temperature is 200C The density of the air is 120 kgm3 at 200C and 100 atm of pressure A small leak occurs in the window seals in the passenger compartment Model the air as an ideal fluid to find the speed of the stream of air flowing through the leak Solution This is simple enough problem that we don t need to draw a gure ldeal uids are incompressible p1 p2 p 120 kg m3 which is surely unrealistic in our case here but it does allow for an estimate of the speed of the leaking gas Use Bernoulli s equation Eq Xl 25 de ne point 1 as the outside of the aircraft and point 2 as the inside and set yl yg 0 with W m 0 inside the aircraft Then solve for m the speed of the leaking gas outside the aircraft 1 1 P1 gm 211 P2 5mg pgyg P1 p 0 P200 gm P2 P1 1g 2P2p P1 U1 2032 131 p Donald G Luttermoser E TSU XI 19 5 2 2100 atm 0287 atm U 1 120 kgmg 144 1 5 N 2 X 0 m 120 x 105 1112 120 kgmg 347 ms 4 Viscosity and Poiseuille s Law a Most uids have internal resistance This internal friction is called Seldomly uids travel in an ideal fashion viscosity i Assume we have a uid owing through a pipe of cross sectional area A 7TB2 where R is the ra dius of the pipe over a length L between to points labeled 1 and 2 ii Using Eq XI 21 can write this volume ow rate Q through a small slice of the pipe of length Age as Ax AV QA AEE XI 27 iii If the uid is not ideal then internal friction of the uid will set up an internal frictional force called the viscous force Fvisc ln uid mechanics the size of this force is measured by coef cient of viscosity often just called the Viscosity 17 XI 20 b Q PHYS 2010 General Physics iv Viscosity is measured with the unit of poise where 1 poise 1 dyne scm2 01Nsm2 01Pas XI 28 The resistive force set up by this internal friction can be determined from a change in pressure along a streamline 862171 7TB4 which is called Poiseuille s law The derivation of this H g mam law requires high level integral calculus so we will just state this law here without proof Whether a uid s ow is laminar or turbulent depends upon the viscosity of the uid The onset of turbulence can be determined by a dimensionless factor called the Reynolds number i va 77 where p is the density of the uid and v the velocity of R mmm the uid owing through a pipe of diameter D i R lt 2000 Flow is laminar ii 2000 lt R lt 3000 Flow is unstable gt ow is laminar but any small disturbance in the ow will cause turbulence to immediately form iii R gt 3000 Flow is turbulent Donald G Luttermoser ETSU Xl 21 Example X176 Problem 959 Page 316 from the Serway amp Faughn textbook A straight horizontal pipe with a diameter of 10 cm and a length of 50 m carries oil with a coefficient of viscosity of 012 Nsm2 At the output of the pipe the flow rate is 86 X 10 5 m3s and the pressure is 10 atm Find the gauge pressure at the pipe input Solution This is simple enough problem that we don t need to draw a gure In Poiseuille s law let P1 P be the pressure of the pipe at input and P2 PO 10 atn1 be the pipe pressure on output Using Eq XI 29 and converting all units to 81 R 05D 05 10 cm x001 rncrn 50 X 10 3 In the gauge pressure P Po is 862171 7TB4 886 x 105 n13s012 N sn1250 m 7T50 X 10 3 n14 21 x 106 Pa 21 atm P Po

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