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# PhysicsEveryday Phenomena I PHYS 103

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This 124 page Class Notes was uploaded by Sonny Breitenberg on Monday September 28, 2015. The Class Notes belongs to PHYS 103 at George Mason University taught by Jason Kinser in Fall. Since its upload, it has received 27 views. For similar materials see /class/215188/phys-103-george-mason-university in Physics 2 at George Mason University.

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Date Created: 09/28/15

Chapter 1 Temperature Two questions that may seem simple 1 What is temperature 2 What is heat We use to common temperature scales Fahrenheit and Celsius The Fahrenheit scale was invented in the early 1700 s and the Celsius scale was invented in 1743 Table 11 shows a few key temperatures Table 111 Important Temperatures Farenheit 32 212 Celsius 0 100 The conversion equations are 9 F 3C 32 11a 0 g F7 32 1 11b 11 Absolute Zero Temperature is the movement of molecules in matter As the temperature lowers the vibrations become less Eventually there is a temperature in which the molecules are motionlessi This is called absolute zero On the Celsius scale absolute zero is at 727320 i Physicists working on matter in this temperature range use a different temperature scaled named Kelvin and temperatures in the scale do not use the word degrees The conversion from degrees Celsius to Kelvin is simply K C 7 273121 12 There is also an absolute temperature scale using Fahrenheit gradationsi Temperatures in this scaled are called Rankinesi 12 Heat There is no such thing as COLDi Objects can have heat or lack thereof Consider the two masses shown in Figure lili In this case object A has more heat than object Bi When in contact heat will ow from A to Bi Object B is not giving cold to object Al Figure 11 Two masses of differing temperature 13 Heat Capacity Each material has an ability to store heati Water for example has a tremendous ability to store heat and therefore it is much harder to change the temperature of water than metali Consider the two examples in Figure 12 In both cases there is a cup of hot tea and in each case a colder object is being placed into the teat Milk 100 g at 5 C lt 0 Metal Ball 100 g at 5 C Tea 90 C a Pouring milk into tea b Dropping a metal ball into tea Figure 12 Two items changing the temperature of a cup of tea The heat from the tea will go into the other object until equilibrium is reached In Figure 112a there is 1009 of milk at 5 C being poured into the tear In Figure 112b there is a metal ball with a mass of 1009 and a temperature of 5 C being poured into the tear Both additions have the same mass and initial temperature The milk is mostly water and therefore it is much harder to heat than a metal balli Even though the two additions have the same mass and initial temperature it will take more heat from the tea to warm up the milk Therefore the cup of tea in Figure 112a will be colder than the tea in Figure 112bi 131 De nition of Heat Capacity The speci c heat capacity is the quantity of heat required to change 1 gram by 1 degree Celsius This is measure in calan39es Soy the 1009 of milk has more calories than does 1009 of metal The speci c heat of some materials is shown in Table 112 Table 112 Speci c Heat Capacities ltem cal 9 C water 110 ice 0 49 steam 0 48 ethyl 0158 steel 0 11 lead 010305 bee s wax 0 82 dry cement 01037 copper 0 09 ammonia 1 1 freon 0 21 milk 094 apples 0 87 brains 0 84 chicken 0 80 There are other units for measuring heat capacity Some tables provide the data in r Aluminum for example is cal J 01215 019007 9 C 9 K Conversions are J kcal Btu 4L1861897K71kg QC 71 oFi 1 3 132 Calories A calorie is a unit of energy Notation for a calorie can be tricky A lowercase c is used for a calorie and an uppercase C is used for a kilocalorie So C 10005 14 In the scienti c elds the use of calorie has been replaced by Joules However calorie is still used to describe food energy A calorie can be measured by burning the food and collecting the heat perhaps by warming an amount of water A calorie is the amount of energy required to raise the temperature of 1 gram of water by 1 C Therefore lcal 419 15 133 Heat Equation The relationship between a quantity of heat the mass and the temperature is Q chT 16 where Q is the quantity of heat m is the mass 5 is the heat capacity of the object and T is the temperature Example 11 Raising the temperature of water How much heat does it take to raise the temperature of 1 gram of water by 1 degree C Q chT lgl00 flow C1cal 9 Example 12 Milk How much heat does it take to raise the temperature of 1 gram of milk by 1 degree C Q chT lg094 flow C 094mi 9 Example 13 Steel How much heat does it take to raise the temperature of 1 gram of steel by 1 degree C Q chT 19mm flow 00 Ollcal 9 134 Equilibrium When two objects are in contact and they are different temperatures then the heat from the hotter object will ow into the colder object until they are both the same temperature This state is called equilibrium Figure 13 shows two masses and the two temperatures are T1 gt T2 In this case heat will ow from object A into object B until both objects are the same temperature T1gtT2 Flgllle 1 3 Two masses of dlelellng lempelalllle Example 1A Walm Apples ln Cold Walel Gwen 5 apples lhal on avezage h v a mass of 1009 and lhese apples ale al loom lelnpelalllle 72mv 2 These a laced lnlo l lllel of walel al am The syslem ls allowed lo Ieachlng a lempelalllle equlllbllum Whal ls lhe nal lempelalulev specl c heal of app es ls 0 Mal9 C The mass of lhe apples ls m 5009 The mass of lhe walel ls m2 10009 Slall by selllng up a heat exchange Q1Q2 m1c1AT1 7m2c2AT2 meal 7 T 7m2c2T2 7 T 021ng 7 mlclT MQCQTQ mmT Solve fol T m1c1T1 7712ch2 mmT mmT M16113 mQCQTQ T 771161 771262 5000 87221000105 5000 87 100010 10 lac 14 Melting and Freezing There are three phases to water It can be ice liquid or steami In any of these phases the speci c heat is different as shown in Table 12 However the transition between the phases is described by a different process lce does not instantly melti Rather it takes heat to convert the ice into water During this process the icewater remains at 0 Ci The heat that is being added is not changing the temperature but it is converting the ice into liquid This process is called a transition This process is described by Q mL 17 where L is the latent heat The process of converting a solid to a liquid uses Lf which is the latent heat offusz39oni The process of converting a liquid to a gas uses L1 which is the latent heat of vapon39zatz39on These are two different values Lf vi For water Lf 800alg and L1 5400algi Clearly it will take 7 times more heat to convert water to steam than it does to convert ice to water Example 15 Melting lce Given a large mixture of ice and water how much ice will melt if 400 calories are added to the mixture A mixture of ice and water at equilibrium will be at T 0 C Use Q mLf to solve for m to determine that mass of ice that goes through a transition Q 4000al i 7 50 m Lf 800alg g So to convert cold ice to very hot steam there are ve steps 1 Raise the temperature of the ice to 0 C 1 Convert the ice to water The temperature remains at 0 C 1 2 3i Raise the temperature of the water from 0 C to 1000 Ci 4 Convert the water to steami The temperature remains at 100 C 5 i Raise the temperature of the steami Consider a case in which there is 1000 grams of ice at 73 C and heat is added until it is steam at 103 Ci How much heat does it take to make this conversion To solve this problem each part is considered in turn The rst step is to raise the temper ature of the ice to 0 Ci ca 0 Q1 cht 10009 049 3 C 14700ali g The second step is to convert the ice to water Q2 mLf 10009 80511 800005ali 9 The third step is to raise the temperature of the water ca 1 9 C Q3 cht 10009 1000 gt 100 C 100000wzi The fourth step is to convert the water to steami cal Q4 mLf 10009 5407 540000wzi 9 The nal step is to raise the temperature of the steami cal 0 Q5 cht 10009 0489 C gt 3 C l4400ali The total heat used is QT Q Q2 Q3 Q4 Q5 7229100ali It is interesting to note that it takes 5 times more heat to convert the water to steam than it does to raise the temperature to the water It is important that water have a very high heat capacity This prevents lake from quickly freezing over The large ocean masses help stabilize the world temperatures as well Most impor tant though is that if water had a low heat capacity then the shower water would be cold by the time it traveled from the hot water heater to the shower 15 First Law of Thermodynamics The rst law of thermodynamics is An increase in the internal energy of a system is equal to the heat added to the system minus the work done by the systemi This is expressed as AU Q 7 W7 18 where U is the internal energy Q is the heat added and W is the work done The internal energy U is the potential energy plus the kinetic energy of the molecules in the systemi 16 Joule s Experiment It is possible to raise the temperature of water by stirring it The stirring motion increases the motion of the molecules and thus the stored heati Given a sample of ice water it is possible to melt the ice by adding heat or by stirring the mixture However the amount of stirring required is a large gure Example 16 Stirring Water Given a mass of icy water sitting on a hotplate The hotplate adds 400 calories to the water and a stirring mechanism adds 500 to the water How much ice melts Lf 800alg 335g 419 Q 4000al 1680 cal AU Q W 1680 500 2180 AU 2180 65 m Lf 335 9 71 ML A g Ellllllliljulllm 3 W l i ii F t a l 1 quot I I l l u y TEquot i ll iilllul llillllllillllilllllh hill 1111111 sllllliilflll quotl a 1121thmmeium Wizarlull39mulilmw39imuu unmm 11 07 11 c l ll u Figure 14 Joule s apparatusWik10 James Prescott Joule 1818 1889 created a machine that could measure mechanical equiv alent of heat A schematic of this machine in shown in Figure 14 It consists of a container of water at a known temperature and inside of this water is paddle wheel The axle of the paddle wheel is wound with a cord that is tied to a weight When the weight is allowed to fall then the cord is pulled and the paddle wheel turns By stirring the water the temperature is increased slightly He was able to measure the work done by the falling weight and assuming negligible friction all of this work is converted into a rise in temperature of the water Since then units of heat have the unit of Joule in his honor The conclusion in modern terms is that it takes 419 to raise the temperature of 1 gram of water by 1 CC 17 Compressing a Gas 171 Gas Law The relationship between pressure volume and temperature of a gas is PV 2 NkT 19 Example 17 Joule s Experiment Given Joule s experiment with a weight of 1kg and 1 cup of water How far would the weight fall in order to raise the temperature of this water by 1 0C A cup of water is 2365907213 The density of water is 1 and therefore by using m pV the mass of the water is calculated to be M 237g The amount of work required to raise the temperature by 1 C is Q chT 237 1 1 237ml 980 Using PE mgh and solving for h the height required to create this amount of energy is PE 980 h 7 7 100m my lt1gtlt978gt 100 meters is about100 yards and so this is the length of a football eld where P is pressure V is volume N is the number of molecules is is the Boltzmann constant and T is the temperature Boltzmann s constant is R k 7 138 1043 K NA where R is Rydberg s constant Given a gas that remains under a constant pressure what is the relationship between volume and temperature Using the gas law of Equation 19 the values of P N and k are constants Therefore for two different scenarios VliNkng TlPT Example 18 Constant Pressure Gas Given 4 liters of a gas at 600Kand the volume of this gas is reduce by a factor of 4 what is the new temperature V2 1 T2 7 in 7 Z600K 7150K Example 18 may seem counter intuitive The volume is reduced and the temperature goes down It may seem that this is the opposite behavior of what we know After all in a piston engine the cylinder is compressed volume goes down and the temperature is increased It should be remembered that in the piston case the pressure is also increased In the example problem the pressure remains a constant In order for this to occur heat must leave the system 172 Work from Compressing a Gas Figure 15 shows a simple piston which is used to compress a gas More compression is equivalent to more work This is expressed as W PAV 110 Flgllle 15 Aslmple plslon 18 Heat Flow Thele 816 1166 melhods by Whlch heal can be Llansfezed 1 Conduellon 2 Conveellon 3 Radlallon Conduellon oceuls whele Lhexe ale lwo masses lhac ale ln wnlael VIth eaeh olhel Heal ows fxom lhe hollel object to lhe wldex obleec Conveellon oeeuls ln an almosphele IL ls lhe well known hot m mes phenomenon la lon oeeuls when heal ls Llansfezed by elecuormagnetlc ux W feel heat fxom che h 11 ls l Rad L e slln because ll ls lmpallng use VIth 11g Le laall allon Not a ladlallon n lhe vlslble speellum A Lxemendous amounl of heal can be Llansfezed ln lhe lnflaled specuum Chapter 1 Rotational Inertia A note on the Change in anglesi This Chapter requires that all angles be computed in radians and not degrees 11 Rotational Inertia In a linear system a mass in motion has inertial The same applies for an object that is rotating or orbitingi It has a rotational inertia A mass with a higher rotational inertia requires more effort to put it into orbit or Change its orbit Figure 11 shows a mass moving in a circular orbit mass Figure 11 An orbiting mass The expression for rotational inertia for a single mass that is in Circular motion is I mrz 11 Where m is the mass r is the radius and I is the rotational inertial 111 Con gurations Equation 11 only applies to a point mass orbiting a center point Consider Figure 12 in which a rod is constructed from a set of masses Each mass moves with the same angular velocity However the radius for each mass is different The total rotational inertia would be the sum of the rotational inertias of the individual massesi Figure 12 A set of masses in rotation In actuality a solid rod is not made up of a set of blocks as shown but rather is a continuum of masses and usually calculus is used to compute items such as rotational inertial Therefore without proof the rotational inertias for many different geometric shapes is given in Figure 13 12 Rotational Force ln linear motion the equation for a force is F mar In rotational motion a rotational force is a torque which is de ned by the Greek letter tau 739 la 12 where I is the rotational inertia and a is the rotational accelerationi Consider the case in Figure 14 in which Ziggy is riding on a MerryGo Roundi In this case there are two masses one from Ziggy and the other from the spinning disc The total rotational inertia is the sum of the rotational inertias of the two objects For this example set the mass Ziggy to 96kg the mass of the disc to 201k and the radius of the disc to 4771 The total rotational inertia is ITIMIZ 1381 IT mMr2 7ner 13b IT 0 520142 9642 13c IT 1608 1536 13d IT 31441694712 1i3e Note that the rotational inertia is not dependent upon the angular velocity This is equivalent to noting that in linear motion the mass is not dependent upon the velocity i 1At non relativistic speeds of course As the speed of a mass approaches the speed of light bizarre effects occur and one of them is that the mass will change a I mrz b I 44 c I mrz d I mrz1izmlz e I mlIZ f Solid sphere Hollow sphere h I mrz I gm I gmrz I gmr Figure 13 Various rotational inertiasi Figure 14 Ziggy on a MerryGo Roundi Now consider a case in which the disc has an initial rotational velocity of 0 and Ezra pushes the disc until it is moving at a rate of 3505 and it takes Ezra 205 to achieve this rate The centripetal acceleration is 9 35 771i75 2 a t 20 S The torque is 739 a 3144x175 5500N mi 13 Rolling Disc Consider Figure 15 in which there is an inclined planer There are two types of discs that the user can roll down the hill Both discs have the same mass and the same radiusi However one disc is made of solid wood and the other is a thin disc of metal Will they roll down the hill at the same rate Figure 15 Two types of discs rolling down a hill The object with the larger rotational inertia will accelerate sloweri The rotational inertia of the wooden disc is and the rotational inertia of the annular ring is I mrzi Since the steel ring has a larger rotational inertia it will accelerate sloweri 14 Angular Momentum and Kinetic Energy In linear motion the momentum is described by p mm In angular motion the angular momentum is described by L wi 14 Linear kinetic energy is described as 1 KE imvz and the equivalent in rotation motion is 1 2 KE 7101 i 15 2 15 Rotational Motion and the World We Know 151 Spinning Athletes such as ice skaters and pool divers use rotational momentum to their advantage An ice skater will push themselves into a spin that lasts for a bit of time During this spinning they do not apply any more forces and friction is small So once they start spinning they establish a certain angular momentumi During the spinning the angular momentum is conserved since there are no outside forceszi How is it that an ice skater can increase their rotational velocityi7 They do this by rst starting with their arms outstretchedi Since the arms have a mass and this mass has radius from the axis of spinning there is a rotational momentumi When the arms are brought close to the body the mass is the same but the radius changes The rotational inertia is decreased but the angular momentum must remain the same Thus the angular velocity is increased by L In 152 Kepler7s Second Law Revisited Kepler s second law was that the planets sweep out equal areas during their elliptical orbit As the planet orbits its angular momentum is conserved For a planet in an orbit the rotational inertia is I mrz so the rotational inertia is L Io mrzwi Since L is constant when r increases 0 must decreasei 16 Momentum as a Vector Angular momentum as a vector is more complicated than linear momentum vectori Figure 16 shows a spinning mass and that the angular momentum vector points along the axis of rotation gt Figure 16 Rotational Inertia as a vector 2Newton s First Law applies to angular motion as well Thls vector obeys the mght hand TulE Cull the the ngezs on the nght axound azound the axis of lotatlon Polnt the thumb outwanls and thls ls the dllectlon of the momentum vectol 161 Balancing a Bicycle Why ls it much easlez to balance a blcycle when 1L ls ln motlen Figuxe 1 7 shows a blcycle and the white vectols are the melha vectols Howevex the blcycle i also ln motlon So thew ls 39a locational momentum fox the Wheels They would pzefex not to tllt zom this momentum It takes a signi cant amount of 0106 to change s lght tllts in the blcycle thus movlng the centezof mass outslole of the tootpnnt ls not sul clent to topp e the blke 00110 Figuxe 1 7 Riding a blcycle 162 Rotational Energy The lotatlonal enezgy IS olescnbeol by Equatlon 15y Consldez the planet Ealth Wlth a mass of 9710941cg and an avezage laollus of 6 378 10 The xoLaLional lneltla ls 2 2 37 2 IEm7 971 10 kgm The tlmelt Lakes fox one levolutlon ls 2393 days whlch ls 85148st Theangulax veloclty is 2 w 7W 729 r 10 5mds 86148 The rotational energy is 1 KE 5le 25810291 17 Gyrobus Figure 18 shows a bus that was built in the 1940s that was powered by a ywheel This is a large wheel that was spun up to 10000 RPM This stored the energy that would be drained o to drive an electric motor when the driver needed propulsion Figure 18 The Gyrobusgyr10 The ywheel weighed 15 tons and could spin at 10 000 RPM This would move a 15 ton bus a total of 35 miles When the ywheel slowed to 5000 RPM the driver would nd a recharging station that would spin up the wheel in about 90 seconds The only data not supplied but the website is the diameter of the wheel which we estimate ate 3 feet First convert the measurements into the metric system 0 Mass 15 ton 1360 kg 0 15 tons 13600 kg 9 10000 RPM 1047 rads The rotational inertia is 1 I Em 0 513600i452 138kg m The angular momentum is L 1w 138kg m210477quotads 1i44105kg mQsi The kinetic energy of the Wheel is KE n 0 5138kg m21047kg 721232 7561071 The kinetic energy of the bus traveling at 30MPH 13i4ms is KE mv2 0 51360013i4 1221051 So the ywheel has 50gtlt more energy than the bus in motion Chapter 1 Rotational Motion Linear motion was discussed in the previous chapters In this cases the motion vector was a straight line In this chapter rotational motion is reviewed 11 Rotational Motion There are equivalents between rotational motion and linear motion 111 Angular Displacement Figure 11 shows at mass blue ball travel in a perfect circlei In the previous chapter the velocity vector is shown as a straight line Another way of viewing the motion is to consider the change in angle The angle here is shown as the Greek symbol 9 thetai As the blue moves in a counter clockwise direction 19 increases a Figure 11 Linear versus rotational motion A scienti c calculator has three different modes for measuring anglesi Two of the methods used in physics are degrees and radiansi Figure 12 shows the relationship between these two methods of measuring anglesi Basically 1 radian is equal to 57296 degrees A full rotation is 380 or 27r radiansi 90 nZ rad 180 7 red 270 3n2 rad Figure 12 Degrees and radians The third type of measurement that a calculator offers is gradians which has been used in some engineering applications In this case a full rotation is 400 gra s 112 Angular Velocity ln linear motion velocity is described by I v 7 t The equivalent in angular measurements is called angular velocity and it is denoted by w omega 7 A19 0 7 1 1 Example 11 Spinning Disc A disc spins at 5 revolutions per second What is its angular velocity7 5 x 36015 1800degs 113 Rotational Acceleration Linear acceleration is described by a Angular accela39ratz39zm is the rate of increase or decrease of angular velocity and is described by Aw 13 12 114 Similarities dvot7at w0t7at 115 Multiple Example An old Subaru is traveling at 607711011 The tires are the car are 22560R16i This means that each tire has a diameter of 266 inches What is the angular velocity of a tirei7 The circumference is c 7rr 7r 3 3 835m 212cm 212ml The speed of the car is 60mph 269ms ln one second the car travels 269m and therefore the wheel must make 269 m 7 127mm i This is equal to 762rpmi Converting this to degrees the angular velocity is 127 360 I 4567degsi S If the car needs 85 to accelerate from 0 to 60 what is the angular acceleration of the tirei7 A 4567 a 7w Y 571idegs 12 Torque and Balance Torque is de ned as a force at a distance from a fulcrumi Figure 13 shows a lever on a fulcrumi The lever is theoretically Inassless and the only weight that exists in this system is the mass as the end of the leveri Figure 13 Torquei The force F is applied at a distance d from the fulcrumi The torque is then de ned as T F d 13 Certainly if F is increased then the lever would have a greater tendency to move clockwisei Likewise if d is increased but F is not the same tendency is experienced Balance occurs when the a lever has equal torques on both sides of the fulcrumi Figure 14 shows a case in which a lever has two different Inasses at different lengths from the fulcrumi The mass on the left is 7711 and it is a distance l1 from the fulcrum The mass on the right is 7712 and is a distance l2 from the fulcrumi If the values given are 7711 3 m2 2 and l1 1 then what is the distance l2 required to balance the lever Figure 141 Balance In this case the two masses create two torques that need to be equal in magnitude in order to balance the leveri The balance is shown in Equation 14a where the two torques are equal but opposite Equation 13 is then infused into both sides of the equation and the force is de ned as a weight or mgr In this case both sides of the equation have 9 term which cancels Equation 1i4e shows the values and it should be noted that since ll was on the left of the fulcrum that is has a negative value 7391 7T2 14a F111 inzz 14b 7711911 4712912 14c mlll 7771212 1i4d 3X4 212 149 3 12 5 14f 13 Center of Gravity The center of gravity or center of mass is a point in a mass in which all torques are balanced If the mass were located on a rod such that the center of mass is directly over the rody then the mass would be balanced This is shown in Figure 15 Balancing is important for many applications especially those in which a wheel spins with a large angular velocity Figure 16 shows an image of a car tire being balancedi If car tires are not balanced then it will oscillation when spinning at a high rate and this oscillation is transfered as Vibration to the passengers and also wears out the tire at a faster rate To make a tire balanced weights are applied to the rim in order to add mass at particular angles from the centeri When the center of mass exceeds the footprint of an object then it will fall overi Figure 17 shows a case in which the center of mass red arrow now exceeds the footprint perimeter of the base and the object falls overi Figure 18 shows a similar case In this case there is a board with a mass and that center of mass is to the left of the edge by a distance L A small lad walks to the right of the edge The board has a clockwise torque from the mass of the board and the boy has a a counterclockwise torque since he is to the right of the edge When the boy s torque exceeds that of the board then he and the board will fall off the edge Figure 15 Center of gravity Figure 16 Tire Balancingi irlo Figure 17 Pisa pis Figure 18 Walking the plank Example 12 Boy walking the plank Consider Figure 18 in Which the board has a mass of 20169 and the boy has a mass of 50169 The center of the mass is 2m to the left of the edge What is the distance d in Which the boy Will cause the board to fall Twood may lew ind mwlw imbd 20kg 72m 50kgd 40 d 7 E 7 08m Homework 01 Phys 103 Name 1 Albert Einstein stood 5 feet 9 inches tall How tail was he in inches Convert to inches H5ftgtlt12m9in 1 H60in9in69in 2 How tail was Albert in centimeters H269inxml75cm 1m 3 It is 20 miles from the Fairfax campus to the Prince William Campus How far is that in kilometers 161km d220migtlt lmz 322km32km 4 A rectangle has the dimensions 13cmgtlt15cm What is the area of this rectangle A 13cm x 15 cm195 cm2 However the measurements are only with two digits of precision and so the correct answer is 20 cmz 5 A new refrigerator has the dimensions of 32gtlt66gtlt28 inches What is the exterior Fluids 2 Jason M Kinser jkinsengmu edu Bioinformatics 8 Comp Biology eorge Mason Univ November 18 2010 149 HHMSZ L Oul ne n Viscosity Archimedes H Weight in the Pool Flow Junkyard Wars Bernoulli39s Principle 249 HHMSZ LArchimedes Outline Archimedes I Floating Objects I Archimedes Golden Crown 349 HuMsZ LArchimedes Archimedes Principle Archimedes Principle states that A buoyant force is equal to the weight of the fluid that is displaced 449 magi 66 SM Figure shows a large ship on the high seas Even though it is huge it floats because the weight of water that is displaced by the part of the ship that is below sea level is equal to the total weight of the ship 5mg HuMsZ LArchimedes Floating Ohjec39s Outline Archimedes I Floating Objects 049 muse l k l mch m g The box shown in Figure 7 has a volume of V X H X L In this example the block is submerged halfway and so the volume of water that is displaced is V gtlt H X L 749 HHMSZ LArchimedes Floating Ohjec39s Buoyant Force The buoyant force can be calculated by obtaining the weight that is displaced Assuming that the box is at sea level the mass is m Thus the mass of the object is WV 1000 gm 549 HuMsZ LArchimedes Floating Ohjec39s Buoyant Force Which is the correct spelling Bouyant force Buoyant force 949 Flukk 2 LArchimedes Floating Objects Question A block of wood shown in Figure T that is 3 X 3 X 2 is floating in water The density of this wood is 500 How much of the wood is submerged 1049 Flukk 2 LArchimedes Floating Objects Question A block of wood shown in Figure T that is 3 X 3 X 2 is floating in water The density of this wood is 500 How much of the wood is submerged 1049 Fluids 2 LArchimedes Floating Ohjec39s A Little Work What is the mass of the wood 1149 Fluids 2 LArchimedes Floating Ohjec39s A Little Work What is the mass of the wood m pBV 50018 9000kg What is the weight of the wood 1149 Fluids 2 LArchimedes Floating Ohjec39s A Little Work What is the mass of the wood m pBV 50018 9000kg What is the weight of the wood W mg 900098 88 2OON 1149 Fluids 2 LArchimedes Floating Ohjec39s A Little More Work The buoyant force is therefore also 88200N The volume of water that also has this weight is 1249 Fluids 2 LArchimedes Floating Ohjec39s A Little More Work The buoyant force is therefore also 88200N The volume of water that also has this weight is m 9000kg 3 7 m pw lOOOkgm3 What does this mean 1249 Fluids 2 LArchimedes Floating Ohjec39s A Little More Work The buoyant force is therefore also 88200N The volume of water that also has this weight is m 9000kg 3 7 m pw lOOOkgm3 What does this mean This is half of the volume of the wood block Therefore half of the block of wood would be submerged 1249 Fluids 2 LArchimedes Floating Ohjec39s Wood from an Apple Tree Consider the previous case but submerge wood from an apple tree instead The density of this wood is pA 700kgm3 1349 Fluids 2 LArchimedes Floating Ohjec39s Wood from an Apple Tree Consider the previous case but submerge wood from an apple tree instead The density of this wood is pA 700kgm3 Since the density is greater the wood will have more weight Therefore it will be necessary for more water to be displace Since the density of the apple wood is 07 of that of water then 70 of the wood will be submerged The geometry of the floating mass is not relevant to the amount of water displaced In the case of the apple wood 70 of the wood will be under water no matter what the shape of the wood is 1349 Fluids 2 LArchimedes Archimedes Golden Crown Outline Archimedes I Archimedes Golden Crown 1449 Fluids 2 LArchimedes Archimedes Golden Crown The Story A goldsmith created a crown for King Hiero II and it was to be made of pure gold But was it 1549 Fluids 2 LArchimedes Archimedes Golden Crown The Story A goldsmith created a crown for King Hiero II and it was to be made of pure gold But was it Or did the goldsmith dilute the gold with a cheaper metal like silver Fluids 2 LArchimedes Archimedes Golden Crown The Story A goldsmith created a crown for King Hiero II and it was to be made of pure gold But was it Or did the goldsmith dilute the gold with a cheaper metal like silver Archimedes was hired to determine if the crown was pure gold How did he do it Fluids 2 LArchimedes Archimedes Golden Crown The Logic Archimedes could not alter the crown and it had a very irregular shape His logic was that if the crown was solid gold then it would have the same density as a block of solid gold The volume and weight of a block of gold was easily measured and the density easily computed 1049 Fluids 2 LArchimedes Archimedes Golden Crown The Logic Archimedes could not alter the crown and it had a very irregular shape His logic was that if the crown was solid gold then it would have the same density as a block of solid gold The volume and weight of a block of gold was easily measured and the density easily computed But how could he measure the volume of the crown 1049 HHMSZ LArchimedes Archimedes Golden Crown The Answer He plunged the crown into a container of water and the volume of water displaced was equal to the volume of the crown Now he could compute the density 1149 HHMSZ LArchimedes Archimedes Golden Crown The Answer He plunged the crown into a container of water and the volume of water displaced was equal to the volume of the crown Now he could compute the density The story also states that Archimedes was so excited about this discovery that he immediately ran into the streets naked yelling quotEurekaquot Greek for quotl have found it 1149 Fluids 2 LArchimedes Archimedes Golden Crown Archimedes You are not to do that during the final 1549 Fluids 2 LFlow Outline I The Shower Flow I Flow in a Pipe 1949 Fluids 2 LFlow LFlow in a Pipe Outline Flow I Flow in a Pipe 2049 Fluids 2 L Flow LFlaw in a Pipe Archimedes Figure shows a fluid in a cylindrical pipe Consider a small portion of this fluid of length L This plug of fluid has an area A a length L and a velocity v The volue of this plug is V AL 21 49 Fluids 2 L Flow LFlaw in a Pipe Archimedes Figure shows a fluid in a cylindrical pipe Consider a small portion of this fluid of length L This plug of fluid has an area A a length L and a velocity v The volumthis plug is V AL The rate of flow is defined as Q VA Unfortunately a common symbol for volume flow rate is Q which is also used for heat It is up to the student to keep these straight 21 49 Fluids 2 L Flow LFlaw in a Pipe Reducer Figure shows a pipe that reduces its diameter V A1 0 The flow rate must remain the same throughout the pipe Therefore A1 V1 A2 V2 2249 Fluids 2 L Flow LFlaw in a Pipe Reducer Figure shows a pipe that reduces its diameter V A o The flow rate must remain the same throughout the pipe Therefore A1 V1 A2 V2 If the diameter of the pipe is reduced then the flow must increase 2249 Fluids 2 LFlow LFlow in a Pipe Flow Example If the water is pumped with a speed of 045 ms under a pressure of 400 torr through a 60 cm diameter pipe what will be the velocity if the pipe is reduced to a diameter of 20 cm 2349 Fluids 2 LFlow LFlow in a Pipe Flow Example If the water is pumped with a speed of 045 ms under a pressure of 400 torr through a 60 cm diameter pipe what will be the velocity if the pipe is reduced to a diameter of 20 cm Use lel VQAQ and solve for v2 7 A1 7 7r6022 7 v2 7 le2 7 045n239O22 7 405ms 2349 Fluids 2 LFlow LThe Shower Outline I The Shower Flow 2449 Fluids 2 LFlow LThe Shower The Story The son of a not so famous physics professor likes to take lengthy showers The maximum flow of the shower is controlled by the shower head and it is rated at 25 gallons per minute 1 gallon 379 liters What is the volume of water used in a 15 minute shower What is the velocity of the water in the pipes Fluids 2 LFlow LThe Shower The Story The son of a not so famous physics professor likes to take lengthy showers The maximum flow of the shower is controlled by the shower head and it is rated at 25 gallons per minute 1 gallon 379 liters What is the volume of water used in a 15 minute shower What is the velocity of the water in the pipes This must be done in parts Fluids 2 LFlow LThe Shower Shower 1 In one minute 379 liters of water are used What is the volume of water that is used in one minute 2549 Fluids 2 LFlow LThe Shower Shower 1 In one minute 379 liters of water are used What is the volume of water that is used in one minute Noting that 1 liter is 0001m3 by definition the volume of water used is 379 0001m3 7 73 3 lgal ll 7 9510 m 25gal 2549 Fluids 2 LFlow LThe Shower Shower 1 In one minute 379 liters of water are used What is the volume of water that is used in one minute Noting that 1 liter is 0001m3 by definition the volume of water used is 379 0001m3 7 73 3 lgal ll 7 9510 m 25gal In a 15 minute shower 0142m3 of water is used 2549 HuMsZ LFlow LThe Shower Shower Two Pipes But there are two pipes feeding the shower Thus in one minute 47 10 3m3 of water is moved 2749 Fluids 2 LFlow LThe Shower Shower 2 Near the source of water most houses have 34 pipe and then nearer the fixtures the pipes are 12 In the metric system the larger pipe has a diameter of 0019m and so the cross sectional area is zapquot Fluids 2 LFlow LThe Shower Shower 2 Near the source of water most houses have 34 pipe and then nearer the fixtures the pipes are 12 In the metric system the larger pipe has a diameter of 0019m and so the cross sectional area is A 7rr2 7r001922 2810 4m3 zapquot Fluids 2 LFlow LThe Shower Shower 3 In one second the amount of water that comes out of the pipe is 78 10 5m3 What is the length of the plug of water 2949 Fluids 2 LFlow LThe Shower Shower 3 In one second the amount of water that comes out of the pipe is 78 10 5m3 What is the length of the plug of water 75 X 7 028m R 11in L A 2810 4 What is the velocity of the plug 2949 Fluids 2 LFlow LThe Shower Shower 3 In one second the amount of water that comes out of the pipe is 78 10 5m3 What is the length of the plug of water 75 X 7 028m R 11in L A 28 104 What is the velocity of the plug 7 028m v dt 15 028ms 2949 Fluids 2 LFlow LThe Shower Shower 3 What is the flow rate all49 Fluids 2 LFlow LThe Shower Shower 3 What is the flow rate Q VA 0282810 4 78 10 5m3s all49 Fluids 2 LFlow LThe Shower Shower 4 What is the cross sectional area of the smaller pipe 025m 63 10 3m 3149 Fluids 2 LFlow LThe Shower Shower 4 What is the cross sectional area of the smaller pipe 025m 63 10 3m A 7rr2 134104m2 What is the flow rate of the smaller pipe 3149 Fluids 2 LFlow LThe Shower Shower 4 What is the cross sectional area of the smaller pipe 025m 63 10 3m A 7rr2 134104m2 What is the flow rate of the smaller pipe 075 v 06ms m 24ins 3149 Fluids 2 LFlow LThe Shower Recap So the velocity in the two pipes are I large 028ms I small 060ms 3249 Fluids 2 LFlow LThe Shower Shower 5 If the son takes a 15 minute 9005 shower what is the volume of water used 3349 Fluids 2 LFlow LThe Shower Shower 5 If the son takes a 15 minute 9005 shower what is the volume of water used V Qt 78 10 5m3s9005 007m3 3349 Fluids 2 LFlow LThe Shower Shower 5 If the son takes a 15 minute 9005 shower what is the volume of water used V Qt 78 10 5m3s9005 007m3 What is this in liters 3349 Fluids 2 LFlow LThe Shower Shower 5 If the son takes a 15 minute 9005 shower what is the volume of water used V Qt 78 10 5m3s9005 007m3 What is this in liters 007m3 70 ll 0001m3 3349 Fluids 2 LFlow LThe Shower Shower 5 If the son takes a 15 minute 9005 shower what is the volume of water used V Qt 78 10 5m3s9005 007m3 What is this in liters 007m3 70 ll 0001m3 Gallons 3349 Fluids 2 LFlow LThe Shower Shower 5 If the son takes a 15 minute 9005 shower what is the volume of water used V Qt 78 10 5m3s9005 007m3 What is this in liters ll 37 007m 039001m3 70 Gallons 1gal 70319 7185gal Recall that there are two pipes and so he is consuming 37 gallons in 15 minutes 3349 Fluids 2 LBernoulli39s Principle Outline Bernoulli39s Principle 3449 Fl ui k 2 LBemouIIi39s Plimiple Reducer The reducer shown in Figure shows a pipe with a reducer and each end of the pipe has two pressure stems 7 3549 Fluids 2 LBernoulli39s Principle Pressure Differential The pressure is different in these two stems because the velocity in the pipe is different This is described by KE P77k where k is a constant 3349 Fluids 2 LBernoulli39s Principle Pressure Differential The pressure is different in these two stems because the velocity in the pipe is different This is described by KE P 7 k V where k is a constant This can also be rewritten as P gov2 k 3349 Outline n Viscosity 3149 Definition Viscosity is a resistance at the perimeter of the pipe as the fluid interacts with the walls of the pipe Thus the velocity near the walls is less gt gt 7 3249 m1 A laminar flow is one in which the fluid flows smoothly A turbulent flow is one in which the flow has more violent swirls and flows in the reverse direction httpenwikipedia orgwikiFile HD Rayleigh Taylorgif 3939 39 40 Fluids 2 LViscosily Samples Table Viscosities Liquid 7 ethyl alcohol 11 blood 3 4 milk 3 olive oil 84 SAE 10 motor oil 65 SAE 40 motor oil 319 water 179 tar 30000 4u49 Fluids 2 I Viscosity Glass Figure shows the viscosity table for various glasses at various temperatures It should be noted that there is no transition phase 12 Common glass I VISCOS39ty Cu Wes Soda lime glass for containers 10 Fiber wool glass for insulation Lowexpansion borosilicate similar to Duran or Pyrex 8 Eglass borosilicate with low alkali Panel glass for traditional color TV 100 silica glass quartz glass Lead crystal example composition 4 V References 1 quotHigh temperature glass melt property database for process log10viscosity Pas CD modelingquot Eds Thomas P Seward Ill and Terese Vascott The American Ceramic Society Westerville Ohio 2005 2 2 SciGlass 67 2006 3 A Fluegel quotGlass Viscosity Calculation Based on a Global Statistical Modeling Approachquot Glass Technol Europ J Glass Sci Technol A vol 48 2007 no 1 p 1330 httpglasspropertiescomviscosityl 0 500 700 900 1 100 1300 Temperature C 4149 Hr LWeiqu in the Pool Outline H Weight in the Pool 4249 HuMsZ LWeight in the Pool Does Ezra Float What is Ezra39s weight in a pool of water I The density of the human body is phuman 945kgm3 I The density of fresh water not pure water is pW 998kgm3 at 20 C I Ezra has a mass of 90kg 4349 HuMsZ LWeight in the Pool Does Ezra Float What is Ezra39s weight in a pool of water I The density of the human body is phuman 945kgm3 I The density of fresh water not pure water is pW 998kgm3 at 20 C I Ezra has a mass of 90kg This has to be done in parts 4349 HuMsZ LWeiqu in the Pool Part 1 What is Ezra39s volume 4449 HuMsZ LWeiqu in the Pool Part 1 What is Ezra39s volume 4449 HuMsZ LWeight in the Pool Part 1 What is Ezra39s volume 90 m 7 0095m3 945 To confirm this consider a rectangle that is 2m high and has a footprint of 03m x 015m This is roughly the shape of a human The volume of this rectangle is 2 X 03 X 015 009 which is about the same as Ezra 4449 HuMsZ LWeiqu in the Pool Part 2 So if Ezra is completely submerged then he will displace 0095m3 of water 4549 HHMSZ LWeiqu in the Pool Part 2 So if Ezra is completely submerged then he will displace 0095m3 of water What is the mass of the water the Ezra displaces 4549 HHMSZ LWeiqu in the Pool Part 2 So if Ezra is completely submerged then he will displace 0095m3 of water What is the mass of the water the Ezra displaces m pV 9980095 948kg 4549 HHMSZ LWeight in the Pool Part 3 Thus the displaced water is 48kg more than Ezra Thus I Ezra weighs 74898N in water which is about 710le I Since he weighs less he will rise to the top 4049 HHMSZ LWeiqu in the Pool Part 4 If Ezra were placed in salt water the mass of the water displaced would be m pV 10270095 976kg He would rise to the top faster 4149 HuMsZ LJunkyard Wars Outline E Junkyard Wars 4349 Chapter 12 lVlOIIientum and Impulse Bounce a ball on the oor It changes direction which is an obvious acceleration Is it possible to compute the force by F ma Even though it looks like the ball instantly changed directions it actually took a nite amount of time In Newtonian physics objects don t change velocities instantly 121 De nitions According to Newton s Second Law F ma Substituting Equation 59 produces AU F 7 121 ma m At Consider then the term of FAt FAtmAU 122 This is the force through a unit of time which is the de nition of impulse JFAtmAU 123 An impulse acting on a mass changes the momentum Using 19 to represent momen tum p mu 124 the impulse is therefore also described as J Ap mAU 125 100 CHAPTER 12 MOMENTUM AND IMPULSE Example 121 Momentum Given a mass m 7kg moving a speed of v 2ms what is the momentum p mu 7kg2ms l4kgms Example 122 Impulse A golf club exerts a force of SOON on a 01kg golf ball for 0013 What is the impulse What is the change in velocity of the ball FAt 500N001s 5N3 Ap 5N3 A Z 01kg mmS Example 123 Bouncing Ball A ball bounces on a oor Just before impact the ball has a velocity of 710msyA and just after the impact the ball has a velocity of 10msy The duration of impact is 013 and the mass of the ball is 01kg What is the force experience by the ball The change in momentum is7 Ap mAU 0120 2N 5 The force is then i Ap i 2 F777720N At 1 121 DEFINITIONS 101 1211 Conservation of Momentum For a collision the total momentum of a system is conserved Consider a case in which there are two train cars on railroad tracks and they are far apart The rst car has a mass of m1 and is moving towards the second car with a velocity of U1 The second car has a mass of mg and is moving towards the rst car with a velocity of 122 When they collide they will couple with each other The momentum before the collision is p1 192 mlvl mgvg where 122 is negative because it is going in the opposite direction as 121 After the collision the momentum is p3 m1 m2v3 because the total momentum is conserved From this it is possible to compute the nal velocity as7 m1111 7712112 U3 126 m1m2 A 22 caliber ri e weighs 751193 and shoots a bullet that weighs 269 with a velocity of 330ms The total momentum before the shooting is 0 since neither the gun nor bullet are in motion After the shooting the bullet is in motion and therefore has a momentum The total momentum is still 0 and therefore the gun must have a velocity in the opposite direction This is the gun s recoil which is usually stopped by the shooter What is the velocity of the gun First the measurements must be converted to the metric system Since the gun is near the surface ofthe Earth the weight can be converted to a mass 34kg The mass of the bullet is 00026kg The momentum of the bullet after the shooting is 193 00026330 086kg ms Since the total momentum is 0 for this system the gun must have the same momentum in the opposite direction p 7086 7 7 7025 127 W mg 34 77187 where the negative sign indicates that the gun is moving in the opposite direction of the bullet An episode of Mythbusters considered the idea of putting bullets in a hot re to see if they will re They did indeed re and the reader should not try this experiment They surrounded the re with planks of wood to absorb the ying bullets However7 they were surprised to see that the damage was less than expected because it wasn t the bullets that were ying into the wood When the bullet red it separated the projectile from the casing Since the casing is much lighter it was the part that ew the farthest and struck the wood planks Again this is a serious dangerous experiment and should not be attempted Recall that in Section 411 that the collision of a Hummer and Smart car were considered Clearly7 if the two cars were traveling at the same speed towards each other the much larger Hummer would survive the collision in much better shape than the Smart car The curb weight of a Hummer H2 is 6614le BOOOkg at sea level and the curb weight 102 CHAPTER 12 MOMENTUM AND IMPULSE of the Smart Fortwo is 16001123 730kg If the collision is to have equal momentum from both cars then ml Ul 777121127 By rearranging this it is possible to determine the difference in speed required to have the same momentum for both vehicles 3000 E i41 128 Ul m2 So the Smart car would have to travel at more than four times the speed of the Hummer to have the same momentum Example 124 lmpulse in Baseball A baseball has a mass of 0145kg and when pitched has a velocity of 745ms After the ball is hit is has a momentum of 55ms The duration of the collision is 2007713 00023 What is the applied force The impulse is the change in momentum Ap mAU 0145100145kgms J F 7 7250N At 1212 A Sticky Pendulum Consider a case in Which there are just two balls When one strikes they stick and both rise together Mass of spheres is 01kg each Height that rst ball is raised to 04m What is the PE when rst ball is lifted PE mgh 019804 0392J What is the KB just before the rst ball strikes KE 0392J Total energy is conserved What is the velocity of the rst ball just before the collision 7 2KE 7 20392 7 V7 01 2398m8 121 DEFINITIONS 103 What is the momentum just before the collision p mu 0128 028kg ms What is the momentum just after the collision p 028kg ms Momentum is conserved What is the velocity of the two balls as they leave the pile p 7 028 7 U2 7 m 7 702 714ms What is the kinetic energy after the collision 1 2 2 KE 7 Emu 7 050214 7 0196 ls KE conserved in this system NO What is the height that the pair reaches PE 0196 779 0298 03917 1213 Vector Representation Figure 121a shows a mass moving downwards with a velocity of my A force from the i direction is applied and the mass now moves with a velocity of 172 which has the initial 37 component plus the additional momentum obtained from the applied force in the i direction Figure 121b shows the change in momentum as a vector in red Chapter 1 Heat Recall the rst law of thermodynamics from Equation 7 which states that an increase in internal energy U is the increase in heat Q minus the work Wi 11 Heat Engine In a heat engine heat will ow from a hot reservoir to a cold reservoiri When this occurs it is possible to perform work This is described by WQHchl 11 This is shown in Figure 11 which shows two heat reservoirs The reservoir marked TH is the hot reservoiri Heat QH can ow from the hot to the cold side and along the way an amount of work can be done Figure 11 Heat exchange and work is done The ef ciency of which this work can be done is described by W QT 12 e Example 11 Work from heat A heat engine takes in 1200 from a hot reservoir and does 400 of work What is the e iciencyi7 W 400 770i3333 e QH 1200 12 Second Law of Thermodynamics The second law of thermodynamics states that heat will ow from a hot source to a cold source In order for heat to ow from the cold to the hot work must be done 121 Carnot Cycle The Carnot cycle describes the processes involved in the ideal cyclical heat enginei Figure 12 shows the four stages of the Carnot cycle with a piston Figure 12 A Carnot cycle The four stages are H i isothermal expansion adiabatic expansion i isothermal compression ewgo adiabatic compression An isothermal expansion is one which the temperature remains constant but the volume is increased In this stage heat is added to the system but the temperature remains constant An adiabatic expansion is the second stage and the volume is increased but there is no heat ow In this case the temperature can change according to the gas lawi Stage three is the isothermal compression in which the volume is decreased and heat ows from the system into an external reservoir The nal stage is adiabatic compression in which heat does not ow but the volume decreases This cycle continues back to the rst stages The e iciency of this system is described by T H 7 To 7 7 1 3 6 1H It should be noted that this computation must be performed in an absolute temperature scale In the metric system the temperature uses Kelvin where K C 273 Figure 13 shows the ow diagram for the Carnot cyclei Figure 13 Carnot cyclei Another manner to examine a Carnot cycle is shown in Figure 14 The top and bottom curves are isotherms In phase I the temperature remains constant but the volume increases and the pressure decreasesi In phase H the adiabatic expansion continues to increase the volume and decrease the pressure but in this phase the temperature is also decreased Phase Ill is the isothermal compression in which the volume is decreased pressure increases along a constant temperature The nal phase is the adiabatic compression in which the volume is decreased and both the pressure and temperature increase The gas law is obeyed in all four phases of this cycle Work is done during the expansion phasesi As the piston is pushed outwards it can do mechanical work In a car engine the piston expands and it turns a crankshaft which through a transmission will turn the car s axlesi 13 The Refrigerator Figure 15 shows the typical refrigerator The blue box represents the inside of the fridge and the two green objects represents the coils which are connected to a pump The coils contain a uid that is pumped through the fridge and its pressures are managed so that heat can be removed from the box V Figure 14 Carnot cyc Example 12 Carnot engine Given a cyclical heat engine that receives stearn at 400 C and dispenses it at 120 C This system takes in 59 of heat in each cycle What is the ef ciency of this system TH e To 673K 7 393K 7 7 0 416 g H 673K What is the maximum arnonnt of work that can be done in a single cycle W eQH 0 416500kf 207k Pump Figure 15 A refrigeratori In this case the purpose is to take heat from the cold air on the inside and move it to the hotter air on the outsider Since heat is being moved from cold to hot work is required to be performed Thus it is necessary to have a pump and pressure regulation All of this requires electricity which is equivalent to stating that work needs to be done The temperature of the box is T1 and the desire is to move heat from there to the outside which has a temperature of T2 and T2 gt Tli The rst step is to provide a uid in the coil that is at a temperature T3 where T3 lt Tli According to the gas law by manipulating the pressure and volume it is possible to manipulate the temperature Heat will naturally ow from the box to the coil since T3 lt Tli The uid is then pumped to the outside and along the way to pressure and volume are adjusted so that the uid heats up so that T4 gt T3 In this case the heat will naturally ow from the uid to the outsider Thus heat has been move from the inside of the box to the outsider However work was required and machines are needed to pump the uid and to adjust the pressures The heat ow is shown in Figure 16 Figure 16 A refrigerator schemei 14 Entropy Entropy is the measures of disorder or randomness in a closed systemi Another way of stating the second law of thermodynamics is that entropy is always increas ing First consider a system in which there is a box that is a divided by a walli Gas molecules are injected into one side of the box There is a bit of order to the systemi Then the wall is removed and the gas naturally expands The gas molecules will stay in motion but the system will never naturally return to the original state The disorder of the system has been increased Now it is possible to put the wall back in place and pump the gas from one side to another In this situation the order has been restored but it took a pump to do it That means that there was work done and the pump needed energy such as electricity Somewhere far away there is a coal red power plant that is generating the electricity by burning the coali In this process the order of the coal is destroyed during the burning process The entropy of the gas in the box is decreased but the entropy of the coal is increased According to the second law the entropy of the entire system must increase Therefore we know the the increase in the entropy of the coal is more than the decrease of entropy of the gas in the box 15 Examples 151 Back to the Future In the movie Back to the Future Doc builds a time machine from a DeLorean car and claims that it takes 1 21GW gigawatts to run the time machine That is a lot of energy Let s assume that the machine has an incredible ef ciency rate of 80i That means that 20 of the energy used in the time machine is converted to waist heat This is a lot of energy Would it melt the DeLorean In order to do this problem a bit more data is needed First the car is constructed mostly from aluminum and so it will be assumed here that that is the only materiali 0 Mass of a DeLorean 1420169 0 Speci c heat of aluminum 902Jkg O Melting temperature of aluminum 661 C 6 Giga 109 The problem states that 20 of the energy is converted to waste heat This is the heat that will be absorbed by the car This is 121 1002 2 42 103 The car is at a normal temperature 20 C Using Q chT the nal temperature of the car is Q 242 108 TT i20 209 C 2 1 me 1420902 This is not the melting temperature of aluminum and so the car survivesi However it would be a bit much for the passenger Chapter 1 Collisions There are three basic types of collisions perfectly inelastic partially inelastic and elastic Each of these will be discussed with examples 11 Overview A collision in this chapter occurs when two masses with at least one mass having a velocity meetl There are a few options that can occur There can be a collision in which the two objects stick together the two objects completely bounce off of each other or the two objects bounce of off each other with a loss of energy due other effects noise heat etc Each of these are de ne l A perfectly inelastic collision is one in which the objects do not bounce off of each other They stick together A large example is that of two railroad carsl One moves towards the other and when they meet they couple together They now move as one mass A partially inelastic collision is one which the masses do not stick together and kinetic energy is lost Actually the energy is not really lost but it is transformed into sound heat light andor deformation of the masses An elastic collision is one in which the objects bounce off of each other and no energy is lost In all collisions momentum is conserved The total momentum before the collision is the same as the total momentum after the collision In an elastic collision kinetic energy is also conserved Kinetic energy is not conserved in an inelastic collision 111 Perfectly Inelastic Collision In an inelastic collision two masses collide and stick together Momentum is conserved Consider two masses both 1kg both moving at lms but towards each other The mo mentum of the rst mass is p1 mlvl lkglms 1kgmsl The momentum of the second mass is the same magnitude but in the opposite direction p2 71kgmsl The total momentum is pT 101 p2 0 After the collision but balls come to a stop and have no velocity The total momentum is still the same Consider a different case in which m1 2kg and 7212 1kg but the velocities are the same as in the previous case Now the total momentum is PT p1 102 2kgmsilkgms lkgmsl After the collision the two objects stick together Thus the velocity of after the collision can be compute 1k 1 3 gS 0 33msl 11 m Example 11 Skateboard A girl mass 700 kg is running 30 ms east when she jumps onto a stationary skateboard mass 20 kg What is the velocity of the girl and skateboard assuming they move off together P1 P2 Pf 700x30 20 0 p3 p3 2100 MW Pf pf 2100 7 7 29 W mf 720 M8 112 Elastic Collisions In an elastic collision the two masses bounce off of each other Both momentum and kinetic energy are conserved Consider two masses The rst has a mass of 1kg and is moving directly towards the second with a velocity of 2mamp The second has a mass of 2kg and it is not moving The two masses collide and bounce off of each other The second mass then has a velocity of 43msl What is the velocity of the rst mass This problem can be solved in two ways The rst solution is to consider the conservation of momentuml The momentum before the collision is mlvil m2vi2 2kg mamp The momentum after the collision must be the same 2kg ms mlvfl mgvfgl All of the variables are known except vfl and so the equation is rearranged to solve for it 12 P Wv 2 243 2 17177 7 fiig m1 This indicates that the rst mass reversed direction The second solution is to use the conservation of kinetic energy The total energy before the collision is KE m1vi21 0 21 The total kinetic energy after the collision must have the same value Thus KE 2111 mgv l Again all of the variables are given in the problem except vfll So the equation is rearranged to provide ig 1 3 113 Two Equations and Two Unknowns In actuality the previous problem can be be solved without knowing the nal velocity vfgl However this math becomes a bit more involved To begin there are two equations with two unknowns The equations are p mlvfl 7212va 14 1 1 KE Emlv l E721212 15 The two unknowns are vfl and vfgl Rearrange Equation 14 to solve for vfl to get p 7 7212va 7 16 f1 ml The expression for vfl is then plugged into Equation 15 to get 1 1 p 7 m 1 2 K EmngQ Eml f2 17a 1 1 K 721212 R p2 7 2177212va mgv 17b 2 2 0 2 E L12 7pm L 7 K 17 v lt 2 T 27211 172 m1 T 27211 C Equation 17c is a quadratic equation of the form 0412 121 c 0 and the solution for this equation is wellknown 7 i b2 7 4ac z 18 a For this example however there is a bit of luch in that 7 K 0 and so it is not necessary to use Equation 18 Instead Equation 17c becomes 0Uf2lt gt TM 19 27211 m1 Plugging in the values for all known terms on the right hand side of Equation 19 the value for vfg is computed to be Plugging vfg into Equation 16 reveals vfl 7 12 Collision Angle The angle of impact and the angle of re ection are the same for an elastic collision as shown in Figure 11 Consider the collision in Figure 12 The rst object m1 is traveling at a speed of v1 and collides with the second mass 7212 which is at rest The balls move off at an angle 9 This angle is dependent upon the angle of collision and therefore can be a given parameter In this problem the total momentum is conserved in both the i and Q directions Given m1 014169 7212 023169 v1 17213 and t9 300 Since 7212 is not moving the total momentum before the collision is the momentum of ml which is 171 721151 014x12 0kg 0142 0kg 1 10 Figure 11 Incident and re ective anglesi Figure 12 A nonsticky collision The momentum in the Q direction was 0 before the collision and must therefore 0 after the collision This means that the 1 component of 103 must be equal and opposite to the 1 component of PE 123 12 0 111 If the velocity after collision of 7711 is measured to be 06 0317715 then the velocity of 7712 after the collision can be computed Using Equation 111 we get 7711 014 A MM Eviy 03 018g ms 1 12 and conserving momentum in the i direction we get 123 122 P17 m10i6mSm2v2 0i14kgms 1 13 Uzi 014kgm32gkgO6ms 024m So the velocity after collision of 7712 is M 024i 7 0 18y7msi This can be con rmed by computing the total momentum after collisioni Actually in an elastic collision both the kinetic energy and the momentum are conserved Using both of these properties and 19 it is possible to calculate the velocity of both balls after the collision However this is a bit involved and is not pursued herei 13 Ballistic Pendulum Given a pendulum in which is made of wood and weighs mm 2169 A bullet weighing 72 0 0051i29 is red into the wood The wood acting as a pendulum swings up to a height of h 00762721 What is the velocity of the bullet The rst item of note is the the weight of the bullet is much less than the weight of the wood In order to properly add the two weights it takes 4 digits of precision Thus this problem will be solved in four digits The gravitational acceleration is 9 9 807ms2i At the full height of the swing the potential energy of the pendulum is PE m9h 2005169 9 807m320i0762m 14981 114 The total energy of the system is the potential energy of the pendulum when it reaches its full height The total energy just after the collision of the bullet and the wood is also the same value From this is it is possible to compute the velocity of the two objects just after the collision 7 ME 7 21i498 7 W m71i222msi 115 This is the velocity of the combined objects after the collision This collision is inelastic which means the KE is not conserved However momentum is The momentum just after the collision is pf m mwvf 2005169li222ms 2450169 mamp 116 The momentum before the equation is the same value pi 2450169 mamp This can be written as Pi Pb 101 mb39Ub 77va 117 where vb is the velocity of the bullet before impact The vw is the velocity of the wood before impact and it is 0 Therefore p5 2450169 mamp From this it is possible to calculate the velocity of the bullet before impact 7 p71 7 2450169ms 490m i 118 vb mb 0005169 8 1 Show that momentum is conserved PbPw Pf 119 7215115 mwvw m v 120 0005 490 0 2005 1222 1 21 245 2 45 122 14 Ballistic Spring Experiment Consider the experiment shown in Figure 13 A bullet with a mass of 269 is red with a velocity of 330ms into a block of wood that has a mass of lilkgi The block of wood is attached to a spring with a spring constant of k 249N in The goal is to determine the how far the block of wood movesi 26 g 330 ms k349 Nm Figure 13 A bullet is red into a block of wood attached to a spring This problem is performed in several parts First the mass of the bullet is converted to kg to be congruent m3 00026in This is an inelastic collision because the bullet sticks into the block of wood Only the bullet is moving before the collision and therefore the total momentum is p vaB 0 0026330 0 858m52i Since it is an inelastic collision the total momentum is conserved The momentum after the collision is also pf 0 85877152 Just after the collision the bullet is stuck in the wood and the two items have a momentum of pf 0 858m52i The velocity just after the collision is then 7 pf 0858 if 0778 i W m 110 M5 Once the block starts moving the spring starts to compress and the blockbullet will slow down The kinetic energy just after the collision and before the spring begins to compress is 1 2 2 KE i771va 0 51i10260i778 03341 Now the spring begins to compress and the kinetic energy becomes potential energy stored in the spring When the movement stops the spring is compressed and the kinetic energy is 0 Therefore all of the energy is now potential energy Since total energy is conserved the potential energy at compression is PEG 03341 The amount of compression can be determined by PE k12 7 2PE 7 20i334 7 I n 4 T 7 MW 7 0 0437mi

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