GEN PHYSICS A STUDIO
GEN PHYSICS A STUDIO PHY 2048C
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Date Created: 09/17/15
Motion in One Dimension The study of motion is called kinematics Displacement of a point particle Ax x2 961 where 961 is the position at time t1 and x2 is the position at time t2 152 gt151 The average velocity Tipler figure 24 is defined by A96 962 961 U av At t2 t1 The average velocity can take positive or negative values The average speed is defined by total distance As gt 0 A e s eed verag p total time At Instantaneous velocity Tipler figure 25 2115 11m g111rn 36 362 d x At gtO At t2 gtt t t2 dt This is the slope of the tangent of the curve ZEt at t and called derivative Note t t1 x 961 of the difference notation The instantaneous speed is the absolute value ivti of the instantaneous velocity Relative VEIOCity If a particle 19 moves with velocity va relative to a coordinate system A which moves with velocity UAB relative to a coordinate system B the velocity ofp with respect to B is UpB UpA i UAB This equation is only valid for velocities M lt C where C is the speed of light Acceleration Acceleration is the rate of change of the instantaneous velocity The average acceleration is defined by Av212 211 At 752 251 aav The instantaneous acceleration is t 1 Av 21 212 d v d a 1m 1m Ala gt0 At t2 gtt t t2 dt dt2 Motion With Constant Acceleration do a aav Integration d vtd voat Here 210 is the velocity at time zero called initial condition Second integration 1 2 05 0 Uot l Eat Here 960 is the second initial condition the position at time zero Illustrations a Air track b constant acceleration of free fall Note 21 210 212 213 2a v 210 1 1 Am avt t 16 x0 21 2 U l 210 2 212 vg l 2an General integration Tipler figure 218 Questions PRS Units The meter is nowadays defined by 1 Scratches on a platiniumiridium alloy kept at the International Bureau of Weights and Measures in S vres France 2 As a fraction of the circumference of the earth around the equator 3 As a distance traveled by the speed of light in vacuum during a certain time Significant figures The experimental measurements 250 m and 2496 m Disagree Are identical Agree within measurement errors Order of magnitude The number 250 has the order of magnitude 10 102 103 25 X 102 Relative velocity A swimmer moves a river upstream with the velocity of 120 kmh relative to the water The speed of the water is 01 ms relative to the ground The velocity of the swimmer over ground positive upstream is 1 0 kmh 2 3000ms 3 084kmh 4 084kmh 5 3000 ms Appendix III Least Squares Adjustment of Data Principle of Least Squares For many experiments one searches for a possible equation that relates one variable call it X to another variable call it Y Experimentally one measures Y for different values ofX plots the results of the measurements and then tries to draw a curve through the data points This is called the regression of Y on X as a simple mathematical model it will be useful as a brief and precise description or as a means of predicting the quantity Y for a given X The following discussion deals with the mathematical formalism used to find the best possible curve through the data The famous least squares criterion for fitting a curve is based on the idea that one would like to minimize the difference between the data and fitted or predicted curve This minimum difference is found by comparing each of the data points Y to their predicted values Y It is not sufficient to simply add up these differences since the positive and negative errors would cancel The accepted practice is to add up the squares of these differences and minimize that sum hence the name least squares Using the absolute values r Y has the disadvantage that it does not always treat each point equally The least squares method of fitting a curve through data points is motivated by the following criteria Squaring the difference between the fitted values and the measured values makes positive and negative errors weighted equally 2 Squaring the difference between the fitted values and the measured values emphasizes large errors and in trying to find the minimum X2 large errors are suppressed as much as possible 3 The algebra ofleast squares is very manageable In all problems simple or complicated the principle of least squares requires the minimizing of the sum of the weighted squares of the residuals N N S E Zial Veiz39daalx2 211 X a2 11 11 where residual Mmveal value 7 tabulated value X 7 a 77 note that we don39t know a yet The value 11 is the weight of the 2M residual For example this could be used when the same residual occurs twice then the residual value is entered once but with a weight of two Also it sometimes occurs that measurements are made with different reliability and then the weight factor is also used We will assume w 1 for all of our examples The method afleaxt iguarex is a set of rules for minimizing S We define a function called Cbzlxaaared by 2253 7 Hm where 7I denotes the standard deviation of the observation making up the measurement saInple If we assume 7I to be a constant X2 is a minimum when X is a minimum Hence the least squares method minimizes X and X2 Least squares also minimizes the external smndard deviation ie the uncertainty in the tted paraIneter In the simplest case only one parameter call it a is varied Its uncertainty is denoted by in S The minimum value off could be found by varying a and checking 3 each time Plotted one obmins To compute the value ofa for which S becomes a minimum L minimum 8 013 one requires E 0 and nds M d N 2 NX a aminimum ltX agt 2 gt 0ltt gt N N 206 dZXZ Na0 11 11 Thus the least squares value ofa is 9 the mean value OfX and a E gives the smallest possible value to the sum ofthe squares of the residuals the X2 and the external standard deviation The standard deviation of the observations 65 is computed as a 71 lt gt2 Xi d N1 and the standard deviation of the mean 7 r 7 a m Least Squares Fit to a Straight Line Consider N data values with ordinate and abscissa X see the sketch below Assume the coordinates but not the X coordinates are subject to uncertainty Although seldom totally true this assumption is reasonable for most of our experiments Suppose the equation to be tted to these data is y a kx Let y denote the y observed ordinate at the 2M point then the residual at that point is residual yam exam M y 751 kg where a and k are to be obtained from the t using the method of least squares This residual is a y residual because the uncertainty in our dam is assumed to be all in Then the sum to be minimized is Residual y ycaic Plotofyabx ya N 2 x1 x2 x3 x4 x x 3 Zia0 PM 11 where w is the weight ie importance assigned toy If we nd values ofa and k which minimize S we will have the least squares result for the best t straight line that describes the data For this example we will assume that the weight w are all equal to unity To nd the minimum S we differentiate X with respect to the unknown parameters in this case a and k and set each of the partial derivatives equal to zero 63 N i 2 k 0 a go a x N 2Xlyl a kxl50 Solving for a and k between the two resulting equations we obtain kzZlel N jZlel N j ZXf NEZ m 1 1 where EEEZXZ jEEZJl and mZXf N 2Allsumsrunfromz391toiN Using these values ofa and k and some elemenmry algebra we nd that the value of the minimum X2 is X 1 N X2 27272ZJG2152 52 a a 11 where 7 the uncertainty in the coordinate of a data point has been assumed to be the same for all points This assumption is usually reasonable for our experiments The Equipartition Theorem Degrees of freedom are associated with the kinetic energy of translations rotation vibration and the potential energy of vibrations A result from classical statistical mechanics is the equipartition theorem When a substance is in equilibrium there is an average energy of krT2 per molecule or RT2 per mole associated with each degree of freedom The Mean Free Path Despite the high average speed of the molecules in a gas the transmission of odor is relatively slow The reason are collisions of the molecules with one another If we have one type of molecule and regard it as a little sphere of diameter d the average time 739 between collisions may be calculated 1 ve nvvtd2 Here vav is the average speed of a molecule and 71 NV is the number of molecules per unit volume The average distance traveled by the molecule between collisions is called the mean free path UaVT The MaxwellBoltzmann Distribution Not all molecules in the gas are at the average speed We would like to calculate the distribution of speeds Familiar examples of distributions are the grades distribution in a class of students or the distribution of the heights of a large number of people The MaxwellBoltzmann speed distribution of molecules in a gas is 4 m 32 771212 fl 2k T U2eXp2kT39 This function is a probability density which fulfills Oodvfv1 O The average of the velocity squared is then a standard integral aw foodvmv 3 O This is our familiar result which gives the rms velocity vrms x3krTm Note that the rms velocity is not the most likely velocity which is given by the velocity vmax for which the probability density takes its maximum A short calculation gives vmax x2k Tm With decreasing temperature fv gets narrower peaked around its maximum values fvmax This is depicted in figure 1716 of Tipler The Energy Distribution The MaxwellBoltzmann speed distribution yields the distribution of kinetic energies For E mUZ2 we get dE mvdv and we can write 4 m 32 2 771212 fwd 7 U exp 2kT d where 32 2 1 E E fenergyw kT eXp kT is the MaxwellBoltzmann energy distribution In the language of statistical physics the factor 32 2 1 E W m is called the density of states seen to be proportional to and the factor E exp is called the Boltzmann factor Heat Capacity and Specific Heat Chapter 19 When heat energy flows into a substance the temperature of the substance rises unless the substance undergoes a phase transition which are treated later The amount of heat energy Q needed to raise the temperature of a substance by a small amount is proportional to the temperature change and to the mass of the substance Q CAT chT where C is called heat capacity and C specific heat The heat capacity is energy needed per one degree temperature change and the specific heat is the heat capacity per unit mass C m The molar specific heat C is defined by where n is the number of moles The calorie cal is the historical unit of heat energy originally defined as the amount of energy needed to raise the temperature of one gram water by one degree more precisely from 200C to 210C but in the range from 00C to 1000C the variations are less than 1 In the same way one one defines the kcal using one kilogram of water Nowadays the calorie is defined in terms of the SI energy unit loal 4184J Another customary unit of heat is the British thermal unit BTU which was originally defined as the amount of energy needed to raise the temperature of one pound of water by on Fahrenheit The BTU is related to the calorie and joule by 1BTU 252 cal 1054 kJ The original definition of the calorie implies that the specific heat of water is Cwater 1ca1gK 1kca1kg K Table 181 of TiplerMosca lists the specific heats of some solids and liquids Calorimetry The specific heat of an object is conveniently measured by heating the object to some temperature placing it in an thermally insulated water bath of known mass and temperature and measuring the final equilibrium temperature This procedure is called calorimetry and the insulated water container is called calorimeter Change of Phase and Latent Heat When heat is added to ice at 00C the temperature of the ice does not change instead it melts This is an example of a phase change Common type of phase changes include 1 2 Fusion liquid to solid Melting solid to liquid Vaporization liquid to vapor or gas Condensation vapor or gas to liquid Sublimation solid to vapor or gas eg for dry ice Various forms of crystallization The fact that the temperature remains constant during a phase change has to do with the potential energy of the molecules For instance the molecules in a liquid are close together and exert attractive forces on each other while the molecules in a gas are far apart Vaporization achieves the corresponding increase in potential energy The heat energy per unit mass required for a change of phase is called the latent heat L Q mL The latent heat for melting is called latent heat of fusion Example For water at a pressure of 1atm the latent heat of fusion is Lf 797 kcalkg 3335 kJkg and the latent heat of vaporization is L 540 kcalkg 226 MJkg Table 192 of Tipler compiles the latent heats of fusion and vaporization for various substances The First Law of Thermodynamics In the 18405 James Joule showed that heat is a form of energy and energy is conserved This is simply energy conservation Joule was the first to demonstrate the conversion of mechanical work into heat His apparatus is shown in figure 182 of Tipler Mosca It is customary to write W for the work done by the system on its surroundings for example if a gas expands against a piston The heat Q is taken positive when it is put into the system Using these conventions and denoting the internal energy of the system by U the first law of thermodynamics reads QAUW The Internal Energy of an Ideal Gas We have already seen that the kinetic energy K of the molecules of an ideal gas is related to the absolute temperature T by K 3nRT2 where n is the number of moles of a gas and R is the universal gas constant If the internal energy of a gas is just the translational kinetic energy then UKgnRT If the molecules have other types of energy in addition to the translational energy such as rotation the internal energy will be larger adding nRT2 for every additional degree of freedom according to the equipartition theorem Again the internal energy depends only on the temperature and not on the volume or the pressure U UT for the ideal gas Joule verified this by finding no temperature decrease in the free expansion experiment of figure 185 of TiplerMosca Precision measurements show a small temperature decrease due to slightly nonideal behavior of real gases Work and the PV Diagram for a Gas In many types of engines work is done by a gas expanding against a movable piston Examples are steam and gasoline engines Quasistatic Processes Figure 186 of TiplerMosca shows an ideal gas confined in a container with a tightly fitting piston assumed to be frictionless We move the piston slowly in small steps so that equilibrium holds after each step Such a process is called quasistatic Let A be the area of the piston If the piston moves a small distance x the work done by the gas on the piston is dWFdxPAdxPdV where dV Adm is the increase in the volume of the gas PV Diagrams We can represent the states of a gas on a diagram P versus V Figure 18 7 of Tipler Mosca shows shows a PV diagram with a horizontal line representing states which all have the same value of P Such a process is called an isobaric expansion The work done by the gas is equal to the area under the P versus V curve WPdV Figure 18 8 of TiplerMosca shows three different paths on a PV diagram for a gas that is initially in the state P1V1T and finally in the state P2V2T As the initial and final temperatures agree P1 V1 P2 V2 nRT holds ideal gas Figure 18 8a The gas is heated at constant pressure and then cooled at constant volume The work done is P1V2 V1 Figure 18 8b The gas is cooled at constant volume and then heated at constant pressure The work done is P2 V2 V1 Clearly P2V2 V1lt P1V2 V1 Figure 188c Path C represents an isothermal expansion meaning that the temperature remains constant We calculate the work done using P nRTV RT dVlisothermal n V2 dV V2 VViso erma R T R T l th 1 n v V n n V1 We see that the amount of work done by the gas is different for each process illustrated dV Since U2 U1 for all these states the net amount of heat added must also be different for each of the processes This illustrates that the work done and the heat added depend on how the system moves from one state to another whereas the change in the internal energy does not Another Projectile Motion Question The initial conditions of two point particles are as follows 20 2x10Ax ASE gt 0 920 910 AC9 Ag gt 0 71mg E 111095 ASE 020 U20y 0 010 gt 07 711 gt 07 There is free fall downwards in y direction Which of the following holds draw the situation and pick one choice 1 Particle one will hit particle two unless it hits ground first 2 Particle one will not hit particle two Solution 96115 21109615 y1t Uloyt g 96215 Ax y2t Ay 3979 At some time t1 1E1t1 21le t1 Ax x2051 391151 71mg 751 975 I 2 010 751 1 Ay Egty2t1 h1t Newton s Laws 1 Law of inertia An object continues to travel with constant velocity including zero unless acted on by an external force 2 The acceleration a of an object is given by m6 net 7 where m is the mass of the object and net the net external force 3 Action Reaction Forces always occur in equal and opposite pairs If object A exterts a force on object B an equal but opposite force is exterted by object B on A Definition of the Mass Mass is an intrinsic property of an object that measures its resistence to acceleration that is the object39s inertia If the same force F produces the acceleration a1 when applied to object 1 and acceleration a2 when applied to object 2 the ratio of their masses is defined to be m2 611 m1 02 By comparing with the 1 kg object kept at S vres we can thus measure the mass of any object Unit of force The force required to produce the acceleration of 1ms2 on the 1 kg standard object is called one Newton 1N1kgmSZ Weight The force due to the gravitational field g is called weight 117 mg Approximately g lgl 981 Nkg 981ms2 An object is said to be in free fall when gravity is the only force acting on it An example of Newton39s 3rd law Block on a table Springs When a spring is compressed or extended by a small amount Am the force it exerts is Hooke39s law sz kAx With Axx x0 Figure 45 of Tipler Mosca Let us choose 9amp0 0 the differential equation for this motion is d FxmamW kx It will be solved later in this course Solving Problems Draw a diagram Isolate the object under investigation Indicate all forces acting on the object Choose a convenient coordinate system Decompose the forces into components along the major axes Use Newton39s laws and solve the resulting equations for the unknowns Example Sledge Tipler Mosca figures 410 and 411 Fnu713 ma Fm max as Fn x 0 mm 0 Fn7ywyFy 0may Example Inclined Plane Tipler Mosca figure 413 0 Fwy Hay max wxwf9mgf9 1 x0vot gf6t2 R A H V PRS 1 f6 sin6 2 f6 3086 Example String Tension Tipler Mosca figure 414 special case and figure 437 ZFx T1 COSO T2 cos 0 ZFy 2 T1 sina T2 sin mg 0 cosa T2 f 1 cosw T1 sina T1W m9 T1 sinoz 008 sin COSO mg 008 T1 smog T1 2mg cosw What happens for 04 gt 0 sina l mm Hg and T2mg 00804 sina l J Example Two Connected Blocks Tipler mosca figure 421 with 6 O For block one gravity is cancelled by the normal force it remains the effect of gravity acting on block two Fm2gm1m2a m2 a g m1m2 1 96115 x0vot at2 1 y2t y0vot at2 Homework Calculate also the tension in the string Questions Assume m1 l m2 1kg and m2 50 g The expected acceleration is pick one 1 981ms2 2 04905ms2 3 0005ms2 4 50ms2 5 4905 ms2 Work and Energy Motion With Constant Force The work W done by a constant Force 13 whose point of application moves through a distance A9 is defined to be W F 3086 Ax where 6 is the angle between the vector 13 and the vector A5 see figure 61 of Tipler Mosca If Azf is along the xaxis ie MAxma then W Fm Ax holds Work is a scalar quantity that is positive if A96 and F96 have the same sign and negative otherwise The Si unit of work and energy is the joule J 1Q1Nm1kgm282 Another energy unit frequently used in physics is the electron volt eV 16V 1602 176 462 63 X 10 19 J is the actual value from the National Institute of Standards and Technology NIST surfe physicsnistgov Often used multiples meV keV MeV and GeV Which of the following choices corresponds respectively to meV keV MeV and GeV 1103eV 104eV 106W 109eV 210 36V 1026V 1036V 1066V 310 3ev 103eV 106eV 109eV 410 66V 1036V 1066V 1096V 510 3ev 1026V 103eV 109eV 610 36V 1026V 1036V 1066V Answer See table 11 of TiplerMosca Power The power P supplied by a force is the rate at which the force does work dW P dt The SI unit of power is called watt W 1W 1J8 1kW h 103 W 36008 36 x106Ws 36MJ mp 505 ft 93 746W 2 0746 kW Work and Kinetic Energy There is and important theorem which relates the total work done on a particle to its initial and final speeds lf Fm is the net force acting on a particle Newton39s second law gives Fm max and we recall the constantacceleration formula TiplerMosca eqn2 17 p28 between initial and final speeds vi UE2axAx Now the total work becomes 1 1 Wm maxAx Emu mv 2 where we substituted am Ax v Uf2 The kinetic energy of the particle is defined by 1 K mv2 2 The workkinetic energy theorem states The total work done on the particle is equal to the change in kinetic energy WW 2 Kf Ki Work Done by a Variable Force Tipler Mosca figures 67 and 68 32 W 11m ZFxAxiz Fmdx 239 1 area under the F96 versus 16 curve Example Work needed to expand a spring from rest When we choose 9amp0 0 for the rest postion of the spring Fxkx x0kx 139 139 1 W dex kxdx kx2 0 0 2 Hence Work and Energy in 3D Figure 612 of TiplerMosca For a small displacement AW 13A F cos A8 FSAS Here 13 A is called the dot product or scalar product of the two vectors For two general vectors A and B it is defined by f g ABCOS where o is the angle between fl and see figure 613 of TiplerMosca Properties of Dot Products Table 61 of TiplerMosca Commutative rule 3314f Distributive rule ff l 6 6 l 3 6 Further the following holds pick one 1 fl and g are perpendicular AB 1 2 fl and g are perpendicular g 3 fl and g are perpendicular 0 1 fland g are parallel AB 1 2 fland g are parallel g 3 fl and g are parallel 0 Cm CM CM CM CM CM CD The General Definition of Work 824 82 W Fd Fsds 31 31 3D Work Kinetic Energy Theorem 82 82 82 Wm asd8m d Ud8m Eds 81 81 dt 8 d8 dt 1 82 d 2 1 1 m v Ud8m vdv mv mv 81 d8 U1 2 2 Example 1 Skier skiing down a hill of constant slope Figure 618 of TiplerMosca Wm mgscos 900 6 h Wmgs sin6 mgsgmgh Final speed 21 1 1 W mgh mv2 mvg where 210 is the initial speed For 210 0 initially at rest we get for the final speed 21 2gh Example 2 Skier skiing down a hill of arbitrary slope Figure 618 of TiplerMosca dWm d mgds COS mgdh s h Wm d mg dhmgh 0 0 independently of the slope of the hill Potential Energy Often work done by external forces on a system does not increase the kinetic energy of the system but is instead stored as potential energy Examples figures 620 and 621 of TiplerMosca 1 Energy stored by lifting a weight 2 Energy stored by a spring Conservative Forces A force is called conservative when its total work done on a particle along a closed path is zero figure 622 of Tipler Mosca Potential Energy Function For conservative forces a potential energy function U can be defined because the work done between two positions 1 and 2 does not depend on the path 82 AUU2 U1 Fd 31 dU 13 d for in nitesimal displacements Example Gravitational potential energy near the earth39s surface dU 13d mg 39 dxidy dzl mgdy y UdUmg dy mgy mgyo yo U U0mgy With U0 mgyo Example Potential energy of a spring with x0 0 dU 13d dex lltxdx krdx 1 UkxdxU0 kx2 We may choose U0 0 such that U becomes a 1 U kxdx kx2 O 2 Nonconservative Forces Not all forces are conservative Friction is an example of a nonconservative force It eats up the energy which is converted as we learn later into heat Momentum Conservation TiplerMosca Chapter 8 The Center of Mass CM The CM 77cm moves as if all the external foces acting on the system were acting on the total mass M of the system located at this point In particular it moves with constant velocity if the external forces acting on the system add to zero Definition 39I L 39I L MfcmZm7F Where Mzzmi 21 21 Here the sum is over the particles of the system m7 are the masses and F are the position vectors of the particles In case of a continuous object this becomes Mfcmde where dm is the position element of mass located at position 7 see figure 84 of Tipler Mosca Example Figures 81 82 and 83 of TiplerMosca Two masses a placed on the xaxis and mm is defined by chm mlxl m2x2 Where Mm1 m2 If the masses are equal 81 the CM is midway between them If the masses are unequal 82 the CM is closer to the heavier mass Let us choose 83 961 0 Then m2 m2 xcm IE2 E2 M m1m2 PRS Assume a 8kg mass is at the origin and a 4kg mass at x06 m The CM is then at pick one 1 72m 2 03m 3 02m 4 04m 5 01m 6 18m Gravitational Potential Energy and CM The gravitational potential energy of a system of particles is given by U Zmighi 927mm z 1 z 1 Therefore by definition of the CM U thcm with thm 2mm 2391 We can use this result to locate the CM experimentally If we suspend an irregular object from a pivot the object will rotate until the CM reaches its lowest point and the CM lies somewhere on the vertical line drawn directly downward from the pivot By repeating this for several pivots we find the CM Demonstration CM of Florida compare figure 811 of TiplerMosca For the mathematicaly ambitious Finding the CM by Integration Example 1 Uniform Stick Figure 812 of Tipler Mosca dx M chmxdmMLxdx x2 E L 0 2L 0 2 L xcm E Example 2 Semicircular Hoop Figure 813 of Tipler Mosca d8 M d M z RdQ m 7TB 7TB x coordinate xcm 0 M W M W x xdm 7TRO deQ 7TRO RCOS6Rd6 MR 7T y coordinate ycm 2 R7r M W M W Mycmydm de6 Rsin6Rd6 7TB 0 7TB 0 W22MR O 71 7r2 2MR sin6d6 2MB 3086 0 7T 7T Motion of the CM Velocity of the CM 39I L dfcm d d772 df dt m1 dt m2 dt dt Differentiating again give the acceleration of the CM TL Micm 771151 m2 2 21 The acceleration of each particle is due to internal and external forces miai Fr Frith Fiext According to Newton39s third law the internal forces come in pairs which cancel and we find 39I L Majom Z Fiext Fnetext 21 The CM moves like a particle of mass M m7 under the influence of the net external force acting on the system Example Tipler Mosca figure 8 18 80 kg and 120 kg sit on a 60 kg rowboat 2 m apart The boat is at rest on a calm lake and 80 kg sits at the center of the boat 80g and 120 kg switch places How far does the boat move Solution As there are no external forces the CM does not change Therefore the difference in the CM position before and after the move is how far the boat moves We choose the center of the boat as the coordinate origin and find Before the move xcm 80 kg 0 60 kg 0 i 120 kg 2 m 0923m 80kg r60kg f120kg After the move 12 k am 0 ggtlt0gtlt60kggtlt0gtlt80kggtlt2mgt 0615m 80kg r60kg f120kg Therefore the boat moves Ax xcm x m 0923m 0615m 0308m Momentum Conservation Definition The mass of a particle times it velocity is called momentum m27 Newton39s second law can be written as d dm m m5 dt d7 net as the masses of our particles have been constant The total momentum 13 of a system is the sum of the momenta of the individual 39I L particles 39I L 21 7L1 Differentiating this equation with respect to time we obtain d d cm M Macm neex dt dt a t t The law of momentum conservation When the net external force is zero the total momentum is constant Fnet xt 0 gt P constant Example An astronaut of ma 60 kg weight is at rest relative to her spaceship She pushes a detached solar pannel of weight ma 80 kg away into space at 03 ms relative to the spaceship What is her subsequent velocity relative to the spaceship Announcements 1 Do not bring the yellow equation sheets to the miderm Idential sheets will be attached to the problems 2 Some PRS transmitters are missing Please bring them back Kinematics Displacement of a point particle A f2 31 where E1 is the position at time t1 and E2 is the position at time t2 152 gt151 Instantaneous velocity dz E This is the slope of the tangent of the curve at at t and called derivative 17 The instantaneous acceleration is dt dt2 i Motion With Constant Acceleration d2 dt average Integration v v d dt 0 Here 270 is the velocity at time zero the first initial condition Second integration 1 If f0170t 5t2 Here ED is the second initial condition the position at time zero Newton s Laws 1 Law of inertia An object continues to travel with constant velocity including zero unless acted on by an external force 2 The acceleration a of an object is given by m6 net 7 where m is the mass of the object and net the net external force 3 Action Reaction Forces always occur in equal and opposite pairs If object A exterts a force on object B an equal but opposite force is exterted by object B on A Friction If an external force acts on a heavy box standing on a floor see figure 51 of Tipler Mosca the box may not move because the external force is balanced by the force f8 of static friction Its maximum value fsamax is obtained when any further increase of the external force will cause the box to slide fsmaac Ms Fn where us is called the coefficient of static friction If the box does not move we have f8 S f8maac Kinetic friction also called sliding friction Once the box slides the opposing force is the force of fk Mk Fn where uk is called the coefficient of kinetic friction Experimentally it is found that uk lt us Example Two Connected Blocks Tipler Mosca figures 59 and 510 PRS How many forces act on block 2 Press the number on your remote How many forces act on block 1 press the number on your remote Moving blocks T m1 a l uk m1 g mgg Tm2a mgg ukm1gzm1am2am1m2a m2g ukm1g or km2g m1m2a m1m2 m19 CL Work and Energy Motion With Constant Force The work W done by a constant Force 13 whose point of application moves through a distance A is defined to be W F 3086 Ax where 6 is the angle between the vector 13 and the vector Af see figure 61 of Tipler Mosca If A3 is along the xaxis ie A53 Ami Am then W Fm Ax holds Work is a scalar quantity that is positive if A96 and F96 have the same sign and negative otherwise The Si unit of work and energy is the joule J 1Q1Nm1kgm282 Work and Kinetic Energy There is and important theorem which relates the total work done on a particle to its initial and final speeds If 13 is the net force acting on a particle Newton39s second law gives 13 me The total work becomes and the mechanical workkinetic energy theorem states The total work done on the particle is equal to the change in kinetic energy Wm Kf Ki Potential Energy Often work done by external forces on a system does not increase the kinetic energy of the system but is instead stored as potential energy Conservative Forces A force is called conservative when its total work done on a particle along a closed path is zero figure 622 of Tipler Mosca Potential Energy Function For conservative forces a potential energy function U can be defined because the work done between two positions 1 and 2 does not depend on the path 82 AUU2 U1 Fd 31 dU 13 d for in nitesimal displacements Example Gravitational potential energy near the earth39s surface dU 13d mg 39 dxidy dzl mgdy y UdUmg dy mgy mgyo y 0 WorkEnergy Theorem with Kinetic Friction Non conservative Forces Not all forces are conservative Friction is an example of a nonconservative force The energy dissipated by friction is thermal energy heat f A8 AEtherm where f is the frictional force applied along the distance A8 The workenergy theorem reads then Wext AEmech l AEtherm Example Assume the block enters a frictionless loop of radius R What is the minimal kinetic energy K7 the block needs to reach the top of the loop without leaving the track with Therefore 1 1 K mvfmg2R l mvt2 2 v 5 2 Compare figure 75 of TiplerMosca Momentum Conservation The Center of Mass CM The CM 77cm moves as if all the external foces acting on the system were acting on the total mass M of the system located at this point In particular it moves with constant velocity if the external forces acting on the system add to zero Definition TL TL MfcmZm F Where Mzzmi 71 71 Here the sum is over the particles of the system m7 are the masses and 7339 are the position vectors of the particles In case of a continuous object this becomes Mfcmde where dm is the position element of mass located at position 77 Momentum The mass of a particle times it velocity is called momentum 13 m Newton39s second law can be written as 4 dz d m2 t dt dt as the masses of our particles have been constant dt me The total momentum 13 of a system is the sum of the momenta of the individual particles TL 77 i1 i1 Differentiating this equation with respect to time we obtain d d cm M CmFneex dt dt a t t The law of momentum conservation When the net external force is zero the total momentum is constant Fnet xt 0 gt P constant Example Inelastic scattering figure 830 of TiplerMosca A bullet of mass m1 is fired into a hanging target of mass m2 which is at rest The bullet gets stuck in the target Find the speed U of the bullet from the joint velocity of of bullet and target after the collision Rotation The angular velocity LU Direction Righthandrule In accordance with the righthandrule the torque is defined as a vector 77x F Angular Momentum Definition E 77X 13 Like the torque angular momentum is defined with respect to the point in space where the position vector 77 originates For a rotation around a symmetry axis we find L LU magnitude L Iw Rotational kinetic energy Kmt Iw2 Energy conservation for an object initially at rest rolling down an inclined plane 1 1 mgh m212 1w2 1 2 mv2 IU Torque and Angular Momentum The net external torque acting on a system equals the rate of change of the angular momentum of the system a dL 7quot Ei zext Conservation of Angular Momentum If the net external torque acting on a system is zero the total angular momentum of the system is constant dt Tnet 0 gt L constant 2O Speci c Heat and Calorimetry Introduction The amount of heat Q required to raise the temperature of a solid body at constant pressure depends on the change in temperature AT of the body its mass m and a characteristic of the material forming the body called its speci c heat C This relationship is expressed by the equatioan MCAT and the dimensions of C are thus heat per unit mass per unit temperature change The values of C do depend on temperature with those of common metals such as aluminum and brass increasing a few percent as the temperature increases from 20 0C to 1000C for example while that for iron or steel increases about 10 over the same range Since these are not large changes average specific heats are often quoted in handbooks for such fairly broad temperature ranges Historically the amount of heat Q was originally expressed in terms of calories The calorie was defined most accurately as the amount of heat required to raise the temperature of 1 gram of water from 1450C to 155 0C at 1 atmosphere pressure With this definition the specific heat of water between 00C and 100 oCis 100 mlgmOC to within better than 1 The use of the calorie began before it was esmblished that heat was a form of energy and that 1 calorie is the equivalent of about 418 joules Thus in the SI system of units specific heats that is the values of C for particular materials are expressed as jkgOC and there is no need for the mlorz39e However since so much work involving heat has used the mlorz39e and since the specific heat of water is unity when it is employed it remains a common unit and will be used in this work The food Calorie with a capiml C is 1000 of these glories or 1 kiloitalorz39e The process of measuring quantities of heat exchanged is called calorimetry In this experiment your objective will be to determine the average specific heat of several metals over a certain temperature range by the calorimeter method ofmixtures Theory We know that when two bodies initially at different temperatures are placed in intimate contact in time they will come to equilibrium at some intermediate temperature Provided no heat is lost to or gained from the surroundings the quantity of heat lost by the hotter body is equal to that gained by the colder body This is the process which occurs in the method of mixtures that you will use The metal sample whose specific heat is to be measured is heated in boiling water to about 100 0C It is then quickly transferred to an aluminum calorimeter cup which contains cold water of known temperature When the metal sample and calorimeter cup come to equilibrium the common temperature is measured with a thermometer It is assumed that the transfer of heat between the thermometer and the system is small enough to be neglected If the net heat exchange with the surroundings can be kept small then the heat lost by the metal sample equals the heat gained by the water and the calorimeter cup Let MI be the mass of the sample whose specific heat is C Let Txbe its temperature before it is placed in the calorimeter Let MW and Cw be the mass and specific heat of the water and let M and Q the mass and specific heat of the calorimeter cup Denote the temperature of the water and calorimeter cup before the sample is added by Tm and the final temperature of the mixture by T Now use these 41 symbols to express mathematically the situation when a hot object the saInple is placed in contact with a cooler one the water and the calorimeter cup and the two are allowed to exchange heat until they reach a common temperature From this equation derive an expression for the specific heat of the sample in terms of the other quantities Procedure Fill a beaker with enough water so that the saInple when placed in it will be covered with some to spare Bring the water to a boil using a bunsen burner Weigh the aluminum sample and the dry inner calorimeter cup Note that the plastic top on the calorimeter is a thermal insulator whose temperature like that of the outer cup is assumed to be unaffected by changes in the temperature of the inner calorimeter cup during the course of experiment Suspend the saInple in the boiling water with string and a glass rod making sure not to touch the sides or bottom While the aluminum saInple is reaching equilibrium fill the calorimeter cup about 23 full of cool water about 5 C below room temperature Cool water can be had from a water cooler Ice is also available Weigh the cup plus water Care should be taken that the water is not so cold as to cause condensation on the outside of the calorimeter If condensation does occur dry the outside of the inner cup before proceeding Place the thermometer in the boiling water near the sample not touching the bottom When equilibrium is reached record the sample temperature Remove the thermometer cool it with tap water wipe it dry and then place it in the calorimeter cup Record the temperature as soon as it is reasonably steady and then quickly transfer the saInple from the boiling water to the calorimeter Care must be taken not to carry any hot water over with the sample nor to splash any cold water out of the calorimeter Stir the water in the calorimeter gently with the glass rod and observe the temperature When equilibrium is reached record the temperature Always read the thermometer as accurately as you can interpolating between the marks Calculate the average specific heat of your aluminum sample as determined by your experiment and specify the temperature range over which your value applies Since the specific heat of the aluminum calorimeter cup is also an unknown approximate its value by assuming it to be the same as the sample even though the cup and the sample are subject to different temperature ranges Later when you check the effect of this approximation you will see that it introduces little error If the value you obtain for the specific heat of your saInple is not between 019 and 025 mlgOC repeat the experiment to improve your technique You may for exaInple want to use a different temperature for the cool water The objective is to have the cool water as far below room temperature initially as the final temperature of the mixture is above room temperature Perform the experiment for the other saInples and determine their specific heats Compare your results with the accepted values which your instructor will furnish and calculate the percentage error and estimate the uncertainty in your experimental technique Questions 1 What do you see as the major sources of error in this experiment Use your calculation for steel as an example and determine the effect on the measured specific heat if the sample were to cool down 30C during the transfer Then calculate the effect if the net uncertainty in T 7 Tm were i 020C 42 Ln These calculations will give you some idea of the sensitivity of the results to some of the measured variables How much difference does it make in your results if the value you use for the speci c heat of the calorimeter cup is offby as much as 20 How would the measured value of the specific heat be affected if some boiling water were carried over with the sample How would the results be affected by splashing water out of the calorimeter or by condensation taking place on the inner cup from water which is too cold Why is it desired to start with the temperature of the water lower than room temperature and end with the temperature about the saIne amount above room temperature
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