GENERAL PHYSICS 1
GENERAL PHYSICS 1 PHYS 2010
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This 90 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 2010 at University of Colorado at Boulder taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/232106/phys-2010-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.
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Date Created: 10/30/15
A particle is moving along the path shown with constant speed Its velocity vector at two different times is shown What is the direction of the acceleration when the particle is at point A pink le lt gt 239 blue green yellow Purple the acceleration is zero CTEnergy O The United States has 5 of the world s population What fraction of the entire world s supply of oil is consumed in the United States A5 B8 C15 D25 Answer 25 CTEnergy l Albert Einstein lowers a book of mass m downward a distance h at constant speed V The work done by the force of gravity is A B C 0 the force of Albert39s hand is A B C 0 the net force on the book is A B C 0 Answers The work done by gravity is since displacement and Fhmd force of gravity are in the same direction The work done by As Al39s hand is 7 since displacement and the force of Al s hand are in the opposite direction The net force is zero so the work done by the net force is zero Fgrav CTEnergy Z A rock of mass m is twirled on a string in a horizontal plane The work done by the tension in the string on the rock is A B C0 Answer The work done by the tension force is zero since in any tiny interval of time the force of the tension in the string is perpendicular to the direction of the displacement CTEnergy 3 A 1 kg mass is moved part way around a square loop as shown The square is 1 meter on a side and the nal position of the mass is 05 m below its original position Assume that g 10 msz What is the work done by the force of gravity during this journey m g Ah 1 kg 10 msz Ah Up 05m 1m start 0 nish A 10JB5 C0J D710J E75J Answer Total work done is 5 J The work done during the horizontal segments is zero since force is perpendicular to displacement there The work during the rst upward vertical segment is negative and this negative work is cancelled by the positive work done during the rst half of the downward vertical segment The total work is that done during the second half of the downward vertical segment where Ah 05 m If instead of moving part way around a square the mass where taken on a long and tortuous journey to the Moon Tibet and Lithuania and then returned to the same nish point as before would the work done by gravity be the A same or B different Answer the same Any journey can be thought of a series of small vertical or horizontal displacements During any horizontal segment the work done by gravity is zero All upward vertical segments are cancelled by corresponding downward vertical segments EXCEPT for the last 05 m between the start and the nish CTEnergy 4 Two marbles one twice as heavy as the other are dropped to the ground from the roof of a building Assume no air resistance Just before hitting the ground the heavier marble has A as much kinetic energy as the lighter one B twice as much kinetic energy as the lighter one C half as much kinetic energy as the lighter one D four times as much kinetic energy as the lighter one E impossible to determine Answer The heavier marble has the same nal speed but twice the mass In so it has twice the KB l2mv2 CTEnergy S A car moves of mass m moving with speed v suddenly applies the brakes and slows to a stop Suppose I ask quotWhat was the work done on the car by the force of frictionquot Do you have enough information to answer the question A Yes B No Answer Yes We can use the WorkEnergy Principle which says that the work done by the net force is the change in kinetic energy Wm AKE In this case the net force is the force of friction The change in KE is KEf 7 KB 0 712mv2 CTEnergy G A mass m is at the end of light massless rod of length R the other end of which has a frictionless pivot so the rod can swing in a vertical plane The rod is initially horizontal and the mass is pushed down with an initial speed v0 What initial kinetic energy is required for the mass to pivot 2700 to the vertical position Va 1 2 KEINITIAL Emvo A ng B mg2R C mg3R D None of these Answer ng by conservation of energy 4 R y Let us put the zero of height at the initial position lt yR KEi PEi KEf PEf KEi0 0ng Instead we could put the zero of height at the bottom of the circle KEi PEi KEf PEf In which case we getthe same answer as before KE ng 0 mg2R KEi ng CTEnergy 7 A stick of dynamite explodes Is energy conserved in this situation A Yes B No Answer Yes The total energy everywhere is always conserved In science the word quotconservedquot means quotthe total amount is constant unchangingquot When the dynamite explodes the chemical potential energy in the molecules of TNT are converted into other forms of energy heat and light CTEnergy S A projectile is red with an initial speed v0 at an angle 9 from the horizontal What is the KB of the projectile when it is on the way down at a height h above the ground Assume no air resistance A l2mv2 mgh B mgh C l2mv2 7 mgh D Impossible to tell Answer Use conservation of energy and write Ei Ef with the initial position where projectile has just left the cannon and the nal position height h KEi PE KEf PEf mvoz 0 mvf2 mgh 1 2 2 zmvf 2mVo mgh CTEnergy 9 A block initially at rest is allowed to slide down a frictionless ramp and attains a speed V at the bottom To achieve a speed 2V at the bottom how many times higher must the new ramp be d A J5 E 14 B 2 C 3 D 4 E none ofthese Answer Using conservation of energy we can write Ei Ef If we place the zero of height at the bottom of the ramp we have KEi PEi KEf PEf 0mgh mV20 h 0c V2 So the initial height h is proportional to the nal speedsquared V2 So to get twice the speed V we need 4 times the height h Since 2V2 4V2 CTEnergy IO A mass slides down a rough ramp with friction of height h Its initial speed is zero Its nal speed at the bottom of the ramp is V As the mass descended its KE A increased B decreased C remained constant As the mass descended its PE A increased B decreased C remained constant As the mass descended its KE PE total mechanical energy A increased B decreased C remained constant Answers KE increased PE decreased and Emech decreased Some of the mechanical energy was converted into thermal energy by the friction Remember KEPE constant ONLY if NO friction and no heat generated If there is friction then KE PE Ememal constant CTEnergy ll A mass is oscillating back and forth on a spring as shown Position 0 is the relaxed unstretched position of the mass WWW At which point is the magnitude of the force on the mass maximum A 0 B M C E At which point is the magnitude of the acceleration of the mass maximum A 0 B M C E At which point is the elastic potential energy maximum A0 BM CE The force on the mass is maximum at point E since the force Esprng ikx is maximum magnitude there The acceleration is maximum at point E Newton39s second law says Fuel m a so the acceleration is max when Fuel is max The elastic potential energy PEelas 2 kxz is maximum at E where x is max CTEnergy IZ A springloaded dart gun shoots a dart straight up into the air and the dart reaches a maximum height of h 3 m The same dart is shot straight up a second time from the same gun but this time the spring is compressed twice as much before firing How far up does the dart go this time neglecting friction and assuming an ideal spring A2h6m B4h12m C8h24m Dh4 075m El 2 15m Answer Use conservation of energy and write Ei Ef with the initial position where the dart is inside the gun with the spring cocked and the nal position top of the trajectory Let s set h 0 at the initial position KEi PE PE KEf PE elas i gravJ gravf PEe1asr 00 kx2 0mgh0 k X2 mgh X2 cx h X2 constantgtlthCh Here C is a constant C 2mgk Since X2 is proportional to h when X is doubled X gt 2X X2 gt 2X2 4 X2 h is increased by a factor of 4 It may be easier to see this by setting up a ratio i 4xf 4 Ihl X12 X12 2 X1 CTEnergy 13 A springloaded dart gun shoots a dart straight up into the air and the dart reaches a maximum height of h 24 m The same gun is reloaded with the spring compressed the same amount but now the gun is aimed at an angle of 450 to the horizontal Will the dart reach the same maximum height of 24m Assume no frictional losses A Yes the dart will reach the same height B No the dart will not reach the same height Answer No We can see why this must be by considering conservation of energy When red at 45 the dart has some horizontal motion when it is at the top of it trajectory So it always has some KB and its initial PEelastic is never fully converted into PEgmv CTEnergy 14 A hockey puck slides without friction along a frozen lake toward an ice ramp and plateau as shown The speed of the puck is 4ms and the height of the plateau is lm Will the puck make it all the way up the ramp A Yes B No C impossible to determine without knowing the mass of the puck Answer We have to do a quick calculation Setting y 0 at the bottom of the ramp we have K13i PE KEf PEf mv20 0ng The nal energy PE would be ng massgtlt10ms2lm massgtlt 10 mzs2 The initial energy KE is l2mv2 massgtlt05l6 mzsz massgtlt8 mzs2 So the initial KE is not large enough the puck will not make it to the top of the hill CTEnergy IS A small mass starting at rest slides without friction down a rail to a loopdeloop as shown The maximum height of the loop is the same as the initial height of the mass Will the ball make it to the top of the loop A Yes the ball will make it to the top ofthe loop B No the ball will not make it to the top C Not enough info to say or don t know Answer No the ball will not make it to the top According to conservation of energy in order for the ball to make it to the top of the loop it must have some nonzero KE at the top of the loop if it did have zero KE at the top of the loop it would have fallen off the track before then because it would have been going too slow Since it must have some KE at the top of the loop some of its initial PEIgrav was converted into that KE so its PEIgrav at the top of the loop must be less than the initial PEng CTEnergy IG A cart rolls without friction along a track The graph of PE vs position is shown The total mechanical energy KE PE is 45k The top of the hill at X 62 m is round and the bottom of the valley near X 130 m is at Etot PEkJ o 20 40 60 80 100 120140 160 mm To within 5k what is the maximum KE over the stretch of track shown A 251d B 7k C 45k D 35 kJ Answer Max KE of35 kJ occurs at about x 130 m When the KB is a minimum over this stretch of track what is the direction of the acceleration A up B down C right D Left E some other direction at an angle or zero Answer KE is min at top of hill near x 62 m At the top the acceleration is down When moving along a circular path at constant or approximately constant speed the acceleration is toward the center of the circle When the KB is max what is the direction of the acceleration A up B down C right D Left E some other direction at an angle or zero Answer Along the at portion of the valley the velocity is constant and the accleration is zero CTEnergy 17 A cart rolls without friction along a track The graph of PE vs position is shown The total mechanical energy KE PE is 0 k 20m 40m 60m What is the MAX KB of the cart during its journey to within SkJ A 351d B 48k C 16k D l6kJ E None of thesedon39t know Answer Max KE is where PE is min Max KE 48 k lt Etotzo PE KE 48kJ 48kJ Suppose the cart is at position X 20m is moving right and has total energy Elm 7 20k Will the cart make it over the hill at X38m A Yes B No Answer No The quotturning pointquot where the cart slows to a stop and turns around The turning point is where the Em line intersects the PE curve This is where KE 0 because Em KE PE Etot 20 M u CTEnergy IS Elevator 1 can ca1ry a load of mass m up a distance h in atime t before the engine overheats Its power output is P1 Elevator 2 can carry the same load up twice the distance 2h in twice the time 2t What is its power output P2 A P2 P1 B P2 2P1 C P2 4P1 D None these Elevator 3 can ca1ry twice the load 2m up twice the distance 2h in the same time t as elevator 1 What is its power output P3 A P3 P1 B P3 2P1 C P3 4P1 D None these 2h 2h time 2t timet h time t m m 2m 1 2 3 Answers P2 worldtime mg2h2t mght P1 P3 worldtime 2mg2ht 4mght 4P1 CTEnergy 19 A hockey puck sliding on an ice rink is moving at 1 ms when it slides onto a ca1pet that someone left on the ice The puck comes to rest after moving lm on the carpet How far along the carpet would the puck go if its initial speed was 2ms A 15m B 2m C 3m D 4m E Impossible to determine Hint Apply the WorkEnergy Theorem If the puck slides twice as far the friction does twice as much negative work Answer 4 m We have to do a calculation to see this This is a case with friction but the workenergy principle still applies Friction does negative work here so we take absolute values just to get rid of the minus signs wmc AKE since APE 0 2 ungAX mv AXocv2 Warmup question How much more KE does the puck have moving at 2 ms compared to moving at 1 ms A xE times as much B twice as much C 3 times as much D 4 times as much Answer 4 times as much CT2 1 A person starts in Boulder drives to Denver 50 km away in 1 hour stays in Denver for 1 hour then speeds back to Boulder in 30 minutes 5min1 Inr sunninvzhou W1 What is the average speed of this round trip A zero B 67 kmhr C 40 kmhr D 75 kmhr E None ofthese What is the average velocity of this round trip same choices Answers Average speed is 100 km 25 hr 40 kmhr X Average veloc1ty 0 Since v Z and X na Xinmal 3 Av 0 t CT2 2A person initially at point 3 on the XaXis stays there 0 l 2 3 for a little while and then strolls I I I I along the XaXis to point 1 stays I I I I there for a moment and then runs to point 2 and remains there 3 JlQJM 1 i351 2 Which of the following graphs correctly depicts this motion Answer D time time E None of these gtlt time time CT2 3 A object starts at X 5 and moves left along the XaXis at constant speed finiShOlt O lt O lt O lt OStart I I X 0 Which graph represents this motion A It X X A B time time A At time time E None of these Answer B CT2 4A train car moves along a long straight track The graph shows the position as a function of time for the train position time The graph shows that the train A speeds up all the time B slows down all the time C speeds up part of the time and slows down part of the time D moves at constant velocity E None of these statements is true Answer B The train is slowing down all the time because the slope of the curve is always decreasing On a graph of X vs t the slope is the velocity CTZ S A train moves along a straight track and its position vs time looks like X t Which graph best depicts the trains velocity vs time M A A v lt B t D C t QUn Answer D CT2 6 The graph show positions pOSlthIl train A as a function of time for two trains running on parallel tracks Which statement is true trall l B A At time tc both trains have the V t tlme same velocity C B Both trains speed up all the time C Both trains have the same velocity at some time before tc D At some time both trains have the same acceleration E None of the above statements is true Answer B Both trains have the same velocity when the slopes of the two curves are equal Notice that train A has constant velocity and therefore has zero acceleration Train B is slowing down all the time Train B has negative velocity CT2 7 An object s velocity vs time is V Which graph best represents the obj ect s acceleration vs time At A a A a B D Answer D The slope ofa curve ofv vs t is the acceleration a CTZ S At the moment that the velocity is zero what is the sign of the acceleration A Positive B negative C zero V a Is the acceleration constant all during the time the velocity is changing A Yes a constant B No a is changing Answers A the acceleration is negative at the moment when the velocity is zero A the acceleration is constant The slope of a graph of v vs t is the acceleration Although v 0 at an instant of time the velocity is in the process of changing from positive v to negative v CT2 9 v gt A A glider on a tilted air track is given a brief push uphill The glider coasts up to near the top end stops and then slides back down When the glider is at the highest point of its path its acceleration is A straight down B downward along the track C upward along the track D no direction the acceleration is zero Answer B the direction of quotDeltavquot is downward along the track Just before the glider reaches the top the velocity v1 is upward along the track Just after the glider reaches the top the velocity v is downward along the track The change in velocity Av is the vector that changes v1 into v2 The direction of the acceleration is the same as the direction of AV CT210 A ball is thrown straight upward At the top of its trajectory its acceleration is A zero B straight up C straight down D depends on the mass of the ball CT2l 1 If you drop an object in the absence of air resistance it accelerates downward at 98ms2 If instead you throw it downward its downward acceleration after release is A less than 98ms2 B 981m2 C more than 98ms2 CT212An object s velocity vs time graph looks like this V What situation produces this kind of motion A A rock is thrown straight up B A rock is dropped from rest C A rock is thrown straight down D A book slides along a table and comes to rest E None ofthese CT2l3 On planet X a cannon ball is red straight upward The position and velocity of the ball at many times are listed below Note that we have chosen up as the positive direction What is the acceleration due to gravity on Planet X A 5ms2 B lOms2 C 15ms2 D 20ms2 E None ofthese CT2l4 A ball is red straight downward out of a special springloaded gun which produces constant acceleration Upward is chosen as the positive direction Assume that air resistance is negligible Which graph properly represents acceleration of the ball Ball released lf rom spring D None of these CT215 A truck traveling at 50 kmhr approaches a car stopped at a red light When the truck is 100m from the back of the car the light turns green and the car immediately begins to accelerate at 200ms2 Which graph belows represents this situation car X car A B X 7 t t truck truc car D None of these CT2 15 A ball is red from a canon straight upward with an initial velocity v0 Assume no air resistance Ifthe initial velocity v0 is doubled the time to reach the apex of the trajectory A doubles B increases by a factor of 4 C Neither ofthese D Impossible to tell from the information given If the initial velocity v0 is doubled the maximum height of the ball A doubles B increases by a factor of 4 C Neither ofthese D Impossible to tell from the information given CT2l7 1 2 X a t The distance x traveled by a ball in a time t1s given by the formula 2 where a is a nonzero positive constant What happens to the distance x when the time t is tripled A X increases by a factor of 3 B X increases by a factor of 92 C X increases by a factor of 9 D X increases by a factor that depends on the value of a Do you have your clicker with you today A Yes B No A trick question Which object A or B has larger velocity A A B B A A better question Which object A or B has higher speed A B 241 SIP Phys 2020 Fa 01 Light waves diffraction and interference In Ch 22 we learned light is a wave and then in Ch 23 we promptly ignored this fact It took people a long time to notice it s a wave because 7 is so small Recall Violet light has 7t400 nm red light is 700 nm Here is a quotCh 23quot picture of light coming in from the left hitting a wall with a hole or slit in it and forming a shadow on a distant wall Dark here gt Big slit in wall quotshadowquot I Bright here gt gt gt gt gt Dark here Incident light quotshadowquot Distant screen This was basically Newton s idea and is quite accurate However if the size of the slight is very small somewhere on the same scale as the 7b of light then the picture is very different Light spreads out Small Sht all over but not equal intensity gt everywhere In iden lig t W veg Distant screen The lines I drew are NOT the same as in the previous picture they are not quotraysquot Now we re trying to indicate the wave nature by drawing quotwavefrontsquot You might think of this as connecting points where E is cresting These lines are now perpendicular to the direction of travel of the waves Think of it as looking down at waves on the ocean and what I m drawing is the lines where the waves are highest In the picture above the waves move right The image on the screen isn t quite right we ll make a more accurate picture soon 242 SIP Phys 2020 Fa 01 Brief review of interference of waves Phys 2010 Giancoli Ch 1111 Imagine two traveling waves same frequency speed and direction C gt C v C If they are quotin phasequot meanng the peaks are lined up and if the waves are traveling at the same place and time they will add up they superpose as waves always do The resulting wave the sum of the two has the same f c and direction but twice the amplitude The waves have constructively interfered Now take those same two waves but let them be exactly quotout of phasequot meaning the peak of one is at the same place as the trough of another Then the result when they superpose is cancellation the quotquot of a peak adds to the quotquot of a trough giving zero no wave at all C 2 gt 1 2 The waves have destructively interfered You can have two waves heading towards a common point At that point the waves will interfere If they are quotin phasequot they will add up to double the amplitude In this figure both waves peak at the point where they meet they add up constructive interference If you were to shift one of the waves 2 a little so one peaks where the other quottroughsquot at the meeting point they would instead be completely quotout of phasequot they would exactly cancel destructive interference at that one point If they are just partly out of phase they will add interfere superpose but not to either extreme Two waves of amplitude A can add up to a wave of anything between 0 and 2A 243 SIP Phys 2020 Fa 01 Huygens a contemporary of Newton came up with a neat way to think about traveling waves Each point on any wavefront can be 7 considered to be the source 0 of a new outgoing spherical 3 wave In this gure only one tiny spherical wave point center of the gap of wave fronts the incoming wavefront is Plane ane allowed to propagate to the right What you get is an outgoing spherical wave with the same f X and c as the incident plane wave The little gap acts as a single quotpoint sourcequot of wave of frequency f With a wider gap each point in the gap acts as a point source Each of 7 those points emits spherical waves These waves are quoton top of each otherquot you have to superpose them add them up to figure out the resulting outgoing wave to wave fronts plane wave The outgoing wave here is not spherical any more nor is it a plane wave It s something inbetween We won t ever do that math in this class it s a little tricky though certainly not impossible Huygens worked it out 350 years ago but the result in the end if often very simple and quite reasonable Giancoli s figures are better than mine check them out 244 SIP Phys 2020 Fa 01 As an example Huygen s principle fully predicts the phenomenon of refraction The details are as I said tricky again Giancoli has somewhat better pictures but the result is straightforward wave fronts Waves always have coming in A f speed so M fl cnl k2 f2 cn2 Now here s an wave fronts important consequence going out 79 of Huygen s ideas fl f2 Why n2gtn1 As the incoming wave reaches the boundary and creates the outgoing wave it is quotjigglingquot with frequency fl and that s the same frequency that the new medium will jiggle with Frequency is just counting it does not depend on what medium you re in But the resulting 7t does depend on the medium The equations above tell us that if fl f2 then M k2 n2nl can you see howl got that That means in the picture above k2 is smaller because n2 is bigger Wavelength depends on speed In a quotfast mediumquot smaller n the wavefronts are more spread out 7c is bigger The picture shows wavefronts remember that the quotraysquot these represent are perpendicular to the wavefronts I have drawn two sample rays in the figure The angle at the boundary is bent Refraction Huygens principle predicts that if light hits small slits you should be more likely to observe the quotwave naturequot of light Look back at the pictures on the previous page the outgoing wave is more noticeably different from the incoming wave when the slit size is small To really see this effect in the lab it helps a lot to have monochromatic light one pure wavelength and also coherent light meaning nice incoming plane waves with all rays quotin synchquot with each other 245 SIP Phys 2020 Fa 01 Young39s Double Slit Experiment The first observation of interference of Visible light Huygen s says each of the U two slits will act as a point x source for outgoing spherical waves emerging heading off in all directions from each slit 4 D C gt On the right light is wave fronts The incoming light is spherical waves Plane ane monochromatlc one 7b and coherent which means that if the E field is peaking at the top slit at some instant in time it is also simultaneously peaking at the bottom slit What will we see on a screen far off to the right The old quotparticlequot or quotrayquot model LI Sha ow predicts this Brlght SpOt h d two bright spots one directly being S a OW each slit ZBUght SpOt shadow This is experimentally dead wrong 39 At least if the slits are very small screen compared to 7b Huygens model of interfering waves LI Bright predicts this We ll see why soon 3 dark And this is what you really see I Bright Never mind Giancoli Fig 249 and 10 3 dark for now For two SMALL slits each Bright bright quotbumpquot in the image is equally quot i screen You see a bunch of bright spots over there We ll explain exactly why you get this pattern on the next page They are not necessarily bright right behind the slits either This is weird quot Shadowsquot are what we re used to but the wave nature of light says it can effectively bend around corners This phenomenon is called diffraction It s not so intuitive for light 246 SIP Phys 2020 Fa 01 How can this be How do you figure out where the bright spots will be What we have is two slits two point sources of light and they are in phase with each other If light goes the same distance from each eg the waves leave the sources and head to the quotmidpoin quot on the screen then the two waves will arrive in phase in synch they will constructively interfere and you get a bright spot on the screen there Yes its bright right smack in center the center just exactly where you d expect a dark shadow from that central wall Are there any other bright spots What really matters is whether the TWO waves reach the screen in phase Draw the rays the light from each slit travels different distances If the two waves BEGIN in phase and travel different distances they might still be in phase or not it all depends on the difference in path lengths This is tricky to visualize Let s look at a spot on the X screen a height quotxquot above the center line a distance L far away from the sources Is it bright or dark there THIS is the extra distance screen the lower ray travels It all depends on the difference in the path lengths of the two rays I m making an assumption here a limit that L is HUGE compared to d X That means that the gure 9 really looks like this d4 L I just exaggerated quotdquot above so we could look at the geometry more easily 247 SIP Phys 2020 Fa 01 The question is is the extra distance EXACTLY a whole number of full wavelengths If it is then the two waves are still in phase Think about the condition for waves to be in phase If a wave travels through exactly ONE full wavelength it s back in phase Same if it travels through exactly TWO full wavelengths So if the extra distance traveled by one ray compared to the other exactly equals X or 27 or 37 or m7t where quotmquot is any integer then the two waves are still back in phase and they constructively interfere which means the spot we labeled quotxquot will be bright If on the other hand one wave travels through HALF a wavelength extra it s exactly out of phase That means the two waves destructively interfere it would be totally dark at point quotxquot Same whether the path length difference is 05 7 or 15 or 25 etc We need to look at the picture again this time a quotzoom inquot right down next to the slits screen is far away T is is d Sime to the right Convince yourself The picture shows that the path length difference is d sin6 The geometry is tricky Think about it compare to the previous pictures see if you can really get it for yourself What we ve just argued is that if that path length difference is an INTEGER number of Ns we get constructive interference and thus the spot x will be bright Constructive d sin6 m 7 where many integer 0 l 2 3 Similarly if the path length difference is a HALF integer number of Ns we get destructive interference the spot x will be dark Destructive d sin6 ml2 7t many integer 0 l 2 3 9 248 SIP Phys 2020 Fa 01 Quick review of radians and the quotsmall angle approximation S R radius A s quotarc lengthquot along circle 6 sR defines the angle in radians Full circumference is s 2 75R That means 6 all the way around sR 2 TERR 2 Tcrad Or 60 deg 2 Tcrad Now look at a small wedge and quotsquare off the curvy side We know 6 sR R 5 X and we know tan 6 XR e s and we know sin 6 XR5 R Those all come from quotsohcahtoaquot But if 6 is very small we see 5 m X roughly just look at the picture and 5 m 0 too Conclusion if 6 is very small 6 m tane m sine That would also mean x R tane m R6 Example try this on your calculator 1 deg 017453 rad sinl deg 017452 and tanl deg 017455 These are very close Finally let s go back to our quot2 slit interferencequot picture I m going to simplify the picture by only drawing ONE quotaveragequot ray not both of them Remember 1 is really tiny compared to L U Bright We already argued there are dark bright spots whenever d sin6 in 7b X Note 6 tan 1XL S Green H Since L is huge we frequently have XL small and then 9 XL So for small X and big L d sin6 m 7 becomes 1 XL z in 7b or BRIGHT SPOTS x mm x L d m 0 1 2 3 DARK SPOTS x m 12 k L ml in 0 1 2 3 If X L is not small just use the correct formula d sin9 In Use this to nd 9 then solve for X R tan9 249 SIP Phys 2020 Fa 01 What ifwe had more than 2 equally spaced slits Draw the picture draw ALL the rays coming from all the slots towards a common point at the screen You should convince yourself that as long as d sin6 in 7b m0l2 then the light from ALL the slots will be in phase constructive interference VERY bright At any other angle you ll get destructive or at least partially destructive interference If 6 is quotslightlyquot off one of those magic bright values then say the rays from slit 1 and slit 2 won t be very much out of phase so if there are only two slits it s darker but only a little if you shift a small angle away from the brightest point But with many slits slit 1 will be way out of phase with some other slit maybe not 2 but one a few further down the row bright So you end up getting VERY bright spots at the places where the formula for 2 slits says it Very dark Should be bright and they re much sharper quotquotquotquotquotquot quot bnght It s very dark almost everywhere else very dark This is handy for clearly separating the bright D bnght spots We call a device like this with many slits a diffraction grating They re very useful Why Notice the positions X m m 7 L d for the bright spots depend linearly on the wavelength 7 Different colors incident on the same 2 slits or grating will make bright spots at different locations The separation of the colored spots X is easy to measure and is proportional to 7b That gives you a very handy and easy way to measure wavelengths of light Think about that a bit how can you measure something so fantastically tiny 7b of light is just a few thousand times larger than an atom 2410 SJP Phys 2020 Fa 01 Example Consider 2 slits a distance d001 mm apart There is a screen L2m away Monochromatic red light X 2700 nm shines on the 2 slits What will the brightness pattern look like over at the screen Lingo X2 is the quotm2quot or quot2nd orderquot line X2 X1 1st bright spot above the center The quotfirst Xl orderquot line or quotm1quot line X1 1 X L d I Here X1 700E9 m2m01E3 m 14 m quotquotquotquotquotquot quot quotzerot quot order or m0 line in middle m0 X0 I X1 1st bright spot below the center quotm1quot line X391 x 17tLd14m Note the symmetry X1 xl X392 Etc Note 61 17td 07 rad 622 7td 14 rad etc If X starts getting big you should go back to the correct formula for the quotangles of maXimaquot namely d sin6 m k o If we d used blue light 500 nm instead of red the pattem is a little different X1 is slightly smaller the bright spots are squeezed together more towards the center which is unchanged 0 If d was bigger the pattern is also squeezed in a similar way That s why it s so hard to notice this pattern in normal life For Visible light 7c is SO small and d is so much bigger that the bright spots all squeeze right next to each other you can t even NOTICE that there are dark and bright spots To make it worse all the colors are normally present with their bright spots in slightly different places so it really just looks bright all over 2411 SJP Phys 2020 Fa 01 What if we used a grating with 1000 linescm instead of 2 slits First we have to think about the spacing between these lines 1000 linescm means lcm 1000 lines or 1 1000 cm per line or 01 mm per line that s the same distance between lines as we had in the previous problem quotlinesquot in a grating can be though of as analogous to quotslitsquot X2 m2 So the pattern is identical to that in the XI mzl previous problem except that the spots are I brighter and sharper quotquotquotquotquotquot quot m0 The X s positions of the bright spots are the I same 391 m239 1 X 2 m2 If you shine white light rather than monochromatic there are many Ns present Each color has bright spots at different positions X So the colors appear to get spread out on the screen You get a little rainbow a spectrum at each order One spectrum for each quotpeakquot in the picture above Since X1 kLd measuring quotXquot for different colors can tell you what the wavelength of each color is Some light sources have different colors but not all colors E g Mercury arclamps used on highways look yellowishwhite but do not contain all wavelengths with anything near equal intensity A diffraction grating which spreads out colors allows you to quickly and easily see which colors are present in the source You will see lines of different colors at each order This is a unique quotfingerprintquot of the light source This technique is called spectroscopy and we use it to determine the chemical composition of materials In this way we can even deduce what elements are present in distant stars by analyzing the light they emit 2412 SJP Phys 2020 Fa 01 Another way to get a spectrum is with a prism How does this work Different colors 7t have slightly different indices of refraction quotnquot in normal glass So they get bent slightly differently red longer 7 n closer to 1 bends less blue shorter 7 n bigger bends more n2 depends on A Notice that the spread of colors is opposite the way it goes with a grating In a grating red has longer wavelength and so bends more In a prism red has smaller quotnquot so it bends less The mechanisms involved are really totally different Rainbows are caused by the prism effect Different colors bend through different angles as they pass from air into water droplets The result is that white light gets dispersed into a spectrum of colors The sun produces all the colors of light so the rainbow has them all See Giancoli Fig 2416 Take a look at a CD It s got a lot of closely spaced lines and is re ective it acts like a grating It s a quotre ectionquot grating rather than a quottransmissionquot grating but the physics is the same Again at different angles you see different colors Check it out hold it at an angle from a single light source Can you see quotROYGBIVquot If you keep tilting can you see the second order How many orders does it have Can you estimate the spacing between the lines MPl SIP Phys 2020 Fa 01 Modern Physics Physics from the 20th century What we ve been studying so far in 20102020 is called classical physics Newtonian mechanics wave phenomena heat electricity and magnetism optics wave optics We have described phenomena of ordinary life and sometimes more exotic things qualitatively and quantitatively Classical physics is a remarkable success story a deep understanding of nature that has also led to a significant fraction of the technology in our modern lives This physics was developed mostly over the period from the late 1600 s Newton through the late 1800 s optics thermodynamics and Maxwell s EM Around 1900 several major revolutions in the way we think about and look at the physical world occurred The new discoveries are called modern physics even though much of it is 70100 years old Specifically I m talking about relativity and quantum mechanics Nuclear and particle physics are spinoffs in a sense new exp tal observables but using the framework of relativity and quantum mech Modern physics doesn t eliminate the need usefulness even in some ways the correctness of classical physics But it extends our understanding to regions where classical ideas simply break down they fail to describe experiment Relativity is particularly important when speed or energy get very high Quantum mechanics is most useful when sizes scales are very small like inside atoms or nuclei In 1890 some physicists admittedly mostly older ones thought everything in nature could be explained quotclassicallyquot It seemed just a question of solving Newton s laws or Maxwell s equations in more complicated situations Was physics quotdonequot they wondered just a matter of measuring quantities to more decimal places to improve accuracy They were quite completely wrong It is impossible to understand the behavior of say atoms from a purely classical or Newtonian approach The world simply does not behave like Newton or really most of us thinks at extreme scales Relativity and quantum mechanics are weird mind boggling Do not keep saying to yourself if you can possibly avoid it But how can it be like that because you will get down the drain into a blind alley from which nobody has yet escaped Nobody knows how it can be like thatquot Richard Feynman one of the greatest physicists of the latter 20th century MPZ SIP Phys 2020 Fa 01 Relativity was pretty much Einstein s idea though Maxwell Michelson Morely Lorentz Mach and many others contributed of course It involves the idea of how we observe physical events in particular how our description of physics depends on our quotreference frame Galileo had thought about this long before Einstein and came up with correct classical expressions relating observables The speed or position of an object depends on the observer in welldefined ways but the equations relating the observables are universal and independent of observer Some things are relative like velocity or value of magnetic field but other things are not like Fma or Maxwell s equations The fundamental principle here the quotrelativity principle was understood by Galileo as well as Einstein the basic laws of physics are the same in all inertial reference ames The extra step that Einstein took was small but had radical implications Einstein argued on the basis of Maxwell s equations and a lot of thinking that the speed of light c is a constant independent of the speed of the source or observer This becomes a basic law of physics with the same status as say Maxwell s equations Galileo would have argued otherwise if you fire a bullet towards me YOU say it has a certain speed say 1000 mihr But if I run away from you at 999 mihr I see it has a different speed in this case a harmless l mihr the difference 1000 999 right This works with bullets but Einstein says that for light that s not correct If you shine a ashlight towards me YOU say it has a speed of c299E8 ms If I run away from you at 298E8 ms I say the light has a speed of drum roll here are you about to say the difference 001E8 ms No the speed I measure is ALSO c299E8 ms That s what Einstein argued and it has been checked and verified in countless experiments including pieces of technology in use every day today We re not going to go much further into the theory of relativity just because we re running out of lectures this semester When you start to think about the consequences of Einstein s postulate your whole way of thinking about the world gets shaken For his postulate to be correct neither time nor space can be absolutes There s not some quotcosmic clockquot clicking off seconds out there Time is relative the passage of time depends on the observer MP3 SIP Phys 2020 Fa 01 Closely tied in to that simultaneity is relative I might see two events say two stars going nova in different parts of the sky and say Hey they happened at the same time But another physicist moving with respect to me perhaps she s an astronomer on quotMars Basequot will disagree She will argue that one happened first Who s right We both are simultaneity is NOT an absolute property of events in different places Weird stuff The consequences continue a particle which decays radioactively has a certain lifetime But if it moves quickly with respect to you it lives longer Time is dilated when things move fast Einstein derived equations which quantify these effects They re not important or noticeable unless the speeds involved are very close to c But we certainly have observed them atomic clocks transported in jets tick at a measurably different rate than reference clocks on the ground And GPS systems must correct for relativistic effects or they won t work the clocks on GPS satellites move fast Einstein does not tell us that everything is relative he s not throwing away science He has unified space and time into a 4dimensional quotspacetimequot that must be considered together to describe these experiments Momentum and energy similarly become connected and the most famous consequence derivable just from the simple principle of relativity described above is that E mc 2 mass and energy are related Energy can be converted into mass and is regularly in particle physics experiments and vice versa mass can be converted into energy and is regularly in the core of our sun nuclear power plants and bombs and in fact when you light a match That s right Emcquot2 is not just about nuclear energy It s about ANY kind of energy including ordinary chemical energy like the burning of a match The equations of relativity are not hard the math involved is nothing worse than the Pythagorean theorem Read Chapter 26 to get a quantitative sense of relativity At this point you are all sophisticated and knowledgeable enough to read about physics on your own even something as intimidating as relativity It is hard to wrap your head around it it s not exactly intuitive But you can understand it There are also lots of great books that you could read eg quotRelativity Simply Explainedquot But we re going to go on to survey a few other topics of modern physics now MP4 SIP Phys 2020 Fa 01 The other part of modern physics is quantum mechanics It is in many ways a far more radical theory than relativity Its consequences are still being studied its philosophical implications still a great source of discussion It was not developed by one person but a large group of really brilliant minds Max Planck started the process Einstein was an active participant though he never really believed in it Neils Bohr Louis de Broglie Werner Heisenberg Erwin Schrodinger Max Born Paul Dirac Wolfgang Pauli Its development wasn t sudden like relativity It began with a series of puzzles experiments whose results were in direct con ict with classical physics predictions E g Maxwell s equations coupled with some classical thermalstatistical physics make a clear prediction about what the spectrum of radiation from hot objects should be socalled quotblackbody radiationquot blackbody just means any ordinary material that absorbs and emits strongly But the data disagreed Not by a little bit either it was totally and completely different Max Planck was able to explain the data by making a wild and crazy hypothesis that light is quantized it comes in bundles in chunks in quotquantaquot In particular if you light of some color frequency there is a minimum chunk ONE quantum of light which we now would call a photon and you can t have any LESS light than that except for none at all of course This is a thoroughly preposterous idea Light is a wave after all and you should be able to make waves with any amplitude any energy you like right You can make sound waves of any loudness can t you However it s not a preposterous idea if light is really quotparticlesquot it doesn t bother you to say you can only go down to one quotquantumquot of electricity namely e an electron does it You can t have any less electric charge than that except for zero Because electricity comes in chunks in quanta But light s supposed to be a wave according to Maxwell not a particle So this was Planck s wild idea He even came up with a formula a very important one The energy of a photon is a constant times the frequency quothquot is a new constant of nature h 667E34 Js h is now called quotPlanck s constantquot This formula is strange not very intuitive MPS SIP Phys 2020 Fa 01 Plank felt this idea of quotquantized lightquot was mathematical fiction a way to describe blackbody data that had nothing to do with reality But Einstein found the idea compelling and asked if there were other consequences of quantizing light He used the idea to explain another totally unrelated puzzle of the day the quotphotoelectric effectquot light shining on metal makes electrons y off We use this effect all the time now in solar cells photodetectors all sorts of applications Even photosynthesis is a complex version of this effect Classical physics says waves transmit energy and momentum and so can quotkick electronsquot but the details are wrong for the photoelectric effect Maxwell says low frequency light should be able to lift electrons out of metal if you wait awhile or if the light is bright enough But Einstein says that if Ehf then a given color light has a minimum amount of energy If that isn t enough to lift an electron out of the quotpotential wellquot of the metal then NOTHING at all will happen If no individual photon can lift out an electron none will ever come out But as you change frequency you reach a critical color f where the energy IS enough and suddenly lots of electrons should come popping out you ll see a quotphotocurren quot Here is Einstein s prediction for kinetic energy KE 0f ejeCted 639 KE of ejected electrons as a function of frequency of light shined on the metal It s a simple consequence of conservation of energy frequency of light f Electrons have a quotbinding energyquot W also called the quotwork functionquot of the metal it s the amount of energy required to lift an electron out of the metal That means the electrons starts off with NEGATIVE potential energy initial electron energy is W it s quotboundquot f min Initial energy hf from the light W for the electron Final energy KE of the electron Conservation of energy says KE hf W the graph shown above The slope should be Planck s constant h which Plank derived in a TOTALLY different way looking at glowing metal in ovens Completely incorrect according to EM theory but absolutely correct in the lab A famous experimentalist Millikan read Einstein s paper and did the experiment He was certain it was all a crock and worked very hard to show that Einstein was wrong But in the end the prediction of Planck and Einstein was exactly correct MP6 SIP Phys 2020 Fa 01 So which is it now is light a particle or a wave Countless interference exp ts cannot be easily explained by a simple particle model they require light to be a wave Yet Einstein and Planck show us these new exp ts that cannot be easily explained by a wave model Light has both characters you simply can t picture what light really quotisquot in any classical way You really have to use the formulas of quantum mechanics tied in with Maxwell s equations A rigorous quotcombined theoryquot was developed in the 1940 s by Richard Feynman and others called quotQEDquot or quotQuantum Electrodynamicsquot but the basic ideas were all in place from earlier in the century The message is profound you can t always impose a simple model of the world on nature Light is complicated It was a revolution in physics and the model that emerged is still unsatisfying to many people It s a model with quotwaveparticle dualityquot We describe it accurately mathematically but it s hard to quotpicturequot in any simple intuitive classical way Classically you either have waves or particles39 you can t have both at the same time But you can and must in the real world of quantum physics There were many more exp ts verifying Ehf and the quantum nature of light The most famous and clinching was in the 30 s by Arthur Compton He managed to shine quanta of light at electrons and watch the electrons quotbouncequot off The light behaved JUST as a particle a billiard ball would The description of the scattering requires relativity but otherwise is just a standard quotparticle collisionquot not at ALL what you d expect if a light wave was quotwigglingquot past the electron So the idea of quotquantizing energyquot and quantizing light in particular was important but this is not the whole story Quantum mechanics goes much further it s not just about photons Louis de Broglie a theorist argued that if light can be both wave and particle why can t electrons also have both characters Physicists were getting very wild in their thinking Come on surely electrons are particles But de Broglie was correct electrons do exhibit wavelike behavior in some cases For example if you beam electrons just like the beam in your TV towards a pair of slits you d expect two bright spots where lots of electrons hit right behind the holes But what you see in reality is a diffractioninterference pattern just like light going through slits Beams of particles interfere in a way that only waves should MP7 SIP Phys 2020 Fa 01 What is the wavelength of a particle De Broglie came up with a formula based on the conclusion of Einstein and Compton that momentum p hlambda for light De Broglie predicted that X hp for particles of momentum p Let s compute X for say me a rather large particle Imagine I m walking through some double doors 2 big slits WillI diffract WillI get quotbentquot off to some funny angle Will there be certain angles the diffraction minima that I will never go to Yikes Let s assume I m walking so V 1 ms and my momentum is p mv 50 kg ms roughly 7t Steve hp 663E34 Js 50 kg ms 13E35 m Now remember you never notice interference effects unless the slit size is roughly as small as wavelength So unless the doors are about lOA35 m apart you ll never notice any effect We re saved from quotquantum weirdnessquot in the ordinary world Atoms are typically lOAlO m apart lOA35 is very small I wouldn t fit through them if they were that close But what about electrons from a low voltage cathode Suppose you let an electron drop through a 100 V difference no big deal a couple of car batteries and a capacitor can do this Let s have them strike a crystal which looks microscopically like a row of atoms about lOAlO m apart A bunch of atoms in a row it s like a grating isn t it The speed of the electron comes from conservation of energy 100 eV 12 m v 2 Solving for v looking up m 91E3l kg I find v 6E6 ms Check this yourself don t forget to convert eV to Joules Do you see why one electron dropping 100V has lOOeV of energy 7telectron hp 663E34 Js9lE3l kg6E6 ms 12E10 m Cool The electron DOES have a wavelength comparable to the spacing of the quotslitsquot between the atoms Electrons should and do diffract when they strike a crystal face The eXperiment was done in 1927 and many more since then Particles have a quotwave naturequot MP8 SIP Phys 2020 Fa 01 This is the principle of the electron microscope You can only quotlookquot at objects if the wavelength of light you are using is smaller than the object If X is bigger than the object the light diffracts and you can t get a good shadow a good image If you want to look at something smaller than Visible light s wavelength a few hundred nm then you need to use a quotlightquot source with smaller wavelength You could still use EM radiation that would mean going into the quotXrayquot region But it s hard to steer X rays there are no good quotlensesquot to focus them Instead you can use electrons They re charged so they re very easy to steer with E and B fields you can build quotlensesquot for electrons with magnets and by varying their energy you can pick wavelengths that are small enough to image individual atoms Very cool practical application of quantum physics It turns out that all particles have wavelike properties Waveparticle duality extends to everything Neutrons e g have MORE mass so more momentum hence small wavelength We use quotneutron interferometersquot if we want interference effects with very low energy particles People have even done experiments where they quotinterferedquot atom beams to get quotbright and dar quot bands The grating was a standing wave of light so the BEAM was particles and GRATING was light exactly backwards of Chapter 24 The equations here are simple nothing new is required d sintheta m lambda for maxima is true for particles too None of these quotquantumquot effects I m discussing were noticed until the 20th century because of the numerical scales the subtlety of the effects as our example of quotmequot diffracting showed Recall the limit of short wavelengths in optics gives quotray opticsquot which is really particlelike behavior We talked about that in Chapter 24 when the slits are big enough you don t NOTICE that light is a wave Neils Bohr came up with an idea called the correspondence principle that basically says any new theory of nature like Einstein s or de Broglie s should still give all the old usual welltested classical results if applied to ordinary situations It s only when you look in new regimes like high speed or tiny systems like atoms that you start to get these new even crazy phenomena Relativity is like that too Newton s laws break down at high speeds but in the limit of speeds much less than c all the formulas reduce to the classical ones You can show this mathematically Classical physics is an approximation to modern physics but an incredibly accurate one in ordinary world situations MP9 SIP Phys 2020 Fa 01 I m afraid quantum mechanics gets much weirder still It s not just that particles have wave character that alone would be something I think we could accept It s much worse their behavior is not fully deterministic I think his is the craziest and most radical break from classical physics of all WhenI describe a particle as a quotwavequot I m really saying that the particle is described by a mathematical quotwave functionquot that tells you the probability of finding the particle somewhere It s not that the electron wiggles like a wave it s more that the location of the electron is not exactly predetermined where the wave function is big the electron is likely to be Where the wave function is small the electron is not likely to be The quotwavequot here is a probability wave This is hard to think about There is an inherent uncertainty in all quantum measurements If you try to measure one aspect of a particle e g its position you can do so but the act of measurement quotdisturbs the wavequot and results in an increased uncertainty about some other aspect eg the momentum Heisenberg39s uncertainty principle quantified this relationship It s not just a statement about our ability to make measurements it s a more fundamental statement about the nature of reality Nature does not know the position and momentum of an electron at the same time because an electron is described only probabilistically There is an intrinsic element of uncertainty in certain observables although not all Einstein helped start the theory of quantum mechanics but he rejected philosophically the conclusions that it inevitably led to quotGod does not play dice with the Universe he said Alas he was fundamentally incorrect about this There is at least as good experimental evidence for the correctness of quantum physics as there is for relativity the consequences of the Heisenberg Uncertainty principle are unavoidable and observable Indeed tunneling electron microscopes make use of it to function Quantum mechanics is a whole new way of thinking about reality The world is not exactly quotclockwor quot there is an element of unpredictability that runs counter to a lot of people s intuitions about science But let s leave this quantum weirdness and look at something a little more concrete atoms MP10 SIP Phys 2020 Fa 01 By the turn of the century people roughly knew the mass of an atom and they had estimated the size of atoms to be about 1 Angstrom lOAIO m from chemistry experiments They know of the existence of electrons Thomson did experiments like our class demo with the CRT to measure em and Millikan did experiments to measure e People had postulated the existence of nuclei Rutherford scattered small atoms off bigger ones Helium from metal and concluded that the nucleus is tiny atoms are mostly empty space He came up with a quotplanetaryquot model a nucleus at the center and an electron orbiting like earth around the sun Only its electricity that holds the electron in not gravity The nucleus is positively charged and massive 999 of the mass of the atom The electron is negatively charged and orbits at a radius of about lOAIO m The nucleus itself is about lOA15 m in size a quotfemtometerquot or quotfmquot also called a Fermi The simplest atom hydrogen is the lightest It consists of one proton in the center and one electron in orbit Helium is the next biggest with 2 protons 2 neutrons to make the mass correct and 2 electrons in orbit And so on up the period table It was a neat simple picture But it was also wrong 1 Electrons in orbit are accelerating centripetally and Maxwell says they should radiate The expected lifetime is calculable a fraction of a second Matter should vanish in a puff of radiation in less than 1 second 2 Hydrogen emits only certain frequencies of light when excited Why The planetary model should allow a continuum of energy electrons can have any energy atoms should just glow with a broad spectrum of energy 2b The light emitted by hydrogen was examined in spectrometers and only a few specific quotlinesquot or colors were seen You ll study this in lab 6 There was even a formula found by Balmer a high school teacher 1gtR122 1 n2 where R was a constant obtained from the hydrogen data and n is any integer You plug in an integer n and out pops an experimentally observed wavelengths The formula had NO theoretical basis it was purely quotempiricalquot MPll SIP Phys 2020 Fa 01 Neils Bohr applied the edgling ideas of quantized energy and photons to understand atoms His model was not fullblown quantum mechanics That took another few years A solid theory of the atom required Schrodinger39s quotquantum wave equationquot which put together de Broglie39s ideas in a more rigorous way and was then improved by Dirac in the late 2039s and nally by QED in the 40 s But Bohr came rst he constructed a quotsemiquantumquot picture a model of atoms that explained some classically inexplicable behavior It paved the way to the full blown quotQuantum mechanicsquot still to come Here were Bohr s quotpostulatesquot 1 Nature only allows certain orbital radii Electrons cannot sit at any random radius with therefore any old energy Phys 2010 will tell us how energy in a circular orbit is determined from radius Bohr didn t know the reason he just guessed the fact that electron radii and thus energies are quantized Only certain values are allowed By the way de Broglie was soon able to explain why If you want a quotwavequot to wiggle in a certain space the wavelength has to t A guitar string only allows certain special wavelengths those that are the length of the string or L2 or L3 They have to t Similarly if an electron is a wave the wavelength hence momentum hence energy is restricted by geometry But never mind for now Bohr didn39t know this 2 Angular momentum rp is ALSO quantized Only certain values of rp are allowed I p n h 2 pi h is Plank s constant and n is any integer Where the heck did that come from Well the units are right quothquot has units of angular momentum and since quothquot quantized light energy maybe it quantizes angular momentum too But why the 2 pi And why quantize angular momentum For Bohr it was purely quotad hocquot it just explained data De Broglie39s idea that quotwavelength has to t into the orbital circumferencequot derives this result too Schrodinger39s wave equation derives it even more rigorously and xes it up a little But like I said Bohr had a crude model not atheory So let39s go on 3 If an electron is in an allowed orbit it won t radiate This is a quotpostulatequot not an explanation It just doesn t radiate says Bohr But if it changes orbits jumps from one allowed radius to another THEN it will radiate It will emit a photon The energy of the photon must conserve total energy E hf of the emitted photon E nal Einitial of the electron in orbit Those are Bohr s postulates Let s look at the consequences of Bohr s model for the simplest atom hydrogen MPl2 SIP Phys 2020 Fa 01 Consider an electron in orbit around a proton electron in orbit Newton says F ma m So Bohr is quotquasiclassicalquot he s still invoking Newton s laws Q Ze with z123 Circular orbits have a v 2r where r is the radius of the electron orbit that s just Phys 2010 kinematics The force is purely electrical Coulomb s law says F k Q1 Q2 rquot2 k Ze e rquot2 Put this together kZe 2r 2 m vA2r Eq n 1 Now Bohr argues angular momentum is quantized rp n h2 pi We will define h h 27 called quoth barquot simply so we don t have to keep writing those 2 pi s all time Anyway Bohr s quantization rule says rmv n h or V nh mr Eq n 2 Plug this into Eq n 1 above and then get quotrquot by itself try it yourself I et r n 2 W N g Z Lkezm Eq n 3 The stuff in parentheses is just a bunch of constants of nature you can look up in Giancoli It s a calculator problem I get 053 ElO m Remember from the figure Z is an integer just the number of protons in the nucleus For hydrogen Zl This formula gives us many different radii you plug in n the quotquantum number of the orbitquot and you ll get out a radius Once you have r you also have a corresponding v from Eq n 2 Bohr s postulate of quantized angular momentum tells us that the radius of the orbit is also quantized But in a funny way it goes like n 2 For hydrogen atoms Zl rl the minimum radius is 53 ElO m This is called the quotBohr radiusquot MP13 SIP Phys 2020 Fa 01 This is a big deal Remember I told you that atoms are experimentally measured to be about lOAIO m in size which is just what the Bohr radius gives us The Bohr model is predicting from fundamental constants what the size of atoms should be Think about that a little Could you have imagined that humans could CALCULATE the size of the building blocks of nature just by knowing what seem to be COMPLETELY unrelated constants like quothquot and quotkquot Remember quothquot is measured from the energy of photons from hot glowing objects or light shining on metal it s a universal constant of nature that has nothing apparently to do with atoms it s about light And quotkquot is the constant from Coulomb s law it tells you about sparks and cat fur It tells you about the forces between charges even just electrons no atoms required to measure it Nature is all tied together the properties of light and electricity are intimately hooked into the objects that make up stuff It s really very cool How about energy Recall from Ch 17 the potential V near a point charge Q is V kQr And potential energy U qV so Etotal PE KE 12 m v 2 k Ze e r Because quotQquot Ze charge of the nucleus quotq e charge of electron Now plug in quotvquot and quotrquot we just found try it if you like algebra puzzles 71 2e lmkz lgetE FZL 2h Once again the mess in parenthesis just involves a bunch of constants of nature m is the electron mass etc Whenl punch in my calculator it looks nicer if I convert to eV instead of Joules I get 136 eV So energy is also quantized again in a funny way Like lnA2 The energy came out negative but that makes sense Electrons are bound in hydrogen you must add energy to get the electron off to infinity Indeed this formula lets us predict that energy the quotbinding energyquot of hydrogen Bohr has concluded that a single integer quotnquot the quotprinciple quantum numberquot tells you the radius speed energy and angular momentum Everything you d want to know about the atom basically MPl4 SIP Phys 2020 Fa 01 Let s examine these quotallowedquot or quotquantizedquot energies noo EOO 0 eV n1 is the lowest orbit the n3 136 eV3A2 15 eV most deeply bound 112 13396 eVzAz 3394 CV You need to add 136 eV to it to make the electron escape to in nity That s the binding energy of hydrogen from chemistry It s bang on correct Another great success of the model the chemical binding energy is calculated exactly And now remember Bohr s ideas 3 that if you re in one of these allowed states called quotorbitalsquot or quotenergy levelsquot you re stable and orbit without radiating but you can also jump to another level If you jump down in energy a photon will be emitted with Ephoton Einit Efinal 136 eV 1ninitquot2 1nfinalquot2 n1 136 eV102 136 eV If you jump down from some excited level to n2 the photon leaves with Ephoton 136 eV12A2 1ninitquot2 Since photons have E hf from Planck and fquot 7tc from Maxwell we have 1lambda fc Ehc 136 eVhc 12A2 1nA2 This is Mr Balmer s formula And Bohr has predicted the factor out front it comes from fundamental constants Think about this Bohr predicts the specific colors of light that hydrogen should emit all from first principles Another remarkable achievement Experimentally the accuracy is stunning Although the theory is crude it works well Now why does nf have to be 2 Can t you jump down to nf1 Sure But if you plug in the numbers you discover that such wavelengths are shorter than visible they re UV So Bohr is now not just matching data he s predicting a whole bunch of additional quotspectral linesquot When you take a UV detector and look for them they re there and exactly predicted by the Bohr formula Bohr also predicts the absorption spectrum For cold hydrogen we expect the electrons to all start in n1 the quot ground statequot lowest possible energy so if you shine light through cold hydrogen you should absorb photons if they have E 136 eV 1n 2 1 1quot2 And that s also right MPlS SIP Phys 2020 Fa 01 Quantum mechanics goes much further than Bohr s idea As I said before Bohr s model is still mostly classical He s thinking of the electron as a small object that obeys Newton s law It has a radius position and velocity momentum But in the end we have learned that s not true The electron s position is uncertain so is its momentum We can derive and calculate a quotwave functionquot for the electron which tells where it is likely to be if you make a measurement The energies that Bohr calculated turn out to be the same in the full quantum calculation but the model the picture of the atom is different I think of it more as a quotclou quot of electron but even that s not right the electron is not fuzzy it s just that no one knows where it is until we look for it It has a probability that is quotspread out like a cloudquot I m sorry it s hard to describe or picture But the mathematics gives spectacular predictions not just the energies of photons but magnetic properties electric properties chemical properties Everything you can measure about hydrogen atoms is calculable from first principles involving only fundamental constants of nature At the same time that physics has become indeterministic in some respects it has also become grandly unified and accurate in others I m afraid a few pages about quantum physics can t possibly do it justice Like relativity you can read Ch 27 and 28 to learn some basic formulas which allow you to understand real life applications tunneling microscopes uorescence how lasers work And why nuclei are radioactive And there are tons of books in any bookstore that try to quotpopularizequot it Unlike relativity which you can really learn and understand on your own I think quantum is much harder A physics major will take three full courses in quantum mechanics as an undergraduate and another 3 as a grad student and spend about a week or two on relativity And then you can use quantum mechanics to make wonderful and accurate predictions but to really start to understand it Well I m still working on that myself CTEnergy l Albert Einstein lowers a book of mass m downward a distance h at constant speed V The work done by the force of gravity is A B C 0 the force of Albert39s hand is A B C 0 the net force on the book is A B C 0 Answers The work done by gravity is since displacement and Fhmd force of gravity are in the same direction The work done by As Al39s hand is 7 since displacement and the force of Al s hand are in the opposite direction The net force is zero so the work done by the net force is zero Fgrav CTEnergy Z A rock of mass m is twirled on a string in a horizontal plane The work done by the tension in the string on the rock is A B C 0 Answer The work done by the tension force is zero since in any tiny interval of time the force of the tension in the string is perpendicular to the direction of the displacement CTEnergy 3 A 1 kg mass is moved part way around a square loop as shown The square is 1 meter on a side and the nal position of the mass is 05 m below its original position Assume that g 10 msz What is the work done by the force of gravity during this journey mgAh 1 kg 10 m2 Ah Up 05m 1m start nish A10J B5J C0J D10J E5J Answer Total work done is 5 J The work done during the horizontal segments is zero since force is perpendicular to displacement there The work during the rst upward vertical segment is negative and this negative work is cancelled by the positive work done during the rst half of the downward vertical segment The total work is that done during the second half of the downward vertical segment where Ah 05 m If instead of moving part way around a square the mass where taken on a long and tortuous journey to the Moon Tibet and Lithuania and then returned to the same nish point as before would the work done by gravity be the A same or B different Answer the same Any journey can be thought of a series of small vertical or horizontal displacements During any horizontal segment the work done by gravity is zero All upward vertical segments are cancelled by corresponding downward vertical segments EXCEPT for the last 05 m between the start and the nish CTEnergy 4 Two marbles one twice as heavy as the other are dropped to the ground from the roof of a building Assume no air resistance Just before hitting the ground the heavier marble has A as much kinetic energy as the lighter one B twice as much kinetic energy as the lighter one C half as much kinetic energy as the lighter one D four times as much kinetic energy as the lighter one E impossible to determine Answer The heavier marble has the same nal speed but twice the mass In so it has twice the KB l2mv2 CTEnergy S A block initially at rest is allowed to slide down a frictionless ramp and attains a speed V at the bottom To achieve a speed 2v at the bottom how many times as high must the new ramp be A 5 214 B 2 C 3 D 4 E none ofthese at rest Answer Using conservation of energy we can write Ei Ef If we place the zero of height at the bottom of the ramp we have KEi PEi KEf PEf Omgh mv20 hocv2 So the initial height h is proportional to the nal speedsquared v2 So to get twice the speed v we need 4 times the height h Since 2v2 4v2 CTEnergy G A mass m is at the end of light massless rod of length R the other end of which has a frictionless pivot so the rod can swing in a vertical plane The rod is initially horizontal and the mass is pushed down with an initial speed v0 What initial kinetic energy is required for the mass to pivot 2700 to the vertical position V0 1 2 KEINITIAL Emvo A ng B mg2R C mg3R D None of these Answer ng Let us put the zero of height at the initial position yR y7R KEi PEi KEf PEf KEi0 0ng Instead we could put the zero of height at the bottom of the circle y2R lt y0 In which case we get the same answer as before KEi PEi KEf PEf KEi ng 0 mg2R KEi ng CTEnergy 7 A mass slides down a rough ramp with friction of height h Its initial speed is zero Its nal speed at the bottom of the ramp is V Vi0 As the mass descended its KB A increased B decreased C remained constant Answer KE increased As the mass descended its PE A increased B decreased C remained constant Answer PE decreased As the mass descended its KE PE total mechanical energy A increased B decreased C remained constant Answer Emech decreased Some of the mechanical energy was converted into thermal energy by the friction Remember KEPE constant ONLY if NO friction and no heat generated CTEnergy S A hockey puck slides without friction along a frozen lake toward an ice ramp and plateau as shown The speed of the puck is 4ms and the height of the plateau is lm Will the puck make it all the way up the ramp A Yes B No C impossible to determine without knowing the mass of the puck Answer We have to do a quick calculation Setting y 0 at the bottom of the ramp we have KEi PEi KEf PEf mvz0 0ng The nal energy PE would be ng massgtlt10ms2lm massgtlt 10 mzs2 The initial energy KE is l2mV2 massgtlt05l6 mzsz massgtlt8 mzs2 So the initial KE is not large enough the puck will not make it to the top of the hill CTEnergy 9 A mass is oscillating back and forth on a spring as shown Position 0 is the relaxed unstretched position of the mass aw At which position is the magnitude of the acceleration of the mass a maximum A 0 At which point is the magnitude of the force on the mass a maximum A 0 At which point is the elastic potential energy a maximum A 0 B M C E Answers The acceleration is maximum at E since the force Fspring ikx is maximum magnitude there Newton s second law says Fuel m a so the acceleration is max when Fuel is max The force on the mass is maximum at E for the same reason that acceleration is max there The elastic potential energy PEelas 12 kxz is maximum at E where x is max Ch6 10 A springloaded dart gun shoots a dart straight up into the air and the dart reaches a maximum height of h 24 m The same dart is shot straight up a second time from the same gun but this time the spring is compressed only half as far before ring How far up does the dart go this time neglecting friction and assuming an ideal spring A 2h 48m B h 24m C m 12m D h4 6m E M 3m Answer Use conservation of energy and write Ei Ef with the initial position where the dart is inside the gun with the spring cocked and the nal position top of the trajectory Let s set h 0 at the initial position KEi PE PE KEf PE gram elasJ PE gravf e1asr 00kx2 0mgh0 k x2 mgh x2 cx h x2 constantgtlthCh Here C is a constant C 2mgk Since x2 is proportional to h when x is cut in half x gt x2 x2 gt x22 x24 h is reduced by a factor of 4 It may be easier to see this by setting up a ratio 2 2 r L 2 39hz X2 2X1 4X1 i 4 2 2 2 39 hr X1 X1 X1 A springloaded dart gun shoots a dart straight up into the air and the dart reaches a maximum height of h 24 m The same gun is reloaded with the spring compressed the same amount but now the gun is aimed at an angle of 450 to the horizontal Will the dart reach the same maximum height of 24m Assume no frictional losses A Yes the dart will reach the same height B No the dart will not reach the same height Answer No We can see why this must be by considering conservation of energy When fired at 45 the dart has some horizontal motion when it is at the top of it trajectory So it always has some KE and its initial PEelastic is never fully converted into PEgmv CTEnergy ll A projectile is red with an initial speed v0 at an angle 9 from the horizontal What is the KB of the projectile when it is on the way down at a height h above the ground Assume no air resistance KE 2 9 hl A l2mv02 mgh B mgh C l2mv2 7 mgh D Impossible to tell Answer Use conservation of energy and write Ei Ef with the initial position where projectile has just left the cannon and the nal position height h KEi PE KEf PEf mvoz 0 mvf2 mgh mvf2 mvoz mgh CTEnergy IZ A small mass starting at rest slides without friction down a rail to a loopdeloop as shown The maximum height of the loop is the same as the initial height of the mass Will the ball make it to the top of the loop A Yes the ball will make it to the top ofthe loop B No the ball will not make it to the top C Not enough info to say or don t know Answer No according to conservation of energy In order for the ball to make it to the top of the loop it must have some nonzero KE at the top of the loop if it did have zero KE at the top of the loop it would have fallen off the track before then because it would have been going too slow Since it must have some KE at the top ofthe loop some of its initial PEIgrav was converted into that KE so its PEIgrav at the top of the loop must be less than the initial PEng CTEnergy 13 A cart rolls without friction along a track The graph of PE vs position is shown The total mechanical energy KE PE is 45k 50 4o Etot PEkJ 20 10 O 0 2O 4O 60 80 100 120140 160 Xm To within 5k what is the maximum KE over the stretch of track shown A 251d B 7k C 45k D 35 kJ Answer Max KE of35 kJ occurs at about X 130 m CTEnergy 14 A cart rolls without friction along a track The graph of PE vs position is shown The total mechanical energy KE PE is 0 k What is the MAX KB of the cart during its journey to within SkJ 20m 40m 60m A 35 k B 48 k C 16 k D 16 k E None of thesedon39t know Answer MaX KE is where PE is min MaX KE 48 kJ lt EtotO PE KE 48kJ 48kJ Suppose the cart is at position X 20m is moving right and has total energy Elm 7 20k Will the cart make it over the hill at X38m A Yes B No Answer No The quotturning pointquot where the cart slows to a stop and turns around The turning point is where the Em line intersects the PE curve This is where KE 0 because Em KE PE Etot 20 M u CTEnergy IS A hockey puck sliding on an ice rink is moving at 1 ms when it slides onto a ca1pet that someone left on the ice The puck comes to rest after moving lm on the carpet How far along the carpet would the puck go if its initial speed was 2ms A 15m B 2m C 3m D 4m E Impossible to determine Hint Apply the WorkEnergy Theorem If the puck slides twice as far the friction does twice as much negative work Answer 4 m We have to do a calculation to see this This is a case with friction Remember that the magnitude of the work done by friction is the amount of thermal energy gained which is the amount of mechanical energy KEPE lost wmc AKE since APE 0 ungAX mv2 AXocv2 So ifwe double v v2 is increased by a factor of 4 and so is AX Warmup question How much more KE does the puck have moving at 2 ms compared to moving at 1 ms A J5 times as much B twice as much C 3 times as much D 4 times as much Answer 4 times as much CTEnergy IG Elevator 1 can carry a load of mass m up a distance h in a time t before the engine overheats Its power output is P1 Elevator 2 can ca1ry the same load up twice the distance 2h in twice the time 2t What is its power output P2 A P2P1 B P22P1 C P2 4P1 DNonethese Answer P2 worldtime mg2h 2t mght P1 CTEnergy 17 Elevator 3 can carry twice the load 2m up twice the distance 2h in the same time t as elevator 1 What is its power output P3 AP3IH BP32P1 CP34P1 DNone wse Answer P3 worldtime 2mg2ht 4mght 4P1 2h 2h me2t met met m m 2m CTTheI39m l A piece of Krell metal is cool to the touch even after a blaster pistol has red several shots at it One shot can vaporize a tiger Compared to water Krell metal has a heat capacity which is very very A small B large C about the same Answer Krell metal has a very large heat capacity AQ m c AT In the clip from the movie quotForbidden Planetquot we saw a blaster pistol red into a Krell metal door The blaster delivers a large amount of heat Q to the door but the metal remained cool to the touch small AT so the speci c heat c must be very large CTTheI39m Z I put a 100W water heater in my cup of water 100W 100 Js 24 cals I have 240 grams of water in my cup How long to raise the temperature of my water by 0 A l s B 10 s C 100 s D None ofthese Answer 10 s 240 grams of water requires 240 cal to raise its temperature by 1 C 1 cal per gram per 0C With ofpower of 24 cals we will need 10 s to get 240 cal CTTheI39m 3 An ice cube is placed in a cup containing some liquid water The water and ice exchange energy with each other but not with the outside world After the water and ice come to the same temperature is it possible the ice could freeze the water rather than the water melt some of the ice A Yes the ice could freeze the water B No the water will always melt some of the ice liquid water Answer Yes the ice could freeze the water If the ice was very cold well below freezing say T 200 C then heat would ow from the liquid water to the cold ice cube The heat required to raise the temperature of the ice cube up to the melting point could be greater than the heat which the liquid water could give up before freezing CTTheI39m 4 Data cice 050 calg 0C cwam 100 calg 0C 05mm 048 calg 0C LiceWater 80 Calg LWatersteam539 calg Which requires the most energy A Raise the temperature of 1 gram of liquid water 50 0C B Melt 1 gram of ice at T0 0C C Raise the temperature of 1 gram of ice from 73K to 273K D Vaporize 05 gram of water at T100 0C Answer Vaporizing 05 g of water requires 5392 268 cal Heating water 50 C requires 50 cal assuming it stays liquid Melting a gram of ice requires 797 cal Heating a gram ice from 73K to 273K requires 100cal CTTheI39m S I put a 100W water heater in my cup of water 100W 100 Js 24 cals I have 240 grams of water in my cup How long to raise the temperature of my water by 1 C A l s B 10 s C 100 s D None ofthese Answer 10 s 240 grams of water requires 240 cal to raise its temperature by 1 C 1 cal per gram per 0C With ofpower of 24 cals we will need 10 s to get 240 cal CTTheI39m G Heat from your house s furnace reaches the rooms in your house primarily by A conduction B convection C radiation D None of these Answer convection If your furnace is quotforced airquot then hot air is blown through the vents into the rooms This is bulk movement of heated matter the hot air is transported from the furnace to the rooms If you have steam heat either inthe oor hot water pipes or radiators then a pump is moving the hot watersteam from the furnace to the rooms It is still bulk movement of hot matter still convection CTThermJ Here m Boulder rt ls eoldrn the wrnter and warm m the summer Why7 Whreh explanatron below ls the most aeeurate7 A In the wrnter there rs less solar radratron per unrt area of the ground B The Sun rs further from the Earth rn the wrnter C Beeause of elrmate patterns rtrs more eloudy than average rn the wrnter therefore less solarradratron than average reaehes the ground ground n r T ka Tnth wtnt r rn the wt r and eaeh square meter ofgroundrecelves fewer Solar rays square meter ofgound Sun over head summer Sun s rays square meter ofgound Sun low rn sky Wrnter Rotational Motion We are going to consider the motion of a rigid body about a fixed axis of rotation The angle of rotation is measured in radians 0rads dimensionless S Notice that for a given angle 0 the ratio sr is A S independent of the size of the circle A Example How many radians in 180 Circumference C 2 11 r S nr snr 0 11rads A r r r 11 rads 180 1 rad 5730 Angle 0 of a rigid object is measured relative to some reference orientation just like 1D position x is measured relative to some reference position the origin Angle 0 is the quotrotational positionquot Definition of angular velocity 03 5 like V E Blg A g Q V V In 1D velocity v has a sign or 7 depending on direction Likewise n has a sign convention depending on the sense of rotation 0 I V i 39 m 7 7r gtllt For rotational motion there is a relation between tangential velocity v velocity along the rim and angular velocity 0 A0 E 3 AsrA0 r Asin A0 timeAt V rrm At At 10302006 University of Colorado at Boulder V A Def1n1tlon of angular acceleration llke a E E 0c rate at which n is changing n constant 3 0c 0 3 speed V along rim constant r 03 De nition of tangential acceleration am rate at which speed V along rim is changing AV Ar n r A03 acan E 3 am r at At At At Z V am 1s d1fferent than the rad1al or centripetal acceleratlon a r ar is due to change in direction of Velocity V am is due to change in magnitude of Velocity speed V am and ar are the tangential and radial components of the acceleration Vector a Angular Velocity 03 also sometimes called angular frequency Difference between angular Velocity n and frequency f radians f reVolutions sec sec T period time for one complete reVolution or cycle or reV 3 T T T T Units of frequency f reVs hertz Hz Units of angular Velocity rad s s391 Example An old Vinyl record disk with radius r 6 in 152 cm is spinning at 333 rpm reVolutions per minute 333 reV 333 reV 60s 60333s o What is the period T 3 180 sreV 1mm 60 s 333reV lreV 10302006 UniVersity of Colorado at Boulder 3 period T 180 s o What is the frequencyf f 1T 1reV180 s 0555 Hz 0 What is the angular velocity 03 n 2 11 f 2 11 0555 s39l 349 rads o What is the speed V of a bug hanging on to the rim of the disk V r w 152 cm349 s39l 530 cms o What is the angular acceleration 0c of the bug at 0 since 03 constant 0 What is the magnitude of the acceleration of the bug The acceleration has only a radial component ar since the tangential acceleration am r at 0 V2 0530 n s2 a a 184 ms2 about 02 g39s r 0152m As we shall see for every quantity in linear translational motion there is an analogous quantity in rotational motion Translation lt gt Rotation X lt gt 0 AX A0 V lt gt n At At AV Am a lt gt 0c At At F lt gt M lt gt 7 F Ma lt gt 0c The rotational analogue of force is torque Force F causes acceleration a lt gt torque 139 causes angular acceleration 0c The torque pronounced quottorkquot is a kind of quotrotational forcequot magnitude of torque HHFFmN 10302006 UniVersity of Colorado at Boulder r quotlever armquot distance from axis of rotation to point of application of force F i component of force perpendicular to lever arm Example Wheel on a fixed axis Notice that only the perpendicular component of the force F will rotate the wheel The component of the FLFsin0 force parallel to the lever arm FH has no effect on the rotation of the F wheel 9 FM If you want to easily rotate an object about an axis you want a large lever arm r and a large perpendicular force F i I bad I better I beSt axis no good Fl 0 no good r 0 Example Pull on a door handle a distance r 08 m from the hinge with a force of magnitude F 20 N at an angle 0 300 from the plane ofthe door like so FLTF trFirFsin0 11mg 9 08 m20 Nsin 30 80 mN Another example a Pulley 1 r F i Fl Torque has a sign or 7 Positive torque causes counterclockwise CCW rotation 1 7 1 Negative torque causes clockwise CW rotation 10302006 University of Colorado at Boulder If several torques are applied the net torque causes angular acceleration In 2 oc 0c To see the relation between torque 1 and angular acceleration 0c consider a mass m at the end of light rod of length r pivoting on an aXis like so aX1s I Apply a force F i to the mass keeping the force perpendicular to the lever arm r FL acceleration awn r X Apply Fuel m a along the tangential direction aXis F i m am m r at Multiply both sides by r to get torque in the game r FL m r 2 0c De ne quotmoment of inertiaquot I mr2 Can generalize de nition of 1 De nition of moment of inertia of an extended object about an aXis of rotation I E 21min2 mlrl2 mzrz2 l i Examples 0 2 small masses on rods of length r r r 0 0 0 2 12mr m aX1s m 10302006 University of Colorado at Boulder o A hoop of total mass M radius R with aXis through the center has Ihoop M R2 mi I Emirf ZmJRZ MR2 since r R for all i In detail Z Z 2 2 2 2 I mlr1 m2r2 m3r3 m1R m2R m3R m1 m2 m3 R2 MR2 0 A solid disk of mass M radius R with aXis through the center mass M 1disk 12 MR2 hard to show Moment of inertia I is a kind of quotrotational massquot Big I 3 hard to get rotating like Big M 3 hard to get moving If I is big need a big torque 1 to produce angular acceleration according to InaIx likanetma Example Apply a force F to a pulley consisting of solid disk of radius R mass M at EIOL RF MRZ0L 2 at Rotational Kinetic Energy How much KE in a rotating object Answer KEmt I 032 like KENquot m V2 10302006 University of Colorado at Boulder Proof KEtot Zemiviz VDI Vi oari aX1s 1 i EI mi KE Zmimriz Zmiri2mz ilmz 9 How much KE in a rolling wheel The formula V r n is true for a wheel spinning about 3 V a xed aX1s or rolling on the ground V center of mass velocity aXiS Stationary point touching ground instantaneously at rest To see why look at situation from the bicyclist39s point of View aXis stationary ground moving gtllt Rolling KE Rolling wheel simultaneously translating and rotating KEtot KEtrans KErot KE M V2 I of 10302006 University of Colorado at Boulder Conservation of energy problems with rolling motion A sphere a hoop and a cylinder each with mass M and radius R all start from rest at the top of an inclined plane and roll down to the bottom Which object reaches the bottom rst Apply Conservation of Energy to determine V nal Largest V nal will be the winner KEi PEi KEf PEf 0 Mgh MV2 032 0 KEtrans KEYOt Value of moment of inertia I depends on the shape of the rolling thing Idisk l2M R2 Ihoop M R2 Isphere 2 5M R2 Computing the coef cient can be messy Let39s consider a disk with I l2MR2 For the disk the rotational KE is V Z Ioa M R2E MV2 usedoaVr 3 Mgh MV2 MV2 MVZ MV2 2 4 ghV V ggh E 1161 gh Notice that nal speed does not depend on M or R Let39s compare to nal speed of a mass M sliding down the ramp no rolling no friction M M gh M V2 M39s cancel D Vi0 3V 2gh E 141 gh Sliding mass goes faster than rolling disk Why 10302006 University of Colorado at Boulder R9 As the mass descends PE is converted into KE With a rolling object KEtm KELmns KEm so some of the PE is converted into KErm and less energy is left over for KEmms A smaller KELrams means slower speed since KEmms 12 M v2 So rolling object goes slower than sliding object because with rolling object some of the energy gets quottied upquot in rotation and less is available for translation Comparing rolling objects Ihoop gt Idisk gt Isphere 3 Hoop has biggest KEm 12 I 032 3 hoop ends up with smallest KEmms 3 hoop rolls down slowest sphere rolls down fastest Angular Momentum quotSpinquot p 1 De nition of angular momentum ofa spinning object like p m V If something has a big moment of inertia I and is spinning fast big n then it has a big quotspinquot big angular momentum Angular momentum is a very useful concept because angular momentum is conserved Conservation of Angular Momentum If a system is isolated from external torques then its total angular momentum L is constant Text 0 3 L101 constant like Fm 0 3 Ptot constant Here is a plausibility argument for conservation of angular momentum proof is a bit too messy First we argue that Enet amp like Fnet E a At At Tnet 10 1 assumingIconst amp At A At This turns out to be true even if I at constant Sonow In 3 if c0then 0 3 L constant At At It turns out that only 4 things are conserved 0 Energy 0 Linear momentum p o Angular momentum L 0 Charge q 10302006 University of Colorado at Boulder R10 Conservation of Angular Momentum is very useful for analyzing the motion of spinning objects isolated from external torques 7 like a skater or a spinning star If Text 0 then L I n constant IfI decreases to must increase to keep L constant Example spinning skater cl Ii Di If 03f 1 big 03 small 1 small 03 big Example rotation of collapsing star A star shines by converting hydrogen H into helium He in a nuclear reaction When the H is used up the nuclear re stops and gravity causes the star to collapse inward l gravity l v H fast 39 f x slow I As the star collapses pulls its arms in the star rotates faster and faster Star radius can get much smaller R m 1 million miles gt Rfm 30 miles f Sphere I M R2 M Ri20 i sz Dr R2 n sz of usingoa2nf2 n f T If Ri gtgt Rf then Ti gtgtgtgtTf The sun rotates once every 27 days quotNeutron starsquot with diameter of about 30 miles typically rotates 100 time per second 10302006 University of Colorado at Boulder Let s review the correspondence between translational and rotational motion Translation X AX V At AV a At F M Fnet M a KEm 12M v2 pmV Fuel Ap At gt gt gt If Fext 0 pm constant 10302006 Rotation 0 A0 3 At Am a At 139 r FL I 2 m r2 Tnet 10 KErm 12 1 of L I n IneLALAt gt If Text 0 L101 constant University of Colorado at Boulder 14 13 12 A11 E10 V3 1m52x10510mls gt L39 7 axt o 6 x10msx1050m 0 5 12axtxt d gt 4 3 2 1 O 012345678910 Time 5 Dis nlacement TOTAL 90 m d Second Method Average velocity Vave 12 4 ms ms 9 ms Vave 12 V0 yt 12 V0 V0 12 a X t Displacement d9msgtlt IOs90m d vmx t v0gtltt12agtltt2 A ball is red straight downward out of a special gun which produces constant acceleration Upward is chosen as the positive direction Assume that air resistance is negligible Which graph properly represents acceleration of the ball Pink ball released a from spring a Yellow t Blue Green Purple None of them A train car moves along a long straight track The graph shows the position as a function of time for the train position time The graph shows that the train Green speeds up all the time Yellow slows down all the time Blue speeds up part of the time and slows down part of the time Pink moves at constant velocity Purple None of these statements is true The X and ycoordinates of a particle as a function of time are xlbcl yld el Where bcd and e are positive constants Question 1 What is the direction of the velocity of the particle Pink y Blue Purple None of these Green Yellow Question 2 What is the direction of the acceleration of the particle Pink y 1 Blue Purple None of these Green Yellow
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