UPGRADE YOUR PHYSICS


 Randall Nigel Holmes
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1 Correctons March 7 UPGRADE YOUR PHYSICS NOTES FOR BRITISH SIXTH FORM STUDENTS WHO ARE PREPARING FOR THE INTERNATIONAL PHYSICS OLYMPIAD, OR WISH TO TAKE THEIR KNOWLEDGE OF PHYSICS BEYOND THE ALEVEL SYLLABI. A. C. MACHACEK Page
2 Correctons March 7 Introducton The Internatonal Physcs Olympad s an annual nternatonal physcs competton for preunversty students. Teams of fve from each partcpatng naton attend, and recently over 6 countres have taken part. Each naton has ts own methods for selectng ts team members. In Brtan, ths s by means of a seres of wrtten and practcal exams. The queston paper for the frst round s crculated to all secondary schools. Once the team has been chosen, t s necessary for ts members to broaden ther horzons. The syllabus for the Internatonal Physcs Olympad s larger than that of the Brtsh Alevel, and ndeed forms a convenent steppngstone to frst year undergraduate work. For ths reason, tranng s provded to help the Brtsh team brdge the gap. The Brtsh Olympad Commttee recognzes the need for teachng materal to help canddates prepare for the nternatonal competton. Furthermore, ths materal ought to have greater potental n the hands of students who wsh to develop ther physcs, even f they have no desre to take part n the examnatons. It s my hope that these notes make a start n provdng for ths need. A.C. Machacek, About the author: Anton Machacek has served on the Brtsh Physcs Olympad Commttee snce 997, and has been nvolved regularly n tranng the Brtsh team and n wrtng Physcs Challenge examnatons. He served on the academc commttee for the Internatonal Physcs Olympad n Lecester n. Anton s Head of Physcs at the Royal Grammar School, Hgh Wycombe, and s an Academc Vstor n the subdepartment of Atomc and Laser Physcs, Unversty of Oxford. Page
3 Correctons March 7 Contents LINEAR MECHANICS...5. MOTION IN A LINE...5. GOING ORBITAL FLUIDS WHEN THINGS GET STICKY QUESTIONS... FAST PHYSICS...4. THE PRINCIPLE OF RELATIVITY...5. HIGH SPEED OBSERVATIONS RELATIVISTIC QUANTITIES THE LORENTZ TRANSFORMS QUESTIONS ROTATION ANGLE ANGULAR VELOCITY ANGULAR ACCELERATION TORQUE ANGULAR FORCE MOMENT OF INERTIA ANGULAR MASS ANGULAR MOMENTUM ANGULAR MOMENTUM OF A SINGLE MASS MOVING IN A STRAIGHT LINE ROTATIONAL KINETIC ENERGY SUMMARY OF QUANTITIES ROTATIONAL MECHANICS WITH VECTORS MOTION IN POLAR COORDINATES MOTION OF A RIGID BODY QUESTIONS VIBES, WIGGLES & LIGHT OSCILLATION WAVES & INTERFERENCE RAYS FERMAT S PRINCIPLE...69 Page 3
4 Correctons March QUESTIONS HOT PHYSICS THE CONSERVATION OF ENERGY THE SECOND LAW HEAT ENGINES AND FRIDGES ENTROPY IRREVERSIBLE PROCESSES AND THE SECOND LAW RESTATEMENT OF FIRST LAW THE BOLTZMANN LAW PERFECT GASES RADIATION OF HEAT QUESTIONS SPARKS & GENERATION ELECTROSTATICS WHEN THINGS ARE STILL MAGNETISM WHEN THINGS MOVE CIRCUITS PUTTING IT TOGETHER QUESTIONS SMALL PHYSICS WAVES AND PARTICLES UNCERTAINTY ATOMS LITTLE NUTS QUESTIONS PRACTICAL PHYSICS ERRORS, AND HOW TO MAKE THEM ERRORS, AND HOW TO MAKE THEM WORSE SYSTEMATIC ERRORS WHICH GRAPH? QUESTIONS APPENDIX MULTIPLYING VECTORS DIMENSIONAL ANALYSIS...35 Page 4
5 Correctons March 7 Lnear Mechancs. Moton n a Lne.. The Fundamentals... Knematcs Mechancs s all about moton. We start wth the smplest knd of moton the moton of small dots or partcles. Such a partcle s descrbed completely by ts mass (the amount of stuff t contans) and ts poston. There s no nternal structure to worry about, and as for rotaton, even f t tred t, noone would see. The most convenent way of labellng the poston s wth a vector r showng ts poston wth respect to some convenent agreed statonary pont. If the partcle s movng, ts poston wll change. If ts speed and drecton are steady, then we can wrte ts poston after tme t as r = s + ut, where s s the startng pont (the poston of the partcle at t=) and u as the change n poston each second otherwse known as the velocty. If the velocty s not constant, then we can t measure t by seeng how far the object goes n one second, snce the velocty wll have changed by then. Rather, we say that u how far the object would go n one second f the speed or drecton remaned unchanged that long. In practce, f the moton remans constant for some small tme (called t), and durng ths small tme, the partcle s poston changes r, then the change n poston f ths were mantaned for a whole second (otherwse known as the velocty) s u = r number of t perods n one second = r t. Smlarly, f the velocty s changng, we defne the acceleraton as the change n velocty each second (f the rate of change of acceleraton were constant. Accordngly, our equaton for acceleraton becomes a = u t. Hopefully, t s apparent that as the moton becomes more complex, and the t perods need to be made shorter and shorter, we end up wth the dfferental equatons lnkng poston, velocty and acceleraton: d u r dt d a u dt r u u dt. a dt Page 5
6 Correctons March 7... Dynamcs Now we have a way of descrbng moton, we need a way of predctng or explanng the moton whch occurs changng our queston from what s happenng? to why? and our explanaton s gong to nvolve the actvty of forces. What do forces do to an object? The frst essental pont s that forces are only needed to change (not mantan) moton. In other words unless there s a change of velocty, no force s needed. But how much force s needed? Newton made the assumpton (whch we fnd to be helpful and true) that the force causes a change n what he called the moton we now call t momentum. Suppose an object has mass m and velocty u (we shall clarfy what we mean by mass later) then ts momentum s equal to mu, and s frequently referred to by physcsts by the letter p. Newton s second law states that f a constant force F s appled to an object for a short tme t, then the change n the momentum s gven by F t. In dfferental notaton d(mu)/dt = F. In the case of a sngle object of constant mass t follows that d mu du F m ma. dt dt Hs next assumpton tells us more about forces and allows us to defne mass properly. Imagne two brcks are beng pulled together by a strong sprng. The brck on the left s beng pulled to the rght, the brck on the rght s beng pulled to the left. Newton assumed that the force pullng the left brck rghtwards s equal and opposte to the force pullng the rght brck leftwards. To use more mathematcal notaton, f the force on block no. caused by block no. s called f, then f =f. If ths were not the case, then f we looked at the brcks together as a whole object, the two nternal forces would not cancel out, and there would be some left over force whch could accelerate the whole object. It makes sense that f the brcks are dentcal then they wll accelerate together at the same rate. But what f they are not? Ths s where Newton s second law s helpful. If the resultant force on an object of If you want to prove that ths s rdculous, try lftng a large bucket whle standng n t. Page 6
7 Correctons March 7 constant mass equals ts mass tmes ts acceleraton, and f the two forces are equal and opposte, we say f m a a a f m m m a, and so the more massve block accelerates less. Ths s the defnton of mass. Usng ths equaton, the mass of any object can be measured wth respect to a standard klogram. If a mystery mass experences an acceleraton of m/s whle pushng a standard klogram n the absence of other forces, and at the same tme the klogram experences an acceleraton of 4m/s the other way, then the mystery mass must be kg. When we have a group of objects, we have the opton of applyng Newton s law to the objects ndvdually or together. If we take a large group of objects, we fnd that the total force F total d F dt m u changes the total momentum (just lke the ndvdual forces change the ndvdual momenta). Note the smplfcaton, though there are no f j n the equaton. Ths s because f j + f j =, so when we add up the forces, the nternal forces sum to zero, and the total momentum s only affected by the external forces F....3 Energy and Power Work s done (or energy s transferred) when a force moves somethng. The amount of work done (or amount of energy transformed) s gven by the dot product of the force and the dstance moved. W = F r = F r cos () where s the angle between the force vector F and the dstance vector r. Ths means that f the force s perpendcular to the dstance, there s no work done, no energy s transferred, and no fuel supply s needed. If the force s constant n tme, then equaton () s all very well and good, however f the force s changng, we need to break the moton up nto lttle parts, so that the force s more or less constant for each part. We may then wrte, more generally, W = F r = F r cos (a) Two useful dfferental equatons can be formed from here. Page 7
8 Correctons March Vrtual Work From equaton (a) t s clear that f the moton s n the drecton of the force appled to the object (.e. =), then W F, r where W s the work done on the object. Accordngly, we can calculate the force on an object f we know the energy change nvolved n movng t. Let s gve an example. An electron (wth charge q) s forced through a resstor (of length L) by a battery of voltage V. As t goes through, t must lose energy qv, snce V s the energy loss per coulomb of charge passng through the resstor. Therefore, assumng that the force on the electron s constant (whch we assume by the symmetry of the stuaton), then the force must be gven by W / d = qv / L. If we defne the electrc feld strength to be the force per coulomb of charge (F/q), then t follows that the electrc feld strength E = V/L. So far, we have gnored the sgn of F. It can not have escaped your attenton that thngs generally fall downwards n the drecton of decreasng [gravtatonal] energy. In equatons () and (a), we used the vector F to represent the externally appled force we use to drag the object along. In the case of lftng a hodful of brcks to the top of a wall, ths force wll be drected upwards. If we are nterested n the force of gravty G actng on the object (whether we drag t or not), ths wll be n the opposte drecton. Therefore F = G, and W = G r, (b) W G. r In other words, f an object can lose potental energy by movng from one place to another, there wll always be a force tryng to push t n ths drecton....5 Power Another useful equaton can be derved f we dfferentate (a) wth respect to tme. The rate of workng s the power P, and so W P t F δr t r F. t As we let the tme perod tend to zero, r/t becomes the velocty, and so we have: Page 8
9 Correctons March 7 P = F u = F u cos () where s now best thought of as the angle between force and drecton of moton. Agan we see that f the force s perpendcular to the drecton of moton, no power s needed. Ths makes sense: thnk of a bke gong round a corner at constant speed. A force s needed to turn the corner  that s why you lean nto the bend, so that a component of your weght does the job. However no work s done you don t need to pedal any harder, and your speed (and hence knetc energy) does not change. Equaton () s also useful for workng out the amount of fuel needed f a workng force s to be mantaned. Suppose a car engne s combatng a frcton force of N, and the car s travellng at a steady 3m/s. The engne power wll be N 3m/s = 6 kw. Our equaton can also be used to derve the knetc energy. Thnk of startng the object from rest, and calculatng the work needed to get t gong at speed U. The force, causng the acceleraton, wll be F=ma. The work done s gven by W P dt F mv dv v dt m v dt U mv mu dv dt (3) although care needs to be taken justfyng the ntegraton stage n the multdmensonal case... Changng Masses The applcaton of Newton s Laws to mechancs problems should pose you no trouble at all. However there are a couple of extra consderatons whch are worth thnkng about, and whch don t often get much attenton at school. The frst stuaton we ll consder s when the mass of a movng object changes. In practce the mass of any selfpropellng object wll change as t uses up ts fuel, and for accurate calculatons we need to take ths nto account. There are two cases when ths must be consdered to get the answer even roughly rght jet aeroplanes and rockets. In the case of rockets, the fuel probably makes up 9% of the mass, so t must not be gnored. The proof s nterestng. It turns out that v dv v dv cos v dv snce the change n speed dv s equal to dv cos where dv (note the bold type) s the vector gvng the change n velocty. Page 9
10 Correctons March 7 Changng mass makes the physcs nterestng, because you need to thnk more carefully about Newton s second law. There are two ways of statng t ether () () Force on an object s equal to the rate of change of ts momentum Force on an object s equal to mass acceleraton The frst says F d( mu) dt mu mu ma m u, whereas the second smply states F=ma. Clearly they can t both be correct, snce they are dfferent. Whch s rght? The frst: whch was actually the way Newton stated t n the frst place! The good old F=ma wll stll work but you have to break the rocket nto parts (say grams of fuel) so that the rocket loses parts, but each part does not lose mass and then apply F=ma to each ndvdual part. However f you want to apply a law of moton to the rocket as a whole, you have to use the more complcated form of equaton. Ths may be the frst tme that you encounter the fact that momentum s a more frendly and fundamental quantty to work wth mathematcally than force. We shall see ths n a more extreme form when lookng at specal relatvty. Let us now try and calculate how a rocket works. We ll gnore gravty and resstve forces to start wth, and see how fast a rocket wll go after t has burnt some fuel. To work ths out we need to know two thngs the exhaust speed of the combuston gas (w), whch s always measured relatve to the rocket; and the rate at whch the motor burns fuel (n kg/s), whch we shall call. We ll thnk about one part of the moton, when the rocket starts wth mass (M+m), burns mass m of fuel, where m s very small, and n dong so ncreases ts speed from U to U+u. Ths s shown below n the dagram. Before M+m m After M U Uw U+u Notce that the velocty of the burnt fuel s Uw, snce w s the speed at whch the combuston gas leaves the rocket (backwards), and we need to take the rocket speed U nto account to fnd out how fast t s gong relatve to the ground. Conservaton of momentum tells us that Page
11 Correctons March 7 (M+m) U = m (Uw) + M (U+u) so m w = M u. (4) We can ntegrate ths expresson for u to evaluate the total change n speed after burnng a large amount of fuel. We treat the u (change n U) as an nfntesmal calculus du, and the m as a calculus dm. Notce the mnus sgn clearly the rocket must lose mass as fuel s burnt. Equaton (4) now tells us Ths can be ntegrated to gve dm w du (5) M U w M w fnal dm ln M U U ntal du M wln M ntal fnal (6) Ths formula (6) s nterestng because t tells us that n the absence of other forces, the gan n rocket speed depends only on the fracton of rocket mass that s fuel, and the exhaust speed. In ths calculaton, we have gnored other forces. Ths s not a good dea f we want to work out the moton at blast off, snce the Earth s gravty plays a major role! In order to take ths, or other forces, nto account, we need to calculate the thrust force of the rocket engne a task we have avoded so far. The thrust can be calculated by applyng F=ma to the (fxed mass) rocket M n our orgnal calculaton (4). The acceleraton s gven by du/t = u/t, where t s the tme taken to burn mass m of fuel. The thrust s T u mw mw m M M w w (7) t Mt t t gven by the product of the exhaust speed and the rate of burnng fuel. For a rocket of total mass M to take off vertcally, T must be greater than the rocket s weght Mg. Therefore for lft off to occur at all we must have w Mg. (8) Ths explans why heavy hydrocarbon fuels are nearly always used for the frst stage of lqud fuel rockets. In the later stages, where absolute Page
12 Correctons March 7 thrust s less mportant, hydrogen s used as t has a better kck per klogram because of ts hgher exhaust speed...3 Fcttous Forces Fcttous forces do not exst. So why do we need to gve them a moment s thought? Well, sometmes they make our lfe easer. Let s have a couple of examples...3. Centrfugal Force You may have travelled n one of those farground rdes n whch everyone stands aganst the nsde of the curved wall of a cylnder, whch then rotates about ts axs. After a whle, the floor drops out and yet you don t fall, because you re stuck to the sde. How does ths work? There are two ways of thnkng about ths. The frst s to look at the stuaton from the statonary perspectve of a frend on the ground. She sees you rotatng, and knows that a centrpetal force s needed to keep you gong round a force pontng towards the centre of the cylnder. Ths force s provded by the wall, and pushes you nwards. You feel ths strongly f you re the rder! And by Newton s thrd law t s equally true that you are pushng outwards on the wall, and ths s why you feel lke you are beng thrown out. Whle ths approach s correct, sometmes t makes the maths easer f you analyse the stuaton from the perspectve of the rder. Then you don t need to worry about the rotaton! However n order to get the workng rght you have to nclude an outwards force to balance the nward push of the wall. If ths were not done, the force from the walls would throw you nto the central space. The outward force s called the centrfugal force, and s our frst example of a fcttous force. It doesn t really exst, unless you are workng n a rotatng reference frame, and nsst that you are at rest. The dfference between the two vewponts s that n one case the nward push of the wall provdes the centrpetal acceleraton. In the other t opposes the centrfugal force  gvng zero resultant, and keepng the rder stll. Therefore the formulae used to calculate centrpetal force also gve the correct magntude for centrfugal force. The two dfferences are: () Centrfugal force acts outwards, centrpetal force acts nwards () Centrfugal force s only consdered f you are assumng that the cylnder s at rest (n the cylnder s reference frame). On the other hand, you only have centrpetal acceleratons f you do treat the cylnder as a movng object and work n the reference frame of a statonary observer. Ths example also shows that fcttous forces generally act n the opposte drecton to the acceleraton that s beng gnored. Here the Page
13 Correctons March 7 acceleraton s an nward centrpetal acceleraton, and the fcttous centrfugal force ponts outward...3. Inertal Force The second example we wll look at s the moton of a lft (elevator) passenger. You know that you feel heaver when the lft accelerates upwards, and feel lghter when t accelerates downwards. Therefore f you want to smplfy your maths by treatng the lft car as a statonary box, you must nclude an extra downward force when the lft s actually acceleratng upwards, and vceversa. Ths fcttous force s called the nertal force. We see agan that t acts n the opposte drecton to the acceleraton we are tryng to gnore. We shall look more closely at ths stuaton, as t s much clearer mathematcally. Suppose we want to analyse the moton of a ball, say, thrown n the ar n a lft car whle t s acceleratng upwards wth acceleraton A. We use the vector a to represent the acceleraton of the ball as a statonary observer would measure t, and a to represent the acceleraton as measured by someone n the lft. Therefore, a = A + a. Now ths ball won t smply travel n a straght lne, because forces act on t. Suppose the force on the ball s F. We want to know what force F s needed to get the rght moton f we assume the lft to be at rest. Newton s second law tells us that F=ma, f m s the mass of the ball. Therefore F=m(A+a ), and so FmA = ma. Now the force F must be the force needed to gve the ball acceleraton a (the moton relatve to the lft car), and therefore F =ma. Combnng these equatons gves F = F ma. (9) In other words, f workng n the reference frame of the lft, you need to nclude not only the forces whch are really actng on the ball (lke gravty), but also an extra force ma. Ths extra force s the nertal force. Let us contnue ths lne of thought a lttle further. Suppose the only force on the ball was gravty. Therefore F=mg. Notce that F = F ma = m (ga) () and therefore f g=a (that s, the lft s fallng lke a stone, because some nasty person has cut the cable), F =. In other words, the ball behaves as f no force (not even gravty) were actng on t, at least from the pont of vew of the unfortunate lft passengers. Ths s why weghtlessness s experenced n free fall. Page 3
14 Correctons March 7 A smlar argument can be used to explan the weghtlessness of astronauts n orbtng spacecraft. As statonary observer (or a physcs teacher) would say that there s only one force on the astronauts gravty, and that ths s just the rght sze to provde the centrpetal force. The astronaut s perspectve s a lttle dfferent. He (or she) experences two forces gravty, and the fcttous centrfugal force. These two are equal and opposte, and as a result they add to zero, and so the astronaut feels just as weghtless as the doomed lft passengers n the last paragraph.. Gong Orbtal.. We have the potental We shall now spend a bt of tme revewng gravty. Ths s a frequent topc of Olympad questons, and s another area n whch you should be able to do well wth your Alevel knowledge. Gravtaton causes all objects to attract all other objects. To smplfy matters, we start wth two small compact masses. The sze of the force of attracton s best descrbed by the equaton F r GMm () R Here G s the Gravtatonal constant ( Nm /kg ), M s the mass of one object (at the orgn of coordnates), and m s the mass of the other. The equaton gves the force experenced by the mass m. Notce the r subscrpt and the mnus sgn the force s radal, and drected nwards toward the orgn (where the mass M s). It s possble to work out how much work s needed to get the mass m as far away from M as possble. We use ntegraton R R R GMm GMm GMm F dx Fr dr dr r r. R Notce the use of F r n the second stage. In order to separate the masses we use a force F whch acts n opposton to the gravtatonal attracton F r. The equaton gves the amount of work done by ths force as t pulls the masses apart. We usually defne the zero of potental energy to be when the masses have nothng to do wth each other (because they are so far away). Accordngly, the potental energy of the masses m and M s gven by R GMm E( R). () R Page 4
15 Correctons March 7 That s, GMm R joules below zero energy. Notce that de( R) F r ( R). (3) dr Ths s a consequence of the defnton of work as W F dx, and s generally true. It s useful because t tells us that a forces always pont n the drecton of decreasng energy. The potental energy depends on the mass of both objects as well as the poston. The gravtatonal potental V(R) s defned as the energy per unt mass of the second object, and s gven by V R E( R, m) GM lm. (4) m R m Accordngly, the potental s a functon only of poston. The zero lmt on the mass m s needed (n theory) to prevent the small mass dsturbng the feld. In practce ths wll not happen f the masses are fxed n poston. To see the consequences of breakng ths rule, thnk about measurng the Earth s gravtatonal feld close to the Moon. If we do ths by measurng the force experenced by a kg mass, we wll be fne. If we do t by measurng the force experenced by a 8 kg planet put n place for the job, we wll radcally change the moton of Earth and Moon, and thus affect the measurement. In a smlar way, we evaluate the gravtatonal feld strength as the force per klogram of mass. Wrtng the feld strength as g gves MG g (5) R and equaton (3) may be rewrtten n terms of feld and potental as dv g( R). (6) dr.. Orbtal trcks There s a useful shortcut when dong problems about orbts. Suppose that an object of mass m s orbtng the centre of coordnates, and n experences an attractve force Fr Ar, where A s some constant. Therefore n= for gravty, and we would have n=+ for moton of a partcle attached to a sprng (the other end fxed at the orgn). Page 5
16 Correctons March 7 If the object s performng crcular orbts, the centrpetal acceleraton wll be u R where R s the radus of the orbt. Ths s provded by the attractve force mentoned, and so: mu R mu F r AR n AR n (7) Now the potental energy E(R) s such that de dr F r AR n, so E( R) n AR n (8) f we take the usual conventon that E(R) s zero when the force s zero. Combnng equatons (7) and (8) gves mu n E( R) (9) so that Knetc Energy = Potental Energy (n+). () Ths tells us that for crcular gravtatonal orbts (where n=), the potental energy s twce as large as the knetc energy, and s negatve. For ellptcal orbts, the equaton stll holds: but now n terms of the average 3 knetc and potental energes. Equaton () wll not hold nstantaneously at all tmes for noncrcular orbts...3 Kepler s Laws The moton of the planets n the Solar system was observed extensvely and accurately durng the Renassance, and Kepler formulated three laws to descrbe what the astronomers saw. For the Olympad, you won t need to be able to derve these laws from the equatons of gravty, but you wll need to know them, and use them (wthout proof).. All planets orbt the Sun n ellptcal orbts, wth the Sun at one focus. 3 By average, we refer to the mean energy n tme. In other words, f T s the orbtal perod, the average of A s gven by T T A t ) ( dt. Page 6
17 Correctons March 7. The area traced out by the radus of an orbt n one second s the same for a planet, whatever stage of ts orbt t s n. Ths s another way of sayng that ts angular momentum s constant, and we shall be lookng at ths n Chapter The tme perod of the orbt s related to the [tme mean] average radus of the orbt: T R 3. It s not too dffcult to show that ths s true for crcular orbts, but t s also true for ellptc ones...4 Large Masses In our work so far, we have assumed that all masses are very small n comparson to the dstances between them. However, ths s not always the case, as you wll often be workng wth planets, and they are large! However there are two useful facts about large spheres and sphercal shells. A sphercal shell s a shape, lke the skn of a balloon, whch s bounded by two concentrc spheres of dfferent radus.. The gravtatonal feld experenced at a pont outsde a sphere or sphercal shell s the same as f all the mass of the shape were concentrated at ts centre.. A sphercal shell has no gravtatonal effect on an object nsde t. These rules only hold f the sphere or shell s of unform densty (strctly f the densty has sphercal symmetry). Therefore let us work out the gravtatonal force experenced by a mner down a very very very deep hole, who s half way to the centre of the Earth. We can gnore the mass above hm, and therefore only count the bt below hm. Ths s half the radus of the Earth, and therefore has one eghth of ts mass (assumng the Earth has unform densty whch t doesn t). Therefore the M n equaton () has been reduced by a factor of eght. Also the mner s twce as close to the centre (R has halved), and therefore by the nversesquare law, we would expect each klogram of Earth to attract hm four tmes as strongly. Combnng the factors of /8 and 4, we arrve at the concluson that he experences a gravtatonal feld ½ that at the Earth s surface, that s 4.9 N/kg..3 Fluds when thngs get stcky Questons about fluds are really classcal mechancs questons. You can tackle them wthout any detaled knowledge of flud mechancs. There are a few ponts you need to remember or learn, and that s what ths secton contans. Perfect gases are also fluds, but we wll deal wth them n chapter 5 Hot Physcs..3. Floatng and... the opposte The most mportant thng to remember s Archmedes Prncple, whch states that: Page 7
18 Correctons March 7 When an object s mmersed n a flud (lqud or gas), t wll experence an upwards force equal to the weght of flud dsplaced. By weght of flud dsplaced we mean the weght of the flud that would have been there f the object was not n poston. Ths upward force (sometmes called the buoyant upthrust) wll be equal to Force = Weght of flud dsplaced = g Mass of flud dsplaced = g Densty of flud Volume of flud dsplaced () For an object that s completely submerged, the volume of flud dsplaced s the volume of the object. For an object that s only partly submerged (lke an ceberg or shp), the volume of flud dsplaced s the volume of the object below the waterlne. Ths allows us to fnd out what wll float, and what wll snk. If an object s completely submerged, t wll have two forces actng on t. Its weght, whch pulls downwards, and the buoyant upthrust, whch pulls upwards. Volume V Mass M Upthrust = V g Object floats f: V > M > M/V Flud Densty Weght = M g Therefore, thngs float f ther overall densty (total mass / total volume) s less than the densty of the flud. Notce that the overall densty may not be equal to the actual densty of the materal. To gve an example a shp s made of metal, but contans ar, and s therefore able to float because ts overall densty s reduced by the ar, and s therefore lower than the densty of water. Puncture the hull, and the ar s no longer held n place. Therefore the densty of the shp = the densty of the steel, and the shp snks. For an object that s floatng on the surface of a flud (lke a shp on the ocean), the upthrust and weght must be equal otherwse t would rse Page 8
19 Correctons March 7 or fall. From Archmedes prncple, the weght of water dsplaced must equal the total weght of the object. There s a branteaser queston lke ths: A boat s floatng n the mddle of a lake, and the amount of water n the lake s fxed. The boat s carryng a large rock. The rock s lfted out of the boat, and dropped nto the lake. Wll the level of water n the lake go up or down? Answer: Level goes down whle the rock was n the boat (and therefore floatng) ts weght of water was beng dsplaced. When t was dropped nto the depths, ts volume of water was dsplaced. Now the densty of rock s hgher than that of water, so the water level n the lake was hgher n the frst case..3. Under Pressure What s the pressure n a flud? Ths must depend on how deep you are, because the deeper you are, the greater weght of flud you are supportng. We can thnk of the pressure (=Force/Area) as the weght of a square prsm of flud above a horzontal square metre marked out n the depths. Pressure = Weght of flud over m square = g Densty Volume of flud over m = g Densty Depth Cross sectonal area of flud (m ) Pressure = g Densty Depth () Of course, ths equaton assumes that there s nothng pushng down on the surface of the lqud! If there s, then ths must be added n too. Therefore pressure m under the surface of the sea = atmospherc pressure + weght of a m hgh column of water. It s wse to take a bt of cauton, though, snce pressures are often gven relatve to atmospherc pressure (.e. MPa more than atmospherc) and you need to keep your wts about you to spot whether pressures are relatve (vacuum =  kpa) or absolute (vacuum = Pa)..3.3 Contnuty Contnuty means conservatsm! Some thngs just don t change lke energy, momentum, and amount of stuff. Ths gves us a useful tool. Thnk about the dagram below, whch shows water n a cm [dameter] ppe beng forced nto a 5cm ppe. Page 9
20 Correctons March 7 cm 5 cm Water, lke most lquds, doesn t compress much so t can t form bottlenecks. The rate of water flow (cubc metres per second) n the bg ppe must therefore be equal to the rate of water flow n the lttle ppe. You mght lke to draw an analogy wth the current n a seres crcut. The lght bulb has greater resstance than the wre but the current n both s the same, because the one feeds the other. How can we express ths mathematcally? Let us assume that the ppe has a cross sectonal area A, and the water s gong at speed u m/s. How much water passes a pont n second? Let us put a marker n the water, whch moves along wth t. In one second t moves u metres. Therefore volume of water passng a pont = volume of cylnder of length u and cross sectonal area A = u A. Therefore Flow rate (m 3 /s) = Speed (m/s) Cross sectonal area (m ). (3) Now we can go back to our orgnal problem. The flow rate n both wde and narrow ppes must be the same. So f the larger one has twce the dameter, t has four tmes the cross sectonal area; and so ts water must be travellng four tmes more slowly..3.4 Bernoull s Equaton Somethng odd s gong on n that ppe. As the water squeezes nto the smaller radus, t speeds up. That means that ts knetc energy s ncreasng. Where s t gettng the energy from? The answer s that t can only do so f the pressure n the narrower ppe s lower than n the wder ppe. That way there s an unbalanced force on the flud n the coneshaped part speedng t up. Let s follow a cubc metre of water through the system to work out how far the pressure drops. The flud n the larger ppe pushes the flud n the cone to the rght. The force = pressure area = P L A L. A cubc metre of flud occupes length /A L n the ppe, where A L s the cross sectonal area of the ppe to the left of the constrcton. Accordngly, the work done by the flud n the wder ppe on the flud n the cone n pushng the cubc metre through s Force Dstance = P L A L /A L = P L. However ths cubc metre does work P R A R /A R = P R n gettng out the other sde. Thus the net energy gan of the cubc metre s P L P R, and ths must equal the change n the cubc metre s knetc energy u R / u L /. Page
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