Chapter 4. Conservation laws for systems of particles

Transcription

1 Chapter 4 Conservaton laws for systems of partcles In ths chapter, we shall ntroduce (but not n ths order) the followng general concepts: 1. The lnear mpulse of a force. The angular mpulse of a force 3. The power transmtted by a force 4. The work done by a force 5. The potental energy of a force. 6. The lnear momentum of a partcle (or system of partcles) 7. The angular momentum of a partcle, or system of partcles. 8. The knetc energy of a partcle, or system of partcles 9. The lnear mpulse momentum relatons for a partcle, and conservaton of lnear momentum 10. The prncple of conservaton of angular momentum for a partcle 11. The prncple of conservaton of energy for a partcle or system of partcles. We wll also llustrate how these concepts can be used n engneerng calculatons. As you wll see, to applyng these prncples to engneerng calculatons you wll need two thngs: () a thorough understandng of the prncples themselves; and () Physcal nsght nto how engneerng systems behave, so you can see how to apply the theory to practce. The frst s easy. The second s hard, and people who can do ths best make the best engneers. 4.1 Work, Power, Potental Energy and Knetc Energy relatons The concepts of work, power and energy are among the most powerful deas n the physcal scences. Ther most mportant applcaton s n the feld of thermodynamcs, whch descrbes the exchange of energy between nteractng systems. In addton, concepts of energy carry over to relatvstc systems and quantum mechancs, where the classcal versons of Newton s laws themselves no longer apply. In ths secton, we develop the basc defntons of mechancal work and energy, and show how they can be used to analyze moton of dynamcal systems. Future courses wll expand on these concepts further Defnton of the power and work done by a force Suppose that a force F acts on a partcle that moves wth speed v. k F(t) P By defnton: The Power developed by the force, (or the rate of work done by the force) s P F v. If both force and velocty are expressed n Cartesan components, then P Fxvx Fyvy Fzvz O Work has unts of Nm/s, or `Watts n SI unts.

2 The work done by the force durng a tme nterval t0 t t1 s t1 W F v dt t0 The work done by the force can also be calculated by ntegratng the force vector along the path traveled by the force, as r1 W F dr where r 0, r 1 are the ntal and fnal postons of the force. r 0 r 0 k O F(t) r 1 P Work has unts of Nm n SI unts, or `Joules A movng force can do work on a partcle, or on any movng obect. stretch a sprng, t s sad to do work on the sprng. For example, f a force acts to 4.1. Defnton of the power and work done by a concentrated moment, couple or torque. Concentrated moment, Couple and `Torque are dfferent names for a generalzed force that causes rotatonal moton wthout causng translatonal moton. These concepts are not often used to analyze moton of partcles, where rotatonal moton s gnored the only applcaton mght be to analyze rotatonal moton of a massless frame connected to one or more partcles. For completeness, however, the power-work relatons for moments are lsted n ths secton and appled to some smple problems. To do ths, we need brefly to dscuss how rotatonal moton s descrbed. Ths topc wll be developed further n Chapter 6, where we dscuss moton of rgd bodes. Defnton of an angular velocty vector Vsualze a spnnng obect, lke the cube shown n the fgure. The box rotates about an axs n the example, the axs s the lne connectng two cube dagonals. In addton, the obect turns through some number of revolutons every mnute. We would specfy the angular velocty of the shaft as a vector ω, wth the followng propertes: 1. The drecton of the vector s parallel to the axs of the shaft (the axs of rotaton). Ths drecton would be specfed by a unt vector n parallel to the shaft. There are, of course, two possble drectons for n. By conventon, we always choose a drecton such that, when vewed n a drecton parallel to n (so the vector ponts away from you) the shaft appears to rotate clockwse. Or conversely, f n ponts towards you, the shaft appears to rotate counterclockwse. (Ths s the `rght hand screw conventon ) n Axs of rotaton

3 Vewed along n Vewed n drecton opposte to n 3. The magntude of the vector s the angular speed d / dt of the obect, n radans per second. If you know the revs per mnute n turned by the shaft, the number of radans per sec follows as d / dt 10 n. The magntude of the angular velocty s often denoted by d / dt The angular velocty vector s then d ω n n. dt Snce angular velocty s a vector, t has components ω x yzk n a fxed Cartesan bass. Rate of work done by a torque or moment: If a moment M M x M y M zk acts on an obect that rotates wth angular velocty ω, the rate of work done on the obect by M s P M ω M M M x x y y z z Smple examples of power and work calculatons Example 1: An arcraft wth mass kg flyng at 00 knots (10m/s) clmbs at 1000ft/mn. Calculate the rate of work done on the arcraft by gravty. The gravtatonal force s vx vy mg, and the velocty vector of the arcraft s v. The rate of work done on the arcraft s therefore Substtutng numbers gves P F v mgv y F T mg F D Example : Calculate a formula for the work requred to stretch a sprng wth stffness k and unstretched length L 0 from length l 0 to length l 1. The fgure shows a sprng that held fxed at A and s stretched n the horzontal drecton by a force F F actng at B. At some nstant the x sprng has length l 0 x l 1. The sprng force law states that the force actng on the sprng at B s related to the length of the sprng x by F Fx k( x L0 ) The poston vector of the force s r x, and therefore the work done s r1 l1 k W F dr k( x L0 ) dx ( l1 L0 ) ( l0 L0 ) r l 0 0 A x F B

4 Example 3: Calculate the work done by gravty on a satellte that s launched from the surface of the earth to an alttude of 50km (a typcal low earth orbt). Assumptons 1. The earth s radus s km. The mass of a typcal satellte s 4135kg - see, e.g The Gravtatonal parameter GM km s (G= gravtatonal constant; M=mass of earth) 4. We wll assume that the satellte s launched along a straght lne path parallel to the drecton, startng the earths surface and extendng to the alttude of the orbt. It turns out that the work done s ndependent of the path, but ths s not obvous wthout more elaborate and sophstcated calculatons. Calculaton: 1. The gravtatonal force on the satellte s. The work done follows as GMm F x Rh GMm 1 1 F dx GMm R h R R x 6 3. Substtutng numbers gves J (be careful wth unts f you work wth klometers the work done s n N-km nstead of SI unts Nm) Example 4: A Ferrar Testarossa skds to a stop over a dstance of 50ft. Calculate the total work done on the car by the frcton forces actng on ts wheels. Assumptons: 1. A Ferrar Testarossa has mass 1506kg (see Testarossa.html). The coeffcent of frcton between wheels and road s of order 0.8 T x F N F 3. We assume the brakes are locked so all wheels skd, and ar resstance s neglected Calculaton The fgure shows a free body dagram. The equaton of moton for the car s ( T T ) ( N N mg) ma F R R F x N N mg 1. The vertcal component of the equaton of moton yelds R F. The frcton law shows that TR NR TF NF TR TF NR NF mg 3. The poston vectors of the car s front and rear wheels are rf x rr ( x L). The work done follows as. We suppose that the rear wheel starts at some pont x0 when the brakes are appled and skds a total dstance d. x0 d x0 d L x0 d x0 d L W FR drr FF drf TR ( dx) TF ( dx) ( TR TF ) d x x L x x L The work done follows as W mg. Substtutng numbers gves W 9 10 J. h 5 R L mg T R N R

5 Example 5: The fgure shows a box that s pushed up a slope by a force P. The box moves wth speed v. Fnd a formula for the rate of work done by each of the forces actng on the box. The fgure shows a free body dagram. The force vectors are 1. Appled force Pcos Psn. Frcton T cos T sn 3. Normal reacton N sn T cos 4. Weght mg The velocty vector s vcos vsn Evaluatng the dot products F v for each formula, and recallng that 1. Appled force Pv. Frcton Tv 3. Normal reacton 0 4. Weght mg sn cos sn 1 gves P P mg N T Example 6: The table lsts the expermentally measured force-v-draw data for a longbow. Calculate the total work done to draw the bow. In ths case we don t have a functon that specfes the force as a functon of poston; nstead, we have a table of numercal values. We have to approxmate the ntegral r1 W F dr r numercally. To understand how to do ths, remember that ntegratng a functon can be vsualzed as computng the area under a curve of the functon, as llustrated n the fgure. We can estmate the ntegral by dvdng the area nto a seres of trapezods, as shown. Recall that the area of a trapezod s (base x average heght), so the total area of the functon s f1 f f f3 f3 f4 f4 f5 W x1 x x3 x You could easly do ths calculaton by hand but for lazy people lke me MATLAB has a convenent functon called `trapz that does ths calculaton automatcally. Here s how to use t >> draw = [0,10,0,30,40,50,60]*0.01; >> force = [0,40,90,140,180,0,70]; >> trapz(draw,force) ans = >> So the soluton s 80.5J F Force N Draw (cm) f 1 f f 3 f 4 f 5 x x 1 x x 3 x 4

6 4.1.4 Defnton of the potental energy of a conservatve force F(t) Preamble: Textbooks nearly always defne the `potental energy of a P force. Strctly speakng, we cannot defne a potental energy of a sngle r 0 k force nstead, we need to defne the potental energy of a par of forces. A force can t exst by tself there must always be an equal and opposte reacton force actng on a second body. In all of the dscusson to be O r presented n ths secton, we mplctly assume that the reacton force s 1 actng on a second body, whch s fxed at the orgn. Ths smplfes calculatons, and makes the dscusson presented here look lke those gven n textbooks, but you should remember that the potental energy of a force par s always a functon of the relatve postons of the two forces. Wth that provso, consder a force F actng on a partcle at some poston r n space. Recall that the work done by a force that moves from poston vector r 0 to poston vector r 1 s r1 W F dr r In general, the work done by the force depends on the path between r 0 to r 1. For some specal forces, however, the work done s ndependent of the path. Such forces are sad to be conservatve. 0 For a force to be conservatve: The force must be a functon only of ts poston.e. t can t depend on the velocty of the force, for example. The force vector must satsfy curl( F) 0 Examples of conservatve forces nclude gravty, electrostatc forces, and the forces exerted by a sprng. Examples of non-conservatve (or should that be lberal?) forces nclude frcton, ar resstance, and aerodynamc lft forces. The potental energy of a conservatve force s defned as the negatve of the work done by the force n movng from some arbtrary ntal poston r 0 to a new poston r,.e. r V ( r) F dr constant r0 The constant s arbtrary, and the negatve sgn s ntroduced by conventon (t makes sure that systems try to mnmze ther potental energy). If there s a pont where the force s zero, t s usual to put r 0 at ths pont, and take the constant to be zero. Note that 1. The potental energy s a scalar valued functon. The potental energy s a functon only of the poston of the force. If we choose to descrbe poston n terms of Cartesan components r x y zk, then V ( r ) V ( x, y, z). 3. The relatonshp between potental energy and force can also be expressed n dfferental form (whch s often more useful for actual calculatons) as F grad( V ) If we choose to work wth Cartesan components, then

7 V V V Fx Fy Fzk k x y z Occasonally, you mght have to calculate a potental energy functon by ntegratng forces for example, f you are nterested n runnng a molecular dynamc smulaton of a collecton of atoms n a materal, you wll need to descrbe the nteratomc forces n some convenent way. The nteratomc forces can be estmated by dong quantum-mechancal calculatons, and the results can be approxmated by a sutable potental energy functon. Here are a few examples showng how you can ntegrate forces to calculate potental energy Example 1: Potental energy of forces exerted by a sprng. A free body dagram showng the forces exerted by a sprng connectng two obects s shown n the fgure. 1. The force exerted by a sprng s F k( x L0 ). The poston vector of the force s r x 3. The potental energy follows as r L 1 V ( r) F dr constant = k( x L0 ) dx k( L L0 ) r L 0 0 where we have taken the constant to be zero. x F Example : Potental energy of electrostatc forces exerted by charged partcles. The fgure shows two charged partcles a dstance x apart. To calculate the potental energy of the force actng on partcle, we +Q +Q place partcle 1 at the orgn, and note that the force actng on 1 F partcle s 1 Q1Q F x 4 x where Q1 and Q are the charges on the two partcles, and s a fundamental physcal constant known as the Permttvty of the medum surroundng the partcles. Snce the force s zero when the partcles are nfntely far apart, we take r 0 at nfnty. The potental energy follows as x Q1Q Q1Q V ( r) dx 4 x 4 x

8 Table of potental energy relatons In practce, however, we rarely need to do the ntegrals to calculate the potental energy of a force, because there are very few dfferent knds of force. For most engneerng calculatons the potental energy formulas lsted n the table below are suffcent. Type of force Force vector Potental energy Gravty actng on a partcle near earths surface F mg V mgy m F y Gravtatonal force exerted on mass m by mass M at the orgn GMm F r 3 r GMm V r r F r m Force exerted by a sprng wth stffness k and unstretched length L 0 r k( r L0 ) r F V k r L 1 F 0 r Force actng between two charged partcles Force exerted by one molecule of a noble gas (e.g. He, Ar, etc) on another (Lennard Jones potental). a s the equlbrum spacng between molecules, and E s the energy of the bond. Q1Q Q1Q F r V 3 4r 4r 13 7 E a a F 1 a r r 1 6 a a E r r F +Q 1 +Q r 1 F 1 r

9 4.1.5 Potental energy of concentrated moments exerted by a torsonal sprng A potental energy cannot usually be defned for concentrated moments, because rotatonal moton s tself path dependent (the orentaton of an obect that s gven two successve rotatons depends on the order n whch the rotatons are appled). A potental energy can, however, be defned for the moments exerted by a torsonal sprng. A sold rod s a good example of a torsonal sprng. You could take hold of the ends of the rod and twst them, causng one end to rotate relatve to the other. To do ths, you would apply a moment or a couple to each end of the rod, wth drecton parallel to the axs of the rod. The angle of twst ncreases wth the moment. Varous torson sprng desgns used n practce are shown n the pcture the mage s from More generally, a torsonal sprng ressts rotaton, by exertng equal and opposte moments on obects connected to ts ends. For a lnear sprng the moment s proportonal to the angle of rotaton appled to the sprng. The fgure shows a formal free body dagram for two obects connected by a torsonal sprng. If obect A s held fxed, and obect B s rotated through an angle about an axs parallel to a unt vector n, then the sprng exerts a moment M n on obect B where s the torsonal stffness of the sprng. Torsonal stffness has unts of Nm/radan. A B n -M The potental energy of the moments exerted by the sprng can be determned by computng the work done to twst the sprng through an angle. 1. The work done by a moment M due to twstng through a very small angle d about an axs parallel to a vector n s dw M dn. The potental energy s the negatve of the total work done by M,.e. 1 V M dn n dn d M A -M M B Defnton of the Knetc Energy of a partcle Consder a partcle wth mass m whch moves wth velocty v v v v k. By defnton, ts knetc energy s x y z 1 1 T m v m vx v y vz r 0 k O P v r 1

10 4.1.7 Power-Work-knetc energy relatons for a sngle partcle Consder a partcle wth mass m that moves under the acton of a force F. Suppose that 1. At some tme t 0 the partcle has some ntal poston r 0, velocty v 0 and knetc energy T 0. At some later tme t the partcle has a new poston r, velocty v and knetc energy T. 3. Let P F v denote the rate of work done by the force t 1 4. Let W Pdt F dr be the total work done by the force t r r 0 0 The Power-knetc energy relaton for the partcle states that the rate of work done by F s equal to the rate of change of knetc energy of the partcle,.e. dt P dt Ths s ust another way of wrtng Newton s law for the partcle: to see ths, note that we can take the dot product of both sdes of F=ma wth the partcle velocty dv d 1 F v ma v m v m ( v v ) dt dt To see the last step, do the dervatve usng the Chan rule and note that a b b a. The Work-knetc energy relaton for a partcle says that the total work done by the force F on the partcle s equal to the change n the knetc energy of the partcle. W T T 0 Ths follows by ntegratng the power-knetc energy relaton wth respect to tme Examples of smple calculatons usng work-power-knetc energy relatons There are two man applcatons of the work-power-knetc energy relatons. You can use them to calculate the dstance over whch a force must act n order to produce a gven change n velocty. You can also use them to estmate the energy requred to make a partcle move n a partcular way, or the amount of energy that can be extracted from a collecton of movng partcles (e.g. usng a wnd turbne) Example 1: Estmate the mnmum dstance requred for a 14 wheeler that travels at the RI speed-lmt to brake to a standstll. Is the dstance to stop any dfferent for a Toyota Echo? Ths problem can be solved by notng that, snce we know the ntal and fnal speed of the vehcle, we can calculate the change n knetc energy as the vehcle stops. The change n knetc energy must equal the work done by the forces actng on the vehcle whch depends on the dstance sld. Here are the detals of the calculaton. Assumptons: 1. We assume that all the wheels are locked and skd over the ground (ths wll stop the vehcle n the shortest possble dstance). The contacts are assumed to have frcton coeffcent The vehcle s dealzed as a partcle. 4. Ar resstance wll be neglected. T A T B TC T D N A N B N C N D mg

11 Calculaton: 1. The fgure shows a free body dagram.. The equaton of moton for the vehcle s T T T T N N N N mg ma A B C D A B C D x N N N N mg. The vertcal component of the equaton shows that A B C D T T T T N N N N mg 3. The frcton force follows as A B C D A B C D 4. If the vehcle skds for a dstance d, the total work done by the forces actng on the vehcle s d W ( TA TB TC TD ) dx ( TA TB TC TD ) d mgd 0 5. The work-energy relaton states that the total work done on the partcle s equal to ts change n knetc energy. When the brakes are appled the vehcle s travelng at the speed lmt, wth speed V; at the end of the skd ts speed s zero. The change n knetc energy s therefore T 0 mv /. The work-energy relaton shows that Substtutng numbers gves mgd mv / d V / g Ths smple calculaton suggests that the brakng dstance for a vehcle depends only on ts speed and the frcton coeffcent between wheels and tres. Ths s unlkely to vary much from one vehcle to another. In practce there may be more varaton between vehcles than ths estmate suggests, partly because factors lke ar resstance and aerodynamc lft forces wll nfluence the results, and also because vehcles usually don t skd durng an emergency stop (f they do, the drver loses control) the nature of the brakng system therefore also may change the predcton. Example 3: Compare the power consumpton of a Ford Excurson to that of a Chevy Cobalt durng stopstart drvng n a traffc am. Durng stop-start drvng, the vehcle must be repeatedly accelerated to some (low) velocty; and then braked to a stop. Power s expended to accelerate the vehcle; ths power s dsspated as heat n the brakes durng brakng. To calculate the energy consumpton, we must estmate the energy requred to accelerate the vehcle to ts maxmum speed, and estmate the frequency of ths event. Calculaton/Assumptons: 1. We assume that the speed n a traffc am s low enough that ar resstance can be neglected.. The energy to accelerate to speed V s mv /. 3. We assume that the vehcle accelerates and brakes wth constant acceleraton f so, ts average speed s V/. 4. If the vehcle travels a dstance d between stops, the tme between two stops s d/v. 5. The average power s therefore mv / 4d. 6. Takng V=15mph (7m/s) and d=00ft (61m) are reasonable values the power s therefore 0.03m, wth m n kg. A Ford Excurson weghs 900 lb (4170 kg), requrng 15 Watts (about that of a lght bulb) to keep movng. A Chevy Cobalt weghs 681lb (116kg) and requres only 36 Watts a very substantal energy savng. Reducng vehcle weght s the most effectve way of mprovng fuel effcency durng slow drvng, and also reduces manufacturng costs and materal requrements. Another, more costly, approach s to use a system that can recover the energy durng brakng ths s the man reason that hybrd vehcles lke the Prus have better fuel economy than conventonal vehcles.

12 Example 4: Estmate the power that can be generated by a wnd turbne. The fgure shows a wnd turbne. The turbne blades deflect the ar flowng past them: ths changes the ar speed and so exerts a force on the blades. If the blades move, the force exerted by the ar on the blades does work ths work s the power generated by the turbne. The rate of work done by the ar on the blades must equal the change n knetc energy of the ar as t flows past the blades. Consequently, we can estmate the power generated by the turbne by calculatng the change n knetc energy of the ar flowng through t. To do ths properly needs a very sophstcated analyss of the ar flow around the turbne. However, we can get a rather crude estmate of the power by assumng that the turbne s able to extract all the energy from the ar that flows through the crcular area swept by the blades. Calculaton: Let V denote the wnd speed, and let denote the densty of the ar. 1. In a tme t, a cylndrcal regon of ar wth radus R and heght Vt passes through the fan.. The cylndrcal regon has mass R Vt 1 3. The knetc energy of the cylndrcal regon of ar s 4. The rate of flow of knetc energy through the fan s therefore T R Vt V dt 3 R V dt 5. If all ths energy could be used to do work on the fan blades the power generated would be dt 3 P R V dt Representatve numbers are () Ar densty 1. kg / m ; () ar speed 5mph (11 m/sec); () Radus 30m Ths gves 1.8MW. For comparson, a nuclear power plant generates about MW. A more sophstcated calculaton (whch wll be covered n EN810) shows that n practce the maxmum possble amount of energy that can be extracted from the ar s about 60% of ths estmate. On average, a typcal household uses about a kw of energy; so a sngle turbne could provde enough power for about 5-10 houses Energy relatons for a conservatve system of partcles. 3 The fgure shows a `system of partcles ths s ust a collecton of obects that we mght be nterested n, whch can be dealzed as partcles. Each partcle n the system can experence forces appled by: Other partcles n the system (e.g. due to gravty, electrc charges on the partcles, or because the partcles are physcally connected through sprngs, or because the partcles collde). We call these nternal forces actng n the system. We wll denote the nternal force exerted by the th partcle on the th partcle by R. Note that, because every acton has an equal and opposte reacton, the force exerted on the th partcle by m 1 ext F m 4 1 R 1 R 13 R 31 R 1 m ext F R 3 R 3 F 3 ext m 3

13 the th partcle must be equal and opposte, to R,.e. R R. Forces exerted on the partcles by the outsde world (e.g. by externally appled gravtatonal or electromagnetc felds, or because the partcles are connected to the outsde world through mechancal lnkages or sprngs). We call these external forces actng on the system, and we wll denote the external ext force on the th partcle by F ( t) We defne the total external work done on the system durng a tme nterval t0 t t0 t as the sum of the work done by the external forces. t0 t ext W ( t) v ( t) dt ext F forces t0 The total work done can also nclude a contrbuton from external moments actng on the system. The system of partcles s conservatve f all the nternal forces n the system are conservatve. Ths means that the partcles must nteract through conservatve forces such as gravty, sprngs, electrostatc forces, and so on. The partcles can also be connected by rgd lnks, or touch one another, but contacts between partcles must be frctonless. If ths s the case, we can defne the total potental energy of the system as the sum of potental energes of all the nternal forces. We also defne the total knetc energy partcles. total T of the system as the sum of knetc energes of all the The work-energy relaton for the system of partcles can then be stated as follows. Suppose that total 1. At some tme t 0 the system has and knetc energy T 0 total. At some later tme t the system has knetc energy T. 3. Let V0 total denote the potental energy of the force at tme t 0 total 4. Let V denote the potental energy of the force at tme t 5. Let Wext denote the total work done on the system between t 0 t t 0 t Work Energy Relaton: Ths law states that the external work done on the system s equal to the change n total knetc and potental energy of the system. W T V T V ext 0 0 We won t attempt to prove ths result - the proof s conceptually very straghtforward: t smply nvolves summng the work-energy relaton for all the partcles n the system; and we ve already seen that the work-energy relatons are smply a dfferent way of wrtng Newton s laws. But when wrtten out the sums make the proof look scary and dffcult to follow so we ll spare you the gory detals. If you are nterested, ask us (or better stll see f you can do t for yourself!) Energy conservaton law For the specal case where no external forces act on the system, the total energy of the system s constant Wext 0 T V T0 V0

14 It s worth makng one fnal remark before we turn to applcatons of these law. We often nvoke the prncple of conservaton of energy when analyzng the moton of an obect that s subected to the earth s gravtatonal feld. For example, the frst problem we solve n the next secton nvolves the moton of a proectle launched from the earth s surface. We usually glbly say that `the sum of the potental and knetc energes of the partcle are constant and f you ve done physcs courses you ve probably used ths knd of thnkng. It s not really correct, although t leads to a more or less correct soluton. Properly, we should consder the earth and the proectle together as a conservatve system. Ths means we must nclude the knetc energy of the earth n the calculaton, whch changes by a small, but fnte, amount due to gravtatonal nteracton wth the proectle. Fortunately, the prncple of conservaton of lnear momentum (to be covered later) can be used to show that the change n knetc energy of the earth s neglgbly small compared to that of the partcle Examples of calculatons usng knetc and potental energy n conservatve systems The knetc-potental energy relatons can be used to quckly calculate relatonshps between the velocty and poston of an obect. Several examples are provded below. Example 1: (Borng FE exam queston) A proectle wth mass m s launched from the ground wth velocty V 0 at angle. Calculate an expresson for the maxmum heght reached by the proectle. If ar resstance can be neglected, we can regard the earth and the proectle together as a conservatve system. We neglect the change n the earth s knetc energy. In addton, snce the gravtatonal force actng on the partcle s vertcal, the partcle s horzontal component of velocty must be constant. Calculaton: 1. Just after launch, the velocty of the partcle s v V0 cos V0 sn. The knetc energy of the partcle ust after launch s 0 /. Its potental energy s zero. 3. At the peak of the traectory the vertcal velocty s zero. Snce the horzontal velocty remans constant, the velocty vector at the peak of the traectory s 0 cos. The knetc energy at ths pont s therefore mv 0 cos / 4. Energy s conserved, so mv0 / mgh mv0 cos / h V0 (1 cos ) / V0 sn Example : You are asked to desgn the packagng for a senstve nstrument. The packagng wll be made from an elastc foam, whch behaves lke a sprng. The specfcatons restrct the maxmum acceleraton of the nstrument to 15g. Estmate the thckness of the packagng that you must use. Ths problem can be solved by notng that () the max acceleraton occurs when d the packagng (sprng) s fully compressed and so exerts the maxmum force on the nstrument; () The velocty of the nstrument must be zero at ths nstant, (because the heght s a mnmum, and the velocty s the dervatve of the heght); and () The system s conservatve, and has zero knetc energy when the package s dropped, and zero knetc energy when the sprng s fully compressed. V 0 h

15 Assumptons: 1. The package s dropped from a heght of 1.5m. The effects of ar resstance durng the fall are neglected 3. The foam s dealzed as a lnear sprng, whch can be fully compressed. Calculatons: Let h denote the drop heght; let d denote the foam thckness. 1. The potental energy of the system ust before the package s dropped s mgh. The potental energy of the system at the nstant when the foam s 1 compressed to ts maxmum extent s kd 3. The total energy of the system s constant, so 1 kd mgh 4. The fgure shows a free body dagram for the nstrument at the nstant of maxmum foam compresson. The resultant force actng on the nstrument s ( kd mg), so ts acceleraton follows as a ( kd / m g). The acceleraton must not exceed 15g, so kd / m 16g 5. Dvdng (3) by (4) shows that d h / 8 The thckness of the protectve foam must therefore exceed 18.8cm. d mg kd h Example 3: The Charpy Impact Test s a way to measure the work of fracture of a materal (.e. the work per unt area requred to separate a materal nto two peces). An example (from s shown n the pcture. You can see one n Prnce Lab f you are curous. It conssts of a pendulum, whch swngs down from a prescrbed ntal angle to strke a specmen. The pendulum fractures the specmen, and then contnues to swng to a new, smaller angle on the other sde of the vertcal. The scale on the pendulum allows the ntal and fnal angles to be measured. The goal of ths example s to deduce a relatonshp between the angles and the work of fracture of the specmen. The fgure shows the pendulum before and after t hts the specmen. 1. The potental energy of the mass before t s released s V1 mgl cos1. Its knetc energy s zero.. The potental energy of the mass when t comes to rest after strkng the specmen s V mgl cos. The knetc energy s agan zero. The work of fracture s equal to the change n potental energy - W V V mgl cos cos F 1 1 l m m

16 Example 4: Estmate the maxmum dstance that a long-bow can fre an arrow. We can do ths calculaton by dealzng the bow as a sprng, and estmatng the maxmum force that a person could apply to draw the bow. The energy stored n the bow can then be estmated, and energy conservaton can be used to estmate the resultng velocty of the arrow. Assumptons 1. The long-bow wll be dealzed as a lnear sprng. The maxmum draw force s lkely to be around 60lbf (70N) 3. The draw length s about ft (0.6m) 4. Arrows come wth varous masses typcal range s between grans (16-38 grams) 5. We wll neglect the mass of the bow (ths s not a very realstc assumpton) Calculaton: The calculaton needs two steps: () we start by calculatng the velocty of the arrow ust after t s fred. Ths wll be done usng the energy conservaton law; and () we then calculate the dstance traveled by the arrow usng the proectle traectory equatons derved n the precedng chapter. 1. Just before the arrow s released, the sprng s stretched to ts maxmum length, and the arrow s 1 statonary. The total energy of the system s T V kl, where L s the draw length and k s 0 0 the stffness of the bow.. We can estmate values for the sprng stffness usng the draw force: we have that F D 1 kl, so k FD / L. Thus T0 V0 LF D. 3. Just after the arrow s fred, the sprng returns to ts un-stretched length, and the arrow has 1 velocty V. The total energy of the system s T1 V1 mv, where m s the mass of the arrow The system s conservatve, therefore T0 V0 T1 V1 LFD mv V LFD / m 5. We suppose that the arrow s launched from the orgn at an angle to the horzontal. The horzontal and vertcal components of velocty are Vx V cos Vy V sn. The poston vector of the arrow can be calculated usng the method outlned n Secton 3.. the result s 1 r Vt cos Vt sn gt We can calculate the dstance traveled by notng that ts poston vector when t lands s d. Ths gves 1 Vt cos Vt sn gt d where t s the tme of flght. The and components of ths equaton can be solved for t and d, wth the result V sn V sn t d g g cos The arrow travels furthest when fred at an angle that maxmzes sn / cos -.e. 45 degrees. The dstance follows as V LF d D g mg

17 6. Substtutng numbers gves 064m for a 50 gran arrow over a mle! Of course ar resstance wll reduce ths value, and n practce the knetc energy assocated wth the moton of the bow and bowstrng (neglected here) wll reduce the dstance. Example 5: Fnd a formula for the escape velocty of a space vehcle as a functon of alttude above the earths surface. The term Escape velocty means that the space vehcle has a large enough velocty to completely escape the earth s gravtatonal feld.e. the space vehcle wll never stop after beng launched. Assumptons 1. The space vehcle s ntally n orbt at an alttude h above the earth s surface. The earth s radus s km 3. Whle n orbt, a rocket s burned on the vehcle to ncrease ts speed to v (the escape velocty), placng t on a hyperbolc traectory that wll eventually escape the earth s gravtatonal feld The Gravtatonal parameter GM km s (G= gravtatonal constant; M=mass of earth) 5. Calculaton 1. Just after the rocket s burned, the potental energy of the system s V GMm / ( R h), whle ts knetc energy s T mv /. When t escapes the earth s gravtatonal feld (at an nfnte heght above the earth s surface) the potental energy s zero. At the crtcal escape velocty, the velocty of the spacecraft at ths pont drops to zero. The total energy at escape s therefore zero. 3. Ths s a conservatve system, so T V mv / GMm / ( R h ) 0 v GM / ( R h) 4. A typcal low earth orbt has alttude of 50km. For ths alttude the escape velocty s 10.9km/sec. h R

18 Force, Torque and Power curves for actuators and motors Concepts of power and work are partcularly useful to calculate the behavor of a system that s drven by a motor. Calculatons lke ths are descrbed n more detal n Secton 4.1. In ths secton, we dscuss the relatonshps between force, torque, speed and power for motors. There are many dfferent types of motor, and each type has ts own unque characterstcs. They all have the followng features n common, however: 1. The motors convert some non-mechancal form of energy nto mechancal work. For example, an electrc motor converts electrcal energy; an nternal combuston engne, your muscles, and other molecular motors convert chemcal energy, and so on. You may be nterested n an extensve dscusson of muscles at A lnear motor or lnear actuator apples a force to an obect (or more accurately, t apples roughly equal and opposte forces on two obects connected to ts ends). Your muscles are good examples of lnear actuators. You can buy electrcal or hydraulcally powered actuators as well. L Force Power Electrcal Electrcal Actuator F A Muscle Muscle - - F B F A F B Contracton rate _ dl/dt Contracton rate _ dl/dt 3. The forces appled by an actuator depend on () the amount of power suppled to the actuator electrcal, chemcal, etc; and () the rate of contracton, or stretchng, of the actuator. The typcal varaton of force wth stretch rate s shown n the fgure. The fgure assumes that the actuator pulls on the obects attached to ts ends (lke a muscle); but some actuators wll push rather than pull; and some can do both. If the actuator extends slowly and s not rotatng t behaves lke a two-force member, and the forces exerted by ts two ends are equal and opposte. motor -M A - M A M B M B Moment Electrc motor Internal Combuston engne Rotaton rate Power Electrcal Internal Combuston engne Rotaton rate 4. A rotary motor or rotary actuator apples a moment or torque to an obect (or, agan, more precsely t apples equal and opposte moments to two obects). Electrc motors, nternal combuston engnes, bacteral motors are good examples of rotary motors. 5. The torques exerted by a rotary actuator depend on () the amount of power suppled to the actuator electrcal, chemcal, etc; and () the relatve velocty of the two obects on whch the forces act. The typcal varaton of force wth external power appled and wth relatve velocty s shown n the fgure. An electrc motor apples a large torque when t s statonary, and ts torque

19 decreases wth rotatonal speed. An nternal combuston engne apples no torque when t s statonary. It s torque s greatest at some ntermedate speed, and decreases at hgh speed. The power curve descrbes the varaton of the rate of work done by an actuator or motor wth ts extenson or contracton rate or rotatonal speed. Note that 1. The rate of work done (or power expended) by a lnear actuator that apples forces F A, FB to the obects attached to ts two ends can be calculated as P FA v A FB v B, where v A and vb are the veloctes of the two ends.. The rate of work (or power expended) by a rotatonal motor that apples moments M A, MB to the obects attached to ts two ends s P M A ω A MB ω B. Typcal power curves for motors and actuators are sketched n the fgures. The effcency of a motor or actuator s the rato of the rate of work done by the forces or moments exerted by the motor to the electrcal or chemcal power. The effcency s always less than 1 because some fracton of the power suppled to the motor s dsspated as heat. An electrc motor has a hgh effcency; heat engnes such as an nternal combuston engne have a much lower effcency, because they operate by rasng the temperature of the ar nsde the cylnders to ncrease ts pressure. The heat requred to ncrease the temperature can never be completely converted nto useful work (EN7 wll dscuss the reasons for ths n more detal). The effcency of a motor always vares wth ts speed there s a specal operatng speed that maxmzes ts effcency. For an nternal combuston engne, the speed correspondng to maxmum effcency s usually qute low 1500rpm or so. So, to mnmze fuel consumpton durng your drvng, you need to operate the engne at ths speed for as much of your drve as possble. There s not enough tme n ths course to be able to dscuss the characterstcs of actuators and motors n great detal. Instead, we focus on one specfc example, whch helps to llustrate the general characterstcs n more detal. Torque Curves for a brushed electrc motor. There are several dfferent types of electrc motor here, we wll focus on the smplest and cheapest the so called `Brushed DC electrc motor. The term `Brushed refers to the fact that the motor s drven by a col of wre wrapped around ts rotatng shaft, and `brushes are used to make an electrcal contact between the power supply and the rotatng shaft. `DC means that the motor s drven by drect, rather than alternatng current. Here s a very bref summary of the basc prncples of ths type of motor. The underlyng theory wll be dscussed n more detal n EN510 An electrc motor apples a moment, or torque to an obect that s coupled to ts output shaft. The moment s developed motor by electromagnetc forces actng between permanent magnets n the housng and the electrc current flowng - M B through the wndng of the motor. In the followng dscusson, we shall assume that the body of the motor s statonary ω 0, and ts output shaft rotates wth angular A velocty ωb n where n s a unt vector parallel to the shaft and s ts angular speed. In addton, we assume that the output shaft exerts a moment MB Tn. n M B

20 Power to drve the motor s suppled by connectng an electrcal power source (e.g. a battery) to ts termnals. The power supply generally apples a fxed voltage to the termnals, whch then causes current to flow through the wndng. The current s proportonal to the voltage, and decreases n proporton to the angular (rotatonal) speed of the output shaft of the motor. The moment appled by the output shaft s proportonal to ths electrc current. The electrc current I flowng through the wndng s related to the voltage V and angular speed of the motor (n radans per second) by I ( V ) / R where R s the electrcal resstance of the wndng, and s a constant that depends on the arrangement and type of magnets used n the motor, as well as the geometry of the wre col. The magntude of moment exerted by output shaft of the motor T s related to the electrc current and the speed of the motor by I T0 0 I T0 / T 0 I T0 / where T 0 and 0 are constants that account for losses such as frcton n the bearngs, eddy currents, and ar resstance. The torque-current and the current-voltage-speed relatons can be combned nto a sngle formula relatng torque to voltage and motor speed V T0 0 V / R T0 T R R V / R T0 0 Ths relatonshp s sketched n the fgure ths s called the torque curve for the motor. The two most mportant ponts on Moment the curve are Stall torque T s 1. The `stall torque Ts ( V / R) T0. The `No load speed nl ( V T0 R) / ( 0R ) No load speed nl The torque curve can be expressed n terms of these quanttes as T Ts 1 nl Rotaton rate Motor manufacturers generally provde values of the stall torque and no load speed for a motor. Some manufacturers provde enough data so that you can usually calculate the values of R,, T0 and 0 for ther product. For example, a very detaled set of motor specs are shown n the table below. The table s from Each row of the table refers to a dfferent motor.

21 Here 1. The `nput voltage s the recommended voltage V to apply to the motor. All other data n the table assume that the motor s used wth ths voltage. If you wsh, you can run a motor wth a lower voltage (ths wll make t run more slowly) and f you are feelng brave you can ncrease the voltage slghtly but ths mght cause t to burn out, nvaldate your warranty and expose you to a potental lawsut from your customers. The `No load speed nl s the speed of the motor when t s spnnng freely wth T=0 3. The `No load current Inl s the correspondng current 4. The `Stall torque Ts s the value of T when the motor s not spnnng 0 5. The Stall current Is s the correspondng current 6. The `rated torque and `rated current are lmts to the steady runnng of the motor the torque and current should not exceed these values for an extended perod of tme. The values n the table can be substtuted nto the formulas relatng current, voltage, torque and motor speed, whch yeld four equatons for the unknown values of R,, T0 and 0 I ( V ) / R I V / R Solvng these equatons gves nl nl s 0 Inl T0 0nl Ts Is T0 0nl V ( I ) / s I R V I nl s Isnl V ( Is Inl ) Ts V ( Is Inl ) T0 Ts 0 nl nl Isnl HEALTH WARNING: When you use these formulas t s crtcal to use a consstent set of unts. SI unts are very strongly recommended! Volts and Amperes are SI unts of electrcal potental and electrc current. But the torques must be converted nto Nm. The converson factor s 1 oz-n s Nm. Not all motor manufacturers provde such detaled specfcatons t s more common to be gven () the stall torque; () the no load speed; and () the stall current. If ths s the case you have to assume I 0 and take 0 0. If you aren t gven the stall current you have to take T 0 0 as well. nl

22 Power Curves for a brushed electrc motor drven from a constant voltage power supply: The rate of work done by the Power motor s P MB ω B T. The power can be calculated n Max Power T s nl / terms of the angular speed of the motor as P Ts 1 nl The power curve s sketched n the fgure. The most No load speed nl Rotaton rate mportant features are () The motor develops no power at both zero speed and the no load speed; and () the motor develops ts maxmum power Pmax T / at speed nl /. s nl Effcency of a brushed electrc motor: The electrcal power suppled to a motor can be calculated from the current I and voltage V as P VI. The effcency can therefore be calculated as n T s (1 / nl ) T0 0 R VI V ( V ) The second expresson can be used to calculate the speed that maxmzes effcency Power transmsson n machnes A machne s a system that converts one form of moton nto another. A lever s a smple example f pont A on the lever s made to move, then pont B wll move ether faster or slower than A, dependng on the lengths of the two lever arms. Other, more practcal examples nclude 1. Your skeleton; used to help convert muscle moton nto varous more useful forms F B. The transmsson of a vehcle used to convert rotatonal F moton nduced by the engne s crank-shaft to translatonal A B moton of the vehcle 3. A gearbox converts rotatonal moton at one angular speed to rotatonal moton at another. 4. A system of pulleys. A Concepts of work and energy are extremely useful to analyze transmsson of forces and moments through a machne. The goal of these calculatons s to quckly determne relatonshps between external forces actng on the machne, rather than to analyze the moton of the system tself. To do ths, we usually make two assumptons: 1. The machne s a conservatve system. Ths s true as long as frcton n the machne can be neglected.. The knetc and potental energy of the mechansm wthn the machne can be neglected. Ths s true f ether () the machne has neglgble mass and s very stff (.e. t does not deform); or () the machne moves at steady speed or s statonary. If ths s the case, the total rate of work done by external forces actng on the machne s zero. Ths prncple can be used to relate external forces actng on the machne, wthout havng to work through the very lengthy and tedous process of computng all the nternal forces.

23 The procedure to do ths s always: 1. Derve a relatonshp between the veloctes of the ponts where external forces act ths s always a geometrc relatonshp.. Wrte down the total rate of external work done on the system 3. Use (1) and () to relate the external forces to each other. The procedure s llustrated usng a number of examples below. Example 1: As a frst example, we derve a formula relatng the forces actng on the two ends of a lever. Assume that the forces act perpendcular to the lever as shown n the fgure. F A F B Calculaton 1. Suppose that the lever rotates about the pvot at some angular rate d / dt.. The ends of the lever have veloctes A v A d A(sn cos ) vb db ( sn cos ) (see the crcular moton example n the precedng chapter f you can t remember how to show ths. 3. The force vectors can be expressed as FA FA (sn cos ) FB FB (sn cos ) 4. The reacton forces at the pvot are statonary and so do no work. 5. The total rate of work done on the lever s therefore F v F v A A B B F (sn cos ) d (sn cos ) F (sn cos ) d ( sn cos ) A A B B FAd A FB db sn cos 6. The rate of external work s zero, so FAd A FB db B Example : Here we revst an EN3 statcs problem. The rabbt uses a pulley system to rase carrot wth weght W. Calculate the force appled by the rabbt on the cable. 1. The velocty of the end of the cable held by the rabbt s dl dt, wth drecton parallel to the cable. The velocty of the carrot s dh ; ts drecton s vertcal dt 3. The total length of the cable (asde from small, constant lengths gong around the pulleys) s l h. Snce the length s constant, t follows that dl dh 0 dt dt 4. The external forces actng on the pulley system are () the weght of the carrot W, actng vertcally; () the unknown force F appled by the rabbt, actng parallel to the cable; and () the dl reacton force actng on the topmost pulley. The rate of work done by the rabbt s F, the rate dt

24 of work done by the gravtatonal force actng on the carrot s dh W dt. statonary, and so does no work. The total rate of work done s therefore dh dl W F 0 dt dt 5. Usng (3) and (4) we see that dh dl dh dh W F W F 0 dt dt dt dt F W / The reacton force s Example 3 Calculate the torque that must be appled to a screw wth ptch d n order to rase a weght W. Here, we regard the screw as a machne subected to external forces. 1. When the screw turns through one complete turn ( radans) t rases the weght by a dstance d. The work done by the torque s T ; the work done by the weght s Wd. The reacton forces actng on the thread of the screw do no work 3. The total work s zero, therefore T Wd 0 T W /. W T d Example 4: A gearbox s used for two purposes: () t can amplfy or attenuate torques, or moments; and () t can change the speed of a rotatng shaft. The characterstcs of a gearbox are specfed by ts gear rato Input specfes the rato of the rotatonal speed of the output shaft k Output shaft to that of the nput shaft r o /. shaft The sketch shows an example. A power source (a rotatonal motor) drves the nput shaft; whle the output M 0 shaft s used to drve a load. The rotatonal moton of the nput and output shafts can be descrbed by angular velocty vectors ω ωo o (note that the shafts don t necessarly have to rotate n the same drecton f they don t then one of or wll be negatve). o A quck power calculaton can be used to relate the moments M T Mo To actng on the nput and output shafts. 1. The gear rato relates the nput and output angular veloctes as ω ωo r. The rate of work done on the gearbox by the moment actng on the nput shaft s T T. The work done by the moment actng on the output shaft s To o Too To r. A reacton moment must act on the base of the gearbox, but snce the gearbox s statonary ths reacton moment does no work. 3. The rate of work done by all the external forces on the gearbox s zero, so T T r 0 T T / r o 0 M 0

25 Example 5: Fnd a formula for the engne power requred for a vehcle wth mass m to clmb a slope wth angle at speed v. Estmate values for engne power requred for varous vehcles to clmb a slopes of 3%, 4% and 5% (these are typcal standard grades for flat, rollng and mountan terran freeways, see, e.g. the Roadway Engneerng Hghway Desgn Manual for Oregon at Also, estmate (a) the torque exerted by the engne crankshaft on the transmsson; and (b) the force actng on the pstons of the engne. Assumptons: 1. The only forces actng on the vehcle are aerodynamc drag, gravty, and reacton forces at the drvng wheels.. Aerodynamc drag wll be assumed to be n the hgh Reynolds number regme, wth a drag coeffcent of order 3. The crankshaft rotates at approxmately 000rpm. 4. As a representatve examples of engnes, we wll consder () a large 6 cylnder 4 stroke desel engne, wth stroke (dstance moved by the pstons) of 150mm; and () a small 4 cylnder gasolne powered engne wth stroke 94mm (see Calculaton. The fgure shows a free body dagram for the truck. It shows () gravty; () Aerodynamc drag; () reacton forces actng on the wheels. 1. The velocty vector of the vehcle s v(cos sn ). The rate of work done by the gravtatonal and aerodynamc forces s P FD (cos sn ) v(cos sn ) mg v(cos sn ) FDv mgvsn F D T N N D C N N B A mg where 1 FD CD Av, wth CD the drag coeffcent, A the frontal area of the vehcle, and the ar densty. 3. The reacton forces actng on the wheels do no work, because they act on statonary ponts (the wheel s not movng where t touches the ground). 4. The knetc energy of the vehcle s constant, so the total rate of work done on the vehcle must be zero. The movng parts of the engne (specfcally, the pressure actng on the pstons) do work on the vehcle. Denotng the power of the engne by P E, we have that PE FDv mgvsn 0 PE FDv mgvsn 5. The torque exerted by the crankshaft can be estmated by notng that the rate of work done by the engne s related to the torque T and the rotatonal speed of the crankshaft by P T. The torque follows as T FDv mgvsn / 6. The force F on the pstons can be estmated by notng that the rate of work done by a pston s related to the speed of the pston v by P Fv. In a four-stroke engne, each pston does work E

26 5% of the tme (durng the frng stroke). If the engne has n cylnders, the total engne power s therefore PE nfv / 4. Fnally, the pston speed can be estmated by notng that the pston travels the dstance of the stroke durng half a revoluton, and so has average speed v d /. Ths shows that P nfd / 4 F 4 P / ( nd). E The table below calculates values for two examples of vehcles Vehcle Grade Weght (kg) 18 wheeler Chevy cobalt Frontal area m Drag Coeft E Speed m/s Engne speed Rad/sec Stroke m Engne power Watts Torque Nm 3% % % % % % Pston force N

27 4. Lnear mpulse-momentum relatons 4..1 Defnton of the lnear mpulse of a force In most dynamc problems, partcles are subected to forces that vary wth tme. We can wrte ths mathematcally by sayng that the force s a vector valued functon of tme F ( t). If we express the force as components n a fxed bass,, k, then F( t) F ( t) F ( t) F ( t) k x y z where each component of the force s a functon of tme. k O F(t) P The Lnear Impulse exerted by a force durng a tme nterval t0 t t0 t s defned as t0 t F ( t) dt The lnear mpulse s a vector, and can be expressed as components n a bass k x y z t0 t0 t t0 t t0 t F ( t) dt F ( t) dt F ( t) dt x x y y z z t0 t0 t0 If you know the force as a functon of tme, you can calculate ts mpulse usng smple calculus. For example: 1. For a constant force, wth vector value F 0, the mpulse s F0 t. For a harmonc force of the form F( t) F0 snt the mpulse s F cos( ( t t)) cos t / It s rather rare n practce to have to calculate the mpulse of a force from ts tme varaton. 4.. Defnton of the lnear momentum of a partcle The lnear momentum of a partcle s smply the product of ts mass and velocty p mv The lnear momentum s a vector ts drecton s parallel to the velocty of the partcle Impulse-momentum relatons for a sngle partcle Consder a partcle that s subected to a force F( t) for a tme nterval t0 t t0 t. Let denote the mpulse exerted by F on the partcle Let p mv( t0 t) mv( t0) denote the change n lnear momentum durng the tme nterval t.

28 The momentum conservaton equaton can be expressed ether n dfferental or ntegral form. 1. In dfferental form d F( t) p k dt Ths s clearly ust a dfferent way of wrtng Newton s law F=ma. P. In ntegral form p Ths s the ntegral of Newton s law of moton wth respect to tme. x O y r z The mpulse-momentum relatons for a sngle partcle are useful f you need to calculate the change n velocty of an obect that s subected to a prescrbed force Examples usng mpulse-momentum relatons for a sngle partcle Example 1: Estmate the tme that t takes for a Ferrar Testarossa travelng at the RI speed lmt to make an emergency stop. (Lke many textbook problems ths one s totally unrealstc nobody n a Ferrar s gong to travel at the speed lmt!) We can do ths calculaton usng the mpulse-momentum relaton for a sngle partcle. Assume that the car has mass m, and travels wth speed V before the brakes are appled. Let t denote the tme requred to stop. 1. Start by estmatng the force actng on the car durng an emergency stop. The fgure shows a free body dagram for the car as t brakes to a standstll. We assume that the drver brakes hard enough to lock the wheels, so that the car skds over the road. The horzontal frcton forces must oppose the sldng, as shown n the pcture. F=ma for the car follows as T T N N mg ma T F T R N R ( ) R F F R x The vertcal component of the equaton of moton shows that NF NR mg. The frcton law then shows that T T N N mg F R F R. The force actng on the car s constant, so the mpulse that brngs the car to a halt s T T t N N mg t mg ( ) R F F R 3. The lnear momentum of the car before the brakes are appled s p0 mv. The lnear momentum after the car stops s zero. Therefore, p mv. 4. The lnear mpulse-momentum relaton shows that p mgt mv t V / ( g) 5. We can take the frcton coeffcent as 0.8, and 65mph s 9m/s. We take the gravtatonal acceleraton g The tme follows as t 3.7s. Note that a testarossa can t stop any faster than a Honda Cvc, despte the prce dfference N F

29 Example : The fgure shows a pendulum that swngs wth frequency f swngs per second. Fnd an expresson for the average reacton force at the pvot of a pendulum durng one cycle of oscllaton. (Ths seems a totally academc problem and ndeed t s but the mportant pont s that certan quanttes, such as average forces, can often be computed wthout solvng the full equatons of moton. These quanttes can be extremely useful for checkng computer smulatons) Calculaton: Consder the pendulum at the extreme lmts of ts swng, at the start and end of one cycle of oscllaton. Note that 1. The velocty of the partcle s zero both at the start and end of the cycle. Its change n momentum s therefore zero, and the total mpulse exerted on the partcle must be zero.. The partcle s subected to two forces () gravty; and () the reacton force exerted on the partcle. Durng a swng of the pendulum these forces exert mpulses,, respectvely. g R 3. The mpulse exerted by the gravtatonal force s g mgt where T 1/ f s the tme for one complete swng of the pendulum. 4. Snce the total mpulse s zero, t follows that g R 0 R mg/ f R mg The MATLAB scrpt below shows how ths result can be used to check the program developed n Secton???? to model the swng of a pendulum. functon pendulum_reactons g = 9.81; l = 1; m = 1; theta = p*45/180; tme = 10; Abstol = 10^(-03)*ones(4,1); optons = odeset('reltol', ,'abstol',abstol,'events',@event); y0 = [l*sn(theta),l*cos(theta),0,0]; % Intal condtons [t,y,tevent,yevent,event] = ode45(@eom,[0,tme],y0,optons);%,optons); plot(t,y(:,1:)); % Calculate and plot the reacton forces as a functon of tme for =1:length(t) [Reacton(,1),Reacton(,)] = reactons(y(,:)); end fgure plot(t,reacton(:,1:)); % Integrate the reacton forces wth respect to tme Ix = trapz(t,reacton(:,1)) Iy = trapz(t,reacton(:,)) AverageRy = trapz(t,reacton(:,))/tme(length(t)) functon dydt = eom(t,y) % The vector y contans [x,y,vx,vy] M = eye(5); % Ths sets up a 5x5 matrx wth 1s on all the dags M(3,5) = (y(1)/m)/sqrt(y(1)^+y()^); M(4,5) = (y()/m)/sqrt(y(1)^+y()^);

30 M(5,3) = y(1); M(5,4) = y(); M(5,5) = 0; f = [y(3);y(4);0;g;-y(3)^-y(4)^]; sol = M\f; % Ths solves the matrx equaton M*[dydt,R]=f for [dydt,r] dydt = sol(1:4); % only need to return tme dervatves dy/dt end functon [Rx,Ry] = reactons(y) % Ths functon calculates the reacton forces Rx, Ry, gven the vector % y contanng [x,y,vx,vy] M = eye(5); % Ths sets up a 5x5 matrx wth 1s on all the dags M(3,5) = (y(1)/m)/sqrt(y(1)^+y()^); M(4,5) = (y()/m)/sqrt(y(1)^+y()^); M(5,3) = y(1); M(5,4) = y(); M(5,5) = 0; f = [y(3);y(4);0;g;-y(3)^-y(4)^]; sol = M\f; % Ths solves the matrx equaton M*[dydt,R]=f for [dydt,r] Rx = -sol(5)*y(1)/sqrt(y(1)^+y()^); Ry = -sol(5)*y()/sqrt(y(1)^+y()^); end functon [eventvalue,stopthecalc,eventdrecton] = event(t,y) % Functon to stop the calculaton after one swng % Note that v_x=0 at the end of the swng and crosses zero from above eventvalue = y(3); stopthecalc = 1; % Ths stops the calculaton eventdrecton = -1; end end 4..5 Impulse-momentum relaton for a system of partcles Suppose we are nterested n calculatng the moton of several partcles, sketched n the fgure. Total external mpulse on a system of partcles: Each partcle n the system can experence forces appled by: Other partcles n the system (e.g. due to gravty, electrc charges on the partcles, or because the partcles are physcally connected through sprngs, or because the partcles collde). We call these nternal forces actng n the system. We wll denote the nternal force exerted by the th partcle on the th partcle by R. Note that, because every acton has an equal and opposte reacton, the force exerted on the th partcle by the th partcle must be equal and opposte, to R,.e. R R. Forces exerted on the partcles by the outsde world (e.g. by externally appled gravtatonal or electromagnetc felds, or because the partcles are connected to the outsde world through m 1 F 1 ext R 1 m m 4 R 1 R 13 R 31 F ext R 3 R 3 F 3 ext m 3

31 mechancal lnkages or sprngs). We call these external forces actng on the system, and we wll ext denote the external force on the th partcle by F ( t) We defne the total mpulse exerted on the system durng a tme nterval t0 t t0 t as the sum of all the mpulses on all the partcles. It s easy to see that the total mpulse due to the nternal forces s zero because the th and th partcles must exert equal and opposte mpulses on one another, and when you add them up they cancel out. So the total mpulse on the system s smply tot partcles t0 t F t0 ext ( t) dt Total lnear momentum of a system of partcles: The total lnear momentum of a system of partcles s smply the sum of the momenta of all the partcles,.e. p m v tot partcles The mpulse-momentum equaton can be expressed ether n dfferental or ntegral form, ust as for a sngle partcle: 1. In dfferental form ext dptot F ( t) partcles dt Ths s clearly ust a dfferent way of wrtng Newton s law F=ma.. In ntegral form tot ptot Ths s the ntegral of Newton s law of moton wth respect to tme. Conservaton of momentum: If no external forces act on a system of partcles, ther total lnear momentum s conserved,.e. p 0 p ( t t) p ( t ) tot 0 tot Examples of applcatons of momentum conservaton for systems of partcles The mpulse-momentum equatons for systems of partcles are partcularly useful for () analyzng the recol of a gun; and () analyzng rocket and et propulson systems. In both these applcatons, the nternal forces actng between the gun on the proectle, or the motor and propellant, are much larger than any external forces, so the total momentum of the system s conserved. Example 1: Estmate the recol velocty of a rfle (youtube abounds wth recol demonstratons see. e.g. for samples. Be warned, however a lot of the vdeos are tasteless and/or sexst ) The recol velocty can be estmated by notng that the m g V m b v

32 total momentum of bullet and rfle together must be conserved. If we can estmate the mass of rfle and bullet, and the bullet s velocty, the recol velocty can be computed from the momentum conservaton equaton. Assumptons: 4 1. The mass of a typcal 0. (.e. 0. dameter) calber rfle bullet s about 7 10 kg (dealzng the bullet as a sphere, wth densty kg / m ). The muzzle velocty of a 0. s about 1000 ft/sec (305 m/s) 3. A typcal rfle weghs between 5 and 10 lb (.5-5 kg) Calculaton: 1. Let mb denote the bullet mass, and let mr denote the mass of the rfle.. The rfle and bullet are dealzed as two partcles. Before frng, both are at rest. After frng, the bullet has velocty v b v ; the rfle has velocty v R V. 3. External forces actng on the system can be neglected, so ptot ( t0 t) ptot ( t0) mb vb mr vr 0. mbv mrv 0 V mbv / mr 4. Substtutng numbers gves V between 0.04 and 0.08 m/s (about 0.14 ft/sec) Example : Derve a formula that can be used to estmate the mass of a handgun requred to keep ts recol wthn acceptable lmts. The precedng example shows that the frearm wll recol wth a velocty that depends on the rato of the mass of the bullet to the frearm. The frearm must be brought to rest by the person holdng t. Assumptons: 1. We wll dealze a person s hand holdng the gun as a sprng, wth stffness k, fxed at one end. The end-pont stffness of a human hand has been extensvely studed see, e.g. Shadmehr et al J. Neuroscence, 13 (1) 45 (1993). Typcal values of stffness durng quas-statc deflectons are of order 0. N/mm. Durng dynamc loadng stffnesses are lkely to be larger than ths.. We dealze the handgun and bullet as partcles, wth mass mb and m g, respectvely. Calculaton. 1. The precedng problem shows that the frearm wll recol wth velocty V m v / m. Energy conservaton can be used to calculate the recol dstance. We consder the frearm and the hand holdng t a system. At tme t=0 t has zero potental energy; and has knetc energy 1 1 T m gv mbv / mg. At the end of the recol, the gun s at rest, and the sprng s fully 1 compressed the knetc energy s zero, and potental energy s kd. Energy conservaton kd m v / m gves b 3. The requred mass follows as 4. Substtutng numbers gves g m m v / kd. g b b g

33 Example 4 Rocket propulson equatons. Rocket motors and et engnes explot the momentum conservaton law n order to produce moton. They work by expellng mass from a vehcle at very hgh velocty, n a drecton opposte to the moton of the vehcle. The momentum of the expelled mass must be balanced by an equal and opposte change n the momentum of the vehcle; so the velocty of the vehcle ncreases. Motor, mass m 0 Rocket, mass M Analyzng a rocket engne s qute complcated, because the propellant carred by the engne s usually a very sgnfcant fracton of the total mass of the vehcle. Consequently, t s mportant to account for the fact that the mass decreases as the propellant s used. Assumptons: 1. The fgure shows a rocket motor attached to a rocket wth mass M.. The rocket motor contans an ntal mass m0 of propellant and expels propellant at rate dm 0 (kg/sec) wth a velocty v 0 v 0 relatve to the rocket. dt 3. We assume straght lne moton, and assume that no external forces act on the rocket or motor. Calculatons: The fgure shows the rocket at an nstant of tme t, and then a very short Rocket, mass M tme nterval dt later. 1. At tme t, the rocket moves at speed v, and the system has momentum p ( M m) v, where m s the motor s mass. v. Durng the tme nterval dt a mass dm dt s expelled from Motor, mass m the motor. The velocty of the expelled mass s v ( v v0 ). v-v 0 Rocket, mass M 3. At tme t+dt the mass of the motor has decreased to m dt. dt 4. At tme t+dt, the rocket has velocty v ( v dv). Motor, mass m-dt 5. The total momentum of the system at tme t+dt s v+dv therefore p ( M m 0dt)( v dv) 0dt( v v0 ) 6. Momentum must be conserved, so ( M m) v ( M m 0dt)( v dv) 0dt( v v0 ) 7. Multplyng ths out and smplfyng shows that ( M m) dv dtv0 0 where the term 0 dtdv has been neglected. 8. Fnally, we see that ( M m) dv v0 dt Ths result s called the `rocket equaton. Specfc Impulse of a rocket motor: The performance of a rocket engne s usually specfed by ts `specfc mpulse. Confusngly, two dfferent defntons of specfc mpulse are commonly used:

34 Isp v0 v I 0 sp g The frst defnton quantfes the mpulse exerted by the motor per unt mass of propellant; the second s the mpulse per unt weght of propellant. You can usually tell whch defnton s beng used from the unts the frst defnton has unts of m/s; the second has unts of s. In terms of the specfc mpulse, the rocket equaton s dv ( M m) Isp gisp dt Integrated form of the rocket equaton: If the motor expels propellant at constant rate, the equaton of moton can be ntegrated. Assume that 1. The rocket s at the orgn at tme t=0;. The rocket has speed v 0 at tme t=0 3. The motor has mass m0 at tme t=0; ths means that at tme t t has mass m0 t Then the rocket s speed can be calculated as a functon of tme: Smlarly, the poston follows as v t dv Ispdt ( M m0 t) Isp dv dt ( M m t) v M m0 v v0 Isp log M m0 t x t dx M m0 M m log 0 v Isp v0 dx Isp log v0 dt dt M m0 t M m t Isp M m x ( v 0 0 Isp ) t M m0 t log M m0 t These calculatons assume that no external forces act on the rocket. It s qute straghtforward to generalze them to account for external forces as well the detals are left as an exercse. Example 5 Applcaton of rocket propulson equaton: Calculate the maxmum payload that can be launched to escape velocty on the Ares I launch vehcle. Escape velocty means that after the motor burns out, the space vehcle can escape the earth s gravtatonal feld see example 5 n Secton Assumptons 1. The specfcatons for the Ares I are at Relevant varables are lsted n the table below. Total mass (kg) Specfc mpulse (s) Propellant mass (kg) Stage I Stage II

35 . As an approxmaton, we wll neglect the moton of the rocket durng the burn, and wll neglect aerodynamc forces. 3. We wll assume that the frst stage s ettsoned before burnng the second stage. 4. Note that the change n velocty due to burnng a stage can be expressed as M0 v v0 Ispg log M0 m where M0 s the total mass before the burn, and m s the mass of propellant burned. 5. The earth s radus s km The Gravtatonal parameter GM km s (G= gravtatonal constant; M=mass of earth) 7. Escape velocty s from the earths surface s v GM / R, where R s the earth s radus. e Calculaton: 1. Let m denote the payload mass; let m1, m denote the total masses of stages I and II, let m01, m0 denote the propellant masses of stages I and II; and let I sp1, Isp denote the specfc mpulses of the two stages.. The rocket s at rest before burnng the frst stage; and ts total mass s m m1 m m01 m0. After burn, the mass s m m1 m m0. The velocty after burnng the frst stage s therefore m m1 m m01 m0 v1 Isp1g log m m1 m m0 3. The frst stage s then ettsoned the mass before startng the second burn s m m m0, and after the second burn t s m m. The velocty after the second burn s therefore log m m m m v Isp g Ispg log m m m m1 m m0 m m 4. Substtutng numbers nto the escape velocty formula gves ve km/sec. Substtutng numbers for the masses shows that to reach ths velocty, the payload mass must satsfy m m log 4.435log m m 04 where m s n kg. 5. We can use MAPLE to solve for the crtcal value of m for equalty

36 so the soluton s 8300kg a very small mass compared wth that of the launch vehcle, but you could pack n a large number of people you would lke to launch nto outer space nonetheless (the entre faculty of the dvson of engneerng, f you wsh) Analyzng collsons between partcles: the resttuton coeffcent The momentum conservaton equatons are partcularly helpful f you want to analyze collsons between two or more obects. If the mpact occurs over a very short tme, the mpulse exerted by the contact forces actng at the pont of collson s huge compared wth the mpulse exerted by any other forces. If we consder the two colldng partcles as a system, the external mpulse exerted on the system can be taken to be zero, and so the total momentum of the system s conserved. The momentum conservaton equaton can be used to relate the veloctes of the partcles before collson to those after collson. These relatons are not enough to be able to determne the veloctes completely, however to do ths, we also need to be able to quantfy the energy lost (or more accurately, dsspated as heat) durng the collson. In practce we don t drectly specfy the energy loss durng a collson nstead, the relatve veloctes are related by a property of the mpact called the coeffcent of resttuton. Resttuton coeffcent for straght lne moton Suppose that two colldng partcles A and B move n a straght lne parallel to a unt vector. Let A0 A0 B0 B0 v vx, v vx denote the veloctes of A and B ust before the collson A1 A1 B1 B1 v vx, v vx denote the veloctes of A and B ust after the collson. The veloctes before and after mpact are related by B 1 A 1 B 0 A v v e v v 0 where e s the resttuton coeffcent. The negatve sgn s needed because the partcles approach one another before mpact, and separate afterwards. v x A0 A v x A1 A * v x B0 v x B1 B B Resttuton coeffcent for 3D frctonless collsons For a more general contact, we defne A 0 B0 v v to denote veloctes of the partcles before collson A 1 B1 v v to denote veloctes of the partcles after collson A v A0 B v B0 In addton, we let n be a unt vector normal to the common tangent plane at the pont of contact (f the two colldng partcles are spheres or dsks the vector s parallel to the lne onng ther centers). n v A1 A B B1 v

37 The veloctes before and after mpact are related by two vector equatons: B1 A1 B0 A0 v v n ev v n B1 A1 B1 A1 B0 A0 B0 A0 v v v v n n v v v v n n To nterpret these equatons, note that 1. The frst equaton states that the component of relatve velocty normal to the contact plane s reduced by a factor e (ust as for 1D contacts). The second equaton states that the component of relatve velocty tangent to the contact plane s unchanged (ths s because the contact s frctonless, and so the reacton forces actng on the partcles durng the collson exert no mpulse s exerted n a drecton tangent to the contact plane). The two equatons can be combned nto a sngle vector equaton relatng veloctes before and after mpact ths form s more compact, and often more useful, but more dffcult to vsualze physcally v B1 A1 B0 A0 B0 A0 v v v (1 e) v v n n It s possble to account for frcton at the contact, but frcton wll always make the colldng partcles rotate, so they cannot be dealzed as partcles. Values of resttuton coeffcent The resttuton coeffcent almost always les n the range 0 e 1. It can only be less than zero f one obect can penetrate and pass through the other (e.g. a bullet); and can only be greater than 1 f the collson generates energy somehow (e.g. releasng a preloaded sprng, or causng an exploson). If e=0, the two colldng obects stck together; f e=1 the collson s perfectly elastc, wth no energy loss. The resttuton coeffcent s strongly senstve to the materal propertes of the two colldng obects, and also vares weakly wth ther geometry and the velocty of mpact. The two latter effects are usually gnored. Collson between a partcle and a fxed rgd surface. The collson formulas can be appled to mpact between a rgd fxed surface by takng the surface to be partcle B, and notng that the velocty of partcle B s then zero both before and after mpact. For straght lne moton, v ev A1 A0 For collson wth an angled wall n s a unt vector perpendcular to the wall. v A1 A0 A0 v (1 ) v n n, where e v x A0 A v x A1 A *

38 A v A0 n v A1 A 4..8 Examples of collson problems Example 1 Suppose that a movng car hts a statonary (parked) vehcle from behnd. Derve a formula that wll enable an accdent nvestgator to determne the velocty of the movng car from the length of the skd marks left on the road. Assumptons: 1. We wll assume both cars move n a straght lne. The movng and statonary cars wll be assumed to have masses m1, m, respectvely 3. We wll assume the cars stck together after the collson (.e. the resttuton coeffcent s zero) 4. We wll assume that only the parked car has brakes appled after the collson Ths calculaton takes two steps: frst, we wll use work-energy to relate the dstance sld by the cars after mpact to ther velocty ust after the mpact occurs. Then, we wll use momentum and the resttuton formula to work out the velocty of the movng car ust before mpact. Calculaton: Let V denote the velocty of the movng car ust before mpact; let v1 denote the velocty of the two (connected) cars ust after mpact, and let d denote the dstance sld. 1. The fgure shows a free body dagram for each of the two cars durng sldng after the collson. Newton s law of moton for each car shows that R N N m g m a x R T3 T4 N3 N4 mg max N N m g ; N3 N4 mg.. The vertcal component of the equatons of moton gve The parked car has locked wheels and skds over the road; the frcton law gves the tangental forces at the contacts as T3 T4 ( N3 N4) m g 4. We can calculate the velocty of the cars ust after mpact usng the work-knetc energy relaton durng skddng. For ths purpose, we consder the two connected cars as a sngle partcle. The T T d m gd. The work done s work done on the partcle by the frcton forces s equal to the change n knetc energy of the partcle, so 1 m gd 0 ( m m ) v v mgd m m Fnally, we can use momentum conservaton to calculate the velocty ust before mpact. The momentum after mpact s h1 ( m1 m ) v1, whle before mpact h0 m1v. Equatng the two gves

39 ( m1 m ) ( m1 m ) mgd V v1 m 1 m1 0 0 Example : Two frctonless spheres wth radus R have ntal velocty A B v v. At some nstant of tme, the two partcles collde. At the pont of collson, the centers of the spheres have poston vectors A B y y. The resttuton coeffcent for the contact s denoted by e. Fnd a formula for the veloctes of the spheres after mpact. Hence, deduce an expresson for the change n knetc energy durng the mpact. Ths s a straghtforward vector algebra exercse. We have two unknown 1 1 velocty vectors: A B v v, and two vector equatons momentum conservaton, and the resttuton coeffcent formula. Calculaton 1. Note that a unt vector normal to the tangent plane can be calculated from the poston vectors of the centers at the mpact A B A B as n y y / y y. It doesn t matter whether you choose to take n to pont from A to B or the other way around the formula wll work ether way.. Momentum conservaton requres that A B B1 A1 B0 A0 mb v mav mb v mav v A1 B1 v 3. The resttuton coeffcent formula gves v B A B A B A v v v (1 e) v v n n B1 4. We can solve () and (3) for v by multplyng (3) by ma and addng the equatons, whch gves B1 B0 A0 B0 A0 B0 A0 mb ma v mb v mav m A v v (1 e) v v n n B1 B0 ma B0 A0 v v (1 e) m m v v n n B A A1 5. Smlarly, we can solve for v by multplyng (3) by mb and subtractng (3) from (), wth the result v A 1 A m 0 B B (1 ) 0 A v e 0 mb m v v n n A 6. The change n knetc energy durng the collson can be calculated as ma A1 A1 mb B1 B1 ma A0 A0 mb B0 B0 T v v v v v v v v 7. Substtutng the results of (4) and (5) for B1 v and A1 v A v A0 and smplfyng the result gves B n v B0

40 A B e 1 B0 A0 m m m m T v v n A B Note that the energy change s zero f e=1 (perfectly elastc collsons) and s always negatve for e<1 (.e. the knetc energy after collson s less than that before collson). Example 3: Ths s ust a borng example to help llustrate the practcal applcaton of the vector formulas n the precedng example. In the fgure shown, dsk A has a vertcal velocty V at tme t=0, whle dsk B s statonary. The two dsks both have radus R, have the same mass, and the resttuton coeffcent between them s e. Gravty can be neglected. Calculate the velocty vector of each dsk after collson. V B Calculaton 1. Before mpact, the velocty vectors are A0 B0 v V v 0. A unt vector parallel to the lne onng the two centers s n (3 7 ) / 4 (to see ths, apply Pythagoras theorem to the trangle shown n the fgure). 3. The veloctes after mpact are B1 B0 ma B0 A0 v v (1 e) m m v v n n B A A1 A0 mb B0 A0 v v (1 e) mb m v v n n A Substtutng the vectors gves B1 1 1 e v (1 e) V (3 7 ) / 4 (3 7 ) / 4 V (3 7 7 ) (1 ) 5 7 e e 3 3 A1 v V (1 e) V (3 7 ) / 4 (3 7 ) / 4 V V Example 4: How to play pool (or snooker, bllards, or your own favorte bar game nvolvng balls, a stck, and a table ). The fgure shows a typcal problem faced by a pool player where should the queue ball ht the eght ball to send t nto the pocket? Ths s easly solved the eght ball s statonary before mpact, and after mpact has a velocty v B1 1 (1 ) A0 e v n n Notce that the velocty s parallel to the unt vector n. Ths vector s parallel to a lne connectng the centers of the two balls at the nstant of mpact. So the correct thng to do s to vsualze an magnary ball ust touchng the eght ball, n lne wth the pocket, and am the queue ball at the magnary ball. Easy! A R 3R/ R/

41 The real secret to beng a successful pool player s not pottng the balls that part s easy. It s controllng where the queue ball goes after mpact. You may have seen experts make a queue ball reverse ts drecton after an mpact (appearng to bounce off the statonary ball); or make the queue ball follow the struck ball after the mpact. Accordng to the smple equatons developed here, ths s mpossble but a pool table s more complcated, because the balls rotate, and are n contact wth a table. By gvng the queue ball spn, an expert player can move the queue ball around at wll. To make the ball rebound, t must be struck low down (below the center of percusson ) to gve t a reverse spn; to make t follow the struck ball, t should be struck hgh up, to make t roll towards the ball to be potted. Gvng the ball a sdeways spn can make t rebound n a controllable drecton laterally as well. And t s even possble to make a queue ball travel n a curved path wth the rght spn. Never let t be sad that you don t learn useful sklls n engneerng classes! Example 5: In ths problem we set up a smple MATLAB program that analyzes moton wth mpacts. The problem to be solved n llustrated n the fgure. A square box contans a number of rgd, frctonless crcular dsks. The dsks all have the same mass. At tme t=0 the dsks are gven some random dstrbuton of veloctes. The goal of the program s to compute and anmate the subsequent moton of the dsks, accountng for all collsons. To do ths, we need to: 1. Derve the equatons of moton for each dsk and ntegrate them;. Detect collsons between partcles and the wall, and change the partcle velocty to account for the collson; 3. Detect collsons between two partcles, and change the veloctes of both colldng partcles to account for the collson. We wll model the system by calculatng the (x,y) coordnates of each dsk. In the followng, we wll use x, y v, v to denote the poston and velocty of the th dsk. x y Equatons of moton : No forces act on the dsks except durng the collsons. Therefore, between collsons, the equatons of moton for the th dsk s d x d y 0 dt dt As always, we turn ths nto MATLAB form by wrtng

42 x1 vx1 y v 1 y1 v 0 x1 d v 0 y1 dt x v x1 y v y1 Collsons wth the walls: The collsons occur at the nstants when the dstance of a dsk to the wall s equal to the radus of the dsk. Therefore 1. Collson wth the wall at x=l occurs when L x R 0. Collson wth the wall at x=-l x R L 0 3. Collson wth the roof at y=l occurs when L y R 0 4. Collson wth the floor at y=-l occurs when y R L 0 A collson can be detected by detectng the pont when d mn{ L x R, x R L, L y R, y R L} crosses zero from above. When a collson occurs, the veloctes must be corrected as follows to account for the collson: Collson wth the walls at x L v ev v v. Collson wth the walls at : x x y y y L v ev v v : y y x x Collsons between partcles The collsons occur at the nstants when the dstance between the centers of any two dsks s equal to the dameter of the dsk. When a collson occurs between the th and th partcles, ther veloctes must be corrected as follows to account for the collson: vx vx (1 e) vn ( x x ) / ( R) vy vy (1 e) vn ( y y ) / ( R) vx vx (1 e) vn ( x x ) / ( R) vy vy (1 e) vn ( y y ) / ( R) vx vx x x vy vy y y vn R To see ths, note that n ( x x ) ( y y ) / ( R) s a unt vector onng the two centers; and let v v v, v v v denote the veloctes of the two partcles, and substtute nto the x y x y general equatons n Example 4. Here s a MATLAB scrpt that mplements these calculatons. It wll produce the anmaton shown n the pcture. You can make varous modfcatons to solve smlar problems. The only new feature of ths code that has not been dscussed before s the ODE solver t turns out that the standard solver ode45 s

43 not accurate enough for ths problem and msses some collsons. The more accurate solver ode113 s used nstead. functon dsk_collsons % Functon to compute and anmate moton of colldng dsks % n a box n_dsks = 3; dsk_radus = 1; box_sze = 10.0; resttuton_wall = 1; resttuton_dsk = 1; % These varables are defned n the outer functon so they are avalable % n all functons dsk_wall_ndex = 1; %specfes the dsk that s closest to a wall colldng_par(1) = 1; %specfes frst dsk n par of closest dsks colldng_par() = 1; %specfes second dsk n par of closest dsks tstart = 0; tstop = 60; count = 0; y0 = [1.,1.,0.,0.,-1.,-1.1,1.,1.,-3.,3.,-0.5,0.5]; whle(tstart<tstop) Abstol = 10^(-05)*ones(4*n_dsks,1); optons = odeset('reltol', ,'abstol',abstol,'events',@detect_collson); [t,y,tevent,yevent,event] = ode113(@eom,[tstart,tstop],y0,optons); end f (~sempty(tevent)) y0 = collde(tevent,yevent,event); end tstart = t(length(t)); for =1:length(t) % The dsks are a sngle pont on an x-y plot hold off plot(y(,1),y(,),'ro','markersze',5,'markerfacecolor','r'); hold on for n = :n_dsks plot(y(,4*(n-1)+1),y(, 4*(n-1)+),'ro','MarkerSze',5,'MarkerFaceColor','r'); end axs([-5,5,-5,5]); axs square pause(0.05) % Ths ust slows down the anmaton a bt end functon dydt = eom(t,y) % Equatons of moton for the dsks % y contans [x,y,vx,vy, ] for each dsk for n=1:n_dsks = 4*(n-1)+1; dydt() = y(+); dydt(+1) = y(+3); dydt(+) = 0; dydt(+3) = 0.;

44 end end dydt = transpose(dydt); functon [eventvalue,stopthecalc,eventdrecton] = detect_collson(t,y) % Functon to detect a collson of a dsk wth a wall or another dsk % % Fnd the shortest dstance between a dsk and a wall dmn = box_sze; for n=1:n_dsks = 4*(n-1)+1; d1 = box_sze/-(y()+dsk_radus); d = (y()-dsk_radus)+box_sze/; d3 = box_sze/-(y(+1)+dsk_radus); d4 = (y(+1)-dsk_radus)+box_sze/; [d,mn_wall] = mn([d1,d,d3,d4]); f (d<dmn) dmn = d; dsk_wall_ndex = n; end end % Frst event s collson of dsk wth a wall eventvalue(1) = dmn; stopthecalc(1) = 1; eventdrecton(1)=-1; dmn = 10*box_sze; % Fnd the mn dst between any par of dsks f (n_dsks>1) for n=1:n_dsks-1 = 4*(n-1)+1; for m = n+1:n_dsks = 4*(m-1)+1; d = sqrt( (y()-y())^ + (y(+1)-y(+1))^)-*dsk_radus; f (d<dmn) dmn = d; colldng_par(1)=n; colldng_par()=m; end end end % Second event s collson of two dsks eventvalue() = dmn; stopthecalc() = 1; eventdrecton()=-1; end end functon y0 = collde(t,y,event) % Functon to change velocty of dsks when a collson occurs y0 = y; f (event==1) %collson wth a wall = 4*(dsk_wall_ndex-1)+1; d1 = box_sze/-(y()+dsk_radus); d = (y()-dsk_radus)+box_sze/;

45 d3 = box_sze/-(y(+1)+dsk_radus); d4 = (y(+1)-dsk_radus)+box_sze/; f (d1< d<0.0001) %collson wth left or rght wall y0(+) = -resttuton_wall*y0(+); end f (d3< d4<0.0001) % collson wth top or bottom wall y0(+3) = -resttuton_wall*y0(+3); end else = 4*(colldng_par(1)-1)+1; = 4*(colldng_par()-1)+1; normal(1) = (y()-y())/(*dsk_radus); normal()= (y(+1)-y(+1))/(*dsk_radus); vn = (y(+)-y(+))*normal(1) + (y(+3)-y(+3))*normal(); y0(+) = y0(+) + (1+resttuton_dsk)*vn*normal(1)/; y0(+3) = y0(+3) + (1+resttuton_dsk)*vn*normal()/; y0(+) = y0(+) - (1+resttuton_dsk)*vn*normal(1)/; y0(+3) = y0(+3) - (1+resttuton_dsk)*vn*normal()/; end end end 4.3 Angular mpulse-momentum relatons Defnton of the angular mpulse of a force The angular mpulse of a force s the tme ntegral of the moment exerted by the force. To make the concept precse, consder a partcle that s subected to a tme varyng force F ( t), wth components n a fxed bass,, k, then F( t) F ( t) F ( t) F ( t) k x y z Let r( t) x( t) y( t) z( t) k k F(t) denote the poston vector of the partcle, and M( t) r( t) F( t) y( t) Fz ( t) z( t) Fy ( t) z( t) Fx ( t) x( t) Fz ( t) x( t) Fy ( t) y( t) Fx ( t) k r(t) O the moment of the force about the orgn. x z The Angular Impulse exerted by the force about O durng a tme y nterval t0 t t0 t s defned as t0 t t0 t M( t) dt r( t) F( t) dt t0 t0 The angular mpulse s a vector, and can be expressed as components n a bass

46 k x y z t0 t x z y t0 t0 t y x z t0 t0 t y( t) F ( t) z( t) F ( t) dt z( t) F ( t) x( t) F ( t) dt x( t) F ( t) y( t) F ( t) dt z y x t0 If you know the moment as a functon of tme, you can calculate ts angular mpulse usng smple calculus. For example for a constant moment, wth vector value M 0, the mpulse s M0t 4.3. Defnton of the angular momentum of a partcle The angular momentum of a partcle s smply the cross product of the partcle s poston vector wth ts lnear momentum h r p r mv The angular momentum s a vector the drecton of the vector s perpendcular to ts velocty and ts poston vectors Angular mpulse Angular Momentum relaton for a sngle partcle Consder a partcle that s subected to a force F( t) for a tme nterval t0 t t0 t. Let r( t) denote the poston vector of the partcle Let M( t) r( t) F( t) denote the moment of F about the orgn Let denote the angular mpulse exerted on the partcle Let η r( t0 t) mv( t0 t) r( t0) mv( t0 ) denote the change n angular momentum durng the tme nterval t. The momentum conservaton equaton can be expressed ether n dfferental or ntegral form. 1. In dfferental form d M h dt k. In ntegral form h Although t s not obvous, these are ust another way of wrtng Newton s laws of moton. To show ths, we ll derve the dfferental form. Start wth Newton s law dv F ma m dt x O y r P z Take the cross product of both sdes wth r

47 dv r F r m dt Note that v mv 0 snce the cross product of two parallel vectors s zero. We can add ths to the rght hand sde, whch shows that r F r m dv v mv r m dv dr mv d r mv dh dt dt dt dt dt Ths yelds the requred relaton. Angular momentum conservaton: For the specal case where the force s parallel to r, the moment of the force actng on the partcle s zero ( r F 0 ), and angular momentum s constant d r mv 0 dt Examples usng Angular Impulse Angular Momentum relatons for a sngle partcle The angular mpulse-angular momentum equatons are partcularly helpful when you need to solve problems where partcles are subected to a sngle force, whch acts through a fxed pont. They can also be used to analyze rotatonal moton of a massless frame contanng one or more partcles. Example 1: Orbtal moton. A satellte s launched nto a geostatonary transfer orbt by the ARIANE V launch faclty. At ts pergee (the pont where the satellte s closest to the earth) the satellte has speed 10.km/sec and alttude 50km. At apogee (the pont where the satellte s furthest from the earth) the satellte has alttude km. Calculate the speed of the satellte at apogee. v a r p r a v a Assumptons: 1. We assume that the only force actng on the satellte s the gravtatonal attracton of the earth. The earth s radus s km Calculaton: 1. Snce the gravtatonal force on the satellte always acts towards the center of the earth, angular momentum about the earth s center s conserved.. At both pergee and apogee, the velocty vector of the satellte must be perpendcular to ts poston vector. To see ths, note that at the pont where the satellte s closest and furthest from the earth, the dstance to the earth s a max or mn, and so the dervatve of the dstance of the satellte from the earth must vansh,.e. d d 1 1 dr dr r r r 0 r r 0 r v 0 dt dt r r dt dt where we have used the chan rule to evaluate the tme dervatve of r r. Recall that f the dot product of two vectors vanshes, they are mutually perpendcular. We take a coordnate system wth and n the plane of the orbt, and k perpendcular to the orbt. 3. We take the satellte orbt to le n the, plane wth k perpendcular to the orbt. The angular momentum at apogee and pergee s then h r mv r mv k p p p p p h r mv r mv k a a a a a

48 where r, r a pergee, and v, v p are the dstance of the satellte from the center of the earth at apogee and a p are the correspondng speeds. 4. Snce angular momentum s conserved t follows that h h r mv r mv v r v / r 5. Substtutng numbers yelds 1.6 km/s Example : More orbtal moton. The orbt for a satellte s normally specfed by a set of angles specfyng the nclnaton of the orbt, and by quotng the dstance of the satellte from the earths center at apogee and pergee r, r. It s possble to calculate the a p speed of the satellte at pergee and apogee from ths nformaton. Calculaton 1. From the prevous example, we know that the dstances and veloctes are related by p a p p a a a p p a r mv r mv p p a a. The system s conservatve, so the total energy of the system s conserved. 3. The potental energes when the satellte s at pergee and apogee are GMm GMm Vp Va r r p 4. The knetc energes of the satellte at pergee and apogee are 1 1 Tp mvp Ta mva 5. The knetc energy of the earth can be assumed to be constant. Energy conservaton therefore shows that 1 GMm 1 GMm mvp mva r r 6. The results of (1) and (5) gve two equatons that can be solved for v, v parameters. For example, (1) shows that v v r / r that p a p p a GMm p p mvp rp ra ra 1 1 mv r a r 1 1 p GMra vp GM vp r a rp r a rp ( ra rp ) Smlarly, at apogee GMrp va r ( r r ) r GMm a a p a a a p n terms of known - ths can be substtuted nto (5) to see