Test1 Derivations

Test1 Derivations - CHAPTER 7 Coordinate Systems 1|...

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Unformatted text preview: CHAPTER 7 Coordinate Systems 1| COORDINATE RDTATIONS Theory Frequently, mere than {the enm'dinate system is needed in erder 1n deserihe the tnnlien ul' u purtiele. lni'ermetidn may be given in time ederdihale s}-‘fi1e111. hill net in the nne that yen wish 1e Use. it may he preferable it] use a refitted entirdinute system heetntse mes: nf‘ the threes in yen: i'ree tied}; diagram line up along its uses, er heeause the metiun is direeteti tilting tine ui‘ its uses. 1When Ihis happens, it beeemes advanta— gemis In transfnrm your quantities frnm the urigina] enurdinide system title the preferred entirdinate system. When yet: have finished. it may lie desirable in transform your quantities heel; tn the erigina! enerdinate systeiit. Referring In Fig. 1H. the linsitinn H} temis ed" the eeurdieates X and F 1.t'ith 11m: ennrtliuete system. in terms ui' the quantities . enurdinute system. We write etnr ni'a paint is expressed in and 1he unit veelurs 1' and J nssneinled The same pnsitieri s'eetnr is aisn espresseti t. _t-_. i. and j tissueiated with the retated r - .11 -'r _'I'j [ii—Ia] 1H" r .-.rt+r.l. LII—Ila] I31 I34 MARKS' MECHANICS FROBLEH-EDLVING CUMFANIDN Figure 3'. I-J Figure 3. f-E {Jur interest lies in expressing X and 1" in terms of x and y, and 1r'iee versa. In order In fit} this. lets first express I and J in terms et'i and j. and viee verse. The retstienships between the unit veetnrs are easy In see if they are celleeted end pltteetI in :I eirele. as shnwn in Fig. ’H-E. We get the unit veeter transfnrmstiens ‘ i=eesI’J‘I—sinUJ, l=eesUi-I-sin[tj. ‘ j; sintt l+eestJJ. .l —. .. sintii-t-eesflj, an”? Next, suhstitut'tnh,Ir Eqs. [1 1 -2b] inter Hq. (T. l-1h} yields .‘ri +_t=j XEeest’i i + sin flj}+ Yt— sinfl i +eestlj] = {Xensti — I“ sin HJirt- {X sin H -t- Fees tJJj. It lhllews that .'t' ees I'i' .lt' sin t} T. _ {ll—3a] .t-' 2 smth t-eestJ 1". Equations [ll-3s} express xsnrlr in terms at" X and tr’. Similarly. X and Y are expressed in terms ef I and _1" by substituting Hqs. [it-23} inte Eq. {ll—ls} to yield X _ ‘u'tt.t'—|—."tntl '. J L 3 “ J {11—3b} } ~- -s|nfl.r--t-eesfl_t'. i Coordlnato Systems I 35 Notice that the transfonnation equations of the unit vectors and the transformation equations of the coordinates. are of the same form. The coordinates .r. y, X. and l" ore interchangeable with the unit vectors i, j. l, and J. respectively. liy interchanging these quantities. Ens. {Tl-23] become Eqs. [ll—3:1}. and Eels. Ell—2h} become Eqs. {'i‘. l—3h}. Further— more, notice that this entire derivation. rather than having started out with Eq. {I 1—lj expressing the position vector 1* in two ways. could have started out with F : fit -t- F rj and F = FNI -t- Fir-J. espressing the force vector P in two ways. or could have started out with expressing any other vector in two ways. In other words, the trans’rormation of coordinates given above is vaIid not only for the components oi" the position vector, but also for the components of any vector. For example. the transforma- tions of the components of force are fr; = {mg [11 FX — 5i“ Ff. [ll-45C! £1. = sinti' F} + cos iilr‘r. [—. _ - F = costi'F + sin [it“s ” ‘ -‘ t?.l4hi FF : siniii’ft +costiF_... Once you know how the tmit vectors translornt. you transform the components of any vector in the same way. to short. the components of all of the vectors transform like the unit vectors. Examples 't‘.l-l Detennine the X and 1’ components of the position vector shown. Solution Let's lirst determine the unit vector transforma— tions. Referring to Fig. 11-2. we get Example; 1L; 1: cos 3G" i + sirtJti"' j. J = - sin flit" i -t- cos 30" j. The coordinates transform like the unit vectors, so X = cos 3D” x + sin 3U" y = Ufififitflfit + 0.3104} = Union], 1" = — sinltfl‘ s; + C0530" _t' = —fl.5{fl.3] + U.Hfiti~l[U.-4] : tilt} in. Coordinate Systems I 45 F; = {3: + Eli + {42‘ id El]. and particle 3 moves along the path rytr} : {IEI — 4H + [2:11]. Partiele A is initially at rest1 and rnr, = 1.1-? Find the path rift]! and veloeityr rift} ot‘a small partiele Flying in the air. The particle has mass m = 0.125 kg and is being influenced by the resultant three P111] : {H.313 — “.4! + ilfili + U. 1!] + [15 eostfihk N. The pnrtiele is initially loenied at the origin and at rest. TJ-lfl Find the relative position vector rmmti‘} and the relative ~.r:;:]:;ir;it:,r veetor vR Mfr} of a model rocket as seen Ii}r Ed. lid is traveling on a Inerry-gorronntl following the path rapt!) = 5t] sini‘i + fill eosr j fl and the roeket blasts oft“ following the path rfiill = Elli-ll + it] + [see — 3a: — 51.31:; a. 1.1-l I Find the pntli ofn srnnil rocket rpm-AI} ns seen by Ed. Ed is now on a train traveling with a veloeit}.r of‘vh-Jfr} — Tfli + 4tlj fifs. The rocket. which was initially loealetl at rflrjtl} = zoom IL is fired upward when Ed is at rid-{ll} ; —ltltltli it. For a short time, the rocket has mass m3 = Ilfltl sing and a eonslnnt thrust of FREE] 2 ll {lfltllt lb. 1.3 POLAR CDOHDINATES Theory We have seen that the standard way of“ expressing the position vector of a point is r =xi l _t'j |- :lt, in whieii i r" {I ti [l], j : [If] 1 U}, and It = {ll {l I] denote standard unit vectors and 1nrhere tar. and :. are called rectangular coordinates. Another convenient was,r ofexp ressing the position vector ot'a point is by writing it as a magnitude times a direetion. This is ealled the polar Fonn otthe position vector. As shown in Fig. 134. the polar form of the position vector is r: rn,.. {Ti-ll in which 1': |r| dennles the magnitude oi" r and nr denotes the unit vector in the direction of r. Polar coordinates are used when the motion MEI HARKS' MECHANICS PROBLEM-SOLVING COMPANION ut‘ the puint is restrieted tu :1 flat plane. They are particularly cunvenient tu use when the puint reeves in a circular ur clreulat‘wlike path. When pelsr cuurdinhtes are used, the vecturs are expressed in terms uf the unit vectur n,.. culled the radial unit vector. and the unit vectur n“. Figure 13-! called the circuml’erential unit vector. The circmrtierentiel unit vectur is perpendicular Lu tt,.. Netiee that the unit vectors It... and n“ are merely rotated unit vecturs. like these we srnv in Seetiun 't'. 1. The I end .I in that sectiun are the same as the n, and n" here. Theref'ure. 11F and n” are related tu i and j by 114!) = eus till} i --'1- sin iltr] j. {13—24:} flair] = — sinil'lt‘} i+ eus tilt} j. t?.3-2h] Netiee that the radial and circumferential unit vecturs are functiuus ut" time. Unlike the standard unit vectors, the rudiul end circumferential unit vecturs depend en the angle l3. whieh in turn depends uu time t. Differentiating Eq. [13-23] with respect tu time using the chain rule yields I'tr : —iisintl l+ ileustl] = lit" -sinti i + eusttj} -—. tin” Similarly, differentiating liq. {13-21:} with respect in time yields [tutu ;_- ---itn,.. Thus1 the derivatives til the ttnit vecturs are given by sr = huh—l {13th m, = 41H tie-set We are ready tu difierentiete r with respect tu time in uhtsin the velucity veetur v, and tu differentiate v tu uhtitin the ueeelemtiun veetur e. We Coordinate System: HT difi‘erenliale Eq. {13-1} with rereel to time using the prhduel rule for difl'eretttiat'ton and [it]. {13—311} tc get V _ fir“, i Pun”, “1'11ch Fr. _ j'. :1” = frilj Differentiating Eq. [13—4) with reepcet to time yields a --. rrrnr. + mitt”. where H... = i" — rtiz. H” -...' r'i'} + 2H}. {13-5} Notice that the radial ccrnpencnt of acceleration a, is made up of twn terms and that the circuttll'crctttialchinphnenl fli'acccicratinlt is made up t'lr twn terms. The radial term “1'”? [which is always negative} is called the centrifugal term. The circumferential ten-n 2H} is caiicrl the Cerinlis term. A graphical rcpresentatien at" each cf the acceleration terms is Hhewn in Fig. 13-2ta—ej. The figures chew hew each hf the terms arises. 1h) % —t.', i? I15: firing in; Muir] Sfi=ufifl l-"It1h Figure fifi-Efa—tt') Coordinate Systems IE] 1.4 CYLINDRICAL CDORDINATES Theelr'gulr {Iylindl'ieal eeerdinates represent an exteneien of polar eenrdinates tn three dimensientt. The extension is directed perpendieulnr tn the pulat eeerdinttte plane in the tilteetien of the :: ttxie. The eemmen error that One makes When thiittg cylindrical eeettttnnles is: te let r dennte the distance between the erigin and the tip of the pesttien veeter, as in pelar eenrdinates. In eyltndrieal enerdinates, hnwever, r dene-tes the distance between the might and the pmjeefinn at" the pettitinn tweeter elite the r—-ti' plane. as Shown in Fig. 7".4-1. In ey1int1rieal enerdinutefi. |r| 7E r. Deneting the prejeetien of r unto the :-—t} plane by rp = r11," then |rP| = r. As shnwn, the pneitten vector in eylindriettl eeerdinatee is. r = rn, + 211;. {14—1} where n: denotes the unit vector In the 3 directinn. Differentiating L-‘q. {144} with respeet te time yieldfi the veleeity veetert and dillerentlatlng the veleeity veetnr with respect tn time yields the acceleration veelnt‘, written as ' 2 WE + Hallie + “:“r l.— [14-3) Figure 34-1 I54 MARKS' MECHANICS FRUBLEH-SDLVING COMPANION Netice that the radial and circumferential eempeiients ef pesitien, velecily, and aeceleratien in cylindrical eeerdinates and the eerrespend— ing einnpnnents in JJUIELT eunrdinates are the same [eee 17.1.15. H.341 and (135)]. The differences lit: in the : directive. Examples 'M-I Determine the unit veeters 11],, tint, and er, and the cylindrieal centpnnents el' the pet-titititt veeler r = 4i + 3] -l- [Eli l'l'l. Salutlan The prejeelidn of the familiar. vector nnle the r—t'} plane il-i r“ = 4i —l— 3]. The magnitude eF the prejeetinn at" r is. r = J43 + 33 = 5. The radial unit veeter is then nr : rpfi' : {4i -l- 3]}!5 : {Lilli -i- llfij. Frem the unit vector tmnathi‘tnatiens, nr 2 [131+ [Hi] T ces ti i + sin ti j, se eestl ; [iii and sini’i : {1.15. Thiis1 n“ = ~511th i + ere-til] : —fi_tii —l— {1.Hj. The unit ‘v'ecttt-r n‘. is the same as the unit 1tweeter k. We new vedfy the cylindrical cempenents til" the pesitien veeter. Taking dat preduets. r— r-nr = til-i+ fij + lEk‘J -{llEi+fl.fij} : 4tfl.$i +3{fl.tii = 5m. ezr-n:=[4i+3j+ lEItJ-ltz |2m_ Fifi r— 5nr+ lln_.m_ 1.44 A head is sliding, dawn a cylindrical Spiral having an incline angle til" ‘3 = iii"~ as shewn. Neglect the Frictien between the head and the Spiral. and assume that the head starts t‘rem rest at the tap at the spiral. Detennine the pesitien 1t'eeter ef the head as a ihnetien at time. Salutian As shewn in the tree tied}; diagram. the head is subjected Le a gi'avitalienal Ierce and a nermal three. 1which lies heen hrekcn Liewri intn lwe eenipenenta that are perpendicular te the tangent til" the spiral. Summing threes along the cylindrical directiants. we get N, -- mar. = int—RIF]. N" cesT — mg = me: = HIE. — N" ain 1| a: mar; = mitt]: where we netiee that i‘ _ ii since r is eenstant. We also get the velneity ‘u'EtIEtJl' in the term 1* T RUan + in: :- +r:- ens }' n“. + eainr n2. taking l‘ Cuerdinaee Systems l5?r 15 TA NGE NTIAL-H CIHHAL CO DRDI H AT ES Th eery When tangential annual ceerdinates am used. the veetnr quantities are expressed in terms at cnn‘tpnnents that act in the directien ef the velneity vceter and in the directien perpendicular, er Henna], tn the velecity veeter. 'l'he directien et' the veleeity vecter is alse the tangent tn the path that tlte paint thllews. This is why the cenrdinates used ttt this sectinn are called tangential—unmtal eeerdinates. These eeerdinates are ttsed when tlte motion is cenlined In a plane, and they are particularly eenvenient te use when the threes in the Free hetly diagram act in the dii'eet'ten at" the velccity veetnr and perpendicular tn the velneity vectnr. This eccurs in aerudynan‘tie and hydredynarnic preblems in which drag threes act. alen g the velocity vectnr and litl threes act perpendicular te the velocity vecter. lt alsu uecurs in certain n‘techanisnis, such as itt guided-tail prehletne, in which nnrtnal reactinns act in the nnnnal dil'ectinn and Frietien threes act in the tangential directinn. Netice, in contrast with pelar cnerd'tnates~ in which the vecters are based en pnsitien, that 1.vith tangeittial—nennal ceerdinates the vecters are Ihased un veleelty. Tangential -nnrmal entirdinales are setup by first writing the velncity veetnr as magnitude times directien. Referring In Fig. 7.5—], we let v 21ml, 5 {IS-ll - . where t' dcnetes the magnitude ef' the velncity vecter and Itr denetes the tangential unit veetnr. The tangential unit veeter is directed aleng the velneity vectnr. We alsu define the ttnit vectel' n”. called the normal unit veetnr. The nnnnal unit vecter is defined in he perpendicular tn the Figure 15-1 tangential unit veetet‘. Just as in the case ni‘ pelar ceerdinates. the unit veeters nr and n,f are retaterl unit vcctnrs, and thus are related to the standard unit vectors tltmugh transl'ermatien equatiens. Frem liq. [lit—EL replacing nr and n” with i1r and n”. we get b: Hint." flu = I53 MARKS” MECHANICS PROBLEM-SOLVIHG COMPANION L. Netti, referring to Fig. 15in we v tH define the radius of curvature p of ' >\ the path at time t. by intersecting the “thee "t lines extending from nxtt} and _ Niuili'tj‘” infirm” nit-lentil in which fit is 11 small WWW increment of time. Over the time nit} f * flttt-nn increment flit. the angle changes an incremental amount eti' and the point mech an incremental utnount m. Notice thttt In dfl' = dot, no dividing by hi and letting the incremental quantities become infu‘tileeitnal yields figure 15-2 pit = in {15—3} in whiclt attifttr = tit and refit! = 1:. Now, diITcrcntittting 1" in liq. {IS—l} with respect to time yields :1 = urn, + ann”, where n, = L'I, an = {715-4} Notice in Eq. [15—4) that |n| gé t}. With tangential normal eeerdintttcit, il in useful to have it formula for the rudiue ol'cur'trnture oftt fitnetitin_t-[.r]. lt cnn be shown From Fig. 15-3 {sec Example 15—4} that f- 2 in, If}? {15-5} It in interesting to notice that when the Slope dyftit'e tent! ot‘ the function fit} in t-i-ITIEIII [when tent? {<1 I]. the first derivative of the Function in uppreximatel‘j' {writ : H. From Eq. {155}, when the slope of the function is small, the second derivative of the function is approximately e'lyfulxz = lip. Coordinate Systems IS'iI rh‘ In!“ (It “’1’ '1 —-: J+ —'— a; t [new] Examples figure 1.113 1.5-] Find the radius: DI" curvature {if a particle that IL'LI'} =3:3 —- 16.x + 5E] m when .r = 4. Solution Diffcrcntiating the functitm fur the path yields tbllnws the path ,3: .- d2 J “T: _—_ 3,11 _ m = 3+2~ Eff = m = 24. so. f'mm Eq. {15-4}, 1 323 “2 = film—J _ = 131513 m. 1.5-: Duturminc the pnlar Form 01' the veEocity vector v = fin, mfg it" the pnsitinu veclur is r = 5a,. winning n 5_ I2_ n 7 H a j r:—|— --, = --_ —_. 13 I31 ’ «ft—:3 «tn—t3 Solutmn From the transformation cquatiuns Fur rutatcd unit mulnrs, n —12I-1£' n —--_-E l+- ?—j “"13 13" “m M' Newton's Laws of Motion I15 the aigdhnl and to solve the equations. suhstituting numbers [or the parameters at the last possible stage of the manipulation. In feet. it" possible, it is heat to express the answers as general expressions. and then to substitute numbers for the parameters to obtain numerical answers. Doing this enables twp important tests to he perlhruted. First. by developing general expressions. the units in your answers can he verified to mateh as a way to test for eareiess errors. Secondly. the general expressions exhibit trends titat can be inspected. The answers will be proportional to some of the parameters in the prohlem. and inverser proportional to ether parameters in the prohiem. These trends are an important part of your comprehensive understanding of the system. ' 6.1 GOVERNING EQUATIONS Theory The lield ot‘ Newtonian meehanies is governed by Newton‘s First. Second, and Third Laws and by Newton‘s Law ot‘tLiraeitatioa. Newton’s First Law states: 21‘ pttt'tt'ei’e mares- ii-t'th constant speed and att'raett'ott when the t'rts'ttt’t'ttttt fittest! tit- :et‘u, Newton‘s Seeond Law states: A partt't'i'e.'r ttc'eetettttt'uti t's ti'tteat'fl' pt‘apttrtt'ttttat to the testtt'tattt three. The ttt'ttttrtt'tt'attatttr r-ttttstattt ts t'ttttrra' tttit ttttt-ttt-tefs- mas-tr. Newton ‘s Third Law states: Tit-rt parades t'titetiart irr'ttt eartt other tiltTJttg-it'fFJt‘t't’s' that are t’tfltr’ttr in magnitude. apposite i'ti atteett'tm. atta‘ eati't'ttattt: Newton‘s Law of Gravitation states: Two pat‘tt'etes flt't.’ attracted to each other tttt‘attgttft'irt'ar tttat are at ttttear proportion to their masses and at tat-arse proportion to the .‘u't’jflfl'f't’ ttt'ttitt' tit-.t'tttttr'tt tte'ftit’att that”. I16 MARKS’ MECHANICS FROBLEH-SDLVING COMPANION Let's new examine a system nfparticlcs. Tlte number of particles is rt. Letting particles he distinguished From one another by subscripts. we pick out the ith particle and tttelfth particle and examine them [see Fig. h.2—t}. Newton‘s Second and Third Laws are expressed mathematically HS l_.. _ _. __| . F.- + H” -i- t}; -|- . . . + I'm} 2 treat. : [til-la] ! [5,. —f_r-l-. {til-lit] rl- x t“ = r! >< f}... tel-1c} where F.- denotes the resultant external {to the system} force acting on the hit particle. 1'” Phi"; t “st-ff“ is the resultant internal [to the system} fi'tt'ce acting on the itlt particle. Itr is the acceleration of tlte t'tlt particle. at,- is the mass of the ith particle. t}!- is the three that thejth particle exerts on the ith particle. t}.- thc force that the ith particle exerts an titcjllt particle. and I; is the position at the i111 particle. 'l‘itc left-hand side of Eq. [n.E-l a] is the resultant Force [external and internal] acting on the i'th particle. Thus. Eq. [sz-lfl} is the mathematical statement of" Newton‘s Second Law [and the mathematical statement ot‘Newtnn '3 First Law it" aJ = it}. Equation tell—th is the mathematical statement of Newlnrt's Thirti Law. expressing that interacting fitrccs are equal in magnitude and oppesite in direction. Hquatinn {oldie} is the mathema- tical statement espressng that the associated interacting moments are Pilgrim tilt-I Newton’s Laws ot‘ Motion I11 '1 too. Although Newton‘s gnitudc and opposite in directiot te that interacting moments are equal in this t‘ollows directly from this law. opposite in direction1 equal in ma Third Law does not explicitly sta magnitude and opposite in direction, which states that the forces are equal in nutgnitudc, and colincar {see L-lsatnple fit-ti}. he equation of motion F : ma for a single particle to at" particles, in which at; denotes the accelera— tion of the system‘s rnass center. We start by delioing the position vector at the mass center o!" a system of particles as the weighted average of the position vectors of the particles. written as Let‘s now extend t to F r: mar- for a systc l -t'mlr1+m3r:+...m,,rnl. m -_: all + an; + . . . + mm. H! r{._. where m denotes the total mass of the particles. Using an indes. notation. we can rewrite these two equations as I... ._ _Jt_ _____H_ l rL- =-. I—El-nl-rl. m = Em]. {fiLZ-E} HI l=1 r—_|_ _.I l___ Differentiating Eq. tel-2‘; twice with respect to time yields the velocity and acceleration vectors of the mass center, written as 1 It!“ 1 n v - —=- my. a - .—— — m a-. {cl-Bali t. mfg-i : r t “I; .' r i' r the equations grwentlng the motion ol‘ each ol'rhc Nest. let‘s add togethe Front the loll-hand side of lids. [oi—la}. we get particles, Eqs. {6.2-1 at. (a, + {j =- r.) + : = F. res—4} _r=t J—.§J—-i r- | :i-l r'—-l internal forces are equal and opposite as stated in Newton's Third Law. Eq. {filth}. and thus cancel tes the resultant external force acting on out. 11' in Ed. raga} deno the system of particles. Substituting Eqs. tel-la] and [oi—4} into Eq. [6.2-3hl. considering Eq. [til—4], and multiplying the result by m yields where the double sum is zero since the it: .—_ mac. i into} L__ _ _| states that the resultant external force acting on a nass of the systetn multiplied by the center. The internal forces are not Equation {oi—5} system of particles is equal to the I acceleration at" the system‘s mass IIB HARHS’ HECHANFC‘S PROBLEM-SDLVING COMPANION present in the etltttrtr'nn. Thus. the hehrreier' rrt‘ the niaernseepie heel}.r eatt be analyzed withfltli regard tn the internal threes acting inside the herd)». Sinee Hq. tel-5] leeks se much like Nentnn’s Seeettrl Late it is ealled Newtnn‘s Heeenr] Law for a System 01‘ Particles. E xamples e.1-l :‘t hie-eh is sliding tn the right en a rettgh stir-thee Iwhile heing stthjeeted tn the three shewn. Determine the aeeeleratien til" the hleelt. Assume that the fl‘ietien between the reuglt sadism and tlte bidet-z is getter'netl by a dry frietint'r ntndel that will he diseusserl shrirtljr. The black has mass tn = 2 slug, the ltinetr'e liietinrt eeei‘tieient is flk : {1.2, F —' I5 ".11 and [J' = It't". Sututten 5'} The talent: enn he regarded as at single particle. Let .T he pnsittve tn the right and t: Erflfltpfg (5,24 he pesitive upward. Unless otherwise stated, the .t axis is always taken pesitive tn the right, and the y axis is always tat-ten pnsttit-e upward. Sttnnning ftnees in the x and y tlireetinrts yields in g t —F', — Fenst‘l‘ 2 mm. N + Fsin ti — nag = es:er where F,- is tlte Frietiett three. heettrtiing tn the kinetic dryr l‘rietinn metiel. the friction three npprrses the directinn of the mefiertt and the magnitude trt' the three is prapartinnat tn the normal t‘uree N between the battles [the hleek and the grennd]. The Itrepertienalit}r constant is called the kinetie erreffieient rrl' frietinn m. The Friet'tt-rn three is F” .— Flr = fixiht‘lj Alsut the problem states that the blue]: is sliding en the heriaentai sur‘t'ttee; there is net mettert in the y direetintt, se the aeeeleratien in the it tlireetitrn is stern1 thtrt is. tr". = t]. Substituting tr}. = fl and lit]. {62—h} ittte ...
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Test1 Derivations - CHAPTER 7 Coordinate Systems 1|...

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