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# Intro to Human Movement EXB 103

UCD

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This 69 page Class Notes was uploaded by Royal Reilly IV on Tuesday September 8, 2015. The Class Notes belongs to EXB 103 at University of California - Davis taught by Staff in Fall. Since its upload, it has received 54 views. For similar materials see /class/187646/exb-103-university-of-california-davis in Exercise Biology at University of California - Davis.

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Date Created: 09/08/15

EXE man Muduie 3 Muscie La mhrTEnsiun LengthTension Relationship in Muscle Part I Introduction Goai Review gEnEYai assumpiiuhsiui Euniiudmg a saisumsis iengihriansiun isiaiiuhship UndEYSiand quaiiiaiwe aspssis unhs iEngihriEnsiun isiaiiuhship This muduiewiii gsi yuuiu bEgin ihihhihg abuui muscie mas gEnEYaiiDn in quantitative isims Tu u thisi We Wiii bEgm Wiih ihs saisumsis w ihs sismshiai uhii ui suhiiamiuh hi paviicuiavi We Wiii expiuve huwihs abiiiiy uia muscie saisumsis in gEnEYaiEiEnsiun mice is dEpEndEM Dquot the iengih unhai saisumsis Keyiu ihis anaiysis isihs assumpiiun ihai saisumsis geomeivy which is known in gisai detaii sissiiuh micmscupy can be diYEEiiy veiaied in ihs iengihriansiun isiaiiahship u a saisumsis This assumpiiun is SUWiEiEMiyvaiid imam puipusss TuunhavsimpiiiyElmanaiysiS WEWiiiuniycunsidEr E ndi i ns ui mahmai isumsiiisishsiuh Thai isJEnSiDn undeviua s 2quot change Thisassumpiiuh iai eiyavuidsiheiimerd siissis nisiisish DY suhiiaiiuh Em ihs iengihriansiun isiaiiuhship s 3 Ana SiveichrshmiEmng sysis Amaiui guai unhis muduie ism gEnEvaiE a uhsiiuh ihaiWiii isiais saisu YE ishgih in ishsia asu mahmai Ducani uh iiu s bEm iack a Van giveihis biack a u uisiihs biack bEIX ya 2 inhaiiuh Wiii Expaci saisumsis iengih as is ihpui And uhss gNEn ihai inpuiiiha inhaiiuh isiuih ui maximaiiEnSiun as is output Tu cveaie Wis mciiun WEWiii E nsidEi Six Awnbased hihiiimns ia mm in the m z iihas Wash wwwmth ia in the cahiaimhaihava wismeim Namemuveviap Thessaiiahgsmshisais shuwn beiuw MuduieUS MuscisjsmiAhmiswmm EgtltE1U3u Mudme 3 Musc e LanmhrTensmn me m and chmamems have nci uvev ap pusmunA be uw Anm engm cvussbndges cannuHuvm and the savcumeve cannm genevam mnsmn Tm zem enscn pmm s knuwn as passwe msummency n s passwe because the acme pvucess c1 muss a ev necessava zem m W5 mudu e we Wm ecu un acme mnsmn and gnuve passwe2as1m cyces Passwe msummency m Name A Thwckmamem Lune Crussbndge formation Helene W a Maximal tensmn W 39 C Tnquot Nament nvellap Thick nlamem commesslon E Acme msummency Madmednam mayhem ecmwmwmmy 194170192 uvev ap Mme uvev ap aHqumvmuve mnsmn Tm cummues um mawma uvev ap uccmsn begmmng at pusmun a abuve mm rmmh y m enscn genevated at pusmun Evs pusmun c MudmeUS Muscxejemmnwevewmm EXB 103q Module 3 Muscle LengthTension Shortening beyond position C thin filaments begin to overlap This has the effect of reducing isometric tension You might think ofthis overlap as hindering cross bridge formation in some way although the hinderance mechanism is not clear Sarcomeres can contract to such short lengths that thick filament ends eventually come into contact with the Zlines anchoring the thin filaments position D Shortening beyond this point deforms the thick filaments The thick filaments resist this compression furthering limiting the active tension generated by cross bridges In theory sarcomere shortening might continue until thin filaments collide with their opposing Z line position E However there is no evidence that real muscle is capable of such extreme shortening Once thick filaments begin to deform active tension decreases rapidly and quotactive insufficiencyquot is quickly reached Like passive insufficiency active insufficiency is a point at which active tension is zero In active insufficiency the force from crossbridges is exactly opposed by hinderance from thin filament overlap position D effects and thick filament compression position E effects Our lengthtension function will be developed piecewise according to the mechanisms outlined above By piecewise we mean that the relationship between length and tension will not be govered by a single expression For example we won39t simply say 2 tension W 100 or something like that Instead the relationship between length and tension will in turn depend on the length of the sarcomere As an analogy imagine you are trying to describe the relationship between speed and distance for participants in a triathalon Because the triathalon involves three qualitatively different modes of tansport ie swimming cycling and running the speed of the athletes is qualitatively dependent on distance Thus the relationship between speed and distance is different for the swimming portion of the race compared to the cycling portion ofthe race This is an example of where a piecewise description would be useful Coming up with a single expresion that is accurate for the entire race would be quite difficult Returning to the muscle between each sarcomere position we will assume that active tension is a 0 t 39 linearfunction of sarcomere length That is a fixed slope in terms of otmaxt ensmn We pmisarcomereilength won39t concern ourselves with the actual tension generated in terms of Newtons but instead generalize ourfunction based on some maximal tension that occurs between positions B and C Module03 3 MuscleSemiAnsweredxmcd EXE mag Megme 3 MuseTe LanmhrTEnsmn Part II Basic Sarcomere Geometry 30515 CVEa E ass1ghmeh1s mhhuwn1ehg1hs uvsayeumeye cumpunems LugmaHy ge1ehh1he1he1ehg1hs assue1a1eg Wuh savcumeve pusmunsAthmugh D Hawm MEMWEd 1he phvs1ea1 Tahgmavhs h 1he savcumeve EVVMHV EHSTDH ve auunshw 1e1 s hegmm guahmwhe veTanehshm bv mahmu assmhmemsvenhe savcumevE geemew These vaTuesweve Tavgew Wuvked eu11h1he1aens bv Hugh E Hux av ahg e1hevs usmg nahsrmssmh eTemvehmmescupv As vuuemevmeseva ues check 1hem aga1ns1wha1 veu see 1h1he e1ee1mh memmaph h a pm eunhe vs12 pages a 1h1s meme Th1 shemg heTp vuu gam seme 1h1u111eh as1e wha1 pans ave wha11h1he savcumeve The 1ehg1h e1 1h1ehmameh1s 1s genevaH assumeg 1 he 16mm Page amp Hux EV 1953 Pevhaps suvphsmgthsvaTUe 1s aheunhe same h hmh mu Wham HuMEv mg h1s uhghaTwmk ahg human WaThev ampSchvud1 1973 H E Mme1924 Ushg1he ass1ghmeh1 upevamv ass1gh LJhick eapuaT L uwevcase 1 1e b216 um heTuw Vuu eah ge11he 11 ehayae1eyhum1he Gveek Symhm Tumhah m anemawe y hy1yp1hg mc1rlog ilhick 1s 2m 141m 1 m Lithnckl xl i m LTheTy yeuw111ge1ah answev e1 1 5 x 111396 m yemmgmg yuu 1ha1 m 1 x 11139 m hh 0111 h emss hhgge hnkages can he wage The Hazune 1s ahuum1Egmw1gewu11ey1953 A5513quot 1h1s va ue 1e L7Hbu1h eapuah heTuw 212 111 e 1 18m The 11712 15 abum n nxmwme Page amp Hux ey 19133 A5513quot 1h1sva1ue 1e Lbu1h eapuah heTuw 172 11 mm MudmeDS MuseTejemTAhwevegxmeg EXE man MudmeS Musc emnmhlensmn The Ls gm mumquot Nememsvanes amass smmsL spemes Ln humans they ave abum 12mm Lung ovstsy L mm 1973 Asngn Lthva uem LJhin sspusL L Luwsmsss L bsLuw 2m Lithm am We mattth SME Lenth Ma smgLs thmmamem pmjecwvg awayhumthe z Lms Tth va ue uppusus sLde uHheZ Lms Wm ave cunsmeved pan ums smassm savcumeve mussLs H s unc eaywhy he chmamems ave svm av m ste between me WEI spemes mm Mm maments ave nm ack m Pan L We Lmvuduced ms andmavk Lengms ums savcumeve m mum Lenthrtensmn ve a mnsmp Wswm nuw get mewsde ssLsuLsLs Muse andmavk Lsnths Theve s a Law New T k n Wm mm mm caHed L m Lengm WE H bsgm by cvea mg m L vectuvwnmwe 22m va ues m n me m y sngn L m be a new Srewmemvemm mm a Srmw summn mam Vuu can u th by usmg ms Venn and Msmrumbsn m sLLsmsmsLy bytypmg cum Ln each sLsmsm mm L mm emev zsm F L um n THLSLSOK MudmeUS Mususjsmmnmvswmm EXE man MudmeS Musc emnmhrTensIun We heee Ic geheIaIe Expvassmnsmv each cIIhe 5 Iahemahh Iethha m L Lefs bEgm WIIh Ihe IchgeaI Iengm e pusmun Aquot Wham Ihe SavcumEYE EMEYS passwe IhaumcIehcy Passwe Insummency mm ame W A II ThIckmamem Zrhne Crussbndge Iarmahon Ihycm h m ahcnenhah pusmunA WW7 Usmg LJhick LJhm aha Liz cIeaIe ah EXpYESSIDn heIcthaI caIcuIaIethe savcumeve Iengm aI pusmun A Because each savcumeve shavesZrhrIesWIm IIa nmghbms asaumeIhaI a thgIe savcumeve EXIerIds hawway IhIc each ems bDYdEYmg Zr 5 T N h appeavm Ihe hghI EMEY um IhIc IhIa pIacetheeI aha mess Emer L7LhnL7LhckL7Lhn 424 m 0IheyeewaIehI EXpYESSIDnS aYE accepIahIe Vuu ahcuIe get ah anWEY cI 4mm DchI EDMWUE MIN yuu geIIhIa answev Havmg cIeaIee ah EXpVESSmn Icy pusmun A Ms pm I IhIc yum L Venn We H have In make a I I I h I I I I I I hh hI huh I Iheex UMununa ENJhIs meahthaIwehe sun cIIeveIthe Ihe 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nverlap D Thick nlamam compressmn nyuuv n N n shunenhan pusmun47 WW7 m Usmg LJhick and LJL name an expvesswun be uwmat ca cu a esthe savcumeve Engm at pusmunA m L7Lhnckl pm Othevequwa em expvessmns ave asssmams Vuu shumd get an answav 152w Dam cummue unm yuu geths answev Cupy yum asswgnmemluv L mm Exevmse 213 abuve and pass A have be uw R ep ace n n We mu m ebmen m be yuuvexpvessmn L z L Wk 68 mm pusmun D savcumeve mm mm Ex21m52215 abuve wmhum an squaws L72 1 Him 7 264 um swan H s OKWyuuv expvessmn duesn Liz 2 Lithn L7H 222 examy match We ans Shawn be uw 5 mm L L L 2 L mm L Wk m as n squaxsw EEum 7 7 7 21a MudmeUS Mususjsmmnmvswmm EgtltE1U3u Mudme 3 Mum Lengtwensmn L7ZL7Lhck 168 L7Z2L7Lhn 264 w LZ2LLhnLH 222 172 2 Lithnn gum 424 Mheuvy as m pusmun Eheve Thick nlamem ccmpressxcn E E Acme Insuf ciency Huwem 01212 5 nu ewdence 0m vea musde s capame munch ememe shunemng Sn mm at cannm be mm by exammmg savcumeve geumehy a une h H w Fm nuw m usjust assume mm passwe msummency uccms at a ength u Wm m Pan w we u 12mm m m ssue and pvuduce a mum accuvam 251mm Cupy yum asswgnmen uv L mm Exevmse 215 abuve and page A have be uw R ep ace 1pm 1 me 22ch m ebment m b21um mm L z L Wk 68 eave uuHhe unns 2 L L7Z2L7Lhnn 254 pm Liz 2 Lithm L7H 2 22 Liz 2 Lithm mm 424 L72 mm 152 L L72 2 Lithm 2 54 pm LZ2LLhnL7H 222 MudmeDS Musuejemmnmvewmm EgtltE1U3u Mudme 3 Musc e LenmhrTensmn Part I Basic LengthTension Goa Undevs and mechamsm Ma pwecewwseWunctmn Beam cunshuclmn uHEnsmn mummy usmg a pwecewwse shaggy n ms Pam vuu Wm hamquot m mea e a mm mm aeeeMs savcumeve enmh as an mum and vemms ensmn a a maman as an uumm Unmnunate w twuu d be mmemnu wme dawn a mathemauca desc mmn uHhe enmmensmn ve atmnshm thatwuvksmv aH savcumeve Enmhs Th y asuMqussma heamuvsa ectmmensmn avE nuamawe vdw eventuvevd levem uHenmhs as vuu saw m Pan H Su m duWnE R e ax H snutasbad asnsuunds m meees neee uHhe mm m use Fm examme Wva We the mummy a enqth an NIH then We mm musmguve um Maths 5 m We mamauensmn vange be ween 12 e 2 54 pm and 13 e 2 22 pm Wheve mnsmn 1DUuVmawmaD Cveate the lenglhj mummy aH uwevcase m be uwbyasswgmngw ubeaErmwveduv g39hlt m cumammg Eun ean s atemems based un yum L T S Mg L Vaclav Be cavemHu am the subsenm numbevmg tuned Alsu nutewheve lt and g ave used L lenglhlt L2 r hr gm L2 s xengm lt L3 L3 s xengm lt L4 L4 xengm 3m lenghlt Ln lzngh MudmeDS Musuejemmnwevewmm EXB 103d Module 3 Muscle LengthTension Test your flength function below First ask Mathcad to display L Then ask Mathcad what f27pm is Ask what f2p1n is 302 0 0 1 W 0 0 168 0 1 L 264 pm f27um 1 f2pm 0 282 0 0 424 Recall that when a Boolean test is quottruequot then Mathcad returns the value 1 one Otherwise Mathcad returns the value 0 zero Are the f results above what you expect based on how you constructed flength and what you see in L How so 303 Create addditional tests for flength below You39ll need to checkthe otherfour length ranges that that didn39t get checked in Exercise 302 above Create separate regions for each test Ifthere39s a problem go backto Exercise 301 and fix it 303 Module03 13 MuscleSemiAnsweredxmcd EgtltE1U3u Mudme 3 Musc e LanmhrTensmn n sum 1 sum mum e 2ngth4ensmn uneuen 1 use n e gate a Vaclav ens uthev expvessmns usmg mumphesuen Tu see ms M s assume mm m ugn nllnnmhl m uwulw m u y 2 34 eng hused meamng 0m 5 nutwha yuu d mherwwse expect the upevatmn s e um pmuuer and w yuu take exememsyyvm ufmuhwphcauun yuu H have m use me weemnze upevamv mme gtlenglhj unmen yeuwememem Exevmse 3 m Uus su yuu dur havemtype M1212 agam Th 1 u w v 1 um eve unze upeva umsm e e man a m hng39hdm Tun bavandthe bunemeekshke r o M V ahemawe y 012CIrloemmuskey eembmsuen 1 s xengm lt L W an 012 same We Vuu mmm have We mus1 1 sueeess bv s1smne m We vectunzauun hke W5 3 L S lengvhlt L2 9 glengvh e mangle u 4 59le and than cveawmpaswm vuuvvecmvs m the empw 5 vemunzed p acehu dev Vuu ma needm exphemv L3 s lenglhlt L4 m pavemheses heme Vuu add the 6 ad an ema se 1 mumpheanun smn Make suve hevectunze anuw ms 1m ex endsacmssbmhvem gm MudmeUS Musexejemmnwevewmm EXB 103d Module 3 Muscle LengthTension gt 7 length lt L0 W L0 S length lt L1 L1 S length lt L2 l 2 3 4 Ltd You39ll get a warning from Mathcad that you39re redefining a builtin unit g is used for grams This is OK We won39t be using grams in this module le th Hg L2SlengthltL3 L3 S length lt L4 i Lglength ll Make a couple tests of your glength function below You should see 6element vectors being returned that contain exactly one nonzero element 305 gl5pm gem OOOONO The glength function demonstrates how we can use our vector of Boolean tests to quotgatequot another vector of expressions But in the end we don39t really want a vector for our answer we want a number a scalar Fortunately this is easily accomplished by adding up all the numbers in the output vector This adding operation is known as a Vector Sum and can be found in the Vector and Matrix Toolbar where its button looks like 2v Ask Mathcad below what Zgl5pm is You should get an answer of 2 Try the other lengths you used in Exercise 305 above ZgUSurn 2 Zglt2umgt 3 Module03 15 MuscleSemiAnsweredxmcd EgtltE1U3u Mudme 3 Musc e LanmhrTensmn Vuu ve nuw seen 3 the echmques necessavym eenaum a mate 52 mm lenglhlt 1T Cumbme these m make a lensinntlenglhj mnean amewevease hketheune unthe 2 Ym mg39hdq quotgm FeeHveem cupy ananem Exevmse n4 5 yuu emu have m vertype evevylhmg t T 3 1 WM 7 quotmm quot g39h g 4 L2 s xengm lt L3 5 L3 s xengm lt L4 6 L xengm lenghlt Ln Lnsxmghu Heme m a e N am 2 Lime LEMMA LASIengh 145 and E numbevswnh numbevs m expvessmns that have bm ugma meamng mam umhe bawe can wme dawn myee e1 012m The m1 Mu ave the pmms atwhmh acme and passwe msummency ave veached m each emeae easesnenamn s equaHu zem seweae ave easy acme and passwe nsummenw WW7 MudmeUS Musuejemmnwevewmm MudmeS Muscxemnmwensmn EXE man mma Eutwheve shumdwe punhanuuw Wha numbevmme e vecmvuHensanengm mm m yepxacgv Vuu H pvubab y mm 5qu backtu make suve yuu knuw Wheve m mum uccm any PM My 5D and75 m m n q We can gvaph mlnn englhj asswgnmem mm Emma 3 U7 and 3512 12 Mg 1 b W Change We e ventuth cumam n ma numbeyshmummg yam Nme aslengvhd mum p acehu d s 25 and 75 av 25 my me upsuuke mum engmmsmn 75 curve wh e 5m 5 an 012 duwnshuke WWW 2 mn L2 s 1engvhlt L3 5 L3 s length lt L4 L4 length m lenghlt Ln n 25 75 Ian z W 1 2 mm 5n n Tudu m vange mama x MudmeDS Musuejemmnmvewmm EgtltE1U3u Mudme 3 Mum Lengtwensmn and We ast shuum be 5pm Dun t eave mm mm m x n mm 5m 1 mm a new gvaph be uw Wm lensinn an m yraws and x an 012 mm m 2 Make suve yuu mm mm axes by appvupnate 8 unns Whmh m m case uHEnsmn s quotIn 3 Because 5 nm ubwuusthat x cunespundsm mm 6 sarcnmerelenglh name an appmpnate Hm abe A Senhe mm vange mm H m 5 2 5 5mm Cmssed Tm mum ehmmate me D 1 2 3 4 5 uppevxraws whmhwmmakethevegmnu x mamauensmn muve c eav When yuu ve dune yum gvaph mum uuk sarcomere lenth sume hmg hke m m an m Wm M m sarcomere length MudmeDS Musuejemmnmvewmm EXB 103q Module 3 Muscle LengthTension Earlier in this Part we said quotwe can use ourvector of Boolean tests to 39gate39 anothervector of expressionsquot using multiplication In your own words explain what we mean by quotgatingquot Why do you thinkthis technique is important in the contruction of a piecewise function 313 Modu e03 19 MuscleSemiAnsweredxmcd EXB 103d Module 3 Muscle LengthTension Part IV Finishing the LengthTension Function Use linear assumptions to finish the lengthtension function The previous two Parts largely established an outline of the lengthtension relationship of a t sarcomere This Part will resolve the mlssmg componen s Let us begin with what happens as active tension decreases between L3 282p1n where it is 100 and L4 424p1n where it is zero As suggested back in Part I we will assume that tension decreases linearly with length overthis region That is we39ll simply draw a straight line between those two points on the curve This linear assumption implies that that the amount of tension that can be generated is proportional to the amount of overlap between thick and thin filaments This assumption is probably quite good and is supported by experimental evidence in frogs Gordon et a 1966 How should we represent this line One ofthe more common expressions for a line is quotyinterceptquot form y mx b Where m is the slope and b is the yintercept However the yintercept of our line is not immediately obvious We could calculate it but let39s try something easier A line in quotxinterceptquot form can be represented as y mx 7 c where c is the xintercept of the line This is a superior stategy in our case because the xintercept is already known What is the xintercept ofthe linejoining L3 282p1nand L4 424p1n That is at what length would tension be zero along that line 401 The xintercept is 424pm the same as L4 To get the slope ofthe line m recall that slope is E or if we explcitly calculate it from two points Iun X1YY1 and Mil2 the y2 T y1 m X2 T X1 Module03 20 MuscleSemiAnsweredxmcd EXE man MudmeS Muscxemnmwensmn Whahsme s upeu he hnejummg L3 222 mm L 424 m7 4m quotmquot 4142 i L47 L3 w Gvand 5m p acehumevwe have nuw 23 mm ememun vep acetheSUWawnmhe expvessmn quotum lmghr L4 L4 L Am I my LB 25 mg L 5 my L2 Ian mm am am 2 D Em L mg n 4 L L3 lenghltLA LASIengh Cupy yum gvaph mm Exevmse 312 abuve m and pas12mheve be uw The gvaph mum amumancauy the mm sumemg hke m gvaph aHhe quotgm mm 6 2 n 1 2 3 4 5 sarcomerelenth 4m MudmeDS Musuejemmnmvewmm EXB 103g Module 3 Muscle LengthTension 1 0 8 tensionx 6 4 2 sarcomere length That takes care of the descending limb of the lengthtension curve Now let39s work on the ascending limb In the region between L2 264p1nand L1 168p1n overlapping of thin filaments causes a decrease in active tension as the sarcomere shortens The rate at which this occurs in frog 0 sarcomeres is known It39s about 4851 Gordon et a 1966 The maximal tension in a frog um sarcomere is likely the same as in a human sarcomere because the length oftheirthick filaments and thus the number of cross bridges are about the same Walker amp Schrodt 1973 Thus while we don39t know much about the mechanism by which thin filament overlap hinders active tension 0 we can reasonably assume that the net effect in terms of A is the same in human as it is in frog um Herzog eta 1992 As before we39ll simply draw a line over this region ofthe lengthtension curve The slope ofthis 0 line is 485 A But we39ll have to do some algebra to get either the yintercept or the xintercept Neither is obvious from our data so I suggest we solve for the xinterceptjust to be consistent with the form of our previous line Because the slope ofthe line is already known we need only evaluate the line at a single point using the equation y mx 7 0 Begin by making an assignment for the slope in this region Let39s call it 3 beta instead of m just so we39re clearthat this is a specific slope in this length region not some general slope Assign p to 0 be 485 1 Don39t leave out the units um Module03 22 MuscleSemiAnsweredxmcd EgtltE1U3u Mudme 3 Musc e Lenmwensmn semi u Tu ubtam the wmevcam eveuy hne M s use Msmesu s svmbuh math espsbmues Tu ue ms 15 set up We wmevcep uvm eveuy hne usmg me Eun ean eeusxs sen mm 7 a L 7 Next pusmumhe mue em hne a SymbeheTumbav thumchckw Vuuv vaesswunshuu d uukhke Hhe av mm euee uWuuv euuauun and chck un sew m 012 ng away type em 32 Msmesu m su ve m e When yuu ve uene W5 mm a L2 7 H salvec 7 dean Msmesu shumd 12mm 3 smme vame m appmwmaxe v 57am weu ue ms wee Mme an e We neman vemen UMmmna eM a hue Wm mavem Vuu mm eemne W5 senswb e answev m Msmesu 14 nsmad Vuu H qe an answev m Mu nans hke n unnnnz A m 7 2 n6185567mn3n927835 um WME W5 s echmcaHV cunem u s a 5qu Mmuvehuub e e eeme Fununatew uvevcummn W5 bug 5 many easy Vuujus1 have m be exphmt abuunhe va ue m L2 12 7 2 54 m 5 We mudmed exmessmn nu avmd We bun uuks hke W5 Ana mm 1264pm7 c salvec 7 mm 7 H17 7 c salvec 7 u uuuuum m 7 2 umsmmuzugzvzzs um mm 7 m2 64m 7 c salvec 7 n 5721442292959n721549 um bemeen H7168 um and 12 7 2 54 um Tu use W5 expvessmn m um 1ensinnuengm uncuun NW H be cunvemen u asswgn W5 expvessmn m be a panmu avvanab e We used Hunhe s upe evms hne su M s use y gamma unhe wmevcep MudmeUS Musexejemmnwevewmm EgtltE1U3u Mudme 3 Musc e LanmhrTensmn w m asswgnmem Wm uuk hke W5 mm 9 L2 7 H salvec gt n5721443292969n721649 um nm y x mm 1264pm7 c salvec gt n5721443292959n721549 um e e mm 7 H17 7 c salvec gt u uuuuum m e 2 umsmmuzugzvzzs um e e mm museum c salvec an5721443292969n721649 um Testyuuvasswgnmembemw kkMathcadwhat ws Vuu shumd getavame e1 nsvxum Vuu H have m theme 012 msmsueu unuswmeh ave me evs bv uevsun 4m VUS78ym Vuu shumd s1up1uva mmu e andthmk abum Wm yuuVe We have m seww yquot We w eyeepue n pamcu av vuuv As an a emawe appvuach we mud have used a peneu and papene sewevm snumen use ms eeusuen m make um assmnmen uv e Fm examme stamnu mu mm p077 c mu 312W a uebva s EpsWe eeum have Shawn M 7 mm 13 Then We eeum have asswgned y based an m Expvesswunmv e Sumethmg hke W5 st 7 mm a That WEIUM ewe usme same anwevmv v u uuum butwe drdn ake ms apwuach nsmad We used Msmesu s abunym an 012 a gebva m us m essence the asswgnmem e y m Exevmse 315 s asswgmng we be sounon uHhe a gebva pvub emma We ve gemng Msmesu m su ve Gwen me u x m us7 Vunctmn MudmeUS Musexejemmnwevewmm EgtltE1U3u Mudme 3 Mum Lengtwensmn Cupy yuuvlemlnntlenglhj asswgnmem mm Exevmse 4 U3 and paste n have be uw mm 2 911 0 mm vep ace me 75 W m expvessmn mm m Cupyyumgyapmmm ExevmseAElA abuve m The gvaph mum amumaucaHy ch 2 mm sumemmg me me gvaph aHhe quotgm 2 sarcomere lenth mmanAx 5 sarcomere length MudmeDS Musuejemmnmvewmm EgtltE1U3u Meeme 3 Musc e LenmhrTensmn We ve annes1 eene We enw have ene vemen e tu ee Thws as1 vemen swheve the savcumeve 5 5e snemnanne chmamems cummess anams he Zrhnes As mennenee bemvE chmamems new 015 cummessmn ane tensmn evens ewew emekwwnn unhev snenenme Tne vate at Whmhtensmn decveases m W5 12mmquot 5 knuwn new expennnen s m nee H s abum maimeveen eraABBE We u assume hws 5012 same mhumans Because thethmk um Namems ave be ween Ln ane H 152 em can be eeseneee by a shang hne Usmg a hne m wmevcep uvm that meanswe ve uukmq tuwme eewn an expvessmn thsmvm Y X E Assmn eue ee 2mi DenHeave eenneenns um eezxei em Fun nnnnqn anlmH H man 2 15 thanhwsva ue uHum s ust a p acehumev Al the nnewe eem have anvwavm umcaHV MEHHE savcumeve enmh atwhmh acme msummencv uccms Su c eav vwe sneexen m1 ese Wmmnhe wmevcem Tha numbev eeesm have an bm mma meamm Heweven we new have a mum en We Enmmensmn ve auunshm m W5 12mmquot Name v we new knuwmat tmamL e 53 44 Thws mne mnsmn savcumeves ave capab e Mpmducmgwhen nn hx m n m can ca cu ate the wmevcem unhws vegmn enne eewe hke we we en y eavhev n mm W5 wmevcem mne pvupevva ue en Ln Su wmh W5 anewen we be ab em eennpme We lensinntlenglhj mummy Unvenenaew at eas m Mamcad M We eannm ese mm lens quotlenglhj mummy m a symbuhc sewe SuJuv examme uHquq smemem vetumsthe enuv nu su mmn wasmund m athcad 14 tmamL nnn a r nnn n Eun ean 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1 Azsszzxnaszzznyszzz w 2511ma12 u 1 Once that s cunecmd we can use 1 mm lensinntlenglhj mummy MudmeDS 27 Musuejemmnmvewmm EgtltE1U3u Mudme 3 Mum Lengtwensmn Cupy yum asswgnmen uv Umm Way back m Emma 215 and page A have be uw thm m mm vep ace me vswa ue1umwnh Lizcllve L acuve 1426 Liz um 152 L72 2 Lithm 254 pm LZ2L7LhnL7H 222 L72 2 Lithm mm 424 m Liam 1 425 L72 L41 152 L L 2 im 2 54 w Liz 2 Lithm L7H 2 22 Liz 2 Lithm L41 4 24 Cupy yuuvlemlnntlenglhj asswgnmem mm Exevmse 4 ma and paste n have be uw mm 2 mm vep ace me 25 W m expvessmn mpmghr Lu W n 2 ng upmam LE Ln 2 mm L Menath L slmghltL2 Ln mm W 91 Z W ngmng quotmghy A ngmwh n 2222 m MudmeDS Musuejemmnmvewmm EXE man MudmeS Muscxemnmwensmn Cupy yum gvaph mm Emma A m abuve m and page M1212 be uw The gvaph mum aummancauy change mm sumemmg hke We 8 gvaph aHhe mm mm 6 2 n 1 2 3 A 5 sarcomere lenth m m 2 mmanAx 5 m sarcomere length Wm a 12W sentences abum Wm yuu 522 m yum engmensmn gvaph Duesthwsmwnh me andmavk engthswe mscussed back m Pan H7 Huw 5m MudmeDS Musuejemmnmvewmm EXB 103d Module 3 Muscle LengthTension Part V A Model Muscle Goal Extrapolate sarcomere lengthtension model to model a full muscle Having completed your function ofthe lengthtension relationship of a single sarcomere it39d be useful if we could relate this to afull muscle For example we could ask What is the isometric lengthtension relationship of the human biceps over its entire range of motion Related to this question we might ask What limits biceps contraction during full elbow flexion Is it the result ofjoin mechanics ie the elbow can39t flex beyond a certain point Or does the biceps enter active insufficiency Or do both effects come into play Similarly when the elbow is fully extended can the biceps generate tension 0r alternatively are they stretched to passive insufficiency at that length Unforunately we can39t answer these questions with certainty However by making some simple assumptions we can arrive at an educated guess At present we have a function that relates maximal sarcomere tension to sarcomere length Let39s imagine that we connect 100000 sarcomeres together endtoend to form a long single myofibril What do you think its lengthtension relationship would look like Why 100000L0 14255cm 100000L4 424cm Module03 30 MuscleSemiAnsweredxmcd EgtltE1U3u Mudme 3 Mum Lengtwensmn Let s mudWy yuuvlensinntlenglhj mummy m 522 Wyuuv mman s tuned Cupy yum lensinntlenglhj mummy mm Emma 415 and pas12n have be uw mnnnn y My T N m sans as an mpm pavememvmm mummy Tu u thaL change We e m mum asswgnmemm be lensinntlenglhszrtsj On We mm m mum asswgnmem yuu H need m mummy au me appeavances u L a 2 LD L1 L2 L3 and LA byszns Vuu need m u m m mm mm venuvs m addmun yuu needtu mump yme appeavance u y byszns AND yuu need m drwdethe appeavances u n and a byszns When yuu ve dune yum asswgnmem shm d uuk sumethmg mm a I length lt sues ITI 5m g39h 5 T 5m 1 s mm was L 5 xengvhi was w 5 1 S WM 5 L2 whmm 2 mm ms L S MW ms L n a lenghlt sarcsLn E lmghr sarcsLn 1 my mm mlmghsucs 5m mm ms L4 s ugh MudmeDS 31 Musuejemmnmvewmm EXB 103g Module 3 Muscle LengthTension As a test of your new function let39s recreate the graph you made in Exercise 416 To do this first copy that graph and paste it here below You39ll get an error That39s OK The reason for this error is that the tension function now expects you to tell it the number of sarcomeres involved Change the yaxis from w to This will tell the tension function that we want 0 A to assume just a single sarcomere That should elminate the error 10 8 tensionx 1 6 4 2 sarcomere length If all has gone well then your graph here in Exercise 503 should look exactly like the graph you made in Exercise 416 above lfthe graphs don39t lookthe same then you likely have a problem with your assignment in Exercise 502 Don39t continue until you get this right Assuming everything went well let39s now scale up the function to include 105 sarcomeres 100000 To graph this we39ll need to change the range variable we use forthe xaxis of the graph Just to minimize the amount of stuff we have to change on our graph let39s reassign the range variable x Assign x below to be a range variable The first value should be 0 zero the second value should be looL4 and the last value should be 115000L4 This will allow us to clearly see both the active and passive insufficiency ranges for our unnaturally long myofibril 504 x 0100L4115000L4 Module03 32 MuscleSemiAnsweredxmcd EgtltE1U3u Mudme 3 Mum Lengtwensmn Cupwummapmmm Exevmse and m Pasta A have be uw Change the y awshum mm 8 tensmn x m 6 5 iv Mamba 2 new We Hm vam s my me maph M s nmbabw SEND venue mu m 5 Cha ng We msmaved um an m Hm m 2 3 4 5 mm um m cm Change We Hm abeHu myuhbm engm m m 2 mmanxm5 6 2 n m 2U 3U 40 SD myofxbnl length abuve mm mm yuu EXpEE Ed m eng mensmn yexanunsmp m uuk hk27 5m MudmeDS Musuejemmnmvewmm EXB 103d Module 3 Muscle LengthTension We should remind ourselves that this myofibril approximation is pretty crude There39s no evidence that a single 30cm myofibril exists anywhere in the human body However it39s not clearthat assembling many myofibrils into a similarlysized structure would have a fundamentally different lengthtension realtionship that what we39ve graphed here That is while this approximation might be crude we have no reason to believe that it39s grossly inaccurate Instead of a single long myofibril let39s now think about a group of many of these long myofibrils assembled in parallel You should convince yourselfthat such a structure would have the same lengthtension relationship as the single myofibril To be sure the group of myofibrils would be able to generate more tension in absolute terms but in terms of maximal tension the two are the same Each would have a lengthtension relationship that peaks at 100 Each lengthtension relationship would have the same shape provided all the sarcomeres move synchronously and vary over the same range of lengths We can use this model of several long myofibrils as an approximation of a real muscle Returning to the question presented at the beginning of this Part What is the isometric lengthtension relationship of the human biceps over its entire range of motion Let39s assume that over its range of motion a human bicep changes in length from 28cm to 36cm Further let us also assume that the bicep generates maximal isometric tension when it39s length is 315cm Let39s use this information and yourtensionlengthsarcs function to make that graph We39ll first need to estimate the number of sarcomeres that would be put endtoend in each ofthe myofibrils making up the bicep We can determine this using two assumptions Th bicep generates maximal isometric tension when it39s length is 315cm this was given 2 Maximal tension is generated when sarcomeres are halfway between L2 264pmand L3 282pmin length This is reasonably inferred from our work in Parts and II We39ll calculate this value ie the number of sarcomeres per myofibril and assign it to be sarcsb for biceps Later we39ll use this value with the tensionengthsarcs function to make our graph Assign sarcsb below to be Below that ask Mathcad what sarcsb is You should 2 get an answer of about 1154 X 105 3150m L2 L3 2 sarcsib 1154 gtlt105 sarcsib Module03 34 MuscleSemiAnsweredxmcd EgtltE1U3u Mudme 3 Musc e LanmhrTEnsmn vanab e WE H WanHD use um emhg x VangEvaHab E meh su M s make a new ehe caHed x m bmeps Assng x n he 3 137132 vanab e be uw The vswa ue shumd he Z cm me SEEund shame he Z 1cm andth2 as shuu d he 36cm xib 22em221em 36cm cepwemeyaphhem Exemse and m Pasta A have be uw 5 mm m Changethe y awshum mmmibmsib 6 ms123d Vuu ng i 2 ah nee maph mmaHv Tm 5 OK change hehahshemlm 22 3D 22 24 m xb ms ead Thashememhe maph 7 chahee he V39aws yahee m 522012 ehme cm Ensmn vanuen um m um chanee We hahsm vahee hum 28m 35 bmps lengvh chahee the ham abeHu bmeps ehmh xib biceps length MudmeUS Musuejemmhwevewmm EgtltE1U3u Mudme 3 Musc e Lsnmwsnsmn y H q unmmvsavmv Amnw Aw Tu mp yuu unusysmu Wha s gumg un have M s add anmhey mum yum gvaph m sumquot mm natuva vange u bmeps muuun M s a su shuwthe enme engtmensmn ysxsuunsmp cupwummspmmm EXEMSESDB m and page M1212 be uw Tu We st add an addmunamace mm b 5m 3 g u L 5 DEME ms st ysnms an 012 maph snsmxsms b ms pmbsbw sum mm mm 28 m 7 35 Tu We vraws add an addmunamance D Ms msms b When vuu ve dune Vuuv maph shuum uuk m m sume hmg hke m une mm mm m smamjsms b 2 39 5 swam smsib bmeps length MudmeUS Mususjsmmnmvswmm EXB 103g Module 3 Muscle LengthTension Write a few sentences below about what the above graph shows At the beginning of this Part on of the questions presented was What limits biceps contraction during full elbow flexion Is it the result ofjoin mechanics ie the elbow can39t flex beyond a certain point Or does the biceps enter active insufficiency Or do both effects come into play Based on what you39ve done in this Part provide your best guess as to the answers these questions below Explain your reasoning 513 Module03 37 MuscleSemiAnswereclxmcd EXB 103g Module 3 Muscle LengthTension We also asked a similar question at the beginning ofthis Part When the elbow is fully extended can the biceps generate tension 0r alternatively are they stretched to passive insufficiency at that length Based on what you39ve done in this Part provide your best guess as to the answers these questions below Explain your reasoning 514 Module03 MuscleSemiAnswereclxmcd EXB 103d Module 3 Muscle LengthTension References Gordon AM Huxley AF Julian FJ 1966 The variation in isometric tension with sarcomere length in vertebrate muscle fibers Journal ofPhysioogy 184170192 Herzog W Kamal 8 Clarke HD 1992 Myofilament lengths of cat skeletal muscle Theoretical considerations and functional implications Journal ofBiornechancs 258945 948 Huxley HE 1963 Electron microscope studies on the structure of natural and synthestic protein filaments from striated muscle Journal of Molecular Biology 71281308 Page SE Huxley HE 1963 Filament lengths in striated muscle Journal of Cell Biology 192369390 Walker SM Schrodt GR 1973 l segment lengths and thin filament periods in skeletal muscle fibers ofthe rheus monkey and the human Anatomical Record 17816382 Module 3 Muscle LengthTension Written by Eric Leaver 2008 Martin VWson Last Modified May 20 2008 644pm Module03 39 MuscleSemiAnsweredxmcd EXB 103d Module 3 Muscle LengthTension Help Improve This Module Completing this section is not required but will be helpful for future development ofthis module Your answers here will not affect your grade About how much time did you spend completing this module Were there any particular parts of this module that you found too difficult too confusing or too tedious How might these sections be improved Were there any particular parts that you found especially useful or helpful in understanding the material from EXB 103 Other suggestions Module03 40 MuscleSemiAnsweredxmcd EXE mad Maddie a Pv ieciiies Projectile Motion Part I Introduction Goai R 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mdude any was The mphcn unus ave 12 but because ODEsuNe masks can hand e unns we wum be exphcn abum 01252 was Ms 5 OK W5 mudu e 2m w demde an 552 Assng and wqu be 5 wnhuut unns 2m and 5 Nuw bsgm yum sum muck be uw Wm M Wm Given 2m Gwzn 3 Mudme Pmmcmesjnmvswmm EXE man We nuw need mma cundnmns m each e 012 um emeyenuax eeuauens That s we n2 make Eun ean expvessm s be uwmv may pvt j velm j and velvt j Ms 5 many 5 nu mue ve umw 5 mm vem and vew 51 um at a zem That eaves mm Wh ch We assumedm be 4D me eys but new memee J mmam has ma uh beams mese unns Makemese em a emems be uw usmg me Eun ean equaxs swan Mudme a Pvmeemes edm mph The zem Funhev we assumed abuve 2D5 pan n M 4n um vewm n Nuw cveam yuuHuuv d levenua eeuauens agam usmg We Eun ean eeuaxs swan a a 4m who 7pm mm en en a a ve t n eve t at 10 at V0 2m 6 i v 3pm em a me a W 6 i v Kim M a v at V0 Nuw and We suwe meek Wow 3 dusmg smemem usmg the odesnlve mummy Vuu may vecugmze thanhese em eweyennax equatmns vepvesem a sys em e1 oDEs That meansmat yuu H have m use the my 5M2 Mudesu ve atemen ha usesveduvs Nme ma yuu may get a gveen undevhne Wammg 0m yuu ve vede mng a bumrm Mame mummy Ms 5 OK IV 5 am a nanma mummy m ca cu atmg pvesem va ue em mves1m2m my man We wem be usmg mm mm Muemena 5 Pvmeemesjnwevewmm Ph Ph m Odesalve W tend 21h velh velv velv EXB 103d Module 8 Projectiles ph ph v BM Odesolve pv tend velh velh Lvele LLvele J Resolve any errors in your ODEsolve statement before moving on Remember that the cause of the error could be in earlier exercises If everything went according to plan you should now have four functions at your disposal 1 pht 2 pvt 3 veht 4 vevt These are the solutions to your system of ODEs Let39s try them out Ask Mathcad below what phO and velh0 are These are the values of ph and velh when t 0 So Mathcad should return your initial conditions both zero Ask Mathcad also what phend and vehend are These are the values of ph and velh when t end presently set to be end 5 These values too should be zero 208 phO 0 ve1h0 0 phend 0 ve1hend 0 Modue08 6 ProjectilesAnsweredxmcd EXB 103d Module 8 Projectiles Describe below what these four values mean in words In each 4 calculations in Exercise 208 give a brief description of how these values relate to the quotrealquot world of the tomato drop 209 phO 0 means that the horizontal position of the tomato is zero right as it39s dropped This is what we assumed at the beginning ofthis Part that we would define the release site as being where ph 0 velvO 0 means that the horizontal velocity of the tomato is zero right as it39s dropped We39re assuming that the tomato isjust dropped not thrown so it39s horizontal velocity should always be zero phend 0 means that at the end ofthe simulation the tomato still hasn39t moved horizontally velhend 0 means that at the end of the simulation the tomato39s horizontal velocity is still zero Ask Mathcad below what pvO and velvO are Mathcad should return your initial conditions 40 and 0 respectively 210 pvO 40 velvO 0 Describe briefly what these two values mean in words pvO 40 means that the vertical position of the tomato is 40 right as it39s dropped This is the assumed hieght of Sproul Hall from where the tomato is dropped velvO 0 means that the vertical velocity of the tomato is 0 right as it39s dropped While gravity will accelerate the tomato initially the tomato isjust dropped so it doesn39t have any velocity to begin with Ask Mathcad what pvend and velvend are 211 pvend 782625 velvend 74905 You should get values of 82625 and 4905 respectively The 981 value for velv makes a fair amount of sense afterfalling for 10 seconds the tomato is descending at high velocity However the 82625 value for pv is harderto tie to reality Modue08 ProjectilesAnsweredxmcd EXE man Mudme a PTuTecTTT es wheTe m 4n SeTheh whsT dues dv whsrs gumg Em heTe 82 525 euTTespuhmuv DeesThTsmahesehse7 EXp am e082 E25 Ts a Teesheh 32 E25 meTeTs heTewThe gmund Th eeeshT YEaHy make sehse The mmam s zeTe Huwevev evame m haHthEvEsnu gTeuhe Th eTeTsTThg she eeeshT undEYgu shmm seeh unh The meeeT The TemsTeTusT keeps eh see Ts as m Teached T eeeeTeTsTTeh ehee m u TuhThTTT msTy YangE Vauab E me use ssThe depEndEMvaHab E ThThe gvaphs Nme shheugh y ex enswew Th yum sehe Muck abuve 1 Tsh T yeT sssTghee Te he shyThThg LeT sTshe esTe enhsT nuw T uuused quotmquot and The I n TsswsTue end ThTsWTTT awe yuu abum 1nn peThTs Th yum gvaphs TegsTeTess evwhsT end Ts sssTehee Te be t em and an cTesTe sepsTsTe gvaphs e1 yeuTTeuT seTmTe heT hs 5n uw eseh agams ohe eTyeuTveuT gvaphs shequ uuk We The ehe snhe TTghT D 0 em 7 mn Muemena PTuTeeTTTesjhweTeexmm EXB 103d Module 8 Projectiles I I I I I u 05 39 0 pht U INC 7 50 7 05 39 I I I I I I I I i l lUU 0 1 2 3 4 5 0 1 2 3 4 5 t t 1 I I I I u I I I I 05 T 10 7 2039 39 velht u velvt 7 30 7 05 7 40 I I I I I I I I i l i U 0 1 2 3 4 5 0 1 2 3 4 5 Focus on your above graphs for pht and velht only These two graphs should look pretty similar Write a couple of sentences below describing what you see How do pht and velht change in time In terms ofthe tomato what does this mean Position horizontal ph and velocity horizontal velh don39t change in time They39re always zero We39re dropping the tomato and no airwind is affecting the tomato on its journey down It starts out with no horizontal velocity doesn39t accelerate horizontally and so never leaves it39s horizontal position of zero Focus on your above graph for velvt only The line describing velvt should have negative slope velvt decreases in time Why is velvt decreasing becoming more negative 216 velvt is decreasing because the tomato39s velocity is becoming large in the downward direction as time progresses The negative velocity indicates that tomato is going down falling Module08 9 ProjectilesAnsweredxmcd EXB 103d Module 8 Projectiles Continue to focus on the plot of velvt only Why is the slope of this line constant That is why is the plot of velvt vs ta straight line and not curvy 217 Gravity accelerates the tomato at a fixed rate per second 981mss in the vertical direction Thus the tomato39s vertical velocity changes at a fixed rate in time that doesn39t depend on anything else in the simulation This means that the line of velvt vs twill have a constant slope will be a straight line In contrast to the velvt plot the plot of pvt vs tis not a straight line It39s curvy Write a few sentences about why this is 218 The tomato39s velocity isn39t constant 80 early in the drop the tomato doesn39t move much but later in the drop it moves quite rapidly This gives pvt a parabolic shape when plotted against time One quirk ofthe model discussed earlier is that there39s nothing special about pv 0 In reality this is the ground where velv would suddenly drop to zero The model doesn39t accounted for this That is in the model there39s nothing special about pv 0 so the tomatojust keeps going and going Module08 ProjectilesAnsweredxmcd Mudme a Pvmecmes EXE man Jus uvlum cvea e a gvaph Wm pm my me Hm 5n and pm my m y m Tm swhanhemmam s dvup wumu uuk hke Wyuuwewed n mm same pmm 4 awaynum 012 mm m 5 H mm mp m suang u n c ange me yraws m vange mu m mo sn 5 We dam havem uuk a aHma 5 y 72 negatwe ms ance mum mudgHumatuUave s m thvuugh 4 ins I I5 1 pm m 5U 40 3 Mo 2n m 4 ins I 5 pm mlt thmw hke We basebaH m We next 52mm Muumena Pmmcmesjnmvewmm EXE man Mudme a Pvmscmss Part III Throwing a Baseball G ua s R ewew sanssms Mpmjecuun spasm meansquot ang e and pymssuan mam MudWy oDEsaws b uckm descnbe a hymn basebaH mm as Pam yuu su ved m sys12m amwsysnnax equauunsma descnbe mmmn Ma mapped amen Thuse umsysnuax equatmns can nuw be apphed as sJu 012 case a a thvuwn amen v n m mvvsysm Whafs Maven s he mmaoondmans uHhe swmu atmn w vslhmvslwny ph j and mm am We same dwleverma equatmns as m Pan H O en the mma cundmuns a1 a ammo am pvujecmewmbehamedusmgmvee Span ML uammes r 1 Pymssuanspssa 2 Pvujec mnang e 3 Pvujecuunhewgm me Hamum Em amacnms Fm exammg as m Pan m was a ways asmsms mma pusmun hunzuma alumna be zsm Su m h we at We thLWheve m be eavesme basebaH p ayev s hand 5 Wham ph u a mev n m mp nH Ms mphesthat We mma cundmmv an W p s mvt j 5 mud a me pymsmmn mam 1 The baH s pymsmmn speed 5 153mm ws u asswgn a me vanams speed 2 The baH s pymsmmn ang e s Z deg whmh ws u asswgn a me vanams n m mn m spsswymg mm mm oDEsaws mask Mudme Pmmcmesjnmvswmm EXE man Mudme a Pvmecmes u u y y n m be Z deg nemmng the ag unns s mpunam nene because mnenwyse Mamcad wm assume we mean vadwansywhmh wm mess up eunneenemeuy Tne a chavactev theta can be eune mme Gveek Symbexreemsn uvgeneva ed by ypmg qmmg 351 speed 15 e maeg ngmmang e mm vebmly seamen r Vemcal Vemcuy Wurnrgwymw WWW gunmen g Hnnzoma Ve oc y Cmvone39u Madmed nam mums Em amneenenes A su vecachat appamz a acent s e an e 4 hypmemse Whmh m mm mphesthat appamz hypmemse we a acem hypmemse ca speedy m n q um yennaynyne mma cundnmns ave vdv l speed me new speed ca Cvea e a new ODEsu ve Muck be uw by 1 Cupymg We vegmns yuu made m Eyeneyses2u4 thvuugh 2 U7 and pas1mg Mam be uw 2 nan eyseussee m W5 Pan Onee yum new ODEsu ve Muck s cump e eJe esen evyewem su utmns by askmg en swab pumts as yuu mu m Eyeneyses2 nay2 my and 211abuy2 m2 Muemena Pmyeemesjnweveewnm EXB 103g Module 8 Projectiles Given Ph0 0 pv0 2 veh0 speed0059 ve1v0 speedsin9 ipha ve1ht ipva velvt dt dt iveht 0 ive1vt g dt dt Bl Ph WV Odesolve pv t end liveh velh MAM velv velv J J ph0 0 phend 64952 PV0 2 pvend 783125 ve1h0 1299 ve1hend 1299 velv0 75 velvend 74155 If everything in your solve block is working correctly you should notice among other things that velh0 is 1299 and velv0 is 75 Before making some graphs let39s doublecheckthis quotanswerquot with the Pythagorean theorem You may remember the Pythagorean theorem as 2 2 2 A B C Where C is the hypotenuse of a right triangle and A and B are its other two sides In turn this implies C IAZ B2 Module08 14 ProjectilesAnsweredxmcd EXB 103d Module 8 Projectiles Use the Pythagorean theorem below to check your values for velh0 and velv0 using the respective solutions from your ODEsolve block in Exercise 302 Hint You can use velh0 and velv0 directly here You don39t have to type in theirvalues 303 yivelh02 ve1v02 15 Does your answer from Exercise 303 make sense How so 304 That looks right The Pathagorean theorem predicts a hypotenuse of length 15 using the vertical and horizontal initial velocities That39s good because 15 is the assumed initial projection speed Basically this confirms that we did our trigonometry correctly Ratherthan comment on more individual values let39s look at your solutions graphically Copy your graphs from Exercise 214 above and paste them here below These graphs should quottake onquot the new solutions that you generated in Exercise 302 Note You don39t need to assign tagain 305 6U I I I I u I I I I 0 60 39 7 20 pht 40 39 pvt 7 40 7 60 20 39 7 80 I I I I I I I I U lUU 0 1 2 3 4 5 0 1 2 3 4 5 t t 1301 I I I I W 0 1339 39 7 10 ve1ht 1299 VSIVU 20 7 30 129839 39 7 40 129 I I I I 7 U I I I I 0 1 2 3 4 5 0 1 2 3 4 5 t t Module08 15 ProjectilesAnsweredxmcd EXE man Mudme a Pvmecmes Wm a aw pavagvaphs be uw abumwhat yuu 522 h yum gvaphs h pamcuxah New humhese gvaphs Mevhum What yuu saw h Pan H ma Vexacny hunzumaHuHhe be 5 Euns am h mam hke WNHME mmam Em h 012 case BMW in Ms vemmy 5 gveatenhan zevu Sn pusmun hunzuma mmases h We Thws cumva swnh themmam whm vemmy hunzuma was zevu and SD pusmun hunzuma mum change Vemmyvemca m hmh We be ahth mmam decveased h mm The Name have mhanhe baH begmswnh a pusmve v2mca vebmty Sn Makes same mm mm tha vemmy becames hegawe The 5 We be s ans um muvmg up hm atev vevevses dwedmn aha s ans muvmg dawn Thws cumvas swnh the tumatu whmh had ah hmax umyvenma mzevu Sn had a hegawe vemmymnhe ahhaa1 au uHhe Shh usmuh vemcax uHuwed a payabuh pa h h mm m hmh the be and We mmam h the case uHhe haw huwevemhe swmu atmn mum hegh aHhE pavebu a s apex R e edmg shun ma bemve muvmg mhe hegawe dwedmn ONCE agamJHE gmund at m n has ha specm mEamng h W5 swmu a mn Sn the be aHs ngh hmugh We gmund have be uw Tu make 012 gvaph a We muve axeah change m the x7 and yraxesm hmh vangE humu D25 mm made any exphcm um s the hhphcu uhus ave m Mo vvheh yuu ve anE yuu gvaph shuum uuk sumemmg hke i the aha aHhE Ham 1 pm 2n Mo m m 2H pm Muumena Pmmcmesjhmvewmm EXB 103q Module 8 Projectiles The graph of pvt vs pht above looks similar to the graph of pvt vs tthat you made in Exercise 305 For example they39re both parabolic Write a few sentences below explaining conceptually what different information these two graphs are showing Also explain conceptually why the traces in the two graphs looks so similar The two graphs are different because they plot different information The pvt vs pht shows the baseball39s path as would be seen by an observerfar away from the thrower looking perpendicular to the path ofthe ball In contrast the pvt vs tgraph shows how the ball quotfallsquot in time irrespective of the ball39s horizontal position The reason the two graphs looks similar is that the velocity horizontal of the baseball is constant through time This means that pht increases at a rate proportional to time overthe course of the throw Thus watching pvt vs pht ort look basically same because pht and tare proportional to one another The motion of a thrown baseball as you modeled in this Part is different from the motion of a dropped tomato which you modeled in Part II However in the beginning of this Part we discussed that the difference is entirely a difference of initial conditions the differential equations describing the motion are identical Imagine you39re trying to explain this point to a fellow student who is having difficulty What might you say to make this concept more accessible 309 Projectile motion neglecting air resistance and the curvature ofthe earth is basicallyjust quotfallingquot That is all projectiles accelerate towards the ground according to gravity and don39t accelerate in the horizontal direction These quotrulesquot are represented in the differential equations in the ODEsolve block and are true for all projectiles Thus the differential equations describing projectile motion are the same for all projectiles The initial conditions ofthe simulation in contrast depend on the particulars ofthat projectile So for example a basketball thrown straight up begins with a positive velocity vertical but the time derivative of that velocity is still goverened by gravity just like a dropped tomato that has zero initial vertical velocity Module08 17 ProjectilesAnsweredxmcd EXE man Muduie a Pv ieciiies Part IV Maximizing Horizontal Distance Javelin Throw Goais Paiameieiize ihe pieieeiiie oDEsehe biuckiu aeeepi n as ah ihpui paiameiei Geheiaie muiiipie seiuiiehs EIva a YangE ei n vaiues Giaph seiuiiehsie Empivicaiiy dEiEYminE n vaiue ihai mahmizes heiizehiai aisiahee Having deveiuped ah ODEsuive hieehihaiwii caicuiaie ihe meiieh ei a pijciiiE iei s haw i a h E pieieeiieh speed iheh whawaiue ei n Wiii mammize ihe heiizehiai disianEE ihai ihe javaimhavais We migm assume ihai a n vaiue uMSdeg Wiii he epiimai Howevev ihat anwev assume ihanheiaveiih has a pieieeiieh height eizeie which is HEIHYUE ieiiheiaveiih ihiew And it unhei assumesihanheiaveiih is hm aiveady nn39l Mining siaii huidmgihejainm Vquot WE39 V quotL WquotltG DU We examvaiue H sime ihaiwe cuuid suive iuiihe anSWEY ahaiwieaiiv Hui VUWE WquEd ham buiiqu VDUY ODEsuive biuck isuueesiwe use ii Reeaii ihai VDUV ODEsuive biuck heuiems aii iesisiahee and MUiWHiEH aie impunam EunsidEYa iunS in piuieeiiie spans We Wiii eumihue in inhuie ihese DYEESiWHiEH meahs BUY ahswei Wiii heeessaiiiv he incumPiEiE Em N be quad Enuuqh iui BUY puipuses heie We ii make iheieiiewihe assu i In 2 The pieieeiieh height enheiaveiih is 2m 3 Theiaveiih is him whiieihe aiheieie is iuhmhg This impaiis ah addiiiunai z ie ihe heiizehiai veiuciiy EIHaVEim Muuuiena Piuieeiiiesjhweiewmm EXB 103d Module 8 Projectiles Because we assume that speed is fixed let39s make an assignment forthat now Assign speed below to be 20 Don39t include explicit units 401 sReed 20 Begin by copying your ODEsolve block from Exercise 302 and paste it here below You can leave the test evaluations behind In addition copy your dvt vs dht graph from Exercise 307 and paste it below your solve block You shouldn39t have any errors in your solve block or the graph at this point 402 Given ph0 0 pv0 2 ve1h0 3 speed0059 ve1v0 speedsin9 iph veht ipv ve1vt ive1ht 0 ive1vt g dt dt ph PV Jave11n9 Odesolv tend Lvelh velv 20 MO 10 I I U 0 10 20 mm 19 Module08 ProjectilesAnsweredxmcd EXE mad Mudme a Pvdecmes Vuu H heed u change the mma eehdmehs 5 they ma ch the assdmphehs gNEn abuve h pamedxah make SUYE the mhmhd speed enhe ame e 5 memded h the summerquot m velh Hm VDU heed m add the mhmhd speed e the mma hunzuma eempehem enhe pyedeeheh speed 4m jwelimm VDU mayWanHu vevwewthwstechmque m the odesdlve hlncls m mddule the vdhehehs phm pvm velhm ahd velvm Tu N h swmp yjavahnw Wheh yedhe dunE yum ass gnmEM shm d uuk sumemmg MEWS meme Odesalve W 1 end mp 39 39 4m vames e n Based eh yum EXpEHEnEE m the odesehe E ucks Mmhmudu e yed may vecugmze ha ms newmndmn jwelint j eahw he used e m make these gvaphs dheeny nsmad yed M k n M W hd Hm Tu gvaph W s emde yed mds1 asswgnthe Du puHD seme mhendheheh m 52 e1 dhehehs Sn as a sun e1 shunwhwe Banjus ass gn the empdne Dn y WD vdhehehs WEI e ph and W vem ahd VENU semhehswm S mp y he mscavded The Tu du thws name We uHuwmg asswgnmem shumd uuk hke W5 pm 5 Javehn35deg pvz pm i h 356 was MM 2 avadame m yum use These vepvesem the hunzuma ahd VEVMEa pesmehs ph and W enhe avehn asmhdmh emme when WNWquot at a 35ded ang E n et stake a uuk atwhat happehsm thejavehn Mudme theemesjhwevedxmcd EXE man Mudme a Pvmeemes phasm be uw abevegmunwmgm Amusnke X39aws vange mu m En Tu make me gvaph uuk mme yeahaneJeex nee m make uwmey 8 e 6 mm 7 A e 2 2m 40 n pmsu Ana vaSU A 2n n phasu Nuwthatwe knuw huwtu make a smg e uaeemy a smg e n va ue M s make a bunch we we eae erem a umeue nameJhen We can p uHhem 3 an 012 same gvaph d deg deg and deg 4m FMD In And FMS h Md phm h Sud W 71mm 2 W45 71mm 2 W 71mm 2 ph h 556 W 71mm 2 Muemena 21 Pmkeemesjnweveekmm EXE man Mudme a Pvmecmes yuu Heated m Exevmse 4 U7 m m gvaph When yuu ve dune yum gvaph 5mm uuk sumemmg m m une 2m 40 n 111350111400ph45 vph5 0ph550 4m 2m 40 n phisu phAuu pmsu mm mm new gvapmu zuum m Wheve mejavehn mmws mach me gmund amund am am Muumena Pmmcmesjnmvewmm EXB 103d Module 8 Projectiles pv35t E 15 5ng pv55t05 I I 48 52 54 ph35t ph40t ph45t ph50t ph55t Write a few sentences below about what you see in the graphs you made in Exercises 408 and 409 What value of 0 seems to maximize horizontal distance 410 The throw where 0 was 45deg had the greatest horizontal distance about 51 75m Angles more shallow than this didn39t go as far And although angles greater than this went higher they didn39t cover as much horizontal distance before reaching the ground Based on your results in Exercise 409 what do you thinkthe optimal value of e is forthis simulated jevelin throw Do you thinkthe optimal 0 is exactly 45deg within 1 deg something less than 45deg or something more than 45deg Why do you think so 411 Ithinkthe optimal value of 0 is something more than 45deg between 45deg and 50deg I say this because the 45deg and 50deg throws each went further than the 40deg throw 45deg went the furthest in Exercise 409 but if 45deg was optimal I39d expect that the distances for 40deg and 50deg would be about the same The fact that they39re different leads me to believe the optimal angle is something more than 45deg but less than 50deg Module08 23 ProjectilesAnsweredxmcd EXB 103d Module 8 Projectiles Determine within ideg what value of 6 maximizes horizontal distance under these assumptions Use any technique you like Support you answer with calculations andor a graphs To be clear write out your answer in words also Does this match your intuition from Exercise 411 above dhl 39 139 46d th 39 139 47 d th 39 139 48d I ave 11 e I avem e I avem e dvl J g dv2 J g dv3 J g 01 i I I dv1t00839 dv20006 dV30004 39 002 U I I I I 515 516 517 518 519 dhl t 11120 11130 A 6 value of 47deg maximizes horizontal distance This matches my intuition from Exercise 411 I guessed the answerwould be between 45deg and 50deg Module08 ProjectilesAnsweredxmcd EXE mad Maddie a Piuieeiiies Part V Maximizing Hang Time American Football Goai Use ihe skiiis yed ve ieained su iaiie maximize hang inne eia pieieeiiie in spans sdeh asiaveim and shed puii ahseidie disianee is key Huwevevi iaiihe hiehewin Ameiiean ieeihaiii baii aiHimE is key Tu enahie ihe deiendeisie gei dewniieidiihe kiEkEY mdsi maximize ihe inne ihanhe baii iemains in he an Disianee is impunam iuui hm uniyiu a eenain pumii inhe baii is kickediuu Vaquot inie ihe endzene w ihe eiiense Wiii simpiy dawn ihe baii and gei a nee 2nyaids a iadehhaeh The hiehew hegins anhe kicking team s 3n yaid imei su ideaiiyihe baii wedid iandidsishyeimyaidsaw yUusi in iieni enhe appesing end and sme adaeis diam 2m7 inihis Paiiwe Wiii assume mainKiiehesseinnaieswzm i The piuiemiun speed unhe iuuihau is xed ai 251 2 Aii vesisiance and m aie negiidihie ihe dimensiunsuiiheiuuibaii eidipiacmgihe baiiimiiaiiyaipHZSD aids Whiieihis p in designaiien is eenvenieniiweWiii have in he eaiemi abuui EILH uniis becausewe eannei mix heeadse oDEsehe eannei handie Expiicii dniis an d hmh a o a m speed and g have meievrbased dniis Begin heiuw by assigning speed in he 25 the impiieii umis heie aie 1 am 25 Cupy yum odesehe aieehhem EXEYEiSE 4 DZ abuve and pasie ii heie beiuw Deni heihei eepying ihe giaph Maddiena Piuieeiiiesjnweiedwned EXB 103d Module 8 Projectiles Given phO aoy d pv0 0 m ve1h0 speed0059 velv0 speedsin9 d d pht ve1ht pvt velvt dt dt iveiha o iveiva g dt dt ph PV k1ckoff9 Odesolve tend Lvelh ml l Your solve block probably has phO set to be 0 as we assumed in Part IV To change it to 30yards you have to be a bit tricky However we don39t have to explicitly know that 30yd 27432m Instead simply change the phO statement to be pho 30y d m Note that this technique does not explicitly associate units with ph0 Indeed 30y d 27432 is In still a unite55 quantity 503 Change the other initial conditions above to suite the present situation of a football kickoff Hint You39ll need to change pv0 and veh0 504 As you did in Exercises 405 and 407 above create particular solutions below for your kickoff ODEsolve block where 6 is 35deg 40deg 45deg 50deg and 55deg 505 Modue08 26 ProjectilesAnsweredxmcd EXB 103d Module 8 Projectiles dh35 k39 k ff35d dh40 k39 k ff40d dh45 k39 k ff45d 1c 0 e 10 o e 1c 0 e dv35 g dv40 g dv45 g dhso k39 k ff50d dhss k39 k ff55d 10 o e 1c 0 e dv50 g dv55 g Copy your graph from Exercise 408 and paste it here below If everything worked out correctly you should see the ball39s flight begin on from the ground just shy of 30 meters along the xaxis the default units ofthe graph are in meters 506 dv35t dv40t 20 dv45t dv50t 10 dv55t 111350 111400 111450 thOt thSt Copy your graph from Exercise 506 above and paste it here below To get the xaxis to display in yards and not meters append the conversion 3d to each of the xaxis plots So for example y dh35t 3d would be your first plot Make the same change to the other four plots on the xaxis When you39re finished with that change the xaxis to range from 30 to 100 This range will cover the whole of the useful flight of the football the endzones are 100 yards apart This will produce the somewhat unusual situation of the yaxis being in meters but the xaxis being in yards Because we39re not really concerned with pv values and we don39t really want to bother entering in more conversion factors I say wejust leave it But to be clearto the observer enter axis labels of quotmetersquot and quotyardsquot where appropriate 507 Module08 27 ProjectilesAnsweredxmcd EXE mad Mudme a Pvmecmes meters 40 6D 1 u amxoiamuiamsmiahsumiahssmi yd yd yd yd yd yards 6U 8n 1 u amslt amo ah45o ahsum ahssm yd yd yd yd yd yards When yuu ve mm yum gvaph 5mm uuk sumethmg m m appeavs upnmaw CavemHy New yum veasumng be uw 5m Muumena Pmtecmesjdnwevedxmm EXB 103d Module 8 Projectiles 55deg seems to be optimal 6 values of 40 45 and 50 all land close to endzone Forthese three 6 values I worry that the ball would continue rolling into the endzone and result in a touchback When 6 is 35deg the distance seems safe but the ball gets there too quickly 6 55deg is a better kickoff because it takes longer forthe ball to arrive which gives the kicker39s teammates time to get downfield You may have mentioned the issue oftime in your answer above Let39s examine that more closely Copy your graph Exercise 506 above and paste it here below Change the xaxis to be simply t Adjust the xaxis so that you see the whole of the flight of the football 509 I l l l dv35t dv40t20 2150 9 dv50t 39 quotquotquotquotquotquotquotquotquotquotquot 10 x dv55t x U I I I I 0 1 2 3 4 5 t Write a couple sentences about what you see in the graph above Is this consistent with your reasoning in Exercise 508 510 The graph shows the vertical position ofthe football relative to the time since it was kicked V th increasing values of 6 the football spends more time in the air hits the ground at a largertime t This graph confirms my assertion above that a kickoff with a 6 value of 55deg takes longer to land than similar kick made with 6 35deg Module 8 Projectiles Written by Eric Leaver 2008 Martin Wilson Last modified September 8 2008 630pm Module08 29 ProjectilesAnsweredxmcd

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