TableOfContents
PartI:Physics
LawOfTheLever
SlidingandOverturning
MaximumCarSpeed
RangeContinued
EscapeVelocity
CoolingandWind-Chill
AdiabaticProcesses
DrainingaTank
Open-ChannelFlow
Wind-DrivenWaves
Sailing
HeatRadiation
MainSequenceStars
ElectricalResistance
StringsandSound
PartII:Mathematics
Cylinders
ArbitraryTriangles
Summation
StandardDeviationandError
ZipfDistribution
PartIII:Appendix
UnitConversion
UnitPrefixes
References
CopyrightandDisclaimer
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PartI:Physics
LawOfTheLever
Oftentimeswhendoingphysicswesimplysay"aforceisactingonabody"
withoutspecifyingwhichpointofthebodyitisactingon.Thisisbasically
point-massphysics.Weignorethefactthattheobjecthasacomplexthreedimensionalshapeandassumeittobeasinglepointhavingacertainmass.
Sometimesthisissufficient,othertimesweneedtogobeyondthat.Andthisis
wheretheconceptoftorquecomesin.
Let'sdefinewhatismeantbytorque.AssumeaforceF(inN)isactingonabody
atadistancer(inm)fromtheaxisofrotation.Thisdistanceiscalledthelever
arm.Takealookattheimagebelowforanexampleofsuchasetup.
Relevantfortherotationofthebodyisonlytheforcecomponentperpendicular
totheleverarm,whichwewilldenotebyF'.IfgiventheangleΦbetweenthe
forceandtheleverarm(asshownintheimage),wecaneasilycomputethe
relevantforcecomponentby:
F'=F·sin(Φ)
Forexample,ifthetotalforceisF=50NanditactsatanangleofΦ=45°to
theleverarm,onlythethecomponentF'=50N·sin(45°)≈35Nwillworkto
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rotatethebody.Soyoucanseethatsometimesitmakessensetobreakaforce
downintoitscomponents.Butthisshouldn'tbecauseforanyworries,withthe
aboveformulaitcanbedonequicklyandpainlessly.
Withthisoutoftheway,wecandefinewhattorqueisinonesimplesentence:
TorqueT(inNm)istheproductoftheleverarmrandtheforceF'acting
perpendiculartoit.Informofanequationthedefinitionlookslikethis:
T=r·F'
Inquantitativetermswecaninterprettorqueasameasureofrotationalpush.If
there'saforceactingatalargedistancefromtheaxisofrotation,therotational
pushwillbestrong.However,ifoneandthesameforceisactingverycloseto
saidaxis,wewillseehardlyanyrotation.Sowhenitcomestorotation,forceis
justonepartofthepicture.Wealsoneedtotakeintoconsiderationwherethe
forceisapplied.
Let'scomputeafewvaluesbeforegoingtotheextremelyusefullawofthelever.
--------------------------We'llhavealookatthewrenchfromtheimage.Supposethewrenchisr=0.2m
long.What'stheresultingtorquewhenapplyingaforceofF=80Natanangle
ofΦ=70°relativetotheleverarm?
Toanswerthequestion,wefirstneedtofindthecomponentoftheforce
perpendiculartotheleverarm.
F'=80N·sin(70°)≈75.18N
Nowontothetorque:
T=0.2m·75.18N≈15.04Nm
--------------------------Ifthisamountoftorqueisnotsufficienttoturnthenut,howcouldweincrease
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that?Well,wecouldincreasetheforceFandatthesametimemakesurethatit
isappliedata90°angletothewrench.Let'sassumethatasameasureoflast
resort,youapplytheforcebystandingonthewrench.Thentheforce
perpendiculartotheleverarmisjustyourgravitationalpull:
F'=F=m·g
Assumingamassofm=75kg,weget:
F'=75kg·9.81m/s²=735.75N
Withthisnotveryelegant,butcertainlyeffectivetechnique,weareableto
increasethetorqueto:
T=0.2m·735.75N=147.15Nm
Thatshoulddothetrick.Ifitdoesn't,there'sstilloneoptionleftandthatisusing
alongerwrench.Withalongerwrenchyoucanapplytheforceatagreater
distancetotheaxisofrotation.Andwithrincreased,thetorqueTisincreased
bythesamefactor.
--------------------------Asyoucansee,calculatingtorqueisnotabigdeal.Butwhat'stheuse?Thelaw
ofthelever,that'swhat.Imagineabeamsittingonafulcrum.Weapplyone
forceF'(1)=20Nontheleftsideatadistanceofr(1)=0.1mfromthefulcrum
andanotherforceF'(2)=5Nontherightsideatadistanceofr(2)=0.2m.In
whichdirection,clockwiseoranti-clockwise,willthebeammove?
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Tofindthatoutwecantakealookatthecorrespondingtorques.Thetorqueon
theleftsideis:
T(1)=0.1m·20N=2Nm
Fortherightsideweget:
T(2)=0.2·5N=1Nm
Sotherotationalpushcausedbyforce1(leftside)exceedsthatofforce2(right
side).Hence,thebeamwillturnanti-clockwise.Ifwedon'twantthattohappen
andinsteadwanttoachieveequilibrium,weneedtoincreaseforce2toF'(2)=
10N.Inthiscasethetorqueswouldbeequalandtheoppositerotationalpushes
wouldcanceleachother.Soingeneral,thisequationneedstobesatisfiedto
achieveastateofequilibrium:
r(1)·F'(1)=r(2)·F'(2)
Thisisthelawoftheleverinitssimplestform.Let'sseehowandwherewecan
applyit.
--------------------------Agreatexamplefortheusefulnessofthelawoftheleverisprovidedbycranes.
Ononeside,let'ssetr(1)=30m,itliftsobjects.Sincewedon'twantittofall
over,westabilizethecraneusinga20,000kgconcreteblockatadistanceofr(2)
=2mfromtheaxis.Whatisthemaximummasswecanliftwiththiscrane?
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Firstweneedtocomputethegravitationalforceoftheconcreteblock.
F'(2)=20,000kg·9.81m/s²=196,200N
Nowwecanusethelawofthelevertofindoutwhatmaximumforcewecan
applyontheoppositesite:
r(1)·F'(1)=r(2)·F'(2)
30m·F'(1)=2m·196,200N
30m·F'(1)=392,400Nm
Divideby30m:
F'(1)=13,080N
Aslongaswedon'texceedthis,thetorquecausedbytheconcreteblockwill
exceedthatoftheliftedobjectandthecranewillnotfallover.Themaximum
masswecanliftisnoweasytofind.Weusetheformulaforthegravitational
forceonemoretime:
13,080N=m·9.81m/s²
Divideby9.81:
m≈1330kg
Toliftevenheavierobjects,weneedtouseeitheraheavierconcreteblockorput
itatalargerdistancefromtheaxis.
--------------------------Thelawofthelevershowswhywecaninterpretaleverasatooltoamplify
forces.SupposeyouwantuseaforceofF'(1)=100Ntoliftaheavyobjectwith
thegravitationalpullF'(2)=2000N.Notpossibleyousay?Withaleveryou
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candothisbyapplyingthesmallerforceatalargerdistancetotheaxisandthe
largerforceatashorterdistance.
Supposetheheavyobjectsitsatadistancer(2)=0.1mtotheaxis.Atwhatwhat
distancer(1)shouldweapplythe100Ntobeabletoliftit?Wecanusethelaw
ofthelevertofindtheminimumdistancerequired.
r(1)·100N=0.1m·2000N
r(1)·100N=200Nm
r(1)=2m
Soaslongasweapplytheforceatadistanceofover2m,wecanlifttheobject.
Weeffectivelyamplifiedtheforcebyafactorof20.Scientistsbelievethatthe
principleofforceamplificationusingleverswasalreadyusedbytheEgyptians
tobuildthepyramids.Givenalongenoughlever,wecouldliftbasically
anythingevenwithamoderateforce.
--------------------------Inthenextsectionwewillseeanotherinterestingapplicationofthetorque
concept,sowe'renotdonewithitjustyet.Itwillleadtoaveryneatformulaall
carmakersmustknow.
SlidingandOverturning
Inthissectionwewilltakealookatcarperformanceincurves.Ofcentral
importanceforourconsiderationsisthecentrifugalforce.Wheneverabodyis
movinginacurvedpath,thisforcecomesintoplay.Youprobablyfeltitmany
timesinyourcar.Itistheforcethattriestopushyououtofacurveasyougo
throughit.
ThecentrifugalforceC(inN)dependsonthreefactors:thevelocityv(inm/s)of
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thecar,itsmassm(inkg)andtheradiusr(inm)ofthecurve.Giventhese
quantities,wecaneasilycomputethecentrifugalforceusingthisformula:
C=m·v²/r
Notethequadraticdependenceonspeed.Ifyoudoublethecar'sspeed,the
centrifugalforcequadruples.Withthisforceacting,theremustbeacounterforcetocancelitforthecarnottoslide.Thisforceisprovidedbythesideways
frictionofthetires.ThefrictionalforceF(inN)canbecalculatedfromtheso
calledcoefficientoffrictionμ(dimensionless),thecarmassmandthe
gravitationalaccelerationg(inm/s²).
F=μ·m·g
Thecoefficientoffrictiondependsmainlyontheroadtypeandcondition.On
dryasphaltwecansetμ≈0.8,onwetasphaltμ≈0.6,onsnowμ≈0.2andon
iceμ≈0.1.Atlowspeedsthefrictionalforceexceedsthecentrifugalforceand
thecarwillbeabletogothroughthecurvewithoutanyproblems.However,as
weincreasethevelocity,sodoesthecentrifugalforceandatacertaincritical
velocitytheforcescanceleachotherout.Anyincreaseinspeedfromthispoint
onwillresultinthecarsliding.
Wecancomputethecriticalspeeds(inm/s)byequatingtheexpressionsforthe
forces:
m·s²/r=μ·m·g
s=sqrt(μ·r·g)
Thisisthespeedatwhichthecarbeginstoslide.Notethatthere'sno
dependenceonmassanymore.Sinceboththecentrifugalaswellasthefrictional
forcegrowproportionallytothecar'smass,itdoesn'tplayaroleindetermining
thecriticalspeedforsliding.Allthat'sleftintermsofvariablesisthecoefficient
offriction(lowerfriction,lowercriticalspeed)andtheradiusofthecurve
(smallerradius,morenarrowcurve,smallercriticalspeed).
However,slidingisnottheonlyproblemthatcanoccurincurves.Undercertain
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circumstancesacarcanalsooverturn.Againthecentrifugalforceistheculprit.
Assumingthecenterofgravity(inshort:CG)ofthecarisataheightofh(inm),
thecentrifugalforcewillproduceatorqueTactingtooverturnthecar:
T=h·C=m·v²·h/r
Ontheotherhand,there'stheweightofthecargivingrisetoanopposingtorque
T'thatgrowswiththewidthw(inm)andmassmofthecar:
T'=0.5·m·g·w
Atlowspeeds,thetorquecausedbythecentrifugalforcewillbelowerthanthe
onecausedbythegravitationalpull.Butatacertaincriticalspeedo(inm/s),the
torqueswillcanceleachotherandanyfurtherincreaseinspeedwillresultinthe
caroverturning.Equatingtheaboveexpressions,weget:
m·o²·h/r=0.5·m·g·w
o=sqrt(0.5·r·g·w/h)
Asidefromthecoefficientoffriction,thedeterminingfactorhereistheratioof
widthtoheight.Thelargeritis,theharderitwillbeforthecentrifugalforceto
overturnthecar.Thisiswhyloweringacarwhenintendingtogofastmakes
sense.IfyoulowertheCGwhilekeepingthewidththesame,theratiow/h,and
thusthecriticalspeedforoverturning,willincrease.
Let'slookatsomeexamplesbeforedrawingafinalconclusionfromthesetruly
greatformulas.
--------------------------Accordingtocaranddriver.comthecenterofgravityofa2014BMW435iish=
0.5mabovetheground.Thewidthofthecarisaboutw=1.8m.Calculatethe
criticalspeedforslidingandoverturninginacurveofradiusr=300monadry
asphaltroad(μ≈0.8).
Nothingtodobuttoapplytheformulas:
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s=sqrt(0.8·300m·9.81m/s²)
s≈49m/s≈175km/h≈108mph
Sowithnormaldrivingbehavioryoucertainlywon'tgetanywherenearsliding.
Butnotethatsuddensteeringinacurvecancausetheradiusoftheyourcar's
pathtobeconsiderablylowerthantheactualcurveradius.
Ontothecriticaloverturningspeed:
o=sqrt(0.5·300m·9.81m/s²·3.6)
o≈73m/s≈262km/h≈162mph
NotevenMichaelSchumachercouldbringthiscartooverturn.
--------------------------Howwouldthecriticalspeedschangeifwedrovethe2014BMW435ithrough
thesamecurveonanicyroad?Inthiscasethecoefficientisconsiderablylower
(μ≈0.1).Forthecriticalslidingspeedweget:
s=sqrt(0.1·300m·9.81m/s²)
s≈17m/s≈62km/h≈38mph
Soeventhissweetsportcarisindangerofslidingrelativelyquicklyunderthese
conditions.Whatabouttheoverturningspeed?Well,ithasnothingtodowiththe
frictionofthetires,soitwillstillbeat73m/s.
--------------------------Slidingdoesnotalwaysendinanaccident.Whilesliding,acarwillusually
increasetheradiusofitspathandatthesametimedecreaseitsspeed.Bothof
theseeffectslowerthecentrifugalforceandthuscanbringthecarbackintoits
normaldrivingmodebeforeanaccidentoccurs.Ontheotherhand,onceacar
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startsoverturning,there'snogoingback.
Thisiswhywhendesigningacar,youwantittoslidebeforeitoverturns,sothe
criticalslidingspeedshouldbelowerthanthecriticalspeedforoverturning.
Usingtheformulas,wecanfindacriticalCGheightH(inm)byequatingthe
criticalspeeds:
s=o
H=0.5·w/μ
AslongastheCGisbelowthisheight,thecarwillstartslidingbefore
overturning.Sincewesawintheexamplethatthedifferencebetweenthecritical
speedsincreasesastheroadconditionsworsen,itissufficienttomakesurethat
theaboveequationholdstruefordryasphalt.Settingμ≈0.8,weget:
H≈0.63·w
Soasaruleofthumb,theheightoftheCGshouldbekeptbelow60%ofthe
width.OfcourseaccuratelydeterminingtheCGnotsomethingyoucandowith
aruler.Butitcanbeestimatedbymeasuringtheweightsoneachtireina
horizontalpositionandtheweightsonthefronttiresafterraisingthefront.With
thisdone,youcaninputthesemeasurementsintotheonlineCGcalculator
featuredonrobrobinette.com.
MaximumCarSpeed
Howdoyoudeterminethemaximumpossiblespeedyourcarcango?Well,one
ratherstraight-forwardoptionistojustgetintoyourcar,goontheAutobahnand
pushdownthepedaluntiltheneedlestopsmoving.Theproblemwiththisoption
isthatthere'snotalwaysanAutobahnnearby.Soweneedtofindanotherway.
Luckily,physicscanhelpusouthere.Youprobablyknowthatwheneverabody
ismovingatconstantspeed,theremustbeabalanceofforcesinplay.Theforce
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thatisaimingtoacceleratetheobjectisexactlybalancedbytheforcethatwants
todecelerateit.Ourfirstjobistofindoutwhatforceswearedealingwith.
Obviouscandidatesfortheretardingforcesaregroundfrictionandairresistance.
However,inourcaselookingatthelatterissufficientsinceathighspeeds,air
resistancebecomesthedominatingfactor.Thismakesthingsconsiderablyeasier
forus.Sohowcanwecalculateairresistance?
Tocomputeairresistanceweneedtoknowseveralinputs.Oneoftheseistheair
densityD(inkg/m³),whichatsealevelhasthevalueD=1.25kg/m³.Wealso
needtoknowtheprojectedareaA(inm²)ofthecar,whichisjusttheproductof
widthtimesheight.Ofcoursethere'salsothedependenceonthevelocityv(in
m/s)relativetotheair.Theformulaforthedragforceis:
F=0.5·c·D·A·v²
withc(dimensionless)beingthedragcoefficient.Thisistheonequantityinthis
formulathatistoughtodetermine.Youprobablydon'tknowthisvalueforyour
carandthere'sagoodchanceyouwillneverfinditoutevenifyoutry.In
general,youwanttohavethisvalueaslowaspossible.
Onecomodder.comyoucanfindatableofdragcoefficientsformanycommon
moderncarmodels.Excludingprototypemodels,thedragcoefficientinthislist
rangesbetweenc=0.25fortheHondaInsighttoc=0.58fortheJeepWrangler
TJSoftTop.Theaveragevalueisc=0.33.Infirstapproximationyoucan
estimateyourcar'sdragcoefficientbyplacingitinthisrangedependingonhow
streamlineditlookscomparedtotheaveragecar.
Withtheequation:powerequalsforcetimesspeed,wecanusetheabove
formulatofindouthowmuchpower(inW)weneedtoprovidetocounterthe
airresistanceatacertainspeed:
P=F·v=0.5·c·D·A·v³
Ofcoursewecanalsoreversethisequation.Giventhatourcarisabletoprovide
acertainamountofpowerP,thisisthemaximumspeedvwecanachieve:
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v=(2·P/(c·D·A))1/3
Fromtheformulawecanseethatthetopspeedgrowswiththethirdrootofthe
car'spower,meaningthatwhenweincreasethepowereightfold,themaximum
speeddoubles.Soevenaslightincreaseintopspeedhastobeboughtwitha
significantincreaseinenergyoutput.
Notethewehavetoinputthepowerinthestandardphysicalunitwattrather
thantheoftenusedunithorsepower.Luckilytheconversionisveryeasy,just
multiplyhorsepowerwith746togettowatt.
Let'sputtheformulatothetest.
--------------------------IdriveatenyearoldMercedesC180Compressor.AccordingtheMercedes-Benz
homepage,itsdragcoefficientisc=0.29anditspowerP=143HP≈106,680
W.Itswidthandheightisw=1.77mandh=1.45mrespectively.Whatisthe
maximumpossiblespeed?
Firstweneedtheprojectedareaofthecar:
A=1.77m·1.45m≈2.57m²
Nowwecanusetheformula:
v=(2·106,680/(0.29·1.25·2.57))1/3
v≈61.2m/s≈220.3km/h≈136.6mph
FrommyexperienceontheAutobahn,thisseemstobeveryrealistic.Youcan
reach200Km/hquitewell,buttheaccelerationisalreadynoticeablylowerat
thispoint.
IfyouevergetthechancetovisitGermany,makesuretorentaridiculouslyfast
sportscar(youcanrentaPorsche911Carreraforaslittleas200$perday)
andfindanicesectionontheAutobahnwithunlimitedspeed.Butremember:
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unlessyou'reovertaking,alwaysusetherightlane.Theleftlanesarereserved
forovertaking.Neverovertakeontherightside,nobodywillexpectyouthere.
Andmakesuretochecktherear-viewmirroroften.Youmightthinkyou'regoing
fast,butthere'salwayssomeonegoingevenfaster.Letthempass.Lastbutnot
least,stayfocusedandkeepyoureyesontheroad.Trafficjamscanappearout
ofnowhereandyoudon'twanttoendupinthebackofatruckatthesespeeds.
--------------------------ThefastestproductioncaratthepresenttimeistheBugattiVeyronSuperSport.
Ishasadragcoefficientofc=0.35,widthw=2m,heighth=1.19mand
powerP=1200HP=895,200W.Let'scalculateitsmaximumpossiblespeed:
v=(2·895,200/(0.35·1.25·2·1.19))1/3
v≈119.8m/s≈431.3km/h≈267.4mph
Doesthisseemunreasonablyhigh?Itdoes.Butthecarhasactuallybeen
recordedgoing431Km/h,sowearerightontarget.Ifyou'dliketopurchasethis
car,makesureyouhave4,000,000$inyourbankaccount.
--------------------------Toconcludethissection,let'sdoaveryroughestimateonhowlongittakesacar
toreachitsmaximumspeedv.Atthisspeedthecar'skineticenergyisE=0.5·
m·v².IfthecarcontinuouslyproducesthemaximumpowerP,thisishowlong
ittakestoprovidetheaboveamountofenergy:
t=0.5·m·v²/P
withtbeinginseconds.Notethatsincethespeedgrowswiththethirdrootofthe
power,thetimeittakestoreachmaximumspeedshouldbeinversely
proportionaltothethirdrootofthepower.Soifyouincreasethepower
eightfold,itwilltakeonlyhalfaslongtogototopspeed.Alsonotethatherethe
massofthecarisalsoafactor.
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WedeterminedthatfortheMercedesC180Compressorthetopspeedisv=61.2
m/s.GiventhepowerP=143HP≈106,680Wandthemassm=1,600kg,
estimatehowlongittakestoreachthemaximumspeed.
Weapplytheformula:
t=0.5·1,600·61.2²/106,680
t≈28.1s
Thisisofcourseassumingthatwecouldconstantlyproducethemaximumpower
(whichisnotthecase).Butthevalueneverthelessseemsquiterealistic
consideringthecarcangoto100Km/h(abouthalfthetopspeed)in9seconds
andtheaccelerationwouldbesignificantlyslowerinthesecondhalf.
--------------------------WhatabouttheBugattiVeyronSuperSport?Howlongdoesittaketogetitupto
itstheoreticaltopspeedofv=119.8m/s?Themassofthecarism=1900kg,
henceweget:
t=0.5·1,800·119.8²/895,200
t≈15.2s
SonotonlydowereachtwicethetopspeedoftheMercedesC180Compressor,
wegettheremorethan10secondsearlier.Bytheway,theaverageacceleration
duringthiswouldbe:
a=v/t=119.8/15.2≈7.9m/s²
whichiscomparabletowhatwe'dexperienceinfreefall.Andthisisjustthe
average.Itactuallygoesfrom0to100Km/h(27.8m/s)inonly2.2seconds,
whichtranslatesintoanaccelerationof:
a=27.8/2.2≈12.6m/s²
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--------------------------Trythesecalculationsforyourowncar.Ifyou'reluckyyou'llfinditsdrag
coefficientonline.Ifnot,justestimateitfromrangeprovidedinthissection.You
shouldn'tbeoffbytoomuch.
RangeContinued
Inthefirstvolumeofthisbookwehadalookathowfaranobjectthrownfrom
thegroundatacertainvelocityandanglegets.We'lltakeyetanotherlookat
range,butthistimeforadifferentsetup.Assumewethrowanobject
horizontallyataspeedofv(inm/s)fromtheheighth(inm)abovetheground.
Whathorizontaldistancewillitcoverbeforehittingtheground?
Well,inx-direction(whichisthehorizontalaxis),theobjectwillkeepon
movingatthevelocityvwhenneglectingairresistance:
x=v·t
Iny-directionitissubjecttotheconstantgravitationalaccelerationg=9.81
m/s².Sotheheightwilldecreaseaccordingtothisformula:
y=h-0.5·g·t²
Wecangetridoftimebyusingt=x/vandpluggingthatintothesecond
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formula.
y=h-0.5·g·x²/v²
Thisisthecoordinateformoftheparabolicpaththeobjecttakes.Wewantto
knowatwhatvaluex=Ritimpactstheground(y=0).Withtheaboveequation
thisiseasytodo.Weset:
0=h-0.5·g·R²/v²
Andsolvefortherange:
R=v·sqrt(2·h/g)
Injustafewstepswederivedagreatformulathatallowsustocomputethe
rangefromtheinitialvelocityoftheobjectandtheinitialheight.Let'stalkabout
thenatureofthedependencies.Therangegrowslinearlywithvelocity.Ifthe
velocitydoubles,sodoestherange.Asfortheheight,quadruplingitwillalso
leadtoadoublinginrange.
Beforewegoontotheexamples,let'stakeacloserlookattheimpactitself.We
mightbeinterestedinknowingatwhatspeedandatwhatangletheobject
impactstheground.Howcanwecomputethesequantities?Againthekeyisto
analyzethex-andy-directionseparately.
Asstated,thespeedinx-directionwillremainequaltotheinitialspeed,sov(x)
=v.Andaccordingtotheformula:speedequalsaccelerationmultipliedbytime,
thespeediny-directionaftertimetwillbe:v(y)=g·t.
Let'sgetridoft.Howlongdoesittaketheobjecttohittheground?Thisiseasy.
Wealreadyknowthatittakestheobjectthetimet=x/vtocoverthehorizontal
distancex.Sothetimetoimpactisjust:
t=R/v=sqrt(2·h/g)
Nowweareabletocomputethespeediny-directionatthetimeofimpact.
Insertingthisexpressionfortintov(y)=g·tleadsto:
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v(y)=sqrt(2·g·h)
Sincethex-andy-velocitiesareatarightangletoeachother(seeimagebelow),
thetotalspeedcanbecomputedusingPythagoras'theorem:
w=sqrt(v(x)²+v(y)²)
Allthat'sleftisinsertingtheappropriateexpressionsandweareleftwitha
handyformulaforcalculatingthetotalimpactspeedw(inm/s).
w=sqrt(v²+2·g·h)
Whatabouttheangleofimpactθ?Forthatwewillneedsometrigonometry.
Rememberthatinthetangentformulaweusetheratio:oppositetoadjacent.In
ourcasetheoppositeisthespeediny-directionandtheadjacentthespeedinxdirection.Hencewecanwrite:
tanθ=sqrt(2·g·h)/v
Notethatthevelocityvisnotpartofthesquareroot.Therelationshipis
somewhatcomplex,butingeneralwecanconcludethatthelargertheratioof
initialheighttoinitialspeedis,thesteepertheimpactwillbe.
Afterallthiswork,let'sgototheexamples.
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Aballrollsovertheedgeofatableofheighth=1mataspeedofv=3m/s.At
whathorizontaldistancetotheedgeofthetablewillitland?Atwhatspeedand
angledoesitimpacttheground?Firstlet'stakealookthetherange:
R=3m/s·sqrt(2·1m/9.81m/s²)
R≈1.35m
Soitwilllandatadistanceof1.35mfromthetable.Nowlet'sturntotheimpact
speed:
w=sqrt((3m/s)²+2·9.81m/s²·1m)
w≈5.35m/s
Sotheball'sspeedincreasessignificantly(byabout80%)duringthefall.The
angleofimpactis:
tanθ=sqrt(2·9.81m/s²·1m)/3m/s
tanθ≈1.48
Usingtheinversefunctionweget:
θ≈arctan(1.48)≈55.9°
--------------------------Weholdagardenhosehorizontallyataheightofh=0.5mandturnonthe
water.ThewaterjethitsthegroundatadistanceofR=1.5m.Atwhatspeedis
thewaterexitingthehose?Fromtherangeformulawecansetupanequation
fortheinitialspeedvusingthegivenvalues(we'llignoretheunitsforsakeof
simplicity):
1.5=v·sqrt(2·0.5/9.81)
4≈v·0.32
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Divideby0.32:
v≈12.5m/s
--------------------------Justafterjumpingoutofthevehicle,theowner'sbrandnewcargoesoveracliff
ofheighth=15m.Thepolicedeterminefromtheimpactcraterthattheangleof
impactwasaboutθ=25°.Theownerinsistthathewasbelowthe100Km/h
limitatthetimeoftheaccident.Ishetellingthetruth?
Let'scalculatetheinitialspeedvandcomparetheresulttothelimit.Fromthe
formulafortheangleofimpactwecansetupthisequation:
tan(25°)=sqrt(2·9.81·15)/v
0.47≈17.16/v
Multiplybyv:
0.47·v≈17.16
Nowdivideby0.47:
v≈36.5m/s
Sothisisthespeedatwhichthecarwentoverthecliff.Howdoesthatcompare
tothespeedlimit?Well,36.5m/sare131.4km/h,sohewassignificantlyover
thelimit.
--------------------------ToconcludethissectionI'llgiveyouanotherrelevantformulayouwillhardly
findinanyotherphysicsbook.Itispossibletocomputethetotaldistance
coveredbytheobject(=thelengthoftheparabolicarc).Sincethederivation
requiresatripdeepintotherealmsofintegralcalculus,I'llskipitentirelyandgo
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righttotheformulaforthearclength.Braceyourself:
s=sqrt(R²+4·h²)+(0.5·R²/h)·arcsinh(2·h/R)
witharcsinhbeingtheinversefunctionofthehyperbolicsinefunction(your
calculatorshouldknowthisone).Granted,thisisamonster.Butitworks.And
onceweknowthetotaldistancecovered,wecangoontocalculatetheaverage
speed<v>:
<v>=s/t=s·v/R
Let'sgiveitatry.
--------------------------Let'sgobacktothesecondexamplewherewehadwaterflowingoutofahoseat
aheightofh=0.5m.TherangewasR=1.5m.Whatisthelengthofthewater
jetfromhosetoground?Weapplythemonsterformula:
s=sqrt(3.25)+2.25·arcsinh(0.67)
s≈3.21m
Onceyoufigureouthowwheretheinversehyperbolicsinefunctionisonyour
calculator,thisisn'tactuallysobad.
--------------------------Asyoucansee,eveninsuchasimpleandbasicsituationasfreefallthere'salot
ofinterestingphysicsandevensomehard-coremathinvolved.Anditgetseven
worseonceyouwanttoincludeairresistance(don'ttrythisathome!).
EscapeVelocity
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Ifyouthrowanobjectupwards,gravitywillforceittocomebackdownquickly.
Afterall,whatgoesupmustcomedown.Unlessit'sgoingfasterthantheescape
velocity,inwhichcaseitindeedwillbeabletoescapeaplanet'sormoon's
gravitationalprison.Inthissectionwewillderiveandapplyaformulathat
allowsustocalculatesaidescapevelocity(undertheassumptionoftherebeing
noatmospherepresent).
Toderiveaformulaweneedtothinkabouthowmuchenergyisneededtomove
anobjectfromonepointtoanotherinagravitationalfield.Mostpeoplehere
wouldsay(ifthey'dsayanything):justusethisformulaforthepotentialenergy.
E(pot)=m·g·h
Fromthiswecancomputehowmuchenergyisneededtoelevateanobjectof
massm(inkg)byaheightofh(inm).Problemsolved?Yesandno.This
formulaworksaslongasthechangeinheightisnottoogreat.Forpeople,
elevators,planesandevenhigh-altitudeballoonsitworksjustfine.Butfor
satellitesandspaceships(aswellasforcalculatingtheescapevelocity)it
unfortunatelyfails.Soweneedsomethingmoreprecise.
Luckily,withNewton'sformulaforthegravitationalforceandsomecalculus,we
canderivethisformulathatallowsustofindouttheenergyneededtobringan
objectfromthesurfaceofacelestialbodytoacertainheight:
E(pot)=G·M·m·(1/R-1/(R+h))
withM(inkg)beingthemassofthecelestialbody,R(inm)itsradiusandGthe
gravitationalconstant.Itsvalueis:G≈6.67·10-11Nm²kg²throughoutthe
universeandforallbodies.Itcan'tgetmoreconstantthanthat,Isuppose.
Tocalculatehowmuchenergyisneededtocompletelyfreeanobjectofa
gravitationalfield,welettheheightgotoinfinity,whichweexpress
symbolicallyassuch:h→∞.Thesecondtermintheaboveformulathenjust
disappears,leavingthismuchnicerexpression:
E(pot,∞)=G·M·m/R
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Thisistheenergyweneedtoprovideanobjectwithsothatitcanleaveaplanet
forgood(assumingnoatmospherepresent).Howdoesthistranslateintoa
velocity?Well,weknowthatanobjectwithvelocityv(inm/s)hasthekinetic
energyE(kin)=0.5·m·v².Sowecanfindtheescapevelocitybyequatingthe
kineticenergywiththepotentialenergy:
E(kin)=E(pot,∞)
0.5·m·v²=G·M·m/R
Solvingforvweget:
v=sqrt(2·G·M/R)
Thisistheescapevelocity.Notethatitdoesnotdependonthemassoftheobject
thatistobeshotintospaceinanyway.Alltheparametersintheformulareferto
thecelestialbodyonly.Anditallcomesdowntotheratioofmasstoradius.The
greaterthisis,thehighertheescapevelocitywillbe.
Let'sderiveanalternateformtotheaboveequationthatwillallowustogo
aroundhavingtodealwithverylow(G)orveryhigh(M)numbers.From
Newton'slawofgravitationwecanquicklyshowthatacelestialbody's
gravitationalaccelerationatthesurface(denotedbygandwell-knownformany
bodieswithinoursolarsystem)is:
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