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# asp.net mvc 4 and the web api pdf free download : Scan multiple pages into one pdf software Library dll winforms asp.net .net web forms mechanics6-part1313

2|OneDimensionalMotion
57
Withthesameinitialpositionasintheprecedingcase,
x
(0)=0,thisdetermines
D
=
mv
0
=b
.
OR,you can do the same calculation more eciently by usingdenite integrals toapply the
initialconditionsdirectly. Whenthetwolimitsonanintegralareequal,sothattheintegrationinterval
is zero,thentheintegraliszero. . That’ssuchasimpleideathatyoumaynotimmediatelyseewhat
gooditis.Repeattheprecedinglinesasdeniteintegrals:
m
dv
x
dt
bv
x
!
Z
v
x
v
0
dv
x
v
x
Z
t
0
b
m
dt
! ln
v
x
ln
v
0
b
m
t
!
v
x
(
t
)=
v
0
e
bt=m
(2
:
15)
When
t
=0the
dt
integralvanishes,andthisinturnrequiresthe
v
x
integraltovanish.Thathappens
ifthetwolimitsthereareequal,andyouhavethenappliedtheinitialconditions,
v
x
(0)=
v
0
. Next,the
integraltoget
x
,applyingthelimitsthesameway:
v
x
=
v
0
e
bt=m
=
dx
dt
!
Z
x
0
dx
=
Z
t
0
v
0
e
bt=m
dt
!
x
(
t
)=
m
b
v
0
e
bt=m
+
mv
0
b
(2
:
16)
Example
(c)Frictionproportionaltospeedsquared:
m
dv
x
dt
bv
2
x
!
Z
v
x
v
0
dv
x
v
2
x
Z
t
0
b
m
dt

1
v
x
+
1
v
0
b
m
t
(2
:
17)
Again,separationofvariableshandledthisequation. This
b
isofcoursedierentfromtheonebefore.
Itdoesn’tevenhavethesamedimensions.Nowtoget
x
,solvefor
v
x
,thenintegrate.
v
x
=
dx
dt
=
v
0
1+
bv
0
m
t
!
Z
x
0
dx
=
Z
t
0
v
0
1+
bv
0
m
t
dt
=
m
b
Z
t
0
bv
0
=m
1+
bv
0
m
t
dt
!
x
(
t
)=
m
b
ln
1+
bv
0
m
t
(2
:
18)
Inallthreecases,starttheanalysisatthestart. Howdothesesolutionsbehaveforsmall(notzero)
time? InEq. . (2.16) for
bv
x
itappears that
x
(
t
) blowsup as
b
! 0. . Doesit? Thepowerseries
expansionofEq.(0.1)(a)iswhatyouneedforthis.
For
F
x
bv
x
,
e
x
=1+
x
+
x
2
=
2!+
x
3
=
3!+
;
soEqs.(2.16)are
v
x
=
v
0
e
bt=m
=
v
0
bt
m
+
1
2
b
2
t
2
m
2

and
x
(
t
)=
m
b
v
0
e
bt=m
+
mv
0
b
m
b
v
0
bt
m
+
1
2
b
2
t
2
m
2

+
mv
0
b
=
v
0
t
bv
0
2
m
t
2
+
The
t
2
terminthisequationfor
x
showsthefamiliarform
a
x
t
2
=
2,soyoucanimmediatelyrecognize
thattheinitialacceleration
a
x
=
F
x
=m
bv
0
=m
,andthatsaysthattheinitialforceis
bv
0
,exactly
asitshouldbe. Theapparentproblemas
b
!0inEq.(2.16)isnowgone. Theseriesexpansionnot
onlyshows thatthissingularitydoesnothappen,itshowsthatthesmalltimebehavior(through
t
2
)
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2|OneDimensionalMotion
58
time
:::
)
Equations(2.18)forthe
bv
2
x
casecallfortwootherseriesfromEqs.(0.1),thegeometricseries
(h)andthelogarithm(d):
1
=
(1+
x
)=1
x
+
x
2
x
3
+
and
ln(1+
x
)=
x
x
2
=
2+
x
3
=
3+
v
x
=
v
0
1+
bv
0
m
t
=
v
0
bv
0
m
t
+
b
2
v
2
0
m
2
t
2

x
=
m
b
ln
1+
bv
0
m
t
=
m
b
bv
0
m
t
b
2
v
2
0
2
m
2
t
2

=
v
0
t
bv
2
0
2
m
t
2
+
The
x
equationshows that the frictional forcestarts outas
bv
2
0
,sothe
t
2
term showsthe initial
acceleration
a
x
=
F
x
=m
,againbecause itisin theform
a
x
t
2
=
2. Also,
x
really doesn’t blowupas
b
!0.
x
v
x
x
v
x
x
v
x
t
(a)
(b)
(c)
Fig.2.2
Whathappensforlargetime?
(a)Equations(2.12)forconstantfrictionshowthatthevelocityreacheszeroinanitetimeand
thepositiongoestoanitelimitatthattime.
(b)Equations(2.16) for
F
x
bv
x
behavedierently. Thevelocity y neverreaches zero,but
evenastimegoestoinnitythepositionapproachesanitevalue. Itmovesnofartherthan
mv
0
=b
andthatisthehorizontalasymptotefor
x
inthegraph.
(c)Equations(2.18)for
bv
2
x
againhavethepropertythatthevelocityneverequalszero,but
forthiscasethepositionfunctionalsogoestoinnityastimeincreases. The
x
-graphkeepsgoingup,
thoughevermoreslowlyastimeapproachesinnity.
Whythedierence? For(a),as
v
x
!0thefrictionalforcedoesnotgotozeroanditmaintains
itsslowingeectallthewaytotheend. Ifyoubrakeacartoastopthisway,youfeelanabrupthalt,
withalargejerkattheendastheaccelerationdropsfrom
F
fr
=m
tozeroinaninstant(jerkbeing
thetime-derivativeofacceleration,andthisequation(2.12)woulddescribeaninnitejerk).
For(b),thefrictiondoesgotozeroasthemasscomestoahalt,andmakesthebrakingmore
gentle. Thevelocity y approacheszerofastenoughthattheintegralfor
x
converges;
e
t
goestozero
very fast.
For(c),asthevelocity dropstowardzerothefrictiondropsreally fast,andastimegoesonit
becomes so small thatit isn’t even capableofbringingthemass toahalt in anitedistance. . But
notice that it takes along time to get very far. . Here e thevelocity approacheszero as 1
=t
,andthe
correspondingintegralfor
x
R
dt=t
isdivergent.
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2|OneDimensionalMotion
59
Don’tskipthisanalysis!
Why spend all thateort to checkand toanalyzethe solution
toafairlysimplesetofproblems? Thesearethetoolsthatyou
will use on every problem that you solve in the future. . If f you
everwanttopublishtheresultsofsomeof yourwork,wouldn’t
itbeniceifthey’recorrect? Wouldn’tyouliketondyourown
mistakesbefore they’reinprint? Ifyoudecidetobecomeasur-
geon,don’tyou wantto be sureyou’reremoving the rightleg?
atleastthat’s the casewith good intuition. . Mostpeople’sintuitionaboutmostthings s comesfrom
casual and unconsideredexperiences. . Ablackcatpasses s in frontofmeandIthen slipandskinmy
knee,soblackcatsareunlucky. Inoticethatwhenabookslidesacrossthetableitsooncomestoa
halt,soforceisalwaysrequiredtosustainmotion.
Howdoyouacquiregoodintuition? Justsolvingaproblemisjustthestart;analyzinghowthe
resultbehavesisthenextstep. Thesubjectofmechanicsisaparticularlygoodoneonwhichtohone
theseskills because you already have ideasabouthowthings move. . Youneedtotiethose e ideas to
describingtherealworld.Intheprocessyouwillprobablyndthatsomeofyourpicturesofrealitywill
When you’veanalyzed the results of yourcalculationsoften enoughyoucanusetheintuition
thatyouranalysisprovidestosolveproblemswithoutsolvingthem! Whatwillhappenifthefrictional
forceis
F
x
F
0
e
v=v
1
(2
:
19)
andthemass
m
startswithvelocity
v
0
asintheprecedingexamples? Youcansolvetheproblemusing
thesolution.
Forlarge
v
thebrakingforceapproachestheconstant
F
0
,sointhatdomainthemotionwilllook
like example(a) above. . Forsmall
v
(whatmeanssmall?) the e behaviorof
F
x
follows from aseries
expansion.
forsmall
v
F
x
F
0
1 [1
v
v
1
+]

F
0
v
v
1
F
x
F
0
v
1
Fig.2.3
andtherstnon-vanishingtermofthislookslikeexample(b).
Putthesetogetherandthey saythatiftheinitialvelocityis wellabove
v
1
,theaccelerationis
constantandthevelocity andpositionlooklikeequations(2.12),with the velocitydroppinglinearly.
Afterthevelocityhasdroppedtotheneighborhoodof
v
1
orso,thebrakingismoreliketheequations
(2.16),sothatthoughthevelocityneverexactlygoestozero,thetotaldistancetravelledwillstillbe
nite.
2.3SolvingF=ma: F(x)
I’velookedatthecasesforwhichtheforceisafunctionof
t
aloneandof
v
x
alone,nowforthecase
thatit’safunctionof
x
alone.
F
x
(
x
)=
m
d
2
x
dt
2
=
m
dv
x
dt
(2
:
20)
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2|OneDimensionalMotion
60
Youcan’tintegratethiswithrespectto
t
because
x
isn’taconstant,makingitimpossibletocarryout
the
R
F
x
(
x
)
dt
untilafter youknowthesolutionfor
x
(
t
x
,
whywouldyouwanttointegrateit
dt
anyway?Youcanhoweverintegratetheleftsideof(2.20)with
respectto
x
,andwiththeproperchangeofvariablesyoucandothesameontheright.
Z
F
x
(
x
)
dx
=
Z
m
dv
x
dt
dx
=
Z
mdv
x
dx
dt
=
Z
mv
x
dv
x
=
1
2
mv
2
x
(2
:
21)
The integral l of
F
x
on the left denes (minus) ) the e potential l energy. . Why y minus? ? It t makes later
manipulationeasiertointerpret. Callitsconstantofintegration
E
,then
F
x
dU
dx
;
so
U
(
x
)+
E
=
1
2
mv
2
x
;
or
1
2
mv
2
x
+
U
(
x
)=
K
+
U
=
E
(2
:
22)
This constant of integration
E
is the total mechanical l energy, , kinetic plus s potential, , and d the e nal
equationisarrangedsothattheintegrationconstantisontherightandeverythingelseontheleft. Itis
intheformofaconservationlaw.Something,
K
+
U
,remainsconstantthroughoutthemotion. This
iswhythere’saminussignindening
U
twoterms. Lookbackatsection1.3foramoredetailedpresentationofthis.
Thisis aconservation law,conservation ofmechanicalenergy,becauseitsays thatthe same
mathematicalexpressionevaluatedattwodierenttimesisguaranteedtogivethesamevalue.
mv
2
=
2
isthekineticenergyand
U
isthepotentialenergy. Thisconservationofenergyequationisarelation
betweenpositionandvelocity,andsometimesthat’sallyouwant. Ifyouneedtogoallthewaytothe
endandtondthepositionasafunctionoftime,thisequationisseparable,asinsection0.8. Solve
for
v
x
=
dx=dt
,thenmoveeverythinginvolving
x
toonesideoftheequationandeverythinginvolving
t
totheother.
v
x
=
dx
dt
=
r
2
m
E
U
(
x
)
;
whichrearrangesto
dx
q
2
E
U
(
x
)

m
=
dt
(2
:
23)
Whetherthisintegralishardoreasywilldependon
U
. Attheveryworstitisreducedtosomething
thatyoucanintegratenumerically,asintheexampleofsection3.12.
Example
Theabsolutesimplestcasetostartwithoughttobeaconstantforce,atleastthat’swhatyoumay
think. Startwith
y
=0at
t
=0,andtakethecommongravitationalforceneartheEarth’ssurface:
F
y
mg
dU=dy
. Thissays
U
(
y
)=
mgy
so
Z
y
0
dy
0
p
2(
E
mgy
0)
=m
=
Z
t
0
dt
0
(2
:
24)
Why didI I startsprinklingprimesaround the
y
and
t
variables? Because e itispoorform to usethe
samesymbolfortwodierentthingsinthesameequation.
y
0
isthevariableofintegrationand
y
isthe
limit;
t
0
istheintegrationvariableandthenalcoordinateisthetime
t
. Ifyouconfusethe(dummy)
integrationvariablewithoneoftheparametersoftheproblemyoucancauseyourselfconfusionandwill
probablymakemistakesthatcanbeveryhardtond. Ifyoulookatpreviouspagesyouwillndthat
Isometimesdon’tobeythisrulemyselfeventhoughIknowbetter. Keepitinmindthough. Perhaps
youwouldliketogobacktoequations(2.15){(2.18)andcorrectmynotation.
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2|OneDimensionalMotion
61
Theintegraliseasy,andifyoudon’tshowanyfear,thealgebraistoo.
1
g
r
2(
E
mgy
0)
m
y
0
1
g
r
2(
E
mgy
)
m
+
1
g
r
2
E
m
=
t
!
y
1
2
gt
2
+
r
2
E
m
t
(2
:
25)
andthelasttermis
v
0
t
.Thisisclearlythehardwaytosolvethisproblem.
Even before doing the integrals there is much more to glean from the conservation equation
(2.22). Youcangetqualitativeinformationevenwithoutndingafullsolution,andsometimesthat’s
U
x
E
a
b c
d
e
f
Fig.2.4
Takeasanexamplethepotentialenergy
U
(
x
)=
kx
2
=
2,with
thecorrespondingforce
dU=dx
kx
(likeasimplespring).The
kinetic energy as afunction of
x
is
K
(
x
) =
E
U
(
x
), and the
totalenergyasafunctionof
x
is
E
=constant. Inthisgraphof
U
versus
x
youcanalsograph
E
versus
x
.Thisisthehorizontalline
representingtheconstanttotalenergy.
This graph has several vertical lines showing the dierence
K
=
E
U
,andtheyrepresentthekineticenergiesatthosevalues
of
x
.Asthemassmovesalongthe
x
-axisitskineticenergyincreases
asitapproaches
x
=0,andthekineticenergydecreasesafterithas
passedthatpointandmovesawayfromtheorigin.As
x
approaches
theintersectionofthe
E
and
U
curves,passingthroughthepoint
eandheadedtowardf,the kineticenergy approaches zero. At the e pointofintersectionf,themass
stops. Evenatthisstoppingpointthereisstillaforceonthemass;
dU=dx
kx
pushesitback
towardthecenterandthisforcecontinuesuntilthemasshitstheotherstoppingpointata.Fromait
ispushedtotherightandrepeatsthecycle.Thisexampleisasimpleharmonicoscillator,andchapter
threeisdevotedtoit.
know it, but pretend d you don’t.) ) On n the right, the slope
dU=dx
is positive,so
F
x

dU=dx
is
negative. Thesamereasoningsaysthat
F
x
ispositiveontheleft. Attheorigintheslopeiszero,so
F
x
iszerothere.Fromlookingatthisgraphof
U
alonecanyoutellthatthegraphof
F
x
isastraight
line?No,justthatitpassesthroughtheoriginandgetsmorenegativeontherightandmorepositive
onthe left. . Nowyou u canrememberthat
F
x
dU=dx
kx
,sothatitreally is astraightline
passingthroughtheorigin.
Intheprecedingexample,
kx
2
=
2,youcanproceedfromthisqualitativeanalysisandndacom-
pletesolutionbecauseyoucandotheintegralinEq.(2.23)withouttoomuchtrouble,(problem2.15).
Iftheforceanditscorrespondingpotentialenergyismorecomplicatedthismaynotbepracticable,but
U
issomefunctionof
x
,andhoweverthemassmoves,thecombinationofkineticandpotential
energiesstaysconstant. Thatmeansthatjustasinthesimplespecialcaseabove,thegraphoftotal
energyversuspositionisjustastraighthorizontalline,and
K
(
x
)=
E
U
(
x
)istheverticaldistance
betweentheline
E
andthecurve
U
inthegraph.Itcannevergonegative;itcanhowevergotozero.
That’swherethemassstops.Portionsofthe
E
=constantgraphsaresolidtoindicatewherethemass
isallowedtobe(
K>
0),andportionsaredashedtoindicatetheforbiddenregions(
K<
0).
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2|OneDimensionalMotion
62
U
x
E
1
E
2
E
3
E
4
E
5
a
A
b
B
c
C d
D
f
F
g
G
F
x
x
c
C
d
a
A
g
G
Fig.2.5
F
x
dU=dx
U
.This
F
x
-graphshowstheforcethatyoundbylookingatthepotentialenergygraphandestimatingnotjust
thesignbutthesizeoftheslope. Thepointsatwhich
F
x
=0correspondtotheminimaandmaxima
ofthepotentialenergy,sothoseareeasytond. BetweencandCforexamplethereisaminimum
ofthepotentialenergy;thatmeansthatthe
F
x
curvepassesthroughzerobetweenthosetwopoints.
Similarly,betweenCanddthereisamaximumof
U
soanotherzerofor
F
x
. Thestoppingpointsfor
theenergies
E
1 4
areindicatedbydotsonthe
F
x
-curve,andafewofthemarelabelled. Forexample,
gandGlabelvaluesof
x
atwhich
U
(
x
)=
E
4
. Atthatenergythemasscanoscillatebetweenthese
points.(AtthesameenergyitcanalsooscillatebetweenpointsfandF.)
K
+
U
=
E
betweenthetotalenergy
E
andthepotentialenergy
U
,andifIknowtheenergythenIcandetermine
where(orwhether)themasswillstop. Thatpointiswhereitskineticenergygoestozero,andthere
itstops.Theslopeof
U
atthesamepointsaysthatitgetspushedbackwhereitcamefrom.Forthe
samepotentialenergyyoucanhaveoscillationbetweenpointsaandAorbetweencandCordandD
orfandForgandG,butapparentlynotwhentheenergyistoobig. Ofcoursetheremaybemoreto
thegraphandtherecouldbeastoppingpointwaytotheright.
Intheexampleof
E
4
thatafteritleavesfitspeedsupuntilitpassesthebottomofthecurve. Itthenslowsdownbutdoesn’t
stopandthenitspeedsupevenmorebeforenallystoppingatF.Thenitreturns. Youcanndthe
totalperiodofthisoscillationfromftoFandbacktofbydoingtheintegralEq.(2.23)from
x
f
to
x
F
andmultiplyingtheresultbytwo. The
H
representsanintegraloverthewholecycle.
Period =
I
dt
=
I
dx
dx=dt
=
I
dx
v
x
=2
Z
x
F
x
f
dx
v
=2
Z
x
F
x
f
dx
q
2
E
4
U
(
x
)

m
Example
For amore e interesting example, use gravity. . The e gravitational eld of the Earth drops s o  with
distance, the e details s of which are discussed d in detail l in chapter r six. . For r now, say y that the e radial
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2|OneDimensionalMotion
63
componentoftheforceonamass
m
is
F
r
mg
0
R
2
=r
2
.Herethesinglecoordinateis
r
,thedistance
fromthecenteroftheEarth,and
g
0
isthegravitationaleldattheEarth’ssurface(atdistance
R
from
thecenter).Ifyoureaprojectilestraightup,howhighwillitgo(ignoringairresistance)?
Thisisstillaone-dimensionalproblemwithcoordinate
r
(
R<r<
1),soitisthesameasthe
equations(2.20)and(2.22)butchangingthenameofthevariable.
F
r
(
r
)=
mg
0
R
2
r
2
and
F
r
dU
dr
!
U
(
r
)=
mg
0
R
2
r
1
2
mv
2
r
+
U
(
r
)=
E
(2
:
26)
Theseequationsrepresentringsomethingstraightupwithaninitialspeed
v
0
. Theyalsopretendthat
thereisnoatmospheretocomplicatethemathematicsandthatheEarthisn’trotating.Howhighwill
itgo?Whereisitsstoppingpoint,tousethelanguageofthissection?
E
=
1
2
mv
2
0
mg
0
R
2
R
(atsurface)
=0
mg
0
R
2
r
stop
(atstop)
U
r
R
E
r
stop
Fig.2.6
Thegraphofpotentialenergybehavesas1
=r
inside,but you donotneedthat becauseyou typicallydon’tre arocket from belowground. . The
equationssimplyevaluatetheenergyattwopoints,at
r
=
R
andat
r
=
r
stop
.
Thisconservationofenergyequationgivesthenalheightintermsoftheinitialspeed.
E
=
1
2
mv
2
0
mg
0
R
2
R
mg
0
R
2
r
stop
!
r
stop
=
R
v
2
0
=
2
g
0
R
(2
:
27)
If
E
islargeenoughthehorizontallinefortheconstant
E
-graphwillneverintersectthepotentialenergy
curve,andtheprojectilewillneverstop(
E>
0). Movethelineofconstant
E
upwardsinthisgraph
andwatchthepointofintersectionwiththe 1
=r
graphmovetotheright. Statedanotherway,if
v
0
isincreasedenoughthedenominatorinthesolutionfor
r
stop
willgotozeroandthestoppingdistance
goestoinnity. Thatdenestheescapespeedfromtheplanet.
STOP!Isthisresultfor
r
stop
plausible?Doesithavethecorrectdimensions? Whathisitifthe
initialspeedissmall? Howbigmust
v
0
be toescapetheEarth? ? Computethenumber, , and isthat
numberreasonable? Canyou u compare it toany othernumberyou’ve seen? ? Howdoes s thenumber
comparetotheorbitalspeedofasatelliteinnear-Earthorbit?[Ifyoudon’tknowthisnumber,maybe
F(v)again
Youcansometimescombinethemethodsofthe precedingtwosectionsbysolvingtheproblemsofa
velocity-dependentforceasafunctionof
x
t
. ThesametrickthatyousawinEq.(2.21)to
solveforforcesdependingon
x
alonealsoworksfor
F
(
v
).
F
x
(
v
x
)=
m
dv
x
dt
=
m
dv
x
dx
dx
dt
=
mv
x
dv
x
dx
!
dx
=
m
v
x
dv
x
F
(
v
x
)
(2
:
28)
2|OneDimensionalMotion
64
Integratingthis gives arelationbetween
x
and
v
x
. Is s this more useful than getting
v
x
(
t
)? Usually
not,butitisanothertoolinyourutilitybelt,andit’ssometimesjustwhatyouneed. Forexamplesee
problem2.53.
2.4Fallingwithresistance
Aplausibleexamplecomesbyassumingthatairresistanceisproportionaltovelocity. It’snotallthat
accurate,butitisafairlysimpleplacetostart.Amoreaccuratemodelwillassumethatairresistance
isproportionalto
v
2
,butthisiseasier.
ma
y
=
m
dv
y
dt
=
m
d
2
y
dt
2
mg
bv
y
mg
b
dy
dt
(2
:
29)
Thereareseveralwaystosolvethisequation.I’lluseseparationofvariablesagain,movingallthe
v
y
’s
toonesideoftheequationandallthe
t
’stotheother.
m
dv
y
dt
mg
bv
y
!
m
dv
y
mg
+
bv
y
dt
ThisissomethingthatIcan integrate. . Use e thesameinitialconditionsas inthe precedingexample,
v
y
(0)=
v
0
,andlet
u
=
bv
0
y
m
Z
v
y
v
0
dv
0
y
mg
+
bv
0
y
Z
t
0
dt
=
m
b
Z
bv
y
bv
0
du
mg
+
u
=
m
b
ln(
mg
+
u
)
bv
y
bv
0
=
m
b
ln
mg
+
bv
y
mg
+
bv
0
t
Solveforthevelocity.
m
b
ln
mg
+
bv
y
mg
+
bv
0
t
!
v
y
(
t
)=
mg=b
+
v
0
+
mg=b
e
bt=m
(2
:
30)
Before proceeding, , does s this s make e sense? ? At t time zero you u have
v
y
=
v
0
as required. . (Look k at
bothequationsinEq.(2.30)tocheck.) Atlargetime,
v
y
approaches
mg=b
,andfromtheoriginal
dierentialequationofmotion,Eq.(2.29),thismaketheaccelerationzero.Thatistheterminalvelocity.
YoucanrewriteEq.(2.30)directlyintermsofthisterminalspeed.;
v
t
=
mg
b
;
then
v
y
(
t
)=
v
t
+(
v
0
+
v
t
)
e
gt=v
t
(2
:
31)
Tond
y
(
t
)youhaveasimpleintegralwithrespecttotime. Usetheboundaryconditionthat
y
(0)=0and
y
(
t
)=
Z
t
0
dt
0
mg=b
+
v
0
+
mg=b
e
bt0=m
mg
b
t
0
bv
0
+
mg
m
b
2
e
bt0=m
t
0
mg
b
t
+
bv
0
+
mg
m
b
2
e
bt=m
v
t
t
+(
v
0
+
v
t
)
v
t
g
e
gt=v
t
(2
:
32)
Dothesemakesense?First,checkthedimensions! Donotassumethatyouwillnevermakeamistake.
Noticethatit’seasiertocheckdimensionsnowthateverythingisexpressedintermsof
v
t
m
,
b
,and
g
. Next,seewhatthebehaviorofthesolutionisinsomespecialcases.Startwiththesmall
timebehavior(notzerotime). Theinitialvelocityofthemassshouldbe
v
y
=
v
0
,anditsacceleration
shouldbe(
mg
bv
0
)
=m
. Whythislastexpression? Lookattheoriginalequation(2.29),andthat
2|OneDimensionalMotion
65
tellsyoutheaccelerationatalltimes,inparticularthetime
t
=0. Nowuseapowerseriesexpansion
on
y
(
t
)tondthesmalltimebehavior.
y
(
t
)=
mg
b
t
+(
bv
0
+
mg
)
m
b
2
bt
m
+
(
bt
)
2
2
m
2
(
bt
)
3
6
m
3
+

v
0
t
(
bv
0
+
mg
)
t
2
2
m
+(
bv
0
+
mg
)
b
6
m
2
t
3
=
v
0
t
1
2
m
(
mg
+
bv
0
)
t
2
+(
bv
0
+
mg
)
bt
3
6
m
2
=
v
0
t
g
t
2
2
v
0
bt
2
2
m
+
g
bt
3
6
m
+
(2
:
33)
WhenImultipliedoutthelastequation,Iwascarefultokeeptermstoaconsistentorderin
b
. That
meansthatwhenIdropa
b
2
terminoneplaceIshoulddropiteverywhere.
Someoftheterms canceled: : Thosethatinvolved d 1
=b
. That’san n importantcheckonthealgebra
becauseiftheydon’tcanceltheneverythinggoestoinnityastheairresistancevanishes.
Sometermsbecomeindependentof
b
astheviscosityapproacheszero,andthosearetheonesthat
reproducewhatyouexpectinthemuchsimplercalculationforzeroviscosity.
Thecorrectionforsmallviscosity comesin the termslinearin
b
. Thesegobeyondtheelementary
calculation, so look at the initial acceleration. . Thatis s in the
t
2
terms,the ordinary
at
2
=
2,so you
can recognizethattheaccelerationhasthecorrectinitial value,thoughitwashiddeninsideamore
complicatedexponentialexpression,waitingtobedugout.Theinitialforceisthesumofgravity(down)
andthefriction(alsodown),producinginitialacceleration
a
y
0
=(
mg
bv
0
)
=m
.
The
t
3
term is startingtogethardertointerpret, , butthesigniseasttounderstand. . Asthe
objectstarttorise,itstartstoslowdown(
gt
).Thisinturnimpliesthattheviscousforceisnotquite
asbigasitwouldhavebeenwithoutgravity,hencethiscorrectionhasapositivesign.
You’re not done. . What t happens in n the opposite e case, forwhich the viscosity is very y large?
Perhapsyou’reringabulletintoabarrelofhoney.If
b
islargetheexponential
e
bt=m
willgotozero
veryquickly. Theviscousforcewillbemuchgreaterthanthegravitationalforce:
bv
0
mg
. Inthis
casethesolution(2.32)willbeapproximately
y
(
t
)
mv
0
b
mgt
b
(2
:
34)
Thismeansthat itvery quicklygoesupadistance proportionalto its initial momentum, stops ata
height
mv
0
=b
,andthenslowlydriftsdownatitsconstantterminalspeed
mg=b
. CanIseewhyanyof
thisistrueindependentofsolvingthewholeequation?Theterminalvelocityiseasytosee. Justgoto
ma
y
=0=
mg
bv
y
!
v
y
mg=b;
theterminalvelocity
Forlargetimes,theexponential
e
bt=m
diesout,andyouareleftwithaconstantvelocity,
y
(
t
)
mg
b
t
+
v
0
+
mg=b
m
b
mg
b
t
m
b
1+
bv
0
mg

(2
:
35)
some timegoingup beforeitstarts downandeventually reachesitsterminalspeed. . Thebiggerthe
initialspeed,themoretimeitspendsgoingup.Didithelptorearrangethefactorsinthelastexpression
andtoputitintheformofsomethinglike(
t
t
0
)?Probablynot,butyoudon’tknowuntilyoutry.
2|OneDimensionalMotion
66
Still anotherway to squeezeinformation outofthisresult: : What t if theairresistance isvery
small? Maybeevenzero? OnethingthatIcannotdoistolet
b
becomesmallinthelastequations,
becauseIwasverycarelessintheprecedingparagraphwhenIsaid\Forlargetimes
:::
"Whatislarge?
That is s meaningless s because time is notdimensionless. . Large e compared to a femtosecond orlarge
comparedtoagalacticyear?WhatIshouldhavesaidisthat
bt=m
1,makingtheexponentialsmall
comparedtoone. Gobacktotheoriginalequation(2.32),and d lookatthatforsmall
b
. ThereIgo
again. Whatis\small
b
"?Thistimeitis
bt=m
1,andIessentiallyanalyzedthiscaseinEq.(2.33),
whereIsaidthat
t
issmall. Ifyouset
b
=0there,youget
v
0
t
gt
2
=
2,andthecorrectionislinearin
thefactor
b
.
Istheremorethatyoucananalyzeinthissimple-lookingproblem. Yes,seeproblem2.14.
Example
someoneelse,orifateachersaysit’sright.Doesyoursolutionmakesense?
v
x
=
v
0
1+
bv
0
m
t
2
?
Wouldthismakesense?Itisdimensionallycorrect;atleastithasthesamecombinationofparameters
bv
0
t=m
thatappearedinEq.(2.18),soifitwascorrecttherethenitisheretoo. As
t
!1thisgoes
tozerofasterthanthe othersolution. . Is s thatbad? ? Notobviously y so. . Before, , the integralof
v
x
dt
wenttoinnityforlargetimebecausethevelocitywenttozerososlowly.Doesthathappenhere? No,
because
Z
1
dt
1+
t
2
convergestoanitevalue
v
x
v
0
h
bv
0
m
t
2
i
forsmall
t
.(thegeometricseries,Eq.(0.1)(h))
This impliesthat attime zero, , theacceleration = 2
b
2
v
3
0
t=m
2
= 0. . Even n withthe air r resistance,
mv
2
,assumedinderivingEq.(2.17)thissaysthatthereisnodecelerationatthebeginning. Thatis
clearlywrongbecause
v
6=0atthattime.
Example
If youhaveasimplemechanicalsystem withamasshangingonastringthatiswrapped
aroundapulley,whatwilltheaccelerationbe?Inchapteronetheequation(1.19)said
dv
y
dt
=
m
1
m
2
m
1
+
m
2
g
=
m
1
+
m
2
m
1
m
2
g
or
=
m
1
m
1
+
m
2
g
?
Intherstcase,lookattheexpressionforthespecialcasethat
m
1
=
m
2
. Thedenominatorvanishes,
a
y
isalwayspositive. Even
thoughthedirectionof
a
y
shouldreverseif
m
2
>m
1
changesto
m
1
>m
2
. Bothoftheseresultsfor
dv
y
=dt
arewrong,andyoucansaythiswhetheryouknowthecorrectsolutionornot.