asp.net mvc 4 and the web api pdf free download : C# merge pdf files into one application software cloud windows html .net class mechanics14-part1284

4|ThreeDimensionalMotion
137
Thispictureshowstwoofthemagneticeldlinesasintheprecedingdrawing,oneofthemalongthe
axis.Forbotheldlinesthechargestartsontheleftsidewithinitialvelocityatsomeangletotheeld,
anditspiralsaroundthedirectionof
~
B
asrepresentedbytheeldline. Astheeldbecomesweaker
towardthemidpointtheorbitalradiusincreases,thetransversecomponentofthevelocitydecreases,
andthelongitudinalcomponentofvelocityincreases. Towardtherightside,thelongitudinalcomponent
ofvelocity decreasestozeroandthechargeisre ectedbackalongitspath. . Thecenteroftheorbit
followstheeldlinetogoodaccuracy.
Howisitpossibletounderstandthismotionwithoutrecoursetoallthisnumericalcomputation?
Therearetwokeyfactsthatprovideinsight,allowingyoutounderstandmostoftheseresultswithout
detailedcalculation. Derivingthemgetsintothesubjectofadiabaticinvariants,whichiswellbeyond
anythinginthisbook,butIwillsimplystatetheresultsandusethem:Asthechargemovesthroughthe
elditcirclesaroundaeldline,andthecenterofthecirclefollowsthateldline. Also,theproduct
ofthelocalmagneticeld andtheareaofthecircularorbitstaysconstant(theadiabaticinvariant).
Bothofthesestatementsareapproximatebutuseful. Ifyouacceptthem,youcanexplaintheresults
oftheprecedingparagraph.Oneassumptionneededtomakethisvalidisthatthemovementalongthe
eldisslowcomparedtotherotationaroundit.
v
k
v
?
. Thesecondassumptionis s thattheeld
doesnotchangemuchoverthediameterofthecircle.
~v
k
~v
?
~
B
r
Fig.4.10
Foralittlecirclearoundtheeldline,take
B
asthevalueatthecenterofthecircle.Thecharge
hasspeed
v
withcomponentsparalleltotheaxisandperpendiculartotheaxis: callthem
v
k
and
v
?
.
The speed is constant
v
2
=
v
2
k
+
v
2
?
. Why y constant? ? Magnetic c forces donowork,Eq.(4.26), so
mv
2
=
2doesn’tchange. Forthiscircle,theradialcomponentof
~
F
=
m~a
is
m
v
2
?
r
=
qv
?
B
!
v
?
r
=
!
=
qB
m
Iclaimedthat
r
2
B
isanadiabaticinvariant,whichmeansthatit’snearly constant. . Combinethat
equationwiththisequationfor
!
.
r
2
B
=
and
v
2
?
r
2
=
q
2
B
2
m
2
imply
v
2
?
=B
=
q
2
B
2
m
2
Rearrangethistoget
v
2
?
=
q
2
m
2
B
(4
:
44)
Thefactorsarelessimportantthantherelationbetween
v
?
and
B
.Itsaysthatasthemagneticeld
gets stronger,the component
v
?
getsbigger. But, thespeedstays s constant,sothat says thatthe
othercomponent,
v
k
,mustgetsmaller. When
B
gets bigenough,thatparallel componentgoes to
zeroandthemotionalongtheeldlinestops. That’sthemagneticmirror.
Youcangetasurprisingresultoutofthisbywritingtheequationforthekineticenergyofthe
charge. (Aconstant,remember.)
K
=
1
2
mv
2
=
1
2
m
v
2
k
+
v
2
?
=
1
2
m
v
2
k
+
q
2
m
2
B
(4
:
45)
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4|ThreeDimensionalMotion
138
Lookatthemotionalongtheaxis,wheretheeldhasonlythe
z
-component. FromEq.(4.42),
B
=
B
0
+
B
0
z
2
=‘
2
!
K
=
1
2
m
v
2
k
+
q
2
m
2
B
0
+
B
0
z
2
=‘
2
Comparethistotheenergyequationforasimpleoscillator,
E
=
mv
2
=
2+
kx
2
=
2.
1
2
mv
2
k
+
1
2
q
2
m
B
0
z
2
2
=aconstant
(4
:
46)
Thatisaharmonicoscillator. Thisresultmeansthatnotonlydoesthemotionalongtheeldlinestop,
itreversesandkeepsbouncingbackandforthinsimpleharmonicmotionuntilithitssomething.
If you u you u take e the Earth’s magnetic c eld d to be that of a magnetic dipole e (a pretty y good
approximation)anddothesameenergyanalysisthatledtoEq.(4.46),youwillndthatthemotion
oftheelectronmatchesthatofapendulum,thesubjectofsections3.4and4.6.
TogetanideaofthefrequencyofoscillationoftheseelectronsintheVanAllenBelt,
!
2
0
=
k
m
=
1
m
.
q
2
B
0
m‘
2
Thefactor=
r
2
B
=
r
2
0
B
0
,where
r
0
isradiusoftheelectron’scircularmotionatthemiddleof
theoscillation;
B
=
B
0
there. Theelectronchargeis
q
e
.
!
2
0
=
e
2
r
2
0
B
2
0
m
2
2
=
eB
0
m
r
0
2
Thequantities
e
and
m
areknown.
B
0
forEarthisabout1/2Gauss.
issomeplausiblepartofthe
Earth’s radius,perhaps 4000km. . Finally,
r
0
isthe least well-known,and will vary greatly from one
electrontothenext. Ifyoumakeaguessofonemillimeterthenyouatleasthaveaplacetostart:
!
0
=
eB
0
m
r
0
=
1
:
610
19
.
0
:
510
4
9
:
110 31
.
1
:
010
3
4106
=0
:
002s
1
sothisperiod,2
=!
0
,isaboutanhour.Ifyounowchange
r
0
from1mmto1m,thisperiodchanges
fromaboutanhourtoabout3seconds.
4.5ApproximateSolutions
Sometimes\
:::
asleazyapproximationthatprovidesgoodphysicalinsightintowhat’sgoingoninsome
system is farmore usefulthan anunintelligibleexactresult." " HereIpresentaparticularmethodby
showingsomeexamples.
Ifyouhaven’treadsection0.11andsolvedprob-
lem0.59(a),thennowisagoodtimeforit.The
restofthissectionwillmakemuchmoresense
after that. . Eq. . (4.11) also used this s method.
Example
To illustrateaparticularmethod, iteration,lookagainat Eqs.(4.1) and(4.2), butthistimesee
whathappensifyouassumefromthebeginningthattheairresistanceissmallorequivalentlythatthe
massislarge. Toarstapproximationthen,Eq.(4.1)is
m~g
mg
^
y
=
m
d
2
~r
0
dt
2
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4|ThreeDimensionalMotion
139
denotingthe approximate solutionbythe subscript zero. . With h thesame initial conditionsas before
(
~r
(0)=0and
_
~r
(0)=
~v
0
),thisis
d
2
~r
0
dt
2
=
~g
!
~r
0
=
~
A
+
~
Bt
+
1
2
~gt
2
!
~r
0
(
t
)=
~v
0
t
+
1
2
~gt
2
(4
:
47)
Sofarthisisalloldmaterial. Nowusethisapproximateresulttoconstructanimprovedresult. Ifthe
b~v
termissmallcomparedtotheothertermsintheequationfor
~r
,thenwrite
d
2
~r
dt
2
~g
b
m
d~r
dt
!
d
2
~r
1
dt
2
~g
b
m
d~r
0
dt
b
m
(
~v
0
+
~gt
)
(4
:
48)
Thisusestheapproximatesolution
~r
0
inthesmallterm 
b~v=m
,andbecausethistermissmallanyway,
usingaslightlywrongvaluefor
~r
(
t
)willintroduceonlyasmallerror.Thefunction
~r
1
(
t
)istheimproved
approximationforthesolution.Thiscanbeintegratedjustaseasilyastheoriginalequationthatdidn’t
includeairresistance:
~r
1
=
~
A
+
~
Bt
+
1
2
~gt
2
b
m
1
2
~v
0
t
2
+
1
6
~gt
3
(4
:
49)
Againyouneedtousetheinitialconditions
_
~r
1
(0)=
~v
0
and
~r
1
(0)=0,toget
~r
1
(
t
)=
~v
0
t
+
1
2
~gt
2
b
m
1
2
~v
0
t
2
+
1
6
~gt
3
(4
:
50)
Justhowgoodisthisapproximation,anddoesitreallygiveinformationabouttheexactresult?
There’snodoubtthatit’ssimpler,butissimplergoodenough? Fortunately,inthiscaseyouhavean
exactsolution,sothisexampleisalaboratoryinwhichtotestthevalidityoftheiterativesolution.Go
backtotheequations(4.3){(4.5). Thoseequationsincludetheexactsolutionandthentheyexpandit
inaseries,keepingonlythelowestorderterm|theequivalentof
~r
0
(
t
)inthelanguageofthissection.
Allthatyouneedtodoistokeepthenextordertermsinthoseequations,onemorepowerof
b
m
,and
youwillseethattheymatchEq.(4.50)perfectly.NoticealsothatderivingEq.(4.50)isfarquickerand
simplerthanallthelaborinsection4.1. Ifyouknow thatyouareaimingforanapproximatesolution
toaproblem,youaregenerallybetteromakingtheapproximationsatthebeginning,notattheend
ofthecalculation.
Inthe lastequation, howdothetwonewterms compare? ? Factorout
1
2
t
from eachand you
have
b
m
t
2
~v
0
t
+
1
3
~gt
2
.Thersttermintheparenthesesisthedistancetheobjectwouldhavegonein
time
t
andthesecondtermis
2
3
thedistanceitwouldhavedropped. Ifyou’redealingwithanobject
thatismovingratherfast,thenthesecondtermcanbemuchsmallerthantherstandtheexpression
for
~r
1
(
t
)wouldstillbenearlyaparabola. Butnotice: thetipofthisparabolaisnotatthetop. The
parabolaistilted,asifthereisaneectivegravitationaleldgivenby
~g
0
=
~g
b
m
~v
0
.Comparethisto
thedrawinginFigure4.1.
Example
What about a problem that you u can’t t solve exactly? ? (That t is, almost all other problems.) ) Air
resistanceismoreaccuratelydescribedasquadraticin
v
,andinchaptertwo,wherethemotionwas
alongastraightline,calculatingthiswaslefttoacoupleofhomeworkproblems: 2.29{2.31.Asyousee
there,theresultsarecomplicatedbutmanageable.Ifyoutrytodothesamethinginthreedimensions,
yougetanequationthatyoucan’tsolveatallwithoutresortingeithertonumericalmethodsorashere,
toaniterativetechnique.
Thedierentialequationforthemotioninthiscase,writtenasvectors,is
m
d
2
~r
dt
2
=
m~g
bv
2
^
v
=
m~g
bv~v
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4|ThreeDimensionalMotion
140
This\
b
"isofcoursedierentfromthepreviousone,andthebeginningoftheproblemisthesame:
d
2
~r
0
dt
2
=
m~g
!
~r
0
=
~
A
+
~
Bt
+
1
2
~gt
2
!
~r
0
(
t
)=
~v
0
t
+
1
2
~gt
2
Thenextstepisthesametoo,atleastinprinciple.
d
2
~r
1
dt
2
~g
b
m
j
_
~r
0
j
_
~r
0
where
_
~r
0
=
~v
0
+
~gt
and j
_
~r
0
j=
v
2
0
+2
~v
0
.
~gt
+
g
2
t
2
1
=
2
Can you u integrate this? ? Yes, , although h you u probably don’t want t to. . The e result is a a complicated
combinationofalgebraicandinversehyperbolic functions. . Itisnotenlightening,andspendingtime
heretocarryoutthealgebrajustisn’tworthit. Ihaveincludedthisexampleonly y toshowthatthis
iterativemethodcanbeusedevenincomplicatedcases,andifyoureallyneedtheresults,youcanget
them.
CurveBalls
Amoreinterestingexample: Whydoesacurveballcurve?Inbaseballthepitcherstands60ft(18.3m)
fromthe batterand throws theball ataspeed that wouldviolatetrac laws anywhere buton the
Autobahn. Askilledpitchercanthrowtheballsothatitcurvesinvariousdirectionsbyascarilylarge
amount(scarilyifyou’rethebatter). Ifyou’remorefamiliarwithsoccer,orinternationalfootball(the
sphericalkind)thenyouknowthataproperlykickedballcanmoveonasurprisinglybentpath.Thisis
thesamephenomenonasthecurvedbaseballpath,soI’lldescribeeverythinginonelanguage:baseball.
Thespinningballismovingthroughtheair,andbecausetheairhassomeviscositysomeofthe
airjustnexttothesurfaceoftheballisdraggedalongwiththatsurface.Thedetailsarecomplicated,
butwhenthedustsettlestheadditionalforceduetotheinteractionoftheball’sspinwiththeaircan
bereasonablewelldescribedbyasimpleexpressioncalledtheMagnusforce.
~
F
M
=
C~!
~v
(4
:
51)
Whatis
C
? Windtunneltestsverifythatthisforceisproportionalto
!v
,so
C
can’tinvolveeitherof
thosefactors. Dimensionalanalysiscanprovidealotofinformation.Whatcan
C
possiblydependon?
;
thedensityofair,dimensionsM
=
L
3
;
theviscosityofair,dimensionsM
=
LT
R;
theradiusoftheball,dimensionsL
C
itselfmusthavedimensionsforceoveracceleration=M. Whatcombinationoffactorscanproduce
thecorrectdimensionsfor
C
?
[
C
]=M=[
]
a
[
]
b
[
R
]
c
=
M
L
3
a
M
LT
b
(L)
c
!
a
+
b
=1
;
3
a
b
+
c
=0
;
b
=0
!
a
=1
; b
=0
; c
=3
Theexpressionfor
C
isthensomedimensionlessfactortimes
R
3
. Avalueof
1
2
R
3
isagoodplace
tostartforthisfactor,butit’sneverquitethatsimple.
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4|ThreeDimensionalMotion
141
Onabaseballthesizeofthisforceissmallcomparedtogravity,sothisiterativemethodshould
workwellhere.FollowthepatternoftheexampleEqs.(4.47).{(4.49).
m
d
2
~r
dt
2
=
m~g
+
~
F
airresistance
+
C~!
~v
(4
:
52)
The airresistance canbe eitherthesimplelinearone,orquadratic,orsomethingworse, butforthe
presentpurposesitdoesn’tmatter.Theiterativemethodallowsyoutosolvethatpartseparatelyfrom
theMagnusforceandthen,forthisrstorderiteration,toaddthetworesults.
DotheMagnusforcealonethen.
m
d
2
~r
dt
2
=
m~g
!
~r
0
(
t
)=
~v
0
t
+
1
2
~gt
2
m
d
2
~r
1
dt
2
=
m~g
+
C~!
_
~r
0
(
t
)=
C~!
~v
0
+
~gt
~r
1
(
t
)=
~
A
+
~
Bt
+
1
2
~gt
2
+(
C=m
)
~!
1
2
~v
0
t
2
+
1
6
~gt
3
=
~v
0
t
+
1
2
~gt
2
+(
C=m
)
~!
1
2
~v
0
t
2
+
1
6
~gt
3
(4
:
53)
Nowforthefunpart: Whereisthecurveballinthisequation?Startbylookingattherelativesizesof
thelasttwoterms. Factora
1
2
t
from(
1
2
v
0
t
2
+
1
6
gt
3
)togettwoterms,
v
0
t
and
1
3
gt
2
. Whentheball
reachestheplate,therstoftheseisjustthedistance
v
0
t
fromthepitchertothebatter. Thesecond
is
2
3
.1
2
gt
2
,andthatis
2
3
theamountthattheballwillnaturallydropbythetimeitreachestheplate.
Inthiscase,thesecondtermisreallygoingtobemuchlessthantherst. Concentratethenon
~v
0
t
+
1
2
t
2
~g
+(
C=m
)
~!
~v
0
(4
:
54)
andpictureitasthebatterwillseeit.Theballiscomingathighspeedalmoststraightatyou.Atleast
youhopethatitis\almost".Assumethattheballisthrownhorizontally,simplybecausethatmakesit
easiertopicturewhatwillhappen.
~v
0
ishorizontal.
~g
isdown.
~!
~v
0
isperpendicularto
~v
0
This means that the ball’s s acceleration is s not necessarily down. It t can be e in n any direction n that t is
perpendiculartotheball’sinitialvelocity. IftheMagnustermislargeenough,itcouldinprincipalbe
up,thoughprobablynohumanpitchercandothis.
WhatistheratioofthesecondterminthebracketsofEq.(4.54)totherst?Take
C
=
1
2
R
3
,
thentheratiois
R
3
!v
0
2
mg
=
(1
:
27kg
=
m
3
)
.
(3
:
7cm)
3
2
.
145gm
.
9
:
8m
=
s2
!v
0
=2
:
310
5
s
2
m
!v
0
If
v
0
is85mph,or38m/s,and
!
is20rev/s,or126rad/s,thenthisratioisabout0.1,andthat’sabig
eect.Itisespeciallybigwhenyourememberthatthebatterhasonlyatime
d=v
0
=18m
(38m
=
s)=
0
:
5sfromwhenthepitcherthrowstheballtowhenthebatterisexpectedtohitit.
Takeleftandrightfromthepitcher’svantagetodescribethis.
If
~!
~v
0
isd
c
own(
~!
left)thentheaddedforceisdown;theballsinksfasterthan
~g
alonewouldcause.
If
~!
~v
0
iscup(
~!
right)thenitwillsinklessthanexpected.
If
~!
~v
0
isl
b
eft(
~!
up)thenyouhavealeftcurvingball. Andofcourse
~!
downimpliesarightcurve.
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4|ThreeDimensionalMotion
142
Whatistheshape ofthiscurveinthelasttwocases? That’seasy. Itjustlooksliketheusual
parabolictrajectoryofaballmovinginagravitationaleld,butthateldisn’tpointingexactlydown;
ithasacomponenteitherleftorright.Insteadof
~g
,youhave
~g
0
.
~r
(
t
)=
~v
0
t
+
1
2
~g
0
t
2
with
~g
0
=
~g
+(
C=m
)
~!
~v
0
(4
:
55)
WillthisbeatiltedparabolaasinFigure4.1orEq.(4.50)?No,because
~!
~v
0
isalwaysperpendicular
to
~v
0
. Instead,thisparabolaisrotatedaroundthe
~v
0
directionasanaxis.
Whydobattersthinkthattheball\breaks"?Whyshoulditappearthattheballiscomingpretty
nearlystraightatyouforawhileandthenittakesasudden detour? ? Thepitcherwillthrowtheball
nearlyhorizontally,maybealittleup. Duringthersthalfoftheball’striptotheplateitwilldropat
mostone-fourthofitstotaldrop,andlessthanthatifitstartsoutwithaslightupwardcomponentof
velocity.Onaveragethedistancefromthebattertotheballisthreetimesaslargeduringthersthalf
ofthetripasinthesecondhalf,sothebatterseesitmovesidewaysthroughan angle thatis about
33=9timesgreaterduringthelasthalfofthetripasduringthersthalf. Hencetheillusionofa
breakingball.
Comparetheeectsforabaseballwithmass145gm,radius3.7cmandtheAssociationfootball
(soccerball)withmass 425gm,radius11cm. . Thecoecient
C=m
inthe lattercaseislargerby a
factor(11
=
3
:
7)
3
(425
=
145)=9. Ofcourseyoustillhavetogetalargeenough
!
. That’stheartand
whyBeckhamearnedthebigbucks.
4.6Pendulum,largeangles_
Thependulumwasanexampleofasimpleharmonicoscillatorinthelastchapter,butitwasn’texactly
harmonic, any y more than any other r oscillation n aboutan equilibrium will be. . In n Eq.(3.23) you had
somethingthat was not asimple harmonicoscillator, and it became simple only in the smallangle
approximation.Whatiftheangleisn’tsmall?
Solving forthe position as afunction of time is dicult,but if allyou wantis the period of
oscillationthen that’snotsohard|allthatis involvedisdoingadicultintegral. . In n theequation
(3.27)youhave
U
E
E
=
1
2
m‘
2
_
2
+
mg‘
(1 cos
)
(4
:
56)
Inthatchapter,Idierentiatedtheenergy,
dE=dt
=0andgotthedierentialequations ofmotion.
Instead,solvethisenergyequationfor
dt
directly.
2
m‘
2
E
mg‘
(1 cos
)
=
d
dt
2
!
dt
=
d
p
(2
E=m‘
2
) (2
g=‘
)(1 cos
)
This separatesthe equation, , and allyou have todo nowis s tointegrate. . Itis s atough integral but
notimpossible,andithelpstodoalittlemanipulationrst. Rememberthetrigonometricidentityfor
half-angles?
sin
2
2
=
1
2
(1 cos
)
4|ThreeDimensionalMotion
143
dt
=
d
q
(2
E=m‘
2) (4
g=‘
)sin
2
(
=
2)
=
s
4
g
d
q
(
E=
2
mg‘
) sin
2
(
=
2)
(4
:
57)
Why manipulate the expressioninthis way? ? Itisnotobvious,andI’m m surethattherstperson to
solvethisproblemdidnotndtherightpaththroughthemazethersttime.
Themotion ofthependulum isbackandforthbetweenthestoppingpoints asin section 2.3.
Thestoppingpointsoccurwhenthekineticenergydropstozero|whenthelineofconstant
E
inthe
graph,Eq.(4.56),crossesthecurve
U
(
).Thatiswhen
_
=0,
_
=0 =)
E
=
mg‘
(1 cos
)=2
mg‘
sin
2
(
=
2)
;
=
max
andthatisthepointwherethedenominatorunder
d
vanishesinEq.(4.57).
Nowyoucantrytodotheintegralof
dt
togettheperiodofthependulum,anduptothispoint
allthemanipulationshavebeenthesamekindthatyouseeina\techniquesofintegration"chapter
inyourintroductorycalculustext. Butnowcomesaclevertrick. . Again,itisnothingbeyondwhatis
doneinbeginningcalculus(achangeofvariables),butitisnotanobviouschange. Asitstandsthe
periodofthependulumisfourtimesthetimeittakestogofromthecentertothestoppingpoint
T
=4
Z
dt
=4
Z
max
0
s
4
g
d
q
(
E=
2
mg‘
) sin
2
(
=
2)
where
max
=2sin
1
s
E
2
mg‘
Theintegralgoesfromzerotoalimitthatdependsontheenergy. Itwouldbemucheasiertohandle
ifthelimitisxed. Thetrick: let
s
E
2
mg‘
sin
=sin
2
(4
:
58)
Thisisdesignedsothatas
goesfromzeroto
=
2,theangle
goesfromzeroto
max
. Whyshould
thishelp?IfitdidnotIwouldn’tbespendingtimeonit. Executethechangeofvariables.
s
E
2
mg‘
cos
d
=
1
2
cos
2
d
d
q
(
E=
2
mg‘
) sin
2
(
=
2)
=
s
E
2
mg‘
cos
d
1
2
cos
2
q
(
E=
2
mg‘
) sin
2
(
=
2)
=2
cos
d
q
1 sin
2
(
=
2)
p
1 sin
2
=2
d
p
k
2sin
2
where
k
2
=
E
2
mg‘
=sin
2
max
2
(4
:
59)
Assembleeverythingtogettheperiodofoscillation
T
=4
Z
dt
=4
s
4
g
Z
=
2
0
d
2
p
k
2
sin
2
=4
s
g
Z
=
2
0
d
1
p
k
2
sin
2
(4
:
60)
Therearealotofmissingstepsinthisderivation,butnotbigones.Youshouldcarryoutthesedetails
foryourself. Why
k
2
? That’ssimplythemostcommonconventioninthetreatmentoftheseElliptic
4|ThreeDimensionalMotion
144
Integralsthoughnottheonlyone. InAbramowitzandSteguntheyusetheletter
m
insteadof
k
2
. The
standardnotationforthis\ellipticintegraloftherstkind"is
K
=
Z
=
2
0
d
1
p
k
2sin
2
;
so
T
=4
K
s
g
(4
:
61)
Insection6.13youwillndabriefintroductiontothesefunctionsandsomeoftheirproperties.
ShouldIbelievetheanswer? Iftheoscillationsaresmallenoughthismustreducetothesolution
of
T
foroscillationswithsmallenoughenergytoallowsmallangleapproximations.Smallenoughenergy
meansthattheparameter
k
2
=
p
E=mg‘
1.Startbymakingitzero.
T
=4
s
g
Z
=
2
0
d
1
1
=4
s
g
2
=2
s
g
andthatis2
=!
0
fromEq.(3.25),soitworks.
Nowwhatiftheenergyis notsosmall?
k
2
canstillbesmallbutnotnegligible. Doapower
seriesexpansionusingthebinomialseriesofEq.(0.1).
(1+
x
)
n
=1+
nx
+
n
(
n
1)
x
2
2!
+
andfor
n
= 1
=
2thisis
1
2
x
+
1
.
3
22
.
2!
x
2
1
.
3
.
5
23
.
3!
x
3
+
T
=4
s
g
Z
=
2
0
d
1+
1
2
k
2
sin
2
+
1
.
3
22
.
2!
k
4
sin
4
+
(4
:
62)
=2
s
g
"
1+
1
2
2
sin
2
max
2
+
1
.
3
2
.
4
2
sin
4
max
2
+
1
.
3
.
5
2
.
4
.
6
2
sin
6
max
2
+
#
(4
:
63)
Youcaneasily dothesecondterm,theoneinsin
2
,yourself,buttherestofthemrequireanother
integral,thatofsin
2
n
d
,andyoucaneitherlearnhowtodoit(probablywithcomplex variables)
orhaveabigenough tableofintegralssuchasthatbyGradshteynandRyzhik. . Theintegralhereis
equationnumber3.621.3intheirbook:
Z
=
2
0
d
sin
2
n
=
(2
n
1)!!
(2
n
)!!
2
Thedoublefactorialnotationmeanstheproductofeveryotherintegeruptothegivenone.Forexample
5!!=15and6!!=48.
Foranangleof
max
=30
thisseriesis
2
s
g
1+0
:
01674+0
:
000005+
;

=1
:
017
(4
:
64)
For90
itis
2
s
g
1+0
:
1250+0
:
0352+0
:
0122+0
:
0003+
;

=1
:
173
4|ThreeDimensionalMotion
145
Youcanseethatevenat90
thisseriesconvergesrapidlyandchangesthesmallangleresultbyabout
17%.Ifyouwanttotry179
youwillquicklyconcludethatthere’sgottobeabetterway,becausethis
timethesineof
max
=
2isclosetooneandtheseriesconvergesveryslowly. Atthispointyoucanlook
upareferenceonellipticintegralstondthatthequantityinbracketsisapproximately3.90,evaluated
usingEq.(4.65)inthenextparagraph.
k
2
=sin
2
(179
=
2)=0
:
9999238  !
K
6
:
128  !

=
2
K
=3
:
90
Thedeningequationforthisellipticintegralcanbeevaluatedbytheseriesshownfornottoo
bigvaluesof
k
,andtheequationfollowingisanapproximateexpressionthatisusefulforlarger
k
(near
one).ItisfromAbramowitzandStegun,equation(17.3.26).
K
=
Z
=
2
0
d
1
p
k
2sin
2
=
2
"
1+
1
2
2
k
2
+
1
.
3
2
.
4
2
k
4
+
1
.
3
.
5
2
.
4
.
6
2
k
6
+
#
1
2
ln
16
k
2
as
k
!1
(4
:
65)
Thoughthisfunctionapproaches innity as
k
!1,itdoes soslowly. . Thelogarithmhasaverymild
sortofinnity.
Exercises
Doproblem0.59.
UsingEq.(4.11),comparethesizeofthecorrectiontermfor
=1
andfor
=89
.
ApplyEq.(4.18)toEq.(4.19).
Whathappenstotheequation(4.1)ifthereisawind,
~v
w
? Thiswindcanbeinanydirection.
DrawseveralequipotentialsforthepotentialenergyinEq.(4.24). Drawsomeforcevectorsinthe
samepicture.
Checktheradialforcefunction
~
F
(
x;y
)=(
k=r
2
)^
r
=(
k=r
3
)
~r
againstEq.(4.18).
Using only the e elementary y methods (before e you u ever heard of f a dierential l equation), Find the
relationsbetweenspeed,radius,angularfrequency,andmagneticeldstrengthforachargedparticle
movinginauniformmagneticeld.
RearrangeEq.(4.46)intermsofthecyclotronfrequencyat
z
=0,
qB
0
=m
,andtheradiusofthe
orbitat
z
=0,
r
0
.
CanyouusethesamemethodasinEq.(4.34)todene
e
~
B
.
?
10 IntheparagraphafterEq.(4.55),itsays\ananglethatisabout33=9timesgreater".Where
dothetwofactorsofthreecomefrom?
11 Ifapendulum clockhad anerrorequivalenttothecorrectioninEq.(4.64),howmuch wouldit
gainorloseinoneday(andwhich)?Andisthenumberinthisequationcorrect?
4|ThreeDimensionalMotion
146
Problems
4.1 Fillinthemissingstepsintakingthelimits,Eqs.(4.4)and(4.5). Also,ndthenextorderterms
in
b
fortheseequations|therstnon-vanishingcorrection.
4.2 DeriveEq.(4.7).
4.3 WhatisthemaximumheightasfoundfromeitherEq.(4.3)or(4.7).
4.4 Intheequations(4.3),ifyouturnogravitytheresultsshouldreducetothoseofEq.(2.16). Do
they?
4.5 ExpandEq.(4.7) for r small
b
,keeping terms upto thoselinearin
b
. Whatisthe e rangeofthe
projectile,assumingalevelsurface,andgettingyouranswercorrecttotermslinearin
b
.Isyouranswer
plausible?
4.6 Amassdropsfromrestwithoutairresistance,startingfromcoordinates
x
=
x
0
,
y
=
y
0
,
z
=0.
Computeitsangularmomentum(Eq.(1.13))abouttheoriginattime
t
;compareittothetorqueon
themass. Repeat t this,butpickyourorigin forthe angularmomentumandtorqueaboutthe point
(
x;y;z
)=(2
x
0
;
0
;
0).
4.7 Theequation(4.7)willtellyouwheretheprojectilehitstheground,therange. Theequationfor
thegroundis
y
=0ifyouassumethatthegroundislevel.Firsttakethecase
b
=0andsolveforthe
range;forthatyouhavethesimpleequations (4.4) and (4.4). . Forsmallairresistance(small
b
)the
rangeisclaimedtobeapproximately(2
v
2
0
=g
)[sin
cos
4
3
(
bv
0
=mg
)sin
2
cos
].Analyzethisclaim
forplausibilityorlackthereof.
4.8Derivetheexpressionintheprecedingproblem. ExpandEq.(4.7)totherstorderin
b
andnd
therstordercorrectiontotherange. Ifyouhaven’tdoneproblem0.59,nowwouldbeagoodtime;
thequadraticequationthatyou gethereisbestsolvedbyiteration. . Ifyoundsomethingnotquite
right about theanswer stated in the precedingproblem,nda betteranswer and then check it for
plausibility.
4.9_Ifyoureaprojectileupahillandwanttonditsrange,theequationforthegroundisnolonger
y
=0. Assumethatthehillisrepresentedby y astraightlineatan angle
from thehorizontal and
writeitsequation.Findtherangeofaprojectileredupsuchahill.Forthisproblem,assumezeroair
resistance. Doyoumeasurethepositionwhereithitshorizontallyoralongthehill? Yourchoice. . At
whatangledoyoureittogetthemaximumrange?(Andisthisanglemeasuredfromthehorizontalor
fromthehillitself?Again,yourchoice.) Doyouchooseyourcoordinatesystem
x
-
y
ashorizontaland
verticalordoyouchooseitparallelandperpendiculartothehill?Again,yourchoice,butIrecommend
thelatter. Doesyouranswerreducetothecorrectvaluesinallthespecialcasesforwhichyoucannd
theanswereasily?E.g.specialvaluesof
oroftheringdirection.
4.10 Theprojectileanalyzedinsection4.1nowhasaheadwind;theairhasahorizontalvelocity
v
air
^
x
.
Solveforthetrajectorynowandanalyzetheresult. Whatwindwillittaketobringthemassbackto
itsstartingpoint?
4.11 Following the methods ofsection 3.11,nd theGreen’sfunction solutionforequationsofthe
formEq.(4.2),butforanarbitraryforceontheright:
m
d2y
dt2
+
b
dy
dt
=
F
(
t
).
4.12_(a)NowusetheformulafromtheprecedingproblemandapplyittoEq.(4.2).
(b)Theresultdoesn’tagreewiththe
y
partofEq.(4.3)becausetheinitialcondition
v
y
(0)=
v
0
sin
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