Modern Classical Mechanics Solution Manual

456 Pages·2020·20.68 MB·English

by  T. M. Helliwell, V. V. Sahakian

Modern Classical Mechanics
Solution Manual
T. HELLIWELL
V. SAHAKIAN
©2020T.HelliwellandV.Sahakian
PublishedbyCambridgeUniversityPress
sahakian.physics.hmc.edu/mcm/
Preamble
Our book “Modern Classical Mechanics” includes more than 500 problems, attached as
usualinafewpagesattheendofeverychapter.Someproblemsareeasy–labeledby(*),
some are challenging tagged with (***), many others are somewhere in between and are
denotedby(**).Truthbetold,ittookusaboutaslongtocollectorinventtheseproblems,
solvethem,andtypesetthem,astowriteandtypesetthemainbodyofthetextbookitself.
Somewehaveassignedalreadymanytimesinourcourses,whileothersarenewlyminted.
It is of course through solving problems that most students learn, really learn, the
material. And for that to succeed it is obviously important to keep the solutions manual
confidential.Aseveryphysicistknows,itisonlythroughgrapplingwithproblemsoneself,
makingmistakes,followingblindalleys,andotherwisechallengingoneself,isthematerial
trulylearned.Havingsolutionswithwhichstudentscancomparetheirworkonlyattheend
thenbecomesaninvaluableresourceforlearningfromtheirmistakes.
Inthisproblemsolutionsmanualwehaveattemptedtosolvealloftheproblemsatthe
endofthechapters.Surprisingly,wehavebeenknowntomakemistakes.Theremayeven
betyposhereandthere.Sowehopethatpeoplewhousethebookandthismanualwilllet
usknowwhentheyfinderrorssothatwecancorrectthem.Wewillkeepanerratawebpage
updatedat:
sahakian.physics.hmc.edu/mcm
You can submit corrections and suggestions by contacting [email protected] via
email; please include the words “MCM errata” in the subject of the email. We will
periodicallyupdatethecontentofthebookletattheCambridgeUniversityPresswebsite.
Finally,wearegratefultoIvyYuanforhelpintypesettingthismanual.
iii

Part I
1 NewtonianParticleMechanics 3
2 Relativity 45
3 TheVariationalPrinciple 82
4 LagrangianMechanics 108
5 FromClassicaltoQuantumandBack 143

1 Newtonian Particle Mechanics
1.1 ProblemsandSolutions
(cid:2) Problem1.1 Ameterstickisatrestinaprimedframeofreference,withoneendattheorigin
andtheotheratx(cid:2) =1.0m.(a)UsingtheGalileantransformationfindthelocationofeach
endofthestickintheunprimedframeataparticulartimet,andthenfindthelengthofthe
meterstickintheunprimedframe.(b)Repeatforthecasethatthestickislaidoutalongthe
positivey(cid:2) axis,withoneendattheoriginandtheotheraty(cid:2) = 1.0m.Whatisthelength
ofthestickintheunprimedframe?
Solution
(a) x = x(cid:2) +vt(cid:2) = x(cid:2) +vt, so the left end has x = 0+vt = vt, and the right end has
l
x =1.0m+vt.Thereforethelength=x −x =1.0m.
r r l
(b) x(cid:2) = 0 for both ends in this case, and y(cid:2) = 0 and y(cid:2) = 1.0 m always. Therefore
x=x(cid:2)+vt=vtforbothends,andy=y(cid:2) =0and1.0mforthetwoends,sothelengthin
theunprimedframeisΔy=Δy(cid:2) =1.0m. (cid:2)
(cid:2) Problem1.2 AriverofwidthDflowsuniformlyatspeedVrelativetotheshore.Aswimmer
swimsalwaysatspeed2Vrelativetothewater.(a)Iftheswimmerdivesinfromoneshore
andswimsinadirectionperpendiculartotheshorelineinthereferenceframeoftheflowing
river,howlongdoesittakehertoreachtheoppositeshore,andhowfardownstreamhasshe
beensweptrelativetotheshore?(b)Ifinsteadshewantstoswimtoapointontheopposite
shoredirectlyacrossfromherstartingpoint,atwhatangleshouldsheswimrelativetothe
directionoftheriverflow,andhowlongwouldittakehertoswimacross?
Solution
(a)Hervelocityperpendiculartotheshorelineis2V,sothetimetoreachtheoppositeshore
ist= D.Duringthistime,sheisalsosweptdownstreamadistanced=Vt= D.
2V 2
(b)ShemusthaveanupstreamcomponentofvelocityVtomakeupfo(cid:2)rtheriverflow.
√FromthePythagoreantheorem,hervelocitycomponentacrosstheriveris (2V)2−V2 =
3Vandtheangle
V V 1
θ =tan−1 √ =sin−1 =sin−1 =30◦
3V 2V 2
3
4 1 NewtonianParticleMechanics
Thereforeheranglerelativetotheflowdirectionis30◦+90◦ = 120◦.Hertimetoswim
acrossist= √D . (cid:2)
3V
(cid:2) Problem1.3 The crews of two eight-man sculls decide to race one another on a river of
widthDthatflowsatuniformvelocityV .ThecrewofscullArowsdownstreamadistance
0
Dandthenbackupstream,whilethecrewofscullBrowstoapointontheoppositeshore
directly across from the starting point, and then back to the starting point. They begin
simultaneously,andeachcrewrowsatthesamespeedVrelativetothewater,withV>V .
0
Whowinstherace,andbyhowmuchtime?
Solution
A:Relativetotheshore,AhasvelocityV +VdownstreamandV−V upstream.Thetime
0 0
spentdownstreamis D andupstream D ,sothetotaltimeforAis
V0+V V−V0
D D D(V−V +V+V ) 2DV
+ = 0 0 = =t
V +V V−V V2−V2 V2−V2 A
0 0 0 0
(cid:3)
B: The velocity of B relative to the shore is V2−V2, so the total time across the
0
streamandbackforBist = √2D .Therefore
B V2−V2
0
⎡ ⎤
(cid:8) (cid:3) (cid:9)
t −t =2D⎣ V − (cid:3) 1 ⎦= 2D V− V2−V2 >0.
A B V2−V20 V2−V2 V2−V20 0
0
√
2D(V− V2−V2)
SoBwinstheracebyΔt= V2−V2 0 . (cid:2)
0
(cid:2) Problem1.4 Passengersstandinginacoastingspaceshipobserveadistantstaratthezenith,
i.e.,directlyoverhead.Ifthespaceshipthenacceleratestospeedc/100wherecisthespeed
oflight,atwhatangletothezenith(tothreesignificantfigures)dothepassengersnowsee
thestar?
Solution
Note that sinθ = c/100 = 1 (cid:3) θ (sinθ (cid:3) θ for θ (cid:4) 1). Alternatively, perhaps the
c(cid:3) 100
hypotenuseshouldbe c2+ c2 ,sotanθ = c/100 (cid:3) θ,soθ (cid:3) 1 ,thesameeitherway
104 c 100
tothreesignificantfigures. (cid:2)
(cid:2)(cid:2) Problem1.5 (a) Snow is falling vertically toward the ground at speed v. (a) A bus driver
is driving through the snowstorm on a horizontal road at speed v/3. At what angle to
the vertical are the snowflakes falling as seen by the driver? (b) Suppose that the large
windshield in the flat, vertical front of the bus has been knocked out, leaving a hole of
area A in the vertical plane. Given that N is the number of falling snowflakes per unit
horizontalareaperunittime,ifthebusmovesatconstantspeedv/3toreachadestination
5 1.1 ProblemsandSolutions
at distance d, how many snowflakes fall into the bus before the destination is reached?
(c) To minimize the total number of snowflakes that fall in, the driver considers driving
fasterorslower.Whatwouldbethebestspeedtotake?
Solution
(a) From the point of view of the ground, snowflakes fall straight down at speed
v, so from the point of view of the bus the snowflakes fall at an angle of
θ =tan−1 V/3 =tan−1(1).
V 3
(b) In a time t the volume swept into the bus is V = A(Vt) = Ad, so the number of
3
snowflakesenteringisNAd,regardlessofspeed.
(c)Thespeedofthebusdoesn’tmatter.Ifthebushashighervelocity,moresnowflakes
comeinperunittime,butthetimetotravelthedistancedisless.Thenumberofsnowflakes
enteringthebusisthesamewhetherthebusmovesfastorslow. (cid:2)
(cid:2)(cid:2) Problem1.6 The jet stream is flowing due east at velocity v relative to the ground. An
J
aircraftistravelingatvelocityv inthenortheastdirectionrelativetotheair.(a)Relative
C
totheground,findthespeedoftheaircraftandtheangleofitsmotionrelativetotheeast.
(b) Keeping the same speed v relative to the air, at what angle would the plane have to
C
move through the air relative to the east so that it would travel northeast relative to the
ground?
Solution
(a)Notethat
◦ ◦
v =v +v cos45 and v =v sin45
net,horizontal J C net,vertical C
√
◦ ◦
Therefore,since cos45 =sin45 =1/ 2,
itfollowsthat
(cid:10) √ (cid:11) √
(cid:4)v = v +v / 2 ˆx+(v / 2)ˆy
net J C C
(ˆx+ˆyareunitvectors).
√
v / 2
θ =tan−1 C √
v +v / 2
J C
(cid:3)
√ √
v = (v +v / 2)2+(v / 2)2
net J C 0
(b)Notethat
v sin45◦ =v sin(π−θ)=v sinθ.
net C C
Also
v =v cos45+v cos(π−θ)=v cos45−v cosθ
J net C net C
√
v =v sinθ−v cosθ =v (sinθ−cosθ)=v 2sin(θ−(π/4))
J c C C C
6 1 NewtonianParticleMechanics
since
1
sin(θ−(π/4))=sinθcos(π/4)−cosθsin(π/4)= √ (sinθ−cosθ).
2
Thus
v
sin(θ−(π/4))= √ J
2v
C
v
θ−(π/4)=sin−1 √ J
2v
C
π v
θ = +sin−1 √ J .
4 2v
C
√
Forexample,supposev =v cos45= 2v ;then
J C C
π π π 3π
θ = +sin−11= + = ,
4 4 2 4
whichiscorrect. (cid:2)
(cid:2) Problem1.7 The earth orbits the sun once/year in a nearly circular orbit of radius
150×106km. The speed of light is c = 3 × 105 km/s. Looking through a telescope,
we observe that a particular star is directly overhead. If the earth were quickly stopped
andmadetomoveintheoppositedirectionatthesamespeed,atwhatangletothevertical
wouldthesamestarnowbeobserved?
Solution
Thespeedoftheearth’sorbitisfoundfrom
−GM m mv2
F = ma: sun =− e
r2 r
(cid:12) e e
GM
⇒ v = sun
e
r
e
HereG=6.67×10−11 m3 ,M =2.0×1030 kg,r =1.5×1011 m.Therefore
kg s2 sun e
(cid:12)
(cid:2)
(2/3)·10−10·2·1030
v = = 8.89×108m/s=2.98×104m/s
e 1.5·1011
Thespeedoflightisc = 3×108 m/s,sosincev /c (cid:4) 1,θ (cid:3) ve = 2.98×104 ≈ 10−4
e c 3.0×108
radians.
Movingintheoriginaldirectionthestar(apparentlyoverhead)isactually10−4 radians
in the forward direction. So if the earth were moving in the opposite direction the star
wouldappeartobeθ (cid:3)2×10−4radiansfromthevertical. (cid:2)