Solution Manual for Fundamentals of Physics Extended 10th Edition - Chapter 2 & Problems

Soluton_ManualFundamentals_of_Physics_Exten j


ded_10th_Edition

1. Thejvelocityjofjthejcarjisjajconstant

vj =j+j(80j km/h)(1000j m/km)(1j h/3600j s)jˆij =j(+22jm s)ˆi,

andjthejradiusjofjthejwheeljisjrj=j0.66/2j=j0.33jm.

(a) Injthejcar’sjreferencejframej(wherejthejladyjperceivesjherselfjtojbejatjrest)jthejroadjisjmo
vingjtowardjthejrearjatjvroadj =j−vj=j−22jm
s,jandjthejmotionjofjthejtirejisjpurelyjrotational.jInjthisjframe,jthejcenterjofjthejtirejisj“fixed”js
ojvcenterj =j0.

(b) Sincejthejtire’sjmotionjisjonlyjrotationalj(notjtranslational)jinjthisjframe,jEq.j10-
18jgivesj vtopj =j(+22 jm/s) jˆi.

(c) Thejbottom-
mostjpointjofjthejtirejisj(momentarily)jinjfirmjcontactjwithjthejroadj(notjskidding)jandjhasjth
ejsamejvelocityjasjthejroad:jvbottomj =j(−22jm s)ˆij.jThisjalsojfollowsjfromjEq.j10-18.

(d) Thisjframejofjreferencejisjnotjaccelerating,jsoj“fixed”jpointsjwithinjitjhavejzerojaccelera
tion;jthus,jacenterj =j0.

(e) Notjonlyjisjthejmotionjpurelyjrotationaljinjthisjframe,jbutjwejalsojhavejj=jconstant,jwhi
chjmeansjthejonlyjaccelerationjforjpointsjonjthejrimjisjradialj(centripetal).jTherefore,jthejma
gnitudejofjthejaccelerationjis

v2 (22jm/s)2 3 2
atopj =j =j j =j1.510j m sj .
rj 0.33jmj j

(f) Thejmagnitudejofjthejaccelerationjisjthejsamejasjinjpartj(d):jabottomj =j1.5jj103jm/s2.

(g) Nowjwejexaminejthejsituationjinjthejroad’sjframejofjreferencej(wherejthejroadjisj“fixed”
jandjitjisjthejcarjthatjappearsjtojbejmoving).jThejcenterjofjthejtirejundergoesjpurelyjtranslatio
naljmotionjwhilejpointsjatjthejrimjundergojajcombinationjofjtranslationaljandjrotationaljmoti
ons.jThejvelocityjofjthejcenterjofjthejtirejisj vj=j(+22jmjs)ˆi.

(h) Injpartj(b),jwejfoundj vtop,carj =j+vj andjwejusejEq.j4-39:

vtop,jgroundj =jvtop,jcarj +jvcar,jgroundj =jvjˆij+jvjˆij =j2vjˆi


522

, 523



whichjyieldsj2vj=j+44jm/s.

(i) Wej canj proceedj asj inj partj (h)j orj simplyj recallj thatj thej bottom-
mostj pointj isj inj firmjcontactjwithjthej(zero-velocity)jroad.jEitherjway,jthejanswerjisjzero.

(j) Thejtranslationaljmotionjofjthejcenterjisjconstant;jitjdoesjnotjaccelerate.

(k) Sincej wej arej transformingj betweenj constant-
velocityj framesj ofj reference,j thejaccelerationsjarejunaffected.jThejanswerjisjasjitjwasjinjp
artj(e):j1.5jj103jm/s2.

(1)jAsjexplainedjinjpartj(k),jaj=j1.5jj103jm/s2.

2. Thejinitialjspeedjofjthejcarjis

vj=j(80jkm/hj)(1000jm/km)(1jh/3600js)j=j22.2jm/sj.

ThejtirejradiusjisjRj=j0.750/2j=j0.375jm.

(a) Thejinitialjspeedjofjthejcarjisjthejinitialjspeedjofjthejcenterjofjmassjofjthejtire,jsojEq.j11-
2jleadsjto
v 22.2jm/sj
j =j comj0j =j =j59.3jrad/s.
0
R 0.375jm

(b) Withjj=j(30.0)(2)j=j188jradjandjj=j0,jEq.j10-14jleadsjto

(59.3jrad/s)2j
 2 j =j02 j + 2  j =j = 9.31jrad/s2j.
2j(188jrad)

(c) Equationj11-1jgivesjRj=j70.7jmjforjthejdistancejtraveled.

3. THINKjThejworkjrequiredjtojstopjthejhoopjisjthejnegativejofjthejinitialjkineticjenergyjofjt
hejhoop.

EXPRESSj Fromj Eq.j 11-5,j thej initialj kineticj energyj ofj thej hoopj isj Kj =jij1j I2 2j+jj1j mv22j,
wherej Ij =j mR2j isj itsj rotationalj inertiaj aboutj thej centerj ofj mass.j Eq.j 11-
2j relatesj thejangularjspeedjtojthejspeedjofjthejcenterjofjmass:jj=jv/R.jThus,
2
Kj =j 1jI  2 j +j 1jmv2j = 1 j(mR2j) j vj  +j 1jmv2j =jmv2
i j Rj 2
2 2 2  

ANALYZEjWithjmj=j140jkg,jandjthejspeedjofjitsjcenterjofjmassjvj=j0.150jm/s,jwejfindjtheji
nitialjkineticjenergyjtojbe

,524 CHAPTERj11


Ki j =jmv2j =j(140jkg)(0.150jm/s) =j3.15jJ
2j




whichjimpliesjthatjthejworkjrequiredjisj Wj =jKj =jKjfj −jKij =j−Kij =j−3.15 j Jj.

LEARNjByjthejwork-
kineticjenergyjtheorem,jthejworkjdonejisjnegativejsincejitjdecreasesjthejkineticjenergy.jAjrol
lingjbodyjhasjtwojtypesjofjkineticjenergy:jrotationaljandjtranslational.

4. Wejusejthejresultsjfromjsectionj11.3.

(a) Wejsubstitutej Ij =jj2j5MjR2j (Tablej10-2(f))jandjaj=j–j0.10gjintojEq.j11-10:

gjsin jgjsin
−0.10gj=j−j =j−j
j


c h
1+ j 2j5 MR2j MR2 7j/j5

whichjyieldsjj=jsin–1j(0.14)j=j8.0°.

(b) Thejaccelerationjwouldjbejmore.jWejcanjlookjatjthisjinjtermsjofjforcesjorjinjtermsjofjener
gy.jInjtermsjofjforces,jthejuphilljstaticjfrictionjwouldjthenjbejabsentjsojthejdownhilljaccelerat
ionjwouldjbejduejonlyjtojthejdownhilljgravitationaljpull.jInjtermsjofjenergy,jthejrotationalj te
rmj inj Eq.j 11-5j wouldj bej absentj soj thatj thej potentialj energyj itj startedj with
wouldjsimplyjbecomej 1j2mv2j(withoutjitjbeingj“shared”jwithjanotherjterm)jresultingjinjajgrea
j



terjspeedj(and,jbecausejofjEq.j2-16,jgreaterjacceleration).

5. Letj Mj bej thej massj ofj thej carj (presumablyj includingj thej massj ofj thej wheels)j andj vj be
itsjspeed.jLetjIjbejthejrotationaljinertiajofjonejwheeljandjjbejthejangularjspeedjofjeachjwhee
l.jThejkineticjenergyjofjrotationjis j1j
K =j4 Ij2j ,
F
G IJ
rot
H2 K
wherej thej factorj 4j appearsj becausej therej arej fourj wheels.j Thej totalj kineticj energyj isjgiv
enjby
Kj=jj1j Mv2j +j4(1jIj2j)j. j

2 2


Thejfractionjofjthejtotaljenergyjthatjisjduejtojrotationjis

Krotjj 4Ij2
fractionj=j j
= .
K Mv2j +j4Ij2j

Forj aj uniformj diskj (relativej toj itsj centerj ofj mass)j Ij =jj1j mR2
2 j (Tablej 10-

2(c)).j Sincej thejwheelsjrolljwithoutjslidingjj=jv/Rj(Eq.j11-
2).jThusjthejnumeratorjofjourjfractionjis

, 525

F1 IFv I
j j j
2


4Ij2j = 4GmR JG J=2mv 2 2
j
H2
KRHK j j



andjthejfractionjitselfjbecomes
2mv2 2m 2j(10) 1
fractionj= = = = =j0.020.
Mv j+j2mv
2 2
Mj +j2m 1000 50

Thejwheeljradiusjcancelsjfromjthejequationsjandjisjnotjneededjinjthejcomputation.

6. Wejplugjaj=j –
j3.5jm/s j(wherejthejmagnitudejofjthisjnumberjwasjestimatedjfromjthej“risejoverjrun”jinjthejg
2

raph),jj=j30º,jMj=j0.50jkg,jandjRj=j0.060jmjintojEq.j11- .
10jandjsolvejforjthejrotationaljinertia.j WejfindjIj=j7.2jj10−4j kg.m2

7. (a)jWejfindjitsjangularjspeedjasjitjleavesjthejroofjusingjconservationjofjenergy.jItsjinitialj
kineticj energyj isj Kij =j 0j andj itsj initialj potentialj energyj isj Uij =j Mghj wherejhj=j6.0sinj
30j=j3.0jmj(wejarejusingjthejedgejofjthejroofjasjourjreferencejleveljforjcomputingjU).jItsjfina
ljkineticjenergyj(asjitjleavesjthejroof)jisj(Eq.j11-5)

Kj j =jj1j Mv2j+jj1j Ij2j.
f 2 2



Herejwejusejvjtojdenotejthejspeedjofjitsjcenterjofjmassjandjjisjitsjangularjspeedj—
jatj thejmomentjitjleavesjthejroof.jSincej(upjtojthatjmoment)jthejballjrollsjwithoutjslidingjwejc

anjsetj vj =jRj =j vjwherej Rj =j0.10j m.j Usingj Ij =jj1j MR22j (Tablej10-2(c)),j conservationj of
energyjleadsjto
1j 1j 1j 1j 3j
Mghj=j Mv2j+j I2j=j MR22j+j MR22j=j MR22.
2 2 2 4 4

ThejmassjMjcancelsjfromjthejequation,jandjwejobtain

1 4j 1
j =j
j

R 3
gh =j
0.10jm 3
c
4j
hb g
9 .8 jmjs2j 3.0j m =j63jradj s.


(b)jNowjthisjbecomesjajprojectilejmotionjofjthejtypejexaminedjinjChapterj4.jWejputjthejorigi
njatjthejpositionjofjthejcenterjofjmassjwhenjthejballjleavesjthejtrackj(thej“initial”jpositionjforj
thisjpartjofjthejproblem)jandjtakej+xjleftwardjandj+yjdownward.jThejresultj ofjpartj(a)jimplie
sjv0j =jRj=j6.3jm/s,jandjwejseejfromjthejfigurejthatj(withjthesejpositivejdirectionjchoices)jitsj
componentsjare

v0xj =jv0jcosj30j=j5.4jm s
v0jyj =jv0jsinj30j=j3.1jm s.

Thejprojectilejmotionjequationsjbecome