Fundamentals of Physics Extended 10th Edition Solution Manual | Halliday, Resnick & Walker – 2025 Step-by-Step Physics Solutions

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ded_10th_Edition

1. Thegvelocitygofgthegcargisgagconstant

vg =g+g(80g km/h)(1000g m/km)(1gh/3600g s)gˆig =g(+22gm s)ˆi,

andgthegradiusgofgthegwheelgisgrg=g0.66/2g=g0.33gm.

(a) Ingthegcar’sgreferencegframeg(wheregthegladygperceivesgherselfgtogbegatgrest)gthegroadgi
sgmovinggtowardgthegreargatgvroadg =g−vg=g−22gm
s,gandgthegmotiongofgthegtiregisgpurelygrotational.gIngthisgframe,gthegcentergofgthegtiregisg“f
ixed”gsogvcenterg =g0.

(b) Sincegthegtire’sgmotiongisgonlygrotationalg(notgtranslational)gingthisgframe,gEq.g10-
18ggivesg vtopg =g(+22 gm/s) gˆi.

(c) Thegbottom-
mostgpointgofgthegtiregisg(momentarily)gingfirmgcontactgwithgthegroadg(notgskidding)gandgh
asgthegsamegvelocitygasgthegroad:gvbottomg =g(−22gm s)ˆig.gThisgalsogfollowsgfromgEq.g10-
18.

(d) Thisgframegofgreferencegisgnotgaccelerating,gsog“fixed”gpointsgwithingitghavegzerogacc
eleration;gthus,gacenterg =g0.

(e) Notgonlygisgthegmotiongpurelygrotationalgingthisgframe,gbutgwegalsoghavegg=gconstant
,gwhichgmeansgthegonlygaccelerationgforgpointsgongthegrimgisgradialg(centripetal).gTherefor
e,gthegmagnitudegofgthegaccelerationgis

v2 (22gm/s)2 3 2
atopg =g =g g =g1.510g m sg .
rg 0.33gmg g

(f) Thegmagnitudegofgthegaccelerationgisgthegsamegasgingpartg(d):gabottomg =g1.5gg103gm/s2.

(g) Nowgwegexaminegthegsituationgingthegroad’sgframegofgreferenceg(wheregthegroadgisg“fi
xed”gandgitgisgthegcargthatgappearsgtogbegmoving).gThegcentergofgthegtiregundergoesgpurely
gtranslationalgmotiongwhilegpointsgatgthegrimgundergogagcombinationgofgtranslationalgandg

rotationalgmotions.gThegvelocitygofgthegcentergofgthegtiregisg vg=g(+22gmgs)ˆi.

(h) Ingpartg(b),gwegfoundg vtop,carg =g+vg andgwegusegEq.g4-39:

vtop,ggroundg =gvtop,gcarg +gvcar,ggroundg =gvgˆig+gvgˆig =g2vgˆi

,522

,
, 523



whichgyieldsg2vg=g+44gm/s.

(i) Weg cang proceedg asg ing partg (h)g org simplyg recallg thatg theg bottom-
mostg pointg isg ing firmgcontactgwithgtheg(zero-
velocity)groad.gEithergway,gtheganswergisgzero.

(j) Thegtranslationalgmotiongofgthegcentergisgconstant;gitgdoesgnotgaccelerate.

(k) Sinceg weg areg transformingg betweeng constant-
velocityg framesg ofg reference,g thegaccelerationsgaregunaffected.gTheganswergisgasgitgwas
gingpartg(e):g1.5gg10 gm/s .
3 2



(1)gAsgexplainedgingpartg(k),gag=g1.5gg103gm/s2.

2. Theginitialgspeedgofgthegcargis

vg=g(80gkm/hg)(1000gm/km)(1gh/3600gs)g=g22.2gm/sg.

ThegtiregradiusgisgRg=g0.750/2g=g0.375gm.

(a) Theginitialgspeedgofgthegcargisgtheginitialgspeedgofgthegcentergofgmassgofgthegtire,gsogEq
.g11-2gleadsgto
v 22.2gm/sg
g =g comg0g =g =g59.3grad/s.
0
R 0.375gm

(b) Withgg=g(30.0)(2)g=g188gradgandgg=g0,gEq.g10-14gleadsgto


2 g =g20g + 2 2g
(59.3grad/s)2g 9.31grad/s .
g = g
=
2g(188grad)

(c) Equationg11-1ggivesgRg=g70.7gmgforgthegdistancegtraveled.

3. THINKgThegworkgrequiredgtogstopgtheghoopgisgthegnegativegofgtheginitialgkineticgenerg
ygofgtheghoop.

EXPRESSg Fromg Eq.g 11-5,g theg initialg kineticg energyg ofg theg hoopg isg K i g =2g
g1g
I2g2+gg1g mv2g,
whereg Ig =g mR2g isg itsg rotationalg inertiag aboutg theg centerg ofg mass.g Eq.g 11-
2g relatesg thegangulargspeedgtogthegspeedgofgthegcentergofgmass:gg=gv/R.gThus,
2
2gg1 gI  +1
g =
2g 1 2g g vg  +g1 gmv2g =gmv2
gmv =
K g (mR )
g
gR g 2
 
i
2 2 2

ANALYZEgWithgmg=g140gkg,gandgthegspeedgofgitsgcentergofgmassgvg=g0.150gm/s,gwegfin
dgtheginitialgkineticgenergygtogbe