Fundamentals of Physics Extended 10th Edition – Complete Solution Manual (Ch. 11 Rolling, Torque, Angular Momentum) – Verified Answers & Step-by-Step Solutions

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ded_10th_Edition
fH




1. The fHvelocity fHof fHthe fHcar fHis fHa fHconstant

v f H = fH+fH(80 f H km/h)(1000 f H m/km)(1 fHh/3600 fH s) fHˆi f H = fH(+22fHm s)ˆi,

and fHthe fHradius fHof fHthe fHwheel fHis fHr fH= fH0.66/2 fH= fH0.33 fHm.

(a) In fHthe fHcar’s fHreference fHframe fH(where fHthe fHlady fHperceives fHherself fHto fHbe fHat fHrest)
fHthe fHroad fHis fHmoving fHtoward fHthe fHrear fHat fHvroad f H = fH−v fH= fH−22 fHm s, fHand fHthe fHmotion


fHof fHthe fHtire fHis fHpurely fHrotational. fHIn fHthis fHframe, fHthe fHcenter fHof fHthe fHtire fHis fH“fixed”

fHso fHvcenter f H = fH0.




(b) Since fHthe fHtire’s fHmotion fHis fHonly fHrotational fH(not fHtranslational) fHin fHthis fHframe, fHEq.
ˆ
fH10-18 fHgives f H vtop f H = fH(+22 fHm/s) fH i.




(c) The fHbottom-most fHpoint fHof fHthe fHtire fHis fH(momentarily) fHin fHfirm fHcontact fHwith fHthe
ˆ
fHroad fH(not fHskidding) fHand fHhas fHthe fHsame fHvelocity fHas fHthe fHroad: fHvbottom f H = fH(−22 fHm s) i


fH. fHThis fHalso fHfollows fHfrom fHEq. fH10-18.




(d) This fHframe fHof fHreference fHis fHnot fHaccelerating, fHso fH“fixed” fHpoints fHwithin fHit fHhave
fHzero fHacceleration; fHthus, fHacenter f H = fH0.




(e) Not fHonly fHis fHthe fHmotion fHpurely fHrotational fHin fHthis fHframe, fHbut fHwe fHalso fHhave fH fH=
fHconstant, fHwhich fHmeans fHthe fHonly fHacceleration fHfor fHpoints fHon fHthe fHrim fHis fHradial

fH(centripetal). fHTherefore, fHthe fHmagnitude fHof fHthe fHacceleration fHis



v2 (22 fHm/s)2 3 2
atop f H = f H = fH fH =fH1.510 f H ms f H .
r fH 0.33 fHm f H fH




(f) The fHmagnitude fHof fHthe fHacceleration fHis fHthe fHsame fHas fHin fHpart fH(d): fHabottom f H = fH1.5 fH fH103
2
fHm/s .




(g) Now fHwe fHexamine fHthe fHsituation fHin fHthe fHroad’s fHframe fHof fHreference fH(where fHthe
fHroad fHis fH“fixed” fHand fHit fHis fHthe fHcar fHthat fHappears fHto fHbe fHmoving). fHThe fHcenter fHof fHthe

fHtire fHundergoes fHpurely fHtranslational fHmotion fHwhile fHpoints fHat fHthe fHrim fHundergo fHa

fHcombination fHof fHtranslational fHand fHrotational fHmotions. fHThe fHvelocity fHof fHthe fHcenter

ˆ
fHof fHthe fHtire fHis f H v fH= fH(+22fHm fHs) i.




(h) In fHpart fH(b), fHwe fHfound f H vtop,car f H = fH+v f H and fHwe fHuse fHEq. fH4-39:

vtop, fHground f H
= fHvtop, fHcar f H + fHvcar, fHground f H
= fHvfHˆi fH+ fHvfHˆi fH = fH2vfHˆi

,522

,
, 523



which fHyields fH2v fH= fH+44 fHm/s.

(i) We f H can f H proceed f H as f H in f H part f H (h) f H or f H simply f H recall f H that f H the f H bottom-most
f H point f H is f H in f H firm fHcontact fHwith fHthe fH(zero-velocity) fHroad. fHEither fHway, fHthe fHanswer

fHis fHzero.




(j) The fHtranslational fHmotion fHof fHthe fHcenter fHis fHconstant; fHit fHdoes fHnot fHaccelerate.

(k) Since f H we f H are f H transforming f H between f H constant-velocity f H frames
f H of f H reference, f H the fHaccelerations fHare fHunaffected. fHThe fHanswer fHis fHas fHit fHwas fHin

fHpart fH(e): fH1.5 fH fH10 fHm/s .
3 2



(1) fHAs fHexplained fHin fHpart fH(k), fHa fH= fH1.5 fH fH103 fHm/s2.

2. The fHinitial fHspeed fHof fHthe fHcar fHis

v fH= fH(80 fHkm/hHf)(1000 fHm/km)(1 fHh/3600 fHs) fH= fH22.2 fHm/s fH.

The fHtire fHradius fHis fHR fH= fH0.750/2 fH= fH0.375 fHm.

(a) The fHinitial fHspeed fHof fHthe fHcar fHis fHthe fHinitial fHspeed fHof fHthe fHcenter fHof fHmass fHof fHthe
fHtire, fHso fHEq. fH11-2 fHleads fHto
v 22.2 fHm/s f H
 f H = fH comfH0 f H = fH = fH59.3 fHrad/s.
0
R 0.375 fHm

(b) With fH fH= fH(30.0)(2) fH= fH188 fHrad fHand fH fH= fH0, fHEq. fH10-14 fHleads fHto

(59.3
2 f H
=fH 2 0f H + 2   2f Hf H= fH 9.31 fHrad/s2 fH.
fHrad/s)
=
2Hf(188 fHrad)

(c) Equation fH11-1 fHgives fHR fH= fH70.7 fHm fHfor fHthe fHdistance fHtraveled.

3. THINK fHThe fHwork fHrequired fHto fHstop fHthe fHhoop fHis fHthe fHnegative fHof fHthe fHinitial
fHkinetic fHenergy fHof fHthe fHhoop.




f H is f H K f H = fH
fH1
EXPRESS f H From f H Eq. f H 11-5, f H the f H initial f H kinetic f H energy f H of f H thei f H hoop
2 2
fH
I2 fH+ fHfH1 f H mv2Hf,
where f H I f H = f H mR2 f H is f H its f H rotational f H inertia f H about f H the f H center f H of f H mass.
f H Eq. f H 11-2 f H relates f H the fHangular fHspeed fHto fHthe fHspeed fHof fHthe fHcenter fHof fHmass: fH fH=

fHv/R. fHThus,


2
2 1 12 f H 1 fH(mR2  f H v f H + fH1 fHmv2
= fHI  + fH
f=H fH = fHmv2
f H
K
fHmv
f H fH
)
i fHR fH 2
2 2 2  

ANALYZE fHWith fHm fH= fH140 fHkg, fHand fHthe fHspeed fHof fHits fHcenter fHof fHmass fHv fH= fH0.150