CHAPTER 6:UNIFORMCIRCULARM OTION ANDGRAVITATION

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NGLEANDANGULARVELOCITY1.Semi- lometersshouldtheodometerread?SolutionGiven:d 1.15 m r 1.15 m2π rad 0.575 m , Δθ 200,000 rot 1.257 10 6 rad21 rotFindΔsusingΔθ Δs,sothatr()Δ s Δ θ r 1.257 10 6 rad (0.575 m)5 7.226 1 0 m 723 ersecond?Whatisthisinrev/min?SolutionGiven:r 0.420 m, v 32.0 m s .v 32.0 m sUse ω 76.2 rad s.r 0.420 mConverttorpmbyusingtheconversionfactor:49

CollegePhysicsStudentSolutionsManualChapter61 rev 2π rad ,1 rev60 s 2π rad 1 min 728 rev s 728 rpmω 76.2 rad s ion(a)Usev rωtofindthelinearvelocity:2 π rad 1 min v rω (0.100 m ) 50,000 rev/min 524 m/s 0.524 km/s1 rev60 s rad1y(b)Given:ω 2π 1.988 10 7 rad s ; r 1.496 1011 m7y 3.16 10 sUsev rωtofindthelinearvelocity:()()v rω 1.496 1011 m 1.988 10 -7 rad s 2.975 10 4 m s 29.7 km 00mradiuscurvebankedata20.0 angle?50

onUsingtan θ v2gives:rgv2tan θ v rgtan θ rg(100 m)(9.8 m s 2 )tan 20.0 18.9 m ttheNorthPoleis9.830 m/s 2 mparethiswiththeacceptedvalueof5.979 10 24 kg .Solution(a)Usingtheequationg GMgives:r22()()6371 10 3 m 9.830 m s 2GMr2gg 2 M 5.979 10 24 kg 1122Gr6.673 10 N m tdistancetoEarth,some6.29 1011 neffect,muchlessthatanunknownforcecausesit.)51

on(a)UseF GMmtocalculatetheforce:r2()GMm 6.673 10 11 N m 2 kg 2 (100 kg )(4.20 kg )Ff 2 7.01 10 7 N2r(0.200 m)(b)ThemassofJupiteris:mJ 1.90 10 27 kgFJ (6.673 10 11)()N m 2 kg 2 1.90 10 27 kg (4.20 kg )(6.29 1011m2) 1.35 10 6 NFf 7.01 10 -7 N 0.521FJ 1.35 10 -6 sactualmass.Usingr3G 2 M ,wecansolvethemassofJupiter:2T4π4π 2 r 3MJ G T23()4π 24.22 10 8 m 6.673 10 -11 N m 2 kg 2 (0.00485 y ) 3.16 10 7 s y[()]2 1.89 10 27 tisinanorbitofthesame52

thatintersectsthesatellite’sorbitatanangleof90 terthantherivet’s.)Solution(a)UseFc mac,thensubstituteusinga v2GmMandF .r2rGmM mv 2 rr2GM Ev rS(6.673 10 11)()N m 2 kg 2 5.979 10 24 kg 2.11 10 4 m s900 10 3 (a):()v tot v 2 v 2 2v 2 2 2.105 10 4 m s 2.98 10 4 m s(c)Usingkinematics:d vtott t (d)F (d3.00 10 3 m 1.01 10 7 s4vtot 2.98 10 m s)()Δp mvtot 0.500 10 3 kg 2.98 10 4 m s 1.48 108 N-7Δtt1.01 10 e’sframeofreference,vi v tot , and vf 0.So,thechangeinthekineticenergyoftherivetis:Δ KE 211122mvtot mvi 0.500 10 3 kg 2.98 10 4 m s 0 J 2.22 10 5 J222()(53)

College"Physics" Student"Solutions"Manual" Chapter"6" " 50" " 728 rev s 728 rpm 1 min 60 s 2 rad 1 rev 76.2 rad s 1 rev 2 rad , π ω π " 6.2 CENTRIPETAL ACCELERATION 18." Verify&that ntrifuge&is&about 0.50&km/s,∧&Earth&in&its& orbit is&about p;linear&speed&of&a .

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