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  • 1. II P6-1 CHAPTER6: MOTION IN TWO DIMENSIONS PLOJECTILE:ANY OBJECTTHAT ISPROJECTEDBY SOME FORCEAND CONTINUESIN MOTIONBY ITS OWNINERTIA (travelsin a curvedpath) TRAJECTORY:PATHOFA PROJECTILE -A curvedpathistheresultof two forces: 1)Theforceyouapplywhenyouthrowit makesit goforwardata constantspeed(theforwardspeeddoesnotchangebecausenootherforces areactingin thisdirectionafteryouthrowtheball) 2) All thetimetheballmoves,gravitykeepspullingontheball(gravity causesit's downwardspeedto keepincreasing) -thecombinationofforwardconstantspeedanddownwardincreasing speedcnusesthepath to curve o WE MUSTLOOK AT THEHORIZONTALAND VERTICAL COMPONENTSOFMOTIONSEPARATELY! PROJECTILELAUNCHEDHORIZONTALLY fra*e, of refuancl (sta^r'tr'ng .P ot i't) V;=inihalvetoct! Vix=Vi Viy=0 (no terttcal Corngonn* o? initial vclocity) tt
  • 2. I I P6-2 PROJECTILENOT LATINCHEDHORIZONTALLY(AT AN ANGLE FROM HORIZONTAL) "F c,?. / a - / l r / l r Ay t., rnaxirrnurnheiql'rt oscrrsin h;lf tha ti nlc (alwaysg) (no.horz.acceleration) dx dx € also catle&t*...e-ttrang" " STEPSFORSOLVINGPROJECTILEMOTIONPROBLEMS Drawyourpicture Establishyoursignconvention Write outyourgivenandwhatyouwantto find (divideit up:andy) Breakyourinitialvelocityintohorizontalandverticalcomponents Apply equationsin onlyonedirection,horizontallyor vertically Lookatwhatyou'resolvingfor to determinewhatto donext. f o':v"'-)s" d -v,*!ot'- Jv ' ' i ' 2 - o - - l a,=ri**lo.t' 2 -no horizontalacceleration,e*:0 -vertically, accelerationis alwaysg! 1 . 2. 3. 4. 5. 6.
  • 3. a a P6-3 EXAMPLE:(Projectilethrownhorizontally) A STONEIS THROWNHORIZONTALLYAT 15M/S. IT WAS THROWNFROMTHETOpOFA CLrFF44l|4HrcH. (A) HOWLONG DOESIT TAKE A STONETOREACHTHEBOTTOMOFTHE CLIFF?(B)HOWFARFROMTHEBASEOFTHECLIFFDOESTHE STONESTRIKE THE GROLTND? Gr-g,' * rnaKe x ' * g 'Y- ftx =O Vix=lsF t tDhere given (,q.) dy A1 LJnKnoarn: (Ot (h) d,x Vi? tSS &--.. $iven u -.#rarne of { rc+uenLa $=-q'8F/ V i $ = O r A9=*44'"'y' +D .firt ? tooKef tha slde of *hc Whuc Yoo aw 3 th rngs =vit + + gt' (viy=o) : i*' Lrgo,w*hj, -l"osofvc-R:t'L) tb) d* - d--r= Vix.::U+t artL [ar= Vlxt =(tstYs.os) O,alr.,4gtg) a d,r=E a
  • 4. I a P6-4 EXAMPLE:(Projectilelaunched atanangle) A GOLFBALL ISHIT AND LEAVESTHETEEWITH A VELOCITY OF25M/SAT 35OWITH RESPECTTOTHEHORIZONTAL.WHAT IS THEHORIZONTALDISPLACEMENTOFTHEBALL? 9iuq' 4 r = o l g r - Q . 8 ? " Vip=th I u,y=fr ,...rSY- -^fl -.. $'{'=o- - fri', *he dx.S €rnfrre +rr,a] -b ((Ak inihd t Velocr inlr hori zon-fzrl and, Vgnt er"t CorvrgoAent-r ' V=ufuT,,,,,,viy--Visrn2s'=(zsgXsrn?J) --,+..lF )r €Ju'I Vix = vrcoszs': LzEbrXco.r3f) a eo.sf* Vix - tl,a +o{"^l {rrne J*he praSachlcis rn rrofion dcpr,rrclsontlnerhifral and ffnat Vcrhcr*l vslo(-tu - or",4orrtd{ eft- it startsr {"o whert tf cnatq #e- Ve,rfical d.isgtaceftLrf ls tefr (er.=o) dy=Viytr t %t?u o ={l.t..t?)c{.t F1,SF)ta; [usa 4bradmficor&c1.r. ouf,t) O = lt.qe - - *{^c hon ?ontzrldrJ?Ilt-emerl- 4'9t = lY't{ L*=V,x1,rt axt.l-D d* =Viri =.lgo.s?Xe"Qrls) . @ l.qtL + af, = ?,1+S a
  • 5. a t P6-5 -uniformcircular motionresultswhena netforce, actingon a mass movingat a constantspeed,changesdirection in sucha way that theforce is alwaysacting at a right angle to thedirection in which themassis moving UNIFORM CIRCIJLAR MOTION: PARTICLESMOVING IN A CIRCULAR PATH AT A CONSTANTSPEED CENTRIPETAL ACCELERATION: ACCELERATION ALWAYS AT RIGHT ANGLES TO THE VELOCITY OF A PARTICLE -centripetalacceleration,like centripetalforce, is alwaysdirectedtoward thecenterof a circle. Thisis a resultof Ittewton'smqws. -"centripetal"meanscenter-seeking a"-centripetalacceleration(m/s2) v- velocity (m/s) r-radiusof circle(m) EXAMPLE:WHAT IS THE CENTRIPETALACCELERATIONOFAN OBJECTMOVINGIN A CIRCULARPATHOF2O.OM RADIUSWITH A SPEEDOF2O.OM/S? Lao.os" sD.oh Given: r=' lD.onn rf - at n ^..r, hh v - 4Ll.J 1 aL-*# J O.on1 I o
  • 6. o t P6-6 *seepage153of textbookfor vectordiagramsdescribingcircularmotion v' (znr' t ' l 4tt2r T2 = 4 , 2nr , = , ) A " : ; ) A " CENTRIPETALFORCE:FORCEDIRECTEDTOWARDTHE CENTEROFA CIRCLE,KEEPSPARTICLESMOVING IN LTNIFORM CIRCULAR MOTION Fr"t = ffiQ, fY'? D,2S Kg r = t,DOrr., v : s . o ? Fc mvz r EXAMPLE:40.25KG MASSISATTACHEDTOA 1.00M LENGTH OFSTRING.THEMASSIS MOVING IN A HORIZONTALCIRCLE AT A SPEEDOF 15.0M/S. WHAT IS THECENTRIPETALFORCE ACTINGON THEMASS? Given: FtnA"Fe e =!1!.t =e r r4 ( o . e s r g X t s . o F )- T b ( l.oonn) w.rL Fc '- sb N Iradlalty tn WAr."L) -properlyengineeredcurvesin aroadarebankedto furnishsuchaforce. -asatellitefollowinganorbitaroundaplanetor aplanetgoingaroundthe sunis heldin orbitby thecentripetalforcefurnishedby the gravitational attraction -astheplanetsmovearoundthesuntheyundergocentripetalacceleration, butthesearenotconstantin magnitude.Theplanetsmovearoundthesun in circularorbits. t a
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