hwsol3 - A Wild Ride A car in a roller coaster moves along...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A Wild Ride A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axis be parallel to the ground and the positive y axis point upward. In the time interval from t = to t= 4s, the trajectory of the car along a certain section of the track is given by r = A(1 m/s)t x + A [(1 m/s )t 3 3 - 6(1 m/s:l)t2] Ji where ftis a positive dimensionless constant. P art A : Att = 2.0 3, t h e ro IIer coaster car ascen d'mg or d escen d'? IS mg, l' t'~ y~ " InC. v(.~ror l'~e roOt,.. r "() t . '~ [3 Yh - ~ Y contjOonenl-s coo..s+er ts ole.sce,noUn~ l il rht.. y-c.cml'0ne;.t ,is ,f-s tAt J..s _ ~ t.ral : _ ...;'1r y - A Part B: Derive a general expression for the speed 'llOfthe car, '1mJ ='1 s" lle8o.+iv(. d.e.s c.SJ'\ 0<.,'1"1 ~ I tt ~!.)1 V -:.. tJ. 12 d:t" ~ .:: A (I 'i's);V:: A ,Itt... Itl(l.~"; +- A of- [3 (l "'Is) )t"J. - 6 J (/ '"Is") t ] ~3 9 (l "t); - A L) t ~- 1.:1 ~ ry l.s : ~ -t- ] 9 rlJeLe... (!)t- -rh- ve.Joc i V =-[A ~ r A? ( 3 t J. I 'J. t )T /s =- A ~-I-+--r-t-2(-t-_-'1~)2,=- T Part C: The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 mfs. Find the maximum value of Aallowed by these regulations. We.. C Clt\ fincJ.. VmD.)( h y re.cfAlI J'" ~ rh~1 mo..X; fll'a. oc.c.uJ' when: olv(t} _ J..t; - 0 I dav(t), tA.t :1,- 0 -~ fA\! Cii =- CD A { I of 'ft"(t f; cJ.. Y tn ,'s UwjO ol -'1tl~ (I r 1t'..t-"I) ~) :z.. 1if; (t - "I) ') (t- -;)) =~ ( 0 ~ J .- _ cJ.~ :::. /BAt (t-J.) (t-'i) _ CIt { / +- ift ~(t--Lt)a;V~ - .,~-t"= () ttl1()..x J..&~r>AIlLJ, ~ .I ~ / { d.~V tifa:" t-.t'A (3 t~):J. 1:,-19")_ A{lk {, of- 'It ~(t-ll , .. ))Y:.. 3-6t2."'Kt))'J. ------~ o-t (I +- 'N:''( /: -4 )/h. (t (s ::t.s ~ yvJ-l( G~::.A <l- S . til .. fi'15 Direction of Acceleration of Pendulum The pendulum shown makes a full swing from -r./4 to +1T/4 . Ignore friction and assume that the string is massless. The eight labeled arrows represent directions to be referred to when answering the following questions. Part A: Which of the following is a true statement about the acceleration of the ii pendulum bob, . ..;>. C E*A B H G ()... ::: tJ... V cLt .> . - the . lYlsfa.nfo..ne.ous r()..r-e.. of c..ha.n -e... Cj c> f fit e.. ve.l 0 cl +-y Part B: What is the direction of iiwhen the pendulum is at position I? H ..:. -Y + h e. fA. c. c. e.l erO.:t id n ~ u../l.s \ \ p(Jil'lr i+ b a..c, I" 1-1> the.. e'OtJ J; briuJ't1 Part C: What is the direction of if at the moment the pendulum passes position 2? OJ- fh IS p (),.. " eire!)..1 ~ +1 +-h e.. ~ )IYI 01-; 0'" is .s; m j Jo...r I---a un r'1o-rh\ ytlot-ion [if] f!!" Part D: What is the direction of awhen the pendulum reaches position 3? m Af fdY' fAe S""Il1e. ;eo.sortS 0-+1 'po./' t- 6 Part E: As the pendulum approaches or recedes from which position(s) is the acceleration vector ;; almost parallel to the velocity vector u. 'pC> S; ti 0 n S 1. o...l1d... o./f 3 V II ~ CA-S wz sn 0 we,cJ.- 1 II .. 'p B 9--tJ .. Problem 3.40 An airplane pilot wishes to fly due west. A wind of 88.0 km/h is blowing toward the south. Part A: If the airspeed of the plane (its speed in still air) is 355 klll/h, in which direction should the -rAe...,.,./ I I\!- pilot head? ...r um SlJ1)\ /'" 01- P t;' - ~ <f-" 11 W s hOLtI ot P 0/11 f d.. ire e+all y t J, - ~ ~ I(/G-S +-, e.- IV Q (' J 0 u...fh or / P IfA Yle J~o ' 1) Ie{ be- 0 0 + ~-~--C:.1 t'w ~. p. -.}/"'"1J S 110(;- E. p .1.5 5 ..fl' YI {) tf 8' WI",..( .::= 0 ---, /1 C/l-' C"J) ~ C355 ;;:.$.= / '1.35 ~ /8 f -rhe pi/of- sAouJc1 Aeexot 1'1.35 Nor-t-A ofU/es f-, Part B: What is the speed of the plane over the ground? 1~ Yl K \ ( \II):>" 0~ -the- +r a j e.., CJ'ori ~s a...t a.. .J-r,' D.. Yl B/e + ( f; +-w) J..::::. p 'J., 1 F t W /.::. l:~55r':..(8'o)~ J WG P-tw tAus ~ Yound the l s.' pilots 9;). sfe-eot. f?efcA.+' \Ie.. +() t-he- v=- Y13. j( YOII ~ yV1/s fOil cou JrJ hfA v- Q/S" Us/n~ b()...S/c, nd r), ,'5 fr,tj 41 A Canoe on a River A canoe has a velocity of 0.400 southeast relative to the earth. The canoe is on a river that is flowing at 0.510 y m/s mls W+E S -> east relati ve to the earth. vl/~~ Part A: Find the magnitude of the velocity Vc/rof the canoe relative to the river. ~1\ x ~ oh 5 er veY' f Y' lo (JA.+ j n j VA ::; ~/ e ~ - - V lYe 11'1 the..... \.41 i V e -r S e. e..s .,..,!; e. VA -.:. c ~ no..e m 0 vi Y1~ i~ 1/e 10 C; -}y -::> -> 0t ::;.. Ye.. V - V Y/e A Fi"cJ. ~ C.Of'YIfJonen-l-s ;:::. O. 'f cos 'i5 ::' - o. tj s i"" ,, ~ 5 V?'e.. =- - Ve. O. 61D X A. .5""'<{ sin i/A=-( o. 'I co f.,'",oJ. i ;t:J lvA/~ ~ s ~ 5 - O. 5 10 ) X - o. WI (;l y .,.It e. j .,;-uct. e. + 0f f his: ve. C /-or : 0101875 o/s ) Part B: Find the direction of the velocity of the canoe relative to the river. IiA -:: ~ {-. ~<7 ;K - ~ .. 28':J..3 A Y f~Y\ -.2gJ.8y &::. -:r~ ;l7A"" ,28 ()vf'C e =- -1-0.. n ( -').2.7 ;l.s') SOlAtA 0';' J,A/ e.s +- Projectile Motion Tutorial A projectile is fired from ground level at time t= 0, at an angle ()with respect to the horizontal. It has an initial speed 1.'0. In this problem we are assuming that the ground is level. Part A: Find the time tllit takes the projectile to reach its maximum height. VXl):; y . - .- - - - - - - ."" i -; -~-';. - - - -- , i i ,, H ,, , , , 1'0/ II I I ( , I I ,, , , I , , \ \ \ , , , \ \ \ \ \ \ V" COSe ttt Vto:=.v,Sif\e , I a.+- x (J:= VoSine-jtH r R -I tH'=~Sine ~ Part B: Find tR, the time at which the projectile hits the ground. 1A e.. +-,. 'Yl e... i /- rCA-Ii e. -t-o V't...o-CA f he. o..f e.,x is ha. If /-CJf-CA-i f-/yV1 e -r), e-re.-f-o,e : if< ::: JtH - ~'!Q Sil1B ~ ---------...- Part C: Find H, the maximum height attained by the projectile :J X Y ::=. ~l-.e~ V{3 ~ (Jf' ~ UJe- o :::y" 2.-..10.. HV .J in ?-<9 ==- ;),[) 2- H H~ v ~ Sj()"e- .;l~ Part D: Find the total distance R(often called the range) traveled in the x direction; in other words, find where the projectile lands. f< ::. Vp'o ~::: t r<. & ~s;,.,e:::: 3 av.'J. Yo c..OS 5 () S'J1(fCCJSeJ . The Archerfish Part A: At what speedvshould an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the 60 pond and spits the water at an angle above the water surface. (roy() -th~ pre.vlofJs prtJ>hlem) 0 we.. JIlf\OW' -rhQ..t:. f< ::;, "lI. 'l. do..!!- e5 i t1 ~ e c. 0 s f!) fot" 110 60 ~ 6. 600 WI SolviY\<j ~.C!J ::::-Vo Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 ill away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60above the surface and with the same initial speed {1 as before. At what height II.above the surface was the insect? S ~f Up e~ ()Q:h'ons of mo fJ'~n -rhe.. X a..nol y d,'re.c+-ial'1 fr:::Xo + In v% t Vy (I) ~ >I.:::. Yo +we.. W~ \/}'= t - ~~ f; ~ Xo=Y()~o 2. (~) O.bDOfY\ J..now Xf= vS J' fl ~ O.60ol"Yl e~ UOw+iOJl\ :L: V C(JJ(60) o oJ 0 Vy=VoJi')'IC60) O. b DOm :::::.Vo GOs (60) fJ/u~ V / f . C-Ct.Y'\ Jve.. f-o r -f: 1;:::::') t O.600~ .:::; ~ co s (6 0) t- in!-o ::=. e~. J. V Slne(!l.600~ ) _ o ~ (Jo.J('o) ~ (0.6001'7\ J. ) v" coS (' 0)/ pI U'j in the. va.l ve. 0 f!. v" .r: ro rn f'o.t' rA Problem 3.56 Don't Do This II. On the day after graduation, you decide to toss a burning match into the top of a cylindrical wastebasket (diameter D and height 2 D) full of old homework papers. To make this event more sporting, the bottom of the wastebasket is at the same height as the point where the match leaves your hand, and the near side of the wastebasket is a horizontal distance 6 Dfrom the point where the 0 match leaves your hand. You launch the match at a 45.0 angle above the horizontal. Part A: Find the minimum value of the launch speed that will make the match enter the top of the wastebasket. You can ignore air resistance and give your answers in terms of gand D Vox == Vo C.OS ('J 5) Vy :; tS i Y\ f:! 5) Va (J ) e.f>. (I) a x.:. v" t-:::. b 0 L:! y.=. Vy f - y~&L t ~::.D (2) J. :_1JJ]JO ~ 6D :> So/vil'l') .f."r r v.!in~ 60 \{, co..s '15 ::: t_ ~ . ==7 r I v <3 CV;c~ ) ; 11 +-0 I e. ~ (.2J / :;! 0 - fuL( / ~ D oS JQJ?f'J)' - /;,.~ LVoco,C'fsJ7 0D1ve.. V 6 Y.J J ;l. o{- course- s,'''' (qS);C-OSC1IS};!J JD=- 6D - t 360~. /).~ v~ o 3 ~ 0 t1( ~ \) o ~ 1-0"" Vo ttr Vo'l..; ~ =/ [Cf 0 3 Part B: Find the maximum value of the launch speed that will make the match enter the top of the wastebasket. Give your answers in terms of gand D. [he s e. t up o LLf fs rA e.. s o.m e.. e-x c V' -t 6 x =- ?0 t- V7!2 GOS 50:=' , 5 ) :1.0 ~ ,x('.sr~).L7~f~).\ '110">~() ~Go?5V ~ ~ Lv;, f.---2Ec..OS('IsJ ) ~ ~;:. J-V-eD ~ Problem 3.66 A 2.7-kg ball is thrown upward with an initial speed of 20.0 mls from the edge of a 45.0 m high cliff. At the instant the ball is thrown, a woman staJ1s running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored. Part A: At what angle above the horizontal should the ball be thrown so that the runner will catch it just before it hits the ground? V.x''::- v 0 co sevy w ('/1-i X:::' f\Ej c1 d W" .J... h~ e ("'e. Ie, vcV\ +- ~~ == ~ e5 j n va-flollS y=- if':, m 'f5~~ __ <;;- +- v" /!) I:- Ii In () -I: - 112. ~ t ~ X::::::6 C, t;,j Ii \I~ ctJ.5 b u.- J- ().,Is" ~/s -f:: 6 % -t' ~ the re /-() e.. r So vtI 6 J"1h ::; v() e- s e.. -e..... ~/s - 6 aO tr'/s -- ::::: ,OJ C e ()::. 0- r c... c. () 5> {,}j ,~ 0..,' n -rnL Part B: How far does the woman run before she catches the ball? b~l\ hih t-Vos;nfi Ct-r y:::::-o -7 haw Lon') i's -I-Jl~ bCA-1I r 7 O~'-I5m lJ S j " ~ &- _~~t::t ,Nj .. 't{'f S ",X- 16.~ +- h~ 1J /.l a.d-~I-i c {-arm tJ 1",- f::. - ~ s j" ~ j: Jo 'Y 0 [)... fA e. C tJ rre.s -n e,e.of.... /-0 -/ l{_ use- V 'sj (l -1-"') -3 p a fl J.s t-O CI'- {>o i n T i t\ S i olL. .;-;., (.,- r ;cJ,..j e. '. C' t--- 5,.5 'tCfs )\IameLYI rlJnS ... So the.. D.X:::" (6 '}'s){5.5ttQs) -- 33. 'i m - Problem 3.69 Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training round with a muzzle speed of vOat an angle f) above the horizontal while advancing toward the second tank with a speed of V1relative to the ground. The second tank is retreating at a speed of "L'1relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at the same height above ground from which it was fired. Part A: Find the distance between the tanks when the round was first fired. V.x::' V V6 CoS & 118 y;;' voSl Fi cL ('l,l ~ .. "t- A 0 r ;z.,or'l"i-oJ op e e.cJ-s i t'\ r=l~ ===,.sl+o.nk ~ :;).nd, +0J1/1 f' Ae.. Shell III.) ,'11 R-e.Ill\.thre. (f \\ +0 ft> +(,&."J'" .1 : c:; ,. () 1I11J.. -ro.n I~ J. : -' V x ::Vo c.os liD f} f h!. a..t r"/ vx;;' vD COS V,x ~ GJ t- V, -rV, J..Jow iDYl' is "to {( "r() in\rio..l JIt\ cose -v~ y~ =-0 = c..c'S f:J t - y~~ 8 t '4.. Th- J.. isl-o.Y\ c.~ hetwe.en Ts _"\ _/ -1. _ 7; - 'J Vo - COS ~ fhe, f-an [.(s -t- i ~ e...s +- ,. eo velocity +-0 C-re..la.+-ive.. t-o.. t\ fI.. ).) e -rv, -v;\) ~ Vo cosG f.or } ni+/cx.1 - ~X'i:::' (III.) COS Part B: Find the distance between the tanks at the time of impact. We.... ho...lIL )'\~ Cln e.-X'pf'e.ssrof\ dJsf-ev1 c,-e...D'xl' d. uri To.. f\ fJ." +-,' mIL t we )4 no H/: Jo'1 1.. c.. los e-s -1-;" e- f) CAp l1 t V.;l To..fl p.. :t ~ i ol. e.IIS T-J., e !jo..p by So 6 -:X f :::.. X 6. t- f - V, t t- V;>. -t 6 X -I '" (Va CO _ S Go :l l' VI - V'l. +- v..- VI) e 2~ !!) c.,~s e, J' OX; -- j V6 ':l. c.. 0 S 'J.. Problem 3.30 A model of a helicopter rotor has four blades, each of length ["'from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of frey/min . . Part A: What is the linear speed of the blade tip? , I ('e"V -(YJ i 1l .--. =--- -P 60 -re..v S 1 r e" v ::::. -;}.17 LT It ~re.. f..a reJ -}- 1) e. WI J( n e o...,r .s.p e. e-cJ..m S is bO + - .:l7Tl- Part B: What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity,9? a.= - -- (&0)'4 v~ -f'" (27{) L- ?.L. ~ . '- J' n f-e r lY\ S a.p ~ L ~ ~= r,:V J. - '5 ~ Problem 3.34 The Ferris wheel in the figure 1jj , which rotates counterclockwise, is just starting up. At a given instant, a passenger on the rim of the wheel and passing through the lowest point of his circular motion is moving at 3.00 mls and is gaining speed at a rate of 0.000 m/82 Part A: Find the magnitude of the passenger's acceleration at this instant. :2- ()..r~J.. ... ( 3 J "'/s.) '1 tyl - o. 6 If 3 ",/s ~ 'J.. (). t(A,n~en + .= o. 5 ""/5 ~ -:2.. ().. ~ fA::: "=' a. rO-J .J- a rAn {O_5)'4 :4. j (O.&<t3)~18 1'-( )')1/s a=-O. Part B: Find the direction of the passenger's acceleration at this instant. t (J. Y\ e- :::. I Col rtV' I , a raJ. I e ::- O-TG f{A h 0.5 ) ( 0- 0 'i3 <A+().t'l () ~ 31.1 () Y' /3 hr oj) verriDo../ Circular Launch A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2 g. y ~L_~ ~ Part A: How far from the bottom of the chute does the ball land ? 2R ac:::-::l~ ~ v-:a. fAer~ fore- v:: / 'J.3 A fro rh a. heiBht :1. ~~ ;;. t , -t ~:;. :r. will ==7 r~i: ~ Ct..d j.J- \AI I 11 fa. Ke.. f: s +-0 ~Il f-r().. ve-l X=:..\/ ,-s f-a n c.. --- b. t= / ::u~/:I..": q.:: ~~ ~ ...
View Full Document

This note was uploaded on 05/22/2008 for the course PHYS 2A taught by Professor Hicks during the Winter '07 term at UCSD.

Ask a homework question - tutors are online