Newtonian Mechanics Single Particle

Transcription

1 CHAPTER Newonian Mechanics Single Paricle -. The basic equaion is F () a) (, ) ( ) ( ) i i i F f g : No inegrable () b) (, ) ( ) ( ) i i i i i i F f g c) (, ) ( ) ( ) d i f d d i ( ) () i i f i g ( ) i ( ) g i d : Inegrable (3) Fi i f i g i i i: No inegrable (4) -. Using spherical coordinaes, we can wrie he force applied o he paricle as F F e + Fe + Fe () r r θ θ φ φ Bu since he paricle is consrained o oe on he surface of a sphere, here us eis a reacion force e ha acs on he paricle. Therefore, he oal force acing on he paricle is F r r Foal Fθeθ + Fφeφ r () The posiion ecor of he paricle is r Re (3) where R is he radius of he sphere and is consan. The acceleraion of he paricle is r a r R e (4) r 9

2 3 CHAPTER We us now epress e r in ers of er, e θ, and e φ. Because he uni ecors in recangular coordinaes, e, e, e 3, do no change wih ie, i is conenien o ake he calculaion in ers of hese quaniies. Using Fig. F-3, Appendi F, we see ha er e sin θ cos φ+ e sin θ sin φ+ e3 cos θ eθ e cos θ cos φ+ e cos θ sin φ e3 sin θ (5) e e sin φ e cos φ φ + Then ( ) ( ) e e φ sin θ sin φ+ θ cos θ cos φ + e θ cos θ sin φ+ φ sin θ cos φ e θ sin θ r e φ sin θ + e θ φ θ 3 (6) Siilarly, e e θ + e φ cos θ (7) θ r φ e e φ sin θ e φ cos θ (8) φ r θ And, furher, ( φ sin θ θ ) ( θ φ sin θ cos θ) ( θφ cos θ φ sin θ) e e + + e + e + (9) r r θ φ which is he only second ie deriaie needed. The oal force acing on he paricle is and he coponens are F θ F φ F r R e () oal R ( θ φ sin θ cos θ) ( θφ cosθ φ sinθ) R + r ()

3 NEWTONIAN MECHANICS SINGLE PARTICLE 3-3. y P α β l The equaion of oion is F a () The graiaional force is he only applied force; herefore, F Fy y g Inegraing hese equaions and using he iniial condiions, ( ) cosα sinα ( ) y () (3) We find ( ) () cos α sin α y g (4) So he equaions for and y are ( ) () cos α (5) sin α y g Suppose i akes a ie o reach he poin P. Then, cos β sin β Eliinaing beween hese equaions, fro which cos α sin α g sinα g + α β g g (6) cos an (7)

4 3 CHAPTER g ( sin α cos α an β) (8) -4. One of he balls heigh can be described by y y + g. The aoun of ie i akes o rise and fall o is iniial heigh is herefore gien by g. If he ie i akes o cycle he ball hrough he juggler s hands is τ.9 s, hen here us be 3 balls in he air during ha ie τ. A single ball us say in he air for a leas 3τ, so he condiion is g 3τ, or 3. s -5.. flighpah N e r plane g poin of aiu acceleraion a) Fro he force diagra we hae ( R) + ( R) r N g e. The acceleraion ha he pilo feels is N g e, which has a aiu agniude a he boo of he aneuer. b) If he acceleraion fel by he pilo us be less han 9g, hen we hae ( 3 33 s ) R 8g s A circle saller han his will resul in pilo blackou. r.5 k () -6. Le he origin of our coordinae syse be a he ail end of he cale (or he closes cow/bull). a) The bales are oing iniially a he speed of he plane when dropped. Describe one of hese bales by he paraeric equaions + ()

5 NEWTONIAN MECHANICS SINGLE PARTICLE 33 y y g () where y 8, and we need o sole for. Fro (), he ie he bale his he ground is τ y g. If we wan he bale o land a ( τ ) 3, hen ( τ ) τ. Subsiuing s and he oher alues, his gies. The rancher should drop he bales behind he cale. b) She could drop he bale earlier by any aoun of ie and no hi he cale. If she were lae by he aoun of ie i akes he bale (or he plane) o rael by 3 in he -direcion, hen 3.68 s. she will srike cale. This ie is gien by ( ) -7. Air resisance is always ani-parallel o he elociy. The ecor epression is: W cwρa c A wρ Including graiy and seing F ne a, we obain he paraeric equaions () b + y () y by + y g (3) where b c ρa w -3. Soling wih a copuer using he gien alues and ρ.3 kg, we find ha if he rancher drops he bale behind he cale (he answer fro he preious proble), hen i akes 4.44 s o land 6.5 behind he cale. This eans ha he bale should be dropped a 78 behind he cale o land 3 behind. This soluion is wha is ploed in he figure. The ie error she is allowed o ake is he sae as in he preious proble since i only depends on how fas he plane is oing. 8 6 y () () Wih air resisance No air resisance

6 34 CHAPTER -8. y P α h Q Fro proble -3 he equaions for he coordinaes are cos α () y g sin α () In order o calculae he ie when a projecie reaches he ground, we le y in (): α (3) sin g g sin α (4) Subsiuing (4) ino () we find he relaion beween he range and he angle as sin α (5) g π π The range is aiu when α, i.e., α. For his alue of α he coordinaes becoe 4 Eliinaing beween hese equaions yields g y g + g (7) We can find he -coordinae of he projecile when i is a he heigh h by puing y h in (7): This equaion has wo soluions: h g + g (8) 4gh g g (9) + 4gh g g (6)

7 NEWTONIAN MECHANICS SINGLE PARTICLE 35 where corresponds o he poin P and o Q in he diagra. Therefore, d 4gh () g -9. a) Zero resising force ( F ): The equaion of oion for he erical oion is: Inegraion of () yields r d F a g () d g+ () where is he iniial elociy of he projecile and is he iniial ie. The ie required for he projecile o reach is aiu heigh is obained fro (). Since corresponds o he poin of zero elociy, we obain ( ) g, (3) b) Resising force proporional o he elociy ( F k) The equaion of oion for his case is: g (4) r : d F g k (5) d where k is a downward force for < and is an upward force for >. Inegraing, we obain For, (), hen fro (6), which can be rewrien as g k + g k () + e (6) k k g k ( e ) (7) k k k ln + g (8) Since, for sall z (z ) he epansion

8 36 CHAPTER 3 ln ( + z) z z + z (9) 3 is alid, (8) can be epressed approiaely as k k g g 3 g + which gies he correc resul, as in (4) for he lii k. () -. The differenial equaion we are asked o sole is Equaion (.), which is k. Using he gien alues, he plos are shown in he figure. Of course, he reader will no be able o disinguish beween he resuls shown here and he analyical resuls. The reader will hae o ake he word of he auhor ha he graphs were obained using nuerical ehods on a 8 copuer. The resuls obained were a os wihin of he analyical soluion. s (/s) (s) s () (s) s (/s) () -. The equaion of oion is d k g d + () This equaion can be soled eacly in he sae way as in proble - and we find

9 NEWTONIAN MECHANICS SINGLE PARTICLE 37 g k log k g k where he origin is aken o be he poin a which so ha he iniial condiion is ( ). Thus, he disance fro he poin o he poin is g k log k g k ( ) s () (3) -. The equaion of oion for he upward oion is Using he relaion we can rewrie () as d k g () d d d d d d d d d d d () d k + g d (3) Inegraing (3), we find log ( k + g) + C (4) k where he consan C can be copued by using he iniial condiion ha when : Therefore, C log ( k + g) (5) k k log + g k k + g (6) Now, he equaion of downward oion is This can be rewrien as d k +g (7) d d + k g d (8) Inegraing (8) and using he iniial condiion ha a (w ake he highes poin as he origin for he downward oion), we find

10 38 CHAPTER g log k g k (9) A he highes poin he elociy of he paricle us be zero. So we find he highes poin by subsiuing in (6): Then, subsiuing () ino (9), h k log + g () k g k + g g log log k g k g k () Soling for, g k g + k We can find he erinal elociy by puing in(9). This gies () Therefore, g (3) k + (4) -3. The equaion of oion of he paricle is d 3 k( + a ) () d Inegraing, and using Eq. (E.3), Appendi E, we find d k d ( + a ) () Therefore, we hae k + ln a a + a + C e A C (3) (4)

11 NEWTONIAN MECHANICS SINGLE PARTICLE 39 where A ak and where C is a new consan. We can ealuae C by using he iniial condiion, a : C a + Subsiuing (5) ino (4) and rearranging, we hae (5) A ace A Ce d (6) d A Now, in order o inegrae (6), we inroduce u e so ha du Au d. Then, A ace a Cu du d A Ce A Cu u a C du A Cu + u (7) Using Eq. (E.8c), Appendi E, we find a sin ( C u) + C A (8) Again, he consan C can be ealuaed by seing a ; i.e., a u : a (9) A C sin ( C ) Therefore, we hae a A sin ( C e + ) sin ( C + ) A Using (4) and (5), we can wrie + a + a sin sin ak a + + a Fro (6) we see ha as. Therefore, Also, for ery large iniial elociies, li sin sin () + a Therefore, using () and () in (), we hae + a π + a π li sin sin ( ) + a () () ()

12 4 CHAPTER and he paricle can neer oe a disance greaer han π ( ) (3) ka π ka for any iniial elociy. -4. α β d y a) The equaions for he projecile are cos α y sin α g Soling he firs for and subsiuing ino he second gies Using d cos β and y d sin β gies g y an α cos α gd cos β dsin β dcos β an α cos α gd cos β d cosβ anα + sinβ cos α Since he roo d is no of ineres, we hae ( ) cosβ anα sinβ cos α d g cos β ( ) cosα sinα cosβ cosα sinβ g cos β ( ) cosα sin α β d () g cos β

13 NEWTONIAN MECHANICS SINGLE PARTICLE 4 b) Maiize d wih respec o α d dα c) Subsiue () ino () sinα sin α β + cosα cos α β cos α β) gcos β ( d ) ( ) ( ) ( ( α β) cos π α β π β α + 4 Using he ideniy we hae d d π β π β cos sin g cos β a sin A sin B cos A+ B sin A B ( ) ( ) π sin sin β sinβ gcos β g sin β a d a g ( + sinβ) -5. g sin θ g θ The equaion of oion along he plane is Rewriing his equaion in he for k d g k d sin θ () d g sin θ k d ()

14 4 CHAPTER We know ha he elociy of he paricle coninues o increase wih ie (i.e., d d > ), so ha ( g k) sin θ > find. Therefore, we us use Eq. (E.5a), Appendi E, o perfor he inegraion. We anh k g g sin θ sin θ k k + The iniial condiion ( ) iplies C. Therefore, ( gk θ ) g sin θ anh sin k C (3) d d (4) We can inegrae his equaion o obain he displaceen as a funcion of ie: Using Eq. (E.7a), Appendi E, we obain ( gk θ ) g sin θ anh sin k d ( gk sin θ ) g ln cosh sin θ k gk sin θ + C (5) The iniial condiion ( ) iplies C. Therefore, he relaion beween d and is Fro his equaion, we can easily find ( gk θ ) d ln cosh sin (6) k ( e dk ) cosh (7) gk sin θ -6. The only force which is applied o he aricle is he coponen of he graiaional force along he slope: g sin α. So he acceleraion is g sin α. Therefore he elociy and displaceen along he slope for upward oion are described by: ( α) where he iniial condiions ( ) and ( ) gsin () ( gsin α ) () hae been used. A he highes posiion he elociy becoes zero, so he ie required o reach he highes posiion is, fro (), A ha ie, he displaceen is g sin α (3)

15 NEWTONIAN MECHANICS SINGLE PARTICLE 43 g sinα (4) For downward oion, he elociy and he displaceen are described by ( sin α) g (5) ( gsin α) (6) where we ake a new origin for and a he highes posiion so ha he iniial condiions are ( ) and ( ). We find he ie required o oe fro he highes posiion o he saring posiion by subsiuing (4) ino (6): Adding (3) and (7), we find g sin α (7) g sin α for he oal ie required o reurn o he iniial posiion. (8) The seup for his proble is as follows: 6 Fence cos θ () y y g + sin θ () where θ 35 and 7. The ball crosses he fence a a ie τ R cosθ, where y. ( ) R 6. I us be a leas h high, so we also need, we obain h y τ θ gτ. Soling for sin gr cos θ Rsin θ ( h y ) cos θ (3) which gies 5. 4 s.

16 44 CHAPTER -8. a) The differenial equaion here is he sae as ha used in Proble -7. I us be soled for any differen alues of in order o find he iniu required o hae he ball go oer he fence. This can be a copuer-inensie and ie-consuing ask, alhough if done correcly is easily racable by a personal copuer. This iniu is 35. s, and he rajecory is shown in Figure (a). (We ake he densiy of air as ρ kg.) y () Wih air resisance No air resisance fence heigh fence range b) The process here is he sae as for par (a), bu now we hae fied a he resul jus obained, and he eleaion angle θ us be aried o gie he ball a aiu heigh a he fence. The angle ha does his is.7 rad 4.7, and he ball now clears he fence by.. This rajecory is shown in Figure (b). () 5 y () Fligh Pah fence heigh fence range ()

17 NEWTONIAN MECHANICS SINGLE PARTICLE The projecile s oion is described by ( cos α) y ( sin α ) g where is he iniial elociy. The disance fro he poin of projecion is () Since r us always increase wih ie, we us hae r > : Using (), we hae r + y () + yy r > (3) r 3 ( sin α ) 3 + yy g g + (4) Le us now find he alue of which yields + yy (i.e., r ): 3 sin α g g ± 9sin α 8 (5) For sall alues of α, he second er in (5) is iaginary. Tha is, r is neer aained and he alue of resuling fro he condiion r is unphysical. Only for alues of α greaer han he alue for which he radicand is zero does becoe a physical ie a which r does in fac anish. Therefore, he aiu alue of α ha insures r > for all alues of is obained fro or, so ha 9sin a 8 α (6) sin α a (7) 3 αa 7.5 (8) -. If here were no reardaion, he range of he projecile would be gien by Eq. (.54): The angle of eleaion is herefore obained fro R sin θ () g

18 46 CHAPTER so ha Rg sin θ ( ) ( 9.8 /sec ) ( 4 /sec).5 () θ 5 (3) Now, he real range R, in he linear approiaion, is gien by Eq. (.55): 4kV R R 3g sin θ 4k sin θ g 3g Since we epec he real angle θ o be no oo differen fro he angle θ calculaed aboe, we can sole (4) for θ by subsiuing θ for θ in he correcion er in he parenheses. Thus, sin θ gr 4k sinθ 3g Ne, we need he alue of k. Fro Fig. -3(c) we find he alue of k by easuring he slope of k N 5 /s. kg/s. The he cure in he iciniy of 4 /sec. We find ( ) ( ) cure is ha appropriae for a projecile of ass kg, so he alue of k is k. sec (6) Subsiuing he alues of he arious quaniies ino (5) we find θ 7.. Since his angle is soewha greaer han θ, we should ierae our soluion by using his new alue for θ in (5). We hen find θ 7.4. Furher ieraion does no subsanially change he alue, and so we conclude ha θ 7.4 If here were no reardaion, a projecile fired a an angle of 7.4 wih an iniial elociy of 4 /sec would hae a range of R ( ) 4 /sec sin /sec (4) (5)

19 NEWTONIAN MECHANICS SINGLE PARTICLE α Assue a coordinae syse in which he projecile oes in he 3 plane. Then, or, cos α r e + e () 3 sin α g 3 3 ( cos α) e + sin α g 3 e () The linear oenu of he projecile is and he angular oenu is ( cos ) e ( sin ) p r α + α g e 3 (3) ( ) ( cos α sin α g ) 3 ( cos α) ( sin α g) L r p + e e e + e 3 (4) Using he propery of he uni ecors ha ei ej e 3 εijk, we find This gies Now, he force acing on he projecile is so ha he orque is ( g cos α) ( cos ) g α e (5) L ( g cos α ) L e (6) F g e (7) N r F ( cos α) e + sin α g e3 ( g) e which is he sae resul as in (6). e 3 3

20 48 CHAPTER -. z B E y Our force equaion is e F q( E+ B ) () a) Noe ha when E, he force is always perpendicular o he elociy. This is a cenripeal acceleraion and ay be analyzed by eleenary eans. In his case we hae also B so ha B B. Soling his for r wihω c qb/. a cenripeal qb () r r (3) qb ω c b) Here we don ake any assupions abou he relaie orienaions of and B, i.e. he elociy ay hae a coponen in he z direcion upon enering he field region. Le r i+ yj+ zk, wih r and a r. Le us calculae firs he B er. B The Lorenz equaion () becoes i j k B y z B y i j) (4) ( ( y ) F r qby i+ q E B j+ qe k (5) Rewriing his as coponen equaions: qb y ωcy (6) qb qey E y ω y + c B (7) qe z z (8) z

21 NEWTONIAN MECHANICS SINGLE PARTICLE 49 The z-coponen equaion of oion (8) is easily inegrable, wih he consans of inegraion gien by he iniial condiions in he proble saeen. qe z() z z + z + (9) c) We are asked o find epressions for and y, which we will call and Differeniae (6) once wih respec o ie, and subsiue (7) for y or y, respeciely. E y ω c y ω c B () E y + ωc ωc () B This is an inhoogeneous differenial equaion ha has boh a hoogeneous soluion (he soluion for he aboe equaion wih he righ side se o zero) and a paricular soluion. The os general soluion is he su of boh, which in his case is Ey C cos ( ωc) + C sin ( ωc) + () B where C and C are consans of inegraion. This resul ay be subsiued ino (7) o ge y ( ) sin ( ) C ω cos ω C ω ω (3) y c c c ( ω ) cos ( ω ) y c c c C sin + C +K (4) where K is ye anoher consan of inegraion. I is found upon subsiuion ino (6), howeer, ha we us hae K. To copue he ie aerages, noe ha boh sine and cosine hae an aerage of zero oer one of heir periods T π ωc. E y, y (5) B d) We ge he paraeric equaions by siply inegraing he elociy equaions. where, indeed, D and C C Ey sin ( ω ) cos ( ωc) ω + + D (6) B c c ωc C C y cos ( ω ) sin ( ω ) ω + + D (7) c c y c ωc D y are consans of inegraion. We ay now ealuae all he C s and A ω E B y A. This gies us D s using our iniial condiions ( ), ( ), y ( ), ( ) c y C D Dy, C A and gies he correc answer A Ey () cos ( ωc) + (8) ω B c

22 5 CHAPTER These cases are shown in he figure as (i) (i) A () sin c (9) ω y ( ω ) c A > Ey B, (ii) A < Ey B, and (iii) A Ey B. (ii) (iii) -3. F ( ) a( ) ke a () wih he iniial condiions () (). We inegrae o ge he elociy. Showing his eplicily, () k α a () d e d () () Inegraing his by pars and using our iniial condiions, we obain k () + e α α α α By siilarly inegraing (), and using he inegral () we can obain (). k α () e α α α α To ake our graphs, subsiue he gien alues of kg, ( ) e k N s, and α.5 s. (5) ( ) 4 ( ) e α + (6) ( ) 6 4 4( 4) e (7) (3) (4)

23 NEWTONIAN MECHANICS SINGLE PARTICLE 5 () () 5 5 a() F f g cos θ g sin θ N g y θ B y F f N d lengh of incline s disance skier raels along leel ground g While on he plane: F N g cosθ y so N gcosθ y F gsin θ Ff ; Ff µ N µ gcos θ g sin θ µ g cos θ So he acceleraion down he plane is: ( ) a g sin θ µ cos θ consan

24 5 CHAPTER While on leel ground: N g; F µ g So F becoes µ g The acceleraion while on leel ground is f a µ g consan For oion wih consan acceleraion, we can ge he elociy and posiion by siple inegraion: Soling () for and subsiuing ino () gies: or a a+ () a + () a ( ) ( ) + a a ( ) a Using his equaion wih he iniial and final poins being he op and boo of he incline respeciely, we ge: ad V V speed a boo of incline B B Using he sae equaion for oion along he ground: Thus So Soling for µ gies ad as as VB a g( sin θ µ cos ) gd (3) θ a ( sin cos ) θ µ θ µ gs d sin θ µ dcos θ + s µ g Subsiuing θ 7, d, s 7 gies Subsiuing his alue ino (3): µ.8

25 NEWTONIAN MECHANICS SINGLE PARTICLE 53 µgs VB VB µ gs V 5.6 /sec B -5. a) A A, he forces on he ball are: N The rack couners he graiaional force and proides cenripeal acceleraion Ge by conseraion of energy: So b) A B he forces are: g N g R E T + U + gh op op op E T U A A + A + E E gh op A N g+ gh R h N g + R N 45 g N R g Ge by conseraion of energy. Fro a), Eoal A B, E + gh + cos 45 R + g () gh.

26 54 CHAPTER R R R 45 Rcos 45 R h So Eoal TB + UB Soling for becoes: R R +h or h R gh gr + Subsiuing ino (): gh gr h 3 N g + R c) Fro b) B h R+ R g d) This is a projecile oion proble ( ) g h R+ R 45 B 45 Pu he origin a A. The equaions: becoe Sole (3) for when y (ball lands). A + y y + y g R + B () B (3) y h + g

27 NEWTONIAN MECHANICS SINGLE PARTICLE 55 g B h B B 8 ± + gh g We discard he negaie roo since i gies a negaie ie. Subsiuing ino (): R B B ± B + 8gh + g Using he preious epressions for B ( ) and h yields 3 R+ h+ h R + R e) U ( ) gy ( ), wih y() h, so U( ) has he shape of he rack. -6. All of he kineic energy of he block goes ino copressing he spring, so ha k, or k 3., where is he aiu copression and he gien alues hae been subsiued. When here is a rough floor, i eers a force µ g k in a direcion ha opposes he block s elociy. I herefore does an aoun of work µ gd k in slowing he block down afer raeling across he floor a disance d. Afer of floor, he block has energy µ gd, which now goes ino copressing he spring and sill oercoing he fricion k on he floor, which is k + µ g. Use of he quadraic forula gies k µ g µ g µ gd + + k k k k () Upon subsiuion of he gien alues, he resul is To lif a sall ass d of rope ono he able, an aoun of work dw ( d) g ( z z) us be done on i, where z 6. is he heigh of he able. The oal aoun of work ha needs o be done is he inegraion oer all he sall segens of rope, giing When we subsiue L ( 4 kg) ( 4 ) ( ) ( ) µ gz W dz g z z z µ () µ., we obain W. 8 J.

28 56 CHAPTER M before collision afer collision The proble, as saed, is copleely one-diensional. We ay herefore use he eleenary resul obained fro he use of our conseraion heores: energy (since he collision is elasic) and oenu. We can facor he oenu conseraion equaion ou of he energy conseraion equaion and ge () (3) 3 This is he conseraion of relaie elociies ha oiaes he definiion of he coefficien of resiuion. In his proble, we iniially hae he superball of ass M coing up fro he ground wih elociy Conseraion of oenu gies 4 gh, while he arble of ass is falling a he sae elociy. and our resul for elasic collisions in one diension gies M+ ( ) M3 + 4 (4) + 3 ( ) + 4 (5) soling for 3 and 4 and seing he equal o gh ie, we obain () h arble 3 α + α h (6) h superball 3α + α h (7) where α M. Noe ha if α < 3, he superball will bounce on he floor a second ie afer he collision.

29 NEWTONIAN MECHANICS SINGLE PARTICLE F f g cos θ g sin θ g N θ y so Inegrae wih respec o ie Inegrae again: θ an F N gcos θ y y N gcos θ F gsin θ F F µ N µ gcos θ f g sin θ µ g cos θ ( sin θ µ cos θ) g ( sin θ µ cos θ) f g + () () + + g ( sin θ µ cos θ) Now we calculae he ie required for he drier o sop for a gien (iniial speed) by soling Eq. () for wih. g ( sin θ µ cos θ) Subsiuing his ie ino Eq. () gies us he disance raeled before coing o a sop. ( ) + g ( sin θ µ cos θ) + g g g ( sin θ µ cos θ) g ( sin θ µ cos θ) ( µ cos θ sin θ)

30 58 CHAPTER We hae θ 4.6, µ.45, g 9.8 /sec. For 5 ph. /sec, 7.4 eers. If he drier had been going a 5 ph, he could only hae skidded 7.4 eers. How fas was he going? Therefore, he was speeding 3 eers gies 3.9 ph. -3. T + () where T oal ie 4. sec. he ie required for he balloon o reach he ground. We can ge fro he equaion When he addiional ie required for he sound of he splash o reach he firs suden. y y + y g ; y y, y h; so (h heigh of building) h g or h g Subsiuing ino (): disance sound raels h speed of sound h h T + or g h h + T g This is a quadraic equaion in he ariable h. Using he quadraic forula, we ge: Subsiuing 4T ± + g g gt h ± + g V V 33 /sec g 9.8 /sec T 4. sec and aking he posiie roo because i is he physically accepable one, we ge:

31 NEWTONIAN MECHANICS SINGLE PARTICLE 59 h 8.46 h 7 eers -3. For, eaple. proceeds as is unil he equaions following Eq. (.78). Proceeding fro here we hae α B α A z so Noe ha z cos sin α α α ( ) α + ( ) y y y z z z cos sin α α α ( ) α + z + + α α ( ) ( z z ) z +. α Thus he projecion of he oion ono he z plane is a circle of radius ( ) So he oion is unchanged ecep for a change in he radius of he heli. The new radius is ( z ) +. qb -3. The forces on he hanging ass are T The equaion of oion is (calling downward posiie) The forces on he oher ass are g T a g or T ( g a) ()

32 6 CHAPTER N T The y equaion of oion gies y θ F f g cos θ g g sin θ N gcosθ y or N gcosθ The equaion of oion gies ( Ff µ k N µ kgcosθ) Subsiuing fro () ino () When θ θ, a. So T gsinθ µ gcosθ a () g g sinθ µ g cosθ a g gsinθ µ gcosθ k k k sin θ + µ k cos θ ( ) sin θ + µ sin θ k Isolaing he square roo, squaring boh sides and rearranging gies Using he quadraic forula gies ( k) + µ sin θ sinθ µ k 4 sin θ ± µ 3+ 4µ k ( + µ k ) k -33. The differenial equaion o sole is cw ρa y g g () where g c w ρa is he erinal elociy. The iniial condiions are y, and. The copuer inegraions for pars (a), (b), and (c) are shown in he figure.

33 NEWTONIAN MECHANICS SINGLE PARTICLE 6 y () (s) 5 5 (s) 5 (s) (/s) (s) 5 5 (s) 5 (s) a (/s ) (s) 5 5 (s) 5 (s) 3 d) Taking ρ 3. kg as he densiy of air, he erinal elociies are 3., 8., and. (all - s ) for he baseball, ping-pong ball, and raindrop, respeciely. Boh he ping-pong ball and he raindrop essenially reach heir erinal elociies by he ie hey hi he ground. If we rewrie he ass as aerage densiy ies olue, hen we find ha ρ aerial R. The differences in erinal elociies of he hree objecs can be eplained in ers of heir densiies and sizes. e) Our differenial equaion shows ha he effec of air resisance is an acceleraion ha is inersely proporional o he square of he erinal elociy. Since he baseball has a higher erinal elociy han he ping-pong ball, he agniude of is deceleraion is saller for a gien speed. If a person hrows he wo objecs wih he sae iniial elociy, he baseball goes farher because i has less drag. f) We hae shown in par (d) ha he erinal elociy of a raindrop of radius.4 will be - larger han for one wih radius. ( 9. s ) by a facor of F R y g Take he y-ais o be posiie downwards. The iniial condiions are y y a.

34 6 CHAPTER a) FR The equaion of oion is g + C α Inegraing gies: ln( α ) Ealuae C using he condiion a : g g α + α So ln( α ) ln( ) α g α α or ln ln g g α d y g α d d d g α ln( g) C α Take he eponenial of boh sides and sole for : Inegrae again: y a, so: Sole (3) for and subsiue ino (4): e g αr α α α e g g ( e α ) () α g α dy ( e ) d α g y+ C + e α α α g C g α α α g y e α α + + α α α α e (3) g ()

35 NEWTONIAN MECHANICS SINGLE PARTICLE 63 α ln α g g α α g α y ln + ln α α α g α g α g α g (4) g α y + ln α α g b) FR β The equaion of oion becoes: d g β d d g β d Inegrae and apply he iniial condiion a : d g β β d Fro inegral ables d anh a a a ; so anh β + C where a a a g β anh C α + so: anh β a a Soling for : aanh a β (5) dy aβ a anh d Fro inegral ables anh udu ln cosh u aβ So y+ C ln cosh β

36 64 CHAPTER Apply he condiions a y and C ln ( cosh ) ln β β So Soling (5) for : Subsiuing ino (6): Use he ideniy: anh u cosh aβ y ln cosh (6) β anh αβ a y ln cosh anh β u, where u <. a (In our case u < as i should be because graiy is sronger han he rearding force, which i us be.) So β ; and he condiion ha u < jus says ha a g y ln cosh cosh ln β β g β ( β g) y ln g β ( β )

37 NEWTONIAN MECHANICS SINGLE PARTICLE y (k) (k) 3 Range (k) We are asked o sole Equaions (.4) and (.4), for he alues k,.5,.,.,.4, and.8 (all in s ), wih iniial speed 6 s and angle of eleaion θ 6. The firs figure is produced by nuerical soluion of he differenial equaions, and agrees closely wih Figure -8. Figure -9 can be os closely reproduced by finding he range for our alues of k, and ploing he s. k. A sooh cure could be drawn, or ore ranges could be calculaed wih ore alues of k o fill in he plo, bu we chose here o jus connec he poins wih sraigh lines. k (/s)

38 66 CHAPTER -36. h Pu he origin a he iniial poin. The equaions for he and y oion are hen y θ R ( θ) cos y g ( sin θ ) Call τ he ie when he projecile lands on he alley floor. The y equaion hen gies Using he quadraic forula, we ay find τ h g ( sin θ ) τ τ sin θ sin θ + gh τ + g g (We ake he posiie since τ >.) Subsiuing τ ino he equaion gies he range R as a funcion of θ. cos sin sin R θ θ + θ + () g where we hae defined equaion we obain is gh. To aiize R for a gien h and, we se dr dθ. The Alhough i can gie ( θ) sin θ cos θ c os θ sin θ sin θ sin θ + + () sin θ +, he aboe equaion canno be soled o gie θ θ( ) in ers of he eleenary funcions. The opiu θ for a gien is ploed in he figure, along wih is respecie range in unis of g. Noe ha, which aong oher hings corresponds o h, gies he failiar resul θ 45 and R g.

39 NEWTONIAN MECHANICS SINGLE PARTICLE θ R/( /g) α d d α Since d d d d hen d d d d d d α α F d d 3 ( ) α F a -38. ( ) n d d d d d d d d n n a) F ( a )( na ) d a d n b) ( ) Inegrae: n d ad ( ) ( ) n F na + n+ a + C n + C using gien iniial condiions

40 68 CHAPTER n ( n ) + + a [( ) ] ( n+ ) n+ a c) Subsiue () ino F(): () ( + ) F na n a { } ( ) ( n+ n+ ) () ( + ) [ ] ( n ) ( n ) F na n a a) F α e β a, so d α e d β α β e d d β α e + C β Soling for gies β e C β β β α ( e e ) β β () + e αβ ln β b) Sole for when αβ + e β αβ e β c) Fro a) we hae αβ ln β d e d β +

41 NEWTONIAN MECHANICS SINGLE PARTICLE 69 a + b a + b d a + b we obain a Using ln( ) ln( ) αβ β αβ β e ln e C β αβ Ealuaing C using a gies C αβ e β So β αβ β αβ β e + e ln e αβ β αβ + + Subsiuing he ie required o sop fro b) gies he disance required o sop + αβ β β β e -4. y ((),y()) a n a Wrie he elociy as () ()T(). I follows ha () d d d T a T+ at+ ann () d d d where N is he uni ecor in he direcion of dt d. Tha N is noral o T follows fro dd( T T ). Noe also is posiie definie. a n a) We hae + y Aα 5 4cos α. Copuing fro he aboe equaion, d a d Aα sin α 5 4cosα () We can ge a fro knowing a in addiion o a. Using n a + y Aα, we ge cosα an a a Aα 5 4cosα (3) b) Graphing an ersus shows ha i has aia a α nπ, where an Aα.

42 7 CHAPTER -4. a) As easured on he rain: T ; i Tf T b) As easured on he ground: Ti u ; T ( ) f + u T + u c) The woan does an aoun of work equal o he kineic energy gain of he ball as easured in her frae. W d) The rain does work in order o keep oing a a consan speed u. (If he rain did no work, is speed afer he woan hrew he ball would be slighly less han u, and he speed of he ball relaie o he ground would no be u +.) The er u is he work ha us be supplied by he rain. W u -4. θ Rθ b R θ Fro he figure, we hae h( θ) ( R+ b )cosθ + Rθ sinθ, and he poenial is U( θ) gh( θ). Now copue: du b g sin θ Rθ cos θ dθ + du b g R cos θ Rθ sin θ dθ () ()

43 NEWTONIAN MECHANICS SINGLE PARTICLE 7 The equilibriu poin (where du dθ ) ha we wish o look a is clearly θ. A ha poin, we hae du d g( R b) θ, which is sable for R> b and unsable for R< b /. We can use he resuls of Proble -46 o obain sabiliy for he case R b, where we will find ha he firs non-riial resul is in fourh order and is negaie. We herefore hae an equilibriu a θ which is sable for R> b and unsable for R b F k+ k α 4 4 U ( ) Fd k k α To skech U(), we noe ha for sall, U() behaes like he parabola behaior is deerined by 4 4 k α U() k. For large, he E E E E 3 E 4 E + U( ) For E E, he oion is unbounded; he paricle ay be anywhere. For E E (a he aia in U()) he paricle is a a poin of unsable equilibriu. I ay reain a res where i is, bu if perurbed slighly, i will oe away fro he equilibriu. du Wha is he alue of E? We find he alues by seing d., ± α are he equilibriu poins 3 k k α U( ± α) E kα kα kα 4 4 For E E, he paricle is eiher bounded and oscillaes beween and ; or he paricle coes in fro ± o ± and reurns o ±. 3

44 7 CHAPTER For E, he paricle is eiher a he sable equilibriu poin, or beyond ±. 3 4 For E, he paricle coes in fro ± o 4 5 ± and reurns T T θ T g g Fro he figure, he forces acing on he asses gie he equaions of oion where is relaed o by he relaion and θ d ( b ) cos alues for he coordinaes g T () g T θ () cos ( b ). A equilibriu, d (3) 4 and T g. This gies as he equilibriu b 4d (4) 4 4 d (5) We recognize ha our epression is idenical o Equaion (.5), and has he sae requireen ha < for he equilibriu o eis. When he syse is in oion, he descripie equaions are obained fro he force laws: ( ) b ( g) ( g 4 ) To eaine sabiliy, le us epand he coordinaes abou heir equilibriu alues and look a heir behaior for sall displaceens. Le ξ and ξ. In he calculaions, ake ers in ξ and ξ, and heir ie deriaies, only up o firs order. Equaion (3) hen becoes ξ ( ) ξ. When wrien in ers of hese new coordinaes, he equaion of oion becoes ( 4 ) 3 g ξ 4 d ξ (7) ( + ) (6)

45 NEWTONIAN MECHANICS SINGLE PARTICLE 73 which is he equaion for siple haronic oion. The equilibriu is herefore sable, when i eiss and -46. Epand he poenial abou he equilibriu poin The leading er in he force is hen i du i U( ) i! d () i i n+ F ( ) ( n+ ) du d U ( n+ ) d n! d The force is resoring for a sable poin, so we need F> ( ) < and ( ) n F< >. This is neer rue when n is een (e.g., U k 3 ( n+ ) ( n+ ) ), and is only rue for n odd when d U d <. () -47. We are gien ( ) U ( a+ a) du d, wih sabiliy deerined by U for >. Equilibriu poins are defined by du d a hose poins. Here we hae du a U d + a which anishes a a. Now ealuae indicaing ha he equilibriu poin is sable. 5 U > du 3 d a a () () U()/U /a

46 74 CHAPTER -48. In he equilibriu, he graiaional force and he eccenric force acing on each sar us be equal 3 G G πd πd τ d d/ d G -49. The disances fro sars o he cener of ass of he syse are respeciely d r + and A equilibriu, like in preious proble, we hae d r + G G πr πd d r d( ) G ( ) 3 τ + + The resul will be he sae if we consider he equilibriu of forces acing on nd sar. -5. a) F F d F () F + d d c c c c c F () d () + F c b) c) Fro a) we find Now if F, hen F c

47 NEWTONIAN MECHANICS SINGLE PARTICLE 75 c when c, we hae.55 year 3 99c when 99% c, we hae 6.67 years a) d d b b d () d b + Now le () /, one finds hours. b b) b + d () ln b We use he alue of found in quesion a) o find he corresponding disance ( ) ln() 6.9 k b -5. a) b) du 4U F ( ) d a a U When F, here is equilibriu; furher when U has a local iniu (i.e. df d < ) i is sable, and when U has a local aiu (i.e. df d > ) i is unsable.

48 76 CHAPTER So one can see ha in his proble a and a are unsable equilibriu posiions, and is a sable equilibriu posiion. c) Around he origin, 4U k 4U F k ω a a d) To escape o infiniy fro, he paricle needs o ge a leas o he peak of he poenial, in e) Fro energy conseraion, we hae U U a in U U in d U + a d a We noe ha, in he ideal case, because he iniial elociy is he escape elociy found in d), ideally is always saller or equal o a, hen fro he aboe epression, 8U a ep d a a+ a ln ( ) U 8U a 8U ep + a a -53. F is a conseraie force when here eiss a non-singular poenial funcion U() saisfying F() grad(u()). So if F is conseraie, is coponens saisfy he following relaions and so on. F y Fy a) In his case all relaions aboe are saisfied, so F is indeed a conseraie force. where f ( y, z) U b F (, ) ayz + b + c U ayz c + f y z () is a funcion of only y and z U F (, ) y az + bz U ayz byz + f z () y

49 NEWTONIAN MECHANICS SINGLE PARTICLE 77 where f (, z) is a funcion of only and z hen fro (), (), (3) we find ha F z where C is a arbirary consan. U ay + by + c U ayz byz + f (, ) 3 y z (3) z b U ayz byz c + C b) Using he sae ehod we find ha F in his case is a conseraie force, and is poenial is U zep( ) yln z+ C c) Using he sae ehod we find ha F in his case is a conseraie force, and is poenial is ( using he resul of proble -3b): U aln r -54. a) Terinal elociy eans final seady elociy (here we assue ha he poao reaches his elociy before he ipac wih he Earh) when he oal force acing on he poao is zero. b) g k and consequenly g k /s. d F d d d ( g+ k) d d d g + k g + k g g + ln k k g+ k a where is he iniial elociy of he poao Le s denoe and y he iniial horizonal and erical elociy of he pupkin. Eidenly, y in his proble. d d d f F k d () d k k f where he suffi f always denoe he final alue. Fro he second equaliy of (), we hae Cobining () and () we hae d () k d f f e k f k k ( f ) e (3)

50 78 CHAPTER Do he sae hing wih he y-coponen, and we hae d dy d g g + k F g k d y d g k k g k k y y yf y yf y y f ln + y + y + y (4) d k d g k g k e f (5) g+ k y and + yf ( + f ) Fro (4) and (5) wih a lile anipulaion, we obain y k gk f f e (6) g + k (3) and (6) are equaions wih unknowns, f and k. We can eliinae f, and obain an equaion of single ariable k. Puing f y k f ( ( g+ ky) ( g) e ) k f 4 and y 38. /s we can nuerically sole for k and obain K.46 s.