C. J. PAPACHRISTOU INTRODUCTION TO

Transcription

1 C. J. PAPACHRISTOU INTRODUCTION TO MECHANICS OF PARTICLES AND SYSTEMS HELLENIC NAVAL ACADEMY

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3 Introducton to MECHANICS of Partcles and Systems Costas J. Papachrstou Department of Physcal Scences Hellenc Naval Academy Hellenc Naval Academy 018

4 Costas J. Papachrstou, 018

5 PREFACE Newtonan Mechancs s, tradtonally, the frst stage of ntaton of a college student nto Physcs. It s perhaps the only truly autonomous subject area of Physcs, n the sense that t can be taught as a self-contaned entty wthout the need for support from other areas of physcal scence. Ths textbook s based on lecture notes (orgnally n Greek) used by ths author n hs two-semester course of ntroductory Mechancs, taught at the Hellenc Naval Academy (the Naval Academy of Greece). It s evdent that no serous approach to Mechancs (at least at the unversty level) s possble wthout the support of hgher Mathematcs. Indeed, the central law of Mechancs, Newton s Second Law, carres a rch mathematcal structure beng both a vector equaton and a dfferental equaton. An effort s thus made to famlarze the student from the outset wth the use of some basc mathematcal tools, such as vectors, dfferental operators and dfferental equatons. To ths end, the frst chapter contans the elements of vector analyss that wll be needed n the sequel, whle the Mathematcal Supplement at the end of the book consttutes a bref ntroducton to the aforementoned concepts of dfferental calculus. The man text may be subdvded nto three parts. In the frst part (Chapters -5) we study the mechancs of a sngle partcle (and, more generally, of a body that executes purely translatonal moton), whle the second part (Chap. 6-8) ntroduces to the mechancs of more complex structures such as systems of partcles, rgd bodes and deal fluds. The thrd part conssts of 60 fully solved problems. I urge the student to try to solve each problem on hs/her own before lookng at the accompanyng soluton. Some useful supplementary materal may also be found n the Appendces. I am ndebted to Arstds N. Magoulas for helpng me wth the fgures. I also thank the Hellenc Naval Academy for publshng the orgnal, Greek verson of the book. Last but not least, I express my grattude and apprecaton to my wfe, Thala, for her patence and support whle ths book was wrtten! Praeus, Greece August 018 Costas J. Papachrstou

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7 CONTENTS 1. VECTORS Basc Notons 1 1. Rectangular Components of a Vector 1.3 Poston Vectors Scalar ( Dot ) Product of Two Vectors Vector ( Cross ) Product of Two Vectors 8. KINEMATICS 11.1 Rectlnear Moton 11. Specal Types of Rectlnear Moton 13.3 Curvlnear Moton n Space 15.4 Change of Speed 16.5 Moton wth Constant Acceleraton 18.6 Tangental and Normal Components 19.7 Crcular Moton 4.8 Relatve Moton 5 3. DYNAMICS OF A PARTICLE The Law of Inerta 8 3. Momentum, Force, and Newton s nd and 3rd Laws Force of Gravty Frctonal Forces Systems wth Varable Mass Tangental and Normal Components of Force Angular Momentum and Torque Central Forces WORK AND ENERGY Introducton Work of a Force Knetc Energy and the Work-Energy Theorem Potental Energy and Conservatve Forces Conservaton of Mechancal Energy Examples of Conservatve Forces Knetc Frcton as a Non-Conservatve Force OSCILLATIONS Smple Harmonc Moton (SHM) 6 5. Force n SHM Energy Relatons Oscllatons of a Mass-Sprng System Oscllaton of a Pendulum Dfferental Equaton of SHM 71

8 6. SYSTEMS OF PARTICLES Center of Mass of a System of Partcles Newton s Second Law and Conservaton of Momentum Angular Momentum of a System of Partcles Knetc Energy of a System of Partcles Total Mechancal Energy of a System of Partcles Collsons RIGID-BODY MOTION Rgd Body Center of Mass of a Rgd Body Revoluton of a Partcle About an Axs Angular Momentum of a Rgd Body Rgd-Body Equatons of Moton Moment of Inerta and the Parallel-Axs Theorem Conservaton of Angular Momentum Equlbrum of a Rgd Body Knetc and Total Mechancal Energy Rollng Bodes The Role of Statc Frcton n Rollng Gyroscopc Moton ELEMENTARY FLUID MECHANICS Ideal Flud Hydrostatc Pressure Fundamental Equaton of Hydrostatcs Unts of Pressure Communcatng Vessels Pascal s Prncple Archmedes Prncple Dynamcs of the Submerged Body Equlbrum of a Floatng Body Flud Flow Equaton of Contnuty Bernoull s Equaton Horzontal Flow 136 APPENDICES 138 A. Composton of Forces Actng n Space 138 B. Some Theorems on the Center of Mass 144 C. Table of Moments of Inerta 148 D. Prncpal Axes of Rotaton 149 E. Varaton of Pressure n the Atmosphere 15 F. Proof of Bernoull s Equaton 153 v

9 SOLVED PROBLEMS 155 MATHEMATICAL SUPPLEMENT 19 BASIC INTEGRALS 5 BIBLIOGRAPHY 6 INDEX 7 v

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11 CHAPTER 1 VECTORS 1.1 Basc Notons Vectors are physcal or mathematcal quanttes carryng two propertes: magntude and drecton. Symbolcally, a vector s usually represented by an arrow: V The magntude of V (proportonal, by conventon, to the length of the arrow) s denoted by V and, by defnton, V 0. In partcular, a vector of zero magntude s called a zero vector, V = 0, and ts drecton s ndetermnate. By defnton, the vector V has the same magntude as V but s orented n the opposte drecton: V V A unt vector (denoted û ) s a vector of unt magntude: u ˆ = 1. A vector V n the drecton of the unt vector û s wrtten V = V uˆ whle a vector W n the opposte drecton s wrtten W = W uˆ. Note that the unt vector û n the drecton of a vector V can be expressed as the quotent V uˆ = (1.1) V (By defnton, the effect of multplyng or dvdng a vector by a postve number s to multply or dvde, respectvely, the magntude of ths vector by ths number wthout alterng the drecton of the vector.) In general, a vector parallel to û s expressed as V = ± V uˆ V uˆ (1.) The quantty V s called the algebrac value of V wth respect to the unt vector û. In partcular, the sgn of V ndcates the orentaton of V relatve to û. 1

12 CHAPTER 1 Example: For the vectors n the followng fgure, û V W we have: V = W =, V = uˆ, W = uˆ, V =, W = (where the last two quanttes represent algebrac values). The sum V = A+ B of two vectors can be represented graphcally n two ways, as shown n the followng fgure: A V θ B B V θ A The dfference W = A B A+ ( B) s found graphcally as follows: A B θ B θ A W W B In the fgure on the rght, note that the arrow of A B s drected toward the tp of A. (Draw the vector B A.) It can be shown that where A= A, B= B, A± B = ( A + B ± AB cos θ ) 1/ and where θ s the angle between A and B. (1.3) 1. Rectangular Components of a Vector An orented straght lne s called an axs. The orentaton s specfed by a unt vector û, parallel to the lne, the drecton of whch vector ndcates the postve drecton on the axs.

13 VECTORS 3 We consder the xy-plane defned by the mutually perpendcular x- and y-axes wth unt vectors u ˆx and u ˆ y, respectvely. Let V be a vector on ths plane. It s often convenent to express V as a sum of two vectors parallel to the correspondng x- and y- axes: y V y uˆ y O uˆx V θ V x x V = V + V x y where V = V uˆ, V = V uˆ x x x y y y The quanttes V x and V y, whch are the algebrac values of the projectons V x and V y of V on the two axes, are the rectangular components of V. These quanttes may be postve or negatve, dependng on the orentatons of V x and V y relatve to u ˆx and u ˆ y, respectvely. (In the above fgure, both V x and V y are postve.) We wrte:. V = V uˆ + V uˆ ( V, V ) x x y y x y (1.4) The magntude of the vector V s V V = V + V x y (1.5) The angle θ n the above fgure s measured relatve to the postve x semaxs and t ncreases counterclockwse. Thus, startng from the x-axs we may form postve or negatve angles by movng counterclockwse or clockwse, respectvely. We note that Vy Vx = V cos θ, Vy = V sn θ, tanθ = (1.6) V In an analogous way one may defne the rectangular components of a vector n 3- dmensonal space. In ths case we use a rght-handed rectangular system of axes x, y, z, wth correspondng unt vectors uˆ, uˆ, u ˆ, as shown n the fgure below: x y z x z uˆz V x uˆx O uˆ y y

14 4 CHAPTER 1 (If we nterchanged, say, the names of the x- and y-axes, the ensung rectangular system xyz would be left-handed, whle the system yxz would now be rght-handed. Can you fnd a practcal way to determne whether a gven system of axes xyz s rghthanded or left-handed? Note the order n whch the names x, y and z are wrtten!) The algebrac values V x, V y, V z of the projectons of V on the three axes consttute the rectangular components of V. As can be shown, V = V uˆ + V uˆ + V uˆ ( V, V, V ) ; x x y y z z x y z V V = V + V + V x y z (1.7) In partcular, f V= 0 then V x =V y =V z = 0 and V= 0. Now, assume that A= A uˆ + A uˆ + A uˆ x x y y z z and B= B uˆ + B uˆ + B uˆ. Then, x x y y z z Moreover, A= B A = B, A = B, A = B x x y y z z A± B= ( A ± B ) uˆ + ( A ± B ) uˆ + ( A ± B ) uˆ x x x y y y z z z ( A ± B, A ± B, A ± B ) x x y y z z (1.8) (1.9) In general, the components of a sum of vectors are the sums of the respectve components of the vectors. Thus, let V = V1 + V + V ( Vx, Vy, Vz ) where V ( V, V, V ). x y z Then, (1.10) V = V, V = V, V = V x x y y z z Example: For the vectors A (1, 1, 0), B (,1, 1), we have: A+ B (3,0, 1), A B ( 1,, 1) and A+ B = ( 1) = 10, A B = ( 1) + ( ) + 1 = 6.

15 VECTORS Poston Vectors A poston vector s a vector used to determne the poston of a pont n space, relatve to a fxed reference pont Ο whch, typcally, s chosen to be the orgn of our coordnate system. For ponts on a plane, we use a system of two axes x, y: y y uˆ y O uˆx r θ P x x The vector r = OP determnes the poston of pont Ρ relatve to O. The components (x, y) of r are the Cartesan coordnates of P. We wrte: r = xuˆ + yuˆ ( x, y) x y (1.11) Alternatvely, we can determne the poston of Ρ by usng polar coordnates (r, θ), where r= r and where 0 θ < π or π < θ π (by conventon, the angle θ ncreases counterclockwse). We notce that x= r cos θ, y= r snθ (1.1) or, conversely, r x y θ y x = +, tan = (1.13) For ponts n space we use a rectangular system of axes x, y, z: z z uˆz r P x x uˆx O uˆ y y y

16 6 CHAPTER 1 We wrte: r = xuˆ + yuˆ + zuˆ ( x, y, z) x y z r = r= x + y + z ; (1.14) The three quanttes (x, y, z) consttute the Cartesan coordnates of pont P n space. (Alternatve systems of coordnates are sphercal and cylndrcal coordnates, whch wll not be used n ths book.) Now, let P 1, P be two ponts n space wth poston vectors r 1, r and wth coordnates (x 1, y 1, z 1 ), (x, y, z ), respectvely. We seek an expresson for the dstance P 1 P between these ponts. P 1 x z r 1 r 1 r O P y We notce that PP 1 = r1 = r r1. But, r = x uˆ + y uˆ + z uˆ, r = x uˆ + y uˆ + z uˆ 1 1 x 1 y 1 z x y z so that Therefore, r = r r = ( x x ) uˆ + ( y y ) uˆ + ( z z ) uˆ x 1 y 1 z. PP = r = ( x x ) + ( y y ) + ( z z ) 1/ (1.15) Example: For the ponts P 1, P, wth respectve coordnates (x 1, y 1, z 1 ) (, 1, 3) and (x, y, z ) (0, 1, ), we have: P1 P = + ( ) + 1 = 3.

17 VECTORS Scalar ( Dot ) Product of Two Vectors Consder two vectors A and B, and let θ be the angle between them, where, by conventon, 0 θ π. B θ A The scalar product (or dot product ) of A and B s a scalar quantty defned by the equaton A B= A B cosθ = AB cosθ (1.16) where A= A, B= B. It s easy to show that ths product s commutatve: A B= B A. In the case where A= B, we have θ=0, cosθ=1 and A A= A = A (1.17) If A and B are mutually perpendcular, then θ=π/, cosθ=0 and therefore A B= 0 A B (1.18) As can be shown, A ( λb ) = ( λ A) B, ( κ A ) ( λb ) = κλ ( A B ) (where κ, λ are scalars) and A ( B+ C) = A B+ A C. For the unt vectors we can show that uˆ uˆ = uˆ uˆ = uˆ uˆ = 1, uˆ uˆ = uˆ uˆ = uˆ uˆ = 0. x x y y z z x y x z y z Thus, f A= A uˆ + A uˆ + A uˆ x x y y z z for the dot product n terms of components: and B= B uˆ + B uˆ + B uˆ, we fnd a useful relaton x x y y z z

18 8 CHAPTER 1 A B= A B + A B + A B x x y y z z (1.19) For two equal vectors we are thus led back to (1.17): A A= A + A + A = A x y z (1.0) 1.5 Vector ( Cross ) Product of Two Vectors A B θ A B B A The vector product (or cross product ) A B of A and B s a vector normal to the plane defned by A and B and n the drecton of advance of a rght-handed screw rotated from A to B. Alternatvely, the drecton of A B may be determned by the rght-hand rule : Let the fngers of the rght hand pont n the drecton of the frst vector, A. Rotate the fngers from A to B through the smaller of the two angles between these vectors. The extended thumb then ponts n the drecton of A B. The magntude of A B s, by defnton, where 0 θ π. We notce that A B = A B snθ = ABsnθ A B= B A (1.1) (hence, the cross product s not commutatve) and that A A= 0. In general, f A and B are parallel vectors, then A B= 0, snce, n ths case, θ=0 or π, so that snθ=0. As can be proven, (where κ, λ are scalars) and A ( λb ) = ( λ A) B, ( κ A ) ( λb ) = κλ ( A B ) A ( B+ C) = ( A B) + ( A C).

19 VECTORS 9 For the unt vectors we can show that uˆ uˆ = uˆ uˆ = uˆ uˆ = 0, uˆ uˆ = uˆ, uˆ uˆ = uˆ, uˆ uˆ = uˆ. x x y y z z x y z y z x z x y Thus, f A= A uˆ + A uˆ + A uˆ, B= B uˆ + B uˆ + B uˆ, we fnd that x x y y z z x x y y z z A B= ( A B A B ) uˆ + ( A B A B ) uˆ + ( A B A B ) uˆ y z z y x z x x z y x y y x z (1.) Ths can be wrtten more compactly n the form of a determnant, uˆ x uˆ y uˆ z A B = A A A x y z B B B x y z (1.3) to be developed wth respect to the frst row.

20 10 CHAPTER 1 x y z Exercses 1. Consder the vectors A ( A, A, A ) and B ( B, B, B ), and let θ be the angle x y z between them. By usng the propertes of the scalar product, show the followng: a. A± B = ( A + B ± A B cosθ) 1/ b. cosθ = A B + A B + A B x x y y z z 1/ 1/ ( Ax + Ay + Az ) ( Bx + By + Bz ) c. If A B, then A+ B = A B = A + B (Pythagorean theorem). Let A ( A, A, A ), x y z B ( Bx, By, Bz ), C ( C, C, C ). x y z a. Show that Ax Ay Az A ( B C) = B B B x y z C C C x y z b. By usng the propertes of determnants, show that A ( B C) = B ( C A) = C ( A B) A ( A B) = 0 3. Fnd the value of α n order that the vectors mutually perpendcular. 1 3 A, α, and B ( 3, 3, 1) 4. Fnd the values of α and β n order that the vectors A (1, α,3) B (, 4, β ) be parallel to each other. be and

21 CHAPTER KINEMATICS.1 Rectlnear Moton Knematcs s the branch of Mechancs that studes moton per se, regardless of the physcal factors that cause or affect t. (The connecton between cause and effect s the subject of Dynamcs, to be dscussed n the next chapter.) The smplest type of moton s rectlnear moton,.e., moton along a straght lne. Such a lne could be, e.g., the x-axs on whch we have defned a postve orentaton n the drecton of the unt vector u ˆx, as well as a pont O (an orgn) at whch x=0. ˆ x O u r A x= 0 x x The poston A of the movng object, at tme t, s specfed by the poston vector OA= r = xu where x and r are functons of t : x= x( t), r = r( t). We note that x>0 or x<0, dependng on whether the object s on the rght or on the left of O, respectvely. The velocty of the object at pont Α, at tme t, s defned as the tme dervatve of the poston vector;.e., t s the vector ˆx dr d dx v = = ( xuˆ ) ˆ x = u dt dt dt (where we have taken nto account that u ˆx s constant). We wrte: x v = vu ˆx where dx v= =± v (.1) dt In the above relaton, v s the algebrac value of the velocty wth respect to the unt vector u ˆx. The magntude of the velocty, equal to v, s called the speed. In general, v and v are functons of t. The sgn of v ndcates the nstantaneous drecton of moton: f v>0, the object s movng n the postve drecton of the x-axs (.e., to the rght, as seen n the fgure), whle f v<0, the object s movng n the negatve drecton (to the left). In S.I. unts, v s expressed n m/s=m.s 1. Gven the functon v=v(t), we can fnd the poston x(t) of the object at all tmes t by ntegratng. From (.1) we have: 11

22 1 CHAPTER dx v= dx= vdt. dt We ntegrate the above dfferental equaton, makng the addtonal assumpton (ntal condton) that, at the moment t=t 0, the nstantaneous poston of the object s x=x 0 : x t t dx= vdt x x = vdt 0 x0 t0 t0 0 t x= x + vdt (.) t0 The dfference x x 0 s called the dsplacement of the object from pont x 0. The acceleraton of the object at tme t s the dervatve of the velocty vector: dv d dv a = = ( vuˆ ) ˆ x = ux. dt dt dt We wrte: a = au ˆx where dv d x a = a dt = dt =± (.3) The unt of acceleraton n the S.I. system s m/s =m.s. The quantty a n (.3) represents the algebrac value of the acceleraton. For a gven functon a=a(t), and by assumng that at the moment t=t 0 the movng object has a velocty v=v 0, we fnd the velocty v(t) at all tmes t by ntegratng: dv v t a= dv= adt dv adt dt = v 0 t0 0 t t v= v + adt (.4) 0 If we know the acceleraton as a functon of x, a=a(x), we can fnd the velocty as a functon of poston, as follows: By dvdng the relatons dv=adt and dx=vdt n order to elmnate t, we get: vdv=adx. We now ntegrate ths relaton, assumng that v=v 0 at the poston x=x 0 : v v vdv= adx = adx v x x 0 v x x v = v + adx (.5) 0 x x 0

23 KINEMATICS 13 Two observatons are n order regardng equaton (.5): 1. In order for (.5) to make sense physcally, the rght-hand sde must be nonnegatve. Ths may place restrctons on the admssble values of x, whch means that the object may not be allowed to move on the entre x-axs.. To fnd v tself (rather than just ts absolute value) from relaton (.5), we must take the square root of the rght-hand sde (the latter assumed nonnegatve). Ths process wll yeld two values for v, wth opposte sgns. One of these values must be excluded, however, snce ts sgn wll not be consstent wth that of v 0. Note also that, n relatons (.) and (.4), the x and v are functons of the varable t that appears on the upper lmts of the correspondng ntegrals. Smlarly, n relaton (.5) the quantty v s a functon of the varable x that appears on the upper lmt of the ntegral. In general, an ntegral wth varable upper lmt s a functon of that lmt.. Specal Types of Rectlnear Moton We now apply the general results of the prevous secton to two famlar cases of rectlnear moton. 1. Unform rectlnear moton: v=constant, a=0 Ths s the moton wth constant velocty (n magntude and drecton). Relaton (.) yelds (by puttng t 0 =0): t x= x + vdt= x + v dt t x= x0+ vt (.6). Unformly accelerated rectlnear moton: a=constant 0 By relatons (.4) and (.) we fnd (puttng t 0 =0): t v= v + adt= v + a dt t v= v0+ at (.7) t t t t x= x + vdt= x + ( v + at) dt= x + v dt+ a tdt Furthermore, relaton (.5) gves: 1 x x v t at = (.8)

24 14 CHAPTER x x 0 0 x 0 x0 v = v + adx= v + a dx v = v + a( x x ) (.9) 0 0 Note that the same result s found by elmnatng t between (.7) and (.8). Note also that, snce the rght-hand sde of (.9) must be nonnegatve, the acceptable values of x may be restrcted; ths, n turn, wll place a restrcton on the possble postons of the movng object. Example: Projectle moton x v> 0 v 0 uˆx v< 0 a O x= 0 At tme t 0 =0, a bullet s fred straght upward from a pont Ο (at whch x 0 =0) of the vertcal x-axs, wth ntal velocty v ˆ 0 = v0 ux ( v0 > 0). We assume that the bullet s subject only to the force of gravty (we gnore ar resstance). Thus, the acceleraton of the projectle equals the acceleraton of gravty, whch s always drected downward, regardless of the drecton of moton (upward or downward) of the projectle. In vector form, a = g uˆ a uˆ a= g x where g s approxmately equal to 9.8 m/s. We notce that a s constant, and therefore the moton s unformly accelerated. The equatons of moton are: x v= v + at= v gt, x= x0+ v0t+ at = v0t gt, v = v + a( x x ) = v gx The projectle wll reach a maxmum heght x=h, where t wll stop momentarly at tme t=t h ; t wll then start movng downward, toward the pont of ejecton O. To fnd h and t h, we use the frst two equatons of moton: v v gt t 0 h = 0 h = 0 h = ; 1 v0 h= v0th gth =. g v g

25 KINEMATICS 15 The maxmum heght h may also be determned by notng that, n order for the rghthand sde of the thrd equaton of moton to be nonnegatve, the value of x must not exceed v 0 /g. Note also that v>0 for 0< t < t h, whle v<0 for t > t h. What s the physcal meanng of ths? (Notce that we chose the postve drecton of the x-axs upward.) Exercse: Fnd the moment at whch the bullet returns to Ο, as well as the velocty of return. What do you observe? Also, show that, n a free fall from heght h, a body acqures a speed v= gh..3 Curvlnear Moton n Space We now consder moton along an arbtrary curve on the plane or n space. The nstantaneous poston of the movng object s determned by the poston vector r wth respect to the orgn Ο of our coordnate system. Ths vector, as well as the correspondng coordnates (x, y, z), are functons of tme t. uˆz z r A a v uˆx O uˆy x y Accordng to (1.14), we may wrte: r = r( t) = xuˆ x+ yuˆ y + zuˆ z where x= x( t), y= y( t), z= z( t). The velocty of the movng object at a pont Α of the trajectory, at tme t, s the tme dervatve of the poston vector: dr v = = d ( xuˆ + yuˆ + zuˆ ) = dx uˆ + dy uˆ + dz uˆ dt dt dt dt dt x y z x y z (.10) We wrte: v = v uˆ + v uˆ + v uˆ x x y y z z dx dy dz vx =, vy =, vz = dt dt dt (.10 )

26 16 CHAPTER As wll be shown analytcally n Sec..6, the velocty vector v s tangent to the trajectory at all ponts and ts drecton s that of the drecton of moton. The magntude of the velocty, called speed, s v = v + v + v x y z (.11) The acceleraton of the object at a pont Α, at tme t, s the tme dervatve of the velocty: We wrte: dv dv dv x y dvz a = = uˆ + uˆ + uˆ dt dt dt dt x y z a= a uˆ + a uˆ + a uˆ x x y y z z dv dv x y dv ax =, ay =, az = dt dt dt z (.1) (.1 ) As wll be shown n Sec..6, the drecton of the acceleraton vector a s toward the concave ( nner ) sde of the trajectory. The magntude of the acceleraton s a = a + a + a x y z (.13) Example: Assume that the coordnates of a movng partcle are gven as functons of tme by the equatons {x=a cos ωt, y=a sn ωt, z=λt}, where Α, ω, λ are postve constants. (What can you say about the trajectory of the partcle?) By usng (.10 ) and (.1 ) we fnd the components of velocty and acceleraton, respectvely, of the partcle: (v x, v y, v z ) ( ωα sn ωt, ωα cos ωt, λ), (a x, a y, a z ) ( ω Α cos ωt, ω Α sn ωt, 0). The magntudes v and a of the correspondng vectors are gven by (.11) and (.13): v= ω A + λ, a= ω A. Note that the speed v of the partcle s constant n tme, as well as that the vectors of velocty and acceleraton are mutually perpendcular [show ths by usng relatons (1.18) and (1.19)]. As wll be seen below, these two facts are closely related..4 Change of Speed Generally speakng, a moton s accelerated f a 0, so that the vector v of the velocty changes wth tme. The term accelerated, however, s often used wth a dfferent meanng n Knematcs, a fact that, f not properly ponted out, may lead to confuson. Thus, a (generally curvlnear) moton s sad to be accelerated or

27 KINEMATICS 17 retarded durng a tme nterval f the speed v = v ncreases or decreases, respectvely, n that nterval. If the speed s constant, the moton s called unform. The knd of moton depends on the angle θ between the vector v of the velocty and the vector a of the acceleraton, where, by conventon, 0 θ π. In general, where a = a. On the other hand, v a= v a cosθ = v a cosθ (.14) dv dv dv d d( v ) d( v ) dv dv ( v a) = v = v + v = ( v v) = = = v dt dt dt dt dt dv dt dt where t should be noted carefully that dv dv v a = v = v dt dt (.15) dv d v dv v v v dt dt dt! By comparng (.14) and (.15), we fnd that dv dt = a cosθ (.16) Gven that a> 0, we note the followng: π dv a. If 0 θ < then > 0 ; v ncreases and the moton s accelerated. dt π dv b. If < θ π then < 0 ; v decreases and the moton s retarded. dt π c. If θ = (that s, f a v dv ) then = 0 ; v s constant and the moton s unform. dt Of specal mportance s the followng concluson: If the acceleraton s perpendcular to the velocty, the speed of the movng object s constant n tme (even though the drecton of the velocty may be changng). Thus, the moton s unform. In the case of rectlnear moton, the angle θ between v and a can only assume two values. If v and a are n the same drecton, then θ=0 and the moton s accelerated, whle f v and a are n opposte drectons, then θ=π and the moton s retarded. Now, v and a are n the same drecton or n opposte drectons f v a > 0 or v a < 0, respectvely. Gven that v = v uˆx and a = a uˆx [see Eqs. (.1) and (.3)],

28 18 CHAPTER where here v and a are algebrac values ( v=± v, a=± a ), we have: v a= v a. We thus conclude that a rectlnear moton s accelerated or retarded, dependng on whether the product of the algebrac values of the velocty and the acceleraton s postve (va>0) or negatve (va<0), respectvely..5 Moton wth Constant Acceleraton We now consder the case where the acceleraton a of the movng object s constant n magntude and drecton. We assume that at tme t=0 the nstantaneous poston vector of the object, relatve to our coordnate system, s r = r, whle the object has 0 ntal velocty v = v. We seek ( ) 0 r t and v( t) for every t >0. Takng nto account that a s a constant vector, we have: dv v t a dv adt dv a dt v v0 at dt = = = = v 0 0 v = v + a t 0 (.17) dr r t t v v at dr v dt atdt dr v dt a tdt dt = = + = + = r r = r0 + v0 t+ a t (.18) We wrte: t r r r 0 t v 0 a. = = + Ths vector relaton s of the form r = κ v 0+ λ a, wth constant v, a and varable 0 κ, λ. Accordng to Analytc Geometry, the vector r = r r les on the constant 0 plane defned by v 0 and a and passng through the pont r 0 (ntal poston of the movng object). On the same plane wll therefore always le the tp of the poston vector r of the object. We conclude that moton wth constant acceleraton (n magntude and drecton) takes place on a constant plane. 1 An example s moton of a body n the gravtatonal feld near the surface of the Earth, n a relatvely small regon of space where ths feld may be consdered unform. If we gnore ar resstance, the body s subject to the constant acceleraton of 1 In the specal case where the acceleraton s zero, the moton s unform rectlnear.

29 KINEMATICS 19 gravty g (drected downward, toward the surface of the Earth) and ts path s confned to the plane defned by the body s ntal velocty v 0 and the acceleraton g. As s easy to see, the plane of moton s perpendcular to the surface of the Earth..6 Tangental and Normal Components The velocty v and the acceleraton a of a movng object are vectors of absolute physcal and geometrcal substance, ndependent of the choce of a system of axes (x, y, z) n our space. If we choose a dfferent set of axes (x, y, z ), wth dfferent orentaton relatve to (x, y, z), the components of v and a wll change but the vectors themselves, as geometrcal quanttes, wll reman the same. We wll now ntroduce a system of components that s assocated wth the trajectory tself of the movng object. As we already know, a pont Α of the trajectory can be determned by ts poston vector r relatve to any reference pont Ο. Ths vector s a certan functon r( t) of tme. An alternatve way of determnng Α s the followng: We choose an arbtrary pont C of the curve representng the trajectory, as well as a postve drecton of moton along the curve (not necessarly concdent wth the actual drecton of moton). The locaton of pont Α on the curve s then gven by the dstance s=ca of A from C, measured along the curve. Note that s may be postve or negatve, dependng on whether Α s located ahead of C or behnd C, respectvely, n the postve sense of traversng the curve,.e., n the drecton of ncreasng s. C s A r r r s A + O To smplfy our analyss, we make the assumpton that the trajectory s a plane curve. The results we wll arrve at, however, wll be vald for moton along any smooth curve n space. We consder the ponts Α and Α of the trajectory, through whch ponts the object passes at tmes t and t, respectvely. Let r and r be the poston vectors of Α and Α relatve to Ο. We call r = r r and t = t t, and we wrte: r = r ( t), r = r ( t ) = r ( t+ t) = r ( t) + r. The velocty at pont Α, at tme t, s dr r( t+ t) r( t) r v = = lm = lm dt t 0 t t 0 t.

30 0 CHAPTER But, r r s = t s t, so that r r s lm = lm lm t s t t 0 s 0 t 0 (snce s 0 when t 0). We thus have: dr dr ds v = = (.19) dt ds dt Ths relaton expresses the dervatve of a composte functon, gven that r= r ( s) and s=s(t), so that r= r( t). We now seek the geometrcal sgnfcance of the vector dr / ds. We notce that ths vector s the lmt of r / s for s 0. As s 0, the vector r / s tends to become tangent to the trajectory at pont Α, whle ts drecton s that of ncreasng s ( s>0); that s, t ponts toward the postve drecton of traversng the curve. Moreover, r / s 1 as s 0. We conclude that dr / ds s a unt vector tangent to the trajectory at Α and orented n the postve drecton of moton on the path. We wrte: Relaton (.19) now takes on the form u ˆT dr = (.0) ds ds v = uˆ ˆ T = v u dt T (.1) where v s the algebrac value of the velocty, wth respect to the unt vector u ˆT : ds v= =± v. dt Thus, f v>0 the moton s n the postve drecton (ncreasng s), whle f v<0 the moton s n the negatve drecton (decreasng s). Note that the unt vector u ˆT s always n the postve drecton, regardless of the drecton of moton! (In the fgure below, the moton s n the postve drecton; thus v>0.) As seen from (.1), the velocty s a vector tangent to the trajectory. A uˆt v +

31 KINEMATICS 1 Our study of acceleraton begns wth an mportant remark: 1. A component of acceleraton parallel to the velocty may alter the magntude but not the drecton of the velocty. (Ths s the case n accelerated rectlnear moton.). A component of acceleraton normal to the velocty may change the drecton but not the magntude of the velocty. (Ths s a drect consequence of the dscusson n Sec..4.) Gven that the velocty s a vector tangent to the trajectory, t s natural to resolve the acceleraton a nto two components: 1. A component a T tangent to the trajectory (tangental acceleraton), whch s responsble for the change of speed (change of magntude of the velocty).. A component a N normal to the trajectory (normal or centrpetal acceleraton), whch s responsble for the change of drecton of the velocty. The total acceleraton of the movng object wll be the vector sum a = a T + a, as N seen n the fgure below: v a T A uˆt uˆn a To evaluate a T and a N, we dfferentate (.1) wth respect to t : a N dv d dv duˆ T a = = ( vuˆ ) ˆ T = ut + v dt dt dt dt (.) The vector duˆt dt s normal to u ˆT. Indeed, duˆ T 1 d 1 d uˆ ( ˆ ˆ ) ( ˆ T = ut ut = ut ) = 0, dt dt dt duˆt gven that u ˆT = 1= constant. Moreover, s drected toward the concave ( nner ) dt sde of the trajectory, snce ths s the case wth the nfntesmal change du ˆT of u ˆT. [By (.), then, the acceleraton a of the movng object s orented toward the

32 CHAPTER concavty of the curve.] We thus consder a unt vector u ˆN normal to the trajectory and drected nward, and we wrte: duˆ ˆ ˆ T du du T T = u ˆ ˆ N = u N (.3) dt dt dt The dfferental du ˆT s approxmately equal to an nfntesmal change of u ˆT when ths unt vector s dsplaced along the curve wthn an nfntesmal tme nterval dt. That s, duˆ = u ˆ uˆ, as seen n the followng fgure. T T T uˆt A ds dϕ A u ˆT dϕ uˆt duˆt ρ dϕ K u ˆT The u ˆT and u ˆT are unt vectors tangent to the curve at the respectve ponts Α and Α. These ponts are an nfntesmal dstance ds apart (measured along the curve), whch dstance the object covers wthn tme dt. Gven that the angle dφ between u ˆT and u ˆT s nfntesmal, we may wrte: duˆ uˆ dϕ = dϕ, where dϕ s n rad. T T Relaton (.3) s thus wrtten (by takng nto account that v=ds/dt): duˆ T dϕ dϕ ds dϕ = uˆ ˆ ˆ N = un = v un (.4) dt dt ds dt ds Gven that ds s nfntesmal, t may approxmately be regarded as an arc of a crcle wth center Κ (the pont of ntersecton of the lnes normal to the curve at Α and Α ) and radus ρ=ακ. The pont Κ s called the center of curvature of the trajectory at A, whle ρ represents the radus of curvature at Α. Thus, ds ρdϕ, so that (.4) becomes: Fnally, (.) yelds: duˆ T v = un ˆ (.5) dt ρ

33 KINEMATICS 3 dv v a= uˆ ˆ T + u dt ρ N (.6) We wrte: a T a = a uˆ + a uˆ T T N N dv d s v = =, a = N dt dt ρ (.7) The magntude of the acceleraton s 4 dv v a = at + an = + dt ρ 1/ (.8) We stress that these results are vald for any curve n space, not just for moton on a plane curve! Let us summarze the ways of resolvng velocty and acceleraton nto components: Note, n partcular, that v = vxuˆ x + vyuˆ y + vzuˆ z = v uˆ T (.9) a= a uˆ + a uˆ + a uˆ = a uˆ + a uˆ x x y y z z T T N N a a a a a a = x + y + z = T + N (.30) Specal cases: 1. Unform curvlnear moton: v=constant. We have: dv v a = = 0, a= a uˆ = uˆ dt ρ T N N N (.31) Note that, n unform moton, the acceleraton s normal to the velocty, n accordance wth the conclusons of Sec..4.. Rectlnear moton: ρ =, s= x, uˆ ˆ T = ux. Hence, ds dx v = uˆ ˆ ˆ T = ux = vux, dt dt v dv a 0, ˆ ˆ N = = a= at ut = ux. ρ dt In unform rectlnear moton ( v = const. ) both a T and a N are zero, so that a= 0.

34 4 CHAPTER.7 Crcular Moton Crcular moton s plane moton wth constant radus of curvature ρ=r. The trajectory of the movng object s a crcle of radus R. + v uˆt A uˆn s θ O C R We choose the postve drecton of moton to be counterclockwse, n whch case the length s of the arc from a reference pont C, as well as the angle θ measured from C (see fgure), ncrease counterclockwse and decrease clockwse. (The fgure represents moton n the postve drecton.) As always, the unt tangent vector u ˆT s orented n the postve drecton, regardless of the actual drecton of moton. As we know, s=rθ, where θ s measured n rad. The velocty of the object s ds dθ v = v uˆ T where v= = R dt dt (.3) (In general, v velocty Then, =± v, dependng on the drecton of moton.) We defne the angular dθ ω= (.33) dt v = Rω (.34) The quantty ω s measured n rad / s = rad.s 1. Note that the sgn of ω s the same as that of v and depends on the drecton of moton. The acceleraton of the object s where a = a uˆ + a uˆ T T N N dv d ω ( ) at R, a v R ω = = N = =. dt dt R R

35 KINEMATICS 5 We defne the angular acceleraton dω α = (.35) dt We thus have: a R a R T = α, N = ω (.36) The magntude of the acceleraton s a = a + a = R α + ω 4 T N (.37) In unform crcular moton the algebrac values v and ω are constant, so that, by (.35) and (.36), α=0 and a T =0. Hence, the acceleraton s purely centrpetal : v a= u = R u v = const R ˆ ˆ N ω N (, ω.) (.38) By (.33) and by takng nto account that ω s constant, we have: θ dθ = ω dt dθ = ω dt= ω dt t θ0 0 0 t θ = θ0+ ω t (.39) The perod Τ (measured n s) and the frequency f = 1/Τ (measured n s 1 or hertz, Hz) of unform crcular moton are defned by the relaton π ω = = π f (.40) T More on these wll be sad n Chapter 5, n the context of smple harmonc moton..8 Relatve Moton Consder two objects Α, Β and an observer Ο who uses the coordnate system (x, y, z). Ths system s called the frame of reference of Ο. The poston vectors of Α and Β relatve to Ο are r A and r B, respectvely, whle the veloctes of Α and Β wth respect to Ο are v A dra drb =, vb =. dt dt We call r BA = r B r A the poston vector of Β relatve to Α.

36 6 CHAPTER z v A r A A r BA B v B v A v BA v B x O r B y The velocty of Β relatve to Α s defned as v BA dr = dt BA (.41) We notce that d drb dra vba = ( rb ra ) = dt dt dt v = v v BA B A (.4) Smlarly, the velocty of Α relatve to Β s v = v v = v AB A B BA In an analogous way we defne the acceleraton of Β relatve to Α: dvba d dvb dva aba = = ( vb va) = dt dt dt dt a = a a = a BA B A AB (.43) (.44) where a A dva dv =, ab = dt dt B are the acceleratons of Α and Β wth respect to Ο. Consder now two observers Ο and Ο, where O s movng at constant velocty V relatve to O ; that s, vo O= V = const. The relatve acceleraton of these observers s, therefore, zero:

37 KINEMATICS 7 dvo O ao O = = 0. dt Assume, further, that these observers record the moton of a partcle Σ : z x x z O O y y r Σ r Σ Σ We denote the velocty and the acceleraton of Σ wth respect to the two observers as follows: v = v, a = a ; v = v, a = a. Σ O Σ O Σ O Σ O We now apply (.4) and (.44) wth Ο n place of Α and Σ n place of Β: v = v v v = v V ; Σ O Σ O O O a = a a a = a. Σ O Σ O O O We thus arrve at an mportant concluson: In partcular, A partcle moves wth the same acceleraton wth respect to two observers that mantan a constant velocty relatve to each other (they do not accelerate wth respect to each other). f the partcle moves wth constant velocty relatve to one observer, t wll also move wth constant velocty relatve to the other observer. Stated dfferently, f the partcle executes unform rectlnear moton relatve to one observer, t wll execute the same knd of moton relatve to the other observer.

38 CHAPTER 3 DYNAMICS OF A PARTICLE 3.1 The Law of Inerta The term pont partcle (or smply partcle) refers to a body whose dmensons are so small that we may gnore ts rotatonal moton (f any). But, even a body of fnte dmensons may be treated as a partcle f ts moton s purely translatonal (that s, f the body s not rotatng). A partcle s sad to be free f (a) t s not subject to any nteractons wth the rest of the world (a case that s rather unrealstc!) or (b) the totalty of ts nteractons somehow sum to zero (.e., they cancel one another) so that the partcle behaves as f t were subject to no nteractons whatsoever. Accordng to the Law of Inerta or Newton s Frst Law, a free partcle moves wth constant velocty (.e., has no acceleraton) relatve to any other free partcle. Therefore, a free partcle ether s n unform rectlnear moton, or s at rest, relatve to another free partcle. Imagne now an observer who herself s a free partcle (ths s approxmately true for someone who s at rest on the surface of the Earth). Such an observer s called an nertal observer and the system of coordnates or axes she uses s called an nertal frame of reference (or, smply, nertal frame). Accordng to the law of nerta, dfferent nertal observers move wth constant veloctes (thus, do not accelerate) relatve to one another. For example, the passenger n a tran that moves wth constant velocty relatve to the ground s (approxmately) an nertal observer and a fxed system of axes (x, y, z) n the tran s an nertal frame of reference. On the bass of the law of nerta we may now gve the followng defnton of an nertal reference frame: An nertal frame of reference s any set of coordnates (or axes) relatve to whch a free partcle ether moves wth constant velocty or s at rest. Thus, n an nertal frame a free partcle does not accelerate. We note that the observer who uses ths frame s, by defnton, at rest relatve to t. A reference frame that accelerates wth respect to an nertal frame s obvously not nertal. Ths s, e.g., the case wth the Earth because of ts daly rotaton as well as ts orbtng moton about the Sun (f we regard the latter as an almost nertal frame). However, snce the acceleraton of the Earth s relatvely small compared to the acceleratons measured n typcal terrestral experments, we may, for practcal purposes, 8

39 DYNAMICS OF A PARTICLE 9 consder the Earth as an almost nertal frame. Hence, any observer movng on the surface of the Earth wth constant velocty wll be regarded as an nertal observer. 3. Momentum, Force, and Newton s nd and 3rd Laws In an abstract sense, force represents the effort necessary n order to alter the state of moton of a body; n partcular, n order to change the body s velocty. The frst dea that comes to mnd s to quanttatvely dentfy force wth acceleraton. We know from our experence, however, that dfferent bodes generally requre a dfferent effort n order to acqure the same acceleraton, or, the same velocty wthn the same perod of tme. (Try, e.g., to produce the same acceleraton on a book and on a truck by pushng them!) Ths happens because dfferent bodes exhbt dfferent nerta, that s, dfferent resstance to a change of ther state of moton. Ths property must therefore be taken nto account n the defnton of force. To ths end, we ntroduce a new physcal quantty called lnear momentum (or smply momentum) of a body: p = mv (3.1) where v s the velocty of the body. The coeffcent m s called mass and s a measure of the body s nerta. Newton s Second Law of Moton (we wll often smply call t Newton s Law ), whch s vald only n nertal frames of reference, n essence defnes the force exerted on a body as the rate of change of the body s momentum at tme t : dp F = dt (3.) If we make the assumpton that the mass m s constant, then dp d ( mv) m dv = =. dt dt dt Hence, F = ma (3.3) where a= dv / dt s the acceleraton of the body at tme t. We stress that the form (3.3) of Newton s law s vald for a body of constant mass, as well as that relatons (3.) and (3.3) are vald on the assumpton that the observer measurng the velocty and the acceleraton of the body s an nertal observer. By usng (3.3) and by takng nto account the results of Sec..8, t s not hard to show that the force on a partcle s the same for all nertal observers.

40 30 CHAPTER 3 The vector equaton (3.3) s equvalent to three algebrac equatons, one for each vector component. We wrte: F = F uˆ + F uˆ + F uˆ, a= a uˆ + a uˆ + a uˆ. x x y y z z x x y y z z Substtutng these expressons nto (3.3) and equatng correspondng components n the two sdes of ths equaton, n accordance wth (1.8), we have: F = ma, F = ma, F = ma (3.4) x x y y z z Now, accordng to the law of nerta, the change of the state of moton of a body relatve to an nertal observer requres an nteracton of the body wth the rest of the world. The force F s precsely a measure of ths nteracton. If the body s not subject to nteractons (.e., s a free partcle ) then F =0 and t follows from (3.3) that the velocty of the body wth respect to an nertal reference frame s constant (snce the acceleraton s zero). We thus conclude that the second law of moton s consstent wth the law of nerta, provded that both these laws are examned from the pont of vew of an nertal frame of reference. It s temptng to argue that, accordng to the above dscusson, the law of nerta s redundant snce t appears to be just a specal case of the second law: Free partcle no nteracton no force no acceleraton constant velocty. There s a subtle pont, however: What knd of observer s enttled to conclude that a partcle that appears to move wth constant velocty (.e., wth no acceleraton) s a free partcle? Answer: Only an nertal observer, who uses an nertal frame of reference! The purpose of the law of nerta s essentally to defne these frames and guarantee ther exstence. So, wthout the frst law of moton, the second law would become ndetermnate, f not altogether wrong, snce t would appear to be vald relatve to any observer regardless of hs or her state of moton. One may say that the frst law defnes the terran wthn whch the second law acqures a meanng. Applyng the latter law wthout takng the former one nto account would be lke tryng to play soccer wthout possessng a soccer feld! Accordng to (3.), f a body s not subject to any force ( F = 0 ) ts momentum p relatve to an nertal frame s constant n tme, snce, n ths case, dp / dt= 0. As wll be seen n Chapter 6, ths s true, more generally, for any solated system of partcles,.e., a system subject to no external nteractons. For such a system, the prncple of conservaton of momentum s vald: The total momentum of a system of partcles subject to no external forces s constant n tme. Ths prncple s ntmately related to a thrd law of moton. Consder a system of two partcles subject only to ther mutual nteracton (there are no external forces). The total momentum of the system at tmes t and t+ t s

41 DYNAMICS OF A PARTICLE 31 P( t) = p1 + p = m1v1 + mv, P( t t) p + = + p = m v + m v By conservaton of momentum, P( t) = P( t+ t) p + p = p + p 1 1 p p ( p = p ) p = p (3.5) Hence, But, by (3.), p1 p p1 p dp1 dp = lm = lm =. t t t 0 t t 0 t dt dt dp dt dp = F, = F dt where F 1 s the nternal force exerted on partcle 1 by ts nteracton wth partcle, whle F 1 s the force on partcle due to ts nteracton wth partcle 1. Thus, fnally, F = F 1 1 (3.6) Relaton (3.6) expresses Newton s Thrd Law or Law of Acton and Reacton. Note that ths law s equvalent to the prncple of conservaton of momentum, whch prncple, n turn, consttutes the generalzaton of the law of nerta for a system of partcles. Takng nto account that Newton s second law (n essence, the defnton of the concept of force) also s a logcal extenson of the frst law, we can apprecate the great mportance of the law of nerta n the axomatc foundaton of Newtonan Mechancs! [See also Papachrstou (01).] You may have notced that we defned momentum, whch depends explctly on mass, wthout prevously gvng a defnton of mass tself. We wll now descrbe a method for determnng mass, based on the prncple of conservaton of momentum. Consder agan an solated system (.e., a system subject to no external forces) consstng of two partcles of masses m 1, m, whch somehow nteract wth each other (e.g., they collde or, f they are electrcally charged, they exert Coulomb forces on each other, etc.). Assume that, wthn a tme nterval t, the momenta of m 1 and m change by p 1 and p, respectvely. Accordng to (3.5), p 1= p or m v = m v 1 1 (3.7) In terms of magntudes,

42 3 CHAPTER 3 m v = m v 1 1 m m v = (3.8) v 1 1 As experment shows, the rato of magntudes on the rght-hand sde of (3.8) always assumes the same value for gven partcles m 1 and m, regardless of the knd or the duraton of ther nteracton. Moreover, the vectors v 1 and v are found to be n opposte drectons, n accordance wth (3.7). These observatons verfy that to each partcle n the system there corresponds a constant quantty m, ts mass, such that the sum m1v 1+ mv retans a constant value when the partcles are subject only to ther mutual nteracton. Ths consttutes an expermental verfcaton of conservaton of momentum. Now, relaton (3.8) allows us to determne the rato m /m 1 expermentally by measurng the rato of magntudes of v 1 and v. Hence, by arbtrarly assgnng unt value to the mass of partcle 1, we can determne the mass of partcle as follows: We allow the two partcles to nteract for a tme nterval t ; then we measure the (vector) changes of ther veloctes wthn ths nterval and we calculate the rato of magntudes of these changes. The result yelds the mass m of partcle numercally (snce, by defnton, m 1 s a unt mass). In a smlar way, we determne the mass m of any partcle by allowng ths partcle to nteract wth a partcle of known mass. By measurng the nstantaneous acceleraton a of m, we then fnd the correspondng nstantaneous force F on ths partcle by usng Newton s law (3.3). In the S.I. system of unts, the unt of mass s 1kg=10 3 g, whle the unt of force s 1Newton=1N=1kg.m.s. Assume now that a body of mass m s subject to varous nteractons wth ts surroundngs, whch nteractons are represented quanttatvely by the forces F1, F,. The vector sum F F = F + F + 1 s the resultant force (or total force) actng on the body. Newton s second law then takes on the form: dp F = = ma dt (3.9) Let a x, a y, a z be the components of the acceleraton a of the body. Accordng to (1.10), the components of the resultant force ΣF are ΣF x, ΣF y, ΣF z, where ΣF x s the sum of the x-components of F1, F,, etc. By equatng correspondng components on the two sdes of (3.9), we have:

43 DYNAMICS OF A PARTICLE 33 F x = ma x, F y = ma y, F ma z = z (3.10) As an example, assume that a body of mass m=kg s subject to the forces F1 (3,1, 1) N and F ( 1, 3, 1) N. The resultant force s F = F1 + F (,4, ) N so that ΣF x =N, ΣF y =4N, ΣF z = N. By (3.10) we fnd the acceleraton of the body: a ( a, a, a ) (1,, 1) m s. x y z A body s sad to be n equlbrum f the total force on t s zero: Σ F = 0. Note that by equlbrum we do not necessarly mean rest ( v = 0 )! Accordng to (3.9), a body s n a state of equlbrum f t does not accelerate ( a = 0) relatve to an nertal observer. Ths state could be nstantaneous, as n the case of an oscllatng pendulum at the moment when the bob reaches the lowest pont of ts path. If, however, the body s ntally at rest at an equlbrum poston, then t remans at rest there. Conversely, a body may be momentarly at rest wthout beng n a state of equlbrum. The total force actng on t wll then cause an acceleraton that wll put the body back n moton at the very next moment. For example, f we throw a stone straght upward, t wll stop nstantaneously as soon as t reaches a maxmum heght and then t wll start movng downward mmedately, under the acton of gravty. Another example s the pendulum bob at the hghest pont of ts path. We noted earler that a body of fnte dmensons can be treated as f t were a pont partcle f ts moton s purely translatonal (.e., the body s not subject to rotaton). Such a moton depends only on the resultant force on the body, regardless of the locaton of the ponts of acton of the varous ndvdual forces that act on ths body. On the contrary, as wll be seen n Chapter 7, the ponts of acton of these forces are mportant when one consders rotatonal moton, as ths moton s determned by the total external torque on the body. 3.3 Force of Gravty Near the surface of the Earth and n the absence of ar resstance, all bodes fall toward the ground wth a common acceleraton g, called the acceleraton of gravty and havng a magntude g 9.8 m s. The force of gravtatonal attracton between a body and the Earth s called the weght w of the body. If m s the mass of the body, then, by Newton s second law, w = mg (3.11) For larger dstances from the surface of the Earth, the value of g (hence also the weght of a body) vares as a functon of the dstance from the Earth. We call M and R the mass and the radus of the Earth, respectvely, and we let h be a gven heght above the surface of the Earth. We would lke to determne the value of g at ths

44 34 CHAPTER 3 heght. Accordng to Newton s Law of Gravty, the magntude of the gravtatonal force on a body of mass m, located at a heght h above the Earth, s Mm w= G ( R + h ) (3.1) where G s a constant, the value of whch s expermentally determned to be G = N.m.kg. Takng nto account that w=mg, we fnd that GM g = ( R+ h) (3.13) Note that the rato w/m, whch represents the gravtatonal feld strength at the consdered locaton, also represents the acceleraton of gravty, g. Accordng to (3.13), ths acceleraton s ndependent of the mass m of the body. Thus, all bodes experence the same acceleraton at any pont n a gravtatonal feld, regardless of the partcular physcal propertes of each body (provded, of course, that no forces other than gravty are present). 3.4 Frctonal Forces Sldng frcton (or smply frcton) s a force that tends to oppose the relatve moton of two surfaces when they are n contact. It s a cumulatve effect of a large number of mcroscopc nteractons of electromagnetc orgn, among the atoms or molecules of the two surfaces. Practcally speakng, these surfaces belong to two bodes that are n contact wth each other (although true contact s never possble at the atomc level!). f s N v= 0 F f k N v 0 F w w Let us consder, for example, a box of weght w lyng on the horzontal surface of a table. The box s at rest under the acton of two forces, namely, ts weght and the normal reacton N from the table. Ths state of equlbrum mples that the resultant force on the box s zero: N + w= 0 N = w. We now push the box to the rght wth a force F that may vary n magntude (see fgure). The box wants to slde to the rght but there s a force f opposng ths moton. Ths force, whch s drected to the left, s the frcton between the box and the surface of the table. If F s not large enough, f manages to balance t and the box