6 Linear and Angular Momentum Conservation of Linear Momentum. If a force F is acting on a particle of mass mm, then according to sn's second law of motion, we have F- ap o (my) ay where p=my is the linear momentum of the particle, Momentum is a ———— vector quantity ; a —TWihe external force acting on the particle is zero, then @ 63 Newte F = op = 0 or p =my =a constant se ay eee Thus, in absence of an external force, the linear momentum of a particle remains constant. In chapter 2, we have already established the law of conservation, of momentum for a system ticles, which ar interacting mutually. There we have shown that in absence of external forces ———_—— 7 | Pi + p2 = Constant or my vy + m2 v2 = Constant «+ Now, let us consider a system of n particles whose masses are mm ,mz,.....,My. The system can be a rigid body in which the particles are in fixed positions with respect to one another, or it can be a collection of particles in which there may be all kinds of internal motion. Suppose that the particles of the system are interacting with each other and are also acted by external forces. If Py = 71 ¥1, Pp = Mp¥2,.-. » Po=/nYq are the momenta of the particles of masses 1,172, -...,My respectively, then the total momentum (P) of the system is the vector sum of the momenta of individual particles i.e., P=pit+po+py+... + Pn 1) i =M1 V+ mzV2 + ... + Min Vy Differentiating it with respect to time ¢, we have 224LINEAR AND ANGULAR MOMENTUM | 225 dp, dor dt or re , de = Fitka+ 0.0 4+ Fr +. (5) where hy, © Fr, ties a rd »F, represent the forces acting on the particles of . cE rs M25 +++ +5!My respectively, 7 Nesoarciitt ea ee and internal forces both. But fccording, v 's third law, the internal es exist in pairs of i : equal and pees a they balance cach other and so do not contribute any te bet : iH Total force. Hence the right hand side sum in eq. (5) pI : nts U © resultant force Fx; only due to the external forces acting on 7 the particles of the system. The internal forces cannot change the otal momentum of the system, because being equal and a they produce equal and opposite changes in the momentum. (Hence, if we want to change the total momentum of the system of the particles, it is necessary to apply external forces on that system. Then the sum of external forces is _aP_d Fea = Gp = qi (Pi + P2 +--+ + Pn) If the resultant external force is zero, then dP as 0 or P = aconstant. ; — ———_ ie, P = pitp2tp3+ -.. + Pn = aconstant .- (6) € hus if the resultant external, force acting on a system of particles is zero, the total linear momentum of the system remains constant. This simple but quite general result is called the law of conservation of linear momentum for a system of particlestt is fo be noted that the momenta of individual particles may change, ‘but their sum i.e. the total linear momentum remains unaltered in the absence of external forces. The law of conservation of momentum is a fundamental and exa: of nature. No violation of it has ever been found and it has been thoroughly checked by all kinds of experiments. We have established this law on the basis of Newton's laws, but this law_holds_true where Newtonian mechanics fails. The law may be proved for the sy of particles (by J ing the case of two interacting particles, chapter 2) with the help of Galilean invariance and the law ‘of conservation of energy Actually, the law of: conservation of linear momentum is one of the sacred laws of neture and is @ prog, in itself. Ex. L. Find the momentum of an electron which is accelerated by @ potential difference of 20 volts._ EECHANTES 226|™ 1 mn?) am (3! mie go 20 ee — —30 1 amet = 2009 * 10720 x 1-6 x 10 24 jg.m/secr ; joactivel @ Ex 2A nucleus, initially at res deca ion A oon of etn electron of e) ight angles : ron nein momenta 1°73 MeVic i a igi ages Oe A ycteus reco peuvino _ In what direction does ¢ pone meme "nd ‘in Sl units. If me mass of the residual nucleus is mom SL. / ‘ 25 eg, calculate its kinetic ener®) jn electron volts ? =amv= Sol. P ing an electron of 224 x10 by emit 3-9 10 Sol. Initial momentum =0 . nave ‘Assuming the direction of the electron as X-axis, we n= 1-731 MeV/c, . momentum of the electro! a =1f Mev/c, momentum of the neutrino momentum of the nucleus p =/"” MeV/c. From the law of conservation of momentum, We have 1.7374 F+mv =0 ormv= -:737 +5 ) MeV/c. Its magnitude = mv = VL “Tay + 12 = 2 MeV/c approx. 6 2x 10' -19 = 2X10 x 1-6 «10 3 x 10' = 10-66 x 1072? kg- m/sec. Direction cosines of p are (-3. -5.0) ie. direction of the recoil of nucleus makes angle of 210° with X-axis. ‘Now, the speed of the residual nucleus 3-9 x 10 KE. of the residual nucleus = 3 mv? =} 3-9 x 107 =! (3-9 x 107) (10-66 x 1077 f 1-6 x10" 3 % = 9-1electron- © 6.2, Centre of Mass. fale haere Let us conside: x etacca| a system of N-particles, having masses 7711 , 72, «+7 aah aie aml ne Vectors ry, r2,.... fy respectively. The contre x this instant is the point whose position vector R is givenMINEAR AND ANGULAR MOMENTUM | 227 MR = mir a mary + + my ry 7% where AFis the total mass of the system ie M=Z n= mits t +m Therefore, m+ Man +. mie Fim -(8) The sum Ent, 5, is called the first moment of mass for the system For a system of two Particles — an + mrp R= my +m TEX, Y, Z) are the coordinates of the centre of Mass, then R=xfs+yfaze +. (9) So that xX a7 t+ may +. myty me my +m + Fy x _ Doky ~ Sin Pry + may2 +... + myyy ~ my Fig t... Fim Fig. 6.1 — =n ++. (10) In z= mi tmz7z2+ . ay tim + A rigid body may be considered as a system of closely packed rarticles, having continuous distribution of mass. The centre of mass of uch a body will be given on replacing the sign of summation (2) by the ign of intergration in eq. (10) ie, st myzy _ ET mzy stmy Sy = Lean 2 peasy = fan = Hy Sv dom mn Zw Led beim, -@)hape, the centre of mass, the eometrical ? For example, centres of th, al s tric ive centres of mass. tr ey. (7) with respect to « arcnante bodies of ach the ular £ Te ayemme : are the FoSPee mass. Differentiating Pa on the f pout ae stick etc isc, MC e of mass. sphere, di es the centre of Motion time ¢, we Bet dry +5. + my “dt dr2 a gm AR yy Gh + 2 ey oa © of mass ang a » centre of 5 and for the velocity dR/dt iu c veloc articles, we ha rd » particles, ce “Tor the velocities of the part ne ret? MV = myyry + mpv2 +--+ ++ (12) MV = Eninyy = P or ++. (13) or This eq gi Ly ass. ion giv of the centre of mas: ati ie Chand side of se) is the total linear momentum of the system Right hand side i is velocity of the centre of ind it is equal to the product of the total mass a eee of ce a In absence of external forces, the total linear remains constant, hence from eq. (12), we have a MV = aconstant or V =a constant Hitt i : This is a remarkable Property of the centre of mass that its ve, locity If the eq. (12) is again differentiated av ap aM with respect to time 4, we get . EF Om) ~ ZF, where F,, is the force acting on any lotal force =F, on all external force Fen, be forces of their mutual int i Particle n of the System. Hence the the particles of the 5) Cause taking all the teraction cancel toy ystem will be equal to the Particles of the system, the gether (Newton’s third law) - (15) Thus, the acceleration of th, : © Centre of Mass is due to onl ! the e 7 OFces and is Btven by Newton § Second law of Mobic : a mass of a eeLINEAR AND ANGULAR MOMENTUM | 229 the total mass of the system, subjected to the external forces applied to the system. Centre of mass frame of reference. A frame attached with the centre of mass of an isolated system* of particles is called the centre of mass frame of referertce or C-frame of reference. in this C-frame of reference, the contre of mass remains at rest i... V=0 . So that in view of eq. (12) the total linear momentum of the system in C-frame of reference is always zero Le., P = MV = Emp va = 0 (in C-frameof reference ) (16) For this reason, the C-frame is also called the zero-momentum frame. Such a frame is sometimes very much convenient for the analysis of many experiments, which are difficult to be analysed in the laboratory system. The centre of mass of a system in absence of external forces moves with constant velocity in an inertial frame, so the centre of mass frame of reference for the isolated system is an inertiai frame. For a system of particles, which is acted by external forces, the C-frame becomes accelerated i.e., non-inertial and is not of much use . Determination of centre of mass of a body. The position of centre of mass of a symmetrical body having uniform density, can easily be known by symmetry. Hence, the centre of mass of a thin unifrom rod is its mid- point; of a circular disc or sphere, its centre and so on. When a body has only one axis of symmetry it may be taken as X-axis and a suitable point on this axis is taken as origin. With the help of the integral calculus, the position of the centre of mass on the X-axis, the axis of symmetry, may be determined. (1) Centre of mass of a thin uniform rod. Let the X-axis lie along the length OA of the rod and O be the origin, If M and / rerepsent its mass FE ae, and length respectively, then mass per 7 k&—- X —41 unit length = M/1. ° A Consider a small clement dr at a Lae distance x from the origin, then its 7 acer Fig. 6.2 dm = Max, Hence, the distance of the centre of mass G of the rod from O is M ya Lean _ Suet ax = "Tan ~ = Hence, the centre of mass of the rod is situated in the middle. ‘© Bolated system means a system free from external forces.far p" esses org shin trianreniar we) re r * 12 Centre of nes a oo simile to KL, each paralle bigest {cach rip Hi an eof mats 0 fe The cent at of ch HT ag similarly, middie point hence the centre Chive plate fies on the median AZ Tp in will ale xs of tl us, the centre of mass oO : ee hae a of intersection G Of the 4 at the poin plate h of ree medians and G ix called the centr iS of three meds ; the triangle. The point G divid 2, 4: at 2, 60 thal ue 4 he on the median ' triangular he tri Let tate. Let the triany Fig 63 3 internally in the ratio T: 2, 74 AG _ 142 i DOK AS oo AG™ 2 DG 1 oe AG 2 AG 2 AG = GAD. (3) Centre of mass of a semi-circular lamina or plate. Let fig, 6. represent a semi-circular lamina with AOB as its diameter and OC a radius perpendicular to it. The raidus OC is evidently the axis of symmetry and, therefore, centre of mass of the lamina lics on it. Let the X-axis be along OC and Y-axis along OA. Ara of the lamina = 1R?/2, If M be the mass of the lamina, its mass per unit area = 2M/nR?, Consider a strip of thickness dx at a distance x from the origin, then Length of the strip = 2 VR? — 2 Area of the strip = 2 VR? x. a Mass of the strip = oa eMTINFAR AND ANGUTAR MOMENTUM [ 111 Hones, x 4 _4R ART “Se _ Hence, the centre of mass of the semi-circular lamina will lie on the line oven 4 distance AR Ax from the centre O of the lar (4 Centre of mass of right circular cone. The fig. 6-5 represents a right circular cone, having circular base AB of radius 8 and centre C Let O, the apex of the cone, be the origin and OC, the X-axis, Here, OC is the axis of symmetry, O< jer two paralicl planes | perpendicular to X-axis and at l———-1r-—— distances x and.x + dv from O, The portion of the cone lying between wo planes is a ciruclar dis Radius of the disc x tang, where ¢ is the semi-vertical angle of the cone. If i be the height of the cone, then tang, = R/h r/x =R/h or Area of the disc = Volume of the disc = = Mass of the disc dm = 222-2 « ax, where p is the density of the material of the cone. Hence, the centre of mass of the cone will lie on OC at a distance X from O, given by pix BER. a fea ayn _ Sxdm 0 n 0 t Lalo 3 X= seca eri A ee me dm eR 2 3 in Re 2. a Six ad io Hence, the centroid of a soild cone is situated at a distance 3 of its vertical height measured from its vertex. (5) Centre of mass of a soid hemisphere. The fig. 6-6 represents 8FINALLY oe ther the shore, ie boat far 10 the other end fowever the centre of mass OF the external forces. Thus em will remain at the same place in absence of total syste When the man moves t » the ced by d. centre of mass of the boat 18 displaced by d centre 0) mo +L+d)+M02+4) +i om +M Subtracting eq. (1) from (ii), we get O=m(L+d)+Md or d= m+M ; Hence, the boat will be displaced towards the shore by the distance, given by mi. 60x10 _ 1200 _4 9) “im +M = 60+ 1400 = 1460 ¥ 63. Collision of Two Particles, Let us consider 1 and ma, moving with 2 respectively before the collision.* If their velocities ‘1 and yp, then according to the la of momentum we have | | w of conservation | . | | muy + mzu) = ™1V1 + mov, (17) | on ger jhe terms ‘before’ and ‘after the collision imeau that the initi nal) positions are i i fend fi Hh y u teraction forces lence the Potential ame. If the isi Collision isLINEAR AND ANGTILAR MOMENTIM | 238 According to the kinetic theory of gases, such clastic collisions ocour iWween the molecules of a gas, This type of collisions mostly tke place tween the atoms, electrons and protons, Tn case of inelastic collisions, a part of kinotic energy is converted in me other forms. This energy appears in the form of heat in larger pacroscopic) particles ie. the collision produces an incrgasc in the brational energy of the constituent atoms of the particles, Ifthe ititial netic energy of an atom or molecule is sufficiently large, then in altision with some other Particle it may absorb some energy. Duc to this bsorption of energy, its inernal structure is changed and we say that the tom has been excited to some higher cncrpy level. The amount of cnergy sed in this Process is called excitation energy *. The final kinetic energy ifthe system is thus reduced by an amount equal to the excitation energy ind then z gm? + dmg? = Amy? +4246... (19) It is also possible that the atom of molecule may be in the excited state before the collision and during the collision its internal excitation energy may change in the form of kinetic energy, so that we have $m? + $m? +E = hm? + zmave? (20) The final velocities yy and v2 of the particles cannot be calculated uniquely only by the knowledge of their initial velocities and conservation laws of momentum and energy. The reason lies in the fact that any velocity has three components and hence for the determination of vy, and Y2, six independent equations are required, But equation (17) gives three equations in component form and equations (18), (19) and (20) give only one equation. In case, if after the collision the direction of movement of one particle is also known, then we can calculate v, and v2 completely, 6.4. Deflection of a Moving Particle by a Particle at Rest. Let us consider a particle of mass mj, colliding clastically with another particle of mass rm, which is initially at rest in the laboratory frame of reference.** Suppose in this frame, the initial velocity of my is u, and after the collision its final velocity be v, in a direction making an angle 6;, with its initial path. This angle 6,, through which the particle of mass m, has been deflected by collision, is called scattering angle. Now, if we take the direction of 0, along X-axis and the plane containing u, and vy, as X-Y plane, then the final velocity vz of m will have no Z-component, because mu, and m ¥ 1%) have no Z-components. If v2 ‘The excitation energy has @ few definite values for each kind of atom or molecule. Generally, this energy is remitted i short interval of time. of light either immediately or after a In case of scattering of a-particles and the laboratory frame. mpton effect initial velocity of m2 is zero inof rt, WC PAVE the path and energy 1m ont 0; 08 Oy my? <1 + f ‘ty, gu 1 OCIS n vnge Oo maker 0 7 equations sio ml 1 neously, WC Can is a little tedious we io we the centre Of mass fy! din the conte My jmore i ve. i 4 more informal and Contre of Mast Frame Noy w I nytt mult jm nin equ sng these three CAN solving these put this ea? mnity of inrerest: BU Ay js IF he bo Peis simpler tic quanti hy. a feng en its refer : m 4 OOM, Y sory Frame 1 ° Laboratory ‘——> —) aH gt Rest Initial Initial 4, i Final Fig. 6.9 f i tre of mass In the laboratory frame, the velocity V of the cen OF the system will be given by (my +m) V= my uy +g X 0 _ muy or “mi tm 2) The centre of mass frame of reference moves with this vel relative to the laboratory frame ie,, centre of mass (C.M.) is at re: centre of mass frame of reference, In this frame, the initial a velocities of the two Particles are given by uy! =ur-V3u2' =~ w'=y- V; v2 The centre of mass is at rest ve equal and opposite m, locity V st in the ind final and +++ (23) =v-V ha m ‘so, Pe m= —mauy! ; MM! =—m, v2" in this frame henc i > € th omentaie, the particles should and anpKy 2 fn aes id with @ 3 metres long wo, nu ig se malsec into the bag and Oey, 10 ke BO nee 0 / si my Asn, ‘as fo ‘wn ifr ty ne bat “ ihe muccirruin displag,)”| s tne bag. Cateniate ti he Tet it CO of the bag (iii) re Phe ia " 0 ts ; Mass of the pullet = 2 ce vetocity of the bullet u= Velocity of the BBY =! From the faw of conservation 0! momentum me om + M)Y 0 sec whence v age a0 a = 0-392 ml aa pendulum. Let the maximup, After collision, the bag will oscillate like @ displacement of the bag bex. fi Restoring force at this displacement is G i Face = mg sin eomg x/h where / is the length of the string. c= oe = 10-2% 9-8 maximum displacement % of the system (bag + bullet) is ae Si cede= 30? . 4c? =} x 10-2 x (0-392) id 0:392¥ 3 9.392 x0-54= 0-21 metre. (i) Energy converted into heat = 4 mu? - 3 heat = 5 mu" — > (m + My the entire kinetic energy At this changed into potential [2* 10-2 x (0-392) ] formie., 2 =3x0-2x20?-2 ; 4 x 10-2 x 0-392)" = 40 - 0-8 = 39-2 joules. ii) 6-5, Impact. When two bodi i ies collide wit impi phenomenon of colliding is calle: diner impinge on, each other, this An impact bet direction of “tween two bodeis is said motion : aid to be di . of each, just before renee ad Aa head on, if the » Is along the common yeor! giee inp t where con! bef thi ma eaaseeeLINEAR AND ANGULAR MOMENTUM | 243 gormal at the point of contact. reocities of the colliding ae oe ea srl anore ta direction as well as in magnitude. o There are two laws, governi joqutotewobodies! a (3) When two bodies collide directly, rere relative velocity after impact is in 7. i instant ratio with their relative velocity pefore impact and is in opposite direction. This constant ratio depends only on the Taterial of bodies and is called the ‘efi - stitution or resilence. It is Fig, 6.10 denoted by the letter e. Thus, if w; and 3 are the velocities of wo bodies pefore the impact and vy; and v2 after the impact, thea C5) 2 Te OF vam se (ui — H2) (The vel ‘are Measur jive T i The quantity has different male for different ae ot) atetances put it always Viog Rewwen 0 and t For perfectly elastic bodies ¢ = 1 and ~ for perfectly plastic ies ¢ = 0. But there are, however. 00 perfectly plastic or elastic pode and actual bodies lie betwen the nwo extremes “ (2) The linear momentum of the system, before and after the collision, is Thus, if the masses of the two bodies are m, ac 72, then conserved. according to this law +++ (29) Momentum before the impact = momennim Ger the impact. athe first law is an experimental fact of ‘Newton’s work and the second tawis the consequence of Newton's third law of motion. Nw using the relations (28) and (29), we can find the final velocities vyand v2. Multiplying €4- (28) by m3, we have getty — migettz = M22 — M2 - (30) (29), we have muy + mye + em2 (HA — u2) _ (my — ema un + (4 2 a my + m2 Subtracting ¢q- (30) from eq. (my + m2) v= whence ‘Adding eq. (29) with eq. (30), We BFF 1 +e) my + (2 = emy) Uz _ ate ears my + MZ Incase, if m, =m and ¢ = 1,weert vps ug andy2 =H1-r PR (the earth and gis an expression for mavinn 00% of escape velocity 1 her fo gravity at earth's surface. W reached by a rocket fired with (Agra (984; Avadh 96; Purvanchal 93; Bundelkhand 95) Sol, As the rocket rises upwards the value of g changes. Fvider 1 veto! a betght will bo given by & changes. Evidently the velocity 2.4 aah v's 2fiman At the earth's surface, x= GMMR? i { aheght A,g'= eM CoM a , Rea Ra aaRYe Ch enRye E a +h/RY c v7 sup? - 2f,— adh. h R *[-rsi, =00 + 28 [reazR ~ X] or wp? Pa yh The escape velocity is given by v2gR. Therefore in this case vo=0- OV2gR, and highest point reached by rocket, »=0, Now substituting the values in the : a2ghR __h O81 x 2gR=fA or 0-81 = Zee whence : = 4-3 R. This is the expression for the maximum height attained by the rocket. 6.7. Angular Momentum and Torque. The angular momentum of a particle is defined as the moment of its linear momentum. Hence the angular momentum J of a particle about a point is defined by 1.89 where r is the vector distance oft the particle from that point and p = my is the momentum in an inertial frame in which the point is stationary. Angular momentum is a vector quantity whose magnitude (p r sin @) is equal to the product of linear momentum and its perpendicular distance from the reference point ; its direction is perpendicular to both r and p (or v). The component of J along any line or axis passing through the fixed reference point is often called the angular momentum of the particle about this axis. SI. units of the angular momentum are kg-m?/sec or J-sec.sanent® : Hatint nvr — ra Oe directo AMET AEN eg, sre mint : wyatt ot, we gel wee ile gaan repent | dp the pine Or gting 04 oGxpren a Cert My dex" dl a de Pee Tie are Aer dy cle ; a id r0¢ apphed on th particle y eap/atis the fore IF is called LOrque OF Momen, sg product OFF A ented by T. Thus, “, The vector P noi and is rept s nowt the reference POre about the Fe [qo derxF T* Gi i ty ito the rate of change of angular MOmentun, Thue sont #8 eg!" 4 Hrs unit snewton-mele ye moving independently OF attache ,, If a number of a i; rid body, then the angular Momentuy, , | ° fy other in the form ¢ ae , ani a : about A point is defined as the vector sum of the angular mome system abo c s oof the separate particles about the Sanne point. “Now, if Jy Ja... are the angular momenta of various Particles y system about a given point, the total angular momentum J of the oa about that point is given by JES + Jat om or T= (11 X MV) (2% V2) F srerseeee N N or J= > (Xm) oF (0X pa) + (3 =1 a= 1 } wheren = 1,2,..,.,N are the particles in the system : ee et & <. Torque Boa HDi X Gn vn) =DraXPn (i * n=1 ts of the internal forces of mutuil | ‘ _ In this summation, the moment interaction balance each other, becai i LINEAR AND ANGULAR MOMENTUM | 253 In a free system left to itself, the external torque T = 0. Thea a3 ad = 9 oF J =constant. Jehehe+ ay (Thus. the total angular momentum of a system of particles is constant if the resultant external torque acting on the system is zero. ‘This is the principle of conservation of angular momentum. In the absence of an external torque, total angular momentum geht ht: of the system remains constant, although the angular momenta of the particles may mutually interchange. Angular momentum (J) and angular momentum about centre of mass Jon). The total angular momentum J of a system of particles can be expressed in a convenient and important form by using the velocity of the centre of mass and velocities of the particles relative to the centre of mass. If R and V are the position vector and velocity of the centre of mass and r,,- and V,- those of a particle of mass mi, relative to the centre of mass, then the position vector and velocity of the particle with respect to the reference point are given by tm =R+ Enc and Vn = V+ Vac Hence the total angular momentum for a system of N particles is N J = Yom (R + rnc) X (V+ vnc) n=1 = Ein (R X V) + Eoin (RX ne) + Din (ne X VY) + Enin (rnc X Vac) -.. (40) But rac=tm—R or mnctnc =Mntn-mnR or | Leintnc = Lentn — LmnR = =n tn — MR where M = =m = mass of all the particles of the system. But according to the property of centre of mass, we know that MR= mri + mor. +... = Zmntn ©. Zitn Enc = 0; similarly, LrtyVnc = 0. Heice, from eq. (37), we have J= RX MV +E (rnc X mnvnc) ... (41) In this equation © (tye X My VYnc) Tepresents the angular momentum of the system about the centre of mass, say Jem, and MV =P is the total linear momentum. The quantity Rx P or Rx MV is the angular momentum of the centre of mass about the reference point. Thus J =Jem + RX P ... (42) ” Tnc = In — R)284 | MECHANICS _ ntral Faves; Areal Velocity, P force, acting, on a puutic hey depends, only joree, ' a inant ite distance from a fined contre, 119 iy the nitude of relative to the Hed conte ion under € 6.8, Motion DOW th LAINIE ving mag NW Nhe oe Wry) > particle osition vector of the par ; force is represented by the relation, F =f (9), where f(r) is a scalar finction of distance » The torque acting on the particle in Ta GaeXB er IF = faery -4) for ond Fe py, Sothat J =r xX mv = constant Thus, the angular momentum of a particle, force, is conserved. Moreover, if J ig constant, it should always be perpendicular to the plane containing r and y. This means that the path ofa particle in the influence of a central force lies in a plane. Now, let O be the centre of force, when the vector r changes to r + Ar, the vector area AS swept by the radius vector in this time is (44) MOVING URAEY Central Of prehen Fig, 6.14 AS =] rx Ar This area is swept in At time, therfore dividing both the equation by Ar and taking limit as At> 0, then dS _ 4 J GI" XVR 5, (44) This expression gives the areal velocity (dS/dt) of the particle. But J is constant for a particle, moving under central force, so from relation (44) we see that the areal velocity remains constant, when the particle moves in the influence of a central force. 6.9. Examples of Conservation of Angular Momentum. sides of this (1) Planetary or satellite motion. The planets move in elliptical orbits about the sun at a focus.* The gravitational force on the planet is directed towards the centre of the sun. In other words, t pked ia central force. Consequently, the angular momentum (J ) e . +H be plane. planet will be constant and so the orbit of motion will be p!AR AND ANGULAR MOMENTUM | 255 por planetary motion, the constancy of J means that the areal city AS/at = (8/27n) is constant. Thus, the rate of sweeping out of the vg by a radial vector from sun to the planet is a constant — this is known aMgepler’s second law of planetary motion. Similarly, artificial satellites # ye in elliptical orbits around the earth and the angular momentum is minserved” : In planetary motion, for the conservation of angular momentum x wv), the planet must move faster at the point of closest approach to sun than af the farthest point. The reason lies in the fact that at these ints the angular momentum is myvr, (because at these positions r is rpendicular to v) and for its constancy the shorter r is associated with jarget If ry,r2 are the distances of closest and farthest points of the planet respectively and 4, v2 are the corresponding velocities, then MVM = MMVI or vr = V2 ... (45) 2) Scattering of protons or a-particles by a heavy nucleus. Suppose that a proton of charge e and mass m is moving towards a heavy nucleus. The charge on the nucleus is where Z is the atomic % ‘umber. Electrostatic force of repulsion will act on the nucleus so that its path willbe +--—~ 4 —-- hyperbola (Fig. 6.15). The , Nucleus perpendicular distance p between the nucleus A and the foot B of the perpendicular drawn from A over the initial direction of the proton is called the impact parameter. If the Fig. 6.15 velocity of the proton at infinite distance is vg, its initial kinetic energy is pimp? and the angular momentum about the nucleus A is mvg p. If the distance of closest approach of the proton from the nucleus is d, then at this distance kinetic energy = 3 vc ?, angular momentum = my, d and Potential energy = Ze?/d. The electrostatic force is a central force, hence for the conservation of ‘ngular momentum and energy, we have mwvop = mved . +. (46) i : aint the artificial satelite comes clase fo the earth, its speed decreases duc to the earth's incre ener and its angular momentum decreases. But the angular momentum of the earth ace in this process and hence the total angular momentum of the system (earth and ite) remains constant,