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Newton Third Law Stadler

Newton Third Law Stadler - Inadequacy of the usual...

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Unformatted text preview: Inadequacy of the usual Newtonian formulation for certain problems in particle mechanics W. Stadler Division of Engineering, San Francisco State University, San Francisco, California 94132 (Received 2 January 1980; accepted for publication 15 September 1981) The problem of the rigid double pendulum is used to show that the Newtonian axioms are inadequate to obtain a solution. It is further shown that an additional independent axiom, such as the law of moment of momentum, must be introduced to make the solution possible. I. INTRODUCTION Although solved correctly some 300 years ago, the prob- lem of the rigid double pendulum is an example which suf- fices to convince the reader of the need for an additional postulate when dealing with constrained particle motion. It will be shown that the problem of the rigid double pendu- lum has no solution when Newton’s third law is taken to include the centrality (or collinearity) of the internal forces. When centrality is not assumed the “principle of angular momentum” no longer follows as a theorem and must be postulated as an independent law. The application of the law of linear momentum and that of moment of momen- tum then leads to a noncentral internal force system. There are a number of other problems which exhibit the same difficulty; e.g., a mass point constrained to slide on the massless rod of a simple pendulum. The analysis here is confined to the rigid double pendulum, for simplicity. II. NEWTONIAN FORMULATION By way of background, the main ingredients of the me- chanics of a system of particles are now summarized. Consider a set (or system) .fl of m particles P,, i = 1,...,m, with masses m,, respectively. The position vector of P, relative to an arbitrary fixed origin in an iner- tial reference frame is r,- =x} il +x§i2 +x§i3, where (x,-',x?,x§) are rectangular Cartesian coordinates and (il ,i2,i3) is a fixed orthonormal basis. The velocity and ac- celeration of P, at the instant t of time are denoted by r..(t) and m t ), respectively. For a system of particles in which all forces are assumed to arise from the interaction of the particles with one an- other, the forces acting on a subsystem JV Q J , of n<m particles, may be subdivided into external and internal forces. The external forces acting on JV are those exerted by particles in % \JV on particles in JV, while the internal forces are those exerted by particles in JV on each other. The external force acting on P, is denoted by F), while the internal force exerted on P, by P]. is denoted by FU. Either of these forces may generally be a function of the current posi- tions and velocities of all n particles as well as time. The usual formulation of Newtonian particle mechan— icsI consists of the following postulatesz: (i) Newton’s second law in the form Fm + (S: Far) = mm) (I) 1:1 for each particle P,- in the system JV. Of course, (1) includes Newton’s first law for a single particle. 595 Am. J. Phys. 50(7), July 1982 (ii) Newton’s third law in the form Fri“) = —— FIJI-(t) and F11“): 0 (2) fort: l,...,n; j= 1,...,n. (iii) The internal forces F” (t) are central 3; that is, [l'.-(t)-l'j(t)]><F.y(t)=0 (3) for i = 1,...,n; j = 1,...,n._ Furthermore, the masses of the particles are always as- sumed to be nonzero. For any given particle P, in the system the problem to be solved in particle mechanics belongs to one of the following categories: (a) The postion r, (t) of a particle P,- is completely speci- fied as a function of the time t. (b) The particle P,- is constrained to move on a specified surface or curve. (c) All of the forces acting on P,- are prescribed as func- tions of the kinematic variables or some of the forces are prescribed together with classes of permissible velocities or accelerations. A problem is considered solved when the motion r, (t) and the external and internal forces have been obtained as functions of time in such a way as to satisfy Newton’s laws (i) and (ii). On the basis of assumptions (i)—(iii) it is easy to establish the following theorem. Theorem I: Moment of momentum. Assume that the forces, masses, and accelerations of 'a system of particles satisfy the postulates (i)—(iii). Then , H0 = M, where H0 is the moment of momentum of the subsystem JV gel! with respect to a point 0 fixed in an inertial reference frame and where M0 is the total moment about 0 of the external forces acting on JV. (See Fig. 1.) Proof. For convenience, let 0 be the origin. For each par- ticle, the application of Newton’s second law (i) implies VF..(z)+ 2 Fat) =m.-i.-(t) (4) i=1 Fig. 1. Central force assumption. © 1982 American Association of Physics Teachers 595 for i = 1, .. .,n. Cross multiplication of each equation by r,- (t ) and a summation over 1' yield n 2 r.(t)><F.(t)+ i )"3 r.(t)><F.-,-(t) 1= 1 l = 1 j = 1 = 2 miriltleiltl- (5) i = l Newton’s third law (ii), together with the assumption that the internal forces are central (iii), then implies that Hom= i m.r.(r)><i.~(t) l=l = 2 r.~(t)><F.-(t) = Mo(t)~ (6) i = I In applications, Theorem 1 is habitually used as if it were an independent axiom; that is, no check is made as to whether the hypotheses of the theorem are satisfied or not. However, the use of the theorem obviously is justified only if all of the hypotheses are met. As will be shown, the hy- potheses are, in fact, not satisfied in the case of the rigid double pendulum. Indeed, the postulates (i)—(iii) are insuffi- cient to solve this problem. Preliminary to the discussion of the rigid double pendu- lum, it is instructive to first analyze the simple pendulum in the light of the above remarks. III. SIMPLE PENDULUM In the standard treatment of the simple pendulum, the freebody diagram of the particle P is drawn as in Fig. 2, with the tensile force T = — Te, and weight W = mg(cos 0 e, — sin 6 es) of the particle. An applica- tion of Newton’s second law yields [9 +(g/a)sin0=0 , (7) as the eo-component equation, and T=mgcos0 +mal92 (8) as the e,-component equation. Equation (7) may be used (subject to apprOpriate initial conditions) to obtain 0 (t ), and (8) then gives the tension required to maintain this motion. It should be emphasized that the tension T was not speci- fied as regards its functional dependence on 49 (t ) but rather that this dependence was derived through the application of postulates and geometrical constraints. Note that, in ad- dition to Newton’s second law, the solution was based on the assumed centrality of the internal force T. The above problem also serves as an illustration of the use of Theorem 1. With 0 as the fixed point and with W as the external force, one again obtains (7). Next, Newton’s second law, together with the central force assumption, is used to calculate T. Thus, in the case of the simple Pendu- lum, the use of Theorem 1 is justified since a central force 596 Am. J. Phys, Vol. 50, No. 7, July 1982 Fig. 2. Standard simple pendulum. \ Fig. 3. pendulum. Variant of the simple does, in fact, sufiice to sustain the motion. Suppose now that one attempted to solve the problem without making use of the central force assumption. To this end, let the internal force on P be represented as F = F,e, + F9 eg. The use of Newton’s second law yields F, = mg sine +maé, (9) F, = —mgcos€—ma€2. (10) One thus has two equations in the three unknowns 0, F9, and F,, so that an additional condition must be introduced before a solution is possible. (See Fig. 3.) As one such condition, the central force assumption, F9 = 0, would again yield (7) and (8). Alternatively, one may assume Ho = M0 as an independent axiom. With F consi- dered as an internal force, and with W as the external force, one obtains man's): —mga sin6. (11) By substituting (1 1) into (9), one deduces that F 9 = 0, and F, may be determined as before. It follows that in this problem, the central force assump- tion and an independent law of moment of momentum are equivalent in the sense that one may be deduced from the other once Newton’s second law is assumed. This equiv- alence does not hold in general, however, as the next exam- ple demonstrates. IV. RIGID DOUBLE PENDULUM In this section, a simple problem in the constrained mo- tion of particles is presented which cannot be solved with the central force assumption but for which an independent- ly postulated axiom of moment of momentum does lead to a solution. It thus appears that the latter postulate allows solutions for a class of problems for which the former pos- tulate fails to provide one. Problem. Consider a system of three particles P0,P,,P2 having masses m0,m,,m2, respectively, and connected by means of a single massless rod. The particle P0 is fixed4 and the rod is hinged at 0 in such a way that the system may move as a pendulum in the xy plane, as illustrated in Fig. 4. The only external forces acting on the system are the weights of the particles and the reaction force at 0. The rod Fig. 4. Rigid double pendulum. W. Stadler 596 is initially at rest in the horizontal (0y) position. Obtain the angular displacement 0 as a function of the time. Strictly speaking, the simple pendulum and the problem as posed above do not conform to the Newtonian formula- tion in Sec. II, because of the inclusion of such concepts as strings and rods. However, these idealizations are artifices which serve to make the pendulum models physically plau- sible; they are not essential to the problem statement. In order to cast the pendulum problems within the framework of particle mechanics, the rods and strings must be re- placed by suitable statements about the intemal forces Fij and by geometrical constraints on the motion. Consider then the rigid double pendulum as posed with- in the framework of Sec. II. It is idealized as a system of three particles P0, P,,P2 aligned along a straight line for all time and moving on concentric circles. The particles are acted upon by the external forces comprised of the weights of the particles, W0,W,,W2, acting vertically downward, and the reaction F0 at 0. In addition, the action of the con- necting members upon the particles is idealized as a set of internal forces {F,0,F0,,F20,F02,F12,F2, }. The unknown re- action, the intemal forces and the motion are to be determined. Ideally, any axioms postulated for the solution of dyna- mical problems provide necessary and sufficient conditions for that solution; that is, for the determination of the mo- tion and the forces (or possibly, their resultants). For a giv- en set of unknowns it is generally accepted that there must be at least as many conditions available for their determina- tion as there are unknowns. With reference to the rigid double pendulum, the process of the introduction of un- knowns and the provision of an equal number of conditions for their determination will now be illustrated. The solution begins with the postulation of Newton’s second law (i) for each particle as the first relationship between the unknowns. Based on the freebodies of Fig. 5, one has the equations of motion given by wo+Fo+F01+F02=mofm W1+F10+F12=mifn (12) W2 + F2: + F20 = ”12.13- The set of vector unknowns in this system of equations is iFo’FonFio,F02:F20:F12F21:1'0sl'1’r2i, so that there are 20 unknown component functions of time for whose determination one has 6 component equations in (12). The geometric constraints are introduced next. For the particle 1"o one has r0 = 0; the presumed circular motion of the particle Pl, namely, r1 = a,e,, yields Fig. 5. Rigid double pendulum as a con- strained particle motion. 597 Am. J. Phys., Vol. 50, No. 7, July 1982' i‘l = a,( — 49 2e, + deg) as the expression for the accelera- tion; finally, since the particle P2 remains aligned with P1, one has r2 = (oz/am]. The use of these 5 conditions leaves the unknowns iFO’FODFIOF’02’F201F12,F2110 } , when suitable initial conditions have been prescribed. Thus there now are 15 unknowns and 6 equations. An application of Newton’s third law (ii) leaves equa- tions of motion of the form wo+Fo—F10_on=0’ wl+F10+F12=mlf1’ (13) W2 — F12 + F20 = m2(aZ/al)i:l 2 with the set of unknowns reduced to {F0,F10,F20,F,2,0}, leaving 9 unknowns and 6 equations. Evidently, in the Newtonian analysis, the effect of the internal forces on a given particle consists of a resultant internal force due to all of the other particles. Conversely, this resultant is all that one may expect to determine when nothing concerning the functional form of the internal forces has been specified. With this in mind, Eq. (13) may be written as W0+Fo—Rl—R2=0, W1+ R1: mlil’ R1: F10 + 1:"12 , (14) W2 + R2 = mziaz/aiifi: R2 = — F12 + F20 - The new set of unknowns is {F0,R,,R2,0 }. With the weights written as W, =m,.g(cos0e, —sint9e9) i=0,l,2, (15) the system (14) of 6 equations in 7 unknowns may now be expressed in the form F0 = " [(mlal + "12%)?2 + (”'0 + m1 + m2)gcos 01% + “"1101 + "1202)0 + (me + "11+ m2)gsin 0199 , . (16) R, = -— m,.(a,-62 +gcos 6)e, + m,-(a,-é +g sin 0)e,9 1': 1,2. An additional condition is needed to solve this system of equations. Usually, this condition is taken to be the as- sumed centrality of the internal forces, condition (iii) above. When used here, this condition comprises two sepa- rate conditions requiring that the «2,9 components of RI and R2 vanish. It then follows from (16) that a = — (g/a,)sin 6 and a = — (g/az)sin a. (17) Since (12 is not equal to a, in general, the es components of R1 and R2 cannot vanish simultaneously and one is led to the conclusion that the internal force system cannot be cen- tral. It follows that this problem has no solution within the traditional framework of Newtonian dynamics. _ Suppose now that instead of assuming centrality of the internal forces, one postulates an independent law of mo- ment of momentum. Thus applying (6), one obtains é: _ mlal +m2a2 mlai + m2“; and hence, by integration, 9 2 = mm mm? + mzafi where the inital conditions 9 (0) = 0 and 6 (0) = 1r/2 have gsin6 (18) gcos 0, (19) W. Stadler 597 been used. The corresponding forces are then given by cos 6 F0: —- g2 2 [(mo‘l‘ml‘l'mzlimlal +m20§l "2101 + mzaz + 2(m101 + mZazlzler + gZSIn 0 2 mlal + mzaz X [(7710 + m1 + mzllmfli + mzagl —‘ (mlal + mzazlzlee (20) and m,g cos 0 2 2 R: = — 2 [mlal +mza2 2 ma, +m2a2 m, sin 6 ‘l‘ 2ailm1al + mzazl]er + 2g 2 mlal + "1202 >< [mla] +m2a§ —a,-(m,al +m2a2)]ea 1’: 1,2. (21) This internal force system clearly is not central. V. CONCLUSIONS The preceding example shows that there are simple problems in constrained particle motion for which the tra- ditional Newtonian axioms are insufficient to obtain a solu- tion. It has also been shown that by invoking the law of moment of momentum (as a separate postulate) the prob- lem of the rigid double pendulum became soluble. It is of interest to note that for the present problem, the postulation of an independent law of moment of momen— tum is equivalent to the following postulates in the sense that Eq. (18) is the consequence of their application: (a) The law of conservation of kinetic and potential ener- gy. A choice of zero potential energy at a vertical distance 02 from 0 yields V(t) = m,g(az ——a1 cos 0) + ngaz(l — cos 0) +m0gaz , T(t)= ma? + magma for the potential and the kinetic energy, respectively. The conservation of the total energy is expressed by V(t ) + T(t ) = const. A differentiation with respect to t re~ sults in Eq. (18). (b) The system of internal forces is in static equilibrium. This, of course, may be expressed in terms of either the virtual work of the forces or in terms of a summation of forces and moments. The total virtual work of the system of internal forces vanishes. With admissible virtual displacements given by 6n, = 0, 5r, = a,60e,,, and 6r2 = a260e9, this statement takes on the form Rl-6l'l + R2'6r2 = [(mlafi + mza§)é +(m,a1+ m2a2)g sin 9 ]60 = 0. 598 Am. J. Phys., Vol. 50, No. 7, July 1982 Since this must be true for any 66, the expression in brack- ets must vanish. The summation of internal forces and the summation of their moments with respect to any fixed point are zero. With R0 = F0l + F02 as the internal force acting on the particle P0 one has R0+R, +R2=0 and 1x710=r,xR,+r2xR,=o. The latter equation again results in Eq. (18). In closing, it may be remarked that within the frame- work of the dynamics of deformable media one does, in fact, postulate Euler’s second law of motion, the law of moment of momentum, as an independent axiom. It is also interesting to note that from a historical viewpoint, the in- ception of the law of moment of momentum preceded the Newtonian postulates. The historical development is traced by Truesdell.2 ACKNOWLEDGMENTS The author is grateful to J. Casey for many discussions and suggestions related to the subject matter of this paper. In addition, he wishes to thank G. Leitmann and R. Rosen- berg for their comments and encouragement. 11. Newton, Philosophiae Naturalis Principia Mathematica. The printing thereof was completed by S. Pepys on 5 July 1686; the book was pub- lished in London in 1687. The most commonly cited English translation is that given in 1729 by Andrew Motte as revised by Florian Cajori: Sir Isaac Newton’s Mathematical Principles of Natural Philosophy and his The System of the World (University of California, Berkeley, CA, 1960). 2Reference 1, Vol. 1, p. 13. Newton’s laws have been stated here in the usual modern form. They cannot be found precisely in this form any- wherein the Princzpia, although all the basic ideas are there. If Newton’s second law is postulated about a system of particles, that is, the total external force is set equal to the total rate of change of the linear momen- tum of the system, and if it is assumed that the law also applies to every subsystem, then Newton’s third law in the form (ii) follows as a theorem. If Newton’s second law is taken as a statement about the motion of an individual particle, as is done here, then the third law does not follow as a theorem but must be independently postulated. For further critiques and historical comments concerning Newton’s laws, the works of I. B. Cohen [Texas Quart. X (3), 127—157 (1967)], C. A. Truesdell [Essays in the His- tory of Mechanics (Springer-Verlag, New York, 1968)] , and I. Szabé [Geschichte der Mechanischen Prinzipien (Birkhiiuser, Stuttgart, 1977)] are recommended. 3Sometimes (ii) and (iii) together are put forward as a strong form of New- ton‘s third law; but it seems to be more in keeping with Newton’s own statement and usage to regard (ii) alone as representing the third law. [See Ref. 1, The System of the World, p. 570.] 4The particle PO has been included only to conform to the Newtonian idea that forces act on masses. W. Stadler 598 ...
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