Atwood s machine with a massive string

Transcription

1 Atwood s mchine with mssive string rxiv: v1 [physics.gen-ph] 1 Oct 2017 Nivldo A. Lemos Instituto de Físic - Universidde Federl Fluminense Av. Litorâne, S/N, Bo Vigem, Niterói, , Rio de Jneiro - Brzil nivldo@if.uff.br October 4, 2017 Abstrct The dynmics of Atwood s mchine with string of significnt mss re described by the Lgrngin formlism, providing n eloquent exmple of how the Lgrngin pproch is gret del simpler nd so much more expedient thn the Newtonin tretment. Either with mssless or mssive pulley, Atwood s mchine [1] is stndrd exmple for the ppliction of the lws of Newtonin mechnics. In its most idelized mnifesttion, the pulley nd the string re supposed to be mssless nd the pulley is ssumed to be mounted on frictionless xle. The mss of the pulley is esily tken into ccount by the dynmics of the pulley s rottionl motion bout its fixed xis [2]. In these two cses treted in the textbooks the string is tken to be mssless. Both the idelized nd the relistic cse, in which ccount is tken of the msses of the string nd the pulley s well s of the friction in the pulley s berings, illustrte the principles involved in the ppliction of Newton s lws [3, 4]. Atwood s mchine is multipurpose mechnicl system, which llows one to investigte from Stokes s lw [5] to vrible-mss rocket motion [6]. Here we wish to determine the effect of the mss of the string on the motion of Atwood s mchine by mens of the Lgrngin formlism. Consider the Atwood s mchine depicted in Fig. 1, in which the string hs uniform liner mss density λ. The mss m s of the string is not supposed to be negligible in comprison with m 1 nd m 2. We ssume tht the string is inextensible nd does not slide on the pulley, which requires enough sttic friction between the string nd the pulley. On the other hnd, we ssume tht the pulley is mounted on frictionless xle. The tretment of this problem by Newtonin mechnics is cumbersome, since it requires the considertion of free-body digrms for the mssesm 1 ndm 2, the pulley together with the segment of the string in contct with it, nd the hnging prts of the string t ech side of the pulley. In order to find the ccelertion of m 1, for instnce, Newton s second lw for ech seprte body must be written down nd the four tensions t the ends of the hnging pieces of the string hve to be eliminted by lgebric mnipultion of the equtions [7]. We tckle the problem by mens of Lgrngin dynmics insted. We hve the constrint x+y = l (1) where l is the length of the hnging prt of the string, tht is, the totl length of the string minus the length of its segment tht touches the pulley. Since the string is inextensible nd does not 1

2 M R x y m 2 m 1 Figure 1: Two msses ttched to mssive string tht goes round pulley. slide on the pulley, the msses m 1, m 2 nd ll points of the string move t the sme speed v = ẋ, while the ngulr speed of the pulley is ω = v/r. Thus, the kinetic energy of the system is T = m 1 2 v2 + m 2 2 v2 + m s 2 v2 + I 2 ω2 = 1 2 (m 1 +m 2 +m s +I/R 2 )ẋ 2, (2) where m s is the mss of the string nd I is the pulley s moment of inerti with respect to the rottion xis. For instnce, if the pulley is homogeneous disk then I = MR 2 /2 Tking the horizontl plne tht contins the pulley s xle s the plne of zero grvittionl potentil energy, the potentil energy of the system is the sum of the grvittionl potentil energies of m 1, m 2, nd of the hnging pieces of the string t ech side of the pulley. Thus V = m 1 gx m 2 gy λxg x 2 λygy 2, (3) where we hve used the fct tht the grvittionl potentil energy of n extended body is determined by the position of its centre of mss. An immteril constnt hs been dropped, nmely the grvittionl potentil energy of the non-hnging prt of the string. With the use of the constrint (1) the potentil energy becomes V = (m 1 m 2 )gx λg 2 x2 λg 2 (l x)2, (4) where the irrelevnt dditive constnt m 2 gl hs been discrded. Thus, the Lgrngin is L = T V = 1 2 (m 1 +m 2 +m s +I/R 2 )ẋ 2 +(m 1 m 2 )gx+ λg 2 x2 + λg 2 (l x)2. (5) Lgrnge s eqution ( ) d L L dt ẋ x = 0 (6) 2

3 yields t once (m 1 +m 2 +m s +I/R 2 )ẍ = (m 1 m 2 )g +2λgx λgl. (7) This eqution predicts tht ẍ > 0 if m 1 +λx > m 2 +λ(l x), tht is, if the totl mss on the left of the pulley is lrger thn the totl mss on the right, which is the correct physicl condition for the mss m 1 to ccelerte downwrd. Furthermore, in the limit of mssless string (λ = 0) we recover the constnt ccelertion ẍ = m 1 m 2 m 1 +m 2 +I/R 2 g (8) tht is found in the textbooks. One cn hrdly fil to pprecite tht the Lgrngin pproch to this problem is substntilly simpler nd much more expedient thn the trditionl Newtonin tretment [7]. When the string is mssive the ccelertion is not constnt. The eqution of motion (7) tkes the form ẍ b 2 x =, (9) where = [ ] 1/2 (m 1 m 2 λl)g m 1 +m 2 +m s +I/R, b = 2λg. (10) 2 m 1 +m 2 +m s +I/R 2 The generl solution to the inhomogeneous liner differentil eqution (9) is the sum of prticulr solution with the generl solution to the homogeneous eqution, nmely where A nd B re rbitrry constnts, with x(t) = +Acoshbt+Bsinhbt (11) b2 The nturl initil conditions x(0) = x 0, ẋ(0) = 0 led to b = m 1 m 2 λl. (12) 2 2λ A = x 0 +, B = 0, (13) b2 whence x(t) = b 2 + ( x 0 + b 2 )coshbt. (14) A prticulrly simple cse ism 1 = m 2 = 0. Then the string moves owing only to the unblnced weights of its hnging prts t ech side of the pulley. In this cse [ ] 1/2 b = l 2 2, b = 2λg, (15) m s +I/R 2 nd it follows tht x(t) = l ( 2 + x 0 l ) cosh bt. (16) 2 3

4 As physiclly expected, if x 0 = l/2 then x(t) l/2, tht is, the string remins t rest in unstble equilibrium. The time scle of the motion is set by the time constnt τ = 1/b. For t bigger thn few time constnts the common speed of the msses nd the string grows exponentilly. This exponentil behvior is due to the fct tht s the string flls to one side more mss is trnsferred to tht side, incresing its weight unblnce with respect to the other side. Suppose m 1 = 0.20 kg, m 2 = 0.10 kg nd consider rope with λ = 0.02 kg/m nd l = 4.0 m. For typicl lbortory pulley we hve I = kg m 2 nd R = 2.5 cm, from which it follows tht m s = λ(l+πr) = kg. From equtions (12) nd (10) we find Therefore, if x 0 = 2.0 m eqution (14) yields b 2 = 0.50 m, b = 1.0 s 1. (17) x(t) = cosht, (18) with x in metres nd t in seconds. If the string s mss were neglected, the mss m 1 would fll with the constnt ccelertion ã = 3.2 m/s 2 given by eqution (8). With the sme initil conditions, the instntneous position of m 1 would be given by x(t) = t 2. (19) Figure 2 shows comprison grph of x(t) (upper line) nd x(t) (lower line) from t = 0 to t = 1.2 s, when x reches its lrgest physiclly llowed vlue, nmely x = 4.0 m. In the present cse the effect of the string s mss is tht of slowing down the motion. If the string s mss is disregrded then x = 4.0 m is reched t t = 1.1 s insted of the correct t = 1.2 s, n 8% error tht certinly requires creful experimentl set up to be detected. If x 0 is smller the discrepncy is lrger nd more esily detected: for x 0 = 0.50 m one hs t = 2.2 s wheres t = 1.5 s. Figure 2: Position, in metres, of m 1 s function of time, in seconds, for mssless string (upper line) nd for mssive string (lower line) for the vlues of the prmeters nd initil conditions given in the text. 4

5 References [1] For historicl ccount of this iconic system, see Greenslde Jr., T B 1985 Atwood s mchine Phys. Tech [2] See, for exmple, Serwy, R A nd Jewett, Jr., J W 2006 Principles of Physics 4th ed. (Belomon, CA: Thomson Brooks/Cole) pp [3] Kofsky, I L 1951 Atwood s mchine nd the teching of Newton s second lw Am. J. Phys. 19, [4] Mrtell, E C nd Mrtell, V B 2013 The effect of friction in pulleys on the tension in cbles nd strings Phys. Tech. 51, [5] Greenwood, M S, Fzio, F, Russotto, M nd Wilkosz, A 1986 Using the Atwood mchine to study Stokes lw Am. J. Phys. 54, ; Lindgren, E R 1988 Comments on Using the Atwood mchine to study Stokes lw [Am. J. Phys. 54, 904 (1986)] Am. J. Phys. 56, 940. [6] Greenwood, M S, Bernett, R, Benvides, M, Grnger, S, Plss, R nd Wlters, S 1989 Using Smrt-pulley Atwood mchine to study rocket motion Am. J. Phys. 57, [7] Trnopolski, M 2015 On Atwood s Mchine with Nonzero Mss String Phys. Tech. 53,