Bulletin of Aichi Univ. of Eduction, 66Nturl Sciences, pp. - 6, Mrch, 07 The Form of Hnging Slinky Kenzi ODANI Deprtment of Mthemtics Eduction, Aichi University of Eduction, Kriy 448-854, Jpn Introduction Slinky is spring toy. If it hngs down, it extends under its own weight. In this pper, we consider the form of Slinky which extends under its own weight. The following is our result. Theorem. Consider spring which extends under the following conditions: i The spring is uniform long the xis, hnging down s the xis to be verticl. ii The grvity cts ech point P of the spring verticlly down, nd its mgnitude is proportionl to the mss locting blow P. iii The spring extends under the locl Hooke s lw, given in Subsection.. iv When no forces ct on the spring, its form is helix of winding number t lest one. v Though the spring extends, it preserves its curvture. Under the bove conditions, the spring lies on one-sheeted hyperboloid of revolution. Though the bove theorem sys so, hnging Slinky seems not to be hyperboloid but cylinder. This is becuse its eccentricity is very lrge. See Subsection 3.. The most importnt ide of this pper is the ssumption v. We will see the vlidity of it in Subsection.3. The uthor got the ide from the textbook of Frnk Morgn []. Figure : Slinky Figure : One-Sheeted Hyperboloid of Revolution Preprtion. Extention of Spring under Its Own Weight Throughout this pper, we denote by g be the ccelertion of grvity, nd by m, k nd L respectively the mss, the spring constnt nd the nturl length of the spring. We lso denote by A nd B the endpoints of the spring. In this subsection, we prove the following well-known lemm. The uthor does not know who got this result first.
Kenzi ODANI Lemm. Consider spring under the conditions i, ii nd iii. Tke ny point P on the spring AB. We denote by h the nturl length between P nd A. After hnging down from A, the length comes to h mg kl Lh. h Especilly, the length of the spring comes to L mg k. Before proving the lemm, we explin locl Hooke s Lw. Consider spring with no other forces cting. We ssume tht the spring pulled with force F extends its length by L. By Hooke s Lw, we hve tht F = k L. The spring constnt k is inversely proportionl to the length L. For exmple, if we rrnge two sme springs in series, then the coupled spring pulled with the force F extends its length by L. Tht is, the spring constnt of the coupled spring is considered s F/ L =k/. Put the spring horizontlly, fix n endpoint A, nd tke two points P, Q on the spring. We denote by h, h δh respectively the distnces of P, Q from A. We ssume tht fter pulling nother endpoint B by the force F in the xil direction, the distnces h, h δh chnge to z, z δz respectively. See Figure 3. Since the spring constnt is inversely proportionl to the length, the spring constnt of the sub-spring PQ comes to kl/δh. Therefore, by Hooke s Lw, we hve tht F = k L δh δz δh. By letting δh 0, we obtin tht We cll it the locl Hooke s Lw. dz F = kl dh.. h δh z δz Figure 3: Derivtion of Locl Hooke s Lw Proof of Lemm. Since the spring is uniform, the mss of sub-spring PB comes to ml h/l. Hooke s Lw., we hve tht By integrting both sides, we hve tht dz kl dh = m L h L g, tht is, dz dh By the locl mg = L h.. kl z = h mg kl Lh h..3 Especilly, by putting h = L, we obtin tht the totl length of the spring L mg k.. Helicl Springs Consider helix represented by position vector x s = cos s b, sin s b, where nd b re positive constnts. By Eq..4, we hve tht bs, 0 s l,.4 b x s =, x s = b..5
The Form of Hnging Slinky The former identity indictes tht s is n rc-length prmeter, nd the ltter indictes tht λ = / b is the curvture of the helix.4. We ssume tht the points A, B, P respectively correspond to the prmetric vlues 0, l, s. Then we hve tht h = bs b, L = bl b..6 Notice tht l is the rc-length though L is the length. They re different notions. By putting Eq..6 into Eq.., we hve tht z s =µ λεl s, where µ = b b, ε = mg kl λ..7 By integrting both sides of Eq..7, we hve tht zs =µs λε ls s..8.3 Preservtion of Curvtures Frnk Morgn [] introduces n interesting story on curvtures of spce curves. There ws compny constructing huge smokestck, which requires the ttchment of spirling strip for structurl support. An engineer imed to cut n nnulus out of flt metl, nd to ttch it to the smokestck long helix. See Figures 4 nd 5. A problem ws wht choice of inner rdius r of the nnulus would mke it fit on the smokestck best? The size of the helix were: =3.75 feet, πb = 3.5 feet. After some tril nd error, the engineer found tht nnulus cut with r 0.5 feet fit well. However, we cn compute the inner rdius r mthemticlly. The wy to compute r is to mke the circle nd the helix to hve the sme curvture. Tht is, It is close to the engineer s experiment. r = λ = b 0.45 feet..9 r Figure 4: Annulus Cut out of Flt Metl Figure 5: A Smokestck Attched with n Annulus 3 Form of Spring Extending under Its Own Weight 3. Finding the Eqution of the Curve Consider spring extending under its own weight. We represent the curve by position vector x s = xs, ys, zs, 3. where zs is function obtined in Eq..8. Since s is n rc-length prmeter, we hve tht x s = x y φ =, 3. 3
Kenzi ODANI where φs =µ λεl s. Since the curvture λ is invrint, we hve tht x s = x y λε = λ. 3.3 So the spring stisfies the differentil equtions 3. nd 3.3. By Eq. 3., the spring must stisfy φs. The restriction comes from the fct tht the length must be smller thn the rc-length. By Eq. 3., we cn put x nd y s follows: By differentiting both sides of Eq. 3.4, we hve tht x = x = φ cos αs, y = φ sin αs. 3.4 λεφ cos α φ α sin α, y = λεφ sin α φ α cos α. 3.5 φ φ By putting them into Eq. 3.3, we hve tht λ ε φ φ φ α λ ε = λ, nd so α = λ φ φ. 3.6 By Eq. 3.6, the spring must stisfy φ 0. The restriction comes from the ssumption tht the spring preserves its curvture. We ssume tht the spring winds counter-clockwise, tht is, α 0. By tking squre root of Eq. 3.6, nd by integrting both sides, we hve tht φ αs =λ φ ds. 3.7 To find this integrl, we introduce new vrible u s follows: φs = cos u, nd so Then Eq. 3.7 comes to αs = sin u ε cos u du = { ε = u ε ε ε tn u ds du = sin u. 3.8 λε ε } cos du u cos du. 3.9 u To find this integrl, we introduce new vrible v by tn u = ε tn v. Then we hve tht cos v = tn v = sin v = cos v = ε cos u ε cos u sin u = ε cos u φ, ε cos u sin u ε cos u φ = sin u φ. 3.0 Then Eq. 3.9 comes to αs = u ε ε ε tn v ε cos v dv = u v c, 3. ε where c is n integrting constnt. By putting Eq. 3. into Eq. 3.4, nd by using Eq. 3.0, we hve tht x s = φ cos ε c v = { } φ cos ε c cos v sin ε c sin v = ε cos ε c cos u sin ε c sin u. 3. 4
The Form of Hnging Slinky Similrly, we hve tht By Eq. 3., we hve tht y s =ε sin ε c cos u cos ε c sin u. 3.3 { x s ds = ε cos = λε = 4λε Similrly, we hve tht y s ds = 4λε } ε c cos u sin ε c sin u sin u du λε } du { ε cos ε c sin u sin ε c cos u { } sin ε u c ε sin ε u c sin ε c du. 3.4 { } cos ε u c ε cos ε u c cos ε c du. 3.5 By integrting both sides of Eqs. 3.4, 3.5, we hve tht { xs = 4λ ε cos ε u c ys = 4λ ε } cos ε u c cos ε c x 0, { ε sin ε u c ε sin ε u c sin ε c } y 0, 3.6 where x 0 nd y 0 re integrting constnts. Eq. 3.6 represents the form of the spring which is extending under its own weight. 3. The Restriction to the Vlue of ε The denomintor of Eq. 3.6 vnishes when ε =/. However, it is not problem. In this subsection, we prove tht the spring must stisfy ε</. Since we hve tht φ 0, we hve tht By solving the qudrtic inequlity 3.7 for ε, we hve tht φ0 = ε µ λεl 0. 3.7 The curvture nd the winding number of the helix re respectively given by ε µ µλl λl µ. 3.8 By combining them, we hve tht λ = b, n = l π b. 3.9 λl =πn b =πn µ. 3.0 By substituting it into Eq. 3.8, we hve tht µ ε πnµ πn < πn. 3. Therefore, spring of winding number n must stisfy ε</π 0.6. It is mthemticl restriction. Physicl conditions my require the vlue of ε much smller. 5
Kenzi ODANI 3.3 Extending Spring Lies on Hyperboloid By Eq. 3.6, we hve tht x x 0 y y 0 = { ε ε ε } 4 6λ cos 4u cos u ε 4ε ε cos u = { ε λ 4ε cos4 u ε 4ε cos u } 4ε } = { λ 4ε ε cos u 7ε 4 λ 4ε. 3. We cn rrnge Eq..8 into By combining Eqs. 3.8 nd 3.3, we hve tht By substituting it into Eq. 3., we hve tht where zs = { φs µ λεl }. 3.3 λε cos u = φs = λεzsµ λεl. 3.4 x x 0 y y 0 = 4ε 4ε z z 0 7ε 4 λ 4ε, 3.5 z 0 = µ λεl / λε. 3.6 By the result of Subsection 3., the constnt ε < /. So Eq. 3.5 represents one-sheeted hyperboloid of revolution. Hence spring extending under its own weight lies on one-sheeted hyperboloid of revolution. The eccentricity of the hyperbol is given by 4ε 4ε = ε. 3.7 Since the constnt ε is very smll, the eccentricity 3.7 is very lrge. So the hyperboloid seems like cylinder. Moreover, by pplying Eq. 3.7 to Eq. 3.6, we hve tht z 0 = µ λεl / λε 3ε < 0. 3.8 4λ Therefore the center of symmetry of the hyperboloid is locted bove the spring. Tht is, the upper prt of the extending spring is thin though the lower prt is thick. References [] F. Morgn, Riemnnin geometry: beginner s guide, nd Ed., Routledge, 009. [] J. D. Sern nd A. Joshi, The center of mss of soft spring, College Mth. J. 4 0, 389 394. [3] https://en.wikipedi.org/wiki/slinky#physicl_properties Received September 0, 06 6