The rolling of a disk on a horizontal plane: I. Stability of equilibria. G. Keady∗ 1 January 2009
Abstract The rolling of a disk (or hoop, etc.) on a horizontal plane – without dissipation – is a classical problem. Many second-year undergraduate students of the author’s generation, studied the stability of the motion in which the disk rolls steadily. Current mathematical software makes work on the problem accessible to present-day second-year engineering mathematics students. This paper (I), and its sequel (II), are partly a composite of assignment and exam questions used between 2006 and 2008. There are research topics surrounding the problem. A catalyst for some recent work on the problem is the toy marketed under the name ‘Euler’s disk’: see (II).
Contents 1 Introduction
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2 The rolling of a disk on a horizontal table 2.1 Motion of a rolling disk: 4th order system . . . . . . . . . . . 2.1.1 Conservation of energy . . . . . . . . . . . . . . . . . . 2.1.2 The position of the centroid . . . . . . . . . . . . . . .
4 4 5 5
3 The equilibria and their stability 3.1 The equilibria in general . . . . . . . . . . . . . . . . . . . . . 3.2 The Jacobian, in general . . . . . . . . . . . . . . . . . . . . .
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∗
School of Maths & Stats, University of Western Australia, Nedlands, Australia. mailto:
[email protected]
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4 The modelling of the ‘table’ is inadequate for some solutions 10 5 Numerical solution
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6 Conclusion
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1
Introduction {sec:intro}
The problem is treated in the text book [10], the author used as an undergraduate 40 years ago. For contemporary students the problem becomes accessible and interesting with some easy use of Computer Algebra (CA). The extra item of interest is the value of CA, for the lecturer, in further – not for teaching – investigation of the problem. In particular, CA helped me with work consequent on my observation that the system of differential equations (d.e.s) – denoted (D) in the following – is integrable. This proved to be a rediscovery: see [12, 3, 8, 2]. Some of this material has been used in two different second-year classes involving engineers. In both of these the following sequence of topics were amongst those treated: • determinants and eigenvalues; • linear systems of constant coefficient d.e.s, exponential matrices; • numerical solution of d.e. initial-value problems (using Matlab); • classification of stability of equilibria for nonlinear systems. The two different classes were as follows. • In 2006 the some of material in this paper was presented to a class of 2nd year engineers, BE students. This class uses Matlab, including its Symbolic Toolbox: a textbook was [9], a recommended reading book is [4]. The matlab m-files associated with the story are made available to the class. The rolling disk problem served to provide several problems for the assignment sheets, sometimes in the ‘Additional Problems’ section where some harder, not-directly-assessed problems are appropriate. I devoted one lecture to the problem, very much emphasising the items listed above. The Euler disk was demonstrated, largely to show that there are equilibria other than rolling in a straight line, and some of them are stable. Advanced topics, e.g. the exact integrability of the system, the special function aspects were not treated. Part I was treated in the 2006 class, but nothing advanced from Part II was. 2
• In 2008-09 the rolling disk problem was treated as a major example in a second-year Applied Maths unit. The vast majority of the students in this unit were BSc/BE double-degree students. This unit was shared with Physics, and is a compulsory unit for any Applied Maths or Physics major. There is a strong emphasis on developing a proficiency in using Mathematica for real problems. Although Matlab is the package used by BE students, UWA’s Mathematica licence is funded by Physics, Mathematics and Mechanical Engineering. In this unit there were weekly computer labs of duration of 2 or 3 hours, and a very substantial amount of the assessment associated with them. There were 6 lectures at the end of the unit, and some carefully prepared printed notes, on the rolling disk problem. The Mathematica code associated with the problem was made available to the class. Material from both Part I and Part II was used. Mathematica has a propensity to use special functions, and, although we do not train the students in properties of these and do not in 2nd year expect the students to do hand calculations involving them (this being treated in a 3rd year ‘Mathematical Methods’ unit taken by Physics and Applied Maths majors), we attempt to ease the students into being relaxed about the formulae involving the named special functions. The last 3 labs of the semester treated the problem of oscillations of a bead on a vertical hoop, rotating about its vertical diameter, and a 2008 exam question on the rolling disk problem used techniques similar to that - applied to the bead on a hoop - in the last of the labs. The exam was held in the lab, and the students’ answers presented electronically in a Mathematica notebook. This 2008 exam question and related material is presented in Part II. For readers of this journal, the easiest route through to the materials and research-related code is via the link at http://www.maths.uwa.edu.au/~keady/papers.html Up until 2006, with my engineering mathematics classes I had treated only simple, perhaps excessively-simple, and, therefore, perhaps unimpressive, examples, both in connection with numerical solution of o.d.e.s and in connection with the classification of equilibria. In particular, the lead-in example is the nonlinear (rigid-bar) pendulum. This is used as an example of numerical solution and also as an example of classifying equilibria as stable or unstable. The second-year engineering students not only treat examples but also
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consider the general setting. In this the d.e. system is written dy = f (y) , dt and we define equilibrium points as vectors ye where f (ye ) = 0. We assess stability by linearizing in the usual way, setting up a problem for the remainder ρ where y = ye + ρ, which is dρ = Jρ where J = Df (ye ) , dt and J is the Jacobian of f evaluated at ye . One classifies the stability of an equilibrium, of course, by determining the signs of the real parts of the eigenvalues of J. If all the real parts are negative, the equilibrium is stable: if all the real parts are nonpositive, with at least one of the real parts zero, we will describe the equilibrium as neutrally stable. If any real part is positive, the equilibrium is unstable. I do not expect modern-day students to be able to do the hand calculations that those of earlier generations would do. However, the CA Systems (CAS, Matlab with its Symbolic Toolbox or Mathematica or Maple, etc.) • give them a tool whereby they can bypass the deficiencies in their stamina and/or skills in hand-calculation, and • enable related numerics and graphics to be completed easily and accurately.
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The rolling of a disk on a horizontal table {sec:rolling}
2.1
Motion of a rolling disk: 4th order system
The first examples, in teaching, concerning the stability of equilibria should be of low order, no higher than 2nd. The nonlinear pendulum is a 2ndorder system: the logistic equation is a 1st-order d.e. It seems educationally worthwhile to give one example which is larger than this, partly to show that Matlab or Mathematica, or whatever package the engineering students are using, can help with engineering sums, or, more accurately for this example, ‘simple’ mechanics sums. We won’t set up the d.e.s here, but take them from mechanics texts: see p402 of [10] for example. We denote the radius of the disk by a, and the acceleration due to gravity by g. Following [10], let θ be the inclination of the plane of the disk to the 4
{subsec:Motion}
vertical. Also following [10], we introduce a rectangular coordinate system moving with the disk and with its origin at the centre of the disk. The first coordinate axes, and associated direction i, is in the direction of the line joining the the centre of the disk to the point of contact. k is normal to the plane of the disk. The vector j so that ijk is an orthogonal triad is therefore in the plane of the disk and horizontal. The angular velocity of the disk is ω = ω1 i + ω2 j + ω3 k. From the equations in [10], we obtain 3 scalar equations each linear in each of (the angular velocities) ω˙ 1 , ω˙ 2 and ω˙ 3 . Solving for ω˙ 1 , ω˙ 2 and ω˙ 3 , we have the following set of four equations for the four unknowns θ, ω1 , ω2 , ω3 . In these, γ = 2/3 for a solid disk and γ = 1/2 for a hoop. θ˙ = −ω2
(0)
ω˙ 1 = −ω2 (tan(θ)ω1 + 2ω3 ) (1 − γ) tan(θ)ω12 + 2ω1 ω3 − 2γ(g/a) sin(θ) ω˙ 2 = (1 + γ) ω˙ 3 = −γω2 ω1
(1)
and
(2) (3)
We denote by (D) the system of four d.e.s (0) (1), (2), (3). 2.1.1
Conservation of energy
The system is autonomous. Invariance of equations under translations in time is associated with the conservation of energy. Hence dE/dt = 0 where E is defined by g (1 − γ)ω12 + (1 + γ)ω22 + 2ω32 = 4γE − 4γ cos(θ) a 2.1.2
The position of the centroid
The preceding 4th-order system for θ, ω1 , ω2 , ω3 , can be solved numerically. So also can the system augmented by equations (??) determining the x, y, z components. This is easy and has been done very many times,e.g. http://demonstrations.wolfram.com/RollingDisk/ The motion of the centroid (xO , yO , zO ) is obtained by integrating the larger system of d.e.s, equations (0)-(6): ω1 φ˙ = − (4) cos(θ) x˙O = aω3 sin(φ) + aω2 cos(φ) cos(θ), (5) y˙O = −aω3 cos(φ) + aω2 sin(φ) cos(θ). (6) 5
(zO = a cos(θ) does not require any further integration.)
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The equilibria and their stability
{sec:equilibriaSt
3.1
The equilibria in general
It is straightforward to solve the equations obtained by putting θ˙ = 0, (so ω2 = 0, ω˙ 1 = 0, ω˙ 3 = 0). We have found it convenient to split these into cases: • rolling in a straight line • spinning about a vertical diameter • the general case, rolling in a circle, and the sub-case where the centroid remains fixed. We begin, though, with the most trivial of the equilibria 0 = ω1 = ω2 = ω3 with θ = 0: this equilibrium is unstable. We note in passing that 0 = ω1 = ω2 = ω3 with θ = π is also an equilibrium of system (D), and it is stable. Clearly, if we are really intending to model the rolling above a table we need to add other constraints, for example that the reaction at the point of contact should point upward. The issue arises also in connection with rolling inside a sphere.
3.2
The Jacobian, in general
Setting ω2 = 0 in the expression for the Jacobian J, required for the analysis of the stability of these equilibria, leads to matrices (which obviously have their trace 0 and determinant 0) of the form 0 0 −1 0 0 0 j23 0 Jgen = j31 j32 0 j34 . 0 0 j43 0 2 Provided Jgen 6= 0, Jgen is rank 2. To solve the d.e.s ρ˙ = Jgen ρ one begins with finding the characteristic polynomial pJgen (λ) = λ2 (λ2 − j34 j43 − j32 j23 + j31 ).
With a similar question having been set on an assignment, we set, as the first question on the 2006 exam paper: 6
{subsec:equilibGe
2006 Exam, Q1. A matrix J (which arose in the ‘rolling disk’ application in lectures) is specified and studied in the following matlab transcript, the last line being the output of its characteristic polynomial. syms j31 j23 j32 j34 j43 x J= sym([0,0,-1,0;0,0,j23,0;j31,j32,0,j34;0,0,j43,0]) charPolJ=factor(poly(J,x)) charPolJ = x^2*(x^2-j34*j43-j32*j23+j31) (a) Given that at least one of j32 or j34 is nonzero, (i) what is the rank of J? (ii) What is det(J)? (b) What is the condition on j31 which ensures that there is a positive eigenvalue of J? (c) If J is such that it has a nonzero eigenvalue, is it diagonalizable?
Not only can the eigenvalues be found explicitly, but there is a simple formula for the exponential matrix. Deriving this is assisted by the observa3 = λ2 Jgen . The following is a (largely unseen) exam question tion that Jgen based on this: 2006 Exam, Q2. (a) Consider any matrix J for which J 3 = λ2 J for some nonzero λ. Verify that the exponential matrix of Jt is given by expm(tJ) = (I −
J2 cosh(λt) sinh(λt) ) + J2 +J . 2 2 λ λ λ
(b) Let u be an eigenvector of matrix A with Au = λu. (i) Show that u is also an eigenvector of expm(At) and find its corresponding eigenvalue. (ii) Given that, for any matrix, • the sum of the eigenvalues equals the trace, • the product of the eigenvalues equals the determinant, show that det(expm(At)) = exp(trace(A)t) (iii) For the matrix J of Q1, what is det(expm(Jt))? (c) Let J0 be the matrix of Q1 with the further restriction that, while it remains rank 2, the only eigenvalue is 0. Is matrix J0 diagonalizable? Given that J03 = 0, use the exponential series to write down a formula for expm(J0 t).
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In any event, the solution of the linearized equations associated with perturbing an equilibrium solution is easy, after the nonzero eigenvalues are obtained. Although caution is needed, some aspects of the solution can be represented by the trajectories in the (θ, ω2 )-plane. We will see, in this plane, either ellipses or hyperbolae according to whether λ2 < 0 or λ2 > 0. We also have det(expm(Jt) = exp(trace(Jt)) = 1 . Stability of the equilibria with θ = 0. The easiest solution has θ = 0 and ω1 = 0 = ω2 and ω3 a constant, and this corresponds to the hoop/disk rolling in a straight line. Perturbing about this, the remainders ρ satisfy ρ˙ = Jρ with 0 0 −1 0 0 0 −2ω3 0 JstLine = − g 2γ 2ω3 0 0 a 1+γ 1+γ 0 0 0 0 The eigenvalues are 0 (with algebraic and geometric multiplicity 2) and ±λstLine where 2( gγ − 2ω32 ) a λ2stLine = . 1+γ We set this as an assignment question, following from an earlier calculation of the eigenvalues of the Jacobian: Assignment Question: Consider the d.e.s describing the rolling of a disk on a horizontal plane. (a) Verify that θ = ω2 = ω1 = 0 and ω3 =constant is an equilibrium solution of these d.e.s. (b) Use the Matlab code given at the links for this assignment to calculate the Jacobian at this equilibrium. Verify that the form of the Jacobian is that given in the previous assignment. (c) Show that the vertically-upright straight-line equilibrium motion of part (a) is neutrally stable only if the angular-velocity ω3 satisfies ω32 > gγ/(2a).
(For readers of this paper, the matlab code and related materials are available from the link at the ‘papers’ page , as specified in the Introduction.) The other, ‘spinning’, equilibrium, in which θ = 0 and ω3 = 0 = ω2 can be analyzed similarly. Stability of the equilibria with 0 < |θ| < π/2.
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In these equilibria, the point of contact of the disk and the plane moves in a circle. 0 0 −1 γ 0 0 − ω1 (2g sin(θ) + ω12 tan(θ)) g g 2 2 Jcircle = 2 a γ cos(θ)−(1−γ)ω1 sec(θ) 2 a γ sin(θ)+(1−γ)ω1 tan(θ) − 0 1+γ (1+γ)ω1 0 0 −γω1 ONLY INCLUDE THE PLOT WITH THE STABILITY BOUNDARY.
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3
W3 0 2 -10 0.0
W1 0.5
1 Th 1.0 1.5
0
fig:thetaW1W3equilib
Figure 1: Equilibrium solutions with 0 < |θ| < π/2. The vertical axis is that of ω3 : the horizontal axes are those of (θ, ω1 ). The solid curve divides the surface into stable and unstable parts: the unstable solutions are those with small θ. The calculations we have lead to results which agree with those in [6]. The equilibria are stable if p −1 + γ 2 + 3(1 + γ) cos(2θ) < 1 + γ + γ2 and, in particular, all equilibria with π/4 < θ < π/2 are stable for all gamma between that of hoop and disk. Specialising this to the γ = 2/3 case of a disk any ‘circle-motion equilibrium’ for which s √ 24 − 9 5 π 0 < cos−1 ( ) 0}. THEOREM.(a) Any solution of the differential equation (9) which starts in the positive half-space {(a, w)|a > 0} of the phase-space remains in that half-space. (b) The trajectory, in the (a, w)-phase-space, of any solution of the differential equation (9) which has a(0) > 0 is a closed convex curve. Proof. The results are immediate from the remarks about E and the fact that E remains constant along a trajectory of solutions of the d.e.. 2008 Exam, Q4. Consider the differential equation a0 = w 1 −1 w0 = a3 (a) Find the equilibrium solutions of the d.e. above, and classify if they are unstable, neutrally stable, or something else. (b) Show that, for any solution of the system, the quantity 1 1 2 E(a, w) = w + 2 +a 2 a is constant along any trajectory. (c) Show a ContourPlot of E(a, w) over an appropriate region of the half-plane {(a, w)|a > 0} showing several closed contours of E. (d) Denote the roots of the cubic for a given by E(a, 0) by as , an and ax . When there are 3 real roots it can be shown (but you need not in this exam) that −1/2 < −as < 0 < an < 1 < ax . The period of the closed orbit for an < as < ax is given by period [as ]where
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period[a_]:= Module[{s1=Sqrt[1-8*a^3],s2}, s2=4*a^3 + s1+1; s3=2*s1/s2; -2*(4*a^3*EllipticK[s3]-s2*EllipticE[s3])/(a*Sqrt[2*s2])]; The parameter as has the physically relevant range 0 < as < 1/2. Furthermore, you are given ax , as and an are related by p p 1 + 1 − 8a3s 1 − 1 − 8a3s ax = , an = 4a3s 4a3s Find as in terms of ax . Find the first two terms in a Series approximation to the period expanding about ax = 1.
The solutions to parts (a) and (b) are easy. The result of the ContourPlot for part (c) is shown in Figure3
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0
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8
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fig:exm08q4cp
Figure 3: Solution to 2008 exam question, part (c). For part (d), use a Solve to find as in terms of ax and then a Series command. These give that the period T satisfies 2π 5π T = √ + √ (ax − 1)2 + smaller . 3 6 3
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With ν = 2π/T this is ν∼
√
5 2 3 1 − (ax − 1) + . . . . 12
fa = ((alphaEq*nu0Sqr)/3)*(alphaEq^3/alpha^3 - 1); ans= Collect[(Normal[Series[fa, {alpha, alphaEq, 3}]] /. \ {alpha -> alphaEq - u}), u, Factor]; coeffList = CoefficientList[ans, u] (* nu0Sqr*{0, 1, 2/alphaEq, 10/(3*alphaEq^2)} *) nu2alphaSmall = Simplify[(9*(coeffList[[4]])*coeffList[[2]] - \ 10*(coeffList[[3]])^2)/(24*coeffList[[2]]^2)] (* -5/(12*alphaEq^2) *)
II.4 Perturbations of the ω3 = 0, θ 6= 0 equilibria The system of d.e.s is integrable and dω1 = tan(θ)ω1 + 2ω3 , dθ
dω3 = γω1 . dθ
These are recorded in the first two lines of the code below. Now treat ω1 (θ) and ω3 (θ) as known. These can be substituted into the right-hand-side of equation (D.2), thereby determining a somewhat elaborate function f , recorded at the third line in the code below. We treat perturbations about the equilibria described at equation (6.0), so we have f (θeq ) = 0, ω3 (θeq ) = 0, ω1 (θeq )2 =
g 2γ cos(θeq ) . a1−γ
These are recorded in the Eq(uilibrium) Rules in the code. W1d= Tan[Th]*W1 + 2*W3; (* Mathematica *) W3d= gamma*W1 f= ((1-gamma)*Tan[Th]*W1^2 +2*W1*W3 -2*gamma*g*Sin[Th])/(1+gamma); derivTh[u_]:= Collect[D[u,Th]+D[u,W1]*W1d+D[u,W3]*W3d,{W1,W3},Factor]; fd= derivTh[f]; nu0SqrAlphaEqSmall= 6*g*gamma/(alphaEq*(1+gamma)); (* Approx *) fdd= derivTh[fd]; fddd= derivTh[fdd]; (* Equilibria with Th(eta)=ThEq, W3=0 and W1^2 as below *) EqRules = {Th-> ThEq, W3 ->0, W1^2 -> 2*g*gamma*Cos[ThEq]/(1-gamma)}; fThEq= Simplify[f /. EqRules ]; (* zero, just a check *) fdThEq= Simplify[fd /. EqRules];
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nu0Sqr= fdThEq; fddThEq= Simplify[fdd /. EqRules]; fdddThEq= Simplify[fddd /. EqRules]; nu2 = Simplify[(9*(fdddThEq/6)*fdThEq -10*(fddThEq/2)^2)/(24*fdThEq^2)]; nu2s = Simplify[nu2 /. {Cos[2*ThEq] -> 2*ct^2 - 1, \ Cos[4*ThEq] -> 8*ct^4 - 8*ct^2 + 1, Sec[ThEq] -> 1/ct, \ Tan[ThEq]^2 -> 1/ct^2 - 1}] (* nu = nu0 * (1 +nu2 *(alphaEq-alphaMin)^2) with nu0=Sqrt[nu0Sqr] *) (* Next treat alphaEq small *) checknu0Sqr = Simplify[Normal[ \ Series[fdThEq /. {ThEq -> Pi/2 - alphaEq}, {alphaEq, 0, 0}]] - \ nu0SqrAlphaEqSmall] (* gives 0 as a check on this term *) (* ct is Cos[ThEq] which, for alphaEq small is such that ct is asymptotic to alphaEq *) (* When alphaEq is small, seek the approximation nu= Sqrt[nu0SqrAlphaEqSmall]*(1 +C2AlphaEqSmall*((alphaEq-alphaMin)/alphaEq)^2) with the C2 term as determined below *) C2AlphaEqSmall= -5/12; checkC2 = Simplify[(Collect[Normal[Series[nu2s, {ct, 0, -1}]], ct, Factor] /. {ct -> alphaEq})*alphaEq^2 - C2AlphaEqSmall] (* gives 0 as a check on C2 *)
What we have done here is check that the expression for ν2 is the same for two different asymptotic approaches. In the preceding section we took the asymptotics in the order first αeq ↓ 0 ,
then
In this section we have taken αeq − αmin first ↓0, αeq
αeq − αmin ↓0. αeq
then αeq ↓ 0 .
It is eminently plausible that the double asymptotics like this yield the same ν2 regardless of the order of the limiting processes. However, both asymptotic approximations are only formal. Given that the trigonometric terms in d.e.s (D) are approximated just by algebraic ones in (A), we expect that at some higher term in the asymptotic expansions for ν, the series will differ.
II.5 Legendre functions and the rolling disk NEEDS DRASTIC REVISION 21
On dividing each of the equations, for ω˙ 1 , and for ω˙ 3 , by θ˙ = ω2 , a homogeneous second-order linear system is obtained for functions ω1 (θ) and ω3 (θ). Matlab’s Symbolic Toolbox (or Maple - which is the package in the Symbolic Toolbox) solves this in terms of conical Legendre functions P 1 +iλ (sin(θ)). 2 There are many equivalent ways to present this. For example V = ω1 cos(θ)/ω3 considered as a function of s = sin(θ) satisfies the Riccati equation C γV 2 dV γV 2 = − = 2 − . ds A 1 − s2 1 − s2 Again, Matlab’s Symbolic Toolbox (or Maple) solves this in terms of conical Legendre functions. After noting this, and deciding that it was worth pursuing, we continued to study the problem, and its literature, eventually reaching the reference [8], which uses the conical Legendre functions. The history is treated in [8]. Several authors, independently, noted it – using hypergeometric functions – in the nineteenth century: see, for example, [12]. The Legendre function form is in a 1904 paper [3]. The problem is old, and the observation not all that deep, so we were not surprised that it proved to be a rediscovery. LESS ON THIS. The readily available code for calculating the conical Legendre functions numerically proved to be rather slow. Hence we chose for most of our checks on the numerics to proceed as follows. The check consists of looking at V against sin(θ), where we have this from two different approaches: 1. from the numerical solution of the 4th-order system (D); 2. from the Legendre d.e., and the Riccati form above. A more important use of the observation it to conclude that the system is integrable, and to find some implications for its global behaviour. The observation that, besides the energy E, there are two other conserved quantities – associated with the conical Legendre solutions – has many implications. [2] state: “ ... for the rolling disk Vierkandt ([12], 1892) showed something very interesting: On an appropriate symmetry-reduced space, ... all orbits of the system are periodic.” Almost all solutions of the system (D) are periodic. A similar statement occurs in the first paragraph of [8], where the ‘quasiperiodic’ motions mentioned there possibly refer to the 7-th order system including φ, xO and yO as well as the other variables. We have yet to devise a method of using the conical Legendre function exact solution to show that the αeq ↓ 0 formal asymptotics are valid approximations for all reasonable initial data, e.g. those, avoiding αmin extremely
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close to 0, with 0 < (αeq − αmin )/αeq < 1 − δ
for some δ = O(1).
Plots of the level curves of the energy for both the formal approximation and the exact solutions make it plausible that something along these lines might be proved.
II.6 Further work? Disks with finite thickness lead to very similar d.e.s which are integrable. The commercial Euler disk is sold with a convex base. The problem of a disk rolling inside a sphere has also been studied. The equilibria, and their stability, can be studied in a similar way to rolling on a flat base. However, this system of d.e.s is 6th order, and is not integrable. The variety of behaviour it exhibits may well be richer than the case of rolling on the plane, but little seems to be known of this.
II.7 Conclusion Realistic problems become accessible to students with limited mathematical manipulation skills, or limited stamina, on using mathematical software on appropriate problems. Furthermore, real problems often have a research element for the lecturer to investigate, if the lecturer so desires.
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1.1
1.2
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1.5
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-4 -1.5
0.0
fig:cpW30ThEq1p5
0.5
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Figure 4: Contours of the energy for the exact solutions perturbing about the ω3 = 0 family of equilibria in the case of a disk γ = 1/2, θeq = 1.5. HAVE TO INCREASE NUMBER OF POINTS IN cpW30ThEq1p5Big
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References [1] Batista, M. The nearly horizontally rolling of a thick disk on a rough plane. Regular and Chaotic Dynamics 13 (2008), no 4, 344-354. [2] Bloch, A.M., Marsden, J.E. and Zenkov, D.V. Nonholonomic dynamics. Notices Amer. Math. Soc. 52 (2005), no. 3, 324–333. [3] Gallop, E.A. On the rise of a spinning top. Trans. Cambridge Philos. Soc. 19 (1904) 356-373. [4] Harman, T.L., Dabney, J. and Richert, N. Advanced Engineering Mathematics Using Matlab (PWS 2nd edition: 1999). [5] G. Keady. The rolling of a disc on a horizontal plane, I. Stability of Equilibria Aust J. Engineering Education., (Submitted 2009). [6] A.S. Kuleshov. The steady rolling of a disc on a rough plane, J. Appl. Maths Mech., 65, 2001, 171–173. [7] Moffatt, H.K. Euler’s disk and its finite-time singularity. Nature 404 (2000) 833. [8] O’Reilly, O.M. The dynamics of rolling disks and sliding disks. Nonlinear Dynamics 10 (1996) 287-305. [9] Strang, G. Linear algebra and its applications. (Brooks/Cole, 3rd ed: 1988; 4th ed: 2006). [10] Synge, J.L. and Griffith, B.A. Principles of Mechanics. (McGraw-Hill 3rd ed: 1959). [11] Thomson, W.T. Introduction to Space Dynamics. (Dover: 1986 reprinting of Wiley 1963). [12] Vierkandt, A. Uber gleitende und rollende Bewungen, Monatshefte der Math. und Phys, 3, 31-54, 1892.
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Bits editted out SCRAP THE Thomson DE EXAMPLE BELOW. There is a closely associated autonomous system of d.e.s, presented in a model form, in an (unseen) exam question for the students: 2006 Exam, Q5. The following autonomous system of d.e.s arises in connection with certain sorts of non-dissipative motions of a rolling disk: α˙ = f1 = ω (1 − α)(1 + α + 2α2 ) ω˙ = f2 = α3 The function f2 is a decreasing function of α on α > 0. (a) Find the equilibrium solution or solutions. (b) Using Matlab one finds that the jacobian of f = [f1 , f2 ]T with respect to y = [α, ω] is 0 1 J = Df = 2 − 3+α 0 α4 Use this to classify any equilibrium point as stable, neutrally stable or unstable. (c) In Figure 5 is given the ‘phase-portrait’ of this system. (i) Mark the equilibrium point with a *. (ii) Sketch, with attention to the main qualitative features, but without too much concern about detailed numerics, the general shape of the trajectory of the solution of of the initial-value problem where the solution starts with α(0) = y1 (0) = 0.8 and ω(0) = y2 (0) = 0.2.
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fig:dirFld06
Figure 5: You are asked in Q5(c) to show, on this phase plot, (i) any equilibrium solution, (ii) a solution to an initial value problem.
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