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I N T E R D I S C I P L I N A R Y L I V E L Y A P P L I C A T I O N S P R O J E C T The Hopping Hoop Interdisciplinary Lively Application Project Title: The Hopping Hoop Authors: Tim Pritchett Department of Physics, United States Military Academy, West Point, New York 10996 Stan Wagon Department of Mathematics and Computer Science, Macalester College, St. Paul, Minnesota 55105 Editor: David C. Arney Mathematics Classifications: Calculus Disciplinary Classifications: Physics, Calculus Prerequisite Skills: 1. Basic Calculus 2. Differential Equations Physical Concepts Examined: 1. Motion 2. Friction Materials: None Computing Requirements: Mathematica Project Intermath 1-43. ©Copyright 2001 by COMAP, Inc. All rights reserved. Permission to make digital or hard copies of part or all of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice. Abstracting with credit is permitted, but copyrights for components of this work owned by others than COMAP must be honored. To copy otherwise, to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP. 2 Project Intermath Contents 1. 2. 3. 4. 5. 6. 7. 8. Introduction Problem Statement and Assumptions Overview of Solution Procedure Constrained Motion: Rolling Without Slipping Some Basic Physics Removing One Constraint: Rolling With Slipping The Hop Postscript: A Final Suprise 1. Introduction Tape a heavy battery or two to a hula hoop and roll the hoop along the floor. Or, attach a weight to the inside of the rim of a bicycle wheel and give the wheel a roll. If you do it right, you’ll discover, as did Littlewood [1, p. 37] (see also [3, 6, 7]), that the hoop/wheel will hop into the air at a certain point. Why does this happen? In this project, we’ll create a mathematical model of a weighted hoop that exhibits this unexpected hopping behavior. In the process, we’ll learn a fair amount of physics. We’ll also develop techniques that are extremely useful in determining the evolution in time of systems subject to contraints. This is a very important dividend, since constrained systems are absolutely everywhere; they are commonplace not only in science and engineering, but in business and finance as well. Figure 1. A stroboscopic photo (created with the help of Dan Schwalbe) that shows the surprising small hop that occurs after about 90 ° of rotation. The Hopping Hoop 3 The mechanical system that is the subject of this project is illustrated in figure 2: a perfectly rigid circular hoop of mass (1 — l)M and radius a with an object of mass l M (the weight) rigidly attached to its rim. Here, M is the total mass of the combined system consisting of the hoop and the object. The parameter l is the ratio of the attached mass to the total; of particular interest will be the limit l Ø 1, in which the mass of the hoop itself is negligible relative to that of the attached object. y lM q x Figure 2. The system: a hoop with an additional weight attached to its rim. Together, the hoop and the weight have mass M; the mass of the hoop alone is (1 — l)M while that of the attached weight is l M. This project is laid out as a series of exercises. Some of these exercises explore aspects of the problem that will be of particular interest to students of physics and engineering, but are not 100% necessary to complete the project. Such exercises, designated by an asterisk, are optional (though all students are strongly encouraged to attempt them). One such exercise follows. 4 Project Intermath * Exercise 1. Center of Mass Consider a system of N particles having masses m1 , m2 , …, mN , and position ÷” ” vectors r1 , ”r2 , …, ”rN . The position R of the center of mass of the system is defined by ÷ ” ⁄Ni=1 mi ”ri R = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ N mÅÅÅÅÅ ⁄i=1 i and it represents — pun very much intended! — a “weighted average” of the positions of the N particles comprising the system. Show that for the system we are considering here, namely, the hoop of mass H1 - lL M with the additional attached mass l M, the center of mass lies along the line connecting the center of the hoop with the attached mass a distance l a from the center of the hoop. 2. Problem Statement and Assumptions To simplify the problem somewhat, we will make two assumptions: 1. We assume that both the hoop and the surface upon which it rolls are perfectly rigid; in other words, neither the hoop nor the surface suffers deformation of any kind as the former rolls along the latter. This is admittedly an idealization — the bottom of a real hoop is slightly flattened while the supporting surface shows a slight depression in the vicinity of the point of contact — but it is an extremely reasonable one. 2. We assume that the hoop rolls without slipping, at least initially. It turns out that this assumption is a little tricky to implement in an actual experiment, at least if you’re interested in seeing a big hop, which requires fairly high speeds. (One way to insure that the hoop initially rolls without slipping in an actual experimental trial is to position the attached weight so that it is not directly above the center of the hoop, and allow it to accelerate from rest entirely under the influence of gravity. This will not, however, result in a very impressive hop.) The problem is then to predict the subsequent motion of the hoop. The Hopping Hoop 5 3. Overview of the Solution Procedure The motion of the hoop consists of three distinct phases: an initial phase in which the hoop rolls entirely without slipping; an intermediate phase in which the hoop slips as it rolls, but has not yet lost contact with the supporting surface; and a final phase in which the hoop has lost contact with the supporting surface and is airborne. Figure 3 shows what we are aiming for: a simulation that yields a graphic display of the hoop and shows the experimentally observed hop. Figure 3. The hopping hoop. The hop is not large, but its existence is very surprising. The trajectory of the center of mass is indicated by the downward sloping curve that appears in the vicinity of the large dots. Exercise 1 formally defines the center of mass. We show, in exercise 2, that the trajectory of the center of mass during the initial rolling-without-slipping phase must be a cycloid. After a brief review of the relevant physics, we proceed to write down, in exercises 3 and 4, the three differential equations of motion for the weighted hoop. During the rolling-without-slipping phase, these three equations can be written (exercises 5, 6, and 8) as a single equation, which we then solve (exercises 9 and 10). This solution is only valid, however, so long as the hoop does not slip, so the next four exercises, 11 through 14, are devoted to determining the critical angle at which slipping begins and the initial phase of the motion terminates. We next turn out attention to the intermediate phase of the motion, in which the hoop slips but maintains contact with the supporting surface. We describe in some detail a procedure for solving the differential equations of motion during this phase and, in exercise 15, ask the student to implement this procedure numerically. Exercise 16 treats the hop itself, the final phase of the motion in which the hoop is actually airborne. The hop ends when the hoop regains contact with the supporting surface, and we determine the time of impact in exercise 17. 6 Project Intermath Exercise 18 assembles the previously obtained solutions for the three phases of the motion into a single simulation. 4. Constrained Motion: Rolling Without Slipping The most general motion of a rigid body in 3-space consists of a translation of the center of mass, combined with a rotation of the body about an axis containing the center of mass. Accordingly, we shall specify the motion of the hoop by the Cartesian coordinates [email protected], [email protected] of the center of mass of the system — indicated by the small dot in figure 2 — and by the angle [email protected] through which the hoop has turned. Of course, [email protected], [email protected], and [email protected] are not the only variables we could choose to specify the configuration of the weighted hoop. Indeed, instead of the horizontal and vertical coordinates of the center of mass of the system, it might seem more natural to choose the horizontal and vertical coordinates of the center of the hoop, quantities that we will denote by [email protected] and [email protected], respectively. However, as we shall see presently, the dynamics of the system will serve to make the choice of coordinates [email protected], [email protected], and [email protected] particularly convenient. Because of our assumptions, the three coordinates [email protected], [email protected], and [email protected] are not independent, but are subject to various constraints: 1. Since the hoop is assumed to be a rigid object that suffers no deformation as it rolls, the center of mass is always a fixed distance from the center of the hoop: It is always true that [email protected] - [email protected] L2 + [email protected] - [email protected] L2 = l2 a2 , or equivalently, [email protected] - [email protected] = a l [email protected] (1) [email protected] - [email protected] = a l [email protected] 2. Since the supporting surface is also assumed to be rigid, the vertical height of the center of the hoop must satisfy [email protected] = a (2) so long as the hoop and the surface are in contact. 3. As long as there is no slipping, the horizontal position of the center of the hoop and the angular coordinate q[t] are related by The Hopping Hoop 7 [email protected] = [email protected] - q0 L where (3) q0 = [email protected] These three constraints together imply that as long as the hoop rolls without slipping along the supporting surface, the center of mass of the system moves along a curtate cycloid (“curtate” means “shortened”), also known as a trochoid [4]. The coordinates of the center of mass are then related to the angular coordinate q[t] by [email protected] = aHq - q0 + l [email protected] (4) [email protected] = aH1 + l [email protected] 2 1 0 0 π 2π 3π 4π Figure 4. The trajectory of the center of mass of an asymmetrically weighted hoop of radius 1 with l = 0.8 is a curtate cycloid. 2 1 0 0 π 2 π 3π 2 2π 5π 2 3π Figure 5. A more detailed depiction of the cycloidal curve that is the trajectory of the center of mass. Also shown are the positions at equal time intervals of the additional weight attached to the rim of the wheel, as well as a spoke connecting the weight to the center. The motion is that of an asymmetrically weighted hoop of unit radius with l = 0.75 rolling without slipping under the influence of gravity with an initial angular speed of 1 radian/second. 8 Project Intermath Exercise 2. Cycloid Constraint Show how the three constraints (1), (2), and (3) combine to give the equations (4), which constrain the center of mass of the weighted hoop to move on a curtate cycloid. Aside for the Advanced Student: Configuration Space NOTE: The material in this section may be useful if this ILAP module is used in conjunction with an advanced course in Langrangian/Hamiltonian mechanics or dynamical systems. This material is entirely supplementary. Constraints have a geometric interpretation. As we have already mentioned, the configuration of our system is specified by a triplet of (generalized) coordinates; we have chosen HX, Y, qL, though this is by no means the only choice. In the absence of constraints, X, Y, and q can assume all possible real values, so the set of all possible configurations of the system, termed the configuration space of the system, is isomorphic to Ñ3 . The constraints restrict the possible configurations of the system to lie on some surface of lower dimension; in the case of the hoop, the configuration space of the system subject to constraints (1), (2), and (3) — which are equivalent to the cycloid constraint, equations (4), as you demonstrated in the preceding exercise — is the onedimensional curve shown in figure 6. The Hopping Hoop 9 Y 0 1 0 X 5 10 10 5 θ 0 Figure 6. Configuration space for the hoop when the cycloid constraint is satisfied. The complete state of a system is specified by the values not only of the generalized coordinates, but of the generalized velocities as well; a complete ° ° ° description of the hoop at an instant in time requires the six values HX, X, Y, Y, q, qL, or, equivalently, HX, PX , Y, PY , q, Pq L, where PX , PY , and Pq are the momenta canonically conjugate to the coordinates X, Y, and q, respectively. The set of all possible states of a classical system is called the phase space of the system. 5. Some Basic Physics Forces Acting on the Hoop The motion of any object is determined by the forces exerted on it by the outside world (external forces). In the case of our weighted hoop, there are two such external forces: the gravitational attraction of the earth for the hoop and the attached object, and the contact force that the supporting surface exerts on the hoop. 10 Project Intermath The gravitational force behaves as if the entire mass of the system were concentrated at the position of the center of mass. This force is directed straight down and has magnitude M g, where M is the combined mass of the hoop and the attached object and g = 9.8 m ê s2 is the constant acceleration imparted by the earth’s gravity to every object near the earth’s surface. The force of contact exerted by the supporting surface on the hoop may be resolved into components normal and tangential to the surface, and these components are conventionally treated as if they were separate forces. The component normal to the surface is commonly known as the normal force, and the component tangential to the surface is friction. q Mg N F Figure 7. Free body diagram of the asymmetrically weighted hoop showing the external forces acting upon it. For later convenience, we define f = F ê M and n = N ê M. These quantities are the forces per unit mass (accelerations) arising respectively from the the x- and ycomponents of the force of contact between the hoop and the supporting surface. (For brevity, we will frequently refer to f and n as “forces” even though they are actually forces per unit mass.) Unlike g, f and n are not constant, but vary as the hoop rolls. Friction, the tangential component of the force of contact between two surfaces, is directed so as to oppose the relative motion of the surfaces, or, in cases where there The Hopping Hoop 11 is no relative motion, the tendency for such motion. In the present case, we assume that the hoop rolls without slipping, and thus there is no relative motion between the hoop and the supporting surface. If, however, the hoop were able to slip freely in the situation depicted in figure 7, then it would rotate clockwise. The friction force in figure 7 is thus directed to the right to oppose this tendency for relative motion. Indeed, it is precisely the action of the friction force that prevents the hoop from sliding; friction is the force responsible for maintaining the “rolling without slipping” constraint expressed mathematically by (3). In much the same way, the normal component of the contact force (the “normal force”) maintains the constraint that the center of the hoop be no closer to the supporting surface than the hoop radius — [email protected] ¥ a — where equality holds when the hoop and the supporting surface are in contact. We shall have more to say about forces of constraint later. Newton’s Second Law Newton’s second law expresses mathematically how the interactions of a body with its environment (manifested by the external forces exerted on the body) determine the body’s subsequent motion (expressed by the acceleration of the body’s ÷” ÷–” center of mass): ⁄ Fexternal = M R Exercise 3. Center of Mass Equations of Motion Review the previous subsection’s discussion of the forces acting on the hoop, and apply Newton’s second law to write down the following pair of second-order differential equations for [email protected] and [email protected], the horizontal and vertical coordinates of the center of mass of the system. – [email protected] = f (5) – [email protected] = n - g (6) We have already observed that an arbitrary motion of a rigid body may be decomposed into a translation of the body’s center of mass combined with a rotation of the body about an axis containing the center of mass. If the direction of the rotation axis does not change with time, the rotational form of Newton’s second law is given by – ⁄ texternal = I q, where q specifies the angle of rotation. The sum on the left-hand side is over the components parallel to the rotation axis of all external torques about the center of mass; I is the moment of inertia of the body about the center of mass. 12 Project Intermath Exercise 4. Equation of Rotational Motion Consider torques about the center of mass, and apply the rotational form of Newton’s second law to obtain the following differential equation for q[t], the angle through which the hoop has rotated. – M [email protected] = ÅÅÅÅ ÅÅ H [email protected] - [email protected] - f [email protected] L I (7) * Exercise 5. Derivation of the Equation of Energy Conservation Consider the differential equations of motion that you obtained in the previous ° ° two exercises. Multiply both sides of equation (5) by M [email protected], of equation (6) by M [email protected], ° and of equation (7) by I [email protected] Eliminate the unknown forces n and f from the resulting three equations using the constraint equations (1), (2), (3), and (4). In this way, obtain a single differential equation for the three coordinates. Integrate this equation with respect to time and so obtain the equation of energy conservation for the weighted hoop. ° 2 ° 2 ° 2 ÅÅÅÅ12 [email protected] + [email protected] M + ÅÅÅÅ12 I [email protected] + M g [email protected] = constant (8) Exercise 6. Energy Conservation Equation in Terms of q Only Using the cycloid constraint (4), eliminate the coordinates [email protected] and [email protected] in favor of q[t] from the equation of energy conservation (8) and so obtain: ° 2 a g M H1 + l [email protected] + ÅÅÅÅ12 @I + a2 M H1 + l2 + 2 l [email protected] LD q HtL = constant (9) The Hopping Hoop 13 * Exercise 7. Moment of Inertia of the Asymmetrically Weighted Hoop The quantity I appearing in (8) and (9) is the moment of inertia of the system consisting of the hoop plus the attached object about the axis that passes through the system center of mass and is perpendicular to the plane of figure 7. Take the attached object to be a mathematical point and show that in such a case I is given by I = M a2 H1 - l2 L. (Hint: You will need to use the parallel axis theorem, which states that the moment of inertia of a rigid body about any axis is equal to the moment of inertia about a parallel axis passing through the center of mass plus the product of the mass of the body and the square of the distance between the two axes.) Of course, in the real world the attached object will not be a mathematical point, but will be of nonvanishing physical dimension, and thus will possess a nonvanishing moment of inertia about its centroid. We will take I = M a2 H1 - l2 + l ¶L (10) where ¶ is a dimensionless parameter (generally quite small) related to the physical dimensions of the attached object. Initial Conditions The coordinate system shown in figure 2 is chosen so that at t = 0 the center of the hoop is situated at (0, a). In general, three parameters are required to specify the initial state of the weighted hoop: one to specify the initial orientation of the hoop, and two more to completely specify its initial motion. One very natural choice would be to specify the angle q0 = [email protected]; the initial translational speed x° 0 of the center of the hoop; ° ° and the initial angular speed of rotation q0 = [email protected] In the present case, however, where we assume that the hoop rolls without slipping, the latter two parameters are not ° independent, but are related by x° 0 = a q0 . We will thus specify the initial state of the ° weighted hoop by giving q0 and q0 . 14 Project Intermath Exercise 8. Mechanical Energy The constant appearing on the right-hand side of equations (8) and (9) is the total mechanical energy E of the weighted hoop, and it may be evaluated from the initial conditions. Take the attached object to be initially located directly above the point of contact (q0 = 0) and show that E = M a g @H1 + lL H1 + cL + ÅÅÅÅ12 ¶ l cD °2 where we have introduced the dimensionless quantity c = a q0 ê g. ° Substitute this result into (9) and solve the resulting equation for [email protected] to obtain the following first-order differential equation describing the time evolution of the weighted hoop constrained to roll entirely without slipping. ° g cH1+ l+ ¶ lê2L+ lH1- [email protected] ######## ####### [email protected] = "##### ÅÅÅÅa "################################ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅL 1 + ¶ lê2+ l [email protected] (11) Exercise 9. Motion on the Cycloid: Numerical Solution for a Typical Case Use a numerical differential equation solver (such as the Mathematica function NDSolve) to obtain a numerical solution to (11) subject to the initial condition [email protected] = 0 for 0 § t § 3 (sec). Use the following values for the parameters: g = 9.8 Hmeters ê sec2 L, a = 1 (meter), l = 0.95, ¶ = 0.01, c = 0.1. To help you visualize your solution, create a plot similar to figure 5 (e.g., with the Mathematica function ParametricPlot.) Exercise 10. Motion on the Cycloid: Analytic Solution for a Special Case Take l = 1 (corresponding to a hoop whose mass is negligible relative to that of the attached object) and ¶ = 0 (effectively treating the attached object as a point mass). For this special case, obtain an analytic solution to the differential equation (11) subject to the initial condition [email protected] = 0. The Hopping Hoop 15 Forces of Constraint We were fortunate to be able to eliminate the forces f and n from the equations of motion. These forces are not known a priori; rather, they are determined by the effects they produce: by the constraints they maintain. In fact, one may even view equations (5) and (6) as defining the forces f and n in the case that the cycloid constraint (4) holds. * Exercise 11. Explicit Expressions for the Forces of Constraint Use the cycloid constraint to obtain the following expressions for the forces of constraint n and f as a function of x = [email protected] 1 f @xD = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Ig l H2+¶ l+2 l xL2 è!!!!!!!!!!!!2 1 - x H2 + l H-2 + ¶ + 6 xL - c H2 + H2 + ¶L lL H1 + ¶ l + l xL + l2 H-2 ¶ - 2 x + 3 ¶ x + 4 x 2 LL (12) 1 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ [email protected] = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H2+¶ l+2 l xL2 H4 g + 4 g ¶ l - 2 g l2 - 2 c g l2 + g ¶2 l2 - 2 g l3 - 2 c g l3 - g ¶ l3 - c g ¶ l3 + 8 g l x - 4 c g l x - 4 g l2 x - 4 c g l2 x + 4 g ¶ l2 x - 4 c g ¶ l2 x - 2 g ¶ l3 x - (13) 2 c g ¶ l3 x - c g ¶2 l3 x + 10 g l2 x 2 - 2 c g l2 x 2 2 g l3 x 2 - 2 c g l3 x 2 + 3 g ¶ l3 x 2 - c g ¶ l3 x2 + 4 g l3 x3 L The fact that n and f are absent from the equation of energy conservation (8) is no accident, but reflects a general property of forces of constraint. Forces that maintain constraints necessarily act perpendicular to the configuration space of the system, and consequently do no work. 16 Project Intermath 6. Removing One Constraint: Rolling With Slipping If the weighted hoop is to become airborne, then its center must acquire a vertical velocity. For this to occur, one of the conditions that constrain the center of mass to move along a cycloid must fail to be satisfied: the hoop must slip. The next exercise examines why. Exercise 12. Why the Hoop Must Slip Use the rigid hoop condition (1) to obtain an expression for y° @tD, the vertical component of the velocity of the center of the hoop. Interpret the two terms in the resulting expression. What values must these terms have (a) if the hoop rolls without slipping, and (b) if the hoop is to become airborne? Having established that there must be some slippage of the hoop relative to the supporting surface just prior to the hoop’s becoming airborne, we now examine at what point this slippage occurs. Equations of Motion for Rolling with Slipping Once the hoop has started to slip, its motion is no longer described by the single equation (11). Rather, the slipping hoop is described by the three equations of motion (5), (6), and (7), subject to the constraints (1) and (2). The friction force is now given by f = -mk n, where the minus sign indicates that the force is directed to the left, a fact that follows from a consideration of the physical particulars of the situation and the observation that [email protected] ¥ 0. The preceding definition of kinetic friction and the rigid hoop constraint (3) allow us to recast the equations of motion in the form: – [email protected] = n - g (14) – [email protected] = -mk n (15) – M [email protected] = ÅÅÅÅ ÅÅ a @l sin @qHtLD + mk H1 + l [email protected] n (16) I The normal force [email protected] must be sufficiently large to ensure that the height of the center of the hoop above the supporting surface is never less than the radius of the The Hopping Hoop 17 hoop, i.e., the point of contact never lies below the x-axis. This constraint is used to dynamically determine the value of n. Solution Procedure: Rolling With Slipping To obtain the motion of the system during the phase in which the hoop rolls with slipping but has not yet lost contact with the supporting surface, one must solve simultaneously the differential equations (14) and (16), subject to the constraint that [email protected] ¥ a. This constraint must be enforced explicitly at each time step of the integration: one checks at each stage whether [email protected] < a, and if this is the case, one adjusts n to give [email protected] = a for that step, then, using the new value of n, one recomputes X and q from (14) and (16), respectively. The integration continues until [email protected] is strictly greater than a. The height of the center of the hoop is [email protected] = [email protected] - a l cos @qHtLD. – – If one differentiates twice with respect to time and substitutes for [email protected] and [email protected] from equations (6) and (16) respectively, one obtains ° 2 M y– @tD = n - g + a lIsin @qHtLD ÅÅÅÅ ÅÅ a nH l sin @qHtLD + mk H1 + l [email protected] L L + [email protected] [email protected] M I (17) In the solution procedure just described, n is updated once every Dt using the constraint [email protected] ¥ a. For the duration of any single time step Dt, n is assumed to be constant. For simplicity, we will take not just n, but also y– @tD to be constant over the interval Dt. This implies that y° @t + DtD = y° @tD + y– @tD Dt and [email protected] + DtD = [email protected] + y° @tD Dt + ÅÅÅÅ12 y– @tD HDtL2 = [email protected] + ÅÅÅÅ12 Dt Hy° @tD + y° @t + DtD L One may view the above expressions for y° @t + DtD and [email protected] + DtD as arising from a slightly modified version of the improved Euler method for numerical solution of differential equations [2]. (The method is “improved” because the term second-order in Dt in the expression for [email protected] + DtD represents a simple but significant improvement over the basic Euler method.) If [email protected] + DtD fails to be greater than or equal to a, then one sets [email protected] + DtD = a and solves the above equation for n. ° 2 [email protected] y° @tD- ÅÅ1ÅÅ D t2 I-g+a l [email protected] [email protected] ME 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ n = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ M a2 Dt2 A1+ ÅÅÅÅÅÅÅÅIÅÅÅÅÅÅÅÅ l [email protected] Hl [email protected]+H1+l [email protected] mk LE The procedure may be represented by the following pseudocode: 18 Project Intermath while (y[t] ≤ a) { if (y[t+∆t] < a) recompute n to give y[t+∆t] = a; use new n to compute X[t+∆t], Y[t+∆t], θ[t+∆t], X [t+∆t], Y [t+∆t], θ[t+∆t]; t = t + ∆t; } end while; A remark is in order about a matter that arises frequently in numerical work: the comparison of floating-point quantities. As we have said, the loop terminates when the height [email protected] of the center of the hoop is strictly greater than a. In order to determine whether or not this is the case, the computer must compare two floating-point values, and these are known to only a finite precision. To insure that the loop is not terminated prematurely, the termination condition is tested only to a certain tolerance d, chosen so as not to exceed the precision of the machine; the while loop ends when [email protected] - a > d. Exercise 13. Numerical Solution: Rolling WITH Slipping Use the programming language of your choice to implement the solution procedure just described. Take mk = 0.7 and use your program to calculate the motion of the hoop from the onset of slipping until time tHop when the hoop loses contact with the supporting surface. To verify that the hoop is indeed slipping, print the values of ° a q and of the horizontal velocity x° of the center of the hoop at selected times during the integration. For your use in later exercises, note the values at t = tHop of X, Y, and q and ° ° ° of their time derivatives X, Y, and q. Aside: Checking for Numerical Correctness Whenever one computes an aproximation to a solution, one must try to ascertain whether the solution is approximately correct. In the present case, the numerical procedure necessitated the introduction of two parameters: the time step Dt and the tolerance d. It is straightforward to assess the effect of the finite step size Dt on the results of the computation: One would expect that for any given value of d, as Dt Ø 0 the numerical solution ought to “converge” to a limiting "numerical function" (i.e., table of values) and the takeoff time tHop , which marks the endpoint of the numerical integration, ought to approach some limiting value. One can verify that this The Hopping Hoop 19 is in fact the case. For instance, let us take d ê a = 10-4 . Dt HsecL tHop HsecL nsteps 0.002 0.0005 1.01507 0.99757 14 21 0.0001 0.00005 0.992471 0.991621 54 91 Interpreting the impact of the tolerance d on the final result demands a certain amount of physical insight. Since floating-point quantities can be computed only to finite precision, we have no choice but to implement the termination condition for the numerical integration, namely, that [email protected] be strictly greater than a, as [email protected] - a > d; in effect, we must consider the hoop to be in contact with the supporting surface until the point of contact clears the surface by at least d. Since the parameter d effectively defines when the hoop is considered to be airborne, the amount of time tHop - tSlip during which the hoop rolls while slipping prior to takeoff depends on d, and so too does the height of the hop. (By the “height of the hop” we mean the maximum height above the supporting surface attained by the center of the hoop; it is max ( [email protected] ) on the interval 0 § t § tImpact.) Roughly speaking, this is because the longer the rolling-with-slipping phase prior to takeoff, the greater the takeoff speed that the hoop can build up; more ° precisely: the greater the amount by which » y° » can exceed » Y ». Numerical simulations confirm that the height of the hop decreases with d. We obtained the following results for Dt = 0.002: dêa tHop nsteps hop height 0.01 1.39 0.001 1.049 0.0001 1.015 76 31 14 0.097 0.0011 0.00012 The observed decrease in the height of the hop with d might lead one to expect in the limit d Ø 0 to observe no hop at all. However, the d Ø 0 limit is not only in conflict with physical reality, it violates even the assumptions of this admittedly idealized problem! The point is simply this: Even if we were able to perform our numerical computations to infinite precision, physical reality will still not let us make d arbitrarily small. It is, after all, microscopic imperfections in the hoop and the supporting surface that give rise to friction in the first place, and the scale of these 20 Project Intermath imperfections provides a physical lower bound to d. In the present case, we began with the assumption that hoop and surface are sufficiently rough that the former would initially roll without slipping over the latter. This precludes d Ø 0. 7. The Hop When the hoop loses contact with the supporting surface, then it is clear that all forces of contact must vanish: [email protected] = f @tD = 0. Exercise 14. Airborne Hoop Assuming that the hoop is flying through the air, write down the equations of motion for the center of mass coordinates [email protected] and [email protected] and for the angular coordinate [email protected] Given that the time at which the hoop loses contact with the supporting surface is tHop, and that the values at that instant of the three coordinates and their time ° ° ° derivatives are XHop , YHop , qHop , XHop, YHop, and qHop , solve the equations of motion to obtain expressions for [email protected], [email protected] and [email protected] valid while the hoop is in the air. The airborne hoop simply rotates about its center of mass at a constant angular speed equal to its angular speed at the instant the contact forces went to zero, while the center of mass itself falls freely under the influence of gravity, describing the characteristic parabolic trajectory. This combined free rotation/free fall continues until the height of the center of the hoop decreases once again to a, at which time [email protected] acquires the positive value required to keep the point of contact between hoop and ground from moving below ground level. Exercise 15. Time of Impact Compute the time tImpact at which the hopping hoop regains contact with the supporting surface. The Hopping Hoop 21 Exercise 16. Putting It All Together Assemble your results from exercises 9, 11, 14, 13, and 14 into a complete solution of the motion of the hoop, and use the resulting combined solution to create a plot similar to the dots in figure 3 showing the motion of the hoop from time t = 0 until tImpact. 8. Postscript: A Final Surprise In computing the motion of the asymmetrically weighted hoop, we took q0 = 0; that is, we assumed that the attached object is initially located at the top of the hoop, directly above the point of contact. Would the behavior of the hoop have been substantially different if the attached object had started out at the bottom of the hoop, i.e., with q0 = p? The answer is a resounding “yes”! We challenge the ambitious among you to examine the behavior of the hoop when the total mechanical energy is the same as in the case which we considered in this module, but with the attached object initially at the bottom instead of the top. Some care is required to modify equations (11–13) to the case where q0 = p (or more generally, to the case where q0 assumes an arbitrary value), but if you do so correctly you will find that the hoop begins to slip at a critical angle qSlip = 210.7 °, at ° which point the angular speed of rotation is qSlip = 10.12 radians/second. You may then proceed as in exercise 13 to compute a numerical solution for the motion of the hoop during the phase in which it rolls with slipping along the supporting surface, and as in exercise 14 to obtain a solution for the motion of the hoop during the phase in which it is airborne. The resulting motion, illustrated in figure 8 and readily observed experimentally, is truly surprising. Note that for the experiment one can start with the weight at the top and ignore the small hop that can occur on the way down. At some point in the upward motion of the weight there should be a very big hop. The electronic version of [5] contains a complete Mathematica package for simulating the hopping hoop with a variety of initial conditions. 22 Project Intermath 3.5 3 2.5 2 1.5 1 0.5 0 -1 0 1 2 3 4 Figure 8. Motion of the hoop for q0 = p. The mechanical energy of the system is identical to that for the case considered throughout this module. The hoop is airborne for most of the motion depicted here; observe that the center of mass (black line) follows the parabolic trajectory characteristic of an object falling freely under the influence of gravity. The Hopping Hoop 23 References [1] B. Bollabás, editor, Littlewood’s Miscellany, Cambridge Univ. Pr., London, 1986. [2] Martin Braun, Differential Equations and their Applications, 2nd ed., Springer, New York, 1975, pp. 104-105. [3] James P. Butler, Hopping hoops don’t hop, American Mathematical Monthly, 106 (1999) 565-568. [4] Shokichi Iyanaga and Yukiyasi Kowada, Encyclopedic Dictionary of Mathematics, MIT Press, Cambridge, Mass., 1977, p. 322. [5] Tim Pritchett, A mechanical system with constraints, Mathematica in Education and Research Vol 8:3-4 (1999) 20-27. [6] Tim Pritchett, The hopping hoop revisited, American Mathematical Monthly 106 (1999) 609-617. [7] T. F. Tokieda, The hopping hoop, American Mathematical Monthly 104 (1997) 152-154. 24 Project Intermath Notes to the Instructor This Interdisciplinary Lively Applications Project is probably best suited for use in a course in mathematical modeling, and it was written with this in mind. We make no assumptions regarding the physics background of the student audience. Those students who have taken the first term of a typical calculus-based general physics course will find themselves in familiar territory. But such a course is by no means prerequisite to this module, since all the necessary physics is fully explained in the module itself; in this sense, the module is self-contained. An instructor who wishes to deemphasize the physics in favor of the mathematics can easily do so simply limiting the number of optional (starred) exercises the students are required to complete. Indeed, certain of the optional exercises (e.g., exercise 11) simply require the student to work through the algebra leading to a given result. Such exercises can probably be skipped with impunity, the stated results being accepted on faith. The module can be implemented in several ways. It could, for instance, form the basis for a semester project by a group of two or three students who would work through it in its entirety. Students with an experimental bent may even wish to construct a model of the hoop and use it to obtain qualitative experimental confirmation of the hop. It is very exciting to see that physical reality matches the model, and building a working model is not difficult: use a hula hoop with holes drilled in the inside to accept four metal spokes, since it is essential that the hoop stay rigid. Students who have an interest in both mathematics and physics might well enjoy this experimental aspect. Alternately, all or a portion of the module could be used as a class project spanning multiple class meetings, with selected exercises assigned as homework. This ILAP project could also serve well in a physics course in classical mechanics, either with individual exercises assigned as homework, or as a semester project by a small group. Prior familiarity with a computer package (such as Mathematica, MathCad, or Maple) offering routines for root-finding, numerical solution of ODEs, and symbolic manipulation will free the students to concentrate on the mathematics and physics of this project, and such familiarity, though not absolutely essential, is very desirable. On the other hand, in order to complete exercise 15, it is essential that students have some experience programming a computer in some language. Absent that knowledge, the computer code necessary to compute the motion of the hoop as it rolls with slipping along the supporting surface will have to be supplied by the instructor. Perhaps the main point of this project is that an undergraduate can handle a simulation where the frictional component is somewhat more complicated The Hopping Hoop 25 than the familiar air resistance models. Creating such a simulation can be a valuable exercise in the art of model-building, and in learning the capabilities (and limits and quirks) of modern software. Sample Solutions to Exercises Exercise 1. Center of Mass In this case, the system consists of two masses: the hoop with mass H1 - lL M and the attached object with mass l M. We can express the position vectors in any coordinate system we choose, so let’s choose a plane polar coordinate system whose origin lies at the center of the hoop. In this system, the position vector of the hoop is the null vector: ”rh = 0, and that of the attached object is given by ”rm = a r̀. (Here, r̀ is the unit vector defined at every point in the plane to point radially away from the origin. Although we do not need it here, we remind the student that the second unit vector in ` plane polar coordinates, q, points in the direction of increasing polar angle q; this is usually taken to be in the sense of counterclockwise rotation about the origin. For a ` more complete discussion of r̀ and q, see, e.g., Daniel Kleppner and Robert J. Kalenkow, An Introduction to Mechanics, McGraw-Hill, 1973.) The position vector of the center of mass is then ÷ ” H1 - lL M ÿ 0 + a M l r̀ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = l a r̀ R = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ M H1 - lL + M l ÷” Since R ∂ r̀, the center of mass of the system lies along the line connecting the center of the hoop (the origin) with the position of the attached object. We observe that for l Ø 1, i.e., when the attached object is much more massive than the hoop, the center of mass coincides with the position of the object. Conversely, for l Ø 0, i.e., when the hoop is much more massive than the attached object, the center of mass lies at the origin (the center of the hoop). Exercise 2. Cycloid Constraint The rigid hoop constraint (1) relates the coordinates of the center of the hoop to the coordinates of the system center of mass. [email protected] - [email protected] = a l sin [email protected] [email protected] - [email protected] = a l cos [email protected] Substituting for [email protected] from the “no slipping constraint,” equation (3), [email protected] = a [email protected] - q0 L, 26 Project Intermath yields the given result for [email protected] : [email protected] = aHq - q0 + l [email protected], while substituting for [email protected] from the “normal force constraint,” equation (2), [email protected] = a, yields the given result for [email protected] : [email protected] = aH1 + l [email protected] Exercise 3. Center of Mass Equations of Motion Referring to the free body diagram, figure 7, we see that the net force acting to – the right is simply F, the tangential component of the contact force, so F = M X, or – – X = F ê M = f . Similarly, the net upward force is N - M g, so N - M g = M Y, or – Y = N ê M - g = n - g. Exercise 4. Equation of Rotational Motion Of three forces shown in the free body diagram, figure 7, only two exert a nonvanishing torque about the center of mass. (The weight M g, which acts at the center of mass, can exert no torque about an axis through the center of mass because the torque arm is zero.) The torque resulting from the normal component of the contact force is N H [email protected] - [email protected] L, and this torque acts to increase q. The torque arising from the tangential component of the contact force is given by -F [email protected], and this torgue acts to decrease the angle q, which is the reason for the minus sign. Summing these two – torques, we obtain, F [email protected] - N [email protected] - [email protected] L = I q, or, with F = M f and N = M n, – q = ÅÅÅÅ1I 8M f [email protected] - M n [email protected] - [email protected] L <, which is (7). Exercise 5. Derivation of the Equation of Energy Conservation Taking our cue from the problem statement, we multiply both sides of (5) by ° ° ° M [email protected], both sides of (6) by M [email protected], and both sides of (7) by I [email protected] This gives – ° ° M [email protected] [email protected] = M f [email protected] – ° ° M [email protected] [email protected] = MHn - gL [email protected] – ° ° I [email protected] [email protected] = M 8 [email protected] - [email protected] L - f [email protected] < [email protected] The left hand side of each of these equations can be written as a total derivative: d M ° 2 ° ÅÅÅÅÅÅÅÅÅ 9 ÅÅÅÅÅÅÅÅÅ [email protected] = = M f [email protected] dt 2 d M ° 2 ° ÅÅÅÅÅÅÅÅÅ 9 ÅÅÅÅÅÅÅÅÅ [email protected] = = M Hn - gL [email protected] (18) dt 2 d I ° 2 ÅÅÅÅÅÅÅÅÅ 9 ÅÅÅÅÅ [email protected] = = M 8 [email protected] - [email protected] - f [email protected] < dt 2 Using the chain rule to differentiate (4) with respect to time, we get The Hopping Hoop 27 ° ° ° ° [email protected] = X£ @[email protected] [email protected] = H1 + l [email protected] L [email protected] = Y [email protected] ° ° ° ° [email protected] = Y£ @[email protected] [email protected] = [email protected] [email protected] = -HX - xL [email protected] (19) We now add the three equations (18) and use (19) to eliminate the cancelling pairs of terms involving n and f . The result is: d M ° 2 d I ° 2 d M ° 2 ° ÅÅÅÅÅÅÅÅÅ 9 ÅÅÅÅÅÅÅÅÅ [email protected] = + ÅÅÅÅÅÅÅÅÅ 9 ÅÅÅÅÅÅÅÅÅ [email protected] = + ÅÅÅÅÅÅÅÅÅ 9 ÅÅÅÅÅ [email protected] = = -M g [email protected] dt 2 dt 2 dt 2 or, d M ° 2 M ° 2 I ° 2 ÅÅÅÅÅÅÅÅÅ 9 ÅÅÅÅÅÅÅÅÅ [email protected] + ÅÅÅÅÅÅÅÅÅ [email protected] + ÅÅÅÅÅ [email protected] + M g [email protected] = = 0 dt 2 2 2 Integration yields the energy conservation equation (8). Exercise 6. Energy Conservation Equation in Terms of q Only Differentiating equations (4) with respect to time, we obtain ° ° [email protected] = a H1 + l [email protected] L [email protected] ° ° [email protected] = -a l [email protected] [email protected] In terms of q, the translational kinetic energy of the center of mass is thus given by ° 2 ° 2 ° 2 ÅÅÅÅ12 M [email protected] + [email protected] M = ÅÅÅÅ12 M a2 H1 + l2 + 2 l [email protected] L [email protected] and the gravitational potential energy is M g [email protected] = M g aH1 + l [email protected] L. Substituting for these terms in (8), one obtains the required result (9). Exercise 7. Moment of Inertia of the Asymmetrically Weighted Hoop The hoop has mass MHoop = H1 - lL M, assumed to be concentrated entirely in the rim, a distance a from the center. Thus, the moment of inertia of the hoop about its center is MHoop a2 , or H1 - lL M a2 . The parallel axis theorem states that the moment of inertia of a rigid body about any axis is equal to the moment of inertia I0 about a parallel axis passing through the center of mass of the body plus the product of the mass m of the body and the square of the distance R between the two axes: I = I0 + m R2 . Now, the center of mass of the hoop alone coincides with the geometric center of the hoop, and, as you showed in Exercise 1, the center of mass of the system consisting of the hoop and the attached object lies a distance R = l a from the center of the 28 Project Intermath hoop. Thus, the moment of inertia of the hoop alone about an axis passing through the center of mass of the hoop-object system is given by IHoop = MHoop a2 + MHoop R2 = H1 - lL M a2 + H1 - lL MHl aL2 = H1 - lL M a2 H1 + l2 L The attached object has mass l M and is situated a distance H1 - lL a from the center of mass of the hoop-object system. If the object is a point mass, then the moment of inertia of the attached object alone about an axis passing through the center of mass of the hoop-object system is IObject = l MH1 - lL2 a2 . We add these two results to obtain the total moment of inertia of the hoop-object system about an axis through its center of mass: I = IHoop + IObject = H1 - lL M a2 H1 + l2 L + l MH1 - lL2 a2 = M a2 H1 + l2 - l - l3 + l - 2 l2 + l3 L = M a2 H1 - l2 L Exercise 8. Mechanical Energy ° ° Setting [email protected] = 0 and [email protected] = q0 in (9) gives immediately ° 2 constant = E = a M I g H1 + lL + a H1 + H1 + ÅÅÅÅ2¶ L lL q0 M °2 One then substitutes g c for a q0 , and so obtains the result sought: E = M a g HH1 + lL H1 + cL + 1 ê 2 ¶ l cL ° Thus, for the weighted hoop rolling without slipping with [email protected] = 0 and [email protected] = è!!!!!!!! g c ê !a!! , the equation of energy conservation (9) takes the form ° 2 a g M H1 + l [email protected]@tDDL + ÅÅÅÅ12 HI + a2 MH1 + l2 + 2 l [email protected] LL [email protected] = M a g @H1 + lL H1 + cL + ÅÅÅÅ12 ¶ l cD ° Solving for [email protected], one obtains a single differential equation for the motion of the hoop, equation (11). Exercise 9. Motion on the Cycloid: Numerical Solution for a Typical Case When Mathematica solves a differential equation numerically it returns an InterpolatingFunction object, which can be treated just as an ordinary function. Using the Mathematica function NDSolve, we obtain the solution q1 in the form of such an interpolating function The Hopping Hoop 29 g = 9.8; a = 1; λ = 0.95; ∂ = 0.01; c = 0.1; t1 = 3.0; θ1 = θ ê. NDSolveA ∂λ c H1 + λ + L + λ H1 − [email protected]θ@tDDL g 2 '''''''''''''''''''''''''''''''' '''''''' '''' , θ@0D == 0=, 9θ @tD == $%%%%%% &'''''''''''''''''''''''''''''''' ∂λ a 1 + + λ [email protected]θ@tDD 2 θ, 8t, 0, t1 <EP1T [email protected], 3.<<, <>D We can then use the function q1 to compute the positions of the center of mass, the center of the hoop, and the attached object. [email protected]_D := a 8θ1 @tD + λ [email protected]θ1 @tDD, 1 + λ [email protected]θ1 @tDD< [email protected]_D := [email protected] − λ a [email protected]θ1 @tDD, [email protected]θ1 @tDD< [email protected]_D := [email protected] + a [email protected]θ1 @tDD, [email protected]θ1 @tDD< The following Mathematica code generates the plot. ntSteps = 30; ∆t = t1 ê ntSteps; ptsCM = [email protected], [email protected], t1 , ∆tDD; ptsCtrHoop = [email protected] , [email protected], t1 , ∆tDD; ptsMass = [email protected] , [email protected], t1 , ∆tDD; lines = 8 [email protected], [email protected], [email protected], ptsMass<DD<, [email protected], [email protected], ptsMassD<, [email protected], [email protected], ptsCtrHoopD<<; [email protected]@[email protected], 8tt, 0, t1 <, Epilog → lines, AspectRatio → Automatic, PlotRange → 8−0.1, 2.2<, Axes → None, Frame −> NoneD; 30 Project Intermath Exercise 10. Motion on the Cycloid: Analytic Solution for a Special Case In the case of a massless hoop (l = 1) with attached point mass (¶ = 0), the equation of motion (9) reduces to "################ ° c + sin2######## A ÅÅ2qÅÅ E### g g 2 c + 1- [email protected]@tDD dq "##### ######## #Å##Å# , or ÅÅÅÅ [email protected] = "##### ÅÅÅÅa "################ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ Å Å = ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅ q 1 + [email protected]@tDD dt a cosA ÅÅ2ÅÅ E where we have made use of the trigonometric identities 1 + [email protected] = 2 cos2 @q ê 2D and 1 - [email protected] = 2 sin2 @q ê 2D. The above expression is separable, so we can straightforwardly integrate. We will use Mathematica to perform these computations, although they can equally well be done by hand. [email protected], a, cD thetaIntegral = ‡ θ@tD 0 [email protected]θ ê 2D !!!!!!!!! θ è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! c + H [email protected]θ ê 2D L2 i j z θ@tD θ@tD %2%%% y è!!! j z $%%%%%%%%%%%%%%%% % %%%%%%%%%%%%%%% −2 j [email protected] c D − LogASinA E + c + SinA E Ez j z j z j z 2 2 k { The other side of the equation is of course just è!!!!!!!! g ê a t. i y g t j thetaIntegral z z $%%%%%% j z integratedEqn = j == êê Simplify j z j z 2 a 2 k { y θ@tD θ@tD 2 1 i j g z j z LogASinA E + $%%%%%%%%%%%%%%%% c + SinA%%%%%%%%%%%%%%%% E%%%% E == j t + [email protected] $%%%%%% j z 2 2 2 a k { eqnsimpler = [email protected]@Exp, integratedEqnDD 1 θ@tD θ@tD 2 SinA E + $%%%%%%%%%%%%%%%% c + SinA%%%%%%%%%%%%%%%% E%%%% == 2 2 2 g# I"##### a [email protected] θ@tD θ@tD sinThetaOver2 = SinA E ê. SolveAeqnsimpler, SinA EEP1T 2 2 g g 1 "###### "###### 1 è!!! c − 2 a t J−1 + a t N 2 simplerSinThetaOver2 = [email protected]@[email protected] The Hopping Hoop 31 è!!! g t è!!! c SinhA E è!!! 2 a θ@tD θ@tD ê. SolveASinA E == simplerSinThetaOver2, θ@tDEP1T 2 è!!! g t è!!! 2 ArcSinA c SinhA EE è!!! 2 a The above expression for [email protected] is valid as long as the hoop remains in contact with the ground and rolls without slipping; in other words, as long as the attached object is constrained to move along a cycloid. Exercise 11. Explicit Expressions for the Forces of Constraint We wish to find the forces of constraint f and n required to keep the asymmetrically weighted hoop rolling without slipping, i.e., the forces required to keep the the center of mass moving along the cycloidal trajectory (4). [email protected]_D := a Hθ@tD + λ [email protected]θ@tDDL [email protected]_D := a H1 + λ [email protected]θ@tDDL ° We’ve already determined that under these conditions, [email protected] must satisfy (11). [email protected], a, λ, ∂, cD; !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!! g è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! "##### 2 c + 2 λ + 2 c λ + c ∂ λ − 2 λ [email protected]θ@tDD a θ = è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!! !!!!!!!! 2 + ∂ λ + 2 λ [email protected]θ@tDD !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!! g è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! "##### 2 c + 2 λ + 2 c λ + c ∂ λ − 2 λ [email protected]θ@tDD a è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!! !!!!!!!!! 2 + ∂ λ + 2 λ [email protected]θ@tDD – d ° We will need an expression for the second derivative [email protected] = ÅÅÅÅ Å [email protected], so we differentiate dt one more time. – θ = [email protected]@θ, tD ê. θ @tD → θD H1 + cL g λ H2 + H2 + ∂L λL [email protected]θ@tDD a H2 + ∂ λ + 2 λ [email protected]θ@tDDL2 2 d From (5) we see that the required force of constraint f @tD is just given by ÅÅÅÅ ÅÅÅÅ [email protected] dt2 forceOfConstraint = [email protected]@tD, 8t, 2<D 32 Project Intermath a Hθ @tD + λ H−[email protected]θ@tDD θ @tD2 + [email protected]θ@tDD θ @tDLL We substitute to eliminate the time derivatives of q and obtain an expression in terms of x = [email protected] (//. abbreviates ReplaceRepeated, which causes the substitutions to be made until there is no further change: twice in this case). forceOfConstraint = [email protected] – forceOfConstraint êê. 8θ @tD → θ, θ @tD → θ, θ@tD → [email protected]ξD<D 1 H2 + ∂ λ + 2 λ ξL2 è!!!!!!!!!!!!! Ig λ 1 − ξ2 H2 + λ H−2 + ∂ + 6 ξL − c H2 + H2 + ∂L λL H1 + ∂ λ + λ ξL + λ2 H−2 ∂ − 2 ξ + 3 ∂ ξ + 4 ξ2 LLM 2 d In the same way, (6) guarantees that [email protected] is just g + ÅÅÅÅ ÅÅÅÅ [email protected] dt2 normalForce = g + [email protected]@tD, 8t, 2<D g + a λ H−[email protected]θ@tDD θ @tD2 − [email protected]θ@tDD θ @tDL normalForce = – [email protected] ê. θ @tD → θ ê. θ @tD → θ ê. θ@tD → [email protected]ξDD 1 H2 + ∂ λ + 2 λ ξL2 H4 g + 4 g ∂ λ − 2 g λ2 − 2 c g λ2 + g ∂2 λ2 − 2 g λ3 − 2 c g λ3 − g ∂ λ3 − c g ∂ λ3 + 8 g λ ξ − 4 c g λ ξ − 4 g λ2 ξ − 4 c g λ2 ξ + 4 g ∂ λ2 ξ − 4 c g ∂ λ2 ξ − 2 g ∂ λ3 ξ − 2 c g ∂ λ3 ξ − c g ∂2 λ3 ξ + 10 g λ2 ξ2 − 2 c g λ2 ξ2 − 2 g λ3 ξ2 − 2 c g λ3 ξ2 + 3 g ∂ λ3 ξ2 − c g ∂ λ3 ξ2 + 4 g λ3 ξ3 L Exercise 12. Why the Hoop Must Slip Rearranging (1), we obtain the following expression for the vertical height of the center of the hoop [email protected] = [email protected] - a l cos @qHtLD Differentiation with respect to time yields ° ° y° @tD = [email protected] + a l sin @qHtLD [email protected] ° The first term, [email protected], is the speed at which the center of mass falls under the ° influence of gravity. The second term, a l sin @qHtLD [email protected], is the speed at which the center of the hoop rises as a result of the rotation of the hoop about the center of mass. If the hoop is to become airborne, then the magnitude of the latter term must exceed that of The Hopping Hoop 33 the former. As long as the hoop rolls without slipping, however, then these two quantities are equal in magnitude, as can be seen by differentiating the second equation of the cycloid constraint (4). Exercise 13. Onset of Slipping for a Massless Hoop With l = 1 (massless hoop) and ¶ = 0 (attached object is a point mass), the expression (12) for the frictional force f required to keep the center of mass of the weighted hoop moving along the cycloidal trajectory required by the constraints becomes g "#########2### 1-x H4 x+4 x2 -4 c H1+xLL g Hx-cL "#########2### 1-x f @xD = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ , H2+2 xL2 x+1 while the corresponding expression (13) for the normal force reduces to 4 g x3 +8 g x2 +4 g x-4 c g-8 c g x-4 c g x2 [email protected] = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = g Hx - cL, H2 x+2L2 °2 where c = a q0 ê g. Alternately, since x = [email protected], f @qD = [email protected] - cL [email protected] ê 2D [email protected] = [email protected] - cL Slipping occurs at the minimum angle q for which » f @qD » = ms » [email protected] ». Since 0 § c < 1, this minimum angle will necessarily occur in a regime where both f and n are nonnegative and the absolute value signs may be dropped: f @qD = ms [email protected] One may see by inspection that this equation has two possible solutions: [email protected] ê 2D = ms and [email protected] - c = 0 Thus qSlip = Min([email protected], 2 [email protected] D) Exercise 14. Onset of Slipping for a Typical Case The hoop begins to slip at the minimum angle for which H f @cos qD L2 = Hms [email protected] qD L2 , where the forces of constraint f and n are given respectively by (12) and (13). So, assuming the computations of exercise 11, we seek the roots of the following equation. forceOfConstraint2 == Hµs normalForceL2 34 Project Intermath 1 H2 + ∂ λ + 2 λ ξL4 Hg2 λ2 H1 − ξ2 L H2 + λ H−2 + ∂ + 6 ξL − c H2 + H2 + ∂L λL H1 + ∂ λ + λ ξL + 1 λ2 H−2 ∂ − 2 ξ + 3 ∂ ξ + 4 ξ2 LL ^ 2L == H2 + ∂ λ + 2 λ ξL4 HH4 g + 4 g ∂ λ − 2 g λ2 − 2 c g λ2 + g ∂2 λ2 − 2 g λ3 − 2 c g λ3 − g ∂ λ3 − c g ∂ λ3 + 8 g λ ξ − 4 c g λ ξ − 4 g λ2 ξ − 4 c g λ2 ξ + 4 g ∂ λ2 ξ − 4 c g ∂ λ2 ξ − 2 g ∂ λ3 ξ − 2 c g ∂ λ3 ξ − c g ∂2 λ3 ξ + 10 g λ2 ξ2 − 2 c g λ2 ξ2 − 2 g λ3 ξ2 − 2 c g λ3 ξ2 + 3 g ∂ λ3 ξ2 − c g ∂ λ3 ξ2 + 4 g λ3 ξ3 L ^ 2 µ2s L The denominators coincide, so we need look at numerators only. g = 9.8; a = 1; λ = 0.95; ∂ = 0.01; c = 0.1; µs = 0.8; slipEqn = [email protected] == Hµs [email protected] 86.6761 H1 − ξ2 L H2 − 0.39095 H1.0095 + 0.95 ξL + 0.95 H−1.99 + 6 ξL + 0.9025 H−0.02 − 1.97 ξ + 4 ξ2 LL ^ 2 == 0.64 H1.53795 + 31.9737 ξ + 68.4348 ξ2 + 33.6091 ξ3 L ^ 2 The relevant root is θSlip = [email protected]@[email protected]ξ ê. [email protected], ξD, _RealDDD; θSlip ê Degree 88.5037 So the hoop begins to slip at q = 88.5037 degrees. We use the solution q1 obtained in exercise 9 to determine the corresponding time tSlip . First we generate a plot to show that the desired time is near 1 second. [email protected]θSlip, θ1 @tD<, 8t, 0, 3<D; The Hopping Hoop 35 7 6 5 4 3 2 1 0.5 1 1.5 2 2.5 3 tSlip = t ê. [email protected]θ1 @tD == θSlip, 8t, 1<DP1T 0.987071 [email protected]θ_D := a 8θ + λ [email protected]θD, 1 + λ [email protected]θD< 8XSlip, YSlip< = [email protected]θSlipD 8XSlip, YSlip< = [email protected]@θ1 @tDD, tD ê. t → tSlip θSlip = θ ê. θ@tD → θSlip 82.49436, 1.02481< 83.34709, −3.10171< 3.26607 Exercise 15. Numerical Solution: Rolling WITH Slipping µk = 0.7; We employ a numerical integration scheme with fixed time step Dt. ∆t = 0.002; (One could devise a more sophisticated scheme utilizing an adaptive step size, but this is by far the simplest approach.) Our solution will consist of a table of values of the three coordinates X, Y, and q computed at the discrete times tSlip , tSlip + Dt, tSlip + 2 Dt, and so on. As a result, it is convenient to parameterize the evolution of the system in terms not of the time t, but rather in terms of the integer variable i, which counts the number of time steps completed. We define new variables relevant to this slipping 36 Project Intermath phase: x[i] and y[i] for the horizontal and vertical coordinates of the center of mass, vx[i] and vy[i] for the horizontal and vertical components of the center of mass velocity, and th[i] and thDot[i] for the rotation angle q and its derivative. The values of this quantities at the onset of slipping (i = 0) are given by [email protected], [email protected]< = 8XSlip, YSlip<; [email protected], [email protected]< = 8XSlip, YSlip<; [email protected], [email protected]< = 8θSlip, θSlip<; We define the height, vertical velocity, and vertical acceleration of the center of the hoop [email protected]_D := [email protected] − a λ [email protected]@iDD [email protected]_D := [email protected] + a λ [email protected]@iDD [email protected] n i [email protected]_D := n − g + a λ j [email protected]@iDD a H1 − λ2 + λ ∂L k y Hλ [email protected]@iDD + µk H1 + λ [email protected]@iDDLL + [email protected]@iDD [email protected] z z { where the expression for the acceleration is simply equation (17). Finally, to be able to start the integration procedure, we require the value of the normal force at the onset of slipping, as well as the value of the vertical acceleration of the center of mass one time step prior to the onset of slipping. Both quantities may be obtained from equation (13). We assume that the quantity is stored in normalForce from the work in exercise 11. normalForce 1.53795 + 31.9737 ξ + 68.4348 ξ2 + 33.6091 ξ3 H2.0095 + 1.9 ξL2 [email protected]ξ_D := [email protected] One begins each step in the integration by checking explicitly to determine whether the constraint is violated: [email protected] < a. If it is, the normal force n to increased to give [email protected] = a for that time step, and, using the revised value of n, the new values x[i + 1], y[i + 1], vx[i + 1], vy[i + 1] , th[i + 1], and thDot[i + 1] are computed for the next time step. The procedure continues until [email protected] is strictly greater than a. Of course, since all of these comparisons involve floating point values, they can be done only to a given tolerance 1 tolerance = ; 102 Thus, the termination condition actually takes the form yctr[i]-a ¥ tolerance The Hopping Hoop 37 At each time step, we compute the values x[i + 1], y[i + 1], vx[i + 1], [email protected] + 1], th[i + 1], and thDot[i + 1] for the next time step by assuming that the corresponding accelerations are constant over a time step. For x and y, whose accelerations depend only on n (which we update only once every time step), this is strictly true, and it is a very good approximation for th. The integration is accomplished in a For loop: n = [email protected]@θSlipDD; [email protected]−1D = [email protected]@θ1 @tSlip − ∆tDDD − g; ForAi = 0, [email protected] − a ≤ tolerance, i ++, H* Compute normal force required to prevent [email protected]+1D < a. *L [email protected] + [email protected] ∆t + [email protected] ∆t2 ê 2 < a, n = MaxA0, 1 i i −j j2 j j−a − a λ [email protected]@iDD + ∆t2 H−g + a λ [email protected]@iDD [email protected] L + 2 k k i 2i yy ∆t Ha λ [email protected]@iDD [email protected] + [email protected] + [email protected] zz zìj j∆t j j1 + λ {{ k k [email protected]@iDD yy Hµk H1 + λ [email protected]@iDDL + λ [email protected]@iDDLz zz zEE; 2 H1 + ∂ λ − λ L {{ H* Recalculate current accelerations to reflect the change in n. *L [email protected] = −µk n; [email protected] = n − g; [email protected] = n Hλ [email protected]@iDD + µk H1 + λ [email protected]@iDDLL; a H1 − λ2 + λ ∂L H* Compute new values of coordinates and their derivatives. *L [email protected] + 1D = [email protected] + [email protected] ∗ ∆t; [email protected] + 1D = [email protected] + [email protected] ∗ ∆t; [email protected] + 1D = [email protected] + [email protected] ∗ ∆t; [email protected] + 1D = [email protected] + [email protected] + [email protected] + 1DL ∗ ∆t ê 2; [email protected] + 1D = [email protected] + [email protected] + [email protected] + 1DL ∗ ∆t ê 2; [email protected] + 1D = [email protected] + [email protected] + [email protected] + 1DL ∗ ∆t ê 2; H* Estimate new value of n. *L n = [email protected], n + [email protected] − [email protected] − 1DLD E For future reference, we save the final values. H* End For loop *L 38 Project Intermath nsteps = i tHop = tSlip + nsteps ∆t; 8XHop, YHop< = [email protected], [email protected]<; 8XHop, YHop< = [email protected], [email protected]<; 8θHop, θHop< = [email protected], [email protected]<; 76 ° Finally, we generate the required table showing t, a q, and vxctr. [email protected]_D := [email protected] − a λ [email protected]@iDD [email protected] [email protected] 8tSlip + j ∗ ∆t, a ∗ [email protected], [email protected]<, 8j, 0, nsteps, 5<D, TableHeadings −> 9None, 9"Time", "a \ .\ θ ", "vx of hoop ctr"==E . Time 0.987071 0.997071 1.00707 1.01707 1.02707 1.03707 1.04707 1.05707 1.06707 1.07707 1.08707 1.09707 1.10707 1.11707 1.12707 1.13707 a θ 3.26607 3.36306 3.46729 3.5784 3.69599 3.81964 3.94886 4.08313 4.22189 4.36451 4.51032 4.65861 4.80861 4.9595 5.20052 8.57936 vx of hoop ctr 3.26607 3.36515 3.47351 3.59222 3.7223 3.86473 4.02037 4.18999 4.37417 4.57334 4.78768 5.01708 5.26115 5.51913 5.82928 7.72878 The Hopping Hoop 39 Exercise 16. Airborne Hoop Setting [email protected] = f @tD = 0 in (5), (6), and (7), we see that for t ¥ tHop , the equations of motion reduce to – [email protected] = 0 – [email protected] = -g – [email protected] = 0 For the given the initial conditions at t = tHop, [email protected] D = XHop [email protected] D = YHop [email protected] D = qHop ° ° [email protected] D = XHop ° ° [email protected] D = YHop ° ° [email protected] D = qHop the equations of motion have the solution ° [email protected] = XHop + XHop Ht - tHop L ° [email protected] = YHop + YHop Ht - tHop L - ÅÅÅÅ12 g Ht - tHop L2 ° [email protected] = qHop + qHop Ht - tHop L Exercise 17. Time of Impact In order to see what’s going on, we start by plotting the height of the center of the hoop. [email protected]_D := 1 YHop + YHop Htt − tHopL − g Htt − tHopL2 − a λ [email protected]θHop + θHop Htt − tHopLD 2 [email protected]@ttD, 8tt, tHop, tHop + 0.1<, AxesLabel → 8"t", "yCtr \\\ Hoop"<D; yCtr Hoop 1.1 1.08 1.06 1.04 1.02 1.14 1.16 1.18 0.98 1.22 1.24 t 0.96 Now we find tImpact. tImpact = tt ê. [email protected]@ttD == a, 8tt, tHop + 0.5, tHop + 1<DP1T 40 Project Intermath 1.22917 Exercise 18. Putting It All Together We combine the results of exercises 9, 13, and 14 into a single comprehensive solution . . . [email protected]θSoln, rCMSolnD [email protected]_ ê; 0 ≤ tt < tSlipD = [email protected]; θ[email protected]_ ê; 0 ≤ tt < tSlipD = θ1 @ttD; [email protected]_ ê; tSlip ≤ tt < tHopD = 8 [email protected] [email protected] ∆t, [email protected], [email protected]<<, 8j, 0, nsteps<[email protected] − tSlipD, [email protected]@8j ∆t, [email protected], [email protected]<<, 8j, 0, nsteps<[email protected] tt − tSlipD<; θ[email protected]_ ê; tSlip ≤ tt < tHopD = [email protected] [email protected] ∆t, [email protected], [email protected]<<, 8j, 0, nsteps<[email protected] − tSlipD; [email protected]_ ê; tt >= tHopD := 8XHop, YHop< + 8XHop, YHop< Htt − tHopL + 80, −g< ê 2 Htt − tHopL2 θ[email protected]_ ê; tt > tHopD := θHop + θHop Htt − tHopL . . . and use it to create a plot. The Hopping Hoop 41 [email protected]_D := [email protected] − λ a [email protected]θ[email protected], [email protected]θ[email protected]< [email protected]_D := [email protected] + a [email protected]θ[email protected], [email protected]θ[email protected]< ntSteps = 40; deltat = tImpact ê ntSteps; ptsCM = [email protected], [email protected], tImpact, deltatDD; ptsCtrHoop = [email protected], [email protected], tImpact, deltatDD; ptsMass = [email protected], [email protected], tImpact, deltatDD; lines = [email protected], [email protected], [email protected], ptsMass<DD<, [email protected], [email protected], ptsMassD<, [email protected], [email protected], ptsCtrHoopD<<; [email protected]@ttD, 8tt, 0, tImpact<, Compiled → False, Epilog → lines, AspectRatio → Automatic, PlotRange → 8−0.2, 2.2<, Frame → True, Axes → 8True, None<D; 2 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 3 42 Project Intermath About the Authors Stan Wagon studied at McGill (BSc) and Dartmouth (PhD), and is now a professor at Macalester College. He has written many articles and books about mathematics, and has received three prizes for his writing: The L. R. Ford award, the Trevor Evans Award, and the Chauvenet Prize. His books include The Banach-Tarski Paradox, Mathematica in Action, and Which Way Did the Bicycle Go?. He recently appeared in Ripley's Believe-It-Or-Not for the square wheel bicycle that he built, and which is on display at Macalester. Every July he teaches a Mathematica course in the mountains of Colorado, and that is how the hopping hoop project got its start: he purchased a hula hoop and he and Tim Pritchett tried to see the bounce. The hoop was too flexible! Other interests include rock-climbing, wilderness skiing, mountain climbing, and snow sculpture. In May 2000, he and three others skied to over 19000 feet on Mt. Logan during a 17-day expedition. In January 2000, his snow sculpture team took second place at an international snow sculpting competition (images at http://www.stanwagon.com). Tim Pritchard graduated with distinction from the Integrated Science Program at Northwestern University. He subsequently attended the Georg-August-Universität The Hopping Hoop 43 Göttingen, where he received a vordiplom in Mathematics, and the University of California Berkeley, from which he holds MA and Ph.D. degrees in Physics. Tim is currently a professor at the U.S. Military Academy at West Point, where he divides his time between teaching physics to cadets and pursuing his own research interests in nonlinear optics. An avid cyclist, Tim recently spent a month touring northern Italy on his (round-wheeled!) bicycle.