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7/27/2019 Hopping Hoop http://slidepdf.com/reader/full/hopping-hoop 1/44 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 Hoo

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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

 TheHopping Hoo

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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.

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Contents

1. Introduction

2. Problem Statement and Assumptions

3. Overview of Solution Procedure

4. Constrained Motion: Rolling Without Slipping5. Some Basic Physics

6. Removing One Constraint: Rolling With Slipping

7. The Hop

8. Postscript: A Final Suprise

1. IntroductionTape 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.

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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 themass of the hoop itself is negligible relative to that of the attached object.

x

y

l M 

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.

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* Exercise 1. Center of Mass

Consider a system of N  particles having masses m1 , m2 , …, mN  , and position

vectors r”1 , r”2 , …, r”N . The position R”÷÷ of the center of mass of the system is defined by

R”÷÷

=⁄i=1

N  mi r”iÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ⁄i=1N  mi

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 inthe 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.

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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 thehoop 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.

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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 HX @tD , Y @tDL of the center of mass of the system — indicated by the small dot

in figure 2 — and by the angle q @tD through which the hoop has turned.

Of course, X @tD , Y @tD , and q @tD 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 tochoose the horizontal and vertical coordinates of the center of the hoop, quantities that

we will denote by x@tD and  y@tD , respectively. However, as we shall see presently, the

dynamics of the system will serve to make the choice of coordinates X @tD , Y @tD , and q @tDparticularly convenient.

Because of our assumptions, the three coordinates X @tD , Y @tD , and q @tD 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 HX @tD - x@tD L2

+ HY @tD - y@tD L2

= l2

 a2

 , or equivalently,

(1)

X @tD - x@tD = a l sin@q HtLD

Y @tD - y@tD = a l cos@q HtLD

2. Since the supporting surface is also assumed to be rigid, the vertical height of the

center of the hoop must satisfy

(2) y@tD = a

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

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(3)x@tD = aHq @tD - q 0L where q 0 = q @0D

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 acurtate 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

(4)

X @q D = aHq - q 0 + l sin@q DL

Y @q D = aH1 + l cos@q DL

0 π 2 π 3 π 4 π

0

1

2

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.

0 π2

π 3 π

2

2 π 5 π

2

3 π

0

1

2

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.

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Exercise 2. Cycloid ConstraintShow 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  , q L , 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 one-dimensional curve shown in figure 6.

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0

5

10

X

0

1

Y

0

5

10

θ

0

1

Y

0

5

10

θ

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 completedescription of the hoop at an instant in time requires the six values HX  , X 

° , Y  , Y 

° , q  , q 

° L , 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 HoopThe 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.

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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.

M g  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  y-

components 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

10 Project Intermath

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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 —  y@tD ¥ 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 LawNewton’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): ⁄ F”÷external = 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 X @tD and Y @tD , the horizontal and vertical coordinates of the

center of mass of the system.

(5)X – @tD =  f 

(6)Y – @tD = n - g

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 rotationaxis 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.

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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 anglethrough which the hoop has rotated.

(7)q –@tD = MÅÅÅÅÅÅÅ

I   H nHX @tD - x@tDL -  f Y @tD L

* 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 X ° @tD , of equation (6) by  M Y 

° @tD ,and of equation (7) by I q 

° @tD. 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.

(8)1ÅÅÅÅ2

  MIX ° @tD2 + Y 

° @tD2M + 1ÅÅÅÅ2

  I q ° @tD2 + M   g  Y @tD = constant

Exercise 6. Energy Conservation Equation in

Terms of q Only

Using the cycloid constraint (4), eliminate the coordinates X 

@t

Dand Y 

@t

Din favor

of q [t] from the equation of energy conservation (8) and so obtain:

(9)a   g   M H1 + l cos@q HtLDL + 1ÅÅÅÅ2

@I + a2 M H1 + l2 + 2 l cos@q HtLD LD q ° HtL2 = constant

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* 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  a2H1 - l2L. (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

(10)I = M  a2H1 - l2 + l ¶L

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 q 0 = q @0D; the initial translational speed x° 0 of the center of the hoop;

and the initial angular speed of rotation q °0 = q 

° @0D. In the present case, however, where

we assume that the hoop rolls without slipping , the latter two parameters are notindependent, but are related by x° 0 = a q 

°0. We will thus specify the initial state of the

weighted hoop by giving q 0 and q °0.

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Exercise 8. Mechanical EnergyThe 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 thepoint of contact (q 0 = 0) and show that E =  M  a   g @H1 + lL H1 + cL + 1ÅÅÅÅ

2  ¶ l cD

where we have introduced the dimensionless quantity c = a q °02 ê g.

Substitute this result into (9) and solve the resulting equation for q ° @tD to obtain

the following first-order differential equation describing the time evolution of the

weighted hoop constrained to roll entirely

without slipping.

(11)q 

°

@tD = "##### g

ÅÅÅÅa   "################################ ###############cH1+ l+¶ lê2L+ lH1-Cos@q HtLD L

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 + ¶ lê2+ lCos@q HtLD

Exercise 9. Motion on the Cycloid: Numerical

Solution for a Typical CaseUse a numerical differential equation solver (such as the Mathematica function

 NDSolve) to obtain a numerical solution to (11) subject to the initial condition q @0D = 0

for 0 § t § 3 (sec). Use the following values for the parameters:  g = 9.8

Hmeters

êsec2

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 ofthe 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 q @0D = 0.

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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 = cos@q D.

(12)

 f @ x D = 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2+¶ l+2 l x L2  I g lè!!!!!!!!!!!!

1 - x 2 H2 + l H-2 + ¶ + 6 x L -

c H2 + H2 + ¶L lL H1 + ¶ l + l x  L + l2 H-2 ¶ - 2 x + 3 ¶ x + 4 x 2LL

(13)

n@ x D = 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2+¶ l+2 l x L2 

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 gl x -

4 c gl x -

4 gl2 x -

4 c gl2 x +

4 g¶ l2 x -

4 c g¶ l2 x -

2 g¶ l3 x -

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 x 2 + 4 g l3 x 3L

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.

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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 verticalcomponent 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 n@tD ¥ 0. The preceding definition of kinetic friction and the rigid

hoop constraint (3) allow us to recast the equations of motion in the form:

(14)Y – @tD = n - g(15)X – @tD = -mk n

(16)q –@tD = MÅÅÅÅÅÅÅ

I   a @l sin @q HtLD + mk H1 + l cos@q HtLDLD n

The normal force n@tD 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

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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

 y@tD ¥ a. This constraint must be enforced explicitly at each time step of the integration:

one checks at each stage whether  y@tD < a , and if this is the case, one adjusts n to give

 y@tD = 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 y@tD is strictly greater than a.

The height of the center of the hoop is y@tD = Y @tD - a l cos @q HtLD.

If one differentiates twice with respect to time and substitutes for Y – @tD and q –@tDfrom equations (6) and (16) respectively, one obtains

(17) y– @tD = n - g + a lIsin @q HtLD MÅÅÅÅÅÅÅI   a nH l sin @q HtLD + mkH1 + l cos@q HtLD L L + cos@q HtLD q 

° @tD2M

In the solution procedure just described, n is updated once every Dt using the

constraint  y@tD ¥ 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

 y@t + DtD = y@tD + y° @tD Dt + 1ÅÅÅÅ2

  y– @tD HDtL2= y@tD + 1ÅÅÅÅ

2Dt H y° @tD +  y° @t + DtD L

One may view the above expressions for y° @t + DtD and y@t + 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  y@t + DtD represents a simple but significant improvement over

the basic Euler method.)

If  y@t + DtD fails to be greater than or equal to a , then one sets  y@t + DtD = a andsolves the above equation for n.

n =2 Aa- y@tD-Dt y° @tD-

1ÅÅÅÅÅ2

D t2 I- g+a l cos@q HtLD q ° @tD2ME

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅDt2 A1+ Ma2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

I l sin@q HtLD Hl sin@q HtLD+H1+l cos@q HtLDL mkLE

The procedure may be represented by the following pseudocode:

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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 y@tD 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  y@t D - 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 ofa 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 Dtand 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

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is in fact the case. For instance, let us take d ê a = 10-4.

Dt  HsecL tHop HsecL nsteps

0.002 1.01507 14

0.0005 0.99757 210.0001 0.992471 54

0.00005 0.991621 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  y@tD be strictly greater than a , as  y@tD - 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 (  y@tD ) 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 withd. We obtained the following resultsfor Dt = 0.002:

d ê a t Hop nsteps hop height

0.01 1.39 76 0.097

0.001 1.049 31 0.0011

0.0001 1.015 14 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 inconflict 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

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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: n@tD =  f @tD = 0.

Exercise 14. Airborne Hoop

Assuming that the hoop is  flying through the air , write down the equations ofmotion for the center of mass coordinates X @tD and Y @tD and for the angular coordinate

q @tD. 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 X Hop , Y Hop , q Hop , X °Hop , Y 

°Hop , and q 

°Hop , solve the equations of motion to

obtain expressions for X @tD , Y @tD and q @tD 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 thecenter 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 n@tDacquires 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 thesupporting surface.

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Exercise 16. Putting It All TogetherAssemble 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 untiltImpact.

8. Postscript: A Final Surprise

In computing the motion of the asymmetrically weighted hoop, we took q 0 = 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 q 0 = 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 q 0 = p (or

more generally, to the case where q 0 assumes an arbitrary value), but if you do so

correctly you will find that the hoop begins to slip at a critical angle q Slip = 210.7 °, atwhich point the angular speed of rotation is q 

°Slip = 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.

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-1 0 1 2 3 4

0

0.5

1

1.5

2

2.5

3

3.5

Figure 8. Motion of the hoop for q 0 = 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 thatthe center of mass (black line) follows the parabolic trajectory

characteristic of an object falling freely under the influence of gravity.

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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.

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Notes to the InstructorThis 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. Thosestudents 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 aclass 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. Onthe 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

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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 MassIn 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: r”h = 0 , and that of the attached object is given by r”m = a r` . (Here, r

` 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 r` 

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

R”÷÷

=H1 - lL M ÿ 0 + a M l r

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ M H1 - lL +  M l

= l a r` 

Since R”÷÷

∂ r`  , 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 ConstraintThe rigid hoop constraint (1) relates the coordinates of the center of the hoop to

the coordinates of the system center of mass.

X @tD - x@tD = a l sin q @tD

Y @tD - y@tD = a l cos q @tDSubstituting for x@tD from the “no slipping constraint,” equation (3),

 x@tD = a Hq @tD - q 0L ,

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yields the given result for X @tD : X @q D = aHq - q 0 + l sin@q DL , while substituting for  y@tDfrom the “normal force constraint,” equation (2),  y@tD = a , yields the given result for

Y @tD : Y @q D = aH1 + l cos@q DL.

Exercise 3. Center of Mass Equations of MotionReferring 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 MotionOf 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 becausethe torque arm is zero.) The torque resulting from the normal component of the contact

force is N H X @tD - x@tD L , and this torque acts to increase q . The torque arising from the

tangential component of the contact force is given by -F  Y @tD , and this torgue acts to

decrease the angle q  , which is the reason for the minus sign. Summing these two

torques, we obtain, F Y @tD - N HX @tD - x@tD L = I q – , or, with F = M f  and N = M n ,

q –

= 1ÅÅÅÅI   8 M f Y @tD -  M  n HX @tD - x@tD L < , which is (7).

Exercise 5. Derivation of the Equation of Energy

ConservationTaking our cue from the problem statement, we multiply both sides of (5) by

 M X ° @tD , both sides of (6) by M Y 

° @tD , and both sides of (7) by I  q ° @tD. This gives

 M X – @tD  X 

° @tD = M   f X ° @tD

 M  Y – @tD Y 

° @tD =  MHn - gL Y ° @tD

I  q –@tD q 

° @tD = M 8 nHX @tD - x@tD L -  f Y @tD < q ° @tD

The left hand side of each of these equations can be written as a total derivative:

(18)

dÅÅÅÅÅÅÅÅÅ

d t 

9 MÅÅÅÅÅÅÅÅÅ

2  X 

°

@t

D2

== M f X 

°

@t

Dd

ÅÅÅÅÅÅÅÅÅd t

 9 MÅÅÅÅÅÅÅÅÅ2

  Y ° @tD 2= = M Hn - gL Y 

° @tDd

ÅÅÅÅÅÅÅÅÅd t

 9 I ÅÅÅÅÅ2

q ° @tD 2= = M 8 nHX @tD - x@tDL -  f Y @tD <

Using the chain rule to differentiate (4) with respect to time, we get

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(19)

X ° @tD = X £@q @tDD q 

° @tD = H1 + l cos@q D L q ° @tD = Y  q 

° @tD

Y ° @tD = Y £@q @tDD q 

° @tD = - lsin@q D q ° @tD = - HX - xL q 

° @tDWe now add the three equations (18) and use (19) to eliminate the cancelling

pairs of terms involving n and  f . The result is:d

ÅÅÅÅÅÅÅÅÅd t

 9 MÅÅÅÅÅÅÅÅÅ2

  X ° @tD 2= +

dÅÅÅÅÅÅÅÅÅd t

 9 MÅÅÅÅÅÅÅÅÅ2

  Y ° @tD 2= +

dÅÅÅÅÅÅÅÅÅd t

 9 I ÅÅÅÅÅ2

q ° @tD 2= = - M g Y 

° @tDor,

dÅÅÅÅÅÅÅÅÅd t

 9 MÅÅÅÅÅÅÅÅÅ2

  X ° @tD 2 +

 MÅÅÅÅÅÅÅÅÅ2

  Y ° @tD 2 +

I ÅÅÅÅÅ2

q ° @tD 2 +  M   g Y @tD = = 0

Integration yields the energy conservation equation (8).

Exercise 6. Energy Conservation Equation in

Terms of q OnlyDifferentiating equations (4) with respect to time, we obtain

X ° @tD = a H1 + l cos@q HtLD L q 

° @tD

Y ° @tD = - a l sin@q HtLD q 

° @tDIn terms of q  , the translational kinetic energy of the center of mass is thus given by

1ÅÅÅÅ2

 M IX ° @tD2 + Y 

° @tD2M = 1ÅÅÅÅ2

 M  a2 H1 + l2 + 2 l cos@q HtLD L q ° @tD2

and the gravitational potential energy is

 M g Y 

@t

D= M g  a

H1 + l cos

@q 

Ht

LD L.

Substituting for these terms in (8), one obtains the required result (9).

Exercise 7. Moment of Inertia of the

Asymmetrically Weighted HoopThe 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 I 0 about a

parallel axis passing through the center of mass of the body plus the product of themass m of the body and the square of the distance R between the two axes:

I = I 0 + 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

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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

I Hoop = MHoop a2 +  MHoop R2

= H1 - lL  M  a2 + H1 - lL  MHl aL2= H1 - lL  M  a2 H1 + l2L

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 I Object = 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 = I Hoop + I Object

= H1 - lL  M a2 H1 + l2L + l MH1 - lL2 a2

=  M a2

H1 + l2

- l - l3

+ l - 2 l2

+ l3

L=  M a2 H1 - l2L

Exercise 8. Mechanical EnergySetting q @tD = 0 and q 

° @tD = q °0 in (9) gives immediately

constant = E = a M I g H1 + lL + a H1 + H1 + ¶ÅÅÅÅ2L lL q 

°02M

One then substitutes g c for a q °02 , and so obtains the result sought:

E =  M a g HH1 + lL H1 + cL + 1 ê2 ¶ l cLThus, for the weighted hoop rolling without slipping with q 

@0

D= 0 and q 

°

@0

D=

è!!!!!!!!!!! g c

êa ,

the equation of energy conservation (9) takes the form

a g M H1 + l Cos@q @tDDL + 1ÅÅÅÅ2

  HI + a2 MH1 + l2 + 2 l Cos@q HtLD LL q ° @tD2 =

 M a g @H1 + lL H1 + cL + 1ÅÅÅÅ2

  ¶ l cDSolving for q 

° @tD , 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 CaseWhen 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 q 1 in the form of such

an interpolating function

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g = 9.8; a = 1; λ = 0.95; ∂ = 0.01; c = 0.1; t1 = 3.0;

θ1 = θ ê. NDSolveA

@t

D==

$%%%%%%g

a

 

&'''''''''''''''''''''''''''''''' '''''''''''''''''''''''''''''''' ''''''''''''c

H1 + λ +

∂ λ

2

L+ λ

H1 − Cos

@t

DDL

1 +∂ λ

2 + λ Cos@θ@tDD, θ

@0

D== 0

=,

θ, 8t, 0, t1<EP1TInterpolatingFunction@880., 3.<<, <>D

We can then use the function q 1 to compute the positions of the center of mass, the

center of the hoop, and the attached object.

rCM @t_ D := a 8θ1@tD + λ Sin@θ1@tDD, 1 + λ Cos@θ1@tDD<rCtrHoop

@t_ 

D:= rCM 

@t

D− λ a

8Sin

@θ1

@t

DD, Cos

@θ1

@t

DD<rMass@t_ D := rCtrHoop@tD + a 8Sin@θ1@tDD, Cos@θ1@tDD<The following Mathematica code generates the plot.

ntSteps = 30;

∆t = t1 êntSteps;

 ptsCM  = Map@rCM, Range@0, t1, ∆tDD;

 ptsCtrHoop = Map@rCtrHoop , Range@0, t1, ∆tDD;

 ptsMass = Map@rMass , Range@0, t1, ∆tDD;

lines =

[email protected], Map@Line, Transpose@8 ptsCtrHoop, ptsMass<DD<,8 AbsolutePointSize@4D, Map@Point, ptsMassD<,8 AbsolutePointSize@2D, Map@Point, ptsCtrHoopD<<;

ParametricPlot@Evaluate@rCM @ttDD, 8tt, 0, t1<,

Epilog → lines,

 AspectRatio → Automatic,

PlotRange → 8−0.1, 2.2<, Axes → None, Frame −> NoneD;

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Exercise 10. Motion on the Cycloid: Analytic

Solution for a Special CaseIn the case of a massless hoop (l = 1) with attached point mass (¶ = 0), the

equation of motion (9) reduces to

q ° @tD = "##### g

ÅÅÅÅa

  "################ ############2 c+ 1- cos@q @tDDÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ1 + cos@q @tDD  , or dq ÅÅÅÅÅÅÅ

dt= "##### g

ÅÅÅÅa

 "################ ######## ###c+ sin2A q 

ÅÅÅÅÅ2E

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅcosA q 

ÅÅÅÅÅ2E

where we have made use of the trigonometric identities 1 + cos@q D = 2 cos2@q ê2D and

1 - cos@q D = 2 sin2@q ê2D. 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.

Clear

@g, a, c

DthetaIntegral = ‡ 0

θ@tD Cos@θê2Dè!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!c + H Sin@θê2D L2

θ

−2ikjjjjjjjLog@è!!!cD − LogASinA θ@tD

2

E + $%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% %%%%c + SinA θ@tD2

E2 Ey{zzzzzzz

The other side of the equation is of course just è!!!!!!!! g ê a t.

integratedEqn =ikjjjjjj

thetaIntegral

2==

$%%%%%%ga

 t2

êêSimplify

y{zzzzzz

LogASinA θ@tD2

E + $%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% %%%%c + SinA θ@tD2

E2 E ==12

ikjjjjj$%%%%%%g

at + Log@cDy{

zzzzzeqnsimpler = Simplify@ Map@Exp, integratedEqnDDSinA θ@tD

2

E + $%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% %%%%c + SinA θ@tD2

E2 == 12I"######g

at+Log@cDM

sinThetaOver2 = Sin

@t

D

2 E ê. Solve

Aeqnsimpler, Sin

@t

D

2 EEP1

T12

è!!!c

−12"######g

a

t J−1 + "######g

a

tNsimplerSinThetaOver2 =

PowerExpand @Simplify@ExpToTrig@sinThetaOver2DDD

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è!!!c SinhA è!!!

g t2è!!!aE

θ

@t

D ê. Solve

ASin

Aθ@tD

2

E== simplerSinThetaOver2, θ

@t

DEP1

T2 ArcSinAè!!!c SinhA è!!!g t2è!!!aEE

The above expression for q @tD 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

ConstraintWe 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).

X @t_ D := a Hθ@tD + λ Sin@θ@tDDLY@t_ D := a H1 + λ Cos@θ@tDDL

We’ve already determined that under these conditions, q ° @tD must satisfy (11).

Clear@g, a, λ, ∂, cD;

θ = "#####ga è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!

2 c + 2 λ + 2 c λ + c ∂ λ − 2 λ Cos

@t

DDè!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!2 + ∂ λ + 2 λ Cos@θ@tDD"#####g

a

è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!2 c + 2 λ + 2 c λ + c ∂ λ − 2 λ Cos@θ@tDD

è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!2 + ∂ λ + 2 λ Cos@θ@tDD

We will need an expression for the second derivative q –@tD = dÅÅÅÅÅÅ

dtq ° @tD , so we differentiate

one more time.

θ–

= Simplify

@D

, t

D ê. θ

@t

D→ θ

DH1 + cL g λ H2 + H2 + ∂L λL Sin@θ@tDDa H2 + ∂ λ + 2 λ Cos@θ@tDDL2

From (5) we see that the required force of constraint  f @tD is just given by d2ÅÅÅÅÅÅÅÅdt2

  X @tDforceOfConstraint = D@X @tD, 8t, 2<D

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a Hθ@tD + λ H−Sin@θ@tDD θ@tD2 + Cos@θ@tDD θ@tDLLWe substitute to eliminate the time derivatives of q  and obtain an expression in terms

of  x = cos@q D (//. abbreviates ReplaceRepeated , which causes the substitutions to

 be made until there is no further change: twice in this case).forceOfConstraint = Simplify@

forceOfConstraint êê. 8θ@tD → θ–

, θ@tD → θ

, θ@tD → ArcCos@ξ D<D1

H2 + ∂ λ + 2 λ ξL2 

Ig λè!!!!!!!!!!!!!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 ξ2LLM

In the same way, (6) guarantees that n@tD is just g +

d2

ÅÅÅÅÅÅÅÅdt2  Y @tDnormalForce = g + D@Y@tD, 8t, 2<Dg + a λ H−Cos@θ@tDD θ@tD2 − Sin@θ@tDD θ@tDLnormalForce =

Together@normalForce ê. θ@tD → θ– ê. θ@tD → θ

ê. θ@tD → ArcCos@ξ DD1

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 ξ3LExercise 12. Why the Hoop Must Slip

Rearranging (1), we obtain the following expression for the vertical height of

the center of the hoop

 y@tD = Y @tD - a l cos @q HtLDDifferentiation with respect to time yields

 y° @tD = Y ° @tD + a l sin @q HtLD q 

° @tDThe first term, Y 

° @tD , is the speed at which the center of mass falls under the

influence of gravity. The second term, a l sin @q HtLD q ° @tD , 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

32 Project Intermath

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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

HoopWith 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

 f @ x D =g"######## ####1-x 2 H4 x+4 x 2-4 c H1+x LL

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2+2 x L2 =g H x-cL"######## ####

1-x 2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

 x+1 ,

while the corresponding expression (13) for the normal force reduces to

n@ x D = 4 g x 3

+8 g x 2

+4 g x-4 c g-8 c g x-4 c g x 2

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2 x+2L2 = g H x - cL ,where c = a q 

°02 ê g. Alternately, since x = cos@q D ,

 f @q D = gHcos@q D - cL tan@q ê 2Dn@q D = gHcos@q D - cL

Slipping occurs at the minimum angle q for which »  f @q D » = ms » n@q D ». Since 0 § c < 1,

this minimum angle will necessarily occur in a regime where both  f  and n are non-

negative and the absolute value signs may be dropped:  f @q D = ms n@q D. One may see by

inspection that this equation has two possible solutions:

tan@q ê2D = ms

and

cos@q D - c = 0

Thus

q Slip = Min(arccos@cD , 2 arctan@ msD)

Exercise 14. Onset of Slipping for a Typical CaseThe hoop begins to slip at the minimum angle for which

H f 

@cos q 

D L2 =

H ms n

@cos q 

D 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

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1H2 + ∂ λ + 2 λ ξL4 

Hg2 λ2 H1 − ξ2L H2 + λ H−2 + ∂ + 6 ξL − c H2 + H2 + ∂L λL H1 + ∂ λ + λ ξL +

λ2

H−2 ∂ − 2 ξ + 3 ∂ ξ + 4 ξ2

LL^2

L==

1

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 ξ3L^2 µs2L

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 =

 Numerator@forceOfConstraintD2

==

s Numerator@normalForceDL2

86.6761 H1 − ξ2L H2 − 0.39095 H1.0095 + 0.95 ξL +

0.95 H−1.99 + 6 ξL + 0.9025 H−0.02 − 1.97 ξ + 4 ξ2LL^2 ==

0.64 H1.53795 + 31.9737 ξ + 68.4348 ξ2 + 33.6091 ξ3L^2The relevant root is

θSlip = Min@ ArcCos@Cases@ξ ê. NSolve@slipEqn, ξ D, _RealDDD;

θSlipêDegree

88.5037

So the hoop begins to slip at q = 88.5037 degrees. We use the solution q 1 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.

Plot@8θSlip, θ1@tD<, 8t, 0, 3<D;

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0.5 1 1.5 2 2.5 3

1

2

3

4

5

6

7

tSlip = t ê. FindRoot@θ1@tD == θSlip, 8t, 1<DP1T0.987071

cycloid @θ _ D := a 8θ + λ Sin@θD, 1 + λ Cos@θD<8X Slip, YSlip< = cycloid @θSlipD8X 

Slip, Y

Slip< = D@cycloid @θ1@tDD, tD ê. t → tSlip

θ

Slip = θ ê. θ@tD → θSlip

82.49436, 1.02481<

83.34709, −3.10171

<3.26607

Exercise 15. Numerical Solution: Rolling WITH 

Slippingµk = 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

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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

8x@0D, y@0D< = 8X Slip, YSlip<;8vx@0D, vy@0D< = 8X 

Slip, Y

Slip<;8th@0D, thDot@0D< = 8θSlip, θ

Slip<;

We define the height, vertical velocity, and vertical acceleration of the center of the

hoop

yctr@i_ D := y@iD − a λ Cos@th@iDDvyctr@i_ D := vy@iD + a λ Sin@th@iDD thDot@iDayctr

@i_ 

D:= n − g + a λ

ikjj

Sin

@th

@i

DD 

na

H1 − λ2 + λ ∂

Hλ Sin@th@iDD + µk H1 + λ Cos@th@iDDLL + Cos@th@iDD thDot@iD2y{zzwhere 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 ξ3H2.0095 + 1.9 ξL2nCyc@ξ  _ D := Evaluate@normalForceD

One begins each step in the integration by checking explicitly to determine whether the

constraint is violated: yctr@iD < a. If it is, the normal force n to increased to give

yctr@iD = 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 yctr@iD is strictly greater than a. Of course, since all ofthese comparisons involve floating point values, they can be done only to a given

tolerance

tolerance =1

102

;

Thus, the termination condition actually takes the form

yctr[i]-a ¥ tolerance

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At each time step, we compute the values x[i + 1],  y[i + 1], vx[i + 1], vy@i + 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 = nCyc@Cos@θSlipDD;

ay@−1D = nCyc@Cos@θ1@tSlip − ∆tDDD − g;

ForAi = 0, yctr@iD − a ≤ tolerance, i++,

H* Compute normal force required to prevent yctr@i+1D < a. *LIfAyctr@iD + vyctr@iD ∆t + ayctr@iD ∆t2 ê2 < a, n = MaxA0,

ikjj2

ikjj−a − a λ Cos

@th

@i

DD+

12

∆t2

H−g + a λ Cos

@th

@i

DDthDot

@i

D2

L+

∆t Ha λ Sin@th@iDD thDot@iD + vy@iDL + y@iDy{zzy{zz ì ikjj∆t2 ikjj1 + λ

Sin@th@iDDH1 + ∂ λ − λ2L Hµk H1 + λ Cos@th@iDDL + λ Sin@th@iDDLy{zzy{zzEE;

H* Recalculate  current  accelerations  to  reflect  the  change  in  n. *Lax@iD = −µk n;

ay@iD = n − g;

thDblDot

@i

D=

na H1 − λ2 + λ ∂L  Hλ Sin@th@iDD + µk H1 + λ Cos@th@iDDLL;

H* Compute new values of coordinates and their derivatives. *Lvx@i + 1D = vx@iD + ax@iD ∗ ∆t;

vy@i + 1D = vy@iD + ay@iD ∗ ∆t;

thDot@i + 1D = thDot@iD + thDblDot@iD ∗ ∆t;

x@i + 1D = x@iD + Hvx@iD + vx@i + 1DL ∗ ∆tê2;

y@i + 1D = y@iD + Hvy@iD + vy@i + 1DL ∗ ∆tê2;

th

@i + 1

D= th

@i

D+

HthDot

@i

D+ thDot

@i + 1

DL∗ ∆t

ê2;

H* Estimate  new  value  of   n. *Ln = Max@0, n + Hay@iD − ay@i − 1DLD E H* End  For  loop *L

For future reference, we save the final values.

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nsteps = i

tHop = tSlip + nsteps ∆t;8X Hop, YHop< = 8x@nstepsD, y@nstepsD<;

8X 

Hop, Y

Hop

<=

8vx

@nsteps

D, vy

@nsteps

D<;

8θHop, θ

Hop< = 8th@nstepsD, thDot@nstepsD<;

76

Finally, we generate the required table showing t , a q ° , and vxctr.

vxctr@i_ D := vx@iD − a λ Cos@th@iDD thDot@iDTableForm ATable@

8tSlip + j ∗ ∆t, a ∗ thDot

@j

D, vxctr

@j

D<,

8j, 0, nsteps, 5<D,

TableHeadings −> 9 None, 9"Time", "a θ\  .\

", "vx of hoop ctr"==ETime a θ

.

vx of hoop ctr

0.987071 3.26607 3.26607

0.997071 3.36306 3.36515

1.00707 3.46729 3.47351

1.01707 3.5784 3.59222

1.02707 3.69599 3.7223

1.03707 3.81964 3.86473

1.04707 3.94886 4.02037

1.05707 4.08313 4.18999

1.06707 4.22189 4.37417

1.07707 4.36451 4.57334

1.08707 4.51032 4.78768

1.09707 4.65861 5.01708

1.10707 4.80861 5.26115

1.11707 4.9595 5.51913

1.12707 5.20052 5.82928

1.13707 8.57936 7.72878

38 Project Intermath

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Exercise 16. Airborne HoopSetting n@tD =  f @tD = 0 in (5), (6), and (7), we see that for t ¥ tHop , the equations of

motion reduce to

X – @tD = 0  Y – @tD = - g

q –@tD = 0

For the given the initial conditions at t = tHop ,

X @tHopD = X Hop X ° @tHopD = X 

°Hop

Y @tHopD = Y Hop Y ° @tHopD = Y 

°Hop

q @tHopD = q Hop q ° @tHopD = q 

°Hop

the equations of motion have the solution

X @tD = X Hop + X °HopHt - tHopL

Y @tD = Y Hop + Y °HopHt - tHopL - 1ÅÅÅÅ2  g Ht - tHopL2

q @tD = q Hop + q °HopHt - tHopL

Exercise 17. Time of ImpactIn order to see what’s going on, we start by plotting the height of the center of

the hoop.

yCH@tt_ D :=

YHop + Y

Hop 

Htt − tHop

L−

12

Htt − tHop

L2 − a λ Cos

@θHop + θ

Hop

Htt − tHop

LDPlot@yCH@ttD, 8tt, tHop, tHop + 0.1<,

 AxesLabel → 8"t", "yCtr \\\ Hoop"<D;

1.14 1.16 1.18 1.22 1.24t

0.96

0.98

1.02

1.04

1.06

1.08

1.1

yCtr Hoop

Now we find tImpact.

tImpact = tt ê. FindRoot@yCH@ttD == a, 8tt, tHop + 0.5, tHop + 1<DP1T

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1.22917

Exercise 18. Putting It All Together

We combine the results of exercises 9, 13, and 14 into a single comprehensivesolution . . .

Clear@θSoln, rCMSolnDrCMSoln@tt_ ê; 0 ≤ tt < tSlipD = rCM @ttD;

θSoln@tt_ ê; 0 ≤ tt < tSlipD = θ1@ttD;

rCMSoln@tt_ ê; tSlip ≤ tt < tHopD = 8Interpolation@

Table@8j ∆t, 8x@jD, vx@jD<<, 8j, 0, nsteps<DD@tt − tSlipD,

Interpolation

@Table

@8j ∆t,

8y

@j

D, vy

@j

D<<,

8j, 0, nsteps

<DD@tt − tSlipD<;

θSoln@tt_ ê; tSlip ≤ tt < tHopD = Interpolation@Table@8j ∆t, 8th@jD, thDot@jD<<, 8j, 0, nsteps<DD@tt − tSlipD;

rCMSoln@tt_ ê; tt >= tHopD := 8X Hop, YHop< +8X 

Hop, Y

Hop< Htt − tHopL + 80, −g< ê2 Htt − tHopL2

θSoln@tt_ ê; tt > tHopD := θHop + θ

Hop Htt − tHopL. . . and use it to create a plot.

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rCtrHoop@t_ D := rCMSoln@tD − λ a 8Sin@θSoln@tDD, Cos@θSoln@tDD<rMass@t_ D := rCtrHoop@tD + a 8Sin@θSoln@tDD, Cos@θSoln@tDD<ntSteps = 40;

deltat = tImpact

êntSteps;

 ptsCM  = Map@rCMSoln, Range@0, tImpact, deltatDD; ptsCtrHoop = Map@rCtrHoop, Range@0, tImpact, deltatDD;

 ptsMass = Map@rMass, Range@0, tImpact, deltatDD;

lines = [email protected],

 Map@Line, Transpose@8 ptsCtrHoop, ptsMass<DD<,8 AbsolutePointSize@4D, Map@Point, ptsMassD<,8 AbsolutePointSize@2D, Map@Point, ptsCtrHoopD<<;

ParametricPlot@rCMSoln@ttD, 8tt, 0, tImpact<,

Compiled → False, Epilog → lines, AspectRatio → Automatic,

PlotRange → 8−0.2, 2.2<, Frame → True, Axes → 8True, None<D;

0 0.5 1 1.5 2 2.5 3

0

0.5

1

1.5

2

The Hopping Hoop 41

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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

42 Project Intermath

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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 onhis (round-wheeled!) bicycle.

The Hopping Hoop 43