Piecewise smooth models of a child's swing · and the system as a whole is piecewise smooth. If Q k...

PIECEWISE SMOOTH MODELS OF PUMPING A CHILD’S SWING∗1

BRIGID MURPHY AND PAUL GLENDINNING†2

Abstract. Some simple models of a child swinging on a playground swing are presented. These3are analyzed using techniques from Lagrangian mechanics with a twist: the child changes the con-4figuration of the system by sudden movements of her body at key moments in the oscillation. This5can lead to jumps in the generalised coordinates describing the system and/or their velocities. Jump6conditions can be determined by integrating the Euler-Lagrange equations over a short time interval7and then taking the limit as this time interval goes to zero. These models give insights into strategies8used by swingers, and answer such vexed questions such as whether it is possible for the swing to9go through a full 360 turn over its pivot. A model of an instability at the pivot observed by Colin10Furze in a rigid swing constructed to rotate through 360 is also described. This uses a novel double11pendulum configuration in which the two components of the pendulums are constrained to move in12orthogonal planes.13

Key words. Lagrangian mechanics, piecewise smooth dynamics, double pendulum, playground14swing.15

AMS subject classifications. 37N05, 70H0316

1. Motivation. Lagrangians and the Euler-Lagrange equations form a central17

part of university mechanics courses. Mathematically they lead directly to variational18

methods, whilst the introduction of generalized coordinates greatly simplifies mod-19

elling. Many examples used in courses are either important but mathematically boring20

unless you want to introduce special function theory (the simple pendulum), math-21

ematically fairly tractable but somewhat artificial (double pendulums and spherical22

pendulums), or positively Victorian (steam engine pressure valves). These examples23

also ignore the role Lagrangian descriptions can play in more modern applications24

such as control theory or hybrid systems. A child’s swing is, admittedly, not the most25

modern of inventions, but it is familiar to most students, it can be analyzed in a26

number of different ways, and it illustrates several extensions of classic Lagrangian27

techniques. The principles or methods described in this paper include28

(a) actuation and control at strategic parts of the motion;29

(b) analysis of jumps due to sudden changes in the configuration (control strate-30

gies); and31

(c) mathematical investigation of effects usually described with words.32

Observations on the design of a rigid swing by Furze [6] leads to33

(d) a variant on the classic double pendulum; and34

(e) a new analysis of instabilities at the pivot of a pendulum.35

Some of the contents of this paper would be suitable for project work or extended36

worksheets related to classical mechanics courses in the second or third year of a37

university course, and variations on this theme could be set for projects at a higher38

degree level (see the discussion at the end of this paper, section 7).39

In section 2 we discuss the models used to describe a child’s swing. In particular40

we revisit some wonderful models introduced by Wirkus et al [16] to describe the child41

and the swing. In section 3 we look at the seated model of [16] using a Lagrangian42

formulation, and then ask whether a swinger on a swing with flexible ropes can swing43

through 360 (which we will refer to as a turnover) without the ropes becoming44

∗Submitted to the editors DATE.†School of Mathematics, University of Manchester, Oxford Road, Manchester M13 9PL, UK

([email protected]).

1

This manuscript is for review purposes only.

mailto:[email protected]

2 BRIGID MURPHY AND PAUL GLENDINNING

slack. This is potentially dangerous, but it turns out that turnover is impossible, so45

seated swinging is safe for this model. In section 4 we do the same for the standing46

and squatting model of [16]. Here we find that a sufficiently tall child could achieve47

turnover without the flexible strings becoming slack, but that swings are still safe for48

smaller children. If the strings are rigid then lack of tension is not an issue and, as49

Colin Furze has so admirably demonstrated [6], turnover is possible. He is forced to50

add a counterweight to his pendulum because of an alarming lateral instability he51

observes (about 1 minute 50 seconds into the video). To model this we introduce a52

variation of the double pendulum in section 5 which is interesting in its own right,53

and then apply this to the rigid pendulum construction in section 6. In the discussion54

of section 7 we describe some of the generalizations of [16] that we would have liked55

to have included in more detail, but which would not fit into this introductory paper.56

We also suggest ideas for future projects.57

2. Strategies and methodology. A typical child’s swing consists of two ropes58

or chains attached to pivots on a horizontal crossbar, and connected together by a59

seat below the bar. The child sits or stands on the seat and moves back and forth to60

initiate the oscillatory motion of the swing. Following most mathematical models we61

will assume that the motion is independent of the direction connecting the ropes, and62

so the model is planar, with the swing swinging from left to right and back again.63

Wirkus et al [16] suggest two related models which take this strategic motion into64

account. These are illustrated in Figure 1 (seated) and Figure 2 (standing). In both65

cases we assume that the child is facing to the left and that the position of the rope66

is determined by the angle θ of the rope to the downward vertical through the pivot;67

positive if the rope is on the right of the pivot negative on the left. The length of the68

ropes is denoted by `.69

2.1. The seated strategy. In the seated strategy, the child is modelled by a70

dumbbell (barbell if American): two equal weights, 12m, separated by a rigid bar of71

length 2a, which is attached to the swing ropes at its centre. In the passive phase of72

the motion from the highest negative angle to the highest positive angle (with θ > 0),73

the child holds his body parallel to the strings, with one mass a distance a above the74

seat and the other a distance a below the seat. At the high point of the backwards75

swing, with θ = 0 and θ > 0, the child throws her body back, so that the rigid rod76

representing her body is at an angle φ to the strings, and holds this position until θ77

is zero again with θ < 0. At this point of the swing the child re-positions his body to78

align with the strings (φ = 0), and swing back to the starting position again.79

2.2. The standing strategy. In the standing strategy, the child’s body is al-80

ways aligned with the ropes as the child stands on the seat. In this case the pumping81

of the swing is achieved by standing and squatting. Suppose that the centre of mass82

of the child is a distance h above her feet when standing. At the highest point of the83

swing θ = 0 the child moves from standing to squatting so the effective length of the84

swing, the distance from the pivot to the centre of mass of the child, changes from85

` − h to ` − αh for some α < 1; we assume that h < `. At the bottom of the swing86

motion, when θ = 0, the child quickly stands up, so the effective length of the swing87

changes abruptly from `− αh back to `− h. This sequence is then repeated.88

2.3. Methodology. Both the strategies described above involve a rapid reposi-89

tioning of the child’s body. Wirkus et al [16] model this as an instantaneous change by90

taking the limit of the Euler-Lagrange equations as the time taken to change position91

tends to zero which provides a nice introduction to piecewise smooth systems.92


SWINGS 3

If a mechanical system can be described by generalized coordinates qk, k =93

1, . . . , N and has kinetic energy T (qk, qk, t) and potential energy V (qk, qk, t) then the94

Lagrangian of the system is95

L(qk, qk, t) = T − V96

and the Euler-Lagrange equations which determine the motion of the system are97

d

dt

(∂L∂qk

)− ∂L∂qk

= Qk, k = 1, . . . , N98

where Qk represents the generalized force on the qk coordinate. In the seated strategy99

the coordinate φ is constant except at the highest points of the swing where for a100

short time ∆t it is a function of time actuated by the child, whilst in the standing101

strategy the effective length of the pendulum is constant except for small times ∆t102

when the child stands or squats and it becomes a function of time. Thus between the103

re-positioning of the child’s body the motion is determined by standard equations,104

and the system as a whole is piecewise smooth. If Qk and ∂L∂qk

are bounded, then105

integrating the Euler-Lagrange equation gives106

∂L∂qk

= C +O(t)107

where C is a constant, and so integrating again from 0 to ∆t gives108 ∫ ∆t

0

∂L∂qk

dt = C∆t+O(∆t2).109

Thus in the limit of ∆t → 0 the right hand side is zero and the left hand side, if110

tractable, becomes a jump condition from just before the change in pose (0−) to just111

after (0+):112

(2.1)[ ∂L∂qk

]0+

0−= 0.113

3. Seated Swinging. During this strategy (section 2.1), the swinger is described114

by two coordinates: θ, the angle of the swing ropes to the vertical, and φ, the angle115

of the body to the ropes. The motion lies in the (x, z)-plane, with x horizontal and z116

vertical, so by elementary geometry, in Cartesian coordinates with the origin at the117

pivot of the pendulum, the positions of the two masses is given by118

(3.1) r± = (` sin θ ± a sin(φ− θ),−` cos θ ± a cos(φ− θ)).119

Wirkus et al [16] derive the equations of motion directly from considerations of the120

change in angular momentum of the system. However, we will use Lagrangian methods121

throughout since this makes generalizations easier.122

3.1. The seated Lagrangian. The kinetic energy of the system is simply the123

sum of the kinetic energies of the masses: 14mr2

± (recall that each mass carries half124

the total mass of the child), and the potential energy is simply 12mgz± for each mass.125

Using (3.1) the Lagrangian for the system is126

(3.2) L =1

2m[(`2 + a2)θ2 − 2a2θφ+ a2φ2] +mg` cos θ,127



aa

m

m

θ

φ

1- 2

1- 2

O

l

(a)

O

l

(b)

Fig. 1. The seated swing strategy in which the child’s body is modelled by a dumbbell with twomasses equal to 1

2m a distance 2a apart. (a) Swinging from right to left (facing left) the child pushes

back making an angle φ with the ropes, and holds this position until θ = 0 with θ < 0; (b) swingingfrom left to right the child aligns her body with the ropes.

and the Euler-Lagrange equation in the θ variable is128

(3.3)d

dt

((`2 + a2)θ − a2φ

)+ g` sin θ = 0.129

There is a similar equation for the φ variable, but since this is being controlled by the130

child the right hand side would be non-zero and unknown.131

During the phase of the swing with θ > 0, φ = 0 and φ = 0, so integrating (3.3)132

once the equation for the energy per unit mass, E, is133

(3.4)1

2θ2 − g `

`2 + a2cos θ = E.134

In the phase of the swing with θ < 0, φ is held at some constant value φ∗ say, and the135

corresponding energy equation is unchanged. The motion can therefore be described136

by phase curves given by (3.4) if θ < 0 or if θ > 0.137

3.2. The seated jump conditions. If θ = 0 the system can jump between138

different energy levels. The jump is determined by the process described in section 2.3.139

Suppose that we are close to the top of the swing phase, with θ close to zero. Since140

g` sin θ is bounded (2.1) implies that141

(3.5) [θ] =a2

`2 + a2[φ]142

describing the instantaneous jump in θ due to an instantaneous change in φ. At θ = 0143

with θ > 0 the angle φ changes from 0 to φ∗, therefore [φ] = φ∗, which increases the144

amplitude of the oscillation. If θ = 0 with θ < 0 then [φ] = −φ∗, so this also increases145

the amplitude of the oscillation.146

3.3. Tension and the no turnover result. If the strings are rigid then since147

the amplitude increases by a fixed amount on each oscillation |θ| will eventually pass148

π and so the swing will turn over a full 360. However, if the strings are flexible then149


SWINGS 5

this model only remains valid whilst there is tension in the strings. If the strings150

lose tension then the swing enters a phase of free fall which needs to be modelled151

separately [9]. It is therefore natural to ask whether the swing can turnover whilst152

always having tension in the strings, and for this we need to calculate the tension153

explicitly. To do this we need to take a more Newtonian approach.154

If φ is fixed then taking moments about O = (0, 0) gives155

(3.6) (`2 + a2)θ = −g` sin θ,156

which also follows directly from from (3.3), with energy equation (3.4). To obtain the157

tension S in the string we will take moments about another point, A = (0,−`). After158

a little manipulation1this gives159

(`2 + a2 − `2 cos θ)mθ +m`2θ2 = −mg` sin θ + S` sin θ.160

Substituting for θ using (3.6) and simplifying gives161

(3.7) S = m`θ2 +mg`2

`2 + a2cos θ.162

The first term of the right hand side of (3.7) can be rewritten using the first integral163

(3.4) as164

(3.8) S = 2E + 3mg`2

`2 + a2cos θ.165

Equation (3.7) shows that if θ = 0 with |θ| < π2 then there is always tension in the166

rope since both terms on the right hand side are positive, but if |θ| ∈ (π2 , π) then the167

tension must have become zero before the maximum angle is reached. This means168

that the jump in θ at an end point must be enough to jump from |θ| < π2 into the169

next oscillation with |θ| > π. The largest change in φ possible when the swinger leans170

back is π2 , with the body perpendicular to the ropes. Equation (3.5) then requires171

a2

`2 + a2

π

2>π

2,172

an obvious contradiction. Hence the seated strategy cannot lead to the full 360173

turnover in this model.174

4. Standing and squatting. In many ways the standing and squatting strategy175

of section 2.2 is easier to deal with mathematically than the seated strategy. The176

turnover result is, however, less clear cut. The simplicity comes from the fact that by177

alternating between squatting and standing, the child changes the effective length of178

the pendulum so it can be modelled as a simple pendulum with variable length L(t).179

Recall that h is the height of the child’s centre of mass above the seat and α ∈ (0, 1)180

the extent to which the child squats, so181

(4.1) L(t) =

Lst = `− h if (θ < 0 and θ < 0) or (θ > 0 and θ > 0)

Lsq = `− αh if (θ < 0 and θ > 0) or (θ > 0 and θ < 0).182

1The algebra gets messy here and it is instructive to walk through several simple cases to besure you have the right methods and ideas. Start by deriving the standard equations for the simplependulum by taking moments about O and then the tension equation S = m`θ2 +mg cos θ by takingmoments about A = (0,−`). Next repeat with the dumbbell of Figure 1 with φ = 0 (aligned alongthe string), noting that internal forces can be ignored). This gives precisely (3.6) and (3.7). Finally,verify that these hold for the dumbbell in general position by taking moments about O and A forthis configuration.



During the transition between poses L is some unknown function of time determined183

by the child, creating potential jumps in the angle θ or the angular velocity θ of the184

pendulum. (There is a slight subtlety here in that the tension acts at the seat of the185

swing, below the centre of mass of the child, but this does not change the moment186

equations.)187

mmθ

Ol

(a)

O

l

(b)

h𝛼h

Fig. 2. The standing swing strategy with body centre of mass a height h above the seat. (a)Swinging from right to left (facing left) with θ > 0 the child squats down so that his centre of massis αh above the seat, 0 < α < 1, and holds this position until θ = 0; (b) continuing to swinging fromright to left the child stands, with her body still aligned with the ropes. He squats again when θ = 0and repeats the strategy on the backwards swing.

4.1. The standing Lagrangian. The velocity of the child is Lθ so the La-188

grangian is189

L =1

2mL2θ2 +mgL cos θ190

with Euler-Lagrange equation191

(4.2)d

dt

(L2θ

)+ gL sin θ = 0.192

During the phases of the motion during which L is constant, solutions move on curves193

of constant energy of the classical simple pendulum194

(4.3)1

2Lθ2 − g cos θ = H.195

4.2. The standing jump conditions. If L changes rapidly over some small196

time interval but remains bounded then the second term in (4.2) is bounded and so197

(2.1) implies that (as ∆t→ 0)198

[L2θ] = 0.199

The change of position from standing to squatting at θ = 0 therefore produces no200

jump in either θ or θ. On the other hand, the change in position at θ = 0 from201

squatting to standing induces a change in θ from θ− to θ+ where202

(4.4) θ+ =

(LsqLst

)2

θ−,203

and since Lsq = `− αh > `− h = Lst the angular speed increases.204


SWINGS 7

4.3. Turnover with rigid ropes. We now need to understand how the jump in205

speed at θ = 0 changes the angle of the swing at θ = 0, i.e. to see how the amplitude206

of the swing changes as a consequence. To do this we use the energy equation (4.3).207

Suppose that the swing starts at an angle θ0 with |θ0| < π2 and θ0 = 0. Then208

it enters a phase in which the child is squatting and the system has energy H0 =209

−g cos θ0. This solution hits θ = 0 for the first time with speed θ−0 given by Lsq θ2−0−210

2g cos 0 = 2H0 = −2g cos θ0, i.e.211

Lsq θ2−0 = 2g(1− cos θ0).212

The speed now jumps to θ0+ given by (4.4) so213

Lstθ2+0 = Lst

(LsqLst

)4

θ2−0 = 2g

(LsqLst

)3

(1− cos θ0).214

The child is now standing and the system has energy H1 = 12Lstθ

2+0 − g cos 0. This215

solution reaches θ = 0 with angle θ1 and H1 = −g cos θ1 so216

2g(1− cos θ1) = Lstθ2+0 = 2g

(LsqLst

)3

(1− cos θ0),217

or, cancelling the factors of 2g,218

(4.5) 1− cos θ1 =

(LsqLst

)3

(1− cos θ0).219

Now, 1− cos θ is an increasing function of θ on [0, π] with a maximum of 2 at θ = π.220

So, taking the positive branch of arccos (in fact, the signs of θk oscillate, but we are221

only interested in the modulus here), the inequality Lsq > Lst implies that provided222

cos θ0 6= 0 repeated iteration of the relationship (4.5) eventually has a right hand side223

which is greater than two after which there is no solution for θ: this corresponds to224

turnover in the model.225

4.4. Tension and a partial no turnover result. If the strings become slack226

then the swing enters a phase of free fall, so the model breaks down. Tension is clearly227

lost if the swing comes to rest momentarily with θ > π2 , and this can be shown to be a228

necessary condition using an argument like that in section 3.3 with a = 0 in (3.7)[9].229

Thus a turnover without free fall is only possible if some initial θ0 <π2 generates a230

solution that does not return to θ = 0, i.e. if there is no solution to (4.5). The limiting231

case is θ0 = π2 , and so since the maximum of the right hand side of (4.5) is two, the232

condition for no turnover without free fall is233

2 >

(LsqLst

)3

.234

Re-arranging this condition using the definition of Lsq and Lst in (4.1) this becomes235

(4.6) h <

(1− 2−

13

1− 2−13α

)`.236

If we assume that a squatting position can half the distance of the centre of mass237

of the child to the seat, i.e. α = 12 then the no turnover condition (4.6) becomes238



h < 0.342`. A human’s centre of mass is about 0.5 of her height, H, (i.e. H = 2h)239

and a typical playground swing has ropes of about 2 m, so this implies that for this240

model, turnover is impossible without free fall, and so the swing is safe, provided the241

height of the child is less than 1.37 m. According to UK health figures [14] this is242

a bit less than the average height of a nine year old boy. Of course, factors such as243

bend in the ropes and friction at the pivot will reduce the efficiency of the swing and244

make it harder to produce the predicted effects.245

4.5. A real experiment. Colin Furze has constructed a 360 pendulum [6]. It246

consists of a single rigid pole attached to a crossbar. He is able to pump the swing247

using a languid version of the standing strategy (the transitions are not abrupt!) so as248

to turn through a full circle. One of the features of his swing is a counterweight on the249

other side of the crossbar. The need for this is illustrated at about 1 minute 50 seconds250

into his YouTube video. Before the counterweight was added lateral strain caused the251

catastrophic buckling of a strut that prevented lateral movement at the pivot. One252

possible explanation for this instability pursued in [12] is that the pendulum had253

enough freedom at the pivot that it acted as a spherical pendulum. A feature of254

the spherical pendulum is that typical oscillations precess: the plane in which they255

oscillate rotates around the pivot. The crossbar prevents this precession of Furze’s256

swing beyond a small angle, and the pressure built up eventually buckled the strut.257

However, the mechanism for the build up of pressure required for this description258

is unclear, and in section 6 we illustrate another mechanism for instability based on259

a double pendulum model in which the two pendulums are constrained to oscillate in260

orthogonal planes. This configuration seems to have some novelty in itself, so we will261

begin this analysis by looking at the orthogonal double pendulum as an object in its262

own right.263

5. Orthogonal double pendulum. Consider a double pendulum in which one264

pendulum, of length `1 and mass (at the end of the string) m1 is pivoted at (0, 0, 0)265

and constrained to move in the (x, z)-plane, whilst the second pendulum, with length266

`2 and mass m2 at the end of the string, is attached to the end of the first pendulum267

and constrained to oscillate in the (y, z)-plane as shown in Figure 3. This is closely268

related to, but different from, the orthogonal double pendulum introduced in [15].269

This configuration can be described in terms of two generalized coordinates θ1270

and θ2, where θ1 (resp. θ2) is the angle between the first (resp. second) pendulum271

rod and the negative z-direction measured anticlockwise in the (x, z)-plane (resp.272

(y, z)-plane). By elementary geometry the positions of the two masses, rk, k = 1, 2,273

are274

(5.1)r1 = (`1 sin θ1, 0,−`1 cos θ1)r2 = (`1 sin θ1, `2 sin θ2,−`1 cos θ1 − `2 cos θ2).

275

The Lagrangian is simply276

L =1

2m1|r1|2 +

1

2m2|r2|2 + (m1 +m2)g`1 cos θ1 +m2`2g cos θ2277

so using (5.1) this can be written as278

(5.2)L = 1

2 (m1 +m2)`21θ21 + 1

2m2`22θ

22 +m2`1`2θ1θ2 sin θ1 sin θ2

+(m1 +m2)g`1 cos θ1 +m2`2g cos θ2.279


SWINGS 9

O

m

m

2

2

111

2θ

θ

l

l

Fig. 3. The orthogonal double pendulum. The upper rod moves in the plane y = 0 and thelower rod lies in the plane through m1 parallel to the plane with x = 0.

5.1. Equations of motion. There are no external forces, so the Euler-Lagrange280

equations for θ1 and θ2 respectively are281

(5.3)ddt

((m1 +m2)`21θ1 +m2`1`2θ2 sin θ1 sin θ2

)= m2`1`2θ1θ2 cos θ1 sin θ2 − (m1 +m2)g`1 sin θ1,

282

and283

(5.4)d

dt

(m2`

22θ2 +m2`1`2θ1 sin θ1 sin θ2

)= m2`1`2θ1θ2 sin θ1 cos θ2 −m2g`2 sin θ2.284

After some manipulation equations for θ1 and θ2 can be written down explicitly:285

(5.5) θ1 =

(m2`1θ

21 cos θ1 sin2 θ2 −m2`2θ

22 cos θ2 −m1g cos2 θ2 −m2g

`1(m1 +m2 −m2 sin2 θ1 sin2 θ2)

)sin θ1286

and287

(5.6) θ2 =

(m2`2θ

22 sin2 θ1 cos θ2 −m2`1θ

21 cos θ1 − (m1 +m2)g cos2 θ1

`2(m1 +m2 −m2 sin2 θ1 sin2 θ2)

)sin θ2.288

Although these equations are very messy, the first thing to notice is that the lineariza-289

tion for small |θ1| and |θ2| reduces to two independent simple pendulums. Moreover,290

since there is no explicit dependence of the Lagrangian on t, the energy is conserved:291

E is constant where292

(5.7)E = 1

2 (m1 +m2)`21θ21 + 1

2m2`22θ

22 +m2`1`2θ1θ2 sin θ1 sin θ2

−(m1 +m2)g`1 cos θ1 −m2g`2 cos θ2.293



The first three terms of (5.7) can be rewritten as294

1

2

(m2(`1θ1 sin θ1 + `2θ2 sin θ2)2 + (m1 +m2 cos2 θ1)`21θ

21 +m2`

22θ

22 cos2 θ2

)295

which is clearly positive and tends to infinity as either |θ1| or |θ2| tends to infinity.296

Since the last two terms of (5.7) are bounded, if E is finite then velocities are bounded.297

5.2. Small `1 and m1. Suppose now that `1 `2 and `2 is O(1). Let ε =298

`1/`2 1 and define ω2 = g/`2. Then the limit ε → 0 of (5.5) is singular (we have299

divided through by `1) and so we need to be more careful with the manipulation of300

(5.3) and (5.4). Dividing (5.4) through by `22 and simplifying gives301

(5.8) θ2 + ω2 sin θ2 +O(ε), or 12 θ

22 − ω2 cos θ2 = E2 +O(ε),302

valid for time scales over which the order ε errors are small and E2 is constant. These303

are, of course, the equations for the standard simple pendulum.304

Dividing (5.3) through by `22 and simplifying gives305

(5.9) εθ1 + θ2 sin θ1 sin θ2 + θ22 sin θ1 cos θ2 = −(m1 +m2)ω2 sin θ1.306

Using (5.8) the θ2 and θ22 terms can be replaced by terms involving only θ2 up to307

order ε:308

(5.10) (m1 +m2)εθ1 + Ω(θ2, ε) sin θ1 = 0309

with310

(5.11) Ω = m1ω2 +m2(3ω2 cos θ2 + 2E2) cos θ2 +O(ε).311

If Ω > 0 varies slowly compared to θ1 then the motion will have approximately312

constant amplitude and frequency 2π/√

Ω. If Ω < 0 then solutions will typically grow313

rapidly for time scales on which the order ε terms are small.314

If m1 is small, order ε say, then, up to terms of order ε the sign of Ω is determined315

by the sign of (3ω2 cos θ2 + 2E2) cos θ2.316

The classic pendulum solutions to (5.8) oscillate without turning through a full317

circle if −ω2 ≤ E2 < ω2. If −ω2 ≤ E2 < 0 then |θ2| < π2 and so cos θ2 > 0.318

Moreover, E2 ≥ −g cos θ2 from (5.8), so 3ω2 cos θ2 + 2E2 ≥ ω2 cos θ2 > 0 and there is319

no instability since Ω > 0.320

Now suppose that 0 < E2 < ω2. In this case 0 < |θ2| ≤ θm where cos θm =321

−E2/ω2 < 0, so θm ∈ (π2 , π). If |θ2| < π

2 then as before cos θ2 > 0 and 3ω2 cos θ2 +322

2E > 0 and so Ω > 0 and the solution has not entered the unstable regime. However, if323

cos θ2 < 0, i.e. |θ2| > π2 (and less than π of course), then Ω < 0 if 3ω2 cos θ2 +2E2 > 0.324

Since E2 > 0 and cos π2 = 0 there must be an interval (π2 , θm) on which Ω < 0, and325

so there will be a region of the swing for θ2 such that θ1 starts growing.326

For completeness, we can calculate θm. It must be less than or equal to θm (the327

height of the swing for θ2), and the limiting case at which 3ω2 cos θ2 + 2E2 changes328

sign is when cos θ = − 23E2/ω

2 > −E2/ω2. Hence θm = arccos(− 2E2

3ω2 ) < θm.329

5.3. Numerical Simulations. Figure 4 shows a time series obtained by numer-330

ical simulation of the orthogonal double pendulum (5.5), (5.6) for two very different331

sets of parameters. In the first, Figure 4a, `1 and `2 are of similar magnitude as332

are m1 and m2 (see figure caption for details). In this case both components of the333


SWINGS 11

t t(b)

θ

1

2

2

1

θ

θ

θ θ

θ

(a)

Fig. 4. Numerical simulation of (5.5), (5.6) showing the angles θ1 and θ2 as a function of timewith g = 9.8, m2 = 80 and `2 = 2.5. (a) m1 = 60, `1 = 2 and initial condition (θ1, θ1, θ2, θ2) =(−0.5222969, 0.047522, 1.2566606,−1.7836296) and 0 ≤ t ≤ 20. (b) m1 = 1, `1 = 0.05 and initialcondition (θ1, θ1, θ2, θ2) = (0.1,−0.01, 1.6, 0) and 0 ≤ t ≤ 8.

pendulum appear periodic and there is a 4 : 5 resonance (anti-phase) between the334

periods.335

Figure 4b has parameter values more relevant for the discussion of section 6: `1 is336

much smaller than `2 and m1 is much smaller than m2. In this case, θ2 undergoes a337

slow oscillation whilst θ1 oscillates when |θ1| is small, but appears to grow in amplitude338

when |θ2| is larger, eventually turning over and causing θ2 to turn over too (these later339

times are not shown). Of course, the energy argument of section 5.1 shows that the340

velocities do not diverge, although they do become large. The oscillations in θ1 when341

|θ2| is small, and growth when |θ2| is larger is clearly vidible in Figure 4b and in the342

impacting solution Figure 6 of section 6.1.343

6. Furze Instability at the pivot. Figure 5 shows a sequence of abstractions344

of the pivot mechanism for a swing. It starts as a sleeve around a crossbar to which345

a single rigid pole is bolted. We assume that the sleeve can rock on the crossbar,346

generating a slight wobble. This is then abstracted to a perpendicular pair of rings,347

the first representing the sleeve (with no width along the crossbar) but with a hole348

slightly bigger than the crossbar, so that a left-right oscillation is possible. The long349

rod is attached to the hole in the perpendicular plate below the plate attached to the350

crossbar. Finally we replace the plates by a rod pivoted at the crossbar and able to351

move left-right, to which is attached to a second rod by a joint that allows it to move352

to and fro. However, since the sleeve/ring attached to the crossbar could only move353

through a small angle in rocking due to the relatively tight fit of the respective holes,354

the short pendulum attached to the crossbar has two buffers on either side to prevent355

it moving through more than a given angle either to the right or to the left.356

Thus the model pivot is an orthogonal double pendulum in which the top pendu-357

lum is smaller and lighter than the bottom. In addition the buffers prevent turnover,358

and we assume that when the smaller pendulum strikes the buffers the effect is a359

classically Newtonian restitution law: if |θ1| = θ∗ then there is an instantaneous jump360

(6.1) θ1 → −rθ1361

for some r ∈ (0, 1) whilst the other variables remain constant (this is the ε→ 0 jump362

condition from (5.3) and (5.4)). The buffers prevent the full instability described in363

section 5.3 from blowing up, but at the same time unless r is small the speed of the364

rocking motion gets large.365



pendulum

sleeve

strut pivot

bufferring

Fig. 5. A sequence of model pivots. From left to right: sleeve around crossbar allowing rockingmotion; rocking ring around crossbar with orthogonal swinging pendulum below; pivoted orthogonaldouble pendulum with buffers to simulate the rocking motion.

We will not attempt to include the actuation methods of section 3 and section 4366

in this analysis. The time scale of the instability is such that unless the change in367

amplitude of the swinger is large, a few oscillations of the swing is sufficient to see368

the effect on the pivot instability. So our model for the Furze instability at the pivot369

is the impact system defined by equations (5.5) and (5.6) if |θ1| < θ∗, with the reset370

(6.1) at θ1 = θ∗.371

6.1. Numerical Simulations. Figure 6 shows the results of a numerical sim-372

ulation of the orthogonal pendulum with buffers and r = 0.3. The parameters used373

were374

g = 9.8, m1 = 1, m2 = 80, `1 = 0.05, `2 = 2.5, θ∗ = 0.5 rad (≈ 28.65),375

A sequence of eight impacts, indicated by arrows, of the upper pendulum with the376

buffers have been computed, and the sequence is more complicated than a simple377

alternation between impacts with the left and right buffers. Note also that there378

is quite a lot of minor oscillation or ‘wobble’ without impact between some of the379

impacts. At values of r above r = 0.9 the upper pendulum does seem to eventually380

settle into a state oscillating directly between the buffers at quite a high frequency381

(an average period of about 0.16 if r = 0.99), whilst the θ2 variable performs its382

slower oscillations for the time scales investigated. We have not investigated the383

longer term effects of the loss of energy from the system (recall that in terms of the384

relevance to the swing strategy we have ignored the changes in energy and amplitude385

due to the swinger’s re-positioning in these simulations). If the amplitude of both386

angles is smaller initially then there are no impacts with the buffers. For example the387

initial conditions (θ1, θ1, θ2, θ2) = (−0.1, 0.3,−0.3, 0.5) leads to oscillations in which388

the maximum amplitude of θ2 (resp. θ1) is approximately 0.4 (resp. 0.1) radians.389

During the fast oscillations between the buffers at higher values of r the speed390

|θ1| of the small pendulum is relatively large at impact as the pendulum oscillates391

through 1 radian in approximately 0.15 seconds. A rather more interesting source392

of fast oscillations, this time from the same threshold, can be observed at r = 0.3393

but with different initial conditions from that shown in Figure 6. Table 1 gives the394


SWINGS 13

θ

θ

θ

t

1

2

Fig. 6. Time series for the two angles θ1 and θ2 of the orthogonal pendulum with buffers. Thecoefficient of elasticity is r = 0.3 and initial conditions (θ1, θ1, θ2, θ2) = (−0.5, 0.659, 1.931,−1.378)and the other parameters are as defined in the text. The impacts were found by integrating asolution and observing when it crosses |θ1| = 0.5 (correct to three decimal places) and then restartingthe trajectory at this new point having applied the reset condition (6.1). These impact events areindicated by the arrows.

Table 1Chattering

Impact 1 2 3 4 5θ1 -0.5 0.5 0.5 0.5 0.5

θ1 0.501 -1.081 -0.570 -0.179 0.054θ2 1.880 2.251 2.045 1.999 1.985

θ2 -1.892 -1.667 -2.176 -2.290 -2.323T 2.2018 0.10792 0.02062 0.005857

times to the next impact, T , for a sequence of impacts on the upper boundary for a395

solution starting with initial conditions on the lower boundary with (θ1, θ1, θ2, θ2) =396

(−0.5, 0.501, 1.880,−1.892), working to three decimal places at impact events. The397

coordinates of subsequent rebounds are given immediately after the impact, i.e. with398

the velocity θ1 reversed and reduced by a factor of 0.3 as in (6.1). It shows a rapid399

decrease in the time between impacts and the corresponding speed of θ1 at impact400

becomes smaller. We were unable to follow the impact beyond the final position shown401

because the solution appeared tangential to the surface θ1 = 0.5. The rapid speed up402

of the time between impact events and the associated near tangency of the solution403

to the impact surface strongly suggests that there is a ‘Zeno’ type singularity in the404

system with infinite chatter in finite time [1, 2]. An analysis of this phenomenon405

goes beyond the scope of this introductory article, but demonstrates the richness of406

behaviour that may be possible in this simple model.407

7. Discussion and other models. The examples in this paper are taken from408

a context almost all students will recognise. Natural piecewise smooth elements have409

been added to the dynamics via control strategies and mechanical impacts, providing410

a good introduction to the different ways that piecewise smooth systems arise in the411



modelling process. In the former case, following [16], we have shown how the instan-412

taneous re-positioning of the body leads to a jump in the configuration of the swing.413

This analysis has been used to provide a mathematical description of the difficulties414

of pumping a swing through a full rotation. The experimental evidence [6] suggests415

that as the swing increases in amplitude it can induce a lateral instability (a Furze416

instability) at the pivot of the swing, and we have introduced one possible mechanism417

which might explain this observation. This mechanism involves the introduction of an418

orthogonal double pendulum. Numerical simulations were performed with standard419

packages (see the comment in the caption of Figure 6. There are specialized software420

packages designed for the analysis of impacts and piecewise smooth systems, and it421

would be useful to obtain more accurate solutions (this could be a project on its own).422

Whilst a swing may seem a childish choice of phenomenon to model, it has many423

interesting features and the ideas used for this analysis have broader application. In424

[8] oscillations induced by the movement of spiders in their webs are described, and425

engineering structures such as cranes swinging heavy loads use time dependent motion426

of the pivot [7]. The swing continues to motivate research [3].427

7.1. Other models. Classic models of actuation use parametric resonance: pe-428

riodic movement of the child changes the effective length of the pendulum in a 2 : 1429

resonance. Thus the models take the form430

θ +(Ω2 + b sinωt

)sin θ = 0.431

If ω ≈ 2Ω this does indeed lead to growing oscillations at small amplitude (sin θ ≈ θ,432

[4, 5]). However, unless the period of the forcing adapts to the period of the swing as433

the amplitude changes (which would require a more complicated model), numerical434

experiments suggest that the swing is damped until it returns to small amplitude and435

the resonance leads to another phase of growth. This does not fit the observations436

either [13].437

7.2. Other strategies. Both the models described here come from [16] and have438

the constraint that when the child changes position on the swing there is no significant439

bending of the ropes. In practice, the change in position of the child is achieved by440

pulling on the ropes, and this creates new configurations in which the ropes are bent441

for some part of the control cycle. These models, and the jumps created by changing442

position, can be a rich source of further problems [12].443

7.3. Pivots. As Furze has demonstrated [6], the pivot itself can be an interest-444

ing (i.e. dangerous) source of dynamics behaviour. We have found very little in the445

mathematical literature about dynamic modelling of pivots. This seems an area that446

would benefit from further study. There are aspects of the orthogonal double pendu-447

lum models of section 5 which are reminiscent of the pony tail instability [10], and a448

classification of lateral instabilities might help identify the mechanism responsible for449

bending the strut in Furze’s pendulum with greater certainty. A careful asymptotic450

analysis of the small ε limit in section 5.2 might be interesting.451

7.4. Role of simple models. The swing models presented here are, of course,452

simplifications of the actual neuro-bio-physical interactions which are going on when453

the child swings. More detailed modelling of the swing and the child’s body can be454

constructed but these rapidly become models that can only be investigated numeri-455

cally; the claim for the simpler models described here is that they make it possible456

to gain insight into the phenomena and the mechanisms at play. Insights which, we457

hope, will be followed up in experiments.458


SWINGS 15

The insight simple models and their mathematical treatment provide is at the459

heart of one extreme of the spectrum of approaches to central societal problems such as460

climate change, in which the more accurate models take vast computers vast amounts461

of time to produce data, and then even longer times to analyse that data [11]. Projects462

associated with the devising and analysis of simpler models help a student to appre-463

ciate the different questions and techniques which can be brought to bear on complex464

problems.465

Acknowledgments. This paper arose from the MSc dissertation of one of us466

(BM) supervised by the other (PG) at the University of Manchester [12]. We are467

grateful to Mark Muldoon for suggesting the orthogonal configuration at the pivot468

and to Alan Champneys and Anne Skeldon for helpful conversations.469

REFERENCES470

[1] M. di Bernardo, C. Budd, A.R. Champneys and P. Kowalczyk, Piecewise-smooth Dynam-471ical Systems, Springer, London, 2008.472

[2] M. di Bernardo and S.J. Hogan, Discontinuity-induced bifurcations of piecewise smooth473dynamical systems, Proc. Roy. Soc. (London) A, 368 (2010) pp. 4915-4935.474

[3] M.V. Berry, Pumping a swing revisited: Minimal model for parametric resonance via matrix475products, Eur. J. Phys., 39 (2018) 055007.476

[4] W.B. Case, The pumping of a swing from the standing position, Am. J. Phys., 64 (1996) pp.477215-220.478

[5] W.B. Case and M.A. Swanson, The pumping of a swing from the seated position, Am. J.479Phys., 58 (1990) pp. 463-467.480

[6] C. Furze, Huge Homemade 360 Swing 2016, https:\\www.youtube.com\watch?v=J9uh-CyBMCs,481accessed 2019-05-29.482

[7] R.M. Ghigliazza and P. Holmes, On the dynamics of cranes, or spherical pendula with483moving supports, Int. J. Non-Linear Mech., 37 (2002) pp. 1211-1221.484

[8] P. Glendinning, Shaking and whirling: dynamics of spiders in their webs, preprint, University485of Manchester, 2018.486

[9] A. Goriely, P. Boulanger and J. Leroy,Toy Models: The Jumping Pendulum, Am. J. Phys.,48774 (2006) pp. 784-788.488

[10] J.B. Keller, Ponytail motion, SIAM J. Appl. Math., 70 (2010) pp. 2667–2672.489[11] P. Maher, E. Gerber, B. Medeiros, T. Merlis, S.C. Sherwood, A. Sheshadri, A. Sobel,490

G.K. Vallis, A. Voigt and P. Zurita-Gotor, The value of hierarchies and simple models491in atmospheric research, preprint, University of Exeter, 2018.492

[12] B. Murphy, Piecewise smooth models of swings, MSc dissertation, University of Manchester,4932018.494

[13] A.A. Post, G. de Groot, A. Daffertshofer and P.J. Beek, Pumping a Playground Swing,495Motor Control, 11 (2007) pp. 136-150.496

[14] Royal College of Paediatrics and Child Health, Boys UK Growth Chart,497https:\\www.rcpch.ac.uk\resources\growth-charts, accessed 2019-05-29.498

[15] A.C. Skeldon, Dynamics of a parametrically excited double pendulum, Physica D, 75 (1994)499pp. 541-558.500

[16] S. Wirkus, R. Rand and A. Ruina, How to pump a swing, College Math. Journal, 29 (1998)501pp. 266-275.502


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