piecewise smooth models of a child's swing · and the system as a whole is piecewise smooth....
TRANSCRIPT
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
1
This manuscript is for review purposes only.
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
This manuscript is for review purposes only.
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
This manuscript is for review purposes only.
4 BRIGID MURPHY AND PAUL GLENDINNING
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
This manuscript is for review purposes only.
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.
This manuscript is for review purposes only.
6 BRIGID MURPHY AND PAUL GLENDINNING
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
This manuscript is for review purposes only.
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
This manuscript is for review purposes only.
8 BRIGID MURPHY AND PAUL GLENDINNING
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
This manuscript is for review purposes only.
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
This manuscript is for review purposes only.
10 BRIGID MURPHY AND PAUL GLENDINNING
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
This manuscript is for review purposes only.
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
This manuscript is for review purposes only.
12 BRIGID MURPHY AND PAUL GLENDINNING
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
This manuscript is for review purposes only.
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
This manuscript is for review purposes only.
14 BRIGID MURPHY AND PAUL GLENDINNING
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
This manuscript is for review purposes only.
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
This manuscript is for review purposes only.