The swing

The swing

Citation for published version (APA):

Harink, J. H. A., Jong, de, P., & Kleijer, H. (1989). The swing. (Opleiding wiskunde voor de industrie Eindhoven : student report; Vol. 8903). Eindhoven University of Technology.

Document status and date: Published: 01/01/1989

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THE SWING

by: drs. J.H.A. Harink , dr=:. P. de Jong

drs. H . K lei j er

January/April 1989

INDUSTRY---CONTENTS

section 1: Preface

section 2: Introduction

section 3: A Mathematical Model section 4: A More Realistic Model

section 5: Air Resistance Included Into The Mathematical Model section 6: Analysis Of A Non Linear Differential Equation

section 7: Application Of The Analysis

section 8: Reduction Of The More Realistic Model To The Mathematical Model

section 1: PREFACE

The so-called Swing Problem has been suggested by dr. S. Howison from Oxford University during his visit at the University of Eindhoven in January 1989. During this visit he gave lectures concerning real-life problems and one of this problems was the swing problem.

This problem is posed as follows:

a child is sitting on a swing and is swinging up and down without help from anybody else. The child is capable to increase its amplitude by moving the legs and upper-body in a certain way.

The question arises what strategy the child should follow to increase its amplitude as efficient as possible.

We have been working on this problem for almost two months with good support from other staff members. for which we would like to thank them.

section 2: INTRODUCTION

We sta~t with investigating a simple mathematical model fo~ the swing. It consists of a single mass point that~ apa~t f~om its swinging motion~ can move up and down in the direction of the cable. If the child (mass point) does not move radially~ ene~gy is

conse~ved. With this obse~vation i t is possible to draw o~bits in the so-called phase plane~ i.e. pictu~es of the amplitude

va~iation ve~sus the amplitude. Using phase plane analysis~ we derive in section 3 a strategy which inc~eases the amplitude of the swing.

In section 4 we analyze a mo~e ~ealistic model consisting of th~ee

mass points. One mass point models the uppe~-body~ one mass point the sitting su~face and the last mass point the legs. Fo~ the mechanics of this model we will use Lagrange's method. This method gives the equation of motion and one could derive~ in the same way as we did fo~ the mathematical model~ a successful strategy.

In section 5 ai~ ~esistance is introduced. It turns out that the influence of ai~ ~esistance leads to a non linea~ diffe~ential

equation~ which we will examine in section 6.

In section 7 we apply the ~esults of section 6.

In section 8 we end with a reduction of the more to the mathematical model and in section 9 the strategy the child should follow to inc~ease formalized.

realistic model conclusion what its amplitude is

section 3: A MATHEMATICAL MODEL

For this model we make two assumptions:

1) the child is represented by a single mass point,

2) the child moves, apart from its swinging motion, only up and down in the direction of the cable.

Picture:

m.

Proposition: Assume the child does not move in the radi al direction, so the length of the cable is constant. Then the energy is constant.

Proof: See standard literature.

In the following, we use the notation:

£: the total energy of the mass point m.. T: the kinetic energy of the mass point m..

U: the potential energy of the mass point m..

The energy is given by:

,where the dot means differentiation with respect to t, i . e. d and the position with 8=0 is taken to be the zero level for U.

From this energy expression we can find 6 as a function of 6. By means of this function we can draw orbits in the phase plane:

(3. 1 ) ~with 6 . == -+ [ Ci 0.5 - 2trCl-cose>/l ] Ci = 2£ / Cm.~ 2.).

Equation (3.1) indicates that

I

de

de

l i s sma I I i f ~ is large and VIce versa.

F't-ogosition: The function 6=/(8) increases (decre:il.ses) fastl'¥, i f ~

is small and the function

(decreases) slowly if 1 is large.

From (3.1) we can find the derivative:

So~ if 1 is small, then 6/1 is large and 6=/(6) increases (decreases) fastly. Look at the pictures below.

6

Z. 1 arge

t

small

We now use phase plane analysis to find a successful strategy. The only way the child can change the amplitude is to sit down or to stand up at well-chosen moments of time. This means in terms of our model that we increase respectively decrease the length

the cable at those moments.

If we change

t

when 8=0, then the total energy

E

equals: E=1n6l(,1-cos8J.

So, to increase E we have to increase

t.

l of

If we change

t

when 8=0, then we have to find out whether we have to increase or decrease

l

to increase the total energy

E.

What happens if we change t? May we use the fact that the angular momentum is conserved?

From standard literature we know that the angular momentum J

eqL\al s:

J=rxp

, I-'Ji th p=m.*r and r=lcos8et. + tsinSe2,

2'

So, thi s yi el ds in our case J=m.t 8 es, wi th es=('eil.Xe2J,

The derivative of J is given by:

J= rxmr + rxmr

=

rxmr .

We change the mass point in the direction of the cable.

Therefore the force to change

t

is also in the direction of the cable.

So, F=Ft.*r and J becomes: J=rxFt.*r=O.

Thus we can conclude that the angular momentum does not change, if the child sits down or stands up!

The question remains whether to increase or decrease L, when 8=0.

2' , 2

We have deduced that J=m.l 8, or equi valentI y that 8=J/Cm.l :>. We have already seen that:

£=0,5m.l2Cf,:>2 + n18lC 1-cos8:>=

=0,5m.l.2C J/Cm.l2:> :>2 + n18l.Cl-cos8:>= =~/C2m.l2:> + n18l.Cl-cos8:>,

So ~ ~o.Jhen 8=0: £=~ /C2m.l2:> ,

The angular momentum is conserved~ so to increase

£

we have to decrease l. when 8=0. At the highest point the kinetic energy is zero~ so the total energy equals the potential energy, so we have increased the amplitude of the swing.

Conclusion:

This leads to the following strategy for the child to follow:

If the child reaches the highest point, i.e. when 8=0~ the child has to sit down and when the child reaches the

section 4: A MORE REALISTIC MODEL

Consider the following model of a child on a swing. Picture:

01

!

i

e

! : ~

I

I

In the above picture are

e

and ~ positive and is

¢

negative.

The three mass points model the upper-body~ the sitting surface and the legs. The three connection rods are rigid and massless. The angles e~

¢

and ~ are defined as in the picture.

The angles

¢

and ~ are arbitrary functions of time. They are prescribed and represent the strategy the child has to follow.

Position of m.i. is xi. ( in terms of the basis: (0.e~.e2.e9) ) ~ so:

x~=l.tCOSeet + l.tsineez.

X2=Cltcose-z.2Cos¢~et + Cz.~Sine-z.2sin¢~e2.

xs=C hCOSe+z.3Cos~,)et + C z.tsine+l.ssin~,)e2.

Differentiating with respect to t gives the velocities in terms of the basis (0.et.e2.es) and their norms~ i.e. IIUi.1\2

=

CUi..ui.,).

We have only one generalised coordinate,

e,

because

¢

and ~ are not: they are known control functions of time.

The kinetic energy of this system is:

,with Hence

T=Tt +T2+Ts

Ti.=·the kinetic energy of mass point mi". Ti.=O. 5mill

vq2.

The potential energy of this system is: U=Ut +U2+Ua

~with LA='the potential energy of mass point mi'

If we take the horizontal axis through e2 as the zero level of potential energy, then we get:

Ut=-1M8 L tcos8.

U2=-m28C l.tcos8-l2COS¢) • US=-m:!J8C l.scos8+lscoslp).

Our system is conservative, because gravitation is a conservative force field. We can apply Lagrange's formalism for 8.

The corresponding Lagrange equation equals:

(4. 1 )

We emphasi ze that we actual 1 y know Xt, X2 and xa and al so their det-ivatives Vt,V2 and va. Therefore we know T as a function of 8,8,¢,¢,lp and lp and also

U

as a function of 8,¢ and lp. So, equation (4.1) is totally determined and is called the equation of motion.

So it is possible to calculate every term of this equation, but this is a very laborious job and that is the reason that we shall only give the general type of the resulting diffential equation in the case that the control angles are chosen to be piecewise constant.

Define 0I.:=n-4>-8 and (3:=11'-8, where n=180 degrees, as Llsual. So a and (3 are prescribed.

Under the assumption th.t a and (3 are piecewise constant, the gener.l type of the resulting differential equation from (4.1) is:

A8 + Bsin8 + CsinC8+a.) + Ds"i.nC8+(3:>

=

O.

By looking . t the phase pl.ne one could derive a successful strategy just . 5 we did with the mathem.tical model.

section 5: AIR RESISTANCE INCLUDED INTO THE MATHEMATICAL MODEL

Here~ we introduce air resistance into the simple mathematical model from section 3.

The swing starts with velocity zero and angle 8CO). So, at time t=O we have the following picture:

8(t) is the angle of the swing at time t.

When m is on the left hand side, 8 is taken negative. Consider the following situation:

When m moves from the left hand side to the right hand side, we allow m to climb a certain distance upwards instantaneously, at a certain angle

¢.

In this way m is able to put extra energy in its motion, as we shall see. So, the length l changes into Al,

Ae[O, 1 J ~ for 8 ( t ) =¢.

The time of this change will be denoted by t~ so 8(t)=¢.

(Note that there is only a slight difference between t and t, the textwriter is to blame for this.)

After this change we have for time t=t: 8betoreCt) =8a.flerCt)=¢

and with the conservation of angular momentum, which is valid, because the force is perpendicular to the motion, i t follows that:

" 2"

We illustrate the described motion by the next picture:

m.

Our aim is to get the amplitude~ i.e. the angle 8, as large as possible at the right hand side.

For a given air resistance and a given X we shall calculate the best angle

¢

for m. to go upwards in order to make the amplitude as large as possible.

Let Fo.i.r be the force appl i ed on m. by the ai r. We will look at two models for the air resistance:

1) The first case: Fo.i.r is proportional to the velocity of m.. 2) The second case: Fo.ir is proportional to the (velocity of m.)2.

The first case (linear friction):

If Fo.i.r=C:ll'Cvelocity of m)=cz.e. c constant~ then the cOLlple the suspension point of the swing is:

2"

-me1.-cl. 8.

The equation of motion for constant 1. becomes: Z. 8 + C I.

e

+ 65 i.

ne

= O.

Assume e is small~ so sine ~ 8.

The equation of motion for small 6 is::

l6 + c l6 + 66 = O.

An e:·:amp1 e: Take c=2/s, l=1 meter, 6eO,)=-0. 1 rod. 6eo.) =0 and

~=O.9, which means that the point mass m climbs 0.1

meter.

The next table gives for these values the time ~ at which m goes upwards, the ma:d mum amp 1 i tude 6mCLX at the ri ght hand si de and the time that the point mass m uses to get from 6eO,) to 6mCLX.

If m does not go up, m reaches the lowest point at time t=0.64 s.

~ (in s) 6max (i n racD time (6eO').6max)

0.0 0.0372 0.99 0.3 0.0414 0.98 0.4 0.0427 0.99 0.48 0.0401 1.00 0.55 0.0428 1.02 0.64 0.0417 1.04 0.7 0.0405 1.05 1 0.0352 1.06

From thi s tab1 e we see that 6mCLX has the rli ghest val ue if m goes up at time t=0.48 s. So the best value for ¢, i.e. the angle at which the mass point climbs up, is 6{0.48).

The main conclusions of the table are

1) m should go up before i t reaches the lowest point,

2) the corresponding swing time is not minimal in that case~

as someone erroneously might expect. The second case (quadratic friction):

If F(1i.r=CM'eve10city of m)2 =

cz.

2e9.)2. then the couple

suspension point of the swing will become:

-~Z.Sin8-clge9.)2si~(6').

Now the equation of motion for constant 1 becomes:

(5. 1 ) 1.6 + cl 2 C6:> . 2 + 8sint3 = O. if m. moves to the right. (Note that this c is not the same c as in the first case.)

With the aid of this equation t3 can be written as a function of t3. (Compare with the method of section

3.>

The ma:dmum amplitude 6mcx at the right hand side can be b'y' solving:

6 (t3mcx) = O.

We wi 11 aSSLlme t3CO:>=O and compute the det-i vati ve of ¢~

compLtted

1. e. the angle at which the mass point climbs LIP~ which eqLlals ¢=t3(¢).

Let ~(t") be the angle as a function of time~ with t"=O is the moment at

~"Jhich

m. climbs up. So,

~O:>=¢

and

~O:>

=

')...-24>

=

x.-zec¢:>. Similarly as for t3~ ~ is a function of ~.

A solution of the equation ~(~)=O is the ma:dmum amplitude ytmax at the right hand side.

So, ytmax is a function of ¢. We shall calculate the optimal value • of ¢ for which ytmaxC¢:> is as large as possible. This means we have to find a • with

But first we have to find 6 as a function of 6.

We can deduce this function starting from the differential equation (5.1). We shall discuss this type of differential equation in section 6 and then apply the result to section 5, equation (5.1) in section 7.

section 6: ANALYSIS OF A NON LINEAR DIFFERENTIAL EQUATION

Equation (4.3):

ACtJ8

=

BCtJC8J2 + Csin8

ACtJ and BCt~ piecewise constant.

We try to find a function f~ such that 8 = fC8~.

Define 8

=

p C

=

pC8~ J

2

and p

=

y (both are functions of 8).

This leads to the following equation: (6. 1 )

,with

O.5yC8J-byC8J=csin8

b=BCtJ/ACtJ en c=C/ACt~.

The solution of the homogeneous equation of (6.1) is: (6.2)

A particulat- solution of (6.1) is:

(6.3)

• with

xC8J

=

d*Ccos8 + 2bsin8J d

=

-c/CO.5+2b2J .

So, the general solution of (6.1) is:

(6.4)

and if yC8COJ~=O. then K.=-d*{cos8COJ + 2bsin8COJ)

*

expC-2b8CO~~.

2 • 2

Finally notice that yC8~=pC8~=C8J , whence:

. + 0.5

(6.5) 8 = - (K.*expC2b8J + d*{cos8 + 2bSine»)

section 7: APPLICATION OF THE ANALYSIS

We continue our discussion of section 5.

STEP 1)

We ended section 5 with the remark that we had to find

e

as a function of

e.

With the results of section 6 i t easily follows that the function we look for is determined by:

,with K given by:

Define s=2cl and assume eco~=o, then:

STEP 2)

We now determine the function ~(~) for the angle ~ after the point mass m has climbed up.

If

m

has climbed up at time t'=(l at the angle

4>,

then ¥O~=4> and

~O~=X

-2ec

4>~

as we have al ready seen.

Just like in step 1) we can derive that:

With the results of step 1) we can express

(~)2

in ¢. From that i t follows that:

STEP 3)

The ma:dmum amplitude ¥'ma.x at the right side is fLllly determined by the equation:

(

~¥'ma.x,)

) 2 =0.

With the equation derived under step 2) we can conclude that is the solution of the next equation:

«

cos¢-s:II'sin¢+<:-cos(:KO')+s:II'sin8<:OX>expCs<:e<:o')-¢::>.:> ) / ( ),..9<:1 +S2,) )

+

<:

-cos¢+sXs i n¢,) /

<:

1 +S2X2,) ) :ll'expC sX<: ¢-¥'ma.x')') = (I.

STEP 4)

The equation in step 3) determines ¥'ma.x as a function of ¢.

Now we want to find the best ¢, denoted by m, for which ¥'ma.x is as large as possible.

We can find

m

by solving the next equation: d

Now apply the operator

~.

I¢=§ onto the equation derived under step 3) and use (7.1).

Doing so~ we get an equation which has to be satisfied for § to be the optimal ¢ ( fortunately the term with ~QXC§> cancels ).

The resulting equation holds. if X=l or if:

().. 2+X+l +sC)" 2+)..»sin§+scos§+sC-cos8CO>+slfsin8eO»lfeXpCsC8eO>-§D> =0.

X=l is an uninteresting case, because m does not climb up at all.

For sand § small ( s=O .. Fa.i.r=O • §=O

written in lowest orders of sand §: § = -slfCl-cos8eO» / eX2+X+l>.

this equation can

So~ for larger Fa.i.r (or larger s) m has to climb up earlier.

section 8: REDUCTION OF THE MORE REALISTIC MODEL TO THE MATHEMATICAL MODEL

Look again at the picture showing the model for a child on a swing , consisting of three point masses (see section 4).

Picture:

We can determine at each moment where the centre of mass is

situated~ given the positions of the three mass points in terms of the basis (O.e~.e2.e9).

Assume that we move mz and ma i nstantaneousl y, such that the distance between the centre of mass and the origin 0 does not change. We should notice that the position of Cm, i.e. the centre of mass~ in the "total system' can not be changed by internal forces.

All the forces which are involved to change the position of mz or ma are internal forces, so these forces do not change the position of em.

There are only two external forces, namely the gravitation force and the reaction force in

O.

But when we rotate mz and ma around mt, the el-:ternal forces do not change either ( the distance between 0 and Cm does not change). So, em remains at its position and there is no change of energy of

In other words, if the child moves its legs or upper-body, under the assumption that ICmOI does not change, its centre of mass does not move in the total system. The child moves the cable,

centre of mass.

not the

Now suppose, that the distance between Cm and 0 changes with Al.

We again assume that the changes are instantaneously. Then we can decompose the motion of Cm in two parts:

1) the motion of Cm in the radial direction until ICmOI=l+Al, 2) the motion of cm around ~, such that the distance between

and Cm does not change.

o

The second part of the motion is the same as described above and does not change the energy of Cm.

The first part of the motion is the same as described in section 3, where we looked at a mathematical model, with only one mass point. The only difference is that the cable of section 3 has to be replaced by the imaginary cable between Cm and O.

change the energy of cm~

Now we can

In fact, by looking at the centre of mass in stead of the three mass points, we get the mathematical model of section 3 back.

section 9: CONCLUSIONS

We end with some conclusions we recover from this report.

The problem to be solved was posed as follows:

A child is sitting on a swing and is swinging up and down without help from anybody else. The child is capable to increase its amplitude by moving its legs and upper-body in a certain way. The question arises what strategy the child should follow to increase its amplitude.

Every successful strategy involves the input of energy into the system one way or the other. The question remains how to do this.

Our first mathematical model shows us that we have to stand up at the lowest point and to sit down at the highest point, or in terms of the centre of mass: we should bring the centre of mass up at the lowest point and bring the centre of mass down at the highest pOint.

Our second model can be brought back to the first model by looking at the centre of mass of the three point masses. The strategy to follow can also be described as follows: when we are at the lowest point we should try to bring our legs and upper-body up as far as we can and when we are at the highest point we should try to bring our legs and upper-body down as far as possible.

When we incorporate the influence of air resistance, the strategy changes a little bit: now we should bring our legs and upper-body up as far as possible just before we reach the lowest pOint.