Physics and Math of Swing Pumping

Introduction

            A child pumping (increasing amplitude) of a swing is a well known example of the way that the amplitude of a harmonic oscillator can be increased.  It is important that the extra force be applied during the middle of the swing. Also, in order to increase the total energy of the swing, the child must act against both gravity and centripetal force.

Math and Physics

            One of the more obvious behaviors in swing pumping is that the child lifts her/his lower legs so that the shins are parallel with the thighs while the swing is going forward and returns them to 90 degrees with respect to the thighs at the height of the backswing. The act of lifting the lower legs at the bottom of the swing trajectory and returning them at the back swing both have the effect of adding gravitational energy to the total energy of the swing. 

            Let the mass of the lower legs be named mL and their length be named L l MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaamiBaaqabaaaaa@37DA@ .  At the low point of the swing, the energy added to the swing by this act alone is then

δE= m L g L l 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjaadw eacqGH9aqpdaWcaaqaaiaad2gadaWgaaWcbaGaamitaaqabaGccaWG NbGaamitamaaBaaaleaacaWGSbaabeaaaOqaaiaaikdaaaaaaa@3F09@  

(1.1)

                                                                       

where the effective height the calves are raised is L l /2 MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadYeadaWgaa WcbaGaamiBaaqabaGccaGGVaGaaGOmaaaa@3953@  and g is the acceleration of gravity.   The centripetal acceleration is a c = ω 2 L MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadggadaWgaa WcbaGaam4yaaqabaGccqGH9aqpcqaHjpWDdaahaaWcbeqaaiaaikda aaGccaWGmbaaaa@3C87@  where ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@37B9@   is the angular rate of the swing and L is the length of the swing rope from the anchor to  the seat.  So when ω MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabeM8a3baa@37B9@  is substantial, then ac should be added to g in equation (1.1).

            A similar argument for energy increase can be made for the use of the hands to bow the rope, thereby lifting the seat at the bottom of the swing arc.   This movement directly increases the angular momentum since the force needed is higher at the middle of the swing arc than at the ends because of larger centripetal force at the middle.

 

At any one time, the total energy of the swing is

 

E t =Mg L s (1−cos( t Amp )] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadweadaWgaa WcbaGaamiDaaqabaGccqGH9aqpcaWGnbGaam4zaiaadYeadaWgaaWc baGaam4CaaqabaGccaGGOaGaaGymaiabgkHiTiGacogacaGGVbGaai 4CaiaacIcacaWG0bWaaSbaaSqaaiaadgeacaWGTbGaamiCaaqabaGc caGGPaGaaiyxaaaa@47E4@  

(1.2)

                                                                       

where M is the mass of the child, Ls is the length of the swing rope, and tAmp is the angle of maximum amplitude of the swing.

Equation (1.2) can be solved for the maximum swing angle:

           

t Amp = cos −1 [ 1− E t Mg L s ] MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadshadaWgaa WcbaGaamyqaiaad2gacaWGWbaabeaakiabg2da9iGacogacaGGVbGa ai4CamaaCaaaleqabaGaeyOeI0IaaGymaaaakmaadmaabaGaaGymai abgkHiTmaalaaabaGaamyramaaBaaaleaacaWG0baabeaaaOqaaiaa d2eacaWGNbGaamitamaaBaaaleaacaWGZbaabeaaaaaakiaawUfaca GLDbaaaaa@48DF@  

(1.3)

We can solve equation (1.3) for the change in amplitude per unit energy input via pumping:

 

                                                                                                                                               

d t Amp d E t = 1 MgL 1− ( MgL− E t MgL ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaam izaiaadshadaWgaaWcbaGaamyqaiaad2gacaWGWbaabeaaaOqaaiaa dsgacaWGfbWaaSbaaSqaaiaadshaaeqaaaaakiabg2da9maalaaaba GaaGymaaqaaiaad2eacaWGNbGaamitamaakaaabaGaaGymaiabgkHi TmaabmaabaWaaSaaaeaacaWGnbGaam4zaiaadYeacqGHsislcaWGfb WaaSbaaSqaaiaadshaaeqaaaGcbaGaamytaiaadEgacaWGmbaaaaGa ayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaaaaaaa@4E40@  

(1.4)

We can then use this derivative with equation (1.1) to find the change in amplitude per swing:

Δ t Amp = d t Amp t E t E= m L L L 2ML 1−( 1− ( E t MgL ) 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabfs5aejaads hadaWgaaWcbaGaamyqaiaad2gacaWGWbaabeaakiabg2da9maalaaa baGaamizaiaadshadaWgaaWcbaGaamyqaiaad2gacaWGWbaabeaaaO qaaiaadshacaWGfbWaaSbaaSqaaiaadshaaeqaaaaakiaadweacqGH 9aqpdaWcaaqaaiaad2gadaWgaaWcbaGaamitaaqabaGccaWGmbWaaS baaSqaaiaadYeaaeqaaaGcbaGaaGOmaiaad2eacaWGmbWaaOaaaeaa caaIXaGaeyOeI0YaaeWaaeaacaaIXaGaeyOeI0YaaeWaaeaadaWcaa qaaiaadweadaWgaaWcbaGaamiDaaqabaaakeaacaWGnbGaam4zaiaa dYeaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGLOa Gaayzkaaaaleqaaaaaaaa@5812@  

(1.5)

We see from equation (1.5) that the amplitude added per swing period is somewhat larger than ½ MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaGqaaKqzagaeaa aaaaaaa8qacaWF9caaaa@3842@  of the mass times length ratio. 

As a numerical example take the following values:

ML=5 kg

LL =0.3 m

L=2.5 m

M=30 kg

tAmp=0.5 radians

Then

Et=90 Joules

Then:

ΔtAmp=0.014 radians per swing period.