The Goldberg Volcano

By Matt, Erik,Scott, and Tom

Track and MarbleTrack and Marble

11stst step step

GIVENGIVEN• Length of Top Track (1)=32.25”=2.688’Length of Top Track (1)=32.25”=2.688’• Length of Bottom Track (2)=33”=2.75’Length of Bottom Track (2)=33”=2.75’• Change in Height of 1 (A)=9.5”=.79’Change in Height of 1 (A)=9.5”=.79’• Change in Height of 2 (B)=11.5”=.958’Change in Height of 2 (B)=11.5”=.958’• Angle of Track 1 to ground (Theta 1) =17.13°Angle of Track 1 to ground (Theta 1) =17.13°• Angle of Track 2 to ground (Theta 2)=20.39°Angle of Track 2 to ground (Theta 2)=20.39°

Rube GoldbergRube Goldberg

Rube GoldbergRube Goldberg

Rube GoldbergRube Goldberg

By using Conservation of Energy, We were able to By using Conservation of Energy, We were able to determine the velocity before hitting the ending determine the velocity before hitting the ending wall and dropping onto the next track.wall and dropping onto the next track.

V1= A*32.2 ft/s²=1/2 v²V1= A*32.2 ft/s²=1/2 v²

=7.14ft/s=7.14ft/s After bouncing back up the track, this is the final After bouncing back up the track, this is the final

velocity at the bottom of the second incline trackvelocity at the bottom of the second incline track V2= .958*32.2=1/2 v²V2= .958*32.2=1/2 v²

=7.85ft/s=7.85ft/s

Coefficient of RestitutionCoefficient of Restitution

Finding the speed of the marble bouncing Finding the speed of the marble bouncing backback

1=32.25”=2.688’1=32.25”=2.688’ 2=5.5”=.4583’2=5.5”=.4583’ V1=7.14ft/sV1=7.14ft/s V2=?V2=?

Rube GoldbergRube Goldberg

Coefficient of RestitutionCoefficient of Restitution

e=-(v1’/v1)e=-(v1’/v1)

-(-4.321/7.14)=e-(-4.321/7.14)=e

=.605=.605

V2=-2.94 ft/s (As V2=-2.94 ft/s (As found by timing and found by timing and measuring the measuring the distance the marble distance the marble traveled.)traveled.)

Dropping WeightDropping Weight

h=8”h=8” w=1.6 ozw=1.6 oz

Weight DroppingWeight Dropping

Weight DroppingWeight Dropping

mgh=(1/2)mv²mgh=(1/2)mv²– The masses cancel out The masses cancel out – gh=(1/2)v²gh=(1/2)v²

32.2(8”*1/12)=(1/2)v² v=6.55ft/s32.2(8”*1/12)=(1/2)v² v=6.55ft/s KE=(1/2)(6.55ft/s)²*(1.6oz)(1lb/16oz)/32.2KE=(1/2)(6.55ft/s)²*(1.6oz)(1lb/16oz)/32.2

=0.66ft-lb=0.66ft-lb

The Domino EffectThe Domino Effect

Domino EffectDomino Effect

Domino EffectDomino Effect

For the first “domino”-For the first “domino”- aann=(6.55 (velocity from PE)/(3.75in*ft/12in) =(6.55 (velocity from PE)/(3.75in*ft/12in)

aann=137.3ft/s²=137.3ft/s² For second “domino”-For second “domino”- Rotates 9.5 times on impactRotates 9.5 times on impact f=9.5 in 5 secf=9.5 in 5 sec f=w*(1/2f=w*(1/2ππ)) w=19w=19ππ radians/5sec =3.8 radians/5sec =3.8ππ v=(3.75/12)(3.8v=(3.75/12)(3.8ππ) v=3.73ft/sec (2)) v=3.73ft/sec (2)

Domino EffectDomino Effect

e=vf/vi= 3.73/6.55 e=.57e=vf/vi= 3.73/6.55 e=.57

3.73^2/(3.75/12)3.73^2/(3.75/12)

Theoretically, since all dominoes are the same, the e Theoretically, since all dominoes are the same, the e remains similar throughout the recation.remains similar throughout the recation.

v=6.55ft/sec av=6.55ft/sec ann=137.3ft/s²=137.3ft/s²

v=3.13ft/sec av=3.13ft/sec ann=44.52ft/s²=44.52ft/s²

v=2.12 av=2.12 ann=14.38=14.38

v=1.21 av=1.21 ann=9.69=9.69

SwingSwing

SwingSwing

SwingSwing

Height the weight is dropped fromHeight the weight is dropped from =10”=.833’=10”=.833’ w=1.32oz=.08125lbsw=1.32oz=.08125lbs m=.0025 slugsm=.0025 slugs Change in height=5”=.417’Change in height=5”=.417’

PE=(.002523)(32.2)(.417)PE=(.002523)(32.2)(.417) =.0338 J=.0338 J

Hot Wheels CarHot Wheels Car

Hot Wheels CarHot Wheels Car

Hot Wheels CarHot Wheels Car

Mass of car=1.1oz=.069lb= .00214slugsMass of car=1.1oz=.069lb= .00214slugs Height=10”=.833’Height=10”=.833’Velocity at BottomVelocity at Bottom

(32.2)(.833)=(1/2)v² v=7.33ft/s(32.2)(.833)=(1/2)v² v=7.33ft/s

PE=(.00214)(.833)(32.2)=.05729JPE=(.00214)(.833)(32.2)=.05729JKE=(1/2)(.00214)(7.33)²= .05749JKE=(1/2)(.00214)(7.33)²= .05749J Impulse at bottom= .015686 slug-ft/sec²Impulse at bottom= .015686 slug-ft/sec²