1. By Matt, Erik, Scott, and Tom
The Goldberg Volcano
By Matt, Erik,
Scott, and Tom

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

3. Rube Goldberg

4. Rube Goldberg

5. Rube Goldberg
By using Conservation of Energy, We were able to determine the velocity before hitting the ending wall and dropping onto the next track.
V1= A*32.2 ft/s²=1/2 v²
=7.14ft/s
After bouncing back up the track, this is the final velocity at the bottom of the second incline track
V2= .958*32.2=1/2 v²
=7.85ft/s

6. Coefficient of Restitution
Finding the speed of the marble bouncing back
1=32.25”=2.688’
2=5.5”=.4583’
V1=7.14ft/s
V2=?

7. Rube Goldberg

8. Coefficient of Restitution
e=-(v1’/v1)
-(-4.321/7.14)=e
=.605
V2=-2.94 ft/s (As found by timing and measuring the distance the marble traveled.)

9. Dropping Weight
h=8”
w=1.6 oz

10. Weight Dropping

11. Weight Dropping mgh=(1/2)mv² 32.2(8”*1/12)=(1/2)v² v=6.55ft/s
The masses cancel out
gh=(1/2)v²
32.2(8”*1/12)=(1/2)v² v=6.55ft/s
KE=(1/2)(6.55ft/s)²*(1.6oz)(1lb/16oz)/32.2
=0.66ft-lb

12. The Domino Effect

13. Domino Effect

14. Domino Effect For the first “domino”-
an=(6.55 (velocity from PE)/(3.75in*ft/12in) an=137.3ft/s²
For second “domino”-
Rotates 9.5 times on impact
f=9.5 in 5 sec
f=w*(1/2π)
w=19π radians/5sec =3.8π
v=(3.75/12)(3.8π) v=3.73ft/sec (2)

15. Domino Effect e=vf/vi= 3.73/6.55 e=.57 3.73^2/(3.75/12)
Theoretically, since all dominoes are the same, the e remains similar throughout the recation.
v=6.55ft/sec an=137.3ft/s²
v=3.13ft/sec an=44.52ft/s²
v= an=14.38
v= an=9.69

16. Swing

17. Swing

18. Swing Height the weight is dropped from =10”=.833’ w=1.32oz=.08125lbs
m=.0025 slugs
Change in height=5”=.417’
PE=( )(32.2)(.417)
=.0338 J

19. Hot Wheels Car

20. Hot Wheels Car

21. Hot Wheels Car Mass of car=1.1oz=.069lb= .00214slugs Height=10”=.833’
Velocity at Bottom
(32.2)(.833)=(1/2)v² v=7.33ft/s
PE=(.00214)(.833)(32.2)=.05729J
KE=(1/2)(.00214)(7.33)²= J
Impulse at bottom= slug-ft/sec²