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By Steven, Jimmy, Joey, and Brad

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1 By Steven, Jimmy, Joey, and Brad
The Goldberg Volcano By Steven, Jimmy, Joey, and Brad

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

22


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