Transcript Example 4

The Goldberg Volcano
By Steven, Jimmy, Joey,
and Brad
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°
Rube Goldberg
Rube Goldberg
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
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=?
Rube Goldberg
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.)
Dropping Weight
 h=8”
 w=1.6 oz
Weight Dropping
Weight Dropping
 mgh=(1/2)mv²
– 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
The Domino Effect
Domino Effect
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)
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=2.12
an=14.38
 v=1.21
an=9.69
Swing
Swing
Swing
 Height the weight is dropped from
=10”=.833’
 w=1.32oz=.08125lbs
 m=.0025 slugs
 Change in height=5”=.417’
PE=(.002523)(32.2)(.417)
=.0338 J
Hot Wheels Car
Hot Wheels Car
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= .05749J
Impulse at bottom= .015686 slug-ft/sec²