rube goldberg project george. team george kyle kirby(electrical engineer) ryan sweet(nuclear...

Rube Goldberg ProjectGeorge

Team George

Kyle Kirby(Electrical Engineer)Ryan Sweet (Nuclear Engineering)Wes Hicks(Civil Engineering)

Project Overview

Our device is set with a loaded spring in a pvc pipe which is released with a mouse trap. It then shoots a marble out of the pipe and lands in a funnel beside the pipe. The marble then lands on a track and does a loop which ends in another funnel. The marble spins around the funnel and triggers the second mouse trap which pulls a weight off the second pvc pipe and raises the banner.

Step One – Launcher and Projectile Motion

Projectile motion(29-1)=tan(1)(-2)-(32.2/2(V)2)(1+tan(1)2)(22)V=34 in/sTheta = 1DegreeY-max=29 in.X-max=2in.

In this portion of the device a launcher shoots the marble out of the pvc pipe and lands into the funnel. In order to figure out how to do this we had to find the projectile motion of the marble. In this step we also encountered one of our issues which is inconsistent shooting of the marble.

Step Two – Loop and Rotational Energy

In this portion of the project the marble rolls down the track and goes through a loop. To do this we had to figure out the velocity of the marble.

Rotational EnergyKE=(1/2)IW2

KE=(1/2)((2/5).005(.005)2)(V/.005)2

Mass of Marble=.005kgRadius of marble=.005mV=34 in/s or 8.636 m/s

Step Three – The Second Funnel Drop and Translational Energy

Translational EnergyStep oneKd2= mgh.456(12)=.005(32.2)(2.833)ftStep twoMgh = mgh + .5mv2

.005(32.2)(2.833)=.005(32.2)(.41666ft)+.5*.005(V)2

V=12.474 ft/s

In this section the marble falls out of the funnel and hits the mousetrap to pull the string. We needed to figure out the velocity of the marble to see if it would set off the mousetrap.

In the transition from step two to this step we came across another issue, which is that occasionally the marble will overshoot the funnel.

Step Four – The Weight and Center of Mass

Center of Mass of the dropping weight(1.25*.0388)/.0388=1.25in.

Here we had to find the center of mass of the weight because we needed to know where to place the weight so it would drop the correct way.