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Rube Goldberg Project

Rube Goldberg Project. George. Kyle Kirby (Electrical Engineer) Ryan Sweet (Nuclear Engineering) Wes Hicks (Civil Engineering). Team George.

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Rube Goldberg Project

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  1. Rube Goldberg Project George

  2. Kyle Kirby (Electrical Engineer) • Ryan Sweet (Nuclear Engineering) • Wes Hicks (Civil Engineering) Team George

  3. 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. Project Overview

  4. 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. Projectile motion (29-1)=tan(1)(-2)-(32.2/2(V)2)(1+tan(1)2)(22) V=34 in/s Theta = 1Degree Y-max=29 in. X-max=2in. Step One – Launcher and Projectile Motion

  5. 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 Energy KE=(1/2)IW2 KE=(1/2)((2/5).005(.005)2)(V/.005)2 Mass of Marble=.005kg Radius of marble=.005m V=34 in/s or 8.636 m/s Step Two – Loop and Rotational Energy

  6. 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. Translational Energy Step one Kd2= mgh .456(12)=.005(32.2)(2.833)ft Step two Mgh = mgh + .5mv2 .005(32.2)(2.833)=.005(32.2)(.41666ft)+.5*.005(V)2 V=12.474 ft/s 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 Three – The Second Funnel Drop and Translational Energy

  7. 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. Center of Mass of the dropping weight (1.25*.0388)/.0388=1.25in. Step Four – The Weight and Center of Mass

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