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The Playground Rube Goldberg Machine

The Playground Rube Goldberg Machine. By Anish Prasanna and Erik Hung. Steps on how it works. 1)The trigger is started by lifting the stick which pushes the ball into the large tube as shown in picture 1.

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The Playground Rube Goldberg Machine

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  1. The Playground Rube Goldberg Machine By Anish Prasanna and Erik Hung

  2. Steps on how it works 1)The trigger is started by lifting the stick which pushes the ball into the large tube as shown in picture 1. 2)The ball then falls into the tube’s inside which is shaped like that of a screw. The ball then falls in free-fall. It is dropped out of the bottom where it starts to roll down the incline as shown in picture 2. 3)Once the ball gains speed, it rolls down the incline. The ball rams into the black roll and knocks down the support of the heavy side of the lever (picture 3). 4) This causes the heavy side to go down and in effect the lighter side to move up and knock the book over as shown in diagram 4. This book then pushes the cup that knocks the heavy side of the pulley over the board. 5) In effect the heavy side pulls downwards on the pulley and the lighter side gets pulled up as in diagram 5. The lighter side is attached to the catapults latch that holds down the catapult. Once the sting pulls upward so does the latch and the catapult is released. Diagram 5 Diagram 4

  3. Energy Transformation calculations of an inclined plane with uncertainty propagation Energy of the ball turns into work to activate the lever by knocking the roll out of the way. Mass of Ball= 5 grams= .005 +/- .001 k.g. Height of incline= 10.5 cm = 0.105 m+/- .001 m Angle of incline= 19 degrees = 19.0+/- .5 degree Distance roll moves= 5.0cm +/- .5 = .05 m +/- .01m (this was found without using the lever) Potential Energy= Mass of ball * gravity * height = (.005kg+/-.001 k.g) * (9.80 m/s^2) * (.105m +/-.001m) Uncertainty Calculations=( (.001 kg/.005kg) + (.001m/.105m)) * (.005kg*.105m) = uncertainty of +/- .00013 J PE= .005145 +/- .00013 J Uncertainty Equation for multiplication= The potential energy of the ball turns into work done by the ball onto the roll. The following calculations find how much force had acted upon the roll. To find how much force was applied we must use the equation W=F*d*cosine(theta) since the displacement is forward however we can ignore cosine since cos(0)=1. We can also change the equation to F=w/d F= (.005145 +/- .00013 J / (.05 m +/- .005m) Uncertainty Equation for division: Uncertainty of F=((.00013 J/ .005145 J) + (.005m/.05m) ) * (.005145J/.05m) = .01055 F= .1029 N +/- .01055 N

  4. Incline Plane/ Power of ball calculations PE=W= .005145 +/- .00013 J To find time taken for ball to roll : First lets calculate the initial velocity on the incline. This can be done by finding the final velocity of the ball in free-fall and then using the horizontal component for the initial velocity. Initial angle launch= -4.2 degrees +/- .2 degrees Height of tube= 23 cm= .23 meters +/- .01 meters Height of drop= 42 cm= .42 meters +/- .05 meters Mass of ball = 5 grams= .005 +/-.001 k.g. Angle of incline=19.0 degrees +/- .5 degree length of hypotenuse of incline: 38.1cm +/- .01cm Visual representation:

  5. Power Cont. Lets find the horizontal initial velocity first. To do this lets use trigonometry. Sin(90)/9.8=sin(4.2)/x Vhi= .7177 m/s sin(90)/9.8=sin(85.8)/x Vvi=9.77 m/s Now that we have the initial velocities, we can use kinematics to find the velocity at the end of the free-fall inside of the tube. Vhf^2=Vhi^2 +2ax Vhf^2=(.7177m/s^2)+2(0)(.42m) Vhf=.7177 m/s Now that we have this we can plug this into the incline and find out the time needed for it to hit the roll. We can do this by finding the acceleration of the plane and then using kinematics to find the time. To find the acceleration we can use trigonometry. Sin(90)/9.8=sin (19)/hvi Hvi=3.19 m/s^2 (as it goes down incline) Now for kinematics- d=vi *t +(1/2)a t^2 .381m = .7177*t+ (1/2)(3.19)t^2 T=.2694956 seconds Finally we can find power. P= .005145 J / .2694956 seconds P=.019091 W Free body of marble

  6. Mechanical Advantages of the 2 simple machines Incline= Fout/fin= 38.1cm/10.5 cm=.381m/.105m= incline multiplies force by factor of 3.6285 lever=li/io= .10875m/.10875m= factor of 1= no advantage

  7. Energy Transformation calculations of a pulley PE of the counterweight of the pulley transforms into work to pull the latch on the catapult Height of counterweight to floor= 1.6m+/- .02m Mass of counterweight=60g= .06kg +/- .01 kg Movement of latch after= 6.875 cm +/- .001 cm Angle to catapult= 120 degrees +/- 5 degrees (projected) PE=mgh= (.06kg)(9.8m/s^2)(1.6m)=.9408 J This PE converts to work done on the latch by the string on the pulley. To find the force done on the latch W=PE= .9408 J=F*d* cos(120) .9408 J= F * .06875m* cos(120) F=w/d F= .9408 J/ .06875m *COS(120) F=-11.142 N

  8. Conservation of momentum calculations for the ball rolling down the incline Lets find the velocity at the end of the incline using conservation of momentum. Initial V of ball= .7177 m/s Mass of ball= 5g= 0.005 kg +/-.001 kg mass of roll= 8g= .008 kg+/- .001kg Height of incline = .105 m +/- .005 m PE1+KE1=PE2+KE2 M1gh+(1/2)m2v^2 = M1gh2 + (1/2)m2v^2 ((.005 kg * 9.8m/s^2 * .105m) + ((1/2)(.008 kg)(0 m/s)= (.005kg)(9.8)(0m) + ((1/2)(.008kg)(v^2) .005145+0=0 + (.004))(v^2) .005145/.004= v^2 Initial velocity of roll= 1.134 m/s

  9. Torque of lever

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