Diary of a Student Engineer Part 8: Rube Goldberg Machine Part 1
So, to clear up the first thing you’re probably wondering about, I should probably give a brief overview as to what exactly a Rube Goldberg Machine actually is? Well, you’ve probably seen one before without knowing what it is called, essentially it is a chain of mechanical reactions designed to perform a task (such as putting a ball in a bucket) in the least efficient way possible. And generally, the result ends up looking pretty cool.
There are some especially impressive Rube Goldberg Machines that you can find online, which must have taken months to build (and a lot of free time) but result in a spectacular ‘domino effect’ incorporating all sorts of triggers and reactions, like this one:
After making my own Rube Goldberg Machine, I have a lot more respect for the amount of work required for even larger scale machines. To say it took a fair amount of time for my comparatively smaller version would be an understatement. However, I do think I was able to incorporate a lot of basic mechanical engineering principles to get the job done.
So, in this post I’m going to discuss the first half of my Rube Goldberg Machine and the engineering design I incorporated to get it to work.
Step 1
As it was required that my Tumbller robot from previous posts played a role in the machine, I decided to place it at the very start given that I could control it and therefore control when to begin the chain reaction. So, the first component consisted of the robot and the principle of friction force:
As you can see, the plan is that as the robot pulls away, the friction between the top of the robot and the bristles of the broomstick will cause the stick to move with the robot. The other end of the broom stick is placed on top of a football on a ramp which keeps it in place until removed. The maths here is really quite simple, as long as the coefficient of friction is greater between the broom and robot than between the broom and the football we can expect the broom to be dragged off the football allowing it to roll away.
Step 2
The next component is a simple one, once the football is released from the top of a ramp, we let gravity play it’s role as the ball rolls down the slope. One design consideration is the placement of guide rails on either side of the ramp so that the ball doesn’t fall off.
A quick calculation done here is to determine the speed of the ball as it reaches the end of the ramp. Approximating the mass of the football to be 0.5kg, the ramp angle, theta, to be 30 degrees and the length of the ramp to be 1 metre. We can calculate the acceleration of the ball as (0.5)g*Sin(30) = 2.45 m/s² and then using the equation v² = 2*a*s we get a final velocity of 2.21 m/s. Therefore the momentum of the ball will be 1.11 kg m/s.
Step 3
Next, my plan was to take advantage of this momentum generated by the ball and use it to knock down a weak structure. Generally, structurally sound designs tend to disperse forces throughout the structure in an even manner. This is generally why triangulation is often used in truss bridges for example and steel is used to absorb external forces. The worst design I could find in my house was a rhombus-shaped cardboard box. As shown below it doesn’t take much deformation for the box to lose much of it’s surface area with the ground and the cardboard itself is rigid enough to not resist deformation in response to an external force. So it collapses. This collapse can then trigger another collapse, in this case a rake, which relies on the box to stand upright.
There was no real calculation I could do here but rather a little knowledge of structural design and trial and error ensured the efficacy of this mechanism.
Step 4
After the rake falls we can then again take advantage of this kinetic energy but this time I wanted to translate the somewhat linear motion to circular motion. I achieved this using an inverted bike. By attaching an extended lever to the bike wheel, I can construct a paddle wheel that translates the momentum of the falling rake to the motion of the wheel.
This is a typical lever calculation. Assuming the mass of 0.5 kg to be at the centre of the rake and the rake to have a length of 1.5 m, we can calculate the final velocity of the rake when it strikes the bike wheel using potential and kinetic energy equations. The centre mass of the rake at rest has potential energy (mgh) of 0.5*g*0.75 = 3.68 J. This energy must be conserved so at 45 degrees, the rake must still have 3.68 but it is dispersed between potential and kinetic energy types. Therefore 3.68 = 0.5*g*(0.75 Sin 45) + 1/2 (0.5)*v². Solve for v to get a velocity of 2.08 m/s. Assuming the rake hits the wheel at it’s midpoint, then we can also take the translation of momentum to be 1.04 kg m/s.
Step 5
With the circular motion now providing a pulling force rather than the pushing forces I have been dealing with in the last few steps, my plan was to leverage this to create a trap door of sorts that would otherwise not be possible with other ‘pushing’ motions.
I kind of achieved this by using a ramp and hinge mechanism. Essentially, using a cord attached to the bike wheel, I could pull a container on a hinge to lift upwards, a bit like opening a door that was facing downwards. By placing this mechanism on a ramp, the opening of the hinge would allow another smaller wheel to be released and travel down the ramp.
Calculations-wise, if we take the cord to be attached at the circumference of the bike wheel then we can roughly calculate the tension force in the cord to be equal to gravitational force (mg) applied by the rake in step 4. In reality the transfer of energy will be very inefficient here so we will assume an efficiency of 50%. Hence: T = 0.5*9.81*0.5 = 2.45N.
Step 6
Time for a bit of reverse engineering! Now that I have a wheel providing linear motion, I thought I’d utilise the other wheel of my bike to translate the linear motion back into circular motion again. This was done by connecting the rolling wheel and the bike wheel using a piece of string. I also attached a small weight made out of cardboard to the bike wheel. This of course would mean that only a small pull force on the bike wheel would trigger the natural motion of the bike wheel to convert the high potential energy level of the weight (when it’s at the top of the wheel’s motion) to a zero-state level, when it reaches the bottom.
Again we can use an energy equation to calculate the velocity (and then the force) of the weight when it reaches the bottom of the bike wheel’s motion.
Again; mgh = 1/2 m v² + mgh. The weight is quite light here so I will take the mass to be 0.1 kg. The wheel diameter is approx. 0.6m. So: 0.1*g*0.6 = 1/2 (0.1)*v² + 0. This gives a velocity of 3.43 m/s. We can then calculate the centripetal force at this moment to be (0.1)*(3.43) / (0.6) = 0.57N.
And so that’s about the halfway point of my Rube Goldberg Machine. If I’ve learned anything so far about the nature of these machines, it is that the time spent designing and adding components to the sequence is absolutely not proportional to the length of time the chain reaction effect gives back. I reckon these 6 steps result in about 10 seconds of motion in total, so fair play to the folk that have the patience to make a machine capable of several minutes in length!
