Due to end-of-semester busyness, I was not able to update over the weekend. But, I want to try and stay on my self-assigned blog schedule, and now I'm just studying for finals, so a day late is better than a week, right?
I already had the idea that science tried to break things apart. But, generally, I always thought this was to find the most basic understanding of the universe -- to be able to explain causation from understanding the way that everything works. I think this is still a part of it, but there's another part to it too.
The human mind can only compute so much.
To demonstrate, see the following physics problem (and if you've had Physics I before, I'm sure you've solved this problem before):
A 1 kg rock is suspended by a massless string from one end of a 1 meter measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the .25 meter mark?
I always find pictures to be useful when solving physics problems, so the first things first:
Really, that's about as complex as I normally draw, just to help me visualize a given scenario. In this problem, there is the word "Massless", which is just a fancy way of saying that the string that connects the rock to the meter stick doesn't have to be accounted for, so we're really just dealing with the rock, the meter stick, and the fact that the meter stick isn't moving even with the rock attached to it and the fact that the balancing point is located .25 meters away from where the rock is connected.
The main concept that needs to be applied here is the concept of Torque. Torque can be written in a number of ways, mathematically, but conceptually it's fairly simple, and related to my previous post talking about Force. When talking about Force, the examples and problems usually use cannon balls, footballs, or cars. That's because they easily relate to something called Translational Motion -- which is just the movement of an object from Point A to Point B. You throw a ball, it goes from your hand, Point A, to some spot on the ground, Point B, and there are a host of equations one can use to predict where that point will be based upon how hard you throw the ball, what angle you throw the ball at, and what the ball interacts with on the way there. These equations all have analogous equations that relate to another type of motion: Rotational Motion. Rotational motion is still motion, but it behaves differently than Translational Motion -- not so different that the Laws of the Universe are different, but we have to model them differently because their Translational motion would be a lot harder to model mathematically than it would if we were to just measure the motion of spinning things by the angles they travel through. So, really, it's still Point A to Point B motion, but instead of measuring things in meters, where the direction would constantly be changing, you measure things in Θ (Theta), a generic symbol meaning "Angle".
Torque is the rotational analogue to Force. But instead of F = ma, you have τ = Iα. τ is the Greek letter Tau, and it stands for Torque, which is rotational Force. α is the Greek letter alpha, and it stands for rotational acceleration (with units of radians/second^2, instead of meters/second^2).
This leaves "I". "I" stands for "Moment of Inertia", which does not explain itself as well as "Mass" does, so it requires a bit of explanation itself. Similarly to mass, if the Moment of Inertia is greater, it takes more Torque to gain a greater angular acceleration. But with rotational motion, you have to take more into account than the mass of an object. You also have to take into account how far away a mass is from the center of rotation. And, as you're actually dealing with a large number of particles all revolving around a single point (we'll call this point the "axle"), all of which may have different masses than each other, and most likely are at different lengths from the axle, this can easily get pretty complex. To be technically correct, you would have to find the distance a single particle is from the axle, find its mass, and compute its individual Moment of Inertia -- which is easy enough when you have only one particle. The equation for the Moment of Inertia of a single particle is "I=mr^2", where m is the mass and r is the distance from the axle. So, you square the distance of the particle from the axle, and multiply it by its mass. But when you're dealing with, say, a wheel, there are a lot of particles.
The way to then tackle a problem like the one above is the realize "Hey, this thing basically has an axle at .25 meters, and it has Torque being applied to that axle due to the Force of gravity. Even better than that, the thing isn't moving, so we know that the Torques are equal on both sides. So, the Torque of Left side is equal to the Torque of Right side, so I'll set their equations equal to one another. I know the mass of the rock, if I can figure out the Moment of Inertia for the Left side and the Moment of Inertia for the Right side, then I can find the mass of the meter stick".
Or, mathematically speaking, Iα(left) = Iα(right) from τ = Iα
This is where I made a mistake in tackling the above problem. There is a way to get around having to add up each individual particle, and in fact this simplification at least makes the moment of inertia calculable by hand. For example, when looking at pulleys (another favorite of physics problems) First, you assume that the particles are, more or less, the same mass, as the object is made of the same material -- a good assumption. Then, because the shape of a wheel is a regular shape where the outside of the wheel is equidistant from the axle, you can actually say "Hey! That pulley's a hellalot like a cylinder!", and make another assumption that is more or less correct: that the pulley will behave as if we had a perfect cylinder. The equation for the Moment of Inertia of a perfect solid cylinder is well known, so you can just plug it into the above equation and work away. It's 1/2mr^2, in case you're curious.
The problem is, the above problem is NOT a perfect cylinder, nor is it anywhere close to one. So, my first instinct was to go back to the basic definition for "I", where you can find "I" for any solid object (as that's what I'm dealing with). This involves integrating the volume of an object with respect to its mass which, quite honestly, is a pain in the ass -- at least for me. And actually, this is what I learned: It's not that there is anything wrong with taking the above approach, but you want to simplify the problem to make it easy, digestible, and understandable. And there is such a solution to the above problem, I just didn't see it initially.
It deals with a concept known as "Center of Mass". Center of Mass lets you treat a whole object as if it were a point particle. You mathematically find some fictional point near the object that the object will follow in motion, so you can use the equations you normally use for a point -- which are easier to deal with than whole objects. It also deals with how you define your system. Before I was looking at the system as "Left Side" and "Right Side", but that combines the mass of the meter stick on the left side, which itself is unknown, and I would have to use more algebra to find the unknown. Instead, if I look at the above problem as "Rock" and "Meter stick", then I at least have less algebra to do.
So, applying the idea of "Center of Mass" to the above problem, I have the rock. The objects weight is concentrated at the left end of the meter stick, due to the massless string, so I can treat the rock like a point at the left end of the meter stick. Then I have the meter stick. By itself, the meter stick, assuming that the mass of the meter stick is more or less spread out evenly (a good assumption), has a Center of Mass at its center. In relation to the axle we're dealing with, that puts the point particle .25 meters away on the right side of the axle, which is the exact distance of the rocks center of mass. So, the above picture can now be drawn as:
I put the circle and arrows in to emphasize the fact that we're really dealing with Torque here, even though this isn't a wheel. What can be seen in the picture is the two torques we're dealing with are in opposite angular direction, and are equidistant from the axle. The beauty to this solution lies in the fact that I is easy to find (mr^2, because now they're points). Also note, because the sum of the Torques are equal to one another, we have no angular acceleration to deal with (which means, technically, we wouldn't have any torque, but the above Torque equation is ACTUALLY written as "The sum of the Torques" with a Σ before τ to denote "Add Torques up" and differs slightly from the mathematical definition of Torque. I just wanted to tie the idea of Torque into Force from before)
So, we substitute "τ" for "Iα", then drop α because there isn't any, and are left with I = I. Substitute mr^2, and you're left with mr^2 = mr^2, and looking at the picture, you see that the Center of Mass is equally distant, so it follows that the masses of the two particles must be the same.
Had I started with Center Mass, I would've realized that the points were equally distant from one another, and that the meter stick wasn't going anywhere, so they'd have to be equal in mass too, and I could have solved this in less than 1 minute.
And that's when it dawned on me -- we can really make things as complex as we want. It's not the complexity in science that we even want. It's the simplicity. We're dealing with a highly complex universe that takes time to understand, and there's no way we'd understand it if all we do is take the clunkiest path to understanding. We want to break things apart and find the root cause of phenomena, sure, but in addition to that we just want to be able to understand the phenomena themselves without going through a huge and sometimes difficult to follow line of thinking (as I did above).
So, that's why you look for the simplest solution -- because we can only compute so much in our head at once, and there's a certain satisfaction that comes with a simple explanation if that little bit explains a whole lot.
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