Hopefully you’re sufficiently enticed to continue to journey with me through the dark-valley of elementary engineering mathematics. Wonderful. I’ll do my best to try to keep it all short and sweet, but be forewarned that this is literally only the beginning. We’re going to drill deeply into the basics here over the next weeks and months (perhaps years) because it’s the basics upon which we’ll eventually have to rebuild after…
Well, let’s not get into that just yet.
OK, so when last we broached this subject we were discussing “motion”. We now have enough information to start to delicately travel into the subject of kinematics – broadly defined as the science of the motion of bodies. Now, we can spend a month of Sundays defining the terms “science” and “bodies”, but let’s put all that off for another day before I start to rant about the differences between statics, dynamics, kinematics, mechanics (a whole-lotta “ick”s there – certainly a topic for another post, methinks – and don’t even get me started on the so-called “debate” between whether the world is linear or nonlinear, I’ll go deeply into that as well). Because we’re going to keep this simple for now, let’s just look at the mathematical kinematics of a particle in 1-dimension.
Given an acceleration a over a time period t, we can utilize the definition of acceleration we introduced to calculate the velocity of that particle through…
v = v0 + a t
… where v0 is the initial velocity of the particle at the start of our time observation. Similarly, we can calculate the new position of the particle given the initial position x0 by
x = x0 +v0 t + 1/2 a t2
This is of course really basic stuff, essentially the integration of the equations defining velocity and acceleration. The relations don’t change as we expand beyond 1D; we only change from working with scalars to working with vectors.
Now, we’ve restricted our discussion so far to what is known as “linear motion” – in other words, motion that does not involve rotation, or angular changes. To understand angular motion, all one needs to do is to grab a yo-yo, run the line out about a foot or so, and start spinning it (make sure nobody is “in range“). Or, just use a yo-yo the way it is typically used – whereby you’ll see both linear and angular motion (2-for-1, in other words).
For each quantity we measure for linear motion, there is an analogue for angular motion. For the angular position of a particle (from a direction in space) we typically use the symbol (the Greek letter theta). Angular displacement therefore can be defined graphically by thinking about a particle rotating about a fixed line; the angular displacement is given by the arc-length swept-out by the particle divided by the radius of the circular travel
= arc-length / radius = s / r
The angular velocity of the particle – the rate at which the angular position changes – is given by
= d / d t (where is the Greek letter omega)
Along the same lines, the angular acceleration is given by the time rate of change of the angular velocity…
= d / d t (where is the Greek letter alpha)
And to make matters even more complicated, using our spinning yo-yo example, there is not only an angular acceleration given by a, above, but there’s a centripetal acceleration for any body undergoing a circular motion, which tends to want to pull the body back toward the center of the circle of travel, given by…
ac = v2 / r
… where, as you may have guessed, the radius of said circular motion is given by r.
Mass, Force, Equilibrium, Work, and Power
We define mass as a quantitative measure of inertia (defined generally as the natural tendency of an object to remain at rest until and unless some external force acts upon it). And at the risk of being accused of circular definitions, a force is defined as any interaction that will work to overcome the inertia of an object, to either put it in motion, or change the motion of a body.
With these definitions in mind we can briefly introduce the concept of equilibrium by introducing one of the fundamental laws of physics – Newton’s Second Law of Motion.
|The vector sum of the external forces F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object: F = ma.|
A body is considered to be in equilibrium if the vector sum of the forces on the body is identically zero; thus, a body in equilibrium has no acceleration (n.b., the body can certainly have velocity, and therefore need not necessarily be at rest – defined as having no motion).
The work done by a force in moving a body is defined as the force multiplied by the distance traveled under the influence of that force…
W = F s
… where F is the average force and s is the magnitude of the displacement, or distance traveled. To understand the difference between Work and Force, let’s take an example picture…
Now, I’m sure nobody is going to argue that this gentleman is not applying substantial force on the barbell in keeping it up over his head right there. But, as the barbell isn’t moving, in the strict sense of the definitions, no work is being done. (I don’t recommend explaining to this gent that he’s not “doing the work” here.)
Energy is equivalent to Work, in the sense that the same units of measure describe them both. Among the energies we’re most familiar with in the mechanical realm are kinetic energy and potential energy. The kinetic energy of a body of mass m and average velocity v is given by…
KE = 1/2 m v2
Whereas the potential energy of a body of mass m that is a distance h over a datum (such as the surface of the Earth) under the effects of a gravitational acceleration g is given by
PE = m g h
We define a conservative force as one that does no net work on an object moving around a closed path when starting and finishing at the same location. As a visual, take a particle under a constant force (say, gravity) around a circle.
So, if Work (and energy) is Force over Distance, we can define Power (in the mechanical sense) as the work done by a force over time
P = W / t
Similarly, we can define the impulse of a force as being the action of a force over time
I = F t
We can now state the impulse-momentum theorem as
F dt = m ( vf – v0 )
Where the momentum p of a particle is given by
p = m v
It’s important to recognize that all of these linear quantities have angular analogues.
Before we go we ought to spend a moment or two defining how all of these quantities are characterized in systems of units. We’re going to stick with SI units on this blog just because, well, that’s the way this all seems to be going. I mean, the British don’t even use “Imperial Units” anymore.
Distance == meters (m)
Mass == kilograms (kg)
Time == seconds (s)
Force == netwon (N, kg*m/s2)
Work/Energy == joule (J, N-m)
Power == watt (W, J/s)
Pressure == pascal (Pa, N/m2)