Fundamental Physics by Ramamurti Shankar at Yale_1_Newtonian Mechanics

Learning Newtonian mechanics certainly ordains the math of differentiation and integration, which belongs to the course of calculus. While the concept of differentiation/derivation is relatively easy – we can make sense of it by thinking of the tangent line on a curve point or velocity of a moving object – the mathematical deduction is interesting and bit challenging.

To intuitively infer the above formula, let’s take the simple steps, first c is constant, obliviously the derivative is 0; second, if n=1, and let’s say f(t) = 3t, it’s a straight line. if the delta t is a vanishingly small one 0.001, the increase/delta of f(t) would be 0.003, hence the first degree of derivative is 3.

What’s if things are more complicated, say f(t) = t^2, referencing from this webiste, “In plain English: We analyzed how f(x)=x2f(x)=x2 changes, found an “imperfect” measurement of 2x+dx, and deduced a “perfect” model of change as 2x.”

The application in physics is about predicting an object trajectory given present status predicated on everything function under gravity. And we all know (based on Newtonian theory) that the gravitational constant g = 9.8, which from the perspective of differentiation in math, it is the acceleration, second degree of derivation.

Professor hence deduced the formula to project an object on earth:

What Physicists differ from pure mathematicians is that they use math as a tool to make sense of the world. So in his lecture, he asked what is the height the object could go, if one stands at a platform 15 meters above the ground, and throw a ball upward with a speed of 10m/s, how tall can the ball go? The tricky part is to figure out that when the ball reaches the top, its speed has to be 0, hence t can be solved, then y can be solved. Or, one can use the bottom equation directly to get the x value.

The other question he asked is the how long the time elapsed when the ball hits the ground, similarly, we figured it is when y(t) = 0, if plug in the equation, we found t could be two values, one is 3, which is sensible, while the other is -1. So what is the -1 mean?

It’s fascinating in physics, that instead of dogmatically disregard this value, physicists contemplate on the legitimacy. Math problem such as this leads to great discovery of anti-particles by Dirac. Back in this case, a time of negative 1 second, simply can be viewed as a symmetrical parabola, it’s the left quarter of the whole picture, since the origin point’s time is set to be 0, the left part time variable has to be designated as -1 second.

Hence, professor Shankar stress the importance of listening to what math equation tells even when it doesn’t seems legitimate at first glance.

Newton’s foundational role in modern science/physics is established by the three Newton’s law:

  1. an object will stay in motion if in motion, stay at rest if at rest unless it is acted on by an outer force
  2. F = ma, the force is the production of the mass and acceleration. Unit of F, since it’s discovered by Newton, will be Newton, Unit of mass is defined in an arbitrary way as how we define what meter is. With the aid of a spring to think: if there is such a standard body, it makes 1 Newton/1 acceleration, then the mass is 1kg, then, replace it with an elephant, or any other objects, we can find the multiple of that standard mass, as the value of it’s mass.
  3. T = -T, caveat is that except gravity, this law ordains that the two objects are tangible in contact.

Prof Shankar deduced a moving object around a circle with radius or r possess an acceleration rate a, in above fomula.

To conduct math calculation, it’s needed to go over the important tiology equations as below, note tan theta = sin theta / cos theta.

Applying Newton’s law into practical problems is fun and useful. For example – two masses linked on an inclined slope problem:

And a rotation problem, which well relates to the amusement park experience:

This explains car racing, and we know to prevent the high-speed racing car from flying off the track, there is a static friction exerting forces. More worth-noting, we pause and ponder ‘is the static friction a must-to-have condition’? Not necessary, if we put the vehicle on a bend or barrel, like what seen in circus, the acrobat riding bicycle round and round without falling down, the surface is so smooth that the friction is negligible, but the clever slope design offers horizontal vector force as static friction serves in flat plane.

Lastly, the climax of Newton problems is the loop-to-loop problem. It is not intuitively understandable even we also can related it to the adrenaline-generating moment when we are pushed on top of roller coaster in amusement park too. The key crux here is that even we were static on top – we don’t drop – the acceleration V^2/R is high. Hence T + mg = ma. Could T be negative, that means V^2 <= Rg, at that exact instant V^2 = Rg, T = 0, but realistically, I don’t think the surface would provide an upward force to the cart.

As is depicted in the right part of the graph, the same is applicable to shooting a bullet into the air, if the bullet went with a fast enough velocity V^2 = Rg, then it will not fall back on earth.

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