Representation of reference systems ... do not laugh, I do what I can

Coriolis, centripeta and vectorial differentiation

Saturday is the day of thinking or the day of mathematics. That is to say it is the day when you will hate me since I will start to put mathematics in this subject by raising the level a little.

Today I want to talk about centripetal and coriolis forces in rotation systems. It’s a very interesting topic. But we are going to see it in a Newtonian way, that is, through the differentiation of the position vectors and their forces. In this way, many of you can see the effort that Newton and like this, dominated mathematics.

The equations we are going to get today are worth a lot, for example, help us physicists to model atmospheric issues such as weather phenomena.

The whole point of all this is to take into account coordinate systems so as not to go crazy when we start to differentiate vectors in a rotating system.

Let us imagine the Earth, round as a sphere, and let us start by considering two Cartesian coordinate systems (X, Y, Z axes), a fixed one that does not rotate (based on vectors I, J, K, vectors remember) and at the same time, another System that rotates with the earth whose base we call {i, j, k} (you see a system of generators and therefore vectors as well). The latter, generated by {i, j, k} we consider it on the surface of the sphere (of Earth) and therefore, rotates at the angular velocity of any point on Earth, pi / 12 radians per time.

As K, the fixed system that does not rotate is in the center, we consider the K axis pointing to the north. And therefore, the angular velocity will be:

$latex \Omega =(\frac{\pi }{12})K$

Now if we take the origin of the coordinate system that rotates from which not, it will do so through a point that we will call $latex P_{0}$ and that will obviously have a vector from the origin of the system that does not rotate until Which rotates we will call $latex R_{0}$. Obviously $latex R_{0}$ will have an angle $latex \theta$ to the axis K so that the angle neither is 0 nor worth pi, that is to say, it is neither in the north pole nor in the south, but in between.

Put all the conditions, we go directly to the “thing”.

As we have to:

$latex \frac{\mathrm{d} r}{\mathrm{d} t}=v(t)=\Omega \times r(t)$

Where we are indicating that the differential of the position vector in time is the velocity (this is also valid for rectilinear movements and is easier to see), except that if we do for circular movements we have that is the vector product of the angular velocity Which is a vector, remember) by the vector of the position in time, which leaves from:

$latex \frac{distance}{time}=\frac{\frac{2\pi }{D}}{\frac{2\pi}{\Omega }}=\Omega D =\left | \Omega \right |\left | r(t) \right | sen\phi = \left | \Omega \times r(t))\right |$

So, we can say that:

$latex \frac{di}{dt} = \Omega\times i;\frac{dj}{dt} = \Omega\times j;\frac{dk}{dt} = \Omega\times k$

And, the most important:

$latex \frac{dR_{0}}{dt} = \Omega\times R_{0}$

Now, by basic algebra we have that any vector can be expressed on any basis with a mere base change. So let R(t) be a position, V(t) velocity and A(t) acceleration with respect to the fixed system (base I, J, K) with respect to the mobile (base i, j, k).

The important thing, and how beautiful the solution will give us is to see what is the relationship between the bases and therefore between the position, the speed and the acceleration.

This come from this equation:

$latex R = R_{0}+r$

And, taking this as a base we can now calculate velocity:

$latex V=\frac{dR}{dt}=\frac{dR_{0}}{dt} + \frac{dx}{dt}i+x\frac{di}{dt}+ \frac{dy}{dt}j+y\frac{dj}{dt} +\frac{dz}{dt}k+z\frac{dk}{dt}=v+\Omega \times R_{0}+x\Omega \times i+y\Omega \times j + z\Omega \times k=v+\Omega \times R_{0}+\Omega \times r=v+\Omega \times R$

Thus, by doing the same for the acceleration knowingly that is the differentiation with respect to the time of the speed, we have:

$latex A=\frac{dV}{dt}=\frac{d}{dt}(v+\Omega \times R)=a+\Omega \times v +\Omega \times (V)= a+2\Omega \times v+\Omega \times (\Omega \times R)$

Where, if we look at the terms (I have not put the double differentiation since I imagine that the reader, seen the above is able to do it for the same) and knowing that they are all vectors and not numbers with Which means that differentiation must be done by the axes, we have to.

Coriolis acceleration $latex 2\Omega \times v$
Centripetal acceleration $latex \Omega \times (\Omega \times R)$

But here the matter does not end, of course since although we have calculated by the movement we have not calculated by the forces, as also did Newton. So let’s see it now with the forces applying the above.

$latex F=mA=ma+2m\Omega \times v+m\Omega \times(\Omega \times R)$

So, letting out the acceleration, we have that:

$latex a=\frac{F}{m}-2\Omega \times v – \Omega \times (\Omega \times R)$

Nice!, we have that the object that is on the surface moving and to which we apply a force F, also has two other components that we will call Coriolis force ($latex -2 \ Omega \ times v$) and the centrifugal force.

In addition, we observe that the forces themselves are not real, but are a consequence of a change of coordinates of two reference systems , a fixed to a moving one. And all thanks to having taken the correct reference systems and to have applied force and trajectory differentiating them as a vector.

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