So in the previous two examples, we have rotated a quasi one dimensional stick, either around its
end or around the middle of the stick.
And in both cases, the numerical solution was really easy.
It was basically just creating this list of our points and then just calculating into with a single
line of code until this one.
And still, you may say it's not really worth doing it because the analytical solution is so simple.
It's just a one dimensional integral where you just have to be careful and what you plug in, but then
you just have to integrate over our square, which is really easy to do.
It's just one-third.
Times are to the power of three.
So and I would agree to this statement, it wasn't really that useful, to be honest, because I would
also say this one is equally as easy.
However, now we're going to more difficult examples where the analytical result will become more and
more difficult and numerical results will basically stay just as easy.
And the first example of we're going to solve here is a sphere which is three dimensional, but it's
still very symmetric, and therefore we can use some tricks and make it a bit more easy.
So the first thing that I want to show you not in detail, but just briefly, is the analytical solution
to calculate the moment of inertia, of rotating a sphere around an axis which I just call the axis,
which goes through the center of the sphere.
And the result will be two over five m times the radius of the sphere squared.
So once again, we start from the same equation as we did before integral over R squared and R is not
the radius.
This is really important to remember.
This is the distance of the point that we are considering with respect to the z axis.
So always the shortest distance.
So it's not the radius, not the distance to the center of the sphere necessarily.
And now, since this time we have a really a three dimensional object, we can take the actual density
instead of the line density.
So I wrote to volume density, which is, of course, mass divided by volume.
And then the is real times.
So we write here that this is a row which is over three times the integral Devi Square.
So I don't know your background.
And if you have done some integration in cylindrical coordinates, but here it's actually pretty easy
to do.
Maybe you ask me now, why are we not using spherical coordinates?
And the reason is that this hour, as I just mentioned, is not the radius of this, you know, of the
sphere, but it's the distance to the axis.
So it's basically a distance in cylindrical coordinates.
And therefore we use cylindrical coordinates and we express this volume element DV in terms of cylindrical
coordinates.
This is something you can look up in any text book.
This is just our times.
DRDO z define where it is.
It's the polar angle and the and then of course, we have to consider also this all square here.
So in total, we get our to the power of three times d.r and easy defy.
And no, basically the only difficult part will be to get the boundaries here, correct.
So for five, this is the angle in the X y plane, basically, which yeah, we in Germany call fi,
but I heard that in other parts of the world, you call this theta.
So please don't be confused here.
So fi goes from zero to two pi to describe a full rotation.
So this is really easy to do.
This is what you see here, basically.
And then once we calculated it's just two pi, and for the other two integrals, it becomes a bit more
difficult because we integrate over our end of a z.
And so first, we have to integrate over our because we have to r here as well and our goes from zero.
So basically, from the Z axes to a certain point, which I called R2, the which is basically the yeah,
the distance from the axis to the shell of the sphere.
And this depends on the Z coordinates.
So if we are at Z equals zero, so this is just in the X Y plane, then this our tilde will be equal
to our.
But if we are at the top of the sphere just at the tip, then it will be zero because they're at a sphere
cuts the z axis.
So when you do some trigonometry or some, yeah, Protagoras.
Then you get that this hour, tilde is square root of our square minus C square, because the radius
of the sphere, the actual radius capital R is equal to z squared plus R to the square.
And now it's basically just mathematics.
You calculate, first of all, the integral of R2 to power three, which is one of four times, are
to the power of four.
Then you use these two boundaries, which are these two here?
The other one is just zero, and the upper one becomes all square minus C square square root to the
poorer, for which you can also write like this.
Then you have to use some binomial formula to get these three terms, and these are also really easy
to integrate.
You basically have.
Here are two to four times Z.
Here you get a term that is that gives one over three times z to the power of three, which in total
gives this one.
With these two factors, and this one just gives one of a five z to the power of five.
Then you use the boundaries Z goes from minus capital R to plus capital R.
And so all of these terms simplify and gives you a power of R2, the power of five times some prefecture.
Now you have to just be careful with the fractions.
And here this one gives you basically 15 over 15.
This one gives you minus 10 over 15.
And this one gives you three over 15.
So if you add all of this up, it gives you eight over 15 together with these twos, which cancel,
actually.
So this will be the results, though we can further simplify this by using the volume of the sphere,
which is, of course, four four three times PI times R two to three.
So you get this fraction here.
And so basically, you just have to divide these two fractions.
So you have eight over 15 times the reciprocal, which is three over four.
So the three and the 15 council, which gives them one over five and the eight and the four cancel,
which gives two.
So maybe this was a bit fast, but as I said, this is not here theoretical physics or mathematics.
So we actually don't want to go through all of this.
I just wanted you to show.
Show you this ones and to explain to you or to make clear to you that when you do this, especially
for the first time, it's pretty difficult that it takes a long time to write this down and to figure
this out.
And maybe you even do mistakes on the way.
And yeah, so all in all, it's just pretty demanding and it would be a good idea to solve this numerically
instead.