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6

The inertia ellipsoid is computed from an integral about an axis - in other words you rotate the object. This will "smooth out" any symmetries and typically increase the symmetry. Sorry this is a "early morning" intuitive explanation - maybe someone else will give you a more formal answer.

1

First, you set up your second integral wrong. The integrand should be $$r\cdot(r^2\cos^2\theta+z^2)$$ since $\sqrt{r^2\cos^2\theta+z^2}$ is the distance of a point from the y-axis. Second, you have used $h$ in two ways. In your picture $h$ measures the vertical distance from the top of the hemisphere whereas in your first iterated integral (the $dr$ ...

4

Use \begin{align} \frac{a_{23}}{a_{33}} & = \frac{ -\cos b \sin a}{\cos a \cos b} = -\tan a \\ \frac{a_{13}}{\sqrt{ a_{23}^2 + a_{33}^2 }} & = \frac{\sin b}{\cos b \sqrt{\sin^2 a +\cos^2 a}} = \tan b \\ \frac{a_{12}}{a_{11}} & = \frac{-\cos b\sin c}{\cos b \cos c} = -\tan{c} \end{align} a = atan2(-a23,a33) b = atan2(a13, ...

1

Now if I consider a particular set of rotation (say X first, then Y , then Z), with the corresponding Tait-Bryan angles --- a,b and c. My rotation matrix will be the following ... Look at that array. It's of the form \begin{array}{ccc} \cos(b) \cos(c) & -\cos(b) \sin(c) & \sin(b) \\ \cdots & \cdots & -\cos(b) \sin (a) \\ \cdots & ...

1

I don't think I fully understand how the Eulers angles were defined, my intuition to these derivation should be defined like this, 1. rotate about z axis 2. rotate about y axis, 3 rotate about x axis. But I can't convince myself why I'm wrong. I hope any experts can point me out. The sequence depicted in the original post involved A rotation by angle ...

1

The rigid body that's being rotated is arbitrary; it need not be rotationally symmetric. To use Euler angles, we assume that rotation is executed by the body being rotated (say, an airplane with flaps, or spacecraft with thrustors) instead of by some external manipulator. So for all intermediate rotations, we have to consider the perspective of the body. ...

0

Compare the distance that your feet move to the distance that the weight moves. Then apply conservation of energy.

0

Ok--I think I understand where my confusion was coming from. The change in basis is simply convenient when attacking problems from a variational point of view. The factors of $\cos \varphi$ and $\sin \varphi$ are simply there to guarantee that the basis is orthonormalized. With Dicke states as eigenstates of the total spin squared ...

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