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Say we have a simple cube, where each vertex of the cube represents the center of a particle.

Let us define a simple rotation matrix, $R$

$$R = \begin{matrix} \cos\theta_x & 0 & -\sin\theta_z\\ \sin\theta_z\sin\theta_x & \cos\theta_x & \cos\theta_z\sin\theta_x\\ \sin\theta_z\cos\theta_x & -\sin\theta_x & \cos\theta_z\cos\theta_x \end{matrix}$$

Where $\theta_x$ and $\theta_z$ represent angles of rotation about the x and z axes, respectively.

$R$, in this sense, is a simple concatenated rotation about the cardinal x and z axes. The use-case for this specific scenario doesn't involve rotation about the y-axis, hence why it's being left out.

If our cube has vertices within the set $S = \{\vec{v} \in \mathbb{R}^3 | \vec{v} = \langle \pm1, \pm1, \pm1 \rangle\}$, where $\vec{v}$ is a position vector vertex of the cube with the respect to the cube's origin, $O$, we can let any particle $p_i$ have its center represented by a member of $S$.

Let $\vec{v_i}$ correspond to the position of particle $p_i$, for all $i$ in [1, 8].

We can then define a newly transformed position for every $p_i$, using $R$:

$$\vec{t_i}=R\vec{v_i}$$

Now, assuming that every particle $p_i$ has a corresponding mass, $m_i$, we can construct the angular momentum, $\vec{a_i}$ from $p_i$'s angular velocity $\vec{\omega_i}$:

$$\vec{a_i} = m_i\left[\vec{t_i} \times \left( \vec{\omega_i} \times \vec{t_i} \right)\right]$$.

With all of this information provided, except for the angular velocity $\vec{\omega_i}$ and angular momentum $\vec{a_i}$, is it possible to use this information to find $\vec{\omega_i}$? Or, must it be given?

If it must be given, what might be a legitimate value of $\vec{\omega_i}$ after the rotation of $R$ has been applied to $\vec{v_i}$?

My understanding is that angular velocity can be thought of as a normalized axis of rotation scaled by the angle with which it is being rotated about. Given that two cardinal axes are accumulated into one transformation (in this case, x and z ), I'm trying to figure out exactly how I can represent these two angles of rotation into an angular velocity vector.

If this doesn't make sense, I clearly need some correction in my reasoning, and would be happy to receive any feedback.

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  • $\begingroup$ I am afraid $R$ is not a rotation matrix. $\endgroup$ – uniquesolution Sep 12 '15 at 8:39
  • $\begingroup$ How? And what is it, then...? $\endgroup$ – about blank Sep 12 '15 at 20:24
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    $\begingroup$ I calculated the determinant of your matrix to be $\cos\theta_x\cos\theta_z+\sin^2\theta_z$, which is not necessarily equal to $\pm 1$, as should be the case with a Rotation matrix. $\endgroup$ – uniquesolution Sep 12 '15 at 23:49

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