homework and exercises - Rotation systems. Problem interpreting an equation

In this equation: $$\mathbf a_i\overset{\rm def}{=}\left(\frac{d^2\mathbf r}{dt^2}\right)_i=\left(\frac{d\mathbf v}{dt}\right)_i=\left[\left(\frac{d}{dt}\right)_r+\boldsymbol\Omega\times\right]\left[\left(\frac{d\mathbf r}{dt}\right)_r+\boldsymbol\Omega\times\mathbf r\right]$$ (from Wikipedia), why is $$\left(\frac{d}{dt}\right)_r \boldsymbol{\Omega} \times \mathbf{r}=\frac{d\boldsymbol{\Omega}}{dt}\times\bf{r}+\bf{\Omega}\times \bf{V_r}$$

In particular, I have qualms with the term $$\frac{d\bf{\Omega}}{dt}\times\bf{r}$$

Why are we deriving the angular velocity? Why is it a derivative not of the rotational type (Namely $(\frac{d}{dt})_r$ )

Other sources do not point out that term I have problems with. In any case, I want to know how you evaluate that derivative.

Answer

Other sources do not point out that term I have problems with.

Other sources explicitly assume a constant angular velocity and thus ignore that component. The wikipedia article you cited is correct.

In any case, I want to know how you evaluate that derivative.

Given any vector quantity $\mathbf q$ that is the same (other than component representation) in the inertial and rotating frame, the time derivative of that vector from the perspective of an inertial versus rotating observer is $$\left(\frac{d\mathbf q}{dt}\right)_I = \left(\frac{d\mathbf q}{dt}\right)_R + \boldsymbol\Omega \times \mathbf q$$ In dynamics, this is sometimes called the transport theorem (but there are a number of other things called the transport theorem).

Applying the transport theorem to the angular velocity vector yields $$\left(\frac{d\boldsymbol\Omega}{dt}\right)_I = \left(\frac{d \boldsymbol\Omega}{dt}\right)_R + \boldsymbol\Omega \times \boldsymbol\Omega = \left(\frac{d \boldsymbol\Omega}{dt}\right)_R$$ In other words, angular acceleration is fundamentally the same vector in the inertial and rotating frame.

Applying the transport theorem instead to angular momentum yields $$\left(\frac{d\mathbf L}{dt}\right)_I = \left(\frac{d\mathbf L}{dt}\right)_R + \boldsymbol\Omega \times \mathbf L$$ The rotational analog of Newton's second law provides an alternative representation of the left-hand side of the above: $\frac {d\mathbf L}{dt} = \boldsymbol \tau_{\text{ext}}$ where the derivative is calculated from the perspective of an inertial frame and $\boldsymbol \tau_{\text{ext}}$ is the external torque on the system. If the system is a rigid body, the angular momentum is given $\mathbf L = \mathrm I \boldsymbol\Omega$ where $\mathrm I$ is the object's inertia tensor. Since the inertia tensor of a rigid body is constant in a frame rotating with the body, the time derivative of the angular momentum vector from the perspective of an observer rotating with the object simplifies to $\left(\frac{d\mathbf L}{dt}\right)_R = \mathrm I\left(\frac{d \boldsymbol\Omega}{dt}\right)_R$. Putting all of the above together yields $$\boldsymbol\tau_{\text{ext}} = \mathbf I \frac{d \boldsymbol\Omega}{dt} + \mathbf \Omega \times (\mathrm I \, \boldsymbol\Omega)$$ or $$\frac {d \boldsymbol\Omega}{dt} = {\mathbf I}^{-1} \left( \boldsymbol\tau_{\text{ext}} - \boldsymbol \Omega \times (\mathrm I \, \boldsymbol \Omega) \right)$$

This yields a way to calculate $\frac {d\boldsymbol \Omega}{dt}$ at any point in time for a rigid body. Whether this is integrable via elementary methods is a different question. In most cases, it isn't. It's rather challenging to find a non-trivial rotational system that has an analytic solution. One typically has to revert to numerical methods to determine the rotational behavior of an object.