Confusions about frames of reference when deriving Euler's equation of rotational motion

Question

I am getting confused about when torques should be frame independent. My understanding is that torque is the same in all frames that are rotating at constant angular velocity. However, this seems to be not completely accurate. For any vector $\def\ba{\mathbf a}\ba$, $$ \left.\frac{d\mathbf a}{dt}\right|_I=\left.\frac{d\mathbf a}{dt}\right|_R+\boldsymbol\omega\times \mathbf a, $$ where $I$ and $R$ denotes taking the derivative in the inertial and rotating frame with angular velocity $\boldsymbol\omega$, respective. My first question is: is this equation true even if $\boldsymbol\omega$ is time-dependent?

If so, then we can continue. Let $\mathbf L$ denote the angular momentum of a rigid body. Then, $$ \left.\frac{d\mathbf L}{dt}\right|_I=\mathbf M\\ \left.\frac{d\mathbf L}{dt}\right|_R+\boldsymbol\omega\times \mathbf L=\mathbf M, $$ Where $M$ is the moment (torque) measured in the inertial reference frame. Until this point, the equation is true for all rotating frame. Let's choose $\boldsymbol\omega$ to be the angular velocity of the rigid body measured in an inertial frame - so we are now in the rotating frame of the body. In the rotating frame, the body is stationary, so $\mathbf L|_R=0$ - but that does not necessarily mean that $\left.\frac{d\mathbf L}{dt}\right|_I=0$, because in the expression $\left.\frac{d\mathbf L}{dt}\right|_I$, the $\mathbf L$ is measured in the inertial frame, although its time derivative is taken in the rotating frame. Am I right so far? All these look strange.

So what we are going to do now is to 1) express $\mathbf L=\mathbf L|_I$ in the basis vectors of rotating frame 2) differentiate each component we get wrt time but assuming those basis vectors are time-independent. Step 1) is easy: $$ \mathbf L|_I=L_i\mathbf e_i=\sum I_i\omega_i\mathbf e_i $$ Moment of inertia is constant here, (Is it constant in all frames or just in the body's frame? Anyway, it doesn't make sense to me to measure the inertia of a rotating body along a fixed axis), so we may safely write $$ \dot {\mathbf L_I}|_R=\mathbf I \dot{\boldsymbol{\omega}} $$ and hence $$ \mathbf{I} \dot{\boldsymbol\omega} + \boldsymbol\omega \times \left( \mathbf{I} \boldsymbol\omega \right) = \mathbf{M} $$ In the above two expressions, $\boldsymbol\omega$ is the angular velocity measured in the inertial frame, but again, its time derivative is taken in the rotating frame. This makes my life extremely difficult - firstly measure in the inertial frame, then take derivative in the rotating frame. Am I really understanding this correctly?

Answer 1

1. Yes, the derivative on the rotating frame is derived using the instantaneous rotational velocity $\boldsymbol{\omega}$ (value at one instant in time) and it is valid for both constant and varying rotation vectors.

2. Your angular momentum equation on the rotating frame is correct.

   $$\mathbf{M}|_I = \frac{d\mathbf L|_I}{dt} = \frac{\partial \mathbf L|_R}{\partial t} +\boldsymbol\omega|_I\times \mathbf L|_I$$ But on the rotating frame $\mathbf L|_R \neq 0$. The rotating frame is just a set of basis vectors on which to resolve the inertial frame vectors. $$ \mathbf L|_I= \mathrm{R}\, \mathbf L|_R$$

   where $\mathrm{R}$ is the rotation vector between body frame and inertial frame.

3. Moment of inertia is constant on the body frame in general unless the body is rotating about a symmetry axis. But to use the MMOI tensor you need to align it into the inertial basis vectors, which is done with the following transformation:

   $$\mathbf{I}|_I = \mathrm{R}\, \mathbf{I}|_R \,\mathbf{R}^\top $$

Now you have everything you need on the same basis vector (the inertial) in order to state the equation of motion

$$ \mathbf{M}|_{I} = \mathbf{I}|_{I}\,\mathbf{\boldsymbol{\dot{\omega}}}|_{I} + \boldsymbol{\omega}|_{I}\times\mathbf{I}_{I}\,\boldsymbol{\omega}|_{I} $$

4. To represent the above on to the rotating frame, use the transformations $\mathbf{M}|_I = \mathrm{R}\, \mathbf{M}|_R$ and $\mathbf{\omega}|_I = \mathrm{R}\, \mathbf{\omega}|_R$. But you realize this is just a change of basis vectors, and not take same as if taking measurements from the moving body (which would result in $\boldsymbol{\omega}|_R = 0$).

   $$\begin{aligned}\mathrm{R}\,\mathbf{M}|_{R} & =\mathrm{R}\,\mathbf{I}_{R}\,\mathrm{R}^{\top}\,\left(\mathrm{R}\,\dot{\boldsymbol{\omega}}|_{R}+\boldsymbol{\omega}|_{I}\times\mathrm{R}\,\boldsymbol{\omega}|_{R}\right)+\left(\mathrm{R}\,\boldsymbol{\omega}|_{R}\right)\times\left(\mathrm{R}\,\mathbf{I}_{R}\,\mathrm{R}^{\top}\,\mathrm{R}\,\boldsymbol{\omega}|_{R}\right)\\ \mathbf{M}|_{R} & =\mathbf{I}_{R}\,\left(\dot{\boldsymbol{\omega}}|_{R}+\mathrm{R}^{\top}\left(\boldsymbol{\omega}|_{I}\times\boldsymbol{\omega}|_{I}\right)\right)+\boldsymbol{\omega}|_{R}\times\mathbf{I}_{R}\,\boldsymbol{\omega}|_{R}\\ \mathbf{M}|_{R} & =\mathbf{I}_{R}\,\dot{\boldsymbol{\omega}}|_{R}+\boldsymbol{\omega}|_{R}\times\mathbf{I}_{R}\,\boldsymbol{\omega}|_{R} \end{aligned}$$

   Notice the transformation of $\boldsymbol{\dot{\omega}}|_I$ onto body coordinates is the derivative of $\boldsymbol{\omega}|_R$ using the rotating frame derivative, but with the convective terms canceling out.

Answer 2

Torque always has a reference frame and has to be kinematically transformed between the inertial frame and the body frame.

Yes, the time derivative equation is an instantaneous differential wrt $\omega$.

By "all rotating frames" I assume you mean all coordinate systems which are fixed wrt the body frame (eg principle moments of inertia, some other global coordinates). Same with "assuming those basis vectors are time-independent". That's fine.

the L is measured in the inertial frame, although its time derivative is taken in the rotating frame

Moment of inertia only makes sense when talking about body coordinates with a fixed point. Remember that for every rotation (matrix), there is a fixed axis, which could be in more than one body-centered or inertial frame directions (ie precession along with rotation).

The Euler equation for rigid body motion with one fixed point, as your last equation states, describes the $\vec{\omega}$ dynamics in the body frame, but you always know the $\omega$ in the laboratory frame.