Derivative of instantaneous rotation vector

Question

Let's say I have a ground (fixed, galilean) $G$ reference frame, and $B$ a moving, rotating (solid) body (or the frame linked to it). Let's define $T_{BG}$ as the transformation matrix such as

$$X_B=T_{BG}X_G$$

Where $X$ is a vector whose coordinates are $X_B$ in the $B$ body frame and $X_G$ in the $G$ ground frame.

Let $\omega$ be the instantaneous rotation vector of the considered body. I can write

$$\frac{d\omega_G}{dt}=\frac{d(T_{BG}^T\omega_B)}{dt}=\frac{dT_{BG}^T}{dt}\omega_B+T_{BG}^T\frac{d\omega_B}{dt}$$

Multiplying both sides by $T_{BG}$, I get

$$[1]:T_{BG}\frac{d\omega_G}{dt}=T_{BG}\frac{dT_{BG}^T}{dt}\omega_B+\frac{d\omega_B}{dt}$$

Now by definition of $\omega$, and if I'm not mistaken, we are supposed to have

$$\frac{dT_{GB}}{dt}=(\omega_G\times)T_{GB}$$

Considering that $(\omega_G\times)=T_{GB}(\omega_B\times)T_{BG}$, the previous equality gives us

$$[2]:\frac{dT_{GB}}{dt}=T_{GB}(\omega_B\times)$$

Using equality [2] in the right member of equality [1], I get

$$T_{BG}\frac{d\omega_G}{dt}=\omega_B\times\omega_B+\frac{d\omega_B}{dt}=\frac{d\omega_B}{dt}$$

Is that right ?

Answer 1

You are correct.

In the world coordinates you have $$\frac{{\rm d}}{{\rm d}t} X_G = \frac{\partial}{\partial t} X_G + \omega_G \times X_G$$

So you differentiate $\omega_G = T_{GB}\, \omega_B$ to get the rotational acceleration transformation

$$\alpha_G = \frac{\rm d}{{\rm d}t} \left( T_{GB}\, \omega_B \right) = \frac{\rm d}{{\rm d}t} \left( T_{GB} \right) \omega_B+T_{GB}\, \frac{\rm d}{{\rm d}t} \left( \omega_B \right) =\omega_G \times T_{GB}\,\omega_B + T_{GB}\,\alpha_B$$

$$\require{cancel} \boxed{ \alpha_G = \cancel{ \omega_G \times \omega_G} + T_{GB}\,\alpha_B } $$

Angular acceleration transforms the same way as rotational velocity when it comes to orientation changes.

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This technique is used to find the kinematics of a pin joint for example. Consider a pin axis ${\bf z}_1$ and angle $\theta$ relating two (rotating) coincident coordinate frames. The axis is in the local coordinate frame of the first body, such that in world coordinate ${\bf z} = T_1 {\bf z}_1$

$$T_2 = T_1 \, {\rm Rot}({\bf z}_1,\theta)$$

The 1st time derivative of the above is

$$\frac{\rm d}{{\rm d}t} T_2 =\frac{\rm d}{{\rm d}t} \left( T_1 \, {\rm Rot}({\bf z}_1,\theta) \right) = \frac{\rm d}{{\rm d}t} \left( T_1 \right) \, {\rm Rot}({\bf z}_1,\theta)+T_1 \,\frac{\rm d}{{\rm d}t} \left( {\rm Rot}({\bf z},\theta) \right)$$

$$ {\boldsymbol \omega}_2 \times T_2 = {\boldsymbol \omega}_1 \times T_1 {\rm Rot}({\bf z}_1,\theta) + T_1 \left( ({\bf z}_1\dot{\theta}) \times {\rm Rot}({\bf z},\theta) \right) $$

_{Use the identity $T({\bf a}\times{\bf b}) = (T {\bf a}) \times (T {\bf b})$}

$$ {\boldsymbol \omega}_2 \times T_2 = {\boldsymbol \omega}_1 \times T_2 + (T_1 {\bf z}_1\dot{\theta}) \times (T_1 {\rm Rot}({\bf z}_1,\theta) )$$

$$ {\boldsymbol \omega}_2 \times T_2 = {\boldsymbol \omega}_1 \times T_2 + \left((T_1 {\bf z}_1)\dot{\theta}\right) \times T_2$$

$${\boldsymbol \omega}_2 = {\boldsymbol \omega}_1 + T_1 {\bf z}_1\dot{\theta} $$

$$\boxed{{\boldsymbol \omega}_2 = {\boldsymbol \omega}_1 + {\bf z} \dot{\theta}} $$

Now take the time derivative of the above

$$ \frac{\rm d}{{\rm d}t}{\boldsymbol \omega}_2 = \frac{\rm d}{{\rm d}t}{\boldsymbol \omega}_1 + \frac{\rm d}{{\rm d}t}({\bf z} \dot{\theta}) $$

$$\boxed{ {\boldsymbol \alpha}_2 = {\boldsymbol \alpha}_1 + {\boldsymbol \omega}_1 \times {\bf z} \dot{\theta} + {\bf z} \ddot{\theta} }$$