# Simultaneous translation and rotation

I would like to describe the motion of a particle in a rigid body by introducing 6 degrees of freedom which are three translational dofs ($u,v,w$ ) in fixed coordinate system $R_0$($X,Y,Z$) and three rotational dofs ($\alpha$, $\beta$ ,$\gamma$ ) about axis of $R_0$ with the order of axis X, then axis Y and finally axis Z, then applying Lagrange's equation to obtain the equation of motion. So the first step is to define the location after displacement and the velocity of this particle. My question is that could I express the location after displacement and the velocity in following procedure? Or is there other approach to introduce these 6 dofs provided that the simultaneous translation and rotation happens?

Here is my trail. The initial particle position vector is $r = [x,y,z]^T$. The translational displacement is $\rho = [u,v,w]^T$, and the rotational displacement is $(R\cdot r - r)$ with the rotation matrix $R$ . When the simultaneous translation and rotation happens, the displaced location of the particle is $r+\rho+ (R\cdot r - r)=\rho+R\cdot r$. Finally, the velocity of this particle takes the form of $\dot{\rho}+\dot{R}\cdot r$. (because the particle is in the rigid body $\dot{r}=0$).

Thank you in advance for taking a look. ;)

What happens simultaneously is the translation and rotation about an arbitrary axis. But parametrizing the axis and angle with 3 variables is problematic so people resort to a sequence of three fixed rotations (Euler angles). These have to happen in sequence and not at the same time as the axis of one depends on the previous rotations. In real life, this is not what is happening.

However you choose to parametrize the rotations you will end up with a 3×3 orientation matrix $\mathtt{E}(\alpha,\beta,\gamma)$ and a location for some given point on the body $\boldsymbol{r}$ such that the location of any other point A on the body is defined by

$$\boldsymbol{r}_A = \boldsymbol{r}(u,v,w) + \mathtt{E}(\alpha,\beta,\gamma) \boldsymbol{A}$$ where $\boldsymbol{A}$ is position vector of A in body coordinates.

You differentiate this to get linear velocity as

$$\boldsymbol{v}_A = \boldsymbol{v} + \dot{\mathtt{E}} A$$

$\mathtt{E}$ depends on $(\alpha,\beta,\gamma)$ and so

$$\dot{\mathtt{E}} = \frac{\partial \mathtt{E}}{\partial \alpha} \dot{\alpha}+ \frac{\partial \mathtt{E}}{\partial \beta} \dot{\beta} + \frac{\partial \mathtt{E}}{\partial \gamma} \dot{\gamma} = \boldsymbol{\omega} \times \mathtt{E}$$

Use this to find out $\boldsymbol{\omega}$ as a function of $(\alpha,\beta,\gamma,\dot{\alpha},\dot{\beta},\dot{\gamma})$. For example if you have a sequence of rotations about XYZ then

$$\mathtt{E} = \mathtt{R}_x(\alpha) \mathtt{R}_y(\beta) \mathtt{R}_z(\gamma) \\ \boldsymbol{\omega} = \hat{i} \dot{\alpha} + \mathtt{R}_x(\alpha) \left( \hat{j} \dot\beta + \mathtt{R}_y(\beta) \hat{k} \dot\gamma\right) \\ \dot{E} = \boldsymbol{\omega} \times E$$

(see https://physics.stackexchange.com/a/243054/392 for an example).