# Differential equation of motion of combined rigid body

I am reading the following paper:
http://tinyurl.com/jv2pu2m

I have a question about (1) and (2):

To (1), I think the following link provides good explanation.

Equation of Motion for Rigid Body Motion

However, I do not know how to do the similar steps to (2), particularly for $$\omega$$.

Specifically, we can write $$v = v_1 +\omega \otimes r$$ and take derivative to it w.r.t $$t$$; how to do this step to $$\omega$$?

I think you are asking about how to transform various vector quantities between points attached to a rigid body. Here are a rundown of the rules of transformation between an arbitrary point A (located at $r_A$) and the center of mass C (located at $r_C$).

\begin{align} v_C & = v_A + \omega \times (r_C - r_A) & & \text{linear velocity at C} \\ \dot{v}_C & = \dot{v}_A + \dot{\omega} \times (r_C - r_A) + \omega \times (v_C - v_A) & & \text{linear acceleration at C} \\ p & = m v_C & & \text{linear momentum}\\ L_C & = \mathtt{J}_C\, \omega & & \text{angular momentum at C} \\ L_A & = L_C + (r_C-r_A) \times p & & \text{angular momentum at A} \\ F & = \frac{{\rm d}}{{\rm d}t} p = m \dot{v}_C & & \text{net force on body} \\ M_C & = \frac{{\rm d}}{{\rm d}t} L_C = \mathtt{J}_C\,\dot{\omega}+ \omega \times \mathtt{J}_C\, \omega & & \text{net torque on body at C} \\ M_A & = M_C + (r_C-r_A) \times F & & \text{net torque at A} \end{align}

Edit 1 If you combine the above you have

\begin{aligned} \begin{pmatrix}v_{C}\\ \omega \end{pmatrix}&=\begin{bmatrix}1 & -[c\times]\\ 0 & 1 \end{bmatrix}\begin{pmatrix}v_{A}\\ \omega \end{pmatrix} \\ \begin{pmatrix}\dot{v}_{C}\\ \dot{\omega} \end{pmatrix}&=\begin{bmatrix}1 & -[c\times]\\ 0 & 1 \end{bmatrix}\begin{pmatrix}\dot{v}_{A}\\ \dot{\omega} \end{pmatrix}+\begin{pmatrix}\omega\times\left(\omega\times c\right)\\ 0 \end{pmatrix} \\ \begin{pmatrix}p\\ L_{A} \end{pmatrix}&=\begin{bmatrix}m [1] & -m[c\times]\\ m[c\times] & \mathtt{J}_{C}-m[c\times] [c\times] \end{bmatrix}\begin{pmatrix}v_{A}\\ \omega \end{pmatrix} \\ \begin{pmatrix}F\\ M_{A} \end{pmatrix}&=\begin{bmatrix}m[1] & -m[c\times]\\ m[c\times] & \mathtt{J}_{C}-m[c\times] [c\times] \end{bmatrix}\begin{pmatrix}\dot{v}_{A}\\ \dot{\omega} \end{pmatrix}+\begin{bmatrix}[1] & [0]\\ [c\times] & [1] \end{bmatrix}\begin{pmatrix}m\left(\omega\times\left(\omega\times c\right)\right)\\ \omega\times\mathtt{J}_{C}\omega \end{pmatrix} \end{aligned}

where $c = r_C-r_A$ is the relative CM location, and $[c\times] = \begin{vmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{vmatrix}$ is the 3×3 cross product operator, and $[0]$, $[1]$ are the 3×3 zero and identity matrices.

• My trouble is how to obtain "$\omega \times J_c \omega$"; could you put more detail here? Commented May 23, 2016 at 22:49
• It is the time derivative of $L_C$ if the rotation was constant. Look up time derivative in a rotating frame in the meantime. Commented May 23, 2016 at 23:20
• Should the fifth be $(r_A-r_C)$? Commented May 24, 2016 at 0:49
• No. Like the torque of force located at $r$ is $r \times F$, the moment of momentum at $r$ is $r \times p$. Commented May 24, 2016 at 2:42