Derivation of Newton-Euler equations I am in search of a simplified version of the derivation of Newton-Euler equations (both translational and rotational) for a rigid body (3D block) that has a body fixed frame and where the center of mass of the body is not at the center of gravity.  I can find elementary derivations for the same system when the center of mass is at the center of gravity, but not for my system in question.  
I am using the derivation as background research for a rotordynamics project that I'm working on.
Any help and or references would be greatly appreciated!  :) 
 A: It depends if you already have the mass moment of inertia tensor defined or not. 
If you know the body inertia is $I_{body}$ and the 3x3 rotation matrix is $E$ then the angular momentum vector at the center of gravity C is
$$ \vec{H}_C = \left( E I_{body} E^\top \right) \vec{\omega} $$
and the linear momentum vector is $$\vec{L} = m \vec{v}_C$$
The mass moment of inertia tensor along the world coordinates on the center of gravity is $I_C = E I_{body} E^\top $ which transforms the rotational velocity $\vec \omega$ into local coordinates, multiplies by $I_{body}$ and transforms backs into world coordinates.
Now the equations of motion on the center of gravity are defined from the sum of forces and moments equals the rate of change of momentum
$$ \sum \vec{F} = \dot{\vec{L}} $$
$$ \sum \vec{M}_C = \dot{\vec{H}}_C $$
or
$$ \sum \vec{F} = m \vec{a}_C  $$
$$ \sum \vec{M}_C = I_C \vec{\alpha} + \vec{\omega} \times I_C \vec{\omega}  $$
since the time derivative of angular momentum on a rotating frame is $\dot{\vec{H}_C} = \frac{\partial \vec{H}_C}{\partial t} + \vec{\omega} \times \vec{H}_C $
Note that $\dot{\vec{v}} = \vec{a} $ and $\dot{\vec{\omega}} = \vec{\alpha} $.
Now to describe the equations on a frame A not on C use the following transformations (with relative position of the c.g. $ \vec{c} =\vec{r}_C - \vec{r}_A $.
$$ \vec{a}_C = \vec{a}_A + \vec{\alpha} \times \vec{c} + \vec{\omega} \times (\vec{\omega} \times \vec{c}) $$
$$ \sum \vec{M}_C = \sum \vec{M}_A - \vec{c} \times \sum \vec{F} $$
So finally the equations of motion of a rigid body, as described by a frame A not on the center of gravity C is (rather messy)
$$ \boxed{ \begin{aligned} 
  \sum \vec{F} &= m \vec{a}_A - m \vec{c}\times \vec{\alpha} + m \vec{\omega}\times(\vec{\omega}\times\vec{c}) \\
\sum \vec{M}_A &= I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A  - m \vec{c} \times \vec{c} \times \vec{\alpha} +\vec{\omega} \times I_C \vec{\omega} + m \vec{c} \times \left( \vec{\omega} \times (\vec{\omega} \times \vec{c}) \right) 
\end{aligned} }
 $$
Which is why people use the spatial notation (look up screw theory) to compact the above into
$$ \sum \bf{f}_A = I_A \bf{a}_A + \bf{p} $$
$$ \begin{pmatrix} \sum \vec{F} \\ \sum \vec{M}_A \end{pmatrix} =
\begin{bmatrix} m & -m \vec{c}\times \\ m \vec{c}\times & I_C - m \vec{c}\times \vec{c}\times \end{bmatrix} \begin{pmatrix} \vec{a}_A \\ \vec{\alpha} \end{pmatrix} + \begin{bmatrix} 1 & 0 \\ \vec{c}\times & 1 \end{bmatrix}
\begin{pmatrix} m \vec{\omega}\times(\vec{\omega}\times\vec{c}) \\ \vec{\omega}\times I_C \vec{\omega} \end{pmatrix}
$$
Notice the above the $0$ and $1$ are 3x3 matrices and $\vec{c}\times$ is the 3x3 cross product operator defined by
$$ \vec{ \begin{pmatrix} x\\y\\z \end{pmatrix} }\times = \begin{vmatrix} 0 & -z & y \\ z & 0  & -x \\ -y & x & 0 \end{vmatrix} $$ 
Now the big 6x6 matrix multiplying the acceleration term is the spatial inertia at A. More here and here.
