The center of mass of a rigid body is given by: $$ \vec{r}_c = \frac{1}{M} \sum_i m_i \cdot \vec{r}_i $$ with $M = \sum_i m_i$ the total mass or $$ \vec{r}_c = \frac{1}{M} \int \vec{r}\ ' \cdot \varrho(\vec{r}\ ')~d^3r' $$ with $M = \int \varrho(\vec{r}\ ')~d^3r'$ for a continious mass distribution. I'll stick to the discrete case for brevity.
Therefore: \begin{align} M~\frac{d^2 \vec{r}_c}{dt^2} = M~\vec{a}_c & = \sum_i m_i~\frac{d^2 \vec{r}_i}{dt^2} \\ & = \sum_i m_i~\vec{a}_i = \sum_i \vec{F}_i \tag1 \end{align} So the center of mass moves due to every force acting on the rigid body's particles. Now, what are these forces? If we set $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} $$ with $\vec{F}_{\text{ext}}$ the external forces acting on a particle, this would not account for the fact, that the body is rigid. There are indeed internal forces that guarantee the body not to deform upon the action of external forces. For exmaple, if an external force is only applied to one particle and there were no internal forces, this would allow the particle to "leave" the body. This is not what one usually means if talking about a rigid body. So one rather has $$ \vec{F}_i = \sum \vec{F}_{\text{ext}} + \sum \vec{F}_{\text{int}} $$, this depends now largely on the specific model of internal forces one chooses (how one "point-force" is propagated through the volume).
Is there a common idealisation for rigid bodies, that allows to calculate or even ignore those internal forces and therefore solve the equation of motion (1)?
