The internal forces come in equal-and-opposite pairs (Newton's 3rd), and therefore result in no net force on the object. If you did take the vector sum of all of them, they would just cancel out. Including them therefore doesn't change the final expression.
There is no need for idealization - just the acceptance of classical mechanics.
Note that this is even true for non-rigid bodies: while the position of their center of mass will change, that center of mass will still move in accordance with the vector sum of the external forces.
You don't have to, but it makes the equations easier to deal with because you don't have to account for the moment of acceleration terms. See the 2nd part this this answer about deriving Newton's laws on an abitrary point not the center of mass.
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) $$ \boxed{ \begin{aligned}
\sum \vec{F} &= m \vec{a}_A - m \vec{c}\times \vec{\alpha} + m \vec{\omega}\times\vec{\omega}\times\vec{c} \\
\sum \vec{M}_A &= I_C \vec{\alpha} + m \vec{c} \times \vec{a}_A - m \vec{c} \times \vec{c} \times \vec{\alpha} +\vec{\omega} \times I_C \vec{\omega} + m \vec{c} \times \left( \vec{\omega} \times \vec{\omega} \times \vec{c} \right)
\end{aligned} } $$
The sum of the forces part equates to mass times acceleration of the center of mass. If the COM is not used, these extra terms appear to account for the change.
To help you the laws of motion can be summarized as follows:
- Linear momentum is defined as mass times the velocity of the center of mass $$\vec{p} = m \vec{v}_{C}$$
- Angular momentum at the center of mass is defined as rotational inertia at the center of mass times angular velocity
$$\vec{L}_C = I_C \vec{\omega}$$
- The net forces acting on a body equal the time derivative of linear momentum
$$ \sum \vec{F} = \frac{{\rm d}}{{\rm d}t} \vec{p} = m \vec{a}_C$$
- The net torques acting on a rigid body about the center of mass equal the time derivative of angular momentum at the center of mass $$\sum \vec{\tau}_C = \frac{{\rm d}}{{\rm d}t} \vec{L}_C = I_C \vec{\alpha} + \vec{\omega} \times I_C \vec{\omega}$$
- To transfer these quantities to a different location A with $\vec{r}=\vec{r}_C -\vec{r}_A$ use the following rules
$$\begin{aligned}
\vec{v}_A & = \vec{v}_C + \vec{r} \times \vec{\omega} \\
\vec{a}_A & = \vec{a}_C + \vec{r} \times \vec{\alpha} + \vec{\omega} \times \left( \vec{r} \times \vec{\omega} \right) \\
\vec{L}_A & = \vec{L}_C + \vec{r} \times \vec{p} \\
\sum \vec{\tau}_A & = \sum \vec{\tau}_C + \vec{r} \times \sum \vec{F} \\
\end{aligned}$$
whereas forces, linear momenta, angular velocity and angular acceleration are shared with the entire rigid body and thus do not change from point to point.
To the above you can add the vector form of the parallel axis theorem with
$$ I_A = I_C - m [\vec{r}\times] [\vec{r}\times]$$
where $[\vec{r}\times]$ is the 3×3 skew symmetric matrix for the cross product operator $\begin{pmatrix}x\\y\\z\end{pmatrix}\times = \begin{vmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{vmatrix}$
This comes out of the momentum transformation from C to A, but it is not the complete picture. To see what happens you have to look at the following 6×6 spatial inertia matrix:
$$\begin{aligned}\vec{p} & =m\vec{v}_{C}=m\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\vec{L}_{A} & =I_{C}\vec{\omega}+m\,\vec{r}\times\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\begin{pmatrix}\vec{p}\\
\vec{L}_{A}
\end{pmatrix} & =\begin{vmatrix}m & -m\,\vec{r}\times\\
m\,\vec{r}\times & I_{C}-m\,\vec{r}\times\,\vec{r}\times
\end{vmatrix}\begin{pmatrix}\vec{v}_{A}\\
\vec{\omega}
\end{pmatrix}
\end{aligned}$$
Best Answer
One way to see a rigid body is a collection of infinity particles such that they preserve the distances between themselves (can only translate and rotate). So, we have a constraint, and then internal forces that maintain this constraint.
This internal forces are dependent on the forces that you apply in the system. If you apply a force in one point of the object (one particle), then all other points would feel a force such that the accelerations of the hole system don't destroy the constraint. Rigid bodies need to have energy stored there to do that.
If you have a static homogeneous rigid body with the center of mass $\vec{r}_{cm}=\vec{0}$, applying a force $\vec{f}(\vec{r}_0)$ in some point $\vec{r}_0$ such that the force is paralel to $\vec{r}_0$, then all the points of the body need to feel this same force to preserve the constraint. The result of this force is an aceleration of the whole system generating a translation.
If the force $\vec{f}(\vec{r}_0)$ is not parallel to $\vec{r}_0$, then all the points need to feel a force $\vec{f}(\vec{r})$ such that
$$ \vec{r}\times\vec{f}(\vec{r})=\vec{r}_0\times\vec{f}(\vec{r}_0) $$.
because the constraint is mathematically translated to $d\vec{r}=d\vec{\theta}_0\times\vec{r}$ for every $\vec{r}$ and the same $d\vec{\theta}_0$ (see here for understand why).
With this we can define a quantity called Torque associated for a force $f$ applied to a point $\vec{r}$ of the rigid body by:
$$ \vec{\tau}=\vec{r}\times\vec{f}(\vec{r}) $$
So, if we have a force $\vec{f}_1$ applied at $\vec{r}_1$ and $\vec{f}_2$ applied at $\vec{r}_2$ we can see that the net force at point $\vec{r}$ need to obey:
$$ \vec{f}(\vec{r})\times\vec{r}=\vec{f}_1\times\vec{r}_1+\vec{f}_2\times\vec{r}_2 $$
Then we may define a net torque:
$$ \vec{\tau}_{net}=\sum_{j}\vec{f}_{j}\times\vec{r}_{j} $$
And, if $\vec{\tau}_{net}=\vec{0}$, then the body does not rotate because at each point $\vec{f}(\vec{r})\times\vec{r}=0$ and for $\vec{r}\neq \vec{0}=\vec{r}_{cm}$ we conclude that the force $\vec{f}(\vec{r})$ is paralell to $\vec{r}$.