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:
$$\begin{aligned}\vec{p} & =m\vec{v}_{C}=m\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\vec{L}_{A} & =I_{C}\vec{\omega}+m\,\vec{r}\times\left(\vec{v}_{A}-\vec{r}\times\vec{\omega}\right)\\
\begin{pmatrix}\vec{p}\\
\vec{L}_{A}
\end{pmatrix} & =\begin{vmatrix}m & -m\,\vec{r}\times\\
m\,\vec{r}\times & I_{C}-m\,\vec{r}\times\,\vec{r}\times
\end{vmatrix}\begin{pmatrix}\vec{v}_{A}\\
\vec{\omega}
\end{pmatrix}
\end{aligned}$$
