Torque equation about center of mass frame

Let say a uniform cylinder is rolling down an incline. Since the center of mass of the cylinder has non-zero acceleration, should it be considered as a non-inertial reference frame?

I know that when we write the torque equation with respect to center of mass, even if we add the pseudo force acting on the center of mass, it will not make any difference because of its torque being zero.

I'm just interested in the concept though. Is the above thinking correct?

The suggestion that the summation point for the net torque is that it is the center of mass is so the equations of motion to assume their simplest form. In general, the COM is not an inertial frame.

In order to understand where the idea of an inertial reference frame comes from, imagine a system of particles, each moving with $$\vec{v}_i$$ and having mass $$m_i$$, and describe the total linear and angular momentum.

First, require that each position $$\vec{r}_i$$ is decomposed into the location of the COM plus a relative component that remains a fixed distance from the COM

$$\vec{r}_i = \vec{r}_C + \vec{d}_i$$

Because the above is described in a reference frame you can differentiate the above as

$$\vec{v}_i = \vec{v}_C + \vec{\omega} \times \vec{d}_i = \vec{v}_C + \vec{\omega} \times (\vec{r}_i - \vec{r}_C)$$

and again

$$\vec{a}_i = \vec{a}_C + \vec{\alpha} \times (\vec{r}_i - \vec{r}_C) + \vec{\omega} \times (\vec{v}_i - \vec{v}_C)$$

without considering the motion of the frame.

Now the total linear momentum is

$$\vec{p} = \sum_i m_i \vec{v}_i = \sum_i m_i \vec{v}_C + \sum_i m_i \vec{\omega} \times (\vec{r}_i - \vec{r}_C)$$

Now consider that $$\sum_i m_i \vec{v}_C = \vec{v}_C \left( \sum_i m_i \right) = m\,\vec{v}_C$$, and that $$\sum_i m_i \vec{r}_i = m\, \vec{r}_C$$ by the definition of center of mass, the above is

$$\vec{p} = m \,\vec{v}_C$$

The total angular momentum about the (moving) center of mass is

$$\vec{L}_C = \sum_i \vec{d}_i \times m_i \vec{v}_i = \sum_i \vec{d}_i \times m_i ( \vec{v}_C + \vec{\omega} \times \vec{d}_i )$$

After further simplifications the above becomes

$$\vec{L}_C = \sum_{i}m_{i}\vec{d}_{i}\times\left(\vec{\omega}\times\vec{d}_{i}\right) = \mathrm{I}_C \vec{\omega}$$

As you can see it is the relative distance of each particle to the COM that counts here $$\vec{d}_i = \vec{r}_i - \vec{r}_C$$.

For dynamics you need the following definitions of force and torque

$$\sum_i \vec{F}_i = \frac{\rm d}{{\rm d}t} \vec{p}$$

$$\sum_i \left( \vec{\tau}_i + \vec{d}_i \times \vec{F}_i \right) = \frac{\rm d}{{\rm d}t} \vec{L}_C$$

again, by definition the center of mass is not an inertial frame location.

There are several different coordinate systems you could use here, but I believe the answer is yes, the are all non-inertial.

For example, a coordinate system "fixed to the disk" rotates as it goes down the ramp, so it has both angular acceleration and linear acceleration.

Another coordinate system you could define is one in which one axis is through the center of the disk, but the other two axis are fixed in the lab frame. This coordinate system accelerates down the ramp, but does not rotate.

Since they are both non-inertial, they both come with ficticious forces.