Pseudo Torque on a Pure Rolling Body If we consider all the torques about instantaneous axis of rotation (which itself is accelerating) of a pure rolling disc, then should we also consider the torque by pseudo force that acts due to acceleration of the body?
 A: Yes, we should. Suppose, we consider the motion from any arbitrary point. About this point, the angular momentum is,
$\vec{L} = \sum_i \vec{r}_i \times \vec{p}_i$.
Then, $\frac{d \vec{L}}{dt} = \sum_i \vec{r}_i \times \frac{d \vec{p}_i}{dt}$.
If the origin itself belongs to a non-inertial frame, then in this frame, $\frac{d \vec{p}_i}{dt}$ is sum of the external forces as well as the pseudo forces on the $i$th particle.
If the origin is the center of mass, then the contribution due to the pseudo forces can be shown to cancel. For any other point, we need to take them into account.
A: Look at this example

The equation of motion about the the instantaneous axis  is:
$$ I\,\ddot\varphi=(F-m\ddot x)\,R+\tau\tag 1$$
where x describe the motion relative to inertial system
now let
$$x\mapsto x+f(t)\\
\Rightarrow\\
\ddot x\mapsto \ddot x+\ddot{f}(t)$$
thus the pseudo force is $m\,\ddot{f}(t)$ and equation (1)
$$ I\,\ddot\varphi=[F-m\,(\ddot x+\ddot{f}(t)]\,R+\tau$$
conclusion : you must consider all the forces  about instantaneous axis of rotation.
