The axis of rotation is the line passing through the centres of the circles in which the particles of the body move during the rotational motion of the body. So, it has nothing to do with the frame of reference or origin. However, when we want to calculate the torque, angular momentum, angular acceleration or other angular variables of the rotating body, it is convenient to consider the origin on the axis of rotation. When we say that a force F    on the ?i?th particle of a rotating body is producing torque ?, we mean that the force is F      and the torque is ? as measured from the axis frame of reference.

        If you consider the origin somewhere else as shown in the figure, and if this 2^nd frame is inertial w.r.t the axis frame, then the force vector remains same but the position vector i.e  r    changes. So, r    X F    also change. Thus ? also change.  Let the torque in the axis frame be ?₁ and that in the 2^nd frame be ?₂

 So, ?₁ is not equal to ?₂

But the moment of inertia and angular acceleration does not change. (Because M.I. does not depend on the origin, it depends only on the axis.)

        As ?₁ is measured in axis frame,

                  ?₁=I?.

             But, ?₁is not equal to ?₂

         ?₂   is not equal to I?.

So, the relation ?=I? will hold only if ? is measured from the axis frame of reference. So, it is most convenient to measure angular quantities of a rotating body from its axis frame . So, henceforth, when we write ? it will mean that the ? is measured from the axis frame.

 

  Now, we come to your question.

  Let the ?i?th particle of a rotating body be acted upon by a force that is measured as F

from the ground frame or any frame other than the axis frame. If the axis frame is translationally accelerating w.r.t. the 2^nd frame (i.e. the ground frame) with acceleration   a       then the force on the ?i?th particle as seen from the axis frame is 

   (F    -  m_ia) 

Then torque of force F about the axis  becomes ?= rX(F    -  m_ia)  .<r,F& a are vectors. I am ommiting the vector sign for the sake of convenience).

 

   Notice that we have to consider the pseudo force ?m_i a     only because the axis frame is accelerating, not because the particles of the body are tangentially or centripetally accelerated. ( Pseudo force is considered due to acceleration of a frame, not due to that of a particle in a frame).

 

        ?_i= rX(F    -  m_ia) 

 or,  total torque ?=? rX(F    -  m_ia) 

If the vector (F    -  m_ia)  makes angle ? with the radial direction, then

   ?= ? r(F    -  m_ia)  sin ?

     = ? rFsin ? ? ? rm_ia sin ?

Thus, if you know the distance of the ?i?th particles from the axis and the forces applied on them you can calculate ? rFsin ?. For example, if a force of x newton acts tangentially on a surface point (e.g. topmost point) of a sphere of radius r then ? rFsin ? = xr.

As for the part ?rm_ia sin ?, you have have to calculate ?m_i r  the same way you do ?m_ir² (i.e. moment of inertia I).

    Once you find out total torque ?, and know I of the body, then angular acceleration produced is given by ?= ?/I

    Note that here ?a? is not equal to ?r, here a is the acceleration of the axis, not the particles.

   There may be an easier way to do the whole thing. If you find such a way, or if you find any errors in this, please let me know.