Rotation of a rigid body combined with translation of the axis parallel to itself:
Let us now introduce translation of the rotational axis parallel to itself - it may even accelerate - and ask what kind on motion is going to follow. So for example there may be a rod on a horizontal table and is hit by an impulse one end, and we may be interested in its subsequent motion. I general it could be a rigid body of general shape on which we apply a force. We split the motion into a translation of the CM of the body and rotation about an axis passing through the CM. By doing so the equation of motion for the translational motion of the CM is very easy. It is

Here is the total momentum of the body; M is its mass;   the acceleration of the CM and the total applied force. With this equation we know how the CM of the body translates. Next we wish to find the rotation of the body with respect to an axis passing through the CM (recall that the most general motion of a rigid body is translation of a point and rotation about that point). But the question is: can we apply

where   is the angular momentum about the CM and is the applied torque about the axis of rotation passing through the CM. I raise this question because in general the CM will also be accelerating and therefore with respect to the CM, there will be a fictitious force that may also give rise to a fictitious torque which is in addition to the applied torque . However, it is easy to see that such a fictitious torque about the CM will always be zero. This is because the fictitious force effectively acts at the CM itself. Because of this reason, there is one more point about which the torque due to the fictitious force vanishes: this is the point that accelerates towards the CM. Thus the equation above can be applied safely about these two points. There is also a third point about which the above equation is valid. This is the point that does not accelerate at all. Let me now prove these statements.

Shown in figure 4 is a rigid body performing a general motion, i.e. it is both translating as well as rotating. For convenience we have shown the body in two dimensions. Two points J and i of the body are also shown. These points are also moving with the body. We now calculate the rate of change of the angular momentum about point J . This is done below.

where all the terms have their standard meaning and the subscript (iJ) denotes the quantity being calculated for point i with respect to point J . Denoting the velocity and acceleration of point i about the origin O as , respectively, and that of J as , we have

With , where is the position vector of the CM with respect to J , we get

If we want the rate of change of the angular momentum to depend only on the applied torque calculated about J, we should have

That will happen under the following three conditions:

I have just shown you that irrespective of the whether point J is accelerating, rotating or performing some general motion, the equation

can be applied about J if it satisfies one of the three conditions obtained above. Notice that in under these conditions the right-hand side has only the externally applied torque. Thus if we choose one of these points to apply the angular momentum-torque equation, we do not have to worry about any fictitious torques arising because we are sitting on an accelerating point. We have been applying the angular momentum-torque equation about points satisfying condition I above; it includes stationary points also. Of the other two points, it is always safer to apply the equation about the CM (condition II ). This is because of the difficulty in ensuring that a point is accelerating towards the CM (condition III ), although in some situations it may be easy. We will discuss one such case below. We now solve some simple examples to illustrate what we have learnt above.