Definition of torque in the (possibly accelerating) centre of mass frame

Question

My lecture note states that the torque in the centre of mass frame ($O^*$) of a rigid body is: $$\bf{G^*}=\sum \bf{r^* \times F}$$ where $\bf{F}$ denotes the real force that is producing the torque and that this is true regardless of whether the centre of mass is accelerating (non-inertial frame).

Does this imply that in an accelerating (but non rotating frame), the perception of "individual" forces acting on the body is the same as that in an inertial frame? By individual I mean forces that contribute to the summation term.

I understand that the inertial term -$M\bf{\ddot{R}}$ is required in the expression for the resultant force when working in such a frame, but this does not need to be considered because we're not interested in the resultant force acting on the body but rather the individual ones that are producing the torque?

How flawed are my arguments?

Answer 1

Individual forces do not change, but you need to add fictitious forces that may or may not contribute to the torque. In the case of an object undergoing linear acceleration, the inertial term can contribute to the torque if your origin of coordinates is not at the center of mass.

Answer 2

For a system of particles, with the origin of the coordinate system taken as the center of mass (CM), $d\vec J/dt = \vec \tau$ where $\vec J$ is the angular momentum with respect to the CM and $\vec \tau$ is the torque from the forces in the inertial frame with respect to the CM. This is true even if the CM is accelerating. The CM of mass is special in this regard. See a good physics mechanics text, such as Symon, Mechanics, for the detailed derivation that proves this relationship.

Also, Euler's equations for the motion of a rigid body expressed in terms of the principal axes of the body that rotate with the body, use the forces in the inertial frame for the torques.

For general three dimensional motion, if the rotation is constrained to be about a fixed point, that point is taken as the origin of the coordinate system for evaluating the rotational motion. If the object is unconstrained and moving freely, the center of mass (even if accelerating) is taken as the origin of the coordinate system for evaluating the rotational motion; translational motion is evaluated for the center of mass using the second law. Only the true forces in the non-inertial frame are considered for these evaluations.

If the body is constrained to rotate about a fixed point and that point is accelerating in an inertial frame, the evaluation must consider the fictitious forces present using that point as the origin. A good intermediate/advanced physics mechanics book addresses the general motion of a rigid body; for example, see Mechanic by Symon, or Classical Mechanics, by Goldstein.

Answer 3

Once the forces and their positions are known, its possible to calculate the torque with respect to the COM. The forces can be affected by the acceleration, but supposing that they are measured correctly (by a load cell for example) that effect is already being taken in consideration.

For example an yoyo toy falls at an acceleration $a$ while I keep the end of the string at a given height. The result is a tension at the string smaller than $mg$.

But if I pull the string up in order to keep the disk in the same height, (instead of let it fall) the tension is now $mg$.

Each situation results in a different torque, and different angular acceleration. But if the tension is properly measured, the torque with respect to the COM is known.