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Assume that we have a free rigid body, its own weight being the only force acting on it. If I want to know its angular acceleration, then I calculate the total moment acting on the body. Now, if I choose to compute the moment wrt the center of mass, then I will get zero (because that's where the weight is being applied) yielding zero angular acceleration, but there are other points for which the moment of the weight is not zero, yielding a nonzero angular acceleration.

Where is this inconsistency coming from?

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  • $\begingroup$ What do you mean by inconsistent? If you change your reference point your observations are bound to change. You are changing your frame from an inertial to a non-inertial frame. Can you please clarify what is it that you found inconsistent? $\endgroup$
    – Tamaghna Chaudhuri
    Commented Jun 23, 2021 at 17:59
  • $\begingroup$ well afaik for any two points $A,B$ in the rigid body it holds $\vec{v}_A = \vec{v}_B + \vec{\Omega} \times \vec{r}_{AB}$ and here $\vec{\Omega}$ is independent of the particular $A,B$ you choose. I assumed that the same should hold for $\vec{\gamma}$ but maybe I got this wrong... $\endgroup$
    – Javi
    Commented Jun 23, 2021 at 18:06
  • $\begingroup$ @TamaghnaChaudhuri Furthermore, why are you assuming that the first frame is inertial? Since there is only one force acting on the body, the center of mass is being accelerated as well, isn't it? $\endgroup$
    – Javi
    Commented Jun 24, 2021 at 17:39

2 Answers 2

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The expression τ = Iα = dL/dt derives from Newton's second and assumes that the chosen axis of rotation is in an inertial frame. It cannot be within a falling object. You can choose an arbitrary horizontal line in the fixed frame as an instantaneous axis. You can then assume that the torque about that axis results from the total weight acting along a vertical line which passes through the the center of gravity (instead of the actual distributed weight). That torque can be shown to be equal to the rate of change of the angular momentum of the falling object about the chosen axis.

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  • $\begingroup$ I see. But if I choose the center of mass as the reference point then $\vec {\tau} = I \, \vec {\alpha} $ works fine, even though the cm is not fixed to an inertial frame (because it is accelerating downwards). So why does it work in this case? $\endgroup$
    – Javi
    Commented Jun 27, 2021 at 4:24
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Angular momentum is not just "rotation". A body moving in a straight line has a non-zero angular momentum about a point not along the line of travel. $L = mvr$, where $r$ is the perpendicular distance between the line of travel and the axis of consideration.

If the velocity changes while the others are constant, this angular momentum will change as well. In your scenario, the body is accelerating downward. If your point of reference is not vertically aligned with the center of mass, then the angular momentum will be changing during the fall.

how do you know if the eq $\tau = I \alpha$ holds for that point?

It holds if the angular momentum change can only go into rotation. If the center of mass can accelerate in a direction that is not in line with your point of consideration, then it does not hold.

For a body falling, then any point in a vertical line from the center of mass is safe.

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  • $\begingroup$ The sum of torques is always equal to the change in angular momentum, but that momentum may not be in rotation. It is always in rotation if you pick an axis that goes through the COM of the body in question. $\endgroup$
    – BowlOfRed
    Commented Jun 27, 2021 at 4:32
  • $\begingroup$ I understand. So when you choose a reference point for computing the torques, how do you know if the eq $\vec{\tau} = I \, \vec {\alpha}$ holds for that point? E.g. in this case it works fine for the cm, even though the cm is not fixed to an inertial frame. $\endgroup$
    – Javi
    Commented Jun 27, 2021 at 4:32

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