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Consider a horizontal long rod that is undergoing free fall. Consider the torque about an axis through the rod (perpendicular to the rod and to the direction of gravitational force), that is a little to the left of the rod's centre of mass. Clearly, there is a net torque about this axis, but we know that the horizontal long rod will only translate downwards.

I suspect it arises from somehow this reasoning being inapplicable for some cases of translating axes about which torque is taken, but in the case of a cylinder rolling down a rough inclined plane, the same paradox is not seen by taking torque about the axis passing through the centre of mass of the cylinder, even though this axis is translating downslope.

What's going on?

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but we know that the horizontal long rod will only translate downwards.

Your problem is that you have not yet realised that constant velocity in a straight line can have constant non-zero angular momentum depending upon where you take reference.

There is no paradox. This is a known and necessary feature of angular momentum.

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  • $\begingroup$ Is there then a way to mathematically derive the equations of motion from taking torque about an axis that doesn't pass through the CoG? Or is taking torque about an axis that doesn't pass through the CoG just foolish and should never be done? $\endgroup$
    – bluesky
    Jun 4, 2023 at 1:37
  • $\begingroup$ The maths just works regardless of where you pick the axis. If you pick somewhere else than the CoG then you just have to take the torque into account. $\endgroup$ Jun 4, 2023 at 5:13
  • $\begingroup$ OP is asking why a net torque about an axis not going through the center of mass doesn't result in a CHANGE in angular momentum about that axis. $\endgroup$ Jun 4, 2023 at 15:05
  • $\begingroup$ @Not_Einstein and my answer directly answers that. $\endgroup$ Jun 5, 2023 at 7:17

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