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Torque, for a system of particles, is defined as:

$$\boldsymbol {\tau}_{net} = \sum_{i=1}^n \mathbf r_i \times \mathbf F_{i,net}$$

Here $\mathbf r_i$ is the position vector of the point from the coordinates axis. Now usually this coordinate axis coincides with the axis of rotation.

  • But what if this coordinate axis doesn't coincided with the axis of rotation then what would the value of torque about that axis mean?
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  • $\begingroup$ Does this answer your question : Is torque independent of choice of the point of rotation? $\endgroup$ Commented Jan 27, 2020 at 17:13
  • $\begingroup$ @sammy can you please tell what is wrong with the concept? $\endgroup$
    – user249968
    Commented Jan 27, 2020 at 17:13
  • $\begingroup$ See also Which is the axis of rotation? $\endgroup$ Commented Jan 27, 2020 at 17:28
  • $\begingroup$ @sammy I have clarified the question. Please do tell if you think it's still not clear. $\endgroup$
    – user249968
    Commented Jan 27, 2020 at 17:29
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    $\begingroup$ Related answer to a question on the nature of torque. $\endgroup$ Commented Jan 27, 2020 at 19:31

2 Answers 2

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Torques exist independent of rotations. Indeed, in statics problems nothing is rotating, yet we can still choose points of reference to calculate torques about (and show that the net torque about any such point is $0$).

You just pick some reference point, then you apply your definition to determine the torque caused by forces about that point. Of course for certain analysis it is smart to choose the point that coincides with the axis of rotation, but this is not a requirement to apply the definition of torque itself.

This is also found in the definition of angular momentum $\mathbf L=\mathbf r\times\mathbf p$. You can choose any reference point you want to calculate angular momentum about. And you can still write out $\boldsymbol\tau=\text d\mathbf L/\text dt$ about that axis, where the torque and angular momentum are relative to the same axis.

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  • $\begingroup$ Can you please continue after: "You just pick some reference point, then you apply your definition to determine the torque caused by forces about that point". So what does the torque about that axis mean? As for now I think that about any arbitrary axis the torque causes increase in angular momentum about that axis and it seems to me that for axis farther from the object the rate of change of angular momentum and is higher(?)). Also can you please tell how any arbitrary axis might be different from the axis of rotation? Thanks (+1 for now) $\endgroup$
    – user249968
    Commented Jan 27, 2020 at 18:01
  • $\begingroup$ As for now I think that about any arbitrary axis the torque causes increase in angular momentum about that axis Yes, that is correct. In terms of what it "means" or how one axis is "different" from the axis of rotation, those questions seem somewhat vague to me, and a more specific answer to them would probably need to assume a specific scenario and axis to calculate torques about. But in any case, the part I quoted will always be true, and I think that is a fine way to think of this. $\endgroup$ Commented Jan 27, 2020 at 18:05
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A fixed rotation axis, such as with a car wheel, can only rotate around the axle. An unrestrained object will only rotate about a line through it's center of mass. Applying a force off-center of it's COM will cause rotation and translation.

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