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Suppose that there is a thin rod $AB$ lying in a gravity-free environment. There are equal and parallel forces which act on it’s ends A and B.

Now w.r.t the COM the net torque is zero. That means that no matter if we look at the rod from the COM or from some random point in space, the rod will not rotate about COM. Now if we look at the rod from A’s or B’s perspective, we see that the torque about A or B is also zero. So that would mean that the body cannot rotate about A or B from their perspectives respectively.

However if we try to calculate the net torque about A from some point in space and not from A’s perspective, we get a net torque about A. But again, the body does not rotate about A.

Why are we getting this torque about A even if the body is not rotating around it?

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  • $\begingroup$ -1, It is not clear what argument you are making. As BowlOfRed suggests, it would be helpful if you gave details for "if we try to calculate the net torque about A from some point in space". $\endgroup$ – sammy gerbil Feb 18 '17 at 6:34
  • $\begingroup$ @sammygerbil, I’ve tried my best to make it as understandable as possible. “if we try to calculate the net torque about A from some point in space” simply means that if we try to calculate the torque about A from an inertial frame of reference and not from the body’s frame itself. $\endgroup$ – Aaryan Dewan Feb 18 '17 at 7:06
  • $\begingroup$ I am asking how you have calculated a non-zero torque about A? I understand what you mean, I don't understand how you have reached this conclusion. $\endgroup$ – sammy gerbil Feb 18 '17 at 7:12
  • $\begingroup$ @sammygerbil , take a look at this question. physics.stackexchange.com/questions/312517/… Here, the OP has calculated a torque about B as viewed from an inertial frame and the torque which he has calculated is non zero. You get different torques about a point if you calculate it from different perspectives. $\endgroup$ – Aaryan Dewan Feb 18 '17 at 7:18
  • $\begingroup$ @AaryanDewan Would you mind actually writing the equation that gives you a non-zero net torque? $\endgroup$ – Steeven Feb 18 '17 at 10:40
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If the body is in equilibrium, it doesn't matter where you situate your axis for taking moments. The moments will sum to zero.

If the center of mass is accelerating, then you need to either (a) take moments about the center of mass or (b) include a pseudo force of -ma at the center of mass and take moments about any arbitrary alternate axis. In both cases, you will get exactly the same net torque.

In the sample problem, the moments about the center of mass are zero. With a pseudo force of $-ma=-2F$ through the center of mass, the moments about point A are $$FL+(-2F)\frac{L}{2}=0$$

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  • $\begingroup$ We add a pseudo when we are trying to torque about A from A’s perspective. What if we try to measure the torque about A from an inertial frame? Then there should be no pseudo force in that case! How will the torque be zero in that case? $\endgroup$ – Aaryan Dewan Feb 18 '17 at 2:29
  • $\begingroup$ I don't understand the question. My answer was regard to how to calculate the torque about the center of mass by taking moments about any arbitrary axis. The pseudo force is something you include mathematically. $\endgroup$ – Chet Miller Feb 18 '17 at 2:44
  • $\begingroup$ You first proved that the torque about the COM is zero. Then the torque about any other point must be zero too. My question is that will the torque about any point be zero w.r.t that point or w.r.t the observer ( inertial frame ) ? If you say that the torque will be zero w.r.t the point then you apply a pseudo force on the COM just like you did in your answer. Now if you want to calculate the torque about that point from an inertial reference frame, then you do not have to apply a pseudo on it! How can you prove without applying pseudo force on the COM, that the torque about any point = 0? $\endgroup$ – Aaryan Dewan Feb 18 '17 at 2:54
  • $\begingroup$ Can this be the case - We know that there is a net torque about any point on the rod except the COM when viewed from an inertial frame. The body does not rotate about that point ( despite the fact that there is a net torque about that point ) because there exists an anti-torque of the same magnitude on some other point on the body w.r.t the inertial frame, so the two torque’s cancel out? $\endgroup$ – Aaryan Dewan Feb 18 '17 at 3:01
  • $\begingroup$ In my judgment, the goal is to determine the torque about the COM. This can then be set equal to the angular acceleration times the moment of inertial about the COM. We can get this torque about the COM using any other point as the axis of rotation by including the pseudo force through the COM and the moment it produces. $\endgroup$ – Chet Miller Feb 18 '17 at 3:17
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However if we try to calculate the net torque about A from some point in space and not from A’s perspective, we get a net torque about A. But again, the body does not rotate about A.

It would be nice if you showed an example. Are you sure that you are including any fictitious forces that appear due to frame acceleration?

A non-zero net torque about A means that the angular momentum about A is changing. But it is possible for that change to be due to a linear acceleration instead of rotation.

If A is at rest in an inertial frame, then the change will have to include rotation. But if A is accelerating, then rotation is not necessary.

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It was not clear in the original question about which point the toque was measured.

Assume the rod to be on a horizontal frictionless table with point $A'$ fixed relative to the table.

enter image description here

If the rod has a mass $m$ then the acceleration of the rod is $a = \dfrac{2F}{m}$.

About point $A'$ there is a clockwise torque of $Fl$.
If the speed of the rod is $v$ then the angular momentum of the rod about $A'$ is $mv\dfrac l 2$.

The rate of change of angular momentum of the rod about $A'$ is $ma\dfrac l 2$ and this must equal the torque about $A$ which is $Fl$.

$Fl = ma \dfrac l 2 =\Rightarrow a = \dfrac{2F}{m}$ as before.


If point $A$ is the end of the rod then a pseudo force must be introduced as per @ChesterMiller 's answer to the question.

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  • $\begingroup$ This is confusing. Is there a non-zero net torque about A or isn't there? Does the OP's question have any meaning? $\endgroup$ – sammy gerbil Feb 18 '17 at 7:20
  • $\begingroup$ @sammygerbil I think that the point $A$ was not defined well enough. $\endgroup$ – Farcher Feb 18 '17 at 9:08

protected by Qmechanic Feb 18 '17 at 8:35

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