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An object in rotational equilibrium has no net external torque:

sum of all external torques = 0

Does it mean "rotational equilibrium" may mean that the object is not rotating ... or it may mean that the object is rotating with constant angular velocity? But when it is rotating ,is it still an equilibrium ?

Rotational equilibrium: $\sum \tau_C=0$

Note that for rotational equilibrium of a rigid body, the net torque about its centre of mass must be zero.

In this quote from a book it says that ror rotational equilibrium, torque about the body's Center of Mass must be zero. But why only about the center of mass? About any other point also it is in equilibrium. It is confusing.

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    $\begingroup$ Possible duplicate of The choice of pivot point in non-equilibrium scenarios $\endgroup$ – ja72 Jan 29 '19 at 14:36
  • $\begingroup$ No,it is different "Does it mean "rotational equilibrium" may mean that the object is not rotating ... or it may mean that the object is rotating with constant angular velocity? But when it is rotating ,is it still an equilibrium ?" $\endgroup$ – user212727 Jan 29 '19 at 15:01
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    $\begingroup$ @ja72 I would say that is a related question, not a duplicate. $\endgroup$ – Aaron Stevens Jan 29 '19 at 15:35
  • $\begingroup$ @TheBroly, just in case, I would edit your question to explain why your question is not a duplicate of the proposed duplicate question $\endgroup$ – Aaron Stevens Jan 29 '19 at 15:37
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Equilibrium can mean different things depending on the context. For example think of a container of gas with a permeable barrier dividing the container in half. At equilibrium you would expect an equal concentration of gas molecules on either side of the partition even though there is still going to be a transfer of gas molecules between each side of the container. The net flow is $0$ though, so we say the system is in equilibrium.

In the context of torques, I would all we need for equilibrium is that the net torque is $0$. Whether or not the system in question has a net rotation depends on your "initial conditions". Typically then you see in many statics scenarios that the system in question is also at rest, but you wouldn't approach the problem any differently if there was rotation (unless the motion changes torques of the system, but then it turns into a dynamics problem instead of a statics problem).

In this snapshot of a book it says that For rotational equilibrium ,torque about the body's Center of Mass must be zero.But why only center of mass? About any other point also it is in equilibrium.It is confusing.

I think this is the distinction between "rotational equilibrium" and more of a complete "mechanical equilibrium". For example, if there is a single force acting on the center of mass of our system, then of course there is no torque about the center of mass of the system and there will be no rotation about the center of mass. However a single force cannot keep a system in mechanical equilibrium, as there will be a net force. But if we pick any other point not along the direction of the force we will find a net torque about that point.

For complete mechanical equilibrium we can choose any point about which to find the torque and it must be $0$. But for just rotational equilibrium we just need to consider torques about the center of mass.

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  • $\begingroup$ So,if the body is rotating with constant angular velocity,is it in equilibrium? $\endgroup$ – user212727 Jan 29 '19 at 15:14
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Equilibrium is generally used to describe systems whose state remains constant with time.

An object rotating with constant angular velocity (even nonzero) is what is meant here. An object that is not rotating (angular velocity of zero) is just a special case.

As for torque, the net torque about any point will be zero if you include pseudoforces as well.

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  • $\begingroup$ As for torque, the net torque about any point will be zero if you include pseudoforces as well. I don't think this is right for just rotational equilibrium. If we push a wheel (in space I guess?) from an axle at its center it will not rotate about its center of mass, so the system is in rotational equilibrium, but there is still a net torque about a point below the axle perpendicular to the applied force. $\endgroup$ – Aaron Stevens Jan 29 '19 at 14:23
  • $\begingroup$ Isn't there also a net acceleration leading to a pseudoforce (also at the center of the wheel) that cancels the torque from the real force you applied? $\endgroup$ – nr2618 Jan 29 '19 at 14:24
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    $\begingroup$ You only consider pseudoforces in accelerating reference frames. I am in an inertial reference frame. It doesn't matter if something is accelerating or not because I am not working in the frame of the accelerating object $\endgroup$ – Aaron Stevens Jan 29 '19 at 15:22
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    $\begingroup$ In any inertial reference frame, if the force applied is through the point $\vec{r}\neq 0$, the torque is nonzero but the wheel also acquires angular momentum according to this reference frame (due to motion of its COM). My statement "the net torque about any point will be zero if you include pseudoforces as well" is in the context of rotational equilibrium. If the wheel is in rotational equilibrium according to your reference frame, the net torque about any point is zero. If your frame happens to be non-inertial, you must include the pseudotorques :) Hope that's clearer now? $\endgroup$ – nr2618 Jan 29 '19 at 15:57
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Aaron Stevens Jan 29 '19 at 19:04
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If the vector sum of all the forces acting on a rigid body is not equal to zero, Then its center of mass has a linear acceleration. That is, $$\sum F ≠ 0\implies a≠0.$$

Similarly, if the vector sum of all torques about the center of mass is not equal to zero, then the body rotate about its center of mass. So for rotational equilibrium torque about center of mass must be zero.

$$\sum T_c ≠0\implies a_c ≠0$$

Torque equation should be applied about center of mass or about axis of rotation.

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