Serway & Jewett's definition of rotational equilibrium

On p. 364 of Physics for Scientists and Engineers (9th ed.), Serway and Jewett define a rigid object to be in rotational equilibrium if it has an angular acceleration of zero. They then state that a necessary condition of rotational equilibrium is that the net torque about any axis must be 0, where "any" does mean "every", as confirmed by a later remark in the text.

I am badly misunderstanding something in the above. Consider the example of a rod of length $$\ell$$ located at time $$t=0$$ from $$(0,0)$$ to $$(0,\ell)$$ in some inertial reference frame. Suppose that two identical and constant forces with magnitudes $$F$$ act on the ends of the rod, wherever they are, and that these forces are directed in the positive $$x$$ direction.

I claim that the rod has zero angular acceleration in this reference frame, and is thus in rotational equilibrium. According to S&J, the net torque on the rod about any axis must be $$0$$. But consider in this reference frame the axis through the origin that is perpendicular to the $$xy$$-plane. Calculating $$\mathbf{r}(t) \times \mathbf{F}$$, isn't the torque about this axis always in the negative $$z$$ direction with constant magnitude $$\ell F \not = 0$$?

[To clarify: the calculation in the previous paragraph is really a calculation of $$\mathbf{r}_A(t) \times \mathbf{F} + \mathbf{r}_B(t) \times \mathbf{F}$$ where $$A$$ and $$B$$ are the ends of the rod.]

• The rod you describe is not in rotational equilibrium. For the reasons you point out, it experiences a net torque about the origin.
– pwf
Commented May 16 at 16:03
• Are you suggesting (1) an object can fail to be in rotational equilibrium even if its angular acceleration is 0 or (2) the rod in question has nonzero angular acceleration? Note that, in the case I've described, the rod is simply experiencing constant acceleration to the right in this reference frame. Commented May 16 at 17:58
• It's (2). It's a bit counterintuitive at first, but an object moving in a straight line has angular momentum around an axis that's not on the line, and if the object is speeding up or slowing down, then it has angular acceleration. The fact that you (correctly) find a nonzero torque about the z axis means its angular momentum is not constant. It's complicated by the fact that its angular velocity and its moment of inertia about that axis are both changing.
– pwf
Commented May 17 at 17:12
• I don't think so, at least as far as Serway defines angular acceleration. Angular displacement is the change in the angle between a reference line on the object and a reference line in the coordinate system. Here, let those be the rod itself (wherever it is in space) and the $x$-axis. The angle never changes, so the angular displacement is always 0, as is the angular acceleration. Right? Commented May 18 at 0:33
• Also, as Serway explicitly points out (pp. 340-1), the relation between net external torque and change in angular momentum will hold for an object with an accelerating COM if one measures from the axis running through the COM. No guarantee it holds otherwise. Commented May 18 at 0:36

For rotational equilibrium the sum of all the torques acting on the object is zero.

In your example the sum of torque about (0,0) due to the force at (0, l) and the torque about (0,l) due to the force at (0,0) is zero.

Hope this helps.

• In my example, the sum of those torques at $t=0$ is indeed $0$ when you choose the axis to be perpendicular to the $xy$-plane and through the point $(0, \frac{\ell}{2})$. But I don't see how that is relevant. That is just a particular axis at a particular time. Commented May 16 at 14:31
• @dontknowphysics It's not just a "particular" axis. It's the all important axis through the COM. My point is if you select any axis through the rod other than the COM there will be a torque about that axis. But for each such axis you can find another on the rod having an equal and opposite torque for a net torque of zero. I don't have Serway and Jewett, but I suspect you might be misinterpreting what they are saying. Commented May 16 at 15:00
• Well, it's only the axis through the COM at time $t=0$, as I'm sure you realize, though the sum of the torques calculated relative to it remains $0$ at all times. Yes, I see what you are saying and that makes sense to me. I have almost quoted S&J verbatim in my post, but I do still suspect I'm misunderstanding something. Could "any" actually have existential import here instead of universal? Is that a standard way of defining rotational equilibrium? Commented May 16 at 15:19
• Here is a different text, Resnik: "The vector sum of all external torques that act on the body, measured about any possible point, must also be zero" (emphasis original). Unless this definition is meant to only capture rotational equilibrium in the case where an object is also in translational equilibrium (and thus we don't have an "independent" definition of rotational equilibrium here), I don't get it. Commented May 16 at 15:32
• @dontknowphysics The Resnik quote is exactly what I'm talking about. Commented May 16 at 16:40

Firstly, the angular momentum of a rigid body about a given point is the sum of orbital and spin angular momentum: $$\mathbf{L}=m\mathbf{r}_\text{CM}\times\mathbf{v}_\text{CM}+I_\text{CM}\boldsymbol{\omega}.$$ The former depends on the choice of origin but the latter doesn't. Secondly, when the point of consideration is changed by $$\mathbf{R}$$, the net torque changes as $$\sum_i(\mathbf{r}_i+\mathbf{R})\times\mathbf{F}_i = \sum_i\mathbf{r}_i\times\mathbf{F}_i + \mathbf{R}\times\sum_i \mathbf{F}_i$$ so, as can be seen from the additional term, the net force is required to be zero in order for the net torque about any point to be the same.

In your example, the net force isn't zero so the net torque will depend on the point chosen. The spin angular momentum is zero but the orbital angular momentum is changing, which is consistent with the nonzero net torque about the origin. The general principle is that the net torque about a point is always equal to the rate of change of the total angular momentum about that point.