Serway & Jewett's definition of rotational equilibrium

Question

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.]

Answer 1

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.

Answer 2

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.