A recent paper in this journal^{1} purports to find that “a truly rigid body model leads to an incorrect statement of the rotational second law” and that “rigid bodies are inconsistent with classical (non-relativistic) physics.” Such a discovery would be profound, and it would be most remarkable if such a fundamental truth about Newtonian mechanics had gone unnoticed for 350 years. However, the paper contains significant flaws, and its conclusion is unwarranted.

The author uses Fig. 1 (redrawn from Fig. 2 of Ref. 1) to analyze two masses connected by massless, rigid rods with the outer mass $m2\u2009$ being acted on by a tangential external force *F*. The author then applies Newton's second law to each mass to obtain

where $\alpha \u2009$ is the object's angular acceleration. This equation is incorrect, as the author notes, because $r2F$ is the torque but $m1r1r2+m2r22$ is not the moment of inertia of the two masses about the pivot point. That is, an analysis using Newton's second law for the individual masses fails to yield the expected rotational second law $\tau =I\alpha $ for the object as a whole. The author concludes that the mistake has arisen from the assumption that the connecting rod is perfectly rigid.

However, the mistake actually arises from a misunderstanding of what forces are being applied and a misapplication of Newton's third law. According to the author, “For the body to remain rigid $m1$ must also accelerate tangentially, which requires a tangential force of magnitude *T* from $m2$, transmitted along the rod, and by the third law there must be a reaction force on $m2$, which is also tangential and of magnitude *T*.” These are the two forces labeled *T* in Fig. 1.

Newton's third law concerns the forces between two interacting objects, but the two masses in Fig. 1 do not interact with each other in any way. The force on $m1$ is not “from $m2$,” and thus any tangential forces on the masses cannot be an action/reaction pair of forces. Instead, each mass interacts with the rod, which does, in some sense, “transmit” the force, but the third law tells us only about the mass-rod interactions.

That the tangential forces on the masses cannot be of equal magnitude can be seen by considering the reaction forces on the rod, assumed to be massless. The rod pushes upward on mass $m1$, so $m1$ exerts a downward reaction force on the rod at distance $r1$ from the pivot. Similarly, $m2\u2009$ exerts an upward reaction force on the rod at $r2$. If these forces were equal in magnitude, as Fig. 1 claims, but applied at different distances from the pivot, they would exert a net torque on the rod. But there can be no net torque on a massless rod.

It's useful to consider the analogy of two masses connected with a massless string. There can be no net force on a massless string because even an infinitesimal net force would cause an arbitrarily large acceleration. The no-net-force constraint requires equal tensions at the ends of a massless string.^{2}

The corresponding constraint for a massless rod is that there can be neither a net force (which would cause an infinite translational acceleration) nor a net torque (which would cause an infinite angular acceleration). To see the implications of this constraint, Fig. 2 redraws Fig. 1 but with: (a) the forces on the masses and on the rod shown separately, (b) no assumption about how $T1$ and $T2$ are related, and (c) the force $Fpivot$ of the pivot on the rod included. The pivot force, which allows the forces on the rod to sum to zero, contributes no torque about the pivot point but does prevent the left end of the rod from moving.

If the rod is massless, the no-net-torque constraint requires

The two tangential forces are not equal-but-opposite forces, as the author assumed. Now one might object that Eq. (2) assumes a definition of torque, whereas one of the author's stated goals is to determine what a proper definition of torque should be. However, Eq. (2) can be found using only Newton's second law for translational motion and the constraint that *all* points on a rigid rod have the same angular acceleration; no prior definition of torque is necessary. In fact, Eq. (2) could be seen as a justification for the conventional definition of torque.

Newton's second law for each mass (with $a1=r1\alpha $ and $a2=r2\alpha $) along with Eq. (2) constraint form three equations in three unknowns. They are easily solved to give

which is exactly the rotational second law $\tau =I\alpha $.

Rather than having the two masses on one rod, suppose mass $m1$ is connected to the pivot by one rigid rod while a second rigid rod connects the two masses. One could imagine that the rods are plugged into sockets drilled into the masses. This is a bit more complex because now, to maintain rigidity, mass $m1$ has to exert torques on the two rods in much that same way that a wall exerts a torque on a stationary pole extending horizontally from a socket in the wall. Even so, the analysis leads to exactly the same conclusion as the above analysis for a single rod; namely, the object rotates as a perfectly rigid body obeying the rotational second law. The author's Fig. 4, which shows the second rod bent at an angle with respect to the first rod, is incorrect.

Now, in the 21st century we know that real objects are not perfectly rigid. One is certainly free to model an object as having spring-like bonds, as the author does in the second part of the paper (although the analysis contains unwarranted assumptions and mathematical errors). By definition, an object modeled with spring-like bonds will not be rigid. But that's not the issue here. The issue is whether we are *compelled* to replace rigid rods with springs; that is, is a truly rigid body—be it a macroscopic object or an atomic model with rigid bonds—incompatible with Newtonian mechanics, as the author claims?

Most certainly not. Rigid bodies, one of the foundational models of classical mechanics, are perfectly consistent with the laws of Newtonian physics and lead to a correct statement of the rotational second law.

The author thanks Thomas Gutierrez and Thomas Bensky for useful discussions.