A recent article in this journal1 focused on the problem of deducing the rotational dynamics equation from Newton's second law without prior definition of torque or angular momentum. The problem in question used a system consisting of two point masses connected by a massless rod and also connected (via a massless rod) to an axis. In the case of a rigid rod, the analysis provided in the article led to an incorrect equation. The model was then modified to allow the rod to bend, which led to the correct equation. Based on this analysis, it was concluded that it is essential to account for the finite deformations of the rod.

Dr. Knight objects to this conclusion by demonstrating that contradictions in the problem with rigid rods are eliminated if one accounts for the torques created in the points of contact between rods and rigid bodies.2 In his response,3 Dr. Cross points out that his conclusion about the rigid body model being incompatible with classical mechanics should only be understood in the context of a model of interactions between different parts of a rigid body.

We deem it reasonable to add the following to this discussion: (i) an analysis of the possibility of reduction from the real two bodies' contact forces to two net forces; (ii) a rigorous solution of the problem with 2 rods and 2 masses; and (iii) a derivation of the equation of rotational dynamics directly from Newton's laws without the introduction of a notion of torque.

Newton's third law in its original form is only applicable to point masses. The following question then arises: can we, in the analysis of the interaction between two bodies consisting of many point masses, use two net forces, F1 and F2, applied to the first and second bodies, respectively? A simple analysis shows that, in general, this would be incorrect. The “strong version” of Newton's third law1 implies that the net torque M of forces in the interaction between point masses is equal to zero. So let us take a point C1 at the center of mass of the first body. Then, calculating torques with respect to C1, we have

M=M12+M21*=0,
(1)

where M12 is the torque of the forces applied to the first body and M21* is the torque of the forces applied to the second body. Keeping in mind that

M21*=M21+r12×F21,
(2)

where M21 is the torque of the forces applied to the second body, calculated with respect to the center of mass of the second body C2, and r12 is the vector from point C1 to point C2. Combining Eqs. (1) and (2), we see that

M12+M21+r12×F21=0,
(3)

where M12 and M21 are calculated for the respective centers of mass.

It is possible to select points other than C1 and C2—those that are not the centers of mass—and Eq. (3) would still hold. This fact can be seen as a consequence of a well-known theorem of statics4 (the Poinsot theorem) that states that any system of forces applied to a rigid body is equivalent to a system of a force and a force couple (a pair of forces). Equation (3) is not essential for solving practical problems, as it follows from Newton's third law, though occasionally it proves to be of utmost use, as is the case with the questions raised in Ref. 1. Equation (3) implies that if forces of interaction between two bodies are not aligned along a straight line, it is essential to account for the resulting torque, since M12+M210 (see Fig. 1). An interaction between a stationary electric dipole and a point charge located on a line perpendicular to the dipole moment provides a vivid example of such a situation.

Fig. 1.

Two interacting masses.

Fig. 1.

Two interacting masses.

Close modal

Let us now show a solution to the problem of two rods and two masses brought up in Ref. 1, but with torques being accounted for. Let us consider the masses to be arbitrary and the rods to be both massless and rigid, thus using them as a way of transferring the interaction between the masses, just as done in Ref. 1. Figure 2 gives an illustration of all the forces and their torques for two masses and a stationary pivot. The torque M21=0 because the masses are considered as mass points (and is thus not shown). For the same reason

M12+M13=0.
(4)
Fig. 2.

Forces and torques in a system of two masses and two rods.

Fig. 2.

Forces and torques in a system of two masses and two rods.

Close modal

The torque M31 acting on the axis is also considered to be zero (and thus not shown), provided that the pivot is ideal.

It follows from Eq. (3) that

M12=(r2r1)F21
(5)

and

M13=r1F13,
(6)

where F21 and F13 are the components of F21 and F13 perpendicular to the rod. If we now write down Newton's second law for the masses, we have

m2(r2α)=FF21
(7)

and

m1(r1α)=F12+F13,
(8)

where α is the angular acceleration. Multiplying the first of these equations by r2 and the second by r1 and then adding, we get the equation of rotational dynamics as

(m1r12+m2r22)α=(FF21)r2+(F12+F13)r1=Fr2M12+M13=Fr2,
(9)

where we have used Eqs. (4)–(6) along with the fact that F21=F12. After finding the angular acceleration, it is possible to express both forces and their torques; for example,

M12=M13=Fm1r12(r2r1)m1r12+m2r22.
(10)

The radial components of the forces depend on the angular velocity of the masses and are responsible for their centripetal accelerations. This problem could be also solved by considering the rods to be part of the system and accounting for the forces and torques applied to them. Since the rods are massless, the net force and torque on them have to be equal to zero. This approach leads to a solution similar to that of Dr. Knight.

Let us now return to the original methodological problem: deriving Newton's second law for rotation without a prior definition of torque.

Let us first note that the torques do not appear if the masses are linked using ideal pivots. In this situation, the interaction forces are aligned along the straight line connecting the pivots. However, a rigid structure cannot be made from only two rods, as suggested in the original article.1 To get a model for a rigid body, it is necessary to add a third rod, as shown in Fig. 3, which completes the rigid triangular truss that is well known to engineers.

Fig. 3.

A rigid truss.

With this model, a derivation of the desired equation is trivial. We need Newton's second law for the two bodies

m2r2α=FT21sinθ2,
(11)
m1r1α=T12sinθ1,
(12)

the law of sines

r2r1=sinθ1sinθ2,
(13)

and the condition of the massless rod being in balance: T12 = T21. Using these equations, we get the correct second rotational law in a few simple steps as

(m1r12+m2r22)α=Fr2.
(14)

Let us also note that any rigid body can be represented as a combination of such rigid triangles.

The inconsistency of a truly rigid-body model with the rotational second law as discussed in Ref. 1 arises from a mistake made in Fig. 2. This figure does not take into consideration the tangential force applied to the body m1 by the rod that connects m2 with the pivot. In his comment,2 Knight gives a correct solution for the problem; we do it here within the framework of the approach suggested in Ref. 1, when the rods just model interactions of the bodies in the system. The discussion2,3 between Dr. Knight and Dr. Cross reveals a disagreement about another important aspect of the model in Ref. 1. Considering the central interaction of bodies, Cross deduces that a two-body system is not able to rotate with angular acceleration as a rigid body because the rotation inevitably causes it to deform. We fully agree with this conclusion; indeed, if the two rods are replaced with flexible ropes, the system with central forces obviously will not be able to rotate with angular acceleration as a rigid body. We have modified Cross's model by adding a third rope (in other words, a third central force), which resolves the controversy. Moreover, the resulting model is a fully correct model of a rigid body.

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D. J.
Cross
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The physical origin of torque and of the rotational second law
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Randall D.
Knight
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Comment on ‘The physical origin of torque and of the rotational second law’ [Am. J. Phys. 83, 121–125 (2015)]
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D. J.
Cross
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Reply to ‘Comment on “The physical origin of torque and of the rotational second law” [Am. J. Phys. 83, 121–125 (2015)]’
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Ferdinand P.
Beer
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