About the Teaching of Plane Motion of Rigid Bodies
Diego Luis Gonzalez, Alejandro Gomez Cadavid, Yeinzon Rodriguez
PPI/UAN-2020-679FT
About the Teaching of Plane Motion of Rigid Bodies
Diego Luis Gonz´alez, ∗ Alejandro Gomez Cadavid, † and Yeinzon Rodr´ıguez
2, 3, 4, ‡ Departamento de F´ısica, Universidad del Valle, A.A. 25360, Cali, Colombia Centro de Investigaciones en Ciencias B´asicas y Aplicadas,Universidad Antonio Nari˜no, Cra 3 Este Escuela de F´ısica, Universidad Industrial de Santander,Ciudad Universitaria, Bucaramanga 680002, Colombia Simons Associate at The Abdus Salam International Centre for Theoretical Physics, Strada Costiera 11, I-34151, Trieste, Italy (Dated: November 9, 2020)The study of the motion of a rigid body on a plane (RBP motion) is usually one of the mostchallenging topics that students face in introductory physics courses. In this paper, we discuss acouple of problems which are typically used in basic physics courses, in order to highlight someaspects related to RBP motion which are not usually well understood by physics students. Thefirst problem is a pendulum composed of a rod and disk. The angular frequency of the pendulumis calculated in two situations: disk fixed to the rod and disk free to spin. A detailed explanationof the change in the angular frequency from one case to another is given. The second problem is aladder which slides touching a frictionless surface. We use this problem to highlight the fact thatthe contact forces applied by the surface perform translational and rotational work despite that thetotal mechanical energy of the ladder is conserved.
I. INTRODUCTION
Two basic objectives that students must achieve inlearning physics are the understanding of fundamentalconcepts, as well as the relationship between them. Abasic diagnostic tool to measure the understanding levelof the students is their performance applying basic con-cepts to solve physical problems. However, in a typicalphysics course, there are often many students who havedifficulty to reach these objectives. For example, it isvery common the misconception that there will alwaysbe a force in the direction of motion. Therefore, it is nota surprise that just a fraction of the students realize thatthe direction of motion is given by the velocity and not bythe net force. One of the most challenging topic in intro-ductory physics courses is the motion of a rigid body ona plane (RBP motion). In the RBP motion, all the par-ticles belonging to the body move in fixed planes whichare all parallel to the frame of reference i.e., the motionis bidimensional. Consequently, the equations related tothe RBP motion can involve up to three coordinates, twofor the position of the center of mass ( cm ) and, an ad-ditional one, for the angular position of the body. Giventhe complexity of the RBP motion, several theoreticalaspects related to the teaching of it have been treatedin previous papers [1–6]. The learning difficulties andtypical misconceptions of the students, associated to theRBP motion, have been reported previously by severalauthors [1, 3, 7–16]. For instance, the students have dif-ficulty in determining the magnitude of the torque andconnecting the total torque to the direction of rotation[17]. It has been also reported that the students typically ∗ [email protected] † [email protected] ‡ [email protected] are in trouble at understanding that, in general, differ-ent points of a rigid body have different velocities andaccelerations [18].In this paper we focus in two situations related to theRBP motion which usually are not well understood bythe students of introductory physics courses, that is whywe restrict to the physical and mathematical tools givenin this kind of courses instead of using more advancedtools such as the Lagrangian formalism.The first situation is related to the contribution of themoment of inertia in the angular frequency of a bodycomposed by several rigid parts. Particularly, we con-sider a compound physical pendulum, which consists ofa disk placed at the end of a rod as shown in Fig. 1(a). In Ref. [19] this system is proposed as a problemthat challenges the students to determine the oscillationfrequency of the system in two cases. In the first case,the disk is fixed to the rod forming a single rigid body.In the second, the disk is mounted to the rod by a fric-tionless bearing so that it is perfectly free to spin. Thisexercise was chosen because it is very illustrative for ourpurpose and surprisingly, a quick search in informal phys-ical forums shows that many of the reported solutions donot clearly explain why the oscillation frequency changesfrom one case to the other even though when it is a com-mon problem that can be found in different course mate-rials [20–23]. Thus, our purpose here is to show that thestudents must realize how the angular momentum is di-vided into two parts and its implications in the oscillationfrequency.On the other hand, there are two basic approaches towrite the work-energy theorem in the RBP motion. Inthe first one, the system is considered as a collection ofparticles and the total work over the body, W , is the sumover the work of each particle belonging to the system. a r X i v : . [ phy s i c s . e d - ph ] N ov (a) (b) (c) (d) FIG. 1: (a) Pendulum composed by a disk and a rod. (b) External forces acting on the rod-disk rigid system. In (c)and (d) the forces acting on the disk and rod are shown, respectively. The interaction between the disk and rod isrepresented by the force (cid:126)N and the torque (cid:126)τ . (a) (b) FIG. 2: (a) A ladder of length 2 (cid:96) starts to slide from rest on a frictionless surface and (b) forces acting on the ladder.Therefore, W is given by: W = (cid:88) i (cid:90) C i (cid:126)F exti · d(cid:126)r i = ∆ E k , (1)where (cid:126)F neti is the total force acting on the i -th particlewhich is the sum of the forces applied by other particlesbelonging to the rigid body ( (cid:126)F inti ) plus forces appliedby particles that are not part of the rigid body ( (cid:126)F exti )[19, 24]. In the second approach for the work-energytheorem, the motion of each particle is decomposed intranslation of the center of mass ( cm ) and rotation aboutthe cm . Consequently, in this last approach the totalwork over the system has two different contributions, onefrom the translation of the cm , W T , and another one fromrotation, W θ .The second situation we want to discuss is related tothe contributions of the external forces to the transla-tional and rotational works. In several cases, the to-tal mechanical energy of a body is conserved, althoughthere are forces which perform translational and rota-tional work. Particularly, we consider another typicalexample: a ladder that slides without friction as in Fig.2 (a) [19, 24–28]. The forces associated to the wall andfloor perform rotational and translational works whichhave the same magnitude but opposite sign. The stu-dents often tend to assume that the translational androtational works associated to these forces are zero in- stead of realizing that the joint contribution to the work, W θ + W T , is the one which vanishes.We choose to discuss the composed pendulum and thesliding ladder because they are classical examples of RBPmotion often used in introductory physics courses to ex-plain several concepts such as torque, angular momen-tum, energy conservation, etc. Additionally, they allowus to illustrate our objectives, so that the solutions pre-sented in this paper are more detailed than those pre-sented in [20–23, 27, 28]. For example, in the ladderproblem we use the integral of the work-energy theoreminstead of writing the energy conservation equation.This paper is divided as follows. In Sec. II we presenta brief description of the basic equations of RBP motion.In Sec. III and IV we apply these equations to the twosystems previously mentioned: the composed pendulumand the sliding ladder. Finally, in Sec. V we providea discussion of the results and give some advices to im-prove the understanding of the students about the RBPmotion. II. PLANE RIGID BODY DYNAMICS
A body with mass M can be considered as a rigid bodyif, for any couple of points p and p (cid:48) of the body, thecondition: || (cid:126)r p − (cid:126)r p (cid:48) || = C pp (cid:48) , (2)is satisfied with C pp (cid:48) being a constant and (cid:126)r p , (cid:126)r p (cid:48) be-ing the positions of the points. Therefore, for a rigidbody, the relative distance between any couple of pointsbelonging to the body remains constant.There exist two simple ways in which a rigid bodywhose motion is restricted to a plane can move, it cantranslate without rotating or rotate about a fixed axis O (cid:48) without translating. In the first one, the motion ofthe body is fully specified by the translational dynamicsof the cm . Consequently, the body motion is completelydescribed by the Newton’s second law, i.e., the net forceover the body, (cid:126)F net , is proportional to the acceleration ofthe cm , whose position, (cid:126)R cm , is defined by the relation M (cid:126)R cm = (cid:82) Ω dm (cid:126)r , where the integral extends over thebody region Ω.The translational work done by the net force movingthe cm along the path C is given by: W T = (cid:90) C d (cid:126)R cm · (cid:126)F net = ∆ E K , (3)where E K = M V cm / (cid:126)V cm are the translational ki-netic energy and the velocity of the cm , respectively. Itis important to emphasize that Eq. (3) is completelygeneral and still valid even if the body rotates on theplane.In the second simple case, the axis may be at rest ormoving with constant velocity without any change in itsorientation, then it is convenient to put the origin of theinertial frame on the rotation axis O (cid:48) . Besides, the ve-locity of an arbitrary point of the body with position (cid:126)r measured from O (cid:48) can be written as (cid:126)v = (cid:126)ω × (cid:126)r (cid:48) where (cid:126)ω is the angular velocity of the rigid body. In general, (cid:126)v is not equal for the different points of the body but (cid:126)ω is the same for all of them. The rotational work W θ isdefined by means of the net torque over the body, whichis proportional to the angular acceleration (cid:126)α . W θ = (cid:90) θ B θ A (cid:126)τ net · d(cid:126)θ = ∆ E K , (4)where E K = I O (cid:48) ω / I O (cid:48) = (cid:82) Ω dm r is the moment of inertia about O (cid:48) . It isimportant to emphasize that, for a rigid body, I O (cid:48) justdepends on the mass distribution relative to O (cid:48) . Thus,it is clear that I (cid:48) O is independent on the motion state ofthe body.The most general case of RBP motion is given whenthe rotation axis O moves on the plane. In this case,it is convenient to decompose the motion of the bodyin displacement of the cm and rotation about the cm .This decomposition can be archived noticing that theposition of any point of the body, (cid:126)r (cid:48) , can be written as the sum of the position of the center of mass (cid:126)R cm and therelative position of the point with respect to to the cm , (cid:126)r , as shown in Fig. 3. Thus, by using the decomposition (cid:126)r (cid:48) = (cid:126)r + (cid:126)R cm , the angular momentum is given by: (cid:126)L = (cid:90) Ω (cid:126)r (cid:48) × d(cid:126)p (cid:48) = (cid:90) Ω dm ( (cid:126)r + (cid:126)R cm ) × ddt ( (cid:126)r + (cid:126)R cm )= (cid:126)R cm × (cid:126)P cm + I cm (cid:126)ω, (5)where I cm is the moment of inertia relative to the cm and (cid:126)P cm the linear momentum of the center of mass. Hence,the net torque can be found by taking the time derivativeon Eq. (5): (cid:126)τ net = (cid:126)R cm × (cid:126)F net + I cm (cid:126)α. (6)If the body describes a pure rotation about the fixed axis O (cid:48) , then (cid:126)V cm = (cid:126)R cm × (cid:126)ω and Eqs. (5) and (6) take theform: (cid:126)L = I O (cid:48) (cid:126)ω and (cid:126)τ net = I O (cid:48) (cid:126)α, (7)where I O (cid:48) is the moment of inertia about O (cid:48) .FIG. 3: (cid:126)R cm and (cid:126)r (cid:48) represent the positions of the cm and of the point p relative to the inertial referencesystem O , respectively. Analogously, (cid:126)r is the position of p relative to the cm .Let d (cid:126)F ext ( (cid:126)r ) be a differential of the total external forceover the body which is applied at position (cid:126)r , and, (cid:126)F net = (cid:82) Ω d (cid:126)F ext ( (cid:126)r ). Consequently (cid:126)τ net can also be calculatedfrom: (cid:126)τ net = (cid:90) Ω (cid:126)r (cid:48) × d (cid:126)F ext ( (cid:126)r ) = (cid:126)τ cm + (cid:126)R cm × (cid:126)F net , (8)which by comparing to Eq. (6) allows us to concludethat, the torque about the cm is given by (cid:126)τ cm = I cm (cid:126)α and is valid regardless the kind of movement of the centerof mass. The work done by the torque (cid:126)τ cm has the sameform as that in Eq. (4) replacing I O (cid:48) by I cm . Finally, thetotal kinetic energy can be written as: E K = 12 (cid:90) Ω dm v (cid:48) = 12 M V cm + 12 I cm ω . (9)From Eqs. (3) and (4) restricted to O (cid:48) → cm , it is pos-sible to conclude that the contributions to the work re-lated to the rotation and translation are separated be-cause they are related to independent degrees of freedom.In contrast, for the case of rotation about a fixed axis,rotation and translation are associated and both contri-butions to the work are completely equivalent.A more complete discussion of the results presented inthis section can be found, for example, in Refs. [19, 24–26]. III. THE DISK AND ROD PROBLEM
Now we apply the results of the previous section to an-alyze the role of the various contributions to the momentof inertia in the dynamic behavior of a system composedby several interacting rigid bodies. In order to accom-plish this, we consider a simple physical system whichhave been widely studied: a compound physical pendu-lum [19–23]. Let’s consider a disk of mass M and ra-dius R fixed to the end of a rod of length (cid:96) and mass m , see Fig. 1 (a). In introductory courses of physics,the students are usually asked to determine the angularfrequency in the limit of small oscillations. Two simplecases can be considered: in the first, the disk is fixed tothe rod in such way that they form a single rigid body; inthe second case, it is considered that the disk is mountedto the rod by a frictionless bearing, so that it is perfectlyfree to spin. A. Case 1: Fixed Disk
In the case 1, the rigid body formed by the rod anddisk rotates around the fixed axis O . Then, from Eq. (7)the angular momentum of the body with respect to therotation axis is given by: (cid:126)L = ( I rod + I disk ) (cid:126)ω = (cid:18) m (cid:96) M (cid:96) + M R (cid:19) (cid:126)ω, (10)where we have used the decomposition shown in Fig. 3to calculate I disk . This result is a particular case of theSteiner’s theorem.On the other hand, taking the whole pendulum as asingle rigid body, the torque with respect to the rotationaxis due to the net force about O depends only on thetotal weight of the body, see Fig.1-(b). The interactionbetween the disk and the rod is represented by the con-tact force (cid:126)N and the torque (cid:126)τ as shown in Figs. 1-(c) and (d). Note that the force (cid:126)N and the torque (cid:126)τ are notzero, but they do not affect the motion of the rod-disksystem. In this way, we have: (cid:126)τ net = − g (cid:18) m (cid:96) M (cid:96) (cid:19) sin θ ˆ k. (11)Taking the time derivative on Eq. (10) and matching theresult with Eq. (11), it is easy to find the equation ofmotion for the pendulum: α = d θdt = − g (cid:96) (cid:0) m + M (cid:1) m (cid:96) + M (cid:96) + M R sin θ. (12)This result can also be obtained easily by using Eq. (3)or (4). Taking Eq. (4) between the trajectory pointsdefined by the angles θ and θ , we have: W θ = − (cid:90) θθ dθg (cid:96) (cid:16) m M (cid:17) sin θ = 12 I (cid:0) ω − ω (cid:1) . (13)After integration, Eq. (13) reduces to: − g (cid:96) (cid:16) m M (cid:17) cos θ + 12 (cid:18) m (cid:96) M (cid:96) + M R (cid:19) ω = E, (14)where E is a constant which represents the total mechan-ical energy, more details can be found in Refs. [20–23].This is nothing more than the conservation energy equa-tion. The first term represents gravitational potentialenergy, while the second the kinetic energy. Taking thetime derivative of Eq. (14) we find: ω (cid:20) g (cid:96) (cid:16) m M (cid:17) sin θ + (cid:18) m (cid:96) M (cid:96) + M R (cid:19) α (cid:21) = 0 , (15)where we have used α = dω/dt and ω = dθ/dt , arrivingto Eq. (12). B. Case 2: Spinning disk
In case 2, the disk can spin freely and the pendulumis not a rigid body any more. However, it is possibleto describe the motion of the rod and disk separately asfollows. From Eq. (7), the angular momentum of the rodrelative to the pivot O (cid:48) can be written as: (cid:126)L rod = m (cid:96) (cid:126)ω. (16)As before, let (cid:126)N be the contact force between the diskand the rod. Nevertheless, now the disk can spin freelyand, therefore, (cid:126)τ = 0. Consequently, the net torque overthe rod about O is given by: (cid:126)τ net = − m g (cid:96) θ ˆ k − (cid:126)τ N , (17)where (cid:126)τ N is the torque applied by (cid:126)N , see Fig. 1-(c). FromEqs. (16) and (17), the motion equation for the rod is: m (cid:96) α = − m g (cid:96) θ − τ N . (18)Now it is necessary to consider the motion of the disk.From the free body diagram it is clear that the torqueabout the cm of the disk is null, see Fig 1 (d). Therefore,the disk rotates about its cm with a constant angular ve-locity, let’s say, (cid:126)β . Therefore, the disk rotates and trans-lates in such way that its angular momentum is given byEq. (5), leading to: (cid:126)L disk = M R (cid:126)β + M (cid:96) (cid:126)ω. (19)The net torque about O (cid:48) can be written as: (cid:126)τ net = − M g (cid:96) sin θ ˆ k + (cid:126)τ N . (20)Consequently, the motion equation for the disk is: M (cid:96) α = − M g (cid:96) sin θ + τ N , (21)where we have used the fact that d(cid:126)β/dt = 0. Note thatEq. (21) does not depend on R . The unknown torque τ N can be eliminated by adding up Eqs. (18) and (21),arriving to the desired motion equation: α = − g (cid:96) m + M m (cid:96) + M (cid:96) sin θ. (22)As we can expect, Eq. (22) can be also obtained applyingEq. (3) for the disk and Eq. (7) for the rod. In this way,we have: w N + M g (cid:96) (cos θ − cos θ ) = 12 M (cid:96) ω (23)and − w N + m g (cid:96) θ − cos θ ) = 16 m (cid:96) ω , (24)respectively. Adding up Eqs. (23) and (24) to eliminatethe work done by the force (cid:126)N , w N , we arrive to the energyconservation equation: − (cid:18) M g (cid:96) + m g (cid:96) (cid:19) cos θ + 12 (cid:18) M (cid:96) + m (cid:96) (cid:19) ω = const , (25)where const is a constant, more details can be found inRefs. [20–23]. Finally, taking the time derivative on Eq.(25) as it was done on Eq. (14), we arrive to Eq. (22).Apart from what the students are usually asked todo in the limit of small oscillations (sin θ ≈ θ ) so thatthe harmonic solution takes place, a comparison betweenthe frequencies of oscillation of the two situations can bemade. Let ϑ be the oscillation frequency, therefore, forthe Case 1, ϑ = g(cid:96) m + M m (cid:96) + M (cid:96) + M R , (26)and for Case 2, ϑ = g (cid:96) m + M m (cid:96) + M (cid:96) . (27)It is not hard to see that ϑ < ϑ regardless the valuesof the parameters. We conclude that, if the disk can spinfreely around its cm , it will not contribute to any changeof the overall angular momentum because it could be atmost spinning at constant angular velocity, so the termrelated to its moment of inertia about its cm will notappear in the overall moment of inertia. Therefore, aphysical reason why the pendulum of case 1 takes longerto complete an oscillation is that it has larger momentof inertia, it is harder to make it rotate. Besides, wecan imagine having to hold the rod from its free end andmake whole pendulum rotate with our own hands. In thefirst case, the torque done is greater because we need tomake the disk rotate exactly like the rod. In contrast, inthe second case we do not need to make the disk rotate,it is free to spin, hence, if we apply the same torque ofthe previous case, it will oscillate fasterIt is important that the students note that, for bothcases, the angular momentum of the disk has two contri-butions as shown in Eq. (19). The first term correspondsto rotation of the disk about its cm while the second oneto rotation of the cm around the pivot. The time deriva-tive of the first term is equal to (cid:126)R cm × (cid:126)F net which in turndepends on (cid:126)N and (cid:126)W d . The time derivative of the secondterm is equal to the internal torque applied by the rod, (cid:126)τ . For Case 1, (cid:126)τ ensures that the disk rotates aroundthe cm with the angular velocity (cid:126)β = (cid:126)ω , just as the roddoes it about the pivot. However, in Case 2, (cid:126)τ = 0 andthe rotation of the disk about its cm occurs at a constantangular velocity (cid:126)β which is not related to (cid:126)ω . An advancephysics student would recognize the presence of a cyclicvariable in the Hamiltonian and, consequently, the ex-istence of a conserved quantity. Equivalently, a studentof an introductory course should realize that for Case 2, (cid:126)τ = 0, implying that the second term of the angular mo-mentum given in Eq. (19) is a constant and does notcontribute to the equation of motion. This result shouldbe used to highlight to novice students the importanceof conserved quantities. IV. THE FREE SLIDING LADDER PROBLEM
In this section we want to illustrate that whenever themechanical energy is conserved, W θ and W T may notvanish individually, though their sum must be zero. As inprevious section, for illustrative purposes we use anotherproblem widely used in introductory physics courses.Let’s consider a ladder of length 2 (cid:96) and mass M whichleans against a wall as shown in Fig. (2)-(a). All surfacesare frictionless. Due to the gravitational field, the ladderstarts to slip downwards from the rest until eventuallyloses contact with the wall. In Ref. [19] the students arechallenged to show that the ladder loses contact whenit is at two-thirds of its initial height. In order to findthis result, the students need to determine the angle forwhich the contact force with the wall becomes zero. How-ever, for the purposes of our discussion we only need tocalculate the rotational equation of motion of the lad-der as follows. The position of the cm can be written as (cid:126)R cm = (cid:96) cos θ ˆ ı + (cid:96) sin θ ˆ . On the other hand, the transla-tional work due to the net force is given by: W T = (cid:90) C ( (cid:126)N w + (cid:126)N f + (cid:126)W ) · d (cid:126)R cm = (cid:90) θθ dθ (cid:96) ( N w ˆ ı + ( N f − M g )ˆ ) · ( − sin θ ˆ ı + cos θ ˆ )= (cid:90) θθ dθ (cid:96) ( − N w sin θ + N f cos θ ) − M g (cid:96) (sin θ − sin θ )= w c − M g (cid:96) (sin θ − sin θ )= 12 M V cm . (28)In Eq. (28), (cid:126)N w and (cid:126)N f are the contact forces applied bythe wall and floor, respectively. Additionally, w c has beendefined as the work due to these contact forces. Thus,from the last two lines of Eq. (28) we conclude: w c = 12 M V cm + M g(cid:96) (sin θ − sin θ ) . (29)The net torque about the cm is just due to (cid:126)N w and (cid:126)N f .Therefore, by using Eq. (4), the rotational work associ-ated to τ cm takes the form: W θ = (cid:90) θθ dθ (cid:96) ( N w sin θ − N f cos θ ) = − w c . (30)Then, replacing I O by I cm as discussed before Eq. (4),we find: − w c = 16 M (cid:96) ω . (31)Note that the rotational work of each contact force can-cels out with its translational counterpart. Thus, thetotal mechanical energy is conserved despite W θ and W T are both different to zero. In fact, adding Eqs. (29) and(31) we arrive to:16 M (cid:96) ω + 12 M V cm + M g (cid:96) sin θ = E, (32)where E is a constant. This result has been previouslyderived in several texts and notes, for example, see Refs.[24, 26, 27]. As expected, Eq. (32) is the conservationenergy equation. Finally, taking the time derivative of (cid:126)R cm it is easy to find V cm = (cid:96) ω . Inserting this result inEq. (32) we find:23 M (cid:96) ω + M g (cid:96) sin θ = E. (33)As before, from the time derivative of Ec. (33) we arriveto the motion equation in terms of the variable θ : α = − g (cid:96) cos θ. (34)In many particle systems, whether they are rigid ornot, it is always possible to describe the motion of anyparticle of the system through the motion of the cm andthe relative motion of the particle about the cm . Thisdecomposition allows us to write the theorem of work andenergy in two separate parts: translation ( W T ) and rota-tion ( W θ ). The former is the work done by the net forcealong the path of the cm . The latter is the work done bythe net torque with respect to the cm . In the simple casewhere the body rotates around a fixed axis, W θ = W T and both parts of the theorem of work and energy arecompletely equivalent. However, in the most general casewhich involves rotation about the cm and translation ofthe cm , W T and W θ , are not necessarily equal. In fact,for systems where the total mechanical energy is con-served, it is satisfied that the sum of the rotational andtranslational works associated to non-conservative forcesis zero. FIG. 4: Sliding ladder with a spring.On the other hand, a deeper observation of Eqs. (29)and (30) shows that the contact forces are completelydefined by the geometrical constraint: x cm + y cm = (cid:96) , (35)in such way that the total work associated to the contactforces is zero regardless the value of these forces. Nat-urally, an advanced physics student would easily realizethat this a consequence of the holonomic nature of theconstraint (35). The total work of the forces associatedto holonomic constraints, as the one given by Eq. (35),is always zero [29, 30]. However, this is not true for allkind of forces. In order to illustrate this idea considerthe sliding ladder with an spring as shown in Fig. (4). Inthis case, the force applied by the spring is (cid:126)F s = − k ∆ x ˆ i ,where ∆ x = 2 (cid:96) (cos θ − cos θ ). Therefore, the rotationaland translational works associated to the force appliedby the spring, w θ and w , satisfy: w θ = w T = (cid:90) θθ k (2 (cid:96) cos θ − l ) (cid:96) sin θ dθ. (36)Assuming that the system is released when the springhas its natural length ( l = 2 (cid:96) cos θ ), we find: w θ = w T = − k(cid:96) (cos θ − cos θ ) , (37)in such way that w θ + w T is minus the potential energyassociated to the spring. V. DISCUSSION
As mentioned in the introduction, the solution of thecompound pendulum problem can be found in severalinformal physics forums. However, in spite that thosesolutions show the correct final result, they usually lackof a rigorous explanation of the main conclusion: ϑ < ϑ given that, in Case 2, the moment of inertia of the diskrelative to its cm does not contribute. In fact, sen-tences such as “The term M R / ϑ > ϑ and accept the re-sult without the need of a formal derivation. However,we strongly suggest to present in introductory physicscourses a complete derivation, as the one provided in thispaper, including an explanation based on the descriptioninteraction between the rod and disk ( (cid:126)τ and (cid:126)N ) and onbasic physical laws. Otherwise, we run the risk to ac-centuate the erroneous idea that, in Case 2, the momentof inertia of the disk becomes zero because it can rotatefreely.On the other hand, from Eqs. (1), (3), (4) and (9) itis clear that: W = W T + W θ = ∆ E k , (38)where the W is the total work due the external forcesover the body given by Eq. (1). On the other hand, thetotal work is the sum of the work due to conservativeand non-conservative forces, W c and W nc , respectively.As usual, W c can be written as the change of the totalpotential energy, − ∆ E p . Thus, it is possible to concludethat: W T + W θ = W nc − ∆ E p . (39)Now, W T and W θ can be also decomposed in the sum ofthe contributions of conservative and non-conservative forces. Thus, we write: W cT + W ncT + W cθ + W ncθ = W nc − ∆ E p . (40)If the total mechanical energy of the body is conservedthe total work due the non-conservative forces satisfy W nc = 0. Given that the potential energy depends onthe parameters that defines the conservative forces suchas the spring constant or the gravity constant, the lastequation implies that: W ncT + W ncθ = 0 and W cT + W cθ = − ∆ E p . (41)Therefore, it is possible to conclude that, as usual, forconservative systems the non-conservative forces do notcontribute to the total work W but they could performtranslational and rotational works as long as the sum ofthem vanish. In introductory courses, the separation ofthe work into two parts improves the understanding ofthe students about the RBP motion. For instance, thisapproach allows students to realize that, in many cases,non-conservative forces could do translational and rota-tional work despite the total energy of the system is con-served. The ladder problem is an example of that. Usu-ally, for the sake of simplicity, in many cases solutionsto the problems solved in class use directly the energyconservation equation. This equation usually is easy touse and simplifies the solution of many problems. How-ever, it can hide some important information about thesystem. For instance, the existence of non-conservativeforces which perform translational and rotational workson a system where the mechanical energy is conserved.Additionally, the conservation energy approach usuallyputs in a second place some concepts which are equallyimportant such as the work and energy theorem. Thistheorem involves an integral, and because of that, forthe students it is more complicated to understand. How-ever, the use of this theorem helps to clarify physicalsituations as the one mentioned here. We recommend tofind the solution of the problems using different methodssuch as the angular momentum/torque, work and energytheorem, energy conservation, etc. This makes easier forstudents the understanding of the relation among dif-ferent physical concepts, for example, helping them tounderstand that the presence of non-conservative forcesdoes not necessarily imply that energy is not conserved. ACKNOWLEDGMENTS
The work of D.L.G was supported by the Vicerrector´ıade investigaciones de la Universidad del Valle C.I. 1164.Y.R. was supported by the following grants: Colciencias-Deutscher Akademischer Austauschdienst Grant No.110278258747 RC-774-2017, Vicerrector´ıa de Ciencia,Tecnolog´ıa, e Innovaci´on - Universidad Antonio Nari˜noGrant No. 2019248, and Direcci´on de Investigaci´on yExtensi´on de la Facultad de Ciencias - Universidad In-dustrial de Santander Grant No. 2460. [1] M. L. L´opez. Angular and linear acceleration in a rigidrolling body: students’ misconceptions. Eur. J. Phys. (2), 189 (2012).[8] L. G. Ortiz, P. R. Heron, and P. S. Shaffer. Studentunderstanding of static equilibrium: Predicting and ac-counting for balancing. American Journal of Physics, (6), 545-553 (2005).[9] L. G. Rimoldini, and C. Singh. Student understanding ofrotational and rolling motion concepts. Physical ReviewSpecial Topics-Physics Education Research, (1), 010102(2005).[10] Y. ¨Unsal. A simple piece of apparatus to aid the under-standing of the relationship between angular velocity andlinear velocity. Physics Education, (3), 265 (2011).[11] K. K. Mashood, and V. A. Singh. An inventory on ro-tational kinematics of a particle: unravelling miscon-ceptions and pitfalls in reasoning. European Journal ofPhysics, (5), 1301 (2012).[12] H. G. Close, L. S. Gomez and P. R. Heron. Student un-derstanding of the application of Newton’s second lawto rotating rigid bodies. American Journal of Physics, (6), 458-470 (2013).[13] N. Khasanah, W. Wartono and L. Yuliati. Analysis ofmental model of students using isomorphic problems indynamics of rotational motion topic. Jurnal PendidikanIPA Indonesia, (2) (2016).[14] I. Rahmawati, S. Sutopo, S. Zulaikah. Analysis of Stu-dents’ Difficulties About Rotational Dynamics Based onResource Theory. Journal Pendidikan IPA Indonesia 95-102 (2017). [15] T. Graham and A. Peek. Developing an approach to theintroduction of rigid body dynamics. International Jour-nal of Mathematical Education in Science and Technol-ogy, , 373-380 (1997).[16] N. Fang and J. Uziak. Student Misconceptions of Gen-eral Plane Motion in Rigid-Body Kinematics. Journal ofProfessional Issues in Engineering Education and Prac-tice..Vol.144