HHuygens’ cycloidal pendulum: anelementary derivation
Riccardo BorghiDipartimento di Ingegneria, Universit`a degli Studi “Roma tre”Via Vito Volterra 62, I-00146 Rome, [email protected]
Abstract
A pedagogical derivation of the Huygens cycloidal pendulum, suitable for high-schoolstudents, is here presented. Our derivation rests only on simple algebraic and geomet-rical tricks, without the need of any Calculus concept.
It is a well known fact that the motion of a simple pendulum is far frombeing rigorously isochrone, as the period of its oscillations depends on theiramplitude, being longer for wider oscillations [1]. For “small” oscillations theswinging time could be assumed approximately independent of the amplitude,and it is within such a regime that simple pendulums are used as clocks. Theisochronism problem was known since Galileo’s time and in 1659 ChristiaanHuygens, “the most ingenious watchmaker of all time”, to use Sommerfeld’swords [2], first found the right vertical trajectory to give a point mass in orderfor it to produce, under the action of gravitational force, a perfect isochrone (ofharmonic type) motion [3, 4]. The curve discovered by Huygens is the celebrated cycloid , namely the trajectory drawn by a typical point on a bicycle wheel whenthe latter moves on the ground without slipping.Now, while the simple pendulum is a topic of pivotal importance in anyundergraduate and/or high-school physics course, the same cannot be said asfar as the cycloidal pendulum is concerned. The latter is rarely offered to stu-dents as an example of application of Newton’s laws of motion, although it wasconsiderably studied already in Newton’s
Principia [5]. A possible reason forthis state of fact is that most treatments of cycloidal pendulum rest on someconcepts of differential geometry and Calculus which could not be yet availableinside the math toolbox of first-year undergraduates, less than ever of high-school students. Moreover, the standard treatment for deriving the pendulumshape could be not easily decodable and make hard grasping the cycloid isochro-nism property. To cover such cultural hole, during last years several didacticalpapers dealing with elementary treatments of the “magic” properties of cycloid,namely brachistochronism and tautochronism, appeared [6, 7, 8, 9, 10].1 a r X i v : . [ phy s i c s . c l a ss - ph ] A p r The scope of the present Letter is to illustrate an elementary derivation ofHuygens’ cycloidal pendulum from basic mechanical principles which could besuitable for high-school students. Differently from other didactical strategies,which are aimed at checking that cycloids are tautochrone (which implies pen-dulum isochronism), within the present approach the cycloid is directly obtainedas a possible answer to the isochronism request.Our derivation rests only on a trivial algebraic trick that, once combinedwith a well known geometrical property of isoscele triangles and translated intothe vector language, leads to the very geometrical definition of cycloids withoutthe need of any Calculus concept.To help teachers, what follows is arranged as a whole didactical unit thatcould be given to students within a standard two-hour lecture. For such reason,some mathematical steps could seem a little bit redundant, but this is done onlyto show a possible way to keep the level suitable for a high-school audience. m g α N α O x y Ps ( ) α τ ^ Fig. 1:
The isochrone pendulum geometry.In Fig. 1 the geometry of the problem is shown: a point mass m is slidingwithin a vertical plane Oxy along a frictionless surface acted on by its weight m g . The vector N describes the surface reaction which is directed along thesurface normal, due to absence of friction. Symbol ˆ τ denotes the tangentialunit vector. The task is to shape the surface in such a way the point motion bepurely harmonic. To this end, Newton’s law is first written m a = m g + N , (1)with a being the point acceleration. Projection of both sides of Eq. (1) alongthe tangential direction ˆ τ gives at once a τ = − g sin α , (2)where the symbol a τ denotes the tangential component of the acceleration and α is the inclination angle of che curve at P with respect the horizontal direction x (see Fig. 1). Now, for the motion along the trajectory to be rigorously harmonicit is necessary that the tangential acceleration a τ be proportional, up to a minussign, to its distance from the origin O measured along the trajectory s = (cid:95) OP , a τ = − ω s , (3)where ω = 2 π/T , with T denoting the oscillation period. From Eqs. (2) and (3)we thus arrive to the following equation: s = (cid:96) sin α , (4)with (cid:96) = g/ω . Equation (4) represents an implicit definition of the trajectoryshape in terms of the curvilinear abscissa s , of the slope α , and of the charac-teristic length (cid:96) . To make explicit what kind of curve is hidden inside Eq. (4),consider a “small” displacement of the point P to another point along the curve,say Q , corresponding to the inclination α + ∆ α , as sketched in Fig. 2. α Q α + Δα ∆ α O s Δ Fig. 2:
Geometrical interpretation of Eq. (7).The displacement length, say ∆ s coincides with the measure of the arc (cid:95) P Q ,i.e., ∆ s = (cid:95) OQ − (cid:95) OP = (cid:96) [sin( α + ∆ α ) − sin α ] , (5)where use has been made of Eq. (4). Now, if Q is sufficiently close to P , i.e., if∆ α is sufficiently small, then it is possible to use the following estimates: sin ∆ α (cid:39) ∆ α , cos ∆ α (cid:39) , (6)so that Eq. (5) can be approximated as∆ s = (cid:96) cos α ∆ α , (7)and the arc (cid:95) P Q can be identified with the segment
P Q , as shown in Fig. 2.
Now the first algebraic trick, which simply consists in recasting Eq. (7) asfollows: ∆ s = 2 (cid:18) (cid:96) α (cid:19) cos α . (8)For what it was said above, it is then clear that ∆ s can be identified with thelength of the small segment P Q , which can also be thought of as directed alongthe curve tangent at P . Consider now the isoscele triangle P RQ , built up insuch a way that
P R = RQ = (cid:96) α , as shown in Fig. 3. P Q α θ =2 α R α Fig. 3:
Geometrical interpretation of Eq. (8).Then, on using trivial geometrical considerations we have that the horizontalinclination of RQ is θ = 2 α and so ∆ θ = 2∆ α . In this way it is possible to write P R = RQ = (cid:96) θ . (9)Now the second trick: to give suitable orientation to the three sides of P RQ in order for them to be interpreted as vectors and to described the trianglethrough the following algebraic equation: −−→
P Q = −→ P R + −−→ RQ , (10)as sketched in Fig. 4. In the same figure it is also shown that, from Eq. (9), theratio RQ ∆ θ = (cid:96) , (11)can be interpreted as the radius of the circumference centred at the point C andhaving radius CR = (cid:96)/ −−→ DC = −→ P R is introduced in such a way thatEq. (10) is written as −−→
P Q = −−→ DC + −−→ RQ , (12)
P Q θ RCD θ r Fig. 4:
Geometrical interpretation of Eq. (10).and the displacement −−→
P Q along the curve can then be viewed as the resultof the composition of two rigid motions of the circumference: (i) an horizontaldisplacement represented by −−→ DC plus (ii) a ∆ θ counterclockwise rotation aroundits centre.However, since DC = RQ , such a composition should immediately be rec-ognized to be equivalent to a perfect (i.e., without slipping) rolling of the cir-cumference along the line r , which then proves that the displacement −−→ P Q layson a cycloid.
Quod Erat Demonstrandum . Acknowledgments
I wish to thank Turi Maria Spinozzi for his invaluable help during the prepara-tion of the work.
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