Apollo's Voyage: A New Take on Dynamics in Rotating Frames
aa r X i v : . [ phy s i c s . c l a ss - ph ] N ov Apollo’s Voyage: A New Take on Dynamics in Rotating Frames
Ujan Chakraborty ∗ and Ananda Dasgupta † Department of Physical Sciences,Indian Institute of Science Education and Research Kolkata (Dated: November 27, 2020)
Abstract
We first demonstrate how our general intuition of pseudoforces has to navigate around severalpitfalls in rotating frames. And then, we proceed to develop an intuitive understanding of thedifferent components of the pseudoforces in most general accelerating (rotating and translating)frames: we show that it is not just a sum of the contributions coming from translation and rotationseparately, but there is yet another component that is a more complicated combination of the two.Finally, we demonstrate using a simple example, how these dynamical equations can be used insuch frames.
The following article has been submitted to the American Journal of Physics. After it is published,it will be found at Link. . INTRODUCTION The diurnal motion of the Sun across the sky from east to west has perhaps been themost significant and most universal natural occurence in the lives of human beings sinceantiquity. There was no lack of theories among our ancients, on the Sun’s apparent motionacross the firmament during the day, and absence at night. The ancient Egyptians picturedthe Sun god Ra sailing his mighty ship across the ethereal waters over the earth at day, andin the underworld at night, while the Greeks believed that Apollo had been tasked withpulling the enchained Sun across the skies everyday, on his golden chariot drawn by fourmighty stallions. The east did not lack theories either: in ancient Hindu mythology, it wasSurya riding across the heavens in his chariot drawn by the seven fiery steeds, with Arunaas his charioteer, while for the Shinto of Japan, it was the goddess Amaterasu coming outof her divine palace to preside over the mortals. It was not until the early Renaissance, thatthe heliocentric model of the solar system was founded:“Nicolaus Copernicus Thorunensis, terrae motor, solis caelique stator”(“Nicolaus Copernicus, son of Toru´n, mover of the earth, stopper of the sun and heavens”). Here, we show that a reflection on the apparent diurnal motion of the Sun might help modifysome of our intuition regarding dynamics in accelerating frames.
II. AN APPARENT PARADOXA. A Problem with our Intuition
We begin with a problem, and an observation that goes against the common intuitionregarding “pseudoforces”. We observe from the Earth, that the Sun appears to move aroundthe Earth once in about 24 hours, but know that actually the Earth rotates on its axis oncein the same time. The Earth also revolves around the Sun in about 365 days, but thismotion due to the gravitational attraction between the two would be ignored at all pointsin our discussion for simplicity. One can see directly that even its inclusion would not affectour conclusion qualitatively, and a back-of-envelope calculation reveals that the quantitativechanges are negligible, so this approximation really isn’t remotely similar to the sphericalchicken approximation, but rather akin to considering the spherical earth (rather than the2blate spheroid earth) approximation. Of course, the entire observation can be explainedquite simply from a space-fixed (inertial) frame. But, suppose an observer on the rotatingEarth (a non-inertial frame) decides to address the situation using “pseudoforces”. Thisshould be possible without much ado using the tools one learns in a basic undergraduatecourse on Newtonian mechanics. So, the following question was posed to quite a largenumber of students who had taken such a course in the same level as the textbook byKleppner and Kolenkow, and also to some students who had also taken a more advancedcourse on the same, based on the celebrated textbook by Goldstein. For such an observer(someone on the Earth), which force on the Sun causes it to revolve around the Earth? Acommon answer received was, the “centripetal” force. This answer is technically correct,but not at all enlightening. It is a bit like the saying: “If you know when a tree was planted,you can determine its age quite accurately”. Because, “centripetal” is a name assigned tojust any force directed centrally inwards: in other words, a force that is responsible for thecentripetal acceleration is the centripetal force. It is the dynamical origin of such a force thatwe are interested in here: how it arises through physical interactions, or as a pseudoforce.However, in spite of being formally trained in the basic physics of non-inertial frames, quitea large number of students failed to correctly identify the origin of this “centripetal” force.The gravitational attraction between the Earth and the Sun is not really relevant in thisparticular problem, since we do not take consider effects due to the rotation of the Earthround the Sun. So, the only forces relevant in this scenario are the pseudoforces. Whileaccounting for the forces of relevance in this scenario, most students successfully identifiedonly the “centrifugal” (pseudo)force. This is perhaps because it is common intuition builtup from day-to-day experience (e.g. when a moving vehicle brakes suddenly), that for anobserver in an accelerated frame, the pseudoforce on an object is equal in magnitude to themass of the object multiplied by the acceleration of the frame, and directed opposite to theacceleration of the frame. And this recipe yields the centrifugal force in this case. However,observe that though the centrifugal force is equal in magnitude to the required “centripetal”force, the direction is opposite. The centrifugal force on the Sun is directed radially outward,so it cannot support the observation of the Sun moving around the Earth once in a day, justby itself. It appears that we have here a paradox: one has an object rotating around theobserver, but, cannot identify the dynamical origin of the centripetal force (the force thatprovides the centripetal acceleration)! 3nother example, of a different physical setting, but of essentially the same nature, wherewe encounter the same “paradox”: an observer sitting at the centre of a rotating merry-go-round (a carousel). Perhaps this might appear more appropriate to some readers, sincewe do not have to make approximations like neglecting the revolution of the Earth aroundthe Sun. For simplicity, let us assume that the origin of the merry-go-round-fixed-frameis a point on its axis of rotation, and that the angular velocity is constant. Our observershall find that the surroundings execute circular motion around him or her. The centrifugalforce would be directed outwards from the axis. Along the same lines as in the previousparagraph, one might think that this is the only force of interest here (neglecting gravityand all other such forces, which are orders of magnitude less). So, the acceleration of thesurrounding bodies can be expected to be horizontal and radially outward from the centreof the merry-go-round. But this is evidently false: in the rotating frame, the surroundingsrevolving around the axis must have an acceleration directed towards the axis itself, and notoutwards. So, who provides the centripetal force?The answer to both problems, is that the pseudoforce in a rotating frame does not consti-tute of the centrifugal force alone. There is yet another force which we usually ignore in mostday-to-day scenarios. It is the “Coriolis force”. It does not occur to us immediately, perhapsbecause in most observations to which we apply our intuition of pseudoforces, Coriolis forcesdo not affect the dynamics significantly over short time scales. For example, in most cases,students imagine Coriolis forces as those which deflect winds and ocean currents, or decidewhich way a whirlpool is going to swirl on different sides of the Equator. It usually does notcross one’s mind that the same Coriolis forces might be crucially relevant for observationsas simple as the diurnal motion of the Sun. In the discussion below, we see how the Coriolisforce resolves the situation completely, and try to build an intuition about the true natureof this force (there is good reason why this intuition is often lacking).
B. The Resolution
As a first step towards a resolution, we look at the following result from Goldstein. Whenwe take the derivative of vector G with respect to time in a rotating frame, we have, (cid:18) d G dt (cid:19) s = (cid:18) d G dt (cid:19) r + ω × G (1)4here s stands for the space-fixed frame, and r for the frame rotating with angular velocity ω .We see that the time derivative in a rotating frame, (cid:0) d G dt (cid:1) r , is different from its counterpartin a non-rotating frame, (cid:0) d G dt (cid:1) s . This can be rationalized in the following fashion: the unitvectors in a rotating frame, unlike their translating counterparts, are not constant. Hence,once has to account for that in the derivative. One also sees that for any translating butnon-rotating frame, ω = , and hence, (cid:0) d G dt (cid:1) translating = (cid:0) d G dt (cid:1) s . Thus, instead of an inertialframe, s can be any non-rotating frame having arbitrary translational motion for the aboveEq. (1) to be satisfied.We address both the merry-go-round problem and the Earth’s rotation problem simul-taneously, as the difference between the two is essentially superficial. Consider two frames,with origins coinciding on the axis of the merry-go-round (or, the axis of the Earth’s rota-tion, which means that our observer is positioned on the North or the South Pole, or thatwe have chosen to ignore the radius of the Earth, an omission the reader will undoubtedlypardon). One is s , non-rotating, the other is r , fixed to the merry-go-round (or the Earth),and rotating with it. Then, let r be the coordinate of some the particle whose motion isunder study (or the Sun, which we treat here as a particle). In both frames r and s , thecoordinate r reads the same, since the origin is the same. Furthermore, let us assume that ω is constant. Then, taking the derivative of r twice, each time making adjustments by Eq.(1), we get the following equation governing the motion in the rotating frame r : F − m ( ω × v r ) − m ω × ( ω × r ) = m a r (2)where v r and a r represent the particle’s velocity and acceleration respectively in therotating frame, and we have made the substituion F = m (cid:16) d r dt (cid:17) s , by Newton’s laws. Adetailed calculation is shown in the appendix (A).Now, if F pseudo be the pseudoforces in the rotating frame r , we should demand F + F pseudo = m a r . Hence, we have: F pseudo = F co + F ce . The second term is the all-too-familiar centrifugal term, F ce = − m ω × ( ω × r ). The first term is the one of greaterinterest here (even if only because it often escapes our intuition): F co = − m ( ω × v r ), theCoriolis term. The Coriolis term is the one responsible for counter-intuitive behaviour. Forthe ‘surroundings’, stationary in the s frame, v r = − ω × r , and hence, the Coriolis termequals twice of the centrifugal term with a reversed sign. This provides a resolution of theproblem. 5hus, we see that the folklore of pseudoforces being equal to the mass of the particletimes the negative of the frame’s acceleration does not really hold true in rotating frames.Why does the Coriolis term not appeal to our intuition as much as the ‘negative of frame’sacceleration term’ ? After all, all of us are in a rotating frame (on the surface of the Earth)!The answer lies in the fact that the Coriolis force is proportional to the cross product of theangular velocity of the frame ω , and the velocity of the observed particle in the frame, v r .For an observer on the surface of the Earth, ω is approximately 7 . × − s − . So, except forbodies moving at very, very high v r , its effects are practically negligible over short intervalsof time. Of course, over long intervals of time, even relatively small v r ’s do cause significantdeviations, depending on the angle between ω and v r , e.g. for ocean currents and winds.But for objects moving at sufficiently high v r , the Coriolis term is definitely not neglible,compared to the centrifugal term. In fact, as demonstrated, for objects fixed respect tothat frame, with respect to which our concerned frame is rotating at angular velocity ω , theCoriolis term is twice the centrifugal term with an opposite sign, and their resultant providesthe centripetal. This is exactly why an observer on the surface of the Earth observes theSun to orbit the Earth diurnally, in spite of the centrifugal term directed radially outward. III. DYNAMICAL EQUATIONS, IN THE MOST GENERAL ACCELERATINGFRAME
Having discussed some of the kinematic consequences of an observer being in a rotatingframe, we now turn to the dynamics in a most general accelerating frame: a frame thatis rotating (no longer at a constant ω ) as well as translating. One may think, shouldn’tthe dynamics in such a frame simply be the sum of the effects due to translation withoutrotation, and due to rotation without translation? Though naively this appears to be aplausible solution, we shall see that this is not entirely correct. The force equation concernedcan still be visualized as a combination of the effects due to translation and rotation, but it isslightly more involved than just their sum. We demonstrate below, that, in the pseudoforce,considering the contributions due to the rotation in absence of translation ( F rot = F ang + F co + F ce ) and the contribution due to translation in absence of translation ( F trans = − m A r ), we still need to add yet another term F comb , which is zero if the frame is eitherstationary (coinciding with s ) or non-rotating.6n what follows, s will denote a stationary / inertial frame, while r will be an acceleratingframe. A. The Force Equations, in the Most General Case
We now can no longer assume that the origins of the s frame and the r frame coincide.Let r s and r r be the coordinates of the particle under study in the s (inertial) and r (non-inertial) frames respectively. Let R the (time-dependent) vector between the origins, thatis, r s = r r + R . Then, R and its derivatives V r = (cid:0) d R dt (cid:1) r and A r = (cid:16) d R dt (cid:17) r capture thetranslational dependence.Let ω r be the angular velocity of the frame, (cid:0) d ω r dt (cid:1) r = (cid:0) d ω r dt (cid:1) s = ˙ ω r .We have, F pseudo = F trans + F rot + F comb (3)where F trans = − m A r (4) F rot = F ang + F co + F ce (5)with F ang = − m ˙ ω r × r r (6) F co = − m ( ω r × v r ) (7) F ce = − m ω r × ( ω r × r r ) (8)and F comb = − m ω r × V r − m ω r × ( ω r × R ) − m ˙ ω r × R (9)Then, F + F pseudo = m a r (10)The derivations are in the appendix (B). B. Considering a Few Special Cases
The last stated Eq. (10) may be considered to be the working form of Newton’s law ina most general accelerating frame. All this might appear a bit cumbersome. It is easier to7nalyze a few simple cases:1. There is a complete agreement with our common intuition when the frame is non-rotating ( ω = = ˙ ω ): F pseudo = − m A r .2. When it is rotating uniformly ( ˙ ω r = ) but not translating, it is reasonable to set R = , and then F pseudo = F ce + F co . We see that this too is in complete agreement,with Eq. (2).3. When it is rotating (perhaps non-uniformly) but not translating, again, it is reasonableto set R = , and then we have F pseudo = F ang + F co + F ce . That is, we have theminimal modification to Eq. (2), simply by addition of the term F ang , just as wemight expect.4. When there is both translation and rotation, the situation is more involved. However,it might be interesting to pose the following questions: given R , V r , does there exist ω r , ˙ ω r such that F comb = , and vice versa? Perhaps this question is similar in spiritsomewhat to Chasles’ theorem.We provide an interesting and instructive example below to demonstrate how these equationsmight be used. C. An Example: the Merry-Go-Round Archer
Suppose an archer decides to try out shooting an arrow from a merry-go-round. Whatadditional factors should be taken into account?The first thing to observe, is that once the arrow leaves the bow (literally), there is norelevant force acting on it (we are only interested in the horizontal motion). Gravity doesnot affect the horizontal motion of the arrow, except for limiting the range, so we leave thatconsideration to the archer’s usual skill. The vertical motion of the arrow (due to gravity /the angle decided upon by the archer) is not affected by the rotation, as ω has a non-zerocomponent only in the vertical direction. Of course, we also ignore the drag force due to theair, and assume the arrow doesn’t go “whichever way the wind doth blow”. So a stationaryobserver should observe the arrow to fly off in a straight line (we ignore the vertical motionof the arrow). But what does the archer see?8uppose, we consider the archer to be sitting on the rotating merry-go-round at a distanceof r from its axis of rotation. And, the merry-go-round is rotating on its axis at a constantangular speed ω , anticlockwise. We consider the inertial frame s to have origin on the axisof the merry-go-round, and for the non-inertial frame r (the archer) to be using cylindricalcoordinates (cid:16) ˆ ρ , ˆ θ , ˆ z (cid:17) where ˆ ρ is directed towards the axis of the merry-go-round, and ˆ z is directed upwards. So, R = − r ˆ ρ , ω r = ω ˆ z , V r = , and A r = . Here, F comb = − m ω r × ( ω r × R ) = − mω r ˆ ρ , F ang = F trans = , F co = 2 mω v r × ˆ z , F ce = mω r r .Hence, we have a r = ω ( r r − r ˆ ρ ) + 2 ω v r × ˆ z (11)Of course, one may solve this second order differential equation with suitable initialconditions to get the full trajectory. However, a qualitative analysis of what happens justat the instant the arrow is shot, is quite easy. At that instant, r r = , and so a r | t =0 = − ω r ˆ ρ + 2 ω v r | t =0 × ˆ z . − ω r ˆ ρ looks the acceleration due to a centrifugal (pseudo)force,but it actually follows from F comb . When r = 0 (that is, the archer is at the center of themerry-go-round), the arrow is observed to be simply deflected to the right by the Coriolisterm 2 ω v r × ˆ z , as expected. IV. CONCLUSION
We have demonstrated some of the limitations of our common intuition when dealingwith accelerating frames, in particular, failure in rotating frames. That being done, we haveprovided a natural way to extend our intuition to such scenarios. In this article, we haverestricted ourselves to dynamics of point particles, that is, equations involving force andacceleration only. It is interesting to note that similar studies can be carried out for systemsof particles, in particular, in the setting of rigid body dynamics. One may study the relationsbetween torque, angular momenta, and angular velocities from general accelerating framesas well, in the same spirit as of this article. We are presently compiling such an article. This leads us to obtain an alternate proof (and more importantly, a new physical realization)of Euler’s equations for rigid body motion, in the same lines as in Mott’s 1966 deduction. ppendix A: Deduction of the Dynamical Equation in a Frame Rotating at a Con-stant Angular Velocity: From Eq. (1), with r as G : (cid:18) d r dt (cid:19) s = (cid:18) d r dt (cid:19) r + ω × r (A1)And again, with (cid:0) d r dt (cid:1) s as G : (cid:18) d r dt (cid:19) s = (cid:18) d r dt (cid:19) r + 2 ω × (cid:18) d r dt (cid:19) r + ω × ( ω × r ) (A2) Appendix B: Deduction of the Dynamical Equation in a Most General AcceleratingFrame:
For a translating-cum-rotating frame, a straightforward calculation by considering thetime derivative of r s in the two frames by Eq. (1) yields: v s = ( v r + V r ) + ω r × ( r r + R ) (B1) a s = ( a r + A r ) + 2 ω r × ( v r + V r ) + ω r × ( ω r × ( r r + R )) + ˙ ω r × ( r r + R ) (B2)Using m a s = F , this can be written as: m a r = F − m A r − m ( ω r × v r ) − m ω r × ( ω r × r r ) − m ˙ ω r × r r − m ( ω r × V r ) − m ω r × ( ω r × R ) − m ˙ ω r × R (B3)We can immediately identify the second term on the right hand side as arising out ofthe translational motion of the frame, the third through fifth terms as arising out of therotational motion of the frame, and the last three terms as a combination of both translationand rotation. 10 CKNOWLEDGMENTS
UC would like to thank the INSPIRE scholarship (2015) and the KVPY scholarship(2016-2020) for support during his degree programme. ∗ [email protected], [email protected] † [email protected] https://publishing.aip.org/resources/librarians/products/journals/ George Hart,
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Classical Mechanics , 3rd ed. (Pearson, 2002) It might appear that the merry-go-round problem is a bit easier to address in detail comparedto the problem of the Sun’s diurnal motion, since one does not have the additional complicationof the translational motion of the frame (one less detail to wrap one’s head around). But it isnot so: as additional translational motions do not contribute anything to the picture, “in theend it doesn’t even matter”. Ujan Chakraborty and Ananda Dasgupta, “Pseudotorques and Euler’s Equations”, manuscriptin preparation David L. Mott, “Another Derivation of Euler’s Equations of Rigid-Body Rotation”, Am. J.Phys. , 1197-1198 (1966), 1197-1198 (1966)