Rattlebacks for the Rest of Us Simon Jonesa) Department of Mechanical Engineering, Rose-Hulman Institute of Technology, Terre Haute, Indiana 47803 Hugh E

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Rattlebacks for the rest of us Simon Jonesa) Department of Mechanical Engineering, Rose-Hulman Institute of Technology, Terre Haute, Indiana 47803 Hugh E. M. Hunt Engineering Dynamics and Vibration, University of Cambridge, Cambridge CB21PZ, United Kingdom (Received 21 November 2018; accepted 18 June 2019) Rattlebacks are semi-ellipsoidal tops that have a preferred direction of spin (i.e., a spin-bias). If spun in one direction, the rattleback will exhibit seemingly stable rotary motion. If spun in the other direction, the rattleback will being to wobble and subsequently reverse its spin direction. This behavior is often counter-intuitive for physics and engineering students when they ﬁrst encounter a rattleback, because it appears to oppose the laws of conservation of momenta, thus this simple toy can be a motivator for further study. This paper develops an accurate dynamic model of a rattleback, in a manner accessible to undergraduate physics and engineering students, using concepts from introductory dynamics, calculus, and numerical methods classes. Starting with a simpler, 2D planar rocking semi-ellipse example, we discuss all necessary steps in detail, including computing the mass moment of inertia tensor, choice of reference frame, conservation of momenta equations, application of kinematic constraints, and accounting for slip via a Coulomb friction model. Basic numerical techniques like numerical derivatives and time-stepping algorithms are employed to predict the temporal response of the system. We also present a simple and intuitive explanation for the mechanism causing the spin-bias of the rattleback. It requires no equations and only a basic understanding of particle dynamics, and thus can be used to explain the intriguing rattleback behavior to students at any level of expertise. VC 2019 American Association of Physics Teachers. https://doi.org/10.1119/1.5115498

I. INTRODUCTION rattleback being particularly confounding. As such, a model of a planar, rocking semi-ellipse is first introduced in Sec. II Rattlebacks, also known as celts or wobblestones, are semi- to acquaint the reader with the necessary concepts used in a ellipsoidal tops that have a preferred direction of spin (i.e., a more complex dynamics analysis. Undergraduates can intui- spin-bias). If spun in, say, the counter-clockwise direction, the tively understand the motion of this body and the equations rattleback will exhibit seemingly stable rotary motion. If spun are less intimidating since they are only two-dimensional. in the clockwise direction, the rattleback’s rotary motion will The dynamics are governed by conservation of linear and transition to a rattling motion and, subsequently, it will reverse angular momenta coupled with basic kinematic constraints, its spin direction resulting in counter-clockwise rotation, as concepts taught in most post-secondary introductory physics depicted in Fig. 1. This counter-intuitive dynamic behavior has classes. Rather than limit the investigation to a roll without long been a favored subject of study in graduate-level dynam- slip condition, which is used in the bulk of the previous liter- ics classes. ature on rattlebacks, the derivation accounts for a conditional Previous literature on rattleback dynamics offer insight stick/slip constraint using a Coulomb friction model. into a myriad of advanced topics, including chaotic motion, The rattleback is introduced in Sec. III, where it is shown energy decay, stability regions, and nondimensionalization the dynamic analysis is virtually identical to that of the rock- 1–11 to name but a few. However, it is the current authors’ ing semi-ellipse. The only significant difference involves the view that focusing on these advanced topics may obscure the use of a mass moment of inertia tensor rather than the scalar understanding of the fundamental kinetics for students new value used in the planar example. Finally, a physical explana- to three-dimensional rigid body dynamics. The goal of this tion for the spin-biased behavior of the rattleback is presented paper is to demonstrate that accurately simulating rattleback in Sec. IV. The explanation is intuitive and can be used to behavior need not be complicated; undergraduate physics accurately explain the spin-bias behavior using a simple dem- and engineering students have all the necessary tools to onstration without the need for complex equations. accurately model the behavior using concepts from introductory dynamics, calculus, and numerical methods courses. II. A ROCKING SEMI-ELLIPSE Furthermore, it appears that no previous literature has provided a simple means of explaining the spin-biased behavior Leaping directly into the seemingly complex dynamics of of the rattleback; the explanations are generally closely linked the rattleback can be intimidating. Many of us do not have to minutia of the complex governing equations. While these reliable intuition when it comes to three-dimensional dynam- complex explanations are accurate, they are too involved to ics, which makes it difficult to assess when simulation results be useful when explaining rattleback behavior to the inter- are correct. As such, a simplified planar analogy to the rattle- ested layman. As such, this article will introduce an intuitive back, the rocking semi-ellipse shown schematically in Fig. 2, explanation for the spin-biased behavior that can be under- will be analyzed first to assist in deriving the governing stood by laymen with only a basic understanding of physics. equations of motion for the rattleback. It is intuitive that if Three-dimensional rigid body dynamics can be a confus- one releases the semi-ellipse from some perturbed state, and ing, counter-intuitive area of study, with the behavior of the one assumes roll without slip and no energy loss terms, the

699 Am. J. Phys. 87 (9), September 2019 http://aapt.org/ajp VC 2019 American Association of Physics Teachers 699 Fig. 1. Schematic showing the spin-bias of a rattleback. semi-ellipse will rock back and forth indefinitely with con- elliptical perimeter of the body such that it is always in stant amplitude. This example will act as a validation case contact with the ground. This style of problem is often and provide confidence in the approach. It should be noted, taught at the end of the introductory dynamics class for though, that the roll without slip condition is too rigid a con- undergraduate engineers as “relative-motion analysis using straint in the current investigation; the effect of sliding on rotating axes.”12 the response of the system is also of interest. Thus, slippage Define the system of Cartesian axes X, Y, Z associated will be allowed at the contact point via a Coulomb friction with the inertial reference frame R and Cartesian axes x, y, z model. While this planar behavior does not seem as exciting associated with the body-fixed reference frame B. In the as that of a spin-biased rattleback, it will be shown that the body-fixed frame, the parametric definition of the semi- vector form of the governing equations is virtually identical elliptical surface is to that of the rattleback, thus this simplified model acts as a useful and intuitive introduction to rattleback dynamics. x2 y2 þ ¼ 1 a x a b y 0; (1) The current authors have found that students, undergradu- a2 b2 ate and graduate alike, often struggle to fully comprehend the derivations presented in previous literature on rattle- where a is the major axis radius and b is the minor axis backs. There can be confusion regarding which reference radius. frame certain vectors are defined in, how to approach derivatives in rotating reference frames, and how to parametrically A. Notation and important kinematic concepts define the geometry of the system. As such, the derivation below is presented in a pedantic manner in order to clarify The section below is included for clarity of notation and to these topics. review important concepts regarding the use of rotating ref- AsshowninFig.2, an inertial reference frame R is erence frames. These concepts can be reviewed in under- defined with origin O. Note, the exact location of O is not graduate level dynamics texts12,13 if further information is important and can be left arbitrary. A body-fixed reference required. frame B is defined with origin C centered on the upper line (1) Define an arbitrary vector function BuðtÞ as follows: of the semi-ellipse, where b^1 is colinear with the top of the ^ ^ semi-ellipse. In this planar example, the k axis of R and b3 X3 B axis of B are parallel. The rotation of the body-fixed uðtÞ¼ ujb^j ¼ u1b^1 þ u2b^2 þ u3b^3; (2) frame relative to the inertial frame is measured via h, j¼1 defined positive in the counter-clockwise direction, as shown in Fig. 3. The center of mass for the semi-ellipse is whose scalar components ujðtÞ¼uðtÞb^j, j ¼ 1, 2, 3 are found at point G. The contact point between the semi- scalar functions of time. The pre-superscript B signifies ellipse and the ground is P, which is the location of the the vector is defined in the rotating, body-fixed reference resultant force vector f. frame. For clarity, bold lowercase letters are used for For clarity, points C and G are fixed in the rigid body, general vectors while a hat over a lowercase letter signi- while point P moves on the body. Point P can be thought of fies a unit vector. as a massless and frictionless slider moving along the Similarly, an equivalent definition could be written for the vector in the inertial reference frame R

R uðtÞ¼uxi^þ uyj^þ uzk^: (3)

Fig. 3. Body-ﬁxed reference frame B rotated by angle h relative to inertial Fig. 2. Schematic of the semi-ellipse showing the body-ﬁxed reference frame. reference frame R.

700 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 700 8 9 B B The scalar components of these two vectors uðtÞ and < x1 = R B uðtÞ will be different, but both Eqs. (2) and (3) describe x ¼ x2 : (11) the same physical vector and thus are equivalent. : ; x3 (2) The deﬁnition of a time derivative du=dt is dependent on the reference frame in which it is being computed, so for (5) Similarly, the angular acceleration of the body-ﬁxed clarity the frame will be included when writing the deriv- frame B in the inertial frame R can be computed by tak- ative as follows: ing the time-derivative of X 2 3 R ðÞB X3 X3 R ^ d u ^ dbj 0 a3 a2 ¼ u_ jbj þ uj ; (4) 6 7 dt dt T T j¼1 j¼1 A ¼ R_ R_ þ R R€ ¼ 4 a3 0 a1 5; (12)

a2 a1 0 where u_ j is the reduced notation for duj/dt. The first term on the right hand side of Eq. (4) accounts for the time which can be used to define the angular acceleration rate of change of the scalar components of the vector, vector where it should be noted that the differentiation of scalar 8 9 13 B functions is always independent of the reference frame. < a1 = B The second term on the right hand side of Eq. (4) a ¼ a2 : (13) accounts for the time rate of change of the unit vectors : ; a3 defining B in the inertial frame R. Since frame B is rotating within frame R, this term is generally non-zero. A fundamental theorem of kinematics12,13 shows that B. Deriving the equations of motion

R ^ The kinetic diagram (KD) and free-body diagram (FBD) for dbj B ^ ¼ x bj; (5) the rocking semi-ellipse are provided in Fig. 4, where a and dt x ay are the translational accelerations of the center of mass in where Bx is the angular velocity vector of frame B in the plane, a is the angular acceleration of the semi-ellipse in frame R, written in terms of frame B the plane, m is the mass of the semi-ellipse, IG is the mass moment of inertia of the semi-ellipse about its center of X3 mass, g is the acceleration due to gravity, and f is the contact B ^ ^ ^ ^ x ¼ xjbj ¼ x1b1 þ x2b2 þ x3b3: (6) force vector comprised of the friction force, Ff, and the j¼1 normal force, N 8 9 Substituting Eq. (5) into Eq. (4) results in the vector R<> Ff => form Rf ¼ N : (14) :> ;> RdðÞBu 0 ¼ Bu_ þ Bx Bu: (7) dt (3) A rotation matrix can be employed to transform a vector 1. Rate form of the law of conservation of linear defined in the body-fixed frame B into one defined in the momentum inertial reference frame R via The general rate form of the law of conservation of linear momentum in the inertial reference frame is commonly R B u ¼ R u; (8) written as

where the bold uppercase R signifies a rotation matrix. RdP ¼ RRF; (15) For the reference frames defined in Fig. 2, the planar dt rotation matrix R is defined as 2 3 where RRF is the sum of all external forces in R acting on cos h sin h 0 6 7 the system and P is the linear momentum of the system, R ¼ 4 sin h cos h 0 5; (9) defined as 001 R B P ¼ m vG or P ¼ m vG; (16) where h is defined as positive for a counter-clockwise rotation, as depicted in Fig. 3. (4) The angular velocity of the body-fixed frame B in the inertial frame R can be computed using the product of the transpose and the time derivative of the rotation matrix 2 3 0 x3 x2 T 6 7 X ¼ R R_ ¼ 4 x3 0 x1 5; (10)

x2 x1 0 Fig. 4. Kinetic diagram (left) and free-body diagram (right) of rocking semi- which can be used to deﬁne the angular velocity vector ellipse.

701 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 701 R where vG is the translational velocity of the center of mass 3. Kinematic equation: Relative position vector B in R, while vG is the translational velocity of the center of B R The relative position vector rP=G in Eq. (23) is computed mass in B. Recall that these absolute velocity vectors vG B via two position vectors measured relative to the origin of and vG are equivalent, even though they are written with respect to different reference frames, so either can be used the semi-ellipse, C depending on which is most convenient. Computing the lin- B B B R rP=G ¼ rP=C rG=C: (24) ear momentum using vG is convenient since it simplifies the time derivative in Eq. (15), resulting in Figure 5 is included to help visualize a convenient method R R for finding Br and Br . In this figure, the inertial frame m aG ¼ f ðmgÞj^; (17) P=C G=C has been visually rotated h in the k^-direction so that the R where aG is the translational acceleration vector of the cen- body-fixed frame B appears square to the page. Recall that ter of mass in R Point C is at the origin (x ¼ 0, y ¼ 0) in the rotating frame B. B 8 9 The position vector rG=C is simply defined by the cen- R 14 < ax = troid of the semi-ellipse R aG ¼ : ay ;: (18) 8 9 B> > az <> 0 => B 4b rG=C ¼ > >: (25) :> 3p ;> 2. Rate form of the law of conservation of angular 0 momentum B The general rate form of the law of conservation of angu- The position vector rP=C requires some calculus to com- lar momentum about the center of mass in the inertial refer- pute. As shown in Fig. 5, the surface of the ground is tangent ence frame is commonly written as to the semi-ellipse at P. This implies a vector normal to the ground surface, nground, must be co-linear with a vector RdH normal to the elliptical curve, nellipse at P,or G ¼ RRM ; (19) dt G nground nellipse ¼ 0: (26) R where R MG is the sum of all external moments acting The vector nground can be computed most easily in the iner- about G in R and HG is the angular momentum of the system about the center of mass, defined as tial frame as 8 9 R R B <> 0 => HG ¼ IG x or HG ¼ IG x; (20) Rn ¼ 1 ; (27) ground :> ;> where IG is the mass moment of inertia tensor of the system 0 about the center of mass 2 3 while the vector nellipse can be computed most easily in the Ixx Ixy Ixz 6 7 body-fixed frame by taking the gradient of the ellipse equa- IG ¼ 4 Iyx Iyy Iyz 5; (21) tion x2=a2 þ y2=b2 ¼ 1 with respect to the body-fixed frame Izx Izy Izz 8 9 B > 2x > > > and Rx is the angular velocity of the body in R, while Bx is <> a2 => the angular velocity of the body in B. It will be convenient to B nellipse ¼ 2y : (28) Bx > > use in this derivation since it can be easily computed > b2 > using Eq. (11). However, since Bx is defined in the rotating :> ;> 0 frame, the concept from Eq. (7) must be employed to take the time derivative in Eq. (19) resulting in The cross product in Eq. (26) must be performed with B B B B R both vectors in the same frame, so Eq. (8) is employed ðI_G x þ IG x_ Þþ x IG x ¼ R MG: (22) resulting in

Note that the first term in Eq. (22) is zero, since the mass R B moment of inertia tensor is independent of time. Also, nground ðR nellipseÞ¼0: (29) B B IG x_ ¼ IG a via the definition in Eq. (12) and the external R R R Evaluating Eq. (29), one finds that two rows provide no use- moment R MG ¼ rP=G f. As shown in the section below, it is more convenient to find the relative position vec- ful information (i.e., 0 ¼ 0), but the final row results in the B first constraint equation required to find the x and y location tor in the body-fixed frame, rP=G, so this vector is transformed using Eq. (8) for the external moment computation of P R B via rP=G ¼ R rP=G. Finally, it is necessary to write all the a2 terms in the same frame, so the terms on the left hand side of x ¼ y tan h: (30) b2 Eq. (22) are transformed using Eq. (8) as well, resulting in Since P lies on the perimeter of the body, the second B B B B R RðIG a þ x IG xÞ¼ðR rP=GÞ f: (23) constraint equation is that of the elliptic curve

702 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 702 R One must now contemplate vP. If the semi-ellipse were rolling without slip and P were fixed in the body then R vP ¼ 0. However, in this case, P can be thought of as a massless and frictionless slider moving along the elliptical perimeter of the body such that it is always in contact with the ground. If the body rolls without slip, this implies there is a moving point on the ground, Pground, which is always in contact with P on the body. As the semi-ellipse rocks back and forth, Pground will move back and forth on the ground under the semi-ellipse. Imagine the movement of P over a small time interval Dt as the semi-ellipse rocks, as depicted in Fig. 6. Since P and Fig. 5. Schematic of the semi-ellipse rocked to an angle of h. Pground must be coincident at both t and t þ Dt, this implies that as the limit of Dt ! 0, the rate that the position of P is changing on the semi-ellipse (i.e., Br_ ) must be equivalent x2 y2 P=G þ ¼ 1; a x a and b y 0: (31) to the rate the position of Pground is changing in R (i.e., a2 b2 R vPground ). Thus Equations (30) and (31) can be used to solve for the x and y R B _ R B vPground ¼ Rð rP=GÞ¼ vP: (36) location of rP=C. This, in conjunction with Eq. (25),is substituted into Eq. (24) to give 8 9 The final consideration is the possibility of sliding; if the B> 2 > semi-ellipse were sliding with respect to the ground then Eq. > a tanðÞh > > pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > (36) would become <> a2 tan2h þ b2 => Br 2 : (32) R B R P=G ¼ > b 4b > vP ¼ Rð r_ P=GÞþ vslip; (37) > pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ > > a2 tan2h þ b2 3p > :> ;> 0 where 8 9 R< = vslipX 4. Kinematic equation: Velocity of the center of mass Rv ¼ v (38) slip : slipY ; v Consider the kinematic equation relating the velocities of slipZ two points, P and G, in an inertial reference frame12,13 would describe the rate and direction of slip between the R R R vP ¼ vG þ vP=G semi-ellipse and the ground. R Substituting the result of Eq. (37) into Eq. (34) provides R d B ¼ vG þ R rP=G ; (33) dt R R B B vG ¼ vslip Rð x rP=GÞ: (39) R where the substitution for vP=G is made since we have defined the relative position vector in the body fixed frame, 5. Kinematic equation: Acceleration of the center of mass B rP=G, in Eq. (32). Making use of the derivative rule defined in Eq. (7), this becomes Equation (17) requires the computation of the acceleration of the center of mass, which can be derived directly from the time R R B B B vP ¼ vG þ Rð r_ P=G þ x rP=GÞ; (34) derivative, in R,ofEq.(39). Making use of the definitions, Rðd=dtÞðRv Þ¼Ra ; Rðd=dtÞðRv Þ¼Ra ,andBx_ ¼ Ba, B _ G G slip slip where rP=G can be computed directly from Eq. (32) as results in 8 9 B> 1 > R R B B B B <> => aG ¼ aslip Rð a rP=G þ x r_ P=G xa2b2 tan hðÞ1 tan2h B þ tan h B B B r_ P=G ¼ : (35) þ x ð x rP=GÞÞ; (40) 2 2 2 3=2 > 1 > ðÞa tan h þ b :> ;> 0 where

Fig. 6. Rotation of the semi-ellipse during a small time interval Dt.

703 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 703 8 9 R< = (17) and (23), and the kinematic constraint equations (32) aslipX R and (40), can be used to solve for: a, F , N, and a . For aslip ¼ aslip (41) f slipX : Y ; _ € aslipZ clarity, in this example: x ¼ h and a ¼ x_ ¼ h. Due to the planar nature of this model, the vector values are thus describes the time rate of change of Rv due to sliding of 8 9 8 9 8 9 slip > > > > > > the semi-ellipse on the ground. Note that if the body rolls B< 0 = B< 0 = R< Ff = without slip then Rv ¼ Ra ¼ 0. Bx ¼ 0 ; Ba ¼ 0 ; Rf ¼ N ; slip slip :> ;> :> ;> :> ;> x a 0 8 9 8 9 C. Solving for the time response of the rocking R> > R> > < vslipX = < aslipX = semi-ellipse Rv ¼ 0 ; Ra ¼ 0 : slip :> ;> slip :> ;> The system parameter values for the current investigation 0 0 are: a ¼ 0.10 m, b ¼ 0.05 m, q ¼ 27:0 kg/m2, g ¼ 9.81 m/s2, ls ¼ 0.3, and lk ¼ 0.2, where the last two parameters are the Finding the analytic solution to this system of ordinary static and kinetic coefﬁcients of friction, respectively. differential equations (ODEs) is impractical, thus a time- From these fundamental parameters, one can compute the stepping algorithm is employed to solve the system numer- mass properties of the system ically. The basic solution scheme is outlined in Fig. 7. 8 9 Since the constraint of this system is state-dependent (i.e., B> > > 0 > sticking or slipping), an algorithm designed for stiff sys- < = 15 pab 4b tems is required; ODE15s is implemented in MATLAB as m ¼ q ; Br ¼ ; 2 G=C > 3p > the time-stepping algorithm. The conditional is described :> ;> 0 below: • 2 2 2 If the semi-ellipse is sliding at the beginning of the time a þ b 16b step (i.e., v 6¼ 0Þ, or the computed friction force Izz ¼ m : slipX 4 9p2 required to prevent sliding during the current time step is larger than the available static friction (i.e., Ff > lsN), Since this is a symmetric, uniform, and planar example, all then the friction force is computed directly from the other entries in the mass moment of inertia tensor deﬁned in Coulomb friction law Eq. (21) can be assumed to be zero without loss of generality. Ff ¼ lkN; (42) For the planar rocking semi-ellipse, the system of equations has three state variables: h, x, and vslipX . Given initial where the direction of the friction force is opposite to the conditions for these state variables, the kinetic equations direction of vslipX .

Fig. 7. Flowchart showing the general solution algorithm for the time response of the rocking semi-ellipse.

704 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 704 • Otherwise, it is assumed the semi-ellipse rolls without slip Mazzoleni et al.16 predicted using a linearized energy formu- for the current time step lation. As further validation, Fig. 9 presents the rocking period for various semi-ellipse geometries plotted against the 16 aslipX ¼ 0 (43) periods predicted by Mazzoleni et al. for small amplitude displacements and roll without slip. These results suggest the and the required friction force is determined from the sys- current numerical simulation is valid and accurate over a tem of ordinary differential equations. wide range of geometries. One may also note in Fig. 8(a) that the normal force, N,is D. Results from rocking semi-ellipse test cases almost constant since angular displacement is small thus inertial effects due to the center of mass accelerating verti- Figure 8 presents the time response for a rocking semi- cally are minimal. The ratio of friction force to normal force, ellipse released from rest, x 0 rad/s and v 0 m/s, at 0 ¼ slipX ¼ F /N, is well below the static friction coefﬁcient, l ¼ 0.3, f s various initial angular displacements: h0 ¼ 1 ; 5 ; 15 , and thus the body is always rolling without slip resulting in zero 25 . Included in each set of results are the angular displace- slip velocity. ment, h(t), the contact forces, Ff (t) and N(t), the ratio of the The initial displacement is increased to h0 ¼ 5 in Fig. friction force to the normal force, Ff ðtÞ=NðtÞ, and the relative 8(b). This deﬂection is no longer in the range of small dis- slip velocity between the semi-ellipse and the ground at the placements, thus the period of rocking increases slightly, as contact point, vslipX ðtÞ. expected. This trend continues as the initial displacements is Figure 8(a) corresponds to the small displacement solution increased in Fig. 8(c) and 8(d). The Ff/N ratio in Fig. 8(b) involving roll without slip. The solution shows steady rock- remains below ls, so again the slip velocity remains zero. ing without decay, as expected since no energy dissipation Slip is initiated when the initial displacement is increased terms are included in the formulation. The rocking period is to h0 ¼ 15 , as shown in Fig. 8(c). The required friction computed to be 0.41 s, which matches the result by force for roll without slip to occur at this exaggerated

Fig. 8. Rocking ellipse response for varying initial angular displacements when released from rest.

705 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 705 Fig. 10. Rendering of a rattleback: uniform semi-ellipsoid with a uniform rectangular prism placed atop, pinned through the centroid of both bodies. Notice the prism has been skewed by 10 about b^3 relative to the semi- ellipsoid, which is sufficient to produce the rattleback behavior. Fig. 9. Comparison of rocking periods for various a/b ratios of a semi- ellipse. axis of the ellipsoid, b^2 is colinear with the minor axis of the ellipsoid, and b^3 is normal to the top surface of the semi- ellipsoid. Note that the uniform prism has been omitted from rocking angle is larger than available given the current coef- the figure for clarity, but its effect on the mass properties of ficient of static friction, thus the conditional switch in the the rattleback is included throughout the derivation. algorithm is triggered to compute the sliding friction force, The geometry of the semi-ellipsoid is defined as Ff ¼ lkN. This can be seen in the plot of Ff/N, where the pla- teaus have magnitude lk ¼ 0.2. Subsequently, the slip veloc- x2 y2 z2 ity is non-zero during these times. This response is also 2 þ 2 þ 2 ¼ 1 a x a b y b visible in Fig. 8(d) to a greater degree. The slippage results a b c in energy loss due to the friction work, thus the magnitude of c z 0 (44) rocking decreases, visible in Fig. 8(c) and more so in Fig. 8(d). and the rectangular prism has length ‘ in the x-direction, Note that the results presented in Fig. 9 are approximate width w in the y-direction, and height h in the z-direction. numerical solutions. The algorithm employed, ODE15s in MATLAB, automatically adapts the time step to improve reso- A. Orienting the body-fixed frame in the inertial frame lution around sharp gradient changes (e.g., the transition between sticking and slipping), but it does not compute the The main difference when deriving the conservation of exact instant such transition occurs. However, the state varia- momenta equations for the rattleback as compared to the bles have converged within each time step to absolute and rocking semi-ellipse is the need for three angles to define the 6 relative tolerances of 1 10 , as defined in the MATLAB rotation state of the rattleback. Rather than using a single documentation.17 Due to these convergence criteria and the rotation angle h, the body-fixed reference frame B for the rat- inherent stability of this system for the given initial condi- tleback is oriented within R using: /1 - roll, /2 - pitch, and tions, small variations to the onset of sticking and/or slipping /3 - yaw, as depicted in Fig. 12. These angles correspond to are deemed to have negligible effect on the outcome, thus Tait-Bryan rotations18 and are commonly used when discus- these approximate solutions are deemed accurate. The same sing an airplane’s orientation. Readers familiar with Euler approach will now be implemented to simulate the spin- angles18 will note the similarity to Tait–Bryan rotations in biased behavior of the rattleback. application. The rotations are applied in the order yaw then pitch then III. RATTLEBACK roll. In this order, the spin angle on the i^ j^ plane will be equal to the yaw, and the elevation angle relative to the i^ j^ The spin-bias of a rattleback results from the planes of geometric symmetry of the semi-ellipsoid and the principal axes of the mass moments of inertia being slightly skewed from one another. A convenient way to accomplish this is to place a uniform rectangular prism atop a uniform semi- ellipsoid, as shown in Fig. 10. Rotating the prism about the b^3 axis atop the semi-ellipsoid, defined by the axis connect- ing the centroid of both bodies and depicted in Fig. 11, allows one to change the amount of “skewness” in the mass moment of inertia tensor and dictate the spin-bias direction. As with the rocking semi-ellipse, a body-fixed reference frame B is attached to the rattleback, with origin at the centroid of the ellipse defining the upper surface of the semi- Fig. 11. Schematic of the rattleback showing the body-fixed reference ellipsoid, as shown in Fig. 11: b^1 is colinear with the major frame.

706 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 706 1. Rate form of the conservation of linear momentum The rate form of conservation of linear momentum developed in Eq. (17) becomes

R R m aG ¼ f ðmgÞk^; (50)

where 8 9 R> > < FfX = Rf ¼ F : (51) :> fY ;> NZ Fig. 12. Three rotations angles to orient the body-fixed frame in the inertial ^ frame: /1 measures roll, /2 measures pitch, and /3 measures yaw. FfX is the friction force in the i direction on the ground, FfY is the friction force in the j^ direction on the ground, and NZ is the normal force perpendicular to the ground in the k^ plane will be equal to the pitch, and rotation about the direction. pitched b^1 axis will be equal to the roll. The rotation matri- ces associated with the three angles are 2 3 2. Rate form of the conservation of angular momentum 10 0 6 7 The rate form of conservation of angular momentum 6 7 R1 ¼ 4 0 cos /1 sin /1 5; (45) developed in Eq. (23) remains B B B B R 0 sin /1 cos /1 R3DðIG a þ x IG xÞ¼ðR3D rP=GÞ f (52) 2 3 cos / 0sin/ though the terms in mass moment of inertia tensor defined in 6 2 2 7 4 5 Eq. (21) as R2 ¼ 010 (46) 2 3 sin / 0 cos / Ixx Ixy Ixz 2 2 6 7 2 3 IG ¼ 4 Iyx Iyy Iyz 5 cos / sin / 0 6 3 3 7 Izx Izy Izz R3 ¼ 4 sin /3 cos /3 0 5: (47) must be computed using superposition since the rattleback is 001 comprised of the two bodies shown in Fig. 10: the semi- ellipsoid (SE) and the rectangular prism (RP). The mass As was possible in the 2D case, a combined rotation matrix properties of the semi-ellipsoid are14 can be employed to transform a vector defined in the body- 8 9 fixed frame, B, into one defined in the inertial frame R via B> 0 > <> => R B 2 B u ¼ R3D u; (48) m ¼q pabc ; r ¼ 0 ; SE SE GSE=C > > 3 > 3b> : ; where 8 2 3 R ¼ R R R : (49) 45b2 3D 1 2 3 6ðÞb2 þc2 007 6 7 6 64 7 This combined rotation matrix R3D is used in conjunction mSE 6 2 7 IG ¼ 6 45b 7 with Eqs. (10)–(13) to compute the angular velocity vector, SE 5 6 0 ðÞa2 þc2 0 7 Bx, and angular acceleration vector, Ba. 4 64 5 B For clarity, one must be diligent when computing the x 00ðÞa2 þb2 and Ba terms. In the planar example of the semi-ellipse, only (53) a single planar rotation occurred, thus Bx ¼ Bh_ . This is no longer the case using the Tait-Bryan angles. Equation (10) and the mass properties of the prism are B B _ _ _ states that x ¼ xð/1; /2; /3; /1; /2; /3Þ and Eq. (12) 8 9 B B _ _ _ € € € B> 0 > states that a ¼ að/1; /2; /3; /1; /2; /3; /1; /2; /3Þ. > > While these relationships may not be as convenient as the < = B 0 planar example, Eqs. (10)–(13) are straightforward to use in mRP ¼ qRPðÞ‘wh ; rG =C ¼ ; RP > > a numerical algorithm and do not hinder the computation. :> h ;> 2 2 3 2 2 B. Governing equations for the rattleback simulation 6 ðÞw þ h 007 6 7 mSE 2 2 Deriving the equations governing the dynamics of the rat- IG ¼ 6 0 ðÞ‘ þ h 0 7; (54) RP 12 4 5 tleback is virtually equivalent to that of the rocking semi- 00ðÞ‘2 þ w2 ellipse. The key equations are repeated below for clarity.

707 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 707 where qSE and qRP are the densities of the semi-ellipse The cross product in Eq. (57) must be performed with and prism, respectively. Terms from Eqs. (53) and (54) both vectors in the same frame, so Eq. (48) is employed can be superimposed to compute the mass properties of resulting in the rattleback as R B B B nground ðR3D nellipseÞ¼0: (60) B mSE rGSE=C þ mRP rGRP=C m ¼ mSE þ mRP; rG=C ¼ 2 mSEmRP3 The first two rows of Eq. (60) provide two of the constraint B 2 equations required to find the location of P; the third row jj rG =Gjj 00 6 SE 7 provides no useful information (i.e., 0 ¼ 0). The final con- 6 B 2 7 IG ¼ IG þ mSE4 5 SE 0 jj rGSE=Gjj 0 straint equation comes from the fact P must also lie on the 000 ellipsoid’s surface 2 3 B 2 jj r jj 00 2 2 2 6 GRP=G 7 x y z T 6 B 2 7 2 þ 2 þ 2 ¼ 1 a x a b y b þ TIGRP T þ mRP4 5; 0 jj rGRP=Gjj 0 a b c 000 c z 0: (61) (55) The three constraint equations from Eqs. (60) and (61) can B where jjrjj represents the Euclidean norm of the position be used to solve for the x, y, and z location of rP=C. Note vectors defined below that the closed-form solution to this system of constraint equations is tedious and thus has not been included for sake of brevity. One would be wise to employ a numerical solver Br ¼ Br Br ; Br ¼ Br Br ; GSE=G GSE=C G=C GRP=G GRP=C G=C (e.g., a multivariate Newton-Raphson15) to compute the B components of rP=C. and T is a transformation matrix to account for the rotation of the prism atop the semi-ellipsoid by angle w, in the b^3 4. Kinematic equation: Velocity of the center of mass direction The velocity of the center of mass is again computed via 2 3 cos w sin w 0 R R B B 6 7 vG ¼ vslip R3Dð x rP=GÞ: (62) T ¼ 4 sin w cos w 0 5: 001 5. Kinematic equation: Acceleration of the center of mass 3. Kinematic equation: Relative position vector The acceleration of the center of mass is computed as B before The relative position vector rP=G is computed via R R B B B B B aG ¼ aslip R3Dð a rP=G rP=G ¼ rP=C rG=C; (56) B B B B B þ x r_ P=G þ x ð x rP=GÞÞ: (63) B where rG=C is defined by the centroid of the rattleback given B B in Eq. (55). The position vector rP=C is found by again not- If rP=G is computed numerically, as recommended above, B ing that the plane of the ground will be tangent to the surface then r_ P=G must be computed numerically by first making of the rattleback at P use of the chain rule nground nellipse ¼ 0: (57) B B B rP=G : rP=G : rP=G : B r_ P=G ¼ @ /1 þ @ /2 þ @ /3 The vector nground for the case of the rattleback is @/1 @/2 @/3 8 9 (64) R< 0 = Rn ¼ 0 ; (58) ground : ; and subsequently computing the partial derivatives using a 1 numerical derivative scheme.15 For example, the first partial derivative in Eq. (64) would be while nellipse is computed from the gradient of the semi- ellipsoid equation x2=a2 þ y2=b2 þ z2=c2 ¼ 1 with respect to Br B B the body-fixed frame P=G rP=GðÞ/ þD/ ;/ ;/ rP=GðÞ/ ;/ ;/ @ ¼ 1 1 2 3 1 2 3 ; 8 9 @/1 D/1 B> 2x > (65) > > > a2 > <> => where Eq. (56) is used to compute Br at the desired rota- B 2y P=G nellipse ¼ : (59) D > b2 > tion angles, and the small change in angle /1 is sufficiently > > small such that the numerical partial derivative has con- > > :> 2z ;> verged to an acceptable tolerance (i.e., a relative tolerance of c2 1 106 is used herein).

708 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 708 C. Solving for the time response of the rattleback a ¼ 0 slipX (67) a ¼ 0 Virtually the same solution scheme outlined for the rock- slipY ing semi-ellipse is employed to solve for the time response and the required friction forces are determined from the of the rattleback; the only significant difference is an system of ordinary differential equations. increase in the number of state variables. The system parameter values for the current investigation Note that the kinetic equation governing rattleback motion are: a ¼ 1.0 10 1 m, b ¼ 2.5 102 m, c ¼ 2.5 102 m, presented by Garcia and Hubbard,3 widely cited as a correct, 3 1 2 qSE ¼ 1020 kg/m , ‘ ¼ 2.0 10 m, w ¼ 2.0 10 m, is reproduced from a combination of Eqs. (50), (52), and 3 3 2 h ¼ 2.5 10 m, qRP ¼ 7900 kg/m , w ¼ 10 , g ¼ 9.81 m/s , (63) when roll without slip is assumed. This validates the ls ¼ 0.3, and lk ¼ 0.2. current rattleback derivation. From these fundamental parameters, one can compute the mass properties of the system D. Results from rattleback test cases 8 9 <> 0 => The properties listed above correspond to a rattleback with spin-bias in the positive b^ direction. Three cases are simu- m 2:1 101kg; Br B 0 m; 3 G=C :> ;> lated to verify the spin-biased behavior of the rattleback is 5:4 103 2 3 being captured. 0:38 0:45 0 (1) Spin in the biased direction is initiated using the follow- 46 7 2 IG 1 10 4 0:45 5:305kg m : ing initial conditions: 8 9 8 9 8 9 8 9 005:5 > > > 3 > > _ > > > < /1 = < 1:0 10 = < /1 = < 0 = / ¼ 1:0 103 rad /_ ¼ 0 rad=s The rattleback simulation has eight state variables: / ; :> 2 ;> :> ;> :> 2 ;> :> ;> 1 _ _ _ _ /3 0 /3 p /2; /3; /1; /2; /3; vslipX ; and vslipY . Given initial conditions 8 9 for these state variables, the kinetic equations (50) and (52), <> 0 => and the kinematic constraint equations (56) and (63), can be R vslip ¼ > 0 > m=s: used to solve for: /€ ; /€ ; /€ ; F ; F , N , a ; and a . : ; 1 2 3 fX fY Z slipX slipY 0 Due to the three-dimensional nature of this model, the vector values are thus (2) Spin against the biased direction, resulting in spin rever- 8 9 8 9 8 9 B B R sal, is initiated using the following initial conditions: > x > > a > > F > < 1 = < 1 = < fX = 8 9 8 9 8 9 8 9 B B R 3 _ x ¼ x2 ; a ¼ a2 ; f ¼ FfY ; >/ > >1:0 10 > >/ > > 0 > :> ;> :> ;> :> ;> < 1 = < = < 1 = < = 3 x3 a3 NZ / ¼ 1:0 10 rad /_ ¼ 0 rad=s 8 9 8 9 :> 2 ;> :> ;> :> 2 ;> :> ;> R> v > R> a > / 0 /_ p < slipX = < slipX = 3 8 9 3 R R vslip ¼ vslip ; aslip ¼ aslip ; <>0=> :> Y ;> :> Y ;> 0 0 Rv ¼ 0 m=s: slip :> ;> 0 where Bx and Ba are defined in Eqs. (10)–(13) as functions of / ; /_ , and /€ for i ¼ 1, 2, 3. i i i (3) Spin is initiated from an initial pitch angle using the fol- The algorithm is again conditional on the state of slip at lowing initial conditions: the contact point. 8 9 8 9 8 9 8 9 • If the rattleback is sliding at the beginning of the time step > > > 3 > > _ > > > R < /1 = < 1:0 10 = < /1 = < 0 = (i.e., jj vslipjj 6¼ 0Þ, or the computed friction force / ¼ 1:7 101 rad /_ ¼ 0 rad=s required to prevent sliding during the current timeqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi step is :> 2 ;> :> ;> :> 2 ;> :> ;> / 0 _ 0 larger than the available static friction (i.e., F2 þ F2 3 8 9 /3 fX fY <> 0 => > lsNZ), then the friction force is computed directly from R the Coulomb friction law, where the magnitude and direc- vslip ¼ > 0 > m=s: R : ; tion of the friction forces are determined from vslip 0 8 9 > n > R < slipX = In all cases, small perturbations in / and / are included to vslip 1 2 n^slip ¼ ¼ nslip help initiate wobble and stabilize the numerical time- jjRv jj :> Y ;> slip 0 stepping solution. Figures 13, 14, and 15 show the response of the rattleback F ¼n l N fX slipX k Z for Cases 1, 2, and 3, respectively. Six sets of results are pre- B FfY ¼nslipY lkNZ: (66) sented for each Case: the angular displacements, /; the R R angular momentum, H; the slip velocities, vslip; the resul- • Otherwise, it is assumed the semi-ellipse rolls without slip tant forces developed at the contact point measured in the for the current time step inertial frame, Rf; those same resultant forces transposed

709 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 709 Fig. 13. Rattleback response to Case 1 initial conditions. into the body-fixed frame via Eq. (48), Bf; and the external by approximately 3 s the rattleback reverses its spin to the R moments acting on the body about the center of mass due to spin-bias direction (i.e., a positive HZ value). Since no slip- the resultant forces, RM. ping occurs there is no energy dissipation, thus the final As shown in Fig. 13, when the rattleback is given an initial momentum about k^ is equal and opposite to the initial value. _ ^ yaw velocity of /3 ¼ pb3 rad/s (i.e., in the spin-bias direc- The cause of the spin reversal is the development of the exter- ^ tion), the rattleback maintains constant angular momentum nal moment about the center of mass in the k direction, MZ. R about the vertical axis, HZ, since no external moments are While not as large in magnitude as the other external moments developed about that axis, MZ, to change the momentum. The developed, the oscillating MZ peaks shown in Fig. 14 are all in R R minor external moments, MX and MY, near the beginning the positive k^ direction, which act incrementally to slow the of the simulation are developed due to wobbling resulting rattleback’s spin and eventually reverse its direction. from the slight initial perturbations in /1 and /2 that are The final case of releasing the rattleback from an initial included in the simulation. pitch angle, /2 ¼ 0.17 rad, is presented in Fig. 15.Thisinitial Conversely, when the rattleback is given an initial yaw pitch is too large to permit roll without slip (i.e., insufficient _ ^ velocity opposite to the spin-bias direction, /3 ¼pb3 rad/s, frictional forces can be developed), thus some slipping is pre- it is evident from Fig. 14 that significant wobbling develops in sent for about the first 0.25 s. However, once sufficient energy the /1 and /2 directions, starting around 1 s. The angular has decayed due to the frictional losses, the rattleback transi- R momentum about the vertical axis, HZ, begins to drop and tions to roll without slip for the remainder of the simulation.

Fig. 14. Rattleback response to Case 2 initial conditions.

710 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 710 semi-ellipsoid (Masses A, B,andC) that are positioned to produce a rattleback with equivalent mass, mass center location, and mass moment of inertia tensor to the original. Mass A and B are anti-symmetrically offset across the major axis to represent the skewed rectangular prism, and mass C represents the center of mass of the semi-ellipsoid.

A. Development of the external moment, MZ, due to wobbling Consider a rattleback released from rest with a pitch angle of 5, as shown in Fig. 17. This is similar to Case 3, but the initial pitch angle has been reduced to ensure roll without slip throughout the response. The ﬁgure includes three views of the rattleback (front, top, and side) at the instant of release. At this instant b^2 and j^ are colinear. The contact

forces FfX ; FfY , and NZ are depicted acting at the contact point on the lower surface of the semi-ellipsoid. The superposition of the body loads, including the weights and the D’Alembert forces, are depicted acting at the lumped-mass locations A, B, and C, as defined in Fig. 16. Recall that a D’Alembert force,12 also known as an inertial force, Fig. 15. Rattleback response to Case 3 initial conditions. accounts for the rate of change of linear momentum for a point mass (i.e., FD0Alembert ¼ma). All forces shown have The question that remains is why does the rattleback reverse been computed from simulation and are displayed at proper its spin? One explanation is to state that the model predicts scale and orientation with one exception: the normal force that the reaction forces at the contact point create an external NZ has been clipped in length so as not to include excessive moment about the center of gravity that drives the rattleback in white-space in the diagrams. the spin-biased direction and that these reaction forces are When the rattleback is released from this position, it will developed due to the wobbling of the rattleback. Since the begin to rock back toward the equilibrium position, thus an wobbling is a direct result of the conservation of momenta angular acceleration is developed. Mass A will accelerate equations, the laws of physics are maintained so the spin- mainly in the þk^ direction, while masses B and C will accel- biased behavior is explained. This is essentially the explanation erate mainly in the k^ direction. Furthermore, the relative that is presented in the bulk of the previous literature on rattle- B position of mass A from the contact point, rA=P, is smaller backs. It is the current authors’ opinion that, though true, this B than that to mass B, rB=P, thus the acceleration of B must be explanation is unsatisfactory. Some cases in physics are too larger than that of A. Equating these accelerations to difficult to comprehend using physical intuition so one must D’Alembert forces, mass A will mainly experience a rela- just trust the equations; this is not one of those cases. Presented tively small force in k^, while mass B will mainly experi- below is an intuitive explanation for why the rattleback ence a relatively large force in þk^. When these D’Alembert reverses its spin that can be explained using a simple demon- forces are superimposed with the weights it results in the stration and common principles from physics. loads depicted in Fig. 17. One can see from the force vectors that the kinetics are more complex than proposed in this ^ ^ IV. A PHYSICAL EXPLANATION FOR THE explanation (i.e., there are also reactions in b1 and b2 due to RATTLEBACK’S SPIN-BIAS the three-dimensional accelerations), but the vertical components play the most significant role at this juncture. The proposed explanation of the mechanism for spin rever- One can see from the front view in Fig. 17 that there is a sal is broken into two components: (A) development of the load imbalance across the major axis of the semi-ellipsoid external moment, MZ, due to wobbling, and (B) wobble devel- (i.e., the equivalent load at A is greater in magnitude than at opment due to the rattleback being spun opposite to the spin- B). Therefore, the rattleback must develop angular accelera- bias. A simplification of the mass properties for the rattleback tion in the b^1 direction. The result can be seen in Fig. 18 at is presented in Fig. 16 to help reduce clutter in the following t ¼ 0.05 s, as the rattleback rocks through the neutral pitch explanatory diagrams. Rather than including the rectangular angle. This body load imbalance during pitching, due to the prism in the diagrams, consider three lumped masses on a skewing of the mass distribution across the major axis of the

Fig. 16. Schematic of the original rattleback with skewed rectangular prism (left) and the simpliﬁed lumped-mass rattleback model with skewed point masses A and B to represent the rectangular prism (right). At the instant shown, the body-ﬁxed frame B is aligned with the inertial frame R.

711 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 711 Fig. 17. The rattleback at t ¼ 0 s, immediately after being released from rest with an initial pitch angle of 5: front view (left), side view (middle), and top view (right). Contact forces included on the bottom surface of the rattleback and body loads included at the lumped-mass locations.

semi-ellipsoid, is the mechanism that couples wobbling in b^1 B. Wobble development due to the rattleback being spun and b^2 directions. opposite to the spin-bias direction Figure 19 depicts the rattleback at t ¼ 0.1 s, when it has ^ essentially reached the limit of its roll for this cycle. One can Assume the rattleback is spun in the b3 direction, oppo- again see the load imbalance across the major axis of the site the spin-bias. Regardless of how carefully one spins the semi-ellipsoid, thus the rattleback must now roll back in the rattleback, inevitably a slight perturbation in the roll and/or pitch direction will be included. Thus, an imperceptibly þb^1 direction. Given the roll without slip condition, this ^ small wobble will exist at t ¼ 0. Consider again Fig. 19, but requires that all of the lumped masses accelerate in the b2 ^ direction. Newton’s second Law states that for this to occur, now assume the rattleback is spinning in the b3 direction (essentially clockwise in the top view). Due to the current a friction force in the b^2 direction is required, which is predicted from the simulation as shown in the top view of Fig. roll and pitch angles of the rattleback, this angular momentum in the b^3 direction results in a slight lifting of the cen- 19. Since the contact point is not directly under the center of ^ mass, this friction force results in an external moment about ter of gravity in the k direction, as a result of roll without the center of mass of the rattleback in the þk^ direction, caus- slip at the contact point. A loose analogy would be that of a ing the rattleback to begin to spin. bicycle rider sharply applying the front brake: if the friction It is the repetition of these two simple steps that results in force between the tire and the ground is sufﬁcient to prevent the spin-biased behavior of the rattleback: (1) the angular sliding, the angular momentum of the rider about the contact acceleration in the pitch direction results in an unbalanced point will cause the system to begin to rotate up and over the load across the major axis of the semi-ellipsoid, and (2) the front wheel. unbalanced load produces an angular acceleration in the roll In other words, the pitch angle is increased during each direction, resulting in a friction force due to the roll without wobble oscillation when the rattleback is spinning opposite slip condition. This friction force produces the external the spin-biased direction. This, in turn, results in the normal moment about the center of mass in the spin-biased direction force at the contact point growing during each wobble oscil- hence the rattleback begins to spin. There are other subtleties lation, which can be seen in Fig. 14. Furthermore, due to the to the dynamics of the rattleback that govern the spin-biased coupling of pitch and roll, explained in Sec. IV A, this also behavior that are being ignored in this explanation, but these increases the roll angle during each wobble oscillation. Since two steps are the dominant mechanism for the development the friction force developed is proportional to the magnitude of spin from wobble. The beauty of this explanation is that of these angles, the external moment about the center of the mechanism is relatively simple and it can be explained mass of the rattleback in the þk^ direction continues to grow ^ using a basic demonstration and simple physical principles until the angular momentum in the b3 direction becomes that are well understood by undergraduate students and lay- zero, as shown in Fig. 14. men alike; no equations are necessary. Once the rattleback transitions to spin in the biased direc- Though this mechanism explains how the spin is devel- tion, the opposite effects occur. The angular momentum in ^ oped due to a wobble, it may not be clear how the wobble the b3 direction is spinning away from the contact point, develops when the rattleback is spun opposite to the spin- thus the normal force decreases in magnitude during each bias direction. This will be explained more thoroughly wobble oscillation, as do the pitch and roll angles, as seen below. in Fig. 14.

Fig. 18. The rattleback at t ¼ 0.05 s: front view (left), side view (middle), and top view (right). The roll angle has been exaggerated to improve clarity, but load magnitudes and angles remain consistent with the simulation predictions.

712 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 712 Fig. 19. The rattleback at t ¼ 0.10 s: front view (left), side view (middle), and top view (right). The rattleback is beginning to roll back toward equilibrium, thus a þb^1 angular acceleration is developed. For this to occur, a friction force in the b^2 direction must also develop.

V. CONCLUSIONS layman or student with only minimal knowledge of particle dynamics. The purpose of this article is to present a straightforward means of deriving the equations of motion that govern the a)Electronic mail: [email protected] spin-biased behavior of a rattleback. The goal is to make the 1J. Walker, “The mysterious rattleback: A stone that spins in one direction complex dynamics accessible to undergraduate physics and and then reverses,” Sci. Am. 241, 172–184 (1979). engineering students using rudimentary conservation of 2H. Bondi, “The rigid body dynamics of unidirectional spin,” Proc. R. Soc. momenta concepts, basic kinematics, and elementary topics London Improv. Nat. Knowl. 405, 265–274 (1986). 3 from calculus and numerical methods. A. Garcia and M. Hubbard, “Spin reversal of the rattleback: Theory and Since not all undergraduate physics and engineering stu- experimentation,” Proc. R. Soc. London Improv. Nat. Knowl. 418, 165–197 (1988). dents are familiar with three-dimensional dynamics, a planar 4L. Franti, “On the rotational dynamics of the rattleback,” Cent. Eur. J. rocking semi-ellipse example is ﬁrst considered to introduce Phys. 11, 162–172 (2013). conservation of momenta equations using a rotating refer- 5S. Rauch-Wojciechowski and M. Przybylska, “Understanding reversals of a rattleback,” Regul. Chaotic Dyn. 22, 368–385 (2017). ence frame. Necessary position, velocity, and acceleration 6 terms are determined from geometry and kinematics, and the W. Case and S. Jalal, “The rattleback revisited,” Am. J. Phys. 82, 654–658 (2014). contact forces are governed using a Coulomb friction model. 7Y. Kondo and H. Nakanishi, “Rattleback dynamics and its reversal time of The temporal response of the rocking semi-ellipse is com- rotation,” Phys. Rev. E 95, 062207 (2017). puted using a standard time-stepping algorithm. Various ini- 8Z. Yoshida, T. Tokieda, and P. J. Morrison, “Rattleback: A model of how tial conditions for the rocking semi-ellipse are considered geometric singularity induces dynamic chirality,” Phys. Lett. A 381, and compared to previously published models to validate the 2772–2777 (2017). 9S. P. Kuznetsov, “On the validity of the nonholonomic model of the modeling approach. All of the required skills for the deriva- rattleback,” Phys.-Usp. 58, 1223–1224 (2015). tions and simulation are commonly taught to physics and 10M. Hanias, S. G. Stavrinides, and S. Banerjee, “Analysis of rattleback cha- engineering students in their ﬁrst two to three years of col- otic oscillations,” Sci. World J. 2014, 1–15. lege, thus it is reasonable for undergraduates to duplicate this 11A. V. Borisov, A. Y. Jalnine, S. P. Kuznetsoz, I. R. Sataev, and J. V. work. Sedova, “Dynamical phenomena occurring due to phase volume compres- The equations governing the rattleback behavior are then sion in nonholonomic model of the rattleback,” Regul. Chaotic Dyn. 17, 512–532 (2012). shown to be virtually identical to those of the planar rocking 12R. C. Hibbeler, Engineering Mechanics Statics and Dynamics, 11th ed. semi-ellipse. Rotations are now in three dimensions and the (Pearson Prentice Hall, NJ, 2007). mass moment of inertia tensor must be computed, but other- 13Roberto A. Tenenbaum, Fundamentals of Applied Dynamics (Springer, wise the previously derived vector equations can be used New York, NY, 2004). 14 directly, enforcing the concept that 3D dynamics need not be J. A. Myers, “Handbook of equations for mass and area properties of various geometric shapes,” Technical Report No. 7827, U.S. Naval Ordinance intimidating. Test Station, China Lake, California, 1962. The simulated results for the rattleback model are shown 15S. C. Chapra and R. P. Canale, Numerical Methods for Engineers, 7th ed. to predict the spin-bias behavior for various initial condi- (McGraw Hill, New York, NY, 2015). tions. Slip is allowed at the contact point via a Coulomb fric- 16M. J. Mazzoleni, M. B. Krone, and B. P. Mann, “Dynamics of rocking tion model rather than requiring the roll without slip semicircular, parabolic, and semi-elliptical disks: Equilibria, stability, and constraint used in the bulk of the previous rattleback litera- natural frequencies,” J. Vibr. Acoust. 137, 1–7 (2015). 17MathWorks, “https://www.mathworks.com/help,” (2019). ture. Finally, a simple, intuitive explanation is presented for 18P. Kim, Rigid Body Dynamics for Beginners: Euler Angles and why the rattleback reverses it spin, which can be used to sat- Quaternions (CreateSpace Independent Publishing Platform, Seoul, South isfactorily explain the rattleback’s intriguing behavior to a Korea, 2013).

713 Am. J. Phys., Vol. 87, No. 9, September 2019 S. Jones and H. E. M. Hunt 713