Motion Analysis of Euler s Disk

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1 Motion Analysis of Euler s Disk Katsuhiko Yaada Osaka University) Euler s Disk is a nae of a scientific toy and its otion is the sae as a spinning coin. In this study, a siple atheatical odel is proposed for the rate daping of the spinning disk. A nuerical siulation is conducted and its results are copared with the experiental ones in order to verify the proposed odel. I Noenclature : inertial coordinates : coordinates fixed on the disk : coordinates where the contact point is attached in y z plane v : expression of vector v in coodinates C C = C I,, ) r ω v v c a : vector fro the center of ass of the disk to the contact point : angular velocity of the disk : velocity of the center of ass of the disk : velocity of the contact point : radius of the disk : halfoftheheightofthedisk 1 Introduction Euler s Disk is a coercial nae of a spinning disk on a flat surface. The contact point of the disk with the surface oves with the very high speed when the disk approaches to the stop condition. This paradoxical otion of the disk has een widely studied1,, 3, 4]. In the previous studies, the otion of the Euler s Disk has een ainly analyzed just efore the disk falls down, and the tie ehavior of the disk is considered when the falling angle of the disk is sall. In this study, the energy dissipation of the disk is focused when the falling angle is relatively large, and the energy dissipation odel of the disk is proposed. The validity of the odel is exained y experients as well as nuerical siulations..1 Slip velocity Constraint Condition Let the attitude of the coordinates with respect to the inertial coordinates e φ,, andψ in 3--3 Euler angles. The angular velocity of the disk is expressed as sin cos ψ φ +sinψ ω = sin sin ψ φ +cosψ 1) φ + ψ Let the vector fro the center of ass of the ody to the contact point on the floor e r. In Fig. 1, the vector ê is a noral vector with the surface which includes the vector r and the vector ê I3 in the vertical rise) direction. The coordinates is defined so that y and z axes correspond to ê and ê I3, respectively. ecause the coordinates is otained y the rotation of the inertial coordinates with angle φ, x axis of the coordinates, ê 1,isexpressedasfollows: ê 1 = cos ψ sin ψ ) 0 The vector r is expressed as follows: a cos ψ r = aê 1 ê 3, r = a sin ψ 3) Let the velocity of the contact point e v c. The slip velocity is otained on the condition that the contact point is fixed on the disk. y defining the velocity of the center of ass as v, v c is expressed as follows: v c = v + ω r 4) If the contact velocity is not 0, the contact point slips. The expressions of v in the coordinates and are given y v x v = v y, 5) v z v = cos ψv x sin ψv y +sinv z sin ψv x +cosψv y sin cos ψv x +sin sin ψv y +cosv z 6)

2 y using the aove equation, the expression of v c in the coordinates ecoes ] cos ψv x sin ψv y +sinv z v c = sin ψv x +cosψv y sin cos ψv x +sinsin ψv y +cosv z a sin + ) a sin ) φ + a ψ 7) a + sin ) The non-slip condition of the disk is assued that the second coponent of v c in the coordinates is 0. Then, the non-slip condition is expressed as follows: sin ψv x +cosψv y +a sin ) φ + a ψ =0 8) In the case where v x and v y are neglected, the slip velocity of the contact point can e evaluated as the second coponent in the coordinates as ê 1 v slip =a sin ) φ + a ψ 9) C ê 3 f N a ê I3 = ê 3 ê Fig 1: Euler s Disk ecause the third coponent of v c in the coordinates is the noral coponent of the contact velocity v c, it ecoes 0 as sin cos ψv x +sin sin ψv y +cosv z + a + sin ) = 0 10) The aove equation expresses a constraint condition aong the velocity coponents. y using this equation, the velocity coponent v z is eliinated and the equations of otion for φ,, ψ, v x,andv y are otained.. Drag force The vector r is expressed in the inertial coordinates as cos φ a sin ) r = sin φ a sin ) 11) I a sin + ) The velocity of the contact point, v c is expressed in the inertial coordinates as ] sin φa sin ) v cos φ c = cos φa sin ) φ sin φ I 0 ) ) cos φ cos ψ cos φ sin ψ sin φ sin ψ v x sin φ cos ψ + v y ) ) + sin φ cos ψ cos φ sin ψ + v x + sin φ sin ψ cos φ cos ψ v y 1) Let the force acting on the disk at the contact point e f c. This force is assued in the following for: f c = k r v c 13) Although the noral reaction also acts on the disk at the contact point, this force is a constraint one and does not appear in the equations of otion..3 Drag torque If the contact point slips, the force f c appears in the equations of otion and dissipates the energy. However, there ust e other energy dissipation sources fro the experiental results. Here, the following drag torque n d is assued aout the center of ass of the disk: ω n d = k d φ ˆω, ˆω = 14) The aove equation assues that the drag torque exerts in the opposite direction of the angular velocity of the disk. The source of this torque is the air drag torque whose agnitude is assued to e proportional to φ ecause it expresses the angular velocity of the contact point in the inertial coordinates. This assuption is appropriate to soe extent fro the experiental results as descried elow. 3 Equations of Motion 3.1 Equations of otion for Forces appeared in the right-hand side of the equations of otion are the generalized forces due to the gravitational force at the center of ass of the disk, the force f C at the contact point, and the drag torque n d. y setting the inertia tensor of the disk as J, itisexpressed as a /4+ /3 J = a /4+ /3 a / 15) Then, the equation of otion for ecoes as follows where v x and v y are regarded as sall quantities and

3 the ters related to v x and v y are oitted: 5a 4 ) a tan + cos 3 + cos ψ a cos 3 cos 3 sin 1 + cos ) ] = k r a sin )sinψ φ + cos ψ ] cos 1) + cos ψ a tan tan ) v x sin ψ a tan tan ) v y ) a + sin 4 3 φ + a sin φ ψ cos 3 a sin ) = g a + sin ) k d φ k r cos ) cos ψv x +sinψv y 16) y sustituting the non-slip condition in Eq. 8) into the aove equation and assuing that φ, k d =0,and k r =0, φ is expressed as follows: φ = 1ga sin ) sin 3a +4 )cos 6a sin ] 17) This is the sae equation in the equation in ]. The disk ehavior where φ in the condition of 0 is explained y this equation. y setting v slip = 0 in Eq. 9), ψ is expressed as follows fro the aove equation: ψ = 1ga sin ) 3 a sin 3a +4 )cos 6a sin ] 3. Equations of otion for φ and ψ 18) The equation of otion for φ isderivedinthesae anner with those of. This is given as follows where the ters related to v x and v y are oitted: a 1 + cos ) 4 1 sin 6 + sin ] φ + a 3 ] 3a 4 )cos φ +3a ψ = k φ d φ +cos ψ) k r a sin ) φ + a sin )sinψv x +cosψv y )] 19) The equation of otion for ψ is also given as a φ + a = k d φ φ + ψ ψ 1 a sin ) φ 3.3 Equations of otion for v x and v y ψ 0) The equations of otion for v x and v y are derived in the sae anner as follows where the ters related to v x and v y are also oitted: cos ψ a tan tan ) sin cos ψ) v x tan sin ψ cos ψ v y + tan sin ψa sin ) φ sin ψ a tan tan ) tan sin ψ cos ψ v x ) 1 + cos tan cos ψ v y + tan cos ψa sin ) sin ψ a cos 3 cos 3 sin 1 + cos ) ] = k r a sin )cosψ φ + sin ψ ] cos ) 4 Siplification of Equations of Motion 4.1 Non-slip condition The vector r fro the center of ass of the disk to the contact point is expressed in the coordinates as a sin r = 0 3) a sin On the other hand, the angular velocity of the disk is expressed in the coordinates as sin ψ ω = 4) φ +cos ψ In the case of v slip = 0, the following equation is otained y sustituting the non-slip condition ψ = a sin ) φ/a into the aove equation: + a sin ) sin φ ω = 5) sin + a ) sin φ Inthecaseof = 0, ω is in the opposite direction of r. Then, the angular oentu of the disk, h, is expressed as h = h 1 h h 3 1 sin 3a +4 ) +6a sin ] φ = 3a +4 ) 1 1 sin 3a +4 ) sin +6a ] φ 6) In this case, the equations of otion of the disk are expressed as follows: ḣ = r f N 7) where ḣ is the tie-derivative of h in the inertial coordinates. The force f N is the noral reaction that acts at the contact point and is approxiately expressed as f N = gê I3 = gê 3. Then, the aove equation is expressed as ḣ = g a sin ) ê = g a sin )ê 3 ê 1) 8)

4 When v slip =0and = 0 in the aove equation, the following equation is otained: ga sin ) ḣ = ê I3 h) 9) h 1 The aove equation eans that the angular oentu h rotates around ê I3 at the rotation speed ga sin )/h 1 in the inertial coordinates. This is the precession y the gravitational torque as is the sae with a general spinning top. On the other hand, this rotation speed is the sae with φ fro the definition of coordinates, the following equation is otained: ga sin ) φ = = h 1 ga sin ) 1 sin 3a +4 )cos +6a sin ] φ 30) This equation is the sae with Eq. 17) that eans the rotation speed φ ecoes large as approaches 0. 5 Coparison etween Experient and Siulation 5.1 Experiental results Figure shows the experiental results of the Euler s Disk. In this figure, a) shows the angular velocity of the disk, and the large values are oserved in x and y coponents in the coordinates. y defining the co- ponents of ω as ω = equations are otained: ω x ω y ω z ] T, the following = ω x sin ψ + ω y cos ψ 31) sin φ = ω x cos ψ + ω y sin ψ 3) φ + ψ = ω z 33) Fro the experiental results gyro data), the agnitude of ω z is uch larger than those of ω x and ω y and the following equations hold: φ, ψ 34) y setting = 0 in Eq. 31), ψ can e coputed. Then, y calculating ψ fro the result, φ is otained fro Eq. 33). y coining the result with Eq. 3), φ and are otained. Then, can e estiated to soe extent. y using the newly estiated and repeating the aove procedure, the estiation accuracy of φ, ψ, and can e iproved. Actually, ecause the estiated is sall enough, the estiated results of φ, ψ, and y repeating the procedure are alost the sae with the firstly estiated results. The estiated results of φ, ψ, and are shown in sufigure ) in Fig.. Fro the estiation results, the slip velocity, v slip,is calculated y Eq. 9) and is shown in sufigure f). As shown in this figure, the slip velocity is sall enough and the non-slip condition alost holds. The estiated φ fro Eq. 17) under the non-slip condition is copared with the estiated φ fro the oservation data and is shown in sufigure c). ecause the oth values alost coincide, it is clear that the non-slip condition alost holds. Sufigure e) shows the echanical energy that is the su of the rotational kinetic energy and the potential energy calculated fro the height of the center of ass of the disk. The translational kinetic energy of the center of ass is not included in the echanical energy ecause the translational velocity cannot e estiated fro the gyro data. y setting the echanical energy as K, K is given y K = 1 ω Jω + g a sin + ) 35) a) angular velocity c) φlue) and its estiate y 17) red) e) echanical energy ) tie-derivative of Euler angles d) inclination Fig : Experiental results 5. Effects of drag force f) slip velocity in Eq. 9) When the drag force f C is given y Eq. 13), the dissipation rate of the echanical energy, K is given as

5 follows: K = k r v C, v C = v C 36) y neglecting ecause of a, the following equation is otained: As shown in Fig. e), the dissipation rate of the echanical energy is alost regarded as constant, and k r vc in Eq. 13) is alost constant. However, even in the case of constant k r vc, is not so uch affected in nuerical siulations. Therefore, the drag torque in the next section is considered as the ain resource of the energy dissipation. K 4g a k d Therefore, in the case of constant k d, K is predicted to e alost constant. As shown in Fig. e), K alost dissipates in proportion to tie in the experiental results. Thus, the drag torque in Eq. 14) is thought to e appropriate. 5.3 Effectsofdragtorque Dissipation of echanical energy ecause in Eq. 31) is an inclination angle of the disk, its rate is regarded as 0. Then, the angle ψ can e defined an angle that satisfies the following equation: sin ψ = ω y ω x + ω y, cos ψ = ω x ω x + ω y 37) y sustituting the aove equation into Eq. 3), the following equation is otained: sin φ = ωx + ωy 38) y neglecting the translational velocity of the center of ass and the slip velocity at the contact point, the following equation is derived fro Eq. 9): a sin ) φ + a ψ = 0 39) y sustituting Eq. 33) into the aove equation, the following equation is otained: aω z sin φ = aω z ωx + ωy = 0 40) Therefore, y setting ω =, the following equation holds: sin φ = ωx + ωy aω = 41) a + ecause φ is otained fro Eq. 17) y neglecting the slip velocity, φ is expressed as follows fro Eqs. 17) and 41): φ = 1g a + a sin ) aω 3a +4 4) )cos 6a sin ] y equating the tie derivative of the work ω n d with that of the echanical energy K, the following equation holds: K = ω n d = k d φ ω 43) y sustituting Eq. 4) into the aove equation, K is given y K = 1g a + a sin ) a 3a +4 )cos 6a sin ] k d 44) 5.3. Coefficient of drag torque Equation 44) expresses the relation aong the dissipative rate of the echanical energy K, k d, and. ecause K and are otained fro the experiental results, the coefficient k d can e calculated at every tie. Although the echanical energy K decreases alost in proportion to tie t as shown in Fig. e), it deflects fro the proportional relation in a strict sense. Therefore, the echanical energy K is approxiated to a quadratic function of tie t unit s]) as follows: K = t t N] The relation is shown in Fig. 3. Fro this relation, K is otained at every tie t, andk d at t is also estiated fro Eq. 44) fro the experiental values of at tie t. The results are shown in Fig. 3. As shown in Fig. 3 ), the experientally estiated values of k d have decreasing tendency with tie t. a) experiental values of echanical energy lue) and its approxiation y a quadratic function red) ) experientally estiated values of k d Fig 3: Experiental results of K and k d

6 5.4 Siulation results Nuerical siulations of the otion of the Euler s Disk are conducted y using the drag torque in Eq. 14) and copared with the experiental results. Siulations are executed under the following conditions: The drag torque coefficient k d is assued to e constant at k d = Ns]. The drag force f C at the contact point is not considered k r =0). Although the drag torque coefficient k d is not regarded constant as shown in Fig. 3 ) and ecoes sall when approaches 0, it is assued here to e constant for the siplicity. Siulations are conducted ased on the equations for φ,, ψ, v x,andv y. Although the echanical energy in the siulation includes the effects of the translational energy y v x and v y, the translational energy contriutes little ecause v x and v y are sall. Siulation results are shown in Fig. 4 as copared with the experiental results. As shown in these figures, the siulation results for the agnitude of the angular velocity, φ, ψ, and the echanical energy coincide well with the experiental results. Therefore, the siulation odel exactly realizes the otion of the Euler s Disk. 6 Conclusions In this study, the otion of the Euler s Disk is analyzed especially when the falling angle is relatively large. The energy dissipation odel of the disk is proposed in this range, and its validity is exained y nuerical siulations and hardware experients. The results show that the proposed odel agrees well with the experiental ehavior of the disk. a) angular velocity siulation) c) tie-derivative of Euler angles solid: experient, dashed: siulation) ) agnitude of angular velocity lue: experient, red: siulation) d) inclination lue: experient, red: siulation) Reference 1] Moffet, H. K., Euler s disk and its finite-tie singularity, Nature, Vol. 404, 0 April, 000, pp ] Petrie, D., Hunt, J. L., and Gray, C. G., Does the Euler Disk slip during its otion?, Aerican Journal of Physics, Vol. 70, No. 10, 00, pp ] Kessler, P. and O Reilly O. M., The Ringing of Euler s Disk, Regular and Chaotic Dynaics, Vol. 7, No. 1, 00, pp ] Caps, H., Dorolo, S., Ponte, S., Croisier, H., and Vandewalle, N., Rolling and slipping otion of Euler s disk, Physical Review E, 69, 004, pp e) echanical energy lue: experient, red: siulation) f) slip velocity in Eq. 9) siulation) Fig 4: Coparison etween siulation and experient

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