{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" 19 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" 19 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" 19 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" 19 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" 19 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 286 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" 19 287 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" 19 288 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Warning" -1 7 1 {CSTYLE "" -1 -1 "Courier " 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Outpu t" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 } 1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 9 "Tippy Top" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 586 "The tippy-top (or Bri tish: tippe top) has fascinated physicists for quite some time. It rep resents a good computational physics problem due to the fascinating mo tion, as well as due to the fact that analytically it is not easy to m ake progress on the problem. A good recipe for numerical work appeared in an article by Richard J. Cohen in American Journal of Physics (AJP 45, 12 (1977)). It did not emphasize the previous attempts at establi shing whether there are conserved quantities in this problem. The latt er are essential at understanding the spherical top's remarkable behav iour." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 619 "What is a tippy top? Think of a sphere which has an axially symmetric mass distribution (such that a rotor axis is defined with associated \+ moment of inertia I3), but whose center of gravity is not at the cente r of the sphere. This can be accomplished by making the sphere out of \+ two parts of different mass density, or by cutting off the sphere and \+ introducing a stem. In the latter version wooden tippy tops are sold i n toy stores, and they resemble mushrooms. What is remarkable about su ch a top? When one twists the top at the stem and puts it down on a no t too smooth surface with the stem straight up (polar angle " }{TEXT 19 7 "theta=0" }{TEXT -1 169 "), it begins after a second or so to pre cess and nutate, i.e., the stem leaves the vertical position, and prov ided the top was spun fast enough it will move towards the " }{TEXT 19 8 "theta=Pi" }{TEXT -1 175 " position and continue spinning on its \+ stem. For the mathematical details that follow Cohen's article, and th en go a bit beyond look at the included lecture notes (jpg-files)." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1164 "Why are physicists so fascinated by the tippy top? Cohen reprints a picture o f Nobel laureates Niels Bohr and Wolfgang Pauli playing with this toy. There are several reasons for this. First of all, experienced mechani cs afficionados have played with normal tops, and have learned about t hem in textbooks. They certainly know about the weird effects, how the gravitational torque makes these precess rather then fall when the to ps are spinning. They know from introductory texts (e.g., Feynman lect ures) how to think about the conservation properties, how the angular \+ momentum vector keeps its magnitude approximately constant and revolve s about the suspension point. Friction forces play a minor role, becau se the precessional angular velocity is small, but they are known to d amp out the nutational oscillation quickly. So the physicist is prepar ed for the phenomenon of gravity being overshadowed by rotational kine tic energy. Slightly more experienced physicists know about the fricti on torque which acts to lift up a normal top: indeed it can be observe d that a top started at a tilt will lift itself up and align the angul ar momentum vector with the vertical." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 177 "What most (even experienced) physici sts are not ready for in the case of a spherical top is the fact that: (i) frictional torques dominate the behaviour of the motion; (ii) the " }{TEXT 19 7 "theta=0" }{TEXT -1 295 " orientation is quite unstable and spinning about the rotor axis gives way to precessional motion fo r any such top, almost irrespective of the mass arrangement provided t he initial spin rate is sufficiently high; (iii) energy dissipation is rather large and leads to tippy-top orientations with " }{TEXT 19 10 "theta>Pi/2" }{TEXT -1 50 ", and for some range of moment of inertia r egimes " }{TEXT 19 5 "I1=I2" }{TEXT -1 8 " versus " }{TEXT 19 2 "I3" } {TEXT -1 28 " to a definite push towards " }{TEXT 19 8 "theta=Pi" } {TEXT -1 100 ". It is the latter behaviour which leads to the phenomen on of the tippy-top standing up on its stem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 242 "To model the entire motion of \+ the tippy-top from beginning to end is complicated, but has recently b een achieved by Christian Friedl and together with Andre Wobst from th e University of Augsburg in Germany turned into virtual reality movies [" }{TEXT 19 42 "www.physik.uni-ausburg.de/~wobsta/tippetop" }{TEXT -1 200 "]. In this worksheet we will restrict the discussion to that o f a spherical top with center of gravity on the rotor axis below the g eometrical center. The sphere is described by two principal moments " }{TEXT 19 5 "I1=I2" }{TEXT -1 5 " and " }{TEXT 19 2 "I3" }{TEXT -1 131 ". Following the work of Friedl, which has been in part inspired b y the interesting paper of Leutwyler [European Journal of Physics " } {TEXT 257 2 "15" }{TEXT -1 216 ", 59 (1994); which most clearly discus ses the stability regimes for the nutation angle theta] we discuss not only the numerical solution of Euler's equations for such a sphere, b ut also illustrate a conservation law." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 235 "The physicist is baffled by the tipp y-top, because the orientation of the top at the beginning is clearly \+ favoured by gravity. Without spinning the gravitational torque lifts t he top into the position with the stem up. Why then is the " }{TEXT 19 7 "theta=0" }{TEXT -1 967 " orientation unstable for high spin rate s? Why in fact, does the top stand up on its stem, obviously going to \+ the most unfavourable orientation from gravity's perspective? Clearly \+ rotational energy is given up during this turning upside-down process. The only torque to remove rotational kinetic energy is the frictional torque which acts about the vertical. Given that friction is importan t to qualitatively determine the motion traditional physicists might b ecome nervous, because most textbook examples deal with energy-cosnser ving systems, or systems where energy is removed slowly (such as in th e nutational motion of the normal top or gyroscope). It turns out that the gravitational potential energy in the tippy-top is dwarfed by the rotational kinetic energy, and can practically be left out of the dis cussion (for high initial spin rates). It is small, because the displa cement of the center of gravity from the center of the sphere (labeled by the variable " }{TEXT 19 1 "a" }{TEXT -1 199 ") is small, i.e., th e gravitational torque is small (one really is interested in the torqu e due to the normal force in response to gravity, because the lever ar m originates in the centre of mass-CM)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 627 "Why is there hope for some analytic al results even though energy is not conserved, and obviously the vert ical component angular momentum is not conserved. The latter point mak es the physicist even more nervous, because conservation of the vertic al component of angular momentum in the laboratory frame provides the \+ main argument for understanding precession, and in particular for unde rstanding why the precession picks up speed when the gyro slows down. \+ The tippy-top spins much more slowly in its upside-down position than \+ in the original starting orientation, i.e., a large amount of kinetic \+ energy has been dissipated. The " }{TEXT 19 2 "z-" }{TEXT -1 252 "comp onent (lab frame) of angular momentum is substantially reduced, but it still points in the same direction. This means that from the body fra me point of view the spin angular velocity about the rotor axis has ch anged sign! Wow, why would it do that?" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 138 "Nevertheless, for the simpler forms \+ of friction forces there remains a conserved quantity. The dot product of the angular momentum vector " }{TEXT 258 1 "L" }{TEXT -1 25 " and \+ the position vector " }{TEXT 259 1 "r" }{TEXT -1 161 " which describes the location of the point of contact with the surface (table) as meas ured from the center of mass (gravity) (CM) of the top (a sphere of ra dius " }{TEXT 19 1 "R" }{TEXT -1 153 ") can be shown to be conserved i n time when the friction is of sliding type. This is done, e.g., in an AJP paper by B.G.Nickel and C.G.Gray [Am. J. Phys " }{TEXT 260 2 "68 " }{TEXT -1 643 ", 821 (2000), where a very comprehensive review of th e history of the subject can be found] although overall the paper is n ot easy to understand, because it presents many results based on energ y conservation (while stressing at the beginning that energy dissipati on is crucial). Leutwyler has probably been the first to intuitively e xplain how the frictional torques work: when the tippy-top goes into s pin+precessional motion the CM executes motion in a circle about the p oint of contact, i.e., there is constantly friction at play. Cohen arg ues that the moment arm for the frictional torque in the spin motion a bout the rotor axis goes like " }{TEXT 19 12 "R*sin(theta)" }{TEXT -1 426 ", i.e., becomes large when the top leaves the vertical orientatio n. This is one reason for the motion favouring the precessional motion (the stem visibly spinning about the vertical, while also nutating). \+ While this is a nice argument (the moment arm remains small for the pr ecessional rotation), one should watch the time evolution of the angul ar velocities, and one will notice that a dramatic change in behaviour occurs at " }{TEXT 19 7 "theta=0" }{TEXT -1 73 " (most simulations st art at a small non-zero value, seem to tend towards " }{TEXT 19 7 "the ta=0" }{TEXT -1 318 ", where a big 'bounce' occurs exchanging the rota tional kinetic energies in the spin and precessional motions. So perha ps more importantly Leutwyler stresses that the motion becomes stable \+ when the frictional torque is minimized. The frictional torque acts su ch as to minimize the velocity of the CM of the top in the " }{TEXT 19 5 "(x,y)" }{TEXT -1 26 " plane. This can occur at " }{TEXT 19 7 "th eta=0" }{TEXT -1 7 " or at " }{TEXT 19 8 "theta=Pi" }{TEXT -1 30 ", bu t also at other values of " }{TEXT 19 5 "theta" }{TEXT -1 199 " when t he moments of inertia are chosen differently. The numerical simulation s of Cohen (and later Friedl) have found examples for this dependence \+ on the relationship of the moment of inertia values " }{TEXT 19 2 "I1 " }{TEXT -1 5 " and " }{TEXT 19 2 "I3" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 237 "Our goal is to presen t the equations of motion for the top without translational motion of \+ the CM. Nevertheless, as explained in the previous paragraph, the CM w ill be moving about the point of contact in a circular fashion (except when " }{TEXT 19 7 "theta=0" }{TEXT -1 4 " or " }{TEXT 19 8 "theta=Pi " }{TEXT -1 454 "), and frictional torques govern the motion. We will \+ follow the Newton-Euler equations of motion as outlined by Cohen rathe r than the Lagrange-Hamilton formalism employed by Friedl (which is co nvenient to observe the conservation law, and to model more complicate d friction torques based on a dissipation function). We cannot derive \+ the equations here, but refer to the paper and to the lecture notes (j pg-files). Nevertheless, we outline some key ideas." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "The laboratory frame is defined by " }{TEXT 19 7 "(x,y,z)" }{TEXT -1 34 ". The table is repre sented by the " }{TEXT 19 5 "(x,y)" }{TEXT -1 70 " plane in which the \+ circular motion of the CM about the contact point " }{TEXT 19 1 "T" } {TEXT -1 38 " occurs. The body frame is denoted by " }{TEXT 19 7 "(1,2 ,3)" }{TEXT -1 6 " with " }{TEXT 19 1 "3" }{TEXT -1 104 " being the ro tor axis. The orientation of the top in the laboratory is described by : the nutation angle " }{TEXT 19 5 "theta" }{TEXT -1 13 " between the \+ " }{TEXT 19 2 "3-" }{TEXT -1 9 " and the " }{TEXT 19 2 "z-" }{TEXT -1 20 "axis; a unit vector " }{TEXT 272 1 "e" }{TEXT 19 2 "_n" }{TEXT -1 32 " formed by the cross product of " }{TEXT 276 1 "e" }{TEXT 19 2 "_z " }{TEXT -1 5 " and " }{TEXT 275 1 "e" }{TEXT 19 3 "_3 " }{TEXT -1 69 "(normalized to unit length) - this vector forms the precession angle \+ " }{TEXT 19 3 "phi" }{TEXT -1 10 " with the " }{TEXT 274 1 "e" }{TEXT 19 2 "_x" }{TEXT -1 28 " unit vector; a unit vector " }{TEXT 273 1 "e " }{TEXT 19 3 "_np" }{TEXT -1 36 " obtained from the cross product of \+ " }{TEXT 271 1 "e" }{TEXT 19 2 "_3" }{TEXT -1 5 " and " }{TEXT 277 1 " e" }{TEXT 19 2 "_n" }{TEXT -1 24 ". The coordinate system " }{TEXT 19 8 "(n,np,3)" }{TEXT -1 97 " defines the orientation of the top, the to p's spinning about the rotor axis is described by the " }{TEXT 19 5 "( 1,2)" }{TEXT -1 31 " axes rotating with respect to " }{TEXT 19 6 "(n,n p)" }{TEXT -1 10 " by angle " }{TEXT 19 3 "psi" }{TEXT -1 110 ". Cohen lists the relationships between these unit vectors in equations (2)-( 9), this is nothing but geometry." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 28 "The angular velocity vector " }{TEXT 261 1 "w" }{TEXT -1 55 " as measured in the laboratory frame has component s in " }{TEXT 19 7 "(n,z,3)" }{TEXT -1 13 " coordinates " }{TEXT 19 29 "(theta-dot, phi-dot, psi-dot)" }{TEXT -1 9 " - where " }{TEXT 19 3 "dot" }{TEXT -1 88 " refers to time derivative. This is simply in ac cord with the definition of the angles (" }{TEXT 19 5 "theta" }{TEXT -1 13 " turns about " }{TEXT 19 1 "n" }{TEXT -1 2 ", " }{TEXT 19 3 "ph i" }{TEXT -1 7 " about " }{TEXT 19 1 "z" }{TEXT -1 6 ", and " }{TEXT 19 3 "psi" }{TEXT -1 7 " about " }{TEXT 19 1 "3" }{TEXT -1 69 "; see F ig. 4 in Cohen). This needs to be expressed in the orthogonal " } {TEXT 19 8 "(n,np,3)" }{TEXT -1 48 " coordinate system, where the comp onents become " }{TEXT 19 59 "(theta-dot, phi-dot*sin(theta), psi-dot+ phi-dot*cos(theta))" }{TEXT -1 6 ". The " }{TEXT 19 8 "(n,np,3)" } {TEXT -1 103 " coordinate system itself has an angular velocity vector in the lab frame which is almost identical to " }{TEXT 262 1 "w" } {TEXT -1 59 " for the top, the only difference being the absense of th e " }{TEXT 19 7 "psi-dot" }{TEXT -1 8 " in the " }{TEXT 19 2 "3-" } {TEXT -1 113 "component. This is the same choice of coordinates as for the discussion of the Euler equations for the gyroscope." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 "In eq. (12) Cohen discusses the simplest-possible frictional torque that can explain th e tippy-top. The argument is that the friction force for rotation of t he CM about T (without translation on the table surface) the friction \+ force points along " }{TEXT 19 1 "-" }{TEXT 267 1 "e" }{TEXT 19 2 "_n " }{TEXT -1 37 ". A rotation of the CM riding on the " }{TEXT 19 2 "3- " }{TEXT -1 51 "axis has a tangential velocity vector aligned with " } {TEXT 268 1 "e" }{TEXT 19 2 "_n" }{TEXT -1 36 ", given that it is perp endicular to " }{TEXT 270 1 "e" }{TEXT 19 2 "_z" }{TEXT -1 5 " and " } {TEXT 269 1 "e" }{TEXT 19 2 "_3" }{TEXT -1 37 ". A kinetic coefficient of friction (" }{TEXT 19 5 "f=0.3" }{TEXT -1 69 " or less) is assumed and the magnitude of the friction force becomes " }{TEXT 19 5 "f*m*g " }{TEXT -1 8 ", where " }{TEXT 19 1 "m" }{TEXT -1 188 " is the mass o f the top. This simplistic friction model does not damp the nutation ( Cohen generalizes it in eq. (16)). The frictional torque is calculated by taking the cross product with " }{TEXT 263 1 "r" }{TEXT 19 18 " = \+ (a*e_3 - R*e_z)" }{TEXT -1 36 ", which results in components along " } {TEXT 266 1 "e" }{TEXT 19 3 "_np" }{TEXT -1 5 " and " }{TEXT 265 1 "e " }{TEXT 19 2 "_3" }{TEXT -1 10 " (but not " }{TEXT 264 1 "e" }{TEXT 19 2 "_n" }{TEXT -1 363 "). We generalize eq. (12) to include damping \+ of the nutation, and also to account for the sign-change when the Cart esian velocity components reverse direction. In a separate section the complete cross product according to eq. (16) is calculated (resulting in messy expressions). The answers can be copy-pasted and used as a r eplacement for the frictional torques." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 356 "The rest is the usual Euler equation treatment: A frame was chosen in which the moments of inertia remain \+ constant (the orientation frame for the top described above). The Eule r equations are straightforward, the rate of change of the angular mom entum vector in the laboratory frame is related to the rate of change \+ in the body frame (where we acquire the " }{TEXT 19 13 "omega-cross-L " }{TEXT -1 317 " term which leads to the nonlinearities of the Euler \+ equation). The friction torque has been translated from the lab frame \+ to the body frame, and the gravitational torque will turn out to be ir relevant (but could be added easily). The equations are then different iated to obtain second-order equations for the angles." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "MOST USERS WILL SKIP THE DERIVATION IN THE INDENTED SECTION BELOW:" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 278 90 "The frictional torque about the CM in (n,n',3) coor dinates according to eq.(16) in R.Cohen" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Calculate the frictional torque by evaluating the cross p roduct." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart; with(li nalg):" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected names no rm and trace have been redefined and unprotected\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "rv:=a*e_3-r*e_z;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#rvG,&*&%\"aG\"\"\"%$e_3GF(F(*&%\"rGF(%$e_zGF(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "sth:=sin(theta(t)): cth:= cos(theta(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "sph:=sin( phi(t)): cph:=cos(phi(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "e_3:=vector([sth*sph,-sth*cph,cth]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e_3G-%'vectorG6#7%*&-%$sinG6#-%&thetaG6#%\"tG\"\"\"-F+6#-%$ph iGF/F1,$*&F*F1-%$cosGF3F1!\"\"-F9F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "e_z:=vector([0,0,1]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e_zG-%'vectorG6#7%\"\"!F)\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalm(rv);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'vect orG6#7%*(%\"aG\"\"\"-%$sinG6#-%&thetaG6#%\"tGF)-F+6#-%$phiGF/F),$*(F(F )F*F)-%$cosGF2F)!\"\",&*&F(F)-F8F,F)F)%\"rGF9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Now enter the friction force in Cartesian coordinates: first do " }{TEXT 19 11 "[-vx,-vy,0]" }{TEXT -1 39 " and multiply on \+ the remaining factors " }{TEXT 19 8 "f*m*g/vR" }{TEXT -1 12 " at the e nd." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 138 "F_f:=-vector([(r*ps idot(t)+a*phidot(t))*sth*cph+thetadot(t)*sph*(a*cth-r),(r*psidot(t)+a* phidot(t))*sth*sph-thetadot(t)*cph*(a*cth-r),0]);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$F_fG,$-%'vectorG6#7%,&*(,&*&%\"rG\"\"\"-%'psidotG6#% \"tGF/F/*&%\"aGF/-%'phidotGF2F/F/F/-%$sinG6#-%&thetaGF2F/-%$cosG6#-%$p hiGF2F/F/*(-%)thetadotGF2F/-F9F?F/,&*&F5F/-F>F:F/F/F.!\"\"F/F/,&*(F,F/ F8F/FEF/F/*(FCF/F=F/FFF/FI\"\"!FI" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "In cartesian coordinates the torque becomes (up to factors -f*m*g/ vR):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "N_fCC:=crossprod(ev alm(rv),F_f);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&N_fCCG-%'vectorG6# 7%,$*&,&*&%\"aG\"\"\"-%$cosG6#-%&thetaG6#%\"tGF.F.%\"rG!\"\"F.,&*(,&*& F6F.-%'psidotGF4F.F.*&F-F.-%'phidotGF4F.F.F.-%$sinGF1F.-FB6#-%$phiGF4F .F7*(-%)thetadotGF4F.-F0FDF.F+F.F.F.F7*&F+F.,&*(F:F.FAF.FJF.F7*(FHF.FC F.F+F.F7F.,&**F-F.FAF.FCF.F8F.F.**F-F.FAF.FJF.FLF.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "expand(N_fCC[3]);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,**,%\"aG\"\"\")-%$sinG6#-%&thetaG6#%\"tG\"\"#F&)-F)6#- %$phiGF-F/F&%\"rGF&-%'psidotGF-F&!\"\"**)F%F/F&F'F&F0F&-%'phidotGF-F&F 8*,F%F&F'F&)-%$cosGF2F/F&F5F&F6F&F8**F:F&F'F&F>F&F;F&F8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "simplify(%,\{cph^2+sph^2=1,cth^2+st h^2=1\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&*(%\"aG\"\"\"%\"rGF'- %'psidotG6#%\"tGF'!\"\"*&)F&\"\"#F'-%'phidotGF+F'F-F')-%$sinG6#-%&thet aGF+F0F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Now we want the comp onents of the torque in the (n,np,3) reference frame. For this purpose we define the transformation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "e_x:=cph*e_n-cth*sph*e_np+sth*sph*e_3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e_xG,(*&-%$cosG6#-%$phiG6#%\"tG\"\"\"%$e_nGF.F.*(-F( 6#-%&thetaGF,F.-%$sinGF)F.%%e_npGF.!\"\"*(-F6F2F.F5F.%$e_3GF.F." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "e_y:=sph*e_n+cth*cph*e_np-st h*cph*e_3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e_yG,(*&-%$sinG6#-%$p hiG6#%\"tG\"\"\"%$e_nGF.F.*(-%$cosG6#-%&thetaGF,F.-F2F)F.%%e_npGF.F.*( -F(F3F.F6F.%$e_3GF.!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "e_z:=sth*e_np+cth*e_3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e_zG,&*& -%$sinG6#-%&thetaG6#%\"tG\"\"\"%%e_npGF.F.*&-%$cosGF)F.%$e_3GF.F." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "N_f:=N_fCC[1]*e_x+N_fCC[2]*e _y+N_fCC[3]*e_z;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$N_fG,(*(,&*&%\" aG\"\"\"-%$cosG6#-%&thetaG6#%\"tGF*F*%\"rG!\"\"F*,&*(,&*&F2F*-%'psidot GF0F*F**&F)F*-%'phidotGF0F*F*F*-%$sinGF-F*-F>6#-%$phiGF0F*F3*(-%)theta dotGF0F*-F,F@F*F'F*F*F*,(*&FFF*%$e_nGF*F**(F+F*F?F*%%e_npGF*F3*(F=F*F? F*%$e_3GF*F*F*F3*(F'F*,&*(F6F*F=F*FFF*F3*(FDF*F?F*F'F*F3F*,(*&F?F*FIF* F**(F+F*FFF*FKF*F**(F=F*FFF*FMF*F3F*F**&,&**F)F*F=F*F?F*F4F*F***F)F*F= F*FFF*FOF*F*F*,&*&F=F*FKF*F**&F+F*FMF*F*F*F*" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 108 "The magnitude of the velocity of the CM about the cont act point T projected onto the x-y plane is given as " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "vR:=sqrt((a*phidot(t)+r*psidot(t))^ 2*sth^2+thetadot(t)^2*(r-a*cth)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%#vRG*$,&*&),&*&%\"rG\"\"\"-%'psidotG6#%\"tGF,F,*&%\"aGF,-%'phidotGF /F,F,\"\"#F,)-%$sinG6#-%&thetaGF/F5F,F,*&)-%)thetadotGF/F5F,),&F+F,*&F 2F,-%$cosGF9F,!\"\"F5F,F,#F,F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "Nn:=factor(simplify(expand(coeff(N_f,e_n)),\{cph^2+sph^2=1,cth ^2+sth^2=1\}))*f*m*g/vR;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NnG*.-% )thetadotG6#%\"tG\"\"\",***\"\"#F*%\"rGF*%\"aGF*-%$cosG6#-%&thetaGF(F* F**$)F.F-F*!\"\"*&)-%$sinGF2F-F*)F/F-F*F**$F " 0 "" {MPLTEXT 1 0 86 "Nnp:=factor(simplify(expand(coeff(N_f,e_np)),\{cph^2+ sph^2=1,cth^2+sth^2=1\}))*f*m*g/vR;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%$NnpG*0-%$sinG6#-%&thetaG6#%\"tG\"\"\",&%\"aG!\"\"*&%\"rGF--%$cosGF (F-F-F-,&*&F2F--%'psidotGF+F-F-*&F/F--%'phidotGF+F-F-F-%\"fGF-%\"mGF-% \"gGF-,&*&)F5\"\"#F-)F&FBF-F-*&)-%)thetadotGF+FBF-),&F2F-*&F/F-F3F-F0F BF-F-#F0FB" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "N3:=factor(si mplify(expand(coeff(N_f,e_3)),\{cph^2+sph^2=1,cth^2+sth^2=1\}))*f*m*g/ vR;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#N3G,$*0%\"rG\"\"\",&*&F'F(-% 'psidotG6#%\"tGF(F(*&%\"aGF(-%'phidotGF-F(F(F(-%$sinG6#-%&thetaGF-\"\" #%\"fGF(%\"mGF(%\"gGF(,&*&)F)F8F()F3F8F(F(*&)-%)thetadotGF-F8F(),&F'F( *&F0F(-%$cosGF5F(!\"\"F8F(F(#FIF8FI" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 220 "Rather than using those messy expressions, we expand a little on the simple frict ion torque modeling introduced in eq. (12). In eq.(12) it was simply a ssumed that the kinetic sliding friction force is along the negative \+ " }{TEXT 279 1 "e" }{TEXT 19 2 "_n" }{TEXT -1 216 " axis. This is vali d when the original spin about the 3-axis is positive, and ignores the velocity component of the CM associated with nutations. If one inputs the torque (12) as is, the tippy-top stands up, reaches " }{TEXT 19 8 "theta=Pi" }{TEXT -1 46 " for many conditions (almost independently \+ of " }{TEXT 19 2 "I1" }{TEXT -1 5 " and " }{TEXT 19 2 "I3" }{TEXT -1 169 "), and then bounces and the friction torque starts dumping energy into the rotational motion. It also gives energy gain when the top is started with a negative value of " }{TEXT 19 9 "psidot(0)" }{TEXT -1 75 ". To obtain a more realistic model we added the torque component a bout the " }{TEXT 280 1 "e" }{TEXT 19 3 "_n'" }{TEXT -1 175 " axis, an d introduced a sign dependence. From the discussion of the velocity, e q. (14) [it is given in the above section], it is obvious that the sig n dependence shoulf be on " }{TEXT 19 11 "thetadot(t)" }{TEXT -1 25 ", and on the combination " }{TEXT 19 25 "(r*psidot(t)+a*phidot(t))" } {TEXT -1 46 ". When we use the discontinuous (unevaluated) " }{TEXT 19 7 "sign(x)" }{TEXT -1 125 " function in Maple, the differential equ ations solver is knocked out. Thus, we use a softened sign function ma de up from the " }{TEXT 19 6 "arctan" }{TEXT -1 87 " function with sca led argument. The dependence on the scale parameter should be tested. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "#plot(2/Pi*arctan(100*x ),x=-1..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "restart; Dig its:=14:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "sth:=sin(theta( t)): cth:=cos(theta(t)):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "We l ist the magnitude of the velocity of the CM projected onto the table s urface. It shows how the components " }{TEXT 19 3 "v_x" }{TEXT -1 5 " \+ and " }{TEXT 19 3 "v_y" }{TEXT -1 32 " change sign when the variables \+ " }{TEXT 19 8 "thetadot" }{TEXT -1 4 " or " }{TEXT 19 17 "a*phidot+r*p sidot" }{TEXT -1 31 " change sign (go through zero)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "vR:=sqrt((a*phidot(t)+r*psidot(t))^2*sth^ 2+thetadot(t)^2*(r-a*cth)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#vR G*$,&*&),&*&%\"aG\"\"\"-%'phidotG6#%\"tGF,F,*&%\"rGF,-%'psidotGF/F,F, \"\"#F,)-%$sinG6#-%&thetaGF/F5F,F,*&)-%)thetadotGF/F5F,),&F2F,*&F+F,-% $cosGF9F,!\"\"F5F,F,#F,F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "# in theta-eq: gravitational (normal force) torque: -m*g*a*sth" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "We introduce the simple friction al torques in the " }{TEXT 19 2 "n'" }{TEXT -1 5 " and " }{TEXT 19 1 " 3" }{TEXT -1 181 "-directions. They are based on sliding kinetic frict ion proportional to the magnitude of the normal force and opposing the velocity, which at this point is not used with known sign." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Sign:=x->2/Pi*arctan(1000*x) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%SignGf*6#%\"xG6\"6$%)operatorG %&arrowGF(,$*(\"\"#\"\"\"%#PiG!\"\"-%'arctanG6#,$*&\"%+5F/9$F/F/F/F/F( F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Nn:=-m*g*a*sth; # t o try eq. (12) & with gravitational torque" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NnG,$**%\"mG\"\"\"%\"gGF(%\"aGF(-%$sinG6#-%&thetaG6# %\"tGF(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "#Nn:=-m*g*a *sth-f*m*g*abs(r*cth-a)*Sign(thetadot(t)); # This friction about n-axi s is too large" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "Nnp:=f*m* g*(r*cth-a)*Sign((r*psidot(t)+a*phidot(t)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$NnpG,$*0\"\"#\"\"\"%\"fGF(%\"mGF(%\"gGF(,&*&%\"rGF(- %$cosG6#-%&thetaG6#%\"tGF(F(%\"aG!\"\"F(%#PiGF7-%'arctanG6#,&*(\"%+5F( F6F(-%'phidotGF4F(F(*(F>F(F.F(-%'psidotGF4F(F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "N3:=-f*m*g*r*sth*Sign((r*psidot(t)+a*phidot (t)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#N3G,$*2\"\"#\"\"\"%\"fGF( %\"mGF(%\"gGF(%\"rGF(-%$sinG6#-%&thetaG6#%\"tGF(%#PiG!\"\"-%'arctanG6# ,&*(\"%+5F(%\"aGF(-%'phidotGF2F(F(*(F;F(F,F(-%'psidotGF2F(F(F(F5" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Manually calculated torques from e q (16): unfortunately " }{TEXT 19 15 "dsolve[numeric]" }{TEXT -1 62 " \+ can't integrate the equations with them beyond a point where " }{TEXT 19 8 "theta(t)" }{TEXT -1 127 " is very small. The answer would be to \+ use the torques Nnp and N3, and keep only the gravitational torque (no rmal force at T). " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "#Nn:= -m*g*a*sth-f*m*g*(r-a*cth)^2*thetadot(t)/vR; # this frictional torque \+ gives trouble." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "#Nnp:=-f* m*g*(a-r*cth)*(r*psidot(t)+a*phidot(t))*sth/vR;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 49 "#N3:=-f*m*g*r*sth^2*(r*psidot(t)+a*phidot(t))/ vR;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "DE1:=diff(theta(t),t )=thetadot(t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "DE2:=diff(phi(t), t)=phidot(t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "DE3:=diff(psi(t),t )=psidot(t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "DE4:=diff(thetadot( t),t)=phidot(t)^2*sth*cth*(1-I3/I1)-(I3*phidot(t)*psidot(t)*sth-Nn)/I1 ;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "DE5:=diff(phidot(t),t)=((I3-2 *I1)*phidot(t)*thetadot(t)*cth+I3*psidot(t)*thetadot(t)+Nnp)/(I1*(sth) );" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "DE6:=diff(psidot(t),t)=phidot (t)*thetadot(t)*sth-cth*diff(phidot(t),t)+N3/I3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$DE1G/-%%diffG6$-%&thetaG6#%\"tGF,-%)thetadotGF+" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$DE2G/-%%diffG6$-%$phiG6#%\"tGF,-%'p hidotGF+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$DE3G/-%%diffG6$-%$psiG6 #%\"tGF,-%'psidotGF+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$DE4G/-%%dif fG6$-%)thetadotG6#%\"tGF,,&**)-%'phidotGF+\"\"#\"\"\"-%$sinG6#-%&theta GF+F3-%$cosGF6F3,&F3F3*&%#I3GF3%#I1G!\"\"F?F3F3*&,&**F=F3F0F3-%'psidot GF+F3F4F3F3**%\"mGF3%\"gGF3%\"aGF3F4F3F3F3F>F?F?" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$DE5G/-%%diffG6$-%'phidotG6#%\"tGF,*(,(**,&%#I3G\"\" \"*&\"\"#F2%#I1GF2!\"\"F2F)F2-%)thetadotGF+F2-%$cosG6#-%&thetaGF+F2F2* (F1F2-%'psidotGF+F2F7F2F2*0F4F2%\"fGF2%\"mGF2%\"gGF2,&*&%\"rGF2F9F2F2% \"aGF6F2%#PiGF6-%'arctanG6#,&*(\"%+5F2FHF2F)F2F2*(FOF2FGF2F?F2F2F2F2F2 F5F6-%$sinGF;F6" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$DE6G/-%%diffG6$- %'psidotG6#%\"tGF,,(*(-%'phidotGF+\"\"\"-%)thetadotGF+F1-%$sinG6#-%&th etaGF+F1F1*&-%$cosGF6F1-F'6$F/F,F1!\"\"*4\"\"#F1%\"fGF1%\"mGF1%\"gGF1% \"rGF1F4F1%#PiGF>-%'arctanG6#,&*(\"%+5F1%\"aGF1F/F1F1*(FKF1FDF1F)F1F1F 1%#I3GF>F>" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "Now set the parame ters I1=9E-7 stable theta around 2.2, tippy doesn't get up. I1=8E-7: gets all the way." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "g:=10; " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "a:=0.3E-2;" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 10 "r:=1.5E-2;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "I1 :=7.5E-7; I3:=7.0E-7;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "m:=6.0E-3; " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "2/5*m*r^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG\"#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG$\" \"$!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\"#:!\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#I1G$\"#v!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#I3G$\"#q!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\" mG$\"#g!\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/++++++a!#?" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "f:=0.3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG$\"\"$!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "IC:=theta(0)=0.1,phi(0)=0,psi(0)=0,thetadot(0)=0,phid ot(0)=0,psidot(0)=150;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6(/-%& thetaG6#\"\"!$\"\"\"!\"\"/-%$phiGF)F*/-%$psiGF)F*/-%)thetadotGF)F*/-%' phidotGF)F*/-%'psidotGF)\"$]\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "sol:=dsolve(\{seq(DE||i,i=1..6),IC\},numeric,method=rosenbrock ,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 " unprotect(Psi):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "Theta:= eval(theta(t),sol): Thetadot:=eval(thetadot(t),sol): Phi:=eval(phi(t), sol): Phidot:=eval(phidot(t),sol): Psi:=eval(psi(t),sol): Psidot:=eval (psidot(t),sol):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 215 "In the follo wing line we test the numerical integration: if it fails to go beyond \+ some time limit due to inability to cope with adaptive error control r equirements we simply choose manually a time integration limit " } {TEXT 19 3 "t_f" }{TEXT -1 35 " that allows us to display results." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "t_f:=1.2; Theta(t_f);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$t_fG$\"#7!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"36i,2H/(*QJ!#<" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "plot([Theta(t)],t=0..t_f,numpoints=500,color=[red,blu e]);" }}{PARA 13 "" 1 "" {GLPLOT2D 896 314 314 {PLOTDATA 2 "6%-%'CURVE SG6$7`jl7$$\"\"!F)$\"/++++++5!#97$$\"/sd)Gkg^#!#;$\"/@e!\\!>#)**!#:7$$ \"/bus$y_q%F0$\"/i5>'e$4**F37$$\"/k9t&ys;(F0$\"/dG'=pQs*F37$$\"/\"=!4& )fX'*F0$\"/)pI3]lR*F37$$\"/YGUR@67F3$\"/_i/rd7*)F37$$\"/y\"*eQ*)R9F3$ \"/(Gx,ivJ)F37$$\"/R$pmymn\"F3$\"/r`h-KdvF37$$\"/#3T&=c@>F3$\"/.%Q,-2j 'F37$$\"/cz(pfc;#F3$\"/!=\"eLz)e&F37$$\"/qYLVu;CF3$\"/IpN=jIWF37$$\"/X B<)**yj#F3$\"/Xx%3N:Q$F37$$\"/HU6&oo)GF3$\"/Z^AMm`AF37$$\"/+*\\\\fo8$F 3$\"/!>`\\>mS\"F37$$\"/)f)H@xxLF3$\"/wOyeDZ8F37$$\"/W;#)Qa'f$F3$\"/1v& 390(=F37$$\"/:rbYocQF3$\"/;6@YaWEF37$$\"/ti8!eq2%F3$\"/Q1!\\IJE$F37$$ \"/h.=YSLVF3$\"/(4h5=Q)QF37$$\"/%)>*z9.c%F3$\"/92C[=HVF37$$\"/EEJ7F4[F 3$\"/jJW+t-ZF37$$\"/alFmLY]F3$\"/FI%\\**R&\\F37$$\"/!)**)\\*o$H&F3$\"/ >*H4\\f7&F37$$\"/.5]m$3_&F3$\"/9\"\\))=$G_F37$$\"/#y!Rb%ew&F3$\"/&zP(Q E<`F37$$\"/DB;4M?gF3$\"/fiGaNPaF37$$\"//=!*)z=C'F3$\"/(HeE2'3cF37$$\"/ FI^![6['F3$\"/*f\"4zf+fF37$$\"/%[TZN$GnF3$\"/>W>2&)RjF37$$\"/aoU,;qpF3 $\"/h:.Z4-pF37$$\"/y)Q%z8/sF3$\"/STj$o!\\vF37$$\"/,++/$RY(F3$\"/?S\\ZB Z$)F37$$\"/KbwaO(p(F3$\"/E9bt5)4*F37$$\"/J\\YsgYzF3$\"/t9l'\\n*)*F37$$ \"/%pYLdC<)F3$\"/5I#[y*e5F,7$$\"/=(e8r$>%)F3$\"/2D#4%)*G6F,7$$\"/3O!y( p^')F3$\"/;JDBz(=\"F,7$$\"/:tlga%*))F3$\"/lAqZ0T7F,7$$\"/nI`Y(>8*F3$\" /#)pkf#\\G\"F,7$$\"/;w!QS0Q*F3$\"/%H[D+HK\"F,7$$\"/#Q!>(Q*>'*F3$\"//U? mC`8F,7$$\"/T*p4eZ')*F3$\"/d4#)>8!Q\"F,7$$\"/%*oC]v55F,$\"/Nr3jG09F,7$ $\"/Ns,^1L5F,$\"/)*p'R>&H9F,7$$\"/apn\\je5F,$\"/h.s#49Y\"F,7$$\"/8kg\\ ]\"3\"F,$\"/`f&y0d\\\"F,7$$\"/j;V\"*)e5\"F,$\"/)H#o()pR:F,7$$\"/W>&y\"F,7$$\"/'pZuliA\"F,$\"/u-q@W_=F,7$ $\"/q^Qrl^7F,$\"/Q@WY$[#>F,7$$\"/V:F,7$$\"/S*p\"o^ )H\"F,$\"/OlJlPZ?F,7$$\"/&[UW=EK\"F,$\"/H'p)Q0-@F,7$$\"/I\\EVqY8F,$\"/ H(\\yH-:#F,7$$\"/B7,;qq8F,$\"/2&GQU?>#F,7$$\"/*=%pav$R\"F,$\"/H\\q0AFA F,7$$\"/Ke6tn=9F,$\"/lD\"*>DhAF,7$$\"/&GUjHCW\"F,$\"/eV6%)o\"H#F,7$$\" /2NNyUn9F,$\"/zC7&zQK#F,7$$\"/U4Fs0!\\\"F,$\"/3fI94bBF,7$$\"/DCTp1::F, $\"/Ervnx$R#F,7$$\"/')H*)=-R:F,$\"/$y-9/jV#F,7$$\"/2MI&=Hc\"F,$\"/9(>? 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*)HD\"\\`?F[z7$Fahu$\"/&f$)zUY5#F[z7$Ffhu$\"/A&))3G%[@F[z7$F[iu$\"/')z hHE&=#F[z7$F`iu$\"/5MFg^:AF[z7$Feiu$\"/5x,ySRAF[z7$Fjiu$\"/HK0\"[QD#F[ z7$F_ju$\"/!*f\"4S6E#F[z-Feju6&FgjuF(FhjuF(-%+AXESLABELSG6$Q\"t6\"Q!Fc hcl-%%VIEWG6$;F(F_ju;$!$]\"F)$\"$]#F)" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "These graphs show clearly that the exchange of rotational energy between the " }{TEXT 19 6 "psidot" } {TEXT -1 5 " and " }{TEXT 19 17 "phidot*sin(theta)" }{TEXT -1 33 " deg rees of freedom happens when " }{TEXT 19 5 "theta" }{TEXT -1 66 " beco mes small, and the reversal occurs when the top 'bounces' at " }{TEXT 19 7 "theta=0" }{TEXT -1 113 ". This makes one nervous naturally, whet her the numerics are all in order. Strange behaviour can happen also f or " }{TEXT 19 8 "theta=Pi" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 "We display the rotational kine tic energy to demonstrate the energy dissipation, and also the total e nergy in order to justify the omission of the gravitational torque (ac tually the torque due to the normal force that acts at the contact poi nt T)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "Trot:=t->(I3*(Ps idot(t)+Phidot(t)*cos(Theta(t)))^2+I1*((sin(Theta(t))*Phidot(t))^2+The tadot(t)^2))/2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%TrotGf*6#%\"tG6 \"6$%)operatorG%&arrowGF(,&*&#\"\"\"\"\"#F/*&%#I3GF/),&-%'PsidotG6#9$F /*&-%'PhidotGF7F/-%$cosG6#-%&ThetaGF7F/F/F0F/F/F/*&F.F/*&%#I1GF/,&*&)- %$sinGF>F0F/)F:F0F/F/*$)-%)ThetadotGF7F0F/F/F/F/F/F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 155 "plot([Trot(t),Trot(t)-m*g*a*cos(Th eta(t))],t=0..t_f,title=\"Rotational Energy (red) and Total Energy (bl ue)\",view=[0..t_f,0..1.1*Trot(0)],color=[red,blue]);" }}{PARA 13 "" 1 "" {GLPLOT2D 933 363 363 {PLOTDATA 2 "6'-%'CURVESG6$7U7$$\"\"!F)$\"/ +++++vy!#;7$$\"/++]#HyI\"!#:$\"/L)e;q$GyF,7$$\"/+++&ech#F0$\"/V^k7D*z( F,7$$\"/+]PPf`PF0$\"/s#[/o_y(F,7$$\"/++v*G:*[F0$\"/e0!)R;vxF,7$$\"/++] L)4X(F0$\"/1mo6GixF,7$$\"/++X.u-5!#9$\"/vZNeBGxF,7$$\"/+]Fy:f7FJ$\"/x- %)fx)o(F,7$$\"/+]d'*)o\\\"FJ$\"/=Ev[DNwF,7$$\"/+](>ZIu\"FJ$\"//*)o%z?d (F,7$$\"/+]xOi(*>FJ$\"/$HfkhK\\(F,7$$\"/+]FPQ^AFJ$\"/SP](RAS(F,7$$\"/+ +IrS7DFJ$\"/JG1#*y*H(F,7$$\"/++&o;Bu#FJ$\"/Y\"z?*f'>(F,7$$\"/++!RS6+$F J$\"/$o,C&>vqF,7$$\"/++]o-hKFJ$\"/vGUL_RpF,7$$\"/++5cZ6NFJ$\"/pT8o*Q!o F,7$$\"/+]xq!*QPFJ$\"/,\\GXRrmF,7$$\"/++!4X$4SFJ$\"/W&oM]&4lF,7$$\"/++ g:WQUFJ$\"/&39trbO'F,7$$\"/+]<_$\\]%FJ$\"/]\"HxWX>'F,7$$\"/++gs#3u%FJ$ \"/Y,i#\\)QgF,7$$\"/+]<#Q'**\\FJ$\"/UkE$fe'eF,7$$\"/+]_u3Y_FJ$\"/gV;w= *p&F,7$$\"/++v8B.bFJ$\"/-9C\\.DbF,7$$\"/+]n(p$RdFJ$\"/6`_WEl`F,7$$\"/+ +Dp2%*fFJ$\"/<*pLac>&F,7$$\"/+]xgkeiFJ$\"/F)f&HF@]F,7$$\"/+]-V&*)['FJ$ \"/H8c,&[([F,7$$\"/++&\\$pPnFJ$\"/@P9Su=ZF,7$$\"/++?am%*pFJ$\"/)HW$oGn XF,7$$\"/++:B1YsFJ$\"/D.5,oAWF,7$$\"/+]PsXSF,7$$\"/++]/;h#)FJ$\"/ ul*z3w$RF,7$$\"/+]P/&f\\)FJ$\"/kG*='[YQF,7$$\"/++5zj_()FJ$\"/=]()eYhPF ,7$$\"/+]<3;%**)FJ$\"/yYuaj*o$F,7$$\"/+]Z=iY#*FJ$\"/&[r]&oCOF,7$$\"/++ ]'[M\\*FJ$\"/AA\"eL_d$F,7$$\"/+]PM&=v*FJ$\"/$Qo#HYINF,7$$\"/++'zs++\"! #8$\"/Z3ZBx)\\$F,7$$\"/+]5Q_D5Fix$\"/+zy*f0[$F,7$$\"/+vxSw]5Fix$\"/4b4 w&eY$F,7$$\"/++is&R2\"Fix$\"/O'Rhc$eMF,7$$\"/+]o#R05\"Fix$\"/$*pN]oeMF ,7$$\"/++KXJC6Fix$\"/i*yuu!fMF,7$$\"/+v@Rm\\6Fix$\"/Lc;f0fMF,7$$\"/+DA l#R<\"Fix$\"/+Ly)R'eMF,7$$\"#7!\"\"$\"/KL\"QU$eMF,-%'COLOURG6&%$RGBG$ \"*++++\"!\")F(F(-F$6$7U7$F($\"/+D]#**ep(F,7$F.$\"/S@gv/\\wF,7$F4$\"/; hk1O>wF,7$F9$\"/8S;sJ0wF,7$F>$\"/)o%o8P&f(F,7$FC$\"/rWVC\")Gm'F,7$Fj p$\"/l)y7Pg`'F,7$F_q$\"/#y6BxAQ'F,7$Fdq$\"/u&eo'yWiF,7$Fiq$\"/6zew<$3' F,7$F^r$\"/B)>h_^$fF,7$Fcr$\"/)p!y^qsdF,7$Fhr$\"/k@y,C:cF,7$F]s$\"/Os \"fOHX&F,7$Fbs$\"/y!4$R@.`F,7$Fgs$\"/W)\\lPF,7$F]x$\"/lEuYJFPF,7$Fbx$\"/-\\?Yi$p$F,7$Fgx$\"/bzW: JpOF,7$F]y$\"/b84S,dOF,7$Fby$\"/xDw0XXOF,7$Fgy$\"/bZ;KNQOF,7$F\\z$\"/9 '=$*\\$QOF,7$Faz$\"/Fq.wMQOF,7$Ffz$\"/>s " 0 "" {MPLTEXT 1 0 100 "Ldotr:=t->I3*(Psidot(t)+cos(Theta( t))*Phidot(t))*(a-r*cos(Theta(t)))-I1*r*sin(Theta(t))^2*Phidot(t);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%&LdotrGf*6#%\"tG6\"6$%)operatorG%&ar rowGF(,&*(%#I3G\"\"\",&-%'PsidotG6#9$F/*&-%'PhidotGF3F/-%$cosG6#-%&The taGF3F/F/F/,&%\"aGF/*&%\"rGF/F8F/!\"\"F/F/**%#I1GF/F@F/)-%$sinGF:\"\"# F/F6F/FAF(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "seq(Ldotr (t_f/10*i)/I1,i=0..10);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6-$!/P3Z(3&p; !#8$!/;jZ(3&p;F%$!/D/Z(3&p;F%$!/R?Z(3&p;F%$!/&znu3&p;F%$!/s`Z(3&p;F%$! /O7Z(3&p;F%$!/!ehu3&p;F%$!/p.X(3&p;F%$!/KM_(3&p;F%$!/T`c(3&p;F%" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "The result is negative, because th e angle between " }{TEXT 281 1 "L" }{TEXT -1 5 " and " }{TEXT 282 1 "r " }{TEXT -1 10 " is above " }{TEXT 19 4 "Pi/2" }{TEXT -1 58 ". If the \+ numbers vary, the plot command below can be used:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 49 "#plot(Ldotr(t)/I1,t=0..t_f,title=\"(L dot r) /I1\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "The conservation of th e dot product of " }{TEXT 287 1 "L" }{TEXT -1 5 " and " }{TEXT 288 1 " r" }{TEXT -1 189 " allows one to calculate the required amount of kine tic energy which has to be dissipated by the frictional torque in orde r to put the tippy top upside down. Assuming that we move from the " } {TEXT 19 7 "theta=0" }{TEXT -1 8 " to the " }{TEXT 19 8 "theta=Pi" } {TEXT -1 56 " configuration we find that the dot product is given by \+ " }{TEXT 19 51 "I3*psidot(0)*(r-a)*(-1) = I3*psidot(t_f)*(r+a)*(-1)" } {TEXT -1 54 ". Therefore the kinetic energy changes by a factor of " } {TEXT 19 15 "((R-a)/(R+a))^2" }{TEXT -1 441 ". This is indeed observed in the energy graph above. Of course, for a more realistic tippy-top \+ the transfer of energy to the translational degrees of freedom plays a lso some role. Nevertheless, the knowledge of this quantity can help i n the design of more successful tippy-tops, namely tops that wind up w ith a more substantial fraction of their original rotational kinetic e nergy in the upside-down configuration. Now it seems that sending " } {TEXT 19 1 "a" }{TEXT -1 271 " to zero would give the best results, in that the least amount of kinetic energy would be needed to put the ti ppy-top upside down (and the gravitational potential to overcome would also be minimized!). It turns out that there is a qualitative change \+ in the solution when " }{TEXT 19 3 "a/r" }{TEXT -1 167 " falls below 0 .1, which makes it unlikely for the tippy-top to work (even though the equations display the instability). In this regime the solution demon strates that " }{TEXT 19 6 "psidot" }{TEXT -1 5 " and " }{TEXT 19 6 "p hidot" }{TEXT -1 115 " can exchange rotational energy without energy d issipation, but that energy dissipation is needed to carry out the " } {TEXT 19 5 "theta" }{TEXT -1 173 "-motion, which happens rapidly at a \+ later time in this case. Also note that in this regime the results are very sensitive to the relationship between the moments of inertia " } {TEXT 19 2 "I1" }{TEXT -1 5 " and " }{TEXT 19 2 "I3" }{TEXT -1 11 ", a nd that " }{TEXT 19 1 "a" }{TEXT -1 5 " and " }{TEXT 19 2 "I1" }{TEXT -1 121 " are not independent of each other when the tippy-top is made \+ of homogeneous material using cut-outs as a method to move " }{TEXT 19 1 "a" }{TEXT -1 22 " along the rotor axis." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "T_fin_over_T0:=((r-a)/(r+a))^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.T_fin_over_T0G$\"/WWWWWWW!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 284 11 " Exercise 1:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 223 "Reduce the initial spin rate by, e.g., a factor of three to 50 rad/sec. What do \+ you observe? How would you explain the difference in behaviour? If you have access to a tippy-top perform experiments to support your findin gs." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 283 11 "E xercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 75 "Model a smoother table surf ace, i.e., reduce the kinetic friction constant " }{TEXT 19 1 "f" } {TEXT -1 161 " by a factor of 2 (to 0.15), and then by another factor \+ of 2 (to 0.075). What do you observe? What does this imply about the i mportance of the frictional torque?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT 285 11 "Exercise 3:" }{TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 268 "Remove the torque due to the normal force (the re action force to gravity), i.e., keep only the frictional torque in the Euler equations. Compare the answers for different initial conditions (initial spin rates) with the full calculations and explain your obse rvations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 286 11 "Exercise 4:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "Cha nge the initial condition such that the tippy-top starts with pure pre cession in the " }{TEXT 19 10 "theta=Pi/2" }{TEXT -1 187 " orientation . Describe your results for different values of the precession rate (e .g., 50, 100, 150 rad/s). If you have access to a tippy-top perform ex periments to support your findings." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT 289 11 "Exercise 5:" }{TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 100 "Change the design parameters of the tippy-top (st art with moments of inertia, then proceed with the " }{TEXT 19 13 "((r -a)/(r+a))" }{TEXT -1 88 " ratio), and observe the changes in the nume rical solution. For example: an increase of " }{TEXT 19 2 "I1" }{TEXT -1 112 " by 20% (while the other parameters remain the same) can have \+ the result that the tippy-top will stabilize at a " }{TEXT 19 5 "theta " }{TEXT -1 27 "-value substantially below " }{TEXT 19 2 "Pi" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 290 11 "Exercise 6:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 146 "The pr esent treatment used a normal force in the frictional torque calculati on which ignored the fact that the CM is accelerated up and down when \+ " }{TEXT 19 5 "theta" }{TEXT -1 43 " changes. The height of the CM is \+ given as " }{TEXT 19 27 "z_CM(t) = R-a*cos(theta(t))" }{TEXT -1 36 ". \+ The second derivative is given as " }{TEXT 19 85 "diff(z_CM(t),t$2) = \+ a*(thetadot(t)^2*cos(theta(t))+diff(thetadot(t),t)*sin(theta(t)))" } {TEXT -1 30 ", and the normal force equals " }{TEXT 19 23 "m*g+m*diff( z_CM(t),t$2)" }{TEXT -1 129 ". Include this correction and find out to what extent it changes the results. What is your expectation before y ou start the work?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "0 0 0" 9 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }