Physics 106a, Caltech 4 December, Lecture 18: Examples on Rigid Body Dynamics. Rotating rectangle. Heavy symmetric top

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1 Physics 106a, Caltech 4 December, 2018 Lecture 18: Examples on Rigid Body Dynamics I go through a number of examples illustrating the methods of solving rigid body dynamics. In most cases, the problem can be worked a variety of ways: Newtonian rate of change of angular momentum equals torque ; Lagrangian approach; Euler equations in body frame; Euler angles etc. Rotating rectangle Even in the absence of external torques, rotation of a body about an axis ω that is not aligned with a principal axis (so that the inertia tensor is non-diagonal and L = T ω is not aligned with ω) will result in internal torques which will cause the rotation axis to precess in space. Hand and Finch problem 8-15: A thin rectangular sheet of dimensions a b and mass M is rotating about a diagonal with constant angular velocity ω. What is the torque? Heavy symmetric top ψ θ l Mg Forces φ-π/2 Left: Heavy symmetric top with torque. Right: Tidal forces and torques acting on the equatorial bulge of the earth from the gravitational pull of the sun or moon. As an application of the Euler angles consider the symmetric top. For the symmetric top θ and φ π 2 give the usual spherical polar coordinates of the symmetry axis and ψ a further rotation about the symmetry axis. Below, will study the motion under a torque tending to rotate the symmetry axis so that there is a potential energy V (θ). One example is gyroscopes and toy tops where gravity acts on the center of mass a distance l from the pivot point so that V (θ) = Mgl cos θ. (1) Thus, ψ is the spin along the symmetry axis, φ is the precession around the vertical due to the torque, and θ is the nutation of the spin axis as it precesses. 1

2 Precession of the equinoxes Another example is the slow precession of the equinoxes of the Earth, due to the average gravitational effect of the moon and the sun acting on the equatorial bulge tending to rotate the symmetry axis of the Earth towards the perpendicular to the orbital plane. The Earth s spin axis, closely (but not exactly) aligned with the principal body axis, is at an inclination angle of θ = 23.4 with respect to the normal to the Ecliptic plane (the plane of the Earth s orbit around the sun. The moon s orbit is almost on the Ecliptic plane, as well). Left: Tidal gravitational forces from the Sun (and moon) on the equatorial bulges of the Earth. Right: The zenith of the North pole (spin axis) relative to the fixed stars and the Ecliptic pole, over the 25,000 years. If the Earth were a perfect sphere, the gravitational torques would cancel and the net force would be the same as the force on the center of mass: F = GM M /R 2, where R is the distance from the center of the Sun to the center of the Earth. Because of the equatorial bulge, and the finite size of the Earth R, there is a net tidal force that isn t aligned with R, roughly: F ( F/ R) R (GM M /R 3 ) R. It s simpler to work with the potential. The gravitational potential energy is ρ( r)d r V = GM R r, (2) where R is the vector between the center of mass of the Earth and the center of the Sun (or moon), r is a vector of integration relative to the center of mass of the Earth, ρ( r) is the mass density of the Earth at r, and the integral d r is over the Earth s volume. Again, if the Earth were a uniform sphere, this integral would be constant, with no dependence on the inclination angle θ. Because of the equatorial bulge, inclined at an angle θ with respect to the Ecliptic plane, there is an additional non-constant term. With a bit more math (see Hand and Finch 8.11), the potential energy takes the form V (θ) = 1 2 ( GM R 3 + GM ) ( 1 3 cos 2 θ moon R 3 (I 3 I ) sun 2 ). (3) The gravitational forces act tidally on the quadrupole moment of the mass distribution of the Earth, and the quantity I 3 I appears in evaluating this quadrupole moment. The quantity GM/R 3 sun = Ω 2 e with 2π/Ω e the orbital period of the Earth (i.e. one year). The additional contribution from the moon is about twice as large. We write the sum as Ω 2 with Ω = π/year = s 1. Then V (θ) = 1 2 I Ω 2 ɛ 1 2 (1 3 cos2 θ) with ɛ = I 3 /I (4) 2

3 Lagrangian analysis In the above two examples, we have a potential V (θ) which depends only on the inclination angle θ resulting in a Lagrangian which depends on ψ, φ, θ, θ) but not φ or ψ. To greatly simplify the math, we will assume, in both cases, that the rigid body has an axis of symmetry, so that the inertia tensor in the body frame is diagonal, with diagonal elements I 1 = I 2 = I and I 3 I. The Lagrangian is L = 1 2 I ( θ 2 + φ 2 sin 2 θ) I 3( ψ + φ cos θ) 2 V (θ). The angles φ, ψ are ignorable so the two conjugate (angular) momenta are constants of the motion p ψ = L ψ = I 3( ψ + φ cos θ) = I 3 ω 3, (5) p φ = L φ = I φ sin 2 θ + I 3 ( ψ + φ cos θ) cos θ. (6) We see that p ψ = I 3 ω 3 is the component of the angular momentum along the symmetry axis in the body frame. Also, p φ is the component of the angular momentum along gravity (for the heavy top) or normal to the Ecliptic plane (for the Earth) in the inertial space frame. Both are conserved. The Euler-Lagrange equation for θ is (remember we must find this before using the constant p φ, p ψ to eliminate ψ, φ) I θ I φ2 sin θ cos θ + I 3 ( ψ + φ cos θ) sin θ φ + V/ θ = 0, (7) and using Eq. (5) in the third term gives I θ I φ2 sin θ cos θ + p ψ sin θ φ + V/ θ = 0, (8) Now we use Eq. (6) to evaluate φ in terms of constants and θ φ = p φ p ψ cos θ sin 2 θ. (9) so that Eq. (8) is the single equation of motion for θ that we must solve. Steady precession Steady precession is given by θ = 0, φ = ω p with ω p the constant precession rate solving (from Eq. (8)) I sin θ cos θ ω 2 p I 3 ω 3 sin θ ω p V/ θ = 0, (10) writing the more intuitive I 3 ω 3 for p ψ. This is a quadratic equation giving two possible steady precession rates for a given θ. (Of course, the top will need to be started with carefully chosen initial conditions to gives this motion, or some dissipation mechanism must damp out other motion see below.) Often we are interested in the case where the top is rapidly spinning compared with the applied torque. The appropriate comparison is the range of the potential energy which we can estimate as V/ θ to eliminate constant parts, and the kinetic energy of the spin 1 2 I 3ω 2 3. For the rapidly spinning top the quadratic equation gives the solutions ω p I 3ω 3 I cos θ, ω 1 p I 3 ω 3 sin θ V θ. (11) 3

4 The first value is ω p ω 3 and is independent of the applied torque in the approximation used. This is just a special case of the Chandler wobble (from the space frame perspective) where the amplitude of the motion is the same as the tilt angle. The second expression is the slow precession ω p ω 3 that is the familiar gyroscopic motion. For the spinning top ω p Mgl/I 3 ω 3, as found from simple arguments. The precession rate of the equinoxes is given using Eq. (4) and is ω p = 3 2 ɛω2 cos θ ω 3, (12) (using I I 3 ). Putting in ɛ = , Ω/ω /365, and θ = 23.4 gives ω p /Ω e = (24891) 1 so that the predicted precession period is years compared with the measured value of years (a 3% difference). Motion from given initial condition V eff E 1 " Oscillations of! (nutation)! E 2 " Steady precession Figure 1: Effective potential for heavy top. More generally we look for the solution from given initial conditions ψ, φ, θ, θ. Use Eqs. (5,6) to fix the constant angular momenta. The Lagrangian is a quadratic form in the angular velocities φ, θ, ψ and has no explicit time dependence so H = T + V = E with E the constant energy. It is convenient to use the constant E and E = E 1 2I 3 p 2 ψ = I 2 θ 2 + V eff (θ) (13) V eff (θ) = 1 (p φ p ψ cos θ) 2 2I sin 2 + Mgl cos θ. (14) θ Remember p φ, p ψ and E (and so E ) are constants of the motion set by the initial conditions. Therefore we have reduced the problem to a one dimensional problem of the dynamics of θ in the effective potential V eff (θ). We can try to integrate the equation for θ or can proceed qualitatively as in Lecture For a fuller understanding, it is useful to introduce u = cos θ and perform the analysis in terms of u. The effective potential for u is a cubic polynomial, which simplifies the work. For this approach see Hand and Finch For simple results on precession and nutation, a direct analysis in terms of θ is sufficient. 4

5 An obvious feature of V eff (θ) is that it diverges positively at the ends of the range θ = 0, π and is bounded below by a finite value Mgl. It therefore has a minimum somewhere between (and maybe more than one). Figure 1 shows a sketch of V eff for some choice of parameters, and examples of two energies E giving oscillations of θ called nutation as well as precession φ 0 (red line), or just steady precession φ 0, θ = 0 (green dot). Since φ is given by Eq. (9), it may or may not change sign during the variation of θ depending on p φ /p ψ and the range of θ, and so the precession may be monotonic, or involve backwards and forwards motion (see Fig. 2). For the common initial condition of θ = φ = 0 the motion involves cusps (third panel). Figure 2: Nutation motion of heavy top. The third panel corresponds to a top released from rest φ = θ = 0. We can investigate the small periodic nutational motion of θ near the uniformly precessing solution by expanding V eff about the minimum to quadratic order. For the large spin case the leading order term is V eff p2 ψ /I. Together with the θ 2 term in the energy E, this gives simple harmonic motion of θ at the frequency p ψ /I = I 3 ω 3 /I. For the Earth example this is the Chandler wobble in the space frame at a frequency close to 2π/day. There are very small O(ε) corrections from the gravitational torque. For given p φ, p ψ and E Eq. (13) for θ can in principal be integrated, although only numerically in general. Some special cases and limits can be investigated analytically. A simple example is an initial condition of θ = θ = 0, i.e. the top spinning about a vertical axis. For this case p φ = p ψ and the effective potential reduces to V eff (θ) = p2 ψ (1 cos θ) 2 2I sin 2 + Mgl cos θ. (15) θ Expanding for small θ V eff (θ) = Mgl + θ2 2 ( ) p 2 ψ Mgl. (16) 4I The θ = 0 equilibrium is stable for p 2 ψ > 4I Mgl and unstable for p 2 ψ > 4I Mgl. A fast spinning top will rotate smoothly about the vertical (called a sleeping top ). Friction will gradually decrease p ψ, and when it reaches the critical value 2 I Mgl the top will begin to wobble and precess. 5

6 Euler s Disk Here s a video of the Euler s Disk in acton: Here is a link to a nice paper on the dynamics: ajw/ph106/files/rollingdisk.pdf In practice, we rest the disk on the horizontal surface, nearly edge on, at some angle slightly less than θ = π/2, then give it a hard spin along ω 2 (perpendicular to the symmetry axis of the disk). Initially, φ = ψ = 0 (it isn t yet rolling and precessing), and θ = 0. Then, torques get it rolling and precessing, and friction causes θ to tend towards zero. Lagrangian approach In this approach I will first look at the general motion, and then specialize to the solution with the center of gravity position constant. The Lagrangian in terms of the Euler angles and the center of mass coordinates is L = 1 2 I ( θ 2 + sin 2 θ φ 2 ) I 3( ψ + φ cos θ) M(Ẋ2 + Ẏ 2 + Ż2 ) MgZ. (17) with I = 1 4 MR2, I 3 = 1 2 MR2 the moments of inertia about the center of the disk. The coin is instantaneously rotating about the contact point. The center of mass motion is therefore R = ω ( R R c ) (18) with R c the position of the contact point. Evaluate this using the geometry of the figure: this gives The components of ω in the space frame are Equation (18) then gives R R c = R( cos θ sin φ, cos θ cos φ, sin θ). (19) ω = ( ψ sin φ sin θ + θ cos φ, ψ cos φ sin θ + θ sin φ, ψ cos θ + φ). (20) Ẋ = Rω 3 cos φ + R sin θ sin φ θ, (21) Ẏ = Rω 3 sin φ R sin θ cos φ θ, (22) with ω 3 = ψ + φ cos θ the component of the angular velocity perpendicular to the plane (i.e. along the ˆk body axis. These are nonholonomic constraints that we will deal with using Lagrange 6

7 multipliers. The third constraint Z = R sin θ is obvious from the geometry of the figure, and is holonomic. Let s use it to eliminate Z from the Lagrangian L = 1 2 I ( θ 2 + sin 2 θ φ 2 ) I 3( ψ + φ cos θ) M(Ẋ2 + Ẏ 2 + R 2 cos 2 θ θ 2 ) MgR sin θ. (23) First, write the rolling constraints in differential form to be included in the Euler-Lagrange equations with Lagrange multipliers δx + R cos φ(cos θ δφ + δψ) R sin θ sin φ δθ = 0, (24) δy + R sin φ(cos θ δφ + δψ) + R sin θ cos φ δθ = 0. (25) The Euler-Lagange equations for θ, φ, ψ, X, Y respectively are d ( I θ + MR 2 cos 2 θ dt θ ) I φ2 sin θ cos θ + I 3 ω 3 cos θ φ + MgR cos θ λ X R sin θ sin φ + λ Y R sin θ cos φ = 0 (26) ( ) I φ sin 2 θ + I 3 ω 3 cos θ + λ X R cos φ cos θ + λ Y R sin φ cos θ = 0 (27) d dt d dt (I 3ω 3 ) + λ X R cos φ + λ Y R sin φ = 0 (28) MẌ + λ X = 0, (29) MŸ + λ Y = 0. For Ẋ = Ẏ = 0 we have λ X = λ Y = 0 and then from Eq. (28) I 3 ω 3 = constant. Assuming θ = 0 we see the constant is zero from the rolling constraint simplified to the case of Ẋ = Ẏ = θ = 0. Thus ω 3 = 0, confirming that ω lies in the plane of the disk if the center of mass is stationary. Explicitly we have ψ + φ cos θ = 0. (31) Equation (27) then tells us φ = Ω is constant, and then Eq. (26) gives Ω 2 = MgR I sin θ = (30) 4g R sin θ. (32) The rotation rate of the rolling motion Ω diverges proportional to sin 1/2 θ as θ 0. The rotation rate of the face on the coin is the space frame component ω z = ω 3. This is ω z = φ + ψ cos θ = Ω sin 2 θ, (33) using Eq. (31). Thus the rate of rotation of the face is proportional to sin 3/2 θ, and goes to zero as θ 0. 7

8 Ball rotating on rotating turntable What is the motion of a solid ball rolling, without slipping, on a rotating turntable? Remarkably, viewed from the space frame, the motion is a circle and the period is always 7/2 times that of the turntable! You can derive this from Newton s law of motion for the horizontal motion of the center of mass and the equation for the angular momentum about the center of mass (only the frictional force maintaining the rolling constraint contributes to both of these the vertical forces give no moment about the center of mass, for example). For movies demonstrating the motion in various cases (center of circle on and off the center of turntable... ) see the references for movies. For details of the calculation see the paper by J. A. Burns, Am. J. Phys. 49, 56 (1981). Top on smooth table Consider the standard heavy symmetric top (e.g. Hand and Finch Fig. 8.14) but when the tip of the top is free to slide on a frictionless, flat surface, rather than being held fixed. The principle components of the moment of inertia of the top about its center of mass are I 3 along the symmetry axis ˆk and I 1 perpendicular to this, and the mass is M with the center of mass a distance h from the tip along the symmetry axis. We use as generalized coordinates the horizontal coordinates of the center of mass X, Y and the Euler angles θ, φ, ψ. The vertical coordinate of the center of mass can be calculated in terms of θ. The expression for the kinetic energy simplifies for rotations about a fixed point, or using translation of and rotation about, the center of mass. For a top with a fixed pivot point we could use either method; for the top on a frictionless table we must use the center of mass method. Since the height of the center of mass above the table is Z = l cos θ the center of mass kinetic energy is T cm = 1 2 M[Ẋ2 + Ẏ 2 + l 2 sin 2 θ θ 2 ]. (34) For the rotational kinetic energy we need the moment of inertia calculated with respect to the center of mass. Using the displaced axis theorem, the components are I 3,cm = I 3 along the top axis and I 1,cm = I 1 Ml 2 perpendicular to this axis, with I 1, I 3 the moment of inertia components about the tip. The rotational kinetic energy is therefore The potential energy is T rot = 1 2 (I 1 Ml 2 )( θ 2 + sin 2 θ φ 2 ) I 3( ψ + φ cos θ) 2. (35) V = Mgl cos θ. (36) The Lagrangian is L = T cm + T rot V. Note that we could also use T cm + T rot to evaluate the kinetic energy for the fixed-tip case, but in that case the kinetic energy of the center of mass would be T cm = 1 2 Ml2 [ θ 2 + sin 2 θ φ 2 ]. (37) Now T cm + T rot just gives the same expression as the kinetic energy for rotation about a fixed tip, as of course it must. The coordinates X, Y are ignorable, so MẊ and MẎ are constants of the motion. This just leaves the angular part of the Lagrangian L rot = 1 2 (I 1 Ml 2 ) θ I 1 sin 2 θ φ I 3( ψ + φ cos θ) 2 Mgl cos θ. (38) This differs from the fixed tip result only in the replacement I 1 I 1 Ml 2 in the θ 2 term. Since the θ 2 only affects the nutational motion, the frequency of uniform precession is the same as in the fixed-tip case calculated in class ω p = Mgl/I 3 ω 3 with ω 3 ψ for large spin. The tip traces out a circle of radius l sin θ about the center of mass, which may be in uniform translational motion or at rest. 8