archive old C4 and C5

Archive of old C.4 and C.5 on Sun Jan 29 PhL 1.29.17 C.4 The Spherical Pendulum We now digress for a subsection on the spherical pendulum, then in the Section C.5 we resume discussion of the use of such a pendulum in its "Foucault pendulum" mode. (a) The equations of motion If we momentarily turn off the rotation of the Earth by setting ω = 0, equations (C.3.9) become 2 + sin2θ 2 = -(g/l)cosθ + T/(ml) - sinθcosθ 2 = - (g/l)sinθ 2cosθ + sinθ = 0 . (C.4.1) These are the equations of motion for a spherical pendulum in the presence of a uniform gravitational field of strength g. It is useful to know something about the solution of these equations before we turn the Earth's rotation back on. Operationally, in the sense of a numerical solution, one can regard the last two equations of (C.4.1) as a pair of coupled second-order non-linear ODE's for functions θ(t) and φ(t) subject to initial conditions θ0, φ0, 0 and 0. Once these equations are solved (physically we know a unique solution must exist), the first equation may be used to determine T(t), the string tension. In practice, there are some difficulties with numerical integration since θ = 0 is a singular point in spherical coordinates (φ is undefined there). This is discussed a bit in Section F.7 for the dumbbell satellite which has similar equations of motion. (b) Lz as constant of the motion The last equation in (C.4.1) may be written in this form d/dt ( sin2θ ) = 0 (C.4.2) which says that sin2θ must be a "constant of the motion". To understand the meaning of this constant, we first compute the torque about the pivot point due to the gravitational force on the mass m, N = r x mg = r x mg = mgl x [cosθ - sinθ ] = -mgl sinθ . (C.4.3) Since this torque lies in a plane normal to the z axis, we may conclude that the Cartesian torque component Nz = 0. The angular version of Newton's Law says N = dL/dt , and therefore we expect that the quantity Lz will be a constant of the motion. Direct calculation shows that Lz = L = m (r x v) = m ( x r) v = m lsinθ (l [ + sinθ ]) = ml2sin2θ (C.4.4) where we have made use of (C.1.4) for and (C.3.7) for v. Thus, we see that our third equation of motion in (C.4.1), rewritten as in (C.4.2), is just the statement that dLz/dt = 0. In the Lagrangian formulation one finds that Lz is one of the canonical momenta which is constant because φ is a cyclic coordinate, meaning it does not appear in the Lagrangian. In terms of the spherical unit vectors, one finds (using (C.3.7) for v) that the vector angular momentum of the pendulum mass is given by L = m r x v = ml2 [ - sinθ ] . (C.4.5) Then application of N = dL/dt, with N as in (C.4.3) and unit-vector change rates as in (C.3.6), simply reproduces the last two equations of motion in (C.4.1). So we conclude that h ≡ Lz/(ml2) = sin2θ (C.4.6) is a constant of the motion of the spherical pendulum. (c) E as another constant of the motion We have shown in (C.4.6) that = h/sin2θ. It this is substituted into the second equation of (C.4.1) one obtains - h2cosθ /sin3θ + (g/l)sinθ = 0 (C.4.7) and we just put this equation on hold for a moment, noticing that it is a second order ODE. Another constant of the motion is the total energy E which may be regarded as E = T + V = kinetic energy + potential energy, E = (1/2)mv2 - mglcosθ = (1/2) m l2(2+ sin2θ2) – mglcosθ (C.4.8) where v2 = l2(2+ sin2θ 2) according to (C.3.7) for v. By rescaling the energy, we can take this constant of the motion to be E = 2/2 + sin2θ2/2 – (g/l)cosθ . E = m l2E (C.4.9) Using (C.4.6) to eliminate gives, E = 2/2 + (h2/2sin2θ) – (g/l)cosθ or (C.4.10) E = 2/2 + Ve(θ) where Ve(θ) ≡ (h2/2sin2θ) – (g/l)cosθ . Unlike (C.4.7), equation (C.4.10) is a first-order ODE so we prefer it to (C.4.7). In fact, if one multiplies (C.4.7) by and notes that = (1/2) ∂t(2) and does a few easy integrals, one obtains an equation of the form ∂t[stuff] = 0 so then stuff = constant. That "stuff" is the right side of (C.4.10) and we have then provided an interpretation for the constant (energy) [ is an "integrating factor" for (C.4.7) ]. (d) Analytic solution to the spherical pendulum problem (outline) Equation (C.4.10) is a first order non-linear ODE for θ(t) which we know how to solve: = (2E - 2Ve(θ) )1/2 => dθ = (2E - 2Ve(θ) )1/2dt => dt = (2E - 2Ve(θ) )-1/2 dθ => t(θ) = !Syntax Error, Idθ' . (C.4.11) Letting z' = cosθ', so dz' = - dθ', this integral can be written as t(θ) = !Syntax Error, Idz' // z = cosθ, z0 = cosθ0 = !Syntax Error, Idz' = !Syntax Error, Idx where a,b,c are the roots of the cubic equation x3 - (El/g) x2 - x + (l/g) (E-h2/2 ) = 0. The dimensionless integral appearing on the last line can be evaluated in closed form using this indefinite integral, . (C.4.12) Here EllipticF is the incomplete elliptic integral of the first kind, EllipticF(sin(φ),k) = !Syntax Error, Idt = F(φ,k) = F(φ|m) = F(φ\α) = sn-1(sinφ,k) (C.4.13) where k = sinα , m = k2 and sn-1 is the inverse Jacobi sn function. Therefore, we have a closed form result for t(θ) which can then be "inverted" to obtain a solution for θ(t). Then from (C.4.6) we get = h/sin2(θ(t)) => φ(t) = φ0 + h !Syntax Error, Idt' /sin2(θ(t')) (C.4.14) and the problem of the spherical pendulum is completely solved more or less in closed form. (e) The nature of the general solution Now recall (C.4.10), which one regards as kinetic plus potential energy for the pendulum mass, E = 2/2 + Ve(θ) where Ve(θ) ≡ (h2/2sin2θ) – (g/l)cosθ . (C.4.10) The effective potential Ve(θ) has the following shape (plotted here for h2/2 = .3 and (g/l) = 1) (C.4.15) Since 2/2 = (E - Ve(θ)) must be positive, one must have Ve(θ) ≤ E which is valid only between θmin and θmax as shown. These are the "turning points" where = 0. Differentiation of (C.4.10) tells us that, 0 = + Ve'(θ) // Ve'(θ) ≡ dVe(θ)/dθ or = – Ve'(θ) . (C.4.16) At the right turning point the slope Ve'(θ) is positive so < 0 which accelerates the particle to the left. At the left turning point the slope Ve'(θ) is negative so > 0 which accelerates the particle to the right. The particle therefore bounces back and forth between these two turning points in some manner. Thus the motion of the pendulum is constrained on the spherical surface r = l between two horizontal circles of angles θmin and θmax and hits both these angles once per "oscillation". Meanwhile, from (C.4.14) the action in the φ dimension of the problem is controlled by = h/sin2(θ(t)) => φ(t) = φ0 + h !Syntax Error, Idt' /sin2(θ(t')) (C.4.14) so as θ(t) does the oscillation just discussed between θmin and θmax, φ(t) increases as shown in its own complicated functional manner. As we shall see in Section C.8, if the spherical pendulum is started in some narrow elliptical orbit, that orbit precesses clockwise roughly in proportion to the area of that ellipse. This "intrinsic apsidal precession" occurs even if the Earth is not rotating. In order to use a spherical pendulum as a "Foucault" pendulum (to be discussed below), one must launch the spherical pendulum with a sufficiently small elliptical area (ideally that area is zero) so the Foucault precession is not masked by the intrinsic precession. Zero area corresponds to h = 0 and Lz = 0 so = 0 in (C.4.14). The Foucault/Intrinsic signal ratio is improved by making the pendulum very long (67 m in Paris). In theory, perhaps by using the "burned string" launch method, one can have h = 0 so = 0 exactly and then the spherical pendulum becomes a plane pendulum moving in the plane φ = φ0. Schumacher and Tarbet (2009) discuss this subject and propose an active electronic device to neutralize the intrinsic precession so that one can have a much shorter length Foucault pendulum. A quick tour of web animations of the spherical pendulum shows the amazing complexity of the possible motions. If the string is replaced with a massless stiff rod, over the top motions are included. Some examples (search youtube if these are dead links) : http://www.youtube.com/watch?v=6hCLkTENfSA . http://www.youtube.com/watch?v=VS1dU5HpfOM&feature=relmfu Both animations demonstrate the θmin ≤ θ ≤ θmax idea, though it takes a while in the first animation. (f) The spherical pendulum for small θ is still complicated unless h is very small Setting sinθ = θ and cosθ = 1, the equations of motion (C.4.1) become 2 + θ2 2 = -(g/l) + T/(ml) - θ 2 = - (g/l) θ 2 + θ = 0 . (C.4.17) Since in general 2 is not small relative to (g/l), one cannot trivially solve the second equation to find simple harmonic motion for θ. If one assumes that 2 << (g/l) ( meaning very small h) and then neglects the 2 term in the second equation, one obtains θ(t) ≈ θ0 cos(Ωt) where Ω = which is the usual angular frequency for a small-angle plane pendulum. (g) The Conical Motion solution The spherical pendulum has an obvious simple solution where θ = θ0 = constant, so the string motion traces out a cone. In this case the equations of motion (C.4.1) become sin2θ0 2 = -(g/l)cosθ0 + T/(ml) cosθ0 2 = (g/l) = 0 . (C.4.18) The second equation says ωφ ≡ = . Since this is a constant, the third equation is satisfied as well, and then the first says T = mg secθ0. In the small angle limit for θ0, ωφ slows down to its smallest possible value ωφ ≡ = Ω which is the frequency of a small-angle plane pendulum. Conversely, as we try to achieve θ0 → π/2, secθ0→ ∞ and both ωφ and tension T become infinite, which seems pretty reasonable. This solution can of course be obtained by elementary methods as well. C.5 The Foucault Pendulum The equations of motion for the spherical pendulum on the rotating Earth were stated in (C.3.9), 2 + sin2θ 2 = -(g/l)cosθ + T/(ml) + 2ω [(cosβsinθ + sinβcosφcosθ )sinθ + sinβsinφ ] - sinθcosθ 2 = - (g/l) sinθ + 2ω [(-cosβcosθ + sinβcosφsinθ)sinθ ] 2cosθ + sinθ = – 2ω [(-cosβcosθ + sinβcosφsinθ) ] . (C.3.9) Recall that the first equation serves only to determine string tension T when θ(t) and φ(t) are known, so we ignore this equation for the time being. We launch the pendulum as a "plane pendulum" with (t=0) = 0 = 0, but right away on the first half-swing the Coriolis deflection generates a small ≠ 0. It is then assumed that remains small enough that we may neglect it in the second equation. In the third equation, we neglect which we expect to be smaller than . We will check these assumptions after the fact. The equations of motion then become, = - (g/l) sinθ = ω (cosβ - sinβcosφtanθ) . (C.5.1) At this point, we have two options, both of which involve approximations. Option 1. (arm waving) We imagine that the pendulum is launched at t = 0 with θ = 0 and some velocity so it reaches a maximum angle θ0 < π and then swings back and forth. The first equation in (C.5.1) is decoupled and is the usual equation which describes a large-swing plane (simple) pendulum. The exact solution θ(t) can be expressed in terms of the Jacobi sn function (see e.g. Lundmark's pendulum section), sin(θ/2) = k sn(Ωt,k) k = sin(θ0/2) Ω = θ(0) = 0 (C.5.2a) or θ(t) = 2 sin-1 [ k sn(Ωt, k)] . (C.5.2b) If θ0 happens to be very small, then k ≈ θ0/2 and (C.5.2a) says (θ/2) ≈ (θ0/2) sn(Ωt, 0) = (θ0/2) sin(Ωt) or θ(t) = θ0 sin(Ωt) which is the expected solution. The quarter period of sn(x;k) is not π/2 but rather K(k) where K is a complete elliptic integral. If we define Ωt' = Ωt - K(k), then at t' = 0 the pendulum is at θ0 instead of 0. The solution for this initial condition is then sin(θ/2) = k sn(Ωt + K(k), k) k = sin(θ0/2) Ω = θ(0) = θ0 (C.5.3a) or θ(t) = 2 sin-1 [ ksn(Ωt + K(k),k)] . (C.5.3b) [ Some sources use sn(z,k) = sn(z | k2) ] The second equation in (C.5.1) is not decoupled, since both θ and φ appear, but we then make the following wobbly argument. The pendulum follows some path as suggested in Fig (C.2.1) which winds around the axis so that angle φ(t) monotonically increases in some non-linear but oscillatory manner. We conjecture that the long term average (many swings) of cosφ is therefore 0. Going even further out on this limb, we might conjecture that the long term time average of the product cosφ tanθ also vanishes as well. This would take some effort to study, so it is just an arm-waving conjecture which is the basis of Option 1. Then one finds from (C.5.1) that <> = ω (cosβ - sinβ <cosφ tanθ>) = ω cosβ (C.5.4) so the effective version of (C.5.1) is this + (g/l) sinθ = 0 <> = ω cosβ (C.5.5) which says that the Foucault pendulum precesses on average at ωcosβ even if it has a large swing angle. Why is sign different from that in (C.5) ?? Option 2. Here we are on firmer ground by assuming that θ0 and therefore θ(t) are small angles. In this case, setting sinθ = tanθ = θ and cosθ = 1, (C.5.1) becomes = - (g/l)θ = ω (cosβ - sinβcosφ θ ) . (C.5.6) Unless we are extremely close to the equator of the Earth where cosβ = 0, the second term in parentheses is much smaller than the first, and we then have + (g/l)θ = 0 = ω cosβ . (C.5.7) The solution of the first equation is θ(t) = θ0cos(Ωt) where Ω = , the usual frequency of the small angle plane pendulum. The second equation then gives the precession rate of the Foucault pendulum. This is the main result of this Appendix. If the pendulum is located at the North Pole, since the Earth rotates counterclockwise when viewed from above the North Pole, we expect the Foucault pendulum to rotate clockwise in non-inertial Frame S. Equation (C.5.6) in this case says = ω. Recall that ω > 0 for the Earth in Fig (C.1.1) so > 0. According to the left side of Fig (C.1.3), > 0 does in fact imply clockwise rotation. If the pendulum is located at the South Pole, since the Earth rotates clockwise when viewed from beneath the South Pole, we expect the Foucault pendulum to rotate counterclockwise in non-inertial Frame S, so we expect < 0. This is the case since cos(π) = -1. In terms of the period of the precession, = ω cosβ may be written as (ignoring direction) (2π/Tprec) = (2π/Tsday) |cosβ| = (2π/Tsday) |sin(θLAT)| => Tprec = Tsday/ |sin(θLAT)| (C.5.8) and this is the fundamental Foucault precession result. The sidereal day (rotation relative to the stars) is 23.93447 hours. The Foucault pendulum at the Pantheon in Paris should have this precession period : Tprec = 23.9344696/ sin(48.846285°) = 31.787731 hours = 31 hours 47.26 minutes (C.5.9) http://www.thegpscoordinates.com/france/paris/pantheon/ // latitude Swing Period = 2π/Ω = 2π/ ≈ 16.4 sec l = 67 m (C.5.10) Justification of assumptions In the second of equations (C.2.9) we neglected both terms. Was this justified? | sinθcosθ 2| ≈ θ 2 ≈ θω2 cosβ ~ θ ω2 |– 2ω [(-cosβcosθ + sinβcosφsinθ)sinθ ] | ≈ |2ω cosβ θ ω (cosβ)| ~ θ ω2 . Both these terms are far smaller than the non-neglected term | (g/l) sinθ | ~ Ω2θ since ω2 << Ω2. For a typical Foucault pendulum Ω = 2π/TΩ ω = 2π/Tday Tday2 >> TΩ2 ? (24*3600)2 >> (16.4)2 ? // Pantheon 7.46 x 109 >> 269 . yes