Question 1: Spherical Pendulum

Transcription

1 Question 1: Spherical Pendulum Consider a two-dimensional pendulum of length l with mass M at its end. It is easiest to use spherical coordinates centered at the pivot since the magnitude of the position vector is constant: r = (lˆr) (lˆr) = l2 (ˆr ˆr) = l. In other words, the mass is restricted to move along the surface of sphere of radius r = l that is centered on the pivot. The motion of the pendulum can therefore be described by the polar angle, the azimuthal angle φ, and their rates of change. (a) The agrangian for a spherical pendulum et s assume that the mass is on bottom half of the sphere, so that the mass has a Cartesian coordinate z = l cos. Since gravity is the only external, non-constraint force acting on the mass, with potential energy U = Mgz = Mgl cos, the agrangian () can be first written as: = T U = 1 2 M v 2 + Mgl cos In spherical coordinates, v = ṙˆr + r ˆ + r φ sin ˆφ, and for the problem under consideration we note that ṙ = l = 0. So we can write as an explicit function of the spherical coordinates, = 1 2 M( l l 2 φ2 sin 2 ) + Mgl cos (1) There are two equations of motion for the spherical pendulum, since in Equation 1 is a function of both and φ; we therefore use the Euler-agrange equation for both coordinates to obtain them. For and φ, ( d dt ( d dt φ ) = 0 = Ml2 Ml 2 2 sin cos + Mgl sin = 0 (2) ) φ = 0 = d ( Ml 2 φ sin 2 ) = 0 (3) dt Equation 3 suggests that the quantity Ml 2 φ sin 2 is a constant in time. This quantity has the same form as the z-component of the angular momentum vector for the mass ( z ), so Equation 3 states that angular momentum is conserved in the z direction for our spherical pendulum. (b) Energy in terms of and z For agrangians that don t explicitly depend on time, and for potentials that only depend on the coordinates (and not explicitly in their rates of change), the total mechanical energy E = T + U is another constant of the motion. Equation 1 satisfies both of these conditions, so E is constant for the spherical pendulum. We can eliminate the φ term in T since, according to Equation 3, z = Ml 2 φ sin 2 is constant in time. So the expression for E becomes E = 1 2 M( l l 2 φ2 sin 2 ) Mgl cos = 1 [ ( ) 2 2 M l l 2 z Ml 2 sin 2 sin ] 2 Mgl cos = 1 2 Ml z Ml 2 sin 2 Mgl cos = E (4) 1

2 Notice that E was re-written as a function of only one coordinate (); one can therefore talk about an equivalent one-body problem, where E = T eff + U eff, T eff = Ml 2 2 /2, and 2 z U eff = Ml 2 sin 2 Mgl cos (5) Given the one-body form of Equation 4, we can technically solve for the solution for as a function of time (or vice versa). Noting that = d/dt and solving for it in Equation 4, we get d 2[E Ueff ()] Ml 2 = d Ml 2[E Ueff ()] t t 0 = t t 0 l M 0 d 2[E Ueff ()] (6) For the full solution of the spherical pendulum, we also need to find the solution for φ. This can be done using the integrated form of Equation 3, z = Ml 2 φ sin 2, and using the chain rule of advanced calculus to put the eventual integral in terms of. In the form of equations, this gives and, solving for dφ/d, we finally get z Ml 2 sin 2 = dφ dt = dφ d d dt = dφ 2[E Ueff ()] d Ml 2 dφ d = z l M 1 φ sin 2 2[E U eff ()] = φ φ 0 = dφ = φ 0 z l 2M 0 d sin 2 E U eff () (7) (c) Max and min values of At the maximum and minimum values of, the mass has no (effective) kinetic energy (i.e. the total mechanical energy at those points is equal to the effective potential energy: = 0) and so E = U eff 2 z = 2Ml 2 sin 2 mgl cos 2 z = 2Ml 2 (1 cos 2 mgl cos (8) ) We can then find an algebraic (cubic) equation for the maximum and minimum values of cos by solving for it in Equation 8: Mgl(cos 3 cos ) + E(cos 2 1) + 2 z 2Ml 2 = 0 (9) 2

3 Question 2: Radial Oscillations In the classic two-body problem, the total mechanical energy for a pair of point-like particles with an interaction potential V (r) can be re-written to reflect an effective one-body problem when using constants of the motion (i.e. conservation laws), as well as a suitable reference frame and coordinate system. Therefore, instead of describing the motion of two particles with masses m 1 and m 2 undergoing motion dictated by the potential V (r), one can instead talk about a particle of reduced mass µ = m 1 m 2 /(m 1 + m 2 ) moving in an effective potential V eff, which has the form V eff = V (r) + 2 2µ 2 r 2 (10) and r is the radial coordinate of the particle with mass µ measured relative to the center of field. Note that Equation 10 yields the potential, which is the potential energy per unit (reduced) mass, and that the potential energy can be computed to be U eff = µv eff. For circular orbits, r = r 0 = constant in time, and for general central-force problems in Newtonian mechanics the orbital angular momentum () is a conserved quantity 1 of the motion: = φ 2. The latter equation can be solved to find the period of orbital motion, φ = dφ = and finally can be put in terms of r 0 and, 2π 0 dφ = 2π = T 0 T T = 2π µr2 0 (a) Newtonian potential for inverse-square-law force (11) Consider the case when the interaction potential V (r) = /r. The effective potential can be explicitly written as a function of r, V eff = r + 2 2µ 2 r 2. (12) A circular orbit corresponds to a system with total mechanical energy E that is equal to U eff at all times. In other words, a circular orbit has a value of r 0 that corresponds to the minimum value of U eff (or, equivalently, V eff ), which can be found to be dv eff dr = 2 2 µ 2 3 = 0 = r 0 = 2 µ 2 (13) We can find the orbital period for this potential by combining the general formula for T (Equation 11) with Equation 13 to eliminate ; this is most easily done by finding T 2 : T 2 = 4π 2 µ2 r = 4π 2 (r 0 µ 2 ) = 4π2 r3 0 (14) If = GM, where G is Newton s gravitational constant and M = m 1 + m 2 is the total (gravitational) mass of the system, then Equation 14 is Kepler s third law of planetary motion. 1 These equations are written for an orbit with its plane embedded in the x y Cartesian plane, so that = π/2 is constant in time. 3

4 For small perturbations in r relative to r 0, we can express V eff as a Taylor expansion in r: V eff (r) V eff (r 0 ) + dv eff dr (r r 0 ) + 1 d 2 V eff 2 dr 2 (r r 0 ) (15) where the first derivative of V eff is equal to 0 by definition. We are therefore left with a constant term and a quadratic term in r for the Taylor-expanded V eff. This has an equivalent form to the potential for a simple harmonic oscillator in one dimension, since the constant term will vanish when we compute the Euler- agrange equation of motion for r. We can therefore find the frequency of oscillations in r by computing the second derivative in V eff, d 2 V eff dr 2 = µ 2 4 = = (r 0µ 2 ) = (16) where the frequency (ω) can be finally found to be: ω = We can compare this to the frequency of orbital motion (Ω) by noting that Ω = 2π/T, and from Equation 14 this can be found to be: Ω = (17) 3 = ω (18) So, in short, the orbital and oscillation frequencies are equal for the case where the interaction potential corresponds to an inverse-square-law force. (b) Two-dimensional harmonic potential Now consider the case when the interaction potential V (r) = kr 2 /2, where k is positive constant. effective potential can be explicitly written as a function of r, The V eff = 1 2 kr µ 2 r 2. (19) We can now apply the same procedures as done for part (a) of this problem to find the various quantities of interest. For starters, we can compute the radius of a circular orbit for this effective potential by setting the first derivative of Equation 19 to 0: dv eff dr ( ) = kr /4 µ 2 3 = 0 = r 0 = µ 2 (20) k We can combine the above result with the general relation for the orbital period T and orbital frequency Ω to find them in terms of r 0 and k: T 2 = 4π2 2 = 4π2 k = 4π2 k 2π = Ω = T = k (21) Using Equation 15, we can find the solution for small oscillations in r about r 0 by computing the second derivative of V eff with respect to r: 4

5 d 2 V eff dr 2 = k and so the frequency of oscillations is found to be: = k + 3 kµ2 r 4 0 = 4k (22) ω = 2 k = 2Ω (23) 5