Homework 4 # 2.18, 2.21, 3.13, 3.14, 3.20

Homework 4: # 2.18, 2.21, 3.13, 3.14, 3.20 2.18 A point mass is constrained to move on a massless hoop of radius a fixed in a vertical plane that rotates about its vertical symmetry axis with constant angular speed ω. Obtain the Lagrange equations of motion assuming the only external forces arise from gravity. What are the constants of motion? Show that if ω is greater than a critical value ω0 , there can be a solution in which the particle remains stationary on the hoop at a point other than the bottom, but if ω < ω0 , the only stationary point for the particle is at the bottom of the hoop. What is the value of ω0 ? Answer: To obtain the equations of motion, we need to find the Lagrangian. We only need one generalized coordinate, because the radius of the hoop is constant, and the point mass is constrained to this radius, while the angular velocity, w is constant as well. 1 2 ˙ ma ( θ2 + ω 2 sin2 θ) − mga cos θ 2 Where the kinetic energy is found by spherical symmetry, and the potential energy is considered negative at the bottom of the hoop, and zero where the vertical is at the center of the hoop. My θ is the angle from the z-axis, and a is the radius. The equations of motion are then: L= d ∂L ∂L = 0 − dt ∂θ˙ ∂θ ma2 ¨θ = ma2 ω 2 sin θ cos θ + mga sin θ We see that the Lagrangian does not explicitly depend on time therefore the energy function, h, is conserved (Goldstein page 61). ∂L h = ˙θ −L ∂θ˙ h = θ̇ma2 θ̇ − 1 ma2 ( θ̇2 + ω 2 sin2 θ) − mga cos θ 2 1 This simplifies to: 1 1 ma2 θ̇ − ( ma2 ω 2 sin2 θ − mga cos θ) 2 2 Because the ‘energy function’ has an identical value to the Hamiltonian, the effective potential is the second term, h= 1 ma2 ω 2 sin2 θ 2 with respect to θ set equal to zero should give us a Vef f = mga cos θ − The partial of Vef f stationary point. ∂Vef f ∂θ = mga sin θ + ma2 ω 2 sin θ cos θ = 0 ma sin θ(g + aω 2 cos θ) = 0 This yields three values for θ to obtain a stationary point, θ= 0 θ=π θ = arccos(− g ) aω 2 At the top, the bottom, and some angle that suggests a critical value of ω . r a g ω0 = The top of the hoop is unstable, but at the bottom we have a different story. If I set ω = ω0 and graph the potential, the only stable minimum is at θ = π, the bottom. Therefore anything ω < ω0 , θ = π is stable, and is the only stationary point for the particle. If we speed up this hoop, ω > ω0 , our angle ω20 ) ω2 is stable and θ = π becomes unstable. So the point mass moves up the hoop, to a nice place where it is swung around and maintains a stationary orbit. θ = arccos(− 2 2.21 A carriage runs along rails on a rigid beam, as shown in the figure below. The carriage is attached to one end of a spring of equilibrium length r0 and force constant k, whose other end is fixed on the beam. On the carriage, another set of rails is perpendicular to the first along which a particle of mass m moves, held by a spring fixed on the beam, of force constant k and zero equilibrium length. Beam, rails, springs, and carriage are assumed to have zero mass. The whole system is forced to move in a plane about the point of attachment of the first spring, with a constant angular speed ω. The length of the second spring is at all times considered small compared to r0 . • What is the energy of the system? Is it conserved? • Using generalized coordinates in the laboratory system, what is the Jacobi integral for the system? Is it conserved? • In terms of the generalized coordinates relative to a system rotating with the angular speed ω, what is the Lagrangian? What is the Jacobi integral? Is it conserved? Discuss the relationship between the two Jacobi integrals. Answer: Energy of the system is found by the addition of kinetic and potential parts. The kinetic, in the lab frame, (x, y), using Cartesian coordinates is 1 T (x, y) = m( ẋ2 + ẏ 2 ) 2 Potential energy is harder to write in lab frame. In the rotating frame, the system looks stationary, and its potential energy is easy to write down. I’ll use (r, l) to denote the rotating frame coordinates. The potential, in the rotating frame is 1 V (r, l) = k(r2 + l2 ) 2 Where r is simply the distance stretched from equilibrium for the large spring. Since the small spring has zero equilbrium length, then the potential 1 2 . energy for it is just2 kl The energy needs to be written down fully in one frame or the other, so I’ll need a pair of transformation equations relating the two frames. That is, relating (x, y) to (r, l). Solving for them, by drawing a diagram, yields x = (r0 + r) cos ωt − l sin ωt y = (r0 + r) sin ωt + l cos ωt Manipulating these so I may find r(x, y) and l(x, y) so as to write the stubborn potential energy in terms of the lab frame is done with some algebra. 3 Multiplying x by cos ωt and y by sin ωt, adding the two equations and solving for r yields r = x cos ωt + y sin ωt − r0 Multiplying x by sin and y by cos, adding and solving for l yields l = −x sin ωt + y cos ωt Plugging these values into the potential energy to express it in terms of the lab frame leaves us with E(x, y) = 1 1 m( ẋ2 + ẏ 2 ) + k((x cos ωt + y sin ωt − r0 )2 + (−x sin ωt + y cos ωt)2 ) 2 2 This energy is explicitly dependent on time. Thus it is NOT conserved in the lab frame. E(x, y) is not conserved. In the rotating frame this may be a different story. To find E(r, l) we are lucky to have an easy potential energy term, but now our kinetic energy is giving us problems. We need 1 E(r, l ) = T (r, l ) + k(l2 + r2 ) 2 Where in the laboratory frame, T (x, y) = 2 1( ẋ2 + ẏ 2 ). Taking derivatives of x and y yield ẋ = −ω(r0 + r) sin ωt + ṙ cos ωt − lω cos ωt −l ˙sin ωt ˙ ωt ẏ = ω(r0 + r) cos ωt + ṙ sin ωt − lω sin ωt −l cos Squaring both and adding them yields ẋ2 + ẏ 2 = ω 2 (r0 + r)2 r˙2 + l2 ω 2 + ˙ l2 + C.T. Where cross terms, C.T. are C.T. = 2ω(r0 + r) ˙l − 2 ṙlω For kinetic energy we know have 1 m(ω 2 (r0 + r)2 r˙2 + l2 ω 2 + ˙ l2 + 2ω(r0 + r) ˙ l − 2 ṙlω) 2 Collecting terms T (r, l) = T (r, l) = ˙ l 1 m(ω 2 (r0 + r + )2 + ( ṙ − lω)2 ) 2 ω Thus 4 ˙ l 2 1 1 ) + ( ṙ − lω)2 ) + k(l2 + r2 ) m(ω 2 (r0 + r + 2 ω 2 This has no explicit time dependence, therefore energy in the rotating frame is conserved. E(r, l) is conserved. E(r, l) = In the laboratory frame, the Lagrangian is just T (x, y) − V (x, y ). L(x, y) = 1 2 m( ẋ2 + ẏ 2 ) − V (x, y ) Where 1 V (x, y) = k((x cos ωt + y sin ωt − r0 )2 + (−x sin ωt + y cos ωt)2 ) 2 The Jacobi integral, or energy function is We have h= h = ẋ Xi q̇i ∂L ∂ q̇i −L ∂L ∂L + ẏ − L(x, y ) ∂ ẋ ∂ ẏ 1 h = ẋm ẋ + ẏm ẏ − m( ẋ2 + ẏ 2 ) + V (x, y ) 2 Notice that V (x, y) does not have any dependence on ẋ or ẏ. Bringing it together h= 1 1 m( ẋ2 + ẏ 2 ) + k((x cos ωt + y sin ωt − r0 )2 + (−x sin ωt + y cos ωt)2 ) 2 2 This is equal to the energy. h(x, y) = E(x, y) Because it is dependent on time, ∂L d h=− 6= 0 dt ∂t we know h(x, y) is not conserved in the lab frame. For the rotating frame, the Lagrangian is 1 L(r, l ) = T (r, l ) − k(r2 + l2 ) 2 5 Where ˙ l 1 m(ω 2 (r0 + r + )2 + ( ṙ − lω)2 ) 2 ω The energy function, or Jacobi integral is T (r, l) = ∂L ˙ ∂L + l ∂ ṙ ∂l˙ − L(r, l) ˙l h(r, l) = ṙm( ṙ − lω) + ˙lmω(r0 + r + ) − L(r, l) ω Collecting terms, with some heavy algebra h(r, l) = ṙ h = ( ṙ−lω)(m ṙ− h = ( ṙ − lω)( 1 2 m( ṙ−lω))+(r0 +r+ ˙ l ˙ ˙ 1 mω 2 (r0 +r+ l ))+ 1 k(l2 +r2 ) )(mω l− 2 ω ω 2 ˙l 1 1 1 m ṙ 1 k(l2 + r2 ) + mlω) + (r0 + r + )( mω l ˙− mω 2 (r0 + r)) + 2 ω 2 2 2 2 More algebraic manipulation in order to get terms that look like kinetic energy, h= 1 1 1 2 2 m( ṙ2 + ˙l2 )+ k(l2 +r2 )+ [ ṙmlω−lωm r−ml ˙ ω +(r0 +r)mω ˙ l−mω 2 (r0 +r)2 −mω ˙ l(r0 +r)] 2 2 2 Yields 1 1 1 mω 2 (l2 + (r0 + r)2 ) m( ṙ2 + l̇2 ) + k(l2 + r2 ) − 2 2 2 This has no time dependence, and this nice way of writing it reveals an energy term of rotation in the lab frame that can’t be seen in the rotating 2 frame. It is of the from E = −21Iω . h(r, l) = ∂L d h=− = 0 ∂t dt We have h(r, l) conserved in the rotating frame. 6 3.13 • Show that if a particle describes a circular orbit under the influence of an attractive central force directed toward a point on the circle, then the force varies as the inverse-fifth power of the distance. • Show that for orbit described the total energy of the particle is zero. • Find the period of the motion. • Find ẋ, ẏ and v as a function of angle around the circle and show that all three quantities are infinite as the particle goes through the center of force. Answer: Using the differential equation of the orbit, equation (3.34) in Goldstein, d2 m d 1 u+u=− 2 V( ) u dθ 2 l du Where r = 1/u and with the origin at a point on the circle, a triangle drawn with r being the distance the mass is away from the origin will reveal r = 2R cos θ 1 2R cos θ Plugging this in and taking the derivative twice, u= 1 sin θ d u= [− cos−2 θ(− sin θ)] = dθ 2R 2R cos2 θ The derivative of this is 1 d sin θ = [sin θ(−2 cos−3 θ)(− sin θ) + cos−2 θ cos θ] dθ 2R cos2 θ 2R Thus 2 d2 1 + sin2 θ 1 2 sin θ cos2 θ + [ ] = u= 2R cos3 θ 2R cos3 θ cos3 θ dθ 2 1 + sin2 θ cos2 θ d2 2 u+u= + = 2R cos3 θ dθ 2 2R cos3 θ 2R cos3 θ That is 8R2 = 8R2 u3 8R3 cos3 θ Solving for V u( )1 by integrating yields, 7 8R2 l2 4 1 u V( ) = − 4m u and we have V (r) = − 2l2 R2 mr 4 with force equal to 8l2 R2 d V (r) = − 5 dr mr This force is inversely proportional to r5 . f (r) = − Is the energy zero? Well, we know V (r), lets find T (r) and hope its the negative of V (r). T = 1 m( ṙ2 + r2 ˙θ2 ) 2 Where r = 2R cos θ ṙ = −2R sin θ ˙θ → ṙ2 = 4R2 sin2 ˙ θ2 r2 = 4R2 cos2 θ So, plugging these in, T = 1 m(4R2 sin2 θ ˙θ2 + 4R2 cos2 θ ˙θ2 ) 2 m4R ˙θ2 = 2mR2 ˙θ2 2 Put this in terms of angular momentum, l, T = l = mr 2 ˙θ l 2 = m2 r 4 ˙ θ 2 T = 2mR2 θ̇2 → T = 2R2 l2 mr 4 Which shows that E =T +V = 2R2 l2 2R2 l2 − = 0 mr 4 mr 4 the total energy is zero. The period of the motion can be thought of in terms of θ as r spans from π θ = − π 2 to θ = 2 . 8 2 Z Z π − π2 P = dt = 2 π − π2 dt dθ dθ This is Z 2 dθ π θ̇ P = 2 ˙ Because θ = l/mr in terms of angular momentum, we have Z P = P = m l P = l And finally, Z π − π2 2 Z l P = 2 mr 2 dθ l 2 π − π2 m From above we have r2 4mR2 −π2 Z 2 π − π2 π − π2 r2 dθ 4R2 cos2 θdθ 4mR2 θ 1 cos θdθ = ( + l 2 2 π 2 π ) = −2 sin 2θ  4 P = 4mR2 π π ( + ) 4 l 4 2mπR2 l For ẋ, ẏ, and v as a function of angle, it can be shown that all three quantities are infinite as particle goes through the center of force. Remembering that r = 2R cos θ, x = r cos θ = 2R cos2 θ y = r sin θ = 2R cos θ sin θ = R sin 2θ Finding their derivatives, ẋ = −4R cos θ sin θ = −2R ˙ θ sin 2θ ẏ = 2Rθ˙ cos 2θ θ p ẏ 2 = 2R ˙ we remember v = ẋ2 + What is θ̇? In terms of angular momentum 9 l = mr 2 ˙θ Plugging in our r, and solving forθ ˙ θ˙ = l 4mR2 cos2 θ As we got closer to the origin, θ becomes close to2 .± π θ = ±( Note that as δ→0 π 2 θ→ ± − δ) π 2 ˙ θ→ ∞ All ẋ, ẏ and v are directly proportional to the θ ˙ term. The ẋ may be questionable at first because it has a sin 2θ and when sin 2θ → 0 as θ → π/2 we ˙ can tell that θ we may be left with ∞ ∗ 0. But looking closely at ẋ = −4Rl cos θ sin θ= − l tan θ mR 4mR2 cos2 θ tan θ → ∞ as θ → ± π 2 2.14 • For circular and parabolic orbits in an attractive 1/r potential having the same angular momentum, show that perihelion distance of the parabola is one-half the radius of the circle. • Prove that in the same central force √ as above, the speed of a particle at any point in a parabolic orbit is 2 times the speed in a circular orbit passing through the same point. Answer: Using the equation of orbit, Goldstein equation 3.55, 1 mk = [1 + ǫ cos(θ − θ′ )] r l2 we have for the circle, ǫ = 0 l2 1 mk = 2 → rc = l rc mk For the parabola, ǫ = 1 l2 1 mk = 2 (1 + 1) → rp = l rp 2mk 10 So rc 2 The speed of a particle in a circular orbit is rp = l2 2 v 2c = r ( m2 r4 ) In terms of k, this is equal to vc2 = r2 ˙θ2 → l → vc = l mr √ mrk r kmr mr =can be found by The speed of a particle in a parabola mr = vp2 = ṙ2 + r2 ˙θ2 ṙ = l2 d l2 ˙θ ( ) = sin θ dt mk(1 + cos θ) mk(1 + cos θ)2 Solving for vp , vp2 = r2 ˙θ2 ( sin2 θ + 1) (1 + cos θ)2 vp2 = r2 ˙θ2 ( vp2 = 2 + 2 cos θ ) (1 + cos θ)2 2r2 ˙θ 1 + cos θ 2 Using r for a parabola from Goldstein’s (3.55), and not forgetting that k = l2 /mr, l2 l2 r= → θ̇2 = 2 4 m r mk(1 + cos θ) we have 2r2 l2 mkr 2k → v 2p = mr m2 r 4 l 2 For the speed of the parabola, we then have vp2 = Thus r √ k mr vp = 2 √ vp = 2vc 11 20. A uniform distribution of dust in the solar system adds to the gravitational attraction of the Sun on a planet an additional force F = −mCr where m is the mas of the planet, C is a constant proportional to the gravitational constant and the density of the dust, and r is the radius vector from the Sun to the planet(both considered as points). This additional force is very small compared to the direct Sun-planet gravitational force. • Calculate the period for a circular orbit of radius r0 of the planet in this combined field. • Calculate the period of radial oscillations for slight disturbances from the circular orbit. • Show that nearly circular orbits can be approximated by a precessing ellipse and find the precession frequency. Is the precession in the same or opposite direction to the orbital angular velocity? Answer: The equation for period is T = 2π θ̇ θ̇ = l mr 2 For a circular orbit, Thus T = 2πmr2 l Goldstein’s equation after (3.58): k l2 = r20 mr 30 In our case, we have an added force due to the dust, mCr0 + l2 k = r20 mr 30 Solving for l yields q l = mr0 k + m2 Cr4 12 0 Plugging this in to our period, 2πmr2 T = → 2π T = 0 p mr0 k + m2 Cr4 Here the orbital angular velocity is q mr3 0 +C k +C r 3 kmr ω = orb This is nice because if the dust was not there, we would have C = 0 and our period would be 2π T0 = q s mr20 k ω0 = → 2 kmr 0 which agrees with l = mr 2 ω and l = √mrk. 0 0 The period of radial oscillations for slight disturbances from the circular orbit can be calculated by finding β. β is the number of cycles of oscillation that the particle goes through in one complete orbit. Dividing our orbital period by β will give us the period of the oscillations. Tosc = T β Equation (3.45) in Goldstein page 90, states that for small deviations from circularity conditions, 1 = u0 + a cos βθ r Substitution of this into the force law gives equation (3.46) u≡ r df  f    dr β2 = 3 + Solve this with f = mCr + k/r 2 r=r0 2k df = − 3 + mC r dr β2 = 3 + r β2 = k + r2 k + 2 r 4mCr mCr − 2kr3 + mC k + mCr r2 → Now 13 β2 = k + mr3 k + mr3 4C C Tosc = T q mr3 + 4C k +C β = q mr3 β k Therefore, our period of radial oscillations is 2π Tosc = q Here mr3 k + 4C + 4C r 3 kmr ω = r A nearly circular orbit can be approximated by a precessing ellipse. The equation for an elliptical orbit is a(1 − e2 ) 1 + e cos(θ − θ0 with e << 1, for a nearly circular orbit, a precessing ellipse will hug closely to the circle that would be made by e = 0. r= To find the precession frequency, I’m going to subtract the orbital angular velocity from the radial angular velocity, ωprec = ωr − ωorb +C r r 3 k kmr ω = + 4C − prec Fixing this up so as to use mr the3 binomial expansion, r r 4Cmr 3 k (1 + ωprec = expansion, 3 Using the binomial mr k r ωprec = 3 kmr ωprec = Therefore, [1 + r − 1 + Cmr 3 2Cmr 3 − (1 + k 4Cmr3 − Cmr 3 r 3 2k k mr r )] = 2k 3Cmr3 = r 2k 14 Cmr 3k 3 mr k [ 3 kmr ) Cmr3 2Cmr 3 − ] k 2k = 3C 2 r mr 3k 3C 2ω0 ωprec = → 3C 4πω0 fprec = Because the radial oscillations take on a higher angular velocity than the orbital angular velocity, the orbit is very nearly circular but the radial extrema comes a tiny bit more than once per period. This means that the orbit precesses opposite the direction of the orbital motion. Another way to do it, would be to find change in angle for every oscillation, ∆θ = 2π − 2π β Using the ratios, 2π Tosc ∆θ With some mean algebra, the period of precession is Tprec = 1 4π Tprec = 1− k +C mr3 k +4C 3 mr q + 4C ) q mr3 k k 4π( + 4C − ( kmr3 mr3 mr3 + 4c k 4π + 4C r 3 kmr Tprec = Because C is very small compared to k, the3C approximation holds, Tprec ≈ Therefore, 4π r = 3 kmr 3C Tprec = q where ω0 = + C) 4π ω0 3C mr3 k → fprec = . 15 3C 4πω0 → ωprec = 3C 2ω0