PHYSICS 200A : CLASSICAL MECHANICS SOLUTION SET #2

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1 PHYSICS 200A : CLASSICAL MECHANICS SOLUTION SET #2 [1] [José and Saletan problem 3.11] Consider a three-dimensional one-particle system whose potential energy in cylindrical polar coordinates {ρ,φ,z} is of the form V ρ,kφ + z), where k is a constant. a) Find a symmetry of the Lagrangian and use Noether s theorem to obtain the constant of the motion associated with it. b) Write down at least one other constant of the motion. c) Obtain an explicit expression for the dynamical vector field see JS eqn. 3.73) and use it to verify that the functions found in a) and b) are indeed constants of the motion. Solution : a) We have L = 1 2 m ρ 2 + ρ 2 φ2 + ż 2) V ρ,kφ + z). Consider now the one-parameter family of coordinate transformations, Clearly φλ) φ + λ zλ) z λk. k φ + z = kφ + z, hence L does not vary with λ, and therefore Q = L q σ = mρ 2 φ mkż q σ σ λ λ=0 is conserved: Q = 0. b) Since L t = 0, the Hamiltonian H is conserved. And since the kinetic energy is homogeneous of degree two in the generalized velocities { ρ, φ,ż}, the Hamiltonian is simply the total energy: H = T + U. Thus, is conserved: Ė = 0. E = 1 2 m ρ 2 + ρ 2 φ2 + ż 2) + V ρ,kφ + z) c) The dynamical vector field is simply the total time derivative, expressed in terms of derivatives with respect to coordinates and velocities: = d dt = q σ q σ + q σ q σ = ρ ρ + φ φ + ż z + ρ ρ + φ φ + z ż 1

2 The generalized accelerations follow from the equations of motion, which yield m ρ = mρ φ 2 ρ d mρ 2 φ) = dt φ m z = z, = k z ρ = 1 m ρ, φ = k mρ 2 z 2 ρ φ ρ, z = 1 m z. Therefore, we have Q = mρ 2 φ mkż ) We also have = ρ 2mρ φ + φ mρ 2 + z mk) = 2mρ ρ φ + k mρ 2 z = 0. E = 1 2 m ρ 2 + ρ 2 φ2 + ż 2) + V ρ,kφ + z) = ρ ρ ) 2 ρ φ mρ ) mk) ρ m z + φ φ + ż z + mρ ρ φ 2 + m ρ ρ + mρ 2 φ φ + mż z = ρ ρ + φ k z + ż z + mρ ρ φ 2 + m ρ 1 ) m ρ ρ φ 2 + mρ 2 φ k mρ 2 z ) 2 ρ φ + mż 1 ) = 0. ρ m z [2] [José and Saletan problem 3.24] Derive the equations of motion for the Lagrangian L = e γt[ 1 2 m q2 1 2 kq2], where γ > 0. Compare with known systems. Rewrite the Lagrangian in terms of the new variable Q q expγt/2), and from this obtain a constant of the motion. 2

3 Solution : We have Mr. Newton then says p = L q = m q eγt, F = L q = kq eγt. ṗ = F m q + γ m q = kq, which is the equation of a damped harmonic oscillator. The phase curves all collapse to the origin, which is a stable spiral if γ < 2 k/m and a stable node if γ > 2 k/m. In general, there is no reason for there to be a conserved quantity in a dissipative system like this... but... consider the coordinate transformation Q q expγt/2), which is inverted trivially to yield q = Q exp γt/2). We have and therefore q = Q 1 2 γ Q) e γt/2 L = 1 2 m Q 1 2 γ Q) kq2 = 1 2 m Q γ m Q Q 1 2 k 1 4 mγ2) Q 2. Since LQ, Q,t) is independent of t, we have that H is conserved: H = Q L Q L = 1 2 m Q k 1 4 mγ2) Q 2. [ = 1 2 m q γ m q q k q2] e γt. [3] A bead of mass m slides frictionlessly along a wire curve z = x 2 /2b, where b > 0. The wire rotates with angular frequency ω about the ẑ axis. a) Find the Lagrangian of this system. b) Find the Hamiltonian. c) Find the effective potential U eff x). d) Show that the motion is unbounded for ω 2 > ω 2 c and find the critical value ω c. e) Sketch the phase curves for this system for the cases ω 2 < ω 2 c and ω2 > ω 2 c. 3

4 f) Find an expression for the period of the motion when ω 2 < ω 2 c. g) Find the force of constraint which keeps the bead on the wire. Solution : We will solve this problem for a general shape zx). Since the curve is rotating, we will use the radial coordinate ρ instead of x, keeping in mind that the wire is a onedimensional object and not a two-dimensional surface. The coordinate ρ then indicates the direction along the wire but perpendicular to the ẑ axis. Note that ρ R may be positive or negative. a) The Lagrangian is This is supplemented by the constraint Lρ,z, ρ,ż) = 1 2 m ρ mż mω2 ρ 2 mgz. Gρ,z) = z zρ) = 0. Of course, we could eliminate z as an independent degree of freedom from the outset, and write Lρ, ρ) = 1 2 m [ 1 + [z ρ)] 2) ρ 2 + ω 2 ρ 2] mgzρ). b) The Hamiltonian is H = p σ q σ L = 1 2 m ρ mż2 1 2 mω2 ρ 2 + mgz = 1 2 m 1 + [z ρ)] 2) ρ 2 + U eff ρ). c) The effective potential is U eff ρ) = mgzρ) 1 2 mω2 ρ 2 = 1 2 m ω2 c ω 2 )ρ 2, where ω c g/b. Note that we do not have m ρ = U eff ρ). This is because p ρ = L ρ = m 1 + [z ρ)] 2) ρ, and thus ṗ ρ = L ρ 1 + [z ρ) ] 2 ) ρ = ω 2 ρ gz ρ) z ρ)z ρ) ρ 2. 4

5 Figure 1: Level sets of the function Cu,v) = 1 + u 2 )v 2 + u 2 superimposed on the phase flow u = v, v = u1 + v 2 )/1 + u 2 ). Note that the phase curves are bounded. d) Since L has no explicit time dependence, H is a constant of the moton: H = 1 2 m 1 + [z ρ)] 2) ρ 2 + U eff ρ) ) = 1 2 m 1 + ρ2 b 2 ρ mω2 c ω2 )ρ 2. Note that if ω 2 > ω 2 c that the level sets of Hρ, ρ) are unbounded. Hence the motion of the system, which takes place along these level sets, is also unbounded. e) Let us define the dimensionless coordinate u ρ/b and dimensionless time variable s ω 2 c ω 2 1/2 t. Then conservation of H means that C = 1 + u 2 )v 2 σu 2 is constant, where v = du ds is the dimensionless velocity, and where σ sgn ω 2 ωc) 2. Setting dc ds = 0, we obtain du ds = v, dv ds = σ v2 )u 1 + u 2. This phase flow has a single fixed point, at u,v) = 0,0), which is either a center ω 2 < ωc 2) or a saddle point ω 2 > ωc). 2 A sketch of the phase flow for ω 2 < ωc 2 is shown in Fig. 1; the 5

6 Figure 2: Level sets of the function Cu,v) = 1 + u 2 )v 2 u 2 superimposed on the phase flow u = v, v = u1 v 2 )/1 + u 2 ). Note that the phase curves are unbounded. flow for ω 2 > ωc 2 is shown in Fig. 2. The Mathematica plot in Fig. 1 was obtained from the following commands: <<Graphics PlotField G1 = ContourPlot[ 1+x^2) y^2 + x^2, {x,-4,4}, {y,-4,4}, PlotPoints -> 50, Contours -> {0.1, 1, 4, 10, 20, 50, 100}, ContourShading -> False]; G2 = PlotVectorField[ {y, -1+y^2) x / 1+x^2)}, {x,-4,4}, {y,-4,4}, PlotPoints -> 30, ColorFunction -> Hue, ScaleFactor -> 0.55]; Show[ {G1, G2} ] It is worthwhile noting that other shapes zρ) may have fixed points for ρ 0. For example, consider the shape zρ) = ρ4 4b 3. If we define u = ρ/b and ω 2 c = g/b as before, but this time write s = ω c t, and define the new dimensionless parameter ε ω 2 /ω 2 c, we have that Cu,v) = 1 + u 6 )v u2 1 2 εu2 6

7 is constant, and the dynamics is given by du ds = v, dv ds = ε u2 6u 4 v 2 )u 21 + u 6 ) This flow, shown in Fig. 3, exhibits a saddle point at u,v) = 0,0) and two centers at u,v) = ± ε,0). The separatrix, which flows through 0,0), has C = 0. All the phase curves are bounded.. Figure 3: Level sets of the function Cu,v) = 1+u 6 )v u4 1 2 εu2 superimposed on the phase flow u = v, v = 1 2 uε u2 6u 4 v 2 )/1 + u 6 ), for ε = 1. There are two centers, at ±1,0), and a saddle at 0,0). All phase curves are bounded. e) The equation of motion can be taken as Ḣ = 0, which yields 1 + [ z ρ) ] ) 2 ρ + z ρ)z ρ) ρ 2 = ω 2 ρ g z ρ). We can expand about an equilibrium solution gz ρ ) = ω 2 ρ, writing ρ = ρ +δρ, in which case δ ρ = Ω 2 δρ, Ω 2 = gz ρ ) ω [ z ρ ) ] 2. Thus, the equilibrium at ρ is stable if ω 2 < gz ρ ) and unstable if ω 2 > gz ρ ). 7

8 We can go even farther in this analysis, using the conservation of H, which allows us to write the motion as a first order ODE, 1 + [ z ρ) ] 2 dt = ± [ H Ueff ρ) ] dρ. 2 m Identifying the turning points as solutions to we have the period for motion TH) is TH) = For the case zρ) = ρ 2 /2b, we have TH) = H = U eff ρ ± ), ρ + H) m 2 dρ ρ H) π/2 4 dθ ω 2 c ω [ z ρ) ] 2 H U eff ρ) H sin2 θ mb 2 ω 2 c ω2 ). g) If we write Gρ,z) = z zρ) = 0 as a constraint, the equations of motion are We now eliminate z = zρ), in which case We may now write m ρ = mω 2 ρ λz ρ) m z = mg + λ. ż = z ρ) ρ, z = z ρ) ρ + z ρ) ρ 2. λ = mg + mz ρ) ρ + mz ρ) ρ 2 and, substituting this into the first of the equations of motion and collecting terms, we find 1 + [z ρ) ] 2 ) ρ = ω 2 ρ gz ρ) z ρ)z ρ) ρ 2. As we have seen above, this result also follows from Ḣ = 0. We may now solve for λ in terms of ρ and ρ: m λ = 1 + [ z ρ) ] 2 g + z ρ) ρ 2 + ω 2 ρz ρ) ). The force of constraint supplied by the wire is Q = Q ˆn = Q ρ ˆρ + Q z ẑ), 8

9 where ˆn = z ρ) ˆρ + ẑ 1 + [ z ρ) ] 2 is the unit vector locally orthogonal to the tangent to the curve. Thus, Q = λ 1 + [ z ρ) ] 2 = m g + z ρ) ρ 2 + ω 2 ρz ρ) ) 1 + [ z ρ) ] 2 We may further eliminate ρ in favor of ρ by invoking conservation of H, which says. ρ 2 = 2H m 2gzρ) + ω2 ρ [z ρ) ] 2. [4] A particle of mass m is embedded, a distance b from the center, in a uniformly dense cylinder of mass M. The mass of the cylinder plus the inclusion is thus M + m.) The cylinder rolls without slipping along a plane inclined at an angle α with respect to the horizontal, under the influence of gravity. The axis of the cylinder remains horizontal throughout the motion. a) Choose an appropriate generalized coordinate and find the Lagrangian. b) Find the equations of motion. c) Under what conditions does a stable equilibrium exist? d) Find the frequency of small oscillations about the equilibrium. Figure 4: A cylinder of radius R with an inclusion rolls along an inclined plane. 9

10 Solution : a) Consulting the diagram in Fig. 4, let q be the distance of the cylinder s point of contact to the bottom of the wedge, and let φ be the angle the inclusion makes with respect to the vertical, with φ = 0 pointing downward. The coordinates of the center of the cylinder, in the plane perpendicular to its symmetry axis, are Thus, the coordinates of the inclusion are We may now write and thus the kinetic energy is x C = q cos α R sin α y C = q sin α + R cos α. x = q cos α R sin α bsinφ y = q sin α + R cos α bcos φ. ẋ C = q cos α ẏ C = q sinα ẋ = q cos α b φ cos φ ẏ = q sinα + b φ sinφ, T = 1 2 Mẋ2 + C ẏ2) + 1 C 2 I φ m ẋ2 + ẏ 2 ) = 1 2 M q I φ m q 2 + b 2 φ2 2b q φcosφ + α) ) I + MR 2 + mr 2 + mb 2 2mbR cosφ + α) φ2, = 1 2 where in the last line we have implemented the holonomic constraint q = d dt Rφ + α) = R φ. We can now write q = q 0 + Rφ, where q 0 is a constant, in which case the potential is U = Mgy C + mgy = M + m)gr φsin α mgbcos φ + U 0, where U 0 is a constant. The remaining generalized coordinate is φ, in which case L = 1 2 I + MR 2 + mr 2 + mb 2 2mbR cosφ + α)) φ2 M + m)gr φsin α + mgbcos φ, up to an irrelevant overall constant. b) The equations of motion, ṗ φ = F φ, are found to be I + MR 2 + mr 2 + mb 2 2mbR cosφ + α)) φ + mbrsinφ + α) φ2 = U φ), 10

11 where Uφ) = M + m)gr φsin α mgbcos φ. c) Equilibrium requires U φ) = 0, which says sin φ = 1 + M m ) R b sin α. Note that sin φ < 0, which means that the inclusion must lie to the right of the vertical midline shown in Fig. 4. In order for a solution to exist, we must have 1 + M ) R m b sin α 1. Thus, no solution exists unless m MR sin α b R sin α, and of course we must have b > R sinα. In fact, there are two solutions: [ φ 1 = π + sin M ) ] [ R m b sin α, φ 2 = 2π sin M m ) ] R b sin α. d) Since U φ) = mgbcos φ, so the solution φ = φ 1 is unstable while φ = φ 2 is stable. The equation for small oscillations is δ φ = Ω 2 δφ, where φ = φ 2 + δφ and Ω 2 = mgbcos φ 2 I + MR 2 + mr 2 + mb 2 2mbR cosφ 2 + α). 11