Goldstein_18_21_13_14_20

Goldstein_18_21_13_14_20 - Homework 4 2.18 2.21 3.13 3.14...

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Homework 4: # 2.18, 2.21, 3.13, 3.14, 3.20 Michael Good Sept 20, 2004 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 ω 0 , there can be a solution in which the particle remains stationary on the hoop at a point other than the bottom, but if ω < ω 0 , the only stationary point for the particle is at the bottom of the hoop. What is the value of ω 0 ? 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. L = 1 2 ma 2 ( ˙ θ 2 + ω 2 sin 2 θ ) - mga cos θ 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: d dt ∂L ˙ θ - ∂L ∂θ = 0 ma 2 ¨ θ = ma 2 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). h = ˙ θ ∂L ˙ θ - L h = ˙ θma 2 ˙ θ - 1 2 ma 2 ( ˙ θ 2 + ω 2 sin 2 θ ) - mga cos θ 1
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This simplifies to: h = 1 2 ma 2 ˙ θ - ( 1 2 ma 2 ω 2 sin 2 θ - mga cos θ ) Because the ‘energy function’ has an identical value to the Hamiltonian, the effective potential is the second term, V eff = mga cos θ - 1 2 ma 2 ω 2 sin 2 θ The partial of V eff with respect to θ set equal to zero should give us a stationary point. ∂V eff ∂θ = mga sin θ + ma 2 ω 2 sin θ cos θ = 0 ma sin θ ( g + 2 cos θ ) = 0 This yields three values for θ to obtain a stationary point, θ = 0 θ = π θ = arccos( - g 2 ) At the top, the bottom, and some angle that suggests a critical value of ω . ω 0 = g a The top of the hoop is unstable, but at the bottom we have a different story. If I set ω = ω 0 and graph the potential, the only stable minimum is at θ = π , the bottom. Therefore anything ω < ω 0 , θ = π is stable, and is the only stationary point for the particle. If we speed up this hoop, ω > ω 0 , our angle θ = arccos( - ω 2 0 ω 2 ) 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. 2
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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 r 0 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
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