Goldstein_13_14 - Homework 3: # 2.13, 2.14 Michael Good...

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Homework 3: # 2.13, 2.14 Michael Good Sept 10, 2004 2.13 A heavy particle is placed at the top of a vertical hoop. Calculate the reaction of the hoop on the particle by means of the Lagrange’s undetermined multipliers and Lagrange’s equations. Find the height at which the particle falls off. Answer: The Lagrangian is L = T - V L = 1 2 m ( ˙ r 2 + r 2 ˙ θ 2 ) - mgr cos θ Where r is the distance the particle is away from the center of the hoop. The particle will eventually fall off but while its on the hoop, r will equal the radius of the hoop, a . This will be the constraint on the particle. Here when θ = 0, (at the top of the hoop) potential energy is mgr , and when θ = 90 o (at half of the hoop) potential energy is zero. Using Lagrange’s equations with undetermined multipliers, ∂L ∂q j - d dt ∂L ˙ q j + X k λ ∂f k ∂q j = 0 with our equation of constraint, f = r = a as long as the particle is touching the hoop. Solving for the motion: d dt ∂L ˙ r = m ¨ r ∂L ∂r = mr ˙ θ 2 - mg cos θ λ ∂f r ∂r = λ * 1 thus - m ¨ r + mr ˙ θ 2 - mg cos θ + λ = 0 solving for the other equation of motion, 1
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d dt ∂L ˙ θ = mr 2 ¨ θ + 2 mr ˙ r ˙ θ ∂L ∂θ = mgr sin θ λ ∂f θ ∂θ = λ * 0 thus - mr 2 ¨ θ - 2 mr ˙ r ˙ θ + mgr sin θ = 0 The equations of motion together are: - m ¨ r + mr ˙ θ 2 - mg cos θ + λ = 0 - mr 2 ¨ θ - 2 mr ˙ r ˙ θ + mgr sin θ = 0 To find the height at which the particle drops off, λ can be found in terms of θ . The force of constraint is λ and λ = 0 when the particle is no longer under the influence of the force of the hoop. So finding λ in terms of θ and setting λ to zero will give us the magic angle that the particle falls off. With the angle
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This note was uploaded on 03/17/2012 for the course PHYS 202 taught by Professor Atkin during the Spring '12 term at Amity University.

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Goldstein_13_14 - Homework 3: # 2.13, 2.14 Michael Good...

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