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goldsteinchap3 - Solutions to Problems in Goldstein...

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Solutions to Problems in Goldstein, Classical Mechanics , Second Edition Homer Reid December 1, 2001 Chapter 3 Problem 3.1 A particle of mass m is constrained to move under gravity without friction on the inside of a paraboloid of revolution whose axis is vertical. Find the one-dimensional problem equivalent to its motion. What is the condition on the particle’s initial velocity to produce circular motion? Find the period of small oscillations about this circular motion. We’ll take the paraboloid to be defined by the equation z = α r 2 . The kinetic and potential energies of the particle are T = m 2 ( ˙ r 2 + r 2 ˙ θ 2 + ˙ z 2 ) = m 2 ( ˙ r 2 + r 2 ˙ θ 2 + 4 α 2 r 2 ˙ r 2 ) V = mgz = mg α r 2 . Hence the Lagrangian is L = m 2 (1 + 4 α 2 r 2 ) ˙ r 2 + r 2 ˙ θ 2 - mg α r 2 . This is cyclic in θ , so the angular momentum is conserved: l = mr 2 ˙ θ = constant . 1
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Homer Reid’s Solutions to Goldstein Problems: Chapter 3 2 For r we have the derivatives L r = 4 α 2 mr ˙ r 2 + mr ˙ θ 2 - 2 mg α r L ˙ r = m (1 + 4 α 2 r 2 ) ˙ r d dt L ˙ r = 8 m α 2 r ˙ r 2 + m (1 + 4 α 2 r 2 r. Hence the equation of motion for r is 8 m α 2 r ˙ r 2 + m (1 + 4 α 2 r 2 r = 4 α 2 mr ˙ r 2 + mr ˙ θ 2 - 2 mg α r or m (1 + 4 α 2 r 2 r + 4 m α 2 r ˙ r 2 - mr ˙ θ 2 + 2 mg α r = 0 . In terms of the constant angular momentum, we may rewrite this as m (1 + 4 α 2 r 2 r + 4 m α 2 r ˙ r 2 - l 2 mr 3 + 2 mg α r = 0 . So this is the di ff erential equation that determines the time evolution of r . If initially ˙ r = 0, then we have m (1 + 4 α 2 r 2 r + - l 2 mr 3 + 2 mg α r = 0 . Evidently, ¨ r will then vanish—and hence ˙ r will remain 0, giving circular motion— if l 2 mr 3 = 2 mg α r or ˙ θ = 2 g α . So if this condition is satisfied, the particle will execute circular motion (assum- ing its initial r velocity was zero). It’s interesting to note that the condition on ˙ θ for circular motion is independent of r .
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Homer Reid’s Solutions to Goldstein Problems: Chapter 3 3 Problem 3.2 A particle moves in a central force field given by the potential V = - k e - ar r , where k and a are positive constants. Using the method of the equivalent one- dimensional potential discuss the nature of the motion, stating the ranges of l and E appropriate to each type of motion. When are circular orbits possible? Find the period of small radial oscillations about the circular motion. The Lagrangian is L = m 2 ˙ r 2 + r 2 ˙ θ 2 + k e - ar r . As usual the angular momentum is conserved: l = mr 2 ˙ θ = constant. We have L r = mr ˙ θ 2 - k (1 + ar ) e - ar r 2 L ˙ r = m ˙ r so the equation of motion for r is ¨ r = r ˙ θ 2 - k m (1 + ar ) e - ar r 2 = l 2 m 2 r 3 - k m (1 + ar ) e - ar r 2 . (1) The condition for circular motion is that this vanish, which yields ˙ θ = k m (1 + ar 0 ) e - ar 0 / 2 r 3 / 2 0 . (2) What this means is that that if the particle’s initial θ velocity is equal to the above function of the starting radius r 0 , then the second derivative of r will remain zero for all time. (Note that, in contrast to the previous problem, in this case the condition for circular motion does depend on the starting radius.)
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