Homework 1 # 1.21, 2.7, 2.12

Homework 1: # 1.21, 2.7, 2.12 1.21. Two mass points of mass m1 and m2 are connected by a string passing through a hole in a smooth table so that m1 rests on the table surface and m2 hangs suspended. Assuming m2 moves only in a vertical line, what are the generalized coordinates for the system? Write the Lagrange equations for the system and, if possible, discuss the physical significance any of them might have. Reduce the problem to a single second-order differential equation and obtain a first integral of the equation. What is its physical significance? (Consider the motion only until m1 reaches the hole.) Answer: The generalized coordinates for the system are θ, the angle m1 moves round on the table, and r the length of the string from the hole to m1 . The whole motion of the system can be described by just these coordinates. To write the Lagrangian, we will want the kinetic and potential energies. T = 1 1 m2 ṙ2 + m1 ( ṙ2 + r2 ˙θ2 ) 2 2 V = −m2 g(R − r) The kinetic energy is just the addition of both masses, while V is obtained so that V = −mgR when r = 0 and so that V = 0 when r = R. 1 1 (m2 + m1 ) ṙ2 + m1 r2 θ̇2 + m2 g(R − r) 2 2 To find the Lagrangian equations or equations of motion, solve for each component: L=T −V = ∂L = 0 ∂θ ∂L = m1 r 2 ˙ θ ∂θ˙ d d ∂L = (m1 r2 ˙θ) = m1 r2 θ̈ + 2m1 r ṙ ˙θ dt dt ∂ θ̇ 1 Thus m1 r(r ¨θ + 2 θ˙ṙ) = 0 and ∂L = −m2 g + m1 r ˙ θ2 ∂r ∂L = (m2 + m1 ) r˙ ∂ ṙ d ∂L = (m2 + m1 )¨ r dt ∂ ṙ Thus r= 0 m2 g − m1 r ˙ θ2 + (m2 + m1 )¨ Therefore our equations of motion are: d (m1 r2 ˙θ) = m1 r(r θ̈ + 2 θ̇ ṙ) = 0 dt r= 0 m2 g − m1 r ˙ θ2 + (m2 + m1 )¨ See that m1 r θ̇ is constant, because ddt (m1 r2 ˙θ) = 0. It is angular momentum. Now the Lagrangian can be put in terms of angular momentum and will lend the problem to interpretation. We have ˙θ = l/m1 r2 , where l is angular momentum. The equation of motion 2 r= 0 m2 g − m1 r ˙ θ2 + (m2 + m1 )¨ becomes l2 + m2 g = 0 m1 r 3 The problem has been reduced to a single non-linear second-order differential equation. The next step is a nice one to notice. If you take the first integral you get r− (m1 + m2 )¨ 1 (m1 + m2 ) ṙ2 + 2 l2 2m1 r2 + m2 gr + C = 0 To see this, check by assuming that C = −m2 gR: l2 d 1 2 − m2 g(R − r)) = (m1 + m2 ) ṙ¨ r− ( (m1 + m2 ) ṙ + 2m1 r2 dt 2 l2 m1 r 3 ṙ + m2 g r˙ = 0 l2 + m2 g = 0 m1 r 3 Because this term is T plus V , this is the total energy, and because its time derivative is constant, energy is conserved. r− (m1 + m2 )¨ 2 2.7 In Example 2 of Section 2.1 we considered the problem of the minimum surface of revolution. Examine the symmetric case x1 = x2 , y2 = −y1 > 0, and express the condition for the parameter a as a transcendental equation in terms of the dimensionless quantities k = x2 /a, and α = y2 /x2 . Show that for α greater than a certain value α0 two values of k are possible, for α = α0 only one value of k is possible, while if α < α0 no real value of k (or a) can be found, so that no catenary solution exists in this region. Find the value of α0 , numerically if necessary. Answer: Starting with Goldstein’s form for a catenary, in section 2.2, not 2.1, y−b a and recognizing by symmetry that the soap film problem and the catenary problem are the same. In Marion and Thorton this is made clear (pg 222). Also, in a similar way to MathWorld’s analysis of a surface of revolution, it is clear that y and x, when interchanged, change the shape of the catenary to be about the x-axis. x = a cosh x−b a To preserve symmetry, x1 = −x2 and y2 = y1 . This switch makes sense because if you hang a rope from two points, its going to hang between the points with a droopy curve, and fall straight down after the points. This shaped revolved around the x-axis looks like a horizontal worm hole. This is the classic catenary curve, or catenoid shape. The two shapes are physically equivalent, and take on different mathematical forms. With this, we see that y = a cosh x2 − b x1 − b y2 = a cosh a a holds. The two endpoints are (x0 , y0 ) and (−x0 , y0 ). Thus y1 = a cosh y0 = a cosh and because x0 − b −x0 − b = a cosh a a cosh(−x) = cosh(x) we have cosh −x0 + b −x0 − b = cosh a a −x0 + b = −x0 − b 3 b = −b b = 0 By symmetry, with the center of the shape or rings at the origin, b = 0, simplifies the problem to a much nicer form: x a y = a cosh Including our end points: x0 a In terms of the dimensionless quantities, y0 = a cosh ρ = x2 /a β = 1 , α = y2 /x2 the equation is y0 = a cosh x0 a y0 a cosh ρ = x0 x0 y0 cosh(ρ) = x0 ρ β= cosh(ρ) ρ The minimum value of β in terms of ρ can be found by taking the derivative, and setting to zero: 0 = d 1 1 cosh ρ ( cosh ρ) = sinh ρ − ρ2 ρ dρ ρ sinh ρ = cosh ρ ρ ρ = coth ρ Thus, solved numerically, ρ ≈ 1.2. Plugging this in to find β0 , the value is: β0 ≈ 1.51 Since β0 = , 1 α0 α0 ≈ .66 4 This symmetric but physically equivalent example is not what the problem asked for, but I think its interesting. If we start at Goldstein’s equation, again, only this time recognize b = 0 due to symmetry from the start, the solution actually follows more quickly. y a x = a cosh xy x = cosh ax a Using, the dimensional quantities defined in the problem, k= x2 a α= y2 x2 we have k = cosh kα Taking the derivative with respect to k, 1 = α0 sinh kα0 Using the hyperbolic identity, cosh2 A − sinh2 A = 1 a more manageable expression in terms of k and α becomes apparent, k2 − 1 = 1 α20 α0 = √ Plug this into k = cosh kα 1 k2 − 1 k = cosh √ Solving this numerically for k yields, k k2 −1 k ≈ 1.81 Since 1 ⇒ α0 ≈ .66 k2 − 1 If α < α0 , two values of k are possible. If α > α0 , no real values of k exist, but if α = α0 then only k ≈ 1.81 will work. This graph is arccosh(k )/k = α and looks like a little hill. It can be graphed by typing acosh(x)/x on a free applet at http://www.tacoma.ctc.edu/home/jkim/gcalc.html. α0 = √ 5 2.12 The term generalized mechanics has come to designate a variety of classical mechanics in which the Lagrangian contains time derivatives of qi higher than the first. Problems for which triple dot x = f (x, x, ˙ ẍ, t) have been referred to as ‘jerky’ mechanics. Such equations of motion have interesting applications in chaos theory (cf. Chapter 11). By applying the mehtods of the calculus of variations, show that if there is a Lagrangian of the form L(qi , q̇iq̈i , t), and Hamilton’s principle holds with the zero variation of both qi and q̇i at the end points, then the corresponding Euler-Lagrange equations are d2 ∂L d ∂L ∂L ( ( )− ) + = 0, ∂qi dt2 ∂q̈i dt ∂ q̇i i = 1, 2, ..., n. Apply this result to the Lagrangian k m qq̈ − q 2 2 2 Do you recognize the equations of motion? L=− Answer: If there is a Lagrangian of the form L = L(qi , q̇i ,q̈i , t) and Hamilton’s principle holds with the zero variation of both qi and q̇i at the end points, then we have: Z I= 2 1 L(qi , q̇i ,q̈i , t)dt and ∂I ∂L ∂qi ∂L ∂ q̇i ∂L ∂q̈i Z 2 X ( ∂qi ∂αi dα + ∂ q̇i ∂αi dα + ∂q̈i ∂αi dα)dt i dα = ∂α 1 To make life easier, we’re going to assume the Einstein summation convention, as well as drop the indexes entirely. In analogy with the differential quantity, Goldstein Equation (2.12), we have δq = ∂q dα ∂α Applying this we have Z δI = 2 1 ( ∂L ∂L ∂ q̇ ∂L ∂q̈ δq + dα + dα)dt ∂ q̇ ∂α ∂q̈ ∂α ∂q 6 The indexes are invisible and the two far terms are begging for some mathematical manipulation. Integration by parts on the middle term yields, in analogy to Goldstein page 44, 2 ∂L ∂ 2 q ∂q d ∂L ∂L ∂q  Z 2 dt = ( )dt  ∂ q̇ ∂α∂t ∂ q̇  − 1 ∂α dt ∂ q̇ 1 ∂α This first term on the right is zero because the condition exists that all the varied curves pass through the fixed end points and thus the partial derivative of q wrt to α at x1 and x2 vanish. Substituting back in, we have: Z 1 2 Z 2 ( ∂L δq − d ∂L δq + ∂L ∂q̈ dα)dt dt ∂ q̇ ∂q̈ ∂α δI = 1 ∂q Where we used the definition δq =∂α ∂q dα again. Now the last term needs attention. This requires integration by parts twice. Here goes: 2 ∂ 2 q d ∂L ∂L ∂ 2 q ∂L ∂q̈  Z 2 dt = ( )dt  ∂q̈ 1 ∂q̈ ∂α  − 1 ∂t∂α dt ∂q̈  ∂t∂α 1 R R again, and we we used are stillvdu left =with by parts Turn the uv −another vdu integration as before. The first problem. term vanishes once Where crank again. Z 2 2 ∂ 2 q d ∂L ∂q d2 ∂L d ∂L ∂q  Z 2 − dt ( )dt =  dt2 ∂q̈ ∂t∂α dt ∂ q̈ ∂ q̈ dt  − 1 ∂α −  ∂α First term vanishes for the third time, and1 we have Z 2 1 d2 ∂L ∂L ∂q̈ Z 2 ∂q dt ∂q̈ ∂α dt2 ∂q̈ 1 dt = ∂α Plugging back in finally, and using the definition of our δq, we get closer Z 2 1 Z 2( 1 d2 ∂L d ∂L ∂L δq + 2 δq − δq)dt ∂q dt ∂q̈ dt ∂ q̇ δI = Gathering δq’s, throwing our summation sign and index’s back in, and applying Hamiliton’s principle: Z δI = 2 d2 ∂L ∂L d ∂L X ( ∂qi − dt ∂ q̇i + dt2 ∂q̈i )δqi dt = 0 i We know that since1 q variables are independent, the variations δqi are independent and we can apply the calculus of variations lemma, (Goldstein, Eq. 2.10) and see that δI = 0 requires that the coefficients of δqi separately vanish, one by one: d2 ∂L ∂L d ∂L + 2 − = 0 dt ∂ q̇i ∂qi dt ∂q̈i 7 i = 1, 2, ...n. Applying this result to the Lagrangian, L=− 1 2 mqq̈ − k 2 q 2 yields 1 ∂L = − mq̈ − kq ∂q 2 − d ∂L = 0 dt ∂ q̇ d2 ∂L d d 1 d 1 1 = ( (− mq)) = (− m q̇) = − mq̈ dt 2 2 dt dt 2 dt2 ∂q̈ Adding them up: −mq̈ − kq = 0 This is interesting because this equation of motion is just Hooke’s Law. This crazy looking Lagrangian yields the same equation for simple harmonic motion using the ‘jerky’ form of Lagrangian’s equations. It’s interesting to notice that if the familiar Lagrangian for a simple harmonic oscillator (SHO) plus an extra term is used, the original Lagrangian can be obtained. L = LSHO + d dt (− mq q̇ ) 2 L= m q̇ 2 kq2 d mq q̇ − + (− ) 2 2 2 dt L= kq2 m q̇ 2 m q̇ 2 mqq̈ − − − 2 2 2 2 L=− mqq̈ 2 − kq2 2 This extra term, ddt (− mq2q̇ ) probably represents constraint. The generalized force of constraint is the Lagrange multipliers term that is added to the original form of Lagrange’s equations. 8