Solution Week 75 (2/16/04) Hanging chain

Transcription

1 Catenary Catenary is idealized shae of chain or cable hanging under its weight with the fixed end oints. The chain (cable) curve is catenary that minimizes the otential energy PHY 322, Sring 208

2 Solution Week 75 (2/6/04) Hanging chain We ll resent four solutions. The first one involves balancing forces. The other three involve various variations on a variational argument. First solution: Let the chain be described by the function y(x), and let the tension be described by the function T (x). Consider a small iece of the chain, with endoints at x and x + dx, as shown. T(x+dx) θ 2 θ T(x) x x+dx Let the tension at x ull downward at an angle µ with resect to the horizontal, and let the tension at x + dx ull uward at an angle µ 2 with resect to the horizontal. Balancing the horizontal and vertical forces on the small iece of chain gives T (x + dx) cos µ 2 = T (x) cos µ, T (x + dx)sinµ 2 = T (x)sinµ + gω dx cos µ, () where Ω is the mass er unit length. The second term on the right is the weight of the small iece, because dx/ cos µ (or dx/ cos µ 2, which is essentially the same) is its length. We must now somehow solve these two differential euations for the two unknown functions, y(x) and T (x). There are various ways to do this. Here is one method, broken down into three stes. First ste: Suaring and adding es. () gives (T (x + dx)) 2 =(T (x)) 2 +2T (x)gω tan µ dx + O(dx 2 ). (2) Writing T (x+dx) º T (x)+t 0 (x) dx, and using tan µ = dy/dx y 0, we can simlify e. (2) to (neglecting second-order terms in dx) Therefore, where c is a constant of integration. T 0 = gωy 0. (3) T = gωy + c, (4)

3 Second ste: Let s see what we can extract from the first euation in es. (). Using cos µ = +(y 0 (x)) 2, and cos µ 2 = +(y 0 (x + dx)) 2, (5) and exanding things to first order in dx, the first of es. () becomes T + T 0 dx +(y 0 + y 00 dx) 2 = T +y 02. (6) All of the functions here are evaluated at x, which we won t bother writing. Exanding the first suare root gives (to first order in dx) T + T 0 dx +y 02 To first order in dx this yields Integrating both sides gives µ y0 y 00 dx +y 02 = T +y 02. (7) T 0 T = y0 y 00. (8) +y02 ln T + c 2 = 2 ln( + y02 ), (9) where c 2 is a constant of integration. Exonentiating then gives where c 3 e c 2. c 2 3T 2 =+y 02, (0) Third ste: We will now combine e. (0) with e. (4) to solve for y(x). Eliminating T gives c 2 3 (gωy + c ) 2 =+y 02. We can rewrite this is the somewhat nicer form, +y 02 = Æ 2 (y + h) 2, () where Æ c 3 gω, and h = c /gω. At this oint we can cleverly guess (motivated by the fact that + sinh 2 z = cosh 2 z) that the solution for y is given by Or, we can searate variables to obtain y(x)+h = cosh Æ(x + a). (2) Æ dx = dy Æ 2 (y + h) 2, (3) and then use the fact that the integral of / z 2 is cosh z, to obtain the same result. The shae of the chain is therefore a hyerbolic cosine function. The constant h isn t too imortant, because it simly deends on where we ick the y = 0 height. Furthermore, we can eliminate the need for the constant a if we ick x =0tobe 2

4 where the lowest oint of the chain is (or where it would be, in the case where the sloe is always nonzero). In this case, using e. (2), we see that y 0 (0) = 0 imlies a = 0, as desired. We then have (ignoring the constant h) thenicesimleresult, y(x) = cosh(æx). (4) Æ We ll show how to determine Æ at the end of the solutions. Second solution: We can also solve this roblem by using a variational argument. The chain will want to minimize its otential energy, so we want to find the function y(x) that minimizes the integral, Z U = (dm)gy = Z Z Ω +y 02 dx gy = Ωg y +y 02 dx, (5) subject to the constraint that the length of the chain is some given length `. That is, Z ` = +y 02 dx. (6) Without this constraint, we could find y(x) by simly using the Euler-Lagrange euation on the Lagrangian y +y 02 given in e. (5). But with the constraint, we must use the method of Lagrange multiliers. This works for functionals in the same way it works for functions. Basically, for any small variation in y(x) near the minimum, we want the change in U to be roortional to the change in `. This means that there exists a linear combination of U and ` that doesn t change, to first order in any small variation in y(x). In other words, the Lagrangian 2 L = y +y 02 + h +y 02 =(y + h) +y 02 (7) satisfis the Euler-Lagrange euation, for some value of h. Therefore, µ 0 d (y + h)y 0 =) = dx 02. (8) +y 02 We must now erform some straightforward (although tedious) differentiations. Using the roduct rule on the left-hand side, and making coious use of the chain rule, we obtain y 02 (y + h)y00 + (y + h)y02 y 00 +y 02 +y 02 ( + y 02 ) 3/2 = +y 02. (9) Multilying through by ( + y 02 ) 3/2 and simlifying gives (y + h)y 00 =(+y 02 ). (20) The reason for this is the following. Assume that we have found the desired function y(x) that minimizes U, and consider two different variations in y(x) that give the same change in `, but different changes in U. Then the difference in these variations will roduce no change in `, while yielding a nonzero first-order change in U. This contradicts the fact that our y(x) yielded an extremum of U. 2 We ll use h for the Lagrange multilier, to make the notation consistent with that in the first solution. 3

5 Having roduced the Euler-Lagrange differential euation, we must now integrate it. If we multily through by y 0 and rearrange, we obtain y 0 y 00 y0 = +y02 y + h. (2) Taking the dx integral of both sides gives (/2) ln( + y 02 )=ln(y + h)+c 4,where c 4 is a constant of integration. Exonentiation then gives (with Æ e c 4 ) in agreement with e. (). +y 02 = Æ 2 (y + h) 2. (22) Third solution: Let s use a variational argument again, but now with y as the indeendent variable. That is, let the chain be described by the function x(y). Then the otential energy is Z Z U = (dm)gy = Ω Z +x dy 02 gy = Ωg y +x 02 dy. (23) The constraint is Z +x ` = 02 dy. (24) Using the method of Lagrange multiliers as in the second solution above, the Lagrangian we want to consider is L = y +x 02 + h +x 02 =(y + h) +x 02. (25) Our Euler-Lagrange euation is then d =) d dy µ (y + h)x 0 =0. (26) +x 02 The zero on the right-hand side makes things nice and easy, because it means that the uantity in arentheses is a constant. Calling this constant /Æ (to end u with the notation in the second solution), we have Æ(y + h)x 0 = +x 02. Therefore, x 0 = Æ 2 (y + h) 2, (27) which is euivalent to e. (3). Fourth solution: Note that our Lagrangian in the second solution above, which is given in e. (7) as L =(y + h) +y 02, (28) is indeendent of x. Therefore, in analogy with conservation of energy (which arises from a Lagrangian that is indeendent of t), the uantity E 0 L = y + h +y 02 (29) 4

6 is indeendent of x. Call it /Æ. Then we have reroduced e. (). Remark: The constant Æ can be determined from the locations of the endoints and the length of the chain. The osition of the chain may be described by giving () the horizontal distance, d, between the two endoints, (2) the vertical distance,, betweenthe two endoints, and (3) the length, `, of the chain, as shown. d l λ -x x = 0 d-x 0 0 Note that it is not obvious what the horizontal distances between the ends and the minimum oint (which we have chosen as the x = 0 oint) are. If = 0, then these distances are simly d/2. But otherwise, they are not so clear. If we let the left endoint be located at x = x 0, then the right endoint is located at x = d x 0.Wenowhavetwounknowns,x 0 and Æ. Our two conditions are 3 y(d x 0 ) y( x 0 )=, (30) along with the condition that the length euals `, which takes the form (using e. (4)) ` = Z d x0 x 0 +y 02 dx = Æ sinh(æx) Ø ØØ d x 0 Writing out es. (30) and (3) exlicitly, using e. (4), we have cosh Æ(d x 0 ) cosh( Æx 0 ) = Æ, and x 0. (3) sinh Æ(d x 0 ) sinh( Æx 0 ) = Æ`. (32) If we take the difference of the suares of these two euations, and use the hyerbolic identities cosh 2 x sinh 2 x = and cosh x cosh y sinh x sinh y = cosh(x y), we obtain 2 2 cosh(æd) =Æ 2 ( 2 `2). (33) We can now numerically solve this euation for Æ. Using a half-angle formula, you can show that e. (33) may also be written as 2sinh(Æd/2) = Æ `2 2. (34) We can check some limits here. If = 0 and ` = d (that is, the chain forms a horizontal straight line), then e. (34) becomes 2 sinh(æd/2) = Æd. The solution to this is Æ = 0, which does indeed corresond to a horizontal straight line, because for small Æ, e. (4) behaves like Æx 2 /2 (u to an additive constant), which varies slowly with x for small Æ. Another limit is where ` is much larger than both d and. In this case, e. (34) becomes 2sinh(Æd/2) º Æ`. The solution to this is a very large Æ, which corresonds to a drooy chain, because e. (4) varies raidly with x for large Æ. 3 We ll take the right end to be higher than the left end, without loss of generality. 5