Solutions to Homework Set 3

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Griffin Griffin
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1 Solutions to Homework Set 3 1 Bent brs: First recll some elementry clculus results: the curvture κ, nd the rdius R of the osculting circle, t point (x, y on curve y(x re given by κ = 1 R = dψ = 1 d tn 1 y 1 + y 2 dx = y (1 + y 2 3/2 Here the rc length s nd the ngle ψ between the curve nd the x xis constitute the intrinisic coordintes of the curve y R ψ x Osculting circle nd intrinsic coordintes For ll prts of the present problem y is ssumed smll enough tht we cn get wy with the pproximtion R 1 = y Euler s problem: We observe tht if rod of fixed length L is bent into curve y(z, we hve 2 = dz 2 + dy 2, dz = ( 1 ẏ 2 = 1 2ẏ2 1 + Here the overdot indictes differentition with respect to the rc-length s For curves with smll y, we my pproximte y = ẏ nd dz in the lst expression The length of the projection of the rod on the z xis is therefore L 1 2 y 2 dz Combining the resulting expresion for the potentil energy of the lod with the bending energy gives the functionl U[y] Inserting the mode expnsion for y(z nd doing the integrl gives U[y] = L { ( Y I n 2 4 π 4 n Mg ( } n 2 π L 4 2 L 2 n=1 The coefficient of 2 n is negtive when Mg > n2 π 2 Y I/L 2 We thus see tht the coefficient of 2 1 is the first to go negtive, mking n = 1 mode unstble, nd this occurs t when Mg = π 2 Y I/L 2 1

b Leonrdo d Vinci s problem: The strength of cntilever ws discussed by Glileo in his Dilogue Concerning Two New Sciences published 1654 His resoning ws not quite complete however Surprisingly, more

2 b Leonrdo d Vinci s problem: The strength of cntilever ws discussed by Glileo in his Dilogue Concerning Two New Sciences published 1654 His resoning ws not quite complete however Surprisingly, more correct nlysis of the problem hd been given by Leonrdo d Vinci more thn century erlier The energy is This is minimum when = δe = The drwing from Glileo s Dilogue E[y] = Y I 2 (y 2 dx + Mgy(L Y Iy (4 δy(x dx + Y I[y δy y δy] L + Mg δy(l for ny vrition δy(x At the wll end, x =, we re told tht y( = y ( = The displcement therefore stisfies y (4 =, with these boundry conditions Thus, we know tht the minimum energy configurtion will be of the form y(x = x 2 + bx 3, but do not yet know the constnts nd b Now consider the integrted-out terms t the weight end: Y Iy (Lδy (L (Y Iy (L Mgδy(L Even though we hve mde Y Iy(4 δy(x dx vnish, we cn still vry y(l nd y (L independently by vrying the s-yet-unknown constnts nd b: δy(l = L 2 δ + L 3 δb, δy (L = 2L δ + 3L 2 δb The vritions re independent becuse given ny desired δy(l nd δy (L, we cn solve for the necessry δ, δb s δ = 3L 2 δy(l 2L 3 δy (L, δb = L 1 δy(l + L 2 δy (L 2

3 Therefore, in the minimum-energy configurtion, the coefficients of δy(l nd δy (L must vnish seprtely, giving the boundry conditions The unique solution is y (L =, Y Iy (L = Mg y(x = Mg Y I Thus y(l = 1 3 MgL3 /Y I Note tht ( 1 6 x3 1 2 Lx2 Y Iy (x = Mg(x L The right-hnd-side of this lst eqution is the bending moment of the weight on the section of the rod t x The mximum bending moment occurs right where the rod enters the wll, nd this is where the cntilever would brek were the weight too lrge Lgrnge multipliers Recll tht finding the sttionry points of x Tx subject to the condition x x = 1 reduces to the problem of finding the normlized eigenvectors of the mtrix T The given qudrtic form is 13x 2 + 8xy + 7y 2 = ( x, y We therefore hve to digonlize the mtrix ( 13 4 T = 4 7 ( ( x y It is possible to digonlize 2 2 mtrix lmost by inspection We begin by observing tht product of the eigenvlues is The sum of the eigenvlues is dett = = 75 trt = = 2 The eigenvlues re therefore λ = 5 nd λ = 15 The esiest wy to find the eigenvectors is to use Dirc s trick (see the ppendix in the notes to decompose the identity mtrix into opertors P λ tht project onto the eigenvectors corresponding to eigenvlue λ: The projection opertors re therefore (T 5I (T 15I I = (5 15 = P 15 + P 5 (1 P 5 = 1 1 (15I T, P 15 = 1 (T 5I 1 3

4 We cn project from lmost ny vector, so let s use v = (, 1 t The eigenvector corresponding to λ = 15 is proportionl to ( ( ( (T 5Iv = = After normlizing we find tht there re sttionry points t x = ±2/ 5, y = ±1/ 5, t which f(x, y = 15 Similrly, the eigenvector corresponding to λ = 5 is proportionl to ( ( ( (15I Tv = =, so the other sttionry points re t x = 1/ 5, y = ±2/ 5, t which f(x, y = 5 Ctenry gin: We seek to mke sttionry { F[x, y] = ρgy + 1 } 2 λ(s(ẋ2 + ẏ 2 1 Prt : Since the en of the chin re fixed, we cn discrd ny integrted out terms, nd so hve { δf = δx(s ( d (λ(sẋ + δy(s (ρg d } (λ(sẏ The potentil energy is therefore sttionry when = δf δx(s = δf δy(s d (λ(sẋ ρg d (λ(sẏ Tking into count the condition ẋ 2 + ẏ 2 = 1, which tells us tht s is the rc-length, we cn set ẋ = cos ψ nd ẏ = sin ψ, where ψ is the intrinsic coordinte Thus = d (λ(s cosψ ρg = d (λ(s sinψ The figure shows tht these re the conditions for the equilibrium of ech infinitesiml segment of the chin, with λ(s T(s being the tension t point s T(s+ ψ(s+ ψ(s ρ g T(s The free body digrm for the forces cting on segment of chin of length 4

5 The horizontl component of the force [T(s cosψ] s+ s must vnish, while the verticl component blnce is [T(s sin ψ] s+ s d (λ(s cosψ d (λ(s sinψ = ρg Prt b: The geometry tells us tht ψ = s/ We know tht T(s cosψ is constnt, C sy, nd so ρ(sg = d ( ( s C tn = C ( s sec2 (2 The constnt C is fixed by mking Mg = π/4 π/4 C ( s sec2 = C[tnψ] π/4 π/4 = 2C Thus ρ(s = M 2 sec2 ( s 5

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space. Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)