z 0 > 0 z = 0 h; (x, 0)

Size: px

Start display at page:

Download "z 0 > 0 z = 0 h; (x, 0)"

Patience Morgan
5 years ago
Views:

3 n = (q 1,..., q n ) T (,, t) V (,, t) L(,, t) = T V d dt ( ) L q i L q i = 0, i = 1,..., n. l l 0 l l 0 l > l 0 {x} + = max(0, x) x = k{l l 0 } +ˆ, k > 0 ˆ

4 (x, z) x z (0, z 0 ) (0, z 0 ) z 0 > 0 x z = 0 h; l; θ; ϕ; k; m; l 0. h m z 0 (x, 0) z 0 = l cos θ + h cos ϕ x = l sin θ + h cos ϕ. (x, ϕ) θ l θ l = l(x, ϕ) = x 2 + z h 2 2h(x sin ϕ + z 0 cos ϕ) = (x h sin ϕ) 2 + (z 0 h cos ϕ) 2. V (x, ϕ) = k(l l 0 ) 2, T = 1 2 mẋ2.

5 L = T V x L = 1 2 mẋ2 k ( (x h sin ϕ)2 + (z 0 h cos ϕ) 2 l 0 ) 2. ( (x ) 2k(x h sin ϕ) h sin ϕ)2 + (z 0 h cos ϕ) 2 l 0 mẍ =. (x h sin ϕ)2 + (z 0 h cos ϕ) 2 ϕ t E 1 ( ) 2 2 mẋ2 + k (x h sin ϕ)2 + (z 0 h cos ϕ) 2 l 0 = E. ẋ 0 ϕ ϕ ϕ + x x ϕ ϕ = ϕ(t) x mẍ + V x = 0, V t V ϕ x t t mẋ = K + O( t) K ẋ O( t) 0 t y 0 t [y] [x] = O( t).

6 t 0 x [x] = 0 ϕ x ϕ = ϕ ± ϕ = ϕ x N j m = N k{ j l 0 } + j, 1 j = j j. j j l 0 + a l 0 a > 0 j = (l 0 + a)(cos 2πj 2πj, sin ), j = 1,..., N. N N X = (0, 0) (0, 0) l 0 + a = ϵ ϵ > 0 l = l 0 + a = ϵl(u, v) j l 0 a ϵl(u cos 2πj N + v sin 2πj N ) + O(ϵ2 ).

7 c j = cos 2πj N s j = sin 2πj N j = (c j, s j ) ϵ (a(u, v) (c j u + s j v)(c j, s j )) + O(ϵ 2 ). (ü, v) = Nk ( 1 + a ) (u, v). 2m l 0 + a θ j = r = (r, 0) = (l 0 + a)(cos θ, sin θ) = (0, 0) l 0 = ((l 0 + a) cos θ r) 2 + (l 0 + a) 2 sin 2 θ l 0 p = l 0 + a l 0 = p 2 + r 2 2rp cos θ l 0. OXZ ϕ x k( p 2 + r 2 2rp cos θ l 0 )( cos ϕ), x 2 = cos ϕ 2r cos ϕ = = 2r 2 2rp cos θ. cos ϕ = r p cos θ x ( ) r p cos θ k r p cos θ l 0. p2 + r 2 2rp cos θ θ 2 l 2 0

8 ψ = 0 ψ 0 ψ π p 2 + r 2 2pr cos ψ = l 2 0. r ( ) π r p cos θ = ˆ 2k r p cos θ l 0 dθ ψ p2 + r 2 2rp cos θ ˆ ( ) π r p cos θ = ˆ 2k (π ψ)r + p sin ψ l 0 p2 + r 2 2rp cos θ dθ. r ψ = 0 ψ

9 ϕ = ϕ = 0 ϕ = ϕ + > 0 x ϕ ϕ ϕ ẋ < 0 ϕ = ϕ 0 ϕ ϕ + ϕ = ϕ ± ϕ = ϕ x 0 ϕ ± < π 2 x = h sin ϕ ± > 0 (x, ẋ) (h sin ϕ ±, 0) h z 0 f a = 1 2k(z 0 l 0 ). 2π mz 0

10 . x x ϕ = ϕ = 0 ẋ < 0 ϕ = ϕ + = 30 ẋ > 0 (x, ẋ) = (0.001, 0) ϕ = 0 1 mm SI m = kg, z 0 = 0.2 m, h = m. l 0 = 1z 2 0 = 0.1 m E 2 GP a Nm 2 3 µm m k = EA/l 0

11 A k = 10EA = π = 18π f a = 1 k 1 2π m = 15 π Hz ϕ + 30 ϕ + 1 mm 10 cm 8.03 Hz ϕ x = h sin ϕ x (x h sin ϕ) (x h sin ϕ) 2 ẋ < 0 ϕ = ϕ = 0 (x, ẋ) = (x, 0) x > 0 ( x, 0) ẋ > 0 ( x, 0) (2h sin ϕ + + x, 0) (x, 0) (x + 2h sin ϕ +, 0) A A = 2h sin ϕ +. ϕ ẋ > 0 ϕ ẋ < 0 ẋ = 0 x = 0 x = h sin ϕ + ẋ = 0

12 ω > 0

13 web b l z y y x body m (a) (b) ( c) x ψ ζ ϕ = ψ + ζ z = 0 P = (b cos ψ, b sin ψ, 0), ψ = ωt b > 0 z = 0 2π y = 0 x > 0 ω 1 ω 2π ω 2π C θ 0 θ π ψ + ζ ζ x C (x, y) ϕ ψ + ζ δ = ζ ϕ b 0 C = (x, y, z) l x = b cos ψ + l sin θ cos(ψ + ζ) y = b sin ψ + l sin θ sin(ψ + ζ) z = l cos θ.

14 L(ζ, θ, t) = 1 2 m(ẋ2 + ẏ 2 + ż 2 ) + mgz (x, y, z) ψ = ωt l θ l(ω + ζ) 2 cos θ sin θ bω 2 cos θ cos ζ + g sin θ = 0, l ζ sin θ + 2l(ω + ζ) θ cos θ + bω 2 sin ζ = 0. b = 0 l 2 θ sin 2 θ = H θ θ z ϕ = ψ + ζ 1 ω 2π 1 g 2π l z θ ζ sin ζ = 0 ζ = 0 ζ = π ζ = 0 θ θ + (0, π) 2 B ζ = π θ ( π, π) 2 b < l [ ( ) ] g b 3 ω > 1. l l g 9.8 ms 2 l 0.05 m g 14 l

15 y x z z y x z x y (x, y, z) 0 < t < 20 ω = 5 (θ, θ, ζ, ζ) = (0.04, 0.01, 0.01, 0.05) ω = 9 (θ, θ, ζ, ζ) = (0.04, 0.01, 0.01, 0.05) ω = 12 (θ, θ, ζ, ζ) = (0.02, 0.07, 0.05, 0.01) ω < 13 g = 9.8, l = 0.05, b = 1 2 l. ω Hz ω = 5 ω = 9 ω z > 0 z = 0

16 ( 3 ) ( 1 = (l 0 + a), 1, = (l 0 + a) ) 3, 1, = (l 0 + a)(0, 1). x = 0 m = 3 1 k{ j l 0 } + ( j ) j. m = , l 0 = 0.17, a = 0.03, k = 18π m a x

17 0.06 (a) y 0 (b) y x (x, y) 0 < t < 3 (x, ẋ, y, ẏ) = (0.02, 0, 0.05, 0) (x, ẋ, y, ẏ) = (0, 0.03, 0.02, 0) r A = 2π k (1 l 0 ) > 0 m 2p r r θ 2 = Ar + O(r 2 ), r 2 θ = H H r 2 r = Ar + H2 r = ( ) 1 3 r 2 Ar2 + H2. 2r 2 V (r) = 1 2 Ar2 + H2 2r 2 r > 0 r 2 = H/ A 1 2π V (r) = 2π 1 2A x

18 H = 0 r = 0

19 ω ω

21 \\ \

22 r I = l 0 π ψ r p cos θ p2 + r 2 2rp cos θ dθ = l π 0 2r ψ 0 2r 2 2rp cos θ p2 + r 2 2rp cos θ dθ p 2 ( I = l π 0 p 2 r 2 ) π 2r ψ p2 + r 2 2rp cos θ dθ + p2 + r 2 2rp cos θ dθ. ψ F (ϕ, k) E(ϕ, k) ϕ dθ ϕ F (ϕ, k) = 1 k2 sin 2 θ, E(ϕ, k) = 1 k 2 sin 2 θ dθ. ϕ = π 2 F ( π 2, k) = π 2 ( k k4 +O(k 6 )), E( π 2, k) = π 2 (1 1 4 k k4 +O(k 6 )). A > B > 0 0 x π 1 A B cos θ dθ = 2 A + B F (δ, R) 0 A B cos θ dθ = 2 A + BE(δ, R) 2B sin θ A B cos θ.

23 δ = sin 1 (A + B)(1 cos θ) 2B, R = 2(A B cos θ) A + B. [ I = l 0 r (p r)f (δ θ, s) (p + r)e(δ θ, s) + 4pr sin θ p2 + r 2 2pr cos θ θ δ δ s = 2 pr. p + r r ψ = 0 ( = ˆ 2k πr + l ) 0 r ((p r)f ( π, s) (p + r)e( π, s)). 2 2 r ( ) = ˆ 2k π l 0 r + O(r 2 ). 2p ] π ψ,

P321(b), Assignement 1

P321(b), Assignement 1 P31(b), Assignement 1 1 Exercise 3.1 (Fetter and Walecka) a) The problem is that of a point mass rotating along a circle of radius a, rotating with a constant angular velocity Ω. Generally, 3 coordinates