質點下滑所需時間的計算(3)

延續第二題 T = ∫ √{ [1 + (y')^2] /(2E/m - 2gy) } dx (0 - >b )...(1)

利用代換z = A + v_0^2/2g - y﹐T = ∫ √{ [1 + (z')^2] /(2gz) } dx (0 - >b )

由Euler's equation,取F = √{ [1 + (z')^2] /z }

z'∂ F/∂ z' - F = 常數。邊界條件z(0) = v_0^2/2g, z(b) = A + z(0)

∂ F/∂ z' = z'/√{ z[1 + (z')^2] }﹐因此

(z')^2/√{ z[1 + (z')^2] } - √{ [1 + (z')^2] /z } = -1/c (c是常數)

整理：z(1 + z') = c^2 或 dz/dx = √(c^2/z - 1)

x = ∫ √[z/(c^2 - z)] dz

代z = c^2(sinϕ)^2

x = 2c^2 ∫ (sinϕ)^2 dϕ = 2c^2 ∫ (1 - cos2ϕ) dϕ

得x = (1/2)c^2(2ϕ - sin2ϕ) + d 及 z = (1/2)c^2(1 - cos2ϕ)

c,d由A,b及v_0決定。而這條方程所表示的正是擺線。

最後求T。用鏈式法則﹐z'(x) = z'(ϕ)/x'(ϕ)

T = ∫ √{ [1 + (z'(ϕ)/x'(ϕ))^2] /(2gz(ϕ)) } x'(ϕ) dϕ (0 - >ϕ(b) )

z'(ϕ)/x'(ϕ) = 2c^2sinϕcosϕ/c^2(1 - 2cosϕ) = 1/tanϕ

所以﹐T = ∫ 2c/√(2g) dϕ (0 - >ϕ(b) ) = √(2A/g) [ϕ(b)/sin(ϕ(b)]

若A = L﹐ϕ(b) = π/2 及 sin(ϕ(b) = 1

則T = (π/2) √(2L/g)

2011-04-17 10:47:11 補充：

brochistonchrone problem 和 tautochrone problem 是兩個不同的問題。 brochistonchrone 是求一條最速下降曲線。tautochrone 是求一條質點在各個起始位置到某一點所需時間相同的曲線。 βραχίστος, brachistos - the shortest, ταὐτό, tauto - the same, χρόνος, chronos - time。雖然兩者都是擺線。

To ： 教書的 ( 專家 5 級 )

The tautochrone problem of cycloid

We will solve it by Abel’s Method for The tautochrone problem.

{* in the following: t for the parameter of the parameter equations [x(t)=t-sint, y(t)=1-cost] of cycloid and T for time.*}

Abel's solution begins with the principle of conservation of energy — since the particle is frictionless, and thus loses no energy to heat, its kinetic energy at any point is exactly equal to the difference in potential energy from its starting point. The kinetic energy is 1/2mv^2, and since the particle is constrained to move along a curve, its velocity is simply v=ds/dT, where s is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height y(t0) to a height y(t) is mg(y(t0)-y(t), thus:

1/2mv^2=1/2m(ds/dT)^2=mg(y(t0)-y(t))

= > ds/dT=√[2g(y(t0)-y(t))]

= > dT=(1/√[2g(y(t0)-y(t))]ds

=(1/√(2g))[1/√(y(t0)-y(t))]ds.

So

dT=(1/√(2g))∫[t0→π] [1/√(y(t0)-y(t))]ds.

For we only interest in the isochronous time of the problem, we omit the constant (1/√(2g)).

T=∫[t0→π] [1/√(y(t0)-y(t))]ds …(*).

For briefly, we also omit the constant L/2, the cycloid can express as:

x(t)=t-sint,

y(t)= cost-1

where 0≦t<π.

We know that

ds=√(dx^2+dy^2

=[√(2-2cost)]dt

=√2[√(1-cost)]dt

And y(t0)-y(t)

=1/[(cos(t0)-1)-(cost-1]

=1/[cos(t0)-cost]

Substitute in (*), we have

T=∫[t0→π]( √2[√(1-cost)]/ [cos(t0)-cost]dt

{let u=t/2, 1-cost=2(sinu)^2, and cos(t0)-cost=cos(t0)-2(cosu )^2+1

=( cos(t0)+1)-2(cosu )^2

=2[(cos(t0)+1)/2-(cosu )^2]

{let k^2= (cos(t0)+1)/2}

=2[k^2- cosu )^2] }

T=∫[t0/2→π/2] √2√(2(sinu)^2)/ √ {2[k^2- cosu )^2]}(2du)

=2√2∫[t0/2→π/2] sinu/√[k^2- cosu )^2]du

{ let w=cosu dw=-sinudu }

=2√2∫[arc cos(t0/2)→0] 1/√[k^2- w^2]dw

=2√2 arc sin(w/k)|[ arc cos(t0/2), 0]

=2√2[arc sin(cos(t0/2)/k)-arc sin(0/k)]

{ since k^2= (cos(t0)+1)/2=(cos(t0/2))^2,

k= cos(t0/2)) }

=2√2[arc sin(1)]

=2√2[π/2]

=√2.

[[Done]]

2011-04-14 19:02:35 補充：

please refer to Tautochrone curve in WIKI.

擺線是最速下降曲線,而且具等時性,即下滑最快的曲線,

個人覺得最特別的是由任意點開始下滑到達終點(a, -L)所需時間是相同的!

2011-04-19 21:08:24 補充：

本題兩提問:(1)由(0,0)下滑至(a, -L), a=πL/2, 問所需時間?

(2)由曲線上任意點(p, q), 0

2011-04-26 19:52:16 補充：

本題特點是任意點下滑所需時間是相同的!

由(0,0)下滑至(a, -L), a=πL/2, 所需時間 T=sqrt(L/(2g))*pi;

由曲線上任意點(p, q)下滑至(a, -L), a=πL/2, 所需時間 T=sqrt(L/(2g))*(pi-t_0), where t_0 is the time at which (x(t_0), y(t_0))=(p,q).

2011-04-11 23:02:48 補充：

大大的(1),(2), and(3)系列是極好的教材. 質點延擺線下滑才是最省時間.

2011-04-13 23:03:28 補充：

很期待"等時性"可以由數學計算裡表現出來.

2011-04-15 08:55:01 補充：

To Sam,

Thank you for the correct answer. I was thinking the brochistonchrone problem and missed the initial condition at the beginning "K.E=P.E."