maths+phy(mechanics2) 22pts

2010-02-21 3:31 am
use the parametrization for a cycloid to show that
"a body
made to oscillate along a cycloid will always take the same time to
cover it"


http://www.scitechantiques.com/cycloidhtml/

The cycloid posses interesting physical properties. It is brachistochronous and tautochronous: brachistochronous, because it represents the path completed in the shortest time between two points for a given type of motion (such
as a fall under the effect of gravity); tautochronous, because a body
made to oscillate along a cycloid will always take the same time to
cover it, whatever the amplitude of the oscillation..........
更新1:

我看書看過的這個theorem的 我學過少少coordinate geometry但證極都證不到 我問過老師他說某國(忘)的科學院以此為公開題 一共有5人交答案 其中一人沒有在答案寫上名字,有人認出了牛頓獨特的符號就說 「雖然魔鬼沒有留下名字,但我們看到魔鬼的爪子」 有無高手可以教路?

更新2:

感謝煩惱即是菩提大師 這個問題我在兩年前已看過 最近偶爾再看到便拿來 我以前嘗試做時都能找到x,y方向的加速度 但要證"最短時間"便完全不知如何入手了

回答 (1)

2010-02-24 7:54 am
✔ 最佳答案
今晚幫您解這兩題

2010-02-23 23:54:34 補充:
已知擺線(x,y)=(θ-sinθ,cosθ-1),t表時間,如圖

圖片參考:http://imgcld.yimg.com/8/n/AE03435620/o/701002200156813873402950.jpg

or http://s585.photobucket.com/albums/ss296/mathmanliu/cy01.gif

設物體(質量m)由A(θ=a, t=0)移動至Q(θ=π, t=T)點,設s(t)=AP之弧長(想像A介於O,P之間)
A移動至P點,物體下降(cosa-1)-(cosθ-1)=cosa-cosθ
由能量守恆: (1/2)m(ds/dt)^2=mg(cosa-cosθ), so ds=√[ 2g (cosa-cosθ)]dt
又弧長導數ds/dθ=√[(dx/dθ)^2+(dy/dθ)^2]=√(2-2cosθ)
so, ∫[a,π]√(2-2cosθ)/√(cosa-cosθ) dθ= ∫[0~T]√( 2g )dt=T√( 2g )
∫[a,π] sin(θ/2)/√{[cos(a/2)]^2-[cos(θ/2)]^2}dθ=T√g
-2arcsin[cos(θ/2)/cos(a/2)] 代θ= a~π
so, T=π/√g
The spending time T is independent on a(initial position). Q.E.D

圖片參考:http://imgcld.yimg.com/8/n/AE03435620/o/701002200156813873402951.jpg

or http://s585.photobucket.com/albums/ss296/mathmanliu/cy02.gif


收錄日期: 2021-05-04 00:47:00
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