[問題] 單擺的邊界值如何找??
y↑
│
│
------------- -------------→x
│↘
│ ↘
│θ ↘ 繩長L 張力T
│ ↘
│ ●
│ ↓ mg
│
從x=Lsinθ
y=-Lcosθ
x'=Lθ'cosθ
y'=Lθ'sinθ
x"=-Lsinθ(θ')^2+Lcosθθ"
y"=Lcosθ(θ')^2+Lsinθθ"
在利用牛頓第二定律F=ma
-Tsinθ=mx"
Tcosθ-mg=my"
最後得到Lθ"+gsinθ=0
我的問題是這種單擺的邊界條件要怎麼找??
y(0)=?? y'(0)=?? 這種的
謝謝大大耐心回答
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 120.107.149.233
→
03/15 12:17, , 1F
03/15 12:17, 1F
推
03/15 12:24, , 2F
03/15 12:24, 2F
→
03/15 12:30, , 3F
03/15 12:30, 3F
→
03/15 12:32, , 4F
03/15 12:32, 4F
若以這題來看 y(0)=-L y'(pi/2)=0 這樣對嗎??
※ 編輯: redwing119 來自: 120.107.149.233 (03/15 16:40)
→
03/15 16:42, , 5F
03/15 16:42, 5F
→
03/15 16:43, , 6F
03/15 16:43, 6F
→
03/15 16:44, , 7F
03/15 16:44, 7F
→
03/15 16:44, , 8F
03/15 16:44, 8F
→
03/15 16:45, , 9F
03/15 16:45, 9F
→
03/15 16:53, , 10F
03/15 16:53, 10F
→
08/13 16:07, , 11F
08/13 16:07, 11F
→
09/17 14:07, , 12F
09/17 14:07, 12F