[請益] 曲線面積的二次矩(慣性矩)
答案:18
小弟的算法:
x = k*y^3
4 = k*27
k = 4/27 = 0.15
x = 0.15*y^3
y = 1.88*x^(1/3)
4
Ix =∫{[(1/2)*y]^2} * dA = ∫{[(1/2)*y]^2}*y*dx = ∫ [(1/4)*y^3]*dx
0
4 4
=∫ 1.66*x*dx = 0.83x^2 | = 13.28
0 0
小弟積分出來的結果為Ix = 13.28,但是答案為Ix = 18
請問版上前輩們,後學哪個地方計算錯誤,還是哪邊的觀念有問題呢?
麻煩不吝嗇指導與告知,謝謝!
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◆ From: 114.35.30.78
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12/16 15:13, , 1F
12/16 15:13, 1F
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12/16 15:14, , 2F
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12/16 15:19, , 3F
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12/16 15:19, , 4F
12/16 15:19, 4F
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12/16 15:20, , 5F
12/16 15:20, 5F
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12/16 15:22, , 6F
12/16 15:22, 6F
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12/16 15:23, , 7F
12/16 15:23, 7F
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這樣積出來的慣性矩應該是矩型的慣性矩,而非曲線的慣性矩耶?
推
12/16 15:25, , 8F
12/16 15:25, 8F
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12/16 15:27, , 9F
12/16 15:27, 9F
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12/16 15:28, , 10F
12/16 15:28, 10F
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12/16 15:29, , 11F
12/16 15:29, 11F
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12/16 15:33, , 12F
12/16 15:33, 12F
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12/16 15:34, , 13F
12/16 15:34, 13F
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12/16 15:37, , 14F
12/16 15:37, 14F
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這題y形心計算方式有錯,應該寫成:∫(1/2)y*y*dx,這樣才是正確的吧?
※ 編輯: mountain161 來自: 114.35.30.78 (12/16 15:45)
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12/16 15:50, , 15F
12/16 15:50, 15F
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12/16 15:52, , 16F
12/16 15:52, 16F
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12/16 15:56, , 17F
12/16 15:56, 17F
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12/16 17:07, , 18F
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12/16 17:08, , 19F
12/16 17:08, 19F
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12/16 17:31, , 20F
12/16 17:31, 20F
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12/16 17:32, , 21F
12/16 17:32, 21F
剛剛仔細翻了一下書籍的解法:
若以(垂直)微小方塊去做慣性矩,(1) Ix = ∫dIx = ∫(dx*y^3)/3
(2) Iy = ∫(x^2)*dA = ∫(x^2)*y*dx
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※ 編輯: mountain161 來自: 114.35.30.78 (12/16 18:50)
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12/16 18:58, , 22F
12/16 18:58, 22F
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12/17 15:25, , 23F
12/17 15:25, 23F
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12/17 15:26, , 24F
12/17 15:26, 24F
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12/17 15:27, , 25F
12/17 15:27, 25F
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12/21 22:47, , 26F
12/21 22:47, 26F