Anonymous
Anonymous asked in 科學數學 · 1 decade ago

轉動慣量的積分

最近學到轉動,有提到轉動慣量,但書上只寫出各式各樣形體的轉動慣量,我想要的是用轉動慣量定義,積出答案來的過程~~(別用平行軸定理)

1.鐵圈,以鐵圈中心為轉軸,距轉軸為R

2.鐵圈,以直徑為轉軸,鐵圈半徑R

3.實心圓柱,底圓半徑R,高L,以圓柱中心為轉軸

4.實心圓柱,底圓半徑R,高L,以高的一半通過直徑為轉軸

5.實心球體,球半徑R,以球直徑為轉軸

6.空心球體,球半徑R,以球直徑為轉軸

7.同心空心圓柱,底大圓半徑R,小圓半徑r,以中心為轉軸

就這樣啦~~@@

1 Answer

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  • 1 decade ago
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    1.

    dm=Rdθ*(M/2πR)=Mdθ/2π

    ∫(0 to 2π)R^2dm

    =(MR^2/2π)∫(0 to 2π)dθ

    =MR^2

    2.

    dm=Mdθ/2π

    r=Rsinθ

    ∫(0 to 2π)r^2dm

    =2∫(0 to π)(R^2)(sin^2θ)Mdθ/2π

    =(MR^2/π)∫(0 to π)[(1-cos2θ)/2]dθ

    =(MR^2/π){π/2-(1/4)[sin2π-sin0]}

    =MR^2/2

    3.

    dm=M*2πrdr/πR^2

    ∫r^2dm

    =∫(0 to R)[2πMr^3/πR^2]dr

    =(2M/R^2)∫(0 to R)r^3dr

    =(2M/R^2)(R^4/4)

    =MR^2/2

    4.

    dm=Mrdθdrdx/πR^2L

    I=∫(-L/2 to L/2)∫(0 to R)∫(0 to 2π)(r^2cos^2θ+x^2) Mrdθdrdx/πR^2L

    =(2M/πR^2L)∫(0 to L/2)∫(0 to R)∫(0 to 2π)(r^2cos^2θ+x^2)rdθdrdx

    =(2M/πR^2L)∫(0 to L/2)∫(0 to R)(πr^2+2πx^2)rdrdx

    =(4M/πR^2L)∫(0 to L/2)(πR^4/4+πR^2x^2)dx

    =(2M/πR^2L)(πR^4L/8+πR^2L^3/24)

    =MR^2/4+ML^2/12

    5.

    dm=Mr^2sinφdrdθdφ/(4/3)πR^3=3Mr^2sinφdθdφ/4πR^3

    I=∫(0 to R)∫(0 to 2π)∫(0 to 2π)r^2sin^2φ*3Mr^2sinφdrdθdφ/4πR^3

    =(3M/4πR^3)*8∫(0 to R)∫(0 to π/2)∫(0 to π/2)r^4sin^3φdθdφdr

    =(6M/πR^3)*(π/2)∫(0 to R)∫(0 to π/2)r^4sin^3φdφdr

    =(3M/R^3)*(2/3)∫(0 to R)r^4dr

    =(2M/R^3)*(R^5/5)

    =2MR^2/5

    6.

    dm=MR^2sinφdθdφ/4πR^2=Msinφdθdφ/4π

    r=Rsinφ

    I=∫(0 to 2π)∫(0 to 2π)R^2sin^2φ*Msinφdθdφ/4π

    =(MR^2/4π)*8∫(0 to π/2)∫(0 to π/2)sin^3φdφdθ

    =(2MR^2/π)*(π/2)∫(0 to π/2)[3sinφ/4-sin3φ/4]dφ

    =MR^2*[3/4-1/12]

    =2MR^2/3

    7.

    dm=M*2πtdt/π(R^2-r^2)

    ∫t^2dm

    =∫(r to R)[2πMt^3/π(R^2-r^2)]dt

    =[(2M/(R^2-r^2)]∫(r to R)t^3dt

    =[(2M/(R^2-r^2)]*(R^4-r^4)/4

    =M(R^2+r^2)/2

    Source(s): 轉至 老王 即下面意見的總和
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