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# 轉動慣量的積分

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

Rating

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