动力学习题

动力学习题

6.1 求一均质的、坐标原点建立在其质心的刚性圆柱体的惯性张量。

解：假设圆柱体的半径为$R$，高为$h$，密度为$\rho$，则其总质量$M=\frac{1}{2}\pi R^{2}h\rho$。

建立柱坐标系，则有
$$\begin{gathered} x=rcos\theta \\ y=rsin\theta \\ z=z \\ dv=rd\theta drdz \end{gathered}$$

I x x = ∫ ∫ ∫ V ( z 2 + y 2 ) ρ d v = ∫ 0 2 π ∫ 0 R ∫ − h / 2 h / 2 ( z 2 + r 2 sin ⁡ 2 θ ) ρ d z d r d θ = ∫ 0 2 π ∫ 0 R ( r 3 h sin ⁡ 2 θ + 1 12 h 3 r ) ρ d r d θ = ∫ 0 2 π ( 1 4 R 4 h sin ⁡ 2 θ + 1 24 R 2 h 3 ) ρ d θ = 1 8 R 4 h ρ ∫ 0 2 π ( 1 − cos ⁡ 2 θ ) d θ + ∫ 0 2 π 1 24 R 2 h 3 d θ = 1 8 π R 4 h ρ + 1 12 π R 2 h 3 ρ = 1 4 M R 2 + 1 12 M h 2 I_{xx}=\int \int \int_V (z^2+y^2)\rho dv\\ =\int_0^{2\pi}\int_0^R\int_{-h/2}^{h/2}(z^2+r^2\sin^2\theta)\rho dzdrd\theta\\ =\int_0^{2\pi}\int_0^R(r^3h\sin^2\theta+\frac{1}{12}h^3r )\rho drd\theta \\ =\int_0^{2\pi}(\frac{1}{4}R^4h\sin^2\theta+\frac{1}{24}R^2h^3)\rho d\theta \\ =\frac{1}{8}R^4h\rho \int_0^{2\pi}(1-\cos2\theta)d\theta+\int_0^{2\pi}\frac{1}{24}R^2h^3d\theta\\ =\frac{1}{8}\pi R^4h\rho+\frac{1}{12}\pi R^2h^3\rho\\ =\frac{1}{4}MR^2+\frac{1}{12}Mh^2 Ixx​=∫∫∫V​(z2+y2)ρdv=∫02π​∫0R​∫−h/2h/2​(z2+r2sin2θ)ρdzdrdθ=∫02π​∫0R​(r3hsin2θ+121​h3r)ρdrdθ=∫02π​(41​R4hsin2θ+241​R2h3)ρdθ=81​R4hρ∫02π​(1−cos2θ)dθ+∫02π​241​R