圆锥体转动惯量

设圆锥体的体积为 $V$，质量为 $M$，底面半径为 $R$，高为 $H$。

体积微元

$$\text dV=r\text dr\text dh\text d\theta$$

体积

$$\begin{aligned} V =& \int_0^{2\pi}\int_0^H\int_0^{r(h)}r\text dr\text dh\text d\theta \\ =& 2\pi\int_0^H\int_0^{\frac RH(H-h)}r\text dr\text dh \\ =& 2\pi\int_0^H\frac 12\left(\frac RH(H-h)\right)^2\text dh \\ \xlongequal[h=Hx]{x=h/H} & \pi R^2\int_0^1(1-2x+x^2)H\text dx \\ =& \pi R^2H\left[x-x^2+\frac{x^3}{3}\right]_0^1 \\ =& \frac 13\pi R^2H \end{aligned}$$

质心高度

$$\begin{aligned} z =& \frac{\int_0^{2\pi}\int_0^{H}\int_0^{r(h)}hr\text dr\text dh\text d\theta} {\int_0^{2\pi}\int_0^{H}\int_0^{r(h)}r\text dr\text dh\text d\theta} \\ =& \frac 1{\frac 13\pi R^2H} 2\pi\int_0^Hh\int_0^{\frac RH(H-h)}r\text dr\text dh \\ =& \frac 6{R^2H}\int_0^H\frac h2\left(\frac RH(H-h)\right)^2\text dh \\ =& \frac 3H\int_0^H\frac {(H-h)^2}{H^2}h\text dh \\ \xlongequal[h=Hx]{x=h/H}& \frac 3H\int_0^1(1-2x+x^2)H^2x\text dx \\ =& 3H\left[\frac{x^2}2-\frac 23 x^3+\frac{x^4}4\right]_0^1 \\ =& \frac H4 \end{aligned}$$

质量微元

$$\text dm=\rho\text dV=\frac M{\frac 13\pi R^2H}r\text dr\text dh\text d\theta$$

绕z轴转动惯量

$$\begin{aligned} J_z =& \int_0^{2\pi}\int_0^H\int_0^{r(h)}r^2\text dm \\ =& \frac M{\frac 13\pi R^2H}\int_0^{2\pi}\int_0^H\int_0^{r(h)}r^2 r\text dr\text dh\text d\theta \\ =& \frac {6M}{R^2H}\int_0^H\int_0^{\frac RH(H-h)}r^3\text dr\text dh \\ =& \frac {6MR^4}{4R^2H}\int_0^H\frac{(H-h)^4}{H^4}\text dh \\ \xlongequal[h=Hx]{x=h/H}& \frac {3MR^2}{2H}\int_0^1(1-x)^4H\text dx \\ =& \frac {3MR^2}2(1-4\frac 12+6\frac 13-4\frac 14+1\frac 15) \\ =& \frac 3{10}MR^2 \end{aligned}$$

计算绕x轴转动惯量时，其内一点到旋转轴的距离为

$$d(r,h,\theta)=\sqrt{(h-\frac H4)^2+r^2\cos^2\theta}$$ $$\begin{aligned} J_x=& \int_0^{2\pi}\int_0^H\int_0^{r(h)}d^2(r,h,\theta)\text dm \\ =& \frac M{\frac 13\pi R^2H}\int_0^{2\pi}\int_0^H\int_0^{\frac RH(H-h)} \left((h-\frac H4)^2+r^2\cos^2\theta\right)r\text dr\text dh\text d\theta \\ =& \frac M{\frac 13\pi R^2H}\left(\pi R^2\int_0^H(h-\frac H4)^2 \frac{(H-h)^2}{H^2}\text dh +\int_0^{2\pi}\cos^2\theta\text d\theta\int_0^H \int_0^{\frac RH(H-h)}r^3\text dr\text dh\right) \\ =& \frac M{\frac 13\pi R^2H}\left(\pi R^2\int_0^1(1-x)^2(x-\frac14)^2H^3\text dx +\int_0^{2\pi}\frac{\cos 2\theta+1}{2}\text d\theta\frac{R^4}4\int_0^H \frac{(H-h)^4}{H^4}\text dh\right) \\ =& \frac M{\frac 13\pi R^2H}\left(\pi R^2H^3 \int_0^1(x^4-\frac 52x^3+\frac{33}{16}x^2-\frac 58x+\frac 1{16})\text dx +\frac{\pi R^4}4\int_0^1(1-x)^4H\text dx\right) \\ =& \frac M{\frac 13\pi R^2H}\left(\pi R^2H^3\frac 1{80} +\frac 1{20}\pi HR^4\right) \\ =& 3M\left(\frac{H^2}{80}+\frac{R^2}{20}\right) \end{aligned}$$

草稿纸

$$(x^2-2x+1)(x^2-\frac12x+\frac1{16})$$ $$\frac15-\frac58+\frac{11}{16}-\frac5{16}+\frac1{16} =\frac15-\frac3{16}=\frac1{80}$$