Center of Mass

 

- for an arc: $\frac{r\sin\theta}{\theta}$ from the centre

moment: $\int_{-\theta}^\theta r\cos\theta\cdot r{\rm d}\theta=r^2\cdot[\sin\theta]|_{-\theta}^\theta=2r^2\sin\theta$

mass: $2r\theta$ so $2r^2\sin\theta\div2r\theta=\frac{r\sin\theta}{\theta}$

- for a sector: $\frac{2r\sin\theta}{3\theta}$ from the centre (分割三角,arc 和 sector 的重心位置的比例为 $\frac{1}{2}:\frac{1}{3}=\frac{2}{3}$)

- for a right circular cone with radius $r$ and height $h$: $\frac{3}{4}h$ from the vertex of the cone

moments: $\int^h_0t\cdot\pi\big({\small\frac{rt}{h}}\big)^2\ {\rm d}t=\pi r^2\int^h_0\frac{r^3}{h^2}\ {\rm d}t=\pi r^2\cdot\Big[{\small\frac{r^4}{4h^2}}\Big]\bigg|^h_0=\frac{\pi r^2 h^2}{4}$

mass: $\frac{\pi r^2h}{3}$ so $\frac{\pi r^2 h^2}{4}\div\frac{\pi r^2h}{3}=\frac{3h}{4}$

- for a right circular cone shell

mass: $\frac{2\pi r\cdot h}{2}=\pi rh$ (or $\int^h_02\pi\big({\small\frac{rt}{h}}\big)\ {\rm d}t=\pi r\cdot\Big[{\small\frac{t^2}{h}}\Big]\bigg|^h_0=\pi rh$)

moments: $\int^h_0t\cdot2\pi\big({\small\frac{rt}{h}}\big)\ {\rm d}t=\pi r\cdot\Big[{\small\frac{2t^3}{3h}}\Big]\bigg|^h_0=\frac{2\pi rh^2}{3}$ so $\frac{2\pi rh^2}{3}\div\pi rh=\frac{2h}{3}$

- similarly for pyramids (棱锥) and their shells

- for a hemisphere with radius $r$: $\frac{3}{8}r$ from the plane it is on

moments: $\int^r_0t\cdot\pi(r^2-t^2)\ {\rm d}t=\pi\int^r_0(r^2t-t^3)\ {\rm d}t=\pi\cdot\big[{\small\frac{r^2t^2}{2}-\frac{t^4}{4}}\big]\Big|^r_0=\frac{\pi r^4}{4}$

mass: $\frac{4\pi r^3}{3}\div2=\frac{2\pi r^3}{3}$ so $\frac{\pi r^4}{4}\div\frac{2\pi r^3}{3}=\frac{3r}{8}$

- for a hemispherical shell with radius $r$: $\frac{1}{2}r$ from the plane it is on

$(h-\frac{3h}{4}):(h-\frac{2h}{3})=\frac{h}{4}:\frac{h}{3}=\frac{3}{4}$ and $\frac{3r}{8}:\frac{r}{2}=\frac{3}{4}$​

posted @ 2024-05-23 17:10  Corylus  阅读(20)  评论(0编辑  收藏  举报