(a) 用代数积分证明：半径为 $r$ 的均匀半圆形薄片的质心到半圆圆心的距离为 $\dfrac{4r}{3\pi}$。

[可直接假设半径为 $r$ 的圆的面积为 $\pi r^2$，无须证明。]

在图 2 阴影所示的模板中，从半圆形薄片 $C_1$ 中挖去半圆形薄片 $C_3$，并将半圆形薄片 $C_2$ 连接于 $C_1$。

• $C_1$ 的半径为 $4a$，直径为 $AB$，圆心为 $O$；
• $C_2$ 的半径为 $2a$，圆心为 $O$；
• $C_3$ 的半径为 $a$，圆心为 $O$。

薄片 $C_1$、$C_2$ 由相同材料制成，并位于同一平面内。

模板从 $A$ 自由悬挂并处于平衡，$AB$ 与竖直向下方向成 $\theta^\circ$ 角。

(b) 求 $\theta$ 的值。

先把半圆放在 $y$ 轴右侧，以圆心为原点。竖条的长度为 $2\sqrt{r^2-x^2}$，质心横坐标就是 $x$；用替换 $u=r^2-x^2$ 完成面积矩积分。

Place the semicircle in $x\ge0$, with centre $O=(0,0)$ and diameter on the $y$-axis. By symmetry, its centre of mass lies on the positive $x$-axis.

A vertical strip has area $2\sqrt{r^2-x^2}\,\mathrm dx$. With uniform surface density cancelling, its first area moment about the $y$-axis is

Using $u=r^2-x^2$, $\mathrm du=-2x\,\mathrm dx$,

Consequently,

The semicircle has area $\pi r^2/2$. Hence its centre of mass is at distance

from $O$.

也可把半圆放在 $x$ 轴上方，用每根竖条的中点计算面积矩。这时被积式成为多项式，特别要保留竖条质心高度中的 $1/2$。

Place the semicircle above the $x$-axis, with centre $O=(0,0)$, so

By symmetry, $\bar x=0$.

A vertical strip has area $y\,\mathrm dx$ and centre of mass at height $y/2$. Its moment about the $x$-axis is therefore $\tfrac12y^2\,\mathrm dx$. Thus

Evaluating the limits,

Divide by the area $\pi r^2/2$:

Since $\bar x=0$, this is the distance from the centre.

第三种方法把半圆分成很窄的扇形。先说明每个微小扇形的质心位于约 $2r/3$ 处，再对这些小扇形的面积矩积分，不能只引用待证明的半圆质心公式。

Place the semicircle above the $x$-axis with centre $O$. Let $\phi$ be the polar angle from the positive $x$-axis, where $0\le\phi\le\pi$.

A narrow sector of angle $\mathrm d\phi$ has area $\tfrac12r^2\,\mathrm d\phi$. In the limit of a narrow sector its centre is at radial distance $2r/3$. This follows by weighting radial distance $s$ with the area element $s\,\mathrm ds\,\mathrm d\phi$:

Its centroid height is therefore $(2r/3)\sin\phi$. The moment of the whole semicircle about the $x$-axis is

By symmetry the centre of mass is on the $y$-axis. Dividing by the area,

which is its distance from $O$.

以直径 $AB$ 为参考线，向大半圆一侧取正。附加的中半圆在负侧，挖掉的小半圆在正侧，两者对总质量矩的贡献都是负的；但总面积是 $16+4-1$。

Take perpendicular distance from $AB$ as positive towards $C_1$. By symmetry the template’s centre of mass $G$ is on the line through $O$ perpendicular to $AB$.

The areas of $C_1,C_2,C_3$ are in the ratio $16:4:1$, and from (a) their signed centroid distances are respectively

The template contains $C_1+C_2-C_3$. Since all pieces have the same surface density, its signed distance $d=OG$ satisfies

Hence

When suspended from $A$, $AG$ is vertically downwards. The triangle $AOG$ is right-angled at $O$, with $AO=4a$, so the required angle between $AB$ and $AG$ is given by

Thus