[ 半径为 $r$ 的半圆弧的质心到圆心的距离为 $\frac{2r}{\pi}$。 ]

图 1 所示的均匀刚性金属丝框架，由圆心均为 $O$ 的两个金属丝半圆弧 $AB$、$CD$，通过两段直金属丝 $AC$、$DB$ 连接而成。

圆弧 $AB$ 的半径为 $2a$，圆弧 $CD$ 的半径为 $a$，$AC=DB=a$，且 $ACDB$ 共线。

框架位于同一平面内。

(a) 证明框架的质心到 $AB$ 的距离为 $\frac{6a}{3\pi+2}$。

$E$ 点位于圆弧 $AB$ 上，且 $\angle AOE$ 为直角。

框架从 $A$ 点自由悬挂。

框架静止悬挂于竖直平面内，$AE$ 与竖直向下方向成 $\theta$ 角。

(b) 求 $\tan\theta$ 的精确值，并化简答案。

这是均匀金属丝框架，质量与长度成正比，不与面积成正比。以连接 $A,B$ 的直线为参考，两半圆弧在相反侧，直线段也贡献质量但力矩为零。

Let the common mass per unit length be $\rho$. Measure signed perpendicular distances from the straight line $AB$ towards $E$ as positive.

The outer semicircular arc has length $2\pi a$ and centroid distance $4a/\pi$. The inner arc has length $\pi a$ and centroid distance $-2a/\pi$.

The straight pieces $AC$ and $DB$ have total length $2a$ and lie on $AB$, so their centroid distance is zero. The total wire length is

By symmetry, the centre of mass $G$ lies on $OE$. Taking moments of masses about $AB$,

Cancelling $\rho$ gives

Hence

悬挂时 $AG$ 竖直。先求 $AG$ 与 $AO$ 的角；由于 $AO=OE$ 且二者垂直，$AE$ 与 $AO$ 成 $45^\circ$。所求角是两角之差，须使用正切差角公式。

Let $\alpha=\angle OAG$. Since $AO=2a$ and $OG=d$,

Triangle $AOE$ is right-angled and isosceles, so

Since $0<d<2a$, $AG$ lies between $AO$ and $AE$. When suspended, $AG$ is vertically downwards, and therefore

Using the tangent subtraction formula,

Thus

and hence

也可将 $AG$ 沿 $AE$ 与垂直 $AE$ 的方向投影。两分量的比直接给角度正切，避免先引入另一个角。

Use the original orientation with coordinates

Then the vector from $A$ to $G$ has components $(d,-2a)$. A unit vector along $AE$ is $(1,-1)/\sqrt2$, so the component of $AG$ along $AE$ is

The magnitude of its perpendicular component is

since $d<2a$.

As $AG$ becomes vertical when suspended, the angle between $AG$ and $AE$ is $\theta$. Thus

Substituting $d=6a/(3\pi+2)$ and cancelling $2a$,

Hence