本题中可直接使用公式手册给出的均匀圆扇形质心公式，无须证明。

图 3 阴影所示的均匀薄片 $ABCDE$ 由一个长方形与一个圆扇形连接而成。

• 长方形 $ABCE$ 满足 $AB=EC=a$，$AE=BC=d$；
• 扇形 $CDE$ 的圆心为 $C$，半径为 $a$；
• $\angle ECD=\frac\pi3$ 弧度。

薄片的质心位于 $EC$ 上。

(a) 证明 $a=\sqrt3d$。

薄片从 $B$ 点自由悬挂并处于平衡状态，$BC$ 与竖直向下方向成 $\beta$ 弧度角。

(b) 求 $\beta$ 的值。

扇形质心位于角平分线上，距圆心的公式须用半角 $\pi/6$。组合质心在 $EC$ 上，意味着长方形和扇形关于 $EC$ 的面积矩大小相等、方向相反。

For a uniform sector of radius $a$ and angle $2\theta$, the centroid is at distance

from the centre along the angle bisector. Here $\theta=\pi/6$, so this distance is

Its perpendicular distance from $EC$ is

The sector has area $\pi a^2/6$. The rectangle has area $ad$, and its centroid is $d/2$ from $EC$ on the opposite side.

Since the combined centre of mass lies on $EC$, moments of areas about $EC$ give

Hence

Since $a,d>0$,

先求组合质心到 $BC$ 的距离。扇形质心距 $BC$ 是沿 $EC$ 方向的投影，与上一问距 $EC$ 的投影不同。悬挂平衡时，悬点到质心的连线竖直向下，所求角等于 $BC$ 与该连线的夹角。

Let $G$ be the centre of mass and let its distance from $BC$ be $y$. The rectangular centroid is at distance $a/2$ from $BC$. The sector centroid is at distance

from $BC$.

Using $d=a/\sqrt3$ and taking moments of areas about $BC$,

Simplifying,

Also $G$ lies on $EC$, so its perpendicular distance from the line $BC$ is $y$, and its distance along $BC$ from $B$ is $d$.

When the lamina hangs freely from $B$, $BG$ is vertically downwards. Therefore

Hence