均匀圆片 $C$ 的圆心为 $X$，半径为 $R$。

从 $C$ 中挖去一个圆心为 $Y$、半径为 $r$ 的圆片，形成图 1 阴影所示的模板，其中 $0<r<R$，且 $XY=R-r$。

模板质心到 $X$ 的距离为 $kr$。

(a) 证明 $r=\frac{k}{1-k}R$。

(b) 由此求 $k$ 的可能取值范围。

$P$ 点位于模板外边缘上，且 $PX$ 垂直于 $XY$。

模板从 $P$ 点自由悬挂并处于平衡状态。

已知 $k=\frac49$，

(c) 求 $XY$ 与竖直方向所成的角。

模板的质量为 $M$。

(d) 用 $M$ 表示可连接于模板、使其从 $P$ 点平衡悬挂且 $XY$ 水平的最轻质点的质量。

孔洞用负质量处理。剩余薄片的质心在远离洞的一侧，以 $X$ 为原点、指向 $Y$ 为正，质心坐标应为 $-kr$。再约去面积公因子，逐步推出给定式。

By symmetry the template’s centre of mass $G$ lies on $XY$, on the opposite side of $X$ from $Y$.

Let the mass per unit area be $\sigma$, and measure signed distances from $X$ towards $Y$ as positive. The complete disc has mass $\sigma\pi R^2$ at coordinate 0; the removed disc has mass $\sigma\pi r^2$ at coordinate $R-r$.

Taking moments of masses about an axis through $X$,

Cancelling $\sigma\pi$ and reversing signs,

Since $0<r<R$, both $r$ and $R-r$ are nonzero. Dividing by $r(R-r)$ gives

Hence

and therefore

利用上一问的关系与 $0<r<R$。先确定分母为正再处理不等式，两个端点都不能取。

From the relation obtained in (a),

Thus $k>0$ and $k<1$, so $1-k>0$. Now $r<R$ and (a) give

Dividing by $R>0$ and multiplying by $1-k>0$,

so

悬挂时 $PG$ 竖直。$PX$ 与 $XG$ 垂直，构成直角三角形；要求的是 $XY$ 与竖直的角，所以正切是 $PX/XG$，不是其倒数。

For $k=4/9$, (a) gives

Hence

Since $PX\perp XY$ and $G$ lies on $XY$, triangle $PXG$ is right-angled at $X$, with $PX=R$.

When suspended, $PG$ is vertically downwards. If $\alpha$ is the smaller angle between $XY$ and the vertical, then

Therefore

让 $XY$ 水平时，薄片重力对悬点的力臂为 $kr$。为使附加质量最小，应在另一侧取得最大的水平力臂 $R$；先说明这个最大值，再列力矩平衡。

With $XY$ horizontal, the vertical through $P$ also passes through $X$. The template’s weight $Mg$ has perpendicular distance

from this vertical.

Let the attached mass be $m_1$, at horizontal distance $b$ on the opposite side of $X$ from $G$. Taking moments about $P$,

Every point of the template is within the original disc, so $b\le R$. To minimise $m_1$, choose the largest possible distance $b=R$, at the farthest outer edge in the direction from $X$ towards $Y$.

Thus

and the minimum mass is