一个物体由半径为 $2a$ 的半球壳连接在一个高为 $h$、底面半径为 $a$ 的封闭空心圆柱体上组成。半球壳和空心圆柱体由相同的均匀材料制成。球壳与圆柱体的对称轴重合。$AB$ 是圆柱体下端的一条直径（见图）。

(a) 求该物体质心到 $AB$ 的距离的表达式，用 $a$ 和 $h$ 表示。

将该物体放置在一个与水平面成 $\theta$ 角的粗糙斜面上，其中 $\tan\theta=\frac23$。物体处于平衡状态，$AB$ 与斜面接触，并沿斜面的一条最大坡度线放置。

(b) 求 $h$ 的可能取值集合，用 $a$ 表示。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

两部分是薄壳，用表面积作质量权重。圆柱封闭，要计入侧面与两个圆形端面。沿对称轴取矩。

The closed cylinder has area $2\pi ah+2\pi a^2$ and centroid at distance $h/2$ above $AB$. The hemispherical shell has curved area $2\pi(2a)^2=8\pi a^2$. Its centroid is a distance $a$ from its rim plane towards the pole, so its distance from $AB$ is $h+a$. Both centroids lie on the common axis. Taking surface-area moments and cancelling $\pi a$ gives

重力作用线投到斜面上的交点必须在底面范围内。沿最大坡度方向，它相对底面圆心向下坡偏移 $d\tan\theta$，不能超过底面半径。

The centre of mass is a perpendicular distance $d$ from the plane. Its vertical line of action meets the plane at distance $d\tan\theta$ downhill from the centre of the base. Equilibrium therefore requires

Using $\tan\theta=2/3$ gives $d\le3a/2$. Since $h>0$ and $a>0$, the denominator $2(h+5a)$ is positive. Substituting for $d$ and multiplying by it gives

Since the cylinder has positive height, the possible heights are $0<h\le a$. The upper endpoint is limiting equilibrium about the downhill edge.