(a) 用代数积分证明：半径为 $a$ 的均匀实心半球 $H$ 的质心到其平面底面圆心 $O$ 的距离为 $\dfrac38a$。

[ 可直接使用半球的体积为 $\dfrac23\pi a^3$。 ]

从 $H$ 中挖去半径为 $\dfrac12a$ 的较小实心半球，形成均匀立体 $S$，其中

• 较小半球的平面底面的圆心为 $C$，且该底面是 $H$ 的平面底面的一部分；
• $OC=\dfrac12a$。

图 5 为 $S$ 的截面图，$S$ 是无阴影部分。

(b) 证明 $S$ 的质心到过 $O$、$C$ 的直线的距离为 $\dfrac{45}{112}a$。

如图 6 所示，立体 $S$ 静止平衡，其曲面与粗糙水平面接触。

$CO$ 与水平面之间的夹角为 $\theta$。

(c) 求 $\tan\theta$ 的精确值。

沿半球对称轴切薄圆盘，用体积矩除以体积。题目是实心半球，圆盘面积含半径平方，不能使用半圆薄片的质心公式。

Let $x$ measure distance from $O$ along the symmetry axis into the hemisphere. At coordinate $x$, the cross-sectional disc has radius $y$ satisfying

For uniform density $\rho$, the disc mass is $\rho\pi(a^2-x^2)\,\mathrm{d}x$. Its first moment about the plane face is

The mass is $M=2\rho\pi a^3/3$. By symmetry the centre of mass lies on the axis, and its distance from $O$ is

体积、质量与半径的立方成正比，因此完整大半球、挖去小半球及余下物体的质量比为 $8:1:7$。以共同平面底面计算垂直方向的质量矩。

Let the removed hemisphere have mass $M_0$. Since the radii are $a$ and $a/2$, the masses of the original, removed and remaining solids are $8M_0$, $M_0$ and $7M_0$ respectively.

Their centres of mass have perpendicular distances $3a/8$, $3(a/2)/8=3a/16$ and $d$ from the common plane face. The remaining centre of mass lies in the cross-sectional plane through $OC$ and the symmetry axes of the two hemispheres. Thus $d$ is also its distance from the line $OC$.

Taking first moments in this perpendicular direction,

Cancelling $M_0$,

挖去左侧小半球后，整体质心还会沿 $CO$ 方向向右偏移。先求这个偏移，再利用平衡时整体质心位于球心 $O$ 的竖直下方求倾角。

Let $q$ be the component of $OG$ parallel to $CO$, measured from $O$ away from $C$. The original hemisphere’s centre of mass has zero component in this direction, whereas the removed hemisphere’s has component $-a/2$. Taking signed first moments gives

Together with part (b), the components of $OG$ are therefore $q=a/14$ parallel to $CO$ and $d=45a/112$ perpendicular to the plane face, into the solid.

At the contact point with the horizontal plane, the sphere’s normal is vertical and passes through $O$. Horizontal equilibrium requires zero friction, so the contact force is vertical. For rotational equilibrium, the weight must have this same line of action. In the illustrated resting position, $G$ is vertically below $O$.

Since $CO$ makes angle $\theta$ with the horizontal, the horizontal components of the two parts of $OG$ cancel:

Consequently