一个固定的光滑球壳的球心为 $O$，半径为 $a$。质量为 $m$ 的质点在球壳光滑内表面上作完整的竖直圆周运动，其中圆周运动所在平面竖直且经过 $O$。当质点位于点 $A$ 时，其速度为 $v$；$OA$ 与经过 $O$ 的竖直向上方向成角 $\theta$，且 $\cos\theta=\frac1{18}$（见图）。

(a) 证明

已知

(b) 用 $m$ 和 $g$ 表示壳与质点之间法向反作用力的最大可能值。

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.

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

需要重力加速度的数值时，取 $g=10\,\mathrm{m\,s^{-2}}$。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。

完整圆周运动要求粒子在最高点仍能贴住内壁。先用最高点向心力求速度下限，再从 $A$ 到最高点用机械能守恒。

Let $w$ be the speed at the top. The inward radial equation there is

The shell can only push, so $R_{\mathrm{top}}\ge0$ and $w^2\ge ag$. The rise from $A$ to the top is $a(1-\cos\theta)$. Conservation of energy gives

Taking the positive square root gives $v\ge\frac13\sqrt{26ag}$.

粒子越低，速度越大，而重力的向心分量越小，故支持力在最低点最大。先由能量求最低点的速度，再列向心力方程。

At a position of angle $\phi$ from the upward vertical, with speed $V$, the inward equation is $R+mg\cos\phi=mV^2/a$. Energy gives $V^2=v^2+2ga(\cos\theta-\cos\phi)$. Therefore $R=mv^2/a+2mg\cos\theta-3mg\cos\phi$. Thus $R$ decreases as $\cos\phi$ increases, and its maximum occurs at the bottom.

The speed squared there is

The inward direction is upward, so $R_{\max}-mg=m(5ag)/a$. Hence $\boxed{R_{\max}=6mg}$.