一个内半径为 $6r$ 的半球形碗被固定，圆形碗口水平。碗口圆心为 $O$，碗面上的 $A$ 点位于 $O$ 正下方。质点 $P$ 在碗的光滑内表面上沿以 $C$ 为圆心的水平圆周运动，角速度恒为 $\sqrt{\dfrac g{4r}}$。如图 1 所示，$C$ 点位于 $OA$ 上。

用 $r$ 表示距离 $OC$。

支持力沿球面法线指向 $O$，其水平分量提供向心力，竖直分量平衡重力。水平圆周的半径是 $CP$，不能直接用碗半径 $6r$。

Let the mass of $P$ be $m$, let $R$ be the normal reaction, and let $\theta=\angle COP$. The only forces on $P$ are its weight $mg$, vertically downwards, and $R$, directed along $PO$ towards $O$.

The radius of the horizontal circle is

and the angular speed satisfies $\omega^2=g/(4r)$. Resolving horizontally towards $C$,

The horizontal circle has non-zero radius, so $\sin\theta>0$. Cancelling it gives

There is no vertical acceleration. Hence

Therefore

还可以在竖直截面内沿球面切线分解，直接消去支持力。注意水平向心加速度在这个切线方向仍有分量，不能把切线方向误当成零加速度。

Let $\theta=\angle COP$, so that the horizontal radius is $6r\sin\theta$. The acceleration is horizontal, towards $C$, with magnitude

In the vertical plane $OCP$, resolve along the tangent to the bowl, directed downwards towards $A$. The normal reaction has no component in this direction. The component of the acceleration is $a_c\cos\theta$, while the component of the weight is $mg\sin\theta$. Thus

Since $\sin\theta>0$,