两个固定点 $A$、$B$ 满足 $AB=0.56\,\mathrm{m}$，且 $A$ 位于 $B$ 正上方。

$AB$ 上的 $C$ 点满足 $AC=0.36\,\mathrm{m}$、$CB=0.2\,\mathrm{m}$。

一根长为 $0.64\,\mathrm{m}$ 的不可伸长轻绳的一端连接于 $A$，另一端连接于 $B$。

质量为 $0.1\,\mathrm{kg}$ 的质点连接于绳上的 $P$ 点，其中 $AP=0.39\,\mathrm{m}$。

如图 2 所示，质点以恒定角速度沿以 $C$ 为圆心的水平圆周运动。

质点的角速度为 $\omega\,\mathrm{rad\,s^{-1}}$。

已知绳的两段均绷紧，

(a) 求 $\omega=3\sqrt5$ 时 $PB$ 中的张力。

(b) 求使 $PB$ 处于松弛状态的 $\omega$ 的最大值，并清楚展示方法。

上、下两段绳的张力不同。先用几何求水平半径及各方向的比例，再分别列竖直平衡和水平向心力方程。

The lower part has length

and the radius of the horizontal circle is

Let the tensions in $PA$ and $PB$ be $T_A$ and $T_B$ respectively. The force directions are along $PA$ towards $A$, along $PB$ towards $B$, and vertically downwards for the weight.

Resolving vertically gives

Resolving horizontally towards $C$, with $\omega^2=(3\sqrt5)^2=45$,

Multiply (1) by $5$ and (2) by $12$:

Subtracting,

Thus

求的是下段绳开始产生张力的临界角速度，因此在临界状态令 $T_B=0$。只有临界状态才使用双绳刚好伸直的几何；真正松弛时，圆心的高度会改变。

At the limiting angular speed $\omega_c$, the lower part is just at zero tension and has its full length. Thus the limiting geometry is still $CP=0.15\ \mathrm m$ and $AC=0.36\ \mathrm m$.

With $T_B=0$, vertical balance and horizontal circular motion give

Dividing the second equation by the first,

Therefore the required upper limiting value is

For clarity, while $PB$ is genuinely slack, let $h$ be the vertical distance of $P$ below $A$. The upper string alone gives $\omega^2=g/h$. With $AP=0.39$ and $AB=0.56$, the straight-line distance from $P$ to $B$ satisfies

Thus $BP<0.25$ requires $h>0.36$, giving $\omega<\omega_c$. The stated limiting value is the boundary where $PB$ first becomes straight with zero tension.