质量为 $m$ 的质点 $P$ 连接于一根长为 $a$ 的不可伸长轻绳的一端，另一端连接于固定点 $O$。如图 4 所示，将质点保持在 $A$ 点，其中 $OA=a$ 且 $OA$ 水平。

将质点以速率 $\sqrt{\dfrac{9ag}5}$ 竖直向下发射。

绳与过 $O$ 的竖直向下方向成 $\theta$ 角且仍然绷紧时，绳中的张力为 $S$。

(a) 证明 $S=\dfrac35mg(5\cos\theta+3)$。

绳变松的瞬间，$P$ 的速率为 $v$。

(b) 证明 $v=\sqrt{\dfrac{3ag}5}$。

(c) 求 $P$ 高于 $O$ 点所在水平面的最大高度。

先用能量把速率写成角度的函数，再代入径向牛顿第二定律。向心方向的重力分量为 $-mg\cos\theta$，张力本身不等于向心合力。

Write $w$ for the speed at a general position while the string is taut. Relative to $A$, the particle has fallen through a vertical distance $a\cos\theta$. Since the tension does no work,

Therefore

Resolving towards $O$, the tension acts inwards and the component of the weight in that direction is $-mg\cos\theta$. Thus

Substituting the energy result,

绳第一次变松时张力降到零。此时速率仍为正，因此是在上升过程中失去绷紧约束，而不是先停下来。

Let $\theta_s$ be the angle when the string first becomes slack. From part (a),

Substituting into the speed relation from part (a),

Taking the positive root,

松绳后作抛体运动。先找松绳点高于 $O$ 的高度，再用竖直速度分量计算额外上升高度；不能把整个速率都当作竖直速率。

At the point of slackening, the height above the level of $O$ is

The particle is moving upwards along the tangent. Since $\cos\theta_s=-3/5$, the corresponding sine is $4/5$, so its upward velocity component is

After slackening, the particle moves as a projectile. At its highest point its vertical velocity is zero. If its additional rise is $h_1$, then

Consequently the maximum height above the level of $O$ is

也可从最初发射点直接写到最高点的能量守恒。最高点只剩水平速度，这个分量从松绳后保持不变，不能把最高点动能写成零。

At slackening, the magnitude of the horizontal component of velocity is

This component is unchanged during the projectile motion. At the highest point the vertical velocity is zero, so its speed there is $v_x$.

Let $H$ be the height above the level of $O$, which is also the initial level of $A$. The tension does no work while the string is taut, and after slackening only gravity acts. Conservation of energy from $A$ to the highest point gives

Cancelling $mg$ and rearranging,