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 5 所示，最初将 $P$ 保持静止，绳绷紧且水平，然后以速率 $u$ 将质点竖直向下发射。

模型中忽略空气阻力。

绳已转过 $\theta$ 角且仍绷紧的瞬间，绳中的张力为 $T$。

(a) 证明 $T=\dfrac{mu^2}a+3mg\sin\theta$。

已知 $u=2\sqrt{\dfrac{3ag}5}$，

(b) 用 $a$ 和 $g$ 表示绳变松的瞬间 $P$ 的速率。

(c) 由此用 $a$ 表示 $P$ 在随后运动中高于 $O$ 的最大高度。

先用能量写出变速率，再列径向方程。角度从初始水平线算起，下降高度为 $a\sin\theta$；两个方程都应保留一致的正负号。

Let the speed at the general position be $w$. The particle has fallen a vertical distance $a\sin\theta$ from its initial level. Since tension does no work, conservation of energy gives

so

Resolving along the string towards $O$,

Using (1),

第一次松绳时张力降至零，此时已经绕过最低点并上升到 $O$ 的水平面以上。将此位置代回能量式求速率。

Let $\theta_s$ be the angle at slackening. Since $u^2=12ag/5$, part (a) with $T=0$ gives

This is first reached on the upward portion of the circular motion, after passing the lowest point. Substituting into (1),

Hence the speed is

松绳后为抛体运动。切向速度的竖直分量只占速率的 $3/5$，算出的额外上升高度还要加上松绳点本来高于 $O$ 的高度。

At slackening, the height above $O$ is

At this first slackening point $\cos\theta_s=-3/5$. The velocity is tangent to the circle and directed upwards, so its upward component is

The subsequent motion is projectile motion. If $h$ is the additional rise before the vertical velocity becomes zero, then

Thus the maximum height above $O$ is

从最初发射点到最高点也可用总能量。松绳后的水平速度不变，因此最高点的动能仍不为零。

At slackening, the horizontal speed is

It stays constant during projectile motion, and is the entire speed at the highest point. Let that point be a height $H$ above $O$, which is also the initial level of projection.

The tension does no work while the string is taut, and only gravity acts after it becomes slack. Conservation of energy from projection to the highest point gives

Cancelling $mg$ and simplifying,