一个质量为 $m$ 的质点连接于一根长为 $l$ 的轻杆的一端，另一端连接于固定点 $O$。杆可绕过 $O$ 的水平轴在竖直平面内自由转动。

如图 2 所示，将质点从 $A$ 点以速率 $u$ 发射，其中 $OA$ 与过 $O$ 的竖直向上方向成 $\alpha$ 角。

质点沿完整的竖直圆周运动。

已知 $\cos\alpha=\dfrac45$，

(a) 证明 $u>\sqrt{\dfrac{2gl}5}$。

杆转动时，杆中的最小张力为 $T$，最大张力为 $4T$。

(b) 证明 $u=\sqrt{\dfrac{17}5gl}$。

轻杆可以提供拉力或压力，因此完成圆周运动的条件来自最高点的速率为正。由初始位置到最高点的高度增加是 $l(1-\cos\alpha)$，据此列能量不等式。

Let $v_t$ be the speed at the highest point. From $A$ to the highest point the increase in height is

Conservation of mechanical energy gives

For complete circles, the particle must pass the highest point with positive speed. Therefore

Equality would give zero speed at the highest point and would not give a completed revolution. The rod can exert compression as well as tension, so no additional positive-tension condition is needed.

先说明张力在最高点最小、最低点最大，再分别写能量方程和径向运动方程。两个位置的速度不同，不能使用同一个速度值。最后才代入张力比。

Let $\theta$ be the angle of the rod from the downward vertical, and let $N$ be the signed inward force exerted by the rod. Energy conservation and radial motion give

Hence

This is greatest at the bottom ($\cos\theta=1$) and least at the top ($\cos\theta=-1$).

Let the corresponding speeds be $v_b$ and $v_t$. At the bottom, energy conservation gives

Resolving upwards towards $O$ at the bottom,

At the top, from (a),

Both gravity and tension act towards $O$ there, so

Now $T_{\max}=4T_{\min}$, giving

Therefore

As a check, this gives $v_t^2=3gl>0$, $T_{\min}=2mg$ and $T_{\max}=8mg$, satisfying both complete rotation and the required ratio.