一位右手高尔夫球手以平面运动挥动高尔夫球杆（下图）。已知高尔夫球的初速度 (v)、手臂长度和球杆长度，球接触点时左肩的角速度是多少？假设左臂和球杆形成一条直线，且球的初速度与击球时球杆头的线速度相同。注意：必须提供国际单位制 (SI) 单位符号，并四舍五入到三位有效数字。

Question

BlackTom AI · Accepted Answer

要理解这个问题，我们要把球棒系统简化为一个绕肩部的单一旋转运动。

- 题设给出：右手高尔夫球手，球接触点时球头的线速度等于球的初速度 v，且左臂和球杆在击球瞬间成一条直线。这意味着球头的线速度就是绕肩部的角速度乘以肩部到球头的距离。
- 需要的距离 r 等于从肩部到球头的直线距离。因为左臂长度为 0.79 m，球杆长度为 1.2 m，且两者在击球时排列成一直线，所以 r = 0.79 m + 1.20 m = 1.99 m。
- 速度关系为 v_ball = ω × r，因此 ω = v_ball / r。
- 将给定数值代入：v_ball = 40 m/s，r = 1.99 m，得到 ω = 40 / 1.99 ≈ 20.1005 rad/s。
- 按要求四舍五入到三位有效数字：ω ≈ 20.1 rad/s。

单位要严格使用国际单位制符号，因此结果应写作 20.1 rad/s。

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