题目

Problem

A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$ . The other end of the string is attached to a fixed point $O$ . The string is held taut with $OP$ making an angle $\alpha$ with the downward vertical, where $\cos\alpha=\frac23$ . The particle $P$ is projected perpendicular to $OP$ in an upwards direction with speed $\sqrt{3ag}$ . It then starts to move along a circular path in a vertical plane.

Find the cosine of the angle between the string and the upward vertical when the string first becomes slack.

(4)

题目中文翻译

一个质量为 $m$ 的质点 $P$ 连接在一根长为 $a$ 的轻质不可伸长细绳的一端。细绳的另一端连接在固定点 $O$ 。保持细绳绷紧，且 $OP$ 与竖直向下方向成 $\alpha$ 角，其中 $\cos\alpha=\frac23$ 。质点 $P$ 以速率 $\sqrt{3ag}$ 沿垂直于 $OP$ 的向上方向射出。随后，它开始在竖直平面内沿圆周运动。

求细绳第一次松弛时，细绳与竖直向上方向夹角的余弦值。

(4)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ( $g$ ) is needed, use $10\,\mathrm{m\,s^{-2}}$ .

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$ 。

解答

解法一

思路

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绳刚松弛时张力为零。先用径向方程把此时的速度平方写成所求余弦，再用能量守恒。

答题过程

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Let $\beta$ be the angle to the upward vertical when the string becomes slack, and $v$ the speed there. With zero tension,

mg\cos\beta=\frac{mv^2}{a}.

The rise from launch is $a(\cos\alpha+\cos\beta)$ . Conservation of energy gives

\begin{align*} \frac12m(3ag)-\frac12mv^2=&\,mga(\cos\alpha+\cos\beta),\\[4mm] v^2=&\,3ag-2ag(\cos\alpha+\cos\beta). \end{align*}

Substituting $v^2=ag\cos\beta$ and $\cos\alpha=2/3$ , and dividing by $ag$ , gives

\begin{align*} \cos\beta=&\,3-2\bigg(\frac23+\cos\beta\bigg),\\[4mm] 3\cos\beta=&\,3-\frac43=\frac53,\\[4mm] \cos\beta=&\,\frac59. \end{align*}

CIE 9231 2022 November Paper 32 Q1

题目

解答

解法一

思路

答题过程