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CIE 9231 2025 November Paper 32 Q1

A Level / CIE / FM

CIE 9231 2025 Nov Paper 32 Paper · Question 1

题目

Problem

A particle PP of mass mm is attached to two light inextensible strings each of length ll. The end of one string is attached to a fixed point AA and the end of the other string is attached to a fixed point BB, with AA vertically above BB. Angle APBAPB is a right angle. The particle PP rotates in a horizontal circle at a constant angular speed ω\omega with both strings taut (see diagram).

Find the tension in string APAP in terms of mm, gg, ll and ω\omega.

(4)
题目中文翻译

质量为 mm 的质点 PP 连接于两根长度均为 ll 的轻质不可伸长细绳。第一根细绳的一端连接于固定点 AA,第二根细绳的一端连接于固定点 BB,其中 AA 在 BB 的正上方。角 APBAPB 为直角。两根细绳均拉紧时,质点 PP 以恒定角速度 ω\omega 作水平圆周运动(见图)。

用 mm、gg、ll 和 ω\omega 表示细绳 APAP 中的张力。

(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.

Where a numerical value for the acceleration due to gravity (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要重力加速度的数值时,取 g=10 m s−2g=10\,\mathrm{m\,s^{-2}}。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。

解答

解法一

思路

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三角形 APBAPB 是等腰直角三角形,所以两根绳都与水平成 45∘45^\circ,圆周半径为 l/2l/\sqrt2。两根绳的张力分别记,列水平及竖直方程。

答题过程

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Let TA,TBT_A,T_B be the tensions in AP,BPAP,BP. Each string makes an angle of 45∘45^\circ with the horizontal, and the radius is l/2l/\sqrt2. Resolving horizontally toward the axis and vertically upward gives

TA+TB2= mω2l2,TA−TB2= mg.\begin{align*} \frac{T_A+T_B}{\sqrt2}=&\,m\omega^2\frac l{\sqrt2},\\[4mm] \frac{T_A-T_B}{\sqrt2}=&\,mg. \end{align*}

Multiplying each equation by 2\sqrt2 and adding gives 2TA=mlω2+2mg2T_A=ml\omega^2+\sqrt2mg. Hence

TA=m2(lω2+2 g).\boxed{T_A=\frac m2(l\omega^2+\sqrt2\,g)}.