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CIE 9231 2022 June Paper 31 Q5

A Level / CIE / FM

CIE 9231 2022 June Paper 31 Paper · Question 5

题目

Problem

A light inextensible string ABAB passes through two small holes CC and DD in a smooth horizontal table where AC=3aAC=3a and DB=aDB=a. A particle of mass mm is attached at the end AA and moves in a horizontal circle with angular velocity ω\omega. A particle of mass 34m\frac34m is attached to the end BB and moves in a horizontal circle with angular velocity kωk\omega. ACAC makes an angle θ\theta with the downward vertical and DBDB makes an angle θ\theta with the horizontal (see diagram).

Find the value of kk.

(7)
题目中文翻译

一根轻质不可伸长细绳 ABAB 穿过光滑水平桌面上的两个小孔 CC 和 DD,其中 AC=3aAC=3a,DB=aDB=a。一个质量为 mm 的质点连接在 AA 端,并以角速度 ω\omega 做水平圆周运动。一个质量为 34m\frac34m 的质点连接在 BB 端,并以角速度 kωk\omega 做水平圆周运动。ACAC 与竖直向下方向成 θ\theta 角,DBDB 与水平方向成 θ\theta 角(见图)。

求 kk 的值。

(7)
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 (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时,取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

解法一

思路

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两段绳子的角度分别与竖直、水平比较。对每个质点,张力的水平分量与圆周半径含有相同的三角因子,约去后直接比较张力。

答题过程

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The smooth holes give the same tension TT in both end segments. The radius of the circle of AA is 3asin⁡θ3a\sin\theta, so

Tsin⁡θ= m(3asin⁡θ)ω2,T= 3maω2.\begin{align*} T\sin\theta=&\,m(3a\sin\theta)\omega^2,\\[4mm] T=&\,3ma\omega^2. \end{align*}

For BB the radius is acos⁡θa\cos\theta because DBDB makes angle θ\theta with the horizontal. Thus

Tcos⁡θ= 34m(acos⁡θ)(kω)2,T= 34mak2ω2.\begin{align*} T\cos\theta=&\,\frac34m(a\cos\theta)(k\omega)^2,\\[4mm] T=&\,\frac34mak^2\omega^2. \end{align*}

The cancellations are valid since 0<θ<90∘0<\theta<90^\circ. Equating the tensions gives

3maω2= 34mak2ω2,k2= 33/4=4.\begin{align*} 3ma\omega^2=&\,\frac34mak^2\omega^2,\\[4mm] k^2=&\,\frac{3}{3/4}=4. \end{align*}

Taking the positive angular-speed ratio gives k=2k=2.

解法二

思路

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也可先用两个质点的竖直平衡求绳的倾角与张力,再分别求角速率。

答题过程

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Vertical equilibrium gives Tcos⁡θ=mgT\cos\theta=mg for AA and Tsin⁡θ=34mgT\sin\theta=\frac34mg for BB. Dividing gives

tan⁡θ=(3/4)mgmg=34.\tan\theta=\frac{(3/4)mg}{mg}=\frac34.

Hence cos⁡θ=4/5\cos\theta=4/5, sin⁡θ=3/5\sin\theta=3/5 and T=mg/(4/5)=5mg/4T=mg/(4/5)=5mg/4. The two radii are 9a/59a/5 and 4a/54a/5. The inward radial equations give

ω2= T(3/5)m(9a/5)=5g12a,(kω)2= T(4/5)(3m/4)(4a/5)=5g3a.\begin{align*} \omega^2=&\,\frac{T(3/5)}{m(9a/5)}=\frac{5g}{12a},\\[4mm] (k\omega)^2=&\,\frac{T(4/5)}{(3m/4)(4a/5)}=\frac{5g}{3a}. \end{align*}

Their ratio is k2=(5g/3a)/(5g/12a)=4k^2=(5g/3a)/(5g/12a)=4, giving the positive ratio k=2k=2.