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CIE 9231 2022 November Paper 32 Q6

A Level / CIE / FM

CIE 9231 2022 Nov Paper 32 Paper · Question 6

题目

Problem

A light inextensible string is threaded through a fixed smooth ring RR which is at a height hh above a smooth horizontal surface. One end of the string is attached to a particle AA of mass mm. The other end of the string is attached to a particle BB of mass 67m\frac67m. The particle AA moves in a horizontal circle on the surface. The particle BB hangs in equilibrium below the ring and above the surface (see diagram).

When AA has constant angular speed ω\omega, the angle between ARAR and BRBR is θ\theta and the normal reaction between AA and the surface is NN.

When AA has constant angular speed 32ω\frac32\omega, the angle between ARAR and BRBR is α\alpha and the normal reaction between AA and the surface is 12N\frac12N.

(a) Show that cos⁡θ=49cos⁡α\cos\theta=\frac49\cos\alpha.

(5)

(b) Find NN in terms of mm and gg and find the value of cos⁡α\cos\alpha.

(4)
题目中文翻译

一根轻质不可伸长细绳穿过固定的光滑圆环 RR,圆环位于光滑水平面上方,高度为 hh。细绳的一端连接在质量为 mm 的质点 AA 上。细绳的另一端连接在质量为 67m\frac67m 的质点 BB 上。质点 AA 在该平面上做水平圆周运动。质点 BB 悬挂在圆环下方、水平面上方,处于平衡状态(见图)。

当 AA 以恒定角速率 ω\omega 运动时,ARAR 与 BRBR 的夹角为 θ\theta,AA 与水平面之间的法向反作用力为 NN。

当 AA 以恒定角速率 32ω\frac32\omega 运动时,ARAR 与 BRBR 的夹角为 α\alpha,AA 与水平面之间的法向反作用力为 12N\frac12N。

(a) 证明 cos⁡θ=49cos⁡α\cos\theta=\frac49\cos\alpha。

(5)

(b) 求 NN,用 mm 和 gg 表示,并求 cos⁡α\cos\alpha 的值。

(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 (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

(a)

解法一

思路

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悬挂粒子静止,因而两种情况下的绳张力相同。对桌面上的质点用半径 htan⁡θh\tan\theta 写向心力式,比较两个角速率。

答题过程

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The ring is smooth and BB is at rest, so the string tension is T=6mg/7T=6mg/7 in both cases. The radii in the two cases are htan⁡θh\tan\theta and htan⁡αh\tan\alpha. The first radial equation is

Tsin⁡θ= mhsin⁡θcos⁡θω2,Tcos⁡θ= mhω2,ω2= 6g7hcos⁡θ.\begin{align*} T\sin\theta=&\,mh\frac{\sin\theta}{\cos\theta}\omega^2,\\[4mm] T\cos\theta=&\,mh\omega^2,\\[4mm] \omega^2=&\,\frac{6g}{7h}\cos\theta. \end{align*}

The cancellation is valid since 0<θ<90∘0<\theta<90^\circ. In the second case,

(32ω)2=6g7hcos⁡α.\bigg(\frac32\omega\bigg)^2=\frac{6g}{7h}\cos\alpha.

Comparing the two relations gives

946g7hcos⁡θ= 6g7hcos⁡α,cos⁡θ= 49cos⁡α.\begin{align*} \frac94\frac{6g}{7h}\cos\theta=&\,\frac{6g}{7h}\cos\alpha,\\[4mm] \cos\theta=&\,\frac49\cos\alpha. \end{align*}

(b)

解法一

思路

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两个情形的竖直加速度都为零,分别列出支持力和绳张力的竖直平衡式。用 (a) 消去 cos⁡θ\cos\theta。

答题过程

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Vertical equilibrium for AA gives

N+67mgcos⁡θ= mg,N2+67mgcos⁡α= mg.\begin{align*} N+\frac67mg\cos\theta=&\,mg,\\[4mm] \frac N2+\frac67mg\cos\alpha=&\,mg. \end{align*}

Writing c=cos⁡αc=\cos\alpha and using cos⁡θ=4c/9\cos\theta=4c/9, the first equation gives N/(mg)=1−8c/21N/(mg)=1-8c/21. Substituting into the second gives

12(1−8c21)+67c= 1,(−421+1821)c= 12,23c= 12.\begin{align*} \frac12\bigg(1-\frac{8c}{21}\bigg)+\frac67c=&\,1,\\[4mm] \bigg(-\frac4{21}+\frac{18}{21}\bigg)c=&\,\frac12,\\[4mm] \frac23c=&\,\frac12. \end{align*}

Thus cos⁡α=3/4\cos\alpha=3/4, and N=mg(1−2/7)=5mg/7N=mg(1-2/7)=5mg/7.