题目

Problem

A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a\sqrt3$ . The other end of the string is attached to a fixed point $A$ . The particle $P$ is also attached to one end of a second light inextensible string of length $a$ . The other end of this string is attached to a fixed point $B$ , where $B$ is vertically below $A$ , with $AB=a$ .

The particle $P$ moves in a horizontal circle with centre $O$ , where $O$ is vertically below $B$ .

The particle $P$ moves with constant angular speed $\omega$ , with both strings taut, as shown in Figure 5.

(a) Show that the upper string makes an angle of $30^\circ$ with the downward vertical and the lower string makes an angle of $60^\circ$ with the downward vertical.

(2)

(b) Show that the tension in the upper string is $\dfrac12m\sqrt3(2g-a\omega^2)$ .

(8)

(4)

题目中文翻译

质量为 $m$ 的质点 $P$ 连接于一根长为 $a\sqrt3$ 的不可伸长轻绳的一端，另一端连接于固定点 $A$ 。质点 $P$ 也连接于第二根长为 $a$ 的不可伸长轻绳的一端，另一端连接于固定点 $B$ ， $B$ 位于 $A$ 正下方，且 $AB=a$ 。

质点 $P$ 沿以 $O$ 为圆心的水平圆周运动， $O$ 位于 $B$ 的正下方。

如图 5 所示，质点 $P$ 以恒定角速度 $\omega$ 运动，两根绳均绷紧。

(a) 证明上方的绳与竖直向下方向成 $30^\circ$ 角，下方的绳与竖直向下方向成 $60^\circ$ 角。

(2)

(b) 证明上方绳中的张力为 $\dfrac12m\sqrt3(2g-a\omega^2)$ 。

(8)

(4)

解答

(a)

解法一

思路

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先在已知三边的三角形 $APB$ 用余弦定理求上绳与竖直线的夹角。下绳的夹角与三角形在 $B$ 的内角互补。

答题过程

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Let $\beta=\angle PAB$ . In triangle $APB$ , $AP=a\sqrt3$ and $AB=BP=a$ . By the cosine rule,

\cos\beta=\frac{AP^2+AB^2-BP^2}{2(AP)(AB)} =\frac{3a^2+a^2-a^2}{2a^2\sqrt3} =\frac{\sqrt3}{2}.

Thus $\beta=30^\circ$ . Since $AB$ is vertically downwards from $A$ , the upper string makes an angle of $30^\circ$ with the downward vertical.

Also $AB=BP$ , so the opposite angles at $P$ and $A$ are equal. Consequently

\angle ABP=180^\circ-30^\circ-30^\circ=120^\circ.

At $B$ , the downward vertical is opposite to $BA$ . Hence the angle between $BP$ and the downward vertical is

180^\circ-120^\circ=\boxed{60^\circ}.

(b)

解法一

思路

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两根绳都向上、向圆心拉质点。先写竖直平衡和水平向心方程，并由(a)求圆周半径；消元时不要把角速度直接当作速率。

答题过程

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Let $T_A$ and $T_B$ be the tensions in the upper and lower strings respectively. From part (a), the horizontal radius is

r=BP\sin60^\circ=\frac{\sqrt3}{2}a.

Resolving vertically, since the acceleration has no vertical component,

T_A\cos30^\circ+T_B\cos60^\circ=mg,

\sqrt3T_A+T_B=2mg.\tag{1}

Resolving horizontally towards $O$ ,

T_A\sin30^\circ+T_B\sin60^\circ=mr\omega^2,

which gives

T_A+\sqrt3T_B=\sqrt3ma\omega^2.\tag{2}

Multiply (1) by $\sqrt3$ and subtract (2):

2T_A=2\sqrt3mg-\sqrt3ma\omega^2.

Therefore

\boxed{T_A=\frac12m\sqrt3(2g-a\omega^2).}

(c)

解法一

思路

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题目要求两绳都有正张力。上绳给角速度平方的上界，下绳给下界；两条条件必须同时满足。

答题过程

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For the upper string to remain taut, $T_A>0$ . By part (b),

2g-a\omega^2>0, \qquad \omega^2<\frac{2g}{a}.

Using the vertical equation (1) from part (b),

\begin{align*} T_B=&\,2mg-\sqrt3T_A\\ =&\,2mg-\frac32m(2g-a\omega^2)\\ =&\,\frac32ma\omega^2-mg. \end{align*}

Thus $T_B>0$ requires

\frac32a\omega^2>g, \qquad \omega^2>\frac{2g}{3a}.

Combining the two strict inequalities,

\boxed{\frac{2g}{3a}<\omega^2<\frac{2g}{a}.}

At either excluded endpoint one of the tensions is zero.

M3 2023 Jan Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程