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 point $A$ which lies above a smooth horizontal table. The particle $P$ moves in a horizontal circle on the table with the string taut. The centre of the circle is the point $O$ on the table, where $AO$ is vertical and the string makes a constant angle $\theta^\circ$ with $AO$ , as shown in Figure 2.

Given that $P$ moves with constant angular speed $\sqrt{\dfrac{2g}a}$ , find the range of possible values of $\theta$

(7)

题目中文翻译

质量为 $m$ 的质点 $P$ 连接于一根长为 $a$ 的不可伸长轻绳的一端，另一端连接于光滑水平桌面上方的 $A$ 点。质点 $P$ 在桌面上做水平圆周运动，绳保持绷紧。如图 2 所示，圆心为桌面上的 $O$ 点， $AO$ 竖直，绳与 $AO$ 的夹角恒为 $\theta^\circ$ 。

已知 $P$ 以恒定角速度 $\sqrt{\dfrac{2g}a}$ 运动，求 $\theta$ 的可能取值范围。

(7)

解答

解法一

思路

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绳的水平张力提供向心力，桌面支持力与绳的竖直分量共同平衡重力。支持力不能为负，并且固定点在桌面上方，所以角度严格小于 90°。

答题过程

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Let the string tension be $T$ and the table’s normal reaction be $R$ . Since $A$ is above the table and the particle describes a nonzero horizontal circle,

0<\theta<90^\circ,\qquad r=a\sin\theta.

Resolving horizontally towards $O$ ,

T\sin\theta=mr\omega^2=m(a\sin\theta)\frac{2g}{a}.

Since $\sin\theta>0$ ,

T=2mg.

Vertical equilibrium gives

T\cos\theta+R=mg,\qquad R=mg(1-2\cos\theta).

The table cannot pull down on the particle, so $R\geq0$ . Thus

1-2\cos\theta\geq0,\qquad \cos\theta\leq\frac12.

Within the geometric range $0<\theta<90^\circ$ , this gives

\boxed{60^\circ\leq\theta<90^\circ}.

At $60^\circ$ , $R=0$ , which is the limiting case of contact. The upper endpoint is excluded because $A$ lies above the table.

M3 2022 Jan UNUSED Q2

题目

解答

解法一

思路

答题过程