题目

Problem

A small smooth ring $R$ of mass $m$ is threaded on to a smooth wire in the shape of a circle with centre $O$ and radius $l$ . The wire is fixed in a vertical plane. The ring $R$ is attached to one end of a light elastic string of natural length $l$ and modulus of elasticity $mg$ . The other end of the elastic string is attached to $A$ , the lowest point of the wire. The point $B$ is on the wire and $OB$ is horizontal.

The ring $R$ is at rest at the highest point of the wire, as shown in Figure 6.

The ring $R$ is slightly disturbed from rest and slides along the wire.

At the instant when $R$ reaches the point $B$ , the speed of $R$ is $v$ and the magnitude of the force exerted on $R$ by the wire is $N$ .

(a) Show that

v^2=2gl\sqrt2

(7)

(b) Show that

N=\frac12mg(5\sqrt2-2)

(7)

题目中文翻译

质量为 $m$ 的光滑小环 $R$ 套在一根光滑圆形金属丝上，圆心为 $O$ ，半径为 $l$ 。金属丝固定在竖直平面内。小环 $R$ 连接于一根原长为 $l$ 、弹性模量为 $mg$ 的弹性轻绳的一端，另一端连接于金属丝的最低点 $A$ 。 $B$ 点位于金属丝上，且 $OB$ 水平。

如图 6 所示，小环 $R$ 静止于金属丝的最高点。

使小环 $R$ 从静止状态受到轻微扰动，它便沿金属丝滑动。

$R$ 到达 $B$ 的瞬间的速率为 $v$ ，金属丝对 $R$ 施加的力的大小为 $N$ 。

(a) 证明

v^2=2gl\sqrt2

(7)

(b) 证明

N=\frac12mg(5\sqrt2-2)

(7)

解答

(a)

解法一

思路

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金属丝的光滑支持力不做功，但弹性绳在最高点已储能，不能漏掉初始弹性势能。在 $B$ 点绳长是直角三角形的斜边。

答题过程

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At the highest point, the string has length $2l$ and extension $l$ . Its elastic potential energy is

E_0=\frac{mg}{2l}l^2=\frac12mgl.

At $B$ , triangle $AOB$ is right-angled with $OA=OB=l$ . Thus

AB=l\sqrt2, \qquad e_B=l(\sqrt2-1).

The ring falls through a height $l$ . The normal force of the smooth wire is perpendicular to the motion and does no work. Treating the small disturbance as giving negligible initial kinetic energy, conservation of energy gives

mgl+\frac12mgl =\frac12mv^2+\frac{mg}{2l}\left[l(\sqrt2-1)\right]^2.

Since $(\sqrt2-1)^2=3-2\sqrt2$ ,

\frac32mgl=\frac12mv^2+\frac12mgl(3-2\sqrt2).

Multiplying by $2/m$ and rearranging,

v^2=3gl-gl(3-2\sqrt2) =\boxed{2gl\sqrt2}.

(b)

解法一

思路

展开

在 $B$ 点向心方向水平指向 $O$ ，重力没有径向分量。绳的张力沿 $BA$ ，须取水平分量；金属丝的力是其余所需向心合力。

答题过程

展开

By Hooke’s law, the tension at $B$ is

T=\frac{mg}{l}\left[l(\sqrt2-1)\right] =mg(\sqrt2-1).

Since $OA=OB$ , the string $BA$ makes $45^\circ$ with the inward horizontal direction $BO$ . Let the signed force of the wire towards $O$ be $N_i$ . Resolving radially inwards at $B$ ,

N_i+T\cos45^\circ=\frac{mv^2}{l}.

The weight has no horizontal component. Using part (a),

\begin{align*} N_i=&\,2mg\sqrt2-mg(\sqrt2-1)\frac{\sqrt2}{2}\\ =&\,mg\left(2\sqrt2-1+\frac{\sqrt2}{2}\right)\\ =&\,\frac12mg(5\sqrt2-2). \end{align*}

This is positive, confirming that the wire acts towards $O$ . Its magnitude is therefore

\boxed{N=\frac12mg(5\sqrt2-2).}

M3 2023 Jan Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程