题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

A uniform lamina $ABCD$ is in the shape of the region $R$

The region $R$ is bounded by the curve with equation $y=\dfrac14(x^2+8)$ , the line with equation $x=4$ , the $x$ -axis and the line with equation $x=-2$ , as shown shaded in Figure 4.

The area of $R$ is $18$

(a) Show that the $x$ coordinate of the centre of mass of the lamina is $\dfrac32$

(4)

(b) Show that the $y$ coordinate of the centre of mass of the lamina is $\dfrac{17}{10}$

(4)

The lamina is freely suspended from $B$ and hangs in equilibrium with $BC$ at an angle $\theta$ to the downward vertical through $B$

(2)

题目中文翻译

本题必须展示所有计算步骤。

不接受完全依赖计算器技术的解法。

均匀薄片 $ABCD$ 的形状为区域 $R$ 。

如图 4 阴影所示，区域 $R$ 由曲线 $y=\dfrac14(x^2+8)$ 、直线 $x=4$ 、 $x$ 轴及直线 $x=-2$ 围成。

$R$ 的面积为 $18$ 。

(a) 证明薄片质心的 $x$ 坐标为 $\dfrac32$ 。

(4)

(b) 证明薄片质心的 $y$ 坐标为 $\dfrac{17}{10}$ 。

(4)

薄片从 $B$ 自由悬挂并处于平衡， $BC$ 与过 $B$ 的竖直向下方向成 $\theta$ 角。

(2)

解答

(a)

解法一

思路

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薄片均匀， $x$ 坐标等于关于 $y$ 轴的面积矩除以面积。题目已给面积 $18$ ，分子必须展示完整的代数积分。

答题过程

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With $y=x^2/4+2$ , the first moment of area about the $y$ -axis is

\begin{align*} Q_y=&\,\int_{-2}^4xy\,\mathrm{d}x =\int_{-2}^4\left(\frac{x^3}{4}+2x\right)\,\mathrm{d}x\\ =&\,\left[\frac{x^4}{16}+x^2\right]_{-2}^4\\ =&\,(16+16)-(1+4)=27. \end{align*}

Since the lamina is uniform and its area is $18$ ,

\boxed{\bar x=\frac{Q_y}{18}=\frac{27}{18}=\frac32.}

(b)

解法一

思路

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一条高度为 $y$ 的竖条，其质心高度为 $y/2$ 。因此关于 $x$ 轴的面积矩含 $y^2/2$ ，不能漏掉二分之一。

答题过程

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The first moment of area about the $x$ -axis is

Q_x=\frac12\int_{-2}^4y^2\,\mathrm{d}x.

Since

y^2=\left(\frac{x^2}{4}+2\right)^2 =\frac{x^4}{16}+x^2+4,

we obtain

\begin{align*} Q_x=&\,\frac12\int_{-2}^4 \left(\frac{x^4}{16}+x^2+4\right)\,\mathrm{d}x\\ =&\,\frac12\left[\frac{x^5}{80}+\frac{x^3}{3}+4x\right]_{-2}^4\\ =&\,\frac12\left(\frac{752}{15}+\frac{166}{15}\right) =\frac{153}{5}. \end{align*}

Dividing by the area,

\boxed{\bar y=\frac{Q_x}{18} =\frac{153}{90}=\frac{17}{10}.}

(c)

解法一

思路

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自由悬挂时，悬点到质心的连线竖直向下。原图中 $BC$ 与 $BG$ 的夹角保持不变，因此用原坐标中的两个距离求正切即可。

答题过程

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In the original coordinates, $B=(4,6)$ , $C=(4,0)$ and $G=(3/2,17/10)$ . When freely suspended from $B$ , the lamina turns until $BG$ is vertically downwards.

Thus the required angle between $BC$ and the downward vertical is the angle between $BC$ and $BG$ . In the original diagram, the component of $BG$ perpendicular to $BC$ is $4-3/2=5/2$ , and the component along $BC$ is $6-17/10=43/10$ .

Therefore

\boxed{\tan\theta=\frac{5/2}{43/10}=\frac{25}{43}.}

M3 2026 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

(c)

解法一

思路

答题过程