题目

Problem

The region $R$ , shown shaded in Figure 3, is the region enclosed between the $x$ -axis and the curve with equation

\frac{x^2}{9a^2}+\frac{y^2}{25a^2}=1

Region $R$ is rotated through $\pi$ radians about the $y$ -axis to form a solid of revolution.

A uniform solid $S$ has the same shape and size as this solid of revolution.

The volume of $S$ is $30\pi a^3$

The distance of the centre of mass of $S$ from $O$ is $d$ .

(a) Use algebraic integration to show that $d=\dfrac{15}8a$ .

(6)

The solid $S$ is joined to a uniform solid right circular cone $C$ that is made of the same material as $S$ to form a model.

The cone has vertex $V$ , height $10a$ and base radius $3a$ .

The plane face of $S$ and the plane face of $C$ coincide.

The model is supported on two light vertical rods.

the first rod has one end attached to the model at $V$ and the other end attached to horizontal ground
the second rod has one end attached to the model at a point on the rim of the common plane face and the other end attached to the horizontal ground

The model rests in equilibrium with its axis of symmetry horizontal and in the same vertical plane as the two rods, as shown in Figure 4.

Given that the weight of the model is $2W$

(b) find, in terms of $W$ , the thrust in the rod attached to the model at $V$ .

(4)

题目中文翻译

图 3 阴影所示的区域 $R$ 是 $x$ 轴与下列曲线围成的区域：

\frac{x^2}{9a^2}+\frac{y^2}{25a^2}=1

将区域 $R$ 绕 $y$ 轴旋转 $\pi$ 弧度，形成旋转体。

均匀立体 $S$ 与此旋转体有相同的形状和大小。

$S$ 的体积为 $30\pi a^3$ 。

$S$ 的质心到 $O$ 的距离为 $d$ 。

(a) 用代数积分证明 $d=\dfrac{15}8a$ 。

(6)

将立体 $S$ 与由相同材料制成的均匀实心正圆锥 $C$ 连接，组成一个模型。

圆锥的顶点为 $V$ ，高为 $10a$ ，底面半径为 $3a$ 。

$S$ 的平面底面与 $C$ 的平面底面重合。

用两根竖直轻杆支撑该模型。

第一根杆的一端连接于模型的 $V$ 点，另一端连接于水平地面；
第二根杆的一端连接于模型公共底面的边缘上一点，另一端连接于水平地面。

如图 4 所示，模型静止平衡，其对称轴水平，并位于两根杆所在的竖直平面内。

已知模型的重量为 $2W$ ，

(b) 用 $W$ 表示连接于模型 $V$ 点的杆中的推力。

(4)

解答

(a)

解法一

思路

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图中是上半椭圆，绕 $y$ 轴旋转 $\pi$ 后已形成完整的圆盘截面。以 $y$ 为积分变量，用 $\pi x^2$ 计算截面积和质量矩，最后除以给定体积。

答题过程

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At height $y$ , where $0\le y\le5a$ , the upper half-ellipse has horizontal coordinate bounds $-x$ and $x$ , with

x^2=9a^2\left(1-\frac{y^2}{25a^2}\right).

Rotating this full segment through $\pi$ about the $y$ -axis forms a complete disc of area $\pi x^2$ .

Let $\rho$ be the density. By rotational symmetry the centre of mass lies on the $y$ -axis. Its first mass moment about the plane $y=0$ is

J=\rho\pi\int_0^{5a}y\,9a^2\left(1-\frac{y^2}{25a^2}\right)\,\mathrm dy.

Algebraic integration gives

J=9\rho\pi a^2\left[\frac{y^2}{2}-\frac{y^4}{100a^2}\right]_0^{5a} =9\rho\pi a^2\left(\frac{25a^2}{2}-\frac{625a^4}{100a^2}\right) =\frac{225}{4}\rho\pi a^4.

Using the mass $\rho(30\pi a^3)$ ,

\boxed{d=\frac{J}{30\rho\pi a^3}=\frac{15a}{8}}.

(b)

解法一

思路

展开

先比较体积，说明圆锥与椭球顶盖的重量各为 $W$ 。绕公共底面边缘的支点取矩，两个重力作用在支点两侧，力矩方向相反。

答题过程

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The cone has volume

V_C=\frac13\pi(3a)^2(10a)=30\pi a^3=V_S.

The solids have equal density and equal volume, so each has weight $W$ .

Let $H$ be the support point on the common rim, and let $T_V$ be the upward thrust at $V$ . Since the symmetry axis is horizontal, the horizontal distance from $H$ to $V$ is $10a$ . The cone’s centre of mass is $10a/4=5a/2$ from its base towards $V$ , whereas the centre of mass of $S$ is $15a/8$ on the opposite side of that base plane.

Taking moments about $H$ ,

W\left(\frac{5a}{2}\right) =W\left(\frac{15a}{8}\right)+T_V(10a).

Hence

10aT_V=\left(\frac{20}{8}-\frac{15}{8}\right)Wa=\frac58Wa, \qquad \boxed{T_V=\frac{W}{16}}.

M3 2026 June A 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