题目

Problem

A square $ABCD$ of side $4a$ is made from thin uniform cardboard. The centre of the square is $O$ . A circle with centre $O$ and radius $\dfrac{7a}4$ is then removed from the square to form a template $T$ , shown shaded in Figure 3.

A right conical shell, with no base, has radius $\dfrac{7a}4$ and perpendicular height $6a$ .

The shell is made of the same thin uniform cardboard as $T$ .

The shell is attached to $T$ so that the circumference of the end of the shell coincides with the circumference of the circle centre $O$ , to form the hat $H$ , shown in Figure 4.

[The surface area of a right conical shell of radius $r$ and slant height $l$ is $\pi rl$ .]

(a) Show that the exact distance of the centre of mass of $H$ from $O$ is

\frac{175\pi a}{63\pi+128}

(8)

A fixed rough plane is inclined to the horizontal at an angle $\alpha$ . The hat $H$ is placed on the plane, with $ABCD$ in contact with the plane, and $AB$ parallel to a line of greatest slope of the plane. The plane is sufficiently rough to prevent the hat from sliding down the plane.

Given that the hat is on the point of toppling,

(b) find the exact value of $\tan\alpha$ , giving your answer in simplest form.

(2)

题目中文翻译

用均匀薄纸板制成边长为 $4a$ 的正方形 $ABCD$ ，其中心为 $O$ 。从正方形中挖去以 $O$ 为圆心、半径为 $\dfrac{7a}4$ 的圆，形成图 3 阴影所示的模板 $T$ 。

一个无底面的正圆锥壳的半径为 $\dfrac{7a}4$ ，垂直高度为 $6a$ 。

圆锥壳由与 $T$ 相同的均匀薄纸板制成。

将圆锥壳连接于 $T$ ，使其端部圆周与以 $O$ 为圆心的圆周重合，形成图 4 所示的帽子 $H$ 。

[ 半径为 $r$ 、斜高为 $l$ 的正圆锥壳的表面积为 $\pi rl$ 。 ]

(a) 证明 $H$ 的质心到 $O$ 的精确距离为

\frac{175\pi a}{63\pi+128}

(8)

一固定粗糙斜面与水平面成 $\alpha$ 角。将帽子 $H$ 放在斜面上， $ABCD$ 与斜面接触， $AB$ 平行于斜面的最大坡度线。斜面足够粗糙，使帽子不会沿斜面下滑。

已知帽子恰好要倾倒，

(b) 求 $\tan\alpha$ 的精确值，以最简形式作答。

(2)

解答

(a)

解法一

思路

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相同纸板意味着质量正比于面积。底板按“正方形减圆洞”处理；圆锥是薄壳，面积用斜高，质心到其底面的距离为高的三分之一。

答题过程

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Let $\sigma$ be the mass per unit area of the cardboard and let $G$ be the centre of mass of the hat. By symmetry, $G$ lies on the cone’s axis through $O$ .

The slant height of the cone is

s=\sqrt{(6a)^2+\left(\frac{7a}{4}\right)^2} =\sqrt{\frac{625a^2}{16}}=\frac{25a}{4}.

Treat the hat as the full square, minus the removed disc, plus the conical shell. Their masses are

M_S=16\sigma a^2, \qquad M_D=\frac{49\pi}{16}\sigma a^2, \qquad M_C=\sigma\pi\frac{7a}{4}\frac{25a}{4} =\frac{175\pi}{16}\sigma a^2.

The square and disc have their centres of mass at $O$ . The centre of mass of a uniform conical shell is one third of its perpendicular height from its base, so its distance from $O$ is $6a/3=2a$ .

Let $OG=d$ . Taking first moments about the plane of the square,

(M_S-M_D+M_C)d=M_C(2a).

The total mass is

M_S-M_D+M_C =\sigma a^2\left(16+\frac{126\pi}{16}\right) =\frac{\sigma a^2}{8}(128+63\pi).

Thus

\frac{\sigma a^2}{8}(128+63\pi)d =\frac{175\pi}{16}\sigma a^2(2a),

and therefore

\boxed{OG=d=\frac{175\pi a}{63\pi+128}.}

(b)

解法一

思路

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临界倾倒时，重力作用线通过下坡侧的支承边。质心沿底面法线离开 $O$ ，把这条距离投影到水平方向，与 $O$ 到下坡边的半边长投影比较即可。

答题过程

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Since $AB$ is parallel to a line of greatest slope, the downhill edge of the square is perpendicular to that line. Let $M$ be its midpoint. Then

OM=2a, \qquad OG=d=\frac{175\pi a}{63\pi+128},

where $OG$ is perpendicular to the plane of the square.

At limiting toppling, the vertical line through $G$ passes through the downhill edge, and hence through $M$ in the central vertical section. The horizontal downhill projections of $OG$ and $OM$ are therefore equal:

d\sin\alpha=2a\cos\alpha.

It follows that

\boxed{\tan\alpha=\frac{2a}{d} =\frac{2(63\pi+128)}{175\pi} =\frac{126\pi+256}{175\pi}.}

M3 2023 Jan Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程