题目

Problem

A uniform solid hemisphere $H$ has radius $r$ and centre $O$

(a) Show that the centre of mass of $H$ is $\dfrac{3r}8$ from $O$

[You may assume that the volume of $H$ is $\dfrac{2\pi r^3}3$ .]

(4)

A uniform solid $S$ , shown below in Figure 3, is formed by attaching a uniform solid right circular cylinder of height $h$ and radius $r$ to $H$ , so that one end of the cylinder coincides with the plane face of $H$ .

The point $A$ is the point on $H$ such that $OA=r$ and $OA$ is perpendicular to the plane face of $H$

(b) Show that the distance of the centre of mass of $S$ from $A$ is

\frac{5r^2+12rh+6h^2}{8r+12h}

(5)

The solid $S$ can rest in equilibrium on a horizontal plane with any point of the curved surface of the hemisphere in contact with the plane.

(2)

题目中文翻译

均匀实心半球 $H$ 的半径为 $r$ ，球心为 $O$ 。

(a) 证明 $H$ 的质心到 $O$ 的距离为 $\dfrac{3r}8$ 。

[ 可直接使用 $H$ 的体积为 $\dfrac{2\pi r^3}3$ 。 ]

(4)

将高为 $h$ 、半径为 $r$ 的均匀实心正圆柱连接于 $H$ ，使圆柱的一端与 $H$ 的平面面重合，形成下方图 3 所示的均匀立体 $S$ 。

$A$ 是 $H$ 上满足 $OA=r$ 且 $OA$ 垂直于 $H$ 的平面面的点。

(b) 证明 $S$ 的质心到 $A$ 的距离为

\frac{5r^2+12rh+6h^2}{8r+12h}

(5)

立体 $S$ 可在水平面上静止平衡，且半球曲面上的任意一点均可与该平面接触。

(2)

解答

(a)

解法一

思路

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沿半球轴从 O 向曲面量 x，把实心半球分成薄圆盘，积分求质量矩，再除以半球质量。

答题过程

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Measure $x$ along the symmetry axis from $O$ into the hemisphere. A disc at $x$ has radius $\sqrt{r^2-x^2}$ , so for uniform density $\rho$ ,

\mathrm{d}m=\rho\pi(r^2-x^2)\,\mathrm{d}x,\qquad 0\leq x\leq r.

The hemisphere’s mass is $M=2\rho\pi r^3/3$ . Its centre of mass lies on the axis by symmetry, and

M\bar x=\rho\pi\int_0^r x(r^2-x^2)\,\mathrm{d}x.

Therefore

M\bar x=\rho\pi\left[\frac{r^2x^2}{2}-\frac{x^4}{4}\right]_0^r=\frac14\rho\pi r^4,

\bar x=\frac{\rho\pi r^4/4}{2\rho\pi r^3/3}=\boxed{\frac{3r}{8}}.

(b)

解法一

思路

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均匀组合立体的质量比等于体积比。半球质心到 A 为 r−3r/8，圆柱质心到 A 为 r+h/2，再用质量加权平均。

答题过程

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Use distance along the axis from $A$ towards the cylinder. The hemisphere’s centre of mass is at distance

r-\frac{3r}{8}=\frac{5r}{8}

from $A$ , while the cylinder’s centre of mass is at $r+h/2$ . Since the solid is uniform, the masses are proportional to the volumes $2\pi r^3/3$ and $\pi r^2h$ . If the combined centre is at distance $\bar y$ from $A$ ,

\left(\frac{2\pi r^3}{3}+\pi r^2h\right)\bar y=\frac{2\pi r^3}{3}\frac{5r}{8}+\pi r^2h\left(r+\frac h2\right).

Cancelling $\pi r^2$ ,

\left(\frac{2r}{3}+h\right)\bar y=\frac{5r^2}{12}+rh+\frac{h^2}{2}.

Multiplying numerator and denominator by $12$ gives

\boxed{\bar y=\frac{5r^2+12rh+6h^2}{8r+12h}}.

(c)

解法一

思路

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任意曲面接触点都能平衡，要求重心在球心 O：这样重力作用线始终与通过球心的支持力共线。将质心距 A 设为 r 后解正根。

答题过程

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At any spherical contact point, the normal reaction passes through $O$ . For equilibrium at every such orientation, the centre of mass must coincide with $O$ ; otherwise some orientation produces a nonzero gravitational moment about the contact point. Thus $\bar y=AO=r$ , giving

r=\frac{5r^2+12rh+6h^2}{8r+12h}.

Therefore

8r^2+12rh=5r^2+12rh+6h^2,

3r^2=6h^2,\qquad r^2=2h^2.

Since $r,h>0$ ,

\boxed{r=\sqrt2\,h}.

M3 2022 Jan UNUSED Q4

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