题目

Problem

(a) Use algebraic integration to show that the centre of mass of a uniform solid hemisphere of radius $r$ is at a distance $\dfrac38r$ from the centre of its plane face.

[You may assume that the volume of a sphere of radius $r$ is $\dfrac43\pi r^3$ .]

(5)

A uniform solid hemisphere of radius $r$ is joined to a uniform solid right circular cone made of the same material to form a toy. The cone has base radius $r$ and height $kr$ . The centre of the base of the cone is $O$ . The plane face of the cone coincides with the plane face of the hemisphere, as shown in Figure 3.

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

(b) Find the exact value of $k$

(5)

题目中文翻译

(a) 用代数积分证明：半径为 $r$ 的均匀实心半球的质心到其平面底面圆心的距离为 $\dfrac38r$ 。

[ 可直接使用半径为 $r$ 的球的体积为 $\dfrac43\pi r^3$ 。 ]

(5)

将半径为 $r$ 的均匀实心半球与同种材料制成的均匀实心正圆锥连接，组成一个玩具。圆锥的底面半径为 $r$ ，高为 $kr$ ，底面圆心为 $O$ 。如图 3 所示，圆锥的平面底面与半球的平面底面重合。

玩具可在水平面上静止平衡，且半球曲面上的任意一点均可与水平面接触。

(b) 求 $k$ 的精确值。

(5)

解答

(a)

解法一

思路

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沿半球的对称轴切成薄圆盘。圆盘半径平方是 $r^2-x^2$ ；用质量对平面底面的矩除以总质量。

答题过程

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Let the density be $\rho$ and measure $x$ along the axis from the centre of the plane face towards the curved surface, so $0\leq x\leq r$ . A thin disc at coordinate $x$ has radius $y$ satisfying

y^2=r^2-x^2.

Its mass is $\mathrm{d}m=\rho\pi(r^2-x^2)\,\mathrm{d}x$ . Therefore the first moment of mass about the plane face is

\begin{align*} \int_0^r x\,\mathrm{d}m =&\,\rho\pi\int_0^r(r^2x-x^3)\,\mathrm{d}x\\ =&\,\rho\pi\left[\frac{r^2x^2}{2}-\frac{x^4}{4}\right]_0^r\\ =&\,\frac{\rho\pi r^4}{4}. \end{align*}

The hemisphere has mass

M=\rho\frac12\left(\frac43\pi r^3\right) =\frac23\rho\pi r^3.

By symmetry the centre of mass lies on the axis. Its distance from the centre of the plane face is

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

(b)

解法一

思路

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任意接触点都能平衡，要求整体质心就在球面球心 $O$ 。两个物体在 $O$ 两侧，以共同底面取带符号的质量矩，密度会约去。

答题过程

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The normal reaction at a point on the hemispherical surface passes through $O$ . For equilibrium at any such contact point, the weight must also act through $O$ . Thus the centre of mass of the whole toy is $O$ .

The two solids have the same density $\rho$ . Their masses are

M_H=\frac23\rho\pi r^3, \qquad M_C=\frac13\rho\pi r^2(kr)=\frac k3\rho\pi r^3.

The hemisphere’s centre of mass is $3r/8$ from $O$ , towards the curved surface. The cone’s centre of mass is one quarter of its height from its base, namely $kr/4$ from $O$ , in the opposite direction.

The signed first moment about the common plane face must be zero:

\frac23\rho\pi r^3\cdot\frac{3r}{8} -\frac k3\rho\pi r^3\cdot\frac{kr}{4}=0.

Dividing by $\rho\pi r^4$ gives

\frac14-\frac{k^2}{12}=0, \qquad k^2=3.

Since the cone has positive height, $k>0$ , so

\boxed{k=\sqrt3.}

M3 2022 June 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