题目

Problem

A uniform solid right circular cylinder has radius $3a$ and height $9a$

A uniform solid hemisphere of radius $3a$ is attached to the cylinder so that the plane face of the hemisphere coincides with one of the plane faces of the cylinder.

A solid right circular cone, of radius $3a$ and height $6a$ , whose plane face is the other plane face of the cylinder, is now removed from the cylinder to form the uniform solid $S$

The point $O$ is the centre of the circular base of $S$ , as shown in Figure 2.

(a) Show that the distance from $O$ to the centre of mass of $S$ is $\dfrac{77}{12}a$

(5)

The solid $S$ is placed with its circular base on a rough horizontal surface.

The point $Q$ , shown in Figure 2, lies on the circumference of the top of the cylinder.

A horizontal force of magnitude $H$ is applied to $S$ at $Q$

The force acts towards the axis of symmetry of $S$

The horizontal surface is sufficiently rough to prevent $S$ from sliding.

The solid $S$ is on the point of tipping.

Given that the weight of $S$ is $W$

(b) find $H$ in terms of $W$

(2)

题目中文翻译

一个均匀实心正圆柱的半径为 $3a$ ，高为 $9a$ 。

将半径为 $3a$ 的均匀实心半球连接于圆柱，使半球的平面底面与圆柱的一个平面端面重合。

现在从圆柱中挖去一个半径为 $3a$ 、高为 $6a$ 的实心正圆锥，其底面与圆柱的另一个平面端面重合，形成均匀立体 $S$ 。

如图 2 所示， $O$ 为 $S$ 的圆形底面的圆心。

(a) 证明 $O$ 到 $S$ 的质心的距离为 $\dfrac{77}{12}a$ 。

(5)

将立体 $S$ 的圆形底面放在粗糙水平面上。

图 2 所示的 $Q$ 点位于圆柱顶面的圆周上。

在 $Q$ 点对 $S$ 施加大小为 $H$ 、指向 $S$ 的对称轴的水平力。

水平面足够粗糙，使 $S$ 不会滑动。

立体 $S$ 恰好要倾倒。

已知 $S$ 的重量为 $W$ ，

(b) 用 $W$ 表示 $H$ 。

(2)

解答

(a)

解法一

思路

展开

质量与体积成正比。将圆柱、上方实心半球相加，再扣去底部挖掉的实心圆锥；各质心距离统一从 $O$ 量起。

答题过程

展开

Let the uniform density be $\rho$ . The masses of the hemisphere, cylinder and removed cone are respectively

M_H=\frac23\rho\pi(3a)^3=18\rho\pi a^3,

M_Y=\rho\pi(3a)^2(9a)=81\rho\pi a^3, \qquad M_C=\frac13\rho\pi(3a)^2(6a)=18\rho\pi a^3.

Thus the remaining mass is $81\rho\pi a^3$ . All centres of mass lie on the common axis. Their heights above $O$ are

h_H=9a+\frac38(3a)=\frac{81a}{8}, \qquad h_Y=\frac{9a}{2}, \qquad h_C=\frac14(6a)=\frac{3a}{2}.

Let the remaining centre of mass be at height $d$ . Taking first moments about the base plane and cancelling $\rho\pi a^3$ gives

81d=18\frac{81a}{8}+81\frac{9a}{2}-18\frac{3a}{2}.

Therefore

81d=\left(\frac{729}{4}+\frac{729}{2}-27\right)a =\frac{2079a}{4},

and hence

\boxed{d=\frac{2079a}{324}=\frac{77a}{12}.}

(b)

解法一

思路

展开

临界倾倒时以底面边缘的接触点取矩。水平力的力臂是施力点高度 $9a$ ，而重力的力臂是底面半径 $3a$ ，不需要使用(a)的质心高度。

答题过程

展开

At limiting tipping, take moments about the edge of the circular base on the side towards which the force pushes the solid. The base reaction acts through this point and has zero moment.

The weight acts on the symmetry axis, with horizontal lever arm $3a$ . The horizontal force at $Q$ has vertical lever arm $9a$ . Thus

H(9a)=W(3a).

Therefore

\boxed{H=\frac W3.}

M3 2026 Jan 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