📚 Centres of Mass | 质心
The concept of centre of mass is fundamental in mechanics. It is the point at which the entire mass of a body can be considered to be concentrated for the purpose of analysing translational motion and equilibrium. Mastering centres of mass is essential for solving problems involving balancing, suspension, and toppling in Edexcel A-Level Mathematics.
质心的概念是力学中的基础。它可以看作是物体全部质量集中用于分析平动和平衡的一个点。掌握质心对于解决 Edexcel A-Level 数学中涉及平衡、悬挂和倾倒的问题至关重要。
1. What is Centre of Mass? | 什么是质心？
The centre of mass of a body is the unique point where the weighted relative position of the distributed mass sums to zero. In a uniform gravitational field, the centre of mass coincides with the centre of gravity, which is the point where the entire weight of the body appears to act.
物体的质心是分布质量的加权相对位置总和为零的唯一点。在均匀重力场中，质心与重心重合，重心是物体全部重力看似作用的那一点。
For a rigid body, the centre of mass is fixed relative to the body, regardless of its orientation. This property is used to determine whether an object will balance or tip when placed on a surface or suspended from a point.
对于刚体，质心相对于物体是固定的，与物体的取向无关。这一性质被用来判断物体放置在表面上或从某点悬挂时是会平衡还是会倾倒。
2. Centre of Mass of a System of Particles | 质点系统的质心
For a discrete set of particles in a plane, the centre of mass (x̄, ȳ) is the mass-weighted average of their individual positions. If particles have masses m₁, m₂, ..., mₙ at coordinates (x₁, y₁), (x₂, y₂), ..., (xₙ, yₙ), the total mass M is m₁ + m₂ + ... + mₙ. The coordinates are given by:
对于平面内的一组离散质点，质心坐标 (x̄, ȳ) 是其各自位置的质量加权平均。若质点质量分别为 m₁, m₂, ..., mₙ，坐标分别为 (x₁, y₁), (x₂, y₂), ..., (xₙ, yₙ)，总质量 M = m₁ + m₂ + ... + mₙ，则坐标公式为：
x̄ = (m₁x₁ + m₂x₂ + ... + mₙxₙ) / M
ȳ = (m₁y₁ + m₂y₂ + ... + mₙyₙ) / M
This can be interpreted as taking moments about the y-axis for x̄ and about the x-axis for ȳ. The same principle extends to three dimensions by including the z-coordinate.
这可以理解为对 y 轴取矩求 x̄，对 x 轴取矩求 ȳ。相同的原理也推广到三维情况，只需加入 z 坐标。
kg, 3 kg and 5 kg are placed at (0,0), (4,0) and (1,3) respectively. Total mass = 10 kg. x̄ = (2×0 + 3×4 + 5×1)/10 = 17/10 = 1.7; ȳ = (2×0 + 3×0 + 5×3)/10 = 15/10 = 1.5. The centre of mass is at (1.7, 1.5).
示例：三个质量分别为 2 kg、3 kg 和 5 kg 的质点位于 (0,0)、(4,0) 和 (1,3) 处。总质量 = 10 kg。x̄ = (2×0 + 3×4 + 5×1)/10 = 17/10 = 1.7；ȳ = (2×0 + 3×0 + 5×3)/10 = 15/10 = 1.5。质心位于 (1.7, 1.5)。
3. Centre of Mass of Uniform Rods and Symmetric Objects | 均匀杆与对称物体的质心
A uniform rod has its centre of mass at its geometrical midpoint. If the rod has length L, the centre of mass lies at a distance L/2 from either end. This is true because the mass is distributed evenly along its length.
均匀直杆的质心位于其几何中点。若杆长为 L，质心到两端的距离均为 L/2。这是因为质量沿杆长均匀分布。
For any uniform object with a plane of symmetry, the centre of mass lies somewhere in that plane. If an object has two orthogonal planes of symmetry, the centre of mass lies along their line of intersection. A uniform sphere's centre of mass is at its geometric centre.
对于任何具有对称平面的均匀物体，质心一定位于该平面内。若物体有两个正交的对称平面，则质心位于它们的交线上。均匀球体的质心在其几何中心。
This symmetry argument simplifies many problems. For a uniform rectangular lamina, the centre of mass is at the intersection of the diagonals, which is also the intersection of the two symmetry axes.
这种对称性论证使许多问题得到简化。对于均匀矩形薄片，质心位于对角线的交点处，也就是两条对称轴的交点。
4. Standard Results for Uniform Laminas | 均匀薄片的标准结果
Many problems can be solved quickly by remembering the positions of centres of mass for standard shapes. The following table summarises the key results for uniform plane laminas (assuming uniform mass per unit area).
记住标准形状的质心位置可以快速解决许多问题。下表总结了均匀平面薄片（假设单位面积质量均匀）的主要结果。
| Shape | Position of Centre of Mass |
|---|---|
| Rectangular lamina | Intersection of diagonals; centre of rectangle |
| Right-angled triangular lamina | Two-thirds of the way from the right-angled vertex to the midpoint of the hypotenuse, or distances h/3 and b/3 from the two perpendicular sides (where h and b are the legs). |
| General triangular lamina | ion of medians (centroid). Distance from any side = one-third of the corresponding height. |
| Semicircular lamina (radius r) | On the axis of symmetry, distance 4r/(3π) from the diameter. |
| Sector of a circle (angle 2α radians, radius r) | On the axis of symmetry, distance (2r sin α)/(3α) from the centre of the circle. |
| Circular arc (wire, angle 2α) | On the axis of symmetry, distance (r sin α)/α from the centre. |
2. For a full circle or full circular wire, the centre of mass is at the geometric centre.
当 α = π/2 时，扇形和圆弧的公式退化为半圆的结果。对于完整的圆或圆形金属丝，质心位于几何中心。
5. Centre of Mass of Composite Bodies | 组合体的质心
To find the centre of mass of a composite body made up of several simple shapes, treat each part as a particle located at its own centre of mass, with its mass (or area, if uniform) acting there. Then apply the particle formula to the whole assembly.
要求由几个简单形状组成的组合体的质心，可将每个部分视为位于其自身质心处的一个质点，其质量（或面积，当密度均匀时）集中在质心处。然后对整个组合应用质点系公式。
For a uniform lamina, we can replace mass by area. Use a table to organise the calculation: list each part, its area A, the coordinates (x, y) of its centre of mass, and the products A x and A y. Sum these and divide by total area.
对于均匀薄片，可以用面积代替质量。可用表格组织计算：列出各部分、其面积 A、其质心坐标 (x, y) 以及乘积 A x 和 A y。求和后再除以总面积。
to its short side along the x-axis. The first rectangle (A=24) has centre (2,3). The second (A=8) if placed with its base from x=4 to 6, has centre (5,2). Total area =32. x̄ = (24×2 + 8×5)/32 = (48+40)/32 = 88/32 = 2.75. ȳ = (24×3 + 8×2)/32 = (72+16)/32 = 88/32 = 2.75. So centre of mass is (2.75, 2.75).
示例：求一个由 4×6 矩形（沿 x 边长 4，沿 y 边长 6）和一个与其短边在 x 轴上相连的 2×4 矩形组成的 L 形薄片的质心。第一个矩形（A=24）中心为 (2,3)。第二个矩形（A=8）若置于 x=4 到 6，底部 y 坐标从0到4，中心为 (5,2)。总面积 =32。x̄ = (24×2 + 8×5)/32 = 88/32 = 2.75；ȳ = (24×3 + 8×2)/32 = 88/32 = 2.75。质心为 (2.75, 2.75)。
6. Centre of Mass of a Body with a Cut-out | 带切口的物体的质心
When a shape has a hole or a piece removed, you can use the negative mass method. Treat the original complete shape as having positive mass (or area) and the removed part as having negative mass. Then find the centre of mass using the standard summation.
当一个形状含有孔洞或被切掉一块时，可以使用负质量法。将原来的完整形状视为具有正质量（或面积），将切除的部分视为具有负质量。然后使用标准求和法求质心。
For a uniform lamina, the mass is proportional to area. So if a circular hole of radius r is cut from a larger rectangular plate, you assign positive area to the rectangle and negative area to the circle. The centre of mass of the hole is at the centre of the circle, taken with a negative sign in the moment sums.
对于均匀薄片，质量与面积成正比。因此，若从一个大矩形板上切去一个半径为 r 的圆孔，给矩形赋予正面积，给圆孔赋予负面积。孔的质心位于圆心，在计算矩时带有负号。
are centre (cut from top right) is at (5,5) with area 4 taken negative. Total area = 32. x̄ = (36×3 - 4×5)/32 = (108 - 20)/32 = 88/32 = 2.75 cm. ȳ similarly 2.75 cm. The centre of mass shifts towards the opposite corner.
示例：一块边长为 6 cm 的均匀正方形薄片，从一个角切去一个边长为 2 cm 的正方形。大正方形中心在 (3,3)，面积 36。被切的小正方形（右上角）中心在 (5,5)，面积为 -4。总面积 = 32。x̄ = (36×3 - 4×5)/32 = 88/32 = 2.75 cm；ȳ 类似为 2.75 cm。质心向对角方向偏移。
7. Equilibrium of Suspended Bodies | 悬挂物体的平衡
When a body is freely suspended from a point, it will come to rest with its centre of mass vertically below the point of suspension. This is because the weight acts through the centre of mass, producing a restoring torque that eliminates any horizontal offset.
当一个物体从某点自由悬挂时，它将静止在质心位于悬挂点正下方的位置。这是因为重力通过质心作用，所产生的恢复力矩会消除任何水平偏移。
This principle is used to locate the centre of mass experimentally. If you hang an irregular shape from different points and draw vertical lines downwards from each pivot, the intersection of these lines gives the centre of mass.
这一原理被用于通过实验定位质心。如果你从不同点悬挂一个不规则形状，并从每个悬挂点向下画垂线，这些垂线的交点即为质心。
In exam problems, you can determine unknown distances or angles by using the fact that the line connecting the pivot and the centre of mass must be vertical in equilibrium.
在考试题目中，可以利用平衡时连接转动轴与质心的直线必定竖直这一事实，来求出未知的距离或角度。
8. Tilting and Toppling | 倾斜与倾倒
A body placed on a rough surface will be in equilibrium as long as the vertical line through its centre of mass falls within its base area. If the centre of mass moves beyond the edge of the base, the body will topple over.
放置在粗糙表面上的物体，只要通过其质心的竖直线落在基底面积之内，就会保持平衡。如果质心超出基底边缘，物体将会倾倒。
ed about one edge, the angle at which it is about to topple occurs when the centre of mass is directly above that edge. At this instant, tan θ = (w/2) / (h/2) = w/h, where θ is the tilt angle from vertical.
考虑一个立放在水平面上的宽为 w、高为 h 的均匀方块，其质心在几何中心。若方块绕一条边缘倾斜，即将倾倒时的临界角发生在质心位于该边缘正上方的瞬间。此时，从竖直线量起的倾斜角 θ 满足 tan θ = (w/2) / (h/2) = w/h。
When analysing more complex shapes on inclined planes, compare the position of the line of action of the weight to the edges of the base. If the line of action falls outside the base, the object will topple, regardless of friction.
在分析斜面上更复杂的形状时，将重力的作用线与基底边缘进行比较。若作用线落在基底之外，无论摩擦力多大，物体都将倾倒。
9. Worked Example: Composite Shape and Tilting | 实例解析：组合形状与倾倒
A uniform solid consists of a cone of height 9 cm on top of a cylinder of height 6 cm, both with base radius 3 cm. The densities are the same. Find the centre of mass relative to the base of the cylinder, and determine the steepest incline on which it can rest without toppling.
一个均匀实心体由一个高 9 cm 的圆锥放在一个高 6 cm 的圆柱上组成，两者底面半径均为 3 cm，密度相同。求质心相对于圆柱底面的位置，并确定它能够不倾倒地放置的最大倾角。
Volumes: Cylinder V
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