这份讲义涵盖主题 3:进阶力学(Further Mechanics)。它把力学扩展到抛射体、刚体、圆周运动、弹性绳、变力和碰撞。取 $g = 10\ \text{m s}^{-2}$。
进阶力学
A-Level 进阶数学 · 第 3 主题
3.1
抛体运动
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| • model the motion of a projectile as a particle moving with constant acceleration and understand any limitations of the model | Vector methods are not required |
| • use horizontal and vertical equations of motion to solve problems on the motion of projectiles, including finding the magnitude and direction of the velocity at a given time or position, the range on a horizontal plane and the greatest height reached | |
| • derive and use the Cartesian equation of the trajectory of a projectile, including problems in which the initial speed and/or angle of projection may be unknown. | Knowledge of the 'bounding parabola' for accessible points is not included. |
来源:剑桥国际大纲
一个抛射体(projectile)在重力下自由运动,所以它有向下的匀加速(constant acceleration)$g$ 且没有水平加速度。把水平和竖直运动分开处理。若它以速率 $u$、角 $\alpha$ 被发射:
$$\text{horizontal: } x = u\cos\alpha\,\cdot t, \qquad \text{vertical: } y = u\sin\alpha\,\cdot t - \tfrac12 g t^2.$$
消去 $t$ 给出轨迹(trajectory,路径)的直角坐标方程(Cartesian equation),它是一条抛物线。对于一个固定的发射速率,所有的轨迹位于一条包络抛物线(bounding parabola,可达点的包络)之下。在水平地面上的射程是 $\dfrac{u^2\sin 2\alpha}{g}$,而最大高度是 $\dfrac{u^2\sin^2\alpha}{2g}$。
例题。 一个球以 $u = 20\ \text{m s}^{-1}$、与水平成 $30^\circ$ 被抛出。求射程和最大高度。
$$\text{range} = \frac{20^2\sin 60^\circ}{10} = 40\times 0.866 = 34.6\ \text{m}, \qquad \text{height} = \frac{20^2\sin^2 30^\circ}{2\times 10} = \frac{400\times 0.25}{20} = 5\ \text{m}.$$
探索
Launch a projectile
Fire the ball, then change the angle and speed. The horizontal motion is steady while gravity pulls it down — together they trace a parabola. Find the angle for the longest range, and try the Moon.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| Further Mechanics | 进阶力学 | jìn jiē lì xué |
| projectile | 抛射体 | pāo shè tǐ |
| constant acceleration | 匀加速 | yún jiā sù |
| Cartesian equation | 直角坐标方程 | zhí jiǎo zuò biāo fāng chéng |
| trajectory | 轨迹 | guǐ jì |
| bounding parabola | 包络抛物线 | bāo luò pāo wù xiàn |
3.2
刚体的平衡
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| • calculate the moment of a force about a point | For questions involving coplanar forces only; understanding of the vector nature of moments is not required. |
| • use the result that the effect of gravity on a rigid body is equivalent to a single force acting at the centre of mass of the body, and identify the position of the centre of mass of a uniform body using considerations of symmetry | |
| • use given information about the position of the centre of mass of a triangular lamina and other simple shapes | Proofs of results given in the MF19 List of formulae are not required. |
| • determine the position of the centre of mass of a composite body by considering an equivalent system of particles | Simple cases only, e.g. a uniform L-shaped lamina, or a uniform cone joined at its base to a uniform hemisphere of the same radius. |
| • use the principle that if a rigid body is in equilibrium under the action of coplanar forces then the vector sum of the forces is zero and the sum of the moments of the forces about any point is zero, and the converse of this | |
| • solve problems involving the equilibrium of a single rigid body under the action of coplanar forces, including those involving toppling or sliding. |
来源:剑桥国际大纲
一个力关于一个点的力矩(moment)是力 $\times$ 垂直距离;它测量转动效果。一个物体的重力作用在它的质心(centre of mass),对一个均匀的平面形状(一个薄片(lamina))你能用对称(symmetry)找到它,或通过把一个复合物体当作一组质点找到它。
一个刚体在共面力(coplanar forces)下,当两个条件都成立时处于平衡(equilibrium):力的向量和为零,而关于任何点的力矩之和为零。一个物体也可能在翻倒(toppling)或滑动(sliding)的边缘。
例题。 一根长 $4\ \text{m}$、重 $100\ \text{N}$ 的均匀梁 $AB$ 放在 $A$ 处的一个支点上。$B$ 处一个竖直力 $F$ 使它保持水平。求 $F$。
关于 $A$ 取力矩(重力作用在中心,距 $A$ $2\ \text{m}$):
$$F\times 4 = 100\times 2 \;\Rightarrow\; F = 50\ \text{N}.$$
探索
Equilibrium of forces
resultant = 0
A body is in equilibrium when the forces add tip-to-tail back to zero.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| moment | 力矩 | lì jǔ |
| centre of mass | 质心 | zhì xīn |
| symmetry | 对称 | duì chèn |
| equilibrium | 平衡 | píng héng |
| toppling | 翻倒 | fān dǎo |
| sliding | 滑动 | huá dòng |
| lamina | 薄片 | báo piàn |
| coplanar forces | 共面力 | gòng miàn lì |
3.3
圆周运动
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| understand the concept of angular speed for a particle moving in a circle, and use the relation $v = r\omega$ | |
| understand that the acceleration of a particle moving in a circle with constant speed is directed towards the centre of the circle, and use the formulae $r\omega^2$ and $\frac{v^2}{r}$. | Proof of the acceleration formulae is not required. |
| solve problems which can be modelled by the motion of a particle moving in a horizontal circle with constant speed | |
| solve problems which can be modelled by the motion of a particle in a vertical circle without loss of energy. | Including finding a normal contact force or the tension in a string, locating points at which these are zero, and conditions for complete circular motion. |
来源:剑桥国际大纲
对一个在半径 $r$ 的圆中运动的质点,角速度(angular speed)$\omega$ 通过 $v = r\omega$ 与速率联系。加速度指向中心——向心加速度(centripetal acceleration)——大小为
$$a = r\omega^2 = \frac{v^2}{r}.$$
在一个水平圆(horizontal circle)中速率是恒定的——如在一个圆锥摆(conical pendulum,一个质量在一根绳上摆动)中。在一个竖直圆(vertical circle)中用能量守恒,因为速率随高度改变;法向接触力(normal contact force)或绳张力提供向心力。
例题。 一个质点在一个半径 $2\ \text{m}$ 的水平圆中以角速度 $3\ \text{rad s}^{-1}$ 运动。求它的速率和加速度。
$$v = r\omega = 2\times 3 = 6\ \text{m s}^{-1}, \qquad a = r\omega^2 = 2\times 3^2 = 18\ \text{m s}^{-2}.$$
探索
Angle in radians
Circular motion is measured in radians: drag the angle θ and radius r to see the arc swept — angular speed ω turns this into v = rω.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| angular speed | 角速度 | jiǎo sù dù |
| centripetal acceleration | 向心加速度 | xiàng xīn jiā sù dù |
| horizontal circle | 水平圆 | shuǐ píng yuán |
| vertical circle | 竖直圆 | shù zhí yuán |
| conical pendulum | 圆锥摆 | yuán zhuī bǎi |
| normal contact force | 法向接触力 | fǎ xiàng jiē chù lì |
3.4
胡克定律
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| use Hooke’s law as a model relating the force in an elastic string or spring to the extension or compression, and understand the term modulus of elasticity | |
| use the formula for the elastic potential energy stored in a string or spring | Proof of the formula is not required. |
| solve problems involving forces due to elastic strings or springs, including those where considerations of work and energy are needed. | e.g. a particle moving horizontally or vertically or on an inclined plane while attached to one or more strings or springs, or a particle attached to an elastic string acting as a 'conical pendulum'. |
来源:剑桥国际大纲
胡克定律(Hooke's law)说一根弹性绳或弹簧中的张力与它的伸长 $x$ 成正比:
$$T = \frac{\lambda x}{L},$$
其中 $L$ 是自然长度而 $\lambda$ 是弹性模量(modulus of elasticity)。拉伸储存弹性势能(elastic potential energy):
$$E = \frac{\lambda x^2}{2L}.$$
例题。 一根自然长度 $2\ \text{m}$、模量 $50\ \text{N}$ 的弹性绳被拉伸 $0.5\ \text{m}$。求张力和储存的能量。
$$T = \frac{50\times 0.5}{2} = 12.5\ \text{N}, \qquad E = \frac{50\times 0.5^2}{2\times 2} = 3.125\ \text{J}.$$
探索
Hooke's law
F = k·x
Force is proportional to extension — the gradient is the stiffness k.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| Hooke's law | 胡克定律 | hú kè dìng lǜ |
| modulus of elasticity | 弹性模量 | tán xìng mó liàng |
| elastic potential energy | 弹性势能 | tán xìng shì néng |
3.5
变力作用下的直线运动
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| solve problems which can be modelled as the linear motion of a particle under the action of a variable force, by setting up and solving an appropriate differential equation. | Including use of $v \frac{\mathrm{d}v}{\mathrm{d}x}$ for acceleration, where appropriate. Calculus required is restricted to content from Pure Mathematics 3 in Cambridge International A Level Mathematics (9709). Only differential equations in which the variables are separable are included. |
来源:剑桥国际大纲
当力取决于位置 $x$ 时,用形式 $a = v\dfrac{dv}{dx}$ 的加速度,它把牛顿定律变成一个联系 $v$ 和 $x$ 的微分方程(differential equation)。
例题。 一个质量 $8\ \text{kg}$ 的质点在一个大小为 $(x^3 + 4x)\ \text{N}$、沿运动方向作用的变力(variable force)下沿一条线运动。当 $x = 0$ 时,$v = 1$。求 $v$ 用 $x$ 表示。
牛顿定律给出 $8v\dfrac{dv}{dx} = x^3 + 4x$。分离并积分:
$$\int 8v\,dv = \int (x^3 + 4x)\,dx \;\Rightarrow\; 4v^2 = \tfrac14 x^4 + 2x^2 + C.$$
在 $x = 0$ 处,$v = 1$ 给出 $C = 4$,所以 $4v^2 = \tfrac14 x^4 + 2x^2 + 4 = 4\left(\tfrac{x^2}{4} + 1\right)^2$。因此 $v = \tfrac14 x^2 + 1$。
探索
Work from a variable force
W = ∫ F dx
When the force changes, the work done is the area under the force–distance graph.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| differential equation | 微分方程 | wēi fēn fāng chéng |
| variable force | 变力 | biàn lì |
| conservation of momentum | 动量守恒 | dòng liàng shǒu héng |
3.6
动量
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| recall Newton’s experimental law and the definition of the coefficient of restitution, the property $0 \leqslant e \leqslant 1$, and the meaning of the terms ‘perfectly elastic’ ($e = 1$) and ‘inelastic’ ($e = 0$) | |
| use conservation of linear momentum and/or Newton’s experimental law to solve problems that may be modelled as the direct or oblique impact of two smooth spheres, or the direct or oblique impact of a smooth sphere with a fixed surface. |
来源:剑桥国际大纲
在一次碰撞中,总动量守恒——动量守恒(conservation of linear momentum)。弹性由牛顿实验定律(Newton's experimental law)测量,它定义恢复系数(coefficient of restitution)$e$:
$$e = \frac{\text{speed of separation}}{\text{speed of approach}}, \qquad 0 \leqslant e \leqslant 1.$$
这里 $e = 1$ 是完全弹性的(没有能量损失)而 $e = 0$ 是非弹性的(物体粘在一起)。在一次斜碰撞(oblique impact)中,把速度沿和垂直于碰撞线分解,沿它应用恢复。
例题。 一个质量 $2\ \text{kg}$、以 $5\ \text{m s}^{-1}$ 运动的球 $A$ 撞上一个静止的质量 $3\ \text{kg}$ 的球 $B$,$e = 0.5$。求之后的速率。
动量:$2(5) = 2v_A + 3v_B$,所以 $2v_A + 3v_B = 10$。恢复:$v_B - v_A = 0.5(5) = 2.5$。一起求解给出 $v_A = 0.5\ \text{m s}^{-1}$ 和 $v_B = 3.0\ \text{m s}^{-1}$。
探索
A collision
Set each mass and speed, then collide them. Total momentum is conserved — see how the velocities come out.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| Newton's experimental law | 牛顿实验定律 | niú dùn shí yàn dìng lǜ |
| coefficient of restitution | 恢复系数 | huī fù xì shù |
| oblique impact | 斜碰撞 | xié pèng zhuàng |
3.6
考试技巧
- 对于抛射体,分解成水平(恒速)和竖直($a = g$)运动,由同一个时间联系。
- 对于一个平衡的刚体,关于一个移除一个未知力的点取力矩。
- 对于圆周运动,用 $F = mv^2/r = m\omega^2 r$ 朝中心;在一个竖直圆中检查顶部的最小速率。
- 用一个变力,用 $a = v\,\frac{dv}{dx}$ 并积分;弹性 PE $= \frac{\lambda x^2}{2L}$。
本主题的互动课程
逐步学习,并即时检测练习。